Introduction to Analysis [5 ed.] 0821847872, 9780821847879

Table of contents :
Cover
Title page
Copyright
Preface
Contents
Chapter 0. Preliminaries
Chapter 1. Sequences
Chapter 2. Limits of functions
Chapter 3. Continuity
Chapter 4. Differentiation
Chapter 5. The Riemann integral
Chapter 6. Infinite series
Chapter 7. Sequences and series of functions
Index
Back Cover

Citation preview

Introduction to Analysis Fifth Edition

Ed.ward D. Gaughan

Introduction to Analysis -

Fifth Edition

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UNDERGRADUATE

ffT'~s · I

Introduction to Analysis Fifth Edition

Edward D. Gaughan

(fiiiit\ American Mathematical Society ~ Providence, Rhode Island

2000 Mathematics Subject Classification. Primary 26-01.

For additional information and updates on this book, visit www .ams.org/bookpages / amstext-1

Library of Congress Cataloging-in-Publication Data Gaughan, Edward. Introduction to analysis-5th ed. / Edward D. Gaughan. p. cm. - (Pure and applied undergraduate texts ; v. 1) Includes index. ISBN 978-0-8218-4787-9 (alk. paper) l. Mathematical analysis. I. Title. QA300.G34 515-dc22

2009 2008047387

Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Acquisitions Department, American Mathematical Society, 201 Charles Street, Providence, Rhode Island 02904-2294, USA. Requests can also be made by e-mail to reprint-permission©ams. org.

©

1968, 1975, 1987, 1993, 1998 held by the American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America.

§ The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability. Visit the AMS home page at http://www. ams. org/ 1098765432

14 13 12 1110 09

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 5th edition contains a number of changes recommended by the reviewers and users of earlier editions of the book. Chapter O contains introductory material on sets, functions, relations, mathematical induction, recursion, equivalent and countable sets, and the set of real numbers. As in the 4th 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 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. At the end of each chapter, you will find several PROJECTS. The purpose of a PROJECT is to give the reader a substantial mathematical problem and the necessary guidance to solve that problem. A PROJECT is distinguished from an exercise in that

I

V

vi Preface

the PROJECT is a multi-step problem that can be very difficult for the beginner without significant assistance. PROJECTS are sometimes used to cover material not included in the chapter discussions or to offer alternative methods for proving important theorems. 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.

TO THE INSTRUCTOR Suppose you have just been assigned to teach the junior-senior level course in analysis this coming semester. The designated text will be Introduction to Analysis, 5th Edition. What should you know before and during the planning of the course you will offer to your students? First of all, you should be aware of some of the special features of the book. Each chapter is divided into sections and the exercises are grouped in a similar fashion. So, for example, if you are discussing Section 2.3 in class, the exercises related to that material will be found in Section 2.3 of the exercise set. As mentioned previously, some exercises are starred to indicate that they contain facts vital to later material. The starred exercises are not necessarily the most difficult ones. An INSTRUCTOR'S MANUAL is available to assist you in planning your course. It contains suggestions for solutions and comments concerning exercises and PROJECTS. It allows you to see at a glance what is needed for the solution of a particular exercise and assist you in deciding when to assign a problem and what hints you might need to offer your students. PROJECTS are multi-stage problems and probably should be assigned over a period of several class meetings. These problems are broken down into small steps that eventually lead to the solution. In the early part of the semester, you may need to lead your class step-by-step, but as students gain confidence, they will find success comes more easily. Some of the PROJECTS are designed to offer you alternatives in approaching certain important theorems. In Chapter 1, the Bolzano-Weierstrass Theorem is proven and used to show that every Cauchy sequence is convergent. PROJECT 1.5 and PROJECT 1.7 offer two ways of obtaining the convergence of Cauchy sequences without the need for the Balzano-Weierstrass Theorem. In Chapter 3, PROJECT 3.1 gives a proof without the use of compact sets that a function continuous on a closed and bounded set is uniformly continuous. In addition, PROJECT 3.1 yields the result that the continuous image of a closed and bounded set is also closed and bounded. As you plan for the coming semester, you will need to make some choices. Unless your class is exceptional, it will be impossible to cover the entire book in one semester. So, what do you cover, and in what order? Most students will have some familiarity

Preface vii with the contents of Sections 0.1, 0.2, and 0.3. How much time you devote to those sections will depend on the extent of that familiarity. Section 0.5 is especially critical as many students have little appreciation of the structure of the system of real numbers as a complete ordered field. For that reason, a comprehensive coverage of Section 0.5 is' crucial. Chapter 1, Sequences, is vital to the study of analysis and deserves careful attention. At this stage, you have some choices. Chapter 6, Infinite Series, is almost independent of Chapters 2-5, enough so that you can go directly from Chapter 1 to Chapter 6 if you so desire . .Otherwise, Chapters 2, 3, 4, and 5, should be covered in that order. This text is designed to offer you flexibility in designing your course and assistance in that effort. I hope you enjoy teaching this course as much as I enjoyed writing the book. If you do, I'm sure your students will share the pleasure with you.

ACKNOWLEDGMENTS I would like to acknowledge the reviewers: Josefina Alvarez, New Mexico State University; J. W. Neuberger, University of North Texas; Alec Norton, University ofTexasAustin; and Richard Thompson, University of Arizona. My sincere appreciation is also extended to Professor Robert J. Wisner, who, some 33 years ago, urged me to write the first edition of this book, and who has made many helpful suggestions relative to each of the first three editions; to Jack N. Thornton, who as managing editor of Brooks/Cole accepted and nurtured a fledgling author through the first and second editions; and to Gary Ostedt, mathematics publisher at Brooks/ Cole, for his support and encouragement in the preparation of this 5th edition. Of course, this book would not be in your hands without the excellent work of Marlene Thom, Brooks/Cole, and Sandra Gormley, University Graphics, who carried it and me through the stages of production. My thanks go to them also. Edward D. Gaughan

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Contents

PREFACE v

Chapter 0.J 0.2 0 .3 0.4 0 .5

1.3 1.4

Preliminaries

1

SETS 2 RELATIONS AND FUNCTIONS 8 MATHEMATICAL INDUCTION AND RECURSION 12 EQUIVALENT AND COUNTABLE SETS 16 REAL NUMBERS 21 EXERCISES 27 PROJECTS 29

Chapter 1.1 1.2

O

J

Sequences

33

SEQUENCES AND CONVERGENCE 33 CAUCHY SEQUENCES 38 ARITHMETIC OPERATIONS ON SEQUENCES 42 SUBSEQUENCES AND MONOTONE SEQUENCES 49 EXERCISES 54 PROJECTS 57 ix

x

Contents

Chapter

2 .J 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

76

EXERCISES 79 PROJECTS 80

Chapter

3 .1 3.2 3.3 3 .4

3

83

Continuity

CONTINUITY OF A FUNCTION AT A POINT 83 ALGEBRA OF CONTINUOUS FUNCTIONS 86 UNIFORM CONTINUITY: OPEN, CLOSED, AND COMPACT SETS 89 PROPERTIES OF CONTINUOUS FUNCTIONS 96 EXERCISES 104 PROJECTS 106

Chapter

4

Differentiation

111

4.1

THE DERIVATIVE OF A FUNCTION 112

4.2 4.3

THE ALGEBRA OF DERIVATIVES 115 ROUE'S THEOREM AND THE MEAN-VALUE THEOREM 119

4 .4

L' HOSPITAL'S RULE AND THE INVERSE-FUNCTION THEOREM 126 EXERCISES 129 PROJECTS 132

Chapter

5

The Riemann Integral

5.J

THE RIEMANN INTEGRAL 138

5 .2 5 .3

RIEMANN SUMS 148

5 .4

THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS 154

CLASSES OF INTEGRABLE FUNCTIONS 146

137

Contents xi

5 .5

ALGEBRA OF INTEGRABLE FUNCTIONS 155

5.6 5 .7

DERIVATIVES OF INTEGRALS 162 MEAN-VALUE AND CHANGE-OF-VARIABLE THEOREMS 162 EXERCISES 165 PROJECTS 169

Chapter 6

In.finite Series

6.J

CONVERGENCE OF INFINITE SERIES 173

6.2 6 .J 6.4 6.5 6.6

ABSOLUTE CONVERGENCE AND THE COMPARISON TEST 178

173

RATIO AND ROOT TESTS 182 CONDITIONAL CONVERGENCE 186 POWER SERIES 194 TAYLOR SERIES 203 EXERCISES 206 PROJECTS 210

Chapter

7

Sequences and Series of Functions

7.J

POINTWISE AND UNIFORM CONVERGENCE 216

7.2 7.J

CONSEQUENCES OF UNIFORM CONVERGENCE 221 UNIFORM CONVERGENCE OF POWER SERIES 225 EXERCISES 232 PROJECTS 234

INDEX 237

215

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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, which is English in this case. In particular, the sentences formed with symbols must make sense when

B

1

2

Chapter O Preliminaries

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. To build the intuitive idea of a set, consider the following examples: 1. 2. 3. 4. 5.

The The The The The

set of all natural numbers set of all the letters of the Greek alphabet set of all rational roots of the equation x 2 + 1 = 0 set consisting of the rational numbers 1, 2, and 3 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.

= B if and only if every object belonging to A also belongs to B and every object belonging to B also belongs to A.

DEFINITION If A and B are sets, then 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 xis an object that belongs to A, we say "xis an element of A" or "xis a member of A" and write x EA. If x does not belong to A, we write x ft_ 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 EA implies x EB, and x EB implies x EA. Let A 1 be the set in example 1, A 2 be the set in example 2, and so on. Now, as observed above, A 4 = A 5 and A 1 A 4, because 13 E A 1 and 13 ft_ A4. Note, however, that each member of A 4 is also a member of A 1 • In a sense, A 4 is a part of A 1•

*

DEFINITION 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. If A C B and A =I= B, then A is a proper subset of B.

0.1

Sets 3

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 CB and B CA, then A = B. To prove the theorem, we must prove that both (I) and (2) hold. Follow along. 0.1 THEOREM B CA.

Let A and B be sets. Then A

=B

if and only if A C B and

Proof Assume A = B. Then, if x EA, then x EB since A = B (see the definition of A = B); hence A CB. Likewise, if x EB, then x EA since A = B, and we have BC A. Thus, if A= B, then AC Band BC A. Assume now that AC Band BC A. We want to show that A = B. Choose any x EA. Since A CB, then x EB. If x EB, then since B CA, we have x EA. 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. ■ Example 0.1 The set of all natural numbers is {x: xis a natural number}. The set of all integers less than 4 and greater than O 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. 2. 3. 4. 5. 6. 7. 8.

J Z

= set of all natural numbers-that is, the set of all positive integers.

= set of all integers.

= the set of all real numbers. Fora P(A) and define C = {x: x E A and x f£. 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, bJ is equivalent to [c, d]. (Hint: Show that [a, b] is equivalent to [0, l] first.)

0.5 REAL NUMBERS x+y *39. If x < y, prove that x < - - < y. 2

*40. If x.?: 0 and y

.?:

0, prove that Vxy s x ; y. (Hint: Use the fact that

(Yx - Vy) 2 .?: 0.)

*41. If 0 < a < b, prove that 0 < a 2 < b2 and 0 < Va < Vb. x a x x+a a 42. If x, y, a, and bare greater than zero and - < - , prove that - < - - < - . y b y y+b b 43. Let A = {r : r is a rational number and r 2 < 2}. Prove the A has no largest member. (Hint: 2If r 2 < 2, and r > 0, choose a rational number 8 such that 0 < 8 < 1 and 8 < - - - . 2r+ 1 Show that (r + 8)2 < 2.) *44. If x = sup S, show that, for each e > 0, there is a E S such that x - e < a 5 x. *45. If y = inf S, show that, for each E > 0, there is a ES such that y s a< y + e.

,.2

PROJECT 0.1 The purpose of this project is to show that the open interval (0, I) is equivalent to the closed interval [O, I]. In the process we will discover that both intervals are equivalent to [O, 1) and (0, I]. It is then easy to generalize to any interval [a, b] with a < b.

30 Chapter O Preliminaries

Define f : (0, I) -

R as follows.

For n E J, n

t(!) = -

2,

2::

1 n - 1

n

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

f(x)

= x.

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

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 xn = p.

Proof 1. Use the factorization

to prove that any positive solution to the equation xn = p is unique. 2. Suppose that p is a positive real number greater than 1 and define A to be the set Iz : z is a positive real number and zn ::s p). Show that A is nonempty and bounded from above. 3. Let x = sup A. Suppose x" < p. Let

i

1

c

=

1

k=O

and choose

11 (

k

)

xk

oto be the smaller of I and p

- xn. Show that C

contrary to x being an upper bound for A. 4. Assume xn > p. Let e = x"-p. Choose y such that O


0, there is at least one number N with the desired property. Follow along to see that {¼1;;"=1 converges to 0. Choose E > 0. Let N be an integer larger than¼- So if n 2: N, we have!; < E. This means that if a,, = !; and A = 0, then for n 2: N, la. - A I = ¼ < E. 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 look for other ways of expressing the notion. Recall that, if a and b are real numbers and E > 0, then la - bl < E iff b - E < a < b + e. Thus, Ia. I;;"=1 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 that, given E > 0, A - E < an < A + E is true e,xcept for a finite set of subscripts-such as, in the case mentioned above, some subsc1ripts from the set { l, . .. , N - I I. E.

t;

st

1. 1 Sequences and Convergence 3S

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 such that (x - E, X + E) C Q.

E

>0

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 Ian },;'. 1 converges to A iffeach neighborhood of A contains all but a finite number of terms of the sequence.

Proof Suppose {anl;:"= 1 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 lanl:=• converges to A and E > 0, there is a positive integer N such that for n ::::: N, Ian - A I < e. In particular, for n 2:: N, we have A - E < an < A + e; hence, an E (A - e, A + e) C Q. Thus, Q contains all terms of the sequence except possibly some of the terms a 1, a 2 , ••• , aN- i ; hence Q contains all but a finite nu1mber of terms of the sequence.

Conversely, suppose each neighborhood of A contains all but a finite number of terms of the sequence Ian I:=•· Choose e > 0. Then Q = (A - e, A + e) is a neighborhood of A and contains all but 21 finite number of terms of the sequence. This means that there is a finite set S = :In., n2 , ••• , n,) of positive integers such that, if an fl. Q, then n E S. Let N = (max S) + 1. Now, if n 2:: N, then n (£. S and, by the definition of S, an E Q; that is, A - E < an < A + E. This last statement is equivalent to the statement Ian - Al < E. Hence, for n::::: N, Ian - Al < E. We have shown that {an),;'. 1 converges to A. We now can answer a question that should be lurking in the back of your mind. Can a sequence lan);:"~1 converge to two different real numbers? Suppose the answer is yes. Then there would be distinct real numbers A and B such that Ian)':=1 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 B respectively, such that P n Q is empty. You aire asked to give a proof of this in Exercise 2. Now, as we have observed above, P contaiins all but a finite number of terms of the sequence lanl:=i• Since lanl:=1 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

36 Chapter 1 Sequences

P n Q is empty, we see that P must contain imfinitely many tenns of the sequence and also must contain at most a finite number of terms of the sequence. Thus, the assumption that (anl:'- 1 converges to both A and B with A :f. 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 tenns. 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 mighit be classified as one that does not seem to have an intuitive proof. Thus, we disting:uish between a difficult theorem and a theorem that is difficult to prove even though the idea of its proof may be intuitively clear. I.I THEOREM

If Ian I.~=• converges t10 A and also to B, then A

= B.

Proof This will be a proof by contradi,ction. We will assume that lanl:=1 con-

*

verges to A and also to B, and A B. We can also assume that A < B. Let E = ½(B - A). Since lanl:. , converges to A, there is N such that for all n 2:: N, we have

A -

E

< an < A +

E.

But A + E = A + ½(B - A) = ½(A + B) = B - ½(B - A) = B - E. This means that for n 2:: N, an < B - E, so there are infinitely many terms of the sequence outside the neighborhood (B - E, B + E). Thus the sequence cannot converge to 8. This contradicts our original hypothesis that {an I;= 1 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 In I:'=1, 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 lanl:=• is bounded from above iff there is a real number M such that ,an s M for all n. A sequence Ian) ; . 1 is bounded from below iff there is a real number P such that P s an for all n. A sequence is bounded iff it is both bounded from above and bounded from below. Note that a sequence I an I;= 1 is bounded iff there are real numbers P and M such that P s an s M for all nor, equivalently, iff there is a real number S such that lanl s S for all n. Let us now suppose that (anl:'=t 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 - I, A + I), there is a positive integer N such that A - l < an < A + l for all n 2:: N. Now we are assured that all terms of the sequence, except possibly the terms a., a:1, ... , aN_., are bounded from below by A - l and from above by A + l. It is now ,easy to find upper and lower bounds for the possibly wayward terms a., ... , aN- t be,cause this is a finite set of real numbers. It should now be clear how upper and lower bounds can be chosen for the sequence

lanl:'=1•

1.1 Sequences and Convergence 37

1.2 THEOREM If lanl::"=i converges tJO A, then lanl::"=i is bounded.

=

Proof Suppose lanl::"= 1 converges to A. Choose e 1. There is a positive integer N such that, if n ~ N, then A - 1 0 there is a positive integer N such that if m, n 2: N, then

Ian - am l
0. There is a positive integer N such that, if n, m 2: N, then A - E < an < A + E and A - E < am < A + E. Thus, for all n, rni 2: N, we find an E (A - E, A + E) and am E (A - E, A + E). The set (A - E, A + e) is an interval of length 2e; hence, the E

difference between an and am is less then 2E.. 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 lanl.:'=i converges to AL Choose E > 0. Then½> 0. There is a positive integer N such that n 2: N implies Ian - A < ;. (The choice of! is not a mere whim. We observed previously that the difference between an and am was less than twice the original choice of E.) Now, if m, n 2: N, then Ian - Al 0 such that (t - E, t + E) C P. Now t - '" is not an upper bound for Q, since tis the least upper bound; hence, there is a pt0sitive integer n such that t - E < a,. .s t . In fact, if m > n, then, by the construction above, t - E < a,. .s am .s t. Each interval [am, bml contains infinitely many points of S; hence, the proof would be complete if m could be found such that t ·- e < am < bm < t + E. As noted above, t - E < a,,, .st form 2::: n; hence, it will suffice to choose m large enough so that m 2::: n and T'"(/3 - a) < E. (2- m(/3 - a) is the length of the interval [am, bm].) To summarize, choose m 2::: n such that 2-'"(/3 - a) < E; then t - E < am .s t s bm = am + T'"(/3 - a) < t + E. Thus, P contains (am, bm]; hence, P contains infinitely many members of S, and tis an accumulation point of S.

42 Chapter 1 Sequences

Suppose now that {a,, I;;_ 1 is a Cauchy sequence. If the range is finite, say {s., ... , s,}, and if we choose E

= min{ Is;

- si l : i

* j, i, j = l , ... , r},

then there is a positive integer N such that n, m :=: N implies that la,, - a,,. 1< E. Since = s1 and a,,. = sk for some j and k amon~: { l, ... , r) and E was chosen to be the least distance between distinct members of the range, we must have an = a,,. for n, m :=: N. Thus, the sequence is constant from some point on and hence converges (see Exercise 11 ). If the range of a Cauchy sequence is in1iinite, then, by Theorem 1.4, the range is an infinite bounded set and hence, by the Bolzano- Weierstrass Theorem, has an accumulation point. We have been searching for a point where the sequence "piles up," and this accumulation point should be it.

a,,

1.7 THEOREM Every Cauchy sequence is convergent.

Proof Let {an);;'= 1 be a Cauchy sequence. By the preceding remarks, if the range is finite, the sequence is constant from some point on; hence it converges. Suppose the range is infinite-call it S. By Theorem 1.4, S is bounded; hence, by the Bolzano-Weierstrass Theorem, S hais an accumulation point-call it a. We shall prove that {anl'::=1 converges to a. Choose E > 0. Since (a - !, a + is a neighborhood of a, it contains ilnfinitely many members of the set S. Since lanl'::=i is Cauchy, there is a positiive integer N such that n, m::: N implies Ian - a,,. I o.

(See the paragraphs preceding Lemma 1.10 to understand the reason for this choice of e'.) There is a positive integer N 2 such that, for n :::= N2 , Ian - Al< e', and a positive integer N3 such that Iii :::: N3 implies lbn - Bl < e'. Let N = max{Ni, N2, N3}. For n :::= N , lbn - Bl< E', Ian - Al< E', and lbnl ~ M. Thus,

~1

an_ = lanB - bnAI = lanB - AB + AB - bnAI B bnB bnB I bn < 1-an_-_A, + '---IA____ llb_n_- Bl bn lbnllBI E'C:nl +


0. Since {an) ;= 1 converges to A , there is a positive integer N such that, for n 2: N, Ian - Al < E. Since Ian,) k= 1 is a subsequence of the given sequence n 1 < n2 < ... , then k s n*. Thus, fork 2: N, nk s N; hence, Ian. - A l < E. Therefore, fan,IZ"- 1 converges to A. Note that if every subsequence of a given seqruence 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 Theiorem 1.14 if we assume the sequence is bounded.

1.15 THEOREM Suppose lxnl:=i is 21 bounded sequence. If all its convergent subsequences have the same limit, then the sequence is convergent.

Proof Assume that all the convergent subsequences of lxnl:=i have the same limit; call it x0 • Exercise 36 guarantees 1that there is at least one convergent subsequence. If lxn );. 1 does not converge to x0 , 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 n 1 > 1 such that lxn, - Xol 2: · E. Likewiise, there is ~ > 2 (and we may choose n2 > n 1) such that lxn, - x0 1 2: E. Continuing in this fashion (we are defining the

1.4 Subsequences and Monotone Sequences 51

sequence In* I;=1 recursively), we find, for each positive integer k, an integer nk such that n1 < n2 < n3 • • • < nk- i < n* and such that lxn, - x0 j 2: e. The sequence lxn,lk=l, being bounded, will have a co111vergent 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 lanl:= i is .increasing iff an $ an+I for all positive integers n. A sequence Ian);= 1 is decreasing iff an 2: an+ 1 for all positive integers n. A sequence is monotone iff it is either increasing or decreasing.

1.16 THEOREM A monotone sequenc,e is convergent iff it is bounded. Proof Suppose I an I; . 1 is a monotone s,equence that is bounded; for definiteness, suppose the sequence is increasing. The set {an : n = l, 2, .. , I is bounded, so let s = suplan: n = I, 2, ... }. It will be shown that Ian}:= • converges to s. Choose e > 0. Since s is the least upper bound of I a" : n = I, 2, ... I, s - e is not an upper bound; hence, there is n0 such that s - e < a"". Now, for n 2: n0 , hence I an} ;e1 converges to s. For the cas:e in which I an);= 1 is decreasing, lets inf Ian : n = I, 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 I;= 1 is indeed monotone, but it does not converge because it is unbounded. The sequence I 0, 1, 0, ... I is bounded, but the theorem does not apply, because it is not monotone; indeed, it fails to converge. ■

Example 1.6 Consider the sequence {s,,};= 1 defined as follows: Si

= \1'2,

Sn=

Y2 + ~

for

n

= 2, 3, ....

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, s1 s 2 and S2

= Y2 + 4¢"2 2: V2 = S1.

We wish to show that Sn s 2 and Sn+i 2: Sn for all n; the preceding statement verifies the truth of this for n = I. Assume the condiltion is satisfied for n = r. Then

S,+I = \/2 + and

Vs, s Y2 + V2 s \/2 + 2 = 2

52 Chapter I Sequences

hence, by the principle of induction, the inequalities hold for all n. By use of Theorem 1.16 we are assured that the sequence converges; call the limit L. Now the subsequence lsn- • };'=2 also converges to L, and since sn > 0 for all n, {~};'=2 converges to VL. Now observe that since sn = v'2 + v:~n- •• the sequence

{s,,} ; =,

= { v'2 + ~ }~=•

converges to v'2 + VL, and also that {snI;=1 converges to L; thus, by the uniqueness of the limit of a sequence, L = v'2 + vf. By eliminating the radicals, we see that L must be a root of the polynomial equation L4 - 4L2 - L + 4 = 0. ■ It is left as an exercise to consider the sequence {(I + ¾YI ;'a 1 • This sequence may be shown to be increasing and bounded, and hence convergent. The limit happens to bee, 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 .17 THEOREM Let E be a set of re.al numbers. Then x0 is an accumulation point of E iff there is a sequence {x,,};'= 1 of members of E, each distinct from x0 , such that {x,, I;= 1 converges to Xo·

Proof Let x0 be an accumulation point of E. Then, for each positive integer n, there is a point x,, EE such that O < Ix,, - .xol < ¾. [The set {Xo - ¾, Xo +¾)is a neighborhood of x 0 , so it contains a member of E distinct from XQ.] It remains to be shown that /xnl:=1 converges to x0 • Choose E > 0. Then there is a positive integer N such that¼ < N ; hence, for n 2= N, Ix,, - x0 1< ¾ s ,¼ s E. Thus, {x,,};'= 1 converges tox0 • Suppose now there is a sequence {x,, );_ , of members of E, each distinct from Xo, that converges to x0 . Since every neighborhood of x0 contains all but a finite number of terms of the sequence, every neighborhood of x0 must contain at least one member of E that is distinct from x0 • Hence, x0 is an accumulation point of E. We close this chapter by examining examples involving techniques that are of sufficient interest to be studied with care.

Example 1.7 Consider a real number O < b < l imd form the sequence {bn);=•• Observe that bn- • - bn = bn- i(l - t,) > O; hence, the sequence is decreasing.

■

In order to show that this sequence converg,~s, it suffices to show that it is bounded from below, an easy task since b" > 0 for all natural numbers n. Although we know now that the sequence converges, we shall seek also to determine its limit. Now Ib" I;= 1 converges; call its limit L. And I b2n) ;:=., being a subsequence of Ib" I; • ., also converges to L. On the other hand, {b2n I;=1, which may be considered as the product of the sequence {b" I;. 1 with itself, converg,~s to L2 • By the uniqueness of the limit, L 2 = L , and so L = 0 or L = l. Clearly, L l since the sequence was decreasing and

*

1.4 Subsequences and Monotone Sequences 53

< 1. Hence, {b"l:= 1 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. ■

b

■

Example 1.8

"\1/c

'Cfc -

< c < 1 and the sequence Cefcl:=1• For all n, = efc(l - n(n -~ ) > 0, Consider 0

since c < l. Hence, the sequence is increasimg, and clearly efc < l for all n. Thus, the sequence converges; call the limit L. By Exercise 28, if {x,.};. 1 converges to x0 , x,, 2: 0 for all n, then IYXnl:= 1 converges to Therefore, the sequence• {\f¼}:= 1 converges to VL, but

v.x;;.

~

= {/c,

and {Vcl:=1 is a subsequence of {efc}:- i andl hence converges toL. By the uniqueness of the limit of a sequence, VL = L; hence L == 0 or L = 1. But c > 0 and the sequence is increasing; therefore L 0. In conclusion, Iefc) 1 converges to 1. (See Exercise 38 for the case c > 1.) ■

:=

*

■ 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 a 1 = 1, and n 2: 2, define a,, = ~ - We shall also us,e induction to attack this sequence. First, look at a few terms of the sequence,

a 1 = 1,

a2 =

\/2,

a3 =

~L

It seems reasonable to expect the sequence tc, be increasing and, hence, convergent if it is bounded. It may not be clear at first glan,ce what to try for an upper bound, so we shall discover this bound by a very useful de.vice. (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,,):= 1 converges to L; then {~):. 11 converges to L2 , but a~ = 2a,._ 1; hence, {a~}:= 1 = {2a,,_ 1 ):=• converges to 2L. Thus, the only candidates for the limit must be the solutions of the equation L2 = 2L; either L = 2 or L = 0. It is easy to rule out the possibility L = 0, so L = 2 is the obvious cho,ice. Note that it has not been proved that Ia,.):= 1 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,. s a,.+ 1 s 2 for all n; hence, the sequence is bounded and monotone and thus is convergent. For n = 1, a 1 = 1 :5 a 2 = V2 :5 2. Assume the statement is true for n = r; in other words, a, s a,+ 1 :5 2. Then a,+1 =

~

:5

~=

a,+ 2

and a,+2

= ~ < \lf':'2 = 2,

54 Chapter J Sequences

and the statement holds for n fanl:'- 1converges to 2.

= r + 1. By induction, the statement holds for all n, and ■

■ Example 1.10 Recall now a word of philosophy mentioned earlier-namely, that determining the limit of a sequence mat be half the battle in showing a sequence to be convergent. Consider the sequence {"O'n I;. 1 • Let us try to guess the limit in advance by reasoning similar to that used in the precc:ding para~ Suppose {'" 0. There is N1 such that for n ~ N1 la. - Al < J, and there is N2 such that for n ~ N2 la. - Bl< J. Use the triangle inequality to show that this implies that IA- BI < E. Argue that A = B. Let x be any positive real number, and define a sequence Ia. I;:'. 1 by

an

+ • • • + [nx) = [x) + [2x] n2

where [x] is the largest integer less than or equal to x. Prove that (a.};:'. 1 converges to x/2.

12 CAUCHY SEQUENCES 14. Prove that every Cauchy sequence is bounded (Theorem 1.4). 15. Prove directly (do not use Theorem 1.8) that, if (a.};:'. 1 and (b.};:'. 1 are Cauchy, so is (a.+ b.};:'. 1• 16. Prove directly (do not use Theorem 1.9) that, if (a.};:'. 1 and {b.};:'. 1 are Cauchy, so is (a.b.};:'. 1 • You will want to use Theorem 1.4. 2 17. Prove that the sequence { n+ l}• is Cauchy. 18. 19. 20. 21. •22.

n n- 1 Give an example of a set with exactly two aocumulation points. Give an example of a set with a countably infinite set of accumulation points. Give an example of a set that contains each of its accumulation points. Determine the accumulation points of the set ( 2" + ¼: n and k are positive integers}. 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.

. · rea1 numbers. Define a. 23. Let a0 and a 1 bed1stmct

= a.-, +2

a•- 2 fior eac h pos1t1ve · · integer ·

n ~ 2. Show that (a.l;:'= 1 is a Cauchy sequence. You may want to use induction to show

that

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

56 Chapter I Sequences 24. Suppose fan):. , converges to A and fan: n E J) is an infinite set. Show that A is an accumulation point of Ian: n E JI.

1.3 ARITHMETIC OPERATIONS ON SEQ!UENCES 25. Suppose {anJ:. 1 and {bnl:-, are sequences such that {anl:=, and {a. + b.);. , converge. Prove that lbn}:. 1 converges. 26. Give an example in which fanJ:. 1 and lbn):. , do not converge but {a. + b.);. , converges. 27. Suppose fa.);. , and {b.l;. 1 are sequences such that {a.1;. 1 converges to A ,t. 0 and {a.b.1:- , converges. Prove that fb.1;. 1 conv1:rges. 28. If {anl:. 1 converges to a with a. 2: 0 for all r.1, show {Vci.:};., converges to Va. (Hint: If .r- . r a-a a> 0, then va. - va = vi_ Va.) a.+ a 29. Prove that

converges to f.., where (

n +

k

k) = (n + k)! . n!k!

30. Prove the following variation on Lemma 1.10. If lbnl:., converges to B ,t. 0 and b. ,t. 0 for all n, then there is M

> 0 such that lbnl 2: M for all n.

31. Consider a sequence {a.);;. , and; for each n, define an

=

n

Prove that if {a.I;;_ 1 converges to A, then (a.I ;;. 1 converges to A. Give an example in which {a.);;_ 1 converges, but lanJ:. 1 does not. 32. Find the limit of the sequences with general tf:nn as given: n 2 + 4n (a) - 2 - - c n - ., cos n (b)n sin n2 (c)

v,j n

(d) n2 - 3 (e)

(R-2)

n

v,j

0. There is a positive integer q0 such that l/q0 < E. There are at most a finite number of rational points in [O, 1) of the fonn p/q where p and q are positive integers with q < q0 , say r1o .•. , '" · We may assume that x0 is deleted from this list if

2.2 Limits of Functions and Sequences 69

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

= minf lxo

8

- r;l:i = 1, ... ,n),

and observe that 8 > 0. Now if O < Ix - x0 I < 8 and x E [0, I], then x is either irrational, in which case f(x) = 0, or x = p/q where p and q are relatively prime with q > q0 , in which case f(x) = liq. In either case,

lf(x) -

1

I

q

Qo

OI :5 if(x)I :5 - < -
0 such that D n (x0 - E, Xo + E] = E n (Xo - E, x0 + E] . If f(x) = g(x) for all x E D n E n [x0 - E, Xo + E] , prove that / Illas a limit at x0 iff g has a limit at x0 .

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 j(: 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.

Project 2.2

81

1. Show that for each positive integer n and each real number x, f(nx) = nf(x). 2. Suppose f is such that there are M > 0 and a > 0 such that if x E [-a, a], then lf(x)I s M . Choose E > 0. 1bere is a positive integer N such that MIN< E. Show that if Ix - YI < a/N,, then lf(x) - f(y)I < E. 3. Prove that if there are M > 0 and .a > 0 such that if x E (-a, a], then lf(x) I s M, then f has a limit at each x in Rand lim,-x f(t) = f(x). 4. Prove that if f has a limit at each x E R, then there are M > 0 and a > 0 such that if x E [-a, a], then lf(x)I s M. You have now proven that if f : 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 lf(x)I s M. In addition, if the condition is satisfied, lim1-+x f1(t) = f(x).

PROJECT 2.2 _ _ _ _ _ _ _ _ _ _ _ _ _ __ In Exercise 39 of Chapter I, 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 sequience for convergence. The theorem you will be asked to prove in this project will ht~ 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 fufither. 1. Prove the following theorem.

n B = 0, x0 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. Prove that h has a limit at x0 if and only if f and g each have a limit at Xo and limx---xo f(x) = limx-Xo g(x).

THEOREM Suppose f : A - R and g : B - R are such that A

You probably recall reading about one-sided liimits in your calculus book. This theorem allows us to examine the relationship betweein one-sided limits and two-sided limits. We need a definition to set the stage. Let f : D - R, and suppose Xo is an accumulation point of {x: x E D, x > x0 ). Then f has a right-hand limit L at Xo iff for each E > 0, there is 8 > 0 such that if Xo < x < x 0 + 8, x ED, then lf(x) - LI < E.

DEFINITION

2. State carefully the definition of left-hcmd 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 hats 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.

82 Chapter 2 Limits of Functions

PROJECT2~--------------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 Suppose f: A - R, g: B -A. Suppose that a is an accumulation point of A, bis an accumulation point of B,

i. lim,_b g(t) = a, ii. there is a neighborhood Q of b siuch that for t E Q iii. f has a limit at a. Then f

O

g has a limit at b and limx-a f(x)

2. Consider f: R\ {0} - R by f(x)

n B, g(t) * a, and

= lim,- f(g(t)).

= !in x. A standard argument in calculus is X

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 : R \ { 'TT(l) - R defined by h(x)

=~ has a limit at f, and find that limit. 'TT - -x 2

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 havin,g a limit "at infinity."

DEFINITION If f is defined on (x: x > a) for some real number a, then f has a limit L at infinity if, for each e > 0, th,ere is M such that for x lf(x) - LI < E. l. Define f(x)

Vx =- for x > 0. Prove: that f x+2

2. Define f(x)

=

Vx

~ r for x 1 - vx

>

> M and x >

a,

has a limit at infinity, and find it.

1. Prove that f has a limit at infinity, and find

it. 3. If f is defined on {x: x > a), define g(x) = f(!) for 0 < x 0. Since (ii) holds, there is 8 > 0 such that if O < Ix - x0 1< 8 and x E £, then

lf(x) - f(.xo) I
0.

lg(x)I ~a.Choose

E

> 0 such that for Ix - x01 < 8' and x E D, lg(x) - g(x0)1 < e'.

There is 8'

Let

8"

= min{ 8',

Now 8"

!

I

8}.

> 0 and for Ix - x0 1 < 8" with x (x) -

g

! g

(x-o)l

=

lg(xo) - g(x) g(x)g(Xo)

ED, we have

I
0 such that if Ix -

lf(x) - f(xo)I

X-O I
.o· Hence there is e > 0 such that (a - e, a + e) C GAo. This means that the interval [a, a + ll can be covered by the open set GAo. Let B = Ix: x E [a, /31 and [a, x] can be covered by a finite subfamily of {G,\l>.eAI- Bis bounded and nonempty since [a, a + l] CB. Let z := sup B. We know that a + ! ~ z and z ~ /3. So, there is A1 E A such that z E GA,• Hence there is 13 > 0 such that (z - S, z + S) C G> "· Now [a, z - ~] can be covered by a finite subfamily of {G" h eA and hence [a, z + ~] can be covered by the subfamily of {G" he A that covers [a, z - ~] plus the open set G,\,• 1fhis contradicts the definition of z unless z = /3. Thus B = [a, /3] and [a, /3] is compact. Now suppose Eis closed and bound,ed. Then there is a closed interval [a, /3] such that E C [a, /31. Let {Ga IaeA be an IDpen cover of E. Then we obtain an open cover of [a, /3) by adding the open set R\E to the family IGalaeA of open sets that cover £. Since [a, /3] is compact, there is a finite set {a., ai, .. . , an} such that Ga ,• Ga2 , • • • , Ga. together with R\E is an open cover of [a, /3). But En R\E = and EC [a, /3], hence Ga,• Ga2, • • • , Ga. is an open cover of E. Thus 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£ is not bounded. For each positive integer n, let Gn = ( - n, n). Now Gn is open for each n and£ C u :a 1G.. , so {G.l:a l is an open cover of E. If n., .. . , n, is any finite set of positive integers, then

Proof First we prove that any closed interval is compact. Let [a,

r

.LJ Gn; = Gn •• I

0

where n0 = max{n., ... , n,), and since Eis unbounded,

is not a cover of E. Thus, (G n I:=1 is an open cover of E with no finite subcover, hence E is not compact. Assume Eis not closed. Then there is an accumulation point of E (call it Xo) such that x0 (£ 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

3.3 Uniform Continuity: Open, Closed, and Compact Sets 9S

Now (x0 - ¾, x0 + ¾l is a closed set for ,each n; hence, Gn is open. Observe that u:=iGn = R\ {x0 1, so {Gnl:= 1 is an open cover of E. Let n., . .. , nr be any finite set of positive integers and n0 = max {ni, ... , nr I. Then r G = Gno = R\ [ U n;

1=1

XO -

-

1

n0

, X0

+ -I ] • n0

In order that {Gn)i=l.....r be a cover of E, we must have EC Gn0

= R\[xo

- ~,Xo +

~l

hence,

En

(xo - ~, xo + ~)

is empty, contrary to x0 being an accumuilation point of E. Thus, {Gn) finite subcover, and E is not compact.

:=has no 1

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 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 ED. Thus, for each x ED, there is Bx> 0 such that if Ix -

YI< 8x and y ED,

then if(x) - f(y) j 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 + I; that is, there is x E ( -1, 1) such that f (x) = 0. This indeed turns out to be the case; in fact, it motivates us to s,tate the following theorem, published by Bernard Bolzano in

1817.

Suppo,se f: [a, b] - R is continuous and f(a) and f(b) have opposite signs. Then there is z E (a, b) such that f(z) = 0.

3.13 BOLZANO'S THEOREM

Proof Consider the case where f(a) < Oi and f(b) > 0, and define three sequences lxnl:=1, IY11l::'=1> and fcnl:= i as follows:

x,

=

a, y 1 = b, c,

= ½(x 1 + y 1),.

and for n;:: 1, 1. if f(cn) < 0, define Xn +, = Cn and Yn+1 = Yn• 2. if f(cn) .:: 0, define Xn+ 1 = Xn and Yn+1 = Cn· 3. Cn+I = ½·

Notice that for each n , c11 , x 11 , and y,, are in the interval [a, b], hence they belong to the domain off. Now, lxnl:=, is an inc:reasing sequence, (Ynl:. 1 is a decreasing sequence, both are bounded and hence they both converge. Moreover, lxn+i - Yn+ il ~ r 11(b - a) for all n. Therefore both sequences converge to the same limit; call it z. Both {f(x11 ) I:=1 and (.f(y11) I;= 1 converge to f(z) since z belongs 'to the domain off and f is continuous at z. But f(y 11) .?! 0 for each n and f(xn) < 0 for each n, hence f(z) = 0. The case for f(a) > 0 and f(b) < 0 follows easily.

3 .4 Properties of Continuous Functions 101

As a beginning student of analysis, you should cultivate the practice of dissecting and examining a proof to determine those facts essential to that proof. Such examination of this proof yields a big payoff in terms of a much more general theorem with almost no effort. Notice that it wasn't important that the domain off be an interval, but rather that any points between a and b belong to thie domain off. We formalize this idea in the following definition.

DEFINITION a

A set A C R is connected if whenever a and b are in A and

< c < b, then c E A .

The connected sets in R are fairly simpl,e to characterize. The following theorem accomplishes that task.

3.14 THEOREM

Let A be a connected subset of R. Then A is one of the

following:

i. Ix : x < a), (x : x > a). (x : x s a), Ix : x ii. [a, b], [a, b), (a, b], (a, b) iii. R

~ a)

Proof It is easy to check that each of the sets listed is connected. So we need to show that if A is connected, then it falls iinto one of the three categories. Note that the empty set is connected and is included in category (ii). Do you see how? Now assume A is nonempty. If A is neither bounded from above nor from below and x ER, then xis neither an upper bound nor a lower bound for A. Hence there are a, b E A such that a < x < b. Since A is connected, x E A. Hence A = R. Assume A is bounded from above, but not from below, and let a = sup A. If x E R, since xis not a lower bound for .A there is y EA such that y < x. If x < a, then there is z EA such that x < z < a.. Thus y < x < z and since A is connected, x EA . Thus A = Is : s < a) or A = {s: s s a). We leave the remaining cases to the reader.

If you have more interest in connected sc!ts, look to Project 3.4. Now we can state a modified version of Bolzano's Theorem. 3.15 MODIFIED BOLZANO'S THEOJrlEM Suppose f: A -+ R is continuous with A connected. Then if a < b, a, b E A with f(a) and f(b) having opposite signs, there is c E (a, b) such that f(c) == 0. Proof The proof of Theorem 3.13 may be used without changes.

Now we are prepared to state and prove the Intermediate-Value Theorem.

102 Chapter 3 Continuity

3.16 INTERMEDIATE-VALUE THEOREM Let f: A - R be continuous with A connected. Further suppose a < b, a, b E A with f(a) < y < f(b) (or f(a) > y > f(b)). Then there is c E (a, b•) such that f(c) = y. Proof Define g: A - R by g(x)

= f(x)

- y. Then g is continuous and g(a) and

g(b) have opposite signs. Consequently, by the modified Bolzano's Theorem, there is c E (a, b) such that g(c) = 0. Therefo1re O = g(c) = f(c) - y, and f(c) = y.

Theorem 3.10 states that if f: D -

R is continuous and D is compact, then

f (D) is compact. In other words, compactness is a property of a set that is preserved by a continuous function. Note that the Int.ennediate-Value Theorem states that if f: A - R is continuous and A is connected, then f(A) is connected. We conclude that connectedness is also a property of a set that :is preserved by a continuous function. The Intennediate-Value Theorem is, of course, the one that is used in trying to find the zeros of a polynomial. If pis 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. 3.17 THEOREM If p is a polynomial of odd degree with real coefficients, then

the equation p(x)

= 0 has at least one re~lll root.

Proof Assume

= Oo + a,x + ... + a,,xn, where n is odd and an * 0. If ao = 0, tlhen p(O) = a0 = 0 and we are through. Assume now that ao * 0, and let p(x)

for all x * 0. Choose E such that O < exists k; > 1 such that

Let K

E

< lanl• For i

= 0, 1, 2, ... , n -

1, there

= max{k0, k,, ... , kn- i ). Now, if lxl;?: K, then

ao +~+ ... +On- II 1

lx"

x"-

X

+ ... +Ian- ii lxl laol + ... + Ian-II < -E + ... + -E :S K K n n

:S~

lx"I

= E.

Thus, for lxl ;?: K, cp(x) has the same sign as a,, . Now p(x) = cp(x)x". So, if an > 0 with x ;?: K, then p(x.) > O; and, if x :S -K, then p(x) < 0. Similarly, if

3.4 Properties of Continuous Functions 103 a" < 0 and x;;: K, then p(x) < 0; and, if x s -K, then p(x) > 0. In any case, we can find points x, and x 2 such that p(x1)i < 0 and p(x2 ) > 0; hence, somewhere between x 1 and x2 , p has a zero.

If we combine Corollary 3.11 and Theo('lem 3.16, we obtain the following result.

3.18 THEOREM If f: [a, b] - R is continuous, there are c and d such that f([a, b])

= [c, d].

Proof

By Corollary 3.11, there atre x, and -½ in [a, b] such that f(x 1) s f (x) s f(x2) for all x E [a, b]. Let c = f(x 1) and d = f(x2 ). By the Intennediate-Value Theorem, if c s y s d, there is x E [a, b] such that f(x) = y. Thus f([a , b]) = [c, d). ■

Example 3.11

As a further application, we shall show that the equation

x = cos x has at least one solution in the closed interval [0, f]. Consider the function f(x) = x - cosxon [0, f]. Clearly, f is continuous, /(0) = -1, andf(r) =¥, Therefore, there is x E [0, ¥J such that f (x) = 0; in otht:r words, x = cos x. ■

3.19 THEOREM Let f: A - R be continuous and 1-1 with A connected. Then f is monotone.

Proof Suppose f is 1-1 and continuous but not monotone. Then one of the following cases applies: i. there are x, y, z E A such that x or ii. there are x, y, z E A such that x

< y < z and f(x) < f(y) and f(z) < f(y)


f(y) and f(z) > f(y).

Let us consider case (ii). We have here x, y, z EA, x < y < z and f(x) > f(y) and f(z) > f( y). Let us now suppose that f (y) < f(z) < f(x). By the IntennediateValue Theorem, there is w E [x, y] such that f(w) = f(z) contrary to f being 1-1. The proof for the other cases is left to the reader. Let us now return to a topic considere 0. 17. Suppose f: D __. R with f(x) is continuous at Xo• 18. Define f : R __. R as follows:

2:

\fl

1 "'.

Show

0 for all x E:: D. Show that, if f is continuous at x0 , then

f(x)

=x

- [x] if [x] is even.

f(x)

=x

- [x

+

= In x. Define h(x) = x·' by x-' =x'

1) if [x) is odd.

Determine those points where/ is continuous:. Justify.

3.3 UNIFORM CONTINUITY: OPEN, CLOSED, AND COMPACT SETS 19. Let f, g : D __. R be uniformly continuous. Pro ve 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 go/ : A __. C? Justify.

21. Define f : [3.4, 5] __. R by f (x)

2 =- . Show that / x- 3

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) __. R by /(x) = x 3 - 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 function f : 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 if f : R __. R is periodic and continuous, then / is uniformly continuous. 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 c:ontinuous. 2S. Give an example of sets A and B and a conti1nuous function f : A U B --+ R such that f is uniformly continuous on A and uniformly continuous on B, but-not uniformly continuous on A U B. *26. Let EC R. Prove that Eis closed if, for every x0 such that there is a sequence {x.};:"= 1 of points of E converging to x0 , it is true that x,0 E E. In other words, prove Eis closed if it contains all limits of sequences of members of E. *27. Prove that every set of the form {x: a < x < bl is open and every set of the form {x: a s x s b I is closed. 28. Let D CR, 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 comtains D, then D C F. D is called the closure of D. 29. If D C R is bounded, prove that D is bounded. 30. Suppose/: R--+ R is continuous and let r0 E: R. Prove that {x: f(x) * r 0 } is an open set. 31. Suppose f : [a, b] __. R and g : [a, b] __. 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 QC D. Define D0 to be the set of interior points of D. Prove that D0 is open and that if Sis any open set contained in D, then SC D0 • D 0 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.

106 Chapter 3 Continuity

*35. Let Ebe compact and nonempty. Prove that E is bounded and that sup E and inf E both belong to E.

36. If Ei, ... , En are compact, prove that E = u:~= 1E; 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 s f(x) s 1) is compact. 39. Suppose that f : R ➔ R is continuous and has the property that for each E > 0, there is M

>

0 such that if lx l 2: M, then 1/(x)I


0 such that D n [Xo - E, Xo + E] = E n [x0 - E, x0 + E] and f (x) = g(x) for all x ED n En [x0 - E, Xo + E]. Prove that f is continuous at Xo iff g is continuous at x0 •

49. Suppose that g: D

PROJECT 3.1 In this project, you will prove a version of Theorem 3.8 and Theorem 3.11 without using the notion of compactness.

3.8 THEOREM Suppose f : A Then f is uniformly continuous.

R ii. continuous with A closed and bounded.

Proof You will be led through the step,s of the proof, but you will need to fill in some missing details and supply reasons. for some of the conclusions. Remember,

Project 3.3

107

you can't use any theorems in the boollc that were proven using the notion of compactness. 1. Assume f is not unifonnly contiinuous on A. Then there is E > 0 such that for each n, there are am b,, E A such that Ian - bnl < 1/n and l f(an) - f(bn)I ~ E. Explain wh:y this must be true if f is not unifonnly continuous on A. 2. There is a sequence of positive iintegers {nk Ik= 1 such that the sequences IanJt"~ 1 and {bn,} ;'s I both conve:rge; 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 COROLLARY If f : A - R is continuous and A is closed and bounded, then there are x., x2 EA such that f(x 1) s f(x) s f(x2) for all x EA.

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 affin.e if f(x + y - z)I = f(x) + f(y) - f(z) for all x, y, z E R. In this project, you will prove that continuou:s affine functions have a special form. A function f: R - R is additive if f(x + y) = f(x) + f(y) for all x, y ER. 1. Assume f is an affine function and m is a real number. Define g: R - R by g(x) = f(x) - [mx + f(O)]. Show thalt g is an additive function. 2. Let m = f (1) - f (0) and show that rthe function defined in item 1 above has the property that g(x + 1) = g (x). 3. Suppose f is affine and bounded on (0, l], and g is defined by g(x) = f(x) - [mx + b] where m = ;f(l) - f(O) and b = f(O}. Show that g is bounded on R. 4. Show that any bounded additive func1tion must be the zero function. 5. Show that if f is affine and continuous, there are real numbers m and b such that f(x) = mx + b for all x E R.

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.

108

Chapter 3 Continuity

In what follows, let R + denote the set of positive real numbers. Assume f: R+ - Risa 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.

*

1. Show that /(1) = 0 and there is a positive real number a such that f(a) = 1.

2. Define g(x) = f(x) - log x where a is such that f(a) = 1. Show that 0

g(xy) = g(x) + g(y) for all x, y ER+. 3. Show that g(ax) = g(x) for every x ER+. Prove that g is bounded. 4. If g is not the zero function, prove th21t g is not bounded. 5. What can you conclude from items 3 and 4?

PROJECT 3.4 After the statement and proof of Bolzano's Theorem, we defined connected set. The definition was motivated by the proof of Bollzano's Theorem and the facts that were used in that proof. Theorem 3.14 characterizes the connected subsets of Rand perhaps explains to the reader the choice of the word ''connected" used in the definition. There are many definitions in this book which have meamng in broader contexts: open set, closed set, compact set, just to mention a fevv. The notion of a connected set belongs to this category. We want the definitions of these concepts to be compatible with the broader usages. Your first task will be to prove the following theorem which describes connected sets in terms of open sets. You may want tc, review the definition of connected sets before going further. Just to simplify the language, we offer a related definition.

DEFINITION A set A is disconnected iff it is not connected. THEOREM A set A C R is disconnected iff there exist open sets P and Q such that i. P () Q ii. P n A

= 0; hence g is differentiable at f(x). Note that g'(u) = 1/2Vu. Thus, by the Chain Rule, g O f = F is differentiable at x and

F '(x) = g'(f(x)) · f'(x) =

4.3

l ----;:==2 · 6x. 2\/1

•

+ 3x

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 .x0 ED is a relative maximum (minimum) off iff there is a neighborhood Q of x0 such that if x E Q

f (x)

s

f (x0 )

(j(x) 2!

n D , then

f (xo)).

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 R such that f(x) = x. Since the domain off is the interval [0, 1), f has a minimum at O and a maximum at I; indeed, 0 is a relative minimum off, and 1 is a relative maximum off. Since f'(x) = 1 for all x E [O, I], 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. ■

f: [0, 1] -

120 Chapter 4 Differentiation

4.6 THEOREM

Suppose f: [a, b] -+ R and suppose f has either a relative maximum or a relative minimum at x 0 E:: (a, b). If f is differentiable at x0 , then f'(xo) = 0.

Proof Assume f has a relative maximum at x0 • Then there is S > 0 such that, if Xo - S < x < Xo + S, then x E [a, b] [since x0 E (a, b)] and f(x) $ f(x0 ). Consider any sequence lxn) ;_ 1 converging to x0 such that Xo - S < Xn < x0. Then, since f is differentiable at x0 ,

converges to f' (Xo), But

f (Xn)

Xn -

f (xo)

:2::

O

Xo

for each n because f (xn) $ f (x0 ) and xn < x 0 , and hence f' (x0 ) ::! 0. Consider now a sequence IYnl:a, such that Xo < Yn < Xo + s. As before, 00

f(Yn) - f (Xo)} { Yn - Xo n=l converges to f' (x0 ), but this time

f(Yn) - f(Xo) Yn - Xo

S

O;

hence, f' (x0) $ 0 . Therefore, f' (Xo) imum at Xo is left to the reader.

= 0. The case in which f

has a relative min-

The reader should now realize that an important fact in this proof is that we are free to choose points in the domain on eithe1r side of Xo as close as we please. Since both the maximum and the minimum occurred at end points in the preceding ex.ample, 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 geo1metric 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 if f(a) == f(b) = 0, there is c E (a, b) such that f'(c) = 0.

Proof If f(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) :I: 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

4.3 Rolle' s Theorem and the Mean-Value Theorem 121

f(c) = 0

a

c

b

Figure 4.2

and minimum in [a, b], say at x 1 and x,!, respectively. Since f is not identically zero on [a, b] and f(a) = f(b) = 0, at least one of x 1 and x2 must belong to (a, b), say x 1 E (a, b). Now, by Theorem 4.6, f"(x 1) = 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 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)1) and (b, f(b)) is f(b) - f(a) b - a

so this is the slope of the tangent line at c; that is, f'( c)

= f(b~

=

~(a).

The precise statement and proof follow.

4.8 MEAN-VALUE THEOREM If f: [a, b]---:> R is continuous on [a, b] and differentiable on (a, b), then there is a c E (a, b) such that 1

f'(c)

= f(b~

- f(a).

-a

122 Chapter 4 Differentiation

I

...I

a

~

b

C

Figure 4.3

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 liniear and satisfy L(a)

= f(a)

and

L(b)

=

f(b).

This is accomplished by choosing L(x)

=

[t(b) - /(a)] (x - a) + f(a), b-a

the equation of the line passing through (a, f(a)) and (b, f(b)). L is continuous and differentiable everywhere and L'(x)

= f(b)

- /(a). 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

= g'(c)

f(b) - f(a) = f'(c) - L'(c) = f'(c) - - - - . b-a

Thus f'(c)

= f(b)

- f(a). b-a

4.3 Roi/e's Theorem and the Mean-Value Theorem 123

The Mean-Value Theorem has some very important applications in relating the behavior off and f'. The next theorem will iilustrate this. 4.9 THEOREM Suppose f is continuotUs on [a, b] and differentiable on (a, b). Under these hypotheses:

i. If f'(x) -:/= 0 for all x E (a, b), thten f is 1- 1. ii. If f'(x) = 0 for all x E (a, b), thten f is constant. iii. If f'(x) > 0 for all x E (a, b), then x < y and x , y E [a, b] imply f(x) < f(y) (that is, f is strictly increasing). iv. If f'(x) < 0 for all x E (a, b), then x < y and x, y E [a, b] imply f(x) > f(y) (that is, f is strictly decreasing).

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 f(x) - f(y) x - y

= f'(c);

that is, f(x) - f(y)

f(x)

= f'(c)(x

- y). With this fact in mind, we shall proceed.

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

< y such that

f'(c) = f(x) - f(y) = O, x-y

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 < y, such that f(x) -:/= f(y). There is c E (a, b) such that f'(c)

= f(x)

- f(y) x-y

*0

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

= f'(c) >

hence f(x) < f(y). (iv) Suppose x f(x) - f(y) x-y

hence f(x)

>

0;

< y and x, y E [a, ho]. There is c E (x, y) such that

= f'(c) < 0;

f(y).

124 Chapter 4 Differentiation Now that the proof has been presented, it is clear that items (iii) and (iv) can be changed as follows to give additional infonnation: iii. f'(x) ?.: 0 for all x E (a, b) implies f is increasing. iv. f'(x) s O 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) = g'(x) for all x E (a, b). Then there is a real number k such that

f(x)

= g(x) + k

for all x E [a, b].

Proof Consider h(x)

= f(x) -

g(x) for all x E [a, b]. Clearly, his continuous on

[a, b] and differentiable on (a, b) and

h '(x)

= f'(x)

- g'(x)

=0

for all x E (a, b). Hence, by Theorem 4.'9, there is a real number k such that

k = h(x) = f(x) - g(x) 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 functiom;. 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, th1:re is t such that O < t < h and

f'(t)

= f(h)

f(h)

= f'(t)(h

- f (0) h- 0

or

Now p

>

f 1

= p(l + t)P- 1h + 1, so

= o + h)P = p(l + w-1h + 1 >

See Exercise 24 for the case of p

1.


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 f'(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 A is a real number such that f'(a) that f'(c) 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 0 = g'(c) = f'(c) - A

so that f'(c)

= A.

Suppose f: [a, b] - R is such that there is a differentiable function g: [a, b] - R such that g'(x) = f(x) for all x E [a, b]. Now f need not be a continuous function, but f must have the inte1111ediate-value property; in other words, if x, y E [a1 b] and f(x) < ,\ < f(y), then there is c between x and y such that f(c) 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 THEOT1~M If f and g are continuous on [a. b] and differentiable on (a, b), then there is c E (a, b) such that [f(b) - f(a)]g'(c)

= [g(b)

- g(a)]lf' (c).

= [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 The:orem. there is c E (a, b) such that

Proof Define h(t)

h'(c)

= [f(b) -

f(a)]g'(c) - [g(b) - g(a)]f'(c)

= 0.

126 Chapter 4 Differentiation

4.4 L'HOSPITAL'S RULE AND THE INVERSE-FUNCTION THEOREM We can use Cauchy's Mean-Value Theorem to prove a familiar rule from calculus, L'Hospital's Rule.

4.13 L'HOSPITAL'S RULE Suppose J 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), ii. f(x0 ) = g(x0 ) = 0; and iii. f' lg' has a limit at Xo, then f lg has a limit at x0 and lim x - .r0

f (x) = g

Jim A-.v0

f:

(x).

g

Proof Let {x,,};;'~ 1 be any sequence converging to x0 with Xn E (a, b)\ {x0 } for all n. By the Cauchy Mean-Value Theorem, there is a sequence {cnl:;"= 1 such that (i) c,, is between Xn and Xo for each n, and (ii) [f(x,,) - f(x0 )]g'(c") = [g(x,,) - · g(.xo)]f'(cn) for each n.

*

*

Since x,, Xo and g'(x) 0 for all x E (a, b), g is l -1, hence g(xn) for all n. Since f R defined by f(.r) = \f2.rJ - 3.r + 6 is differentiable on (0, I) and compute the derivative.

12. Suppose f: la. hi - [c. di. g: [c. di - [p, q], and h: [p, q]-> R, with f differentiable at .r0 E fa. hi. g differentiable at /(.r0 ), and h dif!ferentiable at g(f(: [c. di and g: [c. di -> R are differentiable on [a, h] and [c, dJ, respectively. Suppose

13. Suppose

.f' : la.

hi

->

R

and

g' : [c. di _,, R

are also differentiable on [a, hi and (c. di. respectively. Show that (8 ° f)' : [a. hi -> R is differentiable and find the derivative. 14. Suppose J: R-> R is differentiable and define! g(x) = x 2f(xJ). Show that g is differentiable and compute g '. 15. Define .f(.rl = + + ½ for .r ;z: 0. Determine where f is differentiable and compu te the derivative.

Yx Vx

4.3 ROLLE'S THEOREM AND THE MEAN-VALUE THEOREM 16. Define f: 10. 21-> R by .f(.r) = V2x - .r 2. Show that f satisfies the conditions of Rolle's lhcorcm and find c such that J'(c) = 0. 17. Define .f: R -> R by J(.r) = 1/(1 + x 2). Prove that J has a maximum value and find the point at which that maximum occurs. 18. Prove that the equation .r' - Jr + h = 0 has at most one root in the interval [ - I, I]. 19. Show that cos .1· = .r' + .rJ + 4.r has exactly one root in (0, fl. 20. Suppo~c .f : 10. :21 -> R is differentiable. f(O) = 0. f( I) = 2. and /(2) = 2. Prove that I. there i, c , such that .f'(c1 ) == 0. 2. there i, such that .f'k:l = :?.. and J. there is c., such that J'(c.J = 21. Let f: IO. 11 -+ Rand g: 10. 11 -> R be difforentiable with /(0) = K(O) and f'(x) > >1'(.r) for all .r in JO, 11. Prove that f(.r) > g(.1) for a11l .r in (0. I J.

c,

f

Exercises

131

22. Use the Mean-Value Theorem to prove that nyn-l(X -

J) :S

X" -

yn :S

nxn- l(X -

J)

if n ~ 1 and O :S y s x. 23. Use the Mean-Value Theorem to prove that

1

Vl+h < I + 2 h

for

h

> 0.

24. Generalize Exercise 23 as follows: If O < p < 1 and h > 0, then show that (1

+ h)P < I + ph.

You may assume the usual rules about differentiating powers.

25. Suppose f: (a, b) ➔ R is differentiable and lf'(x)I :SM for all x E (a, b). Prove that f is uniformly continuous on (a, b). Give an exam;ple of a function f: (0, I) ➔ R that is differentiable and uniformly continuous on (0, 1) but such that f' is unbounded.

26. Suppose f is differentiable on (a, b), except possibly at Xo E (a, b), and is continuous on

27.

28. 29. 30. 31.

[a, b }; assume limx- . 0, but there does not exist a neighborhood of a in which f is increasing. Prove that the function f(x) = 2.x3 + 3x2 - :36.x + 5 is 1-1 on the interval [ -1, 1}. Is f increasing or decreasing? Show that the function f(x) = x 3 - 3x2 + 17 is not 1-1 on the interval [-1, I). Give an example of a function f: R ➔ R thait is differentiable and 1-1, but f'(x) = 0 for somexER. If f: [a, b) ➔ R is differentiable at c. a < c < b and f'(c) > 0, prove that there is x, c < x < b, such that f(x) > f(c).

4.4 L'HOSPITAL'S RULE AND THE INVERSE-FUNCTION THEOREM 32. Assume the rules for differentiating the elementary functions, and use L'Hospital's Rule and find the following limits: . In x (a) h m - x-1 X 1 (b) lim-x,-o ex - I . sin x (c) hm - x-o X 33. Use L'Hospital's Rule to find the limit

x2 sin x lim-.----x-o sm X - X cos X 34. Prove the following variant of Theorem 4.14. Suppose f : [a, bJ ➔ R is 1- 1. If f is differ1 entiable at c E [a, b), f'(c) 1' 0, and 1 is continuous at d = f(c), then is differentiable

r

at d and

' 0 such that

41. Suppose f: D---+ R, g: E---+ R, x0 ED

D n [Xo - E, Xo + E]

= E n [Xo -

with f(x) = g(x) for all x E D iff g is differentiable at x0 •

nEn

E,

[x0 -:-

;eo + E,

n £. Suppose

El

x 0 + E]. Prove that f is differentiable at x0

PROJECT4.1 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 conditions, and to give estimates of the speed of convergence. Suppose f: [a, bJ - R is twice diffenentiable; that is, f and f' are both differentiable on [a, b]. Suppose further that there are m and M, both positive, such that lf'(x)I ~ m and 0 < lf'(x)I :5 M for all x Elia, bJ. Finally, suppose that f(a)f(b) < 0. Define a sequence as follows:

a1

= a if f'f'

an

= an - I

-

0. There is Let Q be any refinement of P marked

s U(Q, f) s U(P, f) s

L(P, f)

+

E,

and L(Q, f)

s

J: f

dx s U(Q, f).

ff

dx l s

Hence, IS(Q, f) -

E.

Suppose f: [a, b]-+ R is bounded, and A is a real number such that for each E

> 0 there is a partition P such that if P C Q, where Q is a partition of [a, b]

marked in any fashion, then IS(Q, f) - A l s E. We shall appeal to Theorem 5.2. Choose E > 0. There is a partition P = IXo, Xi, .•. , xnl such that E

IS(P,

n - Al < 4'

regardless of how we choose t1 E [x,- i, x 1]. Recall that M;(f)

= sup{f(x): x E

m1(f)

= inf{f(x) : x

[x,_ ., x;]}

and E [x,- i, x1] J.

150 Chapter 5 The Riemann Integral

Thus, for each i among 1, ... , n, there are ti , s1 E [x,-i, xi] such that E

M;(f) - 4(b - a)< f(t;)

and f(s,)


0 sruch that for any partition P of [a, b) with µ,(P) < 6, we have IS(P, f) - A l ::: E, regardless of how Pis marked. Then, if P is any partition of [a, b) with µ,(P) < 6 and Q is a refinement of P, then µ,(Q) < 6, and hence, regardless of how Q is marked, IS(Q, f) - Al ::: E. By

dx.

Theorem 5.5, then, f E R(x) and A = J!; f Suppose now that f E R(x) on [a, b]. Choose E > 0. Let A = J! f with M a positive real number such that lf(x) I s; M for all x E [a, b) and E' =!-There is a partition P = {x0 , Xi, ••• , Xn} such thatt

U(P, f) - L(P, f) Let




0.

= X- 2-1+-1 is continuous o,n R, so it is integrable on every closed ·

interval. Choose a sequence IPnl:. 1 of partitiions of [a, l] with {µ,(Pnl:. 1 converging to zero. If Pn = Ix0 , x 1, ••• , Xm), choose t 1 E [X;- 1 , X;] such that = x,_ 1x1; that is, t; is the geometric mean of X;- 1 and X;, With each partition so marked, IS(Pn, f));. , converges to

t;

I.

I

a

}

--dx x2 + 1 .

Define a sequence of partitions of (1, ¼1 as foHows: For each Pn define

P'

n

=

{J_ Xm

'

_ I_ Xm- 1

,

...

,

_!_ ' Xi

_!_} •

Xo

= IXo, x 1, x2 , ••• , xml,

154 Chapter 5 The Riemann Integral

and mark P~ with r; = f where the t; are chosen as indicated above. Since P,, is a partition of [a, 1), P~ is~ partition of [l, ¼land µ,(P~) s ~ µ(P,,); hence {µ(P~)}:~ 1 converges to 0. Thus {S(P ~. f) I:- 1 converges to l

11a

1 I

--dx

X2

But S(P,,, f)

+

1

• 1

= S(P~, f) for each n; hence J! X- 2-+-1 dx = Jl'

1

0 -

X

2- -

+ 1

dx

■

5.4 THE FUNDAMENTAL THEOll.EM OF INTEGRAL CALCULUS It is impossible to resist the temptation to use one of these results to prove a very important theorem, the Fundamental Theorem of Integral Calculus. This is indeed an impressive result in the sense that it gives a method for computing certain integrals and reveals the relationship between integration and differentiation. We have already revealed the secret of the proof to be used here in the discussion preceding Theorem 5.5.

5.8 FUNDAMENTAL THEOREM OiP INTEGRAL CALCULUS

Suppose

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

J: f' dx =

f(b) - f(a).

Proof Let P be any partition of [a, b], jp = {Xo, x., ... , x,,}. By the Mean-Value Theorem applied to f on [x,_ ., x1], there is t1 E [x,_i, x1] such that f(x,) - f(x,- 1)

= f'(t,)(x, -

x,-1).

Thus, n

L f'(t,)(x, -

~ I

n

x,- 1) =

L [f(x,) -

/(X;- 1)1 = f(b) - f(a).

~I

This means that any partition P of [a, b] may be marked in a way so that S(P, f') = f(b) - f(a). Choose a sequence {P,,J:. 1 of partitions of [a, b], each marked in this fashion and such that {µ(P,,) } 1 converges to zero. Now, f' E R(x) on [a, b]; hence, by Theorem 5.7 IS(P,,, f')I:=• converges to J! f' dx, but

:=

f') = f(b) - f(a) for each n, so ft f' dx = f(b) - f(a). S(P,,,

It should be pointed out that f' E R(x) iis a very important part of the hypothesis of this theorem. For example, the function f defined on [0, 1) by f(x) = x 2 sin :,\.for x -:I= 0 and /(0) = 0 is differentiable on [0, l], but f' is not Riemann-integrable on [O, 1) because it is unbounded.

5.5 Algebra of Integrable Functions 155

The reader will no doubt recognize this theorem as the basis for many calculations of integrals in early calculus courses. The m:xt example is intended to reinforce that recognition.

Example 5.7 We will use Theorem 5.8 to calculate the integral of F(x) = x 3 - l over the interval [2, 3). First, F is integrable because it is continuous. So we may use Theorem 5.8 if we can find a function f such that f'(x) = F(x) for all x E [2, 3)-that is, find an antiderivative of F. Experience hellps us discover that f (x) = ~ - x serves ■

our purpose. Then

f

3 2

f

3

(x 3

-

I) dx

=

2

f

3

F(x) dx

=

2

f'(x) dx

= /(3)

- f(2)

= -61 . 4

■

Theorem 5.8 does have some limitations. If we wish to use it to find the integral of F, then we must find an antiderivative of F-that is, a function f such that f'(x) = F(x). Under some circumstances, this can be very difficult to do; you may recall quite a bit of time devoted to this problem in your calculus courses. But now we know that it may be impossible! In order for F to have an antiderivative, it must be the derivative off and, hence, must have the intermediate-value property (!;ee Theorem 4.11). If F fails to have the intermediate-value property, it cannot have an antiderivative; therefore, Theorem 5.8 is of no use in finding the integral of F.

5.5 ALGEBRA OF INTEGRABLE FUNCTIONS It is now appropriate to make some observatiions useful in calculating integrals. Since the integral may be considered as a limit of sums in a certain way, the results are derived from similar results for finite sums.

5.9 THEOREM If f., f 2 : [a, b]-+ R :are bounded and /i, f 2 E R(x) on [a, b], then

i. For Ci, c2 any real numbers, cif1

+ cif2 E R(x) on [a, b]

(" (cif 1 + C2f2) dx = C1 1•b,a f I dx + Cz Jarb f2 dx.

Ja

ii. If f 1(x) s fi(x) for all x E [a, b], then

J: f

1(X) dx

$

J:

fz(x) dx.

iii. If m s f 1(x) s M for all x E [a., b], then m(b - a) s

Proof (i) Choose

(le.I +

lc21)E'

E $

f/

1

dx s M(b - a).

> 0. There is E' > 0 such that E.

and

156 Chapter 5 The Riemann Integral There are partitions P, and P2 such that for any refinement Q; of P;, regardless of how Q 1 is marked,

js 0 be such that e'[b - a + 2K] s E. By uniform contin1uity of (f(t)) I < e';

that is, M;( 0. Prove that L(xn) = nL(x) for all x > 0, n E z.

Comment: One possible approach to (3) is to define f(x) = L(ax) and show that f'(x) = L'(x); hence, f(x) = L(ax) = L(x) + k, and then show that k = L(a). A similar approach will supply a proof for (4). 5. Prove that L(i) = L(a) - L(b) for a ,. b > 0. 6. Prove that im L = R. Since L(~) = -- L(x) for 1 < x, anc.l L(l) = 0, it suffices to prove that for any y > 0, there is x > I such that L(x) = y. Since L is continuous, you need to show that thtere 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

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

converges to 1. Do you see why?

= 1.

170 Chapter 5 The Riemann Integral

7. Prove that for each n E J, _n_ s nl(l + ¼) s 1. You will want to write n + 1 L(l + ¼) as an integral and use Theorem 5.9(iii). 8. Finish the proof that L(e) = 1. 9. 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 Eis continuous, differentiable, 1- 1, and increasing. Find E'. You may want to refer to Project

3.5.

10. Prove that E behaves like an exponential function-that is, E(x + y) = E(x)E(y) for all x, y E R. 11. We have some understanding of x' where x is any real number greater than zero and r is any rational number. N,ow 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 = eytoax. Show that f(x) = xx is differentiable and find f' (xis assumed to be a positive real numbe1r).

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 tthe midpoints of the subintervals. Assume that a < band f is Riemann-integrable on [a, b]. For each positive integer n, define Sn(!)

where h

=

ht f ( a + ( Dh). k -

1

= ¼(b -

a). The theorem you wish to prove is the following:

THEOREM Let f be twice-differentialble on [a, b] and f' continuous on [a, b]. Then, given a positive integer n, there is c in [a, b] such that

f

b

f(x) dx - S,.(f)

0

=

(b - a)h2

24

f'i(c)

where h and S n(f) are as described abovt!.

Proof For each k among 1, 2, ... , n, define at h

= ¼(b -

=

a + (k -

½)

h where

a). Define a, +r

gk(t)

f

= a,-, f(x) dx -

21 f(ak)

1. Prove that gk is twice differentiable and gZ(t)

= f'(ak + t)

- f'(at - t).

Project 5.3 171

2. Apply the mean-value theorem to f' on the interval [a1c - t, a1c + t] to find x1c., such that f'(xk.,) = t,(g'ti(t)). Let m = inf(f"(x): x E [a, b]) and M = sup{f"(x): x E [a, b]l. Cornclude that

2tm s gZ(t) s 2tM for all O s t s ½h. 3. Integrate the inequalities in item 2 to obtain the inequality mh3 s g 24 k

4. Show that

24

2

h) s Mh3. 24

J: f(x) dx -

3

nmh

(!

s

{"

J.,

= ;£z_1 g1c(½h). Conclude that

Sn(f)

f(x) dx - Sn(f)

5. Using the fact that h proof.

= ¾(b -

s

nMh 3

24

a) and the continuity off", complete the

You really didn't need the continuity off" in item 5. What theorem could you have used instead of invoking the continuity of f"? Where in the proof did you use the continuity off" (other than in item 5)?

PROJECT 5.3 - - - - - - · - - - - - - - - A method for approximating definite integrals that you have seen before is the Trapezoidal Rule. The basic idea is to use trapezoids to estimate the area under the curve. In this project, you will derive the Trapezoidal Rule and find an upper bound for the error in using the rule to estimate an integral.

1. Let p be the linear function such that p(a) = f(a) and p(b) = f(b). Show that b-a - [p(b) 2

f~ p(x) = -

2

b-a

= - - and define n = a + kh, ... and b = a + nh. Ix:fine

2. Suppose now that h X1c

b-·a

+ p (a)] = - -;:- [f(b) + f(a)].

Tn(f)

h

= 2 [f(a) +

=

a, x 1

2f(x1) + 2f(Xi) + · · · + 2f(Xn-1)

If P1< is the linear function such that p(x1c- 1)

rx·

n

k

x0

1

a + h, ... ,

+ f(b)].

= f(x*_ 1) and P1 0. There is N such that for n and p 2:: 0, Ian + an+I + · · · + an+pl < €.Chooser 2:: n 2:: N. Then

1s, -

Sn l

= lan+I

+ ··· +

a,I
0. By Theorem 6.1, there is N such that, for n 2:: N and p 2:: 0, we have la" + an+t + · · · + an+pl < E. In particular, for p = 0, lan l < e. Hence, I a" I; . 1 converges 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

176 Chapter 6 Infinite Series

~.:'=i an where lanl.:'=i does not converge to ze:ro. Unfortunately, the result is frequently misused, for many students assume that the converse is true-that if {an I;:= 1 converges to zero, then L;'= 1 an converges. This conver:se is definitely false, as we have already seen by considering

+ l) L.. log ( _n_. n

ne J

Here

converges to zero, but the series

..

~1log

7 l)

( +

diverges. The following example emphasizes this fact and will be of use later. ■ Example 63 Consider the infinite series L-;,'. 1~. often referred to as the harmonic series. Since~> 0 for each n, the sequence l[Sn);'. 1 is increasing, and we shall show it is unbounded. If n = 2\ then

l

1

1

2

3

2k

S =1+-+-+ ... +n

! + (! + !) + (! + ! + ! + !) + ... + ( I+ } + .. . + .!..) ~ l + ! + 2 (!) + 4 (!) + · · · + ,zk- (..!..) = 1 + ~

= 1+

2 2

Thus,

3

4

5

4

6

8

7

2k- I

8

.

I

2/c

2/c

2.

•

\Snl:=i is unbounded, so L-;'= 1 ~ 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

The series L;'ao rn is called the geometric series with ratio r. For

n ~ 0,

Sn= 1 + r + r2 + · ·. + rn

l - rn+I

=--l - ir

for all r-:/: 1. Now {rn+ 1_:= 1 converges iff - 1 0 such that IAnl :SM for all n. Choose E > 0. Since f bn};. 1 converges to zero, there is N such that n 2: N implies lbnl

< 2~. Recall now that bn

2:

0 for all n and bn - bn+I 2: 0 for all n.

q- 1

:S

L

n=p

IAn l(bn - bn+.1)

= M2bp < M Thus, by Theorem 6.1 ,

2E

2M

=

+ IAq lbq + IAp-1 lbp

E.

k:. 0 anbn conver_ges.

As an immediate consequence of Theomm 6.12, we have the ''alternating series test."

6.13 THEOREM (Alternating Series Tes.I) Suppose ( bn I:=o is a sequence of real numbers such that i. \bnl:=o converges to zero, and ii. b0 2: b 1 2: b2 2: · · · Then

L;. 0( - ltbn converges.

Proof Set an

= (- 1t

for all n. Then

if n is even and An= 0 if n is odd. Thus, Theorem 6.12 to conclude that

..

L

..

a,.b,.

n• O

converges.

=L

n~O

(-l)"b,.

{Anl:eo is bounded, and we can invoke

188 Chapter 6 Infinite Series In light of Theorem 6.3, one may be tem1pted to conclude that convergent infinite series may be handled in much the same mainner as finite sums. We have postponed the discussion of this problem until now, because an understanding of the distinction between absolute and conditional convergenc,e 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 I;. 0 an and I;.0 b., be two infinite series. For each n, define n

Cn

=

L athn-k· k=O

The infinite series I:=o cn is called the Cauchy product of the two series I;_ 0 an and I:=o bn. This definition may perhaps be best motivated as follows. Suppose that r

and

q(x)

=

1:

bmxm

m='O

are polynomials. Define am = 0 for m > s and bm = 0 for m > r. Then the pointwise product pq is a polynomial, and p(x)q(x)

= (ao + a1x + · · · + a,x')(b0 + b1x + · · · + b,x') = aobo + (aob1 + a1bo)x + · · · + (aob. + a1bn-1 + · · · + anbo)x" + · · · + a,b,x•+r

= Co + C1X + · · · + CnX" + · " · + Cs+rXs+r (ao

+ a 1 + · · · + a,)(bo +

bi

+ ··· +

b,)

= Co + Ct + · · · + C,+r•

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 I;= 1 a. and I;. 1 bn converge to A and B,. respectively, then I;=t Cn converges to AB. It turns out that this is false. In the nex1t few para~phs, we shall explore these questions and prove the main result. ■

Example 6.13 ..

(-l)n

L Yn+i' n + 1

n• O

Consider the series

6.4 Conditional Convergence 189

By the alternating seiies test, this series converges. We shall now consider the Cauchy product I:=o Cn of this series with itself. Now n 1 Cn

For n

= (-Ir ~V(n

===

(n -

- k

+

l)(k

+

1)

0

r-(~l -ky r;

k,

k+ l)(k + 1) = (~ + 1

s (~ + 1

hence, 1 1 -y'-;=(n=-=k=+=l=)(=k=+=l) 2:: -n-

2

= -n -+2.

- + 2

Thus,

lcnl 2::

i -+ 2

kaO

n

-

2

= 2(n + 1), h

+2

{ Cn I;., 0 does not converge to zero, and I;. 0 en fails to converge. As we shall see by the next theorem, I:=o en fails to converge because

f n•O

(-lt

Vn+1

converges conditionally.

■

6.14 THEOREM Suppose I;=o an converges absolutely and I:=o bn converges with

..

and

L

n=O

bn

= B.

Define n

Cn

= kL akbn- k• • O

Then I;. 0

en

converges to AB.

/'roof Define

and

f3n

= Bn ·- B.

190 Chapter 6 Infinite Series We wish to prove that 1Cn};c 1 converges to AB. For each n,

Cn = Co + Ci + C2 + · · · + Cn = aobo + (aob, + a1bo) + · · · + (aobn + a1bn- 1 + · · · + anbo) = aoBn + a,Bn-1 + · · · + anBo, = ao(B + f3n) + a1(B + f3n - ,) + · · · + an(B + /30), Define 'Yn = a0 f3n + a,f3n- i + · · · + a,.{3 0 . Since IAn I :Go converges to A, the theorem will be proved if we can show that {'Yn} converges to zero. Choose E > 0 . Let K = }:;.0 lanl and M = SUPn2:o lf3nl•There is N such that n ;?: N implies that

:so

lf3n l

= /Bn -

and that for q

~

E

Bl < K 2

(We may assume K '1' 0.)

N,

q

E

L laPI < 2M.

(We may assunneM,;, 0.)

psN+ l

Thus, for n

;?:

2N, n

l'Ynl = laof3n + a,f3n- l + ··· + anf3ol S k•LO lakf3n- kl N

= k=O L lakf3n-kl + E

N

n

L

k=N+ I

lakf3n-·kl E

n

E

s 2K k•O L lad+ M k •N+ L I lakl < 2K K + M 2M = E. A related theorem in which no absolute oonvergence is assumed is needed to round out the picture. We shall state it here without proof. (See Project 7.1 for a proof.)

6.15 THEOREM Suppose }:;=o an and :t;=o b,. converge to A and 8, respectively. Define n

Cn

=L

k• O

akbn- k·

If }:;=0 c,. converges to C, then C

= AB.

For finite sums, the order in which the tierms are arranged has no effect upon the sum. We shall now concern ourselves with the similar situation for infinite series.

DEFINITION Let }::'=o a,. be an infinite series. If T is any 1- 1 function from {0, 1, 2, .. . } onto {0, 1, 2, ... }, then the infinite series l:'=o ar is called a rearrangement of l;=o a,..

6.4 Conditional Convergence 191

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 condi1tionally convergent series. Consider the conditionally convergent series

..

1

L (-lf+I -, n

n=I

and denote its sum by S. (S happens to be aJpproximately 0.693; actually, it is log 2. See Project 6.3 for a proof.) Then

i

(-l)n+I_!_

2n

n• I

converges to

f We now consider a new serie:s k;=, an where an = 0 if n is odd and

(-l) 0. There is N such that for n 2: N,

(since k:'t I an converges absolutely). Now Tis a 1-1 function mapping { 1, 2, ... I 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

IS! - Al

s

IS! -

E

co

s

Al
- 1akl

+

~

€

L lakl + -2 < -2 + -2 ·

k• N+I

Thus, (S!):. 1 converges to A, and the theorem is proved. Theorem 6.16 has a converse in a sense. If k:'=i an is an infinite series with the 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 Let {anl:'. 1 be any sequence of real numbers. For each n, define a;!' = an if an 2: 0 and a;!' = 0 if an < 0.. Define a;; = an if an < 0 and a;; = 0 if an 2: 0. Thus, if

(anl:'=1 = { 1, -

11 11 l} 2, 3, - 4, 5, -,;• "· ,

6.4 Conditional Convergence 193

then

{a: );;'=

1,

~ , · · ·}

1

= { 1, 0,

I

= {o, _!2' o, _!4' o' _!6' .. •}.

0,

and {an-,... n•

In any case, an == a:

+ a;;

for all n.

Suppose now that l:;:-_ 1 an converges absolutely. Then for each n, and hence, k;:'= 1 a: and k;:'= 1 a;; converge absolutely. Conversely if k;:'= 1 a; and l:;;'= 1 (-a;;) converge (actually, this is absolute convergence because both are series of nonnegative terms), then since Ian! = a:

- o;;,

converges by Theorem 6.3. This result proves the following theorem. 6.17 THEOREM The infinite series k;:= 1 an converges absolutely iff the series k ;:'= 1 a;; and l:;=1 a;; converge. If k;:'= 1 a: converges with sum A + and l:;;'c I a;; converges with sum A - , then the sum of k.:'= 1 an is A+ + A - , and the sum of k .:'=1 lanl is A+ - A - .

To recap our results, if a series k ;:'. 1 an converges conditionally, then at least one of the two series, L;:'c I a: or L;:'a I a;;' diverg:es. Since we may write and we see that, if L;:'. 1 an converges conditionally, then both L;:'= 1 a-:; and l:;;'= 1 a;; must diverge. Since 2: 0, the partial sums of i:;:-= 1 form an increasing sequence unbounded from above; and since a;; ==: 0, the partial sums of L;;'. 1 a;; must form a decreasing sequence unbounded from below. When we consider rearrangements of series, we shall have recourse to this result. tr we denote the nth partial sum of

a:

a;

and by Sn, s;, ands;;, respectively, then

ISnl::'=1 = {S;; + s;;1:. , converges, whereas 1s: );;'. , and {S;; };;'. 1 do mot. The manner in which the cancellation takes place in the series L;;'= 1 an may be very complicated, and our techniques for

194

Chapter 6 Infinite Series

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~: =1 an converges conditionally, then, given any real number r, there is a rearrangement of ~;=1 a. that converges tor. Proof

First, notice that

1s:1:= 1 and

{S;}:= 1 are unbounded sequences and

Ia" I:= 1 converges to zero. Let n 1 be the least positive integer such that

+ · · · + a;, > r. There is such n 1 because 1s; J;_1 is an increasing sequence not bounded from at

+

a;-

above. Next, let m1 be the least positive integer such that (a t

+ a;- + · · · + a:,) +

(a 1

+ a:2 + · · · +

a;;;,)


r.

As argued before, such an n2 exists. Now choose m2 to be the least positive integer such that (at

+ · · · + a;,) + (a 1 + · · · + a1;;;1) + (a:,+ 1 + · · · + a:2) + (a,:;;,+ 1 + · · · +a;;;,)
b cannot be 2 a lower bound for B. We conclude that b = inf B. Therefore, (ii) implies (i). assumed b

Let us make a few observations about Theorem 6.22. First, the boundedness of lbnl:=, implies that B is nonvoid and bounded, so there is a unique real number b satisfying (i), and hence, (ii) and (itii). Condition (i) is important here in that this description of b was the motiva1ting factor in the investigation of the properties of b. Condition (ii) describes b as that largest subsequential limit of the sequence, and the theorem guarantees that there is a largest subsequential limit if the sequence is bounded. Moreover, if {bn) ;~= 1 converges, then it converges to b. It seems natural that one might be able to prove a similar theorem concerning inf{p : some subsequence of Ibn) 1 converg:es top). See the exercises for some interesting challenges along these lines.

:=

DEFINITION Let lbnl:s i be a sequence of real numbers, and define B = {p : there is a subsequence of Ibn) ;,m I converging top). If B is not void and if f bnl:. 1 is bounded from above, define

= sup B.

lim sup b" n- +•

If Bis not void and if {bnl:=1 is bounded from below, define lim inf bn ■

= inf B.

Example 6.17 Consider the sequence Ibnl:=" where bn lim sup bn

=1

and

lim inf bn n--,ao

= (- lt(l

-

*). Then

= -1.

If bn = n for each n, B is void; hence, we define neither Jim supn_ao bn nor lim inf"_"" bn. If b" = n for n even and b" = l - for n odd, then Jim infn-.ao bn = I, whereas lim sup,,_"" b" is not defined. We have proved that the set of all rational numbers is countable, so we may arrange the rational numbers in (0, l) in a sequence Irn I:. 1• The reader is urged to prove that in this case

*

B

= [0, 1],

lim inf rn

= 0,

and

lim sup r" n-ao

=

■

1.

We now return to the problem that led to, this latest diversion.

6.23 THEOREM Let L:. 0 a,,x" be a power series.

:=o

L:=o

i. If the sequence I ~ I is uinbounded, a,,x" converges only for X = 0. ii. If the sequence {~ ): . 0 converges to zero, then a,,x" converges for all x. iii. If the sequence I ~ ): . 0 is bounded and

:r:=o

a

=

Jim sup ~

* 0,

then ¼is the radius of convergence.

202 Chapter 6 Infinite Series

Proof (i) Choose x

* 0. For each N, there is n :::: N such that

~ > ,:1• Hence, la,.xn l > 1 for some n:::: N. In particular, (a,.xn};:=o does not converge to zero for x i=- 0; hence, ~;:. 0 a,.xn diverge,s for x * 0. (ii) Suppose(~ };;'. 0 converges to zero andx i=- 0. Define E = l~I. There is N such that n :::: N implies that

~ 1/a. Suppose lxl > 1/a. Then a > 1/lxl, and hence, by (iii) of Theorem 6.22, there are infinitely many .n such that

~ > ,!I' So for infinitely many n, la,,xn l > I; hence {a,.xn};:= 1 does not converge to zero. Thus, for !xi > ¼, the series I;:=o a,.x" dilverges. We have as an immediate corollary to TI1eorem 6.23 a slightly different statement of Theorem 6.10.

6.24 THEOREM Let ~;:. 0 a,. be an infinite series. Then i. If { ~ };;'a o is unbounded, the series diverges. ii. If { ~ };:. 0 is bounded, then :~:~oan converges absolutely if lim sup~< 1

n....,.,

6.6 Taylor Series 203

and diverges if lim sup~> 1. Proof The series I;. 0 an converges iff the power series I:'-o a,.xn converges at = 1. The theorem then follows directly from Theorem 6.23.

x

The reader will find it instructive to restalte Theorem 6.21 and 6.8 in tenns of the new concepts just introduced. ■ Example 6.18 Let us reconsider an example mentioned earlier. Define an for n odd and an = 3-n for n even. Then

= r"

for n odd and

for n even. Hence,

has precisely two subsequential limits, l/2 and l/3. Since the larger of the two is lim supn_.., ~ . the radius of convergence is 2, as claimed earlier. ■

6.6 TAYLOR SERIES We now take up a slightly different aspect c,f power series. Recall that for all n and -I< x < 1,

+ x + x 2 + • • • + x"

J-

Xn+I

= ---. 1- X

Since {xn+ 1 }; . 1 converges to zero for - I < x < 1, the power series I:=o xn converges 1 to - - - for -1 < x < 1. We shall now inqlllire what conditions on a function f will 1- x guarantee that there is a power series I:'=o a,.xn such that for all x in some interval, I:'-o a,.xn converges to f(x). Since the point zero is not magic in this discussion, we shall now extend our previous results by a few simple remarks. The function g(x) = ! is not well behaved at zero, but we may still write I

-I g(x) = ~ = 1 - (x + 1)

=-

..

L (x + It n• O

204 Chapter 6 Infinite Series

for Ix

+

11
..

(d)

n-

l

L P-.--., 0< q

0, there is N such that for all n 2: N, a - E < a. 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

I I

lim inf a.- i

n-oo

an

and

I I·

lim sup a.,! !

,,__

an

41. Find the interval of convergence of the power series I;. 1 a.x" where an is as given below. Be sure to check the endpoints. (a) a. = (n + l )(n + 2) (b) a.= sin n (c) a. = 3- n \In (d) a n

= (n!)2

(2n)!

210 Chapter 6 Infinite Series

6.6 TAYWR SERIES 42. Write the Taylor series for log x using powe:rs of x - l. See Project 5.1 in Chapter 5 if you have forgotten the derivative of log x. Prove that this series converges to log x for 1Sx

0.

2. Show, by integration by parts, that I.

=n

-

n

1

1.-2 for all n

3. Prove that l2n+t 2n --=--l2n-l 2n + 1 .

2n 4. Prove that - - 2n + 1

l2n+1 l2n+I =- ::5 - ::s; 1 for all n

l2n-t

/2n

00

{l2n+1}

5. Conclude that - f2n

n=I

= 0, 1, 2, ... , define

converges Ito 1.

~

1.

~ 2.

Project 6 3 211

6. Prove that l2n+i · 4 · 4 · 6 · 6 · · · · (2n)(2n) · 2 - = - - - - 2-·-2 - - - - -- - - - - - - - -

1 • 3 · 3 · 5 • 5 · 7 · 7 · · · · (2n - 1)(2n - 1)(2n + 1)71'

l2n

Items 5 and 6 yield Wallis's fonnula: . 2 • 2 · 4 · 4 • 6 · 6 • • • · (2n)(2n) hm - - - - - - - - - - - - - - - - - - - n-oo 1 · 3 · 3 · 5 · 5 · 7 · 7 · · · · (2n - 1)(2n - 1)(2n + 1)

7T

=- . 2

r

Wallis's fonnula gives 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 paiticular, analysis of complex variablesyou will learn more about infinite products.

PROJECT 6.2 _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ In this project you will construct an increasi1ng function that is discontinuous at each rational point in (0, 1) and continuous at each irrational point in (0, 1). We will need two basic facts: a. The rational numbers in the intervall (0, 1) can be arranged in a sequence {rnl::'=i• 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 tenns with zeros. For each real number x E (0, 1), define K(x) of J, and we define f: (0, I) -+ R by the fonnula

f = I

nEK(x)

= {n: rn s

x}. So K(x) is a subset

1 2. n

1. Prove that f is increasing. 2. Prove that f is discontinuous at r if r is a rational number. 3. Prove 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

L.. (-l)n+I _1 11 -= l

n

212 Chapter 6 Infinite 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.

l. We will need the notation and the result of Exercise 26. If you haven't done that exercise, that is yoUJr first task. To summarize, define 'Yn = 1 + ½ + ½ + · · · + ~ - log n. Prove that {'Yn};. , converges. 2. Let Sn be the nth partial sum of

:=

that {S2n) 3. Show that 2

f (·-

l)n+ 1

! . Your strategy will be to show

n converges to log 2. Expllain why this shows the sum is log 2. 1 n- 1

I (- v+I -k1 = I2n -k1 - I" -k1 == ,'2,. "

k• I

k=I

-y,.

+

log 2.

k= I

Show from this equality that {S2,.};a , converges to log 2.

PROJECT 6.4 _ _ _ _ _ _ _ _ _ _ _ _ _ __ Undoubtedly, you remember the integral test for convergence of infinite series. In this project, you will prove the theorem that gives: this test.

INTEGRAL TEST Let f be continuous on {x: x ative. Then

f

~

1 }, decreasing and nonneg-

f(n) converges if and only if the sequence

n-1

{f

f(x)

dx}

00

n• l

converges. 1. Provide a proof for this theorem. You may want to sketch the graph of f and compare the partial sums of the series with the terms of the sequence. 2. You have probably noted that the convergence of the sequence

{f

00

f(x)

dx},.~, is equivalent to the convergence of the improper integral

J,'° f(x) dx. State and prove a theorem that relates the sum of the series and the limit of the sequence {

ff

(i)

dx}

t,

f(n)

~= 1

3. Use this test to verify the results of EJ 0, f n: [0, l] - R by

f n(x) = xn for each x E [0, l]. For x E [0, 1), {f,,(x)}.:'. ,

converges to zero, and for x

= 1,

lf,,(x)l:-1 converges to 1. Thus, the limit function is no,t continuous, although each f,, is continuous. • ■

Example 7.3 For each positive integer n, define 8n: [0, I] - R by g,,(0)

=0

g,,(x)

=n

for0 0, there is N such that for each positive integer n, n 2: N implies th.at lfn(x) - f(x)I


0, there are positive real numbers N and 8 such that for each positive integer

:=

n, n

2:'. N,

Ix - x0 1 < 8, and x E E imply that

If"(x) - /(x)I


I