IA Analysis I (Full)

Part IA — Analysis I

Based on lectures by W. T. Gowers

Notes taken by Dexter Chua

Lent 2015

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Limits and convergence

Sequences and series in

and

. Sums, products and quotients. Absolute convergence;

absolute convergence implies convergence. The Bolzano-Weierstrass theorem and

applications (the General Principle of Convergence). Comparison and ratio tests,

alternating series test. [6]

Continuity

Continuity of real- and complex-valued functions defined on subsets of

and

. The

intermediate value theorem. A continuous function on a closed bounded interval is

bounded and attains its bounds. [3]

Differentiability

Differentiability of functions from

. Derivative of sums and products. The chain

rule. Derivative of the inverse function. Rolle’s theorem; the mean value theorem.

One-dimensional version of the inverse function theorem. Taylor’s theorem from

R; Lagrange’s form of the remainder. Complex differentiation. [5]

Power series

Complex power series and radius of convergence. Exp onential, trigonometric and

hyperbolic functions, and relations between them. *Direct proof of the differentiability

of a power series within its circle of convergence*. [4]

Integration

Definition and basic properties of the Riemann integral. A non-integrable function.

Integrability of monotonic functions. Integrability of piecewise-continuous functions.

The fundamental theorem of calculus. Differentiation of indefinite integrals. Integration

by parts. The integral form of the remainder in Taylor’s theorem. Improper integrals. [6]

Contents

0 Introduction

1 The real number system

2 Convergence of sequences

2.1 Definitions

2.2 Sums, products and quotients

2.3 Monotone-sequences property

2.4 Cauchy sequences

2.5 Limit supremum and infimum

3 Convergence of infinite sums

3.1 Infinite sums

3.2 Absolute convergence

3.3 Convergence tests

3.4 Complex versions

4 Continuous functions

4.1 Continuous functions

4.2 Continuous induction*

5 Differentiability

5.1 Limits

5.2 Differentiation

5.3 Differentiation theorems

5.4 Complex differentiation

6 Complex power series

6.1 Exponential and trigonometric functions

6.2 Differentiating power series

6.3 Hyperbolic trigonometric functions

7 The Riemann Integral

7.1 Riemann Integral

7.2 Improper integrals

0 Introduction

In IA Differential Equations, we studied calculus in a non-rigorous setting. While

we did define differentiation (properly) as a limit, we did not define what a limit

was. We didn’t even have a proper definition of integral, and just mumbled

something about it being an infinite sum.

In Analysis, one of our main goals is to put calculus on a rigorous foundation.

We will provide unambiguous definitions of what it means to take a limit, a

derivative and an integral. Based on these definitions, we will prove the results

we’ve had such as the product rule, quotient rule and chain rule. We will also

rigorously prove the fundamental theorem of calculus, which states that the

integral is the inverse operation to the derivative.

However, this is not all Analysis is about. We will study all sorts of limiting

(“infinite”) processes. We can see integration as an infinite sum, and differenti-

ation as dividing two infinitesimal quantities. In Analysis, we will also study

infinite series such as 1 +

···

, as well as limits of infinite sequences.

Another important concept in Analysis is continuous functions. In some

sense, continuous functions are functions that preserve limit processes. While

their role in this course is just being “nice” functions to work with, they will be

given great importance when we study metric and topological spaces.

This course is continued in IB Analysis II, where we study uniform convergence

(a stronger condition for convergence), calculus of multiple variables and metric

spaces.

1 The real number system

We are all familiar with real numbers — they are “decimals” consisting of

infinitely many digits. When we really want to study real numbers, while this

definition is technically valid, it is not a convenient definition to work with.

Instead, we take an axiomatic approach. We define the real numbers to be a set

that satisfies certain properties, and, if we really want to, show that the decimals

satisfy these properties. In particular, we define real numbers to be an ordered

field with the least upper bound property.

We’ll now define what it means to be an “ordered field with the least upper

bound property”.

Definition (Field). A field is a set

with two binary operations + and

that

satisfies all the familiar properties satisfied by addition and multiplication in

namely

(i)

(

∀a, b, c

)

+ (

) = (

) +

and

a ×

(

b ×c

) = (

a ×b

)

×c

(associativity)

(ii) (∀a, b) a + b = b + a and a × b = b × a (commutativity)

(iii) ∃0, 1 such that (∀a)a + 0 = a and a × 1 = a (identity)

(iv) ∀a ∃

(

−a

) such that

−a

) = 0. If

a 6

= 0, then

∃a

−1

such that

a×a

−1

= 1.

(inverses)

(v) (∀a, b, c) a × (b + c) = (a × b) + (a ×c) (distributivity)

Example. Q, R, C, integers mod p, {a + b

√

2 : a, b ∈ Z} are all fields.

Definition (Totally ordered set). An (totally) ordered set is a set

with a

relation < that satisfies

(i) If x, y, z ∈ X, x < y and y < z, then x < z (transitivity)

(ii) If x, y ∈ X, exactly one of x < y, x = y, y < x holds (trichotomy)

We call it a totally ordered set, as opposed to simply an ordered set, when

we want to emphasize the order is total, i.e. every pair of elements are related to

each other, as opposed to a partial order, where “exactly one” in trichotomy is

replaced with “at most one”.

Now to define an ordered field, we don’t simply ask for a field with an order.

The order has to interact nicely with the field operations.

Definition (Ordered field). An ordered field is a field

with a relation

that

makes F into an ordered set such that

(i) if x, y, z ∈ F and x < y, then x + z < y + z

(ii) if x, y, z ∈ F, x < y and z > 0, then xz < yz

Lemma. Let F be an ordered field and x ∈ F. Then x

≥ 0.

Proof.

By trichotomy, either

x <

= 0 or

x >

0. If

= 0, then

= 0. So

≥

0. If

x >

0, then

× x

= 0. If

x <

0, then

x − x <

− x

. So 0

< −x

But then x

= (−x)

> 0.

Definition (Least upper bound). Let

be an ordered set and let

A ⊆ X

. An

upper bound for

is an element

x ∈ X

such that (

∀a ∈ A

)

a ≤ x

. If

has an

upper bound, then we say that A is bounded above.

An upper bound

for

is a least upper bound or supremum if nothing

smaller that x is an upper bound. That is, we need

(i) (∀a ∈ A) a ≤ x

(ii) (∀y < x)(∃a ∈ A) a > y

We usually write

sup A

for the supremum of

when it exists. If

sup A ∈ A

then we call it max A, the maximum of A.

Example. Let

. Then the supremum of (0

1) is 1. The set

}

is bounded above by 2, but has no supremum (even though

√

seems like a

supremum, we are in Q and

√

2 is non-existent!).

max

1] = 1 but (0

1) has no maximum because the supremum is not in

(0, 1).

We can think of the supremum as a point we can get arbitrarily close to in

the set but cannot pass through.

Definition (Least upper bound property). An ordered set

has the least upper

bound property if every non-empty subset of

that is bounded above has a

supremum.

Obvious modifications give rise to definitions of lower bound, greatest lower

bound (or infimum) etc. It is simple to check that an ordered field with the least

upper bound property has the greatest lower bound property.

Definition (Real numbers). The real numbers is an ordered field with the least

upper bound property.

Of course, it is very important to show that such a thing exists, or else we

will be studying nothing. It is also nice to show that such a field is unique (up

to isomorphism). However, we will not prove these in the course.

In a field, we can define the “natural numbers” to be 2 = 1 + 1, 3 = 1 + 2

etc. Then an important property of the real numbers is

Lemma (Archimedean property v1)). Let

be an ordered field with the least

upper bound property. Then the set {1, 2, 3, ···} is not bounded above.

Proof.

If it is bounded above, then it has a supremum

. But then

x −

1 is not

an upper bound. So we can find

n ∈ {

, ···}

such that

n > x −

1. But then

n + 1 > x, but x is supposed to be an upper bound.

Is the least upper bound property required to prove the Archimedean prop-

erty? It seems like any ordered field should satisfy this even if they do not have

the least upper bound property. However, it turns out there are ordered fields

in which the integers are bounded above.

Consider the field of rational functions, i.e. functions in the form

P (x)

Q(x)

with

(

)

, Q

(

) being polynomials, under the usual addition and multiplication. We

order two functions

P (x)

Q(x)

R(x)

S(x)

as follows: these two functions intersect only

finitely many times because

(

)

(

) =

(

)

(

) has only finitely many roots.

After the last intersection, the function whose value is greater counts as the

greater function. It can be checked that these form an ordered field.

In this field, the integers are the constant functions 1

, ···

, but it is

bounded above since the function x is greater than all of them.

2 Convergence of sequences

Having defined real numbers, the first thing we will study is sequences. We

will want to study what it means for a sequence to converge. Intuitively, we

would like to say that 1

, ···

converges to 0, while 1

, ···

diverges.

However, the actual formal definition of convergence is rather hard to get right,

and historically there have been failed attempts that produced spurious results.

2.1 Definitions

Definition (Sequence). A sequence is, formally, a function

N → R

(or

Usually (i.e. always), we write

instead of

(

). Instead of

, we usually write

it as (a

), (a

)

∞

or (a

)

∞

n=1

to indicate it is a sequence.

Definition (Convergence of sequence). Let (

) be a sequence and

` ∈ R

. Then

converges to

, tends to

, or

→ `

, if for all

ε >

0, there is some

N ∈ N

such that whenever n > N, we have |a

− `| < ε. In symbols, this says

(∀ε > 0)(∃N )(∀n ≥ N) |a

− `| < ε.

We say ` is the limit of (a

One can think of (∃N)(∀n ≥ N) as saying “eventually always”, or as “from

some point on”. So the definition means, if

→ `

, then given any

, there

eventually, everything in the sequence is within ε of `.

We’ll now provide an alternative form of the Archimedean property. This is

the form that is actually useful.

Lemma (Archimedean property v2). 1/n → 0.

Proof.

Let

ε >

0. We want to find an

such that

/N −

= 1

/N < ε

. So pick

such that

N >

/ε

. There exists such an

by the Archimedean property v1.

Then for all n ≥ N, we have 0 < 1/n ≤ 1/N < ε. So |1/n − 0| < ε.

Note that the red parts correspond to the definition of convergence of a

sequence. This is generally how we prove convergence from first principles.

Definition (Bounded sequence). A sequence (a

) is bounded if

(∃C)(∀n) |a

| ≤ C.

A sequence is eventually bounded if

(∃C)(∃N)(∀n ≥ N) |a

| ≤ C.

The definition of an eventually bounded sequence seems a bit daft. Clearly

every eventually bounded sequence is bounded! Indeed it is:

Lemma. Every eventually bounded sequence is bounded.

Proof.

Let

and

be such that (

∀n ≥ N

)

| ≤ C

. Then

∀n ∈ N

| ≤

max{|a

|, ··· , |a

N−1

|, C}.

The proof is rather trivial. However, most of the time it is simpler to prove

that a sequence is eventually bounded, and this lemma saves us from writing

that long line every time.

2.2 Sums, products and quotients

Here we prove the things that we think are obviously true, e.g. sums and products

of convergent sequences are convergent.

Lemma (Sums of sequences). If a

→ a and b

→ b, then a

+ b

→ a + b.

Proof.

Let

ε >

0. We want to find a clever

such that for all

n ≥ N

−

(

)

| < ε

. Intuitively, we know that

is very close to

and

very close to b. So their sum must be very close to a + b.

Formally, since

→ a

and

→ b

, we can find

, N

such that

∀n ≥ N

− a| < ε/2 and ∀n ≥ N

, |b

− b| < ε/2.

Now let N = max{N

, N

}. Then by the triangle inequality, when n ≥ N ,

|(a

+ b

) − (a + b)| ≤ |a

− a| + |b

− b| < ε.

We want to prove that the product of convergent sequences is convergent.

However, we will not do it in one go. Instead, we separate it into many smaller

parts.

Lemma (Scalar multiplication of sequences). Let

→ a

and

λ ∈ R

. Then

λa

→ λa.

Proof. If λ = 0, then the result is trivial.

Otherwise, let

ε >

0. Then

∃N

such that

∀n ≥ N

− a| < ε/|λ|

. So

|λa

− λa| < ε.

Lemma. Let (a

) be bounded and b

→ 0. Then a

→ 0.

Proof.

Let

C 6

= 0 be such that (

∀n

)

| ≤ C

. Let

ε >

0. Then

∃N

such that

(∀n ≥ N ) |b

| < ε/C. Then |a

| < ε.

Lemma. Every convergent sequence is bounded.

Proof.

Let

→ l

. Then there is an

such that

∀n ≥ N

− l| ≤

1. So

| ≤ |l| + 1. So a

is eventually bounded, and therefore bounded.

Lemma (Product of sequences). Let a

→ a and b

→ b. Then a

→ ab.

Proof. Let a

= a + ε

. Then a

= (a + ε

= ab

+ ε

Since

→ b

→ ab

. Since

→

0 and

is bounded,

→

0. So

→ ab.

Proof.

(alternative) Observe that

−ab

= (

−a

)

−b

)

. We know that

−a →

0 and

−b →

0. Since (

) is bounded, so (

−a

)

+ (

−b

)

a →

So a

→ ab.

Note that in this proof, we no longer write “Let

ε >

0”. In the beginning, we

have no lemmas proven. So we must prove everything from first principles and

use the definition. However, after we have proven the lemmas, we can simply

use them instead of using first principles. This is similar to in calculus, where

we use first principles to prove the product rule and chain rule, then no longer

use first principles afterwards.

Lemma (Quotient of sequences). Let (

) be a sequence such that (

∀n

)

= 0.

Suppose that a

→ a and a 6= 0. Then 1/a

→ 1/a.

Proof. We have

−

a − a

We want to show that this

→

0. Since

a −a

→

0, we have to show that 1

(

)

is bounded.

Since

→ a

∃N

such that

∀n ≥ N

− a| ≤ a/

2. Then

∀n ≥ N

| ≥

|a|/

2. Then

(

)

| ≤

/|a|

. So 1

(

) is bounded. So (

a − a

)

(

)

→

and the result follows.

Corollary. If a

→ a, b

→ b, b

, b 6= 0, then a

= a/b.

Proof.

We know that 1

→

. So the result follows by the product rule.

Lemma (Sandwich rule). Let (

) and (

) be sequences that both converge to

a limit x. Suppose that a

≤ c

≤ b

for every n. Then c

→ x.

Proof.

Let

ε >

0. We can find

such that

∀n ≥ N

−x| < ε

and

−x| < ε

Then ∀n ≥ N, we have x −ε < a

≤ c

≤ b

< x + ε. So |c

− x| < ε.

Example. 1

→

0. For every

n <

. So 0

. The result

follows from the sandwich rule.

Example. We want to show that

+ 3

(n + 5)(2n −1)

→

We can obtain this by

+ 3

(n + 5)(2n −1)

1 + 3/n

(1 + 5/n)(2 −1/n)

→

by sum rule, sandwich rule, Archimedean property, product rule and quotient

rule.

Example. Let k ∈ N and let δ > 0. Then

(1 + δ)

→ 0.

This can be summarized as “exponential growth beats polynomial growth even-

tually”.

By the binomial theorem,

(1 + δ)

≥



k + 1



k+1

Also, for n ≥ 2k,



k + 1



n(n − 1) ···(n −k)

(k + 1)!

≥

(n/2)

k+1

(k + 1)!

So for sufficiently large n,

(1 + δ)

≤

k+1

(k + 1)!

k+1

(k + 1)!

k+1

→ 0.

2.3 Monotone-sequences property

Recall that we characterized the least upper bound property. It turns out that

there is an alternative characterization of real number using sequences, known

as the monotone-sequences property. In this section, we will show that the two

characterizations are equivalent, and use the monotone-sequences property to

deduce some useful results.

Definition (Monotone sequence). A sequence (

) is increasing if

≤ a

n+1

for all n.

It is strictly increasing if

< a

n+1

for all

. (Strictly) decreasing sequences

are defined analogously.

A sequence is (strictly) monotone if it is (strictly) increasing or (strictly)

decreasing.

Definition (Monotone-sequences property). An ordered field has the monotone

sequences property if every increasing sequence that is bounded above converges.

We want to show that the monotone sequences property is equivalent to the

least upper bound property.

Lemma. Least upper bound property ⇒ monotone-sequences property.

Proof.

Let (

) be an increasing sequence and let

an upper bound for (

Then

is an upper bound for the set

n ∈ N}

. By the least upper bound

property, it has a supremum s. We want to show that this is the limit of (a

Let

ε >

0. Since

sup{a

n ∈ N}

, there exists an

such that

> s −ε

Then since (

) is increasing,

∀n ≥ N

, we have

s − ε < a

≤ a

≤ s

. So

− s| < ε.

We first prove a handy lemma.

Lemma. Let (

) be a sequence and suppose that

→ a

. If (

∀n

)

≤ x

, then

a ≤ x.

Proof.

a > x

, then set

a − x

. Then we can find

such that

> x

Contradiction.

Before showing the other way implication, we will need the following:

Lemma. Monotone-sequences property ⇒ Archimedean property.

Proof. We prove version 2, i.e. that 1/n → 0.

Since 1

/n >

0 and is decreasing, by MSP, it converges. Let

be the limit.

By the previous lemma, we must have δ ≥ 0.

δ >

0, then we can find

such that 1

/N <

. But then for all

n ≥

we have 1/n ≤ 1/(4N) < δ/2. Contradiction. Therefore δ = 0.

Lemma. Monotone-sequences property ⇒ least upper bound property.

Proof.

Let

be a non-empty set that’s bounded above. Pick

, v

such that

is not an upper bound for

and

is an upper bound. Now do a repeated

bisection: having chosen

and

such that

is not an upper bound and

is, if (

)

2 is an upper bound, then let

n+1

= (

)

Otherwise, let u

n+1

= (u

+ v

)/2, v

n+1

= v

Then u

≤ u

≤ ··· and v

≥ v

≥ ···. We also have

− u

→ 0.

By the monotone sequences property,

→ s

(since (

) is bounded above by

). Since v

− u

→ 0, v

→ s. We now show that s = sup A.

is not an upper bound, then there exists

a ∈ A

such that

a > s

. Since

→ s

, then there exists

such that

< a

, contradicting the fact that

an upper bound.

To show it is the least upper bound, let

t < s

. Then since

→ s

, we can

find

such that

> t

. So

is not an upper bound. Therefore

is the least

upper bound.

Why do we need to prove the Archimedean property first? In the proof

above, we secretly used the it. When showing that

−u

→

0, we required the

fact that

→

0. To prove this, we sandwiched it with

. But to show

→

we need the Archimedean property.

Lemma. A sequence can have at most 1 limit.

Proof.

Let (

) be a sequence, and suppose

→ x

and

→ y

. Let

ε >

and pick

such that

∀n ≥ N

− x| < ε/

2 and

− y| < ε/

2. Then

|x − y| ≤ |x − a

− y| < ε/

2 +

ε/

2 =

. Since

was arbitrary,

must

equal y.

Lemma (Nested intervals property). Let

be an ordered field with the monotone

sequences property. Let

⊇ I

⊇ ···

be closed bounded non-empty intervals.

Then

∞

n=1

6= ∅.

Proof.

Let

= [

, b

] for each

. Then

≤ a

≤ ···

and

≥ b

≥ ···

For each

≤ b

. So the sequence

is bounded above. So by the

monotone sequences property, it has a limit

. For each

, we must have

≤ a

Otherwise, say

> a

. Then for all

m ≥ n

, we have

≥ a

> a

. This implies

that a > a, which is nonsense.

Also, for each fixed

, we have that

∀m ≥ n

≤ b

. So

a ≤ b

Thus, for all n, a

≤ a ≤ b

. So a ∈ I

. So a ∈

∞

n=1

We can use this to prove that the reals are uncountable:

Proposition. R is uncountable.

Proof.

Suppose the contrary. Let

, x

, ···

be a list of all real numbers. Find

an interval that does not contain

. Within that interval, find an interval that

does not contain

. Continue ad infinitum. Then the intersection of all these

intervals is non-empty, but the elements in the intersection are not in the list.

Contradiction.

A powerful consequence of this is the Bolzano-Weierstrass theorem. This is

formulated in terms of subsequences:

Definition (Subsequence). Let (

) be a sequence. A subsequence of (

) is a

sequence of the form a

, a

, ···, where n

< n

< ···.

Example. 1, 1/4, 1/9, 1/16, ··· is a subsequence of 1, 1/2, 1/3, ···.

Theorem (Bolzano-Weierstrass theorem). Let

be an ordered field with the

monotone sequences property (i.e. F = R).

Then every bounded sequence has a convergent subsequence.

Proof.

Let

and

be a lower and upper bound, respectively, for a sequence

(

). By repeated bisection, we can find a sequence of intervals [

, v

]

⊇

[

, v

]

⊇

[

, v

]

⊇ ···

such that

− u

= (

− u

)

, and such that each

, v

] contains infinitely many terms of (a

By the nested intervals property,

∞

n=1

[

, v

]

∅

. Let

belong to the

intersection. Now pick a subsequence

, a

, ···

such that

∈

[

, v

]. We

can do this because [

, v

] contains infinitely many

, and we have only picked

finitely many of them. We will show that a

→ x.

Let

ε >

0. By the Archimedean property, we can find

such that

−u

(

−u

)

≤ ε

. This implies that [

, v

]

⊆

(

x −ε, x

), since

x ∈

[

, v

Then

∀k ≥ K

∈

[

, v

]

⊆

[

, v

]

⊆

(

x−ε, x

). So

−x| < ε

2.4 Cauchy sequences

The third characterization of real numbers is in terms of Cauchy sequences.

Cauchy convergence is an alternative way of defining convergent sequences

without needing to mention the actual limit of the sequence. This allows us to

say

{

141

1415

, ···}

is Cauchy convergent in

even though the

limit π is not in Q.

Definition (Cauchy sequence). A sequence (

) is Cauchy if for all

, there is

some

N ∈ N

such that whenever

p, q ≥ N

, we have

− a

| < ε

. In symbols,

we have

(∀ε > 0)(∃N )(∀p, q ≥ N) |a

− a

| < ε.

Roughly, a sequence is Cauchy if all terms are eventually close to each other

(as opposed to close to a limit).

Lemma. Every convergent sequence is Cauchy.

Proof.

Let

→ a

. Let

ε >

0. Then

∃N

such that

∀n ≥ N

− a| < ε/

Then ∀p, q ≥ N, |a

− a

| ≤ |a

− a| + |a − a

| < ε/2 + ε/2 = ε.

Lemma. Let (

) be a Cauchy sequence with a subsequence (

) that converges

to a. Then a

→ a.

Proof.

Let

ε >

0. Pick

such that

∀p, q ≥ N

− a

| < ε/

2. Then pick

such that n

≥ N and |a

− a| < ε/2.

Then ∀n ≥ N, we have

− a| ≤ |a

− a

| + |a

− a| <

= ε.

An important result we have is that in

, Cauchy convergence and regular

convergence are equivalent.

Theorem (The general principle of convergence). Let

be an ordered field with

the monotone-sequence property. Then every Cauchy sequence of F converges.

Proof.

Let (

) be a Cauchy sequence. Then it is eventually bounded, since

∃N

∀n ≥ N

− a

| ≤

1 by the Cauchy condition. So it is bounded. Hence by

Bolzano-Weierstrass, it has a convergent subsequence. Then (

) converges to

the same limit.

Definition (Complete ordered field). An ordered field in which every Cauchy

sequence converges is called complete.

Hence we say that R is a complete ordered field.

However, not every complete ordered field is (isomorphic to)

. For example,

we can take the rational functions as before, then take the Cauchy completion

of it (i.e. add all the limits we need). Then it is already too large to be the reals

(it still doesn’t have the Archimedean property) but is a complete ordered field.

To show that completeness implies the monotone-sequences property, we

need an additional condition: the Archimedean property.

Lemma. Let

be an ordered field with the Archimedean property such that

every Cauchy sequence converges. The

satisfies the monotone-sequences

property.

Proof.

Instead of showing that every bounded monotone sequence converges, and

is hence Cauchy, We will show the equivalent statement that every increasing

non-Cauchy sequence is not bounded above.

Let (a

) be an increasing sequence. If (a

) is not Cauchy, then

(∃ε > 0)(∀N )(∃p, q > N) |a

− a

| ≥ ε.

wlog let p > q. Then

≥ a

+ ε ≥ a

+ ε.

So for any N, we can find a p > N such that

≥ a

+ ε.

Then we can construct a subsequence a

, a

, ··· such that

k+1

≥ a

+ ε.

Therefore

≥ a

+ (k − 1)ε.

So by the Archimedean property, (a

), and hence (a

), is unbounded.

Note that the definition of a convergent sequence is

(∃l)(∀ε > 0)(∃N)(∀n ≥ N) |a

− l| < ε,

while that of Cauchy convergence is

(∀ε > 0)(∃N )(∀p, q ≥ N) |a

− a

| < ε.

In the first definition,

quantifies over all real numbers, which is uncountable.

However, in the second definition, we only have to quantify over natural numbers,

which is countable (by the Archimedean property, we only have to consider the

cases ε = 1/n).

Since they are equivalent in

, the second definition is sometimes preferred

when we care about logical simplicity.

2.5 Limit supremum and infimum

Here we will define the limit supremum and infimum. While these are technically

not part of the course, eventually some lecturers will magically assume students

know this definition. So we might as well learn it here.

Definition (Limit supremum/infimum). Let (

) be a bounded sequence. We

define the limit supremum as

lim sup

n→∞

= lim

n→∞



sup

m≥n



To see that this exists, set

sup

m≥n

. Then (

) is decreasing since we

are taking the supremum of fewer and fewer things, and is bounded below by

any lower bound for (a

) since b

≥ a

. So it converges.

Similarly, we define the limit infimum as

lim inf

n→∞

= lim

n→∞



inf

m≥n



Example. Take the sequence

2, −1,

, −

, ···

Then the limit supremum is 1 and the limit infimum is 0.

Lemma. Let (

) be a sequence. The following two statements are equivalent:

– a

→ a

– lim sup a

= lim inf a

= a.

Proof. If a

→ a, then let ε > 0. Then we can find an n such that

a − ε ≤ a

≤ a + ε for all m ≥ n

It follows that

a − ε ≤ inf

m≥n

≤ sup

m≥n

≤ a + ε.

Since ε was arbitrary, it follows that

lim inf a

= lim sup a

= a.

Conversely, if

lim inf a

lim sup a

, then let

ε >

0. Then we can find

such that

inf

m≥n

> a −ε and sup

m≥n

< a + ε.

It follows that ∀m ≥ n, we have |a

− a| < ε.

3 Convergence of infinite sums

In this chapter, we investigate which infinite sums, as opposed to sequences,

converge. We would like to say 1 +

···

= 2, while 1 + 1 + 1 + 1 +

···

does not converge. The majority of the chapter is coming up with different tests

to figure out if infinite sums converge.

3.1 Infinite sums

Definition (Convergence of infinite sums and partial sums). Let (

) be a real

sequence. For each N, define

n=1

If the sequence (S

) converges to some limit s, then we say that

∞

n=1

= s,

and we say that the series

∞

n=1

converges.

We call S

the N th partial sum.

There is an immediate necessary condition for a series to converge.

Lemma. If

∞

n=1

converges. Then a

→ 0.

Proof.

Let

∞

n=1

. Then

→ s

and

n−1

→ s

. Then

− S

n−1

→

However, the converse is false!

Example (Harmonic series). If a

= 1/n, then a

→ 0 but

= ∞.

We can prove this as follows:

− S

n−1

+ 1

+ ··· +

≥

n−1

Therefore S

≥ S

+ n/2. So the partial sums are unbounded.

Example (Geometric series). Let |ρ| < 1. Then

∞

n=0

1 − ρ

We can prove this by considering the partial sums:

n=0

1 − ρ

N+1

1 − ρ

Since ρ

N+1

→ 0, this tends to 1/(1 − ρ).

Example.

∞

n=2

n(n − 1)

converges. This is since

n(n − 1)

n − 1

−

n=2

n(n − 1)

= 1 −

→ 1.

Lemma. Suppose that

≥

0 for every

and the partial sums

are bounded

above. Then

∞

n=1

converges.

Proof.

The sequence (

) is increasing and bounded above. So the result follows

form the monotone sequences property.

The simplest convergence test we have is the comparison test. Roughly

speaking, it says that if 0

≤ a

≤ b

for all

and

converges, then

converges. However, we will prove a much more general form here for convenience.

Lemma (Comparison test). Let (

) and (

) be non-negative sequences, and

suppose that

∃C, N

such that

∀n ≥ N

≤ Cb

. Then if

converges, then

so does

Proof.

Let

M > N

. Also for each

, let

n=1

and

n=1

. We

want S

to be bounded above.

− S

n=N+1

≤ C

n=N+1

≤ C

∞

n=N+1

∀M ≥ N

≤ S

∞

n=N+1

. Since the

are increasing and

bounded, it must converge.

Example.

(i)

converges, since

converges.

(ii)

converges.

n ≥

4, then

n ≤

n/2

. That’s because 4 = 2

4/2

and for

n ≥

(

+ 1)

/n <

√

, so when we increase

, we multiply the right side by a

greater number by the left. Hence by the comparison test, it is sufficient

to show that

n/2

−n/2

converges, which it does (geometric

series).

(iii)

√

diverges, since

√

≥

. So if it converged, then so would

, but

diverges.

(iv)

converges, since for

n ≥

≤

n(n−1)

, and we have proven that

the latter converges.

(v) Consider

∞

n=1

n + 5

− 7n

. We show this converges by noting

−

= n



n −



So if n ≥ 8, then

−

≥

Also, n + 5 ≤ 2n. So

n + 5

− 7n

≤ 4/n

So it converges by the comparison test.

(vi) If α > 1, then

1/n

converges.

Let S

n=1

1/n

. Then

− S

n−1

+ 1)

+ ··· +

)

≤

n−1

)

= (2

n−1

)

1−α

= (2

1−α

)

n−1

But 2

1−α

< 1. So

= (S

− S

n−1

) + (S

n−1

− S

n−2

) + ···(S

− S

) + S

and is bounded above by comparison with the geometric series 1 + 2

1−α

)

+ ···

3.2 Absolute convergence

Here we’ll consider two stronger conditions for convergence — absolute conver-

gence and unconditional convergence. We’ll prove that these two conditions are

in fact equivalent.

Definition (Absolute convergence). A series

converges absolutely if the

series

| converges.

Example. The series

(−1)

n+1

= 1

−

···

converges, but not

absolutely.

To see the convergence, note that

2n−1

+ a

2n − 1

−

2n(2n − 1)

It is easy to compare with 1

to get that the partial sums

converges. But

2n+1

− S

= 1/(2n + 1) → 0, so the S

2n+1

converges to the same limit.

It does not converge absolutely, because the sum of the absolute values is

the harmonic series.

Lemma. Let

converge absolutely. Then

converges.

Proof.

We know that

converges. Let

n=1

and

n=1

We know two ways to show random sequences converge, without knowing

what they converge to, namely monotone-sequences and Cauchy sequences. Since

is not monotone, we shall try Cauchy sequences.

If p > q, then

− S

| =



n=q+1



≤

n=q+1

| = T

− T

But the sequence

converges. So

∀ε >

0, we can find

such that for all

p > q ≥ N , we have T

− T

< ε, which implies |S

− S

| < ε.

Definition (Unconditional convergence). A series

converges uncondition-

ally if the series

∞

n=1

π(n)

converges for every bijection

N → N

, i.e. no

matter how we re-order the elements of a

, the sum still converges.

Theorem. If

converges absolutely, then it converges unconditionally.

Proof. Let S

n=1

π(n)

. Then if p > q,

− S

| =



n=q+1

π(n)



≤

∞

n=q+1

π(n)

Let ε > 0. Since

| converges, pick M such that

∞

n=M+1

| < ε.

Pick N large enough that {1, ··· , M} ⊆ {π(1), ··· , π(N)}.

Then if n > N , we have π(n) > M. Therefore if p > q ≥ N, then

− S

| ≤

n=q+1

π(n)

| ≤

∞

n=M+1

| < ε.

Therefore the sequence of partial sums is Cauchy.

The converse is also true.

Theorem. If

converges unconditionally, then it converges absolutely.

Proof.

We will prove the contrapositive: if it doesn’t converge absolutely, it

doesn’t converge unconditionally.

Suppose that

∞

. Let (

) be the subsequence of non-negative

terms of

, and (

) be the subsequence of negative terms. Then

and

cannot both converge, or else

| converges.

wlog,

∞

. Now construct a sequence 0 =

< n

< ···

such

that ∀k,

k−1

+ b

k−1

+ ··· + b

+ c

≥ 1,

This is possible because the

are unbounded and we can get it as large as we

want.

Let π be the rearrangement

, b

, ···b

, c

, b

, ···b

, c

, b

, ···b

, c

, ···

So the sum up to c

is at least k. So the partial sums tend to infinity.

We can prove an even stronger result:

Lemma. Let

be a series that converges absolutely. Then for any bijection

π : N → N,

∞

n=1

∞

n=1

π(n)

Proof.

Let

ε >

0. We know that both

and

π(n)

converge. So let

be such that

n>M

| <

and

n>M

π(n)

| <

Now N be large enough such that

{1, ··· , M } ⊆ {π(1), ··· , π(N)},

and

{π(1), ··· , π(M)} ⊆ {1, ··· , N}.

Then for every K ≥ N ,



n=1

−

n=1

π(n)



≤

n=M+1

| +

n=M+1

π(n)

| <

= ε.

We have the first inequality since given our choice of

and

, the first

terms of the

and

π(n)

sums are cancelled by some term in the huge

sum.

∀K ≥ N

, the partial sums up to

differ by at most

. So

−

π(n)

| ≤ ε.

Since this is true for all ε, we must have

π(n)

3.3 Convergence tests

We’ll now come up with a lot of convergence tests.

Lemma (Alternating sequence test). Let (

) be a decreasing sequence of

non-negative reals, and suppose that

→

0. Then

∞

n=1

(

−

n+1

converges,

i.e. a

− a

+ a

− a

+ ··· converges.

Proof. Let S

n=1

(−1)

n+1

. Then

= (a

− a

) + (a

− a

) + ··· + (a

2n−1

− a

) ≥ 0,

and (S

) is an increasing sequence.

Also,

2n+1

= a

− (a

− a

) − (a

− a

) − ··· −(a

− a

2n+1

and (S

2n+1

) is a decreasing sequence. Also S

2n+1

− S

= a

2n+1

≥ 0.

Hence we obtain the bounds 0

≤ S

2n+1

≤ a

. It follows from the

monotone sequences property that (S

) and (S

2n+1

) converge.

Since S

2n+1

− S

= a

2n+1

→ 0, they converge to the same limit.

Example.

1 −

√

−

√

+ ···converges.

Lemma (Ratio test). We have three versions:

(i) If ∃c < 1 such that

n+1

≤ c,

for all n, then

converges.

(ii) If ∃c < 1 and ∃N such that

(∀n ≥ N )

n+1

≤ c,

then

converges. Note that just because the ratio is always less than

1, it doesn’t necessarily converge. It has to be always less than a fixed

number c. Otherwise the test will say that

1/n converges.

(iii) If ∃ρ ∈ (−1, 1) such that

n+1

→ ρ,

then

converges. Note that we have the open interval (

−

1). If

n+1

→ 1, then the test is inconclusive!

Proof.

(i) |a

| ≤ c

n−1

. Since

converges, so does

by comparison test.

converges absolutely, so it converges.

(ii)

For all

k ≥

0, we have

N+k

| ≤ c

. So the series

N+k

converges,

and therefore so does

(iii)

n+1

→ ρ

, then

n+1

→ |ρ|

. So (setting

= (1

− |ρ|

)

2) there exists

such that ∀n ≥ N,

n+1

≤

1+|ρ|

< 1. So the result follows from (ii).

Example. If |b| < 1, then

converges, since

n+1

(n + 1)b

n+1



1 +



b → b < 1.

So it converges.

We can also evaluate this directly by considering

∞

i=1

∞

n=i

The following two tests were taught at the end of the course, but are included

here for the sake of completeness.

Theorem (Condensation test). Let (

) be a decreasing non-negative sequence.

Then

∞

n=1

< ∞ if and only if

∞

k=1

< ∞.

Proof.

This is basically the proof that

diverges and

converges for

α < 1 but written in a more general way.

We have

+ a

+ (a

+ a

) + (a

+ ··· + a

) + (a

+ ··· + a

) + ···

≥a

+ a

+ 2a

+ 4a

+ 8a

+ ···

So if

diverges,

diverges.

To prove the other way round, simply group as

+ (a

+ a

) + (a

+ ··· + a

) + ···

≤a

+ 2a

+ 4a

+ ··· .

Example. If a

, then 2

= 1. So

∞

k=1

= ∞. So

∞

n=1

= ∞.

After we formally define integrals, we will prove the integral test:

Theorem (Integral test). Let

: [1

, ∞

]

→ R

be a decreasing non-negative

function. Then

∞

n=1

f(n) converges iff

∞

f(x) dx < ∞.

3.4 Complex versions

Most definitions in the course so far carry over unchanged to the complex

numbers. e.g. z

→ z iff (∀ε > 0)(∃N)(∀n ≥ N) |z

− z| < ε.

Two exceptions are least upper bound and monotone sequences, because the

complex numbers do not have an ordering! (It cannot be made into an ordered

field because the square of every number in an ordered field has to be positive)

Fortunately, Cauchy sequences still work.

We can prove the complex versions of most theorems so far by looking at the

real and imaginary parts.

Example. Let (

) be a Cauchy sequence in

. Let

. Then (

)

and (

) are Cauchy. So they converge, from which it follows that

converges.

Also, the Bolzano-Weierstrass theorem still holds: If (

) is bounded, let

, then (

) and (

) are bounded. Then find a subsequence (

)

that converges. Then find a subsequence of (y

) that converges.

Then nested-intervals property has a “nested-box” property as a complex

analogue.

Finally, the proof that absolutely convergent sequences converge still works.

It follows that the ratio test still works.

Example. If |z| < 1, then

converges. Proof is the same as above.

However, we do have an extra test for complex sums.

Lemma (Abel’s test). Let

≥ a

≥ ··· ≥

0, and suppose that

→

0. Let

z ∈ C such that |z| = 1 and z 6= 1. Then

converges.

Proof. We prove that it is Cauchy. We have

n=M

n+1

− z

z − 1

n=M

n+1

− z

)

z − 1

n=M

n+1

−

n=M

z − 1

n=M

n+1

−

N−1

n=M−1

n+1

z − 1

N+1

− a

N−1

n=M

− a

n+1

We now take the absolute value of everything to obtain



n=M



≤

|z − 1|

+ a

N−1

n=M

− a

n+1

)

|z − 1|

+ a

+ (a

− a

M+1

) + ··· + (a

N−1

− a

))

|z − 1|

→ 0.

So it is Cauchy. So it converges

Note that here we transformed the sum

(

n+1

− z

) into

N+1

−

(

−a

n+1

)

n+1

. What we have effectively done is a discrete analogue

of integrating by parts.

Example. The series

converges if

|z| <

1 or if

|z|

= 1 and

z 6

= 1, and it

diverges if z = 1 or |z| > 1.

The cases

|z| <

1 and

|z| >

1 are trivial from the ratio test, and Abel’s test

is required for the |z| = 1 cases.

4 Continuous functions

4.1 Continuous functions

Definition (Continuous function). Let

A ⊆ R

a ∈ A

, and

A → R

. Then

is continuous at

if for any

ε >

0, there is some

δ >

0 such that if

y ∈ A

is such

that |y − a| < δ, then |f(y) −f(a)| < ε. In symbols, we have

(∀ε > 0)(∃δ > 0)(∀y ∈ A) |y − a| < δ ⇒ |f(y) − f (a)| < ε.

f is continuous if it is continuous at every a ∈ A. In symbols, we have

(∀a ∈ A)(∀ε > 0)(∃δ > 0)(∀y ∈ A) |y − a| < δ ⇒ |f(y) −f(a)| < ε.

Intuitively,

is continuous at

if we can obtain

(

) as accurately as we

wish by using more accurate values of

(the definition says that if we want to

approximate

(

) by

(

) to within accuracy

, we just have to get our

within δ of a for some δ).

For example, suppose we have the function

f(x) =

(

0 x ≤ π

1 x > π

Suppose that we don’t know what the function actually is, but we have a

computer program that computes this function. We want to know what

(

)

is. Since we cannot input

(it has infinitely many digits), we can try 3, and it

gives 0. Then we try 3

14, and it gives 0 again. If we try 3

1416, it gives 1 (since

= 3

1415926

··· <

1416). We keep giving more and more digits of

, but the

result keeps oscillating between 0 and 1. We have no hope of what

(

) might

be, even approximately. So this f is discontinuous at π.

However, if we have the function

(

) =

, then we can find the (approx-

imate) value of

(

). We can first try

(3) and obtain 9. Then we can try

14) = 9

8596,

1416) = 9

86965056 etc. We can keep trying and obtain

more and more accurate values of g(π). So g is continuous at π.

Example.

– Constant functions are continuous.

– The function f(x) = x is continuous (take δ = ε).

The definition of continuity of a function looks rather like the definition of

convergence. In fact, they are related by the following lemma:

Lemma. The following two statements are equivalent for a function

A → R

– f is continuous

– If (a

) is a sequence in A with a

→ a, then f(a

) → f (a).

Proof. (i)⇒(ii) Let ε > 0. Since f is continuous at a,

(∃δ > 0)(∀y ∈ A) |y − a| < δ ⇒ |f(y) −f(a)| < ε.

We want

such that

∀n ≥ N

(

)

− f

(

)

| < ε

. By continuity, it is enough

to find N such that ∀n ≥ N, |a

− a| < δ. Since a

→ a, such an N exists.

(ii)

⇒

(i) We prove the contrapositive: Suppose

is not continuous at

. Then

(∃ε > 0)(∀δ > 0)(∃y ∈ A) |y − a| < δ and |f(y) −f(a)| ≥ ε.

For each

, we can therefore pick

∈ A

such that

− a| <

and

(

)

−

(

)

| ≥ ε

. But then

→ a

(by Archimedean property), but

(

)

6→ f

(

Example.

(i)

Let

(

) =

(

−1 x < 0

1 x ≥ 0

. Then

is not continuous because

−

→

0 but

f(−

) → −1 6= f (0).

(ii) Let f : Q → R with

f(x) =

(

1 x

> 2

0 x

< 2

Then

is continuous. For every

a ∈ Q

, we can find an interval about

which f is constant. So f is continuous at a.

(iii) Let

f(x) =

(

sin

x 6= 0

0 x = 0

Then

(

) is discontinuous. For example, let

= 1

[(2

+ 0

]. Then

→ 0 and f(a

) → 1 6= f (0).

We can use this sequence definition as the definition for continuous functions.

This has the advantage of being cleaner to write and easier to work with. In

particular, we can reuse a lot of our sequence theorems to prove the analogous

results for continuous functions.

Lemma. Let A ⊆ R and f, g : A → R be continuous functions. Then

(i) f + g is continuous

(ii) fg is continuous

(iii) if g never vanishes, then f /g is continuous.

Proof.

(i) Let a ∈ A and let (a

) be a sequence in A with a

→ a. Then

(f + g)(a

) = f(a

) + g(a

But f (a

) → f (a) and g(a

) → g(a). So

f(a

) + g(a

) → f (a) + g(a) = (f + g)(a).

(ii) and (iii) are proved in exactly the same way.

With this lemma, from the fact that constant functions and

(

) =

are

continuous, we know that all polynomials are continuous. Similarly, rational

functions P (x)/Q(x) are continuous except when Q(x) = 0.

Lemma. Let

A, B ⊆ R

and

A → B

B → R

. Then if

and

are

continuous, g ◦ f : A → R is continuous.

Proof. We offer two proofs:

(i)

Let (

) be a sequence in

with

→ a ∈ A

. Then

(

)

→ f

(

) since

is continuous. Then

(

))

→ g

(

)) since

is continuous. So

g ◦ f

is continuous.

(ii)

Let

a ∈ A

and

ε >

0. Since

is continuous at

(

), there exists

η >

0 such

that ∀z ∈ B, |z − f(a)| < η ⇒ |g(z) −g(f(a))| < ε.

Since

is continuous at

∃δ >

0 such that

∀y ∈ A

|y − a| < δ ⇒

|f(y) −f(a)| < η. Therefore |y − a| < δ ⇒ |g(f(y)) − g(f (a))| < ε.

There are two important theorems regarding continuous functions — the

maximum value theorem and the intermediate value theorem.

Theorem (Maximum value theorem). Let [

a, b

] be a closed interval in

and

let

: [

a, b

]

→ R

be continuous. Then

is bounded and attains its bounds, i.e.

f(x) = sup f for some x, and f(y) = inf f for some y.

Proof.

is not bounded above, then for each

, we can find

∈

[

a, b

] such

that f (x

) ≥ n for all n.

By Bolzano-Weierstrass, since

∈

[

a, b

] and is bounded, the sequence

(

) has a convergent subsequence (

). Let

be its limit. Then since

continuous, f(x

) → f (x). But f(x

) ≥ n

→ ∞. So this is a contradiction.

Now let

sup{f

(

) :

x ∈

[

a, b

]

}

. Then for every

, we can find

such that

(

)

≥ C −

. So by Bolzano-Weierstrass, (

) has a convergent

subsequence (

). Since

C −

≤ f

(

)

≤ C

(

)

→ C

. Therefore if

x = lim x

, then f(x) = C.

A similar argument applies if f is unbounded below.

Theorem (Intermediate value theorem). Let

a < b ∈ R

and let

: [

a, b

]

→ R

be continuous. Suppose that

(

)

< f

(

). Then there exists an

x ∈

(

a, b

)

such that f (x) = 0.

Proof. We have several proofs:

(i)

Let

(

)

}

and let

sup A

. We shall show that

(

) = 0

(this is similar to the proof that

√

exists in Numbers and Sets). If

(

)

0, then setting

(

)

in the definition of continuity, we can find

δ >

0 such that

∀y

|y − s| < δ ⇒ f

(

)

0. Then

δ/

∈ A

, so

is not

an upper bound. Contradiction.

(

)

0, by the same argument, we can find

δ >

0 such that

∀y

|y − s| < δ ⇒ f(y) > 0. So s − δ/2 is a smaller upper bound.

(ii)

Let

. By repeated bisection, construct nested intervals

[

, b

] such that

− a

−a

and

(

)

≤ f

(

). Then by the

nested intervals property, we can find

x ∈ ∩

∞

n=0

[

, b

]. Since

−a

→

, b

→ x.

Since

(

)

0 for every

(

)

≤

0. Similarly, since

(

)

≥

0 for every

n, f (x) ≥ 0. So f(x) = 0.

It is easy to generalize this to get that, if

(

)

< c < f

(

), then

∃x ∈

(

a, b

)

such that

(

) =

, by applying the result to

(

)

− c

. Also, we can assume

instead that f(b) < c < f(a) and obtain the same result by looking at −f (x).

Corollary. Let

: [

a, b

]

→

[

c, d

] be a continuous strictly increasing function

with f (a) = c, f (b) = d. Then f is invertible and its inverse is continuous.

Proof.

Since

is strictly increasing, it is an injection (suppose

x 6

. wlog,

x < y

. Then

(

)

< f

(

) and so

(

)

(

)). Now let

y ∈

(

c, d

). By the

intermediate value theorem, there exists

x ∈

(

a, b

) such that

(

) =

. So

is a

surjection. So it is a bijection and hence invertible.

Let

be the inverse. Let

y ∈

[

c, d

] and let

ε >

0. Let

(

). So

(

) =

Let

(

x −ε

) and

(

) (if

, make the obvious adjustments).

Then

u < y < v

. So we can find

δ >

0 such that (

y − δ, y

)

⊆

(

u, v

). Then

|z − y| < δ ⇒ g(z) ∈ (x −ε, x + ε) ⇒ |g(z) −g(y)| < ε.

With this corollary, we can create more continuous functions, e.g.

√

4.2 Continuous induction*

Continuous induction is a generalization of induction on natural numbers. It

provides an alternative mechanism to prove certain results we have shown.

Proposition (Continuous induction v1). Let

a < b

and let

A ⊆

[

a, b

] have the

following properties:

(i) a ∈ A

(ii) If x ∈ A and x 6= b, then ∃y ∈ A with y > x.

(iii) If ∀ε > 0, ∃y ∈ A : y ∈ (x − ε, x], then x ∈ A.

Then b ∈ A.

Proof.

Since

a ∈ A

A 6

∅

is also bounded above by

. So let

sup A

Then ∀ε > 0, ∃y ∈ A such that y > s −ε. Therefore, by (iii), s ∈ A.

If s 6= b, then by (ii), we can find y ∈ A such that y > s.

It can also be formulated as follows:

Proposition (Continuous induction v2). Let A ⊆ [a, b] and suppose that

(i) a ∈ A

(ii) If [a, x] ⊆ A and x 6= b, then there exists y > x such that [a, y] ⊆ A.

(iii) If [a, x) ⊆ A, then [a, x] ⊆ A.

Then A = [a, b]

Proof.

We prove that version 1

⇒

version 2. Suppose

satisfies the conditions

of v2. Let A

= {x ∈ [a, b] : [a, x] ⊆ A}.

Then

a ∈ A

. If

x ∈ A

with

x 6

, then [

a, x

]

⊆ A

. So

∃y > x

such that

[a, y] ⊆ A. So ∃y > x such that y ∈ A

∀ε >

, ∃y ∈

(

x − ε, x

] such that [

a, y

]

⊆ A

, then [

a, x

)

⊆ A

. So by (iii),

[

a, x

]

⊆ A

, so

x ∈ A

. So

satisfies properties (i) to (iii) of version 1. Therefore

b ∈ A

. So [a, b] ⊆ A. So A = [a, b].

We reprove intermediate value theorem here:

Theorem (Intermediate value theorem). Let

a < b ∈ R

and let

: [

a, b

]

→ R

be continuous. Suppose that

(

)

< f

(

). Then there exists an

x ∈

(

a, b

)

such that f (x) = 0.

Proof.

Assume that

is continuous. Suppose

(

)

< f

(

). Assume that

(∀x) f(x) 6= 0, and derive a contradiction.

Let

(

)

}

Then

a ∈ A

. If

x ∈ A

, then

(

)

0, and by

continuity, we can find

δ >

0 such that

|y −x| < δ ⇒ f

(

)

0. So if

x 6

, then

we can find y ∈ A such that y > x.

We prove the contrapositive of the last condition, i.e.

x 6∈ A ⇒ (∃δ > 0)(∀y ∈ A) y 6∈ (x − δ, x].

x 6∈ A

, then

(

)

0 (we assume that

is never zero. If not, we’re done).

Then by continuity, ∃δ > 0 such that |y − x| < δ ⇒ f(y) > 0. So y 6∈ A.

Hence by continuous induction, b ∈ A. Contradiction.

Now we prove that continuous functions in closed intervals are bounded.

Theorem. Let [

a, b

] be a closed interval in

and let

: [

a, b

]

→ R

be continuous.

Then f is bounded.

Proof.

Let

: [

a, b

] be continuous. Let

f is bounded on

[

a, x

]

}

. Then

a ∈ A

. If

x ∈ A, x 6

, then

∃δ >

0 such that

|y − x| < δ ⇒ |f

(

)

− f

(

)

| <

∃y > x

(e.g. take

min{x

δ/

, b}

) such that

is bounded on [

a, y

], which

implies that y ∈ A.

Now suppose that

∀ε >

∃y ∈

(

x, −ε, x

] such that

y ∈ A

. Again, we can

find

δ >

0 such that

is bounded on (

x −δ, x

), and in particular on (

x −δ, x

Pick

such that

is bounded on [

a, y

] and

y > x − δ

. Then

is bounded on

[a, x]. So x ∈ A.

So we are done by continuous induction.

Finally, we can prove a theorem that we have not yet proven.

Definition (Cover of a set). Let

A ⊆ R

. A cover of

by open intervals is a set

: γ ∈ Γ} where each I

is an open interval and A ⊆

γ∈Γ

A finite subcover is a finite subset

, ··· , I

}

of the cover that is still a

cover.

Not every cover has a finite subcover. For example, the cover

{

(

1) :

n ∈ N}

of (0, 1) has no finite subcover.

Theorem (Heine-Borel*). Every cover of a closed, bounded interval [

a, b

] by

open intervals has a finite subcover. We say closed intervals are compact (cf.

Metric and Topological Spaces).

Proof.

Let

γ ∈

}

be a cover of [

a, b

] by open intervals. Let

: [

a, x

]

can be covered by finitely many of the I

Then a ∈ A since a must belong to some I

x ∈ A

, then pick

such that

x ∈ I

. Then if

x 6

, since

is an open

interval, it contains [

x, y

] for some

y > x

. Then [

a, y

] can be covered by finitely

many I

, by taking a finite cover for [a, x] and the I

that contains x.

Now suppose that ∀ε > 0, ∃y ∈ A such that y ∈ (x − ε, x].

Let

be an open interval containing

. Then it contains (

x − ε, x

] for some

ε >

0. Pick

y ∈ A

such that

y ∈

(

x − ε, x

]. Now combine

with a finite

subcover of [a, y] to get a finite subcover of [a, x]. So x ∈ A.

Then done by continuous induction.

We can use Heine-Borel to prove that continuous functions on [

a, b

] are

bounded.

Theorem. Let [

a, b

] be a closed interval in

and let

: [

a, b

]

→ R

be continuous.

Then

is bounded and attains it bounds, i.e.

(

) =

sup f

for some

, and

f(y) = inf f for some y.

Proof. Let f : [a, b] → R be continuous. Then by continuity,

(∀x ∈ [a, b])(∃δ

> 0)(∀y) |y − x| < δ

⇒ |f (y) − f (x)| < 1.

Let

= [

a, b

] and for each

x ∈ γ

, let

= (

x − δ

, x

). So by Heine-Borel,

we can find x

, ··· , x

such that [a, b] ⊆

− δ

, x

+ δ

But

is bounded in each interval (

− δ

, x

) by

(

)

+ 1. So it is

bounded on [a, b] by max |f(x

)| + 1.

5 Differentiability

In the remainder of the course, we will properly develop calculus, and put

differentiation and integration on a rigorous foundation. Every notion will be

given a proper definition which we will use to prove results like the product and

quotient rule.

5.1 Limits

First of all, we need the notion of limits. Recall that we’ve previously had limits

for sequences. Now, we will define limits for functions.

Definition (Limit of functions). Let A ⊆ R and let f : A → R. We say

lim

x→a

f(x) = `,

or “f (x) → ` as x → a”, if

(∀ε > 0)(∃δ > 0)(∀x ∈ A) 0 < |x −a| < δ ⇒ |f(x) − `| < ε.

We couldn’t care less what happens when

, hence the strict inequality

0 < |x −a|. In fact, f doesn’t even have to be defined at x = a.

Example. Let

f(x) =

(

x x 6= 2

3 x = 2

Then lim

x→2

= 2, even though f(2) = 3.

Example. Let f(x) =

sin x

. Then f(0) is not defined but lim

x→0

f(x) = 1.

We will see a proof later after we define what sin means.

We notice that the definition of the limit is suspiciously similar to that of

continuity. In fact, if we define

g(x) =

(

f(x) x 6= a

` x = a

Then f (x) → ` as x → a iff g is continuous at a.

Alternatively,

is continuous at

(

)

→ f

(

) as

x → a

. It follows also

that

(

)

→ `

x → a

iff

(

)

→ `

for every sequence (

) in

with

→ a

The previous limit theorems of sequences apply here as well

Proposition. If f(x) → ` and g(x) → m as x → a, then f(x) + g(x) → ` + m,

f(x)g(x) → `m, and

f(x)

g(x)

→

if g and m don’t vanish.

5.2 Differentiation

Similar to what we did in IA Differential Equations, we define the derivative as

a limit.

Definition (Differentiable function).

is differentiable at

with derivative

lim

x→a

f(x) −f(a)

x − a

= λ.

Equivalently, if

lim

h→0

f(a + h) − f (a)

= λ.

We write λ = f

(a).

Here we see why, in the definition of the limit, we say that we don’t care

what happens when

. In our definition here, our function is 0/0 when

x = a, and we can’t make any sense out of what happens when x = a.

Alternatively, we write the definition of differentiation as

f(x + h) − f (x)

= f

(x) + ε(h),

where ε(h) → 0 as h → 0. Rearranging, we can deduce that

f(x + h) = f(x) + hf

(x) + hε(h),

Note that by the definition of the limit, we don’t have to care what value

takes

when

= 0. It can be 0,

or 10

. However, we usually take

(0) = 0 so that

ε is continuous.

Using the small-

notation, we usually write

(

) for a function that satisfies

o(h)

→ 0 as h → 0. Hence we have

Proposition.

f(x + h) = f(x) + hf

(x) + o(h).

We can interpret this as an approximation of f (x + h):

f(x + h) = f(x) + hf

(x)

| {z }

linear approximation

+ o(h)

|{z}

error term

And differentiability shows that this is a very good approximation with small

o(h) error.

Conversely, we have

Proposition. If

(

) =

(

) +

(

) +

(

), then

is differentiable at

with derivative f

(x).

Proof.

f(x + h) − f (x)

= f

(x) +

o(h)

→ f

(x).

We can take derivatives multiple times, and get multiple derivatives.

Definition (Multiple derivatives). This is defined recursively:

is (

+ 1)-

times differentiable if it is

-times differentiable and its

th derivative

(n)

differentiable. We write

(n+1)

for the derivative of

(n)

, i.e. the (

+ 1)th

derivative of f.

Informally, we will say

-times differentiable if we can differentiate it

times, and the nth derivative is f

(n)

We can prove the usual rules of differentiation using the small

-notation. It

can also be proven by considering limits directly, but the notation will become a

bit more daunting.

Lemma (Sum and product rule). Let

f, g

be differentiable at

. Then

and f g are differentiable at x, with

(f + g)

(x) = f

(x) + g

(x)

(fg)

(x) = f

(x)g(x) + f (x)g

(x)

Proof.

(f + g)(x + h) = f(x + h) + g(x + h)

= f(x) + hf

(x) + o(h) + g(x) + hg

(x) + o(h)

= (f + g)(x) + h(f

(x) + g

(x)) + o(h)

fg(x + h) = f(x + h)g(x + h)

= [f(x) + hf

(x) + o(h)][g(x) + hg

(x) + o(h)]

= f(x)g(x) + h[f

(x)g(x) + f (x)g

(x)]

+ o(h)[g(x) + f(x) + hf

(x) + hg

(x) + o(h)] + h

(x)g

(x)

| {z }

error term

By limit theorems, the error term is o(h). So we can write this as

= fg(x) + h(f

(x)g(x) + f (x)g

(x)) + o(h).

Lemma (Chain rule). If

is differentiable at

and

is differentiable at

(

then g ◦ f is differentiable at x with derivative g

(f(x))f

(x).

Proof.

If one is sufficiently familiar with the small-

notation, then we can

proceed as

g(f(x + h)) = g(f(x) + hf

(x) + o(h)) = g(f (x)) + hf

(x)g

(f(x)) + o(h).

If not, we can be a bit more explicit about the computations, and use

hε

(

)

instead of o(h):

(g ◦ f)(x + h) = g(f(x + h))

= g[f(x) + hf

(x) + hε

(h)

| {z }

the “h” term

]

= g(f(x)) +



(x) + hε

(h)



(f(x))



(x) + hε

(h)



(hf

(x) + hε

(h))

= g ◦ f(x) + hg

(f(x))f

(x)

+ h

(h)g

(f(x)) +



(x) + ε

(h)





(x) + hε

(h)



{z }

error term

We want to show that the error term is

(

), i.e. it divided by

tends to 0 as

h → 0.

But

(

)

(

))

→

(

) is bounded, and

(

hε

(

))

→

because hf

(x) + hε

(h) → 0 and ε

(0) = 0. So our error term is o(h).

We usually don’t write out the error terms so explicitly, and just use heuristics

(

)) =

(

) +

(

);

(

) +

(

) =

(

); and

(

)

·o

(

) =

(

) for any

(bounded) function g.

Example.

(i) Constant functions are differentiable with derivative 0.

(ii) f(x) = λx is differentiable with derivative λ.

(iii)

Using the product rule, we can show that

is differentiable with derivative

n−1

by induction.

(iv) Hence all polynomials are differentiable.

Example. Let f(x) = 1/x. If x 6= 0, then

f(x + h) − f (x)

x+h

−



−h

x(x+h)



−1

x(x + h)

→

−1

by limit theorems.

Lemma (Quotient rule). If

and

are differentiable at

, and

(

)

= 0, then

f/g is differentiable at x with derivative





(x) =

(x)g(x) − g

(x)f(x)

g(x)

Proof.

First note that 1

(

) =

(

)) where

(

) = 1

. So 1

(

) is differ-

entiable at x with derivative

−1

g(x)

(x) by the chain rule.

By the product rule, f/g is differentiable at x with derivative

(x)

g(x)

− f(x)

(x)

g(x)

(x)g(x) − f (x)g

(x)

g(x)

Lemma. If f is differentiable at x, then it is continuous at x.

Proof.

y → x

f(y) −f(x)

y − x

→ f

(

). Since,

y − x →

(

)

− f

(

)

→

0 by

product theorem of limits. So f(y) → f(x). So f is continuous at x.

Theorem. Let

: [

a, b

]

→

[

c, d

] be differentiable on (

a, b

), continuous on [

a, b

and strictly increasing. Suppose that

(

) never vanishes. Suppose further that

(

) =

and

(

) =

. Then

has an inverse

and for each

y ∈

(

c, d

differentiable at y with derivative 1/f

(g(y)).

In human language, this states that if

is invertible, then the derivative of

−1

is 1/f

Note that the conditions will (almost) always require

to be differentiable

on open interval (

a, b

), continuous on closed interval [

a, b

]. This is because it

doesn’t make sense to talk about differentiability at

since the definition of

(a) requires f to be defined on both sides of a.

Proof. g exists by an earlier theorem about inverses of continuous functions.

Let y, y + k ∈ (c, d). Let x = g(y), x + h = g(y + k).

Since

(

) =

, we have

(

). So

(

)

− y

f(x + h) − f (x). So

g(y + k) −g(y)

(x + h) −x

f(x + h) − f (x)



f(x + h) − f (x)



−1

As k → 0, since g is continuous, g(y + k) → g(y). So h → 0. So

g(y + k) −g(y)

→ [f

(x)]

−1

= [f

(g(y)]

−1

Example. Let f(x) = x

1/2

for x > 0. Then f is the inverse of g(x) = x

. So

(x) =

(f(x))

1/2

−1/2

Similarly, we can show that the derivative of x

1/q

1/q−1

Then let’s take x

p/q

= (x

1/q

)

. By the chain rule, its derivative is

p(x

1/q

)

p−1

1/q−1

p−1

−1

5.3 Differentiation theorems

Everything we’ve had so far is something we already know. It’s just that now we

can prove them rigorously. In this section, we will come up with genuinely new

theorems, including but not limited to Taylor’s theorem, which gives us Taylor’s

series.

Theorem (Rolle’s theorem). Let

be continuous on a closed interval [

a, b

] (with

a < b

) and differentiable on (

a, b

). Suppose that

(

) =

(

). Then there exists

x ∈ (a, b) such that f

(x) = 0.

It is intuitively obvious: if you move up and down, and finally return to the

same point, then you must have changed direction some time. Then

(

) = 0

at that time.

Proof. If f is constant, then we’re done.

Otherwise, there exists

such that

(

)

(

). wlog,

(

)

> f

(

). Since

is continuous, it has a maximum, and since

(

)

> f

(

) =

(

), the maximum

is not attained at a or b.

Suppose maximum is attained at x ∈ (a, b). Then for any h 6= 0, we have

f(x + h) − f (x)

(

≤ 0 h > 0

≥ 0 h < 0

since

(

)

−f

(

)

≤

0 by maximality of

(

). By considering both sides as we

take the limit h → 0, we know that f

(x) ≤ 0 and f

(x) ≥ 0. So f

(x) = 0.

Corollary (Mean value theorem). Let

be continuous on [

a, b

] (

a < b

), and

differentiable on (a, b). Then there exists x ∈ (a, b) such that

(x) =

f(b) −f(a)

b − a

Note that

f(b)−f (a)

b−a

is the slope of the line joining f(a) and f(b).

f(x)

f(a)

f(b)

The mean value theorem is sometimes described as “rotate your head and

apply Rolle’s”. However, if we actually rotate it, we might end up with a

non-function. What we actually want is a shear.

Proof. Let

g(x) = f(x) −

f(b) −f(a)

b − a

Then

g(b) − g(a) = f(b) − f(a) −

f(b) −f(a)

b − a

(b − a) = 0.

So by Rolle’s theorem, we can find x ∈ (a, b) such that g

(x) = 0. So

(x) =

f(b) −f(a)

b − a

as required.

We’ve always assumed that if a function has a positive derivative everywhere,

then the function is increasing. However, it turns out that this is really hard to

prove directly. It does, however, follow quite immediately from the mean value

theorem.

Example. Suppose

(

)

0 for every

x ∈

(

a, b

). Then for

u, v

in [

a, b

], we can

find w ∈ (u, v) such that

f(v) −f(u)

v − u

= f

(w) > 0.

It follows that f(v) > f(u). So f is strictly increasing.

Similarly, if

(

)

≥

2 for every

and

(0) = 0, then

(1)

≥

2, or else we

can find x ∈ (0, 1) such that

2 ≤ f

(x) =

f(1) −f(0)

1 − 0

= f(1).

Theorem (Local version of inverse function theorem). Let

be a function with

continuous derivative on (a, b).

Let

x ∈

(

a, b

) and suppose that

(

)

= 0. Then there is an open interval

(

u, v

) containing

on which

is invertible (as a function from (

u, v

) to

((

u, v

))).

Moreover, if g is the inverse, then g

(f(z)) =

(z)

for every z ∈ (u, v).

This says that if

has a non-zero derivative, then it has an inverse locally

and the derivative of the inverse is 1/f

Note that this not only requires

to be differentiable, but the derivative

itself also has to be continuous.

Proof.

wlog,

(

)

0. By the continuity, of

, we can find

δ >

0 such that

(

)

0 for every

z ∈

(

x − δ, x

). By the mean value theorem,

is strictly

increasing on (

x −δ, x

), hence injective. Also,

is continuous on (

x −δ, x

)

by differentiability.

Then done by the inverse function theorem.

Finally, we are going to prove Taylor’s theorem. To do so, we will first need

some lemmas.

Theorem (Higher-order Rolle’s theorem). Let

be continuous on [

a, b

] (

a < b

)

and n-times differentiable on an open interval containing [a, b]. Suppose that

f(a) = f

(a) = f

(2)

(a) = ··· = f

(n−1)

(a) = f(b) = 0.

Then ∃x ∈ (a, b) such that f

(n)

(x) = 0.

Proof. Induct on n. The n = 0 base case is just Rolle’s theorem.

Suppose we have

k < n

and

∈

(

a, b

) such that

(k)

(

) = 0. Since

(k)

(

) = 0, we can find

k+1

∈

(

a, x

) such that

(k+1)

(

k+1

) = 0 by Rolle’s

theorem.

So the result follows by induction.

Corollary. Suppose that

and

are both differentiable on an open interval

containing [

a, b

] and that

(k)

(

) =

(k)

(

) for

= 0

, ··· , n −

1, and also

f(b) = g(b). Then there exists x ∈ (a, b) such that f

(n)

(x) = g

(n)

(x).

Proof. Apply generalised Rolle’s to f − g.

Now we shall show that for any

, we can find a polynomial

of degree at

most

that satisfies the conditions for

, i.e. a

such that

(k)

(

) =

(k)

(

) for

k = 0, 1, ··· , n −1 and p(b) = f(b).

A useful ingredient is the observation that if

(x) =

(x − a)

then

(j)

(a) =

(

1 j = k

0 j 6= k

Therefore, if

Q(x) =

n−1

k=0

(k)

(a)Q

(x),

then

(j)

(a) = f

(j)

(a)

for

= 0

, ··· , n −

1. To get

(

) =

(

), we use our

th degree polynomial

term:

p(x) = Q(x) +

(x − a)

(b − a)



f(b) −Q(b)



Then our final term does not mess up our first

n −

1 derivatives, and gives

p(b) = f(b).

By the previous corollary, we can find x ∈ (a, b) such that

(n)

(x) = p

(n)

(x).

That is,

(n)

(x) =

(b − a)



f(b) −Q(b)



Therefore

f(b) = Q(b) +

(b − a)

(n)

(x).

Alternatively,

f(b) = f(a) + (b − a)f

(a) + ··· +

(b − a)

n−1

(n − 1)!

(n−1)

(a) +

(b − a)

(n)

(x).

Setting b = a + h, we can rewrite this as

Theorem (Taylor’s theorem with the Lagrange form of remainder).

f(a + h) = f(a) + hf

(a) + ··· +

n−1

(n − 1)!

(n−1)

(a)

| {z }

(n−1)-degree approximation to f near a

(n)

(x)

| {z }

error term

for some x ∈ (a, a + h).

Strictly speaking, we only proved it for the case when

h >

0, but we can

easily show it holds for h < 0 too by considering g(x) = f(−x).

Note that the remainder term is not necessarily small, but this often gives

us the best (

n −

1)-degree approximation to

near

. For example, if

(n)

bounded by C near a, then



(n)

(x)



≤

|h|

= o(h

n−1

Example. Let

R → R

be a differentiable function such that

(0) = 1 and

(

) =

(

) for every

(intuitively, we know it is

, but that thing doesn’t

exist!). Then for every x, we have

f(x) = 1 + x +

+ ··· =

∞

n=0

While it seems like we can prove this works by differentiating it and see that

(

) =

(

), the sum rule only applies for finite sums. We don’t know we can

differentiate a sum term by term. So we have to use Taylor’s theorem.

Since

(

) =

(

), it follows that all derivatives exist. By Taylor’s theorem,

f(x) = f(0) + f

(0)x +

(2)

(0)

+ ··· +

(n−1)

(0)

(n − 1)!

n−1

(n)

(u)

for some u between 0 and x. This equals to

f(x) =

n−1

k=0

(n)

(u)

We must show that the remainder term

(n)

(u)

→

0 as

n → ∞

. Note here

that x is fixed, but u can depend on n.

But we know that

(n)

(

) =

(

), but since

is differentiable, it is continuous,

and is bounded on [0, x]. Suppose |f(u)| ≤ C on [0, x]. Then



(n)(u)



≤

|x|

→ 0

from limit theorems. So it follows that

f(x) = 1 + x +

+ ··· =

∞

n=0

5.4 Complex differentiation

Definition (Complex differentiability). Let

C → C

. Then

is differentiable

at z with derivative f

(z) if

lim

h→0

f(z + h) − f(z)

exists and equals f

(z).

Equivalently,

f(z + h) = f(z) + hf

(z) + o(h).

This is exactly the same definition with real differentiation, but has very

different properties!

All the usual rules — chain rule, product rule etc. also apply (with the same

proofs). Also the derivatives of polynomials are what you expect. However, there

are some more interesting cases.

Example. f(z) = ¯z is not differentiable.

z + h −z

(

1 h is real

−1 h is purely imaginary

If this seems weird, this is because we often think of

, but they are

not the same. For example, reflection is a linear map in

, but not in

. A

linear map in

is something in the form

x 7→ bx

, which can only be a dilation

or rotation, not reflections or other weird things.

Example.

(

) =

|z|

is also not differentiable. If it were, then

|z|

would be as

well (by the product rule). So would

|z|

¯z

when

z 6

= 0 by the quotient rule.

= 0, it is certainly not differentiable, since it is not even differentiable on

6 Complex power series

Before we move on to integration, we first have a look at complex power series.

This will allow us to define the familiar exponential and trigonometric functions.

Definition (Complex power series). A complex power series is a series of the

form

∞

n=0

when z ∈ C and a

∈ C for all n. When it converges, it is a function of z.

When considering complex power series, a very important concept is the

radius of convergence. To make sense of this concept, we first need the following

lemma:

Lemma. Suppose that

converges and

|w| < |z|

, then

converges

(absolutely).

Proof. We know that

| = |a

| ·



Since

converges, the terms a

are bounded. So pick C such that

| ≤ C

for every n. Then

0 ≤

∞

n=0

| ≤

∞

n=0



which converges (geometric series). So by the comparison test,

converges

absolutely.

It follows that if

does not converge and

|w| > |z|

, then

does

not converge.

Now let

sup{|z|

converges

}

(

may be infinite). If

|z| < R

then we can find

with

| ∈

(

|z|, R

] such that

∞

converges. So by

lemma above,

converges. If

|z| > R

, then

diverges by definition

of R.

Definition (Radius of convergence). The radius of convergence of a power series

R = sup

|z| :

converges

{z : |z| < R} is called the circle of convergence.

|z| < R

, then

converges. If

|z| > R

, then

diverges. When

|z| = R, the series can converge at some points and not the others.

Example.

∞

n=0

has radius of convergence of 1. When

|z|

= 1, it diverges

(since the terms do not tend to 0).

Example.

∞

n=0

has radius of convergence 1, since the ratio of (

+ 1)th term

to nth is

n+1

/(n + 1)

= z ·

n + 1

→ z.

So if

|z| <

1, then the series converges by the ratio test. If

|z| >

1, then eventually

the terms are increasing in modulus.

= 1, then it diverges (harmonic series). If

|z|

= 1 and

z 6

= 1, it converges

by Abel’s test.

Example. The series

∞

n=1

converges for |z| ≤ 1 and diverges for |z| > 1.

As evidenced by the above examples, the ratio test can be used to find the

radius of convergence. We also have an alternative test based on the nth root.

Lemma. The radius of convergence of a power series

R =

lim sup

Often

| converges, so we only have to find the limit.

Proof.

Suppose

|z| <

/ lim sup

. Then

|z|lim sup

| <

1. Therefore

there exists N and ε > 0 such that

sup

n≥N

|z|

| ≤ 1 −ε

by the definition of lim sup. Therefore

| ≤ (1 −ε)

for every

n ≥ N

, which implies (by comparison with geometric series) that

converges absolutely.

On the other hand, if

|z|lim sup

| >

1, it follows that

|z|

| ≥

1 for

infinitely many

. Therefore

| ≥

1 for infinitely many

. So

does

not converge.

Example. The radius of convergence of

is 2 because

for every

So lim sup

| =

. So 1/ lim sup

| = 2.

But often it is easier to find the radius convergence from elementary methods

such as the ratio test, e.g. for

Note to pedants: yes it is a disc, not a circle

6.1 Exponential and trigonometric functions

Definition (Exponential function). The exponential function is

∞

n=0

By the ratio test, this converges on all of C.

A fundamental property of this function is that

z+w

= e

Once we have this property, we can say that

Proposition. The derivative of e

is e

Proof.

z+h

− e

= e



− 1



= e



1 +

+ ···



But



+ ···



≤

|h|

+ ··· =

|h|/2

1 − |h|/2

→ 0.

z+h

− e

→ e

But we must still prove that e

z+w

= e

Consider two sequences (

)

(

). Their convolution is the sequence (

)

defined by

= a

+ a

n−1

+ a

n−2

+ ··· + a

The relevance of this is that if you take

n=0

and

n=0

and equate coefficients of z

, you get

= a

+ a

n−1

+ a

n−2

+ ··· + a

Theorem. Let

∞

n=0

and

∞

n=0

be two absolutely convergent series, and

let (

) be the convolution of the sequences (

) and (

). Then

∞

n=0

converges (absolutely), and

∞

n=0

∞

n=0

∞

n=0

Proof.

We first show that a rearrangement of

converges absolutely. Hence

it converges unconditionally, and we can rearrange it back to

Consider the series

) + (a

+ a

) + (a

+ a

) + ··· (∗)

Let

n=0

, T

n=0

, U

n=0

|, V

n=0

Also let

→ S, T

→ T, U

→ U, V

→ V

(these exist since

and

converge absolutely).

If we take the modulus of the terms of (

∗

), and consider the first (

+ 1)

terms (i.e. the first

+1 brackets), the sum is

. Hence the series converges

absolutely to UV . Hence (∗) converges.

The partial sum up to (

+ 1)

of the series (

∗

) itself is

, which

converges to ST . So the whole series converges to ST .

Since it converges absolutely, it converges unconditionally. Now consider a

rearrangement:

+ (a

+ a

) + (a

+ a

) + ···

Then this converges to

as well. But the partial sum of the first 1 + 2+

···

terms is c

+ c

+ ··· + c

. So

n=0

→ ST =

∞

n=0

∞

n=0

Corollary.

= e

z+w

Proof. By theorem above (and definition of e

∞

n=0



1 ·

n−1

(n − 1)!

n−2

(n − 2)!

+ ··· +

· 1



∞

n=0





n−1





n−2

+ ··· +





∞

n=0

(z + w)

by the binomial theorem

= e

z+w

Note that if (

) is the convolution of (

) and (

), then the convolution

of (

) and (

) is (

). Therefore if both

and

converge

absolutely, then their product is

Note that we have now completed the proof that the derivative of e

is e

Now we define sin z and cos z:

Definition (Sine and cosine).

sin z =

− e

−iz

= z −

−

+ ···

cos z =

+ e

−iz

= 1 −

−

+ ··· .

We now prove certain basic properties of

sin

and

cos

, using known properties

of e

Proposition.

sin z =

+ ie

−iz

= cos z

cos z =

− ie

−iz

= −sin z

sin

z + cos

z =

2iz

+ 2 + e

−2iz

2iz

− 2 + e

−2iz

−4

= 1.

It follows that if x is real, then |cos x| and |sin x| are at most 1.

Proposition.

cos(z + w) = cos z cos w − sin z sin w

sin(z + w) = sin z cos w + cos z sin w

Proof.

cos z cos w − sin z sin w =

+ e

−iz

)(e

+ e

−iw

)

− e

−iz

)(e

− e

−iw

)

i(z+w)

+ e

−i(z+w)

= cos(z + w).

Differentiating both sides wrt z gives

−sin z cos w −cos z sin w = −sin(z + w).

sin(z + w) = sin z cos w + cos z sin w.

When

is real, we know that

cos x ≤

1. Also

sin

0 = 0, and

sin x

cos x ≤

1. So for

x ≥

sin x ≤ x

, “by the mean value theorem”. Also,

cos

0 = 1,

and

cos x

−sin x

, which, for

x ≥

0, is greater than

−x

. From this, it follows

that when

x ≥

cos x ≥

−

(the 1

−

comes from “integrating”

−x

, (or

finding a thing whose derivative is −x)).

Continuing in this way, we get that for

x ≥

0, if you take truncate the power

series for

sin x

cos x

, it will be

≥ sin x, cos x

if you stop at a positive term,

and ≤ if you stop at a negative term. For example,

sin x ≥ x −

−

11!

In particular,

cos 2 ≤ 1 −

= 1 − 2 +

< 0.

Since

cos

0 = 1, it follows by the intermediate value theorem that there exists

some

x ∈

2) such that

cos x

= 0. Since

cos x ≥

−

, we can further deduce

that x > 1.

Definition (Pi). Define the smallest x such that cos x = 0 to be

Since

sin

cos

= 1, it follows that

sin

1. Since

cos x >

0 on [0

sin

≥ 0 by the mean value theorem. So sin

= 1.

Thus

Proposition.

cos



z +



= −sin z

sin



z +



= cos z

cos(z + π) = −cos z

sin(z + π) = −sin z

cos(z + 2π) = cos z

sin(z + 2π) = sin z

Proof.

cos



z +



= cos z cos

− sin z sin

= −sin z sin

= −sin z

and similarly for others.

6.2 Differentiating power series

We shall show that inside the circle of convergence, the derivative of

∞

n=0

given by the obvious formula

∞

n=1

n−1

We first prove some (seemingly arbitrary and random) lemmas to build up

the proof of the above statement. They are done so that the final proof will not

be full of tedious algebra.

Lemma. Let a and b be complex numbers. Then

−a

−n(b −a)a

n−1

= (b −a)

n−2

+ 2ab

n−3

+ 3a

n−4

+ ···+ (n −1)a

n−2

Proof. If b = a, we are done. Otherwise,

− a

b − a

= b

n−1

+ ab

n−2

+ a

n−3

+ ··· + a

n−1

Differentiate both sides with respect to a. Then

−na

n−1

(b − a) + b

− a

(b − a)

= b

n−2

+ 2ab

n−3

+ ··· + (n − 1)a

n−2

Rearranging gives the result.

Alternatively, we can do

− a

= (b − a)(b

n−1

+ ab

n−2

+ ··· + a

n−1

Subtract n(b −a)a

n−1

to obtain

(b − a)[b

n−1

− a

n−1

+ a(b

n−2

− a

n−2

) + a

n−3

− a

n−3

) + ···]

and simplify.

This implies that

(z + h)

− z

− nhz

n−1

= h

((z + h)

n−2

+ 2z(z + h)

n−3

+ ··· + (n − 1)z

n−2

which is actually the form we need.

Lemma. Let

have radius of convergence

, and let

|z| < R

. Then

n−1

converges (absolutely).

Proof.

Pick

such that

|z| < r < R

. Then

converges, so the terms

are bounded above by, say, C. Now

n|a

n−1

| =

n|a

n−1



|z|



n−1

≤



|z|



n−1

The series



|z|



n−1

converges, by the ratio test. So

n|a

n−1

converges,

by the comparison test.

Corollary. Under the same conditions,

∞

n=2





n−2

converges absolutely.

Proof. Apply Lemma above again and divide by 2.

Theorem. Let

be a power series with radius of convergence

. For

|z| < R, let

f(z) =

∞

n=0

and g(z) =

∞

n=1

n−1

Then f is differentiable with derivative g.

Proof. We want f(z + h) −f(z) −hg(z) to be o(h). We have

f(z + h) − f(z) −hg(z) =

∞

n=2

((z + h)

− z

− hnz

We started summing from

= 2 since the

= 0 and

= 1 terms are 0. Using

our first lemma, we are left with

∞

n=2



(z + h)

n−2

+ 2z(z + h)

n−3

+ ··· + (n − 1)z

n−2



We want the huge infinite series to be bounded, and then the whole thing is a

bounded thing times h

, which is definitely o(h).

Pick

such that

|z| < r < R

. If

is small enough that

h| ≤ r

, then the

last infinite series is bounded above (in modulus) by

∞

n=2

|(r

n−2

+ 2r

n−2

+ ··· + (n − 1)r

n−2

) =

∞

n=2





n−2

which is bounded. So done.

In IB Analysis II, we will prove the same result using the idea of uniform

convergence, which gives a much nicer proof.

Example. The derivative of

= 1 + z +

+ ···

1 + z +

+ ··· = e

So we have another proof that of this fact.

Similarly, the derivatives of sin z and cos z work out as cos z and −sin z.

6.3 Hyperbolic trigonometric functions

Definition (Hyperbolic sine and cosine). We define

cosh z =

+ e

−z

= 1 +

+ ···

sinh z =

− e

−z

= z +

+ ···

Either from the definition or from differentiating the power series, we get

that

Proposition.

cosh z = sinh z

sinh z = cosh z

Also, by definition, we have

Proposition.

cosh iz = cos z

sinh iz = i sin z

Also,

Proposition.

cosh

z − sinh

z = 1,

7 The Riemann Integral

Finally, we can get to integrals. There are many ways to define an integral,

which can have some subtle differences. The definition we will use here is the

Riemann integral, which is the simplest definition, but is also the weakest one, in

the sense that many functions are not Riemann integrable but integrable under

other definitions.

Still, the definition of the Riemann integral is not too straightforward, and

requires a lot of preliminary definitions.

7.1 Riemann Integral

Definition (Dissections). Let [

a, b

] be a closed interval. A dissection of [

a, b

] is

a sequence a = x

< x

< ··· < x

= b.

Definition (Upper and lower sums). Given a dissection

, the upper sum and

lower sum are defined by the formulae

(f) =

i=1

− x

i−1

) sup

x∈[x

i−1

]

f(x)

(f) =

i=1

− x

i−1

) inf

x∈[x

i−1

]

f(x)

Sometimes we use the shorthand

= sup

x∈[x

i−1

]

f(x), m

= inf

x∈[x

i−1

−x

]

f(x).

The upper sum is the total area of the red rectangles, while the lower sum is

the total area of the black rectangles:

a x

···

i+1

···

Definition (Refining dissections). If

and

are dissections of [

a, b

], we say

that D

refines D

if every point of D

is a point of D

Lemma. If D

refines D

, then

f ≤ U

f and L

≥ L

Using the picture above, this is because if we cut up the dissections into

smaller pieces, the red rectangles can only get chopped into shorter pieces and

the black rectangles can only get chopped into taller pieces.

Proof.

Let

< x

< ··· < x

. Let

be obtained from

by the

addition of one point z. If z ∈ (x

i−1

, x

), then

f − U

f =

(z − x

i−1

) sup

x∈[x

i−1

,z]

f(x)

− z) sup

x∈[z,x

]

f(x)

− (x

− x

i−1

But

sup

x∈[x

i−1

,z]

(

) and

sup

x∈[z,x

]

(

) are both at most

. So this is at

most M

(z − x

i−1

+ x

− z − (x

− x

i−1

)) = 0. So

f ≤ U

By induction, the result is true whenever D

refines D

A very similar argument shows that L

≥ L

Definition (Least common refinement). If

and

be dissections of [

a, b

Then the least common refinement of

and

is the dissection made out of

the points of D

and D

Corollary. Let D

and D

be two dissections of [a, b]. Then

f ≥ L

Proof.

Let

be the least common refinement (or indeed any common refinement).

Then by lemma above (and by definition),

f ≥ U

f ≥ L

Finally, we can define the integral.

Definition (Upper, lower, and Riemann integral). The upper integral is

f(x) dx = inf

The lower integral is

f(x) dx = sup

If these are equal, then we call their common value the Riemann integral of

and is denoted

f(x) dx.

If this exists, we say f is Riemann integrable.

We will later prove the fundamental theorem of calculus, which says that

integration is the reverse of differentiation. But why don’t we simply define

integration as anti-differentiation, and prove that it is the area of the curve?

There are things that we cannot find (a closed form of) the anti-derivative of,

−x

. In these cases, we wouldn’t want to say the integral doesn’t exist — it

surely does according to this definition!

There is an immediate necessary condition for Riemann integrability — bound-

edness. If

is unbounded above in [

a, b

], then for any dissection

, there must

be some

such that

is unbounded on [

i−1

, x

]. So

∞

. So

∞

Similarly, if

is unbounded below, then

−∞

. So unbounded functions

are not Riemann integrable.

Example. Let

(

) =

on [

a, b

]. Intuitively, we know that the integral is

(

− a

)

2, and we will show this using the definition above. Let

< ··· < x

be a dissection. Then

f =

i=1

− x

i−1

We know that the integral is

−a

. So we put each term of the sum into the

form

−x

i−1

plus some error terms.

i=1



−

i−1

− x

i−1



i=1

− x

i−1

+ (x

− x

i−1

)

− a

) +

i=1

− x

i−1

)

Definition (Mesh). The mesh of a dissection D is max

i+1

− x

Then if the mesh is < δ, then

i=1

− x

i−1

)

≤

i=1

− x

i−1

) =

(b − a).

So by making δ small enough, we can show that for any ε > 0,

x dx <

− a

) + ε.

Similarly,

x dx >

− a

) − ε.

x dx =

− a

Example. Define f : [0, 1] → R by

f(x) =

(

1 x ∈ Q

0 x 6∈ Q

Let

< x

< ··· < x

be a dissection. Then for every

, we have

= 0 (since

there is an irrational in every interval), and

= 1 (since there is a rational in

every interval). So

f =

i=1

− x

i−1

) =

i=1

− x

i−1

) = 1.

Similarly, L

f = 0. Since D was arbitrary, we have

f(x) dx = 1,

f(x) dx = 0.

So f is not Riemann integrable.

Of course, this function is not interesting at all. The whole point of its

existence is to show undergraduates that there are some functions that are not

integrable!

Note that it is important to say that the function is not Riemann integrable.

There are other notions for integration in which this function is integrable. For

example, this function is Lebesgue-integrable.

Using the definition to show integrability is often cumbersome. Most of

the time, we use the Riemann’s integrability criterion, which follows rather

immediately from the definition, but is much nicer to work with.

Proposition (Riemann’s integrability criterion). This is sometimes known as

Cauchy’s integrability criterion.

Let

: [

a, b

]

→ R

. Then

is Riemann integrable if and only if for every

ε > 0, there exists a dissection D such that

− L

< ε.

Proof.

(

⇒

) Suppose that

is integrable. Then (by definition of Riemann

integrability), there exist D

and D

such that

f(x) dx +

and

f(x) dx −

Let D be a common refinement of D

and D

. Then

f − L

f ≤ U

f − L

f < ε.

(⇐) Conversely, if there exists D such that

f − L

f < ε,

then

inf U

f − sup L

f < ε,

which is, by definition, that

f(x) dx −

f(x) dx < ε.

Since ε > 0 is arbitrary, this gives us that

f(x) dx =

f(x) dx.

So f is integrable.

The next big result we want to prove is that integration is linear, ie

(λf(x) + µg(x)) dx = λ

f(x) dx + µ

g(x) dx.

We do this step by step:

Proposition. Let

: [

a, b

]

→ R

be integrable, and

λ ≥

0. Then

λf

is integrable,

and

λf(x) dx = λ

f(x) dx.

Proof. Let D be a dissection of [a, b]. Since

sup

x∈[x

i−1

]

λf(x) = λ sup

x∈[x

i−1

]

f(x),

and similarly for inf, we have

(λf) = λU

(λf) = λL

So if we choose

such that

f −L

f < ε/λ

, then

(

λf

)

−L

(

λf

)

< ε

. So

the result follows from Riemann’s integrability criterion.

Proposition. Let f : [a, b] → R be integrable. Then −f is integrable, and

−f(x) dx = −

f(x) dx.

Proof. Let D be a dissection. Then

sup

x∈[x

i−1

]

−f(x) = − inf

x∈[x

i−1

]

f(x)

inf

x∈[x

i−1

]

−f(x) = − sup

x∈[x

i−1

]

f(x).

Therefore

(−f) =

i=1

− x

i−1

)(−m

) = −L

(f).

Similarly,

(−f) = −U

(−f) −L

(−f) = U

f − L

Hence if

is integrable, then

−f

is integrable by the Riemann integrability

criterion.

Proposition. Let

f, g

: [

a, b

]

→ R

be integrable. Then

is integrable, and

(f(x) + g(x)) dx =

f(x) dx +

g(x) dx.

Proof. Let D be a dissection. Then

(f + g) =

i=1

− x

i−1

) sup

x∈[x

i−1

]

(f(x) + g(x))

≤

i=1

− x

i−1

)

sup

u∈[x

i−1

]

f(u) + sup

v∈[x

i−1

]

g(v)

= U

f + U

Therefore,

(f(x) + g(x)) dx ≤

f(x) dx +

g(x) dx =

f(x) dx +

g(x) dx.

Similarly,

(f(x) + g(x)) dx ≥

f(x) dx +

g(x) dx.

So the upper and lower integrals are equal, and the result follows.

So we now have that

(λf(x) + µg(x)) dx = λ

f(x) dx + µ

g(x) dx.

We will prove more “obvious” results.

Proposition. Let

f, g

: [

a, b

]

→ R

be integrable, and suppose that

(

)

≤ g

(

)

for every x. Then

f(x) dx ≤

g(x) dx.

Proof. Follows immediately from the definition.

Proposition. Let f : [a, b] → R be integrable. Then |f| is integrable.

Proof. Note that we can write

sup

x∈[x

i−1

]

f(x) − inf

x∈[x

i−1

]

f(x) = sup

u,v∈[x

i−1

]

|f(u) −f(v)|.

Similarly,

sup

x∈[x

i−1

]

|f(x)| − inf

x∈[x

i−1

]

|f(x)| = sup

u,v∈[x

i−1

]

||f(u)| −|f(v)||.

For any pair of real numbers,

x, y

, we have that

||x| − |y|| ≤ |x − y|

by the

triangle inequality. Then for any interval u, v ∈ [x

i−1

, x

], we have

||f(u)| −|f(v)|| ≤ |f(u) −f(v)|.

Hence we have

sup

x∈[x

i−1

]

|f(x)| − inf

x∈[x

i−1

]

|f(x)| ≤ sup

x∈[x

i−1

]

f(x) − inf

x∈[x

i−1

]

f(x).

So for any dissection D, we have

(|f|) −L

(|f|) ≤ U

(f) −L

(f).

So the result follows from Riemann’s integrability criterion.

Combining these two propositions, we get that if

|f(x) −g(x)| ≤ C,

for every x ∈ [a, b], then



f(x) dx −

g(x) dx



≤ C(b − a).

Proposition (Additivity property). Let

: [

a, c

]

→ R

be integrable, and let

b ∈

(

a, c

). Then the restrictions of

to [

a, b

] and [

b, c

] are Riemann integrable,

and

f(x) dx +

f(x) dx =

f(x) dx

Similarly, if

is integrable on [

a, b

] and [

b, c

], then it is integrable on [

a, c

] and

the above equation also holds.

Proof.

Let

ε >

0, and let

< x

< ··· < x

be a dissection of

[a, c] such that

(f) ≤

f(x) dx + ε,

and

(f) ≥

f(x) dx −ε.

Let

be the dissection made of

plus the point

. Let

be the dissection of

[

a, b

] made of points of

from

, and

be the dissection of [

b, c

] made of

points of D

from b to c. Then

(f) + U

(f) = U

(f) ≤ U

(f),

and

(f) + L

(f) = L

(f) ≥ L

(f).

Since

(

)

− L

(

)

, and both

(

)

− L

(

) and

(

)

− L

(

)

are non-negative, we have

(

)

− L

(

) and

(

)

− L

(

) are less than

. Since

is arbitrary, it follows that the restrictions of

to [

a, b

] and [

b, c

] are

both Riemann integrable. Furthermore,

f(x) dx +

f(x) dx ≤ U

(f) + U

(f) = U

(f) ≤ U

(f)

≤

f(x) dx + ε.

Similarly,

f(x) dx +

f(x) dx ≥ L

(f) + L

(f) = L

(f) ≥ L

(f)

≥

f(x) dx −ε.

Since ε is arbitrary, it follows that

f(x) dx +

f(x) dx =

f(x) dx.

The other direction is left as an (easy) exercise.

Proposition. Let f, g : [a, b] → R be integrable. Then f g is integrable.

Proof.

Let

be such that

(

)

|, |g

(

)

| ≤ C

for every

x ∈

[

a, b

]. Write

and

for the

sup

and

inf

in [

i−1

, x

]. Now let

be a dissection, and for each

i, let u

and v

be two points in [x

i−1

, x

We will pretend that

and

are the minimum and maximum when we

write the proof, but we cannot assert that they are, since

need not have

maxima and minima. We will then note that since our results hold for arbitrary

and v

, it must hold when fg is at its supremum and infimum.

We find what we pretend is the difference between the upper and lower sum:



i=1



− x

i−1

)(f(v

)g(v

) − f (u

)g(u

)





i=1

− x

i−1

)



f(v

)(g(v

) − g(u

)) + (f (v

) − f (u

))g(u

)





≤

i=1



C(L

− `

) + (M

− m



= C(U

g − L

g + U

f − L

f).

Since u

and v

are arbitrary, it follows that

(fg) −L

(fg) ≤ C(U

f − L

f + U

g − L

g).

Since

is fixed, and we can get

f − L

and

g − L

arbitrary small

(since

and

are integrable), we can get

(

)

−L

(

) arbitrarily small. So

the result follows.

Theorem. Every continuous function

on a closed bounded interval [

a, b

] is

Riemann integrable.

Proof. wlog assume [a, b] = [0, 1].

Suppose the contrary. Let

be non-integrable. This means that there exists

some

such that for every dissection

− L

> ε

. In particular, for every

n, let D

be the dissection 0,

, ··· ,

Since

− L

> ε

, there exists some interval



k+1



in which

sup f −

inf f > ε

. Suppose the supremum and infimum are attained at

and

respectively. Then we have |x

− y

| <

and f (x

) − f (y

) > ε.

By Bolzano Weierstrass, (

) has a convergent subsequence, say (

). Say

→ x

. Since

− y

| <

→

0, we must have

→ x

. By continuity, we

must have

(

)

→ f

(

) and

(

)

→ f

(

), but

(

) and

(

) are always

apart by ε. Contradiction.

With this result, we know that a lot of things are integrable, e.g. e

−x

To prove this, we secretly used the property of uniform continuity:

Definition (Uniform continuity*). Let

A ⊆ R

and let

A → R

. Then

uniformly continuous if

(∀ε)(∃δ > 0)(∀x)(∀y) |x − y| < δ ⇒ |f(x) −f(y)| ≤ ε.

This is different from regular continuity. Regular continuity says that at any

point

, we can find a

that works for this point. Uniform continuity says that

we can find a δ that works for any x.

It is easy to show that a uniformly continuous function is integrable, since by

uniformly continuity, as long as the mesh of a dissection is sufficiently small, the

difference between the upper sum and the lower sum can be arbitrarily small by

uniform continuity. Thus to prove the above theorem, we just have to show that

continuous functions on a closed bounded interval are uniformly continuous.

Theorem (non-examinable). Let

a < b

and let

: [

a, b

]

→ R

be continuous.

Then f is uniformly continuous.

Proof. Suppose that f is not uniformly continuous. Then

(∃ε)(∀δ > 0)(∃x)(∃y) |x − y| < δ and |f(x) −f(y)| ≥ ε.

Therefore, we can find sequences (x

), (y

) such that for every n, we have

− y

| ≤

and |f (x

) − f (y

)| ≥ ε.

Then by Bolzano-Weierstrass theorem, we can find a subsequence (

) converg-

ing to some

. Since

−y

| ≤

→ x

as well. But

(

)

−f

(

)

| ≥ ε

for every

. So

(

) and

(

) cannot both converge to the same limit. So

is not continuous at x.

This proof is very similar to the proof that continuous functions are integrable.

In fact, the proof that continuous functions are integrable is just a fuse of this

proof and the (simple) proof that uniformly continuously functions are integrable.

Theorem. Let f : [a, b] → R be monotone. Then f is Riemann integrable.

Note that monotone functions need not be “nice”. It can even have in-

finitely many discontinuities. For example, if

: [0

→ R

maps

to the

1/(first non-zero digit in the binary expansion of x), with f(0) = 0.

Proof. let ε > 0. Let D be a dissection of mesh less than

f(b)−f (a)

. Then

f − L

f =

i=1

− x

i−1

)(f(x

) − f (x

i−1

))

≤

f(b) −f(a)

i=1

(f(x

) − f (x

i−1

))

= ε.

Pictorially, we see that the difference between the upper and lower sums is

total the area of the red rectangles.

To calculate the total area, we can stack the red areas together to get something

of width

f(b)−f (a)

and height f (b) − f(a). So the total area is just ε.

Lemma. Let

a < b

and let

be a bounded function from [

a, b

]

→ R

that is

continuous on (a, b). Then f is integrable.

An example where this would apply is

sin

. It gets nasty near

= 0, but

its “nastiness” is confined to

= 0 only. So as long as its nastiness is sufficiently

contained, it would still be integrable.

The idea of the proof is to integrate from a point

very near

up to a

point

n−1

very close to

. Since

is bounded, the regions [

a, x

] and [

n−1

, b

]

are small enough to not cause trouble.

Proof.

Let

ε >

0. Suppose that

(

)

| ≤ C

for every

x ∈

[

a, b

]. Let

and

pick

such that

− x

. Also choose

between

and

such that

b − z <

Then

is continuous [

, z

]. Therefore it is integrable on [

, z

]. So we can

find a dissection D

with points x

< x

< ··· < x

n−1

= z such that

f − L

f <

Let D be the dissection a = x

< x

< ··· < x

= b. Then

f − L

f <

· 2C +

· 2C = ε.

So done by Riemann integrability criterion.

Example.

– f(x) =

(

sin

x 6= 0

0 x = 0

defined on [−1, 1] is integrable.

– g(x) =

(

x x ≤ 1

+ 1 x > 1

defined on [0, 1] is integrable.

Corollary. Every piecewise continuous and bounded function on [

a, b

] is inte-

grable.

Proof.

Partition [

a, b

] into intervals

, ··· , I

, on each of which

is (bounded

and) continuous. Hence for every

with end points

j−1

is integrable on

[

j−1

, x

] (which may not equal

, e.g.

could be [

j−1

, x

)). But then by the

additivity property of integration, we get that f is integrable on [a, b]

We defined Riemann integration in a very general way — we allowed arbitrary

dissections, and took the extrema over all possible dissection. Is it possible to

just consider some particular nice dissections instead? Perhaps unsurprisingly,

yes! It’s just that we opt to define it the general way so that we can easily talk

about things like least common refinements.

Lemma. Let

: [

a, b

]

→ R

be Riemann integrable, and for each

, let

the dissection

< x

< ··· < x

, where

i(b−a)

for each

Then

f →

f(x) dx

and

f →

f(x) dx.

Proof.

Let

ε >

0. We need to find an

. The only thing we know is that

Riemann integrable, so we use it:

Since

is integrable, there is a dissection

, say

< u

< ··· < u

, such

that

f −

f(x) dx <

We also know that f is bounded. Let C be such that |f(x)| ≤ C.

For any n, let D

be the least common refinement of D

and D. Then

f ≤ U

Also, the sums

and

are the same, except that at most

of the

subintervals [x

i−1

, x

] are subdivided in D

For each interval that gets chopped up, the upper sum decreases by at most

b−a

· 2C. Therefore

f − U

f ≤

b − a

2C · m.

Pick n such that 2Cm(b − a)/n <

. Then

f − U

f <

f −

f(x) dx < ε.

This is true whenever

n >

4C(b−a)m

. Since we also have

f ≥

(

) d

therefore

f →

f(x) dx.

The proof for lower sums is similar.

For convenience, we define the following:

Notation. If b > a, we define

f(x) dx = −

f(x) dx.

We now prove that the fundamental theorem of calculus, which says that

integration is the reverse of differentiation.

Theorem (Fundamental theorem of calculus, part 1). Let

: [

a, b

]

→ R

continuous, and for x ∈ [a, b], define

F (x) =

f(t) dt.

Then F is differentiable and F

(x) = f(x) for every x.

Proof.

F (x + h) − F (x)

x+h

f(t) dt

Let

ε >

0. Since

is continuous, at

, then there exists

such that

|y − x| < δ

implies |f (y) − f (x)| < ε.

If |h| < δ, then



x+h

f(t) dt −f(x)



x+h

(f(t) −f(x)) dt



≤

|h|



x+h

|f(t) −f(x)| dt



≤

ε|h|

|h|

= ε.

Corollary. If f is continuously differentiable on [a, b], then

(t) dt = f (b) − f (a).

Proof. Let

g(x) =

(t) dt.

Then

(x) = f

(x) =

(f(x) −f(a)).

Since

(

)

− f

(

) = 0,

(

)

− f

(

) must be a constant function by the mean

value theorem. We also know that

g(a) = 0 = f(a) −f(a)

So we must have

(

) =

(

)

−f

(

) for every

, and in particular, for

Theorem (Fundamental theorem of calculus, part 2). Let

: [

a, b

]

→ R

be a

differentiable function, and suppose that f

is integrable. Then

(t) dt = f (b) − f (a).

Note that this is a stronger result than the corollary above, since it does not

require that f

is continuous.

Proof.

Let

be a dissection

< x

< ··· < x

. We want to make use of this

dissection. So write

f(b) −f(a) =

i=1

(f(x

) − f (x

i−1

)).

For each

, there exists

∈

(

i−1

, x

) such that

(

)

− f

(

i−1j

) = (

−

i−1

) by the mean value theorem. So

f(b) −f(a) =

i=1

− x

i−1

We know that

(

) is somewhere between

sup

x∈[x

i−1

]

(

) and

inf

x∈[x

i−1

]

(

)

by definition. Therefore

≤ f (b) − f (a) ≤ U

Since

is integrable and

was arbitrary,

and

can both get arbitrarily

close to

(t) dt. So

f(b) −f(a) =

(t) dt.

Note that the condition that

is integrable is essential. It is possible to find

a differentiable function whose derivative is not integrable! You will be asked to

find it in the example sheet.

Using the fundamental theorem of calculus, we can easily prove integration

by parts:

Theorem (Integration by parts). Let

f, g

: [

a, b

]

→ R

be integrable such that

everything below exists. Then

f(x)g

(x) dx = f (b)g(b) − f (a)g(a) −

(x)g(x) dx.

Proof. By the fundamental theorem of calculus,

(f(x)g

(x) + f

(x)g(x)) dx =

(fg)

(x) dx = f (b)g(b) − f (a)g(a).

The result follows after rearrangement.

Recall that when we first had Taylor’s theorem, we said it had the Lagrange

form of the remainder. There are many other forms of the remainder term. Here

we will look at the integral form:

Theorem (Taylor’s theorem with the integral form of the remainder). Let

n + 1 times differentiable on [a, b] with with f

(n+1)

continuous. Then

f(b) = f(a) + (b − a)f

(a) +

(b − a)

(2)

(a) + ···

(b − a)

(n)

(a) +

(b − t)

(n+1)

(t) dt.

Proof. Induction on n.

When n = 0, the theorem says

f(b) −f(a) =

(t) dt.

which is true by the fundamental theorem of calculus.

Now observe that

(b − t)

(n+1)

(t) dt =



−(b − t)

n+1

(n + 1)!

(n+1)

(t)



(b − t)

n+1

(n + 1)!

(n+1)

(t) dt

(b − a)

n+1

(n + 1)!

(n+1)

(a) +

(b − t)

n+1

(n + 1)!

(n+2)

(t) dt.

So the result follows by induction.

Note that the form of the integral remainder is rather weird and unexpected.

How could we have come up with it? We might start with the fundamental

theorem of algebra and integrate by parts. The first attempt would be to

integrate 1 to t and differentiate f

(t) to f

(2)

(t). So we have

f(b) = f(a) +

(t) dt

= f(a) + [tf

(t)]

−

(2)

(t) dt

= f(a) + bf

(b) − af

(a) −

(2)

(t) dt

We want something in the form (

b − a

)

(

), so we take that out and see what

we are left with.

= f(a) + (b − a)f

(a) + b(f

(b) − f

(a)) −

(2)

(t) dt

Then we note that f

(b) − f

(a) =

(2)

(t) dt. So we have

= f(a) + (b − a)f

(a) +

(b − t)f

(2)

(t) dt.

Then we can see that the right thing to integrate is (

b −t

) and continue to obtain

the result.

Theorem (Integration by substitution). Let

: [

a, b

]

→ R

be continuous. Let

: [

u, v

]

→ R

be continuously differentiable, and suppose that

(

) =

a, g

(

) =

and f is defined everywhere on g([u, v]) (and still continuous). Then

f(x) dx =

f(g(t))g

(t) dt.

Proof.

By the fundamental theorem of calculus,

has an anti-derivative

defined on g([u, v]). Then

f(g(t))g

(t) dt =

(g(t))g

(t) dt

(F ◦ g)

(t) dt

= F ◦ g(v) −F ◦ g(u)

= F (b) − F(a)

f(x) dx.

We can think of “integration by parts” as what you get by integrating the

product rule, and “integration by substitution” as what you get by integrating

the chain rule.

7.2 Improper integrals

It is sometimes sensible to talk about integrals of unbounded functions or

integrating to infinity. But we have to be careful and write things down nicely.

Definition (Improper integral). Suppose that we have a function

: [

a, b

]

→ R

such that, for every

ε >

is integrable on [

ε, b

] and

lim

ε→0

a+ε

(

) d

exists. Then we define the improper integral

f(x) dx to be lim

ε→0

a+ε

f(x) dx.

even if the Riemann integral does not exist.

We can do similarly for [a, b − ε], or integral to infinity:

∞

f(x) dx = lim

b→∞

f(x) dx.

when it exists.

Example.

−1/2

dx =

−1/2

= 2 − 2ε

1/2

→ 2.

−1/2

dx = 2,

even though x

−1/2

is unbounded on [0, 1].

Note that officially we are required to make

(

) =

−1/2

a function with

domain [0

1]. So we can assign

(0) =

, or any number, since it doesn’t matter.

Example.

dt =



−



= 1 −

→ 1 as x → ∞

by the fundamental theorem of calculus. So

∞

dx = 1.

Finally, we can prove the integral test, whose proof we omitted when we first

began.

Theorem (Integral test). Let

: [1

, ∞

]

→ R

be a decreasing non-negative

function. Then

∞

n=1

f(n) converges iff

∞

f(x) dx < ∞.

Proof. We have

n+1

f(x) dx ≤ f(n) ≤

n−1

f(x) dx,

since

is decreasing (the right hand inequality is valid only for

n ≥

2). It follows

that

N+1

f(x) dx ≤

n=1

f(n) ≤

f(x) dx + f(1)

So if the integral exists, then

(

) is increasing and bounded above by

∞

f(x) dx, so converges.

If the integral does not exist, then

(

) d

is unbounded. Then

n=1

(

)

is unbounded, hence does not converge.

Example. Since

dt < ∞, it follows that

∞

n=1

converges.