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A First Course in Mathematical Analysis
Mathematical Analysis (often called Advanced Calculus) is generally found by students
to be one of their hardest courses in Mathematics. This text uses the so-called sequential
approach to continuity, differentiability and integration to make it easier to understand
the subject.
Topics that are generally glossed over in the standard Calculus courses are given
careful study here. For example, what exactly is a ‘continuous’ function? And how
exactly can one give a careful definition of ‘integral’? This latter is often one of the
mysterious points in a Calculus course – and it is quite tricky to give a rigorous
treatment of integration!
The text has a large number of diagrams and helpful margin notes, and uses many
graded examples and exercises, often with complete solutions, to guide students
through the tricky points. It is suitable for self study or use in parallel with a standard
university course on the subject.
A First Course in
Mathematical
Analysis
DAVID ALEXANDER BRANNAN
Published in association with The Open University
CAMBRIDGE UNIVERSITY PRESS
Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo
Cambridge University Press
The Edinburgh Building, Cambridge CB2 8RU, UK
Published in the United States of America by Cambridge University Press, New York
www.cambridge.org
Information on this title: www.cambridge.org/9780521864398
© The Open University 2006
This publication is in copyright. Subject to statutory exception and to the provision of
relevant collective licensing agreements, no reproduction of any part may take place
without the written permission of Cambridge University Press.
First published in print format 2006
eBook (EBL)
ISBN-13 978-0-511-34857-0
ISBN-10 0-511-34857-6
eBook (EBL)
ISBN-13
ISBN-10
hardback
978-0-521-86439-8
hardback
0-521-86439-9
Cambridge University Press has no responsibility for the persistence or accuracy of urls
for external or third-party internet websites referred to in this publication, and does not
guarantee that any content on such websites is, or will remain, accurate or appropriate.
To my wife Margaret
and my sons David, Joseph and Michael
Contents
Preface
page ix
1
Numbers
1.1 Real numbers
1.2 Inequalities
1.3 Proving inequalities
1.4 Least upper bounds and greatest lower bounds
1.5 Manipulating real numbers
1.6 Exercises
1
2
9
14
22
30
35
2
Sequences
2.1 Introducing sequences
2.2 Null sequences
2.3 Convergent sequences
2.4 Divergent sequences
2.5 The Monotone Convergence Theorem
2.6 Exercises
37
38
43
52
61
68
79
3
Series
3.1 Introducing series
3.2 Series with non-negative terms
3.3 Series with positive and negative terms
3.4 The exponential function x 7! ex
3.5 Exercises
83
84
92
103
122
127
4
Continuity
4.1 Continuous functions
4.2 Properties of continuous functions
4.3 Inverse functions
4.4 Defining exponential functions
4.5 Exercises
130
131
143
151
161
164
5
Limits and continuity
5.1 Limits of functions
5.2 Asymptotic behaviour of functions
5.3 Limits of functions – using " and 5.4 Continuity – using " and 5.5 Uniform continuity
5.6 Exercises
167
168
176
181
193
200
203
vii
Contents
viii
6
Differentiation
6.1 Differentiable functions
6.2 Rules for differentiation
6.3 Rolle’s Theorem
6.4 The Mean Value Theorem
6.5 L’Hôpital’s Rule
6.6 The Blancmange function
6.7 Exercises
205
206
216
228
232
238
244
252
7
Integration
7.1 The Riemann integral
7.2 Properties of integrals
7.3 Fundamental Theorem of Calculus
7.4 Inequalities for integrals and their applications
7.5 Stirling’s Formula for n!
7.6 Exercises
255
256
272
282
288
303
309
8
Power series
8.1 Taylor polynomials
8.2 Taylor’s Theorem
8.3 Convergence of power series
8.4 Manipulating power series
8.5 Numerical estimates for p
8.6 Exercises
313
314
320
329
338
346
350
Appendix 1
Sets, functions and proofs
354
Appendix 2
Standard derivatives and primitives
359
Appendix 3
The first 1000 decimal places of
pffiffiffi
2, e and p
361
Appendix 4 Solutions to the problems
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
363
363
371
382
393
402
413
426
443
Index
457
Preface
Analysis is a central topic in Mathematics, many of whose branches use
key analytic tools. Analysis also has important applications in Applied
Mathematics, Physics and Engineering, where a good appreciation of the
underlying ideas of Analysis is necessary for a modern graduate.
Changes in the school curriculum over the last few decades have resulted
in many students finding Analysis very difficult. The author believes that
Analysis nowadays has an unjustified reputation for being hard, caused by
the traditional university approach of providing students with a highly polished
exposition in lectures and associated textbooks that make it impossible for
the average learner to grasp the core ideas. Many students end up agreeing with
the German poet and philosopher Goethe who wrote that ‘Mathematicians are
like Frenchmen: whatever you say to them, they translate into their own
language, and forthwith it is something entirely different!’
Since 1971, the Open University in United Kingdom has taught Mathematics
to students in their own homes via specially written correspondence texts, and
has traditionally given Analysis a central position in its curriculum. Its philosophy is to provide clear and complete explanations of topics, and to teach these
in a way that students can understand without much external help. As a result,
students should be able to learn, and to enjoy learning, the key concepts of the
subject in an uncluttered way. This book arises from correspondence texts for
its course Introduction to Pure Mathematics, that has now been studied successfully by over ten thousand students.
This book is therefore different from most Mathematics textbooks! It adopts
a student-friendly approach, being designed for study by a student on their own
OR in parallel with a course that uses as set text either this text or another text.
But this is the text that the student is likely to use to learn the subject from. The
author hopes that readers will gain enormous pleasure from the subject’s
beauty and that this will encourage them to undertake further study of
Mathematics!
Once a student has grasped the principal notions of limit and continuous
function in terms of inequalities involving the three symbols ", X and , they
will quickly understand the unity of areas of Analysis such as limits, continuity, differentiability and integrability. Then they will thoroughly enjoy the
beauty of some of the arguments used to prove key theorems – whether their
proofs are short or long.
Calculus is the initial study of limits, continuity, differentiation and integration, where functions are assumed to be well-behaved. Thus all functions
continuous on an interval are assumed to be differentiable at most points in
the interval, and so on. However, Mathematics is not that simple! For example,
there exist functions that are continuous everywhere on R, but differentiable
Johann Wolfgang von Goethe
(1749–1832) is said to have
studied all areas of science of
his day except mathematics –
for which he had no aptitude.
ix
Preface
x
nowhere on R; this discovery by Karl Weierstrass in 1872 caused a sensation in
the mathematical community. In Analysis (sometimes called Advanced
Calculus) we make no assumptions about the behaviour of functions – and
the result is that we sometimes come across real surprises!
The book has two principal features in its approach that make it stand out
from among other Analysis texts.
Firstly, this book uses the ‘sequential approach’ to Analysis. All too often
students starting on the subject find that they cannot grasp the significance of
both " and simultaneously. This means that the whole underlying idea about
what is happening is lost, and the student takes a very long time to master the
topic – or, in many cases in fact, never masters the topic and acquires a strong
dislike of it. In the sequential approach they proceed at a more leisurely pace to
understand the notion of limit using " and X – to handle convergent sequences –
before coming across the other symbol , used in conjunction with " to handle
continuous functions. This approach avoids the conventional student horror at
the perceived ‘difficulty of Analysis’. Also, it avoids the necessity to re-prove
broadly similar results in a range of settings – for example, results on the sum
of two sequences, of two series, of two continuous functions and of two
differentiable functions.
Secondly, this book makes great efforts to teach the "– approach too. After
students have had a first pass at convergence of sequences and series and at
continuity using ‘the sequential approach’, they then meet ‘the "– approach’,
explained carefully and motivated by a clear ‘"– game’ discussion. This
makes the new approach seem very natural, and this is motivated by using
each approach in later work in the appropriate situation. By the end of the book,
students should have a good facility at using both the sequential approach and
the "– approach to proofs in Analysis, and should be better prepared for later
study of Analysis than students who have acquired only a weak understanding
of the conventional approach.
Outline of the content of the book
In Chapter 1, we define real numbers to be decimals. Rather than give a heavy
discussion of least upper bound and greatest lower bound, we give an introduction to these matters sufficient for our purposes, and the full discussion is
postponed to Chapter 7, where it is more timely. We also study inequalities,
and their properties and proofs.
In Chapter 2, we define convergent sequences and examine their properties, basing the discussion on the notion of null sequence, which simplifies
matters considerably. We also look at divergent sequences, sequences
defined by recurrence formulas and particular sequences which converge
to p and e.
In Chapter 3, we define convergent infinite series, and establish a number of
tests for determining whether a given series is convergent or divergent. We
demonstrate the equivalence of the two definitions of the exponential function
x 7! ex , and prove that the number e is irrational.
In Chapter 4, we define carefully what we mean by a continuous function, in
terms of sequences, and establish the key properties of continuous functions.
We also give a rigorous definition of the exponential function x 7! ax .
n
We define ex as lim 1 þ nx
n!1
1
P xn
and as
n! .
n¼0
Preface
In Chapter 5, we define the limit of a function as x tends to c or as x tends
to 1 in terms of the convergence of sequences. Then we introduce the "–
definitions of limit and continuity, and check that these are equivalent to the
earlier definitions in terms of sequences. We also look briefly at uniform
continuity.
In Chapter 6, we define what we mean by a differentiable function, using
difference quotients Q(h); this enables us to use our earlier results on limits to
prove corresponding results for differentiable functions. We establish some
interesting properties of differentiable functions. Finally, we construct the
Blancmange function that is continuous everywhere on R, but differentiable
nowhere on R.
In Chapter 7, we give a careful definition of what we mean by an integrable
function, and establish a number of related criteria for establishing whether
a given function is integrable or not. Our integral is the so-called Riemann
integral, defined in terms of upper and lower Riemann sums. We check the
standard properties of integrals and verify a number of standard approaches for
calculating definite integrals. Then we give a number of applications of
integrals to limits of certain sequences and series and prove Stirling’s Formula.
Finally, in Chapter 8, we study the convergence and properties of power
series. The chapter ends with a marvellous proof of the irrationality of the
number p that uses a whole range of the techniques that have been met in the
previous chapters.
For completeness and for students’ convenience, we give a brief guide to our
notation for sets and functions, together with a brief indication of the logic
involved in proofs in Mathematics (in particular, the Principle of Mathematical
Induction) in Appendix 1. Appendix 2 contains a list of standard derivatives
and primitives and Appendix 3 the first 1000 decimal places in the values of the
pffiffiffi
numbers 2, p and e. Appendix 4 contains full solutions to all the problems set
during each chapter.
Solutions are not given to the exercises at the end of each chapter, however.
Lecturers/instructors may wish to use these exercises in homework assignments.
Study guide
This book assumes that students have a fair understanding of Calculus. The
assumptions on technical background are deliberately kept slight, however, so
that students can concentrate on the newer aspects of the subject ‘Analysis’.
Most students will have met some of the material in the early chapters
previously. Although this means that they can therefore proceed fairly quickly
through some sections, it does NOT mean that those sections can be ignored –
each section contains important ideas that are used later on and most include
something new or have a different emphasis.
Most chapters are divided into five or six sections (each often further divided
into sub-sections); sections are numbered using two digits (such as ‘Section
3.2’) and sub-sections using three digits (such as ‘Sub-section 3.2.4’).
Generally a section is considered to be about one evening’s work for an
average student.
Chapter 7 on Integration is arguably the highlight of the book. However,
it contains some rather complicated mathematical arguments and proofs.
xi
Stirling’s Formula says that,
for
large
pffiffiffiffiffiffiffi
ffi nn, n! is ‘roughly’
2pn ne , in a sense that we
explain.
xii
Therefore, when reading Chapter 7, it is important not to get bogged down in
details, but to keep progressing through the key ideas, and to return later on to
reading the things that were left out at the earlier reading. Most students will
require three or four passes at this chapter before having a good idea of most of it.
We use wide pages with a large number of margin notes in which we place
teaching comments and some diagrams to aid in the understanding of particular
points in arguments. We also provide advice on which proofs to omit on a first
study of the topic; it is important for the student NOT to get bogged down in
a technical discussion or a proof until they have a good idea of the message
contained in the result and the situations in which it can be used. Therefore
clear encouragement is given on which portions of the text to leave till later, or
to simply skim on a first reading.
The end of the proof of a Theorem is indicated by a solid symbol ‘&’ and the
end of the solution of a worked Example by a hollow symbol ‘&’. There are
many worked examples within the text to explain the concepts being taught,
together with a good stock of problems to reinforce the teaching. The solutions
are a key part of the teaching, and tackling them on your own and then reading
our version of the solution is a key part of the learning process.
No one can learn Mathematics by simply reading – it is a ‘hands on’ activity.
The reader should not be afraid to draw pictures to illustrate what seems to be
happening to a sequence or a function, to get a feeling for their behaviour. A
wise old man once said that ‘A picture is worth a thousand words!’. A good
picture may even suggest a method of proof. However, at the same time it is
important not to regard a picture on its own as a proof of anything; it may
illustrate just one situation that can arise and miss many other possibilities!
It is important NOT to become discouraged if a topic seems difficult. It took
mathematicians hundreds of years to develop Analysis to its current polished
state, so it may take the reader a few hours at several sittings to really grasp the
more complex or subtle ideas.
Acknowledgements
The material in the Open University course on which this book is based was
contributed to in some way by many colleagues, including Phil Rippon, Robin
Wilson, Andrew Brown, Hossein Zand, Joan Aldous, Ian Harrison, Alan Best,
Alison Cadle and Roberta Cheriyan. Its eventual appearance in book form
owes much to Lynne Barber.
Without the forebearance of my family, the writing of the book would have
been impossible.
Preface
It is important to read the
margin notes!
This signposting benefits
students greatly in the
author’s experience.
Tackling the problems is a
good use of your time, not
something to skimp.
1
Numbers
In this book we study the properties of real functions defined on intervals of the
real line (possibly the whole real line) and whose image also lies on the real
line. In other words, they map R into R. Our work will be from a very precise
point of view in order to establish many of the properties of such functions
which seem intuitively obvious; in the process we will discover that some
apparently true properties are in fact not necessarily true!
The types of functions that we shall examine include:
exponential functions, such as x 7! ax ; where a; x 2 R;
trigonometric functions, such as x 7! sin x; where x 2 R;
pffiffiffi
root functions, such as x 7! x; where x 0:
The types of behaviour that we shall examine include continuity, differentiability and integrability – and we shall discover that functions with these
properties can be used in a number of surprising applications.
However, to put our study of such functions on a secure foundation, we need
first to clarify our ideas of the real numbers themselves and their properties. In
particular, we need to devote some time to the manipulation of inequalities,
which play a key role throughout the book.
In Section 1.1, we start by revising the properties of rational numbers and
their decimal representation. Then we introduce the real numbers as infinite
decimals, and describe the difficulties involved in doing arithmetic with such
decimals.
In Section 1.2, we revise the rules for manipulating inequalities and show
how to find the solution set of an inequality involving a real number, x, by
applying the rules. We also explain how to deal with inequalities which involve
modulus signs.
In Section 1.3, we describe various techniques for proving inequalities,
including the very important technique of Mathematical Induction.
The concept of a least upper bound, which is of great importance in
Analysis, is introduced in Section 1.4, and we discuss the Least Upper
Bound Property of R.
Finally, in Section 1.5, we describe how least upper bounds can be used to
define arithmetical operations in R.
Even though you may be familiar with much of this material we recommend
that you read through it, as we give the system of real numbers a more careful
treatment than you may have met before. The material on inequalities and least
upper bounds is particularly important for later on.
In later chapters we shall define exactly what the numbers p and e are, and
find various ways of calculating them. But, first, we examine numbers in
general.
For example,pwhat
exactly is
ffiffiffi
the number 2?
You may omit this section
at a first reading.
1
1: Numbers
2
1.1
Real numbers
We start our study of the real numbers with the rational numbers, and investigate their decimal representations, then we proceed to the irrational numbers.
1.1.1
Rational numbers
We assume that you are familiar with the set of natural numbers
N ¼ f1; 2; 3; . . .g;
Note that 0 is not a natural
number.
and with the set of integers
Z ¼ f. . .; 2; 1; 0; 1; 2; 3; . . .g:
The set of rational numbers consists of all fractions (or ratios of integers)
p
: p 2 Z; q 2 N :
Q ¼
q
Remember that each rational number has many different representations as a
ratio of integers; for example
1 2 10
¼ ¼
¼ . . .:
3 6 30
We also assume that you are familiar with the usual arithmetical operations of
addition, subtraction, multiplication and division of rational numbers.
It is often convenient to represent rational numbers geometrically as points
on a number line. We begin by drawing a line and marking on it points
corresponding to the integers 0 and 1. If the distance between 0 and 1 is
taken as a unit of length, then the rationals can be arranged on the line with
positive rationals to the right of 0 and negative rationals to the left.
negative
–3
– 52
–2
positive
–1
0
1
2
1
3
2
2
3
For example, the rational 32 is placed at the point which is one-half of the way
from 0 to 3.
This means that rationals have a natural order on the number-line. For
7
example, 19
22 lies to the left of 8 because
19 76
¼
22 88
and
7 77
¼ :
8 88
If a lies to the left of b on the number-line, then
a is less than b or b is greater than a;
and we write
a<b
or b > a:
For example
19 7
7 19
<
or
> :
22 8
8 22
We write a b, or b a, if either a < b or a ¼ b.
1.1
Real numbers
Problem 1
3
Arrange the following rational numbers in order:
17 45
45
0; 1; 1; 17
20 ; 20 ; 53 ; 53:
Problem 2 Show that between any two distinct rational numbers there
is another rational number.
1.1.2
Decimal representation of rational numbers
The decimal system enables us to represent all the natural numbers using only
the ten integers
0; 1; 2; 3; 4; 5; 6; 7; 8 and 9;
which are called digits. We now remind you of the basic facts about the
representation of rational numbers by decimals.
Definition
A decimal is an expression of the form
a0 a1 a2 a3 . . .;
where a0 is a non-negative integer and a1, a2, a3, . . . are digits.
If only a finite number of the digits a1, a2, a3, . . . are non-zero, then the
decimal is called terminating or finite, and we usually omit the tail of 0s.
Terminating decimals are used to represent rational numbers in the following way
a1
a2
an a0 a1 a2 a3 . . . an ¼ a0 þ 1 þ 2 þ þ n :
10
10
10
It can be shown that any fraction whose denominator contains only powers
of 2 and/or 5 (such as 20 ¼ 22 5) can be represented by such a terminating
decimal, which can be found by long division.
However, if we apply long division to many other rationals, then the process
of long division never terminates and we obtain a non-terminating or infinite
decimal. For example, applying long division to 13 gives 0.333 . . ., and for 19
22
we obtain 0.86363 . . ..
Problem 3
decimals.
Apply long division to 17 and
2
13
For example
0:8500 . . .;
13:1212 . . .;
1:111 . . .:
For example,
0:8500 . . . ¼ 0:85:
For example
8
5
0:85 ¼ 0 þ 1 þ 2
10
10
85
17
¼ :
¼
100 20
to find the corresponding
These non-terminating decimals, which are obtained by applying the long
division process, have a certain common property. All of them are recurring;
that is, they have a recurring block of digits, and so can be written in shorthand
form, as follows:
0:333 . . .
¼ 0:3;
0:142857142857 . . . ¼ 0:142857 . . .;
0:86363 . . .
¼ 0:863:
It is not hard to show, whenever we apply the long division process to a
fraction pq , that the resulting decimal is recurring. To see why we notice that
there are only q possible remainders at each stage of the division, and so one
of these remainders must eventually recur. If the remainder 0 occurs, then the
resulting decimal is, of course, terminating; that is, it ends in recurring 0s.
Another commonly used
notation is
:
:
:
0:3 or 0:142857:
1: Numbers
4
Non-terminating recurring decimals which arise from the long division of
fractions are used to represent the corresponding rational numbers. This
representation is not quite so straight-forward as for terminating decimals,
however. For example, the statement
1
3
3
3
¼ 0:3 ¼ 1 þ 2 þ 3 þ 3
10
10
10
can be made precise only when we have introduced the idea of the sum of a
convergent infinite series. For the moment, when we write the statement 13 ¼ 0:3
we mean simply that the decimal 0:3 arises from 13 by the long division process.
The following example illustrates one way of finding the rational number
with a given decimal representation.
We return to this topic in
Chapter 3.
Example 1 Find the rational number (expressed as a fraction) whose decimal
representation is 0:863:
Solution First we find the fraction x such that x ¼ 0:63:
If we multiply both sides of this equation by 102 (because the recurring block
has length 2), then we obtain
100 x ¼ 63:63 ¼ 63 þ x:
Hence
99x ¼ 63 ) x ¼
63
7
¼ :
99 11
Thus
0:863 ¼
8
x
8
7
95
19
þ
¼
þ
¼
¼ :
10 10 10 110 110 22
&
The key idea in the above solution is that multiplication of a decimal by 10k
moves the decimal point k places to the right.
Problem 4 Using the above method, find the fractions whose decimal
representations are:
(a) 0:231;
(b) 2:281.
The decimal representation of rational numbers has the advantage that it
enables us to decide immediately which of two distinct positive rational numbers
is the greater. We need only examine their decimal representations and notice
the first place at which the digits differ. For example, to order 78 and 19
22 we write
7
¼ 0:875 . . .
8
19
¼ 0:86363 . . .;
22
and
and so
#
#
0:86363 . . . < 0:875 )
19
7
< :
22
8
Problem 5 Find the first two digits after the decimal point in the deci45
mal representations of 17
20 and 53, and hence determine which of these
two rational numbers is the greater.
Warning Decimals which end in recurring 9s sometimes arise as alternative
representations for terminating decimals. For example
1 ¼ 0:9 ¼ 0:999 . . .
and
1:35 ¼ 1:349 ¼ 1:34999 . . .:
Whenever possible, we avoid
using the form of a decimal
which ends in recurring 9s.
1.1
Real numbers
5
You may find this rather alarming, but it is important to realise that this is
a matter of convention. We wish to allow the decimal 0.999 . . . to represent
a number x, so x must be less than or equal to 1 and greater than each of the
numbers
0:9; 0:99; 0:999; . . .:
The only rational with these properties is 1.
1.1.3
Irrational numbers
One of the surprising mathematical discoveries made by the Ancient Greeks
was that the system of rational numbers is not adequate to describe all the
magnitudes that occur in geometry. For example, consider the diagonal of a
square of side 1. What is its length? If the length is x, then, by Pythagoras’
Theorem, x must satisfy the equation x2 ¼ 2. However, there is no rational
number which satisfies this equation.
Theorem 1
x
1
x2 ¼ 12 þ 12 ¼ 2
There is no rational number x such that x2 ¼ 2.
Proof Suppose that such a rational number x exists. Then we can write x ¼ pq.
By cancelling, if necessary, we may assume that p and q have no common
factor. The equation x2 ¼ 2 now becomes
p2
¼ 2;
q2
so
2
ð2rÞ ¼ 2q ;
This is a proof by
contradiction.
p2 ¼ 2q2 :
Now, the square of an odd number is odd, and so p cannot be odd. Hence p is
even, and so we can write p ¼ 2r, say. Our equation now becomes
2
1
so
2
2
q ¼ 2r :
For
ð2k þ 1Þ2 ¼ 4k2 þ 4k þ 1
¼ 4ðk2 þ kÞ þ 1:
Reasoning as before, we find that q is also even.
Since p and q are both even, they have a common factor 2, which contradicts
our earlier statement that p and q have no common factors.
Arguing from our original assumption that x exists, we have obtained two
contradictory statements. Thus, our original assumption must be false. In other
&
words, no such x exists.
Problem 6 By imitating the above proof, show that there is no rational
number x such that x3 ¼ 2.
Since we want equations such as x2 ¼ 2 and x3 ¼ 2 to have solutions, we
must introduce new
We
denote
pffiffiffinumbers
pffiffiffi which are not rational
ffi2
ffiffiffi3 these
pffiffinumbers.
p
new numbers by 2 and 3 2, respectively; thus
2 ¼ 2 and p3ffiffi2ffi p¼
ffiffiffiffiffi2. Of
course, we must introduce many other new numbers, such as 3, 5 11, and
m
so on. Indeed, it can be shown
ffiffiffi that, if m, n are natural numbers and x ¼ n has
p
m
no integer solution, then n cannot be rational. A number which is not rational
is called irrational.
There are many other mathematical quantities which cannot be described
exactly by rational numbers. For example, the number p which denotes the area
of a disc of radius 1 (or half the length of the perimeter of such a disc) is
irrational, as is the number e.
The case m ¼ 2 is treated in
Exercise 5 for this section
in Section 1.6.
Lambert proved that p is
irrational in 1770.
1: Numbers
6
pffiffiffi
It is natural to ask whether irrational numbers, such as 2 and p, can
be represented as decimals. Using your calculator, you can check that
(1.41421356)2 is p
very
approffiffiffi close to 2, and so 1.41421356 is a very
pffiffigood
ffi
ximate value for 2. But is there a decimal which represents 2 exactly? If
such a decimal exists, then it cannot be recurring, because all the recurring
decimals correspond to rational numbers.
In fact, it is possible to represent all the irrational numbers mentioned so
far by non-recurring decimals. For example, there are non-recurring decimals
such that
pffiffiffi
2 ¼ 1:41421356 . . . and p ¼ 3:14159265 . . .:
It is also natural to ask whether non-recurring decimals, such as
0:101001000100001 . . .
and
In fact
ð1:41421356Þ2
¼1:9999999932878736:
pffiffiffi
We prove that 2 has a
decimal representation in
Section 1.5.
0:123456789101112 . . .;
represent irrational numbers. In fact, a decimal corresponds to a rational
number if and only if it is recurring; so a non-recurring decimal must correspond to an irrational number.
We may summarise this as:
recurring decimal , rational number
non-recurring decimal , irrational number
1.1.4
The real number system
Taken together, the rational numbers (recurring decimals) and irrational numbers (non-recurring decimals) form the set of real numbers, denoted by R.
As with rational numbers, we can determine which of two real numbers is
greater by comparing their decimals and noticing the first pair of corresponding digits which differ. For example
#
#
0:10100100010000 . . . < 0:123456789101112 . . .:
We now associate with each irrational number a point on the number-line.
For example, the irrational number x ¼ 0.123456789101112 . . . satisfies each
of the inequalities
0:1 < x < 0:2
0:12 < x < 0:13
0:123 < x < 0:124
..
.
We assume that there is a point on the number-line corresponding to x,
which lies to the right of each of the (rational) numbers 0.1, 0.12, 0.123 . . ., and
to the left of each of the (rational) numbers 0.2, 0.13, 0.124, . . ..
As usual, negative real numbers correspond to points lying to the left of 0;
and the ‘number-line’, complete with both rational and irrational points, is
called the real line.
When comparing decimals in
this way, we do not allow
either decimal to end in
recurring 9s.
1.1
Real numbers
7
There is thus a one–one correspondence between the points on the real line
and the set R of real numbers (or decimals).
We now state several properties of R, with which you will be already
familiar, although you may not have met their names before. These properties
are used frequently in Analysis, and we do not always refer to them explicitly
by name.
Order Properties of R
1. Trichotomy Property If a, b 2 R, then exactly one of the following
inequalities holds
a < b or a ¼ b or a > b:
2. Transitive Property
a<b
If a, b, c 2 R, then
and
3. Archimedean Property
such that
n > a:
b < c ) a < c:
If a 2 R, then there is a positive integer n
4. Density Property If a, b 2 R and a < b, then there is a rational number
x and an irrational number y such that
a < x < b and a < y < b:
Remark
The Archimedean Property is sometimes expressed in the following equivalent way: for any positive real number a, there is a positive integer n such
that 1n < a.
The following example illustrates how we can prove the Density Property.
Example 2
Find a rational number x and an irrational number y satisfying
a < x < b and a < y < b;
where a ¼ 0:123 and b ¼ 0:12345 . . ..
Solution
The two decimals
#
a ¼ 0:1233 . . .
#
and
b ¼ 0:12345 . . .
differ first at the fourth digit. If we truncate b after this digit, we obtain the
rational number x ¼ 0.1234, which satisfies the requirement that a < x < b.
To find an irrational number y between a and b, we attach to x a (sufficiently
small) non-recurring tail such as 010010001 . . . to give y ¼ 0.1234j010010001 . . ..
It is then clear that y is irrational (because its decimal is non-recurring) and that
&
a < y < b.
Problem 7 Find a rational number x and an irrational number y such
that a < x < b and a < y < b, where a ¼ 0:3 and b ¼ 0:3401:
The first three of these
properties are almost selfevident, but the Density
Property is not so obvious.
1: Numbers
8
Theorem 2 Density Property of R
If a, b 2 R and a < b, then there is a rational number x and an irrational
number y such that
a < x < b and
a < y < b:
You may omit the following
proof on a first reading.
Proof For simplicity, we assume that a, b 0. So, let a and b have decimal
representations
a ¼ a0 a1 a 2 a3 . . .
and
b ¼ b0 b1 b2 b3 . . .;
where we arrange that a does not end in recurring 9s, whereas b does not
terminate (this latter can be arranged by replacing a terminating representation
by an equivalent representation that ends in recurring 9s).
Since a < b, there must be some integer n such that
Here a0, b0 are non-negative
integers, and a1, b1, a2, b2, . . .
are digits.
a0 ¼ b0 ; a1 ¼ b1 ; . . .; an1 ¼ bn1 ; but an < bn :
Then x ¼ a0 a1a2a3 . . . an 1bn is rational, and a < x < b as required.
Finally, since x < b, it follows that we can attach a sufficiently small non&
recurring tail to x to obtain an irrational number y for which a < y < b.
Remark
One consequence of the Density Property is that between any two real numbers
there are infinitely many rational numbers and infinitely many irrational
numbers.
Problem 8 Prove that between any two real numbers a and b there are
at least two distinct rational numbers.
1.1.5
Arithmetic in R
We can do arithmetic with recurring decimals by first converting the decimals
to fractions. However, it is not obvious how to perform arithmetical operations
with non-recurring decimals.
pffiffiffi
Assuming that we can represent 2 and p by the non-recurring decimals
pffiffiffi
2 ¼ 1:41421356 . . . and p ¼ 3:14159265 . . .;
pffiffiffi
pffiffiffi
can we also represent the sum 2 þ p and the product 2 p as decimals? Indeed, what do we mean by the operations of addition and multiplication when non-recurring decimals (irrationals) are involved, and do
these operations satisfy the same properties as addition and multiplication
of rationals?
It would take many pages to answer these questions fully. Therefore, we
shall assume that it is possible to define all the usual arithmetical operations
with decimals, and that they do satisfy the usual properties. For definiteness,
we now list these properties.
A proof of the previous
remark would involve ideas
similar to those involved in
tackling Problem 8.
1.2
Inequalities
9
Arithmetic in R
Addition
A1 If a, b 2 R, then
a þ b 2 R.
Multiplication
M1 If a, b 2 R, then
a b 2 R.
A2
If a 2 R, then
a þ 0 ¼ 0 þ a ¼ a.
M2
If a 2 R, then
a 1 ¼ 1 a ¼ a.
A3
If a 2 R, then there is a
number a 2 R such that
a þ (a) ¼ ( a) þ a ¼ 0.
M3
If a 2 R {0}, then there is a
number a1 2 R such that
a a1 ¼ a1 a ¼ 1.
A4
If a, b, c 2 R, then
(a þ b) þ c ¼ a þ (b þ c).
If a, b 2 R, then
a þ b ¼ b þ a.
M4
If a, b, c 2 R, then
(a b) c ¼ a (b c).
If a, b 2 R, then
a b ¼ b a.
A5
D
M5
If a, b, c 2 R, then a (b þ c) ¼ a b þ a c.
CLOSURE
IDENTITY
INVERSES
ASSOCIATIVITY
COMMUTATIVITY
DISTRIBUTATIVITY
To summarise the contents of this table:
Properties A1–A5
R is an Abelian group under the operation of addition þ;
R {0} is an Abelian group under the operation of multiplication ;
These two group structures are linked by the Distributive Property.
Property D
It follows from the above properties that we can perform addition, subtraction
(where a b ¼ a þ (b)), multiplication and division (where ab ¼ a b1) in
R, and that these operations satisfy all the usual properties.
Furthermore, we shall assume that the set R contains the nth roots and
rational powers of positive real numbers, with their usual properties. In
Section 1.5 we describe one way of justifying the existence of nth roots.
1.2
Properties M1–M5
Any system satisfying the
properties listed in the table is
called a field. Both Q and R
are fields.
Inequalities
Much of Analysis is concerned with inequalities of various kinds; the aim of
this section and the next section is to provide practice in the manipulation of
inequalities.
1.2.1
Rearranging inequalities
The fundamental rule, upon which much manipulation of inequalities is based,
is that the statement a < b means exactly the same as the statement b a > 0.
This fact can be stated concisely in the following way:
Rule 1
For any a, b 2 R,
a < b , b a > 0.
Put another way, the inequalities a < b and b a > 0 are equivalent.
There are several other standard rules for rearranging a given inequality into
an equivalent form. Each of these can be deduced from our first rule above. For
Recall that the symbol ‘,’
means ‘if and only if’ or
‘implies and is implied by’.
1: Numbers
10
example, we obtain an equivalent inequality by adding the same expression to
both sides.
Rule 2
For any a, b, c 2 R,
a < b , a þ c < b þ c.
Another way to rearrange an inequality is to multiply both sides by a non-zero
expression, making sure to reverse the inequality if the expression is
negative.
For example
Rule 3
For any a, b 2 R and any c > 0,
For any a, b 2 R and any c < 0,
a < b , ac < bc;
a < b , ac > bc.
253 , 20530 ðc ¼ 10Þ;
253 , 20 > 30
ðc ¼ 10Þ:
Sometimes the most effective way to rearrange an inequality is to take reciprocals. However, in this case, both sides of the inequality should be positive,
and the direction of the inequality has to be reversed.
For example
Rule 4 (Reciprocal Rule)
For any positive a, b 2 R, a5b , 1a > 1b :
Some inequalities can be simplified only by taking powers. However, in order
to do this, both sides must be non-negative and must be raised to a positive
power.
Rule 5 (Power Rule)
For any non-negative a, b 2 R, and any p > 0,
1
2 < 4 , ð¼ 0:5Þ
2
1
> ð¼ 0:25Þ:
4
For example
a < b , a p < b p.
1
4 < 9 , 42 ð¼ 2Þ
1
For positive integers p, Rule 5 follows from the identity
bp ap ¼ ðb aÞ bp1 þ bp2 a þ þ bap2 þ ap1 ;
thus, since the right-hand bracket is positive, we have
< 92 ð¼ 3Þ:
We shall discuss the meaning
of non-integer powers in
Section 1.5.
b a > 0 , bp ap > 0;
which is equivalent to our desired result.
Remark
There are corresponding versions of Rules 1–5 in which the strict inequality
a < b is replaced by the weak inequality a b.
For example,
b3 a3 ¼ ðb aÞ
2
b þ ba þ a2 :
Problem 1 State (without proof) the versions of Rules 1–5 for weak
inequalities.
We shall give one more rule for rearranging inequalities in Sub-section 1.2.3.
1.2.2
Solving inequalities
Solving an inequality involving an unknown real number x means determining those values of x for which the given inequality holds; that is, finding
the solution set of the inequality. We can often do this by rewriting the
inequality in an equivalent, but simpler form, using the rules given in the
last sub-section.
The solution set is the set of
those values of x for which the
inequality holds.
1.2
Inequalities
11
In this activity we frequently use the usual rules for the sign of a product, and
the fact that the square of any real number is non-negative. Also, we need to
remember the difference between the logical statements: ‘implies’, ‘is implied
by’ and ‘implies and is implied by’.
Example 1
þ
þ
þ
þ
x3
Solve the inequality xþ2
xþ4 > 2x1 :
Solution We rearrange this inequality to give a somewhat simpler inequality,
using Rule 1
xþ2
x3
xþ2 x3
>
,
>0
x þ 4 2x 1
x þ 4 2x 1
x2 þ 2x þ 10
>0
,
ðx þ 4Þð2x 1Þ
,
It is a common strategy to
bring all terms to one side.
We bring everything to a
common denominator.
ð x þ 1Þ 2 þ 9
> 0:
ðx þ 4Þð2x 1Þ
Here we complete the square
in the numerator, since we
cannot factorise it.
Now, the numerator is always positive. The denominator vanishes when
x ¼ 4 or x ¼ 12. By examining separately the sign of the denominator when
x < 4, 4 < x < 12 and x > 12, we can deduce that the last fraction is positive
precisely when x < 4 or x > 12. Hence the solution set of the original
inequality is
xþ2
x3
>
x:
&
¼ ð1; 4Þ [ 12 ; 1 :
x þ 4 2x 1
Example 2
Solution
Solve the inequality 2x21þ 2 < 14 :
Since 2x2 þ 2 > 0, we have
1
1
< , 2x2 þ 2 > 4
þ2 4
, x2 þ 1 > 2
2x2
2
,x 1>0
, ðx 1Þðx þ 1Þ > 0:
ðby Rule 4Þ
ðby Rule 3Þ
ðby Rule 1Þ
This last inequality holds precisely when x < 1 or x > 1. It follows that the
solution set of the original inequality is
1
1
x: 2
<
¼ ð1; 1Þ [ ð1; 1Þ:
&
2x þ 2 4
Problem 2 Use each of the following expressions to write down an
inequality with the given expression on its left-hand side which is equivalent to the inequality x > 2:
(a) x þ 3;
Problem 3
(a)
This is because the final
displayed inequality is
equivalent to the inequality
we are solving. The logical
implication symbols between
the displayed inequalities
were all ‘implies and is
implied by’.
4xx2 7
x2 1
(b) 2 x;
(c) 5x þ 2;
(d)
Solve the following inequalities:
3;
(b) 2x2 ðx þ 1Þ2 :
1
5xþ2.
Here we factorise the lefthand side of the inequality
to examine the signs of its
factors.
1: Numbers
12
Warning Great care is needed when solving inequalities which involve
rational powers. In particular, when applying Rule 5 both sides of the inequality must be non-negative.
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Example 3 Solve the inequality 2x þ 3 > x.
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Solution The expression 2x þ 3 is defined only when
2x þ
3 0; that is,
when x 32. Hence we need only consider those x in 3
,
1
.
2
We
can
obtain
an
equivalent
inequality
by
squaring,
provided
that both
pffiffiffiffiffiffiffiffiffiffiffiffiffi
2x þ 3 and x are non-negative. Thus, for x 0, we obtain
pffiffiffiffiffiffiffiffiffiffiffiffiffi
2x þ 3 > x , 2x þ 3 > x2
ðby Rule 5; with p ¼ 2Þ
Note that, for the moment, we
are examining only those x for
which x 0.
, x2 2x 3 < 0
, ðx 3Þðx þ 1Þ < 0:
So the part of the solution set in [0, 1) is [0, 3).
pffiffiffiffiffiffiffiffiffiffiffiffiffi
We now examine
those x for which 32 x < 0. For such x, 2x þ 3 0
pffiffiffiffiffiffiffiffiffiffiffiffiffi
and x < 0, so that 2x þ 3 ð 0Þ > x, for all such x. It follows that all these x,
namely the set ½ 32 , 0Þ, belong to the solution set too.
Combining these results, the solution set of the original inequality is
pffiffiffiffiffiffiffiffiffiffiffiffiffi
3
x : 2x þ 3 > x ¼ ; 0 [ ½0; 3Þ
2
3
&
¼ ;3 :
2
Problem 4
1.2.3
Solve the inequality
Notice the use of the
Transitive Property here.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 2 > x:
Inequalities involving modulus signs
We now turn our attention to inequalities involving the modulus, or absolute value, of a real number. Recall that, if a 2 R, then its modulus jaj is
defined by
a; if a 0,
j aj ¼
a; if a < 0.
y
y= x
It is often useful to think of jaj as the distance along the real line from 0 to a.
x
For example
j3j ¼ j3j ¼ 3:
In the same way, ja bj is the distance along the real line from 0 to a b,
which is the same as the distance from a to b.
We sometimes write
ja bj ¼ dða; bÞ:
Notice also that ja þ bj ¼ ja (b)j is the distance from a to b.
For example, the distance
from 2 to 3 is
jð2Þ 3j ¼ j5j ¼ 5:
1.2
Inequalities
13
We now list some basic properties of the modulus, which follow immediately
from the definition:
Properties of the modulus
For example, if a ¼ 2, b ¼ 1,
then:
For any real numbers a and b:
1. jaj 0, with equality if and only if a ¼ 0;
2. –jaj a jaj;
1. j2j > 0;
2. j2j 2 j2j;
3. jaj2 ¼ a2;
3. j 2j2 ¼ ( 2)2;
4. j(2) 1j ¼ j1 (2)j;
4. ja bj ¼ jb aj;
5. jabj ¼ jaj jbj.
5. j(2) 1j ¼ j 2j j1j.
There is a basic rule for rearranging inequalities involving modulus signs:
Rule 6
For any real numbers a and b, where b > 0: jaj < b , b < a < b.
Note that, in a similar way
jaj b , b a b:
Also, it is often possible, and sometimes easier, to use Rule 5 with p ¼ 2 than to
use Rule 6. The following example illustrates the use of both rules.
Example 4
Solution
Solve the inequality jx 2j < 1.
Using Rule 6, we obtain
We take a ¼ x 2, b ¼ 1 in
Rule 6.
jx 2j < 1 , 1 < x 2 < 1
, 1 < x < 3:
So the solution set of the original inequality is
fx: jx 2j < 1g ¼ ð1; 3Þ:
Alternatively, using Rule 5 (with p ¼ 2), we obtain
jx 2j < 1 , ðx 2Þ2 < 1
For jx 2j2 ¼ (x 2)2.
2
, x 4x þ 3 < 0
, ðx 1Þðx 3Þ < 0:
Again, this shows that the required solution set is (1, 3).
Example 5
Solution
&
Solve the inequality jx 2j jx þ 1j.
Using Rule 5 (with p ¼ 2), we obtain
jx 2j jx þ 1j , ðx 2Þ2 ðx þ 1Þ2
, x2 4x þ 4 x2 þ 2x þ 1
, 3 6x
1
, x:
2
So the solution set of the original inequality is
1
fx: jx 2j jx þ 1jg ¼ ; 1 :
2
&
The inequalities in Examples 4 and 5 can easily be interpreted geometrically.
In Example 4, the inequality jx 2j < 1 holds when the distance from x to 2
is strictly less than 1. So it holds for all points on either side of 2 at a distance
less than 1 from 2 namely, in the open interval (1, 3).
1: Numbers
14
In Example 5, the inequality jx 2j jx þ 1j holds when the distance from x
to 2 is less than or equal to the distance from x to 1, since jx þ 1j ¼ jx (1)j.
The mid-point of 2 and 1 (that is, the point x where the distance from x to 2
equals the distance from x to 1) is 12. So the inequality holds when x lies
in [12 , 1).
Some good ideas when tackling problems involving inequalities of these
types are:
use your geometrical intuition, where possible, to give yourself an idea of
the sets involved;
test one or two values of x in your final solution set to see if they are valid –
this often detects errors in manipulating inequality signs!
Problem 5
Solve the following inequalities:
2
(a) j2x 13j < 5;
1.3
(b) jx 1j 2jx þ 1j.
Proving inequalities
In this section we show you how to prove inequalities of various types. We
shall use the rules for rearranging inequalities given in Section 1.2, and also use
other rules which enable us to deduce new inequalities from old. We have
already met the first rule in Section 1.1, where it was called the Transitive
Property of R.
Transitive Rule
a <b
and
b < c ) a < c.
We use the Transitive Rule when we want to prove that a < c, and we know
that a < b and b < c.
The following rules are also useful:
Combination Rules
If a < b and c < d, then:
Sum Rule
a þ c < b þ d;
Product Rule ac < bd
(provided that a,c 0).
There are also weak and weak/strict versions of the Transitive Rule and
Combination Rules, which we will ask you to work out and use as they arise.
For example, since 2 < 3 and
4 < 5, then
2 þ 4 < 3 þ 5;
2 4 < 3 5:
For example, if a < b and
c d, then
aþc<bþd
Remark
It is important to appreciate that the Transitive Rule and the Combination
Rules have a different nature from Rules 1–6 in Section 1.2. Rules 1–6 tell you
how to rearrange inequalities into equivalent forms, whereas the Transitive
Rule and the Combination Rules enable you to deduce new inequalities which
are not equivalent to the old ones.
and
ac < bd; provided a; c > 0:
1.3
Proving inequalities
1.3.1
15
The Triangle Inequality
If a and b are both vectors in R 2, then the vector a þ b is obtained from the
‘parallelogram construction’:
b
Here we discuss R 2, rather
than R, simply because the
argument is then
geometrically clearer.
In the ‘parallelogram
construction’, you draw the
vector a from the origin to
some point, then the vector b
from that point to a final point.
The vector a þ b is then the
vector from the origin to that
final point.
a+b
a
By elementary geometry, the length of any side of a triangle is less than or
equal to the sum of the lengths of the other two sides. In the special case of R,
when all the vectors lie on a line, this can be interpreted as the Triangle
Inequality, which involves the absolute value of the real numbers a, b and a þ b.
Triangle Inequality
For example, with a ¼ 1 and
b ¼ 3:
If a, b 2 R, then:
1. ja þ bj jaj þ jbj;
2. ja bj jaj jbj (the ‘reverse form’ of the Triangle Inequality).
Proof
1. j1 þ 3j j1j þ j3j;
2. j(1) 3j j j1 j j3j j.
In order to prove part 1, we use Rule 5, with p ¼ 2
ja þ bj jaj þ jbj , ða þ bÞ2 ðjaj þ jbjÞ2
, a2 þ 2ab þ b2 a2 þ 2jajjbj þ b2
, 2ab 2jajjbj:
Remember that jaj2 ¼ a2 :
The final inequality is certainly true for all a, b 2 R, and so the first
inequality must also be true for all a, b 2 R. Hence we have proved part 1.
We prove part 2 by using the same method
ja bj jaj jbj , ða bÞ2 ðjaj jbjÞ2
, a2 2ab þ b2 a2 2jajjbj þ b2
, 2ab 2jajjbj
, 2ab 2jajjbj;
which is again true for all a, b 2 R.
&
Remarks
1. Although we have used double-headed arrows here, the actual proof
requires only the arrows going from right to left. For example, in the
proof of part 1 the important implication is
ja þ bj jaj þ jbj ( 2ab 2jajjbj:
2. Part 1 of the Triangle Inequality can also be proved by using Rule 6, as
follows. By Rule 6
ja þ bj jaj þ jbj , ðjaj þ jbjÞ a þ b ðjaj þ jbjÞ: (1)
Part 2 is sometimes called the
‘cunning form’ or ‘backwards
form’ of the Triangle
Inequality.
1: Numbers
16
Now, we know that |a| a |a| and |b| b |b|; so, by the Sum Rule
ðjaj þ jbjÞ a þ b ðjaj þ jbjÞ:
It follows that the left-hand inequality in (1) must also hold, as required.
3. An obvious modification of the proof in Remark 2 shows that the following
more general form of the Triangle Inequality also holds:
Triangle Inequality for n terms For any real numbers a1, a2, . . ., an
ja1 þ a2 þ þ an j ja1 j þ ja2 j þ þ jan j:
The following example is a typical application of the Triangle Inequality.
Example 1 Use the Triangle Inequality to prove that:
(a) jaj 1 ) j3 þ a3j 4;
(b) jbj < 1 ) j3 bj > 2.
Solution
(a) Suppose that jaj 1. The Triangle Inequality then gives
3 þ a3 j3j þ a3
¼ 3 þ jaj3
3þ1
ðsince jaj 1Þ
Note the use of the Transitive
Rule here.
¼ 4:
(b) Suppose that jbj < 1. The ‘reverse form’ of the Triangle Inequality then
gives
j3 bj j3j jbj
¼ 3 jbj
3 jbj:
Now jbj < 1, so that jbj > 1. Thus
3 jbj > 3 1
¼ 2;
Again, we use the Transitive
Rule.
and we can then deduce from the previous chain of inequalities that
&
j3 bj > 2, as desired.
Remarks
1. The results of Example 1 can also be stated in the form:
(a) j3 þ a3j 4, for jaj 1;
(b) j3 bj > 2, for jbj < 1.
2. The reverse implications
3 þ a3 4 ) jaj 1
and
j 3 bj > 2 ) j bj < 1
are FALSE. For example, try putting a ¼ 32 and b ¼ 2!
Problem 1
Use the Triangle Inequality to prove that:
(b) jbj < 1 ) b3 1 > 7.
(a) jaj 12 ) ja þ 1j 32;
2
8
1.3
Proving inequalities
1.3.2
17
Inequalities involving n
In Analysis we often need to prove inequalities involving an integer n. It is a
common convention in mathematics that the symbol n is used to denote an
integer (frequently a natural number).
It is often possible to deal with inequalities involving n by using the
rearrangement rules given in Section 1.2. Here is such an example.
Example 2
Solution
Prove that 2n2 ðn þ 1Þ2 ;
for n 3:
Rearranging this inequality into an equivalent form, we obtain
2n2 ðn þ 1Þ2 , 2n2 ðn þ 1Þ2 0
, n2 2n 1 0
n
1
2
3
4
2n2
(n þ 1)2
2
4
8
9
18
16
32
25
, ðn 1Þ2 2 0 ðby ‘completing the square’Þ
, ðn 1Þ2 2:
This final inequality is clearly true for n 3, and so the original inequality
&
2n2 (n þ 1)2 is true for n 3.
Remarks
1. In Problem 3 of Section
2x2 (x þ 1)2;
pffiffiffiyou to solve
pffiffiffi the inequality
1.2, we asked
its solution set was 1; 1 2 [ 1 þ 2; 1 . In Example 2, above,
we found those natural numbers n lying in this solution set.
2. An alternative solution to Example 2 is as follows
nþ1 2
2
2
ðby Rule 3Þ
2n ðn þ 1Þ , 2 n
pffiffiffi
1
1
ðby Rule 5, with p ¼ Þ;
, 21þ
n
2
and this final inequality certainly holds for n 3.
Problem 2
1.3.3
Prove that n23nþ2 < 1; for n > 2.
More on inequalities
We now look at a number of inequalities and methods for proving inequalities
that will be useful later on.
2
Example 3 Prove that ab a þ2 b , for a, b 2 R.
Solution We tackle this inequality using the various rearrangement rules and
a chain of equivalent inequalities until we obtain an inequality that we know
must be true
aþb 2
a2 þ 2ab þ b2
ab , ab 2
4
, 4ab a2 þ 2ab þ b2
, 0 a2 2ab þ b2
, 0 ð a bÞ 2 :
This has the following
geometric interpretation: The
area of a rectangle with
sides of length a and b is
less than or equal to the area
of a square with sides of
length aþb
2 .
b
a
1: Numbers
18
This final inequality is certainly true, since
2all squares are non-negative. It
follows that the original inequality ab aþb
is also true, for a, b 2 R. &
2
Remark
pffiffiffiffiffi
In the form ab aþb
2 this
inequality is sometimes called
the Arithmetic–Geometric
Mean Inequality for a, b.
A close examination
of the above chain of equivalent statements shows that in
2
fact ab ¼ aþb
if
and
only if a ¼ b.
2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffi Problem 3 Prove that aþb
a2 þ b2 ; for a, b 2 R.
2
pffiffiffi
Problem 4 Suppose that a > 2. Prove the following inequalities:
2
(a) 12 a þ 2a < a;
(b) 12 a þ 2a > 2:
Hint: In part (b), use the result of Example 3 and the subsequent remark.
Example 4
Prove that
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2 a þ b; for a; b 0:
Solution We tackle this inequality using the various rearrangement rules and
a chain of equivalent inequalities until we obtain an inequality that we know
must be true
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ b2 a þ b , a2 þ b 2 ð a þ bÞ 2
This has the following
geometric interpretation: The
length of the hypotenuse of a
right-angled triangle whose
other sides are of lengths
a and b is less than or equal to
the sum of the lengths of those
two sides.
, a2 þ b2 a2 þ 2ab þ b2
a2 + b2
, 0 2ab:
This final inequality
is certainly
true, since a, b 0. It follows that the
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
&
original inequality a2 þ b2 a þ b is also true, for a, b 0.
Problem
p
ffiffiffi pffiffiffi 5 Use the result of Example 4 to prove that
c þ d; for c; d 0:
pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Example 5 Prove that a b ja bj, for a, b 0.
a
pffiffiffiffiffiffiffiffiffiffiffi
cþd Solution Notice first that interchanging the roles of a and b leaves the
inequality unaltered. It follows that it is sufficient to prove the inequality
under the assumption that a b.
pffiffiffi pffiffiffi
So, assume that a b. Then we know that a b and ja bj ¼ a b:
Hence
pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffi pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi
a b ja bj , a b a b
pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi
, a a b þ b:
This final inequality is certainly true, and is obtained from the result of
Problem 5 by simply substituting a and bffi in place of d. It
pbffiffiffi in place
pffiffiffi ofpcffiffiffiffiffiffiffiffiffiffiffiffiffi
follows that the original inequality a b ja bj is also true, for
&
a, b 0.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi pffiffiffi
Problem 6 Prove that a þ b þ c a þ b þ c; for a; b; c 0:
We often use the Binomial Theorem and the Principle of Mathematical
Induction (see Appendix 1) to prove inequalities.
Example 6 Prove the following inequalities, for n 1:
1
(a) 2n 1 þ n;
(b) 2n 1 þ 1n :
b
This will simplify the details
of our chain of inequalities.
We avoid one modulus as a
result of our simplifying
assumption!
Never be ashamed to utilise
every tool at your disposal!
(Why do the same work
twice?)
n
n
2
1þn
1
2
3
4
2
2
4
3
8
4
16
5
1.3
Proving inequalities
Solution
19
n
(a) By the Binomial Theorem for n 1
nð n 1Þ 2
x þ þ xn
ð1 þ xÞn ¼ 1 þ nx þ
2!
1 þ nx;
for x 0:
Then, if we substitute x ¼ 1 in this last inequality, we get
1 2
1
2n
1 þ 1n
3
4
2 1.41 1.26 1.19
2 1.5 1.33 1.25
We decrease the sum by
omitting subsequent
non-negative terms.
2n 1 þ n; for n 1:
(b) We start by rewriting the required result in an equivalent form
1
1 n
1
n
2 1þ ,2 1þ
ðby the Power RuleÞ:
n
n
Now, if we substitute x ¼ 1n in the Binomial Theorem for (1 þ x)n, we get
n
1 n
1
nð n 1Þ 1 2
1
1þ
¼1þn
þ þ
þ
n
n
2!
n
n
1 þ 1 ¼ 2:
n
Since the inequality 2 1 þ 1n is true, it follows that the original
1
&
inequality 2n 1 þ 1n, for n 1, is also true, as required.
Problem 7
We decrease the sum by
omitting all but the first two
terms.
n
1
Prove the inequality 1 þ 1n 52 2n
; for n 1.
Hint: consider the first three terms in the binomial expansion.
Example 7
Solution
Prove that 2n n2, for n 4.
Let P(n) be the statement
PðnÞ : 2n n2 :
First we show that P(4) is true: 24 42.
STEP 1 Since 24 ¼ 16 and 42 ¼ 16, P(4) is certainly true.
STEP 2 We now assume that P(k) holds for some k 4, and deduce that
P(k þ 1) is then true.
So, we are assuming that 2k k2. Multiplying this inequality by 2
we get
2kþ1 2k2 ;
so it is therefore sufficient for our purposes to prove that 2k2 (k þ 1)2.
Now
2k2 ðk þ 1Þ2 , 2k2 k2 þ 2k þ 1
, k2 2k 1 0 ðby ‘completing the square’Þ
, ðk 1Þ2 2 0:
This last inequality certainly holds for k 4, and so 2kþ1 (k þ 1)2
also holds for k 4.
In other words: P(k) true for some k 4 ) P(k þ 1) true.
It follows, by the Principle of Mathematical Induction, that 2n n2 , for
&
n 4.
Problem 8
Prove that 4n > n4, for n 5.
n
n
2
n2
1
2
3
4
5
2
1
4
4
8
9
16
16
32
25
This assumption is just P(k).
Since P(k þ 1) is:
2kþ1 (k þ 1)2.
1: Numbers
20
Three important inequalities in Analysis
Our first inequality, called Bernoulli’s Inequality, will be of regular use in later
chapters.
Theorem 1 Bernoulli’s Inequality
For any real number x 1 and any natural number n, (1 þ x)n 1 þ nx.
Remark
The value of this result will
come from making suitable
choices of x and n for
particular purposes.
In part (a) of Example 6, you saw that (1 þ x)n 1 þ nx, for x > 0 and n a
natural number. Theorem 1 asserts that the same result holds under the weaker
assumption that x 1.
Proof
Let P(n) be the statement
n
PðnÞ : ð1 þ xÞ 1 þ nx;
for x 1:
STEP 1 First we show that P(1) is true: (1 þ x)1 1 þ x. This is obviously
true.
STEP 2 We now assume that P(k) holds for some k 1, and prove that P(k þ 1)
is then true.
So, we are assuming that ð1 þ xÞk 1 þ kx; for x 1. Multiplying
this inequality by (1 þ x), we get
We prove the result using
Mathematical Induction.
This assumption is P(k).
This multiplication is valid
since ð1 þ xÞ 0.
ð1 þ xÞkþ1 ð1 þ xÞð1 þ kxÞ
¼ 1 þ ðk þ 1Þx þ kx2
1 þ ðk þ 1Þx:
Thus, we have ð1 þ xÞkþ1 1 þ ðk þ 1Þx; in other words the statement
P(k þ 1) holds.
So, P(k) true for some k 1 ) P(k þ 1) true.
We decrease the expression
if we omit the final nonnegative term.
It follows, by the Principle of Mathematical Induction, that ð1 þ xÞn &
1 þ nx, for x 1, n 1:
1
Problem 9 By applying Bernoulli’s Inequality with x ¼ ð2nÞ
, prove
1
1
that 2n 1 þ 2n1, for any natural number n.
You saw in part (b) of
1
Example 6 that 2n 1 þ 1n :
Our second inequality is of considerable use in various branches of
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffi Analysis. In Problem 3 you proved that aþb
a2 þ b2 , for a, b 2 R. We
2
can rewrite this inequality in the equivalent form ða þ bÞ2 2ða2 þ b2 Þ or
ða þ bÞ2 ða2 þ b2 Þð12 þ 12 Þ. The Cauchy–Schwarz Inequality is a generalisation of this result to 2n real numbers.
Theorem 2 Cauchy–Schwarz Inequality
For any real numbers a1 ; a2 ; . . .; an and b1 ; b2 ; . . .; bn ; we have
ða1 b1 þ a2 b2 þ þ an bn Þ2
a21 þ a22 þ þ a2n b21 þ b22 þ þ b2n :
We give the proof of
Theorem 2 at the end of the
sub-section.
1.3
Proving inequalities
Problem 10
21
Use Theorem 2 to prove that
for any positive real numbers
a1 , a2 , . . ., an , then ða1 þ a2 þ þ an Þ
1
a1
þ
1
a2
þ þ
1
an
For example, with n ¼ 3
2
n :
ð1 þ 2 þ 3Þ
1 1 1
11
þ þ ¼6
1 2 3
6
Our final result also has many useful applications. In Example 3 you proved
2
that ab a þ2 b , for a, b 2 R; it follows that, if a and b are positive, then
¼ 11 32 :
1
ðabÞ2 aþb
2 . The Arithmetic Mean–Geometric Mean Inequality is a generalisation of this result for two real numbers to n real numbers.
Theorem 3 Arithmetic Mean–Geometric Mean Inequality
For any positive real numbers a1, a2, . . ., an, we have
1
a1 þ a2 þ þ a n
:
ð a1 a2 . . . an Þ n n
Problem 11 Use Theorem 3 with the n þ 1 positive numbers 1, 1 þ 1n,
1 þ 1n, . . ., 1 þ 1n to prove that, for any positive integer n
nþ1
1 n
1
1þ
1þ
:
n
nþ1
We give the proof of
Theorem 3 at the end of the
sub-section.
For example, with n ¼ 3
1 3 64
¼ 2:37 . . .
1þ
¼
3
27
5 1þ
1
4
4
¼
625
¼ 2:44 . . .
256
Proofs of Theorems 2 and 3
You may omit these proofs at
a first reading.
Theorem 2 Cauchy–Schwarz Inequality
For any real numbers a1, a2, . . ., an and b1, b2, . . ., bn, we have
ða1 b1 þ a2 b2 þ þan bn Þ2 a21 þ a22 þ þ a2n b21 þ b22 þ þ b2n :
Proof If all the as are zero, the result is obvious; so we need only examine
the case when not all the as are zero. It follows that, if we denote the
Pn
2
2
2
2
sum
k¼1 ak ¼ a1 þ a2 þ þ an by A, then A > 0. Also, denote
Pn
2
2
2
2
k¼1 bk ¼ b1 þb2 þþbn
Pn
by B and k¼1 ak bk ¼ a1 b1 þ a2 b2 þþan bn by C.
Now, for any real number l, we have that ðlak þ bk Þ2 0, so that
n n
X
X
l2 a2k þ 2lak bk þ b2k ¼
ðlak þ bk Þ2 0;
k¼1
Note that we will use the notation to keep the argument
brief.
k¼1
which we may rewrite in the form
l2 A þ 2lC þ B 0:
But this inequality is equivalent to the inequality
ðlA þ C Þ2 þ AB C2 ; for any real number l:
Since A is non-zero, we may now choose l ¼ CA. It follows from the last
&
inequality that AB C2, which is exactly what we had to prove.
Remark
If not all the as are zero, equality can only occur if
n
P
k¼1
ðlak þ bk Þ2 ¼ 0; that is,
if all the numbers ak are proportional to all the numbers bk, 1 k n:
A is non-zero, by assumption,
so that 1/A makes sense.
1: Numbers
22
Theorem 3 Arithmetic Mean–Geometric Mean Inequality
For any positive real numbers a1, a2, . . ., an, we have
1
a1 þ a2 þ . . . þan
ð a1 a2 . . . an Þ n :
n
Proof
(2)
Since the ai are positive, we can rewrite (2) in the equivalent form
1
ð a1 a2 . . . an Þ n
1:
ða1 þ a2 þ þan Þ=n
(3)
Now, replacing each term ai by lai for any non-zero number l does not alter
the left-hand side of the inequality (3). It follows that it is sufficient to prove the
inequality (2) in the special case when the product of the terms ai is 1. Hence it
is sufficient to prove the following statement P(n) for each natural number n:
P(n): For any positive real numbers ai with a1a2 . . . an ¼ 1, then
a1 þ a2 þ þ an n.
First, the statement P(1) is obviously true.
Next, we assume that P(k) holds for some k 1, and prove that P(k þ 1) is
then true.
Now, if all the terms a1, a2, . . ., akþ1 are equal to 1, the result P(k þ 1)
certainly holds. Otherwise, at least two of the terms differ from 1, say a1 and a2,
such that a1 > 1 and a2 < 1. Hence
ða1 1Þ ða2 1Þ 0;
which after some manipulation we may rewrite as
a1 þ a 2 1 þ a1 a2 :
(4)
We denote the typical term by
ai rather than ak to avoid
confusion with a different use
of the letter k in the
Mathematical Induction
argument below.
We will prove this by
Mathematical Induction.
The argument is exactly the
same whichever two terms
actually differ from 1.
You should check this
yourself.
We are now ready to tackle P(k þ 1). Then
a1 þ a2 þ þ akþ1 1 þ a1 a2 þ a3 þ a4 þ þ akþ1
By (4).
k þ 1;
since we may apply the assumption that P(k) holds to the k quantities a1a2,
a3, a4, . . ., akþ1. This last inequality is simply the statement that P(k þ 1) is
indeed true.
It follows by the Principle of Mathematical Induction that P(n) holds for all
&
natural numbers n, and so the inequality (2) must also hold.
Remark
A careful examination of the proof of Theorem 3 shows that equality can only
occur if all the terms ai are equal.
1.4
1.4.1
Least upper bounds and greatest lower bounds
Upper and lower bounds
Any finite set {x1, x2, . . ., xn} of real numbers obviously has a greatest element
and a least element, but this property does not necessarily hold for infinite sets.
That is
ða1 a2 Þ a3 a4 . . . akþ1 ¼ 1
) ða1 a2 Þ þ a3 þ a4 þ þ akþ1 k:
1.4
Least upper bounds and greatest lower bounds
23
For example, the interval (0, 2] has greatest element 2, but neither of the sets
N ¼ {1, 2, 3, . . .} nor [0, 2) has a greatest element. However the set [0, 2) is
bounded above by 2, since all points of [0, 2) are less than or equal to 2.
Definitions A set E R is bounded above if there is a real number, M
say, called an upper bound of E, such that
x M; for all x 2 E:
If the upper bound M belongs to E, then M is called the maximum element
of E, denoted by max E.
Geometrically, the set E is bounded above by M if no point of E lies to the
right of M on the real line.
For example, if E ¼ [0, 2), then the numbers 2, 3, 3.5 and 157.1 are all upper
bounds of E, whereas the numbers 1.995, 1.5, 0 and 157.1 are not upper
bounds of E. Although it seems obvious that [0, 2) has no maximum element,
you may find it difficult to write down a formal proof. The following example
shows you how to do this:
Example 1 Determine which of the following sets are bounded above, and
which have a maximum element:
(a) E1 ¼ [0, 2);
(b) E2 ¼ f1n : n ¼ 1; 2; . . .g;
(c) E3 ¼ N.
Solution
(a) The set E1 is bounded above. For example, M ¼ 2 is an upper bound
of E1, since
x 2;
for all x 2 E1 :
However, E1 has no maximum element. For each x in E1, we have x < 2,
and so there is some real number y such that
x < y < 2;
by the Density Property of R.
Hence y 2 E1, and so x cannot be a maximum element.
(b) The set E2 is bounded above. For example, M ¼ 1 is an upper bound of E2,
since
1
1;
n
for all n ¼ 1; 2; . . .:
Also, since 1 2 E2
max E2 ¼ 1:
(c) The set E3 is not bounded above. For each real number M, there is a
positive integer n such that n > M, by the Archimedean Property of R.
Hence M cannot be an upper bound of E3.
&
This also means that E3 cannot have a maximum element.
Problem 1 Sketch the following sets, and determine which are bounded
above, and which have a maximum element:
(a) E1 ¼ (1, 1];
(b) E2 ¼ f1 1n : n ¼ 1; 2; . . .g;
2
(c) E3 ¼ {n : n ¼ 1, 2, . . .}.
For example, y can be of the
form 1.99. . . 9 or y ¼ 12(x þ 2).
2 is not a maximum element,
since 2 2
= E1.
1: Numbers
24
Similarly, we define lower bounds. For example, the interval (0, 2) is
bounded below by 0, since
0 x;
for all x 2 ð0; 2Þ:
However, 0 does not belong to (0, 2), and so 0 is not a minimum element of
(0, 2). In fact, (0, 2) has no minimum element.
Definitions A set E R is bounded below if there is a real number, m say,
called a lower bound of E, such that
m x;
for all x 2 E:
If the lower bound m belongs to E, then m is called the minimum element of
E, denoted by min E.
Geometrically, the set E is bounded below by m if no point of E lies to the
left of m on the real line.
Problem 2 Determine which of the following sets are bounded below,
and which have a minimum element:
(a) E1 ¼ (1, 1];
(b) E2 ¼ f1 1n : n ¼ 1; 2; . . .g;
2
(c) E3 ¼ {n : n ¼ 1, 2, . . .}.
The following terminology is also useful:
Definition
below.
A set E R is bounded if it is bounded above and bounded
For example, the set E2 ¼ f1 1n : n ¼ 1; 2; . . .g is bounded, but the sets
E1 ¼ (1, 1] and E3 ¼ {n2: n ¼ 1, 2, . . .} are not bounded.
Similar terminology applies to functions.
Definitions A function ƒ defined on an interval I R is said to:
be bounded above by M if f(x) M, for all x 2 I; M is an upper bound of ƒ;
be bounded below by m if f(x) m, for all x 2 I; m is a lower bound of ƒ;
have a maximum (or maximum value) M if M is an upper bound of ƒ and
f(x) ¼ M, for at least one x 2 I;
have a minimum (or minimum value) m if m is a lower bound of ƒ and
f(x) ¼ m, for at least one x 2 I.
Example 2 Let ƒ be the function defined by f (x) ¼ x2, x 2 12 , 3 . Determine
whether ƒ is bounded above or below, and any maximum or minimum value
of ƒ.
Solution First, ƒ is increasing on the interval 12 , 3Þ, so that since 12 x < 3 it
follows that 14 f ðxÞ < 9. Hence ƒ is bounded above and bounded below.
Next, since f 12 ¼ 14 and 14 is a lower bound for ƒ on the interval 12 , 3Þ, it
follows that ƒ has a minimum value of 14 on this interval.
Finally,
9 is an upper bound for ƒ on the interval 12 , 3Þ but there is no point
1
x in 2 , 3Þ for which ƒ(x) ¼ 9. So 9 cannot be a maximum of ƒ on the interval.
pffiffiffi pffiffiffi However, if y is any number in (8, 9) there is a number x > y in
y, 3 pffiffiffi 1 2
2 2, 3 2 , 3 such that ƒ(x) ¼ x > y, so that no number in (8, 9) will serve
Strictly speaking, M and m are
the upper bound and lower
bound of the image set
{ƒ(x): x 2 I}.
1
2
2√2
y
3
1.4
Least upper bounds and greatest lower bounds
25
as a maximum of ƒ on the interval. It follows that ƒ has no maximum value
&
on 12 , 3Þ.
Problem 3 Let ƒ be the function defined by f ð xÞ ¼ x12 ; x 2 ½3; 2Þ.
Determine whether ƒ is bounded above or below, and any maximum or
minimum value of ƒ.
1.4.2
Least upper bounds and greatest lower bounds
We have seen that the interval [0, 2] has a maximum element 2, but [0, 2) has
no maximum element. However, the number 2 is ‘rather like’ a maximum
element of [0, 2), because 2 is an upper bound of [0, 2) and any number less
than 2 is not an upper bound of [0, 2). In other words, 2 is the least upper
bound of [0, 2).
Definition A real number M is the least upper bound, or supremum, of a
set E R if:
1. M is an upper bound of E;
2. if M0 < M, then M0 is not an upper bound of E.
In this case, we write M ¼ sup E.
Part 1 says that M is an upper
bound.
Part 2 says that no smaller
number can be an upper
bound.
If E has a maximum element, max E, then sup E ¼ max E. For example,
the closed interval [0, 2] has least upper bound 2. We can think of the least
upper bound of a set, when it exists, as a kind of ‘generalised maximum
element’.
If a set does not have a maximum element, but is bounded above, then we may
be able to guess the value of its least upper bound. As in the case E ¼ [0, 2), there
may be an obvious ‘missing point’ at the upper end of the set. However it is
important to prove that your guess is correct. We now show you how to do this.
Example 3
Solution
Prove that the least upper bound of [0, 2) is 2.
We know that M ¼ 2 is an upper bound of [0, 2), because
x 2;
for all x 2 ½0; 2Þ:
To show that 2 is the least upper bound, we must prove that each number
M0 < 2 is not an upper bound of [0, 2). To do this, we must find an element
x in [0, 2) which is greater than M0 . But, if M0 < 2, then there is a real number
x such that
0
M <x<2
and also
0 < x < 2:
Since x 2 [0, 2), the number M0 cannot be an upper bound of [0, 2). Hence
&
M ¼ 2 is the least upper bound, or supremum, of [0, 2).
Although the conclusion of Example 3 may seem painfully obvious, we have
written out the solution in detail because it illustrates the strategy for determining the least upper bound of a set, if it has one.
For example, x can be of the
form 1.99 . . . 9 for a suitably
large number of digits, or it
can be 12 ðM 0 þ 2Þ since
M 0 < 12 ðM 0 þ 2Þ < 2:
1: Numbers
26
Strategy Given a subset E of R, to show that M is the least upper bound, or
supremum, of E, check that:
1. x M, for all x 2 E;
0
0
2. if M < M, then there is some x 2 E such that x > M .
GUESS
CHECK
the value of M, then
parts 1 and 2.
Notice that, if M is an upper bound of E and M 2 E, then part 2 is automatically satisfied, and so M ¼ sup E ¼ max E.
Example 4
Determine the least upper bound of E ¼ f1 n12 : n ¼ 1; 2; . . .g.
Solution We guess that the least upper bound of E is M ¼ 1. Certainly, 1 is
an upper bound of E, since
1
1 2 1; for n ¼ 1; 2; . . .:
n
To check part 2 of the strategy, we need to show that, if M0 < 1, then there
is some natural number n such that
1
1 2 > M0:
(1)
n
However
1
1
1 2 > M0 , 1 M0 > 2
n
n
1
,
< n2
ðsince 1 M 0 > 0Þ
1rffiffiffiffiffiffiffiffiffiffiffiffiffiffi
M0
1
1
,
>0
<n
ðsince
1 M0
1 M0
and n > 0Þ:
We can certainly choose n so that this final inequality holds, by the
Archimedean Property of R, and so we can choose n so that inequality (1) holds.
&
Hence 1 is the least upper bound of E.
That is, 1 ¼ sup E.
Remark
Although we used double-headed arrows in this solution, the actual proof
required only the implications going from right to left. In other words, the
proof uses only the fact that
rffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
1
0
1 2 >M (
< n:
n
1 M0
Problem 4 Determine sup E, if it exists, for each of the following sets:
(a) E1 ¼ ( 1, 1];
(b) E2 ¼ f1 1n : n ¼ 1, 2, . . .g;
2
(c) E3 ¼ {n : n ¼ 1, 2, . . .}.
Similarly, we define the notion of a greatest lower bound.
Definition A real number m is the greatest lower bound, or infimum, of
a set E R if:
1. m is a lower bound of E;
2. if m0 > m, then m0 is not a lower bound of E.
In this case, we write m ¼ inf E.
Part 1 says that m is a lower
bound.
Part 2 says that no larger
number can be a lower bound.
1.4
Least upper bounds and greatest lower bounds
27
If E has a minimum element, min E, then inf E ¼ min E. For example, the
closed interval [0, 2] has greatest lower bound 0. We can think of the greatest
lower bound of a set, when it exists, as a kind of ‘generalised minimum
element’.
The strategy for establishing that a number is the greatest lower bound of
a set is very similar to that for proving that a number is the least upper bound
of a set.
Strategy Given a subset E of R, to show that m is the greatest lower
bound, or infimum, of E, check that:
1. x m, for all x 2 E;
0
GUESS
CHECK
0
2. if m > m, then there is some x 2 E such that x < m .
the value of m, then
parts 1 and 2.
Notice that, if m is a lower bound of E and m 2 E, then part 2 is automatically
satisfied, and so m ¼ inf E ¼ min E.
Problem 5
Determine inf E, if it exists, for each of the following sets:
(a) E1 ¼ (1, 5];
(b) E2 ¼ fn12 : n ¼ 1; 2; . . .g.
Remarks
1. For any subset E of R, inf E sup E. This follows from the fact that, for
any x 2 E, we have inf E x sup E.
2. For any bounded interval I of R, let a be its left end-point and b its right
end-point. Then inf I ¼ a and sup I ¼ b.
Least upper bounds and greatest lower bounds
of functions
Similar terminology applies to bounds for functions.
Definitions Let ƒ be a function defined on an interval I R. Then:
A real number M is the least upper bound, or supremum, of ƒ on I if:
1. M is an upper bound of ƒ(I);
2. if M0 < M, then M0 is not an upper bound of ƒ(I).
In this case, we write M ¼ sup f or sup f or supf f ð xÞ : x 2 I g or sup f ð xÞ:
I
x2I
A real number m is the greatest lower bound, or infimum, of ƒ on I if:
1. m is a lower bound of ƒ(I);
2. if m0 > m, then m0 is not a lower bound of ƒ(I).
In this case, we write m ¼ inf f or inf f or inf f f ð xÞ : x 2 I g or inf f ð xÞ.
I
There are similar definitions
for the least upper bound and
the greatest lower bound of f
on a general set S in R.
These are really the
definitions for the least upper
bound or the greatest lower
bound of the set {ƒ(x): x 2 I}.
x2I
For
sup f is the least upper bound,
Notice, for instance, that:
if M is an upper bound for f on I, then sup f M,
if m is a lower bound for f on I, then inf f m.
I
I
I
and inf f is the greatest lower
I
bound, for f on I.
1: Numbers
28
The strategies for proving that M is the least upper bound or m the greatest
lower bound of ƒ on I are similar to the corresponding strategies for the least
upper bound or the greatest lower bound of a set E.
Strategies Let ƒ be a function defined on an interval I R. Then:
To show that m is the greatest lower bound, or infimum, of f on I, check
that:
1. f(x) m, for all x 2 I;
2. if m0 > m, then there is some x 2 I such that f(x) < m0 .
To show that M is the least upper bound, or supremum, of f on I, check that:
1. f(x) M, for all x 2 I;
2. if M0 < M, then there is some x 2 I such that f(x) > M0 .
Example 5 Let ƒ be the function defined by f ð xÞ ¼ x2 ; x 2 12 ; 3 . Determine
1 the least upper bound and the greatest lower bound of ƒ on 2 ; 3 .
Solution We have already seen that 9 is an upper bound for ƒ on 12 ; 3 , and
that no smaller number will serve
as
an upper bound. It follows that 9 must be
the least upper bound of ƒ on 12 ; 3 .
Similarly, we have already seen that 14 is the minimum value of ƒ on 12 ; 3 ;
it follows that 14 is the greatest lower bound of ƒ on 12 ; 3 , and this is actually
&
attained ðat the point 12Þ.
You saw this in Example 2.
Example 2.
Problem 6 Let ƒ be the function defined by f ð xÞ ¼ x12 ; x 2 ½1; 4Þ:
Determine the least upper bound and the greatest lower bound of ƒ on ½1, 4).
Remark
For any interval I of R, inf f sup f . This follows from the fact that, for
I
I
any x 2 I, we have inf f f ð xÞ sup f .
I
I
The least upper bound and the greatest
lower bound of a function on an
interval will be particularly significant in our later work on continuity and
integrability of functions.
1.4.3
Chapters 4 and 7.
The Least Upper Bound Property
In the examples in the previous sub-section, it was easy to guess the values of
sup E and inf E. At times, however, we shall meet sets for which these values
are not so easy to determine. For example, if
1 n
E¼
1þ
: n ¼ 1; 2; . . . ;
n
then it can be shown that E is bounded above by 3, but it is not easy to guess the
least upper bound of E.
In such circumstances, it is reassuring to know that sup E does exist, even
though it may be difficult to find. This existence is guaranteed by the following
fundamental result.
We will study this set closely
in Section 2.5.
1.4
Least upper bounds and greatest lower bounds
The Least Upper Bound Property of R Let E be a non-empty subset
of R. If E is bounded above, then E has a least upper bound.
We leave the proof of the Least Upper Bound Property of R to the next subsection. However, the Property itself is intuitively obvious. If the set E lies
entirely to the left of some number M, then you can imagine moving M steadily
to the left until you meet E. At this point, sup E has been reached.
The Least Upper Bound Property of R can be used to show that
pffiffiffi R does
include decimals which represent irrational numbers such as 2, as we
claimed in Section 1.1.
In Sections 1.2 and 1.3 we have taken for granted the existence of rational
powers and their properties, without giving formal definitions.
pffiffiffi How can we
supply these definitions? For example, how can we define 2 as a decimal?
Consider the set
E ¼ x 2 Q : x > 0; x2 < 2 :
29
The Least Upper Bound
Property of R is an example
of an existence theorem, one
which asserts that a real
number exists having a certain
property. Analysis contains
many such results which
depend on the Least Upper
Bound Property of R. While
these results are often very
general, and their proofs
elegant, they do not always
provide the most efficient
methods of calculating good
approximate values for the
numbers in question.
This is the p
setffiffiffi of positive rational numbers whose squares are less than 2.
Intuitively, 2 lies on the number lineptoffiffiffi the right of the numbers in E, but
‘only just’! In fact, we should expect 2 to be the least upper bound of E.
Certainly E has a least upper bound, by the Least Upper Bound Property,
because E is bounded above, by 1.5
pffiffiffifor example. Thus it seems likely that sup E
is the decimal representation of 2. But how can we prove that (sup E)2 ¼ 2?
We shall prove this in Section 1.5, once we have described how to do
arithmetic with real numbers (decimals).
Finally, note that there is a corresponding result about lower bounds.
The Greatest Lower Bound Property of R Let E be a non-empty subset
of R. If E is bounded below, then E has a greatest lower bound.
1.4.4
Proof of the Least Upper Bound Property
You may omit this proof at a
first reading.
We know that E is a non-empty set, and we shall assume for simplicity that E
contains at least one positive number. We also know that E is bounded above.
The following procedure gives us the successive digits in a particular decimal,
which we then prove to be the least upper bound of E.
Procedure to find a ¼ a0 a1a2 . . . ¼ sup E
Choose in succession:
the greatest integer a0 such that a0 is not an upper bound of E;
the greatest digit a1 such that a0 a1 is not an upper bound of E;
the greatest digit a2 such that a0 a1a2 is not an upper bound of E;
..
.
the greatest digit an such that a0 a1a2 . . . an is not an upper bound of E;
..
.
For example, if
E ¼ fx 2 Q : x > 0;
x2 < 2g;
then
a0 ¼ 1; since 12 < 2 < 22 ;
a0 a1 ¼ 1:4; since
1:42 < 2 < 1:52 ;
a0 a1 a2 ¼ 1:41; since
1:412 < 2 < 1:422 ;
..
.
1: Numbers
30
Thus, at the nth stage we choose the digit an so that:
a0 a1a2 . . . an is not an upper bound of E;
a0 a1 a2 . . . an þ 101n is an upper bound of E.
We now prove that the least upper bound of E is a ¼ a0 a1a2 . . ..
First, we have to prove that a is an upper bound of E. To do this, we prove
that, if x > a, then x 2
= E (this is equivalent to proving, that, if x 2 E, then x a).
We begin by representing x as a non-terminating decimal x ¼ x0 x1x2 . . ..
Since x > a, there is an integer n such that
a < x0 x1 x2 . . . xn :
Hence
x0 x1 x2 . . . xn a0 a1 a2 . . . an þ
1
;
10n
and so, by our choice of an, x ¼ x0 x1 x2 . . . xn is an upper bound of E. Since
x > x0 x1x2 . . . xn, we have that x 2
= E, as required.
Next, we have to show that, if x < a, then x is not an upper bound of E. Since
x < a, there is an integer n such that
x < a0 a1 a2 . . . an ;
and so x is not an upper bound of E, by our choice of an.
Thus we have proved that a is the least upper bound of E.
&
Remark
Notice that this proof does not use any arithmetical properties of the real
numbers but only their order properties, together with the arithmetical properties of rational numbers. In the next section, we use the Least Upper Bound
Property to define some of the arithmetical operations on R.
1.5
1.5.1
Manipulating real numbers
Arithmetic in R
At the end of Section 1.1 we discussed the decimals
pffiffiffi
2 ¼ 1:41421356 . . . and p ¼ 3:14159265 . . .;
and asked whether it is possible to add and multiply these numbers to obtain
another real number. We now explain how this can be done, using the Least
Upper Bound Property of R.
pffiffiffi
A natural way to obtain a sequence of approximations to the sum 2 þ p
is to truncate each of the above decimals, and form the sums of the
truncations.
If each of the decimals is truncated at the same decimal place, this gives a
sequence of approximations which is increasing:
pffiffiffi
Here we are assuming that 2
and p can be represented as
decimals.
1.5
Manipulating real numbers
31
pffiffiffi
2
p
pffiffiffi
2þp
1
1.4
3
3.1
4
4.5
1.41
3.14
4.55
1.414
1.4142
..
.
3.141
3.1415
..
.
4.555
4.5557
..
.
pffiffiffi
Intuitively, we should expect that the sum 2 þ p is greater than each of the
numbers in the right-hand column,
pffiffiffi but ‘only just’! To accord with our intuition,
therefore, we define the sum 2 þ p to be the least upper bound of the set of
numbers in the right-hand column; that is
pffiffiffi
2 þ p ¼ supf4; 4:5; 4:55; 4:555; 4:5557; . . .g:
To be sure that this definition makes sense,
pffiffiwe
ffi need to show that this set
is bounded above. But all the truncations of 2 are less than 1.5, and all the
truncations of p are less than, say, 4. Hence, all the sums in the right-hand
column are less than 1.5 þ 4 ¼ 5.5. So, by the Least Upper Bound Property, the
set of numbers
pffiffiffi in the right-hand column does have a least upper bound, and we
can define 2 þ p in this way.
This method can be used to define the sum of any pair of positive real
numbers.
Let us check that this method of adding decimals gives the correct answer
when we use it in a familiar case. Consider the simple calculation
1 2
þ ¼ 0:333 . . . þ 0:666 . . .:
3 3
Truncating each of these decimals and forming the sums, we obtain the set
f0; 0:9; 0:99; 0:999; . . .g:
The supremum of this set is, of course, the number 0.999. . . ¼ 1, which is the
correct answer.
Similarly, we canp
define
the product of any two positive real numbers. For
ffiffiffi
example, to define 2 p, we can form the sequence of products of their
truncations:
pffiffiffi
2
p
pffiffiffi
2p
1
1.4
3
3.1
3
4.34
1.41
1.414
3.14
3.141
4.4274
4.441374
1.4142
..
.
3.1415
..
.
4.4427093
..
.
We do not expect you to use
this method to add decimals!
1: Numbers
32
pffiffiffi
As before, we define 2 p to be the least upper bound of the set of numbers
in the right-hand column.
Similar ideas can be used to define the operations of subtraction and
division.
Thus we can define arithmetic with real numbers in terms of the familiar
arithmetic with rationals, using the Least Upper Bound Property of R.
Moreover, it can be proved that these operations in R satisfy all the usual
properties of a field.
1.5.2
We omit the details.
These properties were listed
in Sub-section 1.1.5.
The existence of roots
Just as we usually take for granted the basic arithmetical operations with real
numbers, so we usually assume that,pgiven
any positive real number a, there is
ffiffiffi
a unique positive real number b ¼ a such that b2 ¼ a. We now discuss the
justification for this assumption.
First, here is a geometrical justification. Given line segments of lengths 1
and a, we can construct a semi-circle with diameter a þ 1 as shown.
For each positive integer
pffiffiffin,
we can also construct n as
follows:
1
1
1
1
b
5
6
a
1
4
3
1
2
7
Using similar triangles, we see that
a b
¼ ;
b 1
and so
1
1
b2 ¼ a:
This shows that there should be a positive real number b such that b2 ¼ a,
so that the length of the vertical
line segment
pffiffiffi
pffiffiffi in the figure can be described
exactly by the expression a. But does b ¼ a exist exactly as a real number?
In fact it does, and a more general result is true.
Theorem 1 For each positive real number a and each integer n > 1, there
is a unique positive real number b such that
bn ¼ a:
ffiffiffi
p
n a. We also define
We
call this number b the nth root of a, and we write bp¼ffiffiffiffiffiffiffiffiffiffi
p
ffiffi
ffi
pffiffiffi
n
0, since 0n ¼ 0, and if n is odd we define n ðaÞ ¼ n a, since
0p¼
ffiffi
ffi
n
ð n aÞ ¼ a if n is odd.
Let us illustrate Theorem 1 with the special case a ¼ 2 and n ¼ 2. In this case,
Theorem 1 asserts the existence of a real number b such that b2 ¼ 2. In other
pffiffiffi
words, it asserts the existence of a decimal b which can be used to define 2
precisely.
Here is a direct proof of Theorem 1 in this special case. We choose the
numbers 1, 1.4, 1.41, 1.414, . . . to satisfy the inequalities
We shall prove Theorem 1 in
Sub-section 4.3.3.
For example,
ffiffiffiffiffiffiffiffiffiffi
p
3
ð8Þ ¼ 2:
1.5
Manipulating real numbers
33
12 < 2 < 22
ð1:4Þ2 < 2 < ð1:5Þ2
ð1:41Þ2 < 2 < ð1:42Þ2
ð1:414Þ2 < 2 < ð1:415Þ2
..
.
(1)
This process gives an infinite decimal
b ¼ 1:414 . . .;
and we claim that
b2 ¼ ð1:414 . . .Þ2 ¼ 2:
This can be proved using our method of multiplying decimals:
b
b
b2
1
1
1
1.4
1.4
1.96
1.41
1.414
..
.
1.41
1.414
..
.
1.9881
1.999396
..
.
Notice that
b ¼ 1:414 . . .
is the decimal that we
obtained as the least upper
bound of the set
fx 2 Q : x > 0; x2 < 2g
in Sub-section 1.5.1.
We have to prove that the least upper bound of the set E of numbers in the righthand column is 2, in other words that
sup E ¼ sup f1; ð1:4Þ2 ; ð1:41Þ2 ; ð1:414Þ2 ; . . .g ¼ 2:
To do this, we employ the strategy given in Sub-section 1.4.2.
First, we check that M ¼ 2 is an upper bound of E. This follows from the lefthand inequalities in (1).
Next, we check that, if M0 < 2, then there is a number in E which is greater
than M0 . To prove this, put
x0 ¼ 1; x1 ¼ 1:4; x2 ¼ 1:41; x3 ¼ 1:414; . . .:
Then, by the right-hand inequalities in (1), we have that
1 2
xn þ n > 2:
10
Also
1 2 2
1
1
xn þ n xn ¼ n 2xn þ n
10
10
10
1
5
5 n ð2 2 þ 1Þ ¼ n ;
10
10
and so
1 2
5
2
xn > xn þ n 10
10n
5
>2 n
10
¼ 1:99 . . . 95:
n digits
For example, if n ¼ 1, then
1 2
¼ 1:52 > 2:
1:4 þ
10
For example, if n ¼ 2, then
ð1:41Þ2 > 1:95:
1: Numbers
34
So, if M0 < 2, then we can choose n so large that xn2 > M0 (while still having
xn 2 E). This proves that the least upper bound of E is 2, and so (1.414. . .)2 ¼ 2.
Thus we can define
pffiffiffi
2 ¼ 1:414 . . .;
pffiffiffi
which justifies our earlier claim that 2 can be represented exactly by a
decimal.
1.5.3
Rational powers
Having discussed nth roots, we are now in a position to define the expression
ax, where a is positive and x is rational.
Definition
If a > 0, m 2 Z and n 2 N, then
pffiffiffim
m
an ¼ n a :
pffiffiffi
1
For example, for a >p0,ffiffiffi with m ¼ 1 we have an ¼ n a, and with m ¼ 2 and
2
2
n ¼ 3 we have a3 ¼ ð 3 aÞ :
This notation is particularly useful, because rational powers (or rational exponents) satisfy the following exponent laws (whose proofs depend on Theorem 1):
Exponent Laws
If a, b > 0 and x 2 Q ,
For example
then axbx ¼ (ab)x.
and x, y 2 Q , then ax ay ¼ axþ y.
and x, y 2 Q , then ðax Þy ¼ axy .
If a > 0
If a > 0
If x and y are integers, these laws actually hold for all non-zero real numbers
a and b. However,
if x and y are not integers, then we must have a, b > 0. For
1
example, ð1Þ2 is not defined as a real number.
m
However, if a is a negative real number, then a n can be defined whenever
m 2 Z, n 2 N and mn is reduced to its lowest terms with n odd, as follows
m
an ¼
p
ffiffiffim
n
a :
Finally, you may have wondered why we did not mention that each positive
number has two nth roots when n is even.
22 ¼ (2)2 ¼ 4. We
p
ffiffiffiFor example,
1
n
shall adopt the convention that, for a > 0, a and an always refer to the positive
nth root of a. If we wish
pffiffiffi to refer to both roots (for example, when solving
equations), we write n a.
1.5.4
Real powers
We conclude this section by briefly discussing the meaning of ax when a > 0
and x is an arbitrary real number. We have defined this expression when x is
rational, but the same definition does not work if x is irrational.
However, it
pffiffiffipffiffi2
is common practice to write down expressions such as 2 , and even to apply
the Exponent Laws to give equalities such as
1
1
1
22 32 ¼ 62 ;
1
1
5
22 23 ¼ 26 ;
1 13
1
22 ¼ 26 :
This extends our
above
m
definition of
a n ; for instance,
1
it defines an whenever n 2 N
and n2 is odd. For example,
ð8Þ3 ¼ 4.
1.6
Exercises
pffiffi p2ffiffi pffiffi pffiffi pffiffiffi 2
pffiffiffi 2 2
pffiffiffi 2
2
¼
2
¼
2 ¼ 2:
Can such manipulations be justified?
In fact, it is possible to define ax, for a > 0 and x 2 R, using the Least Upper
Bound Property of R, but it is then rather tricky to check that the Exponent
Laws work. In Chapters 2 and 3 we shall explain how to define the expression
ex, and in Chapter 4 we use ex to define the real powers in general and show that
the Exponent Laws hold. For the time being, whenever the expression ax
appears, you should assume that x is rational.
1.6
Exercises
Section 1.1
1. Arrange the following numbers in increasing order:
(a)
7 3 1 7 11
36 ; 20 ; 6 ; 45 ; 60 ;
(b) 0:465, 0:465, 0:465, 0.4655, 0.4656.
2. Find the fractions whose decimal expansions are:
(a) 0:481;
(b) 0:481.
3. Let x ¼ 0:21 and y ¼ 0:2. Find x þ y and xy (in decimal form).
4. Find a rational number x and an irrational number y in the interval
(0.119, 0.12).
pffiffiffi
5. Prove that, if n is a positive integer which is not a perfect square, then n
is irrational.
2
Hint: If p, q are positive integers such that pq ¼ n, and k is the positive
integer such that k < pq < k þ 1 (why does such a positive integer k exist?),
p
show that 0 < p kq < q and nqkp
pkq ¼ q, and hence obtain a contradiction.
Section 1.2
1. Solve the following inequalities:
pffiffiffiffiffiffiffiffiffiffiffiffiffi
xþ1
(a) xx1
(b) 4x 3 > x;
2 þ4 < x2 4;
(c) 17 2x4 15;
(d) jx þ 1j þ jx 1j < 4.
Section 1.3
1. Use the Triangle Inequality to prove that
jaj 1 ) ja 3j 2:
2. Prove that (a2 þ b2)(c2 þ d2) (ac þ ad)2, for any a, b, c, d 2 R.
3. Prove the inequality 3n 2n2 þ 1, for n ¼ 1, 2, . . .:
35
1: Numbers
36
(a) by using the Binomial Theorem, applied to (1 þ x)n with x ¼ 2;
(b) by using the Principle of Mathematical Induction.
4. Use the Principle of Mathematical Induction to prove that, for n ¼ 1, 2, . . . :
(a) 12 þ 22 þ 32 þ þ n2 ¼ nðnþ1Þ6ð2nþ1Þ;
qffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffi
5=4
135 ... ð2n1Þ
3=4
(b)
4nþ1
246 ... ð2nÞ
2nþ1.
2
5. Apply Bernoulli’s Inequality, first with x ¼ 2n and then with x ¼ ð3nÞ
to
prove that
2
2
1
3n 1 þ ; for n ¼ 1; 2; . . . :
1þ
3n 2
n
6. By applying the Arithmetic Mean–Geometric Mean Inequality to the n þ 1
positive numbers 1, 1 1n , 1 1n , 1 1n , . . . , 1 1n, prove that
nþ1
n 1
1 1n 1 nþ1
; for n ¼ 1; 2; . . . :
7. Use the Cauchy–Schwarz Inequality to prove that, if a1, a2, . . . , an are
positive numbers with a1 þ a2 þ þ an ¼ 1, then
pffiffiffi
pffiffiffi
pffiffiffi
pffiffiffi
a1 þ a2 þ þ an n:
8. Use the Cauchy–Schwarz Inequality to prove that
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffi
3 4 cos x þ 4 1 cos x 5; for x 2 0; p2 :
Section 1.4
In Exercises 1–4, take E1 ¼ {x: x 2 Q , 0 x < 1} and E2 ¼
n ¼ 1; 2; . . .g:
2
1 þ n1 :
1. Prove that each of the sets E1 and E2 is bounded above. Which of them has
a maximum element?
2. Prove that each of the sets E1 and E2 is bounded below. Which of them has
a minimum element?
3. Determine the least upper bound of each of the sets E1 and E2.
4. Determine the greatest lower bound of each of the sets E1 and E2.
5. For each of the following functions, determine whether it has a maximum
or a minimum, and determine its supremum and infimum:
1
(a) f ð xÞ ¼ 1þx
(b) f ð xÞ ¼ 1 x þ x2 ; x 2 ½0; 2Þ.
2 ; x 2 ½0; 1Þ;
6. Prove that, for any two numbers a, b 2 R
minfa; bg ¼ 12 ða þ b ja bjÞ and
maxfa; bg ¼ 12 ða þ b þ ja þ bjÞ:
2
Sequences
This chapter deals with sequences of real numbers, such as
1 1 1 1 1
2 3 4 5 6
1; ; ; ; ; ; . . .;
Three dots are used to indicate
that the sequence continues
indefinitely.
0; 1; 0; 1; 0; 1; . . .;
1; 2; 4; 8; 16; 32; . . .:
It describes in detail various properties that a sequence may possess, the most
important of which is convergence. Roughly speaking, a sequence is convergent, or tends to a limit, if the numbers, or terms, in the sequence approach
arbitrarily close to a unique real number, which is called the limit of the
sequence. For example, we shall see that the sequence
1 1 1 1 1
2 3 4 5 6
1; ; ; ; ; ; . . .
is convergent with limit 0. On the other hand, the terms of the sequence
0; 1; 0; 1; 0; 1; . . .
do not approach arbitrarily close to any unique real number, and so this
sequence is not convergent. Likewise, the sequence
1; 2; 4; 8; 16; 32; . . .
is not convergent.
A sequence which is not convergent is called divergent. The sequence
0; 1; 0; 1; 0; 1; . . .;
is a bounded divergent sequence. The sequence
1; 2; 4; 8; 16; 32; . . .
is unbounded; its terms become arbitrarily large and positive, and we say that it
tends to infinity.
Intuitively, it seems plausible that some sequences are convergent, whereas
others are not. However, the above description of convergence, involving the
phrase ‘approach arbitrarily close to’, lacks the precision required in Pure
Mathematics. If we wish to work in a serious way with convergent sequences,
prove results about them and decide whether a given sequence is convergent,
then we need a rigorous definition of the concept of convergence.
Historically, such a definition emerged only in the late nineteenth century,
when mathematicians such as Cantor, Cauchy, Dedekind and Weierstrass sought
to place Analysis on a rigorous non-intuitive footing. It is not surprising, therefore, that the definition of convergence seems at first sight rather obscure, and it
may take you a little time to master the logic that it involves.
37
2: Sequences
38
In Section 2.1 we show how to picture the behaviour of a sequence by
drawing a sequence diagram. We also introduce monotonic sequences; that is,
sequences which are either increasing or decreasing.
In Section 2.2 we explain the definition of a null sequence; that is, a
sequence which is convergent with limit 0. We then establish various properties of null sequences, and list some basic null sequences.
In Section 2.3 we discuss general convergent sequences (that is, sequences
which converge but whose limit is not necessarily 0), together with techniques
for calculating their limits.
In Section 2.4 we study divergent sequences, giving particular emphasis to
sequences which tend to infinity or tend to minus infinity. We also show that
convergent sequences are bounded; it follows that unbounded sequences are
necessarily divergent.
In Section 2.5 we prove the Monotone Convergence Theorem, which states
that any increasing sequence which is bounded above must be convergent, and,
similarly, that any decreasing sequence which is bounded below must be convergent. We use this theorem to study simple examples of sequences defined by
recurrence formulas, and particular sequences which converge to p and e.
2.1
2.1.1
Introducing sequences
What is a sequence?
Ever since learning to count you have been familiar with the sequence of
natural numbers
1; 2; 3; 4; 5; 6; . . .:
You have also encountered many other sequences of numbers, such as
2; 4; 6; 8; 10; 12; . . .;
1 1 1 1 1 1
; ; ; ; ; ; . . .:
2 4 8 16 32 64
We begin our study of sequences with a definition and some notation.
Definition
A sequence is an unending list of real numbers
a1 ; a2 ; a3 ; . . .:
The real number an is called the nth term of the sequence, and the sequence
is denoted by
fan g:
In each of the sequences above, we wrote down the first few terms and left
you to assume that subsequent terms were obtained by continuing the pattern in
an obvious way. It is sometimes better, however, to give a precise description
of a typical term of a sequence, and we do this by stating an explicit formula for
the nth term. Thus the expression {2n 1} denotes the sequence
1; 3; 5; 7; 9; 11; . . .;
Alternative notations are
1
fan g1
1 and fan gn¼1 :
2.1
Introducing sequences
39
and the sequence {an} defined by the statement
an ¼ ð1Þn ; n ¼ 1; 2; . . .;
has terms
a1 ¼ 1; a2 ¼ 1; a3 ¼ 1; a4 ¼ 1; a5 ¼ 1; . . .:
Problem 1
(a) Calculate the first five terms of each of the following sequences:
(i) {3n þ 1};
(ii) {3n};
(iii) {( 1)nn}.
(b) Calculate the first five terms of each of the following sequences {an}:
(i) an ¼ n!; n ¼ 1; 2; . . .;
n
(ii) an ¼ 1 þ 1n ; n ¼ 1; 2; . . . (to 2 decimal places).
Sequences often begin with a term corresponding to n ¼ 1. Sometimes,
however, it is necessary to begin a sequence with some other value of n. We
indicate this by writing, for example, fan g1
3 to represent the sequence
For example,
the sequence
1
n!n cannot begin with
n ¼ 1 or n ¼ 2.
a3 ; a4 ; a5 ; . . .:
Sequence diagrams
It is often helpful to picture how a given sequence {an} behaves by drawing a
sequence diagram; that is, a graph of the sequence in R 2. To do this, we mark
the values n ¼ 1, 2, 3, . . . on the x-axis and, for each value of n, we plot the
point
example, the sequence diagrams for the sequences {2n 1},
(n,an). For
n
1
n and {( 1 )} are as follows:
In Figure (a), the points
plotted all lie on the straight
line y ¼ 2x 1.
In Figure (b), they all lie on
the hyperbola y ¼ 1x :
Problem 2 Draw a sequence diagram, showing the first five points, for
each of the following sequences:
n no
n
(a) {n2};
(b) {3};
(c)
1 þ 1n g;
(d) ð1Þ
.
n
(In part (c), use the result of Problem 1, part (b).)
2.1.2
Monotonic sequences
Many of the sequences considered so far have the property that, as n increases,
their terms are either increasing or decreasing. For example, the sequence
{2n 1} has terms 1, 3, 5, 7, . . ., which are increasing, whereas the sequence
2: Sequences
40
1
has terms 1; 12 ; 13 ; 14 ; . . .; which are decreasing. The sequence {( 1)n} is
neither increasing nor decreasing. All this can be seen clearly on the above
sequence diagrams.
We now give a precise meaning to these words increasing and decreasing,
and introduce the term monotonic.
n
Definition
A sequence {an} is:
constant, if
anþ1 ¼ an ;
for n ¼ 1; 2; . . .;
increasing, if
decreasing, if
anþ1 an ;
anþ1 an ;
for n ¼ 1; 2; . . .;
for n ¼ 1; 2; . . .;
monotonic, if
{an} is either increasing or decreasing.
Remarks
1. Note that, for a sequence {an} to be increasing, it is essential that anþ1 an,
for all n 1. However, we do not require strict inequalities, because we
wish to describe sequences such as
1; 1; 2; 2; 3; 3; 4; 4; . . . and 1; 2; 2; 3; 4; 4; 5; 6; 6; . . .
as increasing. One slightly bizarre consequence of the definition is that
constant sequences are both increasing and decreasing!
2. A sequence {an} is said to be:
strictly increasing, if anþ1 > an ;
for n ¼ 1; 2; . . .;
strictly decreasing, if anþ1 < an ; for n ¼ 1; 2; . . .;
strictly monotonic, if {an} is either strictly increasing or strictly
decreasing.
3. A diagram does NOT constitute a proof! In our first example we formally
establish the monotonicity properties of our three sequences.
Example 1
monotonic:
Determine which of the following sequences {an} are
(a) an ¼ 2n 1, n ¼ 1, 2, . . .;
(b) an ¼ 1n , n ¼ 1, 2, . . .;
(c) an ¼ ( 1)n, n ¼ 1, 2, . . .
Solution
(a) The sequence {2n 1} is monotonic because
an ¼ 2n 1 and anþ1 ¼ 2ðn þ 1Þ 1 ¼ 2n þ 1;
so that
anþ1 an ¼ ð2n þ 1Þ ð2n 1Þ ¼ 2 > 0; for n ¼ 1; 2; . . .:
Thus {2n 1} is increasing.
(b) The sequence 1n is monotonic because
1
1
and anþ1 ¼
;
an ¼
n
nþ1
so that
In fact, strictly increasing.
2.1
Introducing sequences
anþ1 an ¼
41
1
1 n ð n þ 1Þ
1
¼
¼
< 0; for n ¼ 1; 2; ...:
nþ1 n
ðn þ 1Þn
ðn þ 1Þn
Thus 1n is decreasing.
Alternatively, since an > 0, for all n, and
anþ1
n
< 1; for n ¼ 1; 2; . . .;
¼
nþ1
an
it follows that
anþ1 < an ; for n ¼ 1; 2; . . .:
In fact, strictly decreasing.
Thus {an} is decreasing.
(c) The sequence {(1)n} is not monotonic. In fact, a1 ¼ 1, a2 ¼ 1 and
a3 ¼ 1.
Hence a3 < a2, which means that {an} is not increasing.
Also, a2 > a1, which means that {an} is not decreasing.
Thus {(1)n} is neither increasing nor decreasing, and so is not mono&
tonic.
A single counter-example is
sufficient to show that
{(1)n} is not increasing;
similarly a (different!) single
counter-example is sufficient
to show that {(1)n} is not
decreasing.
Example 1 illustrates the use of the following strategies:
Strategy To show that a given sequence {an} is monotonic, consider the
expression anþ1 an.
If anþ1 an 0;
for n ¼ 1; 2; . . .;
If anþ1 an 0;
for n ¼ 1; 2; . . .;
then fan g is increasing.
then fan g is decreasing.
If an > 0 for all n, it may be more convenient to use the following version
of the strategy:
Strategy To show that a given sequence of positive terms, {an}, is monotonic, consider the expression aanþ1
.
n
If aanþ1
1;
n
anþ1
If an 1;
for n ¼ 1; 2; . . .;
for n ¼ 1; 2; . . .;
then fan g is increasing.
then fan g is decreasing.
Problem 3 Show that the following sequences {an} are monotonic:
(a) an ¼ n!; n ¼ 1; 2; . . .;
(b) an ¼ 2n ; n ¼ 1; 2; . . .;
(c) an ¼ n þ 1n ; n ¼ 1; 2; . . .:
It is often possible to guess whether a sequence given by a specific formula is
monotonic by calculating the first few terms. For example, consider the
sequence {an} given by
1 n
; n ¼ 1; 2; . . .:
an ¼ 1 þ
n
In Problem 1 you found that the first five terms of this sequence are
approximately
2; 2:25; 2:37; 2:44 and 2:49:
We will study this important
sequence in detail in
Section 2.5.
2: Sequences
42
These terms suggest that the sequence {an} is increasing, and in fact it is.
However, the first few terms of a sequence are not always a reliable guide to
the sequence’s behaviour. Consider, for example, the sequence
an ¼
10n
;
n!
The first two terms certainly
prove that the sequence is
neither decreasing nor
constant.
n ¼ 1; 2; . . .:
The first five terms of this sequence are approximately
10; 50; 167; 417 and 833;
this suggests that the sequence {an} is increasing. However, calculation of
more terms shows that this is not so, and the sequence diagram for {an} looks
like this:
In fact
a6 ’ 1389; a7 ’ 1984;
a8 ’ 2480; a9 ’ 2756;
a10 ’ 2756; a11 ’ 2505;
a12 ’ 2088:
In particular, the fact that
a12 < a11 shows that the
sequence {an} cannot be
increasing.
Simplifying aanþ1
, we find that
n
n
anþ1
10nþ1
10
10
:
¼
¼
nþ1
an
ðn þ 1Þ! n!
10
But nþ1
1, for n ¼ 9, 10, . . ., so that aanþ1
1, for n ¼ 9, 10, . . .; it follows
n
that the sequence {an} is decreasing, if we ignore the first eight terms.
In a situation like this, when a given sequence has a certain property
provided that we ignore a finite number of terms, we say that the sequence
n
eventually has the property. Thus we have just seen that the sequence 10n! is
eventually decreasing.
Another example of this usage is the following statement:
In fact, aa109 ¼ 1 and
for n ¼ 10; 11; . . .:
anþ1
an
< 1;
the terms of the sequence {n2} are eventually greater than 100.
This statement is true because n2 > 100, for all n > 10.
Problem 4 Classify each of the following statements as TRUE or
and justify your answers (that is, if a statement is TRUE, prove it;
if a statement is FALSE, give a specific counter-example):
FALSE,
(a) The terms of the sequence {2n} are eventually greater than 1000.
(b) The terms of the sequence {(1)n} are eventually positive.
1
(c) The terms of the
n are eventually less than 0.025.
n 4sequence
o
n
(d) The sequence 4n is eventually decreasing.
Note this general approach to
‘TRUE’ and ‘FALSE’.
2.2
Null sequences
2.2
43
Null sequences
2.2.1
What is a null sequence?
In this sub-section we give a precise definition of a null sequence (that is, a
sequence which converges to 0) and introduce some properties of null
sequences.
We shall frequently use the rules for rearranging inequalities which you met
in Chapter 1, so you may find it helpful to reread that section quickly before
starting here. We shall also use the following inequalities which were proved in
Sub-section 1.3.3
2n 1 þ n;
Section 1.2
In fact, we use
for n ¼ 1; 2; . . .;
2n n; for n ¼ 1; 2; . . .:
and
n
2
2 n ;
for n 4:
Problem 1 For each of the following statements, find a number X such
that the statement is true:
1
3
(a) 1n < 100
; for all n > X;
(b) 1n < 1000
; for all n > X:
Problem 2 For each of the following statements, find a number X such
that the statement is true:
n
n
Þ
Þ
1
3
(a) ð1
(b) ð1
n2 < 100 ; for all n > X;
n2 < 1000 ; for all n > X:
The solutions of Problems 1 and 2 both suggest that the larger and larger
n we
o
Þn
choose n, the closer and closer to 0 the terms of the sequences 1n and ð1
n2
become.
We can express this in terms of the Greek letter " (pronounced ‘epsilon’),
which we introduce to denote a positive number that may be as small as we
please
given
o particular instance. In terms of the sequence diagrams for
innany
ð1Þn
1
, this means that the terms of these sequences eventually lie
n and
n2
inside a horizontal strip in the sequence diagram from " up to ". However, the
smaller we choose ", the further to the right we have to go before we can be sure
that all the terms of the sequence from that point onwards lie inside the strip.
That is, the smaller we choose " the larger we have to choose X if we wish to have
ð1Þn 1
< "; for all n > X; or 2 < "; for all n > X:
n
n
Definition
A sequence {an} is a null sequence if:
for each positive number ", there is a number X such that
jan j < ";
for all n > X:
(1)
Using this terminology,
n n o it follows from our previous discussion that the
1
Þ
sequences n and ð1
are both null.
n2
We can interpret our finding of a suitable number X for (1) to hold as an
‘" X game’ in which player A chooses a positive number " and challenges
player B to find some number X such that the property (1) holds.
Thus in Problems 1 and 2 we
chose the particular examples
1
3
and 1000
in place of the
of 100
‘general’ positive number ".
ε
X
n
–ε
Note that here X need not be
an integer; any appropriate
real number will serve. X does
NOT depend on n, but does in
general depend on ".
2: Sequences
44
n
Þ
For example, consider the sequence {an} where an ¼ ð1
n2 ; n ¼ 1; 2; . . ..
This sequence is null. For
ð1Þn 1
jan j ¼ 2 ¼ 2 ;
n
n
so that, if we make the choice X ¼ p1ffiffi", it is certainly true that
jan j < ";
for all n > X:
In terms of the " X game, this simply means that whatever choice of " is
made by player A, player B can always win by making the choice X ¼ p1ffiffin !
This is the case because, if
n > p1ffiffi" ½¼ X,
then we have n2 > 1" or " > n12 ;
hence, for n > X, we have
jan j ¼ n12 < X12 ¼ ":
Remarks
n no
Þ
1. In the sequence ð1
, the signs of the terms made no difference to
n2
whether the sequence was null, for they disappeared immediately we took
the modulus of the terms in order to examine whether the definition (1) of a
null sequence was satisfied. Indeed, in general, the signs of terms in a
sequence make no difference to whether a sequence is null. We can express
this formally as follows:
A sequence {an} is null if and only if the corresponding sequence {janj} is null.
2. A null sequence {an} remains null if we add, delete or alter a finite number
of terms in the sequence.
Similarly, a non-null sequence remains non-null if we add, delete or alter
a finite number of terms.
3. If one number serves as a suitable value of X for the inequality in (1) to hold,
then any larger number will also serve as a suitable X. Hence, for simplicity
in some proofs, we may assume if we wish that our initial choice of X in (1)
is a positive integer.
Example 1 Prove that the sequence n13 is a null sequence.
Solution
All that happens is that the
definition (1) remains valid
with a possibly different value
of X having to be chosen.
This is often expressed in the
following memorable way:
‘a finite number of terms do
not matter’.
For example, if 7 þ p will
serve as a suitable value for X,
then so will 12 or 37; but 10
might not.
We have to prove that
for each positive number ", there is a number X such that
1
< "; for all n > X:
n3 (2)
In1 order to find a suitable value of X for (2) to hold, we rewrite the inequality
3 < " in various equivalent ways until we spy a value for X that will suit
n
our purpose. Now
1
< " , 1 < "
n3 n3
1
"
1
ffiffiffi :
, n>p
3
"
, n3 >
So, let us choose X to be p13 ffiffi". With this choice of X, the above chain of
equivalent
inequalities shows us that, if n > X (the last line in the chain),
1
then
3 < " (in the first line). Thus, with this choice of X, (2) holds; so 13 is
n
n
&
indeed null.
The term inside the modulus
is positive.
Here we use the Reciprocal
Rule for inequalities that you
met in Sub-section 1.2.1.
Here we use the Power Rule
for inequalities that you met
in Sub-section 1.2.1.
2.2
Null sequences
Example 2
45
Prove that the following sequence is not null
1; if n is odd;
an ¼
0; if n is even:
Solution To prove that the sequence is not null, we have to show that it does
not satisfy the definition. In other words, we must show that the following
statement is not true:
for each positive number ", there is a number X such that
jan j5 ", for all n 4X:
So, what we have to show is that the following is true:
for some positive number ", whatever X one chooses
jan j5",
6
for all n4X:
which we may rephrase as:
there is some positive number ", such that whatever X one chooses
jan j 5
6 ", for all n4X:
2
1
n
–2
Sequence diagram
1
1/2
–1/2
n
Strip of half-width 12
So, we need to find some positive number " with such a property. The
1
sequence diagram provides
1 1the clue! If we choose " ¼ 2, then the point (n, an)
lies outside the strip 2 ; 2 for every odd n. In other words, whatever X one
then chooses, the statement
jan j < "; for all n > X;
&
is false. It follows that the sequence is not a null sequence.
Note
Notice that any positive value of " less than 1 will serve for our purpose here;
there is nothing special about the number 12.
These two examples illustrate the following strategy:
Strategy for using the definition of null sequence
1. To show that {an} is null, solve the inequality janj < " to find a number X
(generally depending on ") such that janj < ", for all n > X.
2. To show that {an} is not null, find ONE value of " for which there is NO
number X such that janj < ", for all n > X.
However, any value of "
greater than 1 provides
no information!
2: Sequences
46
Problem 3 Use the above strategy to determine which of the following
sequences are null: n
o
n no
1 Þn
Þ
(a) 2n1
;
(c) ðn1
;
(b) ð1
4 þ1 :
10
We now look at a number of Rules for ‘getting new null sequences from old’.
Power Rule If
{an} is a null sequence, where an 0, for n ¼ 1, 2, . . ., and
p > 0, then apn is a null sequence.
1
pffiffi
Thus, for example, the
1sequence n is null – we simply apply the Power
Rule to the sequence n that we saw earlier to be null, using the positive
power p ¼ 12.
We shall prove these Rules
later, in Sub-section 2.2.2.
Recall that apn ¼ ðan Þp :
For p1ffiffin ¼
112
n
:
Remark
Notice that the Power Rule also holds without the requirement that an 0, so
long as apn is defined – for example, if p ¼ m1 where m is a positive integer.
Combination Rules If {an} and {bn} are null sequences, then the following are also null sequences:
Sum Rule
{an þ bn};
Multiple Rule {lan}, for any real number l;
Product Rule {anbn}.
Note that the number l has to
be a fixed number that does
not depend on n.
Thus, for example, we may use known examples of null sequences to verify
that the following sequences are also null:
1 1 þ
– by applying the Sum Rule to the null sequences 1n and n13 ;
n47pn3
1
– by applying the Multiple Rule to the null sequence n3 with
n3
l ¼ 47p;
1
fnð2n1
g – by applying the Product Rule to the null sequences 1n and
1Þ
2n1 :
Problem 4 Use the above rules to show that the following sequences
are null:
n
o
n
o
1
6ffiffi
5
1
p
(a)
þ
;
(b)
(c)
7 ;
3
1 :
5
ð2n1Þ
n
f
ð2n1Þ
3n4 ð2n1Þ3
g
Our next rule, the Squeeze Rule, also enables us to ‘get new null sequences
from old’ – but in a slightly different way. To illustrate
1 this
rule,
we look first at
1pffiffi
p
ffiffi
the sequence diagrams of the two sequences
and 1þ n .
n
1
}√1n }
}1 +1√ n}
...
...
n
In your solution, you may use
any of the sequences that you
have proved to be null so far
in this sub-section.
2.2
Null sequences
47
The points corresponding to the sequence 1þ1pffiffin are squeezed in between
the horizontal axis and the points corresponding to the null sequence p1ffiffin ,
since 1þ1pffiffin < p1ffiffin, for n ¼ 1, 2, . . .. Hence, if from some point onwards all the
points corresponding to p1ffiffin lie in a narrow strip in the sequence diagram of
half-width " about the axis, then (from the same point onwards) all the points
corresponding to 1þ1pffiffin will also lie in the same strip. So, since " may be any
positive number, it certainly looks from this sequence diagram argument that
the sequence 1þ1pffiffin must be a null sequence too.
Squeeze Rule
If {bn} is a null sequence and
jan j bn ; for n ¼ 1; 2; . . .;
then {an} is a null sequence.
That is, {an} is dominated by
{bn}.
The trick in using the Squeeze Rule to prove that a given sequence {an} is
null is to think of a suitable sequence {bn} that
dominates {an}and is itself null.
Thus, for example, since the sequence p1ffiffin is null and 1þ1pffiffin < p1ffiffin, it follows
from the Squeeze Rule that 1þ1pffiffin is also a null sequence.
Proof of the Squeeze Rule We want to prove that {an} is null; that is: for
each positive number ", there is a number X such that
jan j < "; for all n > X:
(3)
We know that {bn} is null, so there is some number X such that
jbn j < "; for all n > X:
(4)
We also know that janj bn, for n ¼ 1, 2, . . ., and hence it follows from (4) that
jan j ð< jbn jÞ < ";
And this is the X that we shall
use to verify (3).
for all n > X:
Thus inequality (3) holds, as required.
&
Remark
In applying the Squeeze Rule, it is often useful to remember the following fact:
The behaviour of a finite number of terms does not matter: it is sufficient to
check that janj bn holds eventually.
n is null.
Example 3 Use the Squeeze Rule to prove that the sequence 12
1n Proof We want to prove that 2
is null. n
What sequence can we find that dominates 12 ? Well, we saw at the start
n
of this sub-section that 2 n, for n ¼ 1, 2, . . ., and it follows from this, by the
Reciprocal Rule for inequalities,
that 21n 1n ; for n¼
1
11;n 2; . . .:
Thus the sequence n dominatesthe
sequence
, and is itself null. It
2
n &
follows from the Squeeze Rule that 12
is null.
n
2
Problem 5 Use the inequality
12n n , for n 4, and the Squeeze Rule
is null.
to prove that the sequence n 2
n
Example 4 Use the Squeeze Rule to prove that the sequence 10n! is null.
n
Solution We want to prove that 10n! is null.
10n With a bit of
linspiration, we guess that the sequence n! is eventually
dominated by n , for some constant l.
That is, that {an} is eventually
dominated by {bn}.
2: Sequences
48
n
Writing out the expression 10n! in full, we see that
10n
10 10
10
10
10
10
...
...
¼
1
2
10
11
n1
n
n!
10
< 3000 ; for all n > 10;
n
30000
:
¼
n
n
In other words, 10n! is eventually dominated by ln , for l ¼ 30000. This
latter
1
sequence is null, by the Multiple Rule applied to thenull sequence
:
n
It follows, fromthe Squeeze Rule and the fact that 30000
is a null sequence,
n
n
&
that the sequence 10n! is also null.
Problem 6 Prove that
n the
ofollowingnsequences
o are null:
1 ð1Þn
sin n2
(a) n2 þn ;
;
(c) n2 þ2n :
(b)
n!
We can now list a good number of generic types of null sequences, of which we
have seen specific instances already in this sub-section. We call these basic
null sequences, since we shall use them commonly together with the
Combination Rules and other rules for null sequences in order to prove that
particular sequences that we meet are themselves null.
Basic null sequences
(a) n1p ;
for p > 0;
(b) fcn g;
for jcj < 1;
p n
(c) fn c g; for p > 0 and jcj < 1;
n
for any real c;
(d) cn! ;
np for p > 0:
(e) n! ;
2.2.2
Instances of these that we
have seen are
1
pffiffiffi ;
n
n 1
;
2 n 1
n
;
2
n
10
;
n!
10 n
:
n!
Proofs
We now give a number of proofs which were omitted from the previous subsection so as not to slow down your gaining an understanding of the key ideas
there. First, recall the definition of a null sequence.
Definition
A sequence {an} is null if
We suggest that you omit
these proofs on a first reading,
and return to them when you
are confident that you
understand the basic ideas.
for each positive number ", there is a number X such that
jan j < ";
for all n > X:
Proofs of the Power Rule and the Combination Rules
Recall that X need not be an
integer; any appropriate real
number will serve.
In the previous sub-section we proved the Squeeze Rule, but did not prove the
Power Rule or the Combination Rules. We now supply these proofs.
Power Rule If{an} is a null sequence, where an 0, for n ¼ 1, 2, . . ., and,
if p>0, then apn is a null sequence.
Recall that anp ¼ (an)p.
2.2
Null sequences
Proof
49
We want to prove that apn is null; that is:
for each positive number ", there is a number X such that
apn < ";
for all n > X:
(5)
For janpj ¼ anp, since an 0.
(6)
Here we use the number "p in
place of " in the definition of
null sequence.
We know that {an} is null, so there is some number X such that
1
an < " p ;
1
for all n > X:
Taking the pth power of both sides of (6), we obtain the desired result (5), with
&
the same value of X.
Remark
1
Notice how we used the (positive) number "p in place of " in (6) in order to
obtain " in the final result (5). In the proofs which follow, we again apply the
definition of null sequence, using positive numbers which depend in some way
on ", in order to obtain " in the inequality that we are aiming to prove.
Sum Rule
sequence.
Proof
If {an} and {bn} are null sequences, then {an þ bn} is a null
We want to prove that the sum {an þ bn} is null; that is:
for each positive number ", there is a number X such that
jan þ bn j < "; for all n > X:
(7)
We know that {an} and {bn} are null, so there are numbers X1 and X2 such that
1
jan j < "; for all n > X1 ;
2
and
1
jbn j < "; for all n > X2 :
2
Hence, if X ¼ max {X1, X2}, then both the two previous inequalities hold; so
if we add them we obtain, by the Triangle Inequality, that
jan þ bn j jan j þ jbn j
1
1
< " þ " ¼ "; for all n > X:
2
2
Thus inequality (7) holds, with this value of X.
We use 12 " here rather than ",
in order to end up with the
symbol " on its own
eventually in the desired
inequality (7).
You met the Triangle
Inequality in Subsection 1.3.1.
&
Before going on, first a comment about the number 12 that appears several
times in the above proof. It is used twice in expressions 12 " at the start of the
proof in order to end up with a final inequality that says that some expression is
‘<"’. While this means that we end up with an inequality that shows at once
that the desired result holds, in fact it is not strictly necessary to end up with
precisely ‘<"’, as the following result shows:
Lemma The ‘Ke Lemma’ Let {an} be a sequence. Suppose that, for
each positive number ", there is a number X such that
jan j < K"; for all n > X;
where K is a positive real number that does not depend on " or X. Then {an}
is a null sequence.
In this case, the result that a
particular sequence is null.
Loosely speaking, we may
express this result as ‘K" is
just as good as "’ in the
definition of null sequence.
For example, K might be 2 or p7
or 259, but it could not be 2n
X
or 259
.
2: Sequences
50
Proof
We want to prove that the sequence {an} is null; that is:
You may omit this proof on a
first reading.
for each positive number ", there is a number X such that
jan j < ";
for all n > X:
Now, whatever this positive number " may be, the number K" is also a
positive number. It follows from the hypothesis stated in the Lemma that
therefore there is some number X such that
"
jan j < K K
¼ "; for all n > X:
This is precisely the condition for {an} to be null.
&
From time to time we shall use this Lemma in order to avoid arithmetic
complexity in our proofs.
Multiple Rule If {an} is a null sequence, then {lan} is a null sequence for
any real number l.
Proof
We want to prove that the multiple {lan} is null; that is:
for each positive number ", there is a number X such that
jlan j < ";
for all n > X:
(8)
If l ¼ 0, this is obvious, and so we may assume that l 6¼ 0.
We know that {an} is null, so there is some number X such that
j an j <
1
";
jlj
for all n > X:
Multiplying both sides of this inequality by the positive number jlj, this gives
us that
jlan j < ";
for all n > X:
Thus the desired result (8) holds.
&
If {an} and {bn} are null sequences, then {anbn} is a null
Product Rule
sequence.
Proof
We use j1lj " here rather than "
in order to end up eventually
with the symbol " on its own
in the desired inequality (8).
We want to prove that the product {anbn} is null; that is:
for each positive number ", there is a number X such that
jan bn j < ";
for all n > X:
(9)
We know that {an} and {bn} are null, so there are numbers X1 and X2 such
that
pffiffiffi
jan j < "; for all n > X1 ;
and
j bn j <
pffiffiffi
";
for all n > X2 :
pffiffiffi
We use " here rather than "
in order to end up eventually
with the symbol " on its own
in the desired inequality (9).
2.2
Null sequences
51
Hence, if X ¼ max {X1, X2}, then both the two previous inequalities hold; so
if we multiply them we obtain that
jan bn j ¼ jan j jbn j
pffiffiffi pffiffiffi
< " " ¼ ";
for all n > X:
Thus inequality (9) holds with this value of X.
&
Basic null sequences
At the end of the previous sub-section we gave a list of basic null sequences.
We end this sub-section by proving that these sequences are indeed null
sequences.
For example
Basic null sequences The following sequences are null sequences:
for p > 0;
(a) n1p ;
(b) fcn g;
for jcj < 1;
(c) fnp cn g;
n
(d) cn! ;
p
(e) nn! ;
for p > 0; jcj < 1;
for any real c;
for p > 0.
1
;
n10
fð0:9Þn g;
3
n ð0:9Þn ;
n
10
;
n!
10 n
n!
Proof
1
(a) To prove that
1 np is null, for p > 0, we simply apply the Power Rule to the
sequence n , which we know is null.
(b) To prove that {cn} is null, for jcj < 1, it is sufficient to consider only the
case 0 c < 1. If c ¼ 0, the sequence is obviously null; so we may assume
that 0 < c < 1.
With this assumption, we can write c in the form
c¼
1
;
1þa
for some number a > 0:
Now, by the Binomial Theorem
n
ð1 þ aÞ 1 þ na
na;
You met the Binomial
Theorem in Sub-section 1.3.3.
for n ¼ 1; 2; . . .;
and hence
1
ð 1 þ aÞ n
1
; for n ¼ 1; 2; . . .:
na
1
Since 1n is null, we deduce that fna
g is also null by the Multiple Rule;
cn ¼
n
hence {c } is null, by the Squeeze Rule.
(c) To prove that {npcn} is null, for p > 0 and jcj < 1, we may again assume
that 0 < c < 1. Hence
c¼
1
;
1þa
for some number a > 0:
Here we take l ¼ 1a in the
Multiple Rule.
2: Sequences
52
First, we deal with the case p ¼ 1. By the Binomial Theorem
1
ð1 þ aÞn 1 þ na þ nðn 1Þa2
2
1
nðn 1Þa2 ; for n ¼ 2; 3; . . .;
2
and hence
ncn ¼
n
ð1 þ aÞn
n
2
¼
; for n ¼ 2; 3; . . .:
2
ð
n
1Þa2
n
ð
n
1
Þa
2
n
o1
2
is null, we deduce that ðn1Þa
is also null, by the
2
1
Since
1 1
n1 2
2
Multiple Rule. Hence {ncn} is null, by the Squeeze Rule. This proves
part (c) in the case p ¼ 1.
To deduce that {npcn} is null for any p > 0 and 0 < c < 1, we note that
np cn ¼ ðnd n Þp ;
for n ¼ 1; 2; . . .;
1
p
where d ¼ c . Since 0 < d < 1, we know that {ndn} is null, and so {npcn} is
null, by the Power Rule.
n
(d) To prove that cn! is null for any real c, we may assume that c > 0. If we
choose any integer m such that m þ 1 > c, then we have, for n > m þ 1, that
cn c c c c
c c ¼
...
...
1 2
m
mþ1
n1 n
n!
c c c c
...
1 2
m
n
c
¼K ;
n
Here we use the special case
of part (c) that we have
already proved,
m
whereK¼ cm! is a constant.
null,
we deduce that Kc
Since 1n is n is also null, by the Multiple Rule; it
n
follows that cn! is also null, by the Squeeze Rule.
p
(e) To prove that nn! is null, for p > 0, we write
p n
np
n
2
¼
; for n ¼ 1; 2; . . .:
n!
2n
n!
p
n
Since n2n and 2n! are
pboth
null sequences, by parts (c) and (d) respec&
tively, we deduce that nn! is null, by the Product Rule.
2.3
2.3.1
Convergent sequences
What is a convergent sequence?
In the previous section we looked at null sequences; that is, sequences which
converge to 0. We now turn our attention to sequences which converge to
limits other than 0.
For c is a constant and m ism
some constant, and hence cm! is
also some constant. Note that
what is varying is n, nothing
else.
2.3
Convergent sequences
Problem 1
53
Consider the sequence an ¼ nþ1
n ; n ¼ 1; 2; . . ..
(a) Draw the sequence diagram for {an}, and describe (informally) how
this sequence behaves.
(b) What can you say (formally) about the behaviour of the sequence
bn ¼ an 1; n ¼ 1; 2; . . .?
The terms of the sequence {an} in Problem 1 appear to get arbitrarily close to 1;
that is, the sequence {an} appears to converge to 1. If we subtract 1 from each
term an to form the sequence {bn}, then we obtain a null sequence. This
example suggests the following definition of a convergent sequence:
Definition The sequence {an} is convergent with limit ł, or converges to
the limit ł, if {an ‘} is a null sequence. In this case, we say that {an}
converges to ł, and we write:
every null sequence converges to 0;
These statements are read as:
‘the limit of an, as n tends to
infinity, is ‘’ and ‘an tends to
‘, as n tends to infinity’.
Often, we omit ‘as n ! 1’.
Do not let this use of the
symbol 1 tempt you to think
that 1 is a real number.
every constantnsequence
{c} converges to c;
o
nþ1
the sequence n is convergent with limit 1.
See Problem 1 above.
lim an ¼ ‘,
EITHER
n!1
an ! ‘ as n ! 1.
OR
The following are examples of convergent sequences:
Problem 2 For each of the following sequences {an}, draw its sequence diagram and show that {an} converges to ‘ by considering an ‘:
(b) an ¼ n
2
(a) an ¼ nn2 1
þ1 ; ‘ ¼ 1;
3
þð1Þn
2n3
; ‘ ¼ 12 :
The definition of convergence of a sequence is often given in the following
equivalent form:
Equivalent definition
The sequence {an} is convergent with limit ł if:
for each positive number ", there is a number X such that
jan ‘j < ";
for all n > X:
(1)
Remarks
1. In terms of the sequence diagram for {an}, this definition states that, for
each positive number ", the terms an eventually lie inside the horizontal
strip from ‘ " up to ‘ þ ".
{an}
+ε
|an – | < ε
–ε
– ε < an < + ε
X
n
Note that here X need not be
an integer; any appropriate
real number will serve.
2: Sequences
54
2. Just as for null sequences, we can interpret ‘an ! ‘ as n ! 1’ as an " X
game in which player A chooses a positive number " and challenges player
B to find some number X such that (1) holds.
For example, consider the sequence {an} where an ¼ 8nþ6
2n , n ¼ 1, 2, . . ..
This sequence is convergent, with limit 4. For
8n þ 6
6
3
4 ¼
¼ ;
j an 4j ¼ 2n
2n n
so that, if we make the choice X ¼ 3", it is certainly true that
jan 4j < ";
for all n > X:
In terms of the " X game, this simply means that whatever choice
of " is made by player A, player B can always win by making the choice
X ¼ 3"!
I'll play
X = 3ε
I'll play
ANY positive
ε
This is the case because, if
3
½¼ X;
"
then we have " > 3n or 3n < ";
hence, for n > X, we have
n>
jan 4j ¼
3 3
< ¼ ":
n X
B wins
A
B
3. If a sequence is convergent, then it has a unique limit. We can see this by
using a sequence diagram. Suppose that the sequence {an} has two limits, ‘
and m, where ‘ 6¼ m. Then it is possible to choose a positive number " such
that horizontal strips from ‘ " up to ‘ þ " and from m " up to m þ " do
not overlap. For example, we can take " ¼ 13 j‘ mj.
m+ε
m
m–ε
The vertical distance between
‘ and m is j‘ mj, so that any
choice of " that is less than
half of this quantity will serve.
–m
+ε
–ε
n
However, since ‘ and m are both limits of {an}, the terms an must eventually
lie inside both strips, and this is impossible.
A formal proof of this fact is given in the Corollary to Theorem 3, later in
this section.
4. If a given sequence converges to ‘, then this remains true if we add, delete
or alter a finite number of terms. This follows from the corresponding result
for null sequences.
5. Not all sequences are convergent. For example, the sequence {(1)n} is not
convergent.
In this section we restrict our attention to sequences which do converge.
In other words ‘altering a
finite number of terms does
not matter’. See Subsection 2.2.1.
We discuss non-convergent
sequences in Section 2.4.
2.3
Convergent sequences
2.3.2
55
Combination Rules for convergent sequences
So far you have tested the convergence of a given sequence {an} by calculating
an ‘ and showing that {an ‘} is null. This presupposes that you know in
advance the value of ‘. Usually, however, you are not given the value of ‘. You
are given only a sequence {an} and asked to decide whether or not it converges
and, if it does, to find its limit. Fortunately many sequences can be dealt with
by using the following Combination Rules, which extend the Combination
Rules for null sequences:
Sub-section 2.2.1.
Theorem 1 Combination Rules
If lim an ¼ ‘ and lim bn ¼ m, then:
n!1
n!1
Sum Rule
Multiple Rule
Product Rule
Quotient Rule
lim ðan þ bn Þ ¼ ‘ þ m;
n!1
lim ðlan Þ ¼ l‘;
n!1
for any real number l;
lim ðan bn Þ ¼ ‘m;
an
‘
lim
¼ ; provided that m 6¼ 0:
n!1 bn
m
n!1
Remarks
1. In applications of the Quotient Rule, it may happen that some of the terms bn
take the value 0, in which case abnn is not defined. However, we shall see (in
Lemma 1) that this can happen only for finitely many bn (because m ¼
6 0), and
so {bn} is eventually non-zero. Thus the statement of the rule does make sense.
2. The following rule is a special case of the Quotient Rule:
If lim an ¼ ‘ and ‘ 6¼ 0, then:
n!1
1
1
Reciprocal Rule lim
¼ .
n!1 an
‘
Corollary 1
We shall prove the Combination Rules at the end of this sub-section, but first
we illustrate how to apply them.
Applying the Combination Rules
Example 1 Show that each of the following sequences {an} is convergent,
and find its limit:
þ 1Þðn þ 2Þ
(a) an ¼ ð2n3n
;
2 þ 3n
2
n
(b) an ¼ 2nn! þþ3n103 .
Solution Although the expressions for an are quotients, we cannot apply the
Quotient Rule immediately, because the sequences defined by the numerators
and denominators are not convergent. In each case, however, we can rearrange
the expressions for an and then apply the Combination Rules.
(a) In this case we divide both the numerator and denominator by n2 to give
2 þ 1n 1 þ 2n
ð2n þ 1Þðn þ 2Þ
¼
:
an ¼
3n2 þ 3n
3 þ 3n
A finite number of terms
do not matter.
2: Sequences
56
Since
1
n
is a basic null sequence, we find by the Combination Rules that
ð2 þ 0Þð1 þ 0Þ 2
¼ :
lim an ¼
n!1
3þ0
3
(b) This time we divide both the numerator and denominator by n! to give
2
n
2n2 þ 10n 2nn! þ 10n!
an ¼
¼
3
n! þ 3n3
1 þ 3nn!
3
2 n
Since nn! , 10n! and nn! are all basic null sequences, we find by the
Combination Rules that
lim an ¼
n!1
0þ0
¼ 0:
1þ0
&
See the list of basic null
sequences (introduced in
Sub-section 2.2.1) which is
repeated in the margin below.
Remark
We simplified each of the above sequences by dividing both numerator and
denominator by the dominant term. In part (a), we divided by n2, which is the
highest power of n in the expression. In part (b), the choice of dominant term was
a little harder, but the choice of n! ensured that the resulting quotients in the
numerator and denominator were all typical terms of convergent sequences.
In choosing the dominant term the following ordering is often useful:
Domination Hierarchy
A factorial term n! dominates a power term cn for any c 2 R.
A power term cn dominates a term np for p > 0, jcj < 1.
The above examples illustrate the following general strategy:
Strategy
Basic
null sequences:
1
; for p > 0;
np
To evaluate the limit of a complicated quotient:
1. Identify the dominant term, bearing in mind the basic null sequences.
2. Divide both the numerator and the denominator by the dominant term.
fcn g;
3. Apply the Combination Rules.
n
c
;
n!p n
;
n!
Problem 3 Show that each of the following sequences {an} is convergent, and find its limit:
3
(a) an ¼ n
þ 2n2 þ 3
2n3 þ 1 ;
2
n
2
(b) an ¼ n3n þ
þ n3 ;
n
ð1Þ
(c) an ¼ n!þ
2n þ 3n! :
Proofs of the Combination Rules We prove the Sum Rule, the Multiple
Rule and the Product Rule by using the corresponding Combination Rules for
null sequences. Remember that
lim an ¼ ‘
n!1
means that
fan ! ‘g is a null sequence:
for jcj < 1;
fnp cn g; for p > 0; c > 1;
for c 2 R;
for p > 0:
Warning Notice that ‘3n!’
means ‘3 (n!)’ and not
‘(3n)!’.
You may omit these proofs at
a first reading.
2.3
Convergent sequences
57
Sum Rule If lim an ¼ ‘ and lim bn ¼ m, then
n!1
n!1
lim ðan þ bn Þ ¼ ‘ þ m:
n!1
Proof
By assumption, {an ‘} and {bn m} are null sequences; since
ðan þ bn Þ ð‘ þ mÞ ¼ ðan ‘Þ þ ðbn mÞ
we deduce that {(an þ bn) (‘ þ m)} is null, by the Sum Rule for null
&
sequences.
Product Rule
If lim an ¼ ‘ and lim bn ¼ m, then
n!1
n!1
lim ðan bn Þ ¼ ‘m:
n!1
Proof
The idea here is to express anbn ‘m in terms of an ‘ and bn m
an bn ‘m ¼ ðan ‘Þðbn mÞ þ mðan ‘Þ þ ‘ðbn mÞ:
Since {an ‘} and {bn m} are null, we deduce that {anbn ‘m} is null, by
&
the Combination Rules for null sequences.
Remark
Note that the Multiple Rule is just a special case of the Product Rule in which
the sequence {bn} is a constant sequence.
To prove the Quotient Rule, we need to use the following lemma, which will
also be needed in the next sub-section:
Lemma 1
If lim an ¼ ‘ and ‘ > 0, then there is a number X such that
n!1
1
an > ‘, for all n > X:
2
Proof Since 12 ‘ > 0, the terms an must eventually lie within a distance 12 ‘ of
the limit ‘.
{an}
3
2
|an – | < 12 1
2
1
< an < 32 2
X
n
In other words, there is a number X such that
1
jan ‘j < ‘; for all n > X:
2
Hence
1
1
‘ < an ‘ < ‘; for all n > X;
2
2
and so the left-hand inequality gives
1
‘ < an ; for all n > X;
2
as required.
&
Here we are taking " ¼ 12 ‘ in
the definition of convergence.
2: Sequences
58
If lim an ¼ ‘ and lim bn ¼ m, then
n!1
n!1
an
‘
lim
¼ , provided that m 6¼ 0:
n!1 bn
m
Quotient Rule
Proof We assume that m > 0; the proof for the case m < 0 is similar. Once
again the idea is to write the required expression in terms of an ‘ and bn m
an ‘ mðan ‘Þ ‘ðbn mÞ
:
¼
bn m
bn m
Now, however, there is a slight problem: {m(an ‘) ‘(bn m)} is certainly a
null sequence, but the denominator is rather awkward. Some of the terms bn
may take the value 0, in which case the expression is undefined.
However, by Lemma 1, we know that that for some X we have
1
bn > m; for all n > X:
2
Thus, for all n > X
an ‘ jmðan ‘Þ ‘ðbn mÞj
¼
b
m
bm
n
n
jmðan ‘Þ ‘ðbn mÞj
1 2
2m
jmj jan ‘j þ j‘j jbn mj
:
1 2
2m
In particular, this implies that
the terms of {bn} are
eventually positive.
Here we use Lemma 1.
Here we apply the Triangle
Inequality to the numerator.
Since this last expression defines a null sequence, it follows by the Squeeze
&
Rule that abnn m‘ is null.
2.3.3
Further properties of convergent sequences
There are several other theorems about convergent sequences which will be
needed in later chapters. The first is a general version of the Squeeze Rule.
Theorem 2
If:
You met the Squeeze Rule
for null sequences in
Sub-section 2.2.1.
Squeeze Rule
1. bn an cn, for n ¼ 1, 2, . . .,
2. lim bn ¼ lim cn ¼ ‘,
n!1
n!1
then lim an ¼ ‘:
n!1
Proof
By the Combination Rules
lim ðcn bn Þ ¼ ‘ ‘ ¼ 0 ;
n!1
so that fcn bn g is a null sequence. Also, by condition 1
0 an bn cn bn ; for n ¼ 1; 2; . . .;
and so fan bn g is null, by the Squeeze Rule for null sequences.
That is, fan bn g is squeezed
by fcn bn g.
2.3
Convergent sequences
59
Now we write an in the form
an ¼ ð an bn Þ þ bn :
Hence by the Combination Rules
lim an ¼ lim ðan bn Þ þ lim bn
n!1
n!1
n!1
&
¼ 0 þ ‘ ¼ ‘:
Remark
In applications of the Squeeze Rule, it is sufficient to check that condition 1
applies eventually. This is because the values of a finite number of terms do not
affect convergence.
A finite number of terms do
not matter.
The following example and problem illustrate the use of the Squeeze Rule and
the Binomial Theorem in the derivation of two important limits.
Example 2
(a) Prove that, if c > 0, then
1
c
ð1 þ c Þn 1 þ ;
n
(b) Deduce that
for n ¼ 1; 2; . . .:
We proved this inequality for
the case c ¼ 1 in Example 6(b)
of Sub-section 1.3.3.
1
lim an ¼ 1; for any positive number a:
n!1
Solution
(a) Using the rules for inequalities, we obtain
1
c
c n
ð1 þ c Þn 1 þ , 1 þ c 1 þ
; since c > 0 :
n
n
The right-hand inequality holds, because
c
c n
1þ
1þn
¼ 1 þ c;
n
n
by the Binomial Theorem. It follows that the required inequality also holds.
(b) We consider the cases a > 1, a ¼ 1, a < 1 separately.
If a > 1, then we can write a ¼ 1 þ c, where c > 0. Then, by part (a)
1
c
1
1 an ¼ ð1 þ cÞn 1 þ ; for n ¼ 1; 2; . . .:
n
Since lim 1 þ nc ¼ 1, we deduce that
All the terms in the Binomial
Theorem expansion of
ð1 þ ncÞn are positive, so we
get a smaller number if we
ignore all the terms after the
first two.
n!1
1
lim an ¼ 1;
n!1
by the Squeeze Rule.
1
If a ¼ 1, then an ¼ 1, for n ¼ 1, 2, . . ., so
1
lim an ¼ 1:
n!1
1
If 0 < a < 1, then 1a > 1, so that lim 1a n ¼ 1 by the first case in part (b).
n!1
Hence, by the Reciprocal Rule
1
1
1
&
lim an ¼
11n ¼ 1 ¼ 1:
n!1
lim
n!1 a
In this application of the
Squeeze Rule, the ‘lower’
sequence is {1}.
2: Sequences
60
Problem 4
(a) Use the Binomial Theorem to prove that
rffiffiffiffiffiffiffiffiffiffiffi
2
1
n
n 1 þ
; for n ¼ 2; 3; . . .:
n1
Þ 2
ð1 þ xÞn nðn1
2! x ;
Hint:
for n 2; x 0:
(b) Use the Squeeze Rule to deduce that
1
lim nn ¼ 1:
n!1
Next we show that taking limits preserves weak inequalities.
Theorem 3 Limit Inequality Rule
If lim an ¼ ‘ and lim bn ¼ m, and also
n!1
n!1
an bn ;
for n ¼ 1; 2; . . .;
then ‘ m.
Proof Suppose that an ! ‘ and bn ! m, where an bn for n ¼ 1, 2, . . ., but
that it is not true that ‘ m. Then ‘ > m and so, by the Combination Rules
lim ðan bn Þ ¼ ‘ m > 0:
n!1
Hence, by Lemma 1, there is an X such that
1
(2)
an bn > ð‘ mÞ; for all n > X:
2
However, we assumed that an bn 0, for n ¼ 1, 2, . . ., so statement (2) is a
contradiction.
&
Hence it is true that ‘ m.
Warning Taking limits does
NOT preserve strict inequalities. That is, if
lim an ¼ ‘ and lim bn ¼ m, and also an < bn, for n ¼ 1, 2, . . ., then it follows
n!1
n!1
from Theorem 3 that ‘ m – but it does NOT necessarily follow that ‘ < m.
1
1
< 1n, for n ¼ 1, 2, . . ., but lim 2n
¼ lim 1n ¼ 0.
For example, 2n
n!1
n!1
We can now give the formal proof, promised earlier, that a convergent
sequence has only one limit.
Corollary 2
If lim an ¼ ‘ and lim an ¼ m, then ‘ ¼ m.
n!1
n!1
Proof Applying Theorem 3 with bn ¼ an, we deduce that ‘ m and m ‘.
&
Hence ‘ ¼ m.
In Sub-section 2.2.1, we saw that a sequence {an} is null if and only if the
sequence fj an jg is null. Our next result is a partial generalisation of this fact.
Theorem 4
If lim an ¼ ‘, then lim jan j ¼ j‘j.
n!1
n!1
This is a ‘proof by
contradiction’.
2.4
Divergent sequences
Proof
Using the ‘reverse form’ of the Triangle Inequality, we obtain
jan j j‘j jan ‘j; for n ¼ 1; 2; . . .:
Since {an ‘} is null, we deduce from the Squeeze Rule for null sequences that
&
fjan j j‘jg is null, as required.
61
You met this in Sub-section
1.3.1, expressed
in the form
ja bj jaj jbj. Here we
are writing an in place of a and
‘ in place of b.
Remark
Theorem 4 is only a partial generalisation of the earlier result about null
sequences, because the converse statement does not hold. That is, if janj ! j‘j,
it does NOT necessarily follow that an ! ‘. For example, consider the sequence
an ¼ (1)n, for n ¼ 1, 2, . . .; in this case, janj ! 1, but {an} does not even
converge.
2.4
2.4.1
Divergent sequences
What is a divergent sequence?
We have commented several times that not all sequences are convergent. We
now investigate the behaviour of sequences which do not converge.
Definition
A sequence is divergent if it is not convergent.
Here are the sequence diagrams for fð1Þn g, {2n} and fð1Þn ng. Each of
these sequences is divergent but, as you can see, they behave differently.
It is rather tricky to prove, directly from the definition, that these sequences
are divergent.
The aim of this section is to obtain criteria for divergence, which avoid
having to argue directly from the definition. At the end of the section, we give
a strategy involving two criteria which deal with all types of divergence.
We obtain these criteria by establishing certain properties, which are necessarily possessed by a convergent sequence; if a sequence does not have these
properties, then it must be divergent.
For example, to show that the
sequence fð1Þn g is
divergent, we have to show
that fð1Þn g is not
convergent; that is, for every
real number ‘, the sequence
fð1Þn ‘g is not null.
2: Sequences
62
2.4.2
Bounded and unbounded sequences
One property possessed by a convergent sequence is that it must be bounded.
Definition
A sequence {an} is bounded if there is a number K such that
jan j K;
for n ¼ 1; 2; . . .:
A sequence is unbounded if it is not bounded.
Thus a sequence {an} is bounded if all the terms an lie on the sequence diagram
in a horizontal strip from K up to K, for some positive number K.
{an}
K
⏐an⏐ ≤ K
n
–K ≤ an ≤ K
–K
For example, the sequence fð1Þn g is bounded, because
jð1Þn j 1;
for n ¼ 1; 2; . . .:
However the sequences {2n} and {n2} are unbounded, since, for each number K,
we can find terms of these sequences whose absolute values are greater than K.
Problem 1
Classify the following sequences as bounded or unbounded:
n ;
(d) 1 1n .
(b) fð1Þn ng;
(c) 2nþ1
n
(a) f1 þ ð1Þn g;
The sequence fð1Þn g shows that:
a bounded sequence is not necessarily convergent.
However we can prove that:
a convergent sequence is necessarily bounded.
Theorem 1 Boundedness Theorem
If {an} is convergent, then {an} is bounded.
Proof We know that an ! ‘, for some real number ‘. Thus {an ‘} is a null
sequence, and so there is a number X such that
jan ‘j < 1;
Take " ¼ 1 in the definition of
a null sequence.
for all n > X:
For simplicity in the rest of the proof, we shall now assume that our initial
choice of X is a positive integer.
Now
jan j ¼ jðan ‘Þ þ ‘j
jan ‘j þ j‘j; by the Triangle Inquality:
It follows that
jan j 1 þ j‘j; for all n > X:
This is the type of inequality needed to prove that {an} is bounded, but it does
not include the terms a1, a2, . . ., aX. To complete the proof, we let K be the
maximum of the numbers ja1j, ja2j, . . ., jaXj, 1 þ j‘j.
K
+1
–1
X
n
–K
That is, K ¼ max{ja1j,
ja2j, . . ., jaXj, 1 þ j‘j}.
2.4
Divergent sequences
63
It follows that
jan j K;
for n ¼ 1; 2; . . .;
&
as required.
From Theorem 1, we obtain the following test for the divergence of a
sequence:
If {an} is unbounded, then {an} is divergent.
Corollary 1
For example, the sequences {2n} and fð1Þn ng are both unbounded, so they
are both divergent.
Problem 2 Classify the following sequences as convergent or divergent, and as bounded
unbounded:
n or
o
pffiffiffi
n
n2 þ n
(a) f ng;
(b) n2 þ 1 ;
(c) ð1Þn n2 ;
(d) nð1Þ :
2.4.3
Sequences which tend to infinity
Although the sequences {2n} and fð1Þn ng are both unbounded (and hence
divergent), there is a marked difference in their behaviour. The terms of both
sequences become arbitrarily large, but those of the sequence {2n} become
arbitrarily large and positive. To make this precise, we must explain what we
mean by ‘arbitrarily large and positive’.
Definition
The sequence {an} tends to infinity if:
for each positive number K, there is a number X such that
an > K;
for all n > X:
In this case, we write
Often, we omit ‘as n ! 1’.
an ! 1 as n ! 1 :
Remarks
1. In terms of the sequence diagram for {an}, this definition states that,
for each positive number K, the terms an eventually lie above the line at
height K.
K
{an}
X
n
2. If a sequence tends to infinity, then it is unbounded – and hence divergent,
by Corollary 1.
2: Sequences
64
3. If a sequence tends to infinity, then this remains true if we add, delete or
alter a finite number of terms.
4. In the definition we can replace the phrase ‘for each positive number K’ by
‘for each number K’; for, if the inequality ‘an > K, for all n > X’ holds for each
number K, then in particular it certainly holds for each positive number K.
A finite number of terms do
not matter.
There is a version of the Reciprocal Rule for sequences which tend to
infinity. This enables us to use our knowledge of null sequences to identify
sequences which tend to infinity.
Theorem 2
Reciprocal Rule
(a) If the sequence {an} satisfies both of the following conditions:
1. {an} is eventually positive,
2. fa1n g is a null sequence,
then an ! 1.
(b) If an ! 1, then a1n ! 0.
Proof of part (a)
To prove that an ! 1, we have to show that:
We prove only part (a): the
proof of part (b) is similar.
for each positive number K, there is a number X such that
an > K;
for all n > X:
(1)
Since {an} is eventually positive, we can choose a number X1 such that
an > 0;
Since
fa1n g
for all n > X1 :
is null, we can choose a number X2 such that
1 1
< ; for all n > X2 :
a K
n
Now, let X ¼ max {X1X2}; then
1
1
0 < < ; for all n > X:
an K
Here we are taking " ¼ K1 in the
definition of a null sequence.
We make this choice of X so
that BOTH of the preceding
inequalities hold for all n > X.
This statement is equivalent to the statement (1), so an ! 1, as required. &
Example 1 Use the Reciprocal Rule to prove that the following sequences
tendnto o
infinity:
3
(a) n2 ;
(c) fn! 10n g
(b) fn! þ 10n g;
Notice that, in parts (b) and
(c), n! is the dominant term.
Solution
3
(a) Each term of the sequence n2 is positive and n31=2 ¼ n23 . Now, n13 is a
2
3
basic null sequence and so n3 is null, by the Multiple Rule. Hence n2
tends to infinity, by the Reciprocal Rule.
(b) Each term of the sequence fn! þ 10n g is positive and
1
1
0
¼ 0;
lim
¼ lim n! 10n ¼
n
n!1 n! þ 10
n!1 1 þ
1
þ
0
n!
by the Combination Rules.
Hence fn! þ 10n g tends to infinity, by the Reciprocal Rule.
Alternatively, since
1
1 , the sequence
n!
þ 101 n n!
n! þ 10n is null, by the
Squeeze Rule for null
sequences.
2.4
Divergent sequences
(c) First note that
65
10n
n! 10 ¼ n! 1 ; for n ¼ 1; 2; . . .:
n!
10n n
Since n! is a basic null sequence, we know that 10n! is eventually less
n
than 1, and so n! 10 is eventually positive. Also
n
1
n!
lim
n!1 n!
1
0
¼ 0;
¼ lim
n ¼
10
n
n!1
10
10
1 n!
by the Combination Rules.
Hence {n! 10n} tends to infinity, by the Reciprocal Rule.
This follows by taking " ¼ 1
in the definition of a null
sequence.
&
There are also versions of the Combination Rules and Squeeze Rule for
sequences which tend to infinity. We state these without proof.
Theorem 3 Combination Rules
If {an} tends to infinity and {bn} tends to infinity, then:
Sum Rule
Multiple Rule
{an þ bn} tends to infinity;
{lan} tends to infinity, for l > 0;
Product Rule
{anbn} tends to infinity.
Theorem 4 Squeeze Rule
If {bn} tends to infinity, and
an bn ;
for n ¼ 1; 2; . . .;
then {an} tends to infinity.
Problem 3
n
(a) 2n ;
For each of the following sequences {an}, prove thatnan ! 1:
o
n
n
2
(b) 2n n100 ;
(c) 2n þ 5n100 ;
(d) 2n10þþnn .
We can also define {an} tends to minus infinity.
Definition The sequence {an} tends to minus infinity if
an ! 1 as n ! 1:
In this case, we write
an ! 1
as n ! 1:
For example, the sequences {n2} and f10n n!g both tend to minus infinity,
because {n2} and fn! 10n g tend to infinity. Sequences which tend to minus
infinity are unbounded, and hence divergent. However, the sequence fð1Þn ng
shows that an unbounded sequence need not tend to infinity or to minus infinity.
2.4.4
Subsequences
We now give two useful criteria for establishing that a sequence diverges; both
of them involve the idea of a subsequence. For example, consider the bounded
divergent sequence fð1Þn g. This sequence splits naturally into two:
This follows from the fact that
sequences which tend to
infinity are unbounded.
2: Sequences
66
the even terms a2, a4, . . ., a2k, . . ., each of which equals 1;
{(–1)n}
the odd terms a1, a3, . . ., a2k 1, . . ., each of which equals 1.
{a2k}
1
Both of these are sequences in their own right, and we call them the even
subsequence {a2k} and the odd subsequence {a2k 1}.
In general, given a sequence {an} we may consider many different subsequences, such as:
1
2
3
4
5
7
n
{a2k – 1}
–1
{a3k}, comprising the terms a3, a6, a9, . . .;
{a4kþ1}, comprising the terms a5, a9, a13, . . .;
6
{a2k!}, comprising the terms a2, a4, a12, . . ..
Definition The sequence fank g is a subsequence of the sequence {an} if
{nk} is a strictly increasing sequence of positive integers; that is, if
n1 < n2 < n3 < . . .:
The sequence fank g is the
sequence
an1 ; an2 ; an3 ; . . .:
For example, the subsequence {a5kþ2} corresponds to the sequence of
positive integers
nk ¼ 5k þ 2; k ¼ 1; 2; . . .:
Note that the sequence {an} is
a subsequence of itself.
The first term of {a5k þ 2} is a7, the second is a12, the third is a17, and so on.
Notice that any strictly increasing sequence {nk} of positive integers must
tend to infinity, since it can be proved by Mathematical Induction that
nk k;
for k ¼ 1; 2; . . .:
Problem 4
(a) Let an ¼ n2, for n ¼ 1, 2, . . .. Write down the first five terms of each
of the subsequences fank g, where:
(i) nk ¼ 2k;
(ii) nk ¼ 4k 1; (iii) nk ¼ k2.
(b) Write down the first three terms of the odd and even subsequences of
n
the following sequence: an ¼ nð1Þ ; n ¼ 1; 2; . . .:
Our next theorem shows that certain properties of sequences are inherited by
their subsequences.
Theorem 5
Inheritance Property of Subsequences
For any subsequence fank g of {an}:
(a) if an ! ‘
as n ! 1, then ank ! ‘
as k ! 1;
(b) if an ! 1
as k ! 1.
as n ! 1, then ank ! 1
A result similar to part (b)
holds for sequences where
an ! 1.
Proof of part (a) We want to show that for each positive number ", there is a
number K such that
jank ‘j < ";
for all k > K:
(2)
However, since {an ‘} is null, we know that there is a number X such that
for all n > X:
jan ‘j < ";
You may omit this proof at a
first reading.We prove only
part (a): the proof of part (b) is
similar.
2.4
Divergent sequences
For simplicity in the rest of the proof, we shall now assume that our initial
choice of X is a positive integer.
So, if we take K so large that
nK X ;
then
nk > nK X; for all k > K;
and so
jank ‘j < "; for all k > K;
&
which proves (2).
67
We will occasionally make
this assumption if we want to
refer to terms such as aX rather
than mess about with nasty
expressions such as a[X],
where [X] is the integral part
of X.
The following criteria for establishing that a sequence is divergent are
immediate consequences of Theorem 5, part (a):
Corollary 2
1. First Subsequence Rule The sequence {an} is divergent if it has two
convergent subsequences with different limits.
2. Second Subsequence Rule The sequence {an} is divergent if it has a
subsequence which tends to infinity or a subsequence which tends to
minus infinity.
Though we do not prove the
fact, ANY divergent sequence
is of one (or both) of these two
types.
We can now formulate the strategy promised at the beginning of this section.
Strategy
To prove that a sequence {an} is divergent:
EITHER
1. show that {an} has two convergent subsequences with different limits
OR
The fact that this strategy will
always apply is simply a
reformulation of the result
mentioned in the above
margin note.
2. show that {an} has a subsequence which tends to infinity or a subsequence which tends to minus infinity.
For example, the sequence fð1Þn g has two convergent subsequences
which have different limits: namely, the even subsequence with limit 1 and
the odd subsequence with limit 1. So the sequence fð1Þn g is divergent, by
the First Subsequence Rule.
Þn
a subsequence (the even
On the other hand, the sequence nð1
ð1Þhas
n
subsequence) which tends to infinity. So n
is divergent by the Second
Subsequence Rule.
Remark
In order to apply the above strategy successfully to prove that a sequence is
divergent, you need to be able to spot convergent subsequences with different
limits or subsequences which tend to infinity or to minus infinity. It is not
always easy to do this, and some experimentation may be required! If the
formula for an involves the expression (1)n, it is a good idea to consider the
odd and even subsequences, although this may not always work. It may be
helpful to calculate the values of the first few terms in order to try to spot some
subsequences whose behaviour can be identified.
Problem 5 Use the above strategy to prove that each of the following
sequences {an} is divergent:
(a) ð1Þn þ 1n ;
(c) n sin 12 np .
(b) 13 n 13 n ;
In part (b) the square brackets
denote ‘the integer part’
function.
2: Sequences
68
We end this section by giving a result about subsequences which will be
needed in later chapters.
Theorem 6 If the odd and even subsequences of {an} both tend to the
same limit ‘, then
lim an ¼ ‘:
n!1
We want to show that:
Proof
for each positive number ", there is a number X such that
jan ‘j < "; for all n > X:
We know that there are integers K1 and K2 such that
ja2k1 ‘j < "; for all k > K1 ;
and
ja2k ‘j < "; for all k > K2 :
So we now let
(3)
X ¼ maxf2K1 1; 2K2 g:
With this choice of X, each n > X is either of the form 2k 1, with k > K1, or of
the form 2k, with k > K2; it follows then that (3) holds with this value of X. &
2.5
The Monotone Convergence Theorem
2.5.1
Monotonic sequences
In Section 2.3, we gave various techniques for finding the limit of a convergent
sequence. As a result, you may be under the impression that, if we know that a
sequence converges, then there is some way of finding its limit explicitly.
However, it is sometimes possible to prove that a sequence is convergent, even
though we do not know its limit. For example, this situation occurs with a given
sequence {an}, which has the following two properties:
1. {an} is an increasing sequence;
2. {an} is bounded above; that is, there is a real number M such that
an M; for n ¼ 1; 2; . . .:
Likewise, if {an} is a sequence which is decreasing and bounded below, then
{an} must be convergent.
We combine these two results into one statement.
Theorem 1
Monotone Convergence Theorem
If the sequence {an} is:
EITHER
1. increasing and 2. bounded above
OR
1. decreasing and 2. bounded below,
then {an} is convergent.
{a2k1} and {a2k} are the odd
and even subsequences of {an}.
2.5
The Monotone Convergence Theorem
69
Proof of the first case Let {an} be a sequence that is increasing and bounded
above. Since {an} is bounded above, the set {an: n ¼ 1, 2, . . .} has a least upper
bound, ‘ say; this is true by the Least Upper Bound Property of R. We now
prove that lim an ¼ ‘.
n!1
We want to show that:
We prove only the increasing
version; the proof of the
decreasing version is similar.
You met this Property in
Sub-section 1.4.3.
for each positive number ", there is a number X such that:
(1)
+ε
We know that, if " > 0, then, since ‘ is the least upper bound of the set
{an: n ¼ 1, 2, . . .}, there is an integer X such that
aX > ‘ ":
jan ‘j < ";
for all n > X:
X
Since {an} is increasing, an aX, for n > X, and so
an > ‘ ";
for all n > X:
for all n > X;
which proves (1).
Hence {an} converges to ‘.
n
This follows from the
definition of least upper
bound.
Thus
jan ‘j ¼ ‘ an < ";
–ε
jan ‘j ¼ ‘ an, because
an ‘.
&
The1Monotone Convergence Theorem tells us that a sequence such as
must
1 n , which is increasing and bounded above (by 1, for example),
be convergent. In this case, of course, we know already that 1 1n is
convergent (with limit 1) without using the Monotone Convergence Theorem.
The Monotone Convergence Theorem is often used when we suspect that a
sequence is convergent, but cannot find the actual limit. It can also be used to
give precise definitions of numbers, such as p, about which we have only an
intuitive idea, and we do this later in this section.
For completeness, we point out that:
We will use the Monotone
Convergence Theorem in
precisely this way in Chapter 3.
Sub-section 2.5.4.
If {an} is increasing but is not bounded above, then an ! 1 as n ! 1.
Indeed, for any real number K, we can find an integer X such that
aX > K;
because {an} is not bounded above. Since {an} is increasing, an aX for n > X,
and so
an > K; for all n > X:
Hence an ! 1 as n ! 1.
Similarly, we have the following result:
If {an} is decreasing but is not bounded below, then an ! 1 as n ! 1.
We summarise these results about monotonic sequences in the following
useful theorem:
Theorem 2
Monotonic Sequence Theorem
If the sequence {an} is monotonic, then:
EITHER
OR
{an} is convergent
an ! 1.
The expression ‘an ! 1’ is
shorthand for ‘either an ! 1
or an ! 1’.
2: Sequences
70
Next we note a consequence of the Monotone Convergence Theorem that is
sometimes useful.
Corollary 1
(a) If a sequence {an} is increasing and bounded above, then it tends to its
least upper bound, sup {an: n ¼ 1, 2, . . .}.
(b) If a sequence {an} is decreasing and bounded below, then it tends to its
greatest lower bound, inf {an: n ¼ 1, 2, . . .}.
Problem 1
Prove part (b) of the Corollary.
We meet applications of these results in the rest of this section, to functions
defined recursively and to the study of the numbers e and p. But our first
application is to a result of considerable value in Analysis, Topology and other
parts of Mathematics.
The proof of part (a) is
contained in the proof of
Theorem 1 above.
This proof is similar to the
earlier discussion.
We shall use it in Section 4.2.
The Bolzano–Weierstrass Theorem
We have seen that if a sequence is convergent, then certainly it must be
bounded. However if a sequence is bounded, it does not follow that it is
necessarily convergent. For example, the sequence {an}, where an ¼ 1n, is
bounded (by 1, since jan j ¼ j 1n j 1, 2, . . .) and is also convergent (to the
n
limit 0). Yet the
sequence {an}, where an ¼ (1) , is also bounded (by 1, since
n
jan j ¼ ð1Þ ¼ 1, for n ¼ 1, 2, . . .) but is divergent (since its odd and even
subsequences tend to different limits).
Our next result shows that, if a sequence is bounded, then, even though it
may diverge, it cannot behave ‘too badly’!
Theorem 3
This was the Boundedness
Theorem – Theorem 1,
Sub-section 2.4.2.
Bolzano–Weierstrass Theorem
A bounded sequence must contain a convergent subsequence.
That is, if a sequence {an} is such that janj M, for all n, then there exist
some number ‘ in [ M, M] and some subsequence fank g such that fank g
converges to ‘ as k ! 1.
For example, the sequence {sin n} is bounded. With the tools at our disposal,
it is not at all clear what the behaviour of this sequence might be as n ! 1!
However, the Bolzano–Weierstrass Theorem asserts that there is at least one
number ‘ in [1, 1] such that some subsequence {sin nk} converges to ‘. This is
itself a surprising result! In fact, however, using quite sophisticated mathematics we can prove that every number ‘ in [1, 1] has this property!
The proof of the Bolzano–Weierstrass Theorem that we give is of interest in
its own right; it depends on the notion of repeated bisection, which is a
standard technique in many areas of Analysis.
Proof We shall assume that the bounded sequence {an} has the property that
janj M, for all n. Then all the terms an lie in the closed interval [M, M],
which we will denote as [A1, B1].
Next, denote by p the midpoint of the interval [A1, B1]. Then at least one of
the two intervals [A1, p] and [p, B1] must contain infinitely many terms in the
For jsin nj 1.
Sadly, this is beyond the range
of this book.
A1 ¼ M and B1 ¼ M.
p ¼ 12 ðA1 þ B1 Þ.
2.5
The Monotone Convergence Theorem
sequence {an} – for otherwise the whole sequence would only contain finitely
many terms. If only one of the two intervals has this property, denote that
interval by the notation [A2, B2]; if both intervals have this property, choose the
left one (that is, [A1, p]) to be [A2, B2].
In either case, we obtain:
1. ½A2 , B2 ½A1 , B1 ;
2. B2 A2 ¼ 12 ðB1 A1 Þ;
3. [A1, B1] and [A2, B2] both contain infinitely many terms in the sequence {an}.
71
If both intervals contain
infinitely many terms in the
sequence, it does not matter
which choice we take; but we
specify the left interval just
for definiteness.
We now repeat this process indefinitely often, bisecting [A2, B2] to obtain
[A3, B3], and so on. This gives a sequence of closed intervals {[An, Bn]} for
which:
1. ½Anþ1 ; Bnþ1 ½An ; Bn , for each n 2 N;
n1
2. Bn An ¼ 12
ðB1 A1 Þ, for each n 2 N;
3. each interval [An, Bn] contains infinitely many terms in the sequence {an}.
Property 1 implies that the sequence {An} is increasing and bounded above by
B1 ¼ M. Hence, by the Monotone Convergence Theorem, {An} is convergent;
denote by A its limit. By the Limit Inequality Rule, we must have that A M.
Similarly, Property 1 implies that the sequence {Bn} is decreasing and
bounded below by A1 ¼ M. Hence, by the Monotone Convergence
Theorem, {Bn} is convergent; denote by B its limit. By the Limit Inequality
Rule, we must have that B M.
We may then deduce, by letting n ! 1 in Property 2 and using the
Combination Rules for sequences, that
Theorem 3, Sub-section 2.3.3.
B A ¼ lim Bn lim An ¼ lim ðBn An Þ
n!1
n!1
n!1
n1
1
¼ lim
ð B1 A 1 Þ
n!1 2
n1
1
¼ ðB1 A1 Þ lim
¼ 0:
n!1 2
In other words, the sequences {An} and {Bn} both converge to a common limit.
Denote this limit by ‘. Since ‘ ¼ A ¼ B, we must have M ‘ M.
We find a suitable subsequence of {an} as follows. Choose any an1 that lies in
[A1, B1]; this is possible since [A1, B1] contains infinitely terms in {an}. Next,
choose a term an2 in the sequence, with n2 > n1, such that an2 2 ½A2 ; B2 ; this is
possible since [A2, B2] contains infinitely terms in {an}, by Property 3. And so on.
In this way, we construct a subsequence fank g of {an}, with nk þ 1 > nk, for
each k. Since Ak ank Bk , it follows from the Squeeze Rule for sequences
&
(that is, by letting k ! 1) that fank g converges to ‘, as required.
2.5.2
Sequences defined by recursion formulas
As we have seen, often sequences are defined by formulas; that is, for a given n,
we substitute that value of n into a formula and obtain the term an in the sequence
{an} Another way of specifying a sequence is to define its terms ‘inductively’, or
‘recursively’; here we specify the first term (or several terms) in the sequence,
and then have a formula that enables us to calculate all successive terms.
That is, ‘ 2 [M, M].
Since nk þ 1 > nk, we must
have nk k, for each k.
For Ak ! A ¼ ‘ and
Bk ! B ¼ ‘.
2: Sequences
72
For example, the following sequences are defined recursively:
{an}, where a1 ¼ 2 and anþ1 ¼ 14 a2n þ 3 , for n 1;
{an}, where a1 ¼ 2, a2 ¼ 4 and an ¼ 12 ðan1 þ an2 Þ, for n 3.
Do these sequences converge? If they do converge, what are their limits? If we
choose a different value for a1 (or for a1 and a2, in the second sequence), do
their behaviours change? We can use the Monotone Convergence Theorem to
answer many such questions.
First, consider the sequence with recursion formula
1
anþ1 ¼ a2n þ 3 ; for n 1:
(2)
4
If indeed {an} is convergent, what value could its limit take? If we let ‘ denote
lim an and let n ! 1 in equation (2), we obtain ‘ ¼ 14 ð‘2 þ 3Þ. This can be
n!1
rearranged as 4‘ ¼ ‘2 þ 3, and so as ‘2 4‘ þ 3 ¼ 0. Hence (‘ 1)(‘ 3) ¼ 0,
so that ‘ = 1 or ‘ = 3.
This suggests that we should look at the differences anþ1 1 and anþ1 3.
Using equation (2), we deduce that
1
1
anþ1 1 ¼ a2n 1 ¼ ðan þ 1Þðan 1Þ; for n 1;
(3)
4
4
1
1
anþ1 3 ¼ a2n 9 ¼ ðan þ 3Þðan 3Þ; for n 1:
(4)
4
4
Of course we have not yet
proved that {an} is
convergent. This has been a
‘what if?’ discussion so far,
but a useful one.
Because 1 and 3 are the
possible limits.
Next, it is useful to examine whether the sequence is monotonic by looking
at the difference anþ1 an
1
1
anþ1 an ¼ a2n 4an þ 3 ¼ ðan 1Þðan 3Þ:
(5)
4
4
So the sign of anþ1 an depends on where an lies on the number-line in relation
to the numbers 1 and 3.
For example, it follows from (5) that
if 1 < an < 3, for some n 1, then anþ1 an < 0,
so that anþ1 < an :
(6)
If an ¼ 1, for some number n 1, it follows from equation (3) that anþ1 ¼ 1.
Consequently, if our initial term is a1 ¼ 1, then it follows that an ¼ 1, for all
n 1; in other words, the sequence is simply the constant sequence {1}.
Similarly, if an ¼ 3, for some number n 1, it follows from equation (3) that
anþ1 ¼ 3. Consequently, if our initial term is a1 ¼ 3, then it follows that an ¼ 3,
for all n 1; in other words, the sequence is simply the constant sequence {3}.
Next, suppose that 1 < an < 3, for some number n 1. It follows from
equation (3) that anþ1 1 is positive, so that 1 < anþ1. On the other hand, it
follows from equation (4) that anþ1 3 is negative, so that anþ1 < 3.
Thus, if 1 < an < 3, for some number n 1, then 1 < an þ 1 < 3.
We can then prove by induction that, if 1 < a1 < 3, then 1 < an < 3, for all
n > 1. (We omit the details.)
Next, it follows, from this last fact and (6) above that, if 1 < a1 < 3, then the
sequence {an} is strictly decreasing.
Hence, if 1 < a1 < 3 the sequence {an} is strictly decreasing and is bounded
below. Hence, by the Monotone Convergence Theorem, {an} is convergent.
We shall use this fact below.
In particular this discussion
covers the case a1 ¼ 2 that we
mentioned at the start of this
sub-section.
Now comes the coup de
grace!
2.5
The Monotone Convergence Theorem
Since {an} is decreasing, whatever its limit might be (and we know that the only
two possibilities for the limit are 1 and 3), its limit must be less than or equal to its
first term a1. It follows that the limit of the sequence must be 1.
73
We saw that 1 and 3 were the
only possible limits at the start
of the discussion.
Problem2 Letthe sequence {an} be defined by the recursion formula
anþ1 ¼ 14 a2n þ 3 , for n 1.
(a) Prove that, if a1 > 3, then an ! 1 as n ! 1.
(b) In the case that 0 a1 < 1, determine whether {an} is convergent
and, if so, to what limit.
(c) Describe the behaviour of the sequence {an} in the case that a1 < 0.
2.5.3
The number e
n We will define e to be the limit of the sequence 1 þ 1n . If we plot the first
few terms of this sequence on a sequence diagram, then it seems that the
sequence is increasing and converges to a limit, which is less than 3.
n is convergent, using the Monotone Convergence
To show that 1 þ 1n
Theorem, we prove that the sequence is increasing and is bounded above.
n 1.
1 þ 1n
is increasing
By the Binomial Theorem
n
1 n
1
nð n 1Þ 1 2 . . .
1
¼1þn
þ
þ
:
þ
1þ
n
n
2!
n
n
The general term in this expansion is
nð n 1Þ . . . ð n k þ 1Þ 1 k
k!
n
1
1
2
k1
(7)
1
¼
1
... 1 :
k!
n
n
n
If k is fixed, then each of the factors
1
2
k1
1 ; 1 ;...; 1 n
n
n
is increasing, and so the general term (7) increases
as n increases. Since this
is true for each fixed k, the sequence ð1 þ 1nÞn is increasing.
n 2.
1 þ 1n
is bounded above
Here, note that the general term (7) satisfies the inequality
1
1
2
k1
1
1
1
... 1 ;
k!
n
n
n
k!
For example, to three decimal
places the first two terms are 2
and 2.25, the sixth term is
2.521, and the hundredth term
is 2.704.
2: Sequences
74
since each of the brackets is at most 1. Hence
1 n
1
1
1
1 þ 1 þ þ þ ... þ
1þ
n
2! 3!
n!
1
1
1
1 þ 1 þ 1 þ 2 þ . . . þ n1 ;
2
2
2
since k! ¼ k (k 1) . . . 2.1 2k1.
Now
n
n 1 12
1
1
1
1
¼2 1
1 þ 1 þ 2 þ þ n1 ¼ 1 1
2
2
2
12
2
1
¼ 2 n1 ;
2
and so
1 n
1
1þ
3 n1 ; for n ¼ 1; 2; . . .:
n
2
n is bounded above, by 3.
Thus the sequence 1 þ 1n
For the sum of the geometric
progression a þn ar þ ar2 þ . . .
þ arn1 is a 1r
1r , if r 6¼ 1;
here we have a ¼ 1, r ¼ 12 :
It follows,
n by the Monotone Convergence Theorem, that the sequence
1 þ 1n
is convergent with limit at most 3. This allows us to make the
following definition:
n
Definition e ¼ lim 1 þ 1n :
n!1
n
For larger and larger values of n, the terms 1 þ 1n give
better
n and better
approximate values for e. Unfortunately, the sequence 1 þ 1n
converges
to e rather slowly, and we need to take very large integers n to get a reasonable
approximation to e ¼ 2.71828. . ..
Now we
like in general to make the definition of ex as
would
x
x n
e ¼ lim 1 þ n , but first we need to know that this limit actually exists.
n!1
For example
1 1000
¼ 2:716 . . .:
1 þ 1000
Then e ¼ e1
Problem 3 Let x > 0.
(a) Prove that ð1 þ nxÞn is an increasing sequence, by adapting the
method above where x ¼ 1.
k
1 k
(b) Verify
that x1 nþ
n ð1 þ nÞ , for k ¼ 1, 2, . . .. Using this fact, prove
that ð1 þ nÞ is bounded above.
(c) Deduce that ð1 þ nxÞn is convergent.
Problem 4 By considering
the product
nof the first n terms of the
, for n ¼ 1, 2, . . ..
sequence ð1 þ 1nÞn , prove that n! > nþ1
e
We now examine the convergence of the sequence ð1 þ nxÞn , for x < 0.
Recall that, when x < 0, then
x > 0; in particular, from the above discussion, the sequence ð1 nxÞn converges. We shall use Bernoulli’s inequality
(1 þ c)n 1 þ nc, for c 1.
In investigating the convergence of the sequence ð1 þ nxÞn , we are
only interested in what happens when n is large. So, we need consider
2
only the situation when n > x. Then n2 > x2, so that n12 < x12 ; thus nx2 < 1,
x2
or n2 > 1.
n
So lim 1 þ nx exists
n!1
for x > 0. Note that the limit is
at least 1, and so cannot be
zero.
You met Bernoulli’s
Inequality in Subsection 1.3.3.
Recall that x > 0.
2.5
The Monotone Convergence Theorem
75
2
It follows that we may substitute nx2 for c in Bernoulli’s Inequality. This
gives that
n
2
x2
x
1 2 1þn 2
n
n
x2
¼ 1 ; for n > x:
n
It follows that
n
x2
x2
1 1 2 1 ; for n > x:
(8)
n
n
n
o
2
Now the sequence 1 xn converges to the limit 1, by the Combination
Rules. Hence, by applying the Squeeze Rule to inequality (8), the sequence
n
n o
2
1 nx2
converges to the limit 1 as n ! 1
Here we use the fact that f1ng is
a basic null sequence.
But
n n
1 þ nx 1 nx
x n
n
¼
1þ
n
1 nx
n
2
1 nx2
n :
¼ 1 nx
(9)
We have just seen that the numerator is convergent, and we saw above that the
denominator is convergent (to a non-zerolimit) forx > 0; it follows from (9),
by the Quotient Rule, that the sequence ð1 þ nxÞn is convergent. So whatever value we choose for the real number x, the sequence ð1 þ nxÞn
is convergent. This means that we can now make the following legitimate
definition:
Definition
For any real value of x,
x n
:
ex ¼ lim 1 þ
n!1
n
Recall that here we are
considering the case x < 0.
Note that when x ¼ 0, the
sequence is simply the
constant sequence {1}.
The function
x 7! ex , for x 2 R, is called
the exponential function.
We can deduce more than this from equation (9), if we rewrite it in the
convenient form
n
x n x n
x2
1
¼ 1 2 :
1þ
n
n
n
Letting n ! 1, we deduce from this last equation that
ex ex ¼ 1:
We have shown that this holds for x < 0; however, by simply interchanging x
and x, it is clear that this holds for x > 0 also.
Here wealso usethe fact that
n
x2
lim 1 2 ¼ 1;
n!1
n
that we proved above.
Theorem 4 Inverse Property of ex
For any real value of x, ex ex ¼ 1
The next property of the exponential function that we need to verify is that
ex ey ¼ ex þ y for any real values of x and y. We shall prove this fact later.
In Sub-section 3.4.3.
2: Sequences
76
2.5.4
The number p
One of the oldest mathematical problems is to determine the area of a disc of
radius r, and the length of its perimeter. It is well known that these magnitudes
are pr2 and 2pr, respectively. But what exactly is p? Is there a real number p
which makes these formulas correct? – and, if so, how is it formally defined?
We now give a precise definition of p as the area of a disc of radius 1, using a
method originally devised by Archimedes to calculate approximate values for
the area of this disc. The idea is to calculate the areas of regular polygons
inscribed in the disc. Archimedes found an easy way to calculate the areas of
such regular polygons with 6 sides, 12 sides, 24 sides and so on. The results of
parts (a) and (b) of the following problem help to give the first two of these areas.
Problem 5
Here ‘inscribed’ means that
all the vertices of the polygon
lie on the circle, so that the
inside of the polygon is
contained in the inside of the
circle.
Verify that the following triangles have the stated areas:
Here p is used only as a
symbol to represent angles; its
value is not required.
2 tan 1#
Hint: In part (d), use the half-angle formula tan # ¼ 1tan22 1# :
2
Let sn denote the number of sides of the nth such inner polygon (so s1 ¼ 6,
s2 ¼ 12, s3 ¼ 24, and, in general, sn ¼ 3 2n) and let an denote the area of the
nth inner polygon. Then
1
2p
an ¼ sn sin
; for n ¼ 1; 2; . . .:
(10)
2
sn
Geometrically, it is obvious that each time we double the number of sides of
the inner polygon the area increases, and so
a1 < a2 < a3 < . . . < an < anþ1 < . . .:
Hence {an} is (strictly) increasing. But is {an} convergent?
For example, to three decimal
places
a1 ¼ 2:598,
a2 ¼ 3,
...
a6 ¼ 3:141:
2.5
The Monotone Convergence Theorem
77
Notice that each of the polygons lies inside a square of side 2, which has
area 4. This means that
an 4;
for n ¼ 1; 2; . . .;
and so {an} is increasing and bounded above (by 4). Hence, by the Monotone
Convergence Theorem, {an} is convergent.
Our intuitive idea of the area of the disc suggests that its area is greater than
each of the areas an, but ‘only just’! Put another way, the area of the disc should
be the limit of the increasing sequence {an}. This leads us to make the
following definition:
Definition
p ¼ lim an :
n!1
We shall explain in a moment how to calculate the terms an without
assuming a value for p.
First, however, we describe how to estimate the area of the disc using outer
polygons. Once again we start with a regular hexagon and repeatedly double
the number of sides. The results of part (c) and (d) of the last problem help to
give the first two such areas.
As before, let sn ¼ 3 2n, for n ¼ 1, 2, . . ., and let bn denote the area of the
nth outer polygon. This nth outer polygon consists of sn isosceles triangles,
each of height 1 and base 2 tanðspn Þ. Thus
p
bn ¼ sn tan
; for n ¼ 1; 2; . . .:
(11)
sn
This will enable us to squeeze
p between the area of the
inner polygons and the area of
the outer polygons.
For example, to three decimal
places
b1 ¼ 3:464;
Geometrically, it is obvious that each time we double the number of sides of
the outer polygon the area decreases, and so
b1 > b2 > b3 > > bn > bnþ1 > :
So the sequence {bn} is (strictly) decreasing and bounded below (by 0,
for example). Thus, by the Monotone Convergence Theorem, {bn} is also
convergent.
Intuitively, we expect that {bn} has the same limit as {an}, which we have
defined to be p. But how can we prove this?
b2 ¼ 3:215;
...
b6 ¼ 3:142:
2: Sequences
78
The terms an and bn can be calculated by using the following equations,
which are known jointly as the Euclidean algorithm
pffiffiffiffiffiffiffiffiffi
anþ1 ¼ an bn ; for n ¼ 1; 2; . . .;
(12)
and
2anþ1 bn
; for n ¼ 1; 2; . . .:
(13)
anþ1 þ bn
pffiffiffi
pffiffiffi
Starting with a1 ¼ 32 3 ¼ 2:598 . . . and b1 ¼p2ffiffiffiffiffiffiffiffiffi
3 ¼ 3:464 . . ., we use
2 b1
these equations iteratively to calculate first a2 ¼ a1 b1 , then b2 ¼ a2a2 þb
, and
1
so on. Here are (approximations to) the first few values of each sequence
obtained in this way:
bnþ1 ¼
sn
6
12
an 2.598 3
24
48
96
Equations (12) and (13)
can be verified from
equations (10) and (11); we
give the details in Subsection 2.5.5 below.
192
3.106 3.133 3.139 3.141
bn 3.464 3.215 3.160 3.146 3.143 3.142
It does appear that both sequences do converge to a common limit, namely p.
Remark
We prove this in Subsection 2.5.5 below.
We have defined p using the areas of approximating polygons. An alternative
approach uses the perimeters of these polygons.
2.5.5
Proofs
You may omit this subsection at a first reading.
We now prove several of the results referred to in the discussion of p in the
previous sub-section.
First, we prove equations (12) and (13)
pffiffiffiffiffiffiffiffiffi
anþ1 ¼ an bn ; for n ¼ 1; 2; . . .;
(12)
and
bnþ1
2anþ1 bn
¼
;
anþ1 þ bn
for n ¼ 1; 2; . . .:
(13)
Using the half-angle formulas for sin # and cos #, it is easy to check that
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
1
1
sin # tan #
sin # ¼
2
2
2
and
1
sin # tan #
:
tan # ¼
2
sin # þ tan #
Hence, since snþ1 ¼ 2sn
1
2p
anþ1 ¼ snþ1 sin
2
snþ1
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
2p
p
sn sin
¼
sn tan
2
sn
sn
pffiffiffiffiffiffiffiffiffi
¼ an bn :
These comprise the Euclidean
algorithm.
We omit the details in both
cases.
This proves equation (12).
2.6
Exercises
79
Similarly
bnþ1 ¼ snþ1 tan
p
snþ1
2sn sin spn sn tan spn
¼
sn sin spn þ sn tan spn
¼
2anþ1 bn
:
anþ1 þ bn
This proves equation (13).
Finally we prove that lim bn ¼ lim an .
n!1
n!1
To do this, we rearrange equation (12) as follows
bn ¼
a2nþ1
:
an
Then, by the Combination Rules, we can deduce that
2
lim anþ1
p2
¼ p:
¼
lim bn ¼ n!1
n!1
lim an
p
n!1
Then, since we defined p to be precisely lim an , it follows that lim bn ¼ lim an ,
n!1
n!1
n!1
as required.
Definition Let an denote the area of the regular polygon with 3 2n sides
inscribed in a disc of radius 1, and bn the area of the regular polygon with
3 2n sides circumscribing the disc. Then
p is the common value of the limits lim an and lim bn .
n!1
n!1
Remark
It can be shown that every two applications of equations (12) and (13) give an
extra decimal place in the decimal expansion for p. This decimal expansion is
now known to many millions of decimal places, using sequences which
converge to p much more rapidly than {an} or {bn}. Since p is irrational,
there is no possibility of the decimal expansion of p recurring.
2.6
Exercises
Section 2.1
1. Calculate the first five terms of each of the following sequences, and draw a
sequence diagram in each case:
n nþ1 o
(b) ð1n!Þ
(a) {n2 4n þ 4};
;
(c) sin 14 np .
We shall return in Section 8.5
to the question of
approximating p, by the use of
power series.
2: Sequences
80
2. Determine which of the following sequences are monotonic:
n no
1
1
ð1Þ
(a) nn þ
;
(c) 2n .
;
(b)
þ2
n
3. Prove that the following sequences are eventually monotonic:
n
(a) 5n! ;
(b) n þ 8n .
Section 2.2
1. For each of the following sequences {an} and numbers ", find a number X
such that janj < ", for all n > X:
n
Þ
, " ¼ 0.001;
(a) an ¼ ð1
n5
(b) an ¼ ð2n þ1 1Þ2 , " ¼ 0.002.
2. Use the definition of null sequence to prove that the two sequences in the
previous exercise are null.
3. Prove that the following sequences are not null:
n
o
n
pffiffiffi
(a) f ng;
(b) 1 þ ð1n Þ .
4. Assuming that 1n is null, deduce that the following sequences are null.
State which rules you use.
n
o
n
o
n 1
o
ð2nÞ
n
(b) nsin
(c) ntan
(a) p2ffiffin þ n37 ;
2 þ1 ;
þ 1 þ sin n .
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi
5. Prove that
n þ 1 n is null.
2
b2
Hint: Use the fact that a b ¼ aa þ
b , for a þ b 6¼ 0.
6. Use the list of basic null sequences to prove that the following sequences are
null. State which rules you use.
10 10 n (b) 6nn! ;
(c) n n!10 .
(a) 43n þ 2n
3n ;
7. Prove that, if the sequence {an} of positive numbers is null, then
null.
pffiffiffiffiffi
an is
Section 2.3
1. Show that the following sequences converge to 1, by calculating an 1 in
each case:
n2 1
(b) n2 þnþ1
.
(a) nn þ3 ;
2. Use the Combination Rules to find the limits of the following sequences:
n
o
þ 5n 10 2n 3
(a) 2n3 þn3n þ 4 ;
;
(c) n5n!
(b) 2nn þn100
100 þ n! .
n
3. (a) Prove that, if the sequence{an}of positive numbers is convergent
pffiffi with
pffiffiffiffiffi
‘ .
limit ‘, where ‘ > 0, then
an is convergent with limit
You met this list in Subsection 2.2.1.
2.6
Exercises
81
2
2
b
Hint: Use the fact that a b ¼ aa þ
b , for a þ b 6¼ 0.
1
(b) Prove that, if the sequence {ak} is convergent with limit ‘, ‘ 6¼ 0, and ank
n 1o
1
is defined, where n 2 N, then ank is convergent with limit ‘n .
Hint: Use the identity aq bq ¼ (a b)(aq1 þ aq2b þ aq3b2 þ þ
bq1) with suitable choices for a, b and q.
Section 2.4
1. Classify the following sequences as convergent or divergent, and as
bounded or unbounded:
n n no
1
(b) ð1Þn!100 .
(a) n4 ;
2. Use the Reciprocal Rule to prove that the following sequences tend to
infinity:
(a) nn!3 ;
(b) n2 þ 2n ;
(c) n2 2n ;
(d) n! n3 3n .
3. Use the Subsequence Rules to prove that the following sequences are
divergent:
n n 2o
1Þ n
(a) fð1Þn 2n g;
(b) ð2n
.
2 þ1
4. Each of the following general statements concerning sequences {an} and {bn}
is true or false. For each statement that is true, prove it. For each statement that
is false, write down examples of sequences to illustrate your assertion:
(a) If {an} and {bn} are divergent, then {an þ bn} is divergent.
(b) If {an} and {bn} are divergent, then {anbn} is divergent.
(c) If {an} is convergent and {bn} is divergent, then {anbn} is divergent.
(d) If a2n is convergent, then {an} is convergent.
2
(e) If an is convergent, and an > 0, then {an} is convergent.
(f) If {bn} is convergent, where
then {an} is convergent.
n bn ¼ anþ1an, o
anþ1 þanþ2
þ
þa2n
ffiffi
p
(g) If an ! 0 as n ! 1, then
is convergent.
n
5. Prove that every sequence has a monotonic subsequence.
Section 2.5
1. Prove that, if {an} is increasing and has a subsequence fank g which is
convergent, then {an} is convergent.
3an þ 1
2. Let fan g1
n¼1 be a sequence for which anþ1 ¼ an þ3 , for n 1.
(a) Prove that, if a1 > 1, the sequence {an} converges.
Hint: consider the cases 1 < a1 < 1, a1 ¼ 1 and a1 > 1 separately.
(b) If a1 1, does the sequence converge or diverge?
The Power Rule in Subsection 2.2.1 and the
subsequent remark there show
that this result holds in the
case that ‘ ¼ 0.
2: Sequences
82
2
¼ 12 an þ a‘ n , for n 1, where
3. Let fan g1
n¼1 be a sequence for which anþ1
a1 > 0 and ‘ > 0.
(a) Prove that {an} is decreasing for n 2, and hence that an ! ‘ as n ! 1.
(b) With ‘2 ¼ 2 and a1 ¼ 1.5, calculate the terms a2, a3, a4 and a5.
4. Obtain formulas for the perimeters of the inner and outer polygons introduced in Sub-section 2.5.4. Prove that these sequences tend to 2p as n ! 1.
5. Bernoulli’s Inequality states that (1 þ x)n 1 þ nx, for x 1, n ¼ 1, 2, . . ..
1
(a) Apply
Bernoulli’s
Inequality, with x ¼ n2 , to give an alternative proof
1 n
is increasing.
that 1 þ n
n
nþ1 o
(b) Apply Bernoulli’s Inequality, with x ¼ n211, to prove that 1 þ 1n
is decreasing.
(c) Deduce from parts (a) and (b) that
1 n
1 nþ1
1þ
e 1þ
;
n
n
You met this in Subsection 1.3.3.
for n ¼ 1; 2; . . .:
6. Let f½an ; bn g1
n¼1 be a sequence of closed intervals with the property that
½anþ1 ; bnþ1 ½an ; bn , for n ¼ 1, 2, . . ..
(a) Prove that there exists some point c that lies in all the intervals [an, bn],
for n ¼ 1, 2, . . ..
(b) Prove that, if bn an ! 0 as n ! 1, then there exists only one such
point c.
This result is sometimes
called The Nested Intervals
Theorem.
3
Series
The Ancient Greek philosopher Zeno of Elea proposed a number of paradoxes
of the infinite, which are said to have had a profound influence on Greek
mathematics. For example, Zeno claimed that it is impossible for an object to
travel a given distance, since it must first travel half the distance, then half
of the remaining distance, then half of what remains, and so on. There must
always remain some distance left to travel, and so the journey cannot be
completed.
This paradox relies partly on the intuitive feeling that it is impossible to add
up an infinite number of positive quantities and obtain a finite answer.
However, the following illustration of the paradox suggests that such an
infinite sum is plausible.
1
4
1
2
0
1
2
1
8
3
4
1
16
7 15
8 16
1
The distance from 0 to 1 can be split up into the infinite sequence of
distances 12 , 14 , 18, . . ., and so it seems reasonable to write
1 1 1 1
þ þ þ þ ¼ 1:
2 4 8 16
This chapter is devoted to the study of such expressions, which are called
infinite series.
The following example shows that infinite series need to be treated with
care. Suppose that it is possible to add up 2, 4, 8, . . ., and that the answer is s
2 þ 4 þ 8 þ 16 þ ¼ s:
If we multiply through by 12, we obtain
1
1 þ 2 þ 4 þ 8 þ ¼ s;
2
which we can rewrite in the form
1
1 þ s ¼ s:
2
It follows from this equation that s ¼ 2, which is obviously non-sensical.
We can avoid reaching such absurd conclusions by performing arithmetical
operations only with convergent infinite series.
In Section 3.1 we define the concept of convergent infinite series in terms
of convergent sequences, and give some examples. We also describe various
properties which are common to all convergent series.
Because the left-hand side of
the previous equation can be
expressed as
1 þ ð2 þ 4 þ 8 þ Þ
¼ 1 þ s:
83
3: Series
84
Section 3.2 is devoted to series with non-negative terms, and we give several
tests for the convergence of such series.
In Section 3.3 we deal with the much harder problem of determining
whether a series is convergent when it contains both positive and negative
terms. We also look at what can happen if we rearrange the terms of a
convergent series – does the new series still converge? This section is quite
long, so you might wish to tackle it in several sessions.
In Section 3.4 we explain how ex can be represented as an infinite series of
powers of x, and we use this representation to prove that the number e is
irrational, and that ex ey ¼ exþy .
3.1
3.1.1
You first met ex in Section 2.5.
Introducing series
What is a convergent series?
We begin by defining precisely what is meant by the statement that
1 1 1 1
þ þ þ þ ¼ 1:
2 4 8 16
Let sn be the sum of the first n terms on the left-hand side. Then
1
s1 ¼ ;
2
1 1 3
s2 ¼ þ ¼ ;
2 4 4
1 1 1 7
s3 ¼ þ þ ¼ ; . . .; and so on:
2 4 8 8
In general, using the formula for the sum of a geometric progression, with
a ¼ r ¼ 12, we obtain
1 1 1
1
1 1
1
1
sn ¼ þ þ þ þ n ¼ þ 2 þ 3 þ þ n
2 4 8
2
2 2
2
2
n
n
1 1 12
1
¼ ¼1
:
1
2
2
12
n is a null sequence, and so
The sequence 12
n 1
lim sn ¼ lim 1 n!1
n!1
2
n
1
¼ 1 lim
¼ 1:
n!1 2
It is the precise mathematical statement that sn ! 1 as n ! 1 which justifies
our earlier (intuitive) statement that
1 1 1 1
þ þ þ þ ¼ 1:
2 4 8 16
We use this approach to define a convergent infinite series.
The sum of a geometric
progression a þ ar þ
n
ar 2 þ þ arn1 is a 1r
1r ,
if r 6¼ 1.
Here we use the Combination
Rules for sequences.
3.1
Introducing series
85
Given a sequence {an} of real numbers, the expression
Definition
a1 þ a2 þ a3 þ is called an infinite series, or simply a series. The nth partial sum of this
series is
sn ¼ a1 þ a2 þ a3 þ þ an :
The behaviour of the infinite series a1 þ a2 þ a3 þ is determined by the
behaviour of {sn}, its sequence of partial sums.
Definition The series a1 þ a2 þ a3 þ is convergent with sum s if the
sequence {sn} of partial sums converges to s. In this case, we say that the
series converges to s, and we write
a1 þ a2 þ a3 þ ¼ s:
The series diverges if the sequence {sn} diverges.
Thus we prove results about series by applying results for sequences proved
in Chapter 2 to the sequence of partial sums {sn}.
Example 1 For each of the following infinite series, calculate its nth partial
sum, and determine whether the series is convergent or divergent:
(a) 1 þ 1 þ 1 þ ;
(b)
1
3
þ 312 þ 313 þ ;
(c) 2 þ 4 þ 8 þ .
Solution
(a) In this case
sn ¼ 1 þ 1 þ þ 1 ¼ n:
The sequence {sn} tends to infinity, and so this series is divergent.
(b) Using the formula for summing a finite geometric progression, with
a ¼ r ¼ 13, we obtain
1 1
1
1
þ þ þ þ n
3 32 33
3
n
n 1 1 13
1
1
1
¼ ¼
:
3
2
3
1 13
sn ¼
Since
1n 3
is a basic null sequence
1
lim sn ¼ ;
2
and so this series is convergent, with sum 12.
(c) In this case
n!1
s n ¼ 2 þ 4 þ 8 þ þ 2n
n
2 1
¼2
¼ 2nþ1 2:
21
The sequence {2nþ1 2} tends to infinity, and so this series is divergent.&
3: Series
86
Sigma notation
Next, we explain how to use the sigma notation to represent infinite series.
Recall that a finite sum such as
1 1 1
1
1
1
1
1
þ þ þ þ
¼ þ þ þ þ 10
2 4 8
1024 21 22 23
2
can be represented using sigma notation as
is the Greek capital letter
‘sigma’.
10
X
1
:
2n
n¼1
This notation can be adapted to represent infinite series, as follows
1
X
an ¼ a1 þ a2 þ a3 þ :
For example
1
X
1
1 1 1
¼ þ þ þ ;
2n 2 4 8
n¼1
1
X
1
n¼1
n
n¼1
Convention When using the sigma notation to represent the nth partial
sum sn of a series, we write
n
X
ak :
s n ¼ a1 þ a2 þ a3 þ þ an ¼
k¼1
We have written k as the ‘dummy variable’ here, to avoid using n for two
different purposes in the same expression. We could have used any letter (other
n
n
P
P
than n) for the dummy variable; for example, the expressions
ai and
ar
i¼1
r¼1
also stand for a1 þ a2 þ a3 þ þ an.
If we need to begin a series with a term other than a1, then we write, for example
1
1
X
X
an ¼ a0 þ a1 þ a2 þ or
an ¼ a3 þ a4 þ a5 þ :
n¼0
n¼3
For any series, the nth partial sum sn is obtained by adding the first n terms. For
example, the nth partial sums of the above two series are
sn ¼ a0 þ a1 þ a2 þ þ an1 ¼
n1
X
ak
and
k¼0
sn ¼ a3 þ a4 þ a5 þ þ anþ2 ¼
nþ2
X
1 1 1
¼ 1þ þ þ þ :
2 3 4
For the above examples
1 1 1
1
sn ¼ þ þ þ þ n
2 4 8
2
n
X
1
¼
;
k
2
k¼1
1 1 1
1
sn ¼ 1 þ þ þ þ þ
2 3 4
n
n
X
1
¼
:
k
k¼1
For example, we write
1
X
1
n!
n
n¼3
for the series
1
1
þ
þ :
3! 3 4! 4
ak :
k¼3
Problem 1 For each of the following infinite series, calculate the nth
partial sum sn, and determine whether the series is convergent or divergent:
1 1
1 n
P
P
P
1 n
13 ;
(b)
ð1Þn ;
(c)
(a)
2 :
n¼1
n¼0
n¼1
Remark
Notice that inserting into a series, or omitting or altering, a finite number of
terms does not affect the convergence of the series, but may affect the sum. For
example, since the series 12 þ 14 þ 18 þ converges with limit 1, it follows that
Recall that the convergence or
divergence of sequences was
not affected by omitting or
altering a finite number of
terms in the sequence.
3.1
Introducing series
87
the series 8 þ 4 þ 2 þ 1 þ 12 þ 14 þ 18 þ ¼ ð8 þ 4 þ 2 þ 1Þ þ 12 þ 14 þ 18 þ converges with limit (8 þ 4 þ 2 þ 1) þ 1 ¼ 16.
3.1.2
For 12 þ 14 þ 18 þ ¼ 1:
Geometric series
All the series considered so far are examples of geometric series. The standard
geometric series with first term a and common ratio r is
1
X
ar n ¼ a þ ar þ ar 2 þ :
n¼0
The following theorem enables us to decide whether any given geometric
series is convergent or divergent.
Theorem 1
Geometric Series
1
P
a
(a) If jrj < 1, then
ar n is convergent, with sum 1r
.
n¼0
1
P
n
(b) If jrj 1 and a 6¼ 0, then
ar is divergent.
n¼0
Proof
(a) If r 6¼ 1, then the nth partial sum sn is given by
1 rn
:
1r
n
Now, if jr j < 1, then fr g is a basic null sequence, and so
1 rn
a
lim ð1 r n Þ
¼
lim sn ¼ lim a
n!1
n!1 1 r
1 r n!1
a
;
¼
1r
1
P
by the Combination Rules for sequences. Thus, if jrj < 1, then
ar n is
a
n¼0
convergent, with sum 1r.
sn ¼ a þ ar þ ar 2 þ þ ar n1 ¼ a
(b) Part (b) can be deduced immediately from the Non-null Test, which
&
appears later in this section. We defer the proof until then.
This value for sn is easily
verified by Mathematical
Induction.
Sub-section 3.1.5, Example 3.
Decimal representation of rational numbers
Geometric series provide another interpretation of the decimal representation
of rational numbers. Recall that rational numbers have terminating, or recurring, decimal representations. For example
3
1
¼ 0:3 and
¼ 0:333 . . . ¼ 0:3:
10
3
Another way to interpret the symbol 0.333 . . . is as an infinite series
3
3
3
þ
þ
þ :
101 102 103
3
1
and common ratio r ¼ 10
.
This is a geometric series, with first term a ¼ 10
1
Since 0 < 10 < 1, this series is convergent, with sum
3
a
3 1
¼ 10 1 ¼ ¼ :
1 r 1 10 9 3
Sub-section 1.1.3.
3: Series
88
Thus we have another method of finding the fraction which is equal to a given
recurring decimal, by calculating the sum of the corresponding geometric series.
Problem 2 Interpret the following decimals as infinite series, and
hence represent them as fractions:
(a) 0.111 . . .;
3.1.3
(b) 0.86363 . . .;
(c) 0.999 . . ..
You should at least read the
solution to part (c), even if
you do not tackle it first.
Telescoping series
Geometric series are easy to deal with because it is possible to find a formula
for the nth partial sum sn. The next problem deals with another series for which
we can calculate sn explicitly.
Problem 3 Calculate the first four partial sums of the following series,
giving your answers as fractions
1
X
1
1
1
1
¼
þ
þ
þ :
n
ð
n
þ
1
Þ
1
2
2
3
3
4
n¼1
The results obtained in Problem 3 suggest the general formula
sn ¼
1
1
1
1
n
þ
þ
þ þ
¼
:
12 23 34
nð n þ 1Þ n þ 1
This formula can be proved by Mathematical Induction, or we can use the identity
1
1
1
¼ ; for n ¼ 1; 2; . . .;
nð n þ 1 Þ n n þ 1
which implies that
1
1
1
1
1
sn ¼
þ
þ
þ þ
þ
12 23 34
ðn 1Þn nðn þ 1Þ
1 1
1 1
1 1
1
1
1
1
¼
þ
þ
þ þ
þ
1 2
2 3
3 4
n1 n
n nþ1
1
¼1
nþ1
n
:
¼
nþ1
Since
n
1
lim sn ¼ lim
¼ lim
¼ 1;
n!1
n!1 n þ 1
n!1 1 þ 1
n
we deduce that the given series is convergent, with sum
1
X
1
¼ 1:
n
ð
n
þ 1Þ
n¼1
Problem 4
fact that
Find the nth partial sum of the series
1
P
n¼1
2
1
1
¼ ;
nðn þ 2Þ n n þ 2
for n ¼ 1; 2; . . .:
Deduce that this series is convergent, and find its sum.
1
nðnþ2Þ,
using the
This cancellation of adjacent
terms explains why this series
is said to be telescoping.
3.1
Introducing series
3.1.4
89
Combination Rules for convergent series
At the start of this chapter we saw that performing arithmetical operations on
the divergent series 2 þ 4 þ 8 þ can lead to absurd conclusions. However,
the following result shows that there are Combination Rules for convergent
series, which follow directly from the Combination Rules for convergent
sequences:
Theorem 2 Combination Rules
1
1
P
P
If
an ¼ s and
bn ¼ t, then:
n¼1
Sum Rule
n¼1
1
P
Multiple Rule
n¼1
1
P
ðan þ bn Þ ¼ s þ t;
lan ¼ ls, for l 2 R.
n¼1
Proof
Consider the sequences of partial sums {sn} and {tn}, where
n
n
X
X
ak and tn ¼
bk :
sn ¼
k¼1
k¼1
By assumption, sn ! s and tn ! t as n ! 1.
Sum Rule
1
P
The nth partial sum of the series
ðan þ bn Þ is
n¼1
n
X
ðak þ bk Þ ¼ ða1 þ b1 Þ þ ða2 þ b2 Þ þ þ ðan þ bn Þ
k¼1
¼ ða1 þ a2 þ þ an Þ þ ðb1 þ b2 þ þ bn Þ ¼ sn þ tn :
By the Sum Rule for sequences
lim ðsn þ tn Þ ¼ lim sn þ lim tn ¼ s þ t;
n!1
n!1
1
P
and so the sequence {sn þ tn} of partial sums of
ðan þ bn Þ has limit s þ t.
n¼1
Hence this series is convergent and
1
X
ðan þ bn Þ ¼ s þ t:
n!1
n¼1
Multiple Rule
1
P
The nth partial sum of the series
lan is
n¼1
n
X
lak ¼ la1 þ la2 þ þ lan
k¼1
¼ lða1 þ a2 þ þ an Þ ¼ lsn :
By the Multiple Rule for sequences
lim ðlsn Þ ¼ l lim sn ¼ ls;
n!1
n!1
We may rearrange the terms
here, because this is the sum
of a finite number of terms.
3: Series
90
1
P
and so the sequence {lsn} of partial sums of
lan has limit ls. Hence this
n¼1
series is convergent and
1
X
lan ¼ ls:
&
n¼1
Example 2
Solution
Prove that the following series is convergent and calculate its sum
1
3
þ
:
2n nð n þ 1Þ
n¼1
1 X
1
P
1
We know that
n¼1
2n
is convergent, with sum 1, and that
1
P
n¼1
1
nðnþ1Þ
is
convergent, with sum 1.
Hence, by the Sum Rule and the Multiple Rule
1 X
1
3
þ
is convergent; with sum 1 þ ð3 1Þ ¼ 4: &
2n nð n þ 1Þ
n¼1
Problem 5
its sum
Prove that the following series is convergent and calculate
1 n
X
3
n¼1
3.1.5
4
2
:
nð n þ 1Þ
The Non-null Test
For all the infinite series we have so far considered, it is possible to derive a
simple formula for the nth partial sum. For most series, however, this is very
difficult or even impossible.
Nevertheless, it may still be possible to decide whether such series are
convergent or divergent by applying various tests. The first test that we give
arises from the following result:
Theorem 3
If
1
P
an is a convergent series, then {an} is a null sequence.
n¼1
Proof
You met the second series in
Sub-section 3.1.3.
Let sn ¼
n
P
ak denote the nth partial sum of
1
P
an . Because
n¼1
k¼1
1
P
n¼1
sn ¼ sn1 þ an ;
and so
an ¼ sn sn1 :
Thus, by the Combination Rules for sequences
lim an ¼ lim ðsn sn1 Þ ¼ lim sn lim sn1
n!1
n!1
n!1
¼ s s ¼ 0;
and so {an} is a null sequence, as required.
1
X
1
n
n¼1
&
1 1
¼ 1 þ þ þ ;
2 3
and
1
X
n¼1
an is
convergent, we know that {sn} is convergent. Suppose that lim sn ¼ s.
n!1
We want to deduce that {an} is null. To do this, we write
n!1
For example, consider the
series
n2
1 4 9
¼ þ þ þ :
þ 1 3 9 19
2n2
3.1
Introducing series
91
The following useful test for divergence is an immediate corollary of
Theorem 3:
Corollary
divergent.
Non-null Test
If {an} is not a null sequence, then
1
P
an is
n¼1
The virtue of the Non-null Test is its ease of application. For example, it
enables us to decide immediately that
1
1
1
X
X
X
1;
ð1Þn and
n are divergent;
n¼1
n¼1
n¼1
because the corresponding sequences {1}, {(1n)} and {n} are not null.
There are various ways to show that a given sequence {an} does not tend to
zero. For example, we can show that
an ! ‘ as n ! 1;
where ‘ 6¼ 0, or that {an} tends to infinity or to minus infinity. More generally,
we can use the result about subsequences which states that
if an ! 0 as n ! 1; then ank ! 0 as k ! 1;
for any subsequence fank g of {an}. This leads to the following strategy:
Strategy
EITHER
To show that
1
P
an is divergent, using the Non-null Test:
You met this Inheritance
Property of subsequences in
Sub-section 2.4.4, Theorem 5,
for any limit ‘.
n¼1
1. show that {an} has a convergent subsequence with non-zero limit;
OR
2. show that {an} has a subsequence which tends to infinity, or a subsequence which tends to minus infinity.
This strategy can be used to prove part (b) of Theorem 1, as follows:
1
P
Example 3 Prove that, if jrj 1 and a 6¼ 0, then
ar n is divergent.
n¼0
Solution We want to show that, if jrj 1 and a 6¼ 0, then {arn} is not a null
sequence. Now, the even subsequence
2k ¼ a; ar 2 ; ar 4 ; ar 6 ; . . .
ar
converges to a 6¼ 0 if jrj ¼ 1, and tends to infinity or to minus infinity if jrj 1.
Thus, {ar2k} is not null, and so {arn} is not null.
1
P
Hence, by the Non-null Test,
ar n is divergent if jrj > 1 and a 6¼ 0. &
It tends to infinity if a > 0 and
to minus infinity if a < 0.
n¼0
Warning
The converse of the Non-null Test is FALSE. In other words:
If {an} is a null sequence, it is not necessarily true that
1
P
an is convergent.
n¼1
1
P
1
1
1
For example, the sequence f1ng is a null sequence, but
n ¼1 þ 2 þ 3 þ n¼1
is divergent.
We shall prove this rather
surprising fact in Subsection 3.2.1.
3: Series
92
The important thing to remember is that you can never use the Non-null Test
to prove that a series is convergent, although you may be able to use it to prove
that a series is divergent.
1
P
n2
Problem 6 Prove that the series
2n2 þ 1 is divergent.
n¼1
3.2
Series with non-negative terms
In this section we consider only series
1
P
an with non-negative terms. In other
n¼1
words we assume that an 0, for n ¼ 1, 2, . . ..
1
P
It follows that the partial sums of
an , which are given by
n¼1
s 1 ¼ a1 ;
s 2 ¼ a1 þ a2 ;
s3 ¼ a1 þ a2 þ a3 ; . . .;
sn ¼ a1 þ a2 þ a3 þ þ an ; . . .;
form an increasing sequence {sn}.
The fact that {sn} is increasing makes it easier to deal with series having
non-negative terms. If we can prove that the sequence {sn} of partial sums is
bounded above, then {sn} is convergent, by the Monotone Convergence
1
P
Theorem, and so
an is convergent. On the other hand, if we can prove
Sub-section 2.5.1, Theorem 1.
n¼1
that the sequence {sn} of partial sums is not bounded above, then {sn} cannot
be convergent, and so we must have that sn ! 1 as n ! 1. We can rephrase
these facts in the following convenient way:
Boundedness Theorem for series
A series
1
P
an of non-negative terms is
n¼1
convergent if and only if its sequence {sn} of partial sums is bounded above.
We shall use this, for example, to prove the Cauchy Condensation Test in
Sub-section 3.2.1.
3.2.1
Tests for convergence
We now explore several tests for the convergence of series with non-negative
terms.
Problem 1 Use your calculator to find the first eight partial sums of
each of the following series
1
1
X
X
1
1
1
1
1 1
¼ 1 þ þ þ ;
¼
1
þ
þ
þ
and
2
2
2
n
2
3
n
2 3
n¼1
n¼1
giving your answers to 2 decimal places. Plot your answers on a
sequence diagram.
Example 1
Prove that the series
1
P
n¼1
1
n2
¼ 1 þ 212 þ 312 þ is convergent.
This is essentially the same
result as that of the
Boundedness Theorem for
sequences, Sub-section 2.4.2,
Theorem 1.
3.2
Series with non-negative terms
93
Solution All the terms of the series are positive, so we shall use the
Monotone Convergence Theorem.
Let sn be the nth partial sum of the series. Then
1
1
1
1
þ þ þ þ 2
22 32 42
n
1
1
1
1
þ
þ
þ þ
<1þ
12 23 34
ð n 1Þ n
1 1
1 1
1 1
1
1
¼1þ
þ
þ
þ þ
1 2
2 3
3 4
n1 n
1
¼ 2 < 2:
n
sn ¼ 1 þ
For any k > 1, k12 <
1
.
ðk1Þk
Here we have telescoping
cancellation.
It follows that {sn} is both increasing and bounded above (by 2), so that by
the Monotone Convergence Theorem {sn} is convergent. Hence the series
&
itself is also convergent.
Remark
In fact the sum of the series is
1
P
n¼1
1
n2
2
¼ p6 , surprisingly. However a proof of this
goes beyond the scope of this book.
The proof depends on use of
trigonometric series or of
Complex Analysis.
Not all series are convergent, though!
Example 2
Prove that the series
1
P
1
n¼1
n
¼ 1 þ 12 þ 13 þ is divergent.
Solution All the terms of the series are positive.
Let sn be the nth partial sum of the series. Then
1 1 1 1 1 1 1 1 1
sn ¼ 1 þ þ þ þ þ þ þ þ þ þ 2 3 4 5 6 7 8 9 10
1
1 1
1 1 1 1
þ
þ þ þ
¼1þ þ
þ
þ 2
3 4
5 6 7 8
1
1
1
þ :
þ k
þ
þ þ k
2 þ 1 2k þ 2
2 þ 2k
It follows that a subsequence of partial sums is
s1 ¼ 1;
1
s2 ¼ 1 þ ;
2 1
1 1
1 1
s4 ¼ 1 þ þ
þ
>1þ þ ;
2
3 4
2 2
1
1 1
1 1 1 1
1 1 1
s8 ¼ 1 þ þ
þ
þ þ þ
þ
>1þ þ þ ;
2
3 4
5 6 7 8
2 2 2
..
.
1 1
1
1
s2k > 1 þ þ þ þ ¼ 1 þ k:
2
2
2
2
|fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}
k terms
This series is often called the
harmonic series, since its
terms are proportional to the
lengths of strings that produce
harmonic tones in music.
We think of n as being fairly
large, in order to see the
pattern.
Here each bracket adds up to
more than 12.
3: Series
94
Hence the sequence {sn} is increasing and not bounded above. Therefore it
cannot be convergent, so that by the Monotone Convergence Theorem the
&
sequence {sn} is divergent. Hence the series itself is also divergent.
We can extend this type of domination approach to test further series for
convergence, in the following way:
Theorem 1
Comparison Test
1
P
(a) If 0 an bn , for n ¼ 1, 2, . . ., and
bn is convergent, then
n¼1
convergent.
(b) If 0 bn an , for n ¼ 1, 2, . . ., and
1
P
an is
n¼1
bn is divergent, then
n¼1
divergent.
1
P
1
P
an is
n¼1
In the proof of part (a), in Subsection 3.2.2, we shall in fact
see that
1
X
an n¼1
1
X
bn :
n¼1
Remark
As with the Squeeze Rule for non-negative null sequences, it is sufficient to
prove that the necessary inequalities in parts (a) and (b) hold eventually.
In applications, we use this result in the following way:
Strategy
To test a series
using the Comparison Test:
EITHER
OR
1
P
an of non-negative terms for convergence
n¼1
1
P
find a convergent series
dominate the an,
find a divergent series
bn of non-negative terms where the bn
n¼1
1
P
That is, where an bn.
bn of non-negative terms where the an
n¼1
dominate the bn.
Example 3
That is, it is sufficient that
0 an bn or 0 bn an for
all n > X, for some number X.
Prove that the series
1
P
n¼1
That is, where bn an.
1
n3
is convergent.
Solution We shall use the Comparison Test.
Let bn ¼ n12 . Then 0 n13 bn , for n ¼ 1, 2, . . ., and we know that the series
1
P
1
n2 is convergent (this was Example 1 above). It follows from part (a) of the
n¼1
1
P
1
&
Comparison Test that the series
n3 is convergent.
For n13 n12 :
n¼1
Example 4
Prove that the series
1
P
n¼1
p1ffiffi
n
is divergent.
Solution We shall use the Comparison Test.
Let bn ¼ 1n. Then 0 bn p1ffiffin, for n ¼ 1, 2, . . ., and we know that the series
1
P
1
2 above). It follows from part (b) of the
n is divergent (this was Example
1
P
n¼1
1
pffiffi is divergent.
&
Comparison Test that the series
n
For 1n p1ffiffin :
n¼1
1
P
ffiffi . Its terms are somewhat similar to
pffiffi 1p
Next, consider the series
nþ3 4 nþ1
1
n¼1
P
p1ffiffi, since we can express pffiffi 1p
ffiffi in the form
those of the series
n
nþ3 4 nþ1
n¼1
pffiffiffi
In other words, the term
ffiffiffi n
p
4
dominates the term 3 n þ 1.
3.2
p1ffiffi
n
Series with non-negative terms
1
1
1 , where the
1þ3n4 þn2
1
P 1
pffiffi is divergent,
series
n
n¼1
95
second fraction tends to 1 as n ! 1. So, since the
it seems likely that the series
1
P
n¼1
pffiffi 1p4 ffiffi
nþ3 nþ1
is also
By Example 4.
divergent.
We can pin down the underlying idea here in the following useful result:
Theorem 2
Limit Comparison Test
1
1
P
P
Suppose that
an and
bn have positive terms, and that
n¼1
n¼1
an
!L
bn
(a) If
1
P
(b) If
n¼1
1
P
as
n ! 1;
bn is convergent, then
bn is divergent, then
n¼1
1
P
where L 6¼ 0:
an is convergent.
In other words, bn behaves
‘rather like an’ for large n.
Note that it is IMPORTANT that
L is non-zero.
n¼1
1
P
an is divergent.
n¼1
So, take an ¼ pffiffinþ31p4 ffiffinþ1 and bn ¼ p1ffiffin, for n ¼ 1, 2, . . .. Then both an and bn
are positive, and
pffiffiffi
pffiffiffi
ð1=ð n þ 3 4 n þ 1ÞÞ
an
pffiffiffi
¼
bn
ð1= nÞ
pffiffiffi
n
ffiffiffi
p
¼ pffiffiffi
nþ34 nþ1
1
¼
1
1 ! 1 as n ! 1:
1 þ 3n 4 þ n2
It then follows from part (b) of the Limit Comparison Test that, since the
1
1
P
P
p1ffiffi is divergent, the series
ffiffi is also divergent.
pffiffi 1p
series
n
nþ3 4 nþ1
n¼1
Since 1 6¼ 0.
By Example 4.
n¼1
Example 5 Use the Limit Comparison Test to prove that the series
1 2
P
n 3nþ4
5n4 n is convergent.
n¼1
2
1
Solution Set an ¼ n 5n3nþ4
4 n and bn ¼ n2 , for n ¼ 1, 2, . . .. Then both an and bn
are positive, and
an
n2 3n þ 4 n2
¼
5n4 n
bn
1
4
3
2
n 3n þ 4n
:
¼
5n4 n
1 3n1 þ 4n2
1
as n ! 1:
¼
!
3
5
5n
1
P
1
Since 15 6¼ 0 and the series
n2 is known to be convergent, then the series
1
n¼1
P n2 3nþ4
&
5n4 n is also convergent.
n¼1
We make this choice for bn
since, for large n, the
2
expression n 5n3nþ4
4 n is ‘more or
n2
1
less the same as’ 5n
4 ¼ 5n2 –
where the factor ‘5’ will not
affect our argument.
Dividing numerator and
denominator by the dominant
term n4.
By Example 1.
3: Series
96
Problem 2 Use the Comparison Test or the Limit Comparison Test to
determine the convergence of the following series:
1
1
1
1
P
P
P
P
cos2 ð2nÞ
1
1pffiffi
nþ4
(a)
(b)
;
(c)
(d)
n3 :
n3 þn;
2n3 nþ1;
nþ n
n¼1
n¼1
n¼1
n¼1
Our next test is motivated in part by the geometric series – recall that the
1
P
ar n , where a 6¼ 0, is convergent if
geometric series a þ ar þ ar 2 þ ¼
n¼0
jrj < 1 but divergent if jrj 1. The reason that this converges, if jrj < 1, is that
the terms are then forced to tend to 0 so quickly that the partial sums sn increase
so slowly as n increases that they remain bounded above. On the other hand, if
jrj 1, the terms do not tend to 0, so that the series must diverge.
Theorem 3
Ratio Test
1
P
an has positive terms, and that abnþ1
! ‘ as n ! 1.
Suppose that
n
n¼1
1
P
(a) If 0 ‘ < 1, then
an is convergent.
n¼1
(b) If ‘ > 1, then
1
P
By the Non-null Test.
Note that with the Ratio Test,
we concentrate on the series
1
P
an itself and do not need to
n¼1
‘think of’ some other series
1
P
bn .
n¼1
an is divergent.
Part (b) includes the case
anþ1
an ! 1.
n¼1
Remark
If ‘ ¼ 1, then the test gives us no information on whether the series converges.
n
For example, if an ¼ 1n, then aanþ1
¼ nþ1
¼ 1þ1 1 ! 1 as n ! 1; we have seen
n
n
1
1
P
P
1
1
an ¼
that the series
n diverges. On the other hand, if an ¼ n2 then
1
1
n¼1
n¼1
P
P 1
anþ1
n2
1
an ¼
an ¼ ðnþ1Þ2 ¼ 1þ2þ 1 ! 1 as n ! 1; but the series
n2 converges.
n
Example 6
series:
1
P
n
(a)
2n ;
n2
n¼1
n¼1
Use the Ratio Test to determine the convergence of the following
(b)
n¼1
1
P
10n
n¼1
n!
.
Solution
(a) Let an ¼ 2nn , n ¼ 1, 2, . . .. Then an is positive, and
anþ1 n þ 1 2n
¼ nþ1 2
an
n
1
1þn
1
as n ! 1:
!
¼
2
2
Since 0 < 12 < 1, it follows from the Ratio Test that the series
1
1
P
P
n
an ¼
2n is convergent.
n¼1
n¼1
n
(b) Let an ¼ 10n! , n ¼ 1, 2, . . .. Then an is positive, and
anþ1
10nþ1
n!
¼
an
ðn þ 1Þ! 10n
10
! 0 as n ! 1:
¼
nþ1
That is, the Ratio Test is
inconclusive if ‘ ¼ 1.
By Example 2.
By Example 1.
3.2
Series with non-negative terms
97
It follows from the Ratio Test that the series
convergent.
1
P
an ¼
n¼1
1
P
n¼1
10n
n!
is
&
Problem 3 Use the Ratio Test to determine whether the following
series are convergent
1 3
1 2 n
1
P
P
P
ð2nÞ!
n
n 2
(a)
(b)
(c)
n! ;
n! ;
nn .
n¼1
n¼1
n¼1
We have now seen examples of many different convergent and divergent
series, including examples of various generic types. We call these basic series,
since we shall use them commonly together with various other tests in order to
prove that particular series are themselves convergent or divergent.
Basic series
1
P
1
(a)
np ;
(b)
n¼1
1
P
(c)
n¼1
1
P
(d)
n¼1
1
P
n¼1
cn ;
np c n ;
The following series are convergent:
for p 2;
Instances of these that we
have seen are
1
P
1
n2 ;
for 0 c < 1;
Geometric Series,
n¼1
1
P
for p > 0; 0 c < 1;
n¼1
cn
n! ;
n
2n
¼
1 P
n
n 12 ;
1 P
1 n
n¼1
2
;
n¼1
1
P
10n ;
for c 0.
n¼1
The following series is divergent:
1
P
1
(e)
for p 1.
np ;
1
P
n¼1
n¼1
n!
p1ffiffi
n
¼
1
P
n¼1
1 .
n1=2
Another test for convergence
When we examined the convergence or divergence of the series
1
P
1
n, we found
n¼1
it convenient to group the terms into blocks that contained in turn 1, 2, 4, 8, . . .
successive individual terms of the series. In fact, a similar technique applies to
a wide range of series, and the following result is thus often invaluable in order
to save repeating that same type of argument:
Theorem 4
You will see this technique
used in the proof of
Theorem 4, in Subsection 3.2.2.
Cauchy Condensation Test
Let {an} be a decreasing sequence of positive terms. Then, if bn ¼ 2n a2n , for
n ¼ 1, 2, . . .
1
1
X
X
an is convergent if and only if
bn is convergent:
n¼1
n¼1
1
P
1
1
For example, consider the series
n. Here, let an ¼ n, so that
n¼1
1
bn ¼ 2n a2n ¼ 2n n ¼ 1:
2
1
P
1
Then the Condensation Test tells us that the series
n is convergent if and
1
1
n¼1
P
P
bn ¼
1 is convergent. This is divergent, by the Nononly if the series
1
n¼1
n¼1
P
1
null Test, so that the original series
n must itself have been divergent.
n¼1
For, if an ¼ 1n, then a2n ¼ 21n .
3: Series
98
This use of the Condensation Test is much simpler than setting out to
perform a grouping and estimation exercise on such occasions! The
Condensation Test is also often useful when the individual terms of a series
include expressions such as loge n.
Example 7 Use the Condensation Test to determine the convergence of the
1
P
1
pffiffiffiffiffiffiffiffi
series
.
n¼2 n
loge n
1
Solution Let an ¼ pffiffiffiffiffiffiffiffi
, n ¼ 2, . . .. Then ann is positive;
and, since
o
n loge n
pffiffiffiffiffiffiffiffiffiffiffiffi
1
n loge n is an increasing sequence, fan g ¼ pffiffiffiffiffiffiffiffi
is a decreasing
n
loge n
sequence.
Next, let bn ¼ 2n a2n ; thus
1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
loge ð2n Þ
1
1
1
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffi 1 :
n loge 2
loge 2 n2
bn ¼ 2n Since
1
P
1
P
1
1
n¼2 n2
2n
is a basic divergent series, it follows by the Multiple Rule that
bn must be divergent. Hence by the Condensation Test, the original series
n¼2
1
P
n¼2 n
1
pffiffiffiffiffiffiffiffi
must also be divergent.
&
loge n
Problem 4 Use the Condensation Test to determine the convergence
of the following series:
1
1
P
P
1
1
(a)
;
(b)
.
n log n
nðlog nÞ2
n¼2
3.2.2
e
n¼2
1
loge 1
loge 1 ¼ 0. Again we are
using the fact that the
convergence of a series is not
affected if we add or alter a
finite of terms.
e
Proofs
You may omit these proofs at
a first reading.
In the previous sub-section we gave a number of tests for the convergence of
series with non-negative terms. We now supply the proofs of these tests.
Theorem 1
Note that here we only sum
from n ¼ 2, since it makes no
1
sense to talk about pffiffiffiffiffiffiffiffi
, as
Comparison Test
1
1
P
P
(a) If 0 an bn, for n ¼ 1, 2, . . ., and
bn is convergent, then
an is
n¼1
n¼1
convergent.
1
1
P
P
(b) If 0 bn an, for n ¼ 1, 2, . . ., and
bn is divergent, then
an is
n¼1
n¼1
divergent.
Proof
(a) Consider the nth partial sums
sn ¼ a1 þ a2 þ þ an ;
for n ¼ 1; 2; . . .;
3.2
Series with non-negative terms
99
and
t n ¼ b 1 þ b2 þ þ bn ;
We know that
for n ¼ 1; 2; . . .:
a1 b1 ; a2 b2 ; . . .; an bn ;
and so
sn tn ;
for n ¼ 1; 2; . . .:
1
P
We also know that
By adding up the previous
inequalities.
bn is convergent, and so the increasing sequence
n ¼1
{tn} is convergent, with limit t, say. Hence
sn tn t;
for n ¼ 1; 2; . . .;
and so the increasing sequence {sn} is bounded above by t. By the Monotone
1
P
Convergence Theorem, {sn} is therefore also convergent, and so
an is a
n ¼1
convergent series.
Note: Moreover, by the Limit Inequality Rule
1
1
X
X
an bn :
n ¼1
n¼1
1
P
an is
(b) We can deduce part (b) from part (a), as follows. Suppose that
1
P
n ¼1
convergent. Then, by part (a),
bn must also be convergent. However,
1
1
n ¼1
P
P
bn is assumed to be divergent, and so
an must also be divergent.&
n ¼1
You met this Rule in
Sub-section 2.3.3, Theorem 3.
This additional inequality will
sometimes be useful.
Such ‘proofs by
contradiction’ can sometimes
save a great deal of detailed
arguments.
n¼1
Theorem 2
Limit Comparison Test
1
1
P
P
Suppose that
an and
bn have positive terms, and that
n ¼1
n¼1
an
! L as n ! 1; where L 6¼ 0:
bn
1
1
P
P
(a) If
bn is convergent, then
an is convergent.
(b) If
n¼1
1
P
n ¼1
bn is divergent, then
n ¼1
1
P
an is divergent.
n¼1
Proof
(a) Because abnn is convergent, it must be bounded. Thus there is a constant K
such that
an
K;
bn
for n ¼ 1; 2; . . .;
and so
an Kbn ;
for n ¼ 1; 2; . . .:
1
P
By the Multiple Rule,
Kbn is convergent. Hence, by the Comparison
n¼1
1
P
an is also convergent.
Test,
n¼1
See Sub-section 2.4.2,
Theorem 1 (the Boundedness
Theorem).
3: Series
100
(b) We can deduce part (b) from part (a), as follows.
1
1
P
P
Suppose that
an is convergent. Then, by part (a),
bn must also be
n¼1
n¼1
convergent, because
bn
1
as n ! 1;
!
L
an
by the Quotient Rule for sequences.
1
1
P
P
However,
bn is assumed to be divergent, and so
an must also be
n¼1
&
divergent. n¼1
We are following a ‘proof by
contradiction’ approach here.
Remember that L 6¼ 0.
Theorem 3
Ratio Test
1
P
Suppose that
an has positive terms, and that aanþ1
! ‘ as n ! 1.
n
n¼1
1
P
(a) If 0 ‘ < 1, then
an is convergent.
n¼1
1
P
(b) If ‘ > 1, then
an is divergent.
Remember that part (b)
includes the case aanþ1
! 1.
n
n¼1
Proof
(a) Because ‘ < 1, we can choose " > 0 so that
r ¼ ‘ þ " < 1:
Hence there is a positive number X, which we may assume to be a positive
integer, such that
anþ1
r; for all n X:
an
Thus, for n X, we have
an
an
an1
aXþ1
¼
...
r nX ;
aX
an1
an2
aX
since each of the n X brackets is less than or equal to r. Hence
an aX r nX ;
for all n X:
For example, take
" ¼ 12 ð1 ‘Þ.
Recall that we occasionally
make the convenient
assumption that X is an integer
to avoid notational
complications.
1
+ε
–ε
(1)
X
n
Now
1
X
aX r nX ¼ aX þ aX r þ aX r 2 þ ;
n¼X
which is a geometric series with first term aX and common ratio r. Since
0 < r < 1, this series is convergent, and so, by inequality (1) and the
1
1
P
P
Comparison Test,
an is also convergent. Hence
an is convergent.
n¼X
n¼1
(b) Since
anþ1
! ‘ as n ! 1;
an
and ‘ > 1, there is a positive number X, which we may assume to be a
positive integer, such that
anþ1
1; for all n X:
an
1
X
n
3.2
Series with non-negative terms
101
(This also holds if aanþ1
! 1 as n ! 1.)
n
Thus, for n X, we have
an
an
an1
aXþ1
¼
...
1;
aX
an1
an2
aX
since each of the brackets is greater than or equal to 1. Hence
an aX > 0;
for all n X;
1
P
and so {an} cannot be a null sequence. Thus, by the Non-null Test,
divergent. Hence
1
P
an is
n¼X
&
an is also divergent.
n¼1
Theorem 4 Cauchy Condensation Test
Let {an} be a decreasing sequence of positive terms. Then, if bn ¼ 2n a2n for
n ¼ 1, 2, . . .
1
1
X
X
an is convergent if and only if
bn is convergent:
n¼1
n¼1
1
1
P
P
Proof Denote by sn and tn the nth partial sums of the series
an and
bn ,
n¼1
n¼1
respectively
s n ¼ a1 þ a2 þ þ an
and tn ¼ 21 a2 þ 22 a4 þ þ 2n a2n :
Since the terms an are decreasing and a1 0, it follows that
The kth bracket contains 2k1
terms.
The kth bracket contains 2k1
occurrences of a2k.
s2n ¼ a1 þ ða2 Þ þ ða3 þ a4 Þ þ þ ða2n1 þ1 þ þ a2n Þ
0 þ ða2 Þ þ ða4 þ a4 Þ þ þ ða2n þ þ a2n Þ
¼ 0 þ ða2 Þ þ ð2a4 Þ þ þ 2n1 a2n
1
¼ tn :
2
ð2Þ
We may also write s2n in another way
s 2 n ¼ a1 þ ð a2 þ a3 Þ þ ð a4 þ a5 þ a6 þ a7 Þ þ þ ða2n1 þ þ a2n 1 Þ þ a2n
a1 þ ð a2 þ a2 Þ
þ ða4 þ a4 þ a4 þ a4 Þ þ þ ða2n1 þ þ a2n1 Þ þ a2n
¼ a1 þ ð2a2 Þ þ ð4a4 Þ þ þ 2n1 a2n1 þ a2n
a1 þ t n :
ð3Þ
Now suppose that
1
P
an is convergent. Then, by the Boundedness Theorem for
n¼1
series, its sequence of partial sums {sn} must be bounded above and so the sequence
of {sn}. It therefore
fs2ng must also be bounded above – since it is a subsequence
follows from inequality (2) that the sequence 12 tn must be bounded above, so the
sequence {tn} must also be bounded above. Another application of the Boundedness
1
P
Theorem shows that therefore the series
bn must be convergent.
n¼1
The kth bracket contains 2k
terms.
The kth bracket contains 2k
occurrences of a2k.
In obtaining (3), we have
replaced the term a2n in the previous line by a larger term 2n a2n.
The Boundedness Theorem
for series appeared in
Section 3.2, above.
3: Series
102
Next, suppose that
1
P
an is divergent. Then, by the Boundedness Theorem
n¼1
for series, its sequence of partial sums {sn} must be unbounded above, and so
the sequence fs2ng must also be unbounded above – since it is a subsequence of
{sn}. It therefore follows from inequality (3) that the sequence {a1 þ tn} must
be unbounded above, so the sequence {tn} must also be unbounded above.
Another application of the Boundedness Theorem shows that therefore the
1
P
&
series
bn must be divergent.
n¼1
We end this section by proving the convergence/divergence of the basic
series given earlier.
The following series are convergent:
Basic series
1
P
1
(a)
np ;
for p 2;
(b)
n¼1
1
P
c ;
for 0 c < 1;
(c)
n¼1
1
P
np c n ;
for p > 0; 0 c < 1;
(d)
n¼1
1
P
cn
n! ;
for c 0.
n¼1
In Sub-section 3.2.1.
In Chapter 7 we shall prove
1
P
1
that
np is convergent, for
n¼1
n
all p > 1.
The following series is divergent:
1
P
1
(e)
for p 1.
np ;
n¼1
Proof
(a) This series is convergent, by the Comparison Test, since, if p 2, then
1
1
2 ; for n ¼ 1; 2; . . .;
p
n
n
1
P
1
and the series
n2 is convergent.
Sub-section 3.2.1, Example 1.
n¼1
(b) The series
1
P
cn is a geometric series with common ratio c, and so it
n¼1
converges if 0 c < 1.
p n
(c) Let an ¼ n c , for n ¼ 1, 2, . . .. Then, for c 6¼ 0
anþ1 ðn þ 1Þp cnþ1
1 p
¼
¼ 1þ
c:
n
an
np c n
Thus, if k is any integer greater than or equal to p, then
anþ1
1 k
c
1þ
c:
n
an
Now, by the Product Rule for sequences
1 k
1þ
! 1 as n ! 1;
n
Sub-section 3.1.2.
If c ¼ 0, the series is clearly
convergent.
We introduce the integer k
here, so that we can use the
Product Rule for sequences.
3.3
Series with positive and negative terms
103
and so, by the Squeeze Rule for sequences
anþ1
! c as n ! 1:
an
1
P
Since 0 c < 1, we deduce, from the Ratio Test, that
np cn is
n¼1
convergent.
n
(d) Let an ¼ cn! , for n ¼ 1, 2, . . .. Then, for c 6¼ 0,
n
anþ1
cnþ1
c
c
¼
:
¼
an
ðn þ 1Þ! n! n þ 1
1 n
P
c
! 0 as n ! 1, and we deduce from the Ratio Test that
Thus aanþ1
n! is
n
n¼1
convergent.
1
P
1
(e) We saw earlier that the series
n is divergent. It follows that the series
1
n¼1
P 1
1
1
np is also divergent, by the Comparison Test, since if p 1 then np n,
If c ¼ 0, the series is clearly
convergent.
In Sub-section 3.2.1,
Example 2.
n¼1
&
for n ¼ 1, 2, . . ..
3.3
Series with positive and negative terms
The study of series
1
P
an , with an 0 for all values of n, is relatively straight-
n¼1
forward, because the sequence of partial sums {sn} is increasing. Similarly, if
an 0 for all values of n, then {sn} is decreasing.
However, it is harder to determine the behaviour of a series with both
positive and negative terms, because {sn} is neither increasing nor decreasing.
However, if the sequence {an} contains only finitely many negative terms, then
the sequence {sn} is eventually increasing, and we can apply the methods of
Section 3.2. Similarly, if {an} contains only finitely many positive terms, then
the sequence {sn} is eventually decreasing, and we can again apply the
methods of Section 3.2, after making a sign change.
In this section we look at series such as
1 1 1 1 1
1
1
1
1
1
1 þ þ þ and 1 þ 2 2 þ 2 þ 2 2 þ ;
2 3 4 5 6
2
3
4
5
6
which contain infinitely many terms of either sign. We give several methods
which can often be used to prove that such series are convergent.
3.3.1
For example, the convergence
of
1þ2þ3
1
1
1
42 52 62
follows from that of
1
P
n¼1
Absolute convergence
Suppose that we want to determine the behaviour of the infinite series
1
X
ð1Þnþ1
1
1
1
1
1
¼ 1 2 þ 2 2 þ 2 2 þ :
2
2
3
4
5
6
n
n¼1
We know that the series
1
X
1
1
1
1
1
1
¼ 1 þ 2 þ 2 þ 2 þ 2 þ 2 þ 2
n
2
3
4
5
6
n¼1
(1)
(2)
is convergent. Does this mean that the series (1) is also convergent? In fact it
does, as we now prove.
The series (2) is a basic
convergent series.
1
n2
:
3: Series
104
Consider the two related series
1
1
1 þ 0 þ 2 þ 0 þ 2 þ 0 þ 3
5
and
1
1
1
0 þ 2 þ 0 þ 2 þ 0 þ 2 þ :
2
4
6
Each of these series is dominated by the series (2), and so they are both
convergent, by the Comparison Test. Applying the Multiple Rule with l ¼ 1,
and the Sum Rule, we deduce that the series
1
1
1
1
1
1 2 þ 2 2 þ 2 2 þ is convergent:
2
3
4
5
6
The argument just given is the basis for a concept called absolute convergence, which we now define.
Definitions
1
1
P
P
A series
an is absolutely convergent if
jan j is convergent.
A series
n¼1
1
P
n¼1
an that is convergent but not absolutely convergent is condi-
If the terms an are all nonnegative, then absolute
convergence and convergence
have the same meaning.
n¼1
tionally convergent.
For example, the series (1) is absolutely convergent, because the series
is convergent.
However, the series
1
X
ð1Þnþ1
n¼1
n
1
P
n¼1
1
n2
We shall examine the
behaviour of conditionally
convergent series later, in
Sub-sections 3.3.2 and 3.3.3.
1 1 1 1 1
¼ 1 þ þ þ 2 3 4 5 6
1
P
1
is not absolutely convergent, because the series
n is divergent; the series is,
n¼1
in fact, conditionally convergent.
Theorem 1 Absolute Convergence Test
1
1
P
P
If
an is absolutely convergent, then
an is convergent.
n¼1
You saw this in Example 2,
Sub-section 3.2.1.
The proofs of the results in
this sub-section appear in
Sub-section 3.3.6.
n¼1
It follows that the series (1) is convergent (as we have already seen), and also
that the series
1þ
1
1
1
1
1
þ þ þ 22 32 42 52 62
is convergent, because the series
1
P
1
n2
is convergent. Indeed, no matter how
we distribute the plus and minus signs among the terms of n12 , the resulting
series is convergent.
However, the Absolute Convergence Test tells us nothing about the
behaviour of the series
n¼1
1 1 1 1 1
1 þ þ þ 2 3 4 5 6
(3)
1 1 1 1 1 1 1 1
1 þ þ þ þ þ þ :
2 3 4 5 6 7 8 9
(4)
and
In fact, series (3) is convergent
(by the Alternating Test,
which we introduce in Subsection 3.3.2), and series (4) is
divergent (see Exercise 2(a)
for Section 3.3).
3.3
Series with positive and negative terms
The series
1
P
n¼1
1
n
105
is divergent, and so the two series (3) and (4) are not
absolutely convergent.
Example 1 Prove that the following series are convergent:
1
P
ð1Þnþ1
1
1
(a)
(b) 1 þ 12 14 þ 18 þ 16
32
.
n3 ;
n¼1
Solution
(a) Let an ¼
1
P
1
for n ¼ 1, 2, . . .; then jan j ¼ n13 . We know that
n3 is
1
n¼1
P ð1Þnþ1
convergent, so it follows that
is absolutely convergent. Hence,
n3
ð1Þnþ1
n3 ,
For
1
P
n¼1
1
n3
is an example of a
basic convergent series.
n¼1
1
P
ð1Þnþ1
is convergent.
by the Absolute Convergence Test,
n3
n¼1
1
P 1
(b) The series
2n is a convergent geometric series, so that the series
n¼0
1 1 1 1
1
1þ þ þ 2 4 8 16 32
is absolutely convergent. Then, by the Absolute Convergence Test, this
&
series is also convergent.
Problem 1 Prove that the following series are convergent:
1
1
P
P
ð1Þnþ1 n
cos n
(a)
;
(b)
3
2n .
n þ1
n¼1
n¼1
P
The Absolute Convergence Test states that, if
jan j is convergent, then
P
an is also convergent, but it does not indicate
P any explicitPconnection
between the sums of these two series. Clearly, an is less than jan j if any
of the terms an are negative.
1
1
P
P
ð1Þnþ1
1
1
1
1
For example,
whereas
¼
2n ¼ 2 þ 4 þ 8 þ ¼ 1,
2n
n¼1
1
2
n¼1
14 þ 18 ¼ 13 :
The following result relates the values of
P
an and
P
jan j:
Triangle Inequality (infinite form)
1 1
1
P P
P
If
an is absolutely convergent, then an jan j.
n¼1
n¼1
n¼1
1
1
1
Problem 2 Show that the series 12 þ 14 18 þ 16
þ 32
64
þ is convergent, and that its sum lies in [1,1]. (You do NOT need to find the sum
of the series.)
3.3.2
The Alternating Test
Suppose that we want to determine the behaviour of the following infinite
series, in which the terms have alternating signs
1
X
ð1Þnþ1
n¼1
n
1 1 1 1 1
¼ 1 þ þ þ :
2 3 4 5 6
Recall that you met the
Triangle Inequality for
a1 ; a2 ; . . .; an
n
n
X
X
a
ja j;
k k¼1 k¼1 k
in Sub-section 1.3.1.
3: Series
106
1
P
1
The Absolute Convergence Test does not help us with this series, because
n
n¼1
is divergent.
1
P
ð1Þnþ1
In fact, the series
is convergent. To see why, we first calculate
n
n¼1
some of its partial sums and plot them on a sequence diagram
s1 ¼ 1;
1
s2 ¼ 1 ¼ 0:5;
2
1 1
s3 ¼ 1 þ ¼ 0:83;
2 3
1 1 1
s4 ¼ 1 þ ¼ 0:583;
2 3 4
1 1 1 1
s5 ¼ 1 þ þ ¼ 0:783;
2 3 4 5
1 1 1 1 1
s6 ¼ 1 þ þ ¼ 0:616:
2 3 4 5 6
The sequence diagram for {sn} suggests that
s1 s3 s5 . . . s2k1 . . .
and
s2 s4 s6 . . . s2k . . . ;
for all k. In other words:
the odd subsequence {s2k1} is decreasing
and:
the even subsequence {s2k} is increasing.
Also, the terms of {s2k1} all exceed the terms of {s2k}, and both subsequences
appear to converge to a common limit s, which lies between the odd and even
partial sums.
To prove this, we write the even partial sums {s2k} as follows
s2k ¼
1
1 1
1
1
þ
þ þ
:
1
2
3 4
2k 1 2k
All the brackets are positive, and so the subsequence {s2k} is increasing.
We can also write
s2k
1 1
1 1
1
1
1
:
¼1
2 3
4 5
2k 2 2k 1
2k
Again, all the brackets are positive, and so {s2k} is bounded above, by 1.
Hence {s2k} is convergent, by the Monotone Convergence Theorem.
Let
lim s2k ¼ s:
k!1
Since
s2k ¼ s2k1 and
1
2k
is null, we have
1
2k
You met subsequences
earlier, in Sub-section 2.4.4.
3.3
Series with positive and negative terms
lim s2k1
k!1
107
1
1
¼ lim s2k þ
¼ lim s2k þ lim
k!1
k!1
k!1
2k
2k
¼ s þ 0 ¼ s;
by the Combination Rules for sequences. Thus the odd and even subsequences
of {sn} both tend to the same limit s, and so {sn} itself tends to s. Hence
1
X
ð1Þnþ1
n
n¼1
1 1 1 1 1
¼ 1 þ þ þ is convergent, with sum s:
2 3 4 5 6
By Theorem 6 of
Sub-section 2.4.4.
In fact, s ¼ loge 2 ’ 0.69.
The same method can be used to prove the following general result:
Theorem 2
This test is sometimes called
the Leibniz Test.
Alternating Test
If
an ¼ ð1Þnþ1 bn ;
n ¼ 1; 2; . . .;
where {bn} is a decreasing null sequence with positive terms, then
1
X
an ¼ b1 b2 þ b3 b4 þ is convergent:
n¼1
When we apply the Alternating Test, there are a number of conditions which
must be checked. We now describe these in the form of a strategy.
Strategy To prove that
check that
1
P
an is convergent, using the Alternating Test,
n¼1
an ¼ ð1Þnþ1 bn ;
n ¼ 1; 2; . . .;
where:
1. bn 0,
for n ¼ 1, 2, . . .;
To show that {bn} is null, use
the techniques introduced in
Sub-section 2.2.1.
2. {bn} is a null sequence;
3. {bn} is decreasing.
Remark
It is often easiest to check that {bn} is decreasing by verifying that
increasing.
1
bn
Here are some examples.
Example 2 Determine which of the following series are convergent:
1
1
1
P
P
P
ð1Þnþ1
ð1Þnþ1
pffiffi ;
(a)
(b)
;
(c)
ð1Þnþ1 .
4
n
n
n¼1
n¼1
n¼1
Solution
n nþ1 o
(a) The sequence ð1pÞffiffin
has terms of the form an ¼ ð1Þnþ1 bn , where
1
bn ¼ pffiffiffi ;
n
n ¼ 1; 2; . . .:
is
3: Series
108
Now:
1. bn ¼ p1ffiffin 0, for n ¼ 1, 2, . . .;
2. fbn g ¼ p1ffiffin is a basic null sequence;
pffiffiffi
3. fbn g ¼ p1ffiffin is decreasing, because b1n ¼ f ng is increasing.
1
P
ð1Þnþ1
pffiffi is convergent.
Hence, by the Alternating Test,
n
(b) The sequence
n
nþ1
ð1Þ
n4
bn ¼
1
;
n4
o
n¼1
has terms of the form an ¼ ð1Þnþ1 bn , where
Alternatively, we can show
that this series is convergent
by the Absolute Convergence
Test.
n ¼ 1; 2; . . .:
Now:
1. bn ¼ n14 0, for n ¼ 1, 2, . . .;
2. fbn g ¼ n14 is a basic null sequence;
3. fbn g ¼ n14 is decreasing, because b1n ¼ n4 is increasing.
1
P
ð1Þnþ1
Hence, by the Alternating Test,
is convergent.
n4
n¼1
(c) The sequence ð1Þnþ1 is not a null sequence. Hence, by the Non-null
Test
1
X
ð1Þnþ1 is divergent.
&
n¼1
Problem 3
(a)
1
P
n¼1
3.3.3
ð1Þnþ1
n3
Notice that the fact that the
nth term includes a factor
(1)nþ1 does not imply at all
that a series is convergent!
Determine which of the following series are convergent:
;
(b)
1
P
n¼1
n
ð1Þnþ1 nþ2
;
(c)
1
P
n¼1
ð1Þnþ1
.
n1=3 þ n1=2
Rearrangement of a series
In the last sub-section we saw that the series 1 12 þ 13 14 þ 15 16 þ is
convergent. Let us denote by ‘ the sum of this series.
If we temporarily ignore the need for careful mathematical proof and simply
‘move terms about’, look what we get
1 1 1 1 1
‘ ¼ 1 þ þ þ 2 3 4 5 6
1 1 1 1 1 1 1
1 1 1
1
¼ 1 þ þ þ þ 2 4 3 6 8 5 10 12 7 14 16
1
1
1 1
1
1 1
1
¼ 1
þ
þ
2
4
3 6
8
5 10
12
1 1
1
þ
þ 7 14
16
In fact this series is
conditionally convergent,
since the harmonic series
1 þ 12 þ 13 þ 14 þ 15 þ 16 þ is
divergent.
(5)
This is the definition of ‘.
ð6Þ
The pattern in (6) is to follow
the next available positive
term by the two next available
negative terms.
Here we insert some brackets.
3.3
Series with positive and negative terms
1 1 1 1 1
1
1
1
þ þ þ þ 2 4 6 8 10 12 14 16
1
1 1 1 1 1 1
¼
1 þ þ þ 2
2 3 4 5 6 8
¼
109
Here we insert the value of
each bracket.
Here we pull out a common
factor of 12.
1
¼ ‘:
2
It follows from the fact that ‘ ¼ 12 ‘ that ‘ ¼ 0. However this is impossible,
since the partial sum s2 of the original series is 12, and the even partial sums s2k
are increasing.
What has gone wrong? We assumed that operations that are valid for sums of
a finite number of terms also hold for sums of an infinite number of terms – we
rearranged (infinitely many of) the terms in the series (5) to obtain the series
(6) – without any justification. This rearrangement operation is not valid in
general!
We now give a precise definition of what we mean by a rearrangement.
1
P
Loosely speaking, the series
bn ¼ b1 þ b2 þ is a rearrangement of the
1
n¼1
P
an ¼ a1 þ a2 þ if precisely the same terms appear in the
series
Recall that, since the series
converges to ‘, then s2k
converges to ‘ also.
In fact the series (6) does
converge; we ask you to
supply a careful proof of this
in Exercise 2 on Section 3.3.
n¼1
sequences {bn} ¼ {b1, b2, . . .} and {an} ¼ {a1, a2, . . .}, though they may occur
in a different order.
1
P
Definition A series
bn ¼ b1 þ b2 þ is a rearrangement of the series
1
n¼1
P
an ¼ a1 þ a2 þ if there is a bijection ƒ such that
n¼1
f: N !N
bn 7! af ðnÞ:
Recall that a bijection is a
one–one onto mapping.
1
P
ð1Þnþ1
Assuming that the sum of the series
¼ 1 12 þ 13 14 þ
n
1
n¼1
P
1
1
an ¼ 1 þ 13 12 þ 15 þ 17 14 þ 19 þ
5 6 þ is loge 2, prove that the series
Example 2
1
11
16 þ converges to 32 loge 2.
n¼1
Solution Let sn and tn denote the nth partial sums of the series
1
1 þ 13 12 þ 15 þ 17 14 þ 19 þ 11
16 þ and
1 12 þ 13 14 þ 15 16 þ ,
respectively.
We now introduce an extra piece of notation, Hn; we denote by Hn the nth
n
P
1
1
1
partial sum
k ¼ 1þ 2 þ þ n of the harmonic series.
k¼1
1
P
an come ‘in a natural way’ in threes, so it
Now, the terms of the series
This will enable us to tackle
things much more easily!
n¼1
seems sensible to look at the partial sums s3n
1 1 1 1 1 1 1 1
s3n ¼ 1 þ þ þ þ þ þ 3 2 5 7 4 9 11 6
1
1
1
þ
þ
4n 3 4n 1 2n
From the definition of s3n.
3: Series
110
1 1
1 1 1
1 1 1
¼ 1þ þ
þ þ
þ þ 3 2
5 7 4
9 11 6
1
1
1
þ
þ
4n 3 4n 1 2n
1 1
1
1 1 1
1
þ þ þ þ
¼ 1þ þ þ þ
3 5
4n 1
2 4 6
2n
1 1
1
1 1 1
1
1
¼
1þ þ þ þ
þ þ þ þ
Hn
2 3
4n
2 4 6
4n
2
1 1
1
1
1 1
1
1
¼
1þ þ þ þ
1þ þ þ þ
Hn
2 3
4n
2
2 3
2n
2
1
1
¼ H4n H2n Hn :
(7)
2
2
We insert some brackets in a
finite sum, for convenience.
We separate out the positive
and negative terms.
We add some terms to the
contents of the first bracket,
then subtract them again; the
last bracket is simply 12Hn.
Furthermore
1 1 1 1 1
1
1
t2n ¼ 1 þ þ þ þ
2 3 4 5 6
2n 1 2n
1 1
1
1 1 1
1
¼ 1þ þ þ þ
2 þ þ þ þ
2 3
2n
2 4 6
2n
1 1
1
¼ H2n 1þ þ þ þ
2 3
n
¼ H2n Hn :
By the definition of t2n.
We insert some positive terms,
then remove them again.
By the definition of H2n.
(8)
We now eliminate the H’s from equations (7) and (8) as follows
1
1
s3n ¼ H4n H2n Hn
2
2
1
¼ ðH4n H2n Þ þ ðH2n Hn Þ
2
1
¼ t4n þ t2n :
2
(9)
But it follows from the hypotheses of the example that tn ! loge 2, and so
t2n ! loge 2 and t4n ! loge 2, as n ! 1. Hence, letting n ! 1 in equation (9),
we see that
1
s3n ! loge 2 þ loge 2
2
3
¼ loge 2:
2
Next
s3n1 ¼ s3n þ
1
2n
3
! loge 2
2
and
s3n2 ¼ s3n þ
1
1
2n 4n 1
3
! loge 2
2
as n ! 1:
It follows from the above three results that sn ! 32 loge 2 as n ! 1:
&
By the definition of Hn.
3.3
Series with positive and negative terms
Problem 4
Prove that the series
1
P
n¼1
111
an ¼ 1 12 14 þ 13 16 18 þ 15 1
10
1
1
1
12
þ 17 14
16
þ converges to 12 loge 2. [You may assume that
1
P
ð1Þnþ1
¼1 12 þ 13 14 þ 15 16 þ is loge 2.]
the sum of the series
n
n¼1
In order to better understand the behaviour of series with positive and
negative terms, we introduce some notation. For a conditionally convergent
series
1
X
an ¼ a1 þ a2 þ ;
For example, for the series
1
X
n¼1
n¼1
we define the quantities aþ
n and an as follows
an ; if an 0,
0;
if an 0,
aþ
¼
¼
and
a
n
n
0;
if an < 0,
an ; if an < 0.
þ
In other words,
the sequence an picks out the non-negative terms and the
sequence an picks out the non-positive terms (and discards their sign).
þ
In particular, an ¼ aþ
n an , and an 0 and an 0.
1
P
Next, since the series
an is convergent, an ! 0 as n ! 1 . It follows that
n¼1
1
aþ
n ¼ ðjan j þ an Þ ! 0
2
as n ! 1
1 1 1
an ¼ 1 þ 2 3 4
1 1
þ þ ;
5 6
we have
1
X
n¼1
1
aþ
n ¼10þ 0
3
1
þ 0þ
5
and
1
X
1
1
a
n ¼0þ þ0þ
2
4
n¼1
1
þ 0 þ :
6
and
1
a
n ¼ ðjan j an Þ ! 0 as n ! 1:
2
1
P
Now, since
an is conditionally convergent, it must contain infinitely
1
1
n¼1
P
P
many negative terms, since otherwise
an and
jan j would be the same
n¼1
n¼1
apart from a finite number of terms. Since altering a finite number of terms
does not affect the convergence of a series, it would then follow that the series
1
P
jan j would be convergent – which we know is not the case.
For the series
1
P
an is
n¼1
assumed to be only
conditionally convergent.
n¼1
1
P
an .
Similarly, there must be infinitely many positive terms in
n¼1
1
P
Finally, we show that, if the series
an is conditionally convergent, then
1
1
1
n¼1
P
P
P
aþ
an ) and
a
the corresponding series
n (the ‘positive part’ of
n (the
1
n¼1
n¼1
n¼1
P
an ) are both divergent.
‘negative part’ of
n¼1 P
1
þ
For, suppose that
aþ
n were convergent. Then, since an ¼ an an , we
1
n¼1
P
þ
have a
a
n ¼ an an ; it follows that the series
n must also be convergent.
1
n¼1
P
And then, since jan j ¼ aþ
jan j would be
n þ an , it would also follow that
n¼1
convergent – which we know is not the case.
1
P
A similar argument shows that
a
n cannot be convergent.
1
1
n¼1
P
P
aþ
a
In particular, it follows from the divergence of the series
n and
n,
n¼1
For the series
1
P
an is
n¼1
assumed to be only
conditionally convergent.
n¼1
whose terms are non-negative, that their partial sums must tend to 1 as
n ! 1.
By the Boundedness Theorem
for series.
3: Series
112
In general it is not true that any rearrangement of a conditionally convergent
series converges. The situation is much more interesting!
Some rearrangements
converge, some diverge.
Theorem 3
Riemann’s Rearrangement Theorem
1
P
Let the series
an ¼ a1 þ a2 þ be conditionally convergent, with nth
n¼1
partial sum sn. Then there are rearrangements of the series that have the
following properties:
1. For any given number s, the rearranged series converges to s;
2. For any given numbers x and y, one subsequence of the sn converges to x
and another subsequence of the sn converges to y;
This is not a complete list of
all the possibilities that can
occur!
3. The sequence sn converges to 1.
4. One subsequence of the sn converges to 1 and another subsequence of
the sn converges to 1.
We will not prove Theorem 3. Instead, we illustrate case (1) using the facts that
we have already discovered about conditionally convergent series. Using
similar ideas, we can prove the various parts of the theorem.
1
P
ð1Þnþ1
Example 3 Find a rearrangement of the series
¼ 1 12 þ 13 14 þ
n
1
1
n¼1
5 6 þ that converges to the sum 3.
1
nþ1
P
Solution Since the ‘positive’ part of the series ð1nÞ has partial sums that
Recall that the sum of this
series is loge 2 or
approximately 0.69.
n¼1
tend to 1, we start to construct the desired rearranged series as follows. Take
enough of the ‘positive’ terms 1, 13 , 15 , . . ., 2N111 so that the sum
1 þ 13 þ 15 þ þ 2N111 is greater than 3, choosing N1 so that it is the first
integer such that this sum is greater than 3. Then these N1 terms will form
the first N1 terms in our desired rearranged series.
Next, take enough of the ‘negative’ terms 12 , 14 , 16 , . . ., 2N1 2 so that the
expression
1 1
1
1 1 1
1
þ þ þ þ
1 þ þ þþ
3 5
2N1 1
2 4 6
2N2
is less than 3, choosing N2 so that it is the first integer such that this sum is less
than 3. These N1 þ N2 terms form the first N1 þ N2 terms in our desired rearranged series.
Now add in some more ‘positive’ terms 2N11þ1 , 2N11þ3 , . . ., 2N311 so that
the sum
1 1
1
1 1 1
1
þ þ þ þ
1 þ þ þþ
3 5
2N1 1
2 4 6
2N2
1
1
1
þ
þ
þþ
2N1 þ 1 2N1 þ 3
2N3 1
is greater than 3, choosing N3 so that it is the first available integer such that
this sum is greater than 3. Then these N2 þ N3 terms will form the first N2 þ N3
terms in our desired rearranged series.
This is possible since
1 1
1þ þ þ ! 1;
3 5
as n ! 1.
3.3
Series with positive and negative terms
113
We then add in just enough ‘negative’ terms to make the next sum of
N3 þ N4 terms less than 3, and so on indefinitely.
In each set of two steps in this process we must use at least one of the
‘positive’ terms and one of the ‘negative’ terms, so that eventually all the
‘positive’ terms and all the ‘negative’ terms of the original series will be taken
1
P
exactly once in the new series, which we denote by
bn . So certainly the
series
At each step there are always
‘positive’ terms or ‘negative’
terms left to choose, since
there are infinitely many of
each.
n¼1
1
P
bn is a rearrangement of the original series
n¼1
1
P
ð1Þnþ1
n¼1
n
.
But how do we know that the sum of the rearranged series is 3? As we keep
1
P
making the partial sums of
bn swing back and forth from one side of 3 to the
n¼1
other, we need the ‘radius’ of the swings to tend to 0. We achieve this by
1
P
ð1Þnþ1
that we choose next, once
always changing the sign of the terms from
n
n¼1
1
P
the partial sum of
bn has crossed beyond the value 3. For this ensures that
n¼1
(from 2N1 2 onwards) the difference between a partial sum and 3 will always be
less than the absolute value of the last term 2Nk11 or 2N1 k used, and we know that
this last term tends to 0 as we go further out along the original series.
1
P
It follows that the rearranged series
bn converges to 3, as required. &
For clearly N1k 1k, since we
use at least 1 term in the
original series at each step of
the rearrangement process.
n¼1
Problem 5 Find a rearrangement of the series
1
1
1
4 þ 5 6 þ whose partial sums tend to 1.
1
P
ð1Þnþ1
n
n¼1
¼1 12 þ 13 The situation is very much simpler, though, in the case of absolutely convergent series.
Let the series
Theorem 4
Then any rearrangement
1
P
bn ¼
n¼1
1
P
1
P
an ¼ a1 þ a2 þ be absolutely convergent.
n¼1
1
P
bn of the series also converges absolutely, and
n¼1
an .
n¼1
For example, the series
1
1
1
1
1
1
1
1
1
þ þ þ þ 22 32 42 52 62 72 82 9 2
For the series
1 1
P
ð1Þnþ1 P
1
n2 ¼
n2 , which is
n¼1
n¼1
a basic convergent series.
is absolutely convergent. It follows from Theorem 4 that the following series is
also absolutely convergent – and to the same sum
3.3.4
1
|{z}
1
1
2 2
2
4
|fflfflfflfflfflffl{zfflfflfflfflfflffl}
1
1
1
þ 2þ 2þ 2
3
5
7
|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}
1 term
2 terms
3 terms
1
1
1
1
2 2 2 2þ
6
8
10
12
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
:
4 terms
Multiplication of series
We now look at the question of how we might multiply together two infinite
1
1
1
P
P
P
series
an and
bn to obtain a single series
cn with the property that
n¼0
n¼0
n¼0
In this sub-section we sum our
series from n ¼ 0 rather than
from n ¼ 1, simply because in
many applications of
Theorem 5 (below) this is the
version that we shall require.
3: Series
114
1
X
cn ¼
n¼0
1
X
!
1
X
an
n¼0
!
bn :
n¼0
1
P
an bn ¼
Now, afirstthought
might be that cn ¼ anbn, so that
n¼0
1
1
P
P
an bn : However this is not the case!
n¼0
n¼0
n
n
For example, let an ¼ 12 and bn ¼ 13 . Then
1
1 n
1
1 n
X
X
X
X
1
1
3
an ¼
¼ 2 and
bn ¼
¼ ;
2
3
2
n¼0
n¼0
n¼0
n¼0
but
1
X
an bn ¼
n¼0
1 n
X
1
n¼0
6
These are both geometric
series.
6
¼ :
5
Note that 65 6¼ 2 32!
We can get a clue to what the correct formula for cn might be by, temporarily, abandoning our rigorous approach and simply ‘pushing symbols about’!
If we multiply out both brackets and collect terms in a convenient way, we get
1
X
n¼0
!
an 1
X
!
bn
n¼0
¼ ða0 þ a1 þ a2 þ a3 þ Þ ðb0 þ b1 þ b2 þ b3 þ Þ
¼ a0 ðb0 þ b1 þ b2 þ b3 þ Þ þ a1 ðb0 þ b1 þ b2 þ b3 þ Þ
þ a2 ðb0 þ b1 þ b2 þ b3 þ Þ þ a3 ðb0 þ b1 þ b2 þ b3 þ Þ þ ¼ a0 b0 þ a0 b1 þ a0 b2 þ a0 b3 þ þ a1 b0 þ a1 b1 þ a1 b2 þ a1 b3 þ þ a2 b0 þ a2 b1 þ a2 b2 þ a2 b3 þ þ a3 b0 þ a3 b1 þ a3 b2 þ a3 b3 þ ¼ a0 b0 þ ða0 b1 þ a1 b0 Þ þ ða0 b2 þ a1 b1 þ a2 b0 Þ
þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þ þ :
Notice that every term that we would expect to be in such a product appears
exactly once in this final expression. We have grouped the terms in the final
expression in a particularly convenient way, so that we can see a pattern
emerging from this non-rigorous argument.
We can formalise this discussion as follows:
Theorem 5
Product Rule
1
1
P
P
Let the series
an and
bn be absolutely convergent, and let
n¼0
n¼0
cn ¼ a0 bn þ a1 bn1 þ a2 bn2 þ þ an b0 ¼
n
X
ak bnk :
k¼0
1
P
cn is absolutely convergent, and
!
!
1
1
1
X
X
X
cn ¼
an bn :
Then the series
n¼0
n¼0
n¼0
Let us now return to the two series
n¼0
1
P
an ¼
n¼0
we considered earlier, whose sums were 2
mula in Theorem 5 for cn, we have
1 P
1 n
2
n¼0
and 32,
and
1
P
n¼0
bn ¼
1 P
1 n
n¼0
3
that
respectively. With the for-
The pattern here is that in the
kth bracket, the sum of the
subscripts of each term is
precisely k.
3.3
Series with positive and negative terms
cn ¼
n
X
ak bnk ¼
k¼0
115
n
X
1
1
nk
k
2
3
k¼0
n k
1X
3
¼ n
3 k¼0 2
nþ1
1 1 32
¼ n
3
1 32
1
¼ n ð2Þ 3
nþ1 !
3
3
2
¼ n n:
1
2
2
3
It follows that
1
X
1
1
X
X
1
1
2
n
n
2
3
n¼0
n¼0
3
¼322
¼ 6 3 ¼ 3:
2
cn ¼ 3 n¼0
Since 2 32 ¼ 3; we see that the sum of the series
predicts it should be.
1
P
cn is what Theorem 5
n¼0
Remark
Notice that the theorem requires that the series
1
P
an and
n¼0
1
P
bn are absolutely
n¼0
convergent. Without this assumption, the result may be false.
1
1
P
P
Þnþ1
pffiffiffiffiffiffi . Then both
an and
bn are converFor example, let an ¼ bn ¼ ð1
nþ1
n¼0
n¼0
gent, by the Alternating Test – but they are not absolutely convergent. Now, by
the formula for cn, we have
cn ¼
n
X
n
X
ð1Þkþ1
ð1Þnkþ1
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
kþ1
nkþ1
k¼0
n
X
1
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi;
¼ ð1Þn
k
þ
1
nkþ1
k¼0
ak bnk ¼
k¼0
so that
n
X
1
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
kþ1 nkþ1
k¼0
n
X
1
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi ¼ 1:
n
þ
1
nþ1
k¼0
jcn j ¼
1
P
Since we therefore do not have cn ! 0 as n ! 1, it follows that the series
cn
n¼0
is divergent.
We shall use Theorem 5 in Section 3.4 to give a definition of the exponential
function x 7! ex in terms of series and to examine its properties, and in
Section 8.4 on power series.
Yet again, this demonstrates
that the hypotheses of our
theorems really do matter!
3: Series
116
3.3.5
Overall strategy for testing for convergence
P
We now give an overall strategy for testing a series an for convergence, as a
flow chart.
We suggest that, given a series which you wish to test for convergence or
absolute convergence, you use the tests in the order indicated in the chart.
Naturally, you may be able to short-circuit this strategy in certain cases.
Is
{an} null?
Σan is
divergent
(Non-null Test)
NO
For example, if you wish to
examine a series whose terms
alternate in sign, it is best to
go straight to the Alternating
Test.
YES
Σ an is
(absolutely)
convergent
YES
Is
Σ⏐an⏐ convergent?
Try using:
Basic series
Ratio Test
Limit Comparison Test
Comparison Test
Condensation Test
NO
or
CAN’T
DECIDE
Σ an is
convergent
(Alternating Test)
YES
Does
Alternating Test
apply?
NO
Try using
First
Principles
Remark
We have labelled the final box ‘First Principles’, to indicate that, if the various
tests do not produce a result, then it may be possible to work directly with the
sequence of partial sums {sn}.
Problem 6 Test the following series for convergence and absolute
convergence:
1
1
1
P
P
P
1
5nþ2n
3
(a)
n;
(b)
;
(c)
n
2
2n3 1;
3
(d)
n¼1
1
P
(g)
n¼1
1
P
n¼1
ð1Þnþ1
1
n3
2n
;
n6
;
(e)
n¼1
1
P
(h)
n¼1
1
P
n¼1
ð1Þnþ1 n2
n2 þ1 ;
(f)
n¼1
1
P
ð1Þnþ1 n
n2 þ2 ;
(i)
n¼1
1
P
ð1Þnþ1 n
n3 þ5 ;
1
3
n¼2 nðloge nÞ4
:
There is a variety of further tests for convergence or divergence which can
be applied to series with non-negative terms. In this book we give only one
further test, called the Integral Test. In particular it will enable us to prove that
1
X
1
is convergent; for all p > 1:
np
n¼1
You will meet the Integral
Test in Section 7.4.
3.3
Series with positive and negative terms
3.3.6
117
Proofs
You may omit these proofs on
a first reading.
We now supply the proofs omitted earlier in this section.
Theorem 1 Absolute Convergence Test
1
1
P
P
If
an is absolutely convergent, then
an is convergent.
n¼1
Proof
n¼1
We know that
1
P
jan j is convergent, and we want to prove that
n¼1
1
P
an
n¼1
is convergent.
To do this, we define two new sequences
an ; if an 0,
0;
if an 0,
þ
and an ¼
an ¼
0;
if an < 0,
an ; if an < 0.
Both the sequences aþ
and a
are non-negative, and
n
n
an ¼ aþ
n an ;
For example, if
1
X
1
1
an ¼ 1 2 þ 2
2
3
n¼1
1
2 þ ;
4
for n ¼ 1; 2; . . .:
then
1
X
Also
aþ
n
jan j;
for n ¼ 1; 2; . . .;
(10)
and
a
n jan j;
for n ¼ 1; 2; . . .:
(11)
1
P
Since
jan j is convergent, we deduce from (10) and (11) that
n¼1
1
P
a
n are convergent, by the Comparison Test. Thus
1
P
n¼1
aþ
n ¼1þ0þ
n¼1
and
1
X
a
n ¼ 0þ
n¼1
aþ
n
and
þ
1
þ 0 þ 32
1
þ0
22
1
þ :
42
n¼1
1
X
an ¼
n¼1
1 X
1
1
X
X
aþ
aþ
a
n an ¼
n n
n¼1
n¼1
(12)
n¼1
is convergent, by the Combination Rules for series.
&
Sub-section 3.1.4.
Triangle Inequality (infinite form) 1
1
1
P
P
P
If
an is absolutely convergent, then an jan j.
n¼1
Proof
n¼1
n¼1
Let
s n ¼ a1 þ a2 þ þ an ;
n ¼ 1; 2; . . .;
and
tn ¼ ja1 j þ ja2 j þ þ jan j;
n ¼ 1; 2; . . .:
Then, by the Absolute Convergence Test
1
X
an exists;
lim sn ¼
n!1
n¼1
also
lim tn ¼
n!1
1
X
n¼1
jan j:
Alternatively, the infinite
form of the Triangle
Inequality can be deduced
from statements (10), (11) and
(12) in the proof of the
Absolute Convergence Test.
3: Series
118
Now, by the Triangle Inequality
In Remark 3 following the
proof of the Triangle
Inequality in Sub-section 1.3.1,
we commented that the
Triangle Inequality for two
numbers, a and b, can be
extended in the obvious way to
any finite sum of numbers.
jsn j ¼ ja1 þ a2 þ þ an j
ja1 j þ ja2 j þ þ jan j ¼ tn ;
and so
tn sn tn :
Thus, by the Limit Inequality Rule for sequences
You met this in Subsection 2.3.3, Theorem 3.
lim tn lim sn lim tn ;
n!1
n!1
n!1
that is
1
X
j an j n¼1
and so
1
X
an n¼1
1
X
jan j;
n¼1
X
X
1
1
an ja j:
n¼1 n¼1 n
Theorem 2
&
Alternating Test
If
an ¼ ð1Þnþ1 bn ;
n ¼ 1; 2; . . .;
where {bn} is a decreasing null sequence with positive terms, then
1
X
an ¼ b1 b2 þ b3 b4 þ is convergent:
n¼1
Proof
We can write the even partial sums s2k of
1
P
an as follows
n¼1
s2k ¼ ðb1 b2 Þ þ ðb3 b4 Þ þ þ ðb2k1 b2k Þ:
Since {bn} is decreasing, all the brackets are non-negative, and so the even
subsequence of partial sums, {s2k}, is increasing.
We can also write the even partial sums s2k as
s2k ¼ b1 ðb2 b3 Þ ðb4 b5 Þ ðb2k2 b2k1 Þ b2k :
Again, all the brackets are non-negative, and so s2k is bounded above, by b1.
Hence {s2k} is convergent, by the Monotone Convergence Theorem.
Now let
lim s2k ¼ s:
k!1
Since s2k ¼ s2k1 b2k so that s2k1 ¼ s2k þ b2k , and {bn} is null, we have
lim s2k1 ¼ lim ðs2k þ b2k Þ
k!1
k!1
¼ lim s2k þ lim b2k ¼ s;
k!1
k!1
3.3
Series with positive and negative terms
119
by the Sum Rule for sequences. Thus the odd and even subsequences of {sn}
both tend to the same limit s, and so {sn} tends to s. Hence
1
X
By Theorem 6, Subsection 2.4.4.
an ¼ b1 b2 þ b3 b4 þ is convergent; with sum s: &
n¼1
Let the series
Theorem 4
1
P
an ¼ a1 þ a2 þ be absolutely conver-
n¼1
1
P
gent. Then any rearrangement
and
1
P
1
P
bn ¼
n¼1
bn of the series also converges absolutely,
n¼1
an :
n¼1
Proof First of all, we prove the result in the special case that ak 0 for
all k.
Choose any positive integer n. Then choose a positive integer N such that
b1, b2, . . ., bn all occur among the terms a1, a2, . . ., aN of the original series.
It follows that
n
X
bk k¼1
N
X
k¼1
1
X
ak
ak :
(13)
k¼1
n
P
Hence the partial sums
bk of the rearranged series
k¼1
1
P
bk are bounded
In particular, N n.
For the right-hand side is
simply the left-hand side with
possibly some more nonnegative terms inserted.
For the partial sums of a series
of non-negative terms are
always less than or equal the
sum of the whole series.
k¼1
above, so that the rearranged series must be convergent. It also follows from
1
1
P
P
bk ak .
(13) that
k¼1
k¼1
By reversing the roles of the terms a1, a2, . . . and b1, b2, . . . in the above
1
1
P
P
argument, the same argument shows that
ak bk . It thus follows that
k¼1
k¼1
1
1
P
P
ak and
bk must in fact have the same sum.
the two series
k¼1
k¼1
To complete our proof, we now drop the condition that ak 0 for all k.
þ
We then define the quantities aþ
k , ak , bk and bk as follows
aþ
k
bþ
k
¼
0;
¼
ak ; if ak 0,
if ak < 0,
bk ; if bk 0,
0; if bk < 0,
a
k
b
k
¼
if ak 0,
ak ; if ak < 0,
¼
0;
0;
if bk 0,
ak ; if bk < 0.
1 1
P
þ
1
a and P a
aþ
k
k
k is convergent, since ak ¼ 2 ðjak j þ ak Þ and both
1
1
k¼1 P
k¼1
k¼1
P
þ
þ
converge.
bk is a rearrangement of
ak , and both series have non-
Now,
1
P
k¼1
k¼1
negative terms. It follows from the first part of the proof that both series
converge, and have the same sum.
1
1
P
P
Similarly,
a
b
k and
k both converge, and have the same sum.
k¼1
k¼1
By the Combination Rules for
series.
3: Series
120
But bk ¼ bþ
k þ bk , and so the series
b converges, by the Sum Rule
k
1
k¼1
P
bk is absolutely convergent.
for series. In other words, the rearranged series
1
P
k¼1
And, in fact
1 1
1
P
b ¼ P bþ þ P b .
k
k
k
k¼1
k¼1
k¼1
Finally
1
X
bk ¼
k¼1
¼
¼
1
X
k¼1
1
X
k¼1
1
X
bþ
k aþ
k 1
X
b
k
k¼1
1
X
a
k
k¼1
ak :
&
k¼1
Theorem 5
Product Rule
1
1
P
P
Let the series
an and
bn be absolutely convergent, and let
n¼0
n¼0
cn ¼ a0 bn þ a1 bn1 þ a2 bn2 þ þ an b0 ¼
n
X
ak bnk :
k¼0
1
P
cn is absolutely convergent, and
!
!
1
1
1
X
X
X
cn ¼
an bn :
Then the series
n¼0
n¼0
Proof
n¼0
n¼0
First, we introduce some notation, for n 0
n
n
n
P
P
P
sn ¼
ak ;
tn ¼
bk ;
un ¼
ck ;
s0n
¼
s¼
s0 ¼
k¼0
n
P
jak j;
k¼0
1
P
ak ;
k¼0
1
P
k¼0
jak j;
tn0
k¼0
n
P
¼
t¼
t0 ¼
jbk j;
k¼0
1
P
u0n
¼
k¼0
n
P
jck j;
k¼0
bk ;
k¼0
1
P
jbk j:
k¼0
In particular, we know that sn ! s, tn ! t, sn0 ! s0 and tn0 ! t0 as n ! 1.
Also, by the Product Rule for sequences, sn tn ! st and sn0 tn0 ! s0 t0 , as n ! 1.
Hence, from the definition of limit, it follows that, for any positive number ",
there is some integer N for which
1
1
jsn tn stj < " and s0n tn0 s0 t0 < "; for all n N:
3
3
It follows that, for n N
0 0
s t s0 t0 ¼ s0 t0 s0 t0 s0 t0 s0 t0 n n
N N
n n
N N
s0 t 0 s0 t 0 þ s0 t 0 s0 t 0 n n
1
1
< "þ "
3
3
2
¼ ":
3
N N
Here we use an integer N
rather than a general number
X and a weak inequality n N
rather than a strict inequality;
this simplifies the notation in
the rest of the argument a
little.
3.3
Series with positive and negative terms
121
Now let n be such that n 2N, and consider the terms ai bj that occur in the
expression un sN tN. Every such term has i þ j n, since un consists of all such
terms; but none of these terms has both i N and j N.
Hence, for every term ai bj in un sN tN, a corresponding term jai bjj
will occur in the expression sn0 tn0 sN0 tN0 – for this last expression consists
of all terms jai bjj with i n and j n, but not with both i N and j N.
(Note that we made the requirement that n 2N in order that every term in
sN tN appears in un.)
N
0
a0b0 a0b1 a0b2
a1b0 a1b1 a1b2
...
a2b0
...
...
a2b1 a2b2
2N
For terms with both i N and
j N are all in sN tN.
n
...
sNtN
N
un – sNtN
2N
n
In the above diagram, we set out the terms ai bj in rows and columns, where
the term ai bj occurs in the (i þ 1)th row and the ( j þ 1)th column. Then the
small square contains all the terms ai bj with 0 i N and 0 j N; these add
up to sN tN.
Hence, for all n 2N, we have
jun stj ¼ jðun sN tN Þ þ ðsN tN stÞj
jun sN tN j þ jsN tN stj
s0 t0 s0 t0 þ jsN tN stj
n n
N N
2
1
< "þ "
3
3
¼ ";
it follows, from this inequality, that un ! st as n ! 1.
Finally, we have
u0n s0n tn0
s0 t 0 :
By the Triangle Inequality.
By the discussions above
concerning un sN tN and
sn0 tn0 sN0 tN0 .
For the product sn0 tn0 contains
more non-negative terms than
does un0 ; and the sequences
{sn0 } and {tn0 } are increasing.
3: Series
122
Hence, the increasing sequence {un0 } is bounded above, and so tends to a
1
P
&
limit as n ! 1. Thus the series
cn is absolutely convergent.
n¼0
3.4
The exponential function x j!ex
Earlier, we defined e ¼ 2.71828 . . . to be the limit
1 n
e ¼ lim 1 þ
;
n!1
n
In Sub-section 2.5.3.
and we defined ex to be the limit
x n
ex ¼ lim 1 þ
; for any real x:
n!1
n
Here we show that the formula
ex ¼
1 n
X
x
n¼0
n!
;
for any real x;
is an equivalent definition of the quantity ex.
Remark
The series
1 n
P
x
n¼0
n!
converges for all values of x. For
This series is a basic series of
type (d), in the case that x 0.
nþ1 n x
x j xj
ðn þ 1Þ! n! ¼ n þ 1
!0
as n ! 1;
so that, by the Ratio Test, the series is absolutely convergent for all x, and so
is convergent for all x.
3.4.1
The definition of ex as a power series, for x > 0
If we plot the partial sum functions of the infinite series of powers of x
1 n
X
x
n¼0
n!
¼1þxþ
x2 x3
þ þ ;
2! 3!
(1)
the resulting graph appears to be that of ex. (A series of multiples of increasing
powers of x is called a power series.) In particular, when x ¼ 1, the sum of
the series
1
X
1
1 1
¼ 1 þ 1 þ þ þ n!
2!
3!
n¼0
is approximately 2.71828 . . ..
3.4
The exponential function x j! ex
123
y
y = ex
3
2
y = 1 + x + 2!x + 3!x
2
y = 1 + x + 2!x
y=1+x
e
y=1
–1
0
1
x
However the fact that the sequence of partial sums of the series (1) appears
to converge to ex does not constitute a proof of this fact. Our first aim is to
supply this proof in the particular case that x > 0.
Theorem 1
If x > 0, then
1 n
P
x
n¼0
n!
¼ ex .
We shall deal with the case
x < 0 in Sub-section 3.4.3.
Hence, for x > 0, the series
1 n
P
x
n! gives an equivalent
n¼0
definition of ex.
Proof
n
We defined ex to be ex ¼ lim 1 þ nx , and so we have to show that
n!1
x n
¼ lim 1 þ
;
n
n! n!1
n¼0
1 n
X
x
for x > 0:
The (n + 1)th partial sum of the above series is
x2 x3
xn
þ þ þ :
2! 3!
n!
By the Binomial Theorem
x nð n 1Þ x 2
x n
x n
1þ
¼1þn
þ þ
:
þ
n
n
2!
n
n
A typical term in this expansion is
nðn 1Þ . . . ðn k þ 1Þ x k xk
1
2
k1
1
¼
1 1 k!
n
n
n
n
k!
xk
;
k!
since each bracket is less than 1.
Thus
x n
x2 x3
xn
1þ
1 þ x þ þ þ þ
n
2! 3!
n!
¼ snþ1 ;
snþ1 ¼ 1 þ x þ
and so, by the Limit Inequality Rule for sequences
x n
lim 1 þ
lim snþ1 ;
n!1
n!1
n
You may omit this proof at a
first reading.
3: Series
124
so that
ex 1 n
X
x
n¼0
n!
:
(2)
On the other hand, for any integers m and n with m n, we have
x nð n 1Þ x 2
x n
nðn 1Þ . . . ðn m þ 1Þ xm
1þn
þ þ
þ
1þ
n
n 2! n
n m!
x2
1
xm
1
2
m1
1
1
¼1þxþ
þ þ
1
... 1 :
n
n
n
n
2!
m!
This follows since {sn} and
{snþ1} both tend to the same
limit, the sum of the infinite
series.
Now keep m fixed and let n ! 1. By the Limit Inequality Rule for
sequences, we obtain
x n
x2 x3
xm
lim 1 þ
1 þ x þ þ þ þ ;
n!1
n
2! 3!
m!
and so
ex smþ1 :
Now let m ! 1; by the Limit Inequality Rule for sequences, we obtain
ex lim smþ1 ;
m!1
so that
ex 1 n
X
x
n¼0
(3)
n!
Combining inequalities (2) and (3), we obtain
1 n
X
x
; for x > 0:
ex ¼
n!
n¼0
&
Problem 1 Estimate e2 (to 3 decimal places) by calculating the seventh
partial sum of the series (1) when x = 2.
3.4.2
Calculating e
The representation of e by the infinite series
1
X
1
1 1
¼ 1 þ 1 þ þ þ e¼
n!
2!
3!
n¼0
(4)
provides a much more efficient way of calculating approximate values for e
n
than the equation e ¼ lim 1 þ 1n . This is illustrated by the following table of
n!1
approximate values:
n
1 þ 1n
n
P
1
k¼0
k!
n
1
2
3
4
5
2
2.25
2.37
2.44
2.49
2
2.50
2.67
2.71
2.717
The calculation of e via the
limit is a new calculation each
time, whereas via the series
involves only adding one
extra term to the previous
approximation.
3.4
The exponential function x j! ex
125
We can estimate how quickly the sequence of partial sums
sn ¼
n1
X
1
1 1
1
¼ 1 þ 1 þ þ þ þ
;
k!
2!
3!
ð
n
1Þ!
k¼0
n ¼ 1; 2; . . .;
converges to e as follows. The difference between e and sn is given by
1
X
1
1
1
1
¼ þ
þ
þ e sn ¼
k!
n!
ð
n
þ
1
Þ!
ð
n
þ
2Þ!
k¼n
1
1
1
1þ
þ
þ ¼
n!
ðn þ 1Þ ðn þ 1Þðn þ 2Þ
1
1 1
<
1 þ þ 2 þ :
n!
n n
The last expression in square brackets is a geometric series with first term
n
1 and common ratio 1n, and so its sum is 11 1 ¼ n1
. Hence
n
1
n
0 < e sn < n! n 1
1
1
(5)
; for n ¼ 1; 2; . . .:
¼
ðn 1Þ! n 1
Here we replace various terms
by larger terms (because their
denominators are smaller); so
the new sum is greater.
For example, this estimate shows that
0 < e s6 <
1 1
1
¼
¼ 0:0016:
5! 5 600
Since the partial sum
1 1 1 1
þ þ þ
2! 3! 4! 5!
1 1 1
1
163
¼
¼ 2:716;
¼1þ1þ þ þ þ
2 6 24 120
60
s6 ¼ 1 þ 1 þ
we deduce very easily that
2:716 < e < 2:716 þ 0:0016 ¼ 2:7183 :
In fact,
e ¼ 2:71828182845 . . .:
Problem 2 Estimate how many terms in the series (4) are needed to
determine e to ten decimal places.
The inequality (5) can also be used to show that e is irrational.
Theorem 2
The number e is irrational.
Proof Suppose that e ¼ mn, where m and n are positive integers.
Now, by the above estimate (5)
0 < e snþ1 <
1 1
;
n! n
and so
1
0 < n!ðe snþ1 Þ < :
n
This is a proof by
contradiction.
3: Series
126
Since e ¼ mn, we have
m
1 1
1
1
0 < n! 1 þ 1 þ þ þ þ
< :
n
2! 3!
n!
n
m But n! n 1 þ 1 þ 2!1 þ 3!1 þ þ n!1 is an integer, since n! times each
expression in the square bracket is itself an integer. So we have found an
integer which lies strictly between 0 and 1. This is clearly impossible, and so
&
e cannot be rational after all.
3.4.3
The definition of ex as a power series, for all real x
Earlier we saw that
1 n
P
x
n¼0
n!
¼ ex , for all x > 0. We now show that this formula is
valid for all real values of x, and we will also verify the fundamental property
of the exponential function, namely that exey ¼ exþy for all real values of x
and y. The Product Rule for series will be our crucial tool in this work.
First we check the following.
For any x 2 R,
Lemma 1
1 n
P
x
n¼0
n!
1
P
ðxÞn
n¼0
Proof By the Product Rule for series,
n 1, we have
cn ¼
¼
n
X
xk
k!
k¼0
n
k
X
n!
1 n
P
x
1
P
ðxÞn
n¼0
n¼0
n! n!
¼
1
P
cn , where, for
n¼0
ðxÞnk
ðn kÞ!
n!
n¼0
are both absolutely
convergent.
¼
n
1X
n!
xk ðxÞnk
n! k¼0 k!ðn kÞ!
¼
1
ðx þ ðxÞÞn ¼ 0:
n!
n¼0
1 n
P
x
n¼0
We may apply the Product
1 n
P
x
Rule since the series
n! and
n¼0
x ðxÞnk
k!ðn kÞ!
k¼0
If x < 0, then
This was Theorem 5 in
Sub-section 3.3.4.
1
P
ðxÞn
&
1 n
P
x
x
We can then use the result in Lemma 1 to prove that
n! ¼ e , for x < 0 as
n¼0
1
P xn
x
well as x > 0. Since, obviously,
n! ¼ e when x ¼ 0, this completes the
n¼0
1
P xn
x
proof that
n! ¼ e , for all x 2 R.
Theorem 3
This was Theorem 1 in
Sub-section 3.4.1.
¼ 1:
Since c0 ¼ 1 1 ¼ 1, the result then follows.
Proof
Here we have simply
substituted mn for e in the
previous expression.
n!
¼ ex .
For x < 0, the following chain of equalities holds
,
1 n
1
X
X
x
ðxÞn
¼1
ðby Lemma 1Þ
n!
n!
n¼0
n¼0
Here we apply the Binomial
Theorem to ðx þ ðxÞÞn .
In other words, the definition
of ex as a power series is
equivalent to its definition
as a limit.
Here the definition of ex on
the right is that x n
.
ex ¼ lim 1 þ
n!1
n
3.5
Exercises
127
¼1
.
xn
lim 1 þ
n!1
n
ðby Theorem 1; since x > 0Þ
¼ 1=ex
¼e
ðthis is the definition of ex Þ
x
ðby the Inverse Property of ex Þ:
&
This completes the proof.
In order to be crystal clear where we have reached, we now combine the results
of Theorem 1 and Theorem 3 into the following:
Theorem 4
For all x 2 R,
1 n
P
x
n¼0
n!
n
¼ lim 1 þ nx ¼ ex .
n!1
Finally, we verify the Fundamental Property of the exponential function,
which was left as unfinished business at the end of Sub-section 2.5.3.
Theorem 5 Fundamental Property of the exponential function
For all x, y 2 R, ex ey ¼ exþy .
Proof
By the Product Rule for series, ex ey ¼
1 n
P
x
n¼0
n!
1 n
P
y
n¼0
n!
¼
1
P
cn ,
n¼0
where, for n 1, we have
cn ¼
¼
n
X
xk
k¼0
n
X
k!
ynk
ðn kÞ!
xk ynk
k!ðn kÞ!
k¼0
¼
n
1X
n!
xk ynk
n! k¼0 k!ðn kÞ!
¼
ðx þ yÞn
:
n!
Since c0 ¼ 1 1 ¼ 1, the result then follows.
3.5
&
Exercises
Section 3.1
1. Prove that
1 P
n¼1
34
n
is convergent, and find its sum.
2. Interpret 0:12 as an infinite series, and hence find the value of 0:12 as a
fraction.
The Inverse Property of ex
was Theorem 4 of Subsection 2.5.3.
3: Series
128
3. Prove that
1
1
1
1
¼
;
4n2 1 2 2n 1 2n þ 1
for n ¼ 1; 2; . . .;
and deduce that
1
X
1
1
¼ :
21
4n
2
n¼1
4. Determine whether the following series converge:
1 1 1 n P
P
P
4 n
4
1
p1ffiffi pffiffiffiffiffiffi
(a)
1 þ 12 ;
(b)
.
(c)
5 þ nðnþ2Þ ;
n
nþ1
n¼1
n¼1
n¼1
Section 3.2
1. Determine whether the following series converge:
pffiffiffiffi
1
1
1
P
P
P
cosð1=nÞ
2n
n2
(b)
(c)
(a)
2n3 n;
2n2 þ3 ;
4n3 þnþ2;
(d)
n¼1
1
P
n¼1
ðnþ1Þ5
2n ;
(e)
n¼1
1
P
n¼1
n2 3n
n! ;
(f)
n¼1
1
P
n¼1
ðn!Þ2
ð2nÞ!.
1 n
P
2 n!
1 n
P
3 n!
2. (a) Use the Ratio Test to prove that
nn converges, but that
nn
n¼1
n¼1
diverges.
(b) For which positive values of c can you use the Ratio Test to prove
1 n
P
c n!
that
nn is convergent?
n¼1
3. Prove that
1
1
1
pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffipffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi pffiffiffi ; for n ¼ 1; 2; . . .;
n
nþ1
n nþ1 nþ1þ n
and use Exercise 4(c) on Section 3.1 to deduce that
1
P
1
3
n¼1 n2
is convergent.
4. Determine whether the following series converge:
pffiffiffiffiffiffi
1
1
P
P
nþ1
1
(a)
.
;
(b)
nðloge nÞðloge ðloge nÞÞ
ð2n2 3nþ1Þðloge nþðloge nÞ2 Þ
n¼3
n¼2
Section 3.3
1. Test the following series for convergence and absolute convergence:
1
1
1
1
P
P
P
P
ð1Þnþ1
ð1Þnþ1 n!
sin n
nþ2n
pffiffi ;
(a)
(b)
(c)
(d)
3n þ5.
n4 þ3 ;
n2 ;
1þ n
n¼1
n¼1
n¼1
n¼1
2. (a) Prove that
1
1
1
1
þ
> ;
3n 2 3n 1 3n 3n
for n ¼ 1; 2; . . .;
and deduce that
1 1 1 1 1
1 þ þ þ þ 2 3 4 5 6
is divergent:
3.5
Exercises
129
(b) Prove that
1 1
1 < 1;
n 2n 2n þ 2 n2
for n ¼ 1; 2; . . .;
and deduce that
1 1 1 1 1 1 1
1 1 1
1
1 þ þ þ þ 2 4 3 6 8 5 10 12 7 14 16
is convergent:
(c) Prove that
1
1
1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ;
2n 1 2n 2n
for n ¼ 1; 2; . . .;
and deduce that
1
1
1
1
1
1 þ pffiffiffi þ pffiffiffi þ 2
3 4
5 6
is divergent:
3. Prove that there exists a rearrangement of the series
1
4
1
P
ð1Þnþ1
n¼1
n
¼ 1 12 þ 13 þ 15 16 þ for which one subsequence of its partial sums tends to 1 and
another subsequence of its partial sums tends to 1. Alternatively, prove that
no such rearrangement exists.
4
Continuity
The graphs of many functions that we come across look to be smooth curves,
without any jumps.
y = ex
1
y = sin x
1
– 12 π
1
π
2
–1
On the other hand, the graphs of some functions that arise naturally, or that
we construct for various purposes, do not appear to be smooth, but to contain
jumps.
y = [x]
y = tan x
1
–2
–1
1
2
– 12 π
1π
2
–1
However there are some functions where the graphs appear to have an
unreasonable number of jumps!
How can we describe this complicated situation? The key underlying idea is
that of continuity. Loosely speaking, a function is said to be continuous if we
can draw its graph without taking our pencil off the page. This is the same as
the graph ‘having no jumps’. The concept is an important one in Analysis
since, in many situations, the crucial step in proving that a function has some
property is to prove that the function is continuous. This is the first of two
chapters that study continuous functions.
In Section 4.1, we define the phrase:
the function f is continuous at the point c,
and we give a number of rules which state, for example, that various combinations and compositions of continuous functions are themselves continuous.
130
4.1
Continuous functions
131
Using these rules, together with a list of basic continuous functions, we can
deduce that many functions are continuous at each point of their domains. For
example, the functions x 7! x þ 1x and x 7! x sin 1x are continuous at each point
of R {0}, and the trigonometric and exponential functions are continuous
at each point of their domains.
Section 4.2 is devoted to the properties of continuous functions. The two
fundamental properties of continuous functions are the Intermediate Value
Theorem and the Extreme Value Theorem. We shall see a useful application of the Intermediate Value Theorem to finding the zeros of various
functions.
In Section 4.3, we discuss the Inverse Function Rule; this rule gives conditions under which a continuous function f has a continuous inverse function
f 1. We then use the Inverse Function Rule to obtain the inverses of standard
functions. Some of these inverse functions will be familiar to you already, but
the Inverse Function Rule enables us to establish their properties by a rigorous
argument.
Finally, in Section 4.4, we shall provide a rigorous definition of the exponential function x 7! ax , where x 2 R and a > 0.
4.1
Continuous functions
4.1.1
What is continuity?
To accord with our intuitive idea of what we should mean by ‘the function f is
continuous at the point c’, we wish to define this concept in such a way that the
following two functions are continuous at the point c:
1.
y
y
2.
f (c)
f (c)
c
x
c
x
On the other hand, we wish to formulate our definition so that the following
two functions are not continuous at the point c:
3.
y
4.
f (c)
y
f (c)
c
x
c
x
In particular, to finding
solutions of polynomial
equations.
4: Continuity
132
So, our definition must say, in a precise way, that:
if x tends to c, then f (x) tends to f (c).
There are several ways of making this idea precise. In this chapter we adopt a
definition which involves the convergence of sequences, as this will enable us
to use the results about sequences that we met in Chapter 2.
In Chapter 5 we shall meet a
different definition, and see
that the two definitions are in
fact equivalent.
Definitions A function f defined on an interval I that contains c as an
interior point is continuous at c if:
for each sequence {xn} in I such that xn ! c, then f (xn) ! f (c).
If f is not continuous at the point c in I, then it is discontinuous at c.
Thus, for example, in graph 3 above, if {xn} is a strictly decreasing
sequence that tends to c as n ! 1, then {f (xn)} is a strictly decreasing
sequence that tends to {f (c)}. On the other hand, if {xn} is a strictly
increasing sequence that tends to c as n ! 1, then {f (xn)} is a strictly
increasing sequence that tends to a limit as n ! 1 – but that limit is some
number less than f (c).
However, in graph 4, if {xn} is a strictly increasing sequence that tends to c
as n ! 1, then {f (xn)} is a strictly increasing sequence that tends to f (c); but, if
{xn} is a strictly decreasing sequence that tends to c as n ! 1, then the
sequence {f (xn)} may not tend to any limit as n ! 1.
On the other hand, in graphs 1 and 2, no matter how the sequence {xn} tends
to c, then for sure we do have that {f (xn)} tends to f (c).
For these examples, then, the definition agrees with what we believe the
essence of continuity should be.
Example 1
Prove that the function f (x) ¼ x3, x 2 R, is continuous at the point 12.
Solution Let {xn} be any sequence in R that converges to 12; that is, xn ! 12.
Then, by the Combination Rules for sequences, it follows that
f ðxn Þ ¼
x3n
3
1
1
!
¼ as n ! 1;
2
8
while f 12 ¼ 18. In other words, {f(xn)} converges to f 12 as n ! 1.
It follows that f is continuous at 12, as required.
Example 2
Prove that the function f ð xÞ ¼
y
y = x3
1
8
x
1
2
y
&
2
1
1; x < 0;
is discontinuous at 0.
2; x 0;
Solution Let {xn} be any sequence in R that converges to 0 from the right. By
looking at the graph y ¼ f(x), it is clear that for such a sequence f(xn) ! 2 ¼ f(0).
This will not help us to show that f is discontinuous at 0!
However, if we let {xn} be any non-constant sequence in R that converges
to 0 from the left, the situation will be very different. By looking at the
graph y ¼ f(x), it is clear that for such a sequence f(xn) ! 1 6¼ f(0). We now
make this precise.
1, x < 0,
y = 2, x ≥ 0.
{
x
To prove that a function is
discontinuous, all we need to
do is to find just one sequence
for which our definition of
continuity at the relevant
point does not hold.
4.1
Continuous functions
Choose the sequence fxn g ¼ 1n ; n ¼ 1; 2; . . .. For this sequence
f ðxn Þ ¼ 1 ! 1 6¼ f ð0Þ as n ! 1:
It follows that the function f cannot be continuous at 0.
133
&
Here we make a specific
choice for the sequence {xn}
to prove that f cannot be
continuous at 0.
Problem 1
(a) Determine whether the function f ð xÞ ¼ x3 2x2 ; x 2 R, is continuous at 2.
(b) Determine whether the function f ð xÞ ¼ ½ x; x 2 R, is continuous at 1.
Problem 2
(a) Prove that the function f ð xÞ ¼ 1; x 2 R, is continuous on R.
(b) Prove that the function f ð xÞ ¼ x; x 2 R, is continuous on R.
Here [] is the integer part
function.
By ‘continuous on R’, we
mean ‘continuous at each
point of R’.
However, as we know, not every function is defined on the whole of R – but we
still want to be able to discuss the continuity of such functions. For example:
pffiffiffi
f ð xÞ ¼ x, where the domain of f is the interval [0, 1);
1
1
f ð xÞ ¼ x2 þ ð1 xÞ2 , where the domain of f is the half-closed interval (0, 1];
f ð xÞ ¼ 1x, where the domain of f is R {0}.
How can we deal with these various different situations?
The first two situations are dealt with by introducing the notion of one-sided
continuity.
Suppose that f is defined on some set S that contains an interval [c, c þ r) for
some r > 0, but where f is not necessarily defined on any interval (c r, c þ r)
that contains c as an interior point. Then it seems reasonable to say that f is
continuous on one side of c. In particular, that f is continuous on the right at the
point c of S.
Similarly, if f is defined on some set S that contains an interval (c r, c] for
some r > 0, but f is not necessarily defined on any interval (c r, c þ r) that
contains c as an interior point, then it seems reasonable to say that f is
continuous on one side of c. In particular, that f is continuous on the left at
the point c of S.
The case of the function f ð xÞ ¼ 1x, where the domain of f is R – {0}, is
different. The domain of f is the whole of R, with just one point omitted. That
is, it is the union of two open intervals (1, 0) and (0, 1) – to both of which
the definition of continuity applies in its originally stated form. So it makes
sense to simply drop the original restriction that the domain of f must be an
open interval in R.
These considerations lead us to making the following less restrictive definition of continuity than our original definition:
y
y=
1
x
x
Definitions A function f defined on a set S in R that contains a point c is
continuous at c if:
for each sequence {xn} in S such that xn ! c, then f(xn) ! f(c).
If f is continuous on the whole of its domain, we often simply say that f is
continuous (without explicit mention of the points of continuity).
If f is not continuous at the point c in S, then it is discontinuous at c.
Here f is continuous on S.
4: Continuity
134
In addition, we have the following related definitions:
Definitions A function f whose domain contains an interval [c, c þ r) for
some r > 0 is continuous on the right at c if:
for each sequence {xn} in [c, c þ r) such that xn ! c, then f(xn) ! f(c).
A function f whose domain contains an interval (c r, c] for some r > 0 is
continuous on the left at c if:
for each sequence {xn} in (c r, c] such that xn ! c, then f(xn) ! f(c).
A function f whose domain contains an interval I is continuous on I if it is
continuous at each interior point of I, continuous on the right at the left endpoint of I (if this belongs to I), and continuous on the left at the right endpoint of I (if this belongs to I).
The connection between the definitions of continuity and of one-sided
continuity is rather obvious.
Theorem 1 A function f whose domain contains an interval I that contains
c as an interior point is continuous at c if and only if f is both continuous on
the left at c and continuous on the right at c.
Example 3 Determine whether the function f given by f ð xÞ ¼
continuous on fx : x 2 R; x 0g.
pffiffiffi
x; x 0, is
Solution The domain of f is the interval I ¼ fx : x 0g.
Thus, we have to show that for each c in I:
pffiffiffi
pffiffiffiffiffi
for each sequence {xn} in I such that xn ! c, then xn ! c.
if c ¼ 0, then we know already that, for any null sequence {xn} in I,
pFirst,
ffiffiffiffiffi
xn is also a null sequence.
let c > 0. We have to prove that if fxn cg is a null sequence, then so
Next,
pffiffiffiffiffi pffiffiffi
is
xn c . Now, from the identity
We omit a proof of this
straight-forward result.
In fact, f is continuous on the
right at 0, and continuous at
all other points of its domain.
y
y = √x
x
0
pffiffiffiffiffi pffiffiffi
xn c
xn c ¼ pffiffiffiffiffi pffiffiffi ;
xn þ c
we see that, since c 6¼ 0
pffiffiffiffiffi pffiffiffi
0
xn c ! pffiffiffi ¼ 0
2 c
In other words,
proof.
as n ! 1:
pffiffiffiffiffi pffiffiffi
xn c is indeed a null sequence. This completes the
&
Problem 3
Prove that the nth root function
1
x 0; if n is even;
n
f ð xÞ ¼ x ; where n 2 N and
x 2 R; if n is odd;
is continuous.
Recall that c 6¼ 0, so that p1ffiffic is
indeed defined.
You may omit this Problem if
you are short of time.
1
We defined an in
Sub-section 1.5.3.
4.1
Continuous functions
Hints: Use the result of Exercise 3(b) on Section 2.3, in Section 2.6, and
the identity aq bq ¼ ða bÞðaq1 þ aq2 b þ aq3 b2 þ þ bq1 Þ
with suitable choices for a, b and q. In your solution, be careful not to
use the letter n for two different purposes.
Example 4
continuous.
135
Example 3 above was the
special case of Problem 3,
where n ¼ 2.
Determine whether the function f ð xÞ ¼ 1x ; x 2 R f0g, is
Solution The domain of f is the set R {0}, the union of the two open
intervals (1, 0) and (0, 1).
Let c be any point of R {0}, and let {xn} be any sequence in R {0} that
converges to c. Then, by the Quotient Rule for sequences, it follows that
ff ðxn Þg ¼ x1n ! 1c as n ! 1; in other words, that f ðxn Þ ! f ðcÞ ¼ 1c as
n ! 1. So f is continuous at c.
Since c is an arbitrary point of R {0}, it follows that f is continuous
&
on R {0}.
Sub-section 2.3.2.
Our work so far on continuity illustrates the following general strategy:
Strategy for continuity
To prove that a function f : S ! R is continuous at a point c of S, prove
that:
for each sequence {xn} in S such that xn ! c, then f(xn) ! f(c).
To prove that a function f : S ! R is discontinuous at a point c of S:
find one sequence {xn} in S such that xn ! c but f(xn) 6! f(c).
Problem 4
Recall that just one such
sequence suffices.
Prove that the following functions are continuous on R:
(a) f(x) ¼ c, x 2 R;
(b) f(x) ¼ xn, x 2 R, n 2 N;
(c) f(x) ¼ jxj, x 2 R.
Problem 5 Determine the points of continuity and discontinuity of the
signum function
8
< 1; x < 0,
x ¼ 0,
f ð xÞ ¼ 0;
:
1;
x > 0.
y
1
y = f(x)
x
–1
Remarks
The definition of function that we are using involves a mapping that we call
f (say) from a set in R, the domain A (say), to another set in R, the codomain
B (say).
1. Let f be continuous at a point c in A, and assume that, for some set A0 A,
c 2 A0 . Suppose that another function g has domain A0 on which g(x) ¼ f(x).
Technically g is a different function from f, for sure. However if f is
continuous at c, then it is a simple matter of some definition checking to
You can think of the mapping
f as being a formula x 7! y that
maps a point x of A onto to a
point y of B, and we write
y ¼ f (x) to indicate this.
4: Continuity
136
verify that g too is continuous at c. So, restriction of a function to a smaller
domain does not affect its continuity at a point.
2. Let f and g be functions defined on sets in R that contain an open interval I,
and c 2 I. Then, if f(x) ¼ g(x) for all x 2 I, f is continuous at c if g is
continuous at c, and f is discontinuous at c if g is discontinuous at c.
Again, we may simply ignore any difference between the domains of the
functions when we are studying continuity at a point.
3. The underlying point here is that continuity at a point is a local property. It
is only the behaviour of the function near that point that determines whether
it is continuous at the point.
4.1.2
Rules for continuous functions
We have seen how to recognise whether a given function is continuous at a
point. However, it would be tedious to have to go back to first principles to
determine on each occasion whether a complicated function is continuous.
As usual, we avoid such problems by having a set of rules that enable us to
construct continuous functions.
Combination Rules for continuous functions
If f and g are functions that are continuous at a point c, then so are:
Sum Rule
f þ g;
Multiple Rule
lf,
Product Rule
Quotient Rule
f
g,
for l 2 R;
fg;
provided that f (c) 6¼ 0.
For example, any polynomial p(x) ¼ a0 þ a1x þ þ anxn, x 2 R, is continuous at all points of R since we can build up the expression for p by successive
applications of the Combination Rules for continuous functions.
Similarly, any rational function rðxÞ ¼ pðxÞ
qðxÞ, where p and q are polynomials
and the domain of r is R minus the points where q vanishes, is continuous at all
points of its domain, since we can build up the expression for r by successive
applications of the Combination Rules for continuous functions.
The Combination Rules above are all natural analogues of the corresponding
results for sequences. However, we can combine functions in more ways than
we can combine sequences – for example, we can compose functions f and g to
obtain the function g f . This approach too will often enable us to obtain new
continuous functions.
Composition Rule Let f be continuous on a set S1 that contains a point c,
and g be continuous on a set S2 that contains the point f(c). Then g f is
continuous at c.
For example, we know that thepfunction
f(x) ¼ x2 þ 1, x 2 R, is continuous on R
ffiffiffi
and that the function gð xÞ ¼ x; x 0, is continuous on I ¼ {xp
: xffiffiffiffiffiffiffiffiffiffiffiffi
0}.ffi It
follows from the Composition Rule that the function g f : x !
7
x2 þ 1 is
For instance, the function
pðxÞ ¼ x17 34x5 þ5x 8
is continuous on R.
For instance, the function
r ð xÞ ¼
1 2x
;
x2 4
x 2 R f2g; is continuous
on its domain.
Functions obtained in this
way are called composite
functions.
4.1
Continuous functions
137
continuous at all points c of R (the domain of f) whose image f(c) lies in I. But
all such points f(c) lie in I; so it follows that in fact the composite g f is
continuous on R.
3
Problem 6 Prove that the function f given by f ð xÞ ¼ x2 ; x 0; is
continuous.
Finally, just as we had a Squeeze Rule for convergent sequences, we have a
corresponding Squeeze Rule for continuous functions.
Theorem 1
Squeeze Rule
Let the functions f, g and h be defined on an open interval I, and c 2 I. If:
1. g(x) f(x) h(x), for all x 2 I,
We state this rule only in the
case of an interior point of an
open interval. However more
general versions also exist!
y
y = h(x)
y = f (x)
y = g (x)
2. g(c) ¼ f(c) ¼ h(c),
3. g and h are continuous at c,
I
then f is also continuous at c.
Example 5 Prove that the function f given by f ð xÞ ¼
is continuous at 0.
2
x sin
0;
1
x
c
;
x¼
6 0,
x ¼ 0,
Solution The diagram in the margin suggests that we should find functions
g and h that squeeze f near 0.
So, we define g(x) ¼ x2, x 2 R, and h(x) ¼ x2, x 2 R. With these two
chosen, we check the conditions of the Squeeze Rule.
The first condition is of course vital! Now we know that
1
1 sin
1;
x
It follows that
x2 x2 sin
We shall study the
trigonometric functions in
detail in Sub-section 4.1.3.
The only property that you
need here is that, for any real
number x, jsin xj 1.
y
y = x2 sin 1x
for any x 6¼ 0:
1
x2 ;
x
x
0
x
for any x 6¼ 0;
so that
gð xÞ ¼ x2 f ð xÞ x2 ¼ hð xÞ;
for any x 2 R:
So condition 1 of the Squeeze Rule is satisfied.
Next, the functions f, g and h all take the value 0 at the point 0. Thus
condition 2 of the Squeeze Rule is satisfied.
Finally, the functions g and h are polynomials, and so in particular they are
continuous at 0. So condition 3 of the Squeeze Rule is satisfied.
It follows then from the Squeeze Rule that f is continuous at 0, as required. &
To test your understanding of these techniques, try the following problems.
Problem 7 Prove that the following function is continuous on R,
stating each rule or fact about continuity that you are using
f ð xÞ ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ x þ 1 5x
; x 2 R:
1 þ x2
These inequalities are
obviously true for x ¼ 0, as
well as for x 6¼ 0.
Recall that all polynomials are
continuous on R.
4: Continuity
138
Problem 8 Using the elementary properties of the trigonometric functions, determine whether the following functions are continuous at 0:
x sinð1xÞ; x 6¼ 0,
sinð1xÞ; x 6¼ 0,
(a) f ðxÞ ¼
(b) f ðxÞ ¼
0;
x = 0;
0;
x = 0.
Proofs
We now give the proofs of the Combination, Composition and Squeeze Rules.
First, recall the Combination Rules.
Combination Rules for continuous functions
If f and g are functions that are continuous at a point c, then so are:
Sum Rule
f þ g;
Multiple Rule
Product Rule
lf, for l 2 R;
fg;
Quotient Rule
f
g,
provided that f(c) 6¼ 0.
The proofs of all these are very similar, and depend on the corresponding
results for sequences. We prove only the Sum Rule.
Proof of the Sum Rule We want to prove that f þ g is continuous at c.
Suppose that f and g have domains S1 and S2, respectively. Then the domain
of f þ g is S1 \ S2, and this set contains the point c.
Thus, we have to show that:
for each sequence {xn} in S1 \ S2 such that xn ! c, then
f ðxn Þ þ gðxn Þ ! f ðcÞ þ gðcÞ:
We know that the sequence {xn} lies in S1 and in S2, and that both functions
f and g are continuous at c. Hence
f ðxn Þ ! f ðc Þ
and gðxn Þ ! gðcÞ;
and it follows from the Sum Rule for Sequences that f(xn) þ g(xn) !
&
f(c) þ g(c), as required.
Next, recall the Composition Rule.
Composition Rule Let f be continuous on a set S1 that contains a point c,
and g be continuous on a set S2 that contains the point f(c). Then g f is
continuous at c.
Proof We want to prove that g f is continuous at c.
Now, we know that g f is certainly defined on the set S ¼ {x : x 2 S1 and
f(x) 2 S2}, and this set contains the point c.
Thus, we have to show that:
for each sequence {xn} in S such that xn ! c, then gðf ðxn ÞÞ ! gðf ðcÞÞ.
We know that the sequence {xn} lies in S1, and that f is continuous at c.
Hence, we have that f(xn) ! f(c).
You may omit the rest of this
Sub-section at a first reading.
4.1
Continuous functions
139
We also know that {f(xn)} lies in S2, and that g is continuous at f(c). It
&
follows that gðf ðxn ÞÞ ! gðf ðcÞÞ, as required.
Finally, recall the Squeeze Rule.
Theorem 1
Squeeze Rule
Let the functions f, g and h be defined on an open interval I, and c 2 I. If:
1. g(x) f(x) h(x), for all x 2 I,
2. g(c) ¼ f(c) ¼ h(c),
3. g and h are continuous at c,
then f is also continuous at c.
Proof We have to show that f is continuous at c.
Thus, we have to prove that:
for each sequence {xn} in the domain of f such that xn ! c, then
f (xn) ! f (c).
Now, since xn ! c there is some number X such that xn 2 I, for all n > X.
Hence, by condition 1 we have that
gðxn Þ f ðxn Þ hðxn Þ;
for all n > X:
(1)
This follows from the
definition of a sequence
converging to c.
So, if we now let n ! 1 and use conditions 2 and 3, we get that
lim gðxn Þ ¼ gðcÞ ¼ f ðcÞ and
n!1
lim hðxn Þ ¼ hðcÞ ¼ f ðcÞ:
n!1
(2)
It follows, from (1), (2) and the Squeeze Rule for sequences, that
Sub-section 2.3.3.
lim f ðxn Þ ¼ f ðcÞ;
n!1
&
as required.
4.1.3
Trigonometric functions and the exponential
function
Trigonometric functions
For the moment we are assuming that you have a knowledge of the trigonometric functions arising from your study of trigonometry.
We will prove that the trigonometric functions are continuous on the
whole of their domains. But, first, we need a basic inequality for the sine
function.
Lemma 1
0 sin x x,
for 0 x p2 :
Proof If x ¼ 0, then sin 0 ¼ 0; so there is an equality.
Suppose next that 0 < x p2, and consider the following diagram, which
represents a quarter circle, centred at the origin, with radius 1.
4: Continuity
140
Since the circle has radius 1, the arc AB has length x and the perpendicular AC
has length sin x. Hence
0 < sin x x;
p
for 0 < x :
2
&
For the shortest distance from
the point A to the line BC is
the perpendicular from A
to BC.
We can now extend the inequality in Lemma 1 to obtain a more general result.
Theorem 2
y
The Sine Inequality
jsin xj j xj; for x 2 R:
y = ⎜x ⎜
1
Proof We saw in Lemma 1 that the inequality holds for 0 x p2 :
For x > p2, we have
–π
⎜sin
y = ⎜sin x ⎜
0
x ⎜≤ ⎜x ⎜, for x ∈
π x
1
jsin xj 1 < p < x ¼ j xj;
2
and so the desired inequality is also true in this case.
Finally, the inequality also holds for x < 0, since
jsinðxÞj ¼ jsin xj
and jxj ¼ j xj:
&
This is the key tool that we need to prove the continuity of the trigonometric
functions.
Theorem 3
continuous.
The trigonometric functions (sine, cosine and tangent) are
Recall that this means they are
continuous at each point of
their domains.
Proof To prove that the sine function is continuous at each point c 2 R, we
need to show that:
for each sequence fxn g in R such that xn ! c; then sin xn ! sin c:
(3)
We use the formula
sin xn sin c ¼ 2 cos
it follows that
1
1
ðxn þ cÞ sin ðxn cÞ ;
2
2
This is a standard
trigonometric formula.
4.1
Continuous functions
141
1
1
jsin xn sin cj ¼ 2cos ðxn þ cÞ sin ðxn cÞ 2
2
1
2sin ðxn cÞ 2
1
2 ðxn cÞ ¼ jxn cj:
2
By taking moduli.
For cos 12 ðxn þ cÞ 1:
Here we use the Sine
Inequality, Theorem 2.
Hence, if {xn c} is null, then {sin xn sin c} is also null, by the Squeeze
Rule for null sequences. In other words, the result (3) holds.
The continuity of the cosine and tangent functions now follows from the
formulas
1
sin x
and tan x ¼
;
cos x ¼ sin x þ p
2
cos x
&
using the Combination Rules and the Composition Rule.
Problem 9 Prove that the following function is continuous on R,
stating each rule or fact about continuity that you are using
f ð xÞ ¼ x2 þ 1 þ 3 sin x2 þ 1 ; for x 2 R:
Problem 10
on R.
Prove that the function f ð xÞ ¼ sin
p
2 cos x
is continuous
The exponential function x 7! ex
We now prove that the exponential function is continuous on R. But first we
start with some inequalities that we will need to do this.
Theorem 4
The Exponential Inequalities
x
for x > 0;
(a) e > 1 þ x,
x
1
1x ;
(b) e 1
(c) 1 þ x ex 1x
;
Proof
and
ex 1 þ x,
for x 0;
for 0 x < 1;
for j xj < 1:
We prove inequalities (a) and (b) using the exponential series
x2 x3
ex ¼ 1 þ x þ þ þ ; for x 0:
2! 3!
2
3
(a) For x > 0, we have x2! > 0, x3! > 0, and so on. Hence
ex > 1 þ x; for x > 0:
It follows from this that ex 1 þ x, for x 0, since e0 ¼ 1
2
3
(b) For x 0, we have x2! x2 , x3! x3 , and so on. Hence
ex 1 þ x þ x2 þ x3 þ :
1
The series on the right is a geometric series; it converges to the sum 1x
,
for 0 x 1. Hence
1
; for 0 x < 1:
ex 1x
You met this series in
Sub-section 3.4.1.
4: Continuity
142
(c) We have just seen that these inequalities hold for 0 x < 1.
For 1 < x < 0, we have that 0 < x < 1, so that, by parts (a) and (b)
1
; for 1 < x < 0:
1 þ ðxÞ ex 1 ðxÞ
By taking reciprocals and reversing the inequalities, we may reformulate
this in the form
1
1 þ x ex ; for 1 < x < 0:
1x
&
This completes the proof of the desired result.
Remark
Notice that, when x 6¼ 0, the results of Theorem 4 hold with strict inequalities.
We are now able to prove the continuity of the exponential function.
Theorem 5
The exponential function x 7! ex, x 2 R, is continuous.
Proof To prove that the exponential function is continuous at each point
c 2 R, we need to show that:
for each sequence fxn g in R such that xn ! c; then exn ! ec :
(4)
We use the formula exn ec ¼ ec ðexn c 1Þ. If we apply Theorem 4, part (c),
with xn c in place of x, we obtain
1
1 þ ðxn cÞ exn c ; for jxn cj < 1:
1 ðx n c Þ
Thus, if {xn c} is null, then jxn cj < 1 eventually, and so exn c ! 1 as
n ! 1, by the Squeeze Rule for sequences. Hence exn ! ec , so that the desired
&
result (4) holds.
Remark
Later we shall give a rigorous definition of the general exponential function
x 7! ax, for x 2 R and a > 0, and we shall show that all exponential functions are
continuous on R.
Section 4.4.
Problem 11 Prove that the following function is continuous on R,
stating each rule or fact about continuity that you are using
2
f ð xÞ ¼ cos x5 5x2 þ 7ex :
We end this section by listing the various types of functions that we have
found to be continuous on their domains.
For example
Basic continuous functions
The following functions are continuous:
polynomials and rational functions;
modulus function;
nth root function;
trigonometric functions (sine, cosine and tangent);
the exponential function.
f ð xÞ ¼ 3x7 4x2 þ 5;
3x
f ð xÞ ¼ 4
;
x 6
f ð xÞ ¼ jxj;
1
f ð xÞ ¼ x2 ; x 0;
1
f ð xÞ ¼ x3 ; x 2 R;
f ð xÞ ¼ sin x; cos x; tan x;
f ð xÞ ¼ ex :
4.2
Properties of continuous functions
4.2
4.2.1
143
Properties of continuous functions
The Intermediate Value Theorem
In Section 4.3 we shall prove that the function f(x) ¼ x5 þ x 1, x 2 R, is a
one-function on R. From the graph of f it certainly looks as though f maps R
onto R; how can we prove this?
For example, is there a value of x for which f(x) ¼ 0? In other words,
is there a root of the equation x5 þ x 1 ¼ 0? The shape of the graph
y ¼ x5 þ x 1 certainly suggests that such a number x exists; since f(0) ¼ 1
and f(1) ¼ 1, we would expect there to be some number x in the interval (0, 1)
such that f(x) ¼ 0. However we do not have a formula for solving the equation
to find x.
Now, we introduced the concept of continuity on the grounds that it would
enable us to pin down precisely the idea that the graph of a ‘well-behaved’
function does not have ‘gaps’ or ‘jumps’, and this is the key to proving that
such a number x exists.
Theorem 1 Intermediate Value Theorem
Let f be a continuous function on [a, b], and let k be any number lying
(strictly) between f(a) and f(b). Then there exists a number c in (a, b) such
that f(c) ¼ k.
This result is illustrated below in the two possible cases:
As the graph above on the right shows, there may be more than one possible value
of c such that f(c) ¼ k. All that we claim is that there is at least one such point c.
The requirement in Theorem 1 that f be continuous at each point of [a, b] is
essential. For example, the function
81
1 x < 0,
<x;
x ¼ 0,
f ðxÞ ¼ 0;
:1
;
0
< x 1,
x
is continuous on [1, 1] except at 0 – where it is discontinuous. For this
function, f(1) ¼ 1 and f(1) ¼ 1, but there is no number c in (1, 1) such that
f ðcÞ ¼ 12 (for example).
The following example shows a typical application of the Intermediate
Value Theorem.
This is one of the main
existence theorems in
Analysis.
Note that f(a) 6¼ f(b) and
a < c < b.
4: Continuity
144
Example 1
Prove that there is a number c in (0, 1) such that c5 þ c 1 ¼ 0.
Solution Consider the function f(x) ¼ x5 þ x 1 on the interval [0, 1]. Then
f is continuous on [0, 1] since it is a basic continuous function, and f(0) ¼ 1
and f(1) ¼ 1.
Since f(0) < 0 < f(1), it then follows from the Intermediate Value Theorem
that there is a number c in (0, 1) such that f(c) ¼ 0; that is, such that
&
c5 þ c 1 ¼ 0.
If f is a function and c is a real number such that f(c) ¼ 0, then c is called a
zero of the function f. We often show that an equation has a solution by proving
that a related continuous function has a zero (by using the Intermediate Value
Theorem with k ¼ 0).
Problem 1
cos c ¼ c.
For f is a polynomial.
The function f is strictly
increasing on [0, 1], and so the
number c must be unique in
this case.
We do not consider the
possibility of complex zeros
in this book.
Prove that there is a real number c in (0, 1) such that
Problem 2 Let the function f: [0, 1] ! [0, 1] be continuous. Prove that
there is a real number c in [0, 1] such that f(c) ¼ c.
Proof of Theorem 1 We use the method of repeated bisection.
We shall assume that f(a) < f(b). If, in fact, f(a) > f(b), the proof is very
similar.
First, denote the closed interval [a, b] as [A1, B1]. Then, denote by p the
midpoint of the interval [A1, B1]. Notice that, if f(p) ¼ k, then the proof is
complete, since we can take c ¼ p.
Otherwise, we define one of the two intervals [A1, p] and [p, B1] to be
[A2, B2] in the following way
½A1 ; p; if f ( p) > 0,
½ A2 ; B2 ¼
½p; B1 ; if f ( p) < 0.
In either case, we obtain:
For example
f ð xÞ ¼ 13 þ 12 x sin p2 x :
You should at least skim this
proof on a first reading, as the
method is an important one.
p ¼ 12 ðA1 þ B1 Þ:
1. [A2, B2] [A1, B1];
2. B2 A2 ¼ 12 ðB1 A1 Þ;
3. f(A2) < k < f(B2).
We now repeat this process indefinitely often, bisecting [A2, B2] to obtain
[A3, B3], and so on. If, at any stage, we encounter a bisection point p such
that f(p) ¼ k, then the proof is complete.
Otherwise, we obtain a sequence of closed intervals {[An, Bn]} with the
following properties:
1. [Anþ1, Bnþ1] [An, Bn],
for each n 2 N;
1n1
2. Bn An ¼ 2
ðB1 A1 Þ, for each n 2 N;
3. f(An) < k < f(Bn),
for each n 2 N.
Property 1 implies that the sequence {An} is increasing and bounded above by
B1 ¼ b. Hence by the Monotone Convergence Theorem, {An} is convergent;
denote by A its limit. By the Limit Inequality Rule for sequences, we must have
that A b.
Similarly, Property 1 implies that the sequence {Bn} is decreasing and
bounded below by A1 ¼ a. Hence by the Monotone Convergence Theorem,
Theorem 3, Sub-section 2.3.3.
4.2
Properties of continuous functions
145
{Bn} is convergent; denote by B its limit. By the Limit Inequality Rule for
sequences, we must have that B a.
We may then deduce, by letting n ! 1 in Property 2 and using the
Combination Rules for sequences, that
B A ¼ lim Bn lim An ¼ lim ðBn An Þ
n!1
n!1
n!1
n1
¼ lim 12
ðB1 A1 Þ
n!1
n1
¼ ðB1 A1 Þ lim 12
¼ 0:
n!1
In other words, the sequences {An} and {Bn} both converge to a common limit.
Denote this limit by c; then we must have a c b.
Now we use the fact that f is continuous at c. It follows that
lim f ðAn Þ ¼ f ðcÞ
n!1
and
That is, c 2 [a, b].
lim f ðBn Þ ¼ f ðcÞ:
n!1
Then, by Property 3, f(An) < k, for n ¼ 1, 2, . . .; so that, by letting n ! 1, we
obtain f(c) k by the Limit Inequality Rule for sequences. Similarly, we can
deduce from the fact that k < f(Bn) that f(c) k. It follows that f(c) ¼ k.
Finally, notice that, since f(a) < k < f(b), we cannot have either c ¼ a or
&
c ¼ b; so that, in fact, a < c < b as required.
Recall that taking limits
‘flattens inequalities’.
That is, c 2 (a, b).
The method of repeated bisection is very powerful, and of wide application
in Mathematics.
Now, we saw above that the continuous
function f(x) ¼ x5 þ x 1, x 2 [0, 1],
1
1
has a zero in (0, 1). Now, since f 2 ¼ 32 þ 12 1 ¼ 15
¼ 1 > 0,
32 < 0 andf(1) 1
we can make the stronger statement that f must have a zero in 2 ; 1 , by the
Intermediate Value Theorem.
Problem 3 Use the method of repeated bisection to find an interval of
length 18 that contains a zero of the function f(x) ¼ x5 þ x 1, x 2 [0, 1].
Antipodal points
Two points on the surface of the Earth are called antipodal points if the line
between them passes through the centre of the Earth. (In this sense, antipodal
points are ‘opposite each other’.)
The following result is a rather interesting application of the Intermediate
Value Theorem! It makes the (physically reasonable) assumption that temperature is a continuous function of position on the Earth’s surface.
Theorem 2
Antipodal Points Theorem
There is always a pair of antipodal points on the Equator of the Earth at
which the temperature is the same.
To prove this, we must set up the situation as a mathematical problem.
Let f() denote the temperature at a point on the Equator at an angle radians
East of Greenwich, for 0 < 2p, and extend f to be defined on [0, 2p] by
In fact the same result holds
for any ‘Great Circle’ on the
Earth’s surface – that is, the
intersection of a plane through
the Earth’s centre with the
Earth’s surface.
This process is often called
Mathematical modelling.
4: Continuity
146
requiring that f(2p) ¼ f(0). Then Theorem 2 can be rephrased in the following
equivalent way:
Theorem 20 Antipodal Points Theorem
Let f : [0, 2p] ! R be continuous, with f(0) ¼ f(2p). Then there exists a
number c in [0, p] such that f(c) ¼ f(c þ p).
Proof of Theorem 20 Notice, first, that if f(0) ¼ f(p) then we can take c ¼ 0.
We now prove the result under the assumption that f(0) < f(p). (The proof in
the case that f(0) > f(p) is very similar.)
Next, define the function g as follows
gðÞ ¼ f ðÞ f ð þ pÞ;
for 2 ½0; p:
Then, since f is continuous on [0, 2p], it follows that g is continuous on [0, p],
by the Combination Rules.
But
gð0Þ ¼ f ð0Þ f ðpÞ < 0
For, if f(c) ¼ f(c þ p), then c
and c þ p are antipodal points
with the same temperature.
Quite often in mathematics
we obtain our results by
applying a standard result to
some cunningly chosen
auxiliary function!
and
gðpÞ ¼ f ðpÞ f ð2pÞ ¼ f ðpÞ f ð0Þ > 0:
It then follows from the Intermediate Value Theorem, with k ¼ 0, that there
exists some number c in (0, p) such that g(c) ¼ 0; in other words, such that
&
f(c) f(c þ p) ¼ 0. This completes the proof.
4.2.2
Zeros of polynomials
You will have already met a standard method for solving a polynomial
equation of degree 2, and possibly equations of degrees 3 and 4 too. However there exists no method of solving a general polynomial equation of
degree 5 or higher by means of formulas. How many zeros can a polynomial
equation have?
In fact, a polynomial equation of degree n can have at most n roots in R.
Theorem 3 Fundamental Theorem of Algebra
Let p(x) ¼ anxn þ an1xn1 þ þ a1x þ a0, x 2 R, where an 6¼ 0. Then
the equation p(x) ¼ 0 has at most n roots in R.
Remark
In fact, in its most general form the Fundamental Theorem of Algebra states
that, if p(x) ¼ anxn þ an1xn1 þ þ a1x þ a0, where the coefficients ak may
These are called quadratic,
cubic and quartic equations,
respectively.
This is a quite difficult result
to prove!
We do not prove this result,
which requires methods
beyond the scope of this book.
4.2
Properties of continuous functions
147
be real or complex numbers and an 6¼ 0, then the equation p(x) ¼ 0 has exactly n
roots in C, the set of all complex numbers.
Now, it is sometimes straight-forward to locate zeros of a polynomial if we
have some idea of where to look for them in the first place.
Problem 4 Let pð xÞ ¼ x6 4x4 þ x þ 1, x 2 R. Prove that p has a
zero in each of the intervals (1, 0), (0, 1) and (1, 2).
However to follow this approach for finding zeros (if any) of a given
polynomial, we need first to have an inkling where to look for them. We
usually start by applying the following result, which gives an interval in
which the zeros must lie:
Theorem 4
Zeros Localisation Theorem
Let pð xÞ ¼ x þ an1 x
þ þ a1 x þ a0 ; x 2 R, be a polynomial. Then
all the zeros of p (if there are any) lie in the open interval ( M, M), where
n
n1
M ¼ 1 þ maxfjan1 j; . . .; ja1 j; ja0 jg:
Notice that the coefficient of
xn, the leading coefficient of p,
has been set as 1.
The proof appears at the end
of the sub-section.
Example 2 Prove that the polynomial pð xÞ ¼ x4 2x2 x þ 1, x 2 R, has
at least two zeros in R.
Solution We will apply the Zeros Localisation Theorem to p, since its
leading coefficient is 1. Since
M ¼ 1 þ maxfj2j; j1j; j1jg
¼ 3;
it follows that all the zeros of p lie in (3, 3).
We now compile a table of values of p(x), for x ¼ 3, 2, 1, 0, 1, 2, 3:
x
p(x)
3
67
2
1
0
1
2
3
11
1
1
1
7
61
We find that p(0) and p(1) have opposite signs, as do p(1) and p(2), so p
must have a zero in each of the intervals (0, 1) and (1, 2), by the Intermediate
Value Theorem.
&
Thus we have proved that p has at least two zeros in R.
Often, it is not necessary to
compute the values of p(x) for
all integers x in [M, M].
In fact this polynomial has
exactly two zeros in R.
Problem 5 Prove that the polynomial pð xÞ ¼ x5 þ 3x4 x 1, x 2 R,
has at least three zeros in R.
Although we cannot at this stage prove the Fundamental Theorem of
Algebra, nevertheless we can use the Zeros Localisation Theorem and the
Intermediate Value Theorem to prove the following:
Theorem 5
Every real polynomial of odd degree has at least one zero in R.
Thus, for example, for sufficiently large x the polynomial pð xÞ ¼
xn þ an1 xn1 þ þ a1 x þ a0 , x 2 R is essentially dominated by its leading
term xn, and so is positive for large positive values of x and is negative for large
negative values of x.
For example,
pð xÞ ¼ x5 þ 3x4 x 1:
Since n is odd.
4: Continuity
148
No corresponding result holds for polynomials of even degree. Thus, for
instance, the polynomial pð xÞ ¼ x2 þ 1 has no zeros in R.
Proofs of Theorems 4 and 5
We supply these proofs which were omitted earlier, so as not to disturb the flow
of the text.
Theorem 4 Zeros Localisation Theorem
Let pð xÞ ¼ xn þ an1 xn1 þ þ a1 x þ a0 ; x 2 R, be a polynomial. Then
all the zeros of p (if there are any) lie in the open interval (M, M), where
M ¼ 1 þ maxfjan1 j; . . .; ja1 j; ja0 jg:
Proof In order to concentrate on the dominating term xn, we define the
function r as follows
pð x Þ
an1
a1
a0
r ð xÞ ¼ n 1 ¼
þ þ n1 þ n ; x 2 R f0g:
x
x
x
x
Then, by using the Triangle Inequality, we obtain that, for jxj > 1
a
a1
a0 n1
þ þ n1 þ n jr ð xÞj ¼ x
x a x a1 a0 n1 þ þ n1 þ n x
x
x
!
1
1
1
þ þ n1 þ n
maxfjan1 j; . . .; ja1 j; ja0 jg j xj
j xj
j xj
!
1
1
1
<M
þ þ n1 þ n þ j xj
j xj
j xj
¼M
1
j xj
1 j1xj
¼
M
:
j xj 1
You may omit these at a first
reading.
Recall that the coefficient of
xn, the leading coefficient
of p, is 1.
Here the modulus of each
coefficient is at most the
maximum value over all
coefficients.
jxj > 1, so that j1xj < 1:
By summing the geometric
series.
It follows that, if j xj M ¼ 1 þ maxfjan1 j; . . .; ja1 j; ja0 jg, then jr ð xÞj < 1:
Now, from the definition of r(x) we see that
pð xÞ ¼ xn ð1 þ r ð xÞÞ;
for j xj M:
But since jr ð xÞj < 1, we certainly have that 1 þ r ð xÞ > 0. It follows from the
above expression for p(x) in terms of r(x) that p(x) must have the same sign as
xn, for j xj M:
&
It follows that any zero of p must lie in (M, M).
Theorem 5
jr ð xÞj51 , 15r ð xÞ51:
Every polynomial of odd degree has at least one zero in R.
Proof By dividing the polynomial by its leading coefficient, we may assume
that the polynomial is of the form
pð xÞ ¼ xn þ an1 xn1 þ þ a1 x þ a0 ; x 2 R, where n is odd.
We can then define M and r(x) as in Theorem 4, and all the statements in the
proof of Theorem 4 then hold here too.
Since n is odd, xn is positive for x M and negative for x M. It follows
from the arguments in the proof of Theorem 4 that p(x) must be positive for
x M and negative for x M. In particular, p(M) > 0 and p(M) < 0.
Then, by the Intermediate Value Theorem, p must have a zero in (M, M). &
So we have certainly got good
value from the arguments in
the previous proof!
4.2
Properties of continuous functions
4.2.3
149
The Extreme Values Theorem
We now look at whether continuous functions on closed intervals are bounded
or unbounded. It turns out that they are bounded, and we use this fact a great
deal throughout Analysis!
Theorem 6
The Extreme Values Theorem
A continuous function on a closed interval possesses a maximum value and
a minimum value on the interval. In other words, if f is a function that is
continuous on a closed interval [a, b], then there exist points c and d in [a, b]
such that
f ðcÞ f ð xÞ f ðdÞ;
We discussed bounds,
suprema, infima, maxima and
minima of functions earlier, in
Sub-section 1.4.2.
We prove this at the end of the
sub-section.
for x 2 ½a; b:
If f is continuous on an interval that is not closed, then it may not be
bounded. For example, the function f ð xÞ ¼ 1x, x 2 ð0; 1, is continuous on (0,1]
but is not bounded above (and so it is not bounded), but it is bounded below
(by 1).
Similarly, if a function is not continuous on a closed interval, then it may not
be bounded. For example, the function
1; x ¼ 0;
f ðxÞ ¼ 1
0 < x 1;
x;
takes the values 1 when x ¼ 0 and 1x when 0 < x 1. f is bounded below, by 0,
but is not bounded above (and so it is not bounded).
Problem 6
(a) Determine the maximum and the minimum of the function
f ð xÞ ¼ x2 , x 2 ½1; 2, on [1, 2]. Specify all points in [1, 2]
where these are attained.
(b) Determine the maximum and the minimum of the function
gð xÞ ¼ sin x, x 2 ½0, 2p, on [0, 2p]. Specify all points in [0, 2p]
where these are attained.
Since a function is bounded if and only it is both bounded above and
bounded below, we sometimes use the following version of Theorem 6 in
applications:
Corollary 1
The Boundedness Theorem
A continuous function on a closed interval is bounded. In other words, if f is
a function that is continuous on the closed interval [a, b], then there exist a
number M such that
jf ð xÞj M; for x 2 ½a; b:
On other occasions, we shall find the following consequence of Theorem 6
and the Intermediate Value Theorem useful:
Corollary 2
The Interval Image Theorem
The image of a closed interval under a continuous function is a closed
interval.
For it will often be sufficient
for our purposes.
4: Continuity
150
Proof of Theorem 6 Without some systematic approach to studying continuous functions, we would never be able to prove this theorem, and would be
reduced to much hand-waving arguments and loose assertions. In this case we
shall use our earlier work on sequences, and the following related result:
Lemma 1
Let f be a function defined on an interval I.
(a) If f is bounded above on I and supff ð xÞ: x 2 I g ¼ M, then there exists
some sequence {xn} in I such that f ðxn Þ ! M as n ! 1.
(b) If f is not bounded above on I, then there exists some sequence {xn} in I
such that f ðxn Þ ! 1 as n ! 1.
Proof
(a) If there is some point, c say, in I for which f(c) ¼ M, then the constant
sequence {c} has the desired property.
Now suppose that no such point c exists. Then, since supff ð xÞ:
x 2 Ig ¼ M, it follows from the definition of supremum that there is
some point, x1 say, in I for which
Notice that this is a result
about sup for any function f on
an interval; no assumption of
continuity is involved.
A similar result holds for
functions that are bounded
below on I or are unbounded
below on I.
This is a proof by
contradiction.
f ðx1 Þ > M 1:
Next, since f(x1) < M (from our assumption that there is no point where
f takes the value M), choose a point, x2 say, in I for which
1
f ðx2 Þ > max f ðx1 Þ; M ;
2
notice, in particular, that this last inequality ensures that x2 6¼ x1.
Continuing this process indefinitely, we obtain a sequence {xn} of
distinct points in I for which
1
f ðxn Þ > max f ðx1 Þ; f ðx2 Þ; . . .; f ðxn1 Þ; M :
n
1
Since the sequence M n converges to M, it follows, by the Squeeze
Rule for sequences, that f ðxn Þ ! M as n ! 1.
(b) It follows from the definition of ‘unbounded above’ that there is some
point, x1 say, in I for which f(x1) > 1. Then, for each n > 1, we can
construct a sequence {xn} of distinct points in I for which
f ðxn Þ > maxff ðx1 Þ; f ðx2 Þ; . . .; f ðxn1 Þ; ng:
Since the sequence {n} tends to 1, it follows, by the Squeeze Rule for
&
sequences, that f(xn) ! 1 as n ! 1.
We are now in a position to prove Theorem 6.
Theorem 6 The Extreme Values Theorem
A continuous function on a closed interval possesses a maximum value and
a minimum value on the interval. In other words, if f is a function that is
continuous on the closed interval [a, b], then there exist points c and d in
[a, b] such that
f ðcÞ f ð xÞ f ðd Þ; for x 2 ½a; b:
So f ðx2 Þ > f ðx1 Þ and
f ðx2 Þ > M 12. Also,
f ðx2 Þ < M:
For M 1n < f ðxn Þ < M:
4.3
Inverse functions
151
Proof First, we shall assume that f is not bounded above, and verify that this
assumption leads to a contradiction with known facts about f.
It follows from part (b) of Lemma 1 that there exists some sequence {xn} in
[a, b] such that f(xn) ! 1 as n ! 1. Then, by the Bolzano–Weierstrass
Theorem, {xn} must contain a convergent subsequence fxnk g; denote by d
the limit of fxnk g. Since all the xnk lie in [a, b], it follows, by the Limit
Inequality Rule for sequences, that d 2 [a, b].
Since f is continuous at d and the sequence fxnk g converges to d, it follows
that f ðxnk Þ ! f ðdÞ. However, ff ðxnk Þg is a subsequence of the sequence {f(xn)}
which tends to 1, so that f ðxnk Þ ! 1:
This is a contradiction. So f must be bounded above on [a, b] after all.
Next, denote by M the number supff ð xÞ: x 2 ½a; bg. It follows, from part (a)
of Lemma 1, that there exists some sequence {xn} in [a, b] such that f(xn) ! M as
n ! 1. Then, by the Bolzano–Weierstrass Theorem, {xn} must contain a convergent subsequence fxnk g; denote by d the limit of fxnk g. Since all the xnk lie in
[a, b], it follows, by the Limit Inequality Rule for sequences, that d 2 [a, b].
Since f is continuous at d and the sequence fxnk g converges to d, it follows
that f ðxnk Þ ! f ðdÞ. Therefore, since ff ðxnk Þg is a subsequence of the sequence
{f(xn)} which tends to M, we have M ¼ f(d).
This complete the proof that there exists a point d in [a, b] such that
f ð xÞ f ðd Þ; for all x 2 ½a; b:
The proof of the existence of a point c in [a, b] such that f(c) f(x), for all
&
x 2 [a, b], is similar; we omit it.
4.3
4.3.1
Inverse functions
Existence of an inverse function
Let f be the function f(x) ¼ 2x, x 2 R. Then, given any number y in R, we can
find a unique number x ¼ 12 y in the domain of f such that y ¼ f(x) ¼ 2x.
The inverse function f 1, defined by f 1 ð yÞ ¼ 12 y; y 2 R, undoes the
‘effect’ of f; that is, f 1( f(x)) ¼ x, x 2 R.
Also, f undoes the effect of f 1; that is, f ( f 1(y)) ¼ y, y 2 R.
Not every function has an inverse function! For example, consider the
function
gð xÞ ¼ x2 ;
x 2 R:
Since g(2) ¼ 4 ¼ g(2), we cannot define g1(4) uniquely. Thus g fails to
have an inverse function, because it is not one–one. However the function
hð xÞ ¼ x2 ; x 2 ½0; 1Þ;
is one–one and has an inverse function
pffiffiffi
h1 ð yÞ ¼ y; y 2 ½0; 1Þ:
In general, if f : A ! R is one–one, then, for each point y in the image f(A),
there is a unique point x in A such that f(x) ¼ y. Thus f is a one–one correspondence between A and f(A); and so we can define the inverse function f 1 by
f 1( y) ¼ x, where y ¼ f(x).
It is also possible to prove
Theorem 6 by the method of
repeated bisection.
Theorem 3, Sub-section 2.5.1.
xnk 2 ½a; b , a xnk b:
Inheritance Property of
Subsequences, Theorem 5,
Sub-section 2.4.4.
M must exist, since we have
shown that the set
ff ð xÞ: x 2 ½a; bg is bounded.
4: Continuity
152
Definition Let f : A ! R be a one–one function. Then the inverse function
f 1 has domain f(A) and is specified by
f 1 ð yÞ ¼ x;
where y ¼ f ð xÞ; x 2 A:
For some functions f, we can find the inverse function f 1 directly, by
solving the equation y ¼ f(x) algebraically to obtain x in terms of y.
1
Example 1 Prove that the function f defined by f ð xÞ ¼ 1x
; x 2 ð1; 1Þ,
has an inverse function defined on (0, 1).
1
Solution First, we solve the equation y ¼ 1x
to give x in terms of y. By
simple manipulation, we obtain
1
1
,x¼1 :
y¼
1x
y
1
Now, for each x 2 (1, 1), we have x < 1, and so f ð xÞ ¼ 1x
> 0; thus
f ðð1,1ÞÞ ð0,1Þ. Also, for each y 2 (0, 1), we have
1
x ¼ 1 2 ð1; 1Þ;
y
and so f is a one–one correspondence between (1, 1) and (0, 1). Hence
1
f 1 ð yÞ ¼ 1 ; y 2 ð0; 1Þ:
&
y
Remark
Usually, when defining a function we write x for the domain variable. To
conform with this practice, we may rewrite the inverse function f 1 in
Example 1 as follows
1
f 1 ð xÞ ¼ 1 ; x 2 ð0; 1Þ:
x
The graph y ¼ f 1 (x) is obtained by reflecting the graph y ¼ f(x) in the line
y ¼ x. This reflection interchanges the x- and y-axes.
Proving that a function f is one–one
We have seen that, if f: A ! R is one–one, then f has an inverse function f 1
with domain f(A). For the function f considered in Example 1, it is possible to
determine f 1 explicitly by solving the equation y ¼ f (x) to obtain x in terms
of y. Unfortunately, it is generally not possible to solve the equation y ¼ f (x) in
this way.
Nevertheless, it may still be possible to prove that f has an inverse function
f 1 by showing that f is one–one in some other way. For example, f is one–one
if it is either strictly increasing or strictly decreasing; that is, if f is strictly
monotonic.
A function f defined on an interval I is:
increasing on I if
x1 5x2 ) f ðx1 Þ f ðx2 Þ; for x1 ; x2 2 I;
strictly increasing on I if x1 5x2 ) f ðx1 Þ5 f ðx2 Þ; for x1 ; x2 2 I;
decreasing on I if
x1 5x2 ) f ðx1 Þ f ðx2 Þ; for x1 ; x2 2 I;
Definitions
y
1
1–x
1
(– ∞, 1)
0
y=
1
x
4.3
Inverse functions
153
x1 5x2 ) f ðx1 Þ > f ðx2 Þ; for x1 ; x2 2 I;
strictly decreasing on I if
monotonic on I if
f is either increasing on I or decreasing on I;
strictly monotonic on I if
f is either strictly increasing on I or strictly
decreasing on I.
The most powerful technique for proving that a function f is strictly monotonic is to compute the derivative f 0 of f and examine the sign of f 0 (x). We shall
not study differentiation in detail until Chapter 6, so here we consider only
functions which can be proved to be strictly monotonic by manipulating
inequalities rather than by Calculus.
For example, if n 2 N, then the function f(x) ¼ xn, x 2 [0, 1), is strictly
increasing; and, if n is odd, the function f(x) ¼ xn, x 2 R, is strictly increasing.
Similarly, if n 2 N, then the function f(x) ¼ xn, x 2 (0, 1), is strictly
decreasing.
Prove that the function f ð xÞ ¼ x5 þ x 1; x 2 R; is one–one.
Example 2
Solution
If x1 < x2, then x51 < x52 . Hence
x51 þ x1 1 < x52 þ x2 1;
so f is strictly increasing, and thus one–one.
Problem 1
Prove that the following functions are one–one:
(a) f ð xÞ ¼ x4 þ 2x þ 3;
2
(b) f ð xÞ ¼ x 4.3.2
&
1
x;
x 2 ½0; 1Þ;
x 2 ð0; 1Þ:
The Inverse Function Rule
If the function f: A ! R is strictly monotonic, then f is one–one, and so f has
an inverse function f 1 with domain f(A). However, it is not always easy to
determine f(A). However, if f is known to be continuous on A, then the
following result simplifies the problem immediately.
Theorem 1
Inverse Function Rule
Let f : I ! J, where I is an interval and J is the image f(I), be a function such that:
1. f is strictly increasing on I;
2. f is continuous on I.
Then J is an interval, and f has an inverse function f 1: J ! I such that:
10 . f 1 is strictly increasing on J;
20 . f 1 is continuous on J.
Remarks
1. The interval I may be any type of interval: open or closed, half-open,
bounded or unbounded.
2. There is a similar version of the Inverse Function Rule with ‘strictly
increasing’ replaced by ‘strictly decreasing’.
We prove the Inverse
Function Rule in
Sub-section 4.3.4.
4: Continuity
154
Example 3 Prove that the function f ð xÞ ¼ x5 þ x 1; x 2 R; has a continuous inverse function, with domain R.
Solution The domain of f is R, which is an interval.
We have already seen, in Example 2, that f is strictly increasing on R and so
has an inverse function f1. Since f is a polynomial, it is a basic continuous
function on R. Thus f satisfies the hypotheses of the Inverse Function Rule.
Then it follows from the Rule that the image J ¼ f(R) is an interval, and that
the inverse function f1: J ! R is strictly increasing and continuous on J.
It remains to check that the image J is the whole of R. Notice that J ¼ f(R)
contains each of the numbers
f ðnÞ ¼ n5 þ n 1;
n 2 Z:
Now, J contains each of the intervals [f (n), f (n)]; and f ðnÞ ¼ n5 n 1 ! 1 as n ! 1, while f ðnÞ ¼ n5 þ n 1 ! 1 as n ! 1. It follows
that, in fact, J ¼ f (R) must be (1, 1) ¼ R.
&
Thus f has a continuous inverse function f 1: R ! R.
As the above example shows, when we apply the Inverse Function Rule the
hardest step is to determine the image J ¼ f (I). Since J is an interval, it is
sufficient to determine the end-points of J, which may be real numbers or one
of the symbols 1 and 1. We must also determine whether or not these endpoints belong to J.
The following diagrams illustrate two examples:
y
y
d
d = f (b)
y = f (x)
J = [c, d )
J = (c, d ]
ð0; 1 has end-points 0
and 1;
½1; 1Þ has end-points 1
and 1:
Do not let this use of the
symbol 1 tempt you to think
that 1 is a real number.
y
y = f (x)
c
c = f (a)
For example:
d
y = f (x)
J = [c, d )
a
I = [a, ∞)
x
a
b
x
I = (a, b]
Notice that, if a is an end-point of I and a 2 I, then c ¼ f(a) is the corresponding end-point of J and c 2 J.
On the other hand, if a is an end-point of I and a 2
= I (this includes the
possibility that a may be 1 or 1), then it is a little harder to find the
corresponding end-point of J. However, it can be shown that, if {xn} is a
monotonic sequence in I and xn ! a, then f(xn) ! c, and c 2
= J: (We prove this
in Sub-section 4.3.4.)
Example 4 Prove that the function f ð xÞ ¼ x4 þ 2x þ 3; x 2 ½0; 1Þ, has a
continuous inverse function with domain [3, 1).
Solution The domain of f is [0, 1), which is an interval.
Also, we know that f is strictly increasing and continuous on [0, 1), and so
conditions 1 and 2 of the Inverse Function Rule hold. It follows that the
image J ¼ f([0, 1)) is an interval, and f has a continuous inverse function f 1:
J ! [0, 1) which is strictly increasing on J.
It remains to check that the image J is [3, 1).
For the end-point 0 of I, we have 0 2 I, so the corresponding end-point of J is
f(0) ¼ 3, and 3 2 J.
c = f (a)
c∈J
a
a∈I
x
I = [a, ∞)
y
d = f (b)
J = (c, d ]
{ f (an)}
y = f (x)
c
c∉J
{an}
a
a∉I
I = (a, b]
b
x
You saw this in Problem 1(a),
above.
4.3
Inverse functions
155
The other end-point of I is 1, so to find the corresponding end-point of J we
choose the monotonic sequence {n}, which lies in I and tends to infinity. Then
f ðnÞ ¼ n4 þ 2n þ 3 ! 1 as n ! 1. It follows that the corresponding end&
point of J is 1. Thus, J ¼ [3, 1), as required.
We now summarise the strategy for establishing that a given continuous
function f has a continuous inverse function.
Strategy To prove that f : I ! J, where I is an interval with end-points a
and b, has a continuous inverse f 1: J ! I:
1. show that f is strictly increasing on I;
2. show that f is continuous on I;
3. determine the end-point c of J corresponding to the end-point a of I as
follows:
if a 2 I, then f(a) ¼ c (and c 2 J);
if a 2
= I, then f(xn) ! c (and c 2
= J),
where {xn} is a monotonic sequence in I such that xn ! a;
4. determine the end-point d of J corresponding to the end-point b of I,
similarly.
Problem 2 Use the above strategy to prove that the function
f ð xÞ ¼ x2 1x ; x 2 ð0; 1Þ, has a continuous inverse function with
domain R.
Hint: Use the result of Problem 1(b) in Sub-section 4.3.1.
4.3.3
Inverses of standard functions
We now use the Inverse Function Rule and the above strategy to define
continuous inverse functions for certain standard functions. Although you
will be familiar with these inverse functions already, we can now prove that
they exist and are continuous. We also remind you of some properties of these
inverse functions.
For each function, we give brief remarks on the four steps of the strategy. In
each case, the continuity of the function f follows directly from the results of
Section 4.1.
The nth root function
We asserted the existence of the nth root function in Sub-section 1.5.2,
Theorem 1. We can at last provide the proof of that assertion!
The nth root function For any positive integer n 2, the function
f ð xÞ ¼ xn ; x 2 ½0; 1Þ;
pffiffiffi
has a strictly increasing continuous inverse function f 1 ð xÞ ¼ n x with
domain ½0; 1Þ, called the nth root function.
There is a corresponding
version of this strategy for the
case that f is decreasing on I.
4: Continuity
156
In this case, the strategy is easy to apply:
1. f is strictly increasing on [0, 1);
2. f is continuous on [0, 1);
3. f(0) ¼ 0;
4. f(k) ¼ kn ! 1 as k ! 1.
It follows, from the Inverse Function Rule, that f has a strictly increasing
continuous inverse function f 1: [0, 1) ! [0, 1).
Remark
If n is odd, then the nth root function can be extended to a continuous function
whose domain is the whole of R.
Inverse trigonometric functions
The function sin1
The function
1 1
f ð xÞ ¼ sin x; x 2 p; p ;
2 2
has a strictly increasing continuous inverse function with domain [1, 1],
called sin1.
In this case:
1. the geometric
definition of f(x) ¼ sin x shows that f is strictly increasing on
12 p; 12 p ;
2. f is continuous on 12 p; 12 p ;
3. sin 12 p ¼ 1;
4. sin 12 p ¼ 1.
It follows that f 12 p; 12 p ¼ ½1; 1. Hence, by the Inverse Function
Rule, f has
increasing continuous inverse function f 1:
1 a 1strictly
½1; 1 ! 2 p; 2 p :
The decreasing version of the strategy can be applied similarly to prove that
the cosine function has an inverse, if we restrict its domain suitably.
The function cos1 The function
f ð xÞ ¼ cos x; x 2 ½0; p;
has a strictly decreasing continuous inverse function with domain [1, 1],
called cos1.
The domain [0, p] of f is chosen here by convention, so that f is a strictly
monotonic restriction of the cosine function.
Similarly, to form an
inverse of the tangent function we must restrict its
domain to 12 p; 12 p , since the tangent function is strictly increasing and
continuous on this interval.
f is a basic continuous
function.
We use {k} rather than {n}
here, to avoid using n for two
different purposes in the same
expression.
4.3
Inverse functions
157
The function tan1
The function
1 1
f ð xÞ ¼ tan x; x 2 p; p ;
2 2
has a strictly increasing continuous inverse function with domain R, called
tan1.
In this case, the image set f 12 p; 12p is R, because (for example) if {xn}
is a monotonic sequence in 12 p; 12 p and xn ! 12 p as n ! 1, then
sin xn
f ðxn Þ ¼ tan xn ¼
! 1 as n ! 1:
cos xn
Remark
Some texts use arc sin, arc cos and arc tan instead of sin1, cos1 and tan1,
respectively.
Problem 3
pffiffiffi
(a) Determine the values of sin1 ðp1ffiffi2Þ, cos1 12 and tan1 3 .
(b) Prove that cos 2 sin1 x ¼ 1 2x2 , for x 2 ½1; 1.
Hint:
Let y ¼ sin1 x:
The function loge
We now discuss one of the most important inverse functions.
The function loge The function
f ð xÞ ¼ ex ; x 2 R;
has a strictly increasing continuous inverse function f 1 with domain
(0, 1), called loge.
In this case:
1. f is strictly increasing on R, since
x1 < x2 ) x2 x1 > 0
) ex2 x1 > 1
) e x2 > e x1 ;
ðsince ex 1 þ x > 1; for x > 0Þ
2. f is continuous on R;
3. f ðnÞ ¼ en ! 1 as n ! 1;
4. f ðnÞ ¼ en ! 0 as n ! 1:
It follows that the image of R under f is f ðR Þ ¼ ð0; 1Þ. Hence, by the
Inverse Function Rule, f has a strictly increasing continuous inverse function
f 1: ð0; 1Þ ! R.
Problem 4 Prove that loge x þ loge y ¼ loge ðxyÞ, for x; y 2 ð0; 1Þ.
Hint: Let a ¼ loge x and b ¼ loge y.
4: Continuity
158
Inverse hyperbolic functions
The function sinh1
The function
1
f ð xÞ ¼ sinh x ¼ ðex ex Þ; x 2 R;
2
has a strictly increasing continuous inverse function f 1 with domain R,
called sinh1.
In this case:
1. f is strictly increasing on R, since
x1 < x2 ) ex1 <ex2
) ex1 < ex2
) ex1 ex1 < ex2 ex2
) sinh x1 < sinh x2 ;
2. f is continuous on R, by the Combination Rules;
3. f ðnÞ ¼ 12 ðen en Þ ! 1 as n ! 1;
4. f ðnÞ ¼ 12 ðen en Þ ! 1 as n ! 1.
It follows that the image of R under f is f (R) ¼ R. Hence, by the Inverse
Function Rule, f has a strictly increasing continuous inverse function
f 1: R ! R.
The function cosh1
The function
1
f ð xÞ ¼ cosh x ¼ ðex þ ex Þ; x 2 ½0; 1Þ;
2
has a strictly increasing continuous inverse function f 1 with domain
[1, 1), called cosh1.
In this case:
1. f is strictly increasing on [0, 1), since
x1 < x2 ) sinh x1 < sinh x2
1 1
) 1 þ sinh2 x1 2 < 1 þ sinh2 x2 2
) cosh x1 < cosh x2 ; since cosh2 x ¼ 1 þ sinh2 x;
2. f is continuous on [0, 1), by the Combination Rules;
3. f ð0Þ ¼ 1;
4. f ðnÞ ¼ 12 ðen þ en Þ ! 1 as n ! 1.
It follows that the image of [0, 1) under f is f ([0, 1)) ¼ [1, 1). Hence, by
the Inverse Function Rule, f has a strictly increasing continuous inverse function f 1: [1, 1) ! [0, 1).
The strategy can be applied in a similar way to show that f(x) ¼ tanh x is
strictly increasing and continuous on R, with f (R) ¼ (1, 1). We omit the
details.
4.3
Inverse functions
159
The function tanh1
The function
sinh x
; x 2 R;
f ð xÞ ¼ tanh x ¼
cosh x
has a strictly increasing continuous inverse function f1 with domain
(1, 1), called tanh1.
The inverse hyperbolic functions can be expressed in terms of loge, as the
following example shows.
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Example 5 Prove that sinh1 x ¼ loge x þ x2 þ 1 , for x 2 R.
Solution
Let y ¼ sinh1 x, for x 2 R. Then
1
x ¼ sinh y ¼ ðey ey Þ;
2
and so
e2y 2xey 1 ¼ 0:
This is a quadratic equation in ey, so that
pffiffiffiffiffiffiffiffiffiffiffiffiffi
e y ¼ x x 2 þ 1:
Since ey > 0, we must choose the þ sign here, so that
pffiffiffiffiffiffiffiffiffiffiffiffiffi
y ¼ loge x þ x2 þ 1 :
Problem 5
4.3.4
&
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Prove that cosh1 x ¼ loge x þ x2 1 , for x 2 ½1; 1Þ.
Proof of the Inverse Function Rule
We now prove the Inverse Function Rule, and justify our strategy for determining the domains of inverse functions.
Theorem 1 Inverse Function Rule
Let f: I ! J, where I is an interval and J is the image f (I), be a function
such that:
1. f is strictly increasing on I;
2. f is continuous on I.
Then J is an interval, and f has an inverse function f 1: J ! I such that:
10 . f 1 is strictly increasing on J;
20 . f 1 is continuous on J.
Proof The proof is in four parts:
J ¼ f(I) is an interval.
Let y1, y2 2 f(I), with y1 < y2, and let y 2 (y1, y2). Now y1 ¼ f (x1) and
y2 ¼ f(x2), for some x1, x2 2 I, with x1 < x2 since f is strictly increasing on I.
It follows, by the Intermediate Value Theorem, that there is some number
x 2 (x1, x2) for which f(x) ¼ y. Hence y 2 f(I).
It follows that f(I) is an interval.
You may omit the proofs in
this sub-section at a first
reading, but you should at
least read the statement of
Theorem 2 below.
4: Continuity
160
The inverse function f 1: J ! I exists.
The function f is strictly increasing and is therefore one–one; also, f maps
I onto J, so the function f 1: J ! I exists, by definition.
f 1 is strictly increasing on J.
We have to show that
y1 < y2 ) f 1 ðy1 Þ < f 1 ðy2 Þ;
for y1 ; y2 2 J:
Notice that
f 1 ðy1 Þ f 1 ðy2 Þ ) f f 1 ðy1 Þ f f 1 ðy2 Þ
) y1 y2 :
Hence y1 < y2 ) f 1 ðy1 Þ < f 1 ðy2 Þ, as required.
f 1 is continuous on J.
Let y 2 J, and (for simplicity) assume that y is not an end-point of J. Then
y ¼ f(x), for some x 2 I, and we want to prove that
yn ! y ) f 1 ðyn Þ ! f 1 ð yÞ ¼ x:
Thus, we want to deduce that:
for each " > 0, there is some number X such that
x " < f 1 ðyn Þ < x þ "; for all n > X:
Since f is strictly increasing, we know that
f ðx "Þ < f ð xÞ < f ðx þ "Þ;
(1)
also, since yn ! y ¼ f(x), there is some number X such that
f ðx "Þ < yn < f ðx þ "Þ;
for all n > X:
If we then apply the strictly increasing function f 1 to these inequalities, we
obtain (1), as required.
This completes the proof of the Inverse Function Rule.
&
In fact, a careful check of the first part of the above proof shows that a slight
change enables us to prove the following result that is of interest in its own right.
Theorem 2
an interval.
The image of an interval under a continuous function is also
Proof Let f be a continuous function on an interval I. We shall assume that f
is non-constant on I, since otherwise the result is trivial.
Let y1, y2 2 f(I), with y1 < y2, and let y 2 (y1, y2). Now y1 ¼ f(x1) and
y2 ¼ f(x2), for some x1, x2 2 I, and x1 6¼ x2. Let I0 denote the interval with endpoints x1 and x2. It follows, by applying the Intermediate Value Theorem to the
function f on I0 , that there is some number x 2 I0 for which f(x) ¼ y. Hence
y 2 f(I).
&
It follows that f(I) is an interval.
Finally, we justify our strategy for determining the end-points of J ¼ f(I).
In Corollary 2,
Sub-section 4.2.3, we showed
that the image of a closed
interval under a continuous
function is a closed interval.
We must introduce the
symbol I0 since we do not
know in general whether
x1 < x2 or x2 < x1.
4.4
Defining exponential functions
161
Strategy for finding the end-points of J If a is an end-point of the
interval I, then we can find the corresponding end-point c of J as follows:
if a 2 I, then f (a) ¼ c (and c 2 J);
if a 2
= I, then f(xn) ! c (and c 2
= J),
where {xn} is a monotonic sequence in I such that xn ! a.
Proof
Here we consider only the
end-point a of I; a similar
result holds for the other
end-point b of I.
For simplicity, we shall suppose that a is the left end-point of I.
y
y
d
d = f (b)
J = [c, d )
J = (c, d ]
y = f (x)
{ f(xn)}
y = f (x)
c = f (a)
c∉J
c∈J
a
I = [a, ∞)
{xn}
x
a
a∉I
a∈I
I = (a, b]
b
x
If a 2 I, then c ¼ f (a) 2 J and
f ð xÞ f ðaÞ ¼ c;
for x 2 I;
and so c is the corresponding left end-point of J.
On the other hand, if a 2
= I, then we select any decreasing sequence {xn} in I
such that xn ! a as n ! 1. Then {f(xn)} is also decreasing; it therefore
follows, by the Monotonic Sequence Theorem for sequences, that
f ðxn Þ ! c as n ! 1;
(2)
where c is a real number or 1.
Now, f(xn) 2 J for n ¼ 1, 2, . . .; therefore, since J is an interval, it follows that
½f ðxn Þ; f ðx1 Þ J;
This was Theorem 2 in Subsection 2.5.1: if the sequence
{an} is monotonic, then either
{an} is convergent or
an ! 1.
for n ¼ 1; 2; . . .:
Hence, by (2), we have that
1
[
ðc; f ðx1 Þ ¼
½f ðxn Þ; f ðx1 Þ J:
n¼1
To deduce, finally, that c is the left end-point of J, we need to show that c 2
= J.
Suppose that in fact c 2 J. Then c ¼ f(x), for some x 2 I; it follows that
f ð xÞ ¼ c < f ðxn Þ;
¼) x < xn ;
¼) x a:
for n ¼ 1; 2; . . .
for n ¼ 1; 2; . . .
Thus x 2
= I. This contradiction completes the proof.
4.4
4.4.1
&
Taking limits ‘flattens’
inequalities.
Defining exponential functions
The definition of ax
pffiffiffi
Earlier, we looked at the definition of the irrational number 2 ¼ 1:4142 . . ..
We have also defined ax for a > 0 when x is rational, but we have not yet
defined ax when x is irrational.
Sub-section 1.1.1.
Sub-section 1.5.3.
4: Continuity
162
pffiffi
2
One possible method for
pffiffiffidefining the irrational powerp2ffiffiffi involves the
decimal representation of 2. Each of the truncations of 2 ¼ 1:4142 . . . is
a rational number, and the corresponding rational numbers
21 ; 21:4 ; 21:41 ; 21:414 ; 21:4142 ; . . .
(1)
are defined, and form an increasing sequence which is bounded above by
22 ¼ 4 (for example). Hence, by the Monotone Convergence Theorem for
sequences, the sequence (1) ispconvergent,
and the limit of this sequence can
ffiffi
be taken as the definition of 2 2 :
We can define ax for a > 0 and x 2 R similarly. However, with this definition
it is difficult to establish the properties of ax, such as the Exponential Laws. It is
more convenient to define ax by using the exponential function x 7! ex, whose
properties we have already discussed.
Recall that
1 n
x n X
x
; for x 2 R;
ex ¼ lim 1 þ
¼
n!1
n
n!
n¼0
ex ¼ ðex Þ1 ;
for x 2 R;
and
exþy ¼ ex ey ;
for x; y 2 R:
(2)
x
Recall too that the function x 7! e is strictly increasing and continuous; and
hence, by the Inverse Function Rule, it has a strictly increasing continuous
inverse function x 7! loge x, x 2 (0, 1). Thus we have
(3)
loge ðex Þ ¼ x; for x 2 R;
and
eloge x ¼ x;
x
for x 2 ð0; 1Þ:
y
Now let a ¼ e and b ¼ e . Then, by equation (2), we have
ab ¼ ex ey ¼ exþy ;
so that, by equation (3), we obtain loge(ab) ¼ logea þ logeb, for a, b 2 (0, 1).
We deduce that, if a > 0 and n 2 N, then
loge ðan Þ ¼ n loge a;
and so
an ¼ en loge a :
With a little more manipulation, we can show that the equation
ax ¼ ex loge a
is true for each rational number x. This suggests that we define ax, for a > 0 and
x irrational, by means of this equation.
Definition
If a > 0, then ax ¼ ex loge a ;
for x 2 R:
For example, 2x ¼ ex loge 2 for x 2 R, so that the graph y ¼ 2x is obtained from
the graph y ¼ ex by a scaling in the x-direction with scale factor log1 2.
e
Sub-section 2.5.1.
Section 3.4.
Sub-section 3.4.3.
4.4
Defining exponential functions
163
This relationship between the graphs y ¼ ex and y ¼ 2x suggests that the
function x 7! 2x must also be continuous. In fact, we can deduce the continuity
of the function x 7! 2x ¼ ex loge 2 ; x 2 R, from the continuity of the function
x 7! ex, by using the Multiple Rule and the Composition Rule for continuity.
Remark
Since x 7! 2x is continuous, we deduce that the sequence
21 ; 21:4 ; 21:41 ; 21:414 ; 21:4142 ; . . .;
where 1, 1.4, 1.41,pffiffi1.414, 1.4142, . . . are truncations
of
pffiffi
does converge to 2 2 , and so both definitions of 2 2 agree.
pffiffiffi
2 ¼ 1:4142 . . .,
In general, we have the following result.
Theorem 1
continuous.
If a > 0, then the function x 7! ax ¼ ex loge a ; x 2 R, is
Prove that the following functions are continuous:
Problem 1
(a) f(x) ¼ x , where x 2 (0, 1) and 2 R;
(b) f(x) ¼ xx, where x 2 (0, 1).
4.4.2
Further properties of exponentials
Our definition of ax enables us to give straight-forward proofs of the following
Exponent Laws.
Exponent Laws
x x
x
If a, b > 0 and x 2 R, then a b ¼ (ab) .
If a > 0 and x, y 2 R, then a a ¼ a
.
If a > 0 and x, y 2 R, then ðax Þy ¼ axy .
x y
xþy
You first met these laws in
Sub-section 1.5.3, but there
x, y 2 Q .
4: Continuity
164
For example, to prove the final Exponent Law, notice that, from the definition
of ax, we have loge(ax) ¼ x loge a; so that
x
ðax Þy ¼ ey loge ða Þ
¼ exy loge a ¼ axy :
Thus manipulations such as
pffiffi p2ffiffi
pffiffiffi 2
pffiffiffiðp2ffiffipffiffi2Þ pffiffiffi2
2
¼ 2
¼
2 ¼2
are indeed justified.
Problem 2
Prove that, if a > 0 and x, y 2 R , then axay ¼ ax þ y.
Finally, our definition of ax enables us to prove the rule for rearranging
inequalities by taking powers.
Rule 5
For any non-negative a, b 2 R , and any p > 0, a < b , ap < bp.
You met this Rule in Subsection 1.2.1.
If a ¼ 0, the result is obvious. In general, since the functions x 7! loge x and
x 7! ex are strictly increasing, we have
a < b , loge a < loge b
, p loge a < p loge b
,e
p loge a
<e
ðsince p > 0Þ
p loge b
, ap < bp :
The following example illustrates the use of Rule 5.
Example 1
Solution
Determine which of the numbers ep and pe is greater.
We use the inequality
x
e > 1 þ x;
for x > 0:
Applying this inequality with x ¼ pe 1, we obtain
p
p
1
e ð e Þ1 > 1 þ
e
p
¼ ¼ pe1 ;
e
p
then, by multiplying through by the positive factor e, we obtain that e e > p.
&
Finally, by applying Rule 5 with p ¼ e, we obtain ep > pe.
Problem 3
4.5
Prove that ex > xe, for x > e.
Exercises
Section 4.1
1. Use the appropriate rules, together with the list of basic continuous
functions, to prove that the following functions are continuous:
This was one of the
Exponential Inequalities:
part (a) of Theorem 4 in
Sub-section 4.1.3.
4.5
Exercises
165
(a) f ð xÞ ¼ expðsinðx2 þ 1ÞÞ; x 2 R;
pffiffi
x 2 ½0; 1Þ:
(b) f ð xÞ ¼ e x þ x5 ;
2. Determine whether the following functions are continuous at 0:
sin x sinð1xÞ; x 6¼ 0,
(a) f ð xÞ ¼
0;
x ¼ 0;
1
1
sin x ;
x 6¼ 0,
(b) f ð xÞ ¼ x
0;
x ¼ 0:
3. Prove that each of the following sequences is convergent, and determine its
limit:
n 1
o
n 1 o
(b) cos 21n 2 :
(a) sin en 1 ;
4. Let f be defined on an open interval I, and c 2 I. Prove that, if f is continuous
at c and f(c) 6¼ 0, then there is an open interval J I such that c 2 J and
f(x) 6¼ 0, for any x 2 J.
1;
x ¼ 0; 1;
5. Determine the points where the function f ð xÞ ¼
is
x þ ½2x; 0 < x < 1;
(a) continuous on the left;
(b) continuous on the right;
(c) continuous.
6. Prove that the following function is continuous on R
8
x 12 p;
< 1;
f ð xÞ ¼ sin x; 12 p < x < 12 p;
:
1;
x 12 p:
7. Determine at which points the following function is continuous
x; 1 x < 0;
f ð xÞ ¼
ex ; 0 x 1:
8. Write down examples of functions with the following properties:
(a) f and g are discontinuous on R, but g f is continuous on R;
(b) f and g are discontinuous on R, but f þ g and fg are continuous on R;
(c) f is continuous
on R 1; 12 ; 13 ; 14 ; . . . but discontinuous at
1; 12 ; 13 ; 14 ; . . . :
Section 4.2
1. The function f is continuous on (0, 1), and takes every real value at most
once. Use the Intermediate Value Theorem to prove that f is strictly monotonic on (0, 1).
2. Give examples of functions f continuous on the half-open interval [0, 1) in
R, to show that f ([0,1)) can be open, closed or half-open.
3. Prove that each of the following polynomials has the stated number of (real)
zeros:
(a) pð xÞ ¼ x4 4x3 þ 3x2 þ 2x 1; 4 zeros;
(b) pð xÞ ¼ 3x3 8x2 þ x þ 3;
3 zeros:
2
4. Prove that
the function f ð xÞ ¼ x sin x 3 p; x 2 R, has a zero in
2
5
3 p; 6 p .
We shall use this result
in Chapter 6.
4: Continuity
166
5. Using the Zeros Localisation Theorem and the Extreme Values Theorem,
prove that every polynomial of even degree n
pð xÞ ¼ an xn þ an1 xn1 þ þ a1 x þ a0 ;
x 2 R;
where an 6¼ 0, has a minimum value on R.
6. Write down an example of a function f(x), x 2 (0, 1), that is continuous on
(0, 1) and for which f ((0, 1)) ¼ (0, 1), but such that there is no point c in
(0, 1) for which f (c) ¼ c.
This result is closely related to
Problem 2 in Subsection 4.2.1, so you might
like to look back at that
problem and compare the two.
Section 4.3
1. For each of the following functions, prove that it has a continuous inverse
function and determine the domain of that inverse function:
(a) f ð xÞ ¼ x3 þ 1 x12 ;
x 2 ð0; 1Þ;
1
ð1þx3 Þ2
(b) f ð xÞ ¼
;
x 2 ð1; 1Þ:
2. Determine whether each of the following statements is true:
(a) sin sin1 x ¼ x, for x 2 ½1; 1;
(b) sin1 ðsin xÞ ¼ x, for x 2 R.
3. (a) Prove that tan1 x þ tan1 y ¼ tan1
tan1 x þ tan1 y lies in 12 p; 12 p .
(b) Use the result in part (a) to evaluate
x þy
1xy
, provided that
tan1 12
þ
tan1 13
.
This is known as the Addition
Formula for tan1.
5
Limits and continuity
In Chapter 4 we made some progress in pinning down the idea of ‘a wellbehaved function’ in precise terms. Using our earlier work on sequences we
defined what was meant by a function being continuous: roughly speaking, its
graph has no jumps or gaps.
However, a number of functions arise quite naturally in mathematics where
we need to handle functions that are already defined near some particular point,
but that are not defined at the point itself.
For example, the function
sin x
f ð xÞ ¼
; x 6¼ 0;
x
which arises when we examine the question of whether the sine function is
differentiable. Clearly from their graphs, the behaviour of f differs significantly
from the behaviour of the function
1
gð xÞ ¼ sin
; x 6¼ 0;
x
that you have already met. It is possible to assign a value (namely, 1) to f at 0 so
that this extension of the domain of f makes f continuous on a domain that is an
interval. On the other hand, it is not possible to assign a value to g at 0 so that this
extension of the domain of g makes g continuous on a domain that is an interval.
In Section 5.1, we discuss limits of functions, and show that the existence of
this helpful value 1 for f can be stated as lim sinx x ¼ 1. The concept of a limit of a
x!0
function is closely related to that of a continuous function, and many of the
rules for calculating limits are similar to those for continuity. We also discuss
one-sided limits.
In Section 5.2, we discuss the behaviour of functions near asymptotes of
their graphs. In particular, we prove that, if n 2 Z, then lim xn ex ¼ 0:
1
y = sin 1x
–1
x!1
Next, in Section 5.3, we introduce a slightly different definition of the limit
lim f ð xÞ of a function f at a point c. Instead of using sequences tending to c to
x!c
define the limit of f at c, we define lim f ð xÞ directly in terms of inequalities
x!c
involving x and f (x), and we verify that this new definition is completely
equivalent to the earlier definition. We also illustrate the changes to proofs of
results about limits that the new definition involves.
In Section 5.4, we introduce a definition for the continuity of a function f at a
point c in terms of inequalities involving x and f (x), rather than in terms of the
behaviour of f on sequences that tend to c. We verify that this new definition is
completely equivalent to the earlier definition of continuity, and illustrate the
changes to proofs of results about continuity that the new definition involves.
Finally, in Section 5.5, we introduce a concept that will be extremely
powerful in your further study of Analysis; namely, that of uniform continuity.
At first sight this new
definition will seem more
complicated. However
throughout the rest of the
book we shall see just how
powerful this new definition
turns out to be!
167
5: Limits and continuity
168
Before starting on Sections 5.3 and 5.4, you may find it useful to quickly
revise the material in Sections 1.2 and 1.3 on inequalities.
5.1
5.1.1
‘may’ means ‘will’!
Limits of functions
What is a limit of a function?
The graph of the function
sin x
; x 6¼ 0;
f ð xÞ ¼
x
shows that, if x takes values which are ‘close to’ but distinct from 0, then f(x)
takes values which are ‘close to’ 1. The closer that x gets to 0, the closer f(x)
gets to 1. We now pin down this idea precisely.
First, we need the following fact.
Theorem 1
Proof
If {xn} is a null sequence whose terms are non-zero, then
sin xn
! 1 as n ! 1:
xn
First we deduce from the inequality
p
sin x x; for 0 < x < ;
2
sin xn
n!1 xn
That is, lim
¼ 1.
Lemma 1, Sub-section 4.1.3.
that
sin x
p
1; for 0 < x < :
x
2
Next, we need to use the formula for the area of a sector of a disc of radius 1
in Figure (a), below. Compare the area of this sector with the area of the
triangle in Figure (b), below.
We find that
p
for 0 < x < :
2
sin x
Since tan x ¼ cos x and cos x > 0 for 0 < x < p2, we obtain
sin x
p
; for 0 < x < :
cos x x
2
Combining this inequality with our earlier upper estimate for sinx x, we find that
sin x
p
1; for 0 < x < :
cos x x
2
x tan x;
We discussed the area and
perimeter of a disc of radius 1
in Sub-section 2.5.4 and in
Exercise 4 on Section 2.5 in
Section 2.6.
5.1
Limits of functions
169
In fact, this inequality holds for 0 < jxj < p2, since
sinðxÞ sin x
¼
:
ðxÞ
x
Now, if {xn} is any null sequence with non-zero terms, then the terms xn
must eventually satisfy the inequality jxn j < p2, and so there is some number X
such that
sin xn
cos xn 1; for n > X:
xn
But cos xn ! 1 as n ! 1, since the cosine function is continuous at 0 and
cos 0 ¼ 1. Hence, by the Squeeze Rule for sequences
sin xn
! 1:
&
xn
cosðxÞ ¼ cos x and
Take " ¼ p2 in the definition of
null sequence.
sin x
This behaviour of cos
x near 0 is an example of a function f tending to a limit as x
tends to a point c.
To define this concept, we need to ensure that the function f is defined near
the point c, but not necessarily at the point c itself. We first introduce the idea
of a punctured neighbourhood of c.
Definitions A neighbourhood of a point c of R is an open interval that
contains the point c, and a punctured neighbourhood of a point c of R is a
neighbourhood of c from which the point c itself has been deleted.
For example, the sets (0, 9), (1, 1) and R are neighbourhoods of the point 2,
and (1, 2) [ (2, 5) and (1, 2) [ (2, 4) are punctured neighbourhoods of the
point 2. In general, a neighbourhood of c is an interval of the form (c r, c þ s),
for some r, s > 0, and a punctured neighbourhood of c is the union of a pair of
intervals (c r, c) [ (c, c þ s), for some r, s > 0. In practice, we often choose as
a neighbourhood of c an open interval (c r, c þ r), r > 0, with centre at c; and
as a punctured neighbourhood of c the union (c r, c) [ (c, c þ r) of two open
intervals of equal length.
We now define the limit of a function in terms of limits of sequences.
Definition Let the function f be defined on a punctured neighbourhood N
of a point c. Then f (x) tends to the limit ł as x tends to c if:
for each sequence fxn g in N such that xn ! c; then f ðxn Þ ! ‘:
(1)
So the ‘puncture’ is at c.
These choices are simply
matters of convenience!
Note that xn 6¼ c, for any n.
We write this as either ‘lim f ð xÞ ¼ ‘’ or ‘f (x) ! ‘ as x ! c’.
x!c
Let us check that this definition holds for f ðxÞ ¼ sinx x at 0. This function f is
defined on the domain R {0}, and so in particular on every punctured
neighbourhood of 0. We have just seen that the statement (1) holds; it follows
then that
sin x
¼ 1:
lim
x!0 x
Since the definition of the limit of a function involves the limit of sequences,
we can use our various Combination Rules for sequences to determine the
limits of many functions.
For example, f is defined on
(1, 0) [ (0, 1).
5: Limits and continuity
170
Example 1 Prove that each of the following functions tends to a limit as x
tends to 2, and determine these limits:
2
3
3x 2
(a) f ð xÞ ¼ xx 24;
(b) f ð xÞ ¼ xx2 3x þ 2 :
Solution
(a) First, notice that f is defined on every punctured neighbourhood of 2.
Next, notice that
x2 4
¼ x þ 2; for x 6¼ 2:
x2
Thus, if {xn} is any sequence that lies in some punctured neighbourhood
N of 2 and xn ! 2, then
f ð xÞ ¼
f ðxn Þ ¼ xn þ 2 ! 4
as n ! 1;
by the Sum Rule for sequences. It follows that
x2 4
¼ 4:
x!2 x 2
(b) The function
We can cancel x 2, since
x 6¼ 2.
For example,
N ¼ (1, 2) [ (2, 3); any other
punctured neighbourhood of
2 would serve our purpose
equally well.
lim
f ð xÞ ¼
x3 3x 2 ðx 2Þðx2 þ 2x þ 1Þ
¼
x2 3x þ 2
ð x 2Þ ð x 1Þ
has domain R {1, 2}, and so f is defined on the punctured neighbourhood
N ¼ (1, 2) [ (2, 3) of 2.
Next, notice that
f ð xÞ ¼
x3 3x 2 x2 þ 2x þ 1
;
¼
x2 3x þ 2
x1
Of course any smaller
punctured neighbourhood of 2
would serve our purpose
equally well.
for x 2 N ¼ ð1; 2Þ [ ð2; 3Þ:
Thus, if {xn} is any sequence that lies in N ¼ (1, 2) [ (2, 3) and xn ! 2, then
x2n þ 2xn þ 1
4þ4þ1
¼ 9 as n ! 1;
!
xn 1
21
by the Combination Rules for sequences. It follows that
f ðxn Þ ¼
lim
x!2
x3 3x 2
¼ 9:
x2 3x þ 2
&
Later in this section we give several further techniques for calculating limits.
Our next example illustrates how to prove that a limit does not exist.
Sub-sections 5.1.2 and 5.1.3.
Example 2 Prove that each of the following functions does not tend to a limit
as x tends to 0:
pffiffiffi
(a) f ðxÞ ¼ 1x ; x 6¼ 0;
(c) f ð xÞ ¼ x; x 0:
(b) f ðxÞ ¼ sin 1x ; x 6¼ 0;
Solution
(a) The function f is defined on
the
punctured neighbourhood N ¼ (2,
0) [ (0, 2)
of 0. The null sequence 1n lies in N and tends to 0, but f 1n ¼ n ! 1.
Hence f does not tend to a limit as x tends to 0.
(b) The function f is defined on the punctured neighbourhood N ¼ (2, 0) [ (0, 2)
of 0.
Any other punctured
neighbourhood of 0 would
serve equally well here, of
course.
5.1
Limits of functions
171
To prove that f(x) does not tend to a limit as x tends to 0, we choose two null
sequences {xn} and {x0n} that lie in N, such that
f ðxn Þ ! 1 and f x0n ! 1:
Since sin 2np þ 12 p ¼ 1 and sin 2np þ 32 p ¼ 1 for n 2 Z, we can choose
1
1
xn ¼
and x0n ¼
; for n ¼ 0; 1; 2; . . .:
2np þ 12 p
2np þ 32 p
Hence f does not tend to a limit as x tends to 0.
pffiffiffi
(c) The function f ð xÞ ¼ x has domain [0, 1), and so f is not defined on any
punctured neighbourhood of zero. Hence f does not tend to a limit as x
&
tends to zero.
In particular, the terms xn and
x n0 will be non-zero.
We collect these techniques together in the form of a strategy.
Strategy To show that lim f ð xÞ does not exist:
x!c
1. Show that there is no punctured neighbourhood N of c on which f is
defined;
OR
2. Find two sequences {xn} and {xn0 } (in some punctured neighbourhood N
of c) which tend to c, such that {f (xn)} and {f (xn0 )} have different limits;
OR
3. Find a sequence {xn} (in some punctured neighbourhood N of c) which
tends to c, such that f (xn) ! 1 or f (xn) ! 1.
Problem 1 Determine whether the following limits exist:
2
(a) lim x xþx ;
(b) lim jxxj :
x!0
5.1.2
x!0
Limits and continuity
Consider the function
1;
f ð xÞ ¼
0;
x 6¼ 0;
x ¼ 0:
Does this function tend to a limit as x tends to zero; and, if it does, what is the
limit?
Certainly, f is defined on any punctured neighbourhood of zero, since the
domain of f is R. Also, if {xn} is any null sequence with non-zero terms, then
f (xn) ¼ 1, for n ¼ 1, 2, . . ., so that f (xn) ! 1 as n ! 1. It follows that lim f ð xÞ ¼ 1.
x!0
This example serves to emphasise that the value of a limit lim f ð xÞ has nothing
x!c
to do with the value of f (c) – even if f happens to be defined at the point c.
However, if f is defined at c, and f is also continuous at c, then the only
possible value for lim f ð xÞ is f (c).
x!c
We put these observations together in the following result.
Theorem 2
Then
Let the function f be defined on an open interval I, with c 2 I.
f is continuous at c , lim f ð xÞ ¼ f ðcÞ:
x!c
Notice that I is a
neighbourhood of c.
5: Limits and continuity
172
Using Theorem 2 and our knowledge of continuous functions, we can evaluate many limits rather easily.
For example, to determine lim 3x5 5x2 þ 1 notice that the function
x!2
f (x) ¼ 3x 5x þ 1 is defined on R and is continuous at 2, since f is a polynomial. Hence, by Theorem 2
lim 3x5 5x2 þ 1 ¼ f ð2Þ ¼ 77:
5
2
Polynomials are basic
continuous functions: see
Sub-section 4.1.3.
x!2
We saw earlier that the following functions are continuous at 0
2 1
x sin x ; x 6¼ 0;
x sin 1x ; x 6¼ 0;
and f ð xÞ ¼
f ð xÞ ¼
0;
x ¼ 0;
0;
x ¼ 0:
Problem 8 and Example 5,
Sub-section 4.1.2,
respectively.
It therefore follows from Theorem 2 that
1
1
2
lim x sin
¼ 0 and lim x sin
¼ 0:
x!0
x!0
x
x
On the other hand, we saw in part (b) of Example 2 that
1
lim sin
¼ 0 does not exist:
x!0
x
Sub-section 5.1.1.
It follows from
Theorem 2 that, no matter how we try to extend the domain
of f ðxÞ ¼ sin 1x to include x ¼ 0, we can never obtain a continuous function.
Remark
If f is defined on an open interval I, with c 2 I, and lim f ð xÞ exists but
x!c
lim f ð xÞ 6¼ f ðcÞ, then f is saidto have
a removable discontinuity at c. For
x!c
x sin 1x ; x 6¼ 0;
has a removable discontinuity
example, the function f ð xÞ ¼
3;
x ¼ 0;
at 0.
This means that, if we
redefine the value of f just at c
itself, the resulting function is
then continuous at c.
Problem 2 Use Theorem 1 to determine the following limits:
pffiffiffiffiffiffiffiffiffi
pffiffiffi
ex
(a) lim x;
(c) lim 1þx
(b) limp sin x;
x!2
x!1
x!2
In the remainder of this chapter we shall frequently use Theorem 2. When the
function f is one of our basic continuous functions, however, we shall not
always refer to the theorem explicitly. For example, f (x) ¼ x2 þ 1 and g(x) ¼
sin x are basic continuous functions, and so we can write limðx2 þ 1Þ ¼ 5 and
x!2
lim sin x ¼ 0 without further explanation.
x!0
Basic continuous functions:
polynomials and
rational functions;
modulus function;
nth root function;
trigonometric functions
(sine, cosine and tangent);
5.1.3
the exponential function.
Rules for limits
As you might expect from your experience with sequences, series and continuous functions, we often find limits by using various rules. First, we state the
Combination Rules.
Theorem 3
Combination Rules
If lim f ð xÞ ¼ ‘ and lim gð xÞ ¼ m, then:
x!c
x!c
Sum Rule
limð f ð xÞ þ gð xÞÞ ¼ ‘ þ m;
Multiple Rule
lim lf ð xÞ ¼ l‘;
x!c
x!c
for l 2 R;
5.1
Limits of functions
173
Product Rule
lim f ð xÞgð xÞ ¼ ‘m;
Quotient Rule
lim gf ððxxÞÞ ¼ m‘ ; provided that m 6¼ 0.
x!c
x!c
For example, since lim sinx x ¼ 1 and limðx2 þ 1Þ ¼ 1; we have
x!0
x!0
2
sin x
þ 2 x þ 1 ¼ 1 þ 2 1 ¼ 3:
lim
x!0
x
The proofs of these rules are all simple consequences of the corresponding
results for sequences. We illustrate this by proving just one rule.
Proof of the Sum Rule Let the functions f and g be defined on a punctured
neighbourhood N of a point c. We want to show that:
for each sequence fxn g in N such that xn ! c, then f ðxn Þ þ gðxn Þ ! ‘ þ m:
There are no new ideas in the
proof. All that we have to do is
to set up things so that we can
use the Sum Rule for
sequences.
Now, since lim f ð xÞ ¼ ‘, we know that, for any sequence {xn} in N for which
x!c
xn ! c, then f (xn) ! ‘. Also, since lim gð xÞ ¼ m, we know that, as xn ! c, then
x!c
g(xn) ! m.
It follows by the Sum Rule for sequences that, as xn ! c, then
&
f (xn) þ g(xn) ! ‘ þ m. This completes the proof.
We can also use the following Composition Rule.
Theorem 4 Composition Rule
If lim f ð xÞ ¼ ‘ and lim gð xÞ ¼ L, then lim gð f ð xÞÞ ¼ L, provided that:
x!c
x!c
x!‘
f (x) 6¼ ‘ in some punctured neighbourhood of c;
g is defined at ‘ and is continuous at ‘.
EITHER
OR
Note that the second limit is a
limit as x ! ‘, not as x ! c.
Remarks on the Composition Rule
(a) In any particular case, the Composition Rule may be FALSE if we do not
ensure that one or other of the two provisos holds! For example, if
f ð xÞ ¼ 0; x 2 R;
sin x
and gð xÞ ¼
x
0;
;
x 6¼ 0;
x ¼ 0;
Here we take c ¼ 0, ‘ ¼ 0 and
L ¼ 1.
then
lim f ð xÞ ¼ 0 and
x!0
lim gð xÞ ¼ 1; but
x!0
lim gð f ð xÞÞ ¼ limð0Þ ¼ 0:
x!0
x!0
So, in this case,
lim gð f ð xÞÞ 6¼ L:
x!0
(b) Suppose the first proviso ‘f (x) 6¼ ‘ in some punctured neighbourhood of c’
in Theorem 4 holds. Then we know that, for each sequence {xn} in a
punctured neighbourhood N of c for which xn ! c, f (xn) does not take the
value ‘ but must lie in a punctured neighbourhood of ‘. In this case, the
desired result will follow from the facts that (i) the sequence {f (xn)}
converges to ‘, (ii) f (xn) lies in a punctured neighbourhood of ‘, and
(iii) lim gð xÞ ¼ L:
x!‘
(c) Suppose the second proviso ‘g is defined at ‘ and is continuous at ‘’
in Theorem 4 holds. In this case, the desired result will follow from
We omit the details of the
proof in this case, which are
now straight-forward to write
down.
5: Limits and continuity
174
the facts that (i) the sequence {f (xn)} converges to ‘, and (ii) g is continuous at ‘.
The following example illustrates the use of the Composition Rule.
Example 3 Determine the
following limits:
2
ðsin xÞ
(a) lim sinsin
(b) lim 1þ sinx x :
x ;
x!0
x!0
Solution
(a) Let f (x) ¼ sin x, x 2 R, and gðxÞ ¼ sinx x , x 6¼ 0: Then
sin x
¼ 1:
x!0
x!0
x!0
x!0 x
Also, f (x) ¼ sin x ¼
6 0 in the punctured neighbourhood ðp; 0Þ [ ð0; pÞ of 0
(for example). It follows, by the Composition Rule, that
sinðsin xÞ
lim gð f ð xÞÞ ¼ lim
¼ 1:
x!0
x!0
sin x
lim f ð xÞ ¼ lim sin x ¼ 0
(b) Let f ðxÞ ¼
and
lim gð xÞ ¼ lim
x 6¼ 0, and gð xÞ ¼ 1 þ x2 , x 2 R: Then
sin x
¼ 1 and lim 1 þ x2 ¼ 2:
lim
x!0 x
x!1
Here we have c ¼ 0, ‘ ¼ 0 and
L ¼ 1.
sin x
x ,
Here we have c ¼ 0, ‘ ¼ 1 and
L ¼ 2.
Also, g is defined and continuous at 1. It follows, by the Composition Rule,
that
!
sin x 2
lim gð f ð xÞÞ ¼ lim 1 þ
¼ 2:
&
x!0
x!0
x
Problem 3 Use the Combination Rules and the Composition Rule to
determine the following limits:
1
sinðx2 Þ
sin x
(b) lim x2 ;
(c) lim sinx x 2 :
(a) lim 2xþx
2 ;
x!0
x!0
x!0
For the functions
8
x ¼ 0;
<0;
f ð xÞ ¼ 2
x ¼ 1;
:
2 þ x; x 6¼ 0; 1;
Problem 4
and gð xÞ ¼
0;
1 þ x;
x ¼ 0;
x 6¼ 0;
determine ð f gÞð xÞ; ð f gÞð0Þ; f lim gð xÞ and lim f ðgð xÞÞ:
x!0
x!0
There is also a Squeeze Rule for limits.
Theorem 5
Squeeze Rule
Let the functions f, g and h be defined on a punctured neighbourhood N of a
point c. If:
1. gð xÞ f ð xÞ hð xÞ; for x 2 N; and
2. lim gð xÞ ¼ lim hð xÞ ¼ ‘;
x!c
x!c
then lim f ð xÞ ¼ ‘:
The proof of Theorem 5 is a
straight-forward application
of the Squeeze Rule for
sequences.
y
y = h(x)
y = f (x)
y = g(x)
x!c
Problem 5 Use
the Squeeze Rule for limits
to prove that:
(a) lim x2 sin 1x ¼ 0;
(b) lim x cos 1x ¼ 0:
x!0
x!0
N
c
x
5.1
Limits of functions
175
The Limit Inequality Rule is another result for limits of functions that is an
analogue of the corresponding result for sequences.
Theorem 3, Sub-section 2.3.3.
Theorem 6 Limit Inequality Rule
If lim f ð xÞ ¼ ‘ and lim gð xÞ ¼ m, and also
x!c
x!c
f ð xÞ gð xÞ, on some punctured neighbourhood of c,
then
‘ m:
5.1.4
One-sided limits
pffiffiffi
In Example 2(c)pabove,
we saw that lim x does not exist, because the
ffiffiffi
x!0
function f ð xÞ ¼ x, x 0, is not defined on any punctured neighbourhood
of zero. Nevertheless, if {xn} is any sequence
pffiffiffi in the domain [0, 1) of f for
pffiffiffiffiffi
which xn ! 0, then xn ! 0. Thus f ð xÞ ¼ x tends to f (0) ¼ 0 as x tends to 0
from the right.
Definition Let f be defined on (c, c þ r), for some r > 0. Then f(x) tends
to the limit ł as x tends to c from the right if:
Note that f need not be defined
at the point c.
for each sequence {xn} in (c, c þ r) such that xn ! c, then f (xn) ! ‘.
We write this either as ‘ limþ f ð xÞ ¼ ‘’ or as ‘f (x) ! ‘ as x ! c þ ’.
For example, limþ
x!0
x!c
pffiffiffi
x ¼ 0:
There is a corresponding definition for limits as x tends to c from the left.
Definition Let f be defined on (c r, c), for some r > 0. Then f(x) tends
to the limit ł as x tends to c from the left if:
for each sequence {xn} in (c r, c) such that xn ! c, then f(xn) ! ‘.
We write this either as ‘ lim f ð xÞ ¼ ‘’ or as ‘f ð xÞ ! ‘ as x ! c ’.
x!c
Sometimes both left and right limits exist. Even when this happens, the two
values need not be equal, as the following example shows.
Example 4 Prove that the function f ð xÞ ¼ jxxj , x 6¼ 0, tends to different limits
as x tends to 0 from the right and from the left.
Solution The function f is defined on (0, 1), and f (x) ¼ 1 on this interval.
Thus, if {xn} is a null sequence in (0, 1), then
lim f ðxn Þ ¼ lim ð1Þ ¼ 1:
n!1
n!1
So limþ f ð xÞ ¼ 1:
x!0
Similarly, f is defined on (1, 0), and f(x) ¼ 1 on this interval. Thus, if {xn}
is a null sequence in (1, 0), then
lim f ðxn Þ ¼ lim ð1Þ ¼ 1:
n!1
So lim f ð xÞ ¼ 1:
x!0
n!1
&
Again, f need not be defined at
the point c.
5: Limits and continuity
176
Problem 6 Write down a function f defined on the interval [1, 3] for
which lim f ð xÞ does not exist but limþ f ð xÞ ¼ 1. Verify that f has the
x!0
x!0
specified properties.
The relationship between one-sided limits and ‘ordinary’ limits is given by
the following result.
Theorem 7
Then
Let f be defined on a punctured neighbourhood of the point c.
We omit a proof of this result.
lim f ð xÞ ¼ ‘
x!c
, limþ f ð xÞ
x!c
and
lim f ð xÞ both exist and equal ‘:
x!c
Remark
If f is defined on an open interval I that contains the point c, and
lim f ð xÞ and lim f ð xÞ both exist; but limþ f ð xÞ 6¼ lim f ð xÞ;
x!cþ
x!c
x!c
x!c
then f is said to have a jump discontinuity at c.
Analogues of the Combination Rules, Composition Rule and Squeeze Rule can
also be used to determine one-sided limits. In the statements of all these rules,
we simply replace lim by limþ or lim , and replace the open interval I containx!c
x!c
x!c
ing c by (c, c þ r) or (c r, c), as appropriate.
Problem 7 Prove
pffiffiffithat:
(a) limþ sinx x þ x ¼ 1;
x!0
Problem 8
prove that
pffiffi
(b) limþ sinpð ffiffix xÞ ¼ 1:
x!0
1
; for jxj < 1, to
Use the inequalities 1þ x ex 1x
limþ
x!0
ex 1
ex 1
¼ lim
¼ 1:
x!0
x
x
Deduce that
ex 1
¼ 1:
x!0
x
lim
5.2
Asymptotic behaviour of functions
In this section we define formally a number of statements that you will have
already met in your study of Calculus, such as
1
! 1 as x ! 0þ and ex ! 1 as x ! 1;
x
and describe the relationship between them.
5.2.1
Functions which tend to infinity
Just as we defined the statement lim f ð xÞ ¼ ‘ in terms of the convergence of
x!c
sequences, so we can define the statement f (x) ! 1 as x ! c.
We verified these inequalities
as part (c) of Theorem 4 (‘The
Exponential Inequalities’) in
Sub-section 4.1.3.
5.2
Asymptotic behaviour of functions
177
Definition Let f be defined on a punctured neighbourhood N of the point c.
Then f (x) ! 1 as x ! c if:
for each sequence {xn} in N such that xn ! c, then f (xn) ! 1.
The statement ‘f (x) ! 1 as x ! c’ is defined similarly, with 1 replaced
by 1.
As with sequences, there is a version of the Reciprocal Rule which relates the behaviour of functions which tend to infinity and functions which
tend to 0.
Theorem 2, Sub-section 2.4.3.
Theorem 1 Reciprocal Rule
(a) If the function f satisfies the following two conditions:
1. f (x) > 0 for all x in some punctured neighbourhood of c, and
2. f (x) ! 0 as x ! c,
then f ð1xÞ ! 1 as x ! c.
(b) If f (x) ! 1 as x ! c, then f ð1xÞ ! 0 as x ! c.
For example,
lim x2 ¼ 0.
1
x2
! 1 as x ! 0, since f (x) ¼ x2 > 0, for x 2 R {0}, and
x!0
The statements
‘f ð xÞ ! 1 or 1 as x ! cþ ’ and
‘f ð xÞ ! 1 or 1 as x ! c ’
are defined similarly, with the punctured neighbourhood of c being replaced by
open intervals (c, c þ r) or (c r, c), as appropriate, for some r > 0.
The Reciprocal Rule can also be applied with ‘x ! c’ replaced by ‘x ! cþ’
or ‘x ! c’, and with the punctured neighbourhood of c being replaced by open
intervals (c, c þ r) or (c r, c), as appropriate, for some r > 0.
For example, 1x ! 1 as x ! 0þ, since f(x) ¼ x > 0, for x 2 ð0; 1Þ, and
limþ x ¼ 0:
x!0
Problem 1 Prove the following:
1
(a) jxj
! 1 as x ! 0;
(b) 1 1 x3 ! 1 as x ! 1þ ;
(c)
sin x
x3
! 1 as x ! 0:
Remark
There are also versions of the Combination Rules and Squeeze Rule for
functions which tend to 1 or 1 as x tends to c, cþ or c; here we state
only the Combination Rules for functions which tend to 1 as x tends to c.
Theorem 2 Combination Rules
If f(x) ! 1 and g(x) ! 1 as x ! c, then:
Sum Rule
f(x) þ g(x) ! 1;
Multiple Rule l f(x) ! 1, for l > 0;
Product Rule f(x) g(x) ! 1.
These are similar to results
stated for sequences in
Theorems 3 and 4,
Sub-section 2.4.3.
5: Limits and continuity
178
Problem 2 Prove that the following statements are false:
(a) If f(x) ! 1 and g(x) ! 1 as x ! 0, then f(x) g(x) ! 1;
(b) If f(x) ! 1 and g(x) ! 0 as x ! 0, then f(x) g(x) ! 0 or
f(x) g(x) ! 1.
5.2.2
Behaviour of f(x) as x tends to 1 or 1
Finally, we define various types of behaviour of real functions f (x) as x tends to
1 or 1. To avoid repetition, here we allow ‘ to denote either a real number
or one of the symbols 1 and 1.
We adopt this convention for
simplicity ONLY in this
sub-section!
Definition Let f be defined on an interval (R, 1), for some real number R.
Then f (x) ! ‘ as x ! 1 if:
for each sequence {xn} in (R, 1) such that xn ! 1, then f (xn) ! ‘.
The statement ‘f (x) ! ‘ as x ! 1’ is defined similarly, with 1 replaced
by 1, and (R, 1) by (1, R). In practice, we usually prove that f (x) ! ‘ as
x ! 1 by showing that f (x) ! ‘ as x ! 1.
When ‘ is a real number, we also use the notation
lim f ð xÞ ¼ ‘ and
lim f ð xÞ ¼ ‘:
x!1
x!1
Once again, we can use versions of the Combination Rules and Reciprocal
Rule to obtain statements about the behaviour of given functions as x tends to 1
or 1. In the statements of these rules, we need only replace c by 1 or 1,
and the punctured neighbourhood of c by (R,1) or (1, R), as appropriate.
For example, for any positive integer n
xn ! 1 as x ! 1 and lim xn ¼ 0:
x!1
More generally, we have the following result for the behaviour of polynomials as x ! 1.
If a0, a1, . . ., an1 are real numbers and
Theorem 3
n
pð xÞ ¼ x þ an1 x
n1
þ þ a1 x þ a0 ;
x 2 R;
then
pð xÞ ! 1 as x ! 1
and
We ask you to prove the first
part of this result in
Section 5.6. The second part
then follows at once.
1
! 0 as x ! 1:
pð x Þ
There are also versions of the Squeeze Rule for functions as x tends to 1,
which have some important applications.
Theorem 4 Squeeze Rule
Let the functions f, g and h be defined on some interval (R, 1).
(a) If f, g and h satisfy the following two conditions:
1. g(x) f(x) h(x), for all x in (R, 1), and
2. lim gð xÞ ¼ lim hð xÞ ¼ ‘;
x!1
x!1
then lim f ð xÞ ¼ ‘:
x!1
We omit the proof of this
theorem, which is straightforward.
5.2
Asymptotic behaviour of functions
179
y
(b) If f and g satisfy the following two conditions:
1. f(x) g(x), for all x in (R, 1), and
l y = f (x)
2. g(x) ! 1 as x ! 1,
then f (x) ! 1 as x ! 1.
y = g(x)
x2
xn
xnþ1
þ þ þ
þ 2!
n! ðn þ 1Þ!
for x 0. Since x 0, all the terms on the right here are non-negative, and so
ex ¼ 1 þ x þ
ex
xn
xnþ1
;
ðn þ 1Þ!
(R, ∞)
R
An important application of this Squeeze Rule is to show that ex tends to 1
faster than any power of x, as x tends to 1. We prove this in the next example.
ex
Example 1 Prove that for each n ¼ 0, 1, 2, . . ., xn ! 1 as x ! 1.
Solution We use the power series expansion
ex y = h(x)
y
x
y = f (x)
y = g(x)
R
(R, ∞)
x
See Section 3.4.
for x 0:
It follows that
ex
x
; for x > 0:
xn ðn þ 1Þ!
x
Since ðnþ1Þ!
! 1 as x ! 1, it follows from part (b) of the Squeeze Rule that
&
! 1 as x ! 1.
Remark
We deduce from Example 1 and the Product Rule that, for any integer n,
xnex ! 1 as x ! 1. Thus, by the Reciprocal Rule, for any integer n,
xnex ! 0 as x ! 1.
Problem 3 Determine the behaviour of the following functions as
x ! 1:
3
(a) f ð xÞ ¼ 2xx3þx ;
(b) f ð xÞ ¼ sinx x.
In our next example, we describe the behaviour of loge x as x ! 1.
Example 2
Prove that loge x ! 1 as x ! 1.
Solution We prove this from first principles.
First, note that the function f (x) ¼ loge x is defined on (0, 1).
Next, we have to prove that:
for each sequence {xn} in (0, 1) such that xn ! 1, then loge xn ! 1.
To prove that loge xn ! 1, we need to show that:
for each positive number K, there is a number X such that
loge xn > K; for all n > X:
(1)
However, since we know that xn ! 1, we can choose X such that
xn > eK ;
for all n > X:
Since the function loge is strictly increasing, the statement (1) is therefore
&
true. It follows that loge x ! 1 as x ! 1, as required.
There is no easy way to prove
this result by using the
Squeeze Rule, since loge x
tends to 1 ‘rather
reluctantly’; that is, more
slowly than any function that
we have considered so far.
5: Limits and continuity
180
5.2.3
Composing asymptotic behaviour
Earlier we gave a Composition Rule for limits. We now describe a more
general Composition Rule which permits the composition of the different
types of asymptotic behaviour that we have now met.
For example, since
1
f ðxÞ ¼ ! 1 as x ! 0þ and gðxÞ ¼ ex ! 1 as x ! 1;
x
we should expect that
1
Sub-section 5.1.3, Theorem 4.
as x ! 0þ :
gð f ð x Þ Þ ¼ e x ! 1
This result is true, as the following Composition Rule shows. To avoid
repetition, we again allow ‘ and L to denote either a real number or one of
the symbols 1 and 1.
We again adopt this
convention for simplicity
ONLY in this sub-section!
Theorem 5 Composition Rule
If:
1. f (x) ! ‘ as x ! c (or cþ, c, 1 or 1), and
We omit the proof of
Theorem 5.
2. g (x) ! L as x ! ‘,
then
If ‘ denotes 1 or 1,
then the first proviso is
automatically satisfied. Note
that we must have conditions
1 and 2 and the first proviso
satisfied or conditions 1 and 2
and the second proviso
satisfied if we are to make any
application of this result.
as x ! c ðor cþ ; c ; 1 or 1; respectivelyÞ;
gð f ð xÞÞ ! L
provided that:
EITHER
OR
f (x) ¼
6 ‘ in some punctured neighbourhood of c (or in (c, c þ r),
(c r, c), (R, 1) or (1, R), respectively, for some r > 0)
‘ is finite, and g is defined at ‘ and is continuous at ‘.
Example 3 Prove that
x
1
(b) xex ! 1 as x ! 0þ ;
(a) ex2 ! 1 as x ! 1;
1
(c) x sin x ! 1 as x ! 1:
Solution
x
(a) Let f ð xÞ ¼ 2x , x 2 R, and gð xÞ ¼ ex , x 2 R f0g, so that
x
g ð f ð x ÞÞ ¼
e2
x
2
x
¼
2e2
,
x
for x 2 R f0g, and so in
particular for x 2 ð0; 1Þ:
Now, by the Multiple Rule, f(x) ! 1 as x ! 1; and, by Example 1,
g(x) ! 1 as x ! 1. It follows, by the Composition Rule, that
x
2e2
gð f ð x Þ Þ ¼
! 1 as x ! 1,
x
x
so that; by the Multiple Rule; we have ex2 ! 1 as x ! 1:
x
(b) Let f ð xÞ ¼ 1x , x 2 R f0g, and gð xÞ ¼ ex , x 2 R f0g, so that
1
g ð f ð x ÞÞ ¼
ex
1
x
1
¼ xex ,
for x 2 R f0g, and so in
particular for x 2 ð0; 1Þ:
5.3
Limits of functions – using " and 181
Now, f (x) ! 1, as x ! 0þ; and, by Example 1, g(x) ! 1 as x ! 1. It
follows, by the Composition Rule, that
1
gð f ð xÞÞ ¼ xex ! 1
as x ! 0þ :
(c) Let f ð xÞ ¼ 1x , x 2 R f0g; and gð xÞ ¼ sinx x ; x 2 R f0g, so that
sin 1x
1
gð f ð xÞÞ ¼ 1 ¼ x sin
; for x 2 R f0g, and so
x
x
In Theorem 5, we have ‘ ¼ 0,
L ¼ 1.
for x 2 ð0; 1Þ:
Now
f ð xÞ ! 0 as x ! 1;
gð xÞ ! 1 as x ! 0; and
This is condition 1.
f ð xÞ 6¼ 0; for x 2 ð0; 1Þ:
So the first proviso is
satisfied.
It follows, by the Composition Rule, that
1
gð f ð xÞÞ ¼ x sin
! 1 as x ! 1:
x
This is condition 2.
&
Problem 4 Prove that:
1
(b) xex ! 0 as x ! 0 .
(a) loge (loge x) ! 1 as x ! 1;
Hint for part (b): Use the fact that 1x ! 1 as x ! 0 .
Problem 5 Give examples of functions f and g (and a specific value for
each of ‘ and m) for which
1. f(x) ! ‘ as x ! 1 and
2. g(x) ! m as x ! ‘,
but for which gð f ð xÞÞ6! m as x ! 1.
5.3
Limits of functions – using e and d
Earlier we gave definitions of the limit of a sequence, continuity of a function
and limit of a function, and strategies for using these definitions. In each case,
the strategy was in two parts:
GUESS
GUESS
that the definition HOLDS: prove that it holds for all cases.
that the definition FAILS: find ONE counter-example.
Each definition is particularly convenient if we wish to prove that the
definition FAILS, but it is not always easy to work with it when we wish to
prove that the definition HOLDS.
For example, when we wish to prove that a given function f is discontinuous
at a point c, then we have to find ONE sequence in a punctured neighbourhood of
c such that
xn ! c
BUT
f ðxn Þ 6! f ðcÞ:
5: Limits and continuity
182
On the other hand, if we wish to use the definition to prove that f is
continuous at c, then we have to show that:
for each sequence {xn} in a punctured neighbourhood of c such that
xn ! c; then f ðxn Þ ! f ðcÞ :
As it is sometimes tricky to determine the behaviour of every such sequence
{xn}, this definition can be very inconvenient to use.
In this section we introduce a definition of the limit of a function that is
equivalent to our earlier definition, but which does not use sequences.
5.3.1
In Section 5.4 we shall
introduce a definition of
continuity which does not
use sequences.
The e – d definition of limit of a sequence
We motivate our discussion by returning to the definition of a convergent
sequence.
Definition The sequence {an} is convergent with limit ł, or converges to
the limit ł, if {an ‘} is a null sequence. In other words, if:
for each positive number ", there is a number X such that
jan ‘j < "; for all n > X:
Sub-section 2.3.1.
(1)
{an}
l+ε
|an – l| < ε
⇔
l – ε < an < l + ε
l
l−ε
n
X
The condition (1) means that, from some point on, the terms of the sequence
all lie in the shaded strip between ‘ " and ‘ þ ", and thus lie ‘close to’ ‘. If we
choose a smaller number ", then the shaded strip in the diagram becomes
narrower, and we may need to choose a larger number X in order to ensure that
the inequality (1) still holds. But, whatever positive number " we choose, we
can always find a number X for which (1) holds.
n
Þ
Problem 1 Let an ¼ ð1
n2 , n ¼ 1, 2, . . .. How large must we take X in
order that:
(a) |an 0| < 0.1,
for all n > X?
(b) |an 0| < 0.01, for all n > X?
(c) |an 0| < ",
for all n > X, where " is a given positive number?
Earlier we described a ‘game’, based on the above definition, in which
player A chooses a positive number " and challenges player B to find a number
X such that (1) holds. If the sequence in question converges, then such a
number X always exists, and player B can always win. If the sequence does
not converge, then, for SOME choices of ", NO number X exists such that (1)
holds, and so player A can always win.
Sub-section 2.2.1.
5.3
Limits of functions – using " and 183
For example, consider the null sequence an ¼ p1ffiffin , n ¼ 1, 2, . . .. In this case,
the game might proceed as follows:
I'll play
X = 100
I'll play
ε = 0.1
B wins
A
B
Here player B wins, since, for n > 100, we have
1
1
pffiffiffi 0 ¼ p1ffiffiffi < pffiffiffiffiffiffiffi
ffi ¼ 0:1:
n
n
100
Here X ¼ 100.
Player A tries again:
I'll play
ε = 0.01
I'll play
X = 10 000
B wins
A
B
Player B again wins, since, for n > 10 000, we have
1
1
pffiffiffi 0 ¼ p1ffiffiffi < pffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 0:01:
n
n
10 000
Here X ¼ 10 000.
But now player B has figured out a winning strategy, and challenges player A
to do his worst!
I'll play
X = 1/ε 2
I'll play
ANY positive
ε
B wins
A
B
This is indeed a winning strategy, since
1
pffiffiffi 1
1
1
for n > 2 , we have n > , and sopffiffiffi 0 ¼ pffiffiffi < " :
"
"
n
n
The reason for introducing " in the definition of convergent sequence is to
formalise the idea of ‘closeness’. The statement (1) means that we can make
the terms an of the sequence as close as we please to ‘ by choosing X large
enough.
Here X ¼ "12 .
5: Limits and continuity
184
Put another way, we can think of a sequence {an} as a function with domain R
n 7! ðn; an Þ:
The number " formalises the idea of closeness of an to ‘ (that is, in the
codomain). The condition ‘for all n > X’ restricts the values n in the domain
to values for which condition (1) holds.
In general, the smaller the
value of ", the larger the X that
we have to choose.
{an}
l+ε
l
{
an close
to l
|an – l| < ε
for all n > X
l–ε
n
X
for all n > X
5.3.2
The e – d definition of limit of a function
The concept of the limit of a function as x ! c also involves the idea of
closeness. To define
f ð xÞ ! ‘
as x ! c;
(2)
we require the statement
j f ð xÞ ‘j < ";
this means that the values of f (x) and ‘ are within a distance " of each other.
Earlier, we formalised the statement (2) by defining the limit of a function in
terms of limits of sequences.
Sub-section 5.1.1.
Definition Let the function f be defined on a punctured neighbourhood N
of a point c. Then f (x) tends to the limit ł as x tends to c if:
for each sequence fxn g in N such that xn ! c, then f ðxn Þ ! ‘: (3)
Note that xn 6¼ c, for any n.
Intuitively, this means that, as xn gets close to c, so f (xn) gets close to ‘. More
precisely, we can make f (xn) as close as we please to ‘ by choosing xn
sufficiently close to c; we can ensure this by considering only xns for sufficiently large n. But the sequence {xn} can be any sequence of points in N that
converges to c, so what we are really saying is that we can make f (x) as close as
we please to ‘ by choosing x sufficiently close to c (but not equal to c).
Thus we now need to specify not only closeness in the codomain
j f ð xÞ ‘j < ";
but also closeness in the domain. To do this, we introduce another small
positive number (which depends on ") and specify closeness in the domain
by a statement of the form
0 < jx cj < :
Actually it is the requirement
j x c j < that specifies the
closeness in the domain. The
requirement 0 < j x c j is
made simply to ensure that
x¼
6 c.
5.3
Limits of functions – using " and y
185
y
y = f (x)
y = f (x)
{ f (xn)}
{
{
f (xn) close
l
to l
f (x) close l
to l
{xn}
c
c
x
x
x close to c
xn close to c
Problem 2 Let f(x) ¼ 2x þ 3, x 2 R. How small must we choose in
order that:
(a) j f (x) 5j < 0.1, for 0 < jx 1j < ?
(b) j f (x) 5j < 0.01, for 0 < jx 1j < ?
(c) j f (x) 5j < ",
number?
for 0 < jx 1j < , where " is a given positive
In general, the statement ‘f (x) ! ‘ as x ! c’ means the following:
for EACH positive number ", there is a
such that
j f ð xÞ ‘j < ";
CORRESPONDING
positive number
for all x satisfying 0 < jx cj < :
(4)
The following diagram illustrates how the numbers " and are used to
formalise the idea of closeness.
y
y
y = f (x)
y = f (x)
l+ε
{
f (x) close
l
to l
l
l–ε
y
c
The same number may serve
for various different values of
"; but, in general, the value of
depends on the value of ".
Condition (4) means that, for
all x with 0 < jx cj < , the
points (x, f (x)) of the graph
lie in the heavily shaded
rectangle, and thus lie ‘close
to’ (c, ‘).
c–δ c c+δ x
x close to c
This leads us to the following definition of the limit of a function.
Definition Let the function f be defined on a punctured neighbourhood N
of a point c. Then f (x) tends to the limit ł as x tends to c if:
for each positive number ", there is a positive number such that
j f ð xÞ ‘j < ";
for all x satisfying 0 < jx cj < :
(5)
Thus
f ðxÞ is ‘close to’ ‘
whenever
x is ‘close to’ c:
5: Limits and continuity
186
Remarks
1. We assume that is chosen sufficiently small in (5), so that the interval
(c , c þ ) lies within N. We also require that 0 < jx cj in (5), since we
are not concerned with the value of f at c, or even whether f is defined at c.
2. There are similar definitions of limits from the left and limits from the right
at c, where we simply replace ‘0 < jx cj < ’ by ‘c < x < c’ and
‘c < x < c þ ’, respectively; and similar definitions of lim f ð xÞ and
x!1
lim f ð xÞ.
For otherwise the values f (x)
may be undefined.
The existence and value of a
limit is a local property, but
the behaviour of the function
AT the point in question is
irrelevant.
x!1
We now formally state the equivalence of the two definitions of ‘limit of a
function’.
Theorem 1 The ‘" definition’ and the ‘sequential definition’ of the
statement lim f ð xÞ ¼ ‘ are equivalent.
x!c
Proof Let the function f be defined on a punctured neighbourhood N of a
point c.
We have to show that:
for each sequence {xn} in N such that xn ! c, then f ðxn Þ ! ‘
(6)
, for each positive number ", there is a positive number such that
(7)
j f ð xÞ ‘j < "; for all x satisfying 0 < jx cj < :
You may omit this proof at a
first reading.
We shall assume, for
convenience, that
N ¼ ðc r; cÞ [ ðc; c þ r Þ
¼ fx : 0 < jx cj < r g:
First, let us assume that (6) holds. Then, for each sequence {xn} in N and
each positive number ", there is a number X such that
( )
j f ðxn Þ ‘j < "; for all n > X:
Here we simply restate (6).
We will now use a ‘proof by contradiction’. So, suppose that (7) does not
hold. Then there must exist some positive number " for which there is no
positive number such that
Here we write down what is
meant by ‘(7) does not hold’.
j f ð xÞ ‘j < ";
for all x satisfying 0 < jx cj < :
(8)
In particular, we can always assume that < r, so that {x : 0 < jx cj < }
N.
It follows that, for each positive integer n, there is some number xn (say) in
N such that
1
(9)
j f ðxn Þ ‘j "; where 0 < jxn cj < :
n
But, from our assumption (6), the sequence {xn} must satisfy (*). Hence, in
addition, in particular, for the positive number " that we have chosen for (*),
there is a number X such that jf (xn) ‘j < ", for all n > X. But this is inconsistent with (9), so that we have a contradiction. It follows, therefore, that (7)
must hold after all!
Next, let us assume that (7) holds. That is, for each positive number ", there
is a positive number such that
j f ð xÞ ‘j < ";
for all x satisfying 0 < jx cj < :
(10)
Recall that
N ¼ fx : 0 < jx cj < r g:
This is possible since (8) does
not hold for any , and since we
may take as successive choices
of the values 1; 12 ; 13 ; . . ..
For all n > X, f (xn) now has to
satisfy the two inconsistent
inequalities
j f ðxn Þ ‘j " and
j f ðxn Þ ‘j < ":
Here we simply restate (7).
5.3
Limits of functions – using " and 187
Now let {xn} be any sequence in N such that xn ! c. In particular, for the
positive number that we have chosen for (10), there is a number X such
that (0 < )jxn cj < for all n > X. It follows from (10) that jf(xn) ‘j < "
for all n > X.
Since this holds for each positive number ", we deduce that f (xn) ! ‘, so that
&
(6) holds, as required.
Remark
Depending on the particular circumstances, we can therefore use whichever
definition of limit is the more convenient for our purposes.
We can think of the choice of in terms of " in the definition of limit of a
function as a game with players A and B, just as we did with the choice of X in
the definition of convergent sequence.
Thus, player A chooses some value for ", and then player B has to try to find a
value of such that
j f ð xÞ ‘j < "; for all x satisfying 0 < jx cj < :
For example, consider how we might show that f (x) ¼ x2 ! 0 as x ! 0.
Here, for the function f (x) ¼ x2, x 2 R, and c ¼ 0, player B has to find a number
such that
j f ð xÞ 0j ¼ 4x2 < "; for all x satisfying 0 < j xj < :
The game might proceed as follows:
I'll play
δ = 0.1
I'll play
ε = 0.2
B wins
A
B
Here player B wins, since
for 0 < j xj < 0:1; we have that 4x2 0:04 < 0:2:
So player A tries again:
I'll play
δ = 0.1
I'll play
ε = 0.02
A wins
A
B
Here player A wins, since
x ¼ 0:09 satisfies 0 < j xj < 0:1
BUT
4 0:092 ¼ 0:0324 6< 0:02:
Player B lost because he did not choose sufficiently small; so he tries again.
We have 0 < jxn cj since xn
lies in a punctured
neighbourhood of c.
5: Limits and continuity
188
I'll play
δ = 0.01
I'll play
ε = 0.02
B wins
A
B
This time player B wins, since
for 0 < j xj < 0:01; we have that 4x2 0:0004 < 0:02:
But now player B has figured out a winning strategy! He challenges player A to
do his worst!
I'll play
ANY positive
I'll play
δ = 12√ε
ε
B wins
A
B
This is indeed a winning strategy, since
2
1 pffiffiffi
1 pffiffiffi 2
" ; we have that 4x < 4 " ¼ ":
for 0 < j xj <
2
2
In general, if f (x) ! ‘ as x ! c, then such a number always exists, and
player B can always win. On the other hand, if f ðxÞ 6! ‘ as x ! c, then for SOME
choices of " no corresponding positive number exists, and so player A can
always win.
Our strategy for using the ‘" definition’ of lim f ð xÞ in particular cases is
x!c
as follows.
Strategy for using the ‘e d definition’ of limit of a function
1. To show that f (x) ! ‘ as x ! c, find an expression for in terms of " such
that:
for each positive number "
j f ð xÞ ‘j < ";
for all x satisfying 05jx cj5: (11)
2. To show that f ðxÞ 6! ‘ as x ! c, find:
positive number " for which there is NO positive number such that
ONE
j f ð xÞ ‘j < ";
for all x satisfying 0 < jx cj < :
Whatever value of " is chosen
by player A, playerpBffiffiffineed
only specify ¼ 12 ".
Of course the choice of may
depend on c too.
5.3
Limits of functions – using " and Example 1
189
Using the strategy, prove that
(a) limð2x þ 3Þ ¼ 5;
x!1
(c) lim x sin 1x ¼ 0;
x!0
(b) limð2x3 Þ ¼ 0;
x!0
(d) lim sin 1x does not exist.
x!0
Solution
(a) The function f (x) ¼ 2x þ 3 is defined on every punctured neighbourhood of 1.
We must prove that:
for each positive number ", there is a positive number such that
jð2x þ 3Þ 5j < "; for all x satisfying 0 < jx 1j < ;
that is
j2ðx 1Þj < ";
for all x satisfying 0 < jx 1j < :
(12)
We choose ¼ 12 "; then the statement (12) holds, since
We could equally well choose
to be any positive number
less than 12 ", since the
statement (12) would still
hold.
for 0 < jx 1j < 12 ", we have j2ðx 1Þj < ":
Hence, limð2x þ 3Þ ¼ 5.
x!1
(b) The function f (x) ¼ 2x3 is defined on every punctured neighbourhood of 0.
We must prove that:
for each positive number ", there is a positive number such that
3
2x 0 < "; for all x satisfying 0 < jx 0j < ;
that is
3
2x < ";
for all x satisfying 0 < j xj < :
p
ffiffi
ffi
We choose ¼ 12 3 "; then the statement (13) holds, since:
pffiffiffi
for 0 < j xj < 12 3 ", we have that
3
pffiffiffi 3
2x ¼ 2j xj3 < 2 1 3 " ¼ 1 " < ":
2
4
(13)
Notice that many other
choices of would also have
served our purpose.
Hence, limð2x3 Þ ¼ 0.
x!0
(c) The function f ð xÞ ¼ x sin 1x is not defined at 0, but it is defined on every
punctured neighbourhood of 0.
We must prove that:
for each positive number ", there is a positive number such that
x sin 1 0 < "; for all x satisfying 0 < jx 0j < ;
x
that is
x sin 1 < ";
x
for all x satisfying 0 < j xj < :
Now, sin 1x 1, for all non-zero x, so that
(14)
5: Limits and continuity
190
x sin 1 j xj;
x
for all non-zero x:
We therefore
choose
¼ "; then the statement (14) holds.
Hence, lim x sin 1x ¼ 0.
x!0
(d) Suppose that the limit exists; call it ‘. Take " ¼ 12, say. Then there is some
positive number , and points xn and yn of the form
1
1
; where n 2 N;
and yn ¼ xn ¼ 1
2n þ 2 p
2n þ 32 p
such that
0 < jxn j < and
0 < jyn j < and
1
j f ðxn Þ ‘j < ;
2
1
j f ðyn Þ ‘j < :
2
and
:
Hence, for sufficiently large n
j f ðxn Þ f ðyn Þj ¼ jð f ðxn Þ ‘Þ ð f ðyn Þ ‘Þj
1 1
þ ¼ 1:
2 2
But f (xn) f (yn) ¼ 1 (1) ¼ 2, which contradicts the previous inequal
ity. It follows that no such numbers ‘ and exist, so that in fact lim sin 1x
x!0
&
does not exist.
j f ðxn Þ ‘j þ j f ðyn Þ ‘j
<
Remarks
1. In Example 1, parts (a) and (b), we can find an appropriate choice of by
‘working backwards’.
For example, in part (b) we have
3
2x < " , x3 < 1 "
2
ffiffiffi
1p
, j xj < 1 3 " ;
3
2
p
ffiffi
ffi
so we can p
choose
¼ 11 3 ". The choice that we made in Example 1 was to
ffiffiffi
23
1 3
take ¼ 2 ", which was a somewhat smaller (but equally valid) value for .
Both possibilities are valid, since we only require to show that
0 < jx 0j < ) 2x3 0 < ";
rather than
0 < jx 0j < , 2x3 0 < ":
On the other hand, in part (c) it is notpossible to ‘solve for ’ in this way –
that is, to solve the inequality x sin 1x < " to obtain a solution such that
x sin 1 < "; for 0 < j xj < :
x
For, if " is sufficiently small, the solution set of the inequality x sin 1x < " is
not even an interval but a union of disjoint intervals.
Of course any choice for smaller than this particular
value will also serve our
1
purposes – such as 17 " or 625
".
5.3
Limits of functions – using " and 191
2. In Example 1, part (d), the point is that, whatever the value of , there are
always points in any punctured neighbourhood of 0 where f takes the values
1 and 1. Thus there cannot exist numbers ‘ and such that
1
j f ð xÞ ‘j < ; for all x satisfying 0 < j xj < :
2
Problem 3 Using the Strategy (except in part (d)), prove that:
(a) limð5x 2Þ ¼ 13;
(b) limð1 7x3 Þ ¼ 1;
x!3
x!0
(d) lim xþxjxj does not exist.
(c) lim x2 cos x13 ¼ 0;
x!0
x!0
Sometimes it is a little tricky to find an expression for in terms of " such that
j f ð xÞ ‘j < ";
for all x satisfying 0 < jx cj < :
The following example illustrates one approach that is sometimes useful.
Example 2
Use the " definition to prove that lim x2 ¼ 4.
x!2
Solution According to the definition, we must show that for x ‘close to 2’ the
following difference is small
2
x 4 ¼ jx þ 2j jx 2j:
For x close to 2, the term jx þ 2j is approximately 4, so that the product
jx þ 2j jx 2j is approximately equal to 4jx 2j. Certainly, if we restrict x
to the punctured neighbourhood (1, 2) [ (2, 3) in which jx þ 2j < 5, we know that
2
x 4 < 5jx 2j:
So, we proceed as follows.
The function f (x) ¼ x2 is defined on every punctured neighbourhood of 2.
We must prove that
for each positive number ", there is a positive number such that
2
x 4 < "; for all x satisfying 0 < jx 2j < ;
that is
jx þ 2j jx 2j < "; for all x satisfying 0 < jx 2j < :
We choose ¼ min 1; 15 " , so that for 0 < jx 2j < we have
1
jx þ 2j < 5 and jx 2j < " :
5
It follows that
1
jx þ 2j jx 2j < 5 " ¼ "; for all x satisfying 0 < jx 2j < :
5
2
&
Hence lim x ¼ 4:
x!2
Use the " definition to prove that lim x3 ¼ 1.
x!1
Try ¼ min 1, 17 " .
Problem 4
Hint:
The discussion so far has been
our motivation.
The formal proof now starts!
5: Limits and continuity
192
Proofs
Since the definition of limit of a function introduced in this section is equivalent to the earlier sequential definition, the rules for combining limits can also
be proved using the " approach. To give you a flavour of what this
involves, we prove here just one of the Combination Rules.
Prove the Sum Rule:
Example 3
If lim f ð xÞ ¼ ‘ and lim gð xÞ ¼ m; then limð f ð xÞ þ gð xÞÞ ¼ ‘ þ m:
x!c
x!c
x!c
Solution Since the limits for f and g exist, the functions f and g must
be defined on some punctured neighbourhoods (c r1, c) [ (c, c þ r1) and
(c r2, c) [ (c, c þ r2) of c. It follows that the function f þ g is certainly defined
on the punctured neighbourhood (c r, c) [ (c, c þ r), where r ¼ min{r1, r2}.
We want to prove that:
Of course f and g may be
defined on some larger sets
too!
for each positive number ", there is a positive number such that
jðf ð xÞ þ gð xÞÞ ð‘ þ mÞj < ";
for all x satisfying 0 < jx cj < : (15)
We know that, since lim f ð xÞ ¼ ‘, there is a positive number 1 such that
x!c
1
j f ð xÞ ‘j < "; for all x satisfying 0 < jx cj < 1 ;
2
similarly, since lim gð xÞ ¼ m, there is a positive number 2 such that
(16)
x!c
1
jgð xÞ mj < ";
2
for all x satisfying 0 < jx cj < 2 :
(17)
We now choose ¼ min{1, 2}. Then both statements (16) and (17) hold
for all x satisfying 0 < |x c| < , so that
jð f ð xÞ þ gð xÞÞ ð‘ þ mÞj ¼ jð f ð xÞ ‘Þ þ ðgð xÞ mÞj
j f ð xÞ ‘j þ jgð xÞ mj
1
1
< " þ " ¼ ";
2
2
so that the statement (15) holds.
Hence limð f ð xÞ þ gð xÞÞ ¼ ‘ þ m.
&
x!c
We now introduce an analogue of a useful tool for writing out proofs that
you met earlier in your work on sequences, the so-called ‘K" Lemma’.
In order to prove that a function f has a limit as x ! c, using the definition of
limit, you need to prove that:
for each positive number ", there is a positive number such that
j f ð xÞ ‘j < ";
for all x satisfying 0 < jx cj < :
It can be tedious in the process of the proof to make a complicated choice for in
order to end up with a final inequality that says that some expression is ‘<"’. While
this means that we end up with an inequality that shows at once that the desired
result holds, in fact it is not strictly necessary to end up with precisely ‘< "’.
Sub-section 2.2.2.
5.4
Continuity – using " and Lemma
The ‘Ke Lemma’
193
For the function f, suppose that:
for each positive number ", there is a positive number such that
j f ð xÞ ‘j < K";
for all x satisfying 0 < jx cj < ;
where K is a positive real number that does not depend on " or X.
Then f (x) ! ‘ as x ! c.
We omit a proof of the Lemma, as it is essentially the same as the previous
proof of the K" Lemma for sequences. There are analogues of the K" Lemma
for the definition of continuity (in terms of " and ), differentiability and
integrability, but we shall not always mention them explicitly every time an
analogue could be stated.
From time to time we shall use this Lemma in order to avoid arithmetic
complexity in proofs.
Problem 5
Use the " definition to prove the Product Rule:
If lim f ð xÞ ¼ ‘ and lim gð xÞ ¼ m; then lim f ð xÞgð xÞ ¼ ‘m:
x!c
x!c
Loosely speaking, we may
express this result as ‘K" is
just as good as "’ in the
definition of limit.
For example, K might be 2 or p7
or 259, but it could not be 2x
or 1".
For example, in Problem 5
below.
You may omit Problem 5 if
you are short of time.
x!c
Problem 6 Use the " definition to prove that, if lim f ð xÞ ¼ ‘ and
x!c
‘ 6¼ 0, then there exists some punctured neighbourhood of c on which f (x)
has the same sign as ‘.
Hint: Take " ¼ 12 j‘j in the definition of limit.
Continuity – using e and d
5.4
In this section we introduce a definition of continuity of a function that is
equivalent to our earlier definition, but which does not use sequences.
5.4.1
The e–d definition of continuity
Recall our earlier definitions of continuity. The first version was for continuity
of a function at an interior point of an interval.
Sub-section 4.1.1.
Definition A function f defined on an interval I that contains c as an
interior point is continuous at c if
for each sequence {xn} in I such that xn ! c, then f (xn) ! f (c).
We then saw how to extend this definition to include continuity of a function at
an end-point of an interval.
Definition A function f defined on a set S in R that contains a point c is
continuous at c if:
for each sequence {xn} in S such that xn ! c, then f (xn) ! f (c).
Sub-section 4.1.1.
Here c may be an interior
point of an interval, or it may
be an end-point of an interval;
both possibilities are covered
by the phrase ‘each sequence
{xn} in S’.
5: Limits and continuity
194
We start by reformulating the first version of this definition in terms of " and .
So, if we replace ‘ by f (c) in the definition of limit of a function, we
immediately obtain the following definition of continuity of a function at an
interior point of an interval.
Definition A function f defined on an interval I that contains c as an interior point is continuous at c if:
for each positive number ", there is a positive number such that
j f ð xÞ f ðcÞj < ";
for all x satisfying jx cj < :
(1)
Remarks
Sub-section 5.3.2.
Note that the inequality
j f (x) f (c)j < " specifies
closeness in the codomain,
whereas the inequality
jx cj < specifies closeness
in the domain.
y
1. Notice that in (1) we write jx cj < rather than 0 < jx cj < , since in
fact (1) is always satisfied when x ¼ c.
2. We may assume that has been chosen sufficiently small so that (c ,
c þ ) lies in I.
3. It is sufficient in (1) to have
j f ð xÞ f ðcÞj < K";
for all x satisfying jx cj < ;
y = f (x)
f (c) + ε
f (c)
f (c) – ε
c–δ c c+δ x
where K is some positive constant. This fact is often referred to as the ‘K"
Lemma’ for continuity.
4. The condition (1) means that, for all x ‘sufficiently close to’ c, the values of f(x)
lie ‘close to’ f(c). If we choose a smaller number ", then we may need to choose
a smaller number in order to ensure that (1) still holds. But, whatever positive
number " we choose, we can always find a number for which (1) holds.
5. We can describe the definition of continuity in terms of an ‘" game’ in
much the same way as we described the definition of limit. WHATEVER choice
of " player A makes, then player B can ALWAYS choose a value of such that
the required condition (1) holds.
6. For a given value of the positive number ", we may need to choose different
values of for different points c.
7. It follows immediately from our definition of continuity in terms of limits
that, if a function f is defined on an open interval containing c as an interior
point, then
We shall return to this point in
Section 5.5.
f is continuous at c , lim f ð xÞ ¼ f ðcÞ.
x!c
8. We can define one-sided continuity in a similar fashion. Thus a function f is
continuous on the left at c if:
for each positive number ", there is a positive number such that
j f ð xÞ f ðcÞj < ";
for all x satisfying c < x < c;
and is continuous on the right at c if:
for each positive number ", there is a positive number such that
j f ð xÞ f ðcÞj < "; for all x satisfying c < x < c þ :
It follows from these definitions and the previous remark that, if a function f
is defined on an open interval containing c as an interior point, then
f is continuous at c , f is continuous on the left and on the right at c.
We shall not spend much
time looking at one-sided
continuity; however there is a
wide range of results similar
to the results for ‘ordinary’
continuity.
5.4
Continuity – using " and 195
We can use the definitions in the last Remark to extend the definition of
continuity of a function f from points of an open interval to all points of a
general interval, in a natural way. We say that f is continuous on an interval I if
f is continuous at all interior points of I, continuous on the right at the left endpoint of I if it belongs to I, and continuous on the left at the right end-point of
I if it belongs to I.
This leads us to extend our initial definition of continuity, as follows.
Definition A function f defined on an interval I that contains a point c is
continuous at c if:
for each positive number ", there is a positive number such that
j f ð xÞ f ðcÞj < ";
for all x in I satisfying jx cj < :
f is said to be continuous on I if it is continuous at each point c of I.
Since the ‘" definition’ and the ‘sequential definition’ of limits are
equivalent, and the two definitions of continuity are simply reformulations
of those definitions, it is obvious that the two definitions of continuity must
also be equivalent. Depending on the particular circumstances, we can therefore use whichever definition of continuity is the more convenient for our
purposes.
This was Theorem 1 in Subsection 5.3.2.
Theorem 1 The ‘" definition’ and the ‘sequential definition’ of the
statement ‘f is continuous at c’ are equivalent.
Our strategy for using the ‘" definition’ of continuity in particular cases
is also similar to the strategy for using the ‘" definition’ of limits.
Strategy for using the ‘e d definition’ of continuity
1. To show that f is continuous at an interior point c of an interval, find an
expression for in terms of " such that:
for each positive number "
j f ð xÞ f ðcÞj5 ";
for all x satisfying jx cj < : (2)
2. To show that f is discontinuous at an interior point c of an interval, find
ONE positive number " for which there is NO positive number such that
j f ð xÞ f ðcÞj < ";
Example 1
for all x satisfying jx cj < :
(3)
Use the Strategy to prove that f (x) ¼ x2 is continuous at 2.
Solution The function f is defined on R.
We must prove that:
for each positive number ", there is a positive number such that
2
x 4 < "; for all x satisfying jx 2j < ;
that is
jx þ 2jjx 2j < ";
for all x satisfying jx 2j < :
Note that, for x near to 2, x þ 2
is near to 4 and x 2 near to 0.
5: Limits and continuity
196
We choose ¼ min 1; 15 " ; then, for jx 2j < , we have
1
jx þ 2j < 5 and jx 2j < ":
5
Hence
2
x 4 < 5 1 " ¼ "; for all x satisfying jx 2j < :
5
&
It follows that f is continuous at 2.
We could equally well choose
to be any positive number
smaller than ¼ min 1; 15 " ,
since the rest of the argument
would still apply. We choose
< 1 to get a bound on the
term x þ 2, and < 15 " as a
small multiple of " such that
the final product is at most ".
Problem 1 Use the Strategy to prove that f (x) ¼ x3 is continuous at 1.
Hint: Try ¼ min 1; 17 " .
Problem 2
uous at 4.
Use the Strategy to prove that f ð xÞ ¼
pffiffiffi
x, x 0, is contin-
Example 2 Use the Strategy to prove that the following function is discontinuous at 0
x; if x 0;
f ð xÞ ¼
1; if x < 0:
Solution The function f is defined on R.
The graph of f suggests that, while f is ‘well behaved’ to the right of 0, we
should examine the behaviour of f to the left of 0. If we choose " to be any
positive number less than 1, then there will always be points x (with x < 0) as
close as we please to 0, where f (x) is not within a distance " of f(0) ¼ 0.
So, take " ¼ 12, say; and let xn ¼ 1n, for n ¼ 1, 2, . . .. Then, for any positive
number 1
1
xn ¼ 2 ð; 0Þ; for all n > X; where X ¼ ;
n
but
y
1
0
For example, the points of
the sequence {xn}, where
xn ¼ 1n, for all n 1, since
then f (xn) ¼ 1.
We make a suitable choice of
" such that no corresponding exists that satisfies the
requirement (3).
1
f ð0Þ
j f ðxn Þ f ð0Þj ¼ f n
¼10
1
¼ 1 6< ¼ ":
2
Thus, with this choice of ", no value of exists such that
j f ð xÞ f ð0Þj < ";
This was requirement (3).
for all x satisfying jx 0j < :
This proves that f is discontinuous at 0.
&
Problem 3 Use the Strategy to prove that the following function is
discontinuous at 2
8
< x 12 ; if x > 2;
f ð xÞ ¼ 1;
if x ¼ 2;
:
x 1; if x < 2:
Finally, we demonstrate a nice application of the " approach to continuity
in order to obtain a result that will be useful later on.
x
5.4
Continuity – using " and Theorem 2 Let f be continuous at an interior point c of an interval I, and
f ðcÞ 6¼ 0. Then there exists a neighbourhood N ¼ (c r, c þ r) of c on
which:
(a) f(x) has the same sign as f(c), and
197
Here r > 0.
(b) j f ð xÞj > 12 j f ðcÞj.
Proof Since f(c) 6¼ 0, we may take 12 j f ðcÞj as the positive number " in the
definition of continuity at c. It follows that there exists some positive number r
such that
1
(4)
j f ð xÞ f ðcÞj < j f ðcÞj; for all x satisfying jx cj < r:
2
We now rewrite (4) in the following equivalent form
1
1
j f ðcÞj < f ð xÞ f ðcÞ < j f ðcÞj; for all x 2 ðc r; c þ r Þ: (5)
2
2
For convenience, here we use
the symbol r rather than in
the definition of continuity –
this makes no real difference.
Suppose, first, that f (c) > 0. It follows from the left inequality in (5) that, for
x 2 (c r, c þ r), we have 12 f ðcÞ < f ð xÞ f ðcÞ – in other words, that
1
2 f ðcÞ < f ð xÞ. Thus, in this case, both (a) and (b) hold for x 2 (c r, c þ r).
Suppose, next, that f (c) < 0. It follows from the right inequality in (5) that,
for x 2 (c r, c þ r), we have f ð xÞ f ðcÞ < 12 j f ðcÞj ¼ 12 f ðcÞ – in other
words, that f ð xÞ < 12 f ðcÞ. Thus, in this case too, both (a) and (b) hold for
x 2 (c r, c þ r).
&
This completes the proof of the theorem.
5.4.2
The Dirichlet and Riemann functions
We can use our new " approach to continuity to tackle two very strange
functions that were devised in the second half of the nineteenth century. We
start with the Dirichlet function.
Definition
Dirichlet’s function is defined on R by the formula
1; if x is rational;
f ð xÞ ¼
0; if x is irrational:
y
f (xn)
1
The graph of f looks rather like two parallel lines, but each line has
‘infinitely many gaps in it’.
f (yn)
c
Theorem 3
The Dirichlet function is discontinuous at each point of R.
Proof Let c be any point of R.Then, for each
n = 1, 2, . . ., by the Density
Property of R, the open interval c 1n , c þ 1n contains a rational number, xn
say, and an irrational number, yn say. Then xn ! c and yn ! c as n ! 1, but
f ðxn Þ ¼ 1 and f ðyn Þ ¼ 0:
Thus f (x) does not tend to a limit as x tends to c, whether c is rational or
irrational.
&
It follows that f is discontinuous at each point c of R.
xn
yn
c + 1n
x
Recall that the Density
Property of R asserts that,
between any two unequal real
numbers, there exists at least
one rational and one irrational
number – Sub-section 1.1.4.
5: Limits and continuity
198
Our next function possesses even stranger behaviour.
Definition
Riemann’s function is defined on R by the formula
1
f ð xÞ ¼
q;
0;
if x is a rational number
if x is irrational:
p
q ; in
lowest terms; with q > 0;
The expression ‘in lowest
terms’ means that p and q
have no common factor.
It is not clear from the graph of f whether f is continuous at any point of R.
Theorem 4 Riemann’s function is discontinuous at each rational point of
R, and continuous at each irrational point of R.
Proof First, let c ¼ pq be any rational point of R, where q > 0 and pq is in its
lowest terms.
Then,
Property of R, for each n = 1, 2, . . ., the open inter by1 the Density
val
c n , c þ 1n
contains an irrational number xn, for which
f ðxn Þ ¼ 0 6¼ 1q ¼ f ðcÞ.
Hence as n ! 1
In this part of the proof, we
use a sequential approach.
Note that xn 6¼ c.
xn ! c but f ðxn Þ 6! f ðcÞ:
Thus f is discontinuous at any rational point c.
Next, suppose that c is any irrational point of R. We must prove that:
for each positive number ", there is a positive number such that
j f ð xÞ f ðcÞj < "; for all x satisfying jx cj < :
In this part of the proof, we
use an " approach.
(6)
Since c is irrational, f (c) ¼ 0; also f (x) 0 for all x. It follows that condition
(6) becomes
f ð xÞ < ";
for all x satisfying jx cj < :
(7)
Our task is therefore to find a value of such that (7) holds.
Now, let N be any positive integer such that N > 1", and let SN be the set of
rational numbers pq (in lowest terms) in the interval (c 1, c þ 1) for which
0 < q < N – that is
p p
SN ¼
: c 1; 0 < q < N :
q q
Since there are only a finite number of points in SN and c 2
= SN, we can define a
positive number as follows
Thus, in particular, N1 < ".
We choose (c 1, c þ 1)
simply so that we are only
looking at points ‘near to c’.
c2
= SN since c is irrational.
5.4
Continuity – using " and 199
¼ minfjx cj : x 2 SN g:
Thus there are NO rational numbers
(c , c þ ).
It follows that, if jx cj < , then:
EITHER
OR
p
q,
with 0 < q < N, that lie in the interval
x is irrational, so that f(x) ( ¼ 0) < ";
x is rational, so that x ¼ pq, with q N, and f ð xÞð¼ 1qÞ N1 < ".
In either case, f(x) < ".
Hence we have proved that (7) holds, and so that f is continuous at c.
5.4.3
is the distance from c to the
nearest point of SN.
&
Proofs
Since the definition of continuity introduced in this section is equivalent to the
earlier sequential definition, the rules for continuity can also be proved using
the " approach. To give you a flavour of what this involves, we first prove
one of the Combination Rules.
Example 3 (Sum Rule) Let f and g be defined on an open interval I containing the point c. Then, if f and g are continuous at c, so is the sum function f þ g.
Solution The functions f, g and f þ g are certainly defined on some neighbourhood (c r, c þ r) I of c, where r > 0.
We want to prove that:
for each positive number ", there is a positive number such that
jð f ð xÞ þ gð xÞÞ ð f ðcÞ þ gðcÞÞj < "; for all x satisfying jx cj < (8)
We know that, since f is continuous at c, there is a positive number 1 (which
we may assume is r) such that
1
(9)
j f ð xÞ f ðcÞj < "; for all x satisfying jx cj < 1 ;
2
similarly, since g is continuous at c, there is a positive number 2 (which we
may assume is also r) such that
1
(10)
jgð xÞ gðcÞj < "; for all x satisfying jx cj < 2 :
2
We now choose ¼ min{1, 2}. Then both statements (9) and (10) hold for all
x satisfying jx cj < , so that
jð f ð xÞ þ gð xÞÞ ð f ðcÞ þ gðcÞÞj ¼ jð f ð xÞ f ðcÞÞ þ ðgð xÞ gðcÞÞj
j f ð xÞ f ðcÞj þ jgð xÞ gðcÞj
1
1
< " þ " ¼ ";
2
2
so that the statement (8) holds.
Hence f þ g is continuous at c.
&
Problem 4 Prove that if f is defined on an open interval I containing
the point c at which it is continuous, and f (c) 6¼ 0, then 1f is defined on
some neighbourhood of c and is also continuous at c.
Hint: Use the result of Theorem 2 in Sub-section 5.4.1.
You might like to compare
this solution with that of
Example 3 in Subsection 5.3.2.
5: Limits and continuity
200
Finally, we give a proof of the Composition Rule for continuity; this is a
particularly pleasing example of the " approach.
Example 4 (Composition Rule) Prove that if f is continuous at c and g is
continuous at f (c), then g f is continuous at c.
Solution
We must prove that:
for each positive number ", there is a positive number such that
jgð f ð xÞÞ gð f ðcÞÞj < ";
for all x satisfying jx cj < :
We do not bother to mention
appropriate neighbourhoods
of c and f (c) on which f and g
are defined, respectively, just
to simplify the statement of
the Rule.
(11)
We know that, since g is continuous at f(c), there is a positive number 1
such that:
jgð f ð xÞÞ gð f ðcÞÞj < ";
for all f ðxÞ satisfying j f ð xÞ f ðcÞj < 1 :
(12)
g
f (c) – δ1
f (c) f (x)
f (c) + δ1
g(f (c)) – ε
g( f (c)) g( f (x)) g( f (c)) + ε
Also, since f is continuous at c, there is a positive number such that
j f ð x Þ f ð c Þ j < 1 ;
for all x satisfying jx cj < :
(13)
f
c–δ
c
x c+δ
f (c) – δ1
f (c) f (x)
f (c) + δ1
Combining (12) and (13), we deduce that:
for all x satisfying jx cj < , then j f ð xÞ f ðcÞj < 1 ,
By (13).
so that
jgð f ð xÞÞ gð f ðcÞÞj < ":
Hence g f is continuous at c.
5.5
By (12).
&
Uniform continuity
We start by reminding you of the " definition of continuity at points of an
interval I.
Definition A function f defined on an interval I that contains a point c is
continuous at c if:
for each positive number ", there is a positive number such that
j f ð xÞ f ðcÞj < ";
for all x in I satisfying jx cj < :
f is said to be continuous on I if it is continuous at each point c of I.
Here the choice of depends,
in general, on both " and c.
5.5
Uniform continuity
201
Earlier we pointed out that, for a given value of the positive number ", we may
need to choose different values of for different points c. This suggests the
following related definition.
Definition
on I if:
Sub-section 5.4.1, Remark 6.
A function f defined on an interval I is uniformly continuous
for each positive number ", there is a positive number such that
(1)
j f ð xÞ f ðcÞj < "; for all x and c in I satisfying jx cj < :
Here the choice of depends
on ", the same choice
whatever c may be.
ONLY
Since the definition of uniform continuity is more restrictive than that of
continuity, it follows that uniform continuity implies continuity.
Theorem 1 If a function f is uniformly continuous on an interval I, then it
is continuous on I.
However the converse result is not true in general.
Example 1 Prove that the function f ð xÞ ¼ 1x , 0 < x 1; is not uniformly
continuous on (0, 1].
Solution
Since f is a rational function, it
is a basic continuous function,
and so is continuous on (0, 1].
To prove that f is NOT uniformly continuous on (0,1], we need to
find ONE positive number " for which there is NO positive number such that
j f ð xÞ f ðcÞj < "; for all x and c in ð0; 1 satisfyingjx cj5:
(2)
We will take as our choice of " the number 1, and prove that there is
corresponding positive number such that
j f ð xÞ f ðcÞj < 1; for all x and c in ð0;1 satisfying jx cj < :
For the statement (2) asserts
that statement (1) is not valid.
NO
(3)
Suppose on the contrary that some number did exist such that (3) was valid.
We will show that this assumption leads us to a contradiction.
Choose a positive integer N with
1
< :
N
Now each point of (0,1] belongs to at least one interval of the form
1 3
k kþ2
2N 2 2N
2N 1
0 2
;
;
;
;1 ;
;
; . . .;
; . . .;
;
;
2N 2N
2N 2N
2N 2N
2N
2N
2N
(4)
and, in addition, all points x and c in any given such interval satisfy the
inequality
2
1
¼
< ;
jx cj <
2N
N
Hence, by (3)
all points x and c in ð0; 1 in any interval in ð4Þ satisfy j f ð xÞ f ðcÞj < 1:
(5)
Now, from the definition of f, we have
2N
2N 1
f ð xÞ <
¼ K; say; for all x in
;1 :
2N 1
2N
This is what we now set out
to do.
For, successive intervals in (4)
overlap.
In particular, f (x) < f (c) þ 1.
For, f is strictly
decreasing
2N
and f 2N1
.
¼ 2N1
2N
5: Limits and continuity
202
Then, since the intervals 2N2N1 ; 1 and 2N2N2 ; 22NN overlap, there is a point c in
2N 2 2N f (c) < K. It follows from (5) that in the second last
2N ; 2N for which
2N 2 2N
interval 2N ; 2N in (4) we have
2N 2 2N
;
:
f ð xÞ < K þ 1; for all x in
2N
2N
Working backwards through the 2N intervals, we can check that
0
2
;
f ð xÞ < K þ ð2N 1Þ; for all x in
:
2N 2N
For f is strictly decreasing.
It follows then, that in fact f(x) < K þ (2N 1), for all x in (0,1]. In other words,
f is bounded in (0,1]. But this is not the case, since f (x) ! 1 as x ! 0þ.
&
We have thus reached the desired contradiction.
Problem 1 Let f be the function f(x) ¼ x2, x 2 [2, 3]. For any given
positive number ", find a formula for in terms of " such that (1) holds.
This proves directly from the definition that f is uniformly continuous on
[2, 3].
Hint:
Use the fact that x2 c2 ¼ (x þ c)(x c).
Notice, though, that the function in Problem 1 is a continuous function on a
closed interval, and recall that we saw earlier that continuous functions on
closed intervals have ‘nice properties’. It turns out that closed intervals are
important too for uniform continuity.
Theorem 2 If a function f is continuous on a closed interval [a, b], then it
is uniformly continuous on [a , b].
Section 4.2.
Example 1 shows that, in
general, f may NOT be
uniformly continuous if its
interval of continuity is a halfclosed interval.
Proof of Theorem 2 We use the Bolzano–Weierstrass Theorem, and a proof
by contradiction.
So, suppose that, in fact, f is not uniformly continuous on [a, b]. Then for
some choice of ", there is NO corresponding positive number such that
You may omit this proof on a
first reading.
j f ð xÞ f ð yÞj < ";
(6)
We use y rather than c in (6),
since it fits in better with the
notation that we will use later
in the proof.
(7)
The choice of x and y will
depend, in general, on the
choice of .
for all x and y in ½a, b satisfying jx yj < :
We can reformulate this statement (6) in the following convenient way:
for any positive number , there are two points x, y in [a, b] such that
j f ð xÞ f ð yÞj "
and jx yj < :
1
n, n
¼ 1, 2, . . ., in turn, we can then define
By applying (7) for the choices n ¼
two sequences {xn} and {yn} in [a,b] such that
1
(8)
j f ðxn Þ f ðyn Þj " and jxn yn j < ; for n 2 N:
n
Now, the sequence {xn} is bounded, so that by the Bolzano–Weierstrass
Theorem it contains a subsequence fxnk g that converges to some point c of
[a, b] as k ! 1. Then, since jxnk ynk j < n1k , it follows that the corresponding
subsequence fynk g must also converge to c as k ! 1.
Since f is continuous at c, we must have
f ðxnk Þ ! f ðcÞ and f ðynk Þ ! f ðcÞ; as k ! 1:
(9)
" is now fixed from here on in
this proof.
Sub-section 2.5.1, Theorem 3.
From (8).
5.6
Exercises
203
Now, it follows from (8) that
j f ðxnk Þ f ðynk Þj ";
for each k 2 N:
(10)
So, if we now let k ! 1 in (10) and use the limits in (9), we deduce that
0 ";
Here we use the Limit
Inequality Rule, Theorem 6
of Sub-section 5.1.3.
which is absurd.
This contradiction shows that (7) and so (6) cannot be valid. It follows
&
that f must be uniformly continuous on [a, b] after all.
We shall use Theorem 2 later, to prove that a function continuous on a closed
interval [a, b] is integrable on [a, b].
Theorem 2 is one of the important theorems in Analysis beyond the scope of
this book.
5.6
Exercises
Section 5.1
1. Determine whether the following limits exist:
3
3
1
1
(a) lim xx1
;
(b) lim jxx1
(c) lim esin x .
j;
x!1
x!1
x!0
2. Determine
limits:
the following
x
(a) lim sin x þ e x1 ;
(b) lim
ex 1
;
x!0 sin x
x!0
(c) lim
x!0
ejxj 1
j xj ;
3
1
(d) limþ jxx1
j.
x!1
3. Write out the proof of
section 5.1.3.
4. For the functions
8
< 0;
f ð xÞ ¼ 1;
:
2;
Theorem 4 (the Composition Rule) in Sub-
x ¼ 0;
x ¼ 1; and
x 6¼0; 1;
gð xÞ ¼
0;
1 þ j xj;
x ¼ 0;
x 6¼ 0;
determine f ðgð0ÞÞ; f lim gð xÞ and lim f ðgðxÞÞ.
x!0
x!0
Section 5.2
1. Prove that
(a) x14 ! 1 as x ! 0;
(c) expðex xÞ ! 1 as x ! 1;
(e) x þ sin x ! 1 as x ! 1;
(b) cot x ! 1 as x ! 0þ ;
(d) loge x ! 1 as x ! 0þ ;
(f) xx ! 1 as x ! 1.
2. Let pð xÞ ¼ xn þ an1 xn1 þ þ a1 x þ a0 , where a0 ; a1 ; . . .; an1 2 R.
Prove that pð xÞ ! 1 as x ! 1.
Hint: Write pð xÞ ¼ xn 1 þ r 1x , for a suitable polynomial r.
3. For a > 0 and n 2 Z, prove that:
(b) eax xn ! 0 as x ! 1.
(a) eax xn ! 1 as x ! 1;
In Theorem 3, Subsection 7.2.2.
5: Limits and continuity
204
4. For a < 0 and n 2 Z, prove that:
(a) eax xn ! 0 as x ! 1;
1; as x ! 1;
(b) eax xn !
1; as x ! 1;
if n is odd;
if n is even:
Section 5.3
1. Use the strategy based on the " definition of limit to prove the following
statements:
pffiffiffiffiffiffiffiffiffiffiffiffiffi
(a) limð3x 4Þ ¼ 5;
(b) limþ x2 4 ¼ 0.
x!3
x!2
2. Use the strategy based on the " definition of limit to prove the following
statements:
x
(a) lim sin
(b) lim cos 1x does not exist.
j xj does not exist;
x!0
x!0
3. Use the strategy based on the " definition of limit to prove that, if
lim f ð xÞ ¼ ‘, where ‘ 6¼ 0, then lim f ð1xÞ exists and equals 1‘ .
x!c
x!c
Section 5.4
1. Use the strategy based on the " definition of continuity to prove that:
(a) f ð xÞ ¼ x5 is continuous at 0;
(b) f ð xÞ ¼ 1x is continuous at 2.
2. Use the strategy based on the " definition of continuity to prove that the
function
sin 1x ; if x 6¼ 0;
f ð xÞ ¼
0;
if x ¼ 0;
is discontinuous at 0.
3. Use the " definition of continuity to prove that, if f and g are continuous
at c, then so is the product fg.
Section 5.5
1. A function f is defined on a bounded interval I, on which it satisfies the
following inequality for some given number K 2 R:
j f ð xÞ f ð yÞj K jx yj; for all x; y in I:
Prove that f is (a) bounded on I, and (b) uniformly continuous on I.
2. Give an example of a function f that is continuous and bounded on R but is
not uniformly continuous on R, and verify your assertions.
3. Prove that, if a function f is continuous on a half-closed interval (a, b], then
it is uniformly continuous on (a, b] if and only if limþ f ð xÞ exists (and is
x!a
finite).
6
Differentiation
The family of all functions is so large that there is really no possibility of
finding many interesting properties that they all possess. In the last two
chapters we concentrated our attention on the class of all continuous
functions, and we found that continuous functions share some important
properties – for example, they satisfy the Intermediate Value Theorem, the
Extreme Values Theorem and the Boundedness Theorem. However, many
of the most interesting and powerful properties of functions are obtained
only when we further restrict our attention to the class of all differentiable
functions.
You will have already met the idea of differentiating a given function f; that
is, finding the slope of the tangent to the graph y ¼ f(x) at those points of the
graph where a tangent exists. The slope of the tangent at the point (c, f(c)) is
called the derivative of f at c, and is written as f 0 (c). But when does a function
have a derivative? Geometrically, the answer is: whenever the slope of the
chord through the point (c, f(c)) and an arbitrary point (x, f(x)) of the graph
approaches a limit as x ! c. In this chapter we investigate which functions are
differentiable, and we discuss some of the important properties that all differentiable functions possess.
In Section 6.1 we give a strategy for determining whether a given function
f is differentiable at a given point c. In particular, we prove that, if f is
differentiable at c, then it is continuous at c; whereas a function f may be
continuous at c, but not differentiable at c. We also consider functions that
possess higher derivatives; that is, functions which can be differentiated more
than once.
In Section 6.2 we obtain various standard derivatives and rules for differentiation, including the Inverse Function Rule.
In Sections 6.3 and 6.4 we study the properties of functions that are differentiable on an interval, and establish some powerful results about derivatives
which are easy to describe geometrically.
In Section 6.5 we meet an important and useful result, called l’Hôpital’s
Rule, which enables us to find limits of the form
In fact there exist functions
that are continuous
everywhere on R, but
differentiable nowhere on R.
This discovery by Karl
Weierstrass in 1872 caused a
sensation in Analysis.
f ð xÞ
;
x!c gð xÞ
lim
in some of the awkward cases when f (c) ¼ g(c) ¼ 0.
In Section 6.6 we construct the Blancmange function, a function that is
continuous everywhere on R, but differentiable nowhere on R. It is related to
certain types of fractals.
In this case, the Quotient Rule
for limits of functions fails.
205
6: Differentiation
206
6.1
6.1.1
Differentiable functions
What is differentiability?
Differentiability arises from the geometric concept of the tangent to a graph.
The tangent to the graph y ¼ f (x) at the point (c, f (c)) is the straight line through
the point (c, f (c)) whose direction is the limiting direction of chords joining the
points (c, f (c)) and (x, f (x)) on the graph, as x ! c.
The following three examples illustrate some of the possibilities that can
occur when we try to find tangents in particular instances.
The function f (x) ¼ x2, x 2 R, is continuous on R, and its graph has a tangent
at each point; for example, the line y ¼ 2x 1 is the tangent to the graph at the
point (1, 1).
On the other hand, although the function g(x) ¼ jx 1j, x 2 R, is continuous
on R, its graph does not have a tangent at the point (1, 0); no line through the
point (1, 0) is a tangent to the graph.
Finally, the signum function
(
1; x < 0,
x ¼ 0,
kð xÞ ¼ 0;
1;
x > 0,
is discontinuous at 0, and no line through the point (0, 0) is a tangent to the
graph.
We now make these ideas more precise, using the concept of limit to pin
down what we meant above by ‘limiting direction’. We define the slope of the
graph at (c, f (c)) to be the limit, as x tends to c, of the slope of the chord through
the points (c, f (c)) and (x, f (x)). The slope of this chord is
f ð xÞ f ðcÞ
;
xc
and this expression is called the difference quotient for f at c. Thus the slope
of the graph of f at (c, f (c)) is
f ð xÞ f ðcÞ
;
(1)
x!c
xc
provided that this limit exists.
Sometimes it is more convenient to use an equivalent form of the difference quotient, particularly when we are examining a specific function for
lim
6.1
Differentiable functions
207
differentiability at a specific point. If we replace x by c þ h, then ‘x ! c’ in (1)
is equivalent to ‘h ! 0’. Thus we can rewrite the difference quotient for f at c as
f ð c þ hÞ f ð c Þ
;
QðhÞ ¼
h
and the slope of the graph of f at (c, f (c)) is then
lim QðhÞ;
(2)
h!0
provided that this limit exists.
We say that f is differentiable at c if the graph y ¼ f (x) has a tangent at the
point (c, f (c)), and that the derivative of f at c is the limit of the difference
quotient given by expression (1) or expression (2). To formalise this concept,
we need to ensure that f is defined near the point c, and so we assume that c
belongs to some open interval I in the domain of f.
Definitions Let f be defined on an open interval I, and c 2 I. Then the
derivative of f at c is
f ð xÞ f ðcÞ
lim
or lim QðhÞ;
x!c
h!0
xc
provided that this limit exists. In this case, we say that f is differentiable at c.
The derivative of f at c is denoted by f 0 (c), and the function f 0: x 7! f 0 ð xÞ
is called the derived function. The operation of obtaining f 0 (x) from f (x)
is called differentiation.
Note that the difference
quotient depends on the
choice of c; for difference
choices of c, the difference
quotient is generally different.
Sometimes f 0 is denoted by Df
and f 0 (x) by Df (x).
For future reference, we reformulate the definition in terms of " and as
follows.
Definition Let f be defined on an open interval I, and c 2 I. Then f is
differentiable at c with derivative f 0 (c) if:
for each positive
number ", there is a positive number such that
f ðxÞf ðcÞ
xc f 0 ðcÞ < ", for all x satisfying 0 < jx cj < .
Remarks
1. In ‘Leibniz notation’, f 0 (x) is written as dy
dx, where y ¼ f (x). As we shall see
later, this notation is sometimes useful; however, it is important to recall
that the symbol dy
dx is purely notation and does NOT mean some quantity dy
‘divided by’ another quantity dx.
2. The existence of the derivative f 0 (c) is not quite equivalent to the existence
Þf ðcÞ
of a tangent to the graph y ¼ f (x) at the point (c, f (c)). For, if lim f ðxxc
x!c
exists (and so is some real number), then the graph has a tangent at the point
(c, f (c)), and the slope of the tangent is the value of this limit. On the other
hand, the graph may have a vertical tangent at the point (c, f (c)). In this
Þf ðcÞ
Þf ðcÞ
! 1 or 1 as x ! c, so that lim f ðxxc
does not exist; thus
case, f ðxxc
x!c
f is not differentiable at c.
3. Since the concept of a derivative is defined in terms of a limit, we shall use
many results obtained for limits of functions to prove analogous results for
derivatives.
This is a simple reformulation,
with nothing else taking place.
We shall use the definition
occasionally in this form,
especially when proving
general results as distinct from
looking at specific functions
and specific points.
For instance, in the statement
of the Composition Rule in
Sub-section 6.2.2.
6: Differentiation
208
Example 1 Prove that the function f (x) ¼ x3, x 2 R, is differentiable at any
point c, and determine f 0 (c).
Solution
At the point c
QðhÞ ¼
f ðc þ hÞ f ðcÞ ðc þ hÞ3 c3
¼
h
h
3c2 h þ 3ch2 þ h3
¼
h
¼ 3c2 þ 3ch þ h2 :
Recall that h 6¼ 0.
Thus, Q(h) ! 3c2 as h ! 0. It follows that f is differentiable at c, and that
&
f (c) ¼ 3c2.
0
Thus, the derived function of
f is f 0 (x) ¼ 3x2, x 2 R.
For comparison, we now prove the same result using the " definition of
differentiability; for simplicity, though, we shall assume that c ¼ 2 and we shall
simply check that f 0 (2) ¼ 12. We think that you will see why the Q(h) method is
often preferred!
Solution We have to show that for each positive number ", there is a positive
number such that
3
x 8
< "; for all x satisfying 0 < jx 2j < ;
12
x2
that is
2
x þ 2x 8 < ";
for all x satisfying 0 < jx 2j < :
Now, x2 þ 2x 8 ¼ (x þ 4)(x 2); so, for 0 < jx 2j < 1, we have x 2 (1, 3)
and jx þ 4j < 7.
Next, choose ¼ min 1; 17 " . With this choice of , it follows that, for all x
satisfying 0 < jx 2j < , we have
2
x þ 2x 8 ¼ jx þ 4j jx 2j
1
< 7 " ¼ ":
7
It follows that f is differentiable at 2, with derivative f 0 (2) ¼ 12.
&
To prove that a function is not differentiable at a point, we use the strategy for
limits that applies in these situations.
Strategy
Here we use the fact that
x3 8 ¼ ðx 2Þðx2 þ 2x þ 4Þ:
To show that lim QðhÞ does not exist:
We arrange for x 2 (1, 3), in
order to concentrate simply on
values of x near to 2. The
bound 7 for jx þ 4j is simply
used in order to keep some
given bound (not necessarily a
small bound) for this term.
It is now clear why we made
that particular choice for ; it
arose from trying various
values and then adjusting the
choice appropriately in order
to end up with a neat ‘"’.
Alternatively, we could have
used a different value for in
terms of ", and the K " Lemma.
h!0
1. Show that there is no punctured neighbourhood of c on which Q is defined;
OR
2. Find
non-zero null sequences {hn} and h0n such that {Q(hn)} and
two
Q h0n have different limits;
OR
3. Find a null sequence {hn} such that Q(hn) ! 1 or Q(hn) ! 1.
Example 2 Prove that the modulus function f(x) ¼ jxj, x 2 R, is not differentiable at 0.
For convenience, we rephrase
this strategy in terms of
lim QðhÞ rather than lim f ð xÞ.
h!0
x!c
6.1
Differentiable functions
209
Solution The graph suggests that the slopes of chords joining the origin to
points on the graph of f to the left and to the right do not tend to the same limit
as these points approach the origin. This suggests that we investigate whether
case 2 of the above strategy may be useful.
At the point 0
f ð0 þ hÞ f ð0Þ jhj j0j
¼
h
h
jhj
¼ :
h
0
Now let {hn} and hn be two null sequences, where hn ¼ 1n and h0n ¼ 1n.
Then
1
1
Q ð hn Þ ¼ Q
¼ n1
n
n
This is motivation for the
approach to the solution rather
than part of the solution itself.
This is the start of the
solution.
Q ð hÞ ¼
y
y = ⏐x⏐
–1
n
0
1
n
x
1
¼ n1 ! 1
as n ! 1;
n
and
1
0
1
Q hn ¼ Q ¼ n1
n
n
¼
1
n
1n
! 1
as n ! 1:
It follows that f is not differentiable at 0.
&
Problem 1
(a) Prove that the function f (x) ¼ xn, x 2 R, n 2 N, is differentiable at
any point c, and determine f 0 (c).
(b) Prove that the constant function on R is differentiable at any point c,
with derivative zero.
In particular, the derivative of
the identity function f (x) ¼ x,
x 2 R, is the constant function
f 0 (x) ¼ 1.
Problem 2 Prove that the function f ð xÞ ¼ 1x ; x 2 R f0g, is differentiable at any point c 6¼ 0, and determine f 0 (c).
Problem 3 Use the " definition of differentiability to prove that the
function f (x) ¼ x4 is differentiable at 1, with derivative f 0 (1) ¼ 4.
Problem 4 Prove that the following functions f are not differentiable
at the given point c:
1
(a) f ð xÞ ¼ j xj2 ; x 2 R, c ¼ 0;
(b) f(x) ¼ [x], x 2 R, c ¼ 1.
Looking back at Example 2, it looks as though chords joining the origin to
points (h, f (h)) have slopes that tend to a limit 1 if h ! 0þ, whereas they have
slopes that tend to a limit 1 if h ! 0. This suggests the concept of one-sided
derivatives that will be useful in our work later on.
Here [x] denotes the integer
part of x.
6: Differentiation
210
y
slope
fL′(c)
y
y = f (x)
y = f (x)
c
x
slope
fR′(c)
c
x
Definitions Let f be defined on an interval I, and c 2 I. Then the left
derivative of f at c is
f ð xÞ f ðcÞ
or fL0 ðcÞ ¼ lim QðhÞ;
x!c
h!0
xc
provided that this limit exists. In this case, we say that f is differentiable on
the left at c.
Similarly, the right derivative of f at c is
f ð x Þ f ðc Þ
or fR0 ðcÞ ¼ limþ QðhÞ;
fR0 ðcÞ ¼ limþ
x!c
h!0
xc
provided that this limit exists. In this case, we say that f is differentiable on
the right at c.
A function f whose domain contains an interval I is differentiable on I if
it is differentiable at each interior point of I, differentiable on the right at the
left end-point of I (if this belongs to I ) and differentiable on the left at the
right end-point of I (if this belongs to I ).
fL0 ðcÞ ¼ lim
All these definitions are
analogous to definitions for
continuity that you met in
Sub-section 4.1.1.
With this notation, the function f in Example 2 is differentiable on the left at
0 and fL0 ð0Þ ¼ 1; it is also differentiable on the right at 0, and fR0 ð0Þ ¼ 1.
The connection between the definitions of differentiability and of one-sided
differentiability is rather obvious.
Theorem 1 A function f whose domain contains an interval I that contains
c as an interior point is differentiable at c if and only if f is both differentiable on the left at c and differentiable on the right at c AND
fL0 ðcÞ ¼ fR0 ðcÞ:
Example 3
We omit a proof of this
straight-forward result.
The common value is
simply f 0 (c).
Determine whether the function
x þ x2 ; 1 x < 0,
f ð xÞ ¼
sin x;
0 x 2p,
y
is differentiable at the points c ¼ 1, 0 and 2p, and determine the corresponding derivatives when they exist.
Solution We investigate the behaviour of the difference quotient Q(h) at
each point in turn.
At 1, the function is not defined to the left of 1; and, for 0 < h < 1,
we have
QðhÞ ¼
f ð1 þ hÞ f ð1Þ
h
y = f (x)
–1
π
2π
x
6.1
Differentiable functions
211
n
o n
o
ð1 þ hÞ þ ð1 þ hÞ2 ð1Þ þ ð1Þ2
¼
h
h þ h2
¼
h
¼ 1 þ h ! 1 as h ! 0þ :
It follows that f is differentiable on the right at 1, and fR0 ð1Þ ¼ 1.
At 0, the function is defined on either side of 0, but by a different formula;
we therefore examine each side separately.
At 0, for 0 < h < 2p, we have
f ðhÞ f ð0Þ sin h sin 0
¼
Q ð hÞ ¼
h
h
sin h
! 1 as h ! 0þ :
¼
h
Also at 0, for 1 < h < 0, we have
h þ h2 fsin 0g
f ð hÞ f ð 0Þ
Q ð hÞ ¼
¼
h
h
h þ h2
¼
h
¼ 1 þ h ! 1 as h ! 0 :
It follows that f is differentiable on the right at 0 and fR0 ð0Þ ¼ 1, and that f
is differentiable on the left at 0 and fL0 ð0Þ ¼ 1. Since the left and right derivatives at 0 are equal, it follows that f is differentiable at 0 and f 0 ð0Þ ¼ 1.
Finally, at 2p, the function is not defined to the right of 2p; and,
for 2p < h < 0, we have
Q ð hÞ ¼
f ð2p þ hÞ f ð2pÞ sinð2p þ hÞ sin 2p
¼
h
h
sin h
! 1 as h ! 0 :
¼
h
It follows that f is differentiable on the left at 2p, and fL0 ð2pÞ ¼ 1.
Problem 5
&
Determine whether the function
8
x2 ; 2 x <0,
>
>
< 4
x ;
0 x < 1,
f ð xÞ ¼
3
>
x
;
1 x 2,
>
:
0;
x > 2,
is differentiable at the points c ¼ 2, 0, 1 and 2, and determine the
corresponding derivatives when they exist.
Hint:
Sketch the graph y ¼ f(x) first.
Remarks
Just as with continuity, the definition of differentiability involves a function
f defined on a set in R, the domain A (say), that maps A to another set in R,
the codomain B (say).
These remarks are analogous
to similar remarks for
continuity in
Sub-section 4.1.1.
6: Differentiation
212
1. Let f and g be functions defined on open intervals I and J, respectively,
where I J; and let f (x) ¼ g(x) on J. Technically g is a different function from f. However, if f is differentiable at an interior point c of J, it
is a simple matter of some definition checking to verify that g too is
differentiable at c. Similarly, if f is non-differentiable at c, g too is nondifferentiable at c.
2. The underlying point here is that differentiability at a point is a local
property. It is only the behaviour of the function near that point that
determines whether it is differentiable at the point.
6.1.2
Differentiability and continuity
All the functions that we have met so far that are differentiable at a particular
point c are also continuous at that point. In fact, this property holds in general.
differentiable ) continuous
Theorem 2 Let f be defined on an open interval I, and c 2 I. If f is differentiable at c, then f is also continuous at c.
There is a similar result for
one-sided derivatives.
Proof
If f is differentiable at c, then there is some number f 0 (c) such that
fð xÞ fðcÞ
¼ f 0 ðcÞ:
x!c
xc
It follows that
lim
f ð xÞ f ðcÞ
ðx cÞ
limf f ð xÞ f ðcÞg ¼ lim
x!c
x!c
xc
n
o
f ð xÞ f ðcÞ
¼ lim
limðx cÞ
x!c
x!c
xc
By the Product Rule for
Limits, Sub-section 5.1.3.
¼ f 0 ðcÞ 0 ¼ 0:
Hence, by the Sum and Multiple Rules for limits
lim f ð xÞ ¼ f ðcÞ:
Sub-section 5.1.3
x!c
Thus f is continuous at c.
&
In fact this gives us a useful test for non-differentiability.
Corollary 1
If f is discontinuous at c, then f is not differentiable at c.
For example, the signum function
(
1; x < 0,
x ¼ 0,
f ð xÞ ¼ 0;
1;
x > 0,
is discontinuous at c, and so cannot be differentiable at c.
It is often worth checking whether a function is even continuous at a point
before setting out on a complicated investigation of difference quotients to
determine whether it is differentiable at that point.
Problem 5, Sub-section 4.1.1.
6.1
Differentiable functions
Problem 6
213
Prove that the function
sin 1x ; x 6¼ 0,
f ð xÞ ¼
0;
x ¼ 0,
is not differentiable at 0.
Example 4
Show that the function
x sin 1x ; x 6¼ 0,
f ð xÞ ¼
0;
x ¼ 0,
is continuous at 0, but not differentiable at 0.
Solution
At 0
We proved earlier that f is continuous at 0.
Sub-section 4.1.2,
Problem 8 (a).
f ð0 þ hÞ f ð0Þ h sin 1h 0
¼
Q ð hÞ ¼
h
h
1
¼ sin
:
h
o
1
n
and x0n ¼ 2nþ1 1 p ;
Thus, if we define two null sequences fxn g ¼ np
ð 2Þ
n 2 N, we have
0
1
1
Qðxn Þ ¼ sinðnpÞ ¼ 0 and Q xn ¼ sin 2n þ p ¼ sin p ¼ 1;
2
2
0 so that {Q (xn)} and Q xn tend to different limits as n ! 1. It follows that
&
f cannot be differentiable at 0.
Problem 7
y
y = x sin
x
y
Prove that the function
x2 sin 1x ; x 6¼ 0,
f ð xÞ ¼
0;
x ¼ 0,
y = x2 sin
1
x
x
is differentiable at 0. Given that, for x 6¼ 0; f 0 ð xÞ ¼ 2x sin 1x cos 1x,
is f 0 continuous at 0?
6.1.3
1
x
The sine, cosine and exponential functions
In order to study the differentiability of each of the sine, cosine and exponential
functions, we need to use the following three standard limits.
Theorem 3
Three standard limits
(a) lim sinx x ¼ 1;
x!0
x
(b) lim 1cos
¼ 0;
x
x!0
x
(c) lim e x1 ¼ 1:
x!0
Proof The limits (a) and (c) have been verified earlier, so we have only to
verify (b) now.
Using the half-angle formula for cosine, we have
2 sin2 12 x
1 cos x
¼ lim
lim
x!0
x!0
x
x
(a) was proved in Subsection 5.1.1; (c) in
Sub-section 5.1.4, Problem 8.
For cos x ¼ 1 2 sin2
1 2x .
6: Differentiation
214
!
sin 12 x
1
¼ lim
sin x
1
x!0
2
x
2
1 !
sin 2 x
1
¼ lim
lim sin x
1
x!0
x!0
2
2x
¼ 1 0 ¼ 0:
By the Product Rule for
limits.
&
Since sine is continuous at 0.
With this tool, we can now tackle the various functions.
Theorem 4
The function f (x) ¼ sin x is differentiable on R, and f 0 (x) ¼ cos x.
Let c be any point in R. At c, the difference quotient for f is
Proof
sinðc þ hÞ sin c
h
sin c cos h þ cos c sin h sin c
¼
h
sin h
1 cos h
sin c ;
¼ cos c h
h
QðhÞ ¼
so that
sin h
1 cos h
sin c lim
h!0
h
h
¼ cos c 1 sin c 0
lim QðhÞ ¼ cos c lim
h!0
h!0
¼ cos c:
It follows that f is differentiable at c, and f 0 (c) ¼ cos c.
&
Problem 8 Prove that the function f (x) ¼ cos x is differentiable on R,
and f 0 (x) ¼ sin x.
Finally we find the derivative of the exponential function. This is an
extremely important result, as the function f (x) ¼ lex, l an arbitrary constant,
is the only function f that satisfies the differential equation f 0 (x) ¼ f (x) on R.
Theorem 5
Proof
The function f (x) ¼ ex is differentiable on R, and f 0 (x) ¼ ex.
Let c be any point in R. At c, the difference quotient for f is
ecþh ec
QðhÞ ¼
h
eh 1
c
;
¼e h
so that
eh 1
h!0
h
c
¼e 1
¼ ec :
lim QðhÞ ¼ ec lim
h!0
It follows that f is differentiable at c, and f 0 (c) ¼ ec.
&
That is, f is its own derived
function.
6.1
Differentiable functions
6.1.4
215
Higher-order derivatives
In the previous sub-section, we saw that sin0 ¼ cos and cos0 ¼ sin. An exactly
similar argument shows that (sin)0 ¼ cos and (cos)0 ¼ sin. Thus sin
and cos are all differentiable on R, and so clearly we can differentiate the
functions sin and cos as many times as we please on R.
In general, when we differentiate any given differentiable function f, we
obtain a new function f 0 (whose domain may be smaller than that of f ). The
notion of differentiability can then be applied to the function f 0 , just as before,
yielding another function f 00 ¼ ( f 0 )0 , whose domain consists of those points
where f 0 is differentiable.
Definitions Let f be differentiable on an open interval I, and c 2 I. If f 0 is
differentiable at c, then f is called twice differentiable at c, and the number
f 00 (c) is called the second derivative of f at c. The function f 00 is called the
second derived function of f.
Provided that the derivatives exist, we can define f 000 or f (3), f (4), . . .,
(n)
f , . . .. The functions f 00 , f (3), f (4), . . ., f (n), . . . are called the higher-order
derived functions of f, whose values f 00 (x), f (3)(x), f (4)(x), . . ., f (n)(x), . . .
are called the higher-order derivatives of f.
f 00 ¼ (f 0 )0 is sometimes written
as f (2).
However, not all derived functions are themselves differentiable. You have
already seen, for example, that the function
x2 sin 1x ; x 6¼ 0,
f ð xÞ ¼
0;
x ¼ 0,
is differentiable at 0. In fact, it has as its derived function
2x sin 1x cos 1x ; x 6¼ 0,
0
f ð xÞ ¼
0;
x ¼ 0,
which is not even continuous at 0.
Example 5
Prove that the function
1 2
x ; x < 0,
f ð xÞ ¼ 1 22
x 0,
2x ;
is differentiable on R, but that its derived function is not differentiable at 0.
For c > 0, the difference quotient for f at c is
1
ðc þ hÞ2 12 c2
Q ð hÞ ¼ 2
h
1
¼ ð2c þ hÞ ! c as h ! 0;
2
so that f is differentiable at c and f 0 (c) ¼ c.
When c ¼ 0 and h > 0, a similar argument shows that f is differentiable on
the right at 0 and fR0 ð0Þ ¼ 0.
For c < 0, the difference quotient for f at c is
Solution
1 ðc þ hÞ2 þ 12 c2
Q ð hÞ ¼ 2
h
1
¼ ð2c þ hÞ ! c
2
as h ! 0;
We assume that h is
sufficiently small that jhj < c,
so that c þ h > 0; and hence
that we are using the correct
value for f at c þ h.
We assume that h is
sufficiently small that
jhj < c, so that c þ h < 0;
and hence that we are using
the correct value for f at c þ h.
6: Differentiation
216
so that f is differentiable at c and f 0 (c) ¼ c.
When c ¼ 0 and h < 0, a similar argument shows that f is differentiable on
the left at 0 and fL0 ð0Þ ¼ 0.
Since f has fL0 ð0Þ ¼ fR0 ð0Þ ¼ 0, it follows that f is differentiable at 0 and
f (0) ¼ 0.
Thus the derived function for f is given by
f 0 ð xÞ ¼ j xj;
x 2 R:
This is the modulus function, which as we showed earlier is not differentiable
&
at 0. It follows that f 0 is not differentiable at 0.
Problem 9
Example 2, Sub-section 6.1.1.
Prove that the function
2
x ; x < 0;
f ð xÞ ¼
x3 ; x 0;
is differentiable on R. Is f 0 differentiable at 0?
6.2
Rules for differentiation
In the last section we showed that the functions sin, cos and exp are differentiable on R, by appealing directly to the definition of differentiability.
However, it would be very tedious if, every time that we wished to prove that
a given function is differentiable and to determine its derived function, we had
to use the difference quotient method described in Section 6.1. Sometimes we
do need to use that method, but usually we can avoid the algebra involved in
the difference quotient method by using various rules for differentiation. In this
section, we introduce the Combination Rules, the Composition Rule and the
Inverse Function Rule for differentiable functions. These are similar to the
rules for continuous functions that you met earlier.
Note that each of the rules for differentiation supplies two pieces of
information:
In Sections 4.1 and 4.3.
1. a function of a certain type is differentiable;
2. an expression for the derivative.
6.2.1
The Combination Rules
The Combination Rules for differentiable functions are a consequence of the
Combination Rules for limits.
Theorem 1 Combination Rules
Let f and g be defined on an open interval I, and c 2 I. Then, if f and g are
differentiable at c, so are:
( f þ g)0 (c) ¼ f 0 (c) þ g0 (c);
Sum Rule
f þ g,
Multiple Rule
l f, for l 2 R,
and
(l f ) (c) ¼ l f (c);
Product Rule
fg,
and
( fg)0 (c) ¼ f 0 (c)g (c) þ f (c)g0 (c);
and
0
0
We differentiate each term
in turn.
We differentiate one term at a
time.
6.2
Rules for differentiation
f
g,
Quotient Rule
217
provided that g(c) 6¼ 0; and
0
gðcÞf 0 ðcÞf ðcÞg0 ðcÞ
f
:
g ðcÞ ¼
ðgðcÞÞ2
Now, we saw in Section 6.1 that, for any n 2 N, the function
x 7! xn ; x 2 R;
is differentiable on R, and that its derived function is
x 2 R:
x 7! nxn1 ;
We can use this fact, together with the Combination Rules, to prove that any
polynomial function is differentiable on R, and that its derivative can be
obtained by differentiating the polynomial term-by-term.
Corollary 1 Let p(x) ¼ a0 þ a1x þ a2x2 þ þ anxn, x 2 R, where a0, a1,
a2, . . ., an 2 R. Then p is differentiable on R, and its derivative is
p0 ð xÞ ¼ a1 þ 2a2 x þ þ nan xn1 ;
x 2 R:
Since a rational function is a quotient of two polynomials, it follows from
Corollary 1 and the Quotient Rule that a rational function is differentiable at
all points where the denominator does not vanish (that is, the denominator
does not take the value 0).
3
Example 1 Prove that the function f ð xÞ ¼ x2x1 ; x 2 R f1g, is differentiable on its domain, and find its derivative.
Solution The function f is a rational function whose denominator does not
vanish on R {1}; hence f is differentiable on this set (the whole of its
domain).
By Corollary 1, the derived function of x 7! x3 is x 7! 3x2 , and the derived
function of x 7! x2 1 is x 7! 2x. It follows from the Quotient Rule that the
derivative of f is
ðx2 1Þ 3x2 x3 ð2xÞ
f 0 ð xÞ ¼
ð x 2 1Þ 2
x4 3x2
&
¼
:
ð x 2 1Þ 2
Problem 1 Find the derivative of each of the following functions:
(a) f (x) ¼ x7 2x4 þ 3x3 5x þ 1, x 2 R;
2
x 2 R f1g;
(b) f ð xÞ ¼ xx3 þ1
1 ;
(c) f (x) ¼ 2 sin x cos x, x 2 R;
x
e
(d) f ð xÞ ¼ 3þsin x2
cos x ;
Problem 2
x 2 R.
x 2 R:
Find the third order derivative of the function f (x) ¼ xe2x,
In the last section we found the derived functions for sin, cos and exp. We
now ask you to find the derived functions for the remaining trigonometric
functions and the three most common hyperbolic functions.
In particular,
0
g0 ðcÞ
1
g ðcÞ ¼ ðgðcÞÞ2 :
6: Differentiation
218
Find the derivative of each of the following functions:
(a) f ð xÞ ¼ tan x; x 2 R 12 p; 32 p; 52 p; . . . ;
(b) f ð xÞ ¼ cosec x; x 2 R f0; p; 2p; . . .g;
(c) f ð xÞ ¼ sec x; x 2 R 12 p; 32 p; 52 p; . . . ;
(d) f ð xÞ ¼ cot x; x 2 R f0; p; 2p; . . .g:
Problem 3
Problem 4
Find the derivative of each of the following functions
(a) f (x) ¼ sinh x, x 2 R;
(c) f (x) ¼ tanh x, x 2 R.
6.2.2
(b) f(x) ¼ cosh x, x 2 R;
The Composition Rule
In the last sub-section we extended our stock of differentiable functions to
include all rational, trigonometric and hyperbolic functions. However, to differentiate many other functions we need to differentiate composite functions,
such as the function f (x) ¼ sin(cos x), x 2 R, which is the composite of two
differentiable functions – namely, f ¼ sin cos. The Composition Rule tells us
that the composite of two differentiable functions is itself differentiable.
Theorem 2 Composition Rule
Let g and f be defined on open intervals I and J, respectively, and let c 2 I and
gðI Þ J. If g is differentiable at c and f is differentiable at g(c), then f g is
differentiable at c and
ð f gÞ0 ðcÞ ¼ f 0 ðgðcÞÞg0 ðcÞ:
This rule is sometimes known
as the Chain Rule.
We give the proof of the Rule
in Sub-section 6.2.4.
Remarks
1. When written in Leibniz notation, the Composition Rule has a form that is
easy to remember: if we put
u ¼ gð xÞ and y ¼ f ðuÞ ¼ f ðgð xÞÞ
then
dy dy du
¼
:
dx du dx
2. We can extend the Composition Rule to a composite of three (or more)
functions; for example
ð f g hÞ0 ð xÞ ¼ f 0 ½gðhðxÞÞg0 ðhð xÞÞh0 ð xÞ:
In Leibniz notation, if we put
v ¼ hð xÞ; u ¼ gðvÞ and
y ¼ f ðuÞ ¼ f ½gðhð xÞÞ;
We frequently use this
extension of the Composition
Rule without mentioning it
explicitly.
6.2
Rules for differentiation
219
then we obtain the chain
dy dy du dv
¼
:
dx du dv dx
Example 2 Prove that each of the following composite functions is differentiable on its domain, and find its derivative:
(a) k(x) ¼ sin (cos
x 2 R; (b) k(x) ¼ cosh (e2x), x 2 R;
x),
1
x2
(c) kð xÞ ¼ tan 4 e þ sin px ; x 2 ð1; 1Þ:
Solution
(a) Here k(x) ¼ sin (cos x), so let
gð xÞ ¼ cos x and
f ð xÞ ¼ sin x;
for x 2 R;
then
g0 ð xÞ ¼ sin x
and f 0 ð xÞ ¼ cos x;
for x 2 R:
By the Composition Rule, k ¼ f g is differentiable on R, and
k0 ð xÞ ¼ f 0 ðgð xÞÞg0 ð xÞ
¼ cosðcos xÞ ð sin xÞ
¼ cosðcos xÞ sin x:
(b) Here k(x) ¼ cosh (e2x), so let
hð xÞ ¼ 2x; gð xÞ ¼ ex and f ð xÞ ¼ cosh x;
then h, g and f are differentiable on R, and
for x 2 R;
h0 ð xÞ ¼ 2; g0 ð xÞ ¼ ex and f 0ð xÞ ¼ sinh x; for x 2 R:
By the (extended form of the) Composition Rule, k ¼ f g h is differentiable on R, and
k0 ð xÞ ¼ f 0 ½gðhð xÞÞg0 ðhð xÞÞh0 ð xÞ
¼ sinh e2x e2x 2
2x
2x
¼
2e2 sinh e :
(c) Here kð xÞ ¼ tan 14 ex þ sin px ; x 2 ð1; 1Þ, so let
1 x2
e þ sin px ; x 2 ð1; 1Þ; and
gð x Þ ¼
4
1 1
f ð xÞ ¼ tan x; x 2 p; p :
2 2
Now, when x 2 (1, 1), we have
1 x2
e þ jsin pxj
j gð x Þ j 4
1
ðe þ 1Þ
4
<1
1
< p;
2
1 1 so that g(x) lies in 2 p; 2 p , the domain of the differentiable function tan.
6: Differentiation
220
Now, by the Composition and Combination Rules, g is differentiable on (1, 1), and
g 0 ð xÞ ¼
1 x2
2xe þ p cos px ; x 2 ð1; 1Þ:
4
Also
f 0 ð xÞ ¼ sec2 x; x 2
1 1
p; p :
2 2
Finally, it follows, by the Composition Rule, that k ¼ f g is differentiable
on (1, 1), and
k0 ð xÞ ¼ f 0 ðgð xÞÞg0 ð xÞ
1 1 x2
2
e þ sin px 2xex þ p cos px ; x 2 ð1;1Þ:
¼ sec2
4
4
&
Problem 5 Find the derivative of each of the following functions:
(a) f(x) ¼ sinh(x2), x 2 R;
(b) f (x) ¼ sin (sinh 2x), x 2 R;
(c) f ð xÞ ¼ sin cosx22x ; x 2 ð0; 1Þ:
6.2.3
The Inverse Function Rule
Earlier we showed that, if a function f with domain some interval I and
image J is strictly monotonic on I, then f possesses a strictly monotonic and
continuous inverse function f 1 on J. In particular, the power function, the
trigonometric functions, the exponential function and the hyperbolic functions all have inverse functions, provided that we restrict the domains, where
necessary.
These standard functions are all differentiable on their domains. Do their
inverse functions also have this property, as their graphs suggest?
As you saw earlier, we obtain the graph y ¼ f 1(x) by reflecting the graph
y ¼ f(x) in the line y ¼ x. This reflection maps a typical point P(c, d) to the
point Q(d, c).
It follows that, if the slope of the tangent to the graph y ¼ f (x) at the point P
is f 0 (c) ¼ m, then the slope of the tangent to the graph y ¼ f 1(x) at Q is
0
ð f 1 Þ ðd Þ ¼ m1 :
However, if the graph of f has a horizontal tangent (m ¼ 0) at a point P, then
the graph of f 1 has a vertical tangent at the corresponding point Q; in this case,
f 1 is not differentiable at Q, since m1 is not defined when m ¼ 0. We therefore
require the condition ‘f 0 (x) is non-zero’ in our statement of the Rule.
Thus, in the Inverse Function Rule we require:
(a) f is strictly monotonic and continuous on an open interval I, so that f has
a strictly monotonic and continuous inverse function on the open interval
J ¼ f(I);
(b) f is differentiable on I and f 0 (x) 6¼ 0 on I, so that f10 ðxÞ is defined.
The Rule tells us that, under these conditions, f 1 is also differentiable, and
there is a simple formula for ( f 1)0 .
Section 4.3.
Recall that strictly monotonic
means that f is either strictly
increasing or strictly
decreasing.
y
y = f (x)
y=x
y = f –1(x)
P
Q
x
y
y=x
y = f (x)
y = f –1(x)
P
Q
x
6.2
Rules for differentiation
221
Theorem 3 Inverse Function Rule
Let f : I ! J, where I is an interval and J is the image f(I), be a function such
that:
1. f is strictly monotonic on I;
2. f is differentiable on I;
y
0
3. f (x) 6¼ 0 on I.
Then f 1 is differentiable on J. Further, if c 2 I and d ¼ f (c), then
1 0
1
f
ðd Þ ¼ 0 :
f ðcÞ
J
d
y
x c
Remark
The Leibniz notation for derivatives can be used to express the Inverse
Function Rule in a form that is easy to remember: if
y ¼ f ð xÞ and
x ¼ f 1 ð yÞ;
1 0
0
dx
and we write dy
dx for f ð xÞ, and dy for ( f ) (y), then
dx 1
¼ :
dy dy
dx
Proof Let F ¼ f1, and let y ¼ f (x), where x 2 I, so that x ¼ F(y), where y 2 J.
Then the difference quotient for F at d is
F ð yÞ F ðd Þ
xc
¼
yd
f ð xÞ f ðcÞ
1
¼ f ðxÞf ðcÞ
xc :
Since f is one–one and differentiable on I, it is necessarily one–one and
continuous on I. Thus F is one–one and continuous on J. It follows that, for
y 6¼ d, we must have x 6¼ c, since x ¼ f 1(y) and c ¼ f 1(d).
Also, since f 1 is continuous, x ! c as y ! d. It follows that
!
F ð y Þ F ðd Þ
1
¼ lim f ðxÞf ðcÞ
lim
x!c
y!d
yd
xc
¼
f ðxÞf ðcÞ
xc
x!c
lim
¼
1
1
f 0 ðcÞ
:
Thus F is differentiable at c, with derivative f 0 ð1cÞ.
&
Example 3 For each of the following functions, show that f 1 is differentiable and determine its derivative:
(a) f (x) ¼ xn, x > 0, n 2 N;
(b) f ð xÞ ¼ tan x; x 2 12 p; 12 p ;
(c) f (x) ¼ ex, x 2 R.
I
x
6: Differentiation
222
Solution
(a) The function
f ð xÞ ¼ xn ; x > 0;
is continuous and strictly increasing on (0, 1), and f ((0, 1)) ¼ (0, 1).
Also, f is differentiable on (0, 1), and its derivative f 0 (x) ¼ nxn1 is nonzero there. So f satisfies the conditions of the Inverse Function Rule.
Hence f 1 is differentiable on (0, 1); and, if y ¼ f (x), then
1 0
1
1
¼
ð yÞ ¼ 0
f
f ð xÞ nxn1
1
1 1n
¼ x1n ¼ y n :
n
n
If we now replace the domain variable y by x, we obtain
1 0
1 1
ð xÞ ¼ xn1 ; x 2 ð0; 1Þ:
f
n
(b) The function
1 1
f ð xÞ ¼ tan x; x 2 p; p ;
2 2
1 1 1 1 is continuous and strictly increasing
R.
1 1on 2 p; 2 p , and f 0 2 p; 2 p ¼
Also, f is differentiable on 2 p; 2 p , and its derivative f ð xÞ ¼ sec2 x is
non-zero there. So f satisfies the conditions of the Inverse Function Rule.
Hence f 1 ¼ tan1 is differentiable on R; and, if y ¼ f (x), then
1 0
1
1
f
¼
ð yÞ ¼ 0
f ð xÞ sec2 x
1
1
¼
:
¼
1 þ tan2 x 1 þ y2
If we now replace the domain variable y by x, we obtain
1 0
1
ð xÞ ¼
; x 2 R:
tan
1 þ x2
(c) The function
f ð xÞ ¼ ex ; x 2 R;
is continuous and strictly increasing on R, and f (R) ¼ (0, 1). Also, f is
differentiable on R, and its derivative f 0 (x) ¼ ex is non-zero there. So
f satisfies the conditions of the Inverse Function Rule.
Hence f 1 ¼ loge is differentiable on (0, 1); and, if y ¼ f (x), then
1 0
1
1
¼
f
ð yÞ ¼ 0
f ð xÞ ex
1
¼ :
y
If we now replace the domain variable y by x, we obtain
1
&
ðloge Þ0 ð xÞ ¼ ; x 2 ð0; 1Þ:
x
Problem 6 For each of the following functions f, show that f1 is
differentiable and determine its derivative:
(a) f(x) ¼ cos x, x 2 (0, p);
(b) f(x) ¼ sinh x, x 2 R.
1
y ¼ xn, so that x ¼ yn :
6.2
Rules for differentiation
223
Most equations of the form y ¼ f (x) cannot be solved explicitly to give x as
some formula involving y alone. The Inverse Function Rule is often used in
such situations to solve problems that would otherwise be intractable. The
following problem illustrates this type of application.
Problem 7 Prove that the function f (x) ¼ x5 þ x 1, x 2 R, has an
inverse function f1 which is differentiable on R. Find the values of
( f1)0 (d ) at those points d corresponding to the points c ¼ 0, 1 and 1,
where d ¼ f (c).
Exponential functions
Earlier we defined the number ax, for a > 0, by the formula
ax ¼ expðx loge aÞ:
Since the functions exp and log are differentiable on R and (0, 1), respectively,
it follows that we can use this formula to determine the derivatives of functions
such as x 7! x ; x 7! ax and x 7! xx .
Example 4 Prove that, for 2 R, the power function f (x) ¼ x, x 2 (0, 1), is
differentiable on its domain, and that f 0 (x) ¼ x1.
Solution
By definition
f ð xÞ ¼ expð loge xÞ:
The function x 7! loge x is differentiable on (0, 1), with derivative x . By the
Composition Rule, f is differentiable on (0, 1) with derivative
f 0 ð xÞ ¼ expð loge xÞ x
1
¼x ¼ x ; x 2 ð0; 1Þ:
&
x
Remark
In the case when is a rational number, ¼ mn say, the result of Example 4 can
also be proved by applying the Composition Rule to the functions x 7! xm and
1
1
x 7! xn . The function x 7! xn is differentiable on (0, 1), by the Inverse Function
Rule, as it is the inverse of the function x 7! xn .
Example 5 Prove that, for a > 0, the function f (x) ¼ ax, x 2 R, is differentiable on its domain, and that f 0 (x) ¼ ax loge a.
Solution
By definition
f ð xÞ ¼ expðx loge aÞ:
The function x 7! x loge a is differentiable on R, with derivative loge a. By the
Composition Rule, f is differentiable on R, with derivative
f 0 ð xÞ ¼ expðx loge aÞ loge a
¼ ax loge a;
x 2 R:
&
Problem 8 Prove that the function f (x) ¼ xx, x 2 (0, 1), is differentiable on its domain, and find its derivative.
Section 4.4.
Section 6.1.
6: Differentiation
224
We end this sub-section with a list of basic differentiable functions.
Basic differentiable functions The following functions are differentiable
on their domains:
polynomials and rational functions
nth root function
trigonometric functions (sine, cosine and tangent)
exponential function
hyperbolic functions (sinh, cosh and tanh).
6.2.4
Proofs
You may omit these proofs at
a first reading.
We now supply the proofs of the Combination Rules and the Composition
Rule, which we omitted earlier. We also illustrate how to prove such results
using the " method.
Theorem 1
Combination Rules
Let f and g be defined on an open interval I, and c 2 I. Then, if f and g are
differentiable at c, so are:
Sum Rule
Multiple Rule
f þ g,
and ( f þ g)0 (c) ¼ f 0 (c) þ g0 (c);
l f, for l 2 R, and (l f )0 (c) ¼ l f 0 (c);
Product Rule
fg,
Quotient Rule
f
g, provided that g (c) 6¼ 0;
and
0
( fg)0 (c) ¼ f0 (c)
g(c) þ f (c) g (c);
and
f
g
0
0
0
ðcÞg ðcÞ
ðcÞ ¼ gðcÞf ððcgÞf
:
ðcÞÞ2
Proof
In this proof we use the
Combination Rules for limits
given in Sub-section 5.1.
Sum Rule
Let F ¼ f þ g. Then
F ð xÞ F ðcÞ
f f ð xÞ þ gð xÞg f f ðcÞ þ gðcÞg
¼ lim
lim
x!c
x!c
xc
xc
f ð x Þ f ðc Þ
gð xÞ gðcÞ
¼ lim
þ lim
x!c
x!c
xc
xc
¼ f 0 ðcÞ þ g0 ðcÞ:
Thus F is differentiable at c, with derivative f 0 (c) þ g0 (c).
Multiple Rule
This is just a special case of the Product Rule, with g(x) ¼ l.
Product Rule
Let F ¼ fg. Then
F ð xÞ F ðcÞ
f ð xÞgð xÞ f ðcÞgðcÞ
¼ lim
x!c
x!c
xc
xc
lim
6.2
Rules for differentiation
225
f f ð xÞ f ðcÞggð xÞ þ f ðcÞfgð xÞ gðcÞg
xc
f ð xÞ f ðcÞ
gð xÞ gðcÞ
¼ lim
gð xÞ þ lim f ðcÞ
x!c
x!c
xc
xc
f ð xÞ f ðcÞ
gð xÞ gðcÞ
¼ lim
lim gð xÞ þ f ðcÞ lim
x!c
x!c
x!c
xc
xc
¼ f 0 ðcÞgðcÞ þ f ðcÞg0 ðcÞ;
¼ lim
x!c
since f and g are differentiable at c, and since g is continuous at c, so that
g(x) ! g(c) as x ! c.
Thus F is differentiable at c, with derivative f 0 (c)g(c) þ f (c)g0 (c).
For, differentiable )
continuous.
Quotient Rule
Let F ¼ gf .
Recall, first, that, since g is continuous at c (as it is differentiable there) and
g (c) 6¼ 0, there exists an open interval J containing c on which g(x) 6¼ 0. Thus
the domain of F contains the open interval J, and c 2 J.
Then
f ð xÞ
f ðcÞ
F ð xÞ F ðcÞ
gð xÞ gðcÞ
¼ lim
x!c
x!c x c
xc
f ð xÞgðcÞ f ðcÞgð xÞ
¼ lim
x!c ðx cÞ gð xÞgðcÞ
ff ð xÞ f ðcÞggðcÞ f ðcÞfgð xÞ gðcÞg
¼ lim
x!c
ðx cÞ gð xÞgðcÞ
f ð x Þ f ðc Þ
f ðcÞ
gð xÞ gðcÞ
lim
¼ lim
x!c ðx cÞ gð xÞ x!c gðcÞ ðx cÞ gð xÞ
f ð x Þ f ðc Þ
1
f ðc Þ
gð x Þ gð c Þ
1
¼ lim
lim
lim
lim
x!c
x!c gð xÞ gðcÞ x!c ðx cÞ
x!c gð xÞ
xc
f 0 ðcÞ f ðcÞg0 ðcÞ f 0 ðcÞgðcÞ f ðcÞg0 ðcÞ
2
¼
;
¼
gð c Þ
g ðcÞ
g2 ð c Þ
lim
since f and g are differentiable at c, and g is continuous at c.
0
ðcÞg0 ðcÞ
Thus F is differentiable at c, with derivative f ðcÞgðcgÞf
:
2 ð cÞ
&
As an illustration of how such results may be proved using the " method
that we introduced in Section 5.4, we prove just the Sum Rule. Notice that in
our proofs we avoid many complications by judicious use of the ‘K" Lemma’.
The ‘K" Lemma’ first
appeared in Sub-section 2.2.2.
Proof of the Sum Rule Let F ¼ f þ g.
In view of the K" Lemma, we must prove that:
As you work through this
proof, compare it with the
earlier proof using limits.
for each positive number ", there is a positive number such that
F ð x Þ F ðcÞ
0
0
< K" for all x satisfying
f
ð
c
Þ
þ
g
ð
c
Þ
f
g
xc
0 < jx cj < :
(1)
First, we write the expression on the left-hand side of (1) in a convenient
form as
Recall that K must NOT depend
on x or ".
6: Differentiation
226
ff ð x Þ þ gð x Þ g ff ð c Þ þ gð c Þ g
ff 0 ðcÞ þ g0 ðcÞg
xc
ff ð xÞ f ðcÞg þ fgð xÞ gðcÞg
ff 0 ðcÞ þ g0 ðcÞg
¼
x
c
f ð xÞ f ðcÞ
gð xÞ gðcÞ
f 0 ðcÞ þ
g0 ð c Þ :
¼
xc
xc
(2)
We now use the information that we already have to tackle each of the terms
in (2) in turn.
Since f is differentiable at c, there exists some positive number 1 such that
f ð x Þ f ðc Þ 0 x c f ðcÞ < "; for all x satisfying 0 < jx cj < 1 : (3)
Also, since g is differentiable at c, there exists some positive number 2
such that
gð x Þ gð c Þ
0
x c g ðcÞ < "; for all x satisfying 0 < jx cj < 2 : (4)
Now let ¼ min{1, 2}. It follows that, for all x satisfying 0 < jx cj < ,
both (3) and (4) hold. Hence, if we apply the Triangle Inequality to (2) and then
use both (3) and (4), we find that, for all x satisfying 0 < jx cj < f f ð xÞ þ gð xÞg f f ðcÞ þ gðcÞg
0
0
f
ð
c
Þ
þ
g
ð
c
Þ
f
g
xc
f ð xÞ f ðcÞ
gð x Þ gð c Þ
f 0 ðcÞ þ g0 ðcÞ
xc
xc
By the Triangle Inequality.
By (3) and (4).
< " þ " ¼ 2":
But this is just the statement (1) that we set out to prove, with K ¼ 2.
This expression appears in (2).
&
Remark
Note that our use of the ‘K" Lemma’ meant that we did not need to know at the
steps (3) and (4) to use expressions like 12 " rather than " in order to end up with
a final conclusion that ‘some expression is <"’! For ‘some expression is <K"’
is sufficient.
We end with the proof of the Composition Rule.
Theorem 2 Composition Rule
Let g and f be defined on open intervals I and J, respectively, and let c 2 I
and gðI Þ J. If g is differentiable at c and f is differentiable at g(c), then
f g is differentiable at c and
ð f gÞ0 ðcÞ ¼ f 0 ðgðcÞÞg0 ðcÞ:
We would only discover the
need for such a choice once
we had made a first attempt
at the proof.
6.2
Rules for differentiation
227
Proof Let F ¼ f g, and let y ¼ g(x) and d ¼ g(c).
The difference quotient for F at c is
F ð x Þ F ð c Þ f ð gð x Þ Þ f ð gð c Þ Þ
¼
:
xc
xc
(5)
Now, the right-hand side of (5) is equal to
f ð yÞ f ðd Þ gð xÞ gðcÞ
;
yd
xc
(6)
provided that y 6¼ d.
Unfortunately, for some choices of g and c it is possible for y ¼ d to occur
when y ¼ g(x) and x is arbitrarily close to c but not actually equal to c. In such
situations the expression (6) will not exist.
To get round this problem, we introduce a carefully chosen auxiliary
function
f ðyÞf ðdÞ
y 6¼ d,
yd ;
hð y Þ ¼
y ¼ d.
f 0 ðdÞ;
Since f is differentiable at d, h(y) ! f 0 (d) as y ! d; and, since h(d ) ¼ f 0 (d ), it
follows that h is continuous at d.
We then apply the Composition Rule for continuous functions, to deduce
that the composite function
(
f ðgðxÞÞf ðgðcÞÞ
gð xÞ 6¼ gðcÞ;
gð xÞgðcÞ ;
h gð x Þ ¼
0
f ðgðcÞÞ;
gð xÞ ¼ gðcÞ;
is continuous at c.
Next, notice that, if g(x) 6¼ g(c), it follows from (5) that
F ð xÞ F ðcÞ
gð x Þ gð c Þ
¼ h gð x Þ ;
xc
xc
(7)
and also that this last statement (7) is also valid when g(x) ¼ g(c), since then
both sides of (7) are zero.
Hence, if we let x ! c in (7) and use the continuity of the function h g, we
obtain
lim
x!c
F ð xÞ F ðcÞ
¼ h gð c Þ g0 ð c Þ
xc
¼ f 0 ðgðcÞÞg0 ðcÞ:
Thus F is differentiable at c, with derivative f 0 (g(c))g0 (c).
&
Remark
Let
gð x Þ ¼
x2 sin
0;
1
x
;
x¼
6 0,
x ¼ 0,
c ¼ 0;
and
d ¼ gð0Þ ¼ 0:
Some texts ignore this
possibility, and so give
incomplete proofs of
Theorem 2! In the Remark
below, we describe one such
situation.
6: Differentiation
228
Since sin(np) ¼ 0 for all n 2 N, it follows that y ¼ g(x) takes the value d ¼ 0
1
that are arbitrarily close to c ¼ 0 but not equal to 0.
at the points np
6.3
This example shows that the
paragraph ‘Unfortunately . . .’
after equation (6) above does
describe a situation that can
occur.
Rolle’s Theorem
In the next two sections we describe some of the fundamental properties of
functions that are differentiable, not just at a particular point but on an
interval. Our results are motivated by the geometric significance of differentiability in terms of tangents, and explain why the graphs of differentiable
functions possess certain geometric properties.
6.3.1
The Local Extremum Theorem
Earlier we described some of the fundamental properties of functions which
are continuous on a closed interval. In particular, we proved the Extreme
Values Theorem, which states that, if a function ƒ is continuous on a closed
interval [a, b], then there are points c, d in [a, b] such that
f ðcÞ f ð xÞ f ðdÞ;
Section 4.2.
Sub-section 4.2.3.
for all x 2 ½a; b;
we called f(c) the minimum of ƒ on [a, b], and f(d) the maximum of ƒ on [a, b].
We used the term extremum to denote either a maximum or a minimum.
But how do we determine the points c, d where these extrema occur? In
general, unfortunately, this is not easy. However, if the function ƒ is differentiable, then we can, in principle, determine c and d by first finding any local
maxima or local minima of the function ƒ on the interval [a, b].
Roughly speaking, for a point c in (a, b), the value f(c) is a local maximum of
ƒ on [a, b] if
f ðxÞ f ðcÞ;
for all x in ½a; b sufficiently close to c;
and a local minimum of ƒ on [a, b] if
f ðxÞ f ðcÞ;
for all x in ½a; b sufficiently close to c:
To make this idea more precise, we use the concept of neighbourhood that you
met earlier.
Definitions
From this point onwards, we
shall commonly use the letter
c to denote an extremum – that
is EITHER a maximum OR a
minimum.
Sub-section 5.1.1.
Let ƒ be defined on an interval I, and let c 2 I. Then:
ƒ has a local maximum f(c) at c if there exists a neighbourhood N of c
such that
f ðxÞ f ðcÞ; for x 2 N \ I;
ƒ has a local minimum f(c) at c if there exists a neighbourhood N of c
such that
f ðxÞ f ðcÞ; for x 2 N \ I;
ƒ has a local extremum at c if f(c) is either a local maximum or a local
minimum.
Notice that N \ I necessarily
includes an interval of the
form (c r, c] or [c, c þ s),
where r, s > 0.
6.3
Rolle’s Theorem
229
When we wish to locate local extrema of a differentiable function ƒ, instead of
using the above definition we usually use the following result, which gives a
connection between local extrema of a function ƒ and the points where f 0 vanishes.
Theorem 1
Local Extremum Theorem
Let ƒ be defined on an interval [a, b]. If ƒ has a local extremum at c, where
a < c < b, and if ƒ is differentiable at c, then
f 0 ðcÞ ¼ 0:
Proof Suppose that ƒ has a local maximum at c.
Since a < c < b, it follows from the definition of local maximum that there
exists a neighbourhood N of c with N ½a, b such that
f ðxÞ f ðcÞ;
for x 2 N:
We now choose numbers r, s > 0 such that N ¼ (c r, c þ s).
First, looking to the left of c, we have
f ðxÞ f ðcÞ 0
and
x c < 0;
for c r < x < c;
so that
f ðxÞ f ðcÞ
0;
xc
for c r < x < c:
Thus
f ðxÞ f ðcÞ
0:
xc
Next, looking to the right of c, we have
fL0 ðcÞ ¼ lim
(1)
x!c
f ðxÞ f ðcÞ 0
and
x c > 0;
for c < x < c þ s;
so that
Thus
f ðxÞ f ðcÞ
0;
xc
for c < x < c þ s:
f ðxÞ f ðcÞ
0:
(2)
xc
Since ƒ is differentiable at c, the left and right derivatives at c exist and are
equal. Hence, in view of inequalities (1) and (2), their common value, f 0ðcÞ,
&
must be 0.
fR0 ðcÞ ¼ limþ
x!c
We prove only the local
maximum version: the proof
of the local minimum version
is similar.
6: Differentiation
230
Remarks
1. The Local Extremum Theorem applies only if the function is differentiable
at a local extremum. For example, the function
f ðxÞ ¼ jxj; x 2 ½1; 1;
has a local minimum 0 at 0, but ƒ is not differentiable at 0.
2. The Local Extremum Theorem does not assert that a point where the derivative vanishes is necessarily a local extremum. For example, the function
f ðxÞ ¼ x3 ; x 2 ½1; 1;
does not have a local extremum at 0, although f 0 (0) ¼ 0.
3. The Local Extremum Theorem does not make any assertion about a
local extremum that occurs at a point c that is one of the end-points of the
interval [a, b].
Find the local extrema of the function
1
1
f ð xÞ ¼ x4 x3 ; x 2 ½1; 2:
4
3
Clearly any extremum of ƒ on [a, b] that occurs at a point other than a or b
must be a local extremum. It follows from Theorem 1 that such a point c must
be a point where f 0 (c) ¼ 0. Thus an immediate consequence of the Local
Extremum Theorem is the following criterion for finding all the extrema of
‘well-behaved functions’ on closed intervals.
Problem 1
Corollary 1 Let ƒ be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then the extrema of ƒ on [a, b] can occur
only at a, at b, or at points c in (a, b) where f 0 (c) ¼ 0.
We now reformulate Corollary 1 as a strategy for locating minima and
maxima.
Strategy To determine the maximum and the minimum of a function ƒ
which is continuous on [a, b] and differentiable on (a, b):
1. Determine the points c1, c2, . . . in (a, b) where f 0 vanishes;
2. Compare the values of f (a), f (b), f (c1), f (c2), . . .;
the least is the minimum, and the greatest is the maximum.
Problem 2 Use the above Strategy to determine the minimum
and the
maximum of the function f ð xÞ ¼ sin2 x þ cos x, for x 2 0; 12 p :
6.3.2
Rolle’s Theorem
In the previous sub-section we saw that, if a function ƒ is continuous on
the closed interval [a, b] and differentiable on the open interval (a, b), then
the extrema of ƒ can occur only at a, at b, or at points c in (a, b) where
f 0 (c) ¼ 0.
6.3
Rolle’s Theorem
231
Now the function
1
f ðxÞ ¼ sin px ;
2
2 2
x2 ;
;
3 3
shows that it can happen that no interior point of (a, b) corresponds to a
maximum or a minimum of ƒ, and that f 0 need not vanish at any interior
point at all.
However, the situation is quite different for the function
1
f ðxÞ ¼ sin px ; x 2½2; 2:
2
That is, the extrema may
not occur at interior points
of (a, b).
Here f(2) ¼ f(2) ¼ 0; on [2, 2] the function f has a maximum at 1 and a
minimum at 1, and at both these interior points f 0 vanishes. This is a special
case of Rolle’s Theorem which asserts that, if f(a) ¼ f(b), then there is at least
one point c strictly between a and b at which f 0 vanishes.
Theorem 2 Rolle’s Theorem
Let ƒ be defined on the closed interval [a, b] and differentiable on the open
interval (a, b). If f(a) ¼ f(b), then there exists some point c, with a < c < b,
for which f 0 (c) ¼ 0.
This is an existence theorem.
Often it is difficult to evaluate
c explicitly.
Remarks
1. This apparently simple result is one of the most important results in
Analysis.
2. In geometric terms, Rolle’s Theorem means that, if the line joining the
points (a, f(a)) and (b, f(b)) on the graph of ƒ is horizontal, then so is the
tangent to the graph at some point c in (a, b).
3. There may be more than one point c in (a, b) at which f 0 vanishes (as in the
diagram in the margin). Rolle’s Theorem simply asserts that at least one
such point c exists.
Proof If ƒ is constant on [a, b], then f 0 (x) ¼ 0 everywhere in (a, b); in this
case, we may take c to be any point of (a, b).
If ƒ is non-constant on [a, b], then either the maximum or the minimum (or
both) of ƒ on [a, b] is different from the common value f (a) ¼ f (b). Since one
of the extrema occurs at some point c with a < c < b, the Local Extremum
&
Theorem applied to the point c shows that f 0 (c) must be zero.
Example 1
function
Verify that the conditions of Rolle’s Theorem are satisfied by the
f ð xÞ ¼ 3x4 2x3 2x2 þ 2x;
x 2 ½1; 1;
and determine a value of c in (1, 1) for which f 0 (c) ¼ 0.
Solution Since f is a polynomial function, f is continuous on [1, 1] and
differentiable on (1, 1). Also, f (1) ¼ f (1) ¼ 1. Thus f satisfies the conditions of Rolle’s Theorem on [1, 1].
It follows that there exists a number c 2 (1, 1) for which f 0 (c) ¼ 0. Now
Since ƒ is continuous on
[a, b], ƒ must have both a
maximum and a minimum on
[a, b], by the Extreme Values
Theorem.
6: Differentiation
232
f 0 ð xÞ ¼ 12x3 6x2 4x þ 2
1
1
2
x ;
¼ 12 x 3
2
so that f 0 vanishes at the points p1ffiffi3, p1ffiffi3 and 12 in (1, 1). Any of these three
&
numbers will serve for c.
Problem 3 Verify that the conditions of Rolle’s Theorem are satisfied
by the function
f ð xÞ ¼ x4 4x3 þ 3x2 þ 2; x 2 ½1; 3;
and determine a value of c in (1,3) for which f 0 (c) ¼ 0.
Problem 4 For each of the following functions, state whether Rolle’s
Theorem applies for the given interval:
(a) f ð xÞ ¼ tan x; x 2 ½0; p;
(b) f ð xÞ ¼ x þ 3jx 1j; x 2 ½0; 2;
(c) f ð xÞ ¼ x 9x17 þ 8x18 ; x 2 ½0; 1;
(d) f ð xÞ ¼ sin x þ tan1 x; x 2 0; 12 p :
6.4
The Mean Value Theorem
Here we continue to study the geometric properties of functions that are
differentiable on intervals, and describe some of their applications.
6.4.1
The Mean Value Theorem
First, recall the geometric interpretation of Rolle’s Theorem: Under suitable
conditions, if the chord joining the points (a, f (a)) and (b, f (b)) of the graph of f
is horizontal, then so is the tangent at some point c of (a, b).
If you imagine pushing the chord (as shown in the margin), always parallel
to itself, until it is just about to lose contact with the graph of f, it looks as
though at this point the chord becomes a tangent to the graph. Similarly, the
‘chord-pushing’ approach suggests that, even if the original chord is not
horizontal (that is, if f(a) 6¼ f(b)), there must still be some point c of (a, b) at
which the tangent is parallel to the chord.
Example 1
Consider the function
f ð xÞ ¼ 3 3x þ x3 ; x 2 ½1; 2:
Find a point c of (1, 2) such that the tangent to the graph of f is parallel to the
chord joining (1, f (1)) to (2, f (2)).
Solution Since f(1) ¼ 3 3 þ 1 ¼ 1 and f(2) ¼ 3 6 þ 8 ¼ 5, the slope of the
chord joining the endpoints of the graph is
f ð 2Þ f ð 1Þ 5 1
¼
¼ 4:
21
21
6.4
The Mean Value Theorem
233
Now, since f is a polynomial, it is differentiable on (1, 2) and its derivative is
qffiffi
f 0 (x) ¼ 3 þ 3x2; hence f 0 (c) ¼ 4 when 3c2 ¼ 7, or c ¼ 73 ’ 1:53. Thus, at
the point (c, f(c)) the tangent to the curve is parallel to the chord joining the
&
end-points.
We now generalise Rolle’s Theorem and assert that there is always a point
where the tangent to the graph is parallel to the chord joining the end-points.
This result is known as the Mean Value Theorem, so-called since
f ð bÞ f ð aÞ
ba
can be thought of as the mean value of the derivative between a and b.
Theorem 1
Mean Value Theorem
Let f be continuous on the closed interval [a, b] and differentiable on the
open interval (a, b). Then there exists a point c in (a, b) such that
f ð bÞ f ð aÞ
f 0 ðcÞ ¼
:
ba
Again, this is an existence
theorem.
Note that when f(a) ¼ f(b), the
Mean Value Theorem simply
reduces to Rolle’s Theorem.
The idea of the proof is as follows. We define h(x) to be the vertical distance
from the chord to the curve; then h(a) and h(b) are both 0; in fact, h satisfies all
the conditions of Rolle’s Theorem. Applying Rolle’s Theorem to h, we obtain
the desired result.
The slope of the chord joining the points (a, f (a)) and (b, f (b)) is
f ðbÞ f ðaÞ
m¼
;
ba
and so the equation of the chord is
y ¼ mðx aÞ þ f ðaÞ:
It follows that the vertical height, h(x), between points with ordinate x on the
graph and those on the chord is given by
hð xÞ ¼ f ð xÞ ½mðx aÞ þ f ðaÞ:
Now h(a) ¼ h(b) ¼ 0, and h is continuous on [a, b] and differentiable on (a, b).
Thus h satisfies all the conditions of Rolle’s Theorem.
It follows from Rolle’s Theorem that there exists some point c in (a, b) for
which h0 (c) ¼ 0. But, since h0 (c) ¼ f 0 (c) m, it follows that
f ð bÞ f ð aÞ
:
&
f 0 ðcÞ ¼ m ¼
ba
Proof
Example 2 Verify that the conditions of the
Mean Value Theorem are
7
satisfied by the function f ð xÞ ¼ x1
;
x
2
1;
xþ1
2 ; and find a value for c that
satisfies the conclusion of the theorem.
Solution
whose denominator
The function f is a rational
function
is non-zero
on 1; 72 ; so f is continuous on 1; 72 and differentiable on 1; 72 : Thus
f satisfies the conditions of the Mean Value Theorem.
Now
5
f 72 f ð1Þ
2
90
¼ ;
¼
7
5
9
21
2
y
y = f (x)
5
9
1
7
2
x
6: Differentiation
234
and
f 0 ð xÞ ¼
2
ðx þ 1Þ2
;
so the Mean Value Theorem asserts that there exists some point c in 1; 72 for
which f 0 (c) ¼ 29. Thus
2
2
¼ ;
2
9
ð c þ 1Þ
and so ðc þ 1Þ2 ¼ 9. It follows that c ¼ 2.
&
This is a quadratic equation
with roots c ¼ 2 and 4; we
ignore the solution c¼ 4,
since it lies outside 1; 72 .
Problem 1 For each of the following functions, verify that the conditions of the Mean Value Theorem are satisfied, and find a value for c that
satisfies the conclusion of the theorem:
(a) f ð xÞ ¼ x3 þ 2x; x 2 ½2; 2;
(b) f ð xÞ ¼ ex ; x 2 ½0; 3:
6.4.2
Positive, negative and zero derivatives
We now study some consequences of the Mean Value Theorem for functions
whose derivatives are always positive, always negative, or always zero.
First, we prove a crucial result about monotonic functions which you use
regularly to sketch the graph of a function f. It concerns the behaviour of f on a
general interval I, so here we denote the interior of I (the set of all interior
points of I ) by Int I.
Theorem 2
For example, if I ¼ [0, 1), then
Int I ¼ (0, 1).
Increasing-Decreasing Theorem
Let f be continuous on an interval I and differentiable on Int I.
(a) If f 0 (x) 0 on Int I, then f is increasing on I.
(b) If f 0 (x) 0 on Int I, then f is decreasing on I.
Proof Choose any two points x1 and x2 in I, with x1 < x2. The function f
satisfies the conditions of the Mean Value Theorem on the interval [x1, x2],
so there exists a point c in (x1, x2) such that
f ðx2 Þ f ðx1 Þ
¼ f 0 ðcÞ:
x2 x1
It follows that f (x2) f(x1) must have the same sign as f 0 (c).
(a) If f 0 (x) 0 on Int I, then f(x2) f(x1) 0, so that f(x2) f(x1). Thus f is
increasing on I.
(b) If f 0 (x) 0 on Int I, then f(x2) f(x1) 0, so that f(x2) f (x1). Thus f is
&
decreasing on I.
Remark
If the inequalities in the statement of Theorem 2 are replaced by strict inequalities, the conclusions of the Theorem become the following:
(a) If f 0 (x) > 0 on Int I, then f is strictly increasing on I;
(b) If f 0 (x) < 0 on Int I, then f is strictly decreasing on I.
The proofs of these assertions
are similar to the proofs in
Theorem 2.
6.4
The Mean Value Theorem
235
Problem 2 For each of the following functions f, determine whether
f is increasing, strictly increasing, decreasing or strictly decreasing:
4
(a) f ð xÞ ¼ 3x3 4x; x 2 ½1; 1Þ;
(b) f ð xÞ ¼ x loge x; x 2 ð0; 1:
Two useful consequences of Theorem 2 are the following corollaries.
Corollary 1
Zero Derivative Theorem
Let f be continuous on an interval I and differentiable on Int I. If f 0 (x) ¼ 0
for all x in Int I, then f is constant on I.
Proof Cases (a) and (b) of Theorem 2 both apply, so that f is both increasing
and decreasing on I.
&
Hence f is constant on I.
As an illustration of the use of this important result, we can now prove the
claim made earlier that the function f (x) ¼ lex, l an arbitrary constant, is the
only function f that satisfies the differential equation f 0 (x) ¼ f (x) on R. For, if
f 0 (x) ¼ f (x), then
d x
ðe f ð xÞÞ ¼ ex f 0 ð xÞ ex f ð xÞ
dx
¼ ex ð f 0 ð xÞ f ð xÞÞ ¼ 0;
it then follows from Corollary 1 that ex f(x) is just some constant l, say, so
that f (x) ¼ lex.
Sub-section 6.1.3.
Corollary 2 Let f and g be continuous on an interval I and differentiable
on Int I. If f 0 (x) ¼ g0 (x) for all x in Int I, then
f(x) ¼ g(x) þ c for all x in I, for some constant c.
Proof This follows immediately by applying Corollary 1 to the function
&
h ¼ f g, since h0 (x) ¼ 0 for all x in the interior of I.
Example 3
Solution
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Prove that sinh1 x ¼ loge x þ x2 þ 1 , for all x 2 R.
Let
f ð xÞ ¼ sinh1 x; x 2 R;
pffiffiffiffiffiffiffiffiffiffiffiffiffi
gð xÞ ¼ loge x þ x2 þ 1 ;
x 2 R:
Then f and g are continuous on R and differentiable on R.
Now
1
f 0 ð xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; x 2 R;
2
x þ1
also, by the Composition Rule for Derivatives
1
1
1
pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 2x x2 þ 1 2
g0 ð x Þ ¼
2
x þ x2 þ 1
x ffi
1 þ pffiffiffiffiffiffiffiffi
x2 þ 1
pffiffiffiffiffiffiffiffiffiffiffiffi
ffi
¼
x þ x2 þ 1
1
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; x 2 R:
x2 þ 1
For h is continuous on I and
differentiable on Int I.
6: Differentiation
236
Hence f 0 (x) ¼ g0 (x) for all x in R.
It follows from Corollary 2 that
pffiffiffiffiffiffiffiffiffiffiffiffiffi
sinh1 x ¼ loge x þ x2 þ 1 þ c;
for some constant c. Putting x ¼ 0 in the above identity, we obtain 0 ¼ loge(1) þ c,
&
so that c ¼ 0.
Problem 3 Use Corollary 1 to prove the following identities:
(a) sin1 x þ cos1 x ¼ 12 p, for x 2 ½1; 1;
(b) tan1 x þ tan1 1x ¼ 12 p, for x > 0.
Next, note that second derivatives can often be used to identify whether a
point c where f 0 (c) ¼ 0 is a local maximum or a local minimum of a function f.
Suppose that f is defined on a neighbourhood of c, and that
f 0 ðcÞ ¼ 0 and
f 00 ðcÞ > 0:
It can be shown that there is some punctured neighbourhood N ¼ ðc r; cÞ [
ðc; c þ sÞ of c such that
f 0 ð xÞ f 0 ðcÞ
> 0;
xc
for x 2 N;
In Exercise 6 on this subsection, in Section 6.7, we ask
you to verify a result that
implies this assertion.
that is
f 0 ð xÞ
> 0; for x 2 N:
xc
We can rewrite this inequality in the form
f 0 ð xÞ < 0;
f 0 ð xÞ > 0;
for x 2 ðc r; cÞ;
for x 2 ðc; c þ sÞ:
It follows from the Increasing–Decreasing Theorem that f has a local
minimum at c.
A similar argument shows that, if
f 0 ðcÞ ¼ 0
and
f 00 ðcÞ < 0;
then f has a local maximum at c.
Thus we have proved the following result.
Theorem 3 Second Derivative Test
Let f be defined on a neighbourhood of c, and f 0 (c) ¼ 0.
(a) If f 00 (c) > 0, then f (c) is a local minimum of f.
(b) If f 00 (c) < 0, then f (c) is a local maximum of f.
Remark
The theorem gives us no information in the case that f 00 (c) ¼ 0.
Problem 4 For the function f(x) ¼ x3 3x2 þ 1, x 2 R, determine those
points c where f 0 (c) ¼ 0. Using the Second Derivative Test, determine
whether these correspond to local maxima, local minima or neither.
The following diagrams are
helpful in remembering this
result.
6.4
The Mean Value Theorem
237
Inequalities
We now demonstrate how the Increasing–Decreasing Theorem can be used to
prove certain inequalities involving differentiable functions.
Prove that, for > 1,
Example 4
ð1 þ xÞ 1 þ x;
This is a generalisation of
Bernoulli’s Inequality
for x 1:
ð1 þ xÞn 1 þ nx;
Solution
Let
f ð xÞ ¼ ð1 þ xÞ ð1 þ xÞ;
for x 1, n 2 N, that you
met in Sub-section 1.3.3.
for x 2 ½1; 1Þ:
The function f is continuous on [1, 1) and differentiable on (1, 1), and
f 0 ð xÞ ¼ ð1 þ xÞ1 h
i
¼ ð1 þ xÞ1 1 :
Firstly, if 1 < x < 0, then
0 < 1 þ x < 1:
Since > 1, we can then take the ( 1)th power of each side of the inequality
1 þ x < 1 to obtain
ð1 þ xÞ1 < 1;
for 1 < x < 0;
Since 1 > 0, we can use
Rule 5 for inequalities, which
you met in Sub-section 1.2.1.
so that
f 0 ð xÞ < 0;
for 1 < x < 0:
Next, if x > 0, then
1 þ x > 1;
so that
ð1 þ xÞ1 > 1;
Since 1 > 0, we can again
use Rule 5 for inequalities.
for x > 0:
Hence
f 0 ð xÞ > 0;
for x > 0:
Bringing together these two arguments, we have
f 0 ð xÞ < 0;
f 0 ð xÞ > 0;
for 1 < x < 0;
for x > 0:
Also, f (0) ¼ 0.
Hence, by the Increasing–Decreasing Theorem
f is decreasing on ½1; 0;
f is increasing on ½0; 1Þ;
so we have
f ð xÞ f ð0Þ ¼ 0;
for x 2 ½1; 0;
f ð xÞ f ð0Þ ¼ 0;
for x 2 ½0; 1Þ:
It follows that
ð1 þ xÞ ð1 þ xÞ 0;
as required.
for x 2 ½1; 1Þ;
&
6: Differentiation
238
Example 4 illustrates the following general strategy.
To prove that g(x) h(x) on [a, b]:
Strategy
1. Let
f ð xÞ ¼ gð xÞ hð xÞ;
and show that f is continuous on [a, b] and differentiable on (a, b);
2. Prove that:
and f 0 ð xÞ 0 on ða; bÞ;
and f 0 ð xÞ 0 on ða; bÞ;
f ðaÞ 0;
f ðbÞ 0;
EITHER
OR
There is a corresponding
version of the Strategy in
which the weak inequalities
are replaced by strict
inequalities.
This also works if b is 1.
This also works if a is 1.
Prove the following inequalities:
(a) x sin x, for x 2 0; 12 p ;
Problem 5
(b)
6.5
2
3x
2
þ 13 x3 ,
for x 2 [0; 1].
L’Hôpital’s Rule
In order to differentiate the functions sin and exp, we needed to use the
following results
sin x
¼1
x!0 x
lim
and
ex 1
¼ 1:
x!0
x
lim
Each of these limits is the limit as x ! 0 of a quotient in which the numerator
and denominator take the value 0 when x is 0. The evaluation of these limits
was not trivial and required considerable care.
In Analysis and in Mathematical Physics we often need to evaluate limits of
the form
lim
f ð xÞ
x!c gð xÞ
;
x!2
cos 3x
sin x ecos x
and
lim
x
x!0 cosh x
The Quotient Rule for limits
of functions states that
f ð xÞ
f ð xÞ lim
;
¼ x!c
x!c gð xÞ
lim gð xÞ
lim
2
1
exist? If they do, what are their values?
Do we have to evaluate all such limits by a direct argument, or is there a handy
rule which we can apply? We shall find that there is, called l’Hôpital’s Rule.
6.5.1
See Sub-section 5.1.1 and
Problem 8 in Sub-section 5.1.4,
respectively.
where f ðcÞ ¼ gðcÞ ¼ 0:
Such limits cannot be evaluated by the Quotient Rule for limits of functions,
because it does not apply in this situation.
For example, do the limits
limp
Sub-section 6.1.3.
x!c
provided that these last two
limits exist.
Cauchy’s Mean Value Theorem
Recall that the Mean Value Theorem asserts that, under certain conditions, the
graph of a function f(x), for x 2 [a, b], has the property that at some intermediate point c the tangent to the graph is parallel to the chord joining the endpoints of the graph. In other words, there exists a point c in (a, b) such that
Theorem 1, Sub-section 6.4.1.
6.5
L’Hôpital’s Rule
f 0 ðcÞ ¼
239
f ð bÞ f ð aÞ
:
ba
(1)
The key tool that we shall need in Sub-section 6.5.2 in the proof of
l’Hôpital’s Rule is the following result.
Theorem 1
Cauchy’s Mean Value Theorem
Let f and g be continuous on [a, b] and differentiable on (a, b). Then there is
some point c in (a, b) for which
f 0 ðcÞ
fgðbÞ gðaÞg ¼ g0 ðcÞ ff ðbÞ f ðaÞg:
(2)
0
In particular, if g(b) 6¼ g(a) and g (c) 6¼ 0, this equation can be written in the
form
f 0 ðcÞ f ðbÞ f ðaÞ
:
(3)
¼
g0 ð c Þ gð bÞ gð aÞ
This form is easier to
remember.
Notice that the Mean Value Theorem is simply the special case when
g(x) ¼ x. For then g0 (c) ¼ 1 and g(b) g(a) ¼ b a, and equation (3) reduces
to equation (1).
However, Theorem 1 is NOT a simple consequence of the Mean Value
Theorem. For, if we apply the Mean Value Theorem separately to the functions
f and g, we establish the existence of two points c1 and c2 in (a, b) for which
f 0 ðc1 Þ ¼
f ðbÞ f ðaÞ
ba
and
g0 ð c 2 Þ ¼
gð bÞ gð aÞ
:
ba
However, since c1 and c2 are usually unequal, we cannot deduce the existence of a single point c satisfying the statement (2).
Our proof of Theorem 1 is similar to that of the Mean Value Theorem: we
choose a suitable ‘auxiliary’ function h on [a, b] for which the conditions of
Rolle’s Theorem are satisfied.
Proof
Consider the function
hð xÞ ¼ f ð xÞ fgðbÞ gðaÞg gð xÞ ff ðbÞ f ðaÞg; for x 2 ½a; b:
By the Combination Rules for continuity and differentiability, h is continuous
on [a, b] and differentiable on (a, b). Also
hðaÞ ¼ f ðaÞ fgðbÞ gðaÞg gðaÞ
¼ f ðaÞgðbÞ gðaÞ f ðbÞ
f f ðbÞ f ðaÞg
Here two terms cancel.
6: Differentiation
240
and
fgð bÞ gð aÞ g gð bÞ
¼ f ðaÞgðbÞ gðaÞf ðbÞ;
hð bÞ ¼ f ð bÞ
f f ð bÞ f ð aÞ g
Again two terms cancel.
so that h(a) ¼ h(b).
Thus h satisfies the conditions of Rolle’s Theorem on [a, b]. Therefore there
exists a point c in (a, b) for which
h0 ðcÞ ¼ 0 ;
The common value of h(a)
and h(b) does not matter; the
important thing is that the two
are equal.
that is, these exists a point c in (a, b) for which
f 0 ðcÞ
fgðbÞ gðaÞg ¼ g0 ðcÞ f f ðbÞ f ðaÞg:
This is precisely equation (2).
Equation (3) follows immediately from equation (2).
&
By applying Cauchy’s Mean Value Theorem to the functions
1
f ð xÞ ¼ x6 1 and gð xÞ ¼ x4 þ 3x3 þ 3x 3 on ½1; 2;
2
5
3
2
prove that the equation 3x 2x 9x ¼ 3 has at least one root in (1, 2).
Example 1
Solution Since f and g are both polynomials, they are continuous on [1, 2]
and differentiable on (1, 2). It follows that Cauchy’s Mean Value Theorem
applies to the functions f and g on [1, 2].
Now, gð2Þ ¼ 12 16 þ 3 8 þ 3 2 3 ¼ 8 þ 24 þ 6 3 ¼ 35 and
gð1Þ ¼ 12 þ 3 þ 3 3 ¼ 3 12, so that g(2) 6¼ g(1). Also, f 0 (x) ¼ 6x5 and
g0 (x) ¼ 2x3 þ 9x2 þ 3 6¼ 0 on (1, 2). It therefore follows from Cauchy’s Mean
Value Theorem that there exists at least one point c in (1, 2) for which
5
6c
63 0
¼
2c3 þ 9c2 þ 3 35 3 12
63
¼ 1 ¼ 2:
31 2
By cross-multiplying, we see that c satisfies the equation 6c5 ¼ 4c3 þ 18c2 þ 6,
or
3c5 2c3 9c2 ¼ 3:
In other words, the equation 3x5 2x3 9x2 ¼ 3 has at least one root in
&
(1, 2).
Problem 1
By applying Cauchy’s Mean Value Theorem to the functions
f ð xÞ ¼ x3 þ x2 sin x and gð xÞ ¼ xcos x sin x on ½0;p;
prove that the equation 3x ¼ (p2 2) sin x xcos x has at least one root
in (0, p).
6.5.2
l’Hôpital’s Rule
We are now in a position to state the key result that we need to evaluate certain
types of limits in a fairly routine way.
Here we use equation (3).
6.5
L’Hôpital’s Rule
241
Theorem 2 l’Hôpital’s Rule
Let f and g be differentiable on a neighbourhood of the point c, at which
f(c) ¼ g(c) ¼ 0. Then
f ð xÞ
f 0 ð xÞ
exists and equals lim 0 ;
x!c gð xÞ
x!c g ð xÞ
lim
provided that this last limit exists.
We assume that
f 0 ð xÞ
lim 0
exists and equals ‘:
(4)
x!c g ð xÞ
Hence there is some punctured neighbourhood N ¼ (c r, c) [ (c, c þ s) of c
0
on which gf 0 is defined and g0 (x) 6¼ 0.
We now prove that our assumption (4) implies that
f ð xÞ
exists and equals ‘:
(5)
lim
x!c gðxÞ
Let y be any specific point in N for which y > c. The functions f and g are
continuous on [c, y] and differentiable on (c, y), so they satisfy the conditions of
Cauchy’s Mean Value Theorem on [c, y].
Now g(y) g(c) 6¼ 0; for otherwise we would have g(c) ¼ 0 (from our
assumptions) and so g(y) ¼ 0; this would imply, by Rolle’s Theorem, that g0
would vanish somewhere in (c, y), which it does not.
It follows from the conclusion (3) of Cauchy’s Mean Value Theorem that
there exists some point z in (c, y) for which
f 0 ðzÞ f ð y Þ f ðc Þ
¼
g0 ðzÞ gð yÞ gðcÞ
f ð yÞ
¼
; since f ðcÞ ¼ gðcÞ ¼ 0:
g ð yÞ
Now let y ! cþ. Since c < z < y, it follows that z ! cþ too.
But we know that
f 0 ðzÞ
limþ 0
exists and has the value ‘:
z!c g ðzÞ
It follows that
f ð yÞ
exists and has the value ‘:
lim
y!cþ gð yÞ
This is exactly the same as the statement that
f ð xÞ
lim
exists and has the value ‘:
x!cþ gð xÞ
A similar argument shows that
f ð xÞ
lim
exists and has the value ‘:
x!c gð xÞ
&
Combining the last two statements, we obtain the desired result (5).
Proof
We now show how we can use l’Hôpital’s Rule to evaluate the two limits that
we mentioned at the start of this section.
You may omit this proof at a
first reading.
Theorem 1, Sub-section 6.5.1.
This is just a special case of
(4), with z in place of x and
z ! cþ in place of x ! c.
Here x is simply a ‘dummy
variable’: it does not matter
what letter we assign to the
variable in the limit.
We simply repeat the whole
argument, starting with a
specific point y in N for
which y < c.
6: Differentiation
242
cos 3x
sin x ecos x
exists and determine its value.
Example 2
Solution
Prove that limp
(6)
x!2
Let
f ð xÞ ¼ cos 3x
and gð xÞ ¼ sin x ecos x ;
for x 2 R:
Then f and g are differentiable on R, and
p
p
f
¼g
¼ 0;
2
2
so that f and g satisfy the conditions of l’Hôpital’s Rule at p2.
Now
f 0 ð xÞ
3 sin 3x
¼
:
g0 ð xÞ cos x þ sin x ecos x
Then, by l’Hôpital’s Rule, the limit (6) exists and equals
f 0 ð xÞ
3 sin 3x
¼ limp
;
x!2 g0 ð xÞ
x!2 cos x þ sin x ecos x
limp
provided that this last limit exists.
0
By the Quotient Rule for continuous functions, we know that gf 0 is continuous
at p2, so that
f 0 ð xÞ f 0 p2
¼ 0 p
limp 0
x!2 g ð xÞ
g 2
3
¼ ¼ 3:
1
It follows, from l’Hôpital’s Rule, that the original limit (6) must also exist,
&
and that its value is 3.
x2
x!0 cosh x 1
exists and determine its value.
Example 3
Solution
Here we are using the fact (see
Remark 7, Sub-section 5.4.1)
that, if f is continuous at c, then
lim f ð xÞ ¼ f ðcÞ:
x!c
(In future, we shall not
mention this fact explicitly in
this particular connection.)
(7)
Prove that lim
Let
f ð xÞ ¼ x2 and gð xÞ ¼ cosh x 1;
Then f and g are differentiable on R, and
for x 2 R:
f ð0Þ ¼ gð0Þ ¼ 0;
so that f and g satisfy the conditions of l’Hôpital’s Rule at 0.
Now, the derivatives of f and g are
f 0 ð xÞ ¼ 2x
and g0 ð xÞ ¼ sinh x;
for x 2 R:
It follows from l’Hôpital’s Rule that the limit (7) exists and equals
2x
(8)
lim
x!0 sinh x
provided that this last limit exists.
Now, both f 0 and g0 are differentiable on R, and f 0 (0) ¼ g0 (0) ¼ 0; thus f 0 and
0
g satisfy the conditions of l’Hôpital’s Rule at 0.
Since
f 00 ð xÞ ¼ 2 and g00 ð xÞ ¼ cosh x; for x 2 R;
We cannot assert that
f 0 ð xÞ f 0 ð0Þ
¼ 0
;
x!0 g0 ð xÞ
g ð0Þ
lim
since f 0 (0) ¼ g0 (0) ¼ 0.
6.5
L’Hôpital’s Rule
243
it follows from l’Hôpital’s Rule that the limit (8) exists and equals
2
;
(9)
lim
x!0 cosh x
provided that this last limit exists.
The function cosh is continuous on R, and cosh 0 ¼ 1, so that lim cosh x ¼ 1.
x!0
It follows, from the Quotient Rule for limits, that the limit (9) does exist and
that its value is
2
2
¼ ¼ 2:
cosh 0 1
Working backwards, we conclude that the limit (8) exists, and equals 2.
Working further backwards, we conclude that the limit (7) also exists and
&
equals 2.
Before applying a theorem, it is important to check that its conditions are
satisfied; for if they are not satisfied you cannot use the theorem! For instance,
a thoughtless application of l’Hôpital’s Rule can give an incorrect answer!
For example, consider the problem of evaluating
2x2 x 1
:
(10)
x!1
x2 x
If we put f(x) ¼ 2x2 x 1 and g(x) ¼ x2 x, we might be tempted to evaluate
(10) as follows
f ð xÞ
f 0 ð xÞ
4x 1
¼ lim 0
¼ lim
lim
(11)
x!1 gð xÞ
x!1 g ð xÞ
x!1 2x 1
f 00 ð xÞ
4
¼ lim 00
¼ lim
x!1 g ð xÞ
x!1 2
Note that the carefully set out
logic of the argument here is
essential, since it is a
consequence of what we
know from Theorem 2.
lim
¼ 2:
In fact, the value of the limit (10) is 3; so let us review the above argument
carefully!
The line (11) in the calculation is valid, since f and g are differentiable on R
and f(1) ¼ g(1) ¼ 0; so the conditions of l’Hôpital’s Rule are satisfied for the
first application of the Rule. However, f 0 (1) ¼ 3 and g0 (1) ¼ 1, so the conditions are not satisfied for the second application of l’Hôpital’s Rule!
In fact, we should have concluded directly from line (11) that
f 0 ð xÞ
4x 1
¼ 3:
lim 0
¼ lim
x!1 g ð xÞ
x!1 2x 1
The moral is that you must apply l’Hôpital’s Rule with care, particularly to
make the proviso ‘provided that this last limit exists’ at the appropriate points. At
the end, you then work backwards through the chain of applications of the Rule
to reach your conclusion about the limit that you set out originally to examine.
Remark
Notice that we cannot apply l’Hôpital’s Rule to evaluate the limits
sin x
ex 1
¼ 1 and lim
¼ 1;
lim
x!0 x
x!0
x
because we used these limits to find the derivatives of sin x and ex, respectively!
Note that f(1) ¼ g(1) ¼ 0.
Here the careful arguments
that we gave in Examples 2
and 3 have been abandoned in
favour of thoughtless
‘formula-pushing’!
6: Differentiation
244
Problem 2
Prove that the following limits exist, and evaluate them.
1
2x
(a) lim sinh
sin 3x ;
x!0
(c) lim
x!0
6.6
sinðx2 þsin x2 Þ
1cos 4x ;
(b) lim
x!0
1
ð1þxÞ5 ð1xÞ5
2
2
ð1þ2xÞ5 ð12xÞ5
;
cos x
(d) lim sin xx
.
x3
x!0
The Blancmange function
We now meet a function, called the Blancmange function, which is continuous
at each point of R but is differentiable nowhere on R. The construction of the
first function with these properties by Karl Weierstrass caused a huge excitement among mathematicians!
You saw earlier that, if a function is differentiable at a point c, then it is also
continuous at c. The Blancmange function shows, in a very striking way, that
the converse result is false!
6.6.1
You may omit this section,
apart from the next two
paragraphs, at a first reading.
Sub-section 6.1.2.
What is the Blancmange function?
We give the formal definition first, and then look at the underlying geometry
of the graph of the function.
Definition
Let f be the function
x ½x;
f ðxÞ ¼
1 ðx ½xÞ;
0 x [x] 12,
1
2 < x [x] < 1,
where [x] denotes the integer part of x. Then the Blancmange function is
defined on R by the formula
1
1
1
BðxÞ ¼ f ðxÞ þ f ð2xÞ þ f ð4xÞ þ f ð8xÞ þ 2
4
8
1
X
1
¼
f ð2n xÞ:
2n
n¼0
So, what does the graph of B look like? And can we obtain any idea why the
function has its strange properties?
The function f is continuous on R, but is not differentiable at the points 12 n,
for any integer n.
This series converges by the
Comparison Test, since
jf ð2n xÞj 12 ; for all x 2 R:
6.6
The Blancmange function
245
Next, we construct the graphs
y ¼ 12 f ð2xÞ; y ¼ 14 f ð4xÞ; y ¼ 18 f ð8xÞ; . . . ;
we obtain each graph by scaling the previous graph by a factor of 12 in both the
x-direction and the y-direction.
Each graph has twice as many peaks in any interval as its predecessor, and
each of these peaks is half the height of the peaks in its predecessor. Thus the
number of points in the interval where the function is not differentiable doubles
(approximately) at each stage.
We obtain the Blancmange function B by adding together all these functions
to form an infinite series Bð xÞ ¼ f ð xÞ þ 12 f ð2xÞ þ 14 f ð4xÞ þ 18 f ð8xÞ þ .
Thus, for example
B 12 ¼ f 12 þ 12 f ð1Þ þ 14 f ð2Þ þ 18 f ð4Þ þ ¼ 12 þ 12 0 þ 14 0 þ 18 0 þ ¼ 12 ;
and
B
1
4
1 1 1 1
Þ þ 18 f ð2Þ þ 4 þ 2 f 2 þ
4 f ð1
¼ 14 þ 12 12 þ 14 0 þ 18 0 þ ¼f
¼ 14 þ 14 ¼ 12 :
To get an idea of the shape of the graph of B, we look at the graphs of
successive partial sum functions of B. To help you follow the construction, we
also draw in the graph at the previous stage (in light dashes) and the function
being added to it (in heavy dashes).
The first few of these graphs
are included in the diagrams
below.
246
6: Differentiation
Eventually, we obtain the following graph of B:
Here the dashes indicate the
graphs at the early stages and
also the graphs of functions
occurring in the summation.
It looks as though B is continuous. However, it is NOT true in general that the
sum of infinitely many continuous function is itself continuous, so a proof of
the continuity of B is necessary.
Similarly, it looks as though B might not be differentiable.
For example, we have marked the points onthe above
graphs corresponding
to
x ¼ 13. At the first stage, 13 lies in the interval 0; 12 , and so the point 13 ; f 13
lies on a line segment of slope 1. At the next stage, 13 lies in the interval 14 ; 12 ,
and so the point 13 ; f 13 þ 12 f 23 lies on a line segment of slope 0. And so on.
At successive stages in the construction of the graph of B, the point corresponding to 13 lies alternately on line segments of slope 0 and 1. Thus it seems
plausible that the slopes of chords joining the points 13 ; B 13 and (x, B(x))
do not tend to any fixed value as x tends to 13, so that B would not be
differentiable at 13.
Notice in the last diagram above that however closely we look at the graph
of the Blancmange function B, it seems to have small blancmanges growing on
it everywhere. At the nth stage, each horizontal line segment has one or two
mini-blancmanges growing on it, and each sloping line segment has one or two
‘sheared’ mini-blancmanges growing on it.
Finally, notice that the Blancmange function B is periodic, with period 1.
This occurs because f is periodic, with period 1.
We give this in
Sub-section 6.6.2.
We prove this in
Sub-section 6.6.3.
6.6
The Blancmange function
6.6.2
247
Continuity of the Blancmange function
To verify that B is continuous on R, we must use the power of the " definition of continuity.
Theorem 1
Proof
The Blancmange function B is continuous at each point c 2 R.
We must show that
for each positive number ", there is a positive number such that
jBðxÞ BðcÞj < ";
for all x satisfying jx cj < :
(1)
Now, it follows, from the definition of B, that Bð xÞ BðcÞ ¼
1
P
1
n
n
2n ð f ð2 xÞ f ð2 cÞÞ; hence, by the infinite form of the Triangle Inequality
Sub-section 3.3.1.
n¼0
j Bð x Þ Bð c Þ j 1
X
1
j f ð2n xÞ f ð2n cÞj:
n
2
n¼0
(2)
For all x and c, both of the numbers f(2nx) and f(2nc) lie in 0; 12 , so that the
modulus of their difference is at most 12; that is
1
j f ð 2n x Þ f ð 2n c Þ j :
2
Now we choose an integer N such that 21N < 12 ". (This choice is possible, since
the sequence 21n is null.) It follows that
1
1
X
X
1
1 1
n
n
f
ð
2
x
Þ
f
ð
2
c
Þ
j
j
n
2
2n 2
n¼N
n¼N
1
2N
1
(3)
< ":
2
Next, since each of the functions x 7! f(2nx), n ¼ 0, 1, . . ., N 1, is continuous,
it follows that
¼
for each n ¼ 0, 1, . . ., N 1, there is a positive number n such that
n
n
j f ð2 xÞ f ð2 cÞj <
1
4 ";
for all x satisfying jx cj < n :
Now we choose ¼ min{0, 1, . . ., N1}. Thus, for each n ¼ 0, 1, . . .,
N 1, we must have
j f ð2n xÞ f ð2n cÞj < 14 ";
for all x satisfying jx cj < :
It follows that
N 1
N 1
X
X
1
1 1
n
n
f
ð
2
x
Þ
f
ð
2
c
Þ
<
"
j
j
n
n
2
2
4
n¼0
n¼0
<
1
X
1
"
nþ2
2
n¼0
1
¼ ":
2
(4)
We apply the definition of
continuity to each function
in turn.
We use 14 " here, in order
to obtain an " in our final
result (1).
6: Differentiation
248
Substituting the upper bounds from (3) and (4) into inequality (2), we deduce
that, for all x satisfying jx cj < , we have
j B ð xÞ B ð cÞ j N 1
1
X
X
1
1
n
n
j
f
ð
2
x
Þ
f
ð
2
c
Þ
j
þ
j f ð 2 n xÞ f ð 2 n cÞ j
n
n
2
2
n¼N
n¼0
1
1
< "þ "
2
2
¼ ":
This is the desired result (1). Hence B is continuous at c.
6.6.3
&
The Blancmange function is differentiable
nowhere
The proof that B is nowhere differentiable on R is more tricky, and the
following lemma about difference quotients in general (which is of interest
in its own right) plays a crucial role.
Lemma Let f be defined on an open interval I and differentiable at the
point c 2 I. Let {xn} and {yn} be sequences in I converging to c such that
xn c yn
and
xn < yn ;
y
y = f (x)
for n ¼ 0; 1; 2; . . .:
Then
lim
n!1
Proof
f ðyn Þ f ðxn Þ
exists and equals f 0 ðcÞ:
yn xn
We must prove that
xn
for each positive number ", there exists some number X such that
f ðy n Þ f ðx n Þ
0
f ðcÞ < "; for all n > X:
y x
n
c
yn
x
(5)
n
Since f is differentiable at c, we know that, for each positive number ", there
exists some positive number such that
f ð x Þ f ðc Þ
1
0
< "; for all x satisfying 0 < jx cj < ;
f
ð
c
Þ
xc
4
so that
1
(6)
j f ð xÞ f ðcÞ f 0 ðcÞðx cÞj "jx cj; for all x satisfying jx cj < :
4
Now, since xn ! c and yn ! c as n ! 1, it follows that there are numbers X1
and X2 such that
jxn cj < for all n > X1 ; and jyn cj < for all n > X2 ;
This is the definition of
differentiability at c, but
with 14 " in place of ", in order
to obtain an " in our final
result (5).
In fact, this is a strict
inequality if x 6¼ c. However
later in the argument we will
need to allow the possibility
that x ¼ c, so we must write
(6) with a weak inequality
sign.
so, if we set X ¼ max{X1, X2}, we certainly have
jxn cj < and jyn cj < for all n > X:
It now follows from (6) that, for all n > X, we have
1
j f ðxn Þ f ðcÞ f 0 ðcÞðxn cÞj "jxn cj and
4
1
0
j f ðyn Þ f ðcÞ f ðcÞðyn cÞj "jyn cj:
4
We put x ¼ xn and then x ¼ yn
into (6).
6.6
The Blancmange function
249
Hence, we can apply the Triangle Inequality to obtain
j f ðyn Þ f ðxn Þ f 0 ðcÞðyn xn Þj
¼ jf f ðyn Þ f ðcÞ f 0 ðcÞðyn cÞg ff ðxn Þ f ðcÞ f 0 ðcÞðxn cÞgj
j f ðyn Þ f ðcÞ f 0 ðcÞðyn cÞj þ j f ðxn Þ f ðcÞ f 0 ðcÞðxn cÞj
1
1
"jyn cj þ "jxn cj
4
4
1
1
"jyn xn j þ "jyn xn j
4
4
1
¼ "jyn xn j; for all n > X:
2
We first insert some terms and
take them away again.
We use the two inequalities that
we have just verified for n > X.
Since xn c yn, both
jyn cj and jxn cj are
|yn xn|.
Since we know that xn 6¼ yn, we can divide this inequality by the non-zero
term yn xn to obtain, for n > X
f ðy n Þ f ðx n Þ
1
0
f ðcÞ " < ":
y x
2
n
n
&
This is the result (5) that we set out to prove.
Remark
The hypothesis that xn and yn must lie on opposite sides of c cannot generally
be omitted. For example, consider the function
x2 sin 1x ; x 6¼ 0,
f ðxÞ ¼
0;
x ¼ 0,
that you saw earlier is differentiable at 0, with f 0 (0) ¼ 0. If we set
1
1
; for n ¼ 0; 1; 2; . . .;
and yn ¼ xn ¼
ð2n þ 1Þp
2n þ 12 p
then
f ðy n Þ f ðx n Þ
¼
yn xn
1
2
ð2nþ12Þ p2
1
ð2nþ12Þp
ð 0Þ
1
ð2nþ1Þp
y = x2
y
2ð2n þ 1Þ
¼
2n þ 12 p
2
!
as n ! 1:
p
However, the value of this limit is not 0, the value of f 0 (0).
We are now ready to prove the principal result in this sub-section.
Theorem 2
c 2 R.
Problem 7, Sub-section 6.1.2.
The Blancmange function B is not differentiable at any point
Proof In order to apply the lemma, we construct two sequences {xn} and
{yn} converging to c such that the corresponding sequence of difference
quotients is not convergent. We use the method of repeated bisection to
construct {xn} and {yn}.
y =f (x)
xn yn x
y = – x2
6: Differentiation
250
Since B is periodic with period 1, we assume for simplicity that c 2 [0, 1].
[x0, y0] ¼ [0, 1]. Then, since c lies in one of the intervals
We
start by defining
0; 12 and 12 ; 1 , we can define
1
0; ;
if c 2 0; 12 ;
1 ½x1 ; y1 ¼ 1 2 if c 2 2 ; 1 :
2; 1 ;
With this definition of [x1, y1], we have that:
1. ½x1 ; y1 ½x0 ; y0 ;
2. y1 x1 ¼ 12 ;
3. x1 c y1;
4. x1 ¼ 12 p1 and y1 ¼ 12 ð p1 þ 1Þ, for some integer p1.
We can then repeat this process, bisecting the interval [x1, y1] to obtain [x2, y2],
and so on. In this way, we obtain a sequence of closed intervals [xn, yn], for
n ¼ 1, 2, . . ., such that:
1. ½xnþ1 ; ynþ1 ½xn ; yn ;
n
2. yn xn ¼ 12 ;
3. xn c yn;
4. xn ¼ 21n pn and yn ¼ 21n ð pn þ 1Þ, for some integer pn.
Properties 2 and 3 imply that both the sequences {xn} and {yn} converge to c,
but we shall show that the sequence of difference quotients {Qn}, where
Bð y n Þ Bð x n Þ
yn xn
n
¼ 2 ðBðyn Þ Bðxn ÞÞ;
Qn ¼
We use Property 2 for the
value of yn xn.
is not convergent. It will then follow, from the lemma, that B is not differentiable at c.
To prove that the sequence {Qn} is divergent, it is sufficient to prove that
Qnþ1 ¼ Qn 1;
for n ¼ 0; 1; 2; . . .;
(7)
since it then follows that {Qn} cannot converge.
To prove (7), we put
1
1
zn ¼ ðxn þ yn Þ ¼ nþ1 ð2pn þ 1Þ:
2
2
Note that
½xnþ1 ; ynþ1 ¼
½xn ; zn ;
½zn ; yn ;
if c 2 ½xn ; zn ;
if c 2 ðzn ; yn :
We now claim that, for n ¼ 0, 1, 2, . . .
1
1
Bðzn Þ ¼ ðBðxn Þ þ Bðyn ÞÞ þ nþ1 :
2
2
It would then follow from (8) that, if [xnþ1, ynþ1] ¼ [xn, zn], then
Qnþ1 ¼ 2
nþ1
n
ðBðzn Þ Bðxn ÞÞ
¼ 2 ðBðyn Þ Bðxn ÞÞ þ 1 ¼ Qn þ 1;
(8)
The structure of the proof is
as follows
(8) ) (7) ) Theorem 2.
6.6
The Blancmange function
251
whereas, if [xnþ1, ynþ1] ¼ [zn, yn], then
Qnþ1 ¼ 2nþ1 ðBðyn Þ Bðzn ÞÞ
¼ 2n ðBðyn Þ Bðxn ÞÞ 1 ¼ Qn 1:
In either case, the required result (7) would then hold. Thus, in order to
complete the proof that (7) holds, it is sufficient to prove that (8) holds.
To verify (8), note that, for k ¼ 0, 1, 2, . . ., we have
9
1
>
2k xn ¼ nk pn ;
>
>
>
2
>
=
1
k
(9)
2 yn ¼ nk ðpn þ 1Þ;
>
2
>
>
>
>
1
2k zn ¼ nkþ1 ð2pn þ 1Þ; ;
2
We shall shortly use the
1
k P
1
expression
2k f 2 x for
k¼0
B(x), hence some ks now
appear.
and f ( p) ¼ 0 for any integer p. It then follows from the definition of B as an
infinite series that
Bðxn Þ þ Bðyn Þ ¼
¼
Bðzn Þ ¼
¼
1
X
1 k f 2 x n þ f 2k y n
k
2
k¼0
n1
X
1 k f 2 x n þ f 2k y n ;
k
2
k¼0
For, f (2kxn) ¼ f (2kyn) ¼ 0 if
k > n 1.
and
1
X
1 k f 2 zn
2k
k¼0
n
X
1 k f 2 zn :
2k
k¼0
For, f (2kzn) ¼ 0 if k > n.
We deduce from (9) that, for k ¼ 0, 1, 2, . . ., n 1, the terms 2kxn, 2kyn and
2 zn all lie in the interval
k
1
1
k
2 xn ; 2 yn ¼ nk pn ; nk ðpn þ 1Þ ;
2
2
1
and so in some interval of the form 2 qk ; 12 ðqk þ 1Þ , where qk is an integer.
Now, the restriction of f to such an interval is linear, so that we have
f 2k zn ¼ 12 f 2k xn þ f 2k yn ; k ¼ 0; 1; 2; . . .; n 1;
k
hence
n1
n1
X
1 k 1X
1 k f
2
z
f 2 x n þ f 2k y n :
¼
n
k
k
2
2
2
k¼0
k¼0
(10)
Finally
1
1
1
n
f
ð
2
z
Þ
¼
f
p
þ
n
n
2n
2n
2
1
¼ nþ1 :
2
If we then add (10) and (11), we obtain (8), as required.
This completes the proof of Theorem 2.
(11)
&
6: Differentiation
252
6.7
Exercises
Section 6.1
1. Determine whether each of the following functions f is differentiable at the
specified point c; if it is, evaluate the derivative f 0 (c).
x2 ; x 0,
(a) f ð xÞ ¼
c ¼ 0;
x2 ;
x > 0,
1;
x < 1,
c ¼ 1;
(b) f ð xÞ ¼
x2 ; x 1,
tan x; 12 p < x < 13 p;
c ¼ 13 p;
(c) f ð xÞ ¼
x2 ;
x 13 p;
1 x;
x < 1,
(d) f ðxÞ ¼
c ¼ 1;
x x2 ; x 1,
1
x sin x2 ; x 6¼ 0;
c ¼ 0;
(e) f ð xÞ ¼
0;
x ¼ 0;
1
sin x sin x ; x 6¼ 0;
(f) f ð xÞ ¼
c ¼ 0:
0;
x ¼ 0;
2. Write down an expression for a function f with domain (1, 2] and the
following properties:
(a) fL0 exists at 1, and fL0 (1) ¼ 1;
(b) fR0 exists at 1, and fR0 (1) ¼ 2.
Verify that f has the properties (a) and (b).
Section 6.2
1. Use the rules for differentiation to verify that the function
2
f ð xÞ ¼ loge ð1 þ xÞ þ ex ; x 2 ð1; 1Þ, is differentiable on its domain,
and determine its derivative.
2. Write down (without justification) the derivatives of the following
functions:
2
þ1
(a) f ð xÞ ¼ xx1
; x 2 ð1; 1Þ;
(b) f ð xÞ ¼ loge ðsin xÞ;
x 2 ð0; pÞ;
(c) f ð xÞ ¼ loge ðsec x þ tan xÞ;
cos xþsin x
(d) f ð xÞ ¼ cos
x 2 0;
xsin x ;
x 2 12 p;
1
4p :
1
2p
;
3. Prove that the function f(x) ¼ tanh x, x 2 R, has an inverse function f 1 that
is differentiable on (1, 1), and find an expression for (f 1)0 (x).
4. Prove that the function f(x) ¼ tan x þ 3x, x 2 12 p; 12 p , has an inverse
function f 1 that is differentiable on R, and find the value of ( f 1)0 (0).
5. Determine the derivatives of the following functions, assuming that they are
differentiable on their domains:
(a) f ð xÞ ¼ coth x; x 2 R f0g;
(b) f ð xÞ ¼ ðloge xÞloge x ; x 2 ð1; 1Þ:
6.7
Exercises
253
Section 6.3
1. The function f has domain [2, 2] and its graph consists of four line
segments, as shown.
Identify (a) the local minima of f, (b) the minima of f, (c) the local maxima
of f, and (d) the maxima of f, and state where these occur,
2. Let f be the function
f ð xÞ ¼ 2x3 3x2 ; x 2 ½0; 2:
Find (a) the local minima and minima of f, (b) the local maxima and
maxima of f and state where these occur.
3. Prove that, of all rectangles with given perimeter, the square has the greatest
area.
4. Verify that the conditions of Rolle’s Theorem are satisfied by the function
f ð xÞ ¼ 1 þ 2x x2 ;
x 2 ½0; 2;
and determine a value of c in (0,2) for which f 0 (c) ¼ 0.
5. Use Rolle’s Theorem to prove that, if p is a polynomial and l1, l2, . . ., ln are
distinct zeros of p, then p0 has at least (n 1) zeros.
6. Use Rolle’s Theorem to prove that, for any real number l, the function
3
f ð xÞ ¼ x3 x2 þ l; x 2 R;
2
never has two distinct zeros in [0, 1].
Hint: Assume that f has two distinct zeros in [0, 1], and obtain a
contradiction.
7. The function f is twice differentiable on an interval [a, b]; and, for some
point c in (a, b), f(a) ¼ f(b) ¼ f(c). Prove that there exists some point d in
(a, b) with f 00 (d) ¼ 0.
8. The function f is differentiable on a neighbourhood N of a point c; f 0 is
continuous at c and f 0 (c) > 0. Prove that f is increasing on some neighbourhood of c.
Harder: Give an example of a function f that is differentiable on R with the
properties that (a) f 0 (0) > 0 and (b) there is NO open neighbourhood of 0 on
which f is increasing.
9. Let the function f : ½0; 1 7! ½0; 1 be continuous on [0, 1] and differentiable
on (0, 1). We know that there is at least one point c in [0, 1] for which
f (c) ¼ c. Use Rolle’s Theorem to prove that, if f 0 ð xÞ 6¼ 1 in (0, 1), then there
is exactly one such point c.
Hint: Consider the function hð xÞ ¼ f ð xÞ x; x 2 ½0; 1; assume that two
such points c exist, and obtain a contradiction.
You saw this in Problem 2
of Sub-section 4.2.1.
6: Differentiation
254
Section 6.4
1. For each of the following functions, verify that the conditions of the Mean
Value Theorem are satisfied, and determine a value of c that satisfies the
conclusion of the theorem:
(a) f ð xÞ ¼ 2x1
(b) f ð xÞ ¼ x3 þ 2x2 þ x; x 2 ½0; 1:
x2 ; x 2 ½1; 1;
2. Use the Mean Value Theorem to prove that
jsin b sin aj jb aj; for a; b 2 R:
3. Use the Zero Derivative Theorem to prove that
pffiffiffiffiffiffiffiffiffiffiffiffiffi
cosh1 x ¼ loge x þ x2 1 ; for x 1:
4. For the function f(x) ¼ x3 2x2 þ x, x 2 R, determine those points c where
f 0 (c) ¼ 0. Using the Second Derivative Test, determine whether these
correspond to local maxima, local minima, or neither.
5. Prove the following inequalities:
(a) loge x 1 1x ; for x 2 ½1; 1Þ;
1
(b) 4x4 x þ 3; for x 2 ½0; 1;
(c) loge ð1 þ xÞ > x 12 x2 ; for x 2 ð0; 1Þ:
6. Let f be defined on a neighbourhood of c, with f 0 (c) > 0. Prove that there is
some punctured neighbourhood N of c such that
f ð xÞ f ðcÞ
> 0; for x 2 N:
xc
pffiffiffi
7. By applying the Mean Value Theorempto
the
function f ð xÞ ¼ x on the
ffiffiffiffiffiffiffi
ffi
1
1
interval [100, 102], prove that 10 11
< 102 < 10 10
:
8. Let f be differentiable on a closed interval [a, b].
(a) Prove that, if the minimum of f on [a, b] occurs at a, then fR0 (a) 0; and
that, if the minimum of f on [a, b] occurs at b, then fL0 (b) 0.
(b) Prove that, if fR0 (a) < 0 and fL0 (b) > 0, then f 0 (x) ¼ 0 for some x 2 (a, b).
Hint: Check that the minimum of f on [a, b] occurs at some point in (a, b).
(c) By considering the function g (x) ¼ f(x) kx, x 2 [a, b], prove that, if
fR0 (a) < k < fL0 (b), then f 0 (x) ¼ k for some x 2 (a, b).
Section 6.5
1. Verify that the following functions satisfy the conditions of Cauchy’s Mean
Value Theorem on [0, 2], and determine a value of c that satisfies the
conclusion of the theorem
f ð xÞ ¼ x4 þ 2x2
and
gð xÞ ¼ x3 þ 3x;
for x 2 ½0; 2:
2. Use l’Hôpital’s Rule to prove that the following limits exist, and evaluate
the limits:
1
1
x þ sin xÞ
(a) lim sinhðsin
;
x
ðxþ3Þ
(b) lim ð5xþ3Þx1
;
x
;
(c) lim 1xcos
2
xx
(d) lim sinh
sinð3x3 Þ :
x!0
x!0
3
x!1
x!0
2
This result is known as
Darboux’s Theorem, and is
an intermediate value theorem
for f 0 .
7
Integration
We have already used the idea of the area of a set in the plane; for example, to
define the number p and to prove that sinx x ! 1 as x ! 0. However, our earlier
discussions begged the question of what exactly we mean by ‘area’.
For simplicity, we shall restrict our attention in this book to defining only the
area between a graph and the x-axis.
In Section 7.1 we use the idea of lower and upper estimates to give a rigorous
definition of such an area. This is done by trapping the desired area between
underestimates and overestimates, each of which is the sum of the areas of
suitably chosen rectangles.
Then the area between the graph y ¼ f (x), x 2 [a, b], and the segment [a, b] of
the x-axis is defined to be A if:
the least upper bound of the underestimates ¼ A,
and:
the greatest lower bound of the overestimates ¼ A.
We call A the integral of f over [a, b], and denote it by
Z b
Z b
f or
f ð xÞdx:
a
a
In practice, it would be inconvenient if we had to revert to the definition in
order to tell whether a given function was integrable or not, and we shall verify
several criteria for integrability.
In Section 7.2, we identify some large classes of integrable functions, and
verify the standard rules for integrable functions.
In Section 7.3 we meet the Fundamental Theorem of Calculus which enables
us to avoid use of the definition of integrability in order to evaluate many
integrals. For instance, we verify the usual methods for integration by parts and
integration by substitution.
It follows from the
Fundamental Theorem of
Calculus that we can think of
integration as the operation
inverse to differentiation.
255
7: Integration
256
Often it is not possible to evaluate an integral explicitly, and the best that we
can do is to obtain upper and lower estimates for its value. In Section 7.4, we
meet a range of inequalities for integrals, including the Triangle Inequality for
integrals
Z b Z b
f j f j; where a < b:
a
a
We use these inequalities to prove Wallis’s Formula for p, namely that
2 2 4 4 6 6
2n
2n
p
: : : : : : :
:
¼ ;
lim
n!1 1 3 3 5 5 7
2n 1 2n þ 1
2
and to establish the Maclaurin Integral Test, which enables us to determine the
1
1
P
P
1
1
convergence or divergence of series such as
np , for p > 0, and
n log n.
n¼1
n¼2
e
Finally, in Section 7.5, we discuss the evaluation of n!. It is easy to evaluate
n! for values of n up to 10, say, by direct multiplication; but for n ¼ 100 or
n ¼ 200, the number n! cannot be evaluated by a standard scientific calculator.
Surprisingly, we can use integration techniques to obtain an excellent estimate
for n!, called Stirling’s Formula
pffiffiffiffiffiffiffiffinn
n! 2pn
as n ! 1:
e
Before starting to read the chapter, we recommend that you refresh your
memory of greatest lower bounds and least upper bounds of functions.
Recall that, if ƒ is a function defined on an interval I R, then:
A real number m is the greatest lower bound, or infimum, of ƒ on I if:
1. m is a lower bound of ƒ(I);
2. if m0 > m, then m0 is not a lower bound of ƒ(I).
At least not in 2005, when
these words are being written.
We shall also explain the
precise meaning of the
symbol tilda: ‘’.
You met sup and inf in Subsection 1.4.2.
We often denote m by
inf { f(x) : x 2 I}, inf f ð xÞ,
x2I
inf f or simply by inf f.
I
M is the least upper bound, or supremum, of ƒ on I if:
1. M is an upper bound of ƒ(I);
2. if M 0 < M, then M 0 is not an upper bound of ƒ(I).
In Sections 7.1 and 7.2 we shall discuss some new properties of infimum and
supremum that are needed for our work on integrability.
Also, we shall recommend that you omit working your way through quite a
number of detailed proofs during your first reading of this chapter. This is not
necessarily because they are particularly difficult, but simply in order to guide
you through the key ideas first before you return to study some of the ‘gory
details’ later on once you have grasped the overall picture.
7.1
We often denote M by
sup { f(x) : x 2 I}, sup f ð xÞ,
x2I
sup f or simply by sup f.
I
The Riemann integral
Earlier we defined the number p to be the area of the unit disc. We obtained
lower and upper estimates for p by trapping the area of the disc between the
area of a 3 2n-sided inner polygon and the area of a 3 2n-sided outer
polygon, and letting n ! 1.
Sub-section 2.5.4.
7.1
The Riemann integral
257
For simplicity, in this book we do not consider general areas in the plane, but
restrict our attention to the area between a graph y ¼ f(x), x 2 [a, b], and the
interval [a, b] of the x-axis. For a continuous function f, we certainly require
such a definition to agree with our intuitive notion of area!
However, it is not obvious that we can always say that the region between a
graph and the x-axis has an area. For example, how can we define such an area
for the following functions?
(a) f(x) =
x2, 0 ≤ x ≤ 1,
2, 1 < x ≤ 2.
{
(b) f(x) =
–2, 0 ≤ x < 1,
3, x = 1.
{
(c) f(x) =
1, 0 ≤ x ≤ 1, x rational,
0, 0 ≤ x ≤ 1. x irrational.
{
Our approach is to find lower and upper estimates for the area, if such a
thing exists, that we can assign to each region by splitting it up into smaller
regions with areas which we can approximate by rectangles, and use the fact
that
Area of a rectangle ¼ base height.
We now illustrate the underlying idea of these estimates by considering the
situation when the function f is positive on [a, b] (see the diagrams below).
First, we divide up the interval [a, b] into a family of smaller intervals, called a
partition of [a, b]. Then we approximate the area by finding two sequences of
rectangles each with one of the subintervals as base. In one sequence, we
choose rectangles as large as possible so that the sum of their individual areas
forms an underestimate for the ‘area’ of the region; in the other, we choose
rectangles as small as possible so that the sum of their individual areas forms an
overestimate for the ‘area’.
Then, if there is a real number A with the properties:
the least upper bound of the underestimates ¼ A,
When defining p, we used
triangles rather than
rectangles.
7: Integration
258
and:
the greatest lower bound of the overestimates ¼ A,
we define A to be the area between the graph and the x-axis.
We find that we can define such an area for the graphs (a) and (b) above, but
not for the graph (c).
7.1.1
The Riemann integral and integrability
We now start on a process leading to the formal definition of the integral.
Definitions A partition P of an interval [a, b] is a family of a finite
number of subintervals of [a, b]
P ¼ f½x0 ; x1 ; ½x1 ; x2 ; . . . ; ½xi1 ; xi ; . . . ; ½xn1 ; xn g;
(1)
where
For brevity, we sometimes
shorten this
expression for P
simply to ½xi1 ; xi gni¼1 or
f½xi1 ; xi : 1 i ng:
a ¼ x0 < x1 < x2 < < xi1 < xi < < xn1 < xn ¼ b:
The points xi, 0 i n, are called the partition points in P.
The length of the ith subinterval is denoted by xi ¼ xi xi1, and the
mesh of P is the quantity kPk ¼ max fxi g.
a = x0 x1 .... xi–1 xi .... xn = b
1in
A standard partition is a partition with equal subintervals.
For example, consider the partition P of [0, 1], where
1 1 3 3 3 3
P¼
0; ; ; ; ; ; ; 1 :
2 2 5 5 4 4
Equivalently
P ¼ f½0; 0:5; ½0:5; 0:6;
½0:6; 0:75; ½0:75; 1g:
Here
1
1
3 1
1
3 3
3
0 ¼ ; x2 ¼ ¼ ; x3 ¼ ¼
2
2
5 2 10
4 5 20
3 1
x4 ¼ 1 ¼ ;
4 4
and the mesh of P is
1 1 3 1
1
¼ :
kPk ¼ max ; ; ;
2 10 20 4
2
x1 ¼
and
P is not a standard partition of [0, 1], since not all its subintervals are of equal
length.
7.1
The Riemann integral
259
Next, we introduce some notation, mi and Mi, associated with the height of
the graph of a bounded function f on its subintervals. Since we are intending to
develop a definition of integral that will apply to discontinuous functions as
well as to continuous functions, we use the concepts of greatest lower bound
and least upper bound of f on subintervals rather than minimum and maximum
of f on the subintervals.
We also introduce some quantities, L( f, P) and U( f, P), that correspond to
underestimates and overestimates of the possible area.
We need to do this, since, if f
is not continuous on a typical
subinterval [xi1, xi], it may
not possess a minimum or a
maximum on the subinterval.
Definitions Let f be a bounded function on [a, b], and P a partition of [a, b]
given by P ¼ {[xi1, xi] : 1 i n}. We denote by mi and Mi the quantities
mi ¼ inf f f ð xÞ : x 2 ½xi1 ; xi g and Mi ¼ supf f ð xÞ : x 2 ½xi1 ; xi g:
Then the corresponding lower and upper Riemann sums for f on [a, b] are
n
n
X
X
mi xi and U ð f , PÞ ¼
Mi xi :
Lð f , PÞ ¼
i¼1
You met infimum and
supremum earlier, in Subsection 1.4.2. The quantities
mi and Mi exist, since f is
bounded.
i¼1
y
y
y = f (x)
m2
m1
x0
x1
m3
x2
x3
...
M2
mn
xn–1 xn x
M3
x0
Example 1 Let f(x) ¼ x, x 2 [0, 1], and let P ¼
partition of [0, 1]. Evaluate L( f, P) and U( f, P).
y
y = f (x)
M1
Mn
x1
x2
x3
...
1
y=x
xn–1 xn x
1 1 1 1 0; 5 ; 5 ; 2 ; 2 ; 1 be a
Solution In this case, the function f is increasing and continuous. Thus, on
each subinterval in [0, 1], the infimum of f is the value of f at the left end-point
of the subinterval and the supremum of f is the value of f at the right end-point
of the subinterval.
Hence, on the three subintervals in P, we have
1
1
1
1
m1 ¼ f ð0Þ ¼ 0;
x1 ¼ 0 ¼ ;
M1 ¼ f
¼ ;
5
5
5
5
1
1
1
1
1 1
3
M2 ¼ f
x2 ¼ ¼ ;
m2 ¼ f
¼ ;
¼ ;
5
5
2
2
2 5 10
1
1
1 1
M3 ¼ f ð1Þ ¼ 1;
m3 ¼ f
x3 ¼ 1 ¼ :
¼ ;
2
2
2 2
It then follows, from the definitions of L( f, P) and U( f, P), that
3
X
L ð f , PÞ ¼
mi xi ¼ m1 x1 þ m2 x2 þ m3 x3
i¼1
1 1 3 1 1
¼0 þ þ 5 5 10 2 2
3 1
31
;
¼0þ þ ¼
50 4 100
0
1
5
1
2
1
x
We use the fact that, if a
function is continuous on a
closed interval, then it attains
its infimum and supremum
there, by the Extreme Values
Theorem in Sub-section 4.2.3.
7: Integration
260
U ð f , PÞ ¼
3
X
Mi xi ¼ M1 x1 þ M2 x2 þ M3 x3
i¼1
1 1 1 3
1
¼ þ þ1
5 5 2 10
2
1
3 1
69
þ þ ¼
:
¼
25 20 2 100
&
Problem 1 Evaluate L ( f, P) and U ( f, P) for the following function
and partition of [0, 1]:
8
1 1
>
>
< 2x; 0 x < 2 ; 2 < x < 1;
f ð xÞ ¼ 0; x ¼ 1 ;
>
>
2
:
1; x ¼ 1;
x 2 ½0; 1; and P ¼ 0; 13 ; 13 ; 34 ; 34 ; 1 :
Hint: Be careful over the values of m2 and M3; you may find it helpful
to sketch the graph of f.
Example 2 Evaluate L( f, Pn) and U( f, Pn) for the following function and
standard partition of [0, 1]
Notice how a careful laying
out of the calculation for the
various terms makes it
straight-forward to calculate
these two sums.
It is quite important that you
tackle this problem, to make
sure that you understand the
notation being used. You
should also read its solution
carefully.
y
1
f ð xÞ ¼ x; x 2 ½0; 1; and
1 1 2
i1 i
1
; ;...; 1 ;1 ;
Pn ¼
0; ; ; ; . . . ;
n n n
n n
n
y=x
and determine lim Lð f ; Pn Þ and lim U ð f ; Pn Þ, if these exist.
n!1
n!1
Solution In this case, the function f is increasing and continuous. Thus, on
each subinterval in [0, 1], the infimum of f is the value of f at the left end-point
of the subinterval and the supremum of f is the value of f at the right end-point
of the subinterval.
i
Hence, on the ith subinterval i1
n ; n in Pn, for 1 i n, we have
i1
i1
i
i
; Mi ¼ f
mi ¼ f
¼
¼ ; and
n
n
n
n
xi ¼
i i1 1
¼ :
n
n
n
It then follows, from the definitions of L( f, Pn) and U( f, Pn), that
n
n
X
X
i1 1
Lð f ; Pn Þ ¼
mi xi ¼
n
n
i¼1
i¼1
1
¼ 2
n
¼
(
n
X
i¼1
i
n
X
)
1
i¼1
1 nð n þ 1Þ
n1
n
¼
;
2
n
2
2n
0
1
n
2
n
...
n –1
n
1
We follow the general
structure of the solution to
Example 1.
x
7.1
The Riemann integral
U ð f ; Pn Þ ¼
n
X
261
Mi xi ¼
i¼1
n
X
i
i¼1
n
1
n
¼
n
1X
i
n2 i¼1
¼
1 nð n þ 1Þ n þ 1
¼
:
n2
2
2n
:
It follows that
n1 1
¼
and
2n
2
nþ1 1
lim U ð f ; Pn Þ ¼ lim
¼ :
n!1
n!1 2n
2
lim Lð f ; Pn Þ ¼ lim
n!1
n!1
&
Problem 2 Evaluate L( f, Pn) and U( f, Pn) for the following function
and standard partition of [0, 1]
f ð xÞ ¼ x2 ; x 2 ½0; 1; and
1 1 2
i1 i
1
; ; . . .; 1 ; 1 ;
Pn ¼
0; ; ; ; . . .;
n n n
n n
n
and determine lim Lð f , Pn Þ and lim U ð f , Pn Þ, if these exist.
n!1
n!1
Properties of Riemann sums
In all the examples that we have seen so far, the lower Riemann sum for f over
an interval has been less than or equal to the corresponding upper Riemann
R1
sum. Also, in Example 2 we found that the value that ‘we hope for’ for 0 xdx,
namely 12, is greater than all the lower Riemann sums and less than all the upper
Riemann sums; it seems reasonable to ask whether such a property holds in
general.
So we now examine some of the key properties of Riemann sums. We shall
need to use some properties of least upper bounds and greatest lower bounds.
Here we are assuming that our
rigorous treatment of
integration will give the same
answers as those that you
obtained in your initial
Calculus course!
We will set these out carefully
as we need them.
Theorem 1 For any function f bounded on an interval [a, b] and any
partition P of [a, b], L( f, P) U( f, P).
Proof Let P ¼ {[xi1, xi] : 1 i n}. Then, on each subinterval [xi1, xi],
1 i n, we have inf{ f(x) : x 2 [xi1, xi]} sup { f(x):x 2 [xi1, xi]} – in other
words, mi Mi.
It follows that
n
n
X
X
mi xi Mi xi ;
i¼1
i¼1
in other words, L( f, P) U( f, P), as required.
&
Next, we need some techniques for comparing the Riemann sums for
different partitions on the interval [a, b]; this will enable us shortly to verify
that, for two partitions P and P0 of [a, b], we must have L( f, P) U( f, P0 ).
For each term in the left-hand
sum is less than or equal to the
corresponding term in the
right-hand sum.
Theorem 3, below.
7: Integration
262
Notice that this fact does NOT follow from Theorem 1, which deals only with
one partition at a time.
Definition Let P ¼ {[x0, x1], [x1, x2], . . ., [xi1, xi], . . ., [xn1, xn]} be a
partition of an interval [a, b]. Then a partition P0 of [a, b] is a refinement of
P if the partition points of P0 include the partition points of P. A partition Q
of [a, b] is the common refinement of two partitions P and P0 of [a, b] if the
partition points of Q comprise the partition points of P together with the
partition points of P0 .
For example, the partition P0 ¼ 0, 12 , 12 , 35 , 35 , 34 , 34 , 1 of [0, 1] is
a refinement of the partition P ¼ 0, 12 , 12 , 34 , 34 , 1 , since it simply has
one additional partition point 35 as compared with P. Similarly,
P00 ¼ 0, 13 , 13 , 12 , 12 , 35 , 35 , 34 , 34 , 1 is also a refinement of P. However the
partition Q ¼ 0, 15 , 15 , 12 , 12 , 1 of [0, 1] is not a refinement of P, since its
partition points do not include all the partition points of P.
Also, the common refinement of the partitions P ¼ 0, 12 , 12 , 34 , 34 , 1
and P0 ¼ 0, 13 , 13 , 23 , 23 , 1 is
1 1 1 1 2 2 3 3
0; ; ; ; ; ; ; ; ; 1 :
3 3 2 2 3 3 4 4
The point 34 is missing from Q.
We shall need the following crucial result in our work on refinements.
I
Lemma 1 For any bounded function f defined on intervals I and J, with
I J, we have
inf f inf f
x2J
Proof
x2I
and
sup f sup f :
x2I
x2J
By definition of greatest lower bound, we know that inf f f ð xÞ, for
x2J
J
Loosely speaking, the larger
interval gives the function
more space to get smaller and
more space to get larger.
x 2 J. It follows, from the fact that I J, that
inf f f ð xÞ; for x 2 I:
x2J
For inf f is the greatest lower
Hence inf f is a lower bound for f on I, so that inf f inf f .
x2J
x2J
The proof that sup f sup f is similar, so we omit it.
x2I
x2I
x2I
&
bound of f on I.
x2J
Problem 3 Prove that, for any bounded function f defined on intervals
I and J where I J; sup f sup f .
x2I
x2J
Lemma 2 Let f be a bounded function on an interval [a, b]. Let P and P0 be
partitions of [a, b], where P0 is a refinement of P that contains just one
additional partition point. Then
Lð f , PÞ Lð f , P0 Þ
Proof
and
U ð f , P0 Þ U ð f , PÞ:
Let P be the partition
P ¼ f½x0 ; x1 ; ½x1 ; x2 ; . . . ; ½xi1 ; xi ; . . . ; ½xn1 ; xn g
of [a, b], and suppose that P0 contains an additional partition point c in the
particular subinterval [, ] of P.
Loosely speaking, the
addition of one partition point
increases L and decreases U.
a = x0 x1 ... xi–1 xi ... xn = b
7.1
The Riemann integral
263
Note that c must be an interior point of [, ]. For we are assuming that all
the partition points of partitions are distinct; and, if c were an end-point or ,
then P0 would contain that point twice in its set of partition points.
Since [, c] [, ], it follows from Lemma 1 that
inf f inf f
x2½; and
x2½;c
inf f inf f :
x2½;
(2)
x2½c; Now, the terms in the lower sums L ( f, P) and L ( f, P0 ) are the same, except that
the contribution to L ( f, P) associated with the interval [, ] is
We use and here rather
than xi1 and xi simply in
order to avoid sub-subscripts.
α
β
c
inf f ð Þ;
x2½; whereas the contribution to L( f, P0 ) associated with the intervals [, c] and
[c, ] is
inf f ðc Þ þ inf f ð cÞ:
x2½; c
x2½c; It then follows, from the inequalities (2), that
inf f ðc Þ þ inf f ð cÞ inf f ðc Þ þ inf f ð cÞ
x2½; c
x2½c; x2½; x2½; ¼ inf f fðc Þ þ ð cÞg
x2½; ¼ inf f ð Þ:
x2½; Since the lower sums L ( f, P) and L ( f, P0 ) are the same, apart from these
contributions to each, it follows that L ( f, P0 ) L ( f, P), as required.
&
The proof that U ( f, P0 ) U ( f, P) is similar, so we omit it.
Lemma 2 shows that the addition of just one point to a partition increases the
lower Riemann sum and decreases the upper Riemann sum. By applying this
fact a finite number of times, we deduce the following general result.
Theorem 2 Let f be a bounded function on an interval [a,b], and let P and
P0 be partitions of [a,b], where P0 is a refinement of P. then
0
Lð f ; P Þ L ð f ; P Þ
0
and U ð f ; P Þ U ð f ; PÞ:
We now compare the Riemann sums for two different partitions, and discover that all the lower Riemann sums of a bounded function on an interval are
less than or equal to all the upper Riemann sums. This is a significant
improvement on the result of Theorem 1 that, for a given partition P of [a, b],
L( f, P) U( f, P).
Theorem 3 Let f be a bounded function on an interval [a, b], and let P and
P0 be partitions of [a, b]. Then
Lð f , PÞ U ð f , P0 Þ:
Proof Let Q be the common partition of P and P0 . Then, since Q is a
refinement of both P and P0 , we have:
L( f, P) L( f, Q),
L( f, Q) U( f, Q),
by Theorem 2,
by Theorem 1,
U( f, Q) U( f, P0 ),
by Theorem 2.
Loosely speaking, refining a
partition increases L and
decreases U.
7: Integration
264
It follows from this chain of inequalities that L( f, P) U( f, P0 ), as
&
required.
This result is exactly what makes our definitions of lower and upper
Riemann sums useful – whatever the integral of f over an interval [a, b]
might be, if it even exists, we are certainly using lower Riemann sums, all of
which provide underestimates, and upper Riemann sums, all of which provide
overestimates.
Definitions
define:
Let f be a bounded function on an interval [a, b]. Then we
the lower integral of f on [a, b] to be
Sometimes written as
Rb
a f ð xÞdx.
Rb
a f ¼ sup Lð f , PÞ,
P
R
the upper integral of f on [a, b] to be ab f ¼ inf U ð f , PÞ,
P
where P denotes partitions of [a, b].
Further, if the
R b lower and upper integrals are equal, we define the integral
of f on [a, b], a f , to be their common value; that is
Z b
Z b Zb
f ¼
f ¼ f:
a
a
Sometimes written as
Rb
a f ð xÞdx.
Often written as
Rb
a
f ðxÞdx.
a
It is all too easy to be lulled into a sense of false security by a firmly stated
definition! So we now prove the following result that assures us that the above
definitions make sense.
Let f be a bounded function on an interval [a, b]. Then:
Rb
R
(a) The lower integral a f and the upper integral ab f both exist;
Rb
R
(b) a f ab f .
Theorem 4
Proof
(a) Since f is bounded on [a, b], there is some number M such that j f (x)j M
on [a, b]; in particular,
f ð xÞ M;
for x 2 ½a; b:
Thus for any partition P ¼ {[xi 1, xi]: 1 i n} of [a, b], we have
f ð xÞ M;
for x 2 ½xi1 ; xi and 1 i n;
in particular, we have
For, inf f ðxÞ f ðxi Þ M:
mi ¼ inf f ð xÞ M:
½xi1 ; xi ½xi1 ; xi It follows that
Lð f ; P Þ ¼
n
X
i¼1
mi xi n
X
Mxi
i¼1
¼M
n
X
xi ¼ M ðb aÞ:
i¼1
Since all the lower sums L ( f, P) are bounded above by M (b a), it follows
that the greatest lower bound of the lower sums, sup Lðf ; PÞ, must exist.
Rb
P
This greatest lower bound is precisely the lower integral a f .
Also,
Rb
a
f M ðb aÞ.
7.1
The Riemann integral
265
The proof of the existence of the upper integral
we omit it here.
Rb
a
f is very similar, so
(b) Let P and P0 be any two partitions of [a, b]. We know, from Theorem 3,
that
Lð f , PÞ U ð f , P0 Þ:
If we fix P0 for the moment, then we know that U( f, P0 ) serves as an upper
bound for all the lower Riemann sums L( f, P), whichever partition P may
be. It follows,
then, that the least upper bound of the L( f, P), the lower
Rb
integral a f , must satisfy the inequality
Z b
f U ð f , P0 Þ:
(3)
a
a
Rb
It follows from the inequality (3) that a f serves as a lower bound for all
the upper Riemann sums U( f, P0 ), whichever partition P0 may be. It
follows, then, that the greatest lower bound of the U( f, P0 ), the upper
Rb
integral a f , must satisfy the inequality
Z b
Z b
f f:
&
a
Remark
It follows, from the definition of the integral (when it exists) as
Z b
f ¼ sup Lð f , PÞ ¼ inf U ð f , PÞ;
P
P
a
that, for any partition P of [a, b]
Z b
Lð f ; P Þ f U ð f ; PÞ:
a
Now,
already
seen
that, if f(x) ¼ x, x 2 [0, 1], and
have
we
Pn ¼ 0; 1n ; 1n ; 2n ; . . . ; 1 1n ; 1 is a standard partition of [0, 1], then
lim Lð f , Pn Þ ¼
n!1
1
2
and
1
lim U ð f , Pn Þ ¼ :
2
n!1
(4)
R1
Since f is bounded on [0, 1], we know that the lower integral 0 f exists. But
R1
Lð f , Pn Þ 0 f , so that, by the Limit Inequality Rule for sequences, it follows
from (4) that
1
2
Z
1
f:
0
(5)
R1
0
Similarly, since f is bounded on [0, 1], we know that the upper integral
f
R1
exists. But 0 f U ð f , Pn Þ, so that, by the Limit Inequality Rule for
sequences, it follows from (4) that
Z 1
1
(6)
f :
2
0
Example 2, above.
7: Integration
266
We then observe that
Z 1
Z 1
1
f f , in light of (5) and (6), and
2
Z 01
Z 1 0
f f , by part (b) of Theorem 4.
0
0
R1
R1
It follows that we must have 0 f ¼ 0 f ¼ 12, so that f is integrable on [0, 1]
R1
and 0 f ¼ 12.
In other words,
R1
0
xdx ¼ 12.
Problem 4 Use the result of Problem 2 to
R 1prove that the function
f(x) ¼ x2 is integrable on [0, 1], and evaluate 0 f .
Problem
5 Let f be the constant
function x 7! k on [0, 1], and
Pn ¼ 0; 1n ; 1n ; 2n ; . . . ; 1 1n ; 1 a standard partition of [0, 1]. Calculate the Riemann sums L( f, Pn)Rand U( f, Pn). Hence prove that f is
1
integrable on [0, 1], and evaluate 0 f .
However, not all bounded functions defined on closed intervals are integrable!
Example 3
Prove that the function
1; 0 x 1;
f ð xÞ ¼
0; 0 x 1;
y
x rational,
x irrational,
y = 1, x ∈
1
is not integrable on [0, 1].
Solution Let P ¼ {[x0, x1], [x1, x2], . . . , [xi1, xi], . . . , [xn1, xn]} be any
partition of [0, 1].
Then
n
n
X
X
Lð f , PÞ ¼
mi xi ¼
0 xi
i¼1
i¼1
¼ 0;
and
U ð f , PÞ ¼
n
X
Mi xi ¼
i¼1
¼
n
X
i¼1
n
X
y = 0, x ∉
0
xi–1 xi
1
x
For, each subinterval [xi1, xi]
contains both rational and
irrational points.
1 xi
xi ¼ 1:
i¼1
Since all the lower Riemann sums are 0, their least upper bound
R1
0
f is also
zero. Similarly, since all the upper Riemann sums are 1, their greatest lower
R1
bound 0 f is also 1.
R1
R1
&
Since 0 f 6¼ 0 f , it follows that f is not integrable on [0, 1].
7.1.2
Criteria for integrability
It would be tedious to have to go through the ab initio discussion of integrability on each occasion that we wished to determine whether a given function
was integrable on a given closed interval. Therefore we now meet three criteria
that we often use to avoid that process.
We suggest that at a first
reading you omit ALL the
proofs in this sub-section but
read the rest of the text.
7.1
The Riemann integral
267
The first criterion is of particular interest in that its statement says nothing
about the value of the integral itself: it mentions only the difference between
the upper and the lower Riemann sums.
Theorem 5 Riemann’s Criterion for integrability
Let f be a bounded function on an interval [a, b]. Then
y
area = U( f, P) – L( f, P)
f is integrable on ½a; b
if and only if:
y = f (x)
for each positive number ", there is a partition P of [a, b] for which
U ð f ; PÞ Lð f ; PÞ < ":
(7)
a
Proof
b
Recall that L( f, P) U( f, P),
for any partition P.
sup Lð f ; PÞ ¼ inf U ð f ; PÞ; over all partitions P of ½a; b;
P
P
denote by I the common value of these two quantities.
It follows that, for any given positive number ", there are partitions Q and Q0
of [a, b] for which
1
1
(8)
Lð f ; QÞ > I " and U ð f ; Q0 Þ < I þ ":
2
2
Now let P be the common refinement of Q and Q0 . It follows from (8) and
Theorem 2 that
1
Lð f ; PÞ Lð f ; QÞ > I ";
2
1
0
U ð f ; PÞ U ð f ; Q Þ < I þ ":
2
Hence, we may deduce from subtracting these inequalities that
1
1
U ð f ; P Þ Lð f ; P Þ < I þ " I "
2
2
That is, that refining a
partition increases L and
decreases U.
¼ ":
This is precisely the assertion (7).
Suppose next that the statement (7) holds; that is, that for each positive
number ", there is some partition P of [a, b] for which
U ð f ; PÞ Lð f ; PÞ < ":
Since
x
Suppose, first, that f is integrable on [a, b]. Then
Rb
a
f U ðf ; PÞ and
from (9) that
Z b
a
f
Z
Rb
a
(9)
f Lðf ; PÞ, whatever partition P is, it follows
b
f U ð f ; P Þ Lð f ; P Þ
a
< ":
(10)
Since the left-hand side of (10) is some non-negative number independent of ",
Rb
Rb
it follows that a f a f ¼ 0: In other words, f is integrable on [a, b], as
required.
&
By (9).
7: Integration
268
Problem 6 Use Riemann’s Criterion to determine the integrability of
the following functions on [0, 1]:
2; 0 x < 1,
(a) f ð xÞ ¼
3; x ¼ 1;
1; 0 x 1, x rational,
(b) f ð xÞ ¼
0; 0 x 1, x irrational.
Our next criterion for integrability is phrased in terms of a sequence of
partitions and its statement involves a value for the integral.
Theorem 6
Common Limit Criterion
Let f be a bounded function on an interval [a, b].
(a) If f is integrable on [a, b], then there is a sequence {Pn} of partitions of
[a, b] such that
Z b
Z b
f and lim U ðf ; Pn Þ ¼
f:
(11)
lim Lð f ; Pn Þ ¼
n!1
a
n!1
a
(b) If there is a sequence {Pn} of partitions of [a, b] such that
lim Lð f ; Pn Þ
n!1
and
lim U ð f ; Pn Þ both exist and are equal;
n!1
(12)
then
R b f is integrable on [a, b] and the common value of these two limits
is a f .
Proof
Rb
(a) Suppose first that f is integrable on [a, b], and let I denote a f .
Then, corresponding to each integer n 1, there exist some partitions
of [a, b], Qn and Qn0 say, for which
1
1
and U f ; Q0n < I þ :
(13)
Lð f ; Qn Þ > I n
n
Now let Pn be the common refinement of Qn and Qn0 . It follows that
1
I < Lð f ; Qn Þ ðby ð13ÞÞ
n
Lð f ; Pn Þ ðsince Pn is a refinement of Qn Þ
U ð f ; Pn Þ
U f ; Q0n ðsince Pn is a refinement of Q0n Þ
1
ðby ð13ÞÞ
<Iþ :
n
Since the sequences I 1n and I þ 1n both converge to I as n ! 1,
it follows from the Squeeze Rule for sequences and the above inequalities that
L ð f ; Pn Þ ! I
and
U ð f ; Pn Þ ! I
as n ! 1:
This is precisely the statement (11).
(b) Suppose next that the statement (12) holds. Let I denote the common value
of the two limits.
7.1
The Riemann integral
269
It follows from (12) that, for any given positive number ", there are
partitions Q and Q0 of [a, b] for which
1
1
Lð f ; QÞ > I " and U ð f ; Q0 Þ < I þ ":
2
2
So, if we let P be the common refinement of Q and Q0 , it follows that
1
Lð f ; PÞ Lð f ; QÞ > I ";
2
1
0
U ð f ; PÞ U ð f ; Q Þ < I þ ":
2
Hence, we may deduce from subtracting these inequalities that
1
1
U ð f ; P Þ Lð f ; P Þ < I þ " I "
2
2
¼ ":
It follows from Theorem 5 that the function f is therefore integrable on [a, b].
Finally, we use the sequence {Pn} of partitions mentioned in the statement of part (b). For them, if we let n ! 1 in the inequality
Z b
f U ð f ; Pn Þ;
Lð f ; Pn Þ a
Rb
and use the assumption
(12), it follows that lim Lð f ; Pn Þ ¼ a f and
Rb
n!1
&
lim U ð f ; Pn Þ ¼ a f . This concludes the proof.
We have just proved that f is
integrable on [a, b]. Here we
determine the value of its
integral over [a, b].
n!1
Now the statement of Theorem 6 can be loosely paraphrased as saying that a
function f is integrable on [a, b] if and only if there is a sequence of partitions
{Pn} such that the corresponding lower and upper Riemann sums tend to a
common value. This gives us no information on what sort of sequences we
should examine to verify that f is integrable. Our next result says that we only
need to examine any sequence of partitions that we please whose mesh tends to
zero. For instance, a function f is integrable on [a, b] if and only if the lower and
upper Riemann sums for the sequence of standard partitions {Pn} of [a, b] tend
to a common value.
Theorem 7 Null Partitions Criterion
Let f be a bounded function on an interval [a, b], and {Pn} any sequence of
partitions of [a, b] with jjPnjj ! 0.
(a) If f is integrable on [a, b], then
Z b
f
lim Lð f ; Pn Þ ¼
n!1
a
and
lim U ðf ; Pn Þ ¼
n!1
Z
b
f:
a
(b) If there is a number I such that lim Lð f ; Pn Þ and lim U ð f ; Pn Þ both
n!1
n!1
Rb
exist and equal I, then f is integrable on [a, b] and I ¼ a f :
Part (b) is simply a special case of part (b) of Theorem 6, for it is true even
without the additional assumption that jjPnjj ! 0.
Part (a) must be less straight-forward, for, if we drop the assumption
Rb
that jjPnjj ! 0, then it is may not be the case that lim Lðf ; Pn Þ ¼ a f or
n!1
Rb
lim U ð f ; Pn Þ ¼ a f . To see this, consider the function f(x) ¼ x, x 2 [0, 1].
n!1
This is a wonderful
simplication!
Often the partitions Pn will be
standard partitions of [a, b].
7: Integration
270
R1
We have already seen that 0 f ¼ 12. However, if we choose for each partition
Pn simply the trivial partition Pn ¼ {[0,1]}, then L( f, Pn) ¼ 0 and U( f, Pn) ¼ 1
for each n.
We have not used a condition such as jjPnjj ! 0 before, so we need to
establish the following preliminary result before we can tackle the proof of
part (a) of Theorem 7.
For inf f ¼ 0 and sup f ¼ 1.
½0;1
½0;1
Lemma 3 Let f be a bounded function on an interval [a, b], and let be
any positive number. Let P be a partition of [a, b] with jjPjj < , and P0 a
partition of [a, b] with the same partition points as P together with at most N
additional partition points. Then
U ð f ; P0 Þ U ð f ; PÞ N ðM mÞ;
There is an analogous result
for lower Riemann sums, but
we do not need it.
where m ¼ inf f ð xÞ and M ¼ sup f ð xÞ.
x2½a;b
x2½a;b
Remark
We already know, since P0 is a refinement of P, that U( f, P0 ) U( f, P).
Lemma 3 gives us some lower bound to how much smaller U( f, P0 ) can be
than U( f, P).
Proof Let c be the first partition point of P0 that is not a partition point of P.
Then, if P ¼ {[x0, x1], [x1, x2], . . . , [xn1, xn]}, there is some integer i for which
xi1 < c < xi. Denote by Q the partition of [a, b] whose partition points are
those of P together with the additional point c.
Now the upper Riemann sums U( f, P) and U( f, Q) are the same, apart from
the contributions
sup f ðxi xi1 Þ to U ð f ; PÞ
Theorem 2, Sub-section 7.1.1.
a
b
xi
xi–1
a
b
xi–1
x2½xi1 ;xi P
c
Q
xi
and
sup f ðc xi1 Þ þ sup f ðxi cÞ to U ð f ; QÞ:
x2½xi1; c
x2½c; xi Now
Lemma 1 (in Sub-section 7.1.1)
said that, for intervals I and J
where I J
sup f M; by Lemma 1;
x2½xi1 ;xi and
inf f inf f
sup f m and sup f m;
x2½xi1; c
x2J
x2½c; xi x2I
and
sup f sup f :
x2I
x2J
since m is a lower bound for f on all of ½a; b:
It follows that
U ð f ; QÞ U ð f ; PÞ ¼ sup f ðc xi1 Þ þ sup f ðxi cÞ
x2½xi1; c
x2½c; xi sup f ðxi xi1 Þ
x2½xi1; xi mðc xi1 Þ þ mðxi cÞ M ðxi xi1 Þ
¼ mðxi xi1 Þ M ðxi xi1 Þ
¼ ðM mÞðxi xi1 Þ
ðM mÞ;
For 0 < xi xi1 .
7.1
The Riemann integral
271
so that
U ð f ; QÞ U ð f ; PÞ ðM mÞ:
We can interpret this inequality as follows: the insertion of one additional point
into the partition P (to obtain a new partition, Q) does not decrease the upper
Riemann sum by more than (M m).
We can repeat this insertion process up to a further N 1 times to end up
with the final partition P0 ; the effect of this is that the upper Riemann sum of P
does not decrease by more than N(M m). In other words
U ð f ; P0 Þ U ð f ; PÞ N ðM mÞ:
&
We now use this result to prove part (a) of Theorem 7.
Theorem 7, part (a) Let f be a bounded function on an interval [a, b], and
{Pn} any sequence of partitions of [a, b] with jjPnjj ! 0. If f is integrable on
[a, b], then
Z b
Z b
f and lim U ðf ; Pn Þ ¼
f:
lim Lð f ; Pn Þ ¼
n!1
n!1
a
a
Proof We are considering a bounded function f integrable on an interval
[a,b], and {Pn} any sequence of partitions of [a, b] with jjPnjj ! 0. We will use
Lemma 3 to prove that
Z b
f:
lim U ð f ; Pn Þ ¼
n!1
a
(The proof that lim Lð f ; Pn Þ ¼
n!1
Rb
a
f is similar, so we omit it.)
First, note that we can assume that f is non-constant on [a, b], since the result
is clearly immediately true in that situation.
Rb
Denote by I the value of the integral a f , and let m ¼ inf f(x) and M ¼
sup f(x).
For each n 1, we can choose a partition Qn of [a, b] for which
Since f is integrable on [a, b].
1
U ð f ; Qn Þ < I þ :
n
Let Nn denote the number of partition points in Qn, and let
¼
1
:
Nn nðM mÞ
Next, choose any partition Pn of [a, b] for which jjPnjj < .
We then apply Lemma 3, with Pn in place of P, and with Pn0 as the common
refinement of Pn and Qn in place of P0 . It follows that
U f ; P0n U ð f ; Pn Þ Nn ðM mÞ
1
¼ U ð f ; Pn Þ :
n
We can now obtain our crucial estimate for U( f, Pn) from this inequality, as
follows
Such a number exist since
M 6¼ m, for we are assuming
that f is non-constant.
Notice that Pn0 has at most
Nn extra points as compared
with Pn.
By Lemma 3.
From the definition of .
7: Integration
272
1
U ð f ; Pn Þ U f ; P0n þ
n
1
U ð f ; Qn Þ þ
n
2
<Iþ :
n
This is the last inequality,
rewritten.
For Pn0 is a refinement of Qn.
By our choice of Qn.
We can therefore deduce that, for each n 1
2
I U ð f ; Pn Þ < I þ :
n
It follows, by the Squeeze Rule for sequences, that U( f, Pn) ! I as n ! 1.&
We shall use these various criteria in the next section.
7.2
Properties of integrals
In Section 7.1 you met the definition of integrability, and saw that the constant
function and the identity function were integrable on any interval [a, b]. But we
want to know that many more functions than that are integrable! In this section
we extend our family of integrable functions to a very large family indeed –
much wider than, for example, just the continuous functions.
7.2.1
Infimum and Supremum of functions (revisited)
In order to capitalise on the definition of integrability via lower and upper
Riemann sums, we need some further properties of the infimum and supremum
of a function. So, first, we remind you of their definitions.
Definitions
Let ƒ be a function defined on an interval I R. Then:
A real number m is the greatest lower bound, or infimum, of ƒ on I if:
1. m is a lower bound of ƒ(I );
2. if m0 > m, then m0 is not a lower bound of ƒ(I ).
You met these earlier, in Subsection 1.4.2 and again in
Section 7.1.
A real number M is the least upper bound, or supremum, of ƒ on I if:
1. M is an upper bound of ƒ(I );
2. if M0 < M, then M0 is not an upper bound of ƒ(I ).
Remark Notice that, if m is a lower bound of ƒ on I, then inf f m;
I
and, if M is an upper bound of ƒ on I, then sup f M.
I
We will also sometimes need a strategy for verifying that a particular
number is the infimum or supremum of a function on a particular interval.
These are the definitions for
the least upper bound and the
greatest lower bound of a
function on an interval I; there
is a similar definition of these
bounds on a general set S in R.
7.2
Properties of integrals
273
Strategies Let ƒ be a function defined on an interval I R. Then:
To show that m is the greatest lower bound, or infimum, of f on I, check that:
1. f (x) m, for all x 2 I;
2. if m0 > m, then there is some x 2 I such that f (x) < m0 .
To show that M is the least upper bound, or supremum, of f on I, check that:
1. f (x) M, for all x 2 I;
2. if M0 < M, then there is some x 2 I such that f (x) > M0 .
Problem 1 Let f be a bounded function on an interval I. Prove that, for
any constant k
inf fk þ f ð xÞg ¼ k þ inf f f ð xÞg and
x2I
When using these strategies,
we shall often use the
numbers m þ " and M " in
place of m0 and M0 , to fit in
with our increased use of " as
a positive number that can be
as small as we please.
x2I
supfk þ f ð xÞg ¼ k þ supf f ð xÞg:
x2I
x2I
In fact you have already met one property of inf f and sup f in your study of
integration.
Lemma 1, Sub-section 7.1.1.
Lemma 1 For any bounded function f defined on intervals I and J where
I J, we have
inf f inf f and sup f sup f :
Loosely speaking, the larger
interval gives the function
more space to get smaller and
more space to get larger.
x2J
x2I
x2I
x2J
A key tool in our work will be the following innocuous-looking result.
Lemma 2 Let f be a bounded function on an interval I. If, for some
number K, f (x) f (y) K for all x and y in I, then sup f inf f K.
I
Proof
I
Note that K must be
independent of the choice
of x and y in I.
Since f (x) f (y) K for all x and y in I, we have
f ð xÞ K þ f ð yÞ; for all x and y in I:
(1)
It follows from (1) that, for any choice whatsoever of y in I, then K þ f (y) serves
as an upper bound for the set of all possible values of f (x) for x in I. It follows that
For supff ð xÞ : x 2 I g is
simply the least upper bound.
(See also the Remark above.)
supf f ð xÞ : x 2 I g K þ f ð yÞ;
which we may rewrite in the form
supf f ð xÞ : x 2 I g K f ð yÞ
(2)
In a similar way, it follows from (2) that supf f ð xÞ : x 2 I g K serves as a
lower bound for the set of all possible values of f(y) for y in I. It follows that
supf f ð xÞ : x 2 I g K inf f f ð yÞ : y 2 I g:
(3)
We may rearrange the inequality (3) in the form supff ð xÞ : x 2 I g
inf f f ð yÞ : y 2 I g K. This is precisely the required result, since the letters
&
x and y in this last inequality are simply ‘dummy variables’.
Remarks
1. If f is bounded on I and f (x) f (y) < K for all x and y in I, then it may not be
true that sup f inf f < K. For example, if f is the identity function on
I
I ¼ (0, 1), then
I
For inf ff ð yÞ: y 2 I g is simply
the greatest lower bound. (See
also the Remark above.)
7: Integration
274
f ð xÞ f ð yÞ < 1, for all x and y in ð0,1Þ, but sup f inf f ¼ 1 0 ¼ 1:
½0,1
½0,1
2. The conclusion of Lemma 2 also holds if we know that, for some number K,
j f (x) f (y)j K, for all x and y in I, since
f ð xÞ f ð yÞ j f ð xÞ f ð yÞj:
In some situations, the following result related to Lemma 2 is also useful.
Let f be a bounded function on an interval I. Then
sup ff ð xÞ f ð yÞg ¼ sup f inf f :
Lemma 3
x; y2I
I
I
Proof We use the strategy for supremum mentioned earlier.
Let x and y be any numbers in I. Then, in particular
f ð xÞ sup f ;
for all x 2 I ;
I
and, since f ð yÞ inf f , for all y 2 I
I
f ð yÞ inf f ;
for all y 2 I:
I
If we add these inequalities, we obtain
f ð xÞ f ð yÞ sup f inf f ;
I
I
for all x; y 2 I;
so that
sup f f ð xÞ f ð yÞg sup f inf f :
x;y2I
By the earlier Remark.
I
I
To prove the desired result, we now need to prove that
for each positive number ", there are X and Y in I for which
f ð X Þ f ðY Þ > sup f inf f ":
I
I
By the strategy for supremum.
(4)
Now, by the definition of infimum and supremum on I, we know that, since
" > 0, there exist X and Y in I such that
1
2
1
f ð X Þ > sup f "
2
I
1
and f ðY Þ < inf f þ ":
I
2
It follows from these two inequalities that
1 1 f ð X Þ f ðY Þ > sup f " inf f þ "
I
2
2
I
¼ sup f inf f ":
I
I
&
This completes the proof.
Problem 2 Let f (x) ¼ x2 on the interval I ¼ (2, 3]. Determine
inf f; sup f; inf f f ð xÞ f ð yÞg and sup f f ð xÞ f ð yÞg:
I
I
x;y2I
x; y2I
7.2
Properties of integrals
275
Theorem 1 Combination Rules
Let f and g be bounded functions on an interval I. Then:
inf ð f þ gÞ inf f þ inf g and supð f þ gÞ sup f þ sup g;
Sum Rule
I
I
I
I I
Multiple Rule inf ðlf Þ ¼ l inf f and supðlf Þ ¼ l sup f ; for l > 0;
I
I
I
I
Negative Rule inf ðf Þ ¼ sup f and supðf Þ ¼ inf f .
I
I
I
I
I
Proof We will prove only the first part of each Rule, for the proofs of the
second parts are similar.
Sum Rule
For any x and y in I
f ð xÞ inf f
gð xÞ inf g;
and
I
I
so that, by adding these two inequalities, we obtain
f ð xÞ þ gð xÞ inf f þ inf g:
I
I
Since inf f þ inf g is a lower bound for f(x) þ g(x) on I, it follows that
I
I
inf ðf þ gÞ inf f þ inf g:
I
I
I
Multiple Rule
For any x in I, f ð xÞ inf f so that
I
lf ð xÞ l inf f ;
for all x 2 I:
I
Since l > 0.
Thus
inf ðlf ð xÞÞ l inf f :
x2I
I
To prove that inf ðlf ð xÞÞ ¼ l inf f , we now need to prove that:
x2I
I
for each positive number ", there is some X in I for which
lf ð X Þ < l inf f þ ":
I
(5)
Now, since " > 0 and l > 0, we have l" > 0. It follows, from the definition of
infimum of f on I, that there exists an X in I for which
"
f ð X Þ < inf f þ :
I
l
Multiplying both sides by the positive number l, we obtain the desired
result (5).
Negative Rule
For any x in I, f ð xÞ sup f so that
I
f ð xÞ sup f;
for all x 2 I :
I
Thus
inf ðf ð xÞÞ sup f :
x2I
I
7: Integration
276
To prove that inf ðf ð xÞÞ ¼ sup f , we now need to prove that:
x2I
I
for each positive number ", there is some X in I for which
f ð X Þ < sup f þ ":
(6)
I
Now, since " > 0 it follows, from the definition of supremum of f on I, that
there exists an X in I for which
f ð X Þ > sup f "
I
so that
f ð X Þ < sup f þ ":
I
&
This is precisely the inequality (6).
Problem 3 Prove that if f is a bounded function on an interval I, then:
(a) supðlf Þ ¼ l sup f ; for l > 0;
(b) supðf Þ ¼ inf f .
I
I
I
I
We will now use these results on greatest lower bounds and least upper bounds
to address the main topic of this section: integration!
7.2.2
Monotonic and continuous functions
We now prove that bounded functions on an interval [a, b] are integrable if they
are either monotonic on [a, b] or continuous on [a, b]. This provides us with
very many integrable functions!
Neither of these two classes of functions includes the other. For example, the
function
f ð xÞ ¼ x x2 ; x 2 ½0; 1;
is continuous on [0,1] but is not monotonic on [0, 1]; while the function
x; 0 x < 1;
f ð xÞ ¼
; x 2 ½0; 1;
2; x ¼ 1;
You should sketch the graphs
of these two functions to
appreciate these statements.
is monotonic on [0,1] but is not continuous on [0, 1].
Theorem 2 Let f be a bounded function on an interval [a, b]. If f is monotonic on [a, b], then it is integrable on [a, b].
Proof We shall assume that f is increasing on [a, b]; if it is decreasing, the
proof is similar. If f is constant on [a, b], it is certainly integrable on [a, b]. So,
we shall assume, in addition, that f is also non-constant on [a, b]; it follows, in
particular, that f (a) 6¼ f (b).
We will prove that:
for each positive number ", there is a partition P of [a, b] for which
U ð f ; PÞ Lð f ; PÞ < ":
It then follows from the Riemann Criterion for integrability that f is integrable
on [a, b].
For any given " > 0, let P be any partition of [a, b] with mesh jjPjj such that
"
kPk < f ðbÞf
ðaÞ. Then, with the usual notation for Riemann sums
Theorem 5, Sub-section 7.1.2.
7.2
Properties of integrals
277
U ðf ; PÞ Lðf ; PÞ ¼
n
X
ðMi mi Þxi ¼
i¼1
n
X
ðf ðxi Þ f ðxi1 ÞÞxi
Since f is increasing on [a, b].
i¼1
kPk n
X
ð f ðxi Þ f ðxi1 ÞÞ
i¼1
¼ kPk ð f ðbÞ f ðaÞÞ
< ":
&
This completes the proof.
Notice that the brevity of the above proof hides the fact that within the proof
we are using quite a lot of work needed to prove the Riemann Criterion!
Theorem 3 Let f be a bounded function on an interval [a, b]. If f is continuous on [a, b], then it is integrable on [a, b].
Proof
We will prove that:
for each positive number ", there is a partition P of [a, b] for which
U ð f ; PÞ Lð f ; PÞ < ":
It then follows from the Riemann Criterion for integrability that f is integrable
on [a, b].
Our key new tool here is the fact that, since f is continuous on an interval
[a, b], which is a closed interval, it is therefore uniformly continuous on [a, b].
"
It follows that, for any given " > 0; 2ðba
Þ > 0, so that there is a positive
number for which
"
; for all x and y in ½a, b satisfying jx yj < : (7)
j f ðxÞ f ð yÞj <
2ðb aÞ
Now, let P ¼ f½xi1 , xi gni¼1 be any partition of [a, b] with mesh jjPjj such
that jjPjj < . Then, for each i we have, for all x and y in [xi1, xi]
You met uniform continuity
in Section 5.5; this assertion is
Theorem 2 there.
Here the choice of depends
ONLY on ", the same choice
whatever x and y may be. We
"
choose 2ðba
Þ rather than ", for
convenience later on.
jx yj xi xi1
kPk < :
It follows, from (7), that
"
; for all x and y in ½xi1 ; xi ;
2ð b aÞ
so that, by Remark 2 following Lemma 2
"
sup f ð xÞ inf f ð xÞ :
2
ð
b
aÞ
x2
½
x
;x
i1 i
x2½xi1 ;xi j f ð x Þ f ð y Þj <
In Sub-section 7.2.1. Recall
that the conclusion of
Lemma 2 is a weak inequality,
not a strong inequality.
Then, with the usual notation for Riemann sums
n
X
U ð f ; P Þ Lð f ; P Þ ¼
ðMi mi Þxi
i¼1
n
X
"
xi
2ðb aÞ i¼1
"
ð b aÞ
¼
2ðb aÞ
1
¼ " < ":
2
This completes the proof.
&
7: Integration
278
7.2.3
Rules for integration
You may omit the details
of all the proofs in this
sub-section at a first reading.
Our first task is to prove the Combination Rules for integrable functions.
Theorem 4
Combination Rules
Let f and g be integrable on [a, b]. Then so are:
Rb
Rb
Rb
Sum Rule
f þ g, and a ð f þ gÞ ¼ a f þ a g;
Rb
Rb
Multiple Rule lf,
and a ðlf Þ ¼ l a f ;
Product Rule fg;
Modulus Rule j f j.
Proof
We use the usual notation for Riemann sums and integrals.
Sum Rule
Let {Pn} be any sequence of partitions of [a, b], where jjPnjj ! 0 as n ! 1.
Since f and g are integrable on [a, b], it follows, from the Null Partitions
Criterion for integrability, that
Z b
Lð f ; Pn Þ and U ð f ; Pn Þ !
f as n ! 1;
Pn ¼ f½xi1 ; xi gni¼1 :
Theorem 7, Sub-section 7.1.2.
a
and
Lðg; Pn Þ
and U ðg; Pn Þ !
Z
b
g
as n ! 1:
a
Then
Lð f ; Pn Þ þ Lðg; Pn Þ ¼
n
X
i¼1
¼
n
X
i¼1
½xi1 ; xi n X
i¼1
inf f xi þ
n
X
n
X
i¼1
i¼1
By the definition of lower
Riemann sum.
inf f þ inf g xi
½xi1 ; xi ½xi1 ; xi inf ð f þ gÞ xi
ð¼ Lðf þ g; Pn ÞÞ By the Combination Rules for
inf and sup, Theorem 1,
Sub-section 7.2.1.
ð¼ U ðf þ g; Pn ÞÞ
½xi1 ; xi sup ð f þ gÞ xi
i¼1 ½xi1 ; xi n
X
inf g xi
½xi1 ; xi (
)
sup f þ sup g
½xi1 ; xi xi
By the Combination Rules for
inf and sup, again.
½xi1 ; xi ¼ U ð f ; Pn Þ þ U ðg; Pn Þ:
We now let n ! 1 in this last set of inequalities. Then, since
Z b
Z b
Z b
Z b
Lð f ; Pn Þ þ Lðg; Pn Þ !
f þ g and U ðf ; Pn Þ þ U ðg; Pn Þ !
f þ g;
a
a
a
a
it follows, from the Limit Inequality Rule for sequences, that
Z b
Z b
Z b
Z b
Lð f þ g; Pn Þ !
f þ g and U ðf þ g; Pn Þ !
f þ g; as n ! 1:
a
a
a
a
Hence, by part (b) of the Null Partitions Criterion for integrability, we deduce
Rb
Rb
Rb
that f þ g is integrable on [a, b] and a ðf þ gÞ ¼ a f þ a g:
Theorem 7, Sub-section 7.1.2.
7.2
Properties of integrals
279
Multiple Rule
Let {Pn} be any sequence of partitions of [a, b], where jjPnjj ! 0 as n ! 1.
Since f is integrable on [a, b], it follows from the Null Partitions Criterion for
integrability that
Z b
Lð f ; Pn Þ and U ð f ; Pn Þ !
f as n ! 1:
Pn ¼ f½xi1 ; xi gni¼1 :
Theorem 7, Sub-section 7.1.2.
a
Then, if l > 0
Lðl f ; Pn Þ ¼
n
X
inf ðlf Þ xi
½x ; x i¼1 i1 i
n
X
¼l
By the Combination Rules for
inf and sup, Theorem 1,
Sub-section 7.2.1.
inf f xi
i¼1
½xi1 ; xi ¼ lLðf ; Pn Þ ! l
Z
b
f
as n ! 1;
a
while, if l < 0
Lðlf ; Pn Þ ¼
n
X
inf ðlf Þ xi
½x ; x i¼1 i1 i
n
X
¼l
By the Combination Rules for
inf and sup.
sup f xi
i¼1 ½xi1 ; xi ¼ lU ðf ; Pn Þ ! l
Z
b
f
a
It follows that, for all real l, Lðlf ; Pn Þ ! l
as n ! 1:
Rb
a
f as n ! 1.
Rb
A similar argument shows that, for all real l, U ðlf ; Pn Þ ! l a f as n ! 1.
Since the limits of the two sequences of Riemann sums are equal, it follows
from part (b) of the Null Partitions Criterion for integrability, that lf is
integrable on [a, b] and
Z b
Z b
ðlf Þ ¼ l f :
a
For, if l ¼ 0, the result is
trivial.
Theorem 7, Sub-section 7.1.2.
a
Product Rule
Since f and g are bounded on [a, b], there is some number M, say, such that
j f ð x Þj M
and
jgð xÞj M;
for all x 2 ½a; b:
Now let {Pn} be any sequence of partitions of [a, b], where jjPnjj ! 0 as
n ! 1. It follows that, if x, y 2 [xi1, xi] for some i, then
j f ð xÞgðxÞ f ð yÞgðyÞj ¼ jff ð xÞgðxÞ f ð xÞgð yÞg þ ff ðxÞgð yÞ f ð yÞgðyÞgj
¼ j f ð xÞfgðxÞ gð yÞg þ gð yÞff ð xÞ f ð yÞgj
M jgðxÞ gðyÞj þ M jf ðxÞ f ð yÞj
(
)
(
)
M
sup g inf g
½xi1 ;xi ½xi1 ;xi þM
sup f inf f ;
½xi1 ;xi ½xi1 ;xi Pn ¼ f½xi1 ; xi gni¼1 :
7: Integration
280
so that, by Lemma 2 in Sub-section 7.2.1 and Remark 2 following that Lemma
(
sup ð fgÞ inf ð fgÞ M
½xi1 ;xi )
(
sup g inf g
½xi1 ;xi ½xi1 ;xi )
þM
½xi1 ;xi (8)
sup f inf f :
½xi1 ;xi ½xi1 ;xi In terms of the Riemann sums for fg, it follows!that
U ð fg; Pn Þ Lð fg; Pn Þ ¼
n
X
i¼1
M
n
X
i¼1
By definition of U and L.
sup ð fgÞ inf ð fgÞ xi
½xi1 ;xi ½xi1 ;xi (
)
sup g inf g xi þ M
n
X
½xi1 ;xi ½xi1 ;xi i¼1
(
)
sup f inf f xi
½xi1 ;xi ½xi1 ;xi By (8).
¼ M fU ðg; Pn Þ Lðg; Pn Þg þ M fU ðf ; Pn Þ Lðf ; Pn Þg
! 0 as n ! 1:
It then follows, from part (b) of the Null Partitions Criterion for integrability,
that fg is integrable on [a, b].
Modulus Rule
Let {Pn} be any sequence of partitions of [a, b], where jjPnjj ! 0 as n ! 1.
Then, if x, y 2 [xi 1, xi] for some i, we have
j f ð xÞj j f ð yÞj j f ð xÞ f ð yÞj;
so that
j f ð xÞj j f ð yÞj sup f inf f
By part (a) of the Null
Partitions Criterion for
integrability (Theorem 7,
Sub-section 7.1.2), since f and
g are integrable on [a, b].
Pn ¼ f½xi1 ; xi gni¼1 :
This follows from the ‘reverse
form’ of the Triangle
Inequality, that you met in
Sub-section 1.3.1.
½xi1 ;xi ½xi1 ;xi and therefore
sup j f j inf j f j sup f inf f :
½xi1 ;xi ½xi1 ;xi (9)
½xi1 ;xi ½xi1 ;xi In terms of the Riemann sums for j f j, it follows that
U ðj f j; Pn Þ Lðj f j; Pn Þ ¼
n
X
!
sup j f j inf j f j xi
½xi1 ;xi i¼1
½xi1 ;xi By definition of U and L.
!
n
X
sup f inf f xi
½xi1 ;xi i¼1
½xi1 ;xi ¼ U ð f ; Pn Þ Lð f ; Pn Þ
!0
By Remark 2 following
Lemma 2, Sub-section 7.2.1.
as n ! 1:
It then follows, from part (b) of the Null Partitions Criterion for integrability,
&
that jfj is integrable on [a, b].
By (9).
By part (a) of the Null
Partitions Criterion for
integrability (Theorem 7,
Sub-section 7.1.2), since f is
integrable on [a, b].
Our next result is often overlooked as obvious – which it is not!
a
Theorem 5
The Sub-interval Theorem
(a) Let f be integrable on [a, Rb], and Rlet a <Rc < b. Then f is integrable on
b
c
b
both [a, c] and [c, b], and a f ¼ a f þ c f :
(b) Let f be integrable on [a, c] and [c, b], where a < c < b. Then f is
integrable on [a, b].
c
b
7.2
Properties of integrals
281
Proof
(a) Let {Pn}, where Pn ¼ f½xi1 , xi gni¼1 , be any sequence of partitions of [a, b]
such that c is a partition point of each Pn, for n 2, and such that jjPnjj ! 0
as n ! 1.
For n 2, let Qn consist of those subintervals in Pn that lie in [a, c], and
Qn0 consist of those subintervals in Pn that lie in [c, b]. Thus {Qn} is a
partition of [a, c] with jjQnjj ! 0 as n ! 1, and {Qn0 } is a partition of [c, b]
with jjQn0 jj ! 0 as n ! 1.
Now, with the usual notation for Riemann sums, we have
n
P
U ð f ; Qn Þ Lð f ; Qn Þ ¼ the sum of those terms in ðMi mi Þxi
i¼1
that correspond to subintervals ½xi1 ; xi that lie in ½a; c
n
X
ðMi mi Þxi
i¼1
¼ U ð f ; P n Þ Lð f ; P n Þ ! 0
as n ! 1:
It follows from the Null Partitions Criterion that f is integrable on [a, c].
A similar argument proves that f is integrable on [c, b].
In particular, it follows that
Z c
Z b
Lð f ; Qn Þ !
f and L f ; Q0n !
f as n ! 1:
a
c
So, if we let n ! 1 in the identity
Lðf ; Pn Þ ¼ Lðf ; Qn Þ þ L f ; Q0n ;
This identity follows from the
definitions of Pn, Qn and Qn0 .
we deduce that
Z c
Z b
Z b
f ¼
f þ f:
a
a
By the definition of integral.
c
(b) In the opposite direction, suppose that f is integrable on [a, c] and [c, b].
Let {Pn}, where Pn ¼ f½xi1 ; xi gni¼1 , be any sequence of partitions of
[a, b] such that c is a partition point of each Pn, for n 2, and jjPnjj ! 0 as
n ! 1.
For n 2, let Qn consist of those subintervals in Pn that lie in [a, c], and
Qn0 consist of those subintervals in Pn that lie in [c, b]. Thus {Qn} is a
partition of [a, c] with jjQnjj ! 0 as n ! 1, and {Qn0 } is a partition of [c, b]
with jjQn0 jj ! 0 as n ! 1.
Then, with the usual notation for Riemann sums, we have
U ð f ; Pn Þ Lð f ; Pn Þ ¼ the sum of those terms in
n
P
ðMi mi Þxi that
i¼1
correspond to subintervals ½xi1 ; xi that lie in ½a; c
n
X
þ the sum of those terms in
ðMi mi Þxi that
i¼1
correspond to subintervals
½xi1 ;xi that
lie in
½c; b
¼ fU ð f ; Qn Þ Lð f ; Qn Þg þ U f ; Q0n L f ; Q0n
! 0 as n ! 1:
It follows, from the Null Partitions Criterion, that f is integrable on
&
[a, b].
7: Integration
282
In light of Theorem 5 we now introduce some notational conventions.
Conventions Let f be integrable on [a, b], where a < b. Then we make the
following definitions
Z a
Z a
Z b
f ¼ 0 and
f ¼ f:
a
b
a
With these conventions, we can assert that if f is integrable on any interval that
Rc
Rb
Rb
contains the points a, b and c, then a f ¼ a f þ c f .
7.3
These conventions apply
where the upper and lower
limits of the integral are the
same, or are in reverse order
on the real line.
To prove this assertion, we
need to look at all the possible
orderings of the three points
on R, which is rather tedious;
so we omit it!
Fundamental Theorem of Calculus
7.3.1
Fundamental Theorem of Calculus
It would be very tedious if, each time that we wished to evaluate an integral, we
had to calculate upper and lower sums and find their greatest lower bound and
least upper bound, respectively. Fortunately, this is usually unnecessary, as
there is a short-cut using the idea of a primitive.
Definition Let f be a function defined on an interval I. Then the function F
is a primitive of f on I if F is differentiable on I and
F0 ¼ f :
For example, if
1
; x 2 ½1; 1;
4 x2
1
2þx
FðxÞ ¼ loge
; x 2 ½1; 1;
4
2x
f ðxÞ ¼
then
is a primitive of f on [ 1, 1], since
1 2 x ð2 xÞ þ ð2 þ xÞ
F 0 ðxÞ ¼ 4 2þx
ð2 xÞ2
1
4
1
¼
¼ :
4 ð2 þ xÞð2 xÞ 4 x2
Problem 1
1 ffi
(a) Prove that the
f ðxÞ ¼ pffiffiffiffiffiffiffi
, x 2 (2, 1), has a primitive
function
x2 4
pffiffiffiffiffiffiffiffiffiffiffiffiffi
F ð xÞ ¼ loge x þ x2 4 .
pffiffiffiffiffiffiffiffiffiffiffiffiffi
(b) Prove that the function f ð xÞ ¼ 4 x2 , x 2 (2, 2), has a primitive
pffiffiffiffiffiffiffiffiffiffiffiffiffi
F ð xÞ ¼ 12 x 4 x2 þ 2 sin1 2x .
The connection between primitives of a function f and the integral of f on an
interval I is given by the following result, which is one of the most important
theorems in Analysis.
Sometimes we denote
a
R
primitive of f byR f , and we
denote F(x) by f ðxÞdx:
7.3
Fundamental Theorem of Calculus
283
Theorem 1 Fundamental Theorem of Calculus
Let f be integrable on [a, b], and F a primitive of f on [a, b]. Then
Z b
f ¼ F ðbÞ F ðaÞ:
Often F(b) F(a) is written
as
½F ðxÞba or F ð xÞjba :
a
Proof
Since f is integrable on [a, b], there is a sequence
Pn ¼ f½x0 ; x1 ; ½x1 ; x2 ; . . .; ½xn1 ; xn g
of partitions of [a, b], with jjPnjj ! 0 as n ! 1, such that
Z b
Z
f and lim U ðf ; Pn Þ ¼
lim Lðf ; Pn Þ ¼
n!1
n!1
a
We use the standard notation
for lower and upper sums and
partitions.
b
f:
(1)
a
Now, the function F satisfies the conditions of the Mean Value Theorem on
each subinterval [xi1, xi], for i ¼ 1, 2, . . ., n. It follows that there exists some
point ci 2 (xi1, xi) such that
F ðxi Þ F ðxi1 Þ ¼ F 0 ðci Þðxi xi1 Þ
¼ f ðci Þxi :
Next, since mi f(ci) Mi, for i ¼ 1, 2 . . ., n, it follows that
n
n
n
X
X
X
mi xi f ðci Þxi Mi xi :
i¼1
i¼1
(2)
For F 0 ¼ f and xi ¼ xi xi1.
Recall that
(3)
mi ¼ inf ff ð xÞ : x 2 ½xi1 ; xi g;
Mi ¼ supff ð xÞ : x 2 ½xi1 ; xi g:
i¼1
n
P
We may now apply (2) to the general term in the sum
f ðci Þxi ; thus it
i¼1
follows from (3) that
n
X
Lð f ; Pn Þ fF ðxi Þ F ðxi1 Þg U ð f ; Pn Þ:
i¼1
Since the middle term is a ‘telescoping’ sum, we deduce that
Lð f ; Pn Þ F ðbÞ F ðaÞ U ð f ; Pn Þ:
(4)
Now let n ! 1 in (4), using the facts in (1); this thus gives
Z b
Z b
f F ð bÞ F ð aÞ f:
a
In other words, FðbÞ FðaÞ ¼
Example 1
Evaluate
R1
0
Rb
a f;
By the Limit Inequality Rule
for sequences.
a
as required.
&
2x dx.
Solution The function f(x) ¼ 2x is continuous on [0, 1] and so is integrable
on [0, 1]; from the Table of Standard Primitives, it has a primitive
x
F ð xÞ ¼ log2 2 on ½0; 1:
e
It follows from the Fundamental Theorem of Calculus that
x 1
Z 1
2
1
:
&
2x dx ¼
¼
loge 2 0 loge 2
0
Problem 2 Using the Table of Standard Primitives, evaluate the
following integrals:
1
R4
Re
(a) 0 ðx2 þ 9Þ2 dx;
(b) 1 loge x dx:
The Table of Standard
Primitives appears in
Appendix 2.
7: Integration
284
7.3.2
Finding primitives
The Fundamental Theorem of Calculus asserts that, if f is integrable on [a, b],
Rb
and F a primitive of f on [a, b], then a f ¼ F ðbÞ F ðaÞ. This does not make it
clear whether a primitive of f is unique.
In fact, for any given function f, a primitive of f is NOT unique. For example,
the functions
x 7! sin1 x and x 7! cos1 x; x 2 ð1; 1Þ;
1 ffi
are both primitives of the function x 7! pffiffiffiffiffiffiffi
; so that, for instance
1x2
Z p1ffi
1
pffi p1ffi
2
1
pffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ sin1 x 0 2 ¼ cos1 x 0 2
1 x2
0
1
¼ p:
4
However, two primitives of a given function f are related; they can only
differ by a constant function.
Theorem 2
Uniqueness Theorem for Primitives
For example
1
sin1 x cos1 x ¼ p:
2
Let F1 and F2 be primitives of a function f on an interval I. Then there exists
some constant c such that
F2 ð xÞ ¼ F1 ð xÞ þ c; for x 2 I:
Since F1 and F2 are primitives of f on I
Proof
F10 ð xÞ ¼ f ð xÞ and
F20 ð xÞ ¼ f ð xÞ;
for x 2 I;
so that
F20 ð xÞ ¼ F10 ð xÞ; for x 2 I:
It follows that there exists some constant c such that
F2 ð xÞ ¼ F1 ð xÞ þ c; for x 2 I:
&
By Corollary 2, Subsection 6.4.2.
Our stock of primitives can be considerably extended by use of the following
Combination Rules. For convenience, we include here an extra rule which
applies to a composite function f g for which the function g is just an x-scaling.
Theorem 3
Combination Rules
Let F and G be primitives of f and g, respectively, on an interval I, and l 2 R.
Then, on I:
Sum Rule
fþg
has primitive F þ G;
Multiple Rule
Scaling Rule
These rules are easily proved
using the corresponding rules
for derivatives.
lf
has primitive lF;
x j! f(lx) has primitive x 7! 1l F ðlxÞ.
For example, it follows from the Table of Standard Primitives and the
Combination Rules that the function x 7! 3 sinhð2xÞ þ 1x, x 2 (0, 1), has a
primitive x 7! 32 coshð2xÞ þ loge x.
Problem 3 Using the Table of Standard Primitives and the Combination Rules, find a primitive of each of the following functions:
2
x 2 ð0; 1Þ;
(a) f ð xÞ ¼ 4 loge x 4þx
2 ;
2x
(b) f ð xÞ ¼ 2 tanð3xÞ þ e sin x; x 2 16 p; 16 p :
In applications of Theorem 3
we do not usually bother to
mention the theorem
explicitly.
7.3
Fundamental Theorem of Calculus
7.3.3
285
Techniques of integration
We are now in a position to use the Fundamental Theorem of Calculus to give
rigorous proofs of some standard techniques for integration.
Theorem 4
Integration by Parts
If f and g are differentiable on an interval [a, b], and f 0 and g0 are continuous
on [a, b], then
Z b
Z b
fg0 ¼ ½ fgba f 0 g:
a
a
We may reformulate the Product Rule for derivatives in the form
Proof
0
0
Theorem 1, Sub-section 6.2.1.
0
ð fgÞ ¼ f g þ fg ;
in other words, that fg is a primitive on [a,b] of f 0 g þ fg0 .
It follows from the Fundamental Theorem of Calculus that
Z b
ð f 0 g þ fg0 Þ ¼ ½ fgba ;
so that
Rb
a
a
0
fg ¼
Rb
½ fgba a f 0 g,
Recall the notation that
&
as required.
Strategy to evaluate an integral using integration by parts
0
1. Write the original function in the form fg , where f is a function that you
can differentiate and g0 a function that you can integrate.
Rb
Rb
2. Use the formula a fg0 ¼ ½ fgba a f 0 g:
Example 2
Evaluate the integral
R1
0
There is a similar strategy for
finding a primitive by
integration by parts.
tan1 xdx:
Solution Here we use a common trick: we consider g0 (x) to be the factor 1,
and use integration by parts. Thus
Z 1
Z 1
1
tan xdx ¼
1 tan1 xdx
0
0
Z 1
1
xdx
1
¼ x tan x 0 1
þ x2
0
1
1
¼ tan1 1 loge 1 þ x2
2
0
1
1
1
1
1
&
¼ p loge 2 þ loge 1 ¼ p loge 2:
4
2
2
4
2
Problem 4
1
(a) Find a primitive of the function f ð xÞ ¼ x3 loge x; x 2 ð0; 1Þ:
Rp
(b) Evaluate the integral 02 x2 cos xdx:
Hint:
½ fgba ¼ f ðbÞgðbÞ f ðaÞgðaÞ:
Use integration by parts twice.
You will meet further instances of integration by parts in the next section.
Here
f ðxÞ ¼ tan1 x; g0 ð xÞ ¼ 1;
so that
f 0 ð xÞ ¼
1
; gðxÞ ¼ x:
1 þ x2
7: Integration
286
Next we look at the method of integration by substitution, and discover that
one needs to be just a little careful in making substitutions.
Theorem 5 Integration by Substitution
If f is continuous on [a, b], g differentiable on [, ], g0 continuous on [, ],
and g([, ]) [a, b], then
Z gðÞ
Z f ð xÞdx ¼
f ðgðtÞÞg0 ðtÞdt:
gðÞ
If, in addition, g() ¼ a, g() ¼ b and g possesses an inverse function g1 on
[a, b], then
Z b
Z g1 ðbÞ
f ð xÞdx ¼
f ðgðtÞÞg0 ðtÞdt:
g1 ðaÞ
a
Proof
Let F be a primitive of f on [a, b], and define the function h by
hðtÞ ¼ F ðgðtÞÞ;
t 2 ½; :
Note that, in this situation, for
g to possess an inverse
function, g must be strictly
increasing.
There is an analogous result in
the situation that g is strictly
decreasing.
(5)
By the Composition Rule for derivatives, applied to the composite
h(t) ¼ F(g(t)), we have that
h0 ðtÞ ¼ F 0 ðgðtÞÞg0 ðtÞ:
Theorem 2, Sub-section 6.2.2.
(6)
By the Fundamental Theorem of Calculus applied to h on [, ], we have
Z h0 ðtÞdt ¼ hð Þ hðÞ:
(7)
If we now substitute for h and h0 from (5) and (6) into the two sides of equation
(7), we obtain
Z F 0 ðgðtÞÞg0 ðtÞdt ¼ F ðgð ÞÞ F ðgðÞÞ:
(8)
Since F is a primitive of f on [a, b], and so a primitive of f on g([, ]), we can
R
rewrite the left-hand side of equation (8) as f ðgðtÞÞg0 ðtÞdt:
Further, since F is a primitive of f on [a, b], and so a primitive of f on
g([, ]), we can apply the Fundamental Theorem of Calculus to the function F
on g([, ]) to express the right-hand side of equation (8) as
Z gðÞ
F ðgðÞÞ F ðgðÞÞ ¼
F 0 ð xÞdx
gðÞ
¼
Z
gð Þ
f ð xÞdx:
gðÞ
So, if we make these substitutions on both sides of equation (8), we deduce that
Z Z gðÞ
f ðgðtÞÞg0 ðtÞdt ¼
f ð xÞdx:
(9)
gðÞ
This is the first part of the theorem.
Suppose next that g possesses an inverse function g1 on [, ], and
a ¼ gðÞ and
then
¼ g1 ðaÞ
b ¼ gðÞ;
and ¼ g1 ðbÞ:
Making these substitutions into equation (9), we obtain the second part of the
&
theorem.
Recall that, by hypothesis
gð½; Þ ½a; b:
7.3
Fundamental Theorem of Calculus
287
b
Strategy to evaluate an integral a f ðgððtÞÞÞg0 ðtÞÞdt using integration by
substitution:
0
1. Choose a function x ¼ g(t) such that dx
dt ¼ g ðtÞ, and express dt in terms of
x and dx.
There is a similar strategy for
finding a primitive by
integration by substitution.
2. Make the necessary substitutions to give an integral in terms of x and dx.
3. Calculate this integral.
Example 3
Evaluate
R3
tþ1
2
0 ð2tþ1Þ12
dt,
Solution Let x ¼ g(t) ¼ 2t þ 1, t 2 R. The function g is one–one on R.
Then dx
dt ¼ 2, so that dx ¼ 2dt. Making the various substitutions into the
integral, we get
Z 41
Z 3
2
tþ1
2 ð x 1Þ þ 1 1
dx
1 dt ¼
1
2
x2
1
0 ð2t þ 1Þ2
Z 41
2 ð x þ 1Þ 1
¼
dx
1
2
x2
1
Z 4
1
1
1
¼
x2 þ x2 dx
4 1
4
1 2 3
1
x2 þ 2x2
¼
4 3
1 1 16
1 2
5
&
þ4 þ2 ¼ :
¼
4 3
4 3
3
Here
when t ¼ 0; x ¼ 1;
when t ¼ 32 ; x ¼ 4:
Also, t ¼ 12 ðx 1Þ.
Problem 5
(a) Find a primitive of the function f (x) ¼ sin(2 sin 3x) cos 3x, x 2 R.
R 1 ex
(b) Evaluate the integral 0 ð1þe
dx.
x Þ2
Remark
If you use the second assertion of Theorem 5, you must verify that the function
g does have an inverse; you may end up with a contradiction if you simply
calculate thoughtlessly.
1
For example, take ¼ 1, ¼ 2, g(t) ¼ t2, and f ð xÞ ¼ x2 ; then a ¼ g(1) ¼ 1
and b ¼ g(2) ¼ 4. Then
4 Z b
Z 4
2 3
16
2
14
1
2
f ð xÞdx ¼
x dx ¼ x2 ¼
¼ ;
3 1
3
3
3
a
1
but
Z 2
Z 2
Z 2
2 12
f ðgðtÞÞg0 ðtÞdt ¼
t
2tdt ¼
2t2 dt
1
1
2
1
2 3
16
2
¼
¼ t
¼ 6:
3 1
3
3
This contradiction arises since the function g does NOT have an inverse on [1, 4]
that maps [1, 4] to [1, 2].
Our second substitution technique applies when we let t ¼ g(x), where the
function g has an inverse, so that x ¼ g1(t), and we express x in terms of t, and
dx in terms of t and dt.
For, if we make the
substitution x ¼ t2, then
t ¼ 1 ) x ¼ 1;
t ¼ 2 ) x ¼ 4:
7: Integration
288
b
Strategy to evaluate an integral a f ðxÞÞdx using integration by
substitution
1. Choose a function t ¼ g(x), where g has an inverse, so that x ¼ g1(t);
express dx in terms of t and dt.
2. Make the necessary substitutions to give an integral in terms of t and dt.
3. Calculate this integral.
Example 4
Evaluate
R loge 5
0
e2x
1
ðex 1Þ2
x
There is a similar strategy for
finding a primitive by
integration by substitution.
dx.
1
Solution Let t ¼ gð xÞ ¼ ðe 1Þ2 , x 2 [0, 1). The function g is one–one
on [0, 1).
Then t2 ¼ ex 1, so that ex ¼ t2 þ 1 and x ¼ loge(t2 þ 1). It follows that
dx
2t
2t
¼ 2
; so that dx ¼ 2
dt:
dt t þ 1
t þ1
Hence
Z loge 5
Z 2 2
2
e2x
ð t þ 1Þ
2t
2
dt
1 dx ¼
x
t
t
þ1
2
0
0
ð e 1Þ
Z 2
¼
2 t2 þ 1 dt
0
2
2 3
¼ t þ 2t
3
0
16
28
þ4 ¼ :
¼
&
3
3
Here
when x ¼ 0; t ¼ 0;
when x ¼ loge 5; t ¼ 2:
Problem 6
1
(a) Find a primitive of the function f ð xÞ ¼
3
1 ; x 2 ð1; 1Þ:
3ðx1Þ2 þ xðx1Þ2
R loge 3 x pffiffiffiffiffiffiffiffiffiffiffiffiffi
(b) Evaluate the integral 0
e 1 þ ex dx:
7.4
Inequalities for integrals and their applications
Often it is not possible to evaluate an integral explicitly, and a numerical estimate
for its value is sufficient for our purposes. This can occur both in applications
of mathematics and in the proofs of theorems that involve integration.
7.4.1
One of the goals of Numerical
Analysis is to obtain good
estimates for integrals.
The key inequalities
Our principal tool connecting inequalities and integrals is the following.
Theorem 1 Fundamental Inequality for Integrals
Rb
If f is integrable on [a, b], and f(x) 0 on [a, b], then a f ð xÞdx 0.
Proof Since f is integrable on [a, b], the sequence {Pn} of standard partitions
of [a, b] has the property that
Z b
lim Lðf ; Pn Þ ¼
f:
n!1
a
Now, for each value of n, we have
By the Null Partitions
Criterion, Theorem 7,
Sub-section 7.1.2.
7.4
Inequalities for integrals
Lð f ; Pn Þ ¼
n
X
mi xi ;
289
where mi ¼ inf f f ð xÞ : x 2 ½xi1 ; xi g:
i¼1
But, since f (x) 0 on [a, b], it follows that f (x) 0 on each subinterval [xi1, xi],
so that 0 is a lower bound for f on [xi1, xi], for each i. Since mi is the greatest
lower bound for f on [xi1, xi], we must therefore have mi 0, for each i.
Since each term mixi in the sum for L(f,Pn) is non-negative, it follows that
Lð f ; Pn Þ ¼
n
X
mi xi 0:
We use the usual notation for
partitions and lower and upper
Riemann sums.
(1)
i¼1
Letting n ! 1 in (1) and using the Limit Inequality Rule for sequences, we
deduce that
Z b
f ¼ lim Lðf ; Pn Þ 0:
&
a
Theorem 3, Sub-section 2.3.3.
n!1
We can now use the Fundamental Inequality for Integrals to prove the most
commonly used inequalities for integrals.
Theorem 2 Inequality Rule for Integrals
Let f and g be integrable on [a, b]. Then:
Rb
Rb
(a) If f(x) g(x) on [a, b], then a f ð xÞdx a gð xÞdx:
Rb
(b) If m f(x) M on [a, b], then mðb aÞ a f ð xÞdx M ðb aÞ:
Recall that earlier we motivated our discussion of integrals in terms of areas.
The following diagrams illustrate the results of Theorem 1 in the special case
that the functions are positive so that we can interpret the integrals as areas
between the curves and the x-axis.
Section 7.1.1.
7: Integration
290
Proof of Theorem 2
(a) Since f and g are integrable on [a, b], so is the function g – f.
Since f (x) g(x) on [a, b], it follows that
gð xÞ f ð xÞ 0; x 2 ½a; b;
By the Combination Rules for
integrals, Theorem 4 of
Section 7.2.3.
so, by Theorem 1, we deduce that
Z b
ðg f Þ 0;
a
from which it follows at once that
Rb
a
f Rb
a
g.
(b) First, the constant function x j! M, x 2 [a, b], is continuous on [a, b] and so
is integrable on [a, b].
We can therefore apply part (a) of the Theorem to the inequality f (x) M
on [a, b] to obtain
Z b Z b
f Mdx ¼ M ðb aÞ:
a
a
Next, the constant function x j! m, x 2 [a, b], is continuous on [a, b] and so
is integrable on [a, b].
We can therefore apply part (a) of the Theorem to the inequality
f (x) m on [a, b] to obtain
Z b
Z b
f mdx ¼ mðb aÞ:
&
a
Example 1
Prove that
a
R1
x3
0 2sin4 x dx
14 loge 2.
Solution Since, by the Sine Inequality, jsin xj jxj, for x 2 R, it follows that
sin4 x x4, for x 2 R. Hence
x3
x3
; for x 2 ½0; 1:
4
2 sin x 2 x4
If we apply part (a) of Theorem 2 to this inequality, we obtain
Z 1
Z 1
x3
x3
dx
dx
4
4
0 2 sin x
0 2x
1
1
4
¼ loge 2 x
4
0
1
1
&
¼ ðloge 1 loge 2Þ ¼ loge 2:
4
4
Example 2
Solution
ffi
Prove that p3ffiffiffi
34
Theorem 2, Sub-section 4.1.3.
R2
dx ffi
pffiffiffiffiffiffiffiffi
1 2 þ x5
3:
pffiffiffiffiffiffiffiffiffiffiffiffiffi
Since the function x 7! 2 þ x5 is increasing on [1, 2], we have
1
1
pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 1;
34
2 þ x5
for x 2 ½1; 2:
If we apply part (b) of Theorem 2 to these inequalities, we obtain
Z 2
3
dx
pffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 3:
34
2
þ x5
1
For the length of [1, 2] is 3.
&
7.4
Inequalities for integrals
291
Problem 1 Use the Inequality Rule for integrals to prove that
1
R3
1 x sin x10 dx 4:
Problem
R 1 2 2 Use the Inequality Rule for integrals to prove that
02 ex dx 23 :
1
2
Hint:
1
, for x 2 [0, 1).
Use the fact that 1 þ x ex 1x
Our final result is of great value in many applications. We saw earlier that, if the
function f is integrable on [a, b], so is the function j f j. We can then use
Theorem 2 to obtain a connection between the values of the integrals of f and j f j.
Theorem 3 Triangle Inequality
If f is integrable on [a, b], then
Z b Z b
f j f j:
a
By the Modulus Rule,
Theorem 4, Sub-section 7.2.3.
The name arises because of
the similarity between this
inequality and the Triangle
Inequality for numbers
X
X
n
n
ai ja j:
i¼1 i¼1 i
a
Furthermore, if jf(x)j M on [a, b], then
Z b f M ðb aÞ:
a
Proof
From the definition of the modulus function, we have
j f ð xÞj f ð xÞ j f ð xÞj:
Thus, since jfj is integrable on [a, b], it follows from part (a) of Theorem 2 that
Z b
Z b
Z b
f jfj j f j:
a
a
a
We can then rewrite this pair of inequalities in the desired form
Z b Z b
f j f j:
a
a
Next, we assume that jf(x)j M on [a,b]. Then, from the definition of the
modulus function, it follows that
M f ð xÞ M;
for x 2 ½a; b:
It follows from part (b) of Theorem 2 that
Z b
M ðb aÞ f M ðb aÞ:
a
We can then rewrite this pair of inequalities in the desired form
Z b f M ðb aÞ:
a
R p
p2
xp
Example 3 Prove that 02 2 þ cos2 x dx 16
.
Solution
By the Triangle Inequality for Integrals
Z p
Z p 2 x p2
2
x p2
dx
dx
0 2 þ cos x
0 j2 þ cos xj
Z p p
2
2x
dx:
¼
2
þ
cos x
0
&
So we must now examine this
last integral.
7: Integration
292
But 2 þ cos x 2, for x 2 0; p2 , so that
h pi
1
1
; for x 2 0; :
2 þ cos x 2
2
By applying this inequality to the integrand in the last integral, we obtain from
the Inequality Rule for Integrals that
Z p
Z p 2 x p2
2 1
p
x dx
dx 0 2 þ cos x
0 2 2
p
1 p
1 2 2
¼
x x
2 2
2
0
2
2
1 p
p
p2
&
¼
¼ :
2 4
8
16
Problem 3 Prove the following inequalities:
R p
R 4 sinð1xÞ
1
x
(a) 1 2 þ cos 1 dx 3;
(b) 04 3 tan
dx
4 loge 2:
2
sinðx Þ
ð xÞ
We end this sub-section with a nice application of the Triangle Inequality,
which is closely related to the Fundamental Theorem of Calculus.
Theorem 4
formula
Theorem 1, Sub-section 7.3.1.
Let f be continuous on [a, b], and F be defined on [a, b] by the
F ð xÞ ¼
Z
x
f ðtÞdt:
a
Then F is differentiable on [a, b], and its derivative is
F 0 ð xÞ ¼ f ð xÞ:
Proof We will prove that F is differentiable at c and F0 (c) ¼ f (c) in the case
that a < c < b. (The cases c ¼ a and c ¼ b are similar, with the appropriate onesided derivatives being used.)
We must prove that:
for each positive number ", there is a positive number such that
F ð x Þ F ðcÞ
f ðcÞ < "; for all x satisfying 0 < jx cj < :
xc
Now
F ð xÞ F ðcÞ
1
¼
xc
xc
¼
and
f ðc Þ ¼
1
xc
Z
1
xc
Z
f ðtÞdt a
Z
x
f ðtÞdt
c
x
f ðcÞdt:
c
x
Z
c
f ðtÞdt
a
You may omit this proof at a
first reading.
We will assume that is
chosen small enough that
(c , c þ ) [a, b].
7.4
Inequalities for integrals
293
It follows that
F ð xÞ F ðcÞ
1
f ðc Þ ¼
xc
xc
so that
Z
x
ff ðtÞ f ðcÞgdt
c
Z x
F ð xÞ F ðcÞ
¼ 1 :
f
ð
c
Þ
f
ð
t
Þ
f
ð
c
Þ
f
gdt
xc
jx cj c
We now suppose that x > c; it then follows from the above equation, using the
Triangle Inequality for integrals, that
Z
F ð xÞ F ðcÞ
1 x
f ðcÞ ¼
ff ðtÞ f ðcÞgdt
xc
xc c
Z x
1
ðsince x > cÞ:
j f ðtÞ f ðcÞjdt
xc c
But, since f is continuous at c, we know that there must exist some positive
number such that
1
j f ðtÞ f ðcÞj < "; for all x satisfying jx cj < :
2
It follows that, for x satisfying c < x < c þ , we have
Z x
F ð xÞ F ðcÞ
1
1
f
ð
c
Þ
"dt
xc
xc
2
c
1
¼ "
2
< ":
A similar argument applies in the case that x < c; so that, for all x satisfying
0 < jx cj < , we have
F ð xÞ F ðcÞ
< ":
f
ð
c
Þ
xc
This completes the proof.
7.4.2
We omit the details.
&
Wallis’s Formula
In this sub-section we use the method of Reduction of Order together with
various inequalities between integrals to establish a remarkable formula for the
number p.
Reduction of Order method
Quite often we need to evaluate an integral In that involves a non-negative
integer n. A common approach to such integrals is to relate the value of In to the
value of In1 or In2 by a reduction formula, using integration by parts.
Rp
Example 4 Let In ¼ 02 sinn xdx, n ¼ 0, 1, 2, . . ..
(a) Evaluate I0 and I1.
(b) Prove that In ¼ n1
n In2 , for n 2.
This is just another name for a
recurrence formula.
7: Integration
294
(c) Deduce from part (b) that, for n 1
I2n ¼
1:3: . . . :ð2n 3Þð2n 1Þ p
2:4: . . . :ð2n 2Þð2nÞ
2
I2nþ1 ¼
and
2:4: . . . :ð2n 2Þð2nÞ
:
3:5: . . . :ð2n 1Þð2n þ 1Þ
Solution
(a) We can evaluate the first two integrals easily
Z p
2
p
I0 ¼
1dx ¼ ; and
2
0
Z p
2
p
I1 ¼
sin xdx ¼ ½ cos x02 ¼ 1:
0
(b) We first write In in the form In ¼
parts, we find that, for n 2
Rp
2
0
sin x sinn1 xdx. Using integration by
Z p
2
p
In ¼ ðcos xÞ sinn1 x 02 ðcos xÞðn 1Þ sinn2 x cos xdx
¼ 0 þ ð n 1Þ
¼ ð n 1Þ
Z
Z
0
p
2
2
cos x sinn2 xdx
0
p
2
1 sin2 x sinn2 xdx
0
¼ ðn 1ÞfIn2 In g:
We can then rewrite this result in the form nIn ¼ (n 1)In2, so that
In ¼
n1
In2 :
n
(c) If we replace n in the formula for In in part (b) by 2n, 2n 2, 2n 4, . . ., in
turn, we obtain
I2n ¼
2n 1
I2n2 ;
2n
I2n2 ¼
2n 3
I2n4 ;
2n 2
I2n4 ¼
Hence
I2n ¼
2n 1
I2n2
2n
¼
ð2n 3Þð2n 1Þ
I2n4
ð2n 2Þð2nÞ
¼
ð2n 5Þð2n 3Þð2n 1Þ
I2n6 ;
ð2n 4Þð2n 2Þð2nÞ
2n 5
I2n6 ; . . .:
2n 4
We will integrate sin x and
differentiate sinn1 x.
7.4
Inequalities for integrals
295
and so on. Continuing this process, we obtain
I2n ¼
¼
1:3: . . . :ð2n 3Þð2n 1Þ
I0
2:4: . . . :ð2n 2Þð2nÞ
1:3: . . . :ð2n 3Þð2n 1Þ p
:
2:4: . . . :ð2n 2Þð2nÞ
2
For I0 ¼ p2 :
Similarly
2n
I2n1
2n þ 1
ð2n 2Þð2nÞ
I2n3
¼
ð2n 1Þð2n þ 1Þ
I2nþ1 ¼
..
.
Problem 4
¼
2:4: . . . :ð2n 2Þð2nÞ
I1
3:5: . . . :ð2n 1Þð2n þ 1Þ
¼
2:4: . . . :ð2n 2Þð2nÞ
:
3:5: . . . :ð2n 1Þð2n þ 1Þ
Let In ¼
R1
0
&
For I1 ¼ 1.
ex xn dx, n ¼ 0, 1, 2, . . ..
(a) Evaluate I0.
(b) Prove that In ¼ e nIn1, for n 1.
(c) Deduce the values of I1, I2, I3 and I4.
Wallis’s Formula
We now use the various results that we have just proved for the integral
Rp
In ¼ 02 sinn xdx to verify some surprising results.
Theorem
5 Wallis’s Formula
2n
2n
(a) lim 21 : 23 : 43 : 45 : 65 : 67 : . . . : 2n1
: 2nþ1
¼ p2 :
n!1
pffiffiffi
n!Þ2 22n
pffiffi ¼
p:
(b) lim ðð2n
Þ! n
Each of the two limits is
called Wallis’s Formula.
n!1
In the next problem, we ask you to establish a number of relationships
between the terms of the two sequences in the statement of Theorem 5.
Problem 5
2 2n
n!Þ 2
pffiffi,
Let an ¼ 21 : 23 : 43 : 45 : 65 : 67 : . . . : 2n2n 1 : 2n2nþ 1 and bn ¼ ðð2n
Þ! n
n 1.
(a) Evaluate an and bn, for n ¼ 1, 2 and 3.
(b) Verify that b2n ¼ 2nþ1
n an , for n ¼ 1, 2 and 3.
2
2nþ1
(c) Prove that bn ¼ n an , for all n 1.
You may omit tackling this
problem and the following
proof at a first reading.
7: Integration
296
Rp
Proof of Theorem 5 Let In ¼ 02 sinn xdx, for all n 0, and let the sequences
{an} and {bn} be as given in Problem 5.
(a) Using the formulas for I2n and I2nþ1 in part (c) of Example 4 above, we
obtain
I2n
1:3:3:5:5: . . .:ð2n 1Þð2n 1Þð2n þ 1Þ p
;
¼
2:2:4:4:6: . . .:ð2n 2Þð2nÞð2nÞ
2
I2nþ1
The numerator of I2nþ1 is
the same as the denominator
of I2n.
which we arrange in the form
an ¼
I2nþ1 p
:
2
I2n
Notice that, in order to complete the proof of part (a), it is sufficient to
show that
I2nþ1
! 1 as n ! 1:
(2)
I2n
We do this as follows.
Since 0 sin x 1 for x 2 0; p2 , we have
h pi
sin2n x sin2nþ1 x sin2nþ2 x; for x 2 0; :
2
It follows, from the Inequality Rule for integrals, that
I2n I2nþ1 I2nþ2 :
Thus
Note that
I2nþ1 I2nþ2
1
I2n
I2n
2n þ 1
:
¼
2n þ 2
I2nþ2 ¼
(3)
2n þ 1
I2n ;
2n þ 2
by the reduction formula
for In.
By letting n ! 1 in (3) and using the Squeeze Rule for sequences, we
deduce that the limit (2) holds, as desired.
(b) We know, from part (c) of Problem 5, that
2n þ 1
an ; for all n 1:
n
We also know, from the proof of part (a) of this theorem, that
p
as n ! 1:
an !
2
b2n ¼
It follows, by applying the Product Rule for sequences to the above
formula for bn2, that
p
b2n ! 2 ¼ p as n ! 1:
2
Hence,pby
the
continuity
of the square root function, we conclude that
ffiffiffi
&
bn ! p as n ! 1.
Unfortunately the sequences in Theorem 5 converge rather slowly as n
pffiffiffi
increases, so they are of little value in determining p or p to any reasonable
degree of approximation. In the next chapter, we shall meet practical ways of
estimating p.
Section 8.5.
7.4
Inequalities for integrals
7.4.3
297
Maclaurin Integral Test
We now introduce a method
for determining the convergence or divergence of
1
P
f ðnÞ, where the terms are positive and decrease to
certain series of the form
1
n¼1
P
1
zero. In particular, we show that the series
np converges if p > 1 and
n¼1
diverges if 0 < p 1.
We use the fact that each term in such a series can be regarded as a
contribution to a lower Riemann sum or upper Riemann sum for a suitable
integral.
Theorem 6 Maclaurin Integral Test
Let f be positive and decreasing on [1, 1), and let f(x) ! 0 as x ! 1.
Then:
1
R n
P
(a)
f ðnÞ converges if the sequence 1 f: n 2 N is bounded above;
(b)
n¼1
1
P
f ðnÞ diverges if the sequence
R n
n¼1
1
f: n 2 N tends to 1 as n ! 1.
You saw in Sub-section 3.2.1
that this series converges if
p 2.
In these results, the number 1
may be replaced by any
convenient positive integer.
Before proving the theorem, we illustrate the underlying ideas.
Example 5
Let Pn1 be the standard partition of [1, n] with n 1 subintervals
f½1; 2; ½2; 3; . . .; ½i; i þ 1; . . .; ½n 1; ng:
(a) Determine the lower and upper Riemann sums, L( f,Pn1) and U( f,Pn1),
for the function f ð xÞ ¼ x12 , x 2 ½1; 1Þ.
1
1
P
P
1
1
(b) Deduce that the series
n2 converges, and that 1 n2 2.
n¼1
n¼1
Solution
(a) Since f is decreasing on [1, 1), it follows that, for i ¼ 1, 2, . . ., n 1,
m i ¼ f ð i þ 1Þ ¼
Mi ¼ f ðiÞ ¼
1
ð i þ 1Þ 2
;
1
:
i2
Also, each subinterval in the partition has length 1.
In2 fact, the sum of this series is
p
6 , but to prove this goes
outside the scope of this book.
7: Integration
298
Hence the lower and upper Riemann sums for f are
n1
X
Lð f ; Pn1 Þ ¼
mi 1 ¼
i¼1
n1
X
1
1
1
þ 2 þ þ 2 ;
2
2
3
n
1
1
1
þ 2 þ þ
:
2
1
2
ðn 1Þ2
i¼1
1
P
1
(b) Let sn denote the nth partial sum of the series
n2 ; thus
U ð f ; Pn1 Þ ¼
Mi 1 ¼
n¼1
1
1
1
1
sn ¼ 2 þ 2 þ þ
þ 2:
2
1
2
n
ð n 1Þ
It follows that
Lð f ; Pn1 Þ ¼ sn 1
and
U ð f ; Pn1 Þ ¼ sn 1
:
n2
Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have
Z n
Lð f ; Pn1 Þ f U ðf ; Pn1 Þ;
1
so that
sn 1 Z
1
n
dx
1
sn 2 :
2
x
n
(4)
Now, the sequence {sn} is increasing, since the series has positive terms.
Also, from (4), we have
Z n
dx
sn þ1
2
1 x
n
1
¼ þ1
x 1
1
1
(5)
þ 1 ¼ 2 2:
¼ 1
n
n
Thus {sn} is bounded above.
Hence, by the Monotone Convergence Theorem for sequences, {sn} is
convergent, so that
1
X
1
is convergent:
2
n
n¼1
Since s1 ¼ 1 and {sn} is increasing, the sum of the series is at least 1.
Finally, we deduce from (5), using the Limit Inequality Rule for
sequences, that lim sn 2; so the sum of the series is at most 2. Hence
n!1
1
Example 6
1
X
1
2:
n2
n¼1
&
Let Pn1 be the standard partition of [1, n] with n 1 subintervals
f½1; 2; ½2; 3; . . .; ½i; i þ 1; . . .; ½n 1; ng:
7.4
Inequalities for integrals
299
(a) Determine the lower and upper Riemann sums, L( f, Pn1) and U( f, Pn1),
for the function f ð xÞ ¼ 1x ; x 2 ½1; 1Þ.
1
P
1
(b) Deduce that the series
n diverges.
n¼1
Solution
(a) Since f is decreasing on [1, 1), it follows that, for i ¼ 1, 2, . . ., n 1
1
;
m i ¼ f ð i þ 1Þ ¼
iþ1
1
Mi ¼ f ðiÞ ¼ :
i
Also, each subinterval in the partition has length 1.
Hence the lower and upper Riemann sums for f are
n1
X
Lð f ; Pn1 Þ ¼
mi 1 ¼
i¼1
1 1
1
þ þ þ ;
2 3
n
n1
X
1 1
1
:
Mi 1 ¼ þ þ þ
1 2
n1
i¼1
1
P
1
(b) Let sn denote the nth partial sum of the series
n; thus
U ð f ; Pn1 Þ ¼
n¼1
1 1
1
1
þ :
sn ¼ þ þ þ
1 2
ð n 1Þ n
It follows that
1
and U ð f ; Pn1 Þ ¼ sn :
n
Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have
Z n
Lð f ; Pn1 Þ f U ð f ; Pn1 Þ;
Lð f ; Pn1 Þ ¼ sn 1
1
so that
sn 1 Z
1
n
dx
1
sn ;
x
n
or
1
sn 1 loge n sn :
(6)
n
In particular, sn loge n. Then, since loge n ! 1 as n ! 1, it follows, from
the Squeeze Rule for sequences that tend to infinity, that sn ! 1 as n ! 1.
1
P
1
&
Hence, the series
n diverges.
n¼1
Remarks
We can in fact get more information from the above argument than just the
result of the example.
1. If we define the sequence { n} by the expression
1 1
1
n ¼ 1 þ þ þ þ loge n
2 3
n
7: Integration
300
then
1
nþ1
loge
nþ1
n
Z nþ1
1
dx
¼
nþ1
x
n
Z nþ1 1
1
dx 0;
¼
nþ1 x
n
nþ1 n ¼
so that { n} is decreasing. Also, it follows from (6), using the fact that
n ¼ sn loge n, that
1
n ;
n
thus the sequence { n} is bounded below by 0.
Hence the sequence { n} tends to a limit, say, as n ! 1. The value of clearly lies between 1 (for 1 ¼ 1) and 0.
This number is called Euler’s constant, and occurs often in Analysis.
In fact,
¼ 0:57721 56649 01532 86060 65120 90082 40243 . . .:
1
2. Although the sequence sn ¼ 1 þ 12 þ þ n1
þ 1n tends to infinity, we can
make more precise statements than that, arising from (6). We can rewrite
the inequalities in (6) in the following form
1
þ loge n sn 1 þ loge n;
n
if we then divide throughout by logen and let n ! 1, we obtain, by the
Limit Inequality Rule for sequences, that
sn
! 1 as n ! 1:
loge n
We can write this last limit in an equivalent form as ‘sn logen as n ! 1’ –
in other words, ‘the behaviour of sn and logen are essentially the same for
large n’.
Problem 6 Use
the Maclaurin Integral Test to determine the behaviour
1
P
1
of the series
np , for p > 0, p 6¼ 1.
n¼1
R
Show that xðlogdx xÞ2 ¼ log1 x, for x > 1, and hence prove
e
1
e
P
1
that the series
2 converges.
nðlog nÞ
Problem 7
n¼2
e
R dx
¼ loge ðloge xÞ, for x > 1, and hence
Problem 8 Show that x log
ex
1
P 1
prove that the series
n log n diverges.
n¼2
e
Sequences revisited
We can apply many ideas similar to those in Example 5 to obtain useful
information about certain types of convergent sequences.
By definition of the sequence
{ n}.
For loge n sn 1n :
We give here just the first 35
decimal places.
We shall address this notation more carefully in
the next section.
7.4
Inequalities for integrals
Example 7
301
1
Prove that n þ1 1 þ n þ1 2 þ þ 2n
! loge 2 as n ! 1.
1
Solution Let f ð xÞ ¼ 1þx
, x 2 [0, 1]; and let Pn ¼ 0; 1n ; 1n ; 2n ; . . .; n1
n ;1
be the standard partition of [0, 1] into n sub-intervals of equal length 1n.
i
Since f is decreasing on [0, 1], it follows that on the ith sub-interval i1
n ; n
we have
i
1
:
mi ¼ f
¼
n
1 þ ni
Recall that
i
mi ¼ inf f ð xÞ : x 2 i1
n ;n :
Thus the lower Riemann sum for f on Pn is
n
X
1
1
i n
1
þ
n
i¼1
n
X
1
1
1
1
¼
¼
þ
þ þ :
nþi nþ1 nþ2
2n
i¼1
L ð f ; Pn Þ ¼
Since f is decreasing on [0, 1], it is integrable on [0, 1]. Hence, as n ! 1, we
have
Z 1
Z 1
1
1
1
dx
þ
þ þ ¼ Lð f ; P n Þ !
f¼
nþ1 nþ2
2n
1
þx
0
0
¼ ½loge ð1 þ xÞ10 ¼ loge 2: &
Example 7 illustrates a general technique that is often useful.
Strategy
n ! 1.
If f is positive and decreasing on [0, 1], then: 1n
Problem 9
Prove that n
1
n2 þ12
n R1
P
f ni ! 0 f as
i¼1
The trick in applying this
strategy is to make a good
choice of f.
1
1
þ n2 þ2
! p4 as n ! 1.
2 þ þ n2 þn2
You have already met the corresponding ideas for divergent sequences in
Example 6 and Remark 2 after that example, so we state without comment
the general strategy illustrated there.
Strategy If f is positive and decreasing on [1, 1), f(x) ! 0 as x ! 1, and
n
Rn
Rn
P
f ðiÞ 1 f as n ! 1.
1 f ! 1 as n ! 1, then
i¼1
Recall that this means that
n
P
f ðiÞ
R n ! 1 as n ! 1:
1 f
i¼1
Remark
In fact under the hypotheses of this strategy
n
P
i¼1
Rn
f ðiÞ ð 1 f Þ þ c, for any fixed
The reason for this will be
explained in Section 7.5.1.
number c; in particular situations we may choose c in any convenient way.
Problem 10
pffiffiffi
Prove that 1 þ p1ffiffi2 þ p1ffiffi3 þ þ p1ffiffin 2 n as n ! 1.
Here you will use both the
strategy and the subsequent
remark.
7: Integration
302
Proof of the Maclaurin Integral Test
Theorem 6 Maclaurin Integral Test
Let f be positive and decreasing on [1, 1), and let f(x) ! 0 as x ! 1. Then:
1
R n
P
(a)
f ðnÞ converges if the sequence 1 f : n 2 N is bounded above;
(b)
n¼1
1
P
f ðnÞ diverges if the sequence
R n
n¼1
1
f : n 2 N tends to 1 as n ! 1.
Rn
Proof Let In denote the integral 1 f , let sn ¼ f(1) þ f(2) þ þ f(n) denote
the nth partial sum of the series, and let Pn1 be the standard partition of [1, n]
with n 1 subintervals
f½1; 2; ½2; 3; . . .; ½i; i þ 1; . . .; ½n 1; ng:
Since f is decreasing on [1, 1), it follows that, for i ¼ 1, 2, . . ., n 1
m i ¼ f ð i þ 1Þ
and Mi ¼ f ðiÞ:
Also, each subinterval in the partition has length 1.
Hence the lower and upper Riemann sums for f are
Lð f ; Pn1 Þ ¼
n1
X
m i 1 ¼ f ð 2 Þ þ f ð 3Þ þ þ f ð n Þ
i¼1
¼ sn f ð1Þ;
U ð f ; Pn1 Þ ¼
n1
X
M i 1 ¼ f ð 1 Þ þ f ð 2Þ þ þ f ð n 1 Þ
i¼1
¼ sn f ðnÞ:
Since f is monotonic on [1, n], it is integrable on [1, n]; hence we have
Lð f ; Pn1 Þ In U ð f ; Pn1 Þ;
so that
sn f ð1Þ In sn f ðnÞ:
(7)
Case 1: {In} is bounded
We are now assuming that, for some M, In M, for all n. It follows from
(7) that
sn f ð1Þ þ M; for all n:
7.5
Stirling’s Formula for n!
303
Thus the increasing sequence {sn} is bounded above, and so, by the Monotone
Convergence Theorem,
it is convergent.
1
P
f ðnÞ is convergent.
Hence the series
n¼1
Case 2: {In} is not bounded
R nþ1
The sequence {In} is increasing, since Inþ1 In ¼ n f 0. Since we are
now assuming that {In} is not bounded, it follows that In ! 1 as n ! 1.
Now, from (7), sn In; so, by the Squeeze Rule for sequences which tend to
infinity,
sn ! 1 as n ! 1:
Hence the series
1
P
&
f ðnÞ is divergent.
n¼1
7.5
Stirling’s Formula for n!
For small values of n, we can evaluate n! directly by multiplication or by using
a scientific calculator.
Problem 1
n
Complete the following table of values of n!
n!
n
n!
n
n!
1
1
6
720
20
2
3
2
6
7
8
5040
40 320
30
40
4
24
9
362 880
50
5
120
10
3 628 800
60
As n increases, n! grows very quickly; for instance:
around 14! seconds have elapsed since the birth of Christ;
around 18! seconds have elapsed since the formation of the Earth.
A calculator soon becomes useless for evaluating n!; thus the author’s
current scientific calculator (2005) gives that 69! ’ 1.7 1098, but an error
message when asked the value of 70!.
Stirling’s Formula gives us a way of estimating n! for large values of n. In order
to state the formula, though, we must first introduce some notation that enables us
to compare the behaviour of positive functions of n for large values of n.
14! ’ 8:7 1010
18! ’ 6:4 1015
7.5.1
The tilda notation
For many numbers that arise in the normal way, we often use the symbol ‘’’ to
pffiffiffi
denote ‘is approximately equal to’. For example, 2 ¼ 1:4142135623730950
4880 . . . (where we have given only the first 20 decimal places); for many
pffiffiffi
pffiffiffi
purposes it is sufficient to use estimates such as 2 ’ 1:414 or even 2 ’ 1:4,
Often an estimate is sufficient
for our purposes.
7: Integration
304
since the error involved is small. But what do we mean by ‘the error involved is
small’? Sometimes we mean that the error, the difference between the exact
value and the approximation, is a small number; sometimes we mean that the
percentage error, namely
actual error
100;
percentage error ¼
exact value
is a small number.
To handle the behaviour of positive functions of n for large values of n, we
introduce a very precise mathematical notation.
Notation
For positive functions f and g with domain N, we write
f ðnÞ gðnÞ as n ! 1;
to mean that gf ððnnÞÞ ! 1
as n ! 1.
2
þ 10
! 1 as
For example, n2 þ 1000n þ 10 n2 as n ! 1, since n þ1000n
n2
1 1
sinð1nÞ
n ! 1; and sin n n as n ! 1, since 1 ! 1 as n ! 1.
n
Notice that, if f(n) g(n) as n ! 1, then g(n) f(n) as n ! 1.
Also, if f(n) g(n) as n ! 1, g(n) ! 1 and c is a given number, then
f(n) þ c g(n) as n ! 1, since, as n ! 1
‘For example, the error is less
than 1’.
For example, ‘the percentage
error is less than 1%’.
The notation is ONLY used
for positive functions.
We say ‘f tilda g’ or ‘f
twiddles g’.
You can think of this as saying
that the percentage error
involved by replacing f(n) by
g(n) is small for large n.
For, lim sinx x ¼ 1:
x!0
For, gf ððnnÞÞ ! 1 , gf ððnnÞÞ ! 1.
f ð nÞ þ c f ð nÞ
c
¼
þ
! 1 þ 0 ¼ 1:
gðnÞ
gðnÞ gðnÞ
Notice, however, that the statement f(n) g(n) as n ! 1 does NOT mean
that f(n) g(n) ! 0 or even that f(n) g(n) is bounded. For example,
n2 þ 1000n þ 10 n2 as n ! 1, yet
2
n þ 1000n þ 10 n2 ¼ 1000n þ 10 ! 1 as n ! 1:
In situations like this example, ‘’ compares the sizes of the dominant terms
in f and g.
Problem 2 Find two pairs from the following functions such that
fi(n) fj(n) as n ! 1
f1 ðnÞ ¼ sinðn2 Þ; f2 ðnÞ ¼ sin n12 ; f3 ðnÞ ¼ 1 cos 1n ;
f4 ðnÞ ¼ n22 ;
f5 ðnÞ ¼ n12 ;
f6 ðnÞ ¼ 1 1n;
f7 ðnÞ ¼ 2n12 :
Remark
If ‘ is finite and positive, then the statements
‘f ðnÞ ! ‘ as n ! 1’ and
‘f ðnÞ ‘ as n ! 1’
are equivalent, since each is equivalent to the statement
‘f ðnÞ
! 1 as n ! 1’:
‘
We can also, in this situation, legitimately say that f(n) ‘ for large n.
The Combination Rules for handling tilda are direct consequences of the
corresponding Combination Rules for sequences.
For example, we may write
1 n
e as n ! 1;
1þ
n
since
1þ
1 n
! e as n ! 1:
n
Note that we only use for
positive funtions.
7.5
Stirling’s Formula for n!
305
For example, if
Combination Rules
f1 ðnÞ ¼ n2 þ n; g1 ðnÞ ¼ n2 ;
f2 ðnÞ ¼ n3 þ n; g2 ðnÞ ¼ n3 ;
Let f1(n) g1(n) and f2(n) g2(n) as n ! 1. Then:
Sum Rule
f1(n) þ f2(n) g1(n) þ g2(n);
Multiple Rule
Product Rule
lf1(n) lg1(n), for any number l > 0;
f1(n) f2(n) g1(n) g2(n);
Reciprocal Rule
1
f1 ðnÞ
Example 1
as n ! 1.
then
2
n þ n þ n3 þ n n2 þ n3 ;
2
2
5 n þ n 5n ðl ¼ 5Þ;
2
n þ n n3 þ n n2 n3 ¼ n5 ;
g11ðnÞ.
1
1
:
n2 þ n n2
1
1
Prove that, if f(n) g(n) as n ! 1, then ð f ðnÞÞn ðgðnÞÞn
Since f(n) g(n) as n ! 1, we have
Solution
f ð nÞ
!1
gð nÞ
as n ! 1;
so that
loge
f ð nÞ
! 0 as n ! 1;
gð nÞ
and therefore
1
f ð nÞ
log
!0
n e gðnÞ
But
Here we use the fact that
the function loge is continuous
at 1.
as n ! 1:
1
1
f ð nÞ
f ð nÞ n
loge
¼ loge
n
gðnÞ
gð nÞ
1
¼ loge
ð f ðnÞÞn
1
;
ðgðnÞÞn
so that
1
loge
ð f ð nÞ Þ n
1
ð gð nÞ Þ n
! 0 as n ! 1:
We now use the fact that the exponential function is continuous at 0; it follows
from the previous limit that
1
ð f ðnÞÞn
1
ðgðnÞÞn
! 1 as n ! 1;
1
1
which is precisely the statement that ð f ðnÞÞn ðgðnÞÞn as n ! 1.
&
Problem
3 Give
specific examples of functions f and g to show that
1
1
ðf ðnÞÞn ðgðnÞÞn as n ! 1 does not imply that f(n) g(n) as n ! 1.
Hint:
7.5.2
Try f(n) ¼ n2 and g(n) ¼ n.
Stirling’s Formula
Stirling’s Formula was discovered in the eighteenth century, in an analysis of
a gambling problem!
7: Integration
306
Theorem 1
Stirling’s Formula
pffiffiffiffiffiffiffiffinn
n! 2pn
as n ! 1:
e
pffiffiffiffiffiffiffiffi n
Problem 4 Use your calculator to evaluate 2pn ne for n ¼ 5. For
this value of n does the expression approximate n! to within 1%?
For small values of n, Stirling’s Formula gives reasonable approximations to
n!, and the percentage error quickly decreases as n increases:
n
10
n!
Stirling’s approximation
3 628 800
3 598 696
18
Error
0.83% (1 in 120)
18
20
52
2.433 10
8.066 1067
2.423 10
8.053 1067
0.42% (1 in 240)
0.16% (1 in 620)
100
9.333 10157
9.325 10157
0.09% (1 in 1170)
We illustrate the use of Stirling’s Formula as follows.
If 200 coins are tossed, then the probability of there being exactly 100 heads
and 100 tails is
200
1
200!
200
¼
:
100
2
ð100!Þ2 2200
It follows from Stirling’s Formula that this probability is
pffiffiffiffiffiffiffiffiffiffi 200200
400p 200!
’ pffiffiffiffiffiffiffiffiffiffi e 2
2
200
100
ð100!Þ 2
200p 100
2200
e
1
pffiffiffi
10 p
1
:
¼
17:724 . . .
In other words, the probability of there being exactly 100 heads and 100 tails
is about 1 in 18 – rather higher than you might expect.
¼
Problem 5
Hint:
1n
nn
n!1 n!
Prove that lim
¼ e.
Use the result of Example 1.
Problem 6 Use Stirling’s Formula to estimate the following numbers
to two significant figures:
300
300
1
300!
1
300
(a)
;
(b)
:
3
150
2
3
ð100!Þ
Problem 7
Use Stirling’s Formula to determine a number l such that
2n
4n
l22n as n ! 1:
n
2n
This fact is easily proved
using Probability Theory.
Here we substitute n ¼ 200
and n ¼ 100 into Stirling’s
Formula.
7.5
Stirling’s Formula for n!
7.5.3
307
Proof of Stirling’s Formula
Theorem 1
Stirling’s Formula
pffiffiffiffiffiffiffiffinn
n! 2pn
as n ! 1:
e
We divide up the proof into a number of steps, for clarity.
Proof
Step 1: Setting things up
We consider the function
f ðxÞ ¼ loge x;
x 2 ½1; n;
and the standard partition Pn1 of the interval [1, n] with n 1 subintervals
f½1; 2; ½2; 3; . . .; ½i; i þ 1; . . .; ½n 1; ng:
We consider also the sequence of numbers fcn g1
n¼2 , where cn is the total area
between the concave curve y ¼ loge x, for x 2 [1, n], and the polygonal line with
vertices (1, 0), (2, loge 2), (3, loge 3), . . ., (n, loge n), as illustrated below. This
consists of n 1 small slivers.
y
y = loge x
total area of
silvers = cn
y = loge x
contribution
to cn
(i + 1,loge(i + 1))
(i, logei)
1
2
3 ... i i + 1
...
n
x
Step 2: Calculating areas
The area between the curve y ¼ loge x and the x-axis, for x 2 [1, n], is
Z n
loge xdx ¼ ½x loge x xn1
1
¼ n loge n ðn 1Þ:
(1)
Next, the area between the polygonal line and the x-axis is
1
2fLð f , Pn1 Þ
þ U ð f ; Pn1 Þg;
(2)
and, since f is increasing, we have
Lð f , Pn1 Þ ¼ loge 1 þ loge 2 þ þ loge ðn 1Þ
¼ loge ðn 1Þ!
¼ loge n! loge n
(3)
and
U ð f , Pn1 Þ ¼ loge 2 þ þ loge n
¼ loge n!:
(4)
Substituting from (3) and (4) into (2), we find that the area between the
polygonal line and the x-axis is
7: Integration
308
loge n! 12 loge n:
It follows from (1) and (5) that
cn ¼ n loge n ðn 1Þ loge n! þ 12 loge n
(5)
1
nnþ2
¼ loge n1 :
e n!
(6)
Step 3: Behaviour of the sequence {cn}
It is obvious from the definition of cn that the sequence {cn} is positive and
increasing. In order to apply the Monotone Convergence Theorem to the
sequence, we must prove next that {cn} is bounded above.
So, let i be an integer with 1 i n 1, and let
A ¼ ði; loge iÞ; B ¼ ði þ 1; loge ði þ 1ÞÞ; and C ¼ ði þ 1; loge iÞ:
Then, as illustrated in the diagram in the margin
BC ¼ loge ði þ 1Þ loge i
1
¼ loge 1 þ :
i
Next, let the tangent at A to the curve meet the line BC at D. Since AC ¼ 1
and the slope of the line AD is 1i , it follows that CD ¼ 1i . Hence
BD ¼ CD BC
1
1
1
¼ loge 1 þ
2:
i
i
2i
Hence the contribution to the area cn of this sliver between the lines x ¼ i and
x ¼ i þ 1 is at most
1
area of ABD ¼ AC BD
2
1
1
1
1 2 ¼ 2:
2
2i
4i
If we now sum such areas over i ¼ 1, 2, . . ., n 1, we obtain
n1
1X
1
cn 4 i¼1 i2
Here we use the inequality
loge ð1 þ xÞ > x 12x2 ;
for x > 0;
that you met in Section 6.7,
Exercise 5(c) on Section 6.4.
1
1X
1
:
4 i¼1 i2
Since this infinite series converges, it follows that the sequence {cn} is
bounded above.
It follows from the Monotone Convergence Theorem that the sequence {cn}
is convergent.
Step 4: Properties of the sequence {an}, where an ¼ ecn
Since {cn} is convergent and the exponential function is continuous, it follows
that, if we set an ¼ ecn , for n ¼ 2, 3, . . ., then the sequence {an} is also
convergent. Thus, using the expression (6) for cn, we have
1
nnþ2
(7)
an ¼ n1 ! L; as n ! 1;
e n!
for some non-zero number L.
The actual value of the bound
does not matter here.
We now introduce {an} so
that we can use Wallis’s
Formula in the next Step.
L 6¼ 0 since the exponential
does not take the value 0.
7.6
Exercises
309
It follows from the formula for an in (7) that
a2n
n2nþ1
e2n1 ð2nÞ!
¼
1
2
a2n e2n2 ðn!Þ
ð2nÞ2nþ2
1
¼
ð2nÞ!n2 e
ðn!Þ2 22nþ2
1
;
(8)
after some cancellation.
Step 5: Proving Stirling’s Formula
We can rewrite (8) in the form
1
ð2nÞ!n2
a2n
e
¼
pffiffiffi :
a2n ðn!Þ2 22n
2
(9)
But, by Wallis’s Formula, the first quotient on the right-hand side of (9) tends
to p1ffiffi as n ! 1. It follows that, if we let n ! 1 on both sides of (9), we obtain
p
L2
1
e
¼ pffiffiffi pffiffiffi ;
L
p
2
so that
e
L ¼ pffiffiffiffiffiffi :
2p
Knowing the value of L, we can then rewrite (7) in the form
1
nnþ2
e
! pffiffiffiffiffiffi ;
n1
e n!
2p
by the Reciprocal Rule for sequences, it follows that
pffiffiffiffiffiffi
2p
en1 n!
pffiffiffi !
:
n
e
n n
. By the definition of tilda, this limit is exactly equivalent to the relation
pffiffiffiffiffiffiffiffinn
n! 2pn
as n ! 1;
e
&
that we set out to prove.
Remark
It is quite remarkable how many of the techniques that we have met so far in the
book are needed in order to prove this apparently straight-forward result.
7.6
Exercises
Section 7.1
1. Sketch the graph of the function
1 j xj; 1 < x < 1;
f ð xÞ ¼
1;
x ¼ 1:
Determine the minimum, maximum, infimum and supremum of f on [1, 1].
You met Wallis’s Formula
in Theorem 5 of Subsection 7.4.2.
ðn!Þ2 22n pffiffiffi
pffiffiffi ¼ p:
n!1 ð2nÞ! n
lim
7: Integration
310
2. Let f be the function
j xj;
f ð xÞ ¼ 1
2;
1 < x < 1;
x ¼ 1:
Evaluate L( f, P) and U( f, P) for each of the following partitions P of [1, 1]:
(a) P ¼ 1; 12 ; 12 ; 0 ; 0; 12 ; 12 ; 1 ;
(b) P ¼ 1; 14 ; 14 ; 13 ; 13 ; 1 :
3. Let f be the function
1 x; 0 x < 1;
f ð xÞ ¼
2;
x ¼ 1:
(a) Using the standard partition Pn of [0, 1] with n equal subintervals,
evaluate L( f, Pn) and U( f, Pn).
R1
(b) Deduce that f is integrable on [0, 1], and evaluate 0 f :
4. Let f be the function f(x) ¼ x3, x 2 [0, 1].
(a) Using the standard partition Pn of [0, 1] with n equal subintervals,
evaluate L( f, Pn) and U( f, Pn).
R1
(b) Deduce that f is integrable on [0, 1], and evaluate 0 f :
5. Let f be the function f (x) ¼ sin x, x 2 0; p2 :
(a) Using the standard partition Pn of 0; p2 with n equal subintervals,
evaluate L ( f, Pn) and U ( f, Pn).
Rp
(b) Deduce that f is integrable on 0; p2 , and evaluate 02 f :
Hint:
Use the formula
sin A þ sin
þ ðn 1ÞBÞ
1 ðA þ BÞþ þ sinðA
sin 2 nB
n1
sin A þ
¼
B ; B 6¼ 0:
2
sin 12 B
6. Prove that the function
1 þ x;
f ð xÞ ¼
1 x;
0 x 1;
0 x 1;
This formula can be proved by
Mathematical Induction.
x rational;
x irrational;
is not integrable on [0, 1].
7. Prove that Dirichlet’s function
1
; if x is a rational number pq, in lowest terms, with q > 0,
f ð xÞ ¼ q
0; if x is irrational.
is integrable on [0, 1].
Hint: Verify that there are at most 12 nðn þ 1Þ points x in [0, 1] for which
f ð xÞ > 1n :
Section 7.2
1. For bounded functions f and g on an interval I, prove that
supð f þ gÞ sup f þ sup g:
I
I
I
For the interval I ¼ [1, 2], write down functions f and g for which strict
inequality holds in this result.
You met this function in Subsection 5.4.2.
7.6
Exercises
311
2. Write down functions f and g and an interval I for which it is not true that
supð fgÞ ¼ sup f sup g:
I
I
I
3. Write down functions f and g on [0, 1] that are not integrable on [0, 1] but
such that f þ g is integrable on [0, 1].
4. (a) Write down functions f and g on [0, 1] such that f is integrable, g is not
integrable, and fg is integrable.
(b) Write down a function f on [0, 1] such that jfj is integrable but f is not
integrable on [0, 1].
5. Prove that, if the functions f and g are integrable on [a, b], then so is the
function max{f, g}.
Hint:
Use the formula maxfa; bg ¼ 12 fa þ b þ ja bjg, for a, b 2 R.
6. Prove that, if f and g are integrable on [a, b], then
Z b 2 Z b Z b fg f2
g2 :
a
a
a
Hint:
This is known as the
Cauchy–Schwarz Inequality
for integrals.
Use the Cauchy-Schwarz Inequality
n
P
2
ai bi
i¼1
n
P
i¼1
a2i
n
P
i¼1
b2i :
Section 7.3
1. Write down a primitive of each of the following functions:
pffiffiffiffiffiffiffiffiffiffiffiffiffi
(a) f ð xÞ ¼ x2 9; x 2 ð3; 1Þ;
(b) f ð xÞ ¼ sinð2x þ 3Þ 4 cosð3x 2Þ;
x 2 R;
(c) f ð xÞ ¼ e sinð3xÞ; x 2 R:
2. Using the result of part (c) of Exercise 1, write down a primitive F of the
function f ð xÞ ¼ e2x sinð3xÞ, x 2 R, for which F(p) ¼ 0.
2x
3. Show that the following
functions are all primitives of the function
f(x) ¼ sech x, x 2 p2 ; p2 :
(b) F2 ð xÞ ¼ 2 tan1 ðex Þ; (a) F1 ð xÞ ¼ tan1 ðsinh xÞ;
1
(c) F3 ð xÞ ¼ sin ðtanh xÞ;
(d) F4 ð xÞ ¼ 2 tan1 tanh 12 x :
Re
4. Let In ¼ 1 xðloge xÞn dx, for n 0:
(a) Prove that In ¼ 12 e2 12 nIn1 , for n 1.
(b) Evaluate I0, I1, I2 and I3.
5. Evaluate each of the following integrals, using the suggested substitution
where given:
Rp
R 1 ðtan1 xÞ2
(a) 02 tanðsin xÞ cos xdx;
(b) 0 1þx2 dx ðtry u ¼ tan1 xÞ;
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R1 (c) 0 x5 þ 2x2 x6 þ 4x3 þ 4dx;
1 R p dx
1 tan2 ð12xÞ
2
(d) 0 2þcos x dx try u ¼ tan 2 x and use the identity cos x ¼ 1 þ tan2 1x ;
ð2 Þ
R e2 loge ðloge xÞ
Re 7
(f) e
dx;
(e) 1 8x loge xdx;
x
R p sinð2xÞ
R 4 dx
pffiffiffi
2
(g) 0 1þ3 cos2 x dx;
(h) 1 ð1þxÞpffiffix ðtry u ¼ xÞ:
You met this in Subsection 1.3.3.
7: Integration
312
Section 7.4
1. Prove the following inequalities:
R 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R 1 x4
1
ffi dx 10
(a) 0 x3 2ð1 þ x99 Þdx 12 ;
(b) 0 pffiffiffiffiffiffiffiffiffiffi
:
1þ3x97
R
30
1
2. Prove that 12 0 1þx
2x99 dx 2:
R 2
2
Þ sinð99xÞ
3. Prove that 0 x ðx3
dx
4:
20
1þx
1
R dx
P
12
1
4. Show that
, and deduce that
is
3 ¼ 2ðloge xÞ
3
xðloge xÞ2
nðloge nÞ2
n¼2
convergent.
1
R dx
P
1
1
2
5. Show that
1 ¼ 2ðloge xÞ , and deduce that
1 is divergent.
xðloge xÞ2
n¼2 nðloge nÞ2
1
1
6. Prove that lim n ðnþ1
þ ðnþ2
þ þ ð2n1Þ2 ¼ 12 :
Þ2
Þ2
n!1
7. Determine the convergence or divergence of the following series:
1
1
P
P
1
1
(a)
(b)
loge n ;
loge ðloge nÞ :
ðlog nÞ
n¼2
8. Prove that
e
n¼2
n
P
loge k
k
k¼2
ðloge nÞ
12 ðloge nÞ2 , as n ! 1.
9. Prove that, if f is integrable on [a, b], where a < b, and f(x) > 0, for all x 2 [a, b],
Rb
then a f > 0:
Hint: Use the following steps:
Rb
(a) Assume that the result is false, so that in fact a f ¼ 0;
Rd
(b) Verify that, if ½c; d ½a; b, then c f ¼ 0;
(c) Prove that there is a partition P of [a, b] with U( f, P) b a, and
deduce that there is some subinterval [a1, b1] of [a, b] with a1 < b1 and
sup f 1;
½a1 ,b1 Rb
(d) Using the fact that a11 f ¼ 0, prove that there is a sequence
1
f½an , bn g1
n ¼ 1 of intervals with ½anþ1 , bnþ1 ½an , bn and sup f n ;
½an ,bn (e) Prove that there is some point c in [a, b] such that an ! c as n ! 1, and
that c 2 [an, bn], for all n 1;
(f) Verify that f(c) ¼ 0.
Section 7.5
ð3nÞ!
3n
n!1 n
1. Prove that lim
1n
¼ 27
e3 :
2. Use Stirling’s Formula to estimate each of the following numbers to two
significant figures:
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1400
400! ð800pÞ
400
;
(b)
:
(a)
200
2
ð100!Þ4 4400
3. Use Stirling’s Formula to determine the number l such that
ð8nÞ!
ðð2nÞ!Þ
4
l
216n
3
n2
as n ! 1:
8
Power series
The evaluation of given functions at given points in their domains is of great
importance. If we are dealing with a polynomial function, the calculation of the
function’s values presents no problem: it is simply a matter of arithmetic. For
example, if
1
1
1
1
f ð xÞ ¼ 1þ x x2 x3 þ x4 ;
2
2
6
4
then
1 1 1 1 13
f ð1Þ ¼ 1þ þ ¼ :
2 2 6 4 12
On the other hand, the sine function is rather different: there is no way of
calculating its values precisely merely by the use of arithmetic.
It is important to be able to estimate values of functions that cannot be
evaluated exactly, and to know how close the estimates are to the actual values
of the function.
In this chapter we are primarily concerned with a procedure for calculating
approximate values of functions, like the sine function, whose values cannot be
calculated easily at all points in their domains. We see how, in principle, we
can use a certain sequence of polynomials to calculate the values of the sine
function, for example, to any desired degree of accuracy; and how we can
represent sin x, for any x, as the sum of a series. We shall see, for instance, that
the polynomial pð xÞ ¼ x 16 x3 approximates f (x) ¼ sin x to within 105 for all
x in the interval [0, 0.1], and that, in general
Section 8.2, Example 1, and
Theorem 3.
x3 x5 x7
þ þ ; for x 2 R:
3! 5! 7!
In Section 8.1 we define the Taylor polynomial Tn(x) and discuss some
particular examples of functions for which the Taylor polynomials appear to
provide useful approximations.
In Section 8.2 we investigate how closely Taylor polynomials approximate a
given function in the neighbourhood of a point a in its domain, and we establish
a criterion for when we can say that
sin x ¼ x f ð xÞ ¼ lim Tn ð xÞ ¼
n!1
1
X
an ð x aÞ n :
n¼0
In this case, we say that f is the sum function of the power series
1
P
an ðx aÞn .
n¼0
In Sections 8.3 and 8.4 we look at the behaviour of power series in their own
right; that is, we consider functions which are defined by power series. In
1
P
particular, in Section 8.3 we see that a power series
an ðx aÞn behaves in
n¼0
one of three ways: it converges only for x ¼ a, or it converges for all x, or it
313
8: Power series
314
converges in an interval (a R, a þ R), where R is a positive number called the
radius of convergence.
In Section 8.4 we discuss various rules for power series, including the Sum,
Multiple and Product Rules. We also find that it is valid to differentiate or
integrate a given power series term-by-term, and that this does not affect the
radius of convergence.
Finally, in Section 8.5 we look briefly at various methods for estimating the
number p, and prove that p is irrational.
Some of the proofs in this chapter are not particularly illuminating, and you
may wish to leave them till a second reading of the chapter. You will need a
calculator handy while you work through Sections 8.1 and 8.2.
8.1
Taylor polynomials
8.1.1
What are Taylor polynomials?
Let f be a continuous function defined on an open interval I containing the
point a. It follows from the definition of continuity that
That is, I is a neighbourhood
of a.
lim f ð xÞ ¼ f ðaÞ:
x!a
Thus we can write
f ð xÞ ’ f ðaÞ;
for x near a:
In geometrical terms, this means that we can approximate the graph y ¼ f(x)
near a by the horizontal line y ¼ f(a) through the point (a, f(a)). For most
continuous functions, this does not give a very good approximation.
However, if the function is differentiable on I, then we can obtain a better
approximation by using the tangent line through (a, f(a)) instead of the
horizontal line. The tangent to the graph at (a, f(a)) has equation
y f ð aÞ
¼ f 0 ð aÞ
xa
or
y ¼ f ðaÞ þ f 0 ðaÞðx aÞ;
so, for x near a, we can write
f ð xÞ ’ f ðaÞ þ f 0 ðaÞðx aÞ:
This approximation is called the tangent approximation to f at a.
Notice that the function f and the approximating polynomial
f ð aÞ þ f 0 ð aÞ ð x aÞ
have the same value at a and the same first derivative at a.
Example 1
Solution
Determine the tangent approximation to the function f(x) ¼ ex at 0.
Here
f ð xÞ ¼ ex ;
f ð0Þ ¼ 1;
f 0 ð xÞ ¼ ex ;
f 0 ð0Þ ¼ 1:
We can think of the tangent at
(a, f(a)) as the line of best
approximation to the graph
near a.
8.1
Taylor polynomials
315
Hence the tangent approximation to f at 0 is
ex ’ f ð0Þ þ f 0 ð0Þðx 0Þ ¼ 1 þ x:
&
Problem 1 Determine the tangent approximation to each of the
following functions f at the given point a:
(a) f ð xÞ ¼ ex ;
a ¼ 2;
(b) f ð xÞ ¼ cos x;
a ¼ 0:
So far we have seen two approximations to f(x) for x near a:
f ð xÞ ’ f ðaÞ
ða constant functionÞ;
0
f ð xÞ ’ f ðaÞ þ f ðaÞðx aÞ ða linear functionÞ:
If the function f is twice differentiable on a neighbourhood I of a, then we can
find an even better approximation to f(x) by considering the expression
f ð x Þ f f ð aÞ þ f 0 ð aÞ ð x aÞ g
ð x aÞ 2
:
Let
F ð xÞ ¼ f ð xÞ ff ðaÞ þ f 0 ðaÞðx aÞg;
2
for x 2 I;
for x 2 I:
G ð x Þ ¼ ð x aÞ ;
Then F and G are differentiable on I, and
F 0 ð xÞ ¼ f 0 ð xÞ f 0 ðaÞ;
G0 ð xÞ ¼ 2ðx aÞ:
Also, F ðaÞ ¼ GðaÞ ¼ 0:
Hence, by l’Hôpital’s Rule, it follows that
F ð xÞ
x!a Gð xÞ
lim
exists and equals
Theorem 2, Sub-section 6.5.2.
F 0 ð xÞ
;
x!a G0 ð xÞ
lim
provided that this latter limit exists.
Now
F 0 ð xÞ
f 0 ð x Þ f 0 ð aÞ
;
¼
lim
x!a G0 ð xÞ
x!a 2ðx aÞ
lim
and, since the function f is twice differentiable at a, this limit exists and equals
1 00
2 f ðaÞ. Hence
lim
x!a
f ð xÞ f f ðaÞ þ f 0 ðaÞðx aÞg
ð x aÞ 2
exists and equals
1 00
f ðaÞ:
2
We can reformulate this result as follows: for x near a
1
f ð xÞ ’ f ðaÞ þ f 0 ðaÞðx aÞ þ f 00 ðaÞðx aÞ2 ða quadratic functionÞ:
2
Notice that the function f and the approximating polynomial f ðaÞ þ
f 0 ðaÞ ðx aÞ þ 12 f 00 ðaÞðx aÞ2 have the same value at a and the same
first and second derivatives at a.
8: Power series
316
This suggests that, if the function is n-times differentiable on I, then we may
be able to find a better approximating polynomial of degree n whose value at a
and whose first n derivatives at a are equal to those of f. This leads to the
following definition.
Definition Let f be n-times differentiable on an open interval containing
the point a. Then the Taylor polynomial of degree n for f at a is the
polynomial
f 0 ð aÞ
ð x aÞ
1!
00
f ð aÞ
f ðnÞ ðaÞ
þ
ð x aÞ 2 þ þ
ðx aÞn :
2!
n!
Tn ð x Þ ¼ f ð a Þ þ
Strictly speaking, we should
use more complicated
notation to indicate that the
Taylor polynomial Tn(x)
depends on n, a and f.
Notice that
Tn ðaÞ ¼ f ðaÞ; Tn0 ðaÞ ¼ f 0 ðaÞ; . . .; TnðnÞ ðaÞ ¼ f ðnÞ ðaÞ;
that is, f and Tn have the same value at a and have equal derivatives at a of all
orders up to and including n.
Example 2 Determine the Taylor polynomials T1(x), T2(x) and T3(x) for the
function f(x) ¼ sin x at each of the following points:
(a) a ¼ 0;
Solution
(b) a ¼ p2
Here
p
¼ 1;
2
p
f 0 ð0Þ ¼ 1;
f0
¼ 0;
f 0 ð xÞ ¼ cos x;
2
p
f 00
f 00 ð xÞ ¼ sin x; f 00 ð0Þ ¼ 0;
¼ 1;
2
p
f 000 ð xÞ ¼ cos x; f 000 ð0Þ ¼ 1; f 000
¼ 0:
2
f ð xÞ ¼ sin x;
f ð0Þ ¼ 0;
f
Hence:
3
(a) T1 ð xÞ ¼ x; T2 ð xÞ ¼ x and T3 ð xÞ ¼ x x3! ;
2
2
(b) T1 ð xÞ ¼ 1; T2 ð xÞ ¼ 1 12 x p2
and T3 ð xÞ ¼ 1 12 x p2 : &
Problem 2 Determine the Taylor polynomials T1(x), T2(x) and T3(x)
for each of the following functions f at the given point a:
(a) f ð xÞ ¼ ex ; a ¼ 2;
(b) f ð xÞ ¼ cos x; a ¼ 0:
Problem 3 Determine the Taylor polynomial of degree 4 for each of
the following functions f at the given point a:
(a) f ð xÞ ¼ 7 6x þ 5x2 þ x3 ; a ¼ 1;
(c) f ð xÞ ¼ loge ð1 þ xÞ; a ¼ 0;
1
(b) f ð xÞ ¼ 1x
; a ¼ 0;
(d) f ð xÞ ¼ sin x; a ¼ p4 ;
(e) f ð xÞ ¼ 1 þ 12 x 12 x2 16 x3 þ 14 x4 ; a ¼ 0:
Problem 4 Determine the percentage error involved in using the
Taylor polynomial of degree 3 for the function f (x) ¼ tan x at 0 to evaluate tan 0.1. (Use your calculator to evaluate tan 0.1.)
We do not usually multiply
out such brackets, since that
would make the results less
clear.
8.1
Taylor polynomials
8.1.2
317
Approximation by Taylor polynomials
We now look at some specific examples of functions to investigate the assertion that Taylor polynomials provide good approximations for a large class of
functions.
The function f ðx Þ ¼ 1 þ 21 x 21 x 2 61 x 3 þ 41 x 4
It follows from the result of Problem 3(e) above that the Taylor polynomials of
degrees 1, 2, 3 and 4 for f at 0 are:
1
1
1
T2 ð xÞ ¼ 1þ x x2 ;
T1 ð xÞ ¼ 1þ x;
2
2
2
1
1
1
T3 ð xÞ ¼ 1þ x x2 x3 ;
2
2
6
1
1
1
1
and T4 ð xÞ ¼ 1þ x x2 x3 þ x4 :
2
2
6
4
Since f ðnÞ ð0Þ ¼ 0 for n 5, it follows that for n 5 the Taylor polynomial of
degree n for f at 0 is just the same as the Taylor polynomial of degree 4 for f at 0.
The graphs of these Taylor polynomials are as follows:
In this case, the polynomials T2(x) and T3(x) provide good approximations to
f(x) near 0, and Tn(x) ¼ f(x) for all n 4.
The function f(x) ¼ sin x
Tn ð xÞ ¼ f ð xÞ
By calculating higher derivatives of the function f(x) ¼ sin x at 0, we can show
that the following are Taylor polynomials for f at 0
T1 ð xÞ ¼ T2 ð xÞ ¼ x;
T5 ð xÞ ¼ T6 ð xÞ ¼ x x3 x5
þ ;
3! 5!
In general, if f is a polynomial
of degree N, then
T 3 ð xÞ ¼ T 4 ð xÞ ¼ x x3
;
3!
T 7 ð xÞ ¼ T 8 ð xÞ ¼ x x3 x5 x7
þ :
3! 5! 7!
for all n N:
8: Power series
318
The following graphs illustrate how the approximation to f(x) given by Tn(x)
gets better as n increases.
For example, the
pgraph
of T5 appears to be very close to the graph of sine
p
over the interval 2 , 2 , so that sin x and T5(x) do not differ by very much for
values of x in this interval. Thus T5(x) seems to be a good approximation to
sin x in this interval.
It appears that, as the degree of the Taylor polynomial increases, so its
graph becomes a good approximation to that of sine over more and more of
R. For instance, in the above diagrams the shaded area covers the interval of the
x-axis on which the Taylor polynomial Tn(x) agrees with sin x to three decimal
places.
1
The function f ðx Þ ¼ 1x
By repeated differentiation it is easy to verify that, for k ¼ 1, 2, . . .
k!
;
f ð k Þ ð xÞ ¼
ð1 xÞkþ1
thus, in particular, f ðkÞ ð0Þ ¼ k!.
Hence the Taylor polynomial of degree n for f at 0 is
Tn ð x Þ ¼
¼
n
X
f ðkÞ ð0Þ
k¼0
n
X
k!
xk
xk ¼ 1 þ x þ x2 þ þ xn :
k¼0
The following diagram shows the graphs of the Taylor polynomials for f at 0 of
degrees 1, 2, 4 and 7.
8.1
Taylor polynomials
319
The graphs show that the nature of the approximation is very different from
the previous examples. For sine, the interval over which the approximation is
good seems to expand indefinitely as the degree of the polynomials increases;
1
the interval of good approximation always seems to be
but for f ðxÞ ¼ 1x
contained in the interval ( 1, 1).
Notice, however, that for this function f, the Taylor polynomials Tn(x) are
1
P
xk ; this series converges
just the nth partial sums of the geometric series
k¼0
1
for jxj < 1, and diverges for jxj 1. It follows that, if jxj < 1, then
with sum 1x
Tn ð xÞ ! f ð xÞ as n ! 1;
so that, if jxj < 1, the polynomials Tn(x) do provide better and better approximations to f(x) as n increases.
Which functions can be approximated by Taylor
polynomials?
In the next section we obtain a criterion for determining those functions f for
which the Taylor polynomials provide useful approximating polynomials, and
For jxj 1, the sequence
{Tn(x)} does not converge,
and so cannot provide an
approximation to f(x).
8: Power series
320
the intervals on which the approximation occurs. We also introduce certain
basic power series which correspond to the functions in the following problem.
Problem 5 Determine the Taylor polynomial of degree n for each of
the following functions at 0:
1
(a) f ð xÞ ¼ 1x
;
(d) f ð xÞ ¼ sin x;
8.2
(b) f ð xÞ ¼ loge ð1 þ xÞ;
(e) f ð xÞ ¼ cos x:
(c) f ð xÞ ¼ ex ;
Taylor’s Theorem
8.2.1
Taylor’s Theorem and approximation
In Section 8.1 we demonstrated how to find the Taylor polynomial Tn(x) of
degree n for a function f at a point a. This polynomial and its first n derivatives
agree with f and its first n derivatives at a, and the polynomial appears to
approximate f at points near a. But how good an approximation is it? What
error is involved if we replace f by a Taylor polynomial at a? The answer to
these questions is given by the following important result.
Theorem 1 Taylor’s Theorem
Let f be (n þ 1)-times differentiable on an open interval containing the
points a and x. Then
f ð xÞ ¼ f ðaÞ þ f 0 ðaÞðx aÞ
f 00 ðaÞ
f ðnÞ ðaÞ
ð x aÞ 2 þ þ
ðx aÞn þ Rn ð xÞ;
þ
2!
n!
where
Rn ð x Þ ¼
f ðnþ1Þ ðcÞ
ðx aÞnþ1 ;
ðn þ 1Þ!
and c is some point between a and x.
Remarks
1. When n ¼ 0, Taylor’s Theorem reduces to the assertion
f ð xÞ ¼ f ðaÞ þ f 0 ðcÞðx aÞ;
which we can rewrite (for x 6¼ a) in the form
f ð x Þ f ð aÞ
¼ f 0 ðcÞ; for some c between a and x:
xa
But this is just the Mean Value Theorem! It follows that Taylor’s Theorem
can be considered as a generalisation of the Mean Value Theorem.
2. The result of Theorem 1 can be expressed in the form
f ð xÞ ¼ Tn ð xÞ þ Rn ð xÞ;
where Rn(x) is thought of as a ‘remainder term’ or ‘error term’ involved in
approximating f(x) by the estimate Tn(x).
Strictly speaking, we should
use more complicated
notation to indicate that the
remainder term Rn(x) depends
on n, a and f.
8.2
Taylor’s Theorem
321
Proof For simplicity, we assume that x > a; the proof is similar if x < a.
We use the auxiliary function
hðtÞ ¼ f ðtÞ Tn ðtÞ Aðt aÞnþ1 ;
t 2 ½a; x;
You may omit this proof at a
first reading.
(1)
where Tn is the Taylor polynomial of degree n for f at a, and A is a constant
chosen so that
hðaÞ ¼ hð xÞ:
(2)
Now
f ðaÞ ¼ Tn ðaÞ; f 0 ðaÞ ¼ Tn0 ðaÞ; . . .;
and f ðnÞ ðaÞ ¼ TnðnÞ ðaÞ;
so that
hðaÞ ¼ 0; h0 ðaÞ ¼ 0; . . .;
and hðnÞ ðaÞ ¼ 0:
The function h is continuous on the closed interval [a, x] and differentiable on
the open interval (a, x); also, h(a) ¼ h(x). Hence, by Rolle’s Theorem, there
exists some number c1 between a and x for which
h0 ðc1 Þ ¼ 0:
Next, we apply Rolle’s Theorem to the function h0 on the interval [a, c1]. The
function h0 is continuous on [a, c1] and differentiable on (a, c1); also,
h0 (a) ¼ h0 (c1) ¼ 0. Hence, by Rolle’s Theorem, there exists some number c2
between a and c1 for which
h00 ðc2 Þ ¼ 0:
In turn, we apply Rolle’s Theorem to the functions
h00 ; h000 ; . . .; hðnÞ
on the intervals
½a; c2 ; ½a; c3 ; . . . ; ½a; cn ;
where c2 > c3 > c4 > > cn > a:
At the last stage, we find that there exists some point c between a and cn for
which
hðnþ1Þ ðcÞ ¼ 0:
(3)
By differentiating equation (1) (n þ 1) times, we obtain
hðnþ1Þ ðtÞ ¼ f ðnþ1Þ ðtÞ Aðn þ 1Þ!:
(4)
From (3) and (4), we deduce that
0 ¼ f ðnþ1Þ ðcÞ Aðn þ 1Þ!;
so that
A¼
f ðnþ1Þ ðcÞ
:
ðn þ 1Þ!
(5)
Finally, it follows from equations (1) and (2), with h(a) ¼ 0, that
f ð xÞ ¼ Tn ð xÞ þ Aðx aÞnþ1 :
Hence, by equation (5)
f ð xÞ ¼ Tn ð xÞ þ
as required.
f ðnþ1Þ ðcÞ
ðx aÞnþ1 ;
ðn þ 1Þ!
&
This choice of A is made so
that we can apply Rolle’s
Theorem to h on [a, x].
8: Power series
322
Problem 1 Obtain an expression for R1(x) when Taylor’s Theorem is
1
applied to the function f ð xÞ ¼ 1x
at a ¼ 0. Calculate the value of c
3
when x ¼ 4.
Problem 2 What can you say about Rn(x) when f is a polynomial of
degree at most n?
Problem 3 By applying Taylor’s Theorem to the function f (x) ¼ cos x
at a ¼ 0, prove that cos x ¼ 1 12 x2 þ R3 ð xÞ, x 2 R, where
1 4
x :
jR3 ð xÞj 24
In most applications of Taylor’s Theorem, we do not know the value of c
explicitly. However, for many purposes this does not
matter, since we can
show that jRn(x)j is small by finding an estimate for f ðnþ1Þ ðcÞ which is valid
for all c between a and x, and then applying the following result.
Corollary 1
Remainder Estimate
Let f be (n+1)-times differentiable on an open interval containing the points
a and x. If
ðnþ1Þ f
ðcÞ M;
for all c between a and x, then
f ð xÞ ¼ Tn ð xÞ þ Rn ð xÞ;
where
jRn ð xÞj Proof
M
jx ajnþ1 :
ðn þ 1Þ!
From Taylor’s Theorem, we have
R n ð xÞ ¼
f ðnþ1Þ ðcÞ
ðx aÞnþ1 ;
ðn þ 1Þ!
where c is some point between a and x. It follows that
ðnþ1Þ f
ðcÞ
jx ajnþ1
jRn ð xÞj ¼
ðn þ 1Þ!
M
jx ajnþ1 :
ðn þ 1Þ!
&
Example 1 By applying the Remainder Estimate to the function f (x) ¼ sin x,
with a ¼ 0 and n ¼ 3, calculate sin 0.1 to four decimal places.
Solution
Here
f ð xÞ ¼ sin x;
f ð0Þ ¼ 0;
f 0 ð xÞ ¼ cos x;
f 0 ð0Þ ¼ 1;
f 00 ð xÞ ¼ sin x;
f 00 ð0Þ ¼ 0;
f 000 ð xÞ ¼ cos x;
f 000 ð0Þ ¼ 1:
Strictly speaking, we should
use more complicated
notation to indicate that the
upper bound M depends on n,
a and f.
8.2
Taylor’s Theorem
323
Hence the Taylor polynomial of degree 3 for f at 0 is
T3 ð xÞ ¼ f ð0Þ þ f 0 ð0Þx þ
f 00 ð0Þ 2 f 000 ð0Þ 3
x þ
x
2!
3!
1
¼ x x3 :
6
Also, f ð4Þ ð xÞ ¼ sin x, so that
ð4Þ f ðcÞ ¼ jsin cj 1;
for c 2 R:
Taking M ¼ 1 in the Remainder Estimate, we deduce that
1
jR3 ð0:1Þj ð0:1Þ4
4!
1
104
¼
24
1
< 105 :
2
Now
sin 0:1 ’ T3 ð0:1Þ
1
¼ 0:1 103
6
¼ 0:1 0:0001666 . . .
¼ 0:0998333 . . .:
It follows from the Remainder Estimate that T3(0.1) gives an estimate for sin 0.1
to four decimal places. Hence sin 0.1 ¼ 0.0998 (to four decimal places). &
Problem 4 By applying the Remainder Estimate to the function
f ð xÞ ¼ loge ð1 þ xÞ, with a ¼ 0 and n ¼ 2, calculate loge 1:02 to four
decimal places.
In many practical situations we do not know how many terms of the power
series are needed in order to calculate the value of a given function to a
prescribed number of decimal places. In such cases, we can use the
Remainder Estimate to determine how many terms are needed.
Example 2 By applying the Remainder Estimate to the function f (x) ¼ ex,
with a ¼ 0, calculate the value of e to three decimal places.
Solution
Since f ðkÞ ð xÞ ¼ ex ;
f
ðk Þ
ð0Þ ¼ 1;
for k ¼ 0; 1; . . .; we have
for k ¼ 0; 1; . . .:
It follows that, for each n
Tn ð xÞ ¼ 1 þ x þ
x2
xn
þ þ :
2!
n!
Also, f ðnþ1Þ ð xÞ ¼ ex , for all x, so that, for all c 2 (0,1), we have
ðnþ1Þ f
ðcÞ e < 3:
It follows from the Remainder Estimate, with x ¼ 1 and M ¼ 3, that
j Rn ð 1 Þ j 3
1nþ1 :
ðn þ 1Þ!
We proved that e < 3 in Subsection 2.5.3.
8: Power series
324
To calculate e to three decimal places, we must choose n so that
3
< 2 104 ; or 15;000 < ðn þ 1Þ!:
ðn þ 1Þ!
Since 7! ¼ 5,040 and 8! ¼ 40,320, we may safely choose n ¼ 7.
It follows that
e ’ T7 ð1Þ
1 1 1 1 1 1 1
¼1þ þ þ þ þ þ þ
1! 2! 3! 4! 5! 6! 7!
¼ 1 þ 1 þ 0:5 þ 0:16 þ 0:1416 þ 0:0083 þ 0:00138
þ 0:000198412 . . .
¼ 2:7182 . . .:
&
Hence, e ¼ 2.718 (to three decimal places).
Problem 5 By applying the Remainder Estimate to the function f(x) ¼
cos x, with a ¼ 0, calculate cos 0.2 rounded to four decimal places.
Our next example illustrates how to obtain an approximation valid over an
interval.
Example 3
1
Calculate the Taylor polynomial T3(x) for f ð xÞ ¼ xþ2
at 1.
Show that T3(x) approximates f (x) with an error less than 5 103 on the
interval [1, 2].
Solution
Here
f ð xÞ ¼
1
;
xþ2
f 0 ð xÞ ¼
1
f ð 1Þ ¼ ;
3
1
;
1
f 0 ð 1Þ ¼ ;
9
3
;
f 00 ð1Þ ¼
4
;
f 000 ð1Þ ¼ ðx þ 2Þ2
f 00 ð xÞ ¼
2
ð x þ 2Þ
f 000 ð xÞ ¼
6
2
;
27
2
:
27
ð x þ 2Þ
Hence the Taylor polynomial of degree 3 for f at 1 is
1 1
1
1
T 3 ð x Þ ¼ ð x 1Þ þ ð x 1Þ 2 ð x 1Þ 3 :
3 9
27
81
24
Also, f ð4Þ ð xÞ ¼ ðxþ2
,
so
that
Þ5
ð4Þ 24
f ðcÞ ; for c 2 ð1; 2Þ:
35
in the Remainder Estimate, we deduce that
Taking M ¼ 24
35
24 ð2 1Þ4
35
4!
1
¼ 5 ¼ 0:0041 . . .; for x 2 ½1; 2:
3
Since the remainder term is less than 0.005, it follows that T3(x) approximates
&
f (x) with an error less than 5 103 on [1, 2].
jR3 ð xÞj 8.2
Taylor’s Theorem
325
Problem 6 Calculate the Taylor polynomial T4(x) for f (x) ¼ cos x at p.
Show that
approximates
f (x) with an error less than 3 103 on the
3 T4(x)
5
interval 4 p; 4 p .
8.2.2
Taylor’s Theorem and power series
From Taylor’s Theorem, we know that, if a function f can be differentiated
as often as we please on an open interval containing the points a and x, then,
for any n
f ð xÞ ¼ Tn ð xÞ þ Rn ð xÞ
n
X
f ðkÞ ðaÞ
ðx aÞk þ Rn ð xÞ;
¼
k!
k¼0
where
Rn ð x Þ ¼
f ðnþ1Þ ðcÞ
ðx aÞnþ1 ;
ðn þ 1Þ!
for some c between a and x. It follows that, if Rn(x) ! 0 as n ! 1, then we can
express f (x) as a power series in (x a).
Theorem 2 Let f have derivatives of all orders on an open interval containing the points a and x. If Rn(x) ! 0 as n ! 1, then
1 ðnÞ
X
f ð aÞ
ð x aÞ n :
f ð xÞ ¼
n!
n¼0
We call f the sum function of the power series
1 ð nÞ
P
f ðaÞ
n¼0
n!
ðx aÞn , and we call
this power series the Taylor series for f at a.
Warning A dramatic example of what happens when the remainder term
does not tend to zero is given by the function
1
2
x
f ð xÞ ¼ e ; x 6¼ 0;
0;
x ¼ 0:
For this function, f (0) ¼ 0, f 0 (0) ¼ 0, f 00 (0) ¼ 0, . . .. Thus the Taylor polynomial Tn(x) is identically zero for each n, although the function f is not identically zero. In this case, Rn(x) ¼ f (x) for all n, and so the Taylor series for f at 0
1 ðnÞ
X
f ð0Þ n
x ¼ 0 þ 0x þ 0x2 þ ;
n!
n¼0
converges to f (x) only at 0!
We can use Theorem 2 to obtain the following basic power series.
Theorem 3
(a)
1
1x
Basic power series
¼1 þ x þ x2 þ x3 þ 2
3
¼
(b) loge ð1 þ xÞ ¼ x x2 þ x3 ¼
1
P
n¼0
1
P
n¼1
xn ;
for j xj < 1;
n
ð1Þnþ1 xn ;
forj xj < 1;
y
1
2
y = e –1/x
–3
–2
–1 0
1
2
3 x
We indicate a proof of this
assertion in Section 8.6,
Exercise 6 on Section 8.2.
8: Power series
326
2
3
(c) ex ¼ 1 þ x þ x2! þ x3! þ ¼
3
5
(d) sin x ¼ x x3! þ x5! ¼
2
1 n
P
x
n¼0
1
P
n¼0
1
P
4
(e) cos x ¼ 1 x2! þ x4! ¼
n¼0
for x 2 R;
n! ;
2nþ1
for x 2 R;
2n
for x 2 R:
x
ð1Þn ð2nþ1
Þ!;
x
ð1Þn ð2n
Þ!;
Proof
(a) Here we use the fact that, for x 6¼ 1
1
xnþ1
¼ 1 þ x þ x2 þ x3 þ þ xn þ
:
1x
1x
n nþ1 o
But the sequence x1x is null, if jxj < 1. Thus, by letting n ! 1, we
This identity is easily proved
by multiplying both sides by
1 x:
obtain
1
X
1
¼
xn ;
1 x n¼0
for j xj < 1:
(b) Similarly
1
ð1Þnþ1 tnþ1
¼ 1 t þ t2 þ ð1Þn tn þ
:
1þt
1þ t
Integrating both sides from 0 to x, where jxj < 1, we obtain
!
Z x
Z x
nþ1 nþ1
dt
ð
1
Þ
t
¼
1 t þ t2 þ ð1Þn tn þ
dt;
1þt
0 1þt
0
so that
nþ1 x
t2 t3
n t
½loge ð1 þ tÞ0 ¼ t þ þ ð1Þ
2 3
nþ1 0
Z x nþ1
t
dt:
þ ð1Þnþ1
0 1þt
x
Hence
x2 x3
xnþ1
þ þ ð1Þn
2
3
nþ1
Z x nþ1
t
nþ1
dt:
þ ð1Þ
0 1þt
loge ð1 þ xÞ ¼ x When 0 x < 1, we have, by the Inequality Rule for integrals
Z x nþ1 Z x nþ1
Z x
t
t
ð1Þnþ1
¼
dt
dt
tnþ1 dt
0 1þt
0 1þt
0
nþ2 x
t
¼
nþ2 0
¼
xnþ2
1
! 0 as n ! 1:
nþ2 nþ2
When 1 < x < 0, we put T ¼ t and X ¼ x, so that 0 < T < 1. Then
Theorem 2, Sub-section 7.4.1.
8.2
Taylor’s Theorem
Z
ð1Þnþ1
0
x
327
Z X nþ1
Z X
tnþ1 T
1
dt ¼
dT T nþ1 dT
1X 0
1þt
0 1T
¼
X nþ2
1
!0
ð1 X Þðn þ 2Þ ð1 X Þðn þ 2Þ
For, 0 < X < 1 and
1
1T
1
1X
for T 2 [0, X].
as n ! 1:
Combining these two results, we deduce that
1
X
xn
ð1Þnþ1 ; for 1 < x < 1:
loge ð1 þ xÞ ¼
n
n¼1
(c) This was one of our definitions of the exponential function.
See Sub-section 3.4.3.
(d) Let f (x) ¼ sin x. Then, by Taylor’s Theorem, we have sin x ¼ Tn ð xÞþ
Rn ð xÞ, where
Rn ð x Þ ¼
f ðnþ1Þ ðcÞ nþ1
x ;
ðn þ 1Þ!
for some number c between 0 and x. We saw earlier that
ðnþ1Þ
cos c;
so that, in particular, we can be sure that f ðnþ1Þ ðcÞ 1. It follows from
the Remainder Estimate, with M ¼ 1, that
f
ðcÞ ¼ sin c
jRn ð xÞj Problem 5(d),
Sub-section 8.1.2.
or
1
j xjnþ1 ! 0 as n ! 1;
ðn þ 1Þ!
so that, in particular, Rn(x) ! 0 as n ! 1.
Hence, by letting n ! 1 in the equation sin x ¼ Tn ð xÞ þ Rn ð xÞ, we obtain
sin x ¼ x 1
X
x3 x5 x7
x2nþ1
þ þ ¼
;
ð1Þn
3! 5! 7!
ð2n þ 1Þ!
n¼0
(e) The proof is similar to that of part (d), so we omit it.
for x 2 R:
&
Remark
Probably the first definitions of sin x and cos x that you met were expressed in
terms
ofpa right-angled triangle, but those definitions only make sense for
x 2 0; 2 . We have now shown that sin x and cos x can be represented by the
power series in parts (d) and (e) of Theorem 3, and we can use these power
series to define sin x and cos x for all x 2 R.
Finally, we use Taylor’s Theorem to prove an interesting limit which is a
generalisation of two limits that you met earlier
1 n
x n
lim 1 þ
¼ e and lim 1 þ
¼ ex ; for any x 2 R:
n!1
n!1
n
n
Example 4
(a) Calculate T1(x) and R1(x) for the function f ð xÞ ¼ loge ð1 þ xÞ; x 2 ð1; 1Þ,
at 0.
(b) By replacing x in part (a) by x , where jxj > jj, prove that, for any real
numbers and Sub-section 2.5.3.
8: Power series
328
x
lim 1 þ
¼ e :
x!1
x
(c) Deduce from part (a) that
1
n loge 1 þ
! 1 as n ! 1:
n
Solution
(a) For the function f ð xÞ ¼ loge ð1 þ xÞ, x 2 ð1; 1Þ, we have
f ð xÞ ¼ loge ð1 þ xÞ;
f ð0Þ ¼ 0;
f 0 ð xÞ ¼
1
;
1þx
f 00 ð xÞ ¼
1
ð1 þ x Þ2
f 0 ð0Þ ¼ 1;
:
Hence
T1 ð xÞ ¼ f ð0Þ þ f 0 ð0Þx ¼ x
and R1 ð xÞ ¼
f 00 ðcÞ 2
x2
x ¼
;
2!
2ð1 þ cÞ2
for some number c between 0 and x.
(b) Replacing x by x in the equation
x2
; for j xj < 1;
loge ð1 þ xÞ ¼ x 2ð1 þ cÞ2
we obtain
2
loge 1 þ
; for j xj > jj:
¼ x
x 2ð 1 þ c Þ 2 x 2
Hence
2 ; for j xj > jj:
¼ x loge 1 þ
x
2ð 1 þ c Þ 2 x
Since 1x ! 0 as x ! 1, it follows that
¼ ;
lim x loge 1 þ
x!1
x
so that
x
¼ :
lim loge 1 þ
x!1
x
Since the exponential function is continuous on R, and
x
x
;
¼ 1þ
exp loge 1 þ
x
x
it follows, from the Composition Rule for limits, that
x
¼ e :
lim 1 þ
x!1
x
x2
,
(c) For x 2 (1, 1), we have, from part (a), that loge ð1 þ xÞ ¼ x 2ð1þc
Þ2
1
for some number c between 0 and x. Putting x ¼ n, we obtain
1
1
1
loge 1 þ
;
¼ n
n 2n2 ð1 þ cn Þ2
for some number cn between 0 and 1n.
8.3
Convergence of power series
329
Hence
1
1
n loge 1 þ
¼1
n
2nð1 þ cn Þ2
! 1 as n ! 1;
since
0
and
8.3
1
2n
1
2nð1 þ cn Þ
2
1
2n
&
is a null sequence.
Convergence of power series
8.3.1
The radius of convergence
In the previous section you saw that certain standard functions can be
expressed as the sum functions of power series; for example
1
X
1
¼
xn
1 x n¼0
and
loge ð1 þ xÞ ¼
1
X
xn
ð1Þnþ1 ;
n
n¼1
for j xj < 1:
These power series are the Taylor series for the given functions at 0.
Conversely, power series can be used to define functions. Thus, for instance,
we defined the exponential function x 7! ex, x 2 R, to be the sum function for
1 n
P
x
the power series
n! .
Section 3.4.
n¼0
But there are many other functions which are defined as the sum functions of
power series (as distinct from a power series obtained as the Taylor series for a
given function). For example, the Bessel function
2n
1
X
ð1Þn 2x
J0 ð xÞ ¼
; for x 2 R;
ðn!Þ2
n¼0
See Exercise 6 on Section 8.4,
in Section 8.6.
arises in analyses of the vibrations of a circular drum and of the radiation from
certain types of radio antenna.
However, all such uses of power series depend on knowledge of those
1
P
numbers x for which a power series
an ðx aÞn converges. In each of the
n¼0
above examples, the series converges on an interval; indeed, all power series
have this property.
Theorem 1
Radius of Convergence Theorem
1
P
an ðx aÞn , precisely one of the following
For a given power series
possibilities occurs:
n¼0
(a) The series converges only when x ¼ a;
(b) The series converges for all x;
We give the proof of
Theorem 1 in Subsection 8.3.3.
8: Power series
330
(c) There is a number R > 0, called the radius of convergence of the power
series, such that the series converges if jx aj < R and diverges if
jx aj > R.
For example:
1
P
(a)
n!xn converges only when x ¼ 0;
(b)
n¼0
1
P
(c)
n¼0
1
P
xn
n!
For all non-zero x, the series
1
P
n!xn diverges, by the Nonn¼0
nð1xÞn o
null Test, as n!x1 n ¼ n!
is
converges for all x;
xn converges if jxj < 1 and diverges if jxj > 1 – its radius of conver-
a basic null sequence.
n¼0
gence is 1.
Remarks
1. Sometimes we abuse our notation by writing R ¼ 0, if a series converges
only for x ¼ a, or R ¼ 1, if a series converges for all x.
2. It is important to remember that Theorem 1, part (c), makes no assertion
about the convergence or divergence of the power series at the end-points
a R, a þ R of the interval (a R, a þ R).
For example, we shall see shortly that the three power series
1
X
n¼1
xn ;
1 n
X
x
n¼1
n
and
Example 2.
1 n
X
x
n¼1
n2
all have radius of convergence 1. However, the intervals on which these
series converge are, respectively, ( 1, 1), [ 1, 1) and [ 1, 1].
The interval of convergence of the power series is the interval (a R, a þ R),
together with any end-points of the interval at which the series converges.
The following diagram illustrates the various types of interval of conver1
P
gence of
an ðx aÞn :
n¼0
Theorem 1 is not concerned with the problem of evaluating the radius of convergence of a power series. However, a power series is simply a particular type of
series, and so all the convergence tests for series can be applied to power series.
In practice, we can tackle most commonly arising power series by using the
following version of the Ratio Test.
Ratio Test for Radius of Convergence
1
P
Suppose that
an ðx aÞn is a given power series, and that
Sections 3.1–3.3.
Theorem 2
n¼0
a nþ1 ! L;
an
We give the proof of
Theorem 2 in Subsection 8.3.3.
as n ! 1:
(a) If L is 1, the series converges only for x ¼ a.
R ¼ 0.
8.3
Convergence of power series
331
(b) If L ¼ 0, the series converges for all x.
R ¼ 1.
(c) If L > 0, the series has radius of convergence
1
L:
R ¼ L1 Example 1 Determine the radius of convergence of each of the following
power series:
1 n
1
P
P
n ðxþ1Þn
ðx2Þn
;
(b)
(a)
n!
n! :
n¼1
n¼1
Solution
n
(a) Here an ¼ nn! for all n, so
a ðn þ 1Þnþ1
n!
1 n
nþ1 ¼ 1þ
;
¼
ðn þ 1Þ!
n
an
nn
Thus
for all n:
a nþ1 ! e as n ! 1:
an
Hence, by the Ratio Test, the radius of convergence is 1e (b) Here an ¼ n!1 , for all n, so
a n!
1
nþ1 ¼
;
¼
ðn þ 1Þ! n þ 1
an
Thus
for all n:
a nþ1 ! 0 as n ! 1:
an
Hence, by the Ratio Test, the power series
1
P
ðx2Þn
n¼1
n!
converges for all x. &
Problem 1 Determine the radius of convergence of each of the following power series:
1
1
P
P
ðn!Þ2 n
(a)
ð2n þ 4n Þxn ;
(b)
ð2nÞ! x ;
(c)
n¼0
1
P
n
ðn þ 2n Þðx 1Þ ;
(d)
n¼1
n¼1
1
P
xn
1:
n
ð
n!
n¼1 Þ
Hint for part (d): Use Stirling’s Formula and the result of Example 1,
Sub-section 7.5.1.
If we wish to find the interval of convergence (as opposed to the radius of
convergence), then we may need to use some of the other tests for series in
order to determine the behaviour at the end-points of the interval.
Example 2 Determine the interval of convergence of each of the following
power series:
1
1 n
1 n
P
P
P
x
x
(a)
xn ;
(b)
(c)
n;
n2 .
n¼1
n¼1
n¼1
Solution
(a) Here an ¼ 1, for all n, so
a nþ1 ¼ 1; for all n:
an
8: Power series
332
Hence, by the Ratio Test, the radius of convergence is 1; in other words
1
X
xn converges for 1 < x < 1:
n¼1
Next, we consider the behaviour of the power series at the end-points of
this interval, namely 1 and 1.1Since the sequences f1n gand fð1Þn g are
P n
x diverges when x ¼ 1, by the Nonboth non-null, it follows that
n¼1
null Test.
1
P
Hence the interval of convergence of
xn is ( 1, 1).
n¼1
1
(b) Here an ¼ n, for all n, so
a n
1
nþ1 ¼
; for all n:
¼
n þ 1 1 þ 1n
an
Thus
a nþ1 ! 1 as n ! 1:
an
Hence, by the Ratio Test, the radius of convergence is 1; in other words
1 n
X
x
converges for 1 < x < 1:
n
n¼1
1
1
P
P
ð1Þn
1
But we know that
n converges. It follows that the
n diverges and
n¼1
n¼1
1 n
P
x
interval of convergence of the power series
n is ½1; 1Þ.
(c) Here an ¼ n12 , for all n, so
a n2
1
nþ1 ¼
;
¼
2
an
ðn þ 1Þ
ð1 þ 1nÞ2
n¼1
By Example 2 of
Sub-section 3.2.1, and
at the start of Subsection 3.3.2, respectively.
for all n:
Thus
a nþ1 ! 1 as n ! 1:
an
Hence, by the Ratio Test, the radius of convergence is 1; in other words
1 n
X
x
converges for 1 < x < 1:
2
n
n¼1
1
P 1
But we know that
n2 converges; so, by the Absolute Convergence Test,
n¼1
1
n
P
ð1Þ
the series
n2 also converges. It follows that the interval of convern¼1
1 n
P
x
&
gence of the power series
n2 is ½1; 1.
n¼1
Now, the Radius of Convergence Theorem tells us that, if a power series
1
P
an ðx aÞn has radius of convergence R, then it converges for jx aj < R. In
n¼0
fact, we can say more than this – namely, that, for all points x with jx aj < R,
the power series is actually absolutely convergent!
Theorem 3
Absolute Convergence Theorem
1
P
Let the power series
an ðx aÞn have radius of convergence R. Then it is
n¼0
absolutely convergent for all x with jx aj < R.
Theorem 1, Sub-section 3.3.1.
8.3
Convergence of power series
333
Proof Let x be a number such that jx aj < R, and chose a number X such
that jx aj < jX aj < R.
1
P
Now, the series
an ðX aÞn is convergent, so that an ðX aÞn ! 0 as
n¼0
n ! 1. In particular, there exists some number N such that jan ðX aÞn j < 1,
for all n > N. It follows that
x a n
n
jan ðx aÞn j ¼ jan ðX aÞ j
X
a
x a n
; for n > N:
Xa
1 1
P
x a n is convergent, so that the series P jan ðx aÞn j is also
But the series
Xa
n¼0
n¼0
convergent, by the Comparison Test for series.
1
P
an ðx aÞn is absolutely convergent, as
It follows that the power series
n¼0
&
required.
Such a choice is always
possible – for example,
choose X to be the midpoint of
x and the nearest end-point of
the interval (a R, a þ R).
For
1 P
x a n
n¼0
Xa
is a geometric
series.
Problem 2 Determine the interval of convergence of each of the
following power series:
1
1
P
P
1 n
(a)
nxn ;
(b)
n3n x :
n¼0
n¼1
Determine the radius of convergence of the power series
1
X
ð 1Þ 2
ð 1Þ . . . ð n þ 1Þ n
1 þ x þ
x þ ¼
x ;
2!
n!
n¼0
Problem 3
where 6¼ 0; 1; 2; . . .:
8.3.2
Abel’s Limit Theorem
We can sometimes use the ideas of power series to sum interesting and
commonly arising series such as
1
1
X
X
ð1Þnþ1
1 1 1
ð1Þn
1 1 1
¼1 þ þ and
¼1 þ þ :
2
3
4
3 5 7
n
2n
þ
1
n¼1
n¼0
A major tool in this connection in the following result.
Theorem 4
Abel’s Limit Theorem
1
P
Let f (x) be the sum function of the power series
an xn , which has radius of
1
n¼0
P
convergence 1; and let
an be convergent. Then
n¼0
1
X
lim f ð xÞ ¼
x!1
an :
n¼0
Proof This proof is a wonderful illustration of the sheer power and magic of
the " ! method for proving results in Analysis! We use the standard
approach for proving that a limit exists as x ! 1 .
1
P
Let sn ¼ a0 þ a1 þ þ an1 be the nth partial sum of the series
an ; and
n¼0
1
P
let s denote the sum
an .
n¼0
A similar result holds for
power series of the form
1
P
an ðx aÞn ; a 6¼ 0.
n¼0
Equivalently
1
1
P
P
n
an x ¼
an :
lim
x!1
n¼0
n¼0
You may omit this proof at a
first reading.
8: Power series
334
Now,
a0 ¼ s1 and an ¼ snþ1 sn; for all n 1:
(1)
It follows that, for j xj51
ð1 x Þ
1
X
snþ1 xn ¼ ð1 xÞ s1 þ s2 x þ s3 x2 þ þ snþ1 xn þ n¼0
¼ s1 þ s2 x þ s3 x2 þ s4 x3 þ snþ1 xn þ
s1 x s2 x2 s3 x3 sn xn snþ1 xnþ1 ¼ s1 þ ðs2 s1 Þx þ ðs3 s2 Þx2 þ þ ðsnþ1 sn Þxn þ 1
X
an xn ¼ f ðxÞ:
¼
Using (1).
n¼0
We now study closely the identity
ð1 xÞ
1
X
snþ1 xn ¼ f ðxÞ:
(2)
n¼0
Next, we want to prove that:
for each positive number ", there is a positive number such that
j f ð xÞ sj < ",
This is simply the definition
of lim f ð xÞ ¼ s:
for all x satisfying 1 < x < 1.
x!1
Now, we use equation (2) in the following way to get a convenient expression for f (x) s
f ð xÞ s ¼
1
X
an x n n¼0
1
X
Here we use the definition
of s, equation (2), and the fact
that the sum of the series
1
P
1
x n is 1x
:
an
n¼0
¼ ð1 xÞ
¼ ð1 xÞ
1
X
n¼0
1
X
snþ1 x n s
n¼0
snþ1 x n ð1 xÞ
n¼0
¼ ð1 xÞ
1
X
1
X
sx n
n¼0
ðsnþ1 sÞx n :
(3)
n¼0
Next, choose a number N such that jsnþ1 sj < 12 " for all n > N. We can
then apply the Triangle Inequality to equation (3), for x 2 ð0; 1Þ, to see that
N
1
X
X
n
n
ðsnþ1 sÞx þ
ðsnþ1 sÞx j f ð x Þ sj ¼ ð 1 x Þ
n¼0
n¼Nþ1
X
X
N
1
n
n
ðsnþ1 sÞx þ ð1 xÞ
ðsnþ1 sÞx ð 1 x Þ
n¼0
n¼Nþ1
1
X
1
xn
ð1 x Þ
jsnþ1 sjx þ ð1 xÞ " 2
n¼0
n¼Nþ1
N
X
n
Such a choice of N is possible,
since snþ1 ! s as n ! 1.
We split the infinite sum into
two parts.
Using the Triangle Inequality.
For jsnþ1 sj < 12 ", for all
n < N; also, we use that
1
P
1
0 < x < 1 so that
xn ¼ 1x
.
n¼0
8.3
Convergence of power series
ð1 xÞ
335
N
X
1
jsnþ1 sjxn þ "
2
n¼0
N
X
1
(4)
jsnþ1 sj þ "
2
n¼0
N
P
Finally, since the linear function x 7!ð1 xÞ
jsnþ1 sj is continuous on
ð1 xÞ
Again we use that 0 < x < 1.
n¼0
R, and takes the value 0 at 1, it follows that we can choose a positive number (with 2 (0, 1)) such that
N
X
1
ð1 x Þ
jsnþ1 sj < "; for all x satisfying 1 < x < 1:
2
n¼0
N
P
If we substitute this upper bound 12 " for ð1 xÞ
jsnþ1 sj into the
We add in the extra
requirement that < 1 in order
to ensure that we are only
considering values of x
in (0, 1).
n¼0
inequality (4), we find that we have proved that
for each positive number ", there is a positive number such that
j f ð xÞ sj < "; for all x satisfying 1 < x < 1:
&
This is what we set out to prove.
As an application of Abel’s Limit Theorem, we use it to evaluate
1
X
ð1Þnþ1
1 1 1
¼ 1 þ þ :
2 3 4
n
n¼1
Let f ð xÞ ¼
1
P
ð1Þnþ1
n¼1
n
xn . We have seen already that this power series converges
in (1, 1) to the sum loge ð1 þ xÞ. Further, we saw earlier that the series
1
P
ð1Þnþ1
is convergent, by the Alternating Test. It follows, from Abel’s
n
Part (b) of Theorem 3, Subsection 8.2.2.
n¼1
Limit Theorem, that
1
X
ð1Þnþ1
n¼1
n
¼ lim
x!1
1
X
ð1Þnþ1
n¼1
n
xn ¼ lim loge ð1 þ xÞ ¼ loge 2:
x!1
Remark
It is always necessary to check that the conditions of Abel’s Theorem apply
before using it. For example, a thoughtless application to the identity
1
X
1
; wherej xj < 1;
1 x þ x2 x3 þ ¼
ð1Þn xn ¼
1
þ
x
n¼0
of taking the limit as x ! 1, would give the following absurd conclusion
1
X
1
1 1 þ 1 1 þ ¼
ð1Þn ¼ !
2
n¼0
We did not check that the
1
P
ð1Þn xn was
series
n¼0
convergent at 1!
Problem 4 Use Abel’s Limit Theorem to evaluate
1
P
ð1Þn
1
1
1
2nþ1 ¼ 1 3 þ 5 7 þ .
n¼0
3
5
7
Hint: Use the facts that tan1 x ¼ x x3 þ x5 x7 þ for jxj < 1, and
that the latter series has radius of convergence 1.
We prove these facts in Subsection 8.4.1.
8: Power series
336
8.3.3
Proofs
You may omit this subsection at a first reading.
We now supply the proofs that we omitted from Sub-section 8.3.1.
Theorem 1
Radius of Convergence Theorem
1
P
For a given power series
an ðx aÞn , precisely one of the following
n¼0
possibilities occurs:
(a) The series converges only when x ¼ a;
(b) The series converges for all x;
(c) There is a number R > 0 such that the series converges if jx aj < R and
diverges if jx aj > R.
Proof For simplicity, we shall assume that a ¼ 0.
Clearly the possibilities (a) and (b) are mutually exclusive; we shall therefore assume that for a given power series neither possibility occurs, and then
prove that possibility (c) must occur.
First, let
(
)
1
X
n
S ¼ x:
an x converges :
For the general proof, replace
x throughout by (x a).
n¼0
Since the possibility (b) has been excluded, we deduce that S must be bounded.
1
P
For, if
an X n diverges, then, in view of the Absolute Convergence Theorem
n¼0
1
P
for power series,
an xn cannot converge for any x with jxj > jXj, since
n¼0
1
P
otherwise
an X n would then have to be convergent – which is not the case.
Theorem 3, Sub-section 8.3.1.
n¼0
Also, S is non-empty, since the power series
1
P
an xn converges at 0.
n¼0
It then follows, from the Least Upper Bound Property of R, that the set S
1
P
an xn is
must have a least upper bound, R say. We now prove that the series
n¼0
convergent if jxj < R and divergent if jxj > R.
First, notice that R > 0. For, since possibility (a) has been excluded, there must
1
P
an xn1 is convergent; hence
be at least one non-zero value of x, x1 say, such that
n¼0
we have x1 2 S. It follows, from the Absolute Convergence Theorem, that
1
P
an xn must converge for those x with jxj < jx1j, so that ðjx1 j; jx1 jÞ lies in S.
n¼0
Now, choose any x for which jxj < R. Then by the definition of R as sup S,
1
P
there exists some number x2 with jxj < x2 < R such that
an xn2 converges. It
n¼0
1
P
an xn converges.
follows, from the Absolute Convergence Theorem, that
n¼0
Finally, choose any x for which j xj > R ¼ sup S. It follows that x 2
= S, so that
1
P
n
&
an x must be divergent.
n¼0
Here we identify a number R
that will be the desired radius
of convergence.
8.3
Convergence of power series
337
In view of the Absolute Convergence Theorem, we can slightly strengthen
the conclusion of the Radius of Convergence Theorem, as follows.
1
P
If the power series
Corollary
an ðx aÞn has radius of convergence
n¼0
R > 0, then it is absolutely convergent if jx aj < R. If the series converges
for all x, then it is absolutely convergent for all x.
Theorem 3, Sub-section 8.3.1.
This result is sometimes also
called the Absolute
Convergence Theorem.
Finally we prove the Ratio Test for Radius of Convergence of power series.
Theorem 2
Ratio Test for Radius of Convergence
1
P
Suppose that
an ðx aÞn is a given power series, and that
n¼0
a nþ1 ! L as n ! 1:
an
(a) If L is 1, the series converges only for x = a.
(b) If L ¼ 0, the series converges for all x.
(c) If L > 0, the series has radius of convergence L1 :
Proof
For simplicity, we shall assume that a ¼ 0.
(a) Suppose that jaanþ1
j ! 1 as n ! 1. Then, for any non-zero value of x, the
n
sequence
ja a xx j ! 1, so that the sequence {anxn} is unbounded. It
nþ1
n
nþ1
n
follows, from the Non-null Test, that the series
1
P
an xn must be divergent.
n¼0
anþ1
an
(b) Suppose that j j ! 0 as n ! 1. Then, for any non-zero value of x
anþ1 xnþ1 anþ1 a xn ¼ a j xj
n
n
! 0 j xj ¼ 0;
1
P
an xn is absolutely convergent, and so is convergent.
(c) Suppose that aanþ1
! L as n ! 1, where L > 0.
n
so that
n¼0
First, suppose that j xj > L1. Then
anþ1 xnþ1 anþ1 a xn ¼ a j xj
n
n
! L j xj > 1;
so that
1
P
an xn is not absolutely convergent. It follows, from the above
n¼0
Corollary to the Radius of Convergence Theorem, that the radius of
1
P
an xn must be less than or equal to L1.
convergence of
n¼0
Next, suppose that j xj < L1. Then
anþ1 xnþ1 anþ1 a xn ¼ a j xj
n
n
! L j xj < 1;
For the general proof, replace
x throughout by (xa).
8: Power series
338
1
P
an xn is absolutely convergent, and so is convergent.
1
P
an xn must be at least equal to L1.
Hence the radius of convergence of
so that
n¼0
n¼0
Combining these two facts, it follows that the desired radius of conver&
gence is exactly L1.
8.4
Manipulating power series
It would be tedious to have to apply Taylor’s Theorem every time that we wished
to determine the Taylor series of a given function. While sometimes this really
has to be done, in many commonly arising situations we can use standard rules
for power series and the list of basic power series to avoid most of the effort.
We now set out to establish the rules for manipulating power series.
8.4.1
Theorem 3, Sub-section 8.2.2.
Rules for power series
Many of the rules for manipulating power series are similar to the corresponding rules for manipulating ‘ordinary’ series.
Theorem 1
Combination Rules
Let
f ð xÞ ¼
gð xÞ ¼
1
X
n¼0
1
X
an ð x aÞ n ;
for jx aj < R;
bn ð x aÞ n ;
for jx aj < R0 :
and
n¼0
Then:
Sum Rule
ðf þ gÞð xÞ ¼
1
P
ðan þ bn Þðx aÞn ;
for jx aj < r;
n¼0
where r ¼ minfR; R0 g;
1
P
lan ðx aÞn ;
Multiple Rule lf ð xÞ ¼
for jx aj < R; where l 2 R:
Thus, if the power series for
both f and g converge at a
particular point x, then so does
the series for f þ g.
n¼0
We do not supply a proof, as Theorem 1 is a simple restatement of the Sum and
Multiple Rules for ‘ordinary’ series.
We can use the Combination Rules to find the Taylor series at 0 for the
function cosh x. We start with the power series for the exponential function
1 n
X
x2 x3
x
;
for x 2 R:
ex ¼ 1 þ x þ þ þ ¼
2! 3!
n!
n¼0
2
3
It follows that ex ¼ 1 x þ x2! x3! þ ¼
1
P
n¼0
ð1Þn xn! ;
n
Theorem 2, Sub-section 3.1.4.
Theorem 3, Sub-section 8.2.2.
for x 2 R. We
may then use the Sum Rule to obtain
1
X
x2 x4
x2n
x
x
;
e þ e ¼ 2 1 þ þ þ ¼ 2 2! 4!
ð2nÞ!
n¼0
for x 2 R;
For the odd-powered terms
cancel.
8.4
Manipulating power series
339
so that, by using the Multiple Rule with l ¼ 12, we obtain
1
X
1
x2 x4
x2n
; for x 2 R:
cosh x ¼ ðex þ ex Þ ¼ 1 þ þ þ ¼
2
2! 4!
ð2nÞ!
n¼0
Problem 1 Find the Taylor series at 0 for each of the following functions, and state its radius of convergence:
(a) f ð xÞ ¼ sinh x; x 2 R;
(b) f ð xÞ ¼ loge ð1 xÞ þ 2ð1 xÞ1 ;
j xj < 1:
Problem 2 Find the Taylor series at 0 for each of the following functions, and state its radius of convergence:
(a) f ð xÞ ¼ sinh x þ sin x; x 2 R;
(b) f ð xÞ ¼ loge 1þx
j xj < 1;
1x ;
1
x 2 R:
(c) f ð xÞ ¼ 1þ2x
2;
Remark
In Theorem 1, the radius of convergence of the power series for f þ g may be
larger than r ¼ minfR; R0 g. For example, we can use the basic power series
and the Combination Rules to verify that the Taylor series at 0 for the functions
1
1
and gðxÞ ¼ 1x
þ 11 x are
f ð xÞ ¼ 1x
2
1
X
xn and
f ð xÞ ¼ 1 þ x þ x2 þ x3 þ x4 þ ¼
n¼0
1 X
1
3 2 7 3 15 4
1 n
gð x Þ ¼ x x x x ¼
1 þ n x ;
2
4
8
16
2
n¼0
each with radius of convergence 1. It follows, from Theorem 1, that the power
series for the function ð f þ gÞð xÞ ¼ 11 x is
2
1
X
1
1
1
1
1 n
ð f þ gÞð xÞ ¼ 1 þ x þ x2 þ x3 þ x4 þ ¼
x ;
n
2
4
8
16
2
n¼0
We simply add the power
series term-by-term.
and that this power series has radius of convergence at least 1. In fact, it has
radius of convergence 2.
x
We can find the Taylor series at 0 for the function f ð xÞ ¼ 11 þ
x, for jxj < 1, by
writing
1þx
1x
function
2
1x
2ð1xÞ
1x
2
¼ 1x
1. Since the Taylor series at 0 for the
1
P
2
3
is 2 þ 2x þ 2x þ 2x þ ¼
2xn , with radius of conver-
in the form
n¼0
gence 1, it follows that the Taylor series for f at 0 is 1 þ 2x þ 2x2 þ
1
P
2x3 þ ¼ 1þ
2xn , with radius of convergence 1.
n¼1
However we could obtain the same result by multiplying together the Taylor
1
, since
series for the functions x 7! 1 þ x and x 7! 1x
ð1 þ xÞ 1 þ x þ x2 þ x3 þ ¼ 1 þ x þ x2 þ x3 þ þ x 1 þ x þ x2 þ x3 þ ¼ 1 þ 2x þ 2x2 þ 2x3 þ :
In fact, we can justify such multiplication together of power series to obtain
further power series.
We now ‘collect together’ the
multiples of successive
various powers of x.
8: Power series
340
Theorem 2
Let
Product Rule
f ð xÞ ¼
gð x Þ ¼
1
X
n¼0
1
X
an ð x aÞ n ;
for jx aj < R;
bn ð x aÞ n ;
for jx aj < R0 :
and
n¼0
Then
ð fgÞð xÞ ¼
1
X
cn ðx aÞn ; for jx aj < r; where r ¼ minfR; R0 g;
n¼0
and
c 0 ¼ a0 b0 ; c 1 ¼ a0 b1 þ a1 b0
As in Theorem 1, the radius of
convergence of the product
1
P
series
cn ðx aÞn may be
n¼0
and
greater than r.
cn ¼ a0 bn þ a1 bn1 þ þ an1 b1 þ an b0 :
Theorem 2 is an immediate consequence of the Product Rule for series, applied
1
1
P
P
to the two series
an ðx aÞn and
bn ðx aÞn , both of which are absolutely
n¼0
n¼0
convergent for jx aj < r – by the Absolute Convergence Theorem.
Example 1
Theorem 5, Sub-section 3.3.4.
Theorem 3, Sub-section 8.3.1.
1þx
Determine the Taylor series at 0 for the function f ð xÞ ¼ ð1x
.
Þ2
Solution We use our knowledge of the power series at 0 for the functions
x
x 7! 1 1 x and x 7! 11 þ
x, both of which have radius of convergence 1. Thus, by
the Product Rule, we have
1þx
1
1þx
¼
2
1x 1x
ð1 xÞ
¼ 1 þ x þ x2 þ x3 þ x4 þ 1 þ 2x þ 2x2 þ 2x3 þ 2x4 þ 1
X
¼
cn xn ;
n¼0
where c0 ¼ 1 1 ¼ 1, c1 ¼ 1 2 þ 1 1 ¼ 3 and cn ¼ 1 2 þ 1 2 þ . . . þ
1 2 þ 1 1 ¼ 2n þ 1.
1
P
&
Thus the required power series for f at 0 is 1 þ x 2 ¼
ð2n þ 1Þxn :
ð1 x Þ
Problem 3
Determine the Taylor series at 0 for:
(a) f ð xÞ ¼ ð1 þ xÞ loge ð1 þ xÞ;
(b) f ð xÞ ¼
n¼0
1þx
ð1xÞ3
;
for j xj < 1;
Also, its radius of
convergence is at least 1.
(In fact it is easy to check that
its radius of convergence is
exactly 1.)
for j xj < 1:
Next, we have seen already that the hyperbolic functions have the following
Taylor series at 0
x3 x5
x2 x4
sinh x ¼ x þ þ þ and cosh x ¼ 1 þ þ þ ;
3! 5!
2! 4!
each with radius of convergence 1. Notice that the derivative of the function
sinh x is cosh x, and that term-by-term differentiation of the series
3
5
2
4
x þ x3! þ x5! þ gives the series 1 þ x2! þ x4! þ . It looks as though we can
At the start of this sub-section.
8.4
Manipulating power series
341
obtain the Taylor series for the derivative of a function f simply by differentiating the Taylor series for f itself.
Our next result states that we can differentiate or integrate the Taylor series
of a function fRterm-by-term to obtain the Taylor series of the corresponding
function f 0 or f , respectively.
Theorem 3
Differentiation and Integration Rules
1
P
Let f ð xÞ ¼
an ðx aÞn ; for jx aj < R. Then:
n¼0
Differentiation Rule f 0 ð xÞ ¼
R
Integration Rule
1
P
n¼1
f ð xÞdx ¼
nan ðx aÞn1 ;
1
P
n¼0
for jx aj < R;
nþ1
aÞ
an ðx n þ 1 þ constant;
To find the constant, put x ¼ a.
for jx aj < R:
They may, however, behave
differently at the end-points of
their intervals of convergence.
All three series have the same radius of convergence.
For example, consider the Taylor series at 0 for the function tan1 . We know that
1
¼ 1 x2 þ x4 x6 þ 1 þ x2
1
X
¼
ð1Þn x2n ; with radius of convergence 1;
This is a geometric series.
n¼0
and that
d
1
tan1 x ¼
; for x 2 R:
dx
1 þ x2
It follows, from the Integration Rule, that
3
5
7
x x x
þ þ þc; for j xj < 1;and some constant c:
3 5 7
Substituting x ¼ 0 into this equation, we find that c ¼ tan1 0 ¼ 0. It follows that
tan1 x ¼ x tan1 x ¼ x Problem 4
x3 x5 x7
þ þ ;
3
5
7
for j xj < 1:
Find the Taylor series at 0 for the following functions:
(a) f ð xÞ ¼ ð1 xÞ2 ; for j xj < 1;
(c) f ð xÞ ¼ tanh1 x; for j xj < 1:
(b) f ð xÞ ¼ ð1 xÞ3 ; for j xj < 1;
Problem 5 Find the first three non-zero terms in the Taylor series at 0
for the function f (x) ¼ ex(1 x)2, jxj < 1. State its radius of convergence.
3
5
x
x
Problem 6 Let f be the function f ð xÞ ¼ x þ 1:3
þ 1:3:5
þ þ
x2nþ1
þ
;
x
2
R.
1:3::ð2nþ1Þ
Determine the Taylor series at 0 for each of the following functions:
(b) f 0 ð xÞ xf ð xÞ:
(a) f 0 ð xÞ;
Problem 7 Determine the Taylor series at 0 for the function
2
f ð xÞ ¼ ex , x 2 R.
R1
2
Þn
1
1
Deduce that 0 ex dx ¼ 1 13 þ 10
42
þ þ ð2nð1
þ 1Þn! þ :
By the Integration Rule, the
final Taylor series must
have the same radius of
convergence as the original
Taylor series, namely 1.
8: Power series
342
We have now found a whole variety of techniques for identifying Taylor series:
Taylor’s Theorem;
Combination, Product, Differentiation and Integration Rules.
But how do we know that different techniques will always give us the same
expression as the Taylor series? The following result states that there is only
one Taylor series for a function f at a given point a – any valid method will give
the same coefficients.
Theorem 4 Uniqueness Theorem
1
1
X
X
If
an ðx aÞn ¼
bn ðx aÞn ; for jx aj < R; then an ¼ bn :
n¼0
Proof
n¼0
Let f ð xÞ ¼
1
P
an ðx aÞn and gð xÞ ¼
n¼0
1
P
bn ðx aÞn , for jx aj <R.
n¼0
If we differentiate both equations n times, using the Differentiation Rule,
and put x ¼ a, we obtain
f ðnÞ ðaÞ ¼ n! an
and gðnÞ ðaÞ ¼ n! bn ;
Since f(x) ¼ g(x), for jx aj < R, we must have f (n)(a) ¼ g(n)(a). It follows that
&
an ¼ bn, for all n 0.
8.4.2
General Binomial Theorem
You will have already met the Binomial Theorem, which states that, for each
positive integer n
n X
nð n 1Þ . . . ð n k þ 1Þ
n!
n k
n
¼
:
ð1 þ xÞn ¼
x ; where
¼
k
k
ð
k!
k!
n
kÞ!
k¼0
See, for example, Appendix 1.
In the Binomial Theorem, the power n is a positive integer and the result is true
for all x 2 R. In fact, a similar result holds for more general values of the power
but with a restriction on the values of x for which it is valid.
Theorem 5 General Binomial Theorem
For any 2 R
1 X
ð 1Þ. . . ð n þ 1Þ
n
ð 1 þ xÞ ¼
¼
x ; where
and j xj < 1:
n
n
n!
n¼0
For example
ð1 þ 2xÞ6 ¼
1 X
6
n¼0
n
ð2xÞn ;
where
6
ð6Þð7Þ . . . ð6 n þ 1Þ
¼
n!
n
1
and j xj < ;
2
By convention,
¼ 1.
0
8.4
Manipulating power series
343
so that
1
for j xj < :
2
Another important type of application occurs when the power is not an
integer. For example
1 1 4 7 1
1 X
3 3 3 ... 3 n þ 1
1
13 n
3
ð1 þ x Þ 3 ¼
;
x ; where
¼
n
n
n!
n¼0
ð1 þ 2xÞ6 ¼ 1 12x þ 84x2 ;
so that
1
1
2
14
ð1 þ xÞ3 ¼ 1 x þ x2 x3 þ ;
3
9
81
for j xj < 1:
Problem 8 Use the General Binomial Theorem to determine the first
four non-zero
terms in the Taylor series at 0 for the function f ð xÞ ¼
1
ð1 þ 6xÞ4 , j xj < 16. State the radius of convergence of the Taylor series.
Problem 9
1
(a) Determine the Taylor series at 0 for the function f ð xÞ ¼ ð1 xÞ2 ,
jxj < 1.
(b) Hence determine the Taylor series at 0 for the function
f ð xÞ ¼ sin1 x, jxj < 1.
In our proof of the General Binomial Theorem, we use the following lemma.
Lemma
For any 2 R and any n 2 N, n
þ ðn þ 1Þ
¼
.
n
nþ1
n
Proof Since
¼ ð1Þ: : :k!: :ðkþ1Þ, for any k 2 N, we may rewrite the
k
left-hand side of the desired identity in the form
ð 1Þ ::: ð nÞ
n
þ ð n þ 1Þ
¼n
þ ðn þ 1Þ
ðn þ 1Þ!
n
nþ1
n
ð 1Þ.. . ð nÞ
¼n
þ
n!
n
ð 1Þ... ð n þ 1Þ
¼n
þ ð nÞ
n!
n
¼n
þ ð nÞ
n
n
&
¼
:
n
Theorem 5
General Binomial Theorem
For any 2 R
1 X
ð 1Þ . . . ð n þ 1Þ
n
ð1 þ xÞ ¼
¼
x ; where
and j xj < 1:
n
n
n!
n¼0
You may omit the remainder
of this sub-section at a first
reading.
Here we cancel the term
(n þ 1).
We now bring the term
( n) in front of the fraction.
The last fraction is just
.
n
8: Power series
344
Proof
Let
1 X
n
x
f ð xÞ ¼
n
n¼0
and
gð xÞ ¼ f ð xÞð1 þ xÞ ;
for j xj < 1:
We want to prove that g(x) ¼ 1, for all x with jxj < 1.
Using the rules for differentiation, we may differentiate the expression for
g to obtain
g0 ð xÞ ¼ f 0 ð xÞð1 þ xÞ f ð xÞð1 þ xÞ1
¼ ½ð1 þ xÞf 0 ð xÞ f ð xÞð1 þ xÞ1 :
Now
Here we differentiate the
power series term-by-term.
Then we split the initial
bracket.
n1
0
n
x
ð1 þ xÞf ð xÞ ¼ ð1 þ xÞ
n
n¼1
1 1 X
n1 X
n
¼
n
x
þ
n
x
n
n
n¼1
n¼1
1
1 X
X
n
n
¼
ðn þ 1Þ
x þ
n
x
n
þ
1
n
n¼0
n¼0
1 X
¼
ð n þ 1Þ
þn
xn
n
þ
1
n
n¼0
1 X
n
¼
x
n
n¼0
1
X
We now replace n by n þ 1 in
the first sum, and check what
the limits of summation are in
both sums.
Now we combine the two
sums into one.
We can apply the result of the
Lemma to this square bracket.
By the definition of f(x).
¼ f ð xÞ:
It follows from the earlier expression for g0 (x) that
g0 ð xÞ ¼ ½ð1 þ xÞf 0 ð xÞ f ð xÞð1 þ xÞ1
¼ 0 ð1 þ xÞ1 ¼ 0:
So, g( x) is a constant. Hence
gð x Þ ¼ gð 0Þ
¼ f ð0Þ ¼ 1;
&
as required.
8.4.3
Proofs
You may omit this Sub-section
at a first reading.
Differentiation Rule The series
1
X
an ðx aÞn and
f ð xÞ ¼
n¼0
1
X
nan ðx aÞn1
n¼1
have the same radius of convergence, R say. Also, f is differentiable on
(a R, a þ R), and
1
X
nan ðx aÞn1 :
f 0 ð xÞ ¼
n¼1
This is one part of Theorem 3,
Sub-section 8.4.1.
8.4
Manipulating power series
345
For simplicity, we assume that a ¼ 0.
1
1
P
P
an xn and
nan xn1 have radii of convergence R and R0 ,
Let the series
Proof
n¼0
n¼1
respectively.
We start by proving that R ¼ R0 .
1
P
nan xn1 is converWe first show that, if jxj < R, then the power series
n¼1
gent. (By the Corollary to the Radius of Convergence Theorem, this shows
that R0 R.)
To prove this, choose a real number c with jxj < c < R. Then the series
1
P
an cn is convergent, and so the sequence {ancn} is a null sequence. Thus
Sub-section 8.3.3.
For example, c ¼ 12 ðj xj þ RÞ.
n¼0
there is a number K such that
jan cn j K;
for n ¼ 0; 1; 2; . . .:
(1)
Then
n1
nan xn1 ¼ jan cn j n x
c
c
xn1
K
n ;
(2)
c
c
1 P
n1
by (1). Since xc < 1, the series
nxc converges. Hence, by the Comparison
n¼1
Test for series, the series
1
P
nan xn1 is absolutely convergent, and so is
Theorem 1, Sub-section 3.2.1.
n¼1
0
convergent. This proves that R R.
Next, suppose that R0 > R. Let c be any number such that R < c < R0 . Then
1
P
nan cn1 is absolutely convergent, by the Absolute Convergence Theorem.
n¼1
For example, c ¼ 12 ðR þ R0 Þ.
But
c
jan cn j ¼ nan cn1 n
c nan cn1 :
Hence, by the Comparison Test for series, the series
1
P
an cn is absolutely
n¼1
convergent, and so is convergent. This contradicts the definition of R, and thus
shows that R0 6> R.
It follows that R ¼ R0 .
Next, we show that f is differentiable on (R, R), and that f 0 is of the stated
form.
So, choose any point c 2 (R, R). Then choose a positive number r < R such
that c 2 (r, r). Then, for all non-zero h such that jhj < r jcj
1
1
f ð c þ hÞ f ð c Þ X
1X
nan cn1 ¼
an ðc þ hÞn cn hncn1 :
h
h n¼1
n¼1
(3)
Now, we apply Taylor’s Theorem to the function x 7! xn on the interval with
end-points c and c þ h. We obtain
1
ðc þ hÞn ¼ cn þ nhcn1 þ nðn 1Þh2 cn2
n ;
2
For example, r ¼ 12 ðjcj þ RÞ.
8: Power series
346
where cn is some number between c and c þ h (and, in particular, with
jcn j r).
Now
ðc þ hÞn cn nhcn1 1 nðn 1Þh2 r n2 :
(4)
2
It follows from (3) and (4) and the Triangle Inequality that
f ð c þ hÞ f ð c Þ X
1 X
1
1
n1 nan c jhj
nðn 1Þjan jr n2 : (5)
h
2
n¼1
n¼2
1
P
Since the series
nan xn1 has radius of convergence R, it follows that so
n¼1
1
1
P
P
does the series
nðn 1Þan xn2 ; we conclude that the series
nð n 1Þ
n¼2
an r n2 is (absolutely) convergent.
Hence, it follows from (5) and the Limit Inequality Rule that
1
f ðc þ hÞ f ðcÞ X
¼
lim
nan cn1 :
h!0
h
n¼1
The series
1
X
an ð x aÞ n
f ð xÞ ¼
n¼2
&
Integration Rule
n¼0
and
1
X
an
ðx aÞnþ1
F ð xÞ ¼
n
þ
1
n¼0
have the same radius of convergence, R say. Also, f is integrable on
(a R, a þ R), and
Z
f ð xÞdx ¼ F ð xÞ:
Proof The two power series have the same radius of convergence, by the
Differentiation Rule for power series, applied to F.
It also follows, from the Differentiation Rule for power series, that F is
&
differentiable and F0 ¼ f, so that F is a primitive of f.
8.5
Numerical estimates for p
One of the most interesting problems in the history of mathematics throughout
the last two thousand years has been the determination to any pre-assigned
pffiffiffi
degree of accuracy of naturally arising irrational numbers such as 2, e and p.
Here we look at various way of estimating p, and prove that p is irrational.
8.5.1
For we have shown that termby-term differentiation does
not alter the radius of
convergence of a power
series.
Calculating p
It is easy to check that p is slightly larger than 3 by wrapping a string round a
circle of radius r and measuring the length of the string corresponding to one
circumference; for this length 2pr is just slightly larger than 6r.
This is the other part of
Theorem 3, Sub-section 8.4.1.
That is, F is a primitive of f on
(a R, a þ R)
8.5
Numerical estimates for p
347
Over hundreds of years mathematicians devised more sophisticated methods
for estimating the value of p. Some of their results are as follows:
the Babylonians
c. 2000 BC
p ¼ 3 18 ¼ 3:125
the Egyptians
the Old Testament
c. 2000 BC
c. 550 BC
p ¼ 3 13
81 ’ 3:160
p¼3
Archimedes
c. 300–200 BC
between 3 17 and 3 10
71, p ’ 3:141
the Chinese
the Hindus
c. 400–500 AD
c. 500–600 AD
p ’ 3:1415926
pffiffiffiffiffi
p ’ 10 ’ 3:16
1 Kings vii, 23;
2 Chronicles iv, 2.
With the development of the Calculus in the seventeenth century, new formulas for estimating p were discovered, including:
Wallis’s Formula
p
2
2n
2n
¼ 21 23 43 45 65 67 2n1
2nþ1
Leibniz’s Series
tan
Gregory’s Series
p
4
1
x¼x
x3
3
þ
x5
5
x7
7
þ ¼ 1 13 þ 15 17 þ However, neither Wallis’s Formula nor Gregory’s Series is very useful for
calculating p beyond a few decimal places, as they converge far too slowly.
However, we can use the Taylor series for tan1 effectively for calculating p
by choosing values of x closer to 0; in fact, the smaller the value of x, the faster
the series converges, and so the fewer the number of terms we need to calculate
p to a given degree of accuracy.
For example, if we choose x ¼ p1ffiffi3 in Leibniz’s Series, we obtain
p
1
1
1 1 3 1 1 5 1 1 7
pffiffiffi þ
pffiffiffi pffiffiffi þ ¼ tan1 pffiffiffi ¼ pffiffiffi 6
5
7
3
3 3
3
3
3
1
1 1
1
þ :
¼ pffiffiffi 1 þ 9 45 189
3
This formula can be used to calculate p to several decimal places without much
effort; we gain about one extra place for every two extra terms.
To obtain series that are more effective for calculating p, we can use the
Addition Formula for tan1
1
1
1 x þ y
tan x þ tan y ¼ tan
;
1 xy
provided that tan1 x þ tan1 y lies in the interval p2 ; p2 .
Problem 1
Theorem 5, Sub-section 7.4.2.
Just before Problem 4, Subsection 8.4.1.
Problem 4, Sub-section 8.3.2.
This is known as Sharp’s
Formula.
Problem 3(a) on Section 4.3,
in Section 4.5.
Use the Addition Formula for tan1 to prove that
p
1 1
1 1
1 2
tan
þ tan
þ tan
¼ :
3
4
9
4
Similar applications of the Addition Formula give further expressions for p4,
such as
1
p
1 1
1 1
1
6 tan
þ 2 tan
þ tan
¼ ;
8
57
239
4
1
1
p
4 tan1
tan1
¼ :
5
239
4
This is known as Machin’s
Formula.
8: Power series
348
Such formulas have been used to calculate p to great accuracy: 1 million decimal
places were achieved in 1974. Recently, improved methods of numerical analysis have been used to calculate p correct to several million decimal places.
In 1770, Lambert showed that p is irrational; that is, p is not the solution of a
linear equation a0 þ a1x ¼ 0, where the coefficients a0 and a1 are integers.
Then, in 1882, Lindemann showed that p is transcendental; that is, p is not a
solution of any polynomial equation a0 þ a1x þ a2x2 þ þ anxn ¼ 0, where
all the coefficients are integers.
We end with a couple of mnemonics that can be used to recall the first few
digits for p; the word lengths give the successive digits:
May I have a large container of coffee?
3:
1 4
1 5
9
2 6
For interest only, we list the
first 1000 decimal places of
p in Appendix 3.
This solved the ancient Greek
problem of finding a ruler and
compass method to construct
a square with an area equal to
that of a given circle.
and
8.5.2
How
3:
I
1
need
4
a
1
drink; alcoholic of
5
9
2
after
5
all those formulas
3
5
8
involving
9
course;
6
tangent
7
functions!
9
Proof that p is irrational
You may omit this at a first
reading.
We now prove that p is irrational. Our proof is rather intricate; and, surprisingly,
it uses many of the properties of integrals that you met in Chapter 7! It depends
on the properties of a suitably chosen integral that we examine in Lemmas 1–3.
R1
n
Lemma 1 Let In ¼ 1 ð1 x2 Þ cos 12 px dx, n ¼ 0, 1, 2, . . .. Then:
(a) In > 0;
(b) In 2.
Proof
n
(a) The function f ð xÞ ¼ ð1 x2 Þ cos 12 px , for x 2 [1, 1], is non-negative,
for n ¼ 0, 1, 2, . . .. It follows, from the Inequality Rule for integrals, that
In 0, for each n.
To prove that In > 0 we need to examine the integral in more detail.
By the Inequality Rule for integrals, we have
Z 1
Z 1 Z 1 Z 1 !
2
2
f ð xÞdx ¼
þ
þ
f ð xÞdx
1
1
Z
1
2
12
12
1
2
f ð xÞdx:
n n
Now, if x 2 12 ; 12 , then ð1 x2 Þ cos 12 px 34 cos 14 p , so that
Z 1
Z 1 n 2
3
1
f ð xÞdx cos p dx
4
1
1 4
2n 3
1
¼
cos p > 0;
4
4
as required.
R 1
For 12 f ð xÞdx 0 and
R1
1 f ð xÞdx 0:
2
By the Inequality Rule for
integrals.
For the length of the interval
of integration is 1.
8.5
Numerical estimates for p
349
n
(b) For x 2 ½1; 1, ð1 x2 Þ cos 12 px 1 1 ¼ 1, so that
Z 1
Z 1
f ð xÞdx 1dx ¼ 2:
1
Problem 2
&
1
By the Inequality Rule for
integrals.
Prove that pI0 ¼ 4 and p3 I1 ¼ 32.
Next, we obtain a reduction formula for In, using integration by parts.
Lemma 2
The integral In satisfies the following reduction formula
p2 In ¼ 8nð2n 1ÞIn1 16nðn 1ÞIn2 :
Proof Using integration by parts twice, we obtain
1
Z
1
2 1
1
2 n2
2 n1
sin px
nð2xÞ 1 x
sin px dx
In ¼ 1 x
p
2
2
1 p 1
Z 1
n1
4n
1
¼
x 1 x2
sin px dx
p 1
2
1
4n 2
1
2 n1
¼
x 1x
cos px
p
p
2
1
Z 1n
n1
n2 o
8n
1
þ 2
1 x2
2x2 ðn 1Þ 1 x2
cos px dx
p 1
2
Z 1
n2
8n
16
1
¼ 2 In1 2 nðn 1Þ
x2 1 x2
cos px dx
p
p
2
1
Z 1n
o
8n
16
1
2 n2
2 n1
¼ 2 In1 2 nðn 1Þ
1x
1x
cos px dx
p
p
2
1
¼
The value of the first term is
zero.
The value of the first term is
zero.
For
n2
x2 1 x2
n2
1 x2
¼ 1 1 x2
n2 n1
¼ 1 x2
1 x2
:
8n
16
In1 2 nðn 1ÞfIn2 In1 g:
p2
p
Multiplying both sides by p2, we obtain
p2 In ¼ 8nIn1 16nðn 1ÞIn2 þ 16nðn 1ÞIn1
¼ 8nð2n 1ÞIn1 16nðn 1ÞIn2 :
&
We now use the result of Lemma 2 to prove the crucial tool in our proof that
p is irrational.
Lemma 3
that
Proof
For n ¼ 0, 1, 2, . . ., there exist integers a0, a1, a2, . . ., an such
!
n
X
p2nþ1
In ¼ a0 þ a1 p þ a2 p2 þ þan pn ¼
ak p k :
n!
k¼0
2nþ1
For simplicity, let Jn ¼ p n! In . It follows from Problem 2 that
J0 ¼ pI0 ¼ 4
and
J1 ¼ p3 I1 ¼ 32:
(1)
8: Power series
350
Next, we rewrite the reduction formula in Lemma 2 in terms of Jn as
p2nþ1
In
Jn ¼
n!
8nð2n 1Þ 2n1
16nðn 1Þ 2n1
p
p
¼
In1 In2
n!
n!
¼ 8ð2n 1ÞJn1 16p2 Jn2 :
(2)
The desired result now follows by Mathematical Induction. We know that
it holds for n ¼ 0 and for n ¼ 1, by the statements (1) above. Using the
reduction formula (2) we can prove that the statement of the Lemma holds
&
for all n 2.
Finally, we can use the fact that, for any integer p, the sequence
null, to prove that p is irrational.
Theorem 1
n
p2nþ1
n!
o
is
The sequence
n
ðp 2 Þ
p is irrational.
where the coefficients ak are integers.
If we now multiply both sides by the expression q2nþ1 , we find that
n
X
p2nþ1
In ¼
ak pk q2nþ1k ;
n!
k¼0
so that
p2nþ1
In is an integer:
n!
(3)
However, we know that the sequence
n
p
o
2nþ1
n!
is a null sequence; and we know,
from Lemma 1, that jInj 2. It follows, from the Squeeze Rule for sequences, that
p2nþ1
In ! 0
n!
as n ! 1:
n!
basic null sequence.
Proof Suppose, on the contrary, that p is rational; so that p ¼ pq, for p, q 2 N.
Then, the conclusion of Lemma 3 can be written in the form
n
X
p2nþ1
pk
I
¼
a
;
n
k
q2nþ1 n!
qk
k¼0
(4)
2nþ1
It follows from (4) that eventually p n! In < 1; and so, from (3), that in fact In
must equal 0. This contradicts the result of Lemma 1, part (a). This contra&
diction proves that, in fact, p must be rational.
8.6
We omit ‘the gory details’!
Exercises
Section 8.1
1. Determine the tangent approximation to the function f(x) ¼ 2 3x þ x2 þ ex
at the given point a:
(a) a ¼ 0;
(b) a ¼ 1.
This is a proof by
contradiction.
is a
8.6
Exercises
351
2. Determine the Taylor polynomial of degree 3 for each of the following
functions f at the given point a:
(a) f(x) ¼ loge(1 þ x), a ¼ 2;
(b) f ð xÞ ¼ sin x, a ¼ p6 ;
(c) f ð xÞ ¼ ð1 þ xÞ2 , a ¼ 12 ;
(d) f ð xÞ ¼ tan x, a ¼ p4 :
3. Determine the Taylor polynomial of degree 4 for each of the following
functions f at the given point a:
(a) f(x) ¼ cosh x, a ¼ 0;
(b) f(x) ¼ x5, a ¼ 1.
4. Determine the percentage error involved in using the Taylor polynomial of
degree 3 for the function f(x) ¼ ex at 0 to evaluate e0.1.
Section 8.2
1. Obtain an expression for the remainder term R1(x) when Taylor’s Theorem is
applied to the function f(x) ¼ ex at 0. Show that, when x ¼ 1 and n ¼ 1, then
the value of c in the statement of Taylor’s Theorem is approximately 0.36.
2. By applying Taylor’s Theorem to the function f(x) ¼ sin x with a ¼ p4, prove
that
1
p 1 p2
sin x ¼ pffiffiffi 1 þ x x
þ R2 ð xÞ; x 2 R;
4
2
4
2
3
where jR2 ð xÞj 16 x p4 :
3. By applying the Remainder Estimate to the function f(x) ¼ sinh x with
a ¼ 0, calculate sinh 0.2 to four decimal places.
x
4. Calculate the Taylor polynomial T3(x) for the function f ð xÞ ¼ xþ3
at 2.
4
Show that T3(x) approximates f(x) to within 10 on the interval 2; 52 :
5. (a) Determine the Taylor polynomial of degree n for the function
f(x) ¼ loge x at 1.
(b) Write down the remainder term Rn(x) in Taylor’s Theorem for this
function, and show that Rn(x) ! 0 as n ! 1 if x 2 (1, 2).
(c) By using Theorem 2 in Section 8.2, determine a Taylor series for f at 1
which isvalid when x 2 (1, 2).
12
x
6. Let f ð xÞ ¼ e ; x 6¼ 0;
0;
x ¼ 0:
(a) Prove that f 0 ð0Þ ¼ 0:
(b) Prove that, for x 6¼ 0, f 0 (x) is of the form
1
(a polynomial of degree at most 3 in 1x)e x2 .
(n)
(c) Prove that, for x 6¼ 0, f (x) is of the form
1
(a polynomial of degree at most 3n in 1x)e x2 .
(d) Prove that, for any positive integer n, f (n)(0) ¼ 0.
Section 8.3
1. Determine the radius of convergence of each of the following power series:
1
1 n
P
P
ð3nÞ!
n
1 n
(a)
ð x þ 1Þ n ;
(b)
n! x e ;
ðn!Þ2
n¼1
n¼1
For this function, the Taylor
polynomial Tn(x) is
identically zero for each n,
although the function f is not
identically zero. In this case,
Rn(x) ¼ f(x), for all n; and
so the Taylor series for f
1 ðnÞ
P
at 0, f n!ð0Þ xn , converges to
n¼0
f(x) only at 0.
8: Power series
352
(c)
1
P
ðn þ 1Þn xn ;
(d)
n¼1
1
2x
2
1:3:5 3
þ 1:3
2:5 x þ 2:5:8 x þ ;
aðaþ1Þ:bðbþ1Þ 2
(e) 1 þ a:b
1:c x þ 1:2:c ðcþ1Þ x þ ,
where a, b, c > 0:
2. Determine the Taylor series for the function f(x) ¼ loge(2 þ x) at the given
point a; in each case, indicate the general term and state the radius of
convergence of the series:
(a) a ¼ 1;
This series is often called the
hypergeometric series.
(b) a ¼ 1.
Hint: Use the Taylor series at 0 for the function x 7! loge ð1 þ xÞ, with
t ¼ x 1 and t ¼ x þ 1, respectively.
3. Determine the interval of convergence of each of the following power
series:
1 n
1
P
P
n
2 n
(a)
(b)
ð1Þn 2n ðx 1Þn :
n! x ;
n¼1
n¼1
4. Give an example (if one exists) of each of the following:
1
P
an xn which diverges at both x ¼ 1 and x ¼ 2;
(a) a power series
n¼0
(b) a power series
1
P
(c) a power series
n¼0
1
P
an xn which diverges at x ¼ 1 but converges at x ¼ 2;
an xn which converges at x ¼ 1 but diverges at x ¼ 2.
n¼0
5. Give an example (if one exists) of each of the following:
1
P
(a) a power series
an xn which converges only if 3 < x < 3;
(b) a power series
n¼0
1
P
(c) a power series
n¼0
1
P
(d) a power series
n¼0
1
P
an xn which converges only if 3 x < 3;
an xn which converges only if 3 < x 3;
an xn which converges only if 3 x 3.
n¼0
Section 8.4
1. Determine the Taylor series for each of the following functions at 0; in each
case, indicate the general term and state the radius of convergence of the
series:
3
x
(a) f ð xÞ ¼ loge ð1 þ x þ x2 Þ ¼ loge 1x
(b) f ð xÞ ¼ cosh
1x ;
1x :
2. Determine the first three non-zero terms in the Taylor series for each of the
following functions at 0; and state the radius of convergence:
(a) f(x) ¼ cos(ex 1);
(b) f(x) ¼ loge(1 þ sin x);
(c) f(x) ¼ ex sin x.
2:4:6 7
5
3. For the function f ð xÞ ¼ x þ 23 x3 þ 2:4
3:5 x þ 3:5:7 x þ , for jxj < 1, prove
that
1 x2 f 0 ð xÞ xf ð xÞ ¼ 1:
4. By using the Integration Rule for power series, prove that
sinh1 x ¼ x 1 x3 1:3 x5 1:3:5 x7
þ
þ ;
2 3 2:4 5 2:4:6 7
for j xj < 1:
1
xffi
.
In fact, f ð xÞ ¼ psinffiffiffiffiffiffiffi
1x2
8.6
Exercises
353
5. By using
R 1 the Integration Rule for power series, find an infinite series with
sum 0 sinðt2 Þdt.
6. The Bessel function J0 is defined by the power series
J0 ð xÞ ¼ 1 x2
22 :ð1!Þ
þ
2
x4
24 :ð2!Þ
2
x6
26 :ð3!Þ
þ þ
2
ð1Þn x2n
22n :ðn!Þ2
þ ; for x 2 R:
(a) Determine power series for J00 ð xÞ and J000 ð xÞ, indicating the general term
in each.
(b) Prove that xJ000 ð xÞ þ J00 ð xÞ þ xJ0 ð xÞ ¼ 0.
J0 is not the Taylor series at 0
of any familiar function.
Appendix 1: Sets, functions and proofs
Sets
A set is a collection of objects, called elements. We use the symbol 2 to mean ‘is a
member of’ or ‘belongs to’; thus ‘x 2 A’ means ‘the element x is a member of the set A’.
Similarly, we use the symbol 62 to mean ‘is not a member of’ or ‘does not belong to’;
thus ‘x 62 A’ means ‘the element x is not a member of the set A’.
We often use curly brackets (or braces) to list the elements of a set in some way. Thus
{a, b, c} denotes the set whose three elements are a, b and c. Similarly, {x : x2 ¼ 2}
denotes the set whose elements x are such that x2 ¼ 2; we would read this symbol in
words as ‘the set of x such that x2 ¼ 2’. When we define a set in this latter way, the
symbol x is a dummy variable; that is, if we replace that symbol x by any other symbol,
such as y, the set is exactly the same set; thus, for instance, the sets {x : x2 ¼ 2} and
{y : y2 ¼ 2} are identical.
Two sets are equal if they contain the same elements. We say that a set A is a subset
of a set B, if all the elements of A are elements of B, and we denote this by writing
‘A B’. We may wish to indicate specifically the possibility that the subset A is equal to
B by writing ‘A B’, or that A is a proper subset of B (that is, A is a subset of B but
A 6¼ B), by writing ‘A B’.
Thus the colon ‘:’ means
‘such that’.
6¼
Notice that, to show that two sets A and B are equal, it is necessary to prove that A is
a subset of B and that B is a subset of A. In symbols
A ¼ B is equivalent to the two properties both holding: A B and B A:
The empty set is the set that contains no elements; it is denoted by the symbol ‘Ø’.
It may seem strange to define such a thing; but, in practice, it is often a convenient set to use.
The union of two sets A and B consists of the set of elements that belong to at least
one of A and B; in symbols
A [ B ¼ fx : x 2 A or x 2 Bg:
The intersection of two sets A and B consists of the set of elements that belong to
both A and B; in symbols
A \ B ¼ fx : x 2 A and x 2 Bg:
We denote the set of elements that belong to A but not to B by A B; that is
A B ¼ fx : x 2 A; x 2
= Bg:
There are some special symbols used that are used to denote commonly arising sets of
real numbers:
N,
Z,
Q,
the set of all natural numbers; thus N ¼ {1, 2, 3, . . .};
the set of all integers; thus Z ¼ {0, 1, 2, 3, . . .};
the set of all rational numbers; that is, numbers of the form pq, where p, q 2 Z
but q 6¼ 0;
R, the set of all real numbers;
R þ, the set of all positive real numbers.
On the real line R, an interval I is a set of real numbers such that, if a and b both lie in I,
then all numbers between a and b also lie in I.
354
Here we also allow x to belong
to both A and B.
Sets, functions and proofs
355
There are nine types of intervals in R:
½a; b ¼ fx : x 2 R; a x bg;
ða; bÞ ¼ fx : x 2 R; a < x < bg;
½a; bÞ ¼ fx : x 2 R; a x < bg;
The first four types of
intervals are bounded
intervals.
ða; b ¼ fx : x 2 R; a < x bg;
½a; 1Þ ¼ fx : x 2 R; x ag;
ða; 1Þ ¼ fx : x 2 R; x > ag;
ð1; b ¼ fx : x 2 R; x bg;
ð1; bÞ ¼ fx : x 2 R; x < bg;
R.
An interval is said to be closed if it contains both of its end-points, open if it contains
neither of its end-points, and half-closed or half-open if it is neither closed nor open.
Functions
A function is a mapping of some element of R onto another element of R. Thus a
function f is defined by specifying:
a set A, called the domain of f;
a set B, called the codomain of f;
a rule x 7! f ð xÞ that associates with each element x of A a unique element f(x) of B.
Symbolically, we write this in the following way:
f: A!B
x 7! f ð xÞ:
The element f(x) is called the image of x under f, and the set f (A) ¼ {y : y ¼ f (x) for
some x 2 A} the image of A under f.
A particularly simple function is the identity function that maps each element of a
set to itself. Thus, the identity function on a set A, iA, is the function
iA : A ! B
x 7! x:
Sometimes every point of the codomain is in the image. We say that a function
f : A ! B is an onto function if f (A) ¼ B.
Sometimes every point of the image is the image of precisely one point of the
domain. We say that a function f : A ! B is a one–one function if:
whenever f(x) ¼ f(y) for elements x, y of A, then necessarily x ¼ y.
Notice that this does NOT mean that f is a one–one mapping of A onto B.
A given function f : A ! B may be onto, one–one, both onto and one–one, or neither
onto nor one–one. A function f : A ! B is called a bijection if it is both onto and
one–one.
If a function f : A ! B is one–one, then it has an inverse function f1 : f (B) ! A with
the defining rule
f 1 ð yÞ ¼ x;
where y ¼ f ðxÞ:
Notice that a function f only has an inverse function f 1 if it is bijective.
Sometimes we want to ‘change the domain’ of a function to a larger set or a
smaller set.
The final five types of
intervals are unbounded
intervals.
Appendix 1
356
If f : A ! B, and C is a subset of A, we say that the function g : C ! B is the restriction
of f to C if
gðxÞ ¼ f ð xÞ; for all x 2 C:
Similarly, if f : A ! B, and A is a subset of C, we say that the function g : C ! B is the
extension of f to C if
gðxÞ ¼ f ð xÞ;
for all x 2 A:
Finally, the composite function or composite g f of two functions
f :A!B
and
Sometimes we use the term
extension to denote a function
g : C ! D where A C,
B D, and g(x) ¼ f(x), for all
x 2 A.
g : C ! D;
where f (A) C, is the function
g
f : A!D
x 7! gð f ð xÞÞ:
Proofs
Logical implications are so important in Mathematics that we use some special symbols, namely ), ( and ,, to denote implications.
Thus, if we are discussing two statements P and Q, we write
P ) Q to mean: ‘if P is true, then Q is true’ or ‘P implies Q’;
P ( Q to mean: ‘P is implied by Q’, or ‘if Q is true, then P is true’ or ‘Q implies P’;
P , Q to mean: ‘P is true if and only if Q is true’;
this is equivalent to the two separate statements
P ) Q and Q ) P:
In this case, P and Q are equivalent statement.
The implications P ) Q and Q ) P are called converse implications.
In order to prove that an assertion ‘P ) Q’ is true, we need to verify that, in all
situations where the statement P holds, then the statement Q also holds. To prove that an
assertion ‘P ) Q’ is not true, all that we need do is to find one example of a situation
where the statement P holds but the statement Q does not. Such a situation is called a
counter-example to the assertion.
Sometimes we prove assertions P by simply checking all possible cases; generally,
though, we devise logical arguments that deal with all cases at the same time.
Sometimes our logical arguments appear slightly convoluted to non-mathematicians;
two examples of these are:
Thus, to prove that P , Q, we
need to prove that P ) Q and
Q ) P.
This is sometimes called
disproof by counter-example.
This approach is sometimes
called proof by exhaustion.
Proof by contradiction: In order to show that a statement P holds, we start by
assuming that P is false; we then look at the consequences of that assumption, and
identify some specific consequence that is untrue or contradictory. It then follows that
P must hold, after all.
Proof by contraposition: In order to show that an implication P ) Q is true, it is
sufficient to prove that
if Q does not hold, then P does not hold.
Sometimes mathematical proofs are long and complicated, and involve the use of a
variety of approaches to prove the desired result.
Notice that not all approaches to proving a result are necessarily valid! Some
examples of such approaches are:
Proof by picture: Here you make a false claim in your argument because it appears
to be true in the particular diagram that you have drawn.
Proof by example: For instance, you prove that some property holds for n ¼ 1 and
n ¼ 2, and then assert that it therefore holds for all positive integers.
This fact is sometimes
(humorously) called proof by
perspiration.
Sets, functions and proofs
357
Proof by omission: For instance, you prove one or two special cases of a result, and
claim that the general proof is ‘similar’.
Proof by superiority: Here you claim that the result is ‘obvious’, rather than sit down
to construct a logical proof of it.
Proof by mumbo-jumbo: Here you write down a jumble of formulas and relevant
words, ending up with the claim that you have proved the result.
Principle of Mathematical Induction
This is a standard method of proving statements involving an integer n, generally a
positive integer.
Principle of Mathematical Induction
Let P(n) denote a statement involving a positive integer n. If the following two
conditions are satisfied:
1. the statement P(1) is true,
2. whenever the statement P(k) is true for a positive integer k, then the statement
P(k þ 1) is also true,
then the statement P(n) is true for all positive integers n.
Sometimes we need to use an equivalent version of the Principle.
(Second) Principle of Mathematical Induction
Let P(n) denote a statement involving a positive integer n. If the following two
conditions are satisfied:
1. the statement P(1) is true,
2. whenever the statements P(1), P(2), . . ., P(k) are true for a positive integer k, then
the statement P(k þ 1) is also true,
then the statement P(n) is true for all positive integers n.
We have stated the Principle when P(n) applies to all integers n 1; analogous
results hold when P(n) applies to all integers n N, whatever integer N may be.
We end with some useful results that can be proved using the Principle of
Mathematical Induction.
Sums
n
X
1 ¼ n;
k¼1
n
X
k¼1
n
X
k2 ¼
nðn þ 1Þð2n þ 1Þ
;
6
n
X
nðn þ 1Þ
;
2
k¼1
n
X
nðn þ 1Þ 2
k3 ¼
;
2
k¼1
k¼
ð2k þ 1Þ ¼ ðn þ 1Þ2 ;
k¼0
sin A þ sinðA þ BÞ þ þ sinðA þ ðn 1ÞBÞ
sin 12 nB
n1
1 sin A þ
¼
B ; B 6¼ 0:
2
sin 2 B
Arithmetic progression
a þ ða þ d Þ þ ða þ 2d Þ þ þ ða þ ðn 1Þd Þ ¼ n a þ n1
2 d :
Thus, in order to use the
Principle, we have a two stage
strategy:
1. check P(1),
2. check that P(k) ) P(k þ 1).
Appendix 1
358
Geometric progression
n
a þ ar þ ar2 þ þ ar n1 ¼ a 1r
r 6¼ 1:
1r ;
Binomial Theorem
n X
nðn 1Þ 2
n k
x þ þ xn ;
ð1 þ xÞn ¼
x ¼ 1 þ nx þ
k
2!
k¼0
ða þ bÞn ¼
n X
n nk k
a b
k
k¼0
¼ an þ nan1 b þ
nðn 1Þ n2 2
a b þ þ bn :
2!
It follows that the sum of the
1
P
geometric series
ar k , for
k¼0
a
jrj < 1,is 1r
.
n
n!
Here,
¼ k!ðnk
Þ! and
k
0! ¼ 1.
Appendix 2: Standard derivatives
and primitives
f(x)
f 0 (x)
Domain
k
0
R
x
xn, n 2 Z {0}
1
nxn1
R
R
x, 2 R
a x, a > 0
x1
ax loge a
(0, 1)
R
xx
xx (1 þ loge x)
(0, 1)
sin x
cos x
R
cos x
tan x
sin x
sec2 x
R
R n þ 12 p : n 2 Z
cosec x
cosec x cot x
sec x
cot x
sec x tan x
cosec2 x
1 ffi
pffiffiffiffiffiffiffi
1x2
1 ffi
pffiffiffiffiffiffiffi
2
R {np : n 2 Z}
R n þ 12 p : n 2 Z
R {np : n 2 Z}
sin1 x
cos1 x
(1, 1)
(1, 1)
tan1 x
1x
1
1þx2
ex
ex
R
loge x
1
x
(0, 1)
sinh x
cosh x
cosh x
sinh x
R
R
tanh x
sinh1 x
sech2 x
1 ffi
pffiffiffiffiffiffiffi
2
R
R
cosh1 x
tanh
1
x
1þx
1 ffi
pffiffiffiffiffiffiffi
x2 1
1
1x2
R
(1, 1)
(1, 1)
359
Appendix 2
360
f (x)
Primitive F(x)
xn, n 2 Z {1}
nþ1
Domain
x
nþ1
xþ1
þ1
ax
loge a
R
sin x
cos x
R
cos x
tan x
sin x
loge (sec x)
R
1 1 2 p; 2 p
ex
ex
loge x
R
(0, 1)
loge x
loge jxj
x loge x x
(1, 0)
(0, 1)
sinh x
cosh x
R
cosh x
tanh x
R
R
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 x 2 , a > 0
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x 2 a2 , a > 0
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a2 þ x 2 , a > 0
ax
e sin bx, a,b 6¼ 0
sinh x
loge (cosh x)
aþx
1
2a loge ax
1
1 x
a tan
a
(
sin1 ax
cos1 ax
(
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
loge x þ x2 a2
1 x
cosh a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
loge x þ a2 þ x2
sinh1 ax
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
2 x2 þ 1 a2 sin1 x
2 xpa
a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2
1
2 a2 1 a2 log x þ
x 2 a2
2 xpx
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 21 2 e pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
2
2
2
2
2 x a þ x þ 2 a loge x þ a þ x
eax
a2 þb2 ða sin bx b cos bxÞ
eax cos bx, a,b 6¼ 0
eax
a2 þb2
x , 2 R {1}
a x, a > 0
1
x
1
x
1
a2 x2 ,
1
a2 þx2 ,
a>0
a>0
1
pffiffiffiffiffiffiffiffiffi
,
a2 x2
a>0
1
pffiffiffiffiffiffiffiffiffi
,
x2 a2
a>0
1
pffiffiffiffiffiffiffiffiffi
,
a2 þx2
a>0
(0, 1)
R
ða cos bx þ b sin bxÞ
(a, a)
R
(a, a)
(a, a)
(a, 1)
(a, 1)
R
R
(a, a)
(a, 1)
R
R
R
Appendix
3: The first 1000 decimal places of
pffiffiffi
2, e and p
More than a million decimal places of these numbers, and various similar commonly
arising numbers, are easily available on the internet.
pffiffiffi
2 ¼ 1.
4142135623
7379907324
9248360558
9505011527
6408891986
6872533965
8355752203
0305440277
0130156185
0146824720
0758474922
8246468157
2636881313
5738108967
4823728997
0249441341
6581726889
7309504880
7846210703
5073721264
8206057147
0955232923
4633180882
3753185701
9031645424
6898723723
7714358548
6572260020
0826301005
7398552561
5040183698
1803268024
7285314781
4197587165
1688724209
8850387534
4121497099
0109559971
0484308714
9640620615
1354374603
7823068492
5288509264
7415565706
8558446652
9485870400
1732204024
6836845072
7442062926
0580360337
8215212822
6980785696
3276415727
9358314132
6059702745
3214508397
2583523950
4084988471
9369186215
8612494977
9677653720
1458398893
3186480342
5091227700
5799364729
9124859052
1077309182
9518488472
7187537694
3501384623
2266592750
3459686201
6260362799
5474575028
6038689997
8057846311
1542183342
2264854470
9443709265
1948972782
2269411275
0607629969
1810044598
8693147101
...
8073176679
0912297024
5592755799
4728517418
5251407989
7759961729
0699004815
1596668713
0428568606
1585880162
9180031138
9064104507
7362728049
4138047565
4215059112
7111168391
e ¼ 2.
7182818284
6277240766
8174135966
8298807531
7614606680
5517027618
7093287091
4637721112
9316368892
6680331825
3012381970
7825098194
5988885193
4841984443
3043699418
7683964243
1718986106
5904523536
3035354759
2904357290
9525101901
8226480016
3860626133
2744374704
5238978442
3009879312
2886939849
6841614039
5581530175
4580727386
6346324496
4914631409
7814059271
8739696552
0287471352
4571382178
0334295260
1573834187
8477411853
1384583000
7230696977
5056953696
7736178215
6465105820
7019837679
6717361332
6738589422
8487560233
3431738143
4563549061
1267154688
6624977572
5251664274
5956307381
9307021540
7423454424
7520449338
2093101416
7707854499
4249992295
9392398294
3206832823
0698112509
8792284998
6248270419
6405462531
3031072085
9570350354
4709369995
2746639193
3232862794
8914993488
3710753907
2656029760
9283681902
6996794686
7635148220
8879332036
7646480429
9618188159
9208680582
7862320900
5209618369
1038375051
...
9574966967
2003059921
3490763233
4167509244
7744992069
6737113200
5515108657
4454905987
8269895193
2509443117
5311802328
3041690351
5749279610
2160990235
0888707016
0115747704
Numerical Analysts enjoy
adding more digits to such
decimals, using ever-more
sophisticated computing
techniques!
For most practical purposes
pffiffiffi
2 ’ 1:414:
For most practical purposes
e ’ 2:718:
A useful fact sometimes is
that e3 ’ 20.
361
Appendix 3
362
p ¼ 3.
1415926535
5923078164
0938446095
6446229489
2712019091
7245870066
0113305305
0921861173
8912279381
1907021798
0005681271
2249534301
8640344181
0597317328
2619311881
5982534904
1712268066
8979323846
0628620899
5058223172
5493038196
4564856692
0631558817
4882046652
8193261179
8301194912
6094370277
4526356082
4654958537
5981362977
1609631859
7101000313
2875546873
1300192787
2643383279
8628034825
5359408128
4428810975
3460348610
4881520920
1384146951
3105118548
9833673362
0539217176
7785771342
1050792279
4771309960
5024459455
7838752886
1159562863
6611195909
5028841971
3421170679
4811174502
6659334461
4543266482
9628292540
9415116094
0744623799
4406566430
2931767523
7577896091
6892589235
5187072113
3469083026
5875332083
8823537875
2164201989
6939937510
8214808651
8410270193
2847564823
1339360726
9171536436
3305727036
6274956735
8602139494
8467481846
7363717872
4201995611
4999999837
4252230825
8142061717
9375195778
...
5820974944
3282306647
8521105559
3786783165
0249141273
7892590360
5759591953
1885752724
6395224737
7669405132
1468440901
2129021960
2978049951
3344685035
7669147303
1857780532
For most practical purposes
p ’ 3:1416:
A useful fact sometimes is
that p2 ’ 10.
Appendix 4: Solutions to the problems
Chapter 1
Section 1.1
45
45
17
1. 1 < 17
20 < 53 < 0 < 53 < 20 < 1:
2. Let a and b be two distinct rational numbers, where a < b. Let c ¼ 12ða þ bÞ; then
c is rational, and
1
c a ¼ ðb aÞ > 0; and
2
1
b c ¼ ðb aÞ > 0;
2
so that a < c < b.
3.
1
7
2
¼ 0:142857142857 . . .; 13
¼ 0:153846153846 . . ..
4. (a) Let x ¼ 0:231.
Multiplying both sides by 103, we obtain
1000x ¼ 231:231 ¼ 231 þ x:
Hence
999x ¼ 231 ) x ¼
231
77
¼
:
999 333
(b) Let x ¼ 0:81.
Multiplying both sides by 102, we obtain
100x ¼ 81:81 ¼ 81 þ x:
Hence
99x ¼ 81 ) x ¼
81
9
¼ :
99 11
Thus
2:281 ¼ 2 þ
5.
17
20
2
9
251
þ
¼
:
10 110 110
45
17
¼ 0:85 and 45
53 ¼ 0:84 . . ., so that 53 < 20.
6. Suppose that there exists a rational number x such that x3 ¼ 2. Then we can write
x ¼ pq. By cancelling, if necessary, we may assume that p and q have no common
factor. The equation x3 ¼ 2 now becomes
p3 ¼ 2q3 :
Now, the cube of an odd number is odd, because
ð2k þ 1Þ3 ¼ 8k3 þ 12k2 þ 6k þ 1
¼ 2 4k3 þ 6k2 þ 3k þ 1;
and so p must be even. Hence we can write p ¼ 2r, say. Our equation now becomes
363
Appendix 4
364
ð2r Þ3 ¼ 2q3 ;
so that
q3 ¼ 4r 3 :
Hence q is also even, so that p and q do have a common factor 2, which contradicts
our earlier statement that p and q have no common factors.
Thus, no such number x exists.
7. We may take, for example, x ¼ 0.34 and y ¼ 0.34001000100001. . ..
8. Let a and b have decimal representations
a ¼ a0 a1 a2 a3 . . .
and
Here a0, b0 are non-negative
integers, and a1, b1, a2, b2, . . .
are digits.
b ¼ b0 b1 b2 b3 . . .;
where we arrange that a does not end in recurring 9s, whereas b does not terminate
(this latter can be arranged by replacing a terminating representation by an equivalent representation that ends in recurring 9s).
Since a < b, there must be some integer n such that
a0 ¼ b0 ; a1 ¼ b1 ; . . . ; an1 ¼ bn1 ; but an < bn :
Then x ¼ a0 a1a2a3 . . . an 1bn is rational, and a < x < b as required.
Next, let c ¼ 12 ðx þ bÞ, so that x < c < b. By repeating the same procedure, this
time to the interval between the numbers c and b, we can find a rational number y
with c < y < b.
We have thus constructed two distinct rational numbers between a and b, as
required.
Section 1.2
1. Rule 1
Rule 2
Rule 3
For any a, b 2 R,
For any a, b, c 2 R,
a b , b a 0.
a b , a þ c b þ c.
For any a, b 2 R and any c > 0,
For any a, b 2 R and any c < 0,
a b , ac bc;
a b , ac bc.
Remark
Note that the following results are NOT true in general:
For any a, b 2 R and any c 0,
For any a, b 2 R and any c 0,
a b , ac bc;
a b , ac bc.
For, if c ¼ 0, then we can make no assertion as to whether a b or a b from
the information that ac bc or ac bc. To see this, take in turn a ¼ 2, b ¼ 3 and
a ¼ 2, b ¼ 1.
Rule 4 (Reciprocal Rule) For any positive a, b 2 R, a b , 1a 1b.
Rule 5 (Power Rule) For any non-negative a, b 2 R, and any p > 0, a b , a p b p.
2. (a) x þ 3 > 5.
(b) 2 x < 0.
(c) 5x þ 2 > 12.
(d)
1
ð5x þ 2Þ
> 1
12 .
3. (a) Rearranging the inequality, we obtain
4x x2 7
4x x2 7
3
,
30
x2 1
x2 1
x2 x þ 1
0
,
x2 1
4x 4x2 4
0
x2 1
2
x 12 þ 34
,
0:
x2 1
,
Note that x does not end in
recurring 9s, from the way in
which it was constructed.
Solutions to the problems
365
2
Since x 12 þ 34 > 0, for all x, the inequality holds if and only if x2 1 < 0.
Hence the solution set is
x:
4x x2 7
3
¼ ð1; 1Þ:
x2 1
(b) Rearranging the inequality, we obtain
2x2 ðx þ 1Þ2 , 2x2 x2 þ 2x þ 1
, x2 2x 1 0
, ðx 1Þ2 2:
Hence the solution set is
n
o n
pffiffiffio n
pffiffiffio
x : 2x2 ðx þ 1Þ2 ¼ x : x 1 2 [ x : x 1 2
pffiffiffii h
pffiffiffi ¼ 1; 1 2 [ 1 þ 2; 1 :
4. We can obtain an equivalent inequality by squaring, provided that both sides are nonnegative. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Now, 2x2 2 is defined when 2x2 2 0; that is, for x2 1. So, 2x2 2 is
defined and non-negative if x lies in (1, 1] [ [1, 1).
Hence, for x 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 2 > x , 2x2 2 > x2
, x2 > 2:
So the part of the solution set in [1, 1) is
Suppose, next, that x 1. Then
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2x2 2 0 > x;
pffiffiffi
2; 1 .
so that the whole of (1, 1] lies in the solution set.
Hence the complete solution set is
n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
o
pffiffiffi x : 2x2 2 > x ¼ ð1; 1 [
2; 1 :
5. (a) 2x2 13 < 5 , 5 < 2x2 13 < 5 ðby Rule 6Þ
, 8 < 2x2 < 18
, 4 < x2 < 9
, 2 < j xj < 3:
Hence the solution set is
2
x : 2x 13 < 5 ¼ ð3; 2Þ [ ð2; 3Þ:
(b) jx 1j 2jx þ 1j , ðx 1Þ2 4ðx þ 1Þ2
, x2 2x þ 1 4x2 þ 8x þ 4
, 0 3x2 þ 10x þ 3
, 0 ð3x þ 1Þðx þ 3Þ:
Hence the solution set is
1
fx : jx 1j 2jx þ 1jg ¼ ð1; 3 [ ; 1 :
3
Appendix 4
366
Section 1.3
1. (a) Suppose that jaj 12.
The Triangle Inequality gives
Hence
ja þ 1j jaj þ 1
1
þ 1 ðsince jaj 12Þ
2
3
¼ :
2
3
ja þ 1j :
2
(b) Suppose that jbj < 12.
The ‘reverse form’ of the Triangle Inequality gives
3
b 1 jjb3 j 1j
¼ jbj3 1
1 jbj3 :
Now
jbj <
1
1
7
) jbj3 < ) 1 jbj3 > ;
2
8
8
so, from the previous chain of inequalities
3
b 1 > 7:
8
2. Rearranging the inequality, we obtain
3n
< 1 , 3n < n2 þ 2
n2 þ 2
, 0 < n2 3n þ 2
, 0 < ðn 1Þðn 2Þ;
and this final inequality certainly holds for n > 2.
3. The result is true for all those a and b for which a þ b 0.
We now consider those a and b for which a þ b > 0. Rearranging the given
inequality, we obtain
a þ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ða þ bÞ2
pffiffiffi a2 þ b2 ,
a2 þ b2
2
2
, a2 þ 2ab þ b2 2a2 þ 2b2
, 0 a2 2ab þ b2
, 0 ða bÞ2 ;
and this final inequality certainly holds.
This completes the proof.
4. (a) Using the rules for rearranging inequalities, we obtain
1
2
1 2
1
aþ
<a, <a a
2
a
2 a
2
1 1
, < a
a 2
, 2 < a2 ðsince a > 0Þ:
Solutions to the problems
367
Since the final inequality is true, the first inequality must also be true. Hence
1
2
aþ
< a:
2
a
(b) In Example 3 and the subsequent remark, we saw that, if a 6¼ b, then
aþb 2
ab <
:
2
()
Now let b ¼ 2a. Then a 6¼ b, since a > 2a (as a2 > 2); so, by (), it follows that
2
1
2 2
aþ
a <
;
a
2
a
which gives the required inequality.
Alternatively, use a direct argument after squaring the expression 12 a þ 2a :
5. Since c, d 0, we can choose non-negative numbers a and b with c ¼ a2 and d ¼ b2.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Substituting into the result a2 þ b2 a þ b, for a, b 0, of Example 4, we obtain
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi
c þ d c þ d:
6. In Problem 5 we saw that
pffiffiffiffiffiffiffiffiffiffiffi pffiffiffi pffiffiffi
cþd cþ d;
for numbers c; d 0:
()
Following the good advice in the margin note before the Problem, we use this!
For a, b, c 0, we deduce from () that
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a þ b þ c ¼ a þ ðb þ c Þ
pffiffiffi pffiffiffiffiffiffiffiffiffiffiffi
aþ bþc
ðby an application of ðÞÞ
pffiffiffi pffiffiffi pffiffiffi
aþ bþ c
ðby a further application of ðÞÞ:
7. If we substitute x ¼ 1n in the Binomial Theorem for (1 þ x)n, when n 2 we get
n
1 n
1
nðn 1Þ 1 2
1
¼1þn
þ þ
1þ
þ
n
n
2!
n
n
n1
1þ1þ
2n
1 1
5 1
¼2þ ¼ :
2 2n 2 2n
When n ¼ 1, the inequality also holds. This completes the proof.
8. Let P(n) be the statement
PðnÞ : 4n > n4 :
STEP 1 First we show that P(5) is true: 45 54.
Since 45 ¼ 1024 and 54 ¼ 625, P(5) is certainly true.
STEP 2 We now assume that P(k) holds for some k 5, and deduce that P(k þ 1) is
then true.
So, we are assuming that 4k > k4. Multiplying this inequality by 4 we get
This assumption is just P(k).
4kþ1 > 4k4 ;
so it is therefore sufficient for our purposes to prove that 4k4 (k þ 1)4.
Now
1 4
4k4 ðk þ 1Þ4 , 4 1 þ
:
()
k
As k increases, the expression 1 þ 1k decreases. Since k 5, 1 þ 1k 1 þ 15 ¼ 65 ; it follows that
Since P(k þ 1) is:
4kþ 1 > (k þ 1)4.
Appendix 4
368
4
1 4
6
1þ
k
5
¼
1296
¼ 2:0736 < 4:
625
Thus the inequality () certainly holds for k 5, and so it follows that
4kþ 1 (k þ 1)4 also holds for k 5.
In other words: P(k) true for some k 5 ) P(k þ 1) true.
It follows, by the Principle of Mathematical Induction, that 4n > n4,
for n 5.
1
9. Substituting x ¼ ð2nÞ
into Bernouilli’s Inequality (1 þ x)n 1 þ nx (which we may
1
do since ð2nÞ
1 for all natural numbers n), we obtain
1 n
1
1n
1
2n
2n
1
¼ :
2
If we take the nth root of both sides of this final inequality (which is permissible, by
the Power Rule), we find that
1
1
1
1;
2n 2n
so that
1
2n 1
2n
¼
1
2n 1
1 2n
1
:
¼ 1þ
2n 1
pffiffiffiffiffi
ffi
10. If we apply the Cauchy–Schwarz Inequality, with ak in place of ak and p1ffiffiffi
ak in
place of bk, we have
1
1
1
ða1 þ a2 þ þ an Þ
þ þ þ
a1 a2
an
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
1
1
1 2
a1 pffiffiffiffiffi þ a2 pffiffiffiffiffi þ þ an pffiffiffiffiffi
a1
a2
an
¼ ð1 þ 1 þ þ 1Þ2
ðwith n terms in the bracketÞ
2
¼n :
11. Applying Theorem 3 to the n þ 1 positive numbers 1; 1 þ 1n ; 1 þ 1n ; . . . ; 1 þ 1n,
we obtain
1
1 n nþ1
1
1
1þ
nþ1þn
n
nþ1
n
¼
nþ2
nþ1
¼ 1þ
1
;
nþ1
taking the (n þ 1)th power of this last inequality, by the Power Rule, we deduce
that
nþ1
1 n
1
1þ
:
1þ
n
nþ1
Solutions to the problems
369
Section 1.4
1. (a)
E1 is bounded above. For example, M ¼ 1 is an upper bound of E1, since
x 1; for all x 2 E1 :
Also, max E1 ¼ 1, since 1 2 E1.
(b)
E2 is bounded above. For example, M ¼ 1 is an upper bound of E2, since
1
1 1; for n ¼ 1; 2; . . .:
n
However, E2 has no maximum element. If x 2 E2, then x ¼ 1 1n for some
1
positive integer n; and so there is another element of E2, such as 1 nþ1
, with
1
1
1
1
since >
:
1 <1
n
nþ1
n nþ1
Hence x is not a maximum element of E2.
(c)
The set E3 is not bounded above, and so it cannot have a maximum element. For
each number M, there is a positive integer n such that n2 > M (for instance, take
n > M, which implies that n2 > n > M).
Hence M cannot be an upper bound of E3.
2. (a) The set E1 ¼ (1, 1] is not bounded below, and so it cannot have a minimum
element. For each number m, there is a (negative) number x such that x < m.
Since x 2 E1, m cannot be a lower bound of E1.
(b) E2 is bounded below by 0, since
1
1
0;
n
for n ¼ 1; 2 . . .:
Also, 0 2 E2, and so min E2 ¼ 0.
(c) E3 is bounded below by 1, since
n2 1;
for n ¼ 1; 2 . . .:
Also, 1 2 E3, and so min E3 ¼ 1.
3. ƒ is increasing on theinterval [3, 2). Since 3 x < 2 so that x2 2 (4, 9], we
have f ð xÞ ¼ x12 2 19 , 14 . Hence ƒ is bounded above and bounded below.
Next, since f ð3Þ ¼ 19 and 19 is a lower bound for ƒ on the interval [3, 2), it
follows that ƒ has a minimum value of 19 on this interval.
Finally, 14 is an upper bound for ƒ on the interval [3, 2) but there is no point x
in [3, 2) for which f ðxÞ ¼ 14. So 14 cannot be a maximum of f on the interval.
However, if y is any number in 19 ; 14 , there is a number x > p1ffiffiy in
p1ffiffiy ; 2 ½3; 2Þ such that f ð xÞ ¼ x12 > y, so that no number in 19 , 14 will
serve as a maximum of f on the interval. It follows that f has no maximum value
on [3, 2).
Appendix 4
370
4. (a) The set E1 ¼ (1, 1] has a maximum element 1, and so
sup E1 ¼ max E1 ¼ 1:
(b) We know that E2 ¼ f1 1n : n ¼ 1, 2, . . .g is bounded above by 1, since
1
1
1;
n
for n ¼ 1; 2 . . .:
To show that M ¼ 1 is the least upper bound of E2, we prove that, if M0 < 1, then
there is an element 1 1n of E2 such that
1
1 > M0:
n
However, since M0 < 1, we have
1
1
1
> M0 , 1 M0 >
n
n
1
,n>
1 M0
ðsince 1 M 0 > 0Þ:
1
1
0
Choosing a positive integer n so large that n > 1M
0 , we obtain 1 n > M , as
required. Hence 1 is the least upper bound of E2.
(c) The set E3 ¼ {n2 : n ¼ 1, 2, . . .} is not bounded above, and so it cannot have a
least upper bound.
5. (a) The set E1 ¼ (1, 5] is bounded below by 1, since
1 x;
for all x 2 E1 :
To show that m ¼ 1 is the greatest lower bound of E1, we prove that, if m0 > 1,
then there is an element x in E1 which is less than m0 . Since m0 > 1, there is a
number x of the form x ¼ 1.00 . . . 01 such that
1 < x < m0 ;
and clearly x 2 E1.
Hence 1 is the greatest lower bound of E1.
(b) The set E2 ¼ fn12 : n ¼ 1, 2, . . .g is bounded below by 0, since
0<
1
;
n2
for n ¼ 1; 2; . . .:
To show that m ¼ 0 is the greatest lower bound of E2, we prove that, if m0 > 0,
then there is an element n12 in E2 such that n12 < m0 . Since m0 > 0, we have
1
1
< m0 , n2 > 0
n2
m
1
, n > pffiffiffiffiffi :
m0
ffi, we obtain
Choosing a positive integer n so large that n > p1ffiffiffi
m0
1
n2
< m0 , as
required. Hence 0 is the greatest lower bound of E2.
6. ƒ is decreasing on the interval [1, 4). Since 1 x < 4 so that x2 2 [1, 16), we have
1 1
, 1 . Hence ƒ is bounded above by 1 and bounded below by 16
.
f ð xÞ ¼ x12 2 16
Next, since f(1) ¼ 1 and 1 is an upper bound for ƒ on the interval [1, 4), it follows
that ƒ has least upper bound 1 on [1, 4).
1
Finally, 16
is a lower bound for ƒ on the interval [1, 4) but there is no point x in
1
.
[1, 4) for which f ðxÞ ¼ 16
1 However, if y is any number in 16
, 1 , there is a number x < p1ffiffiy in [1, 4) such that
1
, 1 will serve as a lower bound of ƒ on [1, 4).
f ð xÞ ¼ x12 < y, so that no number in 16
1
as its greatest lower bound on [1, 4).
It follows that ƒ has 16
Solutions to the problems
371
Chapter 2
Section 2.1
1. (a)
(i) 4, 7, 10, 13, 16;
(ii)
1 1 1 1
1
3 , 9 , 27 , 81 , 243 ;
(iii) 1, 2, 3, 4, 5.
(b) (i) a1 ¼ 1, a2 ¼ 2, a3 ¼ 6, a4 ¼ 24, a5 ¼ 120;
(ii) a1 ¼ 2, a2 ¼ 2.25, a3 ¼ 2.37, a4 ¼ 2.44, a5 ¼ 2.49.
2. (a)
(b)
(c)
(d)
3. (a) n! is monotonic, because
an ¼ n!
and
anþ1 ¼ ðn þ 1Þ!;
Appendix 4
372
so that
anþ1 an ¼ ðn þ 1Þ! n!
¼ n n! > 0; for n ¼ 1; 2; . . .:
Thus {n!} is increasing.
Alternatively, an > 0, for all n, and
anþ1 ðn þ 1Þ!
¼ n þ 1 1;
¼
n!
an
for n ¼ 1; 2; . . .;
so that {n!} is increasing.
(b) {2n} is monotonic, because
an ¼ 2n and anþ1 ¼ 2ðnþ1Þ ;
so that
1
1
2nþ1 2n
1 1
¼ n
1 < 0;
2 2
anþ1 an ¼
for n ¼ 1; 2; . . .:
Thus {2n} is decreasing.
Alternatively, an > 0, for all n, and
anþ1
2n
1
¼ nþ1 ¼ < 1;
2
an
2
for n ¼ 1; 2; . . .;
so that {2n} is decreasing.
(c) n þ 1n is monotonic, because
an ¼ n þ
1
1
and anþ1 ¼ n þ 1 þ
;
n
nþ1
so that
1
1
nþ
nþ1
n
nðn þ 1Þ 1
¼
> 0; for n ¼ 1; 2; . . .:
nðn þ 1Þ
anþ1 an ¼
nþ1þ
Thus n þ 1n is increasing.
4. (a)
(b)
TRUE:
2n > 1000, for n > 9, since {2n} is increasing and 210 ¼ 1024.
FALSE:
All the terms a1, a3, a5, . . . are negative.
1
< 0:025, for all n > 0:025
¼ 40:
anþ1
1 nþ1 4
(d) TRUE: an > 0 for all n, and an ¼ 4 n :
(c)
1
TRUE: n
Now
1 nþ1 4
nþ1 4
1,
4
4
n
n
1 pffiffiffi
, 21
n
So
anþ1
1;
an
for n > 2:
anþ1 an ;
for n > 2;
Hence
so that
n4 4n
is eventually decreasing.
1
1
44
n
1
, n pffiffiffi
’ 2:414:
21
,1þ
Solutions to the problems
373
Section 2.2
1. (a)
1
1
n < 100 , n > 100, by the Reciprocal Rule (for positive numbers). Hence we
may take X ¼ 100.
Any value for X greater than
100 will also be valid.
1
n
Any value for X greater than
333 will also be valid.
3
< 1000
, n > 1000
3 ¼ 333:3333 . . ., by the Reciprocal Rule (for positive numbers). Hence we may take X ¼ 333.
2. (a)
ð1Þn 1
1
1
n2 < 100 , n2 < 100
(b)
, n2 > 100
, n > 10:
Hence we may take X ¼ 10.
ð1Þn 3
1
3
n2 < 1000 , n2 < 1000
1000
, n2 >
3
rffiffiffiffiffiffiffiffiffiffi
1000
’ 18:25:
,n>
3
Hence we may take X ¼ 18.
n
o
3. (a) The sequence 2n 1 1 is a null sequence.
Any value for X greater than 10
will also be valid.
(b)
Any value for X greater than 18
will also be valid.
To prove this, we want to show that:
for each positive number ", there is a number X such that
1 2n 1 < "; for all n > X:
()
We know that
1 1
2n 1 < " , 2n 1 < " ðsince 2n 1 > 0Þ
1
, 2n 1 >
" 1
, n > 12 1 þ ;
"
n
o
1
1
it follows that () holds if we take X ¼ 2 1 þ 1" . Hence 2n1
is null.
n no
Þ
is not a null sequence.
(b) The sequence ð1
10
To prove this, we must find a positive value of " such that the sequence does
not eventually lie in the horizontal strip on the sequence diagram from " up
1
to ". We can take " ¼ 20
, as the following sequence diagram shows:
Appendix 4
374
There is NO value of X such that the following statement holds
ð1Þn 1
10 < 20 ; for all n > X:
n no
Þ
is a null sequence.
(c) The sequence ðn1
4 þ1
To prove this, we want to show that:
for each positive number ", there is a number X such that
ð1Þn n4 þ 1 < "; for all n > X:
()
We know that
ð1Þn 1
n4 þ 1 < " , n4 þ 1 < "
1
, n4 þ 1 >
"
1
4
, n > 1:
"
1
Now, if " 1, then " 1 0, so that
1
n4 > 1; for n ¼ 1; 2; . . .;
"
hence () holds with X ¼ 1.
On the other hand, if 0 < " < 1, then 1" 1 > 0, so that
14
1
1
4
n > 1,n>
1 ;
"
"
1
14
hence () holds with X ¼ " 1 .
n no
Þ
Thus () holds in either case. Hence nð1
is null.
4 þ1
n
o
n
o
1
1
4. (a) We know that 2n1
is null, and so ð2n1
is also null, by the Power Rule.
3
n
o
nÞ o
n
o
1
1
5
(b) The sequences n and 2n1
are null, so p56ffiffin and ð2n1
are also null, by
Þ7
the Power Rule and Multiple
Rule. o
n
5
Hence the sequence p56ffiffin þ ð2n1
is null, by the Sum Rule.
Þ7
o
n
1
1
1
are null, so n14 and
are also null, by
(c) The sequences n and 2n1
1
ð2n1Þ3
the PowerRule.
1
Hence
is also null, by the Product Rule and Multiple Rule.
1
4
3n ð2n1Þ3
5. Here, using the Hint, we have
n
1
1
¼n n
n
2
2
1
1
n 2 ¼ :
n
n
1
n Thus, the sequence n dominates the sequence n 12 , and is itself null. It
n is null.
follows, from the Squeeze Rule, that the sequence n 12
n
o
1
1
6. (a) We guess that n2 þn is dominated by n .
To check this, we have to show that
1
1
; for n ¼ 1; 2; . . .;
n2 þ n n
this certainly holds, because
n2 þ n n; for n ¼ 1; 2; . . .:
n
o
Since 1n is null, we deduce that n21þn is null, by the Squeeze Rule.
Solutions to the problems
(b) We guess that
n
ð1Þ
n!
o
n
375
is dominated by
1
n .
To check this, we have to show that
ð1Þn 1
n! n ; for n ¼ 1; 2; . . . ;
this certainly holds, because
ð1Þn 1
n! ¼ n! ; for n ¼ 1; 2; . . .;
and
n! n; for n ¼ 1; 2; . . .:
Þn is null, by the Squeeze Rule.
Since n is null, we deduce that ð1
n!
n
o
2
n
is dominated by n12 .
(c) We guess that nsin
2 þ 2n
1
To check this, we have to show that
sin n2 1
n2 þ 2n n2 ; for n ¼ 1; 2; . . .;
this certainly holds, because
sin n2 1; for n ¼ 1; 2; . . .;
and
Since
1
n2
n2 þ 2n n2 ; for n ¼ 1; 2; . . .:
n
o
n2
is null, we deduce that nsin
is null, by the Squeeze Rule.
2 þ 2n
Section 2.3
1. (a)
The sequence appears to converge to 1.
(b) Since bn ¼ an 1 ¼ n þn 1 1 ¼ 1n, it follows that fbn g ¼
2. (a)
1
n
is a null sequence.
Appendix 4
376
2
1
2
Here an 1 ¼ nn2 þ 1 1 ¼ n2 þ 1. We know that
n
o
2
n2 þ 1
is a null sequence, so it
follows that {an 1} is also a null sequence.
Hence {an} converges to 1.
(b)
n
3
n
Þ
Þ
Here an 12 ¼ n þ2nð1
12 ¼ ð1
3
2n3 . We know that
follows that an 12 is also a null sequence.
Hence {an} converges to 12.
n
ð1Þn
2n3
3. (a) The dominant term is n3, so we write an as
n3 þ 2n2 þ 3
2n3 þ 1
1 þ 2n þ n33
:
¼
2 þ n13
Since 1n and n13 are basic null sequences
an ¼
lim an ¼
n!1
1þ0þ0 1
¼ ;
2þ0
2
by the Combination Rules.
(b) The dominant term is 3n, so we write an as
an ¼
¼
n2 þ 2n
3n þ n3
2n
n2
3n þ 3
:
3
1 þ n3n
n 2 o n 3o
n
and n3n are basic null sequences
Since n3n , 23
lim an ¼
n!1
0þ0
¼ 0;
1þ0
by the Combination Rules.
(c) The dominant term is n!, so we write an as
an ¼
n! þ ð1Þn
2n þ 3n!
n
1 þ ð1Þ
¼ 2n n! :
n! þ 3
n no
n
Þ
and 2n! are basic null sequences
Since ð1
n!
lim an ¼
n!1
1þ0 1
¼ ;
0þ3 3
by the Combination Rules.
o
is a null sequence, so it
Solutions to the problems
377
4. (a) We know that
rffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffi!n
2
2
,n 1þ
n 1þ
:
n1
n1
qffiffiffiffiffiffi
2
, we obtain
Using the hint, with x ¼ n1
rffiffiffiffiffiffiffiffiffiffiffi!n
rffiffiffiffiffiffiffiffiffiffiffi!2
2
nðn 1Þ
2
1þ
n1
2!
n1
nðn 1Þ
2
¼ n;
¼
2
n1
as required.
1
n
1
(b) Since n 1, we have nn 1. Combining this inequality with that in part (a), we
obtain
rffiffiffiffiffiffiffiffiffiffiffi
2
1
n
; for n 2:
1n 1þ
n
1
nqffiffiffiffiffiffiffiffio1
2
Now,
n 1 n¼2 is a null sequence, by the Power Rule, so that
rffiffiffiffiffiffiffiffiffiffiffi!
2
¼ 1:
lim 1 þ
n!1
n1
1
Hence, by the Squeeze Rule, lim nn ¼ 1:
n!1
Section 2.4
1. (a) {1 þ ( 1)n} is bounded, since the terms 1 þ (1)n take only the values 0 and 2.
Hence
j1 þ ð1Þn j 2; for n ¼ 1; 2; . . .:
n
(b) {( 1) n} is unbounded. Given any number K, there is a positive integer n such
that
jð1Þn nj > K;
for instance, n ¼ jK j þ 1.
2n þ 1
is bounded, since 2n nþ 1 ¼ 2 þ 1n so that
n
2n þ 1
1
n ¼ 2 þ n 3; for n ¼ 1; 2; . . .:
n is bounded, since
(d)
1 1n
1
0 1 1; for n ¼ 1; 2; . . .;
n
so that
n n
1 1 ¼ 1 1 1; for n ¼ 1; 2; . . .:
n
n
pffiffiffi
2. (a) fn ng o
is unbounded, and hence divergent, by Corollary 1.
2
n
(b) nn2 þ
þ 1 is convergent (with limit 1), and hence bounded, by Theorem 1. In fact
n2 þ n 1 þ 1n
1
¼
1 þ 2; for n ¼ 1; 2; . . .:
1
2
n þ 1 1 þ n2
n
(c)
(c) {( 1)nn2} is unbounded, and hence divergent, by Corollary 1.
n
(d) The terms of the sequence nð1Þ are
1
1
1; 2; ; 4; ; 6; . . .;
3
5
so the sequence is unbounded: given any number K, there is an even positive
integer 2n such that 2n > K. Hence the sequence is divergent, by Corollary 1.
Appendix 4
378
n
n
3. (a) Each term of 2n is positive, and 2nn is a basic null sequence. Hence 2n ! 1,
by the Reciprocal Rule.
(b) First, note that
n100
2n n100 ¼ 2n 1 n ; for n ¼ 1; 2; . . .;
2
n 100 o
100
and that n2n is a basic null sequence. It follows that n2n is eventually less
than 1, so that 2n n100 is eventually positive.
Also
1
1
0
n
¼ 0;
¼ lim 2 n100 ¼
lim
n!1 2n n100
n!1 1 n
1
0
2
by the Combination Rules.
Hence 2n n100 ! 1, by the Reciprocal Rule.
n
(c) We know that 2n ! 1, by part (a), and that
2n
2n
þ 5n100 ; for n ¼ 1; 2; . . .:
n
n
n
Hence 2n þ 5n100 ! 1, by the Squeeze Rule.
Remark
You could have used the Reciprocal Rule or the Sum and Multiple Rules.
n n 2o
(d) Each term of the sequence 2n10þþnn is positive, and
n
1
2 þ n2
n10 þ n
¼ lim n
lim 10
n!1 n þ n
n!1 2 þ n2
¼ lim
n10
2n
þ 2nn
2
1 þ n2n
0þ0
¼ 0;
¼
1þ0
n!1
by the Combination Rules.
n
2
Hence 2n10þþnn ! 1, by the Reciprocal Rule.
4. (a)
(i) a2 ¼ 4, a4 ¼ 16, a6 ¼ 36, a8 ¼ 64, a10 ¼ 100;
(ii) a3 ¼ 9, a7 ¼ 49, a11 ¼ 121, a15 ¼ 225, a19 ¼ 361;
(iii) a1 ¼ 1, a4 ¼ 16, a9 ¼ 81, a16 ¼ 256, a25 ¼ 625.
(b) a1 ¼ 1; a3 ¼ 13; a5 ¼ 15;
a2 ¼ 2, a4 ¼ 4, a6 ¼ 6.
5. (a) If an ¼ ð1Þn þ 1n, for n ¼ 1, 2, . . ., then
1
1
a2k ¼ 1 þ
and a2k1 ¼ 1 þ
;
2k
2k 1
so that
lim a2k ¼ 1; whereas lim a2k1 ¼ 1:
k!1
k!1
Hence {an} is divergent, by the First Subsequence Rule.
(b) If an ¼ 13n 13n , for n ¼ 1, 2, . . ., then
1
a3k ¼ 0 and a3kþ1 ¼ ; for k ¼ 1; 2; . . .;
3
so that
lim a3k ¼ 0; whereas lim a3kþ1 ¼ 13:
k!1
k!1
Hence {an} is divergent, by the First Subsequence Rule.
Solutions to the problems
379
(c) If an ¼ n sin 12np , for n ¼ 1, 2, . . ., then
Now
a1 ¼ 1; a2 ¼ 0; a3 ¼ 3; a4 ¼ 0; a5 ¼ 5; a6 ¼ 0; . . .:
1
a4kþ1 ¼ ð4k þ 1Þ sin 2kp þ p
2
¼ 4k þ 1;
for k ¼ 1; 2; . . .;
so that a4kþ1 ! 1.
Hence {an} is divergent, by the Second Subsequence Rule.
Section 2.5
1. Suppose that the sequence {an} is decreasing and bounded below, so that it is
necessarily convergent. We show that the limit of {an} is the number
m ¼ inf {an : n ¼ 1, 2, . . .}.
To see this, let ‘ denote lim an . By the Limit Inequality Rule, since an m we know
n!1
that ‘ m. Now assume that in fact ‘ > m.
Since ‘ > m, it follows that ‘ > 12 ð‘ þ mÞ > m. It follows, from the definition of
greatest lower bound, that there then exists some integer X such that aX < 12 ð‘ þ mÞ;
and so, since {an} is decreasing, that
1
an < ð‘ þ mÞ; for all n > X:
2
We deduce from the Limit Inequality Rule that ‘ 12 ð‘ þ mÞ. We may rearrange this
inequality as 2‘ ‘ þ m, or ‘ m.
This contradicts our assumption that ‘ > m. It follows that, in fact, ‘ ¼ m.
2. We use the ideas in the discussion prior to the problem.
We shall show that this
assumption leads to a
contradiction.
By letting n ! 1.
For we cannot have ‘ < m and
‘ m!
(a) Let a1 > 3. Then, from equation (5) in the discussion before the statement of the
problem
1
a2 a1 ¼ ða1 1Þða1 3Þ
4
> 0;
so that a2 > a1. This suggests that, in general, it might be that anþ1 > an in the
case that a1 > 3; we check this, using Mathematical Induction.
Let P(n) be the statement: If a1 > 3, then an þ 1 > an.
We have just seen that the statement P(1) is true.
Now assume that the statement P(k) is true; that is, that akþ1 > ak for some
k 1. It follows, from equation (5) in the discussion before the statement of the
problem, that
1
akþ1 ak ¼ ðak 1Þðak 3Þ
4
> 0;
so that ak þ 1 > ak; in other words, the statement P(k þ 1) is true.
This proves that P(n) holds for all n 1.
Thus, if a1 > 3, the sequence {an} is increasing.
It follows, from the Monotone Convergence Theorem, that either {an} converges to some (finite) limit ‘ or tends to infinity as n ! 1.
But the only possible limits of the sequence are 1 and 3, so that since a1 > 3
and the sequence is increasing, clearly {an} cannot tend to a limit. It follows that
an ! 1 as n ! 1.
(b) Let 0 a1 < 1. Then, from equation (5) in the discussion before the statement of
the problem,
We need a discussion of this
type in order to identify the
result that we wish to verify
using Mathematical Induction.
Appendix 4
380
1
a2 a1 ¼ ða1 1Þða1 3Þ
4
> 0;
so that a2 > a1. This suggests that, in general, it might be that anþ1 > an in the
case that 0 a1 < 1; we check this, using Mathematical Induction.
Let P(n) be the statement: If 0 a1 < 1, then anþ1 > an.
We have just seen that the statement P(1) is true.
Now assume that the statement P(k) is true; that is, that akþ1 > ak for some
k 1. It follows, from equation (5) in the discussion before the statement of the
problem, that
1
akþ1 ak ¼ ðak 3Þðak 1Þ
4
> 0;
so that ak þ 1 > ak; in other words, the statement P(k þ 1) is true.
This proves that P(n) holds for all n 1.
Thus, if 0 a1 < 1, the sequence {an} is increasing.
It follows, from the Monotone Convergence Theorem, that either {an} converges to some (finite) limit ‘ or tends to infinity as n ! 1.
Now, we can use equation (3) before the problem, with n ¼ 1, to see that the
assumption a1 < 1 implies that
a2 1 ¼
1 2
a 1 < 0;
4 1
so that a2 < 1. We can then prove, by Mathematical Induction, using an argument similar to that in part (a), that, if a1 < 1, then an < 1 for all n 1.
But the only possible limits of the sequence are 1 and 3, so that since an < 1 for
all n 1, clearly {an} cannot tend to 3 or to infinity. It follows that an ! 1 as
n ! 1.
(c) Let a1 < 0. Since anþ1 ¼ 14 a2n þ 3 , it is clear that the behaviour of the sequence
as n ! 1 depends only on the magnitude of a1 and not on its sign.
It follows from this observation that:
if 1 < a1 < 0,
if a1 ¼ 1,
if 3 < a1 < 1,
if a1 ¼ 3,
if a1 < 3,
then an ! 1
then an ! 1
then an ! 1
then an ! 3
then an ! 1
as n ! 1;
as n ! 1;
as n ! 1;
as n ! 1;
as n ! 1.
3. (a) The Binomial Theorem gives
x nðn 1Þ x2
x n
x n
þ
1þ
¼1þn
þ þ
n
n 2! n
n
1
1 2
xn
1 x þ þ n :
¼1þxþ
2!
n
n
The general term in this expansion is
nðn 1Þ . . . ðn k þ 1Þ xk
1
1
2
k1 k
1
1
... 1 x:
¼
k!
n
k!
n
n
n
If k and x
are fixed,
n this general term increases as n increases. Hence the
sequence 1 þ nx
is increasing if x > 0.
(b) By the Binomial Theorem
1 k
1
k
¼1þ ;
1þk
1þ
n
n
n
for k ¼ 1; 2; . . .:
()
Solutions to the problems
381
n is bounded above, we choose an integer k x and use
To prove that 1 þ nx
the inequality (), as follows
x n
k n
1þ
1þ
n
n
!n k
1 k
1 n
1þ
¼
1þ
n
n
1 n
ek ;
n is increasing with limit e. Hence 1 þ nx
is
since the sequence 1 þ n
k
bounded above by e .
n (c) Since 1 þ nx
is increasing and bounded above, it must be convergent, by the
Monotone Convergence Theorem.
n 4. The first n terms of the sequence 1 þ 1n
are
1 2 3
2
3
4
nþ1 n
;
;
;...;
:
1
2
3
n
Taking the product of these terms, each of which is less than e, we obtain
21 32 43 . . . ðn þ 1Þn
< en ;
11 22 33 . . . nn
it follows, by cancellation, that
ðn þ 1Þn
< en :
n!
Hence
n! >
ðn þ 1Þn
¼
en
nþ1 n
;
e
for n ¼ 1; 2; . . .:
5. We use the formulas:
pffiffi
pffiffi
(a) Area ¼ 12 1 1 sin 13p ¼ 12 23 ¼ 43.
(b) Area ¼ 12 1 1 sin 16p ¼ 12 12 ¼ 14.
(c) Area ¼ 12 2 tan 16p 1 ¼ 12 2 p1ffiffi3 ¼ p1ffiffi3.
1 1 (d) Area ¼ 12 2 tan 12
p 1 ¼ tan 12
p .
1 To determine tan 12p , we use the formula
1 2 tan 12
p
1
1 :
tan p ¼
6
1 tan2 12
p
1 1ffiffi
p
Since tan 6p ¼ 3, we obtain
pffiffiffi 1
1
tan2
p 1 ¼ 0;
p þ 2 3 tan 12
12
so that
pffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffi2 ffi
2 3 2 3 þ4
1
p ¼
tan
12
2
pffiffiffi
¼ 3 2:
Appendix 4
382
pffiffiffi
1 1 Since tan 12
p > 0, we must have tan 12
p ¼ 2 3; it follows that
pffiffiffi
Area ¼ 2 3:
Chapter 3
Section 3.1
1. (a) Using the formula for summing a geometric series, with a ¼ r ¼ 13, we obtain
2 3
n
1
1
1
1
þ sn ¼ þ þ þ 3
3
3
3
1n
1
1
3
¼ 3 1 13
n 1
1
1 :
¼
4
3
n Since 13
is a basic null sequence
1
lim sn ¼ ;
n!1
4
and so
1 X
1 n
1
is convergent; with sum :
3
4
n¼1
(b) Here
sn ¼ 1 þ ð1Þ1 þð1Þ2 þ þ ð1Þn1
¼ 1 1 þ 1 1 þ þ ð1Þn1 ;
so that
sn ¼
1; n odd;
0; n even:
Hence s2kþ1 ! 1 and s2k ! 0 as k ! 1, so that {sn} is divergent, by the First
1
P
Subsequence Rule. Thus the series
ð1Þn is divergent.
n¼0
(c) Using the formula for summing a geometric series with a ¼ 2 and r ¼ 12,
we obtain
2
n3
1
1
1
sn ¼ 2 þ 1 þ
þ
þ þ
2
2
2
n
1 12
¼2
1 12
n 1
:
¼4 1
2
n Since 12
is a basic null sequence, lim sn ¼ 4, and so
n!1
1 n
X
1
is convergent; with sum 4:
2
n¼1
2. (a) We interpret 0.111. . . as
1
1
1
þ
þ
þ :
101 102 103
Solutions to the problems
383
1
1
This is a geometric series with a ¼ 10
and r ¼ 10
. Since
convergent with sum
1
10
< 1, the series is
1
a
1
¼ 10 1 ¼ ;
1 r 1 10 9
hence
1
0:111 . . . ¼ :
9
(b) We interpret 0.86363. . . as
8
1
63
63
þ
þ
:
þ
10 10 1001 1002
63
1
The series in the bracket is a geometric series with a ¼ 100
and r ¼ 100
. Since
1
<
1,
this
series
is
convergent
with
sum
100
63
a
63
7
¼ 100 1 ¼
¼ ;
1 r 1 100 99 11
hence
0:86363 . . . ¼
8
7
19
þ
¼ :
10 110 22
(c) We interpret 0.999. . . as
9
9
9
þ
þ
þ :
101 102 103
9
1
and r ¼ 10
. Since
This is a geometric series with a ¼ 10
convergent with sum
1
10
< 1, this series is
9
a
¼ 10 1 ¼ 1;
1 r 1 10
hence
0:999 . . . ¼ 1:
1
12
1
s2 ¼
12
1
s3 ¼
12
1
s4 ¼
12
4. Since
3. s1 ¼
1
;
2
1
1
1
2
þ
¼ þ ¼ ;
23
2
6
3
1
1
2
1
3
þ
þ
¼ þ
¼ ;
23
34
3
12 4
1
1
1
3
1
4
þ
þ
þ
¼ þ
¼ :
23
34
45
4
20
5
¼
1
1 1
1
¼
;
nðn þ 2Þ 2 n n þ 2
we have
for n ¼ 1; 2; . . .;
1
X
1
1 1
1
¼
;
nðn þ 2Þ n¼1 2 n n þ 2
n¼1
1
X
and so
1
1
1
1
1
þ
þ
þ
þ þ
13 24 35 46
nðn þ 2Þ
1
1
1 1
1 1
1 1
1
1
¼
1 þ þ þ þ þ :
2
3
2 4
3 5
4 6
n nþ2
sn ¼
Appendix 4
384
Most of the terms in alternate brackets cancel out, leaving
1
1
1
1
sn ¼
1þ 2
2 nþ1 nþ2
3
1
1
:
¼ 4 2ðn þ 1Þ 2ðn þ 2Þ
n o
n o
Since n þ1 1 and n þ1 2 are null sequences, lim sn ¼ 34, and so
n!1
1
X
1
3
is convergent, with sum :
n
ð
n
þ
2
Þ
4
n¼1
1 P
n
3
3
5. The series
4 is a geometric series, with a ¼ r ¼ 4. Hence, it is convergent, with
n¼1
3
sum 14 3 ¼ 3.
4
The series
1
P
n¼1
1
nðn þ 1Þ
is convergent, with sum 1 (cf. Sub-section 3.1.3). Hence, by
the Combination Rules
1 n
X
3
2
is convergent, with sum 3 ð2 1Þ ¼ 1:
4
nðn þ 1Þ
n¼1
6. By the Combination Rules for sequences
n2
1
1
¼ lim
¼ ;
þ 1 n!1 2 þ n12 2
n 2 o
so that the sequence 2n2n þ 1 is not null.
1
P
n2
Hence, by the Non-null Test,
2n2 þ 1 is divergent.
lim
n!1 2n2
n¼1
Section 3.2
1. Let sn ¼ 1 þ 212 þ 312 þ þ n12 and tn ¼ 1 þ 12 þ 13 þ þ 1n. The values of these
partial sums are as listed below:
n
1
2
3
4
5
6
7
8
sn 1
1.25
1.36
1.42
1.46
1.49
1.51
1.53
tn
1.50
1.83
2.08
2.28
2.45
2.59
2.72
1
2. (a) We use the Comparison Test. Since
n3 þ n n3 ;
for n ¼ 1; 2; . . .;
Solutions to the problems
385
we have
1
1
;
n3 þ n n3
Since
1
P
n¼1
1
n3
for n ¼ 1; 2; . . .:
is convergent, we deduce, from the Comparison Test, that
1
X
1
is convergent:
3þn
n
n¼1
(b) Let
an ¼
1
pffiffiffi
nþ n
1
bn ¼ ;
n
and
for n ¼ 1; 2; . . .;
so that
lim
an
n!1 bn
n
pffiffiffi
þ n
1
¼ lim
n!1 1 þ p1ffiffi
¼ lim
n!1 n
n
¼ 1 6¼ 0:
1
P
1
Since
n is divergent, we deduce, from the Limit Comparison Test, that
n¼1
1
X
1
pffiffiffi is divergent:
n
þ
n
n¼1
(c) Let
an ¼
nþ4
2n3 n þ 1
and
bn ¼
1
;
n2
for n ¼ 1; 2; . . .;
so that
an
n3 þ 4n2
¼ lim 3
n!1 bn
n!1 2n n þ 1
1 þ 4n
¼ lim
n!1 2 12 þ 13
n
n
1
¼ 6¼ 0:
2
is convergent, we deduce, from the Limit Comparison Test, that
1
X
nþ4
is convergent:
3nþ1
2n
n¼1
lim
Since
1
P
n¼1
1
n2
(d) We use the Comparison Test. Since
0 cos2 ð2nÞ 1; for n ¼ 1; 2; . . .;
we have
cos2 ð2nÞ
1
3 ; for n ¼ 1; 2; . . .:
n3
n
is convergent, we deduce, from the Comparison Test, that
0
Since
1
P
n¼1
1
n3
1
X
cos2 ð2nÞ
n3
n¼1
is convergent:
3
3. (a) Let an ¼ nn! , for n ¼ 1, 2, . . ., so that
! anþ1
ðn þ 1Þ3
n!
3
¼
n
an
ðn þ 1Þ!
ðn þ 1Þ2
n3
2
n þ 2n þ 1 1 2
1
¼
¼ þ 2þ 3:
n3
n n
n
¼
Appendix 4
386
Hence, by the Combination Rules for sequences
anþ1
lim
¼ 0;
n!1 an
1 3
P
n
it follows, from the Ratio Test, that
n! is convergent.
n¼1
2 n
(b) Let an ¼ n n!2 , for n ¼ 1, 2, . . ., so that
! anþ1
ðn þ 1Þ2 2nþ1
n!
¼
2 n
n 2
an
ðn þ 1Þ!
2ðn þ 1Þ
2
n
1 1
¼2 þ 2 :
n n
¼
Hence, by the Combination Rules for sequences
anþ1
¼ 0;
lim
n!1 an
1 2 n
P
n 2
it follows, from the Ratio Test, that
n! is convergent.
n¼1
Þ!
(c) Let an ¼ ð2n
nn , for n ¼ 1, 2, . . ., so that
! anþ1
ð2n þ 2Þ!
nn
¼
an
ð2nÞ!
ðn þ 1Þnþ1
¼
ð2n þ 2Þð2n þ 1Þnn
ðn þ 1Þnþ1
2ð2n þ 1Þnn
¼
ðn þ 1Þn
4n þ 2
n :
¼
1 þ 1n
n
n
o
ð1þ1nÞ
1 n
1
We know that lim 1 þ n ¼ e and that 4n þ 2 is null, so that 4n þ 2 is null,
n!1
by the Product Rule for sequences. It follows, from the Reciprocal Rule for
sequences, that
anþ1
! 1 as n ! 1:
an
1
P
ð2nÞ!
It follows, from the Ratio Test, that
nn is divergent.
n¼1
Remark
Notice that
ð2nÞ!
nn
2n
2n 1
nþ1
n
n
n
1;
it follows, from the Non-null Test, that
1
P
ð2nÞ!
n¼1
nn
is divergent.
4. (a) Let an ¼ n log1 n, n ¼ 2, .n. .. Then
o an is positive; and, since {nlogen} is an increase
ing sequence, fan g ¼
1
n loge n
is a decreasing sequence.
Next, let bn ¼ 2n a2n ; thus
1
2n loge ð2n Þ
1
1
1
¼
¼
:
n loge 2 loge 2 n
bn ¼ 2n Solutions to the problems
387
1
P
1
Since
n is a basic divergent series, it follows by the Multiple Rule that
1
n¼2
P
bn must be divergent. Hence, by the Condensation Test, the original series
n¼2
1
P
n¼2
1
n loge n
must also be divergent.
, n ¼ 2, . . .. Then an is positive; and, since {n(loge n)2} is an
n
o
increasing sequence, fan g ¼ nðlog1 nÞ2 is a decreasing sequence.
(b) Let an ¼ nðlog1
e
nÞ2
e
Next, let bn ¼ 2n a2n ; thus
bn ¼ 2n ¼
1
2n ðloge ð2n ÞÞ2
1
ðn loge 2Þ2
¼
1
ðloge 2Þ2
1
:
n2
1
P
1
Since
n2 is a basic convergent series, it follows by the Multiple Rule that
1
n¼2
P
bn must be convergent. Hence by the Condensation Test, the original series
n¼2
1
P
n¼2
1
nðloge nÞ2
must also be convergent.
Section 3.3
nþ1
Þ n
1. (a) Let an ¼ ð1
n3 þ 1 , for n ¼ 1, 2, . . . , so that
jan j ¼
n
;
n3 þ 1
for n ¼ 1; 2; . . .:
Now
n
n
1
¼ ;
n3 þ 1 n3 n2
and
1
P
n¼1
1
n2
for n ¼ 1; 2; . . .;
is a basic convergent series. Hence, by the Comparison Test
1
X
n
is convergent:
3þ1
n
n¼1
It follows, from the Absolute Convergence Test, that
1
X
ð1Þnþ1 n
is convergent:
n3 þ 1
n¼1
n
(b) Let an ¼ cos
2n , for n ¼ 1, 2, . . .; then
1
jan j n ; for n ¼ 1; 2; . . .;
2
since jcos nj 1.
1
P
1
Now
2n is a basic convergent series (in fact, a convergent geometric
n¼1
series). Hence, by the Comparison Test
1
X
cos n
n¼1
2n
is absolutely convergent:
It follows, from the Absolute Convergence Test, that
1
X
cos n
n¼1
2n
is convergent:
Appendix 4
388
2. The series
1 1 1 1
1
1
þ þ þ þ 2 4 8 16 32 64
is absolutely convergent, and hence convergent, by the Comparison Test, since the
1
P
1
series
2n is a convergent geometric series.
n¼1
By the infinite form of the Triangle Inequality, we have
1 1 1 1
þ þ þ 1 1 þ 1 þ 1 þ 1 þ 1 þ 1 þ 1 þ 2 4 8 16 32 64
2 4 8 16 32 64
¼ 1:
It follows that the sum of the original series lies in the interval [1, 1].
n nþ1 o
is of the form ð1Þnþ1 bn , where
3. (a) The sequence ð1nÞ3
bn ¼
1
;
n3
for n ¼ 1; 2; . . .:
Now:
1
n3 0, for n ¼ 1, 2, . . .;
2. n13 is a basic null sequence;
3. n13 is decreasing, because {n3} is increasing.
1
P
ð1Þnþ1
Hence, by the Alternating Test,
is convergent.
n3
n
o n¼1
n
is is not a null sequence, since
(b) The sequence ð1Þnþ1 nþ2
n
¼ 1;
lim
n!1 n þ 2
and so the odd subsequence tends to 1.
Hence, by the Non-null Test
1
X
n
ð1Þnþ1
is divergent:
n
þ
2
n¼1
n
o
Þnþ1
is of the form ð1Þnþ1 bn , where
(c) The sequence ð1
1
1
1.
n3 þn2
bn ¼
Now:
1.
2.
1
1
3
n þn
1
2
1
1
3
1
n þ n2
for n ¼ 1; 2; . . .:
;
0, for n ¼ 1, 2, . . .;
is a null sequence, by the Squeeze Rule, since
1
1
1
n3 þ n2
0
1
1
3
1
2
1
1 ;
n2
n þn
n o
where 11 is a basic null sequence;
n2
n 1
o
1
1
is decreasing, because n3 þ n2 is increasing.
3.
1
1
n3 þ n2
Hence, by the Alternating Test,
1
P
ð1Þnþ1
1
1
n¼1 n3 þ n2
is convergent.
4. Let sn and tn denote the nth partial sums of the series
1 1 1 1 1 1 1
1
1 1 1 1 1
1 þ þ þ and 1 þ þ þ ;
2 4 3 6 8 5 10 12
2 3 4 5 6
n
P
1
1
1
respectively. Denote by Hn the nth partial sum
k ¼ 1þ 2 þ þ n of the harmonic
k¼1
series.
Solutions to the problems
The terms of the series
389
1
P
an come ‘in a natural way’ in threes, so it seems
n¼1
sensible to look at the partial sums s3n
1 1 1 1 1 1 1
1
1
1
1
s3 n ¼ 1 þ þ þ þ
2
4
3
6
8
5
10
12
2n
1
4n
2
4n
1 1
1 1 1
1 1
1
1
1
1
þ
þ
þ þ
¼ 1 2 4
3 6 8
5 10 12
2n 1 4n 2 4n
1 1
1
1 1 1 1
1
1
þ þ þ þ þ
þ
¼ 1 þ þ þ þ
3 5
2n 1
2 4 6 8
4n 2 4n
1 1 1
1
1
1 1
1
þ
þ þ þ
¼ 1 þ þ þ þ þ
2 3 4
2n 1 2n
2 4
2n
1
1 1
1
1 þ þ þ þ
2
2 3
2n
1
1
¼ H2n Hn H2n
2
2
1
¼ ðH2n Hn Þ:
2
But as we saw in Example 2
t2n ¼ H2n Hn ;
so that
1
s3n ¼ t2n :
2
By assumption, tn ! loge 2 and so t2n ! loge 2 as n ! 1. It follows, by the Multiple
Rule for sequences, that
1
s3n ! loge 2 as n ! 1:
2
Finally
1
1
s3n1 ¼ s3n þ ! loge 2 as n ! 1;
4n
2
and
1
1
1
s3n2 ¼ s3n þ
þ ! loge 2 as n ! 1:
4n 2 4n
2
It follows from these three results that sn ! 12 loge 2 as n ! 1.
5. We follow the pattern
of the solution to Example 3, by finding a subsequence of partial
1
P
ð1Þnþ1
sums of the series
¼1 12 þ 13 14 þ 15 16 þ that are greater than 2, 3, . . .
n
n¼1
in turn and ensuring that the other partial sums
are not much less than these values.
1
P
ð1Þnþ1
First, note that all terms of the series
are at most 1 in modulus.
n
1
n¼1 P
ð1Þnþ1
Next, since the ‘positive’ part of the series
has partial sums that tend to
n
n¼1
1, we start to construct the desired rearranged series as follows. Take enough of the
‘positive’ terms 1; 13 ; 15 ; . . .; 2N11 1 so that the sum 1 þ 13 þ 15 þ þ 2N11 1 is greater
than 2, choosing N1 so that it is the first integer such that this sum is greater than 2.
Then these N1 terms will form the first N1 terms in our desired rearranged series.
Next, take one ‘negative’ term 12 and consider the expression
1 1
1
1
:
1 þ þ þ þ
3 5
2N1 1
2
It is certainly less than the partial sum tN1 ¼ 1 þ 13 þ 15 þ þ 2N11 1 of the rearranged
series, for which tN1 > 2. But, since all terms are at most 1 in magnitude, it follows
that tN1 þ1 > 1.
Appendix 4
390
Now add in some more ‘positive’ terms 2N11þ1 ; 2N11þ3 ; . . .; 2N211 so that the sum
1 1
1
1
1
1
1
þ
1 þ þ þ þ
þ
þ þ
3 5
2N1 1
2
2N1 þ 1 2N1 þ 3
2N2 1
is greater than 3, choosing N2 so that it is the first available integer such that this sum
is greater than 3. Then these N2 þ 1 terms will form the first N2 þ 1 terms in our
desired rearranged series.
We then add in just one ‘negative’ term; this makes the next partial sum of the
rearranged series satisfy tN2 þ2 > 2. Then we add in sufficient positive terms to make
the partial sum tN3 þ2 > 4. And so on indefinitely.
In each set of two steps in this process we must use at least one of the ‘positive’
terms and one of the ‘negative’ terms, so that eventually all the ‘positive’ terms and
all the ‘negative’ terms of the original series will be taken exactly once in the new
1
1
P
P
series, which we denote by
bn . So certainly the series
bn is a rearrangement of
n¼1
n¼1
1
P ð1Þnþ1
the original series
n .
n¼1
From our construction, we have ensured that:
1
P
for all n > Nk, for any k, the nth partial sums of
bn are >k.
n¼1
1
P
bn tend to infinity as n ! 1.
It follows, then, that the nth partial sums of
n¼1
1
1 P
1
6. (a) The sequence 2 n is not null; hence, by the Non-null Test, the series
2 n is
n¼1
divergent.
1
P
1
Thus
2 n is neither convergent nor absolutely convergent.
n¼1
(b) We have
n n
5n þ 2n
1
2
¼
5n
þ
; for n ¼ 1; 2; . . .:
3n
3
3
1 1 P
P
n
2 n
Now,
n 13 and
are both basic convergent series; hence, by the
3
n¼1
n¼1
Combination Rules for series
1
X
5n þ 2n
n¼1
3n
(c) The sequence fan g ¼
n
is convergent ðand absolutely convergentÞ:
o
3
2n3 1
is null (so the Non-null Test is inappropriate),
and contains only positive terms.
The Ratio Test fails, since
anþ1
2n3 1
¼ lim
¼ 1:
n!1 an
n!1 2ðn þ 1Þ3 1
lim
Instead, we use the Limit Comparison Test, with
bn ¼
1
;
n3
for n ¼ 1; 2; . . .:
Then
3
an
3
¼ lim 2n 11
n!1 bn
n!1
n3
lim
3n3
n!1 2n3 1
3
¼ 6¼ 0:
2
¼ lim
Since all terms are non-negative.
Solutions to the problems
1
P
Since
n¼1
1
n3
391
is a basic convergent series, we deduce, from the Limit Comparison
Test, that
1
X
3
is convergent:
31
2n
n¼1
Since the terms in the series are non-negative, it follows that it is also absolutely
convergent. n nþ1 o
1
P
1
(d) The sequence ð1Þ1
is null, but
1 is a basic divergent series. Hence
3
n
n¼1 n3
1
nþ1
P
ð1Þ
is not absolutely convergent.
1
n¼1 n3
n nþ1 o
n
o
is of the form ð1Þnþ1 bn , where
However, ð1Þ1
n3
bn ¼
1
1
n3
;
for n ¼ 1; 2; . . . :
Now:
1
for n ¼ 1, 2, . . .;
1 0,
n3
n
o
2. 11 is a null sequence;
3
nn o
1
3. 11 is decreasing, because n3 is increasing.
1.
n3
Hence, by the Alternating Test,
(e) The sequence an ¼
lim
ð1Þnþ1 n2
n2 þ 1
2
n!1 n2
1
P
ð1Þnþ1
1
n¼1
n3
is convergent.
; n ¼ 1; 2; . . ., is not null, since
n
1
¼ lim
¼ 1;
þ 1 n!1 1 þ n12
so that the odd subsequence tends to 1.
1
1
P
P
an are divergent, by the Non-null Test. It follows that
Hence,
jan j and
n¼1
n¼1
1
X
ð1Þnþ1 n2
n¼1
n2 þ 1
is neither convergent nor absolutely convergent:
nþ1
Þ n
(f) Let an ¼ ð1
n3 þ5 ; n ¼ 1; 2; . . .; then
n
; n ¼ 1; 2; . . .;
jan j ¼ 3
n þ5
so that
n
1
jan j 3 ¼ 2 ; for n ¼ 1; 2; . . .:
n
n
1
1
P
P
an
It follows, by the Comparison Test, that
jan j is convergent; that is, that
n¼1
n¼1
is absolutely convergent.
1
P
Hence, by the Absolute Convergence Test,
an is convergent.
n6 n¼1
(g) Since 2n is a basic null sequence of positive terms, it follows, from the
Reciprocal Rule for sequences, that
2n
! 1 as n ! 1:
6
2n n
Hence n6 is not a null sequence; it follows, from the Non-null Test, that
1 n
X
2
is divergent:
n6
n¼1
1 n
P
2
Thus,
n6 is neither convergent nor absolutely convergent.
n¼1
Since
1
P
n¼1
1
n2
is a basic
convergent series.
Appendix 4
392
nþ1
Þ n
(h) Let an ¼ ð1
n2 þ2 , n ¼ 1, 2, . . ., so that
n
; for n ¼ 1; 2; . . .:
jan j ¼ 2
n þ2
Now we try the Limit Comparison Test, with
1
bn ¼ ;
n
for n ¼ 1; 2; . . .:
We obtain
n
jan j
2
¼ lim n 1þ2
n!1 bn
n!1
n
lim
n2
¼ 1 6¼ 0:
þ2
1
1
1
P
P
P
1
Thus,
an is not
jan j is divergent, since
n is divergent; it follows that
¼ lim
n!1 n2
n¼1
n¼1
absolutely convergent.
n
o
However,
ð1Þ n
n2 þ 2
bn ¼
n¼1
n
nþ1
o
is of the form ð1Þnþ1 bn , where
n
;
n2 þ 2
for n ¼ 1; 2; . . .:
Now:
n
n2 þ 2
0, for n ¼ 1, 2, . . .;
o
2. n2 nþ 2 is a null sequence, since
1 n
¼ n 2 ! 0 as n ! 1;
n2 þ 2
1 þ n2
n
o
n2 o
3. n2 nþ 2 is decreasing, since n nþ2 is increasing, because
ðn þ 1Þ2 þ2 n2 þ 2
2
2
¼ nþ1þ
nþ
nþ1
n
nþ1
n
2
¼1
0:
nðn þ 1Þ
1
P
ð1Þnþ1 n
Hence, by the Alternating Test,
n2 þ 2 is convergent.
n¼1
n
o
3
1
(i) Let an ¼
nðloge nÞ4 is an
3 , n ¼ 2, . . .. Then an is positive; and, since
o
n
nðloge nÞ4
1
is a decreasing sequence.
increasing sequence, fan g ¼
3
1.
n
nðloge nÞ4
Next, let bn ¼ 2n a2n ; thus
1
bn ¼ 2n ¼
3
2n ðloge 2n Þ4
1
3
ðn loge 2Þ4
1
1
¼
3 3 :
ðloge 2Þ4 n4
1
P
1
Since
3 is a basic divergent series, it follows (for example, by the Ratio Test)
n¼2 n4
1
P
that
bn must be divergent. Hence, by the Condensation Test, the original
n¼2
1
P
1
series
3 must also be divergent. Hence the series is not convergent, and,
n¼2 nðloge nÞ4
since its terms are non-negative, is not absolutely convergent.
Solutions to the problems
393
Section 3.4
1. Substituting x ¼ 2 into the power series for ex, we find that the seventh partial sum of
the power series for e2 gives
2 22 23 24 25 26
e2 ’ 1 þ þ þ þ þ þ
1! 2! 3! 4! 5! 6!
4 8 16 32
64
þ
¼1þ2þ þ þ þ
2 6 24 120 720
4 2 4
4
¼1þ2þ2þ þ þ þ
3 3 15 45
16
¼ 7 þ ¼ 7:355:
45
Hence this method gives as an estimate for e2 to 3 decimal places the number 7.355.
Remark
In fact, e2 ¼ 7.389, to 3 decimal places.
2. From equation (5) we know that
1
1
;
0 < e sn <
ðn 1Þ! n 1
so that we want n to satisfy the requirement that
1
1
1
< 5 1011 ; or ðn 1Þ! ðn 1Þ > 1011 ¼ 2 1010 :
ðn 1Þ! n 1
5
But 13! 13 ’ 8.095 1010, so n ¼ 14 is sufficient.
In other words, 14 terms will suffice.
Chapter 4
Section 4.1
1. (a) Let {xn} be any sequence in R that converges to 2. Then, by the Combination
Rules for sequences, we have
f ðxn Þ ¼ x3n 2x2n
! 23 2 22 ¼ 8 8 ¼ 0
as n ! 1:
But f(0) ¼ 0, so that f(xn) ! f(0) as n ! 1. Thus f is continuous at 2.
(b) Consideration of the graph of f near 1 suggests that f is not continuous at 1.
So, let {xn} be any sequence in (0, 1) that tends to 1. Then f(xn) ¼ 0, for all n,
so that
f ðxn Þ ! 0 as n ! 1:
But f(1) ¼ 1, so that f(xn) 6! f(1) as n ! 1. Thus f is not continuous at 1.
Remark
If we had chosen {xn} to be any sequence in (1, 2) that tends to 1, we would
have found that f(xn) ! f(1) as n ! 1. However this does NOT prove that f is
continuous at 1, since the definition of continuity requires that f(xn) ! f(1) for
ALL sequences that tend to 1.
2. (a) Let c be any point in R.
Let {xn} be any sequence in R that converges to c. Then f(xn) ¼ 1, for all n,
so that
f ðxn Þ ! 1 as n ! 1:
1
For example, xn ¼ 1 2n
.
Appendix 4
394
But f(c) ¼ 1, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c.
Since c is an arbitrary point of R, it follows that f is continuous on R.
(b) Let c be any point in R.
Let {xn} be any sequence in R that converges to c. Then f(xn) ¼ xn, for all n,
so that
f ðxn Þ ! c as n ! 1:
But f(c) ¼ c, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c.
Since c is an arbitrary point of R, it follows that f is continuous on R.
3. The domain of f is the interval I ¼ {x : x 0} if n is even and R if n is odd.
Thus, we have to show that for each c in I:
1
1
for each sequence {xk} in I such that xk ! c, then xnk ! cn .
First, let c ¼ 0. We have already seen (in the Power Rule for null
and the
n sequences
o
1
subsequent Remark) that, for any null sequence {xk} in I, xnk is also a null
sequence. In other words, f(xk) ! f(0) as k ! 1. Hence f is continuous at 0.
n Next, olet c 6¼ 0. We have to prove that if {xk c} is a null sequence, then so is
1
1
xnk cn .
Now, using the hint, we have
1
xk c
1
xnk cn ¼
n1
n
n2
n
n3
1
n
2
xk þ xk c þ xk n cn þ þ c
n1
n
:
By the result of part (b) of Exercise 3 on Section 2.3, we obtain
n1
n2
n3
1
2
xkn þ xkn cn þ xkn cn þ þ c
!c
n1
n
n2
1
n3
2
n1
n
þ c n cn þ c n cn þ þ c
n1
n
as k ! 1
n1
n
¼ nc ;
since c 6¼ 0, we deduce that
1
0
1
xnk cn ! n1
nc n
as k ! 1
¼ 0:
o
1
In other words, xk cn is null, as required. It follows that f is continuous at c.
n
1
n
4. (a) Let d be any point in R.
Let {xn} be any sequence in R that converges to d. Then f(xn) ¼ c, for all n, so that
f ðxn Þ ! c
as n ! 1:
But f(d) ¼ c, so that f(xn) ! f(d) as n ! 1. Thus f is continuous at d.
Since d is an arbitrary point of R, it follows that f is continuous on R.
(b) Let c be any point in R.
Let {xk} be any sequence in R that converges to c. Then f(xk) ¼ xkn, for all k,
so that, by the Product Rule for sequences
f ðxk Þ ! cn
as k ! 1:
n
But f(c) ¼ c , so that f (xk) ! f (c) as k ! 1. Thus f is continuous at c.
Since c is an arbitrary point of R, it follows that f is continuous on R.
(c) Let c be any point in R.
Let {xn} be any sequence in R that converges to c. Then, by Theorem 4 of
Sub-section 2.3.3, we have
f ðxn Þ ¼ jxn j ! jcj
as n ! 1:
But f(c) ¼ jcj, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c.
Since c is an arbitrary point of R, it follows that f is continuous on R.
Solutions to the problems
395
5. First, let c ¼ 0. Consideration of the graph of f near 0 suggests that f is not continuous at 0.
So, let {xn} be any sequence in (0, 1) that tends to 0. Then f(xn) ¼ 1, for all n, so that
f ðxn Þ ! 1
as n ! 1:
But f(0) ¼ 0, so that f(xn) 6! f(0) as n ! 1. Thus f is not continuous at 0.
Next, let c > 0. We will show that f is continuous at c.
So, let {xn} be any sequence that tends to c. Since c > 0, it follows that eventually
xn > 0; thus xn > 0 for all n > N, for a suitable choice of N. Then f(xn) ¼ 1, for all
n > N, so that
f ðxn Þ ! 1
as n ! 1:
But f(c) ¼ 1, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c.
Finally, let c < 0. We will show that f is continuous at c.
So, let {xn} be any sequence that tends to c. Since c < 0, it follows that eventually
xn < 0; thus xn < 0 for all n > N, for a suitable choice of N. Then f(xn) ¼ 1, for all
n > N, so that
f ðxn Þ ! 1 as n ! 1:
But f(c) ¼ 1, so that f(xn) ! f(c) as n ! 1. Thus f is continuous at c.
It follows that f is continuous on R {0}, and discontinuous at 0.
6. We have already seen, in Example 3 of Sub-section 4.1.1, that the function
1
g : x 7! x2 ;
for x 0;
is continuous on [0, 1). Also, it follows from Problem 4 that the function
h : x 7! x3 ;
for x 2 R;
is continuous on R.
We may apply the Composition Rule to the functions g and h, since
g([0, 1)) R; it follows that
1 3
3
h g : x 7! x2 ¼ x2 ; for x 0;
is continuous on [0, 1).
7. By the Sum and Product Rules, and by using the fact that the constant function and
the identity function are continuous on R, we see that
g : x 7! x2 þ x þ 1;
Equivalently, all polynomials
are continuous on R.
x 2 R;
is continuous on R. Since g(R) (0, 1) and the square root function h is continuous
on R, it follows, by the Composition Rule, that the function
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
h g : x 7! x2 þ x þ 1; x 2 R;
is continuous on R.
Next, by the Sum, Multiple and Product Rules, and by using the fact that the constant
function and the identity function are continuous on R, we see that the functions
k : x 7! 5x
and
l : x 7!1 þ x2 ;
Equivalently, all polynomials
are continuous on R.
x 2 R;
are continuous on R. Since l(x) 6¼ 0, it follows from the Product and Quotient Rules,
that
5x
m : x 7!
; x 2 R;
1 þ x2
is continuous on R.
It then follows, by applying the Sum Rule to h g and m, that the function
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5x
; x 2 R;
f ð xÞ ¼ x2 þ x þ 1 1 þ x2
is continuous on R.
Equivalently, rational functions
are continuous on their
domains.
Appendix 4
396
8. (a) The graph of f suggests that we should find functions g and h that squeeze f near 0.
y
y = x sin
1
x
x
So, we define g(x) ¼ jxj, x 2 R, and h(x) ¼ jxj, x 2 R. With these two chosen,
we check the conditions of the Squeeze Rule.
Now we know that
1
1 sin
1; for any x 6¼ 0:
x
It follows that
1
j xj x sin
jxj;
x
for any x 6¼ 0;
so that
gð xÞ½¼ jxj f ð xÞ ½j xj ¼ hð xÞ;
for any x 2 R:
So condition 1 of the Squeeze Rule is satisfied.
Next, the functions f, g and h all take the value 0 at the point 0. Thus condition
2 of the Squeeze Rule is satisfied.
Finally, the functions g and h are both continuous at 0.
It then follows, from the Squeeze Rule, that f is continuous at 0.
(b) Consideration of the graph of f near 0 suggests that f is not continuous at 0.
1
y = sin 1x
–1
So, let {xn} be the sequence
(
)
1
:
fxn g ¼
2n þ 12 p
Then xn ! 0 as n ! 1, and
1
f ðxn Þ ¼ sin 2n þ p
2
1
¼ sin p ¼ 1
2
for all n, so that
f ðxn Þ ! 1 as n ! 1:
But f(0) ¼ 0, so that f(xn) 6! f(0) as n ! 1. Thus f is not continuous at 0.
Solutions to the problems
397
9. Since the constant function and the identity function are continuous on R, it follows
from the Combination Rules that the function
2
gðxÞ ¼ x þ 1;
x 2 R;
Equivalently, all polynomials
are continuous on R.
is continuous on R.
Next, since the sine function is continuous on R, it follows by the Composition
Rule, that
sin gð xÞ ¼ sin x2 þ 1 ; x 2 R;
is continuous on R.
It then follows from the Multiple and Sum Rules, that the function
f ð xÞ ¼ x2 þ 1 þ 3 sin x2 þ 1 ; x 2 R;
is continuous on R.
10. The cosine function is continuous on R, so that, by the Multiple Rule, the function
p
x 7! cos x; x 2 R;
2
is continuous on R.
Then, since the sine function, also, is continuous on R, we deduce, from the
Composition Rule, that
p
f ð xÞ ¼ sin cos x ; x 2 R;
2
is continuous on R.
11. Since the identity function is continuous on R, it follows, from the Combination
Rules, that the function
gð xÞ ¼ x5 5x2 ;
Equivalently, all polynomials
are continuous on R.
x 2 R;
is continuous on R.
Then, since the cosine function is continuous on R, we deduce, from the
Composition Rule, that
hð xÞ ¼ cos x5 5x2 ; x 2 R;
is continuous on R.
Next, since the identity function is continuous on R, it follows, from the
Combination Rules, that the function
2
kð xÞ ¼ x ;
x 2 R;
is continuous on R.
Then, since the exponential function is continuous on R, we deduce, from the
Composition Rule, that
2
lðxÞ ¼ ex ;
x 2 R;
is continuous on R.
Finally, it follows, from the Combination Rules, that
2
f ðxÞ ¼ cos x5 5x2 þ 7ex ; x 2 R;
is continuous on R.
Section 4.2
1. Consider the function
f ðxÞ ¼ cos x x;
x 2 ½0; 1:
We shall show that f has a zero c in (0, 1).
Equivalently, all polynomials
are continuous on R.
Appendix 4
398
Certainly, f is continuous on [0, 1], and
f ð0Þ ¼ cos 0 0 ¼ 1 > 0;
f ð1Þ ¼ cos 1 1< 0:
Thus, by the Intermediate Value Theorem, there is some number c in (0, 1) such that
f (c) ¼ 0, and so such that
cos c ¼ c:
2. If f (0) ¼ 0 or f (1) ¼ 1, we can take c ¼ 0 or c ¼ 1, respectively.
Otherwise, we have f (0) > 0 and f (1) < 1, since the image of [0, 1] under f lies
in [0, 1].
We now define the auxiliary function
gðxÞ ¼ f ð xÞ x;
x 2 ½0; 1;
and show that g has a zero c in (0, 1).
Certainly, g is continuous on [0, 1], and
gð0Þ ¼ f ð0Þ 0 > 0;
gð1Þ ¼ f ð1Þ 1< 0:
Thus, by the Intermediate Value Theorem, there is some number c in (0, 1) such that
g(c) ¼ 0, and so such that
f ðcÞ ¼ c:
3. For the function f (x) ¼ x5 þ x 1, x 2 [0, 1], we have
f ð0Þ ¼ 1 < 0; f ð1Þ ¼ 1 þ 1 1 ¼ 1 > 0;
1
1 1
15
f
¼ þ 1¼ :
2
32 2
32
and
1 It follows,
1 by the Intermediate Value Theorem applied to 2 ; 1 , that f has a zero
in 2 ; 1 . 3 243
13
Next, f 34 ¼ 1024
þ 34 1 ¼ 1024
< 0. Thus, since
3 f 4 < 0 and f(1) > 0it3 fol
lows, by the Intermediate Value Theorem applied to 4 ; 1 , that f has a zero in 4 ; 1 .
7 16;807 1 12;711
3
7 Finally, f 8 ¼ 32;768 8 ¼ 32;768 > 0. Thus, since f 4 < 0 and f 8 > 0 it fol lows, by the Intermediate Value Theorem applied to 34 ; 78 , that f has a zero in 34 ; 78 .
3 7
This interval 4 ; 8 is of length 18, as required.
4. We compile the following table of values for the continuous function p:
1
0
1
2
p(x) 3
1
1
3
x
Since p(1) < 0 and p(0) > 0, it follows, from the Intermediate Value Theorem, that
p has a zero in (1, 0).
Solutions to the problems
399
Since p(0) > 0 and p(1) < 0, it follows, from the Intermediate Value Theorem,
that p has a zero in (0, 1).
Since p(1) < 0 and p(2) > 0, it follows, from the Intermediate Value Theorem,
that p has a zero in (1, 2).
5. For the polynomial p(x) ¼ x5 þ 3x4 x 1, x 2 R, that is continuous on R, we have
M ¼ 1 þ maxfj3j; j1j; j1jg ¼ 4;
so that, by the Zeros Localisation Theorem, all the zeros of p lie in (4, 4).
We now compile a table of values of p(x), for x ¼ 4, 3, . . ., 4:
x
4
p(x)
253
3 2 1
2
17
0 1
2
3
4
2 1 2 77 482 1,787
By applying the Intermediate Value Theorem to p on each of the three intervals
[4, 3], [1, 0] and [0, 1], we deduce that p has a zero in each of the intervals
ð4; 3Þ; ð1; 0Þ
and
ð0; 1Þ:
6. (a) First, we look separately at the two closed subintervals [1, 0] and [0, 2], on
each of which f is strictly monotonic. The function f is continuous on each of
these subintervals, and so has a maximum and a minimum on each.
Since f is strictly decreasing on [1, 0], it follows that
maxf f ð xÞ : x 2 ½1; 0g ¼ f ð1Þ ¼ 1;
and this is attained in [1, 0] only at 1; also
minf f ð xÞ : x 2 ½1; 0g ¼ f ð0Þ ¼ 0;
and this is attained in [1, 0] only at 0.
Since f is strictly increasing on [0, 2], it follows that
maxf f ð xÞ : x 2 ½0; 2g ¼ f ð2Þ ¼ 4;
and this is attained in [0, 2] only at 2; also
minf f ð xÞ : x 2 ½0; 2g ¼ f ð0Þ ¼ 0;
and this is attained in [0, 2] only at 0.
Combining these results, we see that
maxf f ð xÞ : x 2 ½1; 2g ¼ 4;
and this is attained in [ 1, 2] only at 2; also
minf f ð xÞ : x 2 ½1; 2g ¼ 0;
and this is attained in [1, 2] only at 0.
(b) First, we look separately at the three closed subintervals 0; 12p , 12p; 32p and
3
2p; 2p on each of which f is strictly monotonic. The function f is continuous on
each of these subintervals, and so has a maximum and a minimum on each.
Since f is strictly increasing on 0; 12p , it follows that
max f ð xÞ : x 2 0; 12p ¼ f 12p ¼ 1;
and this is attained in 0; 12p only at 12 p; also
min f ðxÞ : x 2 0; 12p ¼ f ð0Þ ¼ 0;
and this is attained in 0; 12p only at
0. Since f is strictly decreasing on 12p; 32p , it follows that
max f ð xÞ : x 2 12p; 32p ¼ f 12p ¼ 1;
and this is attained in 12p; 32p only at 12p; also
min f ðxÞ : x 2 12p; 32p ¼ f 32p ¼ 1;
Appendix 4
400
1
only at 32p.
Next, since f is strictly increasing on 32p; 2p , it follows that
max f ð xÞ : x 2 32p; 2p ¼ f ð2pÞ ¼ 0;
and this is attained in 32p; 2p only at 0; also
min f ð xÞ : x 2 32p; 2p ¼ f 32p ¼ 1;
and this is attained in 32p; 2p only at 32p.
Combining these results, we see that
and this is attained in
3
2p; 2p
maxf f ð xÞ : x 2 ½0; 2pg ¼ 1;
and this is attained in [0, 2p] only at 12p; also
minf f ð xÞ : x 2 ½0; 2pg ¼ 1;
and this is attained in [0, 2p] only at 32p.
Section 4.3
1. (a) If 0 x1 < x2, then 2x1 < 2x2 and x14 < x24. Hence
x41 þ 2x1 þ 3 < x42 þ 2x2 þ 3;
so that f is strictly increasing on R , and is thus one–one on R .
(b) If 0 < x1 < x2, then x12 < x22 and x11 > x12 , so that x11 < x12 : Hence
1
1
x21 < x22 ;
x1
x2
it follows that f is strictly increasing on (0, 1), and is thus one–one on (0, 1).
2. We perform the three steps of the strategy.
1. We showed, in Problem 1(b) above, that f is strictly increasing on (0, 1).
2. The function
1 x3 1
¼
; x 2 R f0g;
x
x
is a rational function, and therefore is continuous on its domain R {0}. In
particular, f is continuous on (0, 1).
x 7! x2 3. Choose the increasing sequence {n}, which tends to 1, the right-hand end-point
of (0, 1). Then
1
f ðnÞ ¼ n2 ! 1 as n ! 1;
n
by the Reciprocal Rule for sequences. Thus the right-hand end-point of
J ¼ f ðð0; 1ÞÞ is 1.
4. Choose the decreasing sequence 1n , which tends to 0, the left-hand end-point of
(0, 1). Then
1
1
f
¼ 2 n ! 1 as n ! 1;
n
n
by the Reciprocal Rule for sequences. Thus the left-hand end-point of
J ¼ f ðð0; 1ÞÞ is 1.
Hence f has a continuous inverse f 1: R ! (0, 1), by the Inverse Function
Rule.
3. (a) Since sin 14p ¼ p1ffiffi2 and 14p lies in 12p; 12p , we have sin1 p1ffiffi2 ¼ 14p.
1 1
2 1
2
Since
cos
3p ¼ 2, we have cos 3p ¼ 2, and 3p lies in [0, p], so that
1
2
1
cos 2 ¼ 3p.
pffiffiffi
pffiffiffi
Since tan 13p ¼ 3 and 13p lies in 12p; 12p , we have tan1 3 ¼ 13p.
Solutions to the problems
401
(b) Following the Hint, we put y ¼ sin1x. Then
cos 2 sin1 x ¼ cosð2yÞ
¼ 1 2 sin2 y
¼ 1 2x2 ;
since x ¼ sin y.
4. Following the Hint, we put a ¼ logex and b ¼ logey. Then x ¼ ea and y ¼ eb, so that
loge ðxyÞ ¼ loge ea eb
¼ loge eaþb
¼aþb
¼ loge x þ loge y:
5. Let y ¼ cosh1 x, where x 1. Then
x ¼ cosh y ¼ 12ðey þ ey Þ;
so that
e2y 2xey þ 1 ¼ 0:
This is a quadratic equation in ey, and so
pffiffiffiffiffiffiffiffiffiffiffiffiffi
ey ¼ x x2 1:
Both choices of þ or give a positive expression on the right, but recall that y 0
and so ey 1. Since
pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi
x þ x2 1 x x2 1 ¼ 1;
pffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffi
we choose the þ sign, because x þ x2 1 1, whereas x x2 1 1.
Hence
pffiffiffiffiffiffiffiffiffiffiffiffiffi
y ¼ cosh1 x ¼ loge x þ x2 1 :
pffiffiffiffiffiffiffiffiffiffiffiffiffi
(The value y ¼ loge x x2 1 gives the negative solution of the equation
cosh y ¼ x.)
Section 4.4
1. (a) For x > 0, f ð xÞ ¼ x ¼ e loge x .
Now, the functions
x 7! loge x; x 2 ð0; 1Þ;
and
x 7!ex ; x 2 R;
are continuous; hence, by the Multiple Rule and the Composition Rule, f is
continuous.
(b) For x > 0, f ð xÞ ¼ xx ¼ ex loge x .
Now, the functions
x 7! loge x;
x 2 ð0; 1Þ;
and x 7! ex ;
x 7! x;
x 2 R;
x 2 R;
are continuous; hence, by the Product Rule and the Composition Rule, f is
continuous.
2. Since ax ¼ ex loge a and ay ¼ ey loge a , we have
ax ay ¼ ex loge a ey loge a
¼ ex loge aþy loge a
¼ eðxþyÞ loge a
¼ axþy :
Appendix 4
402
x
3. Weuse
the inequality e > 1 þ x, for x > 0. Applying this inequality with x replaced
by xe 1, we obtain
x
x
eðeÞ1 > 1 þ 1 ; for x > e;
e
x
¼ ¼ xe1 :
e
It follows that
x
ee > x;
so that, by Rule 5 with p ¼ e
ex > xe :
Chapter 5
Section 5.1
1. (a) The function
f ð xÞ ¼
x2 þ x
x
has domain R {0}, so that f is defined on any punctured neighbourhood of 0.
Next, notice that f (x) ¼ x þ 1, for x 6¼ 0. It follows that, if {xn} lies in R {0}
and xn ! 0 as n !1, then f(xn) ! 1 as n ! 1.
Hence
lim f ð xÞ ¼ 1:
x!0
(b) First, notice that
f ð xÞ ¼
j xj
¼
x
1;
1;
x > 0,
x < 0.
The domain of f is R {0}, so that f is defined on any punctured neighbourhood of 0.
However, the two null sequences 1n and 1n both have non-zero terms,
but
1
1
¼ 1; whereas lim f ¼ 1:
lim f
n!1
n!1
n
n
Hence, lim f ð xÞ does not exist.
x!0
pffiffiffi
2. (a) The function f ð xÞ ¼ x is defined on [0, 1), which contains the point 2; it is
also continuous at 2.
Hence, by Theorem 2
pffiffiffi pffiffiffi
lim x ¼ 2:
x!2
pffiffiffiffiffiffiffiffiffi
(b) The function f ð xÞ ¼ sin x is defined on [0, p], which contains the point p2; it is
also continuous at p2, by the Composition Rule for continuous functions.
Hence, by Theorem 2
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
p ffi
pffiffiffiffiffiffiffiffiffi
limp sin x ¼ sin
¼ 1:
x!2
2
x
(c) The function f ð xÞ ¼ 1 eþ x is defined on ( 1, 1), which contains the point 1; it is
also continuous at 1, by the Composition Rule for continuous functions.
Hence, by Theorem 2
ex
1
lim
¼ e:
x!1 1 þ x
2
Solutions to the problems
403
3. (a) Since
lim
x!0
sin x
¼ 1 and
x
1
1
¼ ;
þx 2
lim
x!0 2
we deduce from the Product Rule for limits that
sin x
1
¼ :
þ x2 2
lim
x!0 2x
(b) Let f(x) ¼ x2 and gð xÞ ¼ sinx x; then
lim x2 ¼ 0 and
x!0
lim
x!0
sin x
¼ 1:
x
Also, f(x) ¼ x2 6¼ 0 in R {0}; hence, by the Composition Rule
sinðx2 Þ
¼ 1:
x!0
x!0
x2
p
ffiffi
ffi
(c) Let f ð xÞ ¼ sinx x and gð xÞ ¼ x; then f is defined on the punctured neighbourhood
( p, 0) [ (0, p) of 0, and, by the Quotient Rule for limits
sin x 1
lim f ð xÞ ¼ lim
¼ 1:
x!0
x!0 x
lim gð f ð xÞÞ ¼ lim
Also, g is defined on (0, 1), and is continuous at 1; hence, by Theorem 2 and the
Composition Rule
x 12
¼ gð1Þ ¼ 1:
lim gð f ð xÞÞ ¼ lim
x!0
x!0 sin x
4. Here
f ð0Þ;
f ð1 þ xÞ;
8
>
< f ð0Þ;
¼ f ð0Þ;
>
:
f ð1 þ xÞ;
8 0;
<
x ¼ 0,
x 6¼ 0,
x ¼ 0,
x ¼ 1,
x 6¼ 0, 1
x ¼ 0,
ð f gÞðxÞ ¼
¼
¼
(so that 1 + x 6¼ 1, 0)
x ¼ 1,
x 6¼ 0, 1,
x ¼ 0, 1,
0;
:
2 þ ð1 þ xÞ;
0;
3 þ x;
In particular, ( f g)(0) ¼ 0.
Next, lim gð xÞ ¼ 1, so that
x!0
f lim gð xÞ ¼ f ð1Þ ¼ 2:
x 6¼ 0, 1.
x!0
Finally, since f(g(x)) ¼ 3 þ x on the punctured neighbourhood (1, 0) [ (0, 1) of 0
lim f ðgð xÞÞ ¼ limð3 þ xÞ ¼ 3:
x!0
x!0
5. (a) The inequalities
1
x2 ;
x x sin
x
2
2
for x 6¼ 0;
show that condition 1 of the Squeeze Rule holds, with
gð xÞ ¼ x2 and hð xÞ ¼ x2 ;
x 2 R:
Appendix 4
404
Since
lim x2 ¼ 0 and
lim x2 ¼ 0;
x!0
x!0
it follows, from the Squeeze Rule, that
1
¼ 0:
lim x2 sin
x!0
x
(b) The inequalities
j xj x cos
1
j xj;
x
for x 6¼ 0;
show that condition 1 of the Squeeze Rule holds, with
gð xÞ ¼ jxj and hð xÞ ¼ j xj;
x 2 R:
Since
limðj xjÞ ¼ 0 and
x!0
limðj xjÞ ¼ 0;
x!0
it follows, from the Squeeze Rule, that
1
lim x cos
¼ 0:
x!0
x
6. Let
(
1
1 x < 0,
sin ;
x
1 þ x;
0 x 0.
Then, for n ¼ 1, 2, . . ., we have
1
¼ sinðnpÞ ¼ 0 ! 0 as n ! 1;
f np
f ð xÞ ¼
whereas
f 1
¼ sin 2n þ 12 p ¼ 1 ! 1 as n ! 1:
ð2n þ 12Þp
It follows that lim f ð xÞ does not exist.
x!0
However, for any sequence {xn} in ( 0, 3) for which xn ! 0, it follows, from the
fact that the function x 7! 1 þ x is continuous on [0, 3), that
lim f ð xÞ ¼ f ð0Þ ¼ 1:
x!0þ
7. (a) Since lim sinx x ¼ 1, we have, by Theorem 7, that
x!0
sin x
¼ 1:
x
Hence, by the Sum Rule
pffiffiffi
sin x pffiffiffi
sin x
lim
þ x ¼ limþ
þ limþ x
x!0þ
x!0
x!0
x
x
lim
x!0þ
¼1þ0
¼ 1:
(b) Let f ð xÞ ¼
pffiffiffi
x, x 0, and gð xÞ ¼ sinx x, x 6¼ 0. Then
pffiffiffi
sin x
limþ x ¼ 0 and lim
¼ 1:
x!0 x
x!0
pffiffiffi
Also, f ð xÞ ¼ x 6¼ 0 in (0, 1). Hence, by the Composition Rule for limits
pffiffiffi
sinð xÞ
limþ gðf ð xÞÞ ¼ limþ pffiffiffi ¼ 1:
x!0
x!0
x
Solutions to the problems
405
1
8. Since 1 þ x ex 1x
, for jxj < 1, we have
x
; for j xj < 1;
x ex 1 1x
so that
ex 1
1
1
; for 0 < x < 1:
x
1x
()
1
¼ 1, it follows, from the Squeeze Rule, that
Then, since limþ 1 ¼ 1 and limþ 1x
x!0
x!0
limþ
x!0
ex 1
¼ 1:
x
Next, it follows, from the inequalities () above, that
ex 1
1
; for 1 < x < 0;
1
x
1x
so that, by an argument similar to the one above, we have
lim
x!0
ex 1
¼ 1:
x
Hence, by Theorem 7, we may combine these two one-sided limit results to obtain
ex 1
¼ 1:
lim
x!0
x
Section 5.2
1. (a) Let f (x) ¼ jxj. Then f(x) > 0 for x 6¼ 0, and, since f is continuous at 0
limj xj ¼ 0:
x!0
Hence, by the Reciprocal Rule
1
1
¼
! 1 as x ! 0:
f ð xÞ j xj
(b) Let f(x) ¼ x3 1. Then f(x) > 0 for x 2 (1, 1), and, since f is continuous at 1
limþ x3 1 ¼ 0:
x!1
Hence, by the Reciprocal Rule
1
1
¼
!1
f ð xÞ x3 1
as x ! 1þ ;
so that
1
1
¼
! 1 as x ! 1þ :
f ð xÞ 1 x3
1 1 x3
(c) Let f ð xÞ ¼ sin
x. Then f(x) > 0 for x 2 2 p; 2 p , and
2
x3
x
¼ lim sin x
lim
x!0 sin x
x!0
x
lim x2
x!0
¼
lim sinx x
x!0
¼
0
¼ 0;
1
by the Quotient Rule for limits.
Appendix 4
406
Hence, by the Reciprocal Rule
1
sin x
¼ 3 !1
f ð xÞ
x
as x ! 0:
2. (a) Let
f ð xÞ ¼
1
; x 2 R f0g;
x2
and
gð xÞ ¼
1
; x 2 R f0g:
x2
Then f(x) ! 1 as x ! 0, and g(x) ! 1 as x ! 0; but
f ð xÞ gðxÞ ¼ 0 ! 0 as x ! 0:
(b) Let
f ð xÞ ¼
1
; x 2 R f0g;
x2
and
gð xÞ ¼ x2 ; x 2 R f0g:
Then f(x) ! 1 as x ! 1, and g(x) ! 0 as x ! 1; but
f ð xÞgð xÞ ¼ 1 ! 1 as x ! 0:
3. (a) By the Sum Rule for limits as x ! 1, we have
2x3 þ x
1
¼
lim
2
þ
lim
x!1
x!1
x3
x2
1
¼ lim ð2Þ þ lim 2
x!1
x!1 x
¼ 2 þ 0 ¼ 2:
(b) Since 1 sin x 1 for x 2 R, we have
1 sin x 1
;
x
x
x
for x 2 ð0; 1Þ:
Also
1
gð xÞ ¼ ! 0
x
as x ! 1
and
1
! 0 as x ! 1:
x
Hence, by the Squeeze Rule for limits as x ! 1 ,
hð xÞ ¼
f ð xÞ ¼
sin x
! 0 as x ! 1:
x
4. (a) Let f(x) ¼ log ex and g(x) ¼ log ex, for x 2 (0, 1). Then
f ð xÞ ! 1
as x ! 1 and
gð xÞ ! 1 as x ! 1:
Hence, by the Composition Rule for limits
gð f ð xÞÞ ¼ log e ðlog e xÞ ! 1 as x ! 1:
x
(b) Let f ð xÞ ¼ 1x and gð xÞ ¼ ex , for x 2 (0, 1). Then
1
gð f ð xÞÞ ¼ xex ;
for x 2 ð0; 1Þ:
Now
f ð xÞ ! 1
as x ! 0 :
In order to use the Composition rule for limits, we need to determine the
behaviour of g(x) as x ! 1; that is, the behaviour of g(x) as x ! 1.
Solutions to the problems
407
Now
gðxÞ ¼ ex
1
¼ x
xe
x
and
xex ! 1 as x ! 1:
It follows, by the Reciprocal Rule, that
gðxÞ ! 0 as x ! 1:
Hence, by the Composition Rule for limits,
1
gðf ð xÞÞ ¼ xex ! 0 as x ! 0 :
5. Let
for all x 2 R;
f ðxÞ ¼ 1;
and
gðxÞ ¼
2;
3;
x ¼ 1,
x 6¼ 1.
Then
f ðxÞ ! 1 as x ! 1 ðso that ‘ ¼ 1Þ;
and
gð xÞ ! 3 as x ! 1
ðso that m ¼ 3Þ;
however as x ! 1, we have
gðf ð xÞÞ ¼ gð1Þ ¼ 2
! 2 6¼ 3ð¼ mÞ:
Section 5.3
1. (a)
jan 0j < 0:1
n
Þ
, ð1
< 0:1
n2 ,
1
n2
2
, n
< 0:1
> 10:
It follows that j an 0j < 0.1 for all n > X, so long as X (b)
pffiffiffiffiffi
10 ’ 3:16:
ja 0j < 0:01
n n
ð1Þ , n2 < 0:01
, n12
, n2
, n
< 0:01
> 100
> 10:
It follows that jan 0j < 0.01 for all n > X, so long as X 10.
(c)
ja 0j < "
n n
ð1Þ , n2 < "
, n12
, n2
, n
<"
> 1"
> p1ffiffi" :
It follows that jan 0 j < " for all n > X, so long as X p1ffiffi".
Appendix 4
408
2. (a)
j f ð xÞ 5j < 0:1
, j2x 2j < 0:1
, jx 1j
< 0:05:
It follows that j f (x) 5j < 0.1 whenever 0 < jx 1j < , so long as 0.05.
(b)
j f ð xÞ 5j < 0:01
, j2x 2j < 0:01
, jx 1j
< 0:005:
It follows that j f (x) 5j < 0.01 whenever 0 < jx 1j < , so long as 0.005.
(c)
j f ð xÞ 5j < "
, j2x 2j < "
1
, jx 1j
< ":
2
It follows that j f (x) 5j < " whenever 0 < jx 1j < , so long as 12".
3. (a) The function f (x) ¼ 5x 2 is defined on every punctured neighbourhood of 3.
We must prove that:
for each positive number ", there is a positive number such that
jð5x 2Þ 13j < ";
for all x satisfying 0 < jx 3j < ;
that is
1
jx 3j < "; for all x satisfying 0 < jx 3j < :
5
Choose ¼ 15 " ; then the statement () holds.
Hence
()
limð5x 2Þ ¼ 13:
x!3
(b) The function f (x ) ¼ 1 7x3 is defined on every punctured neighbourhood of 0.
We must prove that:
for each positive number ", there is a positive number such that
1 7x3 1 < "; for all x satisfying 0 < jx 0j < ;
that is
3
7x < "; for all x satisfying 0 < j xj < :
qffiffiffiffiffi
Choose ¼ 3 17 " (so that 73 ¼ "); then the statement () holds.
Hence
lim 1 7x3 ¼ 1:
()
x!0
(c) The function f ð xÞ ¼ x2 cos
We must prove that:
1
x3
is defined on every punctured neighbourhood of 0.
for each positive number ", there is a positive number such that
2
x cos 1 0 < "; for all x satisfying 0 < jx 0j < ;
x3
that is
2
x cos 1 < ";
3
x for all x satisfying 0 < jxj < :
()
Solutions to the problems
409
pffiffiffi
Choose ¼ ". Then, since cos x13 1 for all non-zero x, it follows that, for
0 < jxj < 2
x cos 1 x2
x3 < 2 ¼ ";
so that the statement () holds.
Hence
1
lim x2 cos 3 ¼ 0:
x!0
x
(d) We consider the cases that x < 0 and x > 0 separately; that is, the two one-sided
limits.
Since jxj ¼ x when x < 0, we have
lim
x!0
x þ j xj
0
¼ lim ¼ 0;
x!0 x
x
also, since jxj ¼ x when x > 0, we have
lim
x!0þ
x þ j xj
2x
¼ limþ ¼ limþ ð2Þ ¼ 2:
x!0
x!0
x
x
Since the two one-sided limits are not equal, it follows that lim xþxjxj does not
x!0
exist.
3
4. The function f(x) ¼ x , x 2 R, is defined on every punctured neighbourhood of 1.
We must prove that:
for each positive number ", there is a positive number such that
jx3 1j < " for all x satisfying 0 < jx 1j < ;
that is
2
x þ x þ 1 jx 1j < "; for all x satisfying 0 < jx 1j < :
()
Choose ¼ minf1; 17 "g, so that:
(i) for 0 < jx 1j < , we have x 2 (0, 2), and so
2
x þ x þ 1 ¼ x2 þ x þ 1
< 22 þ 2 þ 1 ¼ 7;
(ii) jx 1j < 17 ":
It follows that
2
x þ x þ 1 jx 1j < 7 1" ¼ "; for all x satisfying 0 < jx 1j < ;
7
that is, the statement () holds.
Hence
lim x3 ¼ 1:
x!1
5. Since the limits for f and g exist, the functions f and g must be defined on some
punctured neighbourhoods (c r1, c) [ ( c, c þ r1) and (c r2, c) [ (c, c þ r2) of c, it
follows that the function fg is certainly defined on the punctured neighbourhood
(c r, c) [ ( c, c þ r), where r ¼ min{r1, r2}.
We want to prove that:
for each positive number ", there is a positive number such that
jðf ð xÞgð xÞÞ ð‘mÞj < "; for all x satisfying 0 < jx cj < :
()
Appendix 4
410
In order to examine the first inequality in (), we write
f ð xÞgð xÞ ‘m
as ð f ðxÞ ‘ÞðgðxÞ mÞ þ mð f ðxÞ ‘Þ þ ‘ðgð xÞ mÞ;
so that
j f ð xÞgðxÞ ‘mj ¼ jð f ð xÞ ‘Þðgð xÞ mÞ þ mð f ð xÞ ‘Þ þ ‘ðgð xÞ mÞj
j f ð xÞ ‘j jgð xÞ mj þ jmjj f ð xÞ ‘j þ j‘jjgð xÞ mj:
(1)
We know that, since lim f ð xÞ ¼ ‘, there is a positive number 1 such that
x!c
j f ð xÞ ‘j < ";
for all x satisfying 0 < jx cj < 1 ;
(2)
and a positive number 2 such that
j f ð xÞ ‘j < 1;
for all x satisfying 0 < jx cj < 2 ;
(3)
similarly, since lim gð xÞ ¼ m, there is a positive number 3 such that
x!c
jgð xÞ mj < ";
for all x satisfying 0 < jx cj < 3 :
(4)
We now choose ¼ min {1, 2, 3}. Then the statements (2), (3) and (4) hold for all
x satisfying 0 < jx cj < , so that, from (1), we deduce
j f ð xÞgðxÞ ‘mj j f ðxÞ ‘j jgð xÞ mj þ jmjj f ð xÞ ‘j þ j‘jjgð xÞ mj
1 " þ jmj" þ j‘j"
¼ ð1 þ jmj þ j‘jÞ";
for all x satisfying 0 < jx cj < .
This is equivalent to the statement (*), in which the number " has been replaced
by K", where K ¼ 1 þ jmj þ j‘j. Since K is a constant, this is sufficient, by the
K" Lemma.
It follows that lim f ð xÞgð xÞ ¼ ‘m, as required.
x!c
6. Since lim f ð xÞ ¼ ‘ and ‘ 6¼ 0, we may take " ¼ 12 j‘j in the definition of limit; it
x!c
follows that there is a positive number such that
1
j f ð xÞ ‘j < j‘j for all x satisfying 0 < jx cj < ;
2
in other words, for all x satisfying 0 < jx cj < , we have
1
1
j‘j < f ð xÞ ‘ < j‘j;
2
2
which we may rewrite in the form
1
1
‘ j‘j < f ð xÞ < ‘ þ j‘j:
2
2
()
It follows, from (*), that, if ‘ > 0, then
1
f ð xÞ > ‘ j‘j
2
1
1
¼ ‘ ‘ ¼ ‘ > 0;
2
2
and, if ‘ < 0, then
1
f ð xÞ < ‘ þ j‘j
2
1
1
¼ ‘ ‘ ¼ ‘ < 0:
2
2
In either case, we obtain that f(x) has the same sign as ‘ on the punctured neighbourhood {x: 0 < jx cj < } of c.
Solutions to the problems
411
Section 5.4
1. The function f (x) ¼ x3, x 2 R, is defined on R.
We must prove that:
for each positive number ", there is a positive number such that
3
x 1 < "; for all x satisfying jx 1j < ;
that is
2
x þ x þ 1 jx 1j < ";
Choose ¼ min 1; 17 " , so that
for all x satisfying jx 1j < :
()
(i) for jx 1j < , we have x 2 (0, 2), and so
2
x þ x þ 1 ¼ x2 þ x þ 1
< 22 þ 2 þ 1 ¼ 7;
(ii) jx 1j < 17 ":
It follows that
2
x þ x þ 1 jx 1j < 7 1 " ¼ ";
7
for all x satisfying jx 1j < ;
that is, the statement (*) holds.
Hence f is continuous at 1.
pffiffiffi
pffiffiffi
2. The function f ð xÞ ¼ x is defined on [0, 1), and f ð4Þ ¼ 4 ¼ 2.
We must prove that:
for each positive number ", there is a positive number such that
pffiffiffi
x 2 < "; for all x satisfying jx 4j < :
Now
pffiffiffi
x 2 ¼pxffiffiffi 4 x þ 2
jx 4j
¼ pffiffiffi
xþ2
1
jx 4j;
2
since
ðÞ
pffiffiffi
x 0:
So, choose ¼ ". It follows that, for all x satisfying jx 4j < , we have
pffiffiffi
x 2 1 jx 4j
2
1
< 2
1
¼ "
2
< ":
That is, the statement () holds.
Hence f is continuous at 4.
3. The graph of f suggests that, while f is ‘well behaved’ to the left of 2, we should
examine the behaviour of f to the right of 2. If we choose " to be any positive number
less than 14, say, then there will always be points x (with x > 2) as close as we please to
2 where f (x) is not within a distance " of f (2) ¼ 1.
So, take " ¼ 14. Then, if f were continuous at 2, there would be some positive
number such that
Appendix 4
412
1
j f ð xÞ 1j < ; for all x satisfying jx 2j < :
4
Now let xn ¼ 2 þ 1n, for n ¼ 1, 2, . . .. Then
1
1
xn ¼ 2 þ
2 ð2; 2 þ Þ; for all n > X; where X ¼ ;
n
but
()
1 1
1
j f ðxn Þ 1j ¼ 2 þ n 2
1 1
þ
2 n
1
5
¼ ":
6
4
Thus, with this choice of ", no value of exists such that requirement (*) holds.
This proves that f is discontinuous at 2.
¼
4. Since f is continuous at an interior point c of I and f (c) 6¼ 0, it follows, by Theorem 2
of Sub-section 5.4.1, that there is a neighbourhood N ¼ ( c r, c þ r) of c on which
1
j f ð xÞj > j f ðcÞj:
2
Since f (c) 6¼ 0, it follows from the inequality (1) that
(1)
1
j f ð x Þj > j f ðc Þj
2
> 0; for all x with jx cj < r:
In particular, f (x) 6¼ 0 on N, so that 1f is defined on N.
To prove that 1f is continuous at c, we must prove that,
for each positive number ", there is a positive number such that
1
1 f ðxÞ f ðcÞ < "; for all x satisfying jx cj < :
Now
ðÞ
1
1 f ðcÞ f ð xÞ
¼
f ð xÞ f ðcÞ f ð xÞf ðcÞ ¼
j f ð x Þ f ðc Þj
:
j f ð x Þj j f ðc Þj
(2)
Since f is continuous at c, we know that, for the given value of " in (*), there is a
positive number 1 such that
j f ð xÞ f ðcÞj < ";
for all x satisfying jx cj < 1 :
(3)
Now choose ¼ min{r, 1}, so that both (1) and (3) hold, for all x satisfying
jx cj < . It then follows, from (1), (2) and (3), that
1
1 jf ð xÞ f ðcÞj
f ð xÞ f ðcÞ ¼ j f ð xÞj j f ðcÞj
"
<1
2
2 j f ðcÞj
1
¼1
":
f
ð
c Þj 2
j
2
This is equivalent to the statement (*), in which the number " has been replaced by
K", where K ¼ 1jf ð1cÞj2 . Since K is a constant, this is sufficient, by the K" Lemma.
2
It follows that 1f is continuous at c, as required.
Solutions to the problems
413
Section 5.5
1. Let " be any given positive number. We want to find a positive number such that
j f ð xÞ f ðcÞj < ";
that is, that
2
x c2 < ";
for all x and c in ½2; 3 satisfying jx cj < ;
for all x and c in ½2; 3 satisfying jx cj < :
()
Now, since x2 c2 ¼ (x c)(x þ c), we have
2
x c2 ¼ jx cj jx þ cj:
But, for x and c in [2, 3], we know that
4 x þ c 6:
Hence, in order to prove the statement (*), it is sufficient to prove that there exists a
positive number such that
6jx cj < "; for all x and c in ½2; 3 satisfying jx cj < : ( )
In view of (**), take ¼ 16 " (note that this choice of depends only on ", and
not at all on x or c). With this choice of , it follows that if x, c 2 [2, 3] and satisfy
jx cj < , then
6jx cj < 6
¼ ":
In other words, the statement (**) holds, and hence the statement (*) also holds.
Hence the choice ¼ 16 " serves our purpose.
Chapter 6
Section 6.1
1. (a) Let c be any point of R. Then, for any non-zero h, the difference quotient Q(h)
at c is
f ðc þ hÞ f ðcÞ
h
ðc þ hÞn cn
¼
h
1
1
ncn1 h þ nðn 1Þcn2 h2 þ þ nchn1 þ hn
¼
h
2
ðn terms in the bracketÞ
1
n1
n2
¼ nc
þ nðn 1Þc h þ þ nchn2 þ hn1
2
! ncn1 as h ! 0:
QðhÞ ¼
It follows that f is differentiable at c, and that f 0 (c) ¼ ncn1.
(b) Let f (x) ¼ k, and let c be any point of R. Then, for any non-zero h, the difference
quotient Q(h) at c is
f ðc þ hÞ f ðcÞ
h
kk
¼
h
¼0
QðhÞ ¼
! 0 as h ! 0:
It follows that f is differentiable at c, and that f 0 (c) ¼ 0.
Appendix 4
414
2. Let c be any point of R, with c 6¼ 0. Then, for any non-zero h, the difference quotient
Q(h) at c is
f ðc þ hÞ f ðcÞ
h
1
1
1
¼
h cþh c
c ðc þ hÞ
¼
hðc þ hÞc
1
¼
ðc þ hÞc
1
! 2 as h ! 0:
c
It follows that f is differentiable at c, and that f 0 ðcÞ ¼ c12 .
QðhÞ ¼
3. To prove that f (x) ¼ x4, x 2 R, is differentiable at 1, with derivative f 0 (1) ¼ 4, we
have to show that:
for each positive number ", there is a positive number such that
4
x 1
< "; for all x satisfying 0 < jx 1j < ;
4
x1
that is
3
x þ x2 þ x 3 < ";
for all x satisfying 0 < jx 1j < :
Now, x3 þ x2 þ x 3 ¼ (x 1)(x2 þ 2x þ 3); thus, for 0 <jx 1j < 1, we have
x 2 (0, 2), so that
2
x þ 2x þ 3 < 22 þ 2 2 þ 3 ¼ 11:
1 Next, choose ¼ min 1; 11
" . With this choice of , it follows that, for all x
satisfying 0 <jx 1j< , we have
3
x þ x2 þ x 3 ¼ jx 1j x2 þ 2x þ 3
1
" 11 ¼ ":
11
It follows that f is differentiable at 1, with derivative f 0 (1) ¼ 4.
<
1
4. (a) Let f ð xÞ ¼ jxj2 . The graph of f near 0 suggests that f is not differentiable at 0.
Then, for any non-zero h, the difference quotient Q(h) at 0 is
f ðhÞ f ð0Þ
h
1
jhj2 0
¼
h
1
jhj2
:
¼
h
QðhÞ ¼
For h > 0, it follows that QðhÞ ¼ 11 . So, in particular, if we take h ¼ n12 for n 2 N,
h2
we find that
1
Q 2 ¼ n ! 1 as n ! 1:
n
It follows that f is not differentiable at 0.
(b) Let f(x) ¼ [x], the integer part of x. The graph of f near 1 suggests that f is not
differentiable at 1.
We use the fact that
x4 1 ¼ x2 1 x2 þ 1
2
¼ ðx 1Þðx þ 1Þ x þ 1
3
¼ ðx 1Þ x þ x2 þ x þ 1 :
Solutions to the problems
415
Then, for any h with 1 < h < 0, the difference quotient Q(h) at 1 is
f ð1 þ hÞ f ð1Þ
h
01
¼
h
1
¼ :
h
QðhÞ ¼
It follows that
1
lim QðhÞ ¼ lim h!0
h!0
h
¼ 1:
Hence f is not differentiable at 1.
5.
(a) For 2 > h > 0, the difference quotient for f at 2 is
f ð2 þ hÞ f ð2Þ
h
ð2 þ hÞ2 þ 4
¼
h
¼4h
! 4 as h ! 0þ :
QðhÞ ¼
Hence f is differentiable on the right at 2, and fR0 (2) ¼ 4.
f is not differentiable at 2, since it is not defined to the left of 2.
(b) For 1 < h < 1, the difference quotient for f at 0 is
QðhÞ ¼
¼
f ðhÞ f ð0Þ
( 2h
h 0
h ; h < 0;
¼
h4 0
h ;
h;
h3 ;
h > 0;
h < 0;
h > 0:
Thus
lim QðhÞ ¼ 0 and
h!0
lim QðhÞ ¼ 0;
h!0þ
so that fL0 (0) and fR0 (0) both exist and equal 0.
It follows that f is differentiable at 0, and f 0 (0) ¼ 0.
Appendix 4
416
(c) The difference quotient for f at 1 is
f ð1 þ hÞ f ð1Þ
QðhÞ ¼
h
8
< ð1þhÞ4 1 ;
1 < h < 0,
h
¼
3
ð
1þh
Þ
1
:
;
0 < h < 1,
h
4 þ 6h þ 4h2 þ h3 ; 1 < h < 0,
¼
3 þ 3h þ h2 ;
0 < h < 1.
Thus
lim QðhÞ ¼ 4 and limþ QðhÞ ¼ 3;
h!0
h!0
so that fL0 (1) ¼ 4 and fR0 (1) ¼ 3.
It follows that f is not differentiable at 1.
(d) The difference quotient for f at 2 is
f ð2 þ hÞ f ð2Þ
QðhÞ ¼
h
(
¼
(
¼
ð2þhÞ3 8
;
h
08
h ;
1 < h < 0,
0 < h < 1,
12 þ 6h þ h2 ;
1 < h < 0,
8h ;
0 < h < 1.
Since limþ QðhÞ does not exist, it follows that f is not differentiable at 2.
h!0
However f is differentiable on the left at 2, and fL0 (2) ¼ 12.
6. For n 2 N
!
1
1
p
¼
sin
2n
þ
f 2
2n þ 12 p
¼ 1 ! 1 as n ! 1:
Hence, although
1
ð2nþ12Þp
f ! 0 as n ! 1
!
1
2n þ 12 p
6! 0
¼ f ð0Þ:
It follows that f is not continuous at 0; and so, by Corollary 1, f is not differentiable at 0.
7. For any non-zero h, the difference quotient Q(h) at 0 is
f ðhÞ f ð0Þ
QðhÞ ¼
h h2 sin 1h
¼
h 1
¼ h sin
h
!0
as h ! 0:
It follows that f is differentiable at 0, and that f 0 (0) ¼ 0.
Now, for x 6¼ 0, f 0 ð xÞ ¼ 2x sin 1x cos 1x. Hence, if n 2 N
1
1
f0
¼ sinð2npÞ cosð2npÞ
2np
np
¼01
¼ 1; as n ! 1:
Since f 0 (0) 6¼ 1, it follows that f 0 is not continuous at 0.
Solutions to the problems
417
8. Let c be any point in R. At c, the difference quotient for f is
cosðc þ hÞ cos c
h
cos c cos h sin c sin h cos c
¼
h
cos h 1
sin h
sin c ¼ cos c h
h
! cos c 0 sin c 1; as h ! 0;
QðhÞ ¼
¼ sin c:
It follows that f is differentiable at c, and f 0 (c) ¼ sin c. Since c is an arbitrary point
of R, it follows that f is differentiable on R.
9. For c > 0, the different quotient for f at c is
3
We assume that h is sufficiently
small that jhj < c, so that
c þ h > 0; and hence that we are
using the correct value for f at
c þ h.
3
ðc þ hÞ c
h
1 2
¼ 3c h þ 3ch2 þ h3
h
¼ 3c2 þ 3ch þ h2 ! 3c2
QðhÞ ¼
as h ! 0;
2
0
so that f is differentiable at c and f (c) ¼ 3c .
Next, for c < 0, the different quotient for f at c is
2
We assume that h is sufficiently
small that jhj < c, so that
c þ h < 0; and hence that we are
using the correct value for f at
c þ h.
2
ðc þ hÞ c
h
1
¼ 2ch þ h2
h
¼ 2c þ h ! 2c
QðhÞ ¼
as h ! 0;
0
so that f is differentiable at c and f (c) ¼ 2c.
Finally, the difference quotient for f at 0 is
QðhÞ ¼
¼
f ðhÞ f ð0Þ
( 2 h
h 0
h < 0,
h ;
h3 0
h ;
h;
¼
h2 ;
! 0; as
h > 0,
h < 0,
h > 0,
h ! 0:
It follows that f is differentiable at 0, and that f 0 (0) ¼ 0.
Therefore, the function f 0 is given by the following formula
( 2x;
x < 0,
f 0 ð xÞ ¼
0;
3x2 ;
x ¼ 0,
x > 0.
It follows that f 0 is continuous at 0, since lim f 0 ð xÞ ¼ limþ f 0 ð xÞ ¼ f 0 ð0Þ ¼ 0:
x!0
Section 6.2
1. (a) f 0 ð xÞ ¼ 7x6 8x3 þ 9x2 5;
(b) f 0 ð xÞ ¼
3
x 2 R:
2
ðx 1Þ2x ðx þ 1Þ3x2
ðx3 1Þ2
x 3x2 2x
¼
;
ðx3 1Þ2
4
x 2 R f1g:
x!0
Appendix 4
418
(c) f 0 ð xÞ ¼ 2 cos2 x 2 sin2 x
¼ 2 cos 2x; x 2 R:
ð3 þ sin x 2 cos xÞex ex ðcos x þ 2 sin xÞ
(d) f 0 ð xÞ ¼
ð3 þ sin x 2 cos xÞ2
ex ð3 sin x 3 cos xÞ
¼
; x 2 R:
ð3 þ sin x 2 cos xÞ2
2.
f 0 ð xÞ ¼ e2x þ 2xe2x ¼ e2x ð1 þ 2xÞ;
f 00 ð xÞ ¼ 2e2x ð1 þ 2xÞ þ 2e2x ¼ e2x ð4 þ 4xÞ;
f 000 ð xÞ ¼ 2e2x ð4 þ 4xÞ þ 4e2x ¼ e2x ð12 þ 8xÞ:
3. In each case, we use the Quotient Rule and the derivatives of sine and cosine.
sin x
(a) Here f ð xÞ ¼ cos
x, so that
cos x cos x sin xðsin xÞ
cos2 x
1
¼
cos2 x
¼ sec2 x;
f 0 ð xÞ ¼
on the domain of f.
(b) Here f ð xÞ ¼ sin1 x, so that
cos x
sin2 x
¼ cosec x cot x;
f 0 ð xÞ ¼ on the domain of f.
(c) Here f ð xÞ ¼ cos1 x, so that
sin x
cos2 x
¼ sec x tan x;
f 0 ð xÞ ¼
on the domain of f.
x
(d) Here f ð xÞ ¼ cos
sin x , so that
sin xð sin xÞ cos x cos x
sin2 x
1
¼ 2
sin x
¼ cosec2 x;
f 0 ð xÞ ¼
on the domain of f.
4. (a) Here f ð xÞ ¼ 12 ðex ex Þ; x 2 R; so that
1
f 0 ð xÞ ¼ ðex þ ex Þ
2
¼ cosh x; x 2 R:
(b) Here f ð xÞ ¼ 12 ðex þ ex Þ; x 2 R; so that
1
f 0 ð xÞ ¼ ðex ex Þ
2
¼ sinh x; x 2 R:
(c) Here
f ð xÞ ¼
sinh x
;
cosh x
x 2 R;
Solutions to the problems
419
so that
cosh x cosh x sinh x sinh x
cosh2 x
1
¼
cosh2 x
¼ sech2 x; x 2 R:
f 0 ð xÞ ¼
5. (a) Here
f ð xÞ ¼ sinh x2 ;
so that
x 2 R;
f 0 ð xÞ ¼ 2x cosh x2 ;
x 2 R:
(b) Here
f ð xÞ ¼ sinðsinhð2xÞÞ;
x 2 R;
so that
f 0 ð xÞ ¼ cosðsinhð2xÞÞ2 coshð2xÞ;
x 2 R:
(c) Here
cos 2x
f ð xÞ ¼ sin
;
x2
x 2 ð0; 1Þ;
so that
2
cos 2x
x 2 sin 2x 2x cos 2x
f ðxÞ ¼ cos
;
x2
x4
0
x 2 ð0; 1Þ:
6. (a) The function
f ð xÞ ¼ cos x;
x 2 ð0; pÞ;
is continuous and strictly decreasing on (0, p), and
f ðð0; pÞÞ ¼ ð1; 1Þ:
Also, f is differentiable on (0, p), and its derivative f 0 (x) ¼ sin x is non-zero
there.
So f satisfies the conditions of the Inverse Function Rule.
Hence f 1 ¼ cos1 is differentiable on (1, 1); and, if y ¼ f (x), then
1 0
1
1
:
¼
ð yÞ ¼ 0
f
f ð xÞ
sin x
Since sin x > 0 on (0, p), and sin2 x þ cos2x ¼ 1, it follows that
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi
sin x ¼ 1 cos2 x ¼ 1 y2 :
Hence
0
1
f 1 ðyÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi :
1 y2
If we now replace the domain variable y by x, we obtain
1 0
1
ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; x 2 ð1; 1Þ:
f
1 x2
(b) The function
f ð xÞ ¼ sinh x;
x 2 R;
is continuous and strictly increasing on R, and
f ðR Þ ¼ R:
Appendix 4
420
Also, f is differentiable on R, and its derivative f 0 (x) ¼ cosh x is non-zero there.
So f satisfies the conditions of the Inverse Function Rule.
Hence f 1 ¼ sinh1 is differentiable on R; and, if y ¼ f (x), then
1 0
1
1
:
¼
ð yÞ ¼ 0
f
f ðxÞ cosh x
Since cosh x > 0 on R, and cosh2 x ¼ 1 þ sinh2 x, it follows that
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi
cosh x ¼ 1 þ sinh2 x ¼ 1 þ y2 :
Hence
0
1
f 1 ð yÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi :
1 þ y2
If we now replace the domain variable y by x, we obtain
1 0
1
ð xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ; x 2 R:
f
1 þ x2
7. If x1 < x2, then x15 < x25; it follows that f (x1) < f (x2), so that f is strictly increasing
on R.
Since f is a polynomial function, it is continuous and differentiable on R. Also
f 0 ð xÞ ¼ 5x4 þ 1 6¼ 0 on R:
Thus, f satisfies the conditions of the Inverse Function Theorem.
Now, f (0) ¼ 1, f (1) ¼ 1 and f (1) ¼ 3. Hence, by the Inverse Function Rule
1 0
1
f
¼ 1;
ð1Þ ¼ 0
f ð0Þ
1 0
1
1
f
¼ ;
ð1Þ ¼ 0
f ð1Þ 6
and
0
f 1 ð3Þ ¼
1
1
¼ :
f 0 ð1Þ 6
8. By definition,
f ð xÞ ¼ xx ¼ expðx loge xÞ:
The functions x 7! x and x 7! loge x are differentiable on (0, 1), and exp is differentiable on R. It follows, by the Product Rule and the Composition Rule, that f is
differentiable on (0, 1), and that
1
0
f ð xÞ ¼ expðx loge xÞ loge x þ x x
¼ xx ðloge x þ 1Þ:
Section 6.3
1. Since f is a polynomial function, f is continuous on [1, 2] and differentiable on
(1, 2); also
f 0 ð xÞ ¼ x3 x2 ¼ x2 ðx 1Þ:
Thus f 0 vanishes at 0 and 1.
First, we consider the behaviour of f near 0. For x 2 (1, 1), for example
1
4
f ð xÞ ¼ x3 x 4
3
has the opposite sign to that of x, since x 43 < 0; thus
Solutions to the problems
f ðxÞ > 0 for x 2 ð1; 0Þ
and
f ðxÞ < 0 for x 2 ð0; 1Þ:
Since f (0) ¼ 0, it follows that 0 is not a local extremum of f.
Next, we consider the behaviour of f near 1. Now
1 4 1 3
1 1
x x f ð xÞ f ð1Þ ¼
4
3
4 3
1 3
1 4
¼ x 1 x 1
4
3
1
1
¼ ðx 1Þ x3 þ x2 þ x þ 1 ðx 1Þ x2 þ x þ 1
4
3
1
¼ ðx 1Þ 3x3 x2 x 1
12
1
¼ ðx 1Þ2 3x2 þ 2x þ 1 :
12
It follows that, for x 2 (0, 2), for example
f ðxÞ f ð1Þ 0;
1
so that f has a local minimum at 1, with value f ð1Þ ¼ 12
.
2. Since the sine and cosine functions are continuous and differentiable on R, so also is f.
Now
f 0 ð xÞ ¼ 2 sin x cos x sin x ¼ sin xð2 cos x 1Þ;
thus f 0 vanishes in 0; 12p only when cos x ¼ 12; that is, when x ¼ 13 p.
Since f (0) ¼ 1, f 12 p ¼ 1 and
1
1 pffiffiffi 2 1
3
1
5
p ¼
3 þ ¼ þ ¼ ;
f
3
2
2
4
2
4
1 it follows that, on 0; 2 p :
the minimum of f is 1, and occurs when x ¼ 0 and x ¼ 12 p;
the maximum of f is 54, and occurs when x ¼ 13 p.
3. Since f is a polynomial function, f is continuous on [1, 3] and differentiable on (1, 3).
Also, f (1) ¼ 2 and f (3) ¼ 2, so that f (1) ¼ f (3). Thus f satisfies the conditions of
Rolle’s Theorem on [1, 3].
Now
f 0 ð xÞ ¼ 4x3 12x2 þ 6x ¼ 2x 2x2 6x þ 3 ;
pffiffiffi
so that f 0 (x) ¼ 0 when x ¼ 0 and x ¼ 12 3 3 (the roots of the quadratic equation
0
2x2 6x þ p
3¼
ffiffiffi0). Thus the only value
ofpffiffixffi in (1, 3) such that f (x) ¼ 0 is
1
1
x ¼ 2 3 þ 3 , so we must take c ¼ 2 3 þ 3 ’ 2:37.
NO:
f is not defined at 12 p.
(b)
NO:
f is not differentiable at 1, and f (0) 6¼ f (2).
(c)
YES:
4. (a)
All the conditions are satisfied.
(d) NO: f ð0Þ 6¼ f 12 p .
Section 6.4
1. (a) Since f is a polynomial function, f is continuous on [2, 2] and differentiable on
(2, 2). Thus f satisfies the conditions of the Mean Value Theorem on [2, 2].
421
Appendix 4
422
Now
f 0 ð xÞ ¼ 3x2 þ 2;
so that c satisfies the conclusion of the theorem when
f ð2Þ f ð2Þ 12 ð12Þ
¼
2 ð2Þ
2 ð2Þ
¼ 6;
qffiffi
that is, when 3c2 ¼ 4 so that c ¼ 43 ’ 1:15.
Thus there are two possible values of c.
3c2 þ 2 ¼
(b) The function exp is continuous on [0, 3] and differentiable on (0, 3). Thus f
satisfies the conditions of the Mean Value Theorem on [0, 3].
Now, f 0 (x) ¼ ex, so that c satisfies the conclusion of the theorem when
f ð3Þ f ð0Þ 1 3
¼ e 1 ;
30
3
that is, when c ¼ loge 13 ðe3 1Þ ’ 1:85.
1
1
2. (a) Here f 0 ð xÞ ¼ 4x3 4 ¼ 4 x3 1 , so that f 0 (x) > 0 on (1, 1). Hence, by the
Increasing-Decreasing Theorem, f is strictly increasing on [1, 1).
ec ¼
(b) Here f 0 ðxÞ ¼ 1 1x, so that f 0 (x) < 0 on (0, 1). Hence, by the IncreasingDecreasing Theorem, f is strictly decreasing on (0, 1].
3. (a) Let
f ð xÞ ¼ sin1 x þ cos1 x;
x 2 ½1; 1:
Then f is continuous on [1, 1] and differentiable on (1, 1).
Now
1
1
f 0 ð xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0; for x 2 ð1; 1Þ:
2
1x
1 x2
It follows, from Corollary 1, that
sin1 x þ cos1 x ¼ c;
for x 2 ½1; 1;
for some constant c.
Putting x ¼ 0, we obtain 0 þ 12 p ¼ c, so that c ¼ 12 p. Hence
1
sin1 x þ cos1 x ¼ p; x 2 ½1; 1:
2
(b) Let
1
f ð xÞ ¼ tan1 x þ tan1 ;
x
x 2 ð0; 1Þ:
Then f is continuous and differentiable on (0, 1).
Now
1
1
x2
f 0 ð xÞ ¼
2 ¼ 0; for x 2 ð0; 1Þ:
2
1þx
1þ 1
x
It follows, from Corollary 1, that
1
tan1 x þ tan1 ¼ c; for x 2 ð0; 1Þ;
x
for some constant c.
Putting x ¼ 1, we obtain 14 p þ 14 p ¼ c, so that c ¼ 12 p. Hence
1 1
tan1 x þ tan1 ¼ p; x 2 ð0; 1Þ:
x 2
Solutions to the problems
423
4. We have
f 0 ð xÞ ¼ 3x2 6x ¼ 3xðx 2Þ;
for x 2 R:
0
Thus f (x) ¼ 0 when x ¼ 0 and x ¼ 2, so that c ¼ 0, 2.
Now
f 00 ð xÞ ¼ 6x 6;
for x 2 R;
so that f 00 (0) ¼ 6 < 0 and f 00 (2) ¼ 6 > 0. It follows, from the Second Derivative
Test, that f has a local maximum that occurs at 0 and a local minimum that occurs at 2.
The value of the local maximum is f (0) ¼ 1, and the value of the local minimum is
f (2) ¼ 3.
5. (a) Let
1
x 2 0; p :
2
f ð xÞ ¼ x sin x;
Then f is continuous on 0;
Now
1
2p
and differentiable on 0;
f 0 ð xÞ ¼ 1 cos x > 0;
for x 2
1
2p
.
1
0; p ;
2
and f (0) ¼ 0.
It follows that f (x) 0 for x 2 0; 12 p , so that
1
x sin x; for x 2 0; p :
2
(b) Let
2
1
2
f ð xÞ ¼ x þ x3 ;
3
3
x 2 ½0; 1:
Then f is continuous on [0, 1] and differentiable on (0, 1).
Now
2 2 1
f 0 ð xÞ ¼ x3
3 3
2
1
1 x3 < 0; for x 2 ð0; 1Þ;
¼
3
and f ð1Þ ¼ 23 þ 13 1 ¼ 0.
It follows that f (x) 0 for x 2 [0, 1], so that
2
1
2
x þ x3 ; for x 2 ½0; 1:
3
3
Section 6.5
1. Here
f ðxÞ ¼ x3 þ x2 sin x
and
gð xÞ ¼ x cos x sin x on ½0; p:
Since polynomial functions and the sine and cosine functions are continuous and
differentiable on R, it follows, by the Combination Rules, that f and g are continuous
on [0, p] and differentiable on (0, p). It follows that Cauchy’s Mean Value Theorem
applies to the functions f and g on [0, p].
Now, g(0) ¼ 0 1 0 ¼ 0 and g(p) ¼ pcosp sinp ¼ p, so that g(p) 6¼ g(0).
Also
f 0 ð xÞ ¼ 3x2 þ 2x sin x þ x2 cos x
¼ x2 ð3 þ cos xÞ þ 2x sin x
Appendix 4
424
and
g0 ð xÞ ¼ cos x x sin x cos x
¼ x sin x
on (0, p). In particular, g0 (x) 6¼ 0 on (0, p).
It therefore follows from Cauchy’s Mean Value Theorem that there exists at
least one point c in (0, p) for which
c2 ð3 þ cos cÞ þ 2c sin c
ðp3 Þ ð0Þ
¼
;
c sin c
ðpÞ ð0Þ
so that
cð3 þ cos cÞ þ 2 sin c
¼ p2 :
sin c
By cross-multiplying, we obtain
cð3 þ cos cÞ þ 2 sin c ¼ p2 sin c;
so that
3c ¼ p2 2 sin c c cos c:
Hence the equation 3x ¼ (p2 2)sinx xcosx has at least one root in (0, p).
2. (a) Let
f ð xÞ ¼ sinh 2x
and
gð xÞ ¼ sin 3x;
for x 2 R:
Then f and g are differentiable on R, and
f ð0Þ ¼ gð0Þ ¼ 0;
so that f and g satisfy the conditions of l’Hôpital’s Rule at 0.
Now
f 0 ð xÞ ¼ 2 cosh 2x and g0 ðxÞ ¼ 3 cos 3x;
so that
f 0 ð xÞ 2 cosh 2x
:
¼
g0 ð xÞ
3 cos 3x
It follows, from l’Hôpital’s Rule, that the desired limit lim gf ððxxÞÞ exists and equals
x!0
f 0 ð xÞ
2 cosh 2x
;
¼ lim
lim
x!0 g0 ð xÞ
x!0 3 cos 3x
provided that this last limit exists.
But, by the Combination Rules for continuous functions, the function
x 7!
2 cosh 2x
3 cos 3x
is continuous at 0, so that
f 0 ðxÞ f 0 ð0Þ
¼ 0
x!0 g0 ð xÞ
g ð0Þ
2
¼ :
3
lim
It follows, from l’Hôpital’s Rule, that the original limit exists, and that its
value is 23.
(b) Let
1
1
f ð xÞ ¼ ð1 þ xÞ5 ð1 xÞ5
and
2
2
gð xÞ ¼ ð1 þ 2xÞ5 ð1 2xÞ5 ;
1 1
for x 2 2 ; 2 :
Solutions to the problems
425
Then f and g are differentiable on 12 ; 12 , and
f ð0Þ ¼ gð0Þ ¼ 0;
so that f and g satisfy the conditions of l’Hôpital’s Rule at 0.
Now
4
4
1
1
f 0 ð xÞ ¼ ð1 þ xÞ5 þ ð1 xÞ5
5
5
and
3
3
4
4
g0 ð xÞ ¼ ð1 þ 2xÞ5 þ ð1 2xÞ5 ;
5
5
so that
4
4
1
ð1 þ xÞ5 þ 15 ð1 xÞ5
f 0 ð xÞ
¼ 5
:
0
g ð xÞ 4 ð1 þ 2xÞ35 þ 4 ð1 2xÞ35
5
5
It follows, from l’Hôpital’s Rule, that the desired limit lim gf ððxxÞÞ exists and equals
x!0
0
lim
f ð xÞ
x!0 g0 ð xÞ
¼ lim
þ xÞ
45
þ
þ 2xÞ
35
þ
1
5 ð1
x!0 4 ð1
5
1
5 ð1
4
5 ð1
xÞ
45
3
2xÞ5
;
provided that this last limit exists.
But, by the Combination Rules and the Power Rule for continuous functions,
the functions f 0 and g0 are continuous at 0; hence, by the Quotient Rule
f 0 ð0Þ
g0 ð0Þ
1
þ1
¼ 54 54
5þ5
1
¼ :
4
It follows, from l’Hôpital’s Rule, that the original limit exists, and equals 14.
(c) Let
f ð xÞ ¼ sin x2 þ sin x2
and gð xÞ ¼ 1 cos 4x; for x 2 R:
lim
f 0 ð xÞ
x!0 g0 ð xÞ
¼
Then f and g are differentiable on R, and
f ð0Þ ¼ gð0Þ ¼ 0;
so that f and g satisfy the conditions of l’Hôpital’s Rule at 0.
Now
f 0 ð xÞ ¼ cos x2 þ sin x2 2x þ 2x cos x2
and
g0 ð xÞ ¼ 4 sin 4x;
so that
f 0 ð xÞ cosðx2 þ sinðx2 ÞÞ ð2x þ 2x cosðx2 ÞÞ
¼
:
4 sin 4x
g0 ð xÞ
It follows, from l’Hôpital’s Rule, that the limit lim gf ððxxÞÞ exists and equals
f 0 ð xÞ
lim 0 ;
x!0 g ð xÞ
x!0
()
provided that this last limit () exists.
Next, f 0 (0) ¼ 0 and g0 (0) ¼ 0, and f 0 and g0 are differentiable on R. Thus f 0 and
0
g satisfy the conditions of l’Hôpital’s Rule at 0.
0
It follows, from l’Hôpital’s Rule, that the limit lim gf 0ððxxÞÞ exists and equals
x!0
Appendix 4
426
f 00 ðxÞ
;
x!0 g00 ð xÞ
()
lim
provided that this last limit () exists.
Now
2
f 00 ð xÞ ¼ sin x2 þ sin x2 2x þ 2x cos x2
þ cos x2 þ sin x2 2 þ 2 cos x2 4x2 sin x2
and
g00 ð xÞ ¼ 16 cos 4x:
But, by the Combination Rules for continuous functions, the functions f 00 and g00
are continuous at 0, so that
f 00 ðxÞ f 00 ð0Þ
¼ 00
lim 00
x!0 g ð xÞ
g ð0Þ
4
1
¼
¼ :
16 4
It follows, from l’Hôpital’s Rule, that the limit () exists, and that its value is 14.
It then follows, from a second application of l’Hôpital’s Rule, that the original
limit exists, and that its value is 14.
(d) Let
f ð xÞ ¼ sin x x cos x
and
gð xÞ ¼ x3 ;
for
x 2 R:
Then f and g are differentiable on R, and
f ð0Þ ¼ gð0Þ ¼ 0;
so that f and g satisfy the conditions of l’Hôpital’s Rule at 0.
Now
f 0 ð xÞ ¼ x sin x
and
g0 ð xÞ ¼ 3x2 ;
so that
f 0 ð xÞ x sin x
¼
g0 ð xÞ
3x2
1 sin x
:
¼ 3
x
It follows, from l’Hôpital’s Rule, that the desired limit lim gf ððxxÞÞ exists and equals
x!0
f 0 ð xÞ
1 sin x
lim
¼ lim x!0 g0 ð xÞ
x!0 3
x
1
sin x
;
¼ lim
3 x!0 x
provided that
this
last limit exists.
0
But lim sinx x ¼ 1, so that the limit lim gf 0ððxxÞÞ does exist and equals 13. It follows,
x!0
x!0
from l’Hôpital’s Rule, that the original limit also exists, and that its value is 13.
Chapter 7
Section 7.1
1. In this case, the function f is continuous, except at 12 and 1.
On the three subintervals 0; 13 , 13 ; 34 and 34 ; 1 in P, we have
Solutions to the problems
427
1
2
¼ ;
3
3
3
3
M2 ¼ f
¼ ;
4
2
M3 ¼ 2 ¼ lim f ð xÞ ;
1
1
0¼ ;
3
3
3 1
5
x2 ¼ ¼ ;
4 3 12
3 1
m3 ¼ f ð1Þ ¼ 1;
x3 ¼ 1 ¼ :
x!1
4 4
It then follows, from the definitions of L( f, P) and U( f, P), that
3
X
mi xi ¼ m1 x1 þ m2 x2 þ m3 x3
Lð f ; PÞ ¼
i¼1
1
5
1
¼0 þ0 þ1
3
12
4
1 1
¼0þ0þ ¼ ;
4 4
3
X
U ðf ; PÞ ¼
Mi xi ¼ M1 x1 þ M2 x2 þ M3 x3
i¼1
2 1 3 5
1
¼ þ þ2
3 3 2 12
4
2 5 1 97
¼ þ þ ¼ :
9 8 2 72
m1 ¼ f ð0Þ ¼ 0;
1
¼ 0;
m2 ¼ f
2
M1 ¼ f
x1 ¼
2. In this case, the function f is increasing and continuous on [0, 1]. Thus, on each
subinterval in [0, 1], the infimum of f is the value of f at the left end-point of
the subinterval and the supremum of f is the value of f at the right end-point of the
subinterval.
i
Hence, on the ith subinterval i1
n ; n in Pn, for 1 i n, we have
2
i1
i1 2
i
i
¼
¼
; Mi ¼ f
; and
mi ¼ f
n
n
n
n
i i1 1
¼ :
xi ¼ n
n
n
It then follows, from the definitions of L( f, Pn) and U( f, Pn), that
n
n X
X
i1 2 1
Lð f ; Pn Þ ¼
mi xi ¼
n
n
i¼1
i¼1
n
1 X 2
i 2i þ 1
¼ 3
n i¼1
1 nðn þ 1Þð2n þ 1Þ
nðn þ 1Þ
2
þn
¼ 3
n
6
2
1 ðn þ 1Þð2n þ 1Þ
¼ 2
ðn þ 1 Þ þ 1
n
6
1 ¼ 2 2n2 þ 3n þ 1 6ðn þ 1Þ þ 6
6n
1 1
1 1
¼ 2 2n2 3n þ 1 ¼ þ 2 ;
6n
3 2n 6n
n
n 2
X
X
i
1
Mi xi ¼
U ð f ; Pn Þ ¼
n
n
i¼1
i¼1
n
X
1
¼ 3
i2
n i¼1
1 nðn þ 1Þð2n þ 1Þ
n3
6
2n2 þ 3n þ 1 1 1
1
¼
¼ þ þ 2:
2
6n
3 2n 6n
¼
y
2
y = f (x)
1
0
1 1
3 2
3
4
x
1
y
1
y = f (x)
1
x
Appendix 4
428
It follows that
1 1
1
1
þ 2 ¼
n!1 3
2n 6n
3
lim Lð f ; Pn Þ ¼ lim
n!1
and
lim U ð f ; Pn Þ ¼ lim
n!1
n!1
1 1
1
þ þ
3 2n 6n2
1
¼ :
3
3. By definition of least upper bound, we know that f ð xÞ sup f , for x 2 J. It follows,
x2J
from the fact that I J, that
f ð xÞ sup f ;
for x 2 I :
x2J
Hence sup f is an upper bound for f on I, so that sup f sup f .
x2J
x2I
x2J
4. In Problem 2, we showed that for the function f (x) ¼ x2, x 2 [0, 1], and the partition
i
1
Pn ¼ 0; 1n ; 1n ; 2n ; . . .; i1
of [0, 1]
n ; n ; . . .; 1 n ; 1
1
1
and lim U ð f ; Pn Þ ¼ :
lim Lð f ; Pn Þ ¼
n!1
n!1
3
3
It follows that
Z
1
0
Z 1
1
f 3
and
f 0
1
:
3
But
Z
1
f Z 1
f ;
by part ðbÞ of Theorem 4:
0
0
R1
R1
that we must have 0 f ¼ 0 f ¼ 13, so that f is integrable on [0, 1] and
RIt1follows
1
0 f ¼ 3.
5. Here the function f (x) = k, x 2 [0, 1], isthe constant function on [0, 1].
i
Hence, on the ith subinterval i1
n ; n in Pn, for 1 i n, we have
i i1 1
¼ :
mi ¼ k; Mi ¼ k; and xi ¼ n
n
n
It then follows, from the definitions of L( f, Pn) and U( f, Pn), that
n
n
X
X
1
Lð f ; Pn Þ ¼
m i xi ¼
k
n
i¼1
i¼1
n
X1
¼ k;
¼k
n
i¼1
U ð f ; Pn Þ ¼
n
X
Mi xi ¼
i¼1
n
X
k
i¼1
n
X
¼k
i¼1
1
n
1
¼ k:
n
It follows that
Z
But
f k
and
0
Z
1
Z 1
f k:
0
1
f 0
Z 1
f;
by part ðbÞ of Theorem 4:
0
R1
R1
RIt1follows that we must have 0 f ¼ 0 f ¼ k, so that f is integrable on [0, 1] and
0 f ¼ k:
Solutions to the problems
429
6. (a) Let Pn ¼ f½x0 ; x1 ; ½x1 ; x2 ; . . .; ½xi1 ; xi ; . . .; ½xn1 ; xn g be a standard partition of
[0, 1].
The function f is constant on the interval [0, 1), so that, for this partition Pn,
we have
2
mi ¼ 2 for all i;
2; 1 i n 1;
Mi ¼
3; i ¼ n;
1
xi ¼ :
n
1
1
–1
It follows that
Lð f ; Pn Þ ¼
n
X
mi xi ¼
i¼1
–2
n
X
1
ð2Þ n
i¼1
¼ 2
n
X
1
i¼1
n
¼ 2;
and
U ðf ; Pn Þ ¼
n
X
Mi xi ¼
i¼1
n1
X
Mi xi þ Mn xn
i¼1
¼
n1
X
i¼1
¼ 2
1
1
ð2Þ þ 3
n
n
n1
X
1
i¼1
n
þ
3
n
n1 3
5
þ ¼ 2 þ :
¼ 2
n
n
n
Thus, in particular
5
U ð f ; Pn Þ Lð f ; Pn Þ ¼ :
n
Now let " be any given positive number. It follows that, if we choose n such that
5
5
n < " (that is, if we choose n > " ), then
5
< ":
U ð f ; Pn Þ Lð f ; Pn Þ ¼
n
It follows, from Riemann’s Criterion for integrability, that f is integrable on [0, 1].
Alternatively, let P ¼ {[x0, x1], [x1, x2], . . ., [xi 1, xi], . . ., [xn 1, xn]} be any
partition of [0, 1] . Then, since Mi ¼ mi for 1 i n 1, we have
U ð f ; PÞ Lð f ; PÞ ¼
n
X
i¼1
y
3
ðMi mi Þxi ¼ ðMn mn Þxn
¼ ð3 ð2ÞÞxn ¼ 5xn : ()
Now let " be any given positive number, and choose P to be any partition of [0, 1]
such its mesh, jjPjj, is less than 5"; then, in particular, we have xn < 5". It then
follows, from the inequality (), that
U ð f ; PÞ Lð f ; PÞ ¼ 5xn
"
< 5 ¼ ":
5
It follows, from Riemann’s Criterion for integrability, that f is integrable
on [0, 1].
x
Appendix 4
430
(b) Let P ¼ {[x0, x1], [x1, x2], . . ., [xi 1, xi], . . ., [xn 1, xn]} be any partition of [0, 1].
y
y = 1, x ∈
1
y = 0, x ∉
xi–1 xi
0
1
x
Then, for the partition P, we have
mi ¼ 0 and
Mi ¼ 1;
for all i.
It follows that
Lð f ; PÞ ¼
n
X
mi xi ¼
i¼1
n
X
0 xi ¼ 0;
i¼1
and
U ð f ; PÞ ¼
n
X
Mi xi ¼
i¼1
¼
n
X
i¼1
n
X
1 xi
xi ¼ 1:
i¼1
Thus, for ALL partitions P of [0, 1], we have
U ð f ; PÞ Lð f ; PÞ ¼ 1:
In particular, it follows that, for any positive " for which 0 < " < 1, there is NO
partition P of [0, 1] for which
U ð f ; PÞ Lð f ; PÞ < ":
It follows, from Riemann’s Criterion for integrability, that f is not integrable
on [0, 1].
Section 7.2
1. First, we want to use the Strategy to prove that
inf fk þ f ð xÞg ¼ k þ inf f f ð xÞg:
x2I
x2I
Since f is bounded on I, inf ff ðxÞg exists. In particular
x2I
f ð xÞ inf ff ð xÞg;
x2I
for all x 2 I:
It follows that
k þ f ð xÞ k þ inf f f ð xÞg;
x2I
for all x 2 I;
so that k þ inf ff ð xÞg is a lower bound for k þ f (x) on I.
x2I
Next, let m0 be any number greater than k þ inf f f ð xÞg. It follows that
x2I
m0 k > inf f f ð xÞg;
x2I
Solutions to the problems
431
so that, in particular, there is some number x in I such that
m0 k > f ð xÞ:
We may reformulate this fact as: there is some number x in I such that
m0 > k þ f ð xÞ;
so that m0 is not a lower bound for k þ f (x) on I.
It follows that the greatest lower bound for k þ f (x) on I is k þ inf ff ð xÞg; in other
x2I
words
inf fk þ f ðxÞg ¼ k þ inf f f ðxÞg:
x2I
x2I
Next, we want to use the Strategy to prove that
supfk þ f ð xÞg ¼ k þ supf f ð xÞg:
x2I
x2I
Since f is bounded on I, supf f ð xÞg exists. In particular
x2I
f ðxÞ supf f ð xÞg;
for all x 2 I:
x2I
It follows that
k þ f ð xÞ k þ supf f ð xÞg;
for all x 2 I;
x2I
so that k þ supff ð xÞg is an upper bound for k þ f (x) on I.
x2I
Next, let M0 be any number less than k þ supf f ðxÞg. It follows that
x2I
M 0 k < supf f ð xÞg;
x2I
so that, in particular, there is some number x in I such that
M 0 k < f ð xÞ:
We may reformulate this fact as: there is some number x in I such that
M 0 < k þ f ð xÞ;
so that M0 is not an upper bound for k þ f (x) on I.
It follows that the least upper bound for k þ f (x) on I is k þ supf f ðxÞg; in other
x2I
words
supfk þ f ð xÞg ¼ k þ supf f ð xÞg:
x2I
x2I
2. The function f is decreasing on the interval (2, 0], so that
inf f ¼ f ð0Þ ¼ 0 and
x2ð2;0
sup f ¼ lim þ f ð xÞ ¼ 4;
x2ð2;0
x!2
f is increasing on the interval [0, 3], so that
inf f ¼ f ð0Þ ¼ 0 and
sup f ¼ f ð3Þ ¼ 9:
x2½0;3
x2½0;3
Since I ¼ (2, 0] [ [0, 3], it follows that
inf f ¼ f ð0Þ ¼ 0 and
I
n
o
sup f ¼ max f ð3Þ; lim þ f ð xÞ ¼ 9:
x!2
I
Since 0 f (x) 9 and 0 f (y) 9, so that 9 f (y) 0, we have
9 f ðxÞ f ð yÞ 9:
We will prove that inf f f ð xÞ f ð yÞg ¼ 9 and sup f f ð xÞ f ð yÞg ¼ 9.
x;y2I
x;y2I
First, since f (x) f (y) 9, 9 is a lower bound for f (x) f (y), for x, y 2 I. To
prove that 9 is the greatest lower bound, we now need to prove that:
for each positive number ", there are X and Y in I = (2, 3] for which
f ðX Þ f ðY Þ < 9 þ ":
Appendix 4
432
Now, by the definition of infimum and supremum on I, we know that, since 12 " > 0,
there exist X and Y in (2, 3] such that
f ð X Þ < 12 "
For example, for any number
pffiffiffi X
in (2, 3] with j X j 12 ", we
have
f ðY Þ > 9 12 ":
and
It follows from these two inequalities that
f ð X Þ f ðY Þ < 12 " þ 9 þ 12 "
¼ 9 þ ":
f ðXÞ ¼ X 2 we could,
pffiffiffi for instance, choose
X ¼ 12 ".
It follows that inf f f ð xÞ f ð yÞg ¼ 9.
x;y2I
Finally, since f (x) f (y) 9, 9 is an upper bound for f (x) f (y), for x, y 2 I. To
prove that 9 is the least upper bound, we now need to prove that:
for each positive number ", there are X and Y in I ¼ (2, 3] for which
f ð X Þ f ðY Þ > 9 ":
Now, by the definition of infimum and supremum on I, we know that, since 12 " > 0,
there exist X and Y in (2, 3] such that
f ð X Þ > 9 12 "
f ðY Þ < 12 ":
and
It follows from these two inequalities that
f ð X Þ f ðY Þ > 9 12 " þ 12 "
¼ 9 ":
It follows that sup ff ð xÞ f ð yÞg ¼ 9.
x;y2I
3. (a) For any x in I, f ð xÞ sup f , so that, since l > 0
I
lf ð xÞ l sup f ;
for all x 2 I:
I
Thus
supðlf ð xÞÞ l sup f :
x2I
I
To prove that supðlf ð xÞÞ ¼ l sup f , we now need to prove that:
x2I
I
for each positive number ", there is some X in I for which
lf ð X Þ > l sup f ":
()
I
"
l
Now, since " > 0 and l > 0, we have > 0. It follows, from the definition of
supremum of f on I, that there exists an X in I for which
"
f ð X Þ > sup f :
l
I
Multiplying both sides by the positive number l, we obtain the desired result ().
(b) For any x in I, f ðxÞ inf f so that
I
f ð xÞ inf f ;
for all x 2 I:
I
Thus
supðf ð xÞÞ inf f :
I
x2I
To prove that supðf ð xÞÞ ¼ inf f , we now need to prove that:
I
x2I
for each positive number ", there is some X in I for which
f ð X Þ > inf f ":
I
1
";
4
()
Solutions to the problems
433
Now, since " > 0 it follows, from the definition of infimum of f on I, that there
exists an X in I for which
f ðX Þ < inf f þ ";
I
so that
f ð X Þ > inf f ":
I
This is precisely the desired result ().
Section 7.3
pffiffiffiffiffiffiffiffiffiffiffiffiffi1 1
1
1. (a) F 0 ð xÞ ¼ x þ x2 4
1 þ x2 4 2 ð2xÞ
2
pffiffiffiffiffiffiffiffiffiffiffiffiffi1 1 1 x2 4 2 x2 4 2 þx
¼ x þ x2 4
1
¼ x2 4 2 :
12
1
1 pffiffiffiffiffiffiffiffiffiffiffiffi2ffi 1
1
1
1
4 x þ x 4 x2 2 ð2xÞ þ 2 1 x2
(b) F 0 ð xÞ ¼
2
2
2
4
2
1
1
1
4 x2 x2 þ 2
¼ 4 x2 2
2
2
1
2 2
2
¼ 4x
4x
1
¼ 4 x2 2 :
2. (a) From the Fundamental Theorem of Calculus and the Table of Standard
Primitives, we obtain
Z 4
2
1
1 9
1 4
1 x þ 9 2 dx ¼ x x2 þ 9 2 þ loge x þ x2 þ 9 2
2
2
0
0
9
9
¼ 10 þ loge 9 loge 3
2
2
9
¼ 10 þ loge 3:
2
(b) From the Fundamental Theorem of Calculus and the Table of Standard
Primitives, we obtain
Z e
loge xdx ¼ ½x loge x xe1
1
¼ ðe eÞ ð0 1Þ
¼ 1:
3. Using the Table of Standard Primitives and the Combination Rules, we obtain the
following primitives:
x
;
(a) F ð xÞ ¼ 4ðx loge x xÞ tan1
2
2
1 2x
(b) F ð xÞ ¼ 3 loge ðsec 3xÞ þ 5e ð2 sin x cos xÞ:
4. (a) Here we use integration by parts. Let
f ð xÞ ¼ loge x
and
1
g0 ð xÞ ¼ x3 ;
so that
f 0 ð xÞ ¼
1
x
and
4
gð xÞ ¼ 34x3 :
Appendix 4
434
It follows that
Z
Z
3 4
3 4
1
x3 loge x ¼ x3 loge x x1 x3 dx
4
Z 4
3 4
3
1
3
x3 dx
¼ x loge x 4
4
3 4
9 4
¼ x3 loge x x3 :
4
16
(b) Here we use integration by parts, twice. On each occasion we differentiate
the power function and integrate the trigonometric function.
Hence
Z p
Z p
2
2
p
x2 cos xdx ¼ x2 sin x 02 2x sin xdx
0
Z p 0
2
1 2
¼ p 2
x sin xdx;
4
0
and
Z p
Z p
2
0
p
x sin xdx ¼ ½x ðcos xÞ02 p
2
2
ðcos xÞdx
0
¼ 0 þ ½sin x0
¼ 1:
It follows that
Z
p
2
0
1
x2 cos xdx ¼ p2 2:
4
5. (a) We follow the strategy just before Example 3, with x in place of t and u in
place of x.
Let u ¼ g(x) ¼ 2sin 3x, x 2 R . Then
du
¼ 6 cos 3x;
dx
Hence
Z
so that du ¼ 6 cos 3xdx:
Z
1
sin udu
6
1
¼ cos u
6
1
¼ cosð2 sin 3xÞ:
6
sinð2 sin 3xÞ cos 3xdx ¼
(b) We follow the strategy just before Example 3, with x in place of t and u in place of x.
Let u ¼ gðxÞ ¼ ex , x 2 R. The function g is one–one on R. Then
du
¼ ex ; so that du ¼ ex dx;
dx
also
when x ¼ 0; then u ¼ 1;
when x ¼ 1;
Hence
Z
0
1
ex
ð1 þ
ex Þ
then u ¼ e:
dx ¼
2
Z
e
du
ð1 þ uÞ2
1 e
¼
1þu 1
1
1
þ
¼
1þe 2
e1
:
¼
2ð1 þ eÞ
1
Solutions to the problems
435
6. (a) We follow the strategy just
before Example 4.
1
Let t ¼ gðxÞ ¼ ðx 1Þ2 ; x 2 ð1; 1Þ. The function g is one–one on (1, 1).
Then t2 ¼ x 1, so that x ¼ t2 þ 1. It follows that
dx
¼ 2t; so that dx ¼ 2tdt:
dt
Hence
Z
Z
dx
2tdt
¼
3
1
3
3t þ ðt2 þ 1Þt
3ðx 1Þ2 þxðx 1Þ2
Z
2dt
¼
4t2 þ 1
¼ tan1 ð2tÞ
1
¼ tan1 2ðx 1Þ2 :
(b) We follow the strategy
justffi before Example 4.
pffiffiffiffiffiffiffiffiffiffiffiffi
Let t ¼ gðxÞ ¼ 1 þ ex , x 2 [0, 1). The function g is one–one on [0, 1).
Then t2 ¼ 1 þ ex, so that ex ¼ t2 1 and x ¼ loge (t2 1). It follows that:
dx
2t
2t
¼
; so that dx ¼ 2
dt;
dt t2 1
t 1
also:
pffiffiffi
when x ¼ 0,
then t ¼ 2,
when x ¼ loge 3, then t ¼ 2.
Hence
Z loge 3
Z 2
pffiffiffiffiffiffiffiffiffiffiffiffiffi
ex 1 þ ex dx ¼ pffiffi t2 1 t 0
2
¼
Z
¼
2
pffiffi
2
23
t
3
2t2 dt
2
pffiffi
2
pffiffiffi
16 4 2
¼
:
3
Section 7.4
1. Since
x sin
1
x10
x;
for x 2 ½1; 3;
it follows, from part (a) of Theorem 2, that
Z 3
Z 3
1
x sin 10 dx xdx
x
1
1
3
1
¼ x2
2 1
1
¼ ð9 1Þ ¼ 4:
2
2. Since
1
2
ex 1 x2
1
4
1
¼ ; for x 2 0; ;
1 3
2
1
4
2t
dt
t2 1
Appendix 4
436
it follows, from part (b) of Theorem 2, that
Z 1
2
4 1
2
0
ex dx 3 2
0
2
¼ :
3
(Remark We could obtain a smaller upper estimate for the integral by applying
2
1
part (a) of Theorem 2 to the inequality ex 1x
2 .)
Next, since
1
e 1 þ x 1; for x 2 0; ;
2
it follows, from part (b) of Theorem 2, that
Z 1
2
1
2
0
ex dx 1 2
0
1
¼ :
2
3. (a) Since
sin 1 1; for x 2 ½1; 4;
x x2
and
2 þ cos
2
1
1;
x
it follows that
sin1 x 1;
2 þ cos 1x for x 2 ½1; 4;
for x 2 ½1; 4:
It follows, from Theorem 3, that
Z
4 sin1 x 1 ð4 1Þ
1 2 þ cos 1x ¼ 3:
(b) Since
and
h pi
tan x 0 for x 2 0; ;
4
3 sin x2 2;
h pi
for x 2 0; ;
4
it follows that
tan x
1
tan x;
3 sinðx2 Þ 2
h pi
for x 2 0; :
4
It follows, from Theorems 2 and 3, that
Z p
Z p
4 4
tan x
tan x dx dx
2
2 0 3 sinðx Þ
0 3 sinðx Þ
Z p
4 1
tan xdx
0 2
p4
1
¼ loge ðsec xÞ
2
0
p
1
¼
loge sec
loge 1
2
4
pffiffiffi 1
1
¼ loge 2 ¼ loge 2:
2
4
Solutions to the problems
4. (a)
I0 ¼
¼
Z
437
1
ex dx
0
½ex 10
¼ e 1:
(b) Using integration by parts, we obtain
Z 1
In ¼
ex xn dx
0
Z 1
¼ ½ex xn 10 ex nxn1 dx
0
¼ e nIn1 :
(c) Using the result of part (b), with n ¼ 1, 2, 3 and 4 in turn, we obtain
I1 ¼ e I0 ¼ e ðe 1Þ ¼ 1;
I2 ¼ e 2I1 ¼ e 2;
I3 ¼ e 3I2 ¼ e 3ðe 2Þ ¼ 6 2e;
I4 ¼ e 4I3 ¼ e 4ð6 2eÞ ¼ 9e 24:
5. (a)
2 2 4
¼ ;
1 3 3
2 2 4 4 64
a2 ¼ ¼ ;
1 3 3 5 45
2 2 4 4 6 6 256
a3 ¼ ¼
;
1 3 3 5 5 7 175
a1 ¼
ð1!Þ2 22
pffiffiffi ¼ 2;
2! 1
ð2!Þ2 24 4 pffiffiffi
2;
b2 ¼ pffiffiffi ¼
3
4! 2
b1 ¼
b3 ¼
(b)
ð3!Þ2 26 16 pffiffiffi
pffiffiffi ¼
3:
15
6! 3
b21 ¼ 4
32
9
256
2
b3 ¼
75
(c) We know that
b22 ¼
b2n ¼
and
and
and
ðn!Þ4 24n
ðð2nÞ!Þ2 n
3a1 ¼ 4;
5
32
a2 ¼ ;
2
9
7
256
a3 ¼
:
3
75
:
We now express an in a similar way, tackling the numerator and denominator
separately.
The numerator is a product of 2n even numbers; so, taking a factor 2 from
each term, we obtain
2:2:4:4: . . . :ð2nÞ:ð2nÞ ¼ 22n ð1:1:2:2: . . . :n:nÞ
¼ 22n ðn!Þ2 :
The denominator of an cannot be tackled in quite the same way, as all its factors
are odd. However, we can relate it to factorials by introducing the missing
even factors:
Appendix 4
438
1:3:3:5:5: . . . :ð2n 1Þ:ð2n þ 1Þ ¼
¼
1:2:2:3:3:4:4: . . . :ð2n 1Þ:ð2nÞ:ð2nÞ:ð2n þ 1Þ
2:2: 4:4: . . . :ð2nÞ:ð2nÞ
ðð2nÞ!Þ2 ð2n þ 1Þ
22n ðn!Þ2
:
It follows that
an ¼
22n ðn!Þ2
ðð2nÞ!Þ2 ð2nþ1Þ
22n ðn!Þ2
24n ðn!Þ4
¼
ðð2nÞ!Þ2 ð2n þ 1Þ
n
¼ b2n
;
2n þ 1
so that
b2n ¼
2n þ 1
an :
n
6. Let
1
;
xp
f ð xÞ ¼
for x 2 ½1; 1Þ;
where p > 0 and p 6¼ 1.
Then f is positive and decreasing, and
f ð xÞ ! 0 as x ! 1:
Also,
Z
n
f ¼
1
Z
n
1
dx
xp
¼
n
x1p
1p 1
¼
n1p 1
:
1p
()
Now, if p > 1, then 1 p < 0, so that
Z n
1
n1p
1
<
f ¼
:
p
1
p
1
p
1
1
R n
Since the set 1 f : n 2 N is bounded above, it follows, from the Maclaurin
Integral Test, that the series converges.
Finally, if 0 < p < 1, then 1 p > 0, so that
n1p ! 1
as n ! 1:
It follows, from the Maclaurin Integral Test, that the series diverges.
7. Let t ¼ g(x) ¼ loge x, x 2 (1, 1). The function g is one–one on (1, 1).
Then x ¼ et; also
dt 1
¼
dx x
Hence
Z
and so dt ¼
dx
:
x
Z
dt
¼ 2
t
xðloge xÞ
1
1
¼ ¼
:
t
loge x
dx
2
Solutions to the problems
439
Now let
f ð xÞ ¼
1
xðloge xÞ2
x 2 ½2; 1Þ:
;
Then f is positive and decreasing on [2, 1), and
f ðxÞ ! 0 as x ! 1:
Also
Z
n
f ¼
Z
n
dx
xðloge xÞ2
1 n
¼ loge x 2
1
1
¼
loge 2 loge n
1
:
loge 2
R n
Since the set 2 f : n 2 N is bounded above, it follows, from the Maclaurin
Integral Test, that the series converges.
2
2
8. Let t ¼ g(x) ¼ loge x, x 2 (1, 1). The function g is one–one on (1, 1).
Then x ¼ et; also
dt 1
¼
dx x
Hence
Z
and so dt ¼
dx
:
x
Z
dx
dt
¼
x loge x
t
¼ loge t ¼ loge ðloge xÞ:
Now let
f ð xÞ ¼
1
;
x loge x
x 2 ½2; 1Þ:
Then f is positive and decreasing on [2, 1), and
f ðxÞ ! 0 as x ! 1:
Also
Z
n
2
Z
n
dx
x loge x
¼ ½loge ðloge xÞn2
¼ loge ðloge nÞ loge ðloge 2Þ
! 1 as n ! 1:
f ¼
2
Hence, by the Maclaurin Integral Test, the series diverges.
1
9. Let f ð xÞ ¼ 1þx
2 , x 2 [0, 1].
Then f is positive and decreasing on [0, 1]; it follows, from the strategy, that
Z 1
n
1X
i
f
f:
!
n i¼1 n
0
Now
n
n
n
X
1X
i
1X
1
1
¼
f
i 2 ¼ n
2 þ i2
n i¼1 n
n i¼1 1 þ
n
i¼1
n
1
1
1
;
¼n 2
þ
þ
þ
n þ 12 n2 þ 22
n2 þ n2
Appendix 4
440
and
Z
1
0
Z
1
1
dx
1
þ
x2
0
1 1
¼ tan x 0
f ¼
1
¼ tan1 1 tan1 0 ¼ p:
4
It follows that
1
1
1
þ
þ þ 2
n2 þ 12 n2 þ 22
n þ n2
n
1
! p as n ! 1:
4
10. Let f ð xÞ ¼ p1ffiffix, x 2 [1, 1).
Then f is positive and decreasing on [1, 1), and f (x) ! 0 as x ! 1.
Now
n
n
X
X
1
pffi
f ðiÞ ¼
i
i¼1
i¼1
1
1
1
¼ 1 þ pffiffiffi þ pffiffiffi þ þ pffiffiffi ;
n
3
2
and
Rn
1
f
R n 1
x 2 dx
h1 1 in
¼ 2x2
¼
1
1
¼ 2n2 2 ! 1
as n ! 1:
It follows, from the strategy, that
1
1
1
1
1 þ pffiffiffi þ pffiffiffi þ þ pffiffiffi 2n2 2 as n ! 1;
n
3
2
and so, using the remark before the problem, that
1
1
1
1
1 þ pffiffiffi þ pffiffiffi þ þ pffiffiffi 2n2 as n ! 1:
n
3
2
Section 7.5
1. 20! ’ 2:4329 1018 ;
30! ’ 2:6525 1032 ;
40! ’ 8:1592 1047 ;
50! ’ 3:0414 1064 ;
60! ’ 8:3210 1081 :
2. First, f2(n) f5(n); that is, sin n12 n12 as n ! 1.
We know that
sin x
! 1 as x ! 0;
x
1
since the sequence n2 tends to 0 as n ! 1, it follows that
sin n12
! 1 as n ! 1:
1
2
n
In other words, sin n12 n12 as n ! 1. Second, f3(n) f7(n); that is, 1 cos 1n 2n12 as n ! 1.
We can use the formula cos x ¼ 1 2 sin2 12 x to obtain
Solutions to the problems
441
1 cos x 2 sin2 12x
¼
x2 x2
sin 12 x
1 sin 12 x
¼ 1
1
:
2
2x
2x
Then, since 12 x ! 0 as x ! 0, we deduce, from the fact that sinx x ! 1 as x ! 0, that
1 cos x
1
as x ! 0:
!
x2
2
1
Then, since the sequence n tends to 0 as n ! 1, it follows that
1 cos 1n
1
as n ! 1;
!
1
2
2
n
so that
1 cos
1
n
1
2n2
in other words, 1 cos
! 1 as n ! 1;
1
n
2n12 as n ! 1.
2
3. Using the Hint, let f (n) ¼ n and g(n) ¼ n.
Then
1
1
ð f ðnÞÞn ¼ n2 n
1 2
¼ nn
! ð1Þ2 ¼ 1
as n ! 1;
and
1
1
ðgðnÞÞn ¼ ðnÞn
!1
as n ! 1;
1
n
1
1
so that ðf ðnÞÞ1n ! 1 as n ! 1. In other words, ð f ðnÞÞn ðgðnÞÞn as n ! 1.
ðgðnÞÞ
However
f ðnÞ n2
¼
gðnÞ
n
¼ n 6! 1
as n ! 1;
so that f (n) 6 g(n) as n ! 1.
pffiffiffiffiffiffiffiffi n
4. First, using a calculator, we find that the value of the expression 2pn ne when
n ¼ 5 is approximately 118.019.
Now 5! ¼ 120, so that the error in the Stirling’s Formula approximation is about
120 118.019 ¼ 1.981. The percentage error in this approximation is thus about
1:981
100 ’ 1:651%:
120
This approximation is therefore not within 1% of the exact value.
5. Using Stirling’s Formula, we obtain
nn
nn
pffiffiffiffiffiffiffiffinn
n!
2pn e
en
¼ pffiffiffiffiffiffi ;
2pn
it follows from Example 1 that
n 1n
n
e
1
1
n!
p2n ð2nÞ2n
as n ! 1:
Appendix 4
442
We have seen previously that, for any positive number a
1
1
an ! 1 and
nn ! 1;
as n ! 1:
It follows that
1
p2n ! 1 and
It follows that
n 1n
n
e
n!
1
ð2nÞ2n ! 1;
as n ! 1:
as n ! 1;
that is
6. (a)
(b)
7. Now
so that
n 1n
n
! e as n ! 1:
n!
!
300
1
300!
¼
300
150! 150! 2300
150 2
pffiffiffiffiffiffiffiffiffiffi300300
600p e
’
300
300p 150
2300
e
rffiffiffi
1 6
¼
30 p
¼ 0:046 ðto two significant figuresÞ:
pffiffiffiffiffiffiffiffiffiffi300300
600p e
300!
1
’
300
3
ð100!Þ3 3300 ð200pÞ2 100
3300
e
pffiffiffi
3
¼
200p
¼ 0:0028 ðto two significant figuresÞ:
4n
2n
¼
ð4nÞ!
ðð2nÞ!Þ
2
and
2n
n
4n
2n
ð4nÞ!ðn!Þ2
¼
:
2n
ðð2nÞ!Þ3
n
It follows, from Stirling’s Formula, that
4n
pffiffiffiffiffiffiffiffi4n4n
2n
2n
8pn e 2pn ne
3 6n
2n
ð4pnÞ2 2n
e
n
44n
¼ pffiffiffi
2 26n
22n
¼ pffiffiffi :
2
pffiffiffi
Hence l ¼ 1 2.
¼
ð2nÞ!
ðn!Þ2
;
Solutions to the problems
443
Chapter 8
Section 8.1
1. The tangent approximation to f at a is
f ðxÞ ’ f ðaÞ þ f 0 ðaÞðx aÞ:
f ð xÞ ¼ ex ;
f 0 ð xÞ ¼ ex ;
(a)
f ð2Þ ¼ e2 ;
f 0 ð2Þ ¼ e2 :
Hence the tangent approximation to f at 2 is
ex ’ e2 þ e2 ðx 2Þ:
f ð xÞ ¼ cos x;
(b)
f ð0Þ ¼ 1;
0
f ð xÞ ¼ sin x;
f 0 ð0Þ ¼ 0:
Hence the tangent approximation to f at 0 is
cos x ’ 1 þ 0ðx 0Þ ¼ 1:
2. (a)
f ð xÞ ¼ ex ;
f ð2Þ ¼ e2 ;
f 0 ð xÞ ¼ ex ;
f 0 ð2Þ ¼ e2 ;
f 00 ð xÞ ¼ ex ;
f 000 ðxÞ ¼ ex ;
f 00 ð2Þ ¼ e2 ;
f 000 ð2Þ ¼ e2 :
Hence
f 0 ð2Þ
ðx 2Þ
1!
¼ e2 þ e2 ðx 2Þ;
T1 ð xÞ ¼ f ð2Þ þ
f 0 ð2Þ
f 00 ð2Þ
ðx 2Þ þ
ðx 2Þ2
1!
2!
1
¼ e2 þ e2 ðx 2Þ þ e2 ðx 2Þ2 ;
2
f 0 ð2Þ
f 00 ð2Þ
f 000 ð2Þ
ðx 2Þ þ
ðx 2Þ2 þ
ðx 2Þ3
T3 ð xÞ ¼ f ð2Þ þ
1!
2!
3!
1
1
¼ e2 þ e2 ðx 2Þ þ e2 ðx 2Þ2 þ e2 ðx 2Þ3 :
2
6
T2 ð xÞ ¼ f ð2Þ þ
f ð xÞ ¼ cos x;
(b)
0
f ð0Þ ¼ 1;
f ð xÞ ¼ sin x;
f 0 ð0Þ ¼ 0;
f 00 ð xÞ ¼ cos x;
f 00 ð0Þ ¼ 1;
f 000 ðxÞ ¼ sin x;
f 000 ð0Þ ¼ 0:
Hence
f 0 ð0Þ
x ¼ 1;
1!
0
f ð0Þ
f 00 ð0Þ 2
1
xþ
x ¼ 1 x2 ;
T2 ð xÞ ¼ f ð0Þ þ
1!
2!
2
f 0 ð0Þ
f 00 ð0Þ 2 f 000 ð0Þ 3
1
T3 ð xÞ ¼ f ð0Þ þ
xþ
x þ
x ¼ 1 x2 :
1!
2!
3!
2
T1 ð xÞ ¼ f ð0Þ þ
Appendix 4
444
3. The Taylor polynomial of degree 4 for f at a is
f 0 ða Þ
f 00 ðaÞ
f 000 ðaÞ
f ð4Þ ðaÞ
ðx aÞ þ
ðx aÞ2 þ
ðx aÞ3 þ
ðx aÞ4 :
T4 ð xÞ ¼ f ðaÞ þ
1!
2!
3!
4!
(a)
f ð xÞ ¼ 7 6x þ 5x2 þ x3 ;
f ð1Þ ¼ 7;
f 0 ð xÞ ¼ 6 þ 10x þ 3x2 ;
f 0 ð1Þ ¼ 7;
f 00 ð xÞ ¼ 10 þ 6x;
f 00 ð1Þ ¼ 16;
f 000 ð xÞ ¼ 6;
f 000 ð1Þ ¼ 6;
f ð4Þ ð xÞ ¼ 0;
f ð4Þ ð1Þ ¼ 0:
Hence
T4 ð xÞ ¼ 7 þ 7ðx 1Þ þ 8ðx 1Þ2 þðx 1Þ3 :
(b)
f ð xÞ ¼ ð1 xÞ1 ;
f ð0Þ ¼ 1;
2
f ð xÞ ¼ ð1 xÞ ;
f 0 ð0Þ ¼ 1;
f 00 ð xÞ ¼ 2ð1 xÞ3 ;
f 00 ð0Þ ¼ 2;
f 000 ð xÞ ¼ 3!ð1 xÞ4 ;
f 000 ð0Þ ¼ 3!;
f ð4Þ ð xÞ ¼ 4!ð1 xÞ5 ;
f ð4Þ ð0Þ ¼ 4!:
0
Hence
T4 ð xÞ ¼ 1 þ x þ x2 þ x3 þ x4 :
(c)
f ð xÞ ¼ loge ð1 þ xÞ;
1
f ð0Þ ¼ 0;
f 0 ð xÞ ¼ ð1 þ xÞ ;
f 0 ð0Þ ¼ 1;
f 00 ð xÞ ¼ ð1 þ xÞ2 ;
f 00 ð0Þ ¼ 1;
f 000 ð xÞ ¼ 2ð1 þ xÞ3 ;
f 000 ð0Þ ¼ 2;
f ð4Þ ð xÞ ¼ 3!ð1 þ xÞ4 ;
f ð4Þ ð0Þ ¼ 3!:
Hence
1
1
1
T 4 ð xÞ ¼ x x2 þ x3 x4 :
2
3
4
p
1
¼ pffiffiffi ;
f ð xÞ ¼ sin x;
f
4
2
p
1
f0
¼ pffiffiffi ;
f 0 ð xÞ ¼ cos x;
4
2
1
00
00 p
¼ pffiffiffi ;
f ð xÞ ¼ sin x;
f
4
2
p
1
¼ pffiffiffi ;
f 000
f 000 ð xÞ ¼ cos x;
4
2
1
ð4Þ
ð4Þ p
f
¼ pffiffiffi :
f ð xÞ ¼ sin x;
4
2
(d)
Hence
1
p 1 p2 1 p3 1 p 4
x
x
:
x
þ
T4 ð xÞ ¼ pffiffiffi 1 þ x 4
2
4
6
4
24
4
2
Solutions to the problems
(e)
445
1
1
1
1
f ð xÞ ¼ 1 þ x x2 x3 þ x4 ;
2
2
6
4
1
1
f 0 ð xÞ ¼ x x2 þ x3 ;
2
2
00
f ð xÞ ¼ 1 x þ 3x2 ;
1
f 0 ð0Þ ¼ ;
2
00
f ð0Þ ¼ 1;
f 000 ðxÞ ¼ 1 þ 3!x;
f 000 ð0Þ ¼ 1;
f
ð4Þ
f ð0Þ ¼ 1;
f ð4Þ ð0Þ ¼ 3!:
ð xÞ ¼ 3!;
Hence
1
1
1
1
T4 ðxÞ ¼ 1 þ x x2 x3 þ x4 :
2
2
6
4
f ð xÞ ¼ tan x;
4.
0
f ð0Þ ¼ 0;
f ð xÞ ¼ sec x;
f 0 ð0Þ ¼ 1;
f 00 ð xÞ ¼ 2 sec2 x tan x;
f 00 ð0Þ ¼ 0;
f 000 ðxÞ ¼ 4 sec2 x tan2 x þ 2 sec4 x;
f 000 ð0Þ ¼ 2:
2
Hence
1
T3 ðxÞ ¼ x þ x3 ;
3
so that
1
T3 ð0:1Þ ¼ 0:1 þ 0:001 ¼ 0:1003:
3
Since (by calculator) tan 0.1 ¼ 0.10033467. . ., the percentage error involved is
about
0:10033467 0:10033333
100 ’ 0:001%:
0:10033467
f ð xÞ ¼ ð1 xÞ1 ;
5. (a)
f ð0Þ ¼ 1;
f 0 ð xÞ ¼ ð1 xÞ2 ;
f 0 ð0Þ ¼ 1;
3
00
f ð xÞ ¼ 2 ð1 xÞ ;
f 00 ð0Þ ¼ 2;
..
.
f ðnÞ ð xÞ ¼ ðn!Þ ð1 xÞn1 ;
f ðnÞ ð0Þ ¼ n!:
Hence
1 00
1
f ð0Þx2 þ þ f ðnÞ ð0Þxn
2!
n!
¼ 1 þ x þ x2 þ þ xn :
Tn ð xÞ ¼ f ð0Þ þ f 0 ð0Þx þ
f ð xÞ ¼ loge ð1 þ xÞ;
(b)
0
1
f ð0Þ ¼ 0;
f ð xÞ ¼ ð1 þ xÞ ;
f 0 ð0Þ ¼ 1;
f 00 ð xÞ ¼ ð1Þ ð1 þ xÞ2 ;
f 00 ð0Þ ¼ 1;
f 000 ðxÞ ¼ ð1Þð2Þ ð1 þ xÞ3 ;
f 000 ð0Þ ¼ 2;
..
.
f ðnÞ ð xÞ ¼ ð1Þnþ1 ðn 1Þ! ð1 þ xÞn ;
f ðnÞ ð0Þ ¼ ð1Þnþ1 ðn 1Þ!:
Hence
1 00
1
f ð0Þx2 þ þ f ðnÞ ð0Þxn
2!
n!
1 2 1 3
ð1Þnþ1 n
x :
¼ x x þ x þ
n
2
3
Tn ð xÞ ¼ f ð0Þ þ f 0 ð0Þx þ
Appendix 4
446
f ð xÞ ¼ ex ;
(c)
f ð0Þ ¼ 1;
and, for each positive integer k
f ðk Þ ð xÞ ¼ ex ;
f ðkÞ ð0Þ ¼ 1:
Hence
1 00
1
f ð0Þx2 þ þ f ðnÞ ð0Þxn
2!
n!
x2 x3
xn
¼ 1 þ x þ þ þ þ :
2! 3!
n!
Tn ð xÞ ¼ f ð0Þ þ f 0 ð0Þx þ
f ð xÞ ¼ sin x;
(d)
f ð0Þ ¼ 0;
0
f ð xÞ ¼ cos x;
f 0 ð0Þ ¼ 1;
f 00 ð xÞ ¼ sin x;
f 00 ð0Þ ¼ 0;
f 000 ð xÞ ¼ cos x;
f 000 ð0Þ ¼ 1;
f ð4Þ ð xÞ ¼ sin x;
f ð4Þ ð0Þ ¼ 0;
and, in general, for each positive integer k
8
< 0; if k is even,
ðkÞ
1; if k 1 (mod 4),
f ð0Þ ¼
:
1; if k 3 (mod 4).
Hence
T n ð xÞ ¼ x x3 x5
xn
þ þ þ ð0 or 1 or 1Þ ;
3! 5!
n!
where the coefficient of xk is f
f ð xÞ ¼ cos x;
(e)
0
ðkÞ
ð0Þ
k! ,
and the value of f (k)(0) is as specified above.
f ð0Þ ¼ 1;
f ð xÞ ¼ sin x;
f 0 ð0Þ ¼ 0;
f 00 ð xÞ ¼ cos x;
f 00 ð0Þ ¼ 1;
f 000 ð xÞ ¼ sin x;
f 000 ð0Þ ¼ 0;
f
ð4Þ
ð xÞ ¼ cos x;
f ð4Þ ð0Þ ¼ 1;
and, in general, for each positive integer k
8
< 0; if k is odd;
f ðkÞ ð0Þ ¼
1; if k 0 ðmod 4Þ;
:
1; if k 2 ðmod 4Þ:
Hence
T n ð xÞ ¼ 1 x2 x4
xn
þ þ ð0 or 1 or 1Þ ;
2! 4!
n!
where the coefficient of xk is
above.
f ðkÞ ð0Þ
k! ,
and the value of f (k)(0) is as specified
Section 8.2
1. Here
f ð xÞ ¼
1
1
; f 0 ð xÞ ¼
1x
ð1 x Þ2
Hence
R1 ð xÞ ¼
f 00 ðcÞ 2
x2
:
x ¼
2!
ð1 cÞ3
and
f 00 ð xÞ ¼
2
ð1 xÞ3
:
Solutions to the problems
447
To calculate the value of c, we use Taylor’s Theorem, with n ¼ 1
f ðxÞ ¼ f ðaÞ þ f 0 ðaÞðx aÞ þ R1 ð xÞ:
This gives
f
that is
and so
3
3
3
¼ f ð0Þ þ f 0 ð0Þ þ R1
;
4
4
4
32
3
4
4¼1þ þ
;
4 ð1 cÞ3
32
9
4
¼
:
4 ð1 c Þ3
It follows that ð1 cÞ3 ¼ 14 ; so that 1 c ¼
113
4
’ 0:630. Thus c ’ 0:370:
2. When f is a polynomial of degree n or less
f ðnþ1Þ ðcÞ ¼ 0;
3.
f ð xÞ ¼ cos x;
0
so that Rn ð xÞ ¼ 0:
f ð0Þ ¼ 1;
f ð xÞ ¼ sin x;
f 0 ð0Þ ¼ 0;
f 00 ð xÞ ¼ cos x;
f 00 ð0Þ ¼ 1;
f 000 ðxÞ ¼ sin x;
f 000 ð0Þ ¼ 0;
f ð4Þ ð xÞ ¼ cos x:
Hence, by Taylor’s Theorem with a ¼ 0, f (x) ¼ cos x and n ¼ 3, we have
1
cos x ¼ 1 x2 þ R3 ð xÞ;
2
where
f ð4Þ ðcÞ 4
x :
R3 ð xÞ ¼
4!
Thus
jcos cj 4
x
jR3 ðxÞj 24
1
x4 :
24
4.
f ð xÞ ¼ loge ð1 þ xÞ;
f ð0Þ ¼ 0;
1
;
f 0 ð0Þ ¼ 1;
f 0 ð xÞ ¼
1þx
1
;
f 00 ð0Þ ¼ 1;
f 00 ð xÞ ¼
ð1 þ x Þ2
2
:
f 000 ðxÞ ¼
ð1 þ x Þ3
Hence, by Taylor’s Theorem with x ¼ 0.02
loge 1:02 ¼ 0 þ ð1 0:02Þ 1
0:022 þ R2 ð0:02Þ
2
¼ 0:0198 þ R2 ð0:02Þ:
Now, for c 2 (0, 0.02)
2 j f ðc Þj ¼ 2:
ð1 þ cÞ3 000
Appendix 4
448
Hence, by the Remainder Estimate with M ¼ 2, we obtain
2
jR2 ð0:02Þj 0:023 ¼ 0:0000026:
3!
It follows that
loge 1:02 ¼ 0:0198 ðto four decimal placesÞ:
5. For any positive integer n
f ðnÞ ð xÞ ¼ cos x
hence
ðnÞ f ðcÞ 1;
or
sin x;
for all c 2 R;
so that M ¼ 1.
It follows that the remainder term in the Remainder Estimate satisfies the
inequality
1
ð0:2Þnþ1 :
jRn ð0:2Þj ðn þ 1Þ!
Now
ð0:2Þ4
ð0:2Þ5
¼ 0:00006 and
¼ 0:0000026 . . .;
4!
5!
so we should try n ¼ 4 in the Remainder Estimate, to be safe.
Here
f ð xÞ ¼ cos x;
f ð0Þ ¼ 1;
0
f ð xÞ ¼ sin x;
f 0 ð0Þ ¼ 0;
f 00 ð xÞ ¼ cos x;
f 00 ð0Þ ¼ 1;
f 000 ð xÞ ¼ sin x;
f 000 ð0Þ ¼ 0;
f ð4Þ ð xÞ ¼ cos x;
f ð4Þ ð0Þ ¼ 1
f ð5Þ ð xÞ ¼ sin x:
Hence, by Taylor’s Theorem with f (x) ¼ cos x, a ¼ 0, x ¼ 0.2 and n ¼ 4, we have
1
1
f ð xÞ ¼ 1 x2 þ x4 þ R4 ð xÞ;
2
24
in other words
1
1
cos 0:2 ¼ 1 ð0:2Þ2 þ ð0:2Þ4 þR4 ð0:2Þ
2
24
¼ 1 0:02 þ 0:00006 þ R4 ð0:2Þ
¼ 0:9801 ðrounded to four decimal placesÞ:
6. Using the same function f as in Problem 5, we find that
f ðpÞ ¼ 1;
f 0 ðpÞ ¼ 0;
f 00 ðpÞ ¼ 1;
f 000 ðpÞ ¼ 0;
f ð4Þ ðpÞ ¼ 1:
It follows that
1
1
T4 ð xÞ ¼ 1 þ ðx pÞ2 ðx pÞ4 :
2
24
The difference between T4(x) and f (x) is
f ð5Þ ðcÞ
ðx p Þ5
5!
sin c
¼
ðx pÞ5 ;
120
R4 ð xÞ ¼
Solutions to the problems
449
for some number c between p and x. Hence, for x 2
1 p5
jR4 ðxÞj 120 4
¼ 0:00249 . . . < 3 103 :
3
5
4 p; 4 p
, we have
Section 8.3
1. (a) Applying the Ratio Test with an ¼ 2n þ 4n, we obtain
anþ1 2nþ1 þ 4nþ1 2 þ 2nþ2
a ¼ 2n þ 4n ¼ 1 þ 2n
n
1n
þ2
¼ 2 21n
2 þ1
! 2 2 ¼ 4 as n ! 1:
Hence the series has radius of convergence 14 :
2
ðn!Þ
(b) Applying the Ratio Test with an ¼ ð2nÞ!
, we obtain
1
1
anþ1 ¼ ðn þ 1Þðn þ 1Þ ¼ 1 þ n1 þ n
an ð2n þ 1Þð2n þ 2Þ
2 þ 1n 2 þ 2n
1
as n ! 1:
!
4
Hence the series has radius of convergence 4.
(c) Applying the Ratio Test with an ¼ n þ 2n, we obtain
anþ1 n þ 1 þ 2n1 1 þ 1n þ n21nþ1
¼
a ¼
n þ 2n
1 þ n21n
n
! 1 as n ! 1:
Hence the series has radius of convergence 1.
(d) Applying the Ratio Test with an ¼ 1 1n ; we obtain
ðn!Þ
pffiffiffiffiffiffiffiffi1n n
1
n
anþ1 2pn e
ð
n!
Þ
¼
1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 a nþ1
n
ððn þ 1Þ!Þ
2pðn þ 1Þ nþ1 nþ1
e
1 12
pffiffiffiffiffiffi1n
nn
2p
n
¼ pffiffiffiffiffiffi 1 1 1
n
þ
1
nþ1
2
2p
ððn þ 1Þnþ1 Þ
! 1 as n ! 1:
(Here we have used Stirling’s Formula to estimate n! and (n þ 1)!.)
Hence the series has radius of convergence 1.
2. (a) Applying the Ratio Test with an ¼ n, we obtain
anþ1 n þ 1
a ¼ n ! 1 as n ! 1:
n
Hence the series has radius of convergence 1.
Thus the series converges if x 2 (1, 1), and diverges if jxj > 1.
When x ¼ 1, the sequence {nxn} is non-null; hence, by the Non-null Test, the
series diverges.
Hence the interval of convergence of the series is (1, 1).
(b) Applying the Ratio Test with an ¼ n31n , we obtain
Appendix 4
450
anþ1 n3n
1
an ¼ ðn þ 1Þ3nþ1 ¼ 1 þ 13
n
1
!
as n ! 1:
3
Thus the series converges if x 2 (3, 3), and diverges if jxj > 3.
1
P
1
When x ¼ 3, the series is
n, which is divergent.
n¼1
1
P
When x ¼ 3, the series is
n¼1
ð1Þn
n ,
which is convergent.
Hence the interval of convergence of the series is [3, 3).
3. Applying the Ratio Test with an ¼ ð1Þ: ...n!:ðnþ1Þ, we obtain
anþ1 n n 1
¼
¼
a n þ 1 1 þ 1 n
n
! 1 as n ! 1:
Hence the series has radius of convergence 1.
4. From the Hint, we know that tan1 x is the sum function of the series
1
X
ð1Þn
n¼0
2n þ 1
x2nþ1 ¼ x x3 x5 x7
þ þ ;
3
5
7
for j xj < 1:
()
From the Hint, we also know that the radius of convergence of this power series is 1.
It follows that, by applying Abel’s Limit Theorem to the power series (), we may
obtain the following
1
X
ð1Þn
lim tan1 x ¼
;
x!1
2n þ 1
n¼0
provided that this last series is convergent; it is indeed convergent, by the Alternating
Test for series.
Since the function x 7! tan1 x is continuous at 1, it follows that
1
X
ð1Þn
¼ lim tan1 x
x!1
2n þ 1
n¼0
p
¼ tan1 1 ¼ :
4
Section 8.4
1. (a) We know that
ex ¼ 1 þ x þ
x2 x3
xn
þ þ þ þ ;
2! 3!
n!
for x 2 R;
and so
ex ¼ 1 x þ
x2 x3
xn
þ þ ð1Þn þ ;
2! 3!
n!
for x 2 R:
Hence, by the Sum and Multiple Rules for power series
1
sinh x ¼ ðex ex Þ
2
x3
x2nþ1
þ ;
¼ x þ þ þ
3!
ð2n þ 1Þ!
This series converges for all x.
for x 2 R:
For x 7! tan1 x is continuous
on its domain R.
Solutions to the problems
451
(b) We know that
loge ð1 xÞ ¼ x x2 x3
xn
;
2
3
n
for jxj < 1;
and
1
¼ 1 þ x þ x2 þ x3 þ þ xn þ ;
1x
for j xj < 1:
Hence, by the Sum and Multiple Rules for power series
2
loge ð1 xÞ þ
1
x
x2 x3
xn
¼ x þ 2 þ 2x þ 2x2 þ 2x3 þ þ 2xn þ 2
3
n
3 2 5 3
2n 1 n
x þ ; for jxj < 1:
¼ 2 þ x þ x þ x þ þ
2
3
n
The radius of convergence of this series is 1.
2. (a) We know that
x3
x2nþ1
þ þ
þ ;
3!
ð2n þ 1Þ!
for x 2 R;
x3
ð1Þn x2nþ1
þ þ
þ ;
3!
ð2n þ 1Þ!
for x 2 R:
sinh x ¼ x þ
and that
sin x ¼ x Hence, by the Sum and Multiple Rules for power series
sinh x þ sin x
x3
x2nþ1
x3
ð1Þn x2nþ1
þ þ x þ þ
þ ¼ x þ þ þ
ð2n þ 1Þ!
ð2n þ 1Þ!
3!
3!
x5
x4nþ1
þ ; for x 2 R:
¼ 2 x þ þ þ
ð4n þ 1Þ!
5!
This series converges for all x.
(b) We know that
loge ð1 þ xÞ ¼ x x2 x3
xn
þ þ ð1Þnþ1 þ ;
2
3
n
for j xj < 1;
and
loge ð1 xÞ ¼ x x2 x3
xn
;
2
3
n
for jxj < 1:
Hence, by the Sum and Multiple Rules for power series
1þx
loge
1x
¼ loge ð1 þ xÞ loge ð1 xÞ
x2 x3
xn
x2 x3
xn
¼ x þ þð1Þnþ1 þ x 2
3
n
2
3
n
x3
x2nþ1
¼ 2 x þ þ þ
þ ; for jxj < 1:
3
2n þ 1
The radius of convergence of this series is 1.
(c) We know that
1
¼ 1 þ x þ x2 þ x3 þ þ xn þ ;
1x
for jxj < 1:
Appendix 4
452
It follows, by replacing x by 2x2, that
2 3
n
1
¼ 1 þ 2x2 þ 2x2 þ 2x2 þ þ 2x2 þ 1 þ 2x2
¼ 1 2x2 þ 22 x4 23 x6 þ þ ð1Þn 2n x2n þ ;
for j2x2j < 1; that is, for j xj < p1ffiffi2.
The radius of convergence of this series is p1ffiffi2.
3. (a) We know that
loge ð1 þ xÞ ¼ x x2 x3
xn
þ þ ð1Þnþ1 þ ;
2
3
n
for jxj < 1:
Hence, by the Product Rule
ð1 þ xÞ loge ð1 þ xÞ
x2 x3
xn
¼ x þ þð1Þnþ1 þ 2
3
n
3
4
x
x
xn
þ x2 þ þ ð1Þn
þ 2
3
n1
n
x2 x3
x
þ ; for jxj < 1:
¼ x þ þ ð1Þn
2
6
nðn 1Þ
The radius of convergence of this series is 1.
(b) We know that
1
¼ 1 þ x þ x2 þ x3 þ þ xn þ ;
1x
for j xj < 1;
and (from Example 1) that
1þx
¼ 1 þ 3x þ 5x2 þ þ ð2n þ 1Þxn þ ;
ð1 x Þ2
for j xj < 1:
Hence, by the Product Rule, we obtain
1þx
ð1 x Þ3
1
1þx
¼
1 x ð1 x Þ2
¼ 1 þ x þ x2 þ x3 þ þ xn þ 1 þ 3x þ 5x2 þ þ ð2n þ 1Þxn þ ¼ 1 þ ð3 þ 1Þx þ ð5 þ 3 þ 1Þx2 þ ð7 þ 5 þ 3 þ 1Þx3 þ þ ðð2n þ 1Þ þ ð2n 1Þ þ þ 1Þxn þ ¼ 1 þ 4x þ 9x2 þ 16x3 þ þ ðn þ 1Þ2 xn þ ;
for j xj < 1:
(We can prove, by Mathematical Induction, that
n
X
1 þ 3 þ 5 þ þ ð2n þ 1Þ ¼
ð2k þ 1Þ ¼ ðn þ 1Þ2 ;
for n 2 N:Þ
k¼0
The radius of convergence of this series is 1.
4. (a) We know that
ð1 xÞ1 ¼ 1 þ x þ x2 þ x3 þ þ xn þ ;
for j xj < 1:
Hence, by the Differentiation Rule, we obtain
ð1 xÞ2 ¼ 1 þ 2x þ 3x2 þ 4x3 þ þ nxn1 þ ;
The radius of convergence of this series is 1.
for j xj < 1:
Solutions to the problems
453
(b) If we differentiate the series in part (a), by the Differentiation Rule for power
series, we obtain
2ð1 xÞ3 ¼ 2 þ 6x þ 12x2 þ þ nðn 1Þxn2 þ ;
for jxj < 1:
Hence, by the Multiple Rule, we have
nðn 1Þ n2
x
þ ;
2
The radius of convergence of this series is 1.
(c) Notice that, for the function f (x) ¼ tanh1 x, j x j < 1 , we have
ð1 xÞ3 ¼ 1 þ 3x þ 6x2 þ þ
for j xj < 1:
1
¼ 1 þ x2 þ x4 þ þ x2n þ :
1 x2
Hence, by the Integration Rule for power series, the Taylor series for f at 0 is
f 0 ð xÞ ¼
x3 x5
x2nþ1
þ þ þ
þ ;
3
5
2n þ 1
since f (0) ¼ 0, it follows that c ¼ 0. Hence
tanh1 x ¼ c þ x þ
tanh1 x ¼ x þ
x3 x5
x2nþ1
þ þ þ
þ ;
3
5
2n þ 1
for j xj < 1;
for j xj < 1:
The radius of convergence of this series is 1.
5. We know that
ex ¼ 1 þ x þ
x2 x3
xn
þ þ þ þ ;
2! 3!
n!
for x 2 R;
and we know, from part (a) of Problem 4, that
ð1 xÞ2 ¼ 1 þ 2x þ 3x2 þ 4x3 þ þ ðn þ 1Þxn þ ;
for j xj < 1:
Hence, by the Product Rule for power series, we have
x2 x3
2
x
e ð1 xÞ ¼ 1 þ x þ þ þ 1 þ 2x þ 3x2 þ 4x3 þ 2
6
1
¼ 1 þ ð2 þ 1Þx þ 3 þ 2 þ x2 þ 2
11 2
¼ 1 þ 3x þ x þ ; for jxj < 1:
2
The radius of convergence of this series is 1.
6. We are given that
f ð xÞ ¼ x þ
x3
x5
x2nþ1
þ
þ þ
þ ;
1:3 1:3:5
1:3: ::: :ð2n þ 1Þ
for x 2 R:
(a) By the Differentiation Rule, we can differentiate the power series term-by-term;
this gives
f 0 ð xÞ ¼ 1 þ
x2 x4
x2n
þ
þ þ
þ ;
1 1:3
1:3: ::: :ð2n 1Þ
for x 2 R:
(b) It follows, from the definition of f, that
xf ð xÞ ¼ x2 þ
x4
x6
x2nþ2
þ
þ þ
þ ;
1:3 1:3:5
1:3: ::: :ð2n þ 1Þ
Hence, by the Combination Rules for power series,
f 0 ð xÞ xf ð xÞ ¼ 1;
since all the other terms cancel out.
for x 2 R:
Appendix 4
454
7. Since
ex ¼ 1 þ x þ
x2 x3
xn
þ þ þ þ ;
2! 3!
n!
for x 2 R;
we may deduce that
2
ex ¼ 1 x2 þ
x4 x6
x2n
þ þ ð1Þn
þ ;
2! 3!
n!
for x 2 R:
It follows, from the Integration Rule for power series, that
1
Z 1
x3
x5
x7
x2nþ1
2
þ þ ð1Þn
þ ex dx ¼ x þ
3 5 2! 7 3!
ð2n þ 1Þ n!
0
0
1 1
1
1
n
þ :
¼ 1 þ þ þ ð1Þ
3 10 42
ð2n þ 1Þ n!
8. By the General Binomial Theorem
1 1
X
1
4 xn ;
ð1 þ 6xÞ4 ¼
n
n¼0
where
1
4
n
¼
for j6xj < 1;
1 3 7 1
4 4 4 ... 4 n þ 1
:
n!
It follows that
1 3 7
34
4 4
ð6xÞ2 þ 4
ð6xÞ3 þ 1
2
6
3
27
189 3
1
x ; for jxj < :
¼ 1 þ x x2 þ
2
8
16
6
The radius of convergence of this series is 16.
1
ð1 þ 6xÞ4 ¼ 1 þ
1
4
1
ð6xÞ þ
4
9. (a) By the General Binomial Theorem
1 X
1
12
ðxÞn ;
ð1 xÞ2 ¼
n
n¼0
where
for j xj < 1;
1 3 5 1
2 2 2 ... 2 n þ 1
12
:
¼
n
n!
It follows that
1
1
3
ð1 xÞ2 ¼ 1 þ x þ x2 þ þ ð1Þn
2
8
12 n
x þ ; for j xj < 1:
n
The radius of convergence of this series is 1.
(b) We know that
d
1
sin1 x ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ;
dx
1 x2
for x 2 ð1; 1Þ:
Then, by the result of part (a), with x replaced by x2, we obtain
1
1
1
3
2 2n
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1 þ x2 þ x4 þ þ ð1Þn
x þ ; for j xj < 1:
2
n
2
8
1x
Hence, by the Integration Rule for power series
1
3
sin1 x ¼ c þ x þ x3 þ x5 þ 6
40
1 2nþ1
2 x
þ ; for j xj < 1:
þ ð1Þn
n 2n þ 1
Solutions to the problems
455
Putting x ¼ 0 into this equation, we see that c ¼ 0. It follows that
1
3
sin1 x ¼ x þ x3 þ x5 þ 6
40
1 2nþ1
2 x
þ ;
þ ð1Þn
n 2n þ 1
for j xj < 1:
The radius of convergence of this series is 1.
Section 8.5
1. We use the Addition Formula
tan1 x þ tan1 y ¼ tan1
xþy
;
1 xy
()
which holds provided that tan1 x þ tan1 y lies in the interval p2 ; p2 .
First, we deduce, from the Addition Formula (), that
!
1
1
1 1
1 1
1
3þ4
þ tan
¼ tan
tan
3
4
1 13 14
4þ3
¼ tan1
12 1
7
:
¼ tan1
11
This equation holds, since
1
1
þ tan1
’ 0:3218 þ 0:2450 ¼ 0:5668;
tan1
3
4
and 0:5668 2 p2 ; p2 ’ ð1:5708; 1:5708Þ:
Next, we deduce, from the Addition Formula (), that
1
1
2
7
2
þ tan1
þ tan1
¼ tan1
þ tan1
tan1
3
4
9
11
9
!
7
2
11 þ 9
¼ tan1
7
1 11
29
63 þ 22
¼ tan1
99 14
p
1
¼ tan ð1Þ ¼ :
4
This equation holds, since
7
2
þ tan1
’ 0:5667 þ 0:2187 ¼ 0:7854;
tan1
11
9
and 0:7854 2 p2 ; p2 ’ ð1:5708; 1:5708Þ:
2. First
1
px dx
2
1
1
2
1
4
¼ :
sin px
¼
p
2
p
1
I0 ¼
Z
1
cos
456
It follows that p I0 ¼ 4.
Next, using integration by parts twice, we obtain
Z 1
1
2
1 x cos px dx
I1 ¼
2
1
1 Z 1
2
1
2
1
¼ 1 x2 sin px
ð2xÞ sin px dx
p
2
p
2
1 1
Z
4 1
1
¼
x sin px dx
p 1
2
1
Z 4
2
1
4 1
2
1
x cos px
cos px dx
¼
p
p
2
p
2
1 p 1
1
8 2
1
¼ 2 sin px
p p
2
1
32
¼ 3:
p
It follows that p 3I1 ¼ 32.
Appendix 4
Index
ax, definition and continuity, 161
Abel’s limit theorem, 333
absolute convergence test, 104
absolute convergence theorem, 332, 337
absolute value, 12
absolutely convergent series, 104
addition formula for tan1, 166
alternating test, 107
antipodal points theorem, 145
approximation by Taylor
polynomials, 317
Archimedean property of R, 7
arithmetic in R, 8, 30
arithmetic mean – geometric mean
inequality, 18, 21
asymptotic behaviour of functions, 176
basic
continuous functions, 142
differentiable functions, 224
null sequences, 48
power series, 325
series, 97
Bernoulli’s inequality, 20, 237
Bessel function, 329
Bijection, 355
binomial theorem, 358, 342
blancmange function, 244
Bolzano–Weierstrass theorem, 70
bound, greatest lower, 26
bound, least upper, 25
boundedness theorem, 62, 92, 149
Cauchy condensation test, 97
Cauchy’s mean value theorem, 239
Cauchy–Schwarz inequality, 20, 311
Chain Rule, 218
combination rules
continuous functions, 136
convergent sequences, 55
convergent series, 89
differentiable functions, 216
functions which tend to 1, 177
inequalities, 14
infimum, 275
integrable functions, 278
limits of functions, 172
null sequences, 46
power series, 338
primitives, 284
sequences which tend to 1, 65
series, 89
supremum, 275
tilda, 305
common limit criterion, 268
common refinement, 262
comparison test, 94
composition rule
asymptotic behaviour of functions, 180
continuous functions, 136
differentiable functions, 218
limits of functions, 173
conditionally convergent series, 104
continuity
continuity, 132, 194
limits of functions, 171
one-sided, 134, 194
uniform, 201
continuous functions
basic, 142
boundedness theorem, 149
combination rules, 136
composition rule, 136
extreme values theorem, 169
integrability, 277
intermediate value theorem, 143
inverse function rule, 153
squeeze rule, 137
convergence, interval of, 330
radius of, 330
convergent sequence, 53
combination rules, 55
quotient rule, 55
squeeze rule, 58
convergent series, 85
combination rules, 89
cos1, 156
cosh, power series, 339
cosh1, 158
cosine function
derived function, 214
power series, 326
Darboux’s theorem, 254
decimal representation of numbers, 3, 87
density property of R, 7
derivatives, 207
higher order, 215
one-sided, 210
standard, 359
difference quotient, 206
differentiability of
f(x) ¼ e x, 214
f(x) ¼ a x, 223
f(x) ¼ x, 223
f(x) ¼ x x, 223
trigonometrical functions, 214
differentiable functions
combination rules, 216
composition rules, 218
inequalities involving, 237
inverse function rule, 221
differentiation, 210
rule for power series, 341
Dirichlet’s function, 197, 310
discontinuity, removable, 172
divergent sequence, 61
series, 85
domination hierarchy, 56
e, definition, 74
irrationality, 175
" game, 43, 187
Euclidean algorithm, 78
even subsequence, 66
ex, definition, 75, 122
fundamental property, 127
inverse property, 75
exponent laws, 163
exponential function
continuity, 141
derived function, 214
differentiability, 223
inequalities, 141
power series, 326
extreme values theorem, 149
extremum, 228
field, 9
first subsequence rule, 67
function, nowhere differentiable, 244
fundamental inequality for integrals, 288
fundamental theorem of Algebra, 146
of Calculus, 283, 292
general binomial theorem, 342
geometric series, 87
457
Index
458
Goethe, ix
greatest lower bound, 26, 272
property of R, 29
Gregory’s series for estimating p, 346
harmonic series, 93, 109
higher-order derivatives, 215
hyperbolic functions, differentiability, 224
hypergeometric series, 352
increasing–decreasing theorem, 234
inequalities, power rule, 10
rules for integrals, 289
infimum, 26, 272
infinite series, 85
inheritance property of subsequences, 66
Int I, 234
Integrability, 264
combination rules, 278
continuous functions, 277
integral test, 297
modulus rule, 278
monotonic functions, 276
Riemann’s criterion, 267
integral, 264
additivity, 280
inequality rules, 289
lower, 264
upper, 264
integration by parts, 285
by substitution, 286
integration rule for power series, 341
integration, reduction of order
method, 293
intermediate value theorem, 143
interval image theorem, 149
interval of convergence, 230
inverse function rule for continuous
functions, 153
differentiable functions, 221
inverse hyperbolic functions, 158
inverse trigonometric functions, 156
irrational numbers, 5
K" lemma, 49, 193
least upper bound, 25, 262
property of R, 29
Leibniz notation, 207
Leibniz test, 107
Leibniz’s series for estimating p, 346
l‘Hôpital’s rule, 241
limit comparison test, 95
limit inequality rule for limits, 175
sequences, 60
limit of a function, 169, 177, 185
as x ! 1, 177
one-sided, 175
limit of a sequence, 53, 182
limits of functions and continuity, 171
limits of functions, combination
rules, 172
composition rule, 173
one-sided, 175
squeeze rule, 174
local extremum theorem, 229
loge ð1 þ xÞ, power series for, 325
lower integral, 264
lower Riemann sum, 259
Maclaurin integral test, 297
mathematical induction, 357
maximum, 23
mean value theorem, 233
minimum, 24
modulus function, continuity, 135
modulus rule for integrable
functions, 278
modulus, 12
monotone convergence theorem, 68
monotonic function, 153
integrability, 276
monotonic sequence theorem, 69
multiplication of series, 114
n!, Stirling’s formula for, 306
neighbourhood, 169
non-null test, 91
nth partial sum of series, 85
nth root function, continuity, 155
nth root of a positive real number,
32, 155
null partition criterion, 269
null sequence, 43
odd subsequence, 66
one-one function, 152, 355
one-sided derivative, 210
one-sided limit, 175
onto function, 355
order properties of R, 7
p, 76
irrationality of, 348
numerical estimates, 346
partial sum of series, 85
partition of [a, b], 258
standard, 258
power series
absolute convergence, 332
basic, 325
combination rules, 338
differentiation rule, 341
integration rule, 341
uniqueness theorem, 342
primitives, 282
combination rules, 278
scaling rule, 278
standard, 360
uniqueness theorem, 284
radius of convergence theorem, 329
ratio test, 96
for radius of convergence, 330
rational function, continuity, 136
differentiability, 217
rational power, 34
rearrangement of series, 109
reciprocal rule for
functions which tend to 1, 177
sequences which tend to 1, 64
tilda, 305
recursion formula, 71
reduction of order method, 293
refinement, 262
remainder estimate, 322
removable discontinuity, 172
Riemann’s rearrangement theorem, 112
Riemann’s function, 198
Riemann’s-criterion for integrability, 267
Rolle’s theorem, 231
scaling rule for primitives, 284
second derivative test, 236
second subsequence rule, 67
sequences, 38
basic null, 48
combination rules, 55
convergent, 53
divergent, 61
limit inequality rule, 60
monotonic, 40
null, 43
unbounded, 62
which tend to 1, 63
which tend to 1, 65
squeeze rule, 58
series, 85
absolutely convergent, 104
basic, 97
convergent, 85
divergent, 85
geometric, 87
harmonic, 93, 109
hypergeometric, 352
integral test, 297
multiplication, 114
product rule, 114
Taylor, 325
telescoping, 88
sine function
derived function, 214
power series, 326
sine inequality, 140
sin1, 156
sinh, 158
power series, 339
square root function, continuity, 134
squeeze rule
as x ! 1, 178
continuous functions, 137
Index
convergent sequences, 58
limits of functions, 174
null sequences, 47
sequences which tend to 1, 179
standard derivatives, 359
standard partition, 258
standard primitives, 360
Stirling’s formula, 306
strategy for testing for convergence, 116
sub-interval theorem, 280
subsequence, 66
sum function, 325
supremum, 35
tan1, 157
power series, 341
459
tanh1, 159
power series, 344
tangent approximation, 314
Taylor polynomial, 316
Taylor series, 325
Taylor’s theorem, 320
tilda notation, 300, 304
combination rules, 305
transitive property of R, 7
transitive rule for inequalities, 291
triangle inequality, 15, 16
backwards form, 15
infinite form, 105
integrals, 291
trichotomy property of R, 7
trigonometric functions
continuity, 139
differentiability, 213
unbounded sequence, 62
uniform continuity, 201
uniqueness theorem, power series, 342
primitives, 284
upper integral, 264
upper Riemann sum, 259
Wallis’s formula, 293
Weierstrass, K., x
zero derivative theorem, 235
zero of polynomial, 295
zeros localisation theorem, 147