MAS221 Analysis, Semester 2
Sarah Whitehouse
Contents
0 Revision
3
1 Series
1.1 Convergence and absolute convergence . . . . . . . . . . . . . . .
1.2 Convergence tests . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
4
5
2 Integration
2.1 Integration of step functions . . . . .
2.2 The Riemann integral . . . . . . . .
2.3 The fundamental theorem of calculus
2.4 Improper integrals . . . . . . . . . .
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6
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3 Sequences and series of functions
3.1 Pointwise and uniform convergence
3.2 Continuity under uniform limits . .
3.3 Integration and differentiation . . .
3.4 Uniform continuity . . . . . . . . .
3.5 Series of functions . . . . . . . . .
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15
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4 Applications I
17
4.1 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4.2 e and the exponential function . . . . . . . . . . . . . . . . . . . 18
5 Sequences in Rk
19
5.1 Distances in Rk . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
5.2 Convergence of sequences . . . . . . . . . . . . . . . . . . . . . . 21
5.3 Open sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
6 Functions of several variables
6.1 Images and pre-images . . .
6.2 Continuity . . . . . . . . . .
6.3 Examples . . . . . . . . . .
6.4 Uniform continuity . . . . .
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24
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7 Applications II
27
7.1 Differentiation under the integral sign . . . . . . . . . . . . . . . 27
7.2 The Γ and B functions . . . . . . . . . . . . . . . . . . . . . . . . 28
2
Welcome to Semester 2 of Analysis! This semester we will continue the study
of analysis of functions of one real variable. We will start by studying convergence of series and then move on to integration and its relation to differentiation.
Later on we will generalise to the study of functions of many variables.
0
Revision
Naturally it is important to have a good grasp of the first semester material. In
particular, key to everything is the definition of convergence of a sequence (an )
to a limit l (Definition 2.1 from Semester 1). We revise this here.
Let’s run through how we end up with the formal definition of convergence.
Informally, we might say that the terms of the sequence should get closer and
closer to l. This is certainly too vague! For example, the terms of the sequence
(1 + n1 ) get closer and closer to 0, but we can see that the limit is 1 not 0.
So we realize that we need to say something like the terms of the sequence
should get arbitrarily close to l. The formal definition is designed to capture
this idea.
Definition 0.1 A sequence (an ) is said to converge to a limit l ∈ R if given
any ε > 0, there exists N ∈ N such that for all n > N , we have |an − l| < ε.
In words: no matter what distance ε is specified, there is an index N beyond
which all the terms aN +1 , aN +2 , . . . have a distance smaller than ε to l.
We will now go over an example in detail, first guessing the limit by experimenting with particular values and special cases and then giving a detailed
proof, using the algebra of limits and the sandwich rule.
Example 0.2 Let a, b ∈ R and consider limn→∞ (|a|n + |b|n )1/n .
1. Does this limit exist? What do you think it is? If you don’t know, try
experimenting with some particular values of a and b or try considering
special cases (eg a = 0 or a = b).
2. Prove that your guess is correct.
1
Series
Given
P∞ a sequence (an ), we can try to make sense of the sum of its terms,
n=1 an . We call this a series. We define such an infinite sum via a limit,
when it exists. You have already studied series in MAS110, but without knowing the formal definition of limit. So now we revisit this material in a more
rigorous way.
3
1.1
Convergence and absolute convergence
Definition 1.1 Given a sequence (an ) of real numbers, we consider the associated sequence (sn ) of partial sums, where
sn = a1 + a2 + · · · + an .
We say that the sequence (an ) is summable if the sequence (sn ) of partial
sums converges.
In this case, we write
lim sn =
n→∞
∞
X
an
n=1
or more informally
lim sn = a1 + a2 + a3 + · · ·
n→∞
P∞
If the sequence (an ) is summable, we also say the series P
n=1 an converges.
∞
This is less formal language than the above, as we are using n=1 an for both
the sequence of sums, and the number it converges to (if any). However, it does
not cause problems in practice.
Example 1.2 Consider the geometric series
∞
X
arn
n=0
where a and r are fixed real numbers. For which r does this converge? And
what is the limit in that case?
Note that here the geometric series is indexed so that it begins at n = 0
rather than n = 1.
The proof of the following is left as an exercise (or see the proof given in
MAS110).
P∞
Proposition 1.3 Suppose the series
n=1 an converges. Then an → 0 as
n → ∞.
2
Note
that the converse to the above is not true. For example, the harmonic
P∞
series n=1 n1 does not converge, yet n1 → 0 as n → ∞.
Example 1.4 Show that the series
∞
X
n
n+2
n=1
does not converge.
Definition
1.5 We say a series
P∞
|a
|
converges.
n
n=1
P∞
n=1
an converges absolutely when the series
Proposition 1.6 Any series that converges absolutely also converges.
4
1.2
Convergence tests
In this section we look at some criteria to tell when a series converges. The first
of these is called the comparison test.
Theorem 1.7 (Comparison test) Let (an ) and (bn ) be sequences of real
numbers, where an ≥ 0 and bn ≥ 0. Suppose that (bn ) is summable and an ≤ bn
for all sufficiently large n. Then (an ) is summable.
We can immediately apply this to absolute convergence.
P∞
Corollary 1.8 Let n=1 bn be absolutely convergent. Suppose we have a sequence (an ) where |an | ≤ |bn | for all sufficiently large n. Then the series
P
∞
2
n=1 an is also absolutely convergent.
Example 1.9 Show that the series
∞
X
n3−n
n=1
converges.
Comparing a series to a geometric series, as in the previous example, gives
us the ratio test, perhaps the most useful of any of the tests for convergence.
Theorem 1.10 (Ratio test) Consider a series of non-zero terms
∞
X
an
n=1
and suppose
an+1 = r.
lim
n→∞ an • If r < 1, the series converges absolutely.
• If r > 1, the series does not converge.
Note that if r = 1, we cannot tell by using the ratio test whether or not the
series converges.
Example 1.11 Let x ∈ R. Investigate convergence of the series
∞
X
xn
n
n=1
using the ratio test.
5
The ratio test does not tell us about convergence in the above example for
x = ±1. It turns out that the series converges, but not absolutely, when x = −1
(to − ln(2)), and does not converge when x = 1 (this is the harmonic series).
Our final test for convergence, called the root test, is mainly theoretically
useful. Before stating it, observe that for any bounded sequence of numbers
(bn ), where bn ≥ 0, the sequence (cn ), where
cn = sup{bn , bn+1 , bn+2 , . . .}
is bounded and monotone decreasing, so the sequence (cn ) converges.
If the sequence (bn ) is not bounded, we adopt the convention that the limit
of the sequence (cn ) is ∞.
Theorem 1.12 (Root test) Let
P∞
n=1
an be a series, and set
1
1
n
1
c = lim sup{|an | , |an+1 | n+1 , |an+2 | n+2 , . . .}.
n→∞
• If c < 1, then the series converges absolutely.
• If c > 1, then the series does not converge.
2
Integration
In this section we will go over the rigorous definition of the integral. You have
seen the idea before in MAS110, but now we can make it precise using limits.
2.1
Integration of step functions
We start by considering functions for which it is easy to capture explicitly the
idea of an integral as a (signed) area.
Recall that a bounded interval, I, is a set of one of the forms: (c, d), [c, d],
(c, d] or [c, d). Here c, d ∈ R with c ≤ d. For each of these we define the length
of I to be
length(I) = d − c.
Definition 2.1 Let a, b ∈ R, with a < b. Let A ⊆ [a, b]. Then we define the
characteristic function of A, χA : [a, b] → R, by the formula
1 t ∈ A,
χA (t) =
0 t 6∈ A.
A function s : [a, b] → R is called a step function if we have constants αi ∈ R
and bounded intervals Ii ⊆ [a, b] such that
s(t) =
n
X
for all t ∈ [a, b].
αi χIi (t),
i=1
6
Notice that a step function has only finitely many steps.
The idea of the integral of a function is that it finds the (signed) area under its
graph. Here signed means that area above the x-axis is counted positively, area
under the x-axis is counted negatively. We use this idea to build the definition.
Linking this with the “opposite of differentiation” is a profound theorem, and
not the definition; we will see this theorem soon.
Given a function s(t) = αχI (t), where I is a bounded inteval, the area under
the graph is a rectangle, with width length(I) and height |α|, and the (signed)
area we want is
αlength(I).
The area under a step function comes from adding areas of rectangles togeher, making the following a reasonable definition.
Definition 2.2 Let s : [a, b] → R be a step function. Write
s(t) =
n
X
αi χIi (t).
i=1
Then we define the integral of s to be
Z
b
s(t) dt =
a
n
X
αi length(Ii ).
i=1
The same step function can be written in different ways.
Example 2.3 Define step functions r, s : [0, 5] → R by
r(t) = χ[1,3] (t) + 2χ[2,4] (t),
s(t) = χ[1,2) (t) + 3χ[2,3] (t) + 2χ(3,4] (t).
Notice that in fact r(t) = s(t) for all t. Calculate
R5
0
r(t) dt and
R5
0
s(t) dt.
As illustrated in the example, the definition of the integral for a step function
does not depend on how the step function is written. We will not give the details
of the argument, but it can be done by going via a representation involving
disjoint intervals and then showing that subdivision of those intervals does not
change the integral.
The following is immediate from the definition.
Proposition 2.4 Let r, s : [a, b] → R be step functions, and α, β ∈ R. Then
αr + βs : [a, b] → R is a step function and
Z
b
Z
αr(t) + βs(t) dt = α
b
Z
r(t) dt + β
a
a
b
s(t) dt.
a
2
The next result is clear geometrically; the proof is left as an exercise.
7
Proposition 2.5 Let r, s : [a, b] → R be step functions with r(t) ≤ s(t) for all
t ∈ [a, b]. Then
Z b
Z b
r(t) dt ≤
s(t) dt.
a
a
2
Corollary 2.6 Let s : [a, b] → R be a step function. Then |s(t)| : [a, b] → R is
also a step function and
Z
Z
b
b
|s(t)| dt.
s(t) dt ≤
a
a
The final property we need for later work is also clear from the definition.
Proposition 2.7 Let s : [a, b] → R be a step function. Let a ≤ c ≤ b. Then the
restriction of s to each of [a, c] and [c, b] is again a step function and
Z
b
Z
c
s(t) dt =
a
Z
b
s(t) dt +
a
s(t) dt.
c
2
2.2
The Riemann integral
Let f : [a, b] → R be a bounded function, that is to say we have constants m
and M such that m ≤ f (t) ≤ M for all t ∈ [a, b]. Then if s(t) is a step function,
and s(t) ≤ f (t), then s(t) ≤ M for all t, so
b
Z
s(t) dt ≤ M (b − a).
a
Similarly, if s(t) is a step function, and s(t) ≥ f (t) for all t, then
Z
b
s(t) dt ≥ m(b − a).
a
We want to define the area under the graph of f . We approach this by
approximating f by step functions, since we already know how to treat this
case, by the previous section.
There are two ways we can go about this approximation: approximating
from below, and from above.
Definition 2.8 Let f : [a, b] → R be a bounded function. We define the lower
integral of f :
(Z
)
Z
b
L
b
s(t) dt | s is a step function, s(t) ≤ f (t) for all t .
f (t) dt = sup
a
a
8
We define the upper integral of f :
(Z
Z
U
)
b
b
s(t) dt | s is a step function, s(t) ≥ f (t) for all t .
f (t) dt = inf
a
a
Notice that the set involved in the definition of the lower integral,
)
(Z
b
s(t) dt | s is a step function, s(t) ≤ f (t) for all t ,
a
is a non-empty bounded above subset of R. The integral of a step function is by
definition a real number. If m ≤ f (t) for all t ∈ [a, b], then s = mχ[a,b] : [a, b] →
Rb
R is a step function such that s(t) ≤ f (t) for all t. So a s(t) dt = m(b − a) is
an element of the set. And if f (t) ≤ M for all t ∈ [a, b], then the set is bounded
above by M (b − a). Thus the set has a supremum.
Similarly, the set involved in the definition of the upper integral is a nonempty bounded below subset of R and so has an infimum.
Note that if s is a step function, then
b
Z
b
Z
s(t) dt = L
b
Z
s(t) dt = U
a
a
s(t) dt.
a
Definition 2.9 We call a bounded function f : [a, b] → R Riemann integrable
if
Z b
Z b
L
f (t) dt = U
f (t) dt.
a
a
In this case we define the integral
Z
b
Z
f (t) dt = L
a
b
Z
f (t) dt = U
a
b
f (t) dt.
a
Step functions are clearly Riemann integrable, but not every bounded function is.
Example 2.10 Define f : [0, 1] → R by
0
f (t) =
1
t ∈ Q,
t 6∈ Q.
Show that f is not Riemann integrable.
At the end of the next chapter we will prove the major result that if
f : [a, b] → R is continuous, then f is Riemann integrable (Theorem 3.17).
Propositions 2.4, 2.5, 2.7, and Corollary 2.6 give us the following result.
9
Proposition 2.11 Let the bounded functions f, g : [a, b] → R be Riemann integrable.
• Let α, β ∈ R. Then αf + βg : [a, b] → R is Riemann integrable and
Z b
Z b
Z b
αf (t) + βg(t) dt = α
f (t) dt + β
g(t) dt.
a
a
a
• Suppose f (t) ≤ g(t) for all t ∈ [a, b]. Then
Z b
Z b
f (t) dt ≤
g(t) dt.
a
a
• Let a ≤ c ≤ b. Then
Z b
c
Z
f (t) dt =
a
Z
f (t) dt +
a
b
f (t) dt.
c
• The function |f (x)| : [a, b] → R is Riemann integrable, and
Z
Z
b
b
f (t) dt ≤
|f (t)| dt.
a
a
2
Corollary 2.12 Let the bounded function f : [a, b] → R be Riemann integrable.
Let
m = inf{f (x) | x ∈ [a, b]},
M = sup{f (x) | x ∈ [a, b]}.
Then
Z
m(b − a) ≤
b
f (t) dt ≤ M (b − a).
a
2.3
The fundamental theorem of calculus
Remember that integration was defined in terms of areas. In this section we link
it with the reverse of differentiation. This is known as the fundamental theorem
of calculus. You are already familiar with this statement, but now we are in a
position to give a full proof.
Theorem 2.13 (Fundamental theorem of calculus) Let f : [a, b] → R be
continuous. Define F : [a, b] → R by
Z x
F (x) =
f (t) dt.
a
0
Then F is differentiable, and F (x) = f (x) for all x ∈ [a, b].
The version of the theorem that we use in practice to calculate integrals
follows as a corollary.
10
Corollary 2.14 Let f : [a, b] → R be a continuous function. Suppose we have
a differentiable function F : [a, b] → R such that F 0 (x) = f (x) for all x ∈ [a, b].
Then
Z b
f (t) dt = F (b) − F (a).
a
Note that all of the techniques we already know about finding integrals, such
as substitution and integration by parts, can be proved for continuous functions
with an “antiderivative”.
2.4
Improper integrals
Definition 2.15 Let a ∈ R. We say a bounded function f : [a, ∞) → R is
Riemann integrable if the limit
Z b
lim
f (t) dt
b→∞
exists. We write
Z
a
∞
b
Z
f (t) dt = lim
f (t) dt.
b→∞
a
a
1
Example 2.16 Let f : [1, ∞) → R be given by f (t) =
R ∞t2 . Show that f is
Riemann integrable on the interval [1, ∞), and evaluate 1 f (t) dt.
Similarly, for f : (−∞, b] → R, we define
Z b
Z
f (t) dt = lim
−∞
a→−∞
b
f (t) dt.
a
We also sometimes want to consider integrals of functions f : (a, b] → R
where f (t) → ∞ or f (t) → −∞ as t → a.
Definition 2.17 We say a function f : (a, b] → R is Riemann integrable if the
limit
Z b
Z b
f (t) dt = lim
f (t) dt
a
c↓a
c
exists.
Example 2.18 Let f : (0, 1] → R be given by f (t) = √1t . Show that f is
R1
Riemann integrable on the interval (0, 1] and evaluate 0 f (t) dt.
We similarly define integrals of functions f : [a, b) → R.
We can also define integrals when there are limits involved at both ends of
the interval we integrate over, but we need to evaluate the upper and lower
limits separately.
Example 2.19 Define f : R R→ R by f (t) = e−|t| . Show that f is Riemann
∞
integrable on R and evaluate −∞ f (t) dt.
11
3
Sequences and series of functions
In this section we will study sequences and series of functions. We start with
two different notions of convergence for a sequence of functions. Later we will
consider the interaction between continuity and these notions of convergence,
as well as looking at integration.
3.1
Pointwise and uniform convergence
Consider a sequence of functions, (fn ), where fn : [a, b] → R.
Definition 3.1 We say the sequence (fn ) converges pointwise to a function
f : [a, b] → R if for each t ∈ [a, b], we have
lim fn (t) = f (t).
n→∞
In other words, the sequence (fn ) converges pointwise to f if for each t ∈ [a, b]
and ε > 0, there exists N ∈ N such that |fn (t) − f (t)| < ε whenever n ≥ N .
Note that in this definition, the N can depend not just on the value of ε,
but also on the point t.
Example 3.2 Define fn : [0, 2π] → R by fn (t) = cos(t/n). Show that the
sequence (fn ) converges pointwise and determine the limit function.
Example 3.3 Define fn : [0, 1] → R by fn (t) = tn . Show that the sequence
(fn ) converges pointwise and determine the limit function.
Example 3.4 Define fn : [0, 2] → R by fn (t) = tn . Show that the sequence
(fn ) does not converge pointwise.
Instead of pointwise convergence, we could insist that a sequence of functions converges at the same rate at each point. This concept is called uniform
convergence and it is a much stronger condition. Uniform convergence, as we
shall see, preserves such things as continuity and to an extent differentiation. As
the second of the above examples show, pointwise convergence does not preserve
continuity.
Definition 3.5 Let fn : [a, b] → R. We say the sequence (fn ) converges uniformly to a function f : [a, b] → R when for all ε > 0, we have N ∈ N such that
|fn (t) − f (t)| < ε for all n ≥ N and t ∈ [a, b].
The difference between this definition and pointwise convergence is that the
N does not depend on the point t, only on ε. Note that uniform convergence
of a sequence (fn ) to a function f implies pointwise convergence to the same
function f .
Proposition 3.6 Consider a sequence of functions fn : [a, b] → R.
f : [a, b] → R. Then the following statements are equivalent.
12
Let
• The sequence (fn ) converges uniformly to f .
• Let Mn = sup{|fn (t) − f (t)| | t ∈ [a, b]}. Then Mn → 0 as n → ∞.
The condition on the Mn s in Proposition 3.6 is usually the easiest way to
actually prove uniform convergence in an example.
Example 3.7 For n ≥ 2, define fn : [0, 2π] → R by
fn (t) =
1 − 2n
sin t.
1−n
First show that the sequence (fn ) converges pointwise and determine the limit
function f . Then show that in fact the sequence (fn ) converges uniformly to f .
3.2
Continuity under uniform limits
Theorem 3.8 (Uniform limit theorem) Let fn : [a, b] → R be continuous
for each n ∈ N. Suppose the sequence (fn ) converges uniformly to a function
f : [a, b] → R. Then f is continuous.
The proof is sometimes called the ε/3 proof because of the main trick.
The result also holds, with essentially the same proof, for functions fn : R →
R.
Example 3.9 Let fn : [0, 1] → R be given by fn (t) = tn . Show that the
sequence (fn ) does not converge uniformly.
3.3
Integration and differentiation
We will look at how uniform convergence interacts with integration. The following result is called the uniform convergence theorem for integrals. Many of our
results and examples in the rest of this course are applications of this theorem.
Theorem 3.10 (Uniform convergence theorem) Let fn : [a, b] → R be a
continuous function, for n ∈ N. Suppose the sequence (fn ) converges uniformly
to a function f . Then
Z b
Z b
lim
fn (t) dt =
f (t) dt.
n→∞
a
a
Thus we can, under suitable conditions, swap limits and integral signs. This
is frequently useful.
13
The result actually still holds if each fn is Riemann integrable rather than
continuous, but the proof is much more involved, and the above is enough for
our purposes.
The following example shows that we do need uniform convergence to be
able to swap limits and integrals.
Example 3.11 Let fn : [0, 1] → R, for n ∈ N be defined by
fn (t) = nt(1 − t2 )n .
1. Show that (fn ) converges pointwise to the zero function.
R1
2. Calculate limn→∞ 0 fn (t) dt.
3. Does (fn ) converge uniformly?
Our first immediate application is to differentiation. Indeed, it is natural to
ask at this point about swapping limits and differentiation.
Corollary 3.12 Consider differentiable functions fn : [a, b] → R. Suppose (fn )
converges pointwise to a function f , and the sequence of derivatives (fn0 ) converges uniformly to a function g. Then f is differentiable, and f 0 = g.
Thus we can, under suitable conditions, swap limits and differentiation.
The following example shows that convergence conditions on the sequence
of derivatives are necessary.
Example 3.13 Define fn : [0, 2π] → R by
fn (t) =
1
sin(n2 t).
n
Show that (fn ) converges uniformly to the zero function, but (fn0 ) does not
converge pointwise.
3.4
Uniform continuity
We recall once again what it means for a function to be continuous.
A function f : [a, b] → R is continuous at a point x0 ∈ [a, b] if for all ε > 0
we have δ > 0 such that if |x − x0 | < δ then |f (x) − f (x0 )| < ε.
The function f is continuous on [a, b] if it is continuous at every point x0 ∈
[a, b].
Along similar lines to uniform as opposed to pointwise convergence, we make
the following definition.
Definition 3.14 Let f : [a, b] → R. We say f is uniformly continuous on [a, b]
if for all ε > 0 we have δ > 0 such that if |x − y| < δ then |f (x) − f (y)| < ε for
all x, y ∈ [a, b].
14
The difference here is that the δ can only depend on ε, not on a chosen point
x0 as well. For a given , the same δ has to work across the whole of the interval
[a, b].
We similarly define uniform continuity of a function f : R → R.
It is clear that if f is uniformly continuous then f is continuous. The following example shows that the converse need not hold.
Example 3.15 Define f : R → R by f (x) = x2 . Certainly f is continuous.
Show that f is not uniformly continuous.
In view of the above, the following feels surprising: on a closed bounded
interval there is no difference between continuity and uniform continuity. We
will use some results from semester one to prove this.
Theorem 3.16 Let f : [a, b] → R be continuous. Then f is uniformly continuous.
As an application, we use this result to supply the proof promised earlier
that continuous functions on bounded intervals are Riemann integrable.
Theorem 3.17 Let f : [a, b] → R be continuous. Then f is Riemann integrable.
3.5
Series of functions
We can treat convergence of series of functions as we did for series of real
numbers, via sequences of partial sums.
Definition 3.18 Let (fn ) be a sequence of functions, fn : [a, b] → R. We consider the sequence of partial sums (sn ), where sn : [a, b] → R is defined by
sn (t) = f1 (t) + f2 (t) + · · · + fn (t).
P∞
WePsay the series
n=1 fn converges pointwise if, for each t ∈ [a, b], the
∞
series n=1 fn (t) converges. That is, for each t ∈ [a, b], the sequence of partial
sums (sn (t)) converges, or equivalently, the sequence (sn ) converges pointwise.
P∞We say that the sequence (fn ) is uniformly summable, or that the series
n=1 fn converges uniformly, on [a, b] if the sequence (sn ) of partial sums converges uniformly.
One also makes the same definition for functions defined on R.
15
Theorem 3.8 and Corollary 3.12 immediately give us the following two results.
Theorem 3.19 Let (fn ) be a uniformly summable sequence of continuous functions fn : [a, b] → R. Then the function f : [a, b] → R given by
∞
X
f (t) =
fn (t)
n=1
2
is continuous.
Theorem 3.20 Let (fnP
) be a sequence of differentiable functions, fn : [a, b] →
∞
R, such
that
the
series
n=1 fn converges pointwise to a function f , and the
P∞
series n=1 fn0 is uniformly summable. Then f is differentiable and
∞
X
f 0 (t) =
fn0 (t)
n=1
2
for all t ∈ [a, b].
In order for the above to be useful, we need a criterion for uniform convergence of a series. The following result, called the Weierstrass M -test provides
a handy criterion.
Theorem 3.21 (Weierstrass M -test) Let fn : [a, b] → R be a sequence of
functions. Suppose we have a summable sequence of real numbers (Mn ) such
that |fn (t)| ≤ Mn for all n, and all t ∈ [a, b]. Then the sequence (fn ) is uniformly
summable. Further, for each t ∈ [a, b], the series
∞
X
fn (t)
n=1
converges absolutely.
Note that the Weierstrass M -test also works for functions fn : R → R.
We will see in the next chapter how to use uniform convergence of series in
a class of well-behaved examples. But it can also be used to construct examples
with pathological properties.
Proposition 3.22 There is a function f : R → R which is continuous everywhere, but differentiable nowhere.
16
4
Applications I
4.1
Power series
Definition 4.1 A series of the form
f (x) =
∞
X
an xn
n=0
where an ∈ R are constants is called a power series.
Theorem 4.2 For a power series f (x) =
holds:
P∞
n=0
an xn , one of the following
• The series converges only when x = 0.
• The series converges for all x ∈ R.
• There is a constant R > 0, such that the series converges absolutely if
|x| < R, and does not converge if |x| > R.
Definition 4.3 The constant R appearing in the statement of Theorem 4.2 is
called the radius of convergence of the power series. If the series converges only
when x = 0, we say the radius of convergence is 0. If the series converges for all
x ∈ R, we say the radius of convergence is ∞.
The conventions about 0 and ∞ mean that we don’t have to treat special
cases all the time.
We leave the proof of the following as an exercise, using the ratio test.
Proposition 4.4 Let
f (x) =
∞
X
an xn
n=0
be a power series. Then the radius of convergence is
an R = lim n→∞ an+1 2
if this limit exists.
This result is usually the easiest way to find the radius of convergence in
examples.
17
Example 4.5 Find the radius of convergence of the series
f (x) =
∞
X
nxn .
n=0
For computations using power series, the following allows us to use our results
on uniform convergence.
Lemma 4.6 Let
∞
X
an xn
n=0
be a power series with radius of convergence R. Let 0 < S < R. Then the series
converges uniformly on the interval [−S, S].
Theorem 4.7 (Termwise differentiation of power series) Let
f (x) =
∞
X
an xn
n=0
be a power series with radius of convergence R > 0. Then the function f is
differentiable on (−R, R), with
f 0 (x) =
∞
X
nan xn−1 ,
n=1
where this series also has radius of convergence R.
4.2
e and the exponential function
In the first semester (Example 2.6), you saw that the limit
n
1
lim 1 +
n→∞
n
exists and the limit was taken as a definition of the number e.
Here we will study the exponential function, exp, defined by a power series
and show how to relate the different approaches to e.
Definition 4.8 We define the exponential function, exp : R → R, by
exp(x) =
∞
X
xn
.
n!
n=0
18
Here we take 0! = 1.
If we let an = 1/n!, then
an
1/n!
(n + 1)!
=
=
=n+1
an+1
1/(n + 1)!
n!
so
an =∞
lim
n→∞ an+1 and the series exp(x) converges absolutely for all x ∈ R, by Proposition 4.4.
Thus the exponential function is indeed defined for all x ∈ R.
Using Theorem 4.7, the exponential function is differentiable, and differentiating term by term, we have
exp0 (x) =
∞
∞
∞
X
X
X
nxn−1
xn−1
xm
=
=
= exp(x).
n!
(n − 1)! m=0 m!
n=1
n=1
Proposition 4.9 Let a, b ∈ R. Then
exp(a + b) = exp(a) exp(b).
Now we relate the exponential function to the limit expression for e seen in
semester 1.
Proposition 4.10 Let x ∈ R. Then
x n
lim 1 +
= exp(x).
n→∞
n
In particular,
n
1
exp(1) = e = lim 1 +
.
n→∞
n
Proposition 4.11 e is an irrational number between 2 and 3.
5
Sequences in Rk
We are going to generalise much of what we have done from one variable to many.
There are lots of similarities, so this will involve reviewing and consolidating
what you have seen before.
19
5.1
Distances in Rk
Recall that we define Rk to be the set of k-tuples of real numbers:
Rk = {(x(1) , x(2) , . . . , x(k) ) | x(1) , . . . , x(k) ∈ R}.
Note that we are using the notation x(i) for coordinates in Rk , because we
need lower index notation, xn , for the terms of sequences.
R2 is the two-dimensional plane and R3 is three-dimensional space. An
element x = (x(1) , x(2) , . . . , x(k) ) ∈ Rk is a vector in k-dimensional space. As a
special case, we write 0 = (0, 0, . . . , 0).
Definition 5.1 We define the magnitude of x = (x(1) , x(2) , . . . , x(k) ) ∈ Rk to
be the real number
q
|x| = (x(1) )2 + (x(2) )2 + · · · + (x(k) )2 .
Example 5.2 Let x = (1, −1, 2, 3) ∈ R4 . Calculate |x|.
We would like to understand some basic properties of magnitude and how it
interacts with operations we can carry out in Rk .
Recall from MAS211 that we can multiply a vector by a scalar. In detail,
given α ∈ R and x = (x(1) , x(2) , . . . , x(k) ) ∈ Rk , we define
αx = (αx(1) , αx(2) , . . . , αx(k) ).
Note that |αx| = |α| · |x|.
Example 5.3 Let x = (1, −1, 2, 3) ∈ R4 . Then 3x = (3, −3, 6, 9). Observe
that
p
√
√
√
|3x| = 32 + (−3)2 + 62 + 92 = 135 = 9 × 15 = 3 15 = 3|x|.
Note that |x| ≥ 0 for all x ∈ Rk . Suppose |x| = 0. Then |x|2 = 0, so if
x = (x(1) , x(2) , . . . , x(k) ), then
(x(1) )2 + (x(2) )2 + · · · + (x(k) )2 = 0.
Since (x(i) )2 ≥ 0 for all i (as the square of any real number is non-negative),
it follows that (x(i) )2 = 0 for each i. Thus x(i) = 0, for each i and therefore
x = 0.
We can also add vectors.
(y , y (2) , . . . , y (k) ). We define
Let x = (x(1) , x(2) , . . . , x(k) ) and y =
(1)
x + y = (x(1) + y (1) , x(2) + y (2) , . . . , x(k) + y (k) ).
The next fundamental property we need about the magnitude is the triangle
inequality, which asserts that
|x + y| ≤ |x| + |y|,
for all x, y ∈ Rk . To prove it, we need the following lemma.
20
Lemma 5.4 (Cauchy-Schwarz inequality) Let x = (x(1) , . . . , x(k) ), y =
(y (1) , . . . , y (k) ) ∈ Rk . Then
x(1) y (1) + · · · + x(k) y (k) ≤ |x| · |y|.
Proposition 5.5 (Triangle inquality) For all x, y ∈ Rk ,
|x + y| ≤ |x| + |y|.
To summarise, the magnitude of a vector in Rk has the following properties.
• Let x ∈ Rk and α ∈ R. Then |αx| = |α| · |x|.
• Let x, y ∈ Rk . Then |x + y| ≤ |x| + |y|.
• Let x ∈ Rk . Then |x| ≥ 0, and if |x| = 0, then x = 0.
A function | − | with these properties is called a norm.
Definition 5.6 Let x, y ∈ Rk . Then we define the distance between x and y
by
d(x, y) = |x − y|.
Example 5.7 In R4 , let x = (1, 0, −1, 2) and y = (2, 3, −1, 0). Calculate the
distance between x and y.
Proposition 5.8 The distance has the following properties.
1. Let x, y ∈ Rk . Then d(x, y) = d(y, x).
2. Let x, y, z ∈ Rk . Then d(x, z) ≤ d(x, y) + d(y, z).
3. Let x, y ∈ Rk . Then d(x, y) ≥ 0, and if d(x, y) = 0 then x = y.
A function d with these properties is called a metric. Metrics are examined
in more detail and generality, and surprising applications of the theory developed, in the level 3 Metric Spaces course. Metrics also feature in studies of
mathematical economics.
5.2
Convergence of sequences
We write (xn ) to denote a sequence
x1 , x2 , x3 , . . .
in Rk . The following definition is an immediate generalisation of the definition
of convergence of sequences of real numbers.
Definition 5.9 We say a sequence (xn ) in Rk converges to an element x ∈ Rk
if for all ε > 0 we have N ∈ N such that |xn − x| < ε whenever n ≥ N .
21
If a sequence (xn ) in Rk converges to x ∈ Rk , we write
lim xn = x
n→∞
or
xn → x
as
n → ∞.
Remarks 5.10
1. A sequence (xn ) in Rk converges to x ∈ Rk if and only
if |xn − x| → 0 as n → ∞.
2. For k = 1, this is the familiar definition of convergence of a sequence of
real numbers.
3. We could write the condition |xn − x| < ε as d(xn , x) < ε.
Example 5.11 Let a > 0. In R2 , let
xn =
1
(cos(an), sin(an) + n).
n
Show that (xn ) converges to (0, 1).
Theorem 5.12 Let (xn ) be a sequence in Rk , and let x ∈ Rk . Write
(k)
xn = (x(1)
n , . . . , xn ),
x = (x(1) , . . . , x(k) ).
(i)
Then xn → x as n → ∞ if and only if xn → x(i) as n → ∞ for each i.
Example 5.13 As in the previous example, we let a > 0 and in R2 ,
xn =
1
(cos(an), sin(an) + n).
n
Use Theorem 5.12 to show that (xn ) converges to (0, 1).
Definition 5.14 We call a sequence (xn ) in Rk a Cauchy sequence if for all
ε > 0 there exists N ∈ N such that |xm − xn | < ε whenever m, n ≥ N .
Looking at whether a sequence is Cauchy is a test for whether it converges
without knowing the limit. To be precise, we have the following.
Theorem 5.15 Let (xn ) be a sequence in Rk . Then (xn ) converges to some
limit if and only if (xn ) is a Cauchy sequence.
Recall from semester one that any bounded sequence in R has a convergent
subsequence (the Bolzano-Weierstrass Theorem, 2.4.3). This generalises to Rk .
22
Definition 5.16 A sequence xn in Rk is bounded if there is a real number
C ≥ 0 such that |xn | ≤ C for all n.
Theorem 5.17 (Bolzano-Weierstrass in Rk ) Any bounded sequence in Rk
has a convergent subsequence.
5.3
Open sets
We are going to study open sets in Rk . These generalise the notion of an
open interval in R and they are very useful for the study of functions of several
variables, particularly continuity.
Definition 5.18 Let x ∈ Rk and r > 0. We define the open ball with radius r
around x to be the set
B(x, r) = {y ∈ Rk | |x − y| < r}.
Definition 5.19 We call a subset U ⊆ Rk an open set if for all x ∈ U there is
some δ > 0 such that B(x, δ) ⊆ U .
Example 5.20 Consider the open interval (a, b) = {t ∈ R | a < t < b} in R.
Show that this is indeed open in the sense of Definition 5.19.
One can similarly show that (−∞, b) and (a, ∞) are open sets.
Drawing a picture is often helpful when considering examples of open sets.
Example 5.21 Let x ∈ Rk and r > 0. Show that the open ball B(x, r) is
indeed open in the sense of Definition 5.19.
Example 5.22 Consider the closed interval [a, b] = {t ∈ R | a ≤ t ≤ b} in R.
Show that this is not an open subset of R.
In general, a set which includes “boundary points” is not open for reasons
similar to the above example. If we pick a point x on the boundary, then any
ball B(x, δ) will spill over the boundary and include elements outside the set.
As a general guide, a set defined with “<” tends to be open, though we need
to check the details. On the other hand, a set whose definition involves “≤” will
usually not be open, though again we need to check.
Proposition 5.23 Let U, V ⊆ Rk be open sets. Then U ∪ V and U ∩ V are
also open sets.
We can reformulate convergence of a sequence in terms of open sets. Soon
we will see that this also allows us to study continuity via open sets.
23
Proposition 5.24 Let (xn ) be a sequence in Rk , and x ∈ Rk . Then the sequence (xn ) converges to x if and only if:
(∗) For every open set U containing x, there is some N ∈ N such that xn ∈ U
whenever n ≥ N .
6
Functions of several variables
In this section, we will generalise our study of continuity to functions of several
variables. We begin with a review of some basic terminology for functions.
6.1
Images and pre-images
Let φ : A → B be a function, where A ⊆ Rk and B ⊆ Rl .
Recall from MAS110 that:
• The domain of φ is the set A.
• The range of φ, also called the codomain of φ, is the set B.
The image of φ is the set
{u ∈ B | u = φ(x) for some x ∈ A}.
We write φ(A), or im(φ) to denote the image of φ.
Example 6.1 Define a function φ : R2 → R2 by φ(x, y) = (xy, x + y). Describe
the image of φ.
Definition 6.2 Let φ : A → B be a function, where A ⊆ Rk and B ⊆ Rl . Let
u ∈ B. The pre-image of u under φ is the set
φ−1 (u) = {x ∈ A | φ(x) = u}.
So the pre-image of the point u = (u(1) , . . . , u(l) ) is the set of solutions
(x , . . . , x(k) ) ∈ A, of the equation
(1)
φ(x(1) , . . . , x(k) ) = (u(1) , . . . , u(l) ).
Depending on φ and (u(1) , . . . , u(l) ) this pre-image could consist of a single
element, any number of elements, or be empty.
The notation φ−1 here doesn’t mean an inverse function. Indeed most functions do not have inverses and in general φ−1 does not make sense when written
on its own. But pre-images are defined for any function.
Example 6.3 Define φ : R2 → R2 by φ(x, y) = (xy, x + y). For (u, v) ∈ R2 ,
find φ−1 ((u, v)).
24
Definition 6.4 Let φ : A → B be a function, where A ⊆ Rk and B ⊆ Rl . Let
U ⊆ B. The pre-image of U under φ is the set
φ−1 (U ) = {x ∈ A | φ(x) ∈ U }.
In other words
φ−1 (U ) =
[
φ−1 (u).
u∈U
Notice that if we consider a set with one element, U = {u}, then φ−1 (U ) =
φ−1 ({u}) = φ−1 (u). In other words, we allow ourselves to drop the set brackets
in this case.
6.2
Continuity
The usual definition of a continuous function R to R generalises nicely to functions Rk to Rl .
Definition 6.5 We call a function f : Rk → Rl continuous at a point x0 ∈ Rk if
for all ε > 0 there exists δ > 0 such that if |x − x0 | < δ, then |f (x) − f (x0 )| < ε.
We call f continuous if it is continuous at every point of its domain.
Note that the above definition also works to define continuity of a function
f : A → Rl , where A is a subset of Rk .
Before exploring examples, we will look at some reformulations of the definition.
Proposition 6.6 Let f : Rk → Rl , and let x0 ∈ Rk . The function f is continuous at x0 if and only if the following condition holds.
(∗) If (xn ) is a sequence in Rk converging to x0 , then (f (xn )) is a sequence
in Rl converging to f (x0 ).
It follows that a function f : Rk → Rl is continuous if and only if for every
convergent sequence (xn ) in Rk we have
f lim xn = lim f (xn ).
n→∞
n→∞
We can also formulate the definition in terms of open sets.
Theorem 6.7 A function f : Rk → Rl is continuous if and only if the following
condition holds.
(∗∗) For every open set U ⊆ Rl , the pre-image f −1 (U ) is open in Rk .
The following is very useful in examples, and follows from the above characterisation of continuity.
Proposition 6.8 Let f : Rk → Rl and g : Rl → Rm be continuous. Then g ◦
f : Rk → Rm is continuous.
25
6.3
Examples
In this section we prove that many of “the usual functions” are continuous. We
need to show that sums and products of continuous functions are continuous.
This is sometimes called the algebra of continuous functions.
Proposition 6.9 Let A ⊆ Rk . Let f, g : A → R. If f and g are both continuous
at x0 ∈ A, then f + g and f g are continuous at x0 .
Proposition 6.10 The functions pi : Rk → R defined by the formula
pi (x(1) , . . . , x(k) ) = x(i)
are continuous.
Using the above two propositions, we see that all polynomials in x(1) , . . . , x(k)
are continuous at every point of Rk .
From the previous section, we also know that composites of continuous functions are continuous.
Example 6.11 Define f : R2 → R by f (x, y) = ey sin(x2 y + y 3 ). Show that f
is continuous.
Using functions which are known to be continuous, we can prove that many
sets are open. Recall that an open interval in R is a set of the form (a, b), (a, ∞)
or (−∞, a).
Proposition 6.12 Let I be an open interval in R. Let f : Rk → R be continuous. Then f −1 (I) is open.
Example 6.13 Show that the set
A = {(x, y) ∈ R2 | x2 < y 4 }
is open.
6.4
Uniform continuity
Recall from Section 3.4 that, for functions of one variable, we introduced and
studied a stronger version of continuity, called uniform continuity. This turned
out to have good properties in terms of its interaction with convergence and
integration. We will now do the same in the setting of real-valued functions of
several variables.
Definition 6.14 Let A ⊆ Rk and let f : A → R. We say f is uniformly continuous on A if for all ε > 0 there is some δ > 0 such that if |x − y| < δ then
|f (x) − f (y)| < ε for all x, y ∈ A.
26
The following is proved similarly to the one variable case, Theorem 3.16,
using Theorem 5.17, the Bolzano-Weierstrass theorem. We omit the details.
Theorem 6.15 Let f : [a, b] × [c, d] → R be continuous. Then f is uniformly
continuous.
2
Our next result uses uniform continuity as a way to establish uniform convergence of a sequence of functions.
Theorem 6.16 Let f : [a.b] × [c, d] → R be continuous. Let x ∈ [a, b] and define
f (x, −) : [c, d] → R by
f (x, −)(t) = f (x, t).
Let (xn ) be a sequence in [a, b] converging to x. Then the sequence of functions (f (xn , −)) converges uniformly to the function f (x, −).
Corollary 6.17 Let f : [a, b]×[c, d] → R be continuous. Then for any x0 ∈ [a, b]
we have
Z d
Z b
lim
f (x, t) dt =
f (x0 , t) dt.
x→x0
7
c
a
Applications II
7.1
Differentiation under the integral sign
Consider a function f : [a, b] × [c, d] → R. Let (x, t) ∈ [a, b] × [c, d]. We define
the partial derivatives
f (x + h, t) − f (x, t)
f (x, t + h) − f (x, t)
,
f2 (x, t) = lim
h→0
h
h
if these limits exist. We also sometimes use the notation
∂f
∂f
(x, t) = f1 (x, t),
(x, t) = f2 (x, t).
∂x
∂t
The second notation is perhaps clearer, but depends on how we label our
variables.
f1 (x, t) = lim
h→0
Theorem 7.1 (Differentiation under the integral sign) Let f : [a, b] ×
[c, d] → R be a continuous function such that the partial derivative f1 exists
and is continuous on [a, b] × [c, d]. Then for all x ∈ [a, b],
Z b
Z b
d
∂f
f (x, t) dt =
(x, t) dt.
dx a
a ∂x
27
Taking limits, we have the following result.
Theorem 7.2 Let f : R2 → R be a continuous function where the partial
derivative f1 exists and is continuous on R2 . Then as long as the relevant
integrals exist, we have
Z ∞
Z ∞
∂f
d
f (x, t) dt =
(x, t) dt.
dx −∞
−∞ ∂x
2
A similar result applies when we replace (−∞, ∞) with [0, ∞) or (−∞, 0].
This type of result can sometimes be useful when evaluating tricky integrals.
Example 7.3 Evaluate
∞
Z
0
7.2
sin t
dt.
t
The Γ and B functions
The Γ-function is a differentiable function which extends the factorial to nonintegers. The key to its definition is the following formula.
Lemma 7.4 Let n ∈ N0 and x > 0. Then
Z ∞
n!
=
tn e−xt dt.
xn+1
0
Taking x = 1, we see that
∞
Z
tn e−t dt.
n! =
0
Definition 7.5 The Γ-function, Γ : (0, ∞) → R, is defined by
Z ∞
Γ(s) =
ts−1 e−t dt.
0
By Lemma 7.4, if s ∈ N, then
Γ(s) = (s − 1)!.
Proposition 7.6 For s > 0,
Γ(s + 1) = sΓ(s).
We will cover some more properties of the Γ-function in the exercises. The
Γ-function is related to another function we call the B-function. Note that the
symbol B here is a capital Greek letter beta.
28
Definition 7.7 We define the B-function, B : (0, ∞) × (0, ∞) → R, by
Z
B(x, y) =
1
tx−1 (1 − t)y−1 dt.
0
The proof of the following is left as an exercise.
Proposition 7.8 Let x, y ≥ 0. Then B(x, y) = B(y, x).
The relation with the Γ-function is as follows.
Theorem 7.9 Let x, y > 0. Then
Γ(x)Γ(y) = Γ(x + y)B(x, y).
Using this, we find our first explicit non-integer value of the Γ-function.
Corollary 7.10
Γ
√
1
= π.
2
29
2