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These are Zixuan’s notes for Part IA – Analysis I at the University of Cambridge in 2026. The notes are not endorsed by the lecturers or the University, and all errors are my own.

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Lecture 1 · 2026-01-22

1 Numerical Sequences and Series

1.1 Basics

Definition 1.1 (Sequence)

A sequence on a set is an enumerated list where each element is in the set .

In this section, or . A concrete important case is [ real sequences ].

We shall now consider the issues of

convergence, where converges to , and
divergence of sequences.

On convergence:

we need to be smaller than any given threshold that we choose;
for the comparison, only the tail of the sequence matters, i.e. large behaviour. We can always ignore the first terms of the sequence, for some depending on .

On divergence to infinity:

we need to clear any threshold that we choose;
again, only the tail of the sequence matters. We can always ignore the first terms of the sequence, for some depending on .

Definition 1.2 (Convergence)

We say that converges to some finite if

We write or .

Definition 1.3 (Divergence to Infinity)

We say that a real sequence diverges to (positive) infinity if

We write .

Remark. We can replace by , and replace by , etc. in the definitions above without changing their meanings.

Remark. For complex sequences, we can use analogus definitions of diverging to infinity.

Example 1.4

Consider . Then because ,
Consider . Then because ,
Consider . Then diverges to infinity since ,
Consider . Then does not diverge to , but it does not converge either.

Lemma 1.5

If a sequence converges, then the limit is unique.

Proof. Suppose and . Take , then we have

In particular, for , both inequalities hold. Then

Since is arbitrary, we can conclude that . Hence .

Proposition 1.6 (Sandwich Theorems)

Let , , be real sequences. Then

If for all and , then .
If , , and , then .

Important. The best conclusion in the case that for all and , is still .

Lemma 1.7

Let be a complex sequence. Then if and only if and .

Proof. Recall that for , .

[] We have that and . By the defintiion of convergence, the result follows.

[] Note the inequality Fix . By definiition of convergence,

Hence for all .

Lemma 1.8

Let , . Then and . If for every , then .

Proof. The first and third part is left as an exercise.

We shall prove that . We have

We have

Hence

Lemma 1.9

If , then must be bounded. i.e.

Lecture 2 · 2026-01-24

Proof. Take . Then there exists such that , . Hence for , we have

Let

Then , .

Hence we can replace with and the result follows.

Definition 1.10 (Bounded Sequence)

We say is bounded if such that for all . Equivalently, .

Definition 1.11 (Monotonic Sequence)

We say a real sequence is monotone if either

it is increasing, for every ,
it is decreasing, for every .

Proposition 1.12 (Monotone Convergence Theorem)

Every bounded monotone real sequence converges.

Proof. WLOG suppose is strictly increasing and bounded above. We will use the supremum axiom, that every non-empty set of bounded above has a supremum in .

Refer to IA Numbers and Sets for a full proof of this proposition.

If we drop the monotonicity condition, we may not have convergence. For example, is bounded but does not converge. However, we can still extract convergent subsequences from bounded sequences, e.g. by taking all even terms in .

1.2 Bolzano-Weierstrass Theorem

Theorem 1.13 (Bolzano-Weierstrass Theorem)

If is a real and bounded sequence, then there exists a convergent subsequence.

Definition 1.14 (Subsequence)

A subsequence of a sequence is a sequence of the form where is a strictly increasing sequence of natural numbers.

Lemma 1.15

If , then any subsequence must converge to the same limit.

Proof. Since , by induction, we can show that for all .

Take , then such that , . So if then and hence .

Hence .

Proposition 1.16 (Nested Interval Property)

Take a sequence of nested closed intervals in : , where .

If as , then contains exactly one point.

Proof. This is an application of Monotone Convergence Theorem 1.12.

Since and , we have

Hence,

is increasing and bounded above by ,
is decreasing and bounded below by .

So and converge. Let and . Since limits preserve inequalities, we have . Now we shall prove this proposition by considering two aspects.

Existence. For all , . So . Now, as , we have . Since this is true for all , we have .

Uniqueness. by construction. Since , we must have because limits are unique. Hence .

Thus

Proof. [of Bolzano-Weierstrass Theorem 1.13] We are given and such that for all . We will construt a sequence of nested intervals from which we can sample our subsequence, since that will ensure that our subsequence will converge to the unique intersection point of nested intervals.

Let . Then .

Now take . Then at least one of the intervals and must contain infinitely many terms of the sequence . [If both intervals contained only finitely many terms, then the whole interval would contain only finitely many terms, contradicting the fact that is an infinite sequence.] Take to be a half interval that contains infinitely many terms. Continuing inductively gives a sequence of nested intervals with as , and each contains infinitely many terms of the sequence.

By Nested Interval Property 1.16, . We can now choose as follows: pick such that , then has infinitely many elements of with indices greater than , so pick such that . Continuing in this manner gives a subsequence with for all .

By construction, for every , so , so as .

Lecture 3 · 2026-01-27

Remark. The Bolzano-Weierstrass Theorem also works for complex sequences.

1.3 Cauchy Sequences

Definition 1.17 (Cauchy Sequence)

A sequence is Cauchy if

Example 1.18

. Assume WLOG , Then
is not a Cauchy sequence, because if , for any , then

The definition fails for .
on defined by truncation of decimal expansion of :

This is Cauchy, since for WLOG , we have

This sequence does not converge over , but it does converge over .

Exercise. If satisfies

must be a Cauchy sequence?

Lemma 1.19

If is Cauchy, then it is bounded.

Proof. Take . Then such that ,

Hence for all . Note that is a finite number independent of . So

Lemma 1.20

A complex sequence is Cauchy if and only if and are Cauchy in .

Lemma 1.21

If , then is Cauchy.

Proof. , we have

Consider the converse of Lemma 1.21. Note that Example 1.18 (3) shows that there are Cauchy sequences in that do not converge in . However, we have the following important theorem.

Theorem 1.22 (Completeness Of and )

Every Cauchy sequence in or converges.

Remark. So one can prove convergence of or sequences without having to know the actual limit, by showing that they are Cauchy.

Proof. Recall that a sequence on is Cauchy/convergent if and only if its real and imaginary parts are Cauchy/convergent. So it suffices to prove the result for real sequences.

We have seen that being Cauchy implies that it is bounded by Lemma 1.19. Then by Bolzano-Weierstrass Theorem 1.13, convergent subsequence with limit . We have

More precisely, take ,

since is Cauchy, such that , ,
since , such that , .

We can choose such that , then

1.4 Series and Convergence Tests

Definition 1.23 (Series and Series Convergence)

Let be a sequence over or . We say that is a series.

We say it converges if the sequence of partial sums

converges to some finite or as . In this case, is called the sum of the series,

Example 1.24

does not converge as as .
Geometric series. converges iff . The partial sums for are

Lecture 4 · 2026-01-29

converges to , since we have

Lemma 1.25

Fix . If and converge, then also converges.

Remark. Note that the product of two convergent series need not converge.

As usual, only the tail of the series matters for convergence.

Lemma 1.26

If for all for some , then converges iff converges.

Proof. Let

Note that is a finite sum, so it does not affect convergence. If , then converges iff converges.

Proposition 1.27 ( Term Test)

A necessary condition for to converge is that as .

[i.e., if does not converge to , then diverges.]

Remark. is not a sufficient condition for to converge.

For example, the harmonic series diverges even though its terms converge to . To see that it diverges, note that the partial sums satisfy

Hence is not Cauchy, so it diverges.

Proof.

We shall first focus on tests for convergence of where for all .

Proposition 1.28 (Comparison Test)

If for all sufficiently large , then

Proof. Let and be the partial sums of and . Because , the sequences and are increasing. Since , we have for all . Hence

Example 1.29

converges. This is because

so by Comparison Test 1.28, converges.

The next two tests are about asymptotic comparisons to the geometric series.

Proposition 1.30 (Root Test)

If for all , then consider , and assume such that .

Then

implies converges.
implies diverges.
is inconclusive.

Proof. If , then by the definition of limit,

This implies that for all , so diverges by th term test 1.27.

Now, if , then there is some such that . By the definition of limit,

Hence for all . By Comparison Test 1.28, converges since converges.

Example 1.31

: , so it converges.
: , so it diverges.

Proposition 1.32 (Ratio Test)

If for all , then consider , and assume such that .

Then

implies converges.
implies diverges.
is inconclusive.

Example 1.33

(divergent) and (convergent) are both inconclusive under the root and ratio tests, since both have limit in both tests.
converges, since

Or alternatively, by the root test,

Remark. To show that , write and L’Hospital’s rule shows that .

Exercise. Show that if the ratio test is inconclusive, then so is the root test. Show also that the converse is not true, using

Lecture 5 · 2026-01-31

Proposition 1.34 (Integral Test)

Suppose is a continuous decreasing function (so it is integrable in for each [we will see this later]). Let for each .

Then

Furthermore, as ,

Remark. The RHS is an improper integral, which will be discussed later. The last part tells us that the integral is a good approximation for the series (if it converges), or the rate of divergence (if it diverges).

Proof.

We have

[] since is increasing and converges by assumption. Thus is increasing and bounded above, so it converges.

[] if the integral exists, then is bounded. Hence is a monotone bounded sequence and it converges.

For the last part, let . We have

Hence is decreasing and bounded below, so it converges to some . Also, since

we get .

Example 1.35

converges iff .

Note that

which exists for and diverges for .

Remark. This is a much easier way to see the divergence of the harmonic series. Note a posteriori that the divergence is not surprising, since for to converge we need sufficiently fast to overcome the growth in the number of terms we are adding up.

Rough calculation suggests that for large would be enough for convergence.
diverges since

with the substitution .
converges since

with the substitution .

Proposition 1.36 (Cauchy Condensation Test [Non-Examinable])

Let for all , and suppose that is decreasing. Then

Proof. We have

using the substitution .

From Integral Test 1.34, we have

Hence, letting ,

Thus, converges iff exists, and the result follows.

Proposition 1.37 (Alternating Series Test)

Let be a decreasing sequence with and . Then converges.

Example 1.38

converges though diverges, by the alternating series test. [In Section 5, we will show that it converges to .]

Lecture 6 · 2026-02-03

Proof. Let be the partial sums. Note that

Hence,

so is an increasing sequence.

Also,

and is decreasing.

Moreover, , so we have

Therefore, both and are bounded. By Monotone Convergence Theorem 1.12, both sequences converge. Let and as .

Note that

Lemma 1.39

If the odd and even subsequences of a sequence both converge to the same limit, then the whole sequence converges to the same limit.

Proof. Forall ,

Hence for all .

Therefore, as .

Proposition 1.40 (Dirichlet Test)

Let be a decreasing sequence with and . Let be a sequence such that the sequence of partial sums is bounded.

Then converges.

Definition 1.41 (Absolute Convergence)

A series is said to converge absolutely if converges.

Remark. for all , hence root, ratio, etc. tests can be applied to test for absolute convergence.

Lemma 1.42

If converges absolutely, then it converges.

Remark. The converse is not true by Example 1.38. Hence, absolute convergence is a strictly stronger notion. We call series that converge but not absolutely conditionally convergent.

Proof. Let , and let . Since is convergent, it is Cauchy.

Thus, is Cauchy and hence convergent.

Conditionally convergent series can behave badly under rearrangements.

Example 1.43

Consider

Rearrange terms on the right as

This suggests that the rearranged series sums to half the original series.

In a conditionally convergent series, the order of the sum matters. This is not the case for absolutely convergent series, where any rearrangement converges to the same sum.

Proposition 1.44 (Rearrangements of Absolutely Convergent Series)

Let be a bijection. Let . Then if is absolutely convergent, we have

Proof. Let . By assumption, such that as . For ,

Now, since is a bijection, such that is contained in . Hence, for ,

Hence

2 Continuity of Functions

2.1 Limits of Functions

Take a function with . Consider the meaning of as , even if .

A classic example is at , which is not in the domain of the function. The reason we may think of is that there are points in the domain that are very close to . In other words, is an accumulation point for , since for any threshold , there are points in within of .

Definition 2.1 (Accumulation Point)

Let , and . We say that is an accumulation point for if

If and is not an accumulation point for , we say that is an isolated point of .

Example 2.2

For , is an isolated point of , while any point in is an accumulation point of .

Lecture 7 · 2026-02-05

The points on the circle are accumulation points for .
All the points in are accumulation points for .
For the set , then all the points are accumulation points.

Lemma 2.3

Let , . Then is a accumulation point iff there exists in such that as .

Hence, for as to be meaningful, we need , or if , it should be an accumulation point for . The rationale for our definition are

for any threshold (no matter how small it is), there exist points in which are -close to , i.e.
furthermore, it must contain all points in that are sufficiently close to , i.e.

Definition 2.4 (Limit of a Function)

Let . Take such that is an accumulation point for . We say that as if

is called the limit of as , and we write .

In particular, for where is an accumulation point for , we say that diverges (to ) as if for some , and

Remark. If is an isolated point of , then we can always find sufficiently small that

So the definition of limit can be made for isolated points, but it is not very interesting.

Example 2.5

has fomain . Consider its limit as .

Claim.

Proof. Recall that there is a geometric argument using trignometric circle that shows for all . Hence

Hence , choosing gives

We can also give a sequential characterisation of limits, which is often easier to work with.

Lemma 2.6 (Sequential Characterisation of Limits)

Let . Let where is an accumulation point of . Then

with and is not the constant sequence.

Lemma 2.7

Suppose , and it has a limit at . Then the limit is unique.

Proof. Suppose as , and as are both true.

We have

Note that LHS does not depend on , but RHS does. So taking limit as gives

Hence .

Lemma 2.8

Let . Let where is an accumulation point of . Suppose , . Then,

2.2 Continuity of Functions

From the previous section, we can compute if is a accumulation point for .

Definition 2.9 (Continuity of a Function)

Let . We say that is continuous at every point in .

Take . We say that is continuous at if

Remark. If is an isolated point, then is continuous at . Inf is an accumulation point for , then is continuous at iff .

Example 2.10

is continuous, since given by such that .
is not continuous at , since does not exist.

Lecture 8 · 2026-02-10

Definition 2.11 (Sequential Continuity)

Let . We say that is sequentially continuous at if for every sequence in such that , we have .

Proposition 2.12

Let . Then is continuous at iff is sequentially continuous at .

Proof.

[] By continuity of , , such that . Take on with , then such that for all . Hence for all , so .

[] We shall prove by contradiction. Suppose is not continuous at , then such that , with but .

Take to find with but for all . Hence but does not converge to , contradicting the sequential continuity of at .

Example 2.13

For

consider the sequence with . Then for all and . Hence is not sequentially continuous at , and thus not continuous at .
Consider . is discontinuous at every point :
- if then with but for all and .
- if then with but for all and .
Consider , we shall show that it is continuous at every point . Fix . We can choose ,

Therefore, holds for sufficiently large.

Lemma 2.14

Let be continuous at . Then so are and . If forall , then so is .

Lemma 2.15

Let , , . If is continuous at and is continuous at , then is continuous at .

Proof. Since is continuous at ,

Also, is continuous at . So

Putting everything together,

Hence is continuous at .

Remark. Composition preserves continuity.

2.3 Extreme Value Theorem

Definition 2.16 (Closed Set)

A set is closed if all sequences in which converge in have their limits in .

Example 2.17

is closed,
is not closed,
or are not closed,
is closed,
is not closed.

Definition 2.18 (Bounded Set)

We say is bounded if such that .

In other words, if such that .

Example 2.19

, , are bounded,
, are not bounded.

Proposition 2.20 (Continuity Preserves Closedness and Boundedness)

Let be a closed bounded set. If is continuous then is a bounded closed subset of .

Lecture 9 · 2026-02-12

Proof.

[ is bounded.] Suppose is not bounded, then for each we can find such that . Now, is a sequence in , and it must be bounded since is bounded.

By Bolzano-Weierstrass Theorem 1.13, there exists a convergent subsequence . Let the limit of be . Since is closed, . On the other hand, . So cannot converge as . Then is not continuous at . Therefore must be bounded.

[ is closed.] Take in , suppose that it converges to some . We want to show that . Note that such that . This sequence is inside , hence it is bounded. Copying the argument above to get . Thus . By continuity of ,

Hence .

Theorem 2.21 (Extreme Value Theorem)

Let be a closed bounded set. If is continuous, then there exist with

Remark. We know by Supremum Axiom (from IA Numbers and Sets), that there exist supremum and infimum of . The proof here is that these are always attained, so that they actually maximum and minimum.

Proof. We will focus on the first equality, since the other can be proved similarly.

Let . Hence, is not an upper bound for for all . Hence, we can find a sequence on such that

Note that such that for each of . This gives us a sequence on such that for all with . Now take limits and use the fact that is continuous, we get

By closedness of , is also closed by Proposition 2.20. Hence , so there exists such that . Hence .

2.4 Intermediate Value Theorem

Theorem 2.22 (Intermediate Value Theorem)

If is continuous, then is an interval. Hence, if , then there exists such that .

Remark. The theorem guarantees existence, but not uniqueness.

Proof. If is constant, or if or , then this is trivially true.

If not, WLOG assume , and let . Then , so is non-empty. Also, is bonunded above by , so . We aim to show that .

Suppose . Then . By continuity of , such that , .

This means that , which contradicts the definition of .
Suppose . Then . By continuity of , such that , . But then, , again contradicting the definition of .

Hence .

Example 2.23

We can apply this to show the existence of -th roots (where ). Take , and consider

This is a continuous function, and . By Intermediate Value Theorem 2.22, such that , Hence is a positive -th root of .

Definition 2.24 (Monotone Function)

Consider . We say is (strictly) monotone if either

it is (strictly) increasing, so ,
it is (strictly) decreasing, so .

Proposition 2.25 (Inverse Function Theorem, Version 1)

Let be a continuous function that is strictly monotone. Let and . Then is a bijection and is continuous and strictly monotone.

Lecture 10 · 2026-02-14

Proof.

[Bijiectivity of .] Since is continuous and monotone, so it maps to because its monotonicity implies that the extreme values are attained at the endpoints. Since it is strictly monotone, it is injective. By Intermediate Value Theorem 2.22, it is surjective. Hence is a bijection.

[Monotonicity of .] WLOG take to be strictly increasing. If is not strictly increasing, then such that yet . Since is strictly increasing, , contradicting the choice of . Hence is strictly increasing.

[Continuity of .] Fix and . There are three cases:

if , fix . Then choose sufficiently small that Now, is strictly increasing, so

Then take . We want to prove that for ,

We have
if , fix . Then . Choose and set . Then,

if , the case is similar to the previous one.

3 Differentiation

3.1 Introduction

Definition 3.1 (Differentiability and Derivative)

Let with . We say that is differentiable at if the limit

exists. [We require .]

This limit is called the derivative of at , and is denoted by or .

Remark. At isolated points, the definition of derivative is meaningless. For accumulation points, we can distinguish between

interior points, where we can approach from both sides, and we need the limit to be direction-independent, and
non-interior points, where the limit is domain-restricted.

Example 3.2

is differentiable at every point. This is because we have
is not differentiable at any point. This is because we have

If this limit exists, then it must be the same as the limit along the real axis, which is , and the limit along the imaginary axis, which is .
on is differentiable at every point. We have

We can use results from lectures that and to conclude that .

We can also derive some properties of derivatives from properties of limits.

Lemma 3.3

let be differentiable at . Then so are , and if for all .

Moreover, we have

Proof. These follow from last chapter. The addition rule is left as an exercise. We have

For the last step, we need to show that a function is continuous at if it is differentiable at . This will be dealt with later.

The reciprocal rule is left as an exercise.

Example 3.4

is differentiable with . By induction, we can show that is differentiable with for all .

Hence polynomials are differentiable.
is differentiable on with . By induction, we can show that is differentiable with for all .

Hence rational functions are differentiable on their domains.

Lecture 11 · 2026-02-17

Proposition 3.5 (Chain Rule)

Let and , . Suppose that is differentiable at and is differentiable at . Then is differentiable at , and we have

It will be convenient to have an alternative characterization of derivative to prove this. It is common to see as the slope of the tangent line to the graph of at .

Lemma 3.6

Let . Then is differentiable at iff and function satisfying as such that

Remark. This means for small , and the function is to quantify the error of this approximation. We can equivalently write

Moreover, if is differentiable at , then we must have .

Proof.

[] We have

since as , and is bounded. Hence is differentiable at with .

[] Choose , so that , or equivalently,

Take

Then as , and the required equality holds.

Proof. [of Chain Rule 3.5]

Since are differentiable at respectively, there exists error functions such that

and

So we have

So, we just need to show that as . We have

and since

our conclusion follows.

Example 3.7

Consider the function

At , we have

At ,

which does not exist. Hence is only differentiable on .

Lemma 3.8

If is differentiable at , then is continuous at .

Proof. We have

3.2 Mean Value Theorems

We have concluded that the derivative of a function is the instantaneous rate of change. We want to relate this to the average rate of change.

Theorem 3.9 (Mean Value Theorem)

Let be continuous on and differentiable on . Then such that

We shall consider an easier case first.

Proposition 3.10 (Rolle's Theorem)

Let be continuous on and differentiable on . If , then such that .

Proof. [of Mean Value Theorem 3.9]

Let . Then but also .

By Rolle’s Theorem 3.10, there exists such that , so we have .

Lecture 12 · 2026-02-19

Remark. Mean Value Theorem 3.9 can be rephrased as the followings.

Given such that , such that .

Note that there is no error function in this statement.

Proof. [of Rolle’s Theorem 3.10]

The idea is that we are looking for some that is a local minimum or a local maximum.

By Extreme Value Theorem 2.21, attains its maximum and its minimum on , i.e. such that

If we can show one of , is in , then by the we have our result:

. Hence by taking limits from and , we have .
. Hence by taking limits from and , we have .

Now, suppose is not constant (or otherwise the result is trivial). Either or , so either or is in .

We shall see some applications about Mean Value Theorem 3.9.

Corollary 3.11

Let be continuous on and differentiable on . Then

if on , then is increasing on ;
if on , then is decreasing on ;
if on , then is constant on .

The monotonocity of is strict if the inequalities are strict.

Remark. It is not always possible to replace with some sets . For example, consider the function defined by

Note that by definition, is continuous and differentiable at every point in , and on . However, is not constant on .

Note that we can generalise Corollary of Mean Value Theorem 3.11 (3) to functions defined on .

Lemma 3.12

Let be differentiable in and for all . Then is constant on .

Proof. The idea is to reduce to the case. Fix some , take where . Note that is continuous on and differentiable on , so we can apply Mean Value Theorem 3.9 and its corollaries to and separately. We have

Hence is constant on , so we have .

Since is arbitrary, we have for all , so is constant on .

Theorem 3.13 (Inverse Function Theorem, Version 2)

Let be continuous on and differentiable on . Assume that for all . Then is bijective with inverse continuous and differentiable on with

Proof. By the corollary above, is strictly increasing, so Inverse Function Theorem (Version 1) 2.25 applies, so is a bijection to its image and is continuous.

Now for differentiability, let and . This is unique by bijectivity. Given such that , define such that . Then

So differentiability and the formula holds if we show that implies . This is true because

by continuity of .

Example 3.14

Fix . Define with where . Then is continuous on and differentiable on with for all . Thus by Inverse Function Theorem (Version 2) 3.13, is a bijection to its image with inverse continuous and differentiable on with

We can write , so we have .
is continuous and differentiable on with . Then there exists an inverse

Proposition 3.15 (Cauchy Mean Value Theorem)

Let be continuous on and differentiable on . Then such that

Proof. We aim to reduce to Rolle’s Theorem 3.10. Let

and note that , Hence by Rolle’s Theorem 3.10, there exists such that , so the result follows.

Proposition 3.16 (L'Hôpital's Rule)

Let be continuous on and differentiable on . Assume that for all and that . If exists, then also exists and is equal to .

Example 3.17

We have seen that . With L’Hôpital’s rule, we can also compute this limit as follows:

Behind the scenes in this computation, we are saying that such that

Lecture 13 · 2026-02-21

Corollary 3.18

Let be continuous on and differentiable on . If there exists such that , then locally around , there is an inverse function for which is differentiable. [Refer to Example Sheet 3 for further details.]

3.3 Higher Derivatives and Taylor’s Theorem

Definition 3.19 (Higher Derivatives)

Let be differentiable on . We say that is twice differentiable if

is differentiable. We similarly define thrice differentiable and times differentiable for inductively.

We say that is -times continuously differentiable, and write , if is times differentiable and

is continuous.

Remark. Recall that being differentiable implies its continuity, so if is -times differentiable, then is continuous for all .

Definition 3.20 (Smoothness)

We say is smooth if is -times differentiable for all . We write .

We saw that being differentiable implies that it can be well-approximated by a linear function near :

Meanwhile, if is twice differentiable, then we can do better:

We want to state a general version of this apprixmation, under appropriate conditions, and also quantify the error of this approximation.

Definition 3.21 (Taylor Polynomial)

If is -times differentiable, , then we call

the Taylor polynomial of at of degree . We denote this by .

If is smooth, we call

the Taylor series of at .

Theorem 3.22 (Taylor's Theorem: Lagrange Remainder)

Let be continuous, and assume its first derivatives are continuous as well, and that it is -times differentiable on . [This is all satisfied if where .]

Let the Taylor remainder be

Then there exists such that

Remark. For , this is just Mean Value Theorem 3.9. There is nothing special about ; if and and its derivatives are continuous on and its -th derivative exists on , then we can apply the above to :

Remark. If with , then is continuous and hence bounded on . Hence such that

Hence

Therefore is as . Note that this does not tell use that as , since even if , we do not know how behaves with .

Proof.

Proof 1

WLOG take . [If , apply to instead.] Let be a continuous function, with its first derivatives continuous and the -th derivative existing on , defined by

where we pick such that . Note that and also for all . By Rolle’s theorem,

Now let . Then and we have effective shown that

and hence , so we have

Proof 2

WLOG take . Let

[Note ]

Note that is continuous on , differentiable on with

For , set

Then

i.e. such that

Now, we can choose to get the required result.

The second proof leads to an alternative version of Taylor’s Theorem.

Theorem 3.23 (Taylor's Theorem: Cauchy Remainder)

Let as in Taylor’s Theorem: Lagrange Remainder 3.22 and define similarly. Then there exists such that

Lecture 14 · 2026-02-24

Remark. If with , then as in the previous remark, we can set

Then by Taylor’s Theorem: Cauchy remainder 3.23, we have

It appears that this is not better than the bound in Taylor’s Theorem: Lagrange Remainder 3.22. However, the following example shows that the Cauchy remainder can be better than the Lagrange remainder.

Example 3.24

Consider where , which is smooth on . Clearly

Remark. If , then is exactly its Taylor polynomial of some degree.

Then

We now look at the remainder

There are two ways to estimate this remainder:

Lagrange remainder: such that

If , then for , so for any , we have

Note that argument fails for .
Cauchy remainder: such that

Now , so as long as . Hence

Hence, we can see that the Cauchy remainder estimate gives us that the remainder is exponentially small as for all , while the Lagrange could only guarantee it for .

Definition 3.25 (Analytic Functions)

We say if for every , there exists such that , we have

which is equivalent to saying that

Example 3.26

is an analytic function in , i.e. is analytic in , such that for every .

Remark. If as , it does not necessarily mean that the Taylor series of at does not converge. It instead implies that the Taylor series does not represent near .

4 Integration

4.1 Basics

We want to define as the signed area under the graph of . If the region enclosed by the graph if a simple shape, then we can define the area using geometry. However, in general, the region may be very complicated, and we need to try to approximate.

Definition 4.1 (Upper/Lower Riemann Sums)

Let be bounded. Given a partition, with , we set

where is the lower Riemann sum and is the upper Riemann sum of associated with .

Example 4.2

Consider, on , the function

No matter what we take, we always have rationals and irrationals in each subinterval , hence

Remark. If is bounded, then for some , and

Also, , hence both

are non-empty bounded sets.

Definition 4.3 (Upper/Lower Integral)

Let be bounded. We say

to be the lower integral and upper integral of on respectively.

We say is Riemann integrable on if , and in this case we set

Remark. By the way we set up the sums, .

Example 4.4

Take the function on , defined by

as before, we have , , and is not Riemann integrable.

Lecture 15 · 2026-02-26

The coarsest partition of is . Our intuition is that a finer partition leads to a better estimate for the area.

Proof. We can do this by induction. Let

and choose some for some . Let . Then

So similarly, using instead of , we can show .

This argument can be repeated for any refinement,

and the result follows by induction.

Lemma 4.6

Let be partitions of [not necessarily refinements of each other]. Then

Proof. Note that . By Lemma 4.5, we have

Lemma 4.7

If , then .

Proof. We have

Remark. This means that is integrable iff .

4.2 Integrability Criteria

There is another way to define integrals. No matter how small the threshold is, we can always find a partition of so that the gap between the optimistic and pessimistic estimates is smaller than the threshold. Our definition of integrability is equivalent to this.

Proposition 4.8 (Riemann Integrability Criteria)

Let be bounded. Then is Riemann integrable iff

Proof.

[] We have ,

Hence .

[] From the definition of and , we can always find partitions such that

By assumption, , so . To conclude, we can take , and by Lemma 4.5, we have

It can be shown that this is equivalent to a sequential version.

Proposition 4.9

Let be bounded. Then is Riemann integrable iff there exists a sequence of partitions of such that

Exercise. The proof is left as an exercise.

Example 4.10

Consider on . Let . Then
- With original definition,
  
  So . By , we have , and the function is integrable.
- With Riemann Integrability Criteria 4.8, we have
  
  So is integrable.
Consider on the function

Take as before for . Then

Hence,

By Proposition 4.9, is integrable.

We can try to generalize the above examples to get classes of functions that are integrable.

Proposition 4.11

Let . If is monotone, then is Riemann integrable.

Lecture 16 · 2026-02-28

Proof. If is monotone, then is bounded by .

Since is monotone, and are attained at the endpoints of . WLOG assume is increasing, then

Hence, for any partition of ,

Set , so that for all . Then

Proposition 4.12 (Continuity Implies Integrability)

If is continuous, then it is integrable.

Proof. By Extreme Value Theorem 2.21, is continuous on with , so is bounded.

We will show that contrapositive, that if is not integrable, then is not continuous.

If is not integrable, such that partition of ,

Hence such that .

By Extreme Value Theorem 2.21, such that .

Since the above is true for all partitions, it should also be true for the in the previous proof. So for each , we can get with and .

From this point, we want to show that is not sequentially continuous. The problem is that we do not know if are convergent. However, since and are both bounded, by Bolzano-Weierstrass Theorem 1.13, we can find subsequences that are convergent.

Let and . Then , and by

we have . We can now invoke the sequential continuity: we have

hence and do not converge to the same limit, and is not sequentially continuous, hence not continuous.

Proposition 4.13 (Piecewise Continuity Implies Integrability)

If is piecewise continuous, i.e. suppose there is a partition of such that is continuous and has a finite limit as and , for all , then is integrable, and

where

Proof.

Lemma 4.14

Let be bounded. Suppose is integrable, and the set

is finite, then is integrable and .

Proof. Set . Fix then of such that

The idea is to choose a partition which isolates problematic points, and gives them very little weight. Choose intervals with

Set . Also, let . Then

Try to estimate

Note that if , then , and hence

and if , then we only know that is bounded, and

Hence,

Thus,

For the formula for the integral, we have

Hence

Thus . Similarly, we can show , and the result follows by arbitrariness of .

By Lemma 4.14 we have that is integrable and . To conclude, we just need to show that

Proposition 4.15

Let and . Then is integrable iff and are integrable. Moreover,

Lecture 17 · 2025-03-03

Proof. We will set , , and for brevity.

[] For , then there exists a partition of such that . WLOG let for some [otherwise add to , which can only make smaller]. Then where and are partitions of and respectively. Furthermore,

gives

So and are integrable.

[] For , partitions of and respectively such that

Hence by the equalities in the first part,

Hence is integrable in . Furthermore,

Hence, . Since is arbitrary, the result follows.

Continuous, piecewise continuous and monotone functions are all integrable, we may wonder if these are all the Riemann integrable functions. The answer is negative.

Example 4.16 (Thomae Function)

Consider the function defined by

Since is dense in , for any partition of , hence . We claim that is integrable, then such that .

Pick such that . Set

Define such that

each is some subinterval of
this subinterval has length [we wish to give little weight to the bad points]

Then,

This gives us a different result compared to the Dirichlet function. Despite both functions have infinitely many discontinuities, the integrability properties are fundamentally different.

Proposition 4.17 (Countable Discontinuities Implies Integrability)

If , and , then

is finite implies that is Riemann integrable.
is countable implies that is Riemann integrable.

Proof.

See Example Sheet 3 Q13.
The proof is non-examinable.

Remark. Proposition 4.17 (1) is a stronger version of Proposition 4.13, since we do not require the one-sided limits to exist at the discontinuities. For example, consider oscillating functions like .

Important. If is not Riemann integrable then cannot be countable. However, there are functions which have uncountably many discontinuities but are still Riemann integrable, such as the indicator function of the Cantor set.

4.3 Basic Properties of Integrals

Lemma 4.18

Let be integrable functions. Then

for all implies .
For , .
is integrable and .
is integrable and . [Triangle inequality for integrals]
is integrable but in general.

In order to show the lemma, we first need a few other lemmas, including

an intermediate lemma about upper and lower Riemann sums
an intermediate lemma on and of functions on intervals.

Exercise. Write out the equivalent lemma for the first one, having seen the second one below.

Lemma 4.19

Let be a closed and bounded interval, and let be bounded. Then

If for all , then and preserves this inequality.

2.1. .

2.2. For fixed, .

, and .
.
.

Lecture 18 · 2026-03-05

Proof.

Exercise. 1–3 are exercises.

If on all of , or if on all of , then the result is immediate.

If , then
We have

Proof. [of Lemma 4.18]

Exercise. 1–3 are exercises.

For any partition of ,

Hence if is integrable, then is also integrable.

Since for all ,

by (1) we have

Hence .
We have

Hence we only need to show that being integrable implies that is integrable.

For any partition of ,

Hence if is integrable, then is also integrable.

4.4 Integration and Differentiation

One have probably heard that integration and differentiation are inverse operations. We will make this precise in this section.

We will think about integral of in this section as

If we want to be differentiable, it must be continuous:

Proposition 4.20 (Integration Is Continuous)

Let be Riemann integrable, ans let . Then is continuous in .

Proof.

Theorem 4.21 (Fundamental Theorem of Calculus, Part 1)

If is Riemann integrable and continuous at , then is differentiable at , with

Proof. We will use -definition of integrability, i.e. and we want to show that as .

Estimating numerator, we have

Example 4.22

Take with

is integrable, with

which is not differentiable at .

Hence the condition of continuity at is necessary in Theorem 4.21.

Corollary 4.23

If is continuous on , then

Proof. by Theorem 4.21, so by Mean Value Theorem 3.9, is constant, i.e.

Theorem 4.24 (Fundamental Theorem of Calculus, Part 2)

If is Riemann integrable, and there exists differentiable such that , then

Proof. By assumption, , partition of such that .

Applying Mean Value Theorem 3.9 to on intervals of this partition,

Since is integrable, , hecne .

Lecture 19 · 2026-03-07

Now we shall derive some common consequences.

Proposition 4.25 (Integration by Parts)

Suppose . Then

Proof. By Fundamental Theorem of Calculus, Part 2 4.24 and product rule,

Integrate in and using FTC, the result follows.

Proposition 4.26 (Integration by Substitution)

Let be continuous and let with , and , . Then

Proof. Let , then is well-defined and differentiable by Theorem 4.21. Set . Then is differentiable:

Hence,

Theorem 4.27 (Taylor's Theorem: Integral Remainder)

Suppose . Let be as before. Then

Remark. By Extreme Value Theorem 2.21, . Thus, by Taylor’s Theorem: Integral Remainder 4.27,

Hence as for fixed .

If we knew , then we would get

for all , and this would mean to be analytic at .

Remark. Taylor’s Theorem: Integral Remainder 4.27 generalises to , where contains the line segment (see Example Sheet 3). The reason we don’t further explore this is that in , differentiability implies smoothness, which in turn implies that they are analytic [see IB Complex Analysis]. This comes with estimates on as a function of , that then one can plug into the above to get convergence of to as for some .

Proof. Using Integration by Parts 4.25,

We would use Taylor’s Theorem: Integral Remainder 4.27 to give an alternative proof of Taylor’s Theorem: Cauchy Remainder 3.23 and Taylor’s Theorem: Lagrange Remainder 3.22. To do this, we would need to mean value theorem for integrals.

Proposition 4.28 (Cauchy Mean Value Theorem for Integrals)

Let be continuous, and for all , then such that

Proof. Apply Cauchy Mean Value Theorem 3.15 to and , we get such that

Proof of TT: Lagrange Remainder 3.22. Assuming continuity of , let , then such that

Proof of TT: Cauchy Remainder 3.23. Take .

4.5 Improper Integrals

In this section, we will integrate functions of unbounded domain and unbounded image.

Definition 4.29 (Improper Integrals: Unbounded Domain)

Suppose is integrable on for every , and set

Then we say that exists (converges) if

then we set . Otherwise, we say that does not exist.

If is such that and , then we say exists, and set

Remark. This is different from

See further discussion in Example Sheet 4.

Example 4.30

exists iff .
exists assuming knowledge of normal distribution.

Lecture 20 · 2026-03-10

We need some convergence tests to have a proper proof for the last example.

Proposition 4.31 (Comparison Test for Integrals)

If satisfy for all , then

converges implies converges, and .
diverges (to ) implies diverges to .

Proof.

Let , note that this is increasing since . It is also bounded, since

Since is monotone and bounded, let exists. We claim that

Indeed, by the definition of supremum, such that for ,

Hence,

Taking limits to infinity,
Since , necessarily. Hence , such that . But

Hence diverges to .

Example 4.32

If , . Hence using properties of exponentials. Thus,

Indeed, converges.

Proposition 4.33 (Ratio Test for Integrals)

Let satisfy , and

then

Proof.

Exercise. This is an application of the comparison test. Proof left as an exercise.

Remark. If , then for very large , hence by Comparison Test for Integrals 4.31 from the large onwards,

If , then for very large , and similarly by Comparison Test for Integrals 4.31 from the large onwards,

Example 4.34

For , we have and . By the remark,
For converges since

and by we have .

Exercise. On Example Sheet 4, there are examples of root test, Dirichlet test for integrals. Is a version of the -th term test for integrals true or false?

We shall now consider improper integrals with bounded domain but isolated singularity.

Definition 4.35 (Improper Integrals: Isolated Singularity)

Let be integrable on for any . Set

Then, we say exists (converges) if exists and is finite, and we say

Otherwise, we say it does not exist (converge).

If is such that and , then say

Remark.

In general, this is not equal to

This can be seen by taking .

Example 4.36

converges iff , since as
Consider

which converges iff .

Actually, we can always reduce, by substitution, an improper integral of function with isolated singularities to improper integrals with unbounded domain.

Lemma 4.37

Let be Riemann integrable on for all . Choose on strictly decreasing bijection with , . Then

Lecture 21 · 2026-03-12

5 Sequences and Series of Functions

We now have sufficient tools to revisit and generalise notions in Section 1.

5.1 Introduction

Definition 5.1 (Sequence of Functions)

A sequence of complex-valued functions on a set is an enumerated list where each element is a function .

If, for each , the numerical sequences converges, we can define a function such that

Definition 5.2 (Series of Functions)

Let be a sequence of complex-valued functions. We call the enumerated sum

a series of complex-valued functions on .

If, for each , the numerical series converges, we can define a function such that

Since we have discussed continuity, differentiability and integrability of functions, it is natural to ask whether these properties are preserved under limits.

For the easiest case of continuity we are asking whether

Example 5.3

Swapping limits is not trivial. Consider .

Fixing , and hence = 1.
Fixing , and hence = 0.

Remark. For a general and formal treatment on this issue, see Part IB Analysis II.

5.2 Basics on Power Series

Definition 5.4 (Power Series)

For and , we say

is a power series with centre and coefficients .

Example 5.5

The Taylor series of a smooth function around is a power series, where

Tautologically, any power series converges at its centre. We would like to consider if

has any points other than . [Otherwise we can define a function by only at , which is not very interesting.]

Lemma 5.6

If converges for some and , then converges absolutely.

Proof. Fix such that converges, and let be such that . Then since converges, by -th term test 1.27 we have .

Hence is bounded, so there exists such that

Thus

where . Since converges, by comparison test converges.

Definition 5.7 (Radius of Convergence)

Let be a power series. Then is called the radius of convergence of the power series, if

converges absolutely for all such that .
if , then diverges for all such that .

Remark. On [either 2 points if over , or a circle if over ], this definition does not enforce anything.

Moreover, if , the power series only converges at ; if , the power series converges absolutely for all .

Proposition 5.8

Every power series has a radius of convergence.

Proof. Define

Clearly and is hence non-empty.

If is unbounded, we set ; by Lemma 5.6, not only , there exists with such that converges, but in fact converges absolutely for all .

Otherwise, if is bounded, let . Then

diverges for by the definition of
if , then with . Hence with , such that converges. By Lemma 5.6, converges absolutely.

To compute the radius of convergence, we can use our usual tests for series.

Proposition 5.9 (Root Test for Power Series)

Let be a sequence in such that

exists. Then the power series has radius of convergence

with the convention that if and if .

Proposition 5.10 (Ratio Test for Power Series)

Let be a sequence in such that

exists. Then the power series has radius of convergence

with the convention that if and if .

Exercise. Show that

implies .
implies .

Example 5.11

converges on all of , since
converges only for .
has and converges absolutely at .

At , we have .
has , but behavior at is more subtle.

If , then it diverges as harmonic series.

If , then

Hence

By taking the limit as , observe that converges absolutely, so converges absolutely for all with and .

Lecture 22 · 2026-03-14

Remark. The set

can be bigger or equal to

5.3 Term-by-Term Operations on Power Series

Since are polynomials, they are hence continuous, integrable and differentiable. It is natural to consider whether possible to conclude something about the continuity, integrability, differentiability of

And if so, whether

Example 5.12

converges on all of .

which again converges on all of . This seems to be consistent.
converges on

which does not converge on all of , but it converge on This hints that the argument above is not totally correct.

We need to be careful that term-by-term operations will not hold on the entire set of convergence of power series, but we will show that they do within the radius of convergence.

Proposition 5.13 (Continuity of Power Series)

Let

Let have a radius of convergence . Then

is continuous inside for every .

Proof. [Non-examinable.]

Let

and note the tail estimate

By the definition of radius of convergence, such that

Take . We want to show that it is continuous at . We have

To conclude the proof, choose so that

which is possible because is a polynomial.

Proposition 5.14 (Integration of Power Series)

Let have a radius of convergence . Then

is integrable on for every , and

Proof. [Non-examinable.]

We use the same split

Then

if we choose large enough. Hence

Proposition 5.15 (Differentiation of Power Series)

Let and as in the previous proposition. Then is differentiable on , and

Proof. [Non-examinable.]

If has a radius of convergence , then is continuous on for every by Proposition 5.13, and by Proposition 5.14 we have

Hence, by Fundamental Theorem of Calculus 4.21, we have

It remains to show that new series has radius of convergence . Take , then we have some . THen

Since , by the definition of radius of convergence. Hence such that for all . Thus

Now

Hence,

Example 5.16

has . So
has . For , we have

Lecture 23 · 2026-03-17

5.4 Exponential and Logarithms

Exercise. Using familiar properties of , namely , to show that they Taylor series of at is, with radius of convergence ,

More ambitiously, Taylor’s theorem says that

For fixed ,

Hence for all .

Lemma 5.17

We define

with radius of convergence .

With this definition,

is smooth with ,
,
.

Proof. (1) and (2) are immediate from what we know about power series. For (3), let

Then

Hence is constant on , and thus

Lemma 5.18

Consider , the restirction of to the real axis. Then

is smooth with ,
,
for all ,
is strictly increasing,
as , and as ,
is a bijection.

Proof. (1) and (2) follows immediately from Lemma 5.17.

For , every term in the series is non-negative, so

For , . Hence . Then

It follows that

gives that is strictly increasing, and that is injective.
given = , there are such that . By Intermediate Value Theorem 2.22, such that . Hence is surjective.

Thus we have proved all statements.

Since is a bijection, it must have an inverse. We will call it .

Lemma 5.19

For as defined above, we have

is a bijection with for all , for all .
is smooth and monotone with .
and .
as , and by .

Proof. (1) follows from the definition of inverse. It also gives us that such that

where , and .

Using Inverse Function Theorem 3.13, we get that is smooth, and .

Since in the domain of , is monotone.

Since , by Fundamental Theorem of Calculus 4.21,

Exercise. The final part is left as an exercise.

Remark.

One can push this, in Part IB Analysis II, to show that

We shall define

[The aim is to show that .]

Lemma 5.20

Let , . Then

, = 1.

Proof. (1) and (2) follow by group isomorphism properties of and . (4) is clear from the analogous statements for and . Now for (3),

Corollary 5.21

Take , then

,
by (3),
, and so ,
.

We have that for . Thus

This allows us to identify and use standard notation. We shall write from now on.

Proposition 5.22 (Exponentials, Powers and Logarithms)

as ,
as ,
as .

Proof.

Let ,

for any . Choose . Then
, and any . Pick . Then

Lecture 24 · 2026-03-19

5.5 Trigonometric Functions

With A-Level knowledge about and , we can calculate the Taylor series of them at . Moreover, using Taylor’s theorem, we can show that for all ,

Lemma 5.23

We define

Note that both series have radius of convergence . We will show that and are the complex cosine and sine functions, respectively.

With this definition, for ,

Proof. All statements follow from the definition and properties of . For example, for (4),

Taking , we get (5).

Remark. From (5), ,

Note that this is generally not true for complex .

Proposition 5.24 (Periodicity of Trignometric Functions)

There is a smallest positive such that , and

are periodic with period , i.e.
for all .
for all .

Proof. Once we have existence of smallest such that , the rest of the statements follow from Lemma 5.23.

Let us find the smallest such . Starting from looking at , we have

Thus is strictly decreasing on , and thus it has at most one root in . To see the existence of the root,

By Intermediate Value Theorem 2.22, there is a root in . Now

But since we get , so .

Corollary 5.25

The function for is periodic with period
[c.f. Euler’s identity ]
.

Finally, we need to relate to the more familiar .

Lemma 5.26

is the perimeter of the unit circle,

Proof. If we can show with is a bijection, then we get that

To show that is a surjection, take written as with , and . We have

is continuous and strictly monotone, hence with .
is non0negative, hence . Then if , we get . If , we get .

Hence for every , there is some such that .

To show that is an injection, since has least positive positive , it is injective when restricted to .

Hence , and for all . We can now define etc.

We can further define, for ,

Proposition 5.27

For ,

Exercise. The proof is left as an exercise. It should be similar to the proof of Lemma 5.23.