Analysis I We have been going places in the car of calculus for years, but this analysis course is about how

the car actually works.

Copier’s Message These notes may contain errors. In fact, they almost certainly do since they were just copied

down by me during lectures and everyone makes mistakes when they do that. The fact that I had to type pretty fast to keep up with the lecturer didn’t help. So obviously don’t rely on these notes.

If you do spot mistakes, I’m only too happy to fix them if you email me at mdj27@cam.ac.uk with a message about them. Messages of gratitude, chocolates and job offers will also be gratefully received.

Whatever you do, don’t start using these notes instead of going to the lectures, because the lecturers don’t just write (and these notes are, or should be, a copy of what went on the blackboard) – they talk as well, and they will explain the concepts and processes much, much better than these notes will. Also beware of using these notes at the expense of copying the stuff down yourself during lectures – it really makes you concentrate and stops your mind wandering if you’re having to write the material down all the time. However, hopefully these notes should help in the following ways;

you can catch up on material from the odd lecture you’re too ill/drunk/lazy to go to;

you can find out in advance what’s coming up next time (if you’re that sort of person) and the general structure of the course;

you can compare them with your current notes if you’re worried you’ve copied something down wrong or if you write so badly you can’t read your own handwriting. Although if there is a difference, it might not be your notes that are wrong!

These notes were taken from the course lectured by Dr Paternain in Lent 2010. If you get a different lecturer (increasingly likely as time goes on) the stuff may be rearranged or the concepts may be introduced in a different order, but hopefully the material should be pretty much the same. If they start to mess around with what goes in what course, you may have to start consulting the notes from other courses. And I won’t be updating these notes (beyond fixing mistakes) – I’ll be far too busy trying not to fail my second/third/ th year courses.

Good luck – Mark Jackson

Schedules These are the schedules for the year 2009/10, i.e. everything in these notes that was

examinable in that year. The numbers in brackets after each topic give the subsection of these notes where that topic may be found, to help you look stuff up quickly.

Limits and convergence (1)

Sequences (1.1) and series (1.3) in and . Sums, products and quotients (1.1.2). Absolute convergence (1.5); absolute convergence implies convergence (1.5.1). The Bolzano-Weierstrass theorem (1.1.3) and applications (the General Principle of Convergence) (1.2). Comparison (1.4.1) and ratio (1.4.2) tests, alternating series test (1.4.5).

Continuity (2)

Continuity of real- and complex-valued functions defined on subsets of and (2.1). The intermediate value theorem (2.3). A continuous function on a closed bounded interval is bounded and attains its bounds (2.4).

Differentiability (3)

Differentiability of functions from to (3.1.1). Derivative of sums and products (3.1.2). The chain rule (3.1.3). Derivative of the inverse function (3.1.2). Rolle’s theorem (3.2.1); the mean value theorem (3.2.2). One-dimensional version of the inverse function theorem (3.3). Taylor’s theorem from to ; Lagrange’s form of the remainder (3.5.1). Complex differentiation (3.6). Taylor’s theorem from to (statement only).

Power series (4)

Complex power series and radius of convergence (4.1). Exponential (4.3), trigonometric (4.4) and hyperbolic (4.5) functions, and relations between them. *Direct proof of the differentiability of a power series within its circle of convergence* (4.2).

Integration (5)

Definition and basic properties of the Riemann integral (5.1, 5.3). A non-integrable function (5.1.2). Integrability of monotonic functions (5.2.2). Integrability of piecewise-continuous functions (5.2.4). The fundamental theorem of calculus (5.4.1). Differentiation of indefinite integrals (5.4.2). Integration by parts (5.4.3). The integral form of the remainder in Taylor’s theorem (5.5.1). Improper integrals (5.6).

Contents 1. Limits and Convergence ...................................................................................................................... 3

1.1 Sequences ..................................................................................................................................... 3 1.2 Cauchy sequences ......................................................................................................................... 4 1.3 Series ............................................................................................................................................. 5 1.4 Convergence tests ......................................................................................................................... 6 1.5 Absolute convergence................................................................................................................... 9

2. Continuity .......................................................................................................................................... 10 2.1 Definitions and basic results ....................................................................................................... 10 2.2 Limits of functions ....................................................................................................................... 11 2.3 Intermediate value theorem ....................................................................................................... 12 2.4 Cofu clobi ibaatib ........................................................................................................................ 12 2.5 Inverse functions ......................................................................................................................... 13

3. Differentiability ................................................................................................................................. 13 3.1 Basic calculus .............................................................................................................................. 13 3.2 The mean value theorem ............................................................................................................ 16 3.3 Inverse rule (Inverse function theorem) ..................................................................................... 17 3.4 Cauchy’s mean value theorem .................................................................................................... 18 3.5 Taylor’s theorem ......................................................................................................................... 18 3.6 Some comments on differentiability of functions ..................................................... 21

4. Power series ...................................................................................................................................... 21 4.1 Radius of convergence ................................................................................................................ 21 4.2 Differentiability of power series ................................................................................................. 23 4.3 The standard functions: exponentials......................................................................................... 24 4.4 Trigonometric functions .............................................................................................................. 26 4.5 Hyperbolic functions ................................................................................................................... 28

5. Integration ........................................................................................................................................ 28 5.1 Riemann integration ................................................................................................................... 28 5.2 Integrability of monotonic and continuous functions ................................................................ 29 5.3 Elementary properties of the integral ........................................................................................ 32 5.4 Integration rules and tools .......................................................................................................... 33

5.5 Taylor’s theorem revisited .......................................................................................................... 35 5.6 Infinite integrals (Improper integrals) ......................................................................................... 37

1. Limits and Convergence

1.1 Sequences

1.1.1 Review from Numbers and Sets

Let be a sequence of real numbers.

Definition. as if given ,

Note. in most cases.

We say that is increasing if , that is decreasing if , that is strictly increasing if , and that is monotonic if it is either increasing or decreasing.

1.1.2 The Fundamental Axiom and some properties

Fundamental Axiom of the Real Numbers. Suppose , and , and is increasing. Then as . In other words, an increasing sequence bounded above converges.

Notes. 1) We could have phrased the Axiom with decreasing sequences bounded below.

2) The Axiom is equivalent to the fact that every non-empty set of real numbers bounded above has a least upper bound (supremum).

Lemma 1.1. (i) The limit is unique. That is, if and , then .

(ii) If as and , then as (subsequences

converge to the same limit).

(iii)

(iv) If and , then

(v) If and , then .

(vi) If , and , then .

(vii) If and , then .

Proof. (i) , . Similarly, , . Using the triangle inequality, for , we obtain for . If , just take so which is absurd. Thus .

(ii) , . For , . Thus

,

that is, as .

(v) Definitions as before. By triangle inequality,

If , then

Lemma 1.2. as .

Proof. The sequence is decreasing and bounded below . The Fundamental Axiom . Now we claim that .

by Lemma 1.1(iv). But is a subsequence of . By Lemma 1.1(ii), . By uniqueness of limit,

Remark. If we were instead considering sequences of complex numbers, , we could give essentially the same definition of limit; as if given , , where here stands for the modulus of a complex number.

All properties of Lemma 1.1 carry over to except the last one (we have no order in ).

1.1.3 The Bolzano-Weierstrass Theorem

Theorem 1.3 (Bolzano-Weierstrass Theorem); If and , then we can find and

as . (Every bounded sequence has a convergent

subsequence).

Remark. We say nothing about uniqueness of . E.g. has and .

Proof. Let and . Then either

1) for infinitely many values of , or

for infinitely many values of .

In case 1), set In case 2), set . Proceed inductively to obtain sequences and such that the following holds;

(i)

(ii) for infinitely many values of

(iii)

Then is a bounded increasing sequence and is a bounded decreasing sequence. By the

Fundamental Axiom, and . By property (iii), passing to the limit,

Having selected such that , select such that

. We

can always do this because for infinitely many values of . In other words,

. Since and ,

.

This argument is called the ‘bisection method’ or ‘lion hunting’.

1.2 Cauchy sequences

Definition. A sequence is a Cauchy sequence if given , . As before,

Lemma 1.4. A convergent sequence is a Cauchy sequence.

Proof. and . This means , . . So if we have .

As an application of the Bolzano-Weierstrass theorem, we now show that the converse is true;

Lemma 1.5. A Cauchy sequence is convergent.

Proof. First we’ll prove that if is a Cauchy sequence, then it is bounded.

Cauchy means that , . Choose . Then . In particular, .

.

Take for . For this choice of , .

Now by the Bolzano-Weierstrass theorem, has a convergent subsequence . Now we

show that in fact .

. Choose large enough so that

(since Cauchy, ) and (since

). Thus .

Thus, for sequences of real numbers, convergence and Cauchy property are equivalent. This is called the General Principle of Convergence.

1.3 Series

1.3.1 Convergent and divergent series

We begin with some generalities.

Definition. or . We say that converges to if the sequence of partial sums

tends to as . In that case we wrire

. If does not converge, we

say that diverges.

Remark. Any question about series is really a question about the sequence of partial sums.

Lemma 1.6 (i) If and

converge, then so does

where .

(ii) Suppose . Then either and

both converge or both

diverge (i.e. initial terms do not matter).

Proof. (i)

If and , then clearly by 1.1.

1.3.2 Convergent series and decreasing terms

Lemma 1.7. If the series converges, then .

Proof. . If the sum to infinity converges, then . Therefore .

The converse is not true!

Example. Consider the series

Then , but the series diverges. This can be shown as follows;

so . So if converges, then , but since , in the

limit we get which is a contradiction.

1.3.3 Geometric series

Consider

If ,

If , the series clearly diverges. First note that . So, if ,

In conclusion, converges iff , and if then

.

1.4 Convergence tests The first four tests in this section are for series of positive terms.

1.4.1 Comparison test

The most powerful test for convergence comes out straight from the fundamental axiom.

Theorem 1.8 (Comparison test). Suppose . Then if converges, so does

.

Proof. Let

Now , and the same for , so and are increasing sequences. If , then , so it follows that if

converges, then and thus, since ,

then is bounded above. By the fundamental axiom, has a limit.

Remark. Since initial terms do not matter for convergence, in the theorem it is enough to assume that .

1.4.2 Root and ratio tests

We will now derive two applications of the comparison test, the root and ratio tests.

Theorem 1.9 (Root test, Cauchy). If , suppose

as . Then if ,

converges, and if , diverges.

Remark. If , the test does not give any information. In fact, later on, we’ll see examples with which are convergent and others which are divergent.

Proof. Assume that

with . Take such that . Then

means

that given , ,

, or in other words

. Now take

, so that

. But since the geometric series converges for , the comparison test tells us that converges.

Suppose now that

with . By the definition of a limit, ,

. Thus ; in particular, does not tend to . So by Lemma 1.7, diverges.

Example. Consider

Since , by the root test, converges.

Theorem 1.10 (Ratio test). Suppose and . If , then converges. If , then diverges.

Remark. Again, the test is inconclusive if .

Proof. Suppose with . Take . As in the root test, let , so that , we have . Now write

Using the above result we get for . In other words, there is a constant (independent of ) such that for . Since , the series converges, and by the comparison test, also converges.

Now suppose with . From the definition of limit, , . This is saying that the sequence increases after . In particular, . But this clearly implies that , and thus must diverge.

1.4.3 Some inconclusive examples

Examples. (i) We know

diverges, but both tests are inconclusive.

From the root test, because because . [This can be proved as

follows. Let with , so

by the binomial expansion. Thus

Hence .] Applying the ratio test gives so also.

(ii) Try both tests on

Ratio test;

Root test;

So for this series it also holds that . However, we can show that the series converges.

Thus the series

converges, and by the comparison test, converges.

This shows that if and/or we can’t conclude anything.

(iii)

By the ratio test,

We conclude from the ratio test that the series converges.

1.4.4 Cauchy condensation test

Theorem 1.11 (Cauchy condensation test). Let be a decreasing sequence of positive terms. Then

converges iff

converges.

Proof. Note that for and .

Suppose first that converges. Then

, which by is

, i.e.

Hence

Multiply by to get

because converges. Since the sequence of partial sums is bounded, converges.

Suppose now, conversely, that converges. Then

Thus

because converges. Therefore

is bounded in , so converges. QED.

Example. Consider

for ; if , then . Now is decreasing, so let’s investigate .

which is a geometric series, so converges iff . By the condensation test,

converges iff .

1.4.5 Alternating series test

Theorem 1.12 (Alternating series test). If and , then

converges.

Example. converges, because and is decreasing.

Proof. , so

In both expressions, all terms are since is decreasing. Thus is an increasing sequence bounded above. Now by the fundamental axiom, . But now . Since , .

Since , given , , . And since , given , , . Therefore , . Hence converges.

1.5 Absolute convergence Definition. Take or . If is convergent, then the sequence is called absolutely

convergent.

Note. is convergent but not absolutely convergent.

1.5.1 Absolute convergence implies convergence

Theorem 1.13. If is absolutely convergent, then it is convergent.

Proof. Suppose first that . Introduce

Note that , , and .

Since converges, by the comparison test and also converge, and since , must also converge.

If , write and note that . So, if converges, by the comparison test, and converge. In other words, and converge absolutely, therefore and converge. Since , converges.

Remark. There is an alternative ‘quick’ proof of this using Cauchy sequences. First define

For ,

So convergent Cauchy is Cauchy is convergent.

1.5.2 Conditional convergence and weird sums

Definition. If converges but does not, then we say that is conditionally convergent. The word ‘conditionally’ is used because in this case, the sum to which the series converges is conditional on the order in which the terms are taken. E.g.

converge to different limits ( and respectively).

A rearrangement of is , a bijection, taking .

Theorem 1.14. If is absolutely convergent, then every series consisting of the same term in any order (i.e. a rearrangement) has the same sum.

Proof. We prove this for ; the extension to is an exercise.

Let be a rearrangement of . Let

Suppose we are given a fixed . Then such that contains every term in , so

( ). By the fundamental axiom, with . But by symmetry,

.

If has any sign, consider and from the proof of 1.13, and ,

, . Since

converges, both and converge, and now use the case to conclude that

and , and the result follows. QED.

2. Continuity

2.1 Definitions and basic results

2.1.1 Two definitions of continuity

where and (also applies of course to ). Usually will be some kind of interval. E.g.

Definition 1. For , we say is continuous at if given any sequence , then .

Definition 2. For , we say is continuous at if given , such that if and , then . ( - definition).

Proof that the two definitions are equivalent;

D2 D1; We know that given , such that if and , then . Take with , then it is required to prove that . Since , , . But this implies , i.e. .

D1 D2; Suppose D2 not true. Then we can find with and . Choose , thus there is with and . , but . So does not tend to , which contradicts D1.

2.1.2 Sums, products, multiples and reciprocals

Proposition 2.1. Let , and with both continuous at . Then , , for any constant are also continuous at . In addition, if , then is also continuous at .

Proof. This is a direct consequence of Definition 1 and Lemma 1.1 (about sequences). For example, to show that is continuous at , we take . Since and are continuous at , and . Lemma 1.1 implies ; that is, continuous at . Similarly with the other claims.

Consequence. is clearly continuous. By Proposition 2.1, any polynomial is continuous at every point of . Quotients of polynomials (rational functions) are continuous at every point where the denominator does not vanish.

Terminology. We say that is continuous in if it is continuous at every point in .

2.1.3 Compositions

We now look at compositions.

Theorem 2.2. Let and be two functions such that ( and are subsets of ). Suppose is continuous at and is continuous at , then

is

continuous at .

Proof. Take with . Then it is required to prove that . Since

is continuous at , . Call . Since is continuous at ,

.

Remark. Of course, one can prove this with the - definition.

2.1.4 Examples

Examples. (1)

(assuming continuous). At , is continuous by 2.1 and 2.2. However is not continuous at . Take

so . Also , but . Thus does not tend to zero.

(2)

For , is continuous (same as before). Now let . from definition of . But so so is continuous at .

2.2 Limits of functions . We’d like to make sense of even when may not be in .

E.g.

but . Similarly if e.g. .

Definition. , . Take and assume that there exists a sequence , , . (Note the point may be in but need not be in .) We say that (or as ), if given , such that whenever and , then .

Remarks. 1) iff for every sequence , , we have . The proof of this is exactly as the proof that Definition 1 Definition 2 from the last section.

2) If , then iff is continuous at . There is nothing to prove here really, it follows straight from the definitions.

This limit enjoys the properties which one would expect, i.e.

(i) the limit is unique

(ii) if as and as , then , and if .

We will start by looking at continuous.

2.3 Intermediate value theorem Theorem 2.3 (Intermediate value theorem). Let . Then takes every value which

lies between and .

Proof. Without loss of generality, assume . Take and let . Now , because , so is bounded above (by , in fact). Thus has a supremum, .

If , then obviously holds. If , use the following argument. (Alternatively, by continuity of at , .)

Given a positive integer , is not an upper bound for , otherwise it would contradict the definition of supremum. Then such that , so . Now . So as , so by continuity of , . Thus .

Note that , otherwise , but . Thus, for all sufficiently large, , and . Again by continuity of , . Since , . In the limit, . Hence .

Application. We can show the existence of the th root of a positive number , where is a positive integer. Look at , , which is continuous. We look at in . Now . By the intermediate value theorem, , thus . This means that has a positive th root.

In fact, this th root is unique, since if is another positive th root with , then without

loss of generality , so , so which is a contradiction. So definitely exists.

2.4 Cofu clobi ibaatib If continuous,

Theorem 2.4. such that .

Theorem 2.5. such that .

In words, a continuous function on a closed bounded interval is bounded and attains its bounds.

Note. on is not bounded.

Proof (2.4). Suppose the statement is not true; then for any positive integer , . By Bolzano-Weierstrass, has a convergent subsequence

, with

. Now since is continuous, so

, but

, which is a contradiction.

Proof (2.5). Let . Then by 2.4. Now , so by the definition of supremum, such that . By Bolzano-Weierstrass, has a convergent subseequence

, and . But

. Let , then

, so by continuity of . Similarly

with and the infimum.

Alternative proof. Let . Work by contradiction and suppose . Now consider

in . Now is continuous, since is. By 2.4 applied to , . Since is positive, . Then , i.e. , so that is an upper bound for the set . This contradicts the definition of , since .

2.5 Inverse functions Definition. is said to be increasing if

and strictly increasing if . Similarly for decreasing and strictly decreasing. A function is monotone if it’s either increasing or decreasing.

Theorem 2.6. Let be continuous and strictly increasing. Let and . Then is bijective, and the inverse is also continuous and strictly increasing.

Proof.

is injective; If then . Otherwise, if , , or if , . ( strictly increasing.)

is surjective; take . By the intermediate value theorem, .

So is bijective and has an inverse .

is strictly increasing; take , , . If , then since is increasing, , i.e. .

is continuous; given , let and . Then and for any . Take For this , if then . This was for , but a similar argument gives continuity at the end points also. QED.

3. Differentiability

3.1 Basic calculus

3.1.1 Differentiability

Let be a function. Mostly we’ll be looking at the case where is an interval on and is real-valued.

Take , , .

Definition. is differentiable at with derivative if

is differentiable on if it is differentiable at every point in .

Remarks. (i) We could also write

with .

(ii) Consider

Then . Also . Thus an equivalent way of saying that is differentiable at with derivative is to say that there is a function such that , with . is a linear function in .

Other ways of writing the same thing; , with as . Also with .

(iii) If is differentiable at then it is continuous at ; suppose with . Then

i.e. is continuous at .

Example. Take , . If ,

If , then is differentiable, but . But at , is not differentiable. Does

exist?

is continuous at , therefore .

3.1.2 Constants, sums, products and reciprocals

Proposition 3.1. (i) If , then is differentiable at with .

(ii) If differentiable at , then so is , and .

(iii) If differentiable at , then so is , and .

(iv) If is differentiable at and , then is differentiable at and

Remark. From (iii) and (iv) we get

Proof. (i)

(ii)

(iii) Let . Then

As , and this becomes .

(iv) Let . Then

By continuity, this

Examples. (i) . Then

(ii) , with a positive integer. Write it as . Then

by induction.

(iii) , for a positive integer ( ) . Using 3.1(iv) we get

All polynomials and rational functions are differentiable.

3.1.3 Compositions – the chain rule

Theorem 3.2 (Chain rule). Let which is differentiable at , and ,

and is differentiable at . Then .

Example. , , . Then , so .

Proof. differentiable at means

Similarly, differentiable at means

We substitute to obtain

Let . Then

We need to prove that . Define and . Then are

continuous at respectively. But now is continuous at because it is products, sums and compositions of continuous functions. Thus . QED.

3.2 The mean value theorem

Up to now. everything has worked for . Now we look in more detail at the case of functions .

3.2.1 Rolle’s theorem We first prove the following basic existence result;

Theorem 3.3 (Rolle’s theorem). If continuous and differentiable on , and , then

Proof. Let be the maximum value of in and be the minimum. (By Theorem 2.5, the values and are achieved.) Let . If , then is constant . Otherwise, or .

Suppose (the proof for is similar). We know by Theorem 2.5 that . We’ll prove that .

Suppose first . Since is differentiable at , we can write with . Since , for all sufficiently small. If in addition we take , then . Absurd because is the maximum of .

If , the same argument shows that there are points at left of for which is strictly bigger than , which is again absurd. Thus .

3.2.2 The mean value theorem Theorem 3.4 (Mean value theorem). If continuous and differentiable on ,

then

Remark. We can rewrite this as follows; letting , where

Proof. We consider the auxiliary function . Let’s choose . Then . So

So, for this choice of , satisfies the hypothesis of Rolle’s theorem. Hence . But , so

3.2.3 Corollaries

We have the following important corollary;

Corollary 3.5. continuous and differentiable.

(i) If , then is strictly increasing.

(ii) If , then is increasing.

(iii) If , then is constant.

Proof. (i) with . The mean value theorem for If then .

(ii) If , same argument gives for , .

(iii) Consider on the interval and apply mean value theorem to get , for

3.3 Inverse rule (Inverse function theorem)

3.3.1 Statement and proof

Theorem 3.6. continuous and differentiable in with Let and . The function is a bijection and is continuous on and differentiable on with

Proof. Since , then by Corollary 3.5 we know that is strictly increasing. By Theorem 2.6, exists and is continuous. Let . We need to prove that is differentiable with

If (small) then there is a unique such that . Note; . Writing we get

Note; if , then ( continuous at ).

3.3.2 Differentiating rational powers of

Example. , a positive integer, ( ). Then and

if .

The inverse rule is differentiable in and

Example. , any integer and a positive integer. We find using the chain

rule, as

;

In other words, if where is any rational number, then . Later on we’ll define for real and discover that .

3.3.3 Some lead up to Taylor’s theorem

with continuous on and differentiable on . The Mean Value Theorem for and for . If ,

We’ll see that we can choose .

(Taylor’s theorem).

3.4 Cauchy’s mean value theorem Theorem 3.7 (Cauchy’s mean value theorem) Suppose are continuous and

differentiable on . Then such that

Proof. Consider the function with

Then is continuous on and differentiable on . (It is just a product and sum of functions with the same property.)

because in either case, 2 columns are equal. By Rolle’s theorem . Now expand the determinant and differentiate to see that gives exactly what we want.

Example. Find

Apply CMVT on . Then

Let , then

(L’Hôpital’s rule)

3.5 Taylor’s theorem

3.5.1 Taylor’s theorem with Lagrange’s form of the remainder

Theorem 3.8 (Taylor’s theorem with Lagrange’s remainder). Suppose and its derivatives up

to order are continuous on , and exists for . Then

where .

Notes. 1) is Lagrange’s form for the remainder.

2) For this is the mean value theorem;

Proof. Consider for

is chosen so that . Then

We’ll apply Rolle’s theorem times. We apply it to first, to get an such that . Then

Apply Rolle to in to get such that .

Note that in fact from the definition of we see that

We keep applying Rolle to get at which for .

Now note . Therefore . Write

for to get . Now put this value of in the definition of and the statement is exactly the statement of the theorem. QED.

3.5.2 Taylor’s theorem with Cauchy’s form of the remainder

Theorem 3.9 (Taylor’s theorem with Cauchy’s form of the remainder). Let satisfy the same hypothesis as in Theorem 3.8 and in addition suppose (this is just to simplify matters). Then

where

Proof. Define, for ,

Set

Then

(from definition of ).

Rolle’s theorem applied to gives such that . Now we compute

Rearranging

Set

Note. Setting gives

which is Lagrange’s form of the remainder.

3.5.3 Taylor series

Needs as

Remarks. 1) is just to simplify matters

2) The same result holds in an interval .

3) is the ‘Maclaurin expansion’.

3.5.4 Application to the binomial series

Application. Binomial series; . Then .

Claim;

for , where

is the generalised binomial coefficient. If is a positive integer, then for .

Let’s prove that for the series is absolutely convergent. Ratio test;

So, the series converges absolutely for . In particular, , i.e. as

for .

Now we study the remainders in the Taylor theorem.

missing line

We look at Lagrange’s form;

If , if if . Let

to conclude that .

For , the argument breaks, so Cauchy to the rescue!

Note that

Careful, may depend on . If , . If ,

. In

any case you get

where is a constant that depends on and but not . Therefore as .

3.6 Some comments on differentiability of functions

Standard properties work for both and ; chain rule, the sums and products of differentiable functions are differentiable.

Example. ,

So

does not exist!

, , is differentiable with ‘real glasses’

Complex differentiable functions are called holomorphic.

4. Power series We will look at series of the form

4.1 Radius of convergence

4.1.1 Convergence of power series

Lemma 4.1. If

converges, and , then converges absolutely.

Proof. Since converges,

as . In particular, there is a constant such that

.

since so . Thus the geometric series

converges. By comparison,

converges, i.e. converges absolutely.

Theorem 4.2. A power series either

(i) converges absolutely for all , or

(ii) converges absolutely for all inside a circle and diverges for all outside it, or

(iii) converges for only.

Definition. The circle is called the circle of convergence, and the radius of convergence. In case (i) we agree that , and in case (iii) we agree that .

Proof. Let and converges . Now . If , then by Lemma 4.1, .

If is unbounded, then and we have case (i).

If is bounded, there exists a finite supremum for which we call . If , we’ll prove that if , then

converges absolutely.

Choose such that . Then converges, and by Lemma 4.1,

converges absolutely.

Finally we show that if , then diverges. Take such that . If

converges, again by Lemma 4.1,

converges. But this contradicts the definition of as supremum of . Thus

diverges.

4.1.2 Computing the radius of convergence

The next lemma is useful in computing ;

Lemma 4.3. If as , then .

Proof. By the ratio test we have absolute convergence if , i.e.

. And if , then

and does not tend to zero. Therefore .

Remark. By the root test, if then .

Examples.

(geometric series). We know , but note that if , the series diverges, since which does not tend to .

as , but what if ? If , diverges (Example Sheet 1).

Abel’s test; , , is bounded

bounded in . Therefore converges for and .

has but converges for every .

, but on , we have divergence.

Conclusion In general nothing can be said at .

4.2 Differentiability of power series

Theorem 4.4. Suppose has radius of convergence , so that

for with . Then is differentiable and

Remark. Iterate this theorem, to get that can be differentiated infinitely many times as if it were a polynomial.

Proof (non-examinable). We’ll use two auxiliary lemmas.

Lemma 4.5. If has radius of convergence , then so do and .

Lemma 4.6. (i)

(ii) .

Proof of 4.4. By Lemma 4.5, since has radius of convergence , it defines a function for with . Then we would like to show that

This implies that is differentiable with .

By Lemma 4.6, .

Take . If , we get

By Lemma 4.5, we know that

converges to some number . Therefore

Proof of 4.5. has radius of convergence . Then has radius of convergence and it is required to prove that .

Take with . Choose . Since , as . In

particular, .

But

converges. Ratio test;

By comparison, converges absolutely. Therefore .

But in fact we have equality because

So by comparison if converges absolutely, so does . Therefore .

What we proved also implies that also has radius of convergence of .

Proof of 4.6. (i)

(ii)

by the binomial theorem. Therefore

Example.

We saw last time that it has radius of convergence . Then with defined to be

The theorem we just proved tells us right away that is differentiable and

4.3 The standard functions: exponentials

General remark. Let be differentiable. If , then is constant.

Proof. Let , . Write ,

By the chain rule, . But we also see (Corollary 3.5) constant. Therefore since .

4.3.1 The exponential function

Claim. for .

Proof. ,

By the general remark above, is constant.

From the definition of , , so

Now at , so the claim is proved.

Now we restrict and prove:

Theorem 4.7.

(i) is differentiable everywhere.

(ii)

(iii)

(iv) is strictly increasing

(v) as , as

(vi) is a bijection.

Proof. (i) and (ii) have been done.

(iii) Clearly from the definition of , if and

(iv) is strictly increasing by theorem we proved before

(v) From definition of , for , so as , .

Also, as .

(vi) is strictly increasing so is injective. For surjectivity pick . Since as and as , . The intermediate value theorem says that .

Remark. (vi) and (ii) are saying that is a group isomorphism.

4.3.2 The logarithmic function

Since is a bijection, such that and

.

Theorem 4.8.

(i) is a bijection and ,

(ii) is differentiable and .

(iii)

Proof.

(i) obvious.

(ii) is differentiable by the inverse rule (Theorem 3.6).

(iii) By groups, is a homomorphism.

4.3.3 -functions

Definition. Let , and be any positive number. Then . E.g. .

Theorem 4.9. Suppose and . Then

(i)

(ii)

(iii)

Proof. (i)

(ii)

(iii)

Let be a positive integer. Then

Let be a positive integer, then what is

We can now set for . This definition agrees with the one given before for .

We can now prove that and . Define the real number as

Then , but , so .

can be rewritten as . Then

for , by the chain rule.

Let . Then by the chain rule.

4.4 Trigonometric functions

We define

Both have infinite radius of convergence (check). , . By Theorem 4.4 they’re differentiable, and , .

4.4.1 Relation to exponential function

Now and , so

Now from definitions, and . So . Therefore

It is also obvious that and .

Now we get, using , that

for .

Also, for . If , from which it follows that and . Warning: (or ) are not bounded for , since for , with ,

4.4.2 Periodicity of the trigonometric functions

Proposition 4.10. There is a smallest positive number (where ) such that .

Proof. If , . Then

for and (since and ). Therefore for . But for is a strictly decreasing function in .

We’ll prove that , . Then by the intermediate value theorem, with

.

where each term is

Therefore . Now

where all bracketed terms are for . For ,

Therefore .

Corollary 4.11.

Proof.

But we know .

Definition. We define . Finally,

Theorem 4.12. (i)

(ii)

(iii)

Proof. Immediate from addition formulae and

Note. This implies periodicity of !

4.5 Hyperbolic functions

Definition.

From this we get , . We can check (exercise) that and , and identities such as .

Two final remarks;

1) The other trigonometric functions ( , etc) are defined in the usual way (e.g. ).

2) “Exponentials beat powers”, i.e. as for . To see this, take a positive integer such that and observe (from the definition of as a power series) that

5. Integration

5.1 Riemann integration

bounded, i.e. .

5.1.1 Dissections

Definition. A dissection (or partition) of the interval is a finite subset of which contains and . We’ll write it as

We define the upper and lower sums of with respect to , and , as

Clearly

Lemma 5.1. If and are dissections with , then

Proof. Suppose has one extra point, let’s say for some .

Note that

Then

If now has more than one extra point, just do this argument for each extra point.

; we already noted this.

The proof for is similar to the one for the upper sums. QED.

Lemma 5.2. If and are any two dissections, then

This is a key lemma for integration.

Proof. , so by Lemma 5.1, . QED.

5.1.2 Definitions of integrals and integrability

Definition. The upper integral is defined as

The lower integral is

Remark. These numbers are well-defined because is bounded, so and .

Definition. We say that a bounded function is (Riemann) integrable if . In this case we write

(or just

).

Remark. , thanks to Lemma 5.2.

and

Example. (Dirichlet) A function given by

For any , because every interval contains a rational number. However,

for any partition. Therefore and so is not integrable.

5.2 Integrability of monotonic and continuous functions

5.2.1 Riemann’s theorem

We now prove the following useful criterion for integrability;

Theorem 5.3 (Riemann’s theorem). A bounded function is integrable iff given .

Proof. Let’s assume first that is integrable.

Given , by definition of , . By definition of , . Consider . Then by Lemma 5.2,

because on account of being integrable.

To prove the converse, assume that given , . Then by definitions of and . Since this is true for all . Therefore is integrable. QED.

5.2.2 Integrability of monotonic functions

We now prove that monotonic functions are integrable and continuous functions are integrable.

Remark. Monotonic and continuous functions on are bounded.

Theorem 5.4. If monotonic, then is integrable.

Proof. Assume is increasing (same proof if is decreasing). Take to be any partition of . Then

But is increasing, so

Thus

Now consider, for a positive integer,

For this

When , take large enough so that

Therefore by Theorem 5.3, is integrable. QED.

5.2.3 Uniform continuity

Lemma 5.5 (Uniform continuity). If is continuous, then given and if then .

Remark. Continuity at a point, let’s say , means that given , . The lemma is saying that we can choose a which works for every !

Proof. Suppose claim is not true. Then there exists such that for every , we can find with but .

Take , to get with but . By Bolzano-Weierstrass,

. Then .

Let , then , so

.

On the other hand,

. continuous, therefore ,

. Then, passing to the limit, , but . Absurd. QED.

5.2.4 Integrability of continuous functions

Theorem 5.6. If continuous, then is integrable.

Proof. Consider the partition with points

for a positive integer. Then

Let be given, and consider the given by Lemma 5.5. Choose large enough such that . Then for any , Lemma 5.5 tells us that .

Therefore

By Riemann’s criterion, is integrable.

Example.

We claim that is integrable and

Take any partition, then since every interval contains irrational

numbers where is zero. Clearly .

To prove the claim, it suffices to show that given (by Riemann’s theorem and the fact that ).

Let be given, and take an integer with . Then let . Then and , so . So is a finite set with cardinality .

. Choose a partition such that the points belong to the

intervals with length less than .

First sup is and second sup is (outside ). So . QED.

5.3 Elementary properties of the integral bounded and integrable on then

(1) If on , then

.

(2) is integrable over and

.

(3) For any constant , is integrable and

.

(4) is integrable and

.

(5) The product is integrable.

(6) If except at finitely many points in , then is integrable and

.

(7) If , is integrable over and and

.

Proof (1). If , then integral is upper sum, so;

for any . Now take infimum over all , so

Proof (2). Observe

Now take any two partitions

by Lemma 5.2. Now keep fixed and take infimum on all s. Then

Now take inf over all

since integrable.

A similar argument for the lower sums gives

Putting everything together we get, since , that

.

Proof (3). Exercise.

Proof (4). Consider . Then we claim that is integrable.

(this needs checking). For any partition ,

By Riemann, since is integrable, . By Riemann again and , is integrable. Now note that

since . So ; since and are integrable by (2) and (3), is integrable.

follows from (1) and the fact that . QED.

Proof (5). First we’ll prove that if is integrable, then is integrable. Suppose first that . Since is integrable, given , a partition of such that . Let

since .

bounded, thus where

is integrable, by Riemann’s theorem.

If is now arbitrary (i.e. just integrable) then is integrable by property (4). But since is integrable.

To prove that is integrable, just write

.

Now integrable and integrable and integrable, and we’re done.

Proof (6). Let . Then for every except perhaps a finite set (just

from the definition of the integral) must be integrable with

. But is

integrable with

.

Proof (7). Exercise.

Convention. If , define

Agree that

.

5.4 Integration rules and tools

How do we compute

?

5.4.1 The fundamental theorem of calculus

Let be a bounded Riemann integrable function. For define

Theorem 5.7. is continuous.

Proof. For

by properties of integral, assuming .

is bounded so . Thus

In fact for any , . Let , therefore

Theorem 5.8 (Fundamental Theorem of Calculus). If in addition is continuous, then is differentiable and .

Proof. For , , we’d like to make

small as . We write it as

since

Now

Thus

So we have proved that

Now if , then by continuity of ,

5.4.2 Integration is the “inverse” of differentiation

Corollary 5.9. If is continuous on , then

Proof. From 5.8 we know that . Therefore is constant. Therefore

This gives a way of computing

if we know a “primitive” (anti-derivative) for , that is, a

function such that . Primitives of continuous functions always exist (5.8) and any two primitives differ by a constant.

such that

constant.

5.4.3 Integration by parts

Corollary 5.10. Suppose and exist and are continuous on . Then

Proof. Product rule gives . Therefore

and the rest follows immediately.

5.4.4 Integration by substitution

Corollary 5.11 (Integration by substitution). Let with , , and assume that exists and is continuous on . Let be continuous. Then

Set , to get a more familiar version.

Proof. Set

Let with taking values in . Then by the chain rule,

5.5 Taylor’s theorem revisited

5.5.1 Taylor’s theorem with integral form of the remainder

We now revisit Taylor’s theorem and find an integral form for the remainder.

Theorem 5.12 (Taylor’s theorem with remainder an integral). Let be continuous for , then

where

Remark. Note that here we assume continuity of and not just mere existence (as in our previous versions), so this theorem is a little weaker, but just fine for most practical purposes.

Proof. Let’s make the substitution , so that . Then

By integration by parts this becomes

That is,

Integrate by parts times to get

where the last term is . QED.

5.5.2 Link to Cauchy’s form

This integral form gives back Cauchy’s and Lagrange’s forms.

Previous remarks. continuous, then

The mean value theorem applied to gives . Thus

(mean value theorem for integrals). Let’s apply this to

which is Cauchy’s form of the remainder.

5.5.3 Link to Lagrange’s form

To get Lagrange’s form we use the following proposition. Suppose , then

for some . Then

which is Lagrange’s form of the remainder.

Now we prove the proposition we just used, namely that if continuous and , then such that

If you have then you get the mean value theorem.

Proof. Let

By the Fundamental Theorem of Calculus, and .

Cauchy’s mean value theorem (3.7) applied to and gives

Since , we simplify to get the desired result.

5.6 Infinite integrals (Improper integrals)

We’d like to make sense of

5.6.1 Definition

Suppose we have such that for every , is bounded and integrable on .

Definition. If

we say that

exists and that its value is . In this case we also say that

converges. If

does not tend to a limit, we say that

diverges.

Example.

If , the integral

Let , then we say that the limit is finite only if . If ,

5.6.2 Relation between improper integrals and series

Notice the similarity with the series

Remark. If , for and for constant, then if

converges,

then

converges and

This is the analogue of the comparison test.

Proof.

The function

is an increasing function. Thus if ,

If

converges, then

is bounded above, and so we can consider

We claim

By definition of supremum, given , such that

. If , then

i.e.

, i.e.

Remark. For series, we know that if converges, then . However, it is not true that if

converges, then as .

Example.

converges, so

converges. But , so

as .

5.6.3 The integral test

Theorem 5.13 (Integral test). Let be a positive decreasing function. Then 1)

both converge or both diverge.

2) As ,

tends to a limit as , such that .

Examples. (1)

We just say that

converges iff . We can apply the integral test to deduce that

converges iff .

2)

Let . Then

As , this goes to infinity, so the sum diverges.

Proof (1). Note that since is decreasing, it is Riemann integrable in every interval .

For , since decreasing.

Integrate between and to get

Adding,

Suppose

converges, then

is bounded in . By ,

is bounded,

and therefore it must converge by the fundamental axiom (as it is increasing in ).

Suppose converges, then

is bounded and by

is bounded, thus

since

is increasing, it follows that

converges.

Proof (2). Let

Therefore is decreasing. Also from , . Thus is a bounded decreasing sequence, and therefore it must converge to a limit with . QED.

5.6.4 Euler’s constant

Corollary 5.14 (Euler’s constant).

as with .

Proof. Let and apply the theorem.

Remarks. 1) It is still unknown whether is rational or irrational!

2)

3)

Proof that .

Define

for . We’ll derive estimate for . Let and in the fact from the last section that

for , , .

for . So

But can be easily computed by integration by parts. If you do the calculation, you get

by definition of .

5.6.5 Integrals which are improper at both ends

So far we have looked at

but we can also make perfect sense of

We can also make sense of

.

Definition.

converges if

converges and

converges. Also if

and

, then we let

.

This defintion is independent of ;

so the sum will not change, it will always be equal to .

This is not quite the same as saying that

exists.

Example.

but

does not exist, because neither

or

converge.

Example. on .

Let .

and the same for general functions. This is an improper integral of the second kind.