Introduction to analysis

4 t h Edition Introduction to Analysis 4 t h Edition Introduction to Analysis Edward D. Gaughan New Mexico State U...

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4 t h Edition

Introduction to Analysis

4 t h Edition

Introduction to Analysis

Edward D. Gaughan New Mexico State University

1

I/I

BrookslCole Publishing Company Pacific Grove, California

Consulting Editor: Robert J. Wisner

BrookdCole Publishing Company A Division of Wadsworth, Inc. @ 1993, 1987, 1975, 1968 by Wadsworth, Inc., Belrnont, California 94002. All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transcribed, in any form or by any means-electmnic, mechanical, photocopying, recording, or otherwise-without the prior written permission of the publisher, Brooks/Cole Publishing Company, Pacific Grove, California 93950, a division of Wadsworth, Inc. Printed in the United States of America

Library of Congress Cataloging in Publication Data Gaughan, Edward. Introduction to analysis / Edward D. Gaughan. - 4th ed. p. cm. Includes index. ISBN 0-534-17358-6 1. Mathematical analysis. I. Title. QA300.G34 1992 5 l5-dc2O Sponsoring Editor: Jeremy Hayhurst Editorial Assistant: Nancy Champlin Production Editor: Linda Loba Manuscript Editor: Laurie Vaughan Interior Design: Roy Neuhaus Cover Design: Susan Haberbrn Art Coordinator: Lisa Torri Interior Illustration: Gloria Langer 'I)lpesetting: Archetype Publishing, Inc. Printing and Binding: Arcata Graphics-Faigleld

92- 13531

CIP

Preface

ntroduction to Analysis is designed to bridge the gap between the intuitive calculus normally offered at the undergraduate level and the sophisticated analysis courses the student encounters at the senior or first-year-graduate level. Through a rigorous approach to the usual topics handled in one-dimensional calculus-limits, continuity, differentiation, integration, and infinite series-the book offers a deeper understanding of the ideas encountered in the calculus. Although the text assumes that the reader has completed several semesters of calculus, this assumption is necessary only for some of the motivation (of theorems) and examples. The book has been written with two important goals in mind for its readers: the development of a rigorous foundation for the basic topics of analysis, and the less tangible acquisition of an accurate intuitive feeling for analysis. In the interest of these goals, considerable time is devoted to motivating and developing new concepts. Economy of space is often sacrificed so that ideas can be introduced in a natural fashion. This 4th edition contains a number of changes recommended by the reviewers and users of earlier editions of the book. Chapter 0 contains introductory material on sets, functions, relations, mathematical induction, recursion, equivalent and countable sets, and the set of real numbers. As in the 3rd edition, the set of real numbers is postulated as an ordered field with the least upper bound property. Chapters 1 through 4 contain the material on sequences, limits of functions, continuity, and differentiation. Chapter 5 is devoted to the Riemann integral, rather than the Riemann-Stieltjes integral treated in the first edition. Chapter 6 treats infinite series, and Chapter 7 contains material on sequences and series of functions. The exercise sets have been expanded and offer a selection of exercises with level of difficulty ranging from very routine to quite challenging. The starred exercises are of particular importance, because they contain facts vital to the development of later sections. The star is not used to indicate the more difficult exercises.

vi Preface

At the end of each chapter, you will find several PROJECTS. The purpose of a PROJECT is to give the student the necessary guidance to solve a substantial mathematical problem. A PROJECT is distinguished from an exercise in that the PROJECT is a multi-step problem which can be very difficult for the beginner without significant assistance. PROJECTS are sometimes used to cover material not included in the chapter discussions. For example, PROJECT 2.2 includes left- and right-hand limits. PROJECT 3.1 offers an approach to uniform continuity without the use of compactness. Other PROJECTS are used to generalize theorems and proofs in the text. In the course of this exposition, a number of famous names are mentioned: Cauchy, Bolzano, Weierstrass, Riemann, Caratheodory, and others. A serious student should seek to know something about the persons who have made important contributions to analysis. The reader is urged to indulge in a little historical research when encountering the names of these people. I would like to acknowledge the reviewers: Professor Lew Friedland, State University of New York; Professor Terry L. Herdrnan, Virginia Polytechnic Institute and State University; Professor Charles Hirnrnelberg, University of Kansas; Professor Ken F. Kopfenstein, Colorado State University; and Professor James C. Magee, State University of New York. My sincere appreciation is also extended to Robert J. Wisner, who, some 28 years ago, urged me to write the first edition of this book, and who has made many helpful suggestions relative to each of the three earlier editions; Jack N. Thornton, who as managing editor of Brooks/Cole Publishing Co. accepted and nurtured a fledgling author through the first and second editions; and to Jeremy Hayhurst, mathematics editor at Brooks/Cole Publishing Co., for his support and encouragement in the preparation of the third and fourth edition. Of course, this book would not be in your hands without the excellent work of Linda Loba who carried it and me through the stages of production. My thanks go to her also.

Edward D. Gaughan

Contents

SETS 2 RELATIONS AND FUNCTIONS 8 MATHEMATICAL INDUCTION AND RECURSION 13 EQUIVALENT AND COUNTABLE SETS 17 REAL NUMBERS 22 EXERCISES 29 PROJECT 0.1 32 PROJECT 0.2 32 PROJECT 0.3 33

Chapter

1 Sequences 35

SEQUENCES AND CONVERGENCE 35 CAUCHY SEQUENCES 40 ARITHMETIC OPERATIONS ON SEQUENCES 45 SUBSEQUENCES AND MONOTONE SEQUENCES 52 EXERCISES 57 PROJECT 1.1 60 PROJECT 1.2 61 PROJECT 1.3 62 PROJECT 1.4 62 a

vii

viii Contents

Chapter

2 .I 2.2 2.3 2.4

2

Limits of Functions 63

DEFINITION OF THE LIMIT OF A FUNCTION 63 LIMITS OF FUNCTIONS AND SEQUENCES 69 ALGEBRA OF LIMITS 72 LIMITS OF MONOTONE FUNCTIONS 77 EXERCISES 79 PROJECT 2.1 81 PROJECT 2.2 81 PROJECT 2.3 82 PROJECT 2.4 82

Chapter

3

Continuity 85

3.1 CONTINUZTY OF A FUNCTION AT A POINT 85 3.2 ALGEBRA OF CONTINUOUS FUNCTIONS 88 3.3 UNIFORM CONTINUITY: OPEN, CLOSED, AND COMPACT SETS 3.4 PROPERTIES OF CONTINUOUS FUNCTIONS 100

92

EXERCISES 107 PROJECT 3.1 109 PROJECT 3.2 110 PROJECT3.3 110 PROJECT3.4 111

Chapter

4.1 4.2 4.3 4.4

4

Dwerentiation 113

THE DERNATNE OF A FUNCTION I14 THE ALGEBRA OF DERNATWES I17 R O U S THEOREM AND THE MEAN-VALUE THEOREM I22 L'HOSPITAL'S RULE-AND THE INVERSE-FUNCTION THEOREM, 128 EXERCISES 132 PROJECT 4.1 136 PROJECT 4.2 137 PROJECT 4.3 1;s

Contents ix

Chapter

5

The Riemann Integral 139

5.1 5.2 5.3 5.4 5.5 5.6

THE RZEMANN INTEGRAL 140 CLASSES OF INTEGRABLE FUNCTIONS 149 RIEMANN SUMS 151 THE FUNDAMENTAL THEOREM OF INTEGRABLE CALCULUS 157 ALGEBRA OF INTEGRABLE FUNCTIONS 159 DERIVATWES OF INTEGRALS 164 5.7 MEAN-VALUE AND CHANGE-OF-VARIABLE THEOREMS 166 EXERCISES 170 PROJECT 5.1 174 PROJECT 5.2 175

Chapter 6.1 6.2 6.3 6.4 6.5 6.6

6

Infinite Series

177

CONVERGENCE OF INFINITE SERIES 177 ABSOLUTE CONVERGENCE AND THE COMPARISON TEST 182 RATIO AND ROOT TESTS I86 CONDITIONAL CONVERGENCE 190 POWER SERIES 200 TAYLOR SERIES 209 EXERCISES, 213 PROJECT 6.1 217 PROJECT 6.2 217 PROJECT 6.3 218 t

Chapter

l

'*

7.2 7.3

7

Sequences and Series of Functions 219

CONSEQUENCES OF UNIFORM CONVERGENCE 225 UNIFORM CONVERGENCE OF POWER SERIES 230 EXERCISES 238 PROJECT 7.1 239 PROJECT 7.2 241

INDEX 243

Chapter

0

Preliminaries

efore attempting to study analysis, one must be able to read and communicate mathematics intelligently. This fact is not unique to analysis but is true in all of mathematics. This chapter presents some of the basic vocabulary of mathematics; in fact, its contents, with some rearrangement, may be similar to the beginning chapter of a book at this level in algebra, topology, or other topics in mathematics. This similarity is not accidental. A certain basic vocabulary is common to a good share of mathematics. The chapter does, however, exclude anything unnecessary for the assimilation of the material to come in this book. A few words of both warning and encouragement are in order. First, you should realize that only the fundamentals of your mathematical vocabulary are presented here; the proper usage comes with practice and increasing mathematical maturity. Thus, at the beginning it may seem a bit awkward to use new and possibly unfamiliar ideas in the development of additional concepts. Now the words of encouragement: The initial ideas presented and the theorems proved are quite simple; hence, they will give you many chances to practice your vocabulary in settings where intuition can help to guide your thinking. You are encouraged to play this game quite seriously by giving precise proofs to easy theorems, thus gaining practice in clear and precise mathematical expression-an ability that will be invaluable as the material becomes more difficult in later chapters. One last bit of admonishment is appropriate. Mathematics, by its very nature, begs to be communicated. It is difficult to imagine a mathematician who, upon discovering a new fact or proof, does not have a burning desire to shout it from the rooftops. In fact, any professional mathematician-teacher, researcher, or what have you-must be able to communicate with others. Those who say that mathematics is completely incomprehensible have either failed to learn the language of mathematics or have had the misfortune of trying to learn mathematics from someone who cannot or will not use the language properly. Much symbolism is used in mathematics, but each symbol or set of symbols must stand for a word or phrase in the language used,

2

Chapter 0 Preliminaries

which is English in this case. In particular, the sentences formed with symbols must make sense when translated into words and must impart the meaning intended. A good test for your use of mathematical symbols is to have someone read your writing with a critical eye, translating it aloud to you. Then you can see whether it meets the test for clarity and meaning.

0.1 SETS We shall begin with the naive notion of a set. A simple approach will be sufficient for our purposes in this book. We shall think of a set as a collection of objects. Note that the word collection is as undefined in this setting as is the word set. In order to build the intuitive idea of a set, consider the following examples: 1. 2. 3. 4. 5.

The set of all natural numbers The set of all the letters of the Greek alphabet The set of all rational roots of the equation 2 + 1 = 0 The set consisting of the rational numbers 1, 2, and 3 The set of all integers less than 4 and greater than 0

Note that in examples 1, 2, 3, and 5 the set was described by a rule for determining which objects belong to the set, whereas in example 4 the elements were explicitly named. In particular, observe that those objects that pass the test for membership in the set described in example 5 are precisely the objects listed in example 4. It would be quite disconcerting if these two sets were different-that is, if a set were determined by the method used to describe it rather than by the objects belonging to it. That this will not be the case is made clear by the following definition.

DEFINITION If A and B are sets, then A = B if and only if every object belonging to A also belongs to B and every object belonging to B also belongs to A. This definition makes it obvious that the sets described in examples 4 and 5 are equal. To show that two sets A and B are not equal, it suffices to find an object that belongs to one set and not to the other.

DEFINITION If A is a set and x is an object that belongs to A, we say "x is an element of A" or "x & a member of A" and write x E A. If x does not belong to A, we write x $ A. Let us use this notation to rewrite the definition of equality for sets. If A and B are sets, then A = B if and only if for each object x, x E A implies x E B, and x E B implies x E A. Let A1 be the set in example 1, A2 be the set in example 2, and so on. Now, as observed above, A4 = A5 and Al # A4, because 13 E A1 and 13 4 A4. Note, however, that each member of A4 is also a member of A1. In a sense, A4 is a part of A1.

0.1 Sets 3

D E m T Z O N If A and B are sets such that each member of A is also a member of B, then A is a subset of B or A is contained in B, written A c B. IfA c B and A # B, then A is a proper subset of B.

We are now in a position to state and prove the first theorem of this book. It will be useful later when we wish to prove that two sets are equal. Read and reread the statement of the theorem and reflect on its meaning. Of course, in light of the three definitions just given, it seems obvious that the theorem is true. But let us use the opportunity to present an easy proof and discuss the method of proof to be presented. This format is typical for a theorem in mathematics; it is an "if and only if' theorem. Basically, the theorem has two parts: (1) if A = B, then A C B and B C A; and (2) if A C B and B C A, then A = B. To prove the theorem, we must prove that both (1) and (2) hold. Follow along. 0.1 THEOREM Let A and B be sets. Then A = B if and only if A C B and B CA. Proof Assume A = B. Then, if x E A, then x E B since A = B (see the definition of A = B); hence A c B. Likewise, if x E B, then x E A since A = B, and we have B c A. Thus, if A = B, then A c B and B c A. Assume now that A c B and B c A. We want to show that A = B. Choose a n y x ~ A S. i n c e A C B , t h e n x ~ B I. f x ~ B , t h e n s i n c e BcA, wehavexEA. Thus, by the definition of set equality, A = B. This concludes the proof. It is convenient to have a variety of ways of describing sets. We have already seen sets defined by a rule for membership and by a list of the members of the set. If P(x) is a statement concerning an object x, then we denote {x :P(x)) the set of all objects x such that P(x) is true. The kind of statements that are permissible here is a subject beyond the scope of this book. It suffices to declare that the types encountered in this book are permissible.

Exdmple 0.1 The set of all natural numbers is {x : x is a natural number). The set of all integers less than 4 and greater than 0 is equal to {n : 0 < n < 4 and n is an integer). There are some sets to which we give special names, and we list them below: 1. J = set of all natural numbers-that is, the set of all positive integers. 2. Z = set of all integers. 3. R = the set of all real numbers. 4. F o r a < b, [ a , b ] = { x : x ~ Ra n d a s x s b). 5. For a < b, [a,b)= {x:x E R and a < x < b). 6. F o r a < b , ( a , b ] = { x : x ~ R a n d a < x ~ b ) .

4


7. F o r a < b , ( a , b ) = { x : x ~ R a n d a < x < b ) . 8. We use the symbol 0 for the empty set; that is, 0 is the set that contains no elements.

DEFINITON If A and B are sets, the union of A and B, written A u B, is defined to be the set of all objects that belong to either A or B or possibly both. In the notation used previously, A U B = {x :x E A or x E B). DEFINITION I f A and B are sets, the intersection of A and B, written A n B, is defined to be the set of all objects that belong to both A and B. In other words, AnB={x:x~Aandx~B). To partially digest these new definitions, let us consider a few more examples. Let

NowAnB={a,p),A~B={1,2,4,a,p,~),andAnCisempty. IftwosetsAand B have the property that A n B is empty, we say that they are disjoint. An appropriate theorem to state at this stage contains some parts that are quite obvious from the definitions of union and intersection and some that are not quite so obvious. 0.2 THEOREM Let A, B, and C be sets. Then i. ii. iii. iv. v. vi.

AnB=BnA. AUB=BUA. (AnB)nC=An(BnC). (AUB)UC=AU(BUC). An(BUC)=(AnB)U(AnC). Au(BnC)=(AUB)n(AUC).

Parts (i) through (iv) have very obvious proofs, which will not be supplied here. The proof of (v) follows, and that of (vi) is left as Exercise 3.

Proof In proving (v) we shall use Theorem 0.1 to show that

and

and hence the two sets are equal. Suppose x E A n (B U C). Then x E A and x E B U C. If x E B, then since x E A , x e A n B . I f x e B , t h e n s i n c e x ~ B ~ C , x ~ C a n d a l s o xhence, ~A;

0.1 Sets 5

x E A n c. Thus, in particular, x E A n B or x E A f7 C. In other words, x€(AnB)U(AnC). We have shown that A n (B U C) C (A n B) U (A n C). Suppose

,

x E (AnB)U(AnC). Thenx E A n B o r x A~n C . I f x E A n B , t h e n x ~ Aandx EB; hence,xEA a n d x ~ B U C s i n c e B ~ B U Thus, C. xEAn(BUC). I f x $ A n B , thenx E A ~ C hence,^ ; E A andx E C, but C x E A n (B U C). We have shown

c BUC, so that

and the proof is complete. Some of the reasonably easy relations concerning sets given in the exercises will be used later, arid you are urged to prove them for your own practice-for the simple reason that a proof makes a theorem much easier to remember.

DEFINITION Let A and B be sets. Then the complement of A relative to B, written B \A, is defined to be the set of all objects belonging to B but not to A. In other words,

Let A = {1,2,3,4,5,6) and B = {2,4,6,8,10). Then B \ A = {8,10) and A \ B = {1,3,5). The following form of what are called De Morgan's Laws may look a bit cumbersome because of our decision to speak of the complement of A relative to B rather than the complement of a set relative to some implied universal set. However, this path ieaves little room for misunderstanding. 0.3 THEOREM

Let A, B, and C be sets. Then

Proof Again, the method of proof will be to use Theorem 0.1, showing that the set on the left side of the proposed equality is a subset of the set on the right

6 Chapter 0 Preliminaries

and vice versa. Although this method of proof will be used quite often in such theorems, it will not always be advertised in the future. (i) Let

Then x E A and x $ B n C. Since x $ B n C, either x $ B or x $ C. If x $ B, . then,sincex~A,x~A\B.Ifx$!C,thensincex~A,x~A\C.Hence,

Thus,

Suppose now that

Theneitherx E A \ B o r x ~ A \ C I. ~ x E A \ B t, h e n x ~ Aa n d x $ B, sothat x $ B n C; thus, x E A \ (B f l C). If x E A \ C, then x E A and x $ C, so that x $ B n C ; thus xEA\(BnC). We have shown that (A \ B) U (A \ C) c A \ (B n C). This result, coupled with that of the preceding paragraph, completes the proof of (i). The proof of (ii) is left as Exercise 4.

In order to gain a deeper understanding of Theorem 0.3, let us consider a special case and invent some new notation. Suppose B and C are both subsets of A. If S is a subset of A, let S* denote the complement of S relative to A so that S* = A \ S. In this setting, the theorem states that ( B n C ) * = B * u C * and

(BuC)' = B * n C * .

In cruder terms, the complement of the intersection is the union of the complements, and the complement of the union is the intersection of the complements. The notions of union and intersection can be generalized to unions and intersections of more than two sets. But first some notation is needed. DEFINITION Let A be a set, and suppose for each X E A, a subset Ax of a given set S is specified. The collection of sets Ax is called an indexed family of subsets of S with A as the index set. We denote this by

0.1 Sets 7

DEFINlTZON

If {Ax)xEA is an indexed family of sets, define

and

UAx = {x :x E Ax for some X E A).

.

XEA

The reader is invited to generalize some of the previous theorems in light of these new ideas. There is one logical difficulty here that needs to be pointed out. If A is the empty set, then it is easy to see that UxEAAx is empty; however, it is not clear what to expect of nXEA AX. This could be overcome by insisting that all index sets be nonempty, and this shall be done in some cases. In a context where all sets considered are understood to be subsets of a given set S, the common usage is to let

where 0 is the empty set. The reason for this discussion is to alert the reader of possible problems that might arise from this situation, not to give a solution. In the event that the index set is the set J of positive integers or a finite subset thereof, some specid notation may be adopted. For example, UnEJAn may dso be written U g l A,, and if S = {1,2,. .. ,k), we may write uftl A, instead of UnEsAn. Of course, similar notation may be used for the intersection of a family of sets. Example 0.2

Let A, = (0,

i)for each n E J. Then

n 00

A, = 0 (the empty set).

n= 1

:

Let us see why. If x E ngl A, then x E (0, :) for each n E J; that is, 0 < x < for all n E J. However, if we pick no to be any positive integer greater than then

:,

o < & < x and x 4

0, - , contrary to our assumption. ~ h u ngl s A, is the empty ( k) set since no real number can satisfy the conditions for membership. We are assuming

some prior knowledge of real numbers, integers, and inequalities, but we will discuss these details later in this chapter. Example 0.3

Let B, = ( , 1 ) for each n E J . Then


Again, let us see why. Suppose 0 < x < 1. There is a positive integer m such that < m, hence < x < 1; that is x E (f,I). But then, since x belongs to one set in the indexed family, it must belong to the union of the sets in the indexed family. So (0,l) c Ugl Bn. On the other hand, if x E U g l Bn, then x E Bn for some n E J; hence, 0 < < x < 1. In any case 0 < x < 1. Thus Ugl Bn C (0,l). Therefore UZl Bn = (0,1)*

A generalization of De Morgan's Laws can be stated and proved for 'indexed families of sets, and such a generalization will be useful later.

0.4 THEOREM (De Morgan'shws) dexed family of subsets of S. Then

Let S be a set and {Ax)xEA be an in-

Proof We will prove (i) and leave (ii) as Exercise 11. Assume x E S\(uxEAAx). Then x E S and x 4 UxEAAx. Therefore x E S and x 4 Ax for all X E A. Therefore, x E S \ A x for all X E A. This means that x E nxEA(S\Ax). On the other hand, if x E nxEA(S\Ax), then x E S \ Ax for each X E A. Thus x E S and x 4 Ax for each X E A. This implies that x E S and x 4 UxEAAx. So x E S \ (UxEAAx).We have shown that

Therefore, the two sets are equal.

Example 0.4

Using Example 0.2 and De Morgan's Laws, we see that

0.2 RELATIONS AND FUNCTIONS Given objects x and y, x # y, it is easy to form the set whose only members are x and y: We can write the set as either {x, y} or {y,x} because the method of description doesn't determine the set, but determines only the objects that belong to the set. In

0.2 Relations and Functions 9

this section, however, we are interested in another object one may c o n s ~ - ~from ct two objects, namely an ordered pair. In an ordered pair, each of the objects has a special place-as opposed to the set, in which order is unimportant. This concept is not unfamiliar, for the notion of an ordered pair of real numbers is encountered in algebra at a tender age-for example, as in the case of fractions wherein numerators' and denominators are distinguished.

DEFINITION

The ordered pair (x, y) is defined to be the set {{x), {x, y)).

Give this definition a little time to soak in. The ordered pair (x,y) is the set whose members are the set whose only member is x and the set whose members are x and y. This is our first encounter here with a set that has other sets as members. If x = y, then (x,x) = {{x)). Should this definition seem disconcerting, remember that its purpose is to single out one object as the "first coordinate" and another as the "second coordinate." After proving the following theorem, we shall see that the definition serves the desired purpose.

0.5 THEOREM (x,y) = (u,v) if and only if x = u and y = v.

Proof If x = u and y = v, then

suppose now that (x,y) = (u,v). Then {{x), { x , ~ ) )= {{u), { u , ~ ) ) .I f x = Y , then

Since the set on the left has only one member, the same must be true of the set on the right. This can be true only if {u) = {u, v), which in turn can be true only if u = v. Thus, we have

and so {x) = {u), x = u, and, of course, y = x = u = v. Now suppose x and y are distinct. Then (x) and {x,y) are distinct sets and are the only members of {{x), {x, y)). By the assumed equality,

The case {x) = {u, v) is impossible because u and v must be distinct, and thus {u, v) has two elements, whereas {x) has only one element. Thus {x) = {u), SO x = U. Also


and the only remaining element that can be equal to { x , ~ )is {u, v). But x = u, and by the equality {x,y) = {u, v), we must have y = v. This concludes the proof. The concept of an ordered pair sets the stage for another way of combining two sets to form a new set. In analytic geometry, one considers the set'of all ordered pairs of real numbers-a new set constructed from an old one; a generalization of this will be the next definition.

DEFINITION If A and B are sets, define the Cartesian product of A and B, written A x B, to be the sets of all ordered pairs (a, b) such that a E A and b E B. Thus,AxB={(a,b):a~Aandb~B). W

Example 0.5

As an example, let A = { l,2,3) and B = {2,4,6). Then

.Note that, in general, if A has m elements and B has n elements, then A x B has mn elements.

DEFINITION A relation is a set of ordered pairs. A function is a relation F such that if (x, y) E F and (x, 2 ) E F, then y = z. If F is a function, the domain of F, written dom F, is defined to be dom F = {x : (x, y) E F for some y), and the image of F, written im F, is defined to be im F = {y : (x,y) E F for some x E dom F). Examine the definition of function and reflect on the meaning. For each x E dom F, there is exactly one element y E im F such that (x, y) E F. It is customary to refer to such a y as F(x). Now observe how F works. Choose any x E dom F; then F produces a unique element y in im F. We call it F(x). It is useful to think of F in a dynamic way. Input x and the function F produces an output, F(x). Often F(x) is computed by a formula or rule. Suppose that F is a fundtion with dom F = A and im F C B. We would refer to B as the codomain of F and write this symbolically as:

One reason for specifying the codomain rather than the image of F is that determining the image of F is sometimes very difficult.

Example 0.6 Functions may be defined in a variety of ways. A popular means of defining a function is to give a rule as well as the domain of the function. For

0.2 Relations and Functions 11

example, define a functionf :J + J as follows: f(n) = 2n - 1 for each n E J. Notice that the rule is unambiguous; that is, given n E J , there is no doubt about f (n). Also notice that the domain is specified. Consider now the functions f(x) = 2,x E R, and g(x) = G,x E R. There is a special relationship between f and g that we will discuss shortly. For the moment, notice that f(2) = 8 and g(8) = 2. This means that g "undoes" what f "does" and vice versa. You have seen other pairs of functions with this property before, and we will examine the idea in detail.

DEFINITION by

Let S be a relation. Then the converse of S, written 3, is defined

It would be interesting to discover under what conditions the converse of a function is also a function. Suppose F is a function with (x, y) and (x, z) both belonging to p. In order that jf be a function, it must be true that y = z. Translating back to F, for all (y,x) and (z,x) in F, it must be true that y = z; or, in other words, if F(y) = F(z), then y = z. This idea is important enough to deserve a special definition.

DEFINITION A function f is 1-1 (pronounced "one to one") if and only if, for all y, z in the domain off, f (y) =f(z) implies y = z. In essence, this says that a 1-1 function is one that assumes each value in its image exactly once. The next theorem is another "if and only if" theorem that takes the form: "Property A is satisfied if and only if Property B is satisfied. It has become accepted practice in mathematics literature to use "iff" as an abbreviation for "if and only if." Thus, the theorem could be written: Property A is satisfied iff Property B is satisfied." 0.6 THEOREM Let F be a function. Then p (the converse of F) is a function iff F is a 1-1 functi6n. 1f fi is a function, then dom p = im F and im p = dom F.

Proof Assume k' is a function. Suppose x and y belong to the domain of F with F(x) = F O = z. Then (x,z) E F and (y,z) E F so that (z,y) E p and (z,x) E p. Since P is a hction, y = x. Thus, if E is a function, then F is 1-1. Now assume F is 1-1. Let (u, w) E fl and (u, v) E p. To show that is a function, we must show that w = v. By the definition of p, (w,u) E F and (v, u) E F, making u = F(w) = F(v). By assumption, F is 1-1; hence w = v. The facts that im F = dom and dom F = im fl are immediate.


If F is a 1-1 function, we shall write F-' in place of and call F-' the inverse of F. Furthermore, if F is 1-1, then b = F is a function, so = F-' is also 1-1. Suppose f :A + B and T c A. Define

f (T) is called the image of T under f. In particular f (A) = im f . If C c B, define f-'(c)=

{a € A : f(a) E C).

f -'(c) is called the inverse image of C under f. Note that this definition is given without any assumption that f is 1-1, so it is not assumed here that f is a function. Iff :A + B is a 1-1 function,f is sometimes called an injection or an injective function. When im f = B, we say that f is a function from A onto B and call f a surjection or a surjective function. Iff : A -, B is both 1-1 and onto, f is called a bijection or a bijective function.

DEFINITION Iff : A -,B and g : B + C, the composition of g by f , written g of, is defined to be the set {(x,y) : there is w E B such that (x, w) E f and (w,y) E g).

0.7 THEOREM Iff : A gof : A + C .

-+

B and g : B -, C, then g of is a function and

Proof Suppose (x, u) E g 0 f and (x, v) E g of. We must show that u = v. By the definition of g of there are wlw2 E B such that (x, wl) E f , ( w l ,u) E g, (x, w2) E f , and (w2,v) E g. Since f is a function, wl = WZ;since g is a function, u = v. Thus g of is a function. Letf : A - , B a n d g : B + C . I f x ~ A , t h e n t h e r e i s a y ~ B s u c h t h a t ( x , y ) ~ f , or y = f(x); also, there is z E C such that Q,z) E g, or z = gb). This means that (x,z) E g 0f and z = g o f(x). But

and we see that the composition of g by f is accomplished by "following" f by g. Also, it is clear the dom (g of) = dom f . Though it is defined at length here, the composition of two functions is not a new idea. For example, the function h(x) = sin$ is the composition of the function g(x) = sinx by f (x) = 2, so that h = g 0f. Note that f o g is the function whose value at x is sin^)^; hence f o g # g of.

0.3 Mathematical Induction and Recursion 13

If S is a set, denote by is the function from S onto S defined by ls(s) = s for each s E S. Now, iff is a 1-1 function from A onto B, then we have seen that f - I is a 1-1 function from B onto A. Upon examination of the definition off -I, it becomes clearthatf of-' = IS and f-' of = lA. Exanzple0.7 Let~={x:x~Randx~0).~efinef:~-,~byf(x)=~. f (x) = f (y), then 2 = 3,which means that x = y or x = -y. But, since both x and y must belong to S, then x and y are both zero or both have the same sign-that is, both are positive. In either case x = y. Thus f is 1-1. Let us find f -I. If (x,y) E f then y = x2. Thus, if (y, X) belongs to f - , y = 2 , and because dom f = S, x must be the positive square root of y if y 0, and x must be zero otherwise. Thus f -'(y) = We know that im f = S (previous experience tells us that), so f : S --, S. Notice that

+

a.

-'

and

Example0.8 L e t S = { x : x ~ R a n d x + - 1 ) . Define 2x- 1 f : S --, R by f(x) = x+l. First of all, if f(a) =f(b), then =f(a) =f(b) = 26-' b+l A bit of algebra shows that a = b, hence f is 1-1. Let us find im f and, in the process, we will discover f-'. The number y E im f if and only if there is x E dom f = S such that f (x) = y-that is, y = Solving for x, we obtain x = This means that if y # 2, there is x E S such that f (x) = y. Thus, im f = R \ (2). Also, this means that f = 2-yo

9

3.

g.

*

!

1;

In what is to follow, we will have need of a technique of proof called mathematical induction. In some cases, the use will be rather informal, as in the case of showing that every polynomial function has a limit at every point. On the other hand, in less obvious applications, such as Taylor's Theorem in Chapter 5, the approach will be formal and very precise. Recursion will be used in the definition of certain functions and sequences defined on J. Usually a few initial values of the function are given, and then the value at n is given in terms of the values of the function at k for some k < n. The basis for mathematical induction is the well-ordering principle, which we assume without proof. WELL-ORDERING PRINCIPLE Every nonempty subset of J has a smallest member.

14


0.8 THEOREM (Principle of Mathematical Induction) If P(n) is a statement containing the variable n such that i. P(l) is a true statement, and ii. for each k E J , if P(k) is true, then P(k + 1) is true, then P(n) is true for all n E J. Reflect on the implication of Theorem 0.8. Basically it says that if P(l) is true, and if one can establish that P(k+ 1) is true from the truth of P(k), then P(n) is true for every positive integer n. Intuitively, this means that from the truth of P(l) we can infer that P(2) is true, and from the truth of P(2), we can infer that P(3) is true, and so on. The proof of this theorem is our first example of a proof by contradiction. Usually a theorem is stated in the form "If A is true, then B is true." To prove such a theorem by contradiction, one assumes that A is true and B is false and then shows that this leads to a logical contradiction, often by showing that A is false. Consequently, the assumption "A is true and B is false" can't be true, so the theorem must be true.

Proof

Assume that P(n) is a statement about the variable n such that

(i) P(l) is true, and (ii) for each k E J , if P(k) is true, then P(k + 1) is true. In addition, assume that P(n) is not true for some n E J. (This is the "A is true, B is false" part of the proof by contradiction.) Since P(n) is false for some n E J, the set S = {n : P(n) is false) is a nonvoid subset of J. As such, by the well-ordering principle, S has a smallest member; call it no. Since P(l) is true, no > 1. Since no is the smallest member of S, P(no) is false and P(no - 1) is true. But remember, if P(k) is true, then P(k+ 1) is true! So, if we let k = no - 1, then P(k) is true, but k + 1 = no and P(k + 1) is false. This is a contradiction. Hence, by the method of proof by contradiction, the theorem is true.

In the pages to come, you will see more proofs by contradiction, so now is a good time to examine the process by reading the proof several times for maximum understanding. Example 0.9 Forn E J ,

The following identity will be needed in Chapter 6: 1+x+1+--+x"=

1 - x"+' if x 1-x

+ 1.

0.3 Mathematical Induction and Recursion 15

We will use mathematical induction to verify this identity. Let P(n) be the statement:

a. First check to see that P(l) is true. If n = 1, the left member of this identity is = 1 + x if x # 1. Thus, P(l) is true. 1 + x and the right member is b. Now assume that P(k) is true and try to show that this implies that P(k+ 1) is true. So we have

(because we assumed P(k) is true)

- 1-

i f x # 1. 1-x This last equality asserts that P(k+ 1) is true. Thus, from the truth of P(k), we can prove that P(k + 1) is true. By the principle of mathematical induction, P(n) is true for all n E J. This was our goal. Under some circumstances, we need what is sometimes referred to as the second principle of mathematical induction. 0.9 THEOREM (Second Principle of Mathematical induction) Suppose that P(n) is a statement concerning the variable n. If i. P(l), P(2), . . .,P(m) are true, and ii. for k E J , k 2 m, if P(i) is true for all i E J such that 1 5 i 5 k, then P(k + 1) is true, then P(n) is true for all n E J. Proof We leave the proof as Exercise 23. The proof is very similar to that of Theorem 0.8 and uses the well-ordering principle.

The function defined in Example 0.10 is defined by recursion; that is, some initial values off are given-in this case,f (1) and f (2)-and then, for larger n, f (n) is given in terms of the values f(k) for some k < n. In this case f(n) = f(n-1)ef(n-2) . We will 2 use the second principle of mathematical induction to prove that f (n) = 2 + (for all n E J.

i)

.


Example 0.10

Define f : J --+ R as follows:

3 f ( l ) = 3, f(2) = 2'

and for n 2 3 f (n) = f(n - 1) +f(n - 2) 2

(-i)"-'

for all n E J. To do so, we will use We want to prove that f(n) = 2 + Theorem 0.9. Let P(n) be the statement that f (n) = 2 + (- "-I.

1)

a. We will show that P(l) and P(2) are true. For n = 1 the formula gives

For n = 2, the formula gives

So, for n = 1 and n = 2 , f(n) = 2 + (-$)"-'. b. Assume now that f (i) = 2 + (-

i)'-' for 1 5 i 5 k. Then

Thus, the formula holds for n = k + 1. By the second principle of mathematical induction,f (n) = 2 +

(-i)"-'

for all n E J.

The proofs of both versions of mathematical induction depend on the wellordering principle of the set of positive integers. Both the well-ordering principle and the two principles of mathematical induction can be modified easily as follows.

WELL-ORDERING PRINCIPLE (Modified) If S is any nonempty subset of Z (remember, Z is the set of all integers) such that S has a smallest member, then any nonempty subset 0f-S has a smallest member. Proof Suppose S is a nonempty subset of Z and so is the smallest member of S. Let A be a nonvoid subset of S. If A c J , then the first version of the wellordering principle gives the result. If A is not a subset of J , then so < 1. Let T = {so,so + 1, so + 2, . . .,0). Then A nT is a nonvoid finite set, and the smallest member of A is the smallest member of A n T.

j

If you reexamine the proof of Theorem 0.8, you will see that the argument relied in part on the fact that, if P(n) was false for some n, then there was a smallest such

0.4 Equivalent and Countable Sets

17

n and, since P(1) was true, that particular n was greater than 1. Using the modified well-ordering principle, we can prove the following modification of Theorem 0.8. 0.10 THEOREM (Modifid Principle of Mathematical Induction) P(n) is a statement concerning the variable n. If

Suppose that

i. for some no E 2, P ( 4 ) is true, and ii. for each k € 2 , k 2 4 , if P(k) is true, then P(k+ 1 ) is true, then P(n) is true from all n E 2, n 2 no.

Proof

The proof is left as Exercise 27.

The following example shows how Theorem 0.10 might be applied.

Example O.1I of n:

Consider the statement P(n) : 2n + 1

< 2". Try a few values

n = 1 2 1 + 1 = 3 # 2 = 2l P ( l ) is false. n = 2 2 2 + 1 = 5 # 4 = 22 P(2) is false. n = 3 3 2 + 1 = 7 < 8 = Z3 Aha! P(3) is true. So we will apply the modified version of mathematical induction with

= 3.

a. P(3) is true (see above). b. Assume ke> 3 and P(k) is true. Then

2(k+1)+1= 2 k + 1 + 2 < 2 ~ + 2 < 2k + 2k = 2k+'. Thus, P(k + 1 ) is true. Therefore, P(n) is true for all n 2 3. Exercise 28 asks you to prove a modified version of Theorem 0.9.

Consider now the sets A = {1,2,3,4,5) and B = {2,4,6,8,10). It is clear that A # B, and in fact A gf B and B gf A, so there seems to be no way of comparing the two sets. However, a similarity exists: Both sets have exactly five elements, a result we easily obtain by counting the members of each set. If the sets were quite large, counting elements would be a difficult task; in fact, if the sets were infinite, then we would be out of luck. The purpose of the following discussion is to consider a way of comparing the size of two sets without the necessity of using the natural numbers as a guide for counting. An obvious way of showing that A and B have the same number of elements is to pair each element of A with an element of B and observe


that, when this is accomplished, there are no elements of B remaining. The following is one such way of doing this: (1,2), (2,4), (3,6), (4,8), and (5,lO). The sophisticated reader will recognize that this amounts to defining a 1-1 function from A onto B, in this case the function being easily given by the formula f(x) = 2x for each x E A. There are, of course, many other ways of pairing the elements of A with the elements of B. The reader might be interested in computing how many there are.

-

DEFINITION If A and B are sets, we say that A is equivalent to B, written A B, iff there is a 1-1 function f from A onto B. To the reader who has sampled the flavor of modem mathematics, the word equivalent canies certain implications. Before considering some examples relating to this definition, let us prove a theorem that justifies the usage. 0.11 THEOREM Let A, B, and C be sets. Then

-

i. A A. ii. IfA-B,thenB-A. iii. IfA-BandB-C,thenA-C.

-

Proof (i) To show A A, we must exhibit a 1-1 function f from A onto A. It seems reasonable to try lA.Now if

then

hence, lAis 1-1. It is clear that im lA= A since, for any a E A, lA(a)= a. Thus, A -A. (ii) Suppose A B. Then there is a 1-1 function f from A onto B. To show B A, one must find a 1-1 function g from B onto A. The discerning reader should now observe that f-I is the logical candidate. It has already been shown thatf-I is 1-1, dom f - I = im f = B, and imf-I =dom f =A; hence, f-' is a 1-1 function from B onto A. Therefore, B A. (iii) Assume A B-and B C. There are 1-1 functionsf from A onto B and g from B onto C. We seek a 1-1 function from A onto C. The only reasonable way to obtain a function from A onto C is to consider the composition of g by f , namely g of. We know that dom (g o f ) = A, and it remains to be proved that (g on is 1-1 and that im (g o n = C. This can be done directly, but it's worthwhile to do it via two lemmas.

-

-

-

0.12 l,EMMA

is 1-1.

-

-

Supposef : A + B andg: B 4 C. Iff andg are 1-1, thengof

0.4 Equivalent and Countable Sets

Proof Now

19

Supposef and g are 1-1 and that a1,a2 E A with (g on(al) = (g on(a2).

and, since g is 1-1, f (al) =f (a2) In like fashion, the fact that f is 1-1 guarantees that a1 = a2. Therefore, g of is 1-1. 0.13 LEMMA Suppose f : A then im (g o n = C.

+B

and g : B -, C. If im f = B and im g = C,

Proof Suppose c E C. Since im g = C, there is b E B such that g(b) = c; and, since im f = B, there is an a E A such that f(a) = b. Now

Thus, C c irn (g 0f). Conversely, if a E A, then f(a) E B and

so im ( g o n c C. Thus, im ( g o f ) = C. With these two lemmas we can now complete the proof of Theorem 0.11. Sincef : A + B a n d g : B - + C a r e 1-1,thengof is 1-l;and,sincef isontoB and g is onto C, then g of is onto C. Therefore, A N C. Much more can be said concerning the equivalence of two sets. We shall consider only a few instances that are pertinent to later topics. Let us first consider some examples. Let J denote the set of positive integers, and let E denote the set of even positive integers. Define f :J + E by f (n) = 2n for each n E J. Now, iff (n) =f (m), then

hence n = m, so f is 1-1. Since each even positive integer may be written in the form 2n where n E J , it is clear that f is 1-1 onto E; hence J N E. This may seem to be a peculiar state of affairs-the set J being equivalent to one of its proper subsets. However, this is typical-of such sets; in fact, one may define an infinite set to be any set that is equivalent to one of its proper subsets. Let J denote the set of positive integers, and let Z denote the set of all integers. Define f : J -,Z by the following sets of rules: If n E J and n is even,

and if n E J and n is odd,


To show J 2, we need to show that f is a 1-1 function from J onto 2. By construction off, it is clear that dom f = J and im f c Z. If m E Z and m is positive, then 2m E J and f (2m) = m. If m E Z and m is zero or negative, then 1 - 2m E J and N

since 1 - 2m is odd. Thus, f is onto Z and it remains to show that f is 1-1. For each n E J with n f 1, f(n) > 0 if n is even and f(n) < 0 if n is odd [f(l) = 01. Thus, if f(s) = f (r) = a, then s and r must be both 1, both even, or both odd, as a is zero, positive, or negative. If s and r are both even, then

If s and r are both odd, then

-

Hence f is 1-1. We have shown that J E and J 2, SO,by use of Theorem 0.11, Z equivalent to J are of sufficient importance to deserve a special name. N

N

E. Sets

DEFINITION A set S is countably infinite iff S is equivalent to the set J of positive integers. A set S is finite iff S is empty or there is n E J such that S is equivalent to the set { 1,2,3, . . . ,n). A set is countable if it is either finite or countably infinite. A set is infinite if it is not finite. A set is uncountable if it is not countable. We will see examples of uncountable sets later, but first we consider more examples of countable sets. Three examples of countable sets have been exhibited-namely E, 2,and, of course, J itself since J J. YOUmight find it instructive to stop now and discover a few more examples for yourself. As a start, try to prove that the set of odd positive integers is countable. The following theorem shows that, in a certain sense, countably infinite sets are the smallest infinite sets. N

0.14 THEOREM Any infinite subset of J is countably infinite.

Proof Let S be an infinite subset of J. We will define a function f : J + S recursively as follows: Since S is a lionempty subset of J , it has a least member; define f (1) to be that least member. After f (l),f (2), . . . ,f (k) have been defined, we define f (k + 1) to be the least member of S \ (f(l), f (2), . . . ,f (k)). This is always possible because S \ (f(l), f (2), . . . ,f (k)) is always a nonempty subset of J. The function is certainly 1-1 by its construction. Also, if n E S, then there is m E J such that n is the smallest element of S \ (f(l), f(2), . . .,f (m)),and hence n = f(m + 1). Thus, f is a 1-1 function from J onto S. This means that S is countably infinite.

0.4 Equivalent and Countable Sets 21

0.15 COROLLARY

Any subset of a countable set is countable.

Proof The proof is left as exercise 30. A consequence of Corollary 0.15 is worth noting here. Let S be a set, and suppose there is a 1-1 function f : S -+ J. Then S is equivalent to f (S), a subset of J. But f(S) is countable because it is a countable subset of a countable set; hence S is countable. In other words, to show that a set is countable, it is sufficient to exhibit a 1-1 function from that set into J. That technique is used in the next theorem. 0.16 THEOREM If A and B are countable, then A x B is countable.

Proof Suppose A and B are countable. Then there are 1-1 functions f : A a n d g : B - + J . Defineh:AxB-,Jby

-+

J

Since f and g are 1-1, h must be 1-1 by the Unique Factorization Theorem. By the remarks preceding this theorem, we conclude that A x B is countable. Recall that countable means countably infinite or finite. The fact that S is countable iff it is equivalent to a subset of J allows us to avoid the difficulty of treating the finite and infinite cases separately.

Example 0.12 Let S be the set of all rational numbers. Now each rational number can be written uniquely as 2, where a and b are integers, b > 0, a and b are relatively prime (that is, a and b have no common divisors greater than I), and, if a = 0, b = 1. We define a function f : S -* J x J as follows:

f (r) =f

(i) (a, b) where -b is the unique representation =

a

of the rational number r. This function is 1-1 from S onto a subset of J x J , and, by Theorem 0.16, J x J is countable; hence, any subset of J x J is countable by Corollary 0.15. Therefore, since S is equivalent to a subset of J x J , then S is countable. There are many ways of combining countable sets to obtain countable sets. Theorem 0.16 gives one such way, and the next theorem gives another method. This theorem will be useful in Chapter 2. tI

b# i d

0.17 THEOREM Let S be a nonempty subset of J. Let { A ~ ) ~be~ an s indexed family of countable sets. Then USES As is a countable set.

iU;i


Proof

Since each As is countable, for each s E S, there is a 1-1 function If x E UsEsAs,then there is a smallest member m of S such that x E A,. We will attempt to define a function f : UsEsAs + J x J as follows: f (x) = (m,fm(x)), where m is the least integer in S such that x E A,. Since x E UsEsAs, it must belong to A, for some m E S; hence there is a smallest such m by the well-ordering principle. Thus f is a well-defined function-that is, the rule for finding f (x) is unambiguous. We now need to show that f is 1-1. Suppose

f, : A,

+ J.

Thus, x E A, and y E A,; hence, n = fm(x) = fmO.Because fm is 1-1, x = y. Thus, f is a 1-1 function. We have constructed a 1-1 function f from USESAs onto a subset of J x J; hence, UsEsAsis a countable set.

Our preoccupation with countable sets would seem bizarre if all infinite sets were countable. However, there are many uncountable sets. The set of real numbers is a familiar example, although you may not be familiar with the fact that R is uncountable. In Exercise 37 you are asked to prove that, for any set A, P(A)-the set of all subsets of A-is not equivalent to A. Hence, the set of all subsets of J is another example of an uncountable set. There are several ways to prove that R is uncountable. In one of the projects at the end of Chapter 1, you will have an opportunity to provide such a proof. In anticipation of that project, we next give an example of a set equivalent to R.

Example 0.13 We will show that R is equivalent to (0,l). The easiest way to accomplish this is to use a function familiar to anyone who has studied trigonometry. Define f : (0,l) + R by f (x) = tan (71-x- 5 ) . This gives a 1-1 function from (0,l) to R.

W

See Project 0.1 at the end of this chapter for other sets equivalent to (0,l).

You probably have a good knowledge of the set of rational numbers, the order relation of that set, and the operations of addition and multiplication. However, the set of rational numbers is not sufficient to support all of mathematics. The following theorem shows that an equation as simple as 2 = 2 has no solution in the set of rational numbers. O.I8 THEOREM There is no rational number whose square is 2. Proof Suppose there are positive integers p and q such that 2 = @Iq)*. Assume fuaher that p and q are relatively prime-that is, their greatest common divisor

1

ff

i I

I

0.5 Real Numbers

23

is 1. Thus, we have 2# = p2, so p is even (since p odd implies p2 odd). We may write p = 2r where r is an integer. Then 2$ = p2 = 4?, so = 2?, and hence q is even. Thus, both p and q are even, contrary to the assumption that p and q are relatively prime.

92

Rather than construct the set of real numbers, we shall assume the existence of that set and postulate the properties that we will need. For those who wish a careful and rigorous development of the real numbers, see Edmund Landau's Foundations of Analysis, 2nd ed. (New York: Chelsea, 1960). We assume the existence of a set R, called the set of real numbers, that satisfies the following axioms. There are functions + : R x R 4 R and : R x R + R and a relation < on R such that, for all x, y, z E R, we have

x (y + z) = (X y) + (X 2). There is a unique element 0 E R such that 0 + x = x for all x E R. For each x E R, there is a unique y E R such that x + y = 0, and we write y = -x. There is a unique element 1 E R such that x 1 = x for all x E R. For each x E R with x 0,there is a unique element y E R such that x y = 1, and we write =x-' or y = 21 .

+

x < y and y < z implies x < z. For x,y E R, exactly one of the following is true: x < y, y < x, or x = y. x < y and z > 0 implies xz < yz. This list of properties assures us that the system (R, +, 0,

The rational number

is the required number.

It is worth noting that Theorem 0.22 implies something that sounds stronger. If x and y are real numbers and there are only a finite number of rational numbers between x and y-let rl < 15 < 1-3 < < r, be the list-then there are no rational numbers between rl and r2, for example. Thus, between any two real numbers, there are infinitely many rational numbers. We indicated in Section 0.4 that the set of real numbers is uncountable, whereas the set of rational numbers is countable. This means there are many real numbers that are not rational numbers. A real number that is not a rational number is called an irrational number. As we shall soon see, the set of irrational numbers is also dense in R; that is, every nonempty open interval contains an irrational number. As argued above, this means that every nonempty open interval contains infinitely many

0J Real Numbers 27

irrational numbers. TO prove this theorem, we will need at least one example of an irrational number. Theorem 0.18 asserts that there is no rational number whose square is 2. The usual conclusion is that f i is an irrational number. However, we haven't established that such a number exists. If our set of axioms for R is complete, we should be able to prove from those axioms that the equation a? = 2 has a solution in R; hence, there exists a positive number whose square is 2. The next theorem accomplishes that and more: it establishes for future use the square root function defined on the set of nonnegative real numbers. 0.23 THEOREM If p is any positive real number, there is a positive real number x such that 2 = p.

Proof First we will prove the theorem for p greater than or equal to 1. Define A to be the set {z : z is a positive real number and z 5 p). Now A is nonempty since 1 E A and A is bounded from above. To see the latter, note that if z > p, then 1 > p2 2 p (since p 2 1). Hence z is not in A. Therefore p is an upper bound for A. Let x = supA. We will show that 2 = p. The proof will consist of two parts: in the first part we will show that assuming 2 p leads to a contradiction. The only remaining possibility is that 2 = p, as was to be proven. Suppose 2 < p. Let 6 be the smaller of 1 and !&. Then

Therefore x + 6 belongs to A, and x < x + 6 contrary to x being an upper bound for A. Thus 2 2 p. Then Suppose now that 2 > p. Let 6 =

**.

This means that x -- 6 is less than x but serves as an upper bound for A, contrary to x being the least upper bound. We conclude that a? = p. At the beginning of this proof we assumed that p was greater than or equal to 1. If 0 1, and so the proof applies to 1/p. Thus, there is a positive real number x such that 2 = l/p, and so ( 1 1 ~ = ) ~p; that is, the theorem is proven also for 0 < p < 1. Theorems 0.18 and 0.23 give us f i as an example of an irrational number. Equipped with that knowledge, we proceed.

28 chapter 0 Preliminaries

Om24 THEOREM Between any two distinct real numbers there is an irrational

number.

Proof Let x and y be distinct real numbers with x < y. Now &, the positive square root of 2, is irrational and x f i < y& By Theorem 0.22, there is a rational number r such that x f i < r < y&. We may assume that r 0.-DO you see why? But then x < < y and is an irrational number since 4 2 is irrational. Now do you see why we needed to assume that r # O?

5

5

+

Without using the fact that R is uncountable, Theorem 0.24 guarantees the existence of many irrational numbers. Project 0.3 asks you to prove that if p is a positive integer that is not the square of an integer, then f i is an irrational number. Numbers such as &, fi,and so on, are examples of what are called algebraic numbers. See Exercise 36 for the definition of algebraic number. In that exercise, you are asked to prove that the set of algebraic numbers is countable. There are still many real numbers that are not algebraic; such numbers are called transcendental numbers. The number .rr is an example of a transcendental number, as well as the number e, the base of the natural logarithm function. In the study of analysis, the notation of the distance between two points on the real line-that is, the distance between two real numbers-is of prime importance. The best way to describe this distance is in terms of absolute value. If x is any real number, define 1x1 =

{ -xx iifxf x 2< 0.0

We call 1x1 the absolute value of x. Note that 1x1 = I@ and x 5 1x1 for all x E R. 025 THEOREM Let a and b be any real numbers. Then

i. ii. iii. iv.

la + bl 5 la1 + 1b1. lab1 = la1 lbl. llal - lbll 5 la- bl. If E > 0, then la1 < E iff

-E

< a < E.

Proof Perhaps the easiest way to prove this theorem is to use the fact that 1x1 =

a.

(i) Let a and b be any real numbers. Then

Exercises 29

(ii) ~ e at and b be any two real numbers. Then

lab1 =

= @=

= (a1(bl.

(iii) By (i) above,

Ial = la - b + bl 5 la - b( + (bl; hence

la1 - lbl 5 la - bl*

In similar fashion, Jbl 5 la-bl+lal; hence, lbl-la1 5 la-bl. Since llal-lbll is equal to 1 b 1 - la 1 or 1 a 1 - Ib 1, depending on the relative magnitudes of (a1 and ibl, we have shown that llal - lbll 5 la - bl. Suppose - E < a < E, r > 0. If a > 0, then la1 = a < a If a 5 0, then la1 = -a and - E < a; hence la1 = -a < E. Thus, - E < a < E, E > 0 implies that la1 < E. If la1 < E and a > 0, then - E < 0 < a = la1 < E . If a 5 0, then la1 = -a < r ; hence -r < a 5 0 < E . Thus, la1 < E implies that - E < a < E.

It is to be noted that (iv) implies that the following statements are equivalent.

Again, using the real line as the model for the set of real numbers, la - bl is just the undirected distance fn>m a to b and, for E > 0,

is the set of points that are within a distance E of a. Sets of this type play a major role in the ensuing chapters.

EXERCISES 0.1 SETS 1. List the elements of each of the following sets:

2. Write each of the following in internal notation:

30 Chapter 0 Preliminaries 3. Prove (vi) of Theorem 0.2. 4. Prove (ii) of Theorem 0.3.

6. If A C B, prove that (C \ B) C (C \ A). Either prove the converse is true or give a counterexample.

7. Under what conditions does A \ (A \ B) = B? 8. show that ( A \ B ) u ( B \ A ) = ( A u B ) \ ( A ~ B ) .

9. Look up Russell's paradox and write a brief summary discussing how it relates to Section 0.1. 10. Describe each of the following sets as the empty set, as R, or in interval notation, as appropriate:

00

(c)

0 (-!,I n

(d)

U(-',2n

+ A) n

00

+ 1) n

11. Prove (ii) of Theorem 0.4.

12. Use De Morgan's Laws to give a different and simpler description of the following sets:

0.2 RELATIONS AND FUNCTIONS 13. Define f : J -+ J by f(n) = 2n - 1 for each n E J. What is im f ? Is f 1-l? Is f onto? If f has an inverse, find the domain of the inverse and give a formula for f -*(n). 14. Often, if the domain is not&xified, it is assumed to be the set of all real numbers for which the formula for f(x) defines a real number. What is the domain of the function defined by the function f (x) = &? What is im f ? Is f injective? If so, find the inverse.

For Exercises 15-17, let A = {1,2,3,4,5), B = { 2 , 3 , 4 , 5 , 6 , 7 ) ,and C = { a , b , c , d , e ) . 15. Give an example off :A + B that is not 1-1. #

'

16. Give an example off :A

+B

17. Give an exivnple off : A

-r

that has an inverse, and show the inverse.

B, g : B -r C such that g o f is 1-1 but g is not 1-1.

Exercises 31

*IS. Iff : A -r B is 1-1 and imf = B, prove that (f-' of)(a) = a for all a E A and (f of -l)(b) = b for each b E B.

0.3 MATHEMATICAL INDUCTION AND RECURSION 19. Provethatforalln E J , 1 + 2 + - + n =

,.

n(n+l)

20. Provethat for all n E J , 1 + 3 + 5 + - + ( 2 n -

l)=n2.

21. Prove that n3 + 5n is divisible by 6 for each n E J. 22. Prove that n2 < 2" for n E J , n 2 5. (See Example 0.11.) 23. Prove the second principle of mathematical induction (Theorem 0.9). 24. Define f :J -,J by f(1) = 1, f(2) = 2, f(3) = 3, and f(n) =f(n - l)+f(n - 2)+f(n - 3) for n 2 4. Prove that f(n) 5 2" for all n E J. 25. Definef :J -r J by f(1) = 2 and, for n 2 2, f(n) = d3 +f(n - 1). Prove that f(n) for all n E J. You may want to use your calculator for this exercise.

< 2.4

26. Definef :J + J b y f(1) = 2, f(2) = -8, and, for n 2 3,f(n) = 8f(n-1)-15f(n-2)+6-2". Prove that, for all n E J , f (n) = -5 3" + 5"-' + 2"". 27. Prove Theorem 0.10. *28. Prove the following modified version of the second principle of mathematical induction: Let P(n) be a statement for each n E 2. If (a) P(no),P(no + 1), . . .,P(m) is true, and (b) for k m, if P(i) is true for no 5 i 5 k, then P(k + 1) is true, then P(n) is true for all n no, n E 2.

> >

29. Define f(n) as follows for n E Z, n 2 0. f(0) = 7, f(1) = 4, and, for n 2 2, f(n) = 6f(n - 2) -f(n - 1). Prove that f(n) = 5 2" +2(-3)" for all n E 2, n 2 0.

0.4 EQUIVALENT AND COUNTABLE SETS 30. Prove Corollary 0.15.

31. Find a 1-1 function f from J onto S where S is the set of all odd integers. 32. Let Pn be the set of all polynomials of degree n with integer coefficients. Prove that Pn is countable. (Hint: A proof by induction is one method of approach.) 33. Use Exercise 32 to show that the set of all polynomials with integer coefficients is a countable set. 34. Prove the following generalization of Theorem 0.17: If S is a countable set and {As)sEs is an indexed family of countable sets, then UsEsAs is a countable set. 35. For each p E Pn, define B@) = {x :p(x) = 0). Prove that

U,,p,

B@) is countable.

36. An algebraic number is any number that is a root of a polynomial equation p(x) = 0 where the coefficients of p are integers. Show that the set of algebraic numbers is a countable set.

37. For a set A, let P(A) be the set of all subsets of A. Prove that A is not equivalent to P(A). [Hint: Supposef :A -r P(A) and define C = {x :x E A and x $! f (x)). Show C !$ im f .] 38. Let a, b, c, and d be any real numbers such that a < b and c < d. Prove that [a, b] is equivalent to [c, 4.(Hint: Show that [a, b] is equivalent to [O,1] first.)

32 Chapter 0 Preliminaries 0.5 REAL NUMBERS

*39. If x

< y, prove that x
0, there is a E S such that x - c < a 5 x. If y = inf S, show that, for each e > 0, there is a E S such that y 5 a < y + E.

A t . If x = sup S, show that, for each AS.

E.

PROJECT 0.I The purpose of this project is to show that the open interval (0,l) is equivalent to the closed interval [0, 11. In the process we will discover that both intervals are equivalent to [0, 1) and (0,1]. It is then easy to generalize to any interval [a,b] with a < b. Define f : (0,l) + R as follows

(t)

Forn~J,n22,f

1 = n-1

and for all other x E (0, I),

f ( x ) = x.

1. Prove that f is a 1-1 function from (0,l) into (0, 11. 2. Prove that f is a function from (0,l) onto (0, 11. 3. Find a 1-1 function from [0, 1) onto [0, 11. You might use the function in #2 with some modifications. 4. Prove that [0, 1) is equivalent to (0, 11. 5. Prove that (0,l) is equivalent to [0, I].

PROJECT 0.2 The purpose of this project is to show that every positive real number has a unique nth root. Then Theorem 0.23 follows as a special case. THEOREM If p is any positive real number and n is a positive integer, there is a unique positive real number x such that x" = p.

Projects 33

Proof

1. Use the factorization

to prove that any solution to the equation X = p is unique. 2. Suppose that p is a positive real number greater than 1 and define A to be the set { z : z is a positive real number and f 5 p). Show that A is nonempty and bounded from above. 3. Let x = supA. Suppose X < p. Let

and choose 6 to be the smaller of 1 and $. Show that x + 6 is in A, contrary to x being an upper bound for A. 4. Assume x" > p. Let c = x" -p. Choose y such that 0 < y < x and x-y < ,;;;i;ri. Show that P - f < c, and thus y is an upper bound for A, contrary to x being the least upper bound. 5. Finish the proof.

PROJECT 0.3 The purpose of this project is to use the well-ordering principle to prove that certain square roots are irrational.

THEOREM If p is a positive integer that is not the square of an integer, and 2 = p, then x is irrational. This will be a proof by contradiction using the well-ordering principle. 1. Assume that 2 = p, p is a positive integer that is not the square of an integer, and x is a positive rational number. Define A to be the set {n E J : nx is an integer). Show that A is nonempty, and let m be the smallest member of A. 2. Show that there is k E J such that k < x < k + 1. 3. Let z = mx - mk. Show that 0 < z < m. 4. Show that z and xz ark both integers; hence z E A. Show that this is a contradiction; hence x is not rational.

Chapter

I

Sequences

he aspiring analyst should begin by investigating the folklore of sequences in detail. A vigorous attempt is made in this chapter to remove the cloud of mystery surrounding the inner workings of sequences. A thorough understanding of sequences is invaluable in understanding the rest of this book.

T 1.1

SEQUENCES AND CONVERGENCE

DEFINITION A sequence is a function whose domain is the set of positive integers. If a is a sequence, it is customary to write a(n) = a, for each positive integer n and write a = {an)gl. We call an the nth term of the sequence. If m n, then am and a, are considered to be different terms of the sequence even when a, = am. In this fashion, it is sometimes convenient to describe a sequence by giving a formula for the nth term. For example, the sequence whose nth term is may be written { ) z l ; the fifth term in this sequence is the ninth term is $, and so on. If p, denotes the nth prime, the sequence b ) E , is well defined, although it may be rather time-consuming to determine the 97th term. As a matter of policy, we shall try to use notation that conveys the intended meaning most clearly. The sequences mentioned earlier may also be described as follows:

+

i;

A

This device for describing a sequence seems more useful, for example, in describing the sequence

36 Chapter l

Sequences

From now on, all sequences will be sequences of real numbers unless specified otherwise. This will allow us to dispense with unnecessary language, but will still allow other types of sequences when needed. Before attempting to define the notion of a convergent sequence, let us consider some of the examples we have at hand. Considering the sequence we observe that by "going out far enough" in the sequence-that is, choosing n sufficiently large-we find terms that are very close to zero. Likewise, in the sequence {0,1,0,1,0,1, . . .) we find terms that are very close to zero; in fact many of them are zero. However, in the latter case we discover that however "far out" in the sequence we lookregardless of how large n is chose-there are terms farther out in the sequence that are not close to zero. For example, if we choose n = 1,000,001, then al,m,ool = 0, but al,m,ooz = 1, which is not very close to zero. This defect is not present in the first example. If we decide on a measure of closeness, say within .025 of zero, it is clear that all terms of the sequence beyond the 40th term satisfy this criterion. We shall use this idea as a skeleton upon which to build our definition of convergence. In language less precise than we require, a sequence should converge to a real number A iff, when we settle on how close to A we wish to be, all terms of the sequence from some term on are at least that close to A. Let us now state the fonnal definition of convergence. (Remember, the abbreviation "iff" is used in place of the phrase "if and only if." Such usage occurs in the next definition and in many places in the text to follow.)

{i):!,

{i)zl

DEFINITION A sequence {an)zl converges to a real number A iff for each c > 0 there is a positive integer N such that for all n N we have lan - A1 < E.

>

{k)zl.

It is worthwhile to point out that the choice of N may depend upon the choice of c. Let us return to the sequence Intuition leads us to believe that this sequence should converge to zero. If ths conclusion is correct, then for c > 0, there is N such that, for n 2 N, lan - 01 = 1; 1 - 01 = < E. For example, if E = .025, 1 then, for n 2 41, lan - 01 = ; 5 < .025. Thus, for E = .025, N = 41 satisfies the conditions of the definition. For c = .00025, one may take N = 4001. Note that, if N is such that for n 2 N, lan-A1 < E and N < M, then, for n 2 M, lan-A1 < c. Hence, if { a , ) ~ , converges to A, then for each E > 0, there are many, in fact infinitely many, positive integers N such that for n 2 N, lan - A1 < 6. However, to prove that {an)zl converges to A, it suffices to show that for each E > 0, there is at least one number N with the desired property. ~ o l l o walong to see that converges to 0. Choose E >'o. Let N be an integer larger than So if n 2 N, we have 5 < E. This means that if an = and A = 0, then for n 2 N, lan - A1 = < c. Therefore the sequence converges to zero. By the way, do you see now why we chose N = 41 when e = .025? In order to digest this new idea completely, we shall look for other ways of expressing the notion. Recall that, if a and b are real numbers and E > 0, then )a- b) < c iff b - c < a < b+ e. Thus, {an):, converges to A iff for each e > 0 there is a positive integer N such that for n 2 N we have A - e < an < A + e. This means

&

5

5.

5

A

1 .l

Sequences and Convergence 37

that, given 6 > 0,A - E < an < A + E is true except for a finite set of subscripts-such as, in the case mentioned above, some subscripts from the set { I , . . .,N - 1). To facilitate our study of analysis, we must seek more knowledge of the topology of the set of real numbers. Knowledge will come gradually as concepts and facts are introduced and discussed. At this stage it will help to consider the notion of a neighborhood of a point.

DEFINITION A set Q of real numbers is a neighborhood of a real number x iff Q contains an interval of positive length centered at x-that is, iff there is e > 0 such that (x-e,x+e) C Q. In particular, for each e > 0, (x - e,x + e) is a neighborhood of x. In this setting, a sequence converges to A iff each neighborhood of A contains all but a finite number of terms of the sequence. It is appropriate to state this as an unnumbered lemma and supply a proof. Although the facts are obvious, the proof will further an understanding of the definitions of convergence and neighborhood. As a reminder, this theorem is of the fonn "R iff S," so the proof of the theorem has two parts. We must prove that R implies S and that S implies R.

LEMMA A sequence {an)gl converges to A iff each neighborhood of A contains all but a finite number of terms of the sequence.

Proof Suppose {a,)gl converges to A, and let Q be a neighborhood of A. Then there is E > 0 such that (A - E,A+ e) C Q. Since { a , ) ~ , converges to A and e > 0, there is a positive integer N such that for n 2 N, la, - A1 < E. In particular, for n 2 N, we have A - e < a, < A + e; hence, a, E (A - E,A + E) c Q. Thus, Q contains all terms of the sequence except possibly some of the terms al, a2, . . .,a ~ - 1 ;hence Q contains all but a finite number of terms of the sequence. Conversely, suppose each neighborhood of A contains all but a finite number of terms of the sequence { a , ) ~ , . Choose E > 0. Then Q = (A - E,A+ e) is a neighborhood of A and contains all but a finite number of terms of the sequence. This means that there is a finite set S = {nl, nz, . . . ,n,) of positive integers such that, if a, 4 Q, then n E S. Let N = (max S) + 1. Now, if n 2 N, then n 4 S and, by the definition of S, a, E Q; that is, A - e < a, < A + e. This last statement is equivalent to the statement la, - A1 < e. Hence, for n 2 N, la. - A1 < e. We have shown that {an)gl converges to A. We now can answer a question that should be lurking in the back of your mind. Can a sequence {an)gl converge to two different real numbers? Suppose the answer is yes. Then there would be distinct real numbers A and B such that {a,)E, converges to both A and B. Relying'on our mental picture of the real line, we may easily convince ourselves that there are intervals P and Q, each of positive length, centered at A and

t

38 Chapter l

Sequences

B respectively, such that P n Q is empty. You are asked to give a proof of this in Exercise 2. Now, as we have observed above, P contains all but a finite number of terms of the sequence {an):l. Since {a,);, also converges to B, and since Q is a neighborhood of B, Q contains all but a finite number of terms of the sequence. Recalling that P n Q is empty, we see that P must contain infinitely many terms of the sequence and also must contain at most a finite number of terms of the sequence. Thus, the assumption that { a n ) g l converges to both A and B with A # B leads to a contradiction. We shall state this as a theorem and give a careful proof. Note that the proof is only a restatement of the preceding argument in precise terms. It is worthwhile to point out at this stage that this is the way a good share of mathematics is done. Theorems should follow naturally from definitions in an intuitive way; then, of course, adequate proofs must be given. At this level, a difficult theorem might be classified as one that does not seem to have an intuitive proof. Thus, we distinguish between a difficult theorem and a theorem that is difficult to prove even though the idea of its proof may be intuitively clear. 1.I THEOREM If { a n ) g l converges toA and also to B, thenA = B.

Proof This will be a proof by contradiction. We will assume that {a,):, converges to A and also to B, and A # B. We can also assume that A < B. Let € = L(B - A). Since { a n ) z l converges to A, there is N such that for all n > N , 2 we have

BU~A+E=A+;(B-A)= $ ( A + B ) = B - $ ( B - A ) = B - E . Thismeansthat for n 2 N, a,, < B - 6 , so there are infinitely many terms of the sequence outside the neighborhood (B - 6 , B + c). Thus the sequence cannot converge to B. This contradicts our original hypothesis that { a n ) g l converges to both A and B. You may be able to supply better proofs for this theorem; the proof here is patterned after our initial idea. This is not always the best method of proof, and we shall not adhere rigidly to this procedure. Let US now consider the sequence {n)E,, or the sequence whose nth term is the positive integer n. It seems reasonably clear that this sequence cannot converge because the terms get larger and larger without bound. A sequence { a n ) g l is bounded from above iff there is a real number M such that an 5 M for all n. A sequence { a n ) g l is bounded from below iff there is a real number P such that P 5 a, for all n. A sequence is bounded iff it is both bounded from above and bounded from below. Note that a sequence { a n ) z l is bounded iff there are real numbers P and M such that P 5 an 5 M for all n or, equivalently, iff there is a real number S such that la,l < S for all n. Let us now suppose that { a n ) z l converges to A. If we choose a neighborhood of A, there are, at most, a finite number of terms of the sequence outside this neighborhood. In particular, if we consider the neighborhood (A - 1,A + I), there is a positive integer N such that A - 1 < an < A + 1 for all n 2 N. Now we are

1.1 Sequences and Convergence 39

assured that all terms of the sequence, except possibly the terms al, a2, .. .,ahr- are bounded from below by A - 1 and from above by A + 1. It is now easy to find upper and lower bounds for the possibly wayward terms al, . . . ,a ~ - 1because this is a finite set of real numbers. It should now be clear how upper and lower bounds can be chosen for the sequence {an)gl.

1.2 THEOREM If {an):,

converges to A, then {an)gl is bounded.

Proof Suppose {an)gl converges to A. Choose c = 1. There is a positive integer Nsuchthat,ifn 2 N,thenA-1 < a, 0. Let N be any positive integer greater than Then for n N, we have

a.

>

1

I;-OI=n 0. There is a positive integer N such that, if n, m N, then A - E < an < A + E a n d A - E < a m < A + € . Thus,foralln,m 2 N, we findan E ( A - € , A + ~ ) a n d am E (A - E,A+ E). The set (A - €,A + E) is an interval of length 26; hence, the difference between a, and amis less than 26. We will now state a theorem, the proof of which we have just outlined.

>

1.3 THEOREM Every convergent sequence is a Cauchy sequence.

Proof Suppose {an)gl converges to A. Choose r > 0. Then 5 > 0. There is a positive integer N such that n 2 N implies lan - A1 < 5. (The choice of 5 is

'

1.2 Cauchy Sequences 41

not a mere whim. We observed previously that the difference between a, and am was less than twice the original choice of e.) Now, if m, n 2 N, then la, -A) < 5 and la, -A1 < 5; hence,

Thus, {an)=, is Cauchy. You should digest the basic idea rather than memorize the formal details of the proof. The manner of expressing the idea in a clear, precise fashion will come with experience. Theorem 1.3 gives a necessary condition for convergence. If a sequence is convergent, it must be Cauchy. Equivalently, if a sequence is not Cauchy, then it is not convergent. Consider the sequences { 1,2,3, . . .) and { 1,0,1,0, . . .). It is clear that neither is Cauchy; hence, according to Theorem 1.3, both fail to converge. As mentioned before, the property of being Cauchy is an internal property of a sequence. You might well suspect that this property is not only necessary but also sufficient for convergence. This is true, but a bit of work to prove. The next task will be to prove it. Consider a Cauchy sequence { a n ) z l . Let E = 1. There is a positive integer N such that for n, m 2 N, )an- a,) < 1. In particular, N 2 N. Hence, for n 2 N, laN- anl < 1; that is, a~ - 1 < an < a~ + 1. If you will recall the remarks preceding Theorem 1.2, it will be clear that every Cauchy sequence is bounded, and you should be able to give a nice proof of this theorem for Exercise 14. 1.4 THEOREM Every Cauchy sequence is bounded.

Proof The proof is left as Exercise 14. Consider again a Cauchy sequence {an)El.The terms of this sequence all lie in an interval (C, D), since the sequence is bounded; moreover, the terms of the sequence get closer and closer together as one goes out farther and farther in the sequence. It seems reasonable to suspect that there is a real number A at which the terms of the sequence "pile up." We must find this number A and prove that the sequence converges to A. The existence of such a real number will be proved as a corollary to a theorem that we shall use later. First, the notion of a "pile-up" must be formulated in precise terminology. DEFINITION Let S be a set of real numbers. A real number A is an accumulation point of S iff every neighborhood of A contains infinitely many points of S.

42

Chapter 1 Sequences

A few remarks concerning this definition are in order. First, if A is an accumulation point of S, then every neighborhood of A contains at least one point of S that is different from A. (Indeed, A might not belong to S.) On the other hand, if A is not an accumulation point of S, then some neighborhood of A contains only a finite number of members of S. In this case it is possible to find a smaller neighborhood of A that excludes all points of S different from A. Thus, A is an accumulation point of S iff every neighborhood of A contains a member of S that is different from A. We state this result as follows.

LEMMA Let S be a set of real numbers. Then A is an accumulation point of S iff each neighborhood of A contains a member of S different from A.

{i

Example 1.3 Consider the set S = : n is a positive integer). This set is the range of the sequence {,):! . Earlier, we showed that { converges to zero. Thus, every neighborhood of 0 contains infinitely many terms of the sequence; and, since all terns of the sequence are distinct (that is, if m # n, then a, # an),every neighborhood of 0 contains infinitely many points of the set S. Therefore, 0 is an accumulation point of the set S.

)zl

!

Observe that we did not define an accumulation point of a sequence but, rather, an accumulation point of a set. One might be led to conjecture that the limit of a convergent sequence is always an accumulation point of the range of the sequence. To see that this is false, consider the sequence {an)gl where a, = 1 for all n. This sequence converges to 1, but its range is finite and therefore can have no accumulation points. In Chapter 0, we noted that between any two real numbers there are infinitely many rational numbers and infinitely many irrational numbers. This yields the following theorem. 1.5 THEOREM Every real number is an accumulation point of the set of rational numbers. Every real number is an accumulation point of the set of irrational numbers. Proof Let x be any real number and Q be a neighborhood of x. There is e > 0 such that (x - c, x + e) c Q. There are infinitely many rational numbers between x and x + E; hence, Q contains infinitely many rational numbers. Thus, x is an accumulation point of the set of rational numbers. The proof for the set of irrational numbers is almost identical.

I

You have seen examples of sets with accumulation points, and it is easy to see that finite sets do not have accumulation points. A natural question to pursue is the following: Under what conditions can one guarantee that a set will have at least one accumulation point? Of course, such sets must be infinite. Now consider the set J of all positive integers. Given any real number A, it is easy to find a neighborhood of

1.2 Cauchy Sequences 43

A that will exclude all but a finite number of positive integers (in fact, all if A 6 J ) . Thus J has no accumulation points. The defect here that allows J to avoid possessing accumulation points is that J has plenty of room to "spread out" its members. Perhaps if J is restricted so that it must be contained in an interval of finite length, this restriction will force the existence of accumulation points. This conjecture is true, as we shall now prove.

I .6 BOLZANO-WEZERSTRASS THEOREM Every bounded infinite set of real numbers has at least one accumulation point. We shall preface the proof of this famous theorem with some comments on its meaning. As observed previously, it is impossible for finite sets to have accumulation points; moreover, we have an example-the set of natural numbers-of an infinite set with no accumulation point. Consequently, the boundedness of the set must play some role in forcing the existence of an accumulation point. Note that the theorem does not assert that the accumulation point need belong to the set. As you read the proof, notice how a sequence of intervals {[a,, b,])gl is being defined. The process is the following: The numbers a1 and bl are chosen, then instructions are given for choosing a2 and bz that depend on a1 and bl. Next, the same set of instructions is used to choose a3 and b3 based on the choices of a2 and k. Then come the magic words "Continuing in this fashion"! What all this means is that this sequence is being defined recursively. Usually, when a sequence is defined recursively, the first term is chosen (sometimes several terms are chosen before the recursion instructions are invoked), and then the nth term is chosen based on the choices for the preceding terms. The instructions for choosing the nth term are designed so that the sequence possesses certain properties. Strictly speaking, one should supply a formal proof by mathematical induction that such a definition is possible. In place of that formal proof, however, one usually offers an argument that the first term can be so chosen, and given the fact that the kth term meets the requirements for 1 5 k n - 1, the nth term can be chosen with the desired properties. You will see this method of constructing sequences later in this book and in your further study of analysis.

0;hence, there is a positive integer N1 such that n 2 Nl implies lan - A1 < E' and a positive integer N2 such that n 2 N;? implies lbn - BI < E'. Let N = max{Nl,N2). For n 2 N,

Thus, {anbn)gl converges to AB. Some facts follow immediately from Theorem 1.8 and 1.9. If {an)zl converges to A and a is any real number, then the constant sequence {a)glconverges to a ; hence, the sequence {aan)gl converges to d.In particular, if a = -1, {-an):, converges to -A. Combining this observation with Theorem 1.8, we see that, if {an)gl converges to A and {bn)gl converges to B, then {an - bn)gl converges to A - B. More generally, if a and ,8 are real numbers, den {aan+ ,8bn)zl converges to cuA + PB. Be careful not to read any unintended meaning into Theorems 1.8 and 1.9. If {an}gl and {bn)gl are sequences that do not converge, Theorem 1.8 does not assert

II

1.3 Arithmetic Operations on Sequences 47

that {an+ b n ) z l fails to converge. In the same way, Theorem 1.9 does not assert that {anbn)glfails to converge. Exercises 25, 26, and 27 illustrate this point. The problem with division must be approached with some caution. First of all, consider { a n ) z lconverging to A and { b n ) g lconverging to B. We wish to consider the sequence

and might be led to conjecture that this converges to $. For this even to make sense, we must insist that B $ 0 and that bn # 0 for all n. Let us proceed with these facts in mind.

Considering this result in the same fashion as that preceding Theorem 1.9, we see that the quantities la, - A1 and lB - bnl can be made small and that the constant A poses no difficulty. However, the factor must be bounded somehow; indeed, 1B1 it suffices to keep 1 bn1 away from zero. However, since { b n ) z lconverges to B $0, > 0. The set (B - 5, B + 5) contains this should not be difficult. Since B $ 0, e = all but a finite number of terms of the sequence { b n ) z l ;hence, the choice of e makes it clear that all terms in this neighborhood are "bounded away from zero." In other words, there is a positive number M such that 1b.l 2 M for all but a finite number of terms of the sequence. We shall now prove this in a precise and economical fashion.

F

+

1.10 LEMMA If { b n ) g lconverges to B and B 0, then there is a positive real number M and a positive integer N such that, if n 2 N, then lbn1 2 M.

9

Proof Since B $ 0, = E > 0. There is N such that, if n 2 N, then lbn-Bl k t ~ = Thus for n 2 N,

F.

I

< 6;

Note that, although the idea of this proof is fairly simple, the inequalities used in the proof are a bit tricky. It is tempting to yield to the inclination to give neat, economical proofs like this. In fact, you should try your hand at this pleasant pastime as you gain more confidence and experience and begin to be critical of the proofs appearing in this book. In order to cast aside the veil obscuring the important facts in this proof, we shall give an alternate proof for the case when B > 0. The case B < 0 may be handled similarly.

48 Chapter l

Sequences

Of course, this is more cumbersome than the first proof, but it is much easier to digest. If B > 0, then f > 0. Now there is a positive integer N such that, for n 2 N, S , B - B < bn 0 2 and B < 0 were handled simultaneously by judicious use of inequalities concerning absolute values. The reader will learn a lot by writing out the proof when B < 0 and comparing the two methods of proof. Equipped with this lemma, we are now ready to state and prove a theorem concerning the quotient of two sequences. I.II THEOREM If {an)g, converges to A and {bn)zl converges to B, with B # 0 and bn # 0 for all n, then converges to

(2)n=l

a.

Proof Choose c > 0. By Lemma 1.10, there is a positive real number M and a positive integer Nl such that lbn1 2 M for all n 2 N1. Then

(See the paragraphs preceding Lemma 1.10 to understand the reason for this choice of E'.) There is a positive integer N2 such that, for n 2 N2, lan - A1 < d, and a positive integer N3 such that n 2 N3 implies lbn - BI < E'. Let N = max{N1,N2,N3). For n 2 N, lbn - BI < d, lan -A1 < d, and Ib,l 2 M. Thus,

You have now been initiated into the exclusive club of epsilon pickers. The secrets revealed are representative of the types of epsilon-picking problems to be encountered later in this dook, so you should digest them carefully. As a corollary to Theorem 1.11, observe that, if {bn)gl converges to B with B # 0 and bn # 0 for all n, then { l / b n ) ~ converges , to 1/B. (Let an = 1 for all n, and apply Theorem 1.11.) Exmple 1.4

Consider the sequence


At first glance it would seem there is little use in trying to apply Theorem 1.11 to this sequence because the sequences {n3 - 100n2+ n - 58)z1 and {2n3 + 65n + 6912, are both unbounded and hence are divergent. However, some slight adjustments may improve the situation. Now, for each positive integer n,

converges to 0; hence, by Theorem 1.9, We have already shown that 00 and { $ both converge to 0. If we recall the remarks following Theoconverges to rem 1.9, it is clear that (1 - 100 (!) + $ - 58

}zl

(5

and (2 + 65

($)}zl

)zl

(5)+ 69 ($)

converges to

Now, by Theorem 1.11,

converges to

i.

rn

{Jx

Example 1.5 Consider the sequence - &)El. As we saw in Example 1.4, the form in which this sequence is given does not lend itself to the use of tools at our disposal, but a familiar algebraic device will remedy this problem. Note that, for n > 0,

Now let us refer to some of the homework exercises rather than attack this problem 00 directly. By Exercise 28, converges to 0,since { converges to 0. Now

(5)

n=l

and the sequence

{&)

:}zl

00

n= 1

converges to 0; hence the sequence

50 Chapter l

Sequences

converges to 0 by Exercise 9. A relation between order and convergence remains to be shown. We have a specific relation in mind and shall pursue that exclusively. Suppose { a n ) g l and { b n ) g l are sequences converging to A and B, respectively. Suppose further that a, 5 bn for all n. By this time it should be evident what the conclusion should be-namely, that A 5 B. However, any attempt to improve this result, by asserting that an < bn for all n implies A < B, fails. TO see this, consider the sequences

&

< for all n, but both sequences converge to 0. The anticipated theorem We have will be proved by assuming that A > B and showing that this leads to a contradiction of the hypothesis. If A > B, then it is easy to find neighborhoods P and Q of A and B, respectively, that do not overlap and such that if x E P and y E Q, then x > y. All but a finite number of terms of { a n ) z l belong to the set P. The corresponding terms of the sequence { b n ) g l cannot belong to Q, since an 5 bn for all n. Thus, there are at most a finite number of terms of the sequence { b n ) g lbelonging to the set Q contrary to { b n ) g lconverging to B. 1.12 THEOREM If { a n ) z l converges to A and { b n ) z l converges to B, with a, b, for all n, thenA 5 B.

0. There is a positive integer N l such that n 2 N 1 implies that A - c < a, < A + e. There is a positive integer N2 such that n 2 N2 implies that B - e < bn < B + c. Choose N 2 max{N1,N 2 ) . Then we have

This last inequality is impossible. Note that the contradiction in the proof was not the one mentioned preceding the theorem. But the idea is essentially the same; the choice was motivated by a desire to make the proof simpler to present. A word of friendly advice may help. Contrary to popular belief, mathematics does not consist entirely of formulas and recipes for solving problems. Some of them are convenient in mathematics, but unless they are applied with an understanding of the underlying principles and some intuitive feeling for these principles, the rules and formulas have limited use. You should construct examples other than those given here and examine them critically. The ideas used in formulating a proof may be very


useful in solving a problem in which the theorem may not apply. The next example iflustrates this point. Consider the sequence

This can be viewed as the product of two sequences and {sin F):,. Unfor00 does not converge since sin = 0 if n is divisible tunately, the sequence {sin by 2, and sin is +1 or - 1 if n is odd; thus, we cannot apply Theorem 1.9. However, {sin ?,:) is bounded, and { ,:) converges to 0. Referring to the paragraph preceding Theorem 1.9, let bn = and an = sin y. Then B = 0, and A is unimportant since 1B1 lan - A1 = 0. Thus, it appears that the sequence converges to 0. Let's check this to make sure. Choose E > 0. There is a positive integer N such that < e. Then, for n 2 N, we have

7

:

Our conjecture was correct; in fact, we are led to invent a new theorem. R o facts were used in the discussion-namely, { converges to 0, and {sin ? is bounded.

)zl

)zl

1.I3 THEOREM If { a n ) z lconverges to 0 and { b n ) z lis bounded, then {anbn)gl converges to 0.

Proof Let M be a positive number such that lbnl 5 M for all n. Choose e > 0. Then e' = & > 0. (The reason for this choice of E' will be apparent shortly, but you should have anticipated this choice.) There is a positive integer N such that n 2 N implies lanl = lan - 01 < 6'. Then

Thus, {&bn)zl converges to 0. Now reconsider one of the equivalent formulations of the idea of convergence: a sequence { a n ) g lconverges to A ifY each neighborhood of A contains all but a finite number of terms of the sequence. This statement makes it clear that any finite number of terms of a sequence may be changed without affecting either the convergence of the sequence,or the limit; in fact, a finite number of terms may be deleted from the sequence and the remaining terms relabeled, retaining the original order, without affecting the convergence or limit. The definition of a subsequence of a sequence will facilitiate this discussion.

52 Chapter l

Sequences

DEFINITION Let {an)El be a sequence and {nk)El be any sequence of positive integers such that nl < n2 < n3 . The sequence {ank)& is called a subsequence of {an)El. e e

In less formal terms, we can say that a subsequence is formed by deleting some (or none, since every sequence is a subsequence of itself) of the terms of the sequence and relabeling the remaining terms while retaining the original order of the terms. Of course, there must be infinitely many terms remaining after the deletion. For example, the sequences { 1/k2)El, (1 /2k)El, and { 1 / 2 k=l ~ )are~ subsequences of the sequence {l/n)El formed by setting nk = k2, nk = 2k, and nk = 2k,respectively. The sequences {1/2k)E1 converge to 0 according to our previous remarks. The reader is urged to try to prove that {1/2k)El also converges to 0, although this fact will be evident from some results that follow. Ifa, = 2lf for each positive integer n, and the sequence {nk)El is chosen such that nk = 2k for each positive integer k, then an, = 1 for each k, and the subsequence thus obtained has all terms equal to 1 and, hence, converges to 1. On the other hand, if nk is chosen to be 2k - 1 for each k, then an, = 0 for all k, and the subsequence thus chosen converges to 0. As noted earlier, every sequence is a subsequence of itself, and this sequence does not converge. Reflect a moment on the two examples just given. The first was a convergent sequence, and, of the three subsequences sampled, all converged to the same limit: the limit of the original sequence. Although this evidence is not conclusive, it does give some "statistical support" to lead one to conjecture that every subsequence of a convergent sequence converges and converges to the same limit. This result should not be surprising in light of the discussion preceding the definition of a subsequence. Proceeding with great confidence in the inherent orderliness of mathematics, we consider the second example given-a sequence that does not converge. It has two subsequences with different limits and one subsequence that does not converge at all. By now you should have a clear understanding of how the statements of many theorems can be formulated by considering well-chosen examples. In fact, you should associate with each theorem of importance a stock of examples and counterexamples that illustrate both the necessity of the hypotheses and the strength of the conclusions. A theorem used in this fashion becomes a valued aid and tool to be cherished and exploited, not a tedious fact to be memorized. {

/

1 and

I.I4 THEOREM A sequence converges iff each of its subsequences converges. In fact, if every subsequence converges, then they all converge to the same limit.

Proof The first half of this theorem is very easy to prove because every sequence is a subsequence of itself. Indeed, if every subsequence of a sequence converges, then the sequence is convergent, since the sequence is included among the subsequences, all of which were assumed convergent.

1.4 Subsequences and Monotone Sequences 53

Assume now that { a n ) g l is a sequence that converges to A and that { a n k ) g l is a subsequence. The proof of the theorem will be complete if it can be shown that { a n k ) g l converges to A. (Before going further, read again the paragraph preceding the definition of a subsequence.) Choose E > 0. Since { an )0° converges to A, there is a positive integer N such that, for n 2 N, lan - A( < E. Since { a n k ) g l is a subsequence of the given sequence nl < n2 < *, then k 5 nk. Thus, for k 2 N, nk 2 N ; hence, lank - A1 < E . Therefore, { a n k ) g l converges to A. Note that if every subsequence of a given sequence converges, then the sequence converges, and, by the argument just given, all subsequences converge to the same limit. We can prove a slight variation on Theorem 1.14 if we assume the sequence is bounded. 1.I5 THEOREM Suppose { x n ) Z 1 is a bounded sequence. If all its convergent subsequences have the same limit, then the sequence is convergent.

Proof Assume that all the convergent subsequences of { x n ) Z l have the same limit; call it xo. Exercise 36 guarantees that there is at least one convergent subsequence. If { x n ) z l does not converge to xo, then there is E > 0 for which one cannot find an integer N to satisfy the definition of convergence. If we try N = 1, there is nl > 1 such that (xn,- xo( 2 E. Likewise, there is nz > 2 (and we may choose n2 > n l ) such that 1xn2- xol 2 E . Continuing in this fashion (we are defining the sequence { n k ) z l recursively), we find, for each positive integer k, an integer nk such that nl < n2 < n3 . . . < nk-1 < nk and such that 1xnk - xol 2 e. The sequence { x ~ ~ ) &being , bounded, will have a convergent subsequence that will be a convergent subsequence of the original sequence and that will not converge to xo. This contradicts the hypothesis.

DEFINITION A sequence {a,):, is increasing iff an 5 an+l for all positive integers n. A sequence { a n ) z l is decreasing iff an 2 an+, for all positive integers n. A sequence is monotone iff it is either increasing or decreasing. 1.16 THEOREM

A monotone sequence is convergent iff it is bounded.

Proof Suppose { a n ) g l is a monotone sequence that is bounded; for definiteness, suppose the sequence is increasing. The set {an : n = 1,2, . . .) is bounded, so let s = S U ~ { U ,: n = 1,2,. . .). It will be shown that { a n ) z l converges to s. Choose E > 0. Since s is the least upper bound of {an : n = 1,2, . . .), s - E is not an upper bound; hence, there is no such that s - E < a,. Now, for n 2 no,

54 Chapter 1 Sequences

hence {an)glconverges to s. For the case in which {an)gl is decreasing, let s = inf(% : n = 1,2, . . .), and the details are similar (see Exercise 37). If the sequence is convergent, then by Theorem 1.2 it is bounded.

In keeping with our philosophy of examining examples to illustrate theorems and to show the need for the hypotheses, we shall consider several sequences. First, the sequence {n)gl is indeed monotone, but it does not converge because it is unbounded. The sequence {0,1,0, . . .) is bounded, but the theorem does not apply, because it is not monotone; indeed, it fails to converge. *

W

Example 1.6

Consider the sequence {sn)gl defined as follows:

We shall prove that this sequence is increasing and bounded by 2 and hence, by the preceding theorem, is convergent. Since this sequence is defined recursively, we will use induction to prove the conjecture stated above. Clearly, sl 5 2 and

We wish to show that sn 5 2 and sn+l2 snfor all n; the preceding statement verifies the truth of this for n = 1. Assume the condition is satisfied for n = r. Then

and

hence, by the principle of induction, the inequalities hold for all n. By use of Theorem 1.15, we are assured that the sequence converges; call the limit L. Now the subsequence { ~ ~ -also ~ converges ) 2 ~ to L, and since sn > 0 for all n, {js,_1)z2 converges to &. Now observe that since sn = ,/-, the sequence

{sn)zl

converges to 4 2 + &,and also that cnverges to L; thus, by the uniqueness By eliminating the radicals, we see that L of the limit of a sequence, L = must be a root of the polynomial equation L~ - 4~~ - L + 4 = 0.

d m .

It is left as an exercise to consider the sequence ((1 + f )n)gl. This sequence may be shown to be increasing and bounded, and hence convergent. The limit happens to

1.4 Subsequences and Monotone Sequences 55

be e, the base for the natural logarithm-a number that plays a central role in calculus. See Project 1.1 at the end of the chapter.

I.I7 THEOREM Let E be a set of real numbers. Then xo is an accumulation point of E iff there is a sequence {xn)zl of members of E, each distinct from xo, such that {xn)zl converges to xo.

Proof Let xo be an accumulation point of E. Then, for each positive integer n, 1 there is a point xn E E such that 0 < Gx, -xol < i.[The set (xo - ;,xo + ;)1 is a neighborhood of q,so it contains a member of E distinct from xo.] It remains to be shown that {xn)gl converges to xo. Choose c > 0. Then there is a positive 5 c. Thus, integer N such that $ < N; hence, for n 2 N, Ix, - xol < 1 5 { x n ) ~converges , to xo. Suppose now there is a sequence { x , ) ~ , of members of E, each distinct from xo, that converges to xo. Since every neighborhood of xo contains all but a finite number of terms of the sequence, every neighborhood of xo must contain at least one member of E that is distinct from xo. Hence, xo is an accumulation point of E. We close this chapter by examining several examples involving techniques that are of sufficient interest to be studied with care.

!

I

a

Example 1.7 Consider a real number 0 0; hence, the sequence is decreasing. In order to show that this Sequence converges, it suffices to show that it is bounded from below, an easy task since Z F > 0 for all natural numbers n. Although we know now that the sequence converges, we shall seek also to determine its limit. Now {6")g, converges; call its limit L. And {bZn)zl,being a subsequence of {P)gl,also converges to L. On the other hand, {b2")z1, which may be considered as the product of the sequence {bn)gl with itself, converges to L'. By the uniqueness of the limit, L* = L, and so L = 0 or L = 1. Clearly, L 1 since the sequence was decreasing and b < 1. Hence, {bn)z1 converges to 0. This example is not intended to amaze the reader, but it illustrates a few techniques that are convenient to have at hand.

+

W

Example 1.8

Consider 0 < c < 1 and the sequence {$?)El. For all n,

*

- "-fi= ~ ( - 1""-G) > 0,

since c < 1. Hence, the sequence is increasing, and clearly f i < 1 for all n. Thus, the sequence converges; call the limit L. By Exercise 28, if {xn)gl converges to xo,x, 2 0 for all n, then converges to &. Therefore, the sequence converges to 4,but

{G)z1

56

Chapter l

Sequences

and { m$)gl is a subsequence of {fi):,and hence converges to L. By the uniqueness of the limit of a sequence, & = L; hence L = 0 or L = 1. But c > 0 and the sequence is increasing; therefore L # 0. In conclusion, { if/i}~,converges to 1. (See Exercise 39 for the case c > 1.) I Example 1.9 We shall now consider another sequence that is defined recursively-that is, the first few terms are given, and instructions are provided for computing the nth term in terms of some or all of the preceding terms. Let a1 = 1, and for n 2 2, define an = JG. We shall also use induction to attack this sequence. First, look at a few terms of the sequence,

It seems reasonable to expect the sequence to be increasing and, hence, convergent if it is bounded. It may not be clear at first glance what to try for an upper bound, so we shall discover this bound by a very useful device. (Even if you can guess the bound, play the game to see how the device works.) Since the sequence is increasing, if it converges, the limit will be the least upper bound of the sequence. In an attempt to discover a logical candidate for this limit, let us assume that the sequence converges. Suppose {a,):, converges to L; then {a):, converges to L ~but , a: = 2an-i; hence, {a):, = {2a.-1)21 converges to 2L. Thus, the only candidates for the limit must be the solutions of the equation L~ = 2L; either L = 2 or L = 0. It is easy to rule out the possibility L = 0, so L = 2 is the obvious choice. Note that is has not been proved that {an)El converges to 2. Our only proof so far is that if the sequence converges, then the limit must be 2. We shall now prove by induction that a, an+l5 2 for all n; hence, the sequence is bounded and monotone and thus is convergent. For n = 1, a1 = 1 5 a2 = & 5 2. Assume the statement is true for n = r; in other words, a, I a,+, 5 2. Then

0 such that

Let x and y be distinct real numbers. Prove there is a neighborhood P of x and a neighborhood Q of y such that P n Q = 8. Suppose x is a real number and E > 0. Prove that (x - E,X+ E) is a neighborhood of each of its members; in other words, if y E (x - E,x + E),then there is 6 > 0 such that (y - 6 , y + 6 ) c (x- e , x + ~ ) . Find upper and lower bounds for the sequence

{

00

)*,.

Give an example of a sequence that is bounded but not convergent. Use the definition of convergence to prove that each of the following sequences converges:

(b)

{"Irn n

.=I

Show that {an)z1 converges to A iff {an - A ) z l converges to 0. Suppose {an)zl converges to A, and define a new sequence {bn)zl by bn for all n. Prove that { b n ) z l converges to A.

-- an +2an+i

suppose {an)=,, {bn)El, and {cn)E1 are sequences such that {an):, converges to A, {bn)El converges to A, and an 5 cn 5 bn for all n. Prove that {cn)E1converges to A. Prove that, if {an)El converges to A, then { J a n J ) Econverges l to IAJ. Is the converse true? Justify your conclusion. Let be a sequence such that there exist numbers a and N such that, for n 2 N, an = a. Prove that {an)zl converges to a.

58 Chapter l

Sequences

12. Give an alternate proof of Theorem 1.1 along the following lines. Choose c > 0. There is N1 such that for n 2 N 1 Ian -A1 < 5 , and there is N2 such that for n 2 N2 Ian - BI < 5. Use the triangle inequality to show that this implies that JA- BI < c. Argue that A = B. 13. Let x be any positive real number, and define a sequence { a n ) g l by

where [x] is the largest integer less than or equal to x. Prove that { a n ) g l converges to 42.

1.2 CAUCHY SEQUENCES 14. Prove that every Cauchy sequence is bounded (Theorem 1.4). 15. Prove directly (do not use Theorem 1.8) that, if {an)gl and { b n ) g l are Cauchy, so is {an + bn)El* 16. Prove directly (do not use Theorem 1.9) that, if {an)El and { b n ) g l are Cauchy, so is {anbn)E1.YOUwill want to use Theorem 1.4.

{y)n=l 00

17. Prove that the sequence is Cauchy. 18. Give an example of a set with exactly two accumulation points. 19. Give an example of a set with a countably infinite set of accumulation points. 20. Give an example of a set that contains each of its accumulation points. 21. Determine the accumulation points of the set (2" + : n and k are positive integers). *22. Let S be a nonempty set of real numbers that is bounded from above (below) and let x = sup S (inf S). Prove that either x belongs to S or x is an accumulation point of S. 23. Let a~ and a1 be distinct real numbers. Define an '= "-I"-' for each positive integer

;

n 2 2. Show that { a n ) g lis a Cauchy sequence. You maywant to use induction to show that

and then use the result from Example 0.9 of Chapter 0.

,

24. Suppose {an)gl converges to A and {an : n E J) is an infinite set. Show that A is an accumulation point of {an : n E J ) . I.3 ARITHMETIC OPERATIONS ON SEQUENCES

25. Suppose {an)gl and {bn)gl are sequences such that { a n ) g l and {an+ b n ) g l converge. Rove that {bn)El converges.

26. Give an example in which {an)gl and { b n ) g ldo not converge but {an+bn)Elconverges. 27. Suppose {an)gl and {bn-)gl are sequences such that { a n ) g l converges to A # 0 and {anb n ) g l converges. Prove that { b n ) g lconverges. 28. If { a n ) g lconverges to a with an 2 0 for all n, show {&}El If a > 0,then 6 & = Ffi.]

29. Prove that

converges to

fi. [Hint:

Exercises 59 converges to

i, where

(n + k)! ('ik)= YE-

Rove the following variation on Lemma 1.10. If {bn)Elconverges to B for all n, then there is M > 0 such that IbnJ 2 M for all n.

# 0 and bn # 0

Consider a sequence {an)Zl and, for each n, define

Rove that if {an)gl converges to A, then which { a n ) g l converges, but { a n ) g ldoes not.

converges to A. Give an example in

Find the limit of the sequences with general term as given: n2 + 4n

(a)

n2-5

(c)

sin n2 \/ii

Find the limit of the sequence in Exercise 23 when to look at Example 0.10.

= 0 and a1 = 3. You might want

I .4 SUBSEQUENCES AND MONOTONE SEQUENCES 34. Find a convergent subsequence of the sequence

35. Suppose x is an accumulation point of {a, : n E J). Show that there is a subsequence of

{ a n ) g l that converges to x.

*36. Let {an);, be a bounded sequence of real numbers. Prove that {an)gl has a convergent subsequence. (Hint: You may want to use the Bolzano-Weierstrass Theorem.) *37. Prove that if { a n ) g l is decreasing and bounded, then {an)gl converges. 38. Rove that if c > 1, then { f i ) g l converges to 1. *39. Suppose { x n ) g lconverges to XQ and { y n ) s l converges to XQ.Define a sequence {zn)gl a~ ~ O ~ ~ O W12, S : = xn and 22,-1 = yn. Prove that (zn)El converges to XQ.

40. Show that the sequence defined by a1 = 6 and an = d and find the limit.

z for n > 1 is convergent,

60 Chapter 2

Sequences

41. Let { x n ) z l be a bounded sequence, and let E be the set of subsequential limits of that

sequence. By Exercise 37, E is nonempty. Prove that E is bounded and contains both sup E and inf E. 42. Let { x n ) z 1 be any sequence and T : J --r J be any 1-1 function. Prove that if { x n ) z 1 converges to x, then { x ~ ( ~ ) ) also : ~ converges to x. Explain how this relates to subsequences. Define what one might call a "rearrangement" of a sequence. What does the result imply about rearrangements of sequences? 43. Assume 0 5 a 5 b. Does the sequence {(an + b n ) ' / " ) z l diverge or converge? If the sequence converges, find the limit. 44. Does the seauence

diverge or converge? If the sequence converges, find the limit. 4 5 . Show that if x is any real number, there is a sequence of rational numbers converging to *46. Show that if x is any real number, there is a sequence of irrational numbers converging to x. 47. Suppose that { a n ) z l converges to A and that B is an accumulation point of {an : n E J ) . Prove that A = B.

MISCELLANEOUS 48. Suppose that {an)=, and {bn)El are two sequences of positive real numbers. We say that an is O(bn) (read as an is "big oh" of bn) if there is an integer N and a real number M such that for n 2 N, an M bn. Prove that if { a n / b n ) z , converges to L # 0, then an is O(bn) and bn is O(an). What can you say if L = O? Illustrate with examples.

2,

1 1 2. Prove that 1 + - + - + - + 2 4 3. Prove that the sequence

is bounded. 4. Prove that for all k

1 2n-1

-5 2 for all n E J.

> 2,

5. Prove that the sequence

is increasing. 6. Items 3 and 5 show that the sequence

is bounded and increasing; hence, it is convergent; call the limit e. Compute the 100th, 1,000th, and 100,Oth term of the sequence to approximate e. Compare your results with the number e that your calculator will display.

PROJECT 1.2 The purpose of this project is to establish the convergence of an iterative process for approximating square roots. The process is a familiar one--the divide-and-average process. Define a sequence recursively as follows: a1 = 2, and for n 2 2, define

62 Chapter 1 Sequences

I. Prove that an 2 2

fi for all n.

This is a bit tricky. Use some algebra to write

(

(an + $) - 8 as a square to establish the fact that an +

')* > 8.

2. Prove that the sequence is decreasing. 3. Prove that the sequence converges to 4. Generalize this process to find a sequence that converges to positive real number.

a.

Jii where a is any

PROJECT 1.3 In Exercise 40, you proved that a particular sequence defined recursively was convergent. In this project, you will have an opportunity to generalize that result. i. For what Define a sequence by a1 = a, and for n 2 2, define an = ,/= choices of a (positive, of course) will this sequence converge and to what limit? Prove your conjecture.

PROJECT 1.4 In this project you will prove that the set of real numbers is not a countable set. The common usage is to say that such a set is uncountable. We begin with a few facts about equivalent sets of real numbers. Recall that sets A and B are equivalent if there is a 1-1 function from A onto B. Your strategy will be the following: You will show that [0, 11 is equivalent to R and that [0, 11 is uncountable. Then you can conclude that R is uncountable. First, you will show that [O, 11 is equivalent to R. (You may want to refer to Project 0.1.) 1. Prove that R is equivalent to (0, I), and (0,l) is equivalent to [0, 11. Conclude that R is equivalent to [0, 11. Now your task is to prove that [0, 11 is uncountable. Your strategy will be to consider an arbitrary function f :J + [O, 11 and prove that im f # [O, 11. Thus, there is no function from J onto [0, 11, and so [0, 11 is uncountable. Suppose that T is a function from J to [0, 11. 2. Show that there are sequences {an)gl and {bn)gl such that [al,bl] C [0, 11, and for each n E J, [an+l,bn+lI C [an,bnI and T(n) 6[an,bnI* 3. Show that {an)gl converges; call the limit A. 4. Prove that A E [an,bn] for each n E J . Conclude that A $ im T. 5. Finish the proof that [O,1] is uncountable.

Chapter

2

Limits of Functions

he notion of the existence of a limit of a function at a point underlies the study of calculus. When students fail to understand this idea, their study of calculus becomes a drudgery of juggling formulas. Before considering continuity, differentiation, and integration, one needs a thorough understanding of limits. The material on sequences presented in Chapter 1 will help in achieving this understanding. First we shall consider an example commonly found in calculus books. Let

forx # 1, and let f(1) = 6. We find thatf(x) = x + 1 forx # 1 and f(1) = 6. To the unsophisticated, it may seem objectionable to define a function this way. Since f behaves so well elsewhere, it seems reasonable to define f(1) = 1 + 1 = 2. In this chapter we shall attempt to rearrange any prejudices of the reader concerning the behavior of such functions.

2.1

DEFINITION OF THE LIMIT OF A FUNCTION

First of all, let us exanfine the intuitive idea, presented in elementary calculus, of a limit of a function. Consider a function f : D + R where D C R and L is a real number.' If xo E R, then f has a limit L at xo' if, as x approaches xo, f (x) gets close to L. Of course, the values of x to be considered must belong to the domain of the function. The precise formulation of the idea of getting close no longer poses any problem, since this has already been done in Chapter 1 while dealing with convergent sequences. To summarize, in order to formalize the definition of a limit of a function,

ere

and throughout the remainder of this book, R will be the set of all real numbers, and the domain of any functions considered will be subsets of R.

64

Chapter 2 Limits of Functions

it is necessary to define precisely what is meant by "f(x) gets close to L as x approaches xo." To set the stage for this, it is necessary that there be points in D as close to q as one may wish. Finally, the limit should depend only on the behavior of the function near xo, not on f (xo); indeed, xo need not even belong to D.

DEFINITION Let f :D -.R with xo an accumulation point of D. Then f has a limit L atxo iff for each E > 0 there is a 6 > 0 such that if 0 < Ix-xol < 6 and x E D, then 1 f(x) - LI < E. Observe that the definition is similar to that concerning a sequence converging to a real number L in that it does not specify that if L exists, it is unique. However, this is true, and the proof is left as Exercise 5. Iff has a limit L at xo, we write L = lim,, f (x). It is incorrect for students to think they should memorize all definitions such as the preceding one. That this mistaken impression is widespread among students is evidenced by the fact that many are able to regurgitate the words of the definition, sometimes in the wrong order, without understanding the full meaning. A definition should be carefully examined, dissected, and fully assimilated. The reasons behind it and the full meaning of each part of the statement should be carefully considered. The ideas should become not only a part of one's memory, but a part of one's way of thinking-familiar friends and companions along the journey to enjoyment of mathematics. First, the insistence that xo be an accumulation point of D, the domain off, is necessary in order that f be defined at points near xo. Now, the positive number E is the desired degree of closeness chosen in advance. To fulfill the definition, one must find 6 > 0 (which, of course, will generally depend on E) such that if x E (xo - 6, xo + 6), x E D, and x # q (the behavior off at xo should not affect the limit); then f (x) E (L - E , L + E). This last statement is equivalent to insisting that the graph off for x E Dn(xo-6, xo+6), x # xo, lies in the strip {(x,y) :L-6 < y < L+E). (See Figure 2.1.) The hole in the graph emphasizes that the value off at xo is not considered when determining whether L is the limit off at xo.

Example 2.1 Let us now reconsider the example mentioned at the beginning of the chapter. Let D = R, and define f :D -+ R by

$ 9

= x + 1; hence, f is a linear function, for x # 1 and f(1) = 6. For x # 1, f(x) = and the graph off is a line with slope 1, except for x = 1. As mentioned above, the number f(1) has nothing to do with the existence of a limit at x = 1. Thus, one is led to believe (correctly, by the geometry of the plane) that as x approaches 1, f(x) approaches the value necessary to fill in the gap in the line-namely, 1 + 1 = 2. Let us prove that f has a limit L = 2 at x = 1.

2.1 Definition of the Limit of a Function 65

x0

-6

x0

x,,

+6

Figure 2.1

Choose E > 0. Consider the geometric interpretation of the idea of a limit. It is necessary to choose a neighborhood of 1 such that for x in this neighborhood with x f 1, the corresponding points on the graph off lie in the strip {(x, y) : 2 - E < y < 2 + E). Ignoring the point x = 1, the graph off is a straight line of slope 1. Thus, one is tempted to try E = 6 to obtain the neighborhood (1 - 6, 1 + 6) of x = 1. This result could be obtained by direct computation, but it is worthwhile to take this opportunity to emphasize the geometric aspect of the definition. If 0 < Ix - 1I < 6 = E,then

and the conjecture is shown to be correct. See Figure 2.2.

In the next example, we exhibit a function that fails to have a limit at a point. In order to show that a function does not have a limit at a point, we must show that no number L can meet the requirements of the definition. The strategy will be to assume that the function has a limit and arrive at a contradiction, showing that the limit does not exist. Later we will have simpler methods of showing that a function fails to have a limit at some point.

66 Chapter 2 Limits of Functions

Figure 2.2

Example 2.2 Define f : R\{O) + R by f (x) = H.For x < 0, f (x) = - 1 and for x > 0, f (x) = 1. This suggests that f does not have a k i t at zero. We will proceed by assuming there is a number L that satisfies the requirements of the definition of the limit off at zero and arrive at a contradiction. To the left of zero the values of the function are - 1, and to the right of zero, the values are 1. So if we choose c = 114, there is no number L that differs from - 1 by less than 114 and that differs from 1 by less than 114. Assume L is a limit off at zero. Choose E = 114. There is 6 > 0 such that if 0 < (x - 01 < 6, then 1 f(x) - LI < 114. Choose p = -614 and q = 614. Then f(p) = -1, f(4) = 1, and SO

This is a contradiction. The conclusion is that f does not have a limit at zero.

I

H

In the next example, we will show that the given function fails to have a limit at 0. This time the strategy will be to let L be any real number and exhibit an E > 0 for which no b > 0 can be found that meets the requirements of the definition. Thus, L (an arbitrary real number) cannot be a limit off at 0; hence, f does not have a limit at 0. The differences in the strategies used in Examples 2.2 and 2.3 are somewhat subtle. To comprehend more fully how to show that a function does not have a limit at a point, you might want to use the strategy of Example 2.2 on the function in Example 2.3 and vice versa. Not only will this aid in your understanding of limits of functions, but it will illuminate the differences between the two strategies.

2.1 Definition of the Limit of a Function 67

Figure 2.3

i,

Example 2.3 Define f : (0,l) + R by f (x) = sin and consider the behavior off at x = 0. Although zero is not in the domain off, it is an accumulation point of the domain off; hence, it is reasonable to inquire whether f has a limit at zero. In this example, it is instructive to look at the graph of this function in Figure 2.3 to gain insight into its behavior as x approaches zero. Since f (llna) = 0 for each positive integer n and f (2/n7r) is +1 or -1 for n an odd positive integer, we see that the function oscillates more and more wildly as x becomes close to zero. Consequently, it is reasonable to conjecture that f does not have a limit at zero. Let L be any real number and choose E = 113. To show that L is not a limit off at zero, it suffices to show that for any 6 > 0, there is x E (0,l) such that 0 < )x-01 < 6 and 1f (x)-L1 2 E. By the choice of E,either 1 $! (L - E,L + E)or 0 # (L - E,L + E). Suppose the latter is the case. Then, given 6 > 0, there is a positive integer n such that l/na < 6, and, as observed above, f (llna) = 0 # (L - E, L + E);hence, f (l/na) - 2 6. The case for 1 $! (L - E,L + r) can be handled similarly. Thus, f does not have a limit at zero.

1

LI

Example 2.4 Definef : (0,l) -r R by f (x) = l/x. Now zero is an accumulation point of (0,l); hence, we may inquire as to the existence of the limit off at zero. Again, from a sketch of part of the graph of this function, we can see that the function "blows up" at zero. This is not the sort of behavior one expects a function to have if it is to have a limit at zero, so let us prove that f does not have a limit at zero. Let 1 , L be any real number, and choose E > 0 such that L + 6 > 0. Now, if 0 < x < E then L + c < $ =f(x); hence, 1 f(x) - L( > r. Thus, it is impossible to find a 6 > 0 to fulfill the requirements of the definition; that is, L is not a limit off at zero. Since L rn is any real number, f does not have a limit at zero. (See Figure 2.4.) Example 2.5 Let us now consider a more sophisticated example. Define f : [O,11 + R by f (x) = 0 if x is irrational; and if x is rational, set f (x) = l / q where H


Figure 2.4

x = p/q with p and q nonnegative integers that are relatively prime. Thus, we have

and so on. We shall seek to determine those points at which f has a limit and those at which f does not have a limit. Suppose E [0, 11, and let us examine the behavior off near xo. Since there are irrational points in every neighborhood of that are distinct from a , it is clear that f takes the value 0 in every neighborhood of xo infinitely often. Thus, if there is to be a limit off at a , it is clear that the limit must be zero. It remains to decide if it is reasonable to expect that f has zero as a limit at a. For x E [0, 11,f(x) is small (close to zero) only if x is irrational or if x = p/q and p and q are relatively prime with q large. However, for a fixed positive integer q, there are but a finite number of points in [0, 11 of the form p/q; in fact, for a fixed positive integer qo, there are only a finite number of points in [0, 11 of the form p/q, where q 5 qo with p and q positive integers. Thus, f (x) 2 l/qo at only a finite number of points. In view of these observations, the reader should suspect that indeed f has a limit at xo and the limit is zero. Let us now give a proof of this fact. Choose 6 > 0. There is a positive integer go such that l/qo < 6 . There are at most a finite number of rational points in [O,1] of the form p/q where p and q are positive integers with q < qo, say rl, . ..,r,. We may assume that is deleted from this list if it should happen to be of this form. Now, to guarantee that f(x) is small, it is sufficient to avoid these points. Thus, let

and observe that 6 > 0. Now if 0 < Ix - xol < 6 add x E [O,l], then x is either irrational, in which case f(x) = 0, or x = p/q where p and q are relatively prime with

1 I

2.2 Limits of Functions and Sequences 69

Thus, f has a limit at each x E [0, 11, and that limit is zero. A word of philosophy seems appropriate at this point. The emphasis on rigor, both in definitions and proofs, is intentional because this is the only way mathematics can be communicated intelligently. However, one's feelings and intuition about these matters should not be discarded or considered unimportant. On the contrary, intuition is very important and must be cultivated, but it must also be accompanied by the realization that we must provide rigorous proofs for those facts that our intuition tells us to be true. Finally, the intuition must constantly be readjusted because we a& occasionally led astray by our feelings. In Example 2.3, the oscillatory nature of the function, as evidenced by a sketch of the graph, led our intuition to tell us that the function does not have a limit at zero. The behavior of the function in the preceding example will probably elude the intuition of the novice. However, the facts are now before us and should be considered as a new experience leading to greater mathematical maturity.

2.2

LIMITS OF FUNCTIONS AND SEQUENCES

As mentioned in Chapter 1, a strong relationship exists between limits of sequences and limits of functions. Now is clearly the time to seek out these relationships and exploit them to the fullest. Suppose f : D -r R with Q an accumulation point of D, and suppose f has a limit L at xo. Consider a sequence {xn)El converging to with xn E D, xo # xn, for each positive integer n. Since f has a limit L at xo, as the terms of the sequence get close to xo, the corresponding values off must get close to L; in fact, we must have Cf(xn))z1 converging to L. Fortunately, the converse of this theorem is also true. We summarize these results in the following theorem.

2.1 THEOREM Let f : D + R with xo an accumulation point of D. Then f has a limit at Q iff for each sequence {xn)E1 converging to xo with xn E D and xn # xo for all n, the sequence (f(xn))g1 converges. Before proving this theorem, reflect on one aspect of its content. Consider two Sequences {X,)Z~ and {yn)El such that Xn, yn E D,Xn # Xo, yn # xo for n = 1,2,. . ., and such that both {xn)Zl and {yn)gl converge to xo. If we assume that the latter condition of the theorem holds, then Cf(xn))zl and Cf(yn))gl both converge to, say, L1, and L2 respectively. (Note that the condition does not assert that L1 = h.)Form a new sequence {zn)gl where = Xn and zzn-1 = Yn. This sequence consists of members of D distinct from a,and it converges to xo; hence Cf(zn))zl converges. In particular, Cf(xn)}El and Cf(yn))zl are subsequences of the convergent sequence

70 Chapter 2

Limits of Functions

Cf(zn))gl and hence have the same limit; that is, L1 = L2.NOW,if for every sequence {xn)zl of members of D distinct from xo and converging to xo, the sequence Cf(xn))zl converges, then all such sequences have a common limit by use of the preceding observations. This limit should be the limit of the function, a fact that remains to be seen.

Proof Suppose f has a limit L at xo. Let { x , ) ~be~ a sequence of members of D distinct from xo but converging to xo, and consider the sequence Cf(xn))El. Choose E > 0. There is 6 > 0 such that if 0 < Ix - xol < 6 with x E D, then I f(x) - LI < E. Since {xn)z1 converges to xo, there is N such that for n 2 N, Ixn-xol < 6. Nowforn 2 N,O < IX,-X~~ < 6andx E D; hence, 1f(xn)-Ll < E. Thus, (f(xn))zl converges; indeed, it converges to L. Suppose now that the latter condition is satisfied and so by the remarks following the theorem, all the sequences (f(xn))Zl have a common limit, called (with great originality) L. Suppose that L is not a limit off at xo. (We do not assume anythmg concerning the existence of the limit off at xo; we assume only that L is not a limit off at xo.) Thus, there is c > 0 such that for every 6 > 0, there is x E D, with 0 < Ix - -1 < 6 and such that 1f (x) - LI 2 E . In particular, for each positive integer n, there is xn E D with 0 < Ixn < such that If (x,) - LI 2 E. The sequence { x , ) ~ , converges to xo and is a sequence of members of D distinct from xo; hence, (f(xn))zl converges to L, contrary to the fact that [f(xn)- LI 2 E > 0 for all n. Thus, L must be the limit off at xo. To summarize,f.has a limit of L at xo iff, for each sequence { ~ ~of )members g~ of D distinct from xo converging to xo, the sequence (f(xn))El converges. If the second condition is satisfied, then all the given sequences Cf(xn))zl have a common limit, which is L, the limit of the function f . An example should serve to reveal the usefulness of this equivalence and the related observation that follows it. Let f : (0,l) -+ R satisfy the following condition: there is K > 0 such that, for all X,Y E (0, I), 1 f(x) - f(y)1 5 Klx - yl. Thus, in particular, whenever {xn)zl is a Cauchy sequence in (0, I), then so is (f(xn))El; hence, if {xn)zl is a sequence in (0,l) converging to zero, the sequence {f(xn))zl is convergent. By the theorem, f has a limit at zero; and that may be computed by determining the limit of Cf(xn))Zl for any sequence {xn)El in (0,l) converging to zero. A judicious choice of the sequence {xn)zl may make the task less tedious. See Exercise 10 for an application of this idea. The urge to reconsider the preceding remarks and to generalize the result is irresistible. The conclusion stems from the fact that if is Cauchy, then C f ( ~ ~ ) ) z ~ is Cauchy. The proof of the following theorem is immediate.

{~~)z~

2.2 THEOREM Let f :D -+ R, and suppose xo is an accumulation point of D. If for each sequence { x , ) ~converging ~ to with xn E D\{q) for each n, the sequence (f(xn))zl is Cauchy, then f has a limit at xo.

2.2 Limits of Functions and Sequences 71

Let US now sift through our newly gained knowledge of sequences and discover theorems concerning limits of functions. Many of these theorems can be proved by exploiting Theorem 2.1, but in some cases we will give a direct proof in order to introduce the reader to the act of finding 6's. 2.3 THEOREM Let f : D + R, with xo an accumulation point of D. Iff has a limit at xo, then there is a neighborhood Q of xo and a real number M such that for allx E Q n D , If(x)l 5 M. v

Proof Let E = 1, and let L be the limit off at xo. Then there is 6 > 0 such that forO< lx-xol < b a n d x ~ D, we have if(x)-LI < E = 1. Ifxo ~ D , d e f i n e

otherwise, define

and let Q = (xo - 6, xo + 6). In either case, if x E Q n D, then 1f(x)l 5 M.

In less formal language, iff has a limit at xo, then f is bounded near xo. In particular, this gives another verification that f(x) = does not have a limit at zero, since f is unbounded in every neighborhood of zero. Example 2.6 For each real number x, let [x] denote the largest integer that is less than or equal to x. For example, [n] = 3, = 1, and [2] = 2. For n 5 x < n + 1 with n an integer, [x] = n, so the graph of the function looks like a set of stairs with a "jump" of one unit at each integer. (See Figure 2.5 for a sketch of this function.) If we have been successful in developing the geometric concept of the limit of a function, it should be apparent that f(x) = [x] has a limit at xo iff xo is not an integer. Suppose xo is not an integer, and let 6 be the distance from xo to the nearest integer. Now 6 > 0, and if 0 < (x - xol < 6, then [x] = [xo]; hence, 1f(x) - [xo]l = 0 < E for all E > 0. Thus, f has a limit at xo if xo is not an integer. If xo is an integer, consider the sequence {xo + (-l)":):,. Now, for n odd, n > 1, xo + (-1)": < a,and thus

W

and for n even, xo + (-1)"

[$I

> xo, and thus

Thus, Cf(xo+ (- l)":))~,does not converge, and so f does not have a limit at xo if xo is an integer.


Figure 2.5

:,

:.

Consider the examples f (x) = [x], g(x) = sin and h(x) = Each failed to have a limit at zero for basically different reasons. The function f ( x ) = [x] "jumped" at zero, taking the value -1 to the left of zero and the value 0 to the right of zero; g oscillated too badly at zero to have a limit there; h was unbounded near zero and so could not have a limit there. In a sense, these three examples describe the sort of behavior one may suspect of a function that does not have a limit at a point. We shall comment more on this in Section 2.4.

Following the pattern set in Chapter 1, we shall now determine the relationships between limits of functions and the algebraic operations. I f f : D -,R and g : D -,R, define f + g : D + R by ( f + g)(x) = f ( x ) + g(x) for all x E D. In a similar fashion, define fg : D -, R by (fg)(x) = f(x)g(x). If g(x) 0 for x E D, we may define & g D - - + R ~ ~

+

for all x E D. Now read Theorems 1.8, 1.9, and 1.11, and attempt to formulate corresponding theorems for limits of functions. In each case, the proof can be obtained by use of Theorem 2.1 and the appropriate result from Chapter 1.

2.3 Algebra of Limits 73

2.4 THEOREM Supposef , g :D -+R with xo an accumulation point of D, and fuaher suppose that f and g have limits at xo. Then:

lim (f

x-xo

+ g)(x) = x-xo lim f (x) + lim g(x). x--'xo

ii. fg has a limit at xo and

iii. If g(x)

# 0 for all x E D and limx,

g(x) f 0,then

has a limit at xo and

lim f (x)

x+xo x-xo

x-xo

The proofs of (i) and (iii) will be accomplished by use of sequences, whereas the proof of (ii) will be direct.

Proof (i) Let {xn)gl be any sequence of points in D converging to xo with xn f a for all n. It suffices to show that {(f + g)(xn))gl converges to lim f (x) + lirn g(x).

x+xo

x-xo

By assumption, f and g have limits at xo; hence, Cf(xn))gl converges to limx-+xo f (x) and {g(xn))zl converges to lim,, g(x). By Theorem 1.8,

converges to limx, f (x) + lim,,, g(x). This concludes the proof of (i). (ii) Given e > 0, we must find an appropriate 6 > 0. (Before beginning the proof, see the discussion preceding Theorem 1.9.) As the plan for finding the 6 unfolds, observe the parallel between this search and that for N in Theorem 1.9. Choose c > 0. Let

A = lim f (x) x-xo

We wish to find S

. and

B = lim g(x). x-xo

> 0 such that if 0 < Ix-xol < 6 andx E D, then 6 . By Theorem 2.3, there is 61 > 0 and a real number M > 0

such that for 0 < x - a ( < & , X E D ,we have lf(x)l S M . Let

There is 62 > 0 such that if 0

< lx - xol < &, x E D, then 1f(x) - A1 < e'; and


there is 63 > 0 such that for 0 < Ix - xol < 63, x E D, then lg(x) - BI

Now, if 0

< r'. Let

< Ix - xol < 6 , x E D, then

g ~ of members of D, distinct f'rom xo, con(iii) Suppose { ~ ~is )a sequence verging to xo. Now (f(xn))El converges to lirn,,, f (x) and {g(xn))Zl converges to limx, g(x). Since g(x) # 0 for all x E D, g(xn) # 0 for all n. By assumption, limx, g(x) # 0. Hence,

converges to lim f (x)

x+xo

lim g(x)

'

X+xo

By Theorem 2.1,

$ has the indicated limit at xo.

The adventuresome reader may wish to attempt a direct proof of (iii). A lemma similar to Lemma 1.10 is in order, and then the proof may be patterned after that of Theorem 1.11. See Exercise 2 1. Good luck! 2.5 THEOREM Suppose f : D -. R and g : D -r R, xo is an accumulation point of D, and f and g have limits at xo. If f(x) 5 g(x) for all x E D, then

lim f(x) 5 lim g(x).

x--+xo

Proof

x+xo

The proof is left as Exercise 20.

{i

00

Let us now look at an example suggested by the sequence sin Consider the function f : (0,l) -r R, defined by f(x) = xsin $. It has been shown that sin $ fails to have a limit at zero, so we may not use Theorem 2.4 (ii). However, sin $ is bounded above by 1 and below by - 1. Now it is clear that

I

2.3 Algebra of Limits 75

I

hence, f has a limit at zero; in fact, lim,, f (x) = 0. (We merely choose 6 = c.) A theorem reminiscent of Theorem 1.13 is in order.
0. There is a 61 > 0 and M > 0 such that if x E D and Ix-xol < 61, then lf(x)l 5 M. Let ct=c/M. There is > 0 such that i f x E D a n d o < Ix-xol 0 , there is a neighborhood Q of a such that, for any x, y E Q n D, x # a , y #a,we have lf(x) -f6N
0 such that D n [xo - E,xo + E] = E n [a- E, + E]. If f ( x ) = g(x) for all x E D n E n [xo - E,a +E],prove that f has a limit at xo iff g has a limit at m.

PROJECT 2.1 The purpose of this project is to ascertain under what conditions an additive function has a limit at each point in R. We say that f : R + R is additive if for all x, y E R, f ( x + y) =f (x) +f (y). In what follows, f is an additive function. 1. Show that for each positive integer n and each real number x, f (nr) = nf (x). 2. Suppose f is such that there are M > 0 and a > 0 such that if x E [-a,a], then If (x)l 5 M . Choose E > 0. There is a positive integer N such that M / N < E. Show that if lx - yl < a/N, then 1 f (x) -f 0 1 < c. 3. Prove that if there are M > 0 and a > 0 such that if x E [-a, a], then 1 f(x)l 5 M, then f has a limit at each x in R and lim,, f ( t ) =f (x). 4. Prove that iff has a limit at each x E R, then there are M > 0 and a > 0 such that if x E [-a,a], then 1 f(x)l 5 M . You have now proven that iff : R + R is additive, then f has a limit at each point in R iff there are M > 0 and a > 0 such that if x E [-a,a], then 1 f(x)( 5 M. In addition, if the condition is satisfied, lirn,, f ( t ) =f (x).

PROJ E T 2.2 In Exercise 39 of Chapter 1, you proved what may be called the "shuffle" theorem. In that exercise, you formed a new sequence by putting two sequences together in a special way and then examined the new sequence for convergence. The theorem you will be asked to prove in this project will be similar except that you will deal with functions and limits of functions rather than sequences. You might want to look at Exercise 14 in this chapter before reading further. 1. Prove the following theorem.

THEOREM Suppose f : A + R and g : B -+ R are such that A f l B = 0,xo is an accumulation point of A and also an accumulation point of B. Define h : A U B + R by h(x) = f ( x ) if x E A and h(x) = g(x) if x E B. Rove that h has a limit at xo if and only iff and g each have a limit at xo and limx-f (x) = lim,, g(x).


You probably recall reading about one-sided limits in your calculus book. This theorem allows us to examine the relationship between one-sided limits and twosided limits. We need a definition to set the stage.

DEFINITION Let f : D -+ R, and suppose xo is an accumulation point of {x : x E D, x > xo). Then f has a right-hand limit L at iff for each c > 0, thereis6>Osuchthatif~0< x < x o + 6 , x € D , t h e n I f @ ) - L ( < c 2. State carefully the definition of lef-hand limit for a function. 3. Give an example of a function that has a left-hand limit and a right-hand limit at a point but does not have a limit at that point. 4. Give an example of a function that has a left-hand limit and a right-hand limit at a point and also has a limit at that point. 5. State and prove a theorem that relates the left-hand limit and right-hand limit of a function and the limit of a function.

PROJECT 2.3 When examining a function to determine if it has a limit, it is sometimes convenient to make a change of variable. The theorem you will be asked to prove is designed to assist in using the change of variable in limits of functions. 1. Prove the following theorem:

THEOREM Supposef :A + R, g : B + A. Suppose that a is an accumulation point of A, b is an accumulation point of B, i. limt+b g(t) = a, ii. there is a neighborhood Q of b such that for t E Q nB, g(t) # a, and iii. f has a limit at a. Then f 0 g has a limit at b and lim,, f (x) = f (g(t)). 2. Consider f : R\{O) + R by f (x) = %. A standard argument in calculus is to show that f has a limit 1 at zero. You may use this limit. Use the theorem you just proved to show that the function h :~ \ { a / 2 ) + R defined by h(x) = has 2 a limit at :, and find that limit.

PROJECT 2.4 Let f be a real-valued function defined on {x : x > a) for some real number a. In this project, you will examine the notion of having a limit "at infinity."

Pmiects 83

DEFINITION Iff is defined on { x : x > a) for some real number a, then f has a limit L at infinity if for each c > 0,there is M such that for x > M and x > a, I f @ ) - L I < c.

> 0. Rove that f has a limit at infinity, and find it. Definef ( x ) = for x > 1. Prove that f has a limit at infinity, and find it. Iff is defined on { x :x > a), define g(x) =f (i) for 0 < x < i. Rove that f has

1. Definef (x) =

2. 3.

5,

for x

a limit at infinity if and only if g has a limit at zero. 4. Define what it means for a function to have a limit at minus infinity. Illustrate with an example. 5. State and prove a theorem similar to that in part 3 for a function with a limit at minus infinity.

Chapter

3

Continuity

n the discussion of the limit of a function it was emphasized that the value off at xo has no bearing on the question of the existence of the limit of a function f at point xo. The concept of continuity of a function at a point brings the value off at xo back into the picture. In very imprecise language, a function f will be continuous at a point xo iff (x) lies close to f (xo) whenever x is sufficiently close to xo. Let us examine this notion in a familiar setting. Consider the function f(x) = fi for x 2 0. Because f(2) is an irrational number, it cannot be expressed by any of the conventional means as a decimal. Nevertheless, for purposes of computation, we can attempt to obtain rational numbers fairly close to d. For example, (1.4)~= 1.96, (1.41)~= 1.9881, (1.414)~= 1.999396, and so on. In essence, we are observing that f(1.96) = 1.4, f(l.9881) = 1.41, and f(l.999396) = 1.414, and we assume that 1.414 is close to f i because 1.414 is the value off at a point close to 2. Thus, the assumption is that f(x) = f i is continuous at 2, a fact to be proven later. As with all new ideas, a precise definition is needed.

I

3.1

CONTINUITY OF A FUNCTION AT A POINT

DEFINITION Suppose E c R and f : E -+ R. If xo E E, then f is continuous at xo iff for each r >: 0, there is a 6 > 0 such that if

then

Iff is continuous at x for every x E E, then we say f is continuous. 85

86 Chapter 3 Continuity

x0- 6

xo

xo+ 6

Figure 3.1

Compare this definition with the definition in the previous chapter concerning the limit of a function at a point xo. First, for continuity at xo, the number a must belong to E, but it need not be an accumulation point of E. Indeed, iff :E --,R with xo E E and xo not an accumulation point of E, then there is 6 > 0 such that if lx - xol < 6 and x E E, then x = a;hence for every E > 0. In other words, if xo is not an accumulation point of E and xo E E, then f is continuous at xo by default. Thus, the only interesting case is when a is an accumulation point of E. In this case, the definition requires that f has a limit at xo and that that limit be f (xo). This fact is a part of the next theorem, but before stating and proving that theorem, compare Figure 3.1 with Figure 2.1 of Chapter 2. Thus, f is continuous at xo if and only if for each E > 0 there is a 6 > 0 such that the graph off for xo - 6 < x < xo + 6, x E E, lies in the strip {(x,y) :f (xo)- E < y < f (xo)+ €1. 3.1 THEOREM Let f :E + R with xo E E and Then (i) through (iii) are equivalent: i. f is continuous at Q. ii. f has a limit at a and lim,,

f (x) =f (no).

an accumulation point of E.

3.1 Continuity of a Function at a Point

87

iii. For every sequence {xn)Z1converging to xo with xn E E for each n, {f &))El converges to f (xo). iC

,

This theorem is the first of its type-that is, one that asserts that two or more statements are equivalent. Our method of proof will be to show that (iii) implies (ii), then show that (ii) implies (i), and finally show that (i) implies (iii). This completes the cycle showing that (i), (ii), and (iii) are equivalent-that is, that each implies and is implied by the other. Also, if xo is not an accumulation point of E with xo E E, then (i) is always true, as is (iii). Exercise 4 asks you to prove that (iii) is true in this case.

Proof Assume (iii) holds. In particular, if {xn)El converges to xo with xn # xo and Xn E E for all n, then {f (xn))zl converges to f (xo). Hence, by Theorem 2.1, f has a limit at xo and lim,, f (x) =f (xo). Thus, (iii) implies (ii). Assume (ii) holds, and choose E > 0. Since (ii) holds, there is 6 > 0 such that if 0 < Ix-xol < 6 andx E E, then

The only fact needed to fulfill the definition of continuity at xo is to remove the condition 0 < Ix - xo1. However, when lx - xol = 0, then x = xo; hence

Thus, f is continuous at xo; in other words, (ii) implies (i). Suppose now that (i) holds and that {xn)g1 is a sequence of points in E that converges to XO. Choose E > 0. There is 6 > 0 such that for lx - xol < 6 and x E E, If (x) - f (xo)1 < E. Since {xn)Z1 converges to xo, there is N such that for n 1 N, Ixn - xol < 6. Thus, for n 2 N, I f(xn) - f(xo)l < E. This shows that Cf(xn))z1converges to f (xo). (Does this paragraph look familiar? It should. Compare it with the proof of Theorem 2.1.) It is worthwhile to reflect on this theorem and consider some of its less obvious uses. First, it can be used to show that a given function f is not continuous at a point xo by exhibiting a sequence {xn)zl converging to xo where the sequence { f (xn))zl does not converge at f(m). From a slightly different point of view, this theorem can be used to show that a certain sequence is convergent. For example, assume that the function f (x) = e' is continuous on R. Thus, in particular, if the sequence {xn)El converges to xo, then the continuity of f(x) = e' and Theorem 3.1 guarantee that {P):, converges to @. We shall call attention to this aspect of Theorem 3.1 later. We pause momentarily to reconsider some of the results presented in Chapter 2. Recall the function f :R + R defined by


for x # 1 and f(1) = 6. It has been shown that f has a limit at 1 and lirn,,l f(x) = 2. Unfortunately, lim,,l f (x) = 2 # 6 = f(1); hence, f is not continuous at 1. In this case, f fails to be continuous at 1 because f(1) does not happen to be the limit off at 1. Now consider the function g(x) = sin f for 0 < x < 1, and g(0) = 38. We have seen that g fails to have a limit at zero, so g is not continuous at zero, regardless of what we try for g(0). The function h(x) = f for 0 < x < 1 behaves similarly in the sense that h doesn't have a limit at zero, so it is futile to attempt to define h(0) such that h will be continuous at zero. Recall the function f : [O, 11 + R defined such that f (x) = 0 if x is irrational and f (P/q) = l/q if p and q are positive integers that are relatively prime. It was shown in Chapter 2 that f has a limit at each point of [0, 11 and that its limit is zero at each point. Thus, f is continuous at x E [O,1] iff f(x) = 0 (that is, iff x is irrational).

The next theorem is a natural follow-up to Theorem 2.4. 3.2 THEOREM Suppose f : D Then

+R

and g : D -,R are continuous at xo E D.

i. f + g is continuous at xo. ii. fg is continuous at XO. iii. If g(q) # 0, f /g is continuous at xo. [A word of advice is necessary concerning (iii). The function f/g is defined only where g(x) # 0; hence, when speaking of the function f/g, we assume that the domain consists of those points where g(x) # 0. Lemma 3.3 will shed some light on this situation.]

Proof We shall use a different method of proof for each part of this theorem in order to point out the use of facts previously presented on sequences and limits and also to show a direct proof. Suppose f and g are continuous at xo E D. Let {xn) be any sequence of points in D that converges to xo. Then, by Theorem 3.1, {f(xn)) converges to f (xo) and {g(xn))zl converges to g(xo); hence

{f(xn) + g(xn))Zl converges to f (no)+ g(q). Thus,

converges to f(xo) + g(xo) = (f + g)(-), and hence, by Theorem 3.1, f continuous at XO. This concludes the proof of (i).

+ g is

3.2 Algebra of Continuous Functions 89

Supposef and g are continuous at a E D. If xo is not an accumulation point of D, then fg is continuous at a by our observations following the definition of continuity. Suppose now that xo is an accumulation point of D. By Theorem 3.1, f and g have limits at and

By Theorem 2.4, fg has a limit at xo and

Thus, by Theorem 3.1, fg is continuous at XO. We have purposely elected to give a direct proof of (iii) because it involves some important facts not yet brought to light. We hope the reader has attempted a direct proof of Theorem 2.4(iii) and discovered the need for a lemma similar to the following. The proof of Theorem 3.2(iii) follows this lemma.

+

3.3 LEMMA Let g : D + R be continuous at xo E D with g ( ~ ) 0. Then there are 6 > 0 and a > 0 such that if lx - -1 < 6 and x E D, then lg(x)l 2 a. (Note the similarity between this lemma and Lemma 1.10,the one needed for the theorem concerning the convergence of the quotient of two convergent sequences.)

Proof Choose

There is 6 > 0 such that for lx - xol < 6, x E D, we have

Thus,

Proof of 3.2 continued. Suppose g is continuous at xo and g(xo) # 0. By Lemma 3.3, there are 6 > 0 and a > 0 such that for ix - xol < 6 and x E D, we have ig(x)( 2 a. Choose e > 0. Let


There is 6' > 0 such that for lx - xol < 6' and x E D,

Let

Now 6" > 0 and for lx - -1

< 6" with x E D, we have

5

Thus, is continuous at xo. By part (ii) of this theorem, is then continuous at xo. From our remarks in Chapter 2 and Theorem 3.1, it is clear that every polynomial is continuous. Furthermore, if p and q are polynomials, then is continuous at any point x where q(x) # 0. The case where q(x) = 0 is handled in detail in Exercise 12 at the end of this chapter.

5

Example 3.1 The function f (x) = sinx is undoubtedly familiar to the reader. We observe that, for 0 5 x 5 ./2, 0 5 sinx 5 x. From this and the fact that sin(-x) = - sinx, it is clear that f (x) = sinx has a limit at zero, namely, limx+osinx = 0. From this, we can deduce that the sine function is continuous at every point. Now for all x and y,

W

sinx-siny=2cos

(. ') sin

(X+)

.

Let -Q be any real number and E > 0. Let 6 = min

I sinxo - sin y

= 12cos

Then for lxo - y1 < 6,

(Xy) (xy) sin

Thus, sinx is continu6us at a.For all real x, cos x = sin (q - x ) . In order to verify the continuity of cosx at %, let us proceed directly. Choose c > 0 and let 6 = min{c, T ) . If lx - xol < 6, then

hence,

3.2 Algebra of Continuous Functions 91

Thus, cosx is continuous at q. Now it is clear that the functions tan x, set x, csc x, and ctn x are continuous where defined, since they may be defined by appropriate products and quotients of the functions sinx and cosx. N o points that came to light in the preceding paragraph deserve some special attention. First, a function f : D -, R is continuous at a point xo E D iff for each e > 0, there is 6 > 0 such that for lx -xol < 6 and x E D, we have 1f(x) -f(q)l < e. Note that the choice of 6 > 0 is influenced by two things: the choice of E > 0 and the point in question, q. In the process of investigating the continuity of f(x) = sinx, it was discovered that if c > 0 is given, 6 > 0 could be found that would be suitable for any q. This is not the usual state of affairs and will be considered in some detail very soon. Second, the continuity of the cosine function was deduced from the facts that the sine function is continuous and that cosx = sin (q - x). If we let g(x) = cosx, f (x) = sinx, and h(x) = q - x, then this identity may be written g =f o h. Of course, since h is a polynomial function, it is continuous; indeed, that fact was used to observe that if lx - xol < 6, then

We seek now to generalize this idea and shall attempt to prove that the composition of continuous functions is continuous. 3.4 THEOREM Iff : D -, R and g : Dl + R with im f c D', where f is continuous at xo E D and g is continuous at f (a), then g of is continuous at a.

Proof then

Choose e

There is

> 0. There is b1 > 0 such that if ly -f(xo)l < b1 and y E Dl,

> 0 such that if lx - xol < b2 and x E D, then

Now, since im f C Dl, if lx - xol < b2 and x E D, then f(x) E D', f ( q ) E D', and If(x) -f(xo)l < 61; hence,

Therefore, g 0 f is continuous at xo. This theorem, coupled with previous results, allows us to conclude immediately that functions such as f (x) = c o d and g(x) = sin(cosx) are continuous at q for all real numbers w. Let US now take up the remaining topics that we want to pursue further. We shall give the appropriate definition and then consider some examples.


3.3

UNIFORM CONTINUITY: OPEN, CLOSED, AND COMPACT SETS e

DEFINITION A function f : D + R is uniformly continuous on E C D iff for every c > 0, there is 6 > 0 such that if x,y E E with (x - yl < 6, then [f(x) - f (y)1 < E . If f is uniformly continuous on D, we say f is uniformly continuous. As promised, some examples and nonexamples are forthcoming. Example 3 2 Consider the function f : [2.5,3] -* R defined by f (x) = a.we will show that f is uniformly continuous on [2.5,3]. Choose E > 0. Now

We see no difficulty making the numerator of this last fraction small, but, as we have seen before, the problem of bounding the denominator away from zero must be addressed. However, since x and y must belong to [2.5,3], the smallest nonnegative value for the denominator is 0.25, occurring when x = y = 2.5. With this in mind, we 1 choose 6 = 2 ~Thus, . if lx - yl < 6 and x,y E [2.5,3], then

Example 3.3 Consider the functionf : (0,6) -* R defined by f (x) = 2+2w - 5. We will show that f is uniformly continuous on (0,6). Choose c > 0. Now

To make this expression small, we need to make lx - yl small and bound lx + y + 21. Although 6 is not in the domain off, we can use that number to bound lx + y + 21; that is, lx + y + 21 < 6 + 6 + 2 = 14. With this in mind, we choose 6 = Thus, if IX - yI < 6 and x,y E (0,6), then

he.

The sine function has been shown to be uniformly continuous on the set of all real numbers. Therefore, it is a uniformly continuous function. Similarly, the cosine function is uniformly continuous. This is not surprising in light of Exercise 23.

r

I

Example 3.4 Consider the function f : (0,l) -* R defined by f(x) = l/x. To show that this function is not uniformly continuous, we will choose c = 1 and show

3.3

Uniform Continuity: Open, Closed, and Compact Sets 93

thatforany6 > O,therearexandyin(O,l)suchthat ix-yl < band If@)-f@)l 2 1. Suppose 6 > 0. Let a = min{2,6), x = a/3, and y = a/6. Then lx - yl = 4 5 < 6, 3 and if(x)-f@)l= 2 1.

;1

$1' lo,l

Now consider the function f : (0, a/2) + R defined by f (x) = tanx. Since tanx = sinx/ cos x, for 0 < x < 7r/2, f is a quotient of uniformly cbntinuous functions. [Since sinx and cosx are uniformly continuous on R, they are certainly uniformly continuous on (0, a/2).] Exercises 19 and 20 show under what conditions the sum, product, or composition of uniformly continuous functions yields a uniformly continuous function. We know that the tangent function is unbounded near a/2, so id cannot have a limit there. Let us show that this guarantees that tanx is not uniformly continuous on (0, ~ / 2 ) .

3.5 THEOREM Let f : D --+ R be uniformly continuous. Then, if xo is an accumulation point of D, f has a limit at xo. Proof Let xo be an accumulation point of D, and let {x),:, be any sequence of members of D \ {xo) converging to xo. Recall that it is sufficient to prove that {f (xn))gl is a Cauchy sequence, since every Cauchy sequence of real numbers is convergent. Choose E > 0. Since f is uniformly continuous on D, there is 6 > 0 such that for all x,y E D, lx - yl < 6 implies 1f(x) - f@)l < r. Since {xn)gl converges to xo, it is a Cauchy sequence, and there is N such that for m, n 2 N, Ix, - xml < 6. NOWxn E D for each n; hence, for m, n 2 N, Ix, - xml < 6 and xm,X, E D;hence

Thus, {f (xn))g, is Cauchy. By Theorem 2.1, f has a limit at xo.

Example 3.5 Observe that Theorem 3.5 gives a necessary condition for uniform continuity, but not one that is sufficient. For example, consider g :R + R where g(x) = 2 for all real numbers x. Since g is continuous and dom g = R, the function g has a limit at every accumulation point of R because R contains all its accumulation points. However,

W

Choose 6 > 0 and consider any 6 > 0. Now choose x and y such that lx-yl = 612 < 6 and lx +yl = 3~16.Then

l

I

.

I

In effect, given c > 0, it is impossible to find 6 > 0 such that for (x - y( < 6, we always have lg(x) - g o 1 < c. Thus, g is not uniformly continuous. -

"--h--

-

94 Chapter 3

Continuity

We have seen two examples of continuous functions that were not uniformly continuous; in the first case, the domain was (0,a-); set with accumulation points that did not belong to the set-and the second function had an unbounded domain. The difficulty in the first example comts about near the accumulation points and perhaps may be overcome by considering functions whose domains contain all their accumulation points. Such'sets are of sufficient importance to deserve a name.

DEFINITZON A set E C R is closed iff every accumulation point of E belongs to E. As has been observed previously, a finite set has no accumulation points and, by definition, is a closed set. Also, R, the set of all real numbers, is a closed set, since an accumulation point of R is necessarily a real number, hence a member of R. Suppose E is a closed set and xo E R \ E. Then, by the definition of a closed set, is not an accumulation point of E; hence, there is a neighborhood Q of xo that contains no points of E (since xo E), so Q c R \ E. Thus, if E is closed, there is, for each xo E R \ E, a neighborhood Q of xo such that Q c R \ E. This special type of set comes about in a natural way by considering the complement of a closed set.

DEFINITZON A set A Q of x such that Q c A .

c R is open iff for each x E A there is a neighborhood

The remarks leading up to this definition yield the proof of half of the following theorem. 3.6 THEOREM A set E

c R is closed iff R \ E is open.

Proof Suppose E is closed and xo E R \ E. Then, since E contains all its accumulation points, xo is not an accumulation point of E. Hence, there is a neighborhood Q of xo that contains no points of E. (Recall that xo $! E.) But now Q C R \ E; hence (by definition), R \ E is open. Suppose R \ E is open. To show E is closed, it suffices to show that if xo is an accumulation point of E, then xo E E. Let xo be an accumulation point of E. Then xo E E or xo E R \ E. However, R \ E is open; so if xo E R \ E, there is a neighborhood Q of xo such that Q c R \ E or Q n E is empty, contrary to xo being an accumulation point of E. Thus, xo E E and E is closed.

i

Now let us return to the problem that precipitated this discussion. Consider f : E -,R with f continuous on E. Choose c > 0. Now for each x E E, there is 6, > 0 such that if y E E and lx - yl < 6,, then 1 f ( x ) -fO)< e. In the search for 6 > 0 to guarantee that lx - yl < 6 and x,y E E imply that lf(x) - f(y)1 < e, one might consider 6=inf(6, : x E E ) .

3.3 Uniform Continuity: Open, Closed, and Compact Sets 95

If E is finite, 6 is positive; but if E is infinite, 6 might be zero. Indeed, iff is not uniformly continuous, it may be zero. Consider now the set (x - 6,, x + 6,)n E. If

If one could find xl, . . .,xn such that

there might be some hope of showing that f is uniformly continuous. Note that for each x, (x - 6,, x + 6,) is an open set (see Exercise 27), and E C UXE& - 6,, x + 6,). Here we have a family of open sets whose union contains E, and we wish to choose a finite subfamily with the same property. A new definition is now in order.

DEFINITION A set E is compact iff, for every family {G,),ur of open sets such that E c UaEAGarthere is a finite set {al,.. ., a n ) c A such that E C Ug,G,,. We shall illustrate this concept by giving some examples of sets that are not compact.

Example 3.6 Let E = (0,1] and for each positive integer n, let Gn = ($2). If 0 < x 5 1, there is a positive integer n such that < x; hence, x E Gn, and thus

5

If we choose a finite set nl, . . . ,nr of positive integers, then

where no = max{nl ,. . . ,n,) and

Thus, we have a family of open sets {Gn)nEj such that E C UnEjGn,but no finite subfamily has this property. From the definition, it is clear that E is not compact.


Example 3.7 As a second example, consider the set J of all positive integers. For each n E J, define

Now J c UzlGn, but each Gn contains exactly one member of J , so it is impossible to choose nl ,. . . ,n, such that J C Ur,lGni. Thus, J is not compact. Let US introduce some new terminology to help simplify our language. If E is a set, and {Gcr)aEAis a collection of sets such that E c UaEAGa9then the collection is called a cover of E. If each G, is an open set, then the collection {Ga)aEA is called an open cover of E. If {Go)aEAis a cover of E and B C A such that E C UoEBGa,then the ~01le~ti0n {Ga)aEB is called a subcover of E; and if B is finite, then {Ga)aEB is called a finite subcover of E. In a sense, we are abusing the language, since every subcover of E is also a cover of E; but in light of the notion of compactness, this is a reasonable abuse. With these additions to our vocabulary, we can restate our definition of compactness; A set E is compact iff every open cover of E has a finite subcover. The reader should keep the preceding two examples clearly in mind. The first example, (0,1], is a bounded set that is not closed, since 0 is an accumulation point of (0, 11 that does not belong to (0, 11. The second example is a closed set but is unbounded. As the reader might suspect from our choice of examples, the compact sets on the line are precisely those that are both closed and bounded. This is the content of the next theorem.

3.7 HEZNE-BOREL THEOREM A set E bounded.

cR

is compact iff E is closed and

Proof Suppose that E is closed and bounded and that {Gx)xEAis an open cover of E. Since E is bounded, there is a closed interval [a, P] such that E c [a, PI. Let us suppose that {GX)xEAhas no finite subcover of E (E is not compact). We shall show that this assumption is self-contradictory. Let 'yo be the midpoint of [a, PI. At least one of the two sets [a, yo] n E and ['yo,,8] n E cannot be covered by a finite subfamily of {GX)XEA. Choose one and call it [ a l ,,811; let 71 be the midpoint of [ a l ,PI]. Again, at least one of the two sets [a1,'yl] n E and [yl, Pll n E cannot be covered by a finite subfamily of {GX)XEA; choose one such and call it [ a Z , h ] . (Does this game sound familiar? See the proof of Theorem 1.6, the Bolzan+Weierstrass Theorem.) Continuing in this fashion, we obtain a sequence of closed intervals [an,P,] with the following properties: 1. P n - an= $(,8 - a). 2. [an+l,Pn+llc [an,Pnl for all 3. For each n, the set [an,pn] n E cannot be covered by a finite subfamily of

{GX)XEA-


By (3), [a,, pn],lnE is nonempty for each n; hence, we may choose xn E [aR,P,]nE for n = 1,2, . . .. Consider the set

If P is finite then by (2), there is no such that

for all n. Since {Gx)xEAis an open cover of E,

for some Xo E A; and since Gxo is open, there is c

> 0 such that

Choose n such that

Then, since X, E [an,Pn]f I E,

implies that

IX - x, 1 5 on- an< c.

Hence,

This means that [a,, Pn]n E may be covered by a finite subfamily of {GX)XEA, which is contrary to (3). Suppose P is infinite. Then P is an infinite bounded set and has an accumulation point xo; and since P c E, clearly xo is an accumulation point of E and xo E E since E is assumed closed. There is XI E A such that xo E Gx,; and, since GAlis open, there is E > Osuch that

Choose n such that

Now xo is an accumulation point of P and (xo - 5 , xo + 5 ) is a neighborhood


of Q. Hence, there are infinitely many members of P that belong to (xo - 5, a + f ) . In particular, there is m > n such that

NOWif z E [am,Dm]t7 E, then

Thus, [a,,&] nE c (XO - a xo + E) c GA,,contrary to (3). We have now anived at a contradiction to our original assumption- namely, that E was closed and bounded but not compact. Thus, if E is closed and bounded, E is compact. It now remains to show that every compact set is closed and bounded. We shall accomplish this by showing that if a set is unbounded, it cannot be compact, and if a set is not closed, it cannot be compact. Assume E is not bounded. For each positive integer n, let Gn = (-n,n). Now Gn is open for each n and E c UglGn, SO {Gn)gl is an open cover of E. If nl, . . .,nr is any finite set of positive integers, then

where n.0 = max{nl, .. .,n,), and since E is unbounded,

is not a cover of E. Thus, { G , ) ~ , is an open cover of E with no finite subcover; hence E is not compact. Assume E is not closed. Then there is an accumulation point of E (call it xo) such that xo $! E. In much the same way we did for the set (0,1], we shall construct an open cover of E with no finite subcover. For each positive integer n, define

A,

i]

Now [a- Q + is a closed set for each n; hence, Gn is open. Observe that UglGn = R \ {q),so {Gn)gl is an open cover of E. Let nl, . . ,nr be any finite

.


set of positive integers and no = max{nl, . . .,nr). Then

In order that {Gni)i=l,..., be a cover of E, we must have

hence,

is empty, contrary to xo being an accumulation point of E. Thus, {G.)=, has no finite subcover, and E is not compact. Having read the proof of this theorem, you should go back and see what makes it work and where the hypotheses are used. This time, don't worry about the details. Just seek out the rough idea of the proof. In passing, we give some examples of compact sets. Every finite set is compact because it is necessarily closed and bounded. If a < b, then [a, b] is closed and bounded and thus is compact. It is easy to show that the union of finitely many compact sets is compact, so any union of a finite number of closed, bounded intervals is compact. See Exercise 36. With this knowledge of compact sets in mind, we shall exploit this concept with respect to continuity. The remarks preceding the definition of compactness should make the next theorem no surprise.

3.8 THEOREM Let f : D -r R be continuous with D compact---that is, closed and bounded. Then f is uniformly continuous.

Proof Choose E > 0. Since f is continuous on D, f is continuous at x for each x E D. Thus, for each x E D, there is 6, > 0 such that if lx - yl < 6, and y E D, then 1f (x) -f (y)1 < 5. Consider this family

This is an open cover of D and D is compact, so there is a finite subcover of D. In other words, there are XI,.. .,xn E D such that


Let

Now suppose x, y E D and lx - yl that

< 6. There is an integer i among 1, . . .,n such

Hence,

Therefore, f is uniformly continuous.

In a broad sense, this' theorem has a converse. Suppose f : D + R is uniformly continuous. As has been pointed out,f has a limit at each accumulation point of D. As in Exercise 28, let D' be the set of accumulation points of D and = D u D'. Define g :b -,R by g(x) =f (x) for all x E D; and if xo E b \ D, define g(xo) = lim,, f (x). The function g : B -, R is called an extension off : D + R and indeed turns out to be a continuous function. (The proof of this will be a challenging exercise for the reader.) Thus, iff : D -,R is uniformly continuous with D bounded, then f has a continuous extension g : -,R where is closed and bounded and, hence, compact.

3.4 PROPERTIES

OF CONTINUOUS FUNCTIONS

Some of the reasons for considering continuous functions are the special properties they possess. In the next few pages we consider some of these properties and their applications.

3.9 THEOREM Iff : E -,R is uniformly continuous and E is a bounded set, then f (E) is a bounded set. Proof C h w s e ~= 1. Thereis6 > Osuchthat,if (x-y( < 6,then(f(x) -f(y)l < 1. Since E is a bounded set, there are xl,xz,. . .,x, in E such that

3.4 Properties of Continuous Functions 101

but then

since lx - xi1 < 6, x E E, implies that (f(x) -f (xi)l < 1. Thus f ( E ) is bounded. Supposef : E + R is continuous with E compact. Then f is uniformly continuous and E is bounded, so f(E) is bounded. If we could prove that f(E) was closed, then f(E) would be compact, and we could say that the continuous image of a compact set is compact. Before proceeding, we need the following lemma.

LEMMA Let E be a compact and {xn)g1 be a sequence of points in E. Then there is xo in E and a subsequence {xnk)E1that converges to xo.

Proof Since E is compact, it is both closed and bounded. Thus {xn)E1,being bounded, has a convergent subsequence {xnk)z1.The limit of this subsequence xo belongs to E or is an accumulation point of E. (See Theorem 1.17.) But E is closed, hence xo E E. 3.10 THEOREM Let f : E -+ R be continuous with E compact. Then f(E) is compact.

Proof As pointed out in the preceding paragraph, we need only show that f(E) is closed, since we already know f(E) is bounded. Let yo be an accumulation point off (E). Then there is a sequence {yn)gl of points in f (E) such that y, # yo for all n and {yn)gl converges to yo (see Theorem 1.17.) Since yn E f (E) for each n, there is a sequence {xn)z1 of points in E such that f(xn) = yn for each n. There is a subsequence {xn )El that converges--call the limit xo-and xo E E by the lemma. The sequence ZfCxnk))& converges to f(xo), since f is continuous at xo. But

is a subsequence of {yn)zl and converges to yo, so yo =f (xo) E f (E). Therefore, f (E) is closed. The following important corollary follows from Theorem 3.10. 3.11 COROLLARY Iff : E + R is continuous with E compact, then there are XI ,x2 in E such that, for all x E E,


Proof By Theorem 3.10, f(E) is a compact set, and then, by Exercise 35, both sup f(E) and inf f(E) belong to f(E). Thus, there are xl and x2 in E such that f(xi)=inff(E)

and f(xz)=sup f(E).

Therefore, for all x in E,

In essence, this theorem tells us that iff : E + R is continuous with E compact, then not only is f a bounded function, but f actually assumes a maximum and a minimum value on the set E. Let us explore some examples to illustrate this theorem. Example 3.8 Consider f : (0,l) + R defined by f(x) = :. The function is continuous but not bounded from above. It is bounded from below, but inf f((0,l)) = 1, and there is no x E (0,l) such that f (x) = 1. Example 3.9 The function g : (0,l) -, R defined by g(x) = x is certainly bounded, but inf g((0,l)) = 0 and sup g((0,l)) = 1, neither of which is a value of g W for the reason that (0,l) is not compact.

W

To lay the groundwork for a new theorem worthy of note, suppose f : E -,R is 1-1. Then there is a unique function f - I :f(E) -,E such that f o f-'(x) = x for each x E f(E) and f 0f(y) = y for each y E E. A natural question to raise is the following: Iff is continuous, is f continuous? The next theorem gives a partial answer.

-'

-'

3.12 THEOREM Suppose f : E + R is continuous and 1-1 with E compact. Then f :f (E) + E is continuous.

-'

Proof Let {yn)El be any sequence in f (E) converging to yo E f(E). We will show that {f -'(yn)}zl converges to f (yo). For ease of understanding, we will write xn = f-'(yn) for each n. Thus, f(xn) = yn. Since E is compact, the sequence {xn}Z1 is bounded and Theorem 1.15 applies. Let {xn,)El be any convergent subsequence-of { x , ) ~ , ;call the limit 20. Now 20 belongs to E since E is closed, and so f is continuous at 20; thus, Cf(xnk))E1converges to f(zo). But {f(xnk))E1= {ynk)gl is a subsequence of {yn)gl; hence, it converges to yo = f(x0). Since f is 1-1 and f(xo) =f(zo), xo = 20. By Theorem 1.15, {xn)Z1 converges to xo; or, in other words, {f - (yn))zl converges to f - l (yo). Thus,f is continuous at yo.

-'

'

Example 3.10 To show the need for the hypothesis that E be compact, we shall exhibit an example of a function f : E + R that is 1-1 but such that f :f(E) + E

3.4 Properties of Continuous Functions

103

graph off

Figure 3.2

is not continuous. Of course, E must be noncompact. Let

anddefinef(x)=xforO 5 x 5 1 a n d f ( x ) = 4 - x f o r 2 < x < 3. It islefttothe readers to satisfy themselves that f is indeed 1-1 and continuous. See Figure 3.2 for the graph off. Now f(E) = [0,2] and for 0 5 x 1,f-'(x) = x while for 1 < x 5 2, f-'(x) = 4 - x. See Figure 3.3 for the graph of f-'.

0. An intuitive picture of the geometric nature of the graph of a continuous function leads us to guess that this graph must cross the x-axis somewhere between - 1 and +1; that is, there is x E (- 1,l) such that f (x) = 0. This indeed turns out to be the case; in fact, it motivates us to state the following theorem. 3.13 BOLZANO INTERMEDIATE-VALUE THEOREM Let f : [a, b] 4 R be continuous with f(a) < y < f(b) [or f(b) < y < f(a)J. Then there is c E (a, b) with f (c) = y.

Proof Let A = {x : x E [a, b] and f(x) 5 y). Now A is nonempty (a E A) and bounded. Hence, A has a least upper bound; call it c. Now c either belongs to A, in which case f (c) 5 y, or c is an accumulation point of A. If c is an accumulation point of A, then there is a sequence {an)z, of members of A converging to c. But then, by the continuity off at c, (f(an))El converges to f(c). Since an E A for each n E J , f(an) 5 y, so by Theorem 1.12, f(c) 5 y. Thus, in any case, f (c) l YTo finish the proof, -we need to show that f (c) = y; that is, we must rule out the possibility that f (c) < y. Assume f (c) < 3. Let

Then E > 0 and, by the continuity off at c, there is b > 0 such that lx - cl and x E [a, b] imply that 1f(x) -f(c)l < e; that is,

y and c E [a, b]. So there is x in [a, b] such that c < x < c + 6 and hence f(x) < y. This contradicts the fact that c = supA. Thus, the assumptionf(c) < y must be false since it leads to a contradiction. This theorem is, of course, the one that is used in trying to find the zeros of a polynomial. If p is a polynomial with p(a) < 0 and p(b) > 0, then p has at least one zero between a and b, since every polynomial is continuous. If p is a polynomial of odd degree, then p will have a change of sign, so it necessarily has a zero. Let us prove this fact.

3.14 THEOREM If p is a polynomial of odd degree with real coefficients, then the equation p(x) = 0 has at least one real root.

Proof Assume

where n is odd and an # 0. If a~ = 0, then p(0) = Assume now that a0 # 0, and let

for all x # 0. Choose E such that 0 exists ki > 1 such that

= 0 and we are through.

< 6 < lanl. For i = 0,1,2,. . . , n - 1, there

Let K = max{b, kl, . . .,kn-l). Now, if 1x1 2 K, then

Thus, for 1x1 2 K, +(x) has the same sign as an. Now p(x) = #(x)X. So, if a,, > 0 with x 2 K, then p(x) > 0; and, if x 5 -K, then p(x) < 0. Similarly, if an < 0 and x 1 K, then p(x) < 0; and, if x 5 -K, then p(x) > 0. In any case, we can find points xl and x2 such that p(xl) < 0 and p(x2) > 0; hence, somewhere between xl and x2, p has a zero. If we combine Corollary 3.1 1 and Theorem 3.13, we obtain the following result.


3.15 THEOREM Iff : [a, b] f ([a bl) = [c, dl.

+R

is continuous, there are c and d such that

Proof By Corollary 3.11, there are xl and x2 in [a, b] such that f(x1) I f(x) I f(x2) for all x E [a,& Let c = f(xl) and d = f(x2). By Theorem 3.13, if c f ([a, bl) = [c,dl*

< y 5 d, then there is x E [a, b] such that f (x) = y.

Thus,

W Example 3.11 As a further application, we shall show that the equation x = cosx has at least one solution in the closed interval [0, Consider the function f (x) = x - cos x on [0, . Clearly, f is continuous, f (0) = - 1, and f (5) = Tr . Therefore, there is x E [0, such that f (x) = 0; in other words, x = cos x.

;I.

H] H]

3.16 THEOREM Let f : [a, b] -+ R be continuous and 1-1. Then f is mono-

tone.

Proof Suppose f is 1- 1 and not monotone. Then there are x, y, z E [a, b] such thatx < y < 2, andf(x) f(y). Let us suppose the latter; in fact, let us suppose f (x) > f ( 2 ) > f (y). The other cases may be handled in a similar fashion. By the Intermediate-Value Theorem, there is w E (x,y) such that f (w) =f (z), contrary to f being 1-1. Let us now return to a topic considered at the end of Chapter 2. Let f : [a, P] + R be monotone. Then D = {x : x E (a, P) and f does not have a limit at x) is countable; and iff has a limit at xo E (a, p), then limx, f (x) = f (xo). The discussion of the behavior off at a and at ,6 was left to the reader. It is no betrayal of confidence to state now that f has a limit at a and at p and that lim f (x) = inf {f (x) : a

x--+a!

< x 5 p)

and lim f (x) = sup {f (x) : a 5 x < p)

x-4

iff is increasing. In the case where f is decreasing, both limits still exist, but lim f (x) = sup {f (x) : a

X+CY

< x 5 p)

and lim f(x) = inf {f(x) : a 5 x

x-+P

< p).

Now it is clear that f is continuous at x for each x E (a, p) \ D, and possibly at a and p depending on the definition off at a and ,@. In particular, the set of points at which f is discontinuous is countable.

Exercises 107

EXERCISES 3.1 CONTINUITY OF A FUNCTION AT A POINT Define f : R

-+

R by f(x) = 3 2 - 2x + 1. Show that f is continuous at 2.

Define f : [-4.01 continuous at -3.

+R

by f(x) =

+for x # -3 2a?

18

and f(-3) = -12. Show that f is

Use Theorem 3.1 to prove that {@)gl is convergent, and find the limit. You may assume that the functionf(x) = E is continuous on R. If a E E, a is not an accumulation point of E, and f : E --+ R, prove that, for every sequence {xn)El converging to a with xn E E for all n, {f (xn))Elconverges to f (a). Definef : @,I) -+ R by f (x) = at O? Explain.

5 - E.Can one define f (0) to make f

continuous

Prove that f(x) = 6 is continuous for all x 2 0. Suppose f : R -+ R is continuous and f(r) = 2 for each rational number r. Determine f (fi) and justify your conclusion. Suppose f : (a,b) + R is continuous and f(r) = 0 for each rational number r E (a,b). Prove that f (x)= 0 for all x E (a,b). Define f : ( 0 , l ) + R by f (x) = x sin f . Can one define f (0) to make f continuous at O? Explain. Supposef :E + R is continuous at a, and E F C E. Define g : F --, R by g(x) =f(x) for all x E F. Prove that g is continuous at a. Show by example that the continuity of g at a need not imply the continuity off at a. Define f :R R by f(x) = 8x if x is rational and f(x) = 2? + 8 if x is irrational. Prove from the definition of continuity that f is continuous at 2 and discontinuous at 1. -+

3.2 ALGEBRA OF CONTINUOUS FUNCTIONS be a zero of q of multiplicity m. Rove that p / q can be assigned a value at a such that the function thus defined will be continuous there iff a is a zero of p of multiplicity greater than or equal to m.

42. Let p and q be polynomials and

13. Let f : D Q of

R be continuous at E D. Rove that there is M such that 1 f(x)l 5 M for all x E Q n D. -+

> 0 and a neighborhood

14. I f f : D + R is continuous at E D, prove that the function 1f 1 : D + R such that If l(x) = If(x)l is continuous at a. 15. Suppose f , g : D -+ R are both continuous on D. Define h : D -+ R by h(x) = max {f (x),g(x)). Show that h is continuous on D. 16. Assume the continuity of f(x) = @ and g(x) = In x. Define h(x) = 3 by 3 = 8 '" '. Show h is continuous for x > 0. 17. Supposef :D + R with f(x) 2 0 for all x E D. Show that, iff is continuous at a, then

8is continuous at a.

18. Define f :R + R as follows: f(x) = x - [x]if [x]is even. f (x)= x - [x+ 11 if [x]is odd. Determine those points where f is continuous. Justify.


-

3.3 UNIFORM CONTINUITY: OPEN, CLOSED, AND COMPACT SETS 19. Let f,g : D + R be uniformly continuous. Prove that f + g : D R is uniformly continuous. What can be said about fg? Justify. 20. Let f : A + B and g : B + C be uniformly continuous. What can be said about g of :A + C? Justify.

-

R by f(x) = &. Show that f is uniformly continuous on [3.4,5] without using Theorem 3.8-that is, use the methods of Example 3.2. 22. Define f : (2,7) -t R by f (x) = 2 - x + 1. Show that f is uniformly continuous on (2,7) without using Theorem 3.8---that is, use the methods of Example 3.3. 23. A functionf : R + R is periodic iff there is a real number h # 0 such that f (x + h) =f (x) for all x E R. Prove that iff : R + R is periodic and continuous, then f is uniformly continuous.

21. Define f : [3.4,5]

24. Suppose A is bounded and not compact. Prove that there is a function that is continuous on A, but not uniformly continuous. Give an example of a set that is not compact, but every function continuous on that set is uniformly continuous. 25. Give an example of sets A and B and a continuous function f :A U B + R such that f is uniformly continuous on A and uniformly continuous on B, but not uniformly continuous onAUB. *26. Let E c R. Prove that E is closed if, for every xo such that there is a sequence {xn)gl of points of E converging to m, it is true that xo E E. In other words, prove E is closed if it contains all limits of sequences of members of E.

*27. Rove that every set of the form {x : a {x : a

< x < b)

is open and every set of the form

5 x 5 b) is closed.

28. Let D C R, and let D' be the set of accumulation points of D. Prove that D = D U D' is closed and that if F is any closed set that contains D, then n C F. is called the closure of D. 29. If D C R is bounded, prove that D is bounded. 30. Suppose f : R + R is continuous and let ro E R. Prove that {x :f(x) # ro) is an open set. 31. Suppose f : [a, b] -r R and g : [a,b] -t R are both continuous. Let T = {x :f (x) = g(x)). Prove that T is closed. 32. If D c R, then x E D is said to be the interior point of D iff there is a neighborhood Q of x such that Q c D. Define D o to be the set of interior points of D. Prove that Do is open and that if S is any open set contained in D, then S C Do.Do is called the interior of D. 33. Find an open cover of {x : x > 0 ) with no finite subcover. 34. Find an open cover of (1,2) with no finite subcover. *35. Let E be compact and nonempty. Prove that E is bounded and that sup E and infE both . belong to E. 36. If El, . . .,En are compact, prove that E = uLlEi is compact. 37. Let f : [a, b] + R have a limit at each x E [a,b]. Prove that f is bounded. 38. Suppose f : D + R is continuous with D compact. Prove that {x : 0 5 f(x) 5 1) is compact.

Projects 109

1 I

39. Suppose that f : R + R is continuous and has the property that for each e > 0, there is M > 0 such that if 1x1 2 M, then 1f(x)l < e. Show that f is uniformly continuous. 40. Give an example of a functionf : R -,R that is continuous and bounded but not uniformly continuous.

3.4 PROPERTIES OF CONTINUOUS FUNCTIONS Find an interval of length 1 that contains a root of the equation xd[ = 1. Find an interval of length 1 that contains a root of the equation 2 - &? + 2.826 = 0. Suppose f : [a, b] -,R is continuous and f(b) 5 y f (a). Prove that there is c E [a, b] such that f (c) = y. Suppose that f : [a,b] --, [a, b] is continuous. Prove that there is at least one fixed point in [a, b]--that is, x such that f (x) = x. Iff : [a, b] --, R is 1-1 and has the intermediate-value property-that is, if y is between f(u) and f (v), there is x between u and v such that f (x) = y-show that f is continuous. (Hint: First show that f is monotone.) Prove that there is no continuous function f : R --+ R such that; for each c E R, the equation f (x) = c has exactly two solutions.

0 and a > 0 such that if x E [-a, a], then 1f(x)l 5 M. Rove that f is uniformly continuous. In particular, prove that there is a real number m such that f(x) = mx for all x E R. Let f : [a, b] -r R be continuous, and define g : [a, b] --, R by g(t) = sup {f(x) : a

5 x 5 t).

Prove that g is continuous. (This exercise may be difficult if you didn't cover Section 2.4. In fact, you might also look at Exercise '25 in Chapter 2.) Suppose that g : D -,R is continuous at a and that a is also an accumulation point of D. Define Do = {x : g(x) # 0). If g ( a ) # 0, prove that a is an accumulation point of Do. Suppose f : D -+ R, g : E --, R, and xo E D n E. Suppose further that there is E > 0 such that D n [ a - f , a + € ] =En[%-€, xO+€]and f(x) = g(x) for d l x E DnEn[xo-€, a + € ] . Prove that f is continuous at a iff g is continuous at xo.

PROJECT In this project, you will prove a version of Theorem 3.8 and Theorem 3.1 1 without using the notion of compactness. 3.8 THEOREM Suppose f : A + R is continuous with A closed and bounded. Then f is uniformly continuous.

Proof You will be led through the steps of the proof, but you will need to fill in some missing details and supply reasons for some of the conclusions. Remember, you can't use any theorems in the book that were proven using the notion of compactness.


1. Assume f is not uniformly continuous on A. Then there is c > 0 such that for each n, there are an,bnE A such that lan - bnl < l l n and 1f(an) -f(bn)l 2 E. Explain why this must be true iff is not uniformly continuous on A. 2. There is a sequence of positive integers {nk}gl such that the sequences { a n k } and {bnk)El both converge; call the limits a and b respectively. Explain why this is true. 3. Both a and b belong to A and a = b. Explain why this is true and why it leads to a contradiction. Finish the proof.

3.11 COROUARY Iff : A + R is continuous and A is closed and bounded, then there are XI,X* E A such that f (xl) 5 f (x) 5 f (x2) for all x E A.

Proof This proof has several parts, but the basic idea is to show that f(A) is closed and bounded. 1. f (A) is bounded by Theorem 3.9. This does not use compact sets. 2. f(A) is closed. You will need to modify the statement and proof of Theorem 3.10 to show this is true. 3. You now know that f(A) is both closed and bounded. The rest of the proof is left to you.

PROJECT 3.2 A function f : R + R is afine if f(x+y -2) =f(x)+f(y) -f(z) for allx,y,z E R. In this project, you will prove that continuous affine functions have a special form. A function f : R -,R is additive iff (x + y) =f (x) +f (y) for all x, y E R. 1. Assume f is an affine function and m is a real number. Define g : R + R by g(x) =f (x) - [mr +f (O)]. Show that g is an additive function. 2. Let m = f(1) - f(0) and show that the function defined in Part 1 above has the property that g(x + 1) = g(x). 3. Supposef is affine and bounded on [0, 11, and g is defined by g(x) =f (x) - [mr+b] where m =f(1) -f(0) and b =f(0). Show that g is bounded on R. 4. Show that any bounded additive function must be the zero function. 5. Show that iff is affine &d continuous, there are real numbers m and b such that f (x) = mr + b for all x E R.

I

PROJECT 3.3 In this project you will show that continuous functions that satisfy the functional equation f (xy) = f (x) +f (y) are logarithmic functions. You may assume the usual properties of logarithmic functions in this project.

Projects

111

In what follows, let R+ denote the set of positive real numbers. Assume f : R+ + R is a continuous function, for all positive real numbers x and y, f(xy) =f(x)+f(y), and f is not the zero function; that is, there is u such that f(u) 0.

+

I

I

1. Show that f (1) = 0 and there is a positive real number a such that f (a) = 1. 2. Define g(x) =f(x) - logax where a is such that f(a) = 1. Show that g(xy) = g(x) + g O for all x,y E R+. 3. Show that g ( a ) = g(x) for every x E R+. Prove that g is bounded. 4. If g is not the zero function, prove that g is not bounded. 5. What can you conclude from Parts 3 and 4?

PROJECT 3.4 This is an open-ended project in which you will be asked to make up and prove a theorem of your own; there really isn't a "right" answer. Your purpose will be to modify Theorem 3.12 so it can be applied in some circumstances where the domain of the function is not compact. For example, f : R -, R defined by f(x) = 2 is continuous and 1-1, but Theorem 3.12 cannot be used to prove that the inverse off is also continuous because R is not compact. The inverse is the cube root function and is continuous. Your strategy should be to examine Example 3.10 and the proof of Theorem 3.12 and to find some conditions on the domain off that will guarantee that f-' is continuous. To further focus your efforts, the theorem should take a form similar to the following, and your goal is to fill in the missing words:

THEOREM Suppose dom f = A and A has the property that (till in the missing words). Then iff : A + R is continuous and 1-1, then f-' :f(A) + R is also continuous. Of course, you should then supply a proof of your theorem.

Chapter 4

Differentiation

o describe the direction of a nonvertical line in the x,y plane, it is customary to give the slope of the line. The nonvertical line is the graph of a linear function, and the slope indicates the rate of change of the function. If the slope of the line is positive, the function is increasing; if the slope is negative, the function is decreasing. Moreover, the magnitude of the slope is an indication of how fast the function is increasing or decreasing. Iff is a function whose graph is not a straight line, it would be useful to have a notion of slope for the graph off as a means of determining the rate of change of the function. Fermat and others in the seventeenth century formulated the concept of the derivative for the rate of change of a function. If one draws a smooth curve, the notion of a line tangent to the curve is a reasonably acceptable intuitive idea. The problem is finding the slope of that tangent line. A line tangent to a circle is one that intersects the circle in exactly one point, but that definition cannot be used for the graph of a function except in special cases. (See Exercise 1 for such an example.) The alternative is to use a limiting process to find the slope of the tangent line. The basic idea in calculating the slope of the tangent line at a point Po on the curve is to find the slope of the line through Po and another point P on the curve, and then to take the limit of that slope as P approaches Po. If that limit exists, it is defined to be the slope of the curve at Po. See Figure 4.1 for a sketch of a curve with the line tangent to the curve at Po, as well as a secant line passing through points Po and P. If Po has coordinates (xo,yo),and P has coordinates (x,y), then the slope of the line through P and Po is

or, in terms of the function f ,

114 Chapter 4 Differentiation

Figure 4.1

This means that the slope of the graph of the function f at (xo,yo) is the limit of this quotient as x approaches a.Does this sound familiar? It should. Read on.

DEFINITION Let f : D + R with xo an accumulation point of D and Q E D. For each x E D with x # xo, define

The function f is said to be differentiable at xo (or has a derivative at xo) iff T has a limit at xo, and we write lim,, T(x) =f '(no). The number f '(a) is called the derivative off at xo. Iff is differentiable for each x E E C D , we say f is differentiable on E. The following alternative definition of the derivative may also be familiar. In Exercise 2, you are asked to prove that the two definitions are equivalent.

ALTERNATE DEFINITION Let f : D + R with xo an accumulation point of D and xo E D. For each t E R such that xo + t E D and t # 0, define

4.1 *

The Derivative of a Function 115

The function f is said to be differentiable at xo iff Q has a limit at zero. If this limit exists, it is called the derivative off at xo.

Now we shall state a basic theorem that gives a condition for differentiability in terms of sequences. The definition of differentiability is based on the existence of the limit of a function, so this theorem should come as no surprise. 4.1 THEOREM Suppose f : D + R, xo E D, and xo is an accumulation point

of D. Then f is differentiable at xo iff for every sequence {xn)zl of points of D \ {xo) converging to XO,the sequence

converges. No formal proof will be supplied here because it should be clear from Theorem 2.1 that the condition concerning sequences is equivalent to the condition that the function T given in the definition of differentiability has a limit at xo. Note also that iff is differentiable at xo and {xn)g1 is a sequence of points of D \ {xo) converging to XO,then

converges to f '(xo). The next few examples will further one's understanding of the definition of differentiation. Example 4.1 Consider the function f (x) = 1x1 for all x E R. Set xo = 0, and consider the sequence

this sequence converges to zero, but zero is not a term of this sequence. For n even, (-1 -- -1

and

f

(i)-f(O)

= 1:

for n odd, (-1Y =: -1 -

and

f

(-t) -f@)

= -1.

*

,


so the sequence

does not converge. Thus, by Theorem 4.1, f is not differentiable at zero. The reader should sketch the graph of 1x1 to see why this happens. The graph has a "sharp" comer at zero with the slope suddenly changing from - 1 to +1 as x increases from negative to positive values, and of course the slope is not defined at zero. Example 4.2 For the moment, assume the usual facts about differentiating the product, sum, and composition of the elementary functions. Define f (x) = x sin for x # 0 and f(0) = 0. The fact that f is differentiable at any point other than zero follows from the assumptions made at the beginning of this paragraph. For x # 0,

does not have a limit at zero; hence, f is not differentiable at zero, although it is continuous there. Example 4.3 In an attempt to find a "nicer" function, define f (x) = 2 sin $ for x # 0 and f(0) = 0. Again, assume the basic facts about differentiation of products and composition of elementary functions. Now, f has a derivative at x if x # 0, and f '(x) = 2x sin $ - cos :. So f '(x) is defined for x # 0, but f' does not have a limit at zero. One might suspect that this means f is not differentiable at 0; however, this is false. For x $0,

which has a limit at zero, namely, f'(0) = 0. In this case we have a function with a derivative at every point, but the derivative is not continuous at zero; in fact, it does not have a limit there. Example 4.4 Finally-consider f (x) = I$ sin for x # 0 and f (0) = 0. Again, using the usual rules (to be proven in the next section),f has a derivative everywhere and f is continuous everywhere but fails to be differentiable at one point. Can you guess the point? See Exercise 5.

Let us now return to the initial definition and seek the facts pertinent to functions differentiable on a set. First, consider the difference quotient

4.2

The Algebra of Derivatives 117

For x close to xo, the denominator is close to zero; hence, in order that T have a limit at q , the numerator must be close to zem-otherwise T would not be bounded near xo. We are led to predict that, iff is differentiable at xo, then f must be continuous at xo. The proof may be accomplished by observing that

for x

# xo.

4.2 THEOREM Let f : D + R be differentiable at xo. (It is tacitly assumed that xo E D and that xo is an accumulation point of D.) Then f is continuous at XO.

Proof Let T : D \ {xo) + R be defined by

T has a limit at xo and lim,,,

T(x) =ft(xo). For x

# xo,

Now f is a sum of two functions, one of which is constant and the other the product of two functions, each of which has a limit at xo; so f has a limit at xo and

Thus, f is continuous at xo.

The next theorem presents the usual facts from calculus about the sum, product, and quotient of differentiable functions. 4.3 THEOREM

i. f

Suppose f ,g :D

+R

+ g is differentiable at xo and Cf + g)'(xo) =f t(xo) + gt(xo).

are differentiable at xo. Then


ii. fg is differentiable at xo and

iii. If g ( a ) # 0, then f l g (the domain is the set of all x such that g(x) differentiable at xo and

+ 0 ) is

Proof We shall appeal to Theorem 4.1 to prove this theorem. Let { x n ) g l be any sequence of points in D \ {xo) converging to xo. Since f and g are differentiable at xo,

converges to f '(xo) and

converges to g t ( q ) . Thus,

converges to ft(xo) + gt(xo). This means that f

+ g is differentiable at xo and that

By Theorem 4.2, f is continuous at xo, and so ( f ( x n ) ) z l converges to f ( a ) . Thus,

) . fg is differentiable at xo and converges to f (xo)gt(x0)+ g ( . ~ ' l f ' ( x ~Thus,

Let D' = { x : x E D and g(x) # 0 ) . Then D' is the domain of f / g , and xo is an accumulation point of D' by the continuity of g at XO. Let { x n ) z l be any sequence of points of Dt \ 1x0)converging to q . Then, since g is continuous at

4.2 The Algebra of Derivatives

and g(x0) Now

+ 0, 1/ g is continuous at

XO;

119

hence, { $ ( x , ) ) ~converges ~ to $ (xo).

converges to

Thus, f / g is differentiable at a and

Earlier in this chapter we considered f(x) = sin for x # 0. At that time we assumed some facts concerning the differentiation of the sum, product, and composition of differentiable functions. Now we shall justify our assumptions and consider the differentiability of the composition of differentiable functions. The next theorem is often referred to as the Chain Rule. 4.4 THEOREM (Chain Rule) Suppose f :D + R and g :D' -,R with f ( D ) c D'. Iff is differentiable at xo and g is differentiable at f ( a ) , then g of is

differentiable at xo and

[Remember that the hypothesis that f is differentiable at a implies that xo is an accumulation point of D , and similarly, g differentiable at f(-) implies that f(-) is an accumulation point of D'.] Proof

One might be tempted to prove this theorem by writing


and claiming that the right-hand side has gt(f(~))f'(xo)as a limit, and hence, g of is differentiable at xo as claimed. However, x # X,-J need not imply that f (x) # f (a). Hence, the denominator may well be zero in the first factor of the right-hand side of the equation. We shall seek to circumvent this trap. Let yo =f(xo) and, for each y E Dl, define

and

Since g is differentiable at yo, h has a limit at yo and lim h O = gt(yo)= heo). Y -'Yo

Thus h is conthuous at yo. Since f is differentiable at xo and yo = f(xo), then h 0f is continuous at xo. The function T defined by

has a limit at a , namely, f '(a). Now for

You should check the case where f(x) =f (q). Since both h of and T have limits at xo, g of is differentiable at q and

Let us consolidate a few results at this stage. First, any constant function is differentiable and has derivative zero everywhere. The function f (x) = x is differentiable everywhere and ft(x) = 1 for all x. An easy application of Theorem 4.3 shows that every polynomial function is differentiable and that every rational function (the quotient of two polynomial functions) is differentiable at each point where the denominator is nonzero. Because it is so easy to do and we already know the answer, let us prove now that if f(x) = x" for n an integer, then f '(x) = (Of course, if n < 0, we must assume x # 0, and if n = 0, then ft(x) = 0 for all x.)

4.2 The Algebra of Derivatives 121

425 THEOREM If n is an integer and f (x) = 9 for all x, then f is differentiable

for all x if n > 0 and for all x ft(x) = 0 for all x.

# 0 if n < 0, and fl(x) =

If n = 0, then

Proof For n = 0, f(x) = 1 for all x; hence, f is differentiable for all x and ft(x) = 0, and the theorem is true. We shall now prove the theorem by induction for n > 0 and appeal to Theorem 4.3 to extend to the negative integers. The fact thatf (x) = Y is differentiable for each positive integer n follows from Theorem 4.3 by induction, but this is the type of induction proof that should be unnecessary for one who has progressed this far. We shall now proceed to show that f1(x) = For n = 1, f (x) = x and f '(x) = 1 = lxo, so the theorem holds. Suppose the theorem holds for n = r; that is, iff (x) = Y, then

We may consider this the product of two functions: f (x) = Y Consider g(x) = P1. and h(x) = x. By Theorem 4.3,

Thus, the theorem holds for n = r + 1, and the induction is complete. It now remains to verify the conclusion for f (x) = x" for n a negative integer, x # 0. Now, if n < 0, then -n > 0, and Y = l/x-" is differentiable for x # 0 by our remarks earlier in this proof and in Theorem 4.3. Moreover, if f(x) =x" = l/x-", then

fW =

-(-n)x-n-l

(x- .)*

= nr"-1.

This concludes the proof.

Example 4.5 Let us examine the use of Theorems 4.3, 4.4, and 4.5 in finding the derivative of a function such as F(x) = d m , which is the composition of the functionf (x) = 1+ 3x2 and the function g(u) = fi.For each x, f is differentiable at x and, using Theorems 4.3 and 4.4, ft(x) = 6x. Moreover, for each x, f (x) > 0; hence g is differentiable at f (x). Note that g1(u) = 1/2@. Thus, by the Chain Rule, g 0 f = F is differentiable at x and

122 Chapter 4 Diflerentiation -

-

4.3 ROLLE'S

THEOREM AND THE MEAN-VALUE THEOREM

The classical maxima and minima problems from the calculus are no doubt familiar to the reader. The fact to be applied here is roughly that the graph of a differentiable function has a horizontal tangent at a maximum or minimum point. This is not quite accurate, but let us define some terms before proceeding further. DEFINITION Let f : D + R. A point xo E D is a relative maximum (minimum) off iff there is a neighborhood Q of xo such that if x E Q n D, then

Consider a few examples and reconsider the statement preceding this definition. Example 4.6 Perhaps the point is most easily made by looking at the function f : [0, 1] R such that f (x) = x. Since the domain of f is the interval [0, 11,f has a minimum at 0 and a maximum at 1; indeed, 0 is a relative minimum off, and 1 is a relative maximum off. Since fl(x) = 1 for all x E [0, 11, it is clear what is wrong with our statement concerning the existence of a horizontal tangent at a maximum or minimum point. We shall correct this defect in the following theorem. -+

4.6 THEOREM Suppose f : [a, b] -+ R and suppose f has either a relative maximum or a relative minimum at xo E (a, b). Iff is differentiable at xo, then f /(xo) = 0.

Proof Assume f has a relative maximum at xo. Then there is 6 > 0 such that, if xo - 6 < x < xo + 6, then x E [a, b] [since xo E (a, b)] and f (x) 5 f (xo). Consider any sequence {xn)El converging to such that xo - 6 < xn < xo. Then, since f is differentiable at xo,

converges to f '(xo). But

for each n because f(xn) 5 f(xo) and xn < xo, and hence fl(xo) 2 0. Consider now a sequence {yn)gl such that xo < y < xo + 6. As before,

4.3 Rolle's Theorem and the Mean-Value Theorem 123

converges to ~'(xo),but this time

hence, f '(xo) 5 0. Therefore, f '(xo) = 0. The case in which f has a relative minimum at xo is left to the reader. The reader should now realize that an important fact in this proof is that we are free to choose points in the domain on either side of xo as close as we please. Since both the maximum and the minimum occurred at end points in the preceding example, the theorem did not apply. The following theorem, known as Rolle's Theorem, is an application of Theorem 4.6. We shall postpone a discussion of its geometric interpretation until we have completed the proof.

4.7 ROLLE'S THEOREM Suppose f : [a, b] -+R is continuous on [a, b] and f is differentiable on (a, b). Then iff (a) = f (b) = 0, there is c E (a, b) such that f'(c) = 0.

Proof Iff (x) = 0 for all x E [a, b], then f '(x) = 0 for all x E [a,b], and the theorem is proved. Suppose f(x) # 0 for some x E [a, b]. By Theorem 3.7, [a, b] is compact since it is closed and bounded; hence, by Corollary 3.11, f assumes its maximum and minimum in [a, b], say at xl and x2, respectively. Since f is not identically zero on [a, b] and f(a) =f(b) = 0, at least one of xl and x2 must belong to (a, b), say xl E (a, b). Now, by Theorem 4.6, f'(xl) = 0. In essence, the theorem states that if the graph of a differentiable function touches the x-axis at a and at b, then somewhere between a and b there is a horizontal tangent. See Figure 4.2. Note that the theorem states that there is at least one c such that f'(c) = 0. There may be more such points as indicated in Figure 4.2. If we examine the graph in Figure 4.2 from a geometric point of view-that is, ignoring the choice of coordinate system-it would appear that the tangent line at c

c

Figure 4.2

h


Figure 4 3

is parallel to the line connecting the endpoints of the curve. This is essentially the content of the Mean-Value Theorem to follow. Suppose we have a smooth curve, the graph off, connecting the points (a, f (a)) and (b, f (b)). We can rotate and translate the coordinate axes so that the points lie on the new x-axis. Then Rolle's Theorem will guarantee a point on the curve where the tangent line is parallel to the new x-axis, which contains the segment connecting the points (a, f (a)) and (b, f(b)). (See Figure 4.3.) The slope of the segment joining (a, f(a)) and (b, (b)) is b-a

9

so this is the slope of the tangent line at c; that is,

The precise statement and proof follow. 4.8 MEAN-VALUE THEOREM Iff : [a, b] + R is continuous on [a, b] and differentiable on (a, b), then there is a c E (a, b) such that

. Proof

To prove this theorem, we shall find a linear function L such that f - L satisfies Rolle's Theorem and shall then apply Rolle's Theorem to obtain the desired result. This corresponds to the change of coordinate system mentioned ..before this theorem. The function L must be linear and satisfy

L(a) =f(a)

and L(b) =f (b).


This is accomplished by choosing f(b) -f@) L(x)=[ b - a ](x"a)+f(a), the equation of the line passing through (a, f(a)) and (b, f(b)). L is continuous and differentiable everywhere and Lf(x) = f(b) - f ( 4 b-a Let g =f - L. The function g satisfies the hypotheses of Rolle's Theorem; hence, there is c E (a, b) such that

0 = gf(c) =ff(c) - Lf(c) =ff(c) - f(b) -f@) b-a

Thus

The Mean-Value Theorem has some very important applications in relating the behavior off and f f. The next theorem will illustrate this. I

4.9 THEOREM Suppose f is continuous on [a, b] and differentiable on (a, b). Under these hypotheses: i. If ff(x) # 0 for all x E (a, b), then f is 1-1.. ii. If ff(x) = 0 for all x E (a, b), then f is constant. iii. Iff '(x) > 0 for all x E (a, b), then x < y and x, y E [a, b] imply f (x) (that is, f is strictly increasing). iv. If f f(x) < 0 for all x E (a, b), then x < y and x, y E [a, b] imply f (x) (that is, f is strictly decreasing).

< f (y)

> f (y)

Proof Consider any x, y E [a, b] with x < y. Now f is continuous on [x,y] and differentiable on (x,y); so, by the Mean-Value Theorem, there is c E (x,y) such that

that is, f (x) -f (y) =f (c)(x - y). With this fact in mind, we shall proceed.

126 Chapter 4 Diferentiation

(i) Suppose f is not 1-1. Then there are x,y E [a,b] with x f(x) =f(y). Thus, there is c E (x,y) such that

< y such that

contrary to f' (c) # 0 for all c E [a,b]. (ii) Suppose f is not constant on [a,b]. Then there are x, y E [a,b], x such that f (x) # f (y). There is c E (a, b) such that

< y,

contrary to f' (c) = 0 for all c E [a,b]. (iii)Suppose x < y and x, y E [a,b]. There is c E (x,y) such that

hence f(x) f

0).

Now that the proof has been presented, it is clear that parts (iii) and (iv) can be changed as follows to give additional information: iii. f '(x)2 0 for all x E (a,b) implies f is increasing. iv. f f ( x ) 5 0 for all x E (a,b) implies f is decreasing. The following theorem is an obvious corollary to Theorem 4.9. 4.10 THEOREM Suppose that f and g are continuous on [a,b] and differentiable on (a,b) and that f '(x)= gf(x)for all x E (a,b). Then there is a real number k such that

for all x E [a,b].

Proof Consider h(x) =f (x)- g(x) for all x E [a,b]. Clearly, h is continuous on [a,b] and differentiable on (a,b) and h' (x) =f' (x) - gt(x) = 0


for all x E (a,b). Hence, by Theorem 4.9, there is a real number k such that

for all x E [a, b]. Thus, f(x) = g(x) + k for all x E [a, b]. Example 4.7 The Mean-Value Theorem has a variety of uses, one of which is that of estimating values of certain functions. Suppose p > 1 and h > 0. Define the function f (x) = (1 + x)P. Assuming the basic facts about the differentiation of powers, f is differentiable for x > 0. So, if h > 0, there is t such that 0 < t < h and

Now p

> 1; hence, (1 + t ) ~ '- > 1, SO

See Exercise 24 for the case of p

< 1.

W

Consider a function f differentiable on [a, b]. Iff '(x) # 0 for all x E (a, b), then by Theorem 4.9 f is 1-1 and by Theorem 3.16 f is monotone. Iff is increasing, then, for all x, y E [a, b] with x # y, we have

>

Hence, f '(x) 0 for all x E [a, b]. Similarly, if f is decreasing, f '(x) x E [a, b]. We have thus proved the following theorem.

< 0 for all

LEMMA If f is differentiable on [a, b] and f ' ( x ) # 0 for all x E (a, b), then either ff(x) 2 0 for all x E [a,b] or ff(x) 5 0 for all x E [a,b]. Let us consider this result from a slightly different point of view. Suppose f is differentiable on [a, b], and there are x, y E [a, b] such that f (x) > 0 and f ' ( y ) < 0. If we restrict our attention to [x,y] (assuming x < y), then, by the lemma, there is c E (x,y), such that ft(c) = 0. Thus, we are led to suspect that f has the intermediatevalue property. That this is indeed the case is stated in the following theorem. 4.11 THEOREM Suppose f is differentiable on [a, b] and X is a real number such that f '(a) < X < f '(b) or f '(b) < X < f (a). Then there is c E (a, b) such that ft(c) = A.

128 Chapter 4 Diflerentiation -

Proof Define

for all x E [a,b]. Then g is differentiable on [a,b] and gl(x) = fl(x) - A. If ft(a) < X < f'(b), then g'(a) < 0 and gf(b) > 0; if ff(b) < X < ff(a), then g'(a) > 0 and g'(b) < 0. In either case, g' has opposite signs at a and at b. Hence, by the lemma, there is c E (a, b) such that

so that f '(c) = A. Supposef : [a, b] -,R is such that there is a differentiable function g : [a, b] -,R such that gf(x) = f(x) for all x E [a, b]. Now f need not be a continuous function, but f must have the intermediate-value property; in other words, if x,y E [a, b] and f (x) < X < f (y), then there is c between x and y such that f (c) = A. This means that a function that is a derivative of some function must be special. In particular, f(x) = [x] cannot be the derivative of a function. The following theorem is a generalization of the Mean-Value Theorem. 4.12 CAUCHY MEAN-VALUE THEOREM Iff and g are continuous on [a, b] and differentiable on (a, b), then there is c E (a, b) such that

Proof Define h(t) = [f(b) -f (a)]g(t) - [g(b) - g(a)]f (t) for each t E [a, b]. Note that h is continuous on [a, b] and differentiable on (a, b) and that h(a) = h(b). So, by the Mean-Value Theorem, there is c E (a, b) such that

4.4 L9HOSPZTAL9SRULE AND THE INVERSE-FUNCTION THEOREM We can use Cauchy's Mean-Value Theorem to prove a familiar rule from calculus, L'Hospital's Rule. 4.I 3 L'HOSPZTAL'S RULE Suppose f and g are continuous on [a, b] and differentiable on (a, b). If xo E [a, b] ,

i. g'(x) # 0 for all x E [a,b], x # a; ii. f (a) = g(-) = 0; and iii. f'lg' has a limit at xo,

4.4 L'Hospital's Rule and the lnverse-Function Theorem 129

then f / g has a limit at xo and I lim f-(x) = lim f-$). g X"x0 g

XX 'O

Proof Let { x n ) z l be any sequence converging to xo with xn E [a, b] \ {a)for all n. By the Cauchy Mean-Value Theorem, there is a sequence { c n ) ~such , that

Since xn # xo and gl(x) # 0 for x # a,Rolle's Theorem tells us that g(xn) f g ( q ) = 0 for all n. Since f ( q ) = g(xo) = 0,

and since fl/g' has a limit at q,the sequence

converges; hence, so does the sequence

and both to the same limit. Therefore, f / g has a limit at xo and

Exmple 4.8 To illustrate the use of L'Hospital's Rule, consider the functions f and g defined as follows: f ( x ) = f i - lh+,/=

and

NOWf and g are both continuous on [ 2 , 3 ] and differentiable on ( 2 , 3 ) , and f ( 2 ) = g(2) = 0. Further,

.

130 Chapter 4 Dijgkrentiation

It is clear that fl/gl has a limit at 2 and that limit is 112. So, by L'Hospital's Rule, f /g has a limit at 2 and the limit is 112. The next theorem is a one-dimensional version of the Inverse- Function Theorem. In Chapter 3, we observed that, iff : [a, b] --+ R is continuous and 1-1, then f-' is also continuous. It is now natural to inquire what can be obtained if it is assumed that f is differentiable. Before proceeding further, let's look at the possibilities. Iff is differentiable, then, by the Chain Rule, is 1-1 and differentiable, and iff

but (f

-' of)(x) = x for all x E [a, b]; hence,

for all x E [a, b]. This last equation shows that, for x E [a, b],

and, in particular, f'(x) # 0. These observations give us some insight into what hypotheses are needed and what conclusions might be drawn in the setting mentioned above. 4.14 THEOREM Suppose f : [a, b] --+ R is continuous and differentiable with f '(x) # 0 for all x E [a, b]. Then f is 1-1, f - is continuous and differentiable on f (1%bl), and

'

for all x E [a, b].

Proof Since fl(x) # 0 for all x E [a,b], then, by Theorem 4.9, f is 1-1. Let us suppose that f([a, b]) = [c, dJ. (The fact that f is continuous guarantees that f([a, b]) is indeed a closed interval.) We shall now proceed to show that f-I : [c, dJ + [a, b] is differentiable. Choose yo E [c, dJ and {yn)gl any sequence in [c, 4 \ {yo) converging to yo. Let

4.4 L'HospitalS Rule and the Inverse-Function Theorem 131

-' +

for n = 0,1,2, . . .. BY Theorem 3.12, f is continuous and { ~ ~converges )g~ to xo = f -'(YO); and, since f is 1-1, xn xo for all n. By the differentiability off,

-'

converges to f '(~0).By hypothesis, f '(xo)

+ 0 and, since f is 1-1,

for n = 1,2, . . .. Hence,

converges to

Thus, by Theorem 4.1, f -' is differentiable and fm.

See Exercise 34 for a variation on this theorem.

Example 4.9 We close this chapter with a few applications of our results. Consider the function g(x) = 2 defined. for all real numbers x. Of course, g is differentiable and g' (x) = 2x. We shall restrict our attention to the behavior of g on the interval [8,10]. If x E [8,10], then gf(x) = 2x # 0, and, by the Theorem 4.14, g is 1-1 on [8,10], g- is differentiable on [64, 1001, and (g-')'(g(x)) = 1/g'(x). It is customary to write the inverse of g as & where, of course, & denotes the positive square root of x. Now, g([8, 101) = [64, 1001, and, if h(x) = fi = g-l(x), we have

'

Also,

for each x E [64, 1001, by Theorem 4.14. Suppose one wants to approximate &6 = h(80). Since h(81) = 9 is easy to calculate, we shall try to use this result to come close to By the Mean-Value Theorem,

m.


for some Q E [80, 811. Thus,

Although we do not know the exact value of xo, we do know that 8 < fi < 9. We may conclude that

Example 4.I0 We next consider the functionf (x) = tanx with domain (- 5 , 5) . Assume that f is differentiable,f' ( x ) . sec2x, and sec2x 0 for all x E (- $, 5). Further assume the identity sec2x = l +tan2x. Since fl (x) # 0 for all x E (- $, :), then f is 1-1 and has an inverse g, usually written g(x) = Arctan x. By the Inverse-Function Theorem, g is differentiable and

+

Of course, this is the usual formula for the derivative of Arctan x.

EXERCISES 4.1 THE DERIVATIVE OF A FUNCTION Let (%,yo) be an arbitrary point on the graph of the function f(x) = 2. For # 0, find the equation of the line tangent to the graph off at that point by finding a line that intersects the curve in exactly one point. Do not use the derivative to find this line. Prove that the definition of the derivative and the alternate definition of the derivative are equivalent. Use the definition to find the derivative of f(x) = &, for x zero? Explain. Use the definition to find the derivative of g(x) = 2.

> 0.

Is f differentiable at

Define h(x) = 2 sin $ fork # 0 and h(0) = 0. Show that h is differentiable everywhere and that ht is continuous everywhere but fails to have a derivative at one point. You may use the rules for differentiating products, sums, and quotients of elementary functions that you learned in calculus. Suppose f : (a, b)

-,R

is differentiable at x E (a, b). Prove that

lirn f (x + h) -f (x - h) h+o 2h exists and equals ft(x). Give an example of a function where this limit exists, but the function is not differentiable.

-

Exercises 133

A function f : (a,b) R satisfies a Lipschitz condition at x E (a,b) iff there is M > 0 and e > 0 such that lx - yl < e and y E (a,b) imply that 1f(x) -fQ)l 5 Mlx - yl. Give an example of a function that fails to satisfy a Lipschitz condition at a point of continuity. Iff is differentiable at x, prove that f satisfies a Lipschitz condition at x. A function f : (a,b) + R is said to be uniformly differentiable iff f is differentiable on (a,b) and for each E > 0, there is 6 > 0 such that 0 < Ix - yl < 6 and x, y E (a,b) imply that

Prove that iff is uniformly differentiable on (a,b), then f' is continuous on (a,b). Suppose f : (a,b) -,R is continuous on (a,b) and differentiable at xo E (a,b). Define

Prove that g is continuous on (a,b). Supposef , g, and h are defined on (a,b) and a < a < b. Assumef and h are differentiable at a,f (a)= h(xo),and f (x) 5 g(x) 5 h(x) for all x in (a,b). Prove that g is differentiable at a and f ' ( a ) = g t ( a ) = h t ( a ) . (This problem was suggested by Clyde Dubbs at New Mexico Institute of Mining and Technology in Socorro, New Mexico.)

4.2 THE ALGEBRA OF DERWATZVES Prove f : ( 0 , l ) --+ R defined by f(x) = t/2x2 - 3x + 6 is differentiable on ( 0 , l ) and compute the derivative. Suppose f : [a,b] + [c,4, g : [c,4

--, [p, q], and

h : [p, q] -,R, with f differentiable at

a E [a,b],g differentiable at f (a),and h differentiable at g(f (a)). Prove that h o (g of) is differentiable at a and find the derivative. Suppose f : [ a h ] -, [ c , 4 and g : [ c , 4 respectively. Suppose

+

R are differentiable on [a,b] and [ c , 4 ,

are also differentiable on [a,b] and [c,4, respectively. Show that (g o f l : [a,b] -+ R is differentiable and find the derivative. Supposef :R -,R is differentiable and define g(x) = 2 f (2).Show that g is differentiable and compute g'. Ddme f (x) = J xfi the derivative.

for x 2 0. Determine where f is differentiable and compute

4.3 ROLLE'S THEOREM AND THE MEAN-VALUE THEOREM Definef : [O, 21 + R by f (x) = d m theorem and find c such that ft(c) = 0.

. Show that f

satisfies the conditions of Rolle's

Define f : R + R by f(x) = 1/(1+2).Prove that f has a maximum value and find the point at which that maximum occurs. Rove that the equation 2 - 3x + b = 0 has at most one root in the interval [- 1,1]. Show that cosx = 2 + 2 + 4x has exactly one root in [0,q] .


Supposef : [O, 21 -,R is differentiable,f (0)= 0,f ( 1 ) = 2, and f (2) = 2. Rove that 1. there is cl such that f '(el)= 0, ) 2, and 2. thm is c2 such that f ' ( ~ 2 = 3. there is c3 such that ft(c3) =

i.

Let f : [O, 11 4 R and g : [O,l] -+ R be differentiable with f (0)= g(0) and f '(x)> g t ( ~ ) for all x in [0,1].Rove that f (x) > g(x) for all x in (0,lJ. Use the Mean-Value Theorem to prove that nyn-'(x - y)

5 x" - yn 5 nS-'(x - y)

i f n r landOLy<x. Use the Mean-Value Theorem to prove that 1 d i G < ~ + ~forh h > ~ .

Generalize Exercise 23 as follows: If 0 0, then show that

< 1 +ph.

You may assume the usual rules about differentiating powers.

0 such that

with f (x) = g(x) for all x E D nE n [m - €,a + €1. Prove that f is differentiable at a iff g is differentiable at a.


PROJECT 4.1 The purpose of this project is to examine the Newton-Raphson method for solving an equation of the form f (x) = 0, to establish convergence of the iterative process under certain equations, and to give estimates of the speed of convergence. Suppose f : [a, b] -* R is twice differentiable; that is, f and f f are both differentiable on [a, b]. Suppose further that there are m and M, both positive, such that Ift(x)( ZmandO < (fl'(x)( 5 MforallxE [a,b]. Finally,supposethatf(a)f(b) < 0. Define a sequence as follows: al=aiff'f" 0 on [a, b]. 4. The recurrence relation for {an)zl shows that f (an) =f '(an)(an- an+l). Substitute this in Taylor's formula to obtain the inequality

The following is a slight mhification of the definition of the derivative of a function due to Caratheodory.

DEFINITION Let f : D --+ R with xo an element of D that is also an accumulation point of D. Then f is differentiable at xo iff there is a function k : D + R such that k is continuous at xo and for all x E D,

1. Prove that the preceding definition is equivalent to the one given in this chapter. 2. Use the definition of Caratheodory to prove the chain rule. 3. Use the definition of Caratheodory to prove the inverse function theorem.


PROJECT 4.3 In your calculus course, you learned to graph functions, and symmetry was a useful tool in that task. In this project you will examine the graph of a cubic polynomial and look for symmetry. In what follows, we will assume that every function has domain R; that is, f (x) is defined for all x. DEFINITION The graph of a function f is symmetric with respect to a point (xo,f (xo)) on the graph iff for all t, f (no) = i(f (xo + t) +f (xo - t)). Under these circumstances, we say (xo,f (xo)) is a point of symmetry for the curve. 1. If (0,O) is a point of symmetry for the graph of a function f , show that f is an odd function; that is, f(-x) = -f(x) for all x E R. 2. If (h, k) is a point of symmetry for the graph off, then by translating the axes so that the point of symmetry is the origin in the new coordinate system, the curve is now the graph of an odd function. Write this new function in terms off. 3. Now consider an arbitrary cubic polynomial function p(x) = d +b 2 + cx+d. Find a point of symmetry for the graph of this function by translating the axes in such a way that in the new coordinate system, the curve is the graph of an odd function. 4. Does the point of symmetry you found in part 3 have any other special properties? 5. Suppose (h, k) is a point of symmetry for the graph off, and f is twice differentiable at h. What can you say about ftt(h)? How does this relate to what you discovered in part 4? (Hint: you may want to look at some of the exercises in this chapter before answering this question.)

Chapter S

The Riemann Integral

he concept presented in this chapter is one with which the reader should have some familiarity-that of the Riemann integral. Beginning with an intuitive idea of the area under a curve, we shall progress to a deeper understanding of this concept. By considering a special case at the outset, we can see what is reasonable to try for a definition. Supposef : [a, b] --, R is bounded and f (x) 2 0 for all x E [a, b]. For the purpose of visualizing the notion of the area under the curve, it is worthwhile to assume for the moment that f is continuous. We want to determine some reasonable manner of assigning an area to the portion of the plane bounded by the lines x = a, x = b, y = 0, and the graph off. The approach will be to attempt to approximate the area by use of rectangles, whose areas are easy to compute, and then to use some type of limit process to arrive at our result. Because of the possible pathological nature off (if we don't assume continuity), it is reasonable to expect some difficulty with this limit idea. Choose points XO, xl,xz, . . .,xn such that a = xo < xl < < xn = b. For each i among 1, .. .,n, erect a rectangle Ri with the interval [xi- 1,xi] as base and with altitude such that the portion of the graph off for x E xi] lies in this rectangle. This is possible because f is bounded; in fact, we may choose the altitude of Ri to be

The region in question lies in the union of the rectangles Rl, . . .,R,, and these rectangles are nonoverlapping; hence, the area should be less than or equal to the sum of these areas-that is, less than or equal to

For the more conservative reader, we can try underestimating the area. This time, for each i, erect a rectangle Si with base xi] and with altitude such that

140 Chapter 5 The Riemann Integral

Figure 5.1

the rectangle Si lies entirely inside the region in question. Here we shall choose the altitude of Si to be The rectangles Sl, . . .,Sn are nonoverlapping, and all lie inside the portion of the plane bounded by the lines x = a, x = b, y = 0, and the graph off; hence, the sum of their

must be less than or equal to the sought-after area. See Figure 5.1. Let us temporarily refer to C:=,Mi(xi - xi- 1) as an upper s u m for f and to as a lower s u m for f. The upper sum clearly overestimates the Cy=lmi(xi area, and the lower sum underestimates it. Observe that, as one chooses more points in the interval, the upper sum decreases and the lower sum increases; hence, they should "converge" to the "area" to be found. (The words appear in quotation marks only because these notions have not been formalized in this setting.) Let us now formulate a policy concerning the limiting process to which we have made vague reference. Roughly speaking, we will insist that the upper and lower sums get closer and closer together as we choose more points in [a, b]. The discussion just presented is intended to prepare the reader for the precise formulation needed to give the full meaning of the Riemann integral.

Consider a < b. A partition P of [a, b] is any finite set {-,xl, . . .,xn) such that a = a < x l < * - < x n = b . IfPandQarepartitionsof[a,b] w i t h P ~ Q , t h e n Q i s said to be a refinement of P.

5.1 The Riemann Integral

141

Suppose f : [a, b] + R is a bounded function and is a partition of [a, b]. For each i among 1, . . .,n, define and Also define

and

We shall call U(P,f) an upper sum for f and L(P,f) a lower sum for f. Now f is assumed bounded on [a, b], so there is a real number M such that -M f(x) 5 M for all x E [a, b]. Thus, in particular, for any partition P of [a, b], -M(b - a) L(P,f) 5 U(P,f) M(b - a). Let

0 there is a

We need to show that

By Theorem 5.1,

O
0. There is a partition P of [a,b] such

el = ~ n [ a , ca ]n d Q 2 = ~ n [ c , bThen ].

Thus, by Theorem 5.2,f E R(x) on [a,c] and f E R(x) on [c,b]. We have seen that linear combinations of integrable functions are integrable. Now we seek other operations that preserve integrable functions. As has been shown already, continuous functions seem to behave especially well with respect to integration. The next theorem should help further this opinion.

5.11 THEOREM Suppose f : [a,b] -. S, f E R(x) on [a,b], and 4 : S -,R is continuous with S compact. Then 4 of E R(x) on [a,b].

Proof Choose E > 0. Let K = sup{l4(t)l : t E S ) , and let E' > 0 be such that ~ ' [-ba + 2K] 5 E. By uniform continuity of 4 on S, there is 0 < 6 < E' such that s, t E S and 1s - tl < 6 implies 1 +(s)- &t)l < 8 . By the integrability of f on [a,b], there is a partition P of [a,b] such that

Assume P = {xo,xl,. . . ,x,), and let

and

NOW,for i E A and S , t E [ x ~ - ~ , xK(S) ~ ] ,-f(t)l < 6, SO

,


hence, 6 C ( x i - xi- 1) I C [ M I(f)- mi(f)I(xi - xi- 1)

Thus

CiE-(~i -

5 6. It now ~ O ~ ~ Othat WS

Thus, by Theorem 5.2,

+ of E R(x) on [a,b].

It is interesting to note that this last theorem does not follow the pattern we have observed before. It is true that the composition of continuous functions is continuous and the composition of differentiable functions is differentiable; one might conjecture that, iff : [a,b] + [ c , 4 and g : [c,dJ -, R are such that f E R(x) on [a,b] and g E R(x) on [c, 4 , then g of E R(x) on [a, b]. To see that this is not the case, we shall consider an example. Example 5.8 Define f : [O, 11 -r R by f(x) = 0 if x is irrational, and f(x) = $ if x = 4 with p and q relatively prime nonnegative integers, q # 0. It has already been shown that f E R(x) on [O,l). Define g : [O, 11 -r R by g(x) = 1 if 0 < x 5 1 and g(0) = 0. It is left to the reader to show that g E R(x) on [a, b]. Let h = g 0 f. Then if x is irrational, h(x) = 0, and if x is rational, h(x) = 1. As previously observed, h is not integrable on [O, 11. W

5.12 THEOREM Iff : [a,b] -, R, g : [a,b] -,R, and f , g E R(x) on [a,b], then fg E R(x) on [a, b] and 1f 1 E R(x) on [a, b]. Also,

5.5 Algebra of Integrable Functions 163

Proof Since f and g are integrable, both functions are bounded; hence, there is M > 0 such that im f and irn g are contained in [-M, MI. Define 4 : [-2M, 2M] --+ R by 4(x) = 2.Clearly q5 is continuous, and so 4 o ( f + g) = ( f + g)2 E R(x) on [a, b] by Theorems 5.11 and 5.9. Similarly, 4 o ( f - g ) = ( f - g)2 E R(x) on [a,'b]; hence,

We saw in Chapter 3 that the function #(x) = 1x1 is continuous; hence, q 5 0 f = I f 1 E R(x) on [a,bl. N o w , f ( x ) I Ifl(x) and -f(x) 5 Ifl(x) for all x E [a, b], so [ f d w < [ ~ f / h

and

-[fk=[-fh 0 such that 1x0-yI < 6 with y E [a, b] implies 1f(xo) -f(y)I < 5. Thus, if 0 < 1x0-yl < 6,

Therefore, F is differentiable at xo and F' (xo) =f (xo). We can use Theorem 5.14 to give an easy proof of a slightly weaker version of the Fundamental Theorem of Integral Calculus.

THEOREM Supposef : [a, b] -,R is differentiable with f' continuous on [a,b]. Then f' E R(x) and

Proof Define F : [a, b] -, R by F(x) = J' f '(t) dt. Since f is continuous on [a, b], then F is differentiable on [a, b] and F' (x) = f' (x) for all x E [a, b]. This means there is a constant K such that F(x) = f (x) + K for all x E [a,b]. We evaluate K as follows: 0 = F(a) =f (a) + K, so K = -f (a). Therefore,

r

Example 5.9 Consider the function f(x) = 2 sin for x # 0 and f(0) = 0. Now f is differentiable on [O,1] and f '(x) = 2.x sin - cos for x # 0 and f '(0) = 0. The function f ' is continuous on (O,1] and bounded; hence, by Exercise 24, f ' is integrable on [0, 11. However, the version of the Fundamental Theorem of Integral

166 Chapter 5 The Riernann Integral

Calculus given above cannot be used because f is not continuous on [O,l]. But Theorem 5.8 can be used to determine that

i1

fl(x)dx =f(1) -f(0) = sin 1.

! !.

Example 5.10 For x > 0, define L(x) = [I! dt. Since f (t) = is continuous for t > 0, L is differentiable for t > 0 and Lf(t) = The reader may well remember seeing the natural logarithm function defined in this way. See the project at the end of the chapter for a more detailed examination of this function. Example 5.Il For each x > 0, define f(x) = ~ ~ ~ c dt.o We s ?can consider f as the composition of two functions, g(x) = fi and h(x) = cos? dt. Then f = h o g, and we may use the Chain Rule to find f ':

Sf

= [cos(~(x))~]~'(x) = (cos x)

(a

As pointed out in the discussion of the Fundamental Theorem of Integral Calculus, the reader's experience with integration has been largely that of being given a function f on [a, b] to find a differentiable function g on [a, b] such that f = g'. Iff is defined on [a, b] and g is a differentiable function on [a, b] such that g' =f , then g is called an antiderivative or primitive off. If f is continuous on [a, b], then Theorem 5.4 guarantees that f E R(x) on [a, b]. Then Theorem 5.14 assures us that f has a primitive, namely g(t) = f dx. It is not true that every integrable function has a primitive, since our results from Chapter 4 show that, if f = g' on [a, b] for some differentiable function g, then f must satisfy certain conditions. For example, the function f , defined by f(x) = 0 for 0 5 x 1 , the function f ( x ) = f is integrable on [ l , a ] and on [b,ab]. Use the results of Section 5.3 to show that f dr = J ' f dr.

JP

Exercises 171

for all n. Prove {an)gl converges to

So f dt. 1

Suppose f : [a, b] + R is bounded and for each c > 0 there is a partition P such that for any refinements Qland Q2 of P, regardless of how marked, lS(Ql,f) - S(Q2,f)l < c. Prove that f is integrable on [a, b]. (Note that this applies to different markings of the same partition.) Suppose f is integrable on [-b, b] and f is an odd function--that is, f(-t) = -f(t) for f dr = 0. If f is even-that is, f (-1) = f (t) for all all t E [-b, b]. Rove that t E [-b, b]--prove that

fb

Suppose that f : R -,R is periodic and integrable on every closed interval. If p is the period off, prove that for any a E R,

5.4 THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS use the Fundamental Theorem of Integral Calculus to compute the following: a. b.

C.

r2

x sin x2GX!

~ e f i nfe : [0,2] -r R by f ( x ) = 2 x - 2 for0 s x 5 1 and f ( x ) = ( ~ - 2 ) for ~ 1 < x 5 2. Prove that f is integrable on [O, 21 and find the integral of f over [O, 21. Do not use Theorem 5.10, but rather find the integral by methods similar to those used in the proof of Theorem 5.8.

5.5 ALGEBRA OF INTEGRABLE FUNCTIONS *IS. Suppose f and g are differentiable on [a, b] and f and g' are integrable on [a, b]. Prove that f'g and g'f are integrable on [a,b] and that

Of course, this is the integration-by-partsformula.

19. Suppose f E R(x) on [a, b] and ) is bounded on [a, b]. Prove that ) E R(x) on [a, b].


20. Supposef E R(x) on [a,b] and f (x) 2 0 for all x E [a,b]. Rove that

fl E R(x) on [a,b].

21. Use Exercise 18 to calculate

x cos x dx. You may assume what you need concerning the derivatives of the trigonometric functions.

22. Suppose f is continuous on [0,11. Define gn(x) = f(?)

(5; g n Q dx} 23. Define

for n = 1,2,. . . . Prove that

converges to f(0). n=l

= 1+$+f

+

+ -

slnf

dt. Prove that { 7 n ) z 1converges.

4 4 . Suppose g : [a,b] -t R is bounded and continuous except at X I , . that g E R(x) on [a,b].

. .,xn E [a,b]. Prove

*25. Suppose {an)zl is a sequence of members of [a,b] converging to xt) in [a,b]. I f f is bounded on [a,b] and continuous on [a,b] except at xt) and the points of the sequence {an),"l, prove that f is integrable on [a,b]. 26. Prove that

[Hint: Theorem 5.9(ii) and carefully chosen functions should do the trick.] 27. Supposef and g are integrable on [a,b]. Define h(x) = max{f (x),g(x)). Prove that h is integrable on [a,b].

5.6 DERIVATIVES OF INTEGRALS 28. Supposef : [0,11 + R is continuous and that f(x) = 0 for all x E [0,11.

Sox f (t)dt = Sxl f (t)dt for all x E [0,11. Prove

29. Suppose f and g are continuous on [a,b] and c E [a,b] such that f (c)= g(c).

sab

f (x)dx =

sab

g(x)dx. Prove that there is

30. ~ i n fd where f is defined on [O,1] as indicated:

31. Let f : R + R be continuous and 6 > 0. Define g(t) = that g is differentiable and compute g'.

f (x)dx for all t E R. Prove

*32. Suppose f : [a,b] -t R is continuous and g : [ c , a + [a,b] is differentiable. Define F(x) = )J ': f (t)dt for x E [c,4. Rove that F is differentiable and compute F '.

Exercises 173

-

5.7 MEA~-VALUE AND CHANGE-OF-VARIABLE THEOREMS 33. Suppose f : R R is continuous and has period p, so that f(x + p ) = f(x) for d l x E R. Show that

JXx+'

LX+'

f (t)dt is independent of x in that, for all x, y,

f (t) dt =

f (0dt.

Lp[

f (x + a) -f (x)] dx = 0 for any real number a. Conclude that for any Show, then, that real number a, there is x such that f (x + a) =f (x).

1

I

1

*34. Prove the following variation on Theorem 5.18. Assume that 4 : [a,b] + R is differentiable, 1-1, and increasing with +(a) = c and &b) = d. Iff : [c, dJ + R is integrable on [c, dl and (f o 4)4' is also integrable on [a, b], then

35. Use Theorem 5.18 to evaluate the integrals:

[You may want to consult your calculus book before trying (c) and (d).]

36. Use Theorem 5.18 to establish the result in Exercise 11. 37. Use Theorem 5.18 to establish the result in Example 5.6. 38. Assume f : [a, b] -,R is continuous, f (x) 2 0 for all x E [a, b], and M = sup{f(x) :x E [a, b]). Show that

converges to M.

39. For each positive integer n, define

[(t)'+(f)'+--+

a,, = i n Find the limit of the sequence

(:)2]

174 Chapter 5

The Riernann Integral

40. For each positive integer n, define 7r 27r an = 1[sin - + sin - + n n n

Find the limit of the sequence

+ sin n

{an)zl.

41. Iff and g are integrable on [a, b], show that

[Hint: For all a and

p, sab(af + pg)2 dx 2 0. k

t

=

subdx g2

and

p = - sabfg &.I

42. Use Exercise 41 to prove the Minkowski inequality-that is, iff and g are integrable on [a, b], then

PROJECT 5.1 The purpose of this project is to use the integral to define the natural logarithm function, logx, and to develop some of the properties of that function. Define logx = f dt for 0 < x. For convenience, we will write logx = L(x).

sI[

1. 2. 3. 4.

Prove that L is continuous, differentiable, and 1-1. Prove that L is increasing. Prove that L(ab) = L(a) + L(b) for all a, b > 0. Prove that L(Y) = nL(x) for all x > 0, n E 2.

Comment: One possible approach to (3) is to define f(x) = L(m) and show that ft(x) = L'(x); hence, f (x) = L(m) = L(x) + k, and then show that k = L(a). A similar approach will supply a proof for (4). 5. Prove that L (2) = L(a) - L(b) for a, b > 0.

(i)

6. Prove that im L = R. Since L = -L(x) for 1 < x, and L(l) = 0, it suffices to prove that for any y > 0, there is x > 1 such that L(x) = y. Since L is continuous, you need to show that there is n E J such that L(n) > y and then use the Intermediate-Value Theorem. It might be helpful for you to look ahead to , Example 6.3.

We know that

Projects

175

converges and we choose to call the limit e. Now we will show that L(e) = 1. It suffices to show that

converges to 1. Do you see why?

i)

i)

Prove that for each n E J, 5 5 n~ (1 + 5 1. YOU will want to write L (1 + as an integral and use Theorem 5.9(iii). Finish the proof that L(e) = 1. Let E be the inverse of L (remember, L is 1-1). Since im L = R, then dom E = R and im E = {x : x E R, x > 0). Prove that E is continuous, differentiable, 1-1, and increasing. Find El. Be careful; this is a bit tricky. You cannot use Theorem 3.12 directly because {x : x E R, x > 0) is not compact. However, you can modify the proof of 3.12 to meet your needs. Similar observations are appropriate concerning Theorem 4.14. On the other hand, you might find some assistance in the miscellaneous exercises in Chapters 2, 3, and 4. Prove that E behaves like an exponential function-that is, E(x + y) = E(x)E(y) for all x, y E R. We have some understanding of Y where x is any real number greater than zero and r is any rational number. Now we can define xY for any real number x > 0 and any real number y. Define XY = E(yL(x)) or, in more familiar notation, xY = e~logX. Show that f (x) = X is differentiable and find f (x is assumed to be a positive real number).

PROJECT 5.2 The purpose of this project is to develop the midpoint approximation rule for approximating Riemann integrals. The idea is to approximate the integral by using Riemann sums with the "marking points" chosen to be the midpoints of the subintervals. Assume that a < b and f is Riemann-integrable on [a, b]. For each positive integer n, define

where h = :(b - a). The theorem you wish to prove is the following: THEOREM Let f be twice-differentiable on [a,b] and f I' continuous on [a, b]. Then, given a positive integer n, there is c in [a, b] such that

where h and

S,(nare as described above.

176 Chapter 5 The Riernann Integral

Proo/ Foreachkamong1,2, ...,n , d e f i n e a k = a + ( k - i ) h w h e r e h = i ( b - a ) . Define

Prove that gk is twice differentiable and g;((t) =f '(ak + t ) -f '(ak - t). Apply the mean-value theorem to f on the interval [ak- t , ak + t ] to find 1 rr X ~ Jsuch that ft'(xkJ) = z(gk(t)). Let m = inf(ft'(x) : x E [ a , b ] ) and M = sup{f "(x) :x E [a,b ] ) . Conclude that for all 0 5 t 5 i h . Integrate the inequalities in part 2 to obtain the inequality

Show that

subf

(x) dx - &(f) =

EL, gk ( I h ). Conclude that

Using the fact that h = i ( b - a ) and the continuity off", complete the proof. You really didn't need the continuity of f'' in part 5 . What theorem could you have used instead of invoking the continuity of f''? Where in the proof did you use the continuity of f'' (other than in part 5)?

Chapter

6

Infinite Series

ost students of mathematics who have completed calculus feel competent in the techniques of differentiation and integration, but they may panic at the mention of infinite series. However, the notions of convergence and divergence of infinite series are actually quite easy to assimilate if presented properly. The question of convergence of an infinite series depends on the convergence of a certain sequence, so a brief review of the material concerning sequences, found in Chapter 1, may help you prepare for what is to come. Finite sums of real numbers are familiar objects; our purpose here is to determine when we should assign a real number to a string of symbols such as al+az+a3+. -.The approach is the usual one-that of looking at the partial sums of this "infinite sum."

DEFINITION An infinite series is a pair ({a,)z,, {S,),",) where {an)gl is a. sequence of real numbers and S, = ak for n = 1,2,. . .. The number a, is called the nth term of the series, and S, is called the nth partial sum of the series.

xkl

In keeping with the more customary notation, we may denote the infinite series ({a,),",, {Sn)zl) by a, or by a1 + a2 + + a, + *. It is often convenient to index the terms of an iiifinite series beginning with an integer other than 1. As our discussion unfolds, it will be clear that questions of convergence are independent of whether we index the terms beginning with n = 1 or n = p for some other integer p.

xzl

DEFINITION The infinite series ({a,)z,, {S,)z,) converges if the sequence {S,)z, converges. If {Sn)zl converges to S, we write

178 Chapter 6 Infinite Series

xzl a,.

The number S is called the sum of the infinite series does not converge, we say that the infinite series

{Sn)zl

If the sequence

znzla, diverges.

xE1

Perhaps a word of apology should be offered for using a, as a name both for an infinite series and for the real number that is the limit of the sequence of partial sums when the series converges. However, this abuse conforms to convention and the reader's experiences. Note that, if the series does not converge, we do not use an to denote a real number. Let us first consider an infinite series an where an = 0 for n N + 1. Then, for n 2 N, S, = SN; hence converges to SN = a1 + a2 + + a ~ In. this case, the infinite series is in reality a finite sum, a fact that should certainly be no surprise to the reader. Suppose now that an > 0 for all n; then {Sn)glis increasing. Hence, in order for C z lan to converge, it is necessary that {Sn)glbe bounded. NOWSn+1 = Sn +an+lfor each n; SO for {Sn)glto be bounded, it is necessary that the terms a, get small "fast enough" as n gets large. To illustrate what we mean by "fast enough," we shall consider several examples.

zE1

{Sn)zI

xE1

>

Example 6.1 Consider the infinite series CE1log usual properties of the function logx.) Now

(9) (We. shall assume the

The sequence {Sn)glis unbounded and, hence, is not convergent. Note that converges to 1, so by the continuity of the log function,

{y)zl

converges to log 1 = 0. In this case, the nth term of the series gets small as n gets large, but not "fast enough."

Example 6.2

A.For each n,

Consider now the series CE1

since for all k,

Clearly {Sn)gl converges to 1, and we write

6.1 Convergence of Infinite Series

179

In Example 6.2, the nth term of the series converged to zero "fast enough." Unfortunately, there are no simple criteria for determining whether or not a series converges merely by examining the nth term of the series. In this chapter, we present some standard methods for determining the convergence of infinite series. Suppose now that an is convergent, which means that { S , ) ~is, convergent. By our previous discussion of sequences, { S n ) z l is convergent iff it is Cauchy.

x;,

6.1 THEOREM The infinite series

zzla, converges iff for each

E

> 0 there

is N such that if n 2 N and p 2 0, then

{Sn)zl

Proof Suppose C z la, converges. Then the sequence converges and, hence, is Cauchy. Choose E > 0. There is N such that m, n 2 N implies that ISm-S,I < E. Now ifn ? N + 1 andp 2 0, thenn-1 2 N andn+p 2 N; hence,

Assume the condition holds, and choose E > 0. There is N such that for n 2 N andp 2 0, lan+an+l+*-+a,+,l < E. Choose r 2 n 2 N. Then

Hence, { S n ) E l is Cauchy and the series converges. Let us pause to ponder the content of Theorem 6.1. First, the convergence or divergence of an infinite series is independent of whether we index the terms beginning with n = 1 or with n = p for some other integer p. Second, the convergence is unaffected by changing finitely many terns of the series; although if the series converges, the limit of the partial sums may be altered. If C z lan converges and k is a positive integer, then a, converges; in fact, given E > 0, there is N such that

xEk

for all p 2 0, but

is the mth partial sum for the convergent infinite series CzNan;hence, I 6.2 COROLLARY

If

xzl an converges, then {a,):,

xE4 a, 1 5

E.

converges to zero.

Proof Choose E > 0. By Theorem 6.1, there is N such that, for n 1 N and p 2 0, we have la, + a,+l + + an+pl< E . In particular, for p = 0, lanl < a Hence, { a n ) g lconverges to zero.


Corollary 6.2 is presented with mixed emotions. Its result is very useful because it allows one to observe quite easily that certain series diverge-namely, those series an where { a n ) g l does not converge to zero. Unfortunately, the result is fiequently misused, for many students assume that the converse is true-that if {an):, converges to zero, then an converges. This converse is definitely false, as we have already seen by considering

xGl

xE1

Here

converges to zero, but the series

diverges. The following example emphasizes this fact and will be of use later.

Example 6.3 Consider the infinite series CE15, often referred to as the haris increasing, and we monic series. Since > 0 for each n, the sequence shall show it is unbounded. If n = 2k, then

4

Thus,

{Sn)zl

5

{Sn)glis unbounded, so xE15 is divergent.

In the study of infinite series, it is useful to have many examples at hand. Such examples help to develop the intuition and, as will be seen later, may sometimes be used in testing infinite series for convergence. Before continuing further, we shall present one more classic example.

Example 6.4 For n 0,

>

The series

x> P is called the geometric series with ratio r.

6.1 Convergence of Infinite Series

181

+

for all r 1. Now {P+~):, converges iff -1 < r 5 1; and if lrl < 1, {r+l)El converges to zero. For r = 1, Sn = 1 + 1 + + 1 = n + 1; hence, {Sn)gl is not convergent. To summarize, f l converges to for lrl < 1 and diverges for Irl 1 1-

x>

&

Example 6.5 This example illustrates the fact that we have had experience with infinite series before encountering the idea in calculus. Of course, we are referring to repeating decimals. Some rational numbers have a terminating decimal representation, for example, = 0.25. But for some rational numbers there is no terminating decimal representation, and we commonly write and refer to them as repeating decimals. For example, we would write $ = 0.18 18T8, with the bar over the digits 18 indicating that those digits repeat forever! More compactly, we would write $ = 0.T8, which implies that

a

This means that partial sum is

Since 0 5

$ is the sum of the infinite series czlm. l8 For each n, the nth

& < 1, the sequence of partial sums converges to

and our faith in repeating decimals is reinforced. Notice that the series for the repeating decimal is just a multiple of a geometric series.

In Example 6.5, we were dealing with a multiple of a series already examined in detail, the geometric series. It would have been helpful to have had the following theorem at our disposal.

Proof The proof is left as Exercise 10.


zzl($ + &). NOWthe series xz& converges to $ (see Example 6.4) and zzl converges to 1 (see Example 6.2), so zzl(6+ A), which is zzl(a + 5 A)converges to ($)+5(1) = 6. Notice that we used the fact that xs $ and zzl ,&are the same series. Example 6.6

Consider the series

n(n+ 1)

6.2 ABSOLUTE CONVERGENCE AND THE COMPARISON TEST So far, in most of the examples we have considered, the terns of the infinite series were nonnegative. Let us consider now the infinite series

Shortly it will be shown that this series converges; in fact, this will follow from the alternating series test, which the reader may recall from elementary calculus. It was remarked earlier that if a, 2 0 for all n, the infinite series a, converges if the nth term tends to zero "fast enough." Now is divergent; hence does 1)"; does converge, so it is not converge to zero "fast enough." However, clear that the convergence depends somehow on the fact that the terms alternate in sign. We shall use the following definition to investigate this behavior further.

CzlA

Czl

zgl(-

,:)t{

xzla, converges absolutely iff zzllan1 xEllanl diverges, then xE1a, is said

DEFINITION An infinite series converges. If an converges but to converge conditionally.

xg1

Thus,

czl(- 1 t converges conditionally. The series

converges absolutely since we have already demonstrated that 00

1

C n(n + I) n= 1

converges. Every absolutely convergent infinite series automatically is convergent.

xzlan is > 0. Since xzlan is absolutely convergent, xzllan1 con-

6.4 THEOREM Suppose convergent.

Czla,

is absolutely convergent. Then

Proof Choose E verges. Thus, there is N such that n > N and p 1 0 implies that

6.2 Absolute Convergence and the Comparison Test 183

But la, +an+,+ converges.

+an+,l 5 lan1 +

+ lan+,l < E; hence, by Theorem 6.1, C:,

Some of the distinctions between absolute convergence and conditional convergence shall be brought out as our tale unfolds. Keep in mind that an infinite series CE1an of nonnegative terms converges if the terms get small "fast enough." Thus, the infinite series Czlan converges absolutely if the sequence {lanl)gl converges to zero "fast enough." This notion of "fast enough" is quite vague, and we shall try to make this notion more meaningful. To this end, consider two infinite series Czla,, and bn, where an 2 0 and bn 2 0 for all n. If bn converges, then the sequence {bn)El converges to zero "fast enough." Suppose now that there is N such that for n 2 N, 0 5 an 5 bn. Then it seems plausible to conjecture that {a,):, converges to zero as fast as {bn)El,which is "fast enough." These remarks yield the basis for Theorem 6.5, the "comparison test." In other words, we shall try to determine convergence or divergence of a series by comparing it to one whose behavior is already known. Because the comparison must take place between infinite series of nonnegative numbers, the comparison test will yield information only concerning absolute convergence. The success of this test depends on the user's stock of examples. Therefore, it is important to begin to build this stock. The observant reader will note that the comparison test is the basis for the limit-comparison, ratio, and root tests to be presented later. In the latter two cases, the series to be examined are compared to the geometric series P.

x:,

x:,

I ,

1

xz

6.5 THEOREM (Comparison Test) Suppose series with b, 2 0 for all n. Then

t

4

i. If

+I

a,

xE1an and z;, bn are infinite

x:, bn converges and there is No such that n 2 No implies lanl 5 bn,then

xE1an converges absolutely.

C:, b, diverges and there is No such that n 2 No implies bn 5 lanl,then Czl la, ( diverges.

ii. If

(See Exercise 16 for a generalization of this test.) Proof Suppose the hypotheses of (i) hold. Thus, CE1bn converges and there is No such that n 2 No implies lanl 5 bn. Since bn converges, given E > 0, there is N such that n 2 N and p 2 0 imply that bn+ bn+,+ + bn+, < C. Let

xE1

N1 = max{N, No).

czl

an converges absolutely. Thus, Czl lan1 converges, and . The proof of (ii) can be obtained most easily by using the result just obtained. If la, 1 converges and there is No such that n 2 No implies 1bn1 = bn 5 la. 1,

,:x


then since lan1 2 0, the result of (i) guarantees that to assumption. Thus, (ii) holds.

xzllbn1 converges, contrary

The next two examples illustrate Theorem 6.5.

Example 6.7

Consider the series

We shall choose to compare this series with the series

because this is one of the few examples we know much about and because for all n,

Hence, by Theorem 6.5,

converges since

converges. Note that because convergence. Example 6.8

and

CEI

,h25 0 for all n, convergence is the same as absolute

Consider now the infinite series

diverges. Hence,

a. En=, J;; For all n 2 1, 00

6.2 Absolute Convergence and the Comparison Test 185

does not converge absolutely; but, as we shall see later, it does converge conditionally.

The next few results are intended to populate, as painlessly as possible, our stock of examples of convergent infinite series. 6.6 THEOREM Suppose { a n ) g l is a sequence of nonnegative terms such that an 2 an+lfor all n. Then CE1an converges iff C& 2ka2kconverges.

Proof Define

and

0

Sn is the nth partial sum of Now if n 5 2k,

CE1a,, and Tk is the kth partial sum of C& 2ka2k.

since an 2 an+lfor d l n. On the other hand, if n 2 2k,then

that 2Sn 2 Tk. Now, if an converges, then the sequence {Sn)glis increasing and it converges to S = supsn. Since 2k 0 for all k, it suffices to show that {Tk)& is bounded in order to prove that 2ka2kconverges. Choose k. Then there is n such that n 2 2k;hence Tk 5 2Sn 5 2s. Thus, {Tk)gl is bounded, and hence 2ka2kis convergent. Suppose now that C& zka2k is convergent. Then { T ~ ) E ,converges to T = supTk. As before, it suffices to show that {Sn)glis bounded in order to prove that an converges. Choose n. There is k such that 2k 2 n. Then Sn 5 Tk 5 T. Thus, is bounded and hence Czl an converges. SO

xzl

>

x&

xzl

{Sn)zl


zzl

The infinite series of the form $, p is a real number, are called p-series. Theorem 6.6 allows us to classify these easily.

czl1/nP converges iff p > 1.

6.7 THEOREM

Proof If p 5 0, then I f p > 0, then CEl $ converges iff

5

{$):, does not converge to zero, so Czl $ diverges. > 0 for all n, and Theorem 6.6 may be used. Thus,

converges. But

0

and the series in other words, iff p

is a geometric series that converges iff 2 ' 7

< 1, or,

> 1.

1 . The nth term of this series Examplc 6.9 Consider the series Czl (2n- 1)(2n+1 may be rewritten as ,&, and we want to compare this to a series that is familiar to us. This term almost looks like & but not quite. We suspect the series will converge because of Theorem 6.7 with p = 2. Follow along:

1 4n2 - 1

1 1 5 4n2 - n2 3n2

for all n 2 1.

czl&

converges by Theorems 6.7 and 6.3; hence, so does Czl The series by Theorem 6.5. Since all terms are positive, the convergence is absolute.

6.8 THEOREM (Retio Test) Let Then

xzla, be an infinite series of nonzero terms.

i. If there is a real number q such that 0 that n 2 N implies

1715 then

&

4,

xzla, converges absolutely.

< q < 1 and a positive integer N such

6.3 Ratio and Root Tests 187

ii. If there is No such that n 2 No implies

Proof Suppose the hypothesis of (i) holds. Then

Since 0 < q Thus, for n 2 N, lan1 5 @laNlqmN.

< 1,

x> @ converges, and

xE1

converges. Thus, by the comparison test, an converges absolutely. Suppose the hypothesis of (ii) holds. Since lanl > 0 for all n and a 2 1 1 for n 2 No; hence { a n ) g ldoes not' converge for n No, we have lan1 5 to zero. Thus, an does not converge.

>

I I

xgI

xE1

If anis an infinite series of nonzero terns such that the sequence is convergent, the next theorem follows from Theorem 6.8. This is the form in whiyi the "ratio test" usually occurs in calculus books. 6.9 THEOREM Suppose that the sequence

{I 1}

xzl an is an infinite series of nonzero terms such

00

n=l

converges to L. men,

i. If L < 1, the series converges absolutely. ii. If L > 1, the series diverges. iii. If L = 1, no conclusion concerning convergence can be made. Proof Suppose L

< 1-. Then L
1, then there is N such that n 2 N implies

hence, by Theorem 6.8,

xz,a, diverges.

Example 6.10 To verify that we gain no information when L = 1, we shall give an example of a convergent series for which L = 1 and a divergent series for which L = 1. Consider first the infinite series 1In2. By Theorem 6.7 this series converges and

xgl

converges to 1. The series to 1.

xzIn obviously diverges, but again, { )zl convergesrn

Theorem 6.9 fails to give any information if

fails to converge. In this case, one may have recourse to Theorem 6.8, but that may also fail to shed light on the situation. To illustrate this point, consider the infinite series a, where an = $ for n odd and a, = for n even. For n odd,

xzl

&

and for n even,

Here the sequence

has a subsequence that converges to zero and a subsequence that is unbounded, so neither Theorem 6.8 nor 6.9 applies. The comparison test may be used to verify that this series converges. The "root test," which comes next, is more conclusive than the ratio test in that, if the root test gives no information concerning the convergence of a series, then neither does the ratio test. However, as we shall see by some examples,

6.3 Ratio and Root Tests 189

the root test may be harder to use in some cases. If this were not true, there would be no reason to present the ratio test.

xE1a, is an infinite series. Then i. If there is a real number q such that 0 5 q < 1 and a positive integer N such that qms q for n 2 N, then xE1a, converges absolutely. 2 1, then zzla, diverges. ii. If, for infinitely many n,

6.10 THEOREM (Root Test) Suppose

Proof Suppose there is a real number q such that 0 5 q < 1 and a positive q. Then for n 2 N, 1a.l 5 4". Since integer N such that for n 2 N, 0 < (I < 1, q" converges; hence, by the comparison test, a, converges absolutely. Suppose now that vLl2 1 for infinitely many n. Then for infinitely many n, la,l 2 1, so { a , ) z l does not converge to zero, and a, does not converge.

xzl

xzl

xzl

Example 6.11 Let us now attempt to apply the root test to the earlier e x a m p l e that is, a, where a, = $ for n even and a, = $ for n odd. Now C/ii;; = 21 for n even and C/ii;; = for n odd. Thus, for all n, C/ii;; 5 hence, by Theorem 6.10, the series converges absolutely.

xzl

1;

\

Let us pause to consider a few examples to make use of the root and ratio tests. Example 6.12

Consider the series

If we attempt to apply the root test, we are faced with considering the sequence whose nth term is

Rather than pursue this task further, let us try the ratio test. Now,

In Chapter 1, we observed that the sequence


converges with limit e, 2 < e < 3. Thus,

converges with limit

< 1. Therefore, by the ratio test, this series converges.

.

Let { a n ) s be any sequence of real numbers such that lanl 5 n. Consider -1 < r < 1 and the infinite series C z a,?. Since little is assumed about the sequence {an)%, it seems useless to attempt the ratio test. Rather, let us attempt to use the root test. Now

We appeal to Theorem 6.10. For each n, lanl 5 n; hence,

The sequence { f i ) : , verges to lrl < 1. Let

Then lrl

converges to 1; hence, since - 1 < r

< 1, {Irl C/ii)zlcon-

< q < 1, so there is a positive integer N such that n 2 N implies that

xs

Thus, by Theorem 6.10, anP converges absolutely. If the readers will review the manner of attack, they should observe that we have proved that nP converges absolutely for - 1 < r < 1 and then used the comparison test to show that anP converges absolutely for - 1 < r < 1 if lanl 5 n. In summary, the root test has wider scope than the ratio test, although in practice it may be more difficult to use. Neither is subtle with regard to divergence, since both obtain divergence from the fact that { a n ) z l fails to converge to zero. Neither test will serve to identify series-that converge conditionally. The next few theorems are of some use in studying such series.

xs

Cz

The reader's experience with conditional convergence of series probably relies on the alternating series test. The next two theorems set the stage for that test as well as for more general techniques.

6.4 Conditional Convergence 191

6.11 THEOREM Suppose { a n ) 2and { b n ) zare two sequences of real numbers. Define An = x , a k for n 2 0 and A-1 = 0. Then if 0 5 p 5 q,

Proof

For n = 1,2,. .., we have an = An - An-1;hence,

6.12 THEOREM Assume { a n ) z and { b n ) z are sequences of real numbers such that

xs

i. the partial sums of an are bounded, ii. bo 2 bl 2 -, and iii. { b n ) 2 converges to zero.

>

Then

C z lanbnconverges.

>

Proof As in Theorem 6.11, define An = Elaat for n 0 and A-1 = 0. By (i), {A,):, is a bounded sequence; hence, there is M > 0 such that 1AI. 5 M for all n. Choose E > 0. Since { b n ) g l converges to zero, there is N such that n 2 N implies lbn1 < & . Recall now that bn 2 0 for all n and bn - bn+12 0 for all n. Hence, if N 5 p 5 q,


Thus, by Theorem 6.1,

C z anbnconverges.

As an irnmmediate consequence of Theorem 6.12, we have the "alternating series test." 6.13 THEOREM (Alternating Series Test) Suppose { b n ) z is a sequence of

real numbers such that i. { b n ) z converges to zero, and . ii. bo 2 bl 2 b2 2 Then

cz(-l)"bn converges.

Proof Set an= (- 1)" for all n. Then

if n is even and An = 0 if n is odd. Thus, {An)% is bounded, and we can invoke Theorem 6.12 to conclude that

converges.

In light of Theorem 6.3, one may be tempted to conclude that convergent infinite series may be handled in much the same manner as finite sums. We have postponed the discussion of this problem until now, because an understanding of the distinction between absolute and conditional convergence helps solve the problems to be considered. We have seen that two convergent series may be added term by term, and the resulting series converges to the sum of the two series. Let us now consider multiplication of two infinite series involving what is called the Cauchy product, which is motivated by multiplication of finite sums. DEFINITION Let define

The infinite series and C s bn.

C z an and C s bn be two infinite series.

cz

Cn

For each n,

is called the Cauchy product of the two series

cza,,


This definition may perhaps be best motivated as follows. Suppose that

are polynomials. Define am= 0 for m product pq is a polynomial, and

> s and b, = 0 for m > r. Then the pointwise

akbn-k. In particular, for x = 1,

where cn =

The result one might hope for now is that the Cauchy product of two convergent series is convergent and converges to the product of the sums of the two series. More precisely, if an and bn converge to A and B, respectively, then Czl cn converges to AB. It turns out that this is false. In the next few paragraphs, we shall explore these questions and prove the main result.

xE1

Example 6.13

Ezl

Consider the series

By the alternating series test, this series converges. We shall now consider the Cauchy c, of this series with itself. Now product

cs

For n 2 k,

hence,


Thus,

xs

{cn)% does not converge to zero, and Cn fails to converge. As we shall see by the next theorem, cn fails to converge because

xz

rn

converges conditionally. 6.14 THEOREM

Suppose

xz

anconverges absolutely and

xz

bnconverges

with

C ~ , =andA C b n = ~ . Define

Then

cz-

Cn

converges to AB.

Proof Define

We wish to prove that { C n ) g l converges to AB. For each n,

+an& Since {An)% converges to A, the Define rn= a& + alPn-i + theorem will be proved if we can show that {%)%converges to zero.


Choose E > 0. Let K = n 2 N implies that lPnl=

€

Cn

converges to C, then C = AB.

For finite sums, the order in which the terms are arranged has no effect upon the sum. We shall now concern ourselves with the similar situation for infinite series.

DEFINlTZON Let E > a ,

be an infinite series. If T is any 1-1 function from {0,1,2,. . .) onto {0,1,2,. . .), then the infinite series C z a~c,,is called a rearrangement of C z an.

Of course, we are now concerned with questions about the convergence of the rearrangements of a convergent infinite series and what the sums of such rearrangements might be. Our experience up to now leads us to suspect that all is well with absolutely


convergent series, so let us look at a conditionally convergent series. Consider the conditionally convergent series

and denote its sum by S. (S happens to be approximately 0.693; actually, it is log 2. See Project 6.3 for a proof.) Then

converges to

g.

We now consider a new series

xzl

zzlan where an = 0 if n is odd and

5,

if n is even. a, converges to since all we have done is insert a zero into the series xz,(-l)"+'& between each pair of terms. To see that this is the case, let Sn be the nth partial sum of 1)"' and let A, be the nth partial sum of C z lan. Then Al = 0 , Ah = Sn = A2n+lfor all n 2 1 ; hence, { A n ) z l converges to the same limit as {Sn)gl, which is By adding

xz,(-

&,

3.

y.

Deleting the zeros from this resulting series, we obtain a series that converges to it is easy to see that it is a rearrangement of C z l(- l)"+'!. To clarify this situation, let us write out a few terms of each series in question.

Thus, we have found a rearrangement of limit. Of course,

00

(-

1ln+l

,

that converges to a different


is only conditionally convergent. Let us now attack the problem for absolutely convergent series. If a series converges absolutely, it is easy to see that any rearrangement will converge absolutely, since all one needs to do to verify convergence for series with nonnegative terms is to show that the partial sums are bounded. We seek to solve the more difficult problem of showing that all rearrangements converge to the same limit.

CE1an be an absolutely convergent series converging to A and xzlaT(,) be any rearrangement of CE, an. Then 6.16 THEOREM Let

converges to A.

Proof

For each n, let

Now {Sn)gl converges to A. We seek to prove that Choose r > 0. There is N such that for n 2 N,

{s~)E,also converges to A.

x:,

(since an converges absolutely). Now T is a 1-1 function mapping {1,2, . . .) onto {1,2, . . .); hence, there is K 2 N such that {1,2,3, . . .,N) C {T(l), T(2), . ..,T(K)). Assume n 2 K. Then

~ , to A, and the theorem is proved. Thus, { s ~ ) converges

xz,

an is an infinite series with the Theorem 6.16 has a converse in a sense. If property that every rearrangement converges to the same sum, the series is absolutely convergent. The next few pages will be devoted to this result. But a few preliminary remarks are needed.


DEFINITION k t { a n } g lbe any sequence of real numbers. For each n, define a; = an if an 2 0 and a,+= 0 if a, < 0. Define a; = a, if a, < 0 and a; = 0 if a, 2 0. Thus, if

then

and

In any case, an = a; Suppose now that lall

I Jan1

+ a;

for all n.

zE1a, converges absolutely. Then for each n,

and J a i l 5 lanl;

xEl

xE1

xz,

a; and hence, a: and a; converge absolutely. Conversely, if xzl(-a;) converge (actually, this is absolute convergence because both are series of nonnegative terms), then since lan1 = a,+- a;,

converges by Theorem 6.3. This result proves the following theorem.

xzl xzl zE1

6.17 THEOREM The infinite series an converges absolutely iff the series a; and a; converge. If a; converges with sum A+ and a; converges with sum A-, then the sum of a, is A+ + A-, and the sum of Czl lanl is A+ -A-.

xzl

xgl

xzl

TOrecap our results, if a series Czl a, converges conditionally, then at least one of the two series, a; or a;, diverges. Since we may write

xgl

zgl

i

I

xzl

xzl xzl xE1

we see that, if a, converges conditionally, then both a; and a; must diverge. Since a; 2 0, the partial sums of a; form an increasing sequence unbounded from above; and since a; 5 0, the partial sums of a; must form a decreasing sequence unbounded from below.. When we consider rearrangements of

xgl

6.4 c o n d i t i o ~convergence l 199

series, we shall have recourse to this result. If we denote the nth partial sum of

by Sn, Si, and S i , respectively, then

converges, whereas {%)El and {S;);, do not. The manner in which the cancellation takes place in the series C;, a, may be very complicated, and our techniques for recognizing conditionally convergent series will unfortunately be applicable only to certain special types. We are now equipped to state and prove a partial converse of Theorem 6.16. 6.18 THEOREM If

c:,

an converges conditionally, then, given any real number r, there is a rearrangement of C;, a, that converges to r.

{$)El

and {S;)gl are unbounded sequences and {a,);, converges to zero. Let nl be the least positive integer such that

Proof First, notice that

There is such nl because {Si)zlis an increasing sequence not bounded from above. Next, let ml be the least positive integer such that

As before, such an ml exists because { S ; ) E l is a decreasing sequence bounded from below. Now let n2 be the least positive integer such that

As argued before, such an n2 exists. Now choose m2 to be the least positive integer such that

Again, such an m* exists. Continuing in this way, deleting zeros when they occur, we obtain a rearrangement of the given series. Moreover, the rearrangement converges to r, since {a,):, converges to zero.

200

Chapter 6 Infinite Series

The proof of Theorem 6.18 leaves quite a few details to the reader to verify. However, by this time we assume the reader is sophisticated enough to fill in those details. The intent here is to provide the basic idea that makes the proof work.

We are now going to consider a special type of infinite series called power series. Since the reader has encountered this topic in the study of calculus, many of our results will not be new. However, armed with a firm grasp of the basic facts about infinite series, we should be prepared to gain a much fuller understanding of the inner workings of power series.

DEFINITION Let {an)g, be a sequence of real numbers. For each real number x, we may consider the infinite series

We shall refer to this collection of infinite series as the power series generated by {an)% or, more briefly, as the power series a,#.

xz

xz

a n y , we find that for some values of x, it converges Given a power series (for x = 0, for example), whereas for some other values of x, it may diverge. Our first considerations will be to find the set of points where a given power series converges and to determine the general nature of that set.

Example 6.14 Let us consider several familiar examples. We have already seen that X converges for -1 < x < 1 and diverges elsewhere. On the set (-1, I),

xz

This is, of course, the power series generated by the sequence {an)%, with an = 1 for all n. The power series w 2' a is also familiar, it converges to 8 for all real x. (We shall prove this later.) 1

x,

Intuitively, an infinite series will converge if the nth tern tends to zero "fast enough." Thus, one is led to suspect that if converges and (yl < x, then maybe a n y converges. However, our past experiences with conditionally convergent series should cause us to proceed with caution.

xz

xsa,#

6 5 Power Series 201

xsa n y

6.19 THEOREM Let diverges for x = X I . Then

i.

be a power series that converges for x = xo and

Z> a n y converges absolutely for 1x1 < 1x01; and

ii. C>anx"

diverges for 1x1

> Ixil-

xz

xs

Proof Assume an% converges and 1x1 < 1x01 with XQ $0. Since an$ converges, the sequence {an$)% converges to zero, and hence is bounded; that is, there is M E R such that lan$l 5 M for all n. For all n,

Since

the series

converges, and hence, by the comparison test, En> a n y converges absolutely. If 1x1 > 1x1 1 and a n y converges, then by (I), an$ converges absolutely, contrary to assumption. Hence, a n y diverges.

x>

z>

xz

Let us summarize these results. Suppose C z a n x " is a power series, and let C = { x : C s a n y converges). The possibilities for C are listed below:

z2

a n y converges for all x E R. (i) C = R, in which case (ii) C = {0), in which case C> a n y diverges for all x 0. (iii) There is r > 0 such that (-r, r) C C C [-r, r], in which case absolutely for 1x1 < r and diverges for 1x1 > r.

x> a n y converges

Cases (i) and (ii) are self-explanatory, but we should justify (iii). Suppose C # ( 0 ) and C f R. Since C f R, there is xl E R such that x > a n g diverges. Hence, by Theorem 6.19, for each x E C, 1x1 5 1x1 1. Therefore, C is bounded; let r = sup C. If 1x1 > r, then x $ C, so a n y diverges. If 1x1 < r, then there is a member p of C such that 1x1 0. We cannot assert what happens at r or at -r.

xz

zz

I1

+

xs


Example 6.15

Consider again the power series Cz $. For each n,

hence, by the ratio test, this series converges for all x, as stated earlier. Similarly, the ratio test may be used to show that x z n ! Y converges only for x = 0. The geometric series converges for -1 < x < 1 and diverges for 1x1 2 1. The $ converges for -1 5 x < 1 and diverges for both 1x1 > 1 and power series x = 1.

xzx"

xz

z>

If a n y is a power series, then the set of points at which the series converges is either the set of all real numbers, (01, or an interval of positive finite length centered at zero that may contain all, none, or one of its endpoints. We choose to consider (0) as an interval of zero radius and R as an interval of infinite radius. With this convention, we feel free to speak of the interval of convergence of a power series. In particular, a n y converges absolutely at any point of its interval of convergence that is not an endpoint of that interval. If the interval of convergence is R, the series converges absolutely at each point; and if (-r,r) C C C [-r, r], where C is the interval of convergence, then the series converges absolutely at x for x E (-r, r). As shown by our examples, the behavior at the endpoints is unpredictable. If C z a n y has an interval of convergence C that is different from R and (01, then there is a unique real number r such that

xz

This number r is called the radius of convergence of the power series. We now seek a way to determine r. Since we have already used the ratio test in one case, let's see what we can deduce from this test. Suppose x > a n f l is a power series with an 0 for all n and x # 0. Consider the ratio

+

For the ratio test, we may refer to either Theorem 6.8 or 6.9. For the simplest case, let us assume that the sequence

converges to L $0. Then the sequence

I

i,

converges to x L. Hence, C s a n y converges for 1x1 L < 1, or for 1x1 < and diverges for 1x1 L > 1, or for 1x1 > Thus, the radius of convergence is $ if L $0.

i.

t L

6.5 Power Series 203

If L = 0, then 1x1 L = 0 < 1 for all x, so the series converges for all x. We summarize these results as follows: 6.20 THEOREM Suppose C s a n X " is a power series with an # 0 for all n such that

converges to L. Then i. if L = 0, the series converges for all x; and ii. if L # 0, is the radius of convergence. Theorem 6.20 has the obvious advantage of being easy to apply when it is applicable. However, when

fails to converge, Theorem 6.20 yields no information. At this point, we still have recourse to Theorem 6.8.

6.2I THEOREM Let

C s a n y be a power series with a,, # 0 for all n.

i. If there are real numbers q

Then

> 0 and N such that n 2 N implies that

k;

then C z a n y converges absolutely for 1x1 < ii. If there are real numbers p > 0 and N such that for all n 2 N,

then

cz

a n y diverges for 1x1 >

i.

Proof Suppose the condition of (i) holds and 1x1 < a n y converges absolutely. 6.8, Suppose now the condition of (ii) holds and 1x1 > 6.8, Cz a n y diverges.

c>

k. Then by (i) of Theorem d. Then by (ii) of Theorem

Theorem 6.21 tells us that the radius of convergence r satisfies p and q that satisfy the hypotheses of this theorem. If

d 5r5

for all


converges to L # 0, then p may be chosen to be any number less than L and q any number greater than L; hence, this yields again the fact that 1/ L r 1 /L,which means r = 1/L.

<
1, the series diverges.

Exercises 215 22. If {an)El is a sequence of positive real numbers such that

converges to L, prove that 23. Prove that

{&)El

converges to L.

converges, and find the limit. You might want to look at Example 6.12. 24. Test the following series for convergence:

25. Determine those values of p for which

cE2& converges.

26. For each positive integer n, define

Prove that {%)El converges. (Use the fact that logx = Jlx

$ for x > 0.)

6.4 CONDITIONAL CONVERGENCE 27. If

Cglan converges and { b n ) g l is monotone and bounded, prove that CElanbn con-

verges. 28. Suppose {an)Z1is a sequence of positive real numbers converging to zero such that

xEl(-

an 2 an+lfor all n. Then, by the alternating series test, l)"anconverges; call the sum S. Let Sn be the nth partial sum of (- 1 ) " ~Prove ~ . that ISn - SI 5 an+l.

xz,

xEl an be an infinite series and { n k ) g l a subsequence of the sequence of positive integers. Prove that if XE an converges absolutely, then CElankconverges absolutely.

29. Let

What can be concluded if 30. Prove that

h ~an converges , conditionally?

zEl9converges. (Hint: Use Theorem 6.12.)

31. Test each of the following series for absolute convergence, conditional convergence, or divergence:


00

(c)

C ( f i- 6 ) a nwhere an =

f'"'

(- 1

and f(n) =

I;[

; that is, f (n) is the integer

n part of 2'

6.5 POWER SERIES 32. Suppose that x z a n diverges and that { a n ) z is bounded. Prove that the radius

of convergence of

xzl

C z anx" is equal to 1.

33. Suppose an converges conditionally. Prove that the radius of convergence of a n y is equal to 1.

x,",

34. Show that 35. Show that

CEln!x"converges only for x = 0.

xEl f converges iff -

15x

< 1.

36. Determine the radius of convergence of the power series CEl 9x". 37. Determine the interval of convergence of the power series of p.

xz

xEl $x" for different values

CEl

38. Show that the power series anP and n ~ ~ 2 either - l both converge for all x, both converge only for x = 0, or both have the same finite nonzero radius of convergence. 39. Let {an)El be a sequence of real numbers bounded from below, and let A = @ : there is a subsequence of {an)El converging to p). Suppose A is nonvoid. Define a = infA. Prove that a E A and that, for each E > 0, there is N such that for all n 2 N, a - E < an and there are infinitely many m such that am < a + E .

40. State and prove theorems similar to Theorems 6.8 and 6.21 in terms of

im i n

n+oo

1 1 an

and

I I*

an+1 lim sup n+ oo an

41. Find the interval of convergence of the power series

below. Be sure to check the endpoints.

xE1any where an is

as given

(a) an = (n + 1)(n + 2) (b) an = sin n (c) an = 3-"Jii (4 an = (& (e) an = ( 1 + ;)"* (f) a n = : + $

6.6 TAYLOR SERZES

- 1. See F'roject 5.1 in Chapter 5 if you have forgotten the derivative of logx. Prove that this series converges to logx for

42. Write the Taylor series for log x using powers of x 1 5 ~ ~ 2 .

Projects 217 43. Write the Taylor series for J l - x f o r -1 < x 5 0.

41-x

in powers of x. Prove that the series converges to

45. Show that for 1x1 5 1, Isinx - (x - $ +

&) 1 5 A.

46. Use the Taylor series for d to prove that 1 + x

5 8 for all x > 0.

EL,

47. Suppose that al, a*, . . .,an are positive real numbers, and let A = ak. If we let 5 e(ak/A)-l. Multiply these x = % - 1, the inequality in Exercise 46 shows that l/n < al+oz+.-.+a, inequalities together to prove that (ala2 - an) n 48. Define f(x) = e-'/* if x # 0 and f (0) = 0. Show that f is infinitely differentiable (that is, has derivatives of all orders), but if x # 0,

PROJECT 6.1 The purpose of this project is to develop Wallis's formula. For n = 0,1,2, . . ., define In=":J sinn t dt.

1. Show that 5 I, for n > 0. 2. Show, by integration by parts, that In= 3. Prove that

4. Prove that

for all n 2 2.

& = & 5 '2 5 1 for all n 2 1.

5. Conclude that

converges to 1.

6. Prove that

Parts 5 and 6 yield Wallis's formula: lim

n-rm

2-2°40406-6-00(2n)(2n) 7r =-. 1 3 3 5 5 7 7 (2n - 1)(2n - 1)(2n+ 1 ) 2

Wallis's formula gives 5 as an infinite product, defined as the limit of partial products, in much the same way we defined the infinite sum as the limit of partial sums. If you continue your study of analysis-in particular, analysis of complex variables-you will learn more about infinite products.


PROJECT 6.2 In this project you will construct an increasing function that is discontinuous at each rational point in (0,l) and continuous at each irrational point in (0,l). We will need two basic facts: a. The rational numbers in the interval (0,l) can be arranged in a sequence {rn)gl. This is true because the set of rational numbers is countable. (See Example 0.12 and Corollary 0.15 .) b. Any rearrangement of an absolutely convergent series converges, and any subseries of an absolutely convergent series converges. See Theorem 6.16, and apply the comparison test to the subseries, replacing the missing terms with zeros. For each real number x E (0, I), define K(x) = {n : rn 5 x ) . So K(x) is a subset of J, and we define f : (0,l) -,R by the formula

1. Prove that f is increasing. 2. Prove that f is discontinuous at r if r is a rational number. 3. Rove that f is continuous at x if x is an irrational number. After you finish this project, you may want to refer to Section 2.4 of Chapter 2.

PROJECT 6.3 By Theorem 6.13, the alternating series test, you can prove that the series

converges. In this project you will prove that the sum of this series is log 2, the natural logarithm of 2. We will lead you through the proof, and you will need to supply the missing justification. 1. We will need the notation and the result of Exercise 26. If you haven't done that exercise, that is your first task. To summarize, define y,= 1+ 12 + 13 + *+,1 -log n. Prove that {yn)zl converges. 00 4

2. Let Sn be the nth partial sum of E ( - I ) ~ + A. ' Your strategy will be to show that n n= 1 {Sb)El converges to log 2. Explain why this shows the sum is log 2. 3. Show that

Chapter

7

Sequences and Series of Functions

n Chapter 6 we considered power series but left many rather obvious questions unasked and unanswered. In this chapter we seek out these questions and answer some of them. Suppose a n y is a power series that converges for -r < x < r where r > 0. For each integer n 2 0, define Sn : (-r, r) + R by

x>

for all x E (-r, r). Therefore, for each x E (-r, r) and each integer n 2 0, Sn(x) is the nth partial sum of the convergent series C>anx", so the sequence {Sn(x))% converges to a n y . It is natural to define a function S : (-r, r) + R by defining S(x) to be the limit of the convergent sequence {Sn(x)}% or, equivalently,

x>

for all x E (-r, r). It is customary to denote the limit of a convergent sequence {an)El by limn,, an. We have avoided this usage because of possible conflict with the notation in Chapter 2. However, by now we should not be disturbed by this notation. Thus, we write S(X)= lim Sn(x) n--+w

for each x E (-r, r). Each Sn is a polynomial and hence is continuous everywhere, differentiable everywhere, and Riemann-integrable on every interval of finite length. Does S inherit any of these nice properties? If S is differentiable at x E (-r, r), is it true that

220 Chapter 7 Sequences and Series of Functions

If S is Riemann-integrable on [a, b] C (-r, r), is it true that

Perhaps there lurks in the reader's past some experience with Fourier series-that series of the type

is,

C ( a ncos nx + bn sin nx). Here again, define Tn : R -,R by

for. each integer n 2 0. If this series converges for all x E E, one may define a function T :E + R by T(x) = lirn T&) = c ( a ncos nx + bn sin nx) n+oo

n=o

for all x E E. Again, the same questions may be asked concerning the nature of T. These last remarks lead us to suspect that it may be more profitable to consider the situation in general rather than concentrate our attention on power series alone.

{fn)z

DEFINITION Suppose is a sequence of functions and E is a subset of R such that E C dom fn for each integer n 2 0. Then { converges pointwise on E if for each x E E the sequence

fn)z

{fn(x))Z converges. If

{fn)z converges pointwise on E, define f : E

-+R

by

I

for each x E E. We say that

{fn)s converges pointwise to f on E. It is common practice to call

f the limit function of the sequence {

fn)z and write f = limn,,

f.

*

7.1 Pointwise and Uniform Convergence 221

Example 7.1 As a simple and familiar example, for each integer n 2 0, define f. : (-l,1) + R by

for each x E (- 1,l). Then {fn)glconverges pointwise on (- 1,l) to the function 1 g(x) = Tr;rn We depart from the realm of power series to seek other examples.

Example 7.2

Define for each integer n > 0, fn : [O, 11 --+ R by

for each x E [0, 11. For x E [0, l),

converges to zero, and for x = 1,

converges to 1. Thus, the limit function is not continuous, although each fn is continuous. W

Example 7.3

For each positive integer n, define gn : [0, 11 + R by

Then {gn)gl converges pointwise on [0,1] to the function g, which is zero for all x E [0, 11. Each gn is Riemann-integrable on [0, 11, as is g, but

converges to 1, whereas

g(t) dt = 0.


Example 7.4

Consider the sequence { h n ) z l defined by

The sequence { h n ( x ) ) ~converges , to for 0 < x < 1 and converges to zero at x = 0. Here each h, is continuous, but the limit function is unbounded and, hence, is not Riemann-integrable. Note further that the sequence

is unbounded.

rn

Example 7.5 The sequence {f,)E, given in Example 7.2 yields an example of a sequence of differentiable functions that converges to a function that fails to be differentiable. We now present a sequence of differentiable functions { k , ) Z , such that the limit function k is differentiable, but for some x, k'(x)

# n-oo lim kL(x).

To this end, define, for each positive integer n, k, : [- 1,1] -,R by

for all x E [-I, 11. For each positive integer n, kn is differentiable on [- 1,1] and

From this, it is easy to see that the minimum and maximum values of k, occur at respectively. Hence; and

-1

h,

for all x E [-I, 11 and all positive integers n. It follows quite easily that { k n ( x ) ) z l converges to zero for all x E [-I, 11. The limit function is thus constant on [- 1,1] and hence differentiable, with the derivative equal to zero everywhere on [-I, 11. However G(O) = 1 for all n. I

7.1 Pointwise and Uniform Convergence 223

Rather than be discouraged by this list of examples, we shall examine these examples more carefully with the hope of finding a substitute for pointwise convergence that yields affirmative answers to some of the questions posed earlier. Suppose {fn)gl converges pointwise to a function f on a set E. This means that for each e > 0 and each x E E, there is N (probably depending on both x and e), such that for each positive integer n, n 2 N implies that

Again, for each positive integer n, define fn(x) = Y for each x E [0, 11. Given 0 < x < 1 and e > 0,there is a lower bound for the possible values of N, in fact, N must be larger than llog €1logxl. For x close to 1 but different from 1, N must be quite large, for, given any 0 < e < 1 and any positive integer m, there is 0 < x < 1 such that J? > e. These last remarks are especially meaningful in light of the fact that the discontinuity of the limit function occured at 1. We are led to suspect that a notion of "uniform" convergence, so named because of the similar distinction between continuity and uniform continuity, might be in order.

DEFINITION A sequence {fn)gl of functions is said to converge uniformly on E if there is a function f : E + R such that for each e > 0, there is N such that for each positive integer n, n 2 N implies that

for all x E E. We may also express this by writing "the sequence converges uniformly to f on E."

{fn)zl

Naturally, a series C> fn of functions will converge pointwise or converge uniformly on E as the sequence { S n ) z of partial sums, defined by

converges pointwise or uniformly on E. The following theorem arises naturally after our previous experience with sequences. The proof is left to the reader as Exercise 4. Z

7.1 THEOREM A sequence of functions {fi)glconverges uniformly on E iff for each e > 0 there is a real number N such that for all positive integers m and n,m>Nandn>Nimplythat

for all x E E. For infinite series of functions, there is a very convenient test for uniform convergence, called the Weierstrass M-test.


7.2 THEOREM (Wekrstmss M-Test) Suppose { f * ) g l is a sequence of functions defined on E and { M n ) g l is a sequence of nonnegative real numbers such that

for all x E E and all positive integers n. If converges uniformly on E.

CglM n converges, then Cglfn

xgl

Proof Choose e > 0. Since M n converges, there is a real number N such n 5 m implies that that for all positive integers m and n, N I

(Recall that Mn 2 0 for all positive integers n.) Thus, for all positive integers m and n, N 5 n 5 m implies that

for all x E E. By Theorem 7.1,

zE1./i converges uniformly on E.

We have exhibited remarkable confidence in the definition of uniform convergence by proving the last two theorems before having any concrete evidence that this new concept is really useful. The next few theorems serve to show that this confidence is not unfounded. Suppose { f n ) g l is a sequence of functions converging uniformly to f on E such that for each positive integer n, fn is continuous at xo E E. Choose e > 0. There is N such that for each positive integer n, n 2 N implies that

Ifn(x) -f(x)I

<e

for all x E E. Since fN is continuous at a,there is 6 x E E imply that

IfN(x) - ~ N ( x o ) ) < Hence, for lx - xol

> 0 such that lx - xol < 6 and

€ 0

< 6 and x E E, we have

If(x) - f ( a ) l I I f @ ) -fN(x)l + IfN(x) -fN(xo)I +I fN(x0) -f(-)l < € + €4- € = 3 ~ . It seems we have just proved a theorem. What is really necessary for this proof to yield the desired conclusion-that f is continuous at xo?

7.2 Consequences of Uniform Convergence 225

{fn)zl

First, it is sufficient that infinitely many members of the sequence be continuous at no. Second, uniform convergence on all of E is unnecessary. We now present a slight modification of our original definition of uniform convergence that will suffice for this proof.

DEFINITION Suppose {fn)gl is a sequence of functions converging pointwise to a function f on a set E. Then {fn)gl converges uniformly at xo E E if for each E > 0, there are positive real numbers N and 6 such that for each positive integer n, n 2 N, lx < 6, and x E E imply that

fn)zl

converges uniformly to f on E, then {fn)glconIt is immediate that if { verges uniformly at x for each x € E. Conversely, if E is compact and {fn)gl converges uniformly at x for each x E E, then { converges uniformly on E. Can you prove this?

fn)zl

7.3 THEOREM Suppose {fn)gl converges pointwise to f on E, finitely many members of the sequence are continuous at xo. If { uniformly at xo, then f is continuous at xo.

E E, and in-

fn)zlconverges

Proof Choose c > 0. There are positive real numbers N and 6 such that for each positive integer n 2 N,

for all x E E such that lx - xol < 6. There is a positive integer no 2 N such that f, is continuous at xo; hence, there is 61 > 0 such that lx - xol < 61 and x E E imply that

Let 62 = min{6, 61). Then, for lx - -1

Consequently,f is continuous at xo.

< 62 and x E E, we have


In particular, it follows from Theorem 7.3 that if {f n ) g l is a sequence of functions continuous on E and converging uniformly to f on E, then f is continuous. (Of course, it is implicit that E = domf .) The example presented earlier to convince the reader that the limit of a sequence of continuous functions need not be Riemann-integrable yielded a limit function that was unbounded. The added hypothesis of uniform convergence will eliminate such occurences. 7.4 THEOREM If {f n ) g l is a sequence of functions converging uniformly to f on E, and for each positive integer n, fn is bounded on E, then f is bounded on E.

Proof Let c = 1. There is a real number N such that for each positive integer n, n 2 N implies that

for all x E E. There is M such that 1 fN(x)) 5 M for all x E E. Hence, for all x E E,

One should now hope that, if { f n ) g l converges uniformly to f on [a, b] and for each positive integer n, fn is Riemann-integrable on [a, b], then f is Riemann-integrable on [a, b]. This and more is the content of the next theorem.

7.5 THEOREM Let {f n ) g l be a sequence of functions, each Riemann-integrable on [a, b], converging uniformly to f on [a, b]. Define

for each t E [a, b] and each positive integer n. Then f is Riemann-integrable on [a, b], and {F,):, converges uniformly on [a, b] to the function F defined by ~ ( t=)J al f ( x ) d x

-

for each t E [a, b].

Proof Each fn is bounded on [a, b] so, by Theorem 7.4, f is also bounded. We may suppose a < b. Choose e > 0. There is a real number N such that for each positive integer n, n 2 N implies that

7.2 Consequences of Uniform Convergence 227 -

for all x E [a, b]. There is a partition P of [a, b] such that

For each x E [a, b],

Thus,

Consequently,

Hence, f is Riemann-integrable on [a, b]. Again, choose E > 0 and N such that for each positive integer n, n 2 N implies that

for all x E [a, b]. Then for each positive integer n, n

2 N implies that

for all t E [a, b]. Therefore, {Fn)glconverges uniformly to F on [a, b].

xzf

Soon we shall prove that the power series converges uniformly on any closed interval [a, b] c (- 1,l). (Do you want to try the proof now? Why not?) Assuming this fact, let us see what results we can obtain with the aid of Theorem 7.5. Choose any closed intend [a, b] c (- 1,l). It is no surprise that

is Riemann-integrable on [a, b]. But Theorem 7.5 implies even more; it implies also that


converges uniformly to

[&-- -

St dt/(l - t) on [a, b]. Recall that

log(1 - x),

so we have a power series 00 p+' ;;;r that converges to - log(1 - x) on (- 1,l) and converges uniformly on any closed subinterval of (- 1,l). The concept of uniform convergence does not allow affirmative answers to all the questions posed at the beginning of this chapter. The sequence {kn)zl, defined on [-1,1] by '

for each positive integer n and each x E [- 1,1], converges uniformly to the function that is zero everywhere on [- 1,1]. However, {k; (0))zl converges to 1. Consequently, one must have some additional hypotheses to gain the type of result we have in mind.

{fn)zl

7.6 THEOREM Suppose is a sequence of functions, each of which is differentiable on [a, b]. Suppose further that for some xo E [a, b],

converges and that {fi)zlconverges uniformly to g on [a, b]. Then i. {fn)glconverges uniformly on [a, b] to a function f ; and ii. f is differentiable on [a, b] and f ' ( x ) = g(x) for all x E [a, b].

Proof Assume b > a and choose E > 0. There is a real number N such that for all positive integers m and n, m 2 N and n 2 N imply that

and for all t E [a, b],

For each pair of positive integers m and n, fn -fm is differentiable on [a, b]; hence, for each u E [a, b] and v E [a, b], there is w E [a, b] such that

Therefore, for all positive integers m and n, m 2 N and n 2 N imply that

7.2 Consequences of Uniform Convergence 229

for all x E [a,61. Consequently {f n ) g l converges uniformly on [a,b]. Define

for each x E [a,b]. Choose x E [a,b] and for each positive integer n, define Fn : [a,b] + R by

for all y E [a,b]\{x) and Fn(x)=fi(x). Define F : [a,bl

+

R by

for all y E [a,b]\{x) and F(x) = g(x). The function f will be differentiable at x with f'(x) = g(x) if F is continuous at x. For each positive integer n, Fn is continuous at x by the differentiability of fn at x. If we can show that {Fn)gl converges uniformly to F on [a,b], then F will be continuous at x, and hence f will be differentiable at x with f'(x) = g(x). Choose E > 0. As before, there is a real number N such that for all positive integers m and n, m 2 N and n 2 N imply that for all t E [a,b],

Then, for all positive integers m and n, m 2 N and n 2 N imply that

for all y E [a,b]\{x), and

Thus, by Theorem 7.1, {Fn)gl converges uniformly on [a, b],and it is clear that (F,);, converges to F. This concludes the proof.


7.3

UNIFORM CONVERGENCE OF POWER SERIES

Before examining in more detail the results of Theorems 7.3, 7.5, and 7.6, we return to consider the convergence of power series. The Weierstrass M-test suggests the following theorem.

xs xz

7.7 THEOREM Let a n ybe a power series that converges for -r < x < r, r > 0. Then a n y converges uniformly on -t I x I t for each 0 < t < r. Proof

If 0 < t

< r, then C s ant"converges absolutely. For -t 5 x 5 t,

Thus, by the Weierstrass M-test,

xsa n y converges uniformly on [-t, t].

Consider a power series z z a n x " . To apply the results of Theorem 7.6, one must consider the question of convergence of the series

It is not at all clear from casual observation that this series should converge for any x # 0. Even though a n y converges, it seems reasonable that {nany-')% might not even converge to zero. Indeed, if an = and x = - 1, this is the case. We have, from Chapter 6, a result that allows the direct determination of the interval of convergence of ~ ~ 2 - Perhaps l . considerations of this type will give us the proper perspective.

xz

x>

7.8 LEMMA If { b n ) z lis a bounded sequence of real numbers, then { fibn)gl is a bounded sequence of real numbers and lim sup bn = lim sup fib,,. n+oo

n+oo

Proof Let

In Chapter 1, we proved that be bounded. Let c = lirn sup f i b n . n+oo

{.(%i)glconverges to

1 . Thus, {C/iibn)gl must

7.3 Uniform Convergence of Power Series 231

Let {bnk}glbe a subsequence of {bn)zl converging to b. Then

also converges to b and b 5 c. Suppose

converges to c. Since {

mfi}z converges to 1 $0,we conclude that

converges to c. Consequently, c 5 b. We have shown that b 5 c and c c = b.

5 b, so

Lemma 7.8 and Theorem 6.18 yield the fact that Can."

and

CwnY'

have the same radius of convergence. It is easy to see that this implies that Can." I

1

and

Cna,$-'

also have the same radius of convergence. 7.9 THEOREM Suppose

xz

a n y converges to f on (-r, r) with r

> 0. Then

zs

i. For each 0 < t < r, a n y converges uniformly on [-t, t]. ii. f is n-times differentiable on (-r, r) for each positive integer n. iii. For each 0 < t < r and each positive integer m,

converges uniformly to fm)(x) on [-t, t]. iv. f@)(O) = n!a,. [We make the convention that flo)(x) =f (x) for all x E (-r, r).]

Proof Part (i) of this theorem, a repetition of Theorem 7.7, is included only for the sake of completeness.


The proof of parts (ii), (iii), and (iv) shall be by induction. Choose any xo E (-r,r). There is 0 < t < r such that 1x01 < t. Since a n y converges on (-r, r), Lemma 7.8 shows that

xs

converges on (-r, r). By Theorem 7.7,

converges uniformly on [-t,t]. Define g : [-t,t] -+ R by g(x) = f(x) for all x E [-t,t]. Theorem 7.6 guarantees that g is differentiable on [-t, t] and

for all x E [-t, t]. Thus, since xo E (-t, t), f is differentiable at

and

Since no was any point in (-r, r), we have f differentiable on (-r, r),

xzl

converges uniformly to f '(x) on any closed for all x E (-r, r), and subinterval of (-r, r). In particular, f '(0) = a1 . Assume the theorem holds for n = k. Thus, f is k-times differentiable on (-r, r); for each 0 < t < r,

converges uniformly to flk)(x)on [-t, t]; and

Define h : (-r, r) -+ R by


for each x C (-r, r), and define

for each nonnegative integer n. Then

for all x E (-r, r). By the results in the preceding paragraph, h is differentiable on (-r, r); for each 0 < t < r,

converges uniformly to h'(x) on [-t, t]; and h'(0) = bl . Thus, f k ) is differentiable on (-r, r); for each 0 < t < r,

converges uniformly to fk+l)(x) on [-t, t]; and

This completes the induction. Theorem 7.9 tells us that, inside the interval of convergence (not necessarily at the endpoints), a power series can be differentiated term by term, and the resulting power series will converge to the derivative of the limit function of the original series. Theorem 7.5, coupled with Theorem 7.7, justifies the formula

xz

for -r < a 5 b < r if a n y converges for -r < x < r. Recall that ~ ~ converges x ton & for - 1 < x < 1. Hence, by Theorem 7.9,


converges to

& and

converges to (,_f, o) r, -1<x0

< x < r. The classic example is the function f : R -+ R defined by

for x # 0 and f(0) = 0. In this case, p ( 0 ) exists and equals 0 for each positive integer n, but then

for all x, and f (x) = 0 only for x = 0. Consider the differential equation

{fi)zl

In Chapter 1, we proved that converges to 1 by assuming convergence and showing that then the limit must be 1. Then we proved that the sequence converged to 1. A similar device will be used here. Suppose there is a function f satisfying this differential equation, and suppose that

for -r

and

< x < r with r > 0. Then for -r < x < r,


Consequently, for -r

< x < r,

This equation will be satisfied if {a,)% is chosen so that a1 = a~ and for each integer n 2 2, Thus, we may choose a0 = a1 and, for each integer n 2 2,

Several questions must be considered at this time. First, if a sequence { a , ) z i s chosen so that (2n - l)a,-l - a,-2 ao=al and a,= n2 for each integer n 2 2, is it true that C > a , Y will converge for some x # O? Second, there may be other methods of choosing the sequence {an)% to satisfy the equation above. More to the second point, does there exist a sequence { b n ) z such that C> bnY converges to zero on an interval of positive length with b, # 0 for some integers n? We answer the second question first. 7.10 THEOREM Suppose { a n ) sand { b n ) zare two sequences of real numbers, r > 0, and for all x E (-r, r),

Then for each integer n 2 0, an = b,. Proof

Define f : (-r, r ) --+ R by


for each x E (-r,r). Then, by Theorem 7.9, f has derivatives of all orders on (-r, r) and for each integer n 2 0,

Thus, for each integer n 2 0, an = bn.

In particular, Theorem 7.10 allows one to conclude that if C z a n y converges to zero for all x E (-r, r) with r > 0, then an = 0 for each integer n 2 0. Therefore, any solution to the differential equation that is expressible as a power series radius, must satisfy

xz

a n y , convergent on an interval of positive

The question concerning the convergence of such a power series is still unan1 swered. To simplify matters, choose a0 = a1 = 1. Then a2 = 3, a3 = 61 , . . ., an = z1 , . . .. The solution then is

x,

and this solution is valid for all x because 00 x" a converges for all x. In this case, it is trivial to check that this is a solution, since the function f(x) = 8 is well known to us. Now let us apply the result of Theorem 7.5 to power series. 7.11 THEOREM Suppose with r > 0. Then

xz

an(x - a)" converges to f for -r

<x -a < r

converges to

for -r Proof

< x - a < r. This is an easy application of Theorems 7.9 and 7.5 using the fact that


Theorems 7.9, 7.10, and 7.11 allow us to find the power series for many functions without resorting to Taylor's Theorem.

Example 7.6 We seek a power series for f(x) = Arctan x. Assuming the usual facts about the trigonometric functions and their derivatives, we find that f '(x) = Finding the successive derivatives off will require applications of the quotient rule, and it may be difficult to find a nice formula for f(")(x). Rather, we will use Theorem 7.11. We know that

A.

kctan x =

lxi-;;i 1

dt,

so we will find a power series for Arctan x. From Chapter 6,

-=EX" 1-x 1

for

& and apply Theorem 7.11 to find the series for

- 1 < x < 1.

n=O

Thus,

for - 1 < x2 < 1 or - 1 < x < 1. Theorem 7.11 allows us to find the series for Arctan x by integrating the series just obtained term by term: 00

~ r c t a nx =

(-1Y C2n+ 1

?+I

for

-l<x n, and ifisn.

Show that {f n l g l converges pointwise on [0, 11 to a function f that is not Riemannintegrable.

3. Prove that the sequence {kn)El,defined by k n ( ~ )=

X 1+ n x 2

for all x E R and each positive integer n, converges uniformly on R. 4. Prove Theorem 7.1.

5. Suppose that both {fn)El and {gn)gl converge uniformly on E. Prove that {fn +gn)El converges uniformly on E. What can one prove about the uniform convergence of {fngn)E, ? 6. Let f : R -,R be uniformly continuous, and for each positive integer n and each x E R, define

Prove that {fn)El converges uniformly to f on R. 7. Assume E is compact and {fn)Elis a sequence of functions that converges uniformly on each point of E. Prove that {fn)Elconverges uniformly on E.

7.2 CONSEQUENCES OF UNIFORM CONVERGENCE 8. If A C E C R, then A is dense in E if E = n E. Assume {fn)gl is a sequence of functions continuous on E and converging uniformly on a set A dense in E. Prove that

{fn)El converges uniformly on E. 9. For each positive integer n, define hn : [O, 11 7R by

Show that {hn)gl does not converge uniformly on [O,1] but does converge uniformly on [a, 11 for each 0 < a < 1.

Projects 239

10. Search the literature for an example of a continuous, nowhere-differentiable function. Such an example will probably be more meaningful in light of the contents of this chapter. 11. If {fn)El is a sequence of continuous functions that converges pointwise to f on [a, b], then there is at least one point xo E (a, b) such that Search the literature for a proof of this theorem.

{fn)El converges uniformly at a.

12. Suppose E is compact and {fn)glis a sequence of continuous functions defined on E that converges pointwise to a function f , also continuous on E. If, for each positive integer n, fn(x)

Ifn+l (x)

for all x E E, prove that {fn)glconverges uniformly to f.

7.3 UNIFORM CONVERGENCE OF POWER SERIES 13. Show that the power series

xz 5

converges uniformly on [-a, a] for any real number a.

14. Find the power series for the function f(x) = 1/(2 - 3x) and find the interval in which it

converges to f. You may want to write f in the form o/(l - bx) and use a geometric series. 15. Find the power series for the function log(1 -x) and find the interval in which it converges

to log(1 - x). \

16. Define f(x) =

for x

# 0. Determine whether f has a limit at zero, and if so,

find the limit. 17. Find the first three nonzero terms of the power series for Arcsin x. In what interval does

it converge to Arcsin x? sinx-x

18- Definef (x) = xzl\rcsin

for x

# 0. Determine whetherf has a limit at zero, and if so, find

the limit.

PROJECT 7.1 The purpose of this project is to prove Abel's Theorem concerning convergence of power series at endpoints of the interval of convergence. We know that, given a power series, the interval of convergence is R, {0), or an interval of positive length. If the interval of convergence is (-a, a), we also know that the series converges uniformly on [-s, s] for any 0 < s < a. However, the question of convergence at the endpoints is unanswered. Indeed, we have seen enough examples to expect that the series may converge at all, none, or only one of the endpoints. Suppose that the series in question converges at a. Does the series converge uniformly at a? We haven't addressed that question as yet, and thereby hangs the tale. We will state Abel's Theorem in the most general way and then prove only a special case. You are invited to adapt the proof to the general situation.

x>

ABEL'S THEOREM If a,(x - xo)" converges for xo - R < x < xo + R, and if anRn converges, then the series converges uniformly on [m,xo + R]. Similarly, if an(-R)" converges, the series converges uniformly on 1x0 - R, xol.

xzxz


The version of this theorem that we will prove is the following:

THEOREM If C z anconverges, then

xz a n yconverges uniformly on [0,

11.

The word we used above is a euphemism for "We will give hints and direction and you will write the proof." So get out your pencil and paper and be ready to go to work. 1. Assume

x:

2. Choose E that

-4

an converges and 0 5 x

< 1. Let A, = CE,,ak. Show that

> 0. There is N such that IAnI < for all n 2 N. Then for n 2: N, show

For x = 1, the inequality is just IA, I < E . Thus, the series converges uniformly on [O, 11. (You may want to use the fact that = 1/(1 - x).) As a result of our special case of Abel's Theorem, we can conclude something about the behavior of the function defined by the power series if the series converges at an endpoint. We invite you to state and prove a more general theorem based on Abel's Theorem.

THEOREM Iff is defined by f(x) = C z a n Y for -1 < x < 1, x > a n converges, and f(1) is defined to be an, then f is continuous at 1.

x>

3. Supply a proof for the preceding theorem. (You shouldn't need any hints on this one.) As a result of all this, we can now supply a proof of Theorem 6.15.

xs

THEOREM Suppose C z an converges to A and bn converges to B, and let Cn be the Cauchy product of an and C z bn. If cn converges C, then C = AB.

xz

x>

xz

Supply a proof of the preceding theorem. As a suggestion, define two functions f (x) = a n y and g(x) = bnY, and use some of the results in this project, as well as Theorem 6.14.

zz

Ez

Projects 241

PROJECT 7.2 In this project you will define the trigonometric functions sine and cosine by means of power series and develop some of the properties of those functions from that definition. 1. Show that the power series

" C-"" (2n)!

and

"

C n= 1

n=O

(-ly+l p-1 (2n - I ) !

both converge for every real number x. 2. Define 00

C(x) = n=O

(-1Y' -p (2n)!

00

and S(x) = n=l

(-lY+l p-1 (2n - I ) !

for each x E R. Prove that S and C are both differentiable functions, C = -S, and St = C. 3. Prove that c2(x) + s2(x) = 1 for all x E R. You might want to examine the function F(x) = c2(x)+ s2(x) and prove that it is a constant function. 4. Prove that for all x,y E R, C(x + y) = C(x)C(y) - S(x)S(y) and S(x +y ) = S(x)C(y ) +S(y)C(x). Hint: Define g(x) to be S(x +y) -S(x)C(y) - S(y)C(x). Show that gtt(x) = -g(x), and then

5.

6.

7. 8.

Thus, (g'(x))2 + (g(x))2 is constant. Find that constant and conclude that g(x) = 0 for all x. x4 If 0 5 x 5 2, prove that 1 - 2 T 5 C(x) 5 1 - 2 + G. You might want to group terms of the series to prove these inequalities. Also, show that x - 5 S(x) 5 x for 0 5 x 5 2. Show that the equation C(x) = 0 has exactly one root between 0 and 2; call it 7. Now show that S(27) = 0. Repent for your transgressions, and use the name 7r for 27. Thus, C(a/2) = S(7r) = 0. Prove that S and C are periodic with period 27r; that is, C(x + 27r) = C(x) and S(x + 27r) = S(x) for all x E R. Prove that S and C are cofbnctions; that is, S (5 - x ) = C(x) and C (5 - x ) = S(x).

d

Index

Abel's Theorem, 239 Absolute convergence, 182 Absolute value, 28 Accumulation pint, 4 1 Additive function, 81 continuity of, 109 limits of, 81 M n e function, 110 Algebra: of continuous functions, 88-91 of derivatives, 117-121 of integrable functions, 159-162 of limits of functions, 72-76 of limits of sequences, 45-48 Algebraic number, 28, 31 Alternating series, 192, 215 Antiderivative, 158 Bijection, 12 Bijective, 12 Bolzano Intermediate-Value Theorem, 104 Bolzano-Weierstrass Theorem, 43 Bound: greatest lower, 23 least upper, 23 lower, 23 upper, 23 Bounded: from above, 23, 38 from below, 23, 38 sequence, 38 set, 23 Caratheodory, 137 Cartesian product, 10 Cauchy Mean-Value Theorem, 128 Cauchy product, 192, 194, 240

Cauchy sequence, 40 Chain rule, 119 Change-of-Variable Theorem, 82, 168, 173 Closed interval, 26 Closed set, 94 Closure, 100, 108 Codomain, 10 Compact set, 95 Comparison test, 183, 214 Complement, relative, 5 Complete ordered field, 24 Composition of: continuous functions, 9 1 differentiable functions, 119 functions, 12 Conditional convergence, 182, 190-199 Continuity, uniform, 92 Continuous function, 85 Convergence: absolute, 182 conditional, 182, 190-199 interval of, 202 pintwise, 220 radius of, 202 sequence of functions, 220 sequence of numbers, 36, 39 series, 177 uniform, 223,225 Convergent: sequence, 39, 220 series, 177 Countable set, 20 Countably infinite set, 20 Cover, 96 finite, % open, 96 Decimals, repeating, 181

Decreasing: function, 77 sequence, 53 De Morgan's laws, 5, 8 Dense subset, 26 Derivative: of a function, 114, 137 of an integral, 164 Differentiable: function, 114, 137 uniformly, 133 Disjoint sets, 4 Divergent: sequence, 39 series, 178 Domain, 10 e, 54, 60 Element, 2 Empty set, 4 Equality of sets, 2 Equivalent sets, 18 Extension, 100

Family, indexed, 6 Field, ordered, 23 Finite: cover, % set, 20 Function, 10 additive, 81, 109 m n e , 110 codomain of, 10 composition, 12,91, 119 continuous, 85 dec~easing,77 derivative, 114, 137 differentiable, 114, 137 domain of, 10 extension of, 100 image of, 10, 12 increasing, 77 integrable, 141 inverse, 12, 102, 130 limit, 220 limit of, 64 logarithmic, 111, 174 monotone, 77, 106, 149 one-to-one, 11 periodic, 108 Riemann-integrable, 141 uniformly continuous, 92 Fundamental Theorem of Integral Calculus, 157, 165 Geometric series, 180 'Greatest lower bound, 23

r

Hannonic series, 180 Heine-Bore1 Theorem, 96

Iff, 11 Image, 10, 12 Increasing: function, 77 sequence, 53 Indexed family, 6 Induction, mathematical, 13-1 7 Infimum (id), 23 Infinite: product, 217 series, 177 set, 20 Injection, 12 Injective, 12 Integrable function, 141 Integral: lower, 141 Riemann, 141 upper, 141 Integration by parts, 171 Interior, 108 Intermediate-Value Theorem: for continuous functions, 104 for derivatives, 127 Intersection, 4, 7 Interval: closed, 26 of convergence, 202 notation for, 3 open, 26 Inverse function, 12, 130,202 Inverse-Function Theorem, 130 Inverse image, 12 Irrational number, 26

J, the set of natural numbers, 3 Landau, Edmund, 23 Least upper bound, 23 Least upper bound property, 24 Left-hand limit, 82 L'Hospital's Rule, 128 Lim inf, 203-207 Lim sup, 203-207 Limit: of additive functions, 81 of a function, 64 function, 220 at infinity, 83 left-hand, 82 of monotone functions, 77-79 right-hand, 82 of sequences of functions, 220 of sequences of numbers, 39 Limit comparison test, 214 Lipschitz condition, 133 Logarithmic function, 111, 174 Lower bound, 23 Lower integral, 141 Lower sum, 141

M-test, 224 Marked partitions, 151 Mathematical induction, 13-17 Max, 23 Maximum, relative, 122 Mean-Value Theorem: Cauchy, 128 for derivatives, 124 for integrals, 166, 167 Member, 2 Mesh, 151 Midpoint approximation rule, 175 Min, 23 Minimum, relative, 122 Monotone: function, 77, 106, 149 sequence, 53 Natural number, 3 Neighborhood, 37 Newton-Raphson method, 136 nth roots, 32 nth term: of a sequence, 35 of a series, 177 Number: algebraic, 28, 3 1 irrational, 26 natural, 3 real, 23 transcendental, 28, 31 One-to-one function, 11 Onto function, 12 Open: cover, 96 interval, 26 set, 94 Ordered: field, 23 pair, 9 p-series, 186 Partial sum, 177 Partial summation formula, 191 Partition, 140 marked, 151 mesh of, 151 refinement of, 140 Periodic function, 108 Pointwise convergence, 220 Power series, 200-209 uniform convergence of, 230 Principle: of mathematical induction, 13-17 well ordering, - 13 Product: Cartesian, 10 Cauchy, 192, 194, 240 infinite, 217

R, set of real numbers, 3 Radius of convergence, 200-209 Ratio test, 186, 187 Real number, 23 Rearrangement of: sequence, 60 series, 195 Recursion, 13-1 7 Refinement, 140 Relation, 10 Relative: complement, 5 maximum, 122 minimum, 122 Repeating decimals, 181 Riemann integral, 141 Riemann sum, 152 Right-hand limit, 82 Rolle's Theorem, 123 Root test, 189, 214 Sequences(s), 35 bounded, 38 Cauchy, 40 convergent, 36, 39 decreasing, 53 divergent, 39 increasing, 53 limit of, 39 monotone, 53 nth term, 35 pointwise convergent, 220 product of, 46 quotient of, 48 rearrangement of, 60 subsequence of, 52 sum of, 45 uniform convergence of, 223-225, 230 Series: absolute convergence, 182 alternating, 192, 215 conditionally convergent, 182, 190-1 99 convergent, 177 divergent, 178 geometric, 180 harmonic, 180 interval of convergence, 202 nth term, 177 p-, 186 power, 200-209 product, Cauchy, 192, 194, 240 radius of convergence, 202 rearrangement of, 195 sum, 178 Taylor, 209-2 12 uniformly convergent, 224, 230 Sets, 2 bounded, 23 bounded from above, 23 bounded from below, 23

246 Index Sets, (continued) closed, 94 compact, 95 couhble, 20 countably infinite, 20 dense, 26 disjoint, 4 element of, 2 empty, 4 equality of, 2 equivalent, 18 finite, 20 infinite, 20 intersection, 4, 7 max of, 23 member, 2 min of, 23 open, 94 product of, 10 uncountable, 20, 62 union, 4, 7 Subcover, % finite, % open, 96 Subsequence, 52 Subset, 3 dense, 26 proper, 3 Sum: lower, 141 partial, 177 Riemann, 152 upper, 141

Supremum (sup), 23 Surjection, 12 Surjective, 12 Taylor Series, 209-2 12 Taylor's Theorem, 163, 168, 210 Test: alternating series, 192, 215 comparison, 183, 2 14 limit comparison, 214 M-test, 224 ratio, 186, 187 root, 189, 214 Transcendental number, 28, 31 Uncountable set, 20, 62 Uniform: continuity, 92 convergence, 223-225, 230 Uniformly differentiable, 133 Union, 4, 7 Upper bound, 23 Upper integral, 141 Upper sum, 141 Variable, change of, 82, 168, 173 Wallis formula, 217 Weierstrass M-test, 224 Well-ordering principle, 13, 16 2, set of integers, 3

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