real calculus.pdf

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Analysis I

Course C5

T. W. Korner

March 8, 2003

Small printThe syllabus for the course is defined by the Faculty Board Schedules (which

are minimal for lecturing and maximal for examining). I should very much appreciate

being told of any corrections or possible improvements and might even part with a small

reward to the first finder of particular errors. This document is written in LATEX2e and

should be available from my home page http://www.dpmms.cam.ac.uk/twk in latex, dvi

and ps formats. My e-mail address is twk@dpmms. Some, fairly useless, comments on the

exercises are available for supervisors from me or the secretaries in DPMMS.

Contents

1 Why do we bother? 2

2 The axiom of Archimedes 3

3 Series and sums 4

4 Least upper bounds 7

5 Continuity 8

6 Differentiation 11

7 The mean value theorem 13

8 Complex variable 169 Power series 19

10 The standard functions 20

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11 Onwards to the complex plane 23

12 The Riemann integral 26

13 Some properties of the integral 30

14 Infinite integrals 33

15 Further reading 34

16 First example sheet 35

17 Second example sheet 40

18 Third example sheet 43

19 Fourth example sheet 55

1 Why do we bother?

It is surprising how many people think that analysis consists in the difficultproofs of obvious theorems. All we need know, they say, is what a limit is,the definition of continuity and the definition of the derivative. All the restis intuitively clear.

If pressed they will agree that these definitions apply as much to therationals Q as to the real numbers R. They then have to explain the followinginteresting example.

Example 1.1. Iff : Q Q is given byf(x) =1 ifx2


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interested she will find a complete list of the appropriate axioms in texts likethe altogether excellent book of Spivak [5] and its many rather less excellentcompetitors, but, interesting as such things may be, they are irrelevant toour purpose which is not to consider the shared properties ofR and Q but toidentify adifferencebetween the two systems which will enable us to excludethe possibility of a function like that of Example 1.1 for functions from R toR.

To state the difference we need only recall a definition from course C3.

Definition 1.2. If an R for each n 1 and a R then we say thatana if given >0 we can find ann0() such that

|an a|< for allnn0().

The key property of the reals, the fundamental axiomwhich makes ev-erything work was also stated in the course C3.

Axiom 1.3. [The fundamental axiom of analysis] If an R for eachn 1, A R and a1 a2 a3 . . . and an < A for eachn then thereexists ana R such thatana asn .

Less ponderously, and just as rigorously, the fundamental axiom for thereal numbers says every increasing sequence bounded above tends to a limit.

Everything which depends on the fundamental axiom is analysis, every-thing else is mere algebra.

2 The axiom of ArchimedesWe start by proving the following results on limits, some of which you sawproved in course C3.

Lemma 2.1. (i) The limit is unique. That is, if an a and an b asn thena= b.

(ii) Ifan a asn andn(1) < n(2) < n(3) . . . thenan(j) a asj .

(iii) Ifan = c for alln thenanc asn .(iv) Ifana andbnb asn thenan+bna+b.(v) Ifana andbnb asn thenanbnab.(vi) Ifana asn , an= 0 for eachn anda= 0, thena1n a1.(vii) IfanA for eachn andana asn thenaA.We need the following variation on the fundamental axiom.

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Exercise 2.2. A decreasing sequence of real numbers bounded below tendsto a limit.

[Hint. Ifab thenb a.]Useful as the results of Lemma 2.1 are, they are also true of sequences in

Q. They are therefore mere, if important, algebra. Our first truly analysisresult may strike the reader as rather odd.

Theorem 2.3. [Axiom of Archimedes]

1

n0 asn

Theorem 2.3 shows that there is no exotic real number say (to choosean exotic symbol) with the property that 1/n > for all integers n1 yet >0 (that is, is strictly positive and yet smaller than all strictly positiverationals). There exist number systems with such exotic numbers (the famousnon-standard analysis of Abraham Robinson and the surreal numbers ofConway constitute two such systems) but, just as the rationals are, in somesense, too small a system for the standard theorems of analysis to hold sothese non-Archimedean systems are, in some sense, too big. Archimedesand Eudoxus realised the need for an axiom to show that there is no exoticnumber bigger than any integer1 (i.e. > n for all integers n1; to seethe connection with our form of the axiom consider = 1/). However, inspite of its name, what was an axiom for Archimedes is a theorem for us.

Theorem 2.4. Given any real number K we can find an integer n with

n > K.

3 Series and sums

There is no need to restrict the notion of a limit to real numbers.

Definition 3.1. If an C for each n 1 and a C then we say thatana if given >0 we can find ann0() such that

|an a|< for allnn0().

Exercise 3.2. We work inC.(i) The limit is unique. That is, ifan a andan b asn then

a= b.

1Footnote for passing historians, this is a course in mathematics.

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(ii) Ifan a asn andn(1) < n(2) < n(3) . . . thenan(j) a asj .

(iii) Ifan = c for alln thenanc asn .(iv) Ifana andbnb asn thenan+bna+b.(v) Ifan

a andbn

b asn

thenanbn

ab.

(vi) Ifana asn , an= 0 for eachn anda= 0, thena1n a1.Exercise 3.3. Explain why there is no result in Exercise 3.2 correspondingto part (vii) of Lemma 2.1.

We illustrate some of the ideas introduced by studying infinite sums.

Definition 3.4. We work inF whereF = R orF = C. Ifaj F we saythat

j=1 aj converges to s if

N

j=1

aj

s

asN . We writej=1 aj =s.IfN

j=1 aj does not tend to a limit as N , we say that the sumj=1 aj diverges.

Lemma 3.5. We work inF whereF = R orF = C.(i) Suppose aj , bj F and , F. If

j=1 aj and

j=1 bj converge

then so does

j=1 aj+ bj and

j=1

aj+bj =

j=1

aj+

j=1

bj.

(ii) Suppose aj , bj F and there exists an N such that aj = bj for alljN. Then, eitherj=1 aj andj=1 bj both converge or they both diverge.(In other words, initial terms do not matter.)

Exercise 3.6. Any problem on sums

j=1 aj can be converted into one onsequences by considering the sequencesn=

nj=1 aj. Show conversely that a

sequencesn converges if and only if, when we seta1 = s1 andan = sn sn1[n2] we have

j=1 aj convergent. What can you say aboutlimnn

j=1 ajand limn sn if both exist?

The following results are fundamental to the study of sums.

Theorem 3.7. (The comparison test.) We work in R. Suppose that0bjaj for allj. Then, if

j=1 aj converges, so does

j=1 bj .

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Theorem 3.8. We work inF whereF = R orF = C. If

j=1 |aj | converges,then so does

j=1 aj.

Theorem 3.8 is often stated using the following definition.

Definition 3.9. We work inF whereF = R orF = C. Ifj=1 |aj | con-verges we say that the sumj=1 aj is absolutely convergent.Theorem 3.8 then becomes the statement that absolute convergence im-

plies convergence.Here is a trivial but useful consequence of Theorems 3.7 and 3.8.

Lemma 3.10. (Ratio test.) Supposeaj C and|aj+1/aj| l asj .Ifl 1, then

j=1 aj diverges.

Of course Lemma 3.10 tells us nothing ifl = 1 or l does not exist.Sums which are not absolutely convergent are much harder to deal with

in general. It is worth keeping in mind the following trivial observation.

Lemma 3.11. We work inF whereF = R orF = C. Ifj=1 aj converges,thenaj0 asj .

At a deeper level the following result is sometimes useful.

Lemma 3.12. (Alternating series test.) We work inR. If we have adecreasing sequence of positive numbers an with an 0 as n , then

j=1(1)j+1aj converges.Further

N

j=1(1)j+1aj

j=1(1)j+1aj

|aN+1|

for allN 1.The last sentence is sometimes expressed by saying the error caused by

replacing a convergent infinite alternating sum of decreasing terms by thesum of its first few terms is no greater than the absolute value of the firstterm neglected.

Later we will give another test for convergence called the integral test(Lemma 14.4) from which we deduce the result known to many of you that

n=1 n1 diverges. I will give another proof in the next example.

Example 3.13. (i)

n=11

n

diverges.

(ii)

n=1

(1)nn

is convergent but not absolutely convergent.

(iii) Ifv2n= 1/n, v2n1=1/(2n) then

n=1 vn is not convergent.

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4 Least upper bounds

A non-empty bounded set in R need not have a maximum.

Example 4.1. The setE=

{1/n : n

1

}is non-empty and anye

E

satisfies the inequalities1e0 butEhas no largest member.However, as we shall see every non-empty bounded set in R has a least

upper bound (or supremum).

Definition 4.2. LetEbe a non-empty set inR. We say that is a leastupper bound forE if

(i) e for alleE [that is, is an upper bound for E] and(ii) If e for alleE then [that is, is the least such upper

bound].

IfEhas a supremum we write supeE

e= sup E=.

Lemma 4.3. If the least upper bound exists it is unique.

The following remark is trivial but sometimes helpful.

Lemma 4.4. LetEbe a non-empty set inR. Then is a least upper boundforEif and only if we can findenEwithen andbn such thatbnefor alleE andbn asn .

Here is the promised result.

Theorem 4.5. Any non-empty set in R with an upper bound has a leastupper bound.

We observe that this result is actually equivalent to the fundamentalaxiom.

Theorem 4.6. Theorem 4.5 implies the fundamental axiom.

Of course we have the notion of a greatest lower bound or infimum.

Exercise 4.7. Define the greatest lower bound in the manner of Defini-tion 4.2, prove its uniqueness in the manner of Lemma 4.3 and state andprove a result corresponding to Lemma 4.4.

IfEhas an infimumwe write infeEe= infE= . One way of dealingwith the infimum is to use the following observation.

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Lemma 4.8. LetEbe a non-empty set inR and writeE={e : eE}.Then E has an infimum if and only ifE has a supremum. If E has aninfimum infE=sup(E).Exercise 4.9. Use Lemma 4.8 and Theorem 4.5 to show that any non-empty

set inR with a lower bound has a greatest lower bound.

The notion of a supremum will play an important r ole in our proofs ofTheorem 5.12 and Theorem 9.2.

The following result is also equivalent to the fundamental axiom (that is,we can deduce it from the fundamental axiom and conversely, if we take itas an axiom, rather than a theorem, then we can deduce the fundamentalaxiom as a theorem).

Theorem 4.10. [Bolzano-Weierstrass] If xn R and there exists a Ksuch that|xn| K for alln, then we can findn(1)< n(2)< . . . andx Rsuch thatxn(j)x asj .

The Bolzano-Weierstrass theorem says that every bounded sequence of re-als has a convergent subsequence. Notice that we say nothing about unique-ness; ifxn= (1)n thenx2n1 but x2n+1 1 asn .

We shall prove the theorem of Bolzano-Weierstrass by lion huntingbut your supervisor may well show you another method. We shall use theBolzano-Weierstrass theorem to prove that every continuous function on aclosed bounded interval is bounded and attains its bounds (Theorem 5.12).The Bolzano-Weierstrass theorem will be much used in the next analysiscourse because it generalises to many dimensions.

5 Continuity

We make the following definition.

Definition 5.1. A functionf : R R is continuous atx if given >0 wecan find a(, x)>0 [read a delta depending on epsilon andx] such that

|f(x) f(y)|<

for ally with|

x

y|< (, x).

If f is continuous at each point x R we say that f is a continuousfunction onR.

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I shall do my best to make this seem a reasonable definition but it isimportant to realise that I am really stating a rule of the game (like a knightsmove in chess or the definition of offside in football). If you wish to play thegame you must accept the rules. Results about continuous functions mustbe derived from the definition and not stated as obvious from the notion ofcontinuity.

In practice we use a slightly more general definition.

Definition 5.2. LetEbe a subset ofR. A functionf :E R is continuousatxE if given >0 we can find a(, x)>0 [read a delta depending onepsilon andx] such that

|f(x) f(y)|< for allyE with|x y|< (, x).

If f is continuous at each point x E, we say that f is a continuousfunction onE.

However, it will do no harm and may be positively helpful if, whilst youare getting used to the idea of continuity, you concentrate on the case E= R.

Lemma 5.3. Suppose thatE is a subset ofR, thatxE, and thatf andg are functions fromE to R.

(i) Iff(x) =c for allxE, then f is continuous onE.(ii) Iff andg are continuous atx, then so isf+ g.(iii) Let us definef g : E R byf g(t) = f(t)g(t) for all t E.

Then iff andg are continuous atx, so isf g.(iv) Suppose that f(t)

= 0 for all t

E. If f is continuous at x so is

1/f.

Lemma 5.4. Let U and V be subsets ofR. Suppose f : U R is suchthatf(t)V for all tU. Iff is continuous atxU andg : V R iscontinuous atf(x), then the compositiong f is continuous atx.

By repeated use of parts (ii) and (iii) of Lemma 5.3 it is easy to showthat polynomials P(t) =

nr=0 art

r are continuous. The details are spelledout in the next exercise.

Exercise 5.5. Prove the following results.

(i) Suppose thatE is a subset ofR and thatf :ER is continuous atxE. IfxE E then the restrictionf|E off to E is also continuousatx.

(ii) If J : R R is defined by J(x) = x for all x R, then J iscontinuous onR.

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(iii) Every polynomialP is continuous onR.(iv) Suppose thatP andQ are polynomials and thatQ is never zero on

some subset E ofR. Then the rational function P/Q is continuous on E(or, more precisely, the restriction ofP/Q to E is continuous.)

The following result is little more than an observation but will be veryuseful.

Lemma 5.6. Suppose that E is a subset of R, that x E, and that fis continuous at x. If xn E for all n and xn x as n , thenf(xn)f(x) asn .

So far in this section we have only done algebra but the next result de-pends on the fundamental axiom. It is one of the key results of analysisand although my recommendation runs contrary to a century of enlightenedpedagogy I can see no objections to students learning the proof as a model.

Notice that the theorem resolves the problem posed by Example 1.1 (i).Theorem 5.7. (The intermediate value theorem). If f : [a, b] Ris continuous and f(a) 0 f(b) then there exists a c [a, b] such thatf(c) = 0.

Exercises 5.8 to 5.10 are applications of the intermediate value theorem.

Exercise 5.8. Show that any real polynomial of odd degree has at least oneroot. Is the result true for polynomials of even degree? Give a proof orcounterexample.

Exercise 5.9. Suppose thatg : [0, 1] [0, 1] is a continuous function. Byconsideringf(x) =g(x) x, or otherwise, show that there exists ac[0, 1]with g(c) = c. (Thus every continuous map of [0, 1] into itself has a fixedpoint.) Give an example of a bijective (but, necessarily, non-continuous)

functionh: [0, 1][0, 1] such thath(x)=x for allx[0, 1].[Hint: First find a function H : [0, 1]\ {0, 1, 1/2} [0, 1]\ {0, 1, 1/2}such thatH(x)=x.]Exercise 5.10. Every mid-summer day at six oclock in the morning, theyoungest monk from the monastery of Damt starts to climb the narrow pathup Mount Dipmes. At six in the evening he reaches the small temple at the

peak where he spends the night in meditation. At six oclock in the morningon the following day he starts downwards, arriving back at the monastery atsix in the evening. Of course, he does not always walk at the same speed.Show that, none the less, there will be some time of day when he will be atthe same place on the path on both his upward and downward journeys.

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We proved the intermediate value theorem (Theorem 5.7) by lion hunting.We prove the next two theorems by using the Bolzano-Weierstrass Theorem(Theorem 4.10). Again the results are very important and I can see noobjection to learning the proofs as a model.

Theorem 5.11. Iff : [a, b] R is continuous then we can find anM suchthat|f(x)| M for allx[a, b].

In other words a continuous function on a closed bounded interval isbounded. We improve this result in the next theorem.

Theorem 5.12. If f : [a, b] R is continuous then we can find x1, x2[a, b] such that

f(x1)f(x)f(x2)

for allx[a, b].In other words a continuous function on a closed bounded interval is

bounded and attains its bounds.

6 Differentiation

In this section it will be useful to have another type of limit.

Definition 6.1. LetEbe a subset ofR, f be some function fromE to R,andx some point ofE. If l

R we say thatf(y)

l asy

x [or, if we

wish to emphasise the restriction toE thatf(y)l asyx through valuesy E] if, given > 0, we can find a() > 0 [read a delta depending onepsilon] such that

|f(y) l|<

for allyE with0


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Exercise 6.3. LetEbe a subset ofR, f, g be some functions fromE to R,and x some point of E. (i) The limit is unique. That is, if f(y) l andf(y)k asyx thenl= k.

(ii) If x E E and f(y) l as y x through valuesy E, thenf(y)

l asy

x through valuesy

E.

(iii) Iff(t) =c for alltE thenf(y)c asyx.(iv) Iff(y)l andg(y)k asyx thenf(y) +g(y)l+k.(v) Iff(y)l andg(y)k asyx thenf(y)g(y)lk.(vi) If f(y) l as y x, f(t)= 0 for each t E and l= 0 then

f(t)1 l1.(vii) Iff(t)L for eachtEandf(y)l asyx then lL.We can now define the derivative.

Definition 6.4. LetEbe a subset ofR. A functionf :E R is differen-tiable atx

E with derivativef(x) if

f(y) f(x)y x f

(x)

asyx.Iff is differentiable at each pointxE, we say thatf is a differentiable

function onE.

As usual, no harm will be done if you replace EbyR.Here are some easy consequences of the definition.

Exercise 6.5. LetEbe a subset ofR, fsome function fromE to R, andxsome point ofE. Show that iff is differentiable atx thenf is continuousatx.

Exercise 6.6. LetEbe a subset ofR, f, g be some functions fromE to R,andx some point ofE. Prove the following results.

(i) Iff(t) = c for alltE thenfis differentiable atx withf(x) = 0.(ii) Iff andg are differentiable atx then so is their sumf+ g and

(f+ g)(x) =f(x) +g(x).

(iii) Iff andg are differentiable atx then so is their productf g and(f

g)(x) =f(x)g(x) + f(x)g(x).

(iv) If f is differentiable at x and f(t)= 0 for all t E then 1/f isdifferentiable atx and

(1/f)(x) =f(x)/f(x)2.

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(v) Iff(t) =N

r=0 artr onE thenfis differentiable atx and

f(x) =N

r=1

rarxr1

.The next result is slightly harder to prove than it looks. (we split the

proof into two halves depending on whether f(x)= 0 or f(x) = 0.Lemma 6.7. [Chain rule]LetU andVbe subsets ofR. Supposef :UR is such thatf(t) V for all t U. Iff is differentiable at x U andg: V R is differentiable atf(x), then the compositiong fis differentiableatx with

(g f)(x) =f(x)g(f(x)).

7 The mean value theorem

We have almost finished our project of showing that the horrid situationrevealed by Example 1.1 can not occur for the reals.

Our first step is to prove Rolles theorem.

Theorem 7.1. [Rolles theorem]Ifg : [a, b]Ris a continuous functionwithg differentiable on (a, b) andg(a) = g(b), then we can find ac(a, b)such thatg(c) = 0.

A simple tilt gives the famous mean value theorem.

Theorem 7.2. (The mean value theorem). Iff : [a, b] R is a con-tinuous function withf differentiable on(a, b), then we can find ac(a, b)such that

f(b) f(a) = (b a)f(c).We now have the results so long desired.

Lemma 7.3. Iff : [a, b] R is a continuous function withfdifferentiableon(a, b), then the following results hold.

(i) If f(t) > 0 for all t (a, b) then f is strictly increasing on [a, b].(That is, f(y)> f(x) wheneverby > xa.)(ii) Iff(t)0 for all t(a, b) thenf is increasing on [a, b]. (That is,f(y)f(x) wheneverby > xa.)

(iii) [The constant value theorem] Iff(t) = 0 for allt(a, b) thenf is constant on [a, b]. (That is, f(y) =f(x) wheneverby > xa.)

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Notice that since we deduce Lemma 7.3 from the mean value theorem wecan not use it in the proof of Rolles theorem.

The mean value theorem has many important consequences, some ofwhich we look at in the remainder of the section.

We start by looking at inverse functions.

Lemma 7.4. (i) Supposef : [a, b] R is continuous. Thenf is injectiveif and only if it is strictly increasing (that isf(t)> f(s) wheneveras 0 for allx [a, b]. Letf(a) = c andf(b) = d. Then the mapf : [a, b]

[c, d] is bijective andf1 is differentiable on [c, d] with

f1(x) = 1

f(f1(x)).

In the opinion of the author the true meaning of the inverse rule andthe chain rule only becomes clear when we consider higher dimensions in thenext analysis course.

We now prove a form of Taylors theorem.

Theorem 7.6. [nth mean value theorem] Suppose that b > a and f :[a, b] R isn+ 1 times differentiable. Then

f(b) n

j=0

f(j)(a)

j! (b a)j = f

(n+1)(c)

(n+ 1)! (b a)n+1

for somec witha < c < b.

This gives us a global and a local Taylors theorem.

Theorem 7.7. [Global Taylor theorem] Suppose that b > a and f :[a, b] R isn+ 1 times differentiable. Ifx, t[a, b]

f(t) =

nj=0

f

(j)

(x)j! (t x)j +f(n+1)

(x+(t x))(n+ 1)! (t x)n+1

for some(0, 1).

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Theorem 7.8. [Local Taylor theorem]Suppose that >0 andf : (x , x + ) Risn times differentiable on(x, x + )andf(n) is continuousatx. Then

f(t) =

nj=0

f(j)(x)

j! (t x)j

+ (t)(t x)n

where(t)0 astx.Notice that the local Taylor theorem always gives us some information

but that the global one is useless unless we can find a useful bound on then+ 1th derivative.

To reinforce this warning we consider a famous example of Cauchy.

Exercise 7.9. Consider the functionF : R R defined by

F(0) = 0F(x) = exp(1/x2) otherwise.

(i) Prove by induction, using the standard rules of differentiation, thatFis infinitely differentiable at all pointsx= 0 and that, at these points,

F(n)(x) =Pn(1/x)exp(1/x2)

wherePn is a polynomial which need not be found explicitly.(ii) Explain whyx1Pn(1/x) exp(1/x2)0 asx0.(iii) Show by induction, using the definition of differentiation, thatF is

infinitely differentiable at0 withF(n)(0) = 0 for alln. [Be careful to get thispart of the argument right.]

(iv) Show that

F(x) =

j=0

F(j)(0)

j! xj

if and only ifx = 0. (The reader may prefer to say that The Taylor expansionofF is only valid at0.)

(v) Why does part (iv) not contradict the local Taylor theorem (Theo-

rem 7.8)?Since examiners are fonder of the global Taylor theorem than it deserves

I shall go through the following example.

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Example 7.10. Assuming the standard properties of the exponential func-tion show that

exp x=

j=0xj

j!

for allx.

Please note that in a pure mathematics question many (or even most) ofthe marks in a question of this type will depend on estimating the remainderterm. [In methods questions you may simply be asked to find the Taylorsseries without being asked to prove convergence.]

8 Complex variable

The field C of complex numbers resembles the field R of real numbers inmany ways but not in all.

Lemma 8.1. We can not define an order onC which will behave in the sameway as>forR.

However there is sufficient similarity for us to define limits, continuityand differentiability. (We have already seen some of this in Definition 3.1and Exercise 3.2.)

Definition 8.2. LetEbe a subset ofC, f be some function fromE to C,and z some point of E. If l

C we say that f(w)

l as w

z [or, if

we wish to emphasise the restriction to E thatf(w)l asw z throughvalueswE] if, given >0, we can find a()>0 [read a delta dependingon epsilon] such that

|f(w) l|< for allwE with0


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(ii) Ifz E E andf(w)l asw z through valuesw E, thenf(w)l aswx through valueswE.

(iii) Iff(u) =c for alluE thenf(w)c aswz.(iv) Iff(w)l andg(w)k aswz thenf(w) +g(w)l+k.(v) Iff(w)

l andg(w)

k asw

z thenf(w)g(w)

lk.

(vi) If f(w) l as w z, f(u)= 0 for each u E and l= 0 thenf(w)1 l1.Exercise 8.5. Suppose thatE is a subset ofC, thatzE, and thatf andg are functions fromE to C.

(i) Iff(u) = c for alluE, then f is continuous onE.(ii) Iff andg are continuous atz, then so isf+ g.(iii) Let us definef g :E C byf g(u) =f(u)g(u) for alluE.

Then iff andg are continuous atz, so isf g.(iv) Suppose thatf(u)= 0 for all u E. Iff is continuous atz so is

1/f.(v) IfzE E andf is continuous atz then the restrictionf|E offto E is also continuous atz.

(vi) If J : C C is defined by J(z) = z for all z C, then J iscontinuous onC.

(vii) Every polynomialP is continuous onC.(viii) Suppose thatP andQare polynomials and thatQis never zero on

some subset E ofC. Then the rational function P/Q is continuous on E(or, more precisely, the restriction ofP/Q to E is continuous.)

Exercise 8.6. Let U and V be subsets ofC. Suppose f : U C is suchthatf(z)V for allz U. Iff is continuous atwU andg :VC iscontinuous atf(w), then the compositiong fis continuous atw.Definition 8.7. LetEbe a subset ofC. A functionf :E C is differen-tiable atzEwith derivativef(z) if

f(w) f(z)w z f

(z)

aswz.Iffis differentiable at each pointzEwe say thatfis a differentiable

function onE.

Exercise 8.8. LetEbe a subset ofC, fsome function fromE to C, andzsome point ofE. Show that iff is differentiable atz thenfis continuous atz.

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Exercise 8.9. LetEbe a subset ofC, f, g be some functions fromE to C,andzsome point ofE. Prove the following results.

(i) Iff(u) = c for alluE thenfis differentiable atz withf(z) = 0.(ii) Iff andg are differentiable atzthen so is their sumf+g and

(f+g)(z) =f(z) +g(z).

(iii) Iff andg are differentiable atzthen so is their productf g and

(f g)(z) =f(z)g(z) +f(z)g(z).(iv) If f is differentiable at z and f(u)= 0 for all u E then 1/f is

differentiable atz and

(1/f)(z) =f(z)/f(z)2.

(v) Iff(u) = Nr=0 arur onE thenfis differentiable atz andf(z) =

Nr=1

rarzr1

.

Exercise 8.10. [Chain rule] Let U and V be subsets ofC. Suppose f :U C is such thatf(z)V for all tU. Iff is differentiable atw Uand g : V R is differentiable at f(w), then the composition g f isdifferentiable atw with

(g f)(w) = f(w)g(f(w)).In spite of these similarities the subject of complex differentiable functions

is very different from that of real differentiable functions. It turns out thatwell behaved complex functions need not be differentiable.

Example 8.11. Consider the map : C C given by (z) = z. Thefunction is nowhere differentiable.

Because complex differentiability is so much more restrictive than realdifferentiability we can prove stronger theorems about complex differentiable

functions. For example it can be shown that such functions can be writtenlocally as power series2 (contrast the situation in the real case revealed byExample 7.9). To learn more go to the course P3 on complex methods.

2The syllabus says that this fact is part of this course. In this one instance I adviseyou to ignore the syllabus.

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9 Power series

In this section we work in C unless otherwise stated. We start with a veryuseful observation.

Lemma 9.1. Ifn=0 anzn0 converges and|z| R.

We callRthe radius of convergence of

n=0 anz

n. If

n=0 anz

n convergesfor all z we write R =

. The following useful strengthening is left to the

reader as an exercise.

Exercise 9.3. Suppose that

n=0 anzn has radius of convergenceR. Then

the sequence|anzn| is unbounded if|z| > R and

n=0 anzn converges abso-

lutely if|z|< R.Note that we say nothing about what happens on the circle of conver-

gence.

Example 9.4. (i)

n=1 n2zn has radius of convergence 1 and converges

for allzwith|z|= 1.(ii)n=1 zn has radius of convergence1 and diverges for allzwith|z|=

1.

A more complicated example is given in Exercise 18.16.It is a remarkable fact that we can operate with power series in the same

way as polynomials (within the radius of convergence). In particular we shallshow that we can differentiate term by term.

Theorem 9.5. If

n=0 anzn has radius of convergenceRand we writef(z) =

n=0 anzn thenfis differentiable at all pointsz with|z|< R and

f(z) =

n=1

nanzn1.

The proof is starred in the syllabus. We use three simple observations.

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Lemma 9.6. If

n=0 anzn has radius of convergenceR then given any >0

we can find aK() such that

n=0 |anzn|< K() for all|z| R .Lemma 9.7. If

n=0 anzn has radius of convergenceRthen so do

n=1 nanzn1

andn=2 n(n

1)anz

n2.

Lemma 9.8. (i)

n

r

n(n 1)

n 2r 2

for all2rn.

(ii)|(z+ h)n zn nhzn1| n(n 1)(|z| + |h|)n2|h|2 for allz , h C.In this course we shall mainly work on the real line. Restricting to the

real line we obtain the following result.

Theorem 9.9. (i) Ifaj R there exists a uniqueR (the radius of conver-gence) with 0 R such thatn=0 anxn converges for all real x with|x|< R and diverges for all realx withx > R.

(ii) Ifn=1 anxn has radius of convergence R and we write f(x) =n=0 anxn thenfis differentiable at all pointsx with|x|< R and

f(x) =

n=1

nanxn1.

10 The standard functions

In school you learned all about the functions exp, log, sin and cos and aboutthe behaviour ofx. Nothing that you learned was wrong (we hope) but youmight be hard pressed to prove all the facts you know in a coherent manner,

To get round this problem, we start from scratch making new definitionsand assuming nothing about these various functions. One of your tasks isto make sure that the lecturer does not slip in some unproved fact. On theother hand you must allow your lecturer to choose definitions which allow aneasy development of the subject rather than those that follow some historic,intuitive or pedagogically appropriate path3.

Let us start with the exponential function. Throughout this section weshall restrict ourselves to the real line.

Lemma 10.1. The sum

n=0

xn

n! has infinite radius of convergence.

We can thus define a function e: RR by

e(x) =

n=0

xn

n!.

3If you want to see a treatment along these lines see the excellent text of Burn[3].

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(We usee(x) rather than exp(x) to help us avoid making unjustified assump-tions.)

Theorem 10.2. (i) The function e : R R is everywhere differentiablewithe(x) =e(x).

(ii) e(x+y) =e(x)e(y) for allx, y R.(iii)e(x)>0 for allx R.(iv) e is a strictly increasing function.(v) e(x) asx , e(x)0 asx .(vi) e: R (0, ) is a bijection.It is worth stating some of our results in the language of group theory.

Lemma 10.3. The mapping e is an isomorphism of the group (R, +) andthe group ((0, ), ).

Since e : R (0, ) is a bijection we can consider the inverse functionl: (0, ) R.Theorem 10.4. (i) l : (0, )R is a bijection. We have l(e(x)) =x forallx R ande(l(t)) =t for allt(0, ).

(ii) The function l : (0, )R is everywhere differentiable with l(t) =1/t.

(iii)l(uv) =l(u) +l(v) for allu, v(0, ).We now write e(x) = exp(x), l(t) = log t (the use of ln is unnecessary).

If R and x >0. we write r(x) = exp( log x).

Theorem 10.5. Supposex, y >0 and, R. Then(i) r(xy) =r(x)r(y),(ii) r+(x) =r(x)r(x),(iii)r(r(x)) = r(x)(iv) r1(x) =x.

Exercise 10.6. Use the results of Theorem 10.5 to show that ifnis a strictlypositive integer andx >0 then

rn(x) =xx . . . x

n.

Thus, if we writex =r(x), our new notation isconsistentwith our oldschool notation when is rational but gives, in addition, a valid definitionwhen is irrational.

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Lemma 10.7. (i) If is real, r : (0, )(0, ) is everywhere differen-tiable andr(x) =r1(x).

(ii) Ifx >0 and we definefx() =x thenfx: R (0, )is everywhere

differentiable andfx() = log xfx().

Finally we look at the trigonometric functions.

Lemma 10.8. The sums

n=0(1)nx2n+1

(2n+1)! and

n=0

(1)nx2n

(2n)! have infinite ra-

dius of convergence.

We can thus define functions s, c: R R by

s(x) =

n=0

(1)nx2n+1(2n+ 1)!

and c(x) =

n=0

(1)nx2n(2n)!

Lemma 10.9. (i) The functionss, c: RR are everywhere differentiablewiths(x) =c(x) andc(x) =s(x).

(ii) s(x+y) = s(x)c(y) + c(x)s(y), c(x+y) = c(x)c(y) s(x)s(y) ands(x)2 +c(x)2 = 1 for allx, y R.(iii)s(x) =s(x), c(x) =c(x) for allx R.

Exercise 10.10. Write down what you consider to be the chief properties ofsinhandcosh. (They should convey enough information to draw a reasonablegraph of the two functions.)

(i) Obtain those properties by starting with the definitions

sinh x=

n=0

x2n+1

(2n+ 1)! and cosh x=

n=0

x2n

(2n)!

and proceeding along the lines of Lemma 10.9.(ii) Obtain those properties by starting with the definitions

sinh x=exp x exp(x)

2 and cosh x=

exp x+ exp(x)2

We have not yet proved one of the most remarkable properties of the sineand cosine functions, their periodicity.

Theorem 10.11. Lets andc be as in Lemma 10.9.(i) Ifc(a) = 0 andc(b) = 0 thens(b a) = 0.(ii) We haves(x)>0 for all0< x1.(iii) There exists a unique[0, 2] such thatc() = 0.(iv) s() = 1.(v) s(x+ 4) =s(x), c(x+ 4) =c().(vi) The functionc is strictly decreasing from1 to1 asx runs from0

to 2,

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We nowdefine = 2. Ifxand yare non-zero vectors in Rm we knowby the Cauchy-Schwarz inequality that|x.y| xy and we may definethe angle between the two vectors to be that with 0 which satisfies

cos = x.y

xy.

We can also justify the standard use of polar coordinates.

Lemma 10.12. If (x, y) R2 and (x, y)= (0, 0) then there exist a uniquer >0 and with0


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Theorem 11.3. (i) The function e : C C is everywhere differentiablewithe(z) =e(z).

(ii) e(z+w) =e(z)e(w) for allz, w C.Notice that the remaining parts of Theorem 10.2 are either meaningless

or false (compare Theorem 10.2 (vi) with Lemma 11.4 (iii) which shows thatexp : C C is not injective). We must be very careful in making thetransition from real to complex.

We obtain a series of famous formulae.

Lemma 11.4. (i) If is real

exp i= cos + i sin .

(ii) Ifx andy are real

exp(x+iy) = (exp x)(cos y+i sin y).

(iii)exp is periodic with period2i, that is to say

exp(z+ 2i) = exp z

for allz C.(iv) exp(C) = C \ {0}.

Example 11.5. Observe thatexp0 = exp2i = 1butexp(z) = exp(z)= 0for allz C. Thus the mean value theorem does not hold for differentiablefunctionsf : C C.

It is also possible to extend the definition of sine and cosine to the complexplane but the reader is warned that the behaviour of the new functions maybe somewhat unexpected. Since these extended functions most certainly donot form part of this course (though you will be expected to know them afterthe Complex Methods course) their study is left as an exercise.

Exercise 11.6. (i) Explain why the infinite sums

sin z=

n=0

(1)nz2n+1(2n+ 1)!

and cos z=

n=0

(1)nz2n(2n)!

converge everywhere and are differentiable everywhere with sin z = cos z,cos z= sin z.

(ii) Show that

sin z=exp iz exp(iz)

2i , cos z=

exp iz+ exp(iz))2

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and

exp iz= cos z+i sin z

for allz C.(iii) Show that

sin(z+w) = sin zcos w+ cos zsin w, cos(z+w) = cos zcos w sin zsin w,and(sin z)2 + (cos z)2 = 1 for allz, w C.

(iv) sin(z) = sin z, cos(z) = cos z for allz C.(v) sin andcos are2 periodic in the sense that

sin(z+ 2) = sin zand cos(z+ 2) = cos z

for allz C.(vi) Ifx is real thensin ix= i sinh x andcos ix= cosh x.

(vii) Recover the addition formulae for sinh andcosh by settingz = ixandw= iy in part (iii).

(ix) Show that| sin z| and| cos z| are bounded if|z| K for some Kbut that| sin z| and| cos z| are unbounded onC.

However, as you were already shown in the Algebra and Geometry courseand will be shown again in the Complex Methods course

There is no logarithm defined on all ofC \ {0}.Exercise 11.7. Suppose, if possible, that there exists a continuousL : C \

{0}

C withexp(L(z)) =z for allzC

\ {0}

.(i) If is real, show thatL(exp(i)) = i(+ 2n()) for somen() Z.(ii) Definef : R R by

f() = 1

2

L(exp i) L(1)

i

.

Show thatfis a well defined continuous function, thatf() Z for all R,thatf(0) = 0 and thatf(2) =1.

(iii) Show that the statements made in the last sentence of (ii) are in-compatible with the intermediate value theorem and deduce that no functioncan exist with the supposed properties ofL.

(iv) Discuss informally what connection, if any, the discussion above haswith the existence of the international date line.

A similar argument shows that it is not possible to produce a continuoussquare root on the complex plane.

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We define the upper sumand lower sum associated withDby

S(f, D) =n

j=1

(xj xj1) supx[xj1,xj ]

f(x),

s(f, D) =n

j=1

(xj xj1) inf x[xj1,xj ]

f(x)

[Observe that, if the integralb

af(t) dt exists, then the upper sum ought to

provide an upper bound and the lower sum a lower bound for that integral.]The next lemma is obvious but useful.

Lemma 12.1. IfD andD are dissections withD D thenS(f, D)S(f, D)s(f, D)s(f, D).

The next lemma is again hardly more than an observation but it is thekey to the proper treatment of the integral.

Lemma 12.2. [Key integration property] If f : [a, b] R is boundedandD1 andD2 are two dissections, then

S(f, D1)S(f, D1 D2)s(f, D1 D2)s(f, D2). The inequalities tell us that, whatever dissection you pick and whatever

dissection I pick, your lower sum cannot exceed my upper sum. There is noway we can put a quart in a pint pot5.

SinceS(f, D) (b a)Kfor all dissectionsD we can define the upperintegral as I

(f) = infDS(f, D). We define the lower integral similarly asI(f) = supDs(f, D). The inequalities tell us that these concepts behavewell.

Lemma 12.3. Iff : [a, b] R is bounded, thenI(f)I(f).[Observe that, if the integral

ba

f(t) dt exists, then the upper integralought to provide an upper bound and the lower integral a lower bound forthat integral.]

IfI(f) = I(f), we say that fis Riemann integrable and we write

b

a

f(x) dx= I(f).

The following lemma provides a convenient criterion for Riemann inte-grability.

5Or put a litre in a half litre bottle.

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Lemma 12.4. (i) A bounded functionf : [a, b]R is Riemann integrableif and only if, given any >0, we can find a dissectionD with

S(f, D) s(f, D)< .

(ii) A bounded functionf : [a, b]R is Riemann integrable with integralIif and only if, given any >0, we can find a dissectionD with

S(f, D) s(f, D)< and|S(f, D) I| .Many students are tempted to use Lemma 12.4 (ii) as the definitionof the

Riemann integral. The reader should reflect that, without the inequality ,it is not even clear that such a definition gives a unique value for I. (This isonly the first of a series of nasty problems that arise if we attempt to developthe theory without first proving , so I strongly advise the reader not totake this path.)

We now prove a series of standard results on the integral.Lemma 12.5. (i) The functionJ : [a, b]Rgiven byJ(t) = 1is integrableand b

a

1 dx= b a.

(ii) Iff, g: [a, b] R are Riemann integrable, then so isf+ g and ba

f(x) +g(x) dx=

ba

f(x) dx+

ba

g(x) dx.

(iii) Iff : [a, b] R is Riemann integrable, thenf is and ba

(f(x)) dx= b

a

f(x) dx.

(iv) If R and f : [a, b] R is Riemann integrable, then f isRiemann integrable and b

a

f(x) dx=

ba

f(x) dx.

(v) Iff, g : [a, b]R are Riemann integrable functions withf(t)g(t)for allt[a, b], then ba

f(x) dx b

a

g(x) dx.

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Lemma 12.6. (i) Iff : [a, b]R is Riemann integrable then so isf2.(ii) Iff, g: [a, b]R are Riemann integrable, then so is the productf g.

Lemma 12.7. (i) Iff : [a, b]R is Riemann integrable then so isf+(t) =max(f(t), 0).

(ii) Iff : [a, b]Ris Riemann integrable, then|f|is Riemann integrableand b

a

|f(x)| dx b

a

f(x) dx

.Notice that we often only need the much weaker inequality

ba

f(x) dx

sup

t[a,b]

|f(t)|(b a)

usually stated as

|integral| length sup.

The next lemma is also routine in its proof but continues our programmeof showing that the integral has all the properties we expect.

Lemma 12.8. Suppose thatacb and thatf : [a, b]R is a boundedfunction. Then, f is Riemann integrable on [a, b] if and only iff|[a,c] is Rie-mann integrable on [a, c] andf|[c,b] is Riemann integrable on [c, b]. Further,iffis Riemann integrable on [a, b], then

ba

f(x) dx= c

a

f|[a,c](x) dx+ b

c

f|[c,b](x) dx.

In a very slightly less precise and very much more usual notation we write ba

f(x) dx=

ca

f(x) dx+

bc

f(x) dx.

There is a standard convention which says that, ifbaandfis Riemannintegrable on [a, b], we define

ab

f(x) dx= ba

f(x) dx.

It is, however, a convention that requires care in use.

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Exercise 12.9. Suppose that b a, , R, and f and g are Riemannintegrable. Which of the following statements are always true and which arenot? Give a proof or counter-example. If the statement is not always true,

find an appropriate correction and prove it.

(i) a

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

b f(x) dx+ a

b g(x) dx.

(ii) Iff(x)g(x) for allx[a, b], then a

b

f(x) dx a

b

g(x) dx.

13 Some properties of the integral

Not all bounded functions are Riemann integrable

Lemma 13.1. Iff : [0, 1]R is given by

f(x) = 1 whenx is rational,

f(x) = 0 whenx is irrational,

thenfis not Riemann integrable

This does not worry us unduly, but makes it more important to showthat the functions we wish to be integrable actually are.

Our first result goes back to Riemann (indeed, essentially, to Newton andLeibniz).

Lemma 13.2. (i) Iff : [a, b] R is increasing. thenf is Riemann inte-grable.

(ii)Iff : [a, b]R can be written asf =f1 f2 withf1, f2 : [a, b]Rincreasing, thenf is Riemann integrable.

(iii) If f : [a, b] R is piecewise monotonic, then f is Riemann inte-grable.

It should be noted that the results of Lemma 13.2 do not require f tobe continuous. (For example, the Heaviside function, given by H(t) = 0 fort


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We complete the discussion of integration and this course with a series ofresults which apply only to continuous functions.

Our first result is an isolated, but useful, one.

Lemma 13.4. Iff : [a, b]

R is continuous, f(t)

0 for allt

[a, b] and b

a

f(t) dt= 0,

it follows thatf(t) = 0 for allt[a, b].Exercise 13.5. Let a c b. Give an example of a Riemann integrable

functionf : [a, b] R such thatf(t)0 for allt[a, b] and ba

f(t) dt= 0,

butf(c)= 0.We now come to the justly named fundamental theorem of the calculus.

Theorem 13.6. [The fundamental theorem of the calculus] Supposethatf : (a, b) R is a continuous function and thatu(a, b). If we set

F(t) =

tu

f(x) dx,

thenFis differentiable on(a, b) andF(t) =f(t) for allt

(a, b).

Exercise 13.7. (i) Let H be the Heaviside function H : R R given byH(x) = 0 forx


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We call the solutions ofg (t) =f(t) indefinite integrals(or, simply, inte-grals) off.

Yet another version of the fundamental theorem is given by the nexttheorem.

Theorem 13.9. Suppose thatg : [a, b]R has continuous derivative. Then ba

g(t) dt= g(b) g(a).

Theorems 13.6 and 13.9 show that (under appropriate circumstances)integration and differentiation are inverse operations and the the theoriesof differentiation and integration are subsumed in the greater theory of thecalculus.

We use the fundamental theorem of the calculus to prove the formulaefor integration by substitution and integration by parts.

Theorem 13.10. [Change of variables for integrals]Suppose thatf :[a, b]R is continuous andg : [, ]R is differentiable with continuousderivative. Suppose further thatg([, ]) [a, b]. Then, ifc, d [, ], wehave g(d)

g(c)

f(s) ds=

dc

f(g(x))g(x) dx.

Exercise 13.11. The following exercise is traditional.(i) Show that integration by substitution, usingx= 1/t, gives

b

a

dx

1 + x2

= 1/a

1/b

dt

1 +t2

whenb > a >0.(ii) If we seta=1, b= 1 in the formula of (i), we obtain 1

1

dx

1 +x2?

= 11

dt

1 +t2

Explain this apparent failure of the method of integration by substitution.(iii) Write the result of (i) in terms oftan1 and prove it using standard

trigonometric identities.

Lemma 13.12. Suppose that f : [a, b]

R has continuous derivative and

g: [a, b] R is continuous. LetG: [a, b] R be an indefinite integral ofg.Then, we have b

a

f(x)g(x) dx= [f(x)G(x)]ba b

a

f(x)G(x) dx.

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We obtain the following version of Taylors theorem by repeated integra-tion by parts.

Theorem 13.13. [A global Taylors theorem with integral remain-der] Ifn

1 andf : (

a, a)

R isn times differentiable with continuous

nth derivative, then

f(t) =n1j=0

f(j)(0)

j! tj +Rn(f, t)

for all|t|< a, where

Rn(f, t) = 1

(n 1)! t0

(t x)n1f(n)(x) dx.

In the opinion of the lecturer this form is powerful enough for most pur-poses and is a form that is easily remembered and proved for examination.

14 Infinite integrals

The reader may already be familiar with definitions of the following type.

Definition 14.1. Iff : [a, b] R andM, P 0 let us write

fM,P(x) =

f(x) ifP f(x)MM iff(x)> M

P iff(x) aand

Xa f(x) dxL asX , then we say that

a f(x) dx

exists with valueL.

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It must be said that neither Definition 14.1 nor Definition 14.2 are morethan ad hoc.

In the rest of this section we look at Definition 14.2.

Lemma 14.3. Supposef : [a,

)

R is such thatf

|[a,X] is Riemann inte-

grable on[a, X]for eachX > a. Iff(x)0 for allx, thena f(x) dxexistsif and only if there exists aKsuch that

Xa f(x) dxK for allX.

We use Lemma 14.3 to prove the integral comparison test.

Lemma 14.4. Iff : [1, )R is a decreasing function withf(x)0 asx then

n=1

f(n) and

1

f(x) dx

either both diverge or both converge.

Example 14.5. If >1 then

n=1 n converges. If1 thenn=1 n

diverges.

This is really as far as we need to go, but I will just add one furtherremark.

Lemma 14.6. Suppose f : [a, ) R is such that f|[a,X] is Riemann in-tegrable on [a, X] for each X > a. If

a |f(x)| dx exists, then a f(x) dx

exists.

It is natural to state Lemma 14.6 in the form absolute convergence ofthe integral implies convergence.

Speaking broadly, infinite integrals

a f(x) dx work well when they areabsolutely convergent, that is to say,

a |f(x)| dx


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Spivaks more leisurely. A completely unsuitable but interesting version ofthe standard analysis course is given by BerlinskisA Tour of the Calculus[1] Spivak rewritten by Sterne with additional purple passages by the Ankh-Morpork tourist board. I have writtenA Companion to Analysis [4] whichcovers this course at a higher level together with the next analysis course.It is available off the web but is unlikely to be as suitable for beginners asSpivak and Burkill. If you do download it, remember that you are under amoral obligation to send me an e-mail about any mistakes you find.

References

[1] D. Berlinski A Tour of the CalculusMandarin Paperbacks 1997.

[2] J. C. Burkill A First Course in Mathematical AnalysisCUP, 1962.

[3] R. P. Burn Numbers and FunctionsCUP, 1992.

[4] T. W. KornerA Companion to Analysisfor the moment available via myhome page http://www.dpmms.cam.ac.uk/twk/ .

[5] M. Spivak, CalculusAddison-Wesley/Benjamin-Cummings, 1967.

16 First example sheet

Students vary in how much work they are prepared to do. On the whole,exercises from the main text are reasonably easy and provide good practice

in the ideas. Questions and parts of questions marked with are not intendedto be hard but cover topics less central to the present course.

Q 16.1. Let an R. We say that an as n if, given K >0, wecan find an n0(K) such that anKfor all nn0(K).

(i) Write down a similar definition for an .(ii) Show that an if and only ifan .(iii) Ifan= 0 for all n, show that, ifan , it follows that 1/an 0

asn .(iv) Is it true that, ifan= 0 for all n and 1/an 0, then an as

n

? Give a proof or a counter example.

Q 16.2. Prove that, if

a1> b1 > 0 andan+1=an+bn

2 , bn+1=

2anbnan+bn

,

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then an> an+1> bn+1> bn. Prove that, as n , an andbn both tend tothe limit

(a1b1).

Use this result to give an example of an increasing sequence of rationalnumbers tending to a limit l which is not rational.

Q 16.3. (Exercise 2.2.) Show that a decreasing sequence of real numbersbounded below tends to a limit.[Hint. Ifab thenb a.]Q 16.4. (i) By using the binomial theorem, or otherwise, show that, if >0andn is a positive integer, then

(1 +)n n.

Deduce that (1 +)n asn .(ii) By using the binomial theorem, or otherwise, show that, if > 0,

then

(1 +)n 2 n(n 1)2

.

Deduce that n1(1 +)n asn .(iii) Show that, ifk is any positive integer and a >1, then nkan

asn . [Thus powers beat polynomial growth.](iv) Show that ifk is any positive integer and 1 > a0, thennkan 0

asn .Q 16.5. Ifa

R and a

=

1 describe the behaviour of

an 1an + 1

asn . (That is, for each value ofa say whether the sequence convergesor not, and, if it converges, say what it converges to. For certain values ofayou may find it useful to divide top and bottom by an.)

Q 16.6. (Exercises 3.2 and 3.3.)We work in C.(i) The limit is unique. That is, ifan a and an b as n , then

a= b.(ii) Ifan a as n and n(1) < n(2)< n(3) . . . , then an(j) a asj .

(iii) Ifan= c for all n, thenanc as n .(iv) Ifana andbnb as n , thenan+bna+ b.

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(v) Ifana andbnb as n , thenanbnab.(vi) If an a as n and an = 0 for each n and a= 0, then

a1n a1.(vii) Explain why there is no result in Exercise 3.2 corresponding to part

(vii) of Lemma 2.1.

Q 16.7. [Cesaro summation]Ifaj R we write

n=a1+a2+ +an

n .

(i) Show that, ifan 0, then n 0. [Hint: Given >0 we can findanj0() such that|aj|< /2 for all jj0().]

(ii) Show that, ifana, thenna.(iii) Ifan= (1)n show that an does not converge but n does.(iv) Ifa2m+r = (1)m for 0r2m 1, m0 show that n does not

converge.(v) Ifn converges show that n1an0 asn .

Q 16.8. In this question you may assume the standard properties of theexponential function including the relation exp x1 + x for x0.

(i) Suppose that an0. Show thatn

j=1

ajn

j=1

(1 +aj )exp

nj=1

aj

.

Deduce, carefully, thatn

j=1(1 +aj) tends to a limit as n if and onlyifnj=1 aj does.(ii) Euler made use of these ideas as follows. Letpk be the kth prime.By observing that

1

1 p1k

mr=0

prk

show that

nk=1

1

1 p1k

uS(N,n)

1

u

whereS(N, n) is the set of all integers of the form

u=n

k=1

pmkk

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with 0mkN.By letting N show that

n

k=11

1

p1k

uS(n)

1

u

whereS(n) is the set of all integers whose only prime factors are p1,p2,. . . ,pn. Deduce that there are infinitely many primes (what can could you sayaboutS(n) if there were only nprimes?) and show that

nk=1

1

1 p1k

asn .Conclude that

k=1

11 p1k 1 diverges.Show that|1 (1 p1k )1| 2p1k and deduce that

k=1

1

pkdiverges.

Q 16.9. (i) Ifaj is an integer with 0 aj 9 show from the fundamentalaxiom that

j=1 aj 10j

exists. Show that 0j=1 aj 10j 1, carefully quoting any theorems thatyou use.

(ii) If 0x1, show that we can find integersxj with 0xj9 suchthat

x=

j=1

xj10j.

(iii) Ifaj and bj are integers with 0aj, bj 9 and aj =bj for j < N,aN> bN show that

j=1

aj10j

j=1

bj10j.

Give the precise necessary and sufficient condition for equality and prove it.

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17 Second example sheet

Students vary in how much work they are prepared to do. On the whole,exercises from the main text are reasonably easy and provide good practicein the ideas. Questions and parts of questions marked with are not intended

to be hard but cover topics less central to the present course.

Q 17.1. (Exercises 4.7 and 4.9.)(i)Define the greatest lower bound in the manner of Definition 4.2, prove

its uniqueness in the manner of Lemma 4.3 and state and prove a resultcorresponding to Lemma 4.4.

(ii) Use Lemma 4.8 and Theorem 4.5 to show that any non-empty set inR with a lower bound has a greatest lower bound.

Q 17.2. Suppose thatA and B are non-empty bounded subsets ofR. Showthat

sup{a+b: aA, bB}= sup A + sup B.

The last formula is more frequently written

supaA, bB

a+b= supaA

a+ supbB

b.

Suppose, further that an and bn are bounded sequences of real numbers.For each of the following statements either give a proof that it is always trueor an example to show that it is sometimes false.

(i) supn(an+ bn) = supn an+ supn bn.(ii) supn(an+bn)supn an+ supn bn.(iii) supn(an+bn)supn an+ infn bn.(iv) supaA, bBab= (supaA a)(supbBb).(v) infaA, bBa+b= infaA a+ infbBb.

Q 17.3. (Exercise 5.5) Prove the following results.(i) Suppose that Eis a subset ofR and that f :E R is continuous at

xE. IfxE E then the restriction f|E off to E is also continuousat x.

(ii) IfJ : R R is defined byJ(x) =x for allx R, thenJis continuouson R.

(iii) Every polynomial P is continuous on R.(iv) Suppose that P and Qare polynomials and that Q is never zero on

some subsetEofR. Then the rational functionP /Qis continuous onE(or,more precisely, the restriction ofP /Qto E is continuous.)

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Q 17.4. (Exercises 5.8 to 5.10.)(i) Show that any real polynomial of odd degree has at least one root. Is

the result true for polynomials of even degree? Give a proof or counterex-ample.

(ii) Suppose that g : [0, 1]

[0, 1] is a continuous function. By consid-eringf(x) = g(x) x, or otherwise, show that there exists a c[0, 1] withg(c) =c. (Thus every continuous map of [0, 1] into itself has a fixed point.)Give an example of a bijective (but, necessarily, non-continuous) functionh: [0, 1][0, 1] such that h(x)=x for allx[0, 1].[Hint: First find a functionH : [0, 1] \{0, 1, 1/2} [0, 1] \ {0, 1, 1/2}suchthatH(x)=x.]

(iii) Every mid-summer day at six oclock in the morning, the youngestmonk from the monastery of Damt starts to climb the narrow path up MountDipmes. At six in the evening he reaches the small temple at the peak wherehe spends the night in meditation. At six oclock in the morning on the

following day he starts downwards, arriving back at the monastery at six inthe evening. Of course, he does not always walk at the same speed. Showthat, none the less, there will be some time of day when he will be at thesame place on the path on both his upward and downward journeys.

Q 17.5. (Exercise 6.2.)Let Ebe a subset ofR. Show that functionf :E R is continuous at

xE if and only iff(y)f(x) as yx.Q 17.6. (Exercise 6.3.)

LetEbe a subset ofR,f,g be some functions from E toR, andx some

point ofE. (i) The limit is unique. That is, iff(y) l and f(y) k asyx then l = k.(ii) Ifx E E and f(y) l as y x through values y E, then

f(y)l as yx through values yE.(iii) Iff(t) =c for alltEthenf(y)c as yx.(iv) Iff(y)l andg(y)k as yx thenf(y) + g(y)l+k.(v) Iff(y)l andg(y)k as yx thenf(y)g(y)lk.(vi) If f(y) l as y x, f(t)= 0 for each t E and l= 0 then

f(t)1 l1.(vii) Iff(t)L for each tEandf(y)l as yx then lL.

Q 17.7. (Exercise 6.6.)LetEbe a subset ofR,f,g be some functions from E toR, andx somepoint ofE. Prove the following results.

(i) Iff(t) =c for all tEthenf is differentiable at x with f(x) = 0.

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(ii) Iff and g are differentiable at x then so is their sumf+g and

(f+ g)(x) =f(x) +g(x).

(iii) Iff and g are differentiable at x then so is their product f

g and

(f g)(x) =f(x)g(x) + f(x)g(x).

(iv) If f is differentiable at x and f(t)= 0 for all t E then 1/f isdifferentiable at x and

(1/f)(x) =f(x)/f(x)2.

(v) Iff(t) =N

r=0 artr on E then thenf is differentiable at x and

f(x) =N

r=1

rarxr1

.

Q 17.8. We work in the real numbers. Are the following true or false? Givea proof or counterexample as appropriate.

(i) If

n=1 a4n converges then

n=1 a

5n converges.

(ii) If


n=1 a

4n does.

(iii) Ifan0 for alln and

n=1 an converges then nan0 as n .(iv) Ifan0 for all n and

n=1 an converges then n(an an1)0 as

n .(v)

Ifan is a decreasing sequence of positive numbers andn=1 an con-verges thennan0 asn .(vi) Ifan is a decreasing sequence of positive numbers and nan 0 as

n thenn=1 an converges.[Hint. If necessary, look at Lemmas 14.4 and 14.5.]

(vii) If


n=1 n

1an converges.[Hint: Cauchy-Schwarz]

(viii) If

n=1 an converges then

n=1 n1|an| converges.

Q 17.9. [General principle of convergence]We say that a sequence xnof points in R form a Cauchy sequence if given any > 0 we can find an

n0() such that

|xn xm|< for all n, mn0().

(i) Show that any convergent sequence is a Cauchy sequence.

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(ii) Suppose that the points xn form a Cauchy sequence. Show that, ifwe can findn(j) such thatxn(j)x, it follows that xnx. (Thus, ifany subsequence of a Cauchy sequence converges, so does the sequence.)

(iii) Show that if the pointsxnform a Cauchy sequence the set{xn : n1}

is bounded.(iv) Use the theorem of Bolzano-Weierstrass to show that any Cauchy

sequence converges.[We have thus shown that a sequence converges if and only if it is a Cauchy

sequence. This is the famous Cauchy general principle of convergence. Itsgeneralisation will play an important role in next years analysis course C9.]

Q 17.10. [A method of Abel](i) Suppose that aj and bj are sequences of complex numbers and that

Sn=n

j=1 aj for n1 andS0= 0. Show that, if 1uv thenv

j=uajbj =

vj=u

Sj (bj bj+1) Su1bu+Svbv+1.

(This is known as partial summation, for obvious reasons.)(ii) Suppose now that, in addition, the bj form a decreasing sequence of

positive terms and that|Sn| Kfor all n. Show thatv

j=u

ajbj

3Kbu.(You can replace 3Kbu by 2Kbu if you are careful but there is no advantage

in this.) Deduce that ifbj0 as j thenj=1 ajbj converges.Deduce the alternating series test.(iii) Ifbj is a decreasing sequence of positive terms withbj0 asj

show that

j=1 bjzj converges in the region given by|z| 1 and z= 1.

Show by example that we must have the conditionz= 1. Show by examplethat we must have the condition|z| 1.Q 17.11. Enter any number on your calculator. Press the sin button re-peatedly. What appears to happen? Prove your conjecture using any prop-erties of sin that you need.

18 Third example sheet

Students vary in how much work they are prepared to do. On the whole,exercises from the main text are reasonably easy and provide good practice

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in the ideas. Questions and parts of questions marked with are not intendedto be hard but cover topics less central to the present course. I know thatthere is a general belief amongst students, directors of studies and the FacultyBoard that there is a magic set of questions which is suitable for everybody.If there is one, I will be happy to circulate it. For the moment I remark thatthe unstarred questions on this example sheet represent a good weeks workfor someone who finds the course hard and the whole sheet represents a goodweeks work for someone who finds the unstarred questions boring.

Q 18.1. [Very traditional](i) Consider the function f : RR given byf(t) =t2 sin1/tfor t= 0,f(0) = 0. Show that fis everywhere differentiableand find its derivative. Show that f is not continuous.[Deal quickly with the easy part and then go back to the definition to dealwitht= 0. There are wide selection of counter-examples obtained by lookingat t sin t for various values of and .]

(ii) Find an infinitely differentiable function g : (1, )R

such thatg(t)0 but g (t) 0 as t .Q 18.2. Question 18.1 shows that the derivative of a differentiable functionneed not be continuous. In spite of this the derivative still obeys Darbouxstheorem:- Iff : [a, b] Ris differentiable andk lies betweenf(a) andf(b)then there is a c [a, b] such that f(c) = k . In this question we prove theresult.

Explain why there is no loss of generality in supposing that f(a)> k >f(b). Setg(x) =f(x)kx. By looking atg (a) andg (b) show thatg cannothave a maximum at a or b. Use the method of proof of Rolles theorem to

show that there exists a c (a, b) with g

(c) = 0 and deduce Darbouxstheorem.Give an example of a function f :RR such that there does not exist

a differentiable function F : R R with F =f.Q 18.3. (i)[Cauchys mean value theorem] Suppose thatf, g : [a, b]R are continuous and that f and g are differentiable on (a, b). Supposefurther that g(x)= 0 for all x (a, b). Explain why g(a)= g(b). Byapplying Rolles theorem toF where

F(x) = (g(b) g(a))(f(x) f(a)) (g(x) g(a))(f(b) f(a)),

show that there is a (a, b) such thatf()

g() =

f(b) f(a)g(b) g(a) .

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(ii)[LHopitals rule]Suppose thatf, g: [a, b] R are continuous andthatf andg are differentiable on (a, b). Suppose that f(a) =g(a) = 0, thatg(t) does not vanish near a and f(t)/g(t)l as ta through values oft > a. Show that f(t)/g(t)l as ta through values oft > a.

Q 18.4. (Exercise 7.9.)Consider the functionF : R R defined by

F(0) = 0

F(x) = exp(1/x2) otherwise.

(i) Prove by induction, using the standard rules of differentiation, that Fis infinitely differentiable at all points x= 0 and that, at these points,

F(n)(x) =Pn(1/x)exp(1/x2)

wherePn is a polynomial which need not be found explicitly.(ii) Explain whyx1Pn(1/x)exp(1/x2)0 asx0.(iii) Show by induction, using the definition of differentiation, that F is

infinitely differentiable at 0 withF(n)(0) = 0 for alln. [Be careful to get thispart of the argument right.]

(iv) Show that

F(x) =

j=0

F(j)(0)

j! xj

if and only ifx= 0. (The reader may prefer to say that The Taylor expansionofF is only valid at 0.)(v) Why does part (iv) not contradict the local Taylor theorem (Theo-

rem 7.8)?

Q 18.5. In this question you may assume the standard properties of sin andcos but not their power series expansion.

(i) By considering the sign off1(x), when f1(t) =t sin t, show that

tsin t

for all t

0.(ii) By considering the sign off2(x), whenf2(t) = cos t 1 + t2/2!, show

that

cos t1 t2

2!

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for all t0.(iii) By considering the sign off3(x), whenf3(t) = sin t t + t3/3!, show

that

sin t

t

t3

3!for all t0.

(iv) State general results suggested by parts (i) to (iii) and prove themby induction. State and prove corresponding results for t 0 and f(0) = 0 explain why f(t) > 0 for t > 0.Iff(0) 0 and f(0) = f(1) = 0 what can you say about f and why? Iff(1)0 andf(0) =f(1) = 0 what can you say about fand why?

(ii) Suppose f : [0, 1] R is twice differentiable with f(t) 0 for allt [0, 1] and f(0) = f(1) = 0. Show thatf(t) 0 for all t [0, 1]. [Hint:Consider the sign off.]

(iii) Suppose g : [a, b] R is twice differentiable with g (t) 0 for allt[a, b]. By considering the function f : [0, 1] R defined by

f(t) = g

(1 t)a +tb (1 t)g(a) tg(b)show that

g

(1 t)a +tb (1 t)g(a) +tg(b)for all t with 1t0.

[In other words a twice differentiable function with everywhere positivesecond derivative is convex. Convex functions are considered in the lastpart of the probability course where you prove the very elegant Jensensinequality. You should note, however, that not all convex functions are twice

differentiable (look at x |x|).]Q 18.7. The results of this question are also useful in the probability coursewhen you study extinction probabilities. Notice that the point of this ques-tion is to obtain rigorous proofs of the results stated.

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Suppose a0 > 0, a1, a2, . . . , an 0 andn

j=0 aj = 1. We set P(t) =nj=0 ajt

j.(i) FindP(0),P(1) andP(1). Show thatP(t)0 and P(t)0 for all

t0.(ii) Show that the equation P(t) = t has no solution with 0

t < 1 ifn

j=1jaj1 and exactly one such solution ifn

j=1jaj >1. We write forthe smallest solution ofP(t) =t with 0t1.

(iii) If we set x0 = 0 and xn = P(xn1) for n 1 show, by induction,thatxn1xn.

(iv) Deduce, giving your reasons explicitly, that xn must converge to alimit. Show that 0 andP() =. Deduce that = and so

xn asn .

Q 18.8. The first proof that transcendental numbers existed is due to Li-

ouville. This question gives a version of his proof.Let P(t) =N

j=0 ajtj with the aj integers and aN= 0. Let be a root

ofP.(i) Why can we choose a > 0 such that P(t)= 0 for all t with t

[ , +] and t=.(ii) Let be as in (i). Explain why there exists a K such that|P(t)

P(s)| K|t s| for allt, s[ , +].(iii) Ifp and qare integers such that q 1 and P(p/q)= 0 show that

|P(p/q)| qN. [Hint: Remember that the coefficients ofPare integers.](iv) If p and q are integers such that q 1, P(p/q)= 0 and p/q

[

, +] use (ii) and (iii) to show that pq

K1qN.(v) Show that, ifp and qare integers such that q1 and =p/q, then pq

min(, K1qN).(vi) Show that if is a root of a polynomial with integer coefficients then

there exists a Kand an N (depending on ) such that pq K1qN

whenever p and qare integers withq1 and =p/q.

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(vii) Explain why

n=010n! converges. Call the limit L. Show that

L

mn=0

10n!

2((10)m!)m1.

By taking q= (10)m! and looking at (vi), show that L can not be the rootof a polynomial with integer coefficients.

(viii) By looking at

n=0

n10n!

with n=1 show that there are uncountably many transcendentals.Q 18.9. In this question you may assume standard results on the power

functiontt

.Use the form of Taylors theorem given in Theorem 7.7 to show that

(1 +x) = 1 +x+( 1)

2! x2 + +( 1) . . . ( n+ 1)

n! xn +. . .

for all 0 x < 1. Note that the main part of your task is to estimate theremainder term.

What happens if is a positive integer?Why does your argument fail for1 < x < 0? (The result is true but

our form of the remainder does not give it. This is one of the reasons whythere are so many forms of Taylors with different remainders.)

Q 18.10. (i) Suppose that x0,x1, . . . xn are distinct. Set

ej(x) =k=j

x xkxj xk .

What is the value of ej (xk) if j= k and if j = k? Show that given realnumbers j there is a polynomial of degree n withf(xj) =j.

(ii) Suppose f : [a, b] R is n+ 1 times differentiable, a < x0 < x1 < < xn < b and P is a polynomial of degree n with P(xj) = f(xj) for0

j

n. We are interested in the error

e(x) =f(x) P(x)

at a point x[a, b] with x=xj for all j .

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Set

F(t) =f(t) P(t) e(x)n

k=0

t xkx xk .

Explain whyFvanishes at leastn+2 times on [a, b]. Explain whyF vanishesat least n+ 1 times on [a, b]. By repeating the argument show thatF(n+1)

must vanish at least once (at , say) on [a, b]. Show that

e(x) = 1

(n+ 1)!f(n+1)()

nk=0

(x xk).

Deduce that if|f(n+1)(t)| M for all t[a, b] then

|f(t)

P(t)

|

M

(n+ 1)! |Q(t)

|whereQ(t) =

nk=0(t xk).

[The argument just given should remind you of the proof of Theorem 7.6.]

Q 18.11. Recall from last term (or earlier) that

(cos +i sin )n = cos n+i sin n

for all real and all integers n0.By taking real parts, show that there is a real polynomial Tn of degreen

such that

Tn(cos ) = cos n

for all real .Write down T0(t),T1(t),T2(t) and T3(t) explicitly.Show that, ifn1 then(a)|Tn(t)| 1 for alltwith|t| 1,(b) Tn+1(t) = 2tTn(t) Tn1(t), (why does this result hold for all t?)(c) the coefficient oftn inTn(t) is 2

n1.Explain whyTn+1has n +1 zerosx0,x1,. . . , xnlying in (1, 1). Use (a),

(c) and the final result of question 18.10 to show that if [a, b] = [

1, 1]

and f and Pobey the hypotheses of question 18.10 (so that, in particular,|f(n+1)(t)| M for all t[1, 1]) then

|f(t) P(t)| M(n+ 1)!

|2nTn+1(t)| M2n(n+ 1)!

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The rest of the question just asks for another of useful property of theTchebychev polynomials Tn. (The modern view is that Tchebychev shouldhave called himself Chebychev. He seems to have preferred Tchebycheff.)

(d) Ifn, m0 then 11

Tn(x)Tm(x)

(1 x2)1/2 dx=0 ifn=m

2 ifn = m= 0

What happens ifn= m = 0?

Q 18.12. (Exercise 8.4.)LetEbe a subset ofC,f,g be some functions from EtoC, andzsome

point ofE.(i) The limit is unique. That is, iff(w) l and f(w) k as w z

then l = k.(ii) Ifz

E

E and f(w)

l as w

zthrough values w

E, then

f(w)l as wzthrough values wE.(iii) Iff(u) =c for alluEthenf(w)c as wz.(iv) Iff(w)l and g(w)k as wz then f(w) +g(w)l+k.(v) Iff(w)l andg(w)k as wz then f(w)g(w)lk .(vi) If f(w) l as w z, f(u)= 0 for each u E and l= 0 then

f(w)1 l1.Q 18.13. (Exercise 8.5 and 8.6.) Suppose that E is a subset ofC, thatzE, and that f andg are functions fromE to C.

(i) Iff(u) =c for all uE, then f is continuous on E.(ii) Iff and g are continuous at z, then so is f+g.

(iii) Let us define f g :E C byf g(u) = f(u)g(u) for all uE.Then iff and g are continuous at z, so is f g.

(iv) Suppose that f(u)= 0 for all u E. Iff is continuous at z so is1/f.

(v) IfzE Eand f is continuous at z then the restriction f|E offtoE is also continuous at z.

(vi) IfJ : C C is defined byJ(z) =zfor allz C, thenJis continuouson C.

(vii) Every polynomial P is continuous on C.(viii) Suppose thatP andQ are polynomials and that Q is never zero on

some subsetEofC. Then the rational functionP /Qis continuous onE(or,more precisely, the restriction ofP /Qto E is continuous.)(ix) Let U and V be subsets ofC. Suppose f : U C is such that

f(z) V for all z U. If f is continuous at w U and g : V C iscontinuous at f(w), then the composition g f is continuous at w.

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Q 18.14. (Exercises 8.8 8.9 and 8.10.)LetEbe a subset ofC,f,g be some functions from EtoC, andzsome

point ofE.Show that iff is differentiable at z then f is continuous at z.Prove the following results.(i) Iff(u) =c for all uEthenf is differentiable at zwith f(z) = 0.(ii) Iff and g are differentiable at zthen so is their sum f+g and

(f+ g)(z) =f(z) +g(z).

(iii) Iff and g are differentiable at zthen so is their product f g and

(f g)(z) =f(z)g(z) +f(z)g(z).

(iv) If f is differentiable at z and f(u)= 0 for all u E then 1/f isdifferentiable at zand

(1/f)(z) =f(z)/f(z)2.

(v) Iff(u) =N

r=0 arur on E then thenf is differentiable at zand

f(z) =N

r=1

rarzr1

.(vi) Let U and V be subsets ofC. Suppose f : U C is such that

f(z) V for all t U. Iff is differentiable at w U and g : V R isdifferentiable at f(w), then the composition g f is differentiable at w with

(g f)(w) = f(w)g(f(w)).

Q 18.15. (Exercise 9.3.) Suppose that

n=0 anzn has radius of convergence

R. Then the sequence|anzn| is unbounded if|z| > R and

n=0 anzn con-

verges absolutely if|z|< R.Q 18.16. This question requires Abels test from Exercise 17.10 which isalso starred.

(i) Show that

n=1 zn/nhas radius of convergence 1, converges for all z

with|z|= 1 and z= 1 but diverges ifz= 1.(ii) Let|z1| =|z2| = =|zm| = 1. Find a power series

n=0 anzn

which has radius of convergence 1 and converges for all z with|z| = 1 andz=z1, z2, . . . , z m but diverges ifz= zj for some 1jm.

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Q 18.17. (i) Show that there exist power series of all radii of convergence(including 0 and).

(ii) Suppose the power series

n=0 anzn has radius of convergence R and

the power series

n=0 bnz

n has radius of convergenceS. Show that, ifR=Sthenn=0(an+bn)zn has radius of convergence min(R, S).(iii) Suppose that the conditions of (ii) hold except that R = S. Showthat the radius of convergence T of

n=0(an+bn)z

n satisfies the conditionT R. Show by means of examples that Tcan take any value with T R.[Hint: Start by thinking of a simple relation betweenan and bn which willgive T =.]Q 18.18. (i) Suppose thatxnis a bounded sequence of real numbers. Showcarefully that yn= supmn xm is a bounded decreasing sequence and deducethatyn tends to a limit y say. We write

limsupn

xn= y.

(ii) Suppose that xn is a bounded sequence of real numbers. Prove thefollowing two results.

(A) Given any >0, we can find an M() such that

xnlim supm

xm+

for all nM().(B) Given any > 0 and any N we can find an integer P(N, ) N

such that

xP(N,)lim supm

xm .

(iii) Using (ii) or otherwise show that ifanC and the sequence|an|1/nis bounded and lim supn |an|1/n = 0 then

n=0 anz

n has radius of conver-gence (lim supn |an|1/n)1.

Show also that, if the sequence|an|1/n is unbounded,

n=0 anzn has ra-

dius of convergence 0, and, if lim supn |an|1/n = 0,

n=0 anzn has infinite

radius of convergence.[This result has considerable theoretical significance but I strongly advise

using the definition directly rather than relying on the formula.](iv) Use the formula of this question to obtain the results of Ques-tion 18.17.

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Q 18.19. (i) Starting from the observation that

1 + x+x2 + +xn =1 xn+1

1 x

show that

1

1 x =

n=0

xn

for all|x|< 1. Show that the result is false or meaningless if|x| 1.(ii) Use term by term differentiation to obtain power series expansions

for (1 x)n for all integer n with n1.(iii) Find the radius of convergence of

n=1 x

n/n. Use term by termdifferentiation and the constant value theorem to show that

log(1 x) =

n=1

xn

n.

(iv) Use similar ideas to obtain a power series expansion for tan1 xin arange to be stated.

Q 18.20. (Exercise 10.10.)Write down what you consider to be the chief properties of sinh and cosh.

(They should convey enough information to draw a reasonable graph of thetwo functions.)

(i) Obtain those properties by starting with the definitions

sinh x=

n=0

x2n+1

(2n+ 1)! and cosh x=

n=0

x2n

(2n)!

and proceeding along the lines of Lemma 10.9.(ii) Obtain those properties by starting with the definitions

sinh x=exp x exp(x)

2 and cosh x=

exp x+ exp(x)2

Q 18.21. (Exercise 11.6.)

(i) Explain why the infinite sums

sin z=

n=0

(1)nz2n+1(2n+ 1)!

and cos z=

n=0

(1)nz2n(2n)!

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converge everywhere and are differentiable everywhere with sin z = cos z,cos z= sin z.

(ii) Show that

sin z=exp iz exp(iz)

2i , cos z=

exp iz+ exp(iz)2

and

exp iz= cos z+i sin z

for all z C.(iii) Show that

sin(z+w) = sin zcos w+ cos zsin w, cos(z+ w) = cos zcos w sin zsin w,and (sin z)2 + (cos z)2 = 1 for all z , w C.

(iv) sin(z) = sin z, cos(z) = cos zfor allz C.(v) sin and cos are 2 periodic in the sense that

sin(z+ 2) = sin z and cos(z+ 2) = cos z

for all z C.(vi) Ifx is real then sin ix= i sinh xand cos ix= cosh x.(vii) Recover the addition formulae for sinh and cosh by setting z = ix

andw = iy in part (iii).(viii) Show that| sin z| and| cos z| are bounded if|z| K for some K

but that| sin z| and| cos z| are unbounded on C.Q 18.22. (Exercise 11.7.)

Suppose, if possible, that there exists a continuous L : C \ {0} Cwithexp(L(z)) = zfor all z C \ {0}.

(i) If is real, show that L(exp(i)) = i(+ 2n()) for some n() Z.(ii) Define f : R R by

f() = 1

2

L(exp i) L(1)

i

.

Show thatfis a well defined continuous function, thatf() Z for all R,thatf(0) = 0 and thatf(2) =1.

(iii) Show that the statements made in the last sentence of (ii) are incom-patible with the intermediate value theorem and deduce that no function canexist with the supposed properties ofL.

(iv) Discuss informally what connection, if any, the discussion above haswith the existence of the international date line.

Q 18.23. (Exercise 11.8.) Show by modifying the argument of Exercise 11.7,that there does not exist a continuousS: C C withS(z)2 =zfor allz C.

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19 Fourth example sheet

Students vary in how much work they are prepared to do. On the whole,exercises from the main text are reasonably easy and provide good practicein the ideas. Questions and parts of questions marked with are not intended

to be hard but cover topics less central to the present course.

Q 19.1. (Exercise 12.9.) Suppose that b a, , R, and f and g areRiemann integrable. Which of the following statements are always true andwhich are not? Give a proof or counter-example. If the statement is notalways true, find an appropriate correction and prove it.

(i)

ab

f(x) + g(x) dx=

ab

f(x) dx+

ab

g(x) dx.

(ii) Iff(x)g(x) for allx[a, b], then a

b

f(x) dx a

b

g(x) dx.

Q 19.2. (Exercise 13.5.) Let a c b. Give an example of a Riemannintegrable function f : [a, b] R such that f(t)0 for all t[a, b] and b

a

f(t) dt= 0,

but f(c)= 0.Q 19.3. Define f : [0, 1] R byf(p/q) = 1/qwhen p and qare coprimeintegers with 1p < qandf(x) = 0 otherwise.

(i) Show that fis Riemann integrable and find 1

0 f(x) dx.

(ii) At which points is fcontinuous? Prove your answer.Q 19.4. (Exercise 13.7).

(i) Let Hbe the Heaviside function H : R R given by H(x) = 0 forx


56/62

when b > a >0.(ii) If we set a =1,b= 1 in the formula of (i), we obtain

1

1

dx

1 +x2?

=

1

1

dt

1 +t2

Explain this apparent failure of the method of integration by substitution.(iii) Write the result of (i) in terms of tan1 and prove it using standard

trigonometric identities.

Q 19.6. In this question we give an alternative treatment of the logarithmso no properties of the exponential or logarithmic function should be used.You should quote all the theorems that you use, paying particular attentionto those on integration.

We set

l(x) = x

1

1

t

dt.

(i) Explain whyl : (0, )R is a well defined function.(ii) Use the change of variable theorem for integrals to show that xy

x

1

tdt = l(y)

for all x, y >0. Deduce that l(xy) =l(x) +l(y).(iii) Show that l is everywhere differentiable with l(x) = 1/x.(iv) Show that l is a strictly increasing function.(v) Show that l(x) as x .

Q 19.7. In the lectures we deduced the properties of the logarithm fromthose of the exponential. Reverse this by making a list of the properties ofthe exponential, define exp as the inverse function of log (explaining carefullywhy you can do this) and using the properties of log found in the previousquestion (Question 19.6).

Q 19.8. [The first mean value theorem for integrals] Suppose g :[a, b] R is a continuous function such that g(x) 0 for all x [a, b]. Iff : [a, b]R is continuous explain why we can find k1 and k2 in [a, b] suchthat

f(k1)f(x)f(k2)for all x[a, b]. Deduce carefully that

f(k1)

ba

g(x) dx b

a

f(x)g(x) dxf(k2) b

a

g(x) dx

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and show, stating carefully any theorem that you need, that there exists ac[a, b] such that

b

a

f(x)g(x) dx= f(c) b

a

g(x) dx.

Does this result remain true ifg(x)0 for allx[a, b]? Does this resultremain true if we place no restrictions on g (apart from continuity). In eachcase give a proof or a counter example. (The first mean value theorem forintegrals is used in the numerical analysis course.)

Q 19.9. Use the integral form of the remainder in Taylors theorem (i.e.Theorem 13.13) to obtain the power series expansion for sin.

Q 19.10. [The binomial theorem]If 1> xt0 show thatx t1 +t

x.

If1< xt0 show thatt x1 +t

x.

Use the integral form of the Taylor theorem to show that, if|x|


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In other words ba

f(t)g(t) dt

2 b

a

f(t)2 dt

ba

g(t)2 dt.

(iv) When do we have equality in the inequality proved in (iii).

Q 19.12. Explain why

1

n+ 1

n+1n

1

xdx 1

n.

Hence or otherwise show that if we write

Tn=n

r=1

1

r log n

we have Tn+1 Tn for all n1. Show also that 1Tn 0. Deduce thatTn tends to a limit (Eulers constant) with 10. [It is an indicationof how little we know about specific real numbers that, after three centuries,we still do not know whether is irrational. G. H. Hardy is said to haveoffered his chair to anyone who could prove that was transcendental.]

(ii) By consideringT2n Tn, show that

1 12

+1

31

4+

1

51

6+

1

71

8 = log 2

(iii) By consideringT4n 12T2n 12Tn, show that

1 +131

2+ 1

5+ 1

71

4 = 3

2log 2

The famous example is due to Dirichlet. It gives a specific example whererearranging a non-absolutely convergent sum changes its value.

Q 19.13. [Simple versions of Stirlings formula](The first part of thisquestion is in the probability course.)

(i) Prove that

n

1

log x dxn

r=1 log r n

1

log x dx+ log n.

Computen1

log x dxand hence show that

1

n(n!)1/n 1

e

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asn .(ii) Show that

n+1/2

n1/2

log x dx log n= 1/2

0

log1 + t

n + log1 t

ndt.By using the mean value theorem, or otherwise, deduce that

n+1/2n1/2

log x dx log n 43n2 .

(You may replace 4/(3n2) by An2 with A another constant, if you wish.)

Deduce thatN+1/21/2 log x dx log N! converges to a limit. Conclude that

n!

(n+ 1/2)(n+1/2)en

C

asn for some constant C.(iii) Find a function f(n) not involving factorials such that

f(n)

2n

n

tends to a limit as n . You should try for anf in as simple a form aspossible. (But, of course, different people

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