Calculus II–Math 227.04
David Meredith
Department of Mathematics
San Francisco State University
San Francisco, CA 94132
Fall 2001–MWF 2-3, Tu 12:35-1:25–BUS 110
Contents
1 Introduction
1.1 Instructor . . . . . . . . . . . . .
1.2 Web Site and E-mail . . . . . . .
1.3 Text . . . . . . . . . . . . . . . .
1.4 Software . . . . . . . . . . . . . .
1.4.1 Mathematical Software .
1.4.2 Word Processing Software
1.5 Class procedures . . . . . . . . .
1.5.1 Attendance . . . . . . . .
1.5.2 Schedule . . . . . . . . . .
1.5.3 Homework . . . . . . . . .
1.5.4 Exams . . . . . . . . . . .
1.5.5 Grading . . . . . . . . . .
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3
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5
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8
9
2 Projects
2.1 Learning Mathematica–Due September 28
2.2 Center of a Triangle–Due October 24 . . .
2.3 Measuring a Football–Due November 14 .
2.4 Finding Logarithms–Due December 5 . . .
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10
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3 Lectures
3.1 Introduction to TestGiver . . . . . . . . . . . . .
3.2 Integration . . . . . . . . . . . . . . . . . . . . .
3.2.1 Using Integral Tables . . . . . . . . . . .
3.2.2 Definite Integrals . . . . . . . . . . . . . .
3.2.3 Areas between curves . . . . . . . . . . .
3.2.4 Integration by Substitution . . . . . . . .
3.2.5 Volumes . . . . . . . . . . . . . . . . . . .
3.2.6 Integration by Parts . . . . . . . . . . . .
3.2.7 Work Problems . . . . . . . . . . . . . . .
3.2.8 First Moments and Centers of Mass . . .
3.2.9 Partial Fractions . . . . . . . . . . . . . .
3.2.10 Improper Integrals . . . . . . . . . . . . .
3.2.11 Numerical Integration . . . . . . . . . . .
3.2.12 The Fundamental Theorem of Calculus .
3.2.13 Probability Integrals . . . . . . . . . . . .
3.3 Infinite Series . . . . . . . . . . . . . . . . . . . .
3.3.1 Numerical Series . . . . . . . . . . . . . .
3.3.2 Power Series . . . . . . . . . . . . . . . .
3.3.3 Representing Functions with Power Series
3.3.4 Taylor Polynomials and Remainders . . .
3.3.5 Taylor Series and Convergence . . . . . .
3.3.6 Manipulating Taylor Series . . . . . . . .
3.3.7 Binomial series . . . . . . . . . . . . . . .
3.3.8 Taylor Series and differential equations . .
3.4 Differential Equations . . . . . . . . . . . . . . .
3.4.1 Introduction . . . . . . . . . . . . . . . .
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11
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14
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18
20
25
27
30
33
38
41
48
51
56
56
67
68
72
74
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79
81
81
81
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3.4.2
3.4.3
3.4.4
3.4.5
3.4.6
4 Exams
4.1 Math
4.2 Math
4.3 Math
4.4 Math
Separable Differential Equations . . . . . . . . . . . . . . . . .
Direction Fields . . . . . . . . . . . . . . . . . . . . . . . . . . .
Euler’s Method of Approximating Solutions to Differential Equations y 0 = f (t, y) . . . . . . . . . . . . . . . . . . . . . . . . . .
Exponential Growth and Decay . . . . . . . . . . . . . . . . . .
The Logistic Equation . . . . . . . . . . . . . . . . . . . . . . .
227.04
227.04
227.04
227.04
Midterm
Midterm
Midterm
Midterm
1
1
2
2
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Answers
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Answers
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84
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92
96
. 96
. 97
. 99
. 100
Chapter 1 Introduction
Calculus II divides into two parts: during the first 11 weeks you will study the theory,
methods and applications of integration, including differential equations; during the
last four weeks you will study the theory and application of power series. Along the
way you will learn to use Mathematica as your personal mathematical assistant, and
you will practice your mathematical writing.
Integration is both harder and more useful than differentiation. (Why is that not
surprising?) Since the eighteenth century most scientific laws have been expressed as
differential equations, and integration is the mathematical tool required to use them.
Integration is also widely used in data analysis.
Power series are not as ubiquitous as integration, but they are key to understanding
how calculators and computers do mathematics. Important numerical methods for
evaluating transcendental functions and solving differential equations depend on power
series. Those of you who study advanced mathematics will find that power series play
an important role in advanced analysis.
1.1
Instructor
David Meredith
TH 933
Office Hours: MWF 11-12
(415) 338-2199
E-mail: [email protected]
URL: http://online.sfsu.edu/˜meredith
1.2
Web Site and E-mail
Class notes, handouts, syllabus changes, grade sheet, etc. can be found at the class
web site, which you can access by going to my web site and clicking on Calculus
II. I will not distribute any more paper handouts to you. All information will be
distributed via the class web site.
When you cannot get to my office hour or wait for the next class, you may submit
questions to me by e-mail. Projects may be submitted as e-mail attachments. I can
read Word, Tex, DVI, Adobe Acrobat and Mathematica files.
1.3
Text
Stewart, Calculus: Concepts and Contexts, Brooks/Cole, Pacific Grove CA, 1998.
1.4
Software
1.4.1
Mathematical Software
In this course you will learn to calculate with software. There are several programs
you can use. I will instruct the class to use Mathematica, but I will try to help you
individually if you choose to use another program.
3
1.4.1.1
Maple
Maple is a symbolic and numerical powerhouse similar to Mathematica but not as
widely used outside of academia.
1.4.1.2
Mathematica
P50 1
Mathematica computes numerically and symbolically. You can calculate
i=1 2
i
d 3x
or
e cos 4x. With Mathematica, you can work interactively or you can write
dx
subroutines. The Math Dept. at SFSU has decided to use this program in many of its
courses. Mathematica is freely available M-F, 10-3, in the Math/Stat Computing Lab,
TH404. Buying your own copy is not necessary, but it might be a good investment.
If you learn to use Mathematica effectively, you will find it very valuable in all your
science, engineering and math courses. The bookstore will sell you a copy for $150.
1.4.1.3
Matlab
Matlab is the classic linear algebra program with the best numerical algorithms, and it
can be used with difficulty for calculus. A student version is available at the bookstore.
If you are going into engineering, it might pay to learn this program. Matlab now
includes Maple, so some symbolic calculations can be performed with Matlab. If your
focus is symbolic work, Mathematica or Maple is a better choice.
1.4.1.4
X(PLORE)
Truth in advertising: I wrote X(PLORE). It is freely available from my web site and
in the Math/Stat Computing Lab. You can do most of the calculations for this course
in X(PLORE). The advantage is the cost; the disadvantage is the lack of symbolic
computation.
1.4.1.5
Graphing Calculators
These toys can be used for the simple calculations and graphs. The most recent models will perform symbolic integration and matrix calculations. On the other hand,
calculators are much harder to use than computer software. The keystroke combinations are harder to remember, the screen is much harder to read, and the capabilities
are much less. I have never seen a professional scientist or engineer use a graphing
calculator; they use computers. Maybe you should start learning to use a computer
too. It is possible to do all the computer homework in this course on the most powerful
graphing calculators (TI 89, HP 49), but it won’t be fun.
1.4.2
Word Processing Software
You can write mathematical reports with Mathematica, but not easily. Moreover,
the result is ugly. Much better tools are any standard word processor with equationwriting capability (Word or WordPerfect will do nicely, but you may have to install
the equation module for Word ) or a special scientific work processor like Scientific
WorkPlace ($600) or Scientific Notebook ($75). I highly recommend Scientific Workplace and Scientific Notebook. They include Maple, so you can do simple computations
inside the word processor. The output looks good, much better than the output from
either Mathematica or Word. Sometimes the bookstore stocks Scientific Notebook.
More information about these products can be found at http://www.mackichan.com.
This document was produced with Scientific Workplace.
4
1.5
Class procedures
1.5.1
Attendance
I take attendance every day.
1.5.2
Schedule
Dates
8/29-9/7
9/10-9/14
9/17-9/21
Project 1
9/24-9/28
10/1-10/5
Midterm
10/10-10/12
10/15-10/19
Project 2
10/22-10/26
Midterm
10/31-11/2
11/5-11/9
Project 3
11/12-11/16
11/19-11/21
11/26-11/30
Project 4
12/3-12/7
12/10-12/14
Final
1.5.3
Topics
5.7, 6.1
5.5, 6.2
5.5, 6.2
Due Date
9/10
9/17
9/24
9/28
5.6, 6.5
10/1
App.F, 6.5 10/8
Tuesday
10/9
5.9, 6.5
10/15
5.8, 6.3
10/22
10/24
5.4, 6.7
10/29
Tuesday
10/30
8.1-8.2
11/5
8.2-8.4
11/12
11/14
8.5-8.6
11/19
8.7
11/26
8.8-8.9
12/3
12/5
7.1-7.3
12/10
7.4-7.6
12/17
Monday, 12/17, 1:30-4:00
Homework
You will have two kinds of homework in this course. There will be problems and
projects. Homework is an important part of this course. Forty-five percent of your
grade comes from the homework.
1.5.3.1
Problems
Homework problems will be assigned weekly. They are similar to the exercises in the
text, but the format is new. I’ve developed an electronic homework system called
TestGiver that you will use.
1. You begin by downloading the program TestGiver from the class web site and
installing it on your computer at home. Or you can use TestGiver in the
Math/Stat Lab (TH 404).
(a) TestGiver includes a massive help file, and the folder that contains your
installation also contains a pdf file with a manual for the program. You
should read the section titled “Overview” in the help file or the manual.
2. Then every week, at least seven days before the assignment is due, you
download the weekly assignment file. This will be a file with the extension TGV.
(a) Load your test into TestGiver using the menu item File/Open New Test.
Random items will be given values, and the questions will be displayed on
the Test page.
5
i. If the test is not displayed, click the tab marked ”Test” at the top of
the window.
ii. You should see the questions and spaces for your answers.
iii. You may want to print your test and work away from the computer.
Wise students will keep printed copies of their completed tests. At any
time during the test-taking, a printed copy of the test with all answers
entered so far can be created by selecting the menu item File/Print
Test.
(b) When you enter an answer, it will be checked instantly and you will be told
if it is right or wrong.
i. Most questions allow you to keep trying answers until you get the right
answer. But questions with a yellow background give you only one try.
Think carefully before answering a one-try question.
Usually one-try questions are like true/false questions. They have a
limited number of possible answers, and the one-try rule is there to
keep test takers from just trying all possible answers without actually
doing the problem.
If you enter the wrong answer to a one-try question, you can submit
a written correction that explains the problem thoroughly in complete
sentences. Begin with a statement of the problem.
(c) At any time you can save a partially completed test and reload it later into
TestGiver. All randomly produced parameters and all answers, right and
wrong, are saved. Nothing is changed when the test is reloaded for further
work. Saving is done with the menu item File/Save Test and reloading
is done with the menu item File/Open Old Test. The recommended file
extension for saved tests is TST.
i. WARNING: always reload a TST file that you have saved. If you
reload a TGV file, all answers will be lost and all random parameters
will be recalculated. Loading a TGV file with File/Open New Test
always restarts a test. Loading a TST file with File/Open Old Test
allows you to continue a test you have already started.
ii. If you are doing your work in the lab, you should save your test on a
floppy disk. That way, when you return, you can work at any computer.
(d) When you finish the assignment, send a report of your work to me over the
internet.
i. Get connected to the internet. If you are in the lab, you are already
connected.
ii. Send your report by using the menu item File/Send Report.
iii. Reports must be send on or before the due date. Late papers will be
rejected by the recording software.
A. Falling behind is one of the main reasons students fail calculus.
This rule is designed to help you keep up. It is better to submit a
partially completed assignment and go on to the next one than to
fall behind while trying to complete an old assignment.
B. If you have a good reason for turning in an assignment late, please
see me.
(e) If you have questions about your homework, you can send me an email by
clicking Help/Send e-mail to Dr. Meredith.
1.5.3.2
Projects
1. Projects are formal reports. I will give you a problem, usually with a lot of
parts, and you will write a report based on the problem.
(a) Reports must be written in correct mathematical English. Your textbook
is a model of correct mathematical writing.
6
i. The test of good writing is this: a student in another linear algebra
class at another college should be able to read your paper and understand what problem you were doing and how you answered it.
(b) The most important criteria for a good report is that it is clear and without
errors. It is better to write a grammatical, well-organized report that
correctly answers part of the question than to write a report that purports
to answer the entire question but is either fragmentary or contains mistakes.
i. Think of getting one point for each correct, well-written part and -1
point for each part that is grammatically or mathematically wrong.
You can get a lot of points just by not turning in wrong answers.
ii. There is no partial credit for mistakes.
iii. There is lots of partial credit for correctly stating and answering part
of a problem. If you cannot do a problem as stated, then:
A. do a simpler version. If the problem asks about n × n matrices,
you might solve it for 3 × 3 matrices.
B. do an example. If the problem asks you to prove something about
symmetric matrices, you can show that the property holds for one
symmetric matrix.
C. discuss the problem. Prove what you can and explain why you are
stuck.
The important thing is to write a well-organized report that is internally correct, even if it does not completely answer the problem posed.
(c) You do not have to take up the problem parts in the order I present them.
You can organize your report any way you like, so long as it is clear.
i. You can use a part that you cannot prove to prove another part, so
long as you are clear about what you are doing.
2. Rules for Writing Projects
(a) Things to do
i. Your paper should read from left to right and top to bottom. If your
readers have to follow a wandering path down the page, they will be
confused.
A. Do not put little clusters of words on the page.
B. Start each paragraph at the left margin, but do not start each sentence at the left margin. Write paragraphs, not lists of sentences.
ii. Every word on your page must be part of a sentence.
A. Pick a word at random on your page. Can you find the capital
letter that starts the sentence containing it? Can you find the
period that ends the sentence? Can you find the subject and verb?
iii. Each part of your project should begin with a statement of what you
will prove.
A. Do not assume that your reader has a copy of the problem statement. Although you can refer to theorems in the text by number
(say: “by Lipschutz, 5.3.2”). you have to state completely what
each part of your report proves.
B. Do not simply copy the problem statement. The problem statement asks the reader to do something, which is not what you want.
You want to say what you will calculate or prove, so rewrite the
problem statement as a positive statement of what you will do.
C. If you are only doing part of a problem or a special case, then your
introductory statement should say exactly what you will do (not
what you won’t do). If the problem asks you to prove something for
n points and you do the problem for 3 points, then your statement
should start: “Let M be a set of three points. Then ...”.
7
iv. Finish each part with a conclusion.
v. If you type, you must still use standard mathematical notation. If you
cannot get your word processor to print fractions, exponents, etc., then
write the formulas in by hand.
vi. Good mathematical writing is clear but succinct. Don’t overdo long
calculations. Put in enough steps so another student could repeat your
calculation, but no more.
(b) Things to Avoid
i. NEVER use arrows. They are not well-defined mathematical symbols. They have no meaning. Use words and mathematical formulas
to convey your meaning.
A. If one system of equations is derived from another, don’t just put
an arrow to one from the other. Put in a sentence that says how
the systems are related, for example: “Eliminating x from all but
the first equation, we obtain the system:..”
ii. Never use a triangle of three dots to replace the word “therefore”.
A. Never use a triangle of three dots for anything else either.
iii. Do not use abbreviations or other informal grammatical elements. Prepare a manuscript that you would be proud to see published.
iv. Never turn in a paper that includes unfinished problems, fragments,
scratchwork, or other clutter. The reader will try to read everything
on your page, so make sure that everything on your page is clear and
complete.
A. Do not turn in first drafts. Solve problems first on scratch paper,
then write your final answer on the paper you turn in.
3. Projects may be turned in by groups of up to three students.
(a) If two projects contain significant copied parts, both will receive zeros. I
do not care who copied from whom.
4. Projects will be graded on a scale of 0-4
(a) 0: not submitted, no attempt to write complete sentences, or no correct
parts.
(b) 1: many mathematical or grammatical errors
(c) 2: some parts done correctly, not too many mathematical or grammatical
errors
(d) 3: most parts done correctly and few mistakes, writing is grammatically
correct
(e) 4: almost all parts done correctly and no mistakes, writing is clear and
well-organized
5. Late papers will not be accepted except by prior arrangement.
(a) If you cannot come to class, have a friend bring your work or send it by
email.
(b) Write your work up as you go along. If something happens and you cannot
finish, you will have a partially completed assignment to turn in.
1.5.4
Exams
Exam problems will be similar to the easier homework problems. Exams must be take
individually. Bring plenty of blank 8 12 ” × 11” paper.
8
1.5.4.1
Midterms
There will be two midterms: October 9 and October.30. You may use one 8 12 ” × 11”
page of notes (both sides) for each test.
1.5.4.2
Final
The final will be in the regular lecture room. You may use two 8 12 ” × 11” page of
notes (both sides) for the final. The final exam will cover the entire course.
1.5.5
Grading
Grading System
Attendance 10%
Problems 25%
Projects 20%
Midterms 25%
Final Exam 20%
Final grades will be assigned according to a scale no harsher than: A ≥ 85%,
B ≥ 70%, C ≥ 60%.
9
Chapter 2 Projects
2.1
Learning Mathematica–Due September 28
From the class web site download the Mathematica notebook “Mathematica for Calculus Students”. Do all the “do it yourself” sections, delete all other sections, and
send the resulting notebook to me by e-mail. The usual literacy requirements for
projects are not in force for this part of the assignment.
2.2
Center of a Triangle–Due October 24
Let x, y, z be points in the plane considered as the vertices of a triangle. Show that
x+y+z
is the centroid of the triangle.
3
2.3
Measuring a Football–Due November 14
Carefully measure the circumference of a football at various points along its axis. Use
your measurements to estimate the volume of the football (think Simpson’s method).
Carefully explain your method of computation: exactly what measurements you made
and how you converted them into a volume. Check your work by measuring the volume
some other way, for example by determining how much water the football can displace.
2.4
Finding Logarithms–Due December 5
Assemble the following points into a coherent essay about calculating the logarithm
of any positive number.
1. Using the Maclaurin series for
1
, find the series of ln (1 − x) and ln (1 + x).
1−x
1+x
1+x
2. Use the fact that ln
= ln (1 + x) − ln (1 − x) to find the series for ln
1−x
1−x
.
1+x
1
converges for |x| < 1. Explain why your series for ln
1−x
1−x
converges for |x| < 1.
3. The series for
4. Show if c > 0 then we can find x in the domain −1 < x < 1 such that
1+x
= c.
1−x
5. Conclude that your series will calculate ln c for any positive c. Use the first
ten terms of your series (with Mathematica) to approximate ln (0.5), ln 2, ln 10.
How accurate are your answers?
10
Chapter 3 Lectures
3.1
Introduction to TestGiver
1. I’ve written a computer program TestGiver to “help” you do your homework.
(a) Truth in advertising: TestGiver helps me by automatically grading and
recording your homework
(b) TestGiver helps you in two ways:
i. TestGiver tells you if your answer is right or wrong.
A. Usually you can keep trying for a right answer
ii. TestGiver includes computational and graphical tools
2. You will download TestGiver from the class website and install it on your PC,
or you can use it in the Math/Stat lab in TH 409.
3. You will download weekly homework assignments from the class website.
(a) You can work on an assignment, save your work, and continue at a later
time.
(b) You can print your homework, work out the answers, then go a computer
to enter them. You do not have to do all your work on the computer.
(c) When you have finished your homework, you send it to me over the internet
with one click of the mouse.
4. If using this system causes a hardship for you, please see me. We’ll work out a
different homework system for you.
3.2
Integration
3.2.1
Using Integral Tables
1. Class: define
Z
(a)
f (x) dx
(b)
Z
b
f (x) dx
a
2. Answers
(a) any function whose derivative is f (x)
(b) (informal) the area under the curve y = f (x) between x = a and x = b,
counting area below the x-axis as negative.
(c) F (b) − F (a) is WRONG answer to (b). The fundamental theorem of calculus is a method for evaluating integrals, not the definition of the integral.
i. Lots of functions don’t have antiderivatives, but they can still be integrated.
Z 1p
ii.
x3 + 1dx cannot be evaluated by antiderivatives, but it is still
0
defined.
3. Most problem for the first part of this course reduce to evaluating integrals.
11
4. In back of book is three page table of integrals, which you can copy and bring
to all exams.
(a) All methods of integation are there to change an integral to one in the
table. If the integral is already in the table, then use the table.
Z √
1 + t2
dt
5. Example:
t2
(a) formula 24:
√
Z √ 2
´
³
p
a + u2
a2 + u2
du
=
−
+ ln u + a2 + u2 + C
2
u
u
(b) Let a = 1, u = t, then
√
Z √
³
´
p
1 + t2
1 + t2
2 +C
+
ln
t
+
dt
=
−
1
+
t
t2
t
(c) Check by differentiation:
! p
à √
´
³
p
(1 + t2 )
1 + t2
d
=
−
+ ln t + 1 + t2
dt
t
t2
(d) TestGiver: sqrt(1+t^2)/t^2
(e) Mathematica: Sqrt[1+t^2]/t^2
6. Class: Check answers by differentiation. Write in TestGiver notation.
Z
dx
(a)
4 + x2
³x´
1
i. arctan
+C
2
2
ii. TestGiver: 1/2*atan(x/2)
iii. Mathematica: 1/2*ArcTan[x/2]
Z
(b)
cos2 x dx
x sin 2x
+
+C
2
4
ii. TestGiver: x/2+sin(2*x)/4
iii. Mathematica: x/2+Sin[2x]/4
i.
7.
Z
sin3 x dx
(a) Formula 73.
Z
1
n−1
sin u du = − sinn−1 u cos u +
n
n
n
(b) Let n = 3, u = x. Then
Z
1
sin3 x dx = − sin2 x cos x +
3
1
= − sin2 x cos x −
3
(c) Check:
d
dx
µ
Z
sinn−2 u du
Z
2
sin x dx
3
2
cos x + C
3
¶
2
1
− sin2 x cos x − cos x = sin3 x
3
3
12
(d) TestGiver: -1/3*sin(x)^2*cos(x)-2/3*cos(x)
(e) Mathematica: -1/3 Sin[x]^2 Cos[x]-2/3 Cos[x]
Z
8. Class: cos4 x dx. Check.Write for TestGiver.
1
3
3
cos3 x sin x + cos x sin x + x + C
4
8
8
(b) 1/4*cos(x)^3*sin(x)+3/8*cos(x)*sin(x)+3/8*x
Z √
2 − 3x
9. Class:
dx Check.Write for TestGiver.
x
¯√
¯
¯ 2 − 3x − √2 ¯
√
√
¯
¯
√ ¯+C
(a) 2 2 − 3x + 2 ln ¯ √
¯ 2 − 3x + 2 ¯
(a)
(b) 2*sqrt(2-3*x)+sqrt(2)*ln(abs((sqrt(2-3*x)-sqrt(2))/(sqrt(2-3*x)+sqrt(2))))
Z
1
dx
10. Warning:
9 + 4x2
(a) Formula 17.
Z
1
du
u
du = tan−1 + C
2
+u
a
a
(b) So you might think you could say:
a2
a = 3
u
= 2x
Z
1
2x
1
dx =
tan−1
+C
2
9 + 4x
3
3
But this is wrong because it does not work. Checking by differentiation
shows that this answer is wrong.
µ ¶0
µ
¶
2x
1
1
d 1
−1 2x
³
´
tan
=
dx 3
3
3 1 + ¡ 2x ¢2
3
3
!µ ¶
Ã
1
2
1
=
2
9+4x
3
3
9
18
9 (9 + 4x2 )
2
=
9 + 4x2
In the integration formula #17, the expression that is squared u and the
expression after the differential du are the same. In the problem the expression that is squared 2x and the expression after the differential dx are
not the same.
(c) The trick is to reduce the problem to one with a simple variable squared in
the denominator. This is a special case of the method of substitution.
=
w
dw
Z
1
dw
2
= 2x
= 2dx
= dx
Z
1 1
1
dx
=
dw
2
9 + 4x
9 + w2 2
Z
1
1
dw
=
2
9 + w2
µ
¶
1 1
w
=
tan−1 + C
2 3
3
2x
1
tan−1
+C
=
6
3
13
TestGiver: 1/6*atan(2*x/3)
Mathematica: 1/6 ArcTan[2x/3]
i. Check:
d
dx
µ
1
2x
tan−1
6
3
¶
=
=
µ
¶0
2x
3
3
µ ¶
1
2
1
¡ 2x ¢2
61+
3
1
1
¡ 2x ¢2
61+
3
=
=
2
18
¡ 9+4x2 ¢
9
1
9 + 4x2
11. Class. Check and write for TestGiver.
Z
1
sin 4x + C
cos (4x) dx =
4
= 1/4*sin(4*x)
Z
x
1
1
x sin dx = 4 sin x − 2x cos x + C
2
2
2
= 4*sin(x/2)-2*x*cos(x/2)
Z
1
1
ln (3x + 1) dx =
(3x + 1) ln (3x + 1) − x − + C
3
3
=
1/3*(3*x+1)*ln(3*x+1)-x-1/3
Z
1p
sin−1 5t dt = t arcsin 5t +
(1 − 25t2 ) + C
5
= t*asin(5*t)+1/5*sqrt(1-25*t^2)
3.2.2
Definite Integrals
1. Sample problem: evaluate the integral
Z
Z
π
x sin x dx :
0
π
x sin x dx =
0
−x cos x + sin x|x=π
x=0
= (−π cos π + sin π) − (−0 cos 0 + sin 0)
= (−π (−1) + 0) − (−0 (1) + 0)
= π
TestGiver: pi
Mathematica: Pi
(a) Only put in as many steps as you need to calculate accurately
2. Find the area under the curve y = tan x between x = 0 and x =
1
0.8
0.6
0.4
0.2
0.2
0.4
0.6
tan x
14
0.8
π
4
The answer is the integral
Z
π/4
tan x dx :
0
Z
π/4
tan x dx =
0
x=π/4
− ln cos (x)|x=0
= − ln cos (π/4) + ln cos (0)
µ
¶
1
= − ln √
+ ln (1)
2
³
´
= − ln 2−1/2 + 0
=
1
ln 2
2
TestGiver: 1/2*ln(2)
Mathematica: 1/2 Log[2]
3. Evaluate the integral
Z
π/2
x
e sin 2xdx =
−π/2
=
=
=
R π/2
−π/2
ex sin 2xdx
¯x=π/2
¯
ex
(sin 2x − 2 cos 2x)¯¯
5
x=−π/2
µ π/2
¶ µ −π/2
¶
e
e
(sin π − 2 cos π) −
(sin (−π) − 2 cos (−π))
5
5
µ π/2
¶ µ −π/2
¶
e
e
(0 − 2 (−1)) −
(0 − 2 (−1))
5
5
2 π/2 2 −π/2
− e
e
5
5
TestGiver: 2/5*E^(pi/2)-2/5*E^(-pi/2)
Mathematica: 2/5E^(Pi/2)-2/5E^(-Pi/2)
(a) 97:
Z
eau sin bu du =
eau
(a sin bu − b cos bu) + C
a2 + b2
4. Class: Check and write for TestGiver.
Z π
π
sin2 x dx =
2
0
= pi/2
Z 2
5 6 1
xe3x d =
e +
9
9
0
= 5/9*E^6+1/9
Z 3
√
10 √
2√
x 2 + xdx =
5+
3
3
5
1
= 10/3*sqrt(5)+2/3*sqrt(3)
3.2.3
Areas between curves
1. If you have two curves, say the graphs of f (x) and g (x), and if for all a ≤ x ≤ b
you have g (x) ≤ f (x),
15
fHxL
x=a
x=b
gHxL
then the area between the graph of f (x) and the graph of g (x) for a ≤ x ≤ b is
Z b
(f (x) − g (x)) dx.
a
2. Example: find the area between f (x) = x and g (x) = x2 for 0 ≤ x ≤ 1
(a)
Z
(a)
Z
1
¡
¢
1
x − x2 dx =
6
0
(b) In TestGiver and Mathematica, 1/6
√
3. Class find area between x and x2 for 0 ≤ x ≤ 1.
0
1
¡√
¢
1
x − x2 dx =
3
4. Sometimes you are asked to find the area bounded by two curves.
(a) In the example above we found the area bounded by x and x2
5. Example: find the area bounded by f (x) = sin (πx) and g (x) = 4x2 .
16
(a) Observe that
f (0) = g (0) = 0
f (1/2) = g (1/2) = 1
Therefore the bounded area is
Z
1/2
0
¡
¢
1
1
sin (πx) − 4x2 dx = −
π 6
i. TestGiver: 1/pi-1/6
ii. Mathematica: 1/Pi-1/6
iii. This is plausible–the answer is positive and the area under sin πx =
1
1
, is about twice the area under 4x2 = .
π
6
6. Class find area between 4x2 and x2 + 3.
(a) Answer:
Z
1
−1
¢
¡¡ 2
¢
x + 3 − 4x2 dx = 4
7. Class find area between x + 2 and x2 . Make a graph.
Z 2
¡
¢
9
(a)
(x + 2) − x2 dx =
2
−1
8. Example: Approximate the area bounded by ex and x + 2.to at least four digits
accuracy.
4
3
2
1
-2
x -1
0
1
(a) First we must find the intersections, which can only be approximated.
i. Using TestGiver to find the lower root
Input
FindRoot((x+2)-E^x; x, -2)
Output
-1.84141
The left intersection is at x = −1.84141.
17
ii. Using Mathematica to find the upper root
The right intersection is at x = 1.1462.
(b) The area is
¯x=1.14162
¯
x2
x¯
(x + 2 − e ) =
+ 2x − e ¯
2
−1.8415
x=−1.8415
µ
¶
1.141622
1.14162
=
+ 2 (1.14162) − e
2
Ã
!
2
(−1.8415)
−
+ 2 (−1.8415) − e−1.8415
2
Z
1.1462
x
= (−0.1969) − (−2.1460)
= 1.9491
i. TestGiver calculates the answer as follows:
Input
a=FindRoot((x+2)-E^x; x, -2);
b=FindRoot((x+2)-E^x; x, 1);
c=Integrate((x+2)-E^x; x, a, b)
Output
a : REAL
a = -1.84141
b : REAL
b = 1.14619
c : REAL
c = 1.94909
ii. You could use Mathematica like this:
3.2.4
Integration by Substitution
1. The most important method for integration is substitution. We begin with
indefinite integrals (anti-derivatives).
R
(a) Consider x sin x2 dx
i. You can break this into two factors: sin x2 and x dx. The factors
multiply to the whole thing
ii. Define u = x2 and note that du = 2x dx , so our integral can be
1R
1
1
rewritten as
sin u du = − cos u + C = − cos x2 + C
2
2
2
iii. ALWAYS check by differentiation. In this class you get negative points
for wrong antiderivatives, since they are so easy to check.
18
R
(b) Here is the general idea: given an integral f (x) dx, try to rewrite the
integrand so that g is easier to integrate than f .
f (x) dx = g (u (x)) u0 (x) dx
R
R
i. Then f (x) dx = g (u) du = G (u) + C = G (u (x)) + C
R
2
(c) Example: x e−x dx.
i. I see
Z
Z
2
x e−x dx =
which looks sort of like:
Z
2
e(−x ) (x dx)
eu du
which I can do. The correspondence is not exact, but it is good enough
to encourage me to proceed.
ii. I define
u = −x2
du = −2x dx
1
x dx = − du
2
iii. Then I have
Z
(d) Example:
R ln x
dx.
x
Z
−x2
xe
Z
1
dx = −
eu du
2
1
= − eu + C
2
1 −x2
= − e
+C
2
ln x
dx =
x
≈
Z
Z
µ ¶
1
(ln x)
dx
x
u du
i. I decide to try the substitution
u = ln x
1
du =
dx
x
ii. Then I get:
Z
ln x
dx =
x
Z
u du
1 2
u +C
2
1
=
(ln x)2 + C
2
R
R
R√
R
R √
(e) Class try: cos (5x) dx, cos3 x sin x dx,
2x − 4dx, tan x dx, x x − 1dx,
R
x
√
dx,
2x − 1
=
2. Substitution and definite integrals
19
(a)
R √π
0
x
u
x sin x2 dx
u = x2
1
x dx =
du
2
√
π
π
2
0
0
Z
(b)
Rπ
0
√
x= π
2
x sin x dx =
x=0
R2
sin 5x dx, 0
Z
u=π/2
sin u du
u=0
1
u=π/2
= − cos u|u=0
2
√
1
x= π
= − cos x2 |x=0 BAD IDEA
2
1
π 1
= − cos + cos 0 BETTER IDEA
2
2 2
1
=
2
Rπ
R3 √
dx R π/2 3
, 0 sin x cos x dx, 0 (cos x) esin x dx, 0 x x + 1dx
2x + 3
3. More substitution practice
R π/4
sin 4t dt = 12
0
R 1/2 dx
= 18 π
0
1 + 4x2
R 1 ex + 1
dx = −e−1 + 2
0
ex
R 1/2 sin−1 x
1 2
√
dx = 72
π
0
1 − x2
3.2.5
1
2
R1
cos πt dt = 0
1
dx
√
= 4 − 2 ln 2 e
x ln x
R 1 ex
dx = ln (e + 1) − ln 2
0 ex + 1
r
√
√
R4 1
1
5
1 + dx = − 12
5 + 43 2
1 x2
x
0
R e4
e
R1
100
1
(2x − 1) dx = 101
sin θ
dθ = 1
cos2 θ
2
R 3 3x − 1
dx = 18
2
(x3 − x)2
R π/2 x2 sin x
R 1π
sin x
dx = −2 1 π x2 1+x
6 dx
−π/2 1 + x6
2
0
R π/3
0
Volumes
1. Here is a general procedure for finding the volume of a solid object
(a) Draw a straight line next to the object, called the reference line. Getting
the right reference line is the key to successfully computing the volume.
(b) Pick one point on the line to be 0. Pick a positive direction on the line.
Usually 0 will be at one end of the object, and the object will be on the
positive side of 0.
(c) Consider the cross-section of the object perpendicular to the line at x.If
you choose a good line, this cross-section will have a simple shape and a
computable area. For every point x on the line, find a formula A (x) for
the area of the cross-section.
(d) The volume is the integral of A (x) from the beginning of the object at
x = a to the end at x = b:
Z b
V =
A (x) dx
a
2. Example: consider a circular cone of height 5 and bottom radius 3.
(a) Draw a picture
(b) Let the reference line be parallel to the central axis of the cone.
(c) Let x be the coordinate of the reference line. Put x = 0 at the point on
the line next to the top of the cone. The bottom of the cone is at x = 5.
20
(d) For any point x between 0 and 5, the cross-section of the cone is a circle
3x
9πx2
with radius
. Thus the area of the cross-section is
.
5
25
R 5 9πx2
dx = 15π
(e) The volume is 0
25
1
(f) Check: the formula for the volume of a circular cone is πr2 h where h is
3
the height and r the radius of the bottom.
3. Class find volume of circular cone with bottom radius 2 and height 4.
4. Find volume of truncated cone with height 3 , bottom radius 4, top radius 2.
(a) Draw a picture
(b) Let the reference line be parallel to the central axis of the cone.
(c) Let x be the coordinate of the reference line. Put x = 0 at the point on
the line where the cone would come to a point. The point where the cone
has radius 2 is at a and the point where the cone has radius 4 is at b.Let r
be the radius at x
b−a
b
4
a
b
r
(d) The volume is
Z
3
6
π
µ
= 3
a
=
2
= 3
= 6
2
=
x
3
¶2
2
x dx = 28π
3
5. Class find volume of truncated cone with height 4, bottom radius 3, top radius
1.
6. Volumes of Rotation
(a) Many volumes arise by revolving a curve around a line.Here is an example
π
(b) Consider the curve y = cos x.for 0 ≤ x ≤ . and the region between the
2
curve and the x-axis. Revolve this region around the x-axis. To find the
resulting volume:
i. Draw a picture
Graph of y = cos x rotated about the x-axis
ii. Let the x-axis be the reference line.
21
iii. Use the coordinates on the x-axis for the coordinates of the reference
line.
iv. The cross-section perpendicular to a the x-axis at a point x is a circle
with radius cos x. Thus the area is A (x) = π cos2 x.
v. The volume is
¯π/2
µ
Z π/2
x sin (2x) ¯¯
π cos2 x dx = π
−
2
4 ¯0
0
1 2
=
π
4
by table #64
A. On a test, if you can’t do the problem, at least set up the integral
correctly.
vi. This is called the method of disks because you divide the volume into
a sequence of disks.
(c) Class do: Consider the region between the graph of ex and the x-axis for
1 ≤ x ≤ 3. Revolve the region about the x-axis.
i.
ii.
iii.
iv.
Draw a picture
set up the integral
and find the volume
¡
¢
R3
R3
2
Answer: 1 π (ex ) dx = π 1 e2x dx = π 12 e6 − 12 e2
π
7. Suppose we consider the graph of y = cos x for 0 ≤ x ≤ and rotate the region
2
betwee the curve and the x-axis about the y-axis.
(a) Make a drawing
Graph of y = cos x revolved about the y-axis
(b) Let the reference line be the y-axis.
(c) Use the coordinates on the y-axis for the coordinates of the reference line.
(d) The cross section at point y is a circle of radius arccos y so the area is
2
A (x) = π arccos (y) (that doesnt’ look good)
R1
2
(e) The volume is 0 π (arccos (y)) dy = π (π − 2) by Mathematica .
8. Class do: consider the region bounded by y = 2x, y = 0, x = 1. Rotate the
region about the y-axis.
(a)
i.
ii.
iii.
iv.
Draw a picture
set up the integral
and find
the volume
¡ y ¢2 ´
R2 ³
π
1
−
dy = 43 π
2
0
22
9. There is a better way, the method of shells.
(a) Draw a picture.
(b) Pick a central axis for the object (in this case the y-axis). For this method
to work, the following must be true about your object and the axis. If you
take a cylindrical slice around the axis (take all the points in the volume
a given distance away from the axis), you get a round cylindrical shell of
constant height.
(c) Pick a ray perpendicular to the central axis (in this case the positive xaxis).The coordinates of the ray start at 0 where the ray leaves the axis.
(d) For each point x on the ray, let h (x) be the height of the cylindrical shell
above x. The area of the shell is then A (x) = 2πx h (x).
(e) The volume is the integral of A (x) along the ray from the axis to the edge
of the object.
10. For the example above
(a) Draw a picture
Graph of y = cos x revolved about the y-axis showing a cylindrical shell at x = 1.
(b) Use the y-axis for the central axis.
(c) Use the x-axis for the ray.
(d) The height is h (x) = cos x so the area of a shell is A (x) = 2πx cos x
(e) The volume is
Z
π/2
0
π/2
2πx cos x dx = 2π (cos x + x sin x|0
= π 2 − 2π
by table #81
11. Class try: Find the volume you get revolving the region bounded by y = xex ,
y = 0, x = 2 about the y-axis.
(a) Draw a picture
(b) set up the integral
(c) and find the volume
¡
¢
R2
R2
(d) 0 2πx (xex ) dx = 2π 0 x2 ex dx = 2π 2e2 − 2
12. Variations
(a) Consider the region bounded by y = sin x, x = 0, and y = 1. Rotate the
region about the line y = 1.
23
i. Draw a picture
Graph of y = sin x rotated about y = 1 with cross-section at x = 0.6
i. Use the x-axis as the reference line.
ii. Use the coordinates of the x-axis for the coordinates of the reference
line.
iii. The cross-section at x is a circle of radius 1 − sin x so the area is
2
A (x) = π (1 − sin x) .
iv. The volume is
Z π/2
Z π/2
¢
¡
2
π (1 − sin x) dx = π
1 − 2 sin x + sin2 x dx
0
0
µ
= π x + 2 cos x +
=
3 2
π − 2π
4
¯π/2
x sin 2x ¯¯
−
2
4 ¯0
using table #63 to integrate sin2 x.
(b) Class: region bounded by y = x − x2 and y = 0 about the line y = 1.
i.
ii.
iii.
iv.
Draw a picture
set up the integral
and find the volume
R1 ¡
¡
¢¢2
Answer 0 π 1 − x − x2
dx =
7
10 π
(c) Consider region between the graph of y = ln x, y = 0, and x = 2. Rotate
this region about the y-axis.
i. Draw a picture
Graph of ln x , 1 ≤ x ≤ 2, revolved about y-axis with cylinrical shell at x = 1.5.
ii. Us the y-axis as the central axis.
24
iii. Use the x-axis as the ray.
iv. At distance x the height is ln x and the area of a shell is A (x) =
2πx ln x.
v. The volume
¯2
µ 2
Z 2
¯
x
(2 ln x − 1)¯¯
2πx ln xdx = 2π
4
1
1
3
= 4π ln 2 − π
2
by table #101.
(d) Class: region bounded by y = x − x2 and y = 0 about the line x = 1.
i.
ii.
iii.
iv.
Draw a picture
set up the integral
and find the volume
¢
¡
R1
Answer: 0 2π (1 − x) x − x2 dx = 16 π
13. More volumes
(a) y = sin x, y = 0, 0 ≤ x ≤ π. revolved about the x-axis.
Rπ
i. 0 π sin2 x dx = 12 π2
(b) y = sin x, y = 0, 0 ≤ x ≤ π. revolved about the line y = 1.
Rπ
i. 0 π (1 − sin x)2 dx = 32 π2 − 4π
(c) y = ex , y = 0, x = 0, x = 1; revolved about x-axis.
R1
2
i. 0 π (ex ) dx = 12 πe2 − 12 π
(d) y = x2 , y = 4, x = 0; revolved about y-axis.
R 4 ¡√ ¢2
i. 0 π y dy = 8π
3.2.6
(e) y = x2 , x = 0, x = 2; revolved about y-axis.
¡√ ¢2 ´
R4 ³
y
i. 0 π 22 −
dy = 8π
Integration by Parts
1. differential notation
(a) instead of writing y = f (x),
dy
dx
= f 0 (x), we write dy = f 0 (x) dx
(b) this is called a differential form
(c) for example
i. if y = x2 then dy = 2x dx
ii. if y = sin x then dy = cos x dx
iii. if y = x ln x then dy = (ln x + 1) dx
(d) You can think of this as saying that if you start at x,you have a certain
value for y, and if change x by dx then y changes by approximately dy.
Rb
b
(e) The fundamental theorem of calculus can be written a df = f |a
2. Given a product of functions y = p (x) q (x) we have
Z
dy
= d (pq) = p dq + q dp
Z
Z
d (pq) =
p dq + q dp
Z
Z
Z
p dq =
d (pq) − q dp
Z
= pq − q dp
25
3. How can we use this bit of formal nonsense?
R
(a) xex dx
p=x
dq = ex dx
x
dp
R =xdx q =xe R x
x e dx = xe − e dx
= xex − ex + C
R
(b) x3 ln xdx
p = ln x dq = x3 dx
dx
x4
dp =
q=
x
4
R
x4
1R 3
x
x e dx =
ln x −
x dx
4
4
x4
x4
=
ln x −
+C
4
16
R
(c) sin−1 x dx
dq = dx
p = sin−1 x
dx
q=x
dp = √
1 − x2
R
R
x
sin−1 x dx = x sin−1 x − √
dx
1 − x2
u = 1 − x2 du = −2x dx
R
1 R du
x
√
√
dx = −
2
2
u
1√− x √
2
=
−
u
=
−
1
−
x
√
R
sin−1 x dx = x sin−1 x + 1 − x2
4. class
R
xe−x dx = −xe−x − e−x
R
1
1
(b) x2 ln (2x) dx = x3 ln 2x − x3
3
9
¢
¡
R
1
(c) tan−1 x dx = x arctan x − ln x2 + 1
2
(a)
5. me
(a)
Rπ
x sin x dx
p=x
dq = sin x dx
dp
=
dx
q
= − cos x
Rπ
Rπ
x
sin
x
dx
= −x cos x|π0 + 0 cos x dx
0
= −π (−1) − 0 + 0 = π
Note: we expect a positive result
R π/2
(b) 0 x2 cos x dx
p = x2
dq = cos x dx
dp = 2xdx q = sin x
¯π/2
R π/2 2
R π/2
x cos x dx = x2 sin x¯0 − 2 0 x sin x dx
0
R π/2
π2
=
− 2 0 x sin x dx
4
p=x
dq = sin x dx
dp = dx q = − cos x
R π/2
R π/2
x sin x dx = −x cos x|π/2
+ 0 cos x dx
0
0
=1
R π/2 2
π2
π2
x
cos
x
dx
=
−
2
(1)
=
−2
0
4
4
0
6. class
26
(a)
(b)
(c)
7. me
(a)
R1
0
R3
1
ln x dx = 3 ln 3 − 2
0
x2 ex dx = 2e2 − 2
R2
R π/2
ex sin x
p = sin x
dq = ex dx
dp = cos x dx q = ex
R π/2
R π/2 x
π/2
e sin x dx = ex sin x|0 − 0 ex cos x dx
0
R π/2
= eπ/2 − 0 ex cos x dx
p = cos x
dq = ex dx
dp = − sin x dx q = ex
R π/2
R π/2 x
e cos x dx = ex cos x|π/2
+ 0 ex sin x dx
0
0
R π/2
= −1 + 0 ex sin x dx
³
´
R π/2 x
R π/2
e sin x dx
= eπ/2 − −1 + 0 ex sin x dx
0
R π/2
= eπ/2 + 1 − 0 ex sin x dx
R π/2 x
2 0 e sin x dx = eπ/2 + 1
R π/2 x
eπ/2 + 1
e
sin
x
dx
=
0
2
0
8. class
(a)
x ex dx = 1
Rπ
−π
ex cos x dx =
9. Class
Z
¢
1¡
−1 + e−2π eπ
2
Z
1
x2 e−x dx = −5e−1 + 2
0
π/2
0
x2 sin 2x dx =
Z
1 2 1
π −
8
2
4
√
3
ln x dx = 4 ln 2 −
2
Z1
√
√
√
√
sin xdx = 2 sin x − 2 x cos x
Z 4 √
e x dx = 2e2
1
3.2.7
Work Problems
1. Work is a technical concept from physics. By definition,
work = f orce × distance
2. If distance measured in feet and force in pounds, then work in foot-pounds.
3. If distance in meters and force in Newtons (N), then work in Joules (J).
(a) The force required to lift an object of mass m kilograms is 9.8m Joules at
the Earth’s surface.
4. Example: a bucket of cement weighing 500lbs is lifted from ground level to the
top of a 200’ building. How much work is done?
(a) 500 lbs × 200 ft = 100, 000 ft-lbs.
27
5. Example: a bucket of cement weighing 300 kgs is lifted from ground level to the
top of a 80 m. building. How much work is done?
(a) 300 kg × 80 m × 9.8
J
kg−m
= 2. 352 × 105 J.
6. Class: how much work is required to lift a 50 lb object 75 ft? How much work
is required to lift a 50 kg. object 75 m?
7. Work accumulates as you move an object. How do you measure work when the
force required to move the object is not constant?
8. Example: a bucket of cement weighing 500lbs is lifted from ground level to the
top of a 200’ building, and for every foot lifted 1 lb of cement leaks out of the
bucket. How much work is done?
9. If a constant force F is applied to an object, the work performed in moving the
object a distance x is W (x) = F x.Therefore, dW
dx = F . We can extend this idea
to a variable force.
(a) Alternate definition of work: The rate of change of work with respect to
distance is the force exerted at each point. If an object is being moved
along the x-axis, and a force f (x) is applied at each point x to move the
object, and the work required to move the object from a to x is W (x) then
dW
= f (x)
dx
(b) In other words, if you have to exert a force f (x) at each point x to move an
object along the x-axis from x = a to x = b, then the total work performed
Rb
is W = a f (x) dx.
10. Consider the leaky cement bucket. Coordinatize the situation with a variable x
by setting x = 0 at the bottom of the building and x = 200 at the top. When
the leaky cement bucket is at position x, it weights 500 − x lbs. Therefore the
force required to move the bucket isR 500 − x. The work necessary to move the
200
bucket to the top of the building is 0 (500 − x) dx = 80 000.
11. Class: If the leaky cement bucket starts with 100 kg. of cement and loses 2 kg
per meter lifted, how much work is required to lift the bucket a distance of 50
meters? How much cement reaches the workers at the top?
12. Example. The force required to pull a spring is proportional to the distance
from rest. For a particular spring, suppose the force is 5x, where x is measured
in feet from the resting position and f is measured in pounds.
(a) The spring constant is 5.
(b) The work required to pull the spring out 2 feet is
13. Class: if 4 lbs of force stretch a spring 20 cm.,
R2
0
5x dx = 10 ft-pounds.
(a) what is the spring constant?
(b) how much work is required to stretch the spring from rest to 50 cm?
(c) how much work is required to stretch the spring from 10 cm to 30 cm?
3.2.7.1
Collective Work
1. Work is additive. If a task is divided into two parts, the work required for the
whole is the sum of the work required for each part.
2. Suppose a 200’ chain that weighs 5 lbs/ft is hanging from a 200’ building. How
much work is required to lift the chain to the top of the building?
28
(a) Think of the chain as divided into (infinitessimally) small segments of
length dx. Each segment weighs 5dx lb. Let x measure distance from
the top of the building. The work required to lift the segment of chain at
position x to the top of the building is 5x dx.
(b) The total work required to lift the chain to the top of the building is the
sum of the work required to lift each segment, or
Z 200
5x dx = 100 000 ft-lbs
0
(c) Class: how much work is required if the chain is only 100 feet long?
R 100
5x dx = 25 000 ft-lbs.
0
(d) Suppose the chain is 300 feet long? Divide it into two parts: 100 feet of
chain lying on the ground and 200 feet of chain hanging from the top of
the building. The work required to lift the part on the ground is 100 ft ×5
lbs/ft ×200 ft = 100 000 ft-lbs. The work required to lift the hanging part
has already been computed to be 100 000 ft-lbs. The total work is 200 000
ft-lbs.
3. Suppose a conical water tank is 4 ft tall, and the radius at the top is 3 ft. If the
tank is full of water, how much work is required to pump the water out of the
tank?
(a) This is the work required to lift all the water to the top of the tank.
(b) Divide the water into horizontal slices of thickness dx, where x measures
the vertical distance from the top of the tank. The radius at position x is
¡
¢2
3 − 34 x ft, so the volume of the slice is π 3 − 34 x dx. The weight of the
¡
¢2
slice is 62.4π 3 − 34 x dx. The work required to lift the slice to the top of
¡
¢2
the tank is 62.4πx 3 − 34 x dx, and the total work required to pump the
water out of the tank is:
µ
¶2
Z 4
3
62.4πx 3 − x dx = 2352. 42 ft-lbs
4
0
(c) Class: if the water is only 2 ft deep, how much work is required to pump
it out of the tank?
µ
¶2
Z 4
3
62.4πx 3 − x dx = 735.133
4
2
(d) Class: if the tank is 4 m high and has a top radius of 3 m, how much work
is required to pump out a full tank?
µ
¶2
Z 4
3
9.8 × 1000πx 3 − x dx = 3.69451 × 105 J
4
0
4. Class: A rectangular aquarium 18” long, 12” wide and 12” high is 80% full of
water. How much work is required to pump it out?
Z 1.0
62.5 × 1.5x dx = 32.1563 ft-lbs
0.5
5. Class: A cattle tank is 6’ long with 1/2-round ends of diameter 2’. If the tank
is full, how much work is required to pump it out?
Z 1
p
62.4 × 6 × 2x 1 − x2 dx = 249.6 ft-lbs
0
(a) One cubic foot is 7.481 gallons. If the tank holds 50 gallons, how much
work is required to pump it out?
29
3.2.8
First Moments and Centers of Mass
1. A point-mass is a mathematico-physical fiction consisting of a single point with
weight.
(a) Suppose a point-mass m is placed at a position x1 on the x-axis. The
first moment of the mass about the point x0 is (x1 − x0 ) m. That’s a
definition. You can’t argue with it.
(b) If several masses m1 , . . . , mn are placed at positions x1P
, . . . , xn . The first
moment of the system of masses about the point x0 is ni=1 (xi − x0 ) mi .
(c) Example: a 2 kg mass and a 3 kg mass are 5 m apart. What is the first
moment of the system about a point 2 m from the first mass and 3 m from
the second?
i. Create an x-axis with the 2 kg mass at x = 0 and the 3 kg mass at
x = 5. The first moment about x = 2 is
2 (0 − 2) + 3 (5 − 2) = 5 kg-m
(d) Class: a 4 lb mass is placed 6 ft from a 3 lb mass. What is the first moment
of the system about a point half-way between the masses?
(e) Class: find the point between the masses about which the first moment is
0.
2. The first moment measures how unbalanced the system is (how much torque
it exerts). A system with positive first moment tilts toward the high side; a
system with negative first moment tilts toward the low side; a system with zero
first moment is balanced.
(a) Archimedes discovered this.
(b) Children using a see-saw have intuitive knowledge of this principal.
(c) How can you find the balance point for a system?
3. Theorem: Suppose you have a system of masses placed on the x-axis. Let the
first moment about x = x0 be M, and the let the sum of the masses be m. Then
the first moment about the the point x̄ = x0 + M
m is 0
(a) Corollary: Suppose you have a system of masses placed on the x-axis.
Let the first moment about x = 0 be M, and the let the sum of the masses
be m. Then the first moment about the the point x̄ = M
m is 0
(b) The point about which the first moment is 0 is unique and is called the
center of mass of the system. The system is “balanced” about this point.
(c) Example: a 2 kg mass and a 3 kg mass are 5 m apart. What is the first
moment of the system about a point 2 m from the first mass and 3 m from
the second?
i. Create an x-axis with the 2 kg mass at x = 0 and the 3 kg mass at
x = 5. The sum of the masses is m = 5kg. The first moment about
x = 0 is
M = 2 (0 − 0) + 3 (5 − 0) = 15 kg-m
The center of mass is x̄ =
M
m
=
15 kg-m
5 kg
= 3 m on the x-axis.
(d) Class: a 4 lb mass is placed 6 ft from a 3 lb mass. Where is the center of
mass?
4. Masses are placed at points in the xy-plane:
mass
location
5
(3, 2)
4
(4, −1)
30
3
(−2, 3)
4
(−2, −5)
(a) You can find the moment of the system about the y-axis by ignoring the ymeasurements and pretending that all the masses are on the x-axis. Sometimes this is called the x-moment. Its value is
Mx
= 5 × 3 + 4 × 4 + 3 × (−2) + 4 × (−2)
= 17
(b) Similarly the moment about the x-axis is
My
= 5 × 2 + 4 × (−1) + 3 × 3 + 4 × (−5)
= −5
(c) The sum of the masses is m = 16, and the center of mass is
Mx
m
17
=
16
My
=
m
5
= −
16
x̄ =
ȳ
The system is balanced about this point.
(d) Class: if masses are located at
mass
location
5
(3, 1)
2
(2, −1)
3
(−3, 4)
4
(−2, −6)
find the center of mass.
5. Now we want to extend the concepts of moment and center of mass from pointmasses to extended objects.
6. Suppose a rod has length 3 has density ex/10 where x is the distance from the
left end.
(a) The weight of an (infinitessimal) segment of rod of length dx at position x
is ex/10 dx.
(b) The total weight of the rod is the sum of the infinitessimal weights:
m=
Z
3
ex/10 dx = 3.49859
0
(c) The moment of this segment about the left end x = 0 is xex/10 dx. The
total moment about the point x = 1 is the sum of the moments of all the
infinitessimal segments, or:
Z 3
M=
xex/10 dx = 5.50988
0
(d) The center of mass is x̄ =
rod balances.
M
m
= 1.57489. This is the point about which the
(e) Check that the moment about this point is 0:
Z
0
3
(x − 1.57489) ex/10 dx = −7.9 × 10−6
(f) Class: a rod has density 1 + x2 at distance x from the left end. Find the
center of mass of the rod.
31
7. Finding the center of mass for a 2-dimensional region is more difficult. The
general problem is left for next semester, when you study multiple integration.
We restrict ourselves to regions or lamina of constant density. In this case the
center of mass of the region is called its centroid.
8. Consider the rectangle bounded by (0, 0) , (0, 3) , (3, 4) , (0, 4).
(a) Since we are assuming constant density, we might as well take density equal
to 1 for finding the centroid. The mass is then the same as the area, or
m = 12
(b) To find the first moment in the x-direction, we think of the rectangle as
divided into vertical strips (of infinitessimal width, of course). Each strip
has mass 4dx, and x-moment 4x dx. Therefore the total x-moment is
Z 3
Mx =
4x dx
0
= 18
(c) To find the first moment in the y-direction, we think of the rectangle as
divided into vertical strips again. Each strip has mass 4dx and its center
of mass is at distance 2 from the x-axis. Therefore the y-moment of each
strip is 2 × 4dx, and the y-moment is
Z 3
My =
8 dx
0
= 24
(d) The center of mass is then:
Mx
= 1.5
m
My
=2
m
x̄ =
ȳ
=
which is the answer we expect.
√
9. Consider the half-disk bounded by y = R2 − x2 and the x-axis. By symmetry
x̄ = 0. By geometry,
π
m = R2
2
√
To calculate ȳ, we have divide the half-disk into vertical strips of mass R2 − x2 dx
distance x from the y-axis. The y-moment of each strip is
√
p
R2 − x2
R2 − x2
2
2
R −x
dx =
dx
2
2
and the total y-moment is:
My
=
=
Z
R
−R
R2 − x2
dx
2
2 3
R
3
The y-coordinate of the center of mass is then:
ȳ
=
2 3
3R
π 2
2R
=
4
R
3π
This answer is reasonable because
32
(a) it is a linear multiple of R
(b) it is less than 1/2 way to the top of the circle
10. Class: find the centroid of the region bounded by the x-axis and the graph of
y = x − x2 . Start by setting up the integrals for the the mass and the two
moments.
Z 1
¡
¢
1
m =
x − x2 dx =
6
0
Z 1
¡
¢
1
Mx =
x x − x2 dx =
12
0
¢2
Z 1¡
x − x2
1
My =
dx =
2
60
0
x̄ =
ȳ
=
1
12
1
6
1
60
1
6
=
1
2
=
1
10
Why is your answer reasonable?
11. General formula: given a region bounded by y = f (x), y = g(x), a ≤ x ≤ b, and
assuming f (x) < g(x), we have:
m =
Z
b
(g (x) − f (x)) dx
a
Mx
=
Z
b
x (g (x) − f (x)) dx
a
My
=
x̄ =
Z
(g (x) − f (x))2
dx
2
a
My
Mx
ȳ =
m
m
b
12. Class: Find the center of mass of the triangle bounded by (0, 0), (0, 3), (1, 2).
3.2.9
Partial Fractions
This is really a theorem in algebra that has applications to calculus.
1. Let p (x) = xn + an−1 xn−1 + · · · + a1 x + a0 be a monic polynomial.
(a) monic means leading coefficient is 1.
(b) Carl Friedrich Gauss (1777-1855) proved the Fundamental Theorem of Algebra (1799): p can be factored in linear factors:
p = (x − a1 )
n1
nt
· · · (x − at )
i. the ai are all different
ii. n1 + · · · + nt = n
iii. the √
ai might be complex numbers using the imaginary square root
i = −1.
(c) If the coefficients of the polynomial are real numbers, it is easy to deduce
from Gauss’ theorem that p can be factored into real linear and quadratic
factors:
¡
¢m1
¢ms
¡
p = (x − a1 )n1 · · · (x − as )ns x2 + b1 x + c1
· · · x2 + bt x + ct
where n1 + · · · + ns + 2 (m1 + · · · + mt ) = n
33
i. For the quadratic terms it is always true that b2i < 4ci .so the quadratics
cannot be factored.into real factors.
2. This is not easy to do by hand or by machine.
(a) Even if a polynomial has integer coefficients, it probably does not have
factors with integer coefficients.
√ ¢¡
√ ¢
¡
i. for example x2 − 5 = x − 5 x + 5
(b) Évariste Galois (1811-1832, killed in a duel) proved (1832) that if deg p > 4
then there cannot be a general method for finding the factors.
(c) When a polynomial can be factored into factors with integer coefficients,
Mathematica can sometimes find the factors.
Note 3 + 1 + 2 (1 + 2) = 10
(d) Mathematica can also factor any polynomial approximately into real and
quadratic factors. Just enter the polynomial with floating point coefficients.
3. Here’s the part that helps with calculus.
4. A rational function is a quotient of polynomials
p (x)
.
q (x)
(a) Dividing the numerator and denominator by the leading coefficient of q we
can assume that q is monic.
(b) If deg p ≥ deg q then we can use polynomial division to get
p (x)
r (x)
= a (x) +
q (x)
q (x)
where deg (r) < deg (q)
(c) Every rational function
r (x)
where q is monic and deg r < deg q can be
q (x)
written in a special way.
i. The denominator q (x) can be factored into linear and quadratic factors:
¡
¢m1
¢mt
¡
· · · x2 + bt x + ct
q (x) = (x − a1 )n1 · · · (x − as )ns x2 + b1 x + c1
n
ii. For each linear factor (x − ri ) i we get a sum of terms where numerators are real:
∗
∗
∗
+
+ ··· +
ui =
x − ai
(x − ai )ni
(x − ai )2
iii. For each quadratic factor we get a sum of terms with real linear numerators:
vi =
x2
∗x + ∗
∗x + ∗
∗x + ∗
+
+ ··· + 2
2
2
+ bi x + ci
(x + bi x + ci )mi
(x + bi x + ci )
34
iv.
r (x)
= u1 + · · · + us + v1 + · · · + vt
q (x)
5. Example:
6. Another example:
7. Class: What is the FORM of the partial fraction decomposition of: the following
expressions? Which ones have polynomial parts?
35
(a)
(b)
(c)
1
(x + 1) (x + 2)
3x + 2
2
(x + 1)
x2 − 3x + 2
(x + 1) (x + 2)3
(d)
x2 − 3x + 2
(x + 1) (x + 2)
(e)
x3 − 4x2 + 2x − 1
(x2 − 3x + 4) (x + 2)
(f)
(g)
x3 + 2x2 − x − 1
(x2 − 3x + 4) (x + 2)2
x7 − 6x6 + x5 − x2 + 2
(x2 − 3x + 4)2 (x + 2)2
8. How can we find the constants without using Apart in Mathematica?
(a) By hand. I’ll show you how to do some simple ones:
i. Suppose we have only linear terms appearing to the first power (remember, deg p < u)
a =
=
di
=
p (x)
(x − r1 ) · · · (x − ru )
du
d1
+ ··· +
x − r1
x − ru
p (ri )
(ri − r1 ) · · · (ri−1 − ri ) (ri+1 − ri ) · · · (ru − ri )
ii. Example:
a =
d1
=
=
d2
=
=
d3
=
=
a =
x2 − 2x + 4
x (x − 2) (x + 1)
02 − 2 × 0 + 4
(0 − 2) (0 + 1)
−2
22 − 2 × 2 + 4
(2) (2 + 1)
2
3
2
(−1) − 2 × (−1) + 4
(−1) ((−1) − 2)
7
3
2/3
7/3
−2
+
+
x
x−2 x+1
iii. Class try
3x − 5
A.
(x − 2) (x − 3)
5
B.
(x − 1) (x + 1)
36
(b) How about
3x − 1
=
2
(x − 2)
=
=
a
b
+
x − 2 (x − 2)2
a (x − 2) + b
2
(x − 2)
ax + (b − 2a)
2
(x − 2)
a = 3
b − 2a = −1
b = 5
3x − 1
3
5
=
+
x − 2 (x − 2)2
(x − 2)2
(c) Class try
x+2
(x − 1)2
9. Why do we care:
(a) We can integrate all the terms of a partial fraction decomposition, so we
p (x)
can integrate anything of the form
.
q (x)
(b) We show this by example:
Z
−3/2
3
3 dx =
2 +C
(x − 1)
(x − 1)
Z
3
dx = 3 ln (x − 1) + C
x−1
(c) Class try:
Z
Z
4x − 2
dx =
2
x + 2x + 4
2
dx
(x + 2)2
Z
2 (2x + 2) − 6
dx
x2 + 2x + 4
Z
Z
6
2 (2x + 2)
dx
dx −
=
x2 + 2x + 4
(x + 1)2 + 3
√
√
¢
¡
3
= 2 ln x2 + 2x + 4 − 2 3 arctan
(2x + 2) + C
6
(d) Class try.
R
Z
2x + 1
dx
x2 + 2x + 2
2x + 1
dx . To do these with paper and pencil
+ 2x + 2)2
requires a method we don’t study. Either look up trig substitutions or go
to the lab for Mathematica.
(e) Here’s a hard one:
(x2
37
3.2.10
Improper Integrals
1. Review L’Hôpital’s rule: if limx→a f (x) = ± limx→a g (x) = 0 or ±∞, then
f (x)
f 0 (x)
= lim 0
x→a g (x)
x→a g (x)
lim
and one limit exists if the other does. Note a can be ±∞, or the limit could be
one-sided.
(a) Example:
lim x ln x = lim
x→0+
x→0+
ln x
1/x
Since limx→0+ ln x = − limx→0+ = ∞ we have
lim
x→0+
ln x
1/x
1/x
x→0 −1/x2
= − lim+ x
=
lim+
x→0
= 0
Test: calculate x ln x for x = 1e − 6, 1e − 12
(b) Class do limx→∞ xe−x
Rb
2. Whenever we write a f (x) dx we insist that f (x) be defined for a < x < b.
However, we can sometimes permit integrals where f (a) or f (b) (or both) is
undefined.We all such integrals improper integrals.
(a) An improper integral asks if an area of infinite extent is finite
i. Is the area under the curve y = e−x , x > 0, finite
ii. Is the area under the curve y =
1
, 0 ≤ x ≤ 1, finite?
x
(b) Could these areas possibly be finite? How?
1 1 1
i. Like 1 + + + + · · · = 2
2 4 8
38
(c) Identify which are improper:
R ∞ dx
R3
sin x dx
1 x2
1
√
i. R
R1
x
1
ln x dx
dx
0
0 1−x
R 3 dx
1 x+1
R1
x ln x dx
0
3. One type of improper integral is one with an infinite limit. We define
Z ∞
Z M
f (x) dx = lim
f (x) dx
M→∞
a
a
(a) Example:
Z
∞
1
dx
x3
Z
M
dx
M →∞ 1
x3
¯M
−1 ¯¯
=
lim
M →∞ 2x2 ¯
µ 1
¶
1
1
=
lim
−
M →∞ 2
2M 2
1
=
.
2
=
lim
(b) Not every problem works out nicely
Z M
Z ∞
dx
dx
=
lim
M→∞
x
x
1
1
=
lim ln (M )
M→∞
does not converge
(c) When the integral limit converges to a finite value, we say that the integral
converges or is convergent. Otherwise the integral diverges or fails to converge or is divergent. All improper integrals fall into exactly one of these
categories.
R∞
R ∞ dx
(d) Class try 0 e−x dx
0 1 + x2
4. Another type of improper integral has finite limits but an integrand undefined
at one of the limits
(a) If f is undefined at a we define
Z b
Z
f (x) dx = lim +
M→a
a
b
f (x) dx
M
(b) The defnition is similar if f is undefined at b :
Z M
Z b
f (x) dx = lim−
f (x) dx
M→b
a
a
(c) Example:
Z
1
ln xdx =
0
=
=
lim +
M→0
Z
1
ln xdx
M
1
lim x ln x − x|M
M→0+
lim −1 − M ln M + M
M→0+
= −1
To get this last limit, you have to know that limM→0+ M ln M = 0, which
we proved above using L’Hôpital’s rule
39
i. When you do a problem like this, you have to show all the work.
(d) Not all improper integrals on a finite domain converge either
Z M
Z 1
dx
dx
dx =
lim −
1
−
x
1
−x
M−
→1
0
0
=
=
does not converge because 1 − M
5. Class Try:
R 1 dx R 1 dx R ∞ −x
√ , 0 2 , 0 xe dx,
0
x
x
M
lim
− ln (1 − x)0
lim
− ln (1 − M )
M−→1−
M−→1−
−→ 0 and ln 0 is undefined
6. Warning, not all divergent integrals are infinitely large.
converge, but the limit does not go to infinity either.
R∞
0
sin x dx does not
7. Sometimes all we care about is whether or not an integral converges. We are
not concerned with the exact value of the integral. Here are some ideas you can
use to determine if an integral converges without evaluating it.
Rb
(a) Useful principal of absolute convergence: if a |f (x)| dx converges
Rb
then a f (x) dx converges.
Rb
(b) If a f (x) dx is improper because f (b) is undefined (maybe because b = ∞)
Rb
but f (x) is defined on a ≤ x ≤ c < b then a f (x) dx converges if and only
Rb
if c f (x) dx converges.
i. You just have to look at the bad part.
Z ∞
Z ∞
dx
dx
converges
if
ii.
2
1
+
x
1
+
x2
0
5
R 1 dx
R 0.00001 dx
iii. 0
converges if and only if 0
converges
x
x
(c) The comparison test is useful in some cases
Rb
i. If 0 ≤ f (x) ≤ g (x) for all a < x < b then 0 ≤ a f (x) dx ≤
Rb
g (x) dx.provided the integrals converge
a
Rb
Rb
A. If a g (x) dx converges then a f (x) dx converges
Rb
Rb
B. If a f (x) dx diverges then a g (x) dx diverges.
R ∞ ln x
R ∞ dx
R ∞ dx
ii. 1
dx diverges since 1
diverges so 3
diverges and 0 <
x
x
x
R ∞ ln x
R ∞ ln x
ln x
1
<
for x > 3 so 3
dx diverges so 1
dx diverges.
x
x
x
x
R 1 sin x
iii. 0 √ dx converges because
x
¯
¯
¯ sin x ¯
1
¯
A. 0 ≤ ¯ √ ¯¯ ≤ √
x
x
R 1 dx
B. 0 √ converges, as we have seen
x
¯
¯
R 1 ¯ sin x ¯
C. Therefore by the comparison test, 0 ¯¯ √ ¯¯ dx converges
x
R 1 sin x
D. Therefore, by the absolute convergence test, 0 √ dx converges.
x
8. Class try: Find a comparison to determine the convergence or divergence of
R π/2
R ∞ −x
e sin xdx, 0 arctan x dx.
1
R∞
R1
R∞
dx
dx
dx
√
.Hint: break into two parts 0 √
and 1 √
x (1 + x2 )
x (1 + x2 )
x (1 + x2 )
and do a comparison on each part.
9. Class:
0
40
3.2.11
Numerical Integration
(Warning–this is your teacher’s favorite topic)
1. Two important problems that arise frequently is applied mathematics (mathematics applied at NASA projects, biological research, semiconductor manufacturing, oil refineries, any other area where mathematical models are used to
predict and control some process)
(a) Evaluate an integral you cannot anti-differentiate, for example
Z 100 p
1 + sin4 xdx.
1
(b) If you only know a function through experimental obervations
x
0.32 0.47 · · ·
f (x) 1.36 2.58 · · ·
3.28
1.26
R3
estimate 1 f (x) dx. Example: I knew the amount of ligand absorbed by
cell receptors as a function of the density of free ligand in the bloodstream.
2. Outside university courses, these problems are much more important and much
more frequently encountered than problems using anti-differentiation.
3. Class: without using anti-differentiation, find:
R4
(a) 1 5dx
R4
(b) 1 2xdx
4. The good news is that any integral can be evaluated numerically, and very
efficient methods exist for doing this.
Rb
(a) To numerically evaluate a f dx in Mathematica you use the command
¡ ¢
R4
NIntegrate[f,{x,a,b}]. Here’s how you evaluate 1 sin x2 dx:
(b) To make Mathematica show more digits
There is no point asking for more than 18 digits. To get more, you need to
go deeper into Mathematica.
(c) When you are doing a problem like this and you want lots of accuracy, it
is a good idea to use different software packages to have them check each
other. X(PLORE) says 4.36865542924733e-1 ± 4.4e − 14
(d) X(PLORE) and Mathematica differ in digit 15. After roundoff, Mathematica says 4,
X(PLORE) says 3. But X(PLORE) also warns you not to trust more than 13 digits.
Mathematica does not estimate the precision of its result.
41
5. We are going to study some methods of integrating functions numerically. The
easiest method is the left-hand method
¡ ¢
R4
(a) To evaluate 1 sin x2 dx using 6 intervals,
4−1
= 0.5.
3
ii. make a table for x = 1, 1 + ∆x, . . . , 4 − ∆x
x ¡ ¢ 1
1.5
2
2.5
3
3.5
0.84 0.78 −0.76 −0.03 0.41 −0.31
sin x2
iii. Compute the approximation
¡
¢
¡
¢¢
¡ ¡ ¢
L = sin 12 + sin 1.52 + · · · + sin 3.52 × 0.5
= 0.47
i. Determine the step-size ∆x =
(b) You can do the same thing with one step in Mathematica:
(c) The general formula for the left-hand approximation of
b−a
and then compute
intervals is this: define ∆x =
n
Rb
a
f (x) dx using n
L = (f (a) + f (a + ∆x) + · · · + f (b − ∆x)) ∆x
(d) Class: calculate the left hand approximation of
R2
0
sin x dx using 4 intervals.
6. The right-hand method uses the points x = 1 + ∆x, 1 + 2∆x, . . . , 4:
¢
¡ ¢
¡ ¢¢
¡ ¡
R = sin 1.52 + sin 22 + · · · + sin 42 × 0.5
= −0.0994
(a) In Mathematica:
(b) The general formula for the right-hand approximation if
b−a
n intervals is this: define ∆x =
and then compute:
n
Rb
a
f (x) dx using
R = (f (a + ∆x) + · · · + f (b − ∆x) + f (b)) ∆x
R2
(c) Class: calculate the right hand approximation of 0 sin x dx using 4 intervals.
¡ ¢
R4
7. An even better approximation of 1 sin t2 dt can now be obtained almost for
free. The trapezoidal approximation of the integral is:
L+R
= 0.1829
2
(a) Unfortunately, all the work we’ve done so far is for naught. We have a
lousy approximation.
R2
(b) Class do trapezoidal approximation with 4 intervals for 0 sin x dx
42
8. If we make a table of approximations vs. number of intervals we get (using
Mathematica)
(a) For example, to calculate the left approximation using 130 intervals, I could
have written
(b) The two-in-a-row rule says that if you get the same rounded-off answer
for two different values of n using any method of approximate integration,
then you can trust that rounded-off answer.
i. This is a rule of thumb, a principal of mathematical engineering.
ii. It is not difficult to find an example that contradicts it,
iii. I prefer the three-in-a-row rule and require that the number of intervals be increased by 50% between approxmations.
(c) Looking at these numbers, I believe that the integral is, to 3-digit accuracy,
0.437, since I got that answer two times in a row.
i. From left hand approximations, I could conclude that the integral was
0.4 to one decimal place accuracy
ii. From right hand approximations, I could conclude that the integral
was 0.43 to two decimal place accuracy, and I would be wrong!
(d) To get the table above, I used some more powerful Mathematica commands,
including functions.
9. A direct way to get the trapezoidal approximation is:
µ
¶
f (a)
f (b)
T =
+ f (a + ∆x) + · · · + f (b − ∆x) +
∆x
2
2
(a) Here’s a Mathematica calculation of the 6-interval trapezoidal approxima¡ ¢
R4
tion to 1 sin t2 dt :
43
(b) Class: directly calculate the trapezoidal approximation of
4 intervals.
R2
0
sin x dx using
10. Even better than the trapezoidal approximation is the midpoint approximation.
Rb
(a) The general formula for the midpoint approximation if a f (x) dx using n
b−a
intervals is this: define ∆x =
and then compute:
n
¶
µ
¶
µ
¶¶
µ µ
3∆x
∆x
∆x
+f a+
··· + f b −
∆x
M = f a+
2
2
2
¡ ¢
R4
(b) For example, the 6-interval midpoint approximation to 1 sin t2 dt is:
³ ¡
´
¢
¡
¢
M =
sin 1.252 + sin 1.752 + · · · + sin (3.75)2 × 0.5
= 0.5935
(c) In Mathematica we could compute:
(d) Class: calculate the midpoint approximation of
(e) Using more intervals we get:
R2
0
sin x dx using 4 intervals.
(f) Using the two-in-a-row rule, we could get 3 digit accuracy for our answer:
0.437
i. But I would want a little more evidence.
11. Trapezoidal and midpoint approximations converge faster to the correct answer
than left and right approximations
12. Let’s compare errors in trapezoidal and midpoint approximations, assuming
Mathematica’s answer is correct
44
(a) Class: do you notice a pattern?
13. Why is midpoint better than trapezoidal. See picture.
14. We could average the trapezoidal and midpoint approximations, but it works
better to take a weighted average. If T and M are trapezoidal approximations,
then Simpson’s Approximation is
2M id + T rap
3
R2
(a) Class: calculate Simpson’s approximation of 0 sin x dx using 4 intervals.
simpson =
(b) Simpson’s approximation is much more accurate than even the trapezoidal
or midpoint. Notice how good the values are for even a small number of
intervals:
(c) Why is Simpson’s so much better? Because Simpson’s approximates area
under parabola over interval, not under straight line.
(d) Simpson’s method is often the preferred choice. It is not too complicated,
but fairly accurate.
b−a
(e) If we define ∆x =
then a direct calculation for Simpson’s method is:
2n
µ
¶
∆x
f (a) + 4f (a + ∆x) + 2f (a + 2∆x) + 4f (a + 3∆x) +
· · · + 2f (b − 2∆x) + 4f (b − ∆x) + f (b)
3
We say that this method uses 2n + 1 points.
i. Proof:
2M id + T rap
3


2 (f (a + ∆x) + f (a + 3∆x + · · · + f (b − ∆x)) 2∆x
¶
1 µ

f (a)
f (b)
=
+
+ f (a + 2∆x) + · · · + f (b − 2∆x) +
2∆x
3
2
2
µ
¶
∆x
f (a) + 4f (a + ∆x) + 2f (a + 2∆x) + 4f (a + 3∆x) +
=
· · · + 2f (b − 2∆x) + 4f (b − ∆x) + f (b)
3
45
(f) Class: calculate directly Simpson’s approximation of
points.
R2
0
sin x dx using 5
15. How can you use TestGiver to approximate integrals?
¡ ¢
R4
(a) Let’s approximate 1 sin x2 dx using methods:
i.
ii.
iii.
iv.
v.
left hand using 20 points
right hand using 20 points
trapezoidal using 21 points
midpoint using 20 points
Simpson’s using 41 points
(b) Here is the input to TestGiver’s calculator page.
dx = (4-1)/20;
left = dx*Sum(sin(x^2); x,1,4-dx,dx);
right = dx*Sum(sin(x^2); x,1+dx,4,dx);
trap = (left+right)/2;
mid = dx*Sum(sin(x^2); x,1+dx/2,
4-dx/2,dx);
simpson = 1/3*trap+2/3*mid
(c) Before evaluating the input, choose scientific form for the output.
This will give you lots of digits accuracy in your calculations.
i. Actually calculations are always carried out with lots of precision, but
they can be reported with either 6 or 16 digits accuracy.
(d) Here is TestGiver’s output. You can copy and paste this to the answer
spaces using control-C and control-V
dx : REAL
dx = 1.4999999999999999445E-1
left : COMPLEX
left = 5.047997059245112883E-1
right : COMPLEX
right = 3.3539356070356890482E-1
trap : COMPLEX
trap = 4.2009663331404012431E-1
mid : COMPLEX
mid = 4.45392959589994949991E-1
simpson : COMPLEX
simpson = 4.3696085083134328376E-1
16. But why do all this if you have NIntegrate?
(a) Sometimes I have to imbed an integration routine in a more complicated
algorithm that I’m writing. I have to evaluate many integrals–perhaps
thousands or even millions–and I cannot call on Mathematica for each
one.
(b) Sometimes I have to integrate a function from data.
46
17. Given data
Class
find left hand, right hand, trapezoidal, and Simpson’s approximation for
R2
f
(x)
dx.
0
18. Given data
How could you approximate
R2
0
f (x) dx?
(a) In TestGiver
Input
x = {0,0.4,0.95,1.3,1.7,2};
y = {0,0.151647,0.661645,0.928444,0.983399,0.826822};
Sum((y[j+1]+y[j])/2*(x[j+1]-x[j]);j,1,5)
Output
x : LIST(REAL)
x = {0,0.4,0.95,1.3,1.7,2}
y : LIST(REAL)
y = {0,0.151647,0.661645,0.928444,0.983399,0.826822}
1.18615
19. Errors in Approximations as function of number of points
(a) Rules of thumb
i. If you double the number of points, the error in trapezoidal and midpoint method goes down by 3/4.
ii. If you double the number of points, the error in Simpon’s method goes
down by 15/16.
¡ ¢
R3
(b) Here are some values obtained approximating 0 sin x2 dx.
i. n is the number of points used.
47
3.2.12
The Fundamental Theorem of Calculus
1. How do you set up an integral to represent a quantity like:
(a) an area
(b) a volume
(c) an arc length
(d) a probability
(e) work
2. Integrals represent accumulations
(a) area is an accumulation of strips
(b) volume is an accumulation of slices
(c) arc length is an accumulation of short line segments
(d) probability is the accumulation of unlikely events
(e) work is the accumulation of little pushes (or pulls)
3. The approach is always the same. Find a parameter (represented by a variable,
say x) and a function (say g (x)) such that
(a) the quantity you want to calculate can be represented as a sum of a large
number of small quantities
(b) each of the small quantities can be represented by g (x) ∆x (exactly a product like this)
(c) the quantity you want is the sum of all the small quantities you get taking
all x’s from a to b.
Rb
(d) Then Q = a g (x) dx
4. Let’s look at this for arc length.
(a) A littleqpiece of the graph of q
f (x) above x can be thought of as having
2
length dx2 + (f 0 (x) dx) = 1 + f 0 (x)2 dx.
48
(b) So the entire arc length from x = a to x = b is
Rbq
2
1 + f 0 (x) dx
a
5. There is another approach to creating integrals that model desired quantities
using the Fundamental Theorem of Calculus
6. The Fundamental Theorem of Calculus:
(a) Given a function f (x) (say continuous on [a, b] to be absolutely correct)
Rx
(b) Define a new function F (x) = a f (t) dt
Rb
i. You can’t say a f (x) dx becaus that is a single number
Rx
ii. You can’t say a f (x) dx because the symbol ‘x’ has two different
meaning here
Rx
iii. The value of the integral a f (x) dx is determined by the value of x,
so this is a function of x.
iv. The function F (x) is defined so long as the integral is defined.
A. In particular, the function F (x) is defined on [a, b] and F (a) = 0.
(c) The Fundamental Theorem of Calculus says:
F 0 (x) = f (x) .
7. Let f (x) = sin2 x, and let a = 0. Then F (x) =
Rx
0
sin2 t dt =
1
1
x − sin x cos x.
2
2
(a) This is not an anti-derivative. It is a definite integral:
¯x
Z x
¯
1
1
2
sin t dt =
t − sin t cos t¯¯
2
2
0
0
1
1
=
x − sin x cos x
2
2
µ
¶
1
d 1
x − sin x cos x = sin2 x
(b) Class check that
dx 2
2
R x dt
(c) Class if F (x) = 1
what does the Fundamental Theorem say F 0 (x) =?
t
i. Is this true? Evaluate the integral to find F (x) then differentiate F (x).
49
8. Another version of this theorem is this: if g (x) is a differentiable function on
Rb
[a, b] then a g 0 (x) dx = g (b) − g (a). This is the one you are used to using.
9. So what good is this new version.
10. If you have some quantity F (x) dependent on a parameter x
(a) if F (a) = 0
(b) if you know that F 0 (x) = f (x)
Rx
then F (x) = a f (t) dt.
11. Many scientific laws give you the derivative of the quantity you want.
(a) The law of gravity gives acceleration: we want to know position and speed.
(b) Newton’s law of cooling gives the rate of change of temperature, not the
temperature.
(c) Maxwell’s laws give the derivatives of the electromagnetic waves.
12. For example, let F (x) be the length of the arc of y = f (x) starting at t = a
and ending at t = x.
(a) F (a) = 0 since the arc length from t = a to t = a is 0.
(b)
F 0 (x)
(arc length from 0 to x + ∆x) − (arc length from 0 to x)
∆x
arc length from x to x + ∆x
= lim
∆x→0
∆x
q
2
∆x2 + (f 0 (x) ∆x)
= lim
∆x→0
∆x
q
0
= lim
1 + f (x)2
∆x→0
q
2
=
1 + f 0 (x)
Rxq
Therefore arc length from t = a to t = x is a 1 + f 0 (x)2 dx.
=
lim
∆x→0
13. Exercises:
F (x) =
Z
x
cos t dt
0
F 0 (x) = cos x
F (x) =
Z
2
3t2 dt
x
= −
Z
x
3t2 dt
2
2
F 0 (x) = −3x
F (x) =
Z
x2
et dt
−1
dF dx2
dx2 dx
2
= ex 2x
F 0 (x) =
2
= 2xex
50
14. Class try:
Z
x
et dt
1
Z
2
sin t dt
x2
15. General formula:
d
dx
Z
b(x)
a(x)
f (t) dt = f (b (x)) b0 (x) − f (a (x)) a0 (x)
16. Example
F (x) =
Z
cos x
t3 dt
sin x
3
3
F 0 (x) = (cos x) (− sin x) − (sin x) cos x
³
´
2
2
= − sin x cos x (cos x) + (sin x)
= − sin x cos x
(a) Check:
(cos x)4 − (sin x)4
4
3
0
F (x) = (cos x) (− sin x) − (sin x)3 cos x
F (x) =
17. Class try:
Z
e2x
ex
3.2.13
1
dt
t
Probability Integrals
1. A continuous random variable X can be thought of as a magic box that emits
numbers.
(a) Although no one can predict the next number to appear, it is assumed that
we can make statements about the probability that numbers will occur in
a certain range.
(b) The probability that a ≤ X ≤ b, written P (a ≤ X ≤ b), is the percentage
of emitted numbers that satisfy the condition
(c) A random variable comes with knowledge of all such probabilities.
2. Example: every programming language, including TestGiver and Mathematica,
includes a random function that produces values between 0 and 1.
(a) if 0 ≤ a < b ≤ 1, then P (a ≤ Random ≤ b) = b − a
(b) TestGiver input
x = MakeList(Random;i,1,100);
Sum(Choose(x[i]<1/3,1,0);i,1,100)/100
(c) TestGiver output
x : LIST(REAL)
x = {0.0489052,0.769638,0.0949757,
0.205269,0.800982,0.381896,0.453723,
0.676724,0.498999,0.336422,0.352255,
0.817455,0.535778,0.649806,0.885977,
0.916743,0.169009,0.416907,0.311156,
51
0.915838,0.154497,0.414544,0.250916,
0.375421,0.270999,0.163212,0.191266,
0.893275,0.096042,0.243666,0.251555,
0.894012,0.642095,0.915265.0.463699,
0.962755,0.712461,0.332962,0.0445856,
0.14464,0.0862802,0.909781,0.396014,
0.882005,0.891998,0.621694,0.816544,
0.511161,0.778694,0.301748,0.159643,
0.490599,0.744635,0.220326,0.922486,
0.782521,0.732426,0.498415,0.499808,
0.0595691,0.350143,0.150759,0.587156,
0.658617,0.327814,0.805771,0.967944,
0.453515,0.568628,0.399069,0.16866,
0.0686527,0.904244,0.387533,0.716621,
0.105213,0.913578,0.280736,0.984639,
0.302002,0.414441,0.536631,0.856969,
0.802723,0.487406,0.118382,0.700938,
0.951566,0.141272,0.259189,0.378593,
0.289001,0.979504,0.232006,0.282781,
0.669633,0.47703,0.171012,0.632002,
0.674009}
0.35
3. Example: the time required to wait for phone call to be answered, starting from
the first ring.
(a) That is, the numbers emitted by X have the same statistical distribution
as the waiting times for answering phone calls
4. Example: the weight of a male student
(a) That is, the numbers emitted by X have the same statistical distribution
as the weights of male students
5. What all these examples have in common is that the value can be any real
number in a certain range.
(a) That’s why they are called continuous random variables.
(b) Discrete random variables are random variables that emit integer values
only. We do not study these.
6. The most common way of specifying the probabilities for a random variable X
is to give a function f (x) called the probability density function for X.
(a) The probability density function satisfies
i. f (x) ≥ 0 for all x
R∞
ii. −∞ f (x) dx = 1
(b) The relationship between the probabilities of the random variable X and
the probability density function f (x) is
Z b
f (x) dx
P (a ≤ X ≤ b) =
a
7. Example: the probability density function for the Random function is

 0 if x < 0
1 if 0 ≤ x ≤ 1
f (x) =

0 if x > 1
52
1.2
0.8
0.6
0.4
0.2
0
-0.4 -0.2
-0.2
0.2
0.4
0.6
x0.8
1
1.2
1.4
(a)
(b) The probability that a random number is between 0.4 and 0.7 is
Z
P (0.4 ≤ random ≤ 0.7) =
0.7
f (x) dx
0.4
Z 0.7
=
dx
0.4
= 0.3
(c) The probability that a random number is between −2 and −1 is:
Z
P (−2 ≤ random ≤ −1) =
−1
−2
Z −1
=
= 0
f (x) dx
0dx
−2
8. Example: the probability density function for the time required to wait for a
phone call to be answered is usually thought to be something like:
½
0 if x < 0
f (x) =
1 −x/3
3e
0.3
0.25
0.2
0.15
0.1
0.05
(a)
-2
0
2
4x
6
8
(b) This is an example of an exponential probability distribution
(c) check: f (x) ≥ 0 for all x, and
Z ∞
Z
f (x) dx =
∞
1 −x/3
dx
e
3
0
¯∞
−e−3x ¯
−∞
=
= 1
0
R2
(d) P (0 ≤ wait ≤ 2) = 0 13 e−x/3 dx ≈ 0.487. Almost half the time, the phone
is answered in less than 2 seconds.
53
9. Example: the probability density function for the weight of male students is
approximately
− (x − 160)2
800
f (x) = 0.02e
0.02
0.015
0.01
0.005
(a)
0
120 130
140
150
160
x
170
180
190
200
(b) This is an example of a normal probability distribution
(c) I’m assuming the average weight is 160 lbs and the standard deviation is
10 lbs.
R∞
(d) Clearly f (x) ≥ 0 for all x. It is not obvious, but −∞ f (x) dx ≈ 1
(e) This is only an approximation, since according to this model there is a
small but finite probability that a male student has negative weight.
i. You may have some lightweight friends, but not that lightweight.
2
− (x − 160)
R0
800
ii. P (−10 < weight < 0) = −10 0.02e
dx
iii. Using numerical integration to evaluate this probability:
A. TestGiver input
integrate(0.02*exp(-(x-160)^2/800);
x,-10,0)
B. TestGiver output
6.14241E-16
iv. We can ignore this discrepency from reality.
− (x − 160)2
800
(f) The proportion of students weighing over 180 pounds is 180 0.02e
dx.
R∞
i. Using numerical integration to evaluate this probability:
A. TestGiver input
integrate(0.02*exp(-(x-160)^2/800);
x,180,1000)
B. TestGiver output
0.159076
ii. Sixteen percent of students weigh over 180 lbs.
10. The probability distribution function for a random variable X is F (x) =
(a) F (x) = P (X ≤ x)
(b) F (x) is an increasing function that starts at 0 and goes to 1
(c) The graph of the probability distribution function of random is
54
Rx
−∞
f (t) dt.
1.2
0.8
0.6
0.4
0.2
0
-0.4 -0.2
-0.2
0.2
0.4
0.6
x0.8
1
1.2
1.4
(d) The graph of the probability distribution function of telephone wait is:
0.8
0.6
0.4
0.2
-2
0
2
4
6
x
8
10
(e) The graph of the probability distribution function of male weight is:
1
0.8
0.6
0.4
0.2
0
120 130
140
150
160
x
170
180
190
200
11. Using the probability distribution function, you can answer questions like:
(a) Given a, P (X ≤ a) =?
i. This is the same as F (a) =?
(b) Given b, P (X ≤?) = b
i. This is the same as F (?) = b
(c) the proportion of values less than 1 is P (X ≤ 1) =?
(d) half the values of the random variable are less than ? P (X ≤?) = 0.5
i. This is the median value of the random variable.
12. If F (x) is a probability distribution function for a random variable X, then the
probability distribution function for X is dFdx(x)
13. The long-term average of all values emitted by a random variable with probability density function f (x) is
Z ∞
x f (x) dx
X̄ =
−∞
55
(a) This value is called the mean of X and is denoted µ (X) or X̄.
R1
(b) the mean of the random number function is X̄ = 0 x dx = 0.5
(c) The mean waiting time for telephone answering is
Z ∞
1 −x/3
X̄ =
xe
dx
3
0
= 3
i. Note that almost half the waiting times are less than 2, but a small
proportion of very long waits raises the average to 3
(d) The mean male weight is
X̄
=
Z
∞
− (x − 160)2
800
0.02 x e
dx
−∞
≈ 160
3.3
Infinite Series
3.3.1
Numerical Series
3.3.1.1
Infinite Sequences
1. The theory of infinite series gives ways to better understand and calculate many
functions. To study these methods, we need to begin with numerical series, or
infinte sums of numbers. Before series, we need to consider infinite sequences of
numbers.
(a) Example:
1 1 1
1, , , , . . .
2 3 4
i. What is the generic term sn
A. What is the sequence in Mathematica or TestGiver notation as an
expression in n?
ii. Does the sequence converge to a finite number? What number?
iii. How big does N have to be before all the remaining terms sn , n > N
are within 0.01 of the limit?
(b) Example:
3 5 7
1, , , , . . .
2 3 4
i. What is the generic term sn
A. What is the sequence in Mathematica or TestGiver notation as an
expression in n?
ii. Does the sequence converge to a finite number? What number?
iii. How big does N have to be before all the remaining terms sn , n > N
are within 0.01 of the limit?
(c) Example:
3 7 15
1, , , , . . .
2 4 8
i. What is the generic term sn
A. What is the sequence in Mathematica or TestGiver notation as an
expression in n?
B. Not all sequences start with the same index.
56
ii. Does the sequence converge to a finite number? What number?
iii. How big does N have to be before all the remaining terms sn , n > N
are within 0.01 of the limit?
(d) Example:
1, 1, 1, 1, . . .
i. What is the generic term sn
A. What is the sequence in Mathematica or TestGiver notation as an
expression in n?
ii. Does the sequence converge to a finite number? What number?
iii. How big does N have to be before all the remaining terms sn , n > N
are within 0.01 of the limit?
(e) Example:
1, 2, 3, 4, 5, . . .
i. What is the generic term sn
A. What is the sequence in Mathematica or TestGiver notation as an
expression in n?
ii. Does the sequence converge to a finite number? What number?
iii. How big does N have to be before all the remaining terms sn , n > N
are within 0.01 of the limit?
(f) Example:
1, −1, 1, −1, . . .
i. What is the generic term sn
A. What is the sequence in Mathematica or TestGiver notation as an
expression in n?
ii. Does the sequence converge to a finite number? What number?
iii. How big does N have to be before all the remaining terms sn , n > N
are within 0.01 of the limit?
(g) Example:
sin 1 sin 2 sin 3
,
,
,...
1
2
3
i. What is the generic term sn
A. What is the sequence in Mathematica or TestGiver notation as an
expression in n?
ii. Does the sequence converge to a finite number? What number?
iii. How big does N have to be before all the remaining terms sn , n > N
are within 0.01 of the limit?
2. What does it mean that a sequence converges to a number?
(a) Close books.
(b) Class discuss
3. A sequence converges or is convergent if and only if the numbers get close and
stay close to some value, called the limit of the sequence. If the sequence is
s1, s2 , s3 , . . . and the limit is L, we write:
lim sn = L
n−→∞
(a) We will use this idea informally. In specific cases the convergence or divergence will usually be obvious for at least some components of the problem.
(b) Compare to examples above
(c) Otherwise the sequence is divergent.
57
i. I think it is Chekov who says that all happy families are the same, but
that unhappy families are all unhappy in their own way. Similarly all
convergent sequences converge in the same way, but there are many
ways for a series to be divergent.
(d) Asking if a sequence converges has two answers, one satisfying and the
other not.
i. Sometimes you can prove that the sequence converges to a specific
limit, or that it diverges
ii. Sometimes you can only prove that the sequence converges, but you
cannot find the limit:
A. In these cases the limit may not have a convenient closed form
iii. It may feel unsatisfying to show that a sequence diverges, but the result
can be important.
A. Divergent sequences can also be very useful. See the theory of
assymptotic series
iv. Another example of a sequence. Suppose you have an infinite decimal
like π = 3.14159 . . .. The value is the limit of the finite decimal parts:
s0
s1
s2
s3
π
=
=
=
=
3
3.1
3.14
3.141
..
.
= lim si
i−→∞
A. All terms after the third are within 0.001 of the limit. All terms
after the fifth are within 0.00001 of the limit.
B. You can write a rule for si :
¡
¢
integer part 10i π
si =
10i
v. Here’s a suitably postmodern example (if modern begins about 1880):
s1
s2
s3
s4
s5
s6
s7
s8
s9
s10
s11
=
=
=
=
=
=
=
=
=
=
=
0.0
0.01
0.011
0.0110
0.01101
0.011010
0.0110101
0.01101010
0.011010100
0.0110101000
0.01101010001
Sequence converges to a decimal: 0.011010100010 . . . but what is the
value? What is the rule?
A. Because it is not known which numbers are prime and which are
not, no one can say if some of the digits of this decimal are 1 or 0,
so no one knows what this number is.
4. There are some theorems that help establish convergence or divergence of a
sequence in specific cases.
58
(a) We will assume the obvious: limn −→∞
|a| > 1.
1
n
= 0 and limn −→∞
1
an
= 0 if
(b) If you have one convergent sequence you can construct others: Suppose
limn−→∞ sn = L. Then:
lim (c ± sn ) = c ± L
n−→∞
lim csn
= cL
n−→∞
µ
¶
1
1
i. For example: limn−→∞ 2 −
= 2 − limn−→∞ = 2
n
n
(c) If you have two convergenct sequences you can construct others. Suppose
limn−→∞ sn = L and limn−→∞ tn = M . Then:
lim (sn + tn ) = L + M
n−→∞
lim (sn − tn ) = L − M
n−→∞
lim (sn tn ) = LM
n−→∞
if M 6= 0 then
lim
n−→∞
µ
sn
tn
¶
=
L
M
i. Example:
2n + 1
lim
n−→∞ 3n + 2
1
n
=
lim
2
n−→∞
3+
n
µ
2+
=
=
=
=
¶
1
limn−→∞ 2 +
n
µ
¶
2
limn−→∞ 3 +
n
1
2 + limn−→∞
n
2
3 + limn−→∞
n
2
1
3 + 2 limn−→∞
n
2
3
(d) The squeeze theorem: if you have three convergent sequences {sn } , {tn } , {un }
and for all n, sn ≤ tn ≤ un , and you know that limn−→∞ sn = limn−→∞ un =
L then limn−→∞ tn = L.
sin2 n
1
i. Example: tn = 1 +
. Let sn = 1 and un = 1 + . Then
n
n
sn ≤ tn ≤ un for all n and limn−→∞ sn = limn−→∞ un = 1. so tn = 1
(e) If limn−→∞ |sn | = 0 then limn−→∞ sn = 0
i. But limn−→∞ |sn | = L 6= 0 in general proves nothing about limn−→∞ sn .
Consider sn = (−1)n .
(f) Here’s a really useful one. If sn = f (n) for some continuous function f (x)
and if limx−→∞ f (x) exists, then limn−→∞ sn = limx−→∞ f (x)
59
i. Example:
sn
=
f (x) =
lim sn
n−→∞
=
=
=
=
=
ln (2n + 3)
ln (3n − 2)
ln (2x + 3)
ln (3x − 2)
lim f (x)
x−→∞
2
2x
+ 3 by L’Hôpital
lim
3
x−→∞
3x − 2
2 (3x − 2)
lim
by algebra
x−→∞ 3 (2x + 3)
6
lim
by L’Hôpital
x−→∞ 6
1
(g) A sequence {sn } is increasing if sn ≤ sn+1 for all n. A sequence {sn } is
decreasing if sn ≥ sn+1 for all n. A sequence is monotonic if it is increasing
or decreasing.
i. NOT if is it partially increasing and partially decreasing but if the
whole series is increasing or if the whole series is decreasing.
ii. A sequence {sn } is bounded if there exist numbers BL and BH such
that BL ≤ sn ≤ BH for all n.
iii. A sequnce that is monotonic and bounded converges
A. this theorem does NOT help find the limit
B. The postmodern example above is an increasing sequence that is
bounded below by 0 and above by 1, so it converges.
3.3.1.2
Infinite Series
1. Now we are ready for series.
(a) An infinite series is an infinite sum of numbers:
Pn
(b) The n0 th partial sum is i=? ai
P∞
i=? ai .
(c) Let’s consider some examples.
2. Example:
1+
1 1 1
+ + + ···
2 4 8
(a) What is the generic term sn
(b) What is the series in summation notation
i. What is the series in Mathematica or TestGiver notation?
(c) Partial sum is
1 + ··· +
1
1
=2− n
n
2
2
(d) Does the series converge to a finite number? What number?
(e) How many terms do you have to take before the partial sums are within
0.001 of the limit.
3. Example:
1 − 1 + 1− 1+ ···
(a) What is the generic term sn
(b) What is the series in summation notation.
60
i. What is the series in Mathematica or TestGiver notation?
(c) What are the partial sums?
(d) Does the series converge to a finite number? What number?
(e) How many terms do you have to take before the partial sums are within
0.001 of the limit.
4. Example:
1−
1 1 1
+ − + ···
2 3 4
(a) What is the generic term sn
(b) What is the series in summation notation.
i. What is the series in Mathematica or TestGiver notation?
ii. Not all sums start with the same index.
(c) What are the partial sums?
(d) Does the series converge to a finite number? What number? (ln 2–see
later)
(e) How many terms do you have to take before the partial sums are within
0.001 of the limit.
5. Example: the harmonic series.
1+
1 1 1
+ + + ···
2 3 4
(a) What is the generic term sn
(b) What is the series in summation notation.
i. What is the series in Mathematica or TestGiver notation?
(c) What are the partial sums?
(d) Does the series converge to a finite number? No
i. Hard proof–see integral test below for easier proof.
1 1 1
+ + + ···
2 µ
3 4 ¶
1 1
1
+
+ ···
= 1+ +
2
3 4
¶
¶
µµ
1
1
+
1
+
·
·
·
+
+
2n
2n+1
+···
µ ¶
1
1
+ ···
≥ 1+ +2
2
4
µ
¶
1
+ ···
+2n
n+1
2
1
1
= 1 + + ··· + + ···
2
2
1+
6. Consider the first two examples. What does it mean that a series converges.
under what conditions do we say a series converges
(a) A series converges if the sequence of partial sums converges, gets close
and stays close to some limit.
7. One series that you can always sum exactly is a geometric series. If |x| < 1 and
P∞
1
an = xn then n=0 an =
.
1−x
61
(a) Proof: the partial sums are
PN
n=0 an =
1
.
converges to
1−x
µ ¶n
P∞
1
1
3
=
=
(b)
n=0
1
3
2
1−
3
µ
µ
¶n
¶n
P∞
9 P∞
9
−3
−3
(c)
=
=
n=2
n=0
4
16
4
16
(d) Class do
µ ¶n
1
n=1
2
P∞
1 − xN
and as N −→ ∞ this
1−x
1
9
µ
¶=
.
−3
28
1−
4
8. Interesting aside: repeating decimals
(a)
¡
¢
0.1515151515 . . . = 0.15 1 + .01 + .012 + · · ·
µ
¶
1
= 0.15
1 − .01
0.15
=
0.99
5
=
33
(b)
0.761515151515. . . . = 0.76 + 0.01 (0.15151515 . . .)
µ ¶
5
= 0.76 + 0.01
33
5
76
+
=
100 3300
2513
=
3300
(c) Class do 0.576363636363 . . .
(d) Every eventually repeating decimal (including eventually 0 decimals) is a
fraction, and every fraction can be represented by a repeating decimal.
i. Proof of part 2: when you do the division algorithm it must eventually
repeat.
9. The following properties often simplify problems:
P
P
P
(a) If
an converges then
can converges to c an .
P
P
P
P
P
(b) If
an and
bn converge then
(an ± bn ) converges to
an ± bn
µ
¶
P
1
2
(c) e.g. ∞
+
n=0
2n 3n
(d) Class find limit.
10. We need some tests for convergence.
(a) Often we will not be able to tell what number the series converges to, but
we can tell that it converges.
11. Easiest but most confusing test is the limit test. This either gives no information or says that a series diverges. When it gives no information, the series
might converge or might diverge.
62
P
(a) . If an 9 0 then
an does not converge.
Pn−1
n−1
(b) Example:
diverges since limn−→∞
= 1 6= 0.
n
n
P
an still does not
(c) Converse is false. It is possible that an −→ 0 but
converge.
1 1
i. For example, the harmonic series 1 + + + · · · does not converge.
2 3
P∞ sin (n)
P∞
(d) Class: does n=1 sin (n) converge? How about n=1
?
n
12. The comparison test can tell us if a series comverges without giving much
information about the limit.
P
P
(a) Given an with an ≥ 0 , if P
we can find another series bn that converges
such that 0 ≤ an ≤ bn then
an converges.
P∞ 1
P∞
1
1
1
. Since 0 ≤ n
≤ n and n=0 n converges
(b) Example:
n=0 n
2 +1
2 +1
2
2
P 1
converges to something less than 1.
to 1,
2n + 1
P
P
an diverges then
bn
(c) You can read this backwards: if 0 ≤ an ≤ bn and
diverges.
P
P
1
n
i. example: since ∞
diverges, ∞
diverges
n=2
n=2 2
n
n −1
P∞
1
(d) Class do n=1 n
2 −1
13. More complex but less delicate than the comparison test is the comparison
ratio test.
P
P
an
(a) If lim
converges to a non-zero value then
an and
bn both diverge
bn
or both converge .
P 1
P 1
converges and
converges, because
(b) Example:
n
2 −1
2n
1/2n
n−→∞ 1/ (2n − 1)
lim
2n − 1
n−−→∞
2n
1
=
lim 1 − n
n−→∞
2
= 1
=
lim
both converge or both diverge.because lim
P 1
P 1
(c) Class: compare
and
3n − 1
2n
¯
¯
¯ an+1 ¯
¯ < 1 then the series converges.
14. Another ratio test: if lim ¯¯
an ¯
1
(a) For example, an = n .
2
¯
¯
n+1
¯ an+1 ¯
1
¯ = lim 1/2
= <1
i. lim ¯¯
¯
an
1/2n
2
1
(b) Class do: an =
¯
¯
¯ n! ¯
¯
¯
¯ an+1 ¯
¯ = c < 1 then for n > N , ¯ an+1 ¯ < d = 1 + c < 1.
(c) Proof: If lim ¯¯
¯
¯
an
an ¯
2
n−N
Thus for n > N, |an | < |aN | d
. Therefore (except for the first few
terms)
0 ≤ |an | < Adn
P n
P
A
Since
Ad converges to
,
|an | converges. By the absolute con1
−
d
P
vergence test,
an converges.
63
15. Perhaps the most powerful test.is the integral test
(a) RLet f be a positive continuous decreasing
P∞ function and an = f (n). Then
∞
f
(x)
dx
converges
if
and
only
if
an converges.
M
(b) starting points don’t matter so much.
R∞ 1
P1
(c)
diverges. because 1
dx diverges (Class prove).
n
x
P 1
π2
) .Class set up integral and show it converges.
(d)
converges
(to
n2
6
P 1
√ diverges. Class set up integral and show it diverges.
(e)
n
P1
1
1
. since < √ for
i. Also this series diverges by comparison with
n
n
n
P1
n ≥ 2 and
diverges.
n
P 1
(f) Class try
n3/2
P 1
(g) p-series:
converges if and only if p > 1.
np
R ∞ −p
i. proof: 1 x dx converges if and only if p > 1.
16. Special tests for series with mixed signs
P
P
(a) Absolute Convergence test: if
|an | converges then
an converges
P∞ sin n
i. Example: n=1 n converges by two steps:
¯
¯ 2
¯ sin n ¯
P
1
1
A. 0 ≤ ¯¯ n ¯¯ ≤ n and ∞
n=1 n converges to 1, so by comparison
2
2
2
¯
¯
P∞ ¯¯ sin n ¯¯
converges to something between 0 and 1.
n=1 ¯
2n ¯
P∞ sin n
B. By the absolute convergence test, n=1 n converges to some2
thing between −1 and 1.
P
sin n
converges
ii. Class: show ∞
n=1
n2
¯
¯
¯ an+1 ¯
¯ < 1 then the series converges. If
¯
(b) Another ratio test: if lim ¯
¯
a
n
¯
¯
¯ an+1 ¯
¯ > 1 then the series diverges.
lim ¯¯
an ¯
1
i. For example, an = n .
2
¯
¯
n+1
¯ an+1 ¯
1
¯ = lim 1/2
¯
= <1
A. lim ¯
¯
an
1/2n
2
1
ii. Class do: an =
¯
¯
¯ n! ¯
¯ an+1 ¯
¯ an+1 ¯
¯
¯
¯ < d = 1 + c < 1.
¯
= c < 1 then for n > N , ¯
iii. Proof: If lim ¯
¯
an
an ¯
2
n−N
Thus for n > N, |an | < |aN | d
. Therefore (except for the first few
terms)
0 ≤ |an | < Adn
P n
P
A
Since
Ad converges to
,
|an | converges. By the absolute
1
−
d
P
convergence test,
an converges.
(c) Alternating series test: an easy test when it applies
i. If an ≥ 0 and an ≥ an+1 and limn−→∞ an = 0 then
converges.
64
P
(−1)n an
1 1 1
1
+ − + · · · converges. an =
2 3 4
n
iii. Must check all hypotheses, especially an monotonically converging to
0.
P sin n
1 1 1 1 1
iv. The series 1 − + − + − and
do not satisfy all the
3 2 5 4 7
n
hypotheses
A. the first example is not decreasing monotonically to 0, the signs in
the second example are not alternating.
P2 (−1)n
converges
v. Class show that n=1
ln n
ii. Example: 1 −
17. Rarely useful but always spectacular: telescoping series.
∞
X
1
2+n
n
n=1
1 1
1
+ +
+ ···
2 6 12
¶
∞ µ
X
1
1
=
−
n n+1
n=1
=
= 1
18. A series we have used as an example
µ above ¶whose convergence we could not
P∞ sin (n)
sin 1
determine: n=1
= arctan
n
1 − cos 1
(a) Requires complex analysis to evaluate
(b) Check numerically:
P1000 sin (n)
i.
= 1. 070 694 152
n=1
µn
¶
sin 1
ii. arctan
= 1. 070 796 327
1 − cos 1
(c) Proof:. This is just for fun. You are not expected to understand all of it.
i.
eix = cos x + i sin x
ii.
∞
X
sin (n)
n
n=1
iii.
∞
X
xn
n
n=1
Ã
!
∞
X
ein
= Im
n
n=1
à ∞ ¡ ¢n !
X ei
= Im
n
n=1
=
Z X
∞
Z
xn dx
n=0
dx
1−x
= − ln (1 − x)
=
iv.
∞
X
sin (n)
n
n=1
¢¢
¡
¡
= Im − ln 1 − ei
¡
¢
= − arg 1 − ei
µ
¶
− sin 1
= − arctan
1 − cos 1
µ
¶
sin 1
= arctan
1 − cos 1
65
19. How to use convergence tests. Given a series
P
an
(a) If limn−→∞ an 6= 0 then the series diverges.
K
(b) If it is the form an = Kxn with |x| < 1 then the series converges to
1−x
¯
¯
¯
¯
¯ an+1 ¯
¯
¯
¯ < 1 then the series converges. If lim ¯ an+1 ¯ > 1 then the
(c) If lim ¯¯
¯
¯
an
an ¯
series diverges.
(d) If all the terms are positive:
i. If an = f (n) with f (x) Ra decreasing positive function, then the series
∞
converges if and only if
f (x) dx converges.
P
ii. If you can find a convergent series
bn such that either bn ≥ an or
P
an
limn−→∞
< ∞ then
an converges.
bn
(e) If the terms have mixed signs:
i. If limn−→∞ an 6= 0 and the series is alternating and terms, in absolute
value, are decreasing, then the series converges.
P
P
ii. If
|an | converges, then so does
an .
66
3.3.2
Power Series
1. A power series is an expression of the form::
a0 + a1 x + a2 x2 + · · ·
for some constants an .
(a) When x gets a numerical value, the power series turns into an ordinary
series of numbers
(b) The set of all power series is denoted R [[x]] if you like fancy notation.
(c) We have made a choice of what to study here. Other things we might study
are:
i. polynomials: a0 + a1 x + a2 x2 + · · · where an = 0 for almost all n (all
but a finite number of n). The set of polynomials is R [x].
p (x)
ii. rational functions:
where p, q are polynomials. The set of rational
q (x)
functions is R (x).
iii. Laurent series an xn +an xn+1 + · · · where n can be any integer, positive
or negative. The set of Laurent series is denoted R ((x)). We may show
later that R (x) ⊂ R ((x))
2. For some but not necessarily all values of x, a power series converges to a
value. For these values of x, the power series is a function. For other values of
x, the power series does not converge. The power series, with x an unevaluated
variable, is a mathematical thing that can be studied in its own right.
(a) I’ll try to distinguish when I’m talking about a formal power series and a
series where x has a numerical value.
x2 x3
(b) The series 1 + x +
+
+ · · · converges for all values of x
2!
3!
i. the value converged to is ex
(c) The series 1 + x + x2 + x3 + · · · converges for |x| < 1.
1
i. the value converged to is
1−x
¯
¯
¯ an ¯
¯
¯ = c (which
3. Theorem: if n=0 an x is a power series and limn−→∞ ¯
an+1 ¯
means that the limit has to converge) then the series converges for |x| < c and
diverges for |x| > c. If c = ∞ then the series converges for all x.
P∞
n
(a) c is called the radius of convergence of the series.
(b) If |x| = c then there is no information about the convergence of the series.
It might converge; it might diverge.
(c) If the limit does not converge, then there is no information. More sensitive
tests of this sort are known–try Math 370.
(d) Example: 1 + x + x2 + · · · .
i. an = 1 for all n
1
ii. c = limn−→∞ = 1.
1
x3
x2
+
+ ···.
(e) Example: 1 + x +
2!
3!
1
i. an =
n!
1
n!
1
(n + 1)!
ii. c = limn−→∞
= limn−→∞
= limn−→∞
=0
1
(n + 1)!
n+1
n!
67
¯
¯
¯
¯
¯ an+1 xn+1 ¯
¯ an+1 ¯
¯
¯
¯
¯ = |x| 1 . By the ratio
= |x| limn−→∞ ¯
(f) Proof: limn−→∞ ¯
¯
n
an x
an ¯
c
|x|
|x|
< 1 and diverges for
> 1. That is, the
test, the series converges if
c
c
series converges for |x| < c and diverges for |x| > c
4. Class find radius of convergence.
x2 x3
+
+ ···
2
3
(b) x + 2x2 + 3x3 + · · ·
(a) x +
x2 x3
+
+ ···
4
9
x x2 x3
+
+ ···
(d) 1 + +
2
4
8
nxn
x x2 3x3 x4 5x5
+
+
+
+
+ ··· + n + ···
(e)
2
2
8
16
32
2
(c) x +
5. Example: p = x −
x3 x5
+
− ···
3!
5!
n−1
2
(a) if n is even then an = 0; if n is odd then an =
¯
¯
¯ an+1 ¯
¯
¯ doesn’t exits. What to do?
(b) limn−→∞ ¯
an ¯
(−1)
n!
t
t2
t7
+ − + ···.
3! 5! 7!
(−1)n
an =
(2n + 1)!
c=0
q converges for all t
x2
x4
+
− · · · converges for all x
q1 = 1 −
3!
5!
p = x q1 converges for all x
(c) q = 1 −
i.
ii.
iii.
iv.
v.
6. Class: show carefully that the series 1 −
3.3.3
x2 x4 ‘ x6
+
+
− · · · converges for all x
2
4!
6!
Representing Functions with Power Series
1. We know for |x| < 1,
1
1−x
= 1 + x + x2 + x3 + · · ·
= cxi
2. Powers of x can be substituted for x. Therefore, for |x| < 1
1
1+x
=
∞
X
(−x)
i
i=0
=
∞
X
(−1)i xi
i=0
= 1 − x + x2 − x3 + · · ·
1
= 1 + x2 + x4 + · · ·
1 − x2
68
(a) Class: find power series for
1
1 + x2
3. ¯Multiples
of x can replace x, but that changes the radius of convergence. For
¯
¯ x ¯ < 1 or |x| < 2:
2
1
1−
x
2
2
2−x
x ³ x ´2
= 1+ +
+ ···
2
2
∞
X xi
=
2i
i=0
=
(a) Class: find power series for
1
. What is radius of convergence?
1 − 2x
4. Series can be multiplied by constants, and finite polynomials can be added and
subtracted to series without changing the radius of convergence. For |x| < 1:
3
= 3 + 3x + 3x2 + 3x3 + · · ·
1−x
5. Series can be added and subtracted. If both summands converge, so does the
sum. For |x| < 1:
1
2
+
1 − x 1 + x2
3 − 2x + x2
1 − x + x2 − x3
= 3 + x − x2 + x3
+3x4 + x5 − x6 + x7
+3x8 + · · ·
=
(a) Class: Check previous result by finding power series for
µ
¶
1
1
1
1
=
+
1 − x2
2 1−x 1+x
What is radius of convergence?
1
, you have
6. These operations can be combined. To get a power series for
2−x
a
:
to manipulate the expression into a form
1 − xn /b
1
2−x
=
=
=
1/2
1 − x/2
µ
¶
1
x x2 x3
1+ +
+
+ ···
2
2
4
8
1 x x2 x3
+ +
+
+ ···
2 4
8
16
This series converges for |x/2| < 1 or |x| < 2.
(a) Class: find power series for
2
and find radiusof convergence.
3 − x2
7. Series can also be multiplied and divided.
(a) Multiplication is like polynomial multiplication, but it goes on forever.
¢¡
¢
¡
a0 + a1 x + a2 x2 + · · · b0 + b1 x + b2 x2 + · · ·
= a0 b0 + (a0 b1 + a1 b0 ) x + (a0 b2 + a1 b1 + a2 b0 ) x2 + · · ·
69
(b) The power series for
1
(1 − x)2
is
¡
¢2
1 + x + x2 + · · ·
= 1 + 2x + 3x2 + · · ·
This series converges for |x| < 1.
1+x
can be calculated:
(c)
1−x
1+x
1−x
¢
¡
= (1 + x) 1 + x + x2 + x3 + · · ·
= 1 + 2x + 2x2 + 2x3 + · · ·
This series converges for |x| < 1.
(d) Class: confirm previous result by computing
µ
¶µ
¶
1
1
1
=
1 − x2
1−x
1+x
¢¡
¢
¡
2
1 + x + x + · · · 1 − x + x2 − · · ·
(e) To divide series, set up an equation. Remember, the a’s and b’s are known,
and we are solving for the c’s .We assume b0 6= 0.
a0 + a1 x + a2 x2 + · · ·
= c0 + c1 x + c2 x2 + · · ·
b0 + b1 x + b2 x2 + · · ·
¡
¢¡
¢
b0 + b1 x + b2 x2 + · · · c0 + c1 x + c2 x2 + · · ·
= a0 + a1 x + a2 x2 + · · ·
b0 c0 = a0
a0
c0 =
b0
b0 c1 + b1 c0 = a1
a1 − b1 c0
c1 =
b0
b0 c2 + b1 c1 + b2 c0 = a2
a2 − b1 c1 − b2 c0
c2 =
b0
8. Series can be integrated and differentiated. This is how we represent many
functions.
¶0
µ
1
1
=
1−x
(1 − x)2
= (1 + x + x2 + x3 + · · · )0
1 + 2x + 3x2 + · · ·
Z
dx
1 + x2
Z
¡
¢
=
1 − x2 + x4 − · · · dx
arctan x =
= x−
x3
x5
+
− ···
3
5
Both series converge is |x| < 1
(a) Therefore
π
4
= arctan 1
= 1−
70
1 1 1
+ − + ···
3 5 7
Check:
4
1000
X
i=0
(−1)i
= 3. 142 591 654.
2i + 1
(b) Remarkable: a way to calculate π to any accuracy. And we can do better
tan
π
6
π
=
1
√
3
1
= 6 arctan √
3
!
Ã
1
1
1
= 6 √ −
√ +
√ − ···
3 3 × 33 5 × 35
µ
¶
6
1
1
1
= √
1−
+
−
+ ···
3 × 3 5 × 9 7 × 27
3
∞
n
X
√
(−1)
= 2 3
(2n − 1) 3n
n=0
10
√ X
2 3
(−1)n
= 3. 141 593 305
(2n + 1) 3n
n=0
(c) Class: find series for
Z
dx
1+x
Z
¢
¡
=
1 − x + x2 − · · · dx
ln (1 + x) =
i. What is the radius of convergence of your series.
ii. Show
1 1 1
+ − + ···
2 3 4
∞
n−1
X
(−1)
=
n
n=1
ln 2 = 1 −
= − ln
1
2
1/2 1/22 1/23 1/24
+
+
+
+ ···
1
2
3
4
∞
X 1
=
n2n
n=1
=
Which series converges faster?
ln 2 = 0.693 147 180 6
10
X
(−1)n−1
= 0.645 634 920 6
n
n=1
100
X
(−1)n−1
n
n=1
10
X
1
n
n2
n=1
100
X
1
n2n
n=1
71
= 0.688 172 179 3
= 0.693 064 856 2
= 0.693 147 180 6
3.3.4
Taylor Polynomials and Remainders
1. There are two reasons for studying infinite series
(a) Given a transcendental function like ex or sin x, how do you calculate its
values.
i. How does your calculator or Mathematica or TestGiver know the values?
ii. How did Newton and Euler get the values?
Rx√
(b) How do you represent a function like f (x) = 0 1 + t3 dt which is not
a combination of elementary functions like sine and logarithm and square
root.
2. Here’s a useful fact about integrals. If f (x) is continuous on (a, b) then there is
Rb
a number c between a and b such that a f (x) dx = f (c) (b − a).
3. Here’s a big theorem, the mean value theorem.
(a) For almost any function you will encounter in this course (technically: if f
is continuous on [a, x] and differentiable on (a, x)) then
f (x) = f (a) + f 0 (c) (x − a)
for some c in (a, x).
i. Proof: the theorem follows from the fundamental theorem of calculus,
since
f (x) = f (a) + (f (x) − f (a))
Z x
= f (a) +
f 0 (t) dt
a
= f (a) + f 0 (c) (x − a)
for some c in (a, x).
4. Here is an extension that works for almost all functions we will encounter defined
on an interval [a, x]. (Technically we need f continuous on [a, x] and be n + 1
times differentiable on (a, x)):
f (x) = f (a) + f 0 (a) (x − a) + · · · +
f (n) (a)
f (n+1) (c)
(x − a)n +
(x − a)n+1
n!
(n + 1)!
for some c in (a, x).
f (n) (a)
(x − a)n is called the Taylor
n!
polynomial for f (x) expanded about x = a .
(a) The part f (a) + f 0 (a) (x − a) + · · · +
f (n+1) (c)
n+1
is called the remainder term.
(x − a)
(n + 1)!
(c) n is called the degree of the Taylor polynomial.
(b) The part
π
5. Example: Find the Taylor polynomial of degree 4 around for sin x. Also find
2
the remainder term as best you can.
n
dn
sin x
(a) We need to fill in the table: dxnn
¯
¯
d
sin x¯¯
dxn
x=π/2
72
0
1
2
3
4
sin x
cos x
− sin x
− cos x
sin x
1
0
-1
0
1
Thus the Taylor polynomial is:
³ ´
³ ´
³ ´
(2) π ³
(3) π ³
(4) π ³
³π ´ ³
´
´
´
f
f
f
2
3
π
π
π
π
2
2
2
P4 (x) = f
+
+
+ f0
x−
+
x−
x−
x−
2
2
2
2!
2
3!
2
4!
2
1³
1 ³
π ´2
π ´4
= 1−
+
x−
x−
2
2
24
2
³π ´
and the remainder is:
π ´5
cos c ³
x−
120
2
π
for some value of c between and x.
2
(b) Thus for any x,
π
sin x = p (x) + f (x, c) for some c between and x
2
1 ³
π ´2
π ´4
1³
+
x−
x−
= 1−
2
2
24
2
³
´
cos c
π
π 5
+
for some c between
x−
and x
120
2
2
6. Class find Taylor polynomial of degree 4 and remainder for ex about x = 0.
(a) Write ex = p (x) + f (x, c) for some c between 0 and x.
7. We can use the Taylor polynomial and remainder to estimate sin 2.
(a) This is what is so valuable about Taylor polynomials.
cos c ³
π ´5
sin (2) = P4 (2) +
2−
120
2
1³
1 ³
π ´2
π ´4
P4 (2) = 1 −
+
2−
2−
2
2
24
2
=
0.909
¯
¯
¯ cos c ³
0.55
π ´5 ¯¯
¯
¯ 120 2 − 2 ¯ < 120
= 0.00026
sin 2 = 0.909 ± 0.00026
8. This is revolutionary. We can actually calculate accurate values for sin x. We
can produce estimates and error bounds.
(a) Class use the degree 4 Taylor polynomial for ex to estimate e0.2 and put a
bound on the error.
9. More words. A Taylor polynomial expanded about 0 is called a Maclaurin
polynomial.
10. Here are some common Maclaurin polynomials with remainder:
xn
ec xn+1
+
n!
(n + 1)!
3
2n+1
2n+2
x
n x
n+1 (sin c) x
sin x = x −
+ · · · + (−1)
+ (−1)
3!
(2n + 1)!
(2n + 2)!
2
2n
x
(sin c) x2n+1
x
+ · · · + (−1)n
+ (−1)n+1
cos x = 1 −
2!
(2n)!
(2n + 1)!
n+1
1
x
= 1 + x + · · · + xn +
n+2
1−x
(1 − c)
ex
= 1 + x + ··· +
(a) You are expected to remember these.
73
3.3.5
Taylor Series and Convergence
1. Given a function and points a, x we have the Taylor polynomial and remainder
of any degree:
f (x) = f (a) + f 0 (a) (x − a) + · · · +
f (n) (a)
f (n+1) (c)
(x − a)n +
(x − a)n+1
n!
(n + 1)!
(a) If we let the Taylor polynomial about a go on forever, we call it the Taylor
series about a.
i. If a = 0 we call it the Maclaurin series.
(b) The Taylor series encodes all the Taylor polynomials, but doesn’t tell us
much about the remainder.
f (n) (a)
n
(c) We write f (x) ∼ (a) + f 0 (a) (x − a) + · · · +
(x − a) + · · ·
n!
i. The symbol ∼ means “Taylor series for”.
2. Here are some common Maclaurin series (Taylor series about 0):
ex
∼ 1 + x + ··· +
xn
+ ···
n!
x2n+1
x3
+ · · · + (−1)n
+ ···
3!
(2n + 1)!
2n
x2
n x
+ · · · + (−1)
+ ···
cos x ∼ 1 −
2!
(2n)!
1
∼ 1 + x + · · · + xn + · · ·
1−x
sin x ∼ x −
(a) know these.
3. You cannot evaluate a Taylor series directly, because you cannot add up an
infinite number of terms
4. But you can add up an initial segment of the Taylor series, which is a Taylor
polynomial.
(a) You can hope that the resulting values approximate the function which
gave you the Taylor series.
(b) You can hope that the more terms you add, the more accurate will be your
approximation
(c) You can hope that it will rain nickles tomorrow.
(d) There is a better chance that the Taylor polynomials will give better and
better approximations to the function than that it will rain nickles tomorrow.
(e) However, Taylor polynomials DO NOT always approximate the function.
i. Convergence of series is like convergence of improper integrals.
ii. Sometimes it happens, and sometimes it doesn’t
iii. There are ways to know.
5. When the Taylor polynomials give better and better approximations–when the
difference between the approximations and the function value is eventually less
than any predetermined positive “error”, we say that the Taylor series converges
to the function.
6. We can use the Taylor polynomial and remainder to determine if the series
converges to the function.
74
7. In the last section, we used the remainder to determine the number of terms
required to approximate a function to a certain accuracy. Here we are just
concerned with determining if the series eventually converges, an easier question.
(a) It might appear that there is no need to consider the remainder.
(b) Given a function f (x),
i. construct the Taylor series by evaluating the derivatives of f (x).
ii. use the ratio test to determine the radius of convergence for the series.
iii. that will tell you for which x the series converges to the function
(c) but even if the series converges, how do you know that the series converges
to the function?
i. There are examples of Taylor series that do not converge to their functions
2
ii. The Taylor series about 0 for f (x) = e1/x is 0 + 0x + 0x2 + · · · , which
is not equal to the function for any x 6= 0
(d) So to determine if a Taylor series converges to a function you should examine the remainder term
(e) But the details are best left for a more advanced analysis course, so we will
believe
1
1
i. the series for
converges to
for |x| < 1
1−x
1−x
ii. the series for ex , sin x and cos x converge for all x
iii. any series derived from these by addition, multiplication, integration
and differentiation has the same radius of convergence as the components.
iv. division can change the radius of convergence
A. the series for sin x and cos x converge for all x, but the series for
sin x
π
tan x =
converges for |x| < because tan π2 is undefined.
cos x
2
2 5
17 7
62
x + 315
x + 2835
x9 + · · ·
B. the series is: tan x = x + 13 x3 + 15
there is no apparent pattern
8. Here’s a sample calculation for you to read that we’ll skip in class.
(a) The Maclaurin polynomial of degree n with remainder for ex is:
ex = 1 + x +
x2
xn
ec xn+1
+ ··· +
+
2!
n!
(n + 1)!
for some c between 0 and x.
(b) If we hold x fixed and let n get bigger and bigger, what happens.
i. For each value of n we have a different c, but no matter what the value
of n we can say that the remainder R satisfies
|R| ≤
max (1, ex ) xn+1
n−→∞
(n + 1)!
lim
75
max (1, ex ) xn+1
(n + 1)!
xn+1
n−→∞ (n + 1)!
xn
= max (1, ex ) lim
n−→∞ n!
= max (1, ex ) lim
ii. I want to show that this limit is 0. Choose N > x. Then for n > N
we have
xn
(n)!
=
<
0 ≤
<
=
xN
xn−N
N ! (N + 1) · · · n
µ
¶n−N
x
xN
N! N + 1
xn
lim
n −→∞ (n)!
µ
¶n−N
x
xN
lim
N ! n −→∞ N + 1
0
(c) This means that, no matter what x you pick, the longer the Maclaurin
polynomial for ex , the better the approximation. You can always pick n
so that the nth approximation has error smaller than any desired amount.
(But you might have to pick a very large n.)
(d) In this case we say that the series converges to the function. That means
that we can make the error between the Taylor polynomial value and the
function value as small as we want by taking a sufficiently large Taylor
polynomial.
(e) A similar argument shows that the Maclaurin series for sin and cos converge
for all x, but some series only converge for some values of x.
9. Here’s another sample calculation to be skipped in class. We can show that the
1
series for
converges for −1 ≤ x ≤ 1, but not for other values of x
1−x
(a) In fact the series diverges for other values, since the individual terms are
getting bigger and bigger.
xn
1
is R =
(b) The remainder for
n+2 , and we cannot show that this
1−x
(1 − c)
goes to 0. But we have a more precise remainder:
(1 − x) (1 + · · · + xn ) = 1 − xn+1
1 − xn+1
1 + · · · + xn =
1−x
1
xn+1
= 1 + x + · · · + xn +
1−x
1−x
xn+1
.
1−x
xn+1
ii. If −1 ≤ x ≤ 1 we have limn −→∞
= 0 so the series converges.
1−x
xn
does not converge to 0.
iii. If |x| ≥ 1 we have limn−→∞
1−x
1
is not defined so we cannot say anything about its
A. If x = 1,
1−x
Maclaurin series
xn
1
B. If x = −1 then
= ± and the values do not converge to
1−x
2
anything
xn
C. If x > 1 then
converges to −∞
1−x
n
x
oscillates between values that are larger and
D. If x < −1 then
1−x
larger positive and negative
i. The remainder can be written
76
(c) Therefore
1
≈ 1 + x + x2 + · · · + xn if − 1 < x < 1
1−x
10. We say almost the same thing three ways:
ex
ex
ex
xn
ec xn+1
+
n!
(n + 1)!
xn
∼ 1 + x + ··· +
+ ···
n!
n
x
≈ 1 + x + ··· +
n!
= 1 + x + ··· +
(a) For any x, ex is equal to the right-hand side for some value of c between 0
and x.
(b) The Maclaurin series for ex (Taylor series about 0).
(c) An approximation for ex
11. How good are the approximations:
1
0.5
-4
-2
2
4
x
-0.5
-1
sin x
red
x
black
x−
x3
6
green
x−
x5
x3
+
6
120
purple
x−
x5
x7
x3
+
−
6
120 5040
blue
(a) Here’s another
2
1
-1
-0.8 -0.6 -0.4 -0.2
0
0.2
-1
-2
i.
77
0.4 x 0.6
0.8
1
1
1−x
1+x
1 + x + x2 + x3
1 + x + x2 + x3 + x4 + x5
1 + x + x2 + x3 + x4 + x5 + x6 + x7
red
black
green
purple
blue
(b) Euler, in a moment of weakness, suggested that
3.3.6
1
= 1−1+1−1+1−···
2
Manipulating Taylor Series
1. Finding a Taylor series for each function would be difficult, but we can use the
four basic series to get a lot of others. This is like using differentiation rules
instead of computing the limit for each derivative.
(a) To find the Taylor series of f (x)+g (x) about a of degree n, find the Taylor
series for f and g and add them.
(b) To find the Taylor series of f (x) g (x) about a of degree n, find the Taylor
polynomials for f and g and multiply them up to degree n
(c) Example:
x3 2x4
+
+ ···
3! µ 4! ¶
µ
¶
1
1
1
1
1
x
3
4
−
+x
− +
+ ···
e cos x ∼ 1 + x + x
3! 2
4! 4 4!
1
1
= 1 + x − x3 − x4 + · · ·
3
6
ex + cos x ∼ 2 + x +
(d) Since the components converge for all x, so does the sum and product.
(e) Class do sin x + cos x, sin x cos x, degree 4.
2. When you combine series (or integrate or differentiate), the result converges if
the components do.
x3 x5
+
− ···
3!
5!
converges for all x
sin x ∼ x −
sin x2
x6 x10
+
− ···
3!
5!
converges for all x
= x2 −
cos x = (sin x)0
µ
¶0
x3 x5
∼
x−
+
− ···
3!
5!
2
4
x
x
= 1−
+
− ···
2!
4!
78
(sin x)2
∼
(sin x)2
=
=
µ
Z
x−
x3
x5
+
− ···
3!
5!
¶2
ugh
2 sin x cos x dx
Z
sin 2x dx
!
Z Ã
(2x)3 (2x)5
∼
2x −
+
− · · · dx
3!
5!
23 x4 25 x6
+
− ···
4!
6!
2x4
4x6
8x8
+
−
+ ···
= x2 −
1·3 1·3·5 1·3·5·7
= x2 −
cosh x =
ex + e−x
³ 2
1+x+
∼
x2
2!
´ ³
+ ··· + 1 − x +
x2
2!
− ···
2
´
x2 x4
= 1+
+
+ ···
2!
4!
converges for all x
3. Class find first few terms of Maclaurin polynomials for
(a) cos (x/2)
(b) sin x cos x two ways: by multiplication and by sin x cos x =
1
sin 2x
2
¡ ¢0
4. Class: use Taylor series to show e2x = 2e2x
R
5. Class: use Taylor series to show xex dx = xex − ex
R 0.5
6. Class: use Taylor series to approximate 0 sin t2 dt
7. Class: find the Taylor series for
1
=
(Hint:
2 + 3x2
3.3.7
1
2
)
3
1 + x2
2
1
. What is the radius of convergence?
2 + 3x2
Binomial series
1. We know
(1 − x)−1 ∼ 1 + x + x2 + x3 + · · ·
n
(a) converges for |x| < 1. Actually this is a special case of (1 − x) . We know
what to do if n is a positive integer:
(1 − x)0
(1 − x)
(1 − x)
1
2
∼ 1 + 0x + 0x2 + 0x2 + · · ·
∼ 1 − x + 0x2 + 0x2 + · · ·
∼ 1 − 2x + x2 + 0x2 + · · ·
2. More generally, define the binomial coefficients
³n´
n!
=
i
i! (n − 1)!
n (n − 1) · · · (n − i + 1)
=
i!
which works for any number n so long as i is a non-negative integer. If n is a
non-negative integer we have:
79
(a)
³n´
=
³n´
=1
¶
n
=
=1
1
n−1
µ
¶
³n´
n
=
i
n−i
³n´
= 0 if i > n
i
0
³n´
µn
and most importantly, the binomial theorem
³n´
³n´
³n´
³n´
n
(a + b)
=
an b0 +
an−1 b1 + · · · +
an−i bi + · · · +
a0 bn
0
1
i
n
n (n − 1) n−2 2
= an + nan−1 b +
b + · · · + nabn−1 + bn
a
2
n (n − 1) 2
(1 + x)n = 1 + nx +
x + · · · + nxn−1 + xn
2
(b) Class do (1 + x)4
3. Newton was able to extend this result to all powers, not just integral powers.
In general:
n
(1 + x) ∼ 1 + nx +
³n´
n (n − 1) 2
xi + · · ·
x + ··· +
i
2
(a) This is an infinite series unless n is non-negative integer. It converges for
|x| < 1. For example:
µ
¶µ
¶
1
1
3
−
−
1
1
2
2
2
1/2
∼ 1 + x − x2 +
x3 + · · ·
(1 + x)
2
8
3!
1
1
1
= 1 + x − x2 + x3 + · · ·
2
8
16
2
3
√
(−0.1) (−0.1)
(−0.1)
0.9 ≈ 1 +
−
+
+ ···
2
8
16
= 0. 948 69
Check: 0. 948 692 = . 900 01
0.1
0.05
√
x
1−x≈1−
2
0.01
0.001
0.95
0.975
0.995
0.9995
0.948683
0.974679.
0.994987
0.99949987
(b) How good is the really cheap approximation
i.
x
x
1−
2
√
1−x
4. Class: use the binomial expansion to check the Maclaurin series for
(1 − x)−1 .
1
=
1−x
5. Class
¡
¢−1/2
(a) do four terms for 1 − x2
(b) do four terms for arcsin (x)
(c) If you continued the series, what would the radius of convergence be?
80
3.3.8
Taylor Series and differential equations
1. Series come from functions (Taylor series) and from differential equations
(a) Consider y 0 = y − x, y (0) = 1.
i.
ii.
iii.
iv.
y ∼ a0 + a1 x + a2 x2 + · · ·
y − x ∼ a0 + (a1 − 1) x + a2 x2 + · · ·
y 0 ∼ a1 + 2a2 x + 3a3 x2 + · · ·
y (0) = a0 so a0 = 1 so
a0
a1
a1
2a2
a2
3a3
a3
=
=
=
=
=
=
=
1
a0
1
a1 − 1
0
a2
0
..
.
= an−1
= 0
nan
an
v. solution is 1 + x.
(b) change to y 0 = y − x, y (0) = 0
a0
a1
a1
2a2
a2
3a3
a3
nan
an
y
=
=
=
=
0
a0
0
a1 − 1
−1
=
2
= a2
−1
=
6
..
.
= an−1
−1
=
n!
1
1
−1 2
x − x3 − x4 − · · ·
2!
3!
4!
= 1 + x − ex
∼
(c) Class: use series to solve y 0 = −2xy.
3.4
Differential Equations
3.4.1
Introduction
1. We began by studying integrals, which are useful for calculating quantities that
can be viewed as accumulations
81
(a)
(b)
(c)
(d)
Area is an accumulation of little strips of area
Arc length is an accumulation of little bits of arc length
Volume is an accumulation of little slices or cylinders
Work is an accumulation of little forces
2. There are many ways to calculate integrals, some exact, some approximate
(a) Exact calculations of integrals are carried out with antiderivatives, using
the fundamental theorem of calculus
(b) It turns out that a more general process related to antidifferentiation can
solve a lot of other scientific problems
3. A differential equation is an equation that tells you how a function and its
derivative are related. The solution is a function, not a number.
(a) The differential equation
y0 = y
tells you that you are looking for a function y (t) that equals its derivative.
i. Solutions are y = et , y = −et , y = 3et ,
ii. most general solution y = Cet
iii. If we add an initial condition y (0) = 2 we get a single solution y = 2et
(b) Class solve y 0 = −y, y (1) = 3
(c) When you find the antiderivative of a function f (x) you are solving the
differential equation y 0 = f (x).
(d) In general, differential equations are hard to solve. Imagine:
xy 00 + 2y 0 + xy = 0
One solution is y =
sin x
.
x
i. Class check.
x
d2
dx2
µ
sin x
x
¶
ii. The generic solution is C1
+2
d
dx
µ
sin x
x
¶
+x
µ
sin x
x
¶
=0
sin x
cos x
+ C2
.
x
x
4. Differential equations are the language of science.
(a) The first scientist to use differential equations was Isaac Newton, who
stated and used a differential equation for gravity. If y is the distance
between two bodies, then the distance changes according to the law:
y 00 =
−G
y2
i. Before Newton could do this, he had to invent calculus. Smart guy.
ii. Scientists that followed understood the power of calculus both for expressing scientific laws and using them to predict phenomena.
(b) If a mass M is attached to a spring with spring constant k, let y denote
the position of the spring at time t, with y = 0 corresponding to the rest
position of the spring. Then the acceleration of the spring y 00 is proportional
to the force on the spring, which is −ky, so the position function y for the
mass satisfies the differential equation:
M y 00
M y + ky
00
= −ky
= 0
The solution with initial condition y (0) = 0, y 0 (0) = 1 is y = sin
82
Ãr
!
k
t
M
i. Class check
(c) Population theory. A population is represented by a function p (t), where t
is time measured in years. Suppose the population grows by 2% each year.
Then the rate of change p0 (t) of the population function is a constant
multiple of p (t), or
p0 = kp
(Finding k from the given 2% is indirect, as we will see.) If we assume that
the population at time 0 is p (0) = 10, 000, then
p (t) = Cekt (general solution)
p (0) = C = 10, 000
p (t) = 10000ekt (particular solution)
We know that p (1) = 10, 200, so we can find k:
k
10200 = 10000e
µ
¶
10200
k = ln
10000
= 1. 980 3 × 10−2
So the final solution for the population function is:
p (t) = 10000e1.9803t
(d) Class: find population function if decreases at 3% each year and starts at
500.
(e) Population theory also applies to radioactive decay.
(f) Now for the bad news. We all die. Populations don’t increase forever. Realistic population theory takes account of population growth, stabilization
and decay.
i. Populations grow when they are small (relative to carrying capacity of
environment)
ii. Populations decrease when they exceed carrying capacity
iii. Populations at carrying capacity level stay there (except for random
fluctuations and some other very interesting phenomena leading to
chaos theory–the local expert is Dr. Goetz.)
(g) The general model is is:
P 0 = kP
µ
¶
P
1−
K
Where k and K are positive constants. A “solution” is a function P (t)
that describes population levels at time t.
If 0 < P (t) < K then P 0 (t) > 0 and the population is increasing.
If P (t) > K then P 0 (t) < 0 and the population is decreasing.
If P (t) ≈ K then P 0 (t) ≈ 0 and the population is changing very slowly.
So long as 0 < P (t) << K then P 0 ≈ kP and population grows
exponentially.
v. P (t) = K, is a special solution. One solution is a population that
stays at the same level K for all time.
et
vi. If we set K = k = 1 we get the solutions y = t
. Here are
1
e + y(0)
−1
the graphs of some solutions for y (0) = 0.25, 0.6, 1.25, 2.0.
i.
ii.
iii.
iv.
83
2
1.75
1.5
1.25
1
0.75
0.5
0.25
0.5
1
1.5
2
2.5
3
A. Class check that solution satisfies equation, that initial value correct, that limit is 1 as t −→ ∞.
3.4.2
Separable Differential Equations
1. Some differential equations can be solved exactly and easily. These are the
so-called separable differential equations. They have the form:
y 0 = f (t)g (y)
(a) the solution is
dy
dx
dy
g (y)
Z
dy
g (y)
= f (x) g (y)
= f (x) dx
Z
=
f (x) dx
2. Example:
y0 = y
y (0) = 1
dy
dt
dy
y
Z
dy
y
ln y
y
= y
= dt
Z
=
dt
=
=
=
1 =
y =
t + C1 put arbitrary constant on one side
et+C1
C2 et
C2 e0 put in initial condition
et
3. Another example:
y 0 = ty
y (0) = 2
84
dy
dt
dy
y
Z
dy
y
ln y
y
= ty
= t dt
Z
=
t dt
t2
+ C1
2
2
= et /2+C1
=
2
= C2 et
/2
4. Class try some
3.4.3
(a) y 0 = y/t, y(1) = 1
1
(b) y 0 = , y (0) = 1
y
(c) y 0 = y 2 , y (0) = 1
p
(d) y 0 = 1 − y 2 , y (0) = 0, y (0) = 1
Direction Fields
We are interested in differential equations that have the form y 0 = f (x, y), but not
necessarily separable equations..
1. Example:
y 0 = t2 + y 2
We cannot find exact solution,
(a) We know IF there was a solution y (t) such that y (1) = 2 THEN the slope
of the graph of y (t) at t = 1 would be y 0 (1) = 12 + y (1)2 = 12 + 22 = 5.
(b) More generally, the differential equation tells us the slope of the graph of
the solution function y (t) IF we know a point on the graph.
(c) Could you drive car if the windows were all blacked out, but you had a
compass and you got a constant stream of instructions from the radio: “go
70◦ , go 100◦ , go 120◦ ”? That is what an airline pilot does. Or a boat in
the fog.
(d) If we pick an initial value for our differential equation, say y (0) = 0, then
the differential equation gives is directions how to drive across the ty-plane
to trace out the graph of y (t). We start with slope 0, but as soon as we
have moved a little bit, we have to change our slope and start going up.
(e) The “driving instructions” can be encoded graphically in a direction field.
At representative points (t, y) draw a little line with slope t2 + y 2 .
i. Class do it at (−2, −2) · · · (2, 2).
(f) You can draw the direction field in Mathematica. Here are the commands.
Needs[’’Graphics‘PlotField‘’’]
slopeField[f ] := {1, f}/Sqrt[1 + f^2]
PlotVectorfield[slopeField[tˆ2 + yˆ2],
{t,-2,2},{y,-2,2},Frame->True]
i. Here’s the result.
85
ii. The idea is that at representative points (t, y) we draw a vector of
length 1 with slope y 0 = t2 + y 2 . The formula for the vector is:
¢
¡ 2
1, t + y 2
q
12 + (t2 + y 2 )2
In Mathematica we begin by loading the package for drawing vector
graphics. Don’t forget to use double quotes and backwards single
quotes. Then we plot the desired vector field.
A. Remark: we define
F
G= p 2
F1 + F22
not
F
G= p
1 + (t2 + y 2 )2
Both definitions would be the same for our example, but the first
would work for any vector function F while the second would not.
(For G to work for any F you need the colon-equals instead of
plain equals in Mathematica.)
B. Remark: Graphing vectors of length 1 works better than graphing
the vectors F . Here is a graph of F :
86
2
1
0
-1
-2
-2
-1
0
1
2
Some vectors are much longer than others, and the short ones don’t
show very well.
iii. The field (the picture) represents driving directions in the sense that
if you start at any point, your driving directions would be: “follow the
arrows”.
iv. Here is a trip (trajectory) followiing the directions starting at (0, 0)
2
1.5
1
0.5
-2
-1.5
-1
-0.5
-0.5
0.5
1
1.5
-1
-1.5
-2
2. The direction field for the population equation p0 = 3p (1 − p) is:
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2
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
(a) If, at time 0, the population starts at 0, it stays there.
(b) If, at time 0, the population starts between 0 and 1, it rises to 1.
(c) If, at time 0, the population starts at 1, it stays there.
(d) If, at time 0, the population starts above 1, it declines to 1.
3.4.4
Euler’s Method of Approximating Solutions to Differential Equations y 0 = f (t, y)
1. Ideally, the direction field provides continuous guidance as we trace the solution
to an IVP. In practice we cannot accept or use continuous information. We can
only use discrete bits of information.
(a) Recall: if you have incomplete information about a function y (t), but you
know at t = a the value y (a) and y 0 (a), then for a small value dt you can
approximate y (a + dt) ≈ y (a) + y 0 (a) dt
2. Let’s go back to y 0 = t2 + y 2 , y (0) = 1.
(a) If we start at t = 0, y = 1, the slope of the solution curve is 1. So we
might go to t = 0.1. Then y (0.1) ≈ y (0) + y 0 (0) 0.1 = 1.1. We can go to
(0.1, 1.1) and take another direction reading.
(b) At the point t = 0.1, y = 1.1, the slope is y 0 (0.1) ≈ 0.12 + 1.12 = 1.22.
Then y (0.2) ≈ y (0.1) + y 0 (0.1) 0.1 ≈ 1.1 + 1.22 × 0.1 = 1.222. We can go
to (0.2, 1.222) and take another direction reading.
(c) Let’s make a chart
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