9 This book the is to be returned on or before last date stamped below. NAME CALCULUS Michael Spivak BRANDEIS UNIVERSITY WORLD STUDENT SERIES EDITION A complete and unabridged reprint a/the American textbook, may be this World Student sold only in those countries consigned by Addison-Wesley or distributors. It to which the it may not to its United States of America or is which it is authorized trade be re-exportedfrom the country has been consigned, and The publisher original Series edition pleased to its it may not be sold in possessions. acknowledge the assistance of Wtadistaw Finns, who designed the text and cover, and F. W. Taylor, who produced the illustrations. Copyright reserved. © 1967 by W. A. Benjamin, Inc. All rights of this publication may be reproduced, Ma part stared in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Original edition published in the United States of America, Published simultaneously in Canada. Philippines copyright 1967. Library of Congress Catalog Card Number: 67-20770 Dedicated to the Memory ofY. P. PREFACE / hold every man a debtor to his profession, from the doe seeke which as men of course to receive countenance and profit, endeavour so ought they of duty themselves by way of amends, to be a help to and ornament thereunto. FRANCIS BACON Every aspect of this book was influenced by the desire to present calculus not merely as a prelude to but as the first real encounter with mathematics. Since the foundations of analysis provided the arena in which modern modes of mathematical thinking developed, calculus ought to be the place in which than avoid, the strengthening of insight with logic. In addition to developing the students' intuition about the beautiful concepts of analysis, it is surely equally important to persuade them that precision and rigor are neither deterrents to intuition, nor ends in themselves, but the to expect, rather natural medium in which to formulate and think about mathematical questions. This goal implies a view of mathematics which, in a sense, the entire book No matter how well particular topics may be developed, the goals of this book will be realized only if it succeeds as a whole. For this attempts to defend. reason, it would be of little tion pedagogical practices customarily bestowed on value merely to calculus texts will probably particular aspects of calculus will men- the topics covered, or to and other innovations. Even new the cursory glance tell more than any the teacher with strong feelings about such extended advertisement, and fulfills his list know just where to look to see if this book requirements. A few features do require explicit comment, however. Of the twenty-nine chapters in the book, two (starred) chapters are optional, and the three chapters comprising Part V have been included only for the benefit of those students who might want to examine on their own a construction of the real numbers. Moreover, the appendices to Chapters 3 and 1 1 also contain optional material. The order of the remaining chapters the purpose of the book is is intentionally quite inflexible, since to present calculus as the evolution of one idea, not most exciting concepts of calculus do not and II will probably although the entire book covers require less time than their length suggests a one-year course, the chapters are not meant to be covered at any uniform as a collection of "topics." Since the appear rate. A until Part III, I should point out that Parts I — rather natural dividing point does occur between Parts II and III, so quickly by it is possible to reach differentiation and integration even more detailed more a treating Part II very briefly, perhaps returning later for organization traditional treatment. This arrangement corresponds to the of most calculus courses, but I feel that it will only diminish the value of the book for students who have seen a small amount of calculus previously, and for bright students with a reasonable background. The problems have been designed with this particular audience in mind. They range from straightforward, but not overly simple, exercises- which develop basic techniques and test understanding of concepts, to problems of considerable difficulty and, I hope, of comparable interest. There are about 625 problems in many vii all. Those which emphasize manipulations usually contain examples, numbered with small Roman numerals, while small letters are used to label interrelated parts in other problems. Some indication of is provided by a system of starring and double starring, but relative difficulty there are so many criteria for judging difficulty, and so many hints have been provided, especially for harder problems, that this guide is not completely reliable. Many problems are so difficult, especially if the hints are not con- probably have to attempt only those which difficult problems it should be easy to from the less them; especially interest class busy, but not frustrated. The good will keep a which portion select a answer section contains solutions to about half the examples from an assortment of problems that should provide a good test of technical competence. A sulted, that the best of students will separate answer book contains the solutions of the other parts of these problems, and of all the other problems as well. Finally, there is a Suggested Read- ing list, to which the problems often refer, and an index of symbols. am grateful for the opportunity to mention the many people to whom I owe my thanks. Mrs. Jane Bjorkgren performed prodigious feats of typing^ that compensated for my fitful production of the manuscript. Mr. Richard I Serkey helped collect the material which provides historical sidelights in the problems, and Mr. Richard Weiss supplied the answers appearing in the back of the book. I am especially grateful to my friends Michael Freeman, Jay Goldman, Anthony Phillips, and Robert Wells for the care with which they read, and the relentlessness with which they criticized, a preliminary version of the book. Needless to say, they are not responsible for the deficiencies which remain, especially since I sometimes rejected suggestions which would have made the book appear suitable for a larger group of students. I must express my admiration for the editors and staff of W. A. Benjamin, Inc., who were always eager to increase the appeal of the book, while recognizing the audience for which it was intended. The inadequacies which preliminary editions always involve were gallantly endured by a rugged group of freshmen in the honors mathematics course at Brandeis University during the academic year 1965-1966. About half of this course was devoted to algebra and topology, while the other half covered calculus, with the preliminary edition as the text. It is almost obligatory in such circumstances to report that the preliminary version was a gratifying after all, the class is unlikely to rise up in a body success. This is always safe and protest publically — —but the students themselves, it seems to me, deserve the right to assign credit for the thoroughness with which they absorbed an impressive amount of mathematics. I am content to hope that some other students will be able to use the book to such good purpose, and with such enthusiasm. Waltham, Massachusetts February 1967 MICHAEL SPIVAK PREFACE TO THE WORLD STUDENT SERIES EDITION In American universities, "calculus" courses are so varied in content and viewpoint that only tradition dictates that the same title be given to all. Some of these courses cover only the most rudimentary techniques, sedulously shun- ning all theory. At the other extreme, so-called "honors" courses concentrate on the foundations, and emphasize the theoretical importance of the basic processes of calculus, which have often already been seen by the students in their high school classes. This book was written for just such courses. In the United States, they are so rare that I felt constrained to begin preface with a lengthy defense, or apology for the fact that for the very type of course and I would usually called like to which in it Great Britain and Europe was is my original really written quite standard, "An introduction to Real analysis". From that long prologue salvage only one thought, and once again express my feeling that the foundations of analysis provide one of the most beautiful instances of the welding together of mathematical intuition and mathematical rigor, and that ignoring either one only weakens, rather than strengthens, the other. A few comments about the contents and arrangement should be made, who is learning the material on his own. Of the twenty- especially for the student nine chapters in the book, two (starred) chapters are "optional", and the three V might be thought of in the same way. Moreover, and 1 1 also contain "optional" material. Aside from this, the order of the remaining chapters is intentionally quite inflexible, since the purpose of the book is to present calculus as the evolution of one idea, not as a collection of "topics". Since the most exciting concepts of calculus do not appear until Part III, I should point out that Parts I and II will probably require less time than their length suggests, and although the entire book covers a one-year course, the chapters are not meant to be covered at any uniform rate. A rather natural dividing point does occur between Parts II and III, so it is possible to reach differentiation and integration even more quickly by reading Part II very briefly, perhaps returning later for more detailed study. But I personally feel that the route set forth here, though it requires perseverance at the beginning, will be the most rewarding. The problems, without which any mathematics text is incomplete, range from straightforward (but not overly simple) exercises which develop basic techniques and test understanding of concepts, to problems of considerable difficulty and, I hope, of comparable interest. There are about 625 problems in all. Those which emphasize manipulations usually contain many examples, chapters comprising Part the appendices to Chapters 3 numbered with small Roman numerals, while small interrelated parts in other problems. Some letters are used to label indication of relative difficulty is provided by a system of starring and double starring, but there are so many criteria forjudging difficulty, and so many hints have been provided, especially for harder problems, that are so difficult, especially will this if is not completely reliable. Many problems probably have to attempt only those which especially interest them. From it should be easy to select a portion which will be the less difficult problems ix guide the hints are not consulted, that the best of students * Preface to the World Student Series edition challenging, but not frustrating. The Answer Section contains solutions to about half the examples from an assortment of problems that should provide a good test of technical competence. A separate answer book, Supplement to Calculus, contains the solutions of the other parts of these the other problems as well. Finally, there is problems, and of all a Suggested Reading list, to which and a Glossary of Symbols. Once again I will take the opportunity to mention the many people to whom feats of typing I owe my thanks. Mrs. Jane Bjorkgren performed prodigious Mr. Richard manuscript. the of production fitful my for that compensated in the sidelights historical provided which material the collect helped Serkey problems, and Mr. Richard Weiss supplied the answers appearing in the back of the book. I am especially grateful to my friends Michael Freeman, Jay Goldman, Anthony Phillips, and Robert Wells for the care with which they read, and the relentlessness with which they criticized, a preliminary version the problems often refer, of the book. editions always involve were gallantly in the honors mathematics course at freshmen of group rugged a by endured Brandeis University during the academic year 1965-1966. It is almost obligatory in such circumstances to report that the preliminary version was a The inadequacies which preliminary gratifying success. This is always safe— after — all, the class is unlikely to rise but the students themselves, seerns to up me, in a body and protest publicly deserve the right to assign credit for the thoroughness with which they absorbed an impressive amount of mathematics. I am content to hope that some other students will be able to use the book to such good purpose, and with such it enthusiasm. Berkeley, California MICHAEL SPIVAK CONTENTS PART PART I II Prologue 1 Basic Properties of 2 Numbers Numbers of Various Sorts Foundations 3 Functions 37 Appendix. Ordered Pairs Graphs 5 Limits 6 Continuous Functions 93 Three Hard Theorems 100 1 1 Least Upper Bounds 8 III 52 4 7 PART 3 21 Derivatives 9 54 72 and Integrals Derivatives 125 144 10 Differentiation 11 Significance of the Derivative Appendix. Convexity and Concavity 12 214 Integrals 14 The Fundamental Theorem of Calculus The Trigonometric Functions 256 *16 17 18 IV 191 200 13 15 PART Inverse Functions 163 7r *20 21 Irrational 22 Infinite Series Uniform Complex Complex Complex 25 26 xi 283 Sequences and Infinite Series Approximation by Polynomial Functions 362 e is Transcendental 372 Infinite Sequences 23 24 277 The Logarithm and Exponential Functions Integration in Elementary Terms 302 Infinite 19 is 240 333 388 Convergence and Power Series Numbers 433 Functions 448 Power Series 462 412 xii Contents PART V Epilogue 487 27 Fields 28 Construction of the Real 29 Uniqueness of the Real Numbers Suggested Reading 513 523 Answers {to selected problems) Glossary of Symbols Index 575 571 Numbers 494 507 PART PROLOGUE To you to be conscious that are ignorant is a great step knowledge. BENJAMIN DISRAELI CHAPTER NUMBERS BASIC PROPERTIES OF I The title of this chapter expresses in a few words the mathematical knowledge required to read this book. In fact, this short chapter is simply an explanation addition of what is meant by the "basic properties of numbers/' all of which and multiplication, subtraction and division, solutions of equations and — and factoring inequalities, manipulations— are algebraic other already not a review. Despite the familiarity of the subject, the survey we are about to undertake will probably seem quite novel; it does not aim to present an extended review of old material, but to condense this knowledge into a few simple and obvious properties of numbers. familiar to us. Nevertheless, this chapter is too obvious to mention, but a surprising number of turn out to be consequences of the ones we shall Some may even seem diverse and important facts emphasize. Of the twelve properties which we shall study in this chapter, the first nine are concerned with the fundamental operations of addition and multiplicaperformed tion. For the moment we consider only addition: this operation is —the sum a + b exists for any two given numbers a and b number, of course). It might seem reasonsame (which may possibly be the which can be performed on several operation an as addition regard able to + a n of n numbers numbers at once, and consider the sum a\ + a% as a basic concept. It is more convenient, however, to consider au addition of pairs of numbers only, and to define other sums in terms of sums of this type. For the sum of three numbers a, b, and c, this may be done in twodifferent ways. One can first add b and c, obtaining b + c, and then add a to then this number, obtaining a + (b + c); or one can first add a and b, and compound two the course, Of c. b) (a obtaining + + add the sum a + b to c, sums obtained are equal, and this fact is the very first property we shall list: on a pair of numbers • • • • , (PI) If «, b, and c are a The statement of three • + any numbers, then (b + = c) (a + b) + c. of this property clearly renders a separate concept of the numbers superfluous; we simply agree that = + + a + + b c sum denotes the Addition of four numbers requires b considerations. The symbol a involved, similar, though slightly more mean defined to d is c number a + (b + c) } (a b) c. + + + (1) 3 or (2) or (3) or (4) or (5) + b)+c)+d, + (b + e)) + d a + {(b + c) + d), + d)), a + (b + (a + b) + + d). ((a (a 9 (c (c This definition unambiguous is since these numbers are all equal. Fortunately, need not be listed separately, since it follows from the property PI already listed. For example, we know from PI that this fact (a + b) +c=a+ (b + c), equality of (2) and (3) is a direct consequence of PI, although this may not be apparent at sight (one must let b c play the role of b in PI, and d the role of c). first and it and follows immediately that (1) (2) are equal. The + equalities (3) = (4) = The (5) are also simple to prove. probably obvious that an appeal to PI will also suffice to prove the equality of the 14 possible ways of summing five numbers, but it may not be so clear how we can reasonably arrange a proof that this is so without actually listing these 14 sums. Such a procedure is feasible, but would soon cease to be if we considered collections of six, seven, or more numbers; it would be totally It is inadequate to prove the equality of possible all sums of an arbitrary finite be taken for granted, but an for to worry about the proof (and it is worth worrying about once) a reasonable approach is outlined in Problem 23. Henceforth, we numbers ai, those who would like collection of shall usually + ax • • . , may If a 0 has one property so important that is we list it next: any number, then important role (P3) This fact tacit appeal to the results of this problem and write sums with a blithe disregard for the arrangement of parentheses. + a An . . make a + an The number (P2) . is 0 also played For every number a+ a, = 0 + by 0 there a. in the third property of number —a such a is = (- (-a) = a +a = fl ) our list: that 0. Property P2 ought to represent a distinguishing characteristic of the number and it is comforting to note that we are already in a position to prove 0, this. Indeed, if a number x satisfies a + x = a any one number a, then x = 0 (and consequently this equation also holds numbers a). The proof of this assertion involves nothing more than subtracting a from both sides of the equation, in other words, adding — a to both sides; as the following detailed proof shows, all three properties P1-P3 for for all must be used to justify this operation. If then hence hence hence +x= + + x) = (( — a) + a) + x = 0 + # = a (—a) (a x = <z, {—a) 0; 0; 0. +a = 0; Basic Properties of Numbers As we have just hinted, derived from addition: it is we 5 convenient to regard subtraction as an operation — b). — b to be an abbreviation for a ( + consider a then possible to find the solution of certain simple equations by a series of steps (each justified by PI, P2, orP3) similar to the ones just presented for x = a. For example: the equation a It is + = (* + 3) + (-3) = x + (3 + (-3)) = x + 0 = x — + x If then hence hence hence 3 5, + (-3); 5-3 = 2: 5 2; 2. Naturally, such elaborate solutions are of interest only until you become convinced that they can always be supplied. In practice, it is usually just a waste of time to solve an equation ties Only one other property c so explicitly the reliance and a + the order of (P4) shall of addition remains to be listed. on proper- list). a, b, and When considering + c, + (b a, b, If a we b) only two sums were mentioned: (a if obtained are arrangements c). Actually, several other on equal depends all sums are these That changed. and c is the sums of three numbers + by indicating PI, P2, and P3 (or any of the further properties and any numbers, then b are + a b — b + a. The statement of P4 is meant to emphasize that although the operation of addition of pairs of numbers might conceivably depend on the order of the two numbers, in fact it does not. It is helpful to remember that not all operations are so well behaved. erty: usually a — — b 9± b — For example, subtraction does not have this propa. In passing we might ask just when a — b does amusing to discover how powerless we are if we rely only on properties P1-P4 to justify our manipulations. Algebra of the most elementary variety shows that a — b — b — a only when a = b. Nevertheless, it is impossible to derive this fact from properties P1-P4; it is instructive to examine the elementary algebra carefully and determine which step(s) equal b cannot be detail a, and it is justified when We by P1-P4. indeed be able to justify all steps in listed. Oddly enough, however, the will a few more properties are crucial property involves multiplication. basic properties of multiplication are fortunately so similar to those for addition that little comment will be needed; both the meaning and the conse- The quences should be will (P5) If a, b } * and b> c or simply ab.) are a (P6) If a is elementary algebra, the product of a and clear. (As in be denoted by a any numbers, then ' (b ' c) = (a - b) ' any number, then a ' 1 = 1 • a = a. c. b 1^0. Moreover, 1^0 may seem a strange fact to (The assertion that list it, because there is'no the other properties listed way —these properties would only one number, namely, a (P8) If a and * detail there is no — a~ a~ l l = b ' = means a * b~~ l . if there were a. ' the appearance of the condition is quite necessary; since 0 b = 0 for all numbers -1 = satisfying 0 0 1. This restriction has an impor- 1 • is • in terms of was defined defined in terms of multiplication: the symbol a/b is CT" 1 is Since hold all 1. tant consequence for division. Just as subtraction addition, so division to a number a~ l such that is a • b which deserves emphasis number CP we have any numbers, then b are a 7* 0 in P7. This condition b, there =^ 0, a One but 0.) For every number a (P7) list, could possibly be proved on the basis of it meaningless, a/0 also meaningless is —division by 0 always undefined. is Property P7 has two important consequences. If a = necessarily follow that b matter what from P7 as b and c are. for c; if However, In a l then a~~ hence hence hence (a~ l (a • • if a b • then both a 0, ^ — = — = = b • b) • a) * b 1 • b b a consequence of P7 that then 0, — b b c; = a and this * c, it a ' c does not are no 0, can be deduced if a • a * and c a~ x (a~ l * 1 (a * • ' a) a 0, c)\ • c; c; c. b = 0, then either a = 0 or £ = 0. fact, may happen excluded -1 then a hence hence hence (a~ l that a when we ~ = a) b — 1-^ = = a if (It — follows: If It is also a = (a • * ' 6 6) • • /; and 0 6 say "either a 0 and a ?± 0, 0; 0; 0; 0. = 0 are both true; = 0 or b = 0"; in this possibility is not mathematics "or" is always used in the sense of "one or the other, or both.") This latter consequence of P7 is constantly used in the solution of equations. Suppose, for example, that a number x is known to satisfy ~ 1)(* - 2) - 0. — = 0 or x — 2 ~ (* Then On it follows that either x 1 the basis of the eight properties listed so far 0; hence x it is still — 1 or x = 2. possible to prove Basic Properties of Numbers which combines the operations of addi- Listing the next property, very little. tion and multiplication, (P9) and If a, b, will alter this situation drastically. are any numbers, then c (b + c) = (Notice that the equation (b + c) a • a — b + b ' = a As an example of the usefulness of P9 we — 7 a b will • + a • c. c • a is also true, by P8.) now determine just when - a b a: a — b b) + b If then - (a hence hence Consequently — = = = = — a a a (1 +a + 1) and therefore a — a, - a) + b = b + b — a; {b + b - a) + b (b £ (1 * + b + (b a = b - + a); b. 1), b. 0 = 0 which we have is the justification of the assertion a already made, and even used in a proof on page 6 (can you find where?). This fact was not listed as one of the basic properties, even though no proof A second use of P9 mentioned. With P1-P8 alone a proof was not possible, since the number 0 appears only in P2 and P3, which concern addiP9 the tion, while the assertion in question involves multiplication. With have We obvious: not perhaps though proof is simple, was offered when was it first a as we have already both A sides) that d 0 ' 0 + a = ' ' • will establish the b). To 0) this, may help explain the somewhat is positive. To begin + (-&) Consequently, adding (a ( — a) • b = * = a'b = [(-a) +a]'b = 0-b - (a 0. • b) to both sides) that note that • to 0) note that immediately (by adding (-*) • two negative numbers = — (a-b). Now + easily acceptable assertion that more (-a) 'b It follows (0 0; 0. we prove a a consequences of P9 mysterious rule that the product of — (a = = 0 noted, this immediately implies (by adding (a series of further with, ' + • [-(a 6) to * &)] both = (-a) (-/>) + (-a) = (-«)•[(-*) + *] = (-tf)'O = 0. sides, • we obtain • b (-a) • The fact that the product of two negative numbers is positive is thus a consequence of P1-P9. In other words, if we want PI to P9 to be true, the rule for the product of two negative numbers is forced upon us. various consequences of P9 examined so far, although interesting and important, do not really indicate the significance of P9; after all we could have The each of these properties separately. Actually, P9 is the justification for all algebraic manipulations. For example, although we have shown to solve the equation listed almost how - (x - l)(x = 2) 0, we can hardly expect to be presented with an equation in this form. more likely to be confronted with the equation x2 The 2 "factorization" x - 2>x + - 3x + 2 2 = (x - = We are 0. 1)(* - 2) is really a triple use ofP9: (* - 1) • (* - 2) - x2 + + (-1) (* - 2) (-2) + (-1) x + (-1) x[(-2) + (-l)] + 2 x2 - 3x = x • = = = x • (* + x 2) x • • • + • (-2) 2. A final illustration of the importance of P9 is the fact that this property is actually used every time one multiplies arabic numerals. For example, the calculation 13 X24 52 26 312 is a concise arrangement for the following equations: 13 • 24 = = = 13 • (2 13 -2 26 • • 10 10 + 4) + 13 + 52. 10 • • 4 (Note that moving 26 to the left in the above calculation 26 10.) The multiplication 13 4 = 52 uses P9 also: • • 13-4 = = • (1 10 + 3) 4 + -4-10 + 12 1 • 10 • -4 3 • 4 = 4-10 + 1-10 + = (4 + 1) 10 + 2 = 5-10 + 2 = 52. • 2 is the same as writing 9 Basic Properties of Numbers P1-P9 have descriptive names which are not essential to remember, but which are often convenient for reference. We will take this opportunity to list properties P1-P9 together and indicate the names by which they are commonly designated. The properties (PI) (P2) (Associative law for addition) a (Existence of an additive a + + + (b = 0 = c) + 0 {a + = a b) + c. a. identity) (P3) (Existence of additive inverses) a + + (P4) (Commutative law for addition) a (P5) (Associative law for multiplica- a - (b a - 1 a * <T l a * a • b • + = (-a) (-a) + = b c) = = a 0. a. (a b) ' ' c. tion) (P6) (Existence of a multiplicative = a 1 = 1^0. a; identity) (P7) (Existence of multiplicative = d~ l ' = a 1, tor a 0. inverses) (P8) (Commutative law for multi- = b b ' a. plication) (P9) (Distributive law) (b + c) — a • b +a • c. The three basic properties of numbers which remain to be listed are concerned with inequalities. Although inequalities occur rarely in elementary mathematics, they play a prominent role in calculus. The two notions of inequality, a < b (a is less than b) and a > b (a is greater than b), are intimately two related: a < b means the same as b > a (thus 1 < 3 and 3 > 1 are merely ways of writing the same assertion). The numbers a satisfying a > 0 are 0 are called negative. is possible to reverse it terms of in defined <, be thus can While positivity the procedure: a < b can be defined to mean that b — a is positive. In fact, called positive, while those it is P 9 numbers a satisfying a convenient to consider the collection of as the basic concept, (P10) and all < positive numbers, denoted by state all properties in terms of P: (Trichotomy law) For every number a, one and only one of the following holds: (i) a = (ii) a is (iii) —a 0, in the collection P, is in the collection P. (PI 1) (Closure under addition) If a and (PI 2) (Closure under multiplication) If a in P. b are in P, then a and b + b is in P. are in P, then a b is These three properties should be complemented with the following defini- tions: > < > < a a a a Note, in particular, that a b if a b if b b if a b if a > 0 — > > < b is in • b or a b or a and only if P; a; if a = b\ = b* in P. is however elementary they may seem, are consequences of P10-P12. For example, if a and b are any two numbers, then precisely one of the following holds: All the familiar facts about inequalities, (i) a (ii) a — = b — (a — (iii) Using the definitions 0, b is in the collection P, = b) made, just — b a in the collection P. is follows that precisely one of the following it holds: A more slightly If a < b, < b, then a a so that b + a and £ < c + £ and c so c if (iii) b = > > a < b b, b, a. from the following manipulations. in P, then surely (b is c. + c) Similarly, suppose a — a is in P, — b is in P, — a — (c — b This shows that a a interesting fact results — < c (i) (ii) and b < c, + (b then a < b) — c. < a) — {a b and is + <0 is b < c. in P; thus, if Then in P. (The two inequalities a < b < b < c, which has are usually written in the abbreviated form a the third inequality a < c almost built in.) somewhat less obvious: If a < 0 and b < 0, then ab > 0. The only difficulty presented by the proof is the unraveling of definitions. The symbol a < 0 means, by definition, 0 > a, which means 0 — a — — a is in P. Similarly — b is in P, and consequently, by PI 2, (—a)( — b) — ab is in P. Thus ab > 0. The fact that ab > 0 if a > 0, b > 0 and also if a < 0, b < 0 has one special consequence: a 2 > 0 if a ^ 0. Thus squares of nonzer> numbers are The * There following assertion is one is slightly perplexing feature of the we know usefulness of the symbol some values of a and b, is is < The . statements 1 that bound and 1 1 in the second. This sort of thing > + <3 + <2 1 are both true, even though symbols < to could be replaced by occur when revealed by a statement like < < in the first, and by = used with specific numbers; the Theorem 1 here equality holds for is while inequality holds for other values. — Basic Properties of Numbers 11 have always positive, and in particular we have proved a result which might properties: of list 1 our > 0 seemed sufficiently elementary to be included in (since 1 The = l 2 ). -a > fact that 0 if < a 0 the basis of a concept which will play an For any number a, we define the is role in this book. extremely important absolute value \a\ of a as follows: III ~ \ Note that \a\ 0 when a a a always positive, except is > < a, -a, 0. = For example, we have 0. = + V2 - V3, and — 3j = 3, = 7, Jl + V2 |7J 7 V !^! = VlO — V2 — 1. In general, the most straightforward 1 |1 | any + V2 - approach to problem involving absolute values requires treating several cases sepaapproach since absolute values are defined by cases to begin with. This rately, may THEOREM 1 For values, be used to prove the following very important fact about absolute all numbers a and we have b, +b \a PROOF We < ! + \a\ 1*1- will consider 4 cases: In case (1) we also 0) a a >o, >o, b (2) (3) a < o, b (4) a <o, b have a + b + b\ = a \ > a 0, b > < > < 0; 0; 0; 0. and the theorem +b- + \a\ is obvious; in fact, \b\, so that in this case equality holds. In case we have (4) \a In case (2), + a b\= when a > + b < 0, and again equality holds: + -(a + b) = -a + (-*) = \a\ < 0 and b \a + we must prove 0, b\ < a - a which hand, is if certainly true since b a + b < 0, +b b is <a — < -b, which is certainly true since a b < —a < is + -b is positive a — b, a, and b positive. that —a — i.e., If a > 0, then we b, negative and we must prove that b. This case may therefore be divided into two subcases. must prove that i.e., \b\. —a is negative. On the other may Finally, note that case (3) applying case Although (2) this with a and method b be disposed of with no additional work, by interchanged. | of treating absolute values (separate consideration sometimes the only approach available, there are often simpler methods which may be used. In fact, it is possible to give a much shorter proof of Theorem 1 this proof is motivated by the observation that of various cases) is ; = \a\ (Here, and throughout the book, symbol is when defined only (\a + b\) 2 > x = (a V7 2 . denotes the 0.) + b) 2 square root of x; this positive We may now observe that - a + lab + b < a 2 + 2\a\ \b\ + b 2 = \a\ 2 + 2\a\ \b\ + \b\ = (W + H) 2 2 • • 2 2 . + + 2 we can conclude that \a b\ < \a\ \b\ because x < y 2 implies provided that and x < y, y are both non-negative; a proof of this fact is left to the reader (Problem 5). From this x One final observation may be made about the theorem we have just proved a close examination of either proof offered shows that : + \a a and b have the same sign one of the two is 0, while if (i.e., \a if a and We b are b\ + = \a\ + \b\ are both positive or both negative), or b\< \a\ + if \b\ of opposite signs. this chapter with a subtle point, neglected until now, whose inclusion is required in a conscientious survey of the properties of numbers. After stating property P9, we proved that a — b = b — a implies a = b. The proof began by establishing that will conclude a' (1 + 1) = A - (1 +1), from which we concluded that a = b. This result is obtained from the equation 1. Division by 0 (1 + 1) = b (1 + 1) by dividing both sides by 1 should be avoided scrupulously, and it must therefore be admitted that the validity of the argument depends on knowing that 1 + 1^0. Problem 24 is designed to convince you that this fact cannot possibly be proved from properties P1-P9 alone Once P10, Pll, and PI 2 are available, however, the proof is very simple: We have already seen that 1 > 0; it follows that 1 + 1 > 0, and in particular 1 + 1^0. This last demonstration has perhaps only strengthened your feeling that it is absurd to bother proving such obvious facts, but an honest assessment of our a • + ' ! present situation will help justify serious consideration of such details. In . Basic Properties of Numbers 13 numbers are familiar objects, and that well-known properties of PI -PI 2 are merely numbers. It would be difficult, however, to justify this assumption. Although one learns how to "work with" numbers in school, just what numbers are, remains rather vague. A great deal of this book is devoted to elucidating the concept of numbers, and by the end of the book we will have become quite well acquainted with them. But it will be necessary to work with numbers throughout the book. It is therefore reasonable to admit frankly that we do not yet thoroughly understand numbers; we may still say that, in whatever way numbers are finally defined, they should certainly have properties this we have assumed chapter that explicit statements of obvious, P1-P12. Most of this chapter has been an attempt to present convincing evidence which we should assume in order to deduce other familiar properties of numbers. Some of the problems (which indicate the derivation of other facts about numbers from PI -PI 2) are offered as further evidence. It is still a crucial question whether PI -PI 2 actually account for all properties of numbers. As a matter of fact, we shall soon see that they do not. In the next chapter the deficiencies of properties PI -PI 2 will that PI -PI 2 are indeed basic properties but the proper means for correcting these deficiencies is not so easily discovered. The. crucial additional basic property of numbers which we are seeking is profound and subtle, quite unlike PI -PI 2. The dis- become quite clear, covery of this crucial property will require all the work of Part II of this book. In the remainder of Part I we will begin to see why some additional property is required; in order to investigate this we will have to consider a little more carefully what we mean by "numbers." PROBLEMS 1. Prove the following: (i) If ax (ii) x2 (iii) If x 2 (iv) (v) (vi) = a for -y = 2 = y (x 2 , some number a - then x = ^ 0, then x — 1 + y). y){x y or x = —y. — y){x + xy + y ~ n' x n -y n = (x - y)(x + xy n ~ + y n ). + x n~ y + (There is a particularly x + y = (x + y)(x — xy + y x 3 —y = 3 2 2 (x ). 2 1 3 2 z • l 2 • 2 ). and it will show you how way to do this, using n n torization for x + y whenever n is odd.) (iv), 2. What is wrong with the following "proof? Let -y y(x - y), xy (x +y)(x - y) x +y y> 2 l. 2 , x = y. easy to find a fac- Then 14 Prologue 3. Prove the following: a ad c be ~\- (iii) = (ab)~ L a~ l b~ l 5* 0. if 6, > if a, b , (To do 0. this you must remember the defining property of (ab)~~ l .) (iv) (vi) * d If b 9 0, = then - _ a Find numbers x all for if ad = 5 (vi) (vii) (viii) (ix) 0. (When is a product of two numbers posi- - 2x + 2 > 0. + x + > 2. * - * + 10 > 16. > 0. x + x + (* - 7r)Cr + 5)(* - 3) > (x - \ 2){x - V2) > 0. x2 x2 1 2 2 1 (xi) 2* (xii) * (xiii) - 1 < + 0. 8. < 3* 4. 1 + x 5. when 2 / (x) Also determine 2 tive?) (v) be. which - x < 3 - 2x. - x < 8. 5 - x < -2. (x — \){x - 3) > 4 (ii) (iv) and only a (i) (iii) if b b 4. - d b 1 — > 0. X Prove the following: (i) If a (ii) If a < < b b, and c then < d, then a -b < -a. + c < b + d. Basic Properties of Numbers 6. (iii) If a (iv) If a (v) If a (vi) If a < < < > < < < (vii) If 0 (viii) If 0 .(ix) If 0 (x) If a, £ Prove that and and and b b b > > < c e c then a 0, then ac 0, then ac then a 1 > a. 2 a < 1, then f2 < a. b and 0 < c < b, then a 2 < a < < > 0 and a 2 < 0 < then a if < a b, / b be. 2 , d, + Va6 < < b. bd. (viii).) < then a a < then b. (Use (ix), backwards.) b b. + £)/2 holds for a, b > 0, without generalization of this fact occurs in (a A 2-20. = y. Hint: First explain y and n is odd, then x suffices to consider only the case x, y > 0; then show that Prove that why it x < y and if n x > y = 11 x are both impossible, = y n and is even, then x = or x = —v. Although the basic properties of inequalities were stated in terms of the collection P of all positive numbers, and < was defined in terms of P, this procedure can be reversed. Suppose that P10-P12 are replaced by (b) *8. d. be. b\ (Use — the additional assumption a (a) < > — b < Vab < ~^r— < Notice that the inequality Problem < c 1, a *7. — d, 15 Prove that (P'10) if n x ;? For any numbers a and b and only one, of the one, follow- ing holds: (P'll) « a (ii) a (iii) b (P'13) Show that b, a. < a, b, and < if c, it < a P10-P12 can then be deduced signs. \V2 (ii) \(\a (iii) |(|fl (iv) (v) 7 + V3 - V 5 + + - + +H ~ +b+ b\ |*| |(|V2 \b\)\. \a 6| -_2xy < b and b < c, then if c, then a b, a < b +c<b+ and 0 < c, e. then be. as theorems. Express each of the following with at least one (i) a e. For any numbers <2, b, and c, For any numbers a, b, and ac 9. b, For any numbers a (P'12) = < < c\)\. +/|. + V3|-|V5-V7|)|. less pair of absolute value 16 Prologue 10. Express each of the following without absolute value signs, treating when various cases separately (i) + b\- \a necessary. \b\. (ii) 11. |*| (iv) a Find (i) (ii) 12. - |^|. - \{a - (iii) numbers * all - 3| + 4| |* I* (iii) |* (iv) |* (v) |* (vi) |* (vii) I* (viii) |* \a\)\. - = < < 3| 8. 8. 2. + + + lj 1| lj II- I* • |* 1| which for |* - 2\ |* + 1| |*+ lj > < < + = + 2| = II 1. 2. 1. 0. 3. Prove the following: (i) = |*| = — \xy\ (ii) • > \y\. if x 9* 0. (The best way to do this is to remember what \x\ -1 is.) X if/?* (iii) 0. y Civ) \x (v) \x\ — y\ < + — < \x — y\. |#| jv | I - \y\ (Give a very short proof.) (A very short proof is possible, if you write things in the right way.) (vi) |(|#| — |jy|)| < — y\. (Why \x does this follow immediately from (v)?) (vii) 13. + + + + z\ < \x\ |* y \y\ prove your statement. \z\. when Indicate equality holds, and The maximum of two numbers x and y is denoted by max(^, j). Thus max( — 1, 3) = max (3, 3) = 3 and max( — 1, — 4) = max (—4, —1) = — 1. The minimum of x and y is denoted by min(#, y). Prove that , N min(*, y) = Derive a formula for max(x, max(#, y, z) x +y + \y — x\ x +y — \y — x\ y, z) and min(# y = max ? (at, } z), using, for max(jy, z)). example, b Basic Properties of Numbers 14. (a) Prove that many = \a\ (c) *15. (a) not to become confused by too is \ < > prove the statement for a cases. First obvious for a (b) —a\. (The trick 17 Why 0. is it then 0?) Prove that —b<a<b follows that — < \a\ if < a and only , if < \a\ In particular, b. it \a\. Use this fact to give a new proof that \a + b\ < \a\ + \b\, Use Problem 1 and Problem 7 to prove that if x and j are not both 0, then x1 x + xy 4 z + xy + y ^ 0, + x y + xy + y* 2 2 z 2 9* 0. For every number x 9^ 0, each of these expressions is positive for some number y and also for some negative y (namely, y — ±x); it therefore seems reasonable that the ^ signs can be replaced by > signs. This maneuver is valid, but we are not yet in a position to prove this (see Problem 7-9). Parts fb) and (d) of this problem provide a positive direct demonstration that the (b) Using the > signs hold. fact that + +y = (* + y) > 0, show that the assumption x + xy + y < 0 leads to a contradiction. x2 2 2xy 2 2 (c) Show similarly that 4x 3x 2 **(d) *16. Show that Show (a) (b) **(c) + + 6xy 5xy + + Ay 3y 2 x2 z z x +y +y 2 only z only when when (x x x when to find out b + yY = ^5 17. (a) _|_ \ now be the proof Suppose that -b if b able to is 0. = 0, = 0or# = 0 or y 0or;y (x + yY = 5 - Hint: x* —y. + y\ From the assumption be able to derive the equation 0. This implies that (x + yY = 2 + satisfy the make a good when (x + y) n = guess as to contained in Problem 11-41. — Ac V 2 2 both > 4 2 should +y z 2 — = +y h 2 z n then 0, 2 2 then 0. + x y + x y + xy* + > z 0, 0, *4 + yY = x + y you should xy x + 2x y + 2xy + y = 0, xy xy = + yy xn > > and y are not both (x You 2 x if Use Problem 15 Find out when 2 2 that + y) — = (x + y) (x x and y are not both if 2 2 > 0. - 4c Show that the numbers -b - Vb 2 ' equation x 2 2 + bx +c = 0. 4c Suppose that (b) satisfying x — 2 < Ac Show 0. +c= bx + then x 2 +y xy > 2 numbers x that there are no 2 0; in fact, x + bx + > + bx + c = 0 for all c "Complete the square," i.e., write x 2 Use this fact to give another proof that (c) if x and (x + Hint: x. £/2) + 2 ? are not both 0, jy 0. For which numbers a is it true that x 2 + axy -{- y 2 > 0 whenever x and y are not both 0? Find the smallest possible value of x 2 + bx + c and of ax 2 + bx + c, (d) (e) for a 18. b + 2 (Use the trick in part 0. > (b).) numbers a, elementary as it may seem, is nevertheless the fundamental idea upon which most important in- The fact that a 2 0 for all equalities are ultimately based. is The great-granddaddy of all inequalities the Schwartz inequality: + xiyi The 2 2 2 2 common —their reliance on the Prove the Schwartz inequality by (xi (b) + x V yi + y / 2 2 . three proofs of the Schwartz inequality outlined below have only one thing in (a) < V*! x2y 2 2 + x 2 2 )(yi 2 Prove that +^ ) = if x\ = 2 2 Xy Oiji + first 0 for all a. proving that + x 2y 2 ) 2 and x 2 = X^ 2 x > fact that a 2 for Oij 2 — xoji) 2 . some number X, then equality holds in the Schwartz inequality. Prove the same thing yi — = V2 there is suppose that yi and y 2 are not both no number X such that x\ = \y and x 2 = \y 2 < = (x>-i - \ 9 (yi 2 x x y + (Xv - x y ~ 2X(x iyi + +y* 2 0, if and that Then 2 2 x2y 2) ) + (x, 2 +x 2 2 ). Using Problem 17, complete the proof of the Schwartz inequality. Prove the Schwartz inequality by using 2xy < x 2 + y 2 (how is this derived?) with = x first (d) . x o (c) Now 0. for i — Xi V Xl 2 +X and then 1 = y 2 2 for yi and x 2 = = y = 2 Xy 2 0 or when ; Vy *+y x z — * t 2. Deduce, from each of these three when yi there is proofs, that equality holds only a number X such that Xi — Xyi . In our later work, three facts about inequalities will be crucial. Although proofs will be supplied at the appropriate point in the text, a personal assault on these problems is infinitely more enlightening than a perusal of a completely worked-out proof. The statements of these propositions involve some weird numbers, but their basic message is very simple: if x is close enough to x Q and y is close enough to y Q then x + y will be close to * 0 + ^o, and xy will be , , 19 Basic Properties of Numbers close to x 0 y 0 and , /y will be close to l/y 0 1 these propositions the is fifth letter The symbol "e" which appears . in and of the Greek alphabet ("epsilon"), could just as well be replaced by a less intimidating Roman letter; however, tradition has made the use of £ almost sacrosanct in the contexts to which these theorems apply. 19. Prove that if - \x < Xo\ - and -y < 0 \y -> \ then *20. Prove that \x - then +y) - (x 0 \(x - - (x 0 y) +y -y < < 0 )\ 0 )\ e, 6. if min *o|'< \xy \(x — * 0 .)'o| ^_i^y l) < and \y - ,„| < S. (The notation "min" was defined in Problem 13, but the formula provided by that problem is irrelevant at the moment; the first inequality in the hypothesis just - \x means that e x0 ' < —r~ \ and - \x - x0 \ < 1 , at one point in the argument you need the will first inequality, and at another point you will need the second. One more word of advice: since the hypotheses only provide information about x — x 0 and y — y 0 it is almost a foregone conclusion that the proof will depend upon writing , xy *21. — x 0yo in a Prove that if way 0 y0 0 —y x 0 and y 0 .) and !JV then y — that involves x — / — <mm^— z\yo\ f\yo\ • , , JVo| 2 — \ j> and 1 1 y yo | < e. I *22. i Replace the question marks in the following statement by expressions involving If yQ ^ 0 s, # 0 and y 0 so , 9^ the conclusion will be true: and \y then y that, 0 — Vol < ? and \x and X Xq \y y<> | < e. — xQ \ < ? This problem trivial in is the sense that its from Problems solution follows 20 and 21 with almost no work at all (notice that x/y = x \/y). The crucial point is not to become confused; decide which of the two problems should be used first, and don't panic if your answer looks • unlikely. This problem shows that the actual placement of parentheses in a is irrelevant. The proofs involve "mathematical induction"; sum you are problem, it can if not familiar with such proofs, but still want to tackle this be saved until after Chapter 2, where proofs by induction are explained. Let us agree, for definiteness, that a x a n will denote + + ai Thus ai denotes + (a + _ + + n + (a n _ + a n ))) a a denotes + + a + (a + a and a + a + + (a + (a + # etc. (a 2 * ' ' 2 z 2 2 3 ), a3 2 x +a 4 4 )), 3 Prove that (a) + (ai • ' + • + a k+i ak ) Hint: Use induction on Prove that (b) + (ai (c) ). x x 2 • • • • • if > n k) ai + • + * a k+ x k. (a k+1 + ' ' • + an ) = + ai . . , + * • Hint: Use part (a) to give a proof by induction on Let s(a h a k ) be some sum formed from ai, . . then fc 3 +a + = an . k. . . . ak , Show . that s(a h . . . , ak) + = Hint: There must be two sums s'(a h • . • . • + . , ak at ) . and s ff {a i+h . . . , ak ) such that s(a u . . . , ak ) = s'(a u . . . , ai) + s"(ai+ h . . . , a*). Suppose that we interpret "number" to mean either 0 or 1, and and to be the operations defined by the following two tables. • + 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 Check that properties P1-P9 all hold, even though 1 + 1=0. + CHAPTER *i NUMBERS OF VARIOUS SORTS In Chapter 1 we used word "number" very the with the basic properties of numbers. loosely, despite now It will our concern be necessary to distinguish carefully various kinds of numbers. The simplest numbers are the " counting numbers" ... 1, 2, 3, . significance of this collection of numbers is emphasized by symbol N (for natural numbers). A brief glance at PI -PI 2 will show that our basic properties of "numbers" do not apply to N for example, P2 and P3 do not make sense for N. From this point of view the system N has many deficiencies. Nevertheless, N is sufficiently important to deserve several comments before we consider larger collections of numbers. The most basic property of N is the principle of "mathematical induction." The fundamental its — Suppose P(x) means that the property P holds for the principle of mathematical induction states that P(x) number is Then x. true for the natural all numbers x provided that (1) P(l) (2) Whenever Note that condition assumption that P(k) if is (2) true. P(k) merely is condition (1) also holds. In number P(k) A is is P(h + asserts the truth of fact, if + P{k is true, then 1). Now, since P(2) P(\) = true (using (2) in the special case k will eventually true for all 1) is true. 1) under the true; this suffices to ensure the truth of P(x) for all x, is true (by using (2) in the special case k that P(3) true, = it follows that P(2) is true numbers is follows that each 2). It is clear be reached by a series of steps of this it sort, so that k. favorite illustration of the reasoning behind mathematical induction envisions an infinite line of people, person number 1, person number 2, person number 3, . . . . If each person has been instructed to tell any secret he hears to the person behind him (the one with the next largest number) and a secret is told to person number 1, then clearly every person will eventually learn the secret. If P(x) is the assertion that person number x will learn the secret, then the instructions given (to tell all secrets learned to the next person) assures that and telling the secret to person number 1 makes (1) true. The following example is a less facetious use of mathematical induction. There is a useful and striking formula which expresses the sum of the first n numbers in a simple way: condition (2) is true, f 21 22 Prologue 1+ To prove that for this formula, note some integer k • • first +„ • that + __- n(n = 1) clearly true for n it is = 1. Now assume we have 2 Then k(k = + !) + 1) +*+ +2£ + 1 2 2 *2 = + 3£ + 2 2 (A + + !)(* + 2) the principle of induction this proves This particular example illustrates a phenomenon that frequently occurs, especially in connection with formulas like the one just proved. Although the proof by induction is often quite straightforward, the method by which the formula was discovered remains a so the formula is the formula for also true for k all 1 By . natural numbers mystery. Problems 4 and 5 indicate n. how some formulas of this type may be derived. The principle of mathematical induction may be formulated in an equivway without speaking of "properties" of a number, a term which is sufficiently vague to be eschewed in a mathematical discussion. A more precise formulation states that if A is any collection (or "set"— a synonymous mathealent matical term) of natural numbers and (1) 1 is (2) k + in A, is 1 in A whenever h is in A, then A is the set of all natural numbers. It should be clear that this formulation adequately replaces the less formal one given previously—we just consider the the set A of natural numbers x which satisfy P(x). For example, suppose A is set of natural numbers n for which 1+ Our is • • • it is + „ true that _+ n(n = 1) + 1 previous proof of this formula showed that A contains 1 and that k that the if h is. It follows that A is the set of all natural numbers, i.e., , in A, formula holds for all natural numbers n. There is yet another rigorous formulation of the principle of mathematical induction, which looks quite different. If A is any collection of natural num- ! Numbers of Various bers, it is tempting to say that statement can A must have a smallest member. Actually, this to be true in a rather subtle fail 23 Sorts A way. particularly important numbers is the collection A that contains no natural numbers at all, the "empty collection" or "null set,"* denoted by 0. The null set 0 is a in fact, it has no collection of natural numbers that has no smallest member set of natural — members This at all. set of natural the only possible exception, however; A numbers, then known statement, is as the "well-ordering principle." principle of induction as follows. Suppose that the set Let B be the Clearly 1 numbers 1 were set of natural in is B (because such that n . not in A. This shows that every number n is in B, i.e., number n. Thus ^4 natural is a nonnull 1, can be proved from the A has no least member. . . . , . , k if = + in B, then k is the numbers 1, . . . + . as smallest 1 . , A not in is 1 . k + 1 are in B. It follows that 1 is , n are not in which completes the 0, A. n are all not in k are not in A, surely k member). Moreover, if 1, (otherwise k + 1 would be the smallest member of A), so 1, all A A would have in A, then if . if has a least member. This "intuitively obvious" A for any proof. prove the principle of induction from the well-ordering It is also possible to may principle (Problem 9). Either principle be considered as a basic assump- tion about the natural numbers. There is still another form of induction which should be mentioned. It sometimes happens that in order to prove P(k + 1) we must assume not only P(k), but also P(l) for all natural numbers A "principle of complete induction": If < / k. In this case we rely on the a set of natural numbers and is in A, (1) 1 is (2) k + 1 A in is if 1, . . . , k are in A, A is the set of all natural numbers. Although the principle of complete induction may appear much stronger than the ordinary principle of induction, it is actually a consequence of that principle. The proof of this fact is left to the reader, with a hint (Problem 10). Applications will be found in Problems 6, 16, 19, and 20. Closely related to proofs by induction are "recursive definitions." For then example, the number n natural numbers than or equal to less ! (read "w factorial") n\ = 1 • 2 • . is defined as the product of all the n: . • . {n - 1) - a. This can be expressed more precisely as follows: (1) 1! (2) n\ = = 1, n - (h - 1)!. This form of the definition exhibits the relationship between * Although it may many contexts. satisfying Isome property P; often A might be and (n — 1) not strike you as a collection, in the ordinary sense of the word, the null set arises quite naturally in that nl 0—in fact often We frequently consider the set A, consisting of that P is satisfied always false by showing that we have no guarantee one proves that P is by any all number, A — 0. x so : an in way explicit that is ideally suited for proofs by induction. Problem 21 reviews a definition already familiar to you, which may be expressed more succinctly as a recursive definition; as this problem shows, the recursive definition is really necessary for a rigorous proof of some of the basic properties of the definition. One tion definition which we which may not be familiar involves some convenient nota- will constantly be using. Instead of writing + a\ we employ the Greek will usually • • + an • letter 2 , (capital sigma, for "sum") and write 71 11 In other words, i /=1,2, . . . ^= n. , of the numbers obtained by letting Thus 1 Notice that the sum a { denotes the i letter i +2+ = +n • • n(n + 1) 2 really has nothing to do with the number denoted by n ^ i, and can be replaced by any convenient symbol (except n, of course!): i=i n(n 3 = = + V To define ^ 1) 1 L n= 1) 1 Hi 3 + n = Hi ~ +V 2 1 a precisely really requires a recursive definition { ^ (1) ai = ai = n n ^ (2) °1> —1 ^ ai + an . But only purveyors of mathematical austerity would such precision. In practice, used, it of this too strongly on symbolism are of modifications necessary to add any words of explanation. all sorts and no one ever considers insist Numbers of Various 25 Sorts The symbol t for example, is an obvious way of writing a\ more or + + a2 +a + az i may be set . + ' an, n 1 =5 i numbers which we discovered at the beginremedied by extending this system to the partially . -2, -1, , Z denoted by is PI -PI 2, only P7 A * integers . This = deficiencies of the natural ning of this chapter set of * precisely, 3 The + #6 b (from 0, 1, 2, ... . German "Zahl," number). Of properties Z. fails for numbers is obtained by taking quotients m/n of These numbers are called rational numbers, and the set of all rational numbers is denoted by (for "quotients''). In this system of numbers all of PI -PI 2 are true. It is tempting to conclude that the "properties of numbers," which we studied in some detail in Chapter 1, refer to just one set of numbers, namely, Q. There is, however, a still larger collection of numbers to which properties PI -PI 2 apply the set of all real numbers, denoted by R. The real numbers include not only the rational numbers, but other numbers as well (the irrational numbers) which can be represented still larger system of ^ integers (with n 0). Q — by infinite decimals; proof that tt is w and irrational V2 are both examples of irrational numbers. is not easy The Part III to a proof of this fact. and was known quite simple, irrationality of V2, on The Chapter 16 of shall devote all of the other hand, is Pythagorean theorem of length 1 has a hypotenuse to the Greeks. (Since the shows that an isosceles right triangle, of length it is , —we with sides , not surprising that the Greeks should have investigated this The proof depends on a few observations about the natural numEvery natural number n can be written either in the form 2k for some integer k, or else in the form 2k + 1 for some integer k (this "obvious" fact has a simple proof by induction (Problem 7)). Those natural numbers of the form 2k are called even; those of the form 2k + 1 are called odd. Note that even numbers have even squares, and odd numbers have odd squares: question.) bers. (2k) (2k In particular even; if n 2 is it + l) 2 2 = - 4k 2 = 2 4k 2 + 4k • (2k 2 ), + 1 follows that the converse odd, then n is odd. = 2 must The proof • (2k 2 + 2k) + also hold: if n 2 that is 1. is even, then n irrational is now is quite simple. Suppose that numbers p and We V 2 were rational; that suppose there were natural such that q ^ can assume that p and ? = have no 2. common divisors could be divided out to begin with). P This shows that p 2 for is, is 2 k. Now we common have = 2q\ even, and consequently some natural number divisor (since all p must be even; that is, p = 2k Then so 2k 2 = 2 q . and q 2 This shows that q is even, and consequently that q is even. Thus both/? divisor. This common no have and that fact the q contradicting p are even, contradiction completes the proof. important to understand precisely what this proof shows. We have 2 demonstrated that there is no rational number x such that x = 2. This asserNote, often expressed more briefly by saying that V2 is irrational. It is tion is number however, that the use of the symbol Vl implies the existence of some such a that proved not have We 2. is (necessarily irrational) whose square is proof a present, at that, confidently assert can we exists and number Any impossible for us. proof at this stage would have to be based on P1-P12 have mentioned); since PI -PI 2 are also true for there is a rational number whose Q, the exact same argument would show that (the only properties of R we (Note that the reverse argument will cannot not work our proof that there is no rational number whose square is 2 our because is square 2, be used to show that there is no real number whose every that fact the of property special Q, proof used not only PI -PI 2 but also a square 2, is and this we know is false. — can be written p/% for integers p and q.) could, This particular deficiency in our list of properties of the real numbers existence the asserts which property new adding a by corrected of course, be is, however, of square roots of positive numbers. Resorting to such a measure would still neither aesthetically pleasing nor mathematically satisfactory; we number in Q, know that every number has an nth root if n is odd, and that every positive not number has an nth root if n is even. Even if we assumed this, we could though (even =0 x 1 + prove the existence of a number x satisfying x + the solution there does happen to be one), since we do not know how to write solution the that known it is (in fact, roots of nth of the equation in terms wish to not do certainly we course, And, of form). this in cannot be written x number that all equations have solutions, since this is false (no real not b assume satisfies x 2 + 1 - 0, for example). In fact, this direction of investigation is R distinguishing not a fruitful one. The most useful hints about the property elucidating this of necessity from Q, the most compelling evidence for the — Numbers of Various 27 Sorts come from the study of numbers alone. In order to study the numbers in a more profound way, we must study more numbers. At this point we must begin with the foundations of property, do not properties of the real than the real fundamental concept on which calculus calculus, in particular the is based functions. PROBLEMS 1. Prove the following formulas by induction. (i) 1 (ii) l 2 + • • • 3 + • • • + n(n l)(2n+ 2 _ = n3 = (1 + 3 + 5 + + 32 + 5 +n 1) 6 + Find a formula 2 (i) + • + • • n)\ for ( 2i - *) (2i — I) = 1 + • 2n - V- + (2n - ( n (ii) £= i Hint: What do In and 3. If 0 < k) 2 1 2 l k 2 + < 22 + • * these expressions have to + 32 + + • • (2n) W W W= ( 2 * l) 2 . ) 1. do with 1 +2+3+ * • + • ? the "binomial coefficient" w, k\(n-k)\ (^) = (a) = 2 l ^ is defined by k\ (This becomes a special case of the define 0! = formula first if we 1.) Prove that ("rK:,K> (The proof does not require an induction argument.) "Pascal's triangle" following configuration, rise to the This relation gives —a number not on one of the the two numbers above number in the nth row. it; the binomial coefficient 1 1 13 4 6 1 1 1 2 1 5 10 sides 1 3 1 4 10 1 5 1 is (^j known sum the is as of the kth numbers (b) Notice that all the Use part bers. prove by induction that (a) to number. (Your entire proof by induction up in a glance by Pascal's triangle.) (c) num- in Pascal's triangle are natural Give another proof that always a natural a sense, be will, in summed number by showing a natural is is (^j that (j^j 1, (d) the is (^j . . . , of sets of exactly k integers each chosen number from n. Prove the "binomial theorem": and If a b any numbers and are rcis a natural number, then (a + b) n = a n + (j) Q a»~W +••+(„",) a^b + + *" "SO"a""*. (e) Prove that ®2C)-G) < ii 4. (a) + •••*(;)=»• »s<- K")=(o)-(0 i Prove by induction on 1 +r+ r 2 that + • + • • r = n —- — 1 (if r 1 Derive this 5. = n+1 r sum certainly presents no problem), + r n multiplying by setting S = 1 + r + equation by r, and solving the two equations for S. ifr (b) r 1 evaluating the 1, this result The formula for 2 + (k + l * • • , + • 2 may — 3k* n be derived as follows. We with the formula Writing + (n + (n this _ p = 33 - l) 3 3 - 3 formula for k 23 l) l) 23 n* 1 3 . 12 = 3-2 2 = 3'n = 3[1 2 2 — kz = 1, . . +3 + + 3-2 + + + . i 3-n • • • + + . + , n 3k + l. and adding, we obtain \ 1 1 n 2] + 3[1 + • • • + n] + n. begin Numbers of Various n Thus we can if we already know found in a similar way). Use l (i) (ii) l 3 + • • • 4 + • • • method this £ (which could have been to find 3 . + -!-+•••+ »(« + 1-22-3 1 3 UVJ £ +n + n\ -*- (hi) 29 '* k2 T find Sorts 2« 5 P-2 2 2 ' 2 1) -3 2 » 2 (rc + + l , l) 2 71 Use the method of Problem in the 5 to show Y P can always be written form ^— + An' + En*+ p (The that first 1 + Cn^ + 2 • • • . \ 10 such expressions are n V k = ih 2 + ^ +i/z 2 + W + in n y F = 3 k=l n \ k3 = in' 6 = i* + |« 2 n J= : /c 6 + *« + A« - 8 + 4 5 tVrc 2 1 n =1 : tF = |« fix T F= : V +V + ftll - A" + A" 4 4 +f 7 rc - A« 5 + l« 3 2 - A" —1 n y f= ^«>o + £n« + f 8 rc - -jV 6 + - A" 2 n Notice that the coefficients in the second column are always i, and that decrease after the third column the powers of n with nonzero coefficients 30 Prologue by 2 until n 2 or n is reached. The coefficients in all but the first two columns seem to be rather haphazard, but there actually is some sort of pattern; finding Problem 26-16 may it for the complete story.) 7. Prove that every natural number 8. Prove that A a set either even or odd. is numbers contains of natural w0 and contains + k 1 natural numbers > no. Prove the principle of mathematical induction from the well-ordering whenever 9. if See test. be regarded as a super-perspicacity it contains A then k, contains all principle. 10. Prove the principle of complete induction from the ordinary principle of 1 whenever it coninduction. Hint: If A contains 1 and A contains n + tains 1, . . . n, , consider the set B of all k such that 1, . . . , k are all in A. 11. (a) If a (b) If a and rational is What if is and a + a b is irrational, is b necessarily irrational? are both irrational? b and rational b irrational, is ab necessarily irrational? is (Careful!) there a number (c) Is (d) Are there two 2 a such that a is numbers whose sum irrational but a 4 is rational? and product are both irrational, rational? 12. (a) Prove that V3, V5, and Ve are irrational. example, use the fact that every integer or 3n (b) *13. 2. Why doesn't this proof and Prove that (b) V2 + V3 irrational. V6 - V2 - V3 irrational. (a) Prove that (b) treat V3, for for + 1 and m is a and b. V4? is is if x =p + (a) Prove also that (p Prove that if m and In) 2 1 (m + n) 2 (m + where p and V + = a m = ' q) a > 2; q are rational, for - are natural 72 £ numbers and Prove the same 2 (c) Prove that if results m/n < m'/n' with m/n < m V f V m 2 /n 2 < 2, then 2 ^ 2 with /n' a show, moreover, that 2 (b) some rational Vy. + 2n) _ 2<2 _m n (m + n) (m *16. To are irrational. natural number, then x m 15. work Hint: of the form 3n or 3n Prove that (a) 14. + is all inequality signs reversed. 2, then there < V is another rational number 2. It seems likely that n is irrational whenever the natural number n is not the square of another natural number. Although the method of Problem 1 2 may actually be used to treat any particular case, it is not Numbers of Various advance that clear in always work, and a proof for the general A natural number p is called a will it 31 Sorts case requires .ome extra information. prime number if it is impossible to write p = a and b unless one of these is p, and the other 1 ; ab for natural for numbers convenience we also few prime numbers are 2, = ab, with a and b 3, 5, 7, 11, 13, 17, 19. If n > 1 is not a prime, then n both < n; if either a or 6 is not a prime it can be factored similarly; continuing in this way proves that we can write n as a product of primes. agree that 1 = For example, 28 (a) 4 • = 7 2 2 • first 7. • argument into a rigorous proof by complete induction. (To be sure, any reasonable mathematician would accept the informal argument, but this is partly because it would be obvious Turn to A The a prime number. is not this him how to state rigorously.) it fundamental theorem about states that this factorization is which we integers, will not prove here, unique, except for the order of the factors. Thus, for example, 28 can never be written as a product of primes one is 3, nor can it be written in a way that involves 2 only once of which (now you should appreciate why (b) Using this fact, natural is not allowed as a prime). is irrational unless n = m is irrational unless n — V'n prove that number 1 2 some for m. (c) Prove more generally that (d) No numbers should discussion of prime fail nr. to allude to Euclid's many of them. Prove many prime numbers pi, p pz, beautiful proof that there are infinitely there cannot be only finitely p n by considering pi p 2 - *17. (a) Prove that if x for some integer. (b) first 18. 6 1 integers a n ^ u is this is • pn • • + • that . • , I. V 2 + ^2 powers of this . this trivial if h > The Fibonacci sequence . aQ , +a = Q then x , is 0, irrational unless x is an is irrational. Hint: Start by working out the number. + 0 = = 1, an = a n -i > —1, If h h) n > + 1 then nh. ? ai, a 2 ai ai . • a generalization of Problem 16?) Prove Bernoulli's inequality: Why . . + a n _ lX n - + (1 19. . , x satisfies n (Why Prove that ' 2 , a3 , . . . is defined as follows: 1, + for n > 3. was discovered by This sequence, which begins 1, 1. 2, 3, 5, 8, Fibonacci (circa 1175-1250), in connection with a problem about . . . , 32 Prologue assumed that an initial pair of rabbits gave birth to one new pair of rabbits per month, and that after two months each new pair behaved similarly. The number a n of pairs born in the nth month is # n __i + a M _2, because a pair of rabbits is born for each pair born the previous month, and moreover each pair born two months ago now gives birth to another pair. The number of interesting results about this sequence is truly amazing there is even a Fibonacci Association which rabbits. Fibonacci — publishes a journal, The Fibonacci Quarterly. Prove that V5 One way of deriving this astonishing formula is presented in Problem 23-8. 20. ' The an result in > 0, Problem 1-6 has an important generalization: If ai, . . . , then vai ' . . + ai < an • . . . . + qn n (a) Why aj. happens to ai and If a» aj are = a n ? Suppose not all a { are equal, a^)/2 what both replaced by (a { + mean" A n — (a\ + "geometric mean Gn = Vai the "arithmetic What happens Why = this true if a x is ^ say to the • + * • a n )/ n ? an ? 5 5 ' . . ' . does repeating this process enough times eventually prove G n < A n ? (This is another place where it is a good exercise to provide a formal proof by induction, as well as informal reason.) that The reasoning in the previous proof is closely related to another interest- ing proof. (b) Using the k, (c) that fact that Gn < A n For a general G n < A n when n — 2, prove, by induction on = 2k m m 2 > n. Apply part (b) to the 2 numbers for n n, let a\, . . . . , an , An . , . An , 2 m — n times to 21. The prove that following is Gn < A n . a recursive definition of a n a1 = a, Prove, by induction, that a n+m (a n = a m = ) a n • a m , : Numbers of Various Sorts 33 (Don't try to be fancy: use either induction on n or induction on m, not 22. both at once.) Suppose we know properties PI and P4 for the natural numbers, but that multiplication has never' been mentioned. Then the following can be used as a recursive definition of multiplication: + (a 23. In ' (b a ' 1 a - b + = = b a ' + b = = b9 a- b + b. order suggested!): a c • on (use induction a), a, b ' chapter this — c) -b (in the Prove the following a 1 1) a (you just finished proving the case b we began with the natural = 1). numbers and gradually built A completely rigorous discussion of this process requires a little book in itself (see Part V). No one has ever figured out how to get to the real numbers without going through this process, but up if to the real we do be numbers. accept the real numbers as given, then the natural numbers can 1 1,1 1, etc. The the real numbers of the form 1,1 whole point of matical (a) A way set A + + defined as problem this is to show that there is + a rigorous mathe- of saying "etc." numbers of real (1) 1 is (2) k + is called inductive if in A, is 1 in A whenever k in A. is Prove that R (i) (ii) (iii) (iv) (v) (b) A is real (i) set of positive real set of positive real number inductive (ii) inductive. numbers is inductive. numbers unequal to \ is inductive. set of positive real numbers unequal to 5 is not inductive. If A and B are inductive, then the set C of real numbers which are in both A and B is also inductive. The The The n will be called a natural number if n is in every set. Prove that 1 Prove that k is + a natural number. 1 is a natural number if £ is a natural number. PART FOUNDATIONS The statement is so frequently made that the differential calculus deals with continuous magnitude, and yet an explanation of this continuity nowhere given ; is even the most rigorous expositions of the differential calculus do not base their proofs upon continuity but, with more or less consciousness of the fact, they either appeal to geometric notions or those suggested by geometry, or depend upon theorems which are never established in a purely arithmetic manner. Among these, for example, belongs the above-mentioned theorem, and a more careful investigation me convinced that this theorem, or any one equivalent in some way as a to it, can be regarded sufficient basis for infinitesimal analysis. It then only remained origin in the elements to discover its true- of arithmetic and thus at the to secure a real definition of same time the essence of continuity. I succeeded Nov. 24, 1858, and a few days afterward I communicated the results of my meditations to my dear friend Durege with whom I had a long and lively discussion. RICHARD DEDEKIND > FUNCTIONS CHAPTER that of a Undoubtedly the most important concept in all of mathematics is turn out functions mathematics modern of branch function—in almost every not probably therefore will It investigation. of the central objects to be surprise of great generality. to learn that the concept of a function is one we will be able to will be a relief to learn that, for the present, you Perhaps it restrict our attention even this small class attention for quite our of functions will exhibit sufficient variety to engage For the moment definition. proper some time. We will not even begin with a to functions of a very special kind; and will at length, a provisional definition will enable us to discuss functions mathematicians. by understood as functions, illustrate the intuitive notion of modern mathethe of advantages the discuss Later, we will consider and matical definition. Let us therefore begin with the following: PROVISIONAL DEFINITION A function is a rule which real assigns, to each of certain real numbers, some other number. and amplify this following examples of functions are meant to illustrate clarification. such some definition, which, admittedly, requires The Example 1 The which rule assigns to each number the square of that number. Example 2 The rule which number y the number <? assigns to each y* + 3y + r+ number 5 1 Example 3 The rule which assigns to each + 3c + 2 - 1 c -1 the number x satisfying -17 < ^ 1, 5 The rule which assigns to each number 2 x < 7r/3 the number x Example 5 The rule which assigns to each number a irrational, and the number 1 if a is rational. Example 6 The rule which assigns Example 4 . to 2 the number to 17 the 5, number — 7T 37 the number 0 it a is — the number 28, to 17 36 to — the number 28, , 7T and V2 b any y ^2, to for a, b in 17, 7r 2 /17, or 36/tt, the number 16 if is of the form a + Q number / 3 + x. we are really describing infinitely many different functions, one for each number x.) Example 8 The rule which assigns to each number z the number of 7's in the decimal expansion of z, if this number is finite, and — tt if there are Example 7 The rule which assigns to each (This rule depends, of course, many infinitely One 7's in number 2 on what the number x the decimal expansion of the is, so z. thing should be abundantly clear from these examples —a function is numbers to certain other numbers, not just a rule which can be expressed by an algebraic formula, or even by one uniform condition which applies to every number; nor is it necessarily a rule which you, or anybody else, can actually apply in practice (no one knows, for example, what any rule that assigns tt). Moreover, the rule may neglect some numbers and it not even be clear to which numbers the function applies (try to determine, for example, whether the function in Example 6 applies to tt). The set of rule 8 associates to may numbers to which a function does apply is called the domain of the function. Before saying anything else about functions we badly need some notation. Since throughout this book we shall frequently be talking about functions (indeed we shall hardly ever talk about anything else) we need a convenient way of naming functions, and of referring to functions in general. The standard denote a function by a letter. For obvious reasons the letter "/" a favorite, thereby making "g" and other obvious candidates, but any practice is to is any reasonable symbol, for that matter) will do, not excluding and "y," although these letters are usually reserved for indicating numbers. letter (or If/ by is a function, then the /W—this symbol Naturally, is number which / associates read "/ of x" and is to a number x often called the value "/," leading to the domain and we can symbol *(/)). Note that the symbol /(*) makes sense only of /; for other x the symbol f(x) is not defined. If the functions defined in Examples 1-8 are denoted by /, g h, 9 then f(x) = x2 2 h(c) = rewrite their definitions as follows: for all x. y (3) at x. we denote a function by x, some other letter must be chosen to number (a perfectly legitimate, though perverse, choice would be for x in the (1) denoted off if denote the y, is 2 c' + 1 — T 1 for aI1 ' * 1, -1. r, s, 0, ax , Functions r(x) (4) s{x) (5) = x2 = for all x — 17 < such that 0, x irrational 1, * rational. x < 39 tt/3. 7 x = 2 ' x = 17 17 ' * 17 x — 5, 36 — 7T 7T flW (6) 28 = < '28, 36 = 7T x 16, (7) ax (t) = (8) y(x) _ - t | z + x ^ 17, 2, —j or 17 for all — and * , = + a for a, b in Q. 7T numbers t. exactly «7's appear in the decimal expansion of x n, infinitely I many 7 's appear in the decimal expansion of x. These definitions illustrate the common procedure adopted for defining a function /—indicating what f(x) is for every number x in the domain of /. (Notice that this is exactly the same as indicating f(a) for every number a, or abbreviations are tolerated. f{b) for every number b, etc.) In practice, certain Definition (1) could be written simply f(x) (1) the qualifying phrase "for (4) all It is r( x ) Of x" being understood. the only possible abbreviation (4) = x\ = course, for definition is -17 < x\ x < tt/3. usually understood that a definition such as kix) = - + X x x can be shortened ——— . x * 0, 1 1 to *W = - + -^-7; — 1 a: in other words, unless the domain consist of all numbers for which is explicitly restricted the definition makes any further, it is understood to sense at all. should have little difficulty checking the following assertions about the functions defined above: You /(*+ 1) g(x) r(x + 1) +2x+ 1; x + 3x + 5 = h(x) if -17 r{x) + 2x + =/(*) = = if 3 1 0; < x <| - 1; = + y) s(x s(x) if y is rational; a z (x) = x [f(x) If the expression f(s(a)) looks that s(a) is number a like a matter of fact, f(s(a)) = + 1]; unreasonable to you, then you are forgetting any other number, s(a) for all a. sions than f(s(a)) are, after a so that f(s(a)) more exposure, no first makes sense. As complicated expres- Why? Even more difficult to unravel. The expression formidable as it appears, may be evaluated quite easily with a little patience: f(r(s(0(az(y(i)))))) = = = = f(r(s(6(aM)))) /(rW9(3)))) /(rW16))) /(r(l)) -/(I) = The first 1. few problems at the end of this chapter give further practice manipu- lating this symbolism. The function defined in (1) a rather special example of an extremely function / is a is A important class of functions, the polynomial functions. polynomial function = /(*) a nx n + if there are real a n „ix n~ l + • • • numbers + a 2x 2 a0, + . aiX . + . , a0 , a n such that for all x 0). usually tacitly assumed that a n of/; degree the called The highest power of x with a nonzero coefficient is 6 137* 4 — tt for example, the polynomial function / defined by /(*) = 5x (when is f{x) written in this form it is + has degree 6. functions defined in (2) and (3) belong to a somewhat larger class of functions, the rational functions; these are the functions of the form p/q where p and q are polynomial functions (and q is not the function which is The rational functions are themselves quite special examples of an even larger class of functions, very thoroughly studied in calculus, which are simpler than many of the functions first mentioned in this chapter. The always 0). The following are examples of this kind of function: ro\ (9) f( \ fix) x = ± x2 ; (10) (11) fix) fix) + * sin2 —* + x sm x ; x sin x = = s'mix z 2 ). sin(sin(^ 2 )). 41 Functions „, v , (12) = s /(*) n . «^ sin 2 (sin(sin 2 (* sin 2 * 2 ))) ( \ By what built you may criterion, a monstrosity —+ + fx . sin • sin(# sin #)\ up from a few simple )• sin x / impelled to ask, can such functions, especially feel The answer is that they can be means of combining (9)— (12) we need to start with be considered simple? like (12), - x functions using a few simple functions. In order to construct the functions = and the "sine function" sin, x. The following are some is whose of the important ways in which functions may be combined to produce new the "identity function" which for 7, I(x) x, often written simply sin value sin(#) at x functions. If / the and g are any two functions, we can define a new function / of / and g, by the equation + g, called sum (f + g)(x) =f(x)+g(x). Note that according to the conventions we have adopted, the domain of / + g makes sense, i.e., the set of all x in both consists of all x for which "/(*) + domain /and domain g. If A and B are any two sets, then A C\ B (read "A intersect B" or "the intersection of A and Z?") denotes the set of x in both A and B; this notation allows us to write domain (/ + g) = domain / Pi domain g. we In a similar vein, define the product / • g and the quotient - (or f/g) g of / and g by (/•*)(*)=/(*)•*(*) and Moreover, if g is a function and c is (c a number, g)(x) = c ' we define a new function c g by • g(x). This becomes a special case of the notation / g if we agree that the symbol c should also represent the function / defined by fix) — c; such a function, which has the same value for all numbers x, is called a constant function. • The domain of / g is domain / C\ domain g, and the domain of c g is simply the domain of g. On the other hand, the domain of f/g is rather com* * plicated symbol — general, {x: xz may it \x: g(x) + {x: 3 . {x: x* + 3 < 1 1 } = \x: have been written using y guess that {x than denotes the set of denotes the set of 11} this notation are less domain g C\ {x: g(x) ^0}, the be written domain / 0} denoting the set of numbers x such that g(x) ^ 0. In .} . < consequently just as well ^ < x such that ". numbers . ." is true. 3 x such that x < 8, Thus and Either of these symbols could everywhere instead of x. Variations of x common, but hardly 2} . require any discussion. Any one can 8} denotes the set of positive numbers whose cube is could be expressed more formally as {x: x > 0 and x* < 8}. > 8; it all all 0: *3 < Incidentally, this set is slightly less transparent, equal to the contains just the four numbers x {x: = = or x 1 3 or ^ set < 0 {x: but very standard. The = 2, 1, = 2 or ^ and 3, < x 2}. One variation set {1, 3, 2, 4}, for 4; it is example, can also be denoted by 4). Certain facts about the sum, product, and quotient of functions are obvious facts about sums, products, and quotients of numbers. For consequences of example, very easy to prove that it is (f The proof is ( + k). g =f which demonstrates that two —the two functions must be shown to have the same domain, and the same value + g) + h ) characteristic of almost every proof functions are equal (f + g + h = /+ any number at + (g + domain. For example, in the to prove that note that unraveling the definition of the h), two sides gives Kf = = + g)+h](x) + g)(x) + h(x) + *(*)] + *W (f [/(*) and U+(g + /(*) + + h(x) and =/(*) h)](x) = and the equality of + g(x)] + [/(*) + (g k)(x) UW + Kx)l f(x) + [g(x) + h(x)] is a fact about numbers. In this proof the equality of the two domains was not explicitly mentioned because this is obvious, as soon as we begin to write down these domain of (/ + g) + h and of / + (g + h) is clearly domain / H domain g C\ domain h. We naturally write / + g + h for (/ + g) + h = f + (g + h), precisely as we did for numbers. h = f (g h), and this function is It is just as easy to prove that (/ g) denoted by / g h. The equations / + g = g + / and f'g=g '/should also present no difficulty. in Using the operations +, / we can now express the function / defined (9) by equations; the • • • * • • •, _ + I / It • I We require yet another way If / and g are any two position of / and g, by + I sin sin + / sin sin I sin • ' • • express function (10) this way. of combining functions. This combination, the should be clear, however, that composition of two functions, • is we cannot by functions, far the we (/•*)(*) most important. define a new function / ° g, the com- =/(*«); domain of fog is {x: x is in domain g and g(x) is in domain /}. The symbol u/o g » is often read "/ circle g" Compared to the phrase "the comthere is position of / and g" this has the advantage of brevity, of course, but the another advantage of far greater import: there is much less chance of confusing Functions fog and 43 must not be confused, since they are not usually / and g chosen at random will illustrate this point f = I 1 and g = sin, for example). Lest you become too apprehensive with g o /, these equal; in fact, almost any (try - about the operation of composition, is let us hasten to point out that composition associative: (fog) (and the proof is a triviality) ; o = fo h (g oh) this function is denoted by / ° g We can now ° h. write the functions (10), (11), (12) as (10) (11) (12) / = sine (/•/), / - sin o sine (/•/), / = (sin sin) o sin o • One fact (sin • sin) ° (/ [(sin sin) - o (/ • • /)]) has probably already become clear. Although this method of writing it is hardly short or con- functions reveals their "structure" very clearly, 2 / such that f(x) = sin(x ) for all 2 x unfortunately seems to be "the function / such that f(x) = sin(x ) for all x." The need for abbreviating this clumsy description has been clear for two hundred years, but no reasonable abbreviation has received universal acclaim. At present the strongest contender for this honor is something like venient. The shortest name for the function (read "x goes to sin(# 2 )" or just among x — u x arrow sin(* 2 )"), but sin(* 2 ) writers of calculus textbooks. In this amount popular of is book we and speak of "the function ellipsis, f(x) it is = sin(x 2 )." Even more 2 the quite drastic abbreviation: "the function sin(* )." For the sake speaking, con- we will never use this description, which, strictly number and a function, but it is so convenient that you of precision fuses a hardly popular will tolerate a certain end up adopting it for personal use. As with any convention, will probably utility is the reasonable so long as the slight logical deficiencies cause no confusion. On occasion, confusion will arise unless a 3" is an * more precise description is used. For example, "the function x motivating factor, and this criterion is + ambiguous phrase; x —> x + z t it could i.e., } mean either the function / such that f(x) = x + / = x + t 3 for all x z for all t. or t—>x + t 3 , i.e., the function / such that /(/) many important concepts associated with functions, calculus has a notation which contains the "x —>" built in. By now we have made a sufficiently extensive investigation of functions to warrant reconsidering our definition. We have defined a function as a "rule," but it is hardly clear what this means. If we ask "What happens if you break As we shall see, this rule?" it is however, for not easy to say whether this question is merely facetious or A more substantial objection actually profound. is to the use of the word "rule" that f(x) = x* f(x) = x2 and are certainly for different rules, if determining fix) ; + - x2 fix) = x2 + 3 - 3(* we mean we want by a rule nevertheless, fix) 3x + 1) the actual instructions given and + 3* + 3 - 30 + 1) same function. For this reason, a function is sometimes defined an "association" between numbers; unfortunately the word "association" to define the as escapes the objections raised against "rule" only because it is even more vague. There is, of course, a satisfactory way of defining functions, or we should never have gone to the trouble of criticizing our original definition. But a satisfactory definition can never be constructed by finding synonyms for The definition which mathematicians English words which are troublesome. have finally accepted for "function" is a beautiful example of the means by which intuitive ideas have been incorporated into rigorous mathematics. The correct question to ask about a function is not "What is a rule?" or "What is an association?" but "What does one have to know about a function in order to know all about it?" The answer to the last question is easy for each number x one needs to know the number fix) we can imagine a table which — ; would display /(*)=*»: all the information one could X about the function fix) 1 1 -1 1 2 4 -2 4 V2 V2 It is desire 2 2 7T TT — TT TT 2 2 not even necessary to arrange the numbers in a table (which would if we wanted to list all of them). Instead of a two actually be impossible column array we can consider various (1, 1), (-1, 1), (2, 4), pairs of (-2, numbers 2 4), (t, tt ), (V2, 2), : Functions 45 set.* To find /(I) we simply take the second number of the pair whose first member is 1 to find f(w) we take the second number of the pair whose first member is 7r. We seem to be saying that a func- simply collected together into a ; tion be defined as a collection of pairs of numbers. For example, given the following collection (which contains just 5 pairs) might as well we were if / = then /(l) = {(1,7), (3,7), (5,3), (4,8), (8,4)}, = 7,/(3) = 7,/(5) /= {(1,7), If 8,/(8) = 4 and 1, 3, consider the collection = 3,/(4) only numbers in the domain of/. we (3, 7), (2, 5), 4, 5, 8 are the (1,8), (8, 4)}, decide whether 5, /(8) = 4; but it is impossible to to be any defined be cannot function words, a other In 8. 7 or/(l) /(l) old collection of pairs of numbers; we must rule out the possibility which then /(3) = 7, f{2) arose in this case. DEFINITION A = = = We are therefore led to the following definition. a collection of pairs of numbers with the following property: words, the if (a, b) and {a, c) are both in the collection, then b = c; in other element. first same the with pairs different two contain not collection must function is This is our first full-fledged definition, and illustrates the format we shall always use to define significant new concepts. These definitions are so important (at least as important as theorems) that it is essential to know when one motivating is actually at hand, and to distinguish them from comments, remarks, and casual explanations. They will be preceded by the word DEFINITION, contain the term being defined in boldface letters, and constitute a paragraph unto themselves. There is one more definition (actually defining two things at once) which can now be made rigorously: If / DEFINITION is a function, the domain of / such that (a, b) is in /. If a of a function that there This unique b is is, is is in the the set of domain in fact, a unique all of/, it a for which there is some b from the definition follows number b such that (a, b) is in/. denoted by /(a). we have reached our goal: the important thing about its a function / is that a number fix) is determined for each number x in intuitive an where point the reached also domain. You may feel that we have definition has been replaced by an abstraction with which the mind can With this definition hardly grapple. * The Two consolations may be offered. First, pairs occurring here are often called "ordered pairs," to (2, 4) is not the same pair as in terms of ordered pairs, and an appendix to this (4, 2). It is only fair to warn that although a function emphasize we are going that, for example, to define functions another undefined term. Ordered pairs can be defined, however, chapter has been provided for skeptics. has been defined as a collection of pairs, there is nothing to stop you from thinking of a function as a rule. Second, neither the intuitive nor the formal definition indicates the best way of thinking about functions. The best way to is draw pictures; but this requires a chapter all by itself. PROBLEMS 1. Let/M = is (0 f(f( x )) (f° r which x does this m /©. (hi) f(cx). (iv) f(x+y). (v) /M+/G0. (vi) (vii) = Let g(x) x 2 and let , __ x\ 0, | (1, (ii) For which For which (iii) What (v) 3. _y h(y) jy is w s g(w) is x rational a irrational. y? g(y)? < w? = ^(s)? g(g(e)) is Find the domain of the functions defined by the following formulas. f{x) = (ii) /(*) = (iii) fix) = (i) V - x\ - Vi - ] (iv) /(*) (v) /(*) Let x\ -J- + -L-. — * — 2 x 4. < < is A(jy) is For which For which (iv) sense?). x? fa( (i) make For which numbers c is there a number x such that /(ex) = f(x). Hint: There are a lot more than you might think at first glance. For which numbers c is it true that f(cx) = f(x) for two different numbers 2. +*) What 1/(1 1 = Vj - Vl - x = P(x) 2 a: , let *2 + Vx - 1. + V* -~2. 2 = 2*, and let s(x) = sin a. Find each of the following. In each case your answer should be a number. (J"*) GO. (i) (ii) ,)(,). (5.P.j)(i) (iii) + (jo *)(/). (iv) 5. Express each of the following functions in terms of + - , , and o (for example, the answer to (i) is Po s). P, s, using only In each case your S, Functions 47 answer should be a function. (i) f(x) (ii) f(x) (iii) /(*) (iv) /(*) = = = = (sin x) ( v) f(t) 2 ain *. sin 2*. sin x 2 . sin 2 x (remember that sin 2 x is an abbreviation for 2 ). = 2 2 \ (Note: a bc m this convention means a more simply as a hc .) always b c (vi) because (a ) can be written 2- 2 ). f{u) = sin (2* (vii) f(y) ; is adopted + (viii) 22iin, = = sin(sin(sin(2 2 + sin2a ')))• + sin (a 2 ) 2 sin(a2+sina) . flexible, occupy a favored most investigations of functions. The following two problems illustrate their flexibility, and guide you through a derivation of their most important Polynomial functions, because they are simple, yet role in elementary properties. 6. (a) If x h fi . . . , product of is x n are distinct numbers, find a polynomial function of degree n — which 1 - (x all is and 0 at 1 ^ for ; x<) i, is at Xj for; 0 at xj if ; ^ i. Hint: The (This product usually denoted by n n ; (* - *i)> =1 the symbol II (capital pi) playing the 2 (b) find a where polynomial function / of degree n — 1 such that /(*»•) = a n are given numbers. (You should use the ah . . . , (a). The formula you will obtain is called the "Lagrange interpolation formula.' ) Prove that for any polynomial function /, and any number a, there = is a polynomial function g, and a number b> such that /(*) into (x a) (x fl^W + b for all *. (The idea is simply to divide example, /(*) by long division, until a constant remainder is left. For functions /< from part 5 7. (a) role for products that plays for sums.) Now Gi, same the calculation x2 x - l)x +x x3 -2 -3x 3 + 1 —x 2 x2 —3x x2 —x ^2x -2x + +2 1 -1 48 Foundations - shows that x s proof is + 3* ( +x- l)(x 2 by induction on the degree possible 2) - 1. A formal of/.) (c) (d) Show n roots, i.e., there are at most that for each n there n roots. If h is and if n is odd For which numbers a, b y If A is any = 0. one with only one root. find c, and d will the function cx satisfy /(/(*)) a with f{a) is ax (a) numbers n a polynomial function of degree n with even find a polynomial function of degree n with no roots, 9. - x Prove that if f(a) = 0, then /(*) = (* - a)g(x) for some polynomial function g. (The converse is obvious.) Prove that if/ is a polynomial function of degree w, then /has at most (b) 8. = 1 = * for set of real all x +d +b ? numbers, define a function CA (x) = 1 as follows: A * in , CA # not in A. 0, Find expressions for CA n B and C^ wB and Cr_ a in terms of CA and CB (The symbol A C\ B was defined in this chapter, but the other two may be new to you. They can be defined as follows: , . A VJ B — R —A = (b) {x: x is in A or x {x: x is in R but * Suppose /is a function such that f(x) there is a set A such that / = CA in B}, is = not in A}.) is 0 or 1 for each x. Prove that . (c) 10. (a) (b) *(c) Show that / = / if and only if / = CA for some set A. For which functions / is there a function g such that / = g 2 ? Hint: You can certainly answer this question if "function" is replaced by "number. 55 For which functions / is there a function g such that / = 1 Ig ? For which functions b and c can we find a function x such that 2 OW) + 2 for all *(d) numbers t b(t)x(t) + c(t) = 0 ? What conditions must the functions a a function x such that a{t)x(t) +b{t) = and b satisfy if there is to be 0 numbers ^ ? How many such functions x will there be? Suppose that His a function and y is a number such that H(H(y)) for all 11. (a) y. What is H(H(H(- (H(y) 80 times • • •) ? = 49 Functions (b) (c) Same Same replaced by 81. question if 80 question if H(H{y)) is = H(y). H such that H(H(x)) = H(x) for all numbers x, *(d) Find a function and such that H{\) = 36, H(2) = tt/3, 7/(13) = 47, //(36) = 36, //(tt/3) are - = tt/3, 7/(47) many 47. (Don't try to "solve" for H(x); there = functions //with H{H(x)) The H(x). extra conditions H are supposed to suggest a way of finding a suitable H.) on Find a function //such that H(H(x)) = //(#) for all #, and such that H{\) = 7, //(17) = 18. A function / is even if f(x) = f(—x) and odd if f(x) = —/(—*). For example, / is even if f(x) = x 2 or /(*) = \x\ or /(*) = cos x, while / is *(e) 12. = = odd if f(x) (a) Determine whether / x or /(#) sin +g is even, odd, or not necessarily either, in and g even or odd. (Your answers can most conveniently be displayed in a 2 X 2 the four cases obtained by choosing /even or odd, table.) (c) Do Do (d) Prove that every even function / can be written f(x) (b) same the same the infinitely *13. (a) / for f many • g. ° g. E is Prove that first, even and this — g{\x\), f° r functions g. R can be written/ = E Prove that any function /with domain where (b) for way by "solving" 0 is odd. of writing / for E 4- 0, is unique. and 0 you will (If you do pan; try to (b) probably find the solution to part (a).) 14. any function, define a new function |/| by |/|(*) = /(*)!• ^ /and g are functions, define two new functions, max(/, g) and min(/, g), by If /is | max (/,£)(*) = max (/(*), g(x)), min (/,$)(*) - min(/(*), £(*)). Find an expression max(/, g) and min(/, g) in terms of 0) + min(/, 0). This particular way of useful; the functions max(/, 0) and min(/, 0) are for |. | 15. (a) Show / = max(/, that writing / fairly is and negative parts of /. nonnegative if f(x) > 0 for all x. Prove that any function / can be written / = g — h, where g and h are nonnegative, in infinitely many ways. (The "standard way" is g — max(/, 0) and h = — min(/, 0).) Hint: Any number can certainly be written as the difference of two nonnegative numbers in infinitely called the positive (b) A function many *16. Suppose / (a) / is called ways. satisfies f(x Prove that/fo + y) - f(x) + f(y) and for all x /(*,) + • • • y. +/(*«). 50 Foundations (b) point we're not trying to say anything about fix) Hint: First figure out what c must be. Now prove = that f{x) all If fix) and satisfies fix) (a) (b) = rational = 0 for these such that fix) when x is a natural number, then when x is an when x is the reciprocal of an integer and, finally, for f(x y) = fix) fiy) for all x and suppose that / fiy) for all x and y. properties, but that fix) is not always 0. Prove that all x, • y) two then / ~ fix) + satisfies + Now • all x, as follows: Prove that/(l) Prove that fix) = = > 1. x x if is rational. (c) Prove that fix) 0 if x > 0. (This part is tricky, but if you have been paying attention to the philosophical remarks accompanying (d) Prove that fix) Prove that fix) the problems in the last two chapters, you will (e) > fiy) if x > y. = x for all x. Hint: Use two numbers there *18. is know what to do.) the fact that between any a rational number. what conditions must /, g, h, and k satisfy in order that fix)giy) and y? Prove that there do not exist functions / and g with either of the Precisely = *19. cx for all rational x. also fix x for c cx, first integer, then y, — is (at this for irrational x). *17. some number Prove that there numbers x hix)kiy) for all x (a) following properties: (i) fix) (ii) fix + giy) = xy for all x and y. + y) = gix) — y for all x and y. Hint: Try to get some information about / or g by choosing particular values of x and y, (b) Find functions / and g such that fix y) = gixy) for all x and y. + *20. (a) Find a function /, other than a constant function, such that fiy) /Ml < \y-*V (b) 21. Suppose that fiy) - fix) < iy - x) 2 for all x and y. (Why does this imply that \fiy) - f{x)\ < iy - x) 2 ?) Prove that / is a constant function. Hint: Divide the interval [x,y] into n equal pieces. Prove or give a counterexample (a) (b) — \ + h) =fo g +fo h + h)of = gof+ k of. foig ig . for each of the following assertions: Functions 22. (a) Suppose g = h Prove that ° /. if = /(#) /(j), then £(*) = 51 £(>»). (b) Conversely, suppose that / and g are two functions such that g{x) — g(y) whenever f(x) = f(y). Prove that g = h ° / for some function k. definition will not 23. Suppose that f ° g (a) if x ^ y, (b) every *24. *25. *26. 27. when z is of the form z = f{x) (these and use the hypotheses to show that your Hint: Just try to define h(z) are the only z that matter) = I, run into trouble. where I(x) = x. Prove that then g{x) ^ g(y); b can be written number b = f(a) for some number a. Suppose g is a function with the property that g(x) 5* g(y) if x 9^ y. Prove that there is a function / such that fog = I. (b) Suppose that / is a function such that every number b can be written b — f(a) for some number a. Prove that there is a function (a) g such that fog = I. Find a function / such that g °f = / for some g, but such that there is no function h with / ° h = /. Suppose f ° g = I and h°f = I. Prove that g = h. Hint: Use the fact that composition is associative. (a) Suppose f(x) = x 1. Are there any functions g such that f ° g — + (b) Suppose / is a constant function. For which functions g does f ° g — g°f? (c) Suppose that / o g = identity function, f(x) 28. (a) Let F be the + o = / for all functions g. * Show that / is the x. set of all functions and except P7 hold using g whose domain is R. Prove that, as defined in this chapter, all of properties for F, provided 0 and 1 P1-P9 are interpreted as constant functions, (b) *(c) (d) Show Show P7 does not hold. P10-P12 cannot hold. In other words, show that there is no collection P of functions in F, such that P10-P12 hold for P. (It is sufficient, and will simplify things, to consider only functions which are 0 except at two points x 0 and xi.) that that Suppose we define / < g to mean that f(x) P'lO-P'H (in Problem 1-8) now hold? of (e) If / < g, is h of < h o g ? Is / o h <g o h ? < g(x) for all x. Which 52 Foundations APPENDIX. ORDERED PAIRS Not only well, it is in the definition of a function, but in other parts of the book as necessary to use the notion of an ordered pair of objects. A definition we have never even stated explicitly what properties supposed to have. The one property which we will require formally that the ordered pair (a, b) should be determined by a and b, has not yet been given, and an ordered pair states and the order is which they are given: in if (a, b) — then a d), (c, — c and — b d. Ordered pairs may be treated most conveniently by simply introducing as an undefined term and adopting the basic property as an axiom since this property is the only significant fact about ordered pairs, there is not much point worrying about what an ordered pair "really" is. Those who find this treatment satisfactory need read no further. The rest of this short appendix is for the benefit of those readers who will {a, b) * uncomfortable unless ordered pairs are somehow defined so that this becomes a theorem. There is no point in restricting our attention to ordered pairs of numbers; it is just as reasonable, and just as important, feel basic property have available the notion of an ordered pair of any two mathematical This means that our definition ought to involve only concepts common to all branches of mathematics. The one common concept which pervades all areas of mathematics is that of a set, and ordered pairs (like everything else in mathematics) can be defined in this context; an ordered pair will turn to objects. out to be a The set of a rather special sort. containing the two elements a and b, is an obvious first choice, but will not do as a definition for (a, b), because there is no way of {a, b}, set determining from which of a or {a, b\ more promising candidate is b is meant the rather startling be the to first element. A set: {{a},{a,b}}. This set set {a}, as it has two members, both of which are themselves containing the single may we seem, for this choice are going to define and PROOF <0, b) If (a, b) The = (c, d) hypothesis the other (a, b) to is be sets; one member is the the set {a,b}. Shocking this set. The justification there really isn't anything else worth saying. DEFINITION i a, given by the theorem immediately following the definition is the definition works, THEOREM member 9 then a means = c and = b {{a}, {a,b}). = d. that {{«}, {a,b}} = {{c}, {c,d}\ — { 53 Functions Now only the a = {{a}, {a, b}} contains just two members, c. We and only the proof that 7. b = a. b In this case, one member, namely, \a\, which implies that d {a}. Case not in = = = and a the is d remains. {a, b\ = {{a}, \a,d\\, It is \a\, so the set The same must be = a — convenient to distinguish 2 cases. {a}, {a, b] { true of a\ { , a, { really has only } d \ \ , so { a, d } = b. In this case, b is in one member of {a}, {a, b} but not in the but therefore be true that b is in one member of {{a}, {a, d\ must It the other. This can happen only if ^ is in a, d\ but b is not in {a}; thus 2. b 9* a. other. b {a, b}; therefore have {{a}, \a,b\\ Case and {a} common element of these two members of {{a}, {a, b) }. Similarly, c is unique common member of both members of {{c}, {c, d\\. Therefore a or b ) { } { = d, but b 9^ a; so b = d. | , CHAPTER GRAPHS I Mention the real numbers to a mathematician and the image of a straight probably form in his mind, quite involuntarily. And most likely he neither banish nor too eagerly embrace this mental picture of the real line will will numbers. "Geometric intuition" will allow him to interpret statements about numbers in terms of this picture, and may even suggest methods of proving them. Although the properties of the real numbers which were studied in Part I are not greatly illuminated by a geometric picture, such an interpretation will be a great aid in Part II. You oil -i figure are probably already familiar with the conventional sidering the straight line as a picture of the real numbers, 2 3 to each real number a point on a a point which trarily, i The < label 0, b, to 1, but to the left of 0, To do and a point point twice as far to the right from 0 a we line> is labeled this then the point corresponding to a (Figure 1) of associating we pick, arbi- which we label 1. the point the same distance to the right, 2, labeled —1, etc. is method of con- i.e., lies to With the left this arrangement, if of the point corre- sponding to b. We can also draw rational numbers, such as ^> in the obvious way. It is usually taken for granted that the irrational numbers also somehow fit into this scheme, so that every real number can be drawn as a point on the line. We will not make too. much fuss about justifying this assumption, since method of "drawing" numbers is intended solely as a method of picturing certain abstract ideas, and our proofs will never rely on these pictures (althis though we Because will frequently use this a picture to suggest or help explain a proof). geometric picture plays such a prominent, albeit inessential geometric terminology is frequently employed when speaking of role, numbers number is sometimes called a point, and R is often called the real line. The number \a — b\ has a simple interpretation in terms of this geometric thus a picture: it is the distance between a has a as one end point and ~ a figure ~ 2 £ — . . \ a a v a + £ b as and b, the length of the line segment which the other. This means, to choose an example whose frequent occurrence justifies special consideration, that the set of numbers x which satisfy \x — a\ < £ may be pictured as the collection of points whose distance from a is less than s. This set of points is the "interval" from a — £ to a + £, which may also be described as the points corresponding to numbers x with a — e < x < a + e (Figure 2). g ets Q £ num :)ers which correspond to intervals arise so frequently that it is desirable to have special names for them. The set {x: a < x < b) is denoted by (a, b) and called the open interval from a to b. This notation naturally creates some ambiguity, since (a, b) is also used to denote a pair of numbers, j but in context it is always clear (or can easily be talking about a pair or an interval. 54 Note that if a made clear) whether one is > b, then (a, b) = 0, the set Graphs 55 with no elements; in practice, however, it is almost always assumed (explicitly one has been careful, and implicitly otherwise), that whenever an interval (a, b) is mentioned, the number a is less than b. The set {x: a < x < b] is denoted by [a, b] and is called the closed interval from atob. This symbol is usually reserved for the case a < b, but it is some- if = The and (<z, b) no reasonably accurate picture could ever indicate the difference between the two intervals, various conventions have been adopted. Figure 3 also shows certain "infinite" intervals. The set {x: x > a} is denoted by (a, oo), while the set {x: x > a} is denoted by [a, oo); the sets ( — oo, and (— oo, a] are defined similarly. At this point a standard warning must be issued: the symbols °o and — oo, though usually read "infinity" and "minus infinity," are purely suggestive; there is no number " oc " which satisfies oo > a for all numbers a. While the symbols oo and — oo will appear in many contexts, it is always necessary to define these uses in ways that refer only to numbers. The set R of all real numbers is also considered to be an "interval," and is sometimes denoted by (— oo, oo). Of even greater interest to us than the method of drawing numbers is a method of drawing pairs of numbers. This procedure, probably also familiar to you, requires a "coordinate system," two straight lines intersecting at right angles. To distinguish these straight lines, we call one the horizontal axis, and one the vertical axis. (More prosaic terminology, such as the "first" and "second" axes, is probably preferable from a logical point of view, but most people hold their books, or at least their blackboards, in the same way, so that "horizontal" and "vertical" are more descriptive.) Each of the two axes could be labeled with real numbers, but we can also label points on the horizontal axis with pairs (a, 0) and points on the vertical axis with pairs (0, b), so that the intersection of the two axes, the "origin" of the coordinate system, is labeled (0, 0). Any point (a, b) can now be drawn as in Figure 4, lying at the vertex of the rectangle whose other three vertices are labeled (0, 0), (a, 0), and (0, b). The numbers a and b are called the first and second coordinates, times used for a are [a, b] shown 6, also. determined in respectively, of the point Our real concern, let us recall, function is called the all is just a collection of pairs drawing each of the pairs is usual pictures for the intervals in Figure 3; since graph of this way. method of drawing functions. Since a of numbers, we can draw a function by a in the function. The drawing obtained in this way the function. In other words, the graph of / contains the points corresponding to pairs (*,/(#)). Since most functions contain many pairs, drawing the graph promises to be a laborious under- infinitely taking, but, in fact, Not f(x) = surprisingly, c, tion f(x) from it The lines many functions have have the simplest graphs. = c is graphs which are quite easy to draw. the simplest functions of It is all, the constant functions easy to see that the graph of the func- a straight line parallel to the horizontal axis, at distance (Figure 5). functions f(x) through ~ (0, 0), as in cx also Figure have particularly simple graphs 6. — c straight A proof of this fact is indicated in Figure 7: 56 Foundations Let x be some number not equal to 0, and let L be the straight line which passes through the origin 0, corresponding to (0, 0), and through the point A, corresponding to (x, cx). A point A\ with first coordinate y, will lie on L when the triangle A'B'O similar to the triangle is AB = OB A'B' OB' ABO, thus when c; precisely the condition that A' correspond to the pair (j, cy), A' lies on the graph of /. The argument has implicitly assumed that c this is the other cases are treated easily enough. ratio of the sides of the triangles appearing in the proof, c, is > but 0, which measures the called the slope of the have slope c. This demonstration has neither been labeled nor treated as a formal proof. Indeed, a rigorous demonstration would necessitate a digression which we are not at all prepared to follow. The rigorous proof of any statement connecting geometric and algebraic concepts would first require a real proof (or a pre- straight line, FIGURE The number that i.e., 7 and a line parallel to this line is also said to on a straight line correspond in this, it would be necessary to from Aside an exact way to the real numbers. to develop the properties of intend we precisely as as geometry develop plane real numbers. Now the detailed development of plane geometry is a beautiful cisely stated assumption) that the points length d (a, — b shall it is by no means a prerequisite for the study of calculus. We use geometric pictures only as an aid to intuition; for our purposes (and for most of mathematics) it is perfectly satisfactory to define the plane to be the subject, but b) length c — a set of all pairs of real 8 between (a, b) among others, and (c, d) to the collections f(x) = cx + d ~/~ / / — / / / / r s / / / / / f( x ) - cx motivation for 9 cY + (b- d)\ this definition is been built into our geometry.* Reverting once more to our informal geometric picture, it is not hard to see (Figure 9) that the graph of the function f(x) = cx + d is a straight line with slope c y passing through the point (0, d). For this reason, the functions d are called linear functions. Simple as they are, linear functions /(*) = cx occur frequently, and you should feel comfortable working with them. The + a typical problem whose solution should not cause any trouble. Given two distinct points {a, b) and (c, d), find the linear function / whose following FIGURE . not clear, Figure 8 should serve as adequate explanation— with this definition the Pythagorean theorem has If the (0, d) } } be V( a - // to define straight lines as certain collections To (*, cx):xa real number geometry of structure the all with geometry provide this artificially constructed studied in high school, one more definition is required. If (a, b) and (c, d) are two points in the plane, i.e., pairs of real numbers, we define the distance of pairs, including, FIGURE numbers, and * is might object to this definition on the grounds that nonnegative numbers at the moment are not yet known to have square roots. This objection is really unanswerable —the definition will just have to be accepted with reservations, until this little point is settled. The fastidious reader 57 Graphs graph goes through f(c) = d. If / (a, b) and (c, d). aa ac therefore a = , This amounts to saying that f(a) = ax be of the form f(x) to is - (rf d . c - £)/(<; — — a) + = +0= d c then = b and we must have rf; - /3 = — — b d a c and b f3, b, (3 a + b [(«/ - — — b b)/(c - a)]a, so N a a formula most easily remembered by using the "point-slope form" (see Problem 5). Of course, this solution is possible only account only for the straight if a ^ c; the graphs of linear functions which are not parallel to the vertical axis. The vertical straight lines are not the graph of any function at all; in fact, the graph of a function can never contain even two distinct points on the same vertical line. This conclusion is immediate from the definition of a function two points on the same vertical line correspond to pairs of the form (a, b) and (a, c) and, by definition, a function cannot contain (a, b) and (a, c) if b 9^ c. Conversely, if a set of points in the plane has the property that no two points lie on the same vertical line, then it is surely the graph of a function. Thus the first lines some the graph of a function whose is have no points on them at all. 2 is perhaps the function f(x) = x we draw some of the pairs in /, i.e., some of the pairs of the form (x, x 2 ), we vertical lines After the linear functions the simplest If and the last 2 are; notice domain is not all of R, since 2 sets in Figure 10 are not graphs of functions that the fourth obtain a picture like Figure 1 1 • (2, •I*, f) (-1, !)• •d, 9 (h •<*.*) , (0, 0) FIGURE 11 i) . 1) 4) ! x 2) lie along a curve one shown in Figure 12; this curve is known as a parabola. Since a graph is just a drawing on paper, made (in this case) with printer's ink, the question "Is this what the graph really looks like?" is hard to phrase in any sensible manner. No drawing is ever really correct since the line has thickness. Nevertheless, there are some questions which one can ask: for example, how can you be sure that the graph does not look like one of the drawings in Figure 13? It is easy to see, and even to prove, that the graph cannot look like (a); for if 0 < x < y, then x 2 < y 2 so the graph should be It is not hard to convince yourself that all the pairs (x, like the , higher at y than at x, which is not the case in (a). by drawing a very accurate graph, first plotting It is also many easy to pairs (x, see, simply x 2 ), that the graph cannot have a large "jump" as in (b) or a "corner" as in (c). In order to prove these assertions, however, we first need to say, in a mathematical way, what it means for a function not to have a "jump" or "corner"; these ideas already involve some of the fundamental concepts of calculus. Eventually we will be able to define them rigorously, but meanwhile you may amuse yourself by attempting to define these concepts, and then examining your may be compared with the ones they compare favorably, you are cer- definitions critically. Later these definitions mathematicians have agreed upon. If tainly to be congratulated The called functions f(x) power = xn , for various natural numbers are sometimes functions. Their graphs are most easily compared as in Figure by drawing several at once. The power functions are only special cases of polynomial functions, introduced in the previous chapter. Two particular polynomial functions are 14, \ \ \ ; \ \ \ \ \ \ FIGURE 14 Graphs graphed in Figure 15, while Figure 16 graph of the polynomial function /(*) in the case a n > = a nx + n is an_ xxn meant ~l + • 59 to give a general idea of the • • +a 0, 0. In general, the graph of / will have at most n — 1 "peaks" or "valleys" (a is a point like (*,/(*)) in Figure 16, while a "valley" is a point like "peak" (y> f(y))- power The number of peaks and valleys may actually be much smaller (the have at most one valley). Although these functions, for example, make, we will not even contemplate giving proofs until Part III (once the powerful methods of Part III are available, the proofs will be very easy). assertions are easy to Figure 17 illustrates the graphs of several rational functions. The rational functions exhibit even greater variety than the polynomial functions, but their behavior will also be easy to analyze once tool of Part III. we can use the derivative, the basic Many interesting graphs can be constructed by "piecing together" the graphs of functions already studied. The graph in Figure 18 is made up entirely of straight lines. and is The function / with this graph a linear function on each interval — \l(n + 1)]. (The number 0 is [l/(w + satisfies 1), l/n] not in the domain of /.) Of + and [ — 1/n, course, one can write out an explicit formula for f(x), when x is in [\/{n this is a 1), 1/w] good exercise in the use of linear functions, and will also convince you that a picture is worth a thousand words. n even (a) FIGURE 16 n odd (b) ; 60 Foundations FIGURE 18 Graphs — . 1-1 . = "N. f(x) /-360 61 sin * \ /360 \ /720 --1 It is actually possible to define, in a exhibits this same property In Chapter sine function. much simpler way, a function which of oscillating infinitely often near 0, by using the 15 we will discuss this function in detail, and radian measure in particular; for the time being it will be easiest to use degree measurements for angles. The graph of the sine function is shown in Figure 19 (the scale on the horizontal axis has been altered so that the graph will be clearer; radian measure has, besides important mathematical properties, the additional advantage that such changes are unnecessary). Now consider the function f(x) = sin 1/x. The graph of / is shown in Figure 20. (To draw this graph it helps to first observe that fix) =0 for x = -^—> —!— —180 360 540 f(x) = for x = — 90 fix) 1 = — for x 1 = ) > 90 + Notice that FIGURE 20 when x is * * 360 90 —i —— ^ 270 270 j + 9 720 j + large, so that 1/x * 360 270 is + small, f(x) ) 720 is • * • / also small; when x 62 Foundations f(x) \ = -v sin | / \ / % A u \ / \ / / \ \ / FIGURE / / / / / / \ 1 \ / \ 21 "large negative, 55 that is, when [*| is large for negative x, again f(x) is close although /(*) < 0. An interesting modification of this function is f(x) = x sin 1 fx. The graph of this function is sketched in Figure 21 Since sin 1 /x oscillates infinitely often near 0 between 1 and —1, the function f(x) = x sin 1/x oscillates infinitely is to 0, . often between x FIGURE 22 and —x. The behavior of the graph for x large or large nega- 63 Graphs tive is larger harder to analyze. Since sin 1/x is getting close to 0. while x is getting and larger, there seems to be no telling what the product will do. It is possible to decide, but this III. The graph = of f(x) another question that is is best deferred to Part x 2 sin 1/x has also been illustrated (Figure 22). For these infinitely oscillating functions, it is clear that the graph cannot hope to be really " accurate." The best we can do is to show part of it, and leave out the part near 0 (which much to find The graphs the interesting part). Actually, is simpler functions whose graph cannot be "accurately v I x2 x , < 1 , , f x2 \ , can only be distinguished by some convention similar Our last x < 1 to that used for open intervals (Figure 23). example is a function whose graph spectacularly nondrawable: is 0, x irrational 1, x rational. x rational x irrational 1, 0, FIGURE drawn. of £f and closed easy it is 55 24 The graph of/ must contain infinitely many points on the horizontal axis and but it must not contain either of these lines entirely. Figure 24 shows the usual textbook picture of the graph. To distinguish the two parts of the graph, the dots are placed closer together on the line corresponding to irrational x. (There is also infinitely many points on a line parallel to the horizontal axis, actually a mathematical reason behind this convention, but some sophisticated ideas, introduced in 20-6* Problems and it depends on 20-7.) The peculiarities exhibited by some functions are so engrossing that it is easy to forget some of the simplest, and most important, subsets of the plane, which are not the graphs of functions. The most important example of all is the circle. all A circle the points (#, y) and radius r > 0 contains, by definition, whose distance from (a, b) is equal to r. The circle thus with center (a, b) consists (Figure 25) of all points (x, y) with V( x ~ a) 2 + (y - b) 2 = r or (x - a) 2 + (y - b) 2 = r 2 . The circle copy, is with center and radius (0, 0) called the unit often regarded as a sort of standard 1, circle. A close relative of the circle is the ellipse. This is defined as the set of points, the sum of whose distances from two fixed points fixed points are the same, we points are taken to be (—c, 0) to obtain a and (c, a constant. is circle.) (When the two convenience, the two for If, and the sum of the distances is taken is on the ellipse if 0), be 2a (the factor 2 simplifies some algebra), then (x,y) and only if V( x - {-c)) 2 +y + V( x 2 +y = 2 2 c) 2a or V(* + c) +y = 2 2 - V( x - 2a c) 2 +y 2 or + x2 lex + c +y = 2 2 a 2) = V x - +y -4a V x - + aA = - 4a 2 4a c) or 4{cx - 2 c) +x - 2 2 +y 2 lex + c 2 +y 2 2 or c 2 x2 - lexa 2 a 2 (x 2 - + lex c 2 +y 2 ) or 2 (c - - a 2 )x 2 a 2y 2 - a 2 [c 2 - a 2) or This is usually written simply where a 2 — c c 2 > 0). A 2 (since we must picture of an ellipse intersects the horizontal axis when y = clearly choose a is €, so that *2 —= a2 (0, FIGURE 26 -b) 1 1, shown x = + ±a, > c, it in Figure 26. follows that The ellipse Graphs and it intersects the vertical axis The hyperbola is when = x 0, so 65 that defined analogously, except that we require the difference two distances to be constant. Choosing the points (— c, 0) and (c, 0) once again, and the constant difference as 2a, we obtain, as the condition that (x, y) be on the hyperbola, of the V(x + c y + y - V( x 2 which may be simplified a b = Vc — 2 a 2 , 2 y2 we must then (x,y) is a picture is shown 2 — = c z! = but it c if > a, so that a 2 and only — c 2 < 0. If if i in Figure 27. It contains two different orders. The hyperbola = ± a, ±2a, 1 on the hyperbola ence between the distances of (x,y) from so that x +f = 2 clearly choose The 2 to -+ In this case, however, c) (— c, two pieces, because the differ0) and (e, 0) may be taken in intersects the horizontal axis never intersects the vertical when y = 0, axis. It is interesting to compare (Figure 28) the hyperbola with a = b = V2 and the graph of the function f(x) = 1/x. The drawings look quite similar, and the two sets are actually identical, except for a rotation through an angle of 45° (Problem 21). Clearly no rotation of the plane will change circles or ellipses into the graphs of functions. Nevertheless, the study of these important geometric figures can often be reduced to the study of functions. Ellipses, for example, are FIGURE 28 made up of the graphs of two functions, Vl - ./(*) = g (x ) = -b Vl - b -a<x<a (**/**), and Of course, there are many (x 2 /a 2 ), -a<x<a. other pairs of functions with this same property. For example, we can take b = fix) Vl - (x*/a*), -b Vl - (x 2 /a 2 ), 0 < -a < x x < < a 0 and -b Vl - = g(x) b We Vl - (x 2 /a 2 ), 2 (x /a 2 ), 0<x<a -a<x<0. could also choose = f( x ) — (x /a ), Vl — (x /a 2 ° I b 2 2 x rational, 2 ), x irrational, —a < —a < x x < < a a and ^ = | —b { b But all V y/\ \ — — (x (x 2 2 /a /a 2 ), x rational, 2 x irrational, ), —a < —a < x x < < a a. these other pairs necessarily involve unreasonable functions jump around. A proof, or even a precise statement of this fact, is which too difficult Although you have probably already begun to make a distinction between those functions with reasonable graphs, and those with unreasonable graphs, you may find it very difficult to state a reasonable definition of reasonat present. A mathematical definition of this concept is by no means and a great deal of this book may be viewed as successive attempts to impose more and more conditions that a ''reasonable function must satisfy. As we define some of these conditions, we will take time out to ask if we have really succeeded in isolating the functions which deserve to be called able functions. easy, 5 reasonable. The answer, ' unfortunately, will always be "no," or at best, a qualified "yes." PROBLEMS 1. Indicate on a straight line the set of tions. you Also name each will also (i) |* (ii) \x (hi) j* (iv) |* 2 set, need the KJ - 3| < 1. - 3| < 1. — a\ < e. - 1| < all x satisfying the following condi- using the notation for intervals (in some cases sign). 67 Graphs — (vi) + 1 < a (give > 2. an answer in terms of a, distinguishing various x2 cases). 2. + + (vii) x2 (viii) (x Draw the set of 1 1)(* - 1)(* all - 2) points most cases your picture (*, will > 0. ;y) satisfying the following conditions. (In be a sizable portion of a plane, not just a line or curve.) > x (i) (vi) >y+ y < x 2 y < x |*-j|<1. |*+j|<1. (vii) x +y * +y . (iv) . (v) (vm) (ix) (* < an integer. is an integer. 2 1) + Draw the set of « M+ \y\ k| H (v) (vi) 2) 2 y < 1. all points (x, y) satisfying the following conditions. - 2 2 2 2 2 (vii) (viii) . Draw the set of x=y (iii) x (iv) x Hint: (a) - = 1= i. 1^-11 = 1^-11. 11-^ = 1^-11. * + ^ = 0. xy = 0. - 2* + y = 4. # = y (iii) (i) (y < x\ x (iv) 5. 2 is (x) (ii) 4. b. 2 (iii) 3. y. +a x (ii) = = all points (x, y) satisfying the following conditions: 2 . \y\. sin y. You Show already know the answers when that the straight line through of the function f(x) = m(x — a) + x {a, b) b. and y are interchanged. with slope This formula, m is the graph known as the "point-slope form" more convenient than the equivalent far is expression fix) = mx (b — ma) point-slope form that the slope + a /(*) = c When are the (c) it is immediately clear from the m, and that the value of is / at is b. For a ^ c, show that the straight graph of the function (b) ; — line through (x-a)+ and (a, b) (c, d) is the b. a graphs of fix) = mx + + and g(x) = m'x b b' parallel straight lines? (a) For any numbers A, B, and C, with A and B line is all (x, not both +C= + By y) satisfying Ax (possibly a vertical one). Hint: First decide the set of 0 when show that a vertical straight described. (b) Show (a) can be described as the set of all (x,y) satisfying Ax Prove that the graphs of the functions conversely that every straight are perpendicular if line, including vertical ones, By +C= 0. mn — —1, by computing the squares of the (Why lengths of the sides of the triangle in Figure 29. where the + = mx + b, = nx + c, fix) g(x) case, 0, a straight line is good lines intersect at the origin, as is this special as the general case?) (b) Prove that the two straight lines consisting of all (x,y) satisfying the conditions + By + C = 0, A'x + B'y + C = 0, if and only if A A' + BB' = Ax are perpendicular (a) +yi) (b) 0. Prove, using Problem 1-18, that 2 + (x 2 +y 2 2) < vV + X2 * + VyS+yS. Prove that V(*8 -*i) 2 + (yz ~yi) 2 < ^(x 2 ~ x,Y + Interpret this inequality geometrically inequality"). When does strict + (y 2 y/(x 9 (it is - yi y -xt)*+(y -yt)\ 9 called the "triangle inequality hold? Sketch the graphs of the following functions, plotting enough points to get a good idea of the general appearance. (Part of the make a reasonable decision problem is to "enough"; the queries posed thought will often be more valuable how many below are meant to show that a little than hundreds of individual points.) is . Graphs f(x ) (i) = x (What happens -\ does the graph Why does f(x) = f(x) = (iv) /(*) = (ii) for x 0, and for large x? Where relation to the graph of the identify function? lie in only positive x at suffice to consider it near 69 first?) - x X (iii) 10. * 2 *2 i. + #2 -- - 2 Describe the general features of the graph of / / / / (i) (ii) (iii) is if even. is odd. is nonnegative. = f{x a) for all x (a function with this property periodic, with period a). + (iv) f(x) 11. Graph the way to do is called for m = 1, 2, 3, 4. (There is an easy functions fix) = using Figure 14. Be sure to remember, however, that this, of x when m is even you should also note difference between the graph when m is important that there will be an means the positive mth root even and when 12. m ; odd.) is Graph fix) = |*| and fix) = x\ Graph fix) = |sin*| and fix) = (a) (b) (There is an important which we cannot yet even describe you can discover what it is; part (a) is meant to sin 2 *. difference between the graphs, rigorously. See if be a clue.) 13. 14. Describe the graph of g in terms of the graph of / (i) gix) (ii) gix) (iii) gix) (iv) gix) = = = = (v) gix) =/(lA). (vi) gix) =fi\x\). (vii) gix) (viii) £(*) (ix) £(*) (x) £(*) Draw fix) fix + + c. c). cfix) ( D easy to make ngu sh t he cases (It is i st i i a mistake here.) c = 0, c > 0, c < 0.) ficx). = l/M = max(/, 0). = min(/, 0). = max(/, 1). I the graph of fix) Problem if 1-17. = ax 2 + bx + c. Hint: Use the methods of 15. The symbol = [2] denotes the largest integer which [x] = 2 and [-0.9] functions (they are = -1. Draw [-1.2] is Thus [2.1] = the graph of the following quite interesting, and several will reappear all frequently in other problems). 16. (iv) f{x) = = = = (v) /(*) = [I]- (vi) fix) = 4r (0 /(*) (ii) /(*) (iii) fix) Graph (i) [•*]• x - [v]. VX - fx]. M + Vx - [ X ]. the following functions. /(*) = where {*}, {*} is defined to be the distance from x to the nearest integer. (ii) f(x) (iii) fix) (iv) fix) (v) Many fix) = = = {2*}. + {x} %{2x). {4*}. +i\2x\ +i\4x\. {x} functions may be described in terms of the decimal expansion of a number. Although we will not be in a position to describe infinite decimals rigorously until Chapter 22, your intuitive notion of infinite decimals should suffice to carry you through the following problem, and others which occur before Chapter 22. There is one ambiguity about infinite decimals which must be eliminated: Every decimal ending in a string of 9's is equal to another ending in a string of 0's (e.g., 1.23999 = 1.24000 .). We will always use the one ending in 9's . *17. . . . . Describe as best you can the graphs of the following functions (a comis usually out of the question). plete picture (0 fix) (ii) fix) (iii) fix) — = = the 2nd the number (iv) fix) — finite, ( v) number in the decimal expansion of x. number in the decimal expansion of x. number of 7 s in the decimal expansion the 1st 0 ? is finite, the if and 1 of x if this and 0 otherwise. number of 7's in the decimal expansion of x is otherwise. = the number obtained by replacing all digits in the decimal expansion of x which come after the first 7 (if any) by 0. /(*) (vi) fix) = 0 if 1 first if 1 never appears in the decimal expansion of appears in the 72th place. x, and n 71 Graphs *18. Let x irrational x = - rational in lowest terms. q (A number p/q factor, and q sprinkle points bers with q 19. (a) = is > in lowest 0). Draw terms Hp and q are integers with no common the graph of / as well as you can (don't randomly on the paper; consider 2, = = then those with q first the rational num- 3, etc.). 2 x 2 are the ones of the form (x, x ). points on the graph of f(x) Prove that each such point is equidistant from the point (0, i) and The = -|. (See Figure 30.) Given a point P = (a, fi) and a horizontal line L, the graph of = Y) show that the set of all points (*, y) equidistant from P and g( x ) ax 2 + bx + c. L is the graph of a function of the form f(x) Show that the square of the distance from (c, d) to (x, mx) is the graph of g(x) (b) *20. (a) x 2 (m 2 + 1) + to find the Using Problem 1-17 that the distance from \cm (b) *21. (a) x(-2md - (c, 2c) + minimum d) to the 2 . of these numbers, graph of /(*) - d\/Vm + 2 +c d2 = mx show is 1. Find the distance from (c, d) to the graph of /(*) = mx + (Reduce this case to part (a).) Using Problem 20, show that the numbers x' and / indicated Figure 31 are given by 1 b. in 1 i (b) Show that the set of same as the set of all all (x,y) (x, with (x'/V2)* y) with xy = 1 - (vV^) = 2 1 is the CHAPTER LIMITS The concept of a limit one in difficult we all is surely the most important, of calculus. The and probably the most goal of this chapter is the definition of once more, going to begin with a provisional definition; what we shall define is not the word "limit" but the notion of a function limits, but approaching a PROVISIONAL DEFINITION The are, limit. function / approaches the limit / near a, if we can make fix) as close as / by requiring that x be sufficiently close to, but unequal to, a. we like to Of the six functions graphed Notice that although g(a) it is still true that g is in Figure 1, only the not defined, and h{a) and h approach / near a. is first three approach / at a. wrong way," because we explicitly defined "the This is ruled out, in our definition, the necessity of ever considering the value of the function at a but unequal — it is to a. only necessary that f(x) should be close to / for x close to a, are simply not interested in the value of f(a). or even in We the question of whether f(a) FIGURE is defined. 1 One convenient way of picturing the assertion that / approaches / near a is provided by a method of drawing functions that was not mentioned in Chapter 4. In this method, we draw two straight lines, each representing R, and arrows from a point x in one, to f(x) in the other. Figure 2 illustrates such a picture for two different functions. 72 Limits 73 consider a function / whose drawing looks like Figure 3. Suppose we ask that /(*) be close to /, say within the open interval B which has been drawn in Figure 3. This can be guaranteed if we consider only the numbers x Now in the interval A interval which will 3. (In this diagram we have chosen the largest work; any smaller interval containing a could have been of Figure chosen instead.) If we choose a smaller interval B' (Figure 4) we will, usually, have to choose a smaller A\ but no matter how small we choose the open interval B, there is always supposed to be some open interval A which works. A similar pictorial interpretation is possible in terms of the graph of/, but in this case the interval B must be drawn on the vertical axis, B when x is and the A means set A that the on the horizontal axis. The fact that f(x) part of the graph lying over A is contained in the region which is bounded by the horizontal lines through the end points of B; compare Figure 5(a), where a valid interval A has been chosen, with Figure 5(b), where A is too is in in large. In order to apply our definition to a particular function, let us consider = x sin 1/x (Figure 6). Despite the erratic behavior of this function near 0 it is clear, at least intuitively, that / approaches 0 near 0. and it is certainly to be hoped that our definition will allow us to reach the same con- /(*) clusion. In the case we must ask that x be get x sin if we we can are considering, both a = get /(*) /x within ^ — 1 - succinctly, \x sin l/x\ sin - x 0. To be .11 < — < 10 more / of the definition are 0, so specific, < < x sin x to1, Now if suppose means we want of 0. This — or, ^ sufficiently close to 0, but 1 and x sin 1/x as close to 0 as desired > 10 this is easy. Since for all x ^ 0, we require we wish to 74 Foundations we have s f(x) = * sin - x sin < 1*1, * for all x 0. X This means that 4 / FIGURE / x sin 1/x is \ \ \ is if |*| within 1 y^ < xo an d ^ 0, then — take any positive number £ we can make |* < sin 1/*| y^-; in other words, within yo of 0, but ^ 0. There is just as easy to guarantee that is nothing special about the number y^; it — 0| < j-q-q simply require that |*| \f(x) |*| < £, and * ^ 0. For the function /(*) approaches 0 near 0. If, 6 x of 0 provided that x < y^, \f(x) — but x 0| < £ ^ 0. In fact, if we simply by requiring that = x 2 sin 1/x (Figure 7) for it seems even clearer that / example, we want x £ sin * then we certainly need only require that that |* 2 < yJ~o and consequently < |*| 1 io < y^ and * 9± 0, since this implies | FIGURE 7 (We could do even better, and allow |*| < 1 /Vio and * 5^ 0, but there particular virtue in being as economical as possible.) In general, f(x) = y/\x\ sin ensure that < we need only FIGURE e, require that 8 |*| < £ and * 0, if £ > is no 0, to . Limits FIGURE 75 9 provided that £ < 1 If we are given an £ which is greater than 1 (it might be, even though it is "small" e's which are of interest), then it does not suffice to . require that I < but e, it certainly suffices to require that a third example^consider the function /(*) As iin |*| In order to make |*| if £ < or that 1, < sin l/x\ < |*| 1 < and x £ 5* £ 2 0 = V|*| we can require that and ^ if £ > x 1 < 1 and x 5* 0. 0, (the algebra Finally, let us consider the function f(x) |*| sin \/x (Figure 8)- = sin 1/x is left to you). (Figure 9). For this /approaches 0 near 0. This amounts to saying that it is £ > 0 that we can get \f(x) — 0| < £ by choosing x sufficiently small, and ^ 0. To show this we simply have to find one £ > 0 for which the condition \f(x) - 0| < £ cannot be guaranteed, no matter how function it is false that not true for every small we require number |*| to be. In fact, £ = % will do: it is impossible to ensure that |/(x)| < i no matter how small we require |*| to be; for if A is any interval containing 0, there is some number * = 1/(90 + 360rc) which is in this interval, and for this x we have f(x) = 1 This same argument can be used (Figure 10) to show that / does not approach any number near 0. To show this we must again find, for any particular number /, some number £ > 0 so that \f(x) — /| < £ is not true, no matter how small x is required to be. The choice £ = i works for any number be, we cannot ensure that /; that is, no matter how small we require |*| to _ some and /| #1 also = < 1 The /(90 + reason is, that for any interval A containing 0 there 360«) in this interval, so that some x 2 - 1/(270 + 360m) /(**) in this interval, so that = -I- is 76 Foundations But the interval from total length is only 1 - £ to / + * cannot / - |1 i x, I 0, /(*) no matter what x rational x irrational /| < iand The phenomenon we consider also = exhibited by /(*) the function its - |-1 /[ < no matter what a sin 1/x many near 0 can occur in x irrational /(*) FIGURE since 1, / is. ways. If then, -1 and contain both we cannot have so ; -(?: x rational, / does not approach any number / near a. In fact < \ n o matter how close we bring x to a, because m any interval around a there are numbers x with /(*) = 0, and also numbers x with f(x) = 1, so that we would need |0 - l\ < \ and also |1 - l\ < An amusing variation on this behavior is presented by the function \. shown we cannot make 11 is, - \f{x) l\ ^ in Figure 1 1 = f(x) 0, I c_±J m c f(X) = The behavior of this function is "opposite" to that of g(x) approaches 0 at 0 but does not approach any number at a, if a you should have no difficulty convincing c S — e ) As a contrast . a pathological, FIGURE * ^tional # irrational. *' | If/(*) 12 = c, we = sin 1/x; * 0. it By now yourself that this is true. to the functions considered so far, which have been quite will now examine some of the simplest functions. then /approaches c near a, for every - number a. In that I/O) c\ < s one does not need to restrict x to be near a at tion is automatically satisfied (Figure fact, to all; ensure the condi- 12). As a slight variation, let / be the function shown x x If a 1 ^ /to = i,*> o > 0, then /approaches 1 near certainly suffices to require that \x - b /(*) = -1, x < 0 so . -1 that/W = that |/0) you may - 1. Similarly, (-1)| easily check, The FIGURE 13 function /(*) ensure that |/0) - The < function a, < 0, a <C x 0 < x, a\ x is < £ then / approaches require that |* 2 ^r easily dealt with. Clearly / we just have requires a we must decide - k 2 a2 \ -1 £| near < < 1 e it £. b: to ensure Finally, as 0. |* - a \ < more work. To show to ensure that < 1 approaches a near to require that little how — a / does not approach any number near = /0) = proaches a 2 near if * e it suffices to 0 0. indeed, to ensure that \f(x) a\ < a, since this implies a: or < > in Figure 13: a: to £. that / ap- 77 Limits Factoring looks most promising procedure: we want like the — |* a\ + \x ' a\ < £. Obviously the factor |* a\ is the one that will cause trouble. On the other hand, there is no need to make \x a\ particularly small; as long as we know some bound on the values of \x + a\ we will be in good shape. For example, if \x a\ < 1,000,000, then we will just need to require that |* — a\ < e/1 ,000,000. Therefore, to begin with, let us require that \x — a\ < 1 (any + + + number other than positive ensure that x As a matter of would do 1 just as well) not too large, and consequently that is presumably ; + \x a\ is this will not too large. Problem 1-12 shows that fact, - \x\ \a\ <\x - < a\ 1. so H< + i kl, and consequently + \x Now we a\ < + \x\ < \a\ + 2\a\ need only the additional requirement that 1. — \x a\ < z/{2\a\ + 1). In other words, if \x - < min a\ ( l r \ Naturally, min(l, e/(2\a\ same Precisely the a z near if \x In a. - a\ 1 + + 1)) will just sort of trick will \ then \x 2 - < 2 a \ s. 1/ + be e/(2\a\ show that if f(x) 1) for = small e. x s then / approaches , fact, < min ( 1, \ ' (1 + — ~4 \a\r+ \a\(\ The proof of this assertion will from: If — \x — 2\a\ a\ \x < 2 then 1, + ax + a |*| 2 \ < < < \a\) + nz) 9 then 3 l^ a *\ < e ' \a\*J show where the weird denominator comes + 1, and consequently \a\ \x\ (1 rr, + + + + \a\y+ \a\(l + 2 \a\ ' \a\* \x\ \a\) + \a\\' Therefore _ fl 8| = \ x < n (1 = £. _ . fl | ^2 + ax 4-1\a\) h 2 -Li hi + + + a2 \ + + iui^ \a\) \a\ 2 * [(1 + ^ l)2 + W(1 + M) + W] has now come to point out that of the many demonstrations about which we have given, not one has been a real proof. The fault lies not with our reasoning, but with our definition. If our provisional definition of a function was open to criticism, our provisional definition of approaching a The time limits even more vulnerable. This definition is simply not sufficiently precise is hardly clear how one "makes" f(x) close to / (whatever "close" means) by "requiring" x to be sufficiently close to a (however limit is to be used in proofs. It : 78 Foundations close "sufficiently" close you definition may is supposed to be). Despite the criticisms of our hope you do) that our arguments were feel (I certainly nevertheless quite convincing. In order to present any sort of argument at all, practically forced to invent the real definition. It is possible to we have been arrive at this definition in several steps, each one clarifying phrase which some obscure remains. Let us begin, once again, with the provisional still definition: The function / approaches the limit / near a, if we can make/(*) as close we like to / by requiring that x be sufficiently close to, but unequal to, a. as The very first change which we made in /0) close to The / meant making function The was this definition and small, making to note that and similarly for x a: / approaches the limit / near a, if we can make \f(x) - /| like by requiring that \x - a\ be sufficiently small, and a. second, small as i\ we as small as x 9^ - \f(x) we more crucial, change was means making \f(x) - like" making to note that < l\ any £ for £ > - |/(*) "as l\ 0 that happens to be given us The function / approaches the limit / near a, if for every number £ > 0 we can make \f{x) — /| < e by requiring that \x - a\ be sufficiently small, and x ^ a. There is a common pattern to have given. For each number with the property that 8 say, For the function /(*) = > if x x sin \Jx we the demonstrations about limits which all £ ( 0 we found some a and with a « \x - 0, / < a\ = other positive number, <5, then 0), the — \f(x) number <$ /| < e. was just number £; for f(x) = V\ x sin 1/x, it was £ if £ < 1, and it was 1 if > 1; for/<» = x 2 it was the minimum of 1 and e/(2\a\ + 1). In general, it may not be at all clear how to find the number given £, but it is the condition \x — a\ < 8 which expresses how small " sufficiently" small must be: the 2 \ £ <5, The some function / approaches the limit / near a, 5 > 0 such that, for all *, if \x - a\ < 6 < £. is practically the definition This change, noting that "|* "0 DEFINITION < The is \x - a\ < - a\ < we 8 will adopt. and x ^ for > 0 there is then - l\ every £ ^ a, \f(x) We will make only one trivial a" can just as well be expressed 5." function / approaches the limit I near a 8 > 0 such that, for all *, if 0 < |* some if and x — means a\ < : 8, for every £ then f(x) \ > — 0 there l\ < £. Limits 79 so important (everything we do from now on depends on it) any further without knowing it is hopeless. If necessary memorize it, like a poem! That, at least, is better than stating it incorrectly; A good exercise in if you do this you are doomed to give incorrect proofs. This definition is that proceeding giving correct proofs approaching tions down all to review every fact already demonstrated about funcThis requires writing what you are proving, but not much more work has been done already. When proving that / does not the correct definition of the algebraic approach If is limits, giving real proofs of each. it is / be sure to negate the definition correctly: at a, not true that for every £ > 0 there then |/0) - /| < is some 6 > 0 such that, for all x, if 0 < \x - a\ < 5, e, then there 0 < some £ is \x - < a\ > d 0 such that for but not \f(x) > every 5 - < /| 0 there is some x which satisfies £. = sin 1/x does not approach 0 near 0, we > 0 there is some x with 0 < |* — 0| note that for every 36(k), namely, an x of the form 1/(90 1 /x - 0| < i Thus, to show that the function f(x) consider £ < 8 = \ and but not where n As an limit, is |sin <5 — so large that 1/(90 + + 360n) < 5. illustration of the use of the definition of a function we have reserved the function shown approaching a example, in Figure 14, a standard but one of the most complicated: fix) (Recall that p/q factor and q > is 0, x irrational, 0 Vq, x = p/q in lowest terms < < x 1 in lowest terms, 0 if < x < 1. p and q are integers with no 0.) x irrational (0, h FIGURE 14 x =£ in lowest terms common 80 Foundations For any number prove that be /n 1 with 0 a, consider any this, < Notice that the e. < 1 the function / approaches 0 at a. To > 0. Let n be a natural number so large only numbers x for which \f(x) — 0| < 6 could < a number , £ false are: 112131234 ? ; 2 (If a ; 3 4 3 " 1 9 ; ' ' - n 1 5' 5* s' 5 ; 4'* n * ' ' n rational, then a might be one of these numbers.) However many of numbers there may be, there are, at any rate, only finitely many. There- is these numbers, one one p/q among these numbers. fore, of all these is closest to a\ that is, — \p/q smallest for a\ is one of these numbers, then consider only the values \p/q — a\ for p/q ?± a.) This closest distance may be chosen as the 5. For if 0 < \x — a\ < 8, then x is not one of (If a happens and - n 1 to be 1 — 0| < £ is true. This completes the proof. Note that our which works for a given £ is completely adequate there is no reason why we must give a formula for 8 in terms of £. Armed with our definition, we are now prepared to prove our first theorem; you have probably assumed the result all along, which is a very reasonable thing to do. This theorem is really a test case for our definition: if the theorem could not be proved, our definition would be useless. therefore \f(x) description of the THEOREM l A function cannot approach two different limits near approaches PROOF — <5 Since this / is near our a, m and / approaches theorem about first near limits then a, it In other words, a. if / m. / == will certainly be necessary to translate the hypotheses according to the definition. Since / approaches ber 8 1 > also near a, we know that for any £ - /| > 0 there is some num- 0 such that, for all x, if We / 0< |* - a| < 8h then know, since / approaches m near < e. that there is |/(*) a, some 82 > 0 such that, for all x, if We 0 < |* — a| < have had to use two numbers, 8 2) 8i then |/0) and 82 , - m\ < since there e. no guarantee that is the 8 which works in one definition will work in the other. But, in fact, now £ > 0 there \f(x) - /| easy to conclude that for any is some 8 > all X, if 0 < \x - we simply choose a\ 8 = < 8, then min(6i, S 2 ). < 6 and |/0) - it is 0 such that, for m\ < e; ~ Limits To complete the proof we just have to pick a particular s > 81 0 for which the two conditions - I/O) cannot both hold, / ^ m, so that there is a - |/ > 5 ^ > if / m\ 0 < 0, This implies that for 0 -m = \ - |* |/ < - and 6 < m\ S The proper choice is suggested by Figure 15. If we can choose \l - m\/2 as our e. It follows that m. 0 such that, for if |/ < /| a\ \x all x, < - b, a\ then |/0) " and |/0) - < -H —j— \l < /| m| < ^ " 2 we have <5 - f(x) + fix) - m\ < \l = 1/ - /0)l + - «l 2 2 - l/to m|, a contradiction. | The number / which / approaches near a is denoted by lim fix) (read: the x-*a z—»o approaches a). This definition is possible only because of a: appro; limit mit of fix) f(x) as * ensures that lim }{x) never has to stand for two different Theorem "heorem 1, which ensui numbers. The equation lim/0) = I x-*a has exactly the same meaning as the phrase / approaches The possibility lim fix) = / is / near a. remains that / does not approach / near a, for any /, so that saying false for every number 7. This is usually expressed by still x—*a that "lim fix) does not exist." x, x—>a letter Notice that our new notation introduces an extra, utterly irrelevant already not does which letter which could be replaced by /, y, or any other appear— the symbols lim/O), Hm/(0, lim/0), x-+a t-*a V-*a has denote precisely the same number, which depends on / and a, and at anything denote not do fact, in letters, (these or f, with do to nothing *, y notation, lim/, but this all). A more logical symbol would be something like all a despite its brevity, tried to use it. The is so infuriatingly rigid that almost notation lim f(x) is much more no one has seriously useful because a function / often has no simple name, even though it might be possible by a simple formula involving x. Thus, the short symbol + sin x) 2 lim to express /(*) (.v x-*a could be paraphrased only by the awkward expression lim /, = where f(x) x2 + sin x. a Another advantage of the standard symbolism is illustrated by the expres- sions lim x + J + t\ 3 , ar—>a lim x The first means the number which / approaches near f(x) x + t\ /(*) should have = little difficulty x + when for all x; means the number which / approaches near the second You = a t\ for all (especially if a when t. you consult Theorem 2) proving that lim x + t + t 3 = a + 3 = x +a t\ x~*a lim x z . t->a These examples flexibility. In illustrate the fact, main advantage the notation lim f(x) is of our notation, which so flexible that there is is its some danger x—*a what it really means. Here is a simple exercise in the use of this notation, which will be important later: first interpret precisely, and then of forgetting prove the equality of the expressions lim f(x) and lim f(a x-*a An make important part of it easy to find of inequalities this many + h). h-*0 chapter is the proof of a theorem which will The proof depends upon certain properties values, hardly surprising when one considers the limits. and absolute Although these have already been stated in Problems importance they will be presented once again, in the form of a lemma (a lemma is an auxiliary theorem, a result that justifies its existence only by virtue of its prominent role in the proof of another theorem) The lemma says, roughly, that if x is close to * 0 and y is definition of limit. 1-19, 1-20, and facts 1-21, because of their . close to ^o, then x , +y + be close to x 0 yo, and xy will be close to x 0yo and 1/y will be close to l/y 0 This intuitive statement is much easier to remember than the precise estimates of the lemma, and it is not unreasonable to read will 3 . the proof of Theorem 2 first, in order to see just how these estimates are used. Limits LEMMA (1) If 1* - < *o| | and - y0 < \y |> \ then \(x+y)(2) < +yo)\ (x 0 e. If '* K min * oi " 0' 27i^f+T)) L b ~ y° K 27M M)' and l then — \xy (3) If y 0 ^ proof (i) -j, 0 (2) | < min^— —j-y - *o) » 0 and 9± |o S. 0 and |j then y < Jfo^ol + j) - Since |* — Oo *0 | +^ < 0 )| = 10 < I* - + + *„| - (y yo)\ - yo\ < \y \ +\= e. we have 1 — Ul |*o| < \x — *o| < 1, so that W < 1 + |*o|- Thus |ry - * 0 )>o| = 1*0 - yo) +yo(x < \x\ \y - yo\ + \yo\ • < (i < ~ + M) + 2 (3) We = ~ 2(W + |jy|. > I* - *o| + M 1) 2(W + D s. 2 have W-M< so *o)| i^ol/2. In particular, 0, jy \y - Jo| < ^ and ±<-L. \y\ \yo\ Thus i y _1 y<> \y*-y\ M * M< A l>'o| . _L |>o| = . 2 e . | 83 84 Foundations THEOREM = lim /(*) If 2 and lim / x—* a = m g(x) then 9 x—*a (1) lim (f (2) lim" + g)(x) =l + m; -l-m. (f-g)(x) x—+a Moreover, m ^ if then 0, lim (-) (3) PROOF The means hypothesis = (x) -• m \g/ x->a > that for every e > 0 there are 5 h 8 2 0 such that, for all x, <\x < |* — ifO and This means > 52 <5i, 0 if i(0 and Now 0 < let \x if fi a\ <5 a\ < — <z| < If 0 < |# 2 + m)\ To If £ < - prove > This proves £. 0 if < |* - l\ m\ < e, < £. number) that there are then \f(x) - /| <5 then \g(x) — m\ 2, < a\ then 0 8, < < < " lemma H < Again let 6 = |/W So, 0 if /| < |* < then S l9 > 0 < if (l, < — is a 8 - < 8, |* i a\ - |* — <z| ^^y) 0 there i a\ < Si ((/ and + g)(x) — ~ > , I ° £ < min l\ < £, < m| (l, 2(|m| |, and W this 0 such that, for i / \ then \g(x) - i m\ —- + then 8, and < m\ lemma. all x, - |/(*) then Sa, min(6i, § 2 ), If 0 if £ - similarly, after consulting part (2) of the by the lemma, \(f'g)(x) Finally, \x this implies that 0 such that, for - a\< < min -• | \ and |, (1). > a\ 8h < | and /| we proceed (2) 0 there are 5 h 8 2 — both are true. But by part (1) of the (I \f(x) \g(x) - |* true, so 2 I/O) then then 2, also a positive is <\x - < 0 6i, <5 all x, = min(5i, ). are both < 5 — e/2 (since, after all, 0 such that, for < < a\ a\ < - m\ < proves (2). all x, mm . /\m\ I ^ £|m| 2 \ 2/ \ 1)/ to part (3) of the But according (l/g)(x) makes lemma 85 ^ 0, so that g(x) first, and second that sense, W This proves means, this Limits < e. rn (3). | Using Theorem 2 we can prove, ™ trivially, + 7x = " x* + 5 x* 1 such facts as + la « + r b az 2 without going through the laborious process of finding a must begin with = 7, 1 = 1, lim * = a, lim 7 5, given an 6. We x-*a lim x-+a If we want to find the 5, however, the proof of Theorem 2 amounts to a prescription for doing this. Suppose, to take a simpler example, that we want to find a 5 such that, for all x, but these are easy to prove directly. 0 if < |* - 5% > 0 such that, for and Since how we have to do 5, then Theorem Consulting the proof of and < a\ \x 2 +x- (a 2 + a) I < e. we must 2(1), we < 8i, then < <5 then |* - a| 2 = a2 and lim x - see that if 0 < |* - if 0 < \x - a\ a\ 2, already given proofs that lim a: 2 I* - 2 <z | < < •- mto |- 1 ( '2(HTi)/ Thus we can take « min(5 i? find 8i | this: 8 first all x, <5 2) = min C mm ( yi, I —-— j ,-- y, a, we know If a the 0, same method can be used < 0 if The proof of Theorem find a 8 > - |* < 0 - |* <5, > a 8 then < ^2 v2 0 such that, for all x, 6. 2(3) shows that the second condition will follow 0 such that, for if < a\ to find < a\ if we all x, then 8, \x 2 - a 2 \ < min ^£ Thus we can take 6 = min 2(W + 1) Naturally, these complicated expressions for 8 can be simplified considerably, after they One have been derived. technical detail in the proof of In order for Km f(x) to be defined it is, Theorem 2 deserves some discussion. as we know, not necessary for / to be x—*a defined at a nor y there must be < 8; necessary for / to be defined at is it some > 8 0 such that/(*) points x all ^ a. defined for x satisfying 0 is However, < \x — a\ otherwise the clause "if 0 <\x - < a\ 8, then - \f(x) < l\ s" would make no sense at all, since the symbol f(x) would make no sense for some #'s. If / and g are two functions for which the definition makes sense, it is easy to see that the same is true for / + g and / g. But this is not so clear for 1/g, since 1/g is undefined for x with g(x) = 0. However, this fact gets * Theorem established in the proof of 2(3). There are times when we would like to speak of the limit which / approaches at a, even though there is no 8 > 0 such that f(x) is defined for x satisfying 0 < \x — a\ < 8. For example, we want to distinguish the behavior of the two functions shown in Figure 16, even though they are not defined for numbers less than a. For the function of Figure 16(a) we write = lim f{x) or I x—*a + (The symbols on the left lim f(x) = /. x la are read: the limit of f(x) as x approaches a from above). These "limits from above" are obviously closely related to ordinary limits, e > and the 0 there is definition a 8 > if (The condition "0 x > very similar: lim f(x) is 0 such that, for 0 < < x x — a a < < 8" all 8, / means that for every a*, then f(x) \ is = - l\ < £. equivalent to "0 < \x - a\ < 8 and a.") "Limits from below" (Figure 17) are defined similarly: lim f(x) — I 87 Limits = (or lim f(x) x T I) means that > for every £ 0 there a 5 is > 0 such that, for a all x, if 0 < - a < x 8, then < l\ S. and below even if / is Thus, for the function / quite possible to consider limits from above It is defined for numbers both greater and of Figure 13, less than a. we have = lim f(x) It is - \f(x) and 1 an easy exercise (Problem 27) = — 1. lim f(x) show that lim to f(x) exists and only if x—*a if lim f(x) and lim f{x) both exist and are equal. x—>a + x—*a~ Like the definitions of limits from above and below, which have been smuggled into the text informally there are other modifications of the limit concept which will be found useful. In Chapter 4 it was claimed that if x is ; large, then sin \/x is This assertion close to 0. lim sin l/# FIGURE usually written 0. 18 The symbol lim x—* becomes finity. = is f(x) read "the limit of f(x) as x approaches is infinite," and a limit of the form lim f(x) often called a limit at in- is Figure 18 illustrates a general situation where lim f(x) x— * lim f(x) = I or "as x <*> » means that for every e > 0 there is a = /. Formally, 00 number N such that, for all x, if The analogy with x > N, then f(x) \ l\ < S. the definition of ordinary limits should be clear: whereas < \x — a\ < 5" expresses the fact > N" expresses the fact that x is large. the condition "0 condition "x - that x is close to a, the We have spent so little time on limits from above and below, and at infinity, because the general philosophy behind the definitions should be clear if you understand the definition of ordinary limits (which are by far the most important) Many exercises on these definitions are provided in the Problems, which . also contain several other types of limits which are occasionally useful. PROBLEMS Find the following limits. (These limits all follow, after some algebraic manipulations, from the various parts of Theorem 2 be sure you know which ones are used in each case, but don't bother listing them.) ; x2 (i) lim (ii) lim x* x x—>2 — 1 -8 -2 3 (iii) lim — 2 — v" lim x->y x — y xn - y n lim y-+x x — y X x->3 xn (iv) (v) , (vi) 2. : + h - Va ' lim In each of the following cases, find a 8 such that # satisfying 0 < \x (i) /(*) = (ii) /(*) = -;a = *4 = / ; — a\ a4 < |/(.v) — l\ < £ for all 8. . 1,/ = 1. x + /(*) - *4 (iv) /(*) = —+^ (iii) sin 2 x 1 (v) = 1, / = ; o, 2. / = 0. = VH; = 0 / = 0. = V*; = 1, / = 1. f(x) fl (vi) /(*) 3. = -; a 5 fl For each of the functions in Problem 4-15, decide for which numbers a the limit lim fix) exists. x—>a *4. 5. (a) Do (b) Same problem, the same for each of the functions if we in Problem 4-17. use infinite decimals ending in a string of 0's instead of those ending in a string of 9's. Suppose the functions / and g have the following property: and if 0< |* - 2| < sin 2 if 0 < |* - 2\ < e e > 0 find a 8 > < \x - then For each ii) for all s all x, if 0 2| < 5, 2 , (^j + 6, then \g(x) then |/(*) - < 4| 0 such that, for |/(x) + - 2\ 6. all x, — 6| < e. < 6, > 0 Limits < 0 if (ii) - \x < 2| 1 (iii) if 0 < |* - 2\ < 5, then (iv) if 0 < |* - 2| < j, then _ /to 7. Give an example of a function / false: If \f(x) when 0 (a) < |* - < e when - a\ < 8/2. i lim f(x) If and lim x—>a e. < E. which the following assertion for - |.v do not g(x) < e. 2 < 0 l\ < 8| 1 £to 6. - then f(x)g(x) <5, 89 < a| exist, then 5, - |/(*) < + g(x)] can lim[/(x) x—+a /| is e/2 or x—>a lim f(x)g(x) exist? x—>a (b) If lim f(x) exists and lim [/(#) x—+ a (c) If x— + g(x)] exists, g(x) exist? x—+ a lim /to exists and lim g(x) does not x—*a must lim a exist, can lim x—+o + g(x)] [f(x) x—*a exist? (d) If lim f(x) exists and lim f(x)g(x) x— x— o exists, does follow that lim g(x) it a i —a exists? 8. = Prove that lim f(x) lim f(a x—>a + (This h). mainly an exercise in under- is A—*0 standing what the terms mean.) 9. (a) = Prove that lim /(*) x— a and only / if if lim[/(^) - x— a > = /] 0. (First see why the assertion is obvious; then provide a rigorous proof. In this chapter most problems which ask for proofs should be treated in same way.) Prove that lim f(x) the = (c) lim f(x). x- a ->0 Prove that lim /(*) = lim /(* 3 ). (d) Give an example where lim f(x 2 ) (b) x—*a x—>0 x—»0 exists, x->0 10. but lim f(x) does not. x^O Suppose there is a 8 > 0 such that f(x) = g(x) when 0 < \x — a\ < 8. Prove that lim f(x) = lim g(x). In other words, lim f(x) depends only x— > x— a a * on the values of /(x) for x near a — x—+ a this fact is often expressed that limits are a "local property." (It will clearly help to use other 11. (a) letter, instead of Suppose that f(x) in the definition of limits.) 8, < by saying or some 8', g(x) for all x. Prove that lim f(x) x—ki < lim g(x), x—*a provided that these limits exist. (b) How can the hypotheses be weakened? (c) If f(x) < g(x) for all x, does it necessarily follow that lim f(x) < x—*a lim g(x) ? x—*a 12. Suppose that f(x) < g(x) < h{x) and that lim f(x) that lim g(x) exists, x—*a picture!) and that lim x—>a g(x) = lim f(x) x—»a = lim h{x). Prove x—>a x-+a = lim h(x). (Draw a x—>a . 90 Foundations *13. (a) Prove that = lim /(*)/* if and / b ^ b[f{bx)/bx\ = (b) What happens (c) Part (a) enables us to find lim (sin 2x)/x in terms of lim (sin x)/x. (a) Find this limit in another way. Prove that if lim /(*) = /, then lim if b 0? x—>0 x—*Q 14. (b) Prove that if = lim /(*) and lim / i—»a (a) = |/|. x—*a x—*a 15. = g(x) m, then lim = max(/, /«) and similarly for min. Prove that lim \/x does not exist, false for show that lim \/x = i.e., — (b) Prove that lim \/{x Prove that if = lim f(x) x-+a Af such that 17. + for 0 1 < \x Prove that if f{x) lim f(x) does not /, M < \f(x)\ Why pictoriaily?) Hint: I / is *->0 number every max (/,$)(*) x—»« x—*a x—>0 16. bl. x-+Q = Hint: Write f(bx)/x = then lim f(bx)/x 0, x—*0 - < a\ = /. does not 1) then there if < 0 does a number is — \x exist. a\ < suffice to it d. 5 > 0 and a (What does prove that / — number mean this < 1 f(x) < 8 ? 0 for irrational x exist for any and /(*) = 1 for rational x, then a. x—»o *18. Prove that if }{x) = x for rational *, and f(x) then lim/W does not exist if a ^ 0. = -x x—»a 19. (a) Prove that if = lim g(x) 0, then lim g(x) sin 1 A for irrational x, = 0. x-»0 z—>0 (b) Generalize this fact as follows : lim g (x) If = 0and|AW| < M for all z-+ 0 a-, then lim g(x)h(x) you succeed (b) in may make = 0. (Naturally it is unnecessary to do part (a) if doing part (b); actually the statement of part that's one of the values of easier than (a) — it generalization.) 20. Consider a function / with the following property: which lim g(x) does not exist, then lim[/(x) tion for not exist. This **21. is Prove that this happens if and only if if g is + g(x)] lim f(x) does any funcalso does exist. Hint: actually very easy: the assumption that lim f(x) does not exist leads to an immediate contradiction if you consider the right g. g is replaced This problem is the analogue of Problem 20 when / + by considerably more complex, and the / g. In analysis requires several steps (those in search of an especially challengthis case the situation is • ing problem can attempt an independent solution) (a) Suppose that lim f(x) not exist, exists and is ^ 0. then lim f(x)g(x) also does not x-+0 Prove that exist. if lim g(x) does 91 Limits (b) Prove the same result this sort of limit (c) Prove that lim if (The precise definition of <*>. given in Problem 33.) is neither of these if = \f{x)\ two conditions holds, then there is a function g such that lim g(x) does not exist, but lim f(x)g(x) does Hint: Consider separately the following two cases: (1) For > 0 we have |/(*)| > £ for all sufficiently small x. (2) For exist. some e every 6 > 0, there are arbitrarily small x with |/(*)| second case, begin by choosing points x n with \x n \ < £. In the < 1/n and \f(x n )\<l/n. *22. Suppose that A n is, for each natural number n, some finite set of numbers [0, 1], and that A n and A m have no members in common if m 5* n. in Define / as follows: X = f(x) ( \ Prove that lim /(*) = ln xin An > An x not in 0, 0 for all a in for any n. [0, 1]. x-+a 23. why Explain the following definitions of lim f(x) = are / all correct: x—*a For every *24. 8 > (i) if 0 < i* (ii) if 0 < I* < < (iii) if 0 (iv) if 0 I* I* 0 there - Give examples a\ a\ a\ a\ to < < < < an is £ > 0 such that, for e, then \f(x) s, then \f{x) e, then |/(at) ~ - /| l\ /| e/10, then |/(*) < < < - show that the following all x, 5. 8. 56. /| < «. definitions of lim f(x) = I 5, then are x—*a not correct. (a) For l/W (b) For 0 25. all < £ > < > - a\ all 8 I* /I 0 there £ > 0 such that is a 6 > 0 such that in Problem 4-15 indicate and lim f(x) exist. is an if 0 \x — a\ |/(*) — l\ < < e. 0 there < if < £, then 8. For each of the functions for which numbers a the one-sided limits lim f(x) x—*a + *26. Do same x—*a~ 27. each of the functions in Problem 4-17. (b) Also consider what happens if decimals ending in 0's are used instead of decimals ending in 9's. Prove that lim f(x) exists if lim /(*) = lim /(*). 28. Prove that (a) the for x~*a + x~*a (i) lim f(x) (ii) lim /(H) x~»0 = = lim /(-*). lim/(x). x-*0+ x-*a~ = lim f(x 2 ) (iii) lim /(*). x-»0+ x->0 (These equations, and others like them, are open to several interpretaThey might mean only that the two limits are equal if they both tions. exist; is or that equal to equal to *29. or that < Suppose that lim f(x) < only if — < a m > What n. one easy limit is • • X—> • = • • do some algebra 0; • when m = n? When m > n? Hint: The the limit lim l/x k this some 8 > 0 such that f(x) < f(y) when8 and \y — a\ < 8. Is the converse true? + a 0 )/(b mx m + + b 0) exists if and is < a\ + is suitable.) lim /(*). (Draw a picture to illustrate Prove that there y and \x n Prove that lim (a n x x— 00 ever x and and is the other also exists exists, either limit exists, then the other exists if Decide for yourself which interpretations are it. assertion.) *30. a certain one of the limits if it; so that this is the only oo information you need. 31. Prove that lim f(l/x) = lim x—* x— + Define " lim /(*) = /." 32. X—> (a) — oo Find lim — X—* (b) (a n x n + • + • Prove that lim = /( x) We m + • * • + b 0 ). lim f(-x). x— oo — oo = Prove that lim f(\/x) lim f(x). x— — oo x— 33. a 0 )/(b m x 00 X—+ (c) /(*). oo » define lim f(x) = « mean to that for N all there is a 8 > 0 such x—*a that, for all x, if 0 < — \x < a\ 8, then f(x) > N. (Draw an appropriate picture!) (a) Show that lim \/{x - 3) 2 = oo. x-»3 (b) Prove that if /(*) > > £ 0 for all x, and lim g(x) = then 0, X—Kl 34. (a) Define lim f(x) \imf(x)/g(x) = oo. lim f(x) = oo, = oo ; x—*a + least and lim f(x) = oo. (Or at x—+ oo x-+a~ convince yourself that you could write down the definitions if How many other such symbols can you define?) you had the energy. (b) Prove that lim 1/x = oo. (c) Prove that lim f(x) = oo x->0 + if and only if lim f(\/x) x~+ 00 = oo. CONTINUOUS FUNCTIONS CHAPTER If / is an arbitrary function, it is not necessarily true that lim/W =/(*). x-*a In fact, there are many ways this can fail to be true. For example, / might not even be defined at a, in which case the equation makes no sense (Figure 1). Again, lim f(x) might not exist (Figure 2). Finally, as illustrated in Figure x—*tt 3, even defined at a and lim f(x) exists, the limit might not equal f(a). if / is -A (b) (a) FIGURE (c) 2 We would like to regard all behavior of this type as abnormal and honor, with some complimentary designation, functions which do not exhibit such peculiarities. The term which has been adopted is "continuous." Intuitively, a function / oscillations. is continuous Although whether a function worth cultivating) is it this if the graph contains no breaks, jumps, or wild continuous simply by looking at is you to decide graph (a skill well description will usually enable its easy to be fooled, and the precise definition is very important. DEFINITION The function / is continuous at a if lim/(*) We will have no difficulty finding or are not, continuous at some = /(„). many examples number a of functions which are, —every example involving limits provides an example about continuity, and Chapter 5 certainly provides enough of these. function f(x) = sin 1/x is not continuous at 0, because it is not even defined at 0, and the same is true of the function g(x) — x sin 1/x. On the The other hand, 93 if we are willing to extend the second of these functions, that is, if a 94 Foundations we wish to define a new G function by ^0 = (*sinlA, G(x) = x 0, then the choice of a = G(0) can be made in such a way that G will be continuous at 0 to do this we can (in fact, we must) define G(0) = 0 (Figure 4). — This sort of extension is not possible for F{x) then F will not be continuous at we /; if sin 1/x, x a, x define 0 = 0, no matter what 0, a is, because lim f(x) does x-*0 not exist. The function _ \ f( J x | is not continuous at a, a if rational x ratioi i rrational x irrati 0, \ since lim f(x) does not exist. 0, However, x—»a = lim f(x) 0 = /(0), so / continuous at precisely one point, is 0. x—>0 The functions /(*) numbers a, = = g(x) c, = and x are continuous at all since = lim f(x) = c — c lim x — a lim x—*a f(a), x—>a = lim g(x) x—* x—>a FIGURE *, 2 — lim A(*) lim x 2 = a — 2 g(a), = h{a). 4 Finally, consider the function }W f( In Chapter 5 \ — ~ we showed \ 0> irration x irrational \ l/q, x=p/qiin that lim f(x) = lowest terms. 0 for all a. Since 0 = }(a) only when x—*a a is irrational, this function is continuous at a if a is irrational, but not if a is rational. It is even easier to give examples of continuity if theorems. THEOREM l If / and g are continuous at (1) (2) Moreover, if g(a) ^ 0, then a, / + g i s continuous at a, f g is continuous at a. • then (3) 1/g is continuous at a. we prove two simple Continuous Functions PROOF Since / and g are continuous at lim/O) = By Theorem 2(1) of a, and f(a) +g = Starting with the functions /(*) we can Theorem use b„x = /(*) +g{a) = + g)(x) = }{a) which is just the assertion that / (2) and (3) are left to you. | at a, for every a, (/ + continuous at e and = *, a. proofs of parts The which are continuous to conclude that a function 1 fen-i*" /(*) + • • • +b + 0 c0 X C g(a). is -1 + n = lim g(x) 5 this implies that Chapter lim (/ 95 harder to get much further prove it will be easy to than that. When we discuss the sine function in detail funcA meanwhile. this fact that sin is continuous at a for all a; let us assume is continuous at every point in its domain. But it is tion like f{x) sin 2 x _ " ' + sin 27 x x2 + + 4x sin 2 x x 4 sin x 2 domain. But we are can now be proved continuous at every point in its we of a function like /(*) - sin(* ); still unable to prove the continuity functions. of continuous obviously need a theorem about the composition the definition of conabout point following the theorem, Before stating this Urn f(x) = f(a) according tinuity is worth noting. If we translate the equation to the definition of limits, for every if But in this case, 0 e < we obtain > 0 there |* - where the a\ limit 0 is <5 d, > |/(*) a\ - f(a)\ < e.- phrase f(a), the <\x - all x, 0 such that, for then < d be changed to the simpler condition may \x since if x = a it is - a\ < certainly true that \f(x) 5, - < f(a)\ e. continuous at o continuous at a, and / is continuous at g(a), then / g is a.) at not g(a), at continuous (Notice that / is required to be 2 Ug is PROOF Let £ THEOREM is < > 0. We wish to find a if |* -a\<5, S > 0 such that for then \(fog)(x) i.e., |/UW) - (f all x, ° g)(a)\ < 2. < e, a. We use continuity of / to estimate first how close g(x) that for to g(a) in order must be for this inequality to hold. Since /is continuous atg(a), there is a <5' > 0 such all y, if \y (1) - g(a)\ - f(g(a))\ < < 8', then < 6', then \f(g(x)) \f(y) e. In particular, this means that We now g(a)\ use continuity of g to estimate — the inequality \g{x) any other just like - if \g(x) (2) < g(a)\ positive 8' to - f(g(a))\ < how close x must be The number 8 is a f hold. number; we can therefore take definition of continuity of g at We a. conclude that there to a in order for positive 8' as a 8 is e. the > number £ (!) in the 0 such that, for all x, Combining and (2) can now (3) we < a\ 8, then \g(x) - g(a)\ < - f(g(a))\ < then \f(g(x)) 8'. S. | reconsider the function m_ ,/ x ~ We <5, see that for all x } \x-a\< if We - if \x (3) I \ x sin 1 0 x /x, x 0, = 0. have already noted that / is continuous at 0. A few applications of 1 and 2, together with the continuity of sin, show that / is also con- Theorems tinuous at a, for a ^ 0. Functions like f(x) = sin(* 2 + sin(* + sin 2 (# 3 ))) should be equally easy for you to analyze. The few theorems functions at a been related to continuity of single point, but the concept of continuity doesn't begin to be we some continuous on all points of is called a little differently; all on functions which are continu(a, b), then/ (a, b). Continuity on a closed interval must be defined a function / is called continuous on [a, b] if really interesting until ous at of this chapter have focus our attention interval. If /is continuous at x for all x in (1) /is continuous at x for (2) lim f(x) = f(a) all and lim x in f(x) = (a, b), f(b). £—>&~ x—*a + Functions which are continuous on an interval are usually regarded as might be specified as the first condition which a "reasonable" function ought to satisfy. A continuous function is sometimes described, intuitively, as one whose graph can be drawn without lifting your pencil from the paper. Consideration of the function especially well behaved; indeed continuity f (v /W \ _ | x sin \/x, x i 0, x 0 = 0 97 Continuous Functions shows that this description many that there are is a little too optimistic, but nevertheless true it is which are continu- results involving functions important ous on an interval. These theorems are generally much harder than the ones in this chapter, but there is a simple theorem which forms a bridge between the two kinds of results. The hypothesis of this theorem requires continuity at only a single point, but the conclusion describes the behavior of the function on some interval containing the point. Although this theorem is really a lemma for later THEOREM 3 arguments, included here as a preview of things to come. Suppose /is continuous at that /(*) is PROOF it is a > 0 for number 8 > a\ 0 such that f(x) < 0 for > if \x > 0 we can 0. - A < Then 8. a\ < 8, then take f(a) as the 8y then this last inequality implies f(x) all > \f(x) e. Thus )/(*) number a is if f(a) x satisfying <z, there is \x > if s - f(a)\ < < — > 8 0, a\ 0 there 0 such then there < is 8. a 8 > 0 £. 8 > 0 so that for all x, ~ f(a)\ < /(a), 0. similar proof can be given in the case f(a) can apply the there Similarly, Since/ is continuous at if\x-a\ < and 0. - x satisfying Consider the case f(a) such that, for all x, Since f(a) and f(a) > a, \x all < 0; take £ = —/(a). Or one case to the function —/. | first PROBLEMS 1. For which of the following functions / with domain (i) fix) = (ii) f(x) = R such that F(x) = is there a continuous function f(x) for all x in the F domain of/? X 2. 3. (iii) f{x) (iv) fix) = - 0, x irrational. \/q, x = p/q rational in lowest terms. At which points are the functions of Problems 4-1 5 and 4-17 continuous? Show that (a) Suppose that / is a function satisfying \f(x) < |*| for all x. equal must 0.) that (Notice at 0. /(0) / is continuous at any (b) Give an example of such a function / which is not continuous | a ^ 0. Suppose that g is continuous at 0 and g(0) = 0, and |/(*)| < \g(x)\. Prove that / is continuous at 0. Give an example of a function / such that / is continuous nowhere, but (c) 4. | /| is continuous everywhere. 98 Foundations 5. For each number any other 6. Find a function / which (a) tinuous at continuous at but not at a, is discontinuous at 1, £,. Suppose that . but con- . . other points, all /satisfies f(x at 0. Prove that 8. is Find a function / which is discontinuous at at 0, but continuous at all other points. (b) 7. which find a function 0, points. / + y) = f(x) + f(y), continuous at a for is 1 J, £-, , and that / is . . . , and continuous all a. Suppose that / is continuous at a and f(a) = 0. Prove that if a 9* 0, then a is nonzero in some open interval containing a. / (a) Suppose / is not continuous at a. Prove that for some number £ > 0 there are numbers x arbitrarily close to a with \f(x) — f(a) > 6. + 9. \ Illustrate graphically, (b) Conclude that arbitrarily close to a with f(x) 10. (a) (b) (c) Prove that is continuous at a, then so is there are there are + Prove that / and g are continuous, then if numbers x numbers x |/[ Prove that every continuous / can be written / = .E even and continuous and 0 is odd and continuous. min(/, (d) / if £ > 0 either < f(a) — £ or > f(a) + £. some number for arbitrarily close to a with f(x) so are 0, where E is max(/, g) and g). Prove that every continuous / can be written / = g — h, where g and h are nonnegative and continuous. 11. Prove Theorem f(x) *12. (a) = 1 by using Theorem 2 and continuity of the function (3) l/x. Prove that if /is continuous at / and lim g(x) = L then lim f(g(x)) x—*a /(/). (You can go sider the function (b) Show that ally true if right back to the definitions, but G with G(x) continuity of / at that lim f(g{x)) = / is 13. and/(/) (a) Prove that is if / is = g(x) for x ^ a, it is easier to con- and G(a) not assumed, then it is = L) not gener- /(lim g{x)). Hint: Try /(*) x—*a x^l, = = x—*a = 0 for x— 1. continuous on continuous on R, and which [a, b\ then there satisfies g(x) = is a function g which f(x) for all x in [a, b\ Hint: Since you obviously have a great deal of choice, try making g constant on (— <x> a] and [b, 00). 5 (b) 14. (a) Give an example to show that this assertion is false if [a, b] is replaced by (a, b). Suppose that g and h are continuous at a, and that g{a) = h (a). Define f(x) to be g(x) if x > a and h(x) if x < a. Prove that / is continuous at (b) a. continuous on [a, b] and h is continuous on [b c] and Let f(x) be g(x) for x in [a, b] and h(x) for x in [b, c]. Show that / is continuous on [a, c]. (Thus, continuous functions can be "pasted together.") Suppose g g(b) — is h(b). 9 Continuous Functions 15. (a) Prove the following version of Theorem 3 Suppose that lim nuity'': = f(x) fia), 99 for " right-hand conti- and f(a) > 0. Then there is a x—*a + number 6 > Similarly, fix) (b) < 0 for > 0 such that fix) if fia) all < 0, 0 for then there x satisfying 0 < x all is — a x satisfying 0 number a < 8 < x — a < d. > 0 such that 6. Prove a version of Theorem 3 when lim fix) = fib). x—>b~ 16. If lim fix) x—*a but exists, discontinuity at = sin is ^ fia), then / is 0 and /(0) = said to have a removable a. /x for # ^ does / have a removable (a) If fix) (b) ^ 0, and /(0) = 1? Suppose / has a removable discontinuity at a. Let gix) = /(*) for = lim/(x). Prove that g is continuous at a. * a, and let 1 discontinuity at 0? What if fix) = x sin 1 1 , fx for * x—»o (Don't work very hard; this (c) = 0 if What is Let /(*) terms. a: is quite easy.) let fip/q) = l/q if p/q is in lowest the function g defined by gix) = lim f(y)? is irrational, and y—*x *(d) Let / be a function with the property that every point of discontinuity is a removable discontinuity. This means that lim fiy) exists for y—*x all x, but / numbers is **(e) x. be discontinuous at some (even infinitely many) Define gix) = lim fiy). Prove that g is continuous. (This may not quite so easy as part (b).) and which worth thinking about this problem now, but mainly as a test of intuition; even if you suspect the correct answer, you will almost certainly be unable to prove it Is there a function / which is discontinuous at every point, has only removable discontinuities? at the present time. See (It is Problem 21-24.) CHAPTER THREE HARD THEOREMS This chapter devoted to three theorems about continuous functions, and The proofs of the three theorems themselves will not be given until the next chapter, for reasons which are explained at the end some of is their consequences. of this chapter. THEOREM 1 If / is continuous on such that /(*) - 0. and f(a) [a, b] < 0 < f(b), then there is some x in [a, b] means that the graph of a continuous function which below the horizontal axis and ends above it must cross this axis at some (Geometrically, this starts pointy as in Figure 1.) THEOREM 2 If/ is continuous on some number then /is bounded above on b], [a, N such that /(*) (Geometrically, this theorem < N for means all x in [a, [a, b], that is, there is b]. that the graph of / lies below some line parallel to the horizontal axis, as in Figure 2). THEOREM 3 If / is continuous on [a, f(y) >/(*) for all x in b], [a, then there is some number y in [a, b] such that b] (Figure 3). These three theorems differ markedly from the theorems of Chapter 6. The hypotheses of those theorems always involved continuity at a single point, Three Hard Theorems 101 while the hypotheses of the present theorems require continuity on a whole interval [a, £]— if continuity fails to hold at a single point, the conclusions may fail. For example, let / be the function shown in Figure 4, Then / is continuous but there FIGURE 0-<x<V2 V2 < x < 2. /M V2 is 4 single point at every point of no point x V2 is in [0, 2] - V2, and/(0) < / shown the function is /(*) 0, x ^ 0 x = 0. < 0 f(2), 0; the discontinuity at the sufficient to destroy the conclusion of Similarly, suppose that N except [0, 2] such that /(*) Theorem 1 in Figure 5, but / is not bounded > 0, we have f(£$) = 2N > N. above on [0, 1], In fact for any number This example also shows that the closed interval [a, b] in Theorem 2 cannot be replaced by the open interval (a, b), for the function /is continuous on (0, 1), Then / is continuous at every point of [0, 1] except 0, N • but not bounded there. is Finally, consider the function /« FIGURE 5 On the interval [0, 1] conclusion of Theorem " 2, x 0, x is even though / Theorem fix) for all x in [0,1]; in fact, in Figure 6, x\ the function / not satisfy the conclusion of /GO > shown it is < > 1 1. bounded above, is so / does not continuous on 3— there [0, 1]. satisfy the But / does in [0, 1] such that certainly not true that/(l) > /(*) for is no y we cannot choosey = 1, nor can we choose 0 < y < 1 because 1 f{y) < f(x) if x is any number with y < x < This example shows that Theorem 3 is considerably stronger than Theorem function on 2. Theorem 3 is often paraphrased by saying that a continuous all x in [0, 1] so a closed interval "takes on its maximum value" on that interval. As a compensation for the stringency of the hypotheses of our three theorems, the conclusions are of a totally different order than those of previous theorems. They describe the behavior of a function, not just near a point, but on a whole more interval; such "global" properties of a function are always significantly prove than "local" properties, and are correspondingly of much greater power. To illustrate the usefulness of Theorems 1, 2, and 3, we will soon deduce some important consequences, but it will help to first mention difficult to FIGURE 6 some simple generalizations of THEOREM 4 If / is continuous on such that fix) = c. [a, b] and these theorems. f{a) < c < fib), then there is some x in [a, b] PROOF Let g there THEOREM 5 If / i s = is f Then g c. some x in = /(at) continuous, and g(a) is b] [a, continuous on such that PROOF — such that g{x) [a, b] ' and /(a) > c = 0. > fib), < But 0 gib). By Theorem 1, means that /(*) = c. | < this then there some x is in [a, b] c. The function — / is continuous on [a, b] and —f(a) < —c < —f(b). By Theorem 4 there is some x in [a, b] such that —f(x) = — c, which means that f(x) = c. | Theorems 4 and 5 together show that / takes on any value between f{a) and We can do even better than this: if c and d are in [a, b], then / takes on any value between f(c) and fid). The proof is simple; if, for example, c < d, then just apply Theorems 4 and 5 to the interval [c, d]. Summarizing, if a continuous function on an interval takes on two values, it takes on every value f(b). in between; this slight generalization of Theorem 1 is often called the Inter- mediate Value Theorem. THEOREM 6 PROOF If / is continuous on [a, b], then / is bounded below on some number TV such that f(x) > N for all x in [a, b]. The function — / is continuous on so [a, b], M such that -/(*) < for all x in [a, b]. But = — M. | x in [a, b\ so we can let [a, b\ that is, there is M by Theorem 2 there is a number this means that /(*)" > -AT for all N Theorems 2 and 6 together show that a continuous function / on [a, b] is bounded on [a, b], that is, there is a number N such that \f{x) < N for all x in [a, b]. In fact, since Theorem 2 ensures the existence of a number Ni such that f{x) < Ni for all x in [a, 6], and Theorem 6 ensures the existence of a number \ N THEOREM 7 2 such that fix) If /is > continuous on for all x in [a, N 2 for all x in [a, b], [a, b], then there is we can take somejy in N= [a, b] max(|#i|, \N 2 \). < such that/(j) /W b]. (A continuous function on a closed interval takes on its minimum value on that interval.) PROOF The function —/is continuous on such that all x in —fiy) > b\ —fix) for 8 [a, b] x in by Theorem 3 there is some; which means that f(y) < the trivial consequences of Theorems in [a, b] fix) for 1, 2, and 3, interesting things. Every positive number has a square is \ [a, b], | Now that we have derived we can begin proving a few THEOREM all some number x such that x 2 = root. In other words, if a. a > 0, then there Hard Theorems Three PROOF 103 Consider the function /(*) = * 2 which is certainly continuous. Notice that the statement of the theorem can be expressed in terms of/: "the number a , The proof has a square root" means that / takes on the value a. about / will be an easy consequence of Theorem 4. There number b > 0 such that f(b) > a (as illustrated in > 1 we can take b = a, while if a < 1 we can take < f(b) Theorem 4 applied to [0, b] implies that for we have /(*) = a, i.e., x = a. | obviously a is Figure 7); in fact, if a b = 1. Since /(0) < a some x of this fact y 2 (in [Q, b]), same argument can be used to prove that a positive number has any natural number n. If n happens to be odd, one can do better: every number has an nth root. To prove this we just note that if the posin = n = —a (since n is tive number a has the nth root *. i.e., if x a, then ( — x) odd), so —a has the nth root —x. The assertion, that for odd n any number a Precisely the an nth root, for has an nth root, equivalent to the statement that the equation is xn has a root if n - a = odd. Expressed in this is 0 way the result is susceptible of great generalization. THEOREM 9 If n is odd, then any equation xn + an ^x n ~l + + a0 - 0 has a root. PROOF We obviously want to consider the function fix) = x n + a n ^ 1x n~ 1 + we would like to prove that / is sometimes positive and sometimes negative. The intuitive idea is that for large \x\, the function is very much like g(x) = x n and, since n is large negative The proper fix) = xn odd, this function x. A algebra little is positive for large positive x is all we need analysis of the function + an to make and negative this intuitive / depends on writing ^x n ~ Note that < + [ + • • Igoj n l* Consequently, if we choose (*) then |**| > |*| |*| x satisfying > 1, 2n|a n _i|, 5 2w|a 0 and \a n -k\ |gn-fc| |**| |*| < 2n\a n - k 5 2^ \ | : l for idea work. 1_ xn 2n In 2 n terms In other words, which implies that Therefore, if we choose an 2 > so that f(xi) 0. *i Now < an x in On it then xn/ if < x2 0 satisfies (*), then x 2 n < 0 and *2 / X2 0. 1 to the interval [x 2 x{] , such that /(*) 9 disposes of the frustrating to leave the Some equations, At like x —what more is 2 first — 1 that there is problem of odd degree equations so happily that problem of even degree equations comsight, however, the problem seems insuperable. = 0, have a solution, and some, like x 2 + 1 =0, Theorem pletely undiscussed. we conclude «= 0. | would be do not satisfies (*), xi the other hand, applying Theorem [x2 , xt ] 0 which \ \ so that f(x 2 ) > there to say? If we are willing to consider a more general question, however, something interesting can be said. Instead of trying to solve the equation xn let + a n -ix n -1 + +a « • • 0 ; 0, us ask about the possibility of solving the equations xn + an^x^ + 1 • • • +a 0 = c numbers c. This amounts to allowing the constant term a 0 to information which can be given concerning the solution of these for all possible vary. The equations depends on a fact which is illustrated in Figure 8. ^1 71 an-ix a 0 with n even, The graph of the function /(*) = x n contains, at least the way we have drawn it, a lowest point. In other words, there is a number y such that f(y) < f(x) for all numbers x the function / + + • • • + , — on a minimum value, not just on each closed but on the whole line. (Notice that this is false if n is odd.) The proof depends on Theorem 7, but a tricky application will be required. We can apply Theorem 7 to any interval takes interval, Three Hard Theorems 105 and obtain a point y 0 such that f(y 0 ) is the minimum value of/ on [a, 6]; but if [a, ft] happens to be the interval shown in Figure 8, for example, then the point y 0 will not be the place where / has its minimum value for the whole line. In the next theorem the entire point of the proof is to choose an interval THEOREM io [a, ft], [a> ft] If n is in < As in the that this cannot happen. xn = even and f(x) such that f(y) PROOF way such a + <2 Theorem proof of all + + <z x with 2«|<2 n_i| even, x n provided that such that b n |^| > > 0 for 2 ~ > Af. < xn ?)-/<->. x \ Now consider the number /(0). Let > 0 be a number we have (Figure 9) and also b > M. Then, if * > 2/(0) x number y so all x, ft ft, xn if a 2n|fl 0 |), , x fix) Similarly, is > M, we have |x| 2 is then there 9, if !<!+— + Since n o, f(x) for all x. M — max(l, then for n_1 n -i* b n >|>y >/(0). then —ft, Summarizing: if > ^ ft or * < -ft, Now apply Theorem 7 to the function clude that there is a number y such that if (1) In particular, f(y) (2) THEOREM 11 < if Combining (1) and Theorem 10 now /(0). x < (2) -ft < < ft, /on /(0). the interval then /GO [ > ft, then /(*) see that f(y) < > /(0) > /(>). f(x) for all x. | Consider the equation Xn ft, < /(at). allows us to prove the following result. (*) — Thus -ft or x we x > then /(*) + dn^x"- 1 + + ao = c, ft]. We con- 106 Foundations and suppose n is even. Then there is a number m such that for c > m and has no solution for c < m. PROOF Let /(*) = + an-ix*- + xn + 1 (*) has a solution a 0 (Figure 10). According to Theorem 1 0 there is a number y such that f(y) < f(x) for all Let m — /(>')• If c < m, then the equation (*) obviously has no solution, since the left side always has a value > m. If c — m, then (*) has y as a solution. Finally, suppose c > m. Let b be a number such that b > y and f(b) > c. x. Then f(y) = m < ber x in [y, b] c < f(b) Consequently, by . such that f(x) = c, so x is Theorem a solution of 4, there is some num- (*). | These consequences of Theorems 1,2, and 3 are the only ones we will derive (these theorems will play a fundamental role in everything we do later, however). Only one task remains to prove Theorems 1, 2, and 3. Unfortunately, we cannot hope to do this on the basis of our present knowledge about the real numbers (namely, PI -PI 2) a proof is impossible. There are several ways of convincing ourselves that this gloomy conclusion is actually the case. For example, the proof of Theorem 8 relies only on the proof of Theorem if we could prove Theorem 1, then the proof of Theorem 8 would be com1 plete, and we would have a proof that every positive number has a square root. As pointed out in Part I, it is impossible to prove this on the basis of PI -PI 2. now — — ; Again, suppose we consider the function FIGURE 1 10 x 2 - 2 If there were no number x with x = 2, then / would be continuous, since the denominator would never = 0. But /is not bounded on [0, 2]. So Theorem 2 depends essentially on the existence of numbers other than rational numbers, and therefore on some property of the reals other than PI -PI 2. Despite our inability to prove Theorems 1, 2, and 3, they are certainly results which we want to be true. If the pictures we have been drawing have any connection with the mathematics we are doing, if our notion of continuous function corresponds to any degree with our intuitive notion, Theorems 1, 2, and 3 have got to be true. Since a proof of any of these theorems must require some new property of R which has so far been overlooked, our present diffi2 culties suggest proof of One FIGURE 11 that is, way 1, to discover that property: let us try to construct a for the smallest x in [a, b] is to locate the first point such that /(*) = 0. To where f(x) find this point, first = 0, con- A which contains In Figure all numbers x in [a, b] such that / is negative on such a point, while x' is not. The set A itself is indicated Since / is negative at <2, and positive at b, the set A contains 1 1 , x is by a heavy line. some points greater than (We example, and see what goes wrong. idea which seems promising sider the set [a, x]. a Theorem a, while all points sufficiently close to are here using the continuity of /on [a, b] 9 as well as b are not in A. Problem 6-15.) Hard Theorems Three fix) < 0 for all x in this interval 107 Now suppose a is the smallest number which is greater than all members of < a < b. We claim that /(a) = 0, and to prove this we only have A; clearly a < 0 and /(a) > Then, by Theorem 0. to eliminate the possibilities f(a) Suppose first that /(a) < 0. 6-3, /(*) would be less than 0 for all x in a small interval containing a, in particular for some numbers bigger than a (Figure 12); but this contradicts the fact that a is bigger than A would all every also contain /(a) these points On FIGURE member of A, < 0 since the larger numbers would also be in A. Consequently, is false. the other hand, suppose /(a) > Again applying Theorem 0. 6-3, we would be positive for all a: in a small interval containing a, in particular for some numbers smaller than a (Figure 13). This means that these smaller numbers are all not in A. Consequently, one could have chosen an even smaller a which would be greater than all members of A. Once again we have a contradiction; f(a) > 0 is also false. Hence f(a) = 0 and, we are see that f(x) 12 tempted to say, Q.E.D. We know, however, that something must be wrong, since no new properties or R were ever used, and it does not require much scrutiny to find the dubious than all point. It is clear that we can choose a number a which is greater clear that = so not is but it b), members of A (for example, we can choose a x numbers > 0 all of consists A suppose we can choose a smallest one. In fact, V 2 did not exist, there would not be a least number number greater than all the members of A for any y > V2 we chose, we could such that x 2 < 2. If the ; f{x) > 0 for all in this interval x always choose a Now that still we have smaller one. tional property of the real and use FIGURE 13 it. That is almost obvious what addineed. All we must do is say it properly discovered the fallacy, numbers we it is the business of the next chapter. PROBLEMS 1. For each of the following functions, decide which are bounded above or below on the indicated interval, and which take on their maximum or minimum value. (Notice that/ might have these properties even if/ is not continuous, and even (i) /(*) (ii) f(x) (iii) f(x) (iv) /(*) = = = = x* x2 x2 if the interval is not a closed interval.) on (-1, 1). on (-1, 1). on R. on [0, 00). x x > a on {-a - 1, a + a sary to consider several possibilities for a.) 1). (It will be neces- = sf \ (via) f(x) /•x r, (ix) f = v /(*) xf x )\ 1, x irrational -i/q, x 2. /(*) > # on . * irrational 0, sin 2 (cos on [x] * (i) /(*) (ii) /(*) + V] + rn (iv) /(*) *i79 *5 n [0, a], a 2) on [0, a 3 ]. sin 4 2 If n is = — such n 1. some number x such that + x + sin 2 x This problem (a) an integer 1, 1^3 + x find 5 1 (ii) [0, 1]. *3 Prove that there (i) rrk [0, a]: - x + 3. + 5x + 2x + x + x+ 1. 4* - 4* + = = = = in [0, 1]. on m lowest terms p/q For each of the following polynomial functions /, that f(x) = 0 for some x between n and n + 1 (hi) f{x) 3. = = f(x) (xii) = x rational x, f rn on . in lowest terms f ( (xi) = 1 { = \ f(x ) # irrational x ( I f ( 9 /£, \ k — = n9 # 1 a continuation of Problem 3-7. is is 2 even, and > 0, find a polynomial function of degree n with exactly k roots. (b) A if root a of the polynomial function / is said to have multiplicity m m = (x a) g(x), where g is a polynomial function that does — f(x) have a as a root. Let / be a polynomial function of degree n. Suppose that /has k roots, counting multiplicities, i.e., suppose that not k is sum the of the multiplicities of all the roots. Show that n — k is even. 5, 5. Suppose that / is continuous on [a, b] and that f(x) is always rational. What can be said about / ? Suppose that/is a continuous function on [—1, 1] such that* 2 + (/(*)) 2 = 1 for all x. (This mean s that ( *, f(x)) always lies on the unit circle.) Show . How many for all x i. K V V 2 2 = 1 - x for all x, or else /(*) = 1 - x for continuous functions / are there which satisfy (/(*)) 2 that either f{x) 7 all *. = x2 ? Suppose that / and g are continuous, that f 2 = g 2 and that /(#) ^ 0 for all x. Prove that either f(x) = g (x) for all x, or else /(*) = — g(x) for all #. (a) Suppose that / is continuous, that f(x) = 0 only for x = a, and that , /(*) > 0 for some x > show that x 2 + xy some x < a as well as for about f(x) for all x ^ a ? *(b) Using only the fact that x 2 +y > 2 to consider various fixed y + xy + y ^ 2 0 ^ if 0, x and jy so that 0 if a. What can x and be said are not both 0, (The trick you can define a function.) are not both 0. is ; *(c) Discuss similarly the sign of x not both 10. + 3 x 2y + +y xy 2 109 Hard Theorems Three z when x and y are 0. Suppose / and g are continuous on [a, b] and that f(a) < g(a), but > g(b). Prove that f{x) = g(x) for some x in [a, b). (If your proof f(b) very short, isn't 11. 12. it's not the right one.) Suppose that /is a continuous function on [0, 1] and that /(*) is in [0, 1] for each x (draw a picture). Prove that /(*) = x for some number x. (a) Problem 1 1 shows that / intersects the diagonal of the square in Figure 14 (solid line). Show that / must also intersect the other (dashed) diagonal. Prove the following more general fact: If g and^(0) = 0,g(\) = 1 or^(0) = 1,^(1) = (b) some 13. continuous on is then /(*) 0, = [0, 1] g(x) for x. 0 and let /(0) = 0. Is / continuous on [ 1] ? Show that / satisfies the conclusion of the Intermediate Value Theorem on [— 1, 1]; in other words, if / takes on two values = Let /(*) (a) \jx for x sin ^ — 1, somewhere on [ — 1, 1], it also takes on every value in between. Suppose that / satisfies the conclusion of the Intermediate Value Theorem, and that / takes on each value only once. Prove that / is *(b) continuous. Generalize to the case where /takes on each value only finitely *(c) many times. 14. If / of a continuous function on is !/| on 1], let [0, Prove that for any number c we have < ||/|| *(b) Prove that \\f g\\ (a) \\f (c) *15. + g\\ (b) Hint: *\7. </> 11/11 is + + + hi - /|] < \\k \\cf\\ \\h - g\\ + continuous and lim \\g - cf)(x)/x = Give \c\ • ||/||. an example where = lim f\\. n = 0 <j)(x)/x <f> 0) *16. value n . n Prove that if /us odd, then there is a number* such that x + (x) = n that y such number Prove that if n is even, then there is a y 000 < (*, * Prove that Suppose that (a) maximum be the ||/|| 1]. [0, xn + </>(*) Of which for all 0. + x. proofs does this problem test your understanding? Let / be any polynomial funcdon. Prove that there is some number y such that 1/001 < |/(*)| for all x. Suppose that / is a continuous function with f(x) > 0 for all x, and lim f(x) = 0 = lim f(x). (Draw a picture.) Prove that there is some X— x—* 00 — 00 y such that f(y) > f(x) for all x. (a) Suppose that / is continuous on [a, b], number *18. and let x be any number. Prove that there is a point on the graph of / which is closest to (x, 0) in other words there is some y in [a, b] such that the distance from < (x, 0) to (y, f(y)) [a, b\ (See Figure 15.) is distance from (x, 0) to (z, f(z)) for all z in (b) Show same that this replaced by (c) Show (d) In cases assertion that the assertion and (a) (c), is "19. (a) true if [a. b] is replaced by minimum letg(x) be the a point on the graph of (e) not necessarily true is b] if [a, is throughout. (a> b) Prove that g(y) /. R throughout. distance from < g(x) + \x (*, 0) to — y\, and conclude that g is continuous. Prove that there are numbers x 0 and x x in [a, b] such that the distance from Oo, 0) to Oi,/(*i)) is < the distance from (*</, 0) to (xi\f(xi)) for any *</, x\ in [a, b\ .Suppose that /is continuous on and /(0) = /(l). Let n be any some number x such that [0, 1] natural number. Prove that there is = f(x 1/n), as shown in Figure 16 for n = 4. Hint: Consider the function g(x) = f(x) — f(x 1/n); what would be true if g(x) ^ 0 for all x ? + f(x) + (b) Suppose 0 number satisfies any 20. (a) < < a , but that a not equal to is Why? b). {a, In the than f(a) < implies that f(x) } first any natural case all values close to f(a), but are taken on f(a) for x < a and x > somewhere in (a, b); this b. Refine part (a) by proving that there is no continuous function / which takes on each value either 0 times or 2 times, i.e., which takes on exactly twice each value that it does take on. Hint: The previous hint implies that / has either a value (which must be taken on twice). values close to the maximum maximum or a minimum What can be said about value? Find a continuous function / which takes on every value exactly 3 times. More generally, find one which takes on every value exactly n times, if n (d) /n for x. x in all (c) 1 Prove that there does not exist a continuous function / defined on R which takes on every value exactly twice. Hint: If f(a) — f{b) for a < b, then either f(x) > f(a) for all x in (a, b) or f(x) < f{a) for slightly larger (b) 1 Find a function / which is continuous on [0, 1] and which /(0) = /(l), but which does not satisfy f(x) = f(x + a) for n. Prove that is if odd. n is even, then there is every value exactly n times. Hint: example, all x in two f(x) < 0 for To treat the case n all f{x 2 ) = — 4. for /(* 8 ) = f(x A ). Then either /(*) > 0 for of the three intervals (x h # 2 ), (*2, # 8 ), (#3, ^4), or else let f(x{) = no continuous / which takes on x in two of these three intervals. LEAST UPPER BOUNDS CHAPTER This chapter reveals the most important property of the real numbers. Nevertheless, it is merely a sequel to Chapter 7; the path which must be followed has already been indicated, and further discussion would be useless delay. A DEFINITION set A numbers of real bounded above is > x Such a number x is if there a number x such that for every a in A. a an upper bound called is for A. Obviously A is bounded above if and only if there is a number x which is an upper bound for A (and in this case there will be lots of upper bounds for A) we often say, as a concession to idiomatic English, that "^4 has an upper ; bound" when we mean that there number which is an upper bound for A. now been used in two ways functions, and now in reference to sets. This is a Notice that the term "bounded above" has first, in Chapter 7, in reference to always be clear whether the two definiMoreover, of numbers or a function. dual usage should cause no confusion, since we are talking about a set tions are closely connected: / is bounded above on The if A [a, b] if entire collection R bounded above will the set {f(x): a and only if the set < x < b}, then the function A is bounded above. and the natural numbers N, are bounded above. An example of a set which of real numbers, both examples of sets which are is is it not is A = {x: 0 < x < 1}. bounded above we need only name some upper bound for A, which is easy enough; for example, 138 is an upper bound for A. and so are phrase 2, l£, 1-^, and 1. Clearly, 1 is the least upper bound of A; although the confusion possible avoid any in order to just introduced is self-explanatory, (in particular, to ensure that we all know what the superlative of "less" To show that A is means), we define DEFINITION A number x is a least and Ill this explicitly. upper bound (1) x is (2) if y of A if an upper bound of A, an upper bound of A, then x is < y. The use of the indefinite article "a" in this definition was merely a conces- sion to temporary ignorance. Now that we have made a precise definition, it is if x andjv are both least upper bounds of A, then x = y. Indeed, easily seen that in this case x and y it < < y, since y is an upper bound, and x is a least upper bound, x, since x is an upper bound, and y is a least upper bound; = follows that x For y. The term supremum of reason we speak of the least upper bound of A. synonymous and has one advantage. It abbrevi- this A is ates quite nicely to A sup and (pronounced "soup A") saves us from the abbreviation \\xbA is nevertheless used by some authors). There is a series of important definitions, analogous to those just given, which can now be treated more briefly. A set A of real numbers is bounded (which below if there is a number x such that < x for every a in A. a Such a number x is called a lower bound lower bound of A if and The (1) x is (2) if y greatest lower bound for A. A number x is the greatest a lower bound of A, is of a lower A is bound of A, then x also called the infimum > y. of A, abbreviated inf A; some authors use the abbreviation gib A. One detail has been omitted from our discussion so far —the question of which have at least one, and hence exactly one, least upper bound or greatest lower bound. We will consider only least upper bounds, since the question for greatest lower bounds can then be answered easily (Problem 2). HA is not bounded above, then A has no upper bound at all, so A certainly cannot be expected to have a least upper bound. It is tempting to say that A does have a least upper bound if it has some upper bound, but, like the principle of mathematical induction, this assertion can fail to be true in a rather special way. If A = 0, then A is bounded above. Indeed, any number x is an upper sets bound for 0: x simply because there there is surely no is least noy > y in 0. for every y in 0, Since upper bound number is an upper bound for 0, With this trivial exception however, every for 0. 113 Least Upper Bounds our assertion true^and very important, is warrant consideration of details. enough definitely important to We are finally ready to state the last property numbers which we need. of the real (The (PI 3) A upper bound property) If A is a set of real numbers, and A is bounded above, then A has a least upper bound. least y£ 0, may strike you as anticlimactic, but that is actually one of its To complete our list of basic properties for the real numbers we require Property PI 3 virtues. no particularly abstruse proposition, but only a property so simple that we might feel foolish for having overlooked it. Of course, the least upper bound FIGURE property 1 rational not really so innocent as is numbers | < 2, A and which is satisfying x 2 for all that; after all, it does not hold for the For example, if A is the set of all rational numbers x then there is no rational number y which is an upper bound Q,. an upper bound PI 3 but is. we than or equal to every other rational number which is become clear only gradually how significant less for A. It will are already in a position to demonstrate the proofs which were omitted in Chapter THEOREM 7-1 PROOF If / is continuous on x in [a, Our proof and [a, b] = such that f(x) b] f(a) < 0 < its power, by supplying 7. f(b), then there is some number 0, merely a rigorous version of the outline developed at the end of [a, b] with f(x) = 0. Define the set A, shown in Figure 1, as follows: Chapter Clearly A = {x: a A ^ 0, all points is is 7—we will locate the smallest number x in < < x since a is continuous on is and / A; in x < and f(a) [a, b] > a 8 in < x satisfying a in fact, there b, is negative on the interval fact, there is + a < 8; this some all > x]}. 0 such that A contains Problem 6-15, since/ an upper b^und for A and, follows from 0. Similarly, b is 0 such that 8 [a, points x satisfying b — 8 < x < b are from Problem 6-15, since f(b) > 0. From these remarks it follows that A has a least upper bound a and that a < a < b. We now wish to show that f(a) = 0, by eliminating the possibilities f(a) < 0 and f(a) > 0. Suppose first that f(a) < 0. By Theorem 6-3, there is a 8 > 0 such that upper bounds for A; this also follows < 0 for a -~ 8 < A which satisfies a — f(x) x 8 < a + 8 (Figure 2). Now there is some number x 0 in < x 0 < a (because otherwise a would not be the least upper bound of A). This means that / is negative on the whole interval [a, xo]. But if X\ is a number between a and a + 5, then / is also negative on the whole interval [x 0 x\]. Therefore / is negative on the interval [a, x{], so x\ is in A. But this contradicts the fact that a is an upper bound for A; our original assumption that f(a) < 0 must be false. Suppose, on the other hand, that f(a) > 0. Then there is a number 8 > 0 , such that f(x) that there FIGURE 3 is negative on >0foro; — an x 0 [a, in A 8 < x satisfying x 0 ], which is < a a — + 8 8 (Figure 3). Once again we know < means that / is Thus the assumption x0 < impossible, since f(xo) a; but this > 0. 114 Foundations f(a) > 0 also leads to a contradiction, leaving f(a) = 0 as the only possible alternative. The result, proofs of which Theorems will play 2 and 3 of Chapter 7 require a simple preliminary as Theorem 6-3 played in the previ- much the same role ous proof. THEOREM l If / is continuous at a, then there above on the interval (a — 8, a is + PROOF Since lim f(x) = f(a), there is, number a 8) (see for every s 8 > Figure > 0, 0 such that / is bounded a 8 4). > 0 such that, for all x, x—*a - if |* It is only necessary to apply do), for example, £ = if It follows, in < a\ this 8, then f(a) 1 . e. statement to some particular e (any one will . particular, that + - f(a)\ < We conclude that there is a 8 > 0 such that, for all x, \x-a\< 8, then |/(*) - f(a)\ < 1. 1 if completes the proof: on the interval above by \f(x) \x — a\ < 5, — 8, a + (a then f(x) 8) — f(a) < the function /is 1. This bounded | It should hardly be necessary to add that we can now also prove that / is bounded below on some interval {a — a + 5), and, finally, that / is bounded on some open interval containing a. A more significant point is the observation that if lim f(x) = f(a), then <5, x-*a + 1 1 1 a FIGURE 4 — 5 a a + 5 .' 115 Least Upper Bounds there is > a 6 0 such that / similar observation holds bounded on the set \x\ a < x < a + 8\, and a = f(b). Having made these observations is lim f(x) if x— (and assuming that you theorem. THEOREM 7-2 If / continuous on is supply the proofs), we tackle our second major will then / [a, b], is bounded above on [a, b]. Let A — Clearly A ^ both to the i.e., < x b and / bounded above on is x]}. [a, is in A), and A is bounded above (by 6), so A has a least Notice that we are here applying the term ''bounded above' A which can be visualized as lying on the horizontal axis, and 0 (since a upper bound to /, < [x: a a. set y to the sets {f(y): a < < y x}, which can be visualized as lying on the vertical axis (Figure 5). Our first step is to prove that we actually have a = b. Suppose, instead, that a < 6. By Theorem 1 there is 5 > 0 such that /is bounded on (a — a + 8). Since a is the least upper bound of A there is some 0 in A satisfying a — 8 < x 0 < a. This means that / is bounded on [a, x 0 ]. But if xi is any number with a < xi < a + 8, then / is also bounded on [x 0 xi]. Therefore /is bounded on [a, #i], so xi is in A, contradicting the fact that a is an upper bound for A. This contradiction shows that a — b. One detail should be mentioned: this demonstration implicitly assumed that a < a (so that / would be defined on some interval (a — d, a + 5)); the possibility a = a can be ruled out similarly, <5, ,v , using the existence of a The proof for every x is < 6 > 0 such that /is not quite complete by bounded on —we only know that not necessarily that /is bounded on [a, \x: a < < x a + 8} bounded on [a, x] b\ However, only one / is small argument needs to be added. There A #o in [#o, b], To THEOREM 7-3 If / is is a 8 so / is bounded on We [a, b], continuous on already [a, b], know bounded. This a > x [a, < b} . There is x 0 ] and also on then there is a number y in trick. [a, b] such that /(y) > [a, b]. that / is bounded on {/(*):* in is < | prove the third important theorem we resort to a f(x) for all x in PROOF 0 such that / is bounded on {x: b — 8 b — 8 < x 0 < b. Thus / is bounded on > such that set is f(x) for x in [a, b] obviously not it Suppose instead that a suffices to ^ f(y) 0, b], [a, b]} so it which means that the jv by fix) in set has a least upper bound a. Since show that a for all a - [a, = [a, b]. f(y) for Then some y in the function g [a, b\ denned 116 Foundations continuous on is the other hand, is denominator of the right side since the [a, b], a the least upper bound of { f(x) x in : never is [a 9 b] } ; this 0. On means that > for every £ means This, in turn, means that g this theorem. is x in [a, b] with a — < f(x) e. that > for every £ But 0 there 0 there x in is not bounded on is with g(x) [a y b] > 1 /e. contradicting the previous [a, b], | At the beginning of this chapter the set of natural numbers N was given as an example of an unbounded set. We are now going to prove that N is unbounded. After the difficult theorems proved in this chapter you may be startled to find such an "obvious" theorem winding up our proceedings. If so, you are, perhaps, allowing the geometrical picture of R to influence you too strongly. "Look," you may say, "the real numbers look like — - I 12 — number x 1 0 so every 1 I • * —— + i 1 • 3 is i x n n between two integers n, 1 + n 1 ger)." Basing the argument on a geometric picture (unless x is itself an inte- not a proof, however, and even the geometric picture contains an assumption: that if you place unit segments end-to-end you will eventually get a given segment. This axiom, often omitted from a etry, is segment larger than any first introduction to geom- usually attributed (not quite justly) to Archimedes, and the corre- is sponding property for N numbers, that is not bounded, is called the Archi- consequence of PI -PI 2 (see reference [17] of the Suggested Reading), although it does hold for Q,, of course. Once we have PI 3 however, there are no longer any problems. median property of the reals. This property THEOREM 2 PROOF N is not a not bounded above. is Suppose N were bound a for bounded above. Since N ^ 0, there would be a least upper N. Then a > n ct > n is in N. But for all n in N. for all n in N, Consequently, since + 72 1 is in N if n a — and this that a means that a is — 1 is 1 + > also 1 this means that for all n in N, an upper bound for the least upper bound. | n N, contradicting the fact \ Least Upper Bounds 117 There is a consequence of Theorem 2 (actually an equivalent formulation) which we have very often assumed implicitly. THEOREM 3 PROOF For any > £ 0 there a natural is number Suppose not; then 1/n > £ for all n is an upper bound means that 1/s in n with 1 /n < £. < 1 /s for all n in N. But N, contradicting Theorem 2. | N. Thus n for this A brief glance through Chapter 6 will show you that the result of Theorem 3 was used in the discussion of many examples. Of course, Theorem 3 was not available at the time, but the examples were so important that in order to give them some cheating was honesty but if we can claim tolerated. As partial justification for this dis- that this result was never used in the proof of a theorem, your faith has been shaken, a review of all the proofs given so far order. Fortunately, such deception will not be necessary again. stated every property of the real no more numbers that we We will ever need. is in have now Henceforth, lies. PROBLEMS 1. Find the upper bound and the greatest lower bound (if they exist) sets. Also decide which sets have greatest and least eledecide when the least upper bound and greatest lower bound least of the following ments (i.e., happens to belong (i) |-:rcinNj- (ii) (-: n in to the set). Z and n 0 or x = ^ 0 • \n (iii) {x: x (iv) {x: 0 (v) \x: x2 (vi) \x: x2 (vii) \x: x (viii) 2. (a) |- + = < + + < 1/n for some n in N}. < V2 and > 0}. x + x - 1 < 0}. x x is rational}. 1 0 and x 2 (»l) n : +x- n in 1 < 0}. nJ- Suppose A ^ 0 is bounded below. Let —A denote the set of all —x x in A, Prove that — A ^ 0, that — A is bounded above, and that for — (b) sup ( — ^4) is the greatest lower bound of A. A 7* 0 is bounded below, let B be the set of all lower bounds of A. Show that B 7* 0, that B is bounded above, and that sup B is the If greatest lower bound of A. 118 Foundations 3. Let/ be a continuous The proof (a) function on Theorem of [a> b] showed that there 1 < with }(a) is 0 < f(b). a smallest x in [a, 6] with /(#) = 0. Is there necessarily a second smallest x in [a, with f(x) = 0 ? Show that there is a largest * in [a, 6] with/(x) = (Try to give an easy proof by considering a new 6] 0. function closely related to /.) depended upon consideration of A = {x: a < x < b and / is negative on [a, x]}. Give another proof of Theorem 1, which depends upon consideration of B = {x: a < x < b and /(*) < 0}. Which point x in [a, b] with /(*) = 0 will this proof locate? Give an example where the sets A and B are not the same. The proof (b) 4. Theorem of 1 Suppose that /is continuous on > also that /(*o) and 0 for some x 0 [a, b] and that f(a) = /(£) = 0. Suppose b\ Prove that there are numbers c such that /(c) = f(d) = 0, but/(*) > 0 previous problem can be used to good in [a, a<c<x <d<b with 0 for all x in The Hint: d). (c, advantage. 5. x < < k consider / x < < r x > Hint: Let y. + 1. < y. Suppose x (b) y. Prove that there is an integer £ such that be the largest integer satisfying / < x, and 1. / Prove that there Hint: If 1 /n < y - is x, number a rational then ny — nx (d) from 1. r such that Why (Query: (b) and (c). A set A of real numbers is said to be dense if every open interval contains a point of A. For example, Problem 5 shows that the bers (a) and the set of irrational Prove that if / is (c) If we assume f(x) > continuous and /(#) instead that f(x) g(x) for all x. set of rational num- numbers are each dense. = 0 for all dense set A, then f(x) = 0 for all x. (b) Prove that if / and g are continuous and f{x) dense set A, then f(x) = g(x) for all x. 7. > have parts (a) and (b) been postponed until this problem set?) Suppose that r < s are rational numbers. Prove that there is an irrational number between r and s. Hint: As a start, you know that there is an irrational number between 0 and 1. Suppose that x < y. Prove that there is an irrational number between x and y. Hint: It is unnecessary to do any more work; this follows (c) *6. — Suppose that y (a) Can > be > — numbers x g{x) for all x in a g(x) for all x in A, replaced by > in a show that throughout? Prove that if / is continuous and f(x + y) = f(x) + f(y) for all x and y, then there is a number c such that f(x) = cx for all x. (This conclusion can be demonstrated simply by combining the results of two previous problems.) Point of information: There do exist noncontinuous functions/ prove this satisfying /(# y) = f(x) f(y) for all x andy, but we cannot now; in fact, this simple question involves ideas that are usually never + + Least Upper Bounds mentioned any undergraduate in course. 119 The Suggested Reading con- tains references. *8. Suppose that / (Figure (a) Prove that lim f(x) and lim f(x) both - o i— problem (b) (c) whenever a < b exist. Hint: Why is this in this chapter? Prove that / never has a removable discontinuity (this terminology comes from Problem 6-16). Prove that if / satisfies the conclusions of the Intermediate Value If/is a is continuous. bounded function on = [0, 1], let |||/||| sup {|/(*)|: x in [0, 1]}. Problem 7-14. number x can be written uniquely in every that Prove a 0. Suppose > the form x = ka + x where k is an integer, and 0 < x' < a. is a sequence of positive numbers with (a) Suppose that a h a 2 a z fln+i < a n /2. Prove that for any £ > 0 there is some n with a n < S. Prove analogues of the properties of 10. f{b) x—>a + Theorem, then / *9. < a function such that f(a) is 6). || || in f , 11. , (b) Suppose P is , . . . a regular polygon inscribed inside a inscribed regular polygon with twice as many circle. If sides, P f is the show that the f and the area of P is less than half the difference between the area of the circle and the area difference between the area of the circle P (use Figure 7). Prove that there is a regular polygon P inscribed in a circle with area as close as desired to the area of the circle. In order to do part (c) you will need part (a). This was clear to the Greeks, who used part (a) as the basis for their entire treatment of proportion and of (c) area. By calculating the areas of polygons, this of exhaustion") allows computations of Archimedes used it to show that tt < to ir method ("the method any desired accuracy; < V- But it has far greater theoretical importance: *(d) 12. two regular polygons with the same number of sides have the same ratio as the square of their sides, prove that the areas of two circles have the same ratios as the square of their radii. Hint: Deduce a contradiction from the assumption that the ratio of the areas is greater, or less, than the ratio of the Using the square of the radii by inscribing appropriate polygons. A and B are two nonempty sets of numbers such that x Suppose that for all x in (a) (b) 13. fact that the areas of A and all Prove that sup Prove that sup < y y in B, A < A < y for all y in B. inf B. A and B be two nonempty sets of numbers which are bounded above, let A + B denote the set of all numbers x + y with x in A and y in B, Prove that sup (A + B) = sup A + sup B. Hint: The inequality Let and + sup (A B < sup + (A + B) A — sup FIGURE < B) + suffices to £ for all £ > 0; — i? + + < y e/2. — H 1" 1 8 A prove that sup sup B < sup prove that sup A in 5 with begin by choosing x in A and it s/2 and sup Why? To easy. is B) < x + sup B A sup sup (A I H jj 14. Consider a sequence of closed intervals h (a) that there Show (b) is < and a point x which Suppose that a n <z £ n+ i n+1 that this conclusion is < [a i, in every In b\\,h ~ [a 2, £2], • • • . (Figure 8). Prove . we false if is = 6 n for all n consider open intervals instead of closed intervals. The simple result of Problem It 1 4(a) is called the "Nested Interval may be used to give alternative proofs of Theorems 1 and 2. Theorem. 5 ' The appropriate reasoning, outlined in the next two problems, illustrates a general method, called a "bisection *15. Suppose / f((a + is b) /2) argument." continuous on = 0, [a, b] and f(a) < 0 < Then f(b). or / has different signs at the end points of + [a, (a either + b)/2], b)/2 b]. Why? If / nas different signs at the end points of [(a b)/2) y£ 0, let Ix be one of the two intervals on which / changes f((a or } + Now bisect h. Either / is 0 at the midpoint, or / changes sign on one of the two intervals. Let I2 be such an interval. Continue in this way, to define In for each n (unless / is 0 at some midpoint). Use the Nested Interval Theorem to find a point x where f(x) = 0. Suppose / were continuous on [a, b], but not bounded on [a, b\ Then /would be unbounded on either [a, (a + b)/2] or [(a + b)/2, b]. Why? Let Ii be one of these intervals on which / is unbounded. Proceed as in Problem 15 to obtain a contradiction. (a) Let A — {x: x < a}. Prove the following (they are all easy): sign. *16. 17. If x (i) (b) A in (iii) (iv) If x is A and y < x, in A, then there Suppose, conversely, that {x: *18. is A ?*0. A * R. (ii) x < number finitely then y is is in A. some number A satisfies x' in A such that x Prove that (i)-(iv). < x''. A — sup A}. x is called an almost upper bound many numbers y in A with y > x. An for A if there are only almost lower bound is defined similarly. (a) all almost upper bounds and almost lower bounds of the sets Problem 1. Suppose that A is a bounded infinite set. Prove that the set B of all almost upper bounds of A is nonempty, and bounded below. Find in (b) _ - Least Upper Bounds from part B number 121 (c) It follows (d) and denoted by lim A or lim sup A. Find lim A for each set A in Problem 1. Define lim A, and find it for all A in Problem 1. (b) that inf exists; this called- the is limit superior of A, *19. If A is a bounded prove infinite set (c) A < lim A. lim A < sup A. If lim A < sup A, (d) The analogues lim (a) (b) , A then contains a largest element. of parts (b) and lim (c) for — _— —— ~j . — L -shadow points FIGURE *20. 9 Let / be a continuous function on R. A point x is called a shadow point if there is a number y > x with f(y) > f(x). The rationale for this terminology is indicated in Figure 9; the parallel lines are the rays of of / the sun rising in the east (you are facing north). Suppose that of (a, b) (a) are For x in shadow (a, b), points, but that a prove that f(x) < and f(b). b are not Hint: Let all points shadow points. A — x {y: < y < b and f{x) < f(y) If sup A were less than b, then sup A would be a shadow point. Use this fact to obtain a contradiction to the fact that b is not a shadow point. } (b) Now . prove that f(a) < f(b). (This is a simple consequence of continuity.) (c) Finally, using the fact that a ' f(a) is not a shadow point, prove that =f(b). is known as the Rising Sun Lemma. Aside from serving good illustration of the use of least upper bounds, it is instrumental in proving several beautiful theorems that do not appear in this book; see page 371. This result as a PART DERIVATIVES AND INTEGRALS In 1604, at the height of his scientific career, Galileo argued that for a rectilinear motion in which speed increases proportionally to distance covered, law of motion should be the just that (x = ct 2 ) which he had discovered in the investigation offalling Between 1695 and 1700 bodies. not a single one of the monthly issues of Leipzig's Acta Eruditorum was published without articles of Leibniz, the Bernoulli brothers or the Marquis de I'Hopital treating, with notation only slightly different from that which the we use today, most varied problems of differential calculus, integral calculus and the calculus Thus of variations. in the space of almost precisely one century infinitesimal calculus or, we now call The Calculus, as it in English, the calculating tool par excellence, had been forged; and nearly three centuries of constant use have not completely dulled this incomparable instrument. NICHOLAS BOURBAKI CHAPTER DERIVATIVES The derivative of a function Together with the derives its is integral, particular flavor. the it first of the two major concepts of this section. from which calculus constitutes the source While it is true that the concept of a function is fundamental, that you cannot do anything without limits or continuity, and that least upper bounds are essential, everything we have done until now has been preparation if adequate, this section will be easier than the preceding ones — for — the really exciting ideas to come, the powerful concepts that are truly characteristic of calculus. Perhaps (some would say "certainly") the interest of the ideas to be introduced in this section stems from the intimate connection between the mathematical concepts and certain physical ideas. Many definitions, and even some theorems, may be described in terms of physical problems, often in a revealing the demands of physics were the original inspiration for these way. In fact, fundamental ideas of calculus, and we shall frequently mention the physical interpretations. But we shall always first define the ideas in precise mathematical form, and discuss their significance in terms of mathematical problems. The collection of all functions exhibits such diversity that there is almost no hope of discovering any interesting general properties pertaining to all. Because continuous functions form such a restricted class, we might expect to find some nontrivial theorems pertaining to them, and the sudden abundance of theorems after Chapter 6 shows that this expectation is justified. But the most interesting and most powerful results about functions will be obtained only when we restrict our attention even further, to functions which have even greater claim to be called "reasonable," which are even better behaved than most continuous functions. (a) FIGURE (b) (c) 1 Figure 1 illustrates certain types of misbehavior which continuous functions can display. The graphs of these functions are "bent" at (0, 0), unlike the graph of Figure 2, where it is possible to draw a "tangent line" at each point. The quotation marks have been used to avoid the suggestion that we have 125 126 Derivatives and Integrals defined "bent" or "tangent line/ although we are suggesting that the graph might be "bent" at a point where a "tangent line" cannot be drawn. You have probably already noticed that a tangent line cannot be defined as a line which intersects the graph only once such a definition would be both too restrictive and too permissive. With such a definition, the straight line shown in Figure 3 would not be a tangent line to the graph in that picture, while the parabola would have two tangent lines at each point (Figure 4), and the three functions in Figure 5 would have more than one tangent line at the points where they are "bent." 5 — FIGURE 3 A more promising approach to the definition of a tangent line with "secant lines," and use the notion of limits. points (a y f(a)) whose slope and (a + h,f(a + If h ^ h)) determine, as in 0, might start then the two distinct Figure 6, a straight line is f(a + -/(«) h) h a + >f(a / FIGURE h /(< [a y + h) ) + - h)) f(a) H 6 As Figure 7 illustrates, the "tangent line" at (a, f(a)) seems to be the limit, in some sense, of these "secant lines," as h approaches 0. We have never before talked about a "limit" of lines, but we can talk about the limit of their slopes: 127 Derivatives the slope of the tangent line through f{a \ir* + h) DEFINITION are ready for a definition, The function / and some comments. differentiable at a is ~~ f{a) h h-^o We f(a)) should be (a, if exists. h h-^o In this case the limit a. (We is also say that / denoted by /'(a) and is differentiable if / called the derivative of / at is differentiable at a for every a is domain of /.) in the an addendum; we define the tangent line to the graph of /at (a, f(a)) to be the line through (a, f(a)) with slope f{a). This means that the tangent line at (a, J{a)) is defined only if / is The first comment on our differentiable at definition is really a. refers to notation. The symbol f(a) is certainly reminiscent of functional notation. In fact, for any function/, we denote by/' the function whose domain is the set of all numbers a such that / is differ- The second comment entiable at a, and whose value at such a h^o (To be very for precise: /' which lim [f(a is number is h the collections of + h) - f(a)]/h a pairs all exists.) The function /' is called the derivative of /. comment, somewhat longer than the previous two, refers to the physical interpretation of the derivative. Consider a particle which is moving along a straight line (Figure 8(a)) on which we have chosen an "origin" point 0, Our and a third direction in which distances from 0 shall be written as positive numbers, of points in the other direction being written as negative the distance from 0 numbers. Let denote the distance of the particle from 0, at time s(t) has been chosen purposely; since a distance s{t) suggestive notation FIGURE 8(a) 0 line along which particle is moving t. The s(t) is 128 Derivatives and Integrals determined for each number with a certain function from 0, on the s. /, the physical situation automatically supplies us of s indicates the distance of the particle The graph vertical axis, in terms of the time, indicated on the horizontal axis (Figure 8(b)). "distance" The quotient s(a + h) — s{a) h has a natural physical interpretation. It is the "average velocity" of the partiduring the time interval from a to a h. For any particular a, this average speed depends on k, of course. On the other hand, the limit + cle FIGURE 8(b) lim —+^ ~ W h->0 h J depends only on a (as well as the particular function s) and there are important physical reasons for considering this limit. would like to speak of the We "velocity of the particle at time a" but the usual definition of velocity is really a definition of average velocity; the only reasonable definition of "velocity at time a" (so-called "instantaneous velocity") is the limit Rm s (a + define Notice that - s(a) h h->0 Thus we h) the (instantaneous) velocity of the particle at a to be s'(a). could easily be negative; the absolute value \s\a)\ times called the (instantaneous) speed. s'(a) is some- It is important to realize that instantaneous velocity is a theoretical concept, an abstraction which does not correspond precisely to any observable quantity. While it would not be fair to say that instantaneous velocity has nothing to do with average velocity, remember that s\t) is not s(t + h) - s(t) h any particular k, but merely the limit of these average velocities as h Thus, when velocities are measured in physics, what a physicist really measures is an average velocity over some (very small) time interval; such a procedure cannot be expected to give an exact answer, but this is for approaches really The no 0. defect, because physical velocity of a particle is measurements can never be exact anyway. often called the "rate of change of its position." This notion of the derivative, as a rate of change, applies to any other physical situation in which some quantity varies with time. For example, the "rate of change of mass" of a growing object means the derivative of the function m, where m(t) is the mass at time /. In order to become familiar with the basic definitions of this chapter, we will spend quite some time examining the derivatives of particular functions. Before proving the important theoretical results of Chapter 11, we want to Derivatives 129 have a good idea of what the derivative of a function looks like. The next chapter is devoted exclusively to one aspect of this problem calculating the derivative of complicated functions. In this chapter we will emphasize the concepts, rather than the calculations, by considering a few simple examples. — Simplest of all is = a constant function, f(x) In this case c. = h k a-»o Thus / is differentiable at a for every 0. h fc-*o h-*Q number and a, f'(a) that the tangent line to the graph of / always has slope always coincides with the graph. 0, — 0. This means so the tangent line Constant functions are not the only ones whose graphs coincide with their this happens for any linear function f(x) = cx + d. Indeed tangent lines — f'(a) = lim + h) f(a ' h- = hm -f(a) h o —+ c(a + h) d lim the slope of the tangent line A — = h->o h is c, the same d] as the slope of the = + h) = lim = lim slope 2a = hm f(a • x2 graph of/. Here . — f(a) (a + h) - a a2 + h2 2 + 2ah a— o Some + c\ refreshing difference occurs for f(x) f'(a) [ca ch .. = - h h->o 2 - a2 h = lim {2a = 2a. + h) of the tangent lines to the graph of / are shown in Figure 9. In this pic- ture each tangent line appears to intersect the graph only once, fairly easily: Since the tangent line can be checked it is the graph of the function FIGURE = = g(x) 9 Now, if 2a(x 2ax — a) + —a or In other words, (a, a 2) is x 2 so (x or x = 2ax 2ax = a 2 ) and this fact has slope 2a, a2 . — — a) (a, 2 the graph of / and g intersect at a point x2 through 2 — (x, /(*)) a2 + a2 = 0 = 0; a. the only point of intersection. = 0, g(x)), then 130 Derivatives and Integrals The = function /(#) x2 happens to be quite special in this regard; usually more than once. Consider, a tangent line will intersect the graph = the function f(x) x3 . for example, In this case lim* />(,)- h h->0 + lim^ = - A)3 a5 h h-^o a 3 + = lim = lim = lim 3a 2 = 3a 2 + 3a 2 h + 3a 2h 3ah 2 3ah 2 + +h 3 a 3 h3 h->0 Thus the tangent is of / and g graph of / at has got to be x is easily solved = a, — = + = 3a x — a) + - 2a . This means . (*, = 3a 2x - 2a 3 - 3a 2 x + 2a 3 = f{x)) a) is We a factor of the - ax — 2a 2 also has x a)(x + - g(x)) when solution of the equation left side; the other factor obtain a)(x 2 — 0. we remember that one — - - a3 3 (x (x j{x) 2 x3 if so that (x happens that x 2 3a 2 (x x3 can then be found by dividing. It so 2 a 3 ) has slope 3a (a, intersect at the point or This equation 2 the graph of g(x) The graphs +h 3ah . line to the that the tangent line + ax a)(x + 2a 2 ) — = 0. a as a factor; 2a) = we obtain finally 0. *» Thus, as illustrated in Figure 10, the tangent line through (a, a 3 ) also intersects the graph at the point { — 2a, —8a 3 ). These two points are always dis//slope 3 a 2 -2a 1 Vfa -^H tinct, except when a = 0. We 3 a ) have already found the derivative of sufficiently many functions to illustrate the classical, and still very popular, notation for derivatives. For a given function /, the derivative /' is often denoted by df(x) dx For example, the symbol dx 2 FIGURE 10 dx Derivatives denotes the derivative of the function /(*) parts of the expression = 131 x 2 Needless to say, the separate . df(x) dx 9 are not supposed to have any sort of independent existence— the d s are not numbers, they cannot be canceled, and the entire expression is not the quotient two other numbers "#(*)" and "dx" This notation is due to Leibniz (generally considered an independent co-discoverer of calculus, along with Newton), and is affectionately referred to as Leibnizian notation.* Although of the notation df(x)/dx seems very complicated, in concrete cases it may be 2 shorter; after all, the symbol dx /dx is actually more concise than the phrase *V "the derivative of the function /(*) = The following formulas state in standard Leibnizian notation mation that we have found all the infor- so far: dc = o, = a = 2x, —= 3* 2 dx d(ax + b) 9 dx dx 2 dx dx z . dx Although the meaning of these formulas is clear enough, attempts at literal interpretation are hindered by the reasonable stricture that an equation should not contain a function on one side and a number on the other. For example, if be true, then either df(x)/dx must denote /'(x), rather than/', or else 2x must denote, not a number, but the function whose value at x is 2x. It is really impossible to assert that one or the other of these alternatives /' and sometimes means is intended; in practice df(x)/dx sometimes means the third equation is to denote either a number or a function. Because of this are reluctant to denote f{a) by authors ambiguity, most /'(*), while 2x may —— W, df(x) dx instead /'(a) is usually denoted by the barbaric, but unambiguous, symbol dx * Leibniz was led to this symbol by his intuitive notion of the derivative, + h) - f(x)]/h, which he considered but the "value" of this quotient when h was denoted by dx and the is an "infinitely small" number. This "infinitely small" quantity dx) - /(*) by df(x). Although this point corresponding "infinitely small" difference f(x view is impossible to reconcile with properties (P1)-(P13) of the real numbers, some people to be, not the limit of quotients [f(x + of find this notion of the derivative congenial. In addition to these difficulties, more ambiguity. Although Leibnizian notation the notation dx 2 /dx is associated with one absolutely standard, the is notation df(x)/dx is often replaced by df/dx. This, of course, is in conformity with the practice of confusing a function with its value at x. So strong is this tendency that functions are often indicated by a phrase like the following: We "consider the function^ = x ." will sometimes follow classical practice to the extent of using y as the name of a function, but we will nevertheless carefully distinguish between the function and its values— thus we will always say 2 "consider the function (defined by) y(x) = x 2 ." ambiguities of Leibnizian notation, it is used almost exclusively in older mathematical writing, and is still used very frequently something like Despite the The today. around many staunchest opponents of Leibnizian notation admit that it will be some time, while its most ardent admirers would say that it for quite be around forever, and a good thing too In any case, Leibnizian notation cannot be ignored completely. will ! The policy adopted in this book the text, but to include is to disallow Leibnizian notation within in the Problems; several chapters contain a few (immediately recognizable) problems which are expressly designed to illustrate the vagaries of Leibnizian notation. Trusting that these problems will provide ample practice in this notation, we return to our basic task of examining some simple examples of derivatives. The few it functions examined so far have been differentiable. all To fully appreciate the significance of the derivative it is equally important to know some examples of functions which are not differentiable. The obvious candidates are the three functions Figure Now 1 ; \h\/h if first discussed in this chapter, and illustrated in they turn out to be differentiable at 0 something has clearly gone = 1 for h > 0, \h\/h f(h) - /(0) Jim h-+0 In = - and 1 for h does not < 0. This shows that exist. h fact, lim m -f(Q) = i h and lim Kh) -/(0) h (These two limits are sometimes called the right-hand derivative and the left-hand derivative, respectively, of / at 0.) Derivatives If a 7* Q, then does /''(a) = = -1 1 /'O) The proof of this fact a linear function). is left In fact, exist. f(x) to you The graphs 133 if x if x > < 0, 0. easy if you remember the derivative of and of /' are shown in Figure 11. (it is of / For the function x x2 = fix) i -° > We a similar difficulty arises in connection with f(0). -= - 7(0) /(« lim 0 h = \ Therefore, < A A, have ^- A 1, /(0) = > 0. 0, but Thus /'(0) does not exist; fix) exists for x The graphs of / ^ 0 and — / it is /' are shown Even worse things happen f(h) - not differentiable at is 0. Once again, however, easy to see that /(0) in for /(*) _ ) 2x, x 1, * < > 0 0. Figure 12. = Vjlj. For this function Vh 1 A VA h > 0 h < 0. In this case the right-hand limit 1 lim lim h does not exist; instead And, what's more, 1/Va becomes h-+o+ _ Vh arbitrarily large as h approaches 0. — /V ~k becomes arbitrarily large in absolute value, negative (Figure 13). 1 but . 134 Derivatives and Integrals The better = function f(x) behaved than ~ f(h) /(*) \/x, although not differentiable at 0, this. The <Th /(0) _ h llz = h simply becomes arbitrarily large as h goes to = — /(#) 0. at 0. Geometrically this is at least a little 1 m {<rh y Sometimes one says that /has means that the graph of/ has parallel to the vertical axis (Figure 14). Of course, y/x has the same geometric property, but one would say that / has a derivative of "negative infinity" at FIGURE 1 = h an "infinite" derivative a "tangent line" which is quotient Remember 0. supposed to be an improvement over mere continuity. This idea is supported by the many examples of functions which are continuous, but not differentiable; however, one important point 14 that differentiability is remains to be noted: THEOREM 1 If / is differentiable at PROOF lim /(a + a, then / h) is = f(a) • continuous at lim f(a + = yim f{a As we pointed out in Chapter 5, f(a) + h) ~ f(a) h a-»o f(a) — L h h-+o = = h) a. • lim h h->0 0 0. the equation lim f(a + h) — f(a) = 0 is h->0 equivalent to lim f(x) It is = f(a) ; thus / very important to remember remember that the converse is continuous at is Theorem not true. 1, a. | and just as important to A differentiable function is continuous, but a continuous function need not be differentiable (keep in mind the func= |*|, and you will never forget which statement is true and which tion f(x) false) The continuous functions examined points with at most one exception, but so far it is have been differentiable functions which are not differentiable at several points, even an infinite (Figure 15). Actually, one can do FIGURE 15 at all easy to give examples of continuous much worse than this. There is number a function Derivatives (d) (c) FIGURE 16 which continuous everywhere is 135 and differ entiable nowhere! Unfortunately, the defini- Chapter 23, and I have been unable to persuade the artist to draw it (consider carefully what the graph should look like and you will sympathize with his point of view). It is possible to draw some rough approximations to the graph, however; several successively better approximations are shown in Figure 16. Although such spectacular examples of nondifFerentiability must be postponed, we can, with a little ingenuity, find a continuous function which is not differentiable at infinitely many points, all of which are in [0, 1]. One such function of this function will be inaccessible to us until tion is illustrated in Figure 17. precisely; it is The reader r, . /(*) = [x sin -5 \ I 0, This particular function / ability. is given the problem of defining it a straight line version of the function Indeed, for h ^ 0 is itself x ^ 0 x = 0. x quite sensitive to the question of differenti- we have , /z 1 • 0 sin A = 1 . sin -• h h Long ago we proved that lim sin at 0. Geometrically, one can that the secant line through between —1 and 1, 1 /h does not exist, so f see that a tangent line and (0, 0) no matter how small we require -i 0, not differentiable cannot f{h)) in Figure 18 (h, x sin FIGURE is exist, by noting can have any slope h to be. x ?± 0 x = 0 18 This finding represents something of a triumph; although continuous, the somehow quite unreasonable, and we can now enunciate one mathematically undesirable feature of this function it is not differentiable function /seems — one should not become too enthusiastic about the criterion of differentiability. For example, the function at 0. Nevertheless, , N g(.x) = \ x 2 sin x j differentiable at 0; in fact ^'(0) = = x 0, «• 0 X { 0 0: 1 U2 sin r • = hm lim h-^o h h-+o = lim A sin - = 0. h h-+o The tangent line to the graph of g h at (0, 0) is therefore the horizontal axis (Figure 19). This example suggests that we should seek even more restrictive conditions on a function than mere differentiability. We can actually use the derivative to : Derivatives FIGURE 137 19 formulate such conditions if we introduce another set of definitions, the last of this chapter. For any function we /, by taking the obtain, (whose domain may differentiability can be applied to the function function at The a. whose domain (/')', function derivative of /. a, If (/')' is f t (a) exists, of a particle acceleration at time new function /' The notion of t. or course, yielding another usually written simply /" f and the number/'^) /', consists of all points a such that/' is then / called the In physics the second derivative tion at time derivative, a be considerably smaller than that of /). is and is is differentiate called the second derivative off at a. particularly important. If s(t) is moving along a second said to be 2-times differentiable at straight line, then s"{t) is is the posi- called the Acceleration plays a special role in physics, because, as on a particle is the product of its you can feel the second derivative stated in Newton's laws of motion, the force mass and its acceleration. Consequently when you sit in an accelerating car. There is no reason to stop at the second derivative we can define /'" = (/")'» /"" = (/'")', etc. This notation rapidly becomes unwieldy, so the fol- — lowing abbreviation is usually adopted = /(*+D (it is really a recursive definition) (/<*>)'. Thus / (l > = = /', = (/y, fW =/"' = (/")', /(« = fx" = {f » y> /<« f> etc. various functions f {h \ for k derivatives of /. The Usually, have f for {k) we > resort to the notation 2, f defined for smaller k also. In k) are sometimes called higher-order only for k fact, > 4, but it is convenient to a reasonable definition can be made (0 f \ namely, /CO, = /. Leibnizian notation for higher-order derivatives should also be mentioned. 138 Derivatives and Integrals The is natural Leibnizian symbol for /"(*), namely, abbreviated to W Similar notation The = to - .. following example illustrates the notation how and — x2 . Then, = = /'"W = /<*>(*) = /'(*) /"(*) Figure 20 shows the function - whose graph 2 is shown /, in 2x, 2, 0, if 0, > * 3. together with various derivatives. its by the following function, Figure 21(a): /(*) It is one we have already checked, as A rather more illuminating example is presented /"(*) also shows, in various higher-order derivatives are related to the original function. Let /(*) 2x more frequently used for / (n) (*) is very simple case, 'f'(x) or ' 2 x2 = x>0 , -x\ x < 0. easy to see that (c) /'(«) 2a /'(«) -2a AO) = lim if a > 0, ifa<0. Moreover, /<*>(*) ** 0, k > lim 3 -/(0) f{h) f(h) Now (d) FIGURE h-*0 + 20 and lim ^ = lim h Z*' _ 0) so /'(0)-lim^ = *->o This information can all be summarized as follows: f{x) It follows that /"(0) 0. h = 2|*|. does not exist! Existence of the second derivative is thus a : 139 Derivatives rather strong criterion for a function to some tion like / reveals Even satisfy. "smooth looking" func- a when examined with irregularity the second derivative. This suggests that the irregular behavior of the function sm ~> 0, I, might ^'(O) by the second also be revealed = but 0, we know do not x 7* 0 x = 0 derivative. g'(a) for any a At the moment we know that ^ 0, so it is hopeless to begin computing g"(0). We will return to this question at the end of the next chapter, after we have perfected the technique of finding derivatives. PROBLEMS 1. (a) Prove, working directly from the definition, that = ~\/a\ (b) 2\x\ 2. (a) for a * f(a) Prove that the tangent line to the graph of / at intersect the graph of /, except at {a, I /a). Prove that if /(*) - 1A 2 , then f'(a) which point, (*) = 2, Prove that if/W you obtain x > 0 A, then 1 = ~2/a* (a, for a \/a) does not /at * 0. (a, = 1/2^, for a > 0. (The expression - f(a)]/h will require some algebraic face ^x, then/' (a) for [f(a + h) For each natural number Si'(x) to but the answer should suggest the right lifting, 4. lies = = 1/a 2 ) intersects / at one other on the opposite side of the vertical axis. (b) Prove that the tangent line 3. if /(*) 0. let n, £n (*) = trick.) xw . Remembering that and &'(*) = 3a: 2 conjecture a formula for 1, n Prove your conjecture. (The expression (x + A) may be expanded = = &'(*) 2x , y Sn '(x). by the binomial theorem.) (c) /W = 5. Find/ 6% Prove, starting from the definition (and drawing a picture to illustrate) 7. = /(*) + c, then g'(x) = /'(*); = c/W, then £'(*) = </'(*). Suppose that f(x) — # (a) if if [*]. g(x) (b) if g(x) 3 . (a) (b) (c) What is /'(9), /'(25), /'(36)? What is/'(3 ),/'(5 ),/'(6 2 )? What isf(a 2 ),/'(* )? 2 2 2 you do not find this problem silly, you are missing a very important 2 point: f(x ) means the derivative of / at the number which we happen If 2 to be calling x ; it is not the derivative at x of the Just to drive the point (d) For /(*) = function g(x) = home: x\ compare f'(x 2 ) and g'(x) where g(x) = f{x 2 ). f{x 2 ). 140 Derivatives and Integrals 8. (a) (b) = fix + why + = a) is and differentia ble periodic, with period a Prove that f(x) for all x). /' (i.e., also periodic. is Find fix) and also fix + 3) in the following cases. Be very methodical, or you will surely slip up somewhere. Consult the answers (after you do the problem, naturally). fx) = + f(x + fx (ii) (iii) Find f(x) not 11. a picture to illustrate should be true, also. this Suppose that / (i) 10. Prove (starting from the definition) that this. To do this prob- c). Draw • f(x 9. + f(x c). lem you must write out the definitions of g'{x) and f(x + c) correctly. The purpose of Problem 7 was to convince you that although this problem is easy, it is not an utter triviality, and there is something to prove: you cannot simply put prime marks into the equation To emphasize this point: gi x ) — f( x + o). Prove that if g(x) = f(cx), then g'{x) = c f(cx). Try to see pictorially (c) = Suppose g{x) g'(x) (a) fix) = ix + 5)\ = git + x), and Prove that Galileo was wrong: form f and sit) s = proportional to is ct - if /(*) + git x). The answers will fact will = (i) s"(t) (ii) [/(/)]* if a body falls then cannot be a function of the s, s a distance s(t) in t sec- 2 . Prove that the following first (c) 3) 3)\ = x\ be the same. onds, (b) if + ix 3) about facts are true show why we switched from a (the acceleration = is s if sit) c to = (a/2)t 2 (the a/2): constant). lasit). If s is measured in feet, the value of a is 32. How many seconds do you have to get out of the way of a chandelier which falls from a 400foot ceiling? If you don't make it, how fast will the chandelier be going when it hits you? Where was the chandelier when it was mov- ing with half that speed? 12. Imagine a road on which the speed limit In other words, there is specified at every single point. is a certain function miles from the beginning of the road driving along this road; car ^4's is L such £(*). that the speed limit x Two position at time t is cars, a(t), A and B are and car B's is y bit). (a) What equation expresses the fact that car speed limit? (The answer (b) Suppose that at time t A is ^4's Suppose B not a' it) = always goes at the speed position at time the speed limit at (c) is all t — 1. A always travels at the Lit).) limit, Show and that that B is 2?'s position also going at times. always stays a constant distance behind A. Under what conditions will B still always travel at the speed limit? 141 Derivatives 13. Suppose = that/(<z) g(a) and that the left-hand derivative of/ at a equals < a, and the right-hand derivative of g at a. Define h{x) = fix) for x h(x) = g(x) for x > a. Prove that h is differentiate at a. 14. Let fix) — x2 if x is rational, and fix) = 0 if x is irrational. Prove that /is by this function. Just write out the differentiable at 0. (Don't be scared definition of /'(0).) *15. (a) Let / be a function such that |/(*)| < x 2 for all x. Prove that / is differentiable at 0. (If you have done Problem 14 you should be able do to (b) this.) This result can be generalized x2 if is replaced by \g(x)\> where g has what property? > 16. Let a 17. Let 0 < 0 < 1. Prove that / is not differentiable at 0. *18. Let /(*) 1. prove this for irrational a. = -0.00 Suppose that = f(a) and that f(a) = a , prove that / > is 0 \x\ differentiable at 0, and/(0) = then 0, . + h) . = 0tf n+ ia n+2 . = g(a) If a ... n.aia 2 a 5 — f(a)]/h is for h rational, the decimal and also for .... h(a), that f(x) < g(x) < h(x) for all x, Prove that g is differentiable at a, and that f(a) = g'(a) = h'{a). (Begin with the definition of g'(a).) Show that the conclusion does not follow if we omit the hypothesis f(a) 20. Why? consider [f(a a, h (b) \x\ /satisfies |/(*)| 0 for irrational x, and \/q for x — p/q in lowest terms. Prove not differentiable at a for any a. Hint: It obviously suffices to is (a) if = that / expansion of 19. < If / satisfies \f(x)\ = = g(a) ti(a). h(a). Let / be any polynomial function; we will see in the next chapter that / is differentiable. The tangent line to / at {a, f(a)) is the graph oig(x) = — a) + f(a). Thus f(x) — g(x) is the polynomial function d(x) = — f(a)(x — a) — f(a). We have already seen that if fix) = x then = (x - a) and iff(x) = x\ then dix) = (* - a) (x + 2a). f(a)(x 2 f(x) d(x) , 2 2 , when/(x) = show that it is divisible by (x — a) 2 be some evidence that d(x) is always x 4 and (a) Find (b) There certainly seems to divisible by (x — a) 2 Figure 22 provides an , . argument: graph at two points; the tangent line intersects the graph only once near the 55 point, so the intersection should be a "double intersection. To give . intuitive usually, lines parallel to the tangent line will intersect the a rigorous proof, first d{x) x — note that = a /(*) x -}{a) — _ f(a) a Now answer the following questions. Why is fix) — fia) divisible by (x — a)? Why is there a polynomial function h such that h(x) = a ? Why is lim k(x) = 0 ? Why is h{a) = 0 ? dix) fix - a) for x Why does this solve the problem? 142 Derivatives and Integrals 21. (a) Show = Km that f(a) [/(*) - /(*)]/(* - a). (Nothing deep here.) x—*a (b) Show that derivatives are a "local property": iff(x) = g{x) for all x interval containing a, then f'(a) = g'(a). (This means some open in that in computing/' (a), Of *22. (a) you can ignore f(x) course you can't ignore /(*) for Suppose that / differentiable at is /(,)- 2A Remember an old algebraic trick the same quantity is more added = /(* Iim h,k-^o + Prove that if / is confusion, let g(x) thing g *24. Draw is.) Prove that if / is to + *)-/(«-*), h + k even, then /'(*) = f(—x); find = —/'(—*). (In order to minimize g'(x) = odd, then /'(*) /'(—#). Problems 23 and 24 say that /' is even What can therefore be said about f {k) ? 26. Find /"(*) (ii) (iii) (iv) 27. and then remember what other a picture! 25. (i) — a number is not changed and then subtracted from it. generally, that f{x) *23. a. Prove that x. Hint: **(b) Prove, ^ any particular x + lim fc-o if for such x at once!) all if / Once is again, draw a odd, and odd if / picture. even. is if = x*. = xK f(x) = fix + 3) = f(x) f{x) - If x n and 0 , < £ < n, prove that -"CD*"' *28. (a) (b) Find /'(*) if/(*) Analyze / similarly |*|». Find /"(*). Does /"'(*) if /(*) = x 4 for x > 0 exist for all * ? and /(a:) = —x 4 for * Let fix) = * for * > 0 and let /(*) = 0 for x < 0. Prove that exists (and find a formula for it), but that (n) (0) does not exist. / 30. Interpret the following specimens of Leibnizian notation; each restatement of some fact occuring in a previous problem. —= dxn (i) dx nx n \ < 0. /^"^ *29. n is a — dy ..... (ul) dz . , —IT+ v) c] d[f(x) = dx dy dx* df{x) ST , c = it z = 3a 4 f dx (vi) _ y . dx (Vll) , .... (viu) —7— — rf/(ex) (1X) ^T = C #00 "^~ +c , CHAPTER DIFFERENTIATION The process of rinding the derivative of a function the previous chapter you may have called differentiation. is the impression that this process is From usually laborious, requires recourse to the definition of the derivative, and depends upon successfully recognizing some limit. It is true that such a procedure is often the only possible approach— if you forget the definition of the derivative you are likely to entiate a large be Nevertheless, in this chapter we will learn to differof functions, without the necessity of even recalling the lost. number definition. A few theorems will provide a mechanical process for differentiating a large class of functions, which are formed from a few simple functions by the process of addition, multiplication, division, and composition. This description should suggest what theorems will be proved. We will first find the derivative and then prove theorems about the sum, products, quotients, and compositions of differentiable functions. The first theorem is of a few simple functions, merely a formal recognition of a computation carried out in the previous chapter. THEOREM l / If is = a constant function, f(x) f(a) PROOF — c, 0 then for all numbers /'(*) a. lim h h-^o The second theorem is also a special case of a computation in the last chapter. THEOREM 2 If / PROOF is the identity function, f(x) f'(a) f'{a) - 1 : = x, then for all numbers lim tAjZ h-+o h a lim + — M a h h-+o lim h h a. = l.| h-+0 The derivative of the sum sum of the derivatives. the 144 of two functions is just what one would hope- + 145 Differentiation THEOREM 3 If / and g are differentiable at (/ PROOF (/ + gY(a) = +g then / a, ^+ + lim (/ *)-(/ + g )(g) h ]im ^± ± g(g + a-o = lim = lim r fa /(° = /'W The formula it is A> ~ t/W + g(g)j h h-*o wish, but + — /(g) A) £(a ^ + ^)-/W + h + A) — *(« + THEOREM 4 If / if + *'(«)• for the derivative of a product not as simple as one might and the proof requires only a is nevertheless pleasantly symmetric, the and g are — trick, which we have found useful before a number same quantity is added to and subtracted from it. differentiable at (/"*)'(*) PROOF (f-g)'(a) =lim = *)-*(«) h h->o simple algebraic changed and + *)'(*) = /'(«)+*'(«). fc-+0 = also differentiable at a, is lim (f'g)(* /(<2 + then / g a, • is also differentiable at is not and a, =f(a)'g(a) +f(a)-g'(a). - h) + %(* + (/•*)(«) - f(a)g(a) A) /i->0 lim r f{a + h)[g{a + h) - [/(a g(a)l + A) - /( a )]g(a) | - lim /(a = /(«)•*'(«) + h) lim + lim /(« + *)-/(«) g(* THEOREM 5 If g(x) = cf(x) Theorem and / is If h{x) = c, so that g (-) diflPerentiable at a, = = = = = then g is A) differentiable at =c •/'(*). h-f, then by g'(a) + = /(a)-) 4 simplifies considerably: *'(«) PROOF ^ + /'(«)•*«. (Notice that we have used Theorem 9-1 to conclude that lim /(a In one special case . Theorem 4, (A •/)'(«) A (a) f(a)+k'(a) c-f'(a) + c-f{a).\ 0-f(a) f(a) a, and I 146 Derivatives and Integrals (—f)'{a) = —f(a), and consequently Notice, in particular, that — (/ g)'{a) = (f+l-g]Y(a)=f'(a)-g'(a). To demonstrate what we have already achieved, derivative of THEOREM € If f(x ) = some more x n for some natural number = f(a) PROOF we compute the will special functions. na n~ n, then l for all a, The proof will be by induction on n. For n — 1 this is simply Theorem assume that the theorem is true for n y so that if f(x) = x n then 2. Now , = = Let g(x) * n+1 If I(x) = . x, a. the equation x n+1 = — g(x) £=/•/. thus n • x can be written for all x; I(x) f(x) x from Theorem 4 that It follows g'(a) na*' 1 for all f(a) = U-D'(a) = f(a) 7( ) + /(«) I'(a) = rca w_1 a + a n \ = «fl n + a n = 0 + l)a n for all a. • fl ' ' , This is + precisely the case 1 which we wished Putting together the theorems proved so far to prove. | we can now find /' for / of the form /(*) We a nx n + a n ^ix n ~' 1 + - • + • + a 2x 2 a xx +a 0. obtain = /'(*) We = rc^**- 1 + (n - lK^i*"- + 2 • • + • 2a 2 * + ax . can also find /": /"(*) = »(„ - IK*"" + 2 (« - \){n - 2)a n ^x n ^+ : • • + 2a,. easily. Each differentiation reduces the highest and eliminates one more a,-. It is a good idea to work out the (4) and perhaps / (5) until the pattern becomes quite clear. / This process can be continued power of x by 1 derivatives /"', The for k , , , last interesting derivative is > n Clearly, the next step in our fig. It is /<»>(*) = n\a n /<*>(*) = 0. ; we have program is to find the derivative of a quotient quite a bit simpler, and, because of to find the derivative of 1/g. Theorem 4, obviously sufficient 147 Differentiation THEOREM 7 If g is differentiable at a, and * g(a) 0, 0w PROOF we even Before then 1/g is differentiable at a, and wi? write ©<«+*>-(> we must be (Vg)0* tions. Since g that g there does sure that this expression + A) is is is is, by hypothesis, continuous at some make 8 > a. h. it is necessary to check that This requires only two observa- from Theorem 9-1 from Theorem 6-3 that differentiable at a, irfollows Since g(a) 0 such that g(a sense for small ii„V — makes sense defined for sufficiently small + enough h) h, V_, lim = lim iA 0, it ^ follows 0for|A| lim Therefore (\/g) {a + h) + A)] ~[g(a + -g(a)} h) 1 , A ft-o _ 6. «<i±*L_«<5> a-o _ Um < and we can write -[,(« a-o g(fl)g(a + *)-,(«)] . + h) lim ___L__ fc-»o g(a) A ' g(a + «) [gwr (Notice that we have used continuity of g at a once again.) | The general formula for the derivative of a quotient is now easy to derive. Though not particularly appealing, it is important, and must simply be memorized (I always use the incantation: "bottom times derivative of minus top times derivative of bottom, over bottom squared.") theorem 8 Iff and g are differentiable at a and g(a) 7* 0, then f/g and {a) kg) = sw? is top, differentiable at a y 148 Derivatives and Integrals PROOF Since f/g =/• we have (1/g) /(gKzg^M) 2 teWl /'(«) = i *(«) [*(«)? We • r lf r can now differentiate a few more functions. For example, ( ^~ v /W , fM if /(*) . 1 then * t = /W ( v - 1)(2*) {x*+\)-x(2x) = - ~ = (~1)*~ then /'(*) 1, +1)(2^) _ . (* , , X2 1 -* 2 + i) ' 2 2 . 2 . x2 x Notice that the f(x) example can be generalized: last = x~ n = — for > xn then if some natural number n, n~ l — nx /'W = -fsr- = thus Theorem interpret f(x) 6 actually holds both for positive = then Theorem because it is 6 mean *° to is /(x) = true for n not clear how = 1, and /'(*) and negative = 0 ' * _1 to integers. If mean/'(*) we = 0, word "interpret" is necessary should be defined and, in any case, 0 (T is 0 also. (The 1 0° • meaningless.) Further progress in differentiation requires the knowledge of the derivatives of certain special functions to be studied later. One of these is the sine function. For the moment we shall divulge, and use, the following information, without proof: sin' (a) cos' (a) — cos a — — sin for all a, a This information allows us to differentiate for all a. many other functions. For example, if f(x) = x sin x, then /'(*) /"(*) = x cos x = — * si n = —x sin + x x sin *, + + cos x + 2 cos x; cos x Differentiation 149 if g(x) = sin 2 x = = = = = sin x cos sin x • sin then g"(x) a: 2 sin ^ cos + cos x sin # x, — sin x) + — sin *]; 2[(sin #)( 2[cos 2 ^ cos x cos x] 2 if A(#) = cos 2 x — cos x cos ' a:, then h'(x) /?"(*) = (cos #)( — sin x) + (— = — 2 sin ^ cos = — 2[cos 2 # — sin 2 x\ sin x) cos # Notice that + g'{x) h'(x) = 0, + = 1. As we would have g"{x) The examples above involved only products of two functions. A function involving triple products can be handled by Theorem 4 also; in fact it can be handled in two ways. Remember that / g h is an abbreviation for hardly surprising, we expect, + h)(x) = sin + h"(x) = 0. since also • (/*)•* Choosing the first (f'g' = = = (/ • g)'(x) h{x) • a: cos 2 x • f'(g'h). of these, for example, h)'(x) The or 2 we have + (/ • g)(x)h'(x) [f(x)g(x) +f(x)g'(x)]h(x) +f{x)g(x)h'{x) f(x)g(x)h(x) +f(x)g'(x)h(x) +f(x)g(x)h'(x). choice of / (g h) would, of course, have given the same result, with a different intermediate step. The final answer is completely symmetric and • easily • remembered: (/ g h)' is the sum of the three terms obtained by differentiating each of/, g, and h and multiplying by the other two. * ' For example, if f(x) = x z sin x cos x, then f(x) = 3x 2 sin x cos x +x 3 cos x cos x +x z (sin x)(— sin x). Products of more than 3 functions can be handled similarly. For example, you should have (f-g-h- little difficulty *)'(*) You might even = deriving the formula + f{x)g'{x)h{x)k{x) + f(x)g{x)h'{x)k{x) + f(x)g(x)h(x)k'(x). f'(x)g(x)h(x)k(x) try to prove (by induction) the general formula: n (/x-... •/»)'(*) = J/ito • • • • -/.-iW/.-'W/i+iW • • • • •/.(*)• Differentiating the most interesting functions obviously requires a formula ° gYW in terms of /' and g'. To ensure that / ° g be differentiate at a, one reasonable hypothesis would seem to be that g be differentiate at a. Since the behavior oif°g near a depends on the behavior of/ near g(a) (not near a), it also seems reasonable to assume that / is differentiate at g(a). Indeed we shall prove that if g is differentiate at a and / is differentiate at g(a), then f° r (/ /og differentiate at is a, and = (fog)' (a) g'{a). f'(g(a)) is called the Chain Rule, presumably because a composition of functions might be called a "chain" of functions. Notice that (f°g)' is practically the product off and g', but not quite: /' must be evalu- This extremely important formula ated at g(a) and g' at a. Before attempting to prove this theorem we will try a few applications. Suppose = f(x) Let temporarily, use us, . S to denote the ("squaring") / = Therefore sin x 2 sin function S(x) = x2 . Then o S. we have /'(*) Quite a different result is = = S'(x) S in'(S(x)) cos x 2 obtained if f(x) = • 2x. sin 2 x. In this case / = S o sin, = = ^(sin so /'(*) Notice that this agrees / = sin • x) 2 sin x (as it should) m sin'(*) • cos x. with the result obtained by writing and using the product formula. sin Although we have invented a special symbol, S, to name the "squaring" do problems like this without bothering to write down special symbols for functions, and without even bothering to write down the particular composition which / is one soon becomes accustomed to taking / apart in one's head. The following differif you find it entiations may be used as practice for such mental gymnastics necessary to work a few out on paper, by all means do so, but try to develop the knack of writing /' immediately after seeing the definition of /; problems of this sort are so simple that, if you just remember the Chain Rule, there is no function, it does not take much practice to — — thought necessary. if f(x) f(x) = = sin x 3 sin 3 x then /'(*) /'(*) = = cos x z • 3 sin 2 x 3x 2 • cos x Differentiation fix) — sin «4— — x fix) fix) fix) A = = = sin (sin x) sin(* 3 (* 3 151 + + = = = f{x) 3* 2 ) fix) 3* 2) 53 fix) cos (sin x) cos(* 3 53(* 3 • + + 3* cos x 3* 2) ) (3a: • 52 2 • 2 (3* + 6x) + 6x). 2 function like = /(*) which is sin 2 x 2 = * 2] 2 [sin , the composition of three functions, = S o sin o S, / can also be differentiated by the Chain Rule. It is only necessary to remember that a triple composition f°g°h means (/ o g) o h or / © ig o h). Thus if = fix) we can The sin 2 x 2 write / = (S f= So o sin) o S, (sin o^). by applying the Chain Rule whether the two expressions lead to ec-ually simple calculations; As a matter of fact, as any experienced tlifferentiator knows, it is much better to use the second: derivative of either expression can be found twice; the only doubtful point We can now write down fix) first is f = So in one function to be differentiated is o^). (sin we must put sin * 2 Thus we begin by writing Inside the parentheses function, sin ° S. fix) = To begin with, note that the formula for fix) begins swoop. fell S, so the 2 sin x 2 , the value at x of the second BllifSI • (the parentheses weren't really necessary, after this much involves a composition of two functions, handle. We We must now multiply — S at x; this part is easy which we already know how © it to obtain, for the final answer, fix) The all). of the answer by the derivative of sin following example is = 2 sin x 2 handled f{x) • cos x 2 similarly. = sin (sin • 2x. Suppose x 2 ). Without even bothering to write down / as a composition g o h © k of three we can see that the left-most one will be sin, so our expression for functions, fix) begins /'(*)= cos( ) — : we must put Inside the parenthesis sin (sin x 2 ) (what you get from the value of h by deleting the o k(x); this is simply sin x 2 So our expression for first sin). f(x) begins = fix) cos(sin x 2 ) • We can now forget about the first sin in sin (sin x 2 ) we have to multiply what we have so far by the derivative of the function whose value at x is sin x 2 which is again a problem we already know how to solve: — ; f{x) = cos (sin x 2 ) * cos x 2 2x. • some other functions which are the comsome other triple compositions. You can proba"see" that the answers are correct if not, try writing out / as a Finally, here are the derivatives of position of sin bly just and S as well as y — composition if fx) fx) fx) fx) = — = = /(*) = sin ((sin x) 2 ) [sin (sin x)] f(x) = = = = f(x) — then /'(*) 2 f(x) sin (sin (sin x)) fix) 2 sin (# sin x) cos ((sin x) 2 ) 2 sin (sin x) 2 sin x * cos (sin (sin x)) 2 sin (a* sin x) sin (sin (x sin x)) cos (sin (x • 2 cos x • cos x cos (sin *) * cos x * cos(x sin x) ' • 2 • cos (sin x) ' [sin * + x cos *] sin x)) cos (a; 2 sin x) [2x sin x • + The rule for treating compositions of four (or even more) functions always (mentally) put in parentheses starting from the right, fo(go(hok)) and start smaller x 2 cos is x]. easy 9 reducing the calculation to the derivative of a composition of a number of functions: f(g(h(k( x ))))-z For example, if fx) = 2 sin (sin 2 (*)) [f = So = So sin o So sin (sin o (5 o sin))] then = fix) 2 sin(sin 2 x) cos(sin 2 x) * * 2 sin * • cos if /(*) = sin((sin x 2 ) 2 ) [/ = sin o ^ o sin o = sin © iS o (sin .S* o »S))] then fix) = 2 cos((sin a: 2 ) '• ) 2 sin a 2 cos x 2 • 2a:; if /(a:) = 2 sin (sin(sin [fill a:)) in yourself, if necessary] then /'(*) = 2 sin (sin (sin x)) • cos (sin (sin x)) ' cos (sin x) • cos x. Differentiation 153 With these examples as reference, you require only one thing to become a master differentiator practice. You can be safely turned loose on the exercises — and it is now high time that we proved the Chain Rule. some of the tricks one some of the difficulties encountered. We begin, of course, at the end of the chapter, The might following argument, while not a proof, indicates try, as well as with the definition + *>-(/•*)(«> (/'*)(« (/.,)'(.) h - lim + h)) f(s(a Somewhere in here we would it in by fiat: -f(g(a)) h h-+o like the expression for g'(a). One approach is to put \im f(s(a + A)) h h-+o h-^o This does not look bad, and lim (/°g>( g + *) ~ *-o + h)) - f(g(a)) g(a + h) — g(a) -f^(a)) _ f^(a lim it looks even better if we g(a + h) - g{a) h write (/«*)(«) h = im f(s(a) + l W ++ h)-— g(a h-^o h) g(a)]) - f(g(a)) . g(a) ^ g(a + h) h a—+o The second limit is the factorg'(tf) which we want. Ifwelet^(a (to be precise we should write k (ti)) then the first limit is - g(a) + h) — = k y Hm /kM±*) /i-»o -/(*(«)) ' A; It looks as if this limit should be f(g(a)), since continuity of g at a implies that k goes to 0 as A does. In fact, one can, and we soon will, make this sort of reasoning precise. There is already a problem, however, which you will have noticed if you are the kind of person who does not divide blindly. Even for A^Owe might have g(a + h) — g(a) = 0, making the division and multiplication by g(a + h) — g(a) meaningless. True, we only care about small h, but g(a + h) — g{a) could be 0 for arbitrarily small h. The easiest way this can happen for g to be a constant function, g{x) = Then g(a + h) — is g(a) = so the 0 for all h. c. In this case, / © g is also a constant function, (/ ° g) (x) = f(c), Chain Rule does indeed hold: (fogY (a) = However, there are 0 = f(g(a))-g'(a). also nonconstant functions for arbitrarily small h. For example, if a II = sin -i £ 0, for which g (a (a) the function g might be # 5^ 0 * = 0. * 0, + h) — = 0 154 Derivatives and Integrals = In this case, ^'(O) we must have (f we showed in Chapter 9. 0, as = g)'(0) ° Chain Rule is correct, and this is not exactly If the 0 for any differentiable /, A proof of the Chain Rule can be found by considering such recalcitrant functions separately, but it is easier simply to abandon this approach, and obvious. use a trick. THEOREM 9 (THE CHAIN RULE) If g is differentiable at able at and / is a, differentiable at g(a), then / ° g differenti- is and a, (/•*)'(«) =/'(*(«))•*'(«)• PROOF 0 Define a function + h)) - f(g(a)) g(a + h) — g{a) f(g{a ( 4>{h) as follows. = I /(*(«)), g(a + h) — g{a) is + h) -g(a) g(a + h) -g(a) = if should be intuitively clear that It also small, so if 0 continuous at is + g(a — h) ^0 ilg(a g{a) is 0. When 0: h is small, not zero, then 0(A) will be close to f(g(a)); and if it is zero, then <f>(h) actually equals f(g(a)), which is even better. Since the continuity of is the crux of the whole proof we will provide a careful translation of this intuitive argument. We know that / is differentiable at g(a). This means that <j> M±i^)) =/W) ]im Thus, if £ (1) Now g is > if 0 there 0 < \k\ some number is < . k o <5', f(g(a) then differentiable at a, > 5' 0 such that, for +k)- f(g(a)) all k, -/'(*(«)) hence continuous at a, so - g(a)\ there is < £. > 0 such a 8 that, for all h, Consider (2) if \h\ now any h with 4>{h) it - < 8, \h\ < 8. If k — + h)) - f(g(a)) g(<i + A) — f(g(a follows from (2) that \k\ < + then \g(a 8', g(a the other hand, if g(a + h) = < We 0, f(g(a)) . -f(g(a))\ < (1) that then S. have therefore proved that lim 0(A) =f'(g(a)), A->0 g(a) 5* 0, then 6. surely true that \<t>(h) 8'. A and hence from + h) - g(a) — < _ f(g(a)+k) - \Mh) -f(g(a))\ On h) <t>(h) = f'(g(a)) 9 so it is Differentiation so continuous at is 4> f(g(a even if g(a + A) — {fog y (a) = lim 0. The rest of the + A) ~ f(g(a)) g(#) M = /ok*) = proof = (A) easy. If h ?± 0, then is g(g . + - A) 155 we have g(g) 0 (because in that case both sides are 0). Therefore + h))-f(i(a)) w« • _ Um ^ Hm . ii* + h) — gM i many Now that we can differentiate so another look at the function x 2 sin -j functions so easily x 5* 0 x = 0. we can take /(*) . 0, In Chapter 9 we showed that f(0) = 0, working straight from the definition (the only possible way). For x ^ 0 we can use the methods of this chapter. We have f(x) = + 2x sin x x 2 cos - • x (— \ ^-Y x 2/ Thus 2x sin - = /'(*) — cos -j a: 0, As this it is formula reveals, the first derivative /' not even continuous there. If N /(*) = ( we 5* 0 x = 0. indeed badly behaved at 0 is consider instead x 3 sin -j x 0 * | I x a: * 0, = 0, then , r/ / \ (*) = ( \ 3# 2 sin x — at cos -j I 0, In this case /' sion sion 3a: 2 is continuous at 0, jf ^ 0 ^ = 0. X but /"(0) does not exist (because the expresis differentiable at 0 but the expres- sin 1/x defines a function which — x cos 1/x does not). As you may improvement. suspect, increasing the power of x yet again produces another If x 4 sin -j /« = x 0 x . 0, * = 0, then 1 4# 3 1 x 2 cos sin -> 0, It is x 5* 0 * = 0. easy to compute, right from the definition, that (/')'(O) easy to find for x „, / W . = = 0, and /"(*) is ?± 0: t \2x 2 sin * I 0, ( — — Ax cos x — 2x cos - x sin a: 0 a: * = 0. In this case, the second derivative /" is not continuous at 0. By now you may have guessed the pattern, which two of the problems asks you to establish: if 1 /(*) - i 1 x 10, then/'(0), . . . ,/ (n) (0) exist, 0 x x but f n) is = 0, not continuous at 0; if /(*) k a: = 0, (n) then/'(0), entiable at 0, . n) ,/ (0) exist, and ^discontinuous at 0, but f is not differThese examples may suggest that "reasonable" functions can be . 0. . characterized by the possession of higher-order derivatives hard we try to sufficiently fortunately, mask the infinite oscillation of f(x) = sin 1 —no matter how /x, a derivative of high order seems able to reveal the underlying irregularity. we Un- much worse things can happen. calculations, we will bring this chapter to will see later that After all these involved a close with a minor remark. It is often tempting, and seems more elegant, to write some of the theorems in this chapter as equations about functions, rather than about their values. Thus Theorem 3 might be written (/+£)'=/' Theorem 4 might be written as (/•£)' and Theorem + g', -/•*'+/•*, 9 often appears in the form = <f'g)-g'. may be false, because (fog)' Strictly speaking, these equations left-hand side might have a larger the functions on the domain than those on the right. Nevertheless, this is hardly worth worrying about. If / and g are differentiable everywhere in their domains, then these equations, and others like them, are true, and this is the only case any one cares about. Differentiation 157 PROBLEMS 1. As a warm up exercise, find/'fV) for each of the following/. (Don't worryabout the domain of for /'; just get a formula for f(x) that gives the right answer when it makes sense.) (i) /(*) (ii) f{x) (iii) f{x) (iv) f(x) = = = — (v) /W =sin( -^i). (vi) /to (vii) /(a:) (viii) f(x) +x + sin x 2 sin(* ). sin x 2 . sin (cos x). sin (sin x). C sin (cos x) X = = + sin(* sin x). sin (cos (sin x)). Find f(x) for each of the following functions /. (It took the author 20 minutes to compute the derivatives for the answer section, and it should not take you much longer. Although rapid calculation is not the goal of mathematics, if you hope to treat theoretical applications of the Chain Rule with aplomb, these concrete applications should be child's play mathematicians like to pretend that they can't even add, but most of them can when they have to.) + l)\x + + sin x). ((# + sin x) (i) /to sin((* (ii) /(*) sin (# (iii) /(*) sin (iv) /CO = sin /(*) sin(x sin x) (v) 3 2 2)). 2 2 ). (~^—\ \COS X 3 / + sin (sin x 2 ). 31 '. (vi) /(*) (cos*) (vii) sin 2 x sin x 2 sin 2 x 2 (xiv) /to /to /to /to /to /to /to /to (XV) /to (viii) (ix) (x) (xi) (xii) (xiii) (x + sin 5 #) 6 . sin (sin (sin (sin (sin x)))). + l) (((^ + ^) + ^ + ^) sin(# + sin + sin x )). sin((sin 7 x 1 7 3 4 ). 5 . 2 (a; 2 2 sin (6 cos (6 sin (6 cos 6x))). 1 + sin x 1 (xvi) . sin 3 (sin 2 (sin x)). /to x + sin x 158 Derivatives and Integrals (xvii) /(*) (xviii) /(*) 3. Find the derivatives of the functions tan, cotan, sec, and cosec. (You don't have to memorize these formulas, although they will be needed once in a while; if you express your answers in the right way, they will be simple and somewhat symmetrical.) 4. For each of the following functions /, find f(f(x)) (i) /(*) (ii) f(x) = —+j-. — = = sin x. 1 (hi) f(x) (iv) f{x) 5. /(*) (ii) fix) (iii) fix) (iv) fix) 6. Find = (/•/)'(*)) x x\ 17. For each of the following functions (i) (not /, find f(f(x)). -• X = = = /' in x\ 17. 17*. terms of g' if (0 fix) =g(x+g(a)). (ii) fix) =g(x-g(a)). (iii) fix) =gix + gix)). =g(x)(x-a). fix) = g(a)(x-a). /(* + 3) = gix>). (iv) fix) (v) (vi) 7. (a) A circular object but it is known is that increasing in size in when the radius some unspecified manner, the rate of change of the is 6, Find the rate of change of the area when the radius is 6. (If r(0 and A(t) represent the radius and the area at time t, then the functions r and A satisfy A = rr 2 a straightforward use of the Chain radius is 4. ; (b) Rule is called for.) Suppose that we are now informed that the circular object we have been watching is really the cross section of a spherical object. Find the rate of change of the volume when the radius is 6. (You will clearly need to know a formula for the volume of a sphere; in case you have forgotten, the volume is ^tt times the cube of the radius.) Differentiation (c) Now 159 suppose that the rate of change of the area of the circular is 5 when the radius is 3. Find the rate of change of the cross section 8. volume when the radius is 3. You should be able to do this problem in two ways: first, by using the formulas for the area and volume in terms of the radius; and then by expressing the volume in terms of the area (to use this method you will need Problem 9-3). Let /(*) = x 2 sin 1 fx for x ^ 0, and let /(0) = 0. Suppose also that h and k are two functions such that h'{x) A(0) = sin = 3 2 (sin(* + *'(*) 1)) k(0) = /(* = 0. + 1) Find (/oA)'(0). (i) 9. (ii) (W)'(O). (iii) a'O 2 ), where Find a(x) = h(x 2 ). Exercise great care. /'(0) if \ f(x) = sin -> | g(x) x 9* 0 x = 0, x \ I 0, and *(0) 10. 11. 12. = s'(0) - 0. Using the derivative of /(*) = 1/x, as found in Problem 9-1, find (l/g)'(x) by the Chain Rule. = V} - x 2 (a) Using Problem 9-3, find /'(*) for -1 < x < 1, if fix) 2 (b) Prove that the tangent line to the graph of / at (a, y/\ — a ) interelementary sects the graph only at that point (and thus show that the geometry definition of the tangent line coincides with ours). Prove similarly that the tangent lines to an ellipse or hyperbola intersect . these sets only once. 13. If / If / +g • is differentiable at a, are g and / are differentiable at a 14. (a) (b) (c) / and g necessarily differentiable at a ? what conditions on / imply that g is differentiable at a, ? Prove that if / is differentiable at a, then provided that f(a) ^ 0. Give a counterexample if f(a) = 0. |/| is also differentiable at a, Prove that if / and g are differentiable at a, then the functions max(/, g) and min(/, g) are differentiable at a, provided that f(a) * g(a). Give a counterexample if f(a) = g(a). {n) Suppose that f n) (a) and g (a) exist. Prove (d) 15. Leibniz's formula: 160 Derivatives and Integrals *16. Prove that if frigid)) and £ (n) 0) both exist, then (/ o gY n \a) exists. A experimentation should convince you that it is unwise to seek a formula for (f ° g) (n) (a). In order to prove that (f°g) (n) (a) exists you little have to devise a reasonable assertion about (f°g) {n) (a) which can be proved by induction. Try something like: "(/°£) n (<z) exists and is a sum of terms each of which is a product of terms of the form ~ n (a) If f(x) = a n x + a n -\x n l + + a 0 find a function g such that r g = /. Find another. will therefore 17. • • • , (b) If \ r, find a function (c) Is b b +—+ 2 = fix) g with • • • x* = g' + x—mm b 3 . x2 i f. there a function = /(*) + a n x« - • +a +-+ • • Q • • a: such that /'(*) 18. Show (a) that there = = = = f(x) (b) /'(#) 19. is no x x, if f(x) (a) The number if y n 0 for exactly k = a — is is When (x is even. x, if n — k is odd. double root of the polynomial function / some polynomial function £. Prove that if and only if a is a root of both / and /'. = ax 2 + bx + c {a ^ 0) have a double root? What fl) does fix) w numbers *. called a 2 ^*) a double root of / a numbers 0 for exactly one /'(*) fix) 1 odd. (c ) if — is (d) (b) = 1 fx ? a polynomial function / of degree n such that 0 for precisely n 0 for +^ # for does the condition say geometrically? 20. / If is differentiable at a, let d(x) Problem *21. £i, (a) . . , If #i, bn (b) is a companion to Problem 3-6. Let ... , *22. f(xi). = a\, x n are distinct numbers, prove that there = di and Prove that there /(*,•) a) - /(a). Find another solution for - f'(a)(x - . . . , an and be given numbers. function / of degree In and f(x) 19, this gives 9-20. This problem . = In connection with Problem (a). 0; and is — /'(#») is a polynomial such that f(x 3) = fix 3) = 0 for / b{. Hint: Remember Problem 19. 1, = a polynomial function / of degree In f'ixi) = for all — 1 5^ i, with i. Suppose that a and b are two consecutive roots of a polynomial function but that a and b are not double roots, so that we can write fix) = 5^ 0. 0 and (* — a) ix — b)g{x) where ^(a) /, 161 Differentiation (a) Prove that g{a) and g(b) have the same (b) sign. (Remember that a and consecutive roots.) b are is some number x with a < x < b and /'(*) = 0. draw a picture to illustrate this fact.) Hint: Compare the sign of f{a) and f{b). Now prove the same fact, even if a and b are multiple roots. Hint: m n If f(a) = (x - a) (x - £) £(*) where ^ 0 and g(£) ^ 0, con— m~ — n~ = Prove that there (Also (c) sider the polynomial function h(x) a) f'(x)/(x l {x l b) . This theorem was proved by the French mathematician Rolle, in connection with the problem of approximating roots of polynomials, but the result was not originally stated in terms of derivatives. In fact, Rolle was one of the mathematicians who never accepted the new notions of calcuThis was not such a pigheaded attitude, in view of the fact that for one hundred years no one could define limits in terms that did not verge on the mystic, but on the whole history has been particularly kind to lus. Rolle; his name has become attached to a much more appear in the next chapter, which forms the general result, to basis for the most important theoretical results of calculus. 23. *24. 25. Suppose that f(x) = xg(x) for some function g which is continuous at 0. Prove that / is differentiable at 0, and find /'(0) in terms of g. Suppose / is differentiable at 0, and that /(0) = 0. Prove that f(x) = xg(x) for some function g which is continuous at 0. Hint: What happens if you try to write g(x) = f(x)/x ? n If /W = x~ for n in N, prove that (n + k- 1)! * (*-D! *26. Prove that it differentiable 27. What is *(b) f(x) Let/W = f n) l/(* l/(* 2 - a) n - 1) (n) = ^ and 0. = f(x)g(x) where / and g are Hint: Differentiate. ? is = 0. not continuous at 0. 0, ^ (n) is let /(0) Prove that/' (0), will encounter the . . . , (You Problem 15.) and let /(0) = 0. Prove that /'(0), continuous at 0, and that / (n) is not differentiable basic difficulty as that in = x 2n+l sin \/x Hx (0) exist, that / at 0. ^(0) ? and that / (n) Let f(x) / = x 2n sin 1/xifx (0) exist, same *29. /(0) /<*>(*) if (a) fix) *28. impossible to write x is and 0, . . . , 162 Derivatives and Integrals 30. In Leibnizian notation the Chain Rule ought to read: _ df(g(x)) dg{x) df{y) dx dy v~q(x) dx Instead, one usually finds the following statement: "Let y z = f(y). = g(x) and Then dz = dz^dy" dx dy dx Notice that the z in dz/dx denotes the composite function f ° g, while the z in dz/dy denotes the function /; it is also understood that dz/dy will be "an expression involving y" and that in the final answer g(x) formula; then compare with Problem (i) z (ii) z (iii) z (iv) z — = — = sin y, sin y, cos sin u, v, must be In each of the following cases, find dz/dx by using this substituted for y y u v = = = — x +x cos 2 . x. sin x. cos u, u = sin x. 1. CHAPTER SIGNIFICANCE OF THE DERIVATIVE | I One aim in this chapter derivative of a function. is to justify the time As we we have spent learning to find the about/' tells us a shall see, knowing just a little about/. Extracting information about /from information about/' requires some difficult work, however, and we shall begin with the one theorem which lot is really easy. This theorem is concerned with the value of a function on an maximum Although we have used this term informally while to be precise, and also more general. interval. DEFINITION in Chapter 7, it is worth- of/. Let / be a function and A a set of numbers contained in the domain A point x in A is a maximum point for / on A, if /(*) The number /(*) say that / "has itself is its > for f(y) called the maximum every y in A. maximum value on A value of/ on A (and we also at x"). different x Notice that the maximum value of / on A could be /(*) for several different maximum several have can function a words, / other in (Figure 1); value. Usually we points on A, although it can have at most one maximum interval [a, b]\ if / is continushall be interested in the case where A is a closed have a maximum value ous, then Theorem 7-3 guarantees that / does indeed on b\ [a, The definition of a definition on A We is 1 on A will be left minimum on A at x, of / the following: / has a are to you. if (One -/ has a possible maximum ' at x.) a theorem which does not even depend upon the upper bounds. now ready existence of least THEOREM minimum for Let / be any function defined on (a, b). If x is a maximum (or a then = /'(*) minimum) 0. point for / on (a, b), and / is differentiate at *, continuity, of / at (Notice that we do not assume differentiability, or even other points.) PROOF the simple Consider the case where /has a maximum at x. (Figure 2 illustrates the left of to points idea behind the whole argument— secants drawn through right of the to points through (x, /(*)) have slopes > 0, and secants drawn follows. as proceeds argument this (x)) have slopes < 0.) Analytically, (x[f 163 164 Derivatives and Integrals If h is +h any number such that x >f(x /(*) / has a maximum on since Thus, x + if h > 0 (a, b) in is at then (a, b), + h), This means that *. /(* + A) " /« /(* + *) < 0. we have -f(x) h and consequently FIGURE 2 < lim On the other hand, < if we have 0, /(* o. + *) -/(*) h > ~ nU> so — lim By hypothesis, / is differentiable at fact equal to /'(*). This FIGURE The < 0 it follows that /'(*) = case where / has a 0. so these #, two limits must be equal, in means that f(x) 3 from which > minimum and f(x) > 0, 0. at x is left to you (give a one-line proof). | we cannot replace {a, b) by [a, b] in the statement of we add to the hypothesis the condition that x is in (a, b)). Notice (Figure 3) that the theorem (unless Since f(x) depends only on the values of / near to get a stronger version of Theorem 1 . We x> it is almost obvious how begin with a definition which is illustrated in Figure 4. DEFINITION Let / be a function, and A a A point x in A is a local of /. there AO theorem 2 If / is is (* some - 8, 8 + x defined on > 0 such that You should see numbers contained in the domain 8). (a, b) and has a differentiable at x, then /'(*) PROOF set of maximum [minimum] point for / on A if x is a maximum [minimum] point for / on why this is = local maximum (or minimum) at x, 0. an easy application of Theorem 1 . | and / is 165 Significance of the Derivative The converse of be 0 even if x is Theorem 2 is definitely not true maximum not a local — minimum or it is possible for /'(#) to point for /. The simplest example is provided by the function /(*) = * 3 in this case f(0) = 0, but /has no local maximum or minimum anywhere. Probably the most widespread misconceptions about calculus are concerned with the behavior of a function /near x when f(x) = 0. The point made in the previous paragraph is so quickly forgotten by those who want the world to ; be simpler than true it is, —the condition minimum point of /. we that = /'(*) will repeat Precisely for this adopted to describe numbers x which DEFINITION A critical point of a function / is The /(*) itself is the converse of satisfy the condition /'(*) = the ones which must be considered 2 is not = 0. 0. called a critical value of critical values of /, together Theorem a local maximum or reason, special terminology has been is a number x such that f(x) The number it: 0 does not imply that x /. with a few other numbers, turn out to be in order to find the maximum and mini- /. To the uninitiated, finding the maximum and value of a function represents one of the most intriguing aspects of calculus, and there is no denying that problems of this sort are fun (until you mum of a given function minimum have done your first Let us consider on a closed hundred or so). the problem of finding the maximum or minimum of/ [a, b]. (Then, if / is continuous, we can at least be sure first interval maximum and minimum value exist.) In order to locate and minimum of / three kinds of points must be considered: that a (2) The critical points of / in The end points a and b. (3) Points x in (1) *2 a local minimum FIGURE 4 point maximum point such that / maximum b\ is not differentiable at x. point for / on [a, b], then x must be in one of the three classes listed above: for if x is not in the second or third group, then x is in (a, b) and / is differentiable at x; consequently /'(*) = 0, by Theorem 1, and this means that x is in the first group. If a local [a, b] [a, the x is a maximum If there are minimum point or a minimum many points in these three categories, finding the maximum and may still be a hopeless proposition, but when there are only a of / few critical points, and only a few points where / is not differentiable, the procedure is fairly straightforward: one simply finds /(*) for each x satisfying /'(*) = 0, and/(;t) for each x such that / is not differentiable at x and, finally, and the f(a) and f(b). The biggest of these will be the maximum value of /, smallest will be the minimum. A simple example follows. Suppose we wish to find the maximum and minimum /(*) on the [—1, interval To 2]. so /'(*) = when 3* 0 - = 1 = that 0, = VT73 x — Vl/3 The numbers Vl/3 and candidates for the location of the The third group is - 1, when is, - or both lie VT/3. in [ — 1, maximum and - 2], the so the first minimum group of is V173. contains the end points of the interval -1,2. (2) is x we have 3*' Via (i) The second group — xz begin with, f(x) 2 = value of the function empty, since / is differentiate everywhere. The final step compute to - VT/3 = i VT73 - VI73 = -I VT73, = V173) - (- VT73) = -iVT/3 + VI73 = f VT73, (/(- Vl/3) /(-i) = 0, /(2) = 6. /(VT/3) = (VT/3)» 3 Clearly the mum minimum value is 6, value is occurring at This sort of procedure, — f Vl/3, occurring at VT/3, and the maxi- 2. if feasible, will maximum and always locate the minimum we value of a continuous function on a closed interval. If the function are dealing with is not continuous, however, or if we are seeking the maxi- mum or minimum on an open interval or the whole line, then we cannot even be sure beforehand that the maximum and minimum values exist, so all the may say nothing. Nevertheless, a information obtained by this procedure ingenuity will often reveal the nature of things. In Chapter 7 little just such a problem when we showed that f( x ) has a minimum value on must occur at f(x) n + a n ^x n - + 1 • is • nx l we solved even, then the function • +a 0 the whole line. This proves that the some number x = n~ = 0 =x* if minimum value satisfying + (n - \)a n ^x n ^+ • • • +a Q. If we can solve this equation, and compare the values of f(x) for such x, we can actually find the minimum of /. One more example may be helpful. Sup- pose we wish to find the maximum and minimum, if they exist, of the function 167 Significance of the Derivative on the open (-1, interval We 1). /'(*) have 2x = - (1 so/'(x) = 0 only for x * 2) 2 = O.We can see immediately that for x close to 1 or —1 not have a maxithe values of /(*) become arbitrarily large, so /certainly does minimum at 0. a has that / mum. This observation also makes it easy to show with b, and a numbers be just note (Figure 5) that there will We -1 < such that /(*) > a < 0 and 0 < b < 1, x < a and b < x < 1. /(0) for —1 < minimum of / on [a, b] is the minimum of / on all of only place where (-1,1). Now on |>, b] the minimum occurs either at 0 (the minimum /' = or at a or b, and a and £ have already been ruled out, so the This means that the 0), value is/(0) = 1. _ did not draw the graphs of f(x) graph 2 the draw to cheating z = not is * ), but it 1/(1 x and /(*) x prove anything. (Figure 6) as long as you do not rely solely on your picture to sketching the of method a discuss to going now As a matter of fact, we are in discussused be to information enough gives really that graph of a function local maxima even locate to able be will we fact in minima— and ing maxima In solving these problems we purposely involves consideration of the sign of/'(x), and minima. This method on some deep theorems. and relies which have been proved so far, always This is true even yield information about/' in terms of information about /. to determine used be sometimes can theorem of Theorem 1, although this minima. and maxima of location the namely, certain information about /, that /'(*) is not emphasized we introduced, first was When the derivative numbers as h for any particular h, but only a limit of these The theorems about [f( x + h) derivatives f(x)]/h becomes painfully relevant when one tries to extract and most frusinformation about / from information about /'. The simplest the following by afforded is trating illustration of the difficulties encountered approaches 0; this fact x must*/ be a constant function? It is impossible conviction is strengthened to imagine how / could be anything else, and this velocity of a particle is the if interpretation— by considering the physical it is difficult Nevertheless still! standing be must particle surely the always question: If /'(*) = 0 for all y 0, even to begin a proof that only the constant functions x. The hypothesis /'(*) = 0 only means that lim h-+o and it is not at all obvious /(* + *)-/(*) = satisfy fix) = 0 for all 0> h how one can use the information about the limit to derive information about the function. — 168 Derivatives and Integrals The / is a constant function if /'(*) = 0 for all x, and many other same sort, can all be derived from a fundamental theorem, called the Mean Value Theorem, which states much stronger results. Figure 7 makes it plausible that if/ is differentiable on [a, b], then there is some x in (a, b) such fact that facts of the that Kb) ~f(a) /'(*) this means that some tangent line is parallel to the line between and (b,f(b)). The Mean Value Theorem asserts that this is true some x in (a, b) such that /'(*), the instantaneous rate of change of/ at Geometrically (a,f(a)) there FIGURE is 7 "mean" change of / on [a, b], this average change being [f(b) - f(a)]/[b - a]. (For example, if you travel 60 miles in one hour, then at some time you must have been traveling exactly 60 miles per hour.) This theorem is one of the most important theoretical tools of calculus probably the deepest result about derivatives. From this statement you might x, is exactly equal to the average or — FIGURE conclude that the proof is difficult, but there you would be wrong the hard theorems in this book have occurred long ago, in Chapter 7. It is true that if you try to prove the Mean Value Theorem yourself you will probably fail, but this is neither evidence that the theorem is hard, nor something to be 8 ashamed of. The proof of the theorem was an achievement, but today first can supply a proof which is we quite simple. It helps to begin with a very special case. THEOREM 3 (ROLLE'S THEOREM) If / is there PROOF continuous on is a It follows mum [a, b] number x in and (a, b) differentiable such that/'(*) from the continuity of / on value on [a, b] = on (a, b), and f(a) = f(b), then 0. that / has a maximum and a mini- [a, b]. Suppose first that the maximum value occurs at a point x in (a, b). Then f(x) = 0 by Theorem 1, and we are done (Figure 8). Suppose next that the minimum value of / occurs at some point x in (a, b). Then, again, /'(*) = 0 by Theorem 1 (Figure 9). maximum and minimum values both occur at the end = f(b), the maximum and minimum values of / are equal, Finally, suppose the points. Since f(a) FIGURE 9 so / is a constant function (Figure 10), and for a constant function choose any x in (a, b). | Notice that we really needed the hypothesis that / is differentiable everywhere on (a, b) in order to apply Theorem 1 Without this assumption the theorem is false (Figure 11). You may wonder why a special name should be attached to a theorem as easily proved as Rolle's Theorem. The reason is, that although Rolle's Theorem is a special case of the Mean Value Theorem, it also yields a simple proof of the Mean Value Theorem. In order to prove the Mean Value . FIGURE 10 we can Significance of the Derivative 169 apply Rolle's Theorem to the function which gives the length shown in Figure 12; this is the difference between /(#), and the height at x of the line L between (a, f(a)) and (b, f{b)). Since L is the graph of Theorem we will of the vertical segment we want FIGURE to look at 11 As THEOREM 4 (THE MEAN VALUE it If / THEOREM) turns out, the constant f(a) is continuous on j in [a, b] irrelevant. is and differentiate on /'« = /(*) 6 proof is continuous on = and [a, b] is a number -/(«) - . a differentiable on (a, b), and /(*), AW=/w _[/_M] (4 _ a) /(*)) = Consequently, some x (a, /(*))' in /(«)• we may apply (a, b) Rolle's Theorem = - to h and conclude that there is such that o = *'(*) /'(*) -/(«) /(*) b FIGURE then there Let Clearly, A '(*, (a, b), such that — a so that 12 /'(*) = fib) b Notice that the Mean Value Theorem corollary l If / is is I a still fits into the pattern exhibited by —information about / yields information about /'. This so strong, however, that we can now go in the other direction. previous theorems information -f(a) — defined on an interval and /'(*) = 0 for all x in the interval, then /is constant on the interval. PROOF Let a and b be any two points in the interval with a 9^ b. Then there is some x 170 Derivatives and Integrals in (a> b) such that m = But f(x) FIGURE 0 for _ /(*)- /GO o and consequently /(#) the same, is a b x in the interval, so all 13 interval &L=M. — = = /(£). / i.e., Naturally, Corollary 1 is Thus the value of / at constant on the interval. any two points in the | does not hold for functions defined on two or more intervals (Figure 13). COROLLARY 2 and g are defined on the same interval, and /'(*) = g'(x) is some number c such that / = g + c. If / for all x in the interval, then there PROOF For there 1, The statement A is a of the next corollary requires Figure illustrated in DEFINITION we have (/ — g)'(x) = /'(*) number c such that / — g = c. | x in the interval all Corollary — g'(x) — 0 so, by some terminology, which is 14. / is increasing on an interval if f(a) < f(b) whenever a and b numbers in the interval with a < b. The function / is decreasing on an interval if f(a) > f(b) for all a and b in the interval with a < b. (We often say simply that / is increasing or decreasing, in which case the interval is understood to be the domain of /.) function are two COROLLARY 3 If /'(*) /'(*) PROOF < > 0 for 0 for x in an interval, then / is increasing on the interval; x in the interval, then / is decreasing on the interval. all all Consider the case where /'(*) with a < b. Then there is > Let a and 0. some x in (a, b) b But f(x) > 0 for all x in (a, b), b b — a > 0 it The proof when ~f(a) — follows that f(b) /'(#) < 0 for be two points in the interval with a so fW Since — b all > Q a > * f(a). is left if to you. | 171 Significance of the Derivative are true Notice that although the converses of Corollary 1 and Corollary 2 is increasing, it is If/ true. not is 3 Corollary of converse (and obvious), the hold for some x easy to see that /'(*) > 0 for all *, but the equality sign might (consider /W = x 3 ). graph ot Corollary 3 provides enough information to get a good idea of the the more, once Consider, a function with a minimal amount of point plotting. function /(*) = x* - x. We have = /'(*) 3x 2 — 1. -Vl/3, and have already noted that /'(*) = 0 for x = Vl/3 and x = x. Note that other all for is also possible to determine the sign of /'(*) We it 3x 2 — > 1 0 precisely when > 3x 2 1, x*>h x — thus 3* 2 (a) 1 < > VI73 0 precisely x or when - VT73 < an increasing function < - VT73; Thus / is increasing for x < Vl73, and once again increasing V x < Via 1/3, decreasing for x > V 1/3. - VV3 between Combining this and information with the following facts (1) /(- Vi/3) = f /(Vl/3) = (2) f(x) = 0 for x VTT^ -f Vl/3, = -1, 0, 1, gets large, (3) /(*) gets large as x (b) 14 as x gets large negative, a decreasing function FIGURE and large negative it is to the possible to sketch a pretty respectable approximation graph (Figure 15). on which / increases and decreases to examinejhe sign off. For bothering even could have been found without By the way, notice that the intervals example, since/' is continuous, and vanishes only at - Vl/3 and Vl/3, we Vl/3). Since know that/' always has the same sign on the interval (-Vl/3, y(_VT73) > /(VT73), it follows that / decreases on this interval. Similarly, /' always has the same signon (VlA 00) and /(*) is large for large x, so / If/' is conmust be increasing on (Vl/3, 00). Another point worth noting: points critical adjacent two tinuous, then the sign of/' on the interval between this in x one any for of /'(*) sign the by finding can be determined simply interval. Our sketch of the graph of /(*) = x* -^contains - Vl/3 allow us to say with confidence that VT/3 a local minimum point. In fact, we is sufficient a local information to maximum point, and can give a general scheme for : 172 Derivatives and Integrals deciding whether a critical point, or neither (Figure if/' (1) > 0 in some 1 point if/' < (3) if/' is x, then x is a local minimum is no point point, a local of x and is /' maximum left of x minimum < 0 in some interval some interval point. and minimum has the same sign in some interval to the interval to the right, then x (There left a local 0 in some interval to the to the right of maximum a local interval to the to the right of x, then x (2) is 6) /' > 0 in point. it has in some maximum nor a local left neither a local of x as point. in memorizing these rules —you can always draw the pic- tures yourself.) The polynomial functions can all be analyzed in this way, and graph of such functions. possible to describe the general form of the fig URE 16 it is To even begin, Significance of the Derivative we need a result already mentioned in Problem 3-7: f{x) = + n a nx a n -ix n~ l + • • • 173 If + fl 0 , there are at most n numbers x such that an algebraic theorem, calculus can be used to give an easy proof. Notice that if x\ and x 2 are roots of / (Figure 17), so that /Oi) = f(x 2 ) = 0, then by Rolle's Theorem there is a number x between then / has at most w "roots," fx) = 0. Although this X\ and x 2 such that f(x) = xi < #i x2 < ' ' < ' i.e., really is This means that 0. — x k then /' has at least k , and x 2 one between x 2 and # 3 , now etc. It is , 1 if / has k different roots different roots: one between easy to prove by induction that a polynomial function fx) = has at most n roots that it is true for : n, anx n + a n -ix The statement n~ l + ' • " + = surely true for n is flo 1 , and if we assume then the polynomial = g(x) n+1 £ n+ i* could not have more than n + + b nx + n roots, since if ' 1 it ' + did, g' *0 would have more than n roots. With information this f(x) The n — a = anx + n tf n -i* n_1 + ' ' + ' flo. derivative, being a polynomial function of degree n 1 point if not hard to describe the graph of it is roots. is and Therefore / has at most n negative on 1 maximum not necessarily a local b are — critical points. or minimum — 1, has at most Of course, point, but at a critical any rate, adjacent critical points of /, then /' will remain either positive or since /' (a, b), creasing or decreasing on continuous; consequently, / will be either inThus / has at most n regions of decrease or is (a, b). increase. As a specific example, consider the function - f( x ) = xA f(x) = 4* 3 2x\ Since the critical points of / are — 1, - 4x 0, = and /(-I) /(0) /(I) The behavior Axix 1, - l)ix + 1), and = "I, = 0, = "I. on the intervals between the critical points can be determined by one of the methods mentioned before. In particular, we could determine the sign of /' on these intervals simply by examining the formula for fix). On the other hand, from the three critical values alone we can see (Figure 18) that / increases on ( — 1, 0) and decreases on (0, 1). To determine the sign of —1) and (1, oo) we can compute f on of / 3 ft-!) = 4-(-2) -4-(-2) = -24, f(2) and conclude that / These conclusions large negative is = 4 23 • - 4 2 • = 24, decreasing on (-qo, also follow from the —1) and fact that /(*) is increasing on large for large x (1, <*>). and for x. We can already produce a good sketch of the graph; two other pieces of information provide the finishing touches (Figure 19). First, it is easy to deter- mine /(— x), f(x) = that f(x) so the = x3 — 0 for x graph is = 0, ; symmetric with second, clear that / even, f(x) is is odd, /(*) = The function respect to the vertical axis. already sketched in Figure 15, x, it is = —/(—*), and is consequently symmetric with respect to the origin. Half the work of graph sketching may be saved by noticing these things in the beginning. Several problems in this and succeeding chapters ask you to sketch the graphs of functions. In each case you should determine the critical points of/, (1) (2) the value of / at the critical points, (3) the sign of /' in the regions between critical points (if this is not already clear), numbers x such that f(x) = 0 (if possible), the behavior of f(x) as x becomes large or large negative the (4) (5) Finally, bear in even, may mind (if that a quick check, to see whether the function possible). is odd or save a lot of work. if performed with care, will usually reveal the basic shape of the graph, but sometimes there are special features which require a little more thought. It is impossible to anticipate all of these, but one piece of This sort of analysis, is not defined at certain points (for a rational function whose denominator vanishes at some points), then the behavior of /near these points should be determined. information is often very important. If / example, / is if For example, consider the function = /(*) ff(x) AV2, (-1, -1) FIGURE 19 (1, -1) 0) = - 2* x - x* - + 1 2x 2 2 Significance of the Derivative which is not defined at /'(*) We have (x - l)(2s - 175 1. = - (x» - I) + 2) 2x 2) 2 (* *(« (* - 2) - I) 2 Thus the critical points of / are 0, (1) 2. Moreover, = = (2) /(0) /(2) -2, 2. Because / is not denned on the whole interval (0, 2), the sign of/' must be determined separately on the intervals (0, 1) and (1,2), as well as on the points in intervals (- oo, 0) and (2, oo). We can do this by picking particular /'. Either for formula the at hard staring by simply or intervals, each of these way we find that (3) /'(*) /'(*) f (x). /'(*) > < < > 0 if 0 if 0 0 if 1 0 if x < < < x x < < 2 < x. 0, 1, 2, we must determine the behavior of /(*) as x becomes large or large give us negative, as well as when x approaches 1 (this information will also decreases). and increases which on regions the / determine another way to To examine the behavior as x becomes large we write Finally, *2 clearly /(*) close to x of / near close to x is - 1 1 is -2* + x — 1 - 1 2 = x X — . 1 +x, 1 I i (and slightly larger) (but slightly smaller) when x is 1 when x is large, large negative. and fx) is The behavior also easy to determine; since lim (x 2 -2x + 2) = 1^0, the fraction x2 - 2x + x - 1 2 becomes large as x approaches 1 from above and large negative as x approaches 1 from below. one All this information may seem a bit overwhelming; but there is only account can you that way that it can be pieced together (Figure 20); be sure for each feature of the graph. When this sketch has been completed, we might note that it looks like the 176 Derivatives and Integrals FIGURE 20 graph of an odd function shoved over x2 - 2x x - + 2 1 unit, (x 1 and the expression - l) x - + 2 1 1 shows that this is indeed the case. However, this is one of those special features which should be investigated only after you have used the other information to get a good idea of the appearance of the graph. Although the location of local maxima and minima of a function is always revealed by a detailed sketch of its graph, it is usually unnecessary to do so much work. There is a popular test for local maxima and minima which depends on the behavior of the function only at its critical points. theorem 5 Suppose /'(a) = PROOF By > maximum 0. If f'{a) then / has a local 0, at then /has a local minimum at a; if /"(a) < 0, a. definition, /» = umtk±A^m. h h-^o Since f(a) = 0, this can be written />)=i im fc->o Suppose now that f small h. Therefore: f (a) > 0. Then f(a ^. h + h)/h must be positive for sufficiently 177 Significance of the Derivative f(a and f'(a + + must be positive for sufficiently small h > 0 must be negative for sufficiently small h < 0. h) h) This means (Corollary a \/{x) = -x* and /is decreasing minimum at 3) that some in / is increasing in some interval to the right of interval to the left of a. Consequently, / has a local a. < The proof for the case f'ia) Theorem may be applied to the function fx) = 0 is similar. | (a) 5 been considered. We /"(*) - Vl/3 critical points, - Vl/3 Consequently, mum /(*) - *B *, which has already 3x 2 - 1 6*. and Vl/3, we have is a local maximum 0, point and Vl/3 is a local mini- point. Although Theorem 5 will be the second derivative of many found quite useful for polynomial functions, is so complicated that it is easier to functions a critical point of / it may happen that f'ia) = 0. In this case, Theorem 5 provides no information: minimum point, or it is possible that a is a local maximum point, a local functions neither, as shown (Figure 21) by the consider the sign of the (c) FIGURE = = /"(- Vl/3) = -6 VT/3 < /"(Vl/3) = 6 Vl/3 > 0. (b) - have /'(*) At the *3 first /« 21 in derivative. = ~x\ f{x) Moreover, = x\ if fix) a = is x*; = /"(0) = 0, but 0 is a local maximum point for the first, a point for the second, and neither a local maximum nor minipoint for the third. This point will be pursued further in Part IV. It is interesting to note that Theorem 5 automatically proves a partial coneach case/'(0) minimum local mum verse of THEOREM 6 Suppose }"ia) local PROOF itself. exists. If / maximum at a, has a local then f'ia) < minimum at a, then f'ia) > 0; if / has a 0. a. If f'ia) < 0, then / would also have a /would be constant in some interval Thus local maximum at a, by Theorem = Thus we must have f'ia) > 0. contradiction. a that 0, containing a, so f'ia) Suppose / has a local minimum at 5. The case of a local maximum (This partial converse to and < fix) = signs x* and is handled Theorem cannot be replaced by /(*) The remainder = -* 5 is similarly. | the best > and <, as we can hope for: the > shown by the functions 4 .) of this chapter deals, not with graph sketching, or and minima, but with three consequences of the maxima Mean Value Theorem. The 178 Derivatives and Integrals first is a simple, but very beautiful, theorem which plays an important role in 1 5, and which also sheds light on many examples which have occurred Chapter in previous chapters. theorem 7 Suppose that / is continuous at containing a, except perhaps a, and that f'(x) = for x exists for all x in some interval Suppose, moreover, that lim f'(x) a. x—>a exists. Then f(a) and also exists, f{a) PROOF By lim /'(*). definition, = f{d) For sufficiently small h on differentiable h = < 0). By the (a, a > lim + h) f{a -f(a) + h] and 0 the function / will be continuous on [a, a (a similar assertion holds for sufficiently small + h) Mean Value Theorem there a is number a h in (a, a + h) such that fia + k) -/W =/W h Now a h lim /'(*) approaches a as h approaches exists, it follows = /'(«) (It is + A) ~ /(a) /(a lim a good idea to supply a rigorous we have Even treated if / is 0, because somewhat = e-5 lim /'(a*) ^0 argument an everywhere differentiable function, „ 22 x fix) = ( x 2 sin 7, in («, a + h) ; since = lim /'(*). for this final step, which -> it is still possible for /' to if x 9* 0 a: = 0. * ^ I 0, According to Theorem is informally.) | be discontinuous. This happens, for example, FIGURE ah that however, the graph of /' can never exhibit a discon- shown in Figure 22. Problem 39 outlines the proof of another beautiful theorem which gives further information about the function /'. The next theorem, a generalization of the Mean Value Theorem, is of tinuity of the type interest THEOREM 8 (THE CAUCHY MEAN VALUE THEOREM) If / mainly because of its and g are continuous on number x in (a, b) applications. [a, b] and differentiable on (a, b), such that [fib) ~fia)]g'ix) = [gib) - g(a)]f(x). then there is a Significance of the Derivative (If gib) 9* g(a), and ^ g'(x) 179 equation can be written 0, this fib) -fja) _fjx) g\x) g{b)-g{a) Notice that if = x for all x, then g'(x) = 1, and we obtain the Mean On the other hand, applying the Mean Value Theorem to/ g{x) Value Theorem. and g separately, we find that there are x and M -M _ y in (<z, b) with /'(*). g(b)~g(a) g'iyY but there is no guarantee that the x and y found in this way will be equal. These remarks may suggest that the Cauchy Mean Value Theorem will be quite difficult to prove, but actually the simplest of tricks suffices.) PROOF Let - h(x) =f{x)[g{b) Then h is continuous on It follows - g(x)[f(b) -f(a)]. b\ differentiate on [a, from Rolle's Theorem that h\x) = (a, b), = =f(a)g(b) -g(a)f(b) h(a) means g(a)} and h(b). 0 for some x in (a, b), which that 0 = f(x)[g(b) - - g(a)] The Cauchy Mean Value Theorem theorem which g'(x)[f(b) - (a)]. | the basic tool needed to prove a is facilitates evaluation of limits of the form M Km- x^a g{x) when lim f(x) In this case, = 0 and lim gix) = 0. 5-3 is of no use. Every derivative is a limit of this form, derivatives frequently requires a great deal of work. If some Theorem and computing derivatives are known, however, many limits of this form can now be evalu- ated easily. theorem 9 (L'HOPITAL'S RULE) Suppose that lim f(x) = x—>a and suppose 0 and lim g(x) also that lim /' 1 '(*) g '(*) exists. — hm J x-^agix) Theorem 7 0, is Then lim /(*)/$(*) x->a x->a (Notice that = x—*a = hm — x^agix) a special case.) • exists, and 180 Derivatives and Integrals PROOF The hypothesis that lim f'{x)/g'{x) exists contains two implicit assumptions: (1) there (2) is an interval — {a — a 8, + 8) such that /'(*) and g'(x) exist for + 8) except, perhaps, for x in this interval g'(x) ^ 0 with, once again, the possible exception of x in (a all = x 8, a = a, a. On the other hand, / and g are not even assumed to be defined at a. If we define f(a) = g(a) = 0 (changing the previous values of f(a) and g(a), if necessary), then /and g are continuous at a. If a < x < a 8, then the Mean + Value Theorem and the Gauchy the interval Mean Value Theorem Mean Value Theorem to g, would be some xi in (a, x) with applying the Cauchy Mean Value Theorem applying the there number ax in (a, apply to / and g on — 8 < x < a). First we see that g(x) ^ 0, for if g(x) = 0 g'(x{) = 0, contradicting (2). Now (and a similar statement holds for a [a, x] to / and g, we see that there is a such that x) [/(*) - f - 0]g (ax ) [g(x) - 0]f(ax ) or /(*) Now a x _ f'(ax ) approaches a as x approaches lim f(y)/g'(y) exists, it follows y—>a lim (Once again, the reader argument.) M is a, because ax is in (a, x) ; since that lim f^ lim m. invited to supply the details of this part of the | PROBLEMS 1. For each of the following functions, find the maximum and minimum by finding the points in the interval where the derivative is 0, and comparing the values at these points with the values at the end points. values on the indicated intervals, (i) (ii) /(*) f{x) (hi) f(x) (iv) f(x) (vi) = = x* * 5 x* - + *+ 3* 4 1 = /W = - 8*3 Sx + — Xb +X+ AT — + 1 6* 2 1 on [-2, 2]. on [-1, 1]. on [- J, J]. on [-i, 1]. 1 on 1 [0, 5]. 181 Significance of the Derivative Now sketch the graph of each of the functions in Problem maximum and minimum local 1 , and find all points. n (a) < If ai • < • • an find the , minimum value of f(x) = ^ i — (x ai) 2 . =i n *(b) Now minimum find the = ^ value of f{x) t lem where calculus won't help at on the all: the function /is linear, so that the en's — \x This is a prob- =i intervals between the minimum clearly occurs at one and these are precisely the points where / is not differHowever, the answer is easy to find if you consider how changes as you pass from one such interval to another. of the ai, entiable. f(x) *(c) Let a > Show 0. /<*) that the is (2 + + a)/{\ (— + + |*| l + (i) = / x, x \ 5, x -3, * 9, * { / /• • • (m)\ f r/ /(*) x/ /W v . x c/ \ \ (iv)7 y /(*) = = = W v *| (0, a), all local maximum and mini- // \ JW __ - 0, ( 1/?, ' = = 5 7 9. a: = / * x. x rational 0, x irrational. 1,' * 0, otherwise. = i in lowest terms. \ ( 9 3 ( 1/n for some n in N { I | A straight line then back to 5* 3, 5, 7, * irrational f I , * 7, I (n) - points. fix) /..\ |* (The derivative can be found on each of the and (a, oo) separately.) a). <x, 0), For each of the following functions, find mum value of 1 = i intervals maximum the decimal expansion of x contains a 5 1, if ^ otherwise drawn from the point is (1, b)> (0, a) to the horizontal axis, as in Figure 23. Prove that the total length is and shortest must bring a function where (*, 0) is the point on the horizontal axis. The dashed line in Figure 23 suggests an alternative geometric proof; in either case the problem can be solved without when the angles a and /3 are equal. (Naturally you into the picture: express the length in terms of*, actually finding the point Prove that of greatest area. all (*, 0).) rectangles with given perimeter, the square has the 7. among Find, all volume right circular cylinders of fixed V, the smallest surface area (counting the areas of the faces at top as in Figure 24). 8. one with and bottom, , Figure 25 shows the graph of the derivative of/. Find all local maximum and minimum points of/. FIGURE *9. 25 n~ n l + Suppose that / is a polynomial function, f(x) = x + a n -ix — values critical corresponding and critical points with 4, 3, 1, 2, 1, + 0o, 6, 1, 2, 4, 3. Sketch the graph of /, distinguishing the cases n even and n • odd. *10. (a) ~ • • on the basis of this information. Does there exist a polynomial function with the above properties, except that 3 11. is not a critical point? Describe the graph of a rational function to the text's description of the 12. (a) • n l n + Suppose that the polynomial function f(x) = x + a n -ix = + a 0 has critical points -1, 1,2, 3, and/"(-l) 0,/"(l) > 0, /"(2) < 0, /"(3) = 0. Sketch the graph of/ as accurately as possible • (b) • (in very general terms, similar graph of a polynomial function). Prove that two polynomial functions of degree m and n, respectively, most max(m, n) points, For each m and n exhibit two polynomial functions of degree m and n which intersect max(m, n) times. n ~~ l n + Suppose that the polynomial function f(x) = x + a n -\x critical all for 0 points and ^ critical /"(#) exactly k has a + 0 intersect in at (b) *13. (a) • • • points (b) (c) x. • • • + points. (d) Show — that n k odd. is For each n, show that there is a polynomial function / of degree n with k critical points if n — k is odd. n" 1 n + Suppose that the polynomial function f(x) = x + a n ^\X Let k% = n, maximum a 0 has ki local Show that k 2 — kY + 1 if n even, k h k 2 be three integers with k 2 ki if n is odd, and ki + k2 < n. and k 2 local minimum and k 2 = £i if n is odd. = k\ + 1 if n is even, and points is Show that there is a polynomial function / of degree n, with k\ local maximum points and k 2 local < a kl +k and try minimum points. Hint: Pick a\ < a 2 < • • • t kl+k2 f{x) = n ~ ad ' (1 + * 2 )* for an appropriate number /. Significance of the Derivative 14. Prove that (a) M(b - for x in all Prove that if/'(*) < m for all x in [a, Formulate a similar theorem when (a) Suppose that/'(*) /(*) > g(x) for x > g'(x) > a and Show by an example (b) = the hypothesis f(a) 16. M > /'(*) Find (a) all > then f{b) [a, b], f{a) + - a). a). (c) (b) 15. if 183 functions = sin x. = x\ /'"(*) = x + / such for all fix) a:, < < then /(£) b], |/(*)| < M and that f(a) g(x) for x that these conclusions < /(*) + for all x in = g(a). [>, b]. Show that a. do not follow without g(a). that f(x) (b) /"(*) (c) 17. x\ true that a weight dropped from rest will fall s(t) = 16/ 2 seconds, this experimental fact does not mention the behavior of weights which are thrown upwards or downwards. On the other hand Although feet after it is t = 32 is always true and has just enough ambiguity to account for the behavior of a weight released from any height, with any initial velocity. For simplicity let us agree to measure heights upwards from ground level; in this case velocities are positive for rising bodies and the law s"(t) negative for falling bodies, and s"(t) (a) (b) all bodies fall according to the law = -32. Show that s is of the form s(t) = —l6t 2 + at-\-/3. By setting t = 0 in the formula for s, and then in the formula for /, show that s(t) = — 16t 2 + v 0 t + s 0 where s 0 is the height from which the body is released at time 0, and v 0 is the velocity with which it is , released. (c) A weight ground is level. maximum it thrown upwards with velocity v feet per second, at How high will it go? ("How high" means "what is the height for achieves its all What is its velocity at the moment What is its acceleration at that ground again? What will its velocity times.") greatest height? moment? When will it hit the be when it hits the ground again? 18. A cannon ball is shot from the ground with velocity v at (Figure 26) so that horizontal the law (a) Show v cos a. = — 16t + (v sin v 2 distance Its a)t while 9 an angle a v sin a and a above the ground obeys horizontal velocity remains s(t) its cos a. cannon ball is a parabola (find the position and show that these points lie on a parabola). Find the angle a which will maximize the horizontal distance traveled by the cannon ball before striking the ground. that the path of the at each time (b) has a vertical component of velocity component s(t) constantly it /, *19. (a) Give an example of a function / for which lim X— exists, f(x) but so lim /'(*) does not x—» Prove that (b) exist. °o lim /(*) and lim /'(*) both if x—» oo Prove that (c) a;—* exists, 0. = 0 X— « > = = 00 then lim /"(*) X— W 00 also lim f(x) X—» then \imf(x) exist, oo lim /(*) exists and lim /"(*) if X—> and x—* 0. oo (d) Generalize this result. 20. Suppose that / and g are two differ en tiable functions which satisfy — fg = 0. Prove that if a and b are adjacent zeros of/, and g(a) and (Natug(b) are not both 0, then g(x) = 0 for some * between a and same result holds with / and g interchanged; thus, the zeros of each other.) Hint: Derive a contradiction from the separate / and g assumption that g(x) ^ 0 for all x between a and b: if a number is not 0, rally the there 21. stant 22. is a natural thing to do with Suppose that - \f(x) by considering Prove that /O0| /'. it. < 0 ~ n 3>) for n > 1. Compare with Problem Prove that / is con- 3-20. if ai —+— an ao 12 -r * ' n ' +—7 ; n _ "* u = 0 > 1 then a0 some # for 23. + aix + • • • + an* n in [0, 1]. m never has two Prove that the polynomial function fm (x) = x* - 3x roots in [0, 1], no matter what m may be. (This is an easy consequence of Rolle's Theorem. It is instructive, after giving an analytic proof, to graph and /i, and consider where the graph of fm lies in relation to them.) + f0 24. *25. Suppose that / is continuous and differentiable on [0, 1], that /(*) is in * 1 for all x in [0, 1]. Show that there is [0, 1] for each x, and that/'OO exactly one number x in [0, 1] such that /(*) = x. (Half of this problem has been done already, in Problem 7-11.) Prove that if/ is a twice differentiable function with /(0) = 0and/(l) = 1 an d /'(0) = f{\) = 0, then |/"(jc)| > 4 for some x in [0, 1]. In more A particle which travels a unit distance in a unit time, picturesque terms: and starts and ends with velocity 0, has at some time an acceleration > 4. Hint: Prove that either f"(x) —4 26. for some x > 4 for some x in Suppose that / is a function such that /'(*) = 0. Prove that f(xy) = f(x) + /(>) for /(l) *27. /(*) when^W Suppose that / [0, J], or else /"(*) < in [J, 1]. = = f(xy). satisfies /"(*) +f'(x)g(x) -/W = 1/x for all x, 0 y > all 0. x > 0 and Hint: Find 185 Significance of the Derivative for some function g. Prove that if / is 0 at two points, then / between them. Hint: Use Theorem 6. is 0 on the interval 28 Suppose that / is rc-times differentia ble and that {n) x. Prove that f (x) = 0 for some x. f(x) = 0 for n + 1 different 29. Prove that < V66 - i (without computing 30. V66 <i 8 to 2 decimal places!). Prove the following slight generalization of the Mean Value Theorem: If/ is continuous and differentiable on (a, b) and lim f(y) and lim y exist, then there is some x in b - 31. y—*b~ a+ f(y) — lim f(y) a (Your proof should begin: "This is a Value Theorem because .") . * such that (a, b) lim f(y) — trivial consequence of the Mean . . Prove that the conclusion of the Cauchy written in the form fib) - fjfl) g(b) -g(a) Mean Value Theorem can be f(x) = g'{x) under the additional assumptions that g(b) never simultaneously 0 on (a, b). ^ g(a) and that f(x) and g'(x) are *32. Prove that if/ and g are continuous on [a, b] and differentiable on g'(x) ^ 0 for x in (a, b), then there is some x in (a, b) with and /'(*) g'M Hint: Multiply out 33. What is first, _ m -~ gW to see what — * 2 34. limit is actually —4.) Find the following x (i) lim *-*o (ii) lim +0 this really says. wrong with the following use of FHopital's Rule: x x -2 -— = hm hm — + z-i x - 3x + 2 x~>i (The fia) g(x) tan x cos 2 x x* — 1 limits: 3x* 2x +1 = hm — 6* = - 3 2 , 3. (a, b), 1 186 Derivatives and Integrals 35. Find /'(O) if x = x 10, and 5 (0) = ^(0) - 0 and ^"(O) = 36. 0 a: i = fix) 0, 17. Prove the following forms of l'Hopital's Rule (none requiring any essen- new tially (a) reasoning). = lim f{x) If lim f(x)/g(x) (b) = = (c) — > <z = £— /, then lim x—+a + from below). and lim f(x)/g 0, f (x) = oo, then lim x—*a x—>a — oo (and similarly for oo or s + or x —» - )- if —> x a is replaced by <3 = lim /(*) If = x—*a f(x)/g(x) = f'(x)/g'(x) x—*a + lim g(x) x—*a * and lim 0, (and similarly for limits I lim fix) If = g(x) x—*a + x—*a + - lim x— » f(x)/g(x) = (and I = and lim f(x)/g'(x) 0 x— » flO similarly then lim x—» 00 — oo). for /, oo Consider Hint: lim x-»0 + f(l/x)/g(l/x). (d) X—> lim £(» =0 and lim x—» « oo = fix)/gix) 37. = lim /(*) If x—* f'(x)/g'(x) = », then lim X— oo «0 oo, another form of l'Hopital's Rule which requires more than algebraic manipulations: If lim f(x) = lim g(x) = oo, and lim There is X— > f(x)/g , (x) = X—* oo = then lim f(x)/g(x) X— /, I. Prove X—> 00 oo this as follows. oo (a) For every £ > 0 there m is a - number < a such that for x s > a. *'(*) Apply the Cauchy Mean Value Theorem to /and g on [a, x] to show that fix) g(x) ~ fia) - g(a) (Why can we assume (b) Now g(x) < — £ for x g(a) 9^ 0 > a. ?) write _ f(x) gix) f{x) - gix) -gia) f(x) fja) ~ gja) - fia) gix) — fia) ^ 0 for large x fix) (why can we assume that fix) g(x) that /(*) *(*) - / < 2e for sufficiently large x. ?) and conclude . 187 Significance of the Derivative 38. To complete to prove a many the orgy of variations on l'Hopital's Rule, use Problem 37 few more cases of the following general statement (there are so possibilities that you should select just a few, if any, that interest you): If lim /(*) = f(*)/g(x) = *39. (a) (b) or co in (a, b). oc ( [ + or a~ or oo or ] can be a or a and ( ) can be / or oo or — oo , minimum Hint: Consider the in then lim ), qo , and { } of /on [a, b]; why must it be b)? (a, <c < f'(b), then /'(*) - c for some x in (a, 3). Darboux's Theorem.) Hint: Cook up an appropriate function to which part (b) may be applied. Prove that if (This result *40. - {and lim f(x)/g'(x) = { Suppose that / is differentiable on [a, b]. Prove that if the minimum of/ on is at a, then /'(a) > 0, and if it is at b, then f(b) < 0. (One half of the proof of Theorem 1 will go through.) Suppose that /'(a) < 0 and fib) > 0. Show that /'(*) = 0 for some x somewhere (c) Here ). ( can be 0 or = lim g{x) f(a) is known as It is easy to find a function / such that |/| is differentiable but / is not. For example, we can choose /(*) = 1 for x rational and f(x) = - 1 for x irrational. In this example /is not even continuous, nor is this a mere coincidence: Prove that if |/| is differentiable at a, and/ is continuous at then /is also differentiable at a. Hint: It suffices to consider only a with f(a) = 0. Why? In this case, what must \f\'(a) be? a, *41. (a) Let y ^ when x n 0 and let n be even. Prove that x n + n = (* + only y y) = 0. Hint: If x 0 n + y n = (* + y) n apply Rolle's Theorem n to f(x) = x n +y« - (x + on [0, * 0 y) Prove that if y ^ 0 and n is odd, then x n + n = (x + y) n only if y x = 0 or x = —y. 0 , ], (b) **42 **43. Use the method of Problem 41 to prove that if n is even and f(x) then every tangent line to / intersects / only once. Prove even more generally that intersects *44. is - 0 increasing on Prove that *45. it is essential, as Use and f is increasing. Prove that the function g(x) + Hint: Obviously you should look at g'(x). positive by applying the Mean Value Theorem to / on (it will help to remember that the hypothesis shown by the function /(*) = 1 -f if n > 1, then < x for - 1 (notice that equality holds for x = 0). x) = oo). (0, derivatives to prove that (1 , increasing, then every tangent line the right interval is xn / only once. Suppose that/(0) f(x)/x ; if / is = n > 1 + nx < 0 and 0 < x. /(0) = 0 188 Derivatives and Integrals 46. and x* sin 2 l/x for x 9* 0, a local (a) Prove that 0 (b) Prove that /'(0) is = minimum = 0. let /(0) = point for 0 (Figure 27). /. /"(0) This function thus provides another example to show that Theorem 6 cannot be improved. It also illustrates a subtlety about maxima and minima that often goes unnoticed: a function may not be increasing in any interval to the right of a local minimum point, nor decreasing in any M FIGURE Let/0) = interval to the 27 *47. (a) left. f(a) > 0 and /' is continuous at in some interval containing a. Prove that The next two If g(x) (b) if parts of this = 1 = 0 (see problem show that continuity of/' increasing is essential. 1 by showing that in any interval there are interval containing 0, points x with /'(*) The behavior is /*, = < a < Suppose 0 /(0) then / show that there are numbers x arbitrarily close and also with g'(x) = — 1. 2 1. Let /(*) = ax + x sin l/x for x 9* 0, and let Figure 28). Show that / is not increasing in any open x 2 sin to 0 with g'(x) (c) a, of /for > and 0 a > 1, also points x with which is much more < /'(at) 0. difficult to analyze, is discussed in the next problem. FIGURE *48. Let f(x) 28 = ax +x the sign of /'(*) 1 /x is < -1 for 2 sin l/x for x when a > 1 it is any numbers x if ^00 < -1 0, and let/(0) = close to 0. It - is a In order to find 2x sin l/x - cos more convenient 0. if little cos y for y * 0; we want to the most for large y. This question is quite delicate; to consider the function g(y) know * necessary to decide = 2(sin y)/y — } 189 Significance of the Derivative is - cos y, which does reach the value -1, but happens only when sin y = 0, and it is not at all clear whether g itself can have values < — 1 The obvious approach to this problem is to significant part of g(y) this . minimum find the local values of = solve the equation g'(j) (a) Show that = f if g {y) it is impossible to more ingenuity is required. then 0, = sin y ^ = s'my cos y Unfortunately, g. 0 explicitly, so ~2~/ and conclude that g(y) (b) Now show that if = 0, sin 2 y = g\y) (—2y then Ay 2 + y* 4 and conclude that 1*001 (c) (d) A is +> Va + y 2 2 + 4 +/ 2 fact that (2 > 1, show that y )/VA not increasing in any interval around 0. Using the / = Using the fact that lim (2 then / is increasing in function / is +y some increasing at a 2 interval if + )/V~A 4 >/ = around there is 1, show > f(a) if a < x < a + 5 fix) < fid) if a - 8 < x < a. does not a = that if 1, then a > 1, 0. some number fix) if 5 > 0 such that and Notice that a + 8); for this mean that / is — 8, increasing at 0, increasing in the interval (a example, the function shown in Figure 28 is but is (a) Suppose that /is continuous on [0, 1] and that /is increasing at a for [0, 1]. Prove that /is increasing on [0, 1]. (First convince yourself that there is something to be proved.) Hint: For 0 < b < 1, prove that the minimum of / on [b, 1] must be at b. Prove part (a) without the assumption that / is continuous, by considering for each b in [0, 1] the set Sh = {x: fiy) > fib) for ally in not an increasing function in any open interval containing 0. every a in (b) [b, x] (This part of the problem is not necessary for the other parts.) Hint: Prove that Sb = {x: b < x < 1 } by considering sup Sb If/ is increasing at a and /is differentiable at a, prove that f'ia) > 0 . . (c) (this is easy). 190 Derivatives and Integrals > If f(a) (d) 0, prove that /is increasing at a (go right back to the defi- nition of f(a)). (e) (f) Use parts (a) and (d) to show, without using the Mean Value Theorem, that if / is continuous on [0, 1] and f'(a) > 0 for all a in [0, 1], then / is increasing on [0, 1], Suppose that /is continuous on [0, l]and/'(a) = 0 for all a in (0, 1). Apply part (e) to the function g(x) = f(x) + ex to show that /(l) -/(0) > -e. Similarly, show that /(l) - /(0) < £ by considering h(x) = ex - /(*). Conclude that/(0) = /(l). This particular proof that a function with zero derivative must be constant has many points in common with a proof of H. A. Schwartz, which may be the to think it first rigorous proof ever given. Its discoverer, at least, seemed was. See his exuberant letter in reference [41] of the Suggested Reading. **50. a constant function, then every point is a local maximum point is quite possible for this to happen even if / is not a constant function: for example, if f(x) = 0 for x < 0 and /(*) = 1 for x > 0. Prove, however, that if / is continuous on [a, b] and every point of [a, b] is a If / is for/. It local maximum mum of / occurs at x 0 consider the set {x in point, then / is a constant function. Hint: If the mini- , [#<>, b]\ f(y) — f(x 0 ) for all yin[x Q ,x]\. **51. (a) A point x f(x) > an ordinary maximum point for / on A ^ x (compare with the definition point). A local strict maximum point way. Find all local strict maximum points a strict if A maximum with y of is called f(y) for all y in defined in the obvious is of the function 0, x irrational ~j x — - in lowest terms. seems quite unlikely that a function can have a local strict maximum at every point (although the above example might give one It pause for thought). Prove .(b) this as follows. Suppose that every point is a local strict maximum point for /. Let x\ be any number and choose a\ < Xi < bi with b\ — ai < 1 such that f(xi) > f{x) for all x in [a h b{\. Let x 2 5* *i be any point in (a i, bi) and choose a x < a 2 < x 2 < b 2 < bi with b 2 — a 2 < \ such that f(x 2 ) > f{x) for all x in [a 2 b 2 \ Continue in this way, and , use the Nested contradiction. Interval Theorem (Problem 8-14) to obtain a Significance of the Derivative 191 CONVEXITY AND CONCAVITY APPENDIX. Although the graph of a function can be sketched quite accurately on the by the derivative, some subtle aspects of the graph are revealed only by examining the second derivative. These details were purposely omitted previously because graph sketching is complicated enough without worrying about them, and the additional information obtained is often not worth the effort. Also, correct proofs of the relevant facts are sufficiently difficult to be placed in an appendix. Despite these discouraging remarks, the information presented here is well worth assimilating, because the notions of convexity and concavity are far more important than as mere aids to graph sketching. Moreover, the proofs have a pleasantly geometric flavor not often found in calculus theorems. Indeed, the basic basis of the information provided definition DEFINITION 1 A geometric in nature (see Figure is function / line is convex on an segment joining (a> interval, if for all a and f(a)) (b, The geometric condition appearing in this way that is sometimes more useful in and f(a)) (b, f(b)) = m^m — definition above the graph of / at x b a) g(x) > by . f(x), that is, if +f(a)>f(x) a fib) b 2 if - fja) (x-a)> -a fib) DEFINITION - (x — b We can be expressed in an straight line between The proofs. - a)+ /w {x a b lies the interval, the b in the graph of the function g defined is g(x) This line and above the graph of /. f{b)) lies analytic (a, 1). -fja) — > fix) x a - f(a) f(x) -fja) — _ a therefore have an equivalent definition of convexity. A function / a < x < b is convex on an fix) If the interval if for a, x, and b in the interval with we have word "over" - fja) in Definition 1 < is fjb ) - fja) . replaced by "under" or, equivalently, 192 Derivatives and Integrals if the inequality in Definition 2 /(*) (*, is replaced by -/(«) /(*)) /W > b "/(*) — a is not hard to see that the concave functions are precisely the ones of the form — /, where / is we obtain the definition of a concave function (Figure 2). It convex. For this reason, the next three theorems about convex functions have immediate corollaries about concave functions, so simple that we will not even FIGURE bother to state them. 2 Two Figure 3 shows some tangent lines of a convex function. things seem to be true: The graph (1) point (a, tangent If a (2) < of / lies above the tangent f(a)) itself (this point As a matter of fact THEOREM l PROOF line at (a, f(a)) except at the called the point of contact of the line). b, then the slope of the tangent line at slope of the tangent line at FIGURE is (b, f{b))\ that these observations are true, is, (a, /' f(a)) is is than the less increasing. and the proofs are not difficult. 3 Let / be convex. If / is differentiable at a, then the graph of / lies above the tangent line through (a, /(a)), except at (a, f(a)) itself. If a < b and /is differentiable at a and b, then f(a) < f'(b). If 0 < hi < A 2 then, as Figure 4 indicates, , fja + hQ -f(a) (1) < f(a + h 2) - f(a) hi hi A nonpictorial proof can be derived immediately from Definition 2 a < a + hi < a +h 2 . Inequality (1) shows that the values of /(* + *) -/(*) applied to 193 Significance of the Derivative 1 FIGURE 1 1 a a + hi a -f h 2 4 decrease as h — > 0+ . Consequently, /(„)<MiHW forA>0 h (in fact/'(tf) that for h is > the greatest lower bound of all 0 the secant line through (a, and larger slope than the tangent line, which implies that above the tangent means these numbers). But this f(a)) + h, {a + (a line (an analytic translation of this + h)) f(a + h)) f(a argument is has lies easily supplied). For negative h there is f(a a similar situation (Figure + -f(a) h,) > f(a + h 2) h2 < hi < 0, then - f(a) hi hi This shows that the slope of the tangent line f(° 5): if + »-M is greater than {orh<0 h (in fact f(a) is the least upper above the tangent lies \ 1 a + h2 FIGURE a 5 + line 1 hi a if h bound < 0. of all these numbers), so that f(a This proves the first + h) part of the theorem. 194 Derivatives and Integrals FIGURE Now 6 suppose that a . v ff( / < {a) < Then, b. we have already f(a+(b-a)) -f(a) - b _ as fW b — seen (Figure 6), since b — sincefl -, <0 a > 0 a -f(a\ - a and -»))-/(») yW = /(g) a Combining figure Theorem 7 We 1 if /' is = - /(a) /(*) b a b we these inequalities, < obtain /'(£). I has two converses. Here the proofs will be a begin with a Rolle's -/(*) — Theorem lemma that plays the plays in the proof of the increasing, then the graph of / lies same little more role in the next Mean Value Theorem. below any difficult. theorem that It states that secant line which happens to be horizontal. LEMMA Suppose / /(*) PROOF < is /(«) differentiable = /(*) for a and < x /' is increasing. If a < < b and f(a) = f(b), then b. Suppose first that f(x) > f(a) = f(b) for some x in (a, b). Then the maximum / on [a, b] occurs at some point x 0 in (a, b) with f(x 0) > f(a) and, of course, f(x 0) = 0 (Figure 7). On the other hand, applying the Mean Value Theorem to the interval [a, x 0], we find that there is x\ with a < x\ < #o and of f{xi) = /(*)-/(«) > Xo — a 0) 195 Significance of the Derivative contradicting the fact that /' < for a x for x in < b, and (if it = [xi, x] some x f(a) for in (a, b). < < x and/(*i) < f(a). we conclude that there is x 2 with x\ a x\ X = the other hand, f(x) 0, since contradicts the hypothesis that /' We now attack the general we used that THEOREM 2 PROOF If / is Let a and < b. is case in the proof of the differentiable f(b) < x2 < not constant on is (Figure 8) Mean Value Theorem x and 0. maximum occurs at x. Again this increasing. | by the same sort of algebraic machinations Mean Value Theorem. then / is convex. (x - Define g by fib) = /(*) b It is Applying the a local /' is increasing, g(x) easy to see that g' ing the — also impossible so there [a, x]) - /to) > — X\ /to /'(*2) On is We know that / is were, /' would not be increasing on some #1 with to f(a) f(a) (a, b). Suppose f(x) [a, x] = only remains to prove that f(x) it < increasing. This proves that f(x) is lemma to is a). moreover, g(a) also increasing; = g(b) = f(a). Apply- g we conclude that < g(x) In other words, - f(a) — a if < a < x /to b, - f{a) if < a x < b. then Kb) b - f(a) (x-a)< f(a) —a or /(*) Hence / THEOREM 3 If / is is "/« . Kb) - f(a) convex. | differentiable and the graph of / / is convex. lies above each tangent line except at the point of contact, then PROOF Let a < at /(a)), from Figure 9 that if (b, f(b)) lies above the tangent line above the tangent line at (b, fib)), then the slope of the tangent line at (£, fib)) must be larger than the slope of the tangent line at (a,f(a)). The following argument just says this with equations. Since the tangent line at (a, f(a)) is the graph of the function (a, b. It is and clear (a, f(a)) lies g(x) =f(a)(x-a)+f(a), and since (b, f(b)) lies above the tangent Similarly, since the tangent line at (b, h(x) =f'(b)(x and (a, f(a)) lies above the tangent from a) f(b)) is - +f(b), b) +f(a). the graph of we have line at (b, f(b)), f( a )>f'(bKa-b)+f(b). (2) It follows we have - f(b)> f{a){b (1) line, and (2) that f'(a) < f'(b). from Theorem 2 that / is convex. (1) It now If a function / has a reasonable second derivative, the information given in which / is convex or follows | these theorems can be used to discover the regions in concave. Consider, for example, the function 1 = /(*) + 1 For x2 this function, -2x = /'(*) + (1 Thus /'(*) = = 0 only for x 0, and > < /'(*) /'(*) /(0) x2) 2 = 0 if x 0 if x 1, while < > 0. 0, Moreover, FIGURE /(*) > f(x) —> 0 / even. is 0 for all x, as # — > co or - <x> 10 The graph of / therefore looks something like Figure 10. rod = (1 +x 2 2 ) {-2) + 2x [2(1 + (1 + x Y • 2 2{2>x 2 (i - 1) +x y' 2 x 2) We now compute ; 2* 1 It is Significance of the Derivative 197 = 0 only not hard to determine the sign of /"(*). Note = Vl/3 when x same sign - or Vl/3. Since /" on each of the that /"(*) first clearly continuous, is it must keep the sets (-«,_! ^T/3), (- Vl/3, Vl/3), (Vl/3, Since we oo). compute, for example, that easily *>o, /"(-I) = = -2 < = i> /"(0) /"(I) we conclude /" on (_ > > 0 on (- oo _-^VT/3^and < Oon (- Vl/3, Vl/3). 0 means 5 / _ VT73) 00) o, that /" Since /" 0, is increasing, and (VT73, follows it oo), from Theorem 2 that/ is convex while on <»), (VlA (- V 1/3, V 1/3) / is con- cave (Figure 11). / is convex / is concave - Vl/3 FIGURE to the right, (V left, 1 /3, f) since / is - Vl73 number (a, are inflection points of f(x) = 0 = x\ does not ensure that a then /"(0) = 0, but / lies below the part of the graph and above the part of the graph oo), (-Vl/3, Vl/3); concave on the tangent line to the graph of /at /(*) convex the tangent line since /is convex on (Vl/3, crosses the graph. In general, a f"(a) is 11 Notice that at to the / Vl/3 is is a is f(a)) crosses the graph; thus = an 1/(1 +* 2 ). /, it is right of a. This example Vl/3 and inflection point of /; for example, convex, so the tangent line at an procedure which may if (0, 0) cer- inflection point of necessary that /" should have different signs to the illustrates the if Note that the condition tainly doesn't cross the graph of /. In order for a to be function thus the tangent line called an inflection point of / left a and be used to analyze any function/. After the graph has been sketched, using the information provided \ 198 Derivatives and Integrals by the zeros of /" are computed and the sign of /" is determined on the between consecutive zeros. On intervals where /" > 0 the function is convex; on intervals where /" < 0 the function is concave. Knowledge of the /', intervals and concavity of / can often prevent absurd misinter/. Several functions, which can be analyzed in way, are given in the problems, which also contain some theoretical regions of convexity pretation of other data about this questions. There is one further fact which cannot I resist mentioning here. We have seen that convex and concave functions have the property that every tangent line intersects the graph just once, and a few drawings will probably convince you that no other functions have only proof omit know I this property. many contains so This indeed the case, but the is delicate points that I feel it is wiser to it. PROBLEMS 1. Sketch, indicating regions of convexity the functions in inflection, Problem and concavity and points of 11-1 (consider double as (iv) starred) 2. Figure 25 in Chapter 11 shows the graph of /'. Sketch the graph of/. 3. Find two convex functions / and g such that f(x) = g(x) if and only if x is an integer. Show that / is convex on an interval if and only if for all x and y in the interval we have 4. f(tx 5. *6. + - (1 t)y) < tf(x) + (1 t)f(y), for 0 < t < 1. (This is just a restatement of the definition, but a useful one.) Prove that if / and g are convex and / is increasing, then g o f is convex. Let / be a twice-differentiable function with the following properties: f(x) > 0 for x > 0, and / is = 0 for some x > 0 (so /"(#) is not positive for f(x) = l/(x with/'(*o) + < 1), 0. decreasing, is by all x; and /'(0) = that in reasonable cases an example inflection point at hypothesis in this theorem is given by f(x) = 0. / 1/(1 Prove that have an x 2 )). Every will + shown byf(x) = 1 — * 2 which which is not decreasing; and by essential, as f(x) = x2 , , which does not satisfy /'(0) = 0. Hint: Choose x 0 > 0 cannot have f'(y) < f(x 0) for all y > * 0 Why not? for some xi > * 0 Consider /' on [0, Xi]. We . So f(xi) > f'(xo) (a) Prove that if / is convex, then f([x + y]/2) < [/(*) + f(y)]/2. (b) Suppose that /satisfies this condition. Show that f(kx + (1 — k)y) < kf(x) + (1 — k)f(y) whenever k is a rational number, between 0 and n Hint: Part (a) is the special case n = 1. Use 1, of the form m/2 induction, employing part (a) at each step. (c) Suppose that / satisfies the condition in part (a) and / is continuous. . *7. - . Show that / is convex. 199 Significance of the Derivative *8. Let pi, . . . , p n be numbers with positive i ^= = pi 1. i n (a) For any numbers xi, . . . , x n show that i smallest and the ^= largest n-1 n-1 (b) Show same the between the piXi lies i for (1/t) ^ piXi, where t = ^ i=i t=i n (c) Prove Jensen's inequality: If/ is convex, then/ n ^ A**) < ^ Hint: Use Problem n show that (1/0 ^ t *9. (a) by is /_'(#). and in the — 1 domain of / t. if (Part (b) xu . . . is , needed to * n are.) i /, the right-hand derivative, lim [f(a + h) — if / is convex. Check this assertion, and also show that /_' are increasing, Show that if is tinuous.) Hint: close to **11. is — and that f-'(a) < f+(a), then convex, f+{a) = fJ{a) if and only if /+' is con/ tinuous at a. (Thus / is differentiate precisely when /+' is con- /+' **10. noting that p n denoted by /+'(«), and the left-hand derivative is denoted The proof of Theorem 1 actually shows that /+' and /_' always exist **(b) = For any function f(a)]/h, 4, - pif(xi). i=i i=i <z, and [f(b) /+'(£) — f(a)]/(b — is less than a) is close to /_'(«) for b < a this quotient. Prove that a convex function must be continuous. Prove that if / is convex on R, then either / is decreasing, or else / is increasing, or else there is a number c such that / is decreasing on (-oo t] and increasing on [c, oo ). This problem requires a lot of careful ; bookkeeping, and is perhaps best postponed until after the next chapter, where a similar theorem is proved in detail. An important step for this problem is to establish that if f(a) < f(b) for some a < b, then / is increasing on [b, cc), while / is decreasing on (— oo, a ] if f(a) > f(b). 12 CHAPTER INVERSE FUNCTIONS I We now have at our disposal quite powerful methods for investigating funcis an adequate supply of functions to which these methods be applied. We have studied various ways of forming new functions from tions; may what we lack — addition, old alone, and composition multiplication, division, we can produce only — but using these the rational functions (even the sine function, although frequently used for examples, has never been defined). In the next few chapters we will begin to construct new functions in quite sophisticated ways, but there is one important method which will practically double the usefulness of If we any other method we recall that a function is discover. a collection of pairs of numbers, upon the bright idea of simply reversing we we might hit Thus from the function the pairs. all /= {(1,2), (3, 4), (5, 9), (13, 8)}, = {(2,1), (4, 3), (9, 5), obtain g While /(l) - 2 and /(3) - we have 4, = ^(2) (8,13)}. 1 = and g(4) 3. Unfortunately, this bright idea does not always work. If /= {(1,2), (3, 4), (5, 9), (13, 4)}, then the collection (4,13)} {(2, 1), (4, 3), (9, 5), is not a function at where the trouble sort of thing that since all, lies: contains both it = /(3) /(13), even can go wrong, and it is and (4, 3) (4, 13). It is clear though 3 ^ 13. This is the only worthwhile giving a name to the functions for which this does not happen. DEFINITION A function The / is one-one (read "one-to-one") identity function / is if f(a) ^ f(b) obviously one-one, and so is whenever a ^ b. the following modifi- cation: x The function f(x) = x2 is x 5, x 3, 5 = = 5 3. not one-one, since /( — g{x) (and leave g undefined for x 200 Ix, 3, < 0) = , x\ x then g > is 1) = /(l), but if we define 0 one-one, because g is increasing Inverse Functions (since g'(x) = 2x > 0, for > x - /(*) then / is one-one. If n is is easily generalized: If n one-one (since xn > x , 0, odd, one can do better: the function = f{x) is This observation 0). number and a natural is 201 x n for all x n~ l = nx > 0, for all x ^ 0). easy to decide from the graph of /whether /is one-one: the fib) for a ^ b means that no horizontal line intersects the /'(*) It is particularly condition f{a) ^ graph of / twice (Figure 1). a function which a one one function FIGURE If in we any not one one is (b) (a) 1 reverse all the pairs in (a not necessarily one-one function) some case, collection of pairs. It is / we obtain, popular to abstain from this pro- — cedure unless / is one-one, but there is no particular reason to do so instead of a definition with restrictive conditions we obtain a definition and a theorem. For any function /, the inverse of/, denoted by/- 1 (a, b) for which the pair (b, a) is in /. DEFINITION THEOREM l PROOF / 1 is a function Suppose first and that / (c, a) implies that b = (b, a) if and only is /, Since / is and (*,/(*)) = so a Thus f~ l is -1 Conversely, suppose that/ pairs ib,fib)) _1 is one-one. Let are in c. / if = , is the set of all pairs one-one. and (a, and a = (a, b) fib) be two pairs in f~\ Then f(c); since / is one-one this c) a function. is ic, a function. a function this implies that The graphs off and /_1 b = c. = /(c), then / contains the and (/(*),<;) are in / l Thus / is one-one. | If fib) fib)), so (/(*), b) . are so closely related that it is possible to use the _1 _1 consists of all Since the graph of / graph of / to visualize the graph of / 1 from the graph the off' obtains one graph of /, with (b, a) in the . pairs (a, b) 202 Derivatives and Integrals graph of / by interchanging the horizontal and shown FIGURE 2(a) a "< rs n _ -i C. 0) oq "o tu .3" oq d) / / 2 n C « n 5. ° / / / This procedure it is C C 3 3 C/q_ ST F I(x) = is awkward with books and impossible with blackboards, -1 is another way of constructing the graph of/ The fortunate that there and which is (a, b) x, we merely . are reflections of each other through the graph of -1 called the diagonal (Figure 4). To obtain the graph of/ (b, a) reflect the graph of / through this line (Figure 5). Reflecting through the diagonal twice will clearly leave us right back where we started; this means that (/_1 ) _1 = /, which is also clear from the definition. In conjunction with FIGURE crq 6> dia S°»aI points / !-i oq / so / Q v ^ o • gj crq - (c, 0- f a. to ^ / graph vertical axis. If/ has the in Figure 2(a), 4 Theorem 1, this equation has a significant consequence: )~ l is is also one-one (since a one-one function, then the function /_1 a function). if / is l {f~~ There are a few other simple manipulations with inverse functions of which 203 Inverse Functions you should be aware. Since / Thus j~\b) / / / / / / is in precisely / when (b, a) is in / \ it follows that = b 1/ {a, b) r / f(x) = ber is, x much J , a as ~ number a such that f(a) = b; for example, if unique number a such that a 3 = b, and this num- the (unique) is 1 then/" ^) by The s/ 3 means the same f(a) is the definition, \fb. fact that f 1 (x) is the number y such that f(y) = x can be restated in a more compact form: _1 /(/ W) = *> domain for a11 * in the of Moreover, FIGURE Z" 1 (/(*)) = ^ f° r a ^ * ^ n tlie domain of /; 5 from the previous equation upon replacing / by f~\ These two important equations can be written this follows r i o/ = / (except that the right side will have a bigger not all is of R). Since tions, -1 domain if the domain of /or/ standard functions will be defined as the inverses of other funcquite important that we be able to tell which functions are one-one. many it is already hinted which class of functions are most easily dealt with— is increasing and decreasing functions are obviously one-one. Moreover, if/ -1 -1 decreasis then/ decreasing, if/ is and increasing, is also increasing, then/ We have FIGURE ing (the proof 6 is left to you). In addition, / is increasing decreasing, a very useful fact to remember. It is certainly not true that every one-one function decreasing. Figure One example and only if — / is either increasing or has already been mentioned, and is now graphed in 6: X FIGURE is if 3, x 5, I, x 3, 5 = = 5 3. are Figure 7 shows that there are even continuous one-one functions which you neither increasing nor decreasing. But if you try drawing a few pictures an interval will soon agree that every one-one continuous function defined on the delight; bureaucrat's is either increasing or decreasing. This theorem is a to helps It organization. proof uses no new ideas, but it does require careful 7 begin with two lemmas, the second being a corollary of the LEMMA 1 PROOF first. Let / be a continuous one-one function defined on an interval, and let a, and c be points of the interval with a < c < b. If f(a) < f(b), then f(a) (as in Figure 8). f(c) < f(b); and if }{a) > f(b), then f{a) > f(c) > f(b) Consider first the case where f(a) < f(b). Since c ^ a, b we have /(c) ^ b, < /(*), 204 Derivatives and Integrals ^ h (a) FIGURE and (b) 8 f(c) 7*f(b), so either H FIGURE h f(c) < f(a) or f(a ) <f(c) or f{b) <f(c), <f(b) and it is only necessary to rule out the first and third possibilities. Suppose that f(c) < f(a) were true (Figure 9). Apply the Intermediate Value Theorem to the interval [c, b]\ since f(c) < f(a) < f(b) there is some x in [c, b] such that f(x) = f(a). But this contradicts the fact that / is one-one, 9 since a ^ Thus x. < f{a) is Similarly, suppose that f{c) > f(c) impossible. f(b) were true (Figure 10). Apply the Intermediate Value Theorem to the interval [a, c] since f(a) < f(b) < f(c) there is some x in [a, c] such that f(x) = f(b), again contradicting the fact that / is : one-one. FIGURE Thus f(c) > f(b) is also false, leaving only the possibilitv f(a) < f(b) <f{c). 10 The case where f(a) > f(b) may be handled in precisely the same way. one can be elegant and apply the first case to — /. | LEMMA 2 Or Let / be a continuous one-one function defined on an interval, let a and b be two points of the interval, and suppose that f(a) < f(b). If c is a point of the interval with c < a, then f(c) < f(a); and similarly, if b < c, then f(b) < f(c). (Analogous statements could be made if f(a) > f(b), but we will not need them.) PROOF a fif) < <f{a)<f{b). fQ>). case Finally, 2 < c f(c) The theorem c < a. Suppose first that f(c) > f(b). Then Lemma 1, < b, would imply that f{a) > f(b), which is false. Hence But now Lemma 1 may be applied to c < a < b to conclude that Consider the case applied to If / is b we < c is handled similarly. | are ready for the theorem. continuous and one-one on an interval, then / is either increasing or decreasing on that interval. PROOF Let a and b be two numbers in the interval with a < b. Since f(a) ^ /(£), 205 Inverse Functions < either f(a) > f(b) or f(a) fib). We and show that / is increasing. with c and d be two points in the interval fib), Let follows immediately from b, then it other possibility leads to six cases: equals a or The f{d). (1) c < d < a < b (2) c < a < d < b (3) c < a < b < d (4) a <c <d < (5) a < c < b < d (6) a < b < c < d — — — 1 1 1 c a 1 d b 1 1 — 1 a first then apply b I— a c b 1 1 1 1 1 d 1 bed ! 1 1- 1 cases can be treated easily, using the Lemma apply Lemma 1 to c one (or both) of c, d 1 and 2 that /(c) < —— d c a case (1) d. If 1 \ 1 b Each of these < < —— a c c Lemmas f(a) 1 1 \ on the case will first concentrate 2 to < d c < < a a to < lemmas. For example, in /(c) < f(a), and < }{d). The other conclude that b to conclude that f(c) an easy exercise. prove similarly that /is decreasing. Or one can apply can one If /(a) > f{b), method). | the result just proved to -/ (clearly the preferable cases are left to you, as with continuous Henceforth we shall be concerned almost exclusively If/ is such interval. an on defined are which increasing or decreasing functions 1 will be of/ domain the what precisely quite is possible to say a function, it like. closed interval Suppose first that /is a continuous increasing function on the Value Theorem, / takes on every value [a, b\ Then, by the Intermediate closed interval between f{a) and fib). Therefore, the domain of/" is the b\ then the [a, on Similarly, if / is continuous and decreasing 1 [f(a),f(b)l 1 of f' domain is [f(b)J(a)\ b) the analysis a continuous increasing function on an open interval (a, point c in some choose us let with, becomes a bit more difficult. To begin possiOne on by taken are values / >f(c) (a, b). We will first decide which 1 1). In this case / takes (Figure values large arbitrarily bility is that / takes on the other hand, on all values >f(c) by the Intermediate Value Theorem. If, on c < x < b} is = A then values, {/(*): large arbitrarily on / does not take suppose y is bounded above, so A has a least upper bound a (Figure 12). Now (because any number with /(c) < y < a. Then / takes on some value /(*) > y Theorem, Value Intermediate / a is the least upper bound of A). By the itself; a value the on take cannot that / on the value y. Notice If / is , actually takes for if a = which is /(*) for a impossible. < x < b and we choose t with x < t < b, then /(/) > a, 206 Derivatives and Integrals same arguments work Precisely the on all values than f(c) or there less values between for values less number a is < ($ than f(c) either : / takes such that / takes on f(c) all and f(c), but not fi itself. This entire argument can be repeated if / is decreasing, and if the domain off is R or (a, <x>) or (— oo, a). Summarizing: if / is a continuous increasing, or decreasing, function whose domain is an interval having one of the forms /3 (-oc, {a, b), or R, oo), b), {a, then the domain of f' is also an interval which has one of these four forms. Now that we have completed this preliminary analysis of continuous one1 one functions, it is possible to begin asking which important properties of a one-one function are inherited by its inverse. For continuity there is no problem. THEOREM 3 PROOF If/ is continuous and one-one on an interval, then/ -1 We know by Theorem 2 well assume that / that / by applying the usual We must show that also continuous. either increasing or decreasing. is we can then increasing, since is is We might as take care of the other case — /. trick of considering lim /-*(*) =/-'(*) for each in the b in the domain of/" Such a number b is of the form /(a) for some a For any s > 0, we want to find a 5 > 0 such that, for all x, 1 . domain of/. if - f(a) 8 < Figure 13 suggests the you see the x < way 8, - then a FIGURE + a + e. — < e < a + a Z, - + £); account of the information contained in this picture. choice of Consequently, a < and f(a) - f(a - e). Figure 13 conand what follows is simply a verbal 8 fifl a (x) we let 8 be the smaller of f(a e) f(a) tains the entire proof that this 8 works, Our £ l follows that + a < f~ of finding 8 (remember that by looking sideways /(*-e) </(*) </(* — £ graph of f~ l ): since a it + f{a) ensures that - e) < f(a) - 8 and + f(a) 8 < f(a + if /(a) s - 5 < * < e) < at </( + f{a) 5, then 13 f(a Since / is /~ increasing, x is r U(a - - also increasing, i £)) </-*(*) fl + e). and we obtain </" 1 (/(« + e)), e). Inverse Functions 207 i.e., a which is precisely Having - e < < what we want. + a e, | successfully investigated continuity of f~\ tackle differentiability. Again, a picture indicates just / true. Figure 14 / through shows the graph of a one-one function / with a tangent line If this entire picture is reflected through the diagonal, (a, f(a)). -1 shows the graph of/ V , is only reasonable to result ought to be it is what and the tangent line U through (f(a), the reciprocal of the slope of L. In other words, it a). The L it slope of appears that /(«)) f(a) This formula can directly, for each equally well be written in a b in the domain of f way which expresses (/ *)'(*) 1 : i (r )'w = i Unlike the argument for continuity, this pictorial "proof becomes somewhat involved when formulated analytically. There is another approach which might be it is tried. Since we know that tempting to prove the desired formula by applying the Chain Rule: (rrw - W) _1 is differentiable, since the Chain is not a proof that / Rule cannot be applied unless f~ is already known to be differentiable. But l differentiable, this argument does show what (/""*) 'M wil1 have to be if f~ is information. preliminary and it can also be used to obtain some important Unfortunately, this l 1 THEOREM 4 / is a continuous one-one function defined on an interval and/'Cf" ^)) 1 is not differentiable at a. then If f PROOF We have If/"" 1 aw* differentiable at a, the hence which is absurd. | Chain Rule would imply that = 0, 208 Derivatives and Integrals A Theorem 4 simple example to which = Since /'(0) 0 and = 0 applies f~\0), the function f~ l = #3 is the function f(x) is not differentiable at 0 . (Figure 15). Having decided where an now ready inverse function cannot be differentiable, for the rigorous proof that in all other cases the derivative by the formula which we have already "derived" Notice that the following argument uses in continuity of we is are given two different ways. which we have already proved. THEOREM 5 Let / be a continuous one-one function defined on an interval, and suppose l is differentiable at f"\b), with derivative f(f~~ (b)) ^ 0. Then f~ is l that / differentiable at b, and i f(r\b)) PROOF Let b = f(a). Then lim r\b + = ] f{x) = h) - r\b) f~\b lim + h) - a x* Now every number b + h in the b for a domain of / can be written in the form + h - f(a + k) unique k (we should really write plicity). 1 k(h), but we will stick with k for sim- Then lim }-\b +h)-a h->Q = lim a-*o = We f(a lim ;.-.o + k)) - a + k) — b f'Kfia f(a are clearly on the right track ! + It is k) - f{a) not hard to get an explicit expression for k; since b (b) FIGURE + h = f(a + k) we have 15 f~\b + h) =a + k or k=n(b+h) Now by Theorem 3, _1 the function / is -/-'(*). continuous at b. This means that k Inverse Functions approaches 0 as h approaches f(a lim 209 Since 0. + k)-f(a) = fl{a) = f{nm ^ 0> K k->0 this implies that The work we have done on here inverse functions will be an immediate dividend. For is = /„(#) odd, n x n amply repaid later, but let for all x; for n even, let =x n fn (x) Then fn is x> , 0. a continuous one-one function, whose inverse function = Vx = gn(x) By Theorem 5 we have, for x ^ is 0, 1 /»'(/« "'to) 1 n{fn~~ 1 1 - I 1 Thus, if /(a:) then/'(*) rational ber; = = xa , is an integer or the reciprocal of a natural number, now easy to check that this formula is true if a is any a = and It number: Let a is m/n, where m is n = an integer, and if f( x ) then, = xm ' 1 (j '*)", by the Chain Rule, f(x) = mix _ ^ . 1 '")™- 1 x • i • j™**- 1 [(m/n)-(l/n)l + ((l/n)-ll = ™ _y(m/n)-l_ n is a natural num- Although we now have a formula when for /'(*) ~ f(x) — x a and a is rational, a the treatment of the function f(x) x for irrational a will have to be saved for later at the moment we do not even know the meaning of a symbol 2 like x^ Actually, inverse functions will be involved crucially in the definition — . of x a for irrational Indeed, in the next few chapters several important func- a. tions will be defined in terms of their inverse functions. PROBLEMS 1 . Find / 1 for (i) f{x) (ii) f(X ) each of the following - x*+ = (X - 1. 1)1 x rational x, (iii) f(x) (iv) —x, f{x) x irrational. - \ x\ Ix, = + x fix) (vii) /(0.<zia 2 tf3 < 0. 7* ai 9 x = = an . . a i9 i . = an , 1, . . . , n - 1 . [x]. • • x x x ai, (vi) /. •) = 0.a 2 aia 3 .... (Decimal representation is be- ing used.) (viii) fix) — = -— 1 2. (ii) (iii) (iv) 4. , -1 < / / / / is is is is < x 1. X Describe the graph of/"" 1 (i) 3. — and and decreasing and decreasing and when increasing always positive. increasing always negative. always positive. always negative. Complete the proof of Theorem Prove that if / 2. increasing, then so is is f~\ and similarly for decreasing functions. / and g are increasing, 5. If 6. (a) Prove that (/ (b) 7. ^ 0 Find £ -1 Show if /and g find / Or / g • ? Or/og? then/o^ is also one-one. Find The answer is not f~ o g _1 are one-one, . = — ex _1 +g? terms of f~~ l and g -1 Hint: in terms of that fix) / is in this case. -f- d f~ is l if g{x) = one-one 1 if l . +/(*)• and only if ad — be ^ 0, and : 211 Inverse Functions 8. On (i) which f{x) (ii) f(x) (iii) f(x) (iv) /(*) 9. intervals will the following functions [a, b] be one-one? = x*- 3x = x + x. = (1 + **)"'. 2 . h ^±L = + X2 1 Suppose that / is a one-one function and that Z" 1 has a derivative which nowhere 0. Prove that / is differentiate. Hint: There is a one-step is proof. *10. (a) / such that [f(x)] b + f(x) + x = 0 for all x. Hint: Show that / can be expressed as an And the inverse function. The easiest way to do this is to find easiest way to do this is to set x = f~ iy). Find /' in terms of /, using an appropriate theorem of this chapter. Find /' in another way, by simply differentiating the equation Prove that there a differentiable function is l (b) (c) defining The /. function in Problem 10 by the often said to be defined implicitly is v y + however. As the next problem shows, an equation does not usually define a function implicitly on the whole line, and in some regions more than one func5 equation tion may 11. (a) + = x 0. The situation for this equation be defined implicitly. What are the two differentiable functions / which are defined on implicitly satisfy x 2 + ( — 1, [f(x)] 2 1) = 1 by the equation x 2 + y 2 = 1, i.e., which for all x in (-1, 1)? Notice that there are no solutions defined outside [— 1, (b) *(c) quite special, is It will 1]. + 2 functions / satisfy x 2 [fix)] = -1? differentiable functions / satisfy [fix)]'3 Which Which help to first draw — 3/(#) the graph of the function g(x) — x? Hint: = x — 3x. z In general, determining on what intervals a differentiable function defined implicitly by a particular equation may be a delicate affair, and is is we assume that / is such a differentiable solution, however, then a formula for f(x) can be derived, best discussed in the context of advanced calculus. If exactly as in Problem 10(c), by differentiating both sides of the equation defining / (a process 12. (a) Apply this known method 2 y to the equation [f(x)] 2 + x2 = 1. Notice that only to be expected, since there more than one function defined implicitly by the equation your answer is as "implicit differentiation") + x2 - 1. will involve fix) ; this is 212 Derivatives and Integrals (b) (c) 13. But check that your answer works in Problem 11(a). Apply this same method to [/(*)] 3 Leibnizian notation ation. Because y is both of the functions / found for — 3/(x) = x. particularly convenient for implicit differenti- is so consistently used as an abbreviation for /(*), the equation in x and y which defines / implicitly will automatically stand for the equation which / is supposed to satisfy. How would the following computation be written our notation? in + y* + xy = + ax +, + y* ax dy 14. As long 4 = 0, dx -y _ dx 1, 4y 3 + 3y +x 2 as Leibnizian notation has entered the picture, the Leibnizian notation for derivatives of inverse functions should be mentioned. If _1 dy/dx denotes the derivative of /, then the derivative of / is denoted by dx/dy. Write out Theorem 5 in this notation. The resulting equation will show you another reason why Leibnizian notation has such a large following. It will also explain at which point {f Y is to be calculated l when What using the dx/dy notation. the significance of the following is computation? y = = dx 1 !" _ x y x 1/n , dy dx dx dx ny n— l dy 15. Suppose that / is a differentiate one-one function with a nowhere zero xf~\x) - F(f~ l (x)). Prove that G'{x) — (Disregarding details, this problem tells us a very interesting fact: if we know a function whose derivative is /, then we also know one whose derivative is f~ l But how could anyone ever guess the function G? Two different ways are outlined in Problems 14-10 and and derivative that / = F'. = Let G{x) . 18-13.) 16. Suppose h Find is a function such that 17. sin 2 (sin(# + 1)) and h(0) = 3. l (i) (h~~ )'(?>). (ii) (/?- 1 ) (3), / *18. = h! (x) where Find a formula Prove that if / for (/ = h{x + 1). (/^"M. _1 (fc) (3(x) (*)) exists, and is nonzero, then U~ ik l ) \x ) exists. 213 Inverse Functions 19. (a) Prove that an increasing and a decreasing function intersect at most once. (b) Find two continuous increasing functions / and g such that f(x) g(x) precisely (c) *20. (a) x an geometrically?) 1 Prove that f(x) = if x for / is all x. One possibility is f(x) = — x. an increasing function such that / = /~"\ then Hint: Although the geometric interpretation will be immediately convincing, the simplest proof (about 2 lines) rule out the possibilities f(x) < x and f(x) > x. *21. Which functions have the property that the graph function 22. A when function / more A = integer. Give several examples of continuous / such that / = f~ andf(x) = x for exactly one x. Hint: Try decreasing /, and remember the geometric interpretation. (c) is Find a continuous increasing function / and a continuous decreasing function g, defined on R, which do not intersect at all. If / is a continuous function on R and / = f~\ prove that there is at l least one x such that f(x) = x. (What does the condition / = f~ mean (b) when precise through the graph of reflected is nondecreasing we should is to the graph of a — / (the "antidiagonal") ? < f(y) whenever x < y. (To be that the domain of / be an interval.) if stipulate nonincreasing function is still is f(x) defined similarly. Caution: Some writers use "increasing" instead of "nondecreasing," and "strictly increasing" for (a) our "increasing." Prove that on some is (b) *23. (a) / is nondecreasing, but not increasing, then / is constant (Beware of unintentional puns: "not increasing" not the same as "nonincreasing.") Prove that for all (c) if interval. if / is Prove that if /'(*) Suppose that f(x) there and nondecreasing, then f{x) > 0 differentiate .v. is > > 0 for 0 for all x, all x, then / is nondecreasing. and that /is decreasing. Prove that a continuous decreasing function g such that 0 < g(x) < f{x) for all x. (b) Show that we can even arrange that g will satisfy lim g(x)/f(x) = 0. CHAPTER ^/ I INTEGRALS The derivative does not display its full strength until allied with the "integral," topic may seem to be a comdo not appear even once! The study of integrals does require a long preparation, but once this preliminary work has been completed, integrals will be an invaluable tool for creating new functions, and the derivative will reappear in Chapter 14, more powerful than the second main concept plete digression — of Part III. At first this in this chapter derivatives ever. Although ultimately i to be denned in a quite complicated way, the integral that of area. By now it should come formalizes a simple, intuitive concept as no — surprise to learn that the definition of great difficulties —"area" certainly is an intuitive concept can present no exception. In elementary geometry, formulas are derived for the areas of many plane figures, but a little reflection shows that an acceptable definition of area is seldom given. The area of a region is sometimes defined as the number of 1, which fit in the region. But this definition is hopelessly inadequate for any but the simplest regions. For example, a circle of radius 1 supposedly has as area the irrational number 7r, but it is not at all squares, with sides of length ^ (a, _ 0) 0) what u w squares" means. Even if we consider a circle of radius 1 /vV, which supposedly has area 1 it is hard to say in what way a unit square fits in this circle, since it does not seem possible to divide the unit square into pieces which can be arranged to form a circle. In this chapter we will only try to define the area of some very special those which are bounded by the horizontal axis, the vertiregions (Figure 1) cal lines through (a, 0) and (6, 0), and the graph of a function /such that f(x) > 0 for all x in [a, b]. It is convenient to indicate this region by R(f, a, b). Notice that these regions include rectangles and triangles, as well as many clear FIGURE 1 , — other important geometric figures. area of R(f, a, b) will be b\ Actually, the integral will be defined even for functions / which do not satisfy the condition f(x) > 0 for all x in [a, b). If / is the function graphed in Figure 2, the integral will represent the difference of The number which we called the integral off on will eventually assign as the [a, the area of the lightly shaded region FIGURE 2 interval [a, b] to, a 214 is h, = t2> to t 3, < t\ t± with < h < ts < ti of the heavily shaded indicated in Figure has been divided into four subintervals by means of numbers i and the area region (the "algebraic area" of R(f, a, b)). The idea behind the prospective definition = b 3. The Integrals (the numbering of the will equal the maximum value Mi; m and value of / be s subscripts begins with 0 so that the largest subscript number On the first interval £ = m x (h - of subintervals). h] the function /has the minimum value mi and the ti] let the minimum similarly, on the ith interval [h, let f0 maximum the )'+ rn 2 (t2 - The sum value be Mi. +m h) - (h z +m t%) 4 (*4 - represents the total area of rectangles lying inside the region #(/, the FIGURE 215 h) a, b), while sum 3 S = Mi(h - to) M + 2 (t 2 ~ M (h - h) + h) + z M,(U - h) represents the total area of rectangles containing the region R(f, a, b). The guiding principle of our attempt to define the area A of R(f, a, b) is the obser- vation that A should satisfy < A s A < and S, and that this should be true, no matter how the interval [a, b] is subdivided. It is to be hoped that these requirements will determine A. The following definitions begin to formalize, and eliminate some of the implicit assumptions in, this discussion. DEFINITION Let a [a, b\, The < b. A partition of the interval one of which points in a partition can be a we DEFINITION shall < to ti < • a finite collection of points in is b. numbered t Q, . < < tn • • fw -l . . = 5 tn so that b\ always assume that such a numbering has been assigned. Suppose / [a, b]. = [a, b] is and one of which is a, is bounded on [a, b] and P = {* 0 , • • • , *n} is a partition of Let m = Mi = { The lower sum inf {/(*): sup {/(*): fe-i < < x x of / for P, denoted by L(f, P), < < ti}, h}. defined as is n Uj, P) = The upper sum J rmiu - ti-i). of / for P, denoted by U(f, P), is defined as n U(f, P) = J Miih The lower and upper sums correspond - to the sums s and S in the previous 216 Derivatives and Integrals example; they are supposed to represent the total areas of rectangles lying below and above the graph of /. Notice, however, that despite the geometric motivation, these sums have been defined precisely without any appeal to a concept of "area." Two comment. The requirement that / be rrii and Mi be defined. Note, also, that it was necessary to define the numbers m and Mi as inf s and sup's, rather than as minima and maxima, since / was not assumed continuous. One thing is clear about lower and upper sums: If P is any partition, then details of the definition deserve bounded on [a, b] is essential in order that all the { < L(f, P) £/(/, P), because n Uf, = P) J= i m&t l n U(f, P)= T = i and for each i we have mi(ti On the other hand, FIG URE any 4 two - Mi<M l partitions of - < Mi(k - something [a, b], then less it jf t _i). obvious ought to be true: If Pi and Ps are should be the case that < L(f, Pi) P £/(/, 2 ), because L(f, Pi) should be < area P(/, a, b), and U(f, P 2 ) should be > area b). This remark proves nothing (since the "area of P(/, a, b)" has not even been defined yet), but it does indicate that if there is to be any hope of P(/> defining the area of P(/, first. <z, The proof which we b), a proof that L(f, Pi) < U(f, P2 ) should come are about to give depends upon a lemma which concerns the behavior of lower and upper sums when more points are included in a partition. In Figure 4 the partition P contains the points in black, and Q contains both the points in black and the points in grey. FIGURE 5 that the rectangles region R(f, lemma If Q a, b) contains P drawn for the partition than those (i.e., if all for the original partition P. points of L(/, P) P) £/(/, PROOF Consider first The P picture indicates are a better approximation to the Q To be precise: are also in Q), then < L(/, > U(f 9 Q), Q). the special case (Figure 5) in which Q contains just one point than P: P = Q = {^o, • • . {*o> • • • , tn }, • , • • , t n }, where a = t0 < h < - • • < < m < ffc < • • • < /„ = b. more 217 Integrals Let m! = m" = inf {/(#): tk -i < # < * < inf {/(*): w < u} 9 fe}. Then ^ n = £ L(/, P) ? To = - fe-i), i Jb-1 n t=l i=k+l prove that < P) Q) m k {tk - Now x < < the set {/(#): u and , < a: than or equal - m'iu tk tk + -i) contains ] some smaller possibly first set is less < m"(th u). numbers the all — that in {/(*): ones, so the greatest lower the greatest lower to show therefore suffices to it mk < bound - tk bound -i < of the of the second; thus m! Similarly, m k < m" Therefore, = m k (u — m k {tk — This proves, in > U(f, P) -i) +m k {tk — < u) m'(u special case, that L(f, P) this U(f, Q) tk is similar, and < you left to is - k_ t x) L(f, Q). as + mn (t k — u). The proof that an easy, but valuable, exercise. The now be deduced quite easily. The partition Q can be by adding one point at a time; in other words, there is a general case can P obtained from sequence of partitions P = Ph P such that Pj+ contains just one i = L(f, P) L(f, Pi) 2, . . . Pa = Q , more point than Pjt Then < US, Pi) < U(f, ft) > • • • < US, Pa) = L(f, Q), and U(f, P) = U(f, PO > The theorem we wish THEOREM i Let P and P x [a, b]. 2 to prove be partitions of [a, b], is There • • > U(f, P.) = £/(/, Q). I a simple consequence of this lemma. and be a function which let / is bounded on Then L(f, Pi) PROOF • is a partition in both Pi and P P which 2 ). US, 9 2 ). and lemma, contains both Pi According Pi) < U(f P < US, to the P) < u(f, P) < P 2 (let u(f, P consist of all points p ). 2 | n from Theorem 1 that any upper sum £/(/, P') is an upper bound lower sums L(f, P). Consequently, any upper sum £/(/, P ) is greater than or equal to the least upper bound of all lower sums: It follows f for the set of all ?):?a sup {£(/, partition of for every P set of all upper sums of /. Consequently, f . means This, in turn, that sup {L(f, P)} sup {£(/,/>)} It is clear that sum of / < [a, b]} < U(f, P'), is a lower bound for the inf {£/(/, P)}. both of these numbers are between the lower sum and upper for all partitions: £(/,/*)< sup \L(f, P)} < £/(/, />'), P') <inf ([/(/,?)) < U(f,P'), L(f y for all partitions P'. It may well happen that sup{L(/,P)} in this case, this / for is all partitions, area of R(f, a, b). the <m/y = inf {£/(/, P}; number between the lower sum and upper sum of number is consequently an ideal candidate for the and this On the other hand, sup [L{f,P)\ if < inf \U(f,P)}> then every number x between sup {£(/, P) p')< L{f, < x and } inf {U(f, P)} will satisfy u(f, for all partitions P'. It is The not at all clear just when such an embarrassment of riches will occur. following two examples, although not as interesting as many which will soon appear, show that both phenomena are possible. Suppose first that f(x) = c for any partition of [a, b\ then is all x in rrii — Mi = [a, b] (Figure 6). If P = c, so n L(f, P) = <*< ~ c(h - ^= = <b ~ a ). - a). l i n (/(/, P) = J= t In this case, all = c{b lower sums and upper sums are equal, and sup Now <,_,) 1 P)} - inf {U(f, P)} = c(b consider (Figure 7) the function / defined by v * ' _ ~ I 0, x irrational I 1, x rational. - a). {^o, . • • , tn \ Integrals If P = {fo, • • • > - 0, Af,= 1, rrii any tn] is 219 partition, then since there an is irrational number in [ff-_i, /»•], and since there number a rational is in Therefore, n a = t0 h t% tn-l tn L(/, P) =b = ('«• 1 = - **-0 = °» - fe-i) = b 1 n FIGURE £/(/, 7 P) = i Thus, in The this case (U ' 1 - a. i certainly no/ true that sup {£(/, P) it is upon which principle J= insufficient information to the definition of area was = } inf { U(f, P) }. to be based provides determine a specific area for R(f, 6)—any num- a, — a seems equally good. On the other hand, the region ber between 0 and b refuse to assign it any area at R(J, a, b) is so weird that we might with justice In all. fact, we can maintain, more *inf sup {L(f,P)} the region R(f, whenever generally, that P)l {£/(/, too unreasonable to deserve having an area. As our "unreasonable" suggests, we are about to cloak our a, b) is appeal to the word ignorance in terminology. DEFINITION A function / which sup { L(f, P) : Pa bounded on partition of [a, b] integrable on [a, b] is = } common number In this case, this is is is inf { U(f, P) : [a, b] if P a partition of [a, called the integral of /on [a, b] b] . and denoted by /.'/• (The symbol J is called an "sum;" the numbers a and The in If integral j* f is integral sign an elongated s, for and upper limits of integration.) and was originally b are called the lower also called the area of R(J 9 a, b) when /(*) > 0 for all x [a, b]. / is integrable, then according to this definition, L(f, P) < b J f^ U(f> b Moreover, j f is the unique ^) for a11 Partitions number with P of [a, b]. this property. solve, the problem discussed (nor do we know how integrable are functions which before: we do not know present we know only to find the integral of/ on [a, b] when / is integrable). At This definition merely pinpoints, and does not 220 Derivatives and Integrals two examples: (1) if f(x) = then /is integrable on c, [a, b] and J* f — c • (b — a). (Notice that this integral assigns the expected area to a rectangle.) (2) if /(*) = ( 1, I Several Even ther. * irratlonal x rational, ^' more examples will for these examples, then y not integrable on [s [a, b]. be given before discussing these problems furhowever, it helps to have the following simple criterion for integrability stated explicitly. THEOREM 2 If / S > is bounded on 0 there is j>, b], then / is integrable on P of [a, b] such that U(f, P) PROOF Suppose that for every s first [a, b] if and only if for every a partition > - 0 there - U(f, P) < L(f, P) is s. a partition < L(f, P) P with e. Since vai\U<J,F)\ < U(f,P), supjL(/,P')} >L{f,P), it follows that inf {£/(/, P')} Since this is true for all £ > - sup {£(/,/>')} 0, it follows <£ that sup {L(f,P')} =inf{t/(/,P')}; by definition, then, lar. If / is / is integrable. The proof of the converse = sup \L{f,P)} This means that for each e > P -wS{U{f,T)). 0 there are partitions P', U(f, P") Let assertion is simi- integrable, then - L(f, P') < P" with e. be a partition which contains both P' and P". Then, according to the lemma, U(J, P) L(f, P) < > U(f, P"), L(f, P'); consequently, U(f, P) - L(f, P) < U(J, P") - L(f, P') < s. | Although the mechanics of the proof take up a little space, it should be clear Theorem 2 amounts to nothing more than a restatement of the definition that Integrals of integrability. Nevertheless, 221 a very convenient restatement because there it is no mention of sup's and inf s, which are often difficult to work with. The next example illustrates this point, and also serves as a good introduction to the type of reasoning which the complicated definition of the integral necessi- is tates, even in very simple situations. Let / be defined on by [0, 2] :-\. Suppose P= {*o, • . . , tn a partition of is ] tj-! (see Figure 8). < < 1 t [0, 2] with } Then mi = Mi = mj — 0 if ^ /, i but and 0 Mj = 1. Since t=J+l we have - U(f, P) This certainly shows that / L(f P) 9 tj - h-x. integrable: to obtain a partition is - U(f, P) it is = L(f, P) < P with e, only necessary to choose a partition with < t,-i Moreover, it is / is and tj tj — tj-i < s. clear that < L(J, P) Since < 1 < 0 integrable, there is U(f, P) for all partitions P. only one number between all lower and upper sums, namely, the integral of /, so f/ = o. Jo Although the discontinuity of / was responsible for the difficulties in this example, even worse problems arise for very simple continuous functions. For example, let /(#) = x, and for simplicity consider an interval [0, b], where is a partition of [0, b\ then (Figure 9) tn b > 0. If P = {t^ . . . ) , mi — and Af»- = U and therefore n L(f, P) = 2= = ti-iCh - l t to(h - t 0) + h(t 2 - + h) • • • + *n _i(*„ - f„_i), n U(f, P) = ^= t U(U - Neither of these formulas siderably for partitions case, the length U ti-x) 1 — is Pn = particularly appealing, but both simplify con{^ 0 , • . t , n into ) of each subinterval = to is equal subintervals. In this b/n so f o, h = — ft > n <2 2b — = > etc; n in general, ti ib = n Then n Uf, Pn) = ^ - /,-l(<< /<-l) -m n-l J Remembering —o the formula 1+ ... +A _*C*±2>, 2 this can be written = " n b 1 . ' _l. 2 Integrals 223 Similarly, n U(f, Pn) - = ^ = V+ n(n = - • 1) 2 + * = ^ * 1 . * n 2 rz h l. 2 2 very large, both L(f, Pn ) and U(f, Pn ) are close to b /2, and this remark that first Notice makes it easy to show that / is integrable. If * is Pn ) - U(f, Pn This shows that there are partitions desired. By Theorem / 2 the function be found with only a little L(f, />„)=-• L(f, with is £/(/, Pn ) - Pn < | < ) U(f, Pn) as small as may now Jq f that It is clear, first of all, work. Pn ) L(f, integrable. Moreover, for all n. /2 lies between certain special upper and lower sums, but we have just seen that U(f, P n ) - Uf, Pn) can be made as small as desired, so there is only one number with this property. Since the This inequality shows only that b 2 we can conclude integral certainly has this property, that jT'-r equation assigns area b 2 /2 to a right triangle with base and (Figure 10). Using more involved calculations, or appealing to Notice that altitude b Theorem this 4, it can be shown that 2 2 The function f(x) (Figure 11), if P= mi = = {t 0 , x 2 . /(<*_!) presents even greater difficulties. . . = , tn } is fe_i) 2 a partition of and Choosing, once again, a partition P» so that = Mi = {fo, . - [0, f(ti) . , In this case b\ then = k 2 . **} into « equal parts, the lower and upper sums become n Uf, P.) £<•-<>'-;•n2 Z-/ i = n l n-l n ^ U(f, Pn) k L4 2 * - fa-i) + 1)(2* {U n2 n Recalling the formula 1" from Problem +••*+* — 2-1, these sums can be written + 1) as =V7(»-D(»)(2«-1), rr 6 f/(/,P») =-'^»(«+l)(2« w It is s + l). 6 not too hard to show that L(f, and that U(f, Pn ) sufficiently large. - Pn < y < U(f, ) Pn ), can be made as small as desired, by choosing n The same sort of reasoning as before then shows that L(f, Pn ) J Jo 3 — This calculation already represents a nontrivial result the area of the region hounded by a parabola is not usually derived in elementary geometry. Neverthe theless, the result was known to Archimedes, who derived it in essentially we chapter same way. The only superiority we can claim is that in the next will discover a much simpler way to arrive at this result. 225 Integrals Some of our investigations can be summarized as follows: f = j c ' — (b / = ^2 £2 -— / 3 = *! a) if f(x) = c if /(*) = * for if /(*) = for all *, all *, 3 This ~^ 3 for all x 3 already reveals that the notation j* f suffers from the lack of a con- list venient notation for naming functions defined by formulas. For alternative notation,* analogous to the notation lim /(#), J* means f(x) dx same precisely the is this reason an also useful: j* f. as Thus rb dx = x ax = c c 6 (b • 2 — a), a2 — 2 2 = - x 2 dx 3 3 Notice that, as in the notation lim /(*), the symbol x can be replaced by any x—*a or 6, of course) = p(t) dt = other letter (except * /, /;/(«) <** = /;/« <*» = *. has no meaning in isolation, any more than the symbol has any meaning, except in the context lim /(*). In the equation The symbol = x 2 dx °—i 3 3 the <?nf*re may symbol x 2 dx be regarded as an abbreviation the function / such that /(*) = x 2 for & * The notation J /M dx is actually the older, x—> and was for many for: all x. years the only, symbol for the integral. Leibniz used this symbol because he considered the integral to be the sum of infinitely many rectangles with height /(*) and "infinitely small" width dx. (denoted by J) Later writers used x 0 by Axi. The , integral . . . , xn to denote the points of a partition, and abbreviated Xi was defined as the limit as Ax { approaches 0 of the sums ^= i (analogous to lower and upper sums). f(xi) to /(*), and Axi to dx, delights — n The fact that the limit many people. is /(*»•) Ax { i obtained by changing 2 to /, 226 Derivatives and Integrals This notation for the integral examples may we have made rb / x +y)dx= / Ja fx (y + +^ (1 (3) (fa Ja Ja £ fx xi dX = t + 0 j' f(f (,+,)*)*- £[x = of f* x dx (x - a) t(x - a). dX dx (i+4f-?-«(*-4 = The computations £ a2 ~^ la (1 a2 ---+ + (1 ) = dy 6.* -~--+y(b-a). fx = (4) ydx = / Ja dZ and 5 k2 rb xdx+ jjdy + ^ t)dy= Theorems use of rb ( (2) as flexible as the notation lim fix). Several x—>a aid in the interpretation of various types of formulas which frequently appear; (1) is {d -c)+ <l-l]dx C (~- ^(b-a) + ^ (d-c) f\dx and j* x 2 dx may suggest that evaluating integrals is generally difficult or impossible. As a matter of fact, the integrals of most functions are impossible to determine exactly (although they may be com- puted less, to as any degree of accuracy desired by calculating lower and upper sums). Nevertheshall see in the next chapter, the integral of many functions can be we computed very easily. Even though most integrals cannot be computed exactly, it is important at least to know when a function / is integrable on [a, b\ Although it is possible to say precisely which functions are integrable, the criterion for integrability is a little too difficult to be stated here, and we will have to settle for partial results. The most useful result will be stated now, but proved at the end of the next chapter. THEOREM 3 If / is continuous on (Notice that it is [a, b], then / integrable on is [a, b]. unnecessary to assume that / is bounded on [a, b\ because of continuity.) * Lest chaos overtake the reader when he reads other books, equation (1) requires an impor- tant qualification. This equation interprets / y dx to mean the integral of the function / such f& that each value /(*) is the number^. But classical notation often uses y for_y(x), so might mean the integral of some arbitrary function y. / y dx 227 Integrals Although this theorem of integrals in this book, will provide all the information necessary for the use more it is have a somewhat larger supply satisfying to of integrable functions. Several problems treat this question in detail. It will help to know the following three theorems, which show that / is integrable on [a, b], if it is integrable on [a, c] and [c, b]; that / + g is integrable if /andg are; and that c / is integrable if / is integrable and c is any number. As a simple application of these theorems, recall that if / is 0 except at one point, where its value is 1 then / is integrable. Multiplying this function by c, it follows that the same is true if the value of / at the exceptional point is c. Adding such a function to an integrable function, we see that the value of an integrable function may be changed arbitrarily at one point without destroying integrability. By breaking up the interval into many subintervals, we see that the value can be changed at finitely many points. • , The proofs of these theorems usually use the alternative criterion for inteTheorem 2; as some of our previous demonstrations illustrate, the grability in obscure the point of the proof. It is a good idea to attempt proofs of your own, consulting those given here as a last resort, or as a check. This will probably clarify the proofs, and will certainly give good practice in the techniques used in some of the problems. details of the THEOREM 4 argument often conspire Let a < [c, b]. Conversely, [a, b]. c < b. Finally, If / if is integrable if /is / on b\ then / [a, integrable on integrable on is to integrable is and on [a, c] b], [c, on and on [a, c] then /is integrable on then [a, b], />-/:/+/> PROOF Suppose/ is integrable on of > b].If£ [a, there is P = a partition {t . Q, . . , t n} such that [a, b] - U(f, P) We might as well assume that partition which contains t 0 . , L(f, Q) < Now P U(f, P) = {/ 0 , a partition of y 0, [c, . b] . . L(f, P) . is tj} , = c . < , tn £(/, P) some for tj and < e. then c; (Otherwise, /. Q Q be let the contains P, so £/(/, Q) - e.) a partition of [a, c] and P" = {*,-, . . . , *»} is (Figure 12). Since a/, p) U(J, P) = us, = U(f, n P') + + uf, n, U(J, P"), we have p" P' n a i i i i i n= r i i c i i i m /„ [U(f, P') = b - Uf, P')] + [u(f, p") Since each of the terms in brackets shows that / is integrable on [a, c] Uf,n is and Uf, P")} = u(f, p) nonnegative, each [c, b]. Note Uf, p") < f*f < is less also that < p<u(f,p'), u(f, P"), - Uf, than p) e. < «• This 228 Derivatives and Integrals so that Since this any true for is b < L{f,P) f+ fo fb°f<U(f,P). P, this proves that /:/+/;/-/.*/• Now suppose that / partition P' of integrable on is and a [a, c] partition U(f, P') U{f, P") If P is the partition of [a, b] = = L(f, P) of + + U(f,P') If £ > there 0, is a e. I 6/2, e/2. the points of P' all L(f, P') [c, b]. such that b] [c, - L(f, P') < - L(f, P") < containing U(f,P) and on [a, c] P" and P", then L(f, P"), U(f,P"); consequently, - U(f, P) = L(f, P) Theorem 4 [U(f, P') the basis for is was defined only Ja f for a /;/=0 With even - L(f, P')] a < < c b is [U(f, P") - L(f, />")] some minor notational conventions. The < b. and We now add the ///--/;/ these definitions, the equation if + fcf+ < integral definitions if«>*. = f*f f*f not true (the proof of this assertion holds for is all a, 5 If / and g are integrable on [a, b], then / +g is integrable on [a, b] //(/+*)-/:/+/> PROOF Let P = {*<>, . • - , U be any partition of mi = m/ = m/ = r and define M iy M/, inf {(/ inf \g(x): M" similarly. it is t^i < t^ < x It is * x < < + m/', > m/ + m/'. Mi < Mi + true (Problem 7) that rrii < x Let < ti\, fe}, t-), not necessarily true that — m/ rrii but + g)(x): inf j/0): [>, £]. Similarly, Mi". b a rather tedious case-by-case check). THEOREM c, and Integrals 229 Therefore, L(f,P)+L(g, P)<L(J + g,P) and + g, U(f < U(f,P) P) + U(g, P). Thus, Uf, + P) P) Ug, < Uj + g,P)< U(J + g,P)< Since / and g are integrable, there are partitions P') U(f, P') U(g, P") - Ug, P") US If P contains both P' s/2. + Ug, U(g, P). P" with P', < < + 8/2, and P", then + U(f, P) U(g, P) and consequently - [Uf, P) + g,P) -L(f + g,P)< U(f + g integrable on b\ US P) + Lis, P) ^ L (f + s, P) This proves that / (1) U(f, P) [a, is P)] < «, S. Moreover, </;(/+,) < U(f + and Since US, desired, it Uf, P) ~ P) + Uf, Ug, P) < b J +f g< Wf, \ + is on [a, integrable b] - Ug, - u(g, P) /:</+*> proof + U{g,P); + u(g, P). P) can both be made as small as follows that can also be made as small as desired; If / P) a P) and U{g, p) u(f, p) 6 < U(f,P) also (2) THEOREM g,P) on it [Uf, P) + Ug, P)] therefore follows from (1) and (2) that -/:/+/>• [a, b], then for any number much easier c, the function cf is integrable and The proof (which good idea is than that of Theorem to treat separately the cases c > 0 and c < 5) 0. is left to you. It Why? is a | theorem that / g is (Theorem 6 is just a special case of the more general hard to prove quite is result this but are, integrable on [a, b], if / and g (see Problem 32).) complicated definition, a tew In this chapter we have acquired only one theorem with no proof at all. simple theorems with intricate proofs, and one • 230 Derivatives and Integrals This is not because integrals constitute a more difficult topic than derivatives, but because powerful tools developed in previous chapters have been allowed remain dormant. The most significant discovery of calculus is the fact that and the derivative are intimately related—once we learn the con- to the integral nection, the integral will become The connection between derivatives but the preparations which We first many THEOREM 7 we as useful as the derivative, and and as easy to use. integrals deserves a separate chapter, make in this chapter may serve as a hint. concerning integrals, which plays a role in will state a simple inequality important theorems. Suppose / on integrable is m < and that [a, b] M < f(x) for all x in b]. [a, Then i(b PROOF It is clear - b < a) f<M(b - Ja a). that m(b -a)< and L(f, P) for every partition P. Since J* f = < M(b - U(f, P) sup {£(/, P)\ = a) inf {£/(/, P)}, the desired inequality follows immediately. | on Suppose now that / [a, b] by integrable on is [a, b]. We can define a new function F c+h)-F(c) = /;/ = f{x) c c + dt. Theorem 4.) We have seen that /may be integrable even if not continuous, and the Problems give examples of integrable functions (This depends on h it is FIGURE /;/(*) 13 which are quite pathological. The behavior of F is therefore a very pleasant surprise. THEOREM 8 If / integrable on is [a, b] and F is defined on f(x) F then PROOF is Suppose c on [a, b]; continuous on is in let [a, b\. [a, Since / > 0, by /;/, b]. is integrable on M be a number such that < M [a, b] it is, for all x in |/(*)| If h = [a, b] [a, by definition, b\ then (Figure 13) f(, + a)-f ( ,) = + /; v-/;/ = Since —M < f(x) < M for all x, + /; V. bounded / — t 1 a ^J- F f FIGURE it 14 follows from Theorem 7 that -Af 'A< j C+h f<Mh; c in other words, (1) — Af • h < F(c + h) - < AT- A. If h < 0, a similar inequality can be derived. Note that Applying Theorem 7 to the interval MA < — 1, multiplying by Mh < and Inequalities (1) if £ > 0, + h) - \h\ words F < e/M. is F is true this M - \h\. F(c)\ < e, This proves that continuous at + c. = h) F(c); \ = Figure 14 compares / and F(x) that < -Mh. F{c) < F(c)\ + h) - lim F(c in other we have we have \F{c provided that -Mh- - h) we obtain of length —h, the inequalities, all + F{c c], can be combined: (2) \F{c Therefore, f < f +h Jc which reverses (2) + h, 0 e for various functions /; J* f always better behaved than In the next chapter /. we it appears will see how is. PROBLEMS 1. Prove that J* x z dx = 4 by considering b /4, partitions into n equal sub- n intervals, using the formula for ^ = i i z which was found in Problem 2-5. i This problem requires only a straightforward imitation of calculations in the text, but you should write it up as a formal proof to make certain argument are that all the fine points of the 2. *3. Prove, similarly, that (a) Using Problem 4 J* 2-6, x dx = b /5. n show Y that * to l/(p (b) 4. + = = k p /n p+1 can be b p+l /{p + made n large enough, 1). Decide which of the following functions are integrable on calculate the integral / • \ (0 when you X, 0 < /W = U-2, 1< r, \ ( can. X < X <2. 1 as close i by choosing 1) as desired, Prove that j* x* dx clear. b [0, 2], and 233 Integrals /(*) (iii) /(*) /• /W // \ (1V) (v) - x = + * = * 0, 1 2, * rational [*], # of the form a * not of this form. = / 5. < * * = 0 or * shown the function is < 0 for rational # b and b 1 [i] 0, (vii) + 1, 0, 1 f{x) 1 2. * irrational. f{x) (vi) < < x * [x]. + f ( \ < < 0 x, (ii) > 1. in Figure 15. Find the areas of the regions bounded by = (i) the graphs of fix) (ii) the graphs of fix) x2 x2 = = — and (iii) (iv) (v) 2. = — x 2 and x 2 and g(x) through (-1, 0) and (1, 0). the graphs of fix) — x 2 and g(#) the graphs of /(*) = x 2 and the graphs of /(*) = a: 2 and g(x) + = = = the vertical lines —x - x and hix) = 2. - 2x + 4 and the 2 1 . 2 1 x2 vertical axis. = V*, the graph of fix) (vi) the horizontal axis, 2 through (2, 0). (Don't try to find and the vertical line you should |q see a way of guessing the answer, using only integrals that you already know how to evaluate. The questions that this example should suggest are considered in Problem 16.) 6. Find I!Uc f^siy)dy)dx d b in terms of and f j f a vengeance; 7. (This problem crucial that Prove, using the notation of m.:/ 8. it is g. (a) + Which mi" = inf {/(a) is an exercise in notation, with you recognize a constant when Theorem + g(x 2 ): 5, it appears.) that < *i, *2 < **} <m functions have the property that every lower f. sum equals every upper sum? (b) Which some (c) functions have the property that (other) lower Which continuous are equal? some upper sum equals sum? functions have the property that all lower sums 234 Derivatives and Integrals Which *(d) integrable functions have the property that mind are equal? (Bear in = that one such function lower sums all is = f(x) 0 for x /q for x — p/q in lowest terms.) Hint: You will need the notion of a dense set, introduced in Problem 8-6, as well as irrational, f(x) Problem the results of 9. Ha<b<c<d and / on 10. Prove that (a) in [a, warn / if > f* f Prove that (b) 21. integrable on is d\ prove that / [a, integrable is (Don't work hard.) [b, c]. then 11. 1 if integrable on /and g b\ then that is and [a, b] f(x) > 0 for x in all [a, b], 0. are integrable on > Ja f g> you work hard on part if [a, now (By and f(x) > b~\ g(x) for all x should be unnecessary to it you are wasting time.) (b) Prove that = J* I! X*) f(* - <) d*. (The geometric interpretation should make this very plausible.) Hint: Every partition P = of [a, b] gives rise to a partition tn 0 { P' *12. = {to + c, • • , , . . Ji Hint: P= of can This {t 0 , . [b, ab], . . , and tn I Ji be } + t written b c, + and conversely. c], of [1 , -dt fab = I -dt. Ji t t j"\/tdt = j* a] gives rise to h \/tdt. a partition Every partition P = r {bt 0 , . . . , bt n \ conversely. Prove that b jl c f(t) dt (Notice that Problem 12 = e £ f(ct) dt. a special case.) is Suppose that /is bounded on in [a, b] and that /is continuous with the exception of # 0 in (a, b). Prove that / Hint: Imitate one of the examples in the text. [a, b] [a, b]. 15. + rb\ l ~dt 14. \ , of [a Prove that fa *13. . + c] tn Suppose that / is nondecreasing on [a, b]. Notice that / bounded on [a, b], because f(a) < f(x) < f(b) for x in (a) If P= {fo, ...,*»} is a partition of - = d for [a, b], is what at each point integrable is on automatically [a, is b\ L(f, P) and U(f, P)? (b) Suppose that k each i. Prove that U(f, P) - L(f, P) = (c) 8[f(b) -/(«)]. Prove that / is integrable. (d) Give an example of a nondecreasing function on discontinuous at infinitely many points. [0, 1] which is 235 Integrals might be of It compare interest to problem with the following this extract from Newton's Principia.* LEMMA II AacE, terminated by the right lines there be inscribed any number of parallelograms If in any figure acE, prehended under equal bases &c, bLcm, cMdn, &c, AB, BC, CD, &c, and Aa parallel to one side and their which say, that the ultimate ratios number aKbl, be augmented in infinitum, to the inscribed figure AKbLcMdD, AalbmcndoE, and curvilinear figure AabcdE, circumscribed figure to Bb, Cc, Dd, the sides, of the figure; and the parallelograms are completed: then if the breadth of those parallelograms be supposed to be diminished, / Aa, AE, and the curve Ab, Be, Cd, &c, com- the will have one another, are ratios of equality. For the difference of the inscribed and circumscribed figures is the sum of the parallelograms Kl, Lm, Mn, Do, that is (from the equality of all their bases), the rectangle under one of their bases Kb and the sum of their altitudes Aa, that is, the rectangle ABla. But this recits breadth AB is supposed diminished in infinitum, than any given space. And therefore (by Lem. 1) the figures inscribed and circumscribed become ultimately equal one to the other; and much more will the intermediate curvilinear figure be tangle, because becomes less ultimately equal to either. Q.E.D. *16. Suppose that /is increasing. Figure 16 suggests that fr = rw -ama)(a) If P = f~ {h, . . . {t n )\. L{f~\ P) 17. Now (c) Find j* A P = V* dx s defined on {t Q , . . Prove that if 2 Newton's P' = [f~Kt Q ), . . . . P) = f U(f, bf~Kb) - af~Ka). . / is a < [a, b] is *„} , < of b. may is a parti- constant on each 0 called a step function [a, b] such that s is if there be arbitrary). integrable on - [a, b], then for any Principia, si and s > 0 there is a also a step function 2 2 * [a, b], let < f with f si < e, J* f* — ^ > / with s < e. Ja f Suppose that for all s > 0 there are step functions si < f and > / such that s ^ Si < e. Prove that / is integrable. step function (b) + for 0 (the values of s at h (a) a partition of prove the formula stated above. function tion is ) Prove that, as suggested in Figure 17, l (b) tn , /;;;:/. J* A 2 f* Revision of Mott's Translation, by Florian Cajori. University of Cali- fornia Press, Berkeley, California, 1946. (c) Find a function / which = j f 18. Prove that some L(f, P) for if / <f tinuous function g P partition on integrabie is not a step function, but which is f [a, b]. > then for any £ [a, b], with of — satisfies g < 0 there is a con- Hint: First get a step £. function with this property, and then a continuous one. A picture will si + help immensely. 19. (a) Show that (b) Prove, (c) Use part such that integrabie on is = J* f f. ble to choose x to be in 21. are step functions on The purpose s 2 is also. Si Ja + Ja s2 . Prove that there [a, b]. that in always possi- not it is number x a is (a, b). of this problem P= {/ 0 , < • . — b sup {/(*): choose (c) = s2 ) to is show that many / if points in is integrabie on [a, b], [a, b]. tn be a partition of [0, 1] with U(f, P) Prove that for some i we have Mi — mi < 1. Prove that there are numbers a\ and b\ with a < a\ < b\ < b and Let L(f, P) (b) + (si Show by example then / must be continuous at (a) then [a, b], that 5, 5. Suppose that / [a, b] s2 (and Problem 17) to give an alternative proof of (b) Theorem 20. and if Si without using Theorem < = ai [ai, bi] • , ] < bi} a. * /J — inf {/(#): ai from part < < = x (a) unless i < bi) or 1 two cases a very simple device solves the problem.) Prove that there are numbers a 2 and b 2 with a\ < a 2 < sup {/(*): a 2 (d) Continue in x < b2 ] - inf \f(x): a 2 < x < b2 } n; (You can and in these < b2 < 1. < and bi i. this way to find a sequence of intervals In = [a n b n ] such that sup {/(*): a: in 7n - inf \f(x): x in In < 1/n. Apply the Nested Intervals Theorem (Problem 8-14) to find a point x at which , ) } / (e) 22. is continuous. Prove that / Recall, from (a) is continuous at infinitely Problem f*f > 10, that > Give an example where f(x) b\ and f* f = 0. (b) Suppose f(x) > 0 for 0 many > 0 for if /(*) all x, points in 0 for and f(x) all > [a, b], x in 0 for [a, b\ some x in [a, and f(x 0 ) > 0. all x in lower sum L(f, P) which (c) Suppose / is integrabie on is b Prove that > j f 0. Hint: [«, b] f > Prove that and / is continuous 0. Hint: at * 0 in [a, b] It suffices to find one positive. [a, b] You and will f(x) > 0 for all x in [a, b]. need Problem 21; indeed that was one reason for including Problem 21. Integrals *23. (a) Suppose that / is continuous on and [a, b] = tinuous functions ^ on [a, b]. Prove that / is an obvious g to choose.) (b) Suppose / is continuous on tinuous functions g on = g(a) g(b) = which [a, b] = Prove that / 0. and that [a, b] f* fg 0. = 0 for (This = J fg satisfy the is all 237 con- easy; there 0 for those conextra conditions (This innocent looking fact 0. is an important lemma in the calculus of variations; see the Suggested Reading for references.) Hint: Derive a contradiction from the assumption f(x Q ) > 0 or f(x 0 ) < 0; the g you pick will depend on the behavior of / near x 0 24. *25. — Let f(x) (a) Compute (bj Find inf = Let f(x) that / is 27. L(f, P) for {£/(/, 0 for irrational you must . and all P):Pa integrable on clearly 0; *26. x for x rational = f(x) 0 for x irrational. P partitions of partition of [0, and \/q x x, [0, 1] if and that figure out how to [0, 1]. 1]}. — p/q in lowest terms. Show = 0. (Every lower sum JQ f make upper sums small.) is Find two functions / and g which are integrable, but whose composition g o f is not. Hint: Problem 25 is relevant. for all x in [a, b], then (a) Prove that if m < f(x) < M f*f(x) dx for (b) some number Prove that if / is = - (b a) ii fx < M. continuous on [a, b], jjl with m < then b dx= (b- a)M) jJ{x) some for J in [a, b]; This fact essential. is and show by an example that continuity is called the (First) Mean Value Theorem for Integrals. 28. (a) Suppose that / is continuous on [a, nonnegative on [a, b]. Prove that '/(*)*(*) dx \l for some J in [a, b]. = /(*) This result is b] and that g is integrable and \lg(x)dx called the Second (or Generalized) Mean Value Theorem for Integrals. Hint: The < f(x) < implies that mg(x) < f(x)g(x) < Mg(x), M inequality since g is nonnegative. (b) The Show First that the hypothesis for g is essential. and Second Mean Value Theorems not very essential way) in Chapter 19. The (Tryg(*) = # on [— 1, 1].) be used (in a problem, which for integrals will result of the next I have cribbed from several older calculus texts, is called the Third Mean Value Theorem for Integrals. To be quite frank, I haven't the slightest idea what it is good for. 238 Derivatives and Integrals *29. Suppose that g is increasing on [a, b] and that / is integrable on [a, b]. Prove that there is a number £ in [a, b] such that (a) p(t)g(t) + g(b) g{a) ///(<) dt f*f(t) dt. = a and for £ = b? on g is essential. (Try g(a) = g(b) = 0.) Let / be a bounded function on [a, b] and let P be a partition of [a, b]. Let Af and rrii have their usual meanings, and let M/ and m/ have the corresponding meanings for the function |/|. (b) *30. = dt Hint: What Show that the hypothesis the value of the right side for £ is t (a) (b) (c) (d) Prove that 31. max(|Af<[, m/ = min(|Afi|, \rrii\), |iw,-|). Prove that Af/ — m/ < Mi — Hint: You will have to consider two separate cases, because that's how max and min are defined. Prove that if / is integrable on [a, b], then so is |/|. Prove that if / and g are integrable on [a, b], then so are max(/, g) and min(/, (e) M/ = g). Prove that / is integrable on [a, b] if and only if its "positive part" max(/, 0) and its "negative part" min(/, 0) are integrable on [a, b]. Prove that / if is integrable on [a, b], |/.Vw*|< then Jl i/w dt. i Hint: This follows easily from a certain string of inequalities; Problem 1-14 is relevant. *32. and f{x) > 0 for all x in [a, b]. and m/ denote the appropriate and m/' similarly for g, and define Mi Suppose / and g are integrable on P be a partition of [a, b]. Let Let and inf s for /, and mi similarly for sup's (a) Prove that (b) Show M define M/ fg. < { M" [a, b] M-M" and m > m/m"{ that n - U(fg, P) L(fg, P)<2 i (c) Using the for x in U(fg, P) = [M/M/' - m/m/'^U - / and g are bounded, show that fact that [a, b], - M { so that |/(*)|, < M fc_0 }. L(fg, P) n < /,•_,). l n \ [Mi' - m/](ti - t^) + T t=i (d) Prove that /g (e) Now is [Mi" - m/ ](ti f t=i integrable. eliminate the restriction that fix) > 0 for x in [a, b]. 239 Integrals 33. Suppose that / and g are integrable on [a, b]. The Cauchy -Schwartz inequality states that (/»(/>> (a) Prove the Gauchy-Schwartz inequality by imitating one of the (b) If equality holds, proofs of the Schwartz inequality in is it Problem necessarily true that f (c) / and g are continuous? Show that the Schwartz inequality Schwartz inequality. (d) Prove = 1-18. \g for some X ? What if that^ replaced by a *34. Suppose that / is 1 2 /) and < 2 (J* f Is )- continuous and lim /(*) — f 1 lim - The > some in and 1 are condition lim f(x) x — 00 N. This means that comparison to N, then — a. Prove that oo X f{t) dt / = * t this result true if 0 b ? X Hint: a special case of the Cauchy- is f a = a. implies that fit) is close to a for N+M / Ma/ (N /(f) dt is close to Afa. If + M) is close to a. M is large THE FUNDAMENTAL THEOREM OF CALCULUS CHAPTER From the the hints given in the previous chapter you first fa f theorem of continuous; is original function derivative THEOREM 1 (THE FIRST FUNDAMENTAL THEOREM is / it is only is we fitting that if may / is have already guessed integrable, then F(x) = what happens when the ask F is differentiate (and its especially simple). [a, b], continuous at c in F on and define OF CALCULUS) is that continuous. It turns out that Let / be integrable on Iff We know this chapter. *w = Ij. then F is [a, b], [a, b] by differentiate at c, and F'{c) =/(,). (If c = a or then F'(c) b, mean understood to is the right- or left-hand deriva- tive of F.) PROOF We will assume that in c is supplied by the reader. {a, b) By ; the easy modifications for F'(c)=l^ first that h > 0. mh and M h may h + h)-F(c) = +k f. f; as follows (Figure 1): mh = h = inf {/(*): M It follows a or b Then F(c Define = + h) - F^. F{c A-+0 Suppose c definition, sup < < c \f(x): c x < c x < c + + h], h\. from Theorem 13-7 that h< [c + h f<M h Jc -h. Therefore ^ F(c + ti) — F(c) n If h < 0, only a few details of the argument have to be changed. Let mh = h = M 240 inf {/(*): c sup {/(*): c + + h h < < x x < < c], c\. be The Fundamental Theorem of Calculus 241 A nh c-i-h l FIGURE 1 Then C mh {-h) < f +h - c f<M h -(-h). Since F {c+h)-F(c) = +h f= -f;+ j, f: this yields mh Since h < 0, • h > F{c + h) - F{c) > Mh ' h. dividing by h reverses the inequality again, yielding the same result as before: mh < F{c + - h) F{c) ; < h This inequality / is is continuous at true for c, this h . any integrable function, continuous or not. Since however, lim and M mh = lim M h — f(c), proves that F'(c) = F(c lim + h)-F{c) _ t k h-yO Although Theorem 1 deals only with the function obtained by varying the upper limit of integration, a simple trick shows what happens when the lower limit is varied. If G is defined by CM - /„*/, then = /:/-/:/• 242 Derivatives and Integrals Consequently, if / is continuous at = G\c) The minus Theorem 1 sign appearing here defined even for x < very fortunate, and allows us to extend is if c < a = /;/ we can In this case a. ™ so -/(,). where the function to the situation *w is then c, = - write /;/, we have = F'(c) -(-/(c)) = /(,), exactly as before. F at c is ensured by continuity most interesting when / is con- Notice that in either case, differentiability of of / Theorem at c alone. Nevertheless, tinuous at [a, b] points in all [a, b]. In is 1 this case F is differentiable at all points in and F'=f. In general, it is extremely difficult to decide whether a given function / Theorem derivative of some other function; for this reason Theorem (Problem tain properties problem at all 11-7 is the and Darboux's 11-39) are particularly interesting, since they reveal cer- which / must have. —according to If / Theorem is 1, / continuous, however, there is the derivative of some is no function, namely, the function *w Theorem 1 = /."/• has a simple corollary which frequently reduces computations of integrals to a triviality. COROLLARY If / is continuous on [a, b] /.*/ proof = and / r for g some function g, then = *(*)-*«. Let F{X) Then F' = / = g' on [a, b]. c can be evaluated 0 so c = /;/. Consequently, there F= The number = = g easily: F{a) = +c note that g{a) + c, —g(a); thus F(x) = g(x) is - g(a). a number c such that 243 The Fundamental Theorem of Calculus This is = f*f The proof of for T(b) - g{b) to know /.*/ = gib) = J* f g(a). | ? make the corollary seem that - The point, g(a) of course, is that one might happen quite different function g with this property. For example, x3 = - g(x) then g'{x) = is it what good example, g(x) know a to Thus b. this corollary tends, at first sight, to useless: after all, if g is, — true, in particular, for x — we f(x) so and if = x\ f(x) computing lower and upper obtain, without ever sums: — dx One can x n + l /(n powers similarly; treat other + = then g'(x) 1), x rx For any natural number interval containing 0, but n, if n 3 n dx = and b are is b~~ = f. formula n is a natural number and g(x) = n 1 + 1 = a: n is not bounded on any both positive or both negative, then the function /(*) a -n dx this + n b Naturally if so n+1 -72+1 only true for n -n + 1 — 1 We do not know a simple expression . for a X The problem of computing this integral is discussed later, but it provides a good opportunity to warn against a serious error. The conclusion of Corollary 1 is often confused with the definition of integrals — —many students think that f where g is a function whose derivative is /." This "definition" is not only wrong it is useless. One reason is that a function / may be integrable without being the derivative of another function. For examis defined as: "g(b) g(a), — = = then / is integrable, but / cannot be a is also another reason that is much more imporderivative (why not?). tant: If / is continuous, then we know that / = g' for some function g; but we know this only because of Theorem 1. The function /(*) = \/x provides an excelple, if /(*) 0 for x and There 9* 1 /(I) 1, = 244 1 Derivatives and Integrals lent illustration: > x if 0, The g'(x) 9 f = 8(x) and we know = then f(x) x 1 dt > h i no simpler function g with of where this property. Theorem 1 is so useful that it is frequently called the Second Fundamental Theorem of Calculus. In this book, that name is reserved for a somewhat stronger result (which in practice, however, is not much more useful). As we have just mentioned, a function / might be of the form g' even if / corollary to not continuous. is / If integrable, then is it is still &% - g(a). /„7 = The proof, however, must be entirely different we must return to the definition of integrals. THEOREM 2 (THE SECOND FUNDAMENTAL THEOREM OF CALCULUS) If / integrable on is [a, b] = and / /„7 = PROOF Let P = there is {t 0i . . . a point X{ , tn g, 1, so then gW - g{a). /J such that - g{ti^) g{k) —we cannot use Theorem some function g' for be any partition of } in true that = = By the Mean Value Theorem [a, b]. g'(xt)(ti f(Xi)(ti - t^) - k-l). If = Mi = inf {/(#): U-\ sup {/(#): U-\ rrii < < < < x x k}, t{], then clearly that rriiiti - ti_t) < f( Xi)(ti rriiiti - U-i) < g(ti) - M t ) < - Mi(ti M t ), is, Adding i = ti-i) < these equations for 1, — . g(U-i) . . , n < Mi(U — we obtain n n T % nn(fi - gib) - g(a) < 2 Miih - <,_,) i=l 1 so that Uf, P) < for every partition P. £ut this means gib) We gib) - - g(a) < U(f, P) that gia) = p. | have already used the corollary to Theorem 1 (or, equivalently, 2) to find the integrals of a few elementary functions: Theorem b x n dx = fon+l n+\ n n+l n+ -' 72 7* -1. (a and b both positive or both negative if n < 0.) The Fundamental Theorem of Calculus 245 As we pointed out in Chapter 1 3, this integral does not always represent the area bounded by the graph of the function, the horizontal axis, and the vertical lines through (a, 0) and (b, 0). For example, if a < 0 < b, then does not represent the area of the region shown in Figure instead 2, which is given by Similar care must be exercised in finding the areas of regions which are bounded by the graphs of more than one function —a problem which may frequently involve considerable ingenuity in any case. Suppose, to take a sim- example first, that we wish to find the area of the region, shown between the graphs of the functions ple = f(x) x2 and g(x) = in Figure 3, x3 on the interval [0, I]. If 0 < x < 1, then 0 < x z < x 2 so that the graph of g lies below that of /. The area of the region of interest to us is therefore , area R(f, which 0, 1) - area R(g, 0, 1), is £x 2 dx - \l x 3 dx = i - i = iV This area could have been expressed as /:</-„). If g (x) / and < fix) for all x in g, even if f in Figure 4. If [a, b], i? 2 , [a, b] , then this integral always gives the area bounded by negative. The easiest a number such that f +c and g and g are sometimes c is then the region bounded by / c to see this bounded by / and g, has the same area and g + c. Consequently, R = x area R = 2 = is shown are nonnegative on Rh +c area way + as the region + c) — j* (g + c) - (g + c)] L(f + J* (f b fa c) -/;(/-*)• This observation is useful in the following problem: region bounded by the graphs of f(x) = x3 — x and g(x) = x2 . Find the area of the 246 Derivatives and Integrals The necessity first / and g intersect determine to is region more precisely. this x3 z — — or x or #(# 2 or * = x the interval + (0, [1 f(x) - ([1 x2 , x — x = 0, — x — 1) = 0, + V5 - V5 n 1 1 = 0 j 5 V5]/2, 2 we have 0) — V5]/2) we have * 2 > x 3 the graphs (Figure = g(x) only 5), if x xs — x > x2 and on the interval These assertions are apparent from but they can also be checked easily, as follows. Since = 0, [1 + a:. V5]/2 does not change sign on the intervals it is of 2 2 On The graphs when S or — ([1 Vs]/2, the function / - g V5]/2, 0) and [0, [1 + V5]/2); [1 therefore only necessary to observe, for example, that v (-i) - (-*) - (-*) = i> P _ _ 12 = -1 < 0, s 2 o, 1 to conclude that FIGURE f-g>0 f-g<0 4 The area on on(0, of the region in question Q /l -V5 " * ~ ([1 is [1 V5]/2, + 0), V5]/2). thus ^+/ 0 2 ^ 2 (* 3 - *)1 ^« one of the major problems involved in finding the areas of a region may be the exact determination of the region. There are, however, more substantial problems of a logical nature we have thus far defined the areas of some very special regions only, which do not even include some of the regions whose areas have just been computed! We have simply As this example reveals, — assumed that area made sense for these regions, and that certain reasonable properties of "area" do hold. These remarks are not meant to suggest that you should regard exercising ingenuity to compute areas as beneath you, but are meant to indicate that a better approach to the definition of area is available, although its proper place is somewhere in advanced calculus. The desire to define area was the motivation, both in this book and historically, for the definition of the integral, but the integral does not really provide the best method of defining areas, although it is frequently the proper tool for computing them. be discouraging to learn that integrals are not suitable for the very purpose for which they were invented, but we will soon see how essential they are for other purposes. The most important use of integrals has already been It may The Fundamental Theorem of Calculus 1 FIGURE 247 + V5 5 emphasized: if / is continuous, the integral provides a function y such that /(*) This equation =/«. is the simplest example of a "differential equation" (an equation for a function y which involves derivatives of y). The Fundamental Theorem of Calculus says that this differential equation has a solution, if /is continuous. In succeeding chapters, and in various problems, we will solve more complicated equations, but the solution almost always depends somehow on the integral; in order to solve a differential equation it is necessary to construct a new function, and the integral is one of the best ways of doing this. Since the differentiable functions provided by the Fundamental Theorem of Calculus will play such a prominent role in later work, to realize that these functions to yield still more functions, may be combined, it is very important like less esoteric functions, whose derivatives can be found by the Chain Rule. Suppose, for example, that Although the notation tends to disguise the fact somewhat, / is the composition of the functions C(x) In fact, /(*) = = xs and F(C(x)); in other words, f = Fo Rule, fix) = = F'(C(x)) F'(x*) 1 + • • CM 3x 2 sin 2 xs C. Therefore, by the Chain 248 Derivatives and Integrals If / is defined, instead, as fa = i dt > + sin r* / i Jx* 1 2 • / then If / is defined as the reverse 1 = w /'(*) 7 3* 2 • + 1 . sin 2 x 3 composition, \Ja + 1 sin 2 / 1 then f(x) Similarly, = C(F(x)) ~ 3 (/ F'W • + sin *) 2 1 + sin 1 1 2 x if M x /*sin 1 A/ T + sin : 1 r° 2 £ i J sin x 1 + sin 2 £ X = sin ( [ \Ja + sin 1 dt\ 2 / f then /'(#) g'(x) h'{x) — = + sin 2 (sin x) 1 + sin 2 (sin x) = = • cos ( [* Via The formidable appearing /W also a composition /'« = ; ~~ - + 1 cos x, COS X, dt) i sin 2 • / f 1 + sin +sin't d< ) • * 1 ia in fact, F'(F(*)) + 1 / = • F ° F. sin 2 * Therefore F'M 1 1 0 1 a As 2 function fUal is * 1 1 + sin ^0A 2 1 1 + sin sir 2 x ) the these examples reveal, the expression occurring above (or below) 249 The Fundamental Theorem of Calculus which integral sign indicates the function written as a composition. As a appear on the will example, consider the final j" dt 1 + sin* r(Iaa 1+sin^ f\J t /W = J/ a ) J fl ft 1 | 1 1 + . , sin = g(x) dt, 2 t (/a / i +\imt /al + when / right dt is compositions triple ) sinM J \ / . J° 1 + sm . . dt, 2 t which can be written f = FoFo C and g = F° F° may Omitting the intermediate steps (which you insecure), F. supply, f(x ) = feel still + 1 sin 2 ) 1 1 g'W = + sm 1 A 2 / Ua ,' 3* 2 1 1 \Ja 1 you if we obtain 1 + : sin 2 Via 1 + sin 2 / * J 1 1 1 + sin 2 # Like the simpler differentiations of Chapter 10, these manipulations should become much easier after the practice provided by some of the problems, and, problems of Chapter 10, these differentiations are simply a test of your understanding of the Chain Rule, in the somewhat unfamiliar context provided by the Fundamental Theorem of Calculus. The powerful uses to which the integral will be put in the following chapters like the all depend on the Fundamental Theorem of Calculus, yet the proof of that theorem was quite easy it seems that all the real work went into the definition of the integral. Actually, this is not quite true. In order to apply Theorem 1 to a continuous function we need the one theorem whose proof has not yet been given: If /is continuous on [a, b\ then /is integrable on [a, b\ The proof which will be given depends upon a trick which is not very illuminating, but which at least serves to fill this gap in our knowledge. If / is any bounded function on [a> b], then — sup{L(/,P)} will both exist, even integral of /on will if [a, b] and inf { U(f, P) } / is not integrable. These numbers are called the lower and the upper integral of /on [a, b], respectively, and be denoted by L/;/ The lower and upper integrals both and u/>. have several properties which the integral 250 Derivatives and Integrals < possesses. In particular, if a l /:/= l /:/+ l and if m < fix) < M m{b The < c proofs of these facts are precisely We - M(Z> a). exercise, since they are quite similar to an left as The the corresponding proofs for integrals. corollary of the results for then [a, b], £/ < U £/ < < L a) u /:/= u /:/+ u /> /> for all x in - then b, results for integrals are actually upper and lower integrals, because / is a integrable when prove that a continuous function / is integrable by showing that this equality always holds for continuous functions. It is actually easier to show will that for all x in [a, b]; the trick most of the proof of Theorem to note that is 1 didn't even depend on the fact that / was integrable THEOREM 13-3 PROOF If / is continuous on Define functions b\ then / [a, U on L and integrable on is (a, b). > If h b\ by [a, b] LOO = L J*f and Let x be in [a, = U j* f. U(x) 0 and mh = h = sup < < x inf {/(*): M {/(*): x < < t t x x + h}> + h), then mh < L -h x+h Jx U f < < M k • h, so mh • h < L(x + - h) < U(x L(x) + h) — U(x) < M < M h 'h or 772 a < < ; : lih<0 h- h h and mh = M h = inf {/(f): * sup x {/(*): + + h h < < t t < < *}, x\, one obtains the same inequality, precisely as in the proof of Theorem Since / is continuous at x we have } lim A— 0 mh = lim A->0 M h — /(#), 1 — The Fundamental Theorem and this of Calculus 251 proves that L'(x) This means that there - = U'(x) a number = L(x) is U(x) such that c + for x in {a, b). fix) for all x in c 6]. Since = U(a) the number c must equal = 0, 0, so U(x) = for all x in L(x) [a, b]. In particular, U/ V= 1Kb) U and this means that / is = integrable on = L./ V, a L{b) [a, b]. | PROBLEMS 1. Find the derivatives of each of the following functions (i) F(x) = sin 3 1 dt. J* (ii) (iii) dv) (vi) F(*) F(*) fw F(x) = —+ ^ f* J 15 \Js + 1 + 2 t 1 sin 2 / J* 1 + r + sm Ja 1 + r + sm 2 = sm . sm 2 (vii) F , where F(x) — - / F 1 where F(x) = ^ 1 /(*)? (Caution: it of/. (Find (F' 1 )'^) in terms of 1 , / For each of the following /, / t F-'(x).) - A. VI Vl - Jo 2. dt) dy. 7 / s * n3 Ji (viii) t £ (^J l 2 , & 1 r + shT 6 1 - [*( = 1 1 if F(x) 2 t = J* f, might happen that F'(x) continuous at (i) fix) (ii) /(*) (iii) f{x) = = = 0 if * o if * 0 if a: at < < * 1, /(*) 1, /(*) 1, /(*) = = = 1 if x 1 if x 1 if * > > = 1. 1. 1. which points x = /(*), even is F'(x) if / is = not 252 Derivatives and Integrals FIGURE (vi) = fa) = f(x) = (vii) / (viii) f(x) Find (f (iv) ( 3. 6 /(*) v) 0 if a- 0 if x 0 if a is < < i = 1 if y(0) if (i) /(a) = (ii) /(*) = J* ^ x + 1 = 0, /(*) shown 1 0 * if in lowest terms. a: in Figure 6. = /n for some n in N, /(a) sin (sin sin (sin = = 0, /(*) the function is = \/q if x = p/q > 0. 1/[1 A] if * > 0. irrational, /(*) 0 otherwise. 0 a?/. (Don't try to evaluate / explicitly.) = (The answer you should 4. Find F'(*) 5. perform an obvious manipulation on the integral before trying to find F Prove that if / is continuous, then if F(*) x j0 **6. Ku)(x-u)du = /; (/; /(o *) x *8. is mrt */(*); f(u)( X - uYdu = 2 2 /;(/; f .) du. Hint: Differentiate both sides, making use of Problem Use Problem 5 to prove that jo 7. dt. f*xf(t) 4. (f /(o ^)^)^ 2 + sin 2 x. (This problem is Find a function / such that /'"(*) = 1/Vl "find.") word the misinterpret supposed to be easy; don't = /(*) for all x. }{x if a) period a, with periodic, is function A / + (a) If / is periodic with period a /o°/ = + -// and integrable on 7 ^ all *. [0, a], show that The Fundamental Theorem of Calculus (b) Find a function /such that /is not periodic, but /' periodic g for which is. Hint: Choose a = can be guaranteed that f(x) it 253 J* g not is periodic. (c) 9. Suppose that /' is periodic with period if and only if f{a) = /(0). Find J* Vxdx, by simply a. Prove that / guessing a function / with /'(*) using the Second Fundamental Theorem of Calculus. is = periodic V**, and Then check with Problem 13-16. 10. Use the Fundamental Theorem of Calculus and Problem 13-16 to derive the result stated in Problem 12-15. 11. 12. 13. (a) Find the derivatives of F(x) (b) Now give a new proof = 1/t dt J* Problem for and G(x) = j** 1/t dt. 13-12. Use the Fundamental Theorem of Calculus and Darboux's Theorem (Problem 11-39) to give another proof of the Intermediate Value Theorem. Prove that if h is continuous, / and g are differentiable, and = then F'(x) h(g(x)) - m = g'(x) - hit) dt, /;;? h(f(x)) • f(x). Hint: Try to reduce this two cases you can already handle, with a constant either as the lower or the upper limit of integration. (a) Suppose G' = g and F' = /. Prove that if the function y satisfies the to the *14. differential equation (*) g(y(*)) then there (**) (b) (c) is */W = /(*) a G(y(x)) number = c + F{x) for all x in some interval, such that for all x in this interval. c Show, conversely, that ii y satisfies (**), then y Find what condition y must satisfy if y (In this case g(t) = w >(\ 1 + 1 = + is a solution of (*). *2 l+y(x) t and /(/) = 1 + 2 t .) Then "solve" the resulting equations to find all possible solutions y (no solution will have as its domain), R (d) Find what condition y must /(*) = 1 (An appeal to satisfy if + Problem 12-10 — S[y(x)Y will satisfying the resulting equation.) show that there are functions 254 Derivatives and Integrals (e) Find all functions y satisfying = y(x)y'(x) -x. Find the solution^ satisfying y(0) *15. The limit lim denoted by /, if it exists, is / = — 1. -/° called x r dx> Determine (b) Use Problem 13-12 f What can you Suppose that 0 (d) *16. < < g(x) Explain (b) /(*) dx), say about /(*) > J* 0 for x 1/(1 + 1/x dx / > fix) for all x why it is jj f Explain + J_ why f, aa 1/x dx does not show that to > and 0, then exist. Hint: ? " Jo / and that 0 JJ g exists. Prove that if also exists. x 2 ) dx exists. Hint: Split this integral up at 1 /is denned in the obvious way, as ^lim But another kind of improper integral (a) / Ja < —1 if r The improper integral j"^ way: (or an "improper integral." (a) (c) / / Ja J"^ f is denned in a nonobvious provided these improper integrals both + 1/(1 f. exist. x 2 ) dx exists. J"^ Explain why jf^ x dx does not exist. (But notice that lim j N xdx does exist.) (c) Prove that iff" /exists, then lim [* f exists and equals f"m 1 Show moreover, equal Can you /. J"^ (If you facts? that lim you can't, f and lim f*** f both [ exist f. and state a reasonable generalization of these will have a miserable time trying to do these special cases!) *17. ' There is another kind of 'improper integral" bounded, but the function is unbounded: (a) If a > a 0, find lim [ 1 /Vxdx. This even though the function fix) matter how we define /(0). ^ ^ r — < (b) Find (c) Use Problem 13-12 Ar if 1 to r < limit = 1/V* is is in which the interval denoted by P not bounded on is 1/Vxdx [0, a], 9 no 0. show that x~ dx does not make l J* sense, even as a limit. (d) Invent a reasonable definition of for -1 < r < 0. ja \x\ r dx for a < 0 and compute it 255 The Fundamental Theorem of Calculus (e) Invent a reasonable definition of and show that the of two limits, j (1 + (1 i - x)~ 112 How dx exist? -1 < X 2 )' 112 for It is possible, finally, to < x (1 — x 2) limits exists. does + (1 1/2 dx, Hint: x)~ 112 a as Why compare sum does with 0? combine the two possible extensions of the notion of the integral. (a) If /(*) = \/Vx for 0 < x < 1 and f(x) = \/x 2 for x > find 1, f(x) dx (after deciding what this should mean). (b) JJ Show — 1 < that r < JJ x T dx 0 and other case at qo ; r never makes < — for r 1 . sense. (Distinguish the In one case things go wrong at = — 1 things go wrong cases 0, in at both places.) the THE TRIGONOMETRIC FUNCTIONS CHAPTER The definitions of the functions sin and cos are considerably more subtle than one might suspect. For this reason, this chapter begins with some informal and intuitive definitions, which should not be scrutinized too carefully, as they shall soon be replaced by the formal definitions which we really intend to use. In elementary geometry an angle is simply the union of two half-lines with a common figure More as pairs initial point (Figure 1). 1 useful for trigonometry are "directed angles," (/i, / 2 ) of half-lines with the same which may be regarded initial point, visualized as in Figure 2. ti FIGURE If for h we always choose the positive half of the horizontal axis, a directed described completely by the second half-line (Figure 3). Since each half-line intersects the unit circle precisely once, a directed angle described, even more simply, by a point on the unit circle (Figure 4), that is, angle is 2 is by a point 256 (x, y) with x 2 +y 2 = 1 ; 257 The Trigonometric Functions The and cosine of a directed angle can now be defined 2 is determined by a point (*, y) with x of the angle is defined as y, and the cosine as x. sine (Figure 5) the sine : a directed angle as follows +y = 2 1 Despite the aura of precision surrounding the previous paragraph, we are not yet finished with the definitions of sin and cos. Indeed, we have barely begun. we What we have want to define is defined the sine is and cos x sin x for and cosine each number what of a directed angle; The x. usual procedure for depends on associating an angle to every number. The oldest "measure angles in degrees." An angle "all the way around" is associated to 360, an angle "half-way around" is associated to 180, an angle "a quarter way around" to 90, etc. (Figure 6). The angle associated, in this doing this method to is manner, to the number x, is called "the angle of x degrees." The angle of ambiguity is puralso an angle of is degrees angle of 90 posely extended further, so that an which we will denote function, define now a 360 + 90 degrees, etc. One can 0 degrees by is the same as the angle of 360 degrees, and this sin°, as follows: = sm°(#) sine of the angle of x degrees. There are two difficulties with this approach. Although it may be clear what we mean by an angle of 90 or 45 degrees, it is not quite clear what an angle of V2 degrees is, for example. Even if this difficulty could be circumvented, it is unlikely that this system, depending as it does on the arbitrary choice of 360, will lead to elegant results it would be sheer luck if the func0 tion sin had mathematically pleasing properties. "Radian measure" appears to offer a remedy for both these defects. Given any number x, choose a point P on the unit circle such that x is the length of the arc of the circle beginning at (1, 0) and running counterclockwise to P (Figure 7). The directed angle determined by P is called "the angle of x radians." Since the length of the whole circle is 2tt, the angle of x radians and — the angle of 2ir + x radians are identical. A function sin r can now be defined as follows: = sin r (Ar) sine of the angle of x radians. This same method can easily be adopted to define have sin° 360 = sin r 2w, we can define . sin x = , sin r 2irx —— = . sin r shall soon drop the superscript r in sin only function which will interest us; before we want to • 180 360 We —ttx— sin°; since r we since sin , r (and not sin°) is the do, a few words of warning are advisable. The expressions sin° x and sin r x are sometimes written sin x° sin x radians, but this notation is quite misleading; a number x is simply a number — it does * 258 Derivatives and : Integrals not carry a banner indicating that it is "in degrees" or "in radians." If the the notation "sin x" is in doubt one usually asks: meaning of p "Is x in degrees or radians?" length but what one means is: "Do you mean Even for £ or sin°' 'sin r '?" mathematicians, addicted to precision, these remarks might be it not for the fact that failure to take them into account will dispensable, were lead to incorrect answers to certain problems (an example is given in Problem 18). FIGURE Although the function sin r is the function which we wish to denote simply by sin (and use exclusively henceforth), there is a difficulty involved even in the definition of sin r Our proposed definition depends on the concept of length of a curve. Although we could digress, and define length, at the present time it is much easier to reformulate the definition in terms of areas, which we can treat by means of the integral. (A treatment in terms of length is outlined 7 . Problems 32 to 34.) Suppose that x is the length of the arc of the unit circle from (1 0) to P; this arc thus contains x/2tt of the total length 2w of the circumference of the unit circle. Let S denote the "sector" shown in Figure 8; S is bounded by the unit in , / / Yvlength \ x 11 area circle, the horizontal axis, S should be x/2t times the thus S should have area and the half-line x • With and circle 7T = x — 2 now —more begins. The (This definition is x < through the sum = 2 • jl % it as the area of the unit a semicircle (Figure Vl - 9). x 2 dx. it will be necessary to first define sin x and cos x only tt.) The second of the sector definition identifies not offered simply as an embellishment; to define the trigonometric functions < first precisely, as twice the area of w for 0 P which these remarks as background, the rigorous definition of the functions cos DEFINITION 9 and P. The area of which we expect to be t; (0, 0) We can therefore define cos x and sin x as the coordinates of the point determines a sector of area x/2. 8 sin FIGURE circle, ^ 2tt FIGURE through area inside the unit definition is meant bounded by the unit to describe, for circle, —1 < x < the horizontal axis, 1, the area A(x) and the half-line y/l — x 2 ). If 0 < x < 1, this area can be expressed (Figure 10) as of the area of a triangle and the area of a region under the unit circle (x, The Trigonometric Functions - This same formula happens to work for (x, VI - x 2 < x 0 also. In this case (Figure term 11), the Vl - x is < 1 259 x2 and represents the area of the negative, X triangle which must be subtracted from the term FIGURE Vl - 2 t dt. 10 DEFINITION Notice that if —1 < Fundamental Theorem = A'(x) x < -2x - FIGURE -x 1 (1 - V\ - X2 + 2 2 2 1 - Vi - X2 X 2) - 2x - Vi Vl - X - 2x - 2(1 - x 2 Vl — x 2 1 11 differentiate at x and (using the is + vT 2Vl 2 _ " A then 1, of Calculus), x2 2 2 2 ) 2 -1 2 Vl - x2 Notice also (Figure 12) that on the interval [-1, 1] the function A decreases from 1 a{-\) = to A{\) FIGURE 12 = fact that For 0 P = 0. its < x DEFINITION If 0 < x < + ^ VT^ f 7T 2 t dt This follows directly from the definition of A, and also from the derivative < ir 13). 7r, is we wish (cos x, sin x) x/2 (Figure 0 negative on ( — 1, 1). and sin x as the coordinates of a point which determines a sector whose area is to define cos x on the unit circle In other words: then cos x is the unique A (cos x) number — -; and sin x = Vl — 2 (cos x) . in [ — 1, 1] such that 260 Derivatives and Integrals This definition actually requires a few words of justification. In order to know that there ij a number^ continuous, and that A = satisfying A(y) x/2, we use the fact that A is on the values 0 and w/2. This tacit appeal to the Intermediate Value Theorem is crucial, if we want to make our preliminary definition precise. Having made, and justified, our definition, we can now takes proceed quite rapidly. THEOREM 1 If 0 < x < 7T, then cos'(*) sin'(*) PROOF If B — 2A, then the definition in other words, cos is = — sin = cos x. A (cos x) — x/2 can be written B (cos x) = x; = - have already computed that 1 Vt - 2 (cos x, sin x) We just the inverse of B. A'(x) P ~ x, x2 from which we conclude that = - B\x) Consequently, cos'M = (B-yix) FIGURE 13 1 B'{B-\x)) 1 1 = = — V - 1 (cos x) 2 sin x. Since sin x we = Vl — (cos #) 2 , also obtain . / / \ sin (x) = _ —2 1 2 cos x • . Vl - f • cos (x) =^ (cos x) 2 cos x sin x sin x = The information contained in cos ^. I Theorem 1 can be used to sketch the graphs The Trigonometric Functions of sin and cos on the interval cos'(x) Since [0, ir]. = — = < sin x the function cos decreases from cos 0 quently, cos y 0 for a unique y in 261 = 1 [0, w]. < 0 0, x < x, to cos t = — 1 (Figure 14). ConseTo find y, we note that the definition of cos, A (cos = x) -> 2 means that SO y It is 2 we can Now we 2 dt. t so sin increases The Vi - j* ? dt, have = ' cos ^ I [t/2, = dt t* also write sin'W on ~ £ Vl easy to see that Vi so = 7t] on to sin [0, ir/2] = 7r values of sin x from < = sin 0 < X < 7T, w/2 = 1, 7T/2 0, 0 to sin and then decreases 0 (Figure 15). and cos x for x not in t] are most easily defined by a [0, two-step piecing together process: (1) If 7T < x < 2tt, then sin a* COS X — — sin(27r — x), = cos(27r — x). Figure 16 shows the graphs of sin and cos on (2) If x = 2irk + x' for [0, 2ir]. some integer k and some x f } sin x cos = = sin [0, 2w], then x\ sin cos x\ Figure 17 shows the graphs of sin and cos, Having extended the functions in and now defined on cos to R, all of R. we must now check that the basic properties of these functions continue to hold. In most cases this easy. For example, it is clear that the equation sin 2 x + cos 2 x — 1 is — 262 — Derivatives and Integrals (a) FIGURE 17 holds for not hard to prove that all x. It is also sin' (a:) cos' (a:) x if is not a multiple of 7r. = cos x, = — sin For example, sin x = — sin(27r if ir — x, < x < 2ir, then x), so sin'O) If x is = - sin'(27r — C0S(27T — x) = COS X. x) • (-1) a multiple of t we resort to a trick; it is only necessary to apply 11-7 to conclude that the same formulas are true in this case also. Theorem Jt —— i 1 I 1 1 7T 1 TAP' 1- 1 3tT IT . ' A; \ Ay /esc 1 t i \ cot ' \ | i 1 A FIGURE 18 IT ' i (- 2x Ai ft 1! The Trigonometric Functions The other standard trigonometric functions present no difficulty at 263 all. We define sec x cos x sin x + kw x t± X 9* k-K. tt/2, tan x cos x 1 esc x sin x cos x cot x The graphs are sketched in Figure 18. It is a good idea to convince yourself that the general features of these graphs can be predicted from the derivatives of these functions, which are listed in the next theorem (there is no need to of the theorem, since the results memorize the statement whenever needed.) THEOREM 2 If x ?± kir + 7r/2, then = = sec'(#) tan' (^) If x ^ kir, sec x tan x, sec 2 x. then = — = — csc'(Y) cot'(Y) PROOF can be rederived esc x cot x, esc 2 x, Left to you (a straightforward computation). | The inverses of the trigonometric functions are also easily differentiated. The trigonometric functions are not one-one, so it is first necessary to restrict them to suitable intervals; the largest possible length obtainable is t, and the intervals usually chosen are (Figure 19) [-tt/2, tt/2] for sin, for cos, [0, 7r] (-tt/2, tt/2) for tan. (The inverses of the other trigonometric functions are so rarely used that they not even be discussed here.) will The inverse of the function f{x) = sin x, —tt/2 < x < ir/2 denoted by arcsin (Figure 20); the domain of arcsin is [ — 1, 1]. The -1 has been avoided because arcsin is not the inverse of sin -1 has (which is not one-one), but of the restricted function /; the notation sin _1 the additional disadvantage that sin (*) might be construed as 1/sin x. is notation sin The inverse of the function g(x) — COS X, 0 < X < 7T 264 Derivatives and Integrals denoted by arccos (Figure 21); the domain of arccos is The is [ — 1, 11. inverse of the function h(x) = tan —w/2 < x, x < t/2 denoted by arctan (Figure 22) arctan is one of the simplest examples of a which is bounded even though it is one-one on all of R. is ; differentiable function The derivatives of the inverse trigonometric functions are surprisingly sim- and do not involve trigonometric functions at all. Finding the derivatives a simple matter, but to express them in a suitable form we will have to sim- ple, is plify expressions like cos(arcsin x), sec (arctan x). FIGURE 22 A picture little is the best way to remember the correct simplifications. example, Figure 23 shows a directed angle whose sine is x thus an angle of (arcsin x) radians; consequently cos(arcsin x) theorem FIGURE 3 If will not resort to -1 < x < 1, . then 23 arcsin '(x) arccos Moreover, for all x f (x) 1 — Vi - x2 we have l arctan' (x) 1 PROOF is is the length of x 2 However, in the proof of the next theorem such pictures. the other side, namely, we For —the angle shown + x* arcsin '(#) i i sin' (arcsin x) 1 cos (arcsin x) The Trigonometric Functions 265 Now + 2 [sin(arcsin x)] that 2 [cos(arcsin x)] = 1, is, x2 + = 1; = Vl — x2 [cos(arcsin x)] 2 therefore, cos(arcsin x) (The positive square root cos(arcsin x) > 0.) is . to be taken because arcsin x This proves the is in (— 7r/2, ir/2), so formula. first The second formula has already been established (in the proof of Theorem 1). It is also possible to imitate the proof for the first formula, a valuable exercise if that proof presented any difficulties. The third formula is proved as follows. arctan'M = (/T 1 )'^) " _J h'(h~ l {x)) _ 1 tan '(arc tan x) 1 - sec 2 (arctan x) Dividing both sides of the identity a + cos 2 a tan 2 a + 1 sin 2 by cos a 2 = 1 yields = sec 2 a. It follows that [tan (arc tan x)] 2 + 1 = sec 2 (arc tan x), or x2 + 1 which proves the third formula. The = sec 2 (arctan | traditional proof of the formula the one given here) is sin'(Ar) = cos x (quite different outlined in Problem 26. This proof depends from upon first establishing the limit sin h — hm— h->o = 1, h and the "addition formula" sin(# + y) = sin x cos y + cos x sin y. Both of these formulas can be derived easily now that the derivative of sin and cos are known. The first is just the special case sin'(O) = cos 0. The second depends on a beautiful characterization of the functions sin and cos. In order to derive this result we need a lemma whose proof involves a clever trick; a more straightforward proof will be supplied in Part IV. 266 Derivatives and Integrals LEMMA Suppose / has a second derivative everywhere and that f"+f=0, Then / = PROOF AO) = 0, AO) = o. 0. Multiplying both sides of the equation by first = /'/"+//' /' yields 0. Thus +f 2 [(f) so (/')' +/ 2 is 2 = ]' From a constant function. that the constant is + //') 2(/'/" = /(0) = 0, and 0 /'(0) = 0 it follows 0; thus [/'MP + [/MP = 0 for all This implies that /M = THEOREM 4 If / has a second 0 for all *. | derivative everywhere and r+f-o, /(0) /'(0) = = a, A, then (In particular, /'(0) proof = 0, if then / /(0) = = = sin + a cos. 0 and /'(0) = 1, then / / b ' = sin; if cos.) Let g(x ) ~ /M b sin — x a cos x. Then £'M = /'M — 6 cos * + a sin £"M = /"(*) + * sin * + a cos *. a:, Consequently, g'(0) = = = o, x — a cos g(0) o, o, which shows that 0 THEOREM 5 If = g(x) = f(x) — b sin at, for all #. | * and y are any two numbers, then sin(# cos(* + y) + y) — = j> + cos x sin >>, cos ^ — sin # sin jy. sin x cos cos a- /(0) = 1 and The Trigonometric Functions PROOF 267 For any particular number y we can define a function / by = /(*) sin(x +y). Then = cos(* + y), = - sinO +y). f(x) /"(*) Consequently, r +/ = /(0) /'(0) It follows sin y, cos y. from Theorem 4 that / that o, = = = (cos y) - + sin (sin y) • cos; is, sin (a: Since any + y) = sin x cos + for all sin j cos number^ could have been chosen formula for all x and y. The second formula is proved similarly. to begin with, this proves the first | As a conclusion to this chapter, and as a prelude to Chapter 17, we will mention an alternative approach to the definition of the function sin. Since arcsin'C*) it follows from the = 1 for f Vl - —1 < * < 1, x2 Second Fundamental Theorem of Calculus that arcsin x = — arcsin x arcsin 0 = / Jo This equation could have been taken as the , VI - definition - dt. 2 t of arcsin. It would follow immediately that arcsin (x) = . VI- t 2 -1 the function sin could then be defined as (arcsin) derivative of an inverse function would show that sin^*) which could be defined as cos x. = Vl — and the formula for the sin 2 x, Eventually, one could show that ,4 (cos x) = end of the development the definition with which While much of this presentation would proceed more rapidly, the definition would be utterly unmotivated; the reasonableness of the definitions would be known to the author, but not to the student, for whom it was intended! Nevertheless, as we shall see in Chapter 17, an approach of this sort x/2, recovering at the very we is started. sometimes very reasonable indeed. PROBLEMS Differentiate each of the following functions. (0 /(*) (ii) f{x) (in) f(x) (iv) f(x) = = = arctan(arctan(arctan x)). arcsin (arctan (arccos x)). arctan (tan x arctan 1 = arcsin lim (ii) lim (hi) lim - x +* 3 /6 sin x - * +* 3 /6 cos - 1 +x 2 /2 +x 2 /2 cos lim (v) lim (vi) a: a* - lim (1 *-»o V* - 2 1 — arctan x — at , 1, At in at 4. + 3 a: /3 J-Y sin (a) a: sin x/ I (b) by PHopitaPs Rule, limits sin x X (iv) Y Wl + W ( Find the following (i) x). x*0 a: = 0. Find /'(0). Find /"(0). this point, you will almost certainly have to use PHopitaPs Rule, but Chapter 23 we will be able to find / (0) for all k, with almost no work (fc) all. Graph the following functions. (a) f(x) = = sin 2x. sin(* 2 ). (A pretty respectable sketch of this graph can be obtained using only a picture of the graph of sin. Indeed, pure thought is your only hope in this problem, because determining the (b) f(x) sign of the derivative f'(x) = cos(x 2 ) 2x is no easier than determining the behavior of /directly. The formula for f'(x) does indicate one important fact, however /'(0) = 0, which must be true since / is even, and which should be clear in your graph.) • — (c) f(x) = sin x + sin 2*. (It will the graphs of g(x) = sin probably be instructive x and h(x) = sin 2a: carefully to first draw on the same : 269 The Trigonometric Functions and guess what the sum will look like. You many critical points / has on [0, 2ir] by concan easily find out how sidering the derivative of/. You can then determine the nature of from 0 set of axes, to 2w, by finding out the sign of / these critical points at each point; your sketch will probably suggest the answer.) (d) f(x) = tan x — determine the behavior of /in (— tt/2, tt/2); x. (First (kw in the intervals — tt/2, kir (e) f{x) = sin x — (The material x. + the graph of 7r/2) moved up exactly the same, except a certain amount. in the Appendix to / will look Why?) Chapter 11 will this function.) be particularly helpful for sin x I = x 1, 0. determine approximately where the are located. Notice that / is even and continuous at 0; also you (Part (d) should enable zeros of /' to consider the size of / for large 5. Prove the addition formula 6. (a) From for cos. and the addition formula for sin cos 2x, sin 3x, (b) x.) and cos derive formulas for sin 2x, cos 3*. to find the following values of the trigonometric Use these formulas by geometric arguments functions (usually deduced in elementary trigonometry) 7T sm - = . 7T cos - 4 7T tan 4 = - _ sin ' 4 2 1, 1 2 6 7T cos 6 7. (a) Show A that able a and proof. You sin (a: + + b cos x for suitB) can be written as a sin x in this chapter provides a one-line (One of the theorems should also be able to figure out what a and b are.) b, find numbers A and B such that a sin x (b) Conversely, given a and (c) Use part (a) Prove that + 8. b. ' 2 b cos = A x (b) to sin(* + graph , tan(* B) f{x) for all x. = tan x + y) = x 1 + V3 — sin x + cos x. + tan v tan x tan y x, y, and x y are not of the form addition formulas for sin and cos.) provided that kir + tt/2. (Use the . Derivatives and Integrals (b) Prove that + arc tan x arctan ( * = arc tan y \1 ^ — Y ry/ indicating any necessary restrictions on # andjy. Hint: Replace # 9. arctan # and y by arctan Prove that arcsin + a = arcsin by in part (a) jy — arcsin (aV'l 2 /3 + /3 \/l —a 2 ), a and /5. any numbers, then indicating any restrictions on 10. Prove that if m and mx sin w are sin nx sin rax cos nx cos 11. Prove that cos if rax m and f* /-* / — ^[cos (m n)x 12. cos (m + sin + w)# + cos + i[cos (m sin mx sin nx dx — (m (m 0,3 m m = m TT, 772 ^' | [TT, cos mx cos w# , = ( { ( J* — n)x ^-[sin + — — n)x], n)x], »)*]. n are natural numbers, then /* •/-* — — = sin m# = cos — n n, n n, 0. These relations are particularly important in the theory of Fourier series. Although this topic will receive serious attention only in the Suggested Reading, the next problem provides a hint as to their importance. (a) If / is integrable on [ — ir, x], show that the minimum value of Jl w occurs (/W " a cos nx ^ 2 ^* when a = - \* fix) cos nx and the minimum value dx, of when a = 1 7T r* / fix) sin n# ./ (In each case, bring a outside the integral sign, obtaining a quadratic expression in a,) (b) Define an = — bn = f' f{x) cos nx dx, - [* fix) sin n# i*, n = 0, 1, 2, « = 1, 2, 3, 271 The Trigonometric Functions Show that and if Ci are any numbers, then di N - J ^+ ^ n— + C0S nX Cn dn Sin </* fMf 1 N N = /' [/(*)]* - dx 2t Va + ftp + M.) + * (f + B <„ n=l £ + 2 <„ </» 2 ) n=l AT = £ -* [/(*)]« dx + £ n= + an * *„ 2 ) 1 2 + ((7-2-^) + n= thus showing that the bi = functions = s n (x) integral first d^ In other words, among "- a" )2+( ^- W2)' 1 smallest is when cn = f< and "linear combinations" of the all c n (x) - cos nx for a n cos nx + b n sin and sin nx (c t ir 1 < n < N, the particular function N = g( x) 13. 14. ~+ ^ nx has the "closest fit" to / on [— 7r, 7r]. Find a formula for sin * + sin y and for cos x + cos > Hint: First find a formula for sin {a + b) + sin (a - 6). What good does that do? 2 (a) Starting from the formula for cos 2x, derive formulas for sin x and cos 2 x in terms of cos 2x. (b) Prove that + \ 2^2 x - < x for 0 (c) 15. cos * /l cos v < Use part 16. If x = = x 1 = \ O and sin 2 x and cos(arctan x) trigonometric functions. Hint: y sinjy/cosjy . Sm tt/2. (a) to find Find sin(arctan . and o sin^y/v'l — sin 2 tan u/2, express sin u = expressions as x) cos 2 x ja not arctan x means that x involving = tan ^ = }'. and cos u in terms of x. (Use Problem 15; the answers should be very simple expressions.) 17. (a) Prove that sin(x + the graphs of sin (b) 18. W (a) What is arcsin(cos x) ri Find 1 + dt. 2 t = cos cos as and \ / Jo tt/2) and Hint: if x. along we have been drawing were the case.) (All this arccos(sin x)? The answer is not 45. 272 Derivatives and Integrals Find (b) Jo 19. Find lim x X—* 20. cos°(*) same functions = sin 0 and cos 0 by sin°(» = 0 (sin )' and (cos°)' cos(7r*/180). Find and sin(7r*/180) in terms of these functions. Find lim (b) — X 00 Define (a) sin dt. + 1 Sm - and lim x X x-+0 sin° -• x x->*> 21. Prove that every point on the unit circle is of the form (cos at least one (and hence for infinitely many) numbers 6. 22. (a) 0, sin 6) for is the maximum possible length of an interval on which one-one, and that such an interval must be of the form (2kir - t/2, 2kw tt/2). Prove that w sin is + Suppose we (b) or is 23. Let f(x) 1 = = let g(x) sin x for x in (2/for — t/2, 2kir + tt/2). What )'? < sec x for 0 < x t. _1 Find the domain of / and sketch its graph. 24. Prove that |sin x — sin jy| < |^ — for all numbers x and y. Hint: The same statement, with < replaced by < is a very straightforward consequence of a well-known theorem; simple supplementary considerations then allow < to be improved to <. It is an excellent test of intuition to predict the value of , '25. lim />) X* dx. Continuous functions should be most accessible to intuition, but once you get the right idea for a proof the limit can easily be established for any integrable /. (a) Show (b) Show d lim ( that = \x dx sin 0, computing by the integral explicitly. that if s is a step function on b 13-17), then lim [ s(x) sin \x dx a [a, b] = (terminology from Problem 0. \_>oo J (c) Finally, use for This B= 0 FIGURE 24 (1, 0) 26. any function / which result, like Fourier Problem 13-17 series; it Problem is known 1 1 , is b to show that lim f X->oo J a integrable on f(x) sin \x dx = 0 [a, b]. plays an important role in the theory of as the Riemann-Lebesgue Lemma. This problem outlines the classical approach to the trigonometric functions. The shaded sector in Figure 24 has area x/2. . 273 The Trigonometric Functions (a) By and OCB prove that if 0 < x < then xx < < sin sin x 2 2 (b) OAB considering the triangles 7r/4, 2 cos * Conclude that sin x < cos x < 1, x and prove that hm sin x (c) Use — 1 Using parts (b) sin'(Y), starting cos x x x->0 *27. 1 this limit to find hm (d) = X x->0 and (c), and the addition formula from the definition of the derivative. for sin, find Yet another development of the trigonometric functions was briefly mentioned in the text starting with inverse functions defined by inte- — grals. It for all x. is this function is convenient to begin with arctan, since To do defined problem, pretend that you have never heard of the this trigonometric functions. (a) Let a{x) = f* + (1 2 t )" 1 Prove that a dt. and that lim a(x) and lim a{x) both X— each other. X oo If we is exist, odd and increasing, and are negatives of — »~ define = 7r 2 lim a(x) } then aT 1 is defined on (-x/2,t/2). (b) Show (c) For x (cO'M = that 1 + l [<x~ (x)]*. ^ 7r/2 or — 7r/2, define tan x = aT^*')Then define cos x = /VI + tan x, for x not of the form kir + x/2 or kir 7r/2, and cos {kir ± tt/2) = 0. Prove first that cos'(^) = — tan x cos x, and then that cos"(.x:) = — cos x for all x. = kir + x' with x' 2 1 *28. If we tions, are willing to assume that certain differential equations have solu- another approach to the trigonometric functions pose, in particular, that there and which (a) satisfies Prove thatjy 0 2 ory(O) ^ (b) yo" + +y = 2 is possible. is Sup- not always 0 0. 0 (joO is some function y 0 which constant, and conclude that either ;> 0 (0) ^ 0 +s— + byo'. 0 and s(0) = 0 0. Prove that there and /(O) = is 1. is a function Hint: Try s s satisfying s" of the form ay 0 : 274 Derivatives and Integrals If we define sin = = and cos s trigonometric functions become /, then almost all facts about one point which 7r. This is most done using an exercise from the Appendix to Chapter 1 1 requires work, however easily There trivial. —producing the is number (d) Use Problem 6 of the Appendix to Chapter 1 1 to prove that cos x cannot be positive for all x > 0. It follows that there is a smallest #o > 0 with cos # 0 = 0, and we can define t = 2x 0 Prove that sin ir/2 = 1. (Since sin 2 cos 2 = 1, we have sin x/2 = (e) Find cos (c) . + ±1; the problem 7r, why to decide is and sin w, cos 27r, sin 7r/2 is positive.) sin 2ir. (Naturally sin' (f) 29. (a) = and cos = — cos' Prove that cos and After all the use any we know that sin.) sin are periodic work involved concerting to find that sin (b) may you addition formulas, since these can be derived once with period 2ir. in the definition of sin, is would be it dis- actually a rational function. Prove that it isn't. (There is a simple property of sin which a rational function cannot possible have.) Prove that sin isn't even defined implicitly by an algebraic equation; that is, there do not exist rational functions /0 , (sin x) n + /„_!(*) (sin x) n~ l + • • . . + /„(*) = • . , 0 fn -i such that for all x. Hint: Prove that f0 = 0, so that sin x can be factored out. The remaining factor is 0 except perhaps at multiples of 2ir. But this implies that it is 0 for all x. (Why?) You are now set up for a proof by induction. *30. Suppose that 0i and 0 2 satisfy + £i0i 0i" and that g 2 > (a) Show Show 0, gi. that 0l"02 (b) = that if — 0i(#) 02"01 > (g2 — gl)0102 and 0 2 (*) > 0 0 ,, I* — 02 [0i - 0 2 "0i] = 0. for all x in (a, b), > then 0, and conclude that [0i'W0 2 (c) Show W - 0i'(a)0 2 (<2 )][0 1 (£)0 2 '(£) (d) Show 0i > we cannot have that in this case Consider the sign of <f>i(a) and that the equations <pi(a) 0, 0 < 2 0 or 0! (You should be able - < to 0, do 02 this 0i(*)0 2 'M] 0i(#) = > 0i(£) o. = 0. Hint: <t>i(b). = > 0i(£) 0, = or 0i 0 are also impossible < 0, 0 < 2 0 on with almost no extra work.) if (a, b). 275 The Trigonometric Functions The net result of this problem may be stated as follows: consecutive zeros of <f>i, then </> 2 if a and b are must have a zero somewhere between and b. This result, in a slightly more general form, is known as the Sturm Comparison Theorem. As a particular example, any solution a of the differential equation + y" must have zeros on the + (x - \)y 0 positive horizontal axis which are within w of each other. 31. Prove that + - cos x ^ x/2 if sin + 0, cos 2x then + • • + • sin(n cos nx + ^)x 2 sin At A this stage of the game, problem this is : a mathematical idiot's delight. proof by induction depends on establishing the formula — sin(n + £)* . f- ( cos (« + , 1)# sinO = + £)* 2 sin - 2 sin 2 2 This will require repeated use of addition formulas, in just the right A more reasonable derivation of the formula will occur in Problem 26-13. Like two other results in this probplaces, but very little thinking. lem set, and we this equation also make is very important in the study of Fourier use of it in series, Problems 18-28 and 22-17. The remaining problems in this chapter form a brief introduction to the and show how trigonometric functions can be defined in topic of arc length, terms of arc length. *32. Let /be a continuous function on tion of b], [<z, [a, b\ If P = {^o, . • . ? is tn ] a parti- define n i The number =l t(f P) represents the length of a polygonal curve inscribed 9 graph of / (see Figure 25). We define the length of / on [a, b] to be the least upper bound of all t(f, P) for all partitions P (provided that in the the set of (a) If / is all such If / is [a, b] P) is bounded above). a linear function on tance from (b) -l(f, (a, [a, b], prove that the length of / is the dis- f(a)) to (b, f(b)). not linear, prove that there such that /(/, \bj\b)). (You will is a partition P = P) is greater than the distance from need Problem 4-8.) {a, (a, t9 b] of f(a)) to Derivatives and Integrals 1 1 1 1 1 : a FIGURE = t0 ti t2 (Or, in straight Suppose that (a) b = and f(b) = d, than the length of any other. conventional but hopelessly muddled terminology: "A line is the shortest distance between two points. 55 ) all functions /on [a y b] with f(a) the length of the linear function *33. = 25 Conclude that of (c) t4 t3 /' is bounded on c is less [a, b]. UP any is partition of [a, b] show that L(Vi + cm p) Why Now (c) sup {L(Vl is show that sup integrable on any two [a, b]. *34. Let t{f, P)? /(at) = Vl - /on [x, [a, b] is {/(/, {U( V\ P)}? (This + Vl + f* P ) < ^7(Vl r both P' and P", x 2 for p). (fy, -1 < x < 1. how + is easy.) (f)\ P)\, thereby (f)\ (f) Vl + if (/')« P and P" are P ). If P conf if ;/ 2 , does /(/, P') compare to Define £(x) to be the length of 1]. (a) Show (b) Show (This that is actually Define w as and define sin'(*) an improper integral, as defined in Problem 14-17.) that £'(*) (c) inf + Hint: It suffices to show that partitions, then tains the points of < {/(/, P)\ proving that the length off on is < u(Vi t{L P) Mean Value Theorem. + {f)\ p)\ < sup Hint: Use the (b) < = = 1 for r VI - £( — 1). For 0 < x < = Vl — cos for 0 < x < ir. sin x cos x 2 x. -1 < * < 1. t, define cos * Prove that by «£(cos x) cos'W = = x, —sin x and "CHAPTER I TT IRRATIONAL IS This short chapter, diverging from the main stream of the book, demonstrate that we matics. This entire chapter tional. Like many steps many in 4 is is included to do some sophisticated mathe- are already in a position to devoted to an elementary proof that ir is irraproofs of deep theorems, the motivation for 'elementary' 5 our proof cannot be supplied; nevertheless, it is still quite possi- ble to follow the proof step-by-step. Two observations must be made The before the proof. first concerns the function /»(*) - xY *»(! = ' . nl which clearly satisfies 0< fn (x) < — < for 0 x < 1. nl An important property of the function fn is revealed by considering the expresby actually multiplying out x n {\ — appearing will be n and the highest power will be in the form sion obtained x) n . 2n. The lowest power of x Thus fn can be written 2n u{x) = hX i where the numbers a are integers. It /n <*>(0) =0 if is k CiXi ' =n clear < from n or k > this expression that 2n. Moreover, fn (n) (x) = — [n! cn + terms involving x] nl fn (n+1) (x) =—[(« + 1)! c n +i nl A 277 (2W) W =~[(2n)\c 2n l nl + terms involving x] 278 Derivatives and Integrals This means that /n /n /«< = - (n) (0) (W+1) (0) = 2n) (0) C n) + (» l), n+1 , - (2»)(2« where the numbers on the right are (fc) /n The (0) is «...-(« 1) all integers. an + 1)* 2 «, Thus integer for all £, relation = /»(*) /n(l " *) implies that A W = (-i)Yn tt) (ft) -*); (i therefore, /n The proof that number, and s > x (fc) an (l) is also integer for all k. irrational requires is one further observation: we will have if a is then for sufficiently large n 0, n a n\ To prove this, notice that > n if then 2a, a n+l (n Now let w 0 + a + n 1)! a 1 may (n 0 (n 0 +l) +1)! 2 (no+2) (n 0 (n 0 + + a is a I 2 > 2a. n r! Then, whatever value I < 2)! < k)\ n0 (n 0 )! (n.+ i) J fl A; • have, the succeeding values satisfy a If - — a 1 < n! be any natural number with n 0 (no) n fl - * 2 (» 0 1 a ~7 2* n° (n 0 ) 110 so large that («o)!e < 2*, —TTT + 1)! then < ! J 2 2 a n0 (n 0 )! any 0 x (n 0 for the THEOREM l is the desired result. one theorem in The number x Having made PROOF we these observations, are ready this chapter. x2 irrational. (Notice that the irration- is ality of x 2 implies the irrationality of x, for if 7r would 279 + k)\ irrational; in fact, is Irrational (no+k) a which is were rational, then 7r 2 certainly be.) Suppose x 2 were rational, so that 0 T2 = b for some positive integers a g(x) (i) = - *"[*•'»/»(*) and b. Let T 2n 2/«"to r«»-yw w(*) + + (-l)7n (2n) (*)]. Notice that each of the factors n b TT is an integer. Since 2n ~ 2k fn (k) = (0) b n 2 (TT ) n -k i-Jfe = £ n ^ and fn {k) (^) are integers, this shows that G(0) and G(l) are integers. Differentiating G"(*) (2) The last G - n n b [T* fn "(X ) - T 2n ~ fn ^(x) + 2 term, (-l) n/n (2n+2) (*), (3) Now twice yields G"(x) + H(x) — ir 2 is G(x) b n • + • Thus, adding zero. = • ir (-l) n/n (2n+2) (*)]. (1) and (2) gives = xV/n (*). 2n+2 fn {x) let — G'(x) sin irx xG(x) cos irx. Then #'(*) = xG'O) cos x# + G"0) sin = \G"{x) + x 2 G(*)] sin irx = ir a nfn (x) sin xx, by (3). irx - xG'O) cos irx + ir 2 G(x) sin xx 2 By the Second Fundamental Theorem of Calculus, x 2 f*a nfn (x)simrxdx = = - - H(0) G'{\) sinx x[G(l) + - xG(l) cosx - G'(0) sin 0 G(0)]. Thus x JQ a n fn (x) sin xx </x is an integer. + xG(0) cos — On the other hand, 0 0 < fn (x) < < Ta n fn (x) l/n\ for 0 < sin irx < x < so 1, n ita for 0 < x < 1. nl Consequently, 0 < 7r a f Jo n fn (x) sin wx dx < nl This reasoning was completely independent of the value of n. enough, then 0 < 7r a / Jo n fn (x) sin irx dx < —- < Now if n is large 1. n\ But this is absurd, because the integral is an integer, and there is no integer between 0 and 1. Thus our original assumption must have been incorrect: 7r 2 irrational. | is This proof way is admittedly mysterious; perhaps most mysterious of — we have proved all is the irrational it almost looks as if without ever mentioning a definition of it. A close reexamination of the proof will show that precisely one property of x is essential that 7r enters the proof sin(7r) The proof really depends on = 0. the properties of the function number x with sin x properties of sin are required, namely, irrationality of the smallest positive sin' cos' sin(0) cos(0) Even it = sin, 0. and proves the In fact, very few = cos, = — sin, = 0, = 1. could be shortened; as far as the proof is concerned, cos might be defined as sin'. The properties of sin required in the proof may this list just as well then be written sin" + sin sin(0) sin'(0) = = = 0, 0, 1. this is not really very surprising at all, since, as we have seen in the previous chapter, these properties characterize the function sin completely. Of course, PROBLEMS 1. (a) Prove that the areas of triangles by the equation OAB and OAC in Figure 1 are related 7r is OAC = area 1 - - Vl - jl = Hint: Solve the equations *y (b) Pm be the regular polygon of m A m is the area of Pm show that +/ = 1, for y. sides inscribed in the unit circle. V2 - 2 Vl - = ~ Ai m 281 OAS) 2 2(area OAS), x 2 Let If 16 (area Irrational (2^ m /m) 2 . This result allows one to obtain (more and more complicated) expressions for A 2 % starting with A 4 = 2, and thus to compute t as accurately as desired (according to ods will appear in Chapter 19, a slight a very interesting expression for (a) Using the Although better methvariant of this approach yields Problem 8-11). 7r: fact that area(0A8) = OB, area((MC) show that (b) Show if am is the distance from 0 to one side of Pm , then that 2 A2 (c) Using the «4 ' ' OL S « fact that am = cos 1 and the formula cos x etc. t + 7T — m COS X (Problem 15-14), prove that Together with part "infinite product" to be precise, (b), this shows that 2/tt can be written as an this equation means that the product of the first n faccan be made as close to 2/tt as desired, by choosing n sufficiently large. This product was discovered by Francois Viete in 1579, and is only one of many fascinating expressions for ir, some of which are mentioned later. tors THE LOGARITHM AND EXPONENTIAL FUNCTIONS CHAPTER for a preliminary In Chapter 15 the integral provided a rigorous formulation integral plays a more definition of the functions sin and cos. In this chapter the definition presents essential role. For certain functions even a preliminary difficulties. For example, consider the function = /(*) assumed to be defined for all x and to have an inverse function, 10," positive x, which is the "logarithm to the base This function defined for 10-. is f l = (x) logio x. definition for usually defined only for rational x, while the rational x for definition the of review brief irrational x is quietly ignored. principle important an recall also but will not only explain this omission, In algebra, 10* is A x behind the definition of 10 turns The symbol 10" is first defined for natural numbers n. This notation numbers, large out to be extremely convenient, especially for multiplying very . because x by the desire to extension of the definition of 10 to rational x is motivated the customary us upon preserve this equation; this requirement actually forces The the equation we want definition. Since 10° to be true, we must = define 10° lCT n to be true, we must define lO" 10 1/n . . 71 10 • • . = 10 n • 1 • 10 n 1/n • . m to . 10° = ; = io 1 / the equation we want 1 we want n +"'* +1/n = 10 1 the equation = 10 n times 1/n be true, we must define 10 io = 10 n = l/10 n since n times to 0+n since ; = 1/n 10 • . io = 1/n vT<); and since we want "- +1/n m/n = n+ = 10 m t imes times be true, we must define io 1 / the equation 10 mln = (V 10) m . halt. We have Unfortunately, at this point the program comes to a dead to ensure that as so defined been guided by the principle that 10* should be algebraic way simple any suggest 10*+v = 10*10*; but this principle does not of defining 10* for irrational 283 x. For this reason we will try some more sophisti- 284 Derivatives and Integrals cated ways of finding a function (*) / such + y) - /(*) /(* that for all x /GO ' and y. Of course, we are interested in a function which is not always zero, so we might add the condition /(l) ^ 0. If we add the more specific condition /(l) = 10, then (*) will = imply that /(*) 10* for rational *, and 10 x could be defined as f{x) for other x; in general /(*) will equal [/(l)] x for rational x. One way to more difficult find such a function problem: find a f(* suggested is + y) = /(*) we for all /GO ' = /(I) if try to solve an apparently function / such that differentiable x and jy, io. — Assuming that such a function exists, we can try to find/7 knowing the derivative of / might provide a clue to the definition of / itself. Now h h~+o = lim h h-*o = fix) hm - n h-+o The answer thus depends on /(0) for the moment assume Even a = hm ; and denote this limit exists, fix) =<x-fix) if tive of / If in a The x we examine new light: a. Then for all x. the inverse function/ derivative of f~ l 1 by could be computed, this approach seems self-defeating. has been expressed in terms of / again. is dx is — 1 logio, the 1 1 vfiTKx)) "x the only one which / x n 1 f J x l ~ dt = - is even dx examined previously, the integral we cannot logio x deriva- whole situation appears evaluate. Since logio should have - The about as simple as one could ask! And, what more interesting, of all the integrals / it logio 1 = logio x. 1 = 0 we 285 The Logarithm and Exponential Functions This suggests that a is we unknown. One way of evading this difficulty kg*and hope t~ l define logi 0 x as (\/a) is The dt. difficulty is that to define -dt, I that this integral will be the logarithm to some base, which might be later. In any case, the function defined in this way is surely more determined reasonable, from a mathematical point of view, than logi 0 on the important of logio depends role of the number . The usefulness 10 in arabic notation (and thus ultimately on the fact that we have ten fingers), while the function log provides a notation for an extremely simple integral which cannot be evaluated in terms of any functions already DEFINITION > If x known to us. then 0, dt. The graph and < 0 if x of log < 1, is shown in Figure < then log x x ( < 0, a number log not bounded on l If x, y > Notice Now > 1, then log x > 0, 1/t is l { Jx t -dt<o. t x cannot be defined in this way, because f(t) = log = ,v 0, for the notation "log" first comes from the following theorem. then log(*y) PROOF x 1 The justification theorem if [x, 1]. area FIGURE Notice that since, by our conventions, -dt= - Jl For x 0, 1. that log'(*) choose a = log x 0 = + log y. by the Fundamental Theorem and let 1/x, number y > /(*) = log(*y). of Calculus. 286 Derivatives and Integrals Then = /'(*) — 1 = - 1 = log'foO -y • y x xy Thus /' = log'. This means that there — f(x) that + log x number a is for all x c such that c > 0, is, = log(ry) The number c +c log # for all * > 0. — can be evaluated by noting that when x = — log(l -y) log 1 + 1 we obtain c c. Thus = \og(xy) Since this COROLLARY 1 If n is is true for all y > COROLLARY 2 a natural number and x > 0, then n = ) is log proved. | x. Left to you (use induction). | If x y y > 0, then log0^ = PROOF for all x. the theorem 0, log(# PROOF + log log x log x - logy. This follows from the equations log x Theorem 1 The function = log = m (^ yj lo g + lo S J- 1 provides some important information about the graph of log. log is = clearly increasing, but since \og'(x) 1/x, the derivative becomes large, and log consequently grows more and not immediately clear whether log is bounded or unbounded becomes very small as x more slowly. It is on R. Observe, however, that log(2 it follows that log is, log therefore log takes on is w ) = for a natural log Therefore R important function has a special come clear. bounded above. 1 - not bounded below on all values. (and log 2 n log 2 in fact, not = number log 2 n (0, 1). = -n > n, 0); Similarly, log 2; Since log is continuous, it actually -1 This domain of the function log name, whose appropriateness will soon beis the . The Logarithm and Exponential Functions DEFINITION The "exponential function," exp, 287 defined as log is The graph of exp is shown in Figure 2. Since log x is defined only for x > we always have exp(x) > 0. The derivative of the function exp is easy 0, to determine. THEOREM 2 For all numbers x, = exp'(x) PROOF exp'(*) exp(#). 1 = Gog-yw log'flog-K*)) 1 1 log \x) = A THEOREM 3 If x -1 = expO). <>) second important property of exp is | an easy consequence of Theorem 1 and y are any two numbers, then exp (a: PROOF log Let x f — exp(#) and / = + y) = exp(x) exp(j). • exp (y), so that = log = log/. x y Then x +y = log x' + logy' = log(x'y'). This means that cxp(x + y) = = expO) x'y' exp(jv). | • This theorem, and the discussion at the beginning of FIGURE that exp(l) is particularly important. There is, 2 this number. DEFINITION e This definition is = exp(l). equivalent to the equation 1 = log e = f / Ji e 1 t this chapter, suggest in fact, a special dt. symbol for 288 Derivatives and Integrals As illustrated in Figure 3, [2 i - dt < since 1, 1 — (2 • an upper sum 1) is fpr J = l/*on /(/) [1,2], and / - 7i t </* > since 1, sum £ - (2 • for /(/) +£ 1) = 1/t on -dt < • (4 - 2) = 1 is a lower [1, 4]. Thus FIGURE 3 -dt \ Ji < t Ji t < e -dt, Ji t which shows that 2 In Chapter that 1 9 we < 4. much better approximations for e, and also prove much easier than the proof that tt is irrational !) will find irrational (the proof is e is As we remarked at the beginning of the chapter, the equation + y) = exp(# expO) exp(j) implies that exp(*) = = [exp(l)]* for all rational x. e*, Since exp is defined for all x and expO) = for rational x, it is consistent with our earlier use of the exponential notation to define ex as exp(Y) for all x. DEFINITION For any number x, e The terminology £ x not yet defined a x , a the attempt. If x is rational, for e, but there If a (If a last expression is > = 0, then, for e this any is now be irrational) clear. exponent x. We have We have a reasonable principle to guide us in then a DEFINITION exp(#). an arbitrary (even e But the = 'exponential function" should succeeded in defining if x x = (yoga)* defined for real = ^loga^ all x, so number we can use it to define a x . x, definition clearly agrees with the previous one.) The requirement a > 0 is necessary, in order that log a be defined. This is 289 The Logarithm and Exponential Functions we would not even expect not unduly restrictive since, for example, \ x sense, be denned. (Of course, for certain rational *, the symbol a will make according to the old definition; for example, to = 1 Our definition of a x = = (-1)1/3 /(a) = _1.) to ensure that was designed (f)v e*v and for all x y. equation turns out to be true when e is replaced by any number a > 0. The proof is a moderately involved unraveling of terminology x At the same time we will prove the other important properties of a As we would hope, FIGURE v^T i (-1)1/2 4 THEOREM 4 If a > this then 0, (a (1) b c ) = a bc for all b, c. h (Notice that a will automatically be positive, so a1 (2) = and a a x+v = a x • av (a b c will ) be defined); for all *, y. x old one for (Notice that (2) implies that this definition of a agrees with the rational x.) ^ PROOF ^J>y _ (Each of the steps in or the fact that exp a (2) a l = x+y e lloga = _ ^cloga* e ^clog (e = log = e = l = 1. c (& log a) Suppose ^c6Ioga _ ^ depends upon our last definition, .) a = = a, ^loga-h/loga = on whether > a _ -1 lo * (x+y)loga of the function depends f(x) g this string of equalities Figure 4 shows the graphs of /(*) x = blo «°) 1. ^sloga . * ^ loga a i.e., = fl * . x > 0. | a. 0<l,<z = l,or<z>l. x log a so ^l/loga a* for several different In this case log a if then = The behavior If a = 1, then Thus, < 7, < y log a, < ^ loga < av , . so Thus the function /(*) - a* is increasing. On the other hand, if 0 < a < 1,x is a = function the that /(*) shows that log a < 0, the same sort of reasoning Since one-one. = is a* then/(*) and a 0 * 1, decreasing. In either case, if a > x on every exp takes on every positive value it is also easy to see that a takes FIGURE 5 positive value. Thus and takes on all by loga (Figure the inverse function values. If 5). /W = a x , is numbers, the function usually denoted defined for then f~ l is all positive Just as a* can be expressed in terms of exp, so log a can be expressed in terms of log. Indeed, if = = = y X then log* so or _ y loga X, aV = eV\oga^ y log log * a, loga In other words, log X loga X loga The = derivatives of f(x) /to = «* Iog °, a1 and £(*) = so /'(*) ,(,)-|28f, is also easy to differentiate, if it follows = that the There The g(*) that, by definition, w • Fa'W [*'(*) L + log log^to + first A(*) £to no point in remembering this formula in any specific case that arises; it does is a*, hix) you remember • = £to it • from the Chain Rule that /'to is log a like /to = behind <?^ a = log a * log a more complicated function There loga x are both easy to find: sog'W=_J_. log a A = J —simply apply the principle help, however, to factor in the derivative will be g(x) Hx) remember . one special case of the above formula which is worth remembering. = xa was previously defined only for rational a. We can now function f(x) define and result is find the derivative of the function f(x) just = x a for any number a; the what we would expect: f(x) = xa = a /'(*) = e* 10**, so ?.,«l og x = °L. xa . ax a- K x Algebraic manipulations with the exponential functions will become second nature after a little practice—just remember that all the rules which ought to work actually do. The 2 and 3: basic properties of exp are still those stated in Theorems The Logarithm and Exponential Functions = exp'(*) exp(* - + y) 291 exp(tf), exp(*) • exp(jy). each of these properties comes close to characterizing the function exp. Naturally, exp is not the only function / satisfying /' = /, for if / = cf , then = /(*); these functions are the only ones with this property, f(x) = In fact, however. THEOREM 5 If / differentiable is and /'(*) then there is number a c for all *, /(*) such that /(*) proof = = for all *. ce* Let *w (This is permissible, since Therefore there The second function exp is ** a number ^ — - 0 for all x.) c such that = M. = c Then for all x. | basic property of exp requires a is clearly more involved not the only function / which f(*+y) discussion. The satisfies -/W-/W- any function of the form f(x) = a* also satisfies this equathere are infinitely tion. But the true story is much more complex than this without impossible, is it but many other functions which satisfy this property, one even is there that prove to appealing to more advanced mathematics, In fact, /(*) = 0 or — the function other than those already mentioned! It is for this reason that which functions many infinitely are / 10* there difficult: is so definition of satisfy =/W7W f(*+y) = /(i) but which are not continuous function the function /(*) / = ? io, 10*! One thing is true however—any satisfying f(x+y) must be of the form /(*) = a x or /(*) =/W'/W = 0. (Problem 27 indicates the way to . . 292 Derivatives and Integrals prove and this, few words to say about discontinuous functions with also has a this property.) In addition to the two basic properties stated in Theorems 2 and 3, the function exp has one further property which is very important—exp "grows than any polynomial." In other words, faster THEOREM 6 For any natural number n, lim PROOF The proof consists of several steps. Step 2. > e* x for To prove this = < 1 this is an upper sum > is = 1/t log x on < clear for x for all clearly true, since log x for f(t) e = x oo (this may be con- 0). statement (which x If x oo and consequently lim all x, sidered to be the case n —= [1, x], < x > 0. If x so log x 0) it suffices to show that 0. > < 1, x then (Figure 6) x - 1 < 1, and lim - 1 is = oo x. x Step 2. lim it— To oo prove e — = oo. X this, note that e* 12 e* By Step FIGURE 6 this 1, e x/2 the expression in parentheses x e /x shows that lim lim Step 3, • —= x = is greater than e* 12 oo. oo 11 Note that e_ The _ (f ln y i xln (e \ n expression in parentheses becomes arbitrarily large, power certainly becomes It is now tion: f{x) = possible to e-v*\ x ^ by Step 2, so the wth arbitrarily large. | examine carefully the following very interesting func0. We have A*) X3 ; The Logarithm and Exponential Functions 293 Therefore, < > /'(*) /'(*) FIGURE is large, so — 1 fx e llxl with for x < > 2 is 0, 0, for positive x. close to 0, so e~ lfxi is Moreover, close to 1 if \x\ is (Figure 7). 7 The behavior of/ near so for 0 and increasing so /is decreasing for negative x large, then x 2 x 0 large, so e~ llx2 is e's and 5's, 0 = is more 1/ interesting. If x 1/xi (e ) is small. This small, then 1/x 2 is large, argument, suitably stated is shows that lim e~ l ' xl = 0. x—>0 Therefore, if we define /w 1 then the function / is continuous (Figure FIGURE at 0: = * o, 8). In fact, 8 Indeed /'(0) = lim *->o — h 1 = h lim 0, / is actually differentiable Wc already know that — = lim the it is all more e (x>) = oo, X Z-+00 this ; true that lim and oo X » x—* means that —= lim 0. Thus We can now compute e-U** x 9* 0 0, x = 0. that h *->o 2 = A» lim A-0 an argument similar A r = hm 2^-^ — to the one above shows that /"(0) 4 A = hm —77= x 0, This argument can be continued. In (Problem 29) that (0) = 0 for The \ = function / sin 1/x, ( \ 9 = —2*-; = 0. 0, Thus 0. using induction fact, every k. Lx FIGURE 4 lim is x x it can be shown extremely flat at 0, ^ = 0 0 The Logarithm and Exponential Functions and approaches 0 so quickly that it can mask many example (Figure 9), suppose that 295 irregularities of other functions. For X 5* 0 x = 0. X fix) 0, can be shown (Problem 30) that for this function it is also true that = 0 for all k. This example shows, perhaps more strikingly than any other, just how bad a function can be, and still be infinitely differentiable. In Part IV we will investigate even more restrictive conditions on a function, which will finally rule out behavior of this sort. It /< fc >(0) PROBLEMS 1. Differentiate each of the following functions denotes a m (i) /(*) (ii) f(x) (hi) f(x) (iv) f(x) (v) f(x ) (vi) f(x) 2. (viii) f(x) f{x) (x) f{x) = = = + log(l (sin *) (a) fix) ( c ) fix) (d) fix) (e) fix) = = = = .... (II) + + . [/ = = \~| log (sine*) x -7— II (arcsin *) lo « 3 . (log xY°**. x*. of the following functions. e x+1 e* e ( + e^))e^ + (log(3 . +e ^ _ = ^+ x x . e~\ \ (Compare the graphs with the graphs I 1/exp.) -x x '—_4 = 2X = 1 2 + limits by e~ e * 1 1 x e - 1 - x - * 2 /2 - \ - x - x 2 /2 - 9^ + 1 l'Hopital's Rule. hm hm + e^ *))). log(l sin (sinX) 1 Find the following (I) log(l = eUo'**)= sin x*' **™'. = logc) sin x. Graph each (b) fix) a b * always ). arcsin (ix) (remember that - x 3 /6 of exp and ) 296 Derivatives and Integrals log(l lim (iv) + - x) X log(l (v) lim (vi) lim (cos x, sin x) 4. The + jfl x + x 2 /2 * + * 2/2 2 - log0+^)^^ + ^/2-^/3 functions sinh g = a: x - e~ 2 cosh x = tanh # = e* e x +C + =1 e~x e 2x + 1 1 are called the hyperbolic sine, hyperbolic cosine, and hyperbolic tangent, respectively (but usually read sinch, cosh, and tanch). There are many analogies between these functions and their ordinary trigo- One analogy nometric counterparts. shown that the region is illustrated in Figure 10; a in Figure 10(b) really has area x/2 is proof best deferred when we will develop methods of computing Other analogies are discussed in the following three problems, but the deepest analogies must wait until Chapter 26. If you have not already done Problem 2, graph the functions sinh, cosh, and tanh. until the next chapter, integrals. 5. Prove that (a) cosh 2 (b) tanh 2 (b) FIGURE =1. sinh 2 (d) + 1/cosh = sinh(# + y) = sinh x cosh y + cosh x sinh y. cosh(* + y) = cosh x cosh y + sinh x sinh y. (e) sinh' cosh' = = cosh. (f) (g) tanh ' - 53? 10 (c) 6. — 2 1. sinh. The functions sinh and tanh are one-one; their inverses, denoted by arg and arg tanh (the "argument" of the hyperbolic sine and tangent), are defined on R and ( — 1, 1), respectively. If cosh is restricted to [0, «> it has an inverse, denoted by arg cosh, which is defined on [1, oo). Prove, using the information in Problem 5, that sinh (a) sinh (cosh -1 (b) cosh(sinh(c) 1 x) Ar) (sinh-TM = = = Vx V 1 1 2 - +* 1 2 . . ! The Logarithm and Exponential Functions (d) (e) 7. (a) (b) 1 (cosh-yM = 2 - 1. 1 (tadT^'M = for < L 1*1 1 1 1 Find an explicit formula for sintT cosh" and tanh" (by solving 1 the equation y = sinlT" x for x in terms of y, etc.). Find , , A Vl + H X / ;a dX9 : 2 v> _ —^— dx for fl,^ </x > or 1 K a, 1, i b f for W, 2 |*| Compare your answer — writing 1 8. > for x , Vx 297 - x2 = <1. for the third integral with that obtained by -[— + —} + 211 x x] 1 Find (a) lim ax for 0 X— (b) < a < 1. (Remember the definition!) 00 lim (log *)« (log *)» (c) lim (d) lim *(log x) n Hint: *(log ( (e) . lim n x) -i)^iogiy — = x*. x-»0+ 9. 10. Graph f(x) = (a) Find the x* for x > 0. (Use Problem 8(e).) minimum value of /(*) = ex /x n for > e n /n n for > «. at that /(*) (b) Using the expression /'(*) n+1 for x > n n+1 /(n e l) + + lim f(x) X— = e*(x 1, - and > 0, and conclude n)/x n+1 prove that /'(*) > thus obtain another proof that , co. 00 11. Graph /(*) = *7* n 12. (a) Find lim - log(l + y)/y. l/-»0 would be (b) = x silly.) Find lim x log(l + 1 A). (You can use l'Hopital's Rule, but that 298 Derivatives and Integrals Prove that (c) = ^ lim x—* Prove that (d) e = a + /*)* x (1 + a/x) lim x—» 00 part with just a (c) Prove that log *(e) 13. Graph 14. If a /(*) bank = + (1 = b 1 (1 00 little algebraic fiddling.) - lim x(b 1/x \/x) x derive this from (It is possible to . for * > 1). (Use Problem 12(c).) annum, then an initial investment I If the bank compounds the interest 0. gives a percent interest per + yields 1(1 a/100) after year. 1 (counts the accrued interest as part of the capital for computing interest the next year), then the initial investment grows to 1(1 a/100) n after + n years. Now suppose that interest is given twice a year. The final amount after n years is, although interest is + — a/100) 2n but merely 1(1 a/ 200) 2n awarded twice as often, the interest must be halved not 1(1 alas, + y in each calculation, since the interest is + a/100) larger than 1(1 is a/2 per half year. This amount but not that much larger. Suppose that the 71 , bank now compounds the interest continuously, i.e., the bank considers what the investment would yield when compounding k times a year, and then takes the least upper bound of all these numbers. How much will an investment of 1 dollar yield after 1 year? Let/(*) = log |*| for x ^ 0. Prove that f(x) = 1/x for x ^ 0. Prove that if /' = cf for some number c, then f(x) — ke cz for some initial 15. *16. number *17. k. A radioactive substance diminishes at a present (since total rate proportional to the amount atoms have equal probability of disintegrating, the disintegration is proportional to the number of atoms remaining). all is the amount at time t, this means that A'(t) = cA(t) for some (which represents the probability that an atom will disintegrate). If A(t) (a) Find A(t) (b) Show amount A 0 = A(0) present at time 0. number r (the "half-life" of the radioactive in terms of the that there is a ment) with the property that A(t *18. c + r) = ele- A(t)/2. Newton's law of cooling states that an object cools at a rate proportional to the difference of its temperature and the temperature of the surroundT(t) of the object at time t, in terms of its temperature T 0 at time 0, assuming that the temperature of the surrounding medium is kept at a constant, M. Hint: To solve the differential equation expressing Newton's law, remember that = (T — M)'. ing medium. Find the temperature V *19. Prove that *20. Find 21. (a) all if = j* f(t) /(*) functions / dt, then / - satisfying /'(0 = f(t) + + -* 0. + J* /(/) dt. Prove that 1 +*+ Tx Ti + Hint: Use induction on ' n, ' ' ^ e* and compare for * > 0. derivatives. . 299 The Logarithm and Exponential Functions (b) Give a new proof that lim e x = /x n <*> X-*oo Give yet another proof of this fact, using the appropriate form of PHopital's Rule. (See Problem 11-38.) A point P is moving along a line segment AB of length 10 7 while another point Q moves along an infinite ray (Figure 11). The velocity of P is always equal to the distance from P to B (in other words, if P[t) is the 7 position of P at time t, then P'{i) = 10 - P(t)), while Q moves with 7 = traveled by Q after time t is distance 10 The constant velocity Q'(t) . defined to be the Napierian logarithm of the distance from P to B at time /. Thus 10 7 * = Nap - log[10 7 P(t)l This was the definition of logarithms given by Napier (1550-1617) in his publication of 1614, Mirifici logarithmonum canonis description (A Descrip- work which was done before 10 7 was chosen because Napier's tables (intended for astronomical and navigational calculations), listed the logarithms of sines of angles, for which the best possible available tables extended to seven decimal places, and Napier wanted to Law Wonderful tion of the of Logarithms) the use of exponents was invented! ; The number avoid fractions. Prove that Nap log x - 10 7 log 10 7 Hint: Use the same trick as in Problem 18 to solve the equation for P (a) Sketch the graph of f(x) = (log x)/x (paying particular attention . to the behavior (b) (c) Which is = y x is number y e 0 < < 1, or x = x; but if x if y 5* x **(f) = e, then the only number y satisfying then there is precisely one if x < e, then y > e, moreover, y (Interpret these statements in terms of the 1, x x satisfying xy = x 5* e, \ if x > e, then y < e. graph in part (a) !) Prove that if x and y are natural numbers and x y = — = Show 4, or x 4, y y that the set of all pairs and a straight line or x (e) ? > and (d) oo). ir Prove that xy near 0 and larger, e* or 2, which = = y x > then x = y 2, (x, y) with xy = y x consists of a intersect; find the intersection curve and draw a rough sketch. ix) = For 1 < x < e let g{x) be the unique number > e with x° x Prove that consider g(x) g is differentiate. (It is a good idea to . separate functions, f x (x) 0<x<e = x x 300 Derivatives and Integrals and write g and /2 in terms of fi If . you do part properly you this should be able to show that rt4.JfflL.lzi.) *25. X2 log g(x) 1 / This problem uses the material from the Appendix to Chapter Prove that exp (a) convex and log is is 11. concave. n Prove that (b) £= if t ' = t and 1 all pi > 0, then . . • . + < p lZl z n *>» • + • /> n zn . (Use Problem 8 from the Appendix to Chapter 11.) Deduce another proof that Gn < A n (Problem 2-20). (c) 26. p i Suppose / satisfies /' = / and f(x + y) = f(x)f(y) for all x and y. Prove — exp or / = 0. Prove that if / is continuous and f(x + y) = f(x)f(y) for all x and jy, then that / *27. either / = = 0 or /(*) [/(l)] z for all x. Hint: Show that /(*) = [/(l)]* and then use Problem 8-6. This problem is closely related to Problem 8-7, and the information mentioned at the end of Problem 8-7 can be used to show that there are discontinuous functions / satisfy- for rational ing /(* *28. = f(xy) f(e) log *29. + y) = f(x)f(y). Prove that if x for Prove that / is a continuous function defined on the positive reals, and for all positive x and y, then / = 0 or f(x) = + f(y) f(x) if > = x all f(x) = Hint: Consider g(x) 0. e~ 1 ^ for x ^ and 0, /(0) /(<?*). = 0, then /< fc) (0) = 0 for all k. *30. 31. Prove that 0 for all £. (a) if f(x) Prove that (*) anx n if + = ^^ a is 1 2 sin 1/x for ^ (**) an y 0, and /(0) = then f k \0) 0, = a root of the equation fln-i*"" + 1 • + aix + a = • = then the function y (x) {n) ^ 0 satisfies + a n ^y^ + • • • 0, the differential equation + + - 0. a is a double root of (*), then y(x) = ^ax also satisfies Remember that if a is a double root of a polynomial equation f(x) = 0, then f'(a) = 0. Prove that if a is a root of (*) of order r, then y(x) = x k e ax is a solu- *(b) Prove that if (**). Hint: *(c) tion for 0 If (*) < k < r - has « real numbers as roots (counting multiplicities), part gives n solutions y l9 (d) 1. Prove that in satisfies (**). this . . . , (c) y n of (**). case the function Ciyi + • * + c ny n also 301 The Logarithm and Exponential Functions It is a theorem that in this case these are the only solutions of (**). Problem 16 and the next two problems prove special cases of this theorem, and the general case is considered in Problem 19-17. In Chapter 26 we will see what to do when (*) does not have n real numbers as roots. *32. Suppose that / / = (a) Show that/ 2 - (b) Suppose that constant **(c) /" -/= 0 and /(0) = /'(0) = 0. Prove that 0 as follows. = either f(x) (/') ce x 2 ^ f{x) = 0. 0 for all or else /(#) x in = some interval ce~x for all x in (a, b). Show that for some {a, b) ) c. If/(*o) ^ 0 for x 0 0 < a that Why? Use *33. satisfies < this > 0, say, then there would be a number a such and f(a) = 0, while fix) ^ 0 for a < x < x 0 fact and part (b) to deduce a contradiction. x0 . that if / satisfies /" - / = 0, then f(x) = ae x + be~ x for some and b. (First figure out what a and b should be in terms of /(0) and f(0), and then use Problem 32.) (b) Show also that / = a sinh + b cosh for some (other) a and b. Find all functions / satisfying Show (a) a 34. (a) /<«> (b) /<*> *35. = /(--i). = f n ~*\ This problem, a companion to Problem 15-28, outlines a treatment of the exponential function starting from the assumption that the differential (a) equation f'=f has a nonzero solution. Suppose there is a function / 9^ 0 with /' = /. Prove that f(x) ^ 0 each x by considering the function g(x) = f(x 0 + x)f(x 0 — x), where f(x 0 ) 5* 0. and /(0) = 1 Show that there is a function / satisfying For this / show that f(x + y) = f(x) f(y) by considering the func- for (b) (c) f~f • = f(x+y)/f(x). Prove that / is one-one and that tion g(x) (d) **36. A function / is said to grow faster (/~~ *)'(*) than g if = l/x. lim f(x)/g(x) X— — 00. For 00 example, exp grows faster than any polynomial function. Suppose that are continuous functions. Show that there is a continuous gi, g2, gi, 37. . . • function / which grows faster than each Prove that logi 0 2 is irrational. gi. CHAPTER W7 IS INTEGRATION | ELEMENTARY TERMS IN Every computation of a derivative yields, according to the Second Fundamental Theorem of Calculus, a formula about integrals. For example, if F(x) — — *(log x) F then x, = f (x) log x; consequently, J*\ogxdx = Formulas of F{b) - F(a) = this sort are simplified [a{\og a) - F(b) - we adopt if a\ 0 < a, b. the notation F{a). write J* This evaluation of J* — x(log x) x |\ depended on the lucky guess that log log x dx — f is called = log x dx derivative of the function F(x) satisfying F' - b considerably = F(x) T We may then - b(log b) = x(\og x) — x. Of course, a primitive of /. is the In general, a function F a continuous function / always has a primitive, namely, *w we = /;/> which can be written in terms of familiar functions like sin, log, etc. A function which can be written in this way is called an elementary function. To be precise,* an elementary but in this chapter will try to find a primitive function is one which can be obtained by addition, multiplication, division, and composition from the rational functions, the trigonometric functions and their inverses, and the functions log and exp. should be stated at the very outset that elementary primitives usually cannot be found. For example, there is no elementary function F such that It F'(x) = e~ x * for all x not merely a report on the present state of mathematical ignorance; it theorem that no such function exists) And, what is even worse, you will have no way of knowing whether or not an elementary primitive can be found (you will just have to hope that the problems for this chapter contain (this is * a is (difficult) . The definition which we will give is precise, but not really accurate, or at least not quite standard. Usually the elementary functions are defined to include "algebraic" functions, that is, functions g satisfying an equation (g(x)) n +fn-l(x)(g(*)) n where the /»• 302 -l + • • ' +/oW = 0, are rational functions. But for our purposes these functions can be ignored. — Integration in Elementary Terms 303 no misprints). Because the search for elementary primitives is so uncertain, finding one is often peculiarly satisfying. If we observe that the function F(x) — +* log(l = x arctan x 2 ) satisfies F (just f (x) = arctan x how we would ever be led to such an observation is quite another matter), so that * arctan x ax — log(l x arctan x + x 2) f. then we may feel that we have This chapter consists of "really" evaluated j* arctan x dx. little more than methods primitives of given elementary functions (a process tion"), together with for finding known simply elementary as "integra- some notation, abbreviations, and conventions designed elementary functions can to facilitate this procedure. This preoccupation with be justified by three considerations: (1) (2) (3) Integration is know about it. a standard topic in calculus, Every once in a while you might actually need to evaluate an integral, under conditions which do not allow you to consult any of the standard integral tables (for example, you might take a (physics) course in which you are expected to be able to integrate). The most useful "methods" of integration are actually very important theorems (that apply to all functions, not just elementary ones). Naturally, the last reason how and everyone should is the crucial one. Even if you intend to forget (and you probably will forget some details the first time through), you must never forget the basic methods. These basic methods are theorems which allow us to express primitives of one function in terms of primitives of other functions. To begin integrating we will therefore need a list of primitives for some functions; such a list can be to integrate obtained simply by differentiating various well-known functions. The list given below makes use of a standard symbol which requires some explanation. The symbol //or means "a primitive of/" of /." The symbol jf or, more will often most useful in formulas Jf(x)dx precisely, "the collection of all primitives be used in stating theorems, while //(*) dx is like the following: This "equation" means that the function F(x) = * 4 /4 satisfies F'(x) = # 3 It cannot be interpreted literally because the right side is a number, not a function, but in this one context we will allow such discrepancies; our aim is to . make any the integration process as mechanical as possible, possible device. and we will resort to Another feature of the equation deserves mention. Most people write / 4 emphasize that the primitives of f(x) = x 3 are precisely the functions of the form F{x) = x z + Cfor some number C. Although it is possible (Problem 1 1) to obtain contradictions if this point is disregarded, in practice such difficulties do not arise, and concern for this constant is merely an annoyance. There is one important convention accompanying this notation: the letter appearing on the right side of the equation should match with the letter to appearing after "d" on the left side —thus 4 u z du / — —) 4 *** f j = -, jtxdx / A function in jf(x) dx, integral" of while /, J* i.e., tx dt = xt 2 —2 • a primitive of f(x) dx is called, /, is often called by way of an "indefinite contrast, a "definite integral." This suggestive notation works out quite well in practice, but it is important not to be led astray. At the risk of boring you, the following fact emphasized once again: the integral where Fis an Ja f(x) dx is not you do not defined as "F(b) — is F(a), find this statement repe- of/" time to reread Chapter 13). We can verify the formulas in the following short table of indefinite integrals simply by differentiating the functions indicated on the right side. indefinite integral (if titious, it is / / J adx = x n dx ax = n -dx = + log x , n 9* —1 1 - dx is often written J — for convenience; similar abbreviations are used in the last two examples of this table.) / e x dx = e x Integration in Elementary J sin x dx = — cos x dx — J y sec 2 y sec a: 305 cos x sin x = x dx Terms tan # tan x dx — sec # /* / J 1 + = arctan x *2 / == / VI - ./ Two arcsm * *2 general formulas of the same nature are consequences of theorems about differentiation: = SUM + g(*)] dx jc • f{x) dx //(*) dx = + fg(x) dx, c ff(x) dx. These equations should be interpreted as meaning that a primitive of / + g can be obtained by adding a primitive of / to a primitive of g, while a primitive of c f can be obtained by multiplying a primitive of / by c. - Notice the consequences of these formulas for definite integrals: If / and g are continuous, then j* [/« + g(x)] dx c • = f(x) dx J" = + f{x) dx c ' J* f" g(x) dx, f(x) dx. These follow from the previous formulas, since each definite integral may be written as the difference of the values at a and b of a corresponding primitive. Continuity is required in order to know that these primitives exist. (Of course, the formulas are also true when / and g are merely integrable, but recall how much more difficult the The product formula which THEOREM 1 (INTEGRATION BY PARTS) If /' proofs are in this case.) for the derivative yields a more interesting theorem, be written in several different ways. will f and g are continuous, then ffg'=fg-/ fg, f f(x)g'(x) dx pMg'(x) dx = = f(x)g(x) - / f{x)g(x) dx, - J'{x)g{x) dx. b f{x)g{x) I j (Notice that in the second equation f{x)g{x) denotes the function f g.) 306 Derivatives and Integrals PROOF The formula <Jgy=f'g+fg' can be written fg' = (fg)' - fg. Thus jfg' and fg can be chosen the first as = / (fg)' ~ Jf'g, one of the functions denoted by f(fg)'. This proves formula. is merely a restatement of the first, and the third formula follows immediately from either of the first two. | The second fomula As the following examples illustrate, integration by parts is useful when the function to be integrated can be considered as a product of a function /, whose derivative is simpler than /, and another function which is obviously of the form g'. There are two The first is written U U fg' fg special tricks f g which often work with integration by parts. 1, which can always be to consider the function g' to be the factor in. The second and then trick is by parts to find fh in terms of fh again, simple example is the calculation to use integration solve for fh. A Integration in Elementary Terms 307 or 1, / A more complicated J e* sin f calculation = x dx e x • ( — (log *) 2 2 — J cos x = — e* cos e* ( • — cos x) dx f g = — e* . often required: is cos x) f g' — = . - log xdx x + I + x g cos x dx e* X i u v' • (sin x) \e" u — J v ^(sin x) dx]; u f V therefore, 2 j e x sin = ^(sin # — cos #) xdx — **(sin * — cos x) a: dx or e x sin upon recognizing that a function is of you can already integrate, the greater your chances for success. It is frequently reasonable to do a preliminary integration before tackling the main problem. For example, we can use parts to integrate Since integration by parts depends the form g', the more functions j = j (log x) 2 dx / log x dx = x(log integration by parts) we have if we recall that x) — (log x) (log x) dx x I I f g' (this formula was itself derived by ; j (log *)(log x) dx = (log *)[*(log *)-*]-/ (l/*)|>(log i A i i i i f i' f g f g = = = = (log *)[x(log x) (log x)[x(log x) (log *)[*(log *) x(log *) 2 The most important method The use of this method Rule. integrating - at) - *] <fc - x]- [log x - 1] f — x] — log xdx + 1 dx J J - *] - [*(log x) - x] + x 2*(log *) of integration + 2x. is a consequence of the Chain more ingenuity than by parts, and even the explanation of the method is more difficult. requires considerably 308 Derivatives end Integrals We will therefore nite integrals theorem 2 If / (THE SUBSTITUTION FORMULA) this method in stages, stating the theorem for defiand saving the treatment of indefinite integrals for later. develop first, f and g are continuous, then O(«0«fcPROOF If F is a primitive of /, then the /.*/(*«)*'(*>*• - F{g(b)) left side is F(g(a)). On the other hand, (Fog)'= (F'*g)-g' = (fog)-g', soFo^isa primitive of (fog) - (Fo g)(b) The g' • is facilitated = = j* by the appearance of the g'(x) for g(x) 5 — . a; cos = depend upon recognizing sm 6 a; a tan - , a; 5 , sin x is g(x) tan x dx /(«) / sinx] Uh J f(u)du sin 6 a 6 J / J cob a U a; write . a; = *) for f(u) = cos X = 1 = cos where g'(x) y *= = - / log(cos a) - /(jrMk'M /"cos 6 sin we if / Ja '6 = - (sin x) £ — = — p can then be written /(cos = - will be the factor can be written as which dx can be treated similarly ya this case the factor x, 6 b f / cos x dx^j f(g(fiW(x)dx- / fan In = = !>(*)' , •sin 6 tan 5 factor cos lf(u) J* F(g(a)). | For example, the integration of . (sin *) tf* a: 7a factor 1/cos x g is = b [ sin B5 integration of f • remaining expression, u\ Thus sin x; the f(g(*)), for f(u) / The - F(g(b)) of the form (fog) sin 5 x cos x dx ( (£(.'.')) = (Fog)(a ) right side simplest uses of the substitution formula that a given function is and the cos x; the remaining 1/u. Hence /(«) log(cos * £). Integration in Elementary Terms 309 Finally, to find dx, X lOg X a notice that 1/x f(u) = = = where g{x) g'(x) log and that 1/log x = x, f(g(x)) for Thus 1/w. f a 'g(x) = log x~ M = - X log X /•»(» /* = f(g(*))g'(x) dx / = Jo /(«)& 7<7(o) log 6 =1 1 = log(log 6) - log(log a). U log a Fortunately, these uses of the substitution formula can be shortened conThe intermediate steps, which involve writing siderably. p(g(x))g'(x)dx can easily be eliminated = /*>/(«> by noticing the du, following: To go from the left side to the right side, u for p(x) , ,/\ du for g (x) dx (and change the limits of integration); , ( . substitute _ , ( the substitutions can be performed directly on the original function (accounting for the name of this theorem). For example, f . sm 5 L and /"sin 6 -i 6 x cos x afx u for sin a: r for cos x substitute L I , = u* du, / J sin tfxrj a similarly f b — sin x _ f X^T^r Usually we abbreviate this , w for cos . bStltUtC 1 a: rcoB&i _ du. ^for-sin,^ J J cos a U method even more, and say simply: "Let u du = = g(x) g'(x) dx." Thus let - du 7a X 10 log X In this integrals, if = = log x rioz b = - dx / ./log a i -du. x we are usually interested in primitives rather than definite certainly we can find f(x) dx for all a and b, then we can chapter but w (fx J* find J7(*) dx. For example, since b f it = . sin 5 x cos x dx sin 6 £ sin 6 a 6 6 follows that / sin 6 * = sin 5 x cos * Similarly, y tan x dx /^lcb^ log X It is quite first uneconomical — — = log x cos l0g(l0g to obtain primitives #, ^ from the substitution formula by finding definite integrals. Instead, the two steps can be combined, to yield the following procedure: (1) Let u du (after this = — g(x), g\x) dx; manipulation only the letter u should appear, letter x). (2) (3) Find a primitive (as an expression involving Substitute g(x) back for u. u). Thus, to find J sin 5 x cos x dx, (1) let — — u du so that we sin x, cos x dx, obtain j u*du; (2) evaluate f * u du J (3) remember to substitute sin x f y • 5 sin 5 .*• = u6 V back cos for u, so that j = xdx Similarly, if u = log du — - dx, x #, Sm6 * — - not the Integration in Elementary Terms 311 then f I — fl- 1 becomes dx / x log X J J = du log u, u so that / x log # To dx = log (log x). evaluate / +* 1 2 dx, let a du the factor 2 which has just = = 1 + x\ 2x dx; popped up causes no problem—the integral becomes so I be combined with integration by parts to yield a formula already mentioned: (This result may j 1 • arctan x dx — = x arctan x * arctan * — Jy+ 2 a: — i log(l + x 2 ).) These applications of the substitution formula* illustrate the most straightforward and least interesting types once the suitable factor g'(x) is recognized, the whole problem may even become simple enough to do mentally. The following three problems require only the information provided by the — short table of indefinite integrals at the beginning of the chapter and, of course, the right substitution (the third problem has been disguised a little by some algebraic chicanery). * The substitution formula f is often written in the form f(u)du= This formula cannot be taken f f(g(x))g'(x) dx, u=g(x). literally (after all, //(«) du should mean a primitive of / and not equal). HowIf(gM)g'(x) dx should mean a primitive of (Jog)- g'; these are certainly developed. ever, it may be regarded as a symbolic summary of the procedure which we have well: particularly reads formula fudging, the little and a If we use Leibnitz's notation, du f f{u)du= j m^dx, j sec j 2 x tan 5 x dx, x (cos x)e dx, dx. / you have not succeeded in finding the right substitutions, you should be them from the answers, which are (tan 6 x)/6, e silix and arcsin e x At first you may find these problems too hard to do in your head, but at least when g is of the very simple form g(x) — ax + b you should not have to waste time writing out the substitution. The following integrations should all be clear. (The only worrisome detail is the proper positioning of the constant should the answer to the second be e Zx /3 or 3e 3x ? I always take care of these problems as follows. Clearly Je 3x dx = e Zx (something). Now if I differentiate F(x) = e Zx I get F'(x) = 3e* x so "something" must be i, to cancel the 3.) If able to guess . , • , , /x + / I I log(* 3 dx = cos 4x dx = —3 3), > 4x sin 4 + sin (2x 1) cos (2*.+ 1) dx 2 arctan 2x dx / + 4* 1 More + 2 interesting uses of the substitution formula occur g'(x) does not appear. happens. Consider when the factor There are two main types of substitutions where this first 1 + e x dx. f -e i The prominent appearance of the expression ? suggests the simplifying substitution u du Although the expression J 1 e* = = x e e , x dx. dx does not appear, -? J 1 - e* it e* can always be put in: Integration in Elementary We Terms 313 therefore obtain +u 1 / — 1 u u 1 which can be evaluated by the algebraic [LLa.lto^ J — 1 u trick -l— + \ du= f u . du, J u 1 + io g «, -2iog(i -«) u so that + e / - e 1 1 There is x = -2 dx x - log(l + *») = -2 log an alternative and preferable way of handling does not require multiplying and dividing by u = x — log dx = - du, , we If . x x e e ~ log(l this «*) + x. problem, which write u, then f y Most 1 - " ^ lb _ immediately becomes e* substitution problems are expressing x in terms of hard to see why in terms of x u, this trick is f J and dx much \ one if - * u - afo. u resorts to this trick of in terms of du, instead of vice versa. It all (as substitution u = = g~\u) = (g-V(u) x g(x), dx du to the integral / we On //M(f )'WA. ! the other hand, if we apply the straightforward substitution u du to the same = = g{x) g'(x) dx integral, / /(*(*)) we /(*(*)) dx, obtain (i) J f(g(x)) • g'(x) dx, obtain (2) f /(«) J The integrals (1) and (2) is not long as the function expressing u x under consideration): If we apply the always works one-one for easier j • * ,. y g (g H«) are identical, since _1 (|r du. )'(") = As another concrete example, consider dx. / In we this case by one letter. will + Ve> 1 go whole hog and replace the entire expression Thus we choose the - *2 V> + = = u u2 + e* 1, = log(« 2 dx = — - w2 The integral then - 1), 1 becomes - u2 u J 1 1, * = 1 vV + substitution J 1 3 Thus / A : + is In this case, instead of we will replace x we 1) 3/2 ~ 2[f + 1) 1/2 by Vl - x2 dx. replacing a complicated expression by a simpler one, sin w, because VI — = sin 2 u x dx = = sin u, = cos arcsin x, but which helps us find the expression let [u = u. This really means the expression for x it is to be substituted for dx. arcsin x] cos u du; then the integral becomes j Vl — The . second main type of substitution illustrates the are using the substitution u in terms of u + the integral f that \ {f 3 Another example, which that can occur, = 1 = j sin 2 u cos u du cos 2 w du. evaluation of this integral depends on the equation cos 2 u = 1 + cos 2u 2 (see the discussion of trigonometric functions j cos 2 u du = ~+ J cos 2u below) so that _ </w 2 = a 2 + . sin 2w 4 Thus, Integration in Elementary Terms 315 and //Vl-x*dX ^-+ arcsin x = arcsin x = sin (2 arcsin x) - . 1 . 2 ar^ = . — / N h - sin (arcsin x) 2 V 1 2 * / N cosfarcsm x) 2 2 Substitution and integration by parts are the only fundamental methods which you have to learn; with their aid primitives can be found for a large number of functions. Nevertheless, as some of our examples reveal, success often depends upon some additional tricks. The most important are below. Using these you should be able to integrate all the functions in Problems 1 to 7 (a few other interesting tricks are explained in some of the remaining problems). listed TRIGONOMETRIC FUNCTIONS 1. Since + cos sin 2 x 2 = x 1 and cos 2x we = cos 2 x — sin 2 x, obtain cos 2x cos 2x — — — cos 2 x — (1 sin — cos x) — sin 2 (1 2 x) 2 x = = 2 cos 2 x 1 — — 2 sin 2 1, *, or sin 2 x = 1 — = 1 + cos 2x 2 cos 2 x cos 2x 2 These formulas may be used to integrate J j if n is sin 71 cos n x dx, x dx, even. Substituting (1 - cos 2x) (1 or for sin 2 x or cos 2 x yields a sum + cos 2x) 2 2 of terms involving lower powers of cos. For example, j sin*xdx = j^ -^os2* y dx = l -dx- j cos j 2xdx + cos 2 2^^ and If n is odd, = /z 2k + J sin 2xdx = cos 2 y dx. J 1, n then = x dx J — sin *(1 cos 2 x) k dx; the latter expression, multiplied out, involves terms of the form sin* cos* x, of which can be integrated easily. The integral for cos" x is treated simi- all larly. An integral sin j n x cos m x dx is handled the same way if n or m is odd. If n and formulas for sin 2 x and cos 2 x. A final important trigonometric integral is /—— = f dx cos x sec x dx = m are both even, use the + tan x). log(sec x J Although there are several ways of "deriving" this result, by means of the methods already at our disposal (Problem 10), it is simplest to check this formula by differentiating the right side, and to memorize it. 2. REDUCTION FORMULAS Integration by parts yields (Problem 17) /sin n — — - sin n_1 x dx x cos x + /cos n 1 f J x dx (x 2 + \) n dx = = In + - cos n_1 x sin x n —- * 2 (x f n n 2 + l)"' 1 ^ - In n~ 2 x dx, J n + sin f cos n-2 x dx, J 1 f 2 J (x 2 + l) 71 "" 1 *, ' and many similar formulas. The first two, used repeatedly, give a different method for evaluating primitives of sin n or cos w The third is very important . for integrating a large general class of functions, which will complete our discussion. 3. RATIONAL FUNCTIONS Consider a rational function p/q where = anx n + an -ix n ~ + + 0 m m' + = q(x) b mx + b m^x + bo. We might as well assume that a n = b m — 1. Moreover, we can assume that n < m, for otherwise we may express as a polynomial function plus a p(x) l • 1 • • • • <2 • , J Integration in Elementary which rational function integration of first follows 317 of this form by dividing (for example: is u+l + - u The Terms 1 an arbitrary rational function depends on two facts; the from the "Fundamental Theorem of Algebra" (see Chapter 25, Theorem 2 and Problem 25-3), but the second will not be proved in this book. THEOREM Every polynomial function =X m + q(x) brn^X**- + 1 • • • + bo can be written as a product q( x ) = (x- ai) •-..'(*- r ' a*) r *(* 2 + fa + 7i)'« * (* (where r\ + + rk + 2(jx + + - si) 2 + ft* + 70 -« m). (In this expression, identical factors have been collected together, so that all x — en + fax + and x 2 may 7» be assumed distinct. Moreover, we assume means that that each quadratic factor cannot be factored further. This ft since otherwise 2 a: + ft* + we can = 7,- [* - 47 < 2 0, ; factor ~ ( )] into linear factors). THEOREM If n < m and = + an__i* n-~ + +a + b m_ lXm + — + b 0 ^0) = = (x- ai) r •-..•(*- at) r *(* 1 ^(jf) • • • 0j l * » 2 + ft* + 7i)* • (* 2 + ft* + yi)>i, then p(x)/q(x) can be written in the form q(x) L(* - «i) + (at + ^i I" L(* 2 (* t a*) i-y + is known so complicated that r ~ ... . + J a*) r ' 2 J bi,*i x , ' 71) L(* .(* This expression, ai) (* + fax - + ft* + (* 2 + 7 ,) as the "partial fraction it is + £Mi + 7i) ft* 1 Sl + J (* 2 + fax + 70 s ' decomposition" ofp(x)/q(x), simpler to examine the following example, which 318 Derivatives and Integrals theorem, 2x 7 + how such an expression and shows illustrates to find According to the it. possible to write it is + 13x + 20x + 15* + 16* + 7x + 10 (x + x + 1) (* + 2x + 2)0 - l) b a cx + d ex + / 7 + + * + „ + 2 + 7-T-; (x + x (* - l) +T-7T 1) 4 5 8* 6 2 3 2 2 2 2 7T~n 2 1 o 2 . 2 2a: +k — gx 1 H +*+ 2 (* at I) 2 To find the numbers a, b, c, d, e, f, g, and h, write the right side as a polynomial 2 the over the common denominator (x 2 + x + l) 2 (x + 2x + 3)(* — 1) 2 ; numerator becomes «(* + (cx 1)(* + 2 2x + d)(x- + x + I) + x+\) +(ex+f)(x-\) (x + + 2)(x + x + l) + (gx + h)(x- l) (x + 2x + 2). + 2){x 2 2 2 l) (x +x+ l) + 2 2 £(* + 2x + 2 2 2)(* 2 2 2 2 2x 2 Actually multiplying coefficients are this + the coefficients of 2x 7 coefficient of x 8 out is 8x 6 a, + we obtain 0) we obtain (!) combinations of . . . + 13x 5 a polynomial of degree h. , 2 20x 4 whose 8, Equating these coefficients + + 15x* + 16x 2 Ix unknowns 8 equations in the eight + a, with 10 (the . . . h. 3 After heroic calculations these can be solved to give a < = = 1, b = 2, c 0, /= 0, £ = = 1, d = 3, 0, A = 1. Ix + 4 Thus 2x 7 + 5x« / (* 2 + 13x + 20* + 21x + 16x + + x + 1) + 2x + 2)(x - l) 4 5 3 2 V J 2 r - j jx^T) dx + + i^c^^^^ i We it + , * r example by starting with the into one fraction.) I , obtained partial fraction decomposition will illustrate all the difficulties in integrating rational functions. first two integrals are simple: 1 jX - ' dx = log(x - l) third integration depends x2 +x+ 1 - 1), 1 2 (* The 3 and are already in a position to find each of the integrals appearing in the above expression; the calculations The + /^ + ^ + 2- may actually be feasible. (In simpler cases the requisite calculations this particular converting f /c^H-i + i>»^ 2 x "2 - 1 on " completing the square = (* + i) 2 + f 5 which arise Terms 319 — f instead of f we could not take the square root, but in Integration in Elementary (If we had obtained this case our original quadratic factor could have been factored into linear factors.) We The can now write substitution changes —+ i = x u du = -j^dx, this integral to i f 3 J ( W yl + l) d 2 2 which can be computed using the third reduction formula given above. Finally, to evaluate / we (* x + 3 + 2x + • dx 2) write + 3 dx-'-f 2X + 2 dx+ ^ 2 J x> + 2x + 3 x* + 2x + 2 J The first it du The second integral simply 2 arctan J (x J + l) 2 dx. + 1 on the right side has been purposely constructed by using the substitution u (x> f integral can evaluate f 2 * [ is 2 -* + 3 + 2x + (x on the + * right, 1). If 1 2)» = = + 2x + 3, + 2) dx. x2 (2x which is just the difference of the other two, the original integral were 2* /• 2 J we so that (x> + + 2x + 2Y ^ 2 2 * ^ f [(x J + l) 2 + l] n ' on the right would still be evaluated by the same substitution. would be evaluated by means of a reduction formula. This example has probably convinced you that integration of rational func- the first integral The second tions is integral a theoretical curiosity only, especially since it you can even begin. This factorization of q{x) before is is necessary to find the only partly true. We have already seen that simple rational functions sometimes integration 1 + * dx; j J - e> : 1 arise, as in the another important example - is the integral i_ A -p 1 = ilog(* - 1) - llog(*+l). Z z 1 if a problem has heen reduced to the integration of a rational functhen certain that an elementary primitive exists, even when the difficulty or impossibility of finding the factors of the denominator may preclude Moreover, tion, it is writing this primitive explicitly. PROBLEMS 1. This problem contains some integrals which require little more than algebraic manipulation, and consequently test your ability to discover algebraic tricks, rather than your understanding of the integration processes. Nevertheless, any one of these tricks might be an important preliminary step in an honest integration problem. Moreover, you want to have some feel for which integrals are easy, so that the end of an integration process resort to (v) / it, will tan 2 x is in sight. you can see when section, if you The answer only reveal what algebra you should have used. dx. (Trigonometric integrals are always very touchy, because there are so many trigonometric identities that an easy problem can easily look hard.) Integration in Elementary 2. The Terms 321 following integrations involve simple substitutions, most of which you should be able (i) J J do in your head. e* sin e* dx. xl (ii) to xe~ /log dx. x (In the text this dx. was done by parts.) x r \ (iv) (v) , , x f / J j ^ e**e j (viii) J (ix) J + x 1 dx. ~^dx. xVl - x 2 dx. log (cos x) tan x dx. * log X 7 3. x dx 2e x dx f (vii) + 2 e * Integration by parts. x 2e* dx. (i) J (iii) j (iv) y (v) j (log*) e ax sin bx dx. sin si # ^MA l (vi) (vii) J / dx scc 3 xdx. (This is a tricky and important integral that often If you do not succeed in evaluating it, be comes up. sure to consult the answers.) (viii) j cos (log x) dx. (ix) j V# log x dx. (x) j *(log x) The x = 2 dx. following integrations can = sin u, x cos all be done with substitutions of the form To do some of these you will need to remember u, etc. that / sec x dx = log (sec x as well as the following formula, + tan x) which can also be checked by differ- entiation: J esc x dx = — log (esc x + cot x). In addition, at this point the derivatives of dx f (i) / v the trigonometric func- (You already know this integral, but use the sub— sin u anyway, just to see how it works x ~~ * all handy. tions should be kept stitution x out.) (ii) f ^X V +x dx (iii) , J / (iv) 2 (vi) / (x) The is no = sec 2 u, you want to use the u.) 1 — (The answer will be a certain inverse function that was gi ven short shrift in the text.) 1 dx J x2 , xV\ + X x Vl - x 2 s J - dx f - // V (ix) tan . v x2 /.- Vl J (vii) 1 = dx x w + — Vx - r (Since tan 2 u substitution x 2 1 2 dx. (Yqu wiU need I 1 — Vl + x 2 dx. ( tQ remember the methods and cos.) for integrating powers of sin J x 2 dx. |V?=T*. following integrations involve substitutions of various types. There substitute for cleverness, but there is a general rule to follow: substi- an expression which appears frequently or prominently; if two different troublesome expressions appear, try to express them both in terms of some new expression. And don't forget that it usually helps to tute for Integration in Elementary Terms 323 express x directly in terms of u, to find out the proper expression to substitute for dx. r a) + Vx + 1 ( (H) — — J dx /* ..... 1 Vx+ Vx J /dx (The substitution V +^ w = ^ leads to an ing yet anot her substitution; this 1 integral requir- is all right, but both substitutions can be done at once.) « /it tan # dx (vi) (viii) made 1 + + 4* (vii)J V (Another place where one substitution can be | f X J 2 to do the work of two.) 1 1 y *V»</*. />/ —^=# dx. — Vx 1 (In this case substitution, *(x) two successive substitutions work out two obvious candidates for the first and either will work.) best; there are 1 [J*Z±.Ldx. v J * + *2 1 previous problem provided gratis a haphazard selection of rational functions to be integrated. Here is a more systematic selection. The # y W ( ii) ' (iv) f J (v) +x — 2 A — x *±i + 3x - x 33 4. 7^.2 7x* _ - c„ 5x + (*- i)\x + 2* 2 +x+ 1 + + 4 *. Jx^Ti AT , 1 3x 2 - / (* + 3)(* /• , 3 J x* r v ^ + 7,-1— f (i) 1 5 , - 1) dx. l) 2 *. 324 Derivatives and Integrals (V ° dx 2 (vii) ( ix ) J* - + 2x >+l 3x + 3* + + 2x + Ix + J x* 3 1 2 y (x + x — + l) — 1 2x_ = 2 * 2 3 «. / , 2+ ; +1) 3^ ( *7. (No holds barred.) The following integrations involve methods of the previous problems. Potpourri. » w r*i (iii) (iv) arctan x / /" the . ax. + 1 all *2 — arctan x ; J log Vl + dx. y * log y/\ + * 2 (v) (vi) (vii) y arcsin (viii) (ix) * f J + 1 •am x dx. sin x x cos* . J> ^" dx. __ Vtan /dx * 8. If you 6+1 at — sin X _ . fa m COS^ X a: dx. + (To factor x 6 1, first factor y* + 1, using Problem 14.) have done Problem 1 7-7, the integrals (ii) look very familiar. In general, the substitution x — for integrals involving integrals involving V# + 2 1 . 1 , while x Try — — Problem 4 will cosh u often works is the thing to try for and sinh u (iii) these substitutions in on the other inte- Integration in Elementary grals in Problem (The method 4. is 325 Terms not really recommended; it is easier to stick with trigonometric substitutions.) *9. The world's sneakiest substitution x — t tan — is x = dx = ' undoubtedly 2 arctan L 2 2 + 1 As we found this substitution leads to the expressions Problem 15-16, in ™ 2t X dt. 2 t cosx = = TT?' 1 - 2 t TT7 2 This substitution thus transforms any integral which involves only sin and cos, combined by addition, multiplication, and division, into the integral of a rational function. Find f (j) J (ii) + 1 — f J (iii) 1 (Compare your answer with Problem *L — f / (i n this sin 2 ( *10. ) f J 3 (There + b cos x dx. (An it is better to let x also another is problem —— + t = tan Why?) x. exercise to convince (A way to do this, using 15 . 7> ) should be used only as a * Wv case sin 2 x J a sin x (iv) l(viii).) sin x you that this substitution last resort.) last resort.) 5 sin x Derive the formula for J sec x dx in the following two ways: (a) By writing cos x 1 cos x _ ~ 1 — 1 [" 2 Ll sin 2 x cos X COS X + sin x — 1 ~1 sin *J an expression obviously inspired by partial fraction decompositions. Be sure to note that f cos x/{\ - sin x) dx = - log(l - sin at); the very important. And remember that % log a there on, keep doing algebra, and trust to luck. minus sign From is = log V a. t = tan x/2. Once again, quite a bit of required to put the answer in the desired form; the expression tan x/2 can be attacked by using Problem 15-8, or both answers can be expressed in terms of t. There is another expression By using (b) the substitution manipulation for is which sec x dx, / using Problem 15-8, r sec x dx I — log is less cumbersome than we obtain f l+ tan log (sec x + tan x); t\ = log(tang + ?)} I ( J This last expression was actually the one first discovered, and was due, not to any mathematician's cleverness, but to a curious histori- In accident: cal amounted Wright computed nautical tables that 1599 When the first tables for the logarithms of tangents were produced, the correspondence between the two tables was immediately noticed (but remained unexplained to definite integrals of sec. until the invention of calculus). The derivation of je x sin x dx given in the text seems to prove that the only primitive of f(x) = e x sin x is F(x) = e x (sin x - cos x)/2, whereas F(x) = x (sin x — cos x)/2 C is also a primitive for any number C. Where does C come from? (What is the meaning of the equation + <? xdx = je x sin (a) Find / arcsin x e dx, x sin x — e x cos x — J> sin x dx?) using the same trick that worked for log and arctan. *(b) Generalize this trick: (a) (b) Find $f~ l (x) dx in terms of ff(x) dx. Compare with Problems 12-15 and 14-10. Find / sin 4 x dx in two different ways: first using the reduction formula, and then using the formula for sin 2 x. Combine your answers to obtain an impressive trigonometric identity. Express / log (log x) dx in terms of (log x)~ l J dx. (Neither is expressable in terms of elementary functions.) Express jx 2e~x2 dx in terms of je~ x2 dx. Prove that the function f(x) = e x /(e bx primitive. (Do not try to find it!) Prove the reduction formulas in the dx f J (1 + and work on the tion x = tan u.) X r n 2 ) J (1 last integral. + + ex + text. dx *2)n-l 1) has an elementary For the third one write x 2 dx f J (1 + (Another possibility is x *)n to use the substitu- : Integration in Elementary 18. *19. Find a reduction formula Terms 327 for n (a) fx e°dx. (b) /(log x) n dx. Prove that /—2 Vt A L , \dt = cosh x sinh x x - 2 2 i (See Problem 17-4 for the significance of this computation.) 20. Prove that h ja h = f{x) dx ja f(a + - b x) dx. 21. (A geometric interpretation makes this clear, but it is also a good exerhandling of limits of integration during a substitution.) 2 Prove that the area of a circle of radius r is wr (Naturally you must 22. remember that w is defined as the area Use induction and integration by parts cise in the . of the unit circle.) to generalize Problem 14-6: /;(/;(... (/>„)„,)...)„„. 23. If /'is continuous on Lebesgue Lemma b\ use integration by parts to prove theRiemann- [a, for /: 6 lim f This result is /(/) sin (X*) dt = 0. Problem 15-25, but it can be used to the same way that the RiemannProblem 1 5-25 from the special case in just a special case of much prove the general case (in Lebesgue Lemma was derived in which / 24. is a step function). Prove the following version of integration by parts for improper integrals P u'{x)v{x) dx = u(x)v(x) Ja The first — |o symbol on the right side lim u(x)v(x) X—* *25. f 00 f ; u(x)v'{x) dx. fl means, of course, — u(a)v(a). 96 One of the most important functions in T(x) = jj analysis e-H*- 1 is the gamma function, dt. Prove that the improper integral T(x) is defined if x > 0. (b) Use integration by parts (more precisely, the improper integral version in the previous problem) to prove that (a) I\* (c) Show that T(l) natural numbers = n. 1, + 1) = xY{x). and conclude that I» = (n - 1)! for all 328 Derivatives and Integrals The gamma function thus provides a simple example of a continuous function which 'interpolates" the values of n for natural numbers x. Of course there are infinitely many continuous functions / with ' ! (n — 1 ) there are even infinitely many continuous funcwith f(x 1) = xf(x) for all x > 0. However, the gamma / function has the important additional property that log of is convex, a condition which expresses the extreme smoothness of = f(n) ; ! + tions A beautiful theorem due to Harald Bohr and Johannes this function, Molierup = /(l) + and f(x 1 T states that 1) is the only function = xf(x). See the Suggested Reading for a / with log ° / convex, reference. *26. (a) Use the reduction formula /v/2 sm n x / — n = dx for Now show /2 r^ \ w n x dx to show that 2 sin / Jo (b) sin J n~2 x dx. Jo that 2 sin^,^ = - o .i.^ 3 T/2 ,n sin 2 " . , x <& 7 5 + In x 1 3 5 -•-•-•- = 2 o 4 2 1 - 2» 1 , In 6 and conclude that 72 7r_2244 2 (c) 3 1 Show 3 < This 5 5 sin 1 2n 7 - 2« 2n+1 result, can be 1 + < sin and x 2w * 2n 1 + sin2n * /o" 1 ^ ^Si^n+i ~ ^* 2 2 4 1 3 3 made 271 < x sin 271 "1 x for 0 U < x < tt/2. 2n 2n 6 5*5*7 ' 2n - 1* 2n as close to ir/2 as desired, 7T known = 2 Show is l/2n, starting with the inequalities which shows that the products infinite product, (d) 6 that the quotient of the two integrals in this expression between 0 6 2 2 as Wallis 4 4 6 r3*3*5*5 5 is + 1 usually written as an product: 6 '7* also that the products 1 \Tn 1 • 2 • 3 • 4 5 ; • 6 . 2n • . . • (2n - 1) can be made as close to Vx as desired. (This problem and in Problem 26-18.) fact is used in the next Terms Integration in Elementary 27. It is an astonishing computed where ordinary in cases no elementary formula J improper integrals fact that for can often be f(x) dx cannot. integrals Ja but we can e~ x * dx, ja f(x) dx Jq 329 There is find the value of e~ x dx precisely! There are many ways of evaluating this integral, J but most require some advanced techniques; the following method involves a fair amount of work, but no facts that you do not already * know. (a) Show that ,24 i 3 o 1 i (1 —+ In = (1 —7' 2n 5 1 7T 1 3 2n - 2 2 4 2n - dx + x 2) 71 2 (This can be done using reduction formulas, or by appropriate substitutions, (b) combined with the previous problem.) Prove, using the derivative, that 1 - x2 < xi < ~ e~ (c) e~ x * for 0 < x for 0 < x. 1 1 + < 1. x2 Integrate the nth powers of these inequalities from 0 tol to oo , respectively. Then use the substitution y 3 and from 0 V nx to show that ' In 4 Vn = + 2n 5 1 \fn < e~y* dy j < / e~y" dy q <-Vn2 2 (d) Now (a) 2n 2n - 3 2 use Problem 26(c) to show that I *28. 3 4 rv Use integration by parts f b ' sin .v Ja , f dx / = X and conclude that gate the limit as Jj to dy - T- show that —+ cos a b cos £ a 0 (sin *)/* P cos x dx, / Ja exists. + a -» 0 and the right b f . X2 (Use the left side to investi- side for the limit as b — > co .) 330 Derivatives and / Integrals (b) Use Problem 15-31 to • show that sin (n + \)t dt = 7T i: for (c) any natural number n. Prove that lim sin(X + \)t dt = 0. sin (d) Hint: The term in brackets is bounded by Problem 15-2(vi); the Riemann-Lebesgue Lemma then applies, Use the substitution u = (X + %)t and part (b) to show that sin x ~ ~ X Jo *29. (a) (b) Use the Find = x to t ro) = x Jo substitution u X _ ~ 7T 2 show that r e-« *dx. l PART INFINITE SEQUENCES AND INFINITE SERIES One of the most remarkable algebraic analysis m m(m — +-x+ 1-2 1 of series is the following I) 2 X' 1 m(m - 1) ( m- 2) „. 1-2-3 m(m - + 1) • • - [m - (n 1-2 1)] n + m when which as is a positive whole number sum of the the is is series, then finite, can be expressed, known, by When m is ( 1 + m x) . not an integer, the series goes on to infinity, and it will converge or diverge according as the quantities m and % have this or that value. In this case, one writes the same equality +x) m = (1 1 +™x — A jT2~ m(rn + . . . It is - 1) + "' etc assumed that the numerical equality will whenever the this x always occur series is convergent, but has never yet been proved. NIELS HENRIK ABEL ' APPROXIMATION BY POLYNOMIAL FUNCTIONS CHAPTER There at all. one sense in which the "elementary functions" are not elementary If p is a polynomial function, is p{x) = <2 0 + + aix * + ' • then p(x) can be computed easily for any number At functions like sin, log, or exp. a nx x. n , This present, to find log x not at is = the error involved in accepting such a for log x is true for 1/tdt approxi- J* mately, we must compute some upper or lower sums, and make sum all certain that not too great. Com- log" (a:) would be even more difficult: we would have to computing e pute log a for many values of a until we found a number a such that log a is x approximately x then a would be approximately e In this chapter we will obtain important theoretical results which reduce the computation of/(*), for many functions /, to the evaluation of polynomial — x 1 — . finding polynomial functions which are appropriate, close approximations to/. In order to guess a polynomial which is thoroughly. more themselves it is useful to first examine polynomial functions The method depends on functions. Suppose that p(x) It is at 0. To a0 + + aix ' • + anxn • . our purposes very important, to note that the coeffican be expressed in terms of the value of p and its various derivatives interesting, cients en = and for begin with, note that p(0) = a,. Differentiating the original expression for p(x) yields = p'(x) ax + 2a 2x + ' • • + na n x n '\ Therefore, P'(0) Differentiating again V'(jr) = we 2a 2 =/> (1) (0) =ai. obtain + 3 • 2 • a zx + • • • + n(n - 1) a nx • n "\ Therefore, 2 ^"(0) =/,< >(0) = 2a+ | In general, we will have /><*>(0) agree to define 0! = formula holds for k = 0 also. If wc = 1, kla k and or ak = ~jr' recall the notation (0) /> = p, then this 333 I 334 Infinite Sequences and Infinite Series If we had begun with a (*-«)," p(x) = a0 function p that was written "polynomial in as a + ai(x -*)+'••+ a n (* - a) n , then a similar argument would show that — =P flt Suppose now that / (a) a function (not necessarily a polynomial) such that is /<»(*), all exist. ik) . . ,/»>(«) . Let = a* —— < 0 > k <n, " k\ and define The polynomial Pn.a (Strictly speaking, Pn.aj, to indicate the will be useful.) called the is we should Taylor polynomial of degree n for / at dependence on /; The Taylor polynomial has been Pn/ k) a. more complicated expression, like at times this more precise notation use an even (a) = f k) (a) for 0 < defined so that < k n; it is clearly the only polynomial of degree < n with this property. Although the coefficients of Pn a j seem to depend upon / in a fairly complicated way, the most important elementary functions have extremely simple Taylor polynomials. Consider first the function sin. We have in fact, , sin(O) sin'(0) sin"(0) sin"'(0) sin (4) (0) From this = 0, = cos 0 = 1, = - sin 0 = 0, = - cos 0 = -1, = sin 0 = 0. point on, the derivatives repeat in a cycle of 4. The numbers sin ( *>(0) are 11— 0, 1, 0, > 0, 31 5 , 0, 11— > 7! ! 0, • • • Therefore the Taylor polynomial ftn+i.o of degree 2n Oct X3 , 3! (Of course, P 2 »+i,o = iV^.o). X5 X1 5! 7! • • 9! + 1 for sin at 0 —I— i (2n + 1)1 is — , 335 Approximation by Polynomial Functions The Taylor polynomial P 2n ,o are left of degree In for cos at 0 is (the computations to you) x2 , 2! x4 x6 4! 6! . mxm , , x 2n (2n)l The Taylor polynomial for exp is especially easy to compute. = = exp(O) 1 Taylor polynomial of degree n at 0 for all k, the u3 y ^.o(x) The Taylor polynomial since log 0. The standard • * choice 1 log'W = • +^r- must be computed at some point a for log not even defined at is is yi4 1+- + - + - + -+ = Since exp (fc) (0) —> log'(l) = is a = 1. 0, Then 1; X 1 log"W = X2 —2 log"'W = X3 in general (-l)*- 1 ^ --1)! log<*>(*) 1 ** Therefore the Taylor polynomial of degree n for log at Pn.l(x) = (X ~ — 1) + 2 It is often this case more convenient " n to consider the function f(x) — log(l + x). In We have a = 0. /<*>(*) = log (A) (l /<*>(0) = log<*>(l) we can choose + + 3 1 is + *), so = (-l)*-»(* - 0!. Therefore the Taylor polynomial of degree n for / at 0 -*-- + /»„..(*) 2 — -+ 4 3 ••• + is (-1)"-V /» There is one other elementary function whose Taylor polynomial arctam The computations of the derivatives begin tant is impor- — arctan'O) = — - 1 arctan"(*) = +x ~ 2* +# +* = (1 = 1; = 0; 2 2 ) 2 x arctan"(0) > / (1 arctan"'(*) arctan'(O) > 2 —— 2 ) * (-2) -f (1 — +2* +* 2 -2(1 4 ) +*5_v2* > m = — 2.9 „„ , arctan'"(0) 336 Infinite Sequences and Infinite Series It is clear that this brute force computation will never do. However, the Taylor polynomials of arctan will be easy to find after we have examined the properties of Taylor polynomials more closely although the Taylor poly- — a as /, Pn ,a,/ was simply defined so as to have the same first n derivatives at the connection between / and Pn a j will actually turn out to be much nomial , deeper. One line of evidence for a closer connection between / and the Taylor polynomials for / may be uncovered by examining the Taylor polynomial of degree 1, which is *..(*)-/(«)+/(«)(*-«). Notice that /(*) Now, by FIGURE - PUx) = -f(a) /(*) _ we have the definition of f(a) 1 In other words, as x approaches a the difference /(*) — Pi, a (x) not only small, but actually becomes small even compared to x — a. Figure 1 becomes illustrates the graph of f(x) — e x and of Pi.oM =/(0) +f(P)x which is the Taylor polynomial of degree the graph of *..(*) which is = /(0) + /'(0) + 1 =l+x, for / at 0. ^ x* = — also shows 1+ x + £ 0. As x approaches 0, the seems small to getting even faster than the be i\o(*) the Taylor polynomial of degree 2 for / at difference f(x) The diagram Approximation by Polynomial Functions difference f(x) we are — Pi,o(x). now prepared As it in general i im x—ki = For f(x) e x and a = lim On 0 this is not very precise, but We have just noted that stands, this assertion it to give a definite meaning. ^M — x means = 337 o. a that /w-A,w g = lim --i-, = Q- the other hand, an easy double application of PHopital's Rule shows that eX i Z im z->o " 1 a: * = 2 ± j o. — Pi.oW becomes small compared to x, as x approaches become small compared to x 2 For ft.oW the situation is quite Thus, although /(#) does 0, it not . different; the extra term x 2 /2 provides just the right compensation: e x — 1 — x 2 X a?->0 e = lim — if f{a) Suppose that / is — x 1 = 0. 2 and f'{a) Pn>a exist, then 0; also true. a function for which is /'M, all exist. — lim x— o = in fact, the analogous assertion for l x 1 2X e lim THEOREM — x-+0 = This result holds in general x lim 2 . . ,/->(«) . Let ak = f {k) — (a) 0 ' 7i < A ' ' < n, k\ and define ^n t aW = a0 + ai(x — a) + ' + an {x — Then lim^j ~ *->a (x — ^-W a) n = 0. a) n . 338 Infinite Sequences and Infinite Series PROOF Writing out Pn ,a(x) we explicitly, obtain h - pu*) = n (x - a) zoo i - (x It will help to introduce the new <w) - / a) («) n , n\ functions n-l Q(x) now we must = Z^tp - (* = *W and «)' ( * ~ prove that lim /W - QW _/ <n) («) . n\ g(x) Notice that = /<*>(«), Q(«(a ) £<»(*) = Q(*)] = - n!(* k < - n 1, *)"-*/(» -A)!. Thus - lim [/(*) - /(«) Q(«) = 0, X—>fl lim - [/'(*) lim [/ (B - 2) = /'(*) - Q'OO = Q'(*)] W - = /<"- Q<»- 2) (*)] J) o, - Q (n- (a) 2) (a) = = 0. 0, 2 and limgW = lim*'(*) x—<* x—*a We may therefore Q is a /(*) =/ x-^a and this last limit • • - is (n " 1) (x Q(«) ( "- 2) (*) - (n - 1} _ L/ - 1 times to obtain M - Q^W 1, its (n - . l)st derivative is a constant; Thus (^). - = lim£ x—*a polynomial of degree n in fact, • apply THopitaPs Rule n lim Since = a) n f n) (a)/nl x-~*a by n\(x definition of f n) - a) (a). | One simple consequence of Theorem 1 allows us to perfect the test for local maxima and minima which was developed in Chapter 11. If a is a critical local minipoint of /, then, according to Theorem 11-5, the function / has a mum at a if f"{a) > 0, and a conclusion was possible, but further information; and 0. If/" (a) = 0 no 7 " (a) might give of/ sign that the conceivable local it is if f"(a) maximum at a if/" (a) < - 0, (4) then the sign of/ (a) might be signifi- . Approximation by Polynomial Functions Even more cant. generally, =/» /'(«) (*) we can f n) (a) * The • • =0, (a) • 0. can be guessed by examining the functions situation in this case - (* - «)• = -(*-«)", f(x) *(*) which when ask what happens = 339 Notice (Figure 2) that if « is odd, then a is neither a local nor a local minimum point for / or g. On the other hand, if n is even, then /, with a positive nth derivative, has a local minimum at a, while g, satisfy (*). maximum with a negative wth derivative, has a local maximum at a. Of functions all about the simplest available; nevertheless they indicate the general situation exactly. In fact, the whole point of the next proof is that any function satisfying (*) looks very much like one of these functions, in a satisfying (*), these are sense that is made precise by Theorem 1 Suppose that = f(a) f (1) If n is (2) If n is (3) If n is There is n) • (a) * > < 0, even and f (n) (a) even and f (n) (a) = • then / has a local minimum at a. then / has a local maximum at a. odd, then / has neither a local maximum or minimum at clearly no 0, of generality in assuming that f(a) loss the hypotheses nor the conclusion are affected Then, since the Pn>a of / — n first 1 if / is derivatives of / at a are 0, = a. 0, since neither replaced by / — f(a). the Taylor polynomial is PnM) = f(a) (X 1 (n) f Thus, Theorem 0 = Consequently, if lim x is (x Suppose now that — a) (x • _ a) n n\ states that 1 /M - />„.„(*) _ _ J 1: — a) n —a *-»a sufficiently close to a, /(*) n -«)+• ! (a) (# f(x)/(x 0, 0. n — r L(* f n) /« — - fl) a) n has the same sign as 7f (n) 0 for all (a) n! even. In this case (x has the same sign as f J then a)* is ( nl (n) — a) n > x a. Since (a)/n\ for x sufficiently close to a, it 340 Infinite Sequences and Infinite Series same follows that f(x) itself has the a. lif n) > {a) for x close to a. Consequently, for the case Now f (n) < (a) suppose that n sufficiently close to a, is - a) n > - 0 for x signs for x different mum nor a local for x sufficiently close to > 0 / has a = f(a) minimum local at a. A similar proof 0. The same argument odd. /(*) (x \ as before shows that if xis then (x But n f (a)/n means that 0, this f(x) works sign as — a) > u u the same sign. always has n and a - (x > a and x < a. This minimum at a. \ n < 0 for x < a. Therefore f(x) has proves that / has neither a local maxi- a) Although Theorem 2 will settle the question of local maxima and minima about any function which arises in practice, it does have some theoretical limitations, because f (k) (a) may be 0 for all k. This happens (Figure for just 3(a)) for the function K ) = x \ 0, 0, which has a minimum at 0, and also for the negative of this function (Figure 3(b)), which has a maximum at 0. Moreover (Figure 3(c)), if - /(*) -e~ { then f (k) (0) maximum (c) FIGURE The 3 = 0 for all k, lix \ x = < 0 0, but / has neither a local minimum nor a local at 0. conclusion of Theorem is 1 concept of "order of equality." at a x Jo, often expressed in terms of an important Two functions / and g are equal up to order n if lhn^-y — (x x~+a a) n = 0. Theorem 1 says that the Taylor polynomial The Taylor polynomial might very well have In the language of this definition, Pn.aj equals / up to order n at a. been designed to make this fact true, because there is at most one polynomial < n with this property. This assertion is a consequence of the following elementary theorem. of degree THEOREM 3 PROOF Let P and Q be two polynomials in (x — a), of degree < P and Q are equal up to order n at a. Then P = Q. Let R =P— Q. Since R is a polynomial of degree < n, and suppose that n, it is only necessary to 341 Approximation by Polynomial Functions prove that if R(x) = + b0 + - b n (x a)" satisfies R(x) — x^a (x then R — Now 0. RW lim / x-* a (x For = 2 — = . surely imply that 0 < for 0 = x—>a = + [b 0 < n. = 0; on the other hand, -«)+••• + bi(x - <0 n ] x—*a = b0 lim i ay 0 this condition reads simply lim R(x) x —*a \imR(x) Thus R the hypotheses on n a) 6„. 0 and /?(*) = 4i(* = *i +* -«)+• Therefore, 2 (* - a) + • + • b n (x - a) n~ l and X—>a Thus #i = *(*) Continuing in COROLLARY Let / be (x PROOF — — a this = b 2 (x - a y+ • way, we find that a, and suppose that P is a polynomial in which equals / up to order n at a. Then P = Pn 0 times differentiable at rc- a) of at 0 and degree < n, . , P and Pn a both equal / up to order n at a, it is easy to see that P equals Pnta up to order n at a. Consequently, P = Pn a by the Theorem. | Since , , At first sight this corollary appears to hypotheses; it have unnecessarily complicated might seem that the existence of the polynomial matically imply that /is sufficiently differentiable for this is not so. For example (Figure s( J If P(x) = 0, is undefined. _ { * n+ \ 10, , P would to exist. auto- But in fact suppose that x irrational x rational. P is certainly a polynomial of degree < n which equals / up On the other hand, f(a) does not exist for any a 7± 0, so /"(0) then to order n at 0. \ 4), Pn a 342 Infinite Sequences and Infinite Series V +1 x irrational , x rational 0, FIGURE 4 When useful / does have method that our first however, the corollary may provide a Taylor polynomial of /. In particular, remember find the Taylor polynomial for arctan ended in failure. n derivatives at a, for finding the attempt to The equation = arctan x / Jo suggests a promising 1 by 1 + t 2 , 1 + dt 2 t method of finding a polynomial close to arctan —divide to obtain a polynomial plus a remainder: This formula, which can be checked easily by multiplying both sides by shows that = f / x + 1 . • fx + (-1)V»*+ • / J° JO 3 y 2n+I +(- )"^xr + 2n + + 1 5 (-i) n+1 1 fx / Jo 1 + t 2 , f2n+2 1 1 f + dt t 2 2n+2 hrrt*1 + t 2 According to our corollary, the polynomial which appears here will be the Taylor polynomial of degree In + 1 for arctan at 0, provided that ^2n+2 Jo lim 1 —+ dt 2 2 t' X 2n+1 X-+Q = 0. Since ^2w+2 rx 7o this is clearly true. 2n + 1 i + rx 2 y0 / 2/z + 3 Thus we have found that the Taylor polynomial of degree for arctan at 0 is X =*--X + ---•+ (-1)" 3 ft«+uW 3 h 5 ~2n+l 2n + 1 : ! 343 Approximation by Polynomial Functions By is now the way, that w we have *) and = is, by find arctan a) (0) -... +( -ir 2n I + . i definition, + arctan-(O), + arctan-(O) it find arctan a) (0) for all k: Since *-- + since this polynomial we can discovered the Taylor polynomials of arctan, work backwards and possible to 2 *+ + (2/2 ! by simply equating the + ^ 1) coefficients of x k in these two polynomials arctan a) (0) = . .... 0 if k is even, k\ arctan » +1) ( 0) ( (2/+ = (^}Y arctan <« +l (0) > or 21+1 1)! = . (2 /)|. A much more interesting fact emerges if we go back to the original equation — y3 - * y5 T+?- - « y 2n-fl + • 27+T + - ,r+I fx t 2n+2 ( /. and remember the estimate ^2n f 2 o When < |#| as small as < |*| as we 1, we we can 1 this expression 1 is dt + at 2n + most 1/(2h 3 + 3), and we can make use The most important theorems about Taylor polynomials extend like. this simply by choosing n large enough. In other words, for the Taylor polynomials for arctan to compute arctan x as accurately like this isolated result to other functions, and the Taylor polynomials will soon play quite a new role. The theorems proved so far have always examined the behavior of the Taylor polynomial Pn a for fixed n, as x approaches a. Henceforth we will compare Taylor polynomials P nia for fixed x, and different n. In , anticipation of the coming theorem If / a function for which is we introduce some new notation. exists, we define the remainder term P n a (x) , Rn a{x) by t /(*) - Pn ,a(x) = /GO + Pn.a& +/'(«)(* -*)+•••+ ^-7^ (* ~ *) n + RnA*). n I We would like to have an expression for R n>a (x) whose size is easy to estimate. There is such an expression, involving an integral, just as in the case for arctan. to guess this expression is to begin with the case n = 0: One way /(*) «/(n) +RM- 344 Infinite Sequences and Infinite Series The Fundamental Theorem of Calculus enables us to write /w =/oo + j'mdt, so that AuM = A similar expression for Ri, a {x) x f(t)dt. ja can be derived from this formula using inte- gration by parts in a rather tricky way: Let = "(0 (notice that x represents = v'(t) 1); and f(t) some — v(t) number fixed t — x in the expression for v(t), so then X ( f (t)dt= P/'M-l dt i I u(t) v'(t) = - u(t)v(t)\* r f {t){t-x) u'(t) Since v(x) = dt. v(t) we obtain 0, /w = + j'mdt - «(«>(«) + /(«) = /(«) = f(a) +f'(a)(x - * -a) + /"«(* f' /; /"W(* o t) dt. Thus RU*) = fa* f"(t)(x It is v(t) = hard t. it is any motivation for choosing v(t) = t — x, rather than happens to be the choice which works out, the sort of thing one to give It just might discover ever, now after sufficiently v'(t) — many similar but futile manipulations. easy to guess the formula u (t) then t) dt. (x — t), = f"(t) ( and v(t) so x r f"(tKx Ja How- R 2ta (x). If ~ * - ^\ = for 1) dt = - u(t)v(t)\ \a r Ja -/'(«)('-«>' 2 '(0 + • ~(x-t) — r^ Ja {x dt -tyd, 2 This shows that ./«>(/) (x 2 You should now have little difficulty - 2 O ^- giving a rigorous proof, by induction, 1 345 Approximation by Polynomial Functions that if f (n+1) continuous on is = *»..(*) then [a, x], / ^—t^ " (* 0" dt. form of the remainder, it is expressions for /?„,«(*): important possible (Problem 13) to derive two other remainder, the Cauchy form of the From formula, which this R n aix) = called the integral is - - (* t) n (x - for a) some t in (a, *), n\ and the Lagrange form of the remainder, f(n+l)(A Rn,a(*) = 7 i A (* - SOme f° r we will derive all three One virtue of this proof In the proof of the next theorem (Taylor's Theorem) forms of the remainder in an entirely different way. in 1 Cauchy and Lagrange forms of (aside from its cleverness) is the fact that the the remainder will be proved without assuming the extra hypothesis that (n+1) direct generalis continuous. In this way Taylor's Theorem appears as a / ization of the is = 0, and which suggest a strategy for proving Taylor's Theorem. Mean Value Theorem, to which it reduces for n the crucial tool used in the proof. may These remarks R n a (a) = Since 0, we might try to apply the Mean Value Theorem to the , expression x On is — second thought, however, not at Indeed, clear all if we this how / (n+1) (f) = which is is — a idea does not look very promising, since it ever going to be involved in the answer. take the most straightforward route, of the equation which defines /'(*) x a f(a) +f"(a)(x useless. R n>a we , -*)+•••+ The proper and differentiate both sides obtain (* application of the - a)*" 1 + /?„«'(*), Mean Value Theorem has a This proof lot in common with the integration by parts proof outlined above. was fixed. which number denoted a x which in involved the derivative of a function This THEOREM 4 (TAYLOR'S theorem) is just how x will be treated in the following proof. (n+1) are defined on Suppose that /',..., / by /(*) = f( a ) + f(a )(x _«)+...+ [a, x\ and that (x R n a (x) is denned , - aY + R n ,a(x )- 346 Infinite Sequences and Infinite Series Then /(»+i> (1) ft) Rn,aW = - (x n /) ~ (* some a) for for some in / (a, x). , (2) Moreover, (3) (If a: < f Rnta (x) = if /< n+1) a, (t) - (* ;/ J ; \n -+- is integrable n+1 a) / in (a, x). (n + 1 t ! 1 on then [a, at], /•<»+!) (V) = /?„..(*) (n+1) / I A. ./aft! then the hypothesis should state that /is ) -times differentiable on [x, a]; the number / in (1) and (2) will then be in (x, a)> while remain true as stated, provided that / (n+1) is integrable on [x, a].) PROOF For each number /(*) = fit) t in [a, we have x] +f(t)(x Let us denote the number - + t) R Htt (x) • by • + • ^ S(t); the - (x function + 0" (3) will HA*)- defined on 6" is [a, x], and we have (*) fix) = fit) - +f'(t)(x + 0 ' • + • (* - 0 n + J(0 for all We will now t in [a, *]. differentiate both sides of this equation, function whose value at t is /(#), equals the function which asserts that the whose value at t is {n) f(0 (In + f + ^—y (t) • • (* - ty + sit). common parlance we are considering both sides of (*) Just to make sure that the letter x causes git) = for all /(*) "as a function of /.") no confusion, notice that t, then g'it) and = 0 for all t; if f (k) (t) then g'it) =l-Mk{x-Jf {k) w_ ty-\-\) (t) r x _ t \k-i + L_w f _i_ Z (k+i) (x (t) \ll rx _ _ t) k rt* if 347 Approximation by Polynomial Functions Applying these formulas 0 = f'{i) + [-f(t) + In this beautiful • each term of to + • - (x obtain + [=£@ {x-t)+^(x- 0] + • we (*), C* 0- + - 1 ^ <* - 0-] 02 ] + S'(t). formula practically eveything in sight cancels out, and we obtain = - S>( t ) I -ifi X ( - t )\ n\ Now we can apply the Mean Value Theorem to is some / in (a, x) ™ ~— s(a) at Remember = means (x - S on [a, x] : there o-. n! <2 that S(t) this - - S'(t) the function ^ such that = *„,,(*); in particular that = Rn x (x) = = R„,«(x). S(x) , J(a) 0, Thus o-i M W x — / . <n+1) W a or Rn,*(*) this the is To Cauchy form = f—-^-(x-tnx--a); of the remainder. derive the Lagrange form to the functions S and g(t) = (x we apply — t) n+ l : the / - n\ g(*)-g(a) g'(t) + *...(*) (x w . K S'(t) Thus - a) n+1 or is (n+l) S(a) S(x) which Cauchy Mean Value Theorem is some t in {a, x) such that there the Lagrange form. = / (n+1) (0 (n+1)! _. t)n 348 Infinite Sequences and Infinite Series Finally, if is - S(x) on integrable fx = S(a) fx f(*+i)(A = - S'(t) / then [a, x], Ja ^ I / Ja (x - t) n dt 7l\ or /(n+i)// f\ Cx /(n+i) fx ~ 7 = Rn,a(*) / : Ja - (x t) n dt. | n\ Although the Lagrange and Cauchy forms of the remainder are more than Problem 22-16), the integral form of the remainder will usually be quite adequate. If this form is applied to the functheoretical curiosities (see, e.g., tions sin, cos, and exp, with a = 0, Taylor's Theorem yields the following formulas: sin * = * - —+—3! 5! • • + • (-l) n - v 2 cos x = 1 - X X* + --•••+ 2! ** To 1 (-1)" k ; 4! ^.n X 2n (2« + l)! 1 fi cos (»+DM + — / (x - t) 2n dt, (2n) Jo /*x -+...+ ?L +/ +* + — .»(2n+2) Jo (2n)\ y.2 = + 1)! + {In ' .L (*_,)» evaluate any of these integrals explicitly would be supreme foolishness the answer of course will be exactly the difference of the other terms on the right side ! To and worthwhile. The first two integrals are sin estimate these integrals, left side however, and is especially easy. Since (2n+2) | W | < ! for aU t} we have f oTn? (2n + Jo ~ (x l)\ t)2n+1 dt * ?^4lTT f {2n + l)\\Jo (x I ~ Since ( x -ty n+i dt = -(x o 2rc we conclude 2n+2 t=sx \ 2n + + 2 2 <-o that S in (2n+2) (Y) Jo /; Similarly, t) (2« we can show fx < + l)! (2« + 2)! that coS (2n+1) (/) {X (2,)! -Vndt (2n + l)! <)2 " +1 all the both easy dt — Approximation by Polynomial Functions proved in Chapter 16) interesting, because (as These estimates are particularly any £ > 0 we can make 349 for xn 1 n\ < 8 by choosing n large enough (how large n must be will depend on x). This enables us to compute sin x to any degree of accuracy desired simply by evaluating the proper Taylor polynomial P n ,o(x). For example, suppose we wish to compute sin 2 with an error of less than 10~~ 4 Since . 22n+2 - sin 2 we can P use + /5 2n+i,o(2) 2 n+i,o(2) where R, \R\ —-- < - > as our answer, provided that 22n+2 A it number < + (2n . can be found by a straightforward search n with this property n obviously helps to have a table of values for n and 2 (see page 356) In this ! case it happens that — n sin 2 = is i>n,o(2) +R 3! 5! even easier to calculate sin To sin = P 2n +i,o(l) 1 obtain an error than less . 5 works, so that where It 1(T 4 2)\ e 7! \R\ 11! 9! < 10~ 4 . approximately, since 1 + R, where 1 < (2n we need only _1 (2rc \R\ + < find an n + 2)\ such that £? 2)! only a brief glance at a table of factorials. (Moreover, the P 2n +i,o(l) will be easier to handle.) For very small x the estimates will be even easier. For example, and this requires individual terms of 1 . sin - 10 P2n+M + (l^) \10/ ' + , *' ' lnt where|i?l 1 ' 1 . < 10 2" +2 (2/2 + 2)! < 10~ 10 we can clearly take n = 4 (and we could even get away These methods are actually used to compute tables of sin and cos. A high-speed computer can compute P 2 n+i,o(*) for many different x in almost no time at all. Estimating the remainder for e* is only slightly harder. For simplicity assume that x > 0 (the estimates for x < 0 are obtained in Problem 9). On To obtain with n = \R\ 3). . 350 Infinite Sequences and Infinite Series the interval maximum the [0, x] value of -{x-tYdt<^- Since we since is (»_,)- that * e (« < x pX Y n (n x n+l Once + (n !)! x , < R follows = 4, then (The inequality 0 if n 1) ! sufficiently large. will make x. If < 0 How things more < then x 1, n x x •••+=l+x + -+ + 2! nl In particular, + 1)! again, the estimates are easier for small 2 e +X 4xx n+l < which can be made as small as desired by choosing n large n must be will depend on x (and the factor 4 X difficult). increasing, so is we have 4, + exp A n\ Jo know already l rx pX fx ep t / Jo nl e where 0 /?,9 <R < + (n 1)! immediately from the integral form for R.) 0<R< h < h so ^ = ^1=1 + 1+1+1 + 1 + ^ 2! 65 = Ya 5 where 0 < R < ^~ 10 4! 3! +r -l + '-l + R, ' 24 which shows that < 2 e < 3. (This allows us to improve our estimate of R slightly: 3 x n+i * By taking « = 7 you can compute that the * = 2.718 . x first 3 decimals for e are . . (you should check that n = 7 does give this degree of accuracy, but it would be cruel to insist that you actually do the computations) The function arctan is also important but, as you may recall, an expression for arctan a) (*) is hopelessly complicated, so that the integral form of the remainder is useless. On the other hand, our derivation of the Taylor polynomial for arctan automatically provided a formula for the remainder: arctan x = x x* --+'•' + 3 , , —+— (~l)V> n+1 2n 1 p(-l)"+^+ , h / Jo TT~^ +t 1 2 2 dL 351 Approximation by Polynomial Functions As we have already estimated, dt + 1 + 2n ! t* j 3 For the moment we will consider only numbers x with \x\ < 1. In this case, the remainder term can clearly be made as small as desired by choosing n sufficiently large. In particular, arctan With 1 = 1- this estimate - 1 + -- 3 5 1 it is -- — -+ ±~\}J- + In + n ( where R, 2n for to compute t. Fortunately, there are some surmount these difficulties. Since smaller puting shows 7r is how 3 the remainder less w's will work for only to express arctan this will usually for arctan clever tricks should allow us which enable us somewhat smaller #'s. The comProblem 5 trick for in terms of arctan x for smaller x; 1 to < l**+x..(*)| much + have to be so To obtain a remainder 3)/2. This is really a shame, 4 example, we must take n > (10 = polynomial Taylor 1 the tt/4, so because arctan , make easy to find an n which will than any preassigned number; on the other hand, n large as to make computations hopelessly long. <10~ 4 1 < \R\ 1 can be done in a convenient way. The Taylor polynomial for the function f(x) = log (a: handled in the same manner as the Taylor polynomial + 1) at a = 1 is best Although for arctan. the integral form of the remainder for / is not hard to write down, it is difficult to estimate. On the other hand, we obtain a simple formula if we begin with the equation ' \+t = - + 1 t ~ t* • * + ' (-l)"- 1 *"- 1 + } ^ * 1 + t this implies that 1 ^ y2 1 ,+ ' ) = /.r^'*-'-i jr + xn 3 +( - 1) ""v i--" fx for all x > -1. If x > 0, Jo and there is n then fx / t t + y n ^~l fx fn -=—dt < 1 n t dt = — n Jo + \ when — 1 < x < 0 (Problem remainder term can be made as small as desired by a slightly more complicated estimate 10). For this function the choosing n sufficiently large, provided that — < 1 x < 1 352 Infinite Sequences and Infinite Series The behavior of the remainder terms for arc tan when quite another matter > |*| —+— - 2n = \og(x + 1) is n for arctan, 3 -^7 + < l*».o(*)| f(x) 2n+3 Ui < |#2n+i,o0)| and In this case, the estimates 1. (*>0)for/, 1 m /m become large as m becomes large. This predicament is unavoidable, and is not just a deficiency of our estimates. It is easy to get estimates in the other direction which show that the remainders actually do remain large. To obtain such an estimate for when are of no use, because arctan, note that + t < 2 +x 1 the bounds x 1 in [0, x] (or in if t is 1 > \x\ [x, < 2 < 0] if x 2x\ then 0), > if |*| 1, so 1 *2n+2 fx \ 1 Similarly, if x > Jo 0, 1 + > ~ dt t 2 then for t ± fX 2n \l 2x \J in [0, x] + < +x 1 +2 dt 4n + 6 we have < 1 t t 2 2x, if > x 1, so fx t / Jo t n + i dt > fx - / r dt = 2n 2x Jo 1 + 2 These estimates show that if \x\ > 1, then the remainder terms become large as n becomes large. In other words, for \x\ > 1, the Taylor polynomials for arctan and / are of no use whatsoever in computing arctan x and log(x + 1). This is no tragedy, because the values of these functions can be found for any x once they are known for all x with |#| < 1. This same situation occurs in a spectacular We have already seen that f that the Taylor polynomial (k) (0) Pn , 0 = for for the function 0 for every natural / number k. This means is Pn.oM =/(0)+/'(0)*+ 7f"(0) -^* + Z ! = way 2 in) • • + Jf-^x» 72 (0) I 0. In other words, the remainder term R n ,o(x) always equals f(x), and the Taylor polynomial is useless for computing /(*), except for x — 0. Eventually we will be able to offer some explanation for the behavior of this function, which is such a disconcerting illustration of the limitations of Taylor's Theorem. The word "compute 55 has been used so often in connection with our estimates for the remainder term, that the significance of Taylor's Theorem might < 353 Approximation by Polynomial Functions be misconstrued. It is true tional aid (despite that Taylor's its THEOREM 5 PROOF it has these will be developed will illustrate some ways The first illustration who have waded through the proof, Theorem may be impressive to those an almost ideal computa- consequences. Most of two proofs in succeeding chapters, but 7r is is ignominious failure in the previous example), but equally important theoretical Taylor's Theorem used. in which be particularly in Chapter 16, that will irrational. e is irrational. We know e = el that, for = 1 any n, +h+h+ + h + Rn " • ' Suppose that e were rational, say e = Choose n > b and also n > 3. Then a/b, where 0 *• < > where a and ottoi" b are positive integers. so —= n\ b Every term + + n\ + — 2! ' • • + n\Rn+— n\ than n \R n is an integer (the Consequently, n\R n must be an integer in this equation other integer because n > b). 0 < Rn < 0 <n\R n left side also. is an But 3 (n+l)\ so <-^-<\<\, 4 n which is impossible for an integer. The second illustration proved in Chapter 15: is + 1 | merely a straightforward demonstration of a fact If /"+/=0, f(0) /'(0) then / = 0. To this, observe f* = = / /(5) = (/")' prove (4) etc. (/ (/ 3 )' first = = = that = f k) exists for -/'> =("/')' (4))/ o, o, ^ =-/"=/, every k\ in fact 354 Infinite Sequences and Infinite Series This shows, not only that ones: /, Theorem f all k) exist, = -/, -/'. Since /(0) /', any states, for is — f / is (x - continuous (since f {n+2) (n+1) |/ 0. Now Taylor's < (0l (we can add the phrase "and M all *)" dt. nl M such that number a are 0, all /<*>(()) fx f(n+D(A = Jo there = that n, fix) Each function f {nJrl) but also that there are at most 4 different /'(0) for 0 < exists), so for < * and x, a// any particular x tz n" because there are only four different f k) ). Thus \ Since this is true for every by choosing n M\x\ n+l < Ml ['<*— \f(x)\ Jo n and rc, (»+ nl can be shows that |/(*) since x /n sufficiently large, this ! | 1)1 made < as small as desired s for any £ > 0; conse- quently, f(x) = 0. The other uses to which Taylor's Theorem will be put in succeeding chapcomputational considerations which have concerned us for much of this chapter. If the remainder term R n a (x ) can be made as small as desired by choosing n sufficiently large, then f(x) can be computed to any degree of accuracy desired by using the polynomials Pn ,a( x )- As we require greater and greater accuracy we must add on more and more terms. ters are closely related to the , If we are willing to add up infinitely many terms (in theory at least!), then we ought to be able to ignore the remainder completely. There should be 4 infinite sums" like sm x = xz , = X2 , 1 arctan x + *) 7! X4 X6 4! 6! , * ' ' , xs , = x —— = x Y2 YZ v4 2! 3! 4! ^5 ^7 5 7 h - X^ + X4 - ^ + z 3 4 infinite > • • , • • • if |*| • • • if We are almost completely prepared for this step. necessary definitions. • , H —we have never even defined an • • 8! =\+x + -+-+-+ 3 log(l 5! h 2! e* x7 h 3! COS x x5 x < -1< 1, X < 1. Only one obstacle remains sum. Chapters 21 and 22 contain the Approximation by Polynomial Functions 355 PROBLEMS 1. (of the indicated degree, Find the Taylor polynomials and at the indi- cated point) for the following functions. (i) /(*) = e (ii) /(*) = e* (iii) sin; degree (iv) cos; degree 2n, at t. (v) exp; degree (vi) log; degree n, at 2. (vii) /(*) = x6 (viii) /(*) = xh (ix) f(x) = ~^- /(*) = e degree '; x in at 0. 3, degree 3, at 0. 7T (x) 2. 2n, at -• at n, 1. + x + x; + x + x; 3 degree 4, at 0. z degree 4, at 2 7^t~; x ~t~ -V degree 2n ; degree n, + 1, 1. at 0. at 0. Write each of the following polynomials in x as a polynomial in (x — 3). (It is only necessary to compute the Taylor polynomial at 3, of the same degree as the original polynomial. Why?) 3. - (i) x2 - Ax (ii) x4 - I2x* (iii) x\ (iv) ax 2 + bx 9. + + 44a: 2 + 2x + 1. c. notation) which equals each of the following numbers to within the specified accuracy. To minimize needless comn putation, consult the tables for 2 and n on the next page. Write down a sum (using 2 \ (i) sin 1; error < 1(T 17 (ii) sin 2; error < 1(T 12 (iii) sin J; error < 10~ 20 (iv) e; (v) e error 2 ; error < < 10 -4 . 10" 5 . . . . — Z 71 • 1 2 1 2 4 2 3 8 6 4 16 24 120 720 5 32 6 64 7 128 5,040 8 40,320 9 256 512 362,880 10 1,024 3,628,800 11 2,048 39,916,800 12 4,096 479,001,600 13 8,192 6,227,020,800 14 16,384 87,178,291,200 15 32,768 1,307,674,368,000 16 65,536 20,922,789,888,000 17 131,072 355,687,428,096,000 18 262,144 6,402,373,705,728,000 19 524,288 121,645,100,408,832,000 20 1,048,576 2,432,902,008,176,640,000 This problem is similar to the previous one, except that the errors are so small that the tables cannot be used. You will have to thinking, and in some cases it may be necessary to consult the demanded do a little n Chapter 16, that x /n can be made small by choosing n large the proof actually provides a method for finding the appropriate n. In the previous problem it was possible to find rather short sums; in fact, it proof, in ! was possible to find the smallest n which makes the estimate of the remainder given by Taylor's Theorem less than the desired error. But in this problem finding any specific sum is a moral victory (provided you can demonstrate that the sum works). < 10- (1 ° 10) 10" 1>00 °. (i) sin 1; error (ii) e; error < < 10~ 20 10~ 30 (iii) sin 10; error (iv) e (v) arctan toS error (a) 10 ; error < . . . < 10~ (1 ° 10) . Prove, using Problem 15-8, that 7T - = 4 1 arctan - + 2 * - = 4A arctan 1 4 5 1 arctan -> 3 arctan — 1 239 Approximation by Polynomial Functions Show (b) that = ir 3.14159 . . man (Every young . . 357 should verify a few decimals of ir for himself, but the purpose of this exercise is not to set you off on an immense calculation. If the second expression in part (a) used, the is remarkably decimals for first 5 can be computed with ir work.) little For every number a, and every nonnegative integer "binomial coefficient" 6. \n) a If n > Show a. define the n\ not an integer, then is we n, never is (^j and 0, alternates in sign for that the Taylor polynomial of degree n for f(x) = (1 + x)« n at 0 P ,oW = is x k and that the Cauchy and Lagrange forms ^ rt k , (j^J =0 of the remainder are the following: Cauchy form: _ , N RnAx) = a(a — — 1) (a • • • " a(a — — " n 1) • i • n — n) *(* n - 0 m/A (l + '( 1 + ')-'(rr-;)" , (a • — n) ! + •>(,+ J** + -(" n . . tY 1 ! 1 (rd)"' ' in [0 - or [*• 0I - Lagrange form: _ a(a Ai.oW = . . '= (n — + 1) * • . 1N (a — + ^ (w+1)! J , * n+1 d n) „,., A * n+1 0 a_n_1 ' ' + , N 0° „. n 1 in [°' *1 or & Estimates for these remainder terms are rather difficult to handle, and are postponed to Problem 22-16. 7. Suppose that a of g (i) / and (a)/i\. ai and bi are the coefficients in the Taylor polynomials at g, respectively. Find the In other words, coefficients a + g. (0 f fg. (iii) /'• (iv) h{x) = (v) k(x) = j'f{t)dt. J* f(t) dt. = f {i) (a)/i\ and bi = of the Taylor polynomials at a of the following functions, in terms of the afs (ii) ai and b/s. 358 Infinite Sequences and Infinite Series *8. (a) An + 1 at 0 P is the Taylor polynomial of degree 2n + 1 = P(x) R(x)/x 2n+1 = 0. this (b) Find f (c) In general, gat + imply about lim ik) (0) for all if f(x) where R(x), ^(Ar 2 )A 4n+1 Km if x < What does ? A:. = g(x n ), find f k) terms of the derivatives of (0) in then 0, x - (x - t) n dt 7o « Prove that if -1 < x f.o I (n < + 0, dt (\ t show that (a) to l)\ +x)(n+ Show that if \g (x)\ < M\x - a\ n M\x - a\ n+1 /(n + 1) for \x - a\ < Use part + then f (b) for sin at 0, 0. Prove that (a) of degree v 4n y 10 then sin x Hint: If *11. sin(# 2 ) is y6 10. = Prove that the Taylor polynomial of f(x) for \x - a\ < 8, then n = 0, then \g(x)\ < 8. r if 1) lim g (x)/(x — a) x—*a lim g(x)/(x (c) (d) - a) n+l - 0. n aJ (x),theng'(x) = f'(x) - Pn -i,aAx). Give an inductive proof of Theorem 1, without using PHopitaPs Show that iig{x) =f(x) -P , Rule. 12. Deduce Theorem 13. 14. Theorem, with any form be necessary to assume one as a corollary of Taylor's 1 of the remainder. (The catch is that it will more derivative than in the hypotheses for Theorem 1.) Deduce the Cauchy and Lagrange forms of the remainder from the integral form, using Problems 13-27 and 13-28. There will be the same catch as in Problem 12. (a) Prove that if f f"(a) f (a) exists, - Um-f(a then + h)+ ^ ~ h) - W Hint: Use the Taylor polynomial of degree 2 with x with x — a — h. = a + h and Approximation by Polynomial Functions (b) exists, 15. = Let /(*) > x 2 for x and -x 2 0, for x < Show 0. 359 that even though /"(0) does not. Use the Taylor polynomial Pi, a ,/, together with the remainder, to prove a weak form of Theorem 2 of the Appendix to Chapter 11: If /" > 0, then the graph of / always above the tangent lies line of /, except at the point of contact. *16. ! *17. Problem 17-32 presented a rather complicated proof that / — 0 if /" - / = 0 and /(0) = /'(0) = 0. Give another proof, using Taylor's Theorem. (This problem is really a preliminary skirmish before doing battle with the general case in Problem 1 7, and is meant to convince you that Taylor's Theorem is a good tool for tackling such problems, even though tricks work out more neatly for special cases.) Consider a function / which satisfies the differential equation B /<»> for certain numbers ao, . . f . aj f«\ a n -\. Several special cases , have already received detailed treatment, either in the text or in other problems; in we have found all functions satisfying /' = /, or /" + / = 0, f"—f= 0. The trick in Problem 17-31 enables us to find many solu- particular, or whether these are the only which will be supplied by this some (necessarily sketchy) remarks tions for such equations, but doesn't say solutions. This requires a uniqueness result, problem. At the end you will about the general solution. (a) find Derive the following formula for / (n+1) (let us agree that will beO): n-1 (b) Deduce a formula The formula for in part (b) y is (n + 2) . not going to be used; it was inserted only to convince you that a general formula for f n+k) is out of the question. On the other hand, as part (c) shows, it is not very hard to obtain estimates on the (c) size of (n+A:) (*). Let N this means that = max(l, |<2 0 |, . . . , \a n -i\). Then |<zy_i n-1 = £ bffi$\ where |V| < 2# 2 . + tf n _i<z;| < 2N 2 ; / 360 Infinite Sequences and Infinite Series Show that n-l fn+2) = £ ^yCfl where < |£ y 2| 4 ^ 3j j=0 more and, generally, = ^ fn+k) j (d) bj Conclude from part a number k f Now suppose M < that /(0) - < 2k Nk+1 9 2W* +1 • = /'(0) M < " O+ is /<»> ( /i »(0) =/ (i) 2 T = =f • = 0. Show that 1)! n + k+1 ' 1)! solutions of the differential equation «y/«>, <; < 0 (0) for • + A: + k+ (n for all £. • \2Nx\ • and conclude that / = 0. Show that if /1 and f2 are both and \b*\ any particular number x there (c) that, for I (f) where \ M such that !/<«+*>(*) (e) j =0 - n 1, then x -/ 2. In other words, the solutions of this differential equation are deter- mined by the "initial conditions" (the values f j) (0) for 0 <j < n — 1). This means that we can find all solutions once we can find enough solutions to obtain any given set of initial conditions. If the equation ah has n distinct roots /(*) is . . . a ny then any , = de™ + • function of the form + c n ea x « • a solution, and /'(0) = a+ = OtxCi + /n-i(Q) = /(0) • • ' • + +c + n> • ' - . a^n, + a/-Vn . As a matter of fact, every solution is of this form, because we can obtain any set of numbers on the left side by choosing the c's 361 Approximation by Polynomial Functions properly, but we 18. (a) will not try to prove this last assertion. (It is a which you can easily check for n = 2 or 3.) These remarks are also true if some of the roots are multiple roots, and even in the more general situation considered in Chapter 26. Let/0) = * 4 sin 1/x 2 for x * 0, and/(0) = 0. Show that / = 0 up to order 2 at 0, even though /"(0) does not exist. purely algebraic This example fact, is more complex, but slightly also slightly impressive, than the example in the text, because both f(a) f\a) exist for a ^ 0. Thus, for each number a there is another more and num- ber m(a) such that (*) /(*) = /(«) + mix - = f"(a) for a defined in this way namely, m(a) function (b) m a) - a y + Ra "^>.(x and m(0) 0, is + = 0. ( x) Notice that the not continuous. Let / be a differentiable function. Suppose that there is a function m such that (*) holds for all numbers a, and that m is continuous. Prove that /"(a) exists for all a and equals m(a). Hint: Write (*) for h. x = a h, a = a, and also for x = a, a — a + + *(c) Suppose that there f(x) is a continuous function =f(a)+f'ia)(x-a) + f'ia) 2 ix m - ay + such that mja) (x 3! - ay + Ra(x), where lim Ra(x) (x Does it follow that f"'(a) = m(a)? - ay = 0. * CHAPTER AlV TRANSCENDENTAL IS e The irrationality of e was so easy to prove that in this optional chapter we will attempt a more difficult feat, and prove that the number e is not merely irrational, but actually much worse. Just how a number might be even worse than irrational is suggested by a slight rewording of definitions. A number x irrational is b 7* 0. This if it is is and — solution of = 0, b + aQ - 2 — in this light, the irra- is it not the 0, the solution of the equation = 0, how of one higher degree. Problem 2-17 shows numbers x which satisfy a nx where the is Viewed 0. an equation x2 that with 0 V2 just barely manages to be irrational —although V2 aix it is — a b, any equation does not seem to be such a terrible deficiency; rather, tionality of appears that = except for a b, a/b for any integers a and as saying that x does not satisfy bx for integers a — not possible to write x same the all ai en = is + a n -ix and a0 n~ l ^ + • • produce ao = many irrational • + 0, 0 (this condition rules out the possibility A number which satisfies an called an algebraic encountered n are integers 0). to higher-degree equations number, and 4 4 algebraic" equation of this sort practically every number we have ever (ir and e defined in terms of solutions of algebraic equations are the great exceptions in our limited mathematical experience). All roots, such as V2, ^3, <Tl, are clearly algebraic numbers, and even complicated combinations, like ^3 + V5 + </\ + V2 + </l are algebraic (although we will not try to prove this) Numbers which cannot be obtained by the process of solving algebraic equations are called transcendental; the main result of this chapter states that e is a number of this . anomalous sort. The proof that retically possible this proof, e is is well within our grasp, 19. Nevertheless, and was theo- with the inclusion of we can justifiably classify ourselves as something more than novices mathematics; while many irrationality proofs depend in the study of higher 362 transcendental even before Chapter ; 363 e Is Transcendental only on elementary properties of numbers, the proof that a number is transcendental usually involves some really high-powered mathematics. Even the dates connected with the transcendence of proof that we that e is are impressively recent e transcendental, due to Hermite, dates from 1873. will give —the first The proof a simplification, due to Hilbert. is Before tackling the proof a good idea to itself, it is which depends on an idea used even map in the proof that e out the strategy, is Two irrational. features of the expression e were important = for the proof that e 1 + h +R " --- +h+h+ 1 irrational: is +^+ • • On number the one hand, the +-. • can be written as a fraction p/q with q < n (so that n (/>/ q) is an integer) on the other hand, 0 < R n < 3/(w + 1) (so n\R n is not an integer). These two facts show that e can be approximated particularly well by rational numbers. Of course, every number x can be approximated arbitrarily closely by rational numbers if £ > 0 there is a rational number r with \x — r\ < e; the catch, however, is that it may be necessary to allow a very large denomi! ! ! — nator for r, as large as 1 /s perhaps. For e we are assured that this not the is case: there is a fraction p/q within 3/(n + 1) of e, whose denominator q is at most n\. If you look carefully at the proof that e is irrational, you will see that only this fact about e is ever used. The number e is by no means unique in this respect: generally speaking, the better a number can be approximated by ! rational numbers, the worse Problem in 3). (some evidence for it is The proof that extension of this idea: not only e is but any e, this assertion is presented transcendental depends on a natural finite number of powers e, e 2 , . . . , e can be simultaneously approximated especially well by rational numbers. In our proof we will begin by assuming that e is algebraic, so that n , ane (*) for some integers a 0 find certain €i, . . . , €n , . integers . n + . , M, ' ' + aie + ' a0 = 0, a0 ^ 0 a n In order to reach a contradiction . Mh . . . M , such that e L 9 = + Mi e'i ' M M + M 2 M + M n we will then and certain "small" numbers n e2 €n 364 Infinite Sequences and Infinite Series how Just e's must be will appear when these expressions are subassumed equation (*). After multiplying through by we small the M stituted into the obtain [a 0 The M + aiMi + term first + in brackets an is an M + n] Oitfi integer, and we We be a nonzero integer. will necessarily + + 0. choose the M's so that will manage will also = en a n] it to find e's so small that + |ei<2i * * + * this will lead to the desired contradiction number As a en an —the sum of a nonzero integer and a ^ cannot be zero very reasonable and basic strategy this all is you will function! (This function was introduced in PROOF e is quite straightforward. that the Af's and need to know about the Problem 18-25.) e's are gamma transcendental. Suppose there were integers ane (*) Define numbers M, Mh n a0 + ... xp k . . , . a n ^e M , x*- l [{x M ~ yP-l l an , n' and n unspecified later. number p ^ with a 0 , + x ei, . [(x 0, such that +a = • 0 . . , -!)•••(*(p - 1)! en 0. as follows. n)] p e-* dx, -!)•••(* (P - 1)! »)]' dx, [(*-!)•(*(p The way part of the proof will be the defined. In order to read the proof l ^; of absolute value less than The remarkable THEOREM < \ n)] p e~* dx. - 1)! represents a prime number* which we will choose Despite the forbidding aspect of these three expressions, with a work they will appear much more reasonable. We - n)l concentrate on M little first. If the expression in brackets, is actually multiplied out, [(* - we obtain a polynomial xn * The term "prime number" was numbers will be 1) + • • • • + nl defined in Problem 2-16. used in the proof, although number which does not (* • • divide the integer it is a, An important fact about prime this book: If p is a prime not proved in and which does not divide the integer b, then p also does not divide ab. The Suggested Reading mentions references for this theorem (which is crucial in proving that the factorization of an integer into primes is unique). We will also use the result of is Problem 2-1 6(d), that there are infinitely asked to determine at precisely which points this information is many primes required. —the reader " e Is with integer When coefficients. power raised to the pth Transcendental this 365 becomes an even more complicated polynomial ***+••• ± Thus M can be (n\) p . form written in the np m = Li y Ca - (p a=0 where the — c„ 1 - are certain integers, f Jo l)! and Co = ±{n\) v But . *^ _x / — k\. Jo Thus (,-!+«)! „.£ Now, for a = we 0 c<> . obtain the term We will now consider only primes p > n; then this term is an integer which is not divisible by p. On the other hand, if a > 0, then C* (P 1 = C°(P 7— which by p. « Now divisible + M+ ~ a 1J! (/> M by p. Therefore We consider M^. = , « ^ have -!)••(,- r- (p-i)\ f x»- ~A l [{x 1) (x • • 2) • an integer which itself is )k - - n)]^- (I • is A not divisible dx - A> ^ (p-i)>. This can be transformed into an expression looking very much like M by the substitution u du The changed limits of integration are "+ /•-(« * io ky-*[{u = = x — k, dx. to 0 and °o , and + k-\) •••«••(« + *- n)]V ^ " ~~ (>- D! There is one very significant difference between this expression and that for M. The term in brackets contains the factor u in the kth place. Thus the pth power contains the factor u p This means that the entire expression . (u + k) p ^[(u + *-!)••(« + *- n)] p 1 366 Infinite Sequences and Infinite Series is a polynomial with integer coefficients, has degree at every term of which least/?. Thus np np U(p-\)\ Jo a=l where the a — 1 Da Now it is term in the sum divisible by p. is M divisible is clear that k e M _ _ — + k M ek — k , M + a\M\ + * ' + ' an M + n] I, . . n. , . M we obtain Substituting into (*) and multiplying by [a 0 summation begins with by p. Thus each k is an are certain integers. Notice that the which integer (p-i)\ a= in this case every ; (p-\+a)\ L [aid + * * * + an en ] = 0. In addition to requiring that p > n let us also stipulate that p > \ao\. This and a 0 are not divisible by p, so a Q is also not divisible M M means that both by p. Since each M divisible k is clqM is by not divisible p. + by + a\M\ In particular follows that it p, • + * • an M n a nonzero integer. it is In order to obtain a contradiction to the assumed equation prove that e is transcendental, it is |ai€i + (*), and thereby only necessary to show that • • + an • €n \ can be made as small as desired, by choosing p large enough; it is clearly suffishow that each |€*| can be made as small as desired. This requires nothing more than some simple estimates; for the remainder of the argument remember that n is a certain fixed number (the degree of the assumed polycient to nomial equation (*)). To n \e k \ < e k begin with, \x p~ l [{x be the n, then n)] p \e~ x (/,-!)! np - 1[ \[(x - 1) • maximum of \(x — 1) • • • • (x - < A ~. (P e ~ r- / (x ~~ n)\ for • e~x dx D! n p -i n Ap (P ~ D! cVA p (p-\)\ x 1)! - D! (p --i p n e np n)] p \e~ dx. - (P A < dx f n let k -!)••(*- Q Now < if 1 e n {nAY {p-\)\ x in [0, «]. Then e Is — But n and A are fixed; thus (nA) p /(p by making p sufficiently large. | Transcendental 367 can be made as small as desired 1)1 w is irrational, deserves some philosophic argument seems quite "advanced"—after all, we use integrals, and integrals from 0 to co at that. Actually, as many mathematicians have observed, integrals can be eliminated from the argument completely; the only integrals essential to the proof are of the form This proof, like the At afterthoughts. proof that the first sight, xke~ x dx, fj and these for integral k, Thus M, where for Ca can be replaced by k whenever they occur. integrals ! example, could have been defined are the coefficients of the polynomial [(* If this idea proof that is - 1) • • (* * n)Y. transcendental, depending only on the fact that = +i +i +i + 3! 1! 2! 1 Unfortunately, this "elementary" proof one - developed consistently, one obtains a "completely elementary" e is , original initially as .... harder to understand than the just to eliminate is —the whole structure of the proof must be hidden is by no means peculiar to this specific theorem "elementary" arguments are frequently more difficult than "advanced" ones. Our proof that ir is irrational is a case in point. You probably remember nothing about this proof except that it involves quite a few com- a few integral signs! This situation — more advanced, but much more conwhich is of great historical, as well as intrinsic, interest. One of the classical problems of Greek mathematics was to construct, with compass and straightedge alone, a square whose area is that of a circle of radius 1 This requires the construction of a plicated functions. There is actually a ceptual proof, which shows that 7r is transcendental, a fact . line segment whose length of length 7r is is constructible. VV, which can be accomplished The Greeks were totally if a line segment unable to decide whether such a line segment could be constructed, and even the full resources of modern mathematics were unable to settle this question until 1882. In that year Lindemann proved that w is transcendental; since the length of any seg- ment that can be constructed with straightedge in terms of line +, •, — —, and , segment of length The proof that ir is 7r V , and is and compass can be written therefore algebraic, this proves that a cannot be constructed. transcendental requires a sizable amount of mathematics 368 Infinite Sequences and Infinite Series which not too advanced to be reached in this book. Nevertheless, the proof is much more proof for difficult than the proof that practically the 7r is certainly surprise you. same as the proof for The proof that is transcendental. In fact, the e is e. This last statement should transcendental seems to depend so e is thoroughly on particular properties of e that it is almost inconceivable how any modifications could ever be used for 7r; after all, what does e have to do with 7r? Just wait and see! PROBLEMS 1. (a) Prove that if (b) Prove that if a > 0 is algebraic, then is algebraic. a is algebraic and r is rational, then a + r and ar are algebraic. Part (b) can actually be strengthened considerably: the sum, product, and quotient of algebraic numbers some cult for us to prove here, but 2. Prove that degree (a) a be an diffi- by actually findneed equations of are algebraic, (You satisfy. will number which algebraic is not rational. Suppose that a = anx + a n ^x n ~ + n l • • • +a = 0 0, and that no polynomial function of lower degree has this property. Show that fip/q) 5* 0 for any rational number p/q. Hint: Use Problem 3-7 (b). Now show that \f(p/q)\ > \/q n for all rational numbers p/q with q > 0. Hint: Write f(p/q) as a fraction over the common denominator q\ M = sup {|/'(*) Let to prove that if p/q a \ \a *4. too the polynomial equation /(*) (c) is can be examined: 4.) Let satisfies (b) algebraic. This fact special cases V2 + V3 and V2(l + V3) ing algebraic equations which they *3. is I is - p/q\ > \/Mq n — p/q\ > c/q n for . - \x : a\ < 1}. Use the Mean Value Theorem a rational that for (It follows all rational \a — p/q\ < 1, then max(l, \/M) we have number with c = p/q.) Let a = 0.110010000000000000000001000 where the Ts occur that a is in the n \ place, for each transcendental. (For each equation of degree n, n. Use Problem show that a is 3 to prove not the root of an n.) Although Problem 4 mentions only one specific transcendental number, should be clear that one can easily construct infinitely many other numbers c/q n for a any c and n. Such numbers were considered by Liouville (1809-1882), and the inequality in Problem 3 is which do not first — p/q\ > it satisfy \a ) e Is often called Liouville's inequality. structed in this way happens None Transcendental numbers con- of the transcendental to be particularly interesting, but 369 for a long time numbers were the only ones known. This situation was changed quite radically by the work of Cantor (1845-1918), who showed, without exhibiting a single transcendental number, that most numbers are transcendental. The next two problems provide an introduction to the ideas that allow us to make sense of such statements. The basic definition with which we must work is the following: A set A is called countable if its elements can Liouville's transcendental be arranged in a sequence 01, 02, 03, 04, The set is obvious example (in N, the more fact, set of natural • • • or less the Platonic ideal of) a countable numbers; clearly the set of even natural numbers is also countable: 2, 4, 6, 8, set of all integers (positive, more surprising to learn that Z, the negative, and 0) is also countable, but seeing is It is a little 0, 1, -1, 2, -2, 3, believing: -3, ... . next two problems, which outline the basic features of countable sets, are really a series of examples to show that (1) a lot more sets are countable than one might think and (2) nevertheless, some sets are not countable. The *5. (a) Show or x (b) is Show if A and B are countable, then so is A VJ B = {x: x is in B}. Hint: Use the same trick that worked for Z. that the set of positive rational numbers is countable. (This that A in is really quite startling; use the following look-and-see proof: \ \ \ \ / /f / \ t t (c) Show Ah A2 , . same . . is also countable. Prove that the triple (f) \ i i I • • ' — (/, is countable. (This is as part (b).) are each countable, prove that A (e) \ that the set of all pairs (m, n) of integers practically the (d) If \ X \J A 2 \J A3 KJ ' ' ' (Again use the same trick as in part set of all triples (/, m, n) of integers m, n) can be described by a pair Prove that the set of all rc-tuples (a h you have done part (e), you can do a 2} (/, (b).) countable. (A m) and a number ...,*„) this, is n.) are countable. (If using induction.) 370 Infinite Sequences and Infinite Series Prove that the polynomial functions of degree n is shows that the set of all polynomial functions of degree n can be arranged in a sequence, and each of these functions has at most n roots.) (g) set of all roots of countable. (Part Now use (h) bers *6. Since so is parts (d) is and (g) to prove that the set of all algebraic num- countable. many sets turn out to be countable, numbers between 0 and set of all real there (f) no way of (decimal notation suppose such a listing all these real 1 is it is important to note that the In other words, not countable. numbers in a sequence being used on the right). To prove that were possible and consider the decimal is list this is so, 2 3 4 0.5i 5 2 5 3 5 4 1 where n = n = = n Show and an list, thus obtaining a contradiction. an 5 if a n Problems 5 and 6 can be bers n 5 cannot possibly be in the is summed up 6 if an as follows. 5. The that this number set of algebraic num- countable. If the set of transcendental numbers were also countable, numbers would be countable, by Problem 5(a), and numbers between 0 and 1 would be countable. But this is false. Thus, the set of algebraic numbers is countable and the set of transcendental numbers is not ("there are more transcendental numbers than algebraic numbers")- The remaining two problems illustrate further how important it can be to distinguish between sets which are countable and sets which are not. then the set of all real consequently the *7. set of real Let / be a nondecreasing function on lim f(x) and lim f(x) both exist. x—*a + (a) Recall (Problem 8-8) that x-*a~ For any in (b) [0, 1]. [0, 1] e > 0 prove that there are only finitely many numbers a with lim /(*) — lim /(#) > e. Hint: There are, in fact, at most [/(l) - /(0)]/e of them. Prove that the set of points at which / Hint: If lim f(x) x~*a + number — lim /(*) > 0, then is discontinuous it is > is countable. 1/n for some natural x—*a~ n. This problem shows that a nondecreasing function is automatically continuous at most points. For differentiability the situation is more difficult to analyze and also more interesting. A nondecreasing func- 371 e Is Transcendental tion can able, but at to be differentiable at a set of points fail most points [33] (in Problem is not count- a different sense of the word "most"). Reference of the Suggested Lemma Rising Sun which true that nondecreasing functions are differentiable it is still 9 of the Reading gives a beautiful proof, using the of Problem 8-20. For those Appendix Chapter to 11, it is who have done possible to provide at least one application to differentiability of the ideas already developed in (a) this problem set: If except at those points where its tinuous; but the function /+' is / only a countable bers a x / on and (a x , b x ) bx differentiable is disconis / is . set a constant function. Suppose now dropped. Prove that / takes on of values. Hint: For each x choose rational num- such that a x < x is < Then every value f(x) interval (a x , b x ). (d) is increasing, so a convex function that the hypothesis of continuity (c) convex, then / automatically differentiable except at a countable set of points. Problem 11-50 showed that if every point is a local maximum point for a continuous function /, then (b) is right-hand derivative /+' How many such bx is is a maximum point for maximum value of / on some and x the intervals are there? Now deduce Problem 11-50 as a corollary of part (a). Suppose that every point of / is either a local maximum or a local minimum point for /. Prove that / takes on only a countable set of values. (This is a minor variant of part (a).) Now suppose that / is continuous, and that every point is either a local maximum or a local minimum point for /. stant function. (Although this statement Problem 11-50, the only proof I know is Show that / is a con- only a minor variant of uses part (c), very different ideas than the ones in Chapter 11.) and thus relies on CHAPTER *m I The INFINITE SEQUENCES idea of an infinite sequence is so natural a concept that dispense with a definition altogether. One it is tempting to frequently writes simply "an infinite sequence #2} #3, <Zl) (35, . . . the three dots indicating that the numbers <z» continue to the right "forever." A rigorous definition of an infinite sequence is not hard to formulate, however; the important point about an ber,, ft, there is functions are DEFINITION <Z4 3 An infinite From by a a real meant number infinite sequence is num- that for each natural a n This sort of correspondence . is precisely what to formalize. sequence of numbers real a function whose domain is is N. the point of view of this definition, a sequence should be designated single letter like a, and particular values by a(l), a(2), fl (3), but the subscript notation tfij is #2, a Si . . . itself is usually denoted by a {(-l) n }, and {\/n} denote the sequences a, ft almost always used instead, and the sequence symbol like {a n }. Thus {n}, and 7 defined by fin = ("l) n Tn = - , 1 n A sequence, like any function, can be graphed (Figure 1) but the graph is usually rather unrevealing, since most of the function cannot be fit on the page. 'Hill (c) (a) FIGURE 372 1 (b) 5 373 Infinite Sequences 05 = = 03 01 H = 04 06 : 76 7s 0 FIGURE = 02 2 A more convenient representation of a sequence is obtained by simply as, ... on a line (Figure 2). This sort of picture ai, 55 shows where the sequence 'is going. The sequence {a n "goes out to 55 infinity, the sequence {fi n "jumps back and forth between —1 and 1," and labeling the points 4 } \ the sequence {y n j "converges to 0. Of the three phrases in quotation marks, is the crucial concept associated with sequences, and will be defined 5 the last precisely (the definition a N+ t / FIGURE A DEFINITION illustrated in Figure 3). is /+ a.y 6 3 sequence { an } converges to I (in is a natural number n if N such > /) if that, for all natural N, then \a n — l\ for every g < numbers said to converge if it converges to I for / or has the limit some /, 0 n, s. In addition to the terminology introduced in this definition, say that the sequence {a n } approaches > »o n-~* there = symbols lim a n and to /. we sometimes A sequence diverge if it {a n } is does not converge. To show that the sequence following. If e n > N we > 0, there is = - N converges to 0, it suffices to N such that 1/N < observe the e. Then, if - 0 - Vn - 0 so e, Yn < e. limit lim will } 11 < - < n is yn have Tn The { a natural number Vn + probably seem reasonable after a practically the same as 1 little for large n) , reflection (it just says that + 1 but a mathematical proof might not « » 374 Infinite Sequences and Infinite Series be so obvious. To Vn + f estimate — (Vn /- Vrc= — Vn we 1 + can use an algebraic -V~n ){Vn +7= + +Vn \ 1 trick: Vn) . Vn + 1 + —n Vn + + Vn n 1 1 1 + + Vrc Vn + — Vra by Vrc It is also possible to Theorem estimate 1 applying the 1 Vx on the interval = to the function /(*) Vn + 1 [n, n + Mean Value 1]. We obtain + 1) - Vn 1 for > 2Vx some x in (n, n 1 2 may Either of these estimates proof is left The V^' be used to prove the above but valuable exercise. limit; the detailed to you, as a simple limit 3rc ,™ 4/2 + = + 63 3 + 7n 2 3 - 8n 3 1 4 z should also seem reasonable, because the terms involving n are the most important when n is large. If you remember the proof of Theorem 7-9 you will be able to guess the trick that translates this idea into a proof dividing top — and bottom by n z yields + + 63 3 + In 2 An z - 8n 3rc nz n 1 A 4 63 8 "~ + n T , n3 2 Using this expression, the proof of the above limit one uses the following facts: If lim a n and lim b n both exist, then not difficult, especially if n—* » n— » » lim n— (a n + bn) = lim if lim 71— 6n lim a n + (a n ' b n) = lim a n n—» w • lim » bn ^ lim a n /b n = lim a w /lim 0, 0 for b ni n— then ^ lim n—> n— « cc n—> ao moreover, is all 72 greater than 00 £ ft . some iV, and 375 Infinite Sequences (If we wanted to be utterly precise, the third statement would have to be even more complicated. As it stands, we are considering the limit of the sequence = {a n /b n }, where the numbers c n might not even be defined for certain \c n n < N. This doesn't really matter we could define c n any way we liked for such n because the limit of a sequence is not changed if we change the sequence at a finite number of points.) Although these facts are very useful, we will not bother stating them as a ] — — theorem —you should have no difficulty proving these results for yourself, = because the definition of lim a n / so similar to previous definitions of is n—+ oo X— The = lim f(x) limits, especially /. 00 = between the definitions of lim a n similarity n —* actually closer than mere analogy; it is I and lim x— oo possible to define the f(x) = I is oo first in terms of the second. If/ is the function whose graph (Figure 4) consists of line segments FIGURE 4 joining the points in the graph of the sequence {a n }, so that = f(x) - (fln+i — a n )(x + n) an x < lim f{x) — n , < n + 1, then lim a n This observation 0 < a < 1. is — and only if I prove this frequently very useful. For example, suppose that we — lim a x x. < 0, 0. lim e x— oo = large = note that X—» since log a so that x log a Notice that we is xlo * a oo o, negative and large in absolute value for actually have lim a n - 0 if \a\ < 1; n—* oo for if a < 0 1. Then lim a n To if we can write lim a n = lim (~\) n \a\ n = 0. 376 Infinite Sequences and Infinite Series The behavior becomes of the logarithm function also shows that becomes arbitrarily large as n = lim a n This assertion large. oo, > a if > a 1, then a n often written is 1, n—> oo and sometimes even said that it is {a 71 approaches } <x> We also write equations . like —a n = — lim n— and say that lim a n—* n — an { does not co 5 00 approaches — oo. Notice, however, that even in this extended sense. \ if < — 1, a then exist, oo more important to on a line (Figure 3). There is another connection between limits of functions and limits of sequences which is related to this picture. This connection is somewhat less obvious, but considerably more interesting, than the one previously mentioned Despite this connection with a familiar concept, it is picture convergence in terms of the picture of a sequence as points —instead of defining limits of sequences in terms of limits of functions, it is possible to reverse the procedure. THEOREM 1 Let / be a function defined in an open interval containing at c itself, with - lim /(*) c, except perhaps /. x—*c Suppose that {a n a sequence such that is } each a n each a n (1) (2) lim a n (3) n—* Then the sequence { f(a n ) } is in the ^ c, = domain of /, c. oo satisfies = lim f(a n ) I. n—»oo Conversely, tions, PROOF if this is Suppose first true for every sequence = then lim f(x) { an satisfying the } above condi- /. that lim f(x) = /. Then for every e > 0 there is a 5 > 0 such x—*c that, for all #, if If the 0 sequence {a n } number N such < |* satisfies c\ < d, lim a n then = c, \f(x) - l\ choice of 8, this n > N, then \a n — < 6, means that !/(*») - 1\ < then (Figure that, if By our - c\ < 8. s. 3) there is a natural 377 Infinite Sequences showing that lim f(a n ) n— = 1. *> =/ Suppose, conversely, that lim f(a n ) for every sequence {a n with } n—* *> = lim a n n— c. = If lim f(x) were / not true, there would be some £ > 0 such x—>c <» > that for *z#ry 5 > 0 there < 0 an # with is - |* c\ < but 8 \f(x) - /| > s. In particular, for each n there would be a number x n such that < 0 <r| < - but \f(x n ) - l\ > e. the sequence {# n } clearly converges to c but, since |/(# n ) — l\ > s for the sequence |/(^ n )} does not converge to /. This contradicts the Now all - \x n n, = hypothesis, so lim f(x) Theorem 1 provides the sequences {a n } must be / true. | many examples and {b n } -.in — bn of convergent sequences. For example, defined by (.3+1) cos clearly converge to sin(13) and cos(sin(l)), respectively. It is important, however, to have some criteria guaranteeing convergence of sequences which are not obviously of this sort. There is one important criterion which is very easy to prove, but which the basis for is all other results. This criterion is stated | ai FIGURE ai 03 a* a which therefore apply also to sequences: a sequence {a n is increasing if a n+ i > a n for all rc, nondecreasing if tfn+ > o-n for all n, and bounded above if there is a number such that for all n; there are similar definitions for sequences which are dean < creasing, nonincr easing, and bounded below. in terms of concepts defined for functions, 5 \ M i M THEOREM 2 If {a n is } statement PROOF The set A nondecreasing and bounded above, then is true if {a n is } numbers a n consisting of all has a least upper bound a. if £ > bound 0, there of A. is some Then if n an We > This proves that lim a n is, } converges (a similar by assumption, bounded above, claim that lim a n cin satisfying > [a n nonincreasing and bounded below). = a a — aN — < a — aN < < £, since (Figure a N we have aN, so = | ot. a an £. is 5). In so A fact, the least upper 378 Infinite Sequences and Infinite Series The hypothesis that {a n } 2: if {a n {a n \ is clearly diverges. } should be sequence j is bounded above clearly essential in is not bounded above, then (whether or not {a n } consideration, first Theorem nondecreasing) might appear that there it trouble deciding whether or not a given nondecreasing little an Upon is } bounded above, and consequently whether or not is verges. In the next chapter such sequences j an } con- very naturally and, as will arise we shall see, deciding whether or not they converge is hardly a trivial matter. For the present, you might try to decide whether or not the following (obviously increasing) sequence is bounded above: +* + *, + l, i +*+ i Although Theorem 2 treats only a very special class of sequences, it is more might appear at first, because it is alwayc possible to extract from an arbitrary sequence \a n another sequence which is either nonincreasing or else nondecreasing. To be precise, let us define a subsequence of the sequence {^n} to be a sequence of the form useful than \ \graph ofa R %d where the 2 and nj are 1234 1 1 1 6 are peak points 1 1 56 789 1 1 1 1 10 a na • • j < Hi < 72 3 < * * * h 11 Then every sequence contains a subsequence which or nonincreasing. It is although the proof worth recording as a lemma. 6 LEMMA Any sequence { an } is either nondecreasing become quite befuddled trying to prove very short if you think of the right idea; this possible to assertion, FIGURE 2i natural numbers with ni H an is contains a subsequence which is it is either nondecreasing or nonincreasing. PROOF number Call a natural all m > Case n3 < ' n (Figure • a "peak point" of the sequence {a n \ < if a m a n for 6). The sequence has 1. ' n infinitely many peak are the peak points, then a ni > points. a n2 > In this case, a n} > • * , if Wi so {a Uk < } ft 2 is < the desired (nonincreasing) subsequence. Case 2. The sequence has greater than all such that a n2 greater than > only finitely is peak points) way we many peak m is COROLLARY (THE In this case, not a peak point, there let n\ be is obtain the desired (nondecreasing) subsequence. | If we assume that our original sequence an extra corollary along the way. BOLZANO- WEIERSTRASS THEOREM) points. some n 2 > «i not a peak point (it is greater than m, and hence there is some n 3 > n 2 such that a nt > a nr Con- points. Since a nr Since n 2 all tinuing in this peak {a n \ is bounded, we can pick up Every bounded sequence has a convergent subsequence. Infinite Sequences 379 Without some additional assumptions this is as far as we can go: it is easy having many, evenly infinitely many, subsequences converging to different numbers (see Problem 2). There is a reasonable assumption to add, which yields a necessary and sufficient condition for convergence of any sequence. Although this condition will not be crucial for to construct sequences our work, it does simplify many proofs. Moreover, mental role in more advanced worth stating now. If this condition plays a funda- for this reason alone and investigations, it is a sequence converges, so that the individual terms are eventually all close same number, then the difference of any two such individual terms to the To be should be very small. precise, = lim a n if some for / then for any /, n—* oo £ > 0 there is an and m > N, then \a n This N such that \a n — — am final inequality, can be used < \ \a n \a n — am l\ < \ - £, < l\ + — \l > N; now e/2 for n if both n > N 6 <-£ + -=£• , an 2 2 which eliminates mention of the limit /, Cauchy condition) which is clearly to formulate a condition (the necessary for convergence of a sequence. A sequence DEFINITION number {a n is \ a Cauchy sequence N such that, is > for every £ if \a n usually written lim \a m, n > 0 there is a natural n, N, then if (This condition m and for all — m am — < \ an £. = \ 0.) The beauty of the Cauchy condition is that it is also sufficient to ensure convergence of a sequence. After all our preliminary work, there is very little left to do in order to prove this. THEOREM 3 PROOF A sequence \a n } converges if and only a if it is Cauchy sequence. have already shown that {a n is a Cauchy sequence if it converges. The proof of the converse assertion contains only one tricky feature showing that every Cauchy sequence {a n is bounded. If we take £ = 1 in the definition of We } : \ a Cauchy sequence we find that there \a In particular, this means \a Thus {a m : m > N} the whole sequence is is m m — an \ < 1 is some iV such that for m, n > for all m > N. that — a N +i\ < 1 N. bounded; since there are only bounded. finitely many other ah a 380 Infinite Sequences and » . Infinite Series The Lemma corollary to the thus implies that some subsequence of {a n } converges. Only one point remains, whose proof will be left to you: if a subsequence of a Cauchy sequence converges, then the Cauchy sequence itself converges. | PROBLEMS 1. Verify each of the following limits. lim (i) n-^» n + 3 1. 4 ' + lim V' (iii) = -J- - 1 + 1 = 0. You Hint: should at least be able n—* » vV + to prove that lim vV = - 1 0. n—» = lim (iv) n— for < A: n—* lim (vii) lim = 1, V« = 1. 2 V lim n—* 1) * • • k\ for k < n, in particular, > a 0. n + n + b — n 1 = max{a, b). oo = lim (ix) = »(«— «j (vi) (viii) Hint: «! n/2. lim (v) 0. oo n— » oo where a(n) 0, is The fact that each prime how small a{n) must be. Hint: of the number of primes which divide n. 72 is > 2 gives a very simple estimate i n v^ X n-»oo (a) What can is an Find all an (b) — 1 , limits (c) . . p + 1 be said about the sequence {a n } if it converges and each integer? convergent subsequences of the sequence 1, —1, 1, — 1, 1, (There are infinitely many, although there are only two . . which such subsequences can have.) convergent subsequences of the sequence 1, 2, 1, 2, 3, 1, 2, ... (There are infinitely many limits which Find all 3, 4, 1, 2, 3, 4, 5, . such subsequences can have.) Infinite Sequences (d) Consider the sequence 11 2i_23.i_23_Al 2) 3) 3> 4j 4? 4) 5> 5? 5j 5j 6j 3. 4. 381 • • • • (a) For which numbers a is there a subsequence converging to a? Prove that if a subsequence of a Cauchy sequence converges, then (b) so does the original Cauchy sequence, Prove that any subsequence of a convergent sequence converges. < < (a) Prove that (b) Prove that the sequence if 0 V2, a Vl V2, < Via < then a 2, Vl y/ 2 V2, . . 2. . converges. (c) Find the limit. Hint: Notice that if = lim a n /, then lim n— n—> * oo V2/, by Theorem 1. = Identify the function /(*) lim (lim (cos n n—* oo many tioned \irx) 2k ) . (It has been men- oo times in this book.) Many impressive looking limits can person Vla n = op who makes them be evaluated easily (especially by the up), because they are really lower or upper sums in disguise. With this remark as hint, evaluate each of the following. (Warning: the list contains one red herring which can be evaluated by elementary considerations.) (l) r hm Ve+Ve* + - • • : ; + n ,..v (n) r hm n— (iii) (v) + V? + m + • --- (^fry2+ " lim ( + + ^> oTT2T» + ' 4- - limits like lim n— oo V« + " ' " 7 Although ife" Yl ^(^TT + !l . (vi) V~e oo • • • + (^r^)' " - )• and lim a" can be evaluated using facts n— » * about the behavior of the logarithm and exponential functions, this approach is vaguely dissatisfying, because integral roots and powers can be defined without using the exponential function. Some of the standard "elementary" arguments for such limits are outlined here; the basic tooks are inequalities derived from the binomial theorem, notably (1 + h) n > 1 + nh> for h > 0; 382 Infinite Sequences and Infinite Series and, for part {1 (a) (c), n + hr > l+nh + »±^ h >> ±Z^) h 71—> h > oo if > a by setting a 1, by setting < 1. 1, rh>0. — 1 + h, = 1 + where OO 0. (b) Prove that lim a n (c) Prove that lim n—* estimating (d) = Prove that lim a n {o *, = 0 if = 1 < 0 if < a > a 1. h and oo h. Va = Prove that lim < 0 if 1 a n— > oo (e) = Prove that lim 1. n—* oo 8. (a) (b) Prove that a convergent sequence is always bounded. Suppose that lim a n = 0, and that each a n > 0. Prove that the n~> of 9. (a) all numbers a n actually has a maximum member. Prove that 1 n (b) < + + log (n — 1 ) show = + 1 + - - + = ' is lim ( 1 a refractory; known it is Suppose that / j decreasing, is /(i) + • • + • choose / — = log as Euler's number, has proved * </(2) + and show that rc! < < follows that = lim n^oo <z n > 0. log n) / known whether y is increasing on [1, oo). Show < it n +•••+/(«- ix Now and that each number not even is n, ?? n-^« \ This number, log H ' ' 3 that the sequence \a n It follows that there (b) 1 < n 2 (a) log n 1 If an 10. set oo n e -; to be quite rational. that • • • +/(»). 383 Infinite Sequences Vn This result shows that approximately is \ the ratio of these two quantities conclude that n An close to (n/e) is \ because But we cannot in this sense; in fact, this false. is n\ is theorem which provides the Problem 26-18. which x > 0 the symbol The standard (and tables. n n/e, in the sense that for large n. 1 very desirable, even for concrete computacannot be calculated easily even with logarithm estimate for n\ tions, close to is difficult) right information will be found in *11. This problem investigates for ' x makes sense. In other words, when does b(x) = zX if ' we define ai{x) = 0n x, +iM = x anix \ lim a n (x) exist? n— oo Prove that if b{x) exists, then x b(x) = b(x). (The situation is similar to that in Problem 4.) According to part (a), if b{x) exists, then x can be written in the form lle lly Hint: Consider the for some y. Conclude that 0 < x < e y (a) (b) . graph of f(y) (log y)/y, as in < Suppose, conversely, that 0 (c) since {a n (x) \ also that b(x) is < x Problem 17-24. < (e) b'(x) in (f) terms of (Of course, Prove that if . Show that each a n {x) r b {x) and 1/e < e\ ), find a formula for Problem 17-24(/). by differentiating the equation x h{x) = similar to is method does not prove this = (0, e Hint: This part b{x). lim a n n— lle e). Derive a formula for b(x). e clearly increasing, this proves that b{x) exists (and Find b{Vl) and b(e^). Prove that b is differentiate on (d) *12. = that b is differentiate.) then /, oo lim n—* • • • + = /. n Hint: This problem Problem Suppose that a n + (gi oo is very similar to (in fact it is a special case of) 13-34. *13. ^ 0 for each n and that lim a n +i/a n = L Prove that n— * oo V an — Hint: This requires the same sort of argument that works n /— a = 1, for a > 0. in Problem 12, together with the fact that lim lim n— I. oo V n— > 14. (a) Suppose that {a n Prove that lim a n \ n— (b) a convergent sequence of points is in [0, 1]. * oo {a n } of points all in (0, 1) such that oo Suppose that / all also in [0, 1]. Find a convergent sequence lim a n is not in (0, 1). n—> 15. is is oo continuous and that the sequence *,/«,/(/«),/(/(/(*))), • • • 384 Infinite Sequences and Infinite Series Prove that / is a "fixed point'* for /, i.e., /(/) = /. Hint: have occurred already. Suppose that / is continuous on [0, 1] and that 0 < f(x) < 1 for all x in [0, 1]. Problem 7-1 i shows that / has a fixed point (in the converges to Two 16. (a) /. special cases terminology of Problem 15). If / ment can be made: For any x is increasing, a in [0, 1], the *, /(*), /(/(*)), much stronger state- sequence • • is necessarily a fixed point, by Problem 1 5) Prove by examining the behavior of the sequence for f(x) > x < x, or by looking at Figure 7. A diagram of this sort is has a limit (which FIGURE 7 . this assertion, and f(x) used in Littlewood's Mathematician's Miscellany to preach the value of drawing pictures: "For the professional the only proof needed is Figure]." [this *(b) Suppose that / and g are two continuous functions on [0, 1], with 0 < f(x) < 1 and 0 < g(x) < 1 for all x in [0, 1], which satisfy fog = g o f. Suppose, moreover, that / is increasing. Show that / and g have a common fixed point; in other words, there is a number / such that /(/) = / = g(l). Hint: Begin by choosing a fixed point for ***(c) g. Does the conclusion of part / is (b) hold without the assumption that increasing? Problem 15 is really much more valuable than Problem 15 might suggest, and some of the most important "fixed point theorems" depend upon looking at sequences of the form from x, /(*), /(/(*)), .... A special, but representative, case of one such theorem is treated in Problem 18 (for which the next problem is preparation). The 17. trick in (a) Use Problem 2-4 to show that if c t± 1, + c n then m c = 1 (b) Suppose that \c\ < 1. - c n+1 -c Prove that lim c m + + c n = 0. Suppose that {* n is a sequence with \x n is a Cauchy sequence. c < 1. Prove that {x n Suppose that / is a function on [a, b] such that (c) } — x n +i\ } *18. \f(x ) (*) where (a) (b) c < 1. — /GO ^ I x c ~~ \ (Such a function is jvl> f°r a ^ * ancl y in [a called a contraction.) Prove that / is continuous. Prove that / has at most one fixed point. > < c n > where Infinite Sequences (c) 385 considering the sequence By *,/(*),/(/(*)), for in any x in [a, b], prove that / does have a fixed point. (This result, a more general setting, is known as the "contraction lemma.") Let {x n } be a sequence which — yn (a) bounded, and is SUp {x n * n +l, *n+2, • , Prove that the sequence {y n converges. \ • let •}• The limit lim y n is denoted by Urn * n or lim sup x n and called the limit superior, or upper , n— n—» so *> limit, of the sequence {x n \. (b) Find lim x n each of the following: for n—> (c) (ii) Xn = (-i)"-- (iii) Xn = (iv) *n = vV Define lim x n (or lim inf x n ) and prove that n— n— * oo * lim x n < lim # n (d) Prove that lim x n this case = lim x n (e) and only exists if = lim x n lim * n n— n—* oo n—> oo . n—> «o n—* oo if = lim x n . 00 A Recall the definition, in Problem 8-18, of lim A. Prove that the if numbers lim x n and that in x n are distinct, for a n-> where (a) ^4 = {x n Suppose that : n { n in } is for all n. = set lim 4, oo N}. nonincreasing, with b n ai + aibi + m < bounded then lim x n • • • + an + <z»& n < > n, and that may not look 0 for each M Prove that < bitn This result is known • • as Abel's < 6iM. Lemma. Although it very interesting, it is the basis for several very interesting results which will appear in later problem sets; in addition, the proof 386 Infinite Sequences and Infinite Series depends on a rather tricky way of writing + a\bi • * • + anb n . Deduce that (b) bm < (a trivial + akbk 4 + an b n < * * consequence of part bk M only because (a), inserted it is needed later). *21. The Theorem Bolzano-Weierstrass quite differently than in the text of limit points. there (a) is A \x — < a\ £ but x limit points of the following sets. (i) j-L- tfinNj. (ii) (-- +— {(-1)* (iv) Z. (v) Q. and n : m m in n) A if for every £ > 0 9* a. . J +^J:rcinN). [l a limit point of A many points a of A lim A is the largest Prove that x also proved, statement uses the notion a limit point of the set is with all (hi) (c) point x classical Find \n (b) A a point a in and usually stated, is —the is if and only if for every £ — are infinitely satisfying Prove that limit point of A, \x a\ < > 0 there £. and lim A the smallest. The usual form of the Bolzano-Weierstrass Theorem states that if A an infinite set of numbers contained in a closed interval [a, b], then some point of [a, b] is a limit point of A. Prove this in two ways: Using the form already proved in the text. Hint: Since A is infinite, is (d) there are distinct (e) *22. numbers x h # 2 # 3 , . in A. . Use the Bolzano-Weierstrass Theorem to prove that if / is continuous on [a, b], then / is bounded above on [a, b]. Hint: If / is not bounded above, then there are points x n in **23. . , Using the Nested Intervals Theorem. Hint: If [a, b] is divided into two intervals, at least one must contain infinitely many points of A. (a) Let {a n } h f Suppose that 0 < with f(x n ) f h > n. be the sequence i, gers j [a, b] j is < a f f , < , b n such that aj i, t> < is 1. , Let in % , • . • • be the number of inte(Thus N(2; f) = 2, and A^(rz; a, 6) [a, b]. -g-, Infinite Sequences h f) = 387 Prove that 3.) lim n-+oo (b) A sequence tributed in {a n of } ft numbers in [0, 1] is called uniformly dis- [0, 1] if N(n; a, b) lim = b - a n and b with 0 < a < b < 1 Prove that if s is a step function defined on [0, 1], and {a n is uniformly distributed in [0, 1], then for all a . } (c) Prove that if \a n [0, 1], then ) is uniformly distributed in [0, 1] and /is integrable on (a) Let / be a function defined on a in [0, 1]. points a in set of such that lim f(y) exists for all For any s > 0 prove that there are only finitely many - f(a)\ > £. Hint: Show that the [0, 1] with |lim f(y) such points cannot have a limit point lim f(y) could not (b) [0, 1] x, by showing that exist. Prove that, in the terminology of Problem 20-5, the set of points where / is discontinuous is countable. This finally answers the question of Problem 6-16: If / has only removable discontinuities, then / is continuous except at a countable set of points, and in particular, / cannot be discontinuous everywhere. CHAPTER *m*^ INFINITE SERIES Infinite sequences were introduced + a.\ in this chapter. This of infinitely + a2 + az • • • as yet completely undefined. is sum What can be 5 ' = sn infinite with the specific not an entirely straightforward matter, for the is many numbers defined are the " partial sums and the in the previous chapter "sums" intention of considering their + a\ + • • sum must presumably be an , defined in terms of these partial sums. Fortunately, the mechanism for formulating this definition has already been developed in the previous chapter. puting the infinite represent closer This sum + a\ a2 • 3 • • , is to be any hope of comsums s n should the partial and closer approximations as n is chosen larger and larger. amounts to little more than a sloppy definition of limits: last assertion the "infinite sum" a\ + a2 will necessarily leave the sequence {jn } may + a$ with a n = (— which lim 71— yields the sn ' of * 1, ought ' many have a -1, be lim sn . This approach sequences undefined, since the limit. -1, to . For example, the sequence . . new sequence si = ai ^2 = ai S3 = a\ Si = a\ does not + "sum" easily fail to 1, for If there +a + = 1, + a = 0, +a +a = + a + az + exist. 2 2 2 z 1, ~ 0, Although there happen to be some clever 00 extensions of the definition suggested here (see Problems 8 and 23-11) it seems unavoidable that some sequences will have no sum. For this reason, an acceptable definition of the sum of a sequence should contain, as an essential component, terminology which distinguishes sequences for which sums can be defined from less fortunate sequences. 388 Infinite Series The sequence DEFINITION {a n } summable is — sn In this case, lim » if ai the sequence + ' ' + ' an {s n } 389 converges, where . denoted by s n is n— ^ n= and is an (or, less formally, a\ called the sum of the sequence {a n a2 + as + * ' *) precise expressions; indeed the An infinite \ in this definition The terminology introduced day language. + 1 title sum of this chapter £ n= an is is is usually replaced by less derived from such every- usually called an infinite series, the l word "series" emphasizing the connection with the infinite sequence \a n \. The statement that [a n is, or is not, summable is conventionally replaced by \ the statement that the series £= 71 a n does, or does not, converge. This terminol1 oo ogy is somewhat (so it can't peculiar, because at best the symbol "converge"), and it £ a n denotes a doesn't denote anything at summable. Nevertheless, this informal language unlikely to yield to attacks on logical grounds. is all number unless \a n } is convenient, standard, and direct Certain elementary arithmetical operations on infinite series are if {a that show n \ and consequences of the definition. It is a simple exercise to { bn j are summable, then n = 1 X C ' °n examples of As yet these equations are not very interesting, since we have no terms are the which summable sequences (except for the trivial examples in some sequence, summable eventually all 0). Before we actually exhibit a general conditions for summability will be recorded. can be There is one necessary and sufficient condition for summability which if the sequence only and if summable is {a sequence The n stated immediately. \ only if converges, which happens, according to Theorem 21-3, if and original the terms of s n * 0; this condition can be rephrased in lim sm \s n \ m,n—»*> - 390 Infinite Sequences and Infinite Series sequence as follows. THE CAUCHY CRITERION The sequence {a n summable is } lim m,n— a n +i if and only if +••+«*=•(). 0° Although the Cauchy criterion is of theoretical importance, it is not very useful for deciding the summability of any particular sequence. However, one simple consequence of the Cauchy criterion provides a necessary condition for summability which is too important not to be mentioned explicitly. THE VANISHING CONDITION If \a n } summable, then is = lim a n 0. This condition follows from the Cauchy criterion by taking also be proved directly as follows. If lim s n = I, then n— = lim a n n— » lim n—* » — (s n = Jn -i) lim n Unfortunately, this condition n + sn — lim » * far is from sufficient. is not summable; in numbers \/n shows that the sequence \sn } > > \ (2 terms, each The method never > can jn _i For example, lim fact the following 1 jn = oo grouping of not bounded: is > * (4 terms, each > i) i) it 0. but the sequence {\/n} the ; n—* n— 0, 1 to oo = 1-1 = m = h (8 terms, each > ft) of proof used in this example, a clever trick which one might need for some more standard methods for attacking these problems. These methods shall be developed soon (one of them will give an see, reveals the alternate proof that n= 1/n does not converge) but ^ it will be necessary to first 1 procure a few examples of convergent series. The most important of all infinite series are the "geometric series" |j =0 = r« +r+r + 2 1 r 3 + • • • . n Only the cases approach 0 if \r\ \r\ < 1 > 1. These are interesting, since the individual terms do not sn can be managed because the partial sums series = 1 + + r • • • + r n can be evaluated in simple terms. The two equations sn = rsn = 1 +r+ + r + r + r 2 2 + n + rn + r r n+1 Infinite Series 391 lead to - s n {\ r) = *n = 1 - r» +1 or 1 - r n+1 -r 1 by (division lim r n = 1 — r is 0, since \r\ valid since < Xr 1. It = n we are not considering the case r = 1). Now follows that lim —— 1 — i + n-*<* 1 r 1 [r| , — < 1. r n=0 In particular, n=0 n=l that is, i an infinite FIGURE + i + • • - • 1, sum which can always be remembered from the picture in Figure 1 1 from which Special as they are, geometric series are standard examples derived. will be summability for important tests with each a n > 0; such then sequences are called nonnegative. If \a n is a nonnegative sequence, with combined is clearly nondecreasing. This remark, the sequence {s n summability Theorem 21-2, provides a simple-minded test for shall consider only sequences \a n For a while we \ \ \ : A nonnegative sequence THE BOUNDEDNESS CRITERION sums sn is {a n } is summable if and only if the set of partial bounded. not very helpful— deciding whether or not the set of On the other hand, if some all s n is bounded is just what we are unable to do. this criterion can be comparison, for available already are convergent series By itself, this criterion is used to obtain a result whose simplicity belies almost all other tests). THEOREM 1 (THE COMPARISON TEST) its importance Suppose that 0 < an < bn for all n. (it is the basis for — 392 Infinite Sequences and Infinite Series Then ^ if b n converges, so does n=l PROOF ^ an . n=l If = ai + = bi+ sn tn • ' • • + an +b n • , , then < 0 sn < for all tn n. 00 Now {/ n } is bounded, since ^ n= Therefore £ n converges. {s n } is bounded; l 00 consequently, by the boundedness criterion ^ n= Quite frequently the comparison # n converges. 1 can be used to analyze very com- test plicated looking series in which most of the complication is irrelevant. For example, 2 + sin 2 3 (rc + n + 1) n2 converges because o 2 < ± + sin3 (* 2n + < A, ' 2n n2 and 00 00 X n is a convergent (geometric) = ~ 2n 3 y n tion really means we would expect the series 1 1 + sin 2 n 3 1 term of the series is practically 1 /2 n . This last asser- that Hm n-oo ( n \2 - 1 and this formulation shows us how number c > 1, then it is certainly 0 1 series. Similarly, n L(2 — to converge, since the nth X r* n= 1 * 2 "-l + L_y± W/ +sin 2 to verify 2* = 1 , our conjecture. If true that <C sin*»' h f° r lmge emUgh ^ we choose any 393 Infinite Series This shows that + converges for some N, and of course sin 2 n z this implies convergence of the original series. the comparison test backwards. Consider, for It is also possible to use exam- ple, the series n+ Sn + r n= 1 2 1 We 1 would expect /n for large n. this series to diverge, since (n To prove < 0 this, < - c n2 n This shows that the Y series + choose any number + for large with 0 enough 1) is practically < c < 1. Then n. 1 + (n + l)/(n 2 c \)/{n 2 + 1) diverges, because if it con- 00 verged, then by the comparison test know is \/n would also converge, and this £ n= we 1 false. The comparison analyzed other important test yields tests when we Choosing the geometric series as catalysts. ^= series n vergent series par we excellence, use previously the con " 0 obtain the most important of all tests for summability. THEOREM 2 (THE ratio TEST) Let a n > 0 for all n, and suppose that lim —— n— « <3 = r. n 00 Then a n converges £ n= < if r 1. On the other hand, if r > 1, then the terms 1 so an do not approach 0, so £ n to PROOF compute lim Suppose first a n +i/a n that r < 1 . = <2 n diverges. (Notice that it is therefore essential 1 and not lim a n /a n +\ 0 Choose any number hm —^ == r j < with 1 r < s < 1 . The hypothesis 394 Infinite Sequences and Infinite Series implies that there some is N such ^ < that > N. for n s an This can be written < tfn+i > for n sa n N. Thus Y Since fc 0^ = V ajv =0 fc = j* converges, the comparison test shows that 0 00 Y 00 tf n = Y fltf+fc converges. This implies the convergence of the series ^= n <z n, which has only 1 00 finitely The many case r Y terms not included in > 1 is even easier. If 1 # rt n= iV < j . < a number N such do not approach 0, so {a n } r, then there is that a -^>s for n>N, an which means that aN+k > tfj\rJ* > — k <zn This shows that the individual terms of {# n summable. } 0, 1, ... . is not | oo As a simple application of the ratio test, consider the series <z n = we obtain 1 an+i = (n+l) = »! fln 1 0+1)! \ n\ Thus Iim ^ = 0, = _J » + r 1/n ^ n= 1 !. Letting Infinite Series which shows that the l/»! converges. If £ series n= 395 consider instead the we 1 so series ^ r n /rc!, where r is some number, then fixed positive -J+ = Km lim .R so ^ r n /w = 0, 1 converges. It follows that ! r lim = 0, n->« n! was based on the a result already proved in Chapter 16 (the proof given there the series consider we if Finally, same ideas as those used in the ratio test). ^ nr n we have lim n _>„ since lim (n 7i — verges, + l)/» = nr 1. = n lim = r r, n n->«= This proves that 0 if < r < 1, then rcr* J n= con- 1 and consequently lim nr n = 0. n— » > (This result clearly holds for - 1 < r < 0, also.) It is a useful exercise to proan intermediary. using the ratio test as vide a direct proof of this limit, without as a Although the ratio test will be of the utmost theoretical importance, of the drawback One disappointing. found be practical tool it will frequently determine, and fact that lim a +Ja may be quite difficult to ratio test is the may not even regularity, is is A more exist. will if serious deficiency, which appears with maddening the fact that the limit might equal precisely the one example, n n an show that = £ which is it (1/n) 2 converges, even / i The case lim a n+ i/a n might not be summable might be. In fact, our very next inconclusive: \a n but then again 1. 1 (for \ though y = test 1 396 Infinite Sequences and Infinite Series This test of the comparison THEOREM 3 (THE integral TEST) method provides a quite different divergence of infinite series Suppose that / — but the test, it is chosen for comparison series and decreasing on positive is determining convergence or an immediate consequence for like the ratio test, oo), [1, quite novel. is and that = f(n) a n for 00 Then all n. a n converges if ^ and only if the limit /,•/->.»/// exists. PROOF The existence of lim / / is equivalent to convergence of the series /,'/+/;/+/,'/+••. Now, / since we have decreasing is f{n (Figure 2) + 1)< / < f{n). oo The first double inequality shows that the half of this ^= series n compared to the series converges if lim / ^ /, proving that n=l ^ ^ a n +i may be ^= a n) l a n +i (and hence n=l n exists. as The second half of the inequality shows that the series f may be ^ n= 1 00 FIGURE 2 compared to 00 Y the series A a ni proving that lim f f must Y an settles the exist if converges. | Only one example using the integral test will be given here, but many question of convergence for infinitely vergence of ^ n= \/n p is series at once. If p equivalent, by the integral > it 0, the con- to the existence of test, 1 r Now L (P dx. 1 1 - A*-i 1) log A, + A This shows that lim A— » oo f ^ 1 1 /x p dx exists 1 p - P 1 p if p > 1, but notify = 1. < l.Thus Y J^, l/n p 1 Infinite Series > converges precisely for p 1. ^ V* In particular, n= The tests 397 diverges. l considered so far apply only to nonnegative sequences, but nonmay be handled in precisely the same way. In fact, since positive sequences an I n=l = -( I n= «,), considerations about nonpositive sequences can be reduced to questions involving nonnegative sequences. Sequences which contain both positive and all negative terms are quite another story. ^ If n= an is a sequence with both positive and negative terms, one can con- 1 sider instead the sequence £ \a n all \, whose terms are nonnegative. Cheer- of fully ignoring the possibility that we may have thrown away all the interesting information about the original sequence, we proceed to eulogize those sequences which are converted by this procedure into convergent sequences. 00 oo DEFINITION The series vergent. Y (In summable if an absolutely convergent is more formal language, the sequence the sequence \ Although we have no right is \a n \\ THEOREM 4 is \a n \ to expect this definition to The is con- \ absolutely be of any interest, it following theorem shows that the at least not entirely useless. Every absolutely convergent series is convergent. Moreover, a series is absothe lutely convergent if and only if the series formed from its positive terms and series formed from its negative terms both converge. 00 PROOF is \a n summable.) turns out to be exceedingly important. definition Y the series if If ^ n= \a n Cauchy converges, then, by the criterion, \ l lim |a n+ i| + * * * + \ a m\ = 0. m,n—K*> Since |fl it n +l + ' ' ' + am \ + < • " • follows that lim a n +\ + m,n—> » 00 which shows that ^ a n converges. ' ' ' + am = 0, + km|, 398 Infinite Sequences and Infinite Series To prove the second part of the theorem, An, if an o, if an let > < < > if O-n if so that an~ ^ n= an+ ^ n= is is dn o o, 0 o, the series formed from the positive terms of 1 an ^ n = , and 1 the series formed from the negative terms. l ^ If and n= 1 ^ a n ~ both converge, then n=l OO 00 n = n= 1 00 n= 1 00 n= 1 l 00 also converges, so ^ n= <z converges absolutely. n l 00 On the other hand, ^ if n= |<z n | converges, then, as we have just shown, 1 oo ^= n a n also converges. Therefore 1 t *» + - id * + s m ) and both converge. | It follows from Theorem 4 that every convergent series with positive terms can be used to obtain infinitely many other convergent series, simply by putting in minus signs at random. Not every convergent series can be obtained in this way, however there are series which are convergent but not absolutely convergent (such series are called conditionally convergent). In order to prove this statement we need a test for convergence which applies specifi- — cally to series with positive THEOREM 5 (LEIBNIZ'S THEOREM) and negative terms. Suppose that a\ > a-i > az > • and that lim a n = 0. > 0, H « 399 Infinite Series Then the series 00 £— 71 (-l) n+1 = n <Z — ai + 02 <Z — 3 + <Z4 <Z - 5 * ' * 1 converges. PROOF between the partial sums which we illustrates relationships Figure 3 will establish: (1) S2 (2) Si (3) Sk < > < Si S3 < > k is *S ' • ' ' ' , , —H— / is odd. — I •SlO -^12 3 prove the first two inequalities, observe that (1) (2) To • even and H—t— 56 To S5 Sq if si H FIGURE < > = > = > S 2n +2 S2n+S + S2n S2n, ^n+l " < S2n #2n Since fl2n+l > — #2n+3 #2n + 2 + first S2n-\ ~~ > fl «2n+3- that 2n > since a 2n i-2n-i, ^2nJ Since #2n+2 ^2n+lj prove the third inequality, notice - «2n+l 0. and This proves only a special case of (3), but in conjunction with (1) that general case is easy: if k is even and / is odd, choose n such 2n> k Sk < and In - > 1 (2) the /; then which proves Now, S2n < < S2n-\ ^> (3). the sequence bounded above (by {s 2n s t for } converges, because any odd a = sup /). is it nondecreasing and Let {s 2n } = hm 2n . jr n— » * Similarly, let (S = inf {^2n+l} — nm S2n+1- n— * 00 If follows from (3) that *2n+l a < - 0; since ^2n = «2n+l and UlH fl n = 0 n— » it is actually the case that a = 0. This proves that a = 0 = lim sn , | is 400 Infinite Sequences and Infinite Series The standard example Theorem 5 is -* + *-* + * i which derived from the series , convergent, but not absolutely convergent (since is ^= 71 converge). If the sum of this series denoted by is \/n does not 1 the following manipulations x, lead to quite a paradoxical result: * = i -i + i-i + i -£+ = l- i- i + |- i- i + (the pattern here = (i - i) -i+ • i-T o- TV + + -i r-iV+--1 1 one positive term followed by two negative ones) is - (i • • i) -i+ (* - A) - A+ *- i + *- i + A- A + A- A+ --id - * + *- * + *- * + + -*+•) = so x = ^ 0: x */2, implying that * > that x sum the partial s2 = On (t A - A) - + . . - the other hand, it is easy to see that equals ^, and the proof of Leibniz's Theorem shows 0. sz. This contradiction depends on a step which takes for granted that operations valid for finite sums necessarily have analogues for infinite sums. It is true that the sequence {b n contains {a n = } 1, 1, — i, -h h -h -h h - TV, -i, numbers the all = } -tV, • • • in the sequence h —h t? y ? —h — A> A> — A> i> * ' " • is a rearrangement of {<z n in the following precise sense: each where / is a certain function which "permutes" the natural numbers, that is, every natural number m is f(n) for precisely one n. In our example In fact, \b n bn — /(2m ) } <2/(n) + 1) = 3m + 1 (the terms ... go 1, into the 1st, 4th, 7th, . . . places), = 3m /(4m) /(4m + 2) = 3m — (the terms 9th, + 2 . . — A, -A, (the terms 5th, 8th, • • go into the 3rd, 6th, places), . . . . ... go into the 2nd, places). 00 SB Nevertheless, there is no reason to assume that ^ n= these sums are, by definition, lim b\ n— <*> + • • + b„ should equal bn ^= n l and lim n—* ai + * • an : 1 + an, oo so the particular order of the terms can quite conceivably matter. The series Infinite Series 401 00 2* n = is not special in this regard; indeed, behavior its typical of is 1 which are not absolutely convergent— the following series convergent result (really how bad of a grand counterexample than a theorem) shows more conditionally series are. 00 THEOREM 6 a n converges, but does not converge absolutely, then for any ^ If n= number a l 00 there a rearrangement \b n is of {a n ] \ such that ^ n= = bn a. 1 00 PROOF Let pn denote the ^ n= y series of denote the one of these least positive terms of {a n formed from the series } and let l to converge, for negative terms. It follows does not converge. As a matter of fact, both must fail one had bounded partial sums, and the other had un- series if V partial sums, then the original series bounded n= Now let a be any number. Assume, an would also have un- 1 assumption that partial sums, contradicting the bounded from Theorem 4 that at for simplicity, that it converges. a > 0 (the proof for Pn not convergent, oo a < 0 will be a simple modification). Since the series £ n= there is a number N such that n = We will choose is 1 JVi to 1 be the smallest N with this property. This means that JVi-l (1) 1 n= but (2) < > ^p n n= Then Pn a, l a. l if N^ Si - £ n= j»„, 1 we have This relation, which is clear ^1 from Figure — < 4, — ^ follows immediately /> n = pNi> from equation 402 Infinite Sequences and Infinite Series 0 + pi FIGURE To the 4 sum T\ which for Si we now add on just enough than is less a. negative terms to obtain a we choose In other words, new sum the smallest integer M i which Mi T =Si+ J= q n <Oi. x 71 As before, we have a — We now larger 1 continue procedure indefinitely, obtaining sums alternately this and smaller than < —qM v Ti a, each time choosing the smallest N k or • • M possible. k The sequence • Pu • • • ,pN qi, s9 . . . , qM„pN l+ \, . . . ,pN t, • an The partial sums of this rearrangement increase S u then decrease to T h then increase to S 2 then decrease to T 2 etc. To complete the proof we simply note that \Sk — a\ and Tk — a\ are less than or equal to pN or —qM respectively, and that these terms, being members is a rearrangement of } j . to , , \ k, k of the original sequence {a n must decrease }, to 0, since ^= n a n converges. | 1 Together with Theorem 6, the next theorem establishes conclusively the between conditionally convergent and absolutely convergent distinction series. theorem 7 n= and a n converges absolutely, ^ If {b n } is any rearrangement of {a n }, then l 00 ^ n b n also converges (absolutely), and =1 n=l n=l PROOF Let us denote the partial sums of by {a n } by sn , and the partial sums of tn . 00 Suppose that £ > 0. Since a n converges, there ^ n= | l ^ an — < j-at | e. is some N such that {b n } 403 Infinite Series oe Moreover, since ^= n \a n converges, we can f n= N so that - (W + knl • • + M) < • s, 1 so that i.e., Now choose M so large that each of a u Then whenever m > M, the difference At, also choose \ 1 . . . aw , Thus, if — m m > M, SN \ < • t • aa- , - m appear among b h s N is the sum . . . , b M . of certain a u where Consequently, are definitely excluded. \t • + |flA'+l| |flJV+2| + |flAT +3| + ' * * • then | £ an — / ^an -sN + U-sN = m \ I I n-l n=l 06 < ^ | n= < Since this is true for every £ > s 0, tf — n it \tm — Sn\ J + s. Y the series „ prove that + JA l bn converges to £ an To . n- =1 converges absolutely, notice that the above argument can be applied to the positive terms of shows that the £ <2 n and the negative terms separately; formed from the positive and negative terms of series this £= n bn 1 00 each converge, so ^ n= 6n converges absolutely. | 1 of this chapter have been concerned with summability of reason sequences, but not with the actual sums. Generally speaking, there is no presume that a given infinite sum can be "evaluated' in any simpler terms. The theorems 5 to However, many simple expressions can be equated to infinite sums by using which Taylor's Theorem. Chapter 19 provides many examples of functions for n where lim n—> R n ,a{*) = 0- This is precisely equivalent to oo n-* oo Z—/ i ! • 404 Infinite • Sequences and Infinite Series which means, in turn, that /(*) =0 i As particular examples we have *3 x +-_ 3! 5! sin x X2 cos X 2! e x = 1 + b X1 rL + X* —_ +4! -f . . . . . . X +-+-+ U + 2! 3! 4! X + _ fL + 3 2 X 3 a: 3 a: 4 5 a: arctan x log(l . T X2 + *) xz 2 . --+-+ 4 #4 a: > 1; in addition, when < 5 -1 < , X < 1. 5 + (Notice that the series for arctan * and log(l \x\ M . 7 = — 1, # do not even converge + x) becomes x) for the series for log(l -i-i-i -i which does not converge.) Some pretty impressive results are obtained with particular values of x: 0 = = TZ 7T 3! 5! 5 TT 7 7T + ' * ' , 7! + h + + + ---' h h E-,_i + i_i + ... 4 3 5 7 1,1 ^ = log2 l-- + ---+.... e 1 f More 1 , , , developments may be anticipated if we compare the series and cos x a little more carefully. The series for cos x is just the one we would have obtained if we had enthusiastically differentiated both sides significant for sin x of the equation sin x = x* . xb x • 3! * * 5! we have never proved anything about the if we differentiate both sides of the formula for cos x formally (i.e., without justification) we obtain the formula cos'(*) = — sin x, and if we differentiate the formula for e x we obtain exp'(*) = exp(*). In the next chapter we shall see that such term-by-term term-by-term, ignoring the fact that derivatives of infinite sums. Likewise, differentiation of infinite sums is indeed valid in certain important cases. Infinite Series 405 PROBLEMS 1. Decide whether each of the following infinite series is convergent or divergent. The tools which you will need are Leibniz's Theorem and the comparison, ratio, and integral tests. A few examples have been picked with malice aforethought; two series which look quite similar may require different tests (and then again, they may not). The hint below indicates which tests may be used. eo nd sin <" i-. -* + *-*+ GO 1 (iii) 1 (iv) £(n (v) logn n =i Y— (The summation begins with \ ~ 1 y </n} + 1 — 1 00 (viii) n= 1 (ix) Z-/ log n n=2 eo S =2 (log n)"' 00 (xi) (xii) S Gog nY (-!)»—1_. Y Zv n (log n)» =2 eo (xiii) n3 eo (xiy + 1 n = 2 simply to avoid the meaningless term obtained for n (vii) n - : 1 (x) • 1) ~2 ~^n<1 n * -i+f - i+*-i+t-*+ n (vi) ' = 1.) 406 Infinite Sequences and Infinite Series V-L-. (xv) 1-1 n log n n-2 00 ^ XVi ) / "71 \~2 n—2 00 1 (xvii) _,n 2 (log n) oo (xvii ° s?* = n n l = l Hint: Use the comparison test for (i), (ii), (v), (vi), (ix), (x), (xi), (xiii), (xiv), (xvii); the ratio test for (vii), (xviii), (xix), (xx); the integral test for (viii), (xv), (xvi). The next two problems examine, with hints, some infinite series that require more delicate analysis than those in *2. (a) you have If Problem 1. examples successfully solved (xix) and (xx) from 00 Problem 1, it should be clear that and diverges for a > For a e. = e n ^ n= a n\/n n < converges for a e 1 the ratio test show that fails; 00 ^= 71 e n n\/n n actually diverges, by using Problem 21-10. 1 00 (b) Decide when 21-10 when n n n /a n\ ^ n= converges, again resorting to Problem l the ratio test fails. 00 00 *3. Problem 1 presented the two series ^ (log n)~ k and which the first ^ (log n)~ n of , n=2 n=2 diverges while the second converges. The series 00 1 (log n) los n= which lies between these two, n 2 is analyzed in parts (a) and (b). 00 (a) Show that JJ e v /yv dy exists, by considering the series £ {e/n) n . 407 Infinite Series Show (b) that logn (log n) ; converges, by using the integral substitution Show (c) and part Hint: Use an appropriate test. (a), that V* Z/(log 1 n) l0 « (l0 « n) by using the integral test. Hint: Use the same substitution and show directly that the resulting integral diverges. diverges, as in part (b), 4. Let {a n (a) be a sequence of integers with 0 } a n 10~ n exists (and y {a n < with 0 } < an 1). (This, usually denote by G.aia2a 3 ai number which we (b) Suppose that 0 between 0 and lies < x an < < 1 Prove that there . a n \0~ £ and 9 = n is . . < Prove that 9. of course, the is .) . a sequence of integers Hint: For example, x. n-l ai = [10*] (where [y] denotes the greatest integer which (c) Show that if \a n } is repeating, i.e., is of the form a h a 2 is < . . , y). . , CO ai, 02, • • • ak, ai, a 2 , , . . . , then (and find it). repeating, The same i.e., if a n \0 ^ n= n is a rational result naturally holds if \a n the sequence {a N+k } number l eventually is } repeating for some N. is 00 Prove that (d) x if = a n \0' £ n= n rational, then is {a n eventually is \ 1 repeating. (Just look at the process of finding the decimal expansion of p/q 5. —dividing Suppose that \a n ) proof of Leibniz's | ^— n 6. p by long q into satisfies the hypothesis of Leibniz's Theorem. Theorem (-l) n+1«» - division.) Use the to obtain the following estimate: [«i ~ at + • • ± a N ]\< a N • . l Prove the following theorem (another kind of "comparison test"). If 00 an , bn > 0 and lim a n /b n 00 if ^ converges. = c ^ 0, then T a n converges if and only 408 Infinite Sequences and Infinite Series 7. Prove that an > 0 and lim if r > 1. if and diverges This result known is which the ratio test test shows that the i Va n = r, S then (The proof is very similar as the "root test." It while the root fails, < if r 1, to that of the ratio test.) easy to construct series for is test a n converges works. For example, the root series + + 2 (i) + 2 (i) 3 (i) + 3 (i) + • • • converges, even, though the ratios of successive terms do not approach a limit. Most examples are of this rather artificial nature, but the root test nevertheless quite an important theoretical tool, and if the ratio test works the root test will also (by Problem 21-13). It is possible to elimirnate limits from the root test; a simple modification of the proof shows is 00 that ^ n= a n converges there if is some s < 1 such that all but finitely many l 00 ^/a n are < s, and that ^ n- This result is known a n diverges if infinitely many > are as the "delicate root test" (there ratio test). It follows, using the notation of is a similar delicate Problem 21-19, that if < lim 1 and diverges if 8. possible A sequence \a n ) lim = is <2i + * * • " + • • • + f* = /, if / n oo * sk no conclusion summable, with Cesaro sum called Cesaro is n— (where : lim is sequence 1 1 n—> oo n—* * if > lim an ^ n= converges 1. 1 + Problem 21-12 shows that a summable a k ). sum equal automatically Cesaro summable, with Cesaro sum. Find a sequence which is not to its summable, but which is Cesaro summable. 9. This problem outlines an alternative proof of Theorem 7 which does not rely on the Cauchy (a) Suppose that a n {a n (b) \, and let criterion. > 0 for each = a\ + sn • Show that for each n there Show that ^ an < £ bn is • n. Let + an \b n be a rearrangement of } and some m with ^n t = bi < m n t . + * ' + bn . 409 Infinite Series 00 00 Show (c) that = an ^ ^ bn . n=l n=l 00 Now (d) replace the condition a n > 0 by the hypothesis that n Theorem converges absolutely, using the second part of 10. Prove that (a) a n converges absolutely, ^ if n= and {b n an 1 4. any subse- is } ^= 1 00 quence of {a n then }, £ n= converges (absolutely). 6n 1 00 Show (b) that this n= Prove that *(c) 1 a n converges absolutely, then ^= if a n does not converge absolutely. ^ false if is n l oo n= + = Ol ^ + «3 ^5 + ' ') ' + («a + ^4 + «6 Prove that Problem JQ *13. 18-28 (sin x)/x\ 00 00 Y if an is absolutely convergent, then ^ | shows that j^" (sin x) fx dx an | < "=1 n=l *12. ' ' l 00 11. + £= » converges. \a n \. 1 Prove that dx diverges. | Find a continuous function / with /(*) exists, but lim f(x) does not > 0 for all x, such that J" f(x) dx exist. X-+00 *14. 1 /x for 0 < a: < 1, and let /(0) = 0. Recall the definifrom Problem 15-32. Show that the set of all /(/, P) for P a partition of [0, 1] is not bounded (thus / has "infinite length"). Hint: Try partitions of the form Let f(x) = x sin tion of t(f, P) 15. Let / be the function shown in Figure the shaded region in Figure *16. In this problem we 5. Find J* /, and also the area of 5. will establish the "binomial 5 series' 00 (i + *)« = £(«>*, W<i, fc-0 for any steps, a, by showing that lim Rn ,o(x) = °- The P roof is in several and uses the Cauchy and Lagrange forms as found in Problem 1 9-6. 410 Infinite Sequences and Infinite Series so (a) Use the ratio test to verge for + (1 (b) < |r| a r) ). It Suppose show (this is 1 that the series ^ (^j not to say that it that 0 < x < 1. Show k does indeed con- necessarily converges to follows in particular that lim ( n—>oo \ first r a ) r TL that lim n = 0 for /?„,<>(*) = Lagrange's form of the remainder, noticing that (c) + 1 > < 1. + (1 by using ~ a n~ 0, n—» oo for n \r\ J l t) < 1 a. Now suppose that — 1 < x < 0; the number * in Cauchy's form of the remainder satisfies — 1 < x < t < 0. Show that +'0*" 1 | < M = max(l, where \x\M, + x)*- (1 1 ), and x - 1 + t / Using Cauchy's form of the remainder, and the show that lim /? n(0 (Ar) = fact that 0. n— » oo *17. (a) Suppose that the partial sums of the sequence and that \b n is a decreasing sequence with lim ) {a n bn \ = are bounded, 0. Prove that n—* oo 00 a rt 6„ converges. Hint: ^ n= Use Abel's Lemma (Problem 21-20) to 1 (b) check the Cauchy criterion. Derive Leibniz's Theorem from (c) Prove, using Problem 15-31, that the series this result. ^= n if* is (cos nx)/n converges l not of the form 2kir for any integer £ (in which case it clearly diverges) OO *18. Suppose \a n verges, then ] is decreasing and lim a n 2 n <2 2 » ^ n— also = converges 0. (the Prove that if \ a» con " "Cauchy Condensation l OO Theorem"). Notice that the divergence of 1/n is a special case, for if l 00 00 £ ^ n= 1/n converged, then n= 1 may serve as a hint. ^ n=l 2 n (l/2 n ) would also converge; this remark 411 Infinite Series *19. Prove that (a) n= Prove that (b) an 2 £ if ^ n= £ if and l <z n 2 6n 2 converge, then ^ nA converges, then fl Suppose {a n } > decreasing and each a n is 1 converges. n=l n=l *20. 0 n £ n converges. £ n= l Prove that 0. if ^= n = verges, then lim na n be sure to use the down Hint: Write 0. fact that \a n ] the Cauchy a n con1 and criterion decreasing. is 00 *21. If 0. a n converges, then the partial ; sums sn are bounded, and lim an = =1 n tempting to conjecture that boundedness of the partial sums, It is 00 = together with the condition lim a n " This is will have true, not As a ingenuity. but n a counterexample requires a finding hint, notice that some subsequence of the partial you must somehow allow to converge; \ 0, implies convergence of this to an . =1 little sums happen, without letting the sequence itself converge. oo *22. The divergence of remarkable fact: \/n ^ n= is a particular consequence of the following 1 Any positive rational number x can be written as a numbers of the form \/n. Such a representation can finite always be found by the straightforward method of continually adding on the next number in the sequence which is not too large. For example, sum of distinct the calculation »-i=U < tt f shows that Notice that the numerators 23, 13, 11, 1 are decreasing. Prove that every positive rational x can be written in such a way by showing that the numerators in this sort of calculation must always decrease, so that eventually the remainder has numerator 1. Hint: You just have to check that l(n if + p/q < 1) less is, \/n, but than p. p/q > \/(n + 1), then the numerator of p/q - CHAPTER g^+J 23 The way UNIFORM CONVERGENCE AND POWER SERIES considerations at the end of the previous chapter suggest an entirely of looking at infinite series. infinite sums Our new attention will shift from particular to equations like = e* 1 X x^ 1! 2! + — + f. + . . . which concern sums of quantities that depend on x. In other words, we are interested in functions defined by equations of the form /(*) (in the will =AW previous example +/tW +/aW + n~ fn {x) = x l — /(n we • In such a situation {/„} 1)!). be some sequence of functions; for each x • obtain a sequence of num- bers {/»(*)}, and /(#) is the sum of this sequence. In order to analyze such functions it will certainly be necessary to remember that each sum /iW by is, + /*(*) definition, the limit of the we then define a we can new sequence • • • sequence /iW, /iW +M*)> If + +/•(*) AW +Mx) + /.(*), .... of functions Jn = fi express this fact more /(*) + ' n { ' ' + /», succinctly = lim by } by writing sn (x). n—*to For some time we shall therefore concentrate /(*) = on functions defined as limits, lim /„(*), rather than on functions defined as infinite sums. The total body of results about such functions can be summed up very easily: nothing one would hope to be true actually is instead we have a splendid collection of counterexamples. The first of these shows that even if each fn is continuous, the function / may not be Contrary to what you may expect, the functions fn will be very simple. Figure 1 shows the graphs of the functions — ! /»(*) = , j 1, 412 0 x < > x 1. < 1 413 Uniform Convergence and Power Series These functions are = continuous, but the function f(x) all lim fn (x) is not n—» oo continuous; in fact, hm/n (*) = 0, 0 < 1} x >\ | Another example of this same phenomenon < * 1 m illustrated in Figure 2; the is functions fn are defined by -1, x < - n FIGURE AW 2 = - < nx, 1 - < x < 1 n n i, n In this case, -1, and to if if x < 0, then fn (x) is eventually (i.e., for large enough n) equal > 0, then fn (x) is eventually 1, while n (0) = 0 for all n. Thus x lim/n so, W once again, the function f(x) = = [-1, *<0 (0, x U, ^>0; lim fn (x) = is 0 not continuous. n—> oo By rounding FIGURE off the corners in the previous examples differ entiable functions for produce a sequence of 3 f(x) = lim/n 0) n—* is not continuous. One \jn \ it is even possible to which the function such sequence is easy to define so explicitly: -1, / n 3 - 1, <x. n 2- These functions are differentiable (Figure A 1 1- lim/n (*) = still have < 0 = 0 *>o. -1, x * Continuity and differentiability are, moreover, not the only properties for which problems arise. Another difficulty is illustrated by the sequence {/n } shown in Figure 4; on the interval [0, \/n] the triangle of altitude n y while 4 but we 0, U, \ A l FIGURE 3), defined explicitly as follows: fn (x) = 0 for x > graph of fn forms an \/n. These functions isosceles may be 414 Infinite Sequences and Infinite Series 2n 2x, AW = { 2n 0 < - < ~ x — 2n !<*<! - 2n% 2n n 1 < 0, 1. Because this sequence varies so erratically near 0, our primitive mathematical instincts might suggest that lim fn (x) does not always exist. Nevern—» does exist for theless, this limit continuous. In fact, = moreover, /n (0) if x > 0, 0 for all all x, oo and the function f(x) then /„(*) 0, lim /„(*) n— certainly so lim other words, f(x) = lim fn {x) n— > = 0 for On all x. = dx \q fn(x) 5 fn (x) = 0; = In 0. oo the other hand, the integral oo quickly reveals the strange behavior of this sequence; FIGURE even oo have lim /n (0) n— is oo n— we so that n, eventually is = we have -J", but Thus, lim n—^oo lim/n (*) C fn(x) dx ^ r n— u u •/ » oo This particular sequence of functions behaves in a way that we really never imagined when we first considered functions defined by limits. Although it is true that f(x) = lim fn (x) for n—* oo each x in [0, 1], the graphs of the functions fn do not "approach" the graph of / in the sense of — 5, we draw a strip around /of total width 2e and below), then the graphs of fn do not lie completely within this strip, no matter how large an n we choose. Of course, for each x there is some N such that the point (x, fn (x)) lies in this strip for n > N; this assertion just amounts to the fact that lim fn (x) = f(x). But it is necessary lying close to it if, as in Figure (allowing a width of e above 71—> to choose larger and larger A^'s as x is oo chosen closer and closer to and no one 0, N will work for all x at once. FIGURE 6 The same situation actually occurs, though less blatantly, for each of the other examples given previously. Figure 6 illustrates this point for the sequence 0 X < > x < 1 1. A strip of total width 2s has been drawn around the graph oif{x) = lim /„(*). n—* If e < this strip consists of two pieces, «j which contain no points with second Uniform Convergence and Power Series coordinate equal to of each N such that since each function -5-; within this fn fails to lie strip. takes fn Once on the value 415 the graph -g-, again, for each point x there is fjx)) lies in the strip for n > N; but it is not possible to pick one TV which works for all x at once. It is easy to check that precisely the same situation occurs for each of the other examples. In each case we have a function /, and a sequence of functions {fn }, all defined on some set A, such that some (x, — Km fn {x) f(x) for all x in A. This means that FIGURE 7 for all e then > 0, and -/»(*) \f(x) for all x in A, there < I But in each case different N's must be chosen it is is some N such that if n > N 9 e. for different x% and in each case not true that > for all £ 0 there l/W - AMI < Although is some N such that for all x in A, if n > N, then «. from the this condition differs first only by a minor displacement has a totally different significance. If a sequence {/„} satisfies this second condition, then the graphs of/n eventually lie close to the graph of /, as illustrated in Figure 7. This condition turns out to of the phrase "for all x in A," it be just the one which makes the study of limit functions DEFINITION Let \fn which if ] is feasible. be a sequence of functions defined on A, and let / be a function on A. Then / is called the uniform limit of {/„ on A such that for all x in A, 0 there is some also defined > for every £ } N ifn> N, We also say that {fn } then |/(*) -/«(*)! < 8- converges uniformly to /on A, or that fn approaches / uniformly on A. As a contrast to this definition, f(x) = if we know only lim fn (x) 71—> for that each x in A, 00 we say that {/M converges pointwise to /on A, Clearly, uniform convergence implies pointwise convergence (but not conversely!). Evidence for the usefulness of uniform convergence is not at all difficult to amass. Integrals represent a particularly easy topic; Figure 7 makes it almost then } obvious that made if {fn \ converges uniformly to /, then the integral of fn can be as close to the integral of /as desired. Expressed more precisely, we have the following theorem. THEOREM 1 a sequence of functions which are integrable on [a, b], converges uniformly on [a, b] to a function /which is integrable Suppose that {/„} and that {/n } is — 416 Infinite Sequences and Infinite Series on PROOF Let [a, b]. e > Then 0. There is N such that for all n some ~ /ntol < l/to Thus, if N we have for all * in S [a, b]. N we have > n > f*f(x) dx - | b (x) fa fn dx\ = /; [f(x) - /„(,)] dx * //l/W -/»(*)!<& \ I < f\dx = £(A - a). Since this is true for any £ > 0, it follows that /0V=lim/>| The treatment of continuity is only a little more difficult, involving an "s/3-argument," a three-step estimate of \f(x) - f(x If {fn is a ti)\. sequence of continuous functions which converges uniformly to /, then there is some n such that + Moreover, since fn is -/nto| < (1) l/to (2) |/(*-M) ~/n(* J + A)| <|. continuous, for sufficiently small h (3) l/nto -/n(* \ + *)| we have <|- from (1), (2), and (3) that |/(*) - f(x h)\ < e. In order to however, we must restrict the size of \h\ in a way that cannot be predicted until n has already been chosen; it is therefore quite essential that there be some fixed n which makes (2) true, no matter how small \h\ may be + It will follow obtain it is THEOREM 2 (3), precisely at this point that uniform convergence enters the proof. Suppose that {/n and that {/n } } is a sequence of functions which are continuous on converges uniformly on [a, b] to /. Then /is [a, b] 9 also continuous on [«,*]• PROOF For each x in [a, b] with x in b) fications. (a, ; we* must prove that / the cases x = a and x is = continuous at x. We will deal only b require the usual simple modi- Uniform Convergence and Power Let e > 0. Since {/„} converges uniformly to / on [a, b], Series there is 417 some n such that \f{y) In particular, for Now /„ is all h such that x (1) |/(*) (2) |/(* continuous, so there (3) Thus, |/(* -/.Ml <| if |A| < for all +h -/B (*)| + A) is in is 8 |/nW -/„(* [a, b], [a,*]. we have <|. -/„(* some >in > + A) | <|- 0 such that for + A)| |A| < S we have <|- then 5, + A) -/Ml + A) — /„(* + h) + /„(* + h) - /„(*) + /»(*) - /(at) ~ /(*)| |/(* + A) - /»(* + A) + l/»(* + h) - /»(*)! + |/»to s s s < + + I 3 = < |/(* | I , , 3 This proves that / FIGURE is continuous at *. | 8 After the two noteworthy successes provided by Theorem 1 and Theorem 2, the situation for differentiability turns out to be very disappointing. If each not necesfn is differentiate, and if }/n } converges uniformly to /, it is still there is a that shows Figure 8 sarily true that / is differentiate. For example, sequence of differentiable functions {/n } which converges uniformly to the 418 Infinite Sequences and Infinite Series function f(x) = Even |*|. if / is differentiable, this is not at all surprising if we may reflect that a smooth function can be approxi- oscillating functions. = fn(x) For example (Figure = fn(x) n— x = 9), if -sin(rc 2#), n then {fn } converges uniformly to the function /(*) and lim not be true that = Hm /.'(*); /'(*) mated by very rapidly it = 0, but 2 n cos(n x), 2 n cos(w *) does not always exist (for example, it does not exist if oo 0). n FIGURE 9 Despite such examples, the Fundamental Theorem of Calculus practically guarantees that some sort of theorem about derivatives will be a consequence of Theorem 1; the crucial hypothesis is that {//} converge uniformly (to some continuous function). theorem 3 Suppose that {fn } is a sequence of functions which are differentiable on [a,b\ and that {/n } converges (pointwise) to /. Suppose, moreover, that {fn} converges uniformly on [a, b] to some continuous function g. Then / is differentiable and /'(*) - lim /„'(*). n— PROOF Applying Theorem 1 to the interval = oo [a, x], lim [/„(*) = fix) - we see that for - /„(*)] f(a). each x we have 419 Uniform Convergence and Power Series Since 0p continuous, is = follows that f{x) = g(x) lim fn (x) for x—* the interval [a, b]. x in all *> | that the basic facts about uniform limits have been established, Now clear it how it is to treat functions defined as infinite sums, /« =/iW +/2W + /»(*) + * ' * • This equation means that /(*) = + lim/i(jr) n—> oo • • • +/„W; our previous theorems apply when the new sequence /i,/i+/2,A +/«+/* • we converges uniformly to /. Since this is the only case ested in, we single it out with a definition. DEFINITION The £ series n= is shall ever be inter- (more formally: the sequence {fn /„ converges uniformly \ 1 uniformly summable) to / on A, if the sequence +/»+/• • • • converges uniformly to / on A. We series; convergent can now apply each of Theorems 1, 2, and 3 to uniformly corollary. common in one stated be may results the oo COROLLARY Let ^ (1) (2) fn converge uniformly to /on [a, b]. each fn is continuous on [a, b], then / is continuous on If / and each fn is integrable on [a, b], then If [a, b\ /:/=!/> n= l 00 Moreover, if uniformly on T /n converges (pointwise) to [a, b] to / on fix) - ^ ft], some continuous function, then 00 (3) [a, for aU * in [a ' bl and £ fn converges 420 Infinite Sequences and Infinite Series PROOF If (1) each/n is continuous, then so uniform limit of the sequence f h continuous by Theorem 2. (2) Since /i, /i +/ +/ +/ /i 2, 2 from Theorem follows is +f /i . 3} each /i • • + /2 fi + /n, and /is the • +/s, . . . so /is , converges uniformly to . . 2, + /, it that 1 ///.to /.'</, + ••+/.) -«-(/>+•• +/;/.) 00 Each function (3) ' • ; + /i + /»', • • + /n • and //, with derivative differentiable, is +/ /i' 2 ', / x ' + /,' + /,', . . con- • verges uniformly to a continuous function, by hypothesis. It follows from Theorem 3 that /'(*) = lim [//(*)+ n— • • • +/„'(*)] oo 00 = 2 /.'w. = n At the moment to predict uniformly. i 1 this corollary is not very useful, since it seems quite difficult when the sequence /i, /i + / /i + /2 + /3 will converge The most important condition which ensures such uniform con2 , , . . . is provided by the following theorem; the proof is almost a triviality because of the cleverness with which the very simple hypotheses have been vergence chosen. THEOREM 4 (THE WEIERSTRASS M-TEST) Let {fn be a sequence of functions defined on A, and suppose that sequence of numbers such that \ M < \fn(x)\ {M n } is a for all x in A. n CO Suppose moreover that M T « = n converges. Then for each x in £ A the series i oo 00 fn(x) converges (in fact, it converges absolutely), and n~l uniformly on A fn converges ^ n== 1 to the function 00 /(*) = I= n PROOF For each x in A the series ^ /.(*). 3 |/nW| converges, by the comparison test; Uniform Convergence and Power Series consequently ^= n /„(*) converges (absolutely). Moreover, for all x in 421 ^ we have 1 00 i/w - [/iw + • • • +/-WH = I < n 2 /»w f I 1/nWI = 7?+l 00 < fM = i(2*i M n. 00 00 Y Since I M n converges, the desired, by choosing FIGURE 10 number Y Af n can be made as small as TV sufficiently large. | following sequence {/n } illustrates a simple application of the Weierinteger (the strass M-test. Let {*} denote the distance from x to the nearest define Now Figure = in 10). illustrated is {x} graph of /(*) The (a) /.W -iJ: drawings functions f x and /2 are shown in Figure 11 (but to make the n n has been functions simpler, 10 has been replaced by 2 ). This sequence of clearly applies: defined so that the Weierstrass M-test automatically The /.to = 1 |/«Ml<^ for all*, (b) FIGURE * 00 11 and ^ V10 n converges. Thus £ since each /« fn converges uniformly; continuous, the corollary implies that the function is ! 422 Infinite Sequences and Infinite Series is also continuous. Figure 12 fi + * * • + /n As . shows the graph of the first few partial sums graphs become harder and harder to n increases, the so draw, and the infinite sum ^ n— fn shown by the quite undrawable, as is follow- 1 ing theorem (included mainly as an interesting sidelight, to be skipped if you find the going too rough). THEOREM 5 The function /w = tro"» — n l continuous everywhere and differentiable nowhere is PROOF {10 "* } We have just shown that /is continuous; this is the only part of the proof which We will prove that / is not differentiable at a, for uses uniform convergence. any {h m <z, by the straightforward method of exhibiting a particular sequence for which approaching 0 } f(a lim + hm) m-K» does not < a< 0 exist. It ~ f{a) hm obviously suffices to consider only those numbers a satisfying 1. Suppose that the decimal expansion of a = a Let h m = for these is .... = -10~ m if am = 0.<Zi<22<23tf4 10~ m if a m ^ 4 or 9, but let k m two exceptions will appear soon). Then f(a + h m - f(a) \lO n (a ) + h m )} - ± infinite series i<r-*[{io*(a really a finite is n {10 a} ±10~ m 10 This 4 or 9 (the reason + sum, because h m )} if n - > {I0 n a}l n m, then 10 h m is an integer, so {10 On the other hand, for n n 10 a 10 n (a (in hm n + hm) = = (a + h m )\ - < m we can integer integer + + n \10 a] 9). 0. write 0.a n+ ia n+2 a n +3 . . • 0.a n+ ia n +2a n +3 . . . order for the second equation to be true ~ — 10 -m when a m = - Now it is am (a m • • • . ± • 1) • essential that suppose that «m . 2 • • • we choose 423 Uniform Convergence and Power Series Then we also have O.0 n +i<Z n + 2 0n+3 + (flm - . . 1) • = • ^i m = n + 1 the second equation is true = 4). This means that ±10*-™, {10"(fl + A»)| - {10"*} = (in the special case hm • because we chose -10" m when a m and exactly the same equation can be derived when 0.a n+ ia n -^2a n+ 3 Thus, for n < m we have m- n [{\0*(a + k m)\ - {10M] = ±1\0 . . . h > In other words, + k m - f(a) f(a ) hm is /!+/.+/• to sum of m - 1 numbers, each of which is ± 1. Now adding +1 or The sum of m a number changes it from odd to even, and vice versa. the ±1 numbers each is is therefore an even integer if m is odd, and an odd — 1 - 1 integer if m even. Consequently the sequence of ratios f(a + hm ) -f(a) hm cannot possibly converge, since nately odd and even. a sequence of integers which are alter- it is | Weierstrass M-test is an In addition to its role in the previous theorem, the behaved. We will give well very are ideal tool for analyzing functions which special attention to functions of the form /l+/.+/l+/4 a n (x /(*) which can also fn (x) = a n {x on powers of simplicity, (* we a) n , be described by the equation /to for - sum, of functions which depend only centered at a. For the sake of is called a power series series centered at 0, power on concentrate usually - a) n - a), will = . Such an infinite eo /(*) = £ a n x\ n=0 One especially important group of power n =0 series are those of the form 424 Infinite Sequences and Infinite Series where / is called the some function which has derivatives of all orders at a; this series Taylor series for / at a. Of course, it is not necessarily true that tea (*-«)•; /W- =0 71 this equation holds only when is R n Ax — the remainder terms satisfy lim n—+ We already know that a power series ^ ) 0. as a n x n does not necessarily converge n=0 for all x. For example the power series X3 converges only for |#| < #2 , #3 *4 3 4 2 —1 < converges only for X7 x for x < = 1 . 0, . ^ xh for any x . It is even possible to produce a power For example, the power series series n + l)!(* n+1 = ) + (n nix unbounded ^ 0; indeed, the ratios (n are , 5 i\x does not converge for x , while the power series 1, # which converges only Xb t 9* 0. If \)x a power series ^ a nx n does converge for n=0 00 some for theorem 6 |*| #0^0 < however, then a great deal can be said about the series ^ a nx n |# 0 |. Suppose that the series /Oo) converges, and let <z = ^ ^n^0 be any number with 0 n < a < \x 0 \. Then on [ — a, a] the series 00 /(*) = £= n 0 converges uniformly (and absolutely). Moreover, the same is true for the i Uniform Convergence and Power Series 425 series na n x " «= / Finally, and differentiable is = J n= /'(*) for all x with < |x| nawX ^ 1 1 |* 0 |- 00 PROOF Since £= 71 tf n* 0 n converges, the terms is * in is [ — a, |<z«* n | = < = n | then a], = 0. Hence they are surely \x \On\ • i^l \ < M < | for all n. so \a\, |«n| * |*0] \ \ (this Xq the clever step) is I 0 < Af \a/x 0 n n ' \a n \ < |*| |x 0 • \a n I But approach some number Af such that |a n *o if n 0 bounded: there Now n*0 tf I so the (geometric) series 1, n converges. Choosing Af =0 \a/x 0 n as the number Af n in the Weierstrass Af-test, \ CO it follows that ^ a nx n converges uniformly on [-a, a]. n=0 op To prove the same assertion for g(x) = ^ n= < ._ < Since \a/x 0 \ < 1, «W W Af \a • u . 1 n_1 notice that wa n ^ l n-l| |» „ ±r | a n «!- the series Af = M\ A j n —a h„4 converges (this fact was proved in Chapter 22 as an application of the ratio 426 Infinite Sequences and Infinite Series test). Another appeal to the Weierstrass M-test proves that ^ na n x n 1 con- n-l [— a, verges uniformly on a). Finally, our corollary proves, = /'(*) = £ g(x) n Since that g first = na n x n~ l Theorem \x\ 6, < |x 0 |. for x in [-a, a]. 1 we could have chosen any number for all x with continuous, and then that is < with 0 a < a \x 0 \, this result holds I applied to the infinite series = sin x x7 3! 5! 7! X4 X* 2! 4! 6! X x^ *9 . • • • , , 9! X8 , 1 = eX x^ X2 — cos X x* x x^ +v+v+ 1 — • • ' , 8! x^ + Ti 4i -'-> + which are expected. Each of these converges for any hence the conclusions of Theorem 6 apply for any x: yields precisely the results #o, —+—= = - — + —- —-+••• = + — + — +*•' = = sin'(*) „ cos'(x) , - 1 5x i 3x 2 2x 4x* 6x> 21 41 61 2x exp'(*) 3>x = 1 - = - sin x, 2 - + log(l ~ ^3 x x) the situation is only slightly + x* - —+ x« • ' 7 5 also converges for x2 ^7 j£\> 1 3 arctan'M x, the series arctan x 1, it cos exp(x). For the functions arctan and f(x) = • 1 more complicated. Since converges for x 0 • • < \x\ + • 1, = • and —J— +x 1 In this case, the series happens to converge for x formula for the derivative is not correct for x = = —1 1 for |x| 2 or x also. < 1. However, the = — 1; indeed the series - x + x* - * + = — 1 Notice that this does not contradict Theorem = 1 and x which proves that the derivative diverges for x 6, \x\ < \xo\. 6 2 1 • • • . is given by the expected formula only for . Uniform Convergence and Power Series 427 Since the series + ,)^-| + |-|+| log(l = converges for x 0 1 In = +* 1 1, it also log'(l + x) converges for = - 1 +x 2 x < |#| *3 and 1, + • " • does not converge for x this case, the original series = differentiated series does not converge for x for = —1 power its power series where the derivative derivative, at the points ; < 1. moreover, the 1 All the considerations which apply to a apply to \x\ will automatically is represented by a series. If 00 = /(*) I= 71 converges for all (—R,R), then Theorem x in some interval 6 implies that n na n x ~i = y f{x) a n x» 0 n-l for all * in (—R 9 Theorem R). Applying 6 once again we find that 00 = /"(*) J n= »(» - IK*"-2 , 2 and proceeding by induction we find that oo /<*>(*) = J n(n - 1) Thus, a function defined by a power ( — R,R) is • • (« • series - k + IK**"*. which converges in some interval automatically infinitely differentiable in that interval. Moreover, the previous equation implies that /<*>(0) - k\a k3 so that /<*>(0) A! In other words, a convergent power of the function On this Taylor which it series centered at happy note we could series. A 0 is always the Taylor series at 0 defines. easily end our study of power careful assessment of our situation will reveal and some un- series explained facts however. The Taylor Taylor and exp are as satisfactory as we could and can be differentiated term-by-term for all series of sin, cos, they converge for all #, series of the function f(x) = log(l + x) is slightly less pleasing, desire; x. The because 428 Infinite Sequences and Infinite Series —1 < converges only for it x < but 1, a necessary conse- this deficiency is quence of the basic nature of power series. If the Taylor series for / converged for any x 0 with |# 0 > 1, then it would converge on the interval ( — 1* 0 |*o|), and on this interval the function which it defines would be differentiable, and thus continuous. But this is impossible, since it is unbounded on the interval ( — 1, 1), where it equals log(l + x). The Taylor series for arctan is more difficult to comprehend there seems [, I — be no possible excuse for the refusal of to This mysterious behavior = f(x) + 1/(1 is exemplified even = 7-7— 2 = /(*) +x 1 \x\ > more ~ 1 i ; *2 4 +* - *6 + x* of / ~ all. ' question always dangerous, since is unsympathetic answer: are ! In it happens because it > 1. ' the next best is is given by ' • Why? What unseen obstacle prevents the Taylor series from extending past this sort of \x\ by the function strikingly The Taylor polynomial the Taylor series does not converge at 1 converge when x 2 ), an infinitely differentiable function which thing to a polynomial function. If this series to 1 and —1? Asking to settle for an we may have happens — that's the way things does happen to be an explanation, but this explanation this case there is about real can be answered intelligently only when placed in a broader context. It will therefore be necessary to devote two chapters to quite new material before completing our discussion of Taylor series in Chapter 26. is impossible to give at the present time; although the question numbers, it PROBLEMS 1 For each of the following sequences {/n }, determine the pointwise limit (if it exists) on the indicated interval, and decide whether {fn of {fn } \ converges uniformly to this function. (i) /nW = Vx, (ii) fn(x) x o, (iii) /«(*) (iv) f(x) e on [a, b], and on R. c j x = e e (v) n z —n = < — x —nx* 5 -x* j f(x) on R. n 2- Find the Taylor (i) fix) /(*) = each of the following functions. 1 = x (ii) series at 0 for — a 9^0. a log(* - a), a* 0. 429 Uniform Convergence and Power Series (iii) 1 = /(*) - V — ( 3. v) = /(*) arcsin Find each of the following (ii) _ i - ^! (iii) i v If FT = /W power 5. In _ xs —+— 3-2 (1VJ ^ + . . —+ • • 0 • - +* -* + 3 2 * ? * Problem (See 10.) [ = /(0) 1, find / (A) (0). Hint: Find the series for /. problem we deduce the binomial this < * Hint: Differentiate. 1. series (1 + x) a = without 1 all fact established in part (a) of that problem —the series , =0 work of Problem 22-16, although we the *n ^ n |*| • 1 ' 4 and 1 W < for • L±... 5 ^ for * *)A What is 5-4 J , Hint: . § 4-3 i sums. infinite 3-2^4-3 (sin (Use Problem 19-6.) . y«4 V>3 + *3 *)- 1/2 x. y.2 4. - (1 x 1 will use a x n does ^ n=0 converge for (a) Prove that (b) Now f(x) < |*| (1 1. + *)/'(*) = af(x) for < [x| 1. show that any function / satisfying part (a) is of the form = c(l + x) a for some constant c, and use this fact to establish the binomial series. Hint: Consider g(x) = f(x)/(l + x) a . CO 6. (a) Suppose that /(*) = anX " converS es for £ a11 * in some interval n=0 (-R,R) and <z (b) n = (If 0. that /(*) = 0 for all x in (-/?, R). Prove that each you remember the formula = Suppose we know only that /(*») lim x n = 0. Prove again that each a n 71— for a n this 0 for = is easy.) some sequence 0. {xn } with Hint: First show that 00 f(0) = ao = 0; then that /'(0) = ai = 0, etc. 1/xt sin 1 /x for * 5* 0, then / cannot This result shows that if f(x) = e~ It also shows that a function series. possibly be written as a power be 0 for x < 0 but nonzero for cannot series power defined by a the motion of a particle 0 describe cannot series power thus a x > — which has remained at rest until time 0, and then begins to move 430 Infinite Sequences and Infinite Series = £ n a n x and g(x) = V b n x n converge for all x n=0 n=0 in some interval containing 0 and that f(t m ) = £(/m ) for some Suppose that /(*) (c) = sequence {tm } converging to 0. Show that a n b n for each n. In particular, a function can have only one representation as a power series centered at 0. 00 7. Prove that if = /W ^ n= a n* n an even function, then is <z n = 0 for n odd, 0 *8. and if / is an odd function, then a n = 0 for n even. Recall that the Fibonacci sequence {a n \ is defined by ai tf n+ i = an (a) Show (b) Let + = = a2 1, a n -i. < 2. B_l = that a n+1 /a n 00 /to = 2 n= Use the (c) *n* 1 +*+ if < |4 • • • . |*| < 1/2. = 2 +x- 1 Hint: This equation can be written /(*) - xf(x) - x 2f(x) = 1. Use the partial fraction decomposition for I/O 2 x - 1), and the power series for l/(x - a), to obtain another power series for /. + It follows from Problem 6 that the two power must be the same. Use this fact to show that an 9. + 1/2, then a: (e) 3* 3 ratio test to prove that /(*) converges if Prove that fix) (d) + 2* 2 1 Show that -1 < x < — series obtained for / ~ r V5 power series for /(*) = log(l — x) converges only and that the power series for g (x) = log[(l + x)/(l - the for 1, x)] converges only for a: in ( — 1, 1). * *10. Suppose that £ 00 a n converges. We know that the series f(x) = Y n =o a nx n n =o must converge uniformly on [-a, a] for 0 < a < 1, but it may not converge uniformly on [- 1, 1]; in fact, it may not even converge at the point -1 (for example, if f(x) = log(l x)). However, a beautiful + theorem of Abel shows that the series does converge uniformly on [0, 1]. 00 Consequently, / is continuous on [0, 1] and, in particular, ^ an = Uniform Convergence and Power Km 7 < e, n a nx . then Prove Abel's Theorem by noticing that if \a m \a mx m + • ' + 0n* n < | + Lemma by Abel's e, 431 Series • • • + an \ (Problem 21-20). 11. A sequence {a n } is called Abel summable if hm £ exists; Problem necessarily Abel summable. Find which is not summable. Hint: but a sequence which is Abel summable, find one which does not conyou until Look over the list of Taylor series is continuous at 1. represents it function the verge at 1, even though 10 shows that a 12. summable sequence is Show 3 we assumed only that {/» converges pointwise to/. converges actually that ensure hypotheses n {/ remaining that the uniformly to /. necessarily conis a sequence of bounded (not (a) Suppose that |/n to / on [a 9 b]. uniformly tinuous) functions on [a, b] which converge In Theorem } } 13. J Prove that / is bounded on [a, b\ converge Find a sequence of continuous functions on [a, b] which b]. [a, on function unbounded pointwise to an pointwise Suppose that/is differentiate. Prove that the function/ is the know already we (Since functions. continuous of sequence limit of a (b) *14. 15. 16. example examples of discontinuous derivatives, this provides another continuous.) not where the pointwise limit of continuous functions is to the Find a sequence of integrable functions {/„} which converges irrathe on and rationals 0 the (nonintegrable) function / that is 1 on points. few at a except tionals. Hint: Each fn will be 0 the integral; This problem outlines a completely different approach to learned integrals about facts any use to unfair consequently, it is previously. (a) Let s some be a step function on partition where Si is {* 0> [a, ...,'»} b\ so that of constant on Define [a, b]. the (constant) value of s is on s s as (/ t-_i, ft) - £ f* for i=i (U-i, k). Show that this definition does not depend on the partition {t 0 b] if it is the uniform (b) A function / is called a regulated function on [a, that in this limit of a sequence of step functions {* n on [a, b\ Show m, n for that > we have such case there is, for every e > 0, some , } \s n (d) (e) (x) - Sm(x)\ < S for all x in [a, . . N N (c) . b\ Show that the sequence of numbers f* s n { } will be a Gauchy sequence. on [a, b] Suppose that {*„} is another sequence of step functions there is every 0 £ > for that Show to uniformly /. which converges [a, b\. in x an N such that for n > N we have \sn (x) - t n (x)\ < £ for & we can Conclude that lim f s n - lim /* t n This means that . h define *(f) [* f to be lim f sn for any sequence of The only remaining converging uniformly to /. functions are regulated? Here is step functions question is: {j- n } Which a partial answer. Prove that a continuous function is regulated. Hint: To find a step s on [a, b] with \f(x) — s(x)\ < 6 for all x in [a, b\ consider all y for which there is such a step function on [a, y]. function Find a sequence {fn approaching / uniformly on [0, 1] for which lim (length of fn on [0, 1]) ^ length of /on [0, 1]. (Length is defined in \ n—* oo Problem 15-32, but the simplest example length of whose graphs will be obvious.) will involve functions the COMPLEX NUMBERS CHAPTER With the exception of the last few paragraphs of the previous chapter, this book has presented unremitting propaganda for the real numbers. Neverthenot every polynomial less, the real numbers do have a great deficiency is the fact that no example function has a root. The simplest and most notable 2 that long ago severe is so deficiency =0. This number x can satisfy x + 1 — mathematicians felt the need to "invent" a number i with the property that was quite mysterious: {2 i = o. For a long time the status of the "number" i 2 1 = 0, it is nonsensical to say "let since there is no number x satisfying x 2 Nevertheless, admission of the 1 = 0." the number satisfying i i be + + + simplify greatly i to the family of numbers seemed to bi algebraic computations, especially when "complex numbers" a computation arithmetical (for a and b in R) were allowed, and all the laws of enumerated in Chapter 1 were assumed to be valid. For example, every "imaginary" number + many quadratic equation + ax 2 can be solved formally x = -b +c= bx 0 ^ {a 0) to give + Vb 2 Aac 2 -b - Vb " = x or 2a 2a Ifb 2 — Aac > 0, Aac these formulas give correct solutions; are allowed the formulas seem make to sense in when complex numbers all cases. For example, the equation + x+\ = x2 0 has no real root, since x2 But the formula +x+ 1 = (x for the roots of x + +f 2 i) > 0, for all x. a quadratic equation suggest the "solutions" = and x = ; 2 2 if we understand these to mean , V3. i 2 is 433 • (-1) - V?> V-l • = VI i, then numbers would be 1 It V3 2 . and — 1 V3 — 2 2 . i. not hard to check that these, as yet purely formal, numbers do indeed 434 Infinite Sequences and Infinite Series equation satisfy the +*+ *2 It is 1 = 0. even possible to "solve" quadratic equations whose coefficients are themcomplex numbers. For example, the equation selves +x+ + - x2 1 0 i ought to have the solutions = * ~ -1 ± Vl - — where the symbol square —3 — is 4(1 + 0 = -1 ± V-3 - 4i 2 ' means a complex number a 4i + fit whose In order to have 4i. a2 - P2 + #) = 2 (a + = -3 - 2aPi 4i we need a 2 - 0 2 = -3, 2a0 = -4. These two equations can two possible solutions: be solved for real a and easily a = 1 , 0.-2 Thus the two "square no reasonable way and which real x > to decide — 0, in roots" of — 4z ; which case and in fact, there are a — — /3 —3 — /3; = 4i are 2. — 1 2i and —1 + which one of these should be called the conventional usage of V* 2 i. There V—3 — is 4*', V# makes sense only for denotes the (real) nonnegative root. For this reason, the solution X must be understood ^ With as this — — _+ r as 1 , V-3 - -1 ± where we r is 4i 2 an abbreviation understanding you can — = for: one of the square roots of —3 — 4i. arrive at the solutions + l-2i = * - -l * = — 1 — 21 easily check, these + 2» = 1 numbers do provide formal equation x2 . +x+ + = 1 i 0. solutions for the Complex Numbers For cubic equations complex numbers are equally useful. 435 Every cubic equation ax z with real coefficients divide ax z + bx 2 + + bx 2 a, b, c, cx + + cx + d = and d by x d, 0 (a 5* 0) we know, a real root a, and if we obtain a second-degree polynomial has, as —a we whose roots are the other roots of ax z + bx 2 + cx + d = 0; the roots of this second-degree polynomial may be complex numbers. Thus a cubic equation will have either three real roots or one real root and 2 complex roots. The existence of the real root is guaranteed by our theorem that every odd degree equation has a real root, but it is not really necessary to appeal to this theorem (which is of no use at all if the coefficients are complex); in the case of a cubic we can, with sufficient cleverness, actually find a formula for all the The following derivation is presented not only as an interesting illustra- equation roots. tion of the ingenuity of early mathematicians, but as further evidence for the importance of complex numbers (whatever they To most general cubic equation, only equations of the form solve the x* + bx 2 it + cx + d - may be). obviously suffices to consider 0. 2 even possible to eliminate the term involving x by a let we If forward manipulation. It is , x = y — fairly straight- b -» then so 0 = xz + bx 2 + cx + d The right-hand side now contains no term with^ 2 If we can solve the equation for y we can find x; this shows that it suffices to consider in the first place only . equations of the form xz + px + q - 0. z In the special case p = 0 we obtain the equation x = — q. We shall see later root, in fact it has three, so on that every complex number does have a cube on the other hand, case solutions. The three 0, that this equation has p ^ 436 Infinite Sequences and Infinite Series requires quite an ingenious step. Let P_ 3w Then 0 = +px + *3 w° 3w 2p — +p(w = (w - q +— 3wp 2 2>w p 9w + 27q(w*) q 2 z — bw VF s 27 w* 2 + p ~ 3w V q This equation can be written 27(> 3 ) 2 which is a quadratic equation Thus in w z -p = 3 (!!). = -27? ± V(27)V 2 2 Remember • ^ 4 • Hp* + 27 * 4 + r, where r is a square root of 4 + • 27 can therefore write ^9 this + 27 that this really means: 2 We ~ 0, equation means that square root of q 2 /4 are substituted into +p 3 if is 'A 97' 27 7 some cube root of /27. This — q/2 + r is some when these where r, but allows six possibilities for w, (*), yielding 3/" = ^Zl ± * 2 Jt^I M p 3 27 ^ it _„ 2 " V 27 4 turns out that only 3 different values for x will be obtained surprising feature of this solution arises all when we ! An even more consider a cubic equation of whose roots are real; the formula derived above may still involve com- plex numbers in an essential way. For example, the roots of xz are 4, -2 + - 15* - VJ, and -2 - V3. On 4 = 0 the other hand, the formula derived Complex Numbers above (with p = — 15, x = = — 4) q one solution gives as -15 + V4 - </l 125 3 + = ^2 ^2 + V4 - ' 125 15 + 11/ 437 3 + </i • 11/ Now, (2 so + +3 2 +3 2 + 8 + 122 — 6 — 2 + Hi, + 11 2 + /.Thus, for one solution of the equation = = = 3 /) one of the cube roots of 2 23 2 • • / / • 3 2 / 2 is £ we obtain x = + + 2 i + 6 3/ = + +6 + 2 = / t 3i 3t 45/ + 36 6 3* - 90 + + 2 = - 6 15 9 4(0. found if the other cube roots of 2 + 11/ are The known. The fact that even one of these real roots is obtained from an expression which depends on complex numbers is impressive enough to suggest that the use of complex numbers cannot be entirely nonsense. As a matter of fact, the formulas for the solutions of the quadratic and cubic equations can be other roots can also be interpreted entirely in terms of real numbers. Suppose we agree, writing the real for the number + bi) + (a + bi) (a to write all i 2 as 0 that + + d)h {ad + c) - (ac bd) (b + + If. as a The + bi, laws bc)i. like (1 + 2f) • (3 + = 1+7/ If) for the two equations be regarded as simply an abbreviation = 3 - 2 1-1+2.3 = 1 The / = — 1 show + {a complex numbers number and the Oi relation + di) = + di) - (c (c ' may + a as a and the of ordinary arithmetic Thus, an equation moment, • • 1 2 solution of the quadratic equation ax 1, 7. + bx + c = could be paraphrased as follows: u2 _ v2 UV = 0, If ' (i.e., if = (ii b 2 + _ vi) 4aCj 2 « b 2 - 4*;), 0 with real coefficients 438 Infinite Sequences and Infinite Series + -b «[( •H It is +c= 0, 2a then writing u + -b u not very hard to check this assertion about real numbers without down a single C£ z'," but the complications of the statement itself should convince you that equations about complex numbers are worthwhile as abbreviations for pairs of equations about real numbers. (If you are convinced, try paraphrasing the solution of the cubic equation.) If intend to use complex numbers consistently, however, some reasonable sary to present One it is still we not really going to be neces- definition. been implicit in this whole discussion. All mathematical complex number a + bi are determined completely by the real numbers a and b; any mathematical object with this same property may reasonably be used to define a complex number. The obvious candidate is the ordered pair (a, b) of real numbers; we shall accordingly define a complex number to be a pair of real numbers, and likewise define what addition and multiplication of complex numbers is to mean. possibility has properties of a DEFINITION complex number is an ordered pair of real numbers; if z = (a, b) is a complex number, then a is called the real part of z, and b is called the imaginary part of z. The set of all complex numbers is denoted by C. If (a, b) and (c, d) are two complex numbers we define A b) + (a, b) (The + and while the + - (c, d) (c, d) = — (a + (a ' cy b — c + b d) d, • a + d • b • c). appearing on the left side are new symbols being defined, appearing on the right side are the familiar addition and * and multiplication for real numbers.) When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is taken seriously this all. is (a, 0) (a, this + 0) • is — a real number is not a pair of real numbers, after only a minor annoyance, however. Notice that not true This difficulty (*, 0) (b, 0) = = {a + (a b b, - 0 0 + 0) 0, a = • 0 (a + + shows that the complex numbers of the form same with respect to addition and multiplication b } 0), 0 b) {a • b, 0); 0) behave precisely the complex numbers as real (a, of = Complex Numbers numbers do with their own addition and multiplication. For this reason will adopt the convention that a + bi (a, 0) will be denoted simply by now notation for complex numbers can definition = i (a, b) You may feel if we familiar one more it is (a, 0) (a, 0) + a + + = -1 (the last equality sign (0, b) (*, 0) • (0, 1) bi. was merely an elaborate device for defining form a + bi" That is essentially a firmly established prejudice of modern mathematics that new as "expressions of the must be defined is = = 0) that our definition complex numbers theless, it be recovered The (0, 1). Notice that i 2 = (0, 1) (0, 1) = (-1, depends on our convention). Moreover objects a. made. is DEFINITION correct; 439 as something specific, not as "expressions." Neverwere sincerely worried interesting to note that mathematicians about using complex numbers until the modern definition was proposed. Moreover, the precise definition emphasizes one important point. Our aim in introducing complex numbers was to avoid the necessity of paraphrasing statements about complex numbers in terms of their real and imaginary parts. This means that we wish to work with complex numbers in the same way that we worked with rational or real numbers. For example, the solution of the = w — p/3w, so we want to know that w was found by solving a quadratic equation, cubic equation required writing x 1 /w makes sense. Moreover, 2 which requires numerous other algebraic manipulations. In short, we are likely to use, at some time or other, any manipulations performed on real numbers. We certainly do not want to stop each time and justify every step. Fortunately this is not necessary. Since all algebraic manipulations performed on real numbers can be justified by the properties listed in Chapter 1, it is only necessary to check that these properties are also true for complex numbers. In most cases this is quite easy, and these facts will not be listed as formal theorems. For example, the proof of PI, [(*, b) + (c, d)} + = (e,f) (a, b) + [(c, d) + (ej)] requires only the application of the definition of addition for complex bers. The left side becomes ([a and the right side + c] + e, [b + d]+f) 3 becomes (a + [c + elb + [d+f]); num- is true for real numbers. It is a good idea to check P2-P7 and P9. Notice that the complex numbers playing the role of 0 and 1 in P2 and P6 are (0, 0) and (1, 0), respectively. It is not hard to figure these two are equal because PI out what a — (a, b) but the multiplicative inverse for is, ^ little trickier: if {a, b) + then a 2 (0, 0), b ^ 2 This fact could have been guessed in two ways. it is The = b)-(x,y) (*, (a, b) 0 and To find (x, required in P8 is y) with (1,0) only necessary to solve the equations = solutions are x reason that if \/{a + a Once a/{a 2 bi) + ax — bx + ay 2 b ), = — by 1, 0. = -b/(a y means anything, then 1 1 + + a bi — — a a bi 2 + a a bi 2 It is also possible to ). should be true that it bi __ b 2 — bi + b 2 the existence of inverses has actually been proved (after guessing the by some method), it follows that this manipulation is really valid; it is the easiest one to remember when the inverse of a complex number is actually being sought it was precisely this trick which we used to evaluate inverse — 15 6 + 3* = ~ _ ~ 15 6 # 6 + 90 3i - ' 6 - 3i 3i 45i " 36 + 9 Unlike P1-P9, the rules P10-P12 do not have analogues: it is easy to prove that there is no set P of complex numbers such that P10-P12 are satisfied for all complex numbers. In fact, if there were, then P would have to contain 1 (since 1 = l 2 ) and also -1 (since -1 = 2 z" ), and tHis would contradict P10. The absence of P10-P12 will not have disastrous consequences, but it does mean that we cannot define z < w for complex z and w. Also, you may remember 1^0. Forthat for the real numbers, P10-P12 were used to prove that 1 to this reduced be can tunately, the corresponding fact for complex numbers + one: clearly (1, 0) + (1, 0) ^ (0, 0). Although we will usually write complex numbers in the form a + bi, it is worth remembering that the set of all complex numbers C is just the collection of all pairs of real numbers. Long ago this collection was identified with the plane, and for this reason the plane is often called the "complex plane." The horizontal axis, which consists of all points (a, 0) for a in R, is often called the Complex Numbers and the real axis, vertical axis called the imaginary axis. is Two 441 important definitions are also related to this geometric picture. + If z = x of z is defined as a complex iy is number = z x - and the absolute value or modulus +y 2 > 0, so \z\ iy, of z - x that V# + 7 2 2 is defined unambiguously; +y it 2 .) • Geometrically, z FIGURE defined as is denotes the nonnegative real square root of x 2 z then the conjugate z real), = Vx 2 +y\ \z\ (Notice that x 2 and y (with x 1 distance from z to simply the reflection of z in the real is 0) (Figure (0, 1). while axis, \z\ is the Notice that the absolute value notation for complex numbers is consistent with that for real numbers. The distance between two complex numbers z and w can be defined quite easily as \z — w\. The following theorem lists all the important properties of conjugates and absolute values. THEOREM 1 Let z and w (1) z (2) z = = (3) z be complex numbers. Then z. z and only if number if z real is (i.e., is of the form a + 0/, for some real a). +w= z + w. -(z). (4) w = z w. = (z)~\ if z 9*0. = z-z. \z\ \z-w\ = |*| H|* + H < 1*1 + M. (5) z (6) F - • 1 2 (7) (8) (9) PROOF • Assertions (1) and (2) are obvious. Equations (3) and (5) may be checked by straightforward calculations and (4) and (6) may then be proved by a trick: 0 1 Equations The only (7) = = = J = and difficult 0 (8) z z + (- *) = z_+ -z, = • may 2 • z" 1 so so , -z = = -(*), _1 (*) . proved by a straightforward calculation. is (9). This inequality has, in fact, but the proof will be repeated here, using also be part of the theorem already occurred (Problem 4-8), slightly different terminology. It is clear that equality holds in (9) that (9) is true if z = \w for any real if z = 0'or number X w = 0. It is also easy to see (consider separately the cases 442 Infinite Sequences and Infinite Series X > and X < 0). Suppose, on the other hand, that z ^ Xw X, and that w ^ 0. Then, for all real numbers^ X, 0 for any real number 0 (*) < \z - = = Xw\ 2 — (z (z = + = XV| + zw wz + Notice that wz + 2 — Xw) Xw) — X(wz + -zw) 2 X(wz + zw). \z\ Xw) Xw^ X 2 ww • (z • (z real, since is + zw = wz + zw = wz zw = wz + zw. Thus the right side of (*) is a quadratic equation in X with real coefficients and no real solutions; its discriminant must therefore be negative. Thus (wz it follows, since wz + + zw and zw) 2 \w\ • \z\ + From — 4\w\ 2 \z\ - 2 < 0; are real numbers, < zw) 2\w\ • and \w\ * |z| > 0, that \z\. this inequality it follows that \z + w\ 2 = = < = + w) (z + w) + \w\ + (wf + + |w| + 2\w\ (kl + l^l) (z \z\ \z\ - 2 2 2 2 ' zw) \z\ 2 ? which implies that \z z + IV proof of The then z a a + c this geometric fact is left - w = |w|. | unnecessary is more involved. If z = 0 or w = 0 (a one-line computational proof can be given, but even this —the assertion has already been shown restrict our attention to putting every nonzero complex 2 + to you). interpretation of multiplication we may FIGURE \z\ The operations of addition and multiplication of complex numbers both have important geometric interpretations. The picture for addition is very simple (Figure 2). Two complex numbers z = (a, b) and w = (c, d) determine a parallelogram having for two of its sides the line segment from (0, 0) to z, and the line segment from (0, 0) to w; the vertex opposite (0, 0) is z + w (a c + w\< For any complex number z in this expression, \z\ is a positive real number, while kl is from P1-P9), so nonzero complex numbers. We begin by to follow number into a special form. 0 we can write — 0, = M= kl 1 ' 443 Complex Numbers so that z/\z\ — ber a x is + with some number for = 1 = \a\ = a of 8 radians number a complex iy some but 0 is > r 2 — . + cos 6 i sin 6 Thus every nonzero complex number z can be written 6. = if 0O 0. i sin 0) The number r is unique (it equals \z\), + then the others are do 2kw for k in called an argument of z. Figure 3 shows z in one is + r(cos 6 number 0 and some not unique; +y (cos 0, sin 0) z for Now any complex num1 can be written in the form of absolute value x2 possibility, any one of these numbers is r and 6. (To find an argument terms of — 8 for z x + iy we may note that the equation x means + iy x y If x ?± 0 = 6 ?r/2 Now z ~ = \z\ + i sin 0) sin 0. we can therefore take 6 = arctan y/x, and when y > 0 and 6 = 3?r/2 when y < 0.) the product of two nonzero complex z - (cos 6 \z\ |z|cos 0, == r(cos 0 w — z = = that w = = = + rs(cos 6 i rj[(cos 6 cos + ™[cos (6 s(cos sin 6) (cos 0 — 0) + <j> <t> + + + sin (6 + x = 0 + 0, we can take numbers i sm i sin 0), 6), sin 0) i sin 6 sin 0) i if + z'( s m 6 cos cos 6 sin 0)] 0)]. is the product of the absolute values of of any argument for each of the factors will be an Thus, the absolute value of a product the factors, while the argument sum For a nonzero complex number for the product. z = r(cos 6 + i sin 6) now an easy matter to prove by induction the following very important formula (sometimes known as De Moivre's Theorem): it is z n = n |^| (cos nd + i sin nd), for any argument This formula describes z n so explicitly that z THEOREM 2 n = it is z. easy to decide just when w: Every nonzero complex number has exactly n complex nth roots. More preany complex number w ^ 0, and any natural number n, there are n precisely n different complex numbers z satisfying z = w. cisely, fcr proof 6 of Let w — j-(cos 0 + 2 sin 0) . 1 444 Infinite Sequences and Infinite Series for s = and some number \w\ — 2- satisfies — w zn and only ii r which happens n and only if From the first equation = sin nd) 2 sin 0) i ,r(cos 0 + i sin 0), if + cos nd + r(cos 6 number a complex if + (cos nd Then 0. = sin rc0 i cos + 0 2 sin 0. follows that it = Vs, r where y/s denotes the positive real nth root of s. From the second equation follows that for some integer k we have Conversely, z = r(cos 0 roots of w, Now we choose if + and 6 r it is any integer some integer w. nq + = and some integer q, + i sin 0 k' = fc \J - = z 6k for k it is = 0, then the number v'j (cos + A*, between 0 and = w sin 0*) z k! cos Ov n This shows that every z satisfying z Moreover, some Bk for To can be written Jc cos 6k Vr = = determine the number of rcth therefore only necessary to determine which such z are distinct. sin 0) will satisfy z i h for n it + . . , n — 1 In the course of proving differ by Theorem less 2, — 1. — 1 Then sin 0^. * can be written £ = 0, easy to see that these numbers are . n . . , w all different, than 2w. we have . since any two | actually developed a method for finding the nth roots of a complex number. For example, to find the cube roots of (Figure 4) note that i cube roots of 1 i cos • T + /ir cos ' T ' L and that 1 i sin - 7r/2 is an argument for > 6J 2tt\1 2x\ .../* + t) + sin + t)J Vg . . 1 V6 L 1 — 6 L 1 = |i| are therefore , cos \6 4tt\ 7/ , . . 4ir\~| /tt , sin \6 17 J 5tt = cos T = cos y+ 3tt + . . 1 . ' . sm . sin 5tt T T 3tt i. The . 445 Complex Numbers Since 7T - cos . sin ' 2 6 — — 5tt . sm ' 2 6 — — 3tt . sin i ? 2 = -1, are VI + - VI + i we cannot — i , , 2 In general, 1 2 z 2 z the cube roots of > _ ~ 6 3tt cos = — 2 6 57T cos 1 - 2 expect to obtain such simple results. For exa mpl e, to V 125 and cube roots of 2 + 1 li, note that |2 + 1 li\ = V2 2 + ll 2 = that arctan Af is an argument for 2 + 1 h*. One of the cube roots of 2 + 1 li is find the therefore •\VToe T V125 ^cos (—3—J + Previously 1 2 + i| = can write AM /arctan we noted V2 + 2 this l . . /arctan (—3—jj = VI [cos + is . that 2 = VI, and 2 AMI sm z (««|L*) + ,• sin (5=^)} cube root of 2 also a \ since arctan is + Hi. Since an argument of 2 + i, we cube root as 2 + = i Vs (cos arctan ^ + z sin arctan ^) These two cube roots are actually the same number, because arctan 1 V" = arctan - 2 3 (you can check this by using the formula in Problem 15-8), but this is hardly the sort of thing one might notice! The fact that every complex number has an nth root for all n is just a special The number i was originally introduced in case of a very important theorem. order to provide a solution for the equation x 2 + 1 =0. The Fundamental Theorem of Algebra states the remarkable fact that this one addition automatically provides solutions for all other polynomial equations: every equation zn + an-iz 71 " has a complex root! In the next chapter 1 + we • • • + shall give mental Theorem of Algebra; the a0 = 0 a 0, . . . , a n -i in C an almost complete proof of the Fundagap left in the text can be filled in as an slight 446 Infinite Sequences and Infinite Series The proof exercise (Problem 25-5). come up concepts which on of the theorem will rely quite naturally in a more thorough several new investigation of complex numbers. PROBLEMS 1. 2. Find the absolute value and argument of each of the following. + AL + Ai)'\ (i) 3 (ii) (3 (hi) (1 (iv) a/3 (v) [3 +Q 5 . + + Ai. 4z[. Solve the following equations. + + + + = 0. + = 0. 2ix -1=0. x ix " (1 + l)y = 3 = 4. \(2 + i> + (i) x2 (ii) x* 2 (hi) (iv) K J ix 1 x2 1 ' ( *> (v) 3. — = -z. = z-\ |z — a\ = (i) z (ii) t (iv) |z (v) 5. x2 x — Describe the set of (hi) 4. — #3 — |z| < |z — 2 complex numbers z such that all — — 0. 6|. = a) + |* 1 — real part of z. Prove that \z\ = imaginary part is Prove that + \z 6| c. and that the (z — + w\ 2 real part of z (z is + z)/2, while the z)/2i. - \z = w| 2 2(|z| H + 2 and 2 ), interpret this statement geometrically. 6. What is the pictorial relation between z line goes into the real axis 7. (a) Prove that if a 0, . . . Vi and • z V-f ?Hint: under multiplication by V — , a n -i are real and a + + fo" / Which ? (for a and £ real) - hi n _iz equation. (Thus the nonreal roots of such an equation always occur in pairs, and the number of such roots is even.) 2 n~ l n a 0 is divisible by z a n -iz (b) Conclude that z 2 2 b ) (whose coefficients are real). (a 2az satisfies n the equation z + n_1 <z + • • • a0 = 0, then a also satisfies this + + + *8. (a) Let and a + c b b • • * + + be an integer which are integers y/c, as a we is not the square of another integer. If a define the conjugate of a — bVc Show + b that the conjugate Vc, denoted by is well defined 447 Complex Numbers by showing that a number can be written and b, in only one way. (b) + bVc. for integers a Show that for all a and (3 of the form a + b Ve. we have a = a, a = a if and only if a is an integer, o: + /3 = a + /3, —ol— —a, -1 = a if a 0. and a -1 = (a) a - • (c) a Prove that ~j3, if ao, . . . + n the equation z and z = a n _i are integers , a n _\z n~ x + • • • + a0 = a + then z 0, satisfies £ = — a y/c b also satisfies this equation. 9. the 4th roots of i; express the one having smallest argument form that does not involve any trigonometric functions. Prove that if co is an nth root of 1, then so is u k A number co is called a primitive nth root of 1 if 1, a>, co 2 Find all in a *10. (a) (b) . { co n_1 of is } 1 the set of all are there for n nth roots of = 3, 4, 5, 1. How many . , . . , primitive nth roots 9? n-l (c) Let co be an nth root of 1, with 5* 1. co Prove that V k *11. (a) . . . 0. , clear is if reduce the general case to Show line +^ • • • also easy: the assertion will *12. = lie on one side of some straight line Prove that if zi, ^ 0. Hint: This is obvious from through 0, then zi the geometric interpretation of addition, but an analytic proof is + (b) co* =0 further that Zi through Prove that if \zi\ 0, so — \z 2 -1 that \ = the line ~~ , . . . zC + l \z 3 \ , the real axis, is zk • and z 1 l all lie on one +z~ +z +z • • l k 2 z are the vertices of an equilateral triangle. Hint: that zi is real, and this and a trick this one. can be done with no side of a straight 9* 0. = 0, then z h z 2 and z z It will , help to assume loss of generality. Why? CHAPTER W 25 You COMPLEX FUNCTIONS probably not be surprised will deeper investigation of to learn that a complex numbers depends on the notion of functions. Until now a function was (intuitively) a rule which assigned real numbers to certain other real numbers. But there is no reason why this concept should not be extended; we might just as well consider a rule which assigns complex numbers to certain other complex numbers. A rigorous definition presents no problems (we will not even accord it the honors of a formal definition) a function full : is a collec- complex numbers which does not contain two distinct pairs with the same first element. Since we consider real numbers to be certain complex numbers, the old definition is really a special case of the new one. Nevertion of pairs of we theless, sometimes resort to special terminology in order to clarify is being considered. A function / is called will the context in which a function real-valued iff(z) is a real number for all z in the domain of /, and complexto emphasize that it is not necessarily real-valued. Similarly, we will valued is defined on [a subset of] R in those where the domain of / is [a subset of] R; in other cases we sometimes mention that / is defined on [a subset of] C to emphasize that f{z) is defined for complex z as well as real z. Among the multitude of functions defined on C, certain ones are particularly important. Foremost among these are the functions of the form usually state explicitly that a function / cases = /(*) where a 0) . . . , a n are a nz n + an^z"- 1 + + • • a0 , complex numbers. These functions are called, as in the real case, polynomial functions; they include the function f(z) = number a familiar function Two is the "absolute value function" f(z) and Im Re(# Im(* The "conjugate function" (the is ^ ^ for all z in C. produce new rea j = z = ReW - am (z). R may be combined in complex-valued functions defined on the function ,f(x 448 \z\ defined by Familiar real-valued functions defined on to = numbers are Re (the "imaginary part function"), defined by + iy) — x, + iy) = y, f(z) ways z (the functions of particular importance for complex "real part function") is — some complex and functions of the form f(z) ("constant functions"). Another important generalization of a "identity function") a for + iy) = e v sin (x — y) + ix z cos y. C— an many example . 449 Complex Functions The formula always for this particular function illustrates a possible. Any complex-valued / = function u + decomposition which / can be written is form in the iv — some real-valued functions u and v simply define u(z) as the real part of This decomposition is often very useful, f(z), and v(z) as the imaginary part. be inconvenient to describe a polywould it example, but not always; for for nomial function in this way. other function will play an important role in this chapter. Recall that an argument of a nonzero complex number z is a (real) number 6 such that One z There are infinitely = |*| (cos 0 many arguments + i sin 6). but just one which for z, satisfies unique argument d(z), then 6 is a (real-valued) 0 < 6 < 2tt. If on {z in C: z ^ 0} function") "argument (the function " Graphs" of complex- valued functions defined on C, since they lie in 4-dimensional space, are presumably not very useful for visualization. The alternative picture of a function mentioned in Chapter 4 can be used instead: we draw two copies of C, and arrows from z in one copy, to f(z) in the other we (Figure call this 1). '•/(*) FIGURE The most common pictorial representation of a 1 complex-valued function is produced by labeling a point in the plane with the value /(z), instead of with z Figure (which can be estimated from the position of the point in the picture). Certain features of 2 shows this sort of picture for several different functions. For example, the the function are illustrated very clearly by such a "graph." around circles 0, the funcabsolute value function is constant on concentric respectively, lines, horizontal and tions Re and Im are constant on the vertical 2 and the function /(z) = z wraps the of radius r circle of radius r twice around the circle 2 . functions in Despite the problems involved in visualizing complex-valued properties preimportant of analogues define to possible still is it general, cases these viously defined for real-valued functions on R, and in some the example, For case. properties may be easier to visualize in the complex notion of limit can be defined as follows: lim f(z) = rmmber d / > means that for every (real) 0 such that, for all z, if 0 < number \z - a\ < £ > d, 0 there then \f(z) is a - (real) /| < S. FIGURE 2 451 Complex Functions Although the definition reads precisely as before, the interpretation is slightly different. Since \z — w\ is the distance between the complex numbers z and w the equation lim f(z) = / means that the values of f(z) can be made to lie 9 z—>a inside any given circle sufficiently small circle using trie FIGURE around /, provided that z is restricted to lie inside a around a. This assertion is particularly easy to visualize "two copy" picture of a function (Figure 3). 3 Certain facts about limits can be proved exactly as in the real case. In particular, c = c, lim z = a, lim Z—Kl + g(z)] lim [f(z) = lim /(*) + lim g(z), z—*a 2—* a z—>a lim f(z) g(z) ' lim -\- = z~+a g(z) = lim f(z) • lim g{z) 1 lim g(Z) if > ^ lim g(z) y 0. z-+a upon which these results are based this inequality holds for complex and is + numbers as well as for real numbers. These facts already provide quite a few limits, but many more can be obtained from the following theorem. The essential property of absolute values the inequality THEOREM l < w\ \z + \z\ \w\, Let/(» = u(z) + iv{z) for real- valued functions u and v, and for real numbers a and 0. Then lim f{z) = I if and only if let / = a + z—>a lim u{z) = a, = 0. z-+a lim v(z) 2—>0 proof Suppose first that lim f(z) if The second 0 < = |* /. - If e > 0, < «, then fl| there is |/(*) 6 > - 0 such that, for l\ inequality can be written |[«(z) -a] + i[v(z) -jS]| < e, < e. all iff 452 Infinite Sequences and Infinite Series or - [«(*) Since u{z) — a and v(z) — + a} 2 - [v(z) 0] 2 < e 2 . are both real numbers, their squares are positive; ft this inequality therefore implies that - [u(z) a] 2 < £ 2 and [v{z) - p] and \v{z) - 0\ < 2 £ 2 , which implies that - \u(z) Since this is true for > all £ < a\ £ = a and Now suppose = /3. > 0, there a\ < -> lim v(z) z—* a z—> a that these two equations hold. If £ that, for all z, if 0 < \z |w(z) £. follows that 0, it lim u(z) < - a\ < — a\ < is a 5 > 0 such then <5, £ - and — |z/(z) £ which implies that |/(z) - /| = < - -a\ £ £ , In order to apply the limit lim z = a, = I. + + i[v{z) \i\ • - \v(z) 0]\ - jS| 2 2 This proves that lim f(z) a] \u(z) | Theorem 1 fruitfully, notice we can conclude that that since we already know z—>a lim Re(z) = Re(a), = Im(fl). 2—*a lim Im(z) z—>a A limit like lim sin (Re (z)) = sin (Re (a)) z—»a follows easily, using continuity of sin. Many applications of these principles prove such limits as the following: lim z — a = \a\, } z—>a lim |,?| z->a e y sin lim (x-fi3/)-*a Now + x + ix 3 cos ^ = e b sin a + ia 3 cos fci complex functions, the continuous at a if lim f(z) — that the notion of limit has been extended to notion of continuity can also be extended: / is Complex Functions and /is continuous f(a), if /is continuous at a for previous work on limits shows that f(z) /(') /(*) /(* + zy) all a in domain the of/. 453 The the following functions are continuous: all = a n z n + a n ^z n ~ + = = 14 = ^ sin x + cos l • • • + fl 0 , 3 z'* jy. Examples of discontinuous functions are easy to produce, and certain ones come up very naturally. One particularly frustrating example is the "argument function" 0, which is discontinuous at all nonnegative real numbers (see the "graph" in Figure 2). By suitably redefining 6 it is possible to change f the discontinuities; for example (Figure 4), ment of z with 7r/2 < 0'{z) nonnegative real number tinuities will a. < 57r/2, if 8 (z) denotes the unique argu- 6' is then But, no matter discontinuous at ai for every how 6 is redefined, some discon- always occur. •2A» 7T 2 •2^tt .2f» 7T 2 TT 2 7T 2 7T -2i7T .2|7T 2 •2I7T 7T .2£tt 2 •2i7T 7T 2 •2|tt 7T 2 T T IT IT TT TT TT TT TT 27r 2tt 2tt 2tt 2tt 2t 2ir 2t 2w 3ir 2 2 3tt 2 3tt 2 3tt 2 3tt 2 37T 2 FIGURE The = 0'{z) 4 discontinuity of 0 has important bearing on the problem of defining a a function /such that (/(*)) 2 = zforall^r. For real numbers the function y/ had as domain only the nonnegative real numbers. If complex numbers are allowed, then every number has two square roots "square-root function," that is, 0, which has only one). Although this situation may seem better, it is some ways worse; since the square roots of z are complex numbers, there is no clear criterion for selecting one root to be f(z)> in preference to the other. (except in One way to define / We set /(0) = the following. is and 0, for z 9^ 0 we set /W =VH(cos^+*sinM). Clearly (f(z)) 2 ous. = but the function / z, As a matter of 2 (/(*)) = z for without any all z. is discontinuous, since 6 it discontinu- is the argument, given below, fuss, is impossible to find a continuous / such that Although you may be willing to accept this assertion fact, rather tricky. is We will actually show that a continuous square root function / cannot even be defined on the set of all z with \z\ = 1. Suppose there were such a function /. We must have /(l) = 1 or /(l) = — 1, and we can therefore assume that /(I) = —1, since —/would also be a continuous square root function. In this case, since / does not have the value 1 at 1 it never takes on the value 1 This means that if we define g by , = g{z) for 8(f(z)) |*| = 1, then g will be continuous, since 6 is continuous at all w with w = 1, which is not f(z) for any z. But the equation (f(z)Y = \w\ — 1 except for z implies that 8(z) = 20(f(z)) = 2g(z), which would imply that 0 is continuous at z for all z with \z\ = 1 Since this is false, no such function /can exist. Similar arguments show that it is impossible to define continuous "r/th-root functions" for any n > 2. For continuous complex functions there are important analogues of certain theorems which describe the behavior of real-valued functions on closed intervals. A natural analogue of the interval [a, b] is the set of all complex numbers . z = + x iy < with a < x closed rectangle, and b and c < denoted by is y < d (Figure [a, b] X [c, 5). This set is d]. a continuous complex-valued function whose domain is [a, b] seems reasonable, and is indeed true, that / is bounded on [a, b] If /is then it That is, there is some real |/(<r) | < number M M such that for all z in [a, b] called a X [c, X X [c, d], [c, d]. d]. / has a maximum and a minimum value on [a, b] X [c, d], since there is no notion of order for complex numbers. If / is a real-valued function, however, then this assertion does make sense, and is true. In particular, if /is any complex-valued continuous function on [a, b] X [c, d\ then /| is also continuous, so there is some z 0 in [a, b] X [c, d] such that It does not make sense to say that | \f(zo)\ a similar statement that "/ attains The its is < \f(z)\ for all z in [a, b] X [c, d]; true with the inequality reversed. It maximum and minimum modulus on various facts listed in the previous paragraph will is [a, b] sometimes said X [c, d]." not be proved here, Complex Functions 455 although proofs are outlined in Problem 5. Assuming these facts, however, we can now give a proof of the Fundamental Theorem of Algebra, which is really quite surprising, since we have not yet said much to distinguish polynomial functions from other continuous functions. THEOREM 2 (THE fundamental THEOREM OF ALGEBRA) Let a 0 . , . . , <z n _i Then be any complex numbers. there + PROOF is a complex number , such that = a0 0. Let f(z) Then / is | /| to = l/to + I*" I ^z n ~ + l + +a • • the function is = an defined by |/| tfn-i^" 1 0. + * • ' + a 0 \. based on the observation that a point z 0 with f(z 0 ) = 0 would point for |/|. To prove the theorem we will first show does indeed have a smallest value on the whole complex plane. The proof proof is clearly be a that zn continuous, and so l/l Our = minimum be almost identical to the proof, in Chapter 7, that a polynomial function of even degree (with real coefficients) has a smallest value on all of R; both will depend on the proofs We fact that begin by writing, for z if \z\ ^ is large, then \f(z)\ is large. 0, + ?)' so that + <20 Let M = max(l, Then for all z with \z\ 2w|<2 n _i|, > M, we have \z k \ . . > . \z\ , 2rc|<2 0 |). and 2«|a n _A;| SO <2n-l + + < ' + <1, ~ 2 which implies that + > 1 - + This means that |/to > I for |*| > Af. zn > 456 Infinite Sequences and Infinite Series In particular, if \z\ > M and also \z\ > V2\f(0)\, then I/Ml > l/WI- Now {z: \z\ [a, b] let X [a, b] d] [c, be a closed rectangle (Figure < max(M, X [c, d] is |/(* 0 )| (1) V2\f(0)\)}, and suppose that the attained at z 0 so that if |*| Combining its < X for * in FIGURE > max(Af, (1) minimum minimum < and (2) |/(0)|. ^2|/(0)|), then we [c, |/| on see that |/(z 0 )| d]. Thus \f(z)\ > |/(0)| < |/W| > \f(z 0 )\. for all z, so that value on the whole complex plane at z 0 |/| attains = 0. . 6 To complete the proof of the theorem we now show that f(z Q ) convenient to introduce the function g defined by = g(z) Then g of , It follows, in particular, that |/(^ 0 )| (2) which contains 6) is occurs at fit + a polynomial function of degree 0. We want to show that g(0) = It is ^o). n, whose minimum absolute value 0. Suppose instead that g(0) = a ^ 0. If m is the smallest positive power of* which occurs in the expression for g, we can write g{z) where 0 9* 0. such that = <x + /3z Now, according m to + cm+l z m+l + Theorem 24-2 • • • there + is cn z n , a complex number 7 Complex Functions Then, setting dk = Cky k we have , + (3y mzm + dm+1 zm+1 + m m+1 \a - az + dm+1 z + = = \g(yz)\ \a • = a (l - zm + ^S±! + zm = a (l - a L This expression, so torturously arrived first that at, will < a If we choose, from among enable us to reach a quick we will have 1. z for which this inequality holds, some z which and positive^ then zm \~ *+••]!< < z I Consequently, 1 This all I chosen small enough, if \z\ is —-2- z + is real + dn z»\ ~zm + zm \ d^-z + 1 I contradiction. Notice • -| *+'•])! | \a\- • • • +---)| | = 457 is if 0 < 1 = zm - we have -zm + zm [~*+ * the desired contradiction : = 1 - zm + < 1 - zm + zn = 1. for such a \g(yz)\ < + + ']|<|1 ' \z number • • -]| n z we have \a\, contradicting the fact that ol\ is the minimum of \g\ on the whole plane. Hence, the original assumption must be incorrect, andg(O) = 0. This implies, I finally, that f(z 0 ) = 0. | Even taking into account our omission of the proofs for the basic facts about continuous complex functions, this proof verified a deep fact with surprisingly little work. It is only natural to hope that other interesting developments will arise if we pursue further the analogues of properties of real functions. The next obvious step is to define derivatives: a function /is differentiable at a if r Jf(a - + hm *->o z) z ^ -f(a) . exists, 458 Infinite Sequences and Infinite Series in which case the limit denoted by f'(a). is = = f(a) f(a) It 0 if /(*) 1 if /(^) is = = easy to prove that c, z, + gY(fl) =/'(«) (f (/•*)'(«) =/W«) + /(«)*'(«), =f(g(a))-g'{a); (/«•«)'(«) the proofs of all same these formulas are exactly the as before. It follows, in z n then f'(z) - nz n ~\ These formulas only prove the differentiability of rational functions however. Many other obvious candidates are not differentiate. Suppose, for example, that particular, that if f(z) = + /(* If / is , = iy) iy (i.e., f(z) = z). to be differentiable at 0, then the limit ** + lim exist. »> * (x+t W )->o must - x + -AO), ita (x+iy)->o iy x + iy Notice however, that ify = 0, then^ x + = 1, iy and x if = 0, = -1 then therefore this limit cannot possibly exist, since the quotient has both the values 1 and — 1 for x + iy arbitrarily close to 0. where other differentiable funcand exp, you will see tions are to come that there is no hope at all of generalizing these definitions to complex numbers. At the moment the outlook is bleak, but all our problems will soon be In view of this example, it is not at all clear from. If you recall the definitions of sin solved. PROBLEMS 1. (a) For any real number y, define a (x) = x + iy (so that a is a complexvalued function defined on R). Show that a is continuous. (This follows immediately = that P(y) (b) x + from a theorem iy is in this chapter.) Show similarly continuous, Let / be a continuous function denned on C. For fixed y, let g(x) = iy) Show that g is a continuous function (defined on R) Show f(x iy) is continuous. Hint: Use part (a). similarly that h{y) = f(x + . . + 2. (a) a continuous real-valued function defined on a closed rectangle [a, b] X [c, d]. Prove that if/ takes on the values f(z) Suppose that / is 459 Complex Functions for z and w in [a, b] X [c, d], then / also takes all values between f{z) and f(w). Hint: Consider g(t) = /(£? + (1 - 0«0 for * and f(w) in *(b) If [0, 1]. / is [c, d], a continuous complex-valued function defined on [a, b] the assertion in part (a) no longer makes any sense, since X we cannot talk of complex numbers between f(z) and f(w). We might conjecture that / takes on all values on the line segment between /0) and f(w), but even this is false. Find an example which shows this. (a) Prove that if a 0, . . . any complex numbers, then there are zn (not necessarily distinct) such that a n _ x are , complex numbers z u . . . , n + a n ^z n - + zn 1 • + • • a0 = [I t = i 0" *<)• n~ (b) + n x a0 + a n -i are real, then z + a n _iz Prove that if a 0) quadratic and a z factors linear of can be written as a product factors z 2 az + £ all of whose coefficients are real. (Use Problem . . . ' ' ' , + + 24-7.) this problem we Such a polynomial In ^2 _|_ . . . _|_ will consider is called a only polynomials with real coefficients. of squares if it can be written as sum h ^ n 2 for polynomials with real coefficients. Prove that if / is a sum of squares, then f(x) > 0 for all x, (b) Prove that if / and g are sums of squares, then so is / g. (c) Suppose that/(*) > 0 for all x. Show that / is a sum of squares. Hint: k First write f(x) = x g(x), where g(x) 7* 0 for all x. Then £ must be (a) • > even (why?), and g(x) (a) Let A 0 for all x. be a set of complex numbers. Now use Problem 3(b). A number z is called, as in the real point of the set A if for every (real) £ > 0, there is a point a in A with \z — a\ < £ but z ^ a. Prove the two-dimensional version of the Bolzano- Weierstrass Theorem: If ^4 is an infinite subset Hint: of [a, 6] X [c, d], then ^4 has a limit point in [a, b] X |>, First divide [a, b] X [c, d] ijh half by a vertical line as in Figure 7(a). case, a limit Since A is infinite, at least one half contains infinitely many points of A. Divide this in half by a horizontal line, as in Figure 7(b). Continue in this way, alternately dividing by vertical and horizontal lines. (The two-dimensional bisection argument outlined standard that the the c in this hint is so Bolzano- Weierstrass" often serves to describe in addition to the theorem itself. See, for example, "A Contribution to the Mathematical Theory of Big Game Arner. Math. Monthly, 45 (1938), 446-447.) Prove that a continuous (complex-valued) function on is (c) title method of proof, H. Petard, Hunting," (b) 1 bounded on [a, b] X [c, d]. [a, b] a real-valued continuous function on [a, b] then / takes on a maximum and minimum value on [a, b] (You can use the same trick that works for Theorem 7-3.) Prove that if/ is X [c, d] (Imitate Problem 21-22.) X X [c, d\ [c, d]. 460 Infinite Sequences and Infinite Series *6. The proof Theorem of 2 cannot be considered to be completely ele- mentary because the possibility of choosing y with y m = —a//3 depends on Theorem 24-2, and thus on the trigonometric functions. It is therefore of some interest to provide an elementary proof that there is a solution for the equation z (a) (b) a convex subset of the plane (c) (d) — c = 0. computation to show that solutions of z 2 — c = 0 can be found for any complex number c. Explain why the solution of z n — c — 0 can be reduced to the case Make an where (a) n n explicit is odd. n Let z 0 be the point where the function f(z) = z — c has its minimum absolute value. If z 0 ^ 0, show that the integer m in the proof of Theorem 2 is equal to 1; since we can certainly find y with 7 1 — — a/ /3, the remainder of the proof works for /. It therefore suffices to show that the minimum absolute value of /does not occur at 0. Suppose instead that /has its minimum absolute value at 0. Since n is n —c± 8 n i. Show odd, the points ±5, ± 8i go under /into —c ± 5 , that for small 6 at least one of these points has smaller absolute value — than (b) a nonconvex subset FIGURE of the plane 7. Let f(z) c, thereby obtaining a contradiction. = 0 - z0 Wl •••(*- **)"*. k 8 (a) Show (b) Let^O) = that f{z) = - (z zi) mi ••(*- z k) m" £ m a (z - za )~\ k ^ m a (z - za )~ x . Show that if g(z) = 0, then^i, . . . , zk a= 1 all lie on the same Problem 24-11. cannot (c) through z. Hint: Use K contains the line segment joinFor any set A, there is a smallest convex set containing it, which is called the convex hull of A (Figure 9) if a point P is not in the convex hull of A, then all of A is contained A subset K of the plane ing any two points in ; FIGURE side of a straight line 9 it is convex (Figure if 8). Complex Functions on one side of some straight line through P. prove that the roots of f (z) = 0 lie Using 461 this information, within the convex hull of the set (The definition of convexity in the Appendix to a function / is convex if the set of points lying above or on the graph of / is a convex set, and if the graph contains no straight line segments. Further details about convex sets will be found in reference [19] of the Suggested Reading.) {zi, . . Chapter 8. *9. . , 1 1 zjt}. is related to the one given here Prove that if / is differentiate at z, then / Suppose that / — u + iv where u and v are = + is continuous at z. real- valued functions. (c) and h(x) = v(x + iy 0 ). Show that if and + if$ for real a ft then g' fa) — a and h! (x 0) = ft f(x 0 + iyo) k(y) = u(x 0 + iy) and l{y) = that suppose hand, On the other v(x Q + iy). Show that l'(y 0 ) = a and k'(y 0 ) = -ft Suppose that/(z) = 0 for all z. Show that / is a constant function. (a) Using the expression find f (b) Use this result to find (a) For fixed y 0 \ctg(x) u(x iyo) = a (b) 10. — k) (x) for all L arctan a) (0) for all k. COMPLEX POWER SERIES CHAPTER If you have not already guessed where going to come from, the intend to define functions by means of infinite discussion of infinite complex functions are differentiable of this chapter should give the secret away: title we This will necessitate a sequences of complex numbers, and sums of such series. sequences, but (as was the case with limits and continuity) the basic definitions ai are almost exactly the same as for real sequences and series. # • An sequence of complex numbers is, formally, a complex-valued domain is N; the convenient subscript notation for sequences of real numbers will also be used for sequences of complex numbers. A sequence {a n of complex numbers is most conveniently pictured by labeling as • # infinite function whose • % • a 13 •<2io •«„ an } the points a n in the plane (Figure FIGURE The sequence shown in Figure 1). converges to 1 "convergence' of complex 5 0, 1 sequences being defined precisely as for real sequences: the sequence \a n converges to /, in — lim a n n—» if for every e > 0 there is a natural if n > /, oo N such that, number N, then — \a n l\ < sufficiently large n (Figure 2); expressed eventually inside any circle drawn around Convergence of complex sequences sequences, but can even be reduced to 2 THEOREM 1 is / will contain a n for all colloquially, the sequence is /. not only defined precisely for real this familiar case. Let an and = bn + = 0 + for real b n ic n and cn , let I Then lim an = I if and only lim PROOF more for all n, S. This condition means that any circle drawn around FIGURE ) symbols The proof is left as consult the similar The sum an easy iy for real 0 and y. if — 0 bn and lim exercise. If there Theorem of a sequence \a n 1 is = cn y. any doubt as to \ is defined, once again, as lim s n n— > sn 462 how to proceed, of Chapter 25. | = a\ + ' ' ' + an . oo , where 1 1 Complex Power Sequences which for say that the infinite series ^= n THEOREM 2 summable; this limit exists are an converges alternatively, limit exists, if this Series 463 we may and diverges 1 new unnecessary to develop any otherwise. It is infinite series, because of the following theorem. convergence of tests for Let an Then a n converges if £ n= — + bn and only if £ n= l and for real b n ic n and b* £ n= 1 cn cn . both converge, and in 1 this case an 2 n— PROOF This 00 00 00 = + £ an immediate consequence of Theorem sums of {a n }. | is partial There is ^= an converges absolutely if the series n 1 complex series: the series ^— \a n converges (this is a series of \ l numbers, and consequently one to which our earlier tests may be applied). following theorem is not quite so easy as the preceding two. The 3 applied to the sequence of °o real THEOREM 1 also a notion of absolute convergence for 00 n c n ). n= l Let an = bn + ic n for real b n ^= rt cn . » 00 00 Then and an converges absolutely if and only if ^ n= l bn and ^ n 1 cn both COU- =l verge absolutely. 00 PROOF Suppose n = 6n ^ and n= 1 cn both converge absolutely, 2 and I I that 00 £ n= 1 i.e., 1 00 00 n= 00 £ that first |c n | both converge. It follows that £ |£ n | + \c n converges. \ n- 1 Now, k| = It follows \b n from the comparison + lCn\ test that < \bn\ ^ n= \a n + M» converges (the numbers \a n \ \ 1 00 and \b n \ + \c n \ are real and nonnegative) Thus . £ a n converges absolutely. 464 Infinite Sequences and Infinite Series Now it is suppose that ^ \a n converges. Since \ clear that \b n < \ \a n and \ \c n < \ \a n \. 00 Once again, the comparison test shows that 00 ^ n= \b n \ and ^ n= 1 \c n converge. | \ 1 00 Two consequences of Theorem 3 are particularly noteworthy. ^= If n 00 converges absolutely, then 00 ^ and bn T^l quently ^ and bn n=l Theorem ^= 71 also cn converge absolutely; COnSe- 1 00 00 2. an 1 00 converge, by ^ Theorem ^ 22-4, so a n converges by n=l n—1 In other words, absolute convergence implies convergence. Similar reasoning shows that any rearrangement of an absolutely convergent series has the same sum. These facts can also be proved directly, without using the corresponding theorems for real numbers, by analogue of the Cauchy criterion With power (see Problem these preliminaries safely disposed series, that /0) = ^ first we can now of, establishing an 12). consider complex functions of the form is, - a n (z a) n = a0 + ai(z — a) + — a 2 {z a) 2 + ' ' ' n=0 Here the numbers a and a n are allowed to be complex, interested in the behavior of usually consider power series / for complex z. centered at 0, /w = I n= a n z" As and we are naturally in the real case, we shall 0 in this case, if f(z 0 ) converges, then /(z) will also converge for \z\ < \z 0 The proof of this fact will be similar to the proof of Theorem 23-6, but, for reasons \. that will soon become clear, we will not use all the paraphernalia of uniform convergence and the Weierstrass Af-test, even though they have complex analogues. Our next theorem consequently generalizes only a small part of Theorem THEOREM 4 23-6. Suppose that n ) an z 0 = + aiz + a0 0 a^z 0 2 + • • • nt'o converges for some z Q ^ 0. Then if \z\ < \z 0 \, the two series Series 465 As in the proof of Theorem 23-6, we will need only the fact that the such that numbers a n z 0 n is bounded: there is a number set of Complex Power = n anz ^= + a0 aiz + 2 a 2z + • • 0 71 00 both converge absolutely. PROOF M \a n z 0 We n M < \ for all n then have \a n z n = \ \a n z 0 ^ \ M < and, for z n 0, \na n z Ti n\a n z 0 l \z\ M Since the series ^= 0 n \z/z Q and \ oo n n\z/z 0 converge, this shows that both \ 1 00 CO ^ 71 £= 71 71 n T~\ \z\ <z n£ n Y M n ^ n-1 and =0 71 ^ 0, but this series certainly converges for z = na n z ^ n= =1 assumed that z Theorem 4 converge absolutely (the argument for n~ x 1 0 also). | evidently restricts greatly the possibilities for the set {z: ^= 71 For example, the shaded set A anz n converges}. 0 in Figure 3 cannot be the set of all z where 00 anz ^ n converges, since it contains z y but not the number w satisfying n \w\ < \z\. seems quite unlikely that the set of points where a power series converges could be anything except the set of points inside a circle. If we allow "circles of radius 0" (when the power series converges only at 0) and "circles of radius oo" (when the power series converges at all points), then this assertion is true (with one complication which we will soon mention) the proof requires It ; only Theorem 4 and a knack for good organization. 466 Infinite Sequences and Infinite Series THEOREM 5 For any power series 00 anz n ) = one of the following three + aiz + a0 possibilities a 2z n + must be + a zz 2 • • • true: 00 anz ^= (1) n converges only for z n converges absolutely for = 0. 0 n 00 a n^ ^ (2) z in C. ail n=0 There (3) R> number a is ^ a^ 0 such that 71 converges absolutely if n=0 and diverges if \z\ > R. (Notice that we do not mention w hat |*| < happens when \z\ — R.) r proof Let 00 S — {x in V R: Suppose first £ that <2 n wn converges for some = with a; #}. =0 n is unbounded. Then for any complex number z, there 00 number x in S such that a is |z| < x. By definition of £, this means that ^ 71 w converges for some with = \w\ x > n wn from Theorem 4 that follows It \z\. tf =0 00 ^ <z nz n converges absolutely. Thus, in this case possibility (2) is true. 71=0 Now suppose R = then 0, that ^ tf n£ let R be the least upper bound of S. If converges only for z = 0, so possibility (1) is true. Sup- S is bounded, and n 71=0 on the other hand, that pose, < |z| R, there is a number R > x in 0. Then if < x. £ with \z\ z is a complex Once number with means that again, this 00 00 a nw ^ n converges for some w with < \z\ \w\, so that n=0 71 ^= 0 nz n converges 0 00 absolutely. Moreover, if \z\ > /?, then ^= n not in #nz n does not converge, since |z| is 0 S. | The number R which occurs in case (3) is called the radius of convergence so of ^ anz n . In cases (1) and (2) it is customary to say that the radius of con- n=0 vergence is is 0 and <x> ? called the circle of respectively. When convergence of ^ 0 < R < a nz n . If z <x>, is the circle \z\ \z\ = R\ outside the circle, then, Complex Power of course, anz ^ n does not converge, but actually a much 467 Series stronger statement n=0 can be made: the terms a n z n are not even bounded. To prove this, let w be any number with \z\ > \w\ > R; if the terms a n z n were bounded, then the proof of Theorem 4 would show that anw ^ n converges, which is false. Thus 71=0 00 (Figure 4), inside the circle of convergence the series anz ^ n converges in the n=0 bounded way (absolutely) and outside the circle the series diverges in the way (the terms a n z n are not bounded). What happens on the circle of convergence is a much more difficult question. best possible worst possible We will not consider that question at all, except to mention that there are power series which converge everywhere on the circle of convergence, power series which converge nowhere on the circle of convergence, and power series that do just about anything in between. (See Problem 5.) Remember circle of convergence that our goal in this chapter produce differentiable funcproved for real power series in Chapter 23, that a function defined by a power series can be differentiated term-by-term inside the circle of convergence. At this point we can no longer imitate the proof of Chapter 23, even if we were willing to introduce uniform convergence, because no analogue of Theorem 23-3 seems available. Instead we will use a direct argument (which could also have been used in Chapter 23). Before beginning the proof, we notice that at least there is no problem about the convergence of the series produced by term- by- term differentiation. If the tions. We therefore want is to to generalize the result oo series ^= n a nz n has radius of convergence R, then Theorem 4 immediately 0 implies that the series na n z n ^ ~ l also converges for \z\ < R. Moreover, n=l \z\ > R, so that the terms a n z n are unbounded, then the terms na n z n~ l if are OS surely unbounded, so ^ na n z n~ l does not converge. This shows that the n—1 00 radius of convergence of theorem 6 If the power ^ na n z n ~~ x has radius of convergence < R, also exactly R. series f(z) \z\ is R > 0, = J then / anz is n differentiable at z for all z with and na n z n 468 Infinite Sequences and Infinite Series PROOF We will use another "s/3 argument. for 55 The fact that the theorem is clearly true polynomial functions suggests writing 00 00 /(* + A) -/(*) - (*) ^ na n z n~ l + ((* = £) - n | n=0 n=l AT 00 + ((z — j h) n — z + n ) - A)" ) na n z n k z") \l; n= 0 ((£ + + - A)" z») — N y na n z k » n=1 We will show that for any made < > £ n=»l each absolute value on the right side can be This will 0, N sufficiently large and h sufficiently small. e/3 by choosing clearly prove the theorem. Only the first term in the right side of (*) will present \z\ < \z 0 < The expression {{z + more convenient way if we remember that begin with, choose some z 0 with only h with in a h\ < \z 0 \. *1\.ZJ1 = x n-l \z x Applying + — + yv \ + x n-y x n-2y any _|_ h) - n . . difficulties. we R; henceforth To will consider n z )/h can be written + y«~\ . this to 0+ n h) - zn + h) n - z n (z + h) -z' _ (z . , ; h we obtain (z + h) n - zn = (z + z{z + hy~ + l z(z + h) n ~2 + • • ' + zn ~ l . Since \(z + h) n' 1 + h) n~ 2 + • • • + zn ~l \ < n\zo\ n -\ we have j(z + h) n - g) < n\a n • \ |z 0 n_1 . | 00 But the series ^ n\a n ' \ ko| n_1 converges, so n= 1 ^ n =N+l n\a n \ if N is sufficiently large, then Complex Power 469 Series This means that I V +1 hY - On ((* \l n= V z") + ^— l . v I ccg «• — + *)" - g") 2, v -2/ ^ i 1 n = AT-fl *"> h n=0 0 ~ A )" Kg an - n= JV ((z + A)» - i +1 r h 00 V < In short, (1) N if V sufficiently large, is ((* an l + h) n - for all h with + |* kd— < • v ((* an l + hY - z n < S ) <3 h n=0 h\ f- then zn ) IT n=0 1 «kl \z<\. CO It is easy to deal with the third term on the right side of (*) : Since ^= 71 converges, follows that it if N is sufficiently large, na n z n~ x 1 then IV (2) y^na n z«-i n= ^ na n z n <3 1 Finally, choosing an hm N such that (1) + ((Z ) an h) n ~ and (2) are true, Zn) = £> na n z we note that n J n=l n=0 AT since the polynomial function gO) = ^= 71 <z n 2: n is certainly differentiable. 0 Therefore N AT <3 ^ : >I2 = 71 na n z n ~l 0 for sufficiently small h. As we have already indicated, Theorem 6 has an obvious (1), (2), and (3) prove the theorem. is series is its Taylor series at 0. It follows, in particular, that/ is continuous inside is continuous at z the circle of convergence, since a function differentiable at z (Problem 25-8). by a power and the power corollary: a function represented infinitely differentiable inside the circle of convergence, series | and Infinite Series The continuity of a power series inside its circle of convergence helps explain the behavior of certain Taylor series obtained for real functions, and gives the to the questions raised at the end of Chapter 23. We have already seen that the Taylor series for the function f(z) = 1/(1 z 2 ), namely, promised answers \ + - 1 when converges for real z only vergence J points i \z\ < and consequently has radius of con- 1, no accident that the circle of convergence contains the two and —i at which / is undefined. If this power series converged in a 1. It is circle of radius greater than which was continuous in that circle, in particular at would represent a function i and —i. But this is impossible, since it equals 1/(1 + z 2 ) inside the unit circle, and 1/(1 + z 2 ) does not approach a limit as z approaches i or —i from inside the unit circle. The use of complex numbers also sheds some light on the strange behavior then (Figure 1, 5) it of the Taylor series for the function X = /(*) x 0, 0 = 0. Although we have not yet defined e z for complex that if y is real and unequal to 0, then -l/fiy)« /(«» The interesting fact about this expression small. Thus / numbers, z = so _ presumably be true becomes large it asjy becomes not even be continuous at 0 when defined for complex hardly surprising that it is equal to its Taylor series only for will it is 0. The method by which we will complex z should by now be for * sin cos x = — actually define clear. x^ * X"* 1 3! 5! 2 4 x e x - 1 cos z x , — cos z, cos'(z) • • • , • +" + —+•* • • 2! = z z + -- z 3 5 3! 5! zz z* -++ 2! 4! 1 - 1 +-z + -z 2 = z and cos z) 4! 1! sin' (z) (as well as sin define sin z exp(z) z l 1! For complex z we therefore e For real x we know that 2! Then will p \\y* that is z, it = — sin z, - 2! and exp f (z) = exp(z) by Theorem 6. Complex Power Series Moreover, we if replace z by iz in the series for e z > something 471 particularly- interesting happens: e = lz 1 + iz *2 (this step 2! wr» e iz from the definitions iz' / V — cos z also 3! / 5! n z power — z) = — = series) that sin z, cos z, have g From 1 _|_ i s [ the (i.e., sin( we ' fact that the series converges absolutely), so cos (—z) so ' 5! z4 4! can be justified by the It is clear 4! 3! 21 V ' r -\ the equations for e iz -iz ^ cos z — ' 2 we can and si n Zt derive the formulas sm z = 2 l = cos z 2 The development of complex power series thus places the exponential function at the very core of the development of the elementary funotions — it reveals a connection between the trigonometric and exponential functions which was never imagined when these functions were first defined, and which could never have been discovered without the use of complex numbers. As a by-product of this relationship, we obtain a hitherto unsuspected connection between the numbers e and 7r: if in the formula e we take z = 7r, we iz — CQS z With these remarks And we n z = ir -1. will bring to a close yet there are still have not been mentioned. Thus centered at I gj obtain the remarkable result e functions. _|_ our investigation of complex several basic facts about far, power series we have seldom considered power which series a, 00 f(z) = ^= n except for a — 0. «»(* - «)". 0 This omission was adopted partly to simplify the exposition. 472 Infinite Sequences and Infinite Series For power this centered at a there are obvious versions of all the theorems in series chapter (the proofs require only modifications) there trivial : is number R a 00 (possibly 0 or "oo") such that the series ^= n for z with — \z moreover, for a\ all < — \z <R a\ — unbounded terms R, and has z with a n (z a) n converges absolutely 0 for z with \z — a\ > R; the function 00 f{z) r\ = - a n {z I a)* n=0 has derivative a f(z) - | n It is less tion as a n -\ straightforward to investigate the possibility of representing a func- power centered at FIGURE na n (z-a) =1 series centered at b, if it is already written as a power series a. If 00 6 = f(z) 71 has radius of convergence it is R y and a n (z b is = numbers bn are necessarily „)» a point with 2= n - 0 true that f(z) can also be written as a f(z) (the I= power »•(*- b) — \b a\ series < R (Figure 6), then centered at b, n 0 f n) (b)/n !); moreover, this series has radius of convergence at least R — \b — a\ (it may be larger). We will not prove the facts mentioned in the previous paragraph, and there are several other important facts we shall not prove. If 00 f(z) 00 = 2 a ^z - a) n and *w converge in some circle around / g 1 *»(* a, it is certainly to be converge in a circle around a; 00 and ^(^) = at a, = 2 then it hoped that f moreover, ifg(a) /g should have a power series representation. Finally, f(z) - n=0 n=0 will also = £ ^ 0, +g and the function if 00 ««(* - and g(z) = should be possible to write / J o 6 n (^ - *)», g as a power series centered 6. introducing any basic new / g presents a few problems, and \/g presents even more problems. The possibility of changing a power series centered at a into one centered at b becomes yet more involved, and the treatment offog requires real skill. Rather than end this section with a tour All of these facts could be proved ideas. The treatment off +g is now without easy, while • Complex Power we de force of computations, will instead give a 473 Series 55 preview of " complex analysis, one of the most beautiful branches of mathematics, where all these facts are derived as straightforward consequences of some fundamental results. Power series were introduced in this chapter in order to provide complex functions which are differentiable. Since these functions are actually infinitely differentiable, it is natural to suppose that we have therefore selected only a very special collection of differentiable complex functions. The basic theorems of complex analysis show that If a complex function A, then in A it is is not at is all true: A defined in some region and of the plane is differentiable in automatically infinitely differentiable in A. Moreover, fcr each point a Taylor series for the this at a will converge to f f any in circle A contained in (Figure 7). These among facts are the be proved in complex analysis. It is the methods used are to first impossible to give any idea of the proofs themselves — quite different from anything in elementary calculus. If these are facts granted, however, then the facts mentioned before can be proved very easily. Suppose, for example, that / and g are functions which can be written as power Then, series. we have shown, / and g as are differentiable Appealing differentiable. to the results they can be written as power Similarly, series. = £ = n < a\ centered at b, R}. Thus, which if b is in A, is a)» differentiable in the region A = power series — This possible to write / as a it is converge in the will - a n (z 0 has radius of convergence R, then / — then it ' if /(*) \z\ \z — f + g, f g, 1/g and fog are also from complex analysis, it follows that follows from easy general theorems that circle of radius R — \b a\. 00 series may ^= 0a n (z — actually converge in a larger circle, because a) n may be n the series for a function differentiable in a larger region than A. For example, suppose that f(z) = 1/(1 + z 2 ). Then / is differentiable, except at i and — i, 00 where Y not defined. Thus f(z) can be written as a power series it is anz n n=0 with radius of convergence and ak = 0 if k is odd). It 1 is (as a matter of fact, we know that am — (~~l) n also possible to write 00 f(z) = 2 = n where the numbers bn b n {z are necessarily bn the radius of convergence of this series: to i or —i (Figure 8). - *)", 0 = it is f (n) (j)/n\. Vl + We can easily predict 2 (^-) , the distance from \ 474 Infinite Sequences and Infinite Series As an added incentive complex analysis to investigate mentioned, which result will be quite near the surface, lies found in any treatment of the subject. For real z the values of sin z always lie between z this not at is all true. In = z fact, if gi(iy) sin iy e = large, then sin is iy is — -~i(iy) g—y = 1 and and which 1 , but for complex then for y real, : 2 If y iy, one more will be further, — ey : T l also large in absolute value. This behavior of sin is which are defined and differentiable on the whole complex plane (such functions are called entire). A result which comes quite early in complex analysis is the following: typical of functions Liouville's Theorem: The only bounded entire functions are the constant functions. As a simple application of Liouville's Theorem, consider a polynomial function f(z) where large > n z, 1, = zn + ai z n ~ so that /is not a constant. so Liouville's Theorem tells + • • • +a H9 We already know that f(z) us nothing interesting about is /. large for But con- sider the function g & = 7TT /0) were never 0, then g would be entire; since f{z) becomes large for large would also be bounded, contradicting Liouville's Theorem. Thus f(z) = 0 for some z, and we have proved the Fundamental Theorem of If f(z) z, the function g Algebra. PROBLEMS 1. Decide whether each of the following converges absolutely. n= 1 series converges, and whether it ' Complex Power Series 475 00 (i v ; yiog_« L4 + . B iog». n n n=2 Use the ratio test to show that the radius of convergence of each following power series will to 00 or to a limit is < approach a limit > 1. (In each 1 if \z\ < 1, of the case the ratios of successive terms but for \z\ > 1 the ratios will tend 1.) n-\ » It n = 1 (iii) 00 (iv) 00 (v) 2-")^ n ^(« + _ ^ 2^ Use the root n! . test of the following z .... (») . , (Problem 22-7) power z* z* z* to find the radius of convergence of each series. z6 z* _ _ _ — <?n\z n > . (iv) t**' cv) ^ 2V ! , of root test can always be used, in theory at least, to find the radius convergence of a power series; in fact, a close analysis of the situation The 476 Infinite Sequences and Infinite Series known leads to a formula for the radius of convergence, Hadamard formula." Suppose first as the "Cauchy- numbers V|tf n that the set of is | bounded. 00 Use Problem 22-7 (a) to show that V|^ n z < lim if | 1, Y then n ~*~ n anz n =0 converges. 00 show that (b) Also > lim \Z\a n z if \ "-* Y then 1, 00 n anz n has unbounded =0 terms. 00 (c) Parts (a) and (b) show that the radius of convergence of _ 1/ihn a/I^I (where "1/0" means " oo"). n—* a nz ^ n is n=0 To complete the formula, oo V/ |^ n = define lim oo the set of if | is unbounded. Prove 0, so that the radius of ^/\a n all \ n—+oo 00 anz ^ that in this case, n diverges for z ^ n=0 convergence 5. may 0 (which is Consider the following three be considered as "1/°°"). from Problem series 2: n y^ n= Prove that the n 1 , Vxn= =1 l converges everywhere on the unit circle; that nowhere on the unit circle; and that the second least one point on the unit circle and diverges for first series the third series converges series converges for at on the unit at least one point 6. (a) Prove that e z e w = e zJrW circle. for all complex z and w. (You just have to imitate the relevant parts of Problem 17-35.) (b) Show = 7. (a) Prove iv e (b) 8. (a) (b) 9. (a) cos z sin w and cos(z + w) complex z and w. that every complex number of absolute value 1 can be written some real number y. x x that \e + iy — e for real x and y. that exp takes on every complex value except 0. that sin takes on every complex value. that a complex- valued function / = u + iv satisfies the for Prove Prove Prove Show + that sin(z cos z cos w — w) = sin z sin sin z cos w w + for all \ equation f n) + an-i/^ + 15 (*) if (b) and only Show that a n -iz + e bx n~ l cos a: if u if a * and = • b v do. Hint: + +a 0 ci is = 0, • • • + aof = Use Problem 0, 25-9. n a complex root of the equation z then are both solutions of (*). /(*) = sin cx and /(*) + = Complex Power 10. (a) Show (b) Given «/^0, show that # = that exp log is Series 477 + with one-one on C. not e = w z if and only if z = # (here log denotes the real logarithm function), \w\ iy and j> an argument of w. *(c) Show that there does not exist a continuous function log defined for nonzero complex numbers, such that exp (log (z)) = z for all z ^ 0. (Show that log cannot even be defined continuously for \z\ — 1.) Since there is no way to define a continuous logarithm function we cannot speak of the logarithm of a complex number, but only of "a logarithm for w," meaning one of the infinitely many numbers z with e z = w. (d) Find all logarithms for (e) Find all values of i\ that i. is, of e iz where z is a logarithm of (The i. answers will be real numbers.) (f) Show that many has infinitely (1*)* possible values, while 1*** has only one. 11. (a) For real x show that we can choose log(* (It will (b) The + ^ — arctan log (a; + i) — log(l + x 2) + log(x — i) = log(l + x 2) — — arctan x help to note that t/2 t — i = expression -L-i(J— 1 + 2 2^ \a: a: ? and \og(x i) ^ — be to i) xj, arctan xj. arctan \/x for x ^ 0.) U + z/ yields, formally, ^ / +x 1 Use part 12. (a) (a) to A sequence lim if m,n—> cn \a m {a n — 1 [log(* , 2 check that } an \ 0. ] log(^ + 0]. answer agrees with the usual one. numbers is Suppose that a n called a = bn + Cauchy sequence ic n , where bn and oo is a Cauchy sequence if and only if {a n Cauchy sequences. every Cauchy sequence of complex numbers converges. are real. Prove that \b n this of complex = -0 - 2i and {c n } } are (b) Prove that (c) Give direct proofs, without using theorems about real series, that an absolutely convergent series is convergent and that any rearrangement has the same sum. (It is permitted, and in fact advisable, to use the proofs of the corresponding theorems for real series.) > 478 Infinite Sequences and Infinite Series 13. Prove the formula 1 + - + cos 8 + cos 26 + • • sin (n = cos nd 2 — + i)6 2 sin* 2 by writing cos nd = (^ + e~ )/2, summing the resulting geometric and multiplying top and bottom by e~ i9l K in6 n* series, 14. Let {a n be the Fibonacci sequence, } - (a) If r n (b) Show ^n+iAn, show that that = r lim n— > = r Show (c) + (1 rn a\ 1 + and r = r n +i exists, = — a2 1, a n+2 — an + a n+ \. l/r„. = 1 + 1/r. Conclude that oo V5)/2. ^= that n fl n^ n has radius of convergence 2/(1 + \/S). (Using 1 oo unproved theorems the in this chapter, and the fact that ^ n -\/{z 2 + z - + - z 1 15. — the smallest absolute value of the roots / 0; since the roots are (-1 ± "V 5)/2, the radius = indeed equal to 2/(1 + + Vs)/2. Notice that this number Vs).) can be written as the power series 1 + z/2\ + z /3\ z which is nonzero at 0, it follows that if g(z) = z/(e — 1) for + around 0. circle some = inside 1, then ^ is differentiate z 0, and g(0) there is a that chapter this in theorem It then follows from the unproved Since • (** • power - 1)A 2 • series VK z - Lf = 1 n convergence; bers z = it is 2tt, Ikiri for since this which are called the Bernoulli (a) Clearly b0 = g(0) <?* 1. - + z - e e~ can even predict the radius of the smallest absolute value of the 1 = Now _ _ The numbers 0. bn num- appearing here show that z + z z_ \ 2 2^-1 1 e*+l 1 e z e~ z We is numbers.* = z - n\ 0 with nonzero radius of convergence. * n is of convergence should be (-!_ is anz l from Problem 23-8, we could have predicted that 1) the radius of convergence of z 2 — z - e 1 1 because b n = 0 Sometimes the numbers fl n = (-l)"-^ 2 « are called the Bernoulli numbers, numbers b 2n alternate in sign, although we n is odd and > 1 (see part ((a)) and because the are also in use. will not prove this. Other modifications of this nomenclature if Complex Power Series 479 and deduce that b (b) By l = bn = 0 if is 72 finding the coefficient of z n odd and > n 1. in the right side of the equation z show that SO- 0 for n > 1. This formula allows us to compute any b k in terms of previous ones, and shows that each is rational. Calculate two or three of the following: b2 *(c) Part (a) = = "Mi bi = = 4V5 _i_ " 3 0- shows that b 2n , (In (2n) z e>+\ z 2 e* - 2 2n ! 1 + e 2 ' 2 - ^ 2/2 n=U Replace z by 2iz and show that 00 zcotz= Show *(d) Show n *\n^<Ln„in 2 2 (_1) . • that tan z *(e) b<i A(2T! n=0 = cot z — 2 cot 2z. that 00 tan z = V ^ 2n Li (2ft) (-\)n-i2 2n (2 2n - \)z 2n ~\ ! (This series converges for |*| < 7r/2.) numbers play an important role in a theorem which is denote the best introduced by some notational nonsense. Let us use D to k will mean D Then /'. denotes that so f operator," Df "differentiation The f (k) Bernoulli and e D f will mean (n) £= 0 / A! (of course this series makes no sense in 71 general, but Finally, f( x _|_ i) let it A will make sense if /is a polynomial function, for example). denote the "difference operator" for which Af(x) - f( x y Now Taylor's Theorem implies, = disregarding questions ! 480 Infinite Sequences and Infinite Series of convergence, that or n= we may write this symbolically as e D — 1, — A/ = Even more symbolically "identity operator." A = l where 1 is the can be written 1)/, this which suggests that Thus we obviously ought to have i.e. The (a) beautiful thing about all this nonsense Prove that (**) which case the fk\ (*) to is literally true if infinite sum is / that it works a polynomial function (in By applying really a finite sum). Hint: find a formula for /<*>(* formula in Problem is is + 1) - /<*>(*); 15(b) to find the coefficient of then use the f ij) (x) in the right side of (**). (b) Deduce from (**) that 00 (c) g(0) + (d) Show • • • that for + g(n) Apply any polynomial function g we have = f^git) this to g(x) = dt x v to + ££ show that [*<*-»(!, + 1) - g«-»(0)l Complex Power Using the = fact that bi —h 481 Series show that ten instances of this formula were written out in Problem pattern. 2-6, which offered as a challenge the discovery of the general Berthe but suggestion, This may now seem to be a preposterous After this way! precisely in discovered noulli numbers were actually The first writing out these 10 formulas, Bernoulli claims (in his posthumously printed work Ars Conjectandi, 1713): " Whoever will examine the be able to continue the table. He then writes down the above formula, offering no proof at all, merely noting that the coefficients b k (which he denoted simply by A, B, series as to their regularity C, . . .) satisfy the 55 may equation in Problem 15(b). tween these numbers and the coefficients z/{f — 1) was discovered by Euler. in The relation be- the power series for 16(c) can be generalized to the case where g is the infinite sum must be replaced by a finite function; not a polynomial In order to find an expression for the term. remainder sum plus a The formula remainder, (a) The Problem in it is useful to introduce Bernoulli polynomials some new functions. are defined by <p n 71 The first three are <Pi(x) Show that = bn = b n if > 1, (p n '(x) = n<p n -i(x), *>n« = (-l)Vn(l -*) *>n(0) <Pn(X) , Tl for,* > Hint: Prove the last equation by induction on 1. «, starting with n - 482 Infinite Sequences and Infinite Series (b) R N k (x) Let interval [x, be the remainder term in Taylor's formula for f k \ on the x + so that 1], N /<*>(* (*) + - 1) ^+ U+n)( V ^— fM( X) = Prove that AT = fW [/U>( b ~% i RN + 0 ~ /<4)W] * Xir* - N— (c) Hint: Imitate Problem 16(a). Notice the subscript Use the integral form of the remainder to show that 4 (d) *(*) =0 Deduce the "Euler-Maclaurin Summation Formula": +*(* + 1) + • • + g(x + n) • N +l = k on R. s(t) dt J^" V £ lgV°-»(x + n-l)- + g<*-*>(x)] + SK (x, n), where i (e) Let $n(t) ^n is =o n be the periodic function, with period = <p n (t) \// < for 0 continuous, since even and odd if n is < / <p n 1. (l) odd.) 1, (Part (a) implies that = <p n Show (0), and also that which if n <p n is tions this). become SN (x, n) = 0, satisfies 1, even then if n is that Unlike the remainder in Taylor's Theorem, the remainder Sn(x, does not satisfy lim > n) usually because the Bernoulli numbers and func- large very rapidly (although the first few examples do not suggest Nevertheless, important information can often be obtained from the summation formula. The general situation is best discussed within the context of a specialized study ("asymptotic series"), but the next problem shows one particularly important example. 5 ! Complex Power *18. N= Use the Euler-Maclaurin Formula, with (a) n 1 logtdt--logn / (b) Show (c) Explain + show that [ n Ji — ^(0 dt. that why show that if the improper integral 0 a = g (d) 2, to 483 Series = " |t 2 fa(t)/2t dt exists, and then \a/2» +1 ' V"V Jn It 2 Problem 18-26(d) shows that 0z!) V^r = lim 2 2 2" (2n) Use part (c) to show that V/- a 2n 2n+1 e- 2n 2 2n = 7r lim 7=> a(2n) and conclude that (e) Show that the o: 2w " H/ V~ 2B Vn = V2^. maximum value of *J 4 i/: 2/ 2 < " \(p*(x)\ for x in [0, 1] is £, so that 1 12w Conclude that W »+1/2<T»-1/I2n The final result of Problem / that n is ! approximately V 18, < „! ^+l/V" n+1/12n < . a strong form of Stirling's Formula, shows n+1/2<T n in the sense that this expression differs 2tt w , examfrom n by an amount which is small compared to n when n is large. For <1%. error an with = 3628800, 10 we obtain 3598696 instead of ple, for n " asymptotic the illustrates Formula A more general form of Stirling's used in nature of the summation formula. The same argument which was have we iV > 2 that for show to used be now 18 can Problem ! 5 i ( ni \ = v bk + r ^ dt 484 Infinite Sequences and Infinite Series Since \j/ N is bounded, we can obtain estimates of the form 00 1/ UN n1 N is large, the constant will make MN jMf) N tV! dt < Mn t will also be large; but for very large n the factor the product very small. Thus, the expression N V^r> + l/2,-n may . ( y ** \ be a very bad approximation for n! when * is small, but for large n (how on N) it will be an extremely good one {how good depends on N). large depends There was a most ingenious Architect who had contrived a new Method for building Houses, by beginning at the Roof, and working downwards to the JONATHAN SWIFT Foundation. PART EPILOGUE CHAPTER tff FIELDS I this book a conscientious attempt has been made to define all important concepts, even terms like "function," for which an intuitive definition is often considered sufficient. But Q, and R, the two main protagonists of this story, have only been named, never defined. What has never been defined can never be analyzed thoroughly, and "properties" PI -PI 3 must be con- Throughout sidered assumptions, not theorems, about numbers. Nevertheless, the term "axiom" has been purposely avoided, and in this chapter the logical status more carefully. Like Q, and R, the sets N and Z have also remained undefined. True, some talk about all four was inserted in Chapter 2, but those rough descriptions are far from a definition. To say, for example, that N consists of 1, 2, 3, etc., merely names some elements of N without identifying them (and the "etc." is useless). The natural numbers can be defined, but the procedure is involved and not quite pertinent to the rest of the book. The Suggested Reading list of PI -PI 3 will be scrutinized contains references to this problem, as well as to the other steps that are one wishes to develop calculus from its basic logical starting point. program would proceed with the definition definition in terms of Z. This program results terms of N, and the of required The if further development of this of Z, in Q in a certain well-defined set and properties PI -PI 2 as theorems. struction of R, in terms of + and program is the conconstruction which concerns us. certain explicitly defined operations Q, The •, final step in this Q,. It is this last Assuming that Qhas been defined, and that PI -PI 2 have been proved for Q, we shall ultimately define R and prove all of PI -PI 3 for R. Our intention of proving PI -PI 3 means that we must define not only real numbers, but also addition and multiplication of real numbers. Indeed, the real numbers are of interest only as a set together with these operations: how the real numbers behave with respect to addition and multiplication is crucial; what the real numbers may actually be is quite irrelevant. This assertion can be expressed in a meaningful mathematical way, by using the concept of a "field," which includes as special cases the three important number systems of this book. This extraordinarily important abstraction of modern mathematics incorporates the properties P1-P9 common to Q,, R, and C. A field is a set F (of objects of any sort whatsoever), together with two "binary operations" and defined on F (that is, two rules which associate to elements a and b in F, other elements a b and a b in F) for which the following con- + • + • ditions are satisfied: {a (2) There (i) (ii) 487 + b) + (1) a is c = a + (b + some element 0 +0 = a for all a, c) in F such b, and c in F. that for all a in F, for every a in F, there is some element b in F such that a +b = 0. +b = (3) a (4) (a (5) There b) • a (i) • +a b = c • some element 1 = = (6) a * b (7) <z • (b b + • F with a in = b and in F. c that 1^0 and = a 0, there a is some element • and +a b • 6 in F such F for all a, c 6, and F in c familiar examples of fields are, as already indicated, Q,, R, The b in 1. for all a a c) b, F such 1 in for all a in F, a • b in F. for all a, c) * is that a and for all a (b - For every (ii) + a and C, with + and \ It is probably unnecessary being the familiar operations of to explain why these are fields, but the explanation is, at any rate, quite brief. the rules 1, 3, and and are understood to mean the ordinary When the elements which play 4, 6, 7 are simply restatements of PI, P4, P5, P8, P9; the role of 0 and 1 are the numbers 0 and 1 (which accounts for the choice of and • + + • the symbols 0, 1); and the number b in (2) or (5) is -a or aT\ respectively. (For this reason, in an arbitrary field F we denote by —a the element such that 1 (_ fl ) = 0, and by a -1 the element such that a a" = 1, for a ^ 0.) a + • In addition to R, and C, there are several other Q,, One example is The operations + and the collection described easily. Q for a, b in for real numbers. It is • will, F of Y fields which can be numbers all + b V2 + and a once again, be the usual necessary to point out that these operations really do produce new elements of F\\ {a + b V2) + {a + b V2) Conditions for all real a _|_ & satisfies field are a + 0 in (2) + b V2 = 0. Similarly, verification of (5ii) obvious for Fu Fl since these hold numbers of_the form = 0 + 0 V2_is in Fi number 0 the holds because = (-a) + ( — b) V2 in Fi in Fi the number j3 numbers, they certainly hold a i; is (1), (3), (4), (6), (7) for a = a is c) V2. Condition and, for The + d V2) = (a + + (b + d) V2, which in F (c + d V2) = (ac + 2bd) + (be + ad) V 2, which (c * 1 = 1 for all real + 0 V2 is in Fi, so (5i) is the only slightly difficult point. If a a + b V2 is satisfied. + b V2 ^ 0, then 1 it is therefore necessary to — 7= <z + 6V2 show that \/(a = 1; + b Vl) is in Fi. This is true because 1 aTTVl _ " (The division by a a ( fl - b A - bV2 V2)( V2 is fl + ^ b V~2) a , a*-2b> valid because the relation a J^llL V2 a* - -2b* 6 = 0 could 489 Fields = be true only if hypothesis a + b V2 a b = V2" 0 (since irrational) is which is ruled out by the 0.) The next example of a field, F is considerably simpler in one respect: it con2 two elements, which we might as well denote by 0 and 1. The and are described by the following tables. operations , tains only + The • + 0 1 0 0 1 1 1 0 equation may Our final a in R, and a, b, c — 0 or also be written 1 + (a, a) + (The and 0 0 1 0 1 Notice that in 1. — = this field 1 + 8 equations = 1 0; this 1. silly: F 3 consists of all pairs (0, a) for • (a, a) cation for R.) 0 may be proved by checking the example of a field is rather and are defined by + 1 (l)-(7) are straightforward, case-by-case verification of conditions checks. For example, condition (1) obtained by setting 0 - = = (b, b) (b, b) (a + {a * b, b,a a • + b), b). appearing on the right side are ordinary addition and multipliThe verification that Fz is a field is left to you as a simple exercise. A for detailed investigation of the properties of fields our purposes, fields is a study in itself, but provide an ideal framework in which to discuss the numbers in the most economical way. For example, the consequences of P1-P9 which were derived for "numbers" in Chapter 1 actually hold for any field; in particular, they are true for the fields Q, R, and C. Notice that certain common properties of Q, R, and C do not hold for all properties of fields. fields, a = b. For example, it is For the field C + 1 = 0 to hold in some — a does not necessarily imply that + 1^0 was derived from the explicit possible for the equation 1 and consequently —b a = the assertion description of C; for the fields b 1 Qand R, however, this assertion was derived from further properties which do not have analogues in the conditions for a There is a is field F (with operations + and •) together with a (the "positive" elements) with the following properties: (8) For (i) (ii) (iii) (9) If a (10) If a all a in a a F, one = 0, is in P, —a and and is b and only one of the following in P. are in P, then a b are in P, then a +b • b is is in P. in P. field. An ordered field certain subset P of F a related concept which does use these properties. is true: We have already seen that the field C cannot be made into an ordered field. field F with only two elements, likewise cannot be made into an ordered The 2, field: in fact, then Fh the field be made It applied to (8), + = 1 consisting of all into an ordered positive real into condition (9) implies that 1 numbers an ordered field; 0 is — = 1 numbers + b V2 a P field: let be the the description of P is with in a, b The field left to you. 1 must be in P; On the other hand, + set of all a ordinary sense). (in the shows that 1, in P, contradicting (8). F s Q, certainly can which are can also be made natural to introduce notation for an arbitrary ordered field which is R we corresponds to that used for Q, and a a a a > < < > define : —b b if a b if b b if a b if a > < > in P, is a, b or a b or a Using these definitions we can reproduce, = = for b, b. an arbitrary ordered field F, the definitions of Chapter 7: A set A of elements of Fis bounded above if there is some x in Fsuch that x > a for all a in A. ment x of F is a x < Finally, Any least y for every y in it is such x an upper bound for A. An eleif x is an upper bound for A and an upper bound for A. is called upper bound F which possible to state is A for an analogue of property PI 3 for R; this leads to the last abstraction of this chapter: A complete ordered field is an ordered field in which every nonempty set which is bounded above has a least upper bound. The consideration of fields may seem to have taken us constructing the real numbers. However, intelligible will means of formulating this goal. we are There are two questions which be answered in the remaining two chapters: 1. Is there a complete ordered field? 2. Is there only one complete ordered field? Our field, containing N and Z as certain subsets. be necessary to assume another fact about Q: Let x be an element of Q, with x in assumed to be an At one crucial point it starting point for these considerations will be Q,, ordered will from the goal of provided with an far now N such that nx > > 0, Then for any^ in Q, there is some n y. This assumption, which asserts that the rational numbers have the Archimedian property of the reals, does not follow from the other properties of an ordered field (for the example that demonstrates this conclusively see [17]). The important point for us is that when Q, is explicitly constructed, proper- 491 Fields ties if PI -PI 2 appear as theorems, and so does this additional assumption; really began from the beginning, no assumptions about Q, would be we necessary. PROBLEMS 1. Let F be the set {0, 1, 2} and define operations + following table. (The rule for constructing this table and is • on F as follows: by the add or multiply in the usual way, and then subtract the highest possible multiple of 3; thus 2 = 2 4 = + 0 1 2 0 0 1 2 1 1 2 2 2 0 Show that F is a field, + 3 1, so 2 = 2 • 1.) 0 1 2 0 0 0 0 0 1 0 1 2 1 2 0 2 1 and prove that it cannot be made into an ordered field. 2. Suppose now that we try to construct a field F having elements 0, 1, 2, 3 and defined as in the previous example, by adding or multiplying in the usual way, and then subtracting the highest possible with operations multiple of 3. Let F= 4. + • Show {0, 1, a, that F will not be a field. 0} and define operations + and • on Fby the follow- ing tables. + 0 1 a 0 0 0 1 a 0 1 1 0 /3 a a a 0 Show 4. (a) that F is a OL 0 1 1 0 1 F 0 0 0 0 0 0 1 0 1 a a 0 a 0 0 + 1 = 0. Show that a field in which 1 a). can also be written a = Suppose that a a = 0 for some a ^ 0. Show that consequently £ £ = 0 for all b). Let a 1 a 1 field. be a a (this (b) 0 + + — +a 1 + = 0 1 for all = 0 (and 492 Epilogue 5. (a) Show m mn n times times numbers m and for all natural (b) we have that in any field Suppose that n. F we in the field have 1+^^+1 m times = o times some natural number m. Show that the smallest m with this number (this prime number is called the for property must be a prime characteristic of F) 6. Let F (a) Show be any with only finitely field must be that there m (b) many distinct natural numbers m and n with n times times Conclude that there elements. some natural number is M^_^fi = k with o. k times 7. Let a, b, c, and d be elements of a field F with a — • 6 ^ c • 0. Show that the equations 8. • x £ • at + b2 (a) (b) b • b = Consider an equation x 2 ( In the On b + r)/2 field F 2 is A a, /3, "square root" of a is an element b of 1 + = 1 0, + be incorrect. b b 2 x + —4 = c • • c 0, where b and c has a square root are elements r in F. Show a solution of this equation, of the text, both elements clearly have a square root. the other hand, the equation x 2 Let F. = = which case a has only one. that 10. j; a. of a field F. Suppose that (b) • * How many square roots does 0 have? Suppose a ^ 0. Show that a has two square roots, unless in (a) = £ d y can be solved for x and y. Let a be an element of a field .Pwith 9. a it is +x+ What easy to check that neither element satisfies = 0. Thus some detail in part (a) must 1 is it? F be a field and a an element of F which does not have a square root. This problem shows how to construct a bigger field F\ containing F, in which a does have a square root. (This construction has already been carried through in a special case, namely, F = R and a = — 1 this ; special case should guide you through this example.) . . 493 Fields Let F' consist of are + and 0 O 0, y) (x, y) Prove that Prove that (a) (b) all pairs w) jF', (x, Let + F + (# • 0 0 0) 0) we may agree be the and u, v) • z, (y, w, y • (y, = (x+y, 0) +x z 0 and = O, • is a>) a field. 0), 0), to abbreviate = • (at, 0) by 0) in F' (a, set of all four-tuples (w, x, y, z) of real + u, v) t, (w, x, y, z) (w, x,y,z) — (sw Show that — F satisfies the algebra will plicative inverses (b) • 0) tx = (s + w, — uy — vz, sx + sy (a) + w), y +a z F numbers. Define by • (s, , (x Find a square root of a (c) 0 with the operations (x, so that = = w) (z } and y in F. If the operations on and O on F' as follows: y) with x (x, define operations It is tz, field, except (6). ty — ux). At times quite ornate, but the existence of multi- the only point requiring any thought. customary to denote (0, 1,0,0) byi, 0,1,0) by;, (0, 0,0, (0, Find uw conditions for a all become is + x, u + y, v + z), + tw + uz — vy, + vx — sz + vw + t all 9 products of pairs ticular that condition (6) known i, is as the quarternions. 1) by j, and k. k. The show in par"skew field" F is results will definitely false. This CHAPTER 28 CONSTRUCTION OF THE REAL NUMBERS of drudgery which this chapter necessarily contains is relieved by one truly first-rate idea. In order to prove that a complete ordered field exists we will have to explicitly describe one in detail; verifying conditions (1)— (10) for an ordered field will be a straightforward ordeal, but the description of the The mass the elements in field itself, of is it, ingenious indeed. raw material the real called be ultimately will which it is numbers. To the uninitiated this must seem utterly hopeless if only the rational numbers are known, where are the others to come from? By now we At our disposal is the set of rational numbers, and from this necessary to produce the field have had enough experience to — realize that the situation hopeless as that casual consideration suggests. The may not be quite so strategy to be adopted in our construction has already been used effectively for defining functions and complex numbers. Instead of trying to determine the "real nature" of these concepts, we settled for a definition that described enough about them to determine their mathematical properties completely. A numbers requires similar proposal for defining real numbers in terms of rational numbers. The ought to be determined completely by the suggests a strikingly simple and quite a description of real observation, that a real set of rational numbers less attractive possibility: a real number than it, number might (and in fact eventually will) be described as a collection of rational numbers. In order to make this proposal effective, however, some means must be found for describing "the set of rational numbers less than a real number' without mentioning real numbers, which are still nothing more than heuristic figments of our mathematical imagination. If A is to be regarded as the set of rational numbers which are less than the 5 real is number a, then a rational A ought number If x is in A andy In addition to this have the following property: A should have number x < a, the set A property, the set rational to satisfying y < x, then y is in A. a few others. Since there should be some should not be empty. Likewise, since there should be some rational number x > a, the set A should not be all of Q,. Finally, if x < a, then there should be another rational number y with x < y < a, so A should not contain a greatest member. the real numbers as known, then it is not hard to check (Problem 8-17) that a set A with these properties is indeed the set of rational numbers less than some real number a. Since the real numbers are presently in limbo, your proof, if you supply one, must be regarded only as an unofficial comment on these proceedings. It will serve to convince you, how*If we temporarily regard we have not failed to notice any crucial property of the appears to be no reason for hesitating any longer. ever, that 494 set A. There Construction of the Real DEFINITION A number real is 495 Numbers a set a, of rational numbers, with the following four properties: number withjy < If x (2) OL (3) Q. There is no greatest element in a\ there is some y in a with y > x. (4) The a and^ (1) is * a * in is a rational set of all real numbers is also in a. should be clear that a is if x is then in a, denoted by R. is you of the philosophy behind our example of a real number: a = It thenjy in other words, Just to remind explicit x, 0. \x in Q: # < 0 or x 2 the real < number which definition, here is an 2} will eventually V2, be known is a but it is not an entirely trivial exercise to show that a number. The whole point of such an exercise is to prove this using only facts about Q; the hard part will be checking condition (4), but this has already appeared as a problem in a previous chapter (finding out which one is up to you). Notice that condition (4), although quite bothersome here, is really as actually real essential in order to avoid ambiguity; \ x in without Q: x < 1} < 1} it both and {x in Q/. x would be candidates for the "real number 1." The shift from A to a in our definition indicates both a conceptual and a notational concern. Henceforth, a real number is, by definition, a set of number (a number x has a rational numbers. This means, in particular, that a rational member of Q) is not a real number; instead every rational is a real number, namely, {y in Q: y < x). After completing the construction of the real numbers, we can mentally throw away will henceforth denote these special sets. the elements of and agree that For the moment, however, it will be necessary to work at the same time with natural counterpart which Q Q rational numbers, real real numbers (sets numbers (sets of sets of rational of rational numbers) and even numbers). Some confusion inevitable, but proper notation should keep this to numbers will be denoted by lower case Roman letters (x, y, z, a, b, real numbers by lower case Greek letters (ex, /3, 7) capital Roman ; (A, B, C) will be used to denote c) and letters numbers. devoted to the definition of +, •, and P R, and a proof that with these structures R is indeed a complete ordered The remainder for sets of real sets of perhaps a minimum. Rational is field. of this chapter is 496 Epilogue We we We first define a < P shall define properties for P. we begin with the definition of P, and even here shall actually work backwards. P; later, when +, and 0 are shall available, a with 0 < a, and prove the necessary with the definition of < is the beginning reason for as the set of all The simplicity of this concept in our present setup: Definition. If in a and p a < P means that a is contained an element of/3), but a ^ p. are real numbers, then every element of a also P (that A repetition of the definitions of <, >, is, is > would be but stultifying, it is be expressed more simply than < if a and < only if a is contained in p. and then a if numbers, are real P P If A is a bounded collection of real numbers, it is almost obvious that A interesting to note that < now can ; should have a least upper bound. Each a in A is a collection of rational numbers; if these rational numbers are all put in one collection P, then P is presumably sup A. In the proof of the following theorem we check all the little which have not been mentioned, not least of which is the assertion that (We will not bother numbering theorems in this chapter, P is since they all add up to one big Theorem: There is a complete ordered field.) details a real number. theorem If A least PROOF a set of real numbers and A upper bound. is Let P = {x: x is in some a numbers; the proof that P (1) in A}. is a real ^ 0 and A is bounded above, then A has a Then p is certainly a collection of rational number requires checking four facts. Suppose that x is in P and y < x. The first condition means that x is a for some a in A. Since a is a real number, the assumption y < x in implies that y (2) Since A ^ 0, is in a. Therefore there is some a certainly true that y in A. Since some x in a. This means that x (3) it is is a in P, so is p is in p. a real number, there ^ is 0. bounded above, there is some real number y such that a < 7 for every a in A. Since 7 is a real number, there is some rational number x which is not in 7. Now a < 7 means that a is contained in 7, so it is also true that x is not in a for any a in A. This Since A is that x is not in p; so P ^ Q. Suppose that x is in p. Then x is in a for some a in A. Since a does not have a greatest member, there is some rational number y with x < y and y in a. But this means that y is in p; thus P does not have means (4) a greatest member. is a real number. The proof that P is bound of A is easier. If a is in A, then clearly a is contained in P; this means that a < P, so P is an upper bound for A. On the other hand, if 7 is an upper bound for A, then a < 7 for every a in A; this means These four observations prove that P the least upper Construction of the Real that a contained in 7, for every is a in A, The definition of with a proof that Definition. If a THEOREM PROOF If = ' 'obvious' 5 definition this surely implies that thus 0 is 0 the least upper is con- bound it must be complemented makes any sense at all. are real numbers, then = x {x: y + z for some y in are real numbers, then Once again ; both obvious and easy, but is a and 0 +0 a and 0 + this and < y tained in y. This, in turn, means that 0 of A. I 497 Numbers a +£ is a and some z in 0). a real number. four facts must be verified. some x in a /3. Then a: = jy + 2- for some)' in a which means that w < y + z, and consequently, w — y < z. This shows that w — y is in 0 (since ^ is in 0, and ^ is a real number). Since w = y + (w — y), it follows that 2x> is in a It is clear that a 0. 0, since a 5^ 0 and /3 0 Since a and 0 5* Q, there are rational numbers a and b with a not in a and 6 not in 0. Any x in a satisfies # < a (for if a < *, then condition (1) for a real number would imply that a is in a); similarly any y in 0 satisfies y < b. Thus x + < a + £ for any x in a and ^ in 0. This shows that a + 6 is not in o: /3 ^ Q,. 0, so a If x is in a /3, then a: = y + ^ for y in a and z in 0. There are / in a and in 0 with _y < / and z < z then x < y + z and Thus a is in a 0 has no greatest member. | _/ + Suppose w < x and some z in (1) + for /3, + + (2) (3) + + + (4) f f f ; + + By now you can see how tiresome this whole procedure is going to be. Every we mention a new real number, we must prove that it is a real number; this requires checking four conditions, and even when trivial they require time is really no help for this (except that it will be less you check the four conditions for yourself) Fortunately, however, a few points of interest will arise now and then, and some of our theorems will be easy. In particular, two properties of present no problems. concentration. There boring if . + THEOREM PROOF theorem PROOF and 7 are If a, 0, Since (x + y) + member of (a If a and 0 Left to z real numbers, then (a = + x + 0) + 7 (y is + z) for all rational also a member are real numbers, then you (even easier). | + 0) + 7 a +0 of a = 0 + = a + (0 numbers x, y, + y)> and (0 + a. + 7). and z, every vice versa. | 498 Epilogue To + prove the other properties of = 0 Definition. Q: x < {x in we PROOF If a is +0 a a real number, then define 0. 0}. It is, thank goodness, obvious that 0 theorem is also simple. THEOREM first = a real number, and the following is a. in a and y is in 0, then y < 0, so x + y < x. This implies that x + y is Thus every member of a + 0 is also a member of a. On the other hand, if x is in a, then there is a rational number y in a such that y > x. Since x = y + (x — y), where y is in a, and x — y < 0 (so that * - y is in 0), this shows that x is in a + 0. Thus every member of a is also a If * is in a. member The of + 0. | a —a reasonable candidate for {x in Q,: (since —x not in a means, would seem —x is set not in a} intuitively, that certain cases this set will not even be a real a be the to —x > a, so that x < — a). number. Although a But in number real does not have a greatest member, the set Q— —x is { x in Q,: x is a not in element x 0 when a is a real number of this kind, the set not in a will have a greatest element —x 0 It is therefore necessary may have {x: a — a ; . } to introduce a slight modification into the definition of —a, which comes equipped with a theorem. Definition. If —a = THEOREM PROOF If a is a Q: {x in a real number, then is —* is a real number, then (1) it is is Since in is not the least element of Q— a}. a real number. a ^ Q,, is jy there can assume that y (since —y in is is is — a. some rational number —a Since a ^ —a ^ which not the smallest rational can always be replaced by any y Thus so —a and y < also true that This shows that (3) —a is x. Then — y > —x. Since —x is not not in a. Moreover, it is clear that — y is not the smallest element of Q, - a, since —x is a smaller element. Suppose that x in a, (2) —x not in a, but 1 > y). is Q, We — a in —a. not in a. number in Then —y is 9* 0. 0, Q. there is some x in a. Then — * cannot possibly be in —a, — If (4) x — a, in is y < —x y < < z —.v. element of —a —x then which is Q is number number with a rational also not in a. Let z be a rational Then — not in a, and there is a. z also not in a, is —z So is in — a. and z is Since 499 Numbers Construction of the Real clearly not the smallest —z > shows that x, this does not have a greatest element. | + — a) = 0 is not entirely straightforward. The diffiyou might presume, by the finicky details in the definition of a. Rather, at this point we require the Archimedian property of Q, stated on page 490, which does not follow from PI -PI 2. This property is needed to prove the following lemma, which plays a crucial role in the next The proof a that ( culties are not caused, as — theorem. LEMMA Let a be we may assume over, PROOF Suppose that z first that y 3 then all in a, number w satisfies nz for . . some w, a 490. This contradicts the fact that = — a. z. More- . number would be every rational w < there are x numbers in a. If the is — not the smallest element of Q, is z 2z, 3z, were Then a real number, and z a positive rational number. (Figure 1) rational numbers x in a, and;? not in a, such thatjy is every rational in a, since by the additional assumption on page a real number, so there is some k such not in a. Clearly y — x = z. r Moreover, if y happens to be the smallest element of Q, — a, let x > x be — x (V x). x\ and an element of a, and replace by y by y If z is not in a, there is a similar proof, based on the fact that the numbers that x — kz is in a and y = (k + \)z is + ( — n)z cannot be in a. all fail to | x FIGURE THEOREM If a is a real number, then a PROOF Suppose * x +y < It is —z > a 0. is in a and y +y in 0. 0, so x little more According is is can be written x proves that z is in a + ( 0. —a. Then — y is not in Thus every member of a a). | a, so + go in the other direction. lemma, there (—y) = + (—a) = in difficult to to the + with;/ not the smallest element of tion y 1 Q— z. is some x is If z is is x. in «, Hence in 0. in 0, then and some y not in a, — x = —z. This equaand — y is in «, this in a, a, such that;; Since x —y > (—a) — 500 Epilogue Before proceeding with multiplication, we define the "positive elements" and prove a basic property: Definition. P = + Notice that a THEOREM If a If a /? is 0}. clearly in P if a and 0 are. a real number, then one and only one of the following conditions holds: is (ii) a = a is in P, (iii) —a is (i) PROOF a > {a in R: 0, in P. contains any positive rational number, then numbers, so a contains 0 and a ^ no positive rational numbers, then one of two tive rational (1) (2) a contains certainly contains all nega- a i.e., negative rational numbers; then all is in P. If possibilities a contains must hold: a = 0. some negative rational number x which is not in a; it can be assumed that x is not the least element of Q, — a (since x could be replaced by x/2 > x); then —a contains the positive rational number there — x, is we have so, as This shows that just proved, —a is in P. must hold. If a = 0, it is clearly impossiMoreover, it is impossible that a > 0 and (—a) > 0. | would imply that 0 = a at least one of (i)-(iii) ble for condition —a > a 0, (ii) or (iii) 0 both hold, since to hold. this + but is unequal to a > 0 was defined to mean that a contains This definition was fine for proving completeness, but now we have to show that it is equivalent to the definition which would be made in terms of P. Thus, we must show that a 0 > 0 is equivalent to a > 0. This is clearly a Recall that /5. — consequence of the next theorem. theorem PROOF If a, ft The and y are hypothesis real a > 0 from the definition of a +y > numbers and a implies that + that /3 0 +y is > /3, then +y a contained in a; is contained in a it > 0 + 7- follows immediately + y. This shows that + y. We can easily rule out the possibility of equality, for « + 7 = 0 + 7, if then a = which is false. (a Thus + 7) + (-7) = (0 + 7) + a+y > + y. | ft /3 Multiplication presents difficulties of be defined as follows. ("7) = its own. If a, £ > 0, then a 0 can • a and 0 Definition. If a THEOREM PROOF • j8 = a and 0 If As usual, (1) < z {z: are real numbers 0 or are real 2: = a • 0. a and tive x in < Since 0 (2) (3) a < a; then 0, 0, then 0 with in a 0 is • > x, 0} a real number. in > 0. a 0. If u> < Then z > 0, • positive ^ in 0. then is ze; = so z x • j> automatically for some posi- Now \z z we have z, 0, < zt>/z 1, ) x so (w/z) Thus in a. is u> is 0. • a 0 ^ Clearly • 0. and for all such positive *' and /. Hence ^ > is not in a So ft thus a j3 5^ Q,. Suppose w is in a ft and w < 0. There is some * in a with x > 0 in £ with y > 0. Then z = xy is in a 0 and z > and some Now suppose > 0. Then w = xy for some positive x in a and some positive y in ft Moreover, a contains some x > x; if z = x'.y, then z > xy = and z is in a ft Thus a 0 does not have a greatest If * j > is not in a, and / y' for all is not in 0, then x > *' for all *' in a, in 0. • * (4) 0 > is z in > 0 four conditions. w < z, where z Suppose that a; Suppose in • a, a, xina and y * j for some numbers with we much check and 501 Numbers Construction of the Real • • zt> r • • element. | Notice that a • j8 is clearly in tion of all properties of P. Definition. If a is To P a and 0 if complete the definition of if = a, a and 0 we • first |a| k if If or, 0, \a\. a >0 a < 0. if if - -(|a| • I0|), if a = 0 or 0 = 0 a > 0, 0 > 0 or a < a > 0, 0 < 0 or a < 0, 0, 0 0 < > 0 0. suspect, the proofs of the properties of multiplication usually involve reduction to the case of positive numbers. THEOREM define are real numbers, then 0, As one might This completes the verifica- a real number, then \a\ Definition. If are. and 7 are real numbers, then a • (0 • 7) = (a • 0) • 7- 502 Epilogue PROOF This is clear if a, 0, 7> separate cases (and THEOREM PROOF If a and This (3 THEOREM PROOF If is a Let On a, if = Definition. 1 (It 0 > The proof for is 0. It is • /3 the general case requires considering if one uses the following theorem). = 0 • a. cases are easily checked. | } 1 a = 1 • a. easy to see that every the other hand, suppose x member of a in a. If x is < • 1 is also 0, then x is a member of a. automatically in a 1. If x > 0, then there is some rational number y in a such that Then x = y (x/y), and #/y is in 1, so * is in a 1. This proves that a ifa>0. If a < 0, then, applying the result just proved, we have • a-1 = Definition. If a- 1 = { a theorem is THEOREM PROOF a< If a is 0, then when a = a real number and < x in Q: x a" = 1 > or x 0, = -(H) - -(|a|-|l|) obvious is member if 0 of Q, — (M" a> and 1/* — a 0, is (1) it then x > < 0, so not in a, and in (2) (3) (4) a x, only y. a. 0. | not in a, but 1/x is not the smallest } is a> in 0. a"1 . is a real number. Four conditions must be checked. If y not in a. Since is 1 /jy is = a ). and x 1/* < 1 1 suffices to consider Suppose y x then 1 a real number unequal to 0, then a" Clearly • • • Finally, the | a real number.) a real number, then a > a and the other 0, Q: x < {x in clear that 1 is 0. simplified slightly are real numbers, then clear is is < 0, 1/jy then > 1/x, y is it in a -1 . If j follows that clearly not the smallest element of Q, — > 0, 1 /jy is a, so y is -1 Clearly a" 1 ^ 0. a > 0, there is some positive rational number x in a. Then l/# in a \ so a" not is 1 Suppose x is in a" 1 If x < 0, there is clearly some y in a:"" with -1 contains some positive rationals. If x > 0, then y > x because a 1 /x is not in a. Since 1 /x is not the smallest member of Q, — a, there not in a, with y < 1/x. Choose a rational is a rational number number ^ with y < z < 1/x. Then 1/z is in a, and 1/z > *. Thus a-1 does not contain a largest member. | Since . jy Construction of the Real a" 1 In order to prove that is really the multiplicative inverse of a, have another lemma, which lemma. to LEMMA PROOF is 503 Numbers it helps the multiplicative analogue of our first Let a be a real number with a > 0, and z a rational number with z > Then there are rational numbers x in a, and y not in a, such that y/x = Moreover, we can assume that y is not the least element of Q — a. Suppose first that z in a. Since z is + *»=(! l)) 1 > 0 and n > 1 + n(z - z. 1), numbers follows that the it (*- — 1. z, z 2 z\ . , . . h cannot all be in a. So there is some k such that x = z is in a, and ^ = = to be the least element of happens if z. Moreover, is not in a. Clearly y /x y x' and j byyx'/x. by replace x and a, of element an a, let x' > x be Q, If z is not in a, there is a similar proof, based on the fact that the numbers l/z k cannot all fail to be in a. | number and a ^ THEOREM If a PROOF It obviously suffices to consider only a real is that x or 1 Then in that xy every in 1. is member 1. • a> in a, 0, in and ^ -1 which case a is > Suppose 0. a positive rational number not in a, so \/y > x; consequently xy < 1, which means Since all rational numbers x < 0 are also in 1, this shows that \/y . a a" 1 = then number a positive rational is 0, is a of • a"" 1 in 1. is prove the converse assertion, let z be in 1. If z < 0, then clearly z is in a or1 Suppose 0 < z < 1 According to the lemma, there are positive rational numbers x in a, and y not in a, such that y/x = \/z\ and we can To • . . assume that y is not the smallest element of Q, — a. But this means that 1 where * is in a, and l/> is in or 1 Consequently, z is in a a" | z = x • • . We are almost done The we must consider many cases, but do not despair. numbers are positive contains an interesting point, and the other again all If a, 0, PROOF Assume and y are * • (y a • 03 + 0. • is > 0. A a • (/3 + y) cases is also • • +a • can y. Then both numbers in the equation number in a (/? + also in • 7. a 0 • • in 7. Since x and z a positive element of number is in a 0 + a 7, this = a & positive rational z) for positive x in a, y in 0, • a + 7) y where * y z, element of numbers, then numbers < rational + • real that a, ft first the form x * case be taken care of very neatly. theorem all Once when all the distributive law remains. Only the proof of ! . a +a • • is + z) of = and x z is a positive Thus, every element of 0, 7- (y contain 7) • On form the other hand, a positive rational xi - (*i/*0 ' + y y < x2 ' a in To • (/3 a in • +a j8 7 5 y, so (*i/* 2 ) xi is number z for positive #i * 2 in a, y in ft and z in 7. If x x + 7). Of -jV is - y + in ft x2 * is is of the x 2 then , Thus = + x 2 [(xi/x 2 )y same course, the complete the proof it z < trick works z] if < #2 xi. necessary to consider the cases whema, ft and 7 any one of the three equals 0, the proof is easy and the cases are not all > involving a < 0 can be derived immediately once all the possibilities for 0 and 7 have been accounted for. Thus we assume a > 0 and consider three cases: ft 7 < 0, and 0 < 0, 7 > 0, and 0 > 0, 7 < 0. The first follows immediately from the case already proved, and the third follows from the second by interchanging (3 and 7. Therefore we concentrate on the case 0 < 0, 7 > 0. There are then two possibilities: 0. If (1) + 7 >0. 0 Then a 7 = « • (W • + 7] + = « ' 08 + 7) + « • 101, so «. + 7) 03 = -(a- |/3|) = a (3 + a * (2) 0 a • + 7 < 0. Then = « (10 + 7l + 7) 101 • = ol • |0 + a -7 • 7. + 7I + <* • 7, so a • (0 + 7) = -(« • 10 + 7l) = -(a 101) + a- 7 = <*-0 + <*'7.| This proof completes the work of the chapter. Although long and frequently tedious, this chapter contains results sufficiently important to be read in detail more than once !). For the first time we know there is indeed a complete in a vacuum operating been that we have not ordered field, the theorems of this book are not based on assumptions which can never be realized. One interesting and horrid possibility remains: there may be several complete ordered fields. If this is true, then the theorems of calculus are unexpectedly rich in content, but the properties P1-P13 are at least once (and preferably not disappointingly incomplete. — The last chapter disposes of this possibility; — properties PI -PI 3 completely characterize the real numbers anything that can be proved about real numbers can be proved on the basis of these properties alone. PROBLEMS There are only two problems in this set, but each asks for an entirely different construction of the real numbers The detailed examination of another construction is recommended only for masochists, but the main idea behind these other constructions is worth knowing. The real numbers constructed in this ! Construction of the Real 505 Numbers chapter might be called "the algebraist's real numbers," since they were purposely defined so as to guarantee the least upper bound property, which involves the ordering <, an algebraic notion. The real number system constructed in the next problem might be called "the analyst's real numbers," since they are devised so that Cauchy sequences will always converge. 1. number ought to be the limit of some Cauchy sequence we might try to define a. real number to be a Cauchy sequence of rational numbers. Since two Cauchy sequences might converge to the same real number, however, this proposal requires some Since every real of rational numbers, modifications. (a) Define two Cauchy sequences of rational numbers { a n } and { b n } to be {b n }) if lim (a n - b n ) = 0. Prove that equivalent (denoted by {a n } ~ \a n ) ~ if {a n (b) \ {a n ~ }, {b n that [a n and } Suppose that a is ] ~ {b n ~ n } {b n [c \ the set of all if {b n ~ } {a n }, and that {a n } ~ {c n } \. sequences equivalent to {a n }, and /3 is the set of all sequences equivalent to b n Prove that either a Pi /3 = 0 or a = 13. (If a C\ 0 ^ 0, then there is some [c n in both a and 0. Show } { . \ a and that in this case equivalent to /8 both consist precisely of those sequences {c n }.) Part (b) shows that the collection of all Cauchy sequences can be split up into disjoint sets, each set consisting of all sequences equivalent to (c) some fixed sequence. We define a real number to be such a collection, and denote the set of all real numbers by R. a If a and /3 are real numbers, let \a n be a sequence in a, and {b n ) ) a sequence in 0. Define + j3 to equivalent to the sequence {a n be the collection of + bn show that on the particular sequences \a n and Cauchy sequence and also } Show }. that {a n this definition {£ n } Show that R is a field with tive inverse (e) (f) + bn \ is a is a and fi. Check well defined. these operations; existence of a multiplica- the only interesting point to check. Define the positive real numbers P so that R will be an ordered field. Prove that every Cauchy sequence of real numbers converges. Remember that if { a n } is a sequence of real numbers, then each a n is itself 2. is sequences does not depend chosen for also that the analogous definition of multiplication (d) all a collection of Cauchy sequences of rational numbers. This problem outlines a construction of "the high-school student's real numbers." We define a real number to be a pair (a, {b n }), where a is an integer and \b n is a sequence of natural numbers from 0 to 9, with the proviso that the sequence is not eventually 9; intuitively, this pair repre] sents a + ) b n \0~ n . With this definition, a real number is crete object, but the difficulties involved in defining addition a very con- and multi- plication are formidable (how do you add infinite decimals without worrying about carrying digits infinitely far out?). A reasonable approach is outlined below; the trick to use least is upper bounds right from the start. (a) Define have (a, {b n }) < dn bn < (c, but b s = bound the least upper {dn d3 - }) if for 1 < c, or < j < n. a if a = Using and c some for n we prove this definition, property. k (b) Given a = a + ^ n- bn r and = j8 0. ^ nl0 6 n for = (c, a (the least ak = a + £ n= £ n lCT > 1 all Conversely, given a rational ~n < let « r' < A n £n' = 0 for « intuitively, > r is of the form f (a, where {b n }), Now for a = (a, {b n }) \dn \) define +0 = sup {(a* upper bound exists + &)': A a natural number} by part (a)). If multiplication similarly, then the verification of all conditions for a field forward ak decimal places after number denote th e real number and ; 1 number obtained by changing the rational the kth to {£„}), define (a, task, not highly recommended. is is defined a straight- Once more, however, existence of multiplicative inverses will be the hardest. ^ CHAPTER UNIQUENESS OF THE REAL NUMBERS We shall now revert to the usual notation for real numbers, reserving boldface fields which may turn up. Moreover, we will regard integers numbers as special kinds of real numbers, and forget about the specific way in which real numbers were defined. In this chapter we are interested in only one question: are there any complete ordered fields other than R? The answer to this question, if taken literally, is "yes." For example, the field F z introduced in Chapter 25 is a complete ordered field, and it is certainly not R. This field is a "silly example because the pair {a, a) can be symbols for other and rational 5 ' regarded as just another name + (a, a) (a, a) • for the real number (A, b) = {a + (6, b) « (a • b, a b9 a • + a; the operations b), b)> are consistent with this renaming. This sort of example shows that any intelligent consideration of the question requires some mathematical means of discussing such renaming procedures. If the elements of a field F are going to be used to rename elements of R, R there should correspond a "name" f(a) then for each a in in F. The notation functions. In order f(a) suggests that renaming can be formulated in terms of general than any more much function of to do this we will need a concept general notion most the require will we in fact, now; until which has occurred of "function" used in mathematics. A function, in this general sense, is simply a rule which assigns to some things, other things. To be formal, a function is a collection of ordered pairs (of objects of any sort) which does not contain two distinct pairs with the same first element. The domain of a function / is the set A of all objects a such that (a, b) is in /for some b; this (unique) b is denoted by /(a). Iff (a) is in the to B. For example, = if f(x) sin x for all x in function from if f(z) if f(z) z for s in e z R to R; all R it is all a in (and / 6 is a function from \z in C: z if / is the collection of to is is called a function from A defined only for x in R), then / is from R [— 1, to a function from a function from C is a 1]; to C; C to C; it is also a func- to \z in C: z ?± 0}; from R is z in C, then / for all z in C, then / C A, then / also a function tion from 507 = = B for set F 3. all pairs ^ 0} to {x in R: 0 (a, (a, a)) for a in < x < R, then / 2t}; is a function — : 508 Epilogue Suppose that F\ and F2 are two fields; we will denote the operations in Fi by ©j O, etc. and the operations in F2 by +, •, etc. If F2 is going to be considered as a collection of new names for elements of Fu then there should be a function from Fi to F2 with the following properties: The (1) function / should be one-one, that have fix) ^ f(y); same name. (2) The this function / should be "onto," that some x in Fi such that z there should be (3) is, if x ^ y, then means that no two elements of is, = for every f(x) ; element of F2 is used to name some element of For all x and y in Fi we should have F we should x have the element z in F 2 means that every this Fi. =f(x)+fiy), fixOy) =fix)-f(y}; f(x (By) means that the renaming procedure this is consistent with the opera- tions of the field. If we are also considering Fi and F 2 as ordered fields, we add one more requirement: (4) A lfx©y then /(*)</(>). function with these properties This definition DEFINITION } is so important that is we called an isomorphism from Fi to restate it If Fx and F2 are two fields, an isomorphism from Fi from Fi to F2 with the following properties: (1) Ux^y, (2) If (3) If F 2. formally. to F 2 is a function / then fix) * f(y). = fix) for some x in Fi. 2) then z x and y are in Fi, then z is in F f(x®y) =f(x)+f(y), f(xOy) =f(x)-f(y). If Fx and (4) F If 2 x are ordered fields €> y, then /(*) we also require < fiy). The fields Fi and F2 are called isomorphic if there is an isomorphism between them. Isomorphic fields may be regarded as essentially the same any important property of one will automatically hold for the other. Therefore, we can, and should, reformulate the question asked at the beginning of the chapter; if F is a complete ordered field it is silly to expect F to equal R rather, we would like to know if Fis isomorphic to R. In the following theorem, and "positive elements" P; we and F will be a field, with operations write a < b to mean that b a is in P, and so forth. — — + Uniqueness of the Real Numbers THEOREM PROOF F is If a complete ordered Since two F is then field, isomorphic to R. are defined to be isomorphic fields 509 if there an isomorphism to F which is an is between them, we must actually construct a function /from R isomorphism. We begin by defining / on the integers as follows: M = = f(n) 0, + 1 • +1 • • > for n 0, n times + = -(1 f(n) • • jn| + 1) • < for n 0. times easy to check that It is f(m + n) f(m-n) = = f(m) + f(n), f(m)-f(n), m and n, and it is convenient to denote rational numbers by the on / for all integers define f(m/n) (notice that so n" 1 = • • • • +1^0ifn>0, makes sense because definition m 1+ ft • t" 1 . It is The any definition of real number is /W m/n = then n" 1 • since F is an ordered then ml k/l, + u) = fin) = /(ri) /(ri numbers if We n. = nk so 9 field). m • = I This fc • n, easy to check that f(n for all rational m = m/n = by /(n) ri ' and n) r 2, + /(r • and that for arbitrary x is 2 ), /(n), /(ri) < /(r 2 ) if ri based on the now determined by the rational numbers < r2 . familiar idea that less than it. For any numbers let A x be the subset of F The set A x is certainly not empty, and it is also bounded above, for if >/(r) for all/(r) in A x Since F r 0 is a rational number with r 0 > x, then/(r 0 ) upper bound; we define /(*) least is a complete ordered field, the set A x has a consisting of all f(r), for all rational x in R, r < x. . as sup Ax . have /(*) defined in two different ways, first for rational x, and that these two then for any x. Before proceeding further, it is necessary to show rational number, we definitions agree for rational x. In other words, if x is a We now want to show that sup Ax = f{x), depends where /(x) here denotes m/n, for x = m/n. This is not automatic, but required. thus is digression slight on the completeness of F; a Since F is complete, the elements 1 +1 + n times for natural numbers n which is not bounded above; the proof is exactly the same as the (Theorem 8-2). The consequences of this fact for R have exact analogues in F: in particular, if a and b are elements of F with a < b then form a set proof for R y there is a rational number such that r < /(r) < a Having made we return to the proof that the two definitions a rational number with y < x, then we have Thus every element of A x is < f(x). Consequently, this observation, of f(x) agree for rational already seen that f(y) x. If y is < f(x). sup On sup there But the condition /(r) Ax Ax < /(*) means this clearly contradicts the \ A x < f(x). would be a rational number sup set A x < f(x). we had the other hand, suppose that Then the original assumption b. is false, such that r <f(r) </(*). that < r x, which means that f(r) is in the A x < f{r). This shows that condition sup so sup Ax = /(*). thus have a certain well-defined function / from R to F. In order to verify conditions (l)-(4) of the defini- We show that / is an isomorphism we must We tion. If will begin with (4). x and y are real numbers with x < y, then clearly Ax is contained in Ay . Thus /(*) To r = sup Ax < sup Av = f(y). rule out the possibility of equality, notice that there are rational and s numbers with x We know that f(r) < f(s). < r < s < y. It follows that /w</w</w</w. This proves (4). Condition (1) follows immediately from or y < x; in the either case /(*) first case /(*) < f(y), and (4): If x ^ y, then either x < y in the second case f(y) </(*); in ^ f(y). prove (2), let a be an element of F, and let B be the set of all rational numbers r with f{r) < a. The set B is not empty, and it is also bounded above, because there is a rational number s with f(s) > a, so that/(j) >/(r) for r in B, which implies that s > r. Let x be the least upper bound of B; we claim To Uniqueness of the Real Numbers that f(x) = a. In order to prove this it /(*) < 511 suffices to eliminate the alternatives <h «</(*). In the first would be a rational number case there /(*) But x = this means that x < < and that r This implies that r < x. Since x < a. in B, which contradicts the fact that would be a rational number r with r is sup B. In the second case there a f(r) with r < fir) < fix). = sup B, this means that r < s for some j in 2?. Hence <fis)<a, fir) again a contradiction. Thus f(x) = a, proving (2). To check (3), let x and y be real numbers and suppose that fix /(*) + fiy)- Then In the first this would mean + y) r Then, using the = facts f(r) +> < a contradiction. Finally, if a: ri +r 2, sum where x checked about /for = The number r < /(* + y). such that <m </(*)+/W- could be written as the r + /GO /(*) that ^ Therefore or case there would be a rational fr. But 5* either + y)< /« + /GO /(* + y) fin + n) other case = /(ri) is r. of two rational numbers < ti and rational +/(ri) handled < r2 . numbers, it would follow that > /(*) +/W, similarly. and^ are positive real numbers, the same sort of reasoning shows that /<*-j0 the general case is -/W-/W; then a simple consequence. | This theorem brings to an end our investigation of the real numbers, and any doubts about them: There is a complete ordered field and, up to isomorphism, only one complete ordered field. It is an important part of a mathematical education to follow a construction of the real numbers in detail, but it is not necessary to refer ever again to this particular construction. It is utterly irrelevant that a real number happens to be a collection of rational numbers, and such a fact should never enter the proof of any important resolves theorem about the real numbers. Reasonable proofs should use only the fact numbers are a complete ordered field, because this property of the real numbers characterize them up to isomorphism, and any significant mathematical property of the real numbers will be true for all isomorphic fields. To be candid I should admit that this last assertion is just a prejudice of the author, but it is one shared by almost all other mathematicians. that the real PROBLEMS 1. (a) (b) 2. F Let / be an isomorphism from Fx to 2. Show that /(0) = 0 and /(I) = 1. (Here 0 and 1 on the left denote elements in Fx, while 0 and 1 on the right denote elements of F2 .) Show Here that = /(-a) = -/(a) and /(a- 1 ) f(a)~\ for a * 0. an opportunity to convince yourself that any significant property is shared by any field isomorphic to it. The point of this problem is to write out very formal proofs until you are certain that all statements of this sort are obvious. Fi and F2 will be two fields which are isomorphic; for simplicity we will denote the operations in both by and •. Show that: is of a field + (a) If the in F 2 equation x 2 with a 0y . . equation x n 2 (c) (d) + 1 in F If F Fh 0 has a solution in then it has a solution , <2 * * n _ x in + b n ^x +1 • • xn + <z n __i Fi has a root in ~ l + • • +b • 0 • xn Fu — 0 ~l + +a • • • 0 — 0 then every polynomial with b 0, . . . in , F 2. (summed m times) = 0 in F ly then the same is true 2. and x /(*) 3. . has a root in If = polynomial equation x n (b) If every F +1 . < F 2 are ordered fields (and the isomorphism fiy) f° r x < y) an d Fi is complete, then F 2 is / satisfies complete. Let / be an isomorphism from Fx to F2 and g an isomorphism from F2 to Fz Define the function g of from Fi to F3 by (g © /)(#) = g(f(x)). Show that g o / is an isomorphism. . 4. Suppose that Fis a complete ordered / from R to F. Show that there is field, so is an isomorphism F = R, this is Problem 3-17. Now if / and g are isomorphisms from R to F consider g~ ° f. Find an isomorphism from C to C other than the identity function. to F. Hint: In case l 5. that there actually only one isomorphism from R two SUGGESTED READING A man ought to read just as inclination leads him; for what he reads as a task will do him little SAMUEL JOHNSON good. . purpose of this bibliography is to guide the reader to other sources, but the most important function it can serve is to indicate the variety of mathematical reading available. Consequently, there is an attempt to One achieve diversity, but no pretense of being complete. The present plethora of mathematics books would make such an undertaking almost hopeless in any case, and since I have tried to encourage independent reading, the more standard a text, the less likely it is to appear here. In some cases, this philosophy may seem to have been carried to extremes, as some entries in the list cannot be read by a student just finishing a first course of calculus until several years have elapsed. Nevertheless, there are many selections which can be read now, and I can't believe that it hurts to have some idea of what lies ahead. One of the most elementary unproved theorems mentioned in this book is the fact that every natural number can be written as a product of primes in only one way. A proof of this basic theorem will be found near the beginning of almost any book on elementary number theory. Few books have won so enthusiastic [1] An an audience as Introduction to the Theory of Numbers (third edition), by G. H. Hardy Press, Oxford, 1960. and E. M. Wright; Clarendon The Pergamon Press publishes a series, Popular Lectures in Mathematics, with several worthwhile [2] A Selection of Problems millan (Pergamon), Finally, I will [3] mention a Three Pearls of among them titles, in the New Theory of Numbers, book which little Number by W. Sierpinski; Mac- York, 1964. Theory, I hope is still in print: by A. Khinchin; Graylock Press, Rochester, N.Y., 1952. The subject of irrational analysis. An numbers straddles the fields of excellent introduction will be found [4] Irrational Numbers, by I. M. number theory and in Niven; Wiley, New York, 1956. Together with many historical notes, there are references to some fairly elementary articles in journals. There is also a proof that ir is transcendental (see also [53]) and, finally, a proof of the "Gelfond-Schneider theorem": and If a b are algebraic, with a ^ 0 or 1, and b irrational, is then a h is transcendental All the books listed so far begin witn natural numbers, but whenever necessary take for granted the irrational numbers, not to mention the integers and rational numbers. Several recent books present a construction of the rational numbers from the natural numbers, but it is hard to believe that there can be a more lucid treatment than the one [5] Foundations of Analysis, Incidentally, the original [6] to be found in by E. Landau; Chelsea, German New York, 1951. edition, Grundlagen der Analysis (fourth edition), by E. Landau; Chelsea, New York, 1965. now available in paper back, together with a complete German-English dictionary (of about 300 words) for the whole book an excellent way to begin is — The basic idea for constructing the real derived from Dedekind, whose contributions can be found in reading mathematical German. numbers is Essays on the Theory of Numbers, [7] by R. Dedekind; Dover, New York, 1963. While many mathematicians are content to accept the natural numbers as a natural starting point, numbers can be defined in terms of sets, the most basic starting point of all. A charming exposition of set theory can be found in a sophisticated little book called Naive Set Theory, by P. R. Halmos; [8] Van Nostrand, Princeton, N.J., 1961. Another very good introduction Kamke; Dover, New York, Theory of Sets, by E. [9] is 1950. Perhaps it is necessary to assure some victims of the "new math" that set theory does have some mathematical content (in fact, some very deep theorems). Using these deep results, Kamke proves that there is a disconthose tinuous function / such that f(x y) = /(*) f(y) for all x and y. For + + who first enjoy reading the classics, the most important notions of introduced by Cantor, whose work is reproduced in set theory were [10] Contributions to the Founding of the Theory of Transfinite Numbers, Cantor; Dover, Inequalities, New which were treated actually form a specialized by G. York, 1952. field. an elementary topic in Chapters 1 and 2, good elementary introduction is provided as A by [11] Analytic Inequalities, New by N. Kazarinoff; Holt, Rinehart and Winston, York, 1961. mean is less than or equal to the principle, can be found in the different a each based arithmetic mean, on beginning of the more advanced book Twelve different proofs that the geometric [12] Inequalities, by E. Beckenbach and R. Bellman; Springer, New York, 1961. The classic work on inequalities is [13] Inequalities (second edition), by G. H. Hardy, G. Polya; Cambridge University Press, New J. E. Littlewood, of the authors of this triple collaboration has provided his Each and York, 1952. own con- tribution to the sparse literature about the nature of mathematical thinking, written from a mathematician's point of view. [14] A Mathematician's Apology, by G. Press, New York, 1940. My favorite is H. Hardy; Cambridge University Suggested Reading 517 Littlewood's anecdotal selections are entitled [15] A Mathematician's Miscellany, Polya's contribution is pedagogy by E. Littlewood; J. at the highest level: by G. Polya (Vol. [16] Mathematics and Plausible Reasoning, Analogy in Methuen, 1953. I: Induction and Mathematics; Vol. II: Patterns of Plausible Inference); Princeton University Press, Princeton, N.J., 1954. which can be considered as background a masterful mathematical work, but preparation has been made. Probably some until postponed be perhaps should the best modern work on "classical geometry" is Geometry is the other main field for calculus. Euclid's Elements is still [17] Elementary Geometry from an Advanced Standpoint, by E. Moise; Addison- Wesley, Reading, Mass., 1963. This beautiful book provides excellent historical perspectives and contains a thorough discussion of the role of the "Archimedean axiom" in geometry; in which the Archimedean geometry books, all sorts of addition, Chapter 28 describes an ordered field in axiom does not hold. Speaking of beautiful fascinating things can be found in [18] Introduction to Geometry, by H. S. Coxeter; Wiley, New York, 1961. Almost all treatments of geometry at least mention convexity, which forms another specialized topic. I cannot imagine a better introduction to convexity, or a better mathematical experience in general, than reading and working through [19] Convex Figures, by I. New and Winston, M. Yaglom and W. G. Boltyanskii; Holt, Rinehart York, 1961. This book contains a carefully arranged sequence of definitions and statements of theorems, whose proofs are to be supplied by the reader (worked-out proofs are supplied in the back of the book). Another geometry book has been modeled on the same principle: [20] Combinatorial Geometry in the Plane, Holt, Rinehart and Winston, by H. Hadwiger and H. Debrunner; New York, 1964. Along with these two out-of-the-ordinary books, valuable little book, also of a specialized [21] Counterexamples in Analysis, Day, San Many by B. I might mention an extremely sort, Gelbaum and J. Olmsted; Holden- Francisco, 1964. book come from more advanced topics in few can be appreciated by someone who knows calculus. of the examples in this analysis, but quite a Of calculus books A [22] mention only two, each something I will Course of Pure Mathematics (tenth edition), bridge University Press, New of a classic: by G. H. Hardy; Cam- York, 1952. by R. Courant; Vol. I, second edition, Wiley (Interscience), New York, 1936. [23] Differential and Integral Calculus, 1937; Vol. II, Courant is Volume I first edition, especially strong on applications. In addition, the latter parts of contain material usually found in advanced calculus, including differential equations and Fourier series. An introduction to Fourier series (requiring a little advanced calculus) will also be found in An [24] Introduction to Fourier Series New jamin, The second volume tional material and by R. Seeley; W. A. Ben- Integrals, York, 1966. on of Courant (advanced calculus in earnest) contains addi- differential equations, as well as calculus of variations. an introduction to the not mention any books devoted solely to the calculus of variations, since they are (of necessity) quite difficult. Innumerable I will books on differential equations have been written, but none of the elementary ones seems to generate much enthusiasm. There is a somewhat more advanced book, however, which is universally admired: Ordinary Differential Equations, [25] Lectures on Press, by W. Hurewicz; M.I.T. Cambridge, Mass., 1958. bypass the more or less standard advanced calculus books (which can be found by the reader himself) since nowadays there is a move- I will easily ment to revise the linear algebra. whole presentation of advanced calculus, basing it upon of the first, and still one of the nicest, treatments of One advanced calculus using linear algebra [26] Calculus of Vector Functions, Prentice-Hall, Englewood is by R. H. Crowell and R. E. Williamson; Cliffs, N.J., 1962. Several recent books on advanced calculus attempt to acquaint undergraduates with very large areas of modern mathematics. favorite, of My course, is [27] Calculus on Manifolds, by M. Spivak; W. A. Benjamin, New York, 1965. There are three other topics which are somewhat out of place in this bibliography because they are rapidly becoming established as part of a standard undergraduate curriculum. The purposeful study of fields and related systems is the domain of "algebra." The first undergraduate text on modern algebra was [28] A Survey of Modern Algebra (third edition), by G. BirkhofT and MacLane; Macmillan, New York, 1941. S. 519 Suggested Reading One of the strongest competitors by Topics in Algebra, [29] is now N. Herstein; Ginn I. (Blaisdell), Boston, Mass., 1963. is meant to be intermediate in Modern Algebra and one of the great classics: of Herstein's book [30] difficulty Modern Algebra, by B. L. van der Waerden; Ungar, between New A Survey York, 1953. the way, this book contains a proof of the partial fraction decomposition By of a rational function. One of the standard texts in complex analysis [31] Complex Analysis (second edition), New I am is by L. V. Ahlfors; McGraw-Hill, York, 1966. personally looking forward to the publication of [32] Lectures on the Theory of Functions of a Complex Variable, Van by G. W. Mackey Nostrand, Princeton, N.J., 1967. Finally, the study of topology, although never mentioned before, has really background of many discussions, since it is the natural generalization of the ideas about limits and continuity which play such a prominent role in Part II of this book. There are now many elementary books on topology, but I am dismayed by the prospect of reading and evaluating them all, so I will merely call attention to the fact that they exist, and warn that some are been in the not very good. next few topics, ranging from elementary to very difficult, are included have been alluded to in the text. The proof that a nondecreasing function is differentiable at almost all points (and an explanation of just what this means) receives a beautiful exposition in The in this bibliography because they [33] Functional Analysis, by F. Riesz and B. Sz.-Nagy; Ungar, New York, 1955. book moves on to quite advanced little book devoted entirely to its properties, most of them proved by using the theorem of Bohr and Mollerup which was mentioned in Problem 18-25: (After this elementary beginning, the material.) [34] The gamma The Gamma function has an elegant Function, by E. Artin; Holt, Rinehart and Winston, New York, 1964. The gamma function is only one of several important improper integrals in x% mathematics. In particular, the calculation^ jj e~ dx (see Problem 18-27) is important in probability theory, where the "normal distribution function" : plays a fundamental role. The following book has already become something of a classic: An Introduction to* Probability Theory and Its by W. Feller; Wiley, New York, 1957. [35] Applications (second edition), certain functions in elementary terms (among one of the most esoteric subjects in mathematics. An interesting discussion of the possibilities of integrating in elementary terms, with an outline of the impossibility proofs, and references to the original papers of Liouville, will be found in The impossibility of integrating them = f(x) The [36] e~x *) is Variable (second edition), Integration of Functions of a Single G. H. Hardy; Cambridge University Press, A complete New by York, 1958. presentation of the impossibility proofs will be found in [37] Integration in Finite Terms, by J. Ritt; Columbia University Press, New York, 1948. Oddly enough, a more related but seemingly difficult problem has a much xy — 0 is a example) whose solutions cannot be expressed even in terms of indefinite integrals of elementary functions. This fact is proved on page 43 of the (60-page) book: neater solution. There are simple differential equations (y" + specific An [38] Introduction to Differential Algebra, by I. Kaplansky; Hermann, Paris, 1957. To read (Just this book you will need to know quite a bit of algebra, however. equally algebraic and polished treatments of Liouville's recently, theorems have been obtained, and presumably they will be published soon.) A few words should also be said in defense of the process of integrating in elementary terms, which many mathematicians, look upon as an art (unlike differentiation, which is merely a You skill). are probably already aware that the process of integration can be expedited by tables of indefinite integrals. For those who enjoy perusing tables there is a really beautiful collection, that and a great deal you should ever happen to need the value of the thirty-fourth Bernoulli number, this is the place to look) includes indefinite integrals, definite improper integrals, more besides [39] (if Tables of Integrals, Series, and Products, Ryzhik; Academic Press, For the [40] thrifty, there is New by I. S. Gradschteyn and I. W. York, 1965. a paperback table of integrals: Tables of Indefinite Integrals, by G. Petit Bois; Dover, New York, 1961. 521 Suggested Reading The remaining references are of a three categories, of which the referred to in will different sort. The letter of They fall into H. A. Schwartz be found in Ways of Thought of Great Mathematicians, by H. Meschkowski; HoldenDay, San Francisco, 1964. [41] Some Problem 11-49 somewhat first is historical. historical remarks, and an attempt to incorporate them into the teaching of calculus, will be found in [42] The Calculus: A Genetic Approach, by O. Toeplitz; University of Chicago Press, Chicago, 111., 1963. Most textbooks on the An history of mathematics are both superficial admirable exception [43] An Introduction and Winston, and dull. is to the History of Mathematics, New York, 1964. by H. Eves; Holt, Rinehart Three good scholarly works are [44] History of Analytic Geometry, ' [45] A Press, New York, by C. Boyer; History of the Calculus, and Its Conceptual Development, Dover, [46] by C. Boyer; Academic 1965, New York, 1959. The Mathematics of Great Amateurs, by J. Coolidge; Dover, New York, 1963. A recent book presents philosophical ruminations on the history of matheis all the more impressive matics by an author whose exceptional erudition in a first-rate mathematician: [47] The Role of Mathematics in the Rise of Science, by S. Bochner; Princeton University Press, Princeton, N.J., 1966. Finally, extracts [48] A Source from original sources Book in will be found in Mathematics (2 vols.), by D. Smith; Dover, New York, 1959. large number of books listed here might easily give too hopeful a view of the availability of historical information. Although the very early origins of mathematics are receiving the most careful scrutiny, no one seems to care very The much about the origins of calculus. For example, it is almost impossible to find out who first proved the Mean Value Theorem (according to the Encyhlopddie der Mathematischen Wissenschaften, Volume II, it was O. Bonnet, whose name Gauss-Bonnet Theorem"). Similarly, nearly every history book tells us that Wallis proved Wallis's formula by a "complicated method of interpolation," but almost none bothers to mention what it was, even though it inspired Euler's investiga- is familiar to students of differential geometry from the 6 ' : gamma tions of the function (a description is given in the answer book, along with the solution to Problem 18-26). The second category in this final group of books might be described as "popularizations." The editorials notwithstanding, there are a surprisingly large ones by real mathematicians: [49] [50] New York Times number of first-rate What is Mathematics? (fourth edition), by R. Courant and H. Robbins; Oxford University Press, New York, 1947. Geometry and the Imagination, by D. Hilbert and S. Cohn-Vossen. Chelsea, New York, 1956. The Enjoyment of Mathematics, by H. Rademacher and O. Toeplitz; Princeton University Press, Princeton, N.J., 1957. [52] Famous Problems of Mathematics (second edition), by H. Tietze; Graylock Press, Rochester, N.Y., 1965. [51] One of the most renowned "popularizations" is especially concerned with the teaching of mathematics: [53] Elementary Mathematics from an Advanced Standpoint, Arithmetic, Algebra, Analysis; vol. 2: Geometry)', by Dover, F. Klein (vol. New York, 1 1948. Volume 1 contains a proof of the transcendence of x which, although not so elementary as the one in [4], is a direct analogue of the proof that e is transcendental, replacing integrals with complex line integrals. It can be read as soon as the basic facts about complex analysis are known. The third category is the very opposite extreme original papers. The difficulties encountered here are formidable, and I have only had the courage to list one such paper, the source of the quotation for Part IV. It is not even — you do have a choice of foreign languages. The French is in in English, although in the original [54] Oeuvres New Competes a" Abel; article Christiania. Johnson Reprint Corporation, York, 1965. appeared in a German translation in the Journal fiir die reine und angewandte Mathematik (1) 1826. To compound the difficulties, these references will usually be available only in university libraries. Yet the study of this paper will probably be as valuable as any other reading mentioned here. The reason is suggested by a remark of Abel himself, who attributed his profound knowledge of mathematics to the fact that he read the masters, rather than It first the pupils. ANSWERS TO SELECTED PROBLEMS CHAPTER 1 1. = 1 (i) = 2 If x (iii) 3. One , .y - = a' 1 (ax) = then 0 2 +y or x x that 0, 2 l (a = a)x - j» - 1 • = x x. - (* + y), so either - y = = —^ or x = j. jy) (iv). —y = by x step requires dividing 0. (iii)) = ac/bc. ) (by bc)(b~^) = = bc){bd)+ (ad + + fc)/(W) = a^r + aT = <z/4 + c/d. l ^(a"^" ) « (a a" ) (4 r ) - 1, so a- 6 - («*) = (« ^X* (a lb) I (c/d) = (a/bKc/d)' = = ( r )(c-1 <0 = = M/(4<). (i) */* (ii) (flrf = = ab-i = l (ae)(b^e~ (ac)(bc)^ (by 1 1 1 1 1 • • l 4. (i) *<-l. (iii) * (v) All 2* 7. + 2 = b + - (* l) 2 + 1. (iii) (xiii) (a (v) + (b - ac > c) is in P. Thus, d > a + c. -c < -d; then (i) implies that a + (-*) < b + (-<*). that is, and -c are in />, so -c(b - a ) = ac - be is in P, (ii), fl ) be. a > 0 and a < so a 2 < a. (vii) Using (ix) Substitute a for <;_and 4 for d in (viii). (i) (iii) ( V) (i) (iii) |a (iii) (v) 1, 4| > -4 and 4 > 0; < —4 and b < 0; a + 2b if a > -4 and 4 < 0; -a - 24 if a < -4 and 4 > 0. a if x - a if <z x2 -x (i) (iv), V2 + V3 - V5 + V7. + +>| - |a + * V2 + V3 + V5 - V7. —« 11. - 2 a: 1 Using (xi) 10. V < - a: W ' fl<J (i) (ix) 9. or since *, * * -2 are the roots of x 2 - x - 6 = 0. x > 3 or * < -2, since 3 and * > 7T or -5 < x < 3. x < 3. x > 1 or 0 < x < 1. - * are in P, so (4 - a) + (d - *) = (* + d) & - « and (vii) 5. > Vl - 1 • • fl » • 1 (v) (hi)) 1 1 (iii) 0 a: (* either * is, Replace > by -jy in (vi) 2. = a" 1 a if *2 x > 0; if * < 0. x = 11, -5. -6 < x < -2. No * (the distance from x to plus the distance from x to 1 -1 is at least 2). (vii) 12. (i) = 1, — 1. .1,1111 and 1*1 w = (W since = *V = (M) 2 = = |xy| that proves \y\. |*| are both > 0, this (")) - I^T'I b y ®> = Ml\x\/\y\ = \x\ bl" = |*| Ir x M WW 2 M ' • 1 (iii) 525 1 ( • ! ' 526 Answers to Selected Problems from It follows (v) |*| - \y\ +y + z\ < then \x + y\ = \x same sign 2 1. that Since (i) l \z\ \x\ 2 — \y — (y < x)\ \y\ + \y 1 • + • • • +v+ one (2 • (2) • + 1 l)/6, the formula true for is + *(* + i)( 2 * *) + [k(2k = (A±ll [{k 6 _ + (* l)(* is true for « {k + Then k. + 1) 6 1)2 + 6(* + 1)] + 2 )(2k + 3)] + 2)(2[* + 1] + 1) , 6 so the formula + true for k is 1. (0 (2i - 1) = 1 = 1 +3+5+ • +2+3+ (2n)(2n + 1) - _ • • • • • + (2n - 1) + 2n - 2(1 + n(n 2 = 4. (a) so \z\. \y\ = £ x\, 0). is \y = + (k \x\ 6 2. — [y|, (unless = 2 \x Suppose that the formula i = |*| If equality holds, + y\ + < + + so # and y have the same sign. Moreover, + z must have the same sign as x + y, so x, y, and z must all have the (vii) chapter (iv) <\x- y\. + • • • + n) 1) n\ Since 1 the formula 1 = true for n is +r+ • • • 1. + -r -> —r 2 1 l+r = Suppose that r n = - 1 r n+l ' — \ r Then 1 +r+ * • • + r n + r n+1 = 1 = 1 = — r n +l n+l r 1 — r - r* +1 + r w+1 (l - 0 1 - r - r w+1 \- 1 1 - r = 1. Answers to 527 Selected Problems (b) S = +r+ r + 1 = rS • • • • • • + + r r n + n r n+1 . Thus - J(l r) = J- rS = 1 - r 1 - + 4k + 1, r* +1 , so From (i) + (* we - 4 l) = k4 + 4k 9 6k 2 k = 1, . . . , n obtain n (n + - 1)4 1 = 4 V 1 - 6 n n J P+ 3 + 6 "(" + ^ A: Y 4 * + so n + (B ,_ 1) —r —«2 H " 4 + ^-4^±i>- W 2 «2 3 = 2" 6 , , -* 4 From (iii) 1 1 1 k it + + k{h \ 1) we obtain 1 + n Li 1 + k{k l) even or odd; 1 is either even or odd, in fact it is odd. Suppose n is either = 1 1. In the first case n then n can be written either as 2k or 2k 1) is 1 = 2{k 1 1 = 2A; 1 is odd; in the second case n 2A; 1 is either even or odd. (Admittedly, this looks even. In either case, n + + + + + + + + fishy, Let B Then but set of all natural be the 1 is really correct.) it is in B y and / + 1 is numbers in B if / is / such that n 0 in B, so B — 1 contains + / is all in A. natural numbers, which means that A contains all natural numbers > n 0 b were rational, then b = (a + b) - a would be (a) Yes, for if a rational. If a and b are irrational, then a + b could be rational, for b - + could be (b) If a = 0, r — some a for then ab is rational, for then b rational = (ab) number a. But if a * 0, then ab could not be -1 would be rational. a rational. (c) Yes; for example, \^2. (a) V2 example, (d) Yes; for 12. — V2. and Since 2 (3„ Now /? + + 9n 2 9n follows that \ik 2 it q = = + 1)2 + 2) (3„ suppose that 2 is = x = 3( 3w 2 + 2w ) = 3(3rc + 4rc + + 2 Yin 4 divisible by 3, l ? + 1) 1, then £ must also be divisible by 3. VI were rational, and let V3 = p/q where p and have no common must be. Thus, p quently (3/0 2 + + 6n Then p 2 = 3# so /> is divisible by 3, so 3p' for some natural number p\ and conse- factor. = 3? 2 or 3(//) , = 2 2 ? 2 2 , Thus, 9 - is also divisible by 3, a contradiction. The same + + + + (5n (5 W (5n (Sn 2 l) 2 2) 2 3) 2 4) work proofs = = = = + lOn + = 5(5n + 2«) + + 2(k + 4 = 5(5rc + An) + 4, + 2>0n + 9 = 5(5« + 6n + 1) + 4, + AOn + 16 = 5(5n + + 3) + 2 25rc 2 2 2 and the corresponding equations show that if k 2 is divisible by 5 or V for this a is 4, 1, 2 2Sn 2 25rc 2 1 2 25rc vi because the equations and for because {An + 2) 2 1, then k must be. 6, by The proof fails (For precisely this reason 4. Va is irrational if proof cannot be used to show that in general m) 2 might {an that not a perfect square we have no guarantee — + not be a multiple of a for some m < a. Actually, this assertion but the proof requires the information in Problem 16.) (b) + m, numbers of the form 6n for divisible is 8/2 is true, Since {In + l) 3 = Sn* + \2n 2 + + 6n = 1 2{An s + 6n 2 + 3n) + 1, follows that if k* is even, then k is even. If $2 — p/q where p and had no common factors, then p 3 — 2q z so p z is divisible by 2, so p must be. Thus, /? = 2p' for some natural number //, and conseit # , quently {2p') z = 2q\ or 4(//) 3 = 3 ? - Thus, ? is also even, a contra- diction. The proof (3„ (3„ + + 1)3 2) = If n 18. 3 = = 27n 1, then 27w 3 3 for + + (1 vi 27/2 2 54n + 2 + + A) n is similar, using the equations + 36/z + 1 9rc = 1 + = 3(9« 3 + 9n 2 + 3n) + 1, 2 8 = 3(9rc + 18/z + 12n + 2) 3 Suppose that nA. (1 + h) n > + 1 2. + Then (1 + h )n+i = = (i 1 + h){\ + A) n + (n + \)h + > nh 2 (1 > + 1 A)(l + {n + nh), + l)h. since 1 +h > 0 »A. Answers For h > to Selected 529 Problems the inequality follows directly from the binomial theorem, n are h) the other terms appearing in the expansion of (1 0, + since all positive. chapter 3 1. 0+ (i) x l)/0 + 2); — 1 and x ^ 1/(1 (v) (* Only be 1, since /(*) 0 (iii) 0. (v) -1, (i) {x: (iii) \x: and 0, = rational, or y (for x, y * when and must -1). /(«:) implies that x one x true for at least > (i) c sense only 0). if = makes — 2. + cx) (for x * -l/c +^ + 2)/(* + l)(y + 1) (iii) (vii) the expression /(/(*)) > = cx, this 0. 1 1. —1 < < and x ^ at 1 1 }. a: ^ 2j. (v) 4. 5. 11. (i) 2 2y (iii) 2 2sin< (i) Po (iii) S o S. (v) PoP. (vii) Sos°SoPoPoPos. . + sin(2 f ). S. (a) y. (b) H{y). (c) 12. H(y). even even neither odd neither odd even odd even even odd odd odd even (b) / even / odd g even even even g odd even odd (c) (d) odd even (a) Let g(x) = /(*) for x > 0 and define g arbitrarily for x < 0. . 530 Answers to Selected Problems 21. = 1 and let / be a function for which Then/o(^ + ^fog + foh. = o f]( X = (g + h)(f(x)) = *(/(*)) + *(/(*)) 0 0 o + (A /) = [(5 /) + (* /)] W= Let g(x) (i) /(l) (ii) ° (5 4 1. 2. A) W A V • / Let g(x) = 2 and let/ be a function Then l/(/o^) *fo(l/g). /(£(*)) (i) (2, 4). (iii) [2, 4]. (v) (vii) (-2, 2). (-oo, 1]U[1, (i) All points below the (iii) All points below the graph of fix) (v) All points (vii) A (ix) All points inside the circle of radius g) to/ (i) (iii) 00). = x. = x2 between the graphs of f(x) = x graph of f(x) . + collection of straight lines parallel to the The The (vii) point 1 +y 2 — = * (n, — 1 = —x, 0) for integers n. 1 around 2x +y = 2 (a, b). (The important point about remember the point slope form.) The straight line through (a, b) and so the equation follows m' and b from part b' . (c, d) m(x — a) + b = mx + which goes through the = f(x) a straight line with slope m, When m = — since x 2 (1, 0), is ma) point (c) f(x) -\. 2 Simply observe that the graph of (b (b) and graph of fix) (0, 0). circle of radius ix-\) (a) l//(2). around (1, 2). A square with vertices (1, 0), (0, 1), (-1, 0), and (0, -1). The union of the graph of f(x) = x and of f(x) = 2 — x. (v) 5. ^ which /(£) for intersecting the horizontal axis at the points 3. ^ ) W t /°£ (iv) CHAPTER /) 1- to - 77~rrr = (*M) = 7 (iii) /(2) yi) + [(£ h(x) this exercise has slope (d is — simply to b)/(c — a), (a). In that case, there is clearly with fix) — g(x), while such a number x always namely, x — (b — b)/(m — m'). no number x exists if m m' y f 6. (a) If B = by all 0 and A points (x, then the set is the vertical straight line formed with x = — C/A. If B ^ 0, the set is the graph of y) 5* 0, /to = (~A/B)x (b) The points (x, y) + (-C/A). on the vertical ones which satisfy 1 the graph of f(x) = • +0 mx + b * • jy line with x + — a) ( = 0. — a are precisely the The points (*, y) on are precisely the ones which satisfy i-m)x + 1 y+ {-b) = 0. The graph of / is symmetric with respect to the vertical axis. The graph of / is symmetric with respect to the origin. Equiva- 10. (i) (ii) lently, the part of the graph to the left of the vertical axis is Answers (iii) (iv) 19. (a) obtained by reflecting first through the horizontal axis. through the vertical axis, = x* = (* + which is 2 -+± 16 + i) 2 , y) satisfies this condition (x, - ay + (* - (y 2 /3) over. x the square of the distance from The point 531 and then The graph of / lies above or on the horizontal axis. The graph of / repeats the part between 0 and a over and The square of the distance from (#, x 2 ) to (0, \) is 2 (b) Problems to Selected (x> x 2) to the graph of g. and only if = (y-y) if 2 , or - x2 lax + +y - a2 2 2/3y +£ 2 = y - 2 + a2 - .2/3 2y) - \y y2 /5 -y - 2y 2 20 \ (3/ + 2yy 2 , \ / (3 ^ y, which is just the condition that on L, then the solution is the vertical line (This solution works only for P is not on L. P If is through P.) chapter 5 1. = lim (x + 2x + 4) = —2 x-»2 ~ ^ = lim x n ~ + x n ~ + lim — y 2 lim (ii) = (vi) y = (V a lim It is If |* \x* + ax 2 + 2 a x — + a| = a4 < a*\ 1, < < (* - then (1 h + ^ n" wy"" 1 - . +7^A + h(V a + + h 1 Vfl) Va) by beginning with the equation + a* + a * + a so < + + + + + \a\(l + 2 a)(* 3 1 |*| + + \x\* *)> lim possible to find 8 *4 + k-+o = n~ 2 + • + y n ~* = x A (i) • + y n' + n~l A-V^ Va + lim , 2. 2 l (iv) 12. X x->2 \a\ \a\) |*| s 3 ). \a\, 2 • 2 |a| 2 |*| \a\) 2 |a| \a\ 3 2 (l + \a\) + |a| 3 ; 532 Answers to Selected Problems we can choose therefore S mm 0' = It + (I + \a\(l + \a\y This shows that \x — 4 a < 4 \ is = min 1 ( — By part problem, \x — 1 (v) Let 5 (i) We We f-> -> 1,5 \2 4 = £ need < 0 < 1| e/2. 2 , - i)> — < £ 1| ) + 4 when = -> \2 2/ - 1/x) According - 2| 2| —— - 8 < since 0 i/(x) |* + I |(* i) < 2| £ to parts when |1 and (i) A(ii) 1| < of this happens when this < min 1| 1 (hi) — 4 (a: lemma. 1) i)(H |l/# in < min 1| lemma, (1) of the e/2 and + '4(| a |^ lemma, (3) of the |# (hi) \a\)' to use part (2) of the |«+ + 2(W By part + when true |,- a |< min^l, (ii) \a\) when £ 2(| a which + \a\*(l and probably easier, instructive, is + \a\)> |*| < • 2 < = min ^ £ 2 8. \2 32/ < implies that V|*| - e/2 and \g(x) < min = (\> 2/ • + |, (sin* < 4| ~) = e. e/2, so we need 5. need 2 |£M-4|<min(H^- )> so we need 0 Let = / < \x - < 2| = [min(2, 8e)] 2 lim f(x) and define g(h) = /(a + 5. Then h). for every £ > 0 x—>a there is a 5 Now, if 0 < > 0 such that, for |A| < 5, then 0 all x, if < \(h + can also be written lim f(a that if lim /(a h-+Q + h) — + h) = /. |* - a) — This inequality can be written < 0 /| - < < a\ < a\ £. 5, = then sort of — < - < I \f(x) + h) Thus, lim g(A) The same m, then lim f(x) 5, so \f(a l\ = /, which argument shows m. So either limit exists x—*a other does, and in this case they are equal. e. e. if the Answers 9. (a) we can Intuitively, — can get f(x) / as we like if and only if we we like. The formal proof is so of work to make it look like a proof at all. get f(x) as close to as close to 0 as / that it takes a bit be very precise, suppose lim f(x) trivial To 533 Problems to Selected = and / — let g(x) — f(x) I. x-*a Then — \x < — \g(x) tion (b) > for all £ a\ < 0| 0 there then 8, a 8 > 0 such that, for all x, if 0 < < s. This last inequality can be written = 0. The argument in the other direc- is — \f(x) l\ so lim g(x) £, similarly uninteresting. is making x Intuitively, close to a same as making x — a close /, and let g(x) = f(x — a). the is = Suppose that lim f(x) to 0. Formally: x—> a, Then for < < + so \f(y 8, ~ tion < l\ — lim f(x) < < l\ — a) Now, £. < /| 0 such that, for = < 0 if But £. So lim g(y) £. > a 6 is < \y\ x < 8, |* |*| close to 0 is For every I. have 0 < < | we assume if > l\ so |/0 3 ) 8, and only if £ - then |/0) 3 other hand, then for |/(* |/([\^] (d) Let < m\ - 3 ) = /(a:) > £ all - ) a\ in the reverse direc- there 0 a is < £. Then - l\ < x3 if < 0 if > < |*| £. < m| 1 or £, > for x if 0, - /. 8), we the = 0 < - \f(x) and < |*| 0 Thus lim £. = - /(*) 0 if we have 5 3, < m\ m, x—>0 a 8 such that is if On min(l, that lim f(x z ) exists, say lim /(* 3 ) 0 there Then such that 0 Thus, lim f(x) £. Let Formally: is. 8 x—>0 3 — \x - a\ \(y + a) < inequality can be written this last The argument I. < 0 all x, if then 0 8, similar. is Intuitively, 0 0 there - then \f(x) 5, \g(y) (c) > all £ 1 < for x < < < | = then 5, \Tx\ /(*) 0. < \x\ so 6, m. Then lim f(x 2 ) 2->0 = 15. (a) but lim /(*) does not 1, The = function f(x) 1/* cannot approach a limit at becomes arbitrarily large near be, there is < + (|/| 1/0 - but 5, £)). This a* £)). 0 1) In fact, |#| < 0. some x satisfying 0 < = min(<5, \/{\l\ + No matter what 5 > x (b) exist. 5, This x does not may be, > + there £, |/| is Problem satisfy \\/x some — 9(b): lim 1/x — = 1) = min(l - /| < it < £. — \x 8, 1 £. (It is also 1) + 1 + 1/0 — lim < l\ x satisfying 0 namely, x does not satisfy \\/(x possible to apply since 0, no matter what 8 > 0 may butl/* > |/[ + £, namely, if 1/ the 2-+1 latter exists, so this limit does 23. (i) This is not exist, because of part the usual definition, simply calling the and 8. a minor modification (a).) numbers 8 and £, instead of £ (ii) This 8 0 (in) > < This is of (i) : if the condition 0, then it applies to 8/2, so there |* - < £, is - is an £ > is true for all 0 such that < 8/2 < 8. a similar modification: apply it to 8/5 to obtain a| then I/O) l\ (i). if 534 Answers to Selected Problems (iv) also a modification: This is e/10 > and 0, it same thing says the it only the existence of some £ is as > (i), 0 that since is in question. 27. If lim /(*) = 5 28. < <5 a: (i) then for every /, > £ 0 there are 8 h 8 2 > 0 x—»a such that, for Let = lim /(*) _ x—*a + all x, if a < x if a - 82 = < a or else If / < a + Si, < x < a, min(Si 5 2 ). = \f{x) \f(x) — |* - < a| £ all - £, £. — so \f(x) 5i, > < < /| l\ — then either a 6, a<#<0+5<<z + lim /(*), then for 0 there is /| a 5 > < /| £ for - so |/(-*) < 0 < l\ < x -5 < If 8. x Thus lim /(-*) = e. x is a — e. 0 such that < -x < < 0, /. Similarly, then 0 if and close to 0 and has the same exists positive if and only 5, lim /(*) co- x-^o- then lim /O) exists, < 82 < + x-»0 I/O) < 0 If 3 then then value. (Intuitively, —a: if close to 0 is and negative.) If (ii) / = lim f(x), then for £ all > 0 there is a 8 > 0 such, that x->0 + I/O) < - < /| Thus lim £. (Intuitively, If (iii) / = 0 £ for if at < < = x f(\x\) So 8. if The /. \x\ is > £ all < |*| reverse close to 0, then is lim f(x), then for 0 < close to 0 0 there then 5, |/(|*|) direction is and a 8 > - /| similar. is positive.) 0 such that x-*0 + I/O) - l\ < £ for 0 < x < 8. If 0 < |*| so |/0 2 ) - /| < £. Thus lim /0 2 ) = /. < VS, The < then 0 x2 < reverse direction 5, is x—() if x is close to 0, then x lim /0), then for every £ > 0 there similar. (Intuitively, 2 is close to 0 and positive.) 31. If / = is some N such that |/0) — N < £ for x > N, and we can clearly assume that > 0. Now, if 0 1/N, then 1/x > N, so \f(l/x) - l\ < s. Thus lim /(l/*) = reverse direction chapter CHAPTER 6 7 1 . 1. + FO) = FO) = (i) Bounded above and below; minimum value x x < The similar. is (iii) (i) < /. l\ 2 for 0 for all *. all x. 0; no maximum value. (iii) (v) Bounded below but not above; minimum value 0. Bounded above and below. It is understood that a > -a a + 1 < a + 2 for all * in 1). If -1 < (-a - 1, a a < + \, 1), then a so a + — 1 (so that < -a - 1, so /(*) = 2 is the maximum and Answers —\ < to Selected 535 Problems 0, then /has the minimum value minimum value 0. Since a + 2 > only for [-1 - V5]/2 < a < [1 + V5]/2, when {a + l) a > — i the function / has a maximum value only for a < [1 + \/5]/2 (the maximum value being a + 2). Bounded above and below; maximum value 1 minimum value 0. Bounded above and below; maximum value 1; minimum value minimum 2 a and y if value. If > a < a then / has the 0, 2 (vii) ; (ix) -1. / has a maximum and minimum value, since / is continuous. n = -2, since /(-2) < 0 < /(-l). n = -1, since /( — 1) = — 1 < 0 < /(0). 179 If /(*) = * + 163/(1 + * 2 + sin 2 a:), then / is continuous on R and /(l) > 0, while /(-2) < 0, so f(x) = 0 for some * in (xi) 2. (i) 3. (i) 5. / (iii) (-2, all 7. 1). / took on two different values, then / would take on between, which would include irrational values. constant, for is values in if = x; = -x; = \x\; f(x) fa) = -|4 /(*) (1) (2) f(x) (3) (4) 10. Apply Theorem 11. If /(0) < = /(l) fix) CHAPTER 8 1 . =0 = 1 7(1), * for (i) 1 is to 1 / — g. choose x = 0 or 1. If /(0) > 0 = 7(0) and then Problem 10 applied to / and I implies that - or /(l) 1, some the greatest element, is not in the (iii) 1 is (v) Since and the and the greatest element, {x: +x+ > x2 greatest lower bound is 03 which set. = 0} 1 0 is the least element, R, there is no least upper bound or greatest lower bound, Since (vii) {x: x < greatest lower is 2. (a) 0; neither 0 and * 2 bound x - - + is [ 1 1 belongs to the < V 0} 5]/2, = ([-1 - Vs]/2, and the least 0), the upper bound set. A ^ 0, there is some x in A. Then — x is in — A, so —A ^ 0. A is bounded above, there is some y such that y > x for all x in A. Then -y < -x for all x in 4, so —y < z for all z in -^4, so —4 is bounded below. Let a = sup (--4). Then a is an upper Since Since bound bound for Moreover, for -A, bound 5. (a) If x / is + 1 —A, so, reversing the argument just given, —a is a lower for A. any lower bound for A, then — (3 is an upper bound -/S > a, so 0 < -a. Thus -a is the greatest lower if (3 is so for A. the largest integer with < y. So we can let £ / = < / x, + 1 < then / 1 > x } but / (Proof that a largest such + 1. + 536 Answers to Selected Problems integer exists: / number number Since < of integers ny, + V2 (c) Choose r By part (b), 10. n. —n < < I Pick the largest.) x. - (s is r)/2. number r with x < a rational is number < with x s r < s < y. r < y, and Apply part 0 > < (a) (k x 1 and exists), < x/a (the solution to Problem 5 shows that = x — ka > 0. If x — ka = x > a, then < x/a, contradicting the choice of k. So + l)a, < a. + k so f let x' 1 Since any y in B satisfies y > x for bound for A, so y > sup A. all x in A, any y in Since x < sup + + + 7 sup(A 9 1. B is an upper shows that sup A is a lower bound for B, so sup A < inf i?. A and < sup B for every a: in A, and in it follows that x + y < sup A sup B. Thus, sup ^4 + sup B is an upper bound B so sup (^4 for A B) < sup ^4 sup B. If # and y are chosen in ^4 and B, respectively, so that sup A — x < s/2 and sup B — y < e/2, then sup A + sup B — (x + y) < B. Hence, (b) Part (a) CHAPTER to Let k be the largest integer x 13. there(c) s. such a k 12. is some natural There are consequently only a finite not bounded above, there < with there there fore a rational < / is x some natural number n with 1/n < y — x. there is, by part (a), an integer k with nx < 1, which means that x < k/n < y. (d) r N —n < — x > 0, ny — nx > (b) Since y k Since n with + + B) > x + y > A sup + sup B — £. (a) 1 f {a) = lim = lim fc-+o (b) The tangent line *>-/(«> = l im /fr + a(a + A) a2 g(x) = (a, — 1/a) 1 - (x is a2 If /(*) = g(x), so # = a. +a a then 1 the graph of 1 a) a1 ^ _^ + 2 1 l±A_i — = — through — _ Answers 2. to Selected Problems 537 (a) _1 r(a) , lim = lim /(« The tangent - (-2*A Aa (a line A 2) + A) 2 fc-o (b) + *)-/(«) m through g(x) lim 1/a 2 ) (a, (* " 3 a3 a2 a) +i " —+1 1= the graph of is /7 2 = _2^ *2 ; a3 - ~» /to «*(*), then + *)« 2 = 2 /7* If (, ' a2 a3 or 2x 3 - +a = 3a* 2 3 0, or = 0 So x - (* = a)(2* 2 a or x - ax — —a/2; side of the vertical axis - = a 2) (* - a)(2* the point (—a/2, 4/a 2 ) from = lim &+ lim (Va Vtf)(Va A(V a ^ h + —A + +A+ on the opposite Vfl) = A r lim 2 = Conjecture: = lim S» (X nx n ~\ Proof: + h) ~ S" (X) A h-+o = lim + h h-+o ^- = lim 5 = lim 20--'*'- h-+Q /-I i-i = ^ 11 a: "1 = nx n ~\ — A(VT+A + vi) Va) 1 = 4. ~ lim h~>0 - +A ^° ~ M= A fc-»0 = ~ k) lies a). 1/a 2). (a, 3. f(a) + a)(* - since lim A*" 1 = 0 for ; > 1. 538 Answers to Selected Problems 5 # 6. = f(x) 0 for x not an integer, and /'(*) is not defined if x is an integer. (a) g(x + „, = hm v 2 - h) gjx) = S (x) = h a-*o + + h) - c] [f(x) + c] =m + *)-/(*) /(* lim [fx lim (b) + = hm g(* v g (x) - h) lim cf(x + 2 • . ; . 4 ) (25) 2 2 2 • 2 2 ) ; 4 2 (36) • ; ; 2 (d) 2 • • 2 (c) cfjx) h 2 (b) - n = 3 9 2 /'(25) = 3 (25) /'(36) = 3 2 /'(3 ) = /'(9) = 3 9 ;/'(5 ) = /'(25) = 3 /'(6 ) = /'(36) = 3 (36) 2 = 3a f'(x = 3(x 2 = 3*<. /'(a 2 ) = 3(« = 6* but g(x) = * so /'(* ) = 3x (a) /'(9) h) *->o h-^o 8. = h h-*o 7. g{x) ) 5 6 . , ; (a) „ .. * = lim g(x + /([* + h) *] — g(x) + h + c)- f(x + *)-/(« + '> = f(x +C f(x + c) ). A ji-»o (b) ,'(,) = lim SiX + = hm c[f(cx k) = -lim + fcx)] k) 9. g(x) = f(x + a), = = f is periodic, with period so f{x) + - ~ *) - f(cx)] K k^O then g 9 .. lim /{CX) =c-f'(cx). (Compare the manipulations If = + £*> - *"> k h->o (c) lim /if* Ch /(CX c eix) + ch) - h-*0 = ~ g'(x) in this calculation with g'(x) = fix + + = fix a), a) for all x, Problem by part (a). 5-13.) But which means that /' a. = 5x\ Now fx) = g(x + 5), so by Prob= Six + S)\ = g\x +5) f{x) - 3)\ as in part (i). = = so 5ix (ii) 3) ix fix) fx) (hi) fx) = (* + 2) so /'(*) = 7(* + 2) as in part (i). + *), by Problem 8(a). If ft) = If fx) = git + *), then fix) = so fix) = £'(2*). git + x), then fit) = g\t + *), by Problem 8(a), k such that number no = is = there 2c/, and then s\t) ct (a) If s[t) = all = for '( to is, 2c/ [that ] s ) (i) Ifg(x) lem = x\ then g'(x) 8(a), 5 , 6 7 , 5 10. 11. 2 , 2 t (By the way, at this point t. we do not know any nonzero function / for Answers which to (b) (c) /' proportional to is If s(t) (ii) [/(<)]> The in After Chapter 17 /. determine what the world would be (i) t = = (a/2)t 2 (at) chandelier seconds 2a = = \6t = s'(t) 2 = t 5. it Problems 539 might be amusing were correct.) like if Galileo = so s"(t) a. 2as(t). 16/ 2 feet in or , at, = (a/2)t 2 • falls s(t) 400 if then , = 2 to Selected seconds, so / it falls 400 feet After 5 seconds the velocity will (a) 16 = 80 feet per second. The speed was half amount when 40 = s'(t) = 16*, or t = -f. This is another way of writing the definition (see Problem 5-8). (b) This follows from Problem 5-10, applied to the functions a(h) = be /(5) = 5a 5 • this 21. [f(a 26. 30. /"(*) (hi) f f (x) (ix) (i) (1 (iii) (— (vii) (vii) 2. = = means means means means means (i) (v) 10 h) (i) (iii) CHAPTER + (i) (iii) - + f(a)]/h and = 0(A) [g(a + - h) 6x. 4x 3 . = = that f'(a) that g'(a) the same as that g'(b) 2x) * if 71 a: . + /(*) c (iii). +x cos(x • /'(a) = = /(x) if + = f(b + a) if g(x) = f(x = cf{cb) if g(x) = f(cx). that g'(b) sin x) n~l na a). 2 ). cos (cos x). + sin x)) (1 + cos x). + l) (x + 2))) [2(x + 1)(* + 2) + (x + l) 2(x + sinx)(l + cos *) cos((x + sin x) [2 sin((x + sin x) 2x. (sin x + x cos x) + (cos(sin x )(cos )) (cos(x sin x sin + (2x cos x sin x (2 sin x cos x sin x sin x + (sin x cos sin x sin x 6(x + sin x) (l + 5 sin x cos x). 7x cos x 7(sin x + l) (7 sin x cos(sin x + I) x sin cos(x + sin x sin(x cos(x + + [(2* + )) (2x + 2x cos x ))]. 2 sin x cos x) sin x + sin x (1 + sin x)(2x cos x (cos(x • 2 2 (cos((x ]. • 2 2 )] ) ' 2 2 (v) (vii) a:)) (xi) (xiii) 2 2 2 2 2 2 2 ) ) 2 2 2 ). 5 7 • a* • 2 (ix) = g(a)]/h. 4 5 7 1 7 • 6 6 1 • 2 2 6 7 7 ). 2 2 ) 2 - 2 • 2 2 • — (XV) (1 + sin x) L cos x sin x sin' x 5 540 Answers to Selected Problems 4. (i) {x (iii) 2x*. (i) -x (iii) 17. (i) /'(*) (iii) fix) (v) 7. (a) (b) 2Y 2 . fix) A'(t) + = = = = g'ix gia). = F(0 the is First it 2tt = 4?rr(0 V(0 (c) gia)) g(x)) = 2wr(ty(t). Since r\t) that A'(t) If + + g'ix 4 • = 6 • 4?r 4 • method: Since r(0 f = with then V(t) t, = r(t) 4, it 3 = f — _ — il'W 5 2wr(t) 5 for r{t) when r(/) = 3. 6tt Thus = To apply when 30 /« = ^W 3/2 = = r{t) we the second method, then, using first 3. note that if VA(t)\ Problem 9-3 and the Chain Rule, fit) %A(t) ll2A'(i) (just as we might have Now V(t) = = /3, so • follows that r'H) follows 4. = 47rr(/) 6 = 384tt when r(t) = 4. A'(t) = 2Trr(t)r (t), and A'(t) = volume at time 2 6 for that when 48?r 4wr(t) 3 = 3 3 4[7rr(Q 3tt 2 3/ 2 ] 1/2 4A(t) 312 37T 4*-[r(0»]"» 1/ 2 ' guessed). = 3, Answers to Selected Problems 541 So 8. (i) (iii) 10. = = (/• A)'(0) = a'(x 2 ) A = 2 5 = 30. A'(0) = /'(3) • = sin 2 (sin(x = l = d£. dl = (/•«)'(*) d (i) dx d (iii) ± ^ * . 11 1. (i) 0 = [ (cos ,) • (1 = (- S in w) - 2* - = /'(*) are in 4 + 1)) • 2x 2 . = /'g?w + 2x) = w« (cos(* +x 2 )) • (1 + 2x). • = (- (cos x) (cos x). cos(sin *)) d/x dfo CHAPTER 2 sin (sin 1) dx dy = • implies that Q'w 30. (t) -g-)sin (sin 1). 2x 2 • > 2 9 (sin 4 T u2r(t)A . 3 • f(h(0)) h'ix ) The Chain Rule = — 2, 3x 2 8 for x = 2 = and x both of which 2]; = W; maximum = -gy3 minimum = —11. 2 0 = /'(*) = 12x - 24x + 12x = 12x(x = in is [— only which 0 of and x 1, 2, i]; fi-h) = «,/(*) =H,/(0) =0; maximum — yf minimum = 0. 0 = /'(x) = /(-2) = 5,/(2) = -n,/(-i) -, (iii) 2 3 2x + 1) for x = 0 , (V) + x2 1 (x forx is in = -1 [-1, (i) (iii) (x + + l) l)2x 1 2 (x + V2andx - 2 2x + l) x2 2 -1 - V2, of which only -1 + V2 \]\ = (1 + V2)/2; maximum = (1 + V2)/2, minimum = 0. — is a local maximum point, and 2 is a local minimum point, 0 is a local minimum point, and there are no local maximum /(-I) = 2. 2 0, f{\) = f, /(-l + V2) -g- points. (v) - + V2 1 minimum is a local point. maximum point, and — 1 — V2 is a local 542 Answers to Selected Problems 3. (a) Notice that / actually has a function of even degree. minimum value, The minimum since / is a polynomial occurs at a point x with n = 0 = /'(*) T 2 = so * 4. + (ai and • * + • 7 are local (i) 3 (iii) All irrational x - (* at), =1 1 a n )/n. maximum points, and 5 and 9 are local minimum points. (v) x < x is > 0 are local minimum points, and all irrational maximum points, maximum (minimum) point if the decimal expansion 0 are local a local contains (does not contain) a 5. If /(*) is /(*) The positive lim f(x) = = Vx + 2 a2 oo, and / is V(l - Vx + a = 2 0. Now, (1 - V(l - This equation says that arctan 6. is . since lim f(x) X—* a = /'(#) *) 2 x) = + is = 0 — 0° minimum when Q b2 arctan also possible to notice that f{x) shortest 2 minimum, j3. equal to the sum and the line segment from when the two line segments lie along a of the dashed line segment This +b x) 2 differentiate everywhere, so the _ * 2 is + function / clearly has a occurs at a point * with f'{x) It 5. the total length of the path, then of the lengths (x, 0) to (1, b). line (because of Problem 4-8 (b), if a rigorous reason is required); a little plane geometry shows that this happens when a — /5. If at is the length of one side of a rectangle of perimeter P, then the length of the other side is (P — 2x)/2, so the area is = XJP W 2 So the rectangle with greatest area occurs when x is the maximum point = / on (0, P/2). Since A is continuous on [0, P/2], and A(0) A(P/2) = 0, and A{x) > 0 for x in (0, P/2), the maximum exists. Since A is differentiate on (0, P/2), the minimum point x satisfies for 0 = A'(x) = 1^l2± 2 = PZ 2 so x - P/4. AX : x Answers 7. Let S(r) Problems 543 V be the surface area of the right circular cylinder of volume with radius Since r. V= we have to Selected = h where h >irr?h the height, is F/xr 2 so , S(r) = 2irr 2 + 2wrh = 2wr 2 + 2V — r We want the minimum point of S on «) (0, this exists, since ; lim £(r) = r->0 lim 5(r) = S is Since oo. differentiable on oo), (0, minimum the point r cos x is r- » satisfies = 0 = £'(r) - 4?rr — 2V r 47rr 3 2 - 2V or 3 r = 2r 8. 14. 1 is (a) a local We maximum point, and 3 is minimum a local point. have m —- f(a) = f{x) for some x for some in (a, b) a b > M, so f(b) (b) We - a). have f(b) b - so (c) > M(b - f(a) If |/(*)| ~ — /(«) < /(«) f(a) ^ = a < iti(* - Af(A - a) ( ^ 6) a). for all x in 0, b], - ^ in < f(b) then < -Af < f(a) /(*) M(b - + < M, so «), or |/(*) 16. (a) /(*) = — cos a: ~ /Ml < M(b - + a for some number a a). (because fix) = — one such function, and any two such functions differ by a constant function). 544 Answers to Selected Problems (b) = fix) x 4 /4 + some numbers (c) (a) some number and so f(x) a, = x 5 /20 + + ax b for b. + * 3 /3 + a for some a, so /'(*) = x z /6 + x*/12 + some a and b, so /(*) = * 4 /24 + x 5 /60 + ax 2 /2 + bx + c for some numbers a, b, and £. Equivalently, and more simply, fx) — x 4 /24 + x b /60 + ax 2 + bx + c for some numbers a, b, and c. = -32, we have Since = -32* + a for some a, so 2 j(0 = — \(>t + a* + j8 for some a and Clearly, s(0) = 0 + 0 + 0 and j'(0) = 0 + a. Thus, a: = v 0 and 0 = JoIn this case, j 0 = Oand^o = v, so s(t) = — 16/ 2 + ^. The maximum /'(*) + ax 17, = a for a x 2 /2 b for jff. (b) (c) value of when occurs s maximum value = 0 — zr ~ = At that moment any (as at or t top). v/\6 The (it ~64 same 29. = = -I6t 2 s (t) is < 81, fix) fall • + back down as it was initially = to f(x) V* 8 < < - some x for 7^ 8 2 9 3* 2 lim x-»l +1^0. 0 is —32 when it took to reach the v 9, so 1 < 2 8* • L'H6pitaPs Rule does not lead to the equation because lim 3* 2 > vt, (^) + Vx a/66 :r-+l t 2* moving upward). on [64, 66]: 1 = we have 2 33. which 2 1 ground at time then velocity with — x y/32, so the L hits the takes as long to velocity V66 - ~ V64 = 66 64 < = clearly 0, but the acceleration is Apply the Mean Value Theorem Since 64 * EL. 64* (»/16) = -32 (the or v, 32 The weight 0 = V the velocity time). = — 32/ + s'(t) is +— = 1 — 3 — „ 6x lim x— 1 2 in [64, 66]. . I Answers 34. 545 Problems to Selected (i) x = lim tan x x->o —— 1 lim = — lim cos 2 x sec 2 x x-»o 1. x-*o (ii) — cos 2 x 1 lim x->o CHAPTER 12 1. (i) f-\x) = = - y r (iii) (* = 1 a: - (x 1) 1/3 (If . y (If > = rK*)> = * ±y is * is if „ 1 = fK*)* /+ = /(» = then * 1, so then ^ rati0nal ^ —y, y ( y = 2x -) ,-/(*)-( yw since cos x at lim x->o 1/3 l) /. 2 sin .. = 2 irrational; rational or irrational rational = —* and if if and only if is, y irrational, so x is , an y we have = /(*).) (v) Ix, r f~ f 2. (i) (iii) 4. ^ x = = x an, (vii) x ai, (b), l is 1 is and f~ {x) -1 decreasing and / (^) l increasing a contradiction. 5. —/ Clearly, < • 6. (a) is is If / + g The proof is increasing is . is is , n Then f~ b. < 0. < 0. not defined for x not defined for x or f~ l (a) l (a) > * f~ l since (b) } f'Kb)- But if -1 / is f^a) > = «, similar for decreasing increasing, for /, or one can con- = < f(b) and g(a) < g(b), + g{b) = (/ + g)(b). for example, if /(x) = g(x) = x. if f(a) > 0 for all if a < b, then /(x) if increasing, for (fo g)(x) f(g(y))> so If/(x) . instead. (/o *O0 = r'to, = = so , < >>, since g cx + + 6 d _ ay cy = (But is so f(g(a)) < f(g(b)). = f(g(y)), then g(x) one-one. (/o^)-i(x),thenx = g-KHW). /(>), then ax then x.) so that /(*(*)) g(y) y since / is one-one, so x )-i = ^-i o /-i; for if j (f og 7. . = fU~Kb)) < fiTKa)) = (f+g)(a) = f{a) + g(a) < f(b) f-gis not necessarily increasing; / g fo g 2, then * sider . = ai. -TW l . i -/- 1 Suppose / is increasing. Let a one-one. So either fK*) < f~ . fl t-, + + 6 i = (f * g)(y) = 546 Answers to Selected Problems so + acxy + bd <z<f(# — jy) = adx + bey + ady + flcry + ftc* bd, or If <2^ =^ ftc, If y this implies that x and y for all * = /-1 (V)> men — domain in the = /W? * — bc(x = y 0. jy). (ButifW = ft£, then /(#) = f(y) of /.) so #v + = — x — ft + > so f~ 1 — (x) y —— — = cx 8. Those (i) [2, (iii) since / The formula is for x 9* a/c. a which are contained in (— increasing on (— qo, 0] and [2, [a, / is ing on [0, 2]. Those intervals since 14. intervals oo ), — [a, ft] oo, 0] oo) } or [0, 2] which are contained in (—<*>, 0] or [0, — <», 0] and decreasing on [0, oo). ( ft] or and decreas<*>)> increasing on for the derivative reads: dx = J_ dy dy 1x understood that dx/dy means (f~ l )'(y) while dy/dx an "expression involving and in the final answer x must be replaced (In this formula, is by y, it is 3 by means of the equation y = /(#).) in Problem 14, when completed, shows that The computation dx" n dx nix = is. 16. 1 ^rrw + r w - j 71 ' 1 ) y (l/n)-l - ax) = x(ry(x) + = 1 1 11 *1 F'irK*)) /(/->«) • - (ro'w cryw (i) A\A~ 1 (3)) A'(0) sin 2 (sin 1) Answers 17. to Selected 547 Problems Since (ri) ' w = /lrW we have i crrw CHAPTER 13 1. If Pn ={*„,...,<„} = -/"(r w) (r )'w is i • [/'cnw)] 2 = the partition with ib/n, then n n-1 b* f and (»- *) - 4 (" , i) 3 , - D' l (» similarly n 3 =1 "«44 + Clearly L(f, Pn and ) U(f, Pn + 2 4j" can be made as close ) - to b 4 /4 as desired by Pn ) can be made as £(/, small as desired, by choosing n large enough. This shows that / is integrable. Moreover, there is only one number a with L(/, Pn ) < a ^ choosing /. £/(/, Pn ) = /4 will 6 4 sufficiently large, so £/(/, />n ) for all w; since x 3 dx has this property, the proof that he complete once actually follows from the fact that L(f, 4 as close to 6 /4 as desired were true that U(f, Pn ) V(f, Pn) b 4 /4 < x 3 dx, then as close as desired to b > x3 Jq dx, and AO 4 Pn ) and by choosing n 4 3 * dx that £(/, P„) < £ /4 < £/(/, a straightforward computation, but we show This can be done by for all JQ it £/(/, jPn ) it can be made sufficiently large. In fact, if it would not be possible to make /4 by choosing n large enough, since each similarly we cannot have 6 4 /4 > j* x* dx. 2. We have 30 ly L(f, Pn ) and / (v) = is £ 5 Pn can be made as close to ) enough. As in Problem large loosing n dx U(f, b 1, this /5 as desired implies that (i), (iii), and b A not integrable. (vii) (For a rigorous proof that the functions in integrable, see Problem 15. The (vii) are values of the integrals, which are from the geometric picture, can also be deduced rigorously by using the ideas in the proof of Problem 1 5, together with known clear integrals.) 5. (i) (iii) (v) 6. jl (// ^^ g dy ) dx = f* = // ( ^ g(y) fo g ^ ) dy /; /(*) 24. 27. ^ j g(y) dy Clearly L(f, P) > 0 for every partition P. (b) Apply part (a) to / — g, and use the fact that (a) (a) 0. (b) i (a) Clearly m{b - a) < L(f, P) < iS thC constant) dx (here 10. (herC dx dy U(f, P) < M(b - a) is the constant). Answers P for all partitions m(b of - [a, b]. a) < to 549 Selected Problems Consequently, j* f(x) dx < M(b - a). Thus m < m and satisfies (b) < M. fi M be the minimum and maximum values of /on Let Since / is continuous, it takes on the values quently the number fx of part (a). 31. m and Af, [a, b]. and conse- Since -l/l</<l/l, we have so b (Problem 30 implies that CHAPTER 14 1. (i) (sin 3 x 3) 3* 2 • fx (iii) J K W (v) 2. 1 + t 1 + t / Ja (i) All x 9* (iii) All x (v) All (vii) All + sin 2 2 + sin 2 f t 1. ^ x. 0. 11 (F is not differentiable at 0 because F(x) ~ (/ 1)/(0) > 0 arbitrarily close to 0 with F(x) = = /XrW) = * /;/(<) dt, + 1 » 1 = sense.) 5* 1. (0 F(*) makes -dt. * 4. \ dt. 2 but there are x 3 \f . I 1 / Js f + sin (sin 0)) = sin(sin(/- 1 (0))) 1. so f'w = */« + /;/(<)*. = = 0 for x \.) < 0, 550 Answers to Selected Problems 9. We can choose = /to + 1 1 n Then £<l/n) + l V^rf* = -/(0) = /(*) i+1 n CHAPTER 15 1. (i) 1 + (hi) 1 1 2. (i) 0. (hi) 0. (v) 6. (a) + + (. arctan 2 (arctan x) (tan x arctan 1 arctan 2 x +x 1 2 tan x \ sec 2 x arctan x + + sin a: 1 — 0. sin 2a: 2x cos 2x sin cos 3a: 3a: = = = = = = = + x) sin(# x — sin sin (2a: + a:) cos 2 sin * cos # 2 = x = 2 cos sin a* cos 2 x + (cos 2 x 3 sin a: cos 2 x — sin 3 > (b) Since cos 7r/4 0 = + x) = = = = = cos cos 3 cos 2 = 1, x 4 cos 3 _ 7T 7T - = cos 2 cos 7T cos - = 2 2 sin a;. 2 sin x — sin 2a: sin x — sin x) cos = 2 sin * cos — sin 2 cos x — 2 sin x cos — 3 sin cos x — 3 cos cos x 2 a: 2 a: 2 a: a: a: a: a:. * - = „ 2 cos 2 7T - — It follows, since sin ir/A = V2/2, and 7r/6 > 0 and * 6.6 - 4 cos > 0 and sin 2 + = 1. consequently tan 7r/4 ^^.q^o = cos 3 . 1, 4 that sin 7r/4 = 2a: 2 sin # cos sin 2 x) sin # 4 = V2/2. Similarly, since cos 0 1 + = 0 and cos cos tt/4 = 2 a: cos 3 x 2 we have — a* a:. 2a: (cos 2 x cos — cos x 2a: 2 sin cos (2a: 2 " 71 3 cos -> 6 - Answers we have sinx/6 8. = w/6 cos \^3/2. to Selected since follows, It 551 w/6 > that sin 0, = ^1 - (V3/2) = i 2 (a) + y) = tan(* —+— sin (a: jp - + cos (a: jy) sin # cos cos From part (a) we have + arc tan y) . tan (arc tan x jy sin x sin cos x cos jy cos # cos y — = . + a: tan x tan < + tan (arc tan x) tan (arc tan y) x +y 1 — xy and arctan x y, + } arctan y + provided that arctan x arctan < y 0 jy > and The first < xy = / x arctan ( formula — n)x cos (m + n)x The — = — we must add so that arctan x 1, is arctan y 5* is always 7r + ~\~ y \ - xy) ) (— 7r/2, > 1, which is so that arctan * + 7r/2), to the right side, arctan y < and if x, —ir/2, then we derived by subtracting the second of the following cos (mx — first: nx) = mx cos(-nx) mx sin nx, cos cos nx + sin cos 01* cos nx — sin tw* sin nx. cos + 7r/2, this 7r.) two equations from the cos (m arctan y lies in > 0 and xy (If x, y 1. then 7r/2, ry •> must subtract < arctan x arctan \1 whenever tan(arctan y) when arctan x + arctan^ = ±7r/2, which is equivaFrom this equation we can conclude that arctan x true jy jy tan(arctan x) — j> tan y arctan + 1 sin ^ sin y cos * cos provided that arctan kir 7r/2. Since —7r/ 2 lent to xy — cos # cos y 1 = + cos # sin y cos # cos y . the case except jy cos y cos x sin tan (b) a: sin x cos j 1 10. Problems mx other formulas are derived similarly. — sin mx sin( — nx) . It follows 11. sin mx from Problem 10 that — \ sin nx dx _ — [cos(m — [sin (m 1 m ^ if — n)x m — + cos(m sm(m n)w | 2 IL then n, + «)tt] m+ n n)x] dx n J sin(m — n)ir _ m—n [ = But if sin m = m* The (a) m + n)T + n JJ *l| 0. then n, sin n* </* = ir = = 14. sin(m ~~ 1 J*^ — k{[ir + n)x dx cos(m cos(m + n)7r] — [— x — cos(m + 7T. other formulas are proved similarly. We have cos 2x = = - — cos 2 x sin 2 x 1-2 sin 2 cos 2 * 2 - * 1. So = sin 2 * 1 — cos 2x j 2 = cos 2 * (b) 1 + cos 2x These formulas follow from part sin x/2 > 0 (since 0 < x < ir/2). (a), because cos x/2 > 0 and (c) sin 2 x dx j = [ / h * j cos 2 x dx /* 6 = </x — arctan lim arctan x 1 — * (b — — sin 2a). + -4 (sin 2b — sin 2a). (sin a) — (b — x a) t/4. arctan 0 = tt/2. x—* • 1 9. lim x sin x x->« 20. = lim - sin x x-*0+ = 2b 4 1 2 2 arctan 0 (a) 2 + cos 2x 1 Ja (b) = 2 j Ja 18. dx I Ja Ja 1 x (a)(sinTW=^cos(^)=^cos^). , 1 , . . - v Answers sin°* = S^T < b) 17 1. Si x-^o + e ** ee e (iii) (sin *) • '* • • 8in(sin:r) sin (v) x • 180 • 180 a: a* [(log(sin *)) cos(sin *) • • cos x + ein ?r " „/180 18<)' 553 Problems sin(ir*/180) tt ™ —— = = hm * (i) = i Jim # sin x_>« CHAPTER sin(7T*/180) to Selected *,in *[(log(sin *)) {(log(sin *)) sin ' cos x ' * + (cos */sin x) sin(sin • a:)]. 8in x (cos x/sin x) sin *} • + (cos jc/sin at) • sin * 8in *]. (cos [arcsin (vii) (^J^^fa (arcsin arcsin 3. (i) 0. *) • i ? cos * + log * 4 sin- a: \sm */ * at/ ' ' i i (iii) (v) 5. [logOog sin • | \sm (log *) log * — sin x + log (sin **) (ix) (^))) (a) cosh 2 x — sinh 2 * = — ( — — I J J = 1. (c) sinh x coshjy + cosh gx+y e a: -x-v ir + ~i sinhjy = I ( -r 1 J e -*+v e*-v~\ V ex +y e -*^ ~ ^ e~x+v ^ ^ e"* y -2 '] r + "Tj + L"T + "r + "T""~^J = = smh ! 2 (e) ^ Since sinh x =e + e" (a: + y). 554 Answers to Selected Problems we have = = sinh' (at) K } cosh x. 2 (g) Since . = tanh x sinh x 1 cosh x we have — (cosh x) 2 = tanh'(*) (sinh x) 2 by part (a). cosh 2 x 6. (a) If y = arg cosh x then * x, = cosh^ = > 0 + sinh 2 y/\ and by part ;; (a). So sinh (arg cosh x) — = V* 2 — sinh y 1 > since sinh y (c) 1 = (arg sinh)' (*) sinh '(arg sinh (a:)) 1 cosh (arg sinh(#)) 1 VT+x~ bypart (b) * 2 (arg tanh) (e) 7 l W 7 tanh (arg = X.axih(x)) 2 cosh (arg tanh(#)). Now, tanh 2 jy + —— = - 1 by Problem 5(b), cosh 2 j/ so tanh 2 (arg tanh(*)) + —— — = cosh 2 (arg tanh(#)) or cosh 2 (arg tanh(*)) 1 = 1 7. (a) If y = arg sinh x, then x = sinh = - x2 1, 0 for y > 0. Answers to Selected Problems 555 I so - ev - e™ € = -y 2x, -1=0, v 2xe ± Vax 2x 2 + 4 > 0 so + Vl + ev = = arg sinh x x x2 ey since or y = + Vl + a log (a* 2 ). Similarly, = arg cosh x = i arg tanh x (b) 6 f / J a = —— 1 Vl + = dx log(l arg sinh *2 P , 8. fl . 2 f —-^— r ; <fe 12. (a) (b) 1), ilog(l arg sinh a — *). by Problem 6(c) + Vl + b - log(a + Vl + a + V* - 1) - log(a + Va - 1). = 2 2 ). ) 2 2 log(£ [log(l + - b) - log(l A) - log(l + + log(l - a) «)]. lim x— > e e xlo * a Since log a . < 0, we have ' oo = xlo * a lim x log a x—* = - oo, oe 0. jM^-lim£-0. ^ x-^ (e) = oo so lim (c)1 — — a) 2 lim a* X— 6 + 1 - - 1 (a) & 1 Vx - 2 ~ = logO Ja + V* — \og(x y-* oo lim = lim e * ey zlogX . x->0+ lim log(l +y)/y = lim x log(l x— Now, lim log x a: = 0 by part (d), so lim x* + l/x) = log'(l) = = 1. x->0 + x->0+ x->0+ 1- lim log(l + y)/y = 1. y—*0 + oo (c) « = exp(l) = > (*) = + l/x)) + l/x)) exp(lim x log(l X— lim exp(xlog(l = lim x-+ (1 oo + l/x) x . 00 (The starred equality depends on the continuity of exp at 1, and can be justified as follows. For every £ > 0 there is some 8 > 0 such that \e - exp y\ < £ for \y — 1| < 8. Moreover, there is some N such that|*log(l + l/x) - 1| < 8forx > N.So\e - *log(l + \/x)\ < 6 for x > N. 556 Answers to Selected Problems (d) e = a (1 » X—* = + \/x) f = + /x) ax + a/y) v x [lim lim (1 lim y —* 14. After one year the one dollar will (1 . number of dollars yielded by an initial = + (1 a/100 = x x) e a ' 100 . 00 > 15. Clearly /'(*) 16. f{x) = (_!).!/(-*) = l/x Lctg(x) = f(x)/e cx Then 1/x for x 0. < x If then f(x) 0, so there (a) (b) - ^/(^) - \ some number is UA(t = 16, there . = A(t)/2 9 so so <?<V this r = e ct /2 or does work. T Newton's law * CT — k for all 2)/c. It = c(T - M) is - MY = (i) k = ke*' (V^a = 7(0) 1 (n) , 7 x + (iii) (* (iv) o*/P = 7V + ^c/Vx +* = ( fl 3x — = - c(T y M). some number T(t) and ct . a; - M= So T(t) 1' 10 + aT Vx = )A 4 * == e'* x /*)* = ,*io« («/&>_ + A: such that ke ct , = M+ 7V cf . 1/3 ). 1 - V* + + e~x . easy to check that number which can be written is £<? 2 = ^,sor = - (log T'{t) So by Problem 16 there = c *. states that, for a certain (positive) (T a:. k such that A{t) then 4^ = 1. log(-x), cx some is ^ = - A 0 So + r) f(*)ce such that g(x) A: According to Problem Then k 18 = . . aU CHAPTER investment of be lim 18. ax oo X—> 17. \/x) oo 1 OX—* W = + lira (1 X—* 1 c, Answers = (v) tan 2 x (vi) a+x — sec 2 x 1. + 1/a 1 Va -x 2 —— Vl - (j/*) — sin —^ = = — sin —x 2 - + 1 8a: x : sin x + 2 —+— = + x 4 (iii) (log *) 2 /2. (Let u c 2 2^. = We (iii) e ar ax e . — sec x tan x. +*+ 2 1 e X ( - l) 2 e*.) = log x.) \) = V*.) (Let u = f sec 2 x -(log(cos *)) 2 /2. (Let u = log(cos *).) = x 2ez - J2a:^ </* = x 2e* - [2xex J*V (i) / = (Let u . \ (Let a (vii) (ix) 3. Vl - (i) AT <? = 1 V2* ~ - cos ex (v) sin # cos 2 x 6 -- 8* 1 1 2. — 1 2 1 6* (ix) (x) 2 1 1 .... , (viu) 557 Problems = i (vii) to Selected x 2ex - + 2*** 2e - x fe dx] x . have , , 1 sin 6a afr sin = 6 £a: a J = /" ex / e ax cos bx dx a J sin bx b [e ax cos bx b [ e / ax (- sin bx)dx ] L so y (v) e ax (log x) z " 0 ^2 Using the the text, j ^ sin bx dx dx= + - - az sin 6* c> yo £2" result /(log *) 2 <fe = + 2x]\og x [*(log x) 2 - 2*(log *) A:(log x) 3 - 2x(\og x) 2 *(log a:) 3 = *(log (log b 2 — a:) , „ 6> * 2 ax cos £a\ 2a: (log a*) — 2A*(log x) 2 2 - 3*(log a-) [>(log x) 3 — y ~ [x(log x) 2 + = a: + x) + 2x from we have = " 0 a2 (log x) 2 dx 2x log 2A*(log *) 2 + + 2*] & 2x log x J + 2*ftog *) 6a: 2[x log x - x] - 2x + 2[* log x - x] - 2x a: + log + a* 2x] — 6a*. (vii) sec 3 x dx J* = /(see 2 = = tan x sec # = (sec x) dx a:) tan x sec # — tan — — sec x a; / (tan 2 J sec x(sec # 3 dx sec # J + — #) (sec x tan #) <£* 1) / sec * cfc, so = 3 J sec x dx + log(sec x + %\tan~x sec * //V log x dx — 37 2 2a: = — a: 2 - /f * 3/2 — log x 3 = 4. (i) = Let x = sin w, (iii) = Let # /sec — sec w, (v) Let * = = 1 * 3/2 ^ The . . 1 du * 1/2 / - becomes integral = = u arcsin *. 7 f . The . = sec a du / integral + , . log(sec w becomes >. . tan a) 7 + Vx - = log(* = cos w du. == ~ VI — f / - sec u tan w du. H-l sin u, dx CQ S sin w = <Z* —= Vsec - sin 2 w u tan u du 2 log * = . VI - - cos w du. /cos u du= - log x *)]. - dx x * 3 J — — = tan sin 2 u f 2 The 1). — — esc u du becomes integral log(csc u + cot u) J __ log (i + ^IIi). X \X (vii) J Let x = — sin u, dx (sin 3 u cos w)cos u du = The 2 sin 3 a cos u du J f = cos w du. / , w . o 2 (sin u) (cos w (1 - x 2) 3 ' 2 Let x 2 J sec w sec udu = — = = tan u dx y 3 f sec w = sec 2 u du. (sin a) (1 J ^ cos* (1 3 (ix) becomes integral = — / The - v = . u)du — cos 2 m)cos 2 u du cos 3 a , 1 cos 5 u — * 2) 5 ' 2 5* integral becomes <fo + log (sec u + tan n)] ±[x VTT7 + log(* + Vl + * by Problem 3 (vii) ^[tan w sec w 2 2 )]. Answers 5. = 5 f 6u du J^+^ = = = f / , 6 a u2 ~ u - u* = u 6 , dx 6w 5 du. 2u du. The The integral becomes + Vx + 2 log(l 1). becomes integral \ 1 + = dx 1, + U )=2V, + l- = * 1/6 , x = x 1, 2 log(l J ( -^r )du = 2u^3u^ + 6u-6log(u+l) 1 1 2^-3^ + 6^-6 log(\^ + Let w (v) - 2u Let u (iii) Vx + = Let u (i) 559 Problems to Selected = tan x} x — arc tan a, = </* 1). + </w/(l The u 2). integral becomes f 7 _ du (1 «-2 \ + «y r/_i 1 112 Sj\2 + + « )(2 + u) 2 = = 7 2 + 10 7 - w) d = Let u (vii) = + ^log(2 = 2X x , 2u JL ( +« - log(2 " 1 du 1 f 5 u — 10 tan x) +a 1 + w 2) — f 7 5 + log(l +2 d 2 1 +a - arctan a 5 ^ log(l - = (log a) /(log 2), dx + <fc/( M tan 2 *) log 2). + | *. The becomes u* 1 f log 2 7 + r/ 1 1 " («+l)« 1 - \ g * log2 7 V log 2 7 V a ^+ = 1 [2- log u « " + Let u = V#, x — u 2 , dx r J — + * log 2 - Vl vi 1 2u. The 2w u 22 2x w — w M «/ 1/ 2 log( " + 21og(2* log 2 (ix) 2 integral 1)] + becomes 1)]. integral Now let u ~——~ 2 cos y sin y cosy / . dy = sin^, du = 2 _ sin y 1 cosy integral becomes 2 (1 J , sin y 1 = 2 = 2 y* sin j (1 + sinjy)sin^ </y +^ — cos 2y dy 1 </y ^=— sin 2y cosjy +jy substitution a +y — . 2 cos;; + arcsin u — u Vl — w x + arcsin V* - V* V - = — 2 Vl = -2 Vl The The dy. ^—— —sin fy)sin J-y dy f = —2 = sin^cos.y 2 u2 1 = Vl - *, * = 1 - u 2 dx , x. = —2k <fa leads to -2m 2 du VT and the /— 2 sin 1 2 — y cos y — substitution u </v = — 2. sin ^ —y — . sin y then leads to sin y cos y cosjy = —2u — arcsin u — u Vl — u = -2 Vl - x - arcsin Vl - x - Vl 2 x V*. These answers agree, since arcsin (check x 6. = this V* — — — by comparing arcsin Vl — x their derivatives and 0). In these problems / will denote the original integral, (i) = 2 log(* - 1) x + 1 (iii) _4_ x -iy~"" (* - ( (x + iy 2 1_ 1) J " (* + l) 2 their values for 1 Answers to Selected Problems (v) /= = f-^-dx T + f-^—dx + ; ,* + ilog(* + 1) + 4 arctan *. 1 2 J x2 1 2 (vii) Now f x* + x +i dx J + i; (X FT- I / 4 3' V3 2 -~ (&(• + !)) so / = log(* + 1) + log(* + x + 2 1) - (x arctan (ix) /= 2* + (^+x +d y 2a: + 1 /" j * 2 _ [ j _ 1 16 1 (* 2 +* + /* 1 / Now the substitution V3 2 changes the second integral to VI 16 ' 9 2 f J __du_ ' (u 2 + l) 2 * 2 1) + i)). 561 562 Answers to Selected Problems Using the reduction formula, * [ L2(w 9 = _ 2 + + 1) _ 1 *2 +*+ 2 1 f _*L- 1 2 7 w2 1J V3 can be written ^ = - + l0 ^ [ L 9 Q(* + *+ 2 + 2 V3 The equation J> J sin x dx = e z sin a: — e x A(x + -1\\ / 2 arctan 9 . + 1 arctan J, ) 4 1)) 4 1 1 1 2 log 9 1 this — JV cos # , W3 V I [ ) J. 2// sin x dx means = f sin # can be written Fix) = ^ sin x — any function F with x x e cos x - G(x) where G is another function with G'O) = e sin *. Of but it is not necessarily true that course, G = F + c for some number that F= 12. G. (a) j arcsin x dx = J = 13. 1 — arcsin x dx • x arcsin x — x arcsin x + Vl — x2 . J dx . (a) /sin sin 4 x dx 3 cos x = -—4— + sin 3 = sin x cos * 3 , / sin # cos # 3 f 4 — sin 2x 4 4 . ^_ sin 3a: 2* 4~~ (b) It follows that these same value for x = + 1 Yx + , 1 /* 2i 1A 1 . , J 3 x. 8 cos 4# fa / 2 , 4 L2 2 sin 2x "8 1 _ f 4 J 1 4~ 4 — . 8 x * sin 2 x dx 3 sin x cos # cos # a" /* 3 h - sin 4#~ 4#~l 8 J sin 4# 32 two answers are the same, since they have the 0. + + + x Answers 563 Problems to Selected (a) 17. sin n — /(sin ^(sin*"" — — cos x sin 71-1 = — cos x sin ^ 1 x dx 7 x) dx + + x 1 a: — — (n (n l)/(cos 1)/ (sin a*) (sin n~ 2 71-2 a:) x — f sin sin (cos x) dx n a:) ak, so /sin 71 = — x dx - cos * sin 71 "1 + a: n n 7* -2 a: dx. J (b) n J cos x dx — — = /(cos x)(cos n_1 sin x cos sin x cos 71-1 71 " 1 x dx a:) + + x — — (n (n ~" n 2 a*) sin x dx 1)/ sin #(cos n~ 2 x — cos n x) dx, l)J(cos so /cos n x dx = - sin a* cos 71-1 a* + n cos n f n ~2 x dx. J (c) /dx xy (i « 2 (i f y fx dx [ " j a* )^ " y )- [ y * 1 dx 2 (i a- /dx + (1 1 2 dx + (i ) * xt (i dx a: I" ~ )*" 1 2 A* w 2 - L2(l «)(1 ~h 2 A" 71 -1 " ) - * f - 7 2(1 + n)(l w 2 a: ) 1 ^J so dx r y +* (1 a: 1 n 2 - 2(» ) 1) (a: 2 + l) 71 '1 _ We can also use the substitution x changes the integral to /sec 2 sec u du 271 u f 2 cos 271 2(n - = tan (2w /* 3) 1 y 1) + 2 (a: = a, ^a- l) 71 "1 sec 2 w du, which u du J 2n - cos 271 3 u sin a 2 1 In 2 ' (Vi~+V 2 - 2n - Z ) 1) (1 + ' 2n Vr+T x 2k * 2) n_1 2n 2 — - 3 4 271 f cos u du J x 1 1 2{n +— ~ -2 2w - (1 + a* 2 71 ) dx 3 f 2 y dx f 2 J 2n - ' 1 (1 + 2 a- 71 ) "1 564 Answers to Selected Problems CHAPTER 19 1. (i) p 3i0 ( = x) + ex + ex + 2 e <P2n.iT/2M = (-\) n (x - - — 1 (5e/3\)x\ 2w tt/2) (2n)\ (v) Pn,i(x) = + e(x - e — H 1) • f- H • • 2! 2. (vii) P4,oW = (ix) A H ,,oW If /is Theorem -12 (iii) 243 W (iii) 1 - 1 + *2 - *4 that R„, a (x) • = 0. = Pn a (x). then / n, 0, so f(x) ( * +1) from It follows , . - (x 3) Y 90(* - 3) 17f° r2 " +2 + 2 3) 3 + - 15(* 3) 4 5 . (27^ <10 SinCe =0 (since —— V 2 1 -*' - J"" L(2>(2i+ 1 = + (-i)V. • • 2 t(2^X t' V . + 2(* - 3) + (x - 3) + 405(x - 3) + 270(x - (i) + - = n\ 3 a polynomial function of degree Taylor's 3 +x at 1)! =0 " - 2re+2 (2« < + 1CT 20 ^ for 2n 9 '° r ^ +2> 9 ) 18, 2) > or n 8^ 11 - (since i 7. (i) (iii) =0 a = a = 21 i. ai (t - (i) i (iii) lim n— w + (n + + = (v) chapter \ i\ 10~ 5 for n + > 1 + = 1 ) vV + 1 a = i + + /( w - < 1) = 1 i > e for n (Vn 2 lim n— + > 1/e. + - V^) + 1 lim (V 1 0 = 0. (Each of proved in the same way that lim(V 71—» proved in the 1) = < Vn _+ 2 = n—* » < 0. Vw, n-+ < hese / + ?? « t wo i /? limits - Vn) = can be 0 was = So lim lim *< lo *<*>/* = e o ( by 1. n and lim (Vn) 2 = lim Clearly a(n) t 7 n— » text.) Clearly lim (log a)/n Theorem (ix) ). 0. oo / (vii) 11 / a t-_iA for — V w + l)=0 + (v) > 12, or n ! \)cn. f(t) dt; n / (n < 1) log 2 so Vn + (Vn) 2 < Vl (V n) = 2 1 2 by parts °° n, and lim (log 2 n)/n = 0. n < Vl (Vn) 2 (v) and (vi). , » Answers 4. If (a) < 0 < a (b) Part (a) < then a 2 2, "^2 the 5. If sequence x is same tional, then for all 1 ti\tx is such a multiple of = so f(x) 0, < 2, 2. w = V2/, / w any for or / = is } 2. for sufficiently large w, so 2k so lim(lim(cos n\itx) ) n— 00 k—> 00 n, not a multiple of is = lim(cos n\irx) 2k - - - So the hint shows that as {a n +i}. = 2k < 2. denoted by {a n \, then the sequence {V/2a w is rational, then nlirx (cos n\irx) ^2 V2 V2a < by Theorem so the sequence converges If this < 4, shows that V2 < ^2 V2 < (c) so a < 2a 565 Problems to Selected n, = 1. so |cos x If «!7ta:| is < irra1, so 0. a— * * 6. j (i) (hi) / Jo e x — dx — e (Use partitions of \. —— dx = log 1 + into n equal parts.) [0, 1] 2. * 1 Jo CHAPTER 22 1. (1 + a+*) ^ 2 ~" 2' (i) (Absolutely) convergent, since [(sin nd)/n 2 (iii) Divergent, since the (Leibnitz's Theorem does not apply < \/n 2 sum \ 2n terms have first \ . + " * decreasing in absolute value.) Convergent, since (v) 1 < A for sufficiently large n. Convergent, since (vii) lim (« + !);/(* 2 /Z + !)! . Hm (n_±AV n-*» n-* « \ /tt! (ix) Divergent, since l/(log (xi) Convergent, since l/(log (xiii) Divergent, since «3 for large enough + / _ . n + Q 1 > rc) n2 72 ft n) n > < ^ for « > 9. 1 2n 1 n. Divergent, since (xv) L 3 dx a: log x = log(log AO — log(log 3) * + 1 In. since the terms are not —* oo as N— > oo. . 566 Answers to Selected Problems = (Notice that /(*) m ~ [1 + - Convergent, since (xix) Convergent, since 2n+l n ( < 2 1/n 2 for n 1)n " " 1 + 1 lim n™ a n 10~ ^ so n= < only n 2. 2 r lim = 2 -j n 2 ^10" < The when 1. i(* 1 = = 1, 1 lies + i 1 number is denoted by 0.aifl2«3«4 (Actually, this the sequence {a n } is not eventually The is integral • + - • • . is m ttV] • 0.) - 1] + » - tVi + + •••)- Ui + + + + + 0 + <nr +•••)- id + i + + + tf* 8 n area of the shaded region i(i lCT n ^ 9 1 converges by the boundedness criterion, and w - « + +a - +«A+ + 1. > l between 0 and 23 - N n= CHAPTER oo), since by Problem 17-12. For each we clearly have (a) = = (« [3, + l)nU + l) n+1 2{jl = 2 n «!/fl n 0 15. iorx>e 0 < 2 = 4. decreasing on is x] t l/rc (log n) + n™ lo , (x log X) (xvii) lim l/(x log x) • • •) • •) • •) • • 1 1-1 (0 f(x) = lim n-> « (iii) /nW = * 0 °> ( 11, " < ° X {/n } does not converge uniformly to /. n /(*) = lim fn {x) = 0 (since lim x = n— oo n— > < oo , 1 for * > 1). The se- oo quence {/„} does not converge uniformly to we have /„(*) large for sufficiently large x. /; in fact, for any n Answers ( v) = /(*) !/»(*) 2. (i) - 1 I = lim fn (x) < 1/n for X X a2 az and {fn 0, Selected Problems converges uniformly to /, 567 since all *. -- -- a } to . (iii) k 3. =0 (i) e~* (iii) If yw yZ YA y.'k 2 3-2 4-3 11 ' then = log(l+*) so for (1 4. < |*| 1 we have + *)log(l + = f(x) (1 + *)log(l + x) - Since /(0) = 0, x) - x for \x\ < 1. some number for |*|<1, c. we have c = + x) + c (1 1, so f(x) = Since sin * = — > n =o 0 we have (-l) n x 2n + n=0 (notice that the right side is 1 — for x 1)! 0). n I(-l) (STi)? 0, chapter 24 1. (i) |3 (iii) |(1 + Ai\ +0 ment (v) |(|3 5 for + | = 5; 0 = = (|1 + 1 4t|)| + = i, A; _ = 2n odd. arctan 5 *1) = 1; since tt/4 an argument |5| k So = 5; 6 = 0. for (1 + = i) arctan 1/1 b is 57r/4. is an argu- 568 Answers to Selected Problems 2. (i) X V-l - — -i ± -i ±_ Vb + (-1 (iii) x2 + 2ix s - 2 x (v) x 2 +x+ 2)(x 2 1 ^ 2 2 1 • 2 2 1). All z = All z on the perpendicular bisector of the and No 2 a- = + ry| (z + z)/2 (* - z)/2 \z + w\ 2^w = 2 since 2 j> since [a- 1 * 2 = = + with = —i. solutions are line segment between real. b. z, + iy x is The '* (i) (v) 5. so the only solution , (iii) a 4. L. or z') - (-1 - Vl)i ^— — Vs)i 1 ' 3. i — = (x + - x - 2 = (* 2 4 < — 2x + y [(x [(* \z 2 1 — + 2 + + + |# — 2x < = x2 ] < x2 > 2 + < — 1 that x > which 2^-, and that 1, is impossible, 1. ( (z implies x 2 jv + — y) = +0- (* - = — 1 so , 0 for x z» i» w\ 2 z'jy + — \x iy\. z»]/2 = *. iy)yii = y. w)(z + + fi?) (* — w) (z — w) = 2zz + Geometrically, this says that the sum of the squares of the diagonals of a parallelogram equal the sum of the squares of the sides. CHAPTER 26 1. 2(|z| 2 \w\ ). Converges absolutely, since (i) |(1 + n i) /n\\ < (V2) n /n\, and 00 V (iii) (V2)»/„! converges, Converges, but not absolutely, since the real terms form the series ~i + i ~i+ I" — * and the imaginary terms form the «•(*-* (v) ' ' • series + *- + +•••). Diverges, since the real terms form the series log 3 log 4 log 5 log 7 log 8 log 9 3 4 5 7 8 9 Answers 2. (i) The n+x /{n \z\ / n; n-> • \z\ <1 for |*| < The limit + /n 1, is <1 for |*| < The limit 1 is (v) Problems 569 limit lim (iii) to Selected l) 2 = hm 2 (n + IV, n / >1 but >1 but for for = , f n->« \ > \z\ > \z\ \z\ ) , , \z\ 1. 1. on+l| J(n+l)l lim - 7 0 for is 3. (i) The 2n 3* |*| for <1 for and a/3 |*| |*| > 1. < V2, so h *"Ej V2 V 2 n+1 the series converges absolutely for < y/l. But the series does not converge r /O > V2, so the radius of convergence is 11^ (iii) 2\z\^~* lim *r+* °o fcC- W f |*| but 1, , limits lim are , < |*| = 1 2 n \z\ nl n-- . • i • 1 for Since lim n—> * >_M: = n \ 72 lim !£!i5 <„„!£! = n-> « n— 72 <» the series converges absolutely for all vergence (v) absolutely -v The is 2, so the radius of con- oo limit V 2 n* = / lim n— is », 71 0 for |*| < 1, n! (n 2 lim * n—* oo but oo for |*| > 1 " 1)! oo , so the radius of convergence is 1. GLOSSARY OF SYMBOLS P (— co, a) (-•,*] (- 00, 00) 9 \a\_U Vx 12 max (*,;;) 16 min(x,y) 16 ("epsilon") e N M 19 lim/ 81 lim f(x) * lim /(*) 25 Q lim /(*) 25 x—>a~ R 25, x 38, 45, / 41 f +g lim 41 x— 41 f . . . 42 a„\ , 42 h f-g-h fo g 41 .) . . +g+ 112 112 /' 121 CA 48 *?(*) 130 48 49 max(/, g) m in(/, f<g (a, 6) 137 55 (a, oo) 55 [a, oo) 55 92 52 263 264 265 275 Z(x) 276 log 285 exp 287 287 e 288 e* 288 289 296 296 296 arg sinh 296 arg tanh 296 296 299 arg cosh /"' 137 Nap 137 F(x) log * >» if ' 302 303 If(x)dx 201 rj>) 214 *(/,«,*) P * n,a 215 (a, 6) t/(/,P) 54 263 arcsin /<*> /-» 51 the open interval [a, b] 131 ^ 49 49 the pair cot tanh 48 l/l 92 cosh dx /" R-A 263 sinh * 47 263 esc lo go df(x) a 6' 263 tan a* 127 127 43 47 259, 261 sec *<J,P) 121 f'(a) n « A 4 sin arctan 121 42 *-»/(*) 573 112 infi4 259 259, 261 arccos 112 glb4 lim A lim 43 oo *> lim sup 42 f°g°h Ft i = i oo 254 cos 92 /(*) == sup 4 Iub4 41 {at, 87 lim /(*) — 00 lim /(*) = 41 f-g {*: 87 X— 507 / A(x) 41 AC\B //< 86 lim /(*) 254 257 sin r 257 t 258 a lim /(*) 495 27 G> X / sin° 86 I, Z /(*) 86 x—*a + 24 254 '/(*)<** 78 81 23 »! 254 / 70 lim /(*) 23 0 249 55 ("delta") 5 249 70 W 21 55 55 215 219 334 334 343 357 (:) 225 \a n ) 303 327 372 Glossary of Symbols lim a n n— oo C 373 - > lim a n 71—» = 376 co Ke~ Im A A O 445 T> 382 lim x n 385 a n— > oo lim sup x n 385 479 "70 AA 441 \z\ oo 7 ^ 438 A A A 441 z 11 AAA A 4/V <fn 481 V'n 40Z T 448 4o/, 49/ 44v lim /(^) 4o 449 U /, dUI 4o/, 4vo n—» oo lim * n n— > sin lim inf x n n— > 385 cos exp a, b) 386 bn 00 an 389 Bn Z) i 458 470 470 oo N(n; y /'(a) 385 oo 433, 439 Dk 470 478 479 479 479 —a^ fl" P > < > < |<x| 1 A QQ AClQ 4oo, 4Vo 488, 502 489 490, 496 490, 496 490, 496 490, 496 501 INDEX I I ! 1 AalbmcndoE, 235 Arctan, 264 Abel, Niels Henrik, 332, 430 addition formula Abel summable, 431 Abel's lemma, 385 derivative remainder term Arg Arg Arg Absolutely convergent, 397, 463 Absolutely summable, 397 Acceleration, 137 Acta Eruditorum, 124 Argument Addition, 3 law for, 9 commutative law of, 9 of complex numbers, 438 geometric interpretation Addition formula Argument function, 449 453 of the hyperbolic functions, 296 Arithmetic mean, 32 of, 442 Arrow, "x arrow sin (a: 2 )," 43 Ars conjectandi, 481 Associative law 270 270 for addition, 9 266 for multiplication, 9 Average 266 for sin, 265, for tan, sinh, discontinuities of, associative for cos, 342 351 296 cosh, 296 tanh, 296 Argument, 443 for arctan, for, 335, for, Area, 214, 219 1 of a complex number, 441 for arcsin, 270 for, 264 Taylor polynomials Abel's Theorem, 431 Absolute value, of, 269 velocity, 128 Axis Additive identity, existence of, 9 Additive inverses, existence of, 9 Algebra, Fundamental Theorem horizontal, 55 imaginary, 441 real, of, 317, 445, 455, 474 441 vertical, 55 Algebraic functions, 302 Algebraic number, 362 Bacon, Francis, vi Algebraist's real numbers, 505 Bandes, D., handiwork Almost lower bound, 120 Almost upper bound, 120 Analyst's real numbers, 505 Angle, 256 directed, 256 Antidiagonal, 213 Arabic numerals, multiplication Arccos, 264 Archimedian property for the rational for the real numbers, 490 numbers, 116 Arcsec, 272, 322 8 Bernoulli's inequality, 31 game hunting, mathematical theory 459 Binary operation, 487 Binomial coefficient, 27, Binomial series, 357 409, 429 Binomial theorem, 28 Bisection argument, 120, 459 Bolzano- Weierstrass Theorem, 378, addition formula Taylor Bernoulli polynomials, 481 of, Bohr, Harald, 328 Arcsin, 263 577 Bernoulli numbers, 478 of, Archimedes, 116, 119, 224 derivative "Bent" graphs, 125 Bernoulli, 124, 481 Big Arc length, 275 of, Basic properties of numbers, 3 of, for, 264 series for, 270 386, 459 Bound 429 almost lower, 120 Bound Complex function (Cont.) almost upper, 120 continuous, 453 greatest lower, 112 differentiable, upper, 111, 490 least graph of, 457 449 lower, 112 limit upper, 111, 490 nondifferentiable, 458 Bounded above, Bounded below, 100, 111, 377, 490 449 of, Taylor series for, 469 Complex nth. root, 443 Complex numbers, 433, 438 112, 377 Bourbaki, Nicholas, 124 absolute value addition of, 441 of, 438 Cauchy, 239 Cauchy Condensation Theorem, 410 geometric interpretation Cauchy criterion, 390 Cauchy form of the remainder, imaginary part of, 438 logarithm of, 477 345, modulus 347 Cauchy Mean Value Theorem, 178 Cauchy sequence, 379, 477 Cauchy sequences, equivalence of, 505 Cauchy-Hadamard formula, 476 Cauchy-Schwartz inequality, 239 Cesaro summable, 408 Chain Rule, 150 ff. proof geometric interpretation 154 of, 442 442 of, of, 441 multiplication of, 438 geometric interpretation of, real part of, 438 sequence 462 of, Complex plane, 440 Complex power series, 464 Characteristic (of a field), 492 466 466 Complex sequences, 462-463 Circle, 63 Complex of, Change, rate "/ circle unit, circle of of, 128 of, series, of, 463 Complex-valued functions, 448 Composition of functions, 42 Concave function, 192 64 Circle of convergence, 466 Classical notation Conditionally convergent 142- 143, 162, 212 for integrals, convergence radius of convergence 42 for derivatives, 130-131, 138, 442- 443 225 Cleio, 161 Closed interval, 55 Closed rectangle, 454 Closure under addition, 9 Closure under multiplication, 9 Commutative law 398 Constant function, 41 Construction of the real numbers, 494 Continuous at a, 93, 452 Continuous function, 93, 453 nowhere differentiable, 135, Continuous on (a, Continuous on [a, Contraction, 384 b), 96 b], 96 for addition, 9 Contraction lemma, 385 for multiplication, 9 Converge Comparison test, 391, 407 Comparison Theorem, Sturm, 275 Complete induction, 23 Complete ordered field, 490, 509 Completing the square, 18, 318 Complex analysis, 473 series, Conjugate, 441, 446 pointwise, 415 uniformly, 415, 419 Convergent sequence, 373, 462 Convergent series, 389, 463 absolutely, 397, 463 conditionally, 398 422 Index Convex function, 191 Convex hull, 460 Convex subset of the plane, 460 Cooling, Newton's law of, Deritative, left-hand, 132 Leibniz ian notation right-hand, 132 second, 137 55 Derivative of quotient, incantation for, Coordinate system, 55 147 Diagonal, 202 55 of, Difference operator, 479 "Corner," 58 DifTerentiable, 127, 457 Cos, 256, 259, 273-274, 276, 470 addition formula derivative inverse Deriva- "negative infinity," 134 298 second, 55 origin for, see notation for tive, classical Coordinate first, 579 of, of, see for, 266 Differential equation, 247, 253, 273, 148, 260 275, 301. 359 Arccos initial conditions for, 360 Differentiation, 144 Taylor polynomials for, 334 remainder term for, 348 ff. implicit, 211 Cosh, 296 Differentiation operator, 479 Cosine, hyperbolic, 296 Directed angle, 256 Cot, 263 Discontinuities of a nondecreasing function, 370 Countable, 369 Counting numbers, 21 Discontinuity, removable, 99 Critical point, 165 Disraeli, Critical value, 165 Distance, 56, 441 Csc, 263 Benjamin, 2 between two shortest Cubic equation, general solution, 435 points, 276 Distributive law, 9 Diverge, 373, 463 Division, 6 Darboux's Theorem, 187, 242, 253 Decimal expansion, 70, 407 Division by zero, 6 Dedekind, Richard, 36 Domain, 38, 39, 45, 507 Double intersection, 141 Double root, 160 Defined implicitly, 211 Durege, 36 Decreasing function, 170 Definite integral, 304 DEFINITION, 45 Definition, recursive, 33 e9 Degree (of a polynomial), 40 Degree measurements, 61, 257-258 Delicate ratio test, 408 Delicate root test, 408 De Moivre's Theorem, 443 Dense, 118 Derivative, 125 ff., 127 classical notation for, 130-131, 138, 142-143, 162, 212 a, irrationality of, 353 relation with 368, 471 127 higher-order, 137 "infinite," 134 7r, transcendentality value of, of, 364 288, 350 Elementary function, 302 Ellipse, 64 Empty collection, 23 Entire function, 474 Epsilon, 19 Equal up of/, 127 of/ at 287 to order n, Equality, order of, 340 340 Equations, differential, equations see Differential Equivalent Cauchy sequences, 505 Euler, 481 number, 382 474 Eventually inside, 462 fixed point Exhaustion, method Exp, 287 ff., 470 of, exponential, 287-288 119 graphs 93 of, 384 of, 54-63, 173, 449 hyperbolic, 296 of, Taylor polynomials for, 335 remainder term for, 349-350 Expansion, decimal, 70, 407 increasing, 170 Extension of a function, 93 integrable, 219 identity, 41 imaginary part, 448 integral 219 of, inverse, 201 Factorial, 23 ff. linear, 56 Factorials, table of, 356 local Factorization into primes, 31 Fibonacci, 31 local local Fibonacci Association, 32 logarithm, 285, 289 Fibonacci Quarterly, The, 32 Fibonacci sequence, 31, 430, 478 Field, 487 characteristic of, 492 complete ordered, 490, 509 ordered, 489 First coordinate, 55 First 240 maximum maximum point of, value of, 1 minimum value for Inte- grals, 237 Fixed point of a function, 384 Fourier series, 270, 272 Function, 37, 45 of, 163 63 163 most general definition negative part of, of, nonnegative, 49 43 for, odd, 49, 174 periodic, 69, 140, 252 polynomial, 40 positive part of, 49 power, 58 507 product absolute value, 448 argument, 449, 453 complex- valued, 448 composition of, 42 of, quotient of, rational, 40 real part, 41 41 448 real-valued, 448 concave, 192 conjugate, 448 "reasonable," 96, 125, 156 constant, 41 regulated, 431 continuous, 93 strict critical point of, value 235 step, ff. convex, 191 critical 507 49 nondecreasing, 213 notation Mean Value Theorem to B, maximum point of, 164 minimum point of, 164 strict maximum point of, nonincreasing, 213 Fundamental Theorem of Calcu- from A ff. even, 49, 174 extension lus, 125 elementary, 203 entire, Even function, 49, 174 Even number, 25 First of, differentiate, 127 Euler-Maclaurin Summation Formula, 482 Euler's Function, derivative of, decreasing, 170 1 65 165 sum maximum of, point 41 trigonometric, 256 value of at x, 38 ff. of, 190 190 Index Fundamental Theorem of Algebra, 317, Indefinite integrals, short table 445, 455, 474 of, 304- 305 Fundamental Theorem of Calculus First, 581 Induction, mathematical, 21 240 complete, 23 Second, 244 Inductive set of real numbers, 33 Inequalities, 9 in Galileo, 124, 140 Gamma function, 327, 364 Gauchy-Schwartz, 239 geometric-arithmetic mean, 32 Jensen's, 199 "x goes to sin(x 2 )," 43 Graph sketching, 171-176 to, Liouville's, 369 Schwartz, 18, 239 Graphs, 54-63, 72-73, 173, 449 Greatest lower bound, 112 Grow 490 field, Bernoulli's, 31 Geometric mean, 32 Geometric series, 390 Global properties, 101 Goes an ordered Inequality triangle, 68 Infimum, 112 faster than, 292, 301 "Infinite" derivative, 134 Infinite intervals, 55 Hadamard, 476 Half-life (of radioactive substance), 298 Infinite products, 282, 328 Infinite sequence, 372, 462 Infinite series, 389 Hermite, 363 Infinite High-school student's real numbers, 505 Higher-order derivatives, 137 Infinitely "Infinitely small," 131, 225 Hilbert, 363 Infinity, 55 Horizontal axis, 55 sum, 354, 388 many primes, 31 minus, 55 Hyperbola, 65 Inflection point, 197 Hyperbolic functions, 296 Initial conditions in differential equa- 360 Instantaneous speed, 128 tions, Identity Instantaneous velocity, 128 additive, 9 Integer, 25 multiplicative, 9 Identity function, 41 Identity operator, 480 Imaginary axis, 441 Imaginary part of complex number, 438 Imaginary part function, 448 Integrable, 219 Integral, 219 classical notation for, 225, definite, First Mean Value Theorem Improper integral, 254, 329 Incantation for derivative of quotient, 237 indefinite, 304 short table of, 304-305 Leibnizian notation for, see Integral, classical notation for lower, 249 147 Increasing at for, improper, 254-255, 329 Implicit differentiation, 211 Implicitly denned, 211 226 304 a, 189 Increasing function, 170 Second Mean Value Theorem for, 237 Mean Value Theorem for, Increasing /sequence, 377 Third Indefinite integral, 304 upper, 249 237 Integral form of the remainder, 345, Leibnizian notation for derivatives, 131, 346 142-143, 162, 212 Integral sign, 219 Integral test, for higher-order derivatives, 138 396 Leibniz's formula, 159 Integration 219 by 305 parts, Theorem, 398 Leibniz's limits of, Lemma, ff. substitution, 307 82 Length, 275 of rational functions, 316 by Leibniz, 124, 131, 225 L'Hopital, Marquis de, 124 fT. Interest (finance), 298 L'Hopital's Rule, 179, 186-187 Intermediate Value Theorem, 102, 109, 113, 253 Interpolation, Lagrange, 47 Limit, 72 Intersection of sets, at infinity, 87 of integration, 219 closed, 55 of a sequence, 373 55 uniqueness open, 54 see also Inverse Lindemann, 367 additive, 9 Line, real, 54 multiplicative, 9 Inverse of a function, 201 Line, tangent, ff. Inverses of trigonometric functions, see Trigonometric functions Tangent see line Linear functions, 56 Liouville, 368 Liouville's inequality, 369 Irrational numbers, 25 fields, 80 of, Limit point, 386, 459 Limit superior, 121, 385 Nested Intervals Theorem Isomorphic 449 78, "does not exist," 81 41 Interval, 54 infinite, ff., from above, 86 from below, 86 Theorem, 474 Liouville' s 508 Local Isomorphism, 508 point of function, 164 maximum higher-order derivative second derivative Local Jensen's inequality, 199 Johnson, Samuel, 514 minimum see also test for, test for, point of function, 164, maximum point Local Local property, 89, 101, 142 Local strict maximum point, 190 Jump, 58 Log, 285 Klingenstein, K., trials and tribulations of, Taylor polynomials for, 335 remainder term for, 351 Logarithm to the base 10, Lagrange form of the remainder, 345, 347 Lagrange interpolation formula, 47 Large negative, 62 Least upper bound, 111 ff., 490 Least upper bound property, 113 Lebesgue, see Riemann-Lebesgue Lemma Left-hand derivative, 132 283 of a complex number, 477 Napierian, 299 Lower bound, 112 almost, 120 greatest, 112 Lower integral, 249 Lower limit of integration, 219 Lower sum, 215 Lowest terms, 71 339 176 Index Newton's law of cooling, 298 Maclaurin, 482 Mass, rate of change 128 of, Mathematical induction, 21 Maximum point of a function, 163 local, 164, see also Local maximum point local strict, 190 strict, Mean, of Newton's laws of motion, 137 Nondecreasing function, 213 Nondecreasing sequence, 377 Nondifferentiable complex functions, 458 Nonincreasing function, 213 Nonnegative function, 49 Nonnegative sequence, 391 Notational nonsense, 479 190 Maximum Maximum two numbers, 16 value of function, 163 Nowhere-difTerentiable continuous arithmetic, 32 function, 422 geometric, 32 Mean Value Theorem, nth root, 69, 443 168, 169 Cauchy, 178 existence for integrals (First, Second, and 1 complex, 433, 438 point of a function, local, counting, 21 Minus of Local minimum point two numbers, 16 infinity, Modulus even, 25 imaginary, 433 irrational, 25 55 natural, 25, 33 Mirifici logarithmonum canonis description, 460 algebraic, 362 of function, 163 64, see also Minimum 103, 443, Number of exhaustion, 119 Minimum Minimum of, primitive, 447 Third), 237 Method 583 odd, 25 299 of a complex number, 441 prime, 31 Mollerup, Johannes, 328 rational, 25 Multiplication, 5 real, 25, 441, 495 transcendental, 362 of arabic numerals, 8 associative law for, 9 Numbers, basic properties closure under, 9 Null set, of, 3 23 commutative law of, 9 of complex numbers, 438 geometric interpretation, 442-443 Multiplicative identity, existence of, Multiplicative inverses, existence of, 9 9 Odd function, 49, Odd number, 25 One-one 174 function, 200 Multiplicity (of a root), 108 Open Napier, 299 "Or," 6 Order of equality, 340 Ordered field, 489 complete, 490 Ordered pair, 45, 52 Napierian Logarithm, 299 Natural numbers, 21, 33 Negative, large, 62 "Negative infinity," derivative, interval, 54 Origin (of a coordinate system), 55 134 Negative number, 9 Negative numbers, product of two, 7 Negative part of a function, 49 Nested Intervals Theorem, 120 Newton, 131, 235 Pair, 44 ordered, 45, 52 Parabola, 58 area under, 224 Partial sums, 388 258 Archimedes's approximation Partition, 215 irrationality of, Partial fraction decomposition, 317 7T, relation to "Peak," 59 Peak point, 378 transcendental ity value of, of, 368 357 Period of a function, 69, 140, 252 Vieta's product for 2/x, 282 Periodic function, 69, 140, 252 Wallis's product for tt/2, 328 Perpendicularity of 119 All Pascal's triangle, 27 e, of, 279 68 lines, Petard, H., 459 Pig, yellow, v, Quaternions, 493 314 Quotient of functions, 41 Pigheaded, 161 Plane, 56 complex, 440 Rabbits, growth of population, 32 Point, 54 257-258 Point of contact, 192 Radian measure, Point-slope form of equation of a line, Radioactive decay, 298 Radius of convergence of complex 57, 68 power series, 466 Rate of change of mass, 128 Rate of change of position, 128 Polynomial function, 40 graph of, 58-59, 173 multiplicity of roots, 108, 160 specific ff., see also change Positive elements of field, 489 Positive part of a function, 49 Power Power Power Power of, number, 9 functions, 58 series, series complex, 464 centered at <z, uniqueness 2, table of, 356 Prime number, 31 characteristic of a field, 492 many of, 31 unique factorization into, 31 Primitive, 302 Primitive Principia, nth. root, 447 235 Product, 5 of functions, 41 of 316 Real axis, 441 Real line, 54 Real number (formal definition), 495 Real numbers, 25 algebraist's, 505 505 Archimedian property construction of, 494 ff. of, 116, 490 high-school student's, 505 430 infinitely of, Rational numbers, 25 analysist's, 423, 471 series representation, Powers of 408 integration 128 of, an ordered Positive element of R, 500 Positive 393 Rational functions, 40 functions Position, rate of test, delicate, Ratio Polynomials, Bernoulli, 481 Polynomials, Taylor, 334 61, two negative numbers, 7 Pythagorean theorem, 25, 56 inductive set of, 505 Real part of a complex number, 438 Real part function, 448 Real-valued function, 448 Rearrangement of a sequence, 400 "Reasonable" function, 66, 96, 125, 156 Rectangle, closed, 454 Recursive definition, 23 Reduction formulas, 316 Regulated function, 431 Remainder term for Taylor polynomials, 343 Removable discontinuity, 99 585 Index Riemann-Lebesgue lemma, 272, 327 Right-hand derivative, 132 Rising Sun Lemma, 121 Theorem, 168 Root of a polynomial function, 48 multiplicity test, power, 423, 471 Taylor, 424 108 of, see also nth. roots delicate, convergent, 389 geometric, 390 double, 160 Root conditionally convergent, 398 Fourier, 270, 272 Rolle, 161 Roile's Series (Cont.) 408 408 22 empty, 23 Set, Sets intersection of, 41 notation Shadow Same sign, 12 Sec, 263 inverse Sign, 12 Sin, 41, 256, 259, 273-274, 276, addition formula of, see line, Arcsec 126 Second coordinate, 55 Second derivative, 137 Second derivative test for maxima and minima, 176 Second Fundamental Theorem of Calculus, 244 Second Mean Value Theorem for integrals, 41-42 Sigma, 24 Schwartz, H. A., 18, 190, 239 Schwartz inequality, 18, 239 Secant for, point, 121 237 Sine, hyperbolic, 296 Sine function, 41 Sinh, 296 Sketching graphs, 171-176 Skew-field, 493 Slope of a straight Sequence 470 266 derivative of, 148, 260, 272-273 inverse of, see Arcsin Taylor polynomials for, 334 remainder term for, 348 for, line, 56 Speed, instantaneous, 128 absolutely summable, 397 Cauchy, 379, 505 complex, 477 Square 434 102-103 root, 12, existence of, pointwise, 415 Square root in a field, 492 Square root function, 453 Squaring the circle, 367 Step function, 235 Stirling's formula, 483 uniformly, 415 Straight line (analytic definition), 56 equivalence of, 505 of complex numbers, 462 convergent, 373 decreasing, 377 shortest distance divergent, 373 Fibonacci, 31, 430, 478 increasing, 377 infinite, between two points, 276 slope Strict 372 of, 56 maximum point, 190 Sturm Comparison Theorem, 275 limit of, 373 Subsequence, 378 nondecreasing, 377 Substitution, world's sneakiest, 325 nonincreasing, 377 Substitution formula, 308 rearrangement of, 400 summable, 389 Series absolutely convergent, 397 Subtraction, 5 Sum finite, 3-4 of functions, 41 Sum Two-times (Cont.) infinite, 354, of of an infinite sequence, 389 an infinite sequence of complex numbers, 462 lower, 215 388 partial, sigma notation upper, 215 Sum differentiable, 137 388 for, 24 limit, 415 Uniformly convergent sequence, 415 Uniformly convergent series, 419 Uniformly distributed sequence, 387 Uniformly summable, 419 Uniqueness Uniform of factorization into primes, 31 of squares, 459 of limits, 80 Summable, 389, 463 of power absolutely, 397 Unit circle, Cesaro, 408 Upper bound, Abel, 431 111, 490 490 least, 111, Supremum, 112 Symmetry 430 64 almost, 120 uniformly, 419 Swift, Jonathan, series representations, Upper integral, 249 Upper limit of integration, 219 Upper sum, 215 486 in graphs, 174 Tan, 263 "Valley," 59 Arctan Taylor series for, 497 Tangent, hyperbolic, 296 inverse Tangent of, see line, 125, point of contact "Tangent line," Tanh, 296 Value, absolute, 127 of, average, 128 134 instantaneous, 128 Vertical axis, 55 Taylor polynomial, 334 remainder term Taylor series, Absolute value Velocity 192 vertical, see Value of / at x, 38 Vanishing condition, 390 of, 343, 345, 346 Viete, Francois, 282 424, 469 Taylor's Theorem, 345 Third Mean Value Theorem for inte- grals, 237 Transcendental number, 362 Triangle inequality, 68 Trichotomy law, Cos, Cot, Csc, Sec, Sin, see Bolzano- Weierstrass Theorem Weierstrass M-test, 420 Well-ordering principle, 23 9 Trigonometric functions, 256 Wallis's product, 328 Weierstrass, see also Wright, 326 Tan of, 315-316 263 see also Arccos, Arcsec, Arcsin, Arctan integration inverses of, Zahl, 25 Zero, division by, 6 GLOSSARY OF SYMBOLS P (— .9 oo, w _« (-oo, V* (— 12 max (#,;>) 16 M min(*, ^) 16 {*} 21 0 23 23 55 55 oo \j u/aV 55 oo ) 5 L 70 '/ Km/(x) 81 oo «! lim/ 81 25 R 25,495 lim /(*) 0 27 lim /(*) /(*) 38, 45, 507 lim +* Pi a t — oo 42 an ] , 42 A ^4 121 121 127 127 df(x) 131 47 CA 48 dx 48 R-A 48 49 l/l max(/, £) 49 min (/, 49 5) f<g '» /-» 51 the pair (a, 6) 52 the open interval 55 («, [a> ) 55 oo) 55 00 214 215 215 & Pn,a />»,«,/ Rn,a for 219 /.'/(*) * 302 303 )f(x) dx 303 327 r(x) *(/,«,*) (a, 6) arccos // 201 £/(/,/>) 54 [a, b] 263 F(x) 121 130 a 6' 92 137 112 dx 47 263 arc tan df(x) n cot f» /' *< 92 137 /'(a) f°g°h lim 43 oo 112 4 ihn ^ f-g-h 42 /«| 42 x-*f(x) = 112 lim sup 43 263 arcsin 112 gib 41 . +£ + A n = sup .} . 263 esc /"' infil . 263 tan f" 41 . sec x=a 137 41 . 259, 261 264 265 t(f,P) 275 £(*) 276 log 285 exp 287 287 e e 288 288 a* 289 log„ sinh 296 cosh 296 tanh 296 arg sinh 296 arg tanh 296 arg cosh 296 Nap log 299 A lub4 -g 259, 261 sin 92 /(*) lim /(#) 41 {a h 87 > oo x—+a 41 cos 87 lim f(x) 41 f-g / a: 5 //* {*: a; 86 /(a:) lim 4 254 257 257 258 A(x) 259 o J. Q, / f sin r X 41 254 254 7T 25 / / J 0 sin 86 lim/(;c) Z c 86 /(a:) 24 a< 254 oo J a lim 7 249 f(x) dx 78 ("delta") 5 19 249 70 ' e ("epsilon") N fl ) a] 334 334 343 357 (:) 225 {a n 372 Glossary of Symbols lim a n lim a n 373 = oo Re Im 385 17m sup xn n-» ^ lim x„ * ^ 385 a, b) Um/(z) y Ll an 389 i Printed by Interprint (Malta) Ltd 433,439 + . 449 487,498 1 488, 502 _a 458 470 a-1 cos 470 p 470 > < > < 386 *» D Z>* 478 479 479 479 487, 497 4g7) 501 0 sin bn n = -l 482 448 449 exp N(n; 481 f>n 448 /'(«) 385 479 479 A *-*« 385 lim inf x„ eD 441 \z\ 382 lim x n 438 441 376 n—> oo 7 C |«| 488> 498 488, 502 489 490, 496 490,496 490 496 490, 496 > 501 CDEF798765 SELECTED "WORLD STUDENT SERIES' EDITIONS OF OUTSTANDING UNIVERSITY TEXTBOOKS FOR STUDENTS OF MATHEMATICS •CALCULUS 586 pp, ilius 527pp. 61 ilius By Michael Spivak, 8randeis University •PROS ABILITY WITH STATISTICAL APPLICATIONS, Second By Frederic Mosteller, Harvard Edition University, Rourke. 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