CALCULUS Michael Spivak CALCULUS Fourth Edition Copyright All © 1967, 1980, 1994, 2008 by Michael Spwak rights reserved Library of Congress Catalog Card Number 80-82517 Publish or Perish, Inc. I'M B 377, l302Waugh Drive, louston. Texas 77019 www raathpop com I . Manufactured . in the ( nited States ofAmerica ISBN 978-0-914098-91-1 Dedicated to the Memory of Y. P. PREFACE tan a debi to his profession, from the which as so ought they men of course of duty to endeavour way of amends, and themselves by to be a help ornament thereunto. FRANCIS BACON PREFACE TO THE FIRST EDITION Every aspect of book was influenced by this merely as a prelude to but as the real first the desire to present calculus not encounter with mathematics. the foundations of analysis provided the arena in which Since modern modes of math- ematical thinking developed, calculus ought to be the place in which to expect, rather than avoid, the strengthening of insight with logic. In addition to devel- oping the students' intuition about the beautiful concepts of them equally important to persuade to intuition, nor ends and but the natural in themselves, and think about mathematical that precision analysis, it is surely rigor are neither deterrents medium in which to formulate questions. This goal implies a view of mathematics which, in a sense, the entire book No attempts to defend. how matter well particular topics may be developed, the goals of this book will be realized only if it succeeds as a whole. For this reason, it would be of little value merely to list the topics covered, or to mention pedagogical practices and other innovations. Even the cursory glance customarily bestowed on new calculus texts will probably tell more than any such extended advertisement, and teachers with strong feelings about particular aspects of calculus will know just where to look to see if this book fulfills their requirements. A few features do require explicit ters in the book, two (starred) comment, however. Of the twenty-nine chapchapters are optional, and the three chapters com- V have been included only for the benefit of those students who might examine on their own a construction of the real numbers. Moreover, the appendices to Chapters 3 and 1 1 also contain optional material. prising Part want to The order of the remaining purpose of the book is chapters collection of "topics. " Since the until Part III, less it is intentionally quite inflexible, since the to present calculus as the evolution of one idea, not as a most exciting concepts of calculus do not appear I and II will probably require should be pointed out that Parts time than their length suggests — although the entire book covers a one-year any uniform rate. A rather and III, so it is possible to reach differentiation and integration even more quickly by treating Part II very briefly, perhaps returning later for a more detailed treatment. This arrangement corresponds to the traditional organization 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 course, the chapters are not meant to be covered at natural dividing point does occur between Parts II 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 diffi- culty and, I hope, of comparable interest. There are about 625 problems in all. Those which emphasize manipulations usually contain many examples, numbered viii Preface with small Roman in other problems. many so guide Some and double starring and numerals, while small starring, but there are so hints have been provided, not completely reliable. is letters are used to label interrelated parts indication of relative difficulty Many many is provided by a system of criteria for judging difficulty, especially for harder problems, that this problems are so hints are not consulted, that the best of students will difficult, especially if those which especially interest them; from the less difficult problems should be it The easy to select a portion which will keep a good class busy, but not frustrated. answer section contains solutions to the probably have to attempt only about half the examples from an assortment of problems that should provided a good test of technical competence. A separate answer book contains the solutions of the other parts of these problems, and of all the other problems as well. Finally, there problems often I am grateful and a refer, a Suggested Reading is list, to which the glossary of symbols. mention the many people for the opportunity to to whom I owe my thanks. Jane Bjorkgren performed prodigious feats of typing that compensated for my fitful production of the which provides manuscript. Richard Serkey helped collect the material historical sidelights in the problems, and Richard Weiss supplied the answers appearing in the back of the book. friends Michael I am especially grateful to my Freeman, Jay Goldman, Anthony Phillips, and Robert Wells for and the relentlessness with which they criticized, a the care with which they read, preliminary version of the book. Needles to deficiencies which remain, especially since would have made the book appear my admiration express I say, they are not responsible for the sometimes rejected suggestions which suitable for a larger for the editors and staff of group of students. W A. Benjamin, Inc., I must who were always eager to increase the appeal of the book, while recognizing the audience for which The it was intended. inadequacies which preliminary editions always involve were gallantly en- dured by a rugged group of freshmen in the honors mathematics course University during the academic year 1965-1966. About at Brandeis 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 success. This after all, the class is students themselves, unlikely to rise it seems to up in a body and me, deserve the is always safe- protest publicly —but the right to assign credit for the thor- oughness 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 SECOND EDITION I have often been told that the title of this book should really be something Introduction to Analysis," because the book is usually used in courses students have already learned the mechanical aspects of calculus standard in Europe, and they are becoming more After thirteen years it seems too late to common change the title, like "An where the — such courses are in the United States. but other changes, in addition to the correction of numerous misprints and mistakes, seemed called for. There are now separate Appendices for many topics that were previously slighted: polar coordinates, uniform continuity, parameterized curves, Riemann sums, and the use of integrals for evaluating lengths, volumes and surface areas. A few topics, like manipulations with power series, have been discussed more thoroughly in the text, and there are also more problems on these topics, while other topics, like Newton's method and the trapezoid rule and Simpson's rule, have been developed in the problems. There are in all about 1 60 new problems, many of which are intermediate in difficulty between the few routine problems at the beginning of each chapter and the more difficult ones that occur later. Most of the new problems are the work of Ted Shifrin. Frederick Gordon pointed out several serious mistakes in the original problems, and supplied some non-trivial corrections, as well as the neat proof of Theorem 12-2, which took two Lemmas and two pages in the first edition. Joseph Lipman also told me of this proof, together with the similar trick for the proof of the last theorem in the Appendix to Chapter 1 1 which went unproved in the first edition. Roy O. Davies told me the trick for Problem 1 1-66, which previously was proved only in Problem 20-8 [21-8 in the third edition], and Marina Ratner suggested several interesting problems, especially ones on uniform continuity and infinite series. To all these people go my thanks, and the hope that in the process of fashioning the , new edition their contributions weren't too badly botched. MICHAEL SPIVAK PREFACE TO THE THIRD EDITION The most significant change in this third edition Chapter 17 on planetary motion, in is the inclusion of a which calculus employed is new (starred) for a substantial physics problem. In preparation for dinates are now the old this, three Appendices: the first Appendix to Chapter 4 has been replaced by two cover vectors and conic sections, while polar coor- deferred until the third Appendix, which also discusses the polar coordinate equations of the conic sections. Moreover, the Appendix to Chapter 12 has been extended to treat vector operations on vector-valued curves. Another large change is merely a rearrangement of old material: "The Cos- mopolitan Integral," previously a second Appendix Appendix ter 18, to the chapter now Chapter the material 19); on "Integration in to Chapter 13, is now an Elementary Terms" (previously Chap- moreover, those problems from that chapter which used from that Appendix now appear as problems in the newly placed Appendix. A few other changes and renumbering of Problems result from corrections, and elimination of incorrect problems. and somewhat dismayed when I realized that after allowfirst and second editions of the book, I have allowed another 14 years to elapse before this third edition. During this time I seem to have accumulated a not-so-short list of corrections, but no longer have the original communications, and therefore cannot properly thank the various individuals involved (who by now have probably lost interest anyway). I have had time to make only a few changes to the Suggested Reading, which after all these I ing was both 1 startled 3 years to elapse between the years probably requires a complete revision; this will have to wait until the next edition, which I hope to make in a more timely fashion. MICHAEL SPIVAK PREFACE TO THE FOURTH EDITION I noted that it was 13 years and second editions, and then another 14 years before the third, expressing the hope that the next edition would appear sooner. Well, here it is another 14 years later before the fourth, and presumably final, edition. Although small changes have been made to some material, especially in Chapters 5 and 20, this edition differs mainly in the introduction of additional problems, a complete update of the Suggested Reading, and the correction of numerous errors. These have been brought to my attention over the years by, among others, Nils von Barth; Philip Loewen; Fernando Mejias; Lance Miller, who provided a long list, particularly for the answer book; and Michael Maltenfort, who provided an amazingly extensive list of misprints, errors, and criticisms. Most of all, however, I am indebted to my friend Ted Shifrin, who has been using the book for the text in his renowned course at the University of Georgia for all these years, and who prodded and helped me to finally make this needed revision. I must also thank the students in his course this last academic year, who Promises, promises! In the preface to the third edition between the first served as guinea pigs for the new edition, resulting, in particular, in the current proof in Problem 8-20 for the Rising Sun Lemma, far simpler than Reisz's original proof, or even the proof in [38] of the Suggested Reading, which itself has now been updated considerably, again with great help from Ted. MICHAEL SPIVAK 9 5 CONTENTS PREFACE PART I PART II VI Prologue 1 Basic Properties of 2 Numbers Numbers of Various Sorts 21 Foundations 3 39 Functions 54 Appendix. Ordered Pairs 4 Graphs 56 75 Appendix 1. Vectors Appendix 2. The Conic Appendix 3. Polar Coordinates 5 Limits 90 6 Continuous Functions 7 Three Hard Theorems 8 Least 80 Sections Upper Bounds 1 84 1 122 133 Appendix. Uniform Continuity III 3 144 D< jrivatives and Integrals 9 10 11 Derivatives 149 Differentiation 168 Significance of the Derivative Appendix. Convexity and Concavity 12 Inverse Functions 188 21 230 Appendix. Parametric Representation of Curves 13 Integrals Appendix. Riemann Sums 14 244 253 282 The Fundamental Theorem of Calculus 285 1 XIV Contents 15 The Trigonometric *16 7r *17 Planetary Motion 18 19 is Irrational Functions 303 324 330 The Logarithm and Exponential Functions Integration in Elementary Terms 363 402 Appendix. The Cosmopolitan Integral PART IV Infinite 20 *21 Infinite Series Approximation by Polynomial Functions e is Transcendental 442 Infinite 23 Infinite Series 24 Uniform Convergence and Power Complex Numbers 526 26 27 Sequences Series Epilogue 28 Fields 29 Construction of the Real 30 Uniqueness of the Real Numbers 581 (to selected problems) Glossary of Symbols 669 Numbers 609 Suggested Reading Index 1 471 Complex Functions 541 Complex Power Series 555 Answers 4 452 22 25 PART V Sequences and 339 665 619 588 601 499 CALCULUS PROLOGUE To you to be conscious that are ignorant is a great step knowledge. BENJAMIN DISRAELI . CHAPTER OF NUMBERS BASIC PROPERTIES I The of title required to read what is chapter expresses in a few words the mathematical knowledge this book. In this meant by short chapter and chapter theless, this we of which all —are already —addition and and inequalities, Never- familiar to us. Despite the familiarity of the subject, the not a review. is simply an explanation of division, solutions of equations and other algebraic manipulations factoring is the "basic properties of numbers," multiplication, subtraction survey fact, this are about to undertake will probably seem quite novel; it does not aim an extended review of old material, but to condense this knowledge a few simple and obvious properties of numbers. Some may even seem too to present into number of diverse and important facts turn out to be consequences of the ones we shall emphasize. Of the twelve properties which we shall study in this chapter, the first nine are obvious to mention, but a surprising concerned with the fundamental operations of addition and multiplication. For the moment we of numbers may — consider only addition: sum a + b this operation performed on a pair is any two given numbers a and b (which possibly be the same number, of course). It might seem reasonable to regard the exists for addition as an operation which can be performed consider the more sum a\ + • • • + a„ of n numbers a\, on . . several , convenient, however, to consider addition of pairs of define other sums in terms of sums of this type. numbers at once, a„ as a basic concept. For the numbers sum of and It is and to numbers add b and c, only, three may be done in two different ways. One can first and then add a to this number, obtaining a + (b + c); or one can + first add a and b, and then add the sum a + b to c, obtaining (a + b) + c. Of course, the two compound sums obtained are equal, and this fact is the very first and a, b, c, this obtaining b property (PI) we c, shall list: If a, b, and any numbers, then c are a The statement of three a numbers + (b + c) = more (a this + (b + c) = (a + b) + c. property clearly renders a separate concept of the superfluous; + b) + c. we simply agree that a + b + Addition of four numbers requires involved, considerations. The symbol (1) or (2) or (3) or (4) or (5) a +b+c+d + c) + d, c)) + d, + + a + «b + c) + d), a + (b + (c + d)), (a+b) + (c + d). ((a+b) (a (b c denotes the similar, is sum of number though defined to slightly mean 4 Prologue This definition fact is need not be listed. unambiguous we know from PI For example, (a+b) + and it follows immediately that a direct consequence of P 1 is (one must = (3) It is (4) b let = +c feasible, + a (2) sums of an possible may be proof (and The are equal. may this five equality of (2) and (3) not be apparent at first sight The equalities c). this is it prove the equality will also suffice to numbers, but it may not be so clear how we so without actually listing these 14 sums. would be totally if we considered collections inadequate to prove the equality numbers a\, an This worry about the worth worrying about once) a reasonable approach is outlined in arbitrary finite collection of taken for granted, but for those it is + c), (b but would soon cease to be of six, seven, or more numbers; all this simple to prove. can reasonably arrange a proof that fact = c probably obvious that an appeal to PI is equal. Fortunately, play the role of b in PI, and d the role of (5) are also Such a procedure all from the property PI already that although , follows it and (1) of the 14 possible ways of summing of numbers are since these listed separately, since who would . . . , . like to Problem 24. Henceforth, we shall usually make a tacit appeal to the results of this problem and write sums a\ + a n with a blithe disregard for the arrangement \- of parentheses. The number (P2) If a has one property so important that is important role (P3) list it next: any number, then a An we is For every also played number +0 = + a — a. by in the third a, there a + is (— a) a = property of our number —a such +a = (—a) list: that 0. Property P2 ought to represent a distinguishing characteristic of the and if a it is comforting to note that number x we are already in a position to prove 0, Indeed, satisfies a for number this. any one number a, then x — +x = a (and consequently this equation also holds for all more than subtracting a numbers a). The 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 must be used to justify proof of this assertion involves nothing this operation. a If then hence hence hence +x = + (a + x) = ((—a) + a) + x = +x = x — (-a) a, (-a) 0; 0; 0. + a = 0; Basic Properties ofNumbers 1. As we have just hinted, it is convenient derived from addition: we consider a — b is to regard subtraction as an operation be an abbreviation for a to 5 + (— b). It then possible to find the solution of certain simple equations by a series of steps (each justified by PI, P2, or P3) similar to the ones just presented for the equation a +x = a. For example: If + jc then (jc+3) hence x + (3 + + hence = 3 = = = (-3)) + x hence 5, = 5 + (-3); (-3) x 5 - 3 = 2; 2; 2. Naturally, such elaborate solutions are of interest only until that they can always be supplied. In practice, an equation by indicating so solve you become convinced usually just a waste of time to it is explicitly the reliance on properties PI, P2, and P3 (or any of the further properties we shall list). Only one other property of addition remains to be listed. When considering the sums of three numbers a, b, and c, only two sums were mentioned: {a + b) + c and a + (b + c). Actually, several other arrangements are obtained if the order of a, b, and c is changed. That these sums are all equal depends on If a (P4) and b are any numbers, then a The +b— b + a. 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 statement of For example, subtraction does not have so well behaved. a is — b — a. b 7^ amusing we might ask just when how powerless we are if we In passing to discover — = b — b a only from properties P1-P4; fully it is and determine which be able — when a — this property: usually b does equal b rely only — a, and it on properties P1-P4 Algebra of the most elementary variety shows that to justify our manipulations. a a Nevertheless, b. instructive to step(s) to justify all steps in detail it is impossible to derive this fact examine the elementary algebra carejustified by P1-P4. We will indeed cannot be when a few more properties are listed. Oddly enough, however, the crucial property involves multiplication. The basic properties of multiplication are fortunately so similar to those for ad- dition that should be by a b, (P5) little clear. comment will be needed; both the meaning and the consequences (As in elementary algebra, the product of a or simply ab.) If a, b, and c are any numbers, then a (P6) If a is (b c) = (a b) a = a. any number, then a 1 = 1 • c. and b will be denoted 6 Prologue 1^0. Moreover, (The assertion that because there list it, number, namely, For every (P7) no way is other properties listed may seem ^ 1 — If a / number a 0, there in P7. number we have to hold all if there were only one number a a~ = a • a -1 such that 1. and b are any numbers, then This condition _1 would is = a~ is = b which deserves emphasis detail but 0.) a One list, could possibly be proved on the basis of the it these properties a (P8) a strange fact to b the appearance of the condition is quite necessary; since -1 = a. Ob — for all numbers a^O b, there is no This restriction has an important consequence was defined in terms of addition, so division is Since is defined in terms of multiplication: The symbol o/b means a b meaningless, a/0 is also meaningless division by is always undefined. Property P7 has two important consequences. If a b = a c, it does not necessarily follow that b = c; for if a = 0, then both a b and a c are 0, no matter what b and c are. However, if a ^ 0, then b — c\ this can be deduced from P7 as satisfying • 1. for division. Just as subtraction . — • • follows: a If then a' hence (a~ a) 1 • hence It is also a consequence of P7 that b b = 1 b = c. b) (a hence — — = b • a if 1 then a" hence (a~ b) (a • • hence b a) 1 • b • b hence b (It may happen that a = and b = when we say "either a — or b = and a c a" • (a (a~ = = — — = ^ 0, c); a) c\ c; = 0, a b if a then either a and = or b 0. In fact, a^0, 0; 0; 0; 0. are both true; this possibility 0"; in = mathematics "or" is is not excluded 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 (x Then it follows thai either x — 1 - = 1)(jc is known -2) = or x — 2 — to satisfy 0. 0; hence v = 1 or x = 2. Basic Properties ofNumbers /. On the basis of the eight properties listed so far Listing the next property, little. it is still 7 possible to prove very which combines the operations of addition and multiplication, will alter this situation drastically. (P9) If a, b, and c are any numbers, then a (Notice that the equation (b As an example of the b + c) = a (b • + c) usefulness of = a P9 we b • b a will +a +c c. • a is by also true, P8.) now determine just when — a b — — a: If then (a hence hence Consequently a a — b = - b) + b = a = a +a = b = — b + (1 • 1) and therefore A second use of P9 is a — a, {b - a) + b = b + b + b — a: (b + b — a) + a — • (1 + (b b - + a); b. 1), b. = the justification of the assertion a which we have already made, and even used in a proof on page 6 (can you find where?). This was not fact one of the basic properties, even though no proof was offered listed as when it was first mentioned. With P1-P8 alone a proof was not possible, number appears only in P2 and P3, which concern addition, while the in question involves multiplication. not obvious: We With P9 the proof we have sides) that A series already noted, a = this this, more help explain the somewhat myste- two negative numbers is To begin positive. (—a) b = —(a with, To b). note that b +a follows immediately (by adding Now may easily acceptable assertion that (-a) It immediately implies (by adding —(a -0) to both of further consequences of P9 will establish the prove a-0; 0. rious rule that the product of we assertion simple, though perhaps have = as is since the • b —(a = = = [(-a) + a] • b 0b 0. b) to both sides) that (— a) b • note that (-a) (-b) + [-{a b)\ = (-a) (-b) + (-a) = (-a)-[(-b) + b] = (-a)-0 = 0. b = — (a • b). 8 Prologue Consequently, adding (a we b) to both sides, =a (-a)- (-b) The fact that the obtain b. product of two negative numbers of P1-P9. In other words, if we want PI P9 to positive is is thus a consequence be true, the rule for the product of two to negative numbers is forced upon us. The various consequences of P9 examined so portant, do not although interesting and im- far, really indicate the significance of P9; after each of these properties separately. Actually, P9 is all, we could have we have shown how algebraic manipulations. For example, although listed the justification for almost all to solve the equation (xwe can 0, hardly expect to be presented with an equation in likely to this form. We are more be confronted with the equation x The -2) = l)(x "factorization" (a- - 1) • x (x — - 3.x: 2) 2 - +2= =x — (jc (.v • 3.x - + = 2 — l)(.v 2) + 0. 2) (-1) is really a triple use of P9: - (x 2) = jc.jc+x-(-2) + (-1).jc + = 2 +a-[(-2) + (-1)] + 2 = x 2 - 3x + 2. (-1).(-2) a- 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 = = 13 (Note that moving 26 to the 26 • 10.) The left in multiplication 13 • 4 13-4 • (2 2- 26- 10 10 + 4) 10+ 13-4 + 52. the above calculation = 52 uses P9 also: = 10 + 3) -4 = 10-4 + 3-4 = 4- 10+ 12 = 4- 10+ 10 + = (4+ 1)- 10 + 2 = 5- 10 + 2 = 52. • l . I • 2 is the same as writing Basic Properties of Numbers 1. The ber, list properties P1-P9 have descriptive names which We but which are often convenient for reference. P1-P9 properties together and indicate the are not essential to 9 remem- opportunity to will take this names by which they are commonly designated. law for addition) (PI) (Associative (P2) (Existence of an additive + + a a (b + c) — (a + b) + c. = + a — a. identity) (P3) (Existence of additive inverses) a (P4) (Commutative law a (P5) (Associative for addition) law for multiplica- + (—a) — (—a) + a — 0. + b = b + a. (b c) = (a b) c. a tion) a (Existence of a multiplicative (P6) • 1 = a = a; — a~ a b a. 1 • 1/0. identity) (Existence of multiplicative (P7) a a~ a b x = 1 , for a ^ 0. inverses) (Commutative law (P8) for multi- — plication) The a (Distributive law) (P9) • (b + c) = a b + a c. 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. (a is less than b) and a > b (a is The two greater than notions of inequality, a are intimately related: a b), < < b b means the same as b > a (thus 1 < 3 and 3 > 1 are merely two ways of writing the same assertion). The numbers a satisfying a > are called positive, while those numbers a satisfying a < are called negative. While positivity can thus be defined in terms of <, it is possible to reverse the procedure: a < b can be defined to mean collection of all all — that b a is positive. In fact, numbers, denoted by P, positive it is convenient to consider the as the basic concept, and properties in terms of P: (P10) (Trichotomy law) For every number a, one and only one of the following holds: (PI 1) (PI 2) (i) a = (ii) a is (iii) —a 0, in the collection P, is in the collection P. (Closure under addition) If a and b are in P, then a + b is (Closure under multiplication) If a and b are in P, then a in P. in P. b is state 10 Prologue These three properties should be complemented with the following > a < a > a < a > Note, in particular, that a b if a —b in P; is b if b > a; b if a > b or a b if a < b or a = b* if and only a if definitions: is = b; in P. 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, Using the A < a a-b = 0, (ii) a (iii) —(a definitions just more slightly b b, so that then a (i) +c < — b made, a if in the collection P, Similarly, a in the collection P. is (i) a = (ii) a (iii) b > > b, b, a. from the following manipulations. + c) — and c so c —a= b is in P, is in P, < — b) + (c c, then a (b < c. — a) < The ab > following assertion The 0. The symbol Similarly ab > c almost built c. is in P\ thus, if is If < a b, Then in P. (The two inequalities a < b and < b < c, obvious: If a < b < c are usually written in the abbreviated form a inequality a + c) {a suppose a < b and b < —a —b < b and one of the following holds: follows that precisely it then surely (b in P, is b This shows that is — b) = b — a interesting fact results + c. b — which has the third in.) somewhat is less and b < then 0, only difficulty presented by the proof is the unraveling of definitions. < a —b > a, which means — a = —a is in P. definition, and consequently, by P12, (— a){— b) — ab is in P. Thus means, by in P, is 0. The ab > b > in our has one and also if a < 0, Z? < > if a / 0. Thus squares of nonzero numbers are special consequence: a always positive, and in particular we have proved a result which might have seemed fact that if a > 0, 2 sufficiently * There is elementary to be included one slightly 1 are both true, even though ond. This of the symbol and b, soi is i <>f ihin« <. ; a statemenl like Theorem while inequality holds for other values. 1 > (since 1 = I ). The statements + 1 <3 + 1 <2 we know that < could be replaced l>\ is used with hound to occur when is revealed by of properties: > and perplexing feature of the symbols 1 e< list I in the specific first, and l>\ = in the numbers; the usefulness here equalitj holds for some values of a Basic Properties of Numbers 1. The fact that —a > extremely important role value a < if Note that = number we a, define the absolute of a as follows: \a\ — \a\ 3, |7| the basis of a concept which will play an is book. For any in this 11 \a\ is 7, \\ a, a > -a, a < 0. when a = 0. always positive, except + y/2-V3\ = 1+V2-V3, and \\ we have — 3| = = v/T5-v/ 2-l. For example, + V2 - >/l0| | In general, the most straightforward approach to any problem involving absolute values requires treating several cases separately, since absolute values are defined by cases to important THEOREM 1 For all begin with. This approach fact We numbers a and will be used to prove the following very about absolute values. b, we have \a PROOF may + b\ < \a\ + \b\. consider 4 cases: (3) a>0, a<0, (4) a (2) In case (1) we have a also +b > <0, and the theorem 0, + b\ = \a b>0 b<0 b>0 b<0 >0, a (1) +b= a \a \ + is obvious; in fact, \b\, so that in this case equality holds. In case (4) we have \a In case (2), when a +b< +b\ a > = 0, and again equality -(a +b) and b < \a may This case = -a + (-b) holds: = we must prove 0, + b\ < a — \a\ + \b\. that b. therefore be divided into two subcases. If a +b > 0, then we must prove that a + b < a — b, b i.e., which a certainly true since b is +b < 0, we must prove < and hence —b > a i.e., is On the other hand, if < < a - b, a. > and hence —a < 0. may be disposed of with no certainly true since a Finally, ing case 0. that b which < -b, note that case (2) with a and (3) /; interchanged. | additional work, by apply- 1 2 Prologue Although ious cases) this is method of treating absolute sometimes the only approach may be ods which Theorem 1 In used. fact, possible to give a is it Ifll = Vfl 2 defined only is much shorter proof of . (Here, and throughout the book, sfx denotes the symbol meth- proof is motivated by the observation that this ; values (separate consideration of var- available, there are often simpler (\a when x > We may now 0.) + b\) 2 = (a+ b) 2 = positive square root of x; this observe that 2 + lab + b 2 < a 2 + 2\a\- \b\ + b 2 = \a\ 2 + 2\a\ \b\ + \b\ 2 = (\a + \b\) 2 a \ From we can conclude this that +b\ < \a + \a\ \b\ One close observation final may be made about a and b have the same sign the two + b\ = 0, is (i.e., a and b are of opposite We will inclusion conclude is < implies x y, to the reader we have just proved: a + \a\ 1^1 are both positive or both negative), or if one of this +b\ < + \a\ \b\ signs. chapter with a subtle point, neglected until now, whose required in a conscientious survey of the properties of numbers. After stating property P9, by v is left while \a if the theorem fact examination of either proof offered shows that \a if < because x provided that x and y are both nonnegative; a proof of this (Problem 5). we proved that a —b— b —a implies a = b. The proof began establishing that fl.(l from which we concluded that a + = + +l) = = b. fc.(l + l) This result > obtained from the equation is by dividing both sides by + 1. Division by should it must therefore be admitted that the validity of the argument depends on knowing that 1 + / 0. Problem 25 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 + 1^0. a - (1 1) b • (1 1) 1 be avoided scrupulously, and 1 1 This last demonstration has perhaps only strengthened your feeling that absurd to bother proving such obvious facts, present situation will help justify serious consideration of such details. chapter we have assumed merely explicit statements be difficult, however, to that numbers are familiar objects, deal of this book is and that PI -PI 2 are It how to justify this devoted assumption. Although one learns to elucidating the is In this of obvious, well-known properties of numbers. with" numbers in school, just what numbers it but an honest assessment of our are, remains rather vague. would "work A great concept of numbers, and by the end 1 we of the book . have become quite well acquainted with them. But will necessary to work with numbers throughout the book. It is way numbers say that, in whatever properties PI Most of it will be therefore reasonable admit frankly that we do not yet thoroughly understand numbers; we to 13 Basic Properties of Numbers may still are finally defined, they should certainly have PI 2. chapter has been an attempt to present convincing evidence that this we should assume of numbers. Some of the problems order to deduce PI PI 2 are indeed basic properties which in other familiar properties (which indicate the 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 become quite clear, 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 derivation of other facts about subtle, quite unlike PI -P12. The discovery of this crucial property work of Part II of this book. In the remainder of Part I we why some additional property is required; in order to investigate to consider a little more carefully what we mean by "numbers." the will require all will begin to see this we will have PROBLEMS 1. Prove the following: If (i) (ii) x 2 - 2 = y 3 y (v) x" (vi) x3 + x 3 do x" What is y 2 — - (iv) 2. = x If (iii) ax y" this, + y" a for some number = - (x «^0, + y). y)(x = , then x y or x 2 ). l l ). using (iv), and whenever n wrong with is it will show you how the following "proof"? Let x (x 2 2 -y + y)(x - y) x + v 2y 2 Prove the following: £ = b a (U) i + ~ be c l= if*,c#0. ad + be . . -M--* b -**°- = xy, = xy - y 2 = y(x — y), = y, = y, , = 1. easy way to to find a factorization for odd.) x (i) 1. = —y. = (x — y)(x + xy + y 2 ~ = (x - y)(x n ~ + x n ~ 2 y + + xy"~ 2 + y n ). = (x + y)(x — xy + y 2 (There is a particularly y 2 x 3. = then x = y. Then 1 14 Prologue = 1 (ab) (iii) l a [ b if , / a,b rrs- fM " a (iv) (To do 0. this you must remember the l defining property of (ab)~ .) ac c ft £-g,if». c .^o. 5/ 4. Find a b b a all numbers x for a — c =— b d if (ix) (;c-7z-)(jf+ 5) (jc (x) (jc - (xi) 2* < (xii) x (xiii) -+ (iii) (iv) (v) (vi) (vii) if ad = &c. Also determine when is a product of two numbers positive?) jc 1 + -3) > - V2) > l/2)(x 0. 0. 8. 3' x (xiv) and only which (viii) (ii) 1 < — 4. > 0. x i^l > o. x + Prove the following: (i) If (ii) If (iii) If (iv) If (v) If (vi) If (vii) If (viii) (ix) (x) 6. then 0, 4-jc<3-2jc. 5 - x 2 < 8. 5 - x 2 < -2. (x — l)(x — 3) > 0. (When 2 x - 2x + 2 > 0. 2 x + x + 1 > 2. 2 -x + 10> 16. 2 > 0. x +x + (i) 5. ^ b,d \{ (vi) (a) If If < a < a < fl < a < a > < < < a If a, /? fr, and c < d, then a £ and c b and c > < then a b — < if x then /?c. a. a 2 < a 2 < < b < 2 , if Prove that if x" . (Use then a is y" and n y" and ac < bd. d, then 2 y, then x" « d. a. < c < & Prove that if jc" < ac > > (c) = = < c 0, (b) Prove that — /?c. x < y and (d) + d. then ac 0, < 1, then < Z? and fl a < b, then b > and a 2 a Prove that b —b<—a. then b and c > d, then a 1, +c < /? (viii).) < < b. (Use y", n = < odd, then x" (ix), backwards.) 1, 2, 3 y". = y. x = y is odd, then x is even, then or .v = — v. . 1. 7. Prove that < a < if b, Basic Properties of Numbers 15 then — —+—b < a > a < V ab < - b. 2 Notice that the inequality -Jab alization of this fact occurs in *8. < (a + b)/2 holds for a, b all P of positive all 0. A gener- Problem 2-22. Although the basic properties of inequalities were stated lection > numbers, and < was defined in in terms of the col- terms of P, this procedure can be reversed. Suppose that P10-P12 are replaced by For any numbers a and b one, and only one, of the (P'10) following holds: (P'l (i) a = (ii) a (iii) b < < b, b, a. For any numbers a, 1) < a a +c< b c, if a < b and b b, and c, if a < b, b, and c, if a < b and < c, then < c, then then + c. For any numbers a, (P'l 3) and c. For any numbers a, (P'12) b, ac < be. Show 9. that P10-P12 can then be deduced Express each of the following with as theorems. at least one less pair of absolute value signs. (ii) \V2 + y/3-V5 + y/7\. \(\a+b\-\a\-\b\)\. (iii) \(\a+b\ (i) + \c\ - \a + b + c\)\. \x -2xy + y 2 |(|V2 + V3|-|a/5-a/7|)|. 2 (iv) (v) 10. \. Express each of the following without absolute value cases separately 1 1 when necessary. + b\-\b\. (i) \a (ii) |(|*|-D|. (iii) (iv) l*|-|* fl -|(fl-| fl |)|. Find all 2 |- numbers x - 31 = (i) |x (ii) \x (iii) \x (iv) |jc- (v) |jr -3| < +4\ < - 1| 1 1 + + for which 8. 8. 2. |jc \x -2| > + < 1 1 1. 2. signs, treating various . 1 6 Prologue 12. (vi) |jc-1| (vii) \x - 1 (viii) \x — II + 1 + 1|<1. + 1 = 0. +21 = 3. |jc • \x • \x 1 Prove the following: \xy\ (i) = \x = — x if , (The best way 7^ 0. do to this is to remember what \x\ is. in if y ^ \x\ + \y\. , — y\ < \x (iv) v) ( \ x — \ — \y\ \ x 0. (Give a very short proof.) (A very short proof is possible, ~~ y\- if you write things in the right way.) (vi) |(|.v| — \y\)\ + y + \x (vii) z\ < < — \x \x\ (Why does y\. + + \y\ follow immediately from this when Indicate \z\. equality holds, (v)?) and prove your statement. 13. The maximum of two numbers x and y is denoted by max(x,y). 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(x, y). Prove that max(x, y) = min(jc, y) = x + v + - I v — v — x\ x I L Derive a formula for max(x, y, max(x, 14. (a) Prove that First cases. a ^15. < \a\ = \—a\. y, z) (The —b < a that — |a| < (c) Use this fact to give Prove that if a < Use Problem Show (x , trick is not to > example z)). become confused by too many 0. Why is it then obvious for < b a new proof that if and only if \a\ < b. In particular, |a|. 4 + x3y 0, \a + b\ < |«| + \b\. then 2 + xy + y 2 > 0. + x 2 y 2 + .vy 3 + y 4 > 0. 1 that + y) = x n + y o + y)~ = x + y t (x max(x max(y, x and y are not both x (a) y,z), using, for 0?) Prove that x 46. = prove the statement for a (b) Hint: and min(x, z) when x = only when .v — only or y or y = = 0, or x = —v. it follows 1. (b) Using the fact that 2 x show (c) (d) that Use part 4x 2 0, + 6xy + 4v 2 > when find out (b) to when (x You should now be the proof is + y) 2 > (x unless x (x + 4 y) 0, and y are both 4 =x +y 0. 4 + y) = 5 a 5 +v make able to 5 . . From Hint: the assumption a good guess as to when (x (x+y) + y) n — + y"; x" contained in Problem 11-63. (b) Find the (c) Find the — 3x + 4. Hint: "Complete - 3x + 4 = 2(x - 3/4) 2 + ? i.e., write 2x 2 smallest possible value of x — 3x + 2y + 4y + 2. smallest possible value of x + Axy + 5y — Ax — 6y + 7. (a) Suppose that b~ (a) Find the smallest possible value of 2x 2 the 2 square," 18. + 2xy + y 2 = = +y you should be able to derive the equation x +2x y+2xy +y — 3 = x 2 y + xy 2 — xyix + y). if xy / 0. This implies that (x + y) Find out x 17. 17 Basic Properties of Numbers — -b + Ac > y/b 2 0. Show -b - - Ac numbers that the yjb 2 - Ac ' 2 both (b) 2 + bx + c = equation x satisfy the 2 Suppose that b x 2 + bx + c = 0; — Ac < Show 0. in fact, x~ 0. no numbers x that there are + bx + c. > for all satisfying Complete the x. Hint: square. (c) Use x (d) 2 another proof that this fact to give + xy + y > if x and y are not both 0, then 0. For which numbers a is it true that x~ + axy + y > whenever x and y are not both 0? (e) Find the smallest possible value of x" The fact that a2 nevertheless the ties and of ax + foe + c, for >0. a 19. + bx + c > numbers a, elementary as it may seem, is fundamental idea upon which most important inequalifor all The great-granddaddy of are ultimately based. Schwarz all inequalities is the inequality: x\y\+x2y2 < vx\ (A more general form occurs in + xi vyi +yi Problem 2-2 1 .) • The Schwarz inequality outlined below have only one thing on the fact that a 2 > for all a. three proofs of the in common — their reliance (a) Prove that if x\ = y2 = 0. and X2 — Ay? for some number X > Schwarz inequality. Prove the same thing Xy\ equality holds in the Now suppose that y\ and y2 are not both 0, and 0, then if \| = is no that there 18 Prologue number X such < (Xyi = ^ -x\) 2 2 2 = that x\ (>'i + Using Problem 18, v2 + (Xy 2 2 - ) and x 2 Xy\ -x 2 — Xy 2 Then . 2 ) + x2 y2 + 2X(x iyi ). complete the proof of the Schwarz inequality. < x~+y (how is Prove the Schwarz inequality by using 2xy (b) + x22 (xr ) this derived?) with x Xi = v . , - x\ 2 for first = i 2 and then 1 + x 2 2 )(yi 2 + (vi for = i 2 V2 = ) t 2 V vi -f- v2 2 2. Schwarz inequality by Prove the (c) + x2 y = 2 proving that first +x 2 y 2 2 + (x\y\ (x\y 2 ) - x 2 y\) 2 . Deduce, from each of these three proofs, that equality holds only when (d) yi x2 = y2 = = Xy 2 when or there a is number > A. such that x\ — Xy\ and . In our later work, three facts about inequalities will be crucial. Although proofs will be supplied problems proof. The + y will to 1/yo- is on these enlightening than a perusal of a completely worked-out very simple: if x close is enough and y to xo, is close enough to yo, be close to *o + >'o, and xy will be close to xoyo, and 1/y will be close "e" which appears in these propositions is the fifth letter of the The symbol Greek alphabet Roman more statements of these propositions involve some weird numbers, but their basic message then x at the appropriate point in the text, a personal assault infinitely is and could just ("epsilon"), made however, tradition has letter; as well be replaced by a less intimidating the use of s almost sacrosanct in the contexts to which these theorems apply. 20. Prove that if £ < - xq\ and - \y £ y < \ j, then *21. Prove that jc — jcqI \{x + y) \(x - y) - e ' £. if < min I ,2(|y then \xy - Uo + yo)l < - (xq -yo)\ < x yo\ < | + — , l)' problem just means is I and \y — \q\ J <" 2(|jc irrelevant at the in Problem moment; the 13, but the 1)' first formula provided by inequality in the hypothesis that Ijc |+ £. (The notation "min" was defined thai 1 — *n| < 7- ^ 2(|y l + D and |.v —.vol < 1; Basic Properties of Numbers 1. one point at in the argument you need the will first and inequality, 19 at an- One more word of advice: since the hypotheses only provide information about x — xq and y — yo, it is almost a foregone conclusion that the proof will depend upon writing xy — xq\q in a way that involves x — x$ and y — yo-) other point you will need the second. ^22. Prove that if yo ^ and \y then - y mm / < | I — —^— 2 £|yol Lvol , v/0 and 1 < £. yo ^23. Replace the question marks ing s, xo, If yo / and yo statement by expressions involv- and yol then in the following so that the conclusion will be true: < and ? — \x < xq\ ? _v/0 and xo < £. yo This problem is trivial in the sense that and 22 with almost no work point is be used *24. not to first, its from Problems 21 solution follows at all (notice that x/y = x 1/y). The become confused; decide which of the two problems should and don't panic your answer looks if unlikely. This problem shows that the actual placement of parentheses in a irrelevant. The Chapter Thus a\ (a) + «i (#2 + if you are not fawant to tackle this problem, it can be saved where proofs by induction are explained. 2, (a 2 + (fl3 h (fl„_2 H + ai + «3 denotes + («3 + «4)), etc. + a\ («2 + + + • • • + a„ (a n -\ «3), will denote + an))) ••) and ct\ + a2 + «3 Prove that \-ak ) (a\ H +a k+ = a\ \ H hflt+i- Use induction on k. Prove that if n > k, then Hint: (b) is still Let us agree, for definiteness, that a\ a\ sum proofs involve "mathematical induction"; miliar with such proofs, but until after crucial (a\ H Hint: Use part h fljfc) (a) + (a k+ to give a \ H 1- a„) = a\ V an -\ proof by induction on k. . + «4 denotes . 20 Prologue (c) Let s{a\ . , . . , a k ) be some sum formed from s(a\, ...,a k )=a\-\ Hint: There must be two sums \- a\, ak . . , Show a^. that . s'(a\, ...,«/) and s"(ai + \, . . . , a k ) such that s(a\, ...,a k ) 25. Suppose that we = interpret s'(a\ a "number" to t ) + s"{a t+ mean be the operations defined by the following two + ,ak ). \ or either 1 , and tables. 1 1 1 1 1 Check that properties PI— P9 all hold, even though 1 + 1=0. + and • to , CHAPTER mm NUMBERS OF VARIOUS SORTS In Chapter we used 1 word "number" very the the basic properties of numbers. now be It will our concern with loosely, despite necessary to distinguish carefully various kinds of numbers. The numbers are simplest the "counting numbers" 1,2,3 The fundamental significance of this collection of numbers is emphasized by its natural numbers). A brief glance at P1-P12 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. N symbol (for — N Nevertheless, important to deserve several comments before sufficiently is we consider larger collections of numbers. The most ciple of N basic property of Suppose P(x) means the principle of "mathematical induction." is P that the property mathematical induction number holds for the states that P(x) true for is all Then x. the prin- numbers x natural provided that (1) P(l) (2) Whenever P{k) Note that condition that P(k) (using In P(l) = infinite line k is true. P(k+ 1) under the assumption Now, = 2). it since P(2) is true it clear that each It is sort, so that P(k) then clearly every person 1, person number number .v largest tell 2, condition is true will eventually numbers all person number any secret he hears number) and a secret is will eventually learn the secret. If will learn the secret, the secret to person number 1 makes first n to the . . . be k. . person behind number P(x) is The (2) is true, (to tell all and following example There is a useful and numbers in a simple way: 1 the assertion then the instructions given (1) true. facetious use of mathematical induction. which expresses the sum of the 3, told to person secrets learned to the next person) assures that condition 21 if true (by using is follows that P(3) number true for is x, all follows that P(2) of the reasoning behind mathematical induction envisions one with the next that person then true, person has been instructed to (the 1) of people, person number If each + ensure the truth of P(x) for is of steps of this A favorite illustration him 1). (2) in the special case series P{k true, is merely asserts the truth of fact, if the special case k reached by a an true. true; this suffices to is (1) also holds. (2) in (2) is striking is telling a less formula 22 Prologue + — 2~n(n +-+» = ! To prove this formula, note first that some natural number & we have it l) clearly true for is n = Now 1. assume that for + ... +t = **+i>. i Then 1 + • • • +k+ ( k + 1 ) — = k(k+ 1)- + k+\ k(k _ 2 so the formula the formula for nomenon one also true for k is all natural + + 1) + 2k + 2 A: + 3* + 2 (* + !)(* + 2) \. By n. This particular example numbers the principle of induction this proves illustrates a phe- that frequently occurs, especially in connection with formulas like the just proved. Although the proof by induction is often quite straightforward, method by which the formula was discovered remains a mystery. Problems and 6 indicate how some formulas of this type may be derived. the 5 The principle of mathematical induction may be formulated in an equivalent way 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 mathematical term) of — natural numbers and (2) then A is the set of all k in is 1 (1) + 1 is A, in A whenever natural numbers. It k is in A, should be clear that adequately replaces the less formal one given previously set A of natural numbers x which of natural numbers n for which in A, previous proof of if A' holds for There follows that is. It all natural is this it is —we is numbers A is the set true that __ formula showed that A contains A just consider the P(x). For example, suppose + ... + „ = ! Our satisfy formulation this the set of all natural numbers, 1 , and i.e., that k that the + 1 is formula /;. yet another rigorous formulation of the principle of mathematical in- duction, which looks quite different. If A is am collection of natural numbers, it Numbers of 2. is tempting to say that A must have a smallest member. Actually, can be true to fail numbers is A is that has — the only possible exception, however; A then The 0. no smallest member null set in fact, A if this statement set of natural particularly important no natural numbers that contains denoted by collection" or "null set,"* numbers A in a rather subtle way. the collection is has no it a nonnull is 23 Various Sorts at all, "empty the a collection of natural members at This all. of natural numbers, set has a least member. This "intuitively obvious" statement, known as the "well-ordering principle," can be proved from the principle of induction as follows. Suppose that the set n such that 1, . A would have k k: +1 + 1 are all . A B be the set in B (because has no least member. Let ,n are A. Clearly all not in is 1 1 were in A, then , not in A. This shows that k if then k in B, is number n is in B, i.e., the numbers 1 number n. Thus A = 0, which completes the proof. is if (otherwise k that every It of natural numbers member). Moreover, if I, ... ,k are not in A, surely + 1 would be the smallest member of A), so 1, ... as smallest 1 not in is . A , . . . , may be is 1 n are not in also possible to prove the principle of induction principle (Problem 10). Either principle + A in B. for follows It any natural from the well-ordering considered as a basic assumption about the natural numbers. There is still another form of induction which should be mentioned. + times happens that in order to prove P(k also P(l) for all natural numbers complete induction": A then A the set of is If is (1) 1 is (2) k + : < I In k. in A, is 1 A in if 1 , . . . , be found in fact k are in A, natural numbers. all Although the principle of complete induction will some- numbers and a set of natural the ordinary principle of induction, The proof of this It we must assume not only P(k), but this case we rely on the "principle of 1) is left Problems may appear much to the reader, with a hint 7, 17, stronger than actually a consequence of that principle. it is (Problem Applications 11). 20 and 22. Closely related to proofs by induction are "recursive definitions." For example, the number numbers less n\ (read "n factorial") than or equal to n: n\ = 1 • defined as the product of is 2 • . . . • (n — 1) • all the natural n. This can be expressed more precisely as follows: (1) 1! = (2) n\ =n 1 {n - 1)!. This form of the definition exhibits the relationship between * Although n\ and (n — 1 )! in an may not strike you as a collection, in the ordinary sense of the word, the null set arises many contexts. We frequently consider the set A, consisting of all x satisfying some often we have no guarantee that P is satisfied by any number, so that A might be in it quite naturally in property P; fact often one proves that P — is always false by showing that A = 0. 24 Prologue explicit way that is cursive definition; as this for a rigorous One which may be problem shows, the recursive which will constantly may be will usually employ the more definition succinctly as a reis really necessary definition. not be familiar involves some convenient notation of writing using. Instead a\ H we expressed proof of some of the basic properties of the definition which we by induction. Problem 23 reviews a ideally suited for proofs definition already familiar to you, Greek \-a n letter £ , (capital sigma, for "sum") and write n In other words, i = denotes the /~Jfl/ sum of the numbers obtained by letting Thus 1,2, ... ,n. n(n + A. = + „2+---+n = — Y,i Y-. \) - l /=i n Notice that the letter i really has nothing to do with the number denoted by >_, > > i=\ and can be replaced by any convenient symbol (except A n(n+ . n, of course!): 1) 2 + h= iii t" = 7(7 l) ' 2 + 1) n To define /#/ precisely really requires a recursive definition: (1) ^2a =ai, t n (2) n -i Y< a = J2 ai+an i - /=] But only purveyors of mathematical austerity would precision. In practice, no one ever considers all it too strongly on such symbolism arc used, and add any words of explanation. The symbol sorts of modifications necessary to insist of this , » Numbers of 2. Various Sorts 25 I]«" 7=1 i#4 example, for an obvious way of writing is a or more The of this + (12 + i A3 + « 5 + «6 + «« H precisely, deficiencies of the natural chapter may 3 n /=1 ;=5 numbers which we discovered be partially remedied by extending at the beginning system to the this set of integers ...,-2,-1,0, 1,2 This set is only P7 A denoted by fails Z (from German "Zahl," number). Of properties P1-P12, for Z. numbers is obtained by taking quotients m/n of integers (with n / 0). These numbers are called rational numbers, and the set of all rational numbers is denoted by Q_ (for "quotients"). In this system of numbers all of P1-P12 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 P1-P12 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 larger system of still — numbers) which can be examples of shall of v devote 2, all represented by on the other hand, is quite simple, has a hypotenuse of length v 2, investigated this question.) it and was known is The fact. are both not easy —we irrationality to the Greeks. (Since the isosceles right triangle, with sides of length is 1 not surprising that the Greeks should have The proof depends on number n can be natural numbers. Every natural some irrational is of Chapter 16 of Part III to a proof of this Pythagorean theorem shows that an n and v 2 infinite decimals; numbers. The proof that n irrational + a few observations about the written either in the form 2k some integer k (this "obvious" 8)). 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: for fact integer k, or else in the (2k) (2k In particular if n form 2k 1 for has a simple proof by induction (Problem is it that l) 2 = = 4k 4k 2 = 2- (2k 2 ), 2 +4k +1=2- follows that the converse odd, then n Suppose + 2 is odd. v 2 were must is, 2 also hold: The proof that v2 rational; that (2k is + 2k) if n~ + is irrational 1. even, then n is now is even; quite simple. suppose there were natural numbers p 26 Prologue and q such that We can assume that p and <y have no common divisor could be divided out to begin with). Now we have o 2 This shows that p some natural even, is number k. common (since all divisors 2 and consequently p must be even; that is, p — 2k for Then p 2 = Ak = 2 2q 2 , so T>2 2A: = q 2 . and consequently that q is even. Thus both p and g are even, contradicting the fact that p and q have no common divisor. This This shows that g is even, contradiction completes the proof. It is important to strated that there expressed more is understand precisely what no number x such rational briefly by saying that V2 is this We proof shows. = that x 2. irrational. have demon- This assertion is often Note, however, that the number (necessarily irrational) whose square is 2. We have not proved that such a number exists and we can assert confidently that, at present, a proof is impossible for us. Any proof at this stage would have to be based on PI -PI 2 (the only properties of R we have mentioned); v2 use of the symbol since is PI -PI 2 are a rational reverse square also true for Q the exact same argument would show that there number whose square argument is implies the existence of some will not work — is and 2, this we know our proof that there 2 cannot be used to show that there is no is real (Note that the is false. no rational number whose number whose square is 2, because our proof used not only PI PI 2 but also a special property of Q_, the that every number in can be written p/q for integers p and q.) fact Q This particular deficiency in our of course, be corrected by adding a list of properties of the real numbers could, new property which asserts the existence square roots of positive numbers. Resorting to such a measure aesthetically pleasing nor mathematically satisfactory; is, we would of however, neither know that number has an still not number has an nth root if n is odd, and that every positive nth root if n is even. Even if we assumed this, we could not prove the existence of =0 (even though there does happen to be one), a number x satisfying x 5 + x + since we do not know how to write the solution of the equation in terms of nlh roots (in fact, it is known that the solution cannot be written in this form). And, of course, we certainly do not wish to assume that all equations have solutions, 2 =0, for example). In fact, since this is false (no real number x satisfies x + every 1 1 this direction of investigation of elucidating order to study this tin- R is not a fruitful one. The most useful hints about the most compelling evidence for the necessity Q, property, do not come from the study of numbers alone. In properties of the real numbers in a more profound way, we property distinguishing from the Numbers of 2. must study more than the real numbers. At we must begin point this Various Sorts 27 with the foundations of calculus, in particular the fundamental concept on which calculus is based — functions. PROBLEMS 1. Prove the following formulas by induction. (i) l- (ii) l + -..+/r = + --- + 3 = (l+--- + 3 2. . 2 77) /7 . Find a formula for J2 {2i ~ (i) = 1+3 + 5 + 1) + ••• (2,7- 1). /=] Hint: 2 l 3. 2 < k 3 < + --- + fc!(n 7?\ /n 0/ \n' for k + 32 + (2«) n{n - 2 2 + • • • + - (2/7 - . do with 1) • - {n • k I + I is 1 + 2 + 3 + • • • + 2n and defined by 1) ll , 1 2 l) ? *)! = < 5 "binomial coefficient" n\ £/ 2 l these expressions have to n, the n\ and 2 = 2 1 ) What do +2 + If (a) - ]T(2i (ii) /C 7^ (J, 77 it! (a special case or k > n we just of the first formula if we define 0! = 1), define the binomial coefficient to be 0. Prove that (The proof does not require an induction argument.) This relation gives cal's triangle" rise to the following configuration, — a number not on one of the numbers above it; the binomial coefficient sides is I J in the (77 + l)st row. 1 1 1 1 13 14 1 5 10 2 1 3 1 4 6 10 1 5 1 is the the (k known as "Passum of the two + l)st number 28 Prologue (b) Notice that part (a) entire to all the numbers in Pascal's triangle are natural prove by induction that proof by induction will, in I I is numbers. Use always a natural number. (Your a sense, be summed up in a glance by Pascal's triangle.) (n\ (c) Give another proof that a natural is I number by showing that J ... (d) , n. Prove the "binomial theorem": If a and b are any numbers and n natural number, then '"r b+ W -E("V ;=0 (e) J, Prove that 1 7=0 (.) :3-G)-<»- a D-iy(-) ;: ;=o in £C)-GH>-^ iv / 4. (a) eC-C even + G) + -- 2 «-] Prove that ti\k)\i-k)-\ Hint: (b) Apply the binomial theorem to (1 +.v)"(l +.v)' Prove that n\ k=0 i i In is a 2. 5. (a) 29 + r + r" + • l_ r »+l + r" = .2 • 1 ^ r if = The formula sum evaluating the 1, by this result and r, r (if 1 Derive by 6. Various Sorts Prove by induction on n that 1 (b) Numbers of 1 r certainly presents — +r+ S setting — • • + r" no problem). multiplying this equation , solving the two equations for S. for + 1 • — may \-n be derived as follows. We begin with the formula (k+ Writing formula for k this 2 3 33 3 (n + l) n + 1)3 ( Thus we can - (i) l (ii) l _ 2 = = 3 -k 3 =3k 2 + 3k+ = 33 3 • 1, 2 . 2 • _ = 1 >, ^ find + 3[i2 if we Use . . . 1 + 1 2 + 1 + ,7 already we obtain 1 2j + 3 [! + . . know 2, k method this 1. n and adding, , + 3+3 l 2 . . -/r3 = 3-n 2 + 3-n + . +n + „, ] (which could have been k=\ to find + .-. + n 3 4 + + n\ 3 . • • • —— + —— + 1-2 2-3 1 ... 111 1 1 . • • • + n(n 2 2 22 • + \)' In 5 3 12 ^7. 3 l k=\ in a similar way). found l) 3 2 2 n (« Use the method of Problem 6 to + + 1 l) show 2 that /_^' ; ' can always be written /=l the form P (The first + + An p + Bn p l 10 such expressions are - 1 +Cn p - 2 + --- . in 30 Prologue £* = v + \n k=\ n E* 2 k=] E* 3 k=\ 4 E* w+ =i» + v + = K + v +w = 3 + l» i» 4 2 \» - 30 n k=\ ±* 5 =y + 5 1» + n» 4 - -±n k=\ kb J2 k=\ X> 7 =W +V = +V i„ 8 +I" 5 ~b' + A" + h» 6 - 4 T4" 2 + T2" k=\ +v+i« £*•-*»» 7 -^» 5 +i« 3 -a» A? ^ E/.9 = 1 „ 10 To" 9 1 _, 3 M8 + 2 + + W +l» /7 4 ;/ 7„6, ~ To" - 1 1„4 H + 2 3 ~ 20 „2 n k=\ ]0 =Ti» J2>< n 9 " 7 + 1 " 5 ~V +W». *=1 Notice that the coefficients in the second column are always the third n 2 or n column is the reached. powers of n with nonzero The coefficients in all but the first be rather haphazard, but there actually be regarded as a super-perspicacity i, and that after by 2 until two columns seem to coefficients decrease some sort of pattern; finding it may See Problem 27-17 for the complete is test. story.) 8. 9. Prove that every natural number Prove that whenever 10. if it a set A is either even or odd. of natural numbers contains hq and contains k contains k, then A contains all natural numbers > + 1 hq. Prove the principle of mathematical induction from the well-ordering principle. 11. Prove the principle of complete induction from the ordinary principle of induction. Hint: If /?, 1 12. (a) A contains 1 B all consider the set and b of and A contains k such that If a if a and b arc both irrational? is rational is irrational, is a + 1 // + 1 whenever k are all in it contains A. b necessarily irrational? What 3 2. If a (c) Is there a (d) Are there two (b) and b rational is number irrational, is a such that a ab is Numbers of necessarily irrational? (Careful!) irrational, is 31 Various Sorts but a rational? is numbers whose sum and product are both irrational ra- tional? 13. Prove that V3, v (a) and 5, Why 14. 15. 16. v2 and V3 is treat of the form 3n or 3« V 3, for exam+ or 3« + 2. 1 v 4? are irrational. Prove that (a) V2 + v 6 is irrational. (b) v2 + V3 is irrational. (a) Prove that = p number, then x m = x if (b) Prove also that (a) Prove that (m + In) if /(m (/? + ^fq where p and # are rational, a + b^/q for some rational a and m - m and n are natural + n) > 2; — + 2n)-~2 (m + n)" (b) Prove the same results with (c) Prove that It seems m/n < v 2, m/n < m'/n' < v2. if likely that \fn is m is a natural b. numbers and m</n~ < then 2, show, moreover, that (m with and —a - bjq. y/q) — '17. To are irrational. Hint: doesn't this proof work for Prove that (b) V6 every integer ple, use the fact that all rrr < 2 2 =-. ninequality signs reversed. then there irrational is another rational number m'/n' whenever the natural number // is not method of Problem 1 the square of another natural number. Although the may that actually be used to treat will it information. A natural p = sible to write and the other The first 1 not clear in advance it is number p for natural convenience is prime number called a numbers a and b we also agree that if it is 1 is not < n; if either similarly; continuing in this way a or b is proves that > 1 we can it argument into a rigorous 7 = 2 2 it write n proof by complete induction. this sure, • p, not a is not a prime Turn 4 is a prime number. 2, 3, 5, 7, 11, 13, 17, 19. If n with a and b both impos- unless one of these = 7. (To any reasonable mathematician would accept the informal argu- ment, but this is partly because it would be obvious to her how to state rigorously.) fundamental theorem about that this factorization for case, product of primes. For example, 28 be A for ; — ab, can be factored (a) ab few prime numbers are prime, then n as a any particular always work, and a proof for the general case requires some extra is integers, which we will not prove here, states unique, except for the order of the factors. Thus, example, 28 can never be written as a product of primes one of which . 32 Prologue is 3, nor can appreciate it be written why is 1 (b) Using (c) number m Prove more (d) No discussion prove that this fact, P2 P\ *18. finitely ••• Prove that (a) way • if that involves 2 only irrational unless n *J~n is generally that Zfn infinitely many x + fail 1 . p n by considering , +...+a =0, irrational unless , V6 — v 2 — V3 Prove that v 2 k to allude to Euclid's beautiful p\, p2, P3, ... some integers fln _i, ... «o, then x is (Why is this a generalization of Problem (c) some natural satisfies for Prove that =m for 1. .v"+a„_ix"" (b) = m2 of them. Prove that there cannot be many prime numbers Pi, once (now you should irrational unless n is of prime numbers should proof that there are only a in not allowed as a prime). + v2 is x is an integer. 17?) irrational. is irrational. Hint: Start by working out the first 6 powers of this number. 19. (1 for 20. — 1, Prove Bernoulli's inequality: If h > any natural number n. The Fibonacci sequence 01 fl 2 a„ Why a\, «2> = 1, = = a„-\ 1175-1250), assumed that an in is > 1 +nh a 3? • is 0? defined as follows: + a„_2 1, 1, for n 2, 3, 5, 8, . . . > 3. was discovered by Fibonacci , connection with a problem about rabbits. Fibonacci initial pair of rabbits gave birth to one per month, and that after two months each new number is a„ of pairs > this trivial if h 1, This sequence, which begins (circa +/?)" then born in the nth month new pair of rabbits pair behaved similarly. an -\ + fl„_2, The because a pair of is born for each pair born the previous month, and moreover each born two months ago now gives birth to another pair. The number of there is even a Fiinteresting results about this sequence is truly amazing bonacci Association which publishes a journal, The Fibonacci Quarterly. Prove rabbits pair — that /1+V5Y' a„ /1-V5Y' = v/5 One way of deriving this astonishing formula is presented in Problem 24-16. . 2. 21. The Schwarz Numbers of inequality (Problem 1-19) actually has a Various Sorts 33 more general form: n E* £* 2 Give three proofs of this, analogous 22. The result in Problem 2 - to the three proofs in an important generalization: 1-7 has Problem If a\, . . 1-19. . , an > 0, mean" then the "arithmetic Ya n a\-\ = A„ and "geometric mean" = Va G„ • i • an - satisfy G n < An (a) Suppose say ai > . < A n Then some a, satisfies a, > A„; for convenience, Let a\ = A n and let «2 = «i + «2 — o-\. Show that that a\ A„. . > a\ci2 a\ai. Why does repeating this process enough times eventually prove that G n < An ? (This is another place where a good exercise to provide a formal it is proof by induction, as well as an informal reason.) The (b) reasoning in Using the Gn < A n (c) this proof is related fact that for n Gn < let = . Apply part n. . , an , An . , 2'" The another interesting proof. to A„ when n 2m > a\, 23. does equality 2, prove, by induction on k, that = 2*. For a general n, to When < A„? hold in the formula G„ . -n . , (b) to the 2'" numbers An times prove that G„ < A„. following is a recursive definition of a": a a" 1 +1 = = a, a" -a. Prove, by induction, that a" < +m „ll\l» = a" -a'", „nm (Don't try to be fancy: use either induction on n or induction on m, not both at once.) 34 Prologue 24. Suppose we know properties PI and P4 numbers, but that for the natural multiplication has never been mentioned. Then the following can be used as a recursive definition of multiplication: lb = b, (a Prove the following a-(b a 1 • a 25. b (in + 1) b • In this chapter a -b + b. the order suggested!): + c) = a-b + a-c = a, —b = on (use induction a), = a (you just finished proving the case b we began A the real numbers. 1). with the natural numbers and gradually built up to completely rigorous discussion of this process requires 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 if we do accept the real numbers as given, then the natural numbers can be defined as the real numbers of the form 1, 1 + 1, 1 + 1 + 1, etc. The whole point of this problem is to show that there is a rigorous mathematical way of saying "etc." a little (a) A set A of real numbers (1) 1 is in (2) k + 1 is called inductive if A, is in A whenever k is in A. Prove that (i) R is inductive. The set of positive real numbers is inductive. (iii) The set of positive real numbers unequal to A is inductive. The set of positive real numbers unequal to 5 is not inductive. (iv) If A and B are inductive, then the set C of real numbers which (v) are in both A and B is also inductive. A real number n will be called a natural number if n is in every inductive (ii) (b) set. 26. There (i) Prove that (ii) Prove that k is 1 is + a natural number. 1 is a natural is may be moved from one to spindle the smallest ring first (Figure I ( , I RE is ring. 2 and the next-smallest ring may be moved moves, and that this 1). A For example, is to spindle 3 also, Prove that the entire stack of n rings can be I k a natural number. ring at the top oi a spindle to another spindle, provided that not placed on top of a smaller moved if a puzzle consisting of three spindles, with n concentric rings of decreasing diameter stacked on the stack number cannot be done if the smallest ring moved to spindle 3, then on top of the moved onto in fewer than 2" — next-smallest. spindle 3 in 2" 1 it is moves. — 1 2. *27. Numbers of University B. once boasted 17 tenured professors of mathematics. tion prescribed that at their all 17, any members weekly luncheon meeting, who had 35 Various Sorts Tradi- attended by faithfully discovered an error in their published work make an announcement of this and promptly resign. Such an announcement had never actually been made, because no professor was aware of any errors in her or his work. This is not to say that no errors existed, however. In fact, over the years, in the work of every member of the department at least one error had been found, by some other member of the department. This error had been mentioned to all other members of the department, but the actual author of the error had been kept ignorant of the fact, to forestall any resignations. One fateful year, the department was augmented by a visitor from another university, one Prof. X, who had come with hopes of being offered a permanent position at the end of the academic year. Naturally, he was apprised, by various members of the department, of the published errors which had been discovered. When the hoped-for appointment failed to materialize, Prof. X should obtained his revenge at the last here very much," he said, "but fact, luncheon of the I feel that there year. "I is have enjoyed one thing that I my visit have to tell you. At least one of you has published an incorrect result, which has been discovered by others in the department." **28. What happened the next year? After figuring out, or looking up, the answer to Problem 27, consider the lowing: so how Each member of the department already knew what could his saying it change anything? Prof. fol- X asserted, PART FOUNDATIONS The statement is so frequently made that the differential calculus deals with continuous magnitude, and yet an explanation of this continuity is nowhere given; 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 convinced careful investigation me 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 and thus to at the to discover its true of arithmetic same time secure a real definition of of continuity. I succeeded Nov. 24, 1858, and the essence 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 CHAPTER FUNCTIONS Undoubtedly the most important concept in all of mathematics is that of a function in almost every branch of modern mathematics functions turn out to be the central objects of investigation. It will therefore probably not surprise you to learn that the concept of a function is one of great generality. Perhaps it will — be a we relief to learn that, for the present, will be able to restrict our attention functions of a very special kind; even this small class of functions will exhibit engage our attention for quite some time. cient variety to with a proper definition. For the discuss functions at length, and moment PROVISIONAL DEFINITION modern mathematical A function is will not even begin a provisional definition will enable us to will illustrate the intuitive understood by mathematicians. Later, we of the We to suffi- notion of functions, as consider and discuss the advantages will definition. Let us therefore begin with the following: a rule which assigns, to each of certain real numbers, some other real number. The following examples of functions are nition, which, admittedly, requires Example 1 Example 2 The The some such which assigns to each rule which assigns to 3 + y The which assigns rule c 3 Example 4 the number x Example 5 irrational, The rule which clarification. +5 +l 2 + + -l 3c 2 • each number assigns to each The rule which assigns The rule 1 if which a is c^ 1,-1 the number 5 ' number x to satisfying number a each rational. assigns to 2 the number to 17 the 5, number 36 TV 39 this defi- —17 < x < tt/3 . and the number Example 6 and amplify 3y to c to illustrate number the square of that number. each number y the number rule y Example 3 meant the number if a is 40 Foundations to — the number 28, to 36 — the number 28, TV and any to ^ y 2, 17, jt~/\7, or 36/tt, the number 16 y if is of the form a + bv2 Q. for a, b in The Example 7 which assigns rule to each number t the number 3 / + a\ (This on what the number x is, so 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 — n if there are infinitely many rule depends, of course, .) 7's in the One decimal expansion of z. thing should be abundantly clear from these examples numbers rule that assigns to certain other numbers, not —a function just a rule is any which can be expressed by an algebraic formula, or even by one uniform condition which number; nor applies to every can actually apply is it one knows, in practice (no Moreover, the rule may necessarily a rule 7r). to which numbers the function applies apply is Example 6 applies to it). for example, some numbers and neglect to function in which you, or anybody (try to The set it else, what rule 8 associates may not even be clear determine, for example, whether the of numbers to which a function does called the domain of the function. Before saying anything else about functions throughout this book we shall frequently we badly need some notation. Since be talking about functions (indeed we shall else) we need a convenient way of naming funcand of referring to functions in general. The standard practice is to denote a function by a letter. For obvious reasons the letter "/" is a favorite, thereby making "g" and "h" other obvious candidates, but any letter (or any reasonable symbol, for that matter) will do, not excluding "x" and "y", although these letters hardly ever talk about anything tions, are usually reserved for indicating numbers. If / is number read "/ of a function, then the — which / associates to a number x is denoted by f(x) this symbol is x" and is often called the value of/ at*. Naturally, if we denote a function by x, some other letter must be chosen to denote the number (a perfectly legitimate, though perverse, choice would be "/," leading to the symbol x(f)). Note that the symbol f(x) makes sense only for x in the domain of /; for other x the symbol f(x) not defined. is If the functions defined in we can and y, (1) f(x)=x 2 (2) g(y) then h(c) = = c - 3 1 8 are denoted by /, g, rewrite their definitions as follows: for all x. y (3) Examples — y— + L + —~ c L for all y. 1 — — 3c + 1 5 for all c # 1, -1. /?, r, .v, #, ax , ; 41 3. Functions (4) r(x) = x1 (5) s(x) = for -17 < x < x such that all 0, x irrational 1, x tt/3. rational. x=2 5, 36 x= 17 77 i = 9(x) (6) 28, r • 71 = T7 36 28, 71 36 — 71 x 16. ^ 2, r, or , and x , — a + b\/2 for a, b in Q_. 7T (7) ax (t) (8) v(a-) These tion / = for numbers all 77, exactly —7T, infinitely indicating is t? 7's ?. appear many 7's decimal expansion of x in the appear in the common procedure decimal expansion of adopted x. for defining a func- what f(x) is for every number x in the domain of /. (Notice same as indicating f(a) for every number a, or f(b) for ev- exactly the number ery +x t definitions illustrate the — that this = In practice, certain abbreviations are tolerated. Definition b, etc.) (1) could be written simple the qualifying phrase "for all the only possible abbreviation It is Of course, x" being understood. for definition (4) is r(x)=x 2 (4) = x2 f(x) (1) -17<a<tt/3. , usually understood that a definition such as k(x) I = -v ^ 0, 1 1 can be shortened to *(*) in other all words, unless the numbers for which You should have domain the definition little — 1 1 x — = -+x 1 is explicitly restricted further, makes any sense at difficulty it is understood + 1) g(x) r(x = /(*) + 2x + = h(x)i£ 3 + 3* + 5 = 0; +[) = 1; jc r(x) of checking the following assertions about the func- tions defined above: /(* to consist all. + 2x + 1 if -17 < x < ^- 1; 42 Foundations s{x + y) = six) y if rational; is -G) -MS)-expression /(5(a)) looks unreasonable to you, then you are forgetting that If the s{a) of number a is fact, = f(s(a)) any other number, so that f{s{a)) makes sense. As a matter s(a) for all a. Why? Even more complicated expressions than like f(s(a)) are, after a first more exposure, no The difficult to unravel. expression /(r(s(0(a 3 (j(i)))))), formidable as appears, it may be evaluated quite easily with a little patience: f(r(s(e(a 3 (yQ)))))) = /(r(j(0(a 3 (O))j)) = /(r(*(0(3)))) = /(r(j(16))) = /(r(D) = /(D = The this first few problems at the 1. end of this chapter give further practice manipulating symbolism. The function defined in (1) is a rather special example of an extremely impor- tant class of functions, the polynomial functions. function if = fix) (when f(x) highest numbers there are real + a ,_]A'"~ + an x • • • ; form written in this is flo? it is power of x with a nonzero • • • > A function + ajx" + a\x + ao, usually tacitly functions defined in (2) and (3) is a coefficient is for assumed called the example, the polynomial function / defined by fix) degree 6. The / polynomial a n such that = 5x all that a„ x ^ 0). degree of + 137a belong to a somewhat larger class — The /; for tt has of func- 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 always 0). The rational functions are themselves quite special examples of an even larger class of tions, the functions, very thoroughly studied in calculus, which are simpler than many of the The following are examples of this kind functions mentioned first of function: (9) \x ) Jf(r\ — — X (10) fix) (11) fix) = = -2 A + A"2 + x sin .2 sin a + a sin a 1 1 . \ in this chapter. sin (a 2 ). sin(sin(A )). 43 3. Functions = /(jc) (12) 2 . -22 x ))) -2 • + sin(jc sinx) + sin x /-* • sin (sin (sin (x sin sin • : \ By what criterion, monstrosity like you may jc impelled to ask, can such functions, especially a feel be considered simple? The answer (12), that they can is be built up from a few simple functions using a few simple means of combining functions. we need In order to construct the functions (9)— (12) function" x Z, which I(x) for x, often written simple sin x is which functions If / and g may and the The . be combined and g, to start with the "identity whose value sin(jc) some of the important ways "sine function" sin, following are at in produce new functions. to are any two functions, sum of / the — we can new define a f + function g, called by the equation (f + g)(x) = + g(x). f(x) Note that according to the conventions we have adopted, the domain of / + g consists of all x for which "fix) + gix)" makes sense, i.e., the set of all x in both domain / and domain g. If A and B are any two sets, then A D B (read "A intersect 5" or "the intersection of A and B") denotes the set of x in both A and B; this notation allows us to write domain (/ + g) = domain / Pi domain g. In a similar vein, we product / g and define the the • f — quotient (or f/g) of 8 f and g by (/ • = g)(x) f(x) gix) • and (£\™ ^«== /w g(x) Moreover, g if is a function and c (c is a number, g)(x) = has the same value for / numbers x all define a / g we if • = defined by fix) , is new function c g by g(x). c This becomes a special case of the notation should also represent the function we called a agree that the symbol c c; such a function, which constant function. domain / D domain g, and the domain of c g is simply the domain of f/g is rather complicated it may be written domain / D domain g D {x gix) ^ 0}, the symbol {x gix) ^ 0} denoting the set of numbers x such that gix) / 0. In general, {x denotes The domain of / g • the domain of g. is — On the other hand, : : : . } . . the set of all all jc such that " ..." numbers x such that x~ < 8, is true. Thus {x : and consequently x [x +3 : x < 11} denotes the set of +3 < 11} = {x : x < 2}. Either of these symbols could just as well have been written using y everywhere common, but hardly require any 3 < 8} denotes the set of positive numbers whose cube is less than 8; it could be expressed more formally as {jc and x 3 < 8}. Incidentally, this set is equal to the set {x < jc < 2}. One x > instead of x. discussion. Variations of this notation are Any one can guess that {x > : jt : : 44 Foundations variation {.v = x : 1 or x = = 3 or x The but very standard. slightly less transparent, is example, contains just the four numbers and 2, 3, 1, 4; {1,3,2,4}, for set can also be denoted by it = 4}. 2 or x Certain facts about the sum, product, and quotient of functions are obvious con- sequences of facts about sums, products, and quotients of numbers. For example, very easy to prove that it is (f The proof — same value the any number + g) + h = f + (g + h), (f must be shown to have the same domain, domain. For example, to prove that the two functions at [(/ + h). (g proof which demonstrates that two characteristic of almost every is functions are equal and + g) + h =f + in the note that unraveling the definition of the two sides gives + g) + h](x) = = + g)(x) + h{x) [f(x) + g(x)]+h(x) (f and + [/ and f(x) + [g(x) + h(x)] is a fact about proof the equality of the two domains was not explicitly men- and the equality of [f(x) numbers. In + h)](x) = fix) + (g + h)(x) = f(x)+[g(x) + h(x)], (g this tioned because this is + g(x)] + h{x) obvious, as soon as we domain of (/ + g) + It and of / + (g + h) domain h. We naturally write / + g + h for (/ the we as domain / clearly is + g) + h these equations; —f+ (g Pi domain g + /?), (1 precisely did for numbers. It is by no down begin to write just as easy to prove that (/ / g • - The h. equations f + • = g h g) g =/ +f • and h), (g = and f g • g this • function is denoted should also present f difficulty. Using the operations +, • , / we can now express the function f defined in (9) by /= / + /•/ + /• sin : / sin : +/ • sin • sin : • • sin quire yet another however, that we cannot express function (10) this way. We reway of combining functions. This combination, the composition of two functions, is It should be clear, by far the / and g are any two sition of / and g, by most important. functions, If we (fog)(x) the domain of fog "fog" is and g" this often is {x read "/ - : a is in define a = new is Compared much less o g, the compo- is in domain /}. The symbol to the phrase "the composition of has the advantage of brevity, of course, but there of far greater import: there / f(g(x)): domain g and g(x) circle g." function chance of confusing is /' / another advantage g with go/, and 45 3. Functions these must be confused, since they are not usually equal; not and g chosen at example). Lest let us hasten to point out that composition (and the proof a is triviality); this write the functions (10), (11) (12) / = (10) sin o function oh) (g denoted by f is o o h. g • • /), sin) o sin o (sin sin) o (/ • [(sin • sin) o (/ • /)]) • + / • sin o (/ / become has probably already fact name seems / be "the function to abbreviating / for the function Although clear. functions reveals their "structure" very clearly, shortest now can (12) as (1 1), sin o One We (/•/), sin o sin o (/ (sin associative: is = /o (fog) oh /= /= almost any / random will illustrate this point (try f = I I and g — sin, for you become too apprehensive about the operation of composition, in fact, such that f(x) = sin) + sin method of this writing The hardly short or convenient. it is = such that fix) • sin(jc ) for all 2 sin(x for all ) x unfortunately x" The need for clumsy description has been clear for two hundred years, but no this reasonable abbreviation has received universal acclaim. At present the strongest contender for honor this something is like x -» sin(x ) 2 2 (read "x goes to sin(x )" or just "x arrow sin(x )"), but ellipsis, never use criterion is As with any convention, reasonable so long as the you x — x+ + / 3" is , Even more popular the motivating factor, slight logical deficiencies more the we number and probably end up adopting is is For the sake of precision an ambiguous phrase; i.e., the function / such that f(x) i.e., the function / such that could it = it and for this cause no confusion. precise description J t among a certain amount of speaking, confuses a will utility occasion, confusion will arise unless a example, "the function x sin(jc )." strictly so convenient that is it which, hardly popular it is will tolerate "the function sin(jc )." this description, a function, but personal use. On — and speak of "the function f(x) quite drastic abbreviation: will we In this book writers of calculus textbooks. is used. mean either for all x for all t. For T x +r or t As we -> x + 3 1 , however, for f(t)=x + t 3 many important concepts associated with functions, calculus has a notation which contains the "jc —>" built in. shall see, By now we have made a sufficiently extensive investigation rant reconsidering our definition. hardly clear what is this means. not easy to say whether this If We we of functions to war- have defined a function as a "rule," but ask question "What happens is if you break it is this rule?" it merely facetious or actually profound. 46 Foundations A more substantial objection to the use of the word "rule" is that f(x)=x 2 and f(x) =jc 2 + 3a- + 3-3(jc + 1) by a rule we mean the actual determining f(x); nevertheless, we want are certainly different rules, if f(x) = x /(jt) = jc instructions given for 2 and 2 + 3* + 3 - c(* + 1) same function. For this reason, a function is sometimes defined as an "association" between numbers; unfortunately the word "association" escapes the to define the objections raised against "rule" only because even more vague. it is 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 English words which are troublesome. "function" is The which mathematicians have finally accepted for example of the means by which intuitive ideas have been definition a beautiful 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 for each number x one needs to know the number fix); the last question is easy we can imagine a table which would display all the information one could desire incorporated into rigorous mathematics. — about the function f(x) =x : x fix) 1 1 -1 1 2 4 -2 4 s/2 2 -v/2 2 2 It is IX 7T —n 71 not even necessary to arrange the be impossible if we wanted to list all numbers in a table (which would actually of them). Instead of a two column array can consider various pairs of numbers (1.1). (-1.1). (2.4). (-2.4). 2 (it, it ), (72,2),... we 47 3. Functions simply collected together into a we simply find f(n) we To set.* find f{\) number of the pair whose first member is 1; to number of the pair whose first member is 7r We seem . might as well be defined were given the following / = as a collection of pairs of to take the second take the second be saying that a function numbers. For example, then /(3) or f{\) — = 7, 8. /(2) = we {(1,7), (3,7), (5,3), (4,8), (8,4)}, then f(\) = 7, /(3) = 7, /(5) = 3, /(4) = 8, /(8) = 4 and 1,3, 4, only numbers in the domain of /. If we consider the collection /= if collection (which contains just 5 pairs): 5, 8 are the {(1,7), (3,7), (2,5), (1,8), (8,4)}, 5, = /(8) 4; but it is =7 impossible to decide whether /(l) In other words, a function cannot be defined to be any old collection of pairs of numbers; we must rule out the possibility which arose We in this case. are therefore led to the following definition. DEFINITION A function a collection of pairs of is numbers with (a,b) and (a,c) are both in the collection, then b must not contain two collection the following property: = different pairs with the c; if other words, the in same first element. and illustrates the format we shall always These definitions are so important (at least as important as theorems) that it is essential to know when one is actually at hand, and to distinguish them from comments, motivating 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. This is our first full-fledged definition, use to define significant There If / is (a, b) is domain in /. If of a function that there This unique b With (actually defining two things at once) which can rigorously: a function, the such that concepts. one more definition is now be made DEFINITION new is a is, is of is the set of all a for which there is some b domain of /, it follows from the definition a unique number b such that (a,b) is in /. / in the in fact, denoted by f{a). we have reached our goal: the important thing about a a number f(x) is determined for each number x in its domain. we have also reached the point where an intuitive definition has this definition function You may / is that feel that been replaced by an abstraction with which the mind can hardly grapple. consolations may be offered. First, although a function has been defined *The pairs occurring here are often called "ordered pairs," to not the same pair as ordered this pairs, (4, 2). It is only fair to another undefined term. ( warn that we emphasize that, for example, are going to define functions in Two as a (2. 4) )rdered pairs can be defined, however, and an appendix chapter has been provided lor skeptics. is terms of t< i 48 Foundations collection of pairs, there rule. nothing to stop you from is Second, neither the The of thinking about functions. chapter all by thinking of a function as a nor the formal definition indicates the best way intuitive way best is to draw pictures; but this requires a itself PROBLEMS 1. /(*)= 1/(1+ x). What Let f(f(x)) (i) is which x does (for this make sense?). 'G «> (iii) f(cx). (iv) f(x+y). (v) /(x) (vi) For which numbers c (vii) 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 x? + /(y). there a is number jt such that f(cx) = f(x). Hint: 2. =x Let g(x) and , let x rational 0, *«-!, 3. (i) For which v is /z(v) (ii) For which y is /?(y) (iii) What is (iv) For which (v) For which e is y? g(y)? < g(u;) is irrational. - Mz)? g(A(z)) w < < x if? = g(g(e)) g(£)? Find the domain of the functions defined by the following formulas. f(x) = yi-.v 2 (ii) /(.v) = Vl-\/l-* 2 (iii) /(x)= f(x)= iv /(jc) (v) = 7 y ! >/l = x2 Let S(x) , r — x : + 1 -.v 2 - let . .v «• x — + v/x + P(x) z v/jc = . 2 - -l. 2. 2 V and , let s(jc) = sin*. Find each of the following. In each case you answer should be a number. (i) (SoP)(y). (ii) (So S )(y). (iii) (SoPo S )(t) + (iv) s(t (soP)(t). 3 ). Express each of the following functions +, • , and o (for example, the answer to in (i) terms of is P o s). S, P, S, using only In each case your 1 49 3. Functions answer should be = f{x) — f(x) = /(a) = /(/) = 2s ii) iii) iv) v) S1I > fix) i) a. V I sin2 v . a (remember that sin 2 2 . a sin function. . (Note: a h' sin always a an abbreviation is means a (i) ; this for (sin*) convention is ). adopted because (a b ) c can be written more simply as a bc .) +2" 2 vi) f{u) =sin(2" vii) f{y) = sin (sin (sin (2 viii) f(a) = 2 sin2 ). 2~ ))). + sin (a 2 + 2 sin(a2+sina) " ) . Polynomial functions, because they are simple, yet role in most flexibility, investigations of functions. The flexible, occupy a favored following two problems illustrate their and guide you through a derivation of their most important elementary properties. 6. (a) If x\, ... degree n all (a , — — Xj) x„ are distinct numbers, find a polynomial function 1 which for j / is at x, 1 and at Xj if j /, is ^ at Xj for j / / . i. /) of Hint: the product of (This product is usually denoted by n Y\(x-Xj), the symbol n (capital pi) playing the same role for products that E plays for sums.) (b) Now find a polynomial function / of degree n — 1 such that /(a,) = «,, where a\, ... a n are given numbers. (You should use the functions fi from part (a). The formula you will obtain is called the "Lagrange , interpolation formula.") 7. (a) Prove that for any polynomial function /, and any number a, there is a polynomial function g, and a number b, such that f(x) = (a —a)g(x) + b for all a. (The idea until a constant is simply to divide (a remainder A- A V \ 2 is left. +A X -2 1 —x A2 A2 into f(x) by long For example, the calculation -3a + ),3 3 — a) -3a —A -2a+1 -2a + 2 - division, . 50 Foundations — shows that x 3 + 3.v Prove that Prove that n roots, (d) Show roots, 8. / if If n if odd is For which numbers a, 9. (a) If =x A set any is is /.) (x — a)g(A') for some polynomial / then , = numbers a with f(a) find one with only one and d b, c, has at most 0. for all x +b +d which (for root. will the function ex /(/(*)) formal proof a polynomial function of degree n with is ax satisfy A 1. even find a polynomial function of degree n with no is n — 2) a polynomial function of degree n is there are at most n i.e., and +x— 2 f(a) = 0, then f(x) = (The converse is obvious.) that for each n there n roots. l)(x if function g. (c) — (x on the degree of possible by induction (b) = 1 equation makes sense)? this of real numbers, define a function C\ as follows: 1 C A (x)- a in . A A x not in (J, Find expressions for Cahb and Caub and Cr.^, in terms of Ca and Cg. (The symbol AC\ B was defined in this chapter, but the other two may be new to you. They can be defined as follows: AUB= R—A= (b) Suppose / there 10. is is a set : {x : A (a) For which functions if For which functions all What A .v is in R but x is this / is (x(t))~ for in numbers t f = and only / can certainly answer *(d) is such that / = / Show (b) x or x in B}, is a function such that f(x) (c) that {x is not in A}.) = or 1 each x. Prove that for Caif / = Ca for some set there a function g such that question "function" if is A. f =g Hint: You replaced by "number." there a function g such that / = \/g? + b{t)x(t) + c(t) = ? conditions must the functions a and b satisfy if there is to be a function x such that a(t)x(t) for all 11. (a) numbers t? H is Suppose What that How many a function + b(t) =0 such functions x and v is a number such is H(H(H(-(H(y)-) ? 80 times will there that be? H(H(y)) = v. . 51 3. Functions Same Same (b) (c) question if question if = such that H(l) = 7r/3, //(47) replaced by 8 1 is = H{H{y)) H Find a function *(d) 80 H{y). = such that H{H{x)) = H(2) 36, H{x) = n/3, #(13) 47, for numbers all = H(36) and x, = 36, H(jc/3) 47. (Don't try to "solve" for H(x); there are many func- H{H{x)) = H(x). The extra conditions on // are supposed way of finding a suitable // function H such that H{H{x)) — H{x) for all x, and such that tions // with to suggest a Find a *(e) H(\) 12. A odd (b) if / 13. even is is even —x fix) = fix) or fix) = fix) if — x f(—x) and odd or /(x) Do Prove that every even function the same many for E is Prove that / is / can be O even and with domain / is cosx, while For / is = written fix) g(|x|), for in- for E and O you functions, define R can be written / = E+ O, odd. way of writing / is will unique. (If you try to do part (b) first, probably find the solution to part any function, define a new function and g are = by |/|(x) |/| |/(x)|. two new functions, max(/, g) and min(/, max(/, g)(x) min(/, g)(x) = = (a).) If g), / by max(/(x), g(x)), min(/(x), g(x)). Find an expression for max(/, g) and min(/, g) (a) = — /(— x). = functions g. this by "solving" 15. /(x) fog. Prove that any function where If if or /(x) |x| • (d) (a) = sinx. (c) (b) 14. if 18. Determine whether / + g is even, odd, or not necessarily either, in the four cases obtained by choosing / even or odd, and g even or odd. (Your answers can most conveniently be displayed in a 2 x 2 table.) Do the same for f g. finitely ! H(\l)= l, / function example, (a) = .) in terms of | |. Show that / = max(/, 0) + min(/, 0). This particular way / is fairly useful; the functions max(/, 0) and min(/, 0) are of writing called the positive and negative parts of /. (b) A function / is nonnegative called if fix) > for all x. Prove that any / 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 many ways. function '16. Suppose / (a) (b) satisfies fix + y) = fix) + f(y) for all x and Prove that /(xi +••• + *„) = /(jq) + ••• + /(*„). Prove that there is some number c such that f(x) numbers x (at this irrational x). y. = ex for all rational point we're not trying to say anything about fix) for Hint: First figure out what c must be. Now prove that ) 52 Foundations = fix) then "17. If f{x) and when x first when x = also ex, is the reciprocal of an integer and, / for all x, then fix y) a natural number, then is = these two properties, but that fix) all (a) (b) (c) Prove that /(l) = integer, rational jc . fix + y) = fix) + f(y) for all x and y, all x and y. Now suppose that / satisfies is not always 0. Prove that fix) = x for 1. Prove that fix) — x Prove that fix) > problems (e) an x, as follows: if been paying attention (d) is finally, for all satisfies fiy) for fix) when x in the last Prove that fix) Prove that fix) numbers there is x is if x > (This part 0. to the philosophical two chapters, you > fiy) = rational. x for if x > all will is but tricky, if you have remarks accompanying the know what to do.) y. x. Hint: Use the fact that between any two a rational number. "18. what conditions must f,g,h, and hix)kiy) for all x and y? "19. (a) k satisfy in order that f(x)giy) Precisely Prove that there do not exist functions / and g = with either of the following properties: (i) f(x) (ii) fix) Hint: + giy) = xy for all x and y. giy) = x + y for all x and y. Try to get some information about / or g by choosing particular values of x and y. "20. + = x and (b) Find functions / and g such that fix (a) Find a function /, other than a constant function, such that \fiy) (b) Suppose y) gixy) for all y. — fix)\<\y-x\. that fiy) imply that — fix) < iy \fiy) — fix)\ < — x) 2 for 2 iy — x) ?) all x and Prove that / is y. (Why does this a constant function. Hint: Divide the interval from x to y into n equal pieces. 21. Prove or give a counterexample for each of the following assertions: (a) (b) (c) fo(g + h) = fog + foh. ig + h)of = gof + hof. 7 — = 7°^ f °g (d) f -— = /° f°g 22. (a) (b) Suppose g \8 =ho /. Prove that if fix) — fiy). then g(x) = giy). / and g are two functions such that gix) = giy whenever fix) = fiy). Prove that g = h o f for some function //. Hint: Just try to define hiz) when z is of the form z = fix) (these arc the only z that matter) and use the hypotheses to show that your definition will not Conversely, suppose that run into trouble. . . 53 3. Functions 23. *24. Suppose that / x ^ o = g /, ^ then g(x) Prove that (a) if (b) every (a) Suppose g is a function with the property that g(x) Prove that there is a function / such that / o g = I (b) y, g(y); number b can be Suppose / that is written b *25. Find a function / such that g function h with f o h = I *26. Suppose / o g = is associative. composition (a) (b) and h I — f(a) for some number o a. o f= Prove that there f= I. I for some Prove that g g, = ^ a. g(y) if number b can be a function such that every — f(a) for some number fog = I. b 27. = x. where I(x) is x ^ y. written a function g such that but such that there h. Hint: Use the is no fact that Suppose f(x = x + 1. Are there any functions g such that fog = go f? Suppose / is a constant function. For which functions g does / o g = ) 8 of? (c) Suppose that fog = go f 28. (a) Let and (b) *(c) (d) (e) F be • for all functions g. Show that / is the identity = x. function, f(x) the set of all functions whose domain is as defined in this chapter, all of properties R. Prove that, using + PI— P9 except P7 hold and 1 are interpreted as constant functions. P7 does not hold. that 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 except at two points *o and jci.) Suppose we define / < g to mean that f(x) < g(x) for all x. Which of P'IO-P'13 (in Problem 1-8) now hold? for F, provided Show Show that If / < g, is h o f < h o g? Is / o h < g o h ? 54 Foundations ORDERED PAIRS APPENDIX. Not only it in the definition of a function, but in other parts of the book as well, A necessary to use the notion of an ordered pair of objects. is not yet been given, and ordered pair is supposed definition has we have never even stated explicitly what to have. The one property which we will formally that the ordered pair properties an require states should be determined by a and {a, b) and the b, order in which they are given: if (a, Ordered as may be = b) then a (c, d), = = and b c d. most conveniently by simply introducing {a b) an undefined term and adopting the basic property as an axiom since this pairs treated , — property is the only significant fact about ordered pairs, there worrying about what an ordered pair "really" need read no satisfactory The rest of this short is. Those who is much not point find this treatment further. appendix for the benefit of those readers is uncomfortable unless ordered pairs are somehow defined so that becomes a theorem. There who this basic will feel property no point in restricting our attention to ordered pairs of numbers; it is just as reasonable, and just as important, to have available the notion of an ordered pair of any two mathematical objects. This means that our definition is ought to involve only concepts common to The one common concept which pervades all all branches of mathematics. areas of mathematics is that of a and ordered pairs (like everything else in mathematics) can be defined in this context; an ordered pair will turn out to be a set of a rather special sort. The set {a,b}, 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 determining from {a, b} which of a or b is meant to be the first element. A more promising set, candidate is the rather startling set: {{a},{a,b}}. This {a}, set has two members, both of which are themselves containing the single seem, we member a, the other are going to define (a, b) to be this is set. sets; really isn't anything else DEFINITION theorem l PROOF II' (a, b) The — (c, d), then a hypothesis means — c (a, 6) = and b = The justification {{<-/}, common — the set it may for this choice is the definition works, saying. {{«}, {«,*}}. d. thai {{a}, Now worth member is the set {«, b}. Shocking as given by the theorem immediately following the definition and there one {aJ>}} contains just [a,b}} = {{c], two members, element of these two members of { {c.d}}. {a} and {a.b}; and a {a}, {a, b} }. Similarly, c is is the only the unique 3, Appendix. common member of both members of { {<?}, {c, d) }. Ordered Pairs = Therefore a c. We 55 there- fore have {a,b}} {{a}, and only the proof that b Case 1. = b which implies b 2. other. It d remains. a. In this case, {a,b} member, namely, Case = ^ that a. {a}. d — — = {{a}, It is a — In this case, b ^ a; so b = convenient to distinguish 2 cases. { {a}, {a, b] true of {{a}, {a, } really has only d}}, so {a,d\ = one {a}, b. is in one member of {a}, {a, b} in one member of {a}, must therefore be true that b is This can happen only if b is d, but b {a,d}}, {a}, so the set The same must be in the other. or b = d. | { } { in {a, d}, but b is not in but not in the {a, d) {a}; } but not thus b = a CHAPTER GRAPHS Mention the numbers real probably form in mathematician and the image of a to a her mind, quite involuntarily. banish nor too eagerly embrace And most likely straight line will she will neither mental picture of the real numbers. "Geomet- this allow her to interpret statements about numbers in terms of this 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. ric intuition" will picture, You _, i | ,_ | | -1 \ FIGUR1 2 1 3 method of considering are probably already familiar with the conventional the straight line as a picture of the real numbers, of associating to each real i.e., number a point on a line. To do this (Figure 1) we pick, arbitrarily, a point which we label 0, and a point to the right, which we label 1. The point twice as far to | the right is is same distance from labeled 2, the point the — 1, labeled etc. With arrangement, this to a lies to the left of the point numbers, such as ^, in the a < b, corresponding to obvious way. numbers also somehow can be drawn as a point on the irrational if fit It is to but to the 1, of 0, left then the point corresponding We b. can also draw rational usually taken for granted that the into this scheme, so that every real number We will not make too much fuss about method of "drawing" numbers is intended certain abstract ideas, and our proofs will never line. justifying this assumption, since this method of picturing on these pictures (although we solely as a rely Because explain a proof). inessential role, geometric numbers — thus a number this will frequently use geometric picture plays such a prominent, albeit terminology frequently employed is sometimes called a is a picture to suggest or help point, and R when speaking of is often called the real line. The number \a—b\ has a simple interpretation in terms of this geometric picture: the distance it is between a and end point and b occurrence — a\ < \x than less s s. MCI RE — e a a +£ (Figure pictured as the collection of points whose distance from a of points set as the points —e + e, corresponding to numbers x with a — is the "interval" from a to a is may also < x < a+s which s 2). which correspond called the names for them. open interval from is interval. Note that if to intervals arise so frequently that The a to it is desir- denoted by (a, b) This notation naturally creates some set {x b. : a <x < b) is also used to denote a pair of numbers, but in context always clear (or can easily be 56 satisfy may be ambiguity, since (a, b) an of numbers x which This able to have special and This means, to choose an example whose frequent as the other. Sets of numbers 2 length of the line segment which has a as one justifies special consideration, that the set be described a b, the a > made b, clear) then (a,b) whether one = 0, the set is it is talking about a pair or with no elements; in prac- tice, however, is it almost always assumed whenever an implicitly otherwise), that the open (explicitly if interval ia, b) is 4. Graphs 57 one has been careful, and mentioned, the number a the closed interval [a, b] interval (a, b) is than less a b. ") The the interval (—00, a) set {x from a the interval (a, 00) to b. used for a = : < x < a b) This symbol b, also. The is is denoted by and [a, b] is closed interval it is sometimes and [a, b] are shown called the usually reserved for the case a < b, usual pictures for the intervals (a, b) but in Figure 3; since 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, 00), the interval (—00, a] : the interval [a, 00) FKJURE while the set {x : x > a} is denoted by [a, 00); the sets (—00, a) and (—00, a] are this point a standard warning must be issued: the symbols 00 and —00, though usually read "infinity" and "minus infinity," art purely suggestive; there is no number "00" which satisfies 00 > a for all numbers a. While the symbols 00 and —00 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 (—00, 00). defined similarly. At 3 (a,b) (0, b) 1 -f Of even greater interest to us than the method of drawing numbers is a method 1 of drawing pairs of numbers. This procedure, probably also familiar to you, re- 1 .(l.D (-1.1). 1 To quires a "coordinate system," two straight lines intersecting at right angles. | 1 , (0, 0) distinguish these straight lines, (a, 0) axis. •(1.-1) (-I.-!)' (More prosaic terminology, such horizontal axis, as the "first" and one the and "second" axes, is vertical probably blackboards, in the same way, so that "horizontal" and "vertical" are more we can Each of the two axes could be labeled with real numbers, but also label points on the horizontal axis with pairs (a 0) and points on the vertical axis with pairs (0, /?), so that the intersection of the two axes, the "origin" of the coordinate system, is labeled (0, 0). Any pair ia, 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 4 = 1 , coordinates, respectively, fix) Our tion 5 is real concern, let us recall, is a in this way. method of drawing functions. Since a func- numbers, we can draw a function by drawing The drawing obtained in this way is called the graph of / contains all the points corresponding to pairs ix, fix)). Since most functions contain infinitely many pairs, drawing the graph promises to be a laborious undertaking, but, in fact, many graph of the fix)=x of the point determined just a collection of pairs of each of the pairs fix) one the descriptive.) fix) FIGURE call preferable from a logical point of view, but most people hold their books, or at least their FIGURE we in the function. function. In other words, the functions have graphs which are quite easy to draw. Not surprisingly, the simplest functions of have the simplest graphs. is I I . I RE 6 all, the constant functions fix) easy to see that the graph of the function fix) a straight line parallel to the horizontal axis, at distance c from The I It is functions fix) through (0, 0), — ex also have particularly simple graphs as in Figure 6. A proof of this fact is it (Figure = c, =c 5). straight lines indicated in Figure 7: 58 Foundations number Let x be some not equal to 0, and L be let the straight line which passes through the origin O, 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 BO triangle A' (x, is similar to the triangle ex) ABO, A'B' AB OB' OB when thus = c; precisely the condition that A' corresponds to the pair (y, cy), this is lies on The argument has implicitly assumed that c > 0, The number c, which measures the the graph of /. other cases are treated easily enough. the sides of the triangles appearing in the proof, FIGURE and a line, 7 line parallel to this line is also said to that A' i.e., but the ratio of called the slope of the straight is have slope c. This demonstration has neither been labeled nor treated as a formal proof. we Indeed, a rigorous demonstration would necessitate a digression which not at prepared to all The follow. geometric and algebraic concepts would stated assumption) that the points to the real d-b geometry Now the ia,b) length c — a Aside from numbers. as precisely as detailed we on a this, first require a real proof (or a precisely straight line it correspond would be necessary an exact way in to develop plane intend to develop the properties of real numbers. development of plane geometry no means a prerequisite a beautiful subject, but is for the study of calculus. We shall use it is and FIGURE 8 define the plane to be the set of all pairs of real to define straight lines as certain collections of pairs, including, the collections {(x ex) , geometry with definition is all x 3. real number}. To provide define and (c, d) are two points in the plane, the distance between (a, b) and (c, d) to be If the motivation for this definition —with others, constructed the structure of geometry studied in high school, one y/(a-c) 2 explanation this artificially is this definition the + (b-d) 2 not clear, it is numbers, among required. If (a, b) numbers, we real : by geometric pictures only as an aid to intuition; for our purposes (and for most of mathematics) perfectly satisfactory to are rigorous proof of any statement connecting i.e., more pairs of . Figure 8 should serve as adequate Pythagorean theorem has been built into our geometry* Reverting once more to our informal geometric picture, (Figure 9) that the graph of the function with slope f{x) = ex c, + passing through the point f(x) (0, d). = For ex this it + d is is not hard to see a straight line reason, the functions Simple as they are, linear funcand yon should feel comfortable working with them. The a typical problem whose solution should not cause any trouble. Given points (a,b) and (c,d), find the linear function / whose graph goes b) and (c, d). This amounts to saying that f(a) — b and /(c) = d. II d are called linear functions. tions occur frequently, following two FIG 1 RE 9 distinct through * is The (a, might object to this definition on the grounds that nonncgalive numbers have square roots. This objection is really unanswerable at the moment the fastidious reader are nol yet known del mil ion will just to have — to be accepted with reservations, until this little point is settled. . 4. / is = ax + be of the form f(x) to aa ac = a therefore {d b)/(c d fix) c a formula most easily Of course, and a) ft b — —x+b + + = = = ft fi - b b, d\ b)/(c [{d a c possible only is account only for the straight lines — the "point-slope if ^ a which are not c; conclusion is vertical line function cannot contain (a, b) and (a, c) • r\ J FIGURE 10 if (a, b) b / (see Problem 6). c. and all; the in fact, the The graph of a same vertical line. This two points on the same — (a, c) and, by definition, a Conversely, if a set of points in on the same vertical line, then it is surely the graph of a function. Thus, the first two sets in Figure 10 are not graphs of functions and the last two are; notice that the fourth is the graph of a function whose domain is not all of R, since some vertical lines have no points on them at all. no two points the plane has the property that draw some of the a picture like pairs in /, Figure i.e., is 1 1 '(2,4) '(§-!) (-1,1). (1.1) ,'.. • <v*> (0,0) FKillRK I I > lie If we perhaps the function f(x) — x 2 obtain some of the pairs of the form (x, x ), we After the linear functions the simplest (d) at on immediate from the definition of a function correspond to pairs of the form form" the graphs of linear functions parallel to the vertical axis. function can never contain even two distinct points (b) a graph of any function vertical straight lines are not the a)]a, so b d -b —a = — (x -a) + b, d remembered by using this solution 59 we must have then /3, Graphs . 60 Foundations It is not hard to convince yourself that the one shown is just a the pairs (x, x 2 ) lie along a curve known as a parabola. drawing on paper, made (in this case) with in Figure 12; this Since a graph all curve like is printer's ink, 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 the question "Is this < x < v, then x 2 not the case in plotting first (a) . < y2 It is so the graph should be higher at y than at x, which also easy to see, simply many pairs or a "corner" as in , (x, x ), that the by drawing a very accurate graph, graph cannot have a large "jump" In order to prove these assertions, however, (c). is we as in (b) first need means for a function not to have a "jump" 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 definitions critically Later these definitions may be compared with the ones mathematicians have agreed upon. If they compare favorably, you are certainly to say, in a mathematical way, what it or "corner"; these ideas already involve FIGURE 12 be congratulated! to The = numbers n, are sometimes called power functions. Their graphs are most easily compared as in Figure 14, by functions fix) drawing several The power x", for various natural at once. functions are only special cases of polynomial functions, introduced in the previous chapter. Two particular polynomial functions are fix) (b) I I ( , I l< E 14 = x graphed in Figure 15, while Figure 16 /(x) = x2 +X meant is to give a general idea of the in the case a n > a n x" +a„_\x" is / will have at of peaks and valleys for example, have we will at may 1 , "peaks" or "valleys" \ most one valley). much is a point like (y, smaller (the (a "peak" f(y)). power The functions, Although these assertions are easy will also Many will be very to make, easy). 7 illustrates the graphs of several rational functions. tions exhibit The rational func- even greater variety than the polynomial functions, but their behavior be easy to analyze once we can interesting graphs can be constructed by "piecing together" the graphs of functions already studied. lines. a not even contemplate giving proofs until Part III (once the powerful meth- Figure The function / with The graph this / graph use the derivative, the basic tool of Part in Figure 18 is made up III. entirely of straight satisfies n+\ n n+\ f f{x)=\. 15 most n — actually be ods of Part III are available, the proofs FIGURE \- a point like (x, f(x)) in Figure 16, while a "valley" number 3x H 0. In general, the graph of = x3 - graph of the polynomial function fix) f(x) 61 Graphs 4. > 1. and is a linear function on each interval [\/(n + 1), 1/n] and [—1/n, —\/{n + 1)]. (The number is not in the domain of /.) Of course, one can write out an explicit formula for f(x), when x is in [l/(n + 1), 1/n]; this is a good exercise in the use of linear functions, and will also convince you that a picture is worth a thousand words. FIGURE 16 62 Foundations /(*) = /(*) /(*) = = l (b) (a) /(*) (d) FIGURE 17 FIGURK 18 = l+x 2 . 4. It is this actually possible to define, in a same property of oscillating which we angle of it The graph Now ure 20. Chapter means an angle 0, As by using the usual, "all the we sine function, are using radian way around" a circle, an etc. of the sine function is shown in Figure 19. 19 consider the function fix) To draw this f(x) graph = f{x)= it .v = for x = f(x) Notice that k;i RE 20 = -1 when x is for a- = The graph sin l/x. helps to for 1 — 1 1 t: 1 t7T 1 is shown in Fig- + 1 Z7T ^JT 1 , large, so that \/x / 1 — — of observe that first ^77" I near 15. an angle halfway around (or 180° in layman's terms), an angle of 7r/2 a right angle, FIGURE Itc 63 simpler way, a function which exhibits infinitely often will discuss in detail in measure, so an angle of much Graphs 3 is + A-JT 1 ' 3. small, fix) is also small; when x is 64 Foundations FIGURE 21 "large negative," that although fix) An this < is, when |.v| large for negative x, again f{x) is interesting modification of this function function between 1 x and —x. is and sketched in Figure 2 1 — 1, . the function fix) Since sin = The behavior of the graph x sin is 1 f(x) 2 2 close to 0, jx = x sin l/x. The graph oscillates infinitely often of near l/x oscillates infinitely often between for x large or large negative /(*) FIGURE is 0. = 9 xL • sin 1 is harder to Graphs 4. analyze. Since sin seems be no to /x 1 telling another question that is getting close to 0, while x what the product is is will do. It getting larger and larger, there possible to decide, but this is The graph of f(x) = x best deferred to Part III. 65 sin is i/x has also been illustrated (Figure 22). For these infinitely oscillating functions, The be really "accurate." part near functions (which is is clear that the to graph cannot hope show part of it, and the interesting part). Actually, it is easy to find to leave out the much simpler whose graphs cannot be "accurately" drawn. The graphs of m-\ < x > x x : .2 1 and < X > X g(x) 1 1 1 can only be distinguished by some convention similar to that used for open and (b) FIGURE it is we can do best closed intervals (Figure 23). 2 3 Out last example is a function whose graph x irrational 1, x x irrational 24 The graph / must contain infinitely many points on the many points on a line parallel to the horizontal of also infinitely (x,y) rational. x rational = 0, FIGURE spectacularly nondrawable: 0, fix) /(*) is horizontal axis axis, but it and 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 ical on the line corresponding to irrational x reason behind this convention, but it . (There is actually a mathemat- depends on some sophisticated ideas, introduced in Problems 21-5 and 21-6.) The peculiarities exhibited FIGURE 2 5 some of by some functions are so engrossing that it is easy and most important, subsets of the plane, which are not the graphs of functions. The most important example of all is the circle. A circle with center (a, b) and radius r > contains, by definition, all the points (x, y) whose distance from (a, b) is equal to r. The circle thus consists (Figure 25) to forget the simplest, of all points (x,y) with ^ a) 2 + (v -b) 2 = or (jc - a) 2 + (y - b) 2 r 66 Foundations The circle with center called the unit is A (0,0) and radius close relative of the circle the sum of whose distances we same, foci are the 1 often regarded as a sort of standard copy, , circle. obtain a circle.) defined as the set of points, is is a constant. (When the (c, 0), +y 2 + V(x-(-c)) 2 or ^+ (x c) 2 2 + y two for convenience, the focus points are If, and the sum of the distances is taken to be la some algebra), then (x, y) is on the ellipse if and only if taken to be (— c, 0) and factor 2 simplifies the ellipse. This is from two "focus" points - y/(x =2a- 2 c) + y2 = - y/(x c) 2 + (the la 2 y or 2 + lex + c 2 + y2 = — 4a 4a 2 - \/(x c) 2 + 2 y + x2 - lex + c 2 + y2 or 4(cx - a 2 = -4a V(x ) c) 2 + y2 or L L ,2„2 c x — Icxa^ + a = a (x — lex +c + y ) or / 2 — (c 2\ 2 )jc fl — 2 2 a y~ = 2/2 — a 2\ ) a (c or 11 — a- cr This 2 cz = 1. usually written simply is - + ^=1 bL aL = where b a 2 — c 2 > 0). 2 va — A c2 (since we must picture of an ellipse sects the horizontal axis when y = = FKil'RK 26 is clearly choose shown 0, so that 1. a > in Figure 26. ±a, it follows that The ellipse inter- c, and it when x = intersects the vertical axis 67 Graphs 4. 0, so that „2 b2 The hyperbola is = = ±b. defined analogously, except that and the constant we require the Choosing the points (— c, the two distances to be constant. again, y 1, difference as 2a, we 0) and difference (c, of once 0) obtain, as the condition that (x, y) be on the hyperbola, Ax + cy + y \^ which may 2 —v (x — cy + a-2 — cA 27 In this case, however, /? = vc — a 2 The picture 2 , is then we must (x, y) shown ferent orders. = ±a, but It is it is (x, y) The hyperbola two sets (Problem = 1. clearly choose c 2 2 a2 b2 from (— c, if > a, so that and only — a c2 < 0. If if two pieces, because the difference 0) and (c, 0) may be taken in two dif- intersects the horizontal axis never intersects the vertical interesting to 2 on the hyperbola compare when — l/x. y = 0, so that axis. (Figure 28) the hyperbola with a the graph of the function f(x) the ±2a, in Figure 27. It contains between the distances of x = be simplified to a 2A FIGURE y The drawings = b = v2 and look quite similar, and are actually identical, except for a rotation through an angle of 7r/4 23). Clearly no rotation of the plane will change circles or ellipses into the graphs of functions. Nevertheless, the study of these important geometric figures be reduced to the study of functions. fi guR 1 . : s Ellipses, for example, are can often made up of the 68 Foundations graphs of two functions, /w = Wi-uV), —a < x < a and -bJT~^ (x 2 /a 2 ), g(x) Of course, there are we can example, many <x < -a other pairs of functions with a. this same property. For take b fix) \/l - -W 2 (x /a - 1 2 < x < ), 2 (x /a 2 ), a -a<x<0 and -bf\ = g(x) 2 < I (x /a yrr (x 2 /a We 2\ < x < ), 2 -a < x < , a 0. could also choose f(x) = — b v 1 —b v I (x"/a"), — 1 {x" ja ), x rational, —a < x — irrational, a x < a < x < a and Six) —b v = 1 — 2 b\[\ But (x J a (x /a ), 2 x rational, —a < x irrational, ), a x < <x < a. these other pairs necessarily involve unreasonable functions which all around. A proof, or even a precise statement of this fact, is jump too difficult at present. 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 reasonable functions. by no means easy, and a great deal of this successive attempts to impose more and more conditions mathematical definition of this concept book may be viewed as that a "reasonable" function we will take time out to ask which deserve to A must if is satisfy. we have be called reasonable. As we define some of these really The succeeded conditions, in isolating the functions answer, unfortunately, will always be "no," or at best, a qualified "yes." PROBLEMS 1 . Indicate Also also on a straight line the set of name each set, using need the U sign). (iii) |.v-3|<l. |jc-3| < 1. — a\ < e. Cw) \x (i) (ii) |.v 2 -\\<l all x satisfying the following conditions. the notation for intervals (in some cases you will . 5 M = < 1 1 (vi) +x z 1 2. 2 + (vii) x (viii) (*+ > 1 a (give an answer in terms of a, distinguishing various cases). 2. 1)(jc - -2) > l)(x (a) is a very useful consider the interval First = then x number (b) Now < tb for t? What prove that < t some / x is t < 1 Prove that 0. What . is in [a, b], then x of the interval Prove, conversely, that (d) The = (1 — + t)a if < t < 1, [a, b]? < < t the set of x is in [0, b], tb for then (1 some + t t(b with — a). What is the point 1/3 of the — + tb t)a points of the open interval (a, b) are those of the for if the significance of the Hint: This expression can also be written as a 1. (c) Draw < > b [0, b], for with points of the closed interval [a, b] the mid-point of the interval [0, b]? is if What is the midpoint way from a to b? 3. 0. way of describing the (where we assume, as usual, that a < b). There 69 Graphs 4. is in [a, b]. form (1 — t)a + tb 1 all points (x, y) satisfying the following conditions. (In most cases your picture will be a sizable portion of a plane, not just a line or curve.) (i) x > (ii) x +a > (iii) (iv) (v) (vi) (vii) y < y < 4. (") (iii) (iv) (v) (vi) (vii) (viii) x^-. + is an integer. v (x2 < ;t Draw (i) + b. — y\ < 1. \x + y\ < 1. x + y is an integer. x (x) y x-. \x (viii) (ix) y. l) y 2 + (_v-2) < x the set of |1 Ivl \ jc . < 1. 4 . all points (x, y) satisfying the following conditions: = 1. 1*1 - \y\ = l. \X- 1| =\y- 1|. - x = \y- 1|. 2 x + y 2 = 0. xy = 0. 2 - 2x + v 2 = 4. T 2 x = y 1x1+ 2 70 Foundations 5. Draw the set of all points (x, y) satisfying the following conditions: x (i) y =r 2 (in) x fiv) x Hint: (a) = -> a1 1. b< = = \y\. sin v. You already know Show that is far — mix — (b) is + b. This formula, y are interchanged. c, show m known the graph of the is as the "point-slope the equivalent expression f(x) — immediately clear from the point-slope form that the it is m, and that the value of / / For a a) more convenient than mx + (b — ma); slope when x and the straight line through (a, b) with slope function f(x) form" the answers at a is b. that the straight line through (a,b) and (c, d) is the graph of the function /(*)= — c When (x-a) + b. a = mx + are the graphs of f(x) b and g(x) m'x + b' parallel straight lines? 7. (a) set of all (x, y) satisfying vertical one). Hint: First (b) (1,/n) Show conversely that every straight described as the 8. (a) A and B not both 0, show that the Ax + By + C = is a straight line (possibly a decide when a vertical straight line is described. For any numbers A, B, and C, with set of g(x) are perpendicular if K. i RE in Figure 29. the lines intersect at the origin, as 29 (b) Prove that the two straight good (Why the squares of the lengths this special case, is lines consisting Ax + By + C = 0. +C= 0, A'x are perpendicular (a) if + B'y and only if + A A' of all ix, y) satisfying the BB' = 0. Prove, using Problem 1-19, that >/(*i +y\) 2 + (x2 + yi) 2 < Vxi 2 where as the general case?) ditions 9. 0. = mx + b, = nx + c, mn = — 1, by computing of the sides of the triangle I Ax + By + C = Prove that the graphs of the functions f(x) (1,») including vertical ones, can be line, y) satisfying all (x, +x 2 2 + x/v i + yi con- . Prove that (b) V 2 - (*3 ai ) + 2 - (>'3 < V vi) (x2 - x\) 2 + Interpret this inequality geometrically When ity"). does (it is 2 - (y 2 + 10. 71 Graphs 4. v yi) (-V3 - 2 a- 2 ) + - (V3 2 yz) - called the "triangle inequal- inequality hold? strict Sketch the graphs of the following functions, plotting enough points to get good idea of the general appearance. a how many a reasonable decision meant to show that a little thought is (Part of the problem is to make "enough"; the queries posed below are will often be more valuable than hundreds of individual points.) f(x)=x-\ — does the graph does 1 1 it (What happens . lie suffice to in relation to the graph of the consider only positive x at = x in fix) — X iv fix) = x (ii) (iii) identify function? Why first?) xz 2 1 — ~J1' even. is odd. is nonnegative. = Where 1 1 is f(x) (iv) x? - Describe the general features of the graph of / / / for large 1 fix) (i) and for x near 0, fix + a) for all x (a / if function with this property is called peri- odic, with period a. 12. Graph do the functions fix) this, positive — Xfx for m= 1,2,3, (There 4. is an easy way using Figure 14. Be sure to remember, however, that %fx mth root of x when m is means to the even; you should also note that there will be an important difference between the graphs when m is even and when /;/ is odd.) 13. (a) (b) 14. and f(x) = x sin.v| and f(x) = sin x. (There is an important difference between the graphs, which we cannot yet even describe rigorously. See if you can discover what it is; part (a) is meant to be a clue.) Graph f(x) Graph fix) — = \x\ . | Describe the graph of g in terms of the graph of / if 72 Foundations ij g(x) ii) g(x) = f(x)+c. = f(x + c). IV) g(x) = f{cx). v) g(x) = /(l/x). g(x) = f(\x\). ' . 15. { ' (It is make a mistake here.) = (Distinguish the cases c ' vn) g(jc>=|/(jc)|. viii) g(x) = ix) g(x) = min(/, 0). x) gO) = max(/, Draw easy to max(/, 0, c > 0, c < 0.) 0). 1). the graph of /(x) = ax" + Z?x + c. Use the methods of Prob- Hint: lem 1-18." 16. Suppose A and C that are not both 0. Show that Ax 2 + Bx + Cy 2 + Dy is either a parabola, lines an ellipse, +E= or an hyperbola (or a "degenerate case": two [either intersecting or parallel], one line, a point, or 0). Hint: The and the case A = is just a minor variant. Now consider separately the cases where A and B are both positive or negative, and where one is positive while the other is negative. When do case C = the set of all (x, y) satisfying we have 17. Problem 15, a circle? The symbol all denotes the largest integer which [jc] 2 and [—0.9] (they are essentially is = [—1] = — 1. Draw and quite interesting, is < x. Thus, [2.1] = [2] = the graph of the following functions several will reappear frequently in other problems). f(x) = Jx-[x]. = [x] + Jx~- /(*) = /(*) = fix) iv M [x]. n 1 I 18. Graph (i) the following functions. fix) — {x}, where {x} is defined to be the distance from x to the nearest integer. (ii) f{x) = (iii) fix) (iv) fix) = = (v) /(.v) = {2x}. {x) + \{2x). {4x). {.v} + i{2.v} + i{4.v}. 4. Many ber. may functions Although we Graphs 73 be described in terms of the decimal expansion of a num- not be in a position to describe infinite decimals rigorously will Chapter 23, your intuitive notion of infinite decimals should suffice to carry you through the following problem, and others which occur before Chapter 23. There is one ambiguity about infinite decimals which must be eliminated: Every until decimal ending in a string of 1 s .23999 19. . . = . 1.24000 9's We ). . . equal to another ending in a string of 0's is will always use the one ending in is fix) (ii) fix) (iv) (v) (vi) = = = fix) complete number in the decimal expansion of x. 2nd number in the decimal expansion of x. number of 7's in the decimal expansion of x the 1st the the is finite, and = if fix) and (a usually out of the question). (i) (iii) (e.g., 9's. Describe as best you can the graphs of the following functions picture *20. . number if this otherwise. the number of 7's in the decimal expansion of x is finite, otherwise, 1 — the number obtained by replacing all digits in the decimal fix) expansion of x which come after the first 7 (if any) by 0. if 1 never appears in the decimal expansion of x, and n fix) = first appears in the nth place. if 1 Let /(*) = 0, x irrational 1 x =P q rational in lowest terms. q lowest terms if p and q are integers with no common 0). Draw the graph of / as well as you can (don't sprinkle points randomly on the paper; consider first the rational numbers with q = 2, number p/q factor, and q > (A 21. is in then those with q = (a) The 3, etc.). on the graph of fix) points Prove that each such point (b) from P and L ^22. is equidistant from the point (0, i) ^ fi, show = Show that the square of the distance this set if 2 y + Using Problem 1-18 the distance from (c, 1) +jc(- part (a).) -2md - to find the from 2c) (c, (jc, form fix) - minimum (c, d\/y/m 2 + d) to ix, + d 2 + c2 x 2 ). and the = (a, fi) v) equidistant = ax 2 + bx+c. mx) is . of these numbers, show that d) to the graph of fix) Find the distance from and a point P points (jc, /J? \cm this case to y, all the graph of a function of the is 2 = that the set of What x im (b) x 2 are the ones of the form graph of gix) = —\. (See Figure 30.) Given a horizontal line L, the graph of gix) not on L, so that y l-KH'RE 30 is — = mx is 1. d) to the graph of fix) mx + b. (Reduce . 74 Foundations *23. (a) Using Problem 22, show that the numbers x' and y' indicated in Fig- ure 3 1 are given by x —=y, = —=x + V2 V2 l l s/2 s/l y (b) FIGURE 31 Show that the set of all {x, y) with fx'/v2 as the set of all (jc, y) with xy = 1 ) — (y'/v 2 ) = 1 is the same 4, APPENDIX v a point in the plane; in other words, v is = v For convenience, we (l)\, a pair of numbers is V2). convention that subscripts indicate the will use this second pairs of a point that has been described by a single w and mention the points while z z, it be understood that will w Thus, letter. is first and if we the pair (w\, W2), the pair (z\, zi). is Instead of the actual pair of from the origin O numbers to this point (Figure Of course, in the plane. {v\,v 2 ) 75 Vectors 1. VECTORS 1. Suppose that Appendix we (v\, V2), 1), and we often picture v as an arrow refer to these arrows as vectors new we've haven't really said anything yet, we've simply introduced an alternate term for a point of the plane, and another mental picture. The new terminology real point of the some new V Then we can define a new Notice that sign the left is +W= on the is to do (V\ defined operation, point of the plane) v + + W\, Vl + if by the equation U>2). right side of this equation are numbers, and the On + the other hand, the in the sign on plane wasn't defined, (1) as a definition. might want + w, but there's really no need to some new symbol to use or perhaps on insist possibility of confusion, so Of course, W2). (U>1, for this newly like v we could just points) in the plane, (i.e., new: previously, the sum of two points A very fussy mathematician no new vector (a vectors W = (V\, V2), and we've simply used equation there's emphasize that we are going our usual addition of numbers. just side the letters all we have two = V (1) + to things with points in the plane. For example, suppose that FIGURE is this; since v we might w, v +w as well hasn't been defined before, keep the notation simple. any one can make new notation; for example, since as well have defined v other equally weird formula. The +w + as (v\ real question is, w\ 1U2, does our ^2 it's +w new our 2 \ ), definition, or by some construction have any particular significance? Figure 2 shows two vectors v and w, as well as the point (Ul + W],V2 + W2) (l>l . (l»2 + U>2) - V2 (Vl + Wl) we have simply indicated an arrow. Note that is it as indicated in Figure 2, this slope V\ (t>2 and 2 this, in the usual way, of course, is other words, the line — +Wi) - + W2) l>2 U'2 V\ W\ is just ' the slope of our vector w, from the origin L is without drawing easy to compute the slope of the line L between v and (v\ FIGURE W2), which, for the moment, our new point: - + W\, V2 + parallel to w. O to (w\, W2). In . 76 Foundations M between (w\, W2) Similarly, the slope of the line + Wj) — (V\ + U)\) - (t>2 which v +w M the slope of the vector u; so is lies on the parallelogram having an arrow (Figure 3), it Vl)i t>2 V2 V[ is ' is In short, the parallel to v. w and v and our new point as sides. new point When we draw v + w as points along the diagonal of this parallelogram. In physics, and the sum of two vectors represents the forces are applied simultaneously to the same vectors are used to symbolize forces, resultant force when two different object. FIGURE Figure 4 shows another 3 an arrow +w by Many first + numbers for ordinary +W=W+ also hold for this is the same it w and spanned by as the parallelogram W\, + 1)2 V\ — Wi) + + Similarly, unraveling definitions, [v Figure 5 indicates a The origin O — + w] + (U)\ if we (0, 0) is we V\, we Wi + + for V v v + t>2- + z] [w identity," =v+O = v, (— ui, -V2), ( V = O. also define v =w+ (— v), exactly as with numbers; equivalently, W— V = V2), V[, + w + z. + — v) — — V + w— 5 also easily numbers: have we can It is find the "associative law" +z = v W\ W2+ define also Naturally U>2 an "additive v then = = W\ method of finding O+ and v. states that Vi G UR E new + for V, and thus simply depends on the commutative law FI denote obvious from the geometric picture, since the parallelogram spanned by v and (V\ -f 4 to example, the "commutative law" analytically, since FIGURE "w" is V w If we use and having the same length, but starting at v instead of at the vector from O to the final endpoint; thus we get to following v, and then following w. +w of the properties of vectors. For is sum v + w. the parallel to w, the origin, then v v way of visualizing (W\ — Vl, W2 — V2). checked 4, Just as with numbers, our definition of v w— v + — (w "w — Figure 6(a) shows v and an arrow w— v simply Appendix means Vectors 1. that it 77 satisfies w. v) v" that is parallel to w— v but that starts As we established with Figure 4, the vector from the origin to the endpoint of this arrow is just v + (w — v) — w (Figure 6(b)). In other words, we can picture w — v geometrically as the arrow that goes from itoio (except that it must then be moved back to the origin). There is also a way of multiplying a number by a vector: For a number a and at the endpoint of a vector v = v. (y\, V2), we define a w— v = v (av\, CIV2) (We sometimes simply write av instead of a • v; of course, it is then especially important to remember that v denotes a vector, rather than a number.) vector a direction v points in the when You can FIGURE a easily < same (Figure when direction as v a > and The in the opposite 7). check the following formulas: a 6 • (b v) • = (ab) v, v = v, 0v = O, — 1 v = —v. 1 • • Notice that not defined a we have only defined a product of a number and a vector, we have way of 'multiplying' two vectors, to get another vector.* However, there are various ways of 'multiplying' vectors to get numbers, which are explored in the following problems. PROBLEMS Given a point v of the plane, 1. origin through FIGURE an angle of (Figure 7 Show The aim of this problem is to obtain (b) that = = (cos#, sin 6), Explain why we have R e (\, R 9 (0, 0) 1) [we should really write Re((l, 0)), + w) = Re(a f Re(v) (c) At- Now etc.] (-sin0,cos0). Re (v show that for Re(x, y) = w) = any point (x cosO — + Re(v) a Ro(w), Ro(w). (x, y) we have y sin#, xsmO + y cosO). jump to Chapter 25, you'll find that there is an important way of defining a product, but something very special for the plane it doesn't work for vectors in 3-space, for example, even though the other constructions do. *If you this 8 8). a formula for Rg, with minimal calculation. (a) FIGURE Re(v) be the result of rotating v around the let is — 78 Foundations (d) Use Given v another solution to Problem 4-23. this result to give and w, we define the number V this is w — . + V\W\ V2W2', and w often called the 'dot product' or 'scalar product' of v ('scalar' being a rather old-fashioned word for a number, as opposed to a vector). (a) Given w a vector v, find such that v w — • Now describe 0. the set of all such vectors w. (b) Show that w=W V + z) = v W + V V and (w ' • (v Notice that the product • • last Show = w) (a that v v • > 0, and will w = ' be v • w). (a of a • three products: number and the dot a vector; and that v — v • only when v — O Hence we can . as ||u|| = = only for v y/v V, • What O. the geometric interpretation of is norm? Prove that \\v + and that equality holds some number a > 0. (e) z of two numbers. • ||y|| which v) of two vectors; the product define the norm (d) • of these equations involves the ordinary product the V - that a (c) • Show < w\\ if || + i>|| and only ||w||, v if — or w = or w = a v for that V ' w — + ||l' w\\" — v || U)\\~ 4 3. (a) Let Re be rotation by an angle of 6 (Problem Re(v) (b) Let e let = w = with the (1, 0) axis is is 1 w = 1). • w = ii|| || the angle between v || it • || 1 that first axis, and makes an angle of Calculate that COS0, and w. that pointing along the cos#. • Show w. that in general v where . • a vector of length (compare Problem e Conclude v be the vector of length (cos#, sin 0); this first = Re(w) • 1). 4, 4. Given two vectors v and w, we'd expect the coordinates v\, i^i, u>2, t>2, Appendix 1. Vectors 79 have a simple formula, involving to for the area of the parallelogram they span. Figure 9 indicates a strategy for finding such a formula: since the triangle with vertices the problem u;, + A, v u> is an easier one where one to we can reduce congruent to the triangle 05u, side of the parallelogram lies along the horizontal axis: The (a) L line Conclude and passes through v that the point B parallel to w, so has slope wj/u)]. is has coordinate - V\W2 W\V2 W2 and that the parallelogram therefore has area det(i\ FIGURE w) = — v\u>2 W[V2- This formula, which defines the determinant det, certainly seems to be simple 9 enough, but we all, can't really it clearly be true that det(u, w) always gives the area. After have det(it), v) so sometimes det tion" made all coordinate, etc.) be negative! Indeed, will sorts = — det(u, of assumptions (that Nevertheless, it seems w), easy to see that our "deriva- it is was u>2 B had a positive positive, that likely that det(u, w) is ± the area; the next problem gives an independent proof. (a) If v points along the positive horizontal area of the parallelogram spanned by axis (u>2 (b) If Re is > 0), i; axis, and the negative of the area rotation by an angle of det(/?6iu, for for w (Problem Row) = show w and 1), w that det(u, w) is the above the horizontal below show this axis. that det(i>, w). Conclude that det(i\ w) is the area of the parallelogram spanned by v and w when the rotation from v to w is counterclockwise, and the negative of the area Show and when • sin# 1. det(u, w + z) = det(u, w) det(u + w, z) = det(i>, z) 10 + det(i>, z) + det(u;, z) that = det(a Using the method of Problem 3, w) = det(i', FIGURE clockwise. that a det(u, w) u;|| it is which is also obvious v, w) show || i»|| = det(u, a w). that ||u)|| • sin 0, from the geometric interpretation (Figure 10). 80 Foundations APPENDIX THE CONIC SECTIONS 2. Although we will be concerned almost exclusively with figures defined formally as the set of all pairs of real numbers, in to consider three-dimensional space, real this which we can describe in the plane, Appendix we want in terms of triples of numbers, using a "three-dimensional coordinate system," consisting of three straight lines intersecting at right angles (Figure now mutate to two axes in a horizontal plane, 1). Our horizontal and vertical axes with the third axis perpendicular to both. (1,0,0) One of the simplest subsets of cone around the third illustrated in Figure 2; this FIGURE of slope C say, this may three-dimensional space is the (infinite) cone be produced by rotating a "generating line," axis. 1 slope FIGURE 2 For any given first two coordinates x and + plane has distance v x (x, y, z) (1) We C is y from the on the cone if origin, and thus and only if z can descend from these three-dimensional dimensional one by asking what happens plane y, the point (x, y, 0) in the horizontal P (Figure = ±Cvx" + vistas to the when we y more intersect this . familiar two- cone with some 3). cone plane FIGURE If the P 3 plane is parallel to the horizontal plane, there's certainly intersection is just a circle. Otherwise, the plane P no mystery — the intersects the horizontal plane Appendix 4, We in a straight line. can make things a everything around the vertical axis so that from the plane of the paper, while the The Conic 2. simpler for ourselves lot axis —C (Figure 4) is its view-point the cone If this line itself L happens then equation (1) simply appears as two straight to be vertical, consisting 2 which ^2 2 = r 2 2 — C C a y as the collection of and form we we see from this all axes; lines. points (a,z) for and the plane some a, consists of where , M is third axes, the line first points of the all (x,Mx (x, y, z) Combining (1) and the plane if and only is (2), P in the plane we L can be described + B), the slope of L. For an arbitrary point (x, y, z) (2) if and only see that (x, y, z) is if z follows that it = Mx + B. in the intersection of the cone and if Mx + B = ±C V^ + r (*) Now we first axis, (Figure axes is have to choose coordinate axes in the plane P. measuring distances from the intersection 5); original for the second second axis we just choose If the first coordinate axis. x, then the can be written first in the coordinate of for some a and 6. On respect to these axes Q We can choose L as the with the horizontal plane the line through of a point in P Q parallel to our with respect to these point with respect to the original axes this form ax 5 all that an hyperbola. is Otherwise, in the plane of the FIGURE of says that the intersection of the cone z 4 out points (a, y, z) with all FIGURE rotate in the usual position that is The plane P is thus viewed "straight on," so intersection L with the plane of the first and third are familiar with. slope we if this intersection line points straight first 81 Sections is the other hand, y, then y is + if yS the second coordinate of the point with also the second coordinate with respect to the original axes. Consequently, cone if and only (*) says that the point M(ax + Although this lies on the intersection of the plane and the if B) + B = ±cV(ax + B) looks fairly complicated, after squaring 2 +y 2 . we can write this as C 2 y 2 - a 2 (M 2 - C 2 )x 2 + Ex + F = some E and F that we won't bother writing out. Now Problem 4-16 indicates that this is either a parabola, an ellipse, or an hyperbola. In fact, looking a little more closely at the solution, we see that the for 82 Foundations E and F values of (1) If (2) If M = ±C we obtain a parabola; C > M~ we C < M 2 we (3) If These are irrelevant: obtain an ellipse; obtain an hyperbola. analytic conditions are easy to interpret geometrically (Figure If (1) our plane is parallel to one of the generating lines 6): of the cone we obtain a parabola; our plane slopes If (2) intersection omits (3) If less than the generating line of the cone one half of the cone) we obtain an our plane slopes more than the generating line (so that our ellipse; of the cone we obtain an hyperbola. FIGURE In 6 names of these "conic sections" are comes from a Greek root meaning the very fact, The word parabola related to this description. 'alongside,' the same root mention paradigm, paradox, paragon, paragraph, and even parachute. Ellipse comes from a Greek root that appears in parable, not to paralegal, parallax, parallel, meaning dicate 'defect,' or omission, as in ellipsis (an omission, ... or the dots that in- And hyperbola comes from a Greek root meaning 'throwing beyond,' or With the currency of words like hyperactive, hypersensitive, and hypervennot to mention hype, one can probably say, without risk of hyperbole, that it). excess. tilate, t His. root is familiar to almost everyone.* PROBLEMS 1. I I < . I RE Consider a cylinder with a generator perpendicular (Figure 7 7); the only requirement for a point {x y, z) to , * Although the correspondence between these roots tifully. fi>r io desi iil>< to the horizontal the sake of dull accuracy features of < it lias to and lie on this plane cylinder is die geometric picture correspond so beau- be reported that the ertain equations involving the conic ( heeks originally applied the words set lions. 4, that (jc, y) lies on a 2. The Conic 83 Sections circle: =C y Show Appendix . that the intersection of a plane with this cylinder can be described by an equation of the form (ax What 2. In Figure 8, the sphere S\ has the lies Similarly, the F2 at 8 along the cylinder; equator Ci of S2 z lies it is as the cylinder, so that why the length of the line line L from from z to Fjis P and is plane and a cone is cone. Similarly, use is an an ellipse, F2 ( ; l R E v show that a constant, so . prove geometrically that the intersection of a ellipse (b) to with foci F\ and is when the plane intersects just prove that the intersection is one half of the an hyperbola when the plane intersects both halves of the cone. (a) Explain equal to the length of the vertical a similar fact for the length of the line from z to F2, Similarly, use Figure 9(a) to l P z to C\. that the intersection l its at Fi. tangent to the cylinder. the distance from z to F\ plus the distance from z to F2 3. P also tangent to the plane along the cylinder, and S2 be any point on the intersection of By proving (b) same diameter . Let (a) . possibilities are there? equator C\ FIGURE + /3) 2 +y 2 = C 2 (b) , 84 . Foundations APPENDIX POLAR COORDINATES 3. In this chapter we've been acting all along as points in the plane with pairs of numbers. length r ways, each giving rise to of a point are called its if Actually, there are In many to label different usual coordinates cartesian coordinates, after the French mathematician many situations coordinates (r,9), where r is 1 introduced the idea of first more convenient to introduce polar To the point P we assign the polar is it coordinates, which are illustrated in Figure FIGURE The a different "coordinate system." and philosopher Rene Descartes (1596-1650), who coordinate systems. one way there's only . the distance from the origin O and 9 to P, is the 1 measure, in radians, of the angle between the horizontal axis and the O This 9 to P. is right side of the horizontal axis could have either 9 completely ambiguous at the origin O. So is through the origin if we want to assign a it is line from For example, points on the not determined unambiguously. = or 9 = 2tt ; moreover, 9 necessary to exclude unique pair (r, some ray 9) to each point under consideration. On the other hand, there In (r,9). to (r, fact, when 9) no problem is associating a unique point to any pair it is possible (though not approved of by r < 0, all) according to the scheme indicated in to associate a point Figure 2. Thus, it always makes sense to talk about "the point with polar coordinates (r,9)," (with or without the possibility of r talk P is and < 0), even though there is some ambiguity when we about "the polar coordinates" of a given point. the point with polar coordinates (r,8\) also the point with polar coordinates (-r,92 ). I I CURE It is (r, 2 clear from Figure 1 (and Figure 2) that the point with polar coordinates 6) has cartesian coordinates (x, v) given by x = r cos 9 v = r sin 9 4, Conversely, ordinates Appendix a point has cartesian coordinates (x,y), then (any of) if 85 3. Polar Coordinates polar co- its 9) satisfy (r, =±v? + y tan0 =- 4 if*#0. x Now suppose that / is ordinates we mean the satisfying r = collection of f{9). all FIGURE 3 as opposed all points the P In other words, the graph of graph of f we in polar co- with polar coordinates / in polar coordinates No points with polar coordinates (f(9),9). should be attached to the fact that first, Then by a function. collection of (r, is 9) the special significance are considering pairs (f(0),0), with f(9) graph of /; it is purely a matter of polar coordinate and 9 is considered the to pairs (x, f(x)) in the usual convention that r is considered the first second. The graph of / equation for all 9. r = in polar coordinates often described as "the graph of the is /(#)." For example, suppose that The graph of the equation radius a (Figure 3). This example coordinates are likely to make r = a is a constant function, f(9) = a simply a circle with center O and / illustrates, in is a rather blatant way, that polar symmetry things simpler in situations that involve with respect to the origin O. The graph of the equation r = 9 is shown in Figure 4. The solid line corresponds to all values of 9 > 0, while the dashed line corresponds to values of 9 < 0. FIGURE Spiral of Archimedes 4 As another example involving both the equation r = positive and negative Figure 5(b) shows the part corresponding to n/2 check that no this FIGURE 5 new same graph r, consider the graph of cos9. Figure 5(a) shows the part that corresponds to in points are added < 9 < > n or 9 < for 9 n\ here r 0. It is < < 0. 9 < it/2 You can easy to describe terms of the cartesian coordinates of its points. Since the polar coordinates of any point on the graph satisfy r — cos i 86 Foundations and hence = r cartesian coordinates satisfy the equation its x which describes a circle — cos 2 + y2 = x (Problem 4-16). [Conversely, coordinates of a point satisfy x 2 r r cos 0, + y2 = x, then lies clear that it is if the cartesian on the graph of the equation 0.] Although we've now gotten a two circle in turns out that we different ways, an hesitant about trying to find the equation of it it ellipse in can get a very nice equation if we might polar coordinates. But we choose one of the foci as sum of the the origin. Figure 6 shows an ellipse with one focus at O, with the distances of all the points from O and the other focus f being 2a. We've chosen f of O, with coordinates written as left (-2ea,0). (We have FIGURE The < e < 1, since we must have 2a > distance from f to O). 6 distance r from (x, y) to O r (1) By assumption, is given by =x 2 + y 2 2 the distance from (x, y) to f (2a -r) 2 = - (x . 2a is [-2ea]) — 2 r, hence + y2 , or Aa (2) Subtracting (1) from 2 - bar + r 2 = (2), well be x 2 + 4eax + 4s 2 a 2 + and dividing by 4a, we a — r = ex +e get 9 a, y 2 to Appendix 4, 87 3. Polar Coordinates or r = a — ex — s a — — = which we can write (1 2 e")a EX, as r (3) = A-sx, Substituting r cosO for x, A = for (1 2 — £ )a. we have = A — sr co$0, r(l + e cosO) = A, r and thus r (4) = 1 we found In Chapter 4 + £ COS that 2 x2 y + T, = ab~ (5) is } 1 the equation in cartesian coordinates for an ellipse with la as the distances to the foci, but with the foci at (— c, 0) b = y/a 2 Since the distance between the foci units, so that the focus (c, 0) = take c sa or e = now is by (4), where this ellipse we determining A). (3) and again using we can e (4) left by c when we Conversely, given is A = =-, c = = sa, \l a 2 — we c2 -s 2 get — \l a 2 — s 2a 2 = ay — 1 £2 = yr from moved (3), 1 Thus, is get the ellipse for the corresponding equation (5) the value of a a b when 2c, (c, 0), of the . at the origin, c/a (with equation the ellipse described determined by is -c 2 and sum E' obtain a and b, the lengths of the major and minor axes, immediately and A. The number the eccentricity of the major and minor ellipse, c Va 2 -b 2 a a I (y V determines the "shape" of the axes), while the number A determines its ellipse (the ratio "size," as of the shown by (4). . 88 . Foundations PROBLEMS 1. two points have polar coordinates If d between them tance d What does 2. / / (ii) is odd. = that the dis- 2r\r2cos(0\ — Go). / in polar coordinates if f(9+jz). Sketch the graphs of the following equations. (i) r (ii) r (iii) r (iv) r — a sin#. — a sec9. Hint: — cos 29. Good = cos 39. (v) r = \ cos 29 1 (vi) r = | cos 39 \ It is a very simple graph! luck on one! this Find equations for the cartesian coordinates of points on the graphs and 5. even. show (rj, Oi)-, say geometrically? this is f(9) (iii) 4. 1 + r2 — r\" Describe the general features of the graph of (i) 3. O = and (r\, 9\) given by is (iii) in Problem (i), (ii) 3. Consider a hyperbola, where the difference of the distance between the two foci and choose one focus the constant 2a, is we must have (In this case, > s at Show 1). O that and the other we at (— 2ea, 0). obtain the exact same equation in polar coordinates A = r + 1 as we obtained for an Consider the from the distance to the line is a — r (x, y) cos 9, tion for (4), this any A and s, i i< i ; 7 r(\ + 7). Show that the equation O is equal to that the distance can be written cos 9). which implies (3). Show e )x 2 — 1 , and a hyperbola for e > again of the same form as in (4). polar coordinates of the equa- that the points satisfying this equation satisfy + Using Problem 4-16, show that s is consider the graph 2 . = a (Figure equation for a parabola (1 ( = and conclude a Now, i y (x, y) to =a Notice that i such that the distance (x, v) to the line (x,y) y 9 ellipse. of points set s cos v 2 this 1 = A2 is an 2Asx. ellipse for s < 1, a parabola for . 4, 8. (a) (b) (c) Sketch the graph of the Show Show that it is that it and conclude that 2 it (x 10. + v 2 = y/x 2 + y2 - y, 2 + y 2 + y) 2 = x 2 + 2 y Sketch the graphs of the following equations. (i) r (ii) r (iii) r (a) (b) (c) = l-isin0. = l-2sin0. = 2 + cos#. Sketch the graph of the Find an equation for Show d\d2 (d) 8 sin#. can be described by the equation r FIGURE 89 sin 9. = —1 — r 3. Polar Coordinates can be described by the equation x 9. =1— cardioid r graph of also the Appendix that = Make it is its lemniscate 2 = 2 2a cos26. cartesian coordinates. the collection of all points P in Figure 8 satisfying a~ a guess about the shape of the curves formed by the satisfying d\di = b, when b > a and when b < a 2 . set of all P . CHAPTER LIMITS The concept one in all are, once more, going is PROVISIONAL DEFINITION of a limit not the The of calculus. word surely the The Of the six at a to — a. our it is We the definition of limits, but what we we shall define / near a, is if we can make f(x) sufficiently close to, 1 , only the not defined, and h(a) and h approach / near a. but unequal This is is first as close as three approach defined "the because we to, a. we / at a wrong way," it explicitly ruled definition, the necessity of ever considering the value of the function only necessary that f(x) should be close to / for x close to a, but unequal are simply not interested in the value of f(a), or even in the question of whether f(a) i the limit functions graphed in Figure true that g still out, in is difficult "limit" but the notion of a function approaching a limit. Notice that although g(a) is goal of this chapter by requiring that x be / most important, and probably the most to begin with a provisional definition; / approaches function like to is is defined. i<;rki; One 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 convenient a point x in one, to fix) different functions. 90 in the other. Figure 2 illustrates such a picture for two 5. Limits f(x) (a) FIGURE Now = 91 c 2 consider a function that f(x) be close to /, / whose drawing open say within the looks like Figure 3. B which interval Suppose we ask has been drawn we consider only the numbers x in the A of Figure 3. (In this diagram we have chosen the largest interval which will work; any smaller interval containing a could have been chosen instead.) If we This can be guaranteed in Figure 3. if interval FIGURE FIGURE 3 choose a smaller interval B' (Figure but no matter to how small be some open interval A we choose A which 4) we will, the open usually, 4 have to choose a smaller interval B, there is A', always supposed works. similar pictorial interpretation is possible in terms of the graph of /, but in B must be drawn on the vertical axis, and the set A on the horizontal axis. The fact that / (x is in B when x is in A means that the 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 large. this case the interval ) 92 Foundations To = a take a specific simple example, 5 (Figure be able to f(x) = 3x 5 To be specific, This means that . which we can to 15 — to IT) — <3x < 1 to require that — x < —-, 1 s j 30 or simply \x — 5| < we take any positive — that \x 5 < 1 There 3^; easy to guarantee that if 10' Jo- 1 6 + 15 <3x ~ l5< — wt: < FIGURE we require that x be sufficiently make sure that 3x is within of like if also write as we just have this |3jc — nothing special about the number is 15| number 30 = — how same — function f(x) show sort of argument = — x < a\ works a is first step is \x — a\ < = 3,000,000.v. e/3, 000, 000 in little more Presumably, interesting. we should be This means that we need to show - which gives us the useful 3 or 3,000,000. 3 < x 1 about < 1. +3 < e by requiring e 2 \x = ~9\ — 3| how Once -3| \x — |.v to 3| be small enough. The +3|, \x • Unlike the situation with the previous factor. examples, however, the extra factor here something < 9| to write \x 5 for the function f(x) that f(x) approaches 9 near 3. any given positive number — < a\ to ensure the inequality obvious \x e e. 2 like easy to see that that have to be 1,000,000 times as careful, choosing |.r for < 3a\ 5. It's just as to require that order to ensure that \f(x) The To ensure the limit 3a at a for any a: Naturally, the able to . e/3. \x We just just as l 1 \3x we just have is < LOO simply require that \x 5| < 300 In fact, we can make \3x — 15 < s simply by requiring ; e jq. It — There's also nothing special about the choice a / approaches 3x with —we ought have -Jo To do we we want suppose we want = consider the function f(x) limit 15 near 5 let's Presumably / should approach the f(x) as close to 15 as to get close to 5. 1 6). is which |jc+3|, But the only crucial thing big |jc + 3| So is. the we've specified that 7 and we've guaranteed that 2 \x - 91 = l.v - 31 • is — 3| \x + 3\ + 31 make thing we'll first |.v \x to isn't a convenient constant sure that do is wc can say to require that < 1, or 2 < x < 4, we have < 7. So we now have < 7|.v - 31, 93 5. Limits which shows that we have also required that \x — < min(e/7, 3| The — |jc 3 \x < 1 2 < 9| convenient number. To chose |jc — it is 3 1 < — < 3| was simply made 1 — make how the argument would read we if 10. to show that / approaches a near a / approaches 9 for any again with the expectation that Problem 1-12 shows near 3 + a\. We |jc first ensure that this will work to show worry a bit more will basically we need a, except that about getting the proper inequality for fact, We for convenience. — 3| < 10 or any other |jc 3| jq or \x sure you understand the reasoning in the previous < a good exercise to figure out Our argument that < e/7, provided that we've more official: we require that 3| look it 1). could just as well have specified that paragraph, — £ for \x make Or, to 1. specification \x initial — |jc to require that +a | is |jc — a\ < not too large. 1, In that < \a\ < \x 1. so x\ 1 + \a\ + \a\ < < and consequently \x+a\ < so that we \x\ + 2\a\ 1. then have 2 \x -a 2 = \x < \x \ — — + a\ a\ |* a (2|fl| I + l), — a"\ < s for \x —a\< e/(2\a\ + 1), provided that we — a\ < min(e/(2|a| + 1), 1). also have \x — a\ < 1. Officially: we require that In contrast to this example, we'll now consider the function fix) = \/x (for x ^ 0), and try to show that / approaches 1/3 near 3. This means that we need to show how to guarantee the inequality which shows that we have \x" |.v I < for any given positive number s by requiring £ — 3| \x to be small enough. We begin by writing 1 ~3 = x giving us the nice factor the problem factor small, so that \/\x\ We can first 1/|.\ |. 1 1 = 3x — 3\, and \x 11 ^ _ V 3-x l 1 1 ' • 3 |jc — 3|, \x even an extra i for good measure, along with we In this case, first need to make sure that |.v| isn't won't be too large. require that |jc — 3| < 1, because 111 < 4 < x r this gives 2 <x < 4, so that too , 94 Foundations which not only — < — us that tells order to conclude that — < — Ul . > but also that x , We now which 0, important in is have 2 < -|,-3| which shows that we have — \x 3| < wanted to show < min(6£, 3| If we instead by < To show —a this possible values for x our — (— 3)| < The a < 7 a — \x look 3 < 1 6s, provided that we've official again: we require that 1, works as before. than — It's we proceed in more careful \/a near a for any a that, again, we have be a to little not good enough simply to require that or any other particular number, because 1, that are negative (not to if a is close to mention the embarrassing 0, so that f{x) isn't even defined!). trick in this case is to first require that — \x FIGURE it / approaches — 1/3 near —3, we would begin giving —4 < x < —2, once again implying that initial stipulation. less x s for make / approaches would allow values of x possibility that < 1 that same way, except should be | |jc in general that in formulating 1/3 Or, to . 1/2, so that everything basically the \x 1 1). stipulating that |1/jc| — i/x — also required that \x < a > in other words, You should be we require that x be less than half as far from able to check first that x / and that < l/|.v| as a (Figure 7). 2/|fl|, and then work out the rest of the argument. With quail at really you may have begun to the prospect of tackling even more complicated functions. But that won't all the work required we be necessary, since will eventually rely on. Instead of worrying in functions like f(x) = behavior of = x — = We want it (remember a from consideration, so at 0). to show that \/x we'll turn , — that it is xsinl/x (Figure get f{x) = x sin \x — 0| is 1 as desired In other words, for any I .v < sin e sufficiently small (but < x exempts even defined isn't l/x as close to / I Sill / approaches specifically doesn't matter that this function we can l/U)-0| = = Despite the erratic 8). clear, at least intuitively, that we require that x be sufficiently close to 0, but ^ 0. number e > 0, we want to show that we can ensure that |.v| our attention to some our provisional definition if by requiring that we can be even more frightening. function near this near a / = or f(x) to the function f(x) first have some basic theorems that about the unpleasant algebra that might be involved x examples that might appear Consider for these simple examples, for all x ^ 0, 0). But this is easy. Since 95 5. Limits we have .v so we can make |jc near < x < s simply sin l/.v| = For the function f(x) proaches sin 0. If, for - a - for , all 0, by requiring that l/x (Figure 9) sin / x |a| < and e / 0. seems even clearer that / ap- it example, we want x 1 < sin To then that / ' I ft \ / f(x) = x 2L we \x certainly 2 < \ -4y need only require < \ ^0, and i jq since this implies and consequently x l sin that \x - 2 — 1 . sin < < < \x' 100 x (We could do even better, and allow |jc < 1/V | 10 and x particular virtue in being as economical as possible.) ^ 0, but there In general, if s > is no 0, to ensure that 1 x sm FIGURE we need 9 provided that e < though it that < As a FIGUR E = Vl-vhm is £, "small" but make it If we are e's < s and jc^O, given an £ which which are of interest), is greater than then certainly suffices to require that \y/\x\ sin l/.v| 1 < £ |.v| (the algebra I 1. | third example, consider the function f(x) order to f(x) e, only require that |jc |a| < is left to you) we can < s it |jc — require that and x ^ j 1 does not < x 1 | and sin (it might be, even suffice to require i^O. l/x (Figure 10). In 96 Foundations Finally, let us it is false that / approaches ^ and 0. condition \f(x) |jc| is In to be. matter near l/x (Figure sin For 1 1). This amounts to saying that 0. this function is not true it that we can get \f(x) — 0| < s by choosing x sufficiently e > To show this we simply have to find one e > for which the — 0| < s cannot be guaranteed, no matter how small we require number for every small, = consider the function fix) how = fact, s small some number x 7 will do: we require = \/(\tt |jc it is impossible to ensure that A to be; for if | + Inn) which is |/(jc)| any interval containing is in this interval, and for this < \ no 0, there x we have f{x)=\. FIGURE I 1 This same argument can be used (Figure 12) to show that we must \f(x) — l\ < number near 0. To show some number s > so that any The required to be. small we that for require — \ e is works for any number /; that we cannot ensure that \f(x) — l\ < A containing we can find both x\ and any interval number no matter how small x not true, to be, |.r| not approach again find, for any particular this choice s / does is, 5. /, is no matter how The reason is, X2 in this interval with FI G U RE = /(*i) 12 and l = f(x 2 ) -l, namely X] and Kit for large both 1 , since its 1 no matter what 1 kJT total length — x? + 2nn enough n and m. But the — and 1 1 I = / 1 < I from interval is only and also 1 ; | so — / + 2niJT 7 to / + ^ cannot contain we cannot have — — 1 / 1 < ^, /is. The phenomenon exhibited by f(x) = f we consider the function ,, . fix) = [0. , I sin l/.v near x irrational - . x rational. can occur in many ways, 5. Limits 97 / does not approach any number / near a. In fact, we cannot make \f(x) — l\ < \ no matter how close we bring x to «, because in any interval around a there are numbers x with fix) = 0, and also numbers x with f(x) = 1, so that we would need |0 — /| < j and also |1 — /| < -j. no matter what a then, An amusing .-•'/(*) = x, x rational 0, x irrational is, on variation is presented by the function shown in Figure 13: /(*) The behavior proaches at 0, should have no FIGURE behavior this We 13 of this = function x, x rational 0, .v irrational. but does not approach any difficulty number convincing yourself that - 1 1, x > If a > 0, / approaches then 1 certainly suffices to require that near -a or f{x) = -1, X < 14 — easily check, The time limits 14): — 1 < e it a, since this implies < x < x / approaches —1 near b: to ensure (—1)1 < s it suffices to require that ]x — b\ < —b. Finally, as you may / does not approach any number near 0. so that f{x) that \f(x) FIGURE = you true. indeed, to ensure that \f(x) a: — d\ < |jc a ap- x>0. 1, = is it < x , at a, if this conclude with a very simple example (Figure fix) f(x) = sin l/x; ^ 0. By now "opposite" to that of g(x) is 1. has Similarly, if b now come < 0, then to point out that of the which we have given, not one has been a with our reasoning, but with our definition. If many demonstrations about The real proof. fault lies not our provisional definition of a function was open to criticism, our provisional definition of approaching a limit is even more vulnerable. used in proofs. It is This definition hardly clear is simply not sufficiently precise to be how one "makes" "close" means) by "requiring" x to be sufficiently ciently" close is certainly supposed fix) close to / (whatever close to a (however close "suffi- to be). Despite the criticisms of hope you do) that our (I ing. In order to present any sort of argument at It is may arguments were nevertheless quite convinc- feel to invent the real definition. our definition you all, we have been practically forced possible to arrive at this definition in several steps, each one clarifying some obscure phrase which still remains. Let us begin, once again, with the provisional definition: The (unction / approaches the limit near a. if we can make fix) as close as we like to / by requiring that x be sufficiently close to, but unequal to, a. / The very fix) close change which we made first to / meant making \f(x) — in this definition /| small, and was to note thai making similarly for x and a: 98 Foundations The / approaches function we small as like the limit by requiring that \x near a, / — a\ be The second, more crucial, change was to note we like"" means making \f(x) — l\ < s for any The / approaches the limit can make \f(x) — l\ < s by requiring x function / a. is a There common pattern to number we can make 1/00 —l\ as a. making \f(x) — l\ "as small as > that happens to be given us: e that \x number for every if — a\ be > s we sufficiently small, the demonstrations about limits which all ^ and x that near a, / if sufficiently small, and we have > we found some other positive number, 8 say, with the property that if .r^a and \x — a\ < 8, then \f(x) — 1\ < s. For the function = 0), the number 8 was just the number e; f(x) = .v sin l/.v (with a = 0, for f(x) = yixlsin l/x, it was e for f(x) — x it was the minimum of 1 and s/(2\a\ + 1). In general, it may not be at all clear how to find the number 8, given. For each s / ; given but s, it is \x —a\<8 which expresses how small "sufficiendy" / approaches the limit / the condition small must be: The 8 > This is function such that, for all x, if \x practically the definition change, noting that — \x DEFINITION a The I < — |x a\ < 8 we 8 if and x > for every e ^ there — a, then \f(x) 8 f > is some e. ^ We will make only one trivial a" can just as well be expressed "0 < approaches the limit / near a means: for every e > — a\ < 8, then \f(x) — 1\ < s. < all x, if is so important {everything like a poem! That, at are doomed least, is better to give incorrect proofs. we do from now on depends on it is hopeless. If necessary than stating it incorrectly; A good exercise if formal proofs of each. In most has been done already. conform When cases, this will merely involve a proving that / does not — all bit limits, / is to giving of rewording the algebraic approach that memorize it, this you in giving correct proofs our formal definition to it) you do review every fact already demonstrated about functions approaching the arguments there |.\" proceeding any further without knowing make < will adopt. and x such that, for This definition to l\ 5." function some is u near a, — a\ < at a, work be sure to negate the definition correctly: If it is not true that for every e (hen \f(x) then > — there l\ < s, is some 8 > such that, for all .v, if < a\ < <5, 4 u 5 5. there < — a\ < \x > some £ is 8 such that for but not \f(x) — consider £ but not | = sin and note \ l/x — | so large that l/(n/2 As a limit, < = that for every 8 i there some x which is 99 satisfies s. l/x does not approach sin > there —namely, an x of + 2mt) < the form \/(n/2 + near < some x with is \x — 0, 0| we < 8 2mt), where n is 8. of the use of the definition of a function approaching a final illustration we have < l\ Thus, to show that the function f(x) > every 8 Limits reserved the function shown standard example, but in Figure 15, a one of the most complicated: 0, Jt irrational, \/q, x = /(*) (Recall that and q > p/q is terms in lowest p/q if <x < 1 < x < in lowest terms, p and q are no integers with 1. common factor 0.) x irrational 0, /(*) = — x , = — in lowest terms — j 5 FIGURE 4 3 4 4 5 15 number a, with < a < 1, the function / approaches at a. To prove number £ > 0. Let n be a natural number so large that \/n < £. Notice that the only numbers x for which \f(x) — 0| < s could be false are: For any this, consider any 2 3 1 4' (If a is rational, then a these numbers, one is values \p/q — — a\ if < \x 5, 1 any closest to a; that is, rate, only — a\ \p/q finitely is many of these many. Therefore, of all smallest for one p/q among a happens to be one of these numbers, then consider only the a\ for < — might be one of these numbers.) However be, there are, at (If n 1 5' numbers there may these numbers. 3 p/q then x / is a) This not closest distance one of n- 1 may be chosen as the 8. For — 100 Foundations and therefore \f(x) — 0| < s is true. This completes the proof. Note that our description of the which works for a given e is completely adequate there is no reason why we must give a formula for 8 in terms of e. — <5 Armed with our definition, we are have probably assumed the result This theorem is now prepared to along, which all our really a test case for prove our first theorem; you a very reasonable thing to do. is definition: theorem could not be the if proved, our definition would be useless. THEOREM 1 PROOF A function cannot approach two different limits near a. near a, and approaches / Since our this is / approaches theorem about first m limits near «, then be necessary will certainly it In other words, if / = m. I to translate the hypotheses according to the definition. Since 8\ > / approaches such that, for also < |jc — a\ < then \f(x) 8\, / approaches m near know, since any that for > e there some number is x, all if We we know near a, / — < l\ a, that there s. some is 82 > such that, for all x, < if We |jc — a\ < have had to use two numbers, which works 8\ 82, then \f(x) and 82, since one definition will work in the there is some 8 > for any e > in conclude that if < — \x < a\ then \f(x) 8, we simply choose 8 = min(<$i, dh). To complete the proof we just have — I\ — < m\ there is e. no guarantee other. But, in fact, such < e it is that the 8 now easy to that, for all x, and \f(x) — m\ < > to pick a particular s for e; which the two conditions — |/0r) is 7^ a > and £ \f(x) — m\ < s 0, The proper choice is suggested by Figure 16. If we can choose — m|/2 as our s. It follows that there such that, for all x, \x -a\<8, m, so that 8 < m\ > cannot both hold, / /| |/ — if if / / < w. |/ then |/(jc) - /| < \l — m\ 2 -- in length FI G 1 RE 16 —— - length — and f(x) I —m\ < \l-m\ z - This implies that for |/ -m = \ \l — a\ < < \x - f(x) + 8 we have fix) -m\<\l- f(x)\ \l-m\ = a contradiction. | |/-m|, + |/(.v) \l-m\ - m\ 5. Limits The number which / approaches near a denoted by lim fix) (read: the is / 101 limit of fix) as x approaches a). This definition is possible only because of Theorem 1, which ensures that lim f(x) never has to stand for two different numbers. The x—ni equation = lim f(x) I x—>a has exactly the same meaning as the phrase / approaches The possibility = lim fix) / a. / does not approach / near a, for any /, so number /. This is usually expressed by saying remains that still is near / false for "lim fix) does not every that that exist." x-*a Notice that our new notation introduces an extra, utterly irrelevant letter x, which could be replaced by appear — /, y, or lim fix), x—>a all to any other letter which does not already the symbols lim /(f), lim /(y), t—>a y—>a denote precisely the same number, which depends on / and a, and has nothing do with x, t, or y (these letters, in fact, do not denote anything at all). A more logical symbol would be something like lim /, but this notation, despite its brevity, a is so infuriatingly rigid that almost much more useful because a function lim fix) x—>a is though might be possible it the short no one has seriously to express / tried to use often has it. The notation no simple name, even fix) by a simple formula involving x. Thus, symbol + shut) lim(x x— a could be paraphrased only by the awkward expression lim /, where fix) = x + sin .v. a Another advantage of the standard symbolism lim* + lim* + x— a The first means the means the = x +f 3 by the expressions r\ 3 t . little for , all .v; number which / approaches near a when fit)= x + You should have illustrated number which / approaches near a when f(x) the second is 3 t for all , f. difficulty (especially if you consult lim x +t 3 — a +r\ +f 3 = x +a 3 x -> a Y\mx . Theorem 2) proving that , 102 Foundations These examples In ibility. fact, what forgetting which tation, main advantage of our illustrate the the notation lim f{x) x—>a it will really Here means. be important later: notation, which so flexible that there is is is is its flex- some danger of a simple exercise in the use of this no- first interpret precisely, and then prove the equality of the expressions and lim f(x) An it make to find many limits, as we promised long ago. The proof depends upon properties of inequalities and absolute values, hardly surprising when one important part of easy certain lim f{a +h). chapter this the proof of a theorem which will is considers the definition of limit. Although these facts have already been stated in Problems 1-20, 1-21, and 1-22, because of their importance they be presented will 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 xq, and y is close to yo, then x + y will be close to xo + yo, and xy will be close to xoyo, and 1/y will be once again, in the close to 1/yo- This intuitive statement estimates of the first, LEMMA in lemma, and order to see just it is how much is easier to remember than the precise not unreasonable to read the proof of Theorem 2 these estimates are used. (l)If £ < - and --vol \x \y £ -y < \ -, then \(x + y) - Uo + vo)! < £ (2) If \x — .vol < min 1, ( 2(|yol and +D | — y yo I < 2(1x01 + 1)' then \xy (3) If # y / v — 2 \(x 1 1 y yo + - y) (xo + yo ) I = IU - 2 \x — xq\ < 1 £. + (y - yo)| £ £ + \y - vol <i+2 =£ xo) <\x- .vol Since ' and (1) (2) e\yo\ Vol < min yo| < proof e. I and | then y -xoyo < we have |.V | — |.V()| < \x — xq\ < 1 ' 5. Limits 103 so that < \x\ + 1 |* |. Thus \xy - xoyol = \x(y < \x\ < (1 - y ) \y - vol + + + vol I kol) 2(|.v £ 2(|vol +D have > Ivl Ivol l) 2 - \yo\ so \ + - = £. 2 We -x + + | )\ \x • £ < (3) -x yo(x I ^ vol/2. In particular, v < \y\ -vol < l.v I vol and 0, — 2 1 l.v < l.vol I Thus 1 1 y theorem 2 If lim x—*a f(x) = I lyo vo — \yo\ \y\ and lim g(x) 2 vl = m, l.vol 2 —=r- = f|v 1 _ | 8 - I ivol then x—*a \im(f +g)(x)=-l (1) + m; x—>-a Moreover, PROOF The all if m ^ 0, hypothesis (2) \im(f (3) lim g)(x)=l -m. then means (-)(*) = -• that for every £ > there are <5i,^2 > such that, for x, and This means such < \x if < \x (since, after all, e/2 — — is a\ < 8\, then \f(x) — a\ < 82, then \g(x) — m\ < also a positive l\ number) < s, £. that there are 8\, 82 > that, for all x, and Now < if let |jc — 8 = a\ < if < \x — a\ < <5i, then \f(x) — if < |.v — 52, then \g(x) — min(<5i, #2). 81 are both If true, so a| < < |.v — a\ < 5, < - /| < - and \g(x) £ -, m\ < -. then both £ \f(x) l\ £ - m\ < - < \x — a\ < 8\ and 104 Foundations are true. But by part This proves £ of the (1) lemma this implies that / + g){x) — |( (I + m)| < e. (1). To prove > there we proceed similarly, after consulting such that, for all x, 8\, 82 > (2) are < if — \x < a\ — then \f{x) 8\, part < min 1\ of the lemma. (2) 1, ( 2(|ro|+ and < if — a\ < \x 1) £ — m\ < then \g(x) 82, If 2(|/|+1) Again let 8 = — \f(x) < min(<$i, 82). If / . < nun I\ £ 1, I 2(\m\ So, by the lemma, > Finally, if e |(/ g)(x) • there — a\ < \x —\ > and I — \g(x) ,ov 2(|/| and e, such proves this that, for all < \x — a\ < then \g(x) —m\ < min 1 I'" / I i2\ £\fn\~ 1 2 makes sense, to part (3) of the and second lemma this means, 2 that g(x) /' ^ 0, so (l/g)(x) that - - (x) - < m ,8/ This proves first, \ -r-, I K But according l) ,y, . (5, + (2). /i if £ m\ < l)J —I m\ < a 8 is + then 5, £. (3). | Using Theorem 2 we can prove, trivially, such facts as — + 7x— = — ++la— x^a x- +1 ,. hm x3 5 3 a = a 5 = 2 -, 1 without going through the laborious process of finding a 8, given an e. We must begin with lim 7 = 7, = 1, = a, x—*a lim 1 x—*a lim x x—*a directly. If wc want to find the 8, 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 8 such that, for all x, but these are easy to prove if < |.v - a\ < 8, then \x +x - (a~ + a)\ < £. . 5. Consulting the proof of Theorem such that, for all we have we see that do 9 if < |.v — a\ < <5i, then \x if < \x — a\ < 8i, then |x = already given proofs that lim x —a and > 8i 9^" < — | — a\ < — = a" and lim x x—*a a, we know how this: \ 2 Thus we can take = 8 min(5i, 82) = min min I I 1 2\a\ If find 8\ s x—>a to we must first x, and Since 2(1), 105 Limits a ^ 0, the same method can be used < if I — .V I < <5, then x The proof of Theorem 2(3) shows such that, for all .v, a 8 > 1 > to find a 8 1 a + such that, for Thus we can \x — a\ < 8, < 1 az £. that the second condition will follow then —a \x \ x, 1 9 z i < if all < min i2 or efl i if we find i4 take / I / / . min i2 \ 2 <5 = min fieri |<z| ' 2 1, ' 2|a|+l V Naturally, these complicated expressions for <5 can be simplified considerably, after they have been derived. One technical detail in the proof of order for lim fix) to be defined it is, as x^-a at «, nor is it must be some necessary for 8 > / to Theorem 2 deserves some discussion. In we know, not necessary for / to be defined be defined such that f(x) is at all points x ^ a. defined for x satisfying However, there < \x — a\ < 8; otherwise the clause "if < \x -a\<8, then \f(x) - 1\ < e" would make no sense at all, since the symbol f(x) would make no sense for some x's. If / and g are two functions for which the definition makes sense, it is easy to see that the same is true for f + g and f g. But this is not so clear for i/g, since \/g is undefined for x with g{x) = 0. However, this fact gets established in the proof of Theorem 2(3). 106 Foundations There are times when we would like to speak of the limit which / approaches at a, even though there is no 8 > such that fix) is defined for x satisfying — < \x a\ < 8. For example, we want to distinguish the behavior of the two functions shown in Figure 17, even though they are not defined for numbers less I— than a. we For the function of Figure 17(a) f(x)=l lim x—*a + write or = lim f(x) I. x\.a (a) (The symbols on the These I— "limits definition such left are read: the limit of f(x) as x approaches a from above.) from above" are obviously very similar: is = lim fix) x—>a + / closely related to ordinary limits, means that for every e > there and the a 8 is > that, for all x, < if — a < 8" is equivalent to "0 A' 8, — then \f(x) < l\ e. (b) FIGURE 17 (The condition "0 < x — < a < — \x a\ "Limits from below" (Figure 18) are defined similarly: — lim fix) I) means > that for every e there a 8 is < 8 and x > a.") lim fix) = I (or x—>a~ > such that, for x'fa x, all <a —x < if FIGliRl. 18 It is for 8, then |/(.v) than less < /| e. and below even quite possible to consider limits from above numbers both greater and — if / is defined Thus, for the function / of Figure 14, a. we have lim x^0+ fix) = and 1 an easy exercise (Problem 29) It is lim ' fix) =— 1, jc->0- to show that lim fix) exists if and only if x—*a lim fix) and lim fix) both exist and are equal. x—Ki+ .v—>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 will be found useful. In Chapter 4 (lose to 0. This assertion is it was claimed lim fix) is if x is large, then sin l/.v is usually written lim sin The symbol that which read "the limit 1 /.v = 0. of fix) as x approaches oo," or "as x x—*-oo becomes infinite," and a limit of the form lim fix) is often called a limit at infinity. 5. Limits 107 Figure 19 illustrates a general situation where lim f(x) = I. Formally, lim f{x) AT->00 X—»0O there is a number N such that, for all x, / means that for every £ > if x > N, then |/(jc) - /| < = s. The analogy with the definition of ordinary limits should be clear: whereas the condition "0 < \x — a\ < 5" expresses the fact that x is close to a, the condition > "x ./V" FIGURE We expresses the fact that x is large. 19 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 un- derstand the definition of ordinary limits (which are by far the most important). Many exercises on these definitions are provided in the Problems, tain several other types of limits which are occasionally which also con- useful. PROBLEMS 1. Find the following limits. (These limits all follow, after from the various parts of Theorem nipulations, ones are used in each case, but don't bother — x 1 lim 3 : -8 lim •2 mi LV X lim — x->3 x lim — x-»;y jc lim —x y-*x Va + h — vi sfa lim Find the following .. Inn x-»-l 1 - v7^ limits. 2; listing some algebraic ma- be sure you know which them.) . 108 Foundations - 1 -Vi -v 2 lim A - 3. -v7 -x 2 ! 1 In each of the following cases, determine the limit prove that for all (i) the limit by it is < x satisfying f( x ) = fix) =x 2 + 5x - [3 a- cos(a = a fix) showing how < a\ 2 )], a a = -2, 100 (m) — \x to find a 8 / for the given a, such that \f(x) — 1\ and < e 8. = 0. 2. 1. A" ,4 = x, fix) iv) A4 arbitrary a. vi) = /(*) - f{x) viii) /(a) I. r sin" — 2 vii) + a , =0. a = vM? a = 0. = y^", a = \. For each of the functions in Problem 4-17, decide for which numbers a the imit lim f(x) exists. x—*a *5. Do a) the same for Same problem, b) each of the functions if we in Problem 4-19. 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 all a, if < a — if < I For each £ > \x 21 — 2| < find a 8 (i) if < |a (ii) if < |a - 2| < - 2| < if < |a — (iii) { } < 21 < > + sin' e then |g(A) , such < if |a — 21 < 8, then 5, then |/(x)g(*) <5, then 8, |/(jc) then — — a\ < |a l\ < < s 8/2. when < / \x < 4| — 2| < e > e, s - 6\ < - 8| < £. £. < £. 2 for — < s. 4 : Give an example of a function |/(.v) — +g(Jt) Six) If then |/(a) all that, for all a, g(x) (iv) V ; e, for a\ which the following assertion < 5, then |/(a) — /| is false: < £/2 when 109 5. Limits 8. (a) fix) and lim gix) do not exist, can lim [/(*) x-^a x—>a x—>a lim fix) gix) exist? x—*a (b) If lim If lim + g(x)] Can exist? (c) fix) exists and lim[/(A') + g(x)] exists, must lim g(x) exist? x—*a x—>a x—>a If lim f{x) exists and lim g(x) does not exist, can lim [f(x)-\-g(x)] exist? (d) If lim fix) exists and lim f(x)g(x) x^-a exists, does follow that lim g(x) it x^-a x^>-a exists? 9. = Prove that lim f{x) + lim f(a ' x-+a (This h). is mainly an exercise in under- /,^0 standing what the terms mean.) 10. (a) = = I if and only if lim[/(.t) — /] 0. (First see why x—*a x—»a the assertion is obvious; then provide a rigorous proof. In this chapter Prove that lim f(x) most problems which ask (b) (c) — Prove that lim f(x) v^O Prove that lim f(x) — lim fix X—>0 Give an example where lim Suppose there is that lim f(x) — a 8 > f(x) exists, lim g(x). x—>a are a "local property." 12. (a) same (It but lim f(x) does not. x—>0 g(x) < when \x — a\ < — How If Prove x^>-a this fact is often expressed will clearly help to use 8' that f{x) < g(x) provided that these limits (c) 8. In other words, lim f(x) depends only on the by saying that limits some other letter, or , for all x. Prove that lim f{x) < x—>a (b) way.) the definition of limits.) 5, in Suppose = such that f{x) x—*a values of f(x) for x near a instead of in the ). x—>-0 11. be treated a). x-*a = X—*0 (d) for proofs should lim f(x lim g(x), x—*a exist. can the hypotheses be weakened? < f{x) gix) for all x, does it necessarily follow that lim fix) < x—*a lim gix)? x^a 13. Suppose hix) and that lim fix) = lim hix). Prove that x—>a x—*a and that lim gix) — lim fix) = lim hix). (Draw a picture!) x—>a x-+a x—>a that fix) lim gix) exists, x—>a 44. (a) Prove that gix) < lim fix)/x x—>-0 = / and b Write fibx)/x = b[f(bx)/bx]. (b) What happens if b (c) Part this 15. if < — ^ 0, then lim f{bx)/x bl. Hint: 0? enables us to find lim(sin2x)/x in terms of lim(sinx)/.t. Find x—>0 x—»0 limit in another way. (a) Evaluate the following limits in terms of the number a — lim (sin x) /x. x—*0 (i) = x—>0 lim ™3». x sinajc lim 0 sin bx — 110 Foundations 2a sin lim iii) x x^-0 2a sin lim iv) A^O z a — 1 ,. cos a lim - x^-0 x2 vi) km o vii) lim . tan 2 a + 2x r x+x- a v^O 1 x sin — COS* + /?) — sin (a a sin lim viii) h sin (a lim ix) x- x-»l a 2 (3 lim x^-0 (x 16. a) 1) 1 + sin a) - + sin a) 2 1 — lim (a v-1 xi) — 2 Prove that sin 1) \x , = lim f(x) if 1 b) Prove that |/|. .r—*a = lim f(x) if = then lim |/|(a) l, x^>a ' — and lim g(x) I m, then lim max(/, g)(x) x-*a x—*-a = x-+a max(/, m) and similarly for min. 17. a) Prove that lim \/x does not exist, i.e., show number every b) 18. if such that Hint: 19. *20. l/(.v |/(Jt) < | Why does M Prove that f(x) if (a) Prove that if — x is — a\ < 8. \x prove that /— Generalize a false for exist for any a lim g(x) = ^ > <5 and a number M x and fix) — 1 for rational x, a. and f(x) for rational x, if number (What does this mean pictorially?) < f(x) < /+1 forO < \x—a\ < 8? 1 for irrational = —x for irrational x, then 0. 0, then lim #(a ) sin l/.v = 0. x—*Q x—*0 (b) is exist. then there < if lim f(x) does not exist 21. /, suffice to it does not 1) if f{x) — then lim f(x) does not Prove that — = lim f(x) X—fCl I v^O /. Prove that lim Prove that = that lim 1/a x-^0 this fact as follows: If lim g(x) = and \h(x)\ < M for all a, x—>0 then lim g(x)h(x) — 0. (Naturally it is unnecessary to do part (a) if you x—»o succeed in easier than doing part (a) that's (b); actually the statement of part (b) one of the values of generalization.) may make it 111 5. Limits 22. Consider a function / with the following property: if g is any function for which limgOc) does not exist, then lim[/(x) + g(x)] also does not exist. x—*0 .*—>-0 Prove that happens this and only if Km if f(x) This Hint: does exist. is jc-»-0 an actually very easy: the assumption that lim f(x) does not exist leads to x—*0 immediate contradiction if you consider the right g. **23. This problem when f + g is replaced by f g. more complex, and the analysis the analogue of Problem 22 is In this case the situation considerably is • requires several steps (those in search of an especially challenging problem can attempt an independent Suppose (a) solution). and that lim f(x) exists is ^ 0. = oo. Prove that if x— () exist, then lim f(x)g(x) also does not exist. lim g(x) does not x—»0 x—>0 Prove the same result (b) sort of limit Prove that (c) lim if |,/"(jc)| (The precise definition of given in Problem 37.) is neither of these two conditions holds, then there if g such that lim g(x) does not a function is but lim f(x)g(x) does exist. x—*0 x—>0 Hint: Consider separately the following two cases: (1) for some £ we have > |/(jc)I exist, choosing points x n with Suppose that A n and [0, 1], / > s for all sufficiently small x. (2) For every e < £. In the second < \/n and |/(jc„)| < \/n. are arbitrarily small x with |/(jc)| *24. this \x n \ case, > 0, there begin by number n, some finite set of numbers in A„ and A m have no members in common if m ^ n. Define that for each natural is, as follows: v et f \ Prove that lim f(x) = x 1/", in An x not (J, in A„ lor any n. for all a in [0, 1]. x—*a 25. Explain why For every *26. 8 the following definitions of lim f(x) > there (ii) if < < \x — a\ < — a\ < (iii) if < |jc — (iv) if < \x - a\ (i) if \x Give examples a\ to < < an is e > such e, then \f{x) s, then e, then \f{x) |/(.v) — < - 1\ < — < 1\ I\ I are all correct: jc, 8. S. 58. e/10, then \f(x) -l\ show = x-*a that, for all < 8. that the following definitions of lim f(x) = I are not x—*a correct. (a) For all 8 > there is an e > such that if < \x — a\ < 8, then then < \f(x)-l\ <e. (b) For I jc — all s > a\ < 8. there is a 8 > such that if |/(jc) — /] < s, . 112 . Foundations 27. For each of the functions in Problem 4-17 indicate for which numbers a the one-sided limits lim fix) and lim fix) x—>a + x—*a~ *28. (a) Do (b) Also consider what happens same the each of the functions for decimals ending in 29. Prove that lim fix) in Problem 4-19. decimals ending in if are used instead of O's 9's. exists if lim f{x) x—«+ ' x^-a 30. exist. — lim f(x). x^-a~ Prove that lim (i) = f{x) lim f(—x). lim/(|*|)= lim f(x). (ii) x->0 + x^>0 lim fix (iii) = ) lim f(x). x—»0 + x-+0 (These equations, and others They might mean only if one of the a certain that like that the them, are open to several interpretations. two limits are either limit exists, then the other exists if equal if they both limits exists, the other also exists and is and is equal to exist; or that equal to it. it; or Decide for yourself which interpretations are suitable.) 31. Suppose that lim fix) < lim fix). X—a+ x—>a~ Prove that there sertion.) < x < a 32. y and — \x a\ Prove that lim ia„x" < (Draw a picture to illustrate this as- some 8 > such that fix) < fiy) whenever and |v — a\ < 8. Is the converse true? is 8 h H a )/ib m x'" h bo) (with H an / and bm X-^-OQ exists if and only if m > Hint: the one easy limit n. What lim X—>00 only information you need. 33. Find the following lim x-*oo 1 limits. • , 3 Dx + 6 xsinx 1 *-v-» (ii) (iii) is 1/jc — + sin —x x (i) is c 1 lim X—»00 • Vx2 + x — x. x 2 i\ (iv) lim jr->oo + sin x) ~^— + Sinx) z : ix 34. Prove that lim fi\/x)x^0+ 35. Find the following sin (i) x^-oo (ii) x lim . x lim x sin x-*oo — x limits lim fix). the limit = 0; when m — n? do some algebra ^ When m > so that this is 0) n? the . . 5. 36. Define " lim x-*— oo (a) Find f(x) lim (an x n + • • + a )/(bm x m + = lim f(—x). • (b) Prove that lim f(x) (c) Prove that lim f{\/x) = lim x->—oo x-+0- for define lim f(x) x—*a all x, if (Of course, = +b • ). x—>— 00 X—f-OO We 113 = L" X^>— 00 37. Limits mean oo to f(x). that for all TV there is a 8 > such that, < \x — a\ < then f{x) > N. (Draw an appropriate picture!) we may still say that lim f(x) "does not exist" in the usual sense.) <5, x—>a (a) Show (b) Prove that that lim l/(x x—*-3 if — > f(x) 3) s 2 = > oo. for all x, x—*a lim f(x)/\g(x)\ x—*a 38. (a) = Define lim f(x) x^>-a + self that many 39. = and lim g{x) = oo and lim f{x) 0, then oo. = oo. (Or at least convince your- x^f-a" you could write down the definitions if you had the energy. How other such symbols can you define?) — (b) Prove that lim (c) Prove that lim f(x) Find the following 1 /x limits, oo. = oo when if and only they if lim f{\/x) = oo. exist. 3 ... 1 + 4.y-7 x^-oo l~i lx £ — X TT + 7 lun x(\ + sin"x). a- lim 1 (ii) x—*oc (iii) (iv) lim x sin x x—»oo lim x sin — x .r—>oo vi 2 + (v) lim x—>oo (vi) lim jc(vx x—»oo 2x — +2— x. a/^")- \X Vll lim j:-*oo 40. (a) x Find the perimeter of a regular rc-gon inscribed in a circle of radius [Answer: 2rn sin(7r/n).] (b) (c) What What value does this perimeter approach as n becomes very large? limit can you guess from this? r. a 114 a Foundations 41. For c (a) Hint: More (b) > 1, show Show that generally, that c for c if any > — l/n 0, £ 'l/c > thought of a much then c simpler as n 1 we cannot have Alter sending the manuscript for the I approaches way 1//,? approaches first to 1 c as n edition of this without going through , the factoring tricks all example, that we want to prove that lim x £ > (see a~ — 0, we simply Figure s 19); < x~ < a~ + already been FIGURE 20 is set, \x £, > 1 +e large. for large n. becomes very large. book off to the printer, = a 2 and lim jc 3 = = x—>a on page 92. Suppose, for a", where a > 0. Given minimum of v a 2 + e — a and a — v a 2 — e < 8 implies that \l 2 — e < x < \J 2 + s, so — a < £. It is fortunate that these pages had be the let 8 then becomes very prove that lim x 2 x—>a a l/n — a\ or \x~ so that I couldn't completely fallacious. Wherein \ make lies these changes, because this "proof" the fallacy? CONTINUOUS FUNCTIONS CHAPTER / If is an arbitrary function, not necessarily true that it is lim f(x) = f(a). X—X2 In many ways there are fact, even be defined at a, in this can fail to be true. For example, / might which case the equation makes no sense (Figure not 1). Again, lim f(x) might not even FIGURE / if exist (Figure 2). Finally, as illustrated in Figure 3, x—*a is defined at a and lim f(x) exists, the limit might not equal f(a). 1 S (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 continuous if has been adopted the graph contains and the DEFINITION The at its graph precise definition function / is "continuous." no breaks, jumps, or wild is (a skill well will have no a function oscillations. worth cultivating) it is is / is Although continuous easy to be fooled, important. very continuous at a if lim f(x) We Intuitively, enable you to decide whether a function this description will usually simply by looking is difficulty finding = f{a). many examples of functions which are, or are — some number a every example involving limits provides an example about continuity, and Chapter 5 certainly provides enough of these. not, continuous at I I ( , I K X. 3 The at 0, we function f(x) and the same is = sin \/x is not continuous at 0, because true of the function g(x) = x sin i/x. are willing to extend the second of these functions, that 115 it is On is, if not even defined the other hand, we wish if to define , 116 Foundations a new G function by G(x) x = sin jx 1 a, then the choice of a — at to do extension this = G(0) can be made we can (if fact, not possible for /; is F not be continuous will ^0 x = way such a in 0, we must) define G(0) we define that =0 G will (Figure be continuous This sort of 4). if sin 1/a, x^O a, x F(x) then x = 0, no matter what a at 0, is, because lim f(x) does x^0 not exist. The function /(*) is not continuous = /(0), so The / at a, if is a ^ — c, x, x rational 0, x irrational lim f(x) does not 0, since continuous functions f(x) = at precisely g(x) = x, one point, — and h{x) exist. However, lim f{x) = 0. x are continuous at num- all bers a, since lim fix) = lim c = c = f(a), lim g(x) = lim x = a = g(a), lim = lim x x—*a x^a x^*a /i(a") x-*a Finally, In Chapter 5 but you can x l If is to give continuous a for Since all a). if a a (actually, only for all is examples of continuity irrational, if (2) if g (a) ^ 0, / / +g g • is is continuous continuous at a, at a. then (3) \/g is continuous at a. < a < 1, = f{a) only when a is but not if a is rational. we prove two are continuous at a, then (1) Moreover, at in lowest terms. p/q that lim f(x) x—>a easily see that this is true for even easier / and g = = we showed irrational, this function THEOREM h(a). x irrational 0, = 4 It is = a" consider the function /(*) FIGURE = simple theorems. PROOF / and g Since are continuous at a, — lim f(x) x By Theorem —a ' of Chapter 5 2(1) lim(/ and is (3) left + g)(x) = g{a). + g{a) = f(a) / +g is + g)(a), (f continuous The at a. proofs of parts (2) to you. | = Starting with the functions f(x) for every a, — lim g{x) this implies that just the assertion that are and f{a) .r—>a x—*a which 117 Continuous Functions 6. we can use Theorem which are continuous x, at a, + bn -ix n - +--- + b c m x "+cn,_ x" -' + --- + c bn x n = /(*) = conclude that a function to 1 and f(x) c 1 l 1 l is continuous at every point in than sin When we that. is continuous at domain. But its a for all assume a; let us /(*) now be proved it meanwhile. this fact sin = 2 in its = sin(x"); translate the equation lim f(x) we obtain > But in this case, where the = there — < a\ limit is THEOREM 2 If g if is = x Let s it is continuous (Notice that proof a > 0. / We is worth noting. If 8 > such then \f(x) limits, that, for all x, — < f(a)\ s. \x — a\ < 8 to the simpler condition \x since is /(a), the phrase < may be changed obviously f(a) according to the definition of is 8, we still functions. Before stating this theorem, the following point about the definition of continuity we \x function like But we are domain. f(x) like need a theorem about the composition of continuous < A a unable to prove the continuity of a function if further x continuous at every point for every e much be easy to prove that will x + x suijc +,2,427 i~- 2 sin" x + 4*- sin x • can harder to get is it discuss the sine function in detail — a\ < certainly true that \f(x) at a, and / is continuous S, — f(a)\ at g(a), < s. then /og required to be continuous at g(a), not at wish to find a if \x -a\ < 8 8. > such that for then \(f i.e., o g)( x ) - all (/ is a.) x, 6 g)(a)\ \f(g(x))- f(g(a))\ <e. < e, continuous at a. 118 Foundations We / use continuity of first that for all In particular, this if We now use — g{a)\ < - < g(a)\ (2) and if now 8' to hold. We you order such s. - f(g(a))\ < s. we (3) |.v — if \x < a\ 8, see that for -a\< 8, 8' as the e > then \g(x) — is in the definition (!) such that, for g(a)\ < of all v, 8'. x, all - then \f(g(x)) < f(g(a))\ e. | reconsider the function already noted that fl/0. Functions for < f(g(a))\ a 8 conclude that there / continuous is together with the continuity of 2, > how close x must be to a in order for the The number 8' is a positive number just like x sin \/x, '0. and - 8\ then \f(g(x)) number; we can therefore take positive (3) We have 8' a continuity of g to estimate continuity of g at a. can to g(a) in is that \g(x) if inequality \g{x) Combining \y-g(a)\ < 8\ then |/(j) means (2) We / must be close g(x) continuous at g(a), there is y, (1) any other how to estimate Since for this inequality to hold. = f{x) like sin(.r~ * sin(jc ^ = 0. A few applications of Theorems at 0. show sin, + x that + / is 1 also continuous at a, for should be equally easy sin (x ))) to analyze. The few theorems of this chapter have all been related to continuity of functions a single point, but the concept of continuity doesn't begin to be really interesting at we focus our attention on functions which are continuous at all points of some If / is continuous at x for all x in (a,b), then / is called continuous on (a, b); as a "special case", / is continuous on R = — oo, oo) [see page 57] if until interval. ( it is a continuous little at x for (2) all jc x differently; a function (1) (We all in R. Continuity on a closed interval must be defined / is called /is continuous lim f{x) = continuous on at its all x in (a,b), f(a) and lim f(x) also often simply say that a function in x for is [a, b] if continuous = f(b). if it is continuous at x for domain.) Functions which are continuous on an interval are usually regarded as especially well behaved; indeed continuity might be specified as the "reasonable" function ought to scribed, intuitively, as satisfy. A first continuous function one whose graph can be drawn without from the paper. Consideration of the function ^{X) Y %__ f ~] ' *sinl/*, x#0 0. x = condition which a is sometimes de- lifting your pencil 6. shows that there are this many description is a too optimistic, but little 119 Continuous Functions it is nevertheless true that important results involving functions which are continuous on an There 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 arguments, it is included here as a preview of things to come. interval. THEOREM 3 for all x in some at a, and fia) > 0. Then fix) > more precisely, there is a number > such that fix) > for all x satisfying \x — a\ < 8. Similarly, if f{a) < 0, then there is a number 8 > such that f{x) < for all x satisfying \x — a\ < 8. Suppose / is continuous interval containing a; PROOF <5 Consider the case fia) > 8 > such 0. Since / |.v — a\ < In particular, this must hold for s 8 > so that for —e < + e=\fia)_ fia) fia)-e= \fia) and this itself, A = then -\fia) implies that fix) > \ fia) > is more leading to a proof that similar proof can be given first < f(a)\ fix) — ?/(«), since 8, can apply the FIGURE every s > there is a e, fia) j < s. fia) > (Figure Thus 5). x, - a\ < if \x fia) all — then \fix) 8, i.e., is at a, for that, for all x, if there continuous is 0. < fix) - fia) < \f(a), (We could even have picked elegant, but in the case case to the function fia) — /. more confusing < 0; take s = s to be fia) to picture.) —\f(a). Or one | 5 PROBLEMS 1. For which of the following functions domain R such (i) fix) = (ii) fix) = that Fix) — / fix) for is all there a continuous function x in the F with domain of /? -4 x-2' 2 x \x\ X 2. (iii) fix) (iv) f(x) = = 0, x irrational. l/q, x = p/q rational in lowest terms. At which points are the functions of Problems 4-17 and 4-19 continuous? . 1 20 Foundations 3. Show Suppose (b) / /(0) must equal 0.) Give an example of such a function / which is not continuous ^ a (c) / that a function satisfying \f{x)\ is continuous is for \x\ x. all that at 0. (Notice that any at 0. Suppose that g Prove that / 4. < (a) continuous is continuous is = and g(0) at 0, and < |/(a)| |g(Jt)|. at 0. Give an example of a function / such that / continuous everywhere. continuous nowhere, but is \f\ is 5. number For each which a, find a function continuous is but not at any at a, other points. 6. (a) Find a function / which discontinuous at is 1, \, A, 1, . . . • • but continuous at all other points. (b) Find a function / which continuous 7. Suppose 8. 9. / that / Suppose that /+a nonzero (a) is / continuous — (b) > f(a)\ Conclude > e s. that for trarily close to Prove that every function E (c) £ > 0. / but continuous is Prove that if a / at a. 0, then Prove that for continuous / at a, numbers x arbithere are numbers x arbitrarily or then so continuous on is there are |/|. R can be written f = E + O, O odd and continuous. / and g are continuous, then so are max( if at 0, arbitrarily close to a with either even and continuous and Prove that min(/, (d) is = numbers x < f(a) — s > f(a) + e. (b) where and , Illustrate graphically. Prove that is f(y), and that not continuous is a with f{x) / • interval containing a some number (a) if 1, j? ^, \, all a. a and f(a) there are close to a with f(x) 10. + f(x) a for defined at a but is some number |/(jc) at at some open in = y) continuous is is Suppose / + f(x satisfies Prove that at 0. discontinuous at is other points. at all is /", g) and g). Prove that every continuous / can be written / = g — h, where g and h are nonnegative and continuous. : Theorem /(*)= l/x. 11. Prove 12. (a) Prove that 1(3) /' if is by using Theorem 2 and continuity of the function continuous at / and lim g(x) = x—*a /(/). (b) G Show continuity of that if with G(x) true that lim f(g(x)) x—-a /(0= I- ' = = / g(x) for x at / is ^ a, then lim f(g(x)) it is and G(a) not assumed, then /(lim g(x)). Hint: Try f(x) x—>a = x—*a (You can go right back to the definitions, but the function /, easier to consider = it = I.) is not generally for x ^ I, and . 6. 13. (a) Prove that / if continuous on is continuous on R, and which is then there [a, b], satisfies gix) 121 Continuous Functions — a function g which is fix) for x all in [a,&]. Since you obviously have a great deal of choice, try making g constant on ( — oo,a] and [b, oo). Hint: (b) Give an example to show that this assertion false if [a is b] , replaced is by (a,b). 14. (a) Suppose that g and h are continuous at a, and that g(a) — h(a). Define f(x) to be gix) if jc > a and h(x) if x < a. Prove that / is continuous at a. (b) Suppose g g(b) = Show continuous on [a.b] and h is / that continuous on is [b, c] Let f(x) be g(x) for x in [a,b] and h(x) for x in h(b). continuous on is [a, c]. and [fr,c]. (Thus, continuous functions can be "pasted together".) 15. Prove that whenever 16. (a) if / \x — a\ < continuous is and 8 then for any £ at a, \y — a\ < > 8 if that lim f(x) x—>a + < 0, then there for a is <x —a < 8 version of Theorem there \f(x) number > 8 is a 8 > < s. f(y)\ Then 0. there < x — x satisfying all — so that 3 for "right-hand continuity": f(a), and f(a) > > such that f(x) f(a) = we have Theorem Prove the following version of Suppose 8, > is < a a 8. Similarly, < such that f(x) number for all x satisfying (b) Prove a when 3 = lim f(x) f(b). x—*b~ 17. If lim x—*a f(x) but exists, continuity is ^ f(a), then / removable said to have a is dis- at a. f{x) — \/x for x sin ^ and /(0) = does / have a removable (a) If (b) x sin i/x for x / 0, and /(0) = 1? fix) Suppose / has a removable discontinuity at a. Let gix) = fix) for x ^ a, and let gia) = lim fix). Prove that g is continuous at a. (Don't What discontinuity at 0? 1, = if x-*a work very hard; (c) Let fix) terms. *(d) Let is = What / be if is quite easy.) this is x irrational, is and let fip/q) the function g defined by gix) = = \jq p/q if is in lowest lim fiy)? a function with the property that every point of discontinuity a removable discontinuity. This means that lim fiy) exists for all x, V—KV but / may be Define gix) = discontinuous at v— so easy as part **(e) Is some (even lim fiy). Prove that g * is infinitely many) numbers continuous. (This x. not quite (b).) there a function / which is discontinuous at every point, and which has only removable discontinuities? now, but mainly as a test (It is worth thinking about of intuition; even if Problem 22-33.) this problem you suspect the correct answer, you will almost certainly be unable to prove time. See is it at the present THREE HARD THEOREMS CHAPTER This chapter is devoted to three theorems about continuous functions, and some of their consequences. The proofs of the three theorems themselves will not be given until the next chapter, for reasons which are explained at the end of this chapter. THEOREM l If / is < continuous on [a,b] and f(a) < such that f(x) = fib), then there is some x in [a,b] 0. means that the graph of a continuous function which starts horizontal axis and ends above it must cross this axis at some point, as (Geometrically, this below the theorem 2 in Figure If / is 1.) continuous on [a b] , some number N THEOREM 3 bounded above on then / is < N for such that f(x) (Geometrically, this theorem allel to , means is continuous on [a,&], then there f(y) > fix) for all x , that is, there is graph of / lies below some line par- 2.) / If , x in [«,&]. all that the the horizontal axis, as in Figure [a b] is some number v in [a,b] such that in [a, 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, while FIGl RE FIGURE 122 2 FIGURE * . 7. Three Hard Theorems on a whole the hypotheses of the present theorems require continuity [a, b] — continuity if example, / be let fails hold to the function shown 1, Then / FIGURE continuous at every point of no point x v2 is sufficient to destroy the conclusion Similarly, suppose that / is such that fix) Then / is continuous of Theorem x ^0 0, x = 0. except 0, but [0, 1] < /'(2), 1 in Figure 5, \/x, = every point of at For 0; the discontinuity at the single shown the function /(*) N = is in [0, 2] < except V2, and f(0) [0, 2] but there point 4 is fail. 0<x < Vi V2<x <2 /(*) 2 interval in Figure 4, 1. 4l may a single point, the conclusions at 123 / is bounded above not we have f(\/2N) = 2N > N. 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), but on is 2N [0, 1]. In any number fact, for N > not bounded there. Finally, consider the function shown in Figure 6, X X < 0, X /(*) FIGURE On 5 the interval [0, does not f(y) > fix) for all x in [0, 1] so fiy) < the function if all x x 3 is 1 1. bounded above, is / is Theorem 3 in [0, 1]; in fact, so / does not continuous on — there is no y y < x Theorem 3 is considerably stronger than < nor can we choose 1, < v on its maximum As a compensation < / 1 for because 1. often paraphrased by saying that a continuous function interval "takes > f(x) certainly not true that f{\) it is But such that in [0, 1] = i satisfy the [0, 1]. we cannot choose v s any number with This example shows that Theorem / even though 2, conclusion of satisfy the fi x ) 1] Theorem conclusion of > Theorem 2. on a closed value" on that interval. 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 in- terval; V X to 1 To such "global" properties of a function are always significantly difficult prove than "local" properties, and are correspondingly of much greater power. illustrate the usefulness portant consequences, but FIGURE more of Theorems it 1, 2, will help to first and 3, we will soon deduce some immention some simple generalizations of these theorems. 6 theorem 4 If / is continuous on such that f(x) = c. [a, b] and f(a) < c < f(b), then there is some x in [a, b\ . 124 Foundations PROOF is theorem 5 = f — c. Then Let g some x = such that /(jc) PROOF The continuous, and g(a) is = orem 4 /GO = there c. is this > > c < < means By Theorem g(b). that fix) f(b), then there = is there 1, c. | some x in [a,b] c. —/ function But 0. continuous on [a,b] and f(a) /" is If g in [a,b] such that g(x) continuous on [a,b] and —f(a) some x in [a,b] such that — fix) < — c < —fib). By The- is = — c, which means that I Theorems 4 and 5 together show that / takes on any value between f(a) and f(b). We can do even better than this: if c and d are in [a, ft], then / takes on any value between /(c) and f(d). 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 in between; this slight generalization of Theorem is 1 often called the Intermediate Value Theorem. THEOREM 6 PROOF / If is continuous on [a b] , some number N The —/ function such that in [a, b], , / is > N for such that fix) is bounded below on then continuous on — fix) < M for all x in so we can let N = — M. all [a, b], so [«, b]. But x [a b] , , that is, there is in [a, b]. by Theorem 2 there this means that f(x) is > a M number —M for all x | Theorems 2 and 6 together show that a continuous function / on [a b] is bounded on [a, ft], that is, there is a number TV such that |/(x)| < /V for all x in In fact, since Theorem 2 ensures the existence of a number N\ such that [a b] < N\ for all x in [a, b], and Theorem 6 ensures the existence of a number fix) , , A^2 theorem 7 . such that f(x) / If for is all M > continuous on x in [a , x for all [a, b], in [a, b], then there is we can some y = take iV in [a, b] max(|A^i |, |A^|). such that /(y) < fix) b] (A continuous function on a closed interval takes on its minimum value on that interval.) PROOF The function —/ such that —fiy) all x is continuous on > ~f(x) for all by Theorem 3 there is some y in [a, b] [a,b], which means that /(y) < fix) for [a, b]; x in in [«,/?]. | Now that we have derived the trivial consequences of Theorems 1, 2, and 3, we can begin proving a few interesting things. theorem 8 Every positive number has a square some number x such that x~ = a. root. In other words, if a > 0, then there is Three 7. PROOF Hard Theorems 125 Consider the function fix) = x 2 which is certainly continuous. Notice that the statement of the theorem can be expressed in terms of /: "the number a has a , square root" means that will / on the value takes be an easy consequence of Theorem The proof of this a. fact about / 4. number b > such that f(b)>a (as illustrated in Figure 7); we can take b — a, while if a < 1 we can take b = 1. Since /(0) < a < f(b), Theorem 4 applied to [0, b] implies that for some x (in [0, b]), we have f(x) = a, i.e., x 2 = a. | There obviously a is a > in fact, if 1 same argument can be used Precisely the an nth number a any natural number n. If n happens to be odd, one can do number has an nth root. To prove this we just note that if the positive has the nth root x, has the nth root —x. root, i.e., if The = x" n a, then (— x) assertion, that for = —a (since n is odd), so odd n any number a has an nth equivalent to the statement that the equation is -a = x" has a root FIGURE number has root, for better: every —a to prove that a positive n if is Expressed in odd. way this the result is susceptible of great generalization. 7 THEOREM 9 If n is odd, then any equation x" +a n -\x n ~ x ^ = h«o has a root. PROOF We obviously want to consider the function = f(x) we would like to intuitive idea since n negative x . this A little The proper fix) is that for large is odd, is / prove that function algebra x" _1 + a„_ix"n-\ sometimes \x\, +a H and sometimes positive the function is very is all like = x n + a^i.r"- + 1 • • • + a = xn (1 + — l X Consequently, if Of - X^ we choose x x k \ > \x\ «o + < K-i + ••• satisfying > 1, 2n|fl„_i|, . . , . 2n|aol> and \a n -k\ ; — + X' |jc| (*) \x + < \a n -k\ < |tf#i-*l 2n\a n - k J_ \ In The = and, g(x) x idea work. + + that - + negative. n x and negative for large we need to make this intuitive / depends on writing analysis of the function a, then much positive for large positive is V Note ; l«ol «0\ x") 126 Foundations so a n -\ a n -2 X X «o JL ] l n 2n 2n 2 _ z X In other words. a "- 1 < ~ 2 i x ~ x" 2' which implies that 2 Therefore, if we choose an Ul) 2 so that f(x\) so that f{xi) > < 0. x\ "<u, , On would be ^ more ever, is > y (x 2 1 1 = , I RE 8 x2 < + ••• then (x 2 )" satisfies (*), = + (x 2 to the interval [x 2 x\] , < and f(X2), y we conclude that there is an x 0. | — x 1 =0, have a solution, and some, there to say? If we it problem of even degree equations completely however, the problem seems insuperable. first sight, said. Instead +a n _ + 1 = 0, more general Some do not equa- —what question, how- of trying to solve the equation x"- +...+fl 1 [ x like are willing to consider a something interesting can be =0, us ask about the possibility of solving the equations numbers c. +«„_,*"- depends on a fact which is +a = c This amounts to allowing the constant term ao to vary. The information which can be < if 1 frustrating to leave the for all possible I (x\)" J X] 9 disposes of the problem of odd degree equations so happily that -V" I a° 1 •X? x" let a "- + "fi then satisfies (*), 0. undiscussed. At tions, like which the other hand, such that f(x) Theorem > V Now applying Theorem in [x 2 x\] ) .r" jc given concerning the solution of these equations illustrated in Figure 8. The graph of the function f{x) = Jc"+<r/„_i.Y" _1 + + «(), with n even, contains, In other words, there is a at least the way we have drawn it, a lowest point. < number y such that f(y) the function / takes on a f(x) for all numbers x minimum value, not just on each closed interval, 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 [«,/?], and \a, b] obtain a point yo such that /(>'o) is the minimum value of / on [a, b]; but • • • — if happens to be the interval shown not be the place where / has its in Figure minimum example, then the point yo will value for the whole line. In the next 8, for Three 7. theorem the Hard Theorems proof is to choose an interval entire point of the [a, b] in such a 127 way that this cannot happen. theorem 10 If n is that f(y) PROOF As = even and f(x) in the < f(x) for + a n _\x n x" ~l + proof of Theorem all x with even, x" > for xn provided that such that b" \x\ all ., > M. Now — b, > be a number 2n\ao\), n -\\ + ••• ao a„_i + 5td + ... + ^) = /. consider the number > M. Then, also b if /w . f(0). Let ^ x > y^y^^ > b, we have (Figure 9) ( °)- then x" FIGURE number y such x, so < x .(i > 2/(0) and x < a x /.(*) Similarly, if is > M, we have \x\ 2 is then there 9, if + Since n + ao, • x. all M = max(l, 2n\a then for • - b) " „ /U)>y>-y( n 9 b" ^m = y>/(0). Summarizing: if Now > jc < -b, then f(x) > Theorem 7 to the function / on a number y such that is if (1) In particular, f(y) < (2) Combining (1) and Theorem 10 now li or x apply that there theorem Z? f(0). if x x < b, the interval [—b, b]. then f(y) < f(x). Thus < -b we (2) -b < /(0). or x > then f(x) b, see that /(y) < > /(0) > /(jc) for all x. | allows us to prove the following result. Consider the equation (*) x" +a n _ n-\ 1 ] x' 1 +---+a = c /(v). We conclude 1 28 Foundations and suppose n is even. Then there is a number > m and has no solution for c < m. m such that (*) has a solution for c PROOF Let /(*) = x" According Let m = the left f(y). (Figure 10). Theorem 10 there is a number v such that f(y) < fix) for If c < m, then the equation (*) obviously has no solution, > m. If c = m, then all x. since has y as a solution. (*) suppose c > m. Let b be a number such that b > y and f(b) > c. Then m < c < f(b). Consequently, by Theorem 4, there is some number x in = [y, b] to +oo ... 1 side always has a value Finally, /(v) + a,,-!*"- + = such that f(x) c, so x a solution of (*). is | 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 now — — numbers (namely, PI— PI 2) a proof is impossible. There are several ways of conthis gloomy conclusion is actually the case. For example, the proof of Theorem 8 relies only on the proof of Theorem 1; if we could prove Theorem 1 then the proof of Theorem 8 would be complete, 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 P1-P12. Again, suppose we consider the vincing ourselves that , figure: 10 function = fix) 1 -^r 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 If there . therefore on some property of the Despite our inability to prove which we want to be true. real numbers other than PI -PI 2. Theorems If the pictures nection with the mathematics we 1, 2, and 3, they are certainly results we have been drawing have any are doing, our notion of continuous function if corresponds to any degree with our intuitive notion, Theorems got to be true. property of way Since a proof of any of these R which and 3 have theorems must require some new has so far been overlooked, our present to discover that property: 2, 1, difficulties us try to construct a proof of let con- suggest a Theorem 1 , for example, and see what goes wrong. One I [GURE is, idea which seems promising is to locate the first point the smallest x in [a,b] such that fix) the set A which contains In Figure 11, x heavy line. is Since all numbers x is 0. in [a,b] such a point, while x' / = is not. To find this point, such that The where fix) set / A is first is A indicated by a here using the continuity of / on \a. /?], as well as Problem 6-16.) some (We are contains points sufficiently close to b are not in A. all 0, that consider negative on [o,*]. itself negative at a, and positive at b, the set points greater than a, while = f(x) < for all x in clearly a a < a < is We b. Suppose for all < A < 12); but this 0. prove to this A; to would be less than some numbers bigger 6-3, fix) in particular for contradicts the fact that a, is bigger than every member < is we see that Consequently, /(or) false. these points On the other hand, 12 suppose /(or) > 0. Again applying Theorem 6-3, fix) would be positive for all x in a small interval containing or, 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; is also false. Hence f(a) — and, we are tempted to say, Q.E.D. f{a) > We know, however, that something must be wrong, since no new properties of R were ever used, and it does not require much scrutiny to find the dubious point. It is clear that we can choose a number a which is greater than all members of A (for example, we can choose a = b), but it is not so clear that we can choose a smallest one. In fact, suppose A consists of all numbers x > such that x 2 < 2. could really be number V 2 did the members of A; If the all f(x) >0 for all in this interval x not exist, for any y > there least number greater than we could always choose a still would not be a V2 we chose, smaller one. Now that we have discovered the property of the real numbers 13 and 0, members of we only have all /(or) of A, since the larger numbers would also be in A. also contain only this big. FIGURE greater than is > 0. Then, by Theorem and x in a small interval containing a, than a (Figure FIGURE = claim that f(oc) that f(ct) first number which the smallest eliminate the possibilities /(or) all 129 this interval Now suppose A would Hard Theorems Three 7. That is we fallacy, need. All almost obvious what additional is it we must do say is it properly and use it. the business of the next chapter. PROBLEMS For each of the following functions, decide which are bounded above or below on the indicated interval, value. (Notice that and even if / and which take on might have these properties the interval on (-1,1). fix) fix) =x on /(*) = that —a f(x) (") f(x) (iii) (iv) maximum even if / is or minimum not continuous, not a closed interval.) is =x 2 =x 3 =x 2 W their on (-1,1). onR. [0, oo). x < a x", a 1 + 2, < a on {—a x > a + 1 ; it will — 1,0 + 1). (We assume a > — be necessary to consider several 1. so possibilities for a.) vi fix) = x a vn fix) = x < a , + 2, x > a on 0, x irrational \/q, x = p/q [- a in lowest 1 , a + 1] . on terms (Again assume a [0. 1]. > — 1 .) . 130 Foundations vin fix) = 1, [ { 1/g, [ ,.,,,, rx /(a) = ,./ v (xi) fix) (xii) /(a) a = • , on , terms in lowest p/q x irrational 1, — = = on on a terms in lowest [0, 11. [0, 11. m ^^ r .irrational sin" (cos [a] i , p/g rational Jf f = a l/<y, W /W= {0,*, \ irrational f \ [ / a. 1 + vo + a 2 ) „ , on [0, a']. [0,a]. For each of the following polynomial functions /, find an integer n such that for some a between /7 and n + 1. fix) — /(A)=A 3 -A + 3. 4 /(a) = a 5 + 5a + 2a + /(A) = A 5 +A + 1. /(a) = 4a 2 -4a+ 1. (i) (ii) (iii) (iv) 3. Prove that there *"" + (i) sin 4. a = If n — k — a This problem (a) some number a such is l^-rr- = + a- + sin" A 1 (ii) 1. is 119. 1 a continuation of and > even, is that Problem 0, find 3-7. a polynomial function of degree n with exactly k roots. (b) A root a of the polynomial function — — fix) have a as a root. Let if that / (a Suppose is / is that / is m a polynomial function of degree n. Suppose / be the roots. all multiplicity said to have a polynomial function that does not has k roots, counting multiplicities, of the multiplicities of 5. where g a)'" gix), continuous on [a, b] Show and i.e., suppose that k — that n that fix) is k is the is sum even. What always rational. can be said about /? 6. Suppose for 7. all that / a. (This is a continuous function on [— 1, means that (a, fix)) always either fix) =V — How many continuous functions 1 a 2 for all lies a, or else fix) / 1] such that x~ on the + (/(a))" = unit circle.) = — v — a2 1 for all Show 1 that x. = are there which satisfy (/(a))" x for all*? 8. Suppose that / and g are continuous, all 9. (a) a. Prove that either fix) Suppose that = gix) for that all 2 f = g , and x, or else fix) that fix) = —gix) ^ for for all only for x = a, and / is continuous, that /(a) = some a > a as well as for some x < a. What can be for fix) > about fix) for all x ^ a? v. that said 1 Three 7. some x < that / Now what can be said about fix) for x / a? sign of x + x y + xy + y when x and y are Suppose / and g are continuous on and [a, b] gib). Prove that fix) = gix) for some x in short, it's not the right one.) 12. a continuous function on Suppose that for each x (draw a picture). Prove that fix) / Problem (a) = is a, for a. Discuss the *(c) 11. 131 only for x continuous and that fix) = for some x > a and fix) < but suppose, instead, that fix) > Again assume (b) 10. Hard Theorems is shows that / 11 ure 14 (solid line). Show not both 0. < gia), but fib) > your proof isn't very that fia) [a,/?]. and that fix) is in some number x. [0, 1] =x (If [0, 1] for intersects the diagonal of the square in Fig- that / must also intersect the other (dashed) diagonal. Prove the following more general (b) FIGURE = g (0) 14 13. 0, g(l) = Let fix) (a) = 1 = or giO) x sin l/x for = g(\) 1, ^ fact: and let continuous on If g 0, then fix) is /(0) = 0. = Is and some x. [0, 1] gix) for / continuous on [— 1, 1]? Show that / satisfies the conclusion of the Intermediate Value Theorem on [— 1 1] in other words, if / takes on two values somewhere ; , *(b) on [— 1, 1], Suppose that also takes it on every value in between. / satisfies the conclusion of the Intermediate Value Theo/ takes on each value only once. Prove that / is continuous. to the case where / takes on each value only finitely many rem, and that Generalize *(c) times. 14. If / on (a) *(b) is a continuous function on : 15. maximum value Prove that for any number c Prove that + \\f g\\ that - \\h < ||/|| ||c/|| = \c\ / \\f\\. + \\g\\. Give an example where f\\ < \\h - g\\ + - \\g Prove that if n (b) Prove that if n x" + Hint: ||/ + g\\ / f\\. continuous and lim 0(x)/a" is (a) (a) | llsll. Prove that Suppose + we have = = X—»oo 16. of [0, 1]. 11/11 (c) be the [0, 1], let c()ix) for is is Suppose that / is even, then there a is number x such that x" + </>(.v) = 0. number y such that y" + </>( v) < a x. all Of which odd, then there lim 0(.v)/.v". x—>— oo proofs does this problem test your understanding? is continuous on (a, b) and lim fix) + lim fix) x—*a minimum on all of ia,b). corresponding result when a = — oo and/or = oo. Prove that / has a *17. (I)) Prove the Let / be any polynomial that |/()0| < |/(*)| for function. Prove that there all x. is b = oo some number y such 1 32 Foundations *18. Suppose that lim fix) X—*O0 = / *19. (a) Suppose / that that there Show < is some y is for x, all Prove that there is and some for all x. continuous on is other words there (b) > f{x) picture.) [a, b], a point on the graph of is to (y, /(y)) Figure fix). (Draw a lim X^r—OO that /(y) number y such > a continuous function with fix) is = and let x be any number. Prove / which is closest to (x,0); in from in [a b] such that the distance , distance from (x, 0) to f(z)) for (z, (x 0) , in [a, b]. (See all z 15.) same that this assertion is not necessarily true if [a , b] is replaced by ia,b) throughout. (c) (d) FIGURE 15 (e) is true if [a, b] is replaced by R throughout. and (c), let g(x) be the minimum distance from (jc,0) to a point on the graph of /. Prove that giy) < g(x) + \x — y\, and conclude Show (a) that g continuous. (a) is Prove that there are numbers *o and x\ in [a,b] such that the distance from (jto, 0) to (jci, f(x\)) is < the distance from (xq, 0) to (x\\ f{x\')) any for 20. that the assertion In cases xq' x\ in [a, &]. , Suppose / that continuous on is natural number. Prove that there f(x + \/n), / = g(x) *(b) satisfies FIGURE 16 + any *21. (a) shown < < a = /(0) but that a 1, / which < b, x then either fix) Why? in ia,b). < fia) fix) Refine part takes exist > In the on (c) on each value / (d) continuous on all case all first Hint: twice. fia) for x If in ia,b) or x? [0, 1] = and which fix + a) for / either it What can be fia) fix) = fib) for < fia) for in ia,b); this implies that no continuous function / which times or 2 times, i.e., which takes on exactly generally, find is does take on. Hint: or a minimum The previous hint implies value (which must be taken said about values close to the / which takes maximum on every value value? exactly 3 times. one which takes on every value exactly n times, if odd. Prove that if // is even, then there every value exactly /; times. Hint: is To no continuous / which takes on treat the case n = 4, for example, for all fix\) = fix 2) = /C*3) = /U4). Then either fix) > two of the three intervals ix\,X2), (^2^3)) (^3^x4), or else fix) < all x in two of these three intervals. let R defined on values close to fia), but slightly on somewhere maximum has either a twice). More is is a continuous function < a and x > b. by proving that there Find a continuous function // for all for x (a) twice each value that that / g(x) x. larger than fia), are taken (b) if not equal to \/n for any natural is which takes on every value exactly all Hint: Consider the function /(l), but which does not satisfy fix) Prove that there does not a = 4. what would be true l/«); Find a function n. is in Figure 16 for n — f{x + f(x) Suppose number X X as and /(0) = /(l). Let n be any some number x such that fix) — [0, 1] ,v in for . . LEAST UPPER BOUNDS CHAPTER Never- This chapter reveals the most important property of the real numbers. theless, it is merely a sequel Chapter to path which must be followed has 7; the already been indicated, and further discussion would be useless delay. DEFINITION A set A of real numbers bounded above is x > a Such a number x called an is if there is a number x such that for every a in A. upper bound 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 "A has an upper bound" when we mean that there is a number which is an upper bound for A — now been used in two ways first, in and now in reference to sets. This dual usage Notice that the term "bounded above" has Chapter reference to functions, 7, in should cause no confusion, since it will always be clear whether we are about a set of numbers or a function. Moreover, the two definitions are connected: above on The if A is the set {f(x) and only [a, b\ if entire collection examples of the : which is lj, 1^, A bounded not is {x :0 <x < 1}. is is any possible confusion (in the superlative of "less" means), we self-explanatory, in order to avoid particular, to ensure that we all know what define this explicitly. A number x is a least and 33 is bounded above we need only name some upper bound for A, easy enough; for example, 138 is an upper bound for A, and so are 2, and 1. Clearly, 1 is the least upper bound of A; although the phrase that just introduced DEFINITION / and the natural numbers N, are both bounded above. An example of a set which is A = To show function closely R of real numbers, which are sets bounded above if a < x < b}, then the set A is bounded above. talking upper bound (1) x is (2) if y of A if an upper bound of A, is an upper bound of A, then x < y. . 134 Foundations The to use of the indefinite article "a" in this definition Now temporary ignorance. seen that if that x and v are both least < y < x, we have made was merely a concession a precise definition, = upper bounds of A, then x it easily is Indeed, in v. this case x , and y since v is an upper bound, and x is a least upper bound, since is an upper bound, and y is a least upper bound; jc — y. For this reason we speak of the The term supremum of A synonymous and has one it follows that x is upper bound of A. least advantage. It abbreviates quite nicely to A sup and saves us (pronounced "soup A") from the abbreviation lub (which is There can is more treated number x such a authors). a series of important definitions, analogous to those just given, which is now be there some nevertheless used by A A briefly. set A of real numbers for every a in Such a number x is called a lower bound lower bound of A if ( The 1 ) (2) greatest lower if that x < a and bounded below is x is a lower if y is bound of A A A number for A. x bound of A bound of A, then x > the greatest , a lower is is also called the infimum y. A of , abbreviated inf A; some authors use the abbreviation gib A. One sets detail has have bound. at least We will — been omitted from our discussion so far the question of which one, and hence exactly one, least upper bound or greatest lower consider only least upper bounds, since the question for greatest lower bounds can then be answered easily (Problem If A is cannot be expected to have a have a least upper bound mathematical induction, If A = 0, 2). bounded above, then A has no upper bound not then A is if least it upper bound. It is at tempting has some upper bound, but, this assertion bounded above. can fail to all, be true in so A certainly to say that ,4 does like the principle of a rather special way. Indeed, any number x is an upper bound for 0: x simply because there there is surely no is least > y for every y in no y in 0. Since upper bound for every 0. number With is an upper bound (his trivial for 0, exception however, . 8. our assertion true is —and very important, We consideration of details. definitely 135 Least Upper Bounds important enough to warrant are finally ready to state the last property of the real numbers which we need. (The (PI 3) upper bound property) least /4/0, and A may Property PI 3 is you strike as anticlimactic, To complete our list of basic virtues. A If a set of real numbers, is bounded above, then A has a but that properties for the real least upper bound. particularly abstruse proposition, but only a property so simple that foolish for having overlooked it. Of course, really so innocent as all that; after FIGURE all, it the least does not one of actually is its numbers we require no we might upper bound property is hold for the rational numbers feel not Q A is the set of all rational numbers x satisfying x < 2, then there is no rational number y which is an upper bound for A and which is less than or equal to every other rational number which is an upper bound for A It will become clear only gradually how significant PI 3 is, but we are already in a position to For example, 1 if . demonstrate theorem 7-1 / If is such that f(x) Our proof Chapter 7 fix) < for all this interval 2 fix) > for all in this interval FI GUR E 3 x A ^ < f(b), then there is some number x \x A, shown in Figure : a 0, since <x < a and / b, in is 1, A; in < x < is fact, + number x in [a, b] at the with f(x) = end of 0. as follows: negative there is on the some 8 interval [a, x] > such that }. A contains Problem 6-16, since / is continuous on [a, b] and /(«) < 0. Similarly, b is an upper bound for A and, in such that all points x satisfying b — 8 < x < b are upper fact, there is a 8 > bounds for A; this also follows from Problem 6-16, 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) < and f(a) > 0. Suppose first that /(a) < 0. By Theorem 6-3, there is a 8 > such thai — for a 8 < x < a + 8 (Figure 2). Now there is some number xq in A f(x) < which satisfies a — 8 < xq < a (because otherwise a would not be the least upper bound of A). This means that / is negative on the whole interval [fl,xo]. But if X] is a number between a and a + 8, then /' is also negative on the whole interval [xo,jci]. Therefore / is negative on the interval [a,;q], so x\ is in A. But this contradicts the fact that a is an upper bound for A; our original assumption that must be false. f(cx) < all FIGURE < 7. 0. will locate the smallest set A= Clearly = merely a rigorous version of the outline developed is —we Define the x power, by supplying the proofs which were omitted in Chapter continuous on [a,b] and f(a) in [a, b] PROOF its points x satisfying a a 8; this follows from Suppose, on the other hand, that f(a) > 0. Then there is a number 8 > such that fix) > for a — 8 < x < a + 8 (Figure 3). Once again we know that there is an xq in A satisfying a — 8 < xq < ce; but this means that / is negative on [a, xq], 136 Foundations which is impossible, since /(*o) a contradiction, leaving f(a) The result, proofs of which will > = Thus 0- the assumption f(a) > as the only possible alternative. also leads to | Theorems 2 and 3 of Chapter 7 require a simple preliminary play much the same role as Theorem 6-3 played in the previous proof. THEOREM 1 / If is continuous at a, above on the interval PROOF Since lim f(x) — (a then there — a 8, f(a), there + a is 8) (see number Figure for every s is, > 8 > such that / is bounded 4). > 0, a 8 such that, for all x, x-*-a if \x It is for only necessary to apply example, e = 1 . We if It — a\ < this on the /(«)+!• then \f(x) |.v -a\ < interval {a — — < f(a)\ s. statement to some particular s (any one will do), conclude that there follows, in particular, that if \x the proof: 8, 8, then \f(x) — a\ < 8, is a + 8, a <5 - > such < f(a)\ then f(x) — that, for all , 1. f{a) 8) the function x / is < 1. This completes bounded above by I It should hardly be necessary to add that we can now also prove that / is bounded below on some interval (a — 8, 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 there .v—>a+ 'IGURE 4 . 8. is a 8 > / such that observation holds bounded on is — lim fix) if Least Upper Bounds 137 < x < a + 8}, and a similar Having made these observations (and the set {x f(b). : a x-*-b~ assuming that you THEOREM 7-2 PROOF / If is continuous on [a b] then / , , is our second major theorem. tackle bounded above on [a b] , Let x A / Clearly : a (since a <x < in A), is is bounded above on and A is bounded above to the sets {/(j) : < a < y x}, [a, x]}. A (by b), so has a least are here applying the term "bounded above" both which can be visualized to the set A, (Figure / b and upper bound a. Notice that we {/(y) :a we supply the proofs), will as lying on the horizontal which can be visualized as lying axis, and on the to /, i.e., vertical axis 5). Our first step is to prove that we actually have a = b. Suppose, instead, that a < b. By Theorem 1 there is 8 > such that / is bounded on (a — 8, a +8). Since a is the least upper bound of A there is some xo in A satisfying a— 8 < xq < a. This means that / is bounded on [a, xo]- But if x\ is any number with a < x\ < a + 8, then / is also bounded on [xo, jq]. Therefore / is bounded on [a,xi], so x\ 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 — 8, a + 8)]; the possibility a = a can be ruled out similarly, using the existence of a 8 > such that / is bounded on {x a < x < a + 8}. The proof is not quite complete we only know that / is bounded on [a, x] for every x < b, not necessarily that / is bounded on [a, b]. However, only one small argument needs to be added. There is a > such that / is bounded on {x b — 8 < x < b}. There is xo in A such that b — 8 < xq < b. Thus / is bounded on [a, xo] and also on [xo, b], so / is bounded on [a b] | <y <x : — FIGURE 5 <5 : , To prove theorem 7-3 / If is PROOF We all already theorem we resort the third important continuous on f{x) for . X [a, b], then there number a y in [a, b] such that f(y)> in [a, b]. know / that is bounded on {/(*):* is bounded. This Qf > fix) for is to a trick. x set is in [a, b\ Suppose instead that [a b] , a suffices to ^ f(y) for show all which means that the v in = <x ~ —-, fix) bound a. Since some y in [a, b\. has a least upper it that a. = [a,b]. f(y) x for Then the function by g(x) set [a,b]} in obviously not 0, so it , in [a, b\ g defined 1 38 Foundations is continuous on other hand, a [a, ft], for every e means This, in turn, upper bound of {/(*) x the least is : > //;« means there is x right side in [a, with a in [a, b] — is never ft]); this f(x) < 0. means On the that e. that for every e But denominator of the since the that g is > there is x in [a, bounded on not [a, with g(x) ft] > 1/e. contradicting the previous theo- ft], rem. | 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 "the real say, numbers look 12 n x 3 number x so every like between two integers is + n n, Basing the argument on a geometric picture 1 n is (unless x 1 is itself an integer)." not a proof, however, and even the geometric picture contains an assumption: that end you + if you place unit segments end-to- a segment larger than any given segment. This axiom, introduction to geometry, is usually attributed (not quite will eventually get often omitted from a first N is not Archimedes, and the corresponding property for numbers, that bounded, is called the Archimedean property of the real numbers. This property is not a consequence of PI -P12 (see reference [14] of the Suggested Reading), although justly) to it Q, does hold for of course. Once we have PI 3 however, there are no longer any problems. theorem 2 PROOF N is not bounded above. Suppose N bound a for were bounded above. Since N ^ 0, there would be a least upper N. Then a > for all n in // N. Consequently, a > since n + \ is in N if /; is in means this a the least is thai a- — 1 is also upper bound. -f N. But a and n I 1 > 1 this n for all means for all in // N, that /; an upper bound in N, for N, contradicting the fact that 8. There 3 PROOF For any e Suppose means A often > assumed there that 1/e > is e for number of many will them some cheating was tolerated. can claim that was never used has been shaken, a review of all As we that will < 1/e for n in N. But this all 2. | Theorem 3 Theorem 3 was not that the result of Of course, were so important that order to give in partial justification for this dishonesty proof of a in the the proofs given so far We such deception will not be necessary again. numbers n show you examples. available at the time, but the examples this result Thus s. N, contradicting Theorem for Chapter 6 in the discussion < n with \/n n in N. all an upper bound brief glance through the real 39 implicitly. a natural is not; then 1/n was used 1 a consequence of Theorem 2 (actually an equivalent formulation) which is we have very theorem Least Upper Bounds have now ever need. Henceforth, no theorem, is but if your we faith Fortunately, in order. stated every property of more lies. PROBLEMS 1 . Find the the following (i.e., to upper bound and the greatest lower bound least decide when belong to the — : Also decide which sets. the least upper sets (if have greatest and bound and greatest lower they least exist) of elements bound happens set). n in N n in Z and = or x n — : n ^ n (iv) — \/n for some in < x < V2 and x is rational}. 2 > 0}. x +x + 2 < 0}. :x + xx < and x 2 + x — < 0}. {x : [x : (v) [x (vi) {x (vii) {x A' N}. // 1 : 1 1 : 1 + (viii) 2. (a) (-!)" Suppose A : n ^ for x in A. inN is bounded below. Prove that —A ^ —A denote the set of all — x —A bounded above, and that Let 0, that is — sup (—A) (b) 3. Let (a) is the greatest lower bound of A. A 7^ is bounded below, let B be the set of all lower bounds of A. Show that B ^ 0, that B is bounded above, and that sup B is the greatest lower bound of A. If / be a continuous function on The proof of Theorem with f{x) = 0. is 7-1 If there is [a, b\ with f(a) < < f{b). showed that there is a smallest A" in \a.b\ more than one a in \a,b\ with /(a) = 0, there necessarily a second smallest? Show that there is a largest x in . 140 , . Foundations = \a,b] with f(x) (Try to give an easy proof by considering a 0. new function closely related to /.) (b) The proof of Theorem 7-1 depended upon considering A = ix a < x < b and / is negative on [a, x] }. Give another proof of Theorem 7-1 which depends upon consideration of B = {x a < x < b and f(x)< will this proof locate? Give 0}. Which point x in [a,b] with f(x) = an example where the sets A and B are not the same. : : *4. (a) that Suppose also that f(xo) numbers c / is > some for in [a, b]. xq, a<c<xo<d<b such and d with f(x) > = f(b) = 0. Prove that there are continuous on [a,b] and that f(a) Suppose that /(c) = f(d) = 0, (b) for x in (c, d). Hint: The previous problem can be used good advantage. Suppose that / is continuous on [a, b] and that f(a) < f(b). Prove that there are numbers c and d with a < c < d < b such that f(c) = f(a) and f(d) = f(b) and f(a) < f(x) < f(d) for all x in (c, J). (a) Suppose but all to 5. (b) x < k < v. consider + / < x > Prove that there 1. Let Hint: / an integer k such that is be the largest integer satisfying < I x, and 1 Suppose x < r — that y Prove that there v. is number a rational < y —x, then ny — nx > y. Hint: If \/n r (Query: 1. such that jc < Why have parts 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 and 1 Suppose that x < y. Prove that there is an irrational number between x and v. Hint: It is unnecessary to do any more work; this follows from (b) and (c). (a) (c) (d) *6. A set A of real numbers point of A and the set (a) (b) If / if is A, then f(x) if A, then f(x) all continuous and f(x) = for / and g we assume g(x) for 7. dense if every open interval contains a x. = all g(x) for Can > be if / is is a number for all > g(x) for replaced by + c such that f(x) lems.) Point of information: + y) = f{x) + There f(y) for all all numbers x in a dense = g(x) for all x in a dense x. all x in A, show that f(x) > x and v, > throughout? y) = = f(x) ex for can be demonstrated simply by combining the fact, this of rational numbers x. continuous and f(x Prove that = are continuous and f(x) instead that f(x) then there f(x set of irrational numbers are each dense. Prove that set (c) said to be For example, Problem 5 shows that the . Prove that set is + f(y) for all results x. of two previous prob- do exist noncontinuous functions x and y, but all (This conclusion we cannot prove / satisfying this now; in simple question involves ideas that are usually never mentioned in undergraduate courses (see reference [7] in the Suggested Reading). 8. *8. Suppose / that a function such that f(a) is ure 6). (a) Prove that lim f(x) and lim f{x) both x—>a~ lem / never has from Problem Prove that (c) if If / a is 6 Why is Hint: exist. «+ a removable discontinuity satisfies < b (this terminology comes is bounded function on [0, 1], let sup{ Problem in || || = |||/||| |/(jf)| : x in [0, lj}. 7-14. 11. (a) Suppose &n+\ S a\,a2,a$, that . .. is a sequence of positive > Prove that for any e «/?/2. Suppose P there many inscribed regular polygon with twice as ence between the area of the circle between the area of the difference Prove that there some is need part (a). sides, < n with a n show e. If P' is the that the differ- and the area of P' is less than half the and the area of P (use Figure 7). circle a regular polygon P inscribed in a circle with area as close as desired to the area of the circle. will is in the numbers with a regular polygon inscribed inside a circle. is The- continuous. Suppose a > 0. Prove that every number x can be written uniquely < x' < a. form x — kct + x', where k is an integer, and (c) prob- this 10. (b) (Fig- the conclusions of the Intermediate Value Prove analogues of the properties of FIGURF, a 6-17). / orem, then / *9. < f(b) whenever in this chapter? Prove that (b) jc— 141 Least Upper Bounds In order to do part This was clear to the Greeks, who used part (c) you as the (a) and area. By calculating method of exhaustion") allows basis for their entire treatment of proportion method the areas of polygons, this computations of that =jy *(d) < Using the 7i < tz =f. to ("the any desired accuracy; Archimedes used But it fact that the areas areas of two circles have the 7 Deduce a is to show- has far greater theoretical importance: of two regular polygons with the same ber of sides have the same ratio as the square of their FIGURE it same sides, num- prove that the ratios as the square of their radii. Hint: contradiction from the assumption that the ratio of the areas greater, or less, than the ratio of the square of the radii by inscribing appropriate polygons. 12. Suppose for all 13. that A and B x in A and all are two nonempty sets of numbers such that jc < y v in B. A < sup A < (a) Prove that sup y for (b) Prove that inf B. all y in B. A and B be two nonempty sets of numbers which are bounded above, and let A + B denote the set of all numbers x+y with x in A and y in B. Prove that sup(A + Z?) = sup A+sup B. Hint: The inequality sup(A+5) < sup A+supB is easy. Why? To prove that sup A + sup B < sup(A + B) it suffices to prove that sup A + sup B < sup(/4 + B) + s for all s > 0; begin by choosing x in A Let 5 .. . 142 Foundations and v in FIGURE B with sup 8 (a) 1 1 e/2. 1 1 ai fl3 bj, Consider a sequence of closed intervals Suppose there (b) < e/2 and sup B — y < x 1 | ci\ 14. A — a n+ \ a point x which is Show that a„ < and b n+ is that this conclusion \ < I\ — b„ for b\ £>2 \a\, b\], li all = n (Figure [«2> b{\, . . . Prove that 8). in every /„. is false if we consider open intervals instead of closed intervals. The simple result of Problem 14(a) may be is called the "Nested Interval used to give alternative proofs of Theorems 1 and Theorem." The 2. It appropriate reasoning, outlined in the next two problems, illustrates a general method, called a "bisection argument." *15. Suppose / continuous on [a,b] and f(a) is < < f(b). Then either + b)/2) = 0, or / has different signs at the end points of the interval [a, (a + b)/2], or / has different signs at the end points of [(a + b)/2, b]. Why? If f((a + b)/2) ^ 0, let I\ be the interval on which / changes sign. Now bisect I\. Either / is at the midpoint, or / changes sign on one of the two intervals. Let h be that interval. Continue in this way, to define /„ for f((a is at some midpoint). Use the Nested where f(x) = 0. / each n (unless to find a point x *16. Interval Theorem Suppose / were continuous on [<z,&], but not bounded on [«,/?]. Then / would be unbounded on either [a, (a + b)/2] or [(a +b)/2, b]. Why? Let /] be one of these intervals on which / is unbounded. Proceed as in Problem 1 to obtain a contradiction. 17. (a) Let A = If (i) x < a}. Prove the following (they are is in A and : x (iii) A ^ A ^ (iv) If (ii) (b) {x x v < x, then y is all easy): in A. 0. R. is in A, then there Suppose, conversely, that A is some number satisfies (i)— (iv). x' in A < such that x Prove that A = fv : x' x < sup A). *18. A number finitely x is called an many numbers y almost upper bound for A if there are only A with y > x. An almost lower bound is in defined similarly. (a) Find all Problem (b) almost upper bounds and almosl lower bounds of the sds 1 A is a bounded infinite set. Prove thai the set B of almost upper bounds of A is nonempty, and bounded below. Suppose in that all .. 8. : 19. follows from part that inf (b) B exists; this number called the limit (c) It (d) superior of A, and denoted by lim A or lim sup A. Find lim A set A in Problem 1. Define lim A, and find it for all A in Problem 1. If A is a bounded (a) lim A < (b) lim A < sup A (c) If lim (d) The analogues lim infinite set 143 Least Upper Bounds is for each prove A A < sup A then A , of parts contains a largest element. (b) and (c) for lim . shadow points FIGURE 20. Let of 9 / be a continuous function on R. / if there is a number y > x with terminology is A point /(>') > x is called a fix). The shadow point rationale for this indicated in Figure 9; the parallel lines are the rays of the sun Suppose that all points of (a, b) are and b are not shadow points. Clearly, f(a) > f(b). rising in the east (you are facing north). shadow (a) points, but that a Suppose that f{a) maximum (b) value on Then show f(a) = This little > f(b). [a, b] Show must be that the point where / takes on its a. that this leads to a contradiction, so that in fact we must have fib). result, known proving several beautiful page 450. Sun Lemma, is instrumental in theorems that do not appear in this book; see as the Rising 144 Foundations UNIFORM CONTINUITY APPENDIX. Now come that we've to the end of the "foundations," it might be appropriate to slip in one further fundamental concept. This notion is not used crucially the rest of the book, but it can help clarify many points later on. b 2 We know +e - that the function f{x) x 2 is continuous at a for all a. is some 8 in In other words, a if any number, then for every is such that, for Of course, depends on 8 not work at b (Figure that works - \x But even <5, for but 8, then also depends <5 Indeed, < a\ — a\ < if \x e. 1). for all a, or certainly satisfy x, all it's 0, — In 8 a8 a > s. the 8 that works at a might > there fact, the > a8, is no one number a > 8 +8/2 will then 8 Cl+ - w\ < \x~ clear that given e > a there on a positive a. all if > e 2 + 2 b^8 for b 8 for a 1 !(. i:RE and won't be this < once a > s/8. (This s way of saying tational / that is growing is an admittedly confusing compu- just faster and faster!) On the other hand, for any e > there will be one 8 > that works for all a in any interval [-#,#]. In fact, the 8 which works at N or -N will also work 1 everywhere else in the interval. As a final example, consider the function f{x) = sin l/x, or the function whose graph appears in Figure 18 on page 62. It is easy to see that, so long as £ < 1, that works for these functions at all points a in the there will not be one 8 > open interval (0, 1). These examples illustrate important distinctions between the behavior of various continuous functions on certain intervals, and there is a special term to signal this distinction. DEFINITION The function / is some 8 > there is uniformly continuous on an interval A such if |.r that, for all -y\<8, x and y then |/'U) if for every £ > in A, - f(y)\ < s. We've seen that a function can be continuous on the whole line, or on an open without being uniformly continuous there. On the other hand, the func2 tion f(x) = x did turn out to be uniformly continuous on any closed interval. This shouldn't be too surprising it's the same soil of thing that occurs when we interval, that — bounded on an interval and we would be led to suspect any continuous function on a closed interval is also uniformly continuous on ask whether a function is that interval. In order to prove this, we'll Suppose point b, thai and we have two a function /' that need intervals [a,b] is continuous on to deal and first [b, c] [a, c]. with one subtle point. with the Let £ > common and suppose endthat 145 Appendix. Uniform Continuity 8, the following two statements hold: (i) if (ii) if x and y are in x and y are in and and [a, b] [b, c] — y\ < — y\ < |jc |x 8\, then \f(x) 82, then |/(x) — f(y)\ < — f(y)\ < £, £. We'd like to know if there is some 8 > such that \f(x) — f(y)\ < £ whenever — x and y are points in [a,c] with |jc v < 8. Our first inclination might be to choose 8 as the minimum of 8\ and 8j. But it is easy to see what goes wrong (Figure 2): we might have x in [a,b] and y in \b, c], and then neither (i) nor (ii) tells us anything about |/(jc) — f(y)\. So we have to be a little more cagey, and | <«5 FIGURE also use continuity of 2 LEMMA < b < Let a c and / at b. / be continuous on the interval [a,c]. Let £ > and (ii) hold. Then there is a 8 > such that, let suppose that statements PROOF Since / if x and y are is continuous (i) and in [a, c] at b, there -b\< if \x It a is — \x < y\ > ($3 if \x Choose 8 to — < b\ and 82 minimum be the — \y of < b\ and 8\, 82, suppose that x and y are any two points are both in [a, b], then \f(x) — f(y)\ < £ by then \f(x) — f(y)\ < £ by (ii). The In either case, since \x — / It's is continuous on < y\ |/U)-/(v)|<eby(iii). PROOF < f(b)\ e. -. 8, We and (i); v have also claim that if < b < \x — f(y)\ |jc < e. works. In this 8 — y\ < 8. If x and y x and y are both in is [b, c], that x. b\ < 8 and |v b\ < 8. So I / then [«,/?], is uniformly continuous on the usual trick, but we've got to be a > ^3. in [a, c] with or we — then \f(x) only other possibility x < b < y If - 83, fact, 1 < f(y)\ such that, then \f(x) S3, — then \f(x) 8, follows that 111 THEOREM and 0, little bit careful [a, b], about the mechanism of > such Then A ^ (since a is in A), and A is bounded above (by b), so A has upper bound a. We really should write a B since A and a might depend on we won't since we intend to prove that a = b, no matter what e is. a least the proof. For £ that, for all v and z in [a if Then , b] is £-good on [a, b] if \y-z\ <8, then \f(y)-f(z)\ < Consider any particular {x : £ there is some > 0. 8 , we're trying to prove that A = / say that let's > / 0. a < x is £-good on e. [a, b] for all £ Let < b and / , is £-good on [a, x]}. £. But . 146 Foundations Suppose we had a < that / Since b. v — a\ < continuous is at a, there is some — <5o > — a\ < /(a?) < e/2. Consequently, if \y So <5o, then \f(y) — then < So is surely on the interval e. e-good \z So, / f(z)\ \f(y) [a — So, a +80]. On the other hand, since a is the least upper bound of A, it such that, if — and is | / also clear that e-good on I < ot\ + a [a, e-good on is 80], so + a [a, 8q is a — in 80]. Then the Lemma implies that A, contradicting the fact that a is / is an upper bound. To complete argument <5o > [b — the proof for this such that, But 80, b]. e-good on [a, b~\. we show just have to is practically the same: Since if b / is — 8q also < v < b, e-good on then |/(v) [a, b — that / — is a = b is actually in A. continuous f(b)\ 80], so the < e/2. at b, there So / Lemma is is The some e-good on implies that / is | PROBLEMS 1. (a) For which of the following values of a formly continuous on (b) (c) Find a function Prove that if (b) Prove that if fg (d) / that is — is Show x a uni- (0, 1], but not 1/3, 1/2, 2, 3? continuous and bounded on not uniformly continuous on (a) (c) a = the function f{x) Find a function / that is continuous and bounded on uniformly continuous on (0, 1]. is 2. [0, 00): is [0, 00) but which [0, 00). / and g are uniformly continuous on A, then so is f + g. / and g are uniformly continuous and bounded on A, then uniformly continuous on A. that this conclusion does not hold if one of them isn't bounded. Suppose that / is uniformly continuous on A, that g is uniformly continuous on 5, and that f(x) is in B for all x in A. Prove that g o / is uniformly continuous on A. 3. Use a 4. Derive "bisection Theorem argument" (page 142) to give another proof of Theorem 7-2 as a consequence of Theorem 1 1. 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 bodies. Between 1695 and 1700 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 VHopital 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. of almost precisely in the space one century infinitesimal calculus or, we now call The Calculus, as it in English, the calculating tool had been forged; and nearly three par excellence, centuries of constant use have not completely dulled this incomparable instrument. NICHOLAS BOURBAKI CHAPTER DERIVATIVES The derivative of a function Together with the its integral, particular flavor. While is it it is the first of the two major concepts of constitutes the source true that the concept of a function that you cannot do anything without bounds are essential, everything we have done limits adequate, this section will this section. from which calculus derives is fundamental, or continuity, and that least upper until now has been be easier than the preceding ones — preparation — if 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 cepts and stems from the intimate connection between the mathematical con- certain physical ideas. be described in Many definitions, terms of physical problems, often and even some theorems, may in a revealing way. In fact, the demands of physics were the original inspiration for these 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 functions exhibits such diversity that there all hope of discovering any interesting general properties pertaining to continuous functions form such a restricted class, we might is almost no all. Because expect to find some theorems pertaining to them, and the sudden abundance of theorems Chapter 6 shows that this expectation is justified. But the most interesting nontrivial after and most powerful our attention even results about functions further, to functions will be obtained only when we which have even greater claim to restrict be called "reasonable," which are even better behaved than most continuous functions. = \x\,x>0 f(x)=x 2 ,x<0 f(x) i k, i k i; i Figure display. 1 illustrates certain types The graphs of Figure 2, where FI G UR E 2 it is 149 these functions are "bent" at (0,0), unlike the graph of possible to marks have been used of misbehavior which continuous functions can to draw a "tangent each point. The quotation we have defined "bent" or line" at avoid the suggestion that 150 Derivatives and Integrals "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 FIGURE 3 — such a 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." once FIGURE FIGURE 4 definition 5 A more promising approach to the definition of a tangent line "secant lines," and use the notion of limits. If h (a, f(a)) and (a slope + h, f(a + /?)) ^ 0, might start with then the two distinct points determine, as in Figure 6, a straight line whose is f(a + h)-f(a) h \ FIGURE f(a + h)-f(a) 6 As Figure 7 illustrates, the "tangent some sense, of these "secant lines," talked about a "limit" of lines, but (a,f(a)) seems to be the limit, in approaches 0. We have never before line" at as h we can talk about the limit of their slopes: the « • * 151 9. Derivatives slope of the tangent line through (a, f(a)) should be f(a+h)-f(a) lim h We DEFINITION and some comments. are ready for a definition, The / function differentiable at a is f(a+h)-f(a) lim exists. h h->0 In this case the limit denoted by/' (a) and is (We also say that / domain of /.) a. if differentiable is is called the derivative /" is if of/ at differentiable at a for every a in the The comment on our first tangent line to the definition graph of / really is at (a, f(a)) to an addendum; we define the be the with slope f'(a). This means that the tangent line at / is refers to notation. iniscent of functional notation. function whose 7 through (a, f(a)) (a, f(a)) is defined only The symbol f'(a) is certainly if differentiable at a. The second comment FIGURE line at a, domain and whose value is at In such a any function /, we denote by /'the fact, for the set of numbers a such all number a rem- that / is differentiable is f(a+h)-f(a) lim h ft— (To be very precise: /' is the collection of f(a pairs all + h)-f(a) a, lim h /i— for which lim [f(a + h) — f(a)]/h exists.) The function /' is called the derivative ft—>0 of/. Our comment, somewhat longer than third ical interpretation the previous two, refers to the phys- of the derivative. Consider a particle which on which we have chosen an straight line (Figure 8(a)) direction in which distances from tance from O O shall been chosen purposely; be written as positive numbers, the since a distance s(t) t=0 motion of the t = • particle 5 t < > — =4 • l(, I RE 8 a t. is The dis- >_ ) t = along which particle is moving = 3 — suggestive nota- determined t=\ t • < line I moving along a and a of points in the other direction being written as negative numbers. Let s(t) denote the distance of the particle from O, at time tion s(t) has is "origin" point O, for each 152 Derivatives and Integrals number t, the physical situation automatically supplies us with a certain function s. The graph of s indicates the distance of the particle from (9, on the vertical axis, "distance" graph of s terms of the time, indicated on the horizontal axis (Figure in The 8(b)). quotient s(a + h) — s(a) h has a natural physical interpretation. during the time interval from a to a FIGURE 8(b) depends on of course. h, On It is + h. For any particular a, s{a + h) — h^O h We physical reasons for considering this limit. of the particle at average speed s{a) well as the particular function (as this the other hand, the limit hm depends only on a the "average velocity" of the particle would s) and there are important speak of the "velocity like to 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 s{a + h) — s(a) lim h Thus we define the (instantaneous) velocity of the particle at a to be s'{a). Notice that s'{a) could easily be negative; the absolute value It is important to realize that instantaneous velocity an abstraction which does not correspond precisely While it \s'{a)\ is sometimes (instantaneous) speed. called the would not be with average velocity, fair to to a theoretical concept, is any observable quantity. say that instantaneous velocity has nothing to remember that s'{t) s(t do not is + h)-s(t) h for any particular proaches measures 0. is /z, but merely the limit of these average velocities as h ap- Thus, when velocities are measured in physics, what a an average velocity over some (very small) time cedure cannot be expected to give an exact answer, but physicist really interval; such a pro- this is really no defect, because physical measurements can never be exact anyway. The velocity of a particle is 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) the mass at time In order to /. become familiar with the basic definitions of this chapter, spend quite some time examining the derivatives of particular functions. proving the important theoretical results of Chapter idea of what the derivative of a function looks exclusively i<> one aspect of cated functions. is this In this chapter — like. 1 1, we want to we will Before have a good The next chapter is devoted problem calculating the derivative of compliwill emphasize the concepts, rather than the we 9. Derivatives calculations, by considering a few simple examples. Simplest of function, f(x) = c. f(a+h)-f(a) = hm hm h^O is a constant In this case ,. Thus / all is 153 0. h^O h number differentiable at a for every the tangent line to the graph of c- / always a, h and f'(a) = This means that 0. has slope 0, so the tangent line always coincides with the graph. Constant functions are not the only ones whose graphs coincide with their tangent lines — this happens for f'(a) = any linear function f(x) + d. ex Indeed f(a+h)-f(a) — lim h c(a — + h) + d — [ca + d] lim h —— ch = lim c\ h^Q h the slope of the tangent line A refreshing difference is c, the same occurs for f(x) w, f , (a) slope 2a .. = hm f(a as the slope of the = x. graph of /. Here + h)-f{a) h = (a + h) -a 2 2 lim h = — a~ + 2ah + h - a" lim lim 2a +h h-*0 = Some of the tangent lines to the graph of each tangent checked FIGURE 9 2a. line / are shown in Figure 9. In this picture appears to intersect the graph only once, and fairly easily: Since the tangent line through (a, a ) this fact has slope 2a, can be it is graph of the function g(x) Now, if the graphs of / and = = or ) is . g intersect x In other words, (a, a — a) + a 2ax — a 2a(x x~ so (x or x at a point (x, = 2ax — a~ — 2ax + «" -a) 2 =0 = f(x)) 0; a. the only point of intersection. = (x, g(x)), then the . 1 54 Derivatives and Integrals The function f(x) — x 2 happens to be quite special in more than tangent line will intersect the graph function f(x) = x3 regard; usually a this once. Consider, for example, the In this case . = lim = lim f'(a) f(a+h)-f(a) + h) 3 -a 3 (a h-+0 a 3 3a 2 h+3ah 2 + + h 3 -a 3 lim 3a 2 h = + 3ah 2 = lim 3a = 3a . of / and g = = x or easily solved if is has got to be x = a, so that (x then be found by dividing. so happens that x 2 + ax — (x Thus, as We a — 3a"x — 2a~ - 2 - 2 + la 3 = 3a x 3 - 3a x we remember (x, g(x)) when 3 0. that a factor of the is = one solution of the equation left side; the other factor can obtain a)(x 2 + ax - 2a 2 also has x — 2a a){x 2a — 2 ) —8a 3 ). = 0. a as a factor; — a)(x + 2a) = we obtain finally (a, a 0. through These two points are always have already found the derivative of sufficiendy and the derivative /' still ) also intersects distinct, except is many functions to illustrate very popular, notation for derivatives. For a given function /, often denoted by dfix) dx ki; 10 a~ 0. the classical, figi . illustrated in Figure 10, the tangent line the graph at the point (—2a, when a 3 ) has slope 3a 2 This means that —a) + 3a"(x 3 — a) We (x It at (a, intersect at the point (x, fix)) .v This equation / the graph of is g(x) The graphs 3 + 3ah + h the tangent line to the graph of the tangent line h h h-»0 Thus + lim For example, the symbol dx 2 ~dx 155 9. Derivatives = denotes the derivative of the function f(x) x2 . Needless to say, the separate parts of the expression df(x) dx are not supposed to have any sort of independent existence bers, they cannot be canceled, and the entire expression other numbers "df(x)" and "dx." This notation is — the J's are not not the is due num- quotient of two to Leibniz (generalh considered an independent co-discoverer of calculus, along with Newton), and is Although the notation df{x)/dx may be shorter; after all, the symbol affectionately referred to as Leibnizian notation.* seems very complicated, in concrete cases dx 2 /dx concise than the phrase "the derivative of the function actually is more it f{x)=x 2 r The following formulas state we have found so far: in standard Leibnizian notation all the information that dc = 0, — o. dx d( ax + b) - , dx dx 2 dx dx 3 = 1„ zx — — 1y DX dx Although the meaning of these formulas interpretation are hindered by the reasonable not contain a function on one side and a the third equation than is 2x /', . be true, clear enough, attempts at literal stricture that number on an equation should the other. For example, if then either df(x)/dx must denote f'(x), rather or else 2x must denote, not a number, but the function whose value It is at v one or the other of these alternatives is practice df(x)/dx sometimes means /' and sometimes means f'(.\ really impossible to assert that intended; in while 2x to is is may ), denote either a number or a function. most authors are reluctant to Because of this ambiguity, denote f'(a) by — df(x) —, (a); dx instead f'(a) is usually denoted by the barbaric, but unambiguous, symbol df(x) dx * Leibniz was led to this not the limit of quotients symbol by \ f(x his intuitive + h)- f(x)]/h, notion of the derivative, which he considered to be, small" number. This "infinitely small" quantity was denoted by dx small" difference f(x+dx)—f(x) by when h is an and the corresponding but the "value" of this quotient df(x). Although this poinl of view properties (PI)— (PI 3) ol the real numbers, some people is "inliniteK "infinitely impossible to reconcile with find this notion of the derivative congenial. 1 56 Derivatives and Integrals In addition to these df(x)/dx Leibnizian notation difficulties, Although the notation dx 2 /dx ambiguity. associated with one is often replaced by df/dx. This, of course, is practice of confusing a function with its more absolutely standard, the notation is value at x is conformity with the in So strong . is this tendency that functions are often indicated by a phrase like the following: "consider the function y = x 2 ." as the We name the function sometimes follow will classical practice to the extent and its values function (defined by) y(x) Despite the many — = jc 2 ." ambiguities of Leibnizian notation, mathematical writing, and sively in older is ever, it is used almost exclu- used very frequently today. still staunchest opponents of Leibnizian notation admit that some of using y we will nevertheless carefully distinguish between thus we will always say something like "consider the of a function, but it will The be around for quite time, while its most ardent admirers would say that it will be around forand a good thing too! In any case, Leibnizian notation cannot be ignored completely. The text, policy adopted in this but to include in the it book is to disallow Leibnizian notation within 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 functions examined so far have preciate the significance of the derivative examples of functions which are three functions first first is it discussed in this chapter, f(x) — \x\. and = 1 for h > 0, + h)-f(0) and \h\/h illustrated in Figure 1 ; if they \h\ h = —1 for h < does not lim 0. This shows that exist. h /i— In candidates are the In this case h \h\/h The obvious To fully apknow some something has clearly gone wrong. f(0 Now differentiable. equally important to not differentiable. turn out to be differentiable at Consider been all fact, f(h) .. and - lim // /(0) =— , I. h •() (These two limits are sometimes called the right-hand derivative and the left- hand derivative, respectively, of / at 0.) , 9. Derivatives If a / 0, then f'(a) does In exist. fact, .f(-v)=l = /'(*) -1 if x > 0, if x < 0. The proof of this fact is left to you (it is easy if you remember the linear function). The graphs of / and of /' are shown in Figure /' 2 X > X I), We a similar difficulty arises in connection with f'(0). h 1 1. H ' [ FIGURE derivative of a 1 For the function /M={*X, ./" 157 f(h) I ~ /(0) have 2 = h, h < /? >0. h I = 1, /; Therefore, /i->0- but Thus /'(0) does not exist; exists for x — ^ / /? lim is not differentiable at /'(*) of Even worse / and things 1. 0. Once again, however, f'(x) easy to see that it is f The graphs = /' are = shown happen for 2x, x < 1, ^>0. in Figure 12. f(x) = y/\x\. For this function f'j —=— sfh FIGURE 12 1 h > h <0. f(h)-f(0) I In this case the right-hand limit ,. hm does not exist; what's more, (Figure 13). FIGURE \ } instead 1/v/z f(h) - becomes —l/y/—h becomes /(0) = 1 lim —7= arbitrarily large as h approaches 0. And, arbitrarily large in absolute value, but negative 58 1 Derivatives and Integrals The fix) = function f(x) behaved than better l/x, although not differentiable at 0, this. The - /(0) f(h) 1 ~ h h h "tangent line" which — l/x K.I RE Sometimes one says that / has means that the graph of / has a has the same geometric property, but one would say that Remember 14 0. this (Vh\ parallel to the vertical axis (Figure 14). is of "negative infinity" 1 Geometrically "infinite" derivative at 0. continuity. little I /,2/3 simply becomes arbitrarily large as h goes to an a at least is quotient Of course, / f(x) — has a derivative at 0. that differentiability This idea is is supposed be an improvement over mere to many examples supported by the of functions which are continuous, but not differentiable; however, one important point remains to be noted: THEOREM l / If is / differentiable at a, then PROOF hm / (a + h) — is f(a) continuous at a. = hm • f(a v — hm + h)-f(a) • h-+0 = = As we pointed out to lim x—>a It is f(x) = in Chapter f(a); thus very important to that the converse is / is 5, the /? /'(fl)-0 0. equation lim f(a continuous A hm h-+0 ll + h) — f(a) = is equivalent at a. | and just as important to remember differentiable function is continuous, but a con- remember Theorem not true. // 1, mind the function f(x) and which false). tinuous function need not be differentiable (keep in and you will never forget which statement The continuous functions examined with at most one exception, but which are not Actually, I I ( . i RE it is true so far have been differentiable at all \x\, points easy to give examples of continuous functions differentiable at several points, one can do much worse than 15 is = this. even an There is infinite number a function which (Figure 15). is continuous 9. Derivatives everywhere and differentiable draw it nowhere! Unfortunately, the definition of this function will Chapter 24, and I have been unable to persuade the (consider carefully what the graph should look like and you will be inaccessible to us artist to 159 until sympathize with her point of view). tions to the graph, It is draw some rough approximabetter approximations are shown possible to however; several successively Figure 16. in Although such spectacular examples of nondifferentiability must be postponed, we can, with a at infinitely Figure 17. little ingenuity, find a continuous function = x This particular function ft/0we / have is itself sin — , x ^ x = x 0. Indeed, for not differentiable of the function /(*) 17 is many points, all of which are in [0, 1]. One such function is illustrated in The reader is given the problem of defining it precisely; it is a straight line version FIGURE which 0. quite sensitive to the question of differentiability. 1 60 Derivatives and Integrals f(h) - h sin /(0) = that lim sin 1 h does not /h / exist, so Geometrically, one can see that a tangent line cannot secant line through — and 1 1 , (0, 0) and (h, is not differentiable at exist, 0. by noting that the f(h)) in Figure 18 can have any slope between no matter how small we require h to be. x sin i^ -, x x 0, FIGURE —1 sin h h Long ago we proved i h = 18 This finding represents something of a triumph; although continuous, the function / seems somehow ematically undesirable feature of this ertheless, tiability. and we can now enunciate one mathNevfunction it is not differentiable at 0. quite unreasonable, — one should not become too enthusiastic about the criterion of differen- For example, the function 1 g(x) x = .v^0 sin A=0 0, is differentiable at 0; in fact g'(0) = 0: g(h) - g(0) ,. lim h = lim h-y0 h-* = The ure tangent line to the graph of g at ll lim h sin /i-»0 = sin - — // 0. (0,0) is therefore the horizontal axis (Fig- 19). This example suggests function than mere such conditions if that we should differentiability. we We introduce another seek even more restrictive conditions ean actually use the derivative set of definitions, the last to on a formulate oi this chapter. 9. Derivatives \ 161 / \ / \ \ /I A f . x - sin jc , a; fr U/w IP ^ ,. 1< ^nA 1 J * ^ = 1 \ 1// / \ / \ / \ FIGURE 19 For any function /, we obtain, by taking the derivative, a new function /' (whose domain may be considerably smaller than that of f). The notion of differentiability can be applied whose domain tion (/')' If f"(a) f"(a) is usually written simply /" and is called the / is said to be 2-times second derivative of/ tion at time t acceleration is in sit at time an accelerating There = f"" is no reason (/'")', etc. viation is at a. of a particle moving along a straight acceleration. its differentiable at a. particularly important. line, If s(t) then s"(t) the posi- is is called the Acceleration plays a special role in physics, because, as t. on a stated in Newton's laws of motion, the force and is called the is then In physics the second derivative (/')', The funcsecond derivative of /. differentiable at a, and the number consists of all points a such that /' exists, another function to the function /', of course, yielding Consequently you can particle feel the the product of its is mass second derivative when you car. to stop at the second derivative —we can define /"' = (/")', This notation rapidly becomes unwieldy, so the following abbre- usually adopted (it is really a recursive definition): f(i) f (k+1) _ /', = (f ik) y. Thus HV> /(2) /( 3) W4) =f = f" = ify, = f" = (fy, = (/'")' rllll etc. The various functions (k f \ for k > 2, are sometimes called higher-order > convenient to derivatives of /. Usually, have for / f {k) , we resort to the notation defined for smaller k also. f {k) In only for k fact, namely, f(0) _. r 4, but it is a reasonable definition can be made 162 Derivatives and Integrals Leibnizian notation for higher-order derivatives should also be mentioned. The natural Leibnizian symbol for f"(x), namely, 'df(x) dx dx (fix) = x< is abbreviated to d 2 f(x) or (dx) 2 Similar notation The how Let fix) tion. used for f (k) frequently to dx 2 (x). following example illustrates the notation simple case, (a) is d 2 f(x) more = f {k \ and also shows, in one very various higher-order derivatives are related to the original func- x 2 Then, as we have . already checked, = 2x, /"(*) = 2, f(x) = 0, f(x) fix) = 2x f k) (x) = if*>3. 0, Figure 20 shows the function /, together with A more illuminating example whose graph is shown in Figure 21(a): rather fix) fix) = 2 It is its various derivatives. presented by the following function, is ,2 x >0 —X' x < a ifa >0, <0. - /(0) 0. easy to see that f'ia) = 2a = -la /'(0) = f'(a) if Moreover, (c) lim fih) h->0 ,. lim // fih) /i-»-0 // Now f (k) (x)=0,k>3 = lim /j->o+ and (d) I l(,l Rl. 20 v lim h->0- f(h) v = hm h h->0- so /( 0) = This information can all lim —— ^-*o+ h /? lim h->0 m " /;2 h = o. h be summarized as follows: ./"(-v) = 2|.v|. = n U, 163 9. Derivatives follows that /"(0) does not exist! It Existence of the second derivative rather strong criterion for a function to /(*) = x 2 , x\ Even a "smooth looking" function satisfy. when examined with x>0 like jc<0 suggests that the irregular behavior of the function / some reveals irregularity 1 x g(x) sin 0, the second derivative. This x /0 x = moment we know might also be revealed by the second derivative. At the g'iP) = but 0, we do not know We will return to computing g"(0). after we have any a g'(a) for thus a is ^ 0, so it that hopeless to begin is question at the end of the next chapter, this perfected the technique of finding derivatives. PROBLEMS f\x) = 1. 2\x\ Prove, working directly from the definition, that (a) = -1/a f'(a) (b) 2 fora , ^ Prove that the tangent if = f(x) l/x, then 0. line to the graph of / at (a, I /a) does not intersect the graph of /, except at (a, \/a). 2. Prove that (a) (b) /"(*) 2, l/x 2 then f'(a) f(x)= , Prove that the tangent which point, = if lies = on / line to = -2 /a 3 at (a, I /a ) for a ^ 0. / intersects one other at the opposite side of the vertical axis. = 3. \/(2s/a), for a > 0. (The expression you obtain for [f(a + h) — f(a)]/h will require some algebraic face lifting, but the answer should suggest the right trick.) 4. For each natural x > Prove that S2(x) = f(x) if >/x, then f'(a) number = 2x, and S^'ix) n, let S„(x) 3x (The expression conjecture. = x". Remembering that S\'(x) — 1, conjecture a formula for Sn '(x). Prove your , + (x //)" may be expanded by the binomial theorem.) f"(x) = -2, x < = 5. Find /' 6. Prove, starting from the definition (and drawing a picture to f(x) if [x]. (c) FIGURE 2 if (a) (b) 7. if g(x) g(x) = = + f(x) (a) (b) (c) If What What What is = is f'(x ) calling 2 problem 2 )? you are missing a very important means the derivative of / at the number which we happen x find this 2 ; it is drive the point (d) cf'(x). . f(3 ), /'(5 ), f(6 2 f(a 2 ), f(x )? you do not 2 x /'(*); /'(9), /'(25), /'(36)? 2 is = cf(x), then g'(x) Suppose that f(x) = then g'(x) c, illustrate): For f(x) silly, not the derivative at x of the function g(x) — f(x 2 home: = x3 , compare f'(x ) and g'{x) where g{x) — f(x 2 ). ). point: to be Just to 164 Derivatives and Integrals (a) = f(.x+c). Prove (starting from the definition) that g'{x) — Draw a picture to illustrate this. To do this problem you must c). + f'(x write out the definitions of g'(x) and fix + c) correctly. The purpose Suppose g{x) of Problem 7 was to convince you that although not an utter and there triviality, something is put prime marks into the equation g(x) problem this to prove: = f(x c f'(cx). + is easy, is you cannot simply To emphasize c). it this point: (b) Prove that why (c) f(cx), then g'(x) should be true, this Suppose / that + a) = fix = g(x) if fix) or you will surely and + to see pictorially with period a periodic, Prove that /' all x). Find f'{x) and also f'(x Try also. differentiable is for = (i.e., also periodic. is Be very methodical, up somewhere. Consult the answers (after you do the slip 3) in the following cases. problem, naturally). = (x+3) 5 f(x+3)=x 5 f(x+3) = (x + 5) 7 f(x) (") (in 10. . . Find f'(x) if f(x) = g(t . + x), and fit) if = g(t + x). The answers will not be the same. 11. (a) (b) Prove that Galileo was wrong: and s' s(t) = is — [s'it)] (ii) is then s, s falls a distance s(t) in t seconds, cannot be a function of the form facts are true show why we switched from s"it ) (i) If s a body cf Prove that the following fact will (c) proportional to if 2 a (the acceleration is about = s if s(t) (a/2)t (the first c to a/2): constant). = lasit). measured in feet, the value of a is 32. How many seconds do you way of a chandelier which falls from a 400-foot make it, how fast will the chandelier be going when you? Where was the chandelier when it was moving with half that have to get out of the ceiling? If you don't it hits speed? 12. Imagine a road on which the speed other words, there is from the beginning of the road road; car A's position at time (a) What equation expresses the limit? (The answer Suppose time / is that A\ is specified at every single point. In is t a(t), is = — 1 B, are driving along bit). fi's is A always travels at the speed Lit).) at the / and car fact that car not a' it) A always goes position at time Two cars, A and L(jc). is this (b) limit a certain function L such that the speed limit x miles . speed Show limit, that B and is that ZJ's position at also going at the speed limit at all times. (c) B always stays a constant distance behind A. Under what conditions will B still always travel at the speed limit? Suppose, instead, that 165 9. Derivatives 13. Suppose = g(a) that fia) and the right-hand derivative of g at a. h(x) 14. = g(x) for — Let f{x) / is x > x2 if a. x Prove that h differentiable at 0. (Don't = Define h(x) is at f(x) for x a equals < and a, differentiable at a. = and f(x) rational, is / that the left-hand derivative of be scared by x if Prove that irrational. is Just write out the this function. definition of /'(0).) 15. differentiable at 0. (If do / is you have done Problem 14 you should be able to Prove that for all x. this.) This result can be generalized (b) < x2 a function such at |/(x)| / be Let (a) x if is replaced by |g0c)|, where g has what property? 16. Let a > 17. Let is 18. < / If 1. < Prove that 1. not differentiable at — Let fix) that / is Suppose — Show f(a) 20. , g'(a) = > satisfies |/(jc)| and \/q for = x g(a) - \x\? p/q It = —0.00 . . . 0a n+ \a n+ 2 differentiable at 0. and /(0) = 0, then / in lowest terms. Prove obviously suffices to prove expansion .... g(a) h'(a). that the conclusion does not follow = is = h(a), that f(x) < g(x) < h(x) for all x, Prove that g is differentiable at a, and that h'{a). (Begin with the definition of g'(a).) = = / prove that Why? If a — m.axaia}, ... is the decimal + h) — f(a)]/h for h rational, and also for that f(a) that f'(a) f'(a) (b) a 0. h and / for irrational x, of a, consider [fia (a) if \x\ not differentiable at a for any a. Hint: this for irrational a. 19. < satisfies |/(.v)| if we omit the hypothesis h(a). / be any polynomial function; we will see in the next chapter that / The tangent line to / at (a, /(«)) is the graph of g(x) = f'(a)(x — a) + /(a). Thus f(x) — g(x) is the polynomial function d{x) = 2 fix) — f\a)ix — a) — /(a). We have already seen that if fix) = x then 2 3 then dix) — ix — a) (x + 2a). dix) = (jc — a) and if fix) — x Let is differentiable. , , , (a) Find dix) (b) There ix when fix) — x 4 and show , certainly —a) 2 . seems to 22 it is divisible be some evidence that dix) is intersects the (x — a) 2 . always divisible by graph at two points; the tangent line graph only once near the point, so the intersection should be a "double intersection." To give a rigorous proof, dix) x — a note that fix) — fia) divisible there a polynomial function h such that hix) = answer the following questions. by —a)? Why is first fix)- fia) - fia). x —a Now ix by Figure 22 provides an intuitive argument: usually, lines parallel to the tangent line will intersect the FIGURE that Why is 1 66 Derivatives and Integrals - d(x)/(x does 21. (a) (b) 22. (a) # a) for x that f'(a) lim /?U) is = Why 0? is = ft(a) 0? Why \im[f(x) — — a). f(a)]/(x (Nothing deep here.) x—*-a that derivatives are a "local property": if f(x) = g(x) for all x in = open interval containing a, then some g'(a). (This means that f'(a) in computing f'{a), you can ignore f(x) for any particular x ^ a. Of course you can't ignore f{x) for all such x at once!) Show Suppose / that is differentiable at x. Prove that = f'(x) f(x+h)-f(x-h) lim h->0 Remember an Hint: **(b) = problem? this solve the Show Why a? 2/7 old algebraic trick — a number is same quantity is added to and then subtracted from Prove, more generally, that if the it. + h)-f(x-k) h +k f(x lim fix) not changed Although we haven't encountered something like lim before, its mean- make an appropriate we actually have lim ing should be clear, and you should be able to The important definition. that 23. we Prove that / is = Draw if , so k. —f'(—x). (In order to minimize conand then remember what other thing g find g'(x) f(—x); s-8 = even, then f'(x) a picture! / 24. Prove that 25. Problems 23 and 24 say that /' What can 26. that is are only considering positive h and fusion, let g(x) is.) thing here if is — odd, then f'{x) is therefore be said about Find f"(x) f'{—x). even f (k) if / Once is again, draw a odd, and odd if picture. / is even. ? if = x\ 5 fix) = X f(x)=x 4 fix) ii) iii) iv) 27. . . fix+3) = x 5 If S„(x) — x", and . < k < S„ n, ik) prove that .n-k = ix) (n-k)l' =«, 28. (a) (b) Find /'(*) if fix) = Analyze / similarly if \x\ 3 . ./i-£ . Find f"(x). Does f'"(x) f(x) = 4 .v for x > exist for all and f(x) = — 4 .v for x? x < 0. 9. Derivatives 29. 30. - x n for x > (and find a formula for Let f(x) and it), let = f(x) but that / ( "} for (0) x < 0. does not Prove that dx" „ , nx dx I dy v d[f(x) + c] dfjx) nil dx dx d[cf(x)] = IV 'fix) c- dx dx dz dy dx dx if dx 3 — vi; dx z 3a = v + c. 4 . x=a i - df(x+a) df(x) Vll dx df(cx) = Vlll dx df(cx) IX) = df(x) dx df(y) C dy **-*( "'\k x=b+a C x=b dx dx k dx x=b n-k x=cb (n_1) exists exist. Interpret the following specimens of Leibnizian notation; each ment of some fact occurring in a previous problem. / 167 is a restate- CHAPTER DIFFERENTIATION The process of finding the derivative of a function previous chapter you may have is called the impression that this process requires recourse to the definition of the derivative, limit. approach forget the definition of the derivative if you It is Nevertheless, in this chapter tions, true that such a procedure we is is will learn to differentiate From the usually laborious, and depends upon recognizing some — differentiation. successfully often the only possible you are a large without the necessity of even recalling the definition. A likely to be lost. number of func- few theorems will provide a mechanical process for differentiating a large class of functions, which formed from a few simple functions by the process of addition, multiplication, and composition. This description should suggest what theorems will be proved. We will first find the derivative of a few simple functions, and then prove theorems about the sum, products, quotients, and compositions of differentiable functions. The first theorem is merely a formal recognition of a computation are division, carried out in the previous chapter. THEOREM l If / is a constant function, fix) f'{a) PROOF f (a) = = = for THEOREM 2 If / PROOF is numbers all f(a+h)-f(a) = hm lim h h->0 The second theorem then c, is f'(a) — f'(a) = = x, lim = 0. I h computation in the last chapter. then for 1 c-c /i^o also a special case of a the identity function, f{x) a. numbers all a. f(fl+h)-f(a) h — a + — h h lim - = 1 . h^o h The derivative of the sum of two sum of the derivatives. 168 a lim h^-0 = h functions | is just what one would hope the 10. Differentiation THEOREM 3 / and g If are differentiable at a, then = (f+g)'{a) PROOF (/ + g) +g / f{a) = hm (a) + g'{a). - /(a = hm + h) + g(a +h)- h^O /(fl+/l)-/(fl) but it is the THEOREM 4 / and g If f(a = lim g(fl+/l)-g(fl) = /'(a) h + g\a). added is to (/ . g)'(fl ) = and the proof requires only a simple a number is not changed if and subtracted from = / g is f\a)-g{a) — it. also differentiable at a, + f{a) • and g'{a). + h)-(f-g)(a) (f-g)(a lim h /i^O = not as simple as one might wish, is useful before are differentiable at a, then g(a+h)-g(a) hm | product we have found (/• g)\a) PROOF + h)-f(a) nevertheless pleasantly symmetric, same quantity + g(a)]- | for the derivative of a algebraic trick, which [/(a) h lim The formula and also differentiable at a, is 169 f(a+h)g(a + h)- f(a)g(a} lim h = f(a + h)[g(a + h )-g(a)] lim [f(a+ h) h = hm / (a = (Notice that /i^o + lim h • f(a)-g'(a) f(a)]g(a) h + n) hm /i— - • hm f(a)-g(a). we have used Theorem 9-1 to conclude that lim f(a + h) = f(a).) ft->0 Theorem 4 In one special case THEOREM 5 If g(x) = cf(x) and / is simplifies considerably: differentiable at a, then g g'{a) proof If h(x) = c, so that g = h g{a) h^O h fi^O /; = f, then by g'(a) = = = (h c is differentiable at «, f'(a). Theorem 4, f)'(a) h(a)-f'(a) c-f'(a) = c-f'(a). + h'(a)-f(a) + 0-f(a) | and | 1 70 Derivatives and Integrals = Notice, in particular, that (-f)'(a) (f+[-g])'(a) = —f'(a), and consequently (/ — g)'(a) — f'(a)-g'(a). To demonstrate what we have already we achieved, compute will the derivative of some more special functions. theorem 6 If f(x) = xn some natural number for f'(a)=na PROOF The proof will be by x n+ = If I(x) . =f I follows It . g'{a) This is = = = na n ~ ] = is simply f(x) = .v", for all +l — x" for all I(x) f(x) this 1 if x, the equation x" g(x) thus g for all a. true for n, so that is f'(a) — n- 1 induction on n. For n assume that the theorem Let g(x) then n, Theorem 2. Now then a. x can be written x; from Theorem 4 that (f I)'(a) + precisely the case n = = = = 1(a) f'(a) na"- -a + a" + \)a'\ I'(a) f(a) 1 rw" (n which we wished 1 + + a" 1 for all a. to prove. | we can now Putting together the theorems proved so far find /' for / of the form f(x) We — na„x"~ + — (n n(n - Y)an x n -2 This process can be continued power of x by derivatives /'", H + + a\- 2«2- 1'' + > n \){n - Each easily. , 2)a„_i.v"- 3 + • • • + 2a 2 . differentiation reduces the highest . {4) , , is f for k - (n and eliminates one more a, It is a good idea to work out the and perhaps / (5) until the pattern becomes quite clear. The f 1 last interesting derivative (n) (x) = n\a n ; we have / Clearly, the next step in is \)a„_\x"~ can also find f": fix) = It + ai% + a\x + aq. H obtain f'(x) We = a„x" + a n _\x n ~ quite a bit the derivative <>l u, (x) our program is simpler, and, because of I /g. = 0. of a quotient f/g. 4, obviously sufficient to find to find the derivative Theorem 10. Differentiation THEOREM 7 PROOF If g and g(a) differentiable at a, is Before we even / then \/g 0, is differentiable at a, 171 and write (a+h) (a) h we must be (\/g)(a + h) Since g is, 8 > is at a. differentiable at a, Since g{a) such that g(a for small — it is necessary to check that defined for sufficiently small h. This requires only two observations. by hypothesis, continuous makes sense sure that this expression ^ + h) ^ 0, for it \h\ it follows from < from Theorem 9-1 that g follows Theorem <5. is some make sense 6-3 that there Therefore (\/g)(a + h) does is enough h and we can write , (a 1 + h) (a) g(a lim 1 + h) h^O = — g(a) h g(a) - lim + h) -g(a h[g(a) g(a + h)] ~[g(a + h) -g(a)] g(a)g(a h = -[g(a + h) - g(a)] lim we have used + — h^O h) 1 lim h (Notice that 1 lim g(a) g(a + h) 1 n't n\ 6 . [g(a)] 2 ' 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 top, minus top times derivative of bottom, over bottom squared.") theorem 8 If / and g are differentiable at a and g(a) ^ 0, then f/g and f\ g(a)f(a)-f(a)g'(a) is differentiable at a. , 172 Derivatives and Integrals PROOF Since f/g =f (1/g) we have ^Yw 8 (a) f'(a) + f(a) (a) 8 fja) f(a)(-g'(a)) 8(a) [8(a)] 2 ' f'(a)-g(a)-f(a)-g'(a) [8(a)] 2 We can if/U) now = X -Y if/(*) 2 2 = -l +l few more functions. For example, then f'(x) 2 +l = (x 2 + \)(2x) then fix) 1 then/ r (x) = (x = - - 2 (a- 2 +\) +l)-x(2x) (x 2 +l) 2 (x X = x if/(-v) differentiate a 2 1 ^==(-\)x A - 4x \)(2x) 2 (.v 2 + l) 2. 1-x U2 + D2 2 -2 x Notice that the last example can be generalized: f(x)=x~" = if 1 for some natural number n , then fix) thus Theorem 6 pret f{x) = -nx = -2n = (-n)x-"- actually holds both for positive x° to and negative Theorem 6 is true for n = not clear how 0° should be integers. If we inter- x~ to mean f'(x) = 0, then 1, and f(x) = also. (The word "interpret" is necessary because it is mean f(x) = [ • _1 defined and, in any case, is • meaningless.) Further progress in differentiation requires the knowledge of the derivatives of certain special functions to be studied the moment we shall divulge, and sin' (a) cos' (a) later. One of these = cosrt = — sin« This information allows us to differentiate for all a the sine function. For for all a many without proof: , other functions. For example, f(x) = x sin fix) = x cos a + sin x, = —x sin x + cos a + cosx = —x sin x + 2 cos x A", then f"(x) is use, the following information, ; if , , 10. Differentiation 173 if = g(x) 2 • sin x = a sin sin a, • then = = = = g' (x ) g"(x) + cos x sin x x cos x sin 2 sin x cos a, 2[(sin a)( — sin*) — 2[cos~ a + cos a cosa] a]; sin if h {x ) = cos x = cos x • cos x then h'(x) h"{x) = (cosa)( — sin a) + (— sinA)cosA = —2 sin a cos a, = —2 [cos" a — sin a]. Notice that + h'(x)=0, g'(x) hardly surprising, since (g also have g"(x) = + h"(x) The examples above + h)(x) = Theorem 4 that / g • h (/ The • = = = h)'(x) g / choice of or (g • (f g)'(x) [f'(x)g(x) h(x) + + (/ • g)(x)h'(x) f(x)g'(x)]h(x) answer final can be handled for + f(x)g(x)h'(x) + f(x)g(x)h'(x). h) would, of course, have given the The it f-(g-h). f'(x)g(x)h(x)+ f(x)g'(x)h(x) different intermediate step. A function involv- we have of these, for example, first also; in fact an abbreviation is (f-g)-h Choosing the As we would expect, we . involved only products of two functions. Remember two ways. 1 0. ing triple products can be handled by in + cos 2 a = a sin" is same result, with a completely symmetric and easily remembered: • g, sum of the three terms obtained by differentiating each of /, and h and multiplying by the other two. (f g • h)' For example, is the if f(x) = x sin A COS A then /'(a) = 3a Products of more than should have little (f-g-h- sin a cos a + x~ cos a cos a 3 functions can be = f'(x)g(x)h(x)k(x) + handled (sinA)(— sin a). similarly. For example, you formula difficulty deriving the k)'(x) +a + f(x)g'(x)h(x)k(x) f(x)g(x)h'(x)k(x) + f(x)g(x)h(x)k'(x). . 1 74 Derivatives and Integrals You might even (/l • . try to • . . prove (by induction) the general formula: = J2 h {x) /,)'(*) fi-\(x)fi'(x)fi+i(x) /»(*). l=\ Differentiating the m gY( x ) most interesting functions obviously requires a formula terms of /' and To ensure / (/ reasonable hypothesis would seem to be that g be differentiable ° behavior of also /' o if g that g be differentiable o g near a depends on the behavior of seems reasonable prove that g'. is to assume is Indeed we differentiable at g(a), then is This extremely important formula = (f called the Chain Rule, is practically the product of /' g(a) and at applications. g' at a. and us, g is presumable because Notice that but not quite: /' must be evaluated g', Before attempting to prove this theorem we = sin.v will try a few . temporarily, use S to denote the ("squaring") function S(.x) / = Therefore shall o Suppose f(x) Let / it f'(g(a)).g'(a). a composition of functions might be called a "chain" of functions. is a), and (fo g y(a) §)' Since the at a. near g{a) (not near differentiable at g(a). and / differentiable at a differentiable at a, / that / for one at a, = x 2 . Then sin o S. we have f'(x) = sin'(S(jc)) — Qtiite a different result is obtained • S'(x) 2 cos.t' -2x. if f(x) = sin" X. In this case f = S o sin, so f'{x) Notice that this agrees (as = = ^'(sinA') 2 sin x • sin'(.v) cos x should) with the result obtained by writing it / = sin • sin and using the product formula. Although we have invented a much special symbol, S, to name the "squaring" function, do problems like this without bothering to write and without even bothering to write down the particular composition which / is—one soon becomes accustomed to taking / it does not take down special symbols apart in one's head. practice to for functions, The such mental gymnastics all means do so, but try to following differentiations il yon find it may be used as practice for necessary to work a few out on paper, by develop the knack of writing /" immediately after seeing 175 10. Differentiation the definition of /; problems of this sort are so simple that, the Chain Rule, there if fix) = sinx fix) = sin fix) — sin is = fix) = = fix) A function sin (sin 3 3x~ • 3 /-l=- 1 — = x) fix) — cos (sin x) + 3;r) fix) = cos(jc f(x) = 53(.t + 3x 2 ) 53 — cos • I x \ x- cos* + 3jc 2 3 3 • + 3.t 2 ) (3x • 2 52 • ) (3.x- + 6.x) 2 + 6x). like 9 9 = = sin"x 9 • [sinx 9 ] , the composition of three functions, is f = S o sin o S, can also be differentiated by the Chain Rule. that a triple composition / o g It is = remember Thus if only necessary to means if°g)oh or f h o fix) we can cosjt f (x) /(x) which remember just /'(jc)=3sin x-cosx .1— (a- — then fix) x sinU you no thought necessary. x f(x) if o ig oh). x sin write f = iS o sin) o 5, f = S o (sin o £). The by applying the Chain Rule derivative of either expression can be found twice; the only doubtful point calculations. whether the two expressions lead is As a matter of fact, any experienced as to equally simple differentiator knows, it is much better to use the second: / = S We can now write down fix) function to be differentiated is /'(*) Inside the parentheses sin o S. one in fell S, so the = 2( swoop. To begin with, note that the formula sinjc first fix) begins |. , the value at .v of the second function, writing fix) = 2sinx" (the parentheses weren't really necessary, after much for ) we must put Thus we begin by o (sin o S). of the answer by the derivative of sin o S all). We composition of two functions, which we already know how for the final answer. f{x) = 2 sin* must now multiply at x; this part cos x" 2x. is easy to handle. it this involves a We obtain, . 176 Derivatives and Integrals The following example is handled /(*) Without even bothering we can to write = f'(x) you get from sin (sin * ) can now sin(sin* 2 ). cos( sin, so our expression the value of h °k(x); this is we already we have in sin (sin jc); know how fix) Finally, — to multiply cos(sin* /(jc) = = = f(x) = /(jc) fix) ) • cos*" • is 2jc. 9 sin((sin*) — then ) if You can probably just /'(jc) = cos((sin*) 2sin(sinjc) [sin (sin jc)] /'(jc) sin (sin (sin x)) /'(jc) = = sin /'(jc) = (jc sinjc) / not, try writing out cos (sin (sin 2 sin* • ) as a composition: /(x) = sin (sin (* sin*)) /'(*) = jc)) 2sin(jc sin*) cos(sin(* • start (go cos* • cos(sin;c) sin*)) • cos* • * [sin cos(* [2* sin * + * cos *] sin*) + *~ cos *] more) functions rule for treating compositions of four (or even /o cos* cos(* sin*) • always (mentally) put in parentheses starting from the • cos (sin*) • • and what we —which here are the derivatives of some other functions which are the composition "see" that the answers are correct The sin* is to solve: of sin and S, as well as some other triple compositions. if (what for /'(jc) begins |. ) have so far by the derivative of the function whose value at x again a problem simply sin* So our expression first sin). first sin functions, for f'{x) begins |. ) = cos(sinjc forget about the gohok of three as a composition be by deleting the /'(*) We will we must put Inside the parentheses = down / one see that the left-most Suppose similarly. is easy— right, (h ok)), reducing the calculation to the derivative of a composition of a smaller number of functions: f'(g(h(k(x)))) For example, if /(*) = sin 2 2 (sin (*)) \ f = S = S o sin o S o (sin o o sin (S o sin)) | then /'(*) = 2 sin (sin *) • cos(sin".v) • 2 sin * • cos x : — 177 10. Differentiation if = f(x) sin((sin x ) = = [/ ) sin o 5 o sin o S sin o (S o (sin o 5))] then f'(x) = cos((sin x )-2sinx ) cosx" • 2x; • if /(jc) = f'(x) — sin (sin(sinx)) in yourself, if necessary] [fill then 2 sin (sin (sin x)) With these examples differentiator — of the chapter, and The might as reference, You can be practice. it is now • cos (sin (sin x)) • cos(sinx) you require only one thing to become a master safely turned loose on the exercises at the end we proved high time that the following argument, while not a proof, indicates try, some of as well as cos*. • Chain Rule. some of the We the difficulties encountered. one tricks begin, of course, with the definition = (f °g)'(a) (f lim o g)(a + h))-f(g(a)) f(g(a lim . h /i->o Somewhere it in by in here we would f(g(a+h))-f(g(a)) = h h-+0 One approach expression for g'(a). (fog)(a it + h))-f(g(a)) f(g(a lim to is g(a + h) - g(a) • g(a h^-o This does not look bad, and . + h)—g{a) looks even better if we h write + h)-(fog)(a) h h-+0 = f(g(a) lim + h^O The second (to like the fiat: lim r lim g)(a) o (f h h-*0 = put + h) - limit g(a is g(a+h)-g(a) [g(a+h)-g(a)])-f(g(a)) + h) — g{a) k(h)), lim . h^O we the factor g'{a) which be precise we should write • then the want. If we limit first h g(a let — k a implies that A' + h) — g{a) is f(g(a)+k)-f(g(a)) lim . h^O It looks as goes to precise. if this limit is + h) — g{a) meaningless. one can, and we soon will, make already a problem, however, which you are the kind of person g{a should be f'(g(a)), since continuity of g as h does. In fact, There k = True, for arbitrarily small 0, who does not divide making the division we only care about h. The easiest way blindly. Even at this sort will for h have noticed ^ this can happen + is h) — if you we might have and multiplication by g(a small h, but g(a of reasoning + h) — g(a) g(a) could be for g to be a constant , 1 78 Derivatives and Integrals function, g{x) — Then g(a + c. a constant function, (/ o g){x) = g{a) for /(c), so the = (fog)'(a) However, there are — h) = = Chain Rule does indeed g(x) = if a = x2 sin 1 — x ^0 jc = X we showed 0, as in g is also hold: + h) — g(a) — 0, the function g might be 0. In this case, g'(0) o f'(g(a))-g'(a). nonconstant functions g for which g(a also for arbitrarily small h. For example, / In this case, all h. Chapter 0. 9. If the Chain Rule is correct, we must have (/ o g)'(0) = for any differentiable /, and this is not exactly obvious. 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 use a trick. THEOREM 9 (THE CHAIN RULE) If g is at a, and / differentiable at a, is differentiable at g(a), then Define a function = as follows: </> = 0(A) +h) - g{a + h) — g(a) is also small, so i£ g(a + h) - if gia + h) - gifl) = g(a) j= g(a) should be intuitively clear that g(a differentiable f'{g{a))-g\a). f(g(a+h)) It is and (fog)'{a) PROOF fog if is g(a + continuous — h) g(a) at is 0. When 0: h not zero, then is small, </>(/?) will be close to f'(g(a)); and if it is zero, then (j)(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 f differentiable at g{a). This is + k)-f(g(a)) f(g(a) = lim k^0 Thus, > if s (1) Now for g all is there ifO < |it is | some number <8 > that f'(g(a)). such that, for f(g(a)+k)- f all k, f(g(a)) then , differentiable at a, < -/'(*(«)) hence continuous at a, so there is a 8 > Consider if \h\ now any h with = f(g(a (p(h) \h\ + g(a follows from (2) that |A:| 8'. = g(a + h) — g(a) ^ 0, 8, then \g(a < 8. If h)) - f(g(a)) +h) - < + h) - g(a)\ < < 8', k = f(g(a) + k) - and hence from \(f>ih) - then f(g(a)) ; g(a) k (1) that f\g{a))\ <e. E. such that, h, (2) it 8' means 179 10. Differentiation On the other hand, + g(a if — h) = g(a) = then (p(h) 0, f(g(a)), so it is surely true that <e. \4>(h)- f'{g{a))\ We have therefore proved that = lim 0(A) so continuous at is 0. The /'(*(«)), of the proof is rest f(g(a+h))-f(g(a)) easy. If g(a = if (f (f + h) — g(a w Now that at the = then we have + h)-g(a) h (because in that case both sides are + f(g(a .. = hm . o g) (a) h))-f(g(a)) = - h^o = look g{a) 0, <f>(h) n even / // g(a+h)-g(a) lim lim </>(/?) h^o h Therefore 0). h^o h f(g(a))-g'(a).i we can many differentiate so functions so easily we can take another function x sin — , x yt U x = 0. fix) 0, In Chapter 9 we showed only possible way). For x / = that /'(0) ^ (x) = we can „ 2x — 1 • sin working 0, straight from the definition (the use the methods of this chapter. \- x cos —1 cos — 2 We have at — 1 xz Thus /'(*) 2x sin i^0 , 0. 0, As this formula reveals, the first derivative /' indeed badly behaved is it is not even continuous there. If we consider instead x sm — x : fix) , x/0 x 0, = 0. then fix) 3x x cos sin x sion 3x — xcos sin is continuous at 0, x ^ l) x = 0. but ,/"(0) does not exist (because the expres- 1/x defines a function which 1/x does not). , x 0, In this case /' — is differentiable at but the expression 180 Derivatives and Integrals As you may improvement. power of x suspect, increasing the yet again produces another If x4 — sin /(*) , x ^ x = x 0, 0, then 4x 3 (*) = ./ sin - 1 2 cos a: 0, It is X ^0 x = x 0. easy to compute, right from the definition, that (/')'(O) easy to find for x ^ = 0, and f"(x) is 0: 1 fix) = 2x 2 1 2x cos — 4x cos sin .v/0 sin x 0, In this case, the second derivative /" not continuous at is 0. = 0. By now you may have guessed the pattern, which two of the problems ask you to establish: if 1 /(*) x = sin — x , x 0, then /'(()), . . . / , ( ">(0) exist, but = fix) f jc {n) 0, not continuous at 0; is x 2«+l _ sin f x ^0 jc = 0, X 0, (w) ^ = if and /"" is continuous at 0, but / '" is not differentiable at 0. These examples may suggest that "reasonable" functions can be characterized by the possession of higher-order derivatives no matter how hard then f'(0), ..., / ( (0) exist, — we try to mask the infinite oscillation of / (x ) = sin 1 /x a derivative of sufficiently , high order seems able to reveal the underlying irregularity. Unfortunately, see later that After all much worse It is will things can happen. these involved calculations, a minor remark. we often tempting, we will bring this and seems more chapter to a close with 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 if-gy and Theorem 9 often appears = f-g'+f-g, in the if °g)' form = if °g) -g'- . 10. Differentiation hand is side may equations Strictly speaking, these be false, might have a larger domain than those on the / and g hardly worth worrying about. If domains, then these equations, and others case any because the functions on the 181 left- right. Nevertheless, this are differentiable everywhere in their like them, are true, and this is the only- one cares about. PROBLEMS 1. As a warm up exercise, find fix) for each of the following /. (Don't worry about the domain of / or /'; just get a formula for fix) that gives the right answer when it makes sense.) (i) f(x) (ii) f(x) (iii) f(x) (iv) f(x) = sm(x+x 2 = sinx + sinjc = sin (cos x). = sin(sinx). /(jc) = an(— ). , , ., (vi) 2. • n (v) ). „ /(*) (vii) f(x) (viii) f(x) = — . /cosx\ x) sin (cos - -. X — = + sin x). sin (a sin(cos(sin a)). 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 pretend that they can't even add, but most of them can (i) f(x) (ii) / (iii) f{x) = = = (iv) f(x) = (a- ) (v) f(x) (vi) f(x) / -\ ft \ (vn) /(.v) (viii) f{x) (ix) f(x) (x) fix) (xi) fix) (xii) fix) (xiii) fix) (xiv) fix) + \) 2 (x + 3 2 sin (a + sin x sin ((x + sin a) sm((x 2)). ) ). . / \ r3 sin cos* 3 = sin(x sin x) + sin(sin x~). = (cos a-) 31 -2 2-22 = sin x sm x sin x — sin (sin (sin x)). = (x + sin 5 x) 6 — sin(sin(sin(sin(sin x)))). = sin((sin 7 7 + l) 7 = iHx 2 + x) 3 + x) 4 +x) 5 = sin(x 2 + sin(x 2 + sinjc 2 )). = sin(6cos(6sin(6cos6A))). '. • . . .v ). . when like to they have to.) 182 Derivatives and Integrals sin f{x) (xv) 2 sin + 1 sin 2 x + sin x x „3 / (xvii) = /(*) x 1 = f(x) (xvi) x sin sin y = (xviii)/(*) sin sm \ I sm ^ x 3. — sin .v Find the derivatives of the functions tan, cotan, sec, cosec. (You don't have 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 to symmetrical.) 4. For each of the following functions /, find f'(f(x)) (not (f o f)'(x)). 1 /(*) = 1 5. = = (id f(x) !,11 f(x) (iv /CO = +x simr. x2 . 17. For each of the following functions /, find f(f'(x)). (i) fix) = -. f(x) = = x2 (iii) fix) (iv) f(x)= Find /' in f(x) . 17. 17.v. terms of = = = (") f(x) (iii) f(x) (iv) f(x) = (v) fix) (vi) f(x g(x g' if + g(a)). g(x-g(a)). g(x+g(x)). g(x)(x-a). = g{a)(x -a). + 3) = g(x 2 ). A circular object is known that is when increasing in size in the radius is some 6, the rate unspecified manner, but of change of the radius is it 4. and A(t) 6. Find the rate of change of the area when the radius represent the radius and the area at time t, then the functions r and A satisfy A = nr a straightforward use of the Chain Rule is called for.) is \ (If /•(/) 183 10. Differentiation (b) Suppose that watching we now informed that are when Now is (You 6. will clearly need to a formula for the volume of a sphere; in case you have forgotten, volume the (c) the radius we have been Find the rate really the cross section of a spherical object. is of change of the volume know the circular object is cube of the ^tt times the radius.) suppose that the rate of change of the area of the circular cross section when is 5 when the radius is Find the rate of change of the volume 3. You should be 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). 8. The the radius 3. is area between two varying concentric circles rate of change of the area of the larger circumference of the smaller 9. able to Particle A moves circle circle is changing when along the positive horizontal at all times 10;r in"/sec. is it axis, 9n in has area 16tt in and The . How fast B particle 2 is the ? along the graph of f(x) = -V3i, x < 0. At a certain time, A is at the point (5,0) and moving with speed 3 units/sec; and B is at a distance of 3 units from and moving with speed 4 between A and B changing? the origin 10. Let f(x) =x sin ^ i/x for x and 0, At what units/sec. let f(0) = rate Suppose 0. is the distance also that h and k are two functions such that h'{x) = 2 + sin (sin(jr k\x) 1)) = f(x + 1) k(0)=0. /i(0)=3 Find 11. (i) (foh)'(O). (ii) (k o /)'(0). (iii) a'{x"), Find /'(0) where a(x) — /(*) = h(x~). Exercise great care. if g(.t)sin -, x ^ x = x 0, 0. and g(0) 12. Using the derivative of f{x) by the Chain Rule. 13. (a) (b) = = g'(0) = 0. l/x, as found in Using Problem 9-3, find f'(x) for -1 Prove that the tangent line to the < x Problem < graph of / the graph only at that point (and thus show 1, if 9-1, find (l/g)'(x) f(x) = -x 2 y/\ vl — a 2 . ) intersects that the elementary geometry at (a, definition of the tangent line coincides with ours). 184 Derivatives and Integrals 14. Prove similarly that the tangent 15. an lines to ellipse or hyperbola intersect these only once. sets If / +g If / g and / is differentiable at a, are / and g necessarily differentiable what conditions on / imply are differentiable at a, • at that g a? is differentiable at a? 16. Prove that (a) / if differentiable at a, then |/| is ^ provided that f{a) (b) Give a counterexample (c) Prove that f{a) if / and g if also differentiable at a, is 0. = 0. are differentiable max(/, g) and min(/, g) are a, at then the functions differentiable at a, provided that f(a) ^ 8(a). Give a counterexample (d) 17. f{a) if — g(a). Give an example of functions / and g such that g takes on all values, and fog and g are differentiable, but / isn't differentiable. (The problem becomes we on all values; g could just be a constant function, or a function that only takes on values in some interval (a,b), in which case the behavior of / outside of (a, b) would be irrelevant.) trivial if 18. don't require that g takes = f 2 find a formula for g' (involving /'). = (f) 2 find a formula for g' (involving /"). (a) If g (b) If g (c) Suppose that the function / > , has the property that iff = f + in terms of /. needed at one is If / is three times differentiable at x is defined to be 2>/(*) (a) Show = Show that if f(x) quently, 2)(/ o g) Suppose point.) and f'(x) / 0, the Schwarzian derivative f'{x) 2\f'{x) / 2 (f"(x) 3 of that = ax ex 20. (In addition to simple calculations, f'"(x) ®(fog) = (b) f Find a formula for /" a bit of care 19. 1 that / (n) (fl) = [<2) +b +a with 2 +®g ad - be # 0, then 9/ = 2)g. and g {n) (a) (n fog].g' (f-g) Ha) exist. = J2(i)f k=0 Prow ik) Leibniz's formula: (")-8 (n A, <«> 0. Conse- - . 185 10. Differentiation *21. 22. and g (n) (a) both exist, then (/ o g) (n) (a) exists. A little experimentation should convince you that it is unwise to seek a formula n) (w) for (/ o gY {a). In order to prove that (/ o g) (a) exists you will therefore have to devise a reasonable assertion about (fog)^(a) which can be proved by induction. Try something like: "(/ o g) (n) (a) exists and is a sum of terms each of which is a product of terms of the form ... ." Prove that f if (n) igia)) f{x) = a n x" Find another. (a) If (b) If +a n _\x"~ Is |-ao> find -\ = find a function g with g' (c) { a function g such that g' f. = a„jc +a H H h 1 jc 24. that there = = — = is a polynomial function f'{x) (b) /'(jc) (c) (d) fix) fix) (a) The number jc no for exacdy one x, for exacdy k numbers x, — a x, if n that jc. odd. is n if even. is if n — & is odd. double root of the polynomial function / if some polynomial function g. Prove that a is a called a is 2 a) g(;c) for (jc fix) double root of / of degree n such numbers 1 for = When — for precisely n (a) (b) —w fix)=\/x? such that Show /. there a function /(jc) 23. = / if does fix) — and only ax 2 + bx if a +c is (a a root of both ^ 0) / and /'. have a double root? What does the condition say geometrically? 25. *26. = fix)- f(a)(x -a)- fia). Find d\a). In connection with Problem 24, this gives another solution for Problem 9-20. If / is differentiable at a, let d(x) This problem is a companion to Problem 3-6. Let a\, . . , a n and b\, ... ,bn be given numbers. (a) If x\, tion . , x n are = a, distinct and fixi) Prove that there a, *27. . / of degree 2n — /(jc,) (b) . and /'(jc,) = is &,- numbers, prove that there such that fixf) 1, = b t . Hint: — is fixj) a polynomial func- = for j Remember Problem 24. / of degree In — 1 a polynomial function for all ^ with i, and /(jc,) = i. and b are two consecutive roots of a polynomial function /, but that a and b are not double roots, so that we can write /(jc) = (jc — a)(jc — b)gix) where gia) ^ and gib) ^ 0. Suppose (a) that a Prove that gia) and gib) have the same are consecutive roots.) sign. (Remember that a and b 186 Derivatives and Integrals Prove that there (b) draw a picture and f'(b). Now (c) is some number x with a < x < b and fix) prove the same even fact, if Compare Hint: to illustrate this fact.) = 0. (Also the sign of f'(a) a and b are multiple roots. Hint: If and gib) ^ f{x) = (x —a)'"(x — b)"g(x) where g(a) ^ — = polynomial function h(x) a)'"~ (x — /?)""'. f'(x)/(x 0, consider the [ 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. mathematicians who never accepted the In new fact, Rolle was one of the notions of calculus. This 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 Rolle; his name has become attached to a much more general result, to appear in the next chapter, which forms the 28. 29. Suppose basis for the is Suppose / is differentiable at 0, some function g which If fix) = = x~ n f 32. Prove that for n in {k) <*>,... (x) = What f *(b) Prove that fix) 0. Hint: What happens — xg(x) if you try (-\,x*(» that + *-i)L-«-* ix) n (-!)**! +k- = g(0) = 1 = -n-k for x # 0. f{x)g{x) where / and g are differ- 0. Hint: Differentiate. if /(*)= \/(x-a)"? fix) Let fix) /""(0) same *34. N, prove impossible to write x and /(0) (a) *33. is it {k) continuous at is = 0. that /(0) (n-1)! entiable is and at 0. f(x)/x? = *31. = / to write g(x) results of calculus. xg(x) for some function g which is continuous differentiable at 0, and find /'(0) in terms of g. that f(x) Prove that for 30. most important theoretical Lei = -\)? x 2 "sin exist, and l/.v if that f x # [u) is basic difficulty as that in f(x) /""(0) atO. =\ /(x 2 = x 2n+l exist, that sin l/.v if / (/,) is x 0, and let = /(0) not continuous at Problem 21 ^ 0, and continuous at let 0, 0. 0. Prove that /'(0) (You will encounter the .) /(0) and = that 0. Prow /"" is that ./'(()) not differentiable , . . 10. Differentiation 35. 187 In Leibnizian notation the Chain Rule ought to read: Instead, Z = df(g(x)) df(y) dx dy dgixf -MM dx one usually finds the following statement: f(y). Then dz dz dy dx dy dx "Let y dz/dy denotes the function /; expression involving y," and that for y. In z (ii) z (iii) z (iv) z = siny, = siny, = = also in the final sin u sin v , y y u v 1 — x + x2 = cos*. = sin x = cost/, . u = sin*. fog, while understood that dz/dy will the z be "an answer g(x) must be substituted each of the following cases, find dz/dx by using compare with Problem (i) it is g(x) and „ Notice that the z in dz/dx denotes the composite function in = this formula; then . SIGNIFICANCE OF THE DERIVATIVE CHAPTER One aim in this chapter to justify the time is As we we have spent learning to find the knowing just a little about /' tells us a lot about /. Extracting information about / from information about /' requires some difficult work, however, and we shall begin with the one theorem which is derivative of a function. shall see, really easy. This theorem concerned with the is Although we have used precise, DEFINITION Let and more also this in A is a term informally in Chapter 7, it is A a set worthwhile to be of numbers contained in the domain of /. maximum point for /(*) > /(y) / on A if f° r every v in A. The number f(x) itself is called the maximum value of / on A say that / "has its maximum value on A at x"). maximum (and we also / on A could be f(x) for several different x 1); / can have several different maximum points on A, although it can have at most one maximum value. Usually we shall be interested in the case where A is a closed interval [a b] if / is continuous, then Theorem 7-3 guarantees that / does indeed have a maximum value on [a b] The definition of a minimum of / on A will be left to you. (One possible definition is the following: / has a minimum on A at x, if — / has a maximum on A at x.) We are now ready for a theorem which does not even depend upon the existence Notice that the in other (Figure X2 interval. general. a function and / be A point x maximum value of a function on an X3 value of words, a function , ; , FIGURE 1 of least upper bounds. THEOREM l / be any function defined on (a,b). If x is a maximum (or a minimum) for / on (a,b), and / is differentiate at x, then f'(x) = 0. (Notice that we do not assume differentiability, or even continuity, of / at Let point other points.) PROOF Consider the case where / has a maximum — behind the whole argument have slopes slopes 188 < 0. > secants at x. Figure 2 illustrates the simple idea drawn through points to the left and secants drawn through points to the right of Analytically, this argument proceeds as follows. 0, of (x, f(x)) (x, f(x)) have 11. Significance of the Derivative h If any number such that x is +h in (a, b), is 189 then f(x)>f(x+h), / since maximum on has a This means that (a, b) at x. f(x+h)-f(x)<0. Thus, h if we have > f(x+h)-fjx) a x x + h b <0. h and consequently FIGURE 2 + h)-fix) fix ,. hm On the other hand, if < h < (J. h h->0+ we have 0, f(x + h)- fix) >0, h so /U+/7)-/(x) hm By / hypothesis, is differentiable at x, so these > U. two must be equal, limits in fact equal to fix). This means that FIGURE from which 3 The follows that it Notice (Figure Since < fix) = (unless we add DEFINITION Let fix) depends only on / be is some x is a local then fix) proof is left to you a one-line proof). (give | replace (a, b) by [a, b] in the statement of the x the values of 1. We / near x, it is is in ia,b).) almost obvious begin with a definition which is how to illustrated You should = A <5 is > a local a set of in the domain of /. point for / on A if [minimum] point for / on numbers contained maximum [minimum] such that x is a maximum + 8). A nix -8,x If A a function, and point x in there 2 x Figure 4. A THEOREM at to the hypothesis the condition that get a stronger version of Theorem in 0, 0. we cannot 3) that fix) > and where / has a minimum case theorem f'(x) maximum or minimum for / on (a, b) and / is differentiable at x, 0. see why this is an easy application of Theorem 1 . | 1 90 Derivatives and Integrals The even is converse of Theorem 2 x if is is definitely not true maximum not a local minimum or provided by the function f(x) — x or minimum anywhere. 3 ; — it is possible for f'(x) to be The point for /. = in this case f'(0) simplest example but 0, / has no local maximum Probably the most widespread misconceptions about calculus are concerned with the behavior of a function the previous paragraph simpler than it is, condition f'(x) that — A DEFINITION critical we this / not repeat = The 0. the converse of Theorem 2 it: imply that x condition a local is point of a function / is itself is — fix) a maximum made point or is not true minimum in to be — the point 0. number x such = f'(x) The f'(x) by those who want the world reason, special terminology has been adopted to describe satisfy the The number f(x) when near x so quickly forgotten will does of /. Precisely for numbers x which is that 0. called a critical value of /. values of /, together with a few other numbers, turn out to be the critical maximum and minimum of a maximum and minimum value ones which must be considered in order to find the given function /. To the uninitiated, finding the of a function represents one of the most intriguing aspects of calculus, and there no denying that problems of hundred or so). is Let us consider on a closed that a If a local minimum FIG 1 RE 4 point of / point x (Then, if value three kinds of points / (3) Points x in [a, b] such that / is a is and listed and / is means that .v minimum of above: for is in the first is if at least In order to locate the of / be sure maximum and not differentiable at x. x is [a, £>] , then x must be in one not in the second or third group, then = 0, by Theorem 1, group. in these three categories, finding the a hopeless proposition, but and only a few points where / straightforward: one simply finds f(x) critical points, f(x) for each x such that / The biggest of these will be the maximum minimum. A simple example follows. is minimum or we can differentiable at x; consequentiy f'(x) many points / may still be If there are fairly continuous, maximum point or a minimum point for / on in (a, b) this maximum first must be considered: (2) x is exist.) [a, b]. of the three classes a local maximum interval [a,/?]. problem of finding the The critical points of / in The end points a and b. (1) X2 the maximum and minimum minimum x\ first are fun (until you have done your this sort is when maximum and there are only a few not differentiable, the procedure for not differentiable each x satisfying f'(x) at x and, finally, = 0, is and f{a) and fib). value of /, and the smallest will be the ) 11. Significance of the Derivative Suppose we wish maximum and minimum to find the 191 value of the function f(x)=x 3 -x on the interval [—1,2]. To begin with, =3x 2 - f(x) — so f'(x) when 3x — =0, 1 = x The numbers y that is, y/T/3 1 /lA group is , -VT73: contains the end points of the interval, - (2) third - vT/3. or — y 1/3 both lie in [— 2], so the first group of candidates maximum and the minimum is (1) The second group 1, when 1/3 and for the location of the The we have empty, since / is 1, 2. differentiable everywhere. The final step is to compute f(y/l/3) = 3 (7173 - 1/3 = ivM/3 - vl/3 = — §vT/3, /(- v/r/3) = (- v/T73) 3 /(-D = f(2) = value = + /T73 = § x/Ia -JX/T73 6. occurring at 6, is ) 0. minimum Clearly the (-7173 value is — ^y 1/3, occurring at y 1/3, This sort of procedure, if feasible, will always locate the maximum and minimum value of a continuous function on a closed interval. If the function with is not continuous, however, or on an open /(O) that the this 1 b a interval or the whole if we line, are seeking the then maximum and minimum values exist, so may say nothing. Nevertheless, a we the nature of things. In Chapter 7 maximum we cannot even be procedure all we are dealing or minimum sure beforehand the information obtained by little ingenuity will often reveal solved just such a problem when we showed 1 that FIGURE maximum and the 2. if n is 5 even, then the function f(x)=x"+a n ^x"- +.-.+ao ] minimum value on the whole line. occur at some number x satisfying has a /'(*) we can This proves that the = nx n ~ + (n l l)a„_iJc B -2 +.. minimum value must a\. and compare the values of f(x) for such x, we can actually find the minimum of /'. One more example may be helpful. Suppose we wish to find the maximum and minimum, if they exist, of the function If solve this equation, , 192 Derivatives and Integrals on the open interval (— 1, We 1). have fix) = 2x U-* 2 2 ) = so f'(x) = only for x values of f(x) become We 0. arbitrarily large, so This observation also makes note (Figure can see immediately that for x close to 5) that there will / certainly does not have a easy to show that / has a be numbers a and b, with it — < a 1 < 1 and < b < 1 <x < 1 minimum or — 1 the maximum. We just at 0. such that f(x) > /(0) for — < x < a 1 and minimum of / on [a b\ is the minimum of / on all of on Now [a,b] the minimum occurs either at (the only place where or at a or b, and a and b have already been ruled out, so the minimum This means that the (— 1, 1). = 0), /' value f(x) = 3 jc - x /(0) is b = , 1. In solving these problems we purposely did not draw the graphs of fix) = x —x and f(x) = 1/(1 — x ), but it is not cheating to draw the graph (Figure 6) as long as you do not rely solely on your picture to prove anything. As a matter of fact, we are now going to discuss a method of sketching the graph of a function that really gives enough information to be used in discussing maxima and minima in fact we will be able to locate even local maxima and minima. This method involves consideration of the sign of f'(x), and relies on some deep theorems. — The theorems about which have been proved so far, always yield terms of information about /. This is true even of Theo- derivatives information about /' in rem theorem can sometimes be used to determine certain information about /, namely, the location of maxima and minima. When the derivative was first introduced, we emphasized that f'(x) is not [f(x + h) — f(x)]/h for any particular h, but only a limit of these numbers as h approaches 0; this fact becomes 1 , although this painfully relevant about tered — x' /'. The when one simplest conviction is f'(x) it is = impossible to imagine It is difficult for all x. is always hypothesis /'(jc) h and it is not at = for all x, encoun- must / be a else, and could be anything all must be standing still! — if the Never- even to begin a proof that only the constant functions satisK The f(x Kl. 6 fix) how / 0, surely the particle Inn FIGl / from information strengthened by considering the physical interpretation velocity of a particle theless extract information about frustrating illustration of the difficulties afforded by the following question: If is constant function? this tries to and most obvious = +h)- only means that f(x) = 0. •() how one can use the information about the derive information about the (unction. limit to . 193 11. Significance of the Derivative The fact that of the same / is a constant function can sort, all Value Theorem, which / that if is if /'(jc) = for all jc, and many other be derived from a fundamental theorem, called the states differentiable on much [a, b], stronger results. Figure 7 then there /'(*) fib) = b is some x makes in (a, b) it facts Mean plausible such that - f(a) —a this means that some tangent line is parallel to the line between and (b, there f(b)). The Mean Value Theorem asserts that this is true f(a)) is some x in (a,b) such that fix), the instantaneous rate of change of / at jc, is exactly equal to the average or "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 Geometrically — (a, FIGURE 7 — deepest result about derivatives. proof is From statement you might conclude that the — but there you would be wrong difficult, have occurred long ago, in Chapter Value Theorem yourself you FIGURE this will 7. It is the hard theorems in this true that probably fail, but if this you is try to prove the book Mean neither evidence that the theorem is hard, nor something to be ashamed of. The first proof of the theorem was an achievement, but today we can supply a proof which is quite simple. It 8 helps to begin with a very special case. THEOREM 3 (ROLLE'S THEOREM) If / is there PROOF continuous on [a,b] and differentiable on a is If follows number x in (a, b) such that f'(x) from the continuity of / on [a , = b] that (a, b), and f(a) = f(b), then 0. / has a maximum and a minimum value on [a b] , Suppose — /'(jc) first that the by Theorem 1, maximum value occurs at a point x in (a.b). and we are done (Figure 8). Suppose next that the minimum value of / occurs Then, again, /'(jc) = by Theorem 1 (Figure 9). Finally, points. FIGURE 9 so / any is jc suppose the Since f(a) = Then at some point x in (a,b). maximum and minimum values both occur at the end f(b), the maximum and minimum values of / are equal, a constant function (Figure 10), and for a constant function we can choose in (a,b). | Notice that on (a,b) in we really needed the hypothesis order to apply Theorem 1. that Without / this is differentiable everywhere assumption the theorem is false (Figure 11). FIGURE io 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 Rollc'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 Theorem we will apply Rolle's Theorem to the function which gives the length of the- vertical segment shown in 94 1 Derivatives and Integrals Figure 12; between this (a, is the difference f(a)) and (b, f{b)). between f(x), and the height Since L is the graph of — f(b) gix) b we want FIGURE fib)-f(a) fix) - — b THEOREM 4 (THE MEAN VALUE THEOREM) If it / turns out, the constant f(a) is continuous on in (a b) , (x-a) a and [a, b] is Clearly, h is = continuous on fib)) (x-a)- Consequently, = I ( , I R 1. on then there (a, b), number x a is - fia) —a fib) some x f(b)-f(a) f(x) and h(a) = fia). h(b) = f(b)- differentiable on fjb)-f(a) b -a). (x b-a (a, b), and (x-a) —a fia). we may apply Rolle's Theorem to h and conclude that there is in (a, b) such that = ti(x) = I f(a). irrelevant. differentiable [a, b] = f(a)) + f(a), Let h(x) (a, L a b (b, line such that f'ix) proof x of the to look at 11 As f(a) at fib) f(x) b — fia) a so that 12 fix) fib) = M ; b Mean Notice that the previous theorems mation COROLLARY l If / is is Value Theorem — information about / so strong, however, that — 'fia)' . a still fits into the pattern exhibited yields information we can now go defined on an interval and f'(x) | = about /'. by This infor- in the other direction. for all x in the interval, then / is constant on the interval. PR( )< )i Let a and b be any I \v<> points in the interval with a ^ b. Then there is some x in , 11. Significance of the Derivative 195 (a,b) such that b - fia) —a — a fib) /'(*) But f'(x) i <ig ire — for x all in the interval, so fib) = 13 b = and consequently fia) interval is the same, Naturally, Corollary fib). / i.e., 1 Thus fia) the value of constant on the interval. is / at any two points in the | does not hold for functions defined on two or more in- tervals (Figure 13). corollary 2 / and g If PROOF For all there x The is number c = g'(x) for ) all x in the c. we have (/ — g)\x) = fix — g'ix = / — g = c. | in the interval a is on the same interval, and fix) some number c such that / = g + are defined interval, then there so, ) by Corollary 1 such that statement of the next corollary requires some terminology, which is illus- trated in Figure 14. A DEFINITION function is two numbers an interval if increasing on an in the interval with f(a) often say simply that 3 If fix) > for all PROOF x for all x < a The b. < fib) whenever a and b are function / in an in the interval, interval, then / is then fix) / is = fib) b all x in increasing b - — > it follows that fib) > The proof when fix) < for all x a - fia) —a (a, b), so fib) Since b is on the interval; if fix) < decreasing on the interval. Consider the case where fix) > 0. Let a and b be two points < b. Then there is some x in (a, b) with for case the interval /.) a But fix) > decreasing on with a < b. (We is > fib) for all a and b in the interval / is increasing or decreasing, in which understood to be the domain of corollary interval if fia) fia) >0. a fia). is left to you. | in the interval with 196 Derivatives and Integrals Notice that although the converses of Corollary obvious), the converse of Corollary 3 see that f'{x) f(x) > is 1 and Corollary 2 are not true. If / increasing, is true (and it is easy to might hold for some x (consider for all x, but the equality sign = x\ Corollary 3 provides enough information to get a good idea of the graph of a function with a minimal amount of point plotting. function f(x) = x3 — x. We Consider, once more, the have f'(x) = 3x 2 -\. have already noted that f'(x) = for x = y 1/3 and x = — y^l/3, and it 2 also possible to determine the sign of f'(x) for all other x. Note that 3x — 1 > We precisely when 3x 2 x 2 x > y 1/3 a (a) b — thus 3x is 1 < or > 1 >\. —y x < 1/3; when precisely -7173 < an increasing function x < yi73. Thus / is increasing for x < — yl/3, decreasing between — yl/3 and vl/3, and once again increasing for x > y/i/3. Combining this information with the following facts /(- V/I73) = (1) /(/I73) (b) = / ^ v I73, -|vT73, (2) /(x)=0forx = -l,0, (3) f(x) gets large as x gets large, and large negative as x gets large negative, 1, a decreasing function FIGl RE 14 it is possible to sketch a pretty respectable approximation to the graph (Figure 15). By the way, notice that the intervals on which / increases and decreases could have been found without even bothering to examine the sign of /'. For example, since /' is continuous, and vanishes only at — vl/3 and yl/3, we know always has the same sign on the interval (— \J\/3, vl/3). f(y/\/3 ), it follows that / Since that /' /(— vl/3) > decreases on this interval. Similarly, /' always has the on (y/l/3, oo) and f(x) is large for large x, so / must be increasing on (vl/3, oo). Another point worth noting: If /' is continuous, then the sign of /' on the interval between two adjacent critical points can be determined simply by finding the sign of f'(x) for any one x in this interval. same sign Our to sketch of the graph of f(x) = x3 — allow us to say with confidence that —y/\/3 Vl/3 is a local minimum point. In fact, x contains is we can a local sufficient maximum give a general information point, scheme and that for decid- 197 11. Significance of the Derivative / / increasing decreasing i / increasing / FIGURE 15 ing whether a critical point is a local maximum point, a local minimum point, or neither (Figure 16): (1) if /' > some in the right of x, then x (2) if /' < the right of (3) if some in jc, /' has the is a local in some interval to x and /' > in some interval to maximum interval to the left of then x same < and interval to the left of x is a local sign in interval to the right, then x minimum some is /' point. point. interval to the left of x as has in some it maximum nor a local minimum neither a local point. (There no point is in memorizing these rules —you can always draw the pictures yourself.) The polynomial functions can to describe the general /' > /' (a) FIGURK 16 < all be analyzed in form of the graph of such /' < /' (b) > this way, and functions. To it is even possible begin, /' < we need /' (d) < a 1 98 Derivatives and Integrals result already mentioned Problem in = f(x) / f(x) = — f(x\) Although 0. < JCfc, — f(x2) and X2 such • really is 0, if = that f'(x) — now A3, etc. It is that an algebraic theorem, calculus can be used x\ and X2 are roots of / (Figure 17), so that Theorem This means that 0. numbers x such there are at most n then by Rolle's then /' has at least k between X2 and FIGURE this + <*o. ] i.e., an easy proof. Notice that to give + a„_ hx"- + a„x" has at most n "roots," then 3-7: If if there / number a is A' between has k different roots x\ x\ < a2 < one between x\ and X2, one prove by induction that a polynomial different roots: \ easy to function 17 f(x)=aH x H + an -ix a has at most n roots: it is The statement is - i +---+ao surely true for n = 1 , and if we assume that true for n, then the polynomial = g(x) could not have more than + /; b n+] x" 1 +l +b + b„x" + roots, since if it would have more than n did, g' roots. With information this fix) The n — derivative, 1 a and negative a„x" + fl,,-!*"- 1 + • • • + oo. being a polynomial function of degree n — 1, has at most Therefore / has at most n — 1 critical points. Of course, a not necessarily a local maximum or minimum point, but at any is b are adjacent critical points of /, then /' will on (a,b), since /' specific is continuous; consequently, Thus / has or decreasing on (a,b). As a = roots. cal point if not hard to describe the graph of it is remain either / will rate, positive or be either increasing most n regions of decrease or at criti- increase. example, consider the function f(x)=x 4 -2x 2 . Since f'(x) the critical points of / are = 4a- 3 - 4a = —1,0, and 1 , /(-1) (0,0) 4a(a - 1)(a + 1 ), and = -1, /(0)=0, /(!) (-1,-1) l [GURE (1,-1) 18 The behavior of / 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 be examining the formula for /'(a). On the we can sec (Figure 18) that / To determine the sign of /' on other hand, from the three critical values alone increases on (—1,0) and decreases on (0, 1). 199 11. Significance of the Derivative (—00, —1) and (1, we can compute oo) = 4-(-2) 3 -4-(-2) =4- 2 3 -4-2 = 24, /'(-2) /'(2) and conclude / that -24, decreasing on (— oo, —1) and increasing on is conclusions also follow from the fact that f(x) and large for large x is These (1, oo). for large negative x. We can already produce a good sketch of the graph; two other pieces of mation f(x) provide the finishing touches (Figure 19). First, = graph is for x = 0, ±V2; second, symmetric with respect already sketched in Figure 15, is it is clear that / to the vertical axis. odd, f(x) = it is easy to determine that even, f(x ) = f(—x), so the The function f(x) = x — x, is —f(—x), and is consequently sym- may metric with respect to the origin. Half the work of graph sketching by noticing these things infor- be saved in the beginning. (V2,0) (-1,-1) FIGURE (1,-D 19 Several problems in this and succeeding chapters ask you to sketch the graphs of functions. In each case you should determine (1) the critical points of /, (2) the value of (3) the sign of /' in the regions between critical points / at the critical points, (if this is not already clear), numbers x such that f(x) (4) the (5) the behavior of f(x) as x Finally, even, bear may in mind = becomes (if possible), large or large negative that a quick check, to see (if possible). whether the function is odd or save a lot of work. 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 information is This sort of analysis, if often very important. If rational function of / / is not defined at certain points whose denominator vanishes at some near these points should be determined. For example, consider the function /(*) = 2jc +2 (for example, points), then the if / is a behavior 200 Derivatives and Integrals which is not defined at 1 We . fix) have 2 (jc- 1)(2jc-2)-(jc -2jc = (x — x(x - D 2 of / + 2) 2) * C*-D 2 Thus (1) the critical points are 0, 2. Moreover, = = /(0) (2) f(2) Because / is (—oo, 0) and 2. not defined on the whole interval determined separately on the intervals, or -2, (2, oo). We intervals (0, 1) can do and (0, 2), on the (1,2), as well as by picking particular points this simply by staring hard at the formula for (3) f'{x) the sign of /' must be > x < if f{x) <0 fix) < fix) > if 1 if Either in each of these way we find that 0, <x < if /'. intervals 1, <x < 2 < x. 2, we must determine the behavior of f(x) as x becomes large or large negative, as well as when x approaches 1 (this information will also give us another way to determine the regions on which / increases and decreases). To examine the behavior as x becomes large we write Finally, +2 -2jc jcclearly /(jc) to is jc — 1 is close to jc — 1 (but slightly smaller) = x- 1 + 1 (and slighdy larger) when is jc ] x-V when x is large, and f(x) is close The behavior of / near 1 large negative. also easy to determine; since lim(jc' + 2) = 2jc 1 ^ 0, x->-l the fraction +2 2jc I becomes large as jc approaches 1 from above and large negative as x approaches 1 from below. All this information that it may seem a bit can be pieced together (Figure overwhelming, but there 20); is only one way be sure that you can account for each feature of the graph. When this sketch has been completed, we might note that it looks like the graph 201 11. Significance of the Derivative 2 a- 2 1 FIGURE - 2x + 2 20 of an odd function shoved over x2 1 unit, and the expression (x-!) 2 +l -2.y+2 x- I shows that this indeed the case. However, is which should be investigated only after this 1 is one of those special features 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 vealed by a detailed sketch of work. There is a popular its theorem 5 Suppose f'(a) then PROOF By / = 0. If has a local f"{a) maximum > its it is usually unnecessary to critical points. then 0, / has a local minimum at a. definition, f (a) = hm f(a + h)-f'(a) h /i->0 Since f'(a) — 0, this can be written ,. f (a) always re- do so much maxima and minima which depends on the graph, test for local behavior of the function only at is = hm h^O f'(a + h) h at a; if f"(a) < 0, 202 Derivatives and Integrals Suppose now > that f"(a) Then 0. + f'ia h)/h must be positive for sufficiently small h. Therefore: f'ia and f'ia + + /?) This means (Corollary and / is = - v- must be negative / 3) that for sufficiently small increasing in is Theorem < the case f"(a) some 0. interval to the right of a of a. Consequently, left / has a local similar. | is may be applied to the function f(x) been considered. We have 5 > h < positive for sufficiently small h decreasing in some interval to the minimum at a. The proof for fix) must be /?) = x3 — x, which has already (a) /'(*) /"(*) if(x) = = = 3x 2 - 1 6x. xv4 At the critical points, —y 1/3 and y we have 1/3, = -6/T73<o, / .r(- v T73) f(y/l/3)=6y/t/3>0. —y Consequently, (b) 1/3 is maximum a local point and y 1/3 is a local minimum point. Although Theorem 5 lf(x)=x 5 many functions the second derivative the sign of the that f"{a) that a be found quite useful for polynomial functions, for will is = first maximum is so complicated that Moreover, a if Theorem In this case, 0. a local (Figure 21) derivative. is a critical 5 provides easier to consider / it no information: minimum point, a local it is point of may happen it is possible point, or neither, as shown by the functions (c) FIGURE 2 fix) = -x\ f(x) fix) = x5 ; 1 in each case f'iO) local minimum = /"(0) = but 0, is a local point for the second, and neither maximum point a local maximum point for the third. This point will be pursued further in Part It is of theorem 6 Theorem nor first, a minimum IV 5 automatically proves a partial converse itself. Suppose f"ia) exists. If / has a local < maximum at a, then f'ia) Suppose / has local minimum at by Theorem local PROOF interesting to note that for the local maximum c/, containing a, so that f'ia) The case of a local = at 0, a maximum is minimum at «, then f"(a) > 0; if / has a 0. \\ f'ia) < 0, then / would also have a Thus / would be constant in some interval contradiction. Thus we must have f'ia) > 0. a. 5. handled similarly. | 203 11. Significance of the Derivative Theorem 5 by > and <, (This partial converse to signs cannot be replaced f(x) = -x 4 the best as shown by the for: > and < x 4 and = the functions fix) .) The remainder maxima and of this chapter deals, not with graph sketching, or Mean minima, but with three consequences of the Value Theorem. The theorem which plays an important simple, but very beautiful, and which we can hope is also sheds light first is Chapter role in on many examples which have occurred a 15, in previous chapters. THEOREM 7 Suppose / that is continuous and at a, containing a, except perhaps for x = that f'(x) exists for all x in some interval Suppose, moreover, that lim fix) a. exists. X—fO Then and f'ia) also exists, = /'(a) PROOF By lim /'(*). definition, f(a) = fja+h)- fja) lim h For sufficiently small h differentiable By the Mean on a (a, > the function + h) Theorem Value / there is a be continuous on will (a similar assertion number f(a+h)-f(a) = [a, + a a/, in (a, a + h) and h] holds for sufficiently small h < 0). such that /'(«*)• h Now lim x—>a approaches a as h approaches a/, fix) exists, it ,,, , / ia) = fia .. lim + h )- fia) = lim , / (a/,) a/, is in (a, a + h); since = lim / , (.*). *->a /i—O h a good idea to supply a rigorous s-8 argument for this final step, which have treated somewhat informally.) Even if / is an everywhere differentiable function, x fix) sin — , x ^ x = 0. x 0. to Theorem nuity of the type beautiful theorem which lem 62 uses The 7, shown this result to in it is still possible for /' to be if 2 2 According we | discontinuous. This happens, for example, FIGURE because follows that /j^O (It is 0, U however, the graph of /' can never exhibit a discontiFigure 22. Problem 61 outlines the proof of another gives further information strengthen Theorem next theorem, a generalization of the mainly because of its applications. about the function /', and Prob- 7. Mean Value Theorem, is of interest 204 Derivatives and THEOREM 8 Integrals (THE CAUCHY MEAN VALUE THEOREM) / and g are continuous on number x in (a, b) such that (If g(b) ^ g(a), [f{b) - g'(x) ^ and and [a,b] If f(a)]g'(x) = - f(a) is a g(a)]f(x). f'{x) = g(b)-g(a) = - [gib) (a,b), then there equation can be written 0, this f(b) on differentiable g'(x)- = and we obtain the Mean Value Theorem. On the other hand, applying the Mean Value Theorem to / and g separately, we find that there are x and v in (a, b) with Notice that if g(x) x for all x, then g'{x) f(b)-f(a) but there remarks g'{yY and y found in this way the Cauchy Mean Value Theorem no guarantee may to prove, PROOF is fix) = " g(b)-g(a) 1, that the x suggest that but actually the simplest of will be equal. These will be quite difficult tricks suffices.) Let = h{x) Then h is f(x)[g(b) continuous on follows from Rolle's g(a)\ - = f(a)g{b) Theorem - - g(x)[f(b) [a, b], differentiable h(a) It - on and (a,b), g(a)f(b) = that h'(x) — - g'{x)[f{b) for /(«)]. h(b). some x in (a, b), which means that = f'(x)[g(b) g(a)] The Cauchy Mean Value Theorem which facilitates is - the basic tool - f(a)]. | needed to prove a theorem evaluation of limits of the form lim fix) x^a g( X ) when lim f(x) In this case, Theorem 5-2 is = and lim g(x) = of no use. Every derivative 0. is a limit of this form, and computing derivatives frequently requires a great deal of work. If some derivatives are known, however, many limits of this form can now be evaluated easily. theorem 9 (L'HOPITAL'S rule) Suppose that lim fix) x—*a and suppose — and also that lim f'(x)/g'(x) exists. x—*a * fix) ,. lim x^a g( X ) (Notice that Theorem 7 is a special = case.) lim #(.y) x^a Then 0. lim f(x)/g(x) x—*a fix) lim x^a g'(x) = . exists, and — 11. Significance of the Derivative PROOF The 205 hypothesis that lim f'(x)/g'(x) exists contains two implicit assumptions: x—>a (1) there (a (2) in — is 8, this of x a + — 8, a + 8) such that f'(x) and g'(x) 8) except, perhaps, for x = a, an interval (a / g'(x) interval once again, with, the exist for all possible jc in exception = a. On the other hand, / and g are not even assumed to be defined at a. If we define (changing the previous values of f(a) and g(a), if necessary), f(a) = g(a) = / and g are continuous at a. \{a<x<a+8, then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to / and g on the interval [a, x] (and a similar statement holds for a — 8 < x < a). First applying the Mean Value Theorem to g, we see that g(x) / 0, for if g(x) =0 there would be some x\ in (a, x) with g'(x\ = 0, contradicting (2). Now applying the Cauchy Mean Value Theorem to / and g, we see that there is a number a x in (a, x) such that then ) [f(x)-0]g'(a x = [g(x)-0]f(a x ) ) or Now a x approaches a /(*) /'(«x) g(x) g'iUx) x approaches as assuming that lim f'(v)/g'(y) exists, it because a x a, is in (a,x); since we are follows that y-+a f(x) ,. lim v^a g(x) (Once again, the reader is = f'(a x ) lim x~>a g'(a x = ) lim f'(y) . y^a g'(y) invited to supply the details of this part of the argu- ment.) | PROBLEMS 1. maximum and minimum values on the indicated intervals, 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. For each of the following functions, find the (i) f(x) = x3 - x2 - 8a- f(x)=x +x + \ 4 3 f( x = 3x - 8.v + + 1 5 (ii) (iii) iv (vi) ) /(*) = _-— 6.v 2 on [-2, 2]. on [-1,1]. on [-£,£]. 1 XD + X /W =: +1 Z2—T 2 x +l f{x) = -^— x — x L 1 + on[-i,l], l 1 on[-l,i]. on [0,5]. , 206 Derivatives and Integrals 2. 3. Now sketch the graph of each of the functions in local maximum and minimum 1, and find all Sketch the graphs of the following functions. i) /(*)=* + -. —3 n) f{x) = 111 /(*) = LV 4. Problem points. a) /(*) X - x2 1 = 1+JC < If a\ + x • < 2" a„, find the minimum value of f(x) = /_](x — «,)" is a problem i=\ b) Now find the minimum value of /(jc) = 2, i where calculus won't help at all: on the = \ x ~ a i\- This \ intervals between the a,'s the minimum clearly occurs at one of the a / and these are precisely the points where / is not differentiable. However, the answer is easy to find if you consider how f(x) changes as you pass function is linear, so that the from one such interval Let a > 0. Show (2 + a)/( + a). 1 (—oo, 0), 5. (0, a), maximum value of I = 1 is + + IjcI + 1 \x — a\ (The derivative can be found on each of the intervals and (a, oo) separately.) For each of the following functions, find all local maximum and minimum points. i) ii) iii) iv) fix) fix) /(*) fix) = = = = x, x^3,5,7,9 5, x -3, x 9. x 7. x fix) = 3 5 7 9. x irrational \/q, x = p/q in lowest terms. x, x rational 0. x irrational. x — 0, v) = = = = 0, 1 1, 0, \ In for some n in N otherwise. if , to another. that the /(*) t the decimal expansion of x contains a 5 otherwise. 6. Prove the following (which we 11. Significance of the Derivative 207 on (a, b) often use implicitly): If / is increasing and continuous at a and b, then / is increasing on [«,&]. In particular, if / continuous on [a, b] and /' > on (a,b), then / is increasing on [a, b]. is 7. A straight line then back to 9(1, when (0,a)« drawn from is (1, b), as in the angles a and the point (0, a) to the horizontal axis, Figure 23. Prove that the are equal. (Naturally you must bring a function fi into the picture: express the length in on the horizontal The dashed axis. and total length terms of x, where (x,0) line in shortest is is the point Figure 23 suggests an alternative geometric proof; in either case the problem can be solved without actually ?'' (0,-fl), V (JC,0) finding the point (x, 0).) 8. Let (a) (jto, f(x) FIGURE (Jc, 23 yo) be a point of the plane, = mx + b. /(*)) its let the graph of the function minimizing same square. This may this distance is the as simplify the computations somewhat.] Also find i by noting that the line from (b) L be (xq, yo) to smallest. [Notice that is minimizing and Find the point x such that the distance from (xo, yo) to (x, f(x)) the distance from is perpen- dicular to L. Find the distance from (c) (x, f(x)). = b 0; (*0> yo [It (Ax Surface area sum is + i.e., the computations easier if then apply the result to the graph of f(x) ~ b)-] (Problem 10. make Compare Consider a straight (d) 9. will (xo, yo) to L, Byo 4-7). first = mx assume that and the point with Problem 4-22. line Show described by the equation that the distance + C)/y/A 2 + B 2 from Ax + By + C = (xo, yo) to this line is . The previous Problem suggests the following question: ship between the critical you (xo, yo) to points of / and those of / What is the relation- ? Prove that of all rectangles with given perimeter, the square has the greatest area. the of these areas 11. among Find, all right circular cylinders of fixed volume V, the one with and bottom, as smallest surface area (counting the areas of the faces at top in Figure 24). FKIl'RE 24 12. A right triangle with hypotenuse of length a to generate a right circular cone. is rotated about one of its legs Find the greatest possible volume of such a cone. 13. 14. Show that the sum of a positive number and its reciprocal Find the trapezoid of largest area that can be inscribed is in at least 2. a semicircle of radius a, with one base lying along the diameter. 15. Two is FI G U R E 2 5 hallways, of widths a and b, meet at right angles (Figure 25). What the greatest possible length of a ladder which can be carried horizontally around the corner? 208 Derivatives and Integrals 16. A garden be designed to is R and radius what value of R and 9 shape of a circular sector (Figure in the The garden central angle 6. is to radians) will the length of the fencing (in 26), with have a fixed area A. For around the perimeter be minimized? 17. FIGURE A right angle in Figure 27. 26 by the 18. circle? Ed must Ecological at mph 2 moved along the diameter of a circle of radius a, as shown What is the greatest possible length (A + B) intercepted on it is cross a circular lake of radius What rest (Figure 28). much see as 1 He mile. mph, or he can row or walk around at 4 part can row across way and walk the route should he take so as to scenery as possible? cross as quickly as possible? FIGURE 27 19. (a) Consider points = LAOC of a AAOB the areas of terms of 9 in (b) '20. maximum The lower the and > 3, of all is ABOC a is maximum? Hint: Express things n-gom inscribed in a circle, the regular n-gon area. right-hand corner of a page edge of the paper, as left the page FIGURE Bona circle with center O, subtending an angle How must B be chosen so that the sum of = LAOB. Prove that for n has A and (Figure 29). is folded over so that in Figure 30. If the very long, show that the minimum it just touches width of the paper length of the crease is a and is 3v 3or/4. 2 8 21. Figure 31 shows the graph of the minimum FIGURE FIGURE 2 9 22. 23. 3 Suppose of /. Find derivative all local maximum and points of /. 1 that / a polynomial function, f(x) is — 1, = x" + a„_]x"~ + and corresponding ] ••• +ao, with critical 4, 3. Sketch the graph of /, distinguishing the cases n even and n odd. (a) Suppose a JI 0, points 2, 3, 4, critical values 6, 1,2, = that the critical points of the polynomial function f(x) B- 1 + f'O) = _iJc 1, --- 0. +a are-l, 1,2,3, and /"(-l) Sketch the graph of / = 0, /"(l) > 0, x" + /"(2) < as accurately as possible on the basis of this information. (b) Does there that 3 24. I I', i RE 30 is exist a not a polynomial function with the above properties, except critical point? Describe the graph of a rational function (in very general terms, similar to the text's description of the graph of a polynomial function). 209 11. Significance of the Derivative 25. (a) most max(m,n) intersect in at (b) m For each which 26. and n Suppose / (a) Suppose that the has exactly k (b) + /" +•+f {n) > points critical m and n / > ^ and f"(x) (so n must be 0. — polynomial function f(x) for + an -\x x" all critical + + • ao Show points x. odd. is For each n, show that / (c) —k n that + / /' n, respectively, times. a polynomial function of degree n with is and points. two polynomial functions of degree exhibit max(m,«) intersect even). Prove that *27. m Prove that two polynomial functions of degree of degree n with k — k is n if odd, then there is a polynomial function each of which /" critical points, at non-zero. is Suppose that the polynomial function f(x) — x" + a„-\x"~ + + «o maximum points and ki local minimum points. Show that • • • has k\ local = k\ + 1 &2 (d) Let n, n is k\, =k\ if n ki be three integers with ki — k\ n if odd, and 28. 29. M for M for > < a^^+ki and = f'(x ) then |/'(jc)| for all x in [0, 1]. < Show Find all > (a) fix) (b) f"(x) (c) — / feet after it / g'(x) for all x, and that there = that f(a) — g(a), that f'(x) a. Show > that f(x) g'(x) for ) + M(b-a). [a, £]. is an interval g(a). Show that follow without the > all x, g(x) for all and that /'(xq) > x > ao- such that is . true that a weight dropped from rest will fall s(t) = \6t 2 seconds, this experimental fact does not mention the behavior of weights which are thrown upwards or downwards. law s"(t) +x sin*. = x\ f'"(x)=x+x 2 Although (1 g(a). some xq > functions of > M/4. f{x) > g(x) for x > a and f(x) < g(x) for x < a. Show by an example that these conclusions do not g'(xo) for x in all (b) — / f(b)<f(a) + M(b-a). M for Suppose that f{a) k\ if 1 in [a,b], (a) Suppose (*~ a i) I = x /'(*) that f'{x) — ki local minimum points. ki+k2 all if |/| ki a polynomial function then f{b) > f(a) Prove that > and in [a,b], Formulate a similar theorem when M tr y even, is x (c) > is n if all (b) that f'(x) 1 , f'{x) Suppose + /. if hypothesis f{a) 32. < • odd. and points Prove that (a) (c) 31. • • is that there maximum an appropriate number of length \ on which 30. < ai Show n. k\ local < Hint: Pick a\ for + kj < k\ degree n, with even, and kj is = 32 is On the other hand, the always true and has just enough ambiguity to account for from any height, with any initial velocity. measure heights upwards from ground level; the behavior of a weight released For simplicity let us agree to in this case velocities are positive for rising bodies, and all bodies fall bodies and negative for falling according to the law s"(t) = —32. 210 Derivatives and Integrals Show (a) that s = — \6t 2 +at + of the form s(t) is (3. By setting t = in the formula for s, and then in the formula for s', show that s(t) = —I6t 2 + vtf + sq, where so is the height from which the body is released at time 0, and vq is the velocity with which it is released. (b) A (c) weight thrown upwards with is How ground velocity v feet per second, at ("How high" means "what is the maximum height for all times".) What is its velocity at the moment it achieves its greatest height? What is its acceleration at that moment? When will it hit the ground again? What will its velocity be when it hits the ground level. high will it go? again? 33. A cannon ball zontal component of has a vertical it component — — 16?" + s(t) shot from the ground with velocity v at an angle is ure 32) so that v cos a. Its distance s(t) (vsina)t, while velocity v sin a (Fig- a and a hori- above the ground obeys the law horizontal velocity remains constantly its v cos a. (a) Show (b) 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. (a) Give an example of a function (b) lim f'(x) does not exist. x—>oo Prove that if lim f(x) and lim f'(x) both each time l'Kil'RE 3 2 34. Suppose ~ fg' in f's it is Suppose = (find the position at / for which lim f(x) 0. a. Hint: all On any interval — f(y)\ < \x — y\" for n > f. Compare with Problem 3-20. that \f(x) / is Lipschitz of order = 0. = 0. which functions an v in interval if (*) where f/g defined, interval holds for aatx if there 1. is Prove that / is a constant around all x. The function x and y in the / C show is is is Lipschitz of order Lipschitz of order differentiable at x, then constant by a on an interval. If Lipschitz of order /' is / Lipschitz of order 1 at x. is uniformly at x. Is the converse true? (d) (e) If [f / / x a (c) / all such that (b) / / satisfy for is a > at x, then / is continuous on an interval, then / If is Lipschitz of order a > continuous on this interval (see Chapter 8, Appendix). If - then f(x) 0, \f(x)-f(y)\<C\x-y\ (*) (a) but exists, constant. A function for a parabola / and g are two differentiable Prove that if f(a) = and g(a) ^ that considering 37. is t, an interval around that 36. ball exist, then lim f'(x) x—*oo ;c—>oo Jt—>00 Prove that if lim f(x) exists and lim f"(x) exists, then lim f"(x) -V—>oc x—>oo x—*oo (See also Problem 20-22.) (c) 35. cannon that the path of the on is (li(lcrential)le is Lipschitz of order [<:/./?], is a- > 1 on |<7./;>|, then /' is 1 on [a,fc]? constant on \a,h\. 11 38. Prove that Significance . 211 of the Derivative if an 2 1 n = + 0, 1 then «o for 39. some x + «!•* + • • • + a„x" -0 in (0, 1). 3 fm (a) = x — 3x + m Prove that the polynomial function 40. never has two roots no matter what m may be. (This Theorem. It is instructive, after giving an analytic proof, to graph fo and and consider where the graph of /„, lies in relation to them.) /2, continuous and differentiable on in in [0, 1], Suppose an easy consequence of Rolle's is / that is each x, and that f'(x) [0, 1] for exacdy one number x ^ 1 for x all in [0, 1]. = such that f(x) in [0, 1] [0, 1], that Show (Half of x. f(x) is that there this is problem has been done already, in Problem 7-11.) 41. (a) Prove that the function f(x) two numbers (b) = x — cos a satisfies f(x) = for precisely x. = x — x sin x — cosx. f(x) — 2x — xsinx — cos Prove the same for the function f(x) Prove *(c) this also for the function preliminary estimates will be useful to (Some x. restrict the possible location of the zeros of /.) *42. (a) Prove that /(l) In = / if = and f'(0) 1 = a twice differentiable function with /(0) is more picturesque f'{\) = 0, terms: A particle > 4 then \f"(x)\ which for some x and in (0, 1). travels a unit distance in some time an some x in (0, I), a unit time, and starts and ends with velocity 0, has at > acceleration < —4 or else f"(x) (b) 43. Show Suppose 0. that / is = for = f(x) that / some function interval Suppose for g. \f"(x)\ f(y) for all > 4 = for some x l/.v for all x, y > 0. x in (0, 1). > and /( Hint: Find g'(x) = 1 ) when + f'(x)g(x) - f(x) = Prove that between them. Hint: that / is / is at two Use Theorem 6. if continuous on [«,b], that and that f(x) some x in (a, b). (a, b), in (j, 1). satisfies f\x) 45. + for f(xy). Suppose for some x we must have a function such that f'(x) Prove that f(xy) g(x) 44. that in fact >4 Prove that either f"(x) 4. Hint: = for n+ 1 it points, then is / is on //-times differentiable different x in [a, b]. Prove that / (n) (jc) the on = 212 Derivatives and Integrals 46. Let x\ , . . . x n+ be arbitrary points , = Q(x) / that (n is + let Y\(x-Xi). i Suppose and in [a, b], \ = ] and l)-times differentiable that < n such that P(xj) = f{xi) for Show that for each x in [a,b] there is function of degree Problem 3-6). P = i a is a polynomial n 1 number + 1 (see c in (a,&) such that = f(x)-p(.x) Jf Q(x)- in+l) (» (c) \; + . 1)! Hint: Consider the function F(0 = G(*)[/(0 - P(0] - G(0[/O0 - Show 47. that F is +2 zero at n different points in [«,&], (without computing Prove the following is v 66 66-8 < < decimal to 2 is some x in (a, b) lim /(y) ... - is a trivial on are never simultaneously Prove that g'(x) if ^ / and f(x) " *'(*)' g are continuous on g'ix) Hint: Multiply out What is wrong with 3 Jt ,. lim -^ jc-i (The first, limit is x 2 to see ^ g(a) and that f'(x) and g'(x) (a,b). for x in (a, b), then there fix) 51. Value Mean Value Theorem can be written under the additional assumptions that g(b) and Mean form f(b)-fja) g(b)-g(a) 50. consequence of the ".) Prove that the conclusion of the Cauchy in the lim f(y) v—>a + ' (Your proof should begin: "This Theorem because Mean Value Theorem: If / and lim f(y) and lim /(v) exist, of the (a, b) such that y—>b fix) = o places!). slight generalization continuous and differentiable on then there 49. and use Problem 45. Prove that 9 48. P(x)]. __ " f(x) - and differentiable on some x in (a, b) with [a, b] is f{a) g(b)-g(x)' what this really says. the following use of l'Hopital's Rule: +x-2 -3a- + actually —4.) 2 = ,. lim 3v 2 +l — — = x^l 2jc-3 ,. lim x->1 6.v — - = 2 „ 3. (a,b), . 11. Significance of the Derivative 52. Find the following (i) lim 213 limits: . x^O tan x 2 cos * (11) x .r^O 53. — 1 = lim Find /'(0) l if g(x) , x #0 x = /(*) 0, = and g(0) 54. g'(0) = = and g"(0) 0, 17. Prove the following forms of l'Hopital's Rule (none requiring any essentially new (a) reasoning). If - lim f(x) x — «+ lim f(x)/g(x) = lim g(x) = 0, and lim f'(x)/g'(x) x—>a + x—« + I (and similarly for limits from below). = /, then x—*a + (b) = lim g(x) = 0, and lim f'(x)/g'(x) = oo, then x->a x—>a = oo (and similarly for — oo, or if x —> a is replaced lim f{x) x—>a lim f{x)/g(x) x—*a If by x —> a + or x — a~). »• (c) lim /(*) = X—>00 lim f(x)/g(x) If lim g(x) X—>O0 — I (and = and 0, lim f'{x)/g'(x) X—>00 for — oo). Hint: similarly = /, then Consider lim f(\/x)/g(l/x). (d) lim f(x) = lim g(x) x—>oo x—>oo lim f(x)/g(x) = oo. = If and 0, lim f'(x)/g'(x) = oo, then .*—>oo x -> oo 55. There is another form of l'Hopital's Rule which requires more than algebraic manipulations: If = lim f(x) x-+oo then lim f{x)/g{x) —I. Prove lim g(x) x—>oo = oo, and lim f'(x)/g'(x) = I, x—yoo this as follows. x-»-oo (a) For every e > there is a number a such /'OO / < for e jc that > a *'(*) Apply the Cauchy Mean Value Theorem to / and g on x > that f(x)-f(a) g(x)-g(a) (Why can we assume g(x) — I g(a) < ^ e 0?) for a. [a,x] to show 214 Derivatives and Integrals Now write (b) fix) fix)- fid) fix) g(x) g(x)-g(a) f(x)-f(a) (why can we assume that f(x) — gjx) f(a) - gja) Six) for large x?) 7^ and conclude that — / < 2e for sufficiently y Six) 56. lame * x. To complete the orgy of variations on l'Hopital's Rule, use Problem 55 to prove a few more cases of the following general statement (there are so many you should possibilities that If lim lim g(x) f(x) *-»•[ select just a few, if any, that interest you): = } and lim f'(x)/g'(x) ), ( then lim ] 1 Here [ ] can be a or a + or a or 00 or —00, and ). fix)/gix) — ( can be or 00 or —00, and can be / or 00 or —00. ( 57. If / and g are differentiable { } ) and lim f(x)/g(x) exists, does follow that it x—ya lim f'(x)/g'(x) exists (a converse to l'Hopital's Rule)? X—>fl 58. Prove that of / /' if increasing, then every tangent line of is only once. (In particular, this is / intersects the true for the function f(x) = x" graph if n is even.) 59. Redo Problem 10-18 (c) when 1 iff = f (Why *60. (a) is Suppose (c) / is this chapter?) differentiable on [a b] , . Prove that if the minimum /' part 61. that is 2 on [a, b] is at a, then /'(a) > 0, and if it is at b, then fib) < 0. (One half of the proof of Theorem 1 will go through.) and f'{b) > 0. Show that fix) = for some x Suppose that f'ia) < in (a, b). Hint: Consider the minimum of / on [a, b]; why must it be somewhere in (a, b)? Prove that if f'{a) < c < fib), then fix) = c for some a: in (a, b). (This result is known as Darboux's Theorem. Note that we are not assuming that /' is continuous.) Hint: Cook up an appropriate function to which of (b) problem this T Suppose (b) may that / be applied. is differentiable in some interval containing a, but that /' is discontinuous at a. Prove the following: (a) (b) The one-sided limits lim fix) and lim fix) cannot both x—>« 4 x—ya~ is just a minor variation on Theorem 7.) exist. (This These one-sided limits cannot both exist even if we allow limits with value +00 or —00. Hint: Use Darboux's Theorem (Problem 60). the 11 ^62. For example, x then / is f(a) = 0. (a) that |/| differentiable but is 215 of the Derivative / is not. we can choose f{x) — 1 for a rational and f(x) = —1 for / is not even continuous, nor is this a mere Prove that if |/| is differentiable at a, and / is continuous at a, In this example irrational. coincidence: ^63. / such easy to find a function It is Significance . Hint: also differentiable at a. Why? suffices to It consider only a with In this case, what must \f\'(a) be? / and let n be even. Prove that x" + y" = (x + y)" only when x — 0. Hint: If A'o" + y" = (a'o + y)", apply Rolle's Theorem to f(x) = x" + y" - (x + y)" on [0, x ]. Prove that if y / and n is odd, then x" + y" = (x + y)" only if x = or x = —y. Let y () (b) 64. Suppose f(x)/x that f(0) increasing is Prove that and /' on (it remember as Use derivatives to prove that + a)" > 1 Let f{x) = a 4 sin 2 1 + nx for (a) Prove that (b) Prove that f'(0) is ^ fx for a a local = > n if = (notice that equality holds for a 66. = the function f(x) tial, (1 Hint: Obviously (0, oo). will help to shown by increasing. Prove that the function g(x) is by applying the positive it is right interval 65. = 0, = you should look Value Theorem that the hypothesis /(0) 1 + to = at g'{x). / on is the essen- a"). then 1, - < < a 1 <a and 0). and minimum /"(0) Mean = /(0) let = (Figure 33). point for /. 0. Theorem 6 cannot about maxima and minima that This function thus provides another example to show that be improved. It also illustrates a subtlety often goes unnoticed: a function right of a local minimum may not be increasing in any interval to the point, nor decreasing in 1 any interval to the left. i 1 h \ / / \ \ / \ FIGURE *67. (a) Prove that some The (b) if f'(a) > / 3 3 and /' is continuous at a, then / is increasing in interval containing a. next two parts of this problem show that continuity of /' If #(a) = with g'U) a2 = sin 1/a, 1 and show that there are also with g'(A) =— numbers a 1. is essential. arbitrarily close to 216 Derivatives and Integrals (c) < a < Suppose /(0) = (see interval containing 0, with fix) FIGURE 3 > and = ax + Let f(x) 1. Show that / by showing that in Figure 34). also points jc is sin 1/x for x ^ 0, and let not increasing in any open any interval there are points x x with fix) < 0. 4 The behavior of / for a > 1, which much more is difficult to analyze, is discussed in the next problem. **68. Let f(x) = ax + x 2 s'm the sign of /' (jc ) is < —1 \/x for when a > — g(y) < —1 part of g(y) when < — 1 sin y . = — cos_y, 0, and it is not at The obvious approach (a) Show more that if ingenuity g'(y) let = 0, cosy It — is is a all to this ^ In order to find sin \/x — cos /x 1 more convenient to 0; we want to know — 1, but this happens only whether g itself can have values problem is to find the local minimum impossible to solve the equation g'(y) is is required. then = 2x quite delicate; the most significant 2 - 2 y (sin y) that g(y) 0. if clear it = = little cos y for y 2y and conclude /(0) which does reach the value values of g. Unfortunately, explicitly, so to 0. 2(sin y)/y for large y. This question is and necessary to decide any numbers x close for consider the function g(y) if x^O, it is 1 2+ (sin v) 2y v- — , // Now (b) show = that if g'(y) and conclude 217 then 0, 2 4y sm" of the Derivative Significance . y 4 + 2 +y 4 v that : \g(y)\ V4 + Using the 4 + y 2 )/ v 4 + fact that (2 > y not increasing in any interval around Using the (d) +y lim (2 fact that y->oo / **69. A is increasing in / function some if there a < f(a) if a that / and /(•*) Notice that does this mean not shown a +8); for example, the function is a = that if if 1 , then / is show 1, a > 1, then 0. some number x < —8 < is = v is < if that 0. around f(x) > f(a) show 1, + )/v 4 interval increasing at a is 2 4 v such that +8 a < x > 8 a. increasing in the interval (a Figure 34 in is — 8, increasing at 0, but not an increasing function in any open interval containing 0. Suppose (a) / that every a in continuous on is Prove that [0, 1]. yourself that there prove that the Prove part (b) (a) minimum If (c) / [b, 1] = the set 1] problem {x : b increasing at a is increasing at a for (First [0, 1]. convince < b < For 1 at b. continuous, by consider- Si, = [x f(y) > f(b) for all y in [b, x] }. not necessary for the other parts.) Hint: : < x < l}by and / is is Hint: must be / / on be proved.) to / on of that increasing without the assumption that (This part of the is something is ing for each b in [0, Prove that Sb / and [0, 1] is is considering sup Si,. > differentiable at a, prove that f'(a) (this is easy). If f'{a) (d) > 0, increasing at a (go right back to the definition prove that / and show, without using the is of /'(a)). (e) Use parts that if is / (a) (d) to continuous on is increasing on [0, 1] and f'(a) > /(0) ex - > —e. all Value Theorem, a in [0, 1], then / [0, 1]. Suppose that / is continuous on [0, Apply part (e) to the function g(x) (f) Mean for Similarly, show 1] and f'{a) = f(x) that /(l) f(x). Conclude that /(0) = — = for all a in (0, 1 ). ex to show that /(l) — f(0) < s by considering h(x) = + /(l). This particular proof that a function with zero derivative must be constant has many points in common with a proof of H. A. Schwarz, which first rigorous proof ever given. See his exuberant letter in Its discoverer, at least, seemed may to think reference [54] of the Suggested Reading. be the it was. 218 Derivatives and Integrals ** 70. / If is a constant function, then every point for /. It for example, if f{x) = But prove, using Problem 8-4, function: 0. is maximum a local point happen even if / is not a constant for x < and f{x) — 1 for x > that if / is continuous on [a,b] and quite possible for this to is is a local maximum point, then / is a constant The same result holds, of course, if every point of [a,b] is a local minimum point. Suppose now that every point is either a local maximum or a local minimum point for the continuous function / (but we don't preclude the possibility that some points are local maxima while others are local minima). Prove that / is constant, as follows. Suppose that /(ao) < fibo)- We can assume that /(ao) < /(*) < f(bo) for ao < x < bo. (Why?) Using Theorem 1 of the Appendix to Chapter 8, partition [ao, bo] into intervals on which sup / — inf / < if {bo) — f(ao))/2; also choose the lengths of these intervals to be less than (bo — ao)/2. Then there is one such interval [ai,&l] with ao < a\ < b\ < bo and f(a\) < f{b\). (Why?) Continue inductively and use the Nested Interval Theorem (Problem 8-14) to find a point x that cannot be a local maximum or minimum. every point of [a,b] function. (b) A point x ** 71. for all is v in maximum A ^ with y A local strict all local strict point). obvious way. Find = x = q It maximum maximum point if defined in the is points of the function x irrational 0, f(x) maximum point for / on A f{x) > f(y) x (compare with the definition of an ordinary called a strict P- in lowest terms. q seems quite unlikely that a function can have a at every point (although the maximum local strict above example might give one pause for thought). Prove this as follows. (b) Suppose that every point x\ is a local be any number and choose that f{x\) (a\, b\) > fix) and choose for a\ all < a\ strict < x\ x in [ai,&i]. aj < xj < &2 < maximum < point for /. b\ with b\ / x\ b\ with bi — Let xj — < a\ 1 Let such be any point «2 < i in such that f(xi) > fix) f° r a H x m [ fl 2>^2]- Continue in this way, and use the Nested Interval Theorem (Problem 8-14) to obtain a contradiction. 219 11, Appendix. Convexity and Concavity CONVEXITY AND CONCAVITY APPENDIX. Although the graph of a function can be sketched quite accurately on the basis of the information provided by the derivative, some subde aspects of the graph are revealed only by examining the second derivative. These details were purposely omitted previously because graph sketching complicated enough without wor- is rying about them, and the additional information obtained effort. Also, in often not worth the is correct proofs of the relevant facts are sufficiently difficult to be placed an appendix. Despite these discouraging remarks, the information presented is well worth assimilating, because the notions of convexity and concavity are here far more important than as mere graph sketching. Moreover, the proofs aids to have a pleasantly geometric flavor not often found the basic definition DEFINITION 1 A function / is The geometric analytic way that (a, (b, geometric in nature convex on an segment joining f{a)) and is (a, (b, f{b)) fib)) is line lies and b in the interval, the above the graph of /. an between can be expressed The in proofs. straight line line in the graph of the function g defined by = f(b)-f(a) (x -a) + fia). — / above the graph of / a at x if g(x) > f(x), that is, if f(b)) fib) b (a, theorems. Indeed, 1). this definition sometimes more useful b This lies condition appearing in is in calculus Figure interval, if for all a f(a)) and g(x) (b, (see ~ — fia) (x-a) + fia)> f(x) a fia)) or fib) b — f{a) (x a -a) > fix)- fia) or FIGURE f(b) I b We DEFINITION 2 A a — fia) a x) > f( x — fja) . a therefore have an equivalent definition of convexity. function / is convex on an < x < b we have f(x) x — interval if for a, x, fia) a < fib) - and b fia) in the interval with 220 Derivatives and Integrals If the word "over" in Definition inequality in Definition 2 fix) (b,f(b)) x we replaced by "under" is 1 obtain the definition of a — f{a) f(b) > b a concave — this reason, the FIGURE state 2 the f(a) , function (Figure not hard to see that 2). It is form — /, where / is convex. next three theorems about convex functions have immediate about concave functions, so simple that we corollaries if a the concave functions are precisely the ones of the For equivalently, or, replaced by is will not even bother to them. Figure 3 shows some tangent lines of a convex function. Two things seem to be true: (1) The graph [a, f{a)) (2) If a < b, of itself (this As a matter of fact THEOREM l Let lies above the tangent point is line at (a, line at (b, f{b))\ that is, /' these observations are true, is (a, / be < convex. If / is h\ < ti2, and difficult. b, then f'(a) < graph of / If itself. a lies above < b and / is f'(b). then as Figure 4 indicates, ... a line). than the slope and the proofs are not differentiable at a, then the fia+hO- f(a) (1) A less 3 differentiable at a If f{a)) is increasing. the tangent line through (a, /(«)), except at (a, f(a)) PROOF f(a)) except at the point point of contact of the tangent called the then the slope of the tangent line at of the tangent FIGURE / < f(a + h 2 )-f(a) . nonpictorial proof can be derived immediately from Definition 2 applied to < a + h\ < a + hj. Inequality (1) shows that the values f(a+h)-f(a) h of 11, Appendix. Convexity and Concavity a KKiURE + a + hi Consequently, . f'(a) for h h\ 4 decrease as h (in fact + 221 f'(a) > is < f(a+h) f(fl) for h > bound of all these numbers). But this means that through (a, f{a)) and {a + h, f(a + h)) has larger slope the greatest lower the secant line than the tangent line, which implies line (an analytic translation of this For negative h there is f(a that (a + h, argument is f(a a similar situation (Figure + hi)-f(a) > 5): + h 2 )- f(a This shows that the slope of the tangent line + h)) lies above the tangent easily supplied). if < hi h\ < 0, then f(a) . is greater than f(a+h)-f(a) lor n < U h (in fact f'{a) is above the tangent FIGURE 5 upper bound of all these numbers), so that f(a + h) if h < 0. This proves the first part of the theorem. the least line lies 222 Derivatives and Integrals FIGURE Now 6 suppose that a < j (a) < Then, b. f(a we have as already seen (Figure + (b-a))-f(a) b fib) b — . since b — a > U — b < 6), a - fid) —a and (b) / f(b > + (a-b))-f(b) . since a —b f(a)-f(b) = f(b)-fja) — a b b — a (J a _ Combining a x\ IE these inequalities, we obtain f'{a) < f'(b). | xq Theorem 7 We Theorem is 1 Here the proofs will be a little more difficult. same role in the next theorem that Rolle's proof of the Mean Value Theorem. It states that if /' graph of / lies below any secant line which happens to be has two converses. begin with a lemma plays in the increasing, then the that plays the horizontal. lemma Suppose / is fix) < f{a) proof and /' is a < x < b. differentiable = f(b) for a increasing. If < b and f(a) = f(b), then Suppose that f{x) > fid) = f(b) for some x in (a,b). Then the maximum of / on [a,b] occurs at some point A'o in (a, b) with /(a'o) > fid) and, of course, /'(a'o) = (Figure 7). the interval [a, xq\, we On the other hand, applying the find that there fl( f . U'i ) is x\ with a = /Up) A'o contradicting the fact thai ./ " is increasing. | fja) a < x\ > 0, Mean < Value Theorem xq and to 223 11, Appendix. Convexity and Concavity We now attack we used THEOREM 2 PROOF If / Let a proof of the in the and differentiable is < same sort of algebraic machinations Mean Value Theorem. the general case by the /' f(b)-f(a) = - fix) b the easy to see that g' lemma to g is In other words, if <x < a f(x) x 8 THEOREM 3 If / is convex. differentiable Let a < b. It ia, fia)), clear is and ia, a). = g{b) — f(a). Applying < fia) <x < a if b. fib) b - — fia) fia) —a - ix a) < a < fib) b — fia) fia) a | and the graph of / point of contact, then PROOF - then b, or is f (x a that f(x)~ Hence / — also increasing; moreover, g(a) we conclude g(x) FIGURE convex. is Define g by b. g(x) It is / increasing, then is / above each tangent line except at the convex. is from Figure 8 that fia)) lies lies if ib, above the tangent fib)) lies above the tangent line at ib, fib)), g(x) since ib, fib)) lies line, Similarly, since the tangent line at ib, fib)) and ia, fia)) lies It follows It now from (1) and follows from If a function / (2) is fib), line at ib, fib)), fia)> f'ib)ia-b) + that fia). the graph of = f(b)ix-b) + above the tangent (2) we have fib)> f'ia)ib-a) + h( x ) line at = f\a)(x-a) + f(a), above the tangent (1) line at then the slope of must be larger than the slope of the tangent ia, fia)). The following argument just says this with equations. Since the tangent line at ia, fia)) is the graph of the function the tangent line at ib, fib)) and that we have fib). fia) < fib). Theorem 2 that / is convex. | has a reasonable second derivative, the information given in these theorems can be used to discover the regions in which / is convex or concave. 224 Derivatives and Integrals Consider, for example, the function /(*) For = l+x 2 this function. = fix) Thus f'(x) = only for x = 0, -2x a+x and /(0) = ' 2 2 1, fix) > if x < if a- /'(*) ) while < > 0, 0. Moreover, /U) > f(x) / FIGURE is for all x, —> as x of / — oo, even. (l+.x- 2 2 ) (-2) + 2.v(1+JC 2(3jc 2 - (1+JC 2 [2(1 = y 1/3 or —y/\/3. Since /" is +.v 2 ) - compute 2.vJ 2 4 ) 1) 3 ) not hard to determine the sign of f"(x). Note on each of the We now therefore looks something like Figure 9. fix) = x oo or 9 The graph It is —> first clearly continuous, sets (-00,-/173), (VlAoo). that f"(x) it = only when must keep the same sign Since we compute, for example, that easily r(-D = /"(0) > = -2 <0, 4 we conclude >0, that /" > on (-oo, -yf\j?> ) and /" <Oon(- v/T73, /I73). > Since /" (—oo, means — vl/3 and ) FIGURE 225 and Concavity 11, Appendix. Convexity /' is increasing, it (y/V/3, oo), follows from (vl/3, oo), while on ( Theorem 2 — vl/3, v 1/3 ) f is that / is convex on concave (Figure 10). 10 Notice that at (y 1/3, |) the tangent line right, since / lies below the part of the graph convex on (y 1/3, oo), and above the part of the graph is to the to the left, / concave on (—-y/1/3, vl/3 ); thus the tangent line crosses the graph. In general, a number a is called an inflection point of / if the tangent line to the since is graph of / at (a, points of fix) that a / is is an = 1/(1 +x 2 ). Note convex, so the tangent line at /" has that the condition f"{a) inflection point of /; for order to conclude that a that is an (0, illustrates the example, if f(x) = = are inflection does not x 4 then /"(0) , ensure = 0, but 0) certainly doesn't cross the graph of /. In inflection point of a function /, different signs to the left This example vl/3 and —vl/3 f(a)) crosses the graph; thus and we need to know right of a. procedure which may be used to analyze any func- graph has been sketched, using the information provided by /', the zeros of /" are computed and the sign of /" is determined on the intervals between consecutive zeros. On intervals where /" > the function is convex; tion /. After the on intervals where /" < convexity and concavity of the function is concave. Knowledge of the regions of / can often prevent absurd misinterpretation of other data about /. Several functions, which can be analyzed in this way, are given in the problems, which also contain some theoretical questions. 226 Derivatives and Integrals To round out our discussion of convexity and concavity, we will prove one result that you may already We have begun to suspect. concave functions have the property that every tangent few drawings just once; a this property, THEOREM 4 If / is once, then PROOF /(c)- / either is There are two know on an interval and intersects each of convex or concave on that interval. its tangent lines just parts to the proof. (1) First we Suppose, on the contrary, that some straight / graph rather tricky. is points. of line intersects the probably convince you that no other functions have will but the only proof I differentiable further have seen that convex and claim that no straight line can intersect the graph of at {a, /(a)), (b, f{b)) and < /(c)), with a (c, / in three different line did intersect the < b c (Figure 11). graph Then we would have f(b)-fia) (]) — b f(c)-f(a) a c —a Consider the function g(x) Equation FIGURE 11 (1) = where in (b, c) g'{x), f(a) for = says that g(b) - fix) = So by g{c). Rolle's x in [b, c]. Theorem, there is some number x and thus 0=(x-a)f'(x)-[f(x)-f(a)) or fix) = fix)- — .v But this says (Figure 12) that the f{a) a tangent line at (x, f(x)) passes through (a, f{a)), contradicting the hypotheses. (2) Suppose that ao < bo < .v, y, zt FIGURE 12 Then and = ao an d x\ , step g(t) t)ciQ -t)b -t)c {\ (1 < < b\ + ta\ + tb\ + tc\ c\ are points in the interval. Let 0<t < 1. y, < for 0<t < between oq x, all lie 1. consider the function = , git) By - (1 a\ a\ < x, Now = = = and and (Problem 4-2) the points with analogous statements for y, and z.t- Moreover, xo a\, = co < (1), for g(l) all / ^ in f(y t )-f(x for [0, 1 all ]. / t ) in f(z,)-f(x [0, 1|. So Thus, either / is r ) either g(t) convex or for ()</<!. > lor all / is concave. in / | [0. 1| or 227 11, Appendix. Convexity and Concavity PROBLEMS 1. and concavity and points of inflection, Problem 11-1 (consider (iv) as double starred). Sketch, indicating regions of convexity the functions in 2. Figure 30 in Chapter 3. Show / that interval 4. (a) is (c) convex on an interval is + -t)y) <tf(x) (\ will if be / and g (a) Suppose either (c) + if for all x and y for0<f < -t)f(y), (\ are convex easiest to use / that is and / Problem (a) is in the 1. fog increasing, then is convex. 3.) / by considering second is / is decreasing to the derivatives. and convex on an differentiable is increasing, or else / such that (b) and only Give an example where g o f is not convex. Suppose that / and g are twice differentiable. Give another proof of the result of part 5. if Sketch the graph of /. /'. just a restatement of the definition, but a useful one.) Prove that (It (b) shows the graph of 1 we have f(tx (This 1 interval. decreasing, or else there is of c and increasing to the right of left Use this fact to give another proof of the result in Problem and g are (one-time) differentiable. (You will have to be a when comparing f'(g(x)) and f'igiy)) for x < y.) Prove the result in part (a) Show that a number c without assuming / when / 4(a) little differentiable. c. careful You will have to keep track of several different cases, but no particularly clever ideas are needed. Begin / is by showing that increasing to the right of b; and if if < a fia) > < fib), then is decreasing b and f(a) fib), then / to the left of a. *6. / be f{x) > fix) = Let a twice-differentiable x and / 0, some x > for tion point at in this x > for is is essential, as = function with the following properties: decreasing, (so that in — an example theorem is reasonable cases given by fix) shown by fix) = and f'iO) =1/(1 = 1 / will -\-x )). —x , Prove that 0. have an inflec- Every hypothesis which is not positive = 2 all x; by fix) l/(x + 1), x which is not decreasing; and by fix) with f'ixo) < 0. We which does not satisfy /'(0) = 0. Hint: Choose xq > for , cannot have f'iy) < f'ixo) for some x\ > xq. Consider /' on *7. (a) (b) all y > xo. Why not? So f'(x\ ) > f'ixo) for [0, x\\. / is convex, then fi[x + y]/2) < [fix) + fiy)]/2. Suppose that / satisfies this condition. Show that f(kx + (1 — k)y) < Prove that if and 1, kfix) + (1 — k)fiy) whenever k is a rational number, between = of the form m/2" Hint: Part (a) is the special case n 1. Use induction, . employing part (c) Suppose Show that that / (a) / is at each satisfies convex. step. the condition in part (a) and / is continuous. . 228 Derivatives and . Integrals "8. For n > 1, let p„ be p\ numbers with Y^ positive Pi — 1- n (a) For any numbers x\, and the (b) Show . . show x„ , V] that i=l largest x,. the same (I/O >^ for /?/-*,, where f = >^ p ;=1 (c) Prove Jensen's between the smallest lies /?/•*/ inequality: If / t . i=] is N convex, then /( < J^ /?,a", J Hint: Use Problem = noting that p n 3, 1 — t (Part (b) . p, fjx,). i=\ i=\ needed is show to n-\ that *9. (a) 2, Pi x (1/0 i is i n the domain of / if x\, . . , . xn are.) — f(a)]/h, For any function /, the right-hand derivative, lim [f(a + h) is denoted by f'+ {a), and the left-hand derivative is denoted by fL(a). The proof of Theorem 1 actually shows that f'+ {a) and f'_{a) always exist if / is and (b) convex on some open interval containing also show Conversely, suppose that = with f(b) on [a, c]. (a) is are increasing, convex on < g{b) and f'_ib) and [a, b] and g g' (b) (Figure Hint: Show (b), that compare / if is Hint: [fib) — f'+ {b) than is less the slope of OQ convex, then f'+ia) / is f(a)]/(b f'_{a) if f'+ {a). we If h convex on is ib, [b, c], define h fib)), as in PO with that of = < convex on is 13(a)). + this assertion, that f'_{a) / on [a,b] and g on [b, c], show that Given P and Q on opposite sides of O = tinuous at a. (Thus *10. / f'_ [a,c] to be Figure 13 (c) and that f'+ Check a. and only if f'+ is con- when f'+ is continuous.) for b < a close to a, and differentiable precisely — a) is close to f'_{a) this quotient. Prove that a convex function on R, or on any open interval, must be continuous. (a) (b) Give an example of a convex function on a closed interval that is not continuous, and explain exactly what kinds of discontinuities are possible. 11. Call a function / weakly convex on an interval if for a < b < c in this interval we have (a) Show that a weakly - fib) - fia) convex function is convex f(x) fia) ^ if and only if its graph contains no straight line segments. (Sometimes a weakly convex function is (b) FIG U RE \^ simply called "convex," while convex functions in our sense are called "strictly convex".) (b) Reformulate the theorems of this section for weakly convex functions. 1 1 , Appendix. Convexity 12. 13. and Concavity 229 Find two convex functions / and g such that f(x) = g(x) if and only if x is an integer. Hint: First find an example where g is merely weakly convex, and then modify it, using the result of Problem 9 as a guide. A set A of points in the plane joining any two points in of points (x, y) with y the graph of /. in the > is called convex f(x), so that that if A contains the line segment (Figure 14). For a function /, Af is Af convex is if the set and only Af be the set of points on or above if / let is weakly convex, terminology of the previous problem. Further information on convex sets will (a) Show it be found in reference [18] a convex subset of the plane FIGURE 14 of the Suggested Reading. (b) a non-convex subset of the plane CHAPTER INVERSE FUNCTIONS We now have at our disposal quite powerful methods for investigating functions; what we lack is an adequate supply of functions to which these methods may be applied. We have studied various ways of forming new functions from old addition, multiplication, division, and composition but using these alone, we can produce only the rational functions (even the sine function, although frequently — used for examples, has never been defined). begin to construct new will is one double the usefulness of any other method will practically discover. If we recall that a function is a collection of pairs of numbers, the bright idea of simply reversing we we functions in quite sophisticated ways, but there important method which we In the next few chapters Thus from the pairs. all / = {(1,2), (3,4), (5,9), (13, 8)}, g = {(2,l), (4,3), (9,5), (8, 13)}. we might hit upon the function obtain = While /(l) = 2 and /(3) 4, we have = g(2) and g(4) 1 = 3. Unfortunately, this bright idea does not always work. If / = {(1,2), (3,4), (5,9), (13,4)}, then the collection {(2,1), is not a function at the trouble that can this DEFINITION lies: all, /(3) since = go wrong, and it (4, 3), (9, 5), (4, contains both (4, 3) and /(13), even though it is 13) 3^13. worthwhile giving a (4, 13). This name is It is clear where the only sort of thing to the functions for which does not happen. A function The / is one-one identity function / (read "one-to-one") is if / f(a) obviously one-one, and so is f(b) whenever a ^ the following modifica- tion: ' g (x)= The function f(x) = x2 is x ^3,5 3, x =5 5, x = not one-one, since g(x) 230 x, = x2 , 3. /(— 1) = x>0 b. /(l), but if we define 231 12. Inverse Functions < (and leave g undefined for x = 2x > number and x > 0, for g'(x) 0), then g is one-one, because g This observation 0). is one-one. If n /' is is one-one (since f'(x) condition f(a) / for > nx" ^ / f(b) for a twice (Figure 0, for all ^ x 0). from the graph of / whether / is one-one: the b means that no horizontal line intersects the graph a function that FIGURE not one-one 1 If we reverse all the pairs in (a any case, less / is is (b) (a) some not necessarily one-one function) collection of pairs. It is popular to abstain from is no particular reason we obtain a definition one-one, but there with restrictive conditions l PROOF / ' is which the pair a function if (b, a) and only if is implies that b c. Thus / Conversely, suppose that the pairs (b, f(b)) Since /"' is and (c, is / is the set of all pairs {a, c) = be two pairs /(c); since / in / . Then one-one is this a function. _1 /(c)) , definition one-one. Suppose first that / is one-one. Let (a, b) and (b,a) and (c,a) are in /, so a = f(b) and a = procedure un- this — f obtain, in in /. is / / we do so instead of a and a theorem. to For any function /, the inverse of /, denoted by (a, b) for theorem x all 1). a one-one function DEFINITION a natural is particularly easy to decide It is of — n easily generalized: If odd, one can do better: the function /(*) is increasing (since x>0, f{x)=x\ then is is = a function. (c, If fib) f(b)), so (/(/?), b) a function this implies that b — c. Thus / = /(c), then and (f(b), is one-one. / contains c) are in | f~ l . 232 Derivatives and Integrals The graphs of / and / _1 are so closely related that it is possible to use the graph of / to visualize the graph of / Since the graph of f~ consists of all _1 pairs (a, b) with (b, a) in the graph of /, one obtains the graph of / from the graph of / by interchanging the horizontal and vertical axes. If / has the graph l . shown in Figure 2(a), FIGURE 3" o 3 1-1 CO° 3 3" O 3 "2. oq 1-1 fl> CTQ 3" r-t H o "-) M l"B "rj o f L£ 1 2(a) CL R o re o o 1-1 3 3? re 3" cr O N* o (T) •* 3" C o" 3- n' 3" rn 3 >-! en pj_ re" 3- 5; !-! 1 cr 3 3* CO o 3 CfQ Cfi" 3 3 5" CO O po o o CO O *-! 3~ o c 3 P xs 3" o B. 3 >-i >-*• §" >-s CL O CO re N crq *Tl PJ 3* o 3 o 3" CO re 3- 3* r& "5' O 3* CD O 3 CD So R CO 3 0) o o 3 3 re O >-< This procedure H 3* PJ Qrq^ CfQ po is fortunate that there awkward with books and impossible with blackboards, '. The is another way of constructing the graph of / so it is points 233 12. Inverse Functions (a,b) which and (b,a) are reflections of each other through the graph of I(x) = x, _1 we merely is called the diagonal (Figure 4). To obtain the graph of / reflect the graph of / through this line (Figure 5). (a,b) 'diagonal (c,d) • (b,a) (d, c) FIGURE FIGURE 4 5 Reflecting through the diagonal twice will clearly leave us right back started; this means _1 _1 Theorem conjunction with is that (/ ) 1, = /, which is also clear from the where we equation has a significant consequence: this a one-one function, then the function / is also one-one In definition. / if _1 _1 (since (/ is ) a function). There are a few other simple manipulations with should be aware. Since b Thus f~\b) is is is in / means f{a) the (unique) x 3 then f~(b) , = (a, b) precisely the inverse functions of which when same (b, a) number a such that f(a) — number a such that a' = the unique / in = /~ a as is you follows that (b). b; for b, , it example, and this if f(x) number = by is, definition, \fb. The fact that f~ (x) is the number much more compact form: l (x)) /(/ = x, y such that f(y) for all x in the = x can be restated in a domain of f -l Moreover, fthis follows [ (f(x)) = for x, all x in the domain of /; from the previous equation upon replacing f by f . These two important equations can be written fof~ = of = l r l (except that the right side will have a bigger not all of R). I. i domain if the domain of / or f~ ] is 234 Derivatives and Integrals Since it is many standard functions we be quite important that will be defined as the inverses of other functions, able to tell which functions are one-one. already hinted which class of functions are most easily dealt with decreasing functions are obviously one-one. Moreover, is and also increasing, to you). In addition, One example ing. increasing if f~ and only if [ —increasing and increasing, then is f~ decreasing (the proof is is —/ one-one function certainly not true that every It is 6 / is decreasing, then is / have is l left decreasing, a very useful remember. fact to FIGURE / if if We is has already been mentioned, and r(*) = x, x #3,5 3, x =5 5, x = either increasing or decreasis now graphed in Figure 6: 3. Figure 7 shows that there are even continuous one-one functions which are neither increasing nor decreasing. But if you try drawing a few pictures you suspect that every one-one continuous function defined on an interval will is soon either increasing or decreasing. THEOREM FIGURE 2 If / is continuous and one-one on an interval, then / either increasing or is decreasing on that interval. 7 proof The proof proceeds (1) If a < b < in three easy steps: c are three points in the interval, then either (i) or (ii) < f{b) < f(a) > f(b) > f(a) fie) f(c). Suppose, for example, that fia) < f(c). If we had fib) < fia) (Figure 8), then the Intermediate Value Theorem applied to the interval [b, c] would give an x with FIGURE b < x < c Similarly, fib) > Naturally, the (2) If For a — and fix) < b < c fia), contradicting the fact that / is fie) would lead to a contradiction, so fia) same < d we can apply 8 sort of argument works for the case one-one on < fib) < fia) > /(c). are four points in the interval, then cither (i) fia) < fib) < /(c) < fid) or (ii) fia) > fib) > /(c) > fid). (1) to a < b < c and then to b < c < d. [a.c]. fie). 12. Inverse Functions 235 and suppose that f(a) < f(b). Then / is increasing: For if c and d are any two points, we can apply (2) to the collection {a, b, c, d) (after arranging them in increasing order). | Take any a < b (3) Henceforth we in the interval, shall be concerned almost exclusively with continuous increasing on an interval. If / is such a function, it is possible to say quite precisely what the domain of f~ will be like. Suppose first that / is a continuous increasing function on the closed interval [a, b] Then, by the Intermediate Value Theorem, / takes on every value between f(a) and f{b). Therefore, the domain of f~ is the closed interval [f(a), f(b)]. Similarly, if / is continuous and decreasing on [a, b], then the domain of f~ is or decreasing functions which are defined [ . FIGURE l 9 [ [/(*), /(a)]. If / is a continuous increasing function on an open interval (a, b) the analysis To begin with, let us choose some point c in (a,b). > f(c) are taken on by /. One possibility is that / takes on arbitrarily large values (Figure 9). In this case / takes on all values > f(c), by the Intermediate Value Theorem. If, on the other hand, / does not take on arbitrarily large values, then A = { f (x) c < x < b} is bounded above, so A has a least upper bound a (Figure 10). Now suppose y is any number with f(c) < y < a. Then / takes on some value f(x) > y (because a is the least upper bound of A). By the Intermediate Value Theorem, / actually takes on the value y. Notice that / cannot take on the value a itself; for if a = f(x) for a < x < b and we choose / with x < t < b, then f(t) > a, which is impossible. Precisely the same arguments work for values less than f{c): either / takes on all values less than f(c) or there is a number fi < f(c) such that / takes on all becomes a We bit will first more difficult. decide which values : FIGURE 10 and /(c), but not /3 itself. This entire argument can be repeated if / is decreasing, and if the domain of / is R or (a, oo) or (— oo, a). Summarizing: if / is a continuous increasing, or decreasing, function whose domain is an interval having one of the forms (a,b), (—oo,b), (a, oo), or R, then the domain of f~ is also an interval which has one of these four forms, and we can easily fit the remaining types of intervals, (a, b], [a, b], (—oo, b], and [a, oo), into this discussion also. Now that we have completed this preliminary analysis of continuous one-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. values between ft l theorem 3 PROOF If / is continuous and one-one on an interval, then We know by Theorem 2 that / well assume that / is increasing, is is open, since We it is interval is also continuous. either increasing or decreasing. since applying the usual trick of considering on any / we can then — /. We We might as take care of the other case might as well assume our by interval easy to see that a continuous increasing or decreasing function can be extended to one on a larger open must show that \lmfx—*b l (x) = f-\b) interval. 236 Derivatives and Integrals for each b in the domain of / domain of in the if — f(a) 1 1 see the graph of / suggests the ): way of finding 8 < f s (remember < a + s. (x) by looking sideways you that since a — £ - s) < f(a) < f(a + < < a + s, a e); be the smaller of f(a + e) — f(a) and / (a) — f (a — e) Figure 11 contains the entire proof that this 8 works, and what follows is simply a verbal account of let 8 . the information contained in this picture. b Our — f(a)-S choice of 8 ensures that f(a a +e Consequently, -s)< 1 1 - f(a) 8 + and f(a) < f(a + 8 s). if -8 <x < f{a) FIGURE — then a follows that we f(a)+8 = + 8, < x < f(a) 8 f(a f(a) Such a number b is of the form f(a) for some a such that, for all x, 0, we want to find a 8 > . > /. For any e Figure it ' + f(a) 8. then f(a Since / is increasing, / is — e) < x < f(a + also increasing, e). and we obtain r f-\f(a-s)) < f-\x) < l (f(a + s)), i.e., — a which is precisely Having < f~ s what we want. < (x) | successfully investigated continuity of differentiability. of f{a)). If this entire picture / -1 and is the tangent line L' / it only reasonable to tackle with a tangent it This formula can equally well be written each b in the domain of :•- line L through shows the graph is the reciprocal appears that = -L-. f rectly, for it is through (f(a),a). The slope of L' (/-!)'(/(*,)) 12 , reflected through the diagonal, of the slope of L. In other words, FIGl Kl. / Again, a picture indicates just what result ought to be true. Fig- ure 12 shows the graph of a one-one function (a, + s, a / l\'i in a (a) way which expresses (f~)'(b) di- . ^A f'(f-Hb)) Unlike the argument for continuity, involved when formulated analytically. this pictorial There is "proof" becomes somewhat another approach which might 237 12. Inverse Functions be Since tried. we know that f{f~\x))=x, tempting to prove the desired formula by applying the Chain Rule: it is l f'(f- (x))-(f- y(x)= l l, so I i\', (/"')'(*) f'(f-Hx))' Unfortunately this is f~ is differentiable, since the Chain Rule already known to be differentiable. But this argu- not a proof that cannot be applied unless / _1 is x ment does show what (/)'(*) will have to be if f~ is differentiable, and also be used to obtain some important preliminary information. l fix) = x 3 THEOREM 4 If / is / then PROOF We a continuous one-one function defined on an interval and is can /'(/(#)) = 0, not differentiable at a. have = f(f-\x)) If it / were differentiable at a, the x. Chain Rule would imply 1 f'(f- (a))'(f- )'(a)= that 1 (a) 1, hence 0.(/" 1 ) which absurd. is A simple /'(0) = = to /"'(O), the function Having decided where an 1 f~ following argument uses THEOREM 5 is / be / differentiable at continuity entiable at b, f~ (b), is in all other cases the derivative of / , in two different ways. on an interval, l and 1 )'(b) = — f'(f-Hb))' Let b = f(a). Then f-\b + h)- f~-l (b) l lim fc->0 -1 = lim is . Since (Figure 13). we are now given by the Notice that the which we have already proved. with derivative f'(f~ (b)) (f- proof =x the function f(x) not differentiable at a continuous one-one function defined Let is l inverse function cannot be differentiable, formula which we have already "derived" 3 1, which Theorem 4 applies ready for the rigorous proof that FIGURE (fl)= | example and , f~ l (b + h)-a ^ 0. and suppose Then /~' is that differ- 238 Derivatives and Integrals Now every number b + domain of / h in the + h= b for a can be written ' in the form + k) f(a unique k (we should really write k(h), but we will stick with k for simplicity). Then lim f- [ + h)-a (b = f~\f(a + k))-a f(a + k) — b lim /1-+0 = k lim We are clearly on the right track! It is + f(a ft-»0 k) - f(a) not hard to get an explicit expression for k; since b +h = + k) f(a we have f-\b + h)=a + k or k=f-\b + h)-f-\b). Now by Theorem approaches as h 3, the function approaches 0. this is continuous This means that k at b. Since =/(«) = / lim k^O l / k (/ (ft)) ^ 0, implies that 1 l\'< (f-'Yib) f'(f-Hb)) is The work we have done on inverse functions an immediate dividend. For n odd, let f„(x) = x" will for all be amply repaid later, x; for n even, let fn(x)=x Then /„ is n x>0. , a continuous one-one function, whose inverse function ^= gn (x)= v -v l/ ". is but here 239 12. Inverse Functions By Theorem 5 we ^ have, for x 0, ' 8n (X) " fn'Vn-H*)) 1 n{fn - { {x)) n - { I 1 (A /")"1 77 1 1 -(1/n) X 77 1 =— • X 77 Thus, f(x) if = fix) = ax number: Let a a . x", It is = and a now an integer or the reciprocal of a natural number, then is easy to check that this formula m/n, where m an is integer, fix) =,-"'/" then, by the = and n is is true if a is any rational a natural number; if (x /")"\ ] Chain Rule, f(x) = 1 m(x i/ ") 1 .(!/«)-! n 777 .[(w/n)-(l/n)] + [(l/«)-l] _ ^ x (m/n)-l 77 Although we now have a formula the treatment of the function f{x) for later — at the moment we do Actually, inverse functions will for /'(x) = when f{x) — x a and x" for irrational a not even know the meaning of a be involved crucially irrational a. Indeed, in the next PROBLEMS Find / (i) /(jc)=jc 3 for each of the following /. + 1. (ii) f(x) = (x-\)\ m /(*) = IV fix) = x, x rational —.v, x -x 2 l-x 3 x, (v) fix) = a i+ a (vi) /(*) i \ , =*+[*]. irrational. x > x<0. , a is rational, have to be saved symbol in the definition like a of x few chapters several important functions defined in terms of their inverse functions. 1. will x^a\, a„ x =a /=1, x = t an , . , n : a will . for be ' 240 Derivatives and Integrals /(0.aifl2«3 (vii) = /(*) (viii) • = ) • • 0.fl2 fl l«3 x 1 <x < • • • • (Decimal representation is being used. 1. 1 2. / / / / (i) (ii) (iii) (iv) 3. when / Describe the graph of is increasing and always positive. is increasing and always negative. is decreasing and always positive. is decreasing and always negative. Prove that / if increasing, then so is / is , and similarly for decreasing functions. If / and g are increasing, Prove that (a) o (/ g)~ Find g~ (b) Show l l 7. ' 8. 10. f(x) (ii) f(x) (iii) f(x) Suppose *12. 13. g? l in l . l in 1 —b ax = -\- - ; +d one-one is if and only if ad — be ^ 0, and find intervals [a, b] will the following functions = x 3 -3x 2 be one-one? . = x + x. = (\+x 2 r 5 that = f~ x x + X1 + / is l . 1 1 differentiable with derivative f'(x) 2 satisfies g"(x) = \g(x) = (1 +A" 3 ) -1 ^ 2 . Show . has a derivative which is Suppose that / is a one-one function and that / nowhere 0. Prove that / is differentiable. Hint: There is a one-step proof. As a follow up that 11. o in this case. that g 9. • / and g are one-one, then fog is also one-one. Find terms of / and g~K Hint: The answer is not f~ o g~ terms of f~ if g(x) = + f(x). that f{x) On which (i) f + g? Or f g? Or / if ex f is / is to Problem 10-17, what additional condition on g Find a formula for (/ Prove that if f'(f~ l -1 )"(;c). {x)) ^ and f (k) {f~\x)) The Schwarzian derivative 2)/ was defined (a) Prove that the (b) will insure differentiable? if 2)/(x) exists for domain of / . Find a formula for 2)/~ (a). all x, then in exists, then (/ _1 ) w (jc) exists. Problem 10-19. 2)/ _1 (x) also exists for all v in 241 12. Inverse Functions *14. Prove that there (a) = x The do Show that / way to do this is to Hint: easiest =/ x to set this is such that [/(*)] 5 + /CO + can be expressed as an inverse / find And . the easiest way _1 (v). (b) Find /' in terms of /, using (c) Find /' The / a differentiable function for all x. function. to is an appropriate theorem of this chapter. another way, by simply differentiating the equation defining /. in function in Problem 14 defined implicitly by the equation is quite special, however. As often said to be is equation y +y+x = 0. The situation for this the next problem shows, an equation does not usually define a function implicitly on the whole line, and some regions more than one function may be defined in implicitly. 15. (a) What are the two differentiable functions on (— 1, for 1) x all by the equation (— 1, in jc +y — / which are defined implicitly which x +[f(x)]~ 1, i.e., satisfy = 1 Notice that there are no solutions defined outside 1)? [-1,1]. (b) *(c) Which Which will / functions satisfy x2 + differentiable functions help to first draw [f(x)] / 2 = satisfy -1? [/CO] — 3/(jc) =x the graph of the function g(x) In general, determining on what intervals a differentiable function plicitly by a particular equation may be context of advanced calculus. If however, then a formula for differentiating we and a delicate affair, assume that / It defined im- is best discussed in the is such a differentiable solution, is can be derived, exactly as /'(jc) = x? Hint: — 3x. both sides of the equation defining / in process (a Problem known 14(c), by as "implicit differentiation"): 16. (a) Apply this answer method to the equation [/(.v)] will involve f(x); this is 2 + x2 = Notice that your 1. only to be expected, since there (b) Problem 17. (c) Apply (a) Use this same method to [f(x)]^ more 1. / found in for the functions f — 3f(x)=x. implicit differentiation to find f'(x) One = is 15(a). defined implicitly by the equation x (b) 2 than one function defined implicitly by the equation y + x But check that your answer works for both of the functions 2 of these functions / satisfies and f"(x) + y =7. /(-l) = 2. Find /'(-l) and /"(-l) for this /. 18. The collection of all forms a certain curve this 19. curve at the is in the plane. point (— 1, Leibnizian notation cause y points (x, y) such that 3x 3 is + 4x 2 y — xy 2 + 2y 3 = Find the equation of the tangent 4 line to 1). particularly convenient for implicit differentiation. Be- so consistendy used as and y which defines / an abbreviation for f(x), the equation in v implicitly will automatically stand for the equation , 242 Derivatives and Integrals which / is supposed to How satisfy. would the following computation be written in our notation? 4 + y dy 4y + xy = 3 y + 3y^ dx dx dx -y dy dx 20. As long 1 4v 3 + 3y 2 + ;t as Leibnizian notation has entered the picture, the Leibnizian no- tation for derivatives of inverse functions should be mentioned. If dy /dx l denotes the derivative of /, then the derivative of f~ is denoted by dx/dy. Write out Theorem 5 in this notation. The resulting equation will show you why another reason which point (/ also explain at What notation. Leibnizian notation has such a large following. )' is It will be calculated when using the dx/dy to the significance of the following computation? is y dx x — x 'n dx i/» ' , dy 1 dx dx ny n-\ dy 21. Suppose if whose Two 22. and = f~ G'(x) fact: / that derivative [ (x). we know derivative different Suppose h is is that a differentiable one-one function with a nowhere zero f = — xf Let G(x) F'. (Disregarding details, (x) problem this is / . But how in (x)). Prove that us a very interesting tells /, then we also know one could anyone ever guess the function G? a function whose derivative ways are outlined — F(f~ is Problems 14-14 and 19-16.) a function such that ti{x) = sin (sin(x + 1)) and h(0) = 3. Find l 23. (i) (h- YO). (ii) (£-')'(3), (a) (b) where 0(x) = h(x + 1). Prove that an increasing and a decreasing function intersect at most once. Find two continuous increasing functions / and g such that f(x) when x is an integer. = g(x) precisely (c) Find a continuous increasing function function g, defined *24. (a) If / least is on R, which do not one x such that f(x) x. a continuous decreasing intersect at R all. and /' = / prove that there is at _1 mean (What does the condition / = / a continuous function on = / and , geometrically?) (b) Give several examples of continuous / such that f = f~ and f(x) — x one x. Hint: Try decreasing /, and remember the geometric for exactly interpretation. One possibility is f(x) = —x. 243 12. Inverse Functions (c) Prove that if / then an increasing function such that / = / x. Hint: Although the geometric interpretation will is , fix) = x for all be immediately convincing, the simplest proof (about 2 out the *25. 26. possibilities f(x) < x and f(x) > is to rule x. Which functions have the property that the graph tion when reflected through the graph of — / (the A lines) is still the graph of a func- "antidiagonal")? nondecreasing if fix) < fiy) whenever x < y. (To be more precise we should stipulate that the domain of / be an interval.) A nonincreasing function is defined similarly. Caution: Some writers use function / is ""increasing" instead of "nondecreasing," and our "strictly increasing" for "increasing." (a) Prove that (b) on some interval. (Beware of unintentional puns: "not increasing" the same as "nonincreasing.") Prove that if / is differentiable and nondecreasing, then f'(x) > all *27. Prove that (a) Suppose there (b) / is nondecreasing, but not increasing, then / is constant not is for x. (c) all if is if f'(x) that f(x) > > for all x, then for all x, and / is nondecreasing. that / is decreasing. a continuous decreasing function g such that < g(x) Prove that < f(x) for x. Show that we can even arrange that g will satisfy lim g(x)/f(x) = 0. 244 Derivatives and Integrals PARAMETRIC REPRESENTATION OF CURVES APPENDIX. The (Figure 1). —a perfectly nice looking curve we may In other words, Of course, we (x, fix)). this trick doesn't consisting of all work not be able to describe simplest most cases. one is 1 the set of set of all points 2 points (x ,x). But all 1, or an and ellipse, can't it circle, be used in Figure 2. way of describing curves in the plane in general harks back to the physical conception of a curve as the path of a particle FIGURE as the set of all points it won't allow us to describe the It +y = points (x. v) with x to describe a curve like the The in noticed a need not be the graph of a function might be able to describe the curve as the (f(x),x); for example, the curve in Figure even we material in this chapter serves to emphasize something that long time ago moving in the plane. At 1 each time /, the particle at is a certain point, which has two coordinates; to indicate dependence of these coordinates on the time t, we can call them u(t) and v{t). Thus, we end up with two functions. Conversely, given two functions u and v, we can consider the curve consisting of all points (u(t), v(t)). This curve is said to be represented parametrically by u and v, and the pair of functions u, v is called a the parametric representation of the curve. u and The curve represented parametrically by x v thus consists of all pairs (x, y) with — described briefly as "the curve x function / can u{t), y — = u{t) and y = v(t). often It is Notice that the graph of a v(t)." always be described parametrically, as the curve x — t, y = fit). Instead of considering a curve in the plane as defined by two functions, can obtain a conceptually simpler picture if we broaden our function somewhat. Instead of considering a rule which associates a another number, FIGURE 2 i.e., c(t). numbers Of course, With number this notion, a curve is t, a. point in the plane, just a function which we can from some interval of to the plane. these two different descriptions of a curve are essentially the same: A pair of (ordinary) numbers number with consider a "function c from real numbers to the plane," a rule c that associates, to each denote by real we can we original definition of and functions u to the plane v determines a single function c from the real by the rule c(t) = (u(t),v(t)), and, conversely, given a function c from the real numbers to the plane, each c(t) is a point in the plane, so so that we have unique In Appendix 1 to it is a pair of numbers, which functions u Chapter 4, and we used we can a "vector-valued function." c(t) A = — (c\ (f ), cjit)), (u(t), v(t)) to The but in I ( , I K I . * the term "vector" to describe a point in may also Appendix be called we'll continue to use notation like minimize the use of subscripts. simple example of a vector-valued function that which goes round and round the v(t), conventions of that Appendix would lead us to this e(/) I and v satisfying this equation. the plane. In conformity with this usage, a curve in the plane write c(t) call u(t) = (cos/, sin is /), unit circle (Figure 3). quite useful is 245 12, Appendix. Parametric Representation of Curves / and For two (ordinary) functions by the f(x) + g(x), = f(x) g(x). we have defined a way g)(x) • f + g and / g functions • we can imitate the first of these defwe define the vector-valued function of adding vectors, initions for vector-valued functions c +d new defined + g)(x) = (f c we rules (f Since g, and d: by + d)(t)=c(t) + d(t), (c + where the on the right-hand to saying that side now is the sum of vectors. This simply amounts if c(t) d(t) = = (u(t),v(t)), (w(t),z(t)), then (c + d){t) = we have Recall that extend this to also defined a Then we can = (w(t), z(t)) (u(t) new w(t), v(t) to consider so that for each c, define a + number a and a v for we want vector-valued functions, and a vector-valued function vector c(t). + (u(t), v(t)) we have / + z(t)). a vector To v. an ordinary functio n a a vector-valued function number a c by a(t) and a (a -c)(t) =a(t)-c(t), where the on the right-hand simply amounts to saying that • (a c)(t) = a(t) Notice that the curve a (a is e)(t) side of a in the product of a (u(t), v(t)) = number and u(t), a(t) (a(t) a vector. This v(t)). e, e)(t) already quite general (Figure the point (a is = 4). (a(t)cost, a(t)smt), In the notation of e){t) has polar coordinates a(t) and t , Appendix so that (a 3 to e)(t) is Chapter 4, the "graph polar coordinates." Even more generally, given any vector-valued function functions r and 9 by c, we can define new c(t)=r(t)-e(9(t)). where r{t) is just the distance from the origin the angle of c(t) (as usual, the function 9 FIGURE 4 be careful when using to We we this isn't to c(t), and 9(t) is some choice of defined unambiguously, so one has way of writing an arbitrary curve c). aren't in a position to extend (2) to vector-valued functions in general, since haven't defined the product of two vectors. Appendix 1 to Chapter 4 define two real-valued However, Problems 2 and 4 of products v • w and det(u, w). It 246 Derivatives and Integrals should be given vector-valued functions c and d, clear, how we would define two ordinary (real-valued) functions c Beyond imitating simple more interesting problems, limc(f) (*) Rules = • and d det(c, d). we can consider we can define arithmetic operations on functions, For c(t) like limits. lim(a(f), u(0) = (u(t), v(t)), (\im to De \imv(t)) u(t), . like lim c + lim d, + d = t—>a t—>a lim c t-+a lim a = c follow immediately. lim a(t) lim c t^a t—*a t—t-a Problem 10 shows how an equivalent definition that to give imitates the basic definition of limits directly. Limits lead us of course to derivatives. For c(t) we can define c' (u(t),v(t)) by the straightforward definition c'(a) We = = (u(a), v'(a)). could also try to imitate the basic definition: c(a = c'(a) + h) — c{a) lim // where the fraction on the right-hand -• side [c{a understood to is + h) - mean e{a)}. n As a matter of fact, c(a + h) c(a these + h) — two c(a) lim = definitions are equivalent, because u{a + — h) u(a) v(a u{a+h)-u{a) ,. lim Figure 5 shows c(a + h) we showed in Appendix moved over so that it starts as II < , I !< E 5 (loser and h^Q h (*) h of limits to c(a), as well as the Chapter at c(a). 4, this As h —> 0, this closer to the tangent of our curve, so arrow from c(a) arrow it is c(a it as the set of starts at c(a). all In other words, we +s — to c(a seems reasonable when + h); c(a), except to move to define the c'(a) is moved define the tangent line of c at c(a) points c(a) + h) arrow would appear tangent line of c at c(a) to be the straight line along c'(a), over so that v(a) (u'(a), v'(a)). and 1 + h) - v(a by our definition c(a) v(a) lim , h^O = + h) — lim c\a)\ 12, Appendix. Parametric Representation of Curves = = 247 we get c(a) + c' (a), etc. (Note, however, that this definition does not make sense when c'(a) = (0, 0).) Problem shows that this definition agrees with the old one when our curve c is defined by for s we get the point c(a) itself, for s 1 1 c(t) we simply have Once again, various so that = (t,f(t)), the graph of /. old formulas have analogues. For example, + d)'(a) = (a c)'(a) = (c • + d'(a), a' (a) c(a) + a(a) r c (a) c'(a), equations involving functions, or, as + d)' = c' + d', {ot c) = a c + a. (c • • c . These formulas can be derived immediately from the definition in terms of the component functions. They can also be derived from the definition as a limit, by imitating previous proofs; for the second, we would of course use the standard trick of writing a(a + h)c(a + h) — a(a)c(a) ct(a We = + h) + h) - [c{a c(a)] + [a(a + h) - a(a)] c(a). can also consider the function d(t) where p is now an passes through the = c(p(t)) = (c o p)(t), ordinary function, from numbers to numbers. same points The new curve d p corresponds as c, except at different times; thus to a "reparameterization" of c. For c d we — = (u, v), (u o p, v o p), obtain d'{a) = (O o p)'(a), {v o p)'{a)) = (p'(a)u'(p(a)), p'{a)v\p{a))) = p'{a) (u'{p(a)), v\p(a))) = p'(a) c'(p(a)), • or simply d' Notice that if p (a) time a, then d'(a) of c'(a). = = = a, so p'(a) p' that (c o p). d and c'(a), so that the tangent vector d'(a) This means that the tangent line to the c actually pass through the reparameterized curve d line at to c at c(a) d(a) = c(a). is the same point at is just a multiple same as the tangent The one exception occurs 248 Derivatives and Integrals when = p'(a) tangent line for d 0, since the may be tangent line for c line defined at = t Finally, since 0, even though we can it's then undefined, even though the = 3 ) won't have a tangent c. define real-valued functions det(c, d)(t) = c(t)-d(t), = det(c(f), d(t)), have formulas for the derivatives of these to c(f merely a reparameterization of (C'd)(t) we ought is defined. For example, d(t) new functions. As you might guess, the proper formulas are (c • = = d)'(a) [det(c, d)]'(a) + c'(a) d(a), det(c\ d){a) + det(c\ d')(a). c(a) d'{a) • • These can be derived by straightforward calculations from the definitions in terms of the component functions. But it is more elegant to imitate the proof of the ordinary product rule, using the simple formulas in Problems 2 and 4 of Appendix 1 to Chapter 4, and, of course, the "standard trick" referred to above. PROBLEMS 1. For a function /, the "point-slope form" (Problem 4-6) of the tangent (a) line at (a, — f{a)) can be written as y f(a) = (x — a)f'(a), so that the tangent line consists of all points of the form (x,f(a) Conclude + (x-a)f'{a)). that the tangent line consists of all points of the form (a+s,f(a)+sf(a)). (b) If c (a, = the curve c(t) is (f, new f{a)) [using our f(t)), conclude that the tangent definition] is the same line of c at as the tangent line of / at (a, f(a)). 2. = Let c(t) Show ter 9. = h(x) (f(t), \x\, t ), 3. c' never equal to Suppose and (a) (b) that that u is .v = this u{t), If w differentiable is u{t) = (0, 0). reason, y = v(t) one-one on some that — the function on shown in Figure 21 of Chap- graph of the non-diffcrentiable function In other words, a reparameterization can we are usually only interested in curves c (0, 0). Show x is that c lies along the but that c'(0) "hide" a corner. For with where / a parametric representation of a curve, interval. this interval the on is curve this interval lies and we have v'(t) fix) u'{t) along the graph of //'(/) ^ 0, show / = v ou . that at the point 249 12, Appendix. Parametric Representation of Curves In Leibnizian notation this often written suggestively as is dy dy_ dt dx dx It (c) We also have - u'{t)v"{t) fix) = v'{t)u"{t) (u'(t)Y 4. 5. Consider a function / defined implicitly by the equation x 2/3 Compute f'{x) in two ways: (i) By (ii) By considering Let x = u(t), = y that if the point FIGURE 6 v'(t) the parametric representation x v(t) Q= P = let (xo, yo) from = cos Q be a point on the curve (u(t), v(t)) are not both 0, then the line from 1. y t, = sin t. (Figure 6). closest to (xo, yo)> is P in the plane. to Q is The same Prove an d u'(t) perpendicular to the result holds if Q is (xq, yo). We've seen that the "graph of / (/•e)(0 in other ^ = be the parametric representation of a curve, with tangent line of the curve at furthest 2 y implicit differentiation. u and v differentiable, and and + words, the graph of / = in polar coordinates" is the curve (/(') cos *,/(*) sin 0; in polar coordinates is the curve with the para- metric representation x (a) = f(0)cosG, y Show that for the graph of / line at the point with polar coordinates (f(0),9) -/(6>) sin Show that if the origin f(0) sinO. in polar coordinates the slope /(#)cos(9 (b) = f(0) = (9 and / of the tangent is + / (6>)sin<9 + /'(#) cos 6>' / is differentiable at 9, then the line making an angle of 9 with the through positive horizontal axis is a / in polar coordinates. Use this result to add some details to the graph of the Archimedean spiral in Appendix 3 of Chapter 4, and to the graphs in Problems 3 and 10 of that Appendix tangent line of the graph of as well. (c) Suppose that the point with polar coordinates (f(9), 9) the origin O than any other point on the graph of /. say about the tangent line to the graph at this point? Problem 5. is further from What can you Compare with 250 Derivatives and Integrals Suppose that the tangent (d) line to the graph of / at the point with po- (f(0),0) makes an angle of a with the horizontal axis is the angle between the tangent line and the that a — lar coordinates (Figure 7), ray from O so to the point. Show tan (or - that 0) f(0) = f'(0) 7. In Problem 8 of (a) r = 1 — sin 6 is Appendix 3 to also described Find the slope of the tangent (ii) FIGURE Check be 7 + y 2 + y) = x2 +y . a point on the cardioid in two ways: that at the origin the tangent lines are vertical, as they appear to in Figure 8. FIGURE The line at (x By implicit differentiation. By using the previous problem. (i) (b) Chapter 4 we found that the cardioid by the equation 8 next problem uses the material from Chapter 15, in particular, radian measure, and the inverse trigonometric functions and their properties. 8. A cycloid is defined as the path traced out by a point on the rim of a rolling wheel of radius You can a. see a beautiful cycloid by pasting a reflector the edge of a bicycle wheel and having a friend ride slowly in front on of the headlights of your car at night. Lacking a car, bicycle, or trusting friend, you can settle instead for Figure 9. FI GURE 9 12, Appendix. Parametric Representation of Curves 251 Let u{t) and v(t) be the coordinates of the point on the rim after the (a) wheel has rotated through an angle of arc of the wheel rim from P wheel is rolling, at is to Q in / This means that the (radians). Figure 10 has length at. Since the also the distance from O to Q. Show that we have the parametric representation of the cycloid — — u{t) v(t) Figure 1 1 shows the curves we obtain center of the wheel FIGURE 10 is (a) is FIGURE In Figure 9 need to is (b) from the point to the greater than the radius. not the graph of a function; the wheel is at certain times moving forwards! 1 we drew check that that the distance moving backwards, even though Compute (b) if than the radius or less In the latter case, the curve the point — sin t) a(\ — cos/). a{t the cycloid as the graph of a function, but this is we really the case: and conclude that u is increasing. Problem 3 then shows the cycloid is the graph of / = v o u~\ and allows us to compute u'{t) fit). Show (c) It isn't (d) that the tangent lines of the cycloid at the "vertices" are vertical. possible to get an explicit formula for /, but Show that u{t) Hint: we can come first = a arccos solve for / in a - v(t) ±y/[2a-v(t)]v(t). terms of v(t). close. 252 Derivatives and Integrals The (e) half of the first g(y) 9. = first arch of the cycloid — a v aarccos yj(2a is the graph of g - y)y. v give a parametric representation of a curve Q = Geometrically, it point on the curve the tangent line is to {u(b), v(b)). Prove to Q. Hint: this analytically. assume that we don't have u'(x) Problem 2). 10. The = from P seems clear (Figure parallel to the line This problem interpretation for one of the theorems in Chapter 12 , where Let u and v be continuous on [a,b] and differentiable on (a,b); then u and FIGURE l v'(x) — 1 1 You . (u(a),v(a)) 12) that at some segment from P will give any x for = a geometric will also in (a, b) following definition of a limit for a vector-valued function is need to (compare the direct analogue of the definition for ordinary functions: limc(0 = means / that for every e > there some is 8 > such that, for t->a < all t, if Here / = || || \t is (/i,/2), - a\ < 8, then < -l\\ e. Problem 2 of Appendix the norm, defined in 1 to Chapter 4. If then ||c(/)-/|| (a) \\c(t) Conclude = HO-/il 2 + MO-/2l 2 - that \u(t)-h\ < and show 2 that and \\c(t)-l\\ if lim c{t) = I \v(t) - l2 \ < ||c(0-/||, according to the above definition, then we t—>a also have lim u{t) so that lim c(t) functions, (b) = I = l\ lim v(t) according to our definition on page 246. show that if lim Conversely, and c(t) = I h, (*) in terms of component according to the definition in terms of component functions, then also lim c(t) t—*a definition. — = I according to the above CHAPTER INTEGRALS The derivative does not display second main concept of Part digression — strength until allied with the "integral," the its full At III. first this completed, integrals Although ultimately to Chapter difficulties 14, be defined malizes a simple, intuitive concept 1 to once The given. reflection — that of area. an ever. complicated way, the integral By now intuitive should it 1, which number as area the irrational Even if we many iz , but unit square In this chapter (Figure 1) we will — those which x in [a, b]. It is as circle all is in this circle, since fig- seldom hopelessly inadequate of radius clear is number of squares, with as the this definition not at plane what "jt 1 supposedly has squares" means. it 1, it is hard does not seem possible which can be arranged to form a circle. bounded by the horizontal axis, the vertical lines and the graph of a function / such that f(x) > are convenient to indicate these regions include rectangles geometric come only try to define the area of some very special regions through (a,0) and (b,0), all it is fits to divide the unit square into pieces for But consider a circle of radius l/y/rr, which supposedly has area what way a to say in 2 sometimes defined is in the region. any but the simplest regions. For example, a for FIGURE fit for- concept can present great shows that an acceptable definition of area area of a region sides of length study of preliminary work has been In elementary geometry, formulas are derived for the areas of little The no exception. certainly is this more powerful than in a quite to learn that the definition of — "area" ures, but a be a complete be an invaluable tool for creating new functions, and the will derivative will reappear in FIGURE may seem chapter derivatives do not appear even once! in this integrals does require a long preparation, but no surprise topic and this region by /?(/, a, triangles, as well as many b). Notice that other important figures. The number which we will eventually assign as the area of R(f,a,b) will be / on [a b] Actually, the integral will be defined even for for all x in [a, b]. If / is functions / which do not satisfy the condition f(x) > called the integral of . , the function graphed in Figure 2, the integral will represent the difference of the area of the lightly shaded region and the area of the heavily shaded region (the "algebraic area" of R(f,a, The [a, b] b)). idea behind the prospective definition by means of numbers 3 numbering of equal the 253 = [fe.fc] [/l.fe] to, t\, ti, a FIGURE indicated in Figure 3. The interval has been divided into four subintervals [fO.'l] (the is t?,, to < t\ t\ with < tj < ?3 the subscripts begins with number of subintervals). [*3,*4] <U =b so that the largest subscript will 254 Derivatives and Integrals On the maximum interval first / be w, and of / the function [to, t\] = s maximum the let m\(t\ - to) [f,-_i, tf\ let + niT,(tT, - t\) value m\ and the minimum the value The sum value be M\. + ni2(t2 - minimum has the value M\; similarly, on the zth interval + m\(t\ - tn) £3) represents the total area of rectangles lying inside the region R(f, a, b), while the sum = S M x (t\ - + M 2 (t2 to) M + ~t[) - 3 (t3 + t2 ) M4 (t 4 - t3 ) represents the total area of rectangles containing the region R{f, a, b). A ing principle of our attempt to define the area that A should of R(f, a,b) is The guid- the observation satisfy s < A 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 A b. partition of the interval one of which is we DEFINITION and one of which a, points in a partition can be a shall [a, b] = to < t\ < a finite collection of points in b. is numbered < is tn so that = b; to /„_] < tn always assume that such a numbering has been assigned. Suppose / is bounded on [a, b] m, = and P = inf{/(.v) Mi =sup{/(x) The lower sum of / for P, : : {to t„ /,_i <x < <x r,_i } is a partition of [« b}. Let ti), < \ti). denoted by L(f, P), defined as is n L(f,P) = J^m (fi-ti- it- t i—\ The upper sum of / for P, denoted by U(f,P), is = YlM {t -t - l). U(f,P) i i i defined as 1=1 The lower and upper sums correspond to thr example; they are supposed to represent the sums total areas s and S in the previous 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." 255 13. Integrals Two The requirement of the definition deserve comment. details that / be and M, be defined. Note, also, that it was necessary to define the numbers m, and M, as inf's and sup's, rather than as minima and maxima, since / was not assumed continuous. bounded on One thing [a,b] order that essential in is clear about lower is all the and upper sums: ra, P If any is partition, then L(f,P)<U(f,P), because n n and for each i we have -ti-\) mtiti FIGURE On 4 any MM -ti-i). < the other hand, something less obvious ought to be true: If P\ two partitions of [a b] , , then it and Pi are should be the case that L(f,Pi)<U(f,P2 ), because L(f, P\) should be < area R(f,a,b), and U(f, Pi) should be > area R(f,a,b). This remark proves nothing (since the "area of R(f,a, by has not even been defined yet), but does indicate that it if there be any hope of defining the to is area of R(f,a,b), a proof that L(f, P\) < U(f, P2) should come first. The proof which we are about to give depends upon a lemma which concerns the behavior of lower and upper sums the partition P when more black and the points in grey. the partition FIGURE Q lemma If Q contains P Q The contains both the points in picture indicates that the rectangles drawn for are a better approximation to the region R(f, a,b) than those for the original partition P. 5 points are included in a partition. In Figure 4 contains the points in black, and To be (i.e., if all precise: points of P are also in Q), then L(f,P)<L(f,Q), U(f,P)>U(f,Q). PROOF Consider first the special case (Figure 5) in which Q contains just one than P. P = Q= {to,...,tn ), Uo tk -i,u, tk — ,tn ), where a — to < t\ < • • • < tk-] < u < tk < • • - < tn = b. more point 256 Derivatives and Integrals Let = m" = m' inf{/(.v) t^] < x < : inf {f(x) < x < w : "}- /^}. Then L(/,P)=£m f (f;-f;-i), fe-I £(/, = (?) y^m,-(fj ( To prove /i - Now is less that L(f, P) < L(f, Q) tk _\ : tk < < m\u - -\) < x therefore suffices to it tk } to tk -\) contains and possibly some smaller ones, than or equal ^ + m"(tk -u)+ _i) m,-ft - f,-_i). i=Jt+l the set {f(x) u}, A =1 m k (tk x < + m'(« -r r,-_i) the greatest lower show + m"(tk - all the u). numbers in {f(x) bound of the : so the greatest lower bound of the second; mk that tk -\ < first set thus <tn'. Similarly, mk < m". Therefore, m k (t k This proves, U(f, Q) The is tk = m k (u - _\) tk -\) + m k (t k < m'(u - u) tk _\) + m"(tk - u). P) < L(f, Q). The proof that U(f, P) > you as an easy, but valuable, exercise. in this special case, that L(f, and similar, to is left now be deduced quite easily. The partition Q can be P by adding one point at a time; in other words, there is a sequence general case can obtained from of partitions P such that Pj + \ = P ,P2 ,...,Pa = Q l contains just one L(f, P) = L(f, Pi ) more < L(/, point than Pj. P2 ) < • • - < Then L(f, Pa = ) L(/, Q), and £/(/, P) = U(f, Pi) The theorem we wish THEOREM l Let P\ on and P2 be [<7, /?]. to > U(f, P2 ) > prove is •• • > £/(/, Pa = U(f, ) Q). | a simple consequence of this lemma. partitions of [a,b], and let / be Then U.f\P\)< U(f,P2 ). a function which is bounded . 257 13. Integrals PROOF There in is a partition both P\ and Pi). P which According L(f, It contains both P\ and Pj PO < (let P consist of all points lemma, to the < U(f, P) < U(f, P2 ). L(f, P) | from Theorem follows 1 that any upper sum U(f, P') is an upper bound for sums L(f, P). Consequently, any upper sum U(f, P') is greater the least upper bound of all lower sums: the set of all lower than or equal to sup{L(/, P) for every P' . This, in turn, P : a partition of [a, b]} means that sup{L(/, P)} < U(f, P'). a lower bound for the is set of all upper sums of /. Consequently, sup{L(/,P)} <inf {£/(/, P)}. It is clear that both of these / of numbers are between the lower sum and upper sum for all partitions: L(f, P') < sup{L(/, P)} < U{f, P'), L(f,P')<in£{U(f,P)} <U(f,P'), for all It partitions P' may well happen that = inf {U(f, P}; sup{L(/, P)} in this case, this for all partitions, of R(f,a.b). and On number between the lower sum and upper sum of / number is consequently an ideal candidate for the area the only is this the other hand, if sup{L(/,P)} then every f(x) = number* between < inf {£/(/, />)}, sup{L(/, P)} and inf {£/(/, P)} will satisfy L{f,P')<x<U(f,P') c for all partitions P' show that when such an embarrassment of riches will occur. The many which will soon apboth phenomena are possible. Suppose first that f(x) It is a = t t] t2 tn-\ tn =b pear, FIGURE not at all clear just following two examples, although not as interesting as 6 any partition of [a b] , , = c for * all in [a, b] (Figure 6). then irij — Mi = c, so L(f, P) = J2 c ( fi ~ ti-l) = c(b - a), (=1 U(f,P) = J2c(ti-ti-i)=c(b-a). i=l If P = {to, . . . , /„} is 258 Derivatives and Integrals In this case, lower sums and upper sums are equal, and all sup{L(/, P)} Now = inf {£/(/, P)} / defined by 0, x irrational 1, x rational. | P = {to, . . . ,tn } any is - c(ft consider (Figure 7) the function f If = a). partition, then m, = 0, since there is an irrational number Mj = \, since there is a rational in [/,_], /,], and a = FIGURE 7 ?o t\ t2 tn-\ h = b number in [/,-_ i, ?,]. Therefore, n L(f,P) = J^0-(t -t . )=0, i i 1 /=i /? C/(/,P) = £)l.(f --r _i)=fe-a. i I i=\ Thus, in principle this case it is upon which certainly not true that sup{L(/, P)} the definition of area was On more generally, that peal to the is insufficient any area at all. In fact, so is we can whenever sup{L(/, P)} the region R(f,a,b) it The — any number between the other hand, the region R(f, a, b) weird that we might with justice refuse to assign maintain, inf {(/(/, P)}. be based provides to information to determine a specific area for R(f, a, b) and b — a seems equally good. = # inf {£/(/, />)}, too unreasonable to deserve having an area. word "unreasonable" we suggests, As our ap- are about to cloak our ignorance in terminology. DEFINITION A function / which sup{L( /, P) : P In this case, this is bounded on a partition of [a, b] [a, b]} common number is = is integrable on inf{c/(/, P) : P [a, b] if a partition of called the integral of / on [a, b]}. [a, b] and is denoted by / f. (The symbol J is called an integral sign and was originally an elongated s, for "sum;" the numbers a and b are called the lower and upper limits (>/ integration.) The integral in [a, ft]. f f is also called the area of R(f, a, b) when f(x) > for all .v 13. Integrals If / 259 integrable, then according to this definition, is < L(f, P) f < U(f, P) for partitions all P of [a, b]. Ja Moreover, f f is the unique number with this property. This definition merely pinpoints, and does not solve, the problem discussed we do not know which functions are integrable (nor do we know how to / on [a, b] when / is integrable). At present we know only before: find the integral of two examples: b f(x) if (1) = / then c, is integrable on and [a, b] f / f = c • (b — a). r ,, Ja (Notice that this integral assigns the expected area to a rectangle.) ,~ N - (z) Several Even r r it , x j (x) = f x irrational 0. . II, x more examples . then / < . . , . not integrable on [a,b\. is rational, will be given before discussing these problems for these examples, however, further. it helps to have the following simple criterion is integrable for integrability stated explicitly. THEOREM 2 If £ / > is bounded on there is [a, b], a partition P then / of [a , b] on [a, b] if and only if for every such that U(f,P)-L(f,P) <s. PROOF Suppose first > that for every s there £/(/, P) - is a partition L(f, P) < P with < e. e. Since mr{U(f,P')}<U(f,P), sup{L(f,P')}>L(f,P), it follows that inf {[/(/, P')} Since this is true for all e > 0, it - follows that sup{L(/, P')} by definition, then, If / is / is sup{L(/, P')} integrable. = inf {U(f, />')}; The proof of the converse assertion integrable, then sup{L(/, P)} This means that for each e > = inf {U(f, P)}. there are partitions P', U(f,P")-L(f,P') <e. P" with is similar: . 260 Derivatives and Integrals Let P be a partition which contains both P' and P" . Then, according to the lemma, U(f,P) <U{f,P"), L(f,P) >L(f,P'); consequently, U(f, P) - L(f, P) < U(f, P") - L(f, P') Although the mechanics of the proof take up a that Theorem 2 amounts little < e. | space, it should be clear nothing more than a restatement of the definition to of integrability. Nevertheless, a very convenient restatement because there is 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 necessitates, even in very simple situations. Let / be defined on by [0, 2] ,, fW= , 10, x # 1. x = \ Suppose P = tn } {/(> a partition of is tj-i tj-i 1 tj 2 (see Figure 8). 8 with [0, 2] 1 < l. tj Then = mi FIGURE < 1 M, = if /' ^ j, but = nij Mj and = 1 Since 7-1 L(f, P) = ^m,(f, - tt -i) + mjOj - tj-i) + J2 nii(t ' ~ f'-i)« (=7+1 i=] /-i C/(/,P) ^ = 5^M ft--r -_i) + M/(f -r _i)+ i I / / Mi(ti-ti-i), t=j+\ /=i we have U(f,P)-L(f,P)=tj-tj- h This certainly shows that / is integrable: to obtain a partition U(f,P)-L(f,P)<s, it is only necessary to choose a partition with < tj-i Moreover, it is 1 < tj and tj — tj^\ < e. clear that L(f, P) < < U(f, P) for all partitions P. P with 13. Integrals Since / is integrable, there is number between only one 261 lower and upper sums, all namely, the integral of /, so = 0. /„' 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 fix) P = = x, and /„} is {to for simplicity consider a partition of = mi [0, b], tt -\ an interval [0, &], then (Figure 9) and Mi = + *„_! (tn where b > 0. If tt and therefore Lif,P) = J^t i -. l (ti -t - 1 ) - 1 + nit2 i i=\ Uif,P) = t = Y,ti(ti~ti-l) it\ ) -?!) + •• -t + --- • -r„_i), 1=1 = FIGURE 9 t\it\ Neither of these formulas for partitions t, — tj_\ Pn — {to, is , of each subinterval + -to) t2 it 2 l ) + tl ,it„ -r„_i). particularly appealing, but both simplify considerably tn ) is into n equal subintervals. b/n, so fo = 0, t\ = b -, n t2 = 2b — , etc; n in general ib n Then Lif,Pn ) = Y,h-\iU-t,- l ) i=l A \{i-i)b\ i=\ 1 n 1 u2 Etf-D "-/=1 n-\ b2 E^'b y=o b In this case, the length 262 Derivatives and Integrals Remembering the formula ! this k(k+\) + ••• + * can be written b2 (n-l)(n) n-1 £ « 2 2 Similarly, £/(/, />„) = ]£*,&-?,-!) ^ i = n n \ + l) b 2 ~T~ "n 2 n+\ b 2 n(n If n is makes very large, both L(f, P„) and £/(/, it show easy to that / is Pn ) are close to & /2, integrable. Notice and this remark that first 2 U(f,Pn )-L(f,Pn ) 2 b = ---. 1 n Pn This shows that there are partitions By Theorem 2 with only a the function little work. / is with U(f, P„) — L(f, It is clear, first ) as small as desired. fQ f may now be found Moreover, integrable. Pn of all, that h2 L(f,Pn )< -j<U(f,Pn ) This inequality shows only that b /2 sums, but fix) = desired, so there xy has area we have only one number with we can conclude that — 2 Jo Notice that I H.I K I. Ml tude b (Figure it this 10). between certain just seen that £/(/, P„) is this property, lies for all h. — L(f, Pn this property. ) special can be upper and lower made as small as Since the integral certainly "f^ equation assigns area b /2 to a right triangle with base and Using more involved calculations, or appealing can be shown that h I b2 to alti- Theorem 4, . 13. Integrals The ure function f(x) 1 1), if P = {to = x tn ) m = { presents even greater a partition of is = /fa_i) 2 Pn = Choosing, once again, a partition the lower M, . . = tn , f(tt ) = 2 t, . into n equal parts, so that ) -b i = ti {to, In this case (Fig- difficulties. then [0, b], and (/,_i) 263 and upper sums become E, n b ib n* n (=i FIGURE 11 ; = 1 A .b ; = /? 77 1 b - 7? 1 3 " ./ = ! Recalling the formula 2 l from Problem 2- 1 , these + + ^2 = --- sums can be written L(f, Pn ) = l/C/,/3,,) = and + l) as — .-(n - l)(n)(2n 6 n 1), — .-(n+ 6 l). J 77 It is ±*(*+l)(2Jfc l)(n)(2/7 + •* not too hard to show that that U(f, P„) sufficiently large. — L(f, b 3 L(f,Pn )< J Pn made The same ) can be sort <U(f,Pn ), as small as desired, of reasoning as before then shows that b i 'o /= b3 -*3 This calculation already represents a nontrivial result bounded by a parabola theless, the result is by choosing n — the area of the region not usually derived in elementary geometry. was known to Archimedes, who derived it Never- in essentially the same . . 264 Derivatives and Integrals The only superiority we can claim is that in the next chapter we much simpler way to arrive at this result. Some of our investigations can be summarized as follows: way. a if = / = i: h c b — a) (b 2 a f{x) — if fix) =x if / (x ) = x c for all x. 2 y-y b3 if will discover a3 for all x, for all x i. This list already reveals that the notation notation for naming f f suffers from the lack of a convenient functions defined by formulas. For this reason an alternative notation,* analogous to the notation lim f(x), x—>a also useful: is rh I r>b f{x)dx means same precisely the as Ja /. / Ja Thus /' cdx = c (b • — a), Ja L XdX= Ja u2 b a 2 2 2 3 3 Notice that, as in the notation lim f(x), the symbol x can be replaced by any x—>a f,a,orb, of course): other letter (except pb \ Ja f(a)da= f(t)dt= Ja The symbol dx pb pb nb f(x)dx= rb fiy)dy= Ja Ja f{c)dc. Ja has no meaning in isolation, any more than the symbol x ->• has any meaning, except in the context lim f(x). In the equation /' 2 3 b> fl Ja *The notation / f(x)dx is actually the older, and was for many years the only, symbol for the symbol because he considered the integral to be the sum (denoted by /) of infinitely many rectangles with height f(x) and "infinitely small" width dx. Later writers used xq, x n to denote the points of a partition, and abbreviated x\ — .v,_i by A.v,. The integral was integral. Leibniz used this . . , 77 defined as the limit as Ax, approaches of the sums Y^ / '( v, ) , ./ A,v, (analogous to lower and upper (v, to fix), 7=1 sums). The many people. fact that the limit is obtained by changing >_] to / ) and A v, todx, delights 265 13. Integrals the entire symbol x dx may be regarded the function This notation for the integral amples may appear; for all x. Several ex- as flexible as the notation lim f{x). use of Theorems 5 and 6.* h b b f / is =x that f(x) for: aid in the interpretation of various types of formulas which frequently we have made (1) / such an abbreviation as (x + y) dx f = f + x dx 2 2 b a — - —+y = y dx - (b a). J (2) / (y + t)dy= Ja ydy+ / Ju (3) l!(J! — %- + t(x-a). Ja dx= a+,)dt ) (1 +t)(x -a)dx I. = (1+/) = (4) X = tdy / (x-a)dx / I b 2 (1+/) L\Ox+y)dy ) dx =l The computations of f x dx and f x dx may suggest is — a) a(b generally difficult or impossible. tions are impossible to As a matter of fact, determine exactly {although they that evaluating integrals may be computed of accuracy desired by calculating lower and upper sums). Nevertheless, as the next chapter, the integral of Even though most to know when precisely / is Theorem cannot be computed integrable be stated here, and we on [a b] , . will have we shall see in computed very exactly, easily. important it is Although any degree to it is at least possible to say for integrability is to settle for partial results. a little The too next most useful result, but the proof given here uses material from Chapter 8. If you prefer, you can wait until the end of the next gives the the Appendix to when a totally different *Lest chaos overtake the reader when qualification. This equation interprets value f(x) functions can be which functions are integrable, the criterion difficult to chapter, integrals a function many most func- the integrals of is the number v. But proof will be given. consulting other books, equation f ydx classical integral of some arbitrary Junction y. to mean (1) requires the integral of the function notation often uses y for v(.v), so / an important such that each f ydx might mean the 266 Derivatives and Integrals THEOREM 3 PROOF If / continuous on is Notice, every e / > / that first, prove that is / then [a, b], on integrable is [a, b]. bounded on [a,/?], because it is continuous on [a,£]. To on [a, b], we want to use Theorem 2, and show that for is integrable there is P of a partition such that [a b] , U(f,P)-L(f,P) <e. Now we Theorem know, by continuous on [a b] , . if \x of the Appendix to Chapter 1 So there — y\ < The trick is simply to choose 8. Then for each we have is some > 8 such that for — then \f{x) 8, a partition P = 8, that all / is x and y uniformly in [a , b] , £ < f(y)\ {to tn 2{b-a) such that each ] \t, — /,_i < | i £ \f(x)-f(y)\< and [f,-_i, f,-], follows easily that it Since for all x, y in 2{b-a) this is true for M - we then have all /, [/(/, - P) >"/ < 2(b-a) < b-a = ^(M,- - L(f, P) < b — a = which is what we wanted. Although this theorem integrals in this book, it — ti-i) £>-',-! 1 £ b - m,)(t, b — a a | will is provide more all the information necessary for the use of satisfying to have a somewhat larger supply of integrable functions. Several problems treat this question in detail. It will know on is the following three theorems, which integrable on [a, c] and [c, b]; that show /+g is that / is integrable integrable if / and g help to [«./?], if are; and it that f is integrable if / is integrable and c is any number. except at one As a simple application of these theorems, recall that if / is 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 c such a function to an integrable function, function may be changed arbitrarily By breaking up changed The in the interval into at finitely many at many we see that the value of an integrable one point without destroying subintervals, we integrability. see that the value can be points. proofs of these theorems usually use the alternative criterion for integrability Theorem 2; as some of our previous demonstrations illustrate, the details of the . 267 13. Integrals argument often conspire to 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 used in the techniques THEOREM 4 < Let a on < c b. / If Conversely, [c, b\. [a b] , is on integrable is Suppose / [a b] , is integrable / on Ja and on [a, c] / then [c, b], integrable is Jc > [a, b]. If £ on integrable is integrable Ja PROOF then [a, b], on [a, c] and on integrable on [#,£], then / / is if Finally, if . some of the problems. in 0, there is P = a partition [to, . . . , tn ) of such that U(f,P)-L(f,P) <e. We might which contains to,...,t„ and U(f,P)-L(f,P) <s.) Now of P' [c, b] = {to, = assume that c as well . . . ,tj} is c; some for tj Q then a partition of j (Otherwise, . let Q be the partition — contains P, so U(f, Q) [a, c] = and P" {tj, . . /„} is , L(/, Q) < a partition (Figure 12). Since L(f,P)=L(f,P') U(f,P) = U(f,P') + L(f,P"), + U(f,P"), we have U =b [U(f, P') - L(f, P')] + Since each of the terms in brackets FIGURE 12 that / is integrable on - [U(f, P") [a, c] and is L(f, P")] = U(f, P) nonnegative, each [c, b]. Note is less - L(f, P) than e. < s. This shows also that L(f,P')< f f<U(f,P'), L(f,P")< / f<U{f,P"), so that L(f, Since this is P)< f f+ < f f U(f, P). true for any P, this proves that ff + Jcff=ff. J (i Ja Now suppose that partition P' of [a, c] on [a, and a partition P" of / is integrable c] and on [c, b] [c, b]. such that U{f,P')-L{f,P') <e/2, U(f,P")-L(f,P")<e/2. If e > 0, there is a 268 Derivatives and Integrals If P the partition of [a, b] containing is the points of P' all L(f,P)=L(f,P') + L(f,P") U(f,P) = + U(f,P') and P", then ) U(j\P"); consequently, U(f, P) - L(f, P) Theorem 4 f f was is = - [U(f, P') some minor the basis for defined only for a L(f, P')] < [U(f, P") Ja With these f f Ju definitions, the a < c < b is The e. | integral if > a b. Jb =f f f +f f equation f' t if < pa and = L(f, P")] the definitions ph f - notational conventions. We now add b. + not true (the proof of this assertion is holds for all a, c, b check). THEOREM 5 If / and g are integrable on [a, b], then PROOF Let P — {to tn ) (/+*)= (f + g) be any partition of m, and define M,, = inf {(/ = inf {fix) m," = inf{g(x) M,', M,-" similarly. it is /+ : It is f,_i on [a, b] and / Ja [a, b\. :r,_i integrable pt rb / Ju + g)(x) mi rm but is no pb .& no / Ja /+g : Let ?;_] < x < /,}, < x <t }, < x < t,}, t not necessarily true that = m/ + m/', true (Problem 10) that m, > m{ + mj". Mj < M/ + Mi". Similarly, Therefore, L(f,P)+L(g,P) <L(f + g,P) and U(f + g,P) < U(f,P) + U(g,P). Thus, L(f, P) Since + / and g L(g, P) < L(f + g, P) <U(f + g, are integrable, there are partitions even a rather tedious case-by-case P) < U(f, P\ P" with U(f,P')-L(f,P') <e/2, U(g, P")-L(g,P") <e/2. P) + U(g, P). 269 13. Integrals P If and P", then contains both P' + P) [/(/, - U(g, P) + [L(f, P) L(g, P)] < e, and consequently U(f + g,P)-L(f + g,P) <s. /+g This proves that L(f, P) (1) on integrable is + [a b\ , P)<L{f + g, L{g, < pb rb L(f.P) /+/ + L(g.P)< Ja Since U(f, P) desired, — L(g, P) can both be - [L(f, P) made P) + U(g, P) as small as desired; pb (/ / + *)= on [a , integrable b] as small as on [a, b], L(g, P)] Ja then for any (1) and (2) that fb /+/Ja / Ja is + therefore follows from it pb / made follows that it can also be If <U(f,P) + U(g,P). g Ja — L(f, P) and U(g, P) £/(/, 6 < u(f,P) + U( g ,py, + g ,P) also (2) THEOREM P) + *) (/ J < u(f and Moreover, . number g-l c, the function cf integrable is and f- Ja PROOF The proof (which much is Ja easier than that of Theorem 5) idea to treat separately the cases c (Theorem 6 is on [a Problem 38).) integrable In this chapter theorems with Appendix to > and c just a special case of the , b] , if / and g are, we have acquired intricate proofs, Chapter 8. This but < 0. Why? is left to you. It is a good | more general theorem this result is / that • g is quite hard to prove (see only one complicated definition, a few simple and one theorem which required material from the is not because integrals constitute a more difficult topic than derivatives, but because powerful tools developed in previous chapters The most have been allowed to remain dormant. is the fact that the integral and the derivative are intimately related learn the connection, the integral will easy to use. The connection between chapter, but the preparations hint. We first state many important significant discovery of calculus become derivatives which we will — once we as useful as the derivative, and make and as integrals deserves a separate in this chapter may serve as a a simple inequality concerning integrals, which plays a role in theorems. ) 270 Derivatives and Integrals THEOREM 7 Suppose / is on [a b] and m < fix) < integrable , that M for all x in [a,b]. Then m(b-a)< PROOF It is -a)< for every partition P. fa f = Since Suppose now that / [a,b] by + h) - F(c) is and L(f, P) inequality follows immediately. F(c a). clear that m(b area f < M(b - / [/(/, M(b - P) < = sup{L(/, P)} a) m£{U(f, P )}, the desired | on integrable F(x)= We [a, b]. f= f f can define a new function / on f(t)dt. on Theorem 4.) We have seen that / may be integrable even if it not continuous, and the Problems give examples of integrable functions which (This depends FIGURE is 13 are quite pathological. THEOREM 8 If f is integrable on The behavior [a, b] F and is of F is therefore a very pleasant surprise. defined on by [a, b] Fix) Ja F then PROOF is continuous on [a, b]. Suppose c is in [a, b]. Since / is integrable on on [«,&]; let be a number such that [a, b] it is, by definition, M < |/U)| If h >0, then (Figure M for all x in [a, b]. 13) rc+h Fic + h)- re +h nc F{c) f~ I Ja f = Ja f•/<' Since -M it follows from Theorem 7 < fix) < M for all x, that /c+h f < Mh: in other words, ( If h < 0, - 1 M -h < Fic + h)- Fie) < M a similar inequality can be derived: Note that ., Fic + h)- t/, Fie) Jc J c+h • /?. bounded 13. Integrals Applying Theorem 7 to the interval [c + h, Mh < c], of length —h, we 271 obtain f < -Mh; Jc+h — multiplying by 1 which reverses , Mh (2) Inequalities (1) and (2) if e > 0, the inequalities, < F(c +h)- + h) - \h\ words F < s/M. is this F is F(c)\ < + h) FIGUKI. I 4 s, = F(c); f for various functions /; at c. | / and F(x) — x f' always better behaved than /. In the next chapter is. / / \h\. This proves that continuous Figure 14 compares that M we have lim F(c in other < -Mh. F(c) < F(c)\ \F(c+h)provided that we have can be combined: \F(c Therefore, all / we it will see appears how true 272 Derivatives and Integrals PROBLEMS 1. 4 3 fQ x dx — b /4, by Prove that considering partitions into n equal subin- n Tjr using the formula for tervals, which was found in Problem 2-6. This i=\ problem requires only a straightforward imitation of calculations but you should write it up as a formal proof to make certain that all the fine Using Problem 2-7, show that the sum >^ k p /n p+ can be made as close argument are points of the 2. *3. Prove, similarly, that (a) in the text, clear. 4 5 f x dx = b /5. k=l + to \/(p (b) 1) as by choosing n large enough. desired, = Prove that j^x p dx b p+l /\p + l) . b *4. way This problem outlines a clever P = = a result for /„} for {to partitions for (a) will Show which then follow by which that for such a partition (b) If f(x) = — f, — P we in i/n , U(f,P) =a p+ \\ - C - 1/ is ")^](c , ' ,+ _ 1 £P+l )c (P+l)/« 1 = (b p+] Problem 2-5, that 1,/ ")' -r_ c (p+l)/n 1 (c) and find a similar formula Conclude that L p dx = (You might find Problem 5-41 b p+l_ a p+l - p + useful.) Evaluate without doing any computations: /.I (i) / A-Vl (ii) f (x 5 + ~X 2 dx. 3)y/\ -x 2 dx. /" for L(f, P). I' K +c [ 1 b. (The are equal, instead of using 1 -a p+l )c p/n < to use partitions |/,! p+l ( fl a have x p show, using the formula c trick < are equal. = =a The ti/tj_\ £;_] for c ti x p dx for / Ja continuity.) all ratios r differences all f to find H he/'/" , . 13. Integrals 273 Prove that sin J dt f Jo 7TT for all x > > o 0. Decide which of the following functions are integrable on the integral (i) /(*) when you 0<x x-2, 1 x /(*) x-2, * (iv) f(x) 1 ./•(.*•) .v = + [.r] = / is < jc < 1 < a < 2. irrational. + bv 2 for rational a and b x not of this form. < A < 1 = the function FIGURE 8. 2. x of the form a 1, x 0, (vii) 1 x rational , 1 f(x) < < a < x 0, (vi) calculate + W0, (v) and can. x, = [0, 2], = shown or X in > 1 Figure 15. 15 Find the areas of the regions bounded by —+ (i) the graphs of f(x) =x and g(x) = (ii) the graphs of f(x) = and g(x) = —a - and g(x) and #(a) = = and g(x) =x x 2. and the vertical lines through (-1,0) and (1,0). (iii) the graphs of /(a) (iv) the graphs of f(x) (v) the graphs of f(x) =a —x —x 1 1 — jc . — x and h(x) = 2. — 2x + 4 and the vertical axis. 274 Derivatives and Integrals = the graph of f(x) (vi) through (Don't try to find (2, 0). and s/x, the horizontal axis, s/x dx\ / the vertical line you should see a way of Jo know how guessing the answer, using only integrals that you already evaluate. in 9. The Problem 2 questions that this example should suggest are considered 1 .) Find [(j:< (x)g(y)dy in fa f and fc terms of vengeance; 10. g. crucial that it is dx ) (This problem is an exercise in notation, with a you recognize a constant when it appears.) Prove, using the notation of Theorem 5, that +m" - m,-' 11. to (a) Which + inf{/(.xi) gixj) : < < x\,xj <t /,_i t ] ///,-. sum functions have the property that every lower equals every upper sum? (b) Which (c) (other) lower sum? Which continuous functions have the property that some upper sum equals some functions have the property that all lower sums are all lower sums are equal? *(d) Which integrable functions have the property that equal? (Bear in = f(x) mind Problem 12. set, Ua<b<c<d and / (a) (b) work / p/q / /> 0. Ja Prove that if is / and g then 14. on integrable integrable is Hint: for x irrational, You will need the [a, d], prove that / is integrable on on and f(x) > [a, b] are integrable on [a, b] for all x and f(x) > g(x) in [a, b], for all x rh / > / Ja that if in lowest terms.) = introduced in Problem 8-6, as well as the results of pb in [a, b], f(x) is hard.) if Prove that then one such function 30. [6,c]. (Don't 13. = \/q for x notion of a dense that / g. now (By it should be unnecessary to warn Ja you work hard on part (b) you are wasting time.) Prove that no / b+c fb+c f(x)dx = / f(x-c)dx Ja+c Ju (The geometric interpretation should make partition ...,/„+ P — c} {/(),...,/„} of \a + c, b of + c\, this very plausible.) Hint: Every \ci,b] gives rise to a partition and conversely. P' — [tQ + c, c 15. For a, b > / r 1 -dt + t J\ l -dt t J\ r*abw = 16. pan ...,/„} of [1 = \/tdt. I = a] gives rise to a partition P' , [bt$, . . . , btn ) of [b, ab], conversely. Prove that f f(t)dt =c J cu Given ellipse a special case.) is by the unit circle, described by the equation use Problem 16 to show that the area enclosed by the that the area enclosed +y = x f f(ct)dt. Ja Problem 15 (Notice that 17. P = t Jx l/tdt : Every partition 1 -dt. /a and 275 prove that \ r {to, 13. Integrals 7i, is 1, +y described by the equation x /a = /b nab. is 1 h 18. This problem way outlines yet another to f compute / x" dx; it was used by Ja Cavalieri, one of the mathematicians working just before the invention of calculus. = Let c n (b) Jo Problem 14 shows that / formula to (x 19. Suppose Problem 2-3 / that to in [a,b] with the exception of [a b] , 20. . . + a)"dx. £ QW prove that bounded on is n+ prove that k Now use cn a J-a 2" +1 c„a"+'=2a" +1 (c) = x" dx / pa x"dx = Jo this that Jo p la \ Use show x" dx. Use Problem 16 to (a) [a, b] .\"o even c„ and = \/{n that / + is 1). continuous Prove that in (a,b). at each point on / is integrable / is automatically Hint: Imitate one of the examples in the text. Suppose / that bounded on P = (a) If (b) Suppose nondecreasing on is [a, b], {r , . . that . , f, [a,fr]. Notice that because f(a) < f(x) < f{b) tn } is a partition of [a, b], —t{-\ =8 for each i. for what x is Prove that in [a, b\. L(f, P) and U(f, P)? U (f, P) — L(f, P) = 8[f(b)-f(a)]. (c) Prove that (d) Give an example of a nondecreasing function on /' is integrable. tinuous at infinitely many points. [0, 1 ] which is discon- 276 and Derivatives Integrals It might be of interest from Newton's to compare this problem with the following extract Principia* LEMMA II If in any under equal bases AB, BC, CD, &c, and Aa one side of the figure; and D E number their and to Bb, Cc, Dd, &c, the sides, the inscribed figure AKbLcMdD, AabcdE, / say, sum be diminished, to the circumscribed figure sum area b total f l (b) Lm, Mn, Do, their bases), the rectangle of their altitudes Aa, that because its breadth AB is is, that under one of AalbmcndoE, "21. f- l Suppose that / area a FIGURE f (a) f~ t,,} , Prove (b) Now prove (c) Find 22. : Suppose / that that for a, b af~\a) is and a K I. /. that, as let P' = ] {f~ (t ), suggested in Figure 17, > l , P) + U(f, P') = bf~\b) - = /o t] t2 ti t„ = b I 7 af~\a). dx is < a < for b. a continuous increasing function with f(0) Youngs inequality, — 0. Prove we have that equality holds if and only if l I b f~ (x)dx, — f{a). Hint: Draw a picture like Figure 16! Newton's I (b) / a partition of [a,b], ab< f f{x)dxJo . and much the formula stated above. & [ Ja i the figures inscribed 16 L(f~ I less Jf-Ha) {t (/„)}. and the becomes to the other; - Ja I 1) l P = Kb in infinitum, ff- l (from the equality increasing. Figure 16 suggests that is /' /"• =bf~\b) If the curvilinear figure be ultimately equal to either. (a) (a) is the rectangle ABla. But this rectangle, supposed diminished And therefore (by Lem. and circumscribed become ultimately equal one more will the intermediate QE.D. is their bases than any given space. f~\b) are will have to one another, are ratios of equality. of the parallelograms Kl, all parallel that the ultimate ratios For the difference of the inscribed and circumscribed figures of acE, aKbl, bLcm, cMdn, &c, the parallelograms be augmented in infinitum, curvilinear figure the curve Cd, &c, comprehended completed: then if the breadth of those parallelograms be supposed which B F C Aa, AE^ and any number of parallelograms Ab, Be, and A terminated by the right lines there be inscribed to M AacE, figure Principia, A Revision ofMott's Translation, by Florian Cajori. Press, Berkeley, California, 1946. University of California ) . . 211 13. Integrals 23. Prove that (a) [a, b], if / m < on [a,b] and integrable is M < f(x) for all x in then /" f{x)dx = (b — a)ii Ja for some number Prove that (b) if / is with /i m < < M. i± continuous on [a , b] then , /' f(x)dx = (b-a)f($) for some £ in [a b] , (c) Show by an example (d) More that continuity essential. is continuous on [a b] and that 8 integrable and nonnegative on [a b] Prove that generally, suppose that / is , . , /' f(x)g(x)dx = /($) some f g(x)dx / Ja Ja for is £ in [a b] , . This result is called the Mean Value Theorem for Integrals. FIGURE 18 24. (e) Deduce (f Show same the that angle 9. 4, we is integrable 6 is consider the region and nonpositive on for g 3). measured A shown area Let [a , / be b] , 19 in radians, the area of this sector where the curve in Figure 19, A = 1 f Show is r~ — . Now is the graph in is a partition of that Hl « / z J0o a continuous function on 2 f(0) d0. [a, b]. If P — {to /„ define £(/, P) b] with central circle, n 'IGURK , essential. Figure 18 shows a sector of a polar coordinates of the continuous function /. *25. is [a consider the graph of a function in polar coordinates Appendix When g one of these two hypotheses In this problem (Chapter result if = J2 V^-f;-,) 2 +[/(*;) -/(f;_i)] 2 . } 278 Derivatives and Integrals The number £(f, P) represents the length of a polygonal curve inscribed in the graph of / the least of all such £(f, P) If / If (b) — to t\ h ti ?4 = b / (You FIGURE will {provided that the set P = a partition is is the {a,t,b} of [a, b] greater than the distance from (a, f{a)) to is / prove that the length of , prove that there linear, P be [a, b] to to (b, f(b)). f (a)) such that £(/, P) partitions above). , not is all a linear function on [a b] is length of / on define the i(f, P) for bounded is distance from (a, a We Figure 20). (see upper bound of all need Problem (b, f{b)). 4-9.) and f(b) = d, the length of the linear function is less than the length of any other. (Or, in conventional but hopelessly muddled terminology: "A straight line is Conclude 20 that of / on functions all = with f(a) [a, b] c the shortest distance between two points".) Suppose that (d) show /' bounded on is [a b] , P If . Use Hint: (e) Why is (f) Now show +(f)\ the Mean + sup{l(\/l (/') grable on [#,&]. Hint: 2 , how < P)} < 2 (f') P). , sup{£(/, P)}? (This inf \U (V + 1 2 (./") , is easy.) P)j, thereby proving 2 v 1 if vl / if') Ja suffices to show that if P' and [a,b] It partitions, then £(/, P') + is 2 < U{J\ + (./") , P P"). If , + 2 (f) P" are any two contains the points does i(f, P') compare to €(/, P)? Let i£(x) be the length of the graph of / on v if 1 + lim f'{x) (/') = integrable is inte- is and [a, x], d(x) be the let Show length of the straight line segment from (a, f{a)) to (x,f(x)). if , Value Theorem. / on of both P' and P", U(Vl + P) < i(f, P) < that sup{£(/, P)} that the length of that [a b] that L(\/\ (g) any partition of is on [a, b] and /' is continuous at a (i.e., f(a)), then = lim 1. x^a+ d{x) Hint: In Figure 21, the part of the (h) size of the part between 5 and the size of the part I !'. i RE part 21 26. A {/<) of at (a) .v s defined on tn tj between does not hold for (g) function P — Mean Value Theorems. graph of / between ^ and ^ is just help to use a couple of It will ] of may be Prove that if \a, b] [a, b\ 1 , the part and ^ 5, etc. between Show g and 4 is half the just half that the conclusion of this /. is step function if there a constant on each (f,-_] called a such that 5' is , is a partition /, ) (the values arbitrary). / is integrable on [a, b], then for any £ > there is a step . 279 13. Integrals pb pb function < f s\ with /— I pb Ja f < I such that — S2 / > there are step functions < s\ I / Prove that e. < f and s\ si > f f" / = integrable. is Ja Ja / which Find a function (c) > f also a step function S2 s. Ju that for all s pb pb Suppose (b) and e, pb — S2 I < Ja Ja with s\ / not a step function, but which is satisfies / Ja P L(f, P) for some partition *27. Prove that / if integrable is on of [a, b] pb < f < functions g h with 28. Show (a) that and if s\ > then for any e / — // there are continuous / £ g < Hint: First get step functions Ja and then continuous property, this , pb Ja with [a, b\. A picture will ones. S2 are step functions on [a, b], then Prove, without using Use part (c) 29. Suppose [a b] , / integrable is such that / / = *30. (51+52)= / on / Show by example that choose x to be in Let b (b) — P = tn {to ) be a partition of Prove that for some a. it is is not is integrable we have M, — m, < / < a\ 1. [«!,/?]] = S2- number x 5. in always possible < x < [ti_i ti\ , b\) inf{/(jc) from part (a) < a\ : unless i x = 1 on < [a b] , - , then L(f, P) < 1. < a\ a with (/(/, P) [a, b] — : I Ja proof of Theorem Prove that there are numbers a\ and b\ with a sup{/(A") + {a, b). The purpose of this problem is to show that if / / must be continuous at many points in [a, b]. (a) s\ / also. is pb Ja Prove that there [a,b]. + S2 JX «/</ to / 5, that Ja 26) to give an alternative (and Problem (b) that Theorem s\ pb pb (b) help immensely. b\) < < b and b\ (You can choose or n; and in these two cases a very simple device solves the problem.) (c) /?2 with a\ < 02 < ^2 < b\ and — < inf < x < ^2} < hci2 «2 S ^2} [fix) Continue in this way to find a sequence of intervals /„ = [a„, b n ] such that sup{/(x) x in /„} — inf{/(x) x m I,,} < l/n. Apply the Nested Intervals Theorem (Problem 8-14) to find a point x at which / is con- Prove that there are numbers a2 and sup{/Cv) (d) x : : : : tinuous. (e) Prove that Let / be / is continuous at infinitely many points in [a b] , b *31. integrable on [#,&]. Recall, from Problem 13, that f / / > if Ja fix) > (a) for all x in [a,b]. Give an example where fix) > [a,b], and j Ja f = 0. for all x, and fix) > for some .v in a 280 Derivatives and . Integrals Suppose f{x) > (b) and f(xo) > for x all f" Prove that 0. and / in [a, b] > j / Hint: 0. continuous is at xo in [a,b] to find It suffices one lower Ja sum L(f, P) which Suppose f (c) positive. is on [a, b] You will integrable is and f(x) > x for all in [a, b\. Prove rb that / > / Hint: 0. need Problem 30; indeed that was one Ja reason for including Problem 30. b *32. Suppose (a) / that continuous on is f and [a, b] fg=0 for all continuous / Ja on functions g [a, b]. f —0. Prove that (This easy; there is is an obvious g to choose.) h Suppose / (b) f continuous on [a,b] and that is / = fg for those con- J tinuous functions g on [a,b] which satisfy the extra conditions g(a) gib) = Prove that 0. lemma / = (This innocent looking fact 0. is an important in the calculus of variations; see reference [22] of the Suggested Reading.) Hint: Derive a contradiction from the assumption /(xq) < or f(xo) 33. *34. — Let f(x) Compute L( f, P) (b) Find inf {£/(/, P) that / is depend on the behavior of / near will = x for x rational and f(x) (a) = Let f{x) g you pick 0; the P a partition of on [0, 1] and that of > *o- irrational. [0,1]. [0, 1]}. and \/q for irrational x, integrable P for all partitions : x for = x if — f / = p/q in lowest terms. (Every lower 0. sum is Show clearly Jo you must 0; *35. o Let / is to make upper sums Problem 34 not. Hint: / be how / and g which Find two functions g *36. figure out is are integrable, but corresponding meanings for the function (a) Prove that M (b) Prove that if Prove that if (c) min(/, (d) ' t / / and g is M, - a partition of [a,/?]. M/ let and m,' have the |/|. m,. on integrable / is [a b] , are integrable if / is its integrable , then so on [a, b], is | / 1 then so are max(/, g) and on [a b] , if and only if its "positive part" "negative part" min(/, 0) are integrable on [a, b]. integrable / Hint: P be g). Prove that Prove that - m,-' < is max(/, 0) and 37. let have their usual meanings, and m,- whose composition relevant. a bounded function on [a,b] and Let Mi and small.) on [a, b], f{t)dt < / then \fit)\dt. Phis follows easily from a certain string of inequalities; relevant. Problem 1-14 N . 281 13. Integrals "38. for all x in Suppose / and g are integrable on [a,b] and f(x),g(x) > appropriate M,-' and m,' denote the [a, b]. Let P be a partition of [a, b]. Let sup's and inf 's for /, define M," and m," and define M, and m, similarly for g, similarly for fg. (a) Prove that (b) Show < Mi'M" and t > mi'm" ra, that Using the x in [a L(fg, P) < Y,[ / and fact that b] , - F) tf (/*, (c) M show , M M >' >' ~ mi'mflte - ft_i). < g are bounded, so that \f(x)\, \g(x)\ M for that U(fg,P)-L(fg,P) n In^[M/ - m/](/, - r,-_i) 39. Prove that fg (e) Now Suppose m,"]{U U-l) is \ integrable. eliminate the restriction that f(x),g(x) that - i=\ i=\ (d) 1 + ^[M," - / and g are integrable on > The [a, b]. for x in [a, b]. Cauchy-Schwarz inequality states that ([>)<{ >%!> (a) Show that the Schwarz inequality is a special case of the Cauchy-Schwarz inequality. (b) Give three proofs of the Cauchy-Schwarz inequality by imitating the proofs of the Schwarz inequality in Problem 2-2 1 take (c) some If equality holds, / and g (d) (The . one last will imagination.) is it / = Xg necessarily true that for some What A.? if are continuous? Prove that j I / < / I /" I • Is this result true and if 1 are J replaced by a and b? *40. Suppose that / is integrable on [0, x] for all x > and lim f(x) = a. Prove that lim Hint: The condition r f{t)dt = a. i ,. x ^°° - / x Jo lim f(x) x—>oo = a implies that /(/) is close to a for pN + M t > some N. This means that / f(t)dtis close to Ma. J comparison to N, then Ma/(N + M) is close to a. If M is large in 282 Derivatives and Integrals RIEMANN SUMS APPENDIX. Suppose P = that choose some point t„] {to a partition of is Then we x, in [f,_i, t{\. and [a, b], that for each we i have clearly n - L{f, P) <Y^f(xt){t t < U(f, tt-i) P). i=\ n Any sum / — ^/CxtXfr called a Riemann sum of f,-_i) is / for P. Figure 1 shows Riemann sum; it is the total area of n rectangles that lie partly below the graph of / and partly above it. Because of the arbitrary way in which the heights of the rectangles have been picked, we can't say for sure whether a particular Riemann sum is less than or greater than the integral the geometric interpretation of a a = to FIGURE X\ t\ x2 tn =b f(x)dx. But J J 1 does seem that the overlaps shouldn't matter too much; it bases of all the rectangles are narrow enough, then the The close to the integral. THEOREM 1 the if CI Suppose / that such that, if {to, . . , . be to following theorem states this precisely. on integrable is P= Riemann sum ought tn } is [a b] , Then . > for every s any partition of [a, b] with there all is lengths some /, 8 > — /,_i < 8, then " no >b ^/(jC/)(ft -/,_!)- for PROOF any Riemann sum formed by choosing x Since the Riemann sum and amounts this by choosing a tt showing that to { in (7,_i, the integral both B, — /,]. between L(/, P) and U(f, P), lie we can make U(f, P) — L(f, P) < e P) < £ for any partition with all lengths any given for such that U(f, P) 8 < f(x),\dx / Ja i=\ L(f, s -ti-i <8. The M for / being definition of some M. on integrable [a, b] includes the choose some particular partition P* First condition that = [uq, , . . |/| < .,Uk) for which U(f,P*)-L(f,P*)<B/2, and then choose a such that 8 8 For any partition P with £/(/, P) all - — t, < AMK < f,-_i L(f, P) 8, we can break - YlS Mi ~ m,){ti " the sum r'- l) /=i into two sums. The first pletely contained within U(f, P*) — L(f, P*) < some j = 1, K — 1, sum for these terms is < . . . , involves those one of the s/2. For / for (A' 1 ) • interval [/,_i,/,] intervals \uj-\,Uj\. other all so there are at most — which the 2M • 8 < i we K — s/2. | will 1 This have sum < f,_] is Uj is com- clearly < tj < for of them. Consequently, the . 13, Appendix. The moral to of an integral this tale really that anything is provided that is, Some partition are small enough. this message with even greater all which looks the lengths /, like 283 Riemann Sums a good approximation — ?;_i of the intervals in the home of the following problems should bring force. PROBLEMS 1. Suppose P = / and g that {to of tn } of points are continuous functions [a, b] choose on For a partition [a,b]. a set of points x, in [?,_] and another /,] , set Consider the sum Uj in [f;_i, tf\. Y^f(.Xi)g(Ui)(ti -ti-l). (=1 Notice that this is not a Riemann sum of fg for P. Nevertheless, show that b all lengths /, sum and 8, 2. — a f; f fg provided that the partition P has all Ja _i small enough. Hint: Estimate the difference between such a such sums will be within e of Riemann sum; you / will need to use uniform continuity (Chapter Appendix). similar to, but somewhat harder than, the previous one. and / g are continuous nonnegative functions on [a b] For a P, consider sums This problem Suppose is that partition , . n Y2 y/f(Xi)+g(Uj) (ti - f,-_i). i=\ Show that these sums will fVf + g be within s of if all — f,-_i t, are small Ja enough. Hint: Use the fact that the square-root function tinuous on a closed interval 3. Finally, For a partition (u(t ),v(t )) uniformly con- M] (Compare Problem 13-25.) given parametrically by two functions u and v on [a, b]. we're ready to tackle something big! Consider a curve (wfa), v(ti)) [0, is £(c P) c P = {to /„} of = J2 \/W) - [a, b] «^-i)] we 2 + define b>di) - wfe-i)] 2 ; i=\ this IK.l'RE 2 represents the length of an inscribed polygonal curve (Figure define the length of c to be the least upper bound of all £(/, P), 2). We if it exists. 284 Derivatives and Integrals Prove that if u' and v' are continuous f 4. y/u' 2 Let /' be continuous on the interval polar coordinates on this interval on + v' [a, b], then the length of c is 2 . [Go, 0\]. Show that the graph of / in has the length .ft 5. Using Theorem is 1, show that the Cauchy-Schwarz inequality (Problem 13-39) a consequence of the Schwarz inequality. THE FUNDAMENTAL THEOREM OF CALCULUS CHAPTER From continuous; is We know that theorem of this chapter. first may the hints given in the previous chapter you it is continuous. only It fitting that we / when differentiable (and is =f f is the original function / integrable, then is ask what happens F turns out that if have already guessed the its F(x) derivative is especially simple). THEOREM 1 (THE FIRST / be Let integrable on [a b] , , F on and define FUNDAMENTAL THEOREM OF CALCULUS) = F(x) [a, b] by f. f Ja If / is continuous at c in [a, b], F then = F'ic) c (If — a or b, then F'(c) is understood differentiable at c, is and f{c). mean to the right- or left-hand derivative of F.) PROOF We will assume that c supplied by the reader. in (a, b); the easy modifications for c is By = a or b definition, — F'(c) + F(c h) - F(c) lim h Suppose first that h > Then 0. c+h + h) - F(c Define nth and M\ x as follows (Figure mi, Mi, It follows from Theorem = = F(c) f- -I. 1): inf \f{x) : sup{/(x) c < x < c + h}, c < c + h}. : x < 13-7 that c+h h nu, f < < >L Mi, h Therefore mh < F(c + h)-F(c) <M h . h If h < 0, only a few details of the argument have to be changed. Let mi, M 285 h = — inf {f(x) sup{/(jc) + h < x <c}, c + h < x < c}. c \ : may be 286 Derivatives and Integrals FIGURE 1 Then mh {-h) < f <M - h i-h). Jc+h Since pc+n •C+/1 F(c + h) - F(c) = pc f = - f Jc+h Jc this yields mh Since h < -h > F(c + h)- F(c) > M h. h dividing by h reverses the inequality again, yielding the 0, same result as before: mh < F(c + h)-F(c) < M h . h This inequality continuous at is c, true for any integrable function, continuous or not. Since lim this /»/, = lim Mi, = h^O f(c), proves that F (c) = hm F(c - — + h)-F(c) Although Theorem 1 = /(c). h h-*0 limit is however, h->0 and / deals only with the function obtained by varying the upper of integration, a simple trick shows what happens varied. If G is | defined by G{x) = f f, when the lower limit is 287 The Fundamental Theorem of Calculus 14. then = G(x) f~ Jf f Jd Consequently, / if continuous at is /• a then c, = G\c) -/(c). The minus sign appearing here is very fortunate, and rem 1 to the situation where the function defined even for <a jc . we can In this case if c < write -£> F(x) so Theo- = f f Ja (*) is allows us to extend we have a = F'(c)=-(-f(c)) f(c), exactiy as before. Notice that in either case, differentiability of at c alone. Nevertheless, all points in [a b] , . Theorem F In this case is 1 Problems it is extremely of some derivative 1 1 = is ensured by continuity of when / is / continuous at and f. decide whether a given function function; for Theorem reason this / 1 1 -7 is the and -60 and 11-61 are particularly interesting, since they reveal certain properties which / must have. at all at c , difficult to other F interesting differentiable at all points in [a b] is F' In general, most —according to Theorem / If 1 , / is is continuous, however, there the derivative of some is function, no problem namely the function Fix) = -r / /. Ja Theorem 1 has a simple corollary which frequently reduces computations of integrals to a triviality. corollary If / is continuous on [a, b] and f I proof = g' for some function g, then f = g(b)-g(a). Let Fix) Then F' =f— g' on [a, ft]. = f ./'. Consequently, there F = g + c. is a number c such that 288 Derivatives and Integrals The number c can be evaluated 0= so c — = g(a) + c, F(a) —g(a); thus = Fix) This note that easily: is x true, in particular, for = g(x) - g(a). Thus b. f=F(b) = g(b)-g(a).l f Ja The proof of this after all, corollary tends, at what good is it know to for if g know is, example, g(x) the corollary seem useless: that f=g(b)-g(a) I to make first sight, to = f f ? The point, of course, is that one might happen a quite different function g with this property. For example, if ~3 *(*) then g'ix) = fix) so we = and = f{x) obtain, without ever x , computing lower and upper sums: /' x z dx = Ja One x n+] can treat other powers /in + similarly; if n is a natural gix) = then g'ix) =x", so 1), ,n+l ,"+1 /' = x" dx For any natural number n, containing 0, but if n 1 the function fix) — + x~" is 1 not bounded on any interval a and b are both positive or both negative, then f x dx b -n+\ — -n Ja formula this + n Ja Naturally number and is only true for n a -n+\ -n + +1 /— 1 . We 1 do not know a simple expression for b 1 — I, The problem of computing opportunity to warn against a dx. x this integral is discussed later, but it provides a good The conclusion of Corollary many students think that f f is serious error. confused with the definition of integrals — 1 is often defined "gib) — gia), 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 as: without being the derivative of another function. x ^ 1 There and /( is 1 ) = 1 , then /' is is if fix) = for / cannot be a derivative (why not?). much more important: If / is continuous, integrable, but also another reason that For example, The Fundamental Theorem of Calculus 1 4. we know that f — g' for some Theorem 1. The function f(x) = \/x then then f{x) = g'(x), function g; but x > of 0, -dt, / t and we know of no simpler function g with this property. Theorem 1 is so useful that it is Fundamental Theorem of Calculus. In this book, corollary to frequently called the Second that somewhat stronger result (which in practice, however, is As we have just mentioned, a function / might be of the / if where J\ continuous. If this only because provides an excellent illustration: *(*)= The we know 289 is integrable, then is still it name is reserved for a not much more useful). form g' even if / is not true that = g(b)-g{a). f f The must return THEOREM 2 (THE SECOND FUNDAMENTAL THEOREM OF CALCULUS) If / must be proof, however, is entirely different Let is on integrable P = {to, a point [a, b] . . . ,tn ) x, in [/,_i and / = g' for , t{\ some function g, so we then [a, b]. Mean By the )(t i -tt-i) Value Theorem there such that i )-g(t i _ l = = ) g'(x i f(Xi)(ti -tt-i). If mt =inf{/(.v) Mi =sup{f(x) :/,_! :f,-_i < x < <x < tt }, tt), then clearly ntiifi - < ff_i) f(Xi)(ti - r,_i) < Mtiti - < Mtiti - tt -i), is, mtiti Adding , f = g(b)-g(a). be any partition of g(t that 1 to the definition of integrals. I PROOF —we cannot use Theorem - u-i) these equations for / < = g{tt) 1 - gfa-i) « we tt-i). obtain n n ^]w,(r, -/,_,) <g(b)-g(a) < ^M,(/ i=l i=i so that L(f, P) for every partition P. But this <g(b)-g(a)<U(f,P) means that *(*)-*( /'•I ( -r,_,) 290 Derivatives and Integrals We have already used the corollary to x" dx ,«+] b"~ = + n Ja (or, 1 equivalently, Theorem 2) few elementary functions: to find the integrals of a /' Theorem n+\ 1 n , # and b both (a -1. both negative positive or if n > 0). As we pointed out in Chapter 13, 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 < < b, then x 3 dx does not represent the area of the region shown in Figure instead which is given by -(f*H + FIGURE 2, v /o 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 frequentiy involve considerable ingenuity in any case. Suppose, to take a simple example first, that 2 — we wish to find the area of the region, shown in Figure 3, between the graphs of the functions f(x) = x2 and g(x) = x3 < x 3 < x 2 so that the graph of g on the interval [0,1]. If < x < 1, then below that of /. The area of the region of interest to us is therefore , /(*) = area 1)- area R(g,0, /?(/, 0, lies 1), x^ which is l 1 r I o x dx — Jo r , x dx I 3 Jo 4 This area could have been expressed as /' (f-8)Ja If g(x) < f(x) by / and in K, I RJ Figure for all x in [a, b], then g, even if f and g 4. If c a number such that bounded by /' is then thr icgion R\, this integral are sometimes negative. / +c and always gives the area bounded The and g g, has the easiest +c way to see this is shown are nonnegative on [a,&], same area as the region #i. The Fundamental Theorem of Calculus 14. +c bounded by / and g + c. area R[ = 291 Consequently, pb pb area R 2= (f+c)- / = (g (g + c)] + c) [(f f + C) / *b (f-8). Ja This observation bounded by useful in the following problem: Find the area of the region is the graphs of The first and g necessity intersect is to determine and g(x) region this more 1 - = X2 X or or x v5]/2) we have x graphs (Figure (b) FIGURE 4 — v5]/2, the interval ([1 + only if on the x — 0, The graphs of / , = n + ^5 1-75 \ 0, 2 (0, [1 precisely. 2 3 — x — X 0, 2 -* - 1) =0, jc(jc x or cA On = x' when x /+ —X x fix) 5), 0) intervals ([1 ] /2, or we have x > x — 3 but they can also + V5 [ 1 2 [ 1 2 —x > x and on the interval These assertions are apparent from the be checked easily, as follows. Since f(x) = g(x) x. — V5 — V5]/2, 0) and ] /2, the function + V5 ]/2); (0, [1 f — g does it is not change sign therefore only necessary to observe, for example, that (-^) 3 -(-^)-(-^) 2 = |>o. 1- I 2 1 = -1 <0, to conclude that f-8>0 /-g<0 The area of the region in question on([l-V5]/2,0), on is + V5]/2). thus r—Z— p0 I (0, [1 (x — x — x~) dx + [x I — (x~ — x)] dx. example reveals, 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 he 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 assumed that area made As this — t 292 Derivatives and Integrals FIGURE 5 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 sense for these regions, to the definition of area The is available, although advanced calculus. book and historically, for the definition really provide the best proper place its is somewhere was the motivation, both desire to define area in in this of the integral, but the integral does not method of defining areas, although it is frequently the proper tool for computing them. It may be discouraging to learn that integrals are not suitable for the very pur- pose for which they were invented, but we The most important other purposes. if / is soon see will how essential they are for use of integrals has already been emphasized: continuous, the integral provides a function y such that /(*) This equation is = /(*) the simplest example of a "differential equation" (an equation The Fundamental Theorem of Calculus says that this differential equation has a solution, / 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, for a function y which involves derivatives of y). if and the integral is one of the best ways of doing this. Since the differentiable functions provided by the Fundamental Calculus will play such a prominent role in later work, realize that these functions still more functions, whose may be combined, 1 1 + very important to by the Chain Rule. Suppose, for example, that -[ is of like less esoteric functions, to yield derivatives can be found fix) it Theorem sin dt. z t ~ The Fundamental Theorem of Calculus 14. Although the notation tends to disguise the fact somewhat, / is 293 the composition of the functions C{x)=x 3 and L-^-dt. fr+ r- = = F(x) j Ja In fact, f(x) — F(C(x)); / = F other words, in sin 1 t C. Therefore, by the Chain o Rule, = fix) F'{Cix))-C'ix) =F f (x 2 -3x ) 2 1 3x + 1 If /is defined, x sin 2 . i instead, as r l fix)-- Jx 3 + 1 sin dt, 2 t then 1 fix) + sin 1 If /is defined 3x 2 -- . 3 x as the reverse composition, fix) = — or f/ + sin i 1 / r then fix) = C'(F( x)) F\x) = 2 1 ' ' <i. 2 + sin" 1 t *)) 1 + sin x Similarly, if /> fix) = sin a i r^T dt / Ja s(*) = 1 r = sin > t l —r + Sin / dt * t 1 ./sin.* h(x) + sin" 1 Ja 1 + sin 2 7 r then f'(x\ \X ) J — 2 1 \X ) Se'(r\ = A , -1 — — 1 /z'OO V-U.3 + sin" (sin x) 2 + cos ' • sin (sin ( / \Ja *) 1 1 + 21 sin \ / / 1 , 1 -2 x + sin 294 Derivatives and Integrals The formidable appearing function yJ - fix) ti 1 -Hsin— 1 J i Ja is / = F also a composition; in fact, + 1 o sin dt 2 t F. Therefore = F\F{x))-F\x) f'{x) 1 1 1 1 As these examples + "Of 1 + siiT which appear on the will x sin t reveal, the expression occurring sign indicates the function + 1 dt sin above right below) the integral (or when / is written as a composition. As a final example, consider the triple compositions (r-^-dt) Ja l+sin" / \ -dt L L +sirT r 1 /(*) = / Ja 1 + sin =/ g(x) dt, f 1 1 t dt. 1 + sin / which can be written f = F o FoC and g may Omitting the intermediate steps (which you we = FoF o F. supply, you if still feel insecure) obtain fix) 1 = 1 . 1 1 + sin' -o-t 1 + sin" x 3 1 + sin' 3x\ • dt + 1 sin t g'ix) 1 + sin" tecM I 1 1 dt 1 + sin r Ja dt 1 + sin" t t I l+sin x Like the simpler differentiations of Chapter 10, these manipulations should be- come much easier after the practice provided by some of the problems, and, like the problems of Chapter 10, these differentiations are simply a test of your under- standing 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 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 continuous function integrable on [«, b\. is we need not quite true. to know that if In order to apply / is continuous on Theorem [«,/?], Although we've already offered one proof of this 1 then to a / is result, there is a more elementary argument arguments, it's quite tricky, but proof of Theorem If / sup{L(/, will both exist, even integral of / on [a will be denoted by has the virtue that if , [a, b], is not integrable. These numbers are and the upper integral of / on [a b] / b] , The lower and upper U and / pb and if m< < c /+L / if a / Ja fix) /. < b, then all jc f integrals. Ja f < M{b - /, Jc a). exercise, since they are quite similar to the The results for integrals are actually a corollary ff-vf will / Ja of the results for upper and lower integrals, because We U an as left pb /+ / pb <U/ Ja corresponding proofs for U then in [a, b], pb proofs of these facts are / = / Ja m(b-a)<L The pc pb U and / Jc M for < and Ja pb pc =L lower respectively, both have several properties which the integral integrals possesses. In particular, called the , pb Ja / Ja a review of the inf {!/(/, P)} pb / will force it then and />)} L L Like most "elementary" prefer. 1. any bounded function on is you might that it 295 The Fundamental Theorem of Calculus 1 4. prove that a continuous function / is integrable precisely when /• integrable by showing that this is equality always holds for continuous functions. / It is actually easier to show that f Ja Ja for all x in [a b] , ; the trick even depend on the theorem 13-3 PROOF If / is to note that is / was fact that continuous on [«,/?], Define functions L and U L(x) then on =L / [a b] , f f most of the proof of Theorem integrable! integrable is on by and U(x) =U f Ja Ja Let x be in (a,b). If h > [a, b\. and mh - inf {f{t) M sup{/(0 h = x < : : x < t t <x + /?}, < x +h), f. 1 didn't . 296 Derivatives and . Integrals then /x+h rx+h /<U/ f<M h -h, so mh < L(x + h - h) L(x) < U(x + - U(x) < h) h M,, or mi, L(x+h)-L(x) < U(x+h)-U(x) < If h < / inf{/(0 Mh = sup{/(f) : x + h < < x), x +h < t < t x], proof of Theorem we have at x, = lim nih this : inequality, precisely as in the continuous is mh = = lim Mi, h^O and f(x), /i-vO proves that = L'{x) This means that there is U'(x) number a U (x = = for f(x) x in (a, b). c such that L(x) ) +c for all x in [a b] , Since U(a) the number c must equal = L(a) = 0, 0, so U(x)= L(x) x for all = L(b) in [a,b]. In particular, U and this means / that is f f = U(b) on integrable =L f f, [a ,b\. | PROBLEMS 1 . Find the derivatives of each of the following functions. 3 (i) F(x)= r* / sin 3 tdt. Ja (ii) F{x) = Jf sixth dt) J / + sin 6 t + t 2 1 (iii) nrdt F(x) J 15 VA +t 2 +sin z 1 1 (iv) . // and one obtains the same Since <M h h F(x) -I dt. 1 + t 2 + sin 2 / ? dy. 1 The Fundamental Theorem of Calculus 14. (v) F{x) dt. -f Jla a (vi) F(x) (vii) F = sin 297 1 + t sin / ( + sin" 2 / sin / ( 3 1 dt c 1 x _, where F(x) , f = 1 -dt. I [Find (F T F -i where F CO = (viii) For each of the following /, f{x)? (Caution: uous at x.) 3. (i) fix) (ii) /(jc) (iii) /(jc) (iv) /(x) (v) /(jc) (vi) f(x) (vii) / (viii) fix) = = = = = = 1 x < 1, f(x) if .v < 1, /(*) if x ^ 1, /(x) if jc is if jc FIGURE 6 if = = = irrational, if x > 1. 1 if x > 1. if jc 1 1 fix) = 1. = \/q f, at which points x — fix), even if x x = shown in Figure io sill / + 1 f \/n for some n in N, fix) ! ho l dt, x +r dt. I vT p/q if if / is is F'(x) in lowest terms. < x < in (0,7r/2). — — not contin- 1. 6. that the values of the following expressions 2 ° t = <0, fix) =x if x >0. or jc> 1, fix) = l/[l/x] x < Wx (ii) — fQ F(x) might happen that F'(x) the function is if if if = Show it dt vT^72 Jo 2. F-'U).) ! / , terms of )'(;0 in otherwise. do not depend on x: 298 Derivatives and Integrals 4. Find (i) (./"-' )'(0) if f(x )= I + sm(sin t)dt. 1 Jo (ii) — f( x ) cos(cosf) dt. / / (Don't try to evaluate 5. explicitly.) Find a function g such that (i) + x2 tg{t)dt=x / Jo . 2 (ii) = x+x 2 tg{t)dt / . Jo (Notice that g 6. (a) Find not assumed continuous is continuous functions all / = / Jo (/(*))" assuming that / has (b) > at on is [a b] Use Problem 13-23 for most one some constant C / 0. on an is and any , satisfying interval (— oo, b] < < < with b, but b. C = Finally, for that 7. +C Also find a solution that non-zero for x (c) / at 0.) b, find a solution but non-zero elsewhere. , to interval [a, b] with a prove that 1 -dx < Isfl '/ 2 3 — " < Find F'(x) i 1 F(.x) if ^ + /l-x i;V 8 8. 7 J\ ! v Jo +x 1 -. v2 v/3 J ^-Y = / - ^r4 xf{t)dt. (The answer perform an obvious manipulation on the 9. Prove that if / f(u)(x-u)du= Hint: Differentiate both sides, *10. Use Problem 9 Find a function supposed to xf(x); you should integral before trying to find F' .) f{t)dt\du. I making use of Problem 8. to prove that f f(u)(x-u) 2 du 11. wo/ continuous, then is / is / = 2 f (f such that f'"(x) ' = IT f(t)dt) duA du 1 / v 1 be easy; don't misinterpret the word + sin x. "find.") 2 (This . problem is 12. A / function (a) If / is periodic, with period / Find a function such that / a periodic g for which it + a) = fix on fix) for [0, a], show that not periodic, but /' is. Hint: can be guaranteed that fix) = fQ is x. all for all b. f f'-7 (b) a, if periodic with period a and integrable is 299 The Fundamental Theorem of Calculus 1 4. Choose g is not periodic. (c) If /' periodic with period a and f(a) is = /(0), then / is also periodic with period a. "(d) Conversely, if /' is periodic with period a = period not necessarily 13. Find fQ l/xdx, by simply then f(a) a), — and / guessing a function / with fix) = y/x, Then check 14. Use the Fundamental Theorem of Calculus and Problem 13-21 result stated in Problem 12-21. 15. Let C\, C and some /(0). the Second Fundamental Theorem of Calculus. lem 13-21. : periodic (with is Ci be curves passing through the origin, as and using with Prob- to derive the shown in Figure 7. Each point on C can be joined to a point of C\ with a vertical line segment and to a point of C^ with a horizontal line segment. We will say that C bisects C\ and Cj if the regions A and B have equal areas for every point on C. (a) (b) and C is the graph of fix) = 2x 2 x > 0, find C2 so that C bisects C\ and CiMore generally, find C2 if C\ is the graph of fix) = x m and C is the graph of fix) — ex" for some c > 1. If C\ — x2 the graph of fix) is x > , , , 1 FIGURE 7 16. (a) Find the derivatives of Fix) (b) Now give a = ff \/t dt and Gix) new proof for Problem = fb * \/t dt. 13-15. 17. Use the Fundamental Theorem of Calculus and Darboux's Theorem (Problem 1 1-60) to give another proof of the Intermediate Value Theorem. 18. Prove that if h is continuous, F(x) / and g are differentiable, = h(t)dt, / and Jfix) then F\x) = - fix). Hint: Try to reduce this to the two cases you can already handle, with a constant either as the lower or h(g(x)) • g'ix) hifix)) the upper limit of integration. 19. Let / be integrable on [a, b], let c Fix) = I Ja be in a f, (a, b), and let <x <b. For each of the following statements, give either a proof or a counterexample. F (a) If / is differentiable at c, then (b) If / is differentiable at c, then F' (c) If /' is continuous at c, then F' is is is differentiable at c. continuous at continuous at c. c. 300 Derivatives and Integrals *20. Let r 1 /(*) — cos = x^O , x 0, Is 21. the function F(x) Suppose [0, 1] that /' = on integrable is *22. and /(0) [0, 1] = 0. page 179. Prove that for all x in we have uy- < l/(*)l Show 0. differentiable at 0? Hint: Stare at f Jq = A" = also that the hypothesis /(0) Suppose that / Prove that for a differentiable function with /(0) is x all needed. Hint: Problem 13-39. is > = and < /' < 1. we have />(/,'')' *23. (a) Suppose G' differential = #(>'(-*)) (*) then there is a • y'(x) number G{y{x)) (**) = g and F' equation = — f. Prove that if fix) for all x in (c) Find what condition y must if y all interval, 1 for all x in this interval. then v satisfies (**), y\x) equations to find +c F(x) Show, conversely, that =1+ some c such that (b) (In this case g(t) the function y satisfies the is a solution of (*). satisfy if = l+x 2 1+yU) and f{t) = + 1 t~.) Then "solve" the resulting possible solutions y (no solution will have R as its domain). (d) Find what condition y must satisfy if -1 y'ix) 1 (An appeal to Problem 12-14 +5[v(jc)] show will 4* that there are functions satisfying the resulting equation.) (e) Find all functions y satisfying y(x)y\x) = Find the solution y satisfying y(0) -x. = — 1. 24. Problem 10-19 we found In f" (b) / 2 3 /f' (f'{x) f"\x) 3 f'(x) 2\f'(x) d). Now suppose that / any function is a constant function. is form f(x) the is Schwarzian derivative that the was for f(x) = (ax + b)/(cx + whose Schwarzian derivative is 0. (a) 301 The Fundamental Theorem of Calculus 14. = + (ax b)/(cx + Hint: d). Consider u — f and apply the previous problem. 25. The lim limit called f x r dx, Determine (b) Use Problem 13-15 Jj if r can you say about Suppose < [0, 26. (or f(x)dx), and f to < — 1. show that Jj On (d) f an "improper integral." (a) (c) denoted by f if it exists, is f, < f(x) g(x) N], then Explain What Hint: > for for all x x > > 0, and that f f exists. Prove that if and g is integrable on each interval also exists. fQ g why exist. \/xdx? Jj that /(jc) l/x dx does not +x f^° 1/(1 )dx exists. Hint: Split this integral Decide whether or not the following improper integrals up at 1. exist. 1 dx. I Jo v/l+.v 3 X L Jo f°° m / —-=^ dx 1 (this is really a type considered in Problem 28). xJ\ +x ./0 27. dx. l+.t 3 / 2 The improper integral f j°_ is defined in the obvious way, as But another kind of improper integral f_ it is Jq /+ provided these /.qo /, f improper (a) Explain why f_ 1/(1 + x~)dx (b) Explain why J^xdx does not is lim f° f. defined in a nonobvious way: integrals both exist. exists. exist. (But notice that lim f_ x N dx does exist.) (c) Prove that Show f™ f exists, then moreover, that lim f_ N J-oo /• can't, if Can you you will state a lim J_ N /"and lim f exists f_ n2 f and equals both exist to do f. and equal reasonable generalization of these facts? have a miserable time trying f™ (If you these special cases!) / 302 Derivatives and Integrals 28. There is another kind of "improper bounded, but the function is unbounded: (a) > If a lim f" l/^/xdx. This limit 0, find = even though the function f(x) matter how we define /(0). (b) (c) Find ( JV </* integral" Use Problem 13-15 is interval denoted by f^ l/^/xdx, not bounded on is is [0, a], no <r <0. -1 if \/y/x which the in to show _1 that /q jc dx does not make sense, even as a limit. (d) Invent a reasonable definition of r dx for a — x~)~ < Invent a reasonable definition of j_\(\ limits, exist? and show How does +x) (1 ^~ Why that the limits exist. Hint: _1//2 ] compare with (1 (a) If / is continuous on *(b) If / is integrable on [0, 1], [0, 1] compute lim x and continuous lim x I *^0+ Jx 30. It is —^2 dx, as a does /_](1 2 _1//2 —jc ) /"' 29. and compute it for <r <0. -1 (e) \x\ f for sum of two + x)~ 1 dx ^ —1 < x < 0? fit) dt. I at 0, compute dt. t combine the two possible extensions of the notion of possible, finally, to the integral. (a) If f(x) = 1/Vjc for < x < 1 and f(x) = l/x 2 for x > 1, find poo f(x)dx I Jo (b) this should mean). poo Show that < and r deciding what (after at oo; for r r x dx never makes / r < — —— 1 1 . sense. (Distinguish the cases In one case things go wrong wrong places.) things go at both at 0, in the — 1 < other case CHAPTER THE TRIGONOMETRIC FUNCTIONS The definitions of the functions sin one might suspect. For intuitive definitions, and cos are considerably more subtle than some informal and reason, this chapter begins with this which should not be scrutinized too soon be replaced by the formal definitions which we In elementary geometry an angle common FIGURE More pairs point (Figure they shall simply the union of two half-lines with a 1). 1 useful for trigonometry are "directed angles," (l\, FIGURE 2 If for l\ is initial is carefully, as really intend to use. h) of half-lines with the same we always choose initial which may be regarded the positive half of the horizontal axis, a directed angle described completely by the second half-line (Figure 3). Since each half-line intersects the unit circle precisely once, a directed angle described, even FIGURE I more point (x, v) with x 303 simply, + y2 = as point, visualized as in Figure 2. 1. by a point on the unit circle (Figure 4), that is, is by a 304 Derivatives and Integrals The and cosine of a directed angle can now be defined as follows (Figure 5): is determined by a point (x, y) with x + y 2 = 1; the sine of the defined as y, and the cosine as x. sine a directed angle angle is we Despite the aura of precision surrounding the previous paragraph, and yet finished with the definitions of sin What we have to define this defined is depends on associating an angle "measure angles in degrees." an angle "half-way around" to 90, etc. (Figure 6). An 4 of 360 degrees, and of 90 degrees function, The is The which we ambiguity this also There are two mean by an denote by = difficulties example. Even that this system, to elegant results is to associated to 360, an angle "a quarter way around" manner, this degrees + 90 degrees, is number to the same the x, is as the angle One etc. can now define a sin°, as follows: sine of the angle of with is purposely extended further, so that an angle is an angle of 360 will what we want usual procedure for doing way around" 80, 1 angle; are not barely begun. number. The oldest method angle of this x degrees. approach. Although angle of 90 or 45 degrees, for is, to every angle associated, in sin°(x) degrees The angle "all the associated to is called "the angle of x degrees." FIGURE and cosine of a directed the sine sin* and cos* for each number x. is we have Indeed, cos. it is it may be clear what we not quite clear what an angle of v 2 if this difficulty could be circumvented, it is unlikely as it does on the arbitrary choice of 360, will lead would be sheer luck if the function sin° had mathematically depending — it pleasing properties. "Radian measure" appears number x choose , a point 5 remedy for both these the unit circle such that x the length of the and running counterclockwise The P directed angle determined by radians are identical. A r 2tt is = sin' can now be 360 = sin r x = 2ttx sin soon drop the superscript function which will interest us; before The 7). +x and the angle of 2n x radians. sine of the angle of sin°; since we want to have 7TX = sm T80 360 shall (Figure 2n, we can define sin We P defined as follows: This same method can easily be adopted to define sin° to called "the angle of x radians." Since the the angle of x radians , function sin (x) is Given any defects. is arc of the circle beginning at (1,0) length of the whole circle K,L RE T 1 to offer a P on r in sin' we , since sin do, a few r (and not sin ) is words of warning are the only advisable. r expressions sin° x and sin x are sometimes written sinjc sin.v radians, I K. i RE 6 but this notation is quite misleading; a carry a banner indicating that ii is number x is simply a number "in degrees" or "in radians." it If the does not meaning 15. of the notation "sinx" in is doubt one usually x "Is ngth x but what one means in asks: you mean or sin sin r Even for mathematicians, addicted to precision, these able, were Although the function sin' . remarks might be dispens- not for the fact that failure to take them into account will lead to it sin' is Our proposed example the function (and use exclusively henceforth), there of 305 degrees or radians?" incorrect answers to certain problems (an 7 Trigonometric Functions is: Do FIGURE The definition given in Problem which we wish to 19). denote simply by sin a difficulty involved even in the definition is depends on the concept of the length of a curve. Although the length of a curve has been defined to reformulate the definition in is in several problems, it is also easy terms of areas. (A treatment in terms of length is outlined in Problem 28.) length x Suppose that x is the length of the arc of the unit circle from (1, 0) to P; this arc thus contains x/2jt of the total length 2jt of the circumference of the unit circle. Let S denote the "sector" shown in Figure horizontal axis, and the half-line through x/2jt times the area inside the unit area circle, bounded by the unit circle, the and P. The area of S should be which we expect to be jt; thus S should 5 8; is (0, 0) have area x x 71 2t7 FIGURE We 8 can therefore define cos x and sin = 2- x as the coordinates of the point P which determines a sector of area x/2. With these remarks and cos now more as The begins. background, the rigorous definition of the functions first definition identifies precisely, as twice the area jt sin as the area of the unit circle- of a semicircle (Figure 9). DEFINITION (This definition is onometric functions area it will be necessary to < x < jr.) The second definition is meant to the sector bounded by the unit circle, (*, V 1 — x2 ). If < * < the area of a triangle FIGURE not offered simply as an embellishment; to define the 1, this define siiut first and —1 < x < 1, horizontal axis, and the describe, for the area A(x) of the half-line through area can be expressed (Figure 10) as the and the area of a region under the unit 9 Vl -a 2 trig- cosjc only for vT t 2 dt. circle: sum of 306 Derivatives and Integrals This same formula happens to work for —1 <x < also. In this case (Figure 1 1), the term Xyfl negative, is and represents the area of the which must be subtracted from triangle the term FIGURE iy \-t 10 DEFINITION If - 1 <x < 1 if — 1 xfT-x 2 = < x < 1 then , A is A'W = [^ t 2 dt. differentiable at x -2x \ and (using the Funda- -/T1^ + ^l-A- 2 2y/\-x 2 -x 2 + (\-x 2 J\-x I + mental Theorem of Calculus), (x,Vl-x 2 ) FIGURE dt. then , A(x) Notice that 2 ) Vl-x 2 2 1 -x 2 2>/l 1 -2.v 2 -2(l 2s/\-x -x 2 ) 2 -1 2>/1-jc 2 Notice also (Figure 12) that on the interval [— 1, from „ Al.(-1) I I < . 1 to K E 12 A(l) that its For P = = 0. 1] the function If | This follows directly from the definition of A, and also from the derivative is negative on (—1, < x < n we wish to on the unit (cosx, sin x) < X < decreases i = 0+/' 7l-f2^ = define cos x circle 7T, then cos X is and sin x as the coordinates of a point which determines a sector whose area the unique number A (cos A in [- -1, 1] X and sin x = fact 1). (Figure 13). In other words: DEFINITION A vl — (cosjc) 2 . such that is x/2 . 77z£ Trigonometric Functions 307 This definition actually requires a few words of justification. In order to know 75. that there and that is A a number y takes Value Theorem on the values is crucial, if Having made, and justified, THEOREM l < If x < jt, = x /2, we use the fact that A A(y) satisfying and to then = — sinx, = cos*. sin'U) B = If continuous, make our preliminary definition precise. our definition, we can now proceed quite rapidly. we want cos'(jc) PROOF is tt/2. This tacit appeal to the Intermediate = 2 A, then the definition A(cosx) x/2 can be written = B(cosx) in other words, cos is We just the inverse of B. = A\x) x; have already computed that 1 ' 2V\-x 2 P = (cosjc, sinx) from which we conclude that x area — 1 = - B'(x) 2 ' Vl-x 2 Consequently, FIGURE (#)'(*) cos' (*)= 1 3 1 B'(B-Hx)) 1 v^l - [^U)] 2 = —V — = — sin x (COSX) 2 1 Since sin we x =v — 1 (cos*) 2 , also obtain . 1 , sin (x) = - — 2 cos* VI - 1 • — cos'C*) (cosx) 2 cos x sin x sinx cosx. The information contained in | Theorem 1 can be used to sketch the graphs of , 308 Derivatives and Integrals sin and cos on the interval [0, = cosj unique y for a Since . = — sinjc < cos'(jt) the function cos decreases tt] = from cosO To in [0, tt]. 1 < x < 0, =— to cos7r we find y, A(cosx) 1 tt, (Figure 14). Consequently, note that the definition of cos, =— 2 means that A(0) = |, so dt. Jo It is easy to see that /o J\-t 2 dt = FIGURE r \ Vi-t 2 dt / 15 so we can Now we also write have = sin (x ) cos x I so sin increases on [tt/2, tt] to sin7r The [0, tt/2] from sinO =0 (Figure > 0, < 0, < x < tt/2 f \ = <x < tt/2 to sin7r/2 = TT, 1, and then decreases on 15). values of sin x and cos x for x not in [0, tt] are most easily defined by a two-step piecing together process: (1) If tt < x < 27r, then sin = — sin(2;r — x), — cos(2tt — x). sinx cosjc Figure 16 shows the graphs of sin and cos on (a) (2) If x — 2nk + x' for some integer k, cos sin* cosjc 2tt (b) FIG! RE 16 the functions sin and some x' in [0. 2tt], now and cos to defined on all R, we must of R. now check that the basic properties of these functions continue to hold. In most eases this example, it is clear that the equation sin"" x then = sin a', = cosx'. Figure 17 shows the graphs of sin and cos, Having extended [0, 2tt]. + cos x = 1 is easy. For ) . 15. The Trigonometric Functions 309 (b) FIGURE 17 holds for all x. It is also not hard to prove that sin' (a) cos' (a) if x is not a multiple of tt. = cos a, = — sin For example, sin a = if A', n < x < 2n, then —sin (2jt —a), so sin'(A) If a rem is a multiple of n we resort to a trick; 11-7 to conclude that the I'KiURK 18 = — sin'(27r — a) = COS(27T — A) = COS A it is (— • 1 only necessary to apply Theo- same formulas are true in this case also. 310 Derivatives and Integrals The other standard trigonometric functions present no difficulty We at all. define 1 = secx COSJC tan x = cos -1 x + kix x ^ kir. sinx + 7T/2, cos* cscx — 1 = x sin COSJC sinx The graphs t 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 which are functions, listed in the statement of the theorem, since the results theorem FIGURE 2 If x £ kit + tt/2, = = sec' (jc) tan'(jc) x ^ kjr, 1- - The y you sec x. = — cscjc cotx, = — csCx. cot'(;c) Left to sec x tan x, then csc'OO PROOF is then 19 If no need to memorize the can be rederived whenever needed.) next theorem (there (a straightforward computation). | The inverses of the trigonometric functions are also easily differentiated. trigonometric functions are not one-one, so it is first necessary to restrict to suitable intervals; the largest possible length obtainable is jt, and the them intervals 1 usually chosen are (Figure 19) 1 -71 rt 2 .-^ y-\- 2 [— 7r/2, 7r/2] for cos, [0, 7r] (— tt/2, for sin, tt/2) for tan. (The inverses of the other trigonometric functions are so rarely used that they will not even be discussed here.) an The sin inverse of the function f(x) is — denoted by arcsin (Figure sin 20); the has been avoided because arcsin one), but of the restricted function /; I M. I RE 20 and tt/2 sin x, arcsin by Sin . <x < domain of is tt/2 arcsin is — | 1, lj. The notation not the inverse of sin (which sometimes this function / is is not onc- denoted by Sin, The 15. The inverse of the function = g(x) is denoted by arccos (Figure is denoted by Cos, and arccos The JT arccos is [— 1, Sometimes g 1]. . = tan x The is < x < -jr/2 , 22); arctan is tt/2 one of the simplest examples of a bounded even though it is one-one on denoted by Tan, and arctan by Tan" which Sometimes the function h 2 domain of by Cos" 21); the denoted by arctan (Figure differentiable function FIGURE <X < COS*, inverse of the function h (x ) is 311 Trigonometric Functions all is of R. . derivatives of the inverse trigonometric functions are surprisingly simple, and do not involve trigonometric functions at all. Finding the derivatives is a simple matter, but to express them in a suitable form we will have to simplify expressions 1 like sec (arctan*). cos(arcsinx), FIGURE A little ple, 2 2 picture is the best way to remember the correct simplifications. For Figure 23 shows a directed angle whose sine is x — angle of (arcsinx) radians; consequently cos(arcsin x) side, namely, v 1 — x 2 . However, in the 2 3 THEOREM 3 If —1 < x < 1 , then arcsin'U) = 1 J\-x 2 arccos' (x) = -1 ' Vl-x 2 Moreover, for all x we have arctan' (x) = is shown is \+x 2 exam- thus an the length of the other proof of the next theorem we resort to such pictures. FIGURE the angle will not 312 Derivatives and Integrals PROOF = arcsin'(x) (/ )'{x) 1 f'if-Hx)) l sin'(arcsin x) cos(arcsinx) Now + [sin(arcsin x)\ that [cos(arcsinx)]" = 1, is, + x = [cos(arcsinx)] 1; therefore, = cos(arcsin x) (The positive square root cos(arcsinx) > 0.) x2 . first any proof for the The difficulties. arctan'(x) is in (—7t/2,7t/2), so formula. has already been established also possible to imitate the that proof presented — 1 be taken because arcsinA' to This proves the The second formula It is is v = first (in Theorem the proof of formula, a valuable exercise third formula is proved 1). if as follows. (h~ )'(x) 1 h'(h- l (x)) 1 tan' (arctan x) 1 2 sec (arctan x) Dividing both sides of the identity sin by cos a + a + cos" a = 1 yields tan It a 2 = 1 sec" a. follows that [tan (arctan*)]" + 1 = sec (arctan .v). or x which proves the third formula. The traditional given here) is + 1 = sec (arctan x), | proof of the formula sin'(x) = cosx (quite different outlined in Problem 27. This proof depends upon from the one first establishing The 15. 313 Trigonometric Functions the limit h sin hm = 1. h h-+0 and the "addition formula" sin(x + y) = sin Both of these formulas can be derived are known. The first is + cos x siny. x cos y now that easily just the special case sin'(O) = the derivative of sin cosO. and cos The second depends on a beautiful characterization of the functions sin and cos. In order to derive this we need a lemma whose proof involves a clever trick; a more straightforward result proof will be supplied LEMMA in Part IV Suppose / has a second derivative everywhere and that + f = 0, = 0, /'(0) = 0. f" /(0) = 0. Then / PROOF Multiplying both sides of the first equation by /' yields f'f" + ff' = 0. Thus 2 , [(/ so (f) 2 + f2 the constant is is ) + / 2 = 2(/7" + //') = 0, , ] a constant function. From /(0) = and /'(0) = it follows that 0; thus !-'/„vi2 [f(x)Y+[f(x)] 2 = for all*. This implies that f(x)=0 THEOREM 4 If / for all*. | has a second derivative everywhere and r+/= /(0) ./" (0) = = o, a, 6, then /=b (In particular, if then / = cos.) /(0) = and /'(0) = sin +a 1, then cos. / = sin; if /(0) = 1 and /"(0) = 0, . 314 Derivatives and . Integrals proof Let g{x) = fix) = = fix) — bcosx f"(x) + bsinx — bsinx — a cosx. Then g'ix) g'ix) + a sinx, + acosjc. Consequently, = = g(0) g'iO) o, 0, which shows that = THEOREM 5 If lix) = a cosx, for x. | all x and y are any two numbers, then + y) = + y) = sin(x cosix PROOF — bsinx — fix) For any particular x cos y + cos x cos x cos y — sin jc sin number y we can fix) sin y sin y define a function = , / by + y). sin(x Then = cosU +y) fix) — -sin(x + y). f'ix) Consequently, +/= /(0) = /'(0) = f" It follows from Theorem 4 siny, cosy. that / = that 0, (cos y) • sin +(sin y) cos; • is, sin(x Since any + = y) cos y sin x + sin y cos number y could have been chosen formula for all x and for , all x to begin with, this proves the first y. The second formula As a conclusion jc is proved to this chapter, an alternative approach similarly. | and as a prelude to to the definition 1 for arcsin'(x) nTr / v Chapter of the function — 1 sin. < x < 1, 1 8, we Since will mention . . The 15. it follows Trigonometric Functions from the Second Fundamental Theorem of Calculus fX = arcsin x — arcsin x = arcsin This equation could have been taken as the 315 that 1 -dt. / h J\~-t 2 definition of arcsin. It would follow immediately that . , arcsin (x) 1 = ; Vl-x 2 the function sin could then be defined as (arcsin) tive -1 and the formula for the deriva- of an inverse function would show that sin'(.v) = v 1 — x, sin one could show that A(cosx) = x/2, recovering at the very end of the development the definition with which we started. While much of this presentation would proceed more rapidly, the definition would which could be defined be utterly as cosa". Eventually, unmotivated; the reasonableness of the definitions would be whom the author, but not to the student, for we shall see in Chapter an approach of this 18, indeed. PROBLEMS 1. Differentiate each of the following functions. (i) f(x ) (ii) f(x) (iii) f(x) = = = arctan (arctan (arctan x / (iv) = f(x) ) ) arcsin(arctan(arccosx)). arctan (tan x arctan x ) \ 1 arcsin v/T+^2 Find the following sin.t limits by l'Hopital's Rule. — x + * 3 /6 lim x3 sin n x lim x4 .0 cosx in — x + x 3 /6 — x ( lv ) h m cosx — + x/2 1 . X* arctan x — lim lim x ; x-+0 fvi) 2 4 ,v^0 (v) + x/2 1 lim .V [ — \ x sin 3 x + x 3 /3 known to was intended! Nevertheless, as sort is sometimes very reasonable it — 316 Derivatives and Integrals a sin 3. Let f{x) = x , #0 x=0. 1 (a) Find/'(0). (b) Find/"(0). At this point, you will almost certainly have to use l'Hopital's Rule, but in Chapter 24 we will be able to find f'iO) for all k, with almost no work at 4. all. Graph the following functions. (a) f(x) (b) f(x) = sin2x. = sin (a 2 (A pretty respectable sketch of ). tained using only a picture of the graph of your only hope is of / cos(a = Indeed, pure thought sin. problem, because determining the sign of the -2x ) The formula directly. however (c) = fix) derivative in this no is easier than determining the behavior fix) does for should be clear in fix) = sin a + graphs of gix) sin* and axes, from easily find 0, sin 2a. = (It how many = /z(a) sin even, critical points / sum points by finding out the sign of / at draw first [0, the set of You can will look like. has on fact, and which 2x carefully on the same You can then determine the derivative of /. is probably be instructive to will 2n, and guess what the to out one important indicate which must be true since / your graph.) f'(0) graph can be ob- this 2n] by considering the nature of these critical each point; your sketch will probably suggest the answer.) fix) (d) — — x. tan a fix) = n/2, moved up same, except (e) (First — the intervals ikn — x. sinjr determine the behavior of / in (— n/2, n/2); in kit + n/2) the graph of / will look exactly the a certain amount. (The material in the Why?) Appendix to Chapter 1 1 will be particularly helpful for this function.) x sin fix) (0 X * 1, = 0. (Part (d) should enable you to of /' are located. Notice that the size of f(x) (g) The = x / sin to / is even and continuous at 0; also consider for large x.) .v. hyperbolic spiral nates (Chapter 4, its determine approximately where the zeros is the graph of the function f(0) Appendix behavior for 3). Sketch close to 0. Prove the addition formula for cos. this = a/0 in polar coordi- curve, paying particular attention , . The 15. (a) From (b) Use and cos derive formulas the addition formula for sin cos2jc, sin3x, Trigonometric Functions 317 for sin2x, and cos3x. these formulas to find the following values of the trigonometric func- tions (usually deduced by geometric arguments elementary trigonom- in etry): —= cos 4 4 re tan V2 — = ~2~' 71 71 . sin —= . 1 4 7X . 1 Sm = 6 V3 7X COS 8. (a) Show A that sin(x + r 6=^- B) can be written as a sinjc + b cos x for suitable a and b. (One of the theorems in this chapter provides a one-line You should also be able to figure out what a and b are.) (b) Conversely, given a and b, find bcosx 9. = A sin(.x + (c) Use part (a) Prove that (b) to B) for all + y, v) + tan y tan x = + and x + x. 1 provided that x, that a sin x = V3 sin x + cosx. graph f(x) tan(x numbers A and B such proof. — tan x tan y y are not of the form kit + 7r/2. (Use the addition formulas for sin and cos.) (b) Prove that x ' arctan x + arctan y = arctan 1 and y by arctan y 10. — xy y. Hint: (a). Prove that arcsin a + arcsin indicating any restrictions 11. in part v on x and indicating any necessary restrictions arctan x + Prove that if m and a? fi — on arcsin(av or and 1 — fi~ + ^v 1 —a /?. are any numbers, then mx sin «jc = sin mx cos «jc = = cos mx cos sin a?.v j[cos(w — ri)x — cos(w +/?)-*], + n)x + sin (aw — n)x], — «)jc j [cos(m + n )x + cos(m ^[sin(m ] ), Replace x by , 318 Derivatives and ' Integrals Prove that 12. if m and n are natural numbers, then sin i: cos sin These /// 7r, m = n. 7^ n, /; 0. / If only in the Suggested Reading this topic will receive serious attention (see reference [26]), the (a) 0, n relations are particularly important in the theory of Fourier series. Al- though 13. mx cos nx dx = mx cosnx dx = n, m ^ m = 0, mx sin nx dx = is next problem provides a hint as to their importance. on [— n, integrable r show tt], (f(x) minimum that the value of — a cosnx) dx J —It occurs when a i r * J-TT — —\ f(x) cosnx dx, and the minimum value of — a s'mnxy dx (f(x) J —TT /; when * (In r = — TT / Show = J obtaining a quadratic r =— m * that f(x) cosnx dx, n —0, f(x)s'mnxdx, n = I + 1,2,3 J-tt and d are any numbers, then if c, t -iv fo 1,2, ... J-TT i bn = sign, a.) i ( nxdx. Define an £ sin each case, bring a outside the integral expression in (b) f(x) J-TT 2_. c " cos [f(x)fdx-2iz( nx + d„ sin 2 dx nx + ^2 °» c » + b » d» I +n co [ ~y~ 2 + 5Z r " + ^" n=] j\f {x) f dx -J'^-+jr a,l+l,,r +n ((vl " 7i) + ^ "" (r " " )2 + h {d" ~ "j . . . The 15. thus showing that the among all In other words, sinnjc and = cn (x) integral first cosnx 319 Trigonometric Functions smallest is . when a { = c and b t { = dj. "linear combinations" of the functions s n (x) for 1 < < n TV, = the particular function N = g(x ) — + v^ > a„ cos nx + b„ ao sin nx n=\ has the "closest 14. Find a formula for (a) x sin — sin v.) good does 15. sin x + sin y. tt]. (Notice that this also gives a formula for + b) + sin(a — b). What Hint: First find a formula for sin(a that do? + cos y — and cos x Also find a formula for cos x (a) Starting from the formula for cos 2x derive formulas for sin x (b) Prove that cos v , and cos jc terms of cos2x. x = COS 2 <x < for /l+cosx x . V-— and Sm = 2 — I— V- cos* tv/2. (c) Use part (d) Graph f(x) (a) to find fa = x. sin x dx and f cos 2 x dx sin Find sin(arctanx) and cos(arctanx) as expressions not involving trigonometric functions. Hint: y sin 17. / on [— n, (b) in 16. to fit" If v/v = x 1 — sin = arctanx means that x = tan v = sin v/ cos v = v. tanw/2, express sinw and cosw in terms of x. (Use Problem 16; the answers should be very simple expressions.) 18. (a) graphs of 19. (b) What is (a) Find / sin ~dt. Hint: (All along were the we have been drawing the case.) The answer is not 45. l jdt. / l+f Jo . 1 — 20. Find lim x 21. (a) Define functions (b) Find lim sin and cos by sin°(x) = sin(7rx/180) and cos°(x) = cos(7rx/180). Find (sin )' and (cos )' in terms of these same functions. x-s-0 22. cos*. as if this l+t 2 r°° Find and cos arcsin(cosx) and arccos(sinx)? Jo (b) + n/2) — Prove that sin(x sin° and lim x X sin° Prove that every point on the unit least one (and hence for infinitely x .v->-oo circle is of the form (cos many) numbers 0. 0, sin 0) for at . 320 Derivatives and Integrals 23. (a) Prove that which [2kn (b) sin - tz 25. Prove that + tt/2] or g(x) = sin a tt/2, 2kit = Let f{x) graph. maximum the | let < x < sec a for — x sin statement, with length possible of an interval < — replaced by < x for — (Ikn in y\ for all + 1 )n tt/2, numbers x - 2kn / / tt/2] + _1 y. . tt/2). What and sketch *26. It is to < an excellent The same Hint: f" sin fix) / get the right idea for a proof the (b) A.jc dx limit can easily to intuition, but be established for once you any inte- /. Show that lim Show that lem (c) to of intuition to predict the value of test Continuous functions should be most accessible (a) its . lim grate is a very straightforward consequence of a is , 2(k Find the domain of tt. \x < sin y\ + tt/2, [2A-jr well-known theorem; simple supplementary considerations then allow < be improved on one-one, and that such an interval must be of the form is Suppose we 24. is if s f is sin = by computing the integral 0, explicitly. a step function on [a b] (terminology from Prob, 13-26), then lim Finally, use kx dx f s sin (jc ) Problem 13-26 Xx dx show to = that 0. lim f fix)sir\Xxdx — for any function / which is integrable on [a, b]. This result, like Problem 12, plays an important role in the theory of Fourier series; it is known as the Riemann-Lebesgue Lemma. 27. This problem outlines the The shaded (a) By approach classical sector in Figure 24 has area a/2. considering the triangles OCB prove that OAB and x — 2 2 cos x then fi = sin (l,0) x < — < 2 FIGl RI". 24 (b) Conclude sin < sm < A = 1. A *-^0 Use x that hm (c) sin a that cosjc and prove to the trigonometric functions. this limit to find 1 lim — COS A 1. if < x < 7r/4, . The 15. (d) Using parts starting ^28. and (b) from the and the addition formula (c), for sin, find sin'(x), definition of the derivative. This problem gives a treatment of the trigonometric functions in terms of = yl — x 2 and uses Problem 13-25. Let f(x) Define %(x) to be the length of / on [x, 1]. length, Show (This is an improper integral, as defined in then prove that i£(x) (b) Show x < Define Problem 14-28, so you must the length on [x, 1 — s] and + the limit of these lengths as e -> is =- as i£(— 1). tt = define sin* v for , < x < For — 1 < x < cos x for .) cos 2 - < x < 1 define cos* by i£(cosx) tt, Prove that cos'(x) .v. 1 = — sin* and = x, tioned in the text — this = tt. starting with inverse functions defined convenient to begin with arctan, since To do and sin'C*) Yet another development of the trigonometric functions was briefly is 1. that £'{x) (c) —1 < —L=dt. prove the corresponding assertion for first for that ££(*)= f ^29. 321 Trigonometric Functions function this is by men- integrals. defined for all It x. problem, pretend that you have never heard of the trigonometric functions. Let <x(x) (a) = lim a(x) and x-+oo we (b) (c) define + f") -1 Jq{\ tt dt. = is odd and increasing, and that and are negatives of each other. If Prove that a lim a(x) both x—*— oo exist, 2 lim a(x), then oc~ X—>00 l is defined on (—tt/2, tt/2). Show that {a~ )'{x) = 1 + [cy-'U)] 2 For — tt/2 < x < tt/2, define tan.v = a~ l . tan;c/\/l + tan 2 x. lim (i) sin x Show = [ (x), and then define sinx = that 1 x—>n/2- lim (ii) sin x =— 1 X^-7t/2+ sin* 111 sin'Cr) tt/2 = tan ^30. If we sin"(x) tt/2 and x ^ x=0 1, IV < x < jc = — sinx for — tt/2 < jc < tt2. are willing to assume that certain differential equations have solutions, another approach to the trigonometric functions particular, that there satisfies yo" + vo = 0. is some function yo which is is possible. not always Suppose, in and which : 322 Derivatives and Integrals (a) Prove that (b) + yo°- ^ or yo'(0) Prove that there = s'(0) If we — = 0, become There trivial. x > all 0. problem Find cos7r, is = and cos' to decide why (a) After > a smallest xo is ? = we have 1, with = sin7r/2 ±1; positive.) is sin 27T. (Naturally you may use any addi- can be derived once we know that the sin are periodic work involved certing to find that sin is prove that cosx cannot to 1 1 = sin' cos —sin.) Prove that cos and (There about trigono- 1 1 +cos sin tt/2 and and 2*o2 (Since sin 1. sin7r, cos 2tt, (f) all — tt all facts = one point which requires work, is most easily done using an is Chapter to and s(0) This follows that there It tion formulas, since these = tt. Chapter to and we can define Prove that sin7r/2 the / that either yo(0) = s then almost s', Use Problem 6 of the Appendix cos*o 31. — and cos s + + byo'. s satisfying s" form ayo producing the number be positive for (e) a function from the Appendix exercise (d) is — define sin however and conclude constant, Hint: Try s of the 1. metric functions (c) is (joO 0. with period 2tt. of sin, in the definition it would be discon- actually a rational function. Prove that is it isn't. a simple property of sin which a rational function cannot pos- sibly have.) (b) Prove that is, sin isn't do not there + (sin x)" even defined implicitly by an algebraic equation; that exist rational functions /o, /„_! UMsinx)"- = Hint: Prove that /o |: 32. is x. Suppose that (a) (b) that #2 Show Show • • +f so that sin x can (Why?) You are (f>\ and now > set up . (x) . , /„_] such that = for all x. be factored tt. for a But out. this The remaining implies that proof by induction. satisfy (j>2 0l" 02 " and • except perhaps at multiples of factor for all 5 + 1 . +£101 = 0' = 0' + 5202 8\- that that 01 "02 - if (p\(x) > / 02"0l - (£2 - and 02 U) > [0| "02 - £l)0102 = for x all 0. in (a, b), then 02"0l| > 0, Ja and conclude [4>l(b)4>2(b) that - 0i / (a)02 (a)] - \4>\(b)<f> 2 '(b) - 0,(«)02'(«)] > 0. it is The 15. (c) Show we cannot have that in this case and sider the sign of 4>\'{a) (d) Show that 02 < < 02 > 0, 4>\(a) = = (p\(b) 0. Con- Hint: <p\'{b). the equations 0i (a) or 0i 323 Trigonometric Functions = 0i (b) < 0, or 0i = are also impossible 02 < 0, on (a, b). if 0i > 0, (You should be able to do this with almost no extra work.) problem may be stated as follows: if a and b are consecutive zeros of 0i, then 02 must have a zero somewhere between a and b. This result, in a slightly more general form, is known as the Sturm Comparison Theorem. As a particular example, any solution of The net result of this the differential equation must have 33. (a) at least (x + one zero in any interval Using the formula sin (A' (b) Conclude + for sin ^)x — a: — sin(/c l)y — + («7r, (n 1 )tt). show derived in Problem 14, sin v j)x = 2 sin — that coskx. that 1 — =Q + y" + cos x + cos 2x + + cos nx = • sin(n + l)x . x 2 2 sin — 2 Like two other results in in the study of Fourier this series, problem set, this equation is very important and we also make use of it in Problems 19-43 and 23-22. (c) Similarly, derive the formula sin sin x + sin 2x + + sin nx I -y-A-jsin(-.rj = . . sin x — 2 (A more natural derivation of these formulas will be given in Prob- lem 27-14.) nb (d) Use parts (b) and (c) to find / Jo the definition of the integral. rb sin x dx and / Jo cos x dx direcdy from *CHAPTER ^T I IRRATIONAL 71 IS This short chapter, diverging from the main stream of the book, demonstrate that ics. we are already in a position to This entire chapter many "elementary" is devoted to an elementary proof that it is proofs of deep theorems, the motivation for proof cannot be supplied; nevertheless, is still it is included to do some sophisticated mathematirrational. Like many steps in our quite possible to follow the proof step-by-step. Two observations must be made before the proof. The first concerns the func- tion -A MX)= nl which clearly satisfies < f„(x) < - forO < < jc 1. n\ An important property of the function /„ obtained by actually multiplying out x"(\ will be n and the highest power is HI where the numbers c, are integers. —x) n The be 2n will It is /„(*)(())= revealed by considering the expression . Thus . power of 1 from <nork this expression that >2n. Moreover, (x) fn — — [n\c + n terms involving x] n\ /;/" + l| an fn 324 (x) = -[(« + l)!c„ + \x)=- [(2n)\c2n y ]. jc i + terms involving x] appearing in the t— clear it'k /„ lowest can be written form 16. n is Irrational 325 This means that (n) (0)=Cn fn /n (n+1) fn (0) = (n (0) = (2n)(2n {2n) where the numbers on the + l)cn+1 - 1) • . . right are all integers. {k) fn The , (0) is an • . + (n l)c 2n , Thus integer for all k. relation =/n (l ~X) fn(x) implies that /B (k) = (x) (-i)*/« w (i-*); therefore, (1) /« The proof positive that tt is number, and e an also is irrational requires > this, notice that n if r" (n Now let hq be any natural one further observation: then for sufficiently large n 0, — To prove integer for all k. > will <£. a n 1)! a + \ n n\ number with no > 1 a" 2 n\ < Then, whatever value 2a. r ,"0 a (no)! may have, the succeeding values satisfy fl («o+D 2" (^)! («o+2) (n + 2)\ (n + k)\ a »o 1 < (no+D! fl fl J < — 2 <" • (n 2* «"° If /: is so large S that < 2 » then , (no)! 8 have 2a, then +1 + we if («o+D (n )!' 1 1 r— < — +l)! 2 • — 2 a "o (n )! a is any 326 Derivatives and Integrals +k) ,(n which THEOREM 1 one theorem in this chapter. The number tt of irrational; in fact, is implies the irrationality of tt~ e, k)\ Having made these observations, we are ready the desired result. is < + (no tt, irrational. is jr for if were tt for the (Notice that the irrationality rational, then tt certainly would be.) PROOF Suppose were tt~ rational, so that 2 TT a = b for some G(x) (1) and positive integers a = b"[TT 2 "fn (x) ~ 2 TT b. Let 2 "- 4 + "-\fn "(x) TT fn {4) ( X) - (- D" fn^ix)]. + Notice that each of the factors b"TT is an integer. Since fn 2 "- 2k (k) (0) = 2 k b"(TT )"- = (|V~ - b" and fn ik \l) are integers, this G(0) and G(l) are Differentiating G"{X) (2) The last G = b"[TT 2 G"(x) Now k shows that integers. twice yields "fn "{x) - term, (—l) n fn (3) k a"- b 2 "- 2 TT (,x), is fn zero. + tt 2 G(x) = b"TT {4) ( X) + (-\Tfn a " + 2) (x)]. + Thus, adding (1) 2n+2 2 fn (x) = TT and (2) gives a"f„(x). let — — ttG(x) cos7r.v. = ttG'(x) cos ttx + G"{x) sinTr.r — = [G"U) + 2 GU)]sin7TJc = tt a n fn (x)sinTtx, by (3). 7rG'U)cos7r;t H(x) G'(x) sin7r.v Then H'{x) + tt"G(x) sin ttx 77- By the Second Fundamental Theorem of Calculus, / Jo a" 2 tt /;, (.v ) sin ttx dx = = = // ( 1 ) - H (0) G'(l)siii7r 7r[G(l) + -ttGCDcostt - G'(0) sinO G(0)]. Thus * f a"fn (x) Jo sin ttx dx is an integer. + ttG(0) cosO — 16. On < the other hand, < < f„(x) Tta" fn {x) < x < \/n\ for 327 so 1, jza" < sin7r.v Irrational it is < x < for 1 //! Consequently, 1 f < n I a 7r «" n f„(x) smTtxdx < , n\ Jo This reasoning was completely independent of the value of n . Now if n is large enough, then /* 1 < 7r / tl fl"/„(jc)sin < ixxdx < But this and 1 absurd, because the integral is an integer, and there is no integer between Thus our original assumption must have been incorrect: ti is irrational. | is . This proof that it 1. n\ Jo is admittedly mysterious; perhaps most mysterious of — enters the proof it almost looks as ever mentioning a definition of that precisely one property of ti it A . is really we have proved it the all is way irrational without close reexamination of the proof will show essential sin(7r) The proof if = 0. depends on the properties of the function irrationality of the smallest positive number x with sin x — sin, 0. and proves the In fact, very few properties of sin are required, namely, sin' cos' sin(0) cos(0) Even this list as well = cos, = — sin, = 0, = 1. could be shortened; as far as the proof be defined as sin'. The is concerned, cos might just properties of sin required in the proof may then be written + sin = sin(0) = sin'(0) = sin" Of course, this is 0, 0. 1. not really very surprising at all, since, as we have seen in the previous chapter, these properties characterize the function sin completely. PROBLEMS 1. (a) For the areas of triangles show that we have OAB and O AC in Figure 1, with LAOB < tt/4, 328 Derivatives and Integrals /l -v/l Ari „ AC = -W O 16(area 1 area (b) Let Am (x, y) Pm be is = xy Hint: Solve the equations Pm show the area of . 2(area the regular polygon of OAB) 2 m OAB), x 2 + y 2 = 1, for y. sides inscribed in the unit circle. If that =A A 2m = / l-(2A,„/m) 2 y V 2 - 2v . This result allows one to obtain (more and more complicated) expressions (1.0) =C A4 — 2, and thus to compute n as accurately as desired (according to Problem 8-11). Although better methods will appear in Chapter 20, a slight variant of this approach yields a very for A 2«, starting with interesting expression for FIGURE 2. (a) Using the tt : fact that 1 area(OAfl) = OB, area(OAC) show that if a,„ is = A 2n (b) Show to one side of P,„ , then a,, that CK4 L (c) O the distance from Using the • org • . . . Q?2*-i 2A fact that am = 7T cos—, m yl "I - COS X (Problem 15-15), prove that a4 "8 = Of|6 = 1 V2 1 1 2V2' 11/11 + + \J etc + 2 2V 2 2V 2' 16. Together with part (b), this 7i is Irrational shows that 2/tt can be written as an 329 "infinite product" 7T to be precise, can be made this equation means that the product of the as close to 2/n as desired, by choosing n fascinating expressions for jt , some of which n factors sufficiently large. This product was discovered by Francois Viete in 1579, and many first is only one of are mentioned later. PLANETARY MOTION *CHAPTER Nature and Nature's Laws lay hid God Newton said "Let be," and in night all was light. Alexander Pope Unlike Chapter 16, a short chapter diverging from the main stream of the book, long chapter diverges from the main stream of the book to demonstrate that this we are already in a position to do some In 1609 Kepler published his first real physics. two laws of planetary motion. The first law describes the shape of planetary orbits: The planets move The second law in ellipses, with the sun at one focus. involves the area swept out by the segment from the sun to the planet (the 'radius vector from the sun to the planet') in various time intervals (Figure 1): Equal areas swept out FIGURE are swept out by the radius vector in equal times. in time t is proportional to t (Equivalently, the area .) Kepler's third law, published in 1619, relates the motions of different planets. If a 1 is the major axis of a planet's elliptical orbit and T is its period, the time it takes the planet to return to a given position, then: The ratio a /T is the same for all planets. Newton's great accomplishment was force on a body is its mass times its to show (using his general law that the acceleration) that Kepler's laws follow assumption that the planets are attracted to the sun by a force from the (the gravitational toward the sun, proportional to the mass of the and satisfying an inverse square law; that is, by a force directed toward the sun whose magnitude varies inversely with the square of the distance from the sun to the planet and directly with the mass of the planet. Since force is mass times acceleration, this is equivalent simply to saying that the magnitude of the acceleration is a constant divided by the square of the distance from the sun. force of the sun) always directed planet, Newton's analysis actually established three individual laws. was 330 The first actually discovered results that correlate with Kepler's of Newton's results concerns Kepler's second law (which first, nicely preserving the symmetry of the situation): 17. Planetary Motion Kepler's second the law is true preciselyfor 'centralforces', sun and the planet always Although Newton is lies i.e., 331 if and only theforce between along the line between the sun and the planet. revered as the discoverer of calculus, and indeed invented calculus precisely in order to treat such problems, his derivation hardly seems to use calculus at planet moves, Instead of considering a force that varies continuously as the all. Newton a momentary force To be line imagine that during the , where the length of and the same case where the force is move along to as the triangle this just says that during the next equal If, this line, it would end up at This would imply that P?, SP1P3 (since they have equal Kepler's law holds in the special 0. moment at the the planet arrives at P2 it experi- which by itself would cause the planet the point Q. Combined with the motion that the planet already has, sides are S the linefrom move causes the planet to whose — suppose (Figure 2b) that move this same area height) ences a force exerted along to that time interval the planet moves Pi Pi equals the length of Pj the triangle SP\ Pj has the bases, Now first P\Pi, with uniform velocity (Figure 2a). time interval, the planet continued to f*3 and assumes exerted at the ends of each of these intervals. is specific, let us along the considers short equal time intervals first P2, to to R, the vertex opposite Pj in the parallelogram P2P3 and P%Q. Thus, the area swept out in the second time interval actually the triangle is SPiR. But the area of triangle SP2R is equal to the area of triangle SP3P2, since they have the same base SP2, and the same heights (since RP2 is parallel to SP2). Hence, finally, the area of triangle SP2R is the same as the area of the original triangle SP\ P2 ! Conversely, if SRP2 the triangle has the same area as SP\ Pj, and hence the same area as SP3P2, then RP3 must be parallel FIGURE 2 Q must lie along SP2. Of course, this isn't quite to SP2, and this implies that modern book, but in its argument one would expect to find in a it shows physically just why the result the sort of own charming way should be true. To analyze planetary motion Chapter Chapter 12, 4. we will Appendix Problem 4 of Appendix 1 be using the material and the "determinant" det defined in describe the motion of the planet by the parameterized curve = r(0 (cos 0(0, so that r always gives the length of the line gives the angle, then follows which we easily). will It will assume is c(0 from the sun increasing (the case be convenient e(0 just the to sin 0(0), to the planet, while where is decreasing to write this also as = r(0 • e(0(O), where is to We c(t) (1) in the = (cosf, sin t) parameterized curve that runs along the unit e'(t) = (— sinf, cosO circle. Note that 332 Derivatives and Integrals is also a vector of unit length, but perpendicular to e(0, det(e(f),e'(f)) (2) Differentiating = c'{t) and combining with to Chapter 4, we get det(c(0, c'(t)) since det(u, v) r\t) r(t)r'(t) det(e(0(f)), e(0(f))) = 2 r(f) 0'(Odet(e(0(f)),e'(0(O)), always Using 0. that A{t) formula is segment between = 1 and to e'(0(t)), from time the area swept out c(t + h). It is of the triangle A(/z) with vertices O, and 5 of Appendix obtain Chapter area(A(/j)) c(t shows 1 e'(0(f))) Newton, and area t we'll We want begin by making (Figure 3). A(t), together with a straight line easy to write c(t), 4, the = + h) — to c(t down + h): a formula for the area according to Problems 4 is + £det(c(f), c(t h) - c{t)). Since the triangle A(/z) has practically the same area as the region A(t this Appendix 2 r 0'. for A'(t), and, in the spirit of c(t ) we then get c') an educated guess. Figure 4 shows A(t 3 have out to have another important interpretation. will see, r &' turns to get a FIGURE we (2) det(c, Suppose also + r(t) 2 0'(t) det(e(0(f)), = is we together with the formulas in Problem 6 of (1), (4) As we + r(t)6'{t) e(0(t)) that 1. using the formulas on page 247, (1), (3) = and + h) — A(t), (or practically shows) that + h) A'(t) A(t = lim = lim + h) - A(t) area A(h) h = c(t 4det c(0, lim + h)-c(t) h-+0 = \det(c(t),c'(t)). A rigorous derivation, establishing more in the process, lem 13-24, which gives a formula for the area of a region of a function in polar coordinates. According to this can be made using Prob- determined by the graph Problem, we can write 0(0 I A(0= if our parameterized curve c(t) — polar coordinates (here we've used o r(t) / P(<P)~d<t> e(0(t)) for the is the graph of the function p in angular polar coordinate, to avoid confusion with the function 9 used to describe the curve c). 1 7. Planetary Now the function p just is = p [for any particular angle # </>, _1 (0) (t)) r the time at which the curve c has angular is l polar coordinate 0, so r{0~ the radius coordinate corresponding to 0]. is Although the presence of the inverse function might look a Applying the actually quite innocent: and the Chain Rule to 333 Motion (*) A\t) 2 = \p{e{t)) = 2 \r(t) 6'(t), = \r it's Fundamental Theorem of Calculus First we immediately bit forbidding, get -0'(t) since = p r o >-l Briefly, A' Combining with (4), we e'. thus have l„2/i/ A'=$ det(c, c') = (5) Now is 2 \r we're ready to consider Kepler's second law. Notice that Kepler's second law equivalent to A" = saying that A' \ is constant, = [det(c, c')]' = and thus \ det(c', J + c) it is equivalent to A" \ det(c, c") (see = 0. But page 248) det(c, c"). So Kepler's second law Putting this THEOREM l all we together Kepler's second law planetary path c(t) true is — r(t) if and only Saying that the force c"(t) is And if det(c, this the force c) means of the force, that in the direction points along c{t). = central just is is is Moreover, this this is is central, — 0. and in this case each constant. that it always points along c(t). Since equivalent to saying that c"(t) always equivalent to saying that det(c, c") We've just seen that = e(0(t)) satisfies the equation .2/1/ PROOF equivalent to det(c, c") have: r~0' (Ki) is = we always have 0. equivalent to Kepler's second law. equation implies that [det(c, c')] = 0, which by (5) gives (Ki). | 334 Derivatives and Integrals Newton next showed and that if the gravitational force of the sun an inverse square also satisfies conic section having the sun at one focus. case where the conic section an is is a central force then the path of any object in law, will be a Planets, of course, correspond to the and ellipse, it this is also true for comets that visit and hyperbolas represent objects that come from and eventually continue on their merry way back outside the sun periodically; parabolas outside the solar system, the system. THEOREM 2 If the gravitational force then the path of any body in law, focus (more PROOF of the sun precisely, either an is it a central force that satisfies an inverse square be a conic section having the sun will one at parabola, or one branch of an hyperbola). ellipse, Notice that our conclusion specifies the shape of the path, not a particular parameterization. But this parameterization essentially is determined by Theorem hypothesis of a central force implies that the area A{t) (Figure 5) to so determining c(t) t, on the points mined c(t) 0~ r{t) essentially equivalent to determining : is proportional A for arbitrary This means that e(0(t)) without finding we have its to determine the shape of the path parameterization! Since it is the function which actually describes the shape of the path in polar coordinates, shouldn't be surprised to find 6~ entering into the proof. r o the Unfortunately, the areas of such segments cannot be deter- ellipse. explicitly* = is 1 l we x FIGURE 5 By Theorem 1 , the hypothesis of a central force implies that 2 r 0' [K, for = det(c,c') = some constant M. The hypothesis of an M inverse square law can be written H e(0(t)) c"{t) r(ty for some constant H . Using (K2), this can be written H -— = -—e(6{t)). c"{t) M 0'(t) Notice that the left-hand side of this equation [c'o0- So if we x ]'{Q{t)). let D (this is is the main trick — "we consider C o c' as a function of 0"), then the equation can be written as D'(0(t)) *More prn an sin, etc. iscly. we = can't write H M down e(0(O) = a solution in H M (cos 0(f), sin 0(f)), v terms of familiar "standard functions," like sin, 1 7. Planetary and we can write D [for all u H = M , that components of D, each of which we can H for two constants for c'\ A and c [Here M — B. Letting u H = sin 6 Although we can't get with c = H + A, thus find that cos w B M we 9{t) again H we cos M thus have an explicit formula +B we will use throughout.] if we substitute this equa- an explicit formula for c itself, r(cos6, sin#), into the equation det(c, c') = M (equation (#2)), get h — M cos which 2 H M ~> . B H 2 Problem 15-8 shows some constants M A cos 6 = M, M C and We D. is form can let D — 1, 0, since this simply amounts to which ray corresponds to 6 = 0), write, finally, r\\ L this 1 —i+CcosieiO + D) = rotating our polar coordinate system (choosing we can . sin that this can be written in the r(t) But — A sin 9 sin simplifies to r- for a + B cos 6 H — M so 0. simply a pair of equations, for the is easily solve individually; + A, M . sin u sin 6 really stands for sino#, etc., abbreviations that tion, together we sin u • M , completely eliminating t], just obtained D(u) cos u M ( \ we have H H ( = . (cos u sin u ) of the form 6{t) for some The equation 335 simply as this (u) Motion +ecos01J = H = A. the formula for a conic section derived in (together with Problems 5, 6, In terms of the constant and 7 of that Appendix). M in the equation r d' = M 2 and the constant A in the equation of the orbit /• 1 I + s cos 0] = A Appendix | 3 of Chapter 4 - 336 Derivatives and Integrals we can the last equation in our proof shows that ** c"(t) M— A =- rewrite while the minor axis b as I r(t) 1 e(0(r)). Recall (page 87) that the major axis a of the ellipse is given by A = a (*) \-e 2 ' given by is A = b (b) y/l-e 2 Consequently, = a. \ Remember that equation (5) gives l 9' A'{t) = \r A(t) = \Mt. = \M, and thus We is, can therefore interpret by M in terms of the period T of the orbit. definition, the value of complete t for which we have 6{t) = 2tt, so that This period T we obtain the Hence ellipse. area of the ellipse = A(T) = ^MT, or M= Hence 2(area of the ellipse) lizab T T the constant M"/A in (**) M 2 by Problem 13-17. is 2 2 4jT a b 2 T2 A 4jT 2„3 cl using (c) T< This completes the theorem 3 Kepler's third law moving on is final step true if of Newton's analysis: and only if the accelerations c"{t) of the various planets, ellipses, satisfy c "( f ) = _G.— e(0(O) r~ for a constant G that does not depend on thr planet. 1 7. Planetary It this, should be mentioned that the converse of we want first Theorem 2 is 337 Motion also true. To prove one further consequence of Kepler's second to establish law. Recall that for e(t) = (cos?, sin e'(/) = (— = (—cost, t) we have sin t, cos /). Consequently, e Now differentiating = c"(t) r"{t) = e"(t) e(0(f)) c"(t) = we —e(t) [r"(t) - — sinf) = —e(t). gives (3) + r\t)6'(t) Using (t) + r'(t)6'(t) e'(6(t)) e'(0(f)) + r{t)0"{t) + e'(0(f)) • r(t)9\t)G\t) get r(t)6'(t) 2 ] e(0(O) + [2/(00' (0 + r(f)0"(*)] Since Kepler's second law implies central forces, hence that c"(t) tiple of [as (A^)] • Thus 4 c"{t) If the = As in 1 path of a planet moving under a central gravitational force Theorem 2, notice that the we have knowing what to obtain information lies on a conic an inverse square law. equivalent to Kepler's second law, we can't write down an explicit about the acceleration without actually again, the hypothesis of a central force implies that l ) for some constant M, and r G' it satisfies r[l (A) some entiating = M, the hypothesis that the path the sun as focus implies that First satisfy it is. (K 2 for is determines the parameterization. But solution, so Once must hypothesis on the shape of the path, together with the hypothesis of a central force, which essentially always a mul- [r"{t)-r(t)e'{t) ]-e{6{t)). section with the sun as focus, then the force PROOF is Kepler's second law implies that (6) THEOREM e'(0(f)). and thus always a multiple of e(0(O), the coefficient of e'(6(t)) a matter of fact, we can see this directly by taking the derivative of c(t), must be formula e"(0(/)). +ecos0] = A, and A. For our (not especially illuminating) and substituting from these two equations. proof, differentiate (A) to obtain /•'[l Multiplying by r this +ecos0] -er0'sin0 =0. becomes rr' \ 1 on a conic section with the equation s we lies + s cos 0] - 2 er 0' sin = 0. we will keep differ- 338 Derivatives and Integrals Using both (A) and (K2), this becomes Ar' -eAfsinfl Differentiating again, we = 0. get -e MO' cos 9 = Ar" 0. Using (K2) we get eMcos# Ar' and then using (A) we =0, get A M- = Ar' 0. r Substituting from (K2) yet again, we get A[r"-r(tf') 2 — 2 M- + iV=(X r- ] or r r(6') = M 2 " Ar 2 Comparing with (6), we obtain M c"(t) 2 Ar 2 e(0(O), what we wanted to show: the force is the square of the distance from the sun to the planet. | which is precisely inversely proportional to CHAPTER THE LOGARITHM AND ^^ I EXPONENTIAL FUNCTIONS In Chapter 15 the integral provided a rigorous formulation for a preliminary def- and inition of the functions sin In this chapter the integral plays a cos. even a preliminary definition presents essential role. For certain functions more difficul- For example, consider the function ties. = /(*) assumed to be defined for all x and to have an inverse function, positive x, which is the "logarithm to the base 10," This function defined for is l f~ In algebra, 10* rational x 10*. is is (x) =log 10 x. usually defined only for rational x, while the definition for A quietly ignored. ir- brief review of the definition for rational x will not only explain this omission, but also recall an important principle behind the definition of 10*. The symbol to 10" 10" The definition. Since we want true, we must be true, we must define 10 1/w • 0* to rational x . . • _" 10 = 10" • be true, we must define 10 10 1/w be true, we must = I" • m to 10 A 1 339 v 1 ; 10° = 10 10" = the equation 1 we want +l/ " l/,?+ ly v 10 and define 10'" 7 " " = 10 1 /"" m = since ; times Unfortunately, at this point the guided by motivated by the desire is = 10 the equation 1 = 10 n times 1//,! • = 1/10"; since n times to 0+ " we want since 1; = 10 = 10" • = define 10° 10 1 10"+'". the equation 10-" to This notation turns out n. requirement actually forces upon us the customary 10° be = 10'" • extension of the definition of to preserve this equation; this to numbers defined for natural is first be extremely convenient, especially for multiplying very large numbers, because ( \/70 — 1 we want hI /" = 10"' the equation '" / times )'"• program comes to a dead halt. We have been the principle that 10* should be defined so as to ensure that 10* +v but this principle does not suggest any simple algebraic way of = defining . 340 Derivatives and Integrals we 10* for irrational x. For this reason finding a function / f{x (*) Of course, we + y) = for other x; in general One way more / f(x) = \0 X suggested is fix) for f(y) fix lim we try / such if all moment assume / that — knowing the derivative Now fih) - fix) lim • h^0 h^0 =a h and denote for fix) all approach seems this / the inverse function log '»' (x) 1 lim this limit exists, a could be computed, we examine a new light: x. self-defeating. f~ l — interest in g, of all derivative log 10 the whole situation appears in , J_ of/(/-'(.r)) " Vx about as simple as one could is The again. 1 derivative of / by a. Then it = 7(7^0) " more an apparently x and y, itself. fih)- = has been expressed in terms of If The to solve thus depends on fix) if 10, f(x) h fix) f'iO) / defined as fih) = Even — lim fc-+0 of f{\) +h)- fix) fix) = for the and 10 could be try to find /' a clue to the definition of fix) The answer v we might 10. we can that such a function exists, / might provide specific condition function f(\)= of ways of y. not always zero, so is more the differentiable + y) = fix Assuming sophisticated equal [/(l)]* for rational x. will problem: find a x and all for rational x, to find such a function difficult for f(y) • we add If 0. imply that f(x) (*) will f(x) are interested in a function which add the condition /(l) then some more will try such that r x" dx the integrals examined / ask! And, what is even previously, the integral Ja x~ dx is the only one which we cannot evaluate. Since log 1() Ja have 1 /' / oi J] v I -dt t = log | () x - log 1 , () = log to x 1 = we should 18. we This suggests that define log 10 * as unknown. One way of evading 341 The Logarithm and Exponential Functions (1 /or) dt. t / n difficulty is that a is define this difficulty is to log* The dt, and hope that this integral will be the logarithm to some base, which might be determined later. In any case, the function defined in this way is surely more reasonable, from a mathematical point of view, than logi The usefulness of logjQ depends on the important role of the number 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 known to us. . DEFINITION If x > 0, then log X = / - dt. The graph of log is shown in Figure 1 Notice that if x > < x < 1, then log* < 0, since, by our conventions, if . 1 , > 0, \/t is then log x and -dt / J\ =- - dt Jx t < 0, a number log* cannot be defined not bounded on [x, 1]. For x area FIGURE l If x. 0. in this way, because f(t) = log x 1 The justification THEOREM = < t y > 0, for the notation "log" comes from then log(.vv) = log.v + log >'. the following theorem. 342 Derivatives and Integrals PROOF Notice number choose a = that log'(x) first > y by the Fundamental Theorem of Calculus. l/x, and = f(x) log(xy). 11 Then / Thus /' = log'. (x) = log (xy) This means that there f{x) that Now let — y is for = y • -. x number a +c log* — xy = c such that > x all 0, is, = log(;ty) The number c can be evaluated +c log* for by noting log(l y) • = = that log 1 x > all 0. when x = 1 we obtain +c c. Thus = log(xy) Since COROLLARY 1 If n this is true for is a natural > y all theorem 0, the number and x > + logy log* COROLLARY 2 Let to you (use induction). If x, y > 0, x. proved. | = n logx. | then lo g( PROOF all then 0, log(jc") PROOF is for ~ =logx -logy. ) This follows from the equations log x Theorem provides 1 function log is very small as jc It is log -= — y I = log I — ) + log y some important information about clearly increasing, but since log'C*) becomes large, = . the graph of log. follows that log is, in fact, = n , n log 2 (and log 2 > 0); not bounded above. Similarly, log(-) The becomes and log consequently grows more and more slowly. is bounded or unbounded on R. Observe, number log(2") | l/x, the derivative not immediately clear whether log however, that for a natural it ( :logl -bg 2" /; log 2; - ally takes on is all not bounded below on R important function has a special Since log (0, 1). is continuous, domain name, whose appropriateness Therefore values. 343 The Logarithm and Exponential Functions 18. therefore log . of the function log" the is will it actu- . This soon become clear. DEFINITION The "exponential function," exp, The graph of exp always have exp(x) THEOREM 2 For all numbers x is > shown 0. The in defined as log" is Figure Since log x 2. is defined only for x derivative of the function exp is > 0, we easy to determine. , = exp'(x) PROOF exp'(x) = exp(x). lv, (log = )'U) — 1 -1 log (log 00) 1 log-'(x) = A second THEOREM 3 If = (x) important property of exp is exp(x). | an easy consequence of Theorem 1 x and v are any two numbers, then exp(x PROOF -1 log Let x' = exp(x) and y' = + = expU) y) • exp(y). exp(y), so that = = x v log*', log y'. Then x +y = log*' + logy' = log(x'y'). This means that exp(x + y) = x'y' This theorem, and the discussion FIGURE exp(l) 2 DEFINITION is particularly important. exp(x) at the There e = = is, • exp(y). | beginning of this chapter, in fact, a special exp(l). symbol suggest that for this number. . 344 Derivatives and Integrals This definition equivalent to the equation is 1 As = f = loge 1 / - J\ t dt. illustrated in Figure 3, — dt < I J] 1 since , 1 — (2 • % = f(t) an upper sum 1 ) is for l/ton[l,2], and n FIGURE > dt since \ 1, - (2 • sum +\ 1) • (4 = for /(r) - 2) = 1/r on 1 is a lower [1,4]. Thus 3 ,2 h -4 ,, j j j -dt < -dt < / t Ji t -dt, t Ji which shows that < we In Chapter 20 e is will find proof is irrational (the As we remarked e < much better approximations for e, and also prove much easier than the proof that tt is irrational!). that beginning of the chapter, the equation at the exp(x + expU) = = = v) exp(x) • exp(v) implies that Since exp our DEFINITION is defined for all [exp(l)]* e x , for all rational x and exp(x) = e x for rational x, earlier use of the exponential notation to define e For any number The terminology in defining e yet defined a attempt. If x x , is if a last x = it is consistent with as exp(jc) for all x. exp(x). now be "exponential function" should x an arbitrary (even for ^ rational, e, but there expression is irrational) clear. exponent x. We have sucWe have not a reasonable principle to guide us then a But the x x, e ceeded x is x = (e defined for l ° sa x ) all — x, so e x[oga . we can use it to define a x . in the DEFINITION If a > any o, then, for real number x ax = (If a = e = this definition clearly The requirement a > unduly f(x) = is restrictive since, for 345 The Logarithm and Exponential Functions 1 8. x log a agrees with the previous one.) necessary, in order that logo be defined. This we would example, is not not even expect 10* <-i) to 1/2 = be defined. (Of course, for certain rational x, the symbol a x make will sense, according to the old definition; for example, (_i)i/3 definition of a x Our was designed = (e*) y As we would hope, FIGURE the 4 THEOREM 4 If a > > this a 0, 1) 0. to ensure that xy x and for all y. when e replaced by any is a moderately involved unraveling of terminology. At the other important properties of a x . then {a ] a =a bc b c ) (Notice that a (2) e _i.) equation turns out to be true The proof is same time we will prove number = ^cr = (Notice that b = will for all c. automatically be positive, so (a b ) c will be defined); a and a x+y (2) ft, = ax ay for all x, y. implies that this definition of a x agrees with the old one for all rational x.) PROOF (a (1) b Y= (Each of the steps in the fact that exp (2) a 1 = e c]ogah — this string log" ' £ c lo gO Mog ") = e _ c(bloga) e _ ^c cb\oga of equalities depends upon our last definition, or .) = e ll°z a = e °z a = a, f,x+y l _ Ax+y)\oga __ xloga+yloga __ Figure 4 shows the graphs of f(x) = of the function depends on whether a a < x vloga _ _ vloga x for several different a. 1, a = \, or a > 1 . _ y | The behavior If«= 1, then , 346 Derivatives and Integrals f(x)= \ x = Suppose a > 1. x < if then e < e x < a a the function f(x) < so that log a is decreasing. 0, the — ax same > a if Thus takes on (Figure all it is the inverse function values. If f(x) = a x then , . / the other hand, then f(x) — also easy to see that x ^ and a Since exp takes on every positive value positive value. , v is 1, defined for _1 is < a < if shows that the function f(x) sort of reasoning In either case, , ylo % a On increasing. is Thus, y, xl °z a i.e., 0. < y log a x log a so Thus > In this case log a 1. all a positive a x is = 1, a x one-one. on every numbers, and takes the function usually denoted by log a 5). Just as a x can be expressed in terms of exp, so log a can be expressed in terms of log. Indeed, = log a x, if y then x-= a y = e ylosa .. = y log a so log x or y = x = log* loga In other words, lo g„ FIGURE The 5 derivatives of f(x) f{x) = e — a x and g{x) x[oga f(x) = so g'(x) = so , log* g(x) more complicated loga — log a x are both easy to find: loga also easy to differentiate, if follows -= loga a\ function like you remember f(x ) it xlo ^ a 1 f(x)=g(x)" is e xloga loga A tog* from the Chain Rule fix) _ e (x) that, by definition, =e hM\o S g(x). that h(x) \ogg(x) h'(x)logg(x) + h(x) 8(x)j g(x) h(x) h'(x)logg(x) + h(x) g'(x) g(x)j There is no point in remembering this formula simply apply the principle behind it in any specific case that arises; it does help, however, to remember that the first factor in the derivative will be g(x) h{x) . There The one is = function fix) and define result is above formula which special case of the a x 347 The Logarithm and Exponential Functions 18. was previously defined only worth remembering. for rational a. = find the derivative of the function fix) is x a for We can any number now a; the what we would expect: just _ a fix) „a\ogx SO = - fix) a log* ax X become second naall the rules which ought to work those stated in Theorems 2 and 3: Algebraic manipulations with the exponential functions will —just remember that ture after a little practice actually do. The basic properties of exp fix) THEOREM 5 If / = is ce = and a close to characterizing the function exp. / — satisfying /' = fa) number f, for if / = ce x , then for all x, c such that fix) PROOF exp(y). .; ; fix) is • fix); these functions -are 'the only ones with this property, however. clifTerentiable then there exp(x) not the only function is still =:exp(x), (.x) comes In fact, each of these properties x ; + y) — exp(x Naturally, exp exp are = ce x for all x. Let *(*) (This is permissible, since e x ^ 8 ix) for = fix) = - e*f'{x) is a number The second is = fact, fix) = = ex c clearly not the only function = But the true story for all x. more + y) = / which much more complex than functions which satisfy this property, but advanced mathematics, to involved discussion. The satisfies fix)-f(y). or any function of the form f(x) is 0. y basic property of exp requires a fix In x c such that gix) function exp f(x)e* -^- ie Therefore there Then all x.) it is prove that there this = a x also satisfies this equation. —there are infinitely many other impossible, without appealing to is more even one function other than those 348 Derivatives and Integrals already mentioned! for this reason that the definition of 10 x It is many there are infinitely / which functions f(x + y) = /d)= continuous function / f{x) f(y), 10, = but which are not the function f(x) so difficult: is satisfy 10 v One ! thing is —any however true satisfying f(x + y) = f(x)-f(y) must be of the form f(x) = a x or f(x) = 0. (Problem 38 indicates the way to prove this, and also has a few words to say about discontinuous functions with this property.) Theorems 2 and In addition to the two basic properties stated in exp has one further property which is very important — exp "grows 3, the function faster than any polynomial." In other words, THEOREM 6 number For any natural n. lim x-»oo proof The proof consists e Step 1. x > x of several for oo. x" steps. and consequently lim x, all —= x = 0) it e oo (this may be considered X—>-00 to be the case n — To prove statement (which this 0). x If jc < 1 this upper sum is > log* clearly true, since for f{t) = \/t on x < clear for is for all log* < [1, jc], x > show that 0. .v > <x — If 0. so log* suffices to then (Figure 1, 1 < 6) x — 1 is an x. x Step 2. lim x-»oo To prove e — = oo. x this, note that e er x/2 x . e x/2 1 21 2 \ 2 By Step 1 I ( , I R 1 . 1 , (e x ' r the expression in parentheses is 2 greater than shows that lim x e /x — oo. X—*oo Note 1 , and lim X—>00 6 Step 3. S/2 x —= lim x—*oo x n oo. that e x (e /gx/n x/ ")" 1 ' ' " x" G)" n" e xl = oo; this 1 8. The 349 The Logarithm and Exponential Functions expression in parentheses becomes arbitrarily large, by Step 2, so the nth power certainly It is f(x) now = { e- becomes possible to 'x \x # 0. arbitrarily large. | examine We carefully the following very interesting function: have Xi Therefore, < f'(x)>0 x < 0, forx>0. for f'{x) so / is decreasing for negative x and increasing for positive x. Moreover, large, then FIGURE x2 is large, so — \/x 2 so e e's and close to 0, so e~ is I* large, so e <5's, = is more \/(e l/x ") interesting. is Therefore, is close to 1 if (Figure |jc| is 7). x is small, then l/x is large, argument, suitably stated with shows that if we l/x2 =0. define /(*) = r 1 /* 2 , then the function / is continuous (Figure 1 x^O x 0, 8 If small. This \ime- FIGURE l/x ~ 7 The behavior of / near x,x ~ is 8). In = 0. fact, / is actually differentiable — 350 Derivatives and Integrals Indeed at 0: l/h 2 = /'(0) \/h = lim lim and —r— h^o+ lim -y e We already (X / = h)1 know x ^°° e all the more =— lim oo z ^U 2^ * ) true that l ) = x-*oo means t — = oo; lim this e {] /^ that ,(x and 1//? A -0- (x2) lim it is nm while lim OO, x that lim 2 X->-0O ^(JC ) = 0. Thus e- l 'x = fix) x#0 -3, jc-' x 0, We can now compute 0. that 2 1/^2 h3 lim 2- e -lA 2 lim lim an argument similar = to the /"(*) one above shows — ,4 t- = that f"(0) lim = .^.4 + ^^.4 *6 X = 4 ! 0, lem 40) that approaches f {k) (0) — 9), using induction it = it 0. can be shown (Prob- The function / is extremely flat can mask many irregularities of other for every k. so quickly that For example (Figure fact, Thus ,#0 .v This argument can be continued. In 0. suppose that ,-l/.v /(-*) 2 sin — , .v 7^ x =0. x 0, at 0, and functions. 18. It The Logarithm and Exponential Functions can be shown (Problem 41) that for for all k. this function it is f {k) (0) This example shows, perhaps more strikingly than any other, just bad a function can investigate be, even more out behavior of this and still be IV we In Part infinitely differentiable. restrictive conditions on a function, which = how will will finally rule sort. e FIGURE also true that 351 1/jr sin 1/jt, x ^ x = 9 PROBLEMS 1. Differentiate each of the following functions notes a (remember that a b " always de- (b'h 1) f(x) ii) fix) iii) fix) = e e° . = log(l + log(l + log(l + e [+e]+ '))). = (smxY m(smx) . iv) v) f(x) vi) fix) = = vii) f(x) = viii) 2. f(x) = = ,smv)Mn> (sinA-) log (£jV) sinx. + i\ogx) ^ l f(x) fix)=x x xi) f(x) Check V sin x f . 4 4x e ))e (log(3 x) (a) x-ilogCsin^) X / arcsin ix) = . + (arcsin jc) 1o s 3 . x . . sm(x sm ^ mt) ). that the derivative of This expression is logo/ is /"'//• called the logarithmic derivative of /. It is often easier products and powers in the expression for / compute than become sums and products in the expression for logo/. The deriva/', since to tive /' (b) can then be recovered simply by multiplying by /; called logarithmic differentiation. Use logarithmic differentiation to find (i) /(*) = (] xl +*)0 +e ). fix) this process for each of the following. is 352 Derivatives and Integrals f .. s e (1 f(x) (iii) = (3-x) ]/3 x 2 — x)(3 + x) l (sinx) C0SJC + ' i (cosx) si Find *i b f(t) (It / fit) a (for 4. / > Oon [a,b]). Graph each of the following functions. (a) f(x) (b) f(x) = e x+1 = e sinx f(x) — (d) f(x) = e -e~ (e) f(x) = (c) e x . . + e~ x x (ii) lim (iii) lim l (Compare x - .\ limits lx by V e v - 1 + 1 = 1 ,2, + l'Hopital's Rule. - x - x 2 /2 1 "' ~x^ 6" the graph with the graphs of exp 1/exp.) e 2x e lim . e Find the following (i) x - 1 - - 1 -x -x 2 /2 .v - 2 x /2 - 3 a- /6 x->0 log(l (iv) +Jt)-;t +* 2 /2 lim x->-0 2 log(l +jc)-.y+.x- /2 (v) lim x^O log(l (vi) + x) - x + x 2 /2 - x 3 /3 lim 73 x^O 6. Find the following -x) l/x limits (i) lim(l lim (tanjc) tan 1> (ii) (iii) Um(cosje) /* . . by ' l'Hopital's Rule. ! and . . The Logarithm and Exponential Functions 18. 7. The functions e :x 2 + y 2 = 353 sinh — x e -x .y 2 1] e x -X cosh Jt 2 e x, sin x) tanh x — x e~ X = + e~ ex 1 e 2x + \ hyperbolic sine, hyperbolic cosine, and hyperbolic (but usually read 'sinch,' 'cosh,' and 'tanch'). There are many analogies between these functions and their ordinary trigonometric are called the tangent, respectively One counterparts. shown ter, analogy illustrated in Figure 10; a is in Figure 10(b) really has when we area x/2 is proof that the region best deferred until the next chap- develop methods of computing integrals. Other analogies will are discussed in the following three problems, but the deepest analogies must wait until Chapter 27. If you have not already done Problem 4, graph the and tanh. functions sinh, cosh, 8. Prove that 2 (a) cosh" — 2 • sinh" = 1 2 2 +1/ cosh = (b) tanh (c) sinh(x 1. (f) + y) = sinh cosh y + cosh x sinh y. cosh(x + y) = cosh x cosh y + sinh x sinh y. sinh' = cosh. cosh' = sinh. (g) tanh' (d) (e) jc 1 = cosh The FIGURE 10 - - and tanh and tanh are one-one; their inverses sinh are defined on R and (— 1 1), respectively. These inverse functions are sometimes denoted by arg sinh and arg tanh (the "argument" of the hyperbolic sine and tangent). If cosh is restricted to [0, oo) it has an inverse, denoted functions sinh , , by arg cosh, or simply cosh" which information in Problem 8, that , sinh (cosh x) = jc) = Vl +x 2 -1 (b) cosh (sinh (c) (sinh- )'(x) , 1 = \f x 2 — is 1. . 1 yr + .V (d) (cosh _1 )'(jc) = 1 for x > 1 v^^T (e) (tanh i '/(Jt) = i-x 2 for Ul < 1. defined on [1, oo). Prove, using the . 354 Derivatives and Integrals 10. Find an (a) (b) formula for sinh explicit , cosh , and tanh (by solving the - = equation y x for x sinh terms of in y, etc.). Find b f 1 dx, Ja + s/\ K 2 b r l dx for a b , > or a,b 1 < — 1 1 r b l d for fif.v 1--V 2 a < \a\, \b\ 1. Compare your answer for the third integral with that obtained 1 1 1 l-x< 1 1 . Show 1 1 + - x +x \ — that F(x) by writing f logr dt = J2 is 12. not bounded on Let / be [2, oo). a nondecreasing function on F(x) ) = 13. / bounded on is x dt. / and define > 1. t J\ Prove that [1, oo), [1, oo) if and only if F / log is bounded on [1 , oo). Find (a) lim a x for < a < 1. (Remember the definition!) .V—>00 (b) lim x^oo (logJC)" (c) lim x—>oo (log*)" x (-l)Mlog (d) lim x(logx) n (e) lim x . IV Hint: x(logx)" x x . x->0 + 14. Graph f(x) = x x 15. (a) Find the fix) (b) > minimum n e /n" for > 0. +i /in + lim fix) Graph /(*) 1)" +1 = x > for oo. = <?7* w . (Use Problem — value of fix) 13(e).) x e /x" for x > 0, and conclude that n. Using the expression e" 16. for x fix) x > n = + x 1, — n+l prove that fix) > and thus obtain another proof that e ix n)/x , The Logarithm and Exponential Functions 18. 17. (a) + y)/y. Find lim log(l \ —>o 355 (You can use l'Hopital's Rule, but that would be silly.) (b) Find lim jclog(l +l/x). (c) Prove that e (d) Prove that e a x—*oo with just a *(e) 18. 19. — = . = lim x(b [ ^x — x> 0. l/x) x for +cr/100) after 1 possible to derive this (Use Problem year. If the then the initial suppose that interest + not 7(1 alas, is, 17(c).) computing interest the next investment grows to 7(1 +«/100)" after n years. is The given twice a year. 2 but merely 7(1 <r//100) ", (c) annum, then an initial investment / bank compounds the interest (counts the accrued interest as part of the capital for year), from part 1). a bank gives a percent interest per If yields 7(1 is . + (1 x x little = \/x) lim (1 +a/x) (It X—»00 algebraic fiddling.) Prove that log/? Graph f(x) + lim (1 x—>oo amount final + a/200) 2" Now after n years — although interest is awarded twice as often, the interest must be halved in each calculation, since the interest is a/2 per half year. This amount is larger than 7(1 + a/ 100)", but not that much larger. Suppose that the 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 How much these numbers. 1 20. 21. = # (a) Let f{x) (b) If f(x) Suppose (a) (b) (c) log (d) A all on some that investment of initial dollar yield after 1 0. Prove that f'(x) = l/x for x x, prove that (log |/|)' = /'//. interval the function / satisfies /' = ^ 0. cf for some Assuming that / is never 0, use Problem 20(b) to prove that |/(jc)I = le cx cx for some k. for some number / (> 0). It follows that f(x) = ke Show that this result holds without the added assumption that / is at the endpoint of an open never 0. Hint: Show that / can't be interval on which it is nowhere 0. Give a simpler proof that f(x) Suppose — that /' f(x)/e = = ke cx for some k by considering the cx . fg' for some g. Show that f(x) — ke K{x) for some radioactive substance diminishes at a rate proportional to the disintegration the amount all is at atoms have equal probability of proportional to the time t, represents the probability (a) # c. present (since is for x |jc| for function g{x) 22. an year? number K will Find A(t) in remaining). means that A'(t) = cA(t) that an atom will disintegrate). this — amount disintegrating, the total number of atoms terms of the amount Aq k. A(0) present for at some time 0. If A(t) c (which 356 Derivatives and Integrals (b) Show that there is number a with the property that A(t 23. + = r) 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 surrounding medium. Find the temperature T(t) of the object at time temperature 7b at medium 24. r (the "half-life" of the radioactive element) time /, in terms of its assuming that the temperature of the surrounding 0, expressing Newton's law, M. Hint: To solve the differential equation remember that T — (T — M)'. Prove that fit) dt, then is kept at a constant, if f(x) f = 0. Jo 25. Find all ff = e Jo (i) / continuous functions satisfying x . f f=\-e 2x\ (ii) Jo 26. Find all / functions — satisfying f'{t) + f(t) f(t)dt. / Jo 27. Find all (fix)) 28. (a) / which continuous functions satisfy the t 2 f = Jo fit) +f l / and g be continuous functions on pose that for some C we have Let fix) Prove Gronwall's equation <C+ , inequality: f(x)<CeJ« g . Hint: Consider the derivative of the function h{x ) (b) Let / and g be nonnegative Suppose that f'{x) (Compare Problem 2 tiable. 1 29. fa) = = (C functions with g continuous g(x)f(x) and /(0) = 0. + f* fg)e and / 8 •><> . differen- Prove that / = 0. .) Prove that 1+ * + X2 2! + X3 + + <e" for x > 0. 3! Hint: Use induction on (b) g nonnegative. Sup- a<x<b. fg, f [a b] with //, and compare x Give a new proof that lim e /x n = derivatives. oo. x—>oo 30. Give yet another proof of this Rule. (See Problem 11-56.) fact, using the appropriate form of l'Hopital's . The Logarithm and Exponential Functions 18. 31. (a) Km Evaluate 2 x e 357 be able (You should (Vc dt. e' I f to make an educated Jo guess before doing any calculations.) (b) Evaluate the following lim e~ (i) limits. r x+a/x) x~ x^oo r e / Jx dt. px + (\ogx)/x lim e~ (ii) x e'' dt. / x^oo Jx />.v+(log.v)/2x 2 lim e~ ' e / x^co dt. Jx This problem outlines the 32. r2 approach classical to logarithms and exponentials. we will simply assume that the function f(x) = a x defined in an elementary way for rational x can somehow be extended to a continuous To begin with, , , one-one function, obeying the same algebraic rules, on the whole line. (See Problem 22-29 for 'a direct proof of this.) The inverse of / will then be denoted by log v (a) ( , Show, direcdy from the definition, that l/h , log a (x) = limlog h+ fl -J \. lo g;(linj(l +«'/*). X Thus, the whole problem has been reduced to the determination of + lim(l h—»0 — • h) {,h If . \og e e.— we can show — and , log/U) = has derivative exp'(x) = that this has a limit e, then consequently exp = log" exp(x). (b) Let a„ show = ( 1 I —IV 1 for natural n. Using the binomial theorem, + Ei( -i)( -=)....( 1 A! Vx k=2 l n 7I I n \x < a ll+ \. Using the fact that \/k\ < 1/2 A_1 for k > 2, show that all a„ < 3. Thus, the set of numbers {a\, ai, «3, is bounded, and therefore has a least upper bound e. Show that for any e > we have e — a n < s for large Conclude that a„ . enough (d) numbers that — (c) H If n n. < x < n + 1 , then . . } 358 Derivatives and Integrals Conclude that lim and conclude *33. A point Q point H 1 I x—»oo B time 1 lim 1 x-+—oo = + e. e. ' is = then P'(t) t, 10 The . P to B 7 — P(t)), while 10 (in other words, distance traveled by Napierian logarithm of the distance Q FIGURE Also show that e. moving along a line segment AB of length 10 7 while another moves along an infinite ray (Figure 1 1). The velocity of P is always P = Q'(t) = I I + //) 1// = that lim(l h-*Q equal to the distance from at — x \ from P Q P(t) if Q moves after B to at the position of is P with constant velocity time t time t. is defined to be the Thus 1 10 =Naplog[10 7 -P(/)]. 7 / This was the definition of logarithms given by Napier (1550-1617) in publication of 1614, Mirifici logarithmonum canonis the Wonderful Law description of Logarithms); work which was done exponents was invented! The number 10 his (A Description of before the use of 7 was chosen because Napier's taand navigational calculations), listed the logafor which the best possible available tables extended and Napier wanted to avoid fractions. Prove that bles (intended for astronomical rithms of sines of angles, decimal places, to seven Nap logx = Use the same Hint: ^34. (a) trick as in Sketch the graph of f{x) and behavior near (b) Which (c) Prove that xy = larger, e is x is = x; Problem 23 = equation for P. to solve the (log*)/* (paying particular attention to the oo). or 7r < x < if y n 10 7 10 log but if e ? 1, x = or x > 1, x ^ e, e, then the only then there is number precisely y satisfying one number y y moreover, if x < e, then y > e, and if x > e, y t^ x satisfying x — y then y < e. (Interpret these statements in terms of the graph in part (a)!) x ; (d) Prove that x (e) = y 2, Show = x and y are natural numbers and x y 4, or x = 4, y — 2. if that the set of a straight line which all pairs (x, y) with x y = x y = y x , then x = y or and and draw a rough consists of a curve intersect; find the intersection sketch. ** (f) For 1 < x < Prove that g e let g(x) is be the unique number > differentiable. (It is fl(x) = = logx < x < X logx e g{x) = g(x) x . a good idea to consider separate functions. f\(x) e with x < x e and write g terms of in ,, *35. you do this and f\ You should be fa. [g(x)] 2 able to show -logx 1 xL -logg(x) part properly.) This problem uses the material from the Appendix to Chapter (a) that , 1 if 359 The Logarithm and Exponential Functions 18. Prove that exp is convex and log 1 1. concave. is n (b) Prove that if = y_. Pi 1 and all > /?, then for 0, all Zi > we have i=\ Z\ Pi ...-Zn P" < + PlZl ••• + PnZn- (Use Problem 8 from the Appendix to Chapter (c) 36. (a) Deduce another proof that Let / be Gn < An (Problem 2-22). a positive function on [a, b], and Use Problem 2-22 into n equal intervals. 11.) let Pn be the show to partition of [a, b] that -—— L(f, P )) 1 b (b) — L(logf, a Pn ) < log ( Use the Appendix to Chapter 13 n \b conclude that for to . a all integrable / > we have log f / -aJa A more (c) direct approach is <log( \b-aJa/ / illustrated in the next part: In Problem 35, Problem 2-22 was deduced as a special case of the inequality n for pi > 0, y. Pi = 1 an d 8 convex. For g concave we have the reverse inequality ^Pigixt) Sg l^PiXi) \i=\ i=\ Apply this with g = I log to prove the result of part (b) directly for any integrable /. (d) 37. State a general theorem of which part Suppose / satisfies /' = / and f(x that / = exp or f = 0. + y) (b) is = just a special case. f(x)f(y) for all jc and y. Prove 360 Derivatives and Integrals Prove that "38. either / if / = continuous and f(x is = or f(x) x [f(l)] + y) = f(x)f(y) Show for all x. Hint: for all x and y, then = that /(*) [f(l)] x 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 / satisfying f(x + y) = for rational x, f(x)f(y). Prove that "39. / if a continuous function defined on the positive real numbers, is = = and f(xy) f(x) + f(y) for all positive x and y, then / f(e)\ogx for all x > 0. Hint: Consider g(x) = f(e x ). all if = Hint: Consider functions g(x) Prove that "41. = l/x2 for x # 0, and /(0) = 0, then / (*>(0) = for f(x) = e~ k (you will encounter the same sort of difficulties as in Problem 10-21). Prove that "40. or f(x) = f(x) if e~ l/x2 e~ l ^x ' \/x for x sin P(\/x) for a polynomial function P. / = and /(0) 0, then 0, / ,A) (0) = for all k. 42. (a) Prove that a if is a root of the equation n + a n _ix"~ an x (*) = then the function y(x) a„y (**) *(b) Prove that a " /U) = if < a - ]) if a 0, the differential equation + +a x y' +a y — = 0. xe ax also satisfies (**). a double root of a polynomial equation is =0. (*) of order r, then y(x) —x e ax is a solution for < If (*) has n real numbers as roots (counting multiplicities), part k r 1. n solutions y\, ... (d) that a root of is — {n satisfies a double root of (*), then y(x) is then f'{a) 0, Prove that -\ ax +a„_iy ) Remember Hint: *(c) if i e +a\x + a = l Prove that in , (c) gives y n of (**). this case the function + c\ y\ • • • + cn yn also satisfies (**). a theorem that in this case these are the only solutions of (**). Problem 21 and the next two problems prove special cases of this theorem, and the general case is considered in Problem 20-26. In Chapter 27 we will see what to do when (*) does not have n real numbers as roots. It is Suppose "43. that / satisfies /" - / = and /(0) = /'(0) 0. Prove that / = as follows. (a) Show (b) Suppose f(x) ** c) If = a < this fact 2 if') that f(x) ce f(xo) < 2 f - that x ^ ^ = or else f(x) for X() > 0. for all x in = ce~ 0, say, x some for all x interval (a,b). Show that either some constant c. number a such that in (a,b), for then there would be a for a < x < jco and f(a) = 0, while f(x) ^ and part (b) to deduce a contradiction. xq. Why? Use *44. Show (a) that / if satisfies - f = f" = then f(x) 0, some a and b. (First figure out what a and b should be and /'(0), and then use Problem 43.) Show (b) 45. Find {,,) (other) a in and + be' x for terms of /(0) b. satisfying . { (b) *46. some for ae x ~ (n X) f = f "- 2) (n) f = f (a) / functions all / = a sinh +&cosh also that 361 The Logarithm and Exponential Functions 18. . This problem, a companion to Problem 15-30, outlines a treatment of the exponential function starting from the assumption that the differential equation =f f' has a nonzero solution. Suppose there (a) is a function / / f= with = each x by considering the function g(x) f. Prove that fix) /(xo ^ *)/(*0 — •*)> and /(0) = 1. + for where /(*o)#0. Show (b) For (c) / show this g(x) 47. that there = + f(x / is a function f(x that Let / and g be continuous / is =f f(y) by considering the function f(x) l )'(x) = l/x. functions such that lim f(x) *—>-oo grows faster than g (f ^> g) if hm and we say = one-one and that (f~ Prove that say that v) satisfying f' y)/f(x). (d) We + / that / and g lim grow = lim g(x) oo. JC—>-oo — — = oo, - at the same exists and : = rate is (f / For example, for any polynomial function ~ g) if 0, oo. P with lim P(x) — oo (i.e., P X—>00 is non-constant and has positive leading coefficient) P ^> log" for any positive integer n. (a) Given / and with lim f(x) X—>0O that one of the three conditions g, (b) BF/»g,then/+*~/. (c) If tog/ = = oo, is it S> P and necessarily true .V—»oo / » g or g S> / or / ~ g holds? c > . > lim g(x) we have exp , 1 logg for sufficiently large x, then (d) If / that ;§> F g and F(jc) » G? = / Jo / ^> g. f,G(x)= I JO g, does it necessarily follow . 362 Derivatives and Integrals (e) 48. 49. Arrange each of the following growth (for value at x): convenience, we x 3 ex , x 3 (ii) x log x, e 5x log(x v ), ex (iii) e + log(x 3 ), log4.v, (logx)*, , Suppose that x , x e x x ex , , g\, gi, g3, . , . , , Prove that log 10 2 irrational. , x 3 logx. , . are continuous functions. / which grows + e~ 5x x x x logx (logx) A , continuous function is xx x its '. , 2X e x/2 (\ogx) 2x , . of functions in increasing order of indicate each function simply by giving (i) , sets faster than each g, Show that there is a <%^ CHAPTER INTEGRATION Every computation of a derivative Theorem ELEMENTARY TERMS IX according to the Second Fundamental yields, of Calculus, a formula about integrals. For example, if = x(logx) — x F(x) = then F\x) log x\ consequently, b f \ogx dx I = F(b) — F(a) = — b(\ogb) b — [a(loga) — 0<a,b. a], Ja Formulas of this sort are simplified considerably if = F(x) We may Fib) - we adopt the notation F{a). then write b is fa of the function Fix) called x(logx) — x dx depended on the lucky guess that log is the derivax(\ogx)—x. In general, a function F satisfying F' = f a primitive of /. Of course, a continuous function f always has a This evaluation of tive = \ogxdx L log x = primitive, namely, Fix) Ja but in this chapter we will try to find familiar functions like sin, log, etc. is called an elementary function. a primitive which can be written in terms of A function which can be written in this To be precise,* one which can be obtained by addition, multiplication, division, from the rational functions, the trigonometric functions and functions log and exp. It should be stated at the is no F'{x) a * The and composition and the their inverses, elementary function = e~ x2 for all F such that x not merely a report on the present state of mathematical ignorance; (difficult) theorem that no such function definition which we will give is exists). And, what is 1 where 363 the /, + it is even worse, you precise, but not really accurate, or at least not quite standard. Usually the elementary functions are defined to include "algebraic" functions, that satisfying an equation (g(x))' is very outset that elementary primitives usually cannot be found. For example, there (this is way an elementary function /„_,(x)U(.v))"- 1 + • •• +f (x) = is, functions g 0, are rational functions. But for our purposes these functions can be ignored. . 364 Derivatives and Integrals 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 no misprints). Because the search for elementary primitives so uncertain, finding is one is often peculiarly satisfying. If we observe that the function log(l = x arctan x F(x) + x2 ) satisfies F'{x) how we would (just = arctan jc ever be led to such an observation is quite another matter), so that /' arctan x dx — log(l — x arctan x +x 2 ) Ja then we may feel that we have This chapter consists of itives little arctan x f dx more than methods for finding elementary prim(a process known simply as "integration"), of given elementary functions together with this "really" evaluated some notation, abbreviations, and conventions designed to facilitate procedure. This preoccupation with elementary functions can be justified by three considerations: (1) Integration (2) Every once about and everyone should know a standard topic in calculus, is it. 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). (3) The most orems useful (that "methods" of integration are actually very important apply to all Naturally, the last reason to integrate (and is you probably Even the crucial one. will forget some function in terms of primitives of other functions. list if you intend to express primitives of one we will To begin of primitives for some functions; such a simply by differentiating various well-known functions. The use of a standard symbol which requires some explanation. OT /' means "a primitive of The symbol f f /" or, will often useful in formulas more / in stating A x I list can be obtained given below makes The symbol of all primitives of /." theorems, while like the following: x dx integrating list f(x)dx precisely, "the collection by used how to forget time through), you details the first must never forget the basic methods. These basic methods are theorems which allow us therefore need a the- functions, not just elementary ones). / f(x)dx is most 19. Integration in Elementary Terms = This "equation" means that the function F(x) cannot be interpreted but in because the right side literally one context we this is F'(x) satisfies and we = x3 It . a number, not a function, allow such discrepancies; our aim will integration process as mechanical as possible, device. x 4 /4 365 is will resort to to make the any possible Another feature of the equation deserves mention. Most people write / .3 x'dx X4 —4 +C = emphasize that the primitives of f(x) — x are precisely the functions of the form F(x) — x 4 /4 + C for some number C. Although it is possible (Problem 14) to to obtain contradictions if this point not arise, There and concern is disregarded, in practice such difficulties for this constant is merely an annoyance. one important convention accompanying is 'W on the side left letter appearing — thus u / / u du ' tx dx 4 - -= ' 4 2 tx ~2~' xt / A the letter ap- this notation: pearing on the right side of the equation should match with the after the do 2 tx dt - f f(x)dx, i.e., a primitive of /, is often called an "indefinite while / f(x) dx is called, by way of contrast, a "definite integral." function in integral" of /, This suggestive notation works out quite well in practice, but be led /" integral of Chapter risk / f{x)dx the integral We At the astray. (if of boring you, the following fact is not you do not defined as "Fib) — is F(a), where find this statement repetitious, a dx = F is is an indefinite time to reread = — dx = n^-\ , + log x 1 I / — dx table.) e x dx sin = x dx side. ax is often written / — abbreviations are used in the / it can verify the formulas in the following short table of indefinite integrals n / to 13). x"dx I important not emphasized once again: simply by differentiating the functions indicated on the right I it is e x — — cos x for convenience; similar last two examples of this 366 Derivatives and Integrals I cos x - = x dx sec I dx = sin* tanx x tan x dx sec = sec x dx ' I \+x = 2 dx I arctan x — arcsin x •A Two general formulas of the same nature are consequences of theorems about differentiation: f[f(x) + g(x)]dx = j f(x)dx + j g(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 / 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 rb nb nb f(x)dx+ [f(x)+g(x)]dx= / Ja Ja f f(x)dx c = g(x)dx, J c ci f 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 (Of course, the that these primitives exist. when / and g are merely integrable, but recall how much more difficult the proofs are in this case.) The product formula for the derivative yields a more interesting theorem, which formulas are also true will THEOREM l (integration BY PARTS) be written If /' and in several different ways. g' are continuous, then / fg' = fs-f f'8> f f{x)g\x)dx = f(x)g{x) nb / Jn b f(x)g'(x)dx = f(x)g(x) j f'(x)g(x)dx, /. /> / a f'(x)g(x)dx. Ja (Notice that in the second equation f(x)g(x) denotes the Junction f g. 19. Integration in Elementary Terms PROOF 367 The formula = (/*)' + fg' ~ f'g- f'g can be written = fg' (fg)' Thus j fg'=f(fgy-f fg, and fg can be chosen first The second formula As is merely a restatement of the from either of the follows immediately two. first is be considered as a — xe x dx I a = xsinx dx I ^ x • xe x — (— cosx) — i f f g g' I e i is useful when the func- 1 obviously of the form is e • x g'. dx J = xe x - | | formula product of a function /, whose deriva- 44 fg fg' third | simpler than /, and another function which J and the first, the following examples illustrate, integration by parts tion to be integrated can tive one of the functions denoted by f(fg)'. This proves the as formula. i i f g x / (— cosx)dx • 1 J 4, i r g = — x cos* + sin x There are two first is which often work with integration by parts. The be the factor 1 which can always be written special tricks to consider the function g' to , in. log -Y / dx = / The second and then trick is solve for / which implies 1 J J • log x | 4, g' f to use integration fh A . dx dx = x log x = g f x(\ogx) is — 1414f by parts simple example log x — I (\/x) dx x g — to find X. fh in terms of the calculation log x log x — (l/x) log x -X- -X- -X- "X -X -X g' f g f f g (1 /x) - I that 2 f / J • — losxdx = x (logx) 2 dx , /h again, 368 Derivatives and Integrals or 1 - /x A more complicated I e sin x calculation dx = £ r 4- 4- / 2 often required: is (— cos x) - A • 4- 4- / 8 = —e 2 (log*) = ax log 5x e / x (— cosa)o?a • f g + je< cos a dx cos x : = - e -cosx + I i u v' [,<.( 51 „.v,-/^( S in,»^]; V u' therefore, 2 e / / x sin x dx = x e (sin x — cos x) or — x / e x sin = x dx form success. the the It is more functions you can already integrate, main problem. For example, we can use / (log*) dx we = / J /logx dx by parts); we have (\ogx)(logx)dx I is of the the greater your chances for tackling parts to integrate (logx)(logx) dx J recall that gration that a function do a preliminary integration before frequently reasonable to 3 if cosx) upon recognizing Since integration by parts depends g', e {?,mx = = x(logx) —x I I f g' (this (\ogx)[x(\ogx) - formula was - x] derived by inte- itself (\/x)[x(logx)- x]dx / J I I i i I I f g' f g f g — (log x) [x (log x) — x] — = (log x)[x (log x) —a] — = = (log A) [A (log A) — x (log A ) — A] j / — 1] dx \ogxdx + / [log [A(logA) — 2a (log A + 2a ) A" — A'] 1 dx +A . The most important method of integration is a consequence of the Chain Rule. The use of this method requires considerably more ingenuity than integrating by parts, and even the explanation of the method is more difficult. We will therefore 369 19. Integration in Elementary Terms develop this method theorem in stages, stating the saving the treatment of indefinite integrals for THEOREM 2 If / and g' are for definite integrals first, and later. continuous, then (THE SUBSTITUTION FORMULA) J/g(a) /= rb ?(») f{u)du= If F Ja a primitive of /, then the is f(g(x))-g'(x)dx. \ Jg(a] a) PROOF (fog)-g' / Ja left side — F(g(b)) is F(g(a)). On the other hand, = (Fo g y so F o g is The (fog) a primitive of (F o (F'og).g' g)(b) - (F and the g' o g)(a) (fog).g>, right side F(g(b)) of the form (fog)- is I J x cos x dx sin a facilitated for g(x) = - F(g(a)). | K cosxdx (sinx) by the appearance of the factor cosx, which = /' u . be the factor g'(x) = = sin x f(u)=w g(x) sin x cos x dx ro b = 'g(b) rg\o) f(g(x))g'(x)dx= / Ja tan x sin dx can be treated h f I b tanxdx — sinx is a sin g'(x), similarly = \/u. f(g(x))g'(x)dx =- g(x) = f(u) = - write dx. = dx we COSX where g(x) for f(u) if - sin ^ I Ja 1/cosx can then be written /(cosx) tan x f =— Jo In this case the factor b u du -L f f(u)du / J'g(c via) sin b /' will can be written as (g(x)) 5 , Thus Ja integration of that a For example, the integration of g'. sinx; the remaining expression, (sin.v) f(g(x)), for f(u) The is depend upon recognizing simplest uses of the substitution formula given function is = = cosx; the remaining factor Hence cosx 1 Ja 'g(b) -L /•cos/j J i. as, i / Jg{a) f(u)du i - du / ft " = log(cosa) — log(cos/?). 370 Derivatives and Integrals Finally, to find I /' dx, *log* JIa = notice that \/x = f(u) = where g(x) g'(x) = logx, and that l/logjc f(g(x)) for Thus 1/m. g(x) = logx dx L x\ogX a /•gtfo = ^S b 1 <iw f = f(u)du / log(logb) — log(loga). u 'log a Fortunately, these uses of the substitution formula can be shortened considerably. The intermediate steps, which involve writing rg(b) >b f f(g(x))g'(x)dx= Ja can easily f(u)du, f J g(a) be eliminated by noticing the following: To go from the left side to the right side, u for g{x) substitute du g'(x)dx for (and change the limits of integration); performed the substitutions can be for the name directly b f sin I original function (accounting x cos x dx />sin u for sinx substitute du Ja and on the of this theorem). For example, = cosx dx for b / u du, J sin a similarly rb — r sin x Ja Usually we dx COS X abbreviate this /•cos/? u for cosx substitute du for — sin x dx XJ method even more, and "Let u du = = = / J cos a 1 — du " say simply: g(x) g'(x)dx" Thus u let = log L a In this chapter tegrals, but if we we can dx *logX 1 du /•log* ,v I , — — dx = / j -du. x are usually interested in primitives rather than definite infind fa f(x)dx for all a and b, then we can certainly find — , 19. Integration f f(x)dx. For example, Elementary Terms 371 since b • . sin 5 — x cos x ax — sin 6 6 • i o a — 6 sin - 6 1. it in follows that f sm 5 x cos jc a jc 6 —- * sin = . / Similarly, tan x dx = — log cos x : dx = / / J It is — xlogx , log(logx). quite uneconomical to obtain primitives from the substitution formula by finding definite integrals. two Instead, the can be combined, steps first 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). Thus, primitive (as an expression involving (2) Find (3) Substitute g(x) back for u. . \ to find /-sin cos x jc dx , 1) let u du so that we — = sinx, cosx dx obtain / (2) evaluate (3) remember to substitute sin du; x back for u so that /sin 5 x ' " , cos x dx = — sin - x u). not the , 372 Derivatives and Integrals Similarly, if = u log x 1 = — dx, du x then /* / J — /" 1 dx - becomes 1 - du / xlogx = log u, u J so that 1 To xlogx = dx log(logjc). evaluate j / l+.v 2 dx, let u du the factor 2 which has just = = 1 + 2 .v , 2x dx; popped up causes no problem 1 1 1 j/ u o I /" 1 ~ du = lo S M — the integral becomes - so / l+x (This result may 2 "" f/x = 1 -io g (i 2 9 + .A be combined with integration by parts I 1 • arctan x dx to yield x = x arctan x = x arctan x -I — l+.v 2 1 j log(l dx +x / ), a formula that has already been mentioned.) 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 *The beginning of the chapter and, of course, the right substitution substitution formula is I often written in the form fiu)du= j f(g(x))g'(x)dx, u=g(x). literally (after all, / f(u)du should mean a primitive of / and the symbol f f(g(x))g'(x)dx should mean a primitive of (fog)- g'\ these are certainly not equal). However, it may be regarded as a symbolic summary of the procedure which we have developed. If This formula cannot be taken we use Leibniz's notation, and a little fudging, the formula reads particularly well: I fundu = . j f(u)-£dx. , ) , . 19. Integration in Elementary Terms (the third problem has been disguised a / by some algebraic chicanery). little x dx sec x tan / (cosx)e 373 smx dx. dx. / -e 2x y/\ you have not succeeded in finding the right substitutions, you should be able to guess them from the answers, which are (tan 6 x)/6, e smx 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 If . , substitution. detail The ix I (something). - following integrations should the proper positioning of the constant 3x /3 or 3e ? e e is be all (The only worrisome clear. — should the answer to the second be always take care of these problems as follows. Clearly Now "something" must be differentiate if I cancel the 3, to = F(x) e 3x get F'(x) I , = fe 3e ix 3x , dx = so the 3.) logU+3), / +3 .v , 3x / / cos Ax sin(2.t / dx = dx = e e sin 4; 4 + 1) dx + 1 2 / does ' cos(2a' = dx More 3x — "3 arctan 2x l+4.r 2 interesting uses of the substitution formula occur not appear. Consider There are two main types of when substitutions the factor g'(x) where this happens. first +e x 1 t -e dx. 1 The prominent appearance of the expression e x suggests the simplifying substitu- tion u du Although the expression e /1 We x = e\ = e dx. x dx does not appear, - ex dx = / • J - 1 ex therefore obtain L +u 1 • I 11 — du. u can always be put it — ex e x dx in: 374 Derivatives and Integrals which can be evaluated by the algebraic / — u u 1+e / \-e dx 1 2 f — du = 1 + - du = - 21og(l — - / J - \ trick u) + logM, u u so that There is = -2 log(l - e x + log e x = -2 log( - an alternative and preferable way of handling not require multiplying and dividing by e x — a e 1 ) . ) + x. problem, which does If we write x = logM, dx = — du, e this x 1 u then 1+e - dx / -e immediately becomes J 1 Most substitution ing x in terms of u, see is why this trick one-one for all problems are much easier and dx in if one I — u u resorts to this trick of express- terms of du, instead of vice versa. always works It is long as the function expressing u in terms of x (as x under consideration): If we apply the substitution u x=g-\u) dx = (g- Y(u)du =g(x), l to the integral / we f (1) the other hand, if we apply u to the same l f(u]<)(g- )'(u)du. the straightforward substitution du = g(x) = g\x)dx integral, j f(g(x))dx we f(g(x))dx, obtain On = j f (g(x)) g'(x)dx, g'(x) obtain (2) The integrals (1) and //<- )• du. g'(g-Hu)) arc identical, since (g~ )'(u) [ (2) not hard to As another concrete example, consider 1 vV+T dx. = \/g'(g~ '(«)). r , 19. Integration in Elementary Terms the entire expression y/e x we will go the whole hog and replace by one letter. Thus we choose the substitution In this case = = u u u 2 e -l=e\ 2 x x = integral then f / J + 1 1 + 1, = log(w 2 - dx The + y/e x 375 1), 2w du. —x uz — 1 2 — [ becomes (ir-D 2 — • 2u «2 n — du -i f l 1 u du 2M — — 3 zw. 3 y Thus e t Ve 2x Jx x + Another example, which occur, is 2 = -(e A + " l) 3/2 -2(e v + 1/2 l) . 3 1 illustrates the second main type of substitution that can the integral / Vl — x 2 dx. In this case, instead of replacing a complicated expression by a simpler one, will replace — x by sinw, because vl are using the substitution u = which helps us find the expression x let dx = = = « sin arcsinx, but it is cost/. This really means that sin u cos = [u , Thus, . arcsin x] udu; then the integral becomes / The v 1 — udu — sin~ u cos cos u du. I evaluation of this integral depends on the equation 2 cos u + cos 1u 1 = 2 (see the discussion of trigonometric functions below) so that I cos u du = 1 + cos 2u u du I 2 = - + 2 sin 2w 4 and r / 7 Vl-x' dx = arcsin x sin (2 arcsin jc) 1 arcsin x 1 h - . . sin(arcsinx) —2~x + r^^7 arcsin 1 / we the expression for x in terms of u be substituted for dx to we r - • cos(arcsin x) , 376 Derivatives and Integrals Substitution you have and integration by parts are the only fundamental functions. some of our examples Nevertheless, as reveal, success often The most important additional tricks. you should be able to integrate all the functions in depends are listed below. Using these Problems 1 to 10 (a few other some of the remaining problems). interesting tricks are explained in 1 can be found for a large number of to learn; with their aid primitives upon some methods which TRIGONOMETRIC FUNCTIONS . Since sin x + = cos x cos x 1 and cos o Lx we -2 — sin x 2 , obtain — = cos lx cos 2.* — cos x (1 — 7 sin — cos" x = 2 cos x — -9 .9 — 2 sin x) — sin"x = x, (1 1 ) 1 or — 1 sin" cos 2x x ? + cos lx 1 cos X 2 These formulas may be used to integ rate cos" x dx, / if n is x dx, sin" / even. Substituting (1- cos lx ) ( 1 + cos 2x) or 2 for sin x or cos 2 jc yields a 2 sum of terms involving lower powers of cos. For example, 2 1 - dx 4 C / cos 2x 2 and / If n is odd, n = 2k + / I, cos" 2x = dx 1 + cos Ax / then s'm" x dx = / sin.vd — k cos x) dx\ dx H —/ / cos 2x dx 19. Integration in Elementary Terms 377 the latter expression, multiplied out, involves terms of the form sinjccos'x, all of which can be integrated easily. The integral for cos"x is treated similarly. An integral sin" / handled the same way if n or formulas for sin 2 x and cos 2 *. is A final m x cos is " = dx secxdx I jc odd. important trigonometric integral / w dx If n and m are both even, use the is = log(sec;c +tanjc). Although there are several ways of "deriving" this result, by means of the methods already at our disposal (Problem 13), it is simplest to check this formula by differentiating the right side, and to memorize it. 2. REDUCTION FORMULAS Integration by parts yields (Problem 21) sm n xdx / = --sm 'f / cos" x dx and many U +l)« J n l 2 + ^—— f sin"- 2 xdx, n xcosx = - cos" -1 x sin x + ^—— J f J l n J dx * n / cos" -2 x dx J x In-?, r 1 1 _ ~2n-2(x 2 +\r-i + 2^2j~(xTTTy^ dX similar formulas. The first two, used repeatedly, give a different for evaluating primitives of sin" or cos" . The a large general class of functions, which will third method very important for integrating complete our discussion. is RATIONAL FUNCTIONS 3. Consider a rational function p/q where p(x) q(x) We = = + an _ix n ~ H m b m x +b m ^x m + an x which is l l might as well assume that a„ for otherwise n y ... a + , b . =bm = we may express p/q as a \. Moreover, we can assume that n < m, polynomial function plus a rational function of this form by dividing (the calculation u2 u — 1 = u+l + \ u - 1 is a simple example). The integration of an arbitrary rational function depends on two facts; the first follows from the "Fundamental Theorem of Algebra" (see Chapter 26, Theorem 2 and Problem 26-3), but the second will not be proved in this book. ) 378 Derivatives and Integrals theorem Every polynomial function q{x)=x m +b m - l x m -l + +b can be written as a product q(x) (where r\ = (x- + a,)" \- • + rk . . • . (x - ak ) + 2(s\ rk (x h Si) = (In this expression, identical factors x — c, and + fax + y, may be x 2 + + fox S| 2 (x ]/,) + frx + yiy m). have been collected together, so that assumed all Moreover, we assume that each distinct. quadratic factor cannot be factored further. This means that 2 A- since otherwise + Aa + x we can factor 2 -A + Va -4 k = Yi -4y; <0, -A , 2 - >/A- -4y, into linear factors.) THEOREM < m and If n p(x) q(x) = x" + a„_]X n ~ H \-clq, m = x + b m _ x"- + ---+b 2 = {x- a,)'' (x - a k ) (x + fox + n l ] l • . . . • then /?(a)/<7(a) can be written p(x) q(x) . -a U + ••• + - in the + _{x + . . • . (a 2 + A* + YiY\ form +• — «1 n -a b u x+c ]A k _(a 2 (x ) + fox + + -ak Y k r known it is _ (x 2 yi) */ IX + + pix + n) + si ••• _ C/.1 f fax so complicated that ) (x -cci)'i_ i This expression, S] rk 1 + as the "partial fraction y /) u 2 + A-v + y/)". decomposition" of p(x)/q(x), simpler to examine the following example, which illustrates such an expression and shows how to find it. According to the theorem, it possible to write 2.v 7 + 8.v ft + 20a- 4 + 15a 3 + 16a 2 + 7a + 10 +a + 1) 2 (a 2 + 2a + 2)(a- l) 2 b a ex + d ex + f —, + + + —r + A-l (a-1) 2 a 2 + 2a+2 a 2 +a+1 + (a 2 13a is 5 +h +a + gx (a 2 1) 2 is . 379 19. Integration in Elementary Terms To find the numbers common denominator over the and a, b, c, d, e, f, g, (a 2 2 +x + polynomial h, write the right side as a 1 ) (x 2 + 2.x + 3) (a - — 1 ), the numerator becomes - a(x + (ex 1)(a 2 + 2x + 2)(x 2 + x + + d)(x - 2 \) {x 2 +x + l) 2 1 ) 2 + + b(x 2 + 2x + 2)(x 2 + x + + (ex + Actually multiplying out this ficients are combinations of a, cients of 2.v 7 we 6 + 8.v + 1 3.r 5 we (!) . . 2 - f)(x ( <? 1) (a 2 + 2x .v+/2)U- l) + 2)(x 2 2 1) (a- obtain a polynomial of degree h. , 4 Equating these + 20.x- + 15.x- + 3 16.y 2 + + 8, +x+ + 2.v whose 10 (the coefficient of unknowns a ,...,/? obtain 8 equations in the eight 2 1) 2). coef- coefficients with the coeffi- + 7.x 2 . 8 is jc 0) After heroic calculations these can be solved to give a=l, = 0, e = 2, f = 0, b c=l, d = h g 0, = 3, = \. Thus f 2x 7 + J = + 13.v 5 + 20a- 4 + 17.x- 3 + 16x 2 + 2 2 (x 2 + x+\) 2 (x + 2x + 2)(x-\) 6 77 / (X-I) J dx + / (X J 77T 2 -l) dx + / J —i2 (A- simpler cases the requisite calculations (In particular ing it We into 7x 5.x- example by starting with the one fraction.) +.X + 7 dx dx + / 1)+ T^ J A + ~ A + 2 may actually be feasible. partial fraction 2.v I 3 + dx. 2 obtained this decomposition and convert- are already in a position to find each of the integrals appearing in the above expression; the calculations will illustrate all the difficulties which arise in integrat- ing rational functions. The first two integrals are simple: 1 / x - dx — dx = 2 / The log (a — 1), 1 (X-l) 1 -2 A-l depends on "completing the square": third integration A 2 +A+l --=(X+W~ + 1 2 (If we had obtained — ^ instead of | we could not take the square root, but in this case our original quadratic factor could have been factored into linear factors.) We 380 Derivatives and Integrals now can write 16 1 / The (a 2 +a + 1) dx 2 h ~9~ j dx. + x + 1 substitution changes u = du = + x dx, this integral to 16 ~9~ du, + D2 J (w which can be computed using the third reduction formula given above. Finally, to evaluate l we (a 2 x + + 2x 3 + dx 2) write x+3 / The first evaluate 2 .v + 2.t integral it + on 2 2 J x -\l the right side has u The second integral simply 2arctan(A a J : (x< 2 dx + / J (a- + 1 ) 2 + Ja. 1 been purposely constructed so that we can by using the substitution du 2 +2 + 2x + 2 2a dx + + 2x + on the + 2a f \ , dx 2)" which is just the difference of the other two, If the original integral 1). 3 right, = x + 2a + 2, = (2a + 2) dx -i2 J (a 2 + + 2a is were 2 + 2)" dx + 2 f J [(a [U + 2 1 ) + dx. 1]" on the right would still be evaluated by the same substitution. The second integral would be evaluated by means of a reduction formula. This example has probably convinced you that integration of rational functions the is first integral a theoretical curiosity only, especially since of q{x) before you can even begin. This that simple rational functions sometimes l another important example 2 J x -l J x it is necessary to find the factorization only partly true. We have already seen arise, as in the integration 1 +e x 1 -e* dx; the integral is - is 1 A+l log(A 2 - 1) - -log(A+ 1) . 381 19. Integration in Elementary Terms Moreover, it is if a problem has been reduced to the integration of a rational function, then certain that an elementary primitive impossibility of finding the factors of the even exists, denominator when may the difficulty or preclude writing this primitive explicitly. PROBLEMS 1. This problem contains some integrals which require braic manipulation, and consequently test your little more than alge- ability to discover algebraic rather than your understanding of the integration processes. Never- tricks, theless, any one of these might be an important preliminary step tricks an honest integration problem. Moreover, you want which you can integrals are easy, so that process The answer in sight. is see section, if when you the in have some feel for end of an integration to resort to it, will only reveal what algebra you should have used. Vx* + yz / dx. x dx / - + v7 * + + e 2x + e 3> yjx 1 dx. iin e 4x iv) 1 — dx fI x J b I tan" x VI dx (Trigonometric . +x 2 2 integrals are always very touchy, because there are so many problem can easily look hard.) trigonometric identities that an easy ' dx Vll 2 -x 2 dx 1 . ix) 1 + sin x 8jc / I 2 + 6.v+4 x+ j dx. 1 dx, J 2. The Jlx - x 2 following integrations involve simple substitutions, most of which you should be able to do in your head. . / e x sin e x dx " 382 Derivatives and . .. " Integrals x~ dx. / in iv f log* dx. /' e 2x + eX x e x (In the text this was done by parts. dx 2e* + 1 e dx. e J x dx vi vn I V 7 -a- 4 ! f e^* — x2 viii) / x\J 1 ix) / log(cosx) tan.vJx. Jjc. r log(iogx) ^ x\ogx Integration by parts dx. (ii) (iii) (iv) /VV /" /x (v) / w /" J vn e / v sin sin bx dx x dx (log^) 3 6?JC. —- log(logx) sec x dx . (This up. is If a tricky and important integral that often comes you do not succeed consult the answers.) Vlll / cos(logx)dx. ix) / s/x log x x) / x(\ogx) dx. dx in evaluating it, be sure to ) ' 19. Integration in Elementary Terms 4. The x = following integrations can = sinw, x cosw, all be done with substitutions of the form To do some of these you etc. 383 will need to remember that sec x / = dx as well as the following formula, / cscxdx + log (sec x which can tan x also = — log(cscx be checked by differentiation: +cotx). In addition, at this point the derivatives of all the trigonometric functions should be kept handy. dx - / -x (You already know = x sin u dx / (Since tan v + 1 x tution x this integral, anyway, just to see 2 = u + 1 = sec 2 u, but use the substitution how it works you want out.) to use the substi- tan u dx m 1 J y.v 2 - /dx —-j=^ iv x\x^— (The answer . given short 1 will be a certain inverse function that was the text.) shrift in dx r ' xs/l-x 2 J dx r VI ' J xVl+x 2 x- dx. Vll You will need integrating vm 5. 1 — 2 remember powers of sin the and methods / 71 (x) f y/x 2 cos. dx. - \dx. following integrations involve substitutions of various types. no substitute for cleverness, but there for an expression which appears frequently or prominently; troublesome expressions appear, new in expression. terms of / dx 1 + v/.t dx (u) / And u, to find f (i) 1 for +x 2 dx. LX The .v to + ex ' + 1 is it is a general rule to follow: substitute try to express don't forget that There them both in if two different terms of some usually helps to express x directly out the proper expression to substitute for dx. 384 Derivatives and Integrals dx C mi J \fx + 1/x Li A. IV! I (The substitution u . v + 1 e* = e x leads to an integral requir- ing yet another substitution; this substitutions can be done is all right, but both at once.) dx / 2 + tan x dx VI; (Another place where one do the work of two fJ v S7* y/x + 4' + r 4 substitution can be made to V 1 Vll J 2* +T dx. Vlll dx ix) (In this case . Vx- there are two obvious candidates for the tion, The + Ix - 1 x+A-2 is 2x z2 J * 3 h 3 .v I 3x 2 1) 2 dx. + 3x - + 7.v 2 - (X- dx. 1 -hi 2jc 2x 2 (A- + 5.v + +x + 1 1) 5 dx. 3 1 / (A+3)(A- l) 2 Ja. +4 flA. + 3 A +A + 2 rfjc. / A 4 + 2a 2 + 3a 2 + 3 a + It, + 2a 2 + 2a + A 1 VI 1 1 Vll dx Vlll substitu- x7 ' 1 be integrated. Here f LV first either will work.) previous problem provided gratis a haphazard selection of rational func- tions to in and best; dx. + J V-v two successive substitutions work out I a 4 + 1 2a IX + A+ l) 2 dx. 3a J (a^ + a- I) 3 dx. dx. 1 a more systematic selection. . . .. . 385 19. Integration in Elementary Terms /dx - which looks a , little from any of the previous different vV - x 2 problems. Hint: Use a It helps to write (x" form u substitution of the Extra Hint Use a 2: 2 = . — x2 . l/2 ) = x(x n to obtain . substitution of the form v ~2 — 1) 1/2 . Extra Hint 1: an answer involving arctan. = xa to obtain an answer involving arcsin. *8. (No holds barred.) Potpourri. The following integrations involve all the methods of the previous problems arctan x / dx. 2 +x T 1 arctan x dx. ft + *2)2 in / logy/l+X 2 dx. IV I xlog v r 2 - + x 1 dx. 1 1 1 = dx. + VI vii) viii) ix) / / f / e x cos3 x sin v dx vtan x dx ft' (To factor x b +l 6 + following two problems provide (and can bear Problem 9 it). and integration by many -s'mx = 2 cos x Ix it dx : + sin x 1 dx The dx arcsin ^/x / I yr » 1, first still factor v more practice +x 1 I integrals. )dx. + cos X . 2 sin" /A.x + T" dx. x 11 dx. y/4-X 2 (iv) / x arctan x dx using Problem 1-1. at integration, if you need and trigonometric manipulations Problem 10 involves substitutions. (Of course, in cases the resulting integrals will require log(a" 1, involves algebraic parts, while Find the following + still further manipulations.) . 386 Derivatives and . . Integrals (v) VI /sin 3 /• 3 cos z x 2 Vll / dx. arctan * dx. .v x dx Vlll 10. / - 2* + s]x 2 2 x tanx dx. (ix) I sec (x) I x tan x dx Find the following integrals. I&T&- (i) (ii) / in I iv v 1 — x dx sin arctan >fx Vx + I sin I log(.v dx \ dx. (v) There are obvious VI \/x 2 + — 1 ) dx. but integration by parts Comparing Vll I + log (x substitutions to y/x)dx. is much try, easier. the answers obtained is, perhaps, instructive. /dx Vlll 11. x-x ' 3 /5 ix / (arcsin x)" dx. (x) / x arctan (x )dx. If you have done Problem 18-10, the look very familiar. integrals involving \/x 2 involving lem 4. yi + 2 1. integrals (ii) In general, the substitution x — 1, while x Try these (The method is — sinhw is and = (iii) The world's sneakiest substitution t = tan — , x = is will on the other integrals in Probrecommended; it is easier to stick with trigonometric substitutions.) 12. Problem 4 the thing to try for integrals substitutions not really in cosh u often works for undoubtedly 2 arctan /, dx = 2 \+r dt. 387 19. Integration in Elementary Terms As we found Problem 15-17, in = sinx expressions this substitution leads to the It 1+r = cos* ' 2 1 -V ' l+t 2 This substitution thus transforms any integral which involves only cos, combined by and addition, multiplication, and sin division, into the integral of a rational function. Find dx / Compare your answer + sin x 1 with Problem dx / J IV / + b cos x si a sin x sin" (In this case it is better to let dx (An jc (There - (A . + 5 sin x Derive the formula for a) By Why?) tanjc. to do this, using 15-8.) should be used only as a 13. = way also another is Problem exercise to convince . /i— — 3 t x dx I- mi 2 sin l(viii).) you that substitution this last resort.) last resort.) f sec x dx in the following two ways: writing 1 cos* cos* cos 2 x COS* — 1 COS X 1 ~ x sin : 2 _ 1 + sin x + an expression obviously inspired by sure to note that is 1 : — sin x _ partial fraction decompositions. — sin x dx = — log (1 — sin x ) f cos x/(l And remember very important. and on, keep doing algebra, (b) cos X - that ^ log a — ) ; the log y/a. minus From Be sign there trust to luck. = One manip- By using the substitution ulation required to put the answer in the desired form; the expression is t tanx/2. again, quite a bit of tanx/2 can be attacked by using Problem 15-9, or both answers can be expressed in terms of t. There is another expression for fsecxdx, which is less cumbersome than log(secx + tan*); using Problem 15-9, we obtain /l / sec x dx = last \ = log log (tan ( 1 This + tan - tan + ;)) J expression was actually the one first discovered, and was due, not to any mathematician's cleverness, but to a curious historical acci- . 388 Derivatives and Integrals computed dent: In 1599 Wright When integrals of sec. the first nautical tables that amounted to definite tables for the logarithms of tangents were produced, the correspondence between the two tables was immediately noticed (but remained unexplained until the invention of calculus). 14. The derivation of = primitive of fix) e*(sinjt /e e Suppose that /" is is (a) *(b) is is seems in the text — F{x) to prove that the only — cosx)/2, whereas Fix) any number C. Where does e*(sinjc also a primitive for = C meaning of the equation continuous and that — Given that fin) x dx given sinx the 1, + f"(x)]smxdx = 2. [f(x) Jo / 16. sin x — cosx)/2 + C come from? (What 15. x compute /(0). Find f arcsinx dx, using the same trick that worked for log and arctan. _1 Generalize this trick: Find / / (x) dx in terms of f fix) dx. Compare with Problems 12-21 and 14-14. 17. (a) (b) Find f sin x dx in two different ways: and then using the formula for sin x Combine your answers to obtain first using the reduction formula, an impressive trigonometric 18. Express j \og(\ogx)dx in terms of /(log a)" dx. (Neither terms of elementary functions.) 19. Express 20. Prove that the function fix) (Do not try to find it!) 21. Prove the reduction formulas 1 x f x e~ dx in and work on the .v 22. = + = x e /ie 2 (x ix last integral. 1) has an elementary primitive. For the third one write dx = J l)» +e x + in the text. f ~~ 5x f +l)"+ l)"- " J 1 x~ dx (jc (Another possibility 2 + is 1)" to use the substitution tan u .) Find a reduction formula for (a (b) <23. j (log x)"dx. Prove that />cosh. sit J] (See Problem 1 expressible in terms of f e~ x dx. /dx (x 2 is identity. 1 - cosh x sinh x 1 x dt 2 8-7 for the significance of this computation. 19. Integration in Elementary Terms 24. 389 Prove that pb pb f(x)dx= / + b — x)dx. f(a / Ja Ja (A geometric interpretation makes but this clear, it is also a good exercise in the handling of limits of integration during a substitution.) 25. Prove that the area of a ber that 26. n and such that Show (b) circle of radius Let that / be 0=1. / 0/,(jc) = (Naturally you . = T (j)(x/h). h |jc Show 1] h and that / (j) h = J-h and continuous at f= lim for \x\ > 1 \. 0. Show that f fc /= /(O). that l h-+Q+ final part analogue of part Let — rh cf> h v lim The 4>{x) 0, let > | must remem- circle.) 4>hix) on [—1, J h / be of (b), but integrable Hint: If h is f J_ h —1 / x +x h ^dx = TT. problem might appear, this lim The nr > 1 (d) is For h for integrable lim (c) r defined as the area of the unit be a nonnegative integrable function such that Let (a) is it actually requires on [—1, / 1] more 2 + x2 ) will be an exact careful argument. and continuous TT-^f(x)dx = small, then h/(h at first sight, to at 0. Show that jrf(0). be small on most of [—1, 1]. next two problems use the formula •0i If 2 J 2 f(0) d0, On derived in Problem 13-24, for the area of a region bounded by the graph of / in polar coordinates. 27. For each of the following functions, find the area bounded by the graphs in polar coordinates. (Be careful about the proper range for 0, or you will get nonsensical results!) a (i) f(0)= (ii) /(0) (iii) f(G) (iv) /(0)=acos20. sin 9. = 2 + cos0. = 2a 2 cos 20. 2 . 390 Derivatives and Integrals 28. Figure 1 has area r = f(0) shows the graph of / - Now f(6) dO. / the ordinary graph of in suppose that some function AOx\B + area polar coordinates; the region g I Then g. — area this graph also happens OAB the region OAB thus to be also has area AOxqA. JX] Prove analytically that these two numbers are indeed the same. Hint: function g niU'RE is The determined by the equations x = f (6) cosO, = g(x) f(0)smO. I The next four problems use the formulas, derived in Problems 3 and 4 of the Appendix to Chapter in particular, as the 29. 13, for the length of a curve represented parametrically (and, graph of a function in polar coordinates). Let c be a curve represented parametrically by u and v on be an increasing function with h(a) functions u curve (a) — uoh,v — voh c; intuitively, c is just the = a and h(b) = b. and [a, b], Then on [a, let h b] the give a parametric representation of another same curve c traversed at a different rate. Show, directly from the definition of length, that the length of c on [a, b] equals the length of c on [a b ] , (b) Assuming differentiability of any functions required, show that the lengths are equal by using the integral formula for length, and the ap- propriate substitution. 30. Find the length of the following curves, tions, except for which (iii), f(x) = -(x 2 fix) = x3 + 2) 3 all described as the graphs of func- represented parametrically. is <x < /2 , 1. 1 + 1 < x < 2. 12a in (iv) (v) (vi) 31. = a cos f{x) = log(cosA-), f{x) = logx, x fix) = arcsin e x t. = a 3 sin 3 • V <X < 1 ', <x < < t /, t < In. 7T/6. e. — log 2 < x < 0. For the following functions, find the length of the graph in polar coordinates. (i) fie) = a cose. (ii) /(0) =a(l-cos0). (iii) (iv) (v) fie)=asm 2 ie/2). 0<0<2tt. f(0) = 0<6>< /(0) = 3sec<9 7r/3. . 391 19. Integration in Elementary Terms 32. Problem 8 of the Appendix In to Chapter 12 we described the cycloid, which has the parametric representation — x = u{t) — a(t y sin t), = v(T) = — a(\ cost). (a) Find the length of one arch of the cycloid. [Answer: 8a.] (b) Recall that the cycloid is the graph of v o u Find the area under . one arch of the cycloid by using the appropriate substitution in f f and evaluating the resultant integral. [Answer: 3jza-.] 33. Use induction and integration by za f 34. If /' is Problem 14-10: parts to generalize ^*-jf(f(-(f «*)*)•••)*• continuous on Lemma Lebesgue use integration by parts to prove the [a, b], Riemann- for /: lim f(t)sm(Xt)dt / = 0. Problem 15-26, but it can be used to prove the general case (in much the same way that the Riemann-Lebesgue Lemma was derived in Problem 15-26 from the special case in which / is a step This result is just a special case of function). 35. The Mean Value Theorem for Integrals was introduced in Problem 13-23. The "Second Mean Value Theorem for Integrals" states the following. Supis either nondecreasing or pose that / is integrable on [a b] and that nonincreasing on [a b] Then there is a number £ in [a b] such that , . , f f(x)<j>(x)dx In this problem, tiable, will = <Ka) f f(x)dx assume that with a continuous derivative Prove that (a) we , if the result is / + <Kb) f f{x)dx. continuous and that is 4> is differen- 4>' true for nonincreasing 0, then it is also true for nondecreasing 0. (b) Prove that then Thus, we have it is if the result true for all we can assume is true for nonincreasing satisfying (p(b) — 0, nonincreasing 0. that is nonincreasing and (f)(b) = 0. In this case, to prove that f f(x)<P(x)dx=(f)(a) f f(x)dx. Ja by using integration by (c) Prove this (d) Show that the hypothesis that is needed. Ja is parts. either nondecreasing or nonincreasing . 392 Derivatives and Integrals Second Mean Value Theorem for Integrals, the general case could be derived by some approximation arguments, just as in the case of the Riemann-Lebesgue Lemma. But there is a more instructive way, outlined From in the next 36. (a) case of the this special problem. Given a\ ... , ,an and b\, . bn , . = let Sk , = s\(b\ - \-a n bn a\b\ H (*) . b£i + a\ • + s 2 (b 2 - + • Show cik- b?) h s n _\{b n -\ H that - bn ) + sn bn is sometimes called "Abel's formula for summation regarded as an analogue for sums of the integration by parts formula This disarmingly simple formula by parts." may be It rb rb f'(x)g(x)dx I = f(b)g(b)-f(a)g(a)- f(x)g'(x)dx, Ja especially P = {to, . J(i we if . tn , ) Riemann sums (Chapter use of [a, b], the left side 13, Appendix). In fact, for a partition approximately is n £ (1) - f'(tk)g ftfc_i)ftfc fjfc_l), k=\ while the right side approximately is n f(b)g(b) - - J2 f(a)g(a) f<fk)g'(?k)<!k ~ fc-l) Jfc=l which is approximately f(b)g(b) - f(a)g(a) - J^ f(r k ) 8W ~ 8{tk ~° Uk ~ fc-l) 'k k=\ = - f(b)g(b) f(a)g(a) ~tk-\ + J2 f('k)[8('k-0- 8(t k )] k=\ n = f(b)g(b) - f(a)g(a) + £[/(**) - f(a)} [g(fft _i) - g(tk )] k=\ n + f(a)J2s(t k -\)-g('k)k=\ sum Since the right-most is just g(a) — g(b), this works out to be [8(fk-l) ~ n - [f(b) (2) f(a)]g(b) + £[/(fc) - /(«)] ' *('*)] k=\ If we choose «k = /'('*)('* h -**-l). =g(fk-i) then (1) is k=\ which is the left side of (*), while k k sk = ^/'(/,)(/,( = - f;_i) is approximately ]T f(tj) - /(f,-_i) /=1 1 so (2) is approximately sn bn + /^k&k ~ k=\ which is the right side of (*). h k-\)> = f(t k ) - f(a), . 19. Integration in Elementary Terms This discussion is 393 not meant to suggest that Abel's formula can actually be derived from the formula for integration by parts, or vice versa. But, as we formula can where the functions in shall see, Abel's often be used as a substitute for integration by parts in situations question aren't differentiable. (b) Suppose that {b n } m < Prove Abel's for all n. > nonincreasing, with b„ is a\ -\ each n, and that M an < \- for Lemma: + b\m < a\b\ • • + a„b n < • b\M. (And, moreover, bk m < akbk a formula which only looks (c) / be Let — cp(b) integrable 0. P = Let on a n bn V -\ more [a b] , < b k M, general, but really and let isn't.) be nonincreasing on ...,/„} be a partition of [«,&]. {to, [a b] with , Show sum n ^/fe-iWfe-iXfi-fi-i) /=1 lies between the smallest and the largest of the sums A: d>(a)^2f(tt.i){ti-ti Conclude i). that f{x)(p{x)dx I, lies maximum between the minimum and the 0(a) f of f(t)dt, Ja and that it therefore equals </>(«) f(t)dt for some § / in [a, b]. Ja 37. (a) Show (i) (ii) (b) that the following f Jo sin sii ( x H : — improper ) integrals dx ^(x + I)*, Decide which of the following improper (i) (ii) both converge. fsing / sin" I - dx. I d: integrals converge. that the . 394 Derivatives and Integrals 38. Compute (a) the (improper) integral \ogxdx. / Jo Show (b) that the improper integral log(sinx)Jx converges. / Jo Use the (c) substitution x — 2w to show that '7t/2 / = log(sinx) dx 2 '71/2 Jo + \og(sinx)dx I 2 Jo log(cosjc) I dx + tt log2. Jo nn/2 Compute (d) log(cos.v) dx. / Jo Using the relation cosx (e) = r log(sinjt)<fo. — x), compute sin(7r/2 / Jo 39. Prove the following version of integration by parts for improper \oo /"OO u'(x)v(x)dx / = Ja The right side means, of course, lim u(x)v(x) *40. One u(x)v'(x)dx. / Ja in symbol on the first /»00 — u(x)v(x)\ — u(a)v(a). of the most important functions in analysis roo= integrals: e~'t f x - is the gamma function, ] dt. Jo Prove that the improper integral V(x) is defined if x > 0. Use integration by parts (more precisely, the improper integral version (a) (b) previous problem) to prove that in the ru+ Show (c) that T(l) = 1, = \) x r(x). and conclude that V(n) = (n — 1)! for all natural numbers n The gamma function thus provides a simple example of a continuous function which "interpolates" the values of x > 0. However, the o and f(x (a) Of course there — 1)!; there are r is A states that + 1) gamma for all function has the important additional property convex, a condition which expresses the extreme smoothness this function. Mollerup HI, n. infinitely that log of numbers many continuous functions / with f{n) — (n many continuous functions / with f(x + 1) = xf(x) are infinitely even n\ for natural = Harold Bohr and Johannes the only function / with logo/ convex, /(l) = 1 beautiful V is theorem due to xf(x). See reference [43] of the Suggested Reading. Use the reduction formula for i.JT/2 / $m n xdx n-\ - 1 sin" Jo x dx r / " Jo to show that 12 Sill " x dx. ) ' 19. Integration in Elementary Terms (b) Now show 395 that / • sin I 2n+l L =— , x ax 3 Jo I 71 ' 2 -In sin" / • 3 5 4 6 1 2« - 1 • ' 2 Jo and conclude + 2n 7 5 * —i = — * dx 2n 6 ^ — • 2 2/2 ' that I 7T 2 2 4 4 6 6 2 13 3 5 5 7 In ln - In 2 In sin " x dx Jo + 2/i 1 71 ' / /, C* 12 1 sin / 2 " +1 x^x Jo (c) Show 1 and that the quotient of the + 1 < This 2 " +1 < x 2 sin " < x 2 4 4 6 6 13 3 5 5 7 can be made sin 2 " -1 x <x < for 2n n/2. 2n - In as close to 7r/2 as desired, known In 1 + 1 an usually written as is infinite as Wallis' product: n 4 4 6 6 3 '3 "5 '5 '7 2 2 ~I" 2 Show between which shows that the products 2 (d) is , result, product, integrals in this expression /2n starting with the inequalities 1 sin two ' " "" also that the products 1 v^ can be made problem and Wallis' 2-4-6-...1 3 • 5 • . . . • 2n (2n - as close to y/n as desired. 1 (This fact is used in the next Problem 27-19.) in procedure was quite different! He worked 1 with the integral / () ( 1 — x 1 )" dx (which appears in Problem 42), hoping to recover, from the values obtained for natural numbers /;, a formula for j=/"(l-.v 2 )'/4 A complete account can be found following summary 1 f , ( in reference [49] gives the basic idea. Wallis in /,. Jo 2 first of the Suggested Reading, but the obtained the formula In 4 (2-4---2n) 2 2-3-4---2n(2n He 2" + 1) 2n + then reasoned that n/4 should be ?1 (1-1)- I Jo d-, 2 1/2 ) ^=y^f =(^!) 2 . (n!) 1 2 (2/i)!' ' 396 Derivatives and Integrals we interpret ^! to mean T(l + \), this agrees with Problem 45, but Wallis did not know of the gamma function (which was invented by Euler, guided principally by Wallis' If Since (2n)!/(n!) work). finding for ) ( the binomial coefficient is =q= p Wallis ), I hoped to find \\ by Now 1/2. {p+qHp+q- \)-(p+\, rand this makes sense even p if not a natural number. Wallis therefore decided that is +q (h (*+«)' 1 2 With +q p of this interpretation for = p 1/2, true that it is still P p+q + l q+\ 'P (p P j Denoting +q P by W(q) W(q + 1) equation can be written this + 2q + 3 -r—W(q) = ^—-W{q), + 2q + 2 + ^7 = 1 </ q which leads \ to the table q 2 1 3 1 ' 2 But, since W(^) should be 4/n, Next Wallis notes (this 3), that appearing if 4 2 rt 3 a4 a\ en «3 logo(l/W) ' 6 2 3 of these 4 successive values W(g), W(g tables, 2q 2^ 5 3 ;r since is 7 4 S 7T tt in either 5 ' 2 a\, ai, 03, 04 are a2 says that 3 I 2 wu'' 5 Wallis also constructs the table I a '/ W(q + +q + 1), W(<7 + 2), then +3 > +2 7 2«/ z<? -t- <+ zcy +o convex, compare the remarks before Problem 41), which implies that V Wallis then argues that this should combined values in a table 4 In +4 + jr 3 5 2« 4 4 6 in " \| 6 ' ' ir 3 ' 5 2/;+ 3 4 jr' 3 2 4 be true 6 5 7 ' when a\, a-i, 03, a$ are four successive given both integer and half-integer values! Thus, \), 6 W{n), W(n 2n +2 2/! + ' 5 ' is W(n + 4 3 1 still where q taking as the four successive values 4 V a2 ai a\ ln+ + 3 2 1 3 1 2n 2 |), 5 ' 4 In 4 In 7 5 ' W(n + ' + + 2/; 6 2a; +4 +3 4 7T 2-4-4-6-6---(2«)(2n)(2« 3-3-5-5---(2« + l)(2n from which Willis" producl follows immediately. 3 2 + 2#i which yields simply 2/1 1), + 2) 2n I) 2n +3 +2 1 he obtains 397 19. Integration in Elementary Terms />oo 42. It is an astonishing fact that improper integrals f(x)dx can / often be Jo b computed where ordinary in cases f integrals f(x)dx cannot. There / is no Ja h f elementary formula for e / _x 2 dx but we can find the value of ... , Ja f°° / e _x 2 dx J° There are many ways of evaluating this integral, but most require some advanced techniques; the following method involves a fair amount of precisely! work, but no (a) Show you do not already know. facts that that f -x 2 (1 n dx ) = 2 4 3 5 Jo 1 (l+.t I,o 2 )" = dx In + 2« 7T In 3 1 2 1 "" 4 2 2n-2' done using reduction formulas, or by appropriate combined with the previous problem.) (This can be tions, (b) Prove, using the derivative, that 1 e (c) substitu- x _x < e~ x 2 2 for <x < for < 1 < ~ r +x 2 1 1. x. Integrate the nth powers of these inequalities from oo, respectively. use the substitution y 4 ,-2 sjn Then - y/nx to and from show that 1 to In In 5 3 = to + 1 < dy < e~ / : ' dy Jo n TT In 3 1 2^2-4" (d) Now use Problem 4 1 (d) I **43. (a) Use to show f sin.v dx Ja — that " y= e 2 integration by parts to I show cosfl that cos 6 and conclude that j the limit as a —> f / a x In -3 -2 (sinx)/x dx b exists. J cosx r— dx — x l tl (Use the + and the right side for the , left limit as side to investigate b —> oo.) 398 Derivatives and Integrals (b) Use Problem 15-33 show to T A)* dt Jo (c) + sin(/z /' for that = Ti sin any natural number n. Prove that f + sm(X lim A—>-00 Jo \)t = rff sin 0. 2J The term in brackets is bounded by Problem Riemann-Lebesgue Lemma then applies. Hint: (d) Use the substitution u = + (X ^)t and part C°° sin X show (b) to 15-2(vi); the that 71 x JO /•OO 44. Given the value of (sin / x)/xdx from Problem 43, compute ./O sin x dx by using integration by play an important *45. (a) Use the parts. (As in Problem 38, the formula for sin 2x will role.) substitution u = x t show to ^ 1 = - r(x) that -" Ux du. / x Jo 46. (b) Find r(i). (a) Suppose and fix) that is that lim f(x) integrable = A on every and lim f(x) = < interval [a, b] for B. Prove that for all a, a yS < /?, > we have dx f Jo Hint: To — (A - B) log x estimate /' ' f(ax)-f(Bx) — a dx use two different substitutions. OO (b) Now suppose instead that lim f{x) — A. Prove that r Jo fiotx ) f°° f(x) / Ja ~ x dx converges X f(M ax = A log 8 — a for all a > and that 399 19. Integration in Elementary Terms Compute (c) the following integrals: 10 e (i) -<*x f — e ~P * Jo f°° cos(ax) — cosifix) dx. X Jo In Chapter 13 x we said, rather blithely, that integrals may be computed to any degree of accuracy desired by calculating lower and upper sums. But an applied mathematician, doing may it, an integral needed who really has to to three ought decimal places, say in [tj-\, (a to mention computing upper and lower sums might might not be possible to compute the quantities m, it [/,-_i, ti\. n It is and consider >^ /(*;) (U far — more reasonable simply This represents the ?j_i). i=\ of certain rectangles which partially overlap the graph of Appendix to Chapter the trapezoids FIGURE shown 13. But we will get a much / — better result to pick points sum we t of the areas see Figure if x 1 in the instead choose in Figure 2. 2 Suppose, in particular, that we divide [a,b] into n equal intervals, by the points tj Then refined at the outset that for each interval ti\ about degree of accuracy that might easily be The next three problems show how more calculations much more efficient. not even be practical, since and Mi calculation, rather than just talking in certain circumstances). methods can make the We do the not be overjoyed at the prospect of computing lower sums to evaluate = a + the trapezoid with base [f,_i, b —a . a i tj] + ih. has area (t< -ti-l) means of 400 Derivatives and Integrals and the sum of all these areas S„ f{t\ =h ) + is simply f(a) f(t2 ) 2 + f(ti) fQ>) + 2 ./"(/„-!) 2 n-\ — b = h a i=] This method of approximating an integral obtain to Tj2,i Yj2„ is from £„ just isn't it So 2^«- called the trapezoid is rule. necessary to recompute the old /(?;); their contribution in practice it is compute best to Y,i, £4, £85 /.In / the next problem we will estimate Ja 47. (a) Suppose • • to get / / — E„. Ja that /" continuous. Let is P, be the linear function which agrees / at ti-\ and /,. Using Problem minimum and maximum of /" on with the • pb pb approximations to Notice that to 1 -f / (x - r,_i)(.v 1-46, show that and if n, and jV, are [f,-_i, t{\ - tj)dx then ml > (b) Evaluate Nil ['' / to get nth* [' Nik > 3 l2~ (c) Conclude that there is some c in (a, b) with b (b-a) 3 „ 1 Notice that the "error term" (/? — the error obtained using ordinary We can obtain still more accurate I. 3 the interval [a, b] 48. (a) is Suppose We that a < 2 which = and b = agrees with 2. / we approximate / by first Let Conclude P be (while quadratic i^4-^) when 3). the polynomial function and 2 (Problem /(2)]. 3 that in the general case P = Ja 2 consider what happens at 0, 1, p = -[/(0) 4-4/(1) + /o varies as \/n varies as 1/n). that (b) 2 divided into two equal intervals (Figure first of degree sums results if functions rather than by linear functions. FIGUR a) 3 f"{c)/\2n 3-6). Show 19. Integration in Elementary Terms pb pb Naturally (c) P = J The a quadratic polynomial. is But, re- J ci same relation holds when / is a cubic polynomial! using Problem 11-46; note that f" is a constant. markably enough, this, f when / j Ja Prove 401 this we do previous problem shows that new not have to do any calculations b to f compute a+b we Q when Q / a is cubic polynomial which agrees with 49. (a) much more lee-way in Show Q(a) that there = is Q(b) f(a), + A(x - a)(x - b) defined on is P(x) [a, b], — - Q(x) = (x-a)(x 2 a+b (x ) ) for some A. [a, b\ we have x then for every x in a+b\ 2 / f(x) / f / 1 Prove that ' satisfying ,(a-\ -r- = I Q use to our advantage: ^) = Q\ f(b), = ( we can C = Hint: Clearly Q(x) / , choosing Q, which ,(a+b\ if a+b a cubic polynomial function <2 (b) - byf {4) (^) 4! some ^ in (a,b). Hint: Imitate the proof of Problem 11-46. is continuous, then Conclude that if / for (c) ( ' b-a f = /a r - f f(a)+4f n + bi.\ . fib) Ja for (d) some Now (b-a) 5 2880 H4) r\c) c in (a,b). divide [a b] into 2n intervals , t; — a + ih, h by means of the points = —a b 2n Prove Simpson's b 6n I. rule: n-\ " b-a - ( /(«) +4 £ f(to-i) (b-a) 5 2880h for and + 4f -=-. + /(*) , f(a) / is at a, b, have still b-a But there / some c in (a, b). 4 / (4) + 2Y. fted + f {b) (c) 402 Derivatives and Integrals THE COSMOPOLITAN INTEGRAL APPENDIX. We under the graph of originally introduced integrals in order to find the area a function, but the integral is considerably more than versatile For example, that. Problem 1 3-24 used the integral to express the area of a region of quite another sort. Moreover, Problem 13-25 showed that the integral can also be used to ex- press the lengths of curves lot of work ably a little may be more — though, as we've seen in Appendix surprising, since the integral seems, at two-dimensional creature. Actually, the integral makes few geometric formulas, which we formulas to Chapter 13, a necessary to consider the general case! This result was prob- we will assume some present in will this first its blush, to be a very appearance in quite a Appendix. To derive these from elementary geometry (and allow a results little fudging). Instead of going down three-dimensional ones. be expressed by to one-dimensional objects, we'll begin by tackling some There are some very The integrals. special solids V simplest such solid a "solid of revolution," is obtained by revolving the region under the graph of / when we regard the horizontal axis, FIGURE whose volumes can > on [a b] . around the plane as situated in space (Figure 1). 1 If p = {to,...,tn } is any partition of [a,&], and w, and M, have their usual meanings, then 7tmi is the volume of a irMi~{tj —/,_]) and /,. is that lies disc the volume of a 2 (ti -r,_i) the inside solid (Figure disc that contains the part of Similarly, 2). V between t(-\ Consequently, n n tt FIGURE V y^mrOi - *i-i) 2 S volume V < n ^PM, (f, - ;=1 2 f,_i). 1=1 But the sums on the ends of this inequality are just the lower and upper sums for 2 f on [a, b]: 7T L(f 2 , Consequently, the volume of P) < volume V < n U(f 2 V must be volume V , P). given by = n /' I f(x)dx. Ja This method of finding volumes Figure 3 shows a FIGURE < is affectionately referred to as the "disc more complicated solid V method/' obtained by revolving the region under the graph of / around the vertical axis (V is the solid left over when we start with the big cylinder of radius b and take away both the small cylinder of radius a and the solid V\ sitting right on top of it). In this case we assume a > as well . 19, Appendix. as / > Figures 4 and 5 indicate 0. FIGURE some other The Cosmopolitan possible shapes for For a partition P = rectangle with base shells we ...,/„} {to, [f,-_i, t{\ we tx and height m, or M-, (Figure 2 - 2 r,_i ) < volume V < n ^M^f, 2 - 2 f,-_i ), /=i as n /; ^mi{ti +ti-\)(tt - tt-i) < volume V < Ti^M^t, (=1 Now the volumes n y]m,-(f,- which we can write 6 Adding 6). obtain i=l FIGURE 5 consider the "shells" obtained by rotating the /; 7i V 4 FIGURE of these 403 Integral these Appendix +f,-_i)(r,- -/,_i). /=l sums are not lower or upper sums of anything. But Problem Chapter 13 shows that each sum 1 of the to and ^m,7,(f, -tt-i) can be made as close as desired to J^mjfj_i(fj -/,-i) xf(x)dx by choosing / the lengths f, — f,_i Ja small enough. The same is true of the sums on the right, so we find that b volume V — f 7.71 \ xf(x)dx; Ja this is The the so-called "shell method" of finding volumes. surface area of certain curved regions can also be expressed in terms of inte- we tackle complicated regions, a little review of elementary geometric formulas may be appreciated here. Figure 7 shows a right pyramid made up of triangles with bases of length / and altitude s The total surface area of the sides of the pyramid is thus grals. Before . 1 FIGURE ps, 7 where p is the perimeter of the base. By choosing the base to be a regular polygon 404 Derivatives and Integrals with a large number of sides we see that the area of a right circular cone (Figure 8) must be = — (27tr)s where FIGURE s height s 8 is and and radii r\ ri frustum of a cone with slant Finally, consider the the "slant height." in Figure 9(b), Ttrs, shown Completing in Figure 9(a). this to a cone, as we have s\ + s\ .v r? r\ so s\ r\s = >"2 Consequently, the surface area 7zri(s\ + s) — s\ _ r2 ''1 Now consider = 7ir\s\ r2 2 tts cones, as in Figure 10. £[/ft_l) + JT — ~ r\ r P = The f(ti)]y/(ti {tQ tn total surface - r\ + r,-!) 2 * £[/fe-i) + \ 2 = ns{r\ + ri). \ formed by revolving the surface zontal axis. For a partition ~ is r2 (a) r->s +s = graph of / around the hori- the we can inscribe a series of frustums of area of these frustums - [f(t,) f(U)\\ 1 /ft_l)] + : f{t'- fit,) ( t, is -t,-\ l) ) i=l By the Mean Value Theorem, this is (b) n FIGURE 9 JT £[/fe-rl) + /U/)]>/l+/'(jC 2 ; ) a - fi-l) 1=1 for some we conclude that the surface area x, in (f,-_i, ?j). Appealing to Problem 1 of the Appendix to Chapter 13, is 2n I f(x)\/\ + fix) 2 dx. Ja PROBLEMS 1. (a) Find the volume of the solid obtained by revolving the region bounded by the graphs of f(x) (b) i< i. in 2. and f(x) = x around the horizontal Find the volume of the solid obtained by revolving around the iii.i =x vertical axis. Find the volume of a sphere of radius r. this axis. same region 3 19, Appendix. 3. When the ellipse consisting of all FIGURE 4. 1 we volume of the enclosed (x a) 2 405 + solid. 1 Find the volume of the "torus" (Figure — Integral y /b — 1 iIN obtain an "ellipsoid of revolution" points (x, y) with x l /a~ rotated around the horizontal axis (Figure 11). Find the The Cosmopolitan +y = 2 b 2 (a > b) around the 12), obtained by rotating the circle vertical axis. length b FIGURE 5. A 12 cylindrical hole of radius a radius 2a (Figure [GURE 1 13). is bored through the center of a sphere of Find the volume of the remaining solid. 406 Derivatives and Integrals 6. For the solid shown in Figure 14, find the volume by the (a) FIGURE shell 14 This volume can also be evaluated by the disc method. (b) method. must be evaluated the integral which complicated. The Write in this case; notice that next problem takes up a question which it is this down more might suggest. 7. Figure 15 shows a cylinder of height b and radius f(b), divided into three solids, one of which, V\, is a cylinder of height a and radius f(a). If / one-one, then a comparison of the disk method and the shell method of is computing volumes leads us nbf{b)~ — 7zaf{a) —n to believe that J f(x) dx = volume Vi Ja rfib) f(b) l ' yf~ (y)dy. I Jf(a fu>) Prove this analytically, using the formula for more simply by going through I I ( , I RE I 5 / / from Problem 19-16, or the steps by which this formula was derived. 19, Appendix. Figure The Cosmopolitan Integral 6 shows a solid with a circular base of radius a 1 perpendicular to the diameter AB . 407 Each plane intersects the solid in a square. Using arguments similar to those already used in this Appendix, express the volume of the (b) 9. Same problem if an integral, and evaluate A of it. each plane intersects the solid in an equilateral triangle. Find the volume of a pyramid (Figure 17) area FIGURE solid as in terms of its height h and the base. its 16 FKiURE 10. 17 Find the volume of the solid which is the intersection of the two cylinders Hint: Find the intersection of this solid with each horizontal in Figure 18. plane. FIGURE 11. 18 (a) Prove that the surface area of a sphere of radius (b) Prove, more in Figure 19 is 2nrh. (Notice that position of the planes.) FIGURE 1 I * r is Ativ . shown not on the generally, that the area of the portion of the sphere this depends only on h, 408 Derivatives and Integrals A circular mud boards of length that it puddle can just be covered by a parallel collection of of the at least the radius circle, as in cannot be covered by the same boards non-parallel configuration, as in if Figure 20(a). Prove they are arranged in any (b). (b) FIGURE 12. 13. 2 (a) Find the surface area of the ellipsoid of revolution (b) Find the surface area of the torus The graph of f(x) — l/x, x > 1 in is Problem in Problem 19-3. 19-4. revolved around the horizontal axis (Figure 21). (a) Find the volume of the enclosed (b) Show (c) that the surface area surface FIGURE we "infinite trumpet." infinite. fill up the trumpet with the finite amount of paint found would seem that we have thereby coated the infinite inside area with only a finite amount of paint. How is this possible? Suppose in part is that (a). 2 It PART INFINITE SEQUENCES AND INFINITE SERIES One of the most remarkable m x+ m(m-l)„ + 1 1 m(m - 5 -2 l)(m - 2) ^ [m - (n -2-3 1 m(m - 1) • • • 1-2 - the n is is • • series, then finite, can be expressed, known, by When m • a positive whole number sum of the which as is ^ 1)] + When m of is the following: algebraic analysis 1 series is (I + x) m . not an integer, and the series goes on to infinity, it will converge or diverge according as the quantities m and x have this or that value. In this case, one writes the same equality + (1 x) m = 1 + . . . It is x H m(m — — 1) ^ ' 1-2 „ x2 + • • • etc. assumed that the numerical equality will always occur whenever the ill is series is convergent, but has never yet been proved. NIELS HENRIK ABEL APPROXIMATION BY POLYNOMIAL FUNCTIONS mm ^/ CHAPTER There is If a polynomial function, p is one sense in which the "elementary functions" are not elementary p(x) at all. =a + aix + ---+a n x", then p(x) can be computed easily for any number This x. functions like sin, log, or exp. At present, to find log* is not at = fi 1/t dt all true for approximately, we must compute some upper or lower sums, and make certain that the error involved in accepting such a sum for log* is not too great. Computing e x = log" (x) would be even more difficult: we would have to compute log a for many values of a until we found a number a such that log a is approximately x then a would be approximately e — x . we will obtain important theoretical results which reduce the computation of f(x), for many functions /, to the evaluation of polynomial functions. The method depends on finding polynomial functions which are close approximations to /. In order to guess a polynomial which is appropriate, In this chapter it is useful examine polynomial functions themselves more thoroughly. Suppose that to first p(x) and = + a\x a H a n x". \- our purposes very important, to note that the coefficients a, can be expressed in terms of the value of p and its various derivatives at 0. To begin with, note that It is interesting, for p(0) =a . Differentiating the original expression for p(x) yields = a\ + 2a 2 x + p'(x) \- na n x"~ l . Therefore, =p (» 0)=a p '(0) Differentiating again p"(x) we = ( 1 . obtain 2a 2 +3 • 2 • a3x + + n{n - 1) • a n x"- 2 . Therefore, p"(0) In general, we will = p a \0) = 2a 2 . have P WI (0)=*! ai . or ak = ^. k\ If we agree to define holds for k 411 = also. 0! = 1, and recall the notation p (0) = p, then this formula 412 Infinite Sequences and Infinite Series If we had begun with a function p that was written = ao + a\(x - a) H p(x) as a "polynomial in (x — a)," a n (x -a)", h then a similar argument would show that p Suppose now that / is a) (a) = ~kT- ak a function (not necessarily a polynomial) such that fQ\a),...,f M (a) all exist. Let ak : —— = 7] 0<k<n, , and define Pn ,a(x) — The polynomial Pnfl (Strictly speaking, we to indicate the «o +a\(x — a) H a)". called the has been defined so that PIIM ik) (a) = f {k) (a) it is - fl„(.x- Taylor polynomial of degree n for/ at a. should use an even more complicated expression, like Pn ,a,f, dependence on /; at times this more precise notation will be useful.) is The Taylor polynomial in fact, h clearly the only Although the for < k < n\ polynomial of degree < n with coefficients of Pn ,a,f seem to this properly. depend upon / in a fairly compli- cated way, the most important elementary functions have extremely simple Taylor polynomials. Consider first the function sin. sin(O) sin'(O) sm'(0) sin"'(0) sin From this (4 '(0) We have = 0, =cos0= 1, = -sin = 0, = -cos0 = -l, = sin = 0. point on, the derivatives repeat in a cycle of 4. The numbers (A) av = sin (0) k\ are 0,1.0,-1.0.1.0.-1.0,1 Therefore the Taylor polynomial P2n+1,0 of degree 2n 3 P2n+ i,o(*) = x (Of course, P2ll , i.o = 5 + 1 x x x -_ + ---- + •• + (- P2«+2,0)- 1 for sin at v is 2 " +1 )" - 1 (2|I+1)! 20. Approximation by Polynomial Functions The Taylor polynomial left Pi n ,Q of degree 2n for cos at to you) „2„4„6 P2n.0(x)= - 1 — + -JT--ZT + 4! 2! The Taylor polynomial exp(0) = for = The Taylor polynomial not even defined at is exp is + l + (-1)" for log X2 compute. Since exp (k) (0) log (x) is n must be computed = = X X - + X_ + _+... + -. The standard 0. at 4 3 " (2n)! 6! especially easy to X + computations are x 2n -'- Taylor polynomial of degree n for all k, the 1 p„„(.v) log (the is 413 choice a is at some point a — 1. log'(l) = 1; = -\, z log"(l) = -l; = ^, log" (l) -, / 0, since Then x \og"(x) x / log "U) # = 2; in general log w (-l)*- 1 ^-!)! (jc) x Therefore the Taylor polynomial of degree n for log PnA (x) It is case often = (jc - 1) 15 (x-1) 2 - - more convenient we can choose a = f {k) = (x) + , + • • • is 1 (-\)"- + l (x - If . n to consider the function We 0. U-l) 3 at f(x) = log(l +jc). In this have \og {k) a+x), so /«(0)=tog cfc) = (l) 1 (-l)*- (*-l)!. Therefore the Taylor polynomial of degree n for / PBi0 (x) There arctan. is x2 4 3 x x =x- — + — - — + 2 4 ••• + is (-l)"- 1 *" . n 3 one other elementary function whose Taylor polynomial The computations arctan'U) = arctan" (x) (x) + xl -2x «-=-, =— 2 2 (1 = +.v (\+x 2 is important— of the derivatives begin x 1 arctan at arctan'(O) = 1; arctan" (0) = 0; arctan = —2. ) 2 ) — -(-2) (1 + 2x.2(\+x 2 )-2x ^—. 4 +x 2 ) , (()) , 414 Infinite Sequences and Infinite Series It is computation clear that this brute force polynomials of arctan be easy will to find after of Taylor polynomials more closely One line mials for which Pn ,a,f W U1 we have examined the properties — although simply defined so as to have the same between / and never do. However, the Taylor will the Taylor polynomial PnM .f was n derivatives at a as /, the connection first much actually turn out to be deeper. of evidence for a closer connection between / may be uncovered by examining / and the Taylor polyno- the Taylor polynomial of degree 1 is P La (x) = f(a) + f(a)(x-a). Notice that f(x)-P lM (x) — x Now, by x a — f(a). a we have the definition of f'{a) f(x)-Pha (x) lim x^-a FIGURE f(x)-f(a) x — = (J. a 1 — P\ >a (x ) not to x — a. Figure In other words, as x approaches a the difference f(x) small, but actually becomes small even graph of fix) = e x is 1 only becomes illustrates the and of Pl,0(x) which compared = f(0) + f(0)x= the Taylor polynomial of degree 1 for / 1 +x, The diagram at 0. also shows As x approaches 0, the the graph of ft.o(*) which is = /<P) + /'(0) + ^r* the Taylor polynomial of degree 2 for difference f(x) — P2,o(x) seems to 2 / = l +x + at 0. y • be getting small even faster than the difference 20. Approximation by Polynomial Functions f(x) — P\o(x). As it stands, this assertion is not very precise, but we are prepared to give it a definite meaning. We have just noted that in general /w-ft,w Bm x^-a For f(x) — e x — and a this e x -\-x = x^o Thus, although /(a) become = Q X an easy double application of l'Hopitafs Rule shows that hm not «--!-« ]im jc->-0 ,. does =ft that x x-»0 the other hand, means — small x l n 1 - ^ 0. 2 P[,o(x) becomes small compared to x, as * approaches 0, it compared toi". For P2.0C*) the situation is quite different; the extra term x /2 provides just the right compensation: e'-\-x-hm e'-l-x = hm X2 x->0 2x x^-0 = ^— e = lim x^o This result holds in general — f'(a) if ,. hm . THEOREM l (x analogous assertion for Suppose that / is — PnM is 1 0. exist, = then U; also true. a function for which f "\a) ( f'(a) all exist. a) 1 x 2 and f"(a) f(x)-P2 a~ (x) x-*a in fact, the now a /U)-/YoU) = |im On — x 415 Let ai = f (k) (a) n < k < n. H a n (x . ~ , k\ ~ and define Pn ,a(x) — ao + a\(x - a) \- Then hm *-*•« f(x)-Pn — a (x) , (x — a)" —= _ U. - a)". . 416 Infinite Sequences and Infinite Series PROOF Writing out PnM {x) explicitly, we obtain fM-T^—^ix-ay f(x)-P„ M (x) {x - a)" It will ' i=0 - (x new help to introduce the f a) (n) n {a) n\ functions " _1 Q( x ) now we must Y = ( ' ] (a) Lf——(x-a) 1 and = g(x) - (x a) n ; prove that hm f(x)-Q(x) x^a /W( = fl ) . g(x) n\ Notice that Q {k ik) g \a) =f {k k<n-\, \a), k =n\(x -a)"- (x) /{n - k)\ Thus = lim[/U) - Q(x)] - Q(a) = f(a) 0, x -> a - lim[/'U) lim[/ ( Q'(x)] = - Q "- 2) "- 2, (.t) ( f'ia) (x)] and Km g(x) = We may lim g'(x) - Q'{a) = 0, = f u '- 2) (a) - Q == — therefore apply l'Hopital's Rule n lim g ( "~ 2) u, - 2) = (x) '- u f(x)-Q(x) f"- (x)-Q = lim lim x^a x^a {x — a) n n\{x—a) l) 0. l) (x) . ; Q is a polynomial of degree {a). Thus \x) = f Q fact, (n {n { ,. lim x-*a and this last limit is f n — f{x)-Q(x) = (x — a)" un (a)/n\ ( of /, then, according to > 0, and a "- ,, ./• lim l)st derivative is a constant; in U)-/ ", by definition of f {n) (a). Theorem local ,, («) n\(x—a) x-*-a 1 /"(a) — 1, its (n {) One simple consequence of Theorem maxima and minima which was developed at a if 0. times to obtain 1 ,. Since = (a) allows us to perfect the test for local in Chapter 11-5, the function maximum | at a if 11. / f"(a) < If a is a critical has a local 0. If point minimum f"(a) = no — . 417 20. Approximation by Polynomial Functions conclusion was possible, but and further information; significant. Even more nor a ask =/ --. f(x) = g(x) = -(x-a) n (M - 1) = 0, (a) (x-a)'\ 2) that if n minimum local might be (a)^0. Notice (Figure satisfy (*). maximum "} (4) / (tf) = what happens when then the sign of can be guessed by examining the functions situation in this case which { 0, /» = = /'(a) f = we can generally, (*) The conceivable that the sign of f'"{a) might give is it f"'{a) if / point for , is odd, then a On or g. is neither a local the other hand, if n is minimum at a, while maximum at a. Of all functions even, then /, with a positive nth derivative, has a local with a negative nth derivative, has a local g, satisfying these are about the simplest available; nevertheless they indicate the (*), In fact, the whole point of the next proof general situation exactly. function satisfying is theorem 2 made much looks very (*) by Theorem precise one of these functions, in that any a sense that 1 Suppose that /» = f is If n is (3) If n is (2) There / odd, then no clearly is (n) {n) even and f (a) > even and f (n \a) < n (1) If PROOF like is ••• (n ~ l) 0, n first — 0, has a local of generality in assuming that f(a) / derivatives of \ P„ a( x ) = 0, minimum at a. has a local maximum at a. local maximum nor a local minimum / then / then hypotheses nor the conclusion are affected since the (a) (a)^0. has neither a loss =f at / if a are = f(a)+ Lj^(x -«) + 0, since neither the replaced by is 0, the = / — f{a). Then, Taylor polynomial •••+ at a. —^(x J P, ua of / is - af ' (a) nodd = Thus, Theorem n 1 = f^Ha) if x , is (x figure 2 Suppose now f(x)/(x — a) n — lim that n - n (n) {a) n\ then , . (x-a) x^t-a Cl)" _ f fix) ,. sufficiently close to a, (x n even . f(x)-Pn a (x) = ,. lim fix) (b) -a) states that x-*-a Consequently, (x ; . has the same sign as (n f \a) a)" is even. In this case has the same sign as f {n) (.v — a)" > for all x ^ {a)/n\ for x sufficiendy close to a, a. it Since follows 418 Infinite Sequences and Infinite Series same that f(x) itself has the (n) f > ia) means 0, this sign as f"(a)/n\ for = /(*) > for x close to for the case Now e~ l/x x#0 \ a. f {n) / Consequently, < ia) is signs for a local for x > a and as before shows that a. always has the same y sign. 5 — a) n < Therefore f{x) has (jc < for x / This proves that a. has neither a local if x is different maximum nor because f {k) ia) maxima and minima question of local will settle the about any function which limitations, A similar proof works at a. | Although Theorem 2 just at a. then x > a and x < minimum minimum The same argument odd. n — a)" > fia) has a local -™(x-a) ix If 0. suppose that n sufficiently close to a, But sufficiently close to a. jc that arises in practice, may be for all k. This /(*) = e~ for does have some theoretical it happens (Figure 3(a)) for the function (b) /x [ x^O \ jc=0, 0. which has a minimum which has a and also for the negative of this function (Figure 0. Moreover (Figure 3(c)), if at 0, maximum at x e- l ' /(*) = x 0, -e~ X > fix) -e- = then x<0 atO. x 0, l/x \ The (c) FIGURE {k f \0) = for all k, but conclusion of Theorem at a \ 1 is = x <0, has neither a local Two cept of "order of equality." 3 / x > , l/x 3(b)), minimum nor a local maximum often expressed in terms of an important con- functions / and g are equal up to order n if m x-*a (x / up been designed degree < to to order n at make a. this fact true, n with this property. — = o. a)" Theorem 1 says that the Taylor polynomial The Taylor polynomial might very well have In the language of this definition, Pn.a.f equals -* (t) /(t) ii because there This assertion is is at most one polynomial of a consequence of the following elementary theorem. THEOREM 3 PROOl Let P and Q be two polynomials and Q Let R = P — Q. are equal up in (x to order n at a. Since R is a — a), of degree < n, and suppose that P Then P = Q. polynomial of degree < //, it is only necessary to . 20. Approximation by Polynomial Functions prove that 419 if = b + ... + bn (x-a) n R(x) satisfies lim-^= - x^a ( x R = then Now the 0. P(x) hm For = i this R hypothesis on *-+a (x — a) n surely imply that =0 < for condition reads simply lim R(x) = x^-a lim R{x) = Thus = bo +b lim [b =b < i n. a)' (x { - a) 0; on the other hand + bn (x - a) n + ' ] . and R(x) = - b (x l a) + - a) + bn (.x -a)". Therefore, —— =bi + R(x) X b 2 (x + b n {x - a) n - + x CI and lim x ^a Thus = Z?i Let / be of degree PROOF = bj. a in this way, we =b2 (x - a) 2 + • • • + b n (x - af find that b COROLLARY — and R(x) Continuing x = = bn = 0. | n -times differentiable at a, and suppose that P is a polynomial in (x < n, which equals / up to order n at a. Then P P = naf P and />„,«,/ both equal / up to order n at a, is easy to see that P Pn ,aj up to order n at a. Consequently, P = Pnaf by the Theorem. Since -a) . it equals | it At first sight might seem that / is this corollary appears to have unnecessarily complicated hypotheses; polynomial P would automatically imply that the existence of the sufficiently differentiable for Pn aJ , example (Figure 4), /(*) = x" +l 0, IfP(x) = 0, then order n at 0. On undefined. P to exist. But in fact this is not so. For suppose that , x irrational x rational. < n which equals / up to the other hand, f'{a) does not exist for any a 0, so /"(0) is is certainly a polynomial of degree # — 420 Infinite Sequences and Infinite Series x irrational = /(*) x rational 0, FIGURE 4 When / useful does have n derivatives at a, however, the corollary method that our r* arctan x — method of suggests a promising + t~, to obtain a 1 ^ dt \+t 2 finding a polynomial close to arctan — divide 1 polynomial plus a remainder: l-? 2 + / 4 -/ 6 + \+f in failure. \ I Jo 1 provide a attempt to find the Taylor polynomial for arctan ended first The equation by may Taylor polynomial of /. In particular, remember for finding the + --- (-l)"r" + /i+l, 2/1 + 2 (-l) n+l t 1+r This formula, which can be checked easily by multiplying both sides by 1 + t , shows that px arctan x -f l-t' + + t' (-\)"r"dt + (-\)" +l Jo x3 = According to v - x5 x -y y + / Jo px 2n+\ 2n+2 — { 1 + dt { ~ 2n+2 { 2/i + Jo 1 1 + t l dt. our corollary, the polynomial which appears here will be the Taylor polynomial of degree In + for arctan at 0, provided that 1 r x lim zn+ +2 ,2/1 t - r+i Jo dt .2«+l x-+0 Since \2n+3 l this is 2«+l clearly true. for arctan at + r 2 dt 2 < dt 2/Z+3 ^0 Thus we have found that the Taylor polynomial of degree is ^2/i+i.oU) = - 1 —+X X x 5 ,.2/i+l + (-D" 2//+ I 421 20. Approximation by Polynomial Functions By now the way, possible to we have that P2n+l,0(x) and discovered the Taylor polynomials of arctan, work backwards and since this polynomial /m,_ arctan (0) (0) find arctan (/:) (0) for all k: Since X5 A3 =X- — +— 3 5 by is, X 1 definition, ——a arctan (2) (0) + arctan*mn (0) + ,_ find arctan* (0) 2 " +l + (-1)"——, 2n + , 2 + " arctan (2 +1) (0) + by simply equating the 2 „. —x 2n+l + 1)! coefficients of xk (2n 2! we can it is , , in these two polynomials: ik) arctan (0) - — (J it A: even, is k\ arctan (2/+1) (0) (-1)' (2/+1)! 2/+1 A much more interesting fact „3 arctan a- emerges „5 =x- — + — 3 arctan or + 5 if (2/+1) (0) we go back x 2n+\ M (-1) 2n + r = (-!)' -(2/)!. to the original equation p\ + (-1)" t k 1 2n+2 *dt, / 1 + ^ 2 and remember the estimate Un+l l+t 2 Jo When x \2n+3 ~ In + 3 < 1, this expression is at most 1/(2/7 + 3), and we can make this as we like simply by choosing n large enough. In other words, for \x\ < |a| small as we can \ dt 1 arctan use the Taylor polynomials for to compute arctan a as accurately as we The most important theorems about Taylor polynomials extend like. this isolated result and the Taylor polynomials will soon play quite a new role. so far have always examined the behavior of the Taylor polynomial Pna for fixed n, as a approaches a. Henceforth we will compare Taylor polynomials PnM for fixed a, and different n. In anticipation of the coming theorem we introduce some new notation. to other functions, The theorems proved If / is a Rn ,a(x) by function for which f(x) = P„, a (x) = f(a) + Pn>a {x) + RnAx) f'(a)(x we define the — a)" exists, -«) + •••+ Jf [n) (a) K -L(x remainder term + R nM (x). nl We There would is One way like to have an expression for R„, a (x) whose such an expression, involving an integral, just as to guess this expression f(x) is size f(a) + R iU (x). easy to estimate. in the case for arctan. to begin with the case n = is = 0: 422 Infinite Sequences and Infinite Series The Fundamental Theorem of Calculus f(x) enables us to write = f(a)+ f'(t)dt, f so that Ja A similar expression for tion by parts R\ a (x) in a rather tricky can be derived from and u(t)-f'(t) (notice that formula using integra- this way: Let = t-x v(t) x represents some fixed number in the expression for v(t), so v'(t) = 1 ); then / f\t)dt= Ja f{t)- Jf a dt 1 I I u\t) v'{t) X = n\ f"(t){t-x)dt. u(t)v(t) Ja a Since v(x) = 0, we f(x) I | u'(t) v(t) obtain = f(a)+ Jf a f'(t)dt = f(a)-u(a)v(a)+ f f"(t)(x-t)dt Ja = + f'(a)(x-a)+ f f(a) f"(t)(x - t)dt. Ja Thus Rl,a(x)= It is It just hard to give any motivation for choosing happens to many v'{t) = f Jo - (x — — x, t t), = rather than v(t) sort of thing similar but futile manipulations. easy to guess the formula for u(t) then v(t) be the choice which works out, the discover after sufficiendy now f"(t){x-t)dt. J f f"it) /?2, fl U). and — t. one might However, it is If 2 = ~(x-t) t/(0 , so f"(t)(x-t)dt = -(x-tyv2 u(t)v(t)\ \a f - /'"(/) dt z J<> f"{a){x-aV C h 2 / x —=— f'"(t) (.v - 2 /) dt. I. This shows that fO)(t) -^—{x-n-dt. Ja You should now have little difficulty giving a rigorous proof, by induction, that 20. Approximation by Polynomial Functions /'" +1) if is 423 continuous on [a,x], then fx f(n+l)(f\ Rn, a (x)= L---^-(x-trdt. / Ja n. This formula is called the integral form of the remainder, and we can easily estimate it in terms of estimates for p^/nl on [a, *]. If m and are the minimum and maximum of f»^Jn\ on [a,x], then R (x) satisfies na M /// we can so (x j - t) n < Rna (x)<M f dt t) dt, write -a) n+l (x = Rn,a(x) a n for - (jc + 1 some number a between m and M. Since we've assumed means that for some t in {a, x) we can also write that tinuous, this n n\ which called the is familiar to those + (n \ + Lagrange form of the remainder who have done Problem a) iy} / ( " +1 > is con- ' (these manipulations will look 13-23). The Lagrange form of the remainder is the one we and we can even give a proof that doesn't require will need in almost all cases, f^"+^ to be continuous refinement admittedly of little importance in most applications, where we often assume that / has derivatives of all orders). This is the form of (a we LEMMA the remainder that will choose in our statement of the next theorem (Taylor's Theorem). Suppose R that the function (n is + l)-times differentiable {k R \a) = Then for any x in (a, b] R (x-ay+i For n = for n by induction on n. all 0, this is = 0,l,2,...,ii. we have R(x) PROOF forJfc on [a,b], and "+ ] \t) for -"^nyr Mean just the { some ' Value Theorem, and To do we this use the m («>*)• we will prove the theorem Cauchy Mean Value Theorem to write *(*) (x - fl )«+2 = " R'iz) (n + 2 )(z n and then apply the induction hypothesis R(x) (x -a)"+ 2 (R') 1 ' n R'(Z ) 1 - a)»+i ~ + { to "+ l (z (/!+ + 2)! I ^ S R' on the interval [a, ° me 1)! some t Z in (a ' z] to \t) for 2 R {n+2) (t) (n +2 - a)»+l in (a, z) get x) ' 424 Infinite Sequences and Infinite Series THEOREM 4 (TAYLOR'S THEOREM) Suppose y<"+i> that /' denned on [a,x], and are that Rn a (x) is , = f(x) f(a) + Rn,a(x) = f'(a){x - a) + • • defined by + ^-^-(x - a)" + Rn a (x) • , Then f(n+\) (t) (n+ n+\ ~ 7zr(x : a )' some f° r t in (a, x) 1) (Lagrange form of the remainder). PROOF The R na function Lemma the conditions of the satisfies by the very definition of the Taylor polynomial, so - (jr for some / a)" +l R na — f ' 1)! (n+1) _- /' /•(»+D = a polynomial of degree is + (n (0 But in (a, x). Rn since (/l+l) Rn Rn.a(x) Applying Taylor's Theorem /;. | — 0, we remainder terms for sin and exp, with a to the functions sin, cos, obtain the following formulas: x3 2 " +1 5 - + x- smx=x- + (-D" (2n 4 2 cos* = x x l-+- +( _ ir , + sin l <? (of course, and = 1 + we could A" + —+ • • 2' ' ^l + r — + (2n n\ actually go (/? + -x first sin (2n+2) (0l < easy. in the Since for all 1 f, we have • (2/1 sin {In Similarly we can show + 2),. |2/i+. > (/) + x 9 „ +2 < (2n+2)\ 2)! that COS (2 " +1) (0 (2/j + l)! |2/i+l »+i (0 ,^- + ,. l)! !)! one power higher two are especially „2«+2 + 2)! n+1 cos). Estimates for the (/) (2n l)! (2n)! (2 " +2, < (2«+ 1)! 20. Approximation by Polynomial Functions These estimates are particularly any s > we can make because interesting, (as proved in Chapter 425 16) for xn n\ by choosing n large enough (how large n must be will depend on x). This enables us to compute sin* to any degree of accuracy desired simply by evaluating the proper Taylor polynomial P (*). For example, n suppose we wish to compute sin 2 with an error of less than J0~ 4 Since , . sin 2 = P2n+ \,o{2) + where 7?, < \R\ (2/i we can use P2n+l,o(2) as our answer, provided that 2 2«+2 (2/i A number n with this < + 2)! 10 sin -4 property can be found by a straightforward search— it ob- viously helps to have a table of values for n\ happens that n = 5 works, so that 2= Pu,o(2) and 2" (see + 5! where even easier to calculate sin 1 sin = P2n+l page 432). In 7! 91 < 10~ 4 \R\ ~ U! + error less than e it *' . approximately, since 1 (1) + r, wriere \R < I : (2a? To obtain an this case +R 3! It is + 2)f we need + 2)!' only find an n such that 1 + (2/i and this < £ ' 2)! requires only a brief glance at a table of factorials. vidual terms of P2n+l,o(l) will For very small x the estimates = sin To To obtain jP2 " +1 ° ( lo (Moreover the indi- be easier to handle.) will be even +R ) easier. For example, I where ' |/?l < 10 2 "+2(2/ 2 + 2)! \R\ < lO"^ we can clearly take n = 4 (and we could even get away with n ==3). These methods can actually be used to compute tables of sin and cos; a high-speed computer can compute P2 „ + i.oU) for many different x in almost no time at all. Nowadays, computers, and even cheap calculators, determine the values of such functions "on-the-fly", though by specialized methods that arc even I aster. 426 Infinite Sequences and Infinite Series Estimating the remainder for e x > that x the [0, x] x < (the estimates for maximum value of only slightly harder. For simplicity assume is x e e' is exp since , x e x" R n.O < Since we already know < that e 4, Once increasing, so D! n+\ A\ < as small as desired + (n * 1)! 1)!' by choosing n depend on x (and the 4 factor X In particular, if n - + v = 4, ^ + '" + ^ + *' How large sufficiently large. make things more < x < 1, then will again, the estimates are easier for small x. If 1+ the interval we have (n+ n must be will is On 15). +] («+ ex,n+. which can be made Problem are obtained in difficult). < R < where (/i+ D! then 4 0<R< 1 < IO- 5i SO 1 1 + + 1 2! = 2 + 1 1 + + +/? < R < where - — 4! 3! 17 24 which shows that 2 < < e 3. (This then shows that x < R < 3 a (n+ R allowing us to improve our estimate of compute that the first (you should check that n The arctan you function arctan (/r) (x) is D! By taking slightly.) n — 1 you can 3 decimals for e are e cruel to insist that »+i = 1 does give actually is = 2.718... degree of accuracy, but this would be do the computations). important but, as you also it may recall, an expression for hopelessly complicated, so that our expressions for the remainder are pretty useless. On the other hand, our derivation of the Taylor polynomial for arctan automatically provided a formula for the remainder: + arctan x 3 •+ (_\\n v 2n+\ (-l)";r" 2n + l rx /-i\n+lf2n+2 + (it. ./o 1+/ 2 . t 20. Approximation by Polynomial Functions 427 As we have already estimated (_ x \"+l +2n+2 1 L l moment we For the |2n+3 +v r < dt in+i dt 2/2 Jo only numbers x with will consider remainder term can clearly be made as small as desired \x | + < 1 3 . In this case, the by choosing n sufficiently large. In particular, arctan (-1)" 1 1 1=1 — + + 5" 3 + 2/2 1 + where R, \R\ < In 1 + 3 With this estimate it is easy to find an n which will make the remainder less than any preassigned number; on the other hand, n will usually have to be so large as to make computations hopelessly long. To obtain a remainder < 10~ 4 for example, we must take n > (10 — 3)/2. This is really a shame, because arctan 1 = 7r/4, , compute n. Fortunately, surmount these difficulties. Since so the Taylor polynomial for arctan should allow us to there are some clever tricks which enable us to |2n+3 |/?2n+l,o(*)| < 2/ ? much + 3 work for only somewhat smaller x 's. The trick for computing is to express arctan 1 in terms of arctan x for smaller x; Problem 6 shows how this can be done in a convenient way. smaller n 's will 72" From on page 413, we the calculations logO X = X2 X + X3 X see that for x 4 (-!)" -4 + T T + > we have v (-1)" It+1 where ,n+\ (-1)" n and there For this is another matter 22+1 1 more complicated n sufficiently large, provided that The behavior < when — 1 < x < (Problem 16). remainder term can be made as small as desired by choosing a slighdy function the + B+l estimate —1 < x < 1 of the remainder terms for arctan and f(x) when \x I > 1 . = log(x + 1) is quite In this case, the estimates |2«+3 l^2«+1.0(*)l < 2/2 +3 for arctan, „/i+i \R„.o(x)\ < (x 22+1 > 0) for /, bounds x'"/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 arctan, note are of no use, because when \x\ > 1 the 428 Infinite — — . ' Sequences and Infinite Series that if t is in [0, x] (or in [x, 0] if +r 1 jc < 0), then < 1+jc 2 <2x 2 > ]f\x\ , I, so 2n+2 rx f ITT1 Jo To Ix 2 get a similar estimate for log(l I 1 +t = An Jo + x), we -t+r 1 \2n+\ r r" +2 dt l > dt + +6 can use the formula (-l)"-V'- + 1 (-l)V \+t to get 2 1 1 log(l+*)= f + (-1) n-\ Jo + i-iy r x Jo If x > 0, then for t in [0, x] 1 so t These estimates show that +t < r + 1 1 +x are of no use whatsoever if \x\ in < dt> w~ " 2* > +t 2x, if x > 1, dt t" 2« J + 2 \x\ > 1, the Taylor polynomials for arctan computing arctan x and log(.v + 1 ). This is no tragedy, because the values of these functions can be found for any x once they are for all x with |jc | < known 1 This same situation occurs in a spectacular fix) = have already seen that /^'(O) that the Taylor polynomial Pn,0(x) = Pn /(0) $ for + way -l/x 2 - 0, We dt. then the remainder terms become large as 1, n becomes large. In other words, for and / 1 we have — / Jo t" - / / /'(0)JC for the function .v #0 x = 0. for every natural number k. This means is + f"(0) ^^X 2 H x : 2! 0. remainder term R„q(x) always equals f(x), and the Taylor useless for computing f(x), except for x = 0. Eventually we will be In other words, the polynomial is able to offer some explanation for the behavior of this function, disconcerting illustration of the limitations of Taylor's The word "compute" for the has been used so often in which is such a Theorem. connection with our estimates remainder term, that the significance of Taylor's Theorem might be mis- construed. It is true that Taylor's Theorem can be used as a computational aid 20. Approximation by Polynomial Functions (despite ignominious failure its previous example), but in the it has even 429 more im- portant theoretical consequences. Most of these will be developed in succeeding two proofs chapters, but be used. The first waded through THEOREM 5 PROOF e is will illustrate some ways which Taylor's Theorem in be particularly impressive illustration will the proof, in Chapter 16, that tt is to those who may have irrational. irrational. We know = e any that, for e l = n, — + •••+ — + R +-+ 1 l I Suppose Choose that e n were > b and = rational, say e also n > n < Rn < where , ni 2! 1! 3. («+ 1)! a/b, where a and b are positive integers. Then a -= + + rr + b 2! , i 1 1 - + Rn l ! + , n\ so — n\a =»! + + »! n\ + nl n\R„. n\ Every term because n in this > b). equation other than n\R n is an integer Consequently, n\R n must be an integer <R n < (n + (the left side also. is an integer But 1)!' so 3 < n\R„ < + n which is impossible for an integer. The second proved in illustration Chapter is merely a straightforward demonstration of a 15: If +/= / (0) = /'(0) = / = 0. 1, 4 1 | f" then 3 < - < T To prove this, /( 3) /( 4) /(5) etc. observe that first 0, o, o. f ik) exists for = {f y = _ / = (/ (3)y = _ f y = _ f = = (/ (4)y = r> / i ( fi every k; in fact fact , I 430 Infinite Sequences and Infinite Series This shows, not only that ones: /, /', Theorem all {k) f states, for any = t Each function / in [0, x]. any particular jc there is ( / 0, all ( *>(0) are 0. Now Taylor's n, that ( " (» some = .f (0) f for but also that there are at most 4 different exist, -/, -/'. Since /(0) number a * +1) (0l |/ (we can add the phrase "and + 1)! " +1) ( continuous (since is f M such that m < +u (t) < for < t and jc, all (n+2) exists), so for n n" because there are only four different f {k) ). all Thus M|.v|" s ,/U)l Since this is = The J^Tv: and since |jc|"/"! can be made as small as desired by this shows that |/(x)| < £ for any e > 0; consequently, true for every n, choosing n sufficiently large, /(*) +1 0. Theorem other uses to which Taylor's be put will in succeeding chapters are closely related to the computational considerations which have concerned us for much of RnM (x) remainder term If the this chapter. can be made as small as computed any degree desired by choosing n sufficiently large, then f(x) can be to Pna (x). As we require greater and greater accuracy we must add on more and more terms. If we are willing to add up infinitely many terms (in theory at least!), then we ought to be able to ignore of accuracy desired by using the polynomials There should be the remainder completely. 1 sin x =x— JC — JC (- 3! 5! like • • • , 7! 4 2 JC , sums" X7 "S — "infinite 6 JC 8 JC .v COSX=1 -2! + 4!-6! + 8!-'--' 2 cX = 1 + A:+ JC 3 JC + 2! jc = JC jc 2 log(l We +*) = *- x 5 J x 4 + -- 71 —H it •• "if + are almost completely prepared for this step. we have never even necessary definitions. defined an |jc | < 1 4 3 y+y V 4! 3! —— + — 3 arctan + infinite sum. if-l<v<l. Only one obstacle remains— Chapters 22 and 23 contain the . . 20. Approximation by Polynomial Functions 431 PROBLEMS 1 Find the Taylor polynomials (of the indicated degree, and at the indicated point) for the following functions. i) f(x) — e ii) f(x) = e iii) sin; degree iv) cos; degree 2n, v) exp; degree n vi) log; vii) f(x) — x + x + x; viii) f(x) — x +jr + ix) f(x) = e ; degree 2/7, at , degree n, f(x) at -. 1 at 2. ; degree 4, at 0. x\ degree 4, at 1. degree In + xl l+x — at n. r-; — 3, at 0. smx degree 3, at 0. 1 2. * degree n, s 1, at 0. at 0. Write each of the following polynomials in x as a polynomial in (x is only necessary to compute the Taylor polynomial at as the original polynomial. 3. + (i) x2 - Ax - 9. (ii) x4 - +44.r 2 (iii) x (iv) ax 12.Y 3). (It of the same degree Why?) + 2x + l. . + bx + c. down Write 3 3, — numbers sum a (using 2, to within the specified accuracy. tion, consult the tables for 2" (i) sin 1; error < 10~ 17 (ii) sin 2; error < lO" (iii) sin ~; error < 10~ (iv) e; (v) e error ; notation) which equals each of the following error < 10~ 4 < 10 . . 1 . ^. . and n\ To minimize on the next page. needless computa- 432 Infinite Sequences and Infinite Series 2" n 1 2 1 2 4 2 3 8 6 4 16 24 5 32 120 6 64 720 7 128 5,040 8 256 512 362,880 9 *4. n\ 40,320 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 19 524,888 121,645,100,408,832,000 20 1,048,576 2,432,902,008,176,640,000 This problem similar to the previous one, except that the errors is are so small that the tables cannot be used. thinking, Chapter and some in it it x n /n\ can be 16, that actually provides a problem cases method may made You be necessary will demanded have to do a little to consult the proof, in small by choosing n large — the proof In the previous for finding the appropriate n. was possible to find rather short sums; in fact, it was possible which makes the estimate of the remainder given by to find the smallest n Taylor's specific Theorem less than the sum is a moral victory desired error. But in this problem, finding any (provided you can demonstrate that the sum works). (i) sin 1; error (ii) <?; (iii) < error sin 10; ,10. (iv) 10 ll0 10" 1000 < . < < 10 error error I (v) 5. (a) '. arctan y^; error 10~ 20 . -30 < r\— I0'°) '. lU u 1 ( cisely you showed that the equation x = cos* has pretwo solutions. Use the third degree Taylor polynomial of cos to show that the solutions arc approximately In Problem the error. 1 1 -41 Then use the fifth ±^2/3, and degree Taylor polynomial find bounds on to get a better approximation. (b) Similarly, estimate the solutions of the equation 1x = x sin* + cos" x. , 433 20. Approximation by Polynomial Functions 6. Prove, using (a) Problem n — = 15-9, that 1 arctan — 4 — ify off that = it a arctan — 4 l arctan 239' (Every budding mathematician should ver- 3.14159 on an immense first — 5 a few decimals of used, the — 3 ] 4 Show arctan 2 71 (b) 1 + but the purpose of this exercise tt, If the calculation. 5 decimals for is not to set second expression in part n can be computed you (a) is with remarkably little work.) 7. Suppose that a and b, t and g, respectively. In other the coefficients in are the coefficients in the Taylor polynomials at a of terms of the c, words, a s = f U) (a)/il of the Taylor polynomials a,'s and and = /?, g u) (a)/i\ . / Find a of the following functions, at Z?,'s. + g- (i) f (ii) fg- (iii) /'• (iv) h(x)= [ f{t)dt. Ja k(x)= f (v) f(t)dt. Jo 8. (a) = Prove that the Taylor polynomial of f(x) sin(x 2 ) of degree An + 2 at is Hint: If sinx — P is P(x) v6 „10 3! 5! v 4»+2 (2« + 1)!' the Taylor polynomial of degree 2n + R(x), where lim R(x)/x^" + ^ — 0. + 1 for sin at 0, then What does this implv about lim R(x 2 )/x 4n+2 ? (b) Find f a) (0) for (c) In general, if all k. f(x) = g(x m ), find f ik) (0) in terms of the derivatives of g atO. The ideas in this in the next three 9. (a) problem can be extended significantly, in problems. Problem 7 (i) amounts to the *n,a.f+g = equation *n.a,f "t *n,a,g- Give a more direct proof by writing /(*)= Pn,a,f(x) + Rn,a,f(x) g(x)= Pn ,a,g(x) + R„ M ,g(x), and using the obvious fact about Rn a ,f + , Rn,a,g- ways that are explored . 434 Infinite Sequences and Infinite Series (b) Problem 7 Similarly, could be used to show that (ii) — *n,a,fg I'n.a.f ^n,a,g\ni ' where [P] n denotes the truncation of P P terms of < of degree P n [with sum of written as a polynomial in x — degree n, the to Again, give a more direct proof, using obvious Rn volving terms of the form (c) Prove that about products in- . p and q are polynomials if facts all a] in x—a and lim R(x)/(x — a)" — 0, then + R(x)) = p(q(x)) + R(x) p(q(x) where if > a polynomial in x and q n, is terms of p(q(x (d) If = a The same If g{a) then 'n, a, fog — Y*n,b,f ° *n,a,g\n- — f(x+g(a)), G(x) = C*W For the following applications of Problem 9, = i [i i + = (a) For fix) (b) For the same (c) Find e x and g(x) / and + p„,,., fl..«.« and just write Pn j instead of • • • + = sin*, find g, find Psjg /^.tanG*)- Hint: First use find ^.l/cosU)- (Answer: * we assume Problem 9 +y+ 9 (f) e (e) Find P5J for = sin;c/cos2x. (Answer: a Calculations of this sort cos*. (Answer: + x 2 )e*]. x 3 /[0 may be to y=-) = = for simplicity, and the value of Ps.cosOO for /(jc) for /(a-) = 5 P4;/ P6J . . Find f{x) a n Psj +g (x). (d) 2x ewr] P„, a ,f- 3 11. and any value of g(a). Hint: g(x+a) and H(x) = G(x) — g(a). + Find all case.) 2 (Pn,,. e :' (f) then n. 0, g(a) 0, = 0, then **„.«. 10. > is result actually holds for all a Consider F(x) (f) having only terms of degree a whose constant term = a special is — a)) are of degree = and b (Problem 8 (e) — is 0. —a a polynomial in x Also note that p = -a)" lim R(x)/(x 1 + 2x + + y.v 3 (Answer: a 3 used to evaluate \x 2 - + + a limits that 4 ±x 3 - ^a 4 ) 5 f|^A ) - \x 5 + |a 6 we might ) other- wise try to find through laborious use of l'Hopital's Rule. Find the following: , N (a) hm X- 0 ,. e x -\-x-\x— 2 a - sin Hint: First find a Pt, Nix) = lim x-»-0 o ,/v(a) nator Nix) and D(a). D(x) and /^.o.oU) for the numerator and denomi- 20. Approximation by Polynomial Functions 435 -x x2 2 { (b) lim 1+a — sm x x term Hint: For the 1 (c) +x), first write 1/(1 — — x+x~— a 3 + +x) 1 - 1 lim 2 x2 Hsin * 1 (d) x e /(I — COS(a ) lim r2« x x^sin ' 1 (e) 1 lim *->0 sin 2 sin(A 2 ) a (sinx)(arctan.v) (f) lim 1 12. — — x cos(a 2 ) Let sin /(*) a A" = 7^0 A=0. 1. Starting with the Taylor polynomial of degree In the estimate for the remainder ' w= 1 + -3T 3T + 1 for sin a together with , term derived on page 424, show + - + ( - 1) + "(25i)i that /?2*.0,/00 W here |2n+l #2«.0./'U)| < (2n and use this to conclude that .4 l 1 ^0 Jo -^ with an error of less than 10 13. + 2)!' + 1703 ^j^' .946 = 3 Let e /(*) (a) (b) = x - 1 , x #0 A = 0. Find the Taylor polynomial of degree n for / at give an estimate for the remainder term R„,o,f. Compute / f with an error of less than 10 0, compute f {k) (0). and . JO A 14. Estimate f° / Jo 2 exp(x )dx with an error of less than 10" 436 Infinite Sequences and Infinite Series 15. Prove that if x < 0, then the remainder term R„q for e x kl" 16. Prove that —1 < x < if +1 + (n satisfies 1)! then the remainder term R„q for log(l 0, -f a ) satisfies \ S l*».°l *17. (a) a+xKn+1 y that if \g'(x)\ < M\x - a\" for -a\" +] /(n l)fbr\x-a\<8. Show - \x < a\ then \g(x) 8, - g(a)\ < + M\x (b) x \n+l Use part to (a) show that = 0, then then g'{x) = fix) - Pn _ haJ ,(x). lim g'(x)/(x if -a)" x->a gM-M = x^a (c) (d) 18. Show that if = g(x) ( f(x) X -a) n+] - Pn . a ,f(x), Give an inductive proof of Theorem Deduce Theorem the remainder. 7 (The catch is that it will Theorem Lagrange's method for proving Taylor's We consider a fixed f{x) (*) - Theorem, with any form of be necessary to assume one more as a corollary of Taylor's 1 derivative than in the hypotheses for 19. without using l'Hopital's Rule. 1, number x and + f(t) /'(*)(* - 1 .) Theorem used the following device. write + • • • + — f {n) (t) ^-(x - + t)" Sit) IV. for Sit) = R nt ix). The notation right side as giving the value of down is a tip-off that some function the fact the derivative of this function function of x and whose value t, is check that always fix). To is make we are going to consider the for a given 0, since sure it /, equals the constant you understand the if git) / w (0, = ^—ix-t)\ then git) = £^k(x-t) k -\-l) + f £-^(x-t) kl kl {k) (n_ U , ,._, _,)*-! + , f k+]) ^ix-t) ^ it) and then write k k roles 20. Approximation by Polynomial Functions Show (a) = 437 that + /'(*) f'(0 + QP-V 1! no (x-t) + /w(o (3) + (.y - ty 1! + -/"'<"„ _, (n - 1) + r (n+l), + . /^£) u _„. /?! + S'(0, and notice that everything collapses to /(«+!) 5'(r) = -^— --(.v-/)". (f ) Noting that S(x) 5(a) apply the Cauchy (x — /)" +1 Mean on [a,x] to = = *„.„ CO, Value Theorem to the functions S and h{t) = obtain the Lagrange form of the remainder (Lagrange actually handled (b) = 0, /?„,*(*) this Similarly, apply the regular part of the argument differently). Mean Value Theorem to S to obtain the strange hybrid formula Rn.a(x) (n+l) (t) = f (x-t) n (x-a). n\ This 20. is Cauchy form of the remainder. called the Deduce the Cauchy and Lagrange forms of the remainder from the integral form on page 423, using Problem 13-23. There will be the same catch as in Problem 18. know of only one situation where The next problem is preparation for I 21. For every number Cauchy form of the coefficient" a (a — 'a\ and we define and for , = (1 • as usual. If alternates in sign for n f{x) 1) . . . • n, —n+ (a we + x) a at > is a. used. define the "binomial l) a 1 is that eventuality. and every natural number a, the remainder a a is not an integer, then Show that ( ) is never 0, the Taylor polynomial of degree n Pn ,o(x ) = ^2\ v , ) ^ and Lagrange forms of the remainder arc the following: tnat me Cauchy and 438 Infinite Sequences and Infinite Series Cauchy form: Rn.0(x) — a(a — = 1) • . . . — n) (a • —T^ ~ tTO + O a -"~ -*(* , l IV. a{a- -n) (a ... 1) +t) a , -x{\ _\ n\ U = (H - t \ f t 1 X ~ 1-1 + +'t) a""' ((fx VI +t l \x(\ l.V" + 1/ // (x 1 M , (0,x)or (.v,0). Lagrange form: + (« " .V^Ci + O""" + // 1)! -1 Mn(0,.x)orU,0). , 1/ Estimates for these remainder terms are rather postponed 22. (a) Suppose that for all x > / twice differentiable is while \f"(x)\ 0, — ^2 on for all Taylor polynomial to prove that for any x < \f'(x)\ (b) Show handle, difficult to and are Problem 23-21. to that for all x 2 and that |/(x)| < Mo x > 0. Use an appropriate (0, oo) > we have h -M + -Mi h 2 > for all h > 0. we have L/"(.v)| <2 V / MM 2 . Hint: Consider the smallest value of the expression appearing in (c) If / is twice differentiable proaches (d) If lim 0. 23. (a) as f(x) f The (b) exists if limit and lim f"(x) f"(a) (a) = on the is exists, bounded, and f(x) apas x then lim f"(x) — — oo. lim fix) = f(a then + h) + f(a-h)-2f(a) lim . is called the Schwarz second derivative of / at a. Hint: Use the Taylor polynomial P2, a (x) with x = a + h and with x Let fix) = x 2 for x > 0, and — x 2 for x < 0. Show that exists, (a). 11-34.) exists, right lim (c) (0, oo), x —> oo, then also f'(x) approaches (Compare Problem Prove that on /" f(0 = a — h. + h) + fi0-h)-2fi0) = even though /"(()) does not. Prove that derivative / has a local maximum at is < 0. of / at a exists, then if it a, and the Schwarz second 439 20. Approximation by Polynomial Functions (d) Prove that if f'"(a) exists, then — — = hm /'"(a) 3 24. fifl ,. r + h)- f(a-h)- 2hf'{a) . 5 h h-+Q 3 Use the Taylor polynomial P\ M ,f, together with the remainder, to prove a weak form of Theorem 2 of the Appendix to Chapter 1 1 If f" > 0, then : the graph of / always above the tangent lies line of /, except at the point of contact. *25. Problem 18-43 presented a rather complicated proof that / = if f"—f = 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 26, and meant is to convince you that Taylor's Theorem a good tool for tackling such problems, even though tricks work out is more neatly for special cases.) **26. Consider a function / which for certain numbers ao, . . . , satisfies fl„_i. the differential equation Several special cases have already received we detailed treatment, either in the text or in other problems; in particular, have found all functions satisfying /' 0, or f" — f = 0. The solutions for such equations, = /, or f" + f = Problem 1 8-42 enables us to find many but doesn't say whether these are the only solutions. This requires a uniqueness result, which will be supplied by this problem. At the end you will find some trick in (necessarily sketchy) (a) remarks about the general solution. Derive the following formula for ( / " +1) (let us agree that "a_i" will be 0): 7=0 (b) Deduce a formula for The formula in part (b) is f {n+2 \ not going to be used; vince you that a general formula for other hand, as part size of (c) Let f {n+k) /V means = (c) shows, it is f { " +k) is it was inserted only out of the question. to con- On the not very hard to obtain estimates on the (x). max(l, |oo|, l«„-il)- Then |a -_j / + an _ia/| < that /'" +1 = J2bj f u \ l » ,/=0 where l \bj \ < IN 2 . 2N 2 ; this ) 440 Infinite Sequences and Infinite Series Show that n-\ z " +2) ( = J2 b 2 f (j) ^ where b/\ < 4A < J \ 7=0 and, more f generally, = J2 bj k f u \ („+k) where \bf\ < 2 k N k+l . 7=0 (d) Conclude from part number (n+k \f (e) Now suppose " and conclude Show that = /i any particular number x, there y) (0) = = • is a / = and /2 0) for all*. = / (n_1).(0) = • • M | " " (n+k+l)\ that if /i N k+] n+* +1 JC | " k /'(()) 2*+ 1 /y*+ 2 \2Nx\" Show 0. that +M ' (n+k+\)\ 0. both solutions of the /? are / and that, for <M -2 \x)\ that /(0) M l/U)l (f (c) M such that (M) = X>/ < (0) for j < differential equation 0) , - n then 1, /i = /2 . In other words, the solutions of this differential equation are determined by the "initial means that to obtain we can find all solutions any given x" has n distinct roots is a solution, / (7) (0) for once we can conditions" (the values set < j find < n — enough This 1). solutions of initial conditions. If the equation -a n _\x n - x an then any o-i =0 a , function of the form and + •••+<?„, /(0) =ci f'(0) =a\C] /(«-D(0) =a M- 1 i \-a„cn H , ci+...+an M -1 cn . 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 properly, but we will not try to prove you can easily this last assertion. (It check for n = 2 or 3.) is a purely algebraic These remarks are fact, also true which if some 20. Approximation by Polynomial Functions 441 of the roots are multiple roots, and even in the more general situation considered in Chapter 27. ** 27. (a) Suppose and that that a continuous function on [a,b] with f(a) = f(b) Schwarz second derivative of / at x is (Problem 23). Show that / is constant on [a,b]. Hint: Suppose that fix) > f{a) for some x in {a, b). Consider the function / for all is x in (a, b) the = g(x) = with g(a) g(x) > Problem its (b) (a) = g 23(c) (the For sufficiendy small e > have a maximum point y in {a, Schwarz second derivative of ordinary second derivative). is a continuous function on at all points of - a )(b-x) f(a). will If/ is *28. g(b) g(a), so f( x )-s(x (a, b), then [a, b] / is (x we will - a)(b - x) whose Schwarz second have Now b). is use simply derivative linear. Let /(*) = x 4 sin 1 /x 2 for x # 0, and /(0) = 0. order 2 at 0, even though /"(0) does not exist. Show /= that up to This example is slightly more complex, but also slightly more impressive than the example in the text, because both f'(a) and f"{a) exist for a # 0. Thus, for each number a there is another number m(a) such that f(x) (*) = fi a) + f (a )ix -a) + ^f {x - a f + j^), Ra (x) —=O - a) 2 .. , where lim x^a (X namely, m(a) m (b) defined in = f"{a) for a this way is ± 0, — and m(0) = 0. Notice that the function not continuous. Suppose that / is a differentiable function such with mia) = 0. Use Problem 27 to show that that (*) holds for /» = m (a ) = all a for all a. c Now suppose that that for all (*) holds for all a, and that m is continuous a the second derivative f"(a) exists and equals m(a). Prove ^chapter mm The irrationality of e attempt a more but actually is TRANSCENDENTAL e IS was and b, = a/b a any integers a and for except for a V2 does not seem to of how as saying that x does not satisfy for integers a be such a barely manages to be irrational = = 0. = 0, b, with b / 0. This is the any equation bx - a b Viewed in this light, the irrationality terrible deficiency; rather, — although V2 C1]X it is will and prove worse. Just suggested by a slight we that the number e is not merely irrational, number might be even worse than irrational rewording of definitions. A number x is irrational if it is difficult feat, much not possible to write x same so easy to prove that in this optional chapter + (70 = is appears that it V2 just not the solution of an equation 0, the solution of the equation x 2 -2 = of one higher degree. Problem 2-18 shows 0, how to produce many irrational num- bers x which satisfy higher-degree equations a„x" where the a, are integers not tion of this sort is called an have ever encountered and is + a n _ix"- +-"+flo = 1 all 0. A number which 0, satisfies an "algebraic" equa- algebraic number, and practically every number we defined in terms of solutions of algebraic equations e are the great exceptions in our limited mathematical experience). All (tt roots, such as are clearly algebraic numbers, 7 73, v/2, v ?. and even complicated combinations, like P + VE+ V1 + V2+4/6 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 The proof that e is transcendental is e is a number of this anomalous well within our grasp, and was sort. theoretically Chapter 20. Nevertheless, with the inclusion of this proof, we can justifiably classify ourselves as something more than novices in the study of higher mathematics; while many irrationality proofs depend only on elementary possible even before properties of numbers, the proof that a 442 number is transcendental usually involves . 21. some really 443 Transcendental e is high-powered mathematics. Even the dates connected with the tran- scendence of e are impressively recent — the first Hermite, dates from 1873. The proof that due to due to Hilbert. Before tackling the proof itself, depends on an idea used even is it in the proof that e we a good idea to proof that e is will give map transcendental, is a simplification, is out the strategy, which irrational. Two features of the expression e= l + were important for the Ij +_+ proof that e 1 is .. . On irrational: +-+ number the one hand, the - + --- nl 1! can be written as a fraction p/q with q < other hand, < R n < 3/(« + 1)! (so n\R„ that e can be - + *„ + n\ (so that n\ is (p/q) is an integer); on approximated particularly well by rational numbers. Of course, — number x can be approximated arbitrarily closely by rational numbers there is a rational number r with \x —r\ < e; the catch, however, is that it s > may be necessary to allow a very large denominator for r, as a fraction p/q within 3/(« not the case: there is is every if we are assured that this the not an integer). These two facts show large as \/e perhaps. For e + 1)! whose denominator q is at most n\ If you look carefully at the proof that e 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 it is (some evidence for this assertion is presented in Problem 3). The proof that e is transcendental depends on a natural 2 e", extension of this idea: not only e, but any finite number of powers e, e can be simultaneously approximated especially well by rational numbers. In our of is e, . irrational, , proof we will begin by assuming that e a„e" (*) for some integers certain integers ciq, ... M, M\, . • a„. In , . + . , M n • • 7 c" how , ^ «o we certain "small" numbers €\, . . will . , than find € n such that = e" into the 0. order to reach a contradiction e Just . algebraic, so that is + a\e + ciq = and . M M + €2 = M 2 = M„ + €„ M small the e's must be will appear when these expressions are substituted assumed equation (*). After multiplying through by we obtain M [aoM + a\M\ H \-a„M„] + [€\a\ H \- €„a n ] = 0. 444 Infinite Sequences and Infinite Series The first term in brackets an is and we integer, We necessarily be a nonzero integer. \-€ n a n lead to the desired contradiction this will number of absolute As a value less basic strategy this < \ — the \\ sum of a nonzero integer and a The very reasonable and quite straightforward. remarkable part of the proof will be the way that the M's and order to read the proof you will it to find e's so small that than ^ cannot be zero! all is choose the M's so that manage will also \€\a\ H will need will know about to the e's are defined. In gamma function! (This function was introduced in Problem 19-40.) THEOREM l proof e transcendental. is Suppose there were integers a„e" (*) Define numbers M, M\, ao, . . . . . . + a n -.ie"~ + " x p - [(x l M - €\, . . . \) . . . -L (Pex xP- - Y [(x (p xp - '[U- - unspecified such that • 0. €„ as follows: , (x 1) - n)]Pe~ x J I (x -n)ye~x dx, — 1)-. D! • (.v - -n)Ye~ x dx. €k The • -0- M, k • 0, +a = l M„ and , ^ a„, with «o , (p-iy. number p represents a prime number* which we Despite the forbidding aspect of these three expressions, with a appear much more reasonable. We concentrate on M first. will little choose later. work they will If the expression in brackets, [(x-1).... •(*-»)], is actually multiplied out, we obtain a polynomial x" *The term "prime number" was denned in + ---±nl Problem 2-17. An important fact about prime numbers not proved in this book: If p is a prime number which does not divide the integer a, and which does not divide the integer b, then p also does not divide ab. will be used in the proof, although The Suggested Reading mentions it is references for this theorem (which factorization of an integer into primes that there are infinitely information is many required. primes is unique). —the reader is We will also is crucial in proving that the use the result of Problem 2- 17(d), asked to determine at precisely which points this 21. When with integer coefficients. power raised to the pth Transcendental e is 445 becomes an even this more complicated polynomial x Thus np + • • ± • {n\) p . M can be written in the form "P ,-00 1 a=0 where the Ca = ± («!)''. and Cq are certain integers, k x e r x = dx But k\ Jo Thus ^ Now, for = a we obtain the term ±(n\f We now will divisible by consider only primes On p. Ca which is Now divisible (P-D! the other hand, (p — r—- = consider M*. We M =e (p - ^ =±(n\) p . 1)! p > n; then this term if a > 0, then Caip Therefore p. ~ +a)\ 1 ; by iP M +a- itself is is an integer which + a - 2) \)(p . an integer which . . is • not by p. p, not divisible have k k • is \ 3 - dx (p-iy (x - ri)¥e-<*-Q , xP- l [(x- 1) • . . . , • dx. -I. This can be transformed into an expression looking very much like M by the substitution u du The limits = = of integration are changed to Mt= f°° (u+k) p -l [(u L + k- x — k dx. and oo, 1)....-M- o^ and ...- (u + k-n)]''e-" '"'• 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)"- \(u+k- \)-...-(u+k [ -/?)]'' a 446 Infinite ; Sequences and Infinite Series is a polynomial with integer coefficients, of which has degree at least p. every term Thus "/' M k "P /-00 1 = Y" Da np / - x+a = Y" D„ e-" du m—i Da where the term by divisible sum in the Substituting into (*) = K \-an H > In addition to requiring that p Mk M and flo divisible not divisible M + n] let by p, n. M we obtain [a\€\ -\ n en ] \- us also stipulate that p are not divisible by p, so a$M is > =0. |«ol- also not divisible + a\M\ by p. In particular Yan H is by p. Since M n a nonzero integer. it is In order to obtain a contradiction to the assumed equation prove that e This means follows that it ciqM is n k=\, , M and multiplying by M +a\M\ [a is divisible is p. e each summation begins with a = 1 by p. Thus each Mk is an integer clear that it is that both + <*)! are certain integers. Notice that the in this case every Now 1 (p-l)\ a=l which is (p- transcendental, (*), and thereby only necessary to show that it is h«„e„| |«iei H made as small as desired, by choosing p large enough; it is clearly sufficient show that each \€k\ 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 polynomial equation (*)). To can be to begin with, if 1 < k < n, then - /* \xP~ [(x LV l l€kl - eK i ' ,n np -l \(x - 1 A be the )•...• (x -n)\ p e -x -dx. (p-iy. ^o let dx V-dI (p 1)! -"/' Now x -n)y\e~ ' l)-...-(.v " maximum of 1**1 |(.v < — 1) n nP~ l e . . n np ~l Jo A p [™ e / n np ~l xr dx Jo Ap (P- D! n e npAp e"(nA) p ~~ : n)\ for r (p-l)\ e — AP (p-iy. e (x . (p-l)\ = (p-l)\ x in [0, n]. Then 21. A But n and making p — are fixed; thus (nA) p /(p 1)! can be e is made $A1 Transcendental as small as desired by some philosophic af- sufficiently large. | This proof, proof that like the tt irrational, deserves is — argument seems quite "advanced" after all, we use integrals, and integrals from to oo 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 terthoughts. At the first sight, 00 .k / JO and these for integral k, Thus M, where for Ca can be replaced by integrals example, could have been defined A:! whenever they occur. initially as are the coefficients of the polynomial [(X-1).... •(*->!)]'. If this idea that e is is developed one obtains a "completely elementary" proof consistently, transcendental, depending only on the fact that e=l+ ] + , v. Unfortunately, this "elementary" proof one — { ! + v is this actually a This situation n irrational is is by no means peculiar is proof except that it a case in point. difficult classical a fact which is to this specific theorem— involves quite a few complicated functions. of great Our than "advanced" ones. You probably remember nothing more advanced, but much more conceptual transcendental, of the ---- harder to understand than the original "elementary" arguments are frequendy more proof that is + the whole structure of the proof must be hidden just to eliminate a few integral signs! about v proof, There which shows that historical, as well as intrinsic, interest. problems of Greek mathematics was to construct, is it One with compass whose area is that of a circle of radius 1 This requires the construction of a line segment whose length is y/n, which can be accomplished if a line segment of length n is constructible. The Greeks were totally 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 and straightedge alone, a square In that year until 1882. . Lindemann proved that n is transcendental; since the length of any segment that can be constructed with straightedge and compass can be written that a line in terms of +, •, — -=-, and y/~, and is therefore algebraic, segment of length n cannot be constructed. , The proof that which is it is transcendental requires a sizable this proves amount of mathematics too advanced to be reached in this book. Nevertheless, the proof much more difficult than the proof that e is transcendental. is not In fact, the proof . 448 Infinite Sequences and Infinite Series for it is practically the on same e 7r; after This e. last statement should certainly transcendental seems to depend so thoroughly is particular properties of e that could ever be used for proof for as the The proof that surprise you. almost inconceivable it is all, what does e have to how any do with 7r? modifications and Just wait see! PROBLEMS 1. (a) Prove that (b) Prove that a > is algebraic, then >J~a. is algebraic. a is algebraic and r is rational, then a if if + r and ccr are algebraic. Part can actually be strengthened considerably: the sum, product, (b) and quotient of algebraic numbers is algebraic. This fact is too for us to prove here, but some special cases can be examined: 2. *3. V2 + V3 and v2(l + v3) are algebraic, by actually finding algebraic equations which they satisfy. (You will need equations of degree 4.) Prove that (a) Let be an algebraic number which a. satisfies and not rational. = an x n + a n _ lX "- + Suppose that • no polynomial function of lower degree has for any rational number p/q. f(p/q) / Show this property. Use Prob- Hint: 3-7(b). (b) Now show that (c) Let \f{p/q)\ > \/q" for all rational numbers p/q with q > Hint: Write f(p/q) as a fraction over the common denominator q" to a + oo = 0, 1 that that lem is the polynomial equation f{x) M — sup{ prove that |/'(jc)| if \x : p/q — < a\ a rational is 1 }. . 0. Use the Mean Value Theorem number with a ~ P/q\ > \/Mq n (It follows that \a — p/q\ > c/q" for all rational p/q.) for c \ *4. difficult = — p/q\ < min(l, \/M) |a 1, then we have Let a = 0.1 10001000000000000000001000 where the 1 's occur a is transcendental. of degree in the n ! place, for each n (For each n, show that a . Use Problem 3 is to prove that not the root of an equation n.) Although Problem 4 mentions only one specific transcendental number, it should many other numbers a which do Such numbers were first considered by Liouville (1 809 882), and the inequality in Problem 3 is often called Liouville's inequality. None of the transcendental numbers constructed in this way happens to be particularly interesting, but for a long time Liouville's transcendental numbers were the only ones known. This situation was changed quite radically by the work be clear that one can not satisfy \a — easily construct infinitely n p/q\ > c/q for any c and n. 1 of Cantor (1845 1918), who showed, without exhibiting a single transcendental 21. e is 449 Transcendental number, that most numbers are transcendental. The next two problems provide an introduction to the ideas that allow us to definition with if its which we must work elements can be arranged in a is make sense of such statements. the following: A A set is called The basic countable sequence a\, Q2, #3, #4, .... The obvious example is N, the set more or (in fact, less the Platonic ideal of) a countable set of natural numbers; clearly the of even natural numbers set also is countable: 2,4,6,8 It is a and 0) little is more surprising to learn that Z, the set of all integers (positive, negative also countable, but seeing 0, The believing: 1,-1,2,-2,3,-3 next two problems, which outline the basic features of countable examples might think and (2) (a) Show x (b) is in that B }. Show that Show A and B if are countable, then so Use the same Hint: that the set of same If A\, A2, A3, . . is x : x is countable. (This (m,n) of integers U A2 U A3 U of all Prove that the (f) Prove that the set done part you can do set in A or is is really to enlightenment.) countable. (This is triples (/, ... same trick as in part m, n) of integers (l,m,n) can be described by a pair Prove that the numbers { for Z. are each countable, prove that . (e) (g) AUB = (b).) also countable. (Again use the (e), is worked below indicates the path pairs all as part A\ is trick that the set of positive rational practically the (d) are to quite startling, but the figure (c) sets, show that (1) a lot more sets are countable than one nevertheless, some sets are not countable. really a series of *5. is (/, set integer coefficients of is all countable. (A triple m) and a number of all ^-tuples (a\, ai this, is (b).) an ) is n.) countable. (If you have using induction.) roots of polynomial functions of degree n with countable. (Part (f ) shows that the set of all these . . 450 Infinite . Sequences and Infinite Series polynomial functions can be arranged most n (h) Now Since so set of there use parts and (d) many to (g) prove that the is turn out to be countable, sets numbers between real all no way of listing all ct\ 0t 2 «3 (decimal notation such a list is these set of all algebraic numbers where a„„ important to note that the it is and 1 is not countable. In other words, real numbers in a sequence = O.auanauau = 0.a21fl22«23«24 = 0.a3ifl32fl33fl34 being used on the . • • • . . . To prove right). so, suppose that this number that this is were possible and consider the decimal 1^22^33^44 O.flj = 5 if ann cannot possibly be / 5 in the Problems 5 and 6 can be is at countable. is *6. and each has in a sequence, roots.) and list, = a, m 6 = ann if Show 5. thus obtaining a contradiction. summed up The as follows. set numbers were countable. If the set of transcendental set of all real numbers would be countable, by Problem set of real numbers between of algebraic numbers also countable, then the 5(a), and consequently the and 1 would be countable. But this is false. Thus, numbers is countable and the set of transcendental numbers are more transcendental numbers than algebraic numbers"). The the set of algebraic is not ("there remaining two problems between *7. sets / be Let sets x—>a + important which are a nondecreasing function on lim fix) and lim fix) both (a) how illustrate further which are countable and it can be to distinguish not. Recall (Problem 8-8) that [0, 1]. exist. x-*a~ For any e [0, 1] > prove that there are only with lim fix) — lim fix) > £. finitely Hint: many numbers There are, in fact, at a in most [/(l)-/(6)]/e ofthenT" (b) Prove that the Hint: If number set lim fix) x—>« + n of points — at lim fix) x—>a~ which / > 0, discontinuous is then it is > This problem shows that a nondecreasing function \/n for is to and also be differen liable more at a interesting. set A some natural is more difficult nondecreasing function can of points which is not countable, but true that nondecreasing functions are differentiable at word countable. automatically con- tinuous at most points. For differentiability the situation to analyze is fail it is still most points (in a Reference [38] of the Suggested Reading gives a proof using the Rising Sun Lemma of Problem 8-20. different sense of the "most"). 21. who have done Problem For those is possible to provide at least e is Transcendental 451 9 of the Appendix to Chapter 11, one application it to differentiability of the problem set: If / is convex, then / is differentiable except at those points where its right-hand derivative /+' is discontinuous; but the function /+' is increasing, so a convex function ideas already developed in this is (a) automatically differentiable except at a countable Problem -70 showed that if a continuous function /, then / 1 1 the hypothesis of continuity countable set and b x such (a x ,b x ). of values. that ax Then every point is Hint: is interval (ax b x ). , How many (c) Prove the result of Problem maximum a constant function. Suppose < x < bx and x Deduce Problem a local is the point for now that is a maximum point for / on maximum value of / on some such intervals are there? 11 -70(a) as a corollary. 1 1 of points. dropped. Prove that / takes on only a For each x choose rational numbers a x every value f{x) (b) is set -70(b) similarly. CHAPTER INFINITE SEQUENCES The idea of an infinite sequence is so natural a concept that One dispense with a definition altogether. it is tempting to frequently writes simply "an infinite sequence " «1, «2, «3, «4, «5, the three dots indicating that the numbers rigorous definition of an infinite sequence important point about an there are DEFINITION a real is meant An number a n This sort of . is sequence • 3 continue to the right "forever." A not hard to formulate, however; the is that for each natural correspondence is precisely number, n, what functions to formalize. infinite From infinite a, •• sequence of real numbers is a function whose domain is N. the point of view of this definition, a sequence should be designated by a single letter like a, and particular values by a(l),a(2),a(3), but the subscript notation a\, ai, #3, is almost always used instead, and the sequence like {a n }. Thus {n}, {(—1)"}, itself is usually denoted by a symbol and {\/n} denote the sequences or, /3, and y defined by ot n = n, fin = (-1)", Yn = -• 1 A sequence, like any function, can be graphed (Figure 1) rather unrevealing, since most of the function cannot be • {<*«} {ft; (a) FIGURE 452 I but the graph fit (b) {Yn\ (c) on the page. is usually 22. «2 <X\ • = • $5 = ^3 = = fh £l = At = /?6 • • 453 Sequences 0f5 0(4 <*3 Infinite • K2 K4 1 1 K5 Yi IGURE y\ 2 A more convenient the points a\, ci2, representation of a sequence «3, • • on a • is obtained by simply labeling This sort of picture shows where line (Figure 2). The sequence {or,,} "goes out to infinity," the sequence "jumps back and forth between —1 and 1," and the sequence {y„} "converges the sequence "is going." {/3„} to 0." Of the three phrases in quotation marks, the last associated with sequences, and will be defined precisely is the crucial concept (the definition illustrated is in Figure 3). • • l-e FIGURE DEFINITION a/v+5 > " N+ +e A3 «2 symbols lim a„ = \ l a natural {«„} converges to number / (in such that for Af natural all , if > N, then n \a„ converge if it approaches converges to / for 1) some /, and > there n, this definition, to every e £. or has the limit / if for numbers -l\ < In addition to the terminology introduced in that the sequence {a n } to <2l 3 A sequence is • °/v+3 A /. diverge if we sometimes sequence it } is say said does not converge. To show that the sequence {yn converges to 0, it suffices to observe > 0, there is a natural number N such that \/N < s. Then, If s {a n } the following. if n > N we have 1 Yn . The will =- < — 1 < e, £, lim yjn + N - 0| < reflection (it so \yn £. limit probably seem reasonable practically the same after a 1 — little yjn as y/n for large n), but a — just says that vn + 1 is mathematical proof might not be so , 454 Infinite Sequences and Infinite Series obvious. To + estimate \Jn yjn + — 1 — 1 (s/n+ = \jn we can \Jn 1 use an algebraic - y/n)(y/n 7 + v " + n — 1 + + y/n 1 + 1 + 1 + trick: yfn) \fn n 7 V" 1 7 v It is = >Jx 7 v" 1 — : 1 y n + — \Jn by applying the Mean Value Theorem on the interval [//,/? + 1]. We obtain also possible to estimate to the function f(x) + + /? /'(*) == I for some x in (n, n + 1) 1 < 2V^' Either of these estimates is left may be used to prove the above limit; the detailed to you, as a simple but valuable exercise. The limit + 7« 2 +l - 8// + 63 3n 3 4« n-><x> 3 3 4 should also seem reasonable, because the terms involving n tant when n is remember large. If you the proof of Theorem are the most impor- 7-9 you will be able —dividing top and bottom proof to guess the trick that translates this idea into a by « 3 proof yields o 3n 4/7 3 3 2 +7n + l -8/7 + 63 this expression, the 1 « «3 8 63 " 4 " Using 7 proof of the above — + "n3 n L 2 limit is not difficult, especially if uses the following facts: If lim a„ and lim b n both exist, then n—+oo n—>-oo lim (a„ n—+ 00 + lim (a„ • bn ) = b„) = lim a„ n—+oo moreover, if lim £„ n—+00 ^ 0, lim a„ «—+00 + • /i—+oo then lim n—+00 /?„ <:/„//?„ ^ = for all lim h—+00 a,,/ lim b„ n—+ oo lim b n n—>oo // \ greater than lim &„. n—>oo some tV, and one . 22. Infinite 455 Sequences 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 {c n = [a„/b n }, where the numbers c„ might not even be defined for certain n < N This doesn't really matter we could define c„ 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 (If } — theorem —you should have no = the definition of lim a n proving these results for yourself, because difficulty so similar to previous definitions of limits, especially / is n-><x> = lim f{x) The /. between the similarity mere analogy; it is function whose graph actually closer than second. If / is the (3,fl 3 ) = = lim a„ and lim fix) I is 1 .v—>oo n—»oo possible to define the first in terms of the definitions of (Figure 4) consists of line segments joining a5 ) (5, (l,«l) FIGURE 4 the points in the graph of the sequence f(x) = \ - a n ){x = I if (a n+ - {a,,}, n) so that + a„ n < x < n + 1, then lim a„ Conversely, if f satisfies lim fix) This second observation < a < 1. and only is = /, and we this we = /. fin), then lim a n = I. For example, suppose that = 0. note that lim a -^ < 0, so that x logfl Notice that we actually have since log a is x = we can lim e x—>00 a negative lim a" n—>oo < an = Then \ for if a set frequently very useful. lim a" To prove lim fix) if =0 xlo z a and if \a\ = 0, large in absolute value for large x. < 1; write lim a" n—»oo = lim(-l)>r = h—<-oo 0. 456 Infinite Sequences and Infinite Series The behavior comes of the logarithm function also shows that arbitrarily large as n becomes = lim a" n—>-oo and sometimes even said that it is large. This assertion > a oo, 1 , then a" be- often written 1, We approaches oo. {«"} is > a if also write equations like —a" lim n—»oo and say — oo. that {—a"} approaches = — oo, Notice, however, that if < —1, then a lim a" n—>oo does not exist, Despite this even in this extended sense. connection with a familiar concept, it is more important convergence in terms of the picture of a sequence as points on a There which another connection between limits of functions and limits of sequences is is related to siderably this picture. This connection more interesting, than is somewhat less one previously mentioned the of sequences in terms of limits of functions, limits to visualize line (Figure 3). obvious, but con- — instead of defining possible to reverse the pro- it is cedure. THEOREM 1 Let / be itself, a function defined in an open interval containing c, except perhaps at c with lim x—*c Suppose that Then {a n } is /(*)=/. a sequence such that (1) each a n is (2) each a n (3) lim a„ ^ c, = c. domain of in the /, the sequence {f(a n )} satisfies lim f(a„) = I. n—>O0 Conversely, if this is then lim f(x) PROOF Suppose first = true for every sequence {a n } satisfying the above conditions, I. that lim f(x) = I. Then < \x — c\ for every s > there is a 8 > such that, x—yc for all x, if If the ber N sequence {a n } such satisfies < lim a„ 8, = then \f(x) c, — 1\ then (Figure n-+oc that, if By our choice of 8, this n means > N, then \a„ — c\ that \f(a„)-l\ <e, < 8. < 3) e. there is a natural num- 22. Infinite 457 Sequences showing that = lim f(a„) = Suppose, conversely, that lim / (a„ ) I. for every / sequence n—X30 c. 8 If lim x—>c > f(x) there is = were / would be some not true, there {a n = n—»oo such that for every > s with lim a„ } an x with < — c\ < |.v In particular, for each n there < — \x„ \f(x) —l\ but 8 > e. would be a number x n such c\ < — but |/(jc„) — /| that > s. n Now the sequence clearly converges to c but, since \f(x n {.x,,} the sequence {/(-f„)} does not converge to — lim f(x) must be I Theorem 1 many examples {b,,} —» -• 02 Cl\ FIGURE #3 » have some > e for all n, This contradicts the hypothesis, so /. of convergent sequences. For example, the I a„ = b„ =cos sin and (13 + (sin 1 + (-1)" cos(sin(l)), respectively. It important, however, is guaranteeing convergence of sequences which are not obvi- criteria ously of this sort. There is which other the basis for is l\ defined by clearly converge to sin(13) to — true. | provides sequences {a n } and ) all one important criterion which results. This criterion is is very easy to prove, but stated in terms of concepts ««»» » defined for functions, which therefore apply also to sequences: a sequence {«„} ^4 increasing 5 if a n+ bounded above > an \ if there for is a nondecreasing if a„ + > a n number M such that a„ < M for all n all n, \ ; for all n, is and there are sim- sequences which are decreasing, nonincreasing, and bounded ilar definitions for below. THEOREM 2 If {a n ment PROOF The } is is set nondecreasing and bounded above, then true A if [a n is Then if bound {a,,} converges (a similar state- nonincreasing and bounded below). numbers consisting of all a least upper there ] is a. We some a^ satisfying n > N we have a„ is, by assumption, bounded above, so A has claim that lim a„ n—>oo a — a^ < a>i > tf/v, This proves that lim a„ = a. | so £, since a — a„ = a (Figure a is the least upper < a — «/v < s. 5). > 0, bound of A. In fact, if e 458 Infinite Sequences and Infinite Series The {a n is bounded above Upon diverges. first consideration, Theorem clearly essential in is not bounded above, then (whether or not is } hypothesis that {a n } {a,,} is nondecreasing) {a„) clearly might appear that there should be it trouble deciding whether or not a given nondecreasing sequence {«„} above, and consequendy whether or not sequences they converge hardly a is trivial we +l \ Although Theorem 2 useful than might you might matter. For the present, + 2 at first, j, +i + 1 + i is try to decide bounded above: •••• 5< very special class of sequences, because it is more always possible to extract from an it is another sequence which {a,,} To be nondecreasing. + treats only a appear arbitrary sequence 1 little bounded deciding whether or not shall see, whether or not the following (obviously increasing) sequence 1, is converges. In the next chapter such {a,,} very naturally and, as will arise 2: if is either nonincreasing or else subsequence of the sequence precise, let us define a {a,,} be a sequence of the form to jraph of a "71 where the rij are natural 89 "I Then 10 " ' ' " ' < < n3 «2 ' ' " every sequence contains a subsequence which 11 nonincreasing. It is possible to although the proof sertion, figure "3 ' numbers with 2 and 6 are peak points 1234 567 "71T ) | is become very short either nondecreasing or is quite befuddled trying to prove this asif you think of the right idea; it is worth recording as a lemma. 6 LEMMA Any sequence {a„} contains a subsequence which is either nondecreasing or non- increasing. PROOF number Call a natural > in n (Figure Case < «3 • • if am < if n\ a„ for all 6). The 1. • n a "peak point" of the sequence {«„} sequence has infinitely are the peak many peak points, then a,n > a„ 2 points. > a ny In this case, > • • • , so {a„ k } < n-± < the desired is (nonincreasing) subsequence. Case than a,,-, all > than all sequence has only finitely . Since 112 peak points) many peak points. In this case, let \ is not a peak point, there COROLLARY (THE 1 be greater . 3 that our original sequence {a,,} is | bounded, we can pick up an extra corollary along the way. BOLZANO-WEIERSTRASS THEOREM) \ is obtain the desired (nondecreasing) subsequence. we assume n some nj > n\ such that is not a peak point (it is greater than n\, and hence greater there is some ut, > ni such that a„ > a„ 2 Continuing in this peak points. Since n a, n way we If The 2. a cry bounded sequence has a convergent subsequence. . 22. Without some additional assumptions this is we can as far as 459 Sequences Infinite go: it is easy to construct sequences having many, evenly infinitely many, subsequences converg- Problem 3). There is a reasonable assumption to a necessary and sufficient condition for convergence of any se- numbers ing to different add, which yields quence. Although many investigations, If a the and N worth it is To be such that \a n A precise, —l\< - — lim a n if I e/2 for n > N; < ar , - \a„ + l\ — am < now \I if -a, both n > N and < = E. sequence N {a,,} is a is all A sequence {a,,} and condition all 2 condition) which is limit /, can clearly necessary converges {a n } is bounded. find that there if and only {a n } is some is means \am : m > N} whole sequence is there is a natural that a„\ it is = e. 0.) also sufficient to ensure converis very little left to do is it is a Cauchy sequence. Cauchy sequence a one if it = 1 converges. The proof of showing that every Cauchy tricky feature: in the definition of a Cauchy sequence such that Qn | < for 1 m, n > N that \a m Thus if If we take e /V \Q m In particular, this \ — \a m is — am < \a n our preliminary work, there the converse assertion contains only sequence > for every e if this. We have already shown that we N, then n, lim usually written The beauty of the Cauchy order to prove m m, n > N, then gence of a sequence. After PROOF Cauchy Cauchy sequence such that, for (This condition 3 m > convergence of a sequence. if theorem close to two such individual terms should be for some /, then for any e > there is which eliminates mention of the s, \ to formulate a condition (the number in all the difference of any final inequality, \a„ be used DEFINITION now stating 2 for does simplify sequence converges, so that the individual terms are eventually \a„ This it more advanced condition plays a fundamental role in this for this reason alone same number, then very small. an condition will not be crucial for our work, this Moreover, proofs. (see -a N +\\ < bounded; bounded. 1 for all m > N. since there are only finitely many other a,'s the . , 460 Infinite Sequences and Infinite Series The Lemma corollary to the some subsequence of thus implies that {a n } con- verges. 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 Verify each of the following limits. 1. (i) lim-^- = n->oo (ii) lim „_>oo 111 n + ^- = ni + — v n2 lim n—>oo 0. _|_4 + Z/n 2 —= lim n—>-oo n" 1 - Ifn 2 a > = Hint: You should at least be able to prove 0. 0. lim «—>oo '{fa = 1 VI lim '{fn = 1 Vll lim \fn 2 +n = lim Vfl" + b" = (viii) + 1=0. \Jn 1 that lim IV l. 1 0. 1. max(a, a, b b), > 0. n—>-oo oiifi) (ix) = lim The Hint: how where a(n) 0, (x) small a(n) must be. lim j- nP +i n-+oo = Find the following lim + n + 1 lim n — n 1 >/« 2" in 1 . n n-+oo + limits. fl (i) -. p lim n^-oo 2" +1 + + + aV" + b. (-1)" (-1)" +1 ' (-l) n v^sin(/i") (iv) lim 77^-00 (v) lim /?->-oo (vi) // -f 1 . a" + lim nc'\ /?" \c\ < the each prime fact that n ? is 1. is number of primes which > 2 divide n. gives a very simple estimate of — . . 22. Infinite 461 Sequences 2 (vii) —-. lim n—»oo n\ 3. (a) What can be about the sequence said {a n } if converges and each a„ it is an integer? (b) Find convergent subsequences of the sequence all 1 — 1, , 1 many, although there are only two .... (There are infinitely , — 1, 1 , — 1, which limits such subsequences can have.) (c) Find all 3, 4, 1, convergent subsequences of the sequence 2, 3, 4, 5, ... (There are . 1, many infinitely 2, 1, 2, 3, 1, 2, which such limits subsequences can have.) (d) Consider the sequence 11212312341 2 3 ' 3 ' ' 4 For which numbers a 4. (a) Prove that if 4 ' 4 ' 5 ' 5 ' 5 ' 5 ' ' 6 ' there a subsequence converging to is Cauchy sequence a subsequence of a a? converges, then so does the original Cauchy sequence. 5. (b) Prove that any subsequence of a convergent sequence converges. (a) Prove that (b) Prove that the sequence < a < if yjlsfl, y/2, < sjla < then a 2, 2. V2V2V2, converges. (c) Find the limit. by Theorem 6. < Let a\ Hint: Notice that — /, then lim n—>oo <b\ and define \ — I — s/an bn , b„ + , i —+— an = - bn (a) Prove that the sequences {a n } and {b n ) each converge. (b) Prove that they have the same In Problem 2-16 we saw that limit. any rational approximation k/ 1 replaced by a better approximation (k with k — = I 1 , we Prove that this + 21) /{k + /). obtain I 1 ' 2' 5 is given recursively by i (a) y 1a n = V2/, 1 an+ 7. lim a„ n—>oo if sequence a\ = 1, a n+ \ = 1 + — 1 1 + a„ . to V2 can be In particular, starting — — 462 Infinite Sequences and Infinite Series (b) Prove that — lim a„ n—xx This gives the so-called continued V2. fraction expansion / v 2= + 1 . 2 + 2 + ... Hint: Consider separately the subsequences {ai n } (c) Prove that for any natural numbers a and +b= V a2 many 9. = Identify the function /(.v) times in lim {fl2«+l b, a H . 2a 8. and lim ( + la H (cosrclTTjt)*-*). (It has been mentioned book.) this Many impressive looking limits can be evaluated easily (especially by the person who makes them With disguise. ing: the list remark this because they are really lower or upper sums in up), as hint, evaluate each of the following. (Warn- contains one red herring which can be evaluated by elementary considerations.) l/e .. + + lfe 2 --- + Ve" lim . lim lim iii) «->-oo 1 ( / lim iv) « + y n + + ... 1 1 H V n + (n hm vi) 10. Although \(n In + + —7T H \) 2 n n-V5b 1 1 —z2 lim — -, (2ny n ] ) 2 n 2) lim dissatisfying, 2 (n ' ++ n -j-— + limits like + (n n '±fn + n) 2 n 2 and lim a" can be evaluated using the behavior of the logarithm vaguely — and exponential functions, this Some out using the exponential function. and, for part (1 approach 1 is of the standard "elementary" ar- +/?)" > 1 lor// +////. > tools are inequalities 0; (e), +h)" > about because integral roots and powers can be defined with- guments for such limits are outlined here; the basic derived from the binomial theorem, notably (1 facts +/;// + —- -h 2 > —^--/* 2 , for// > 0. . . , . 22. (a) Prove that Km = 11 a oo > a if by 1, =1+ a setting n—>00 (b) Prove that Km = a" < a < if Infinite re, 463 Sequences where h > 0. 1 n—*oo (c) Prove that lim %/a = (d) Prove that lim y/a = 1 — 1 > a if 1 by 1, < a < if setting = '±J~a 1+re and estimating re. 1 /;—>-oo 11. (e) Prove that lim (a) Prove that a convergent sequence (b) Suppose all 12. (a) lim a„ that numbers "-Jn < + some a„ > Prove that the 0. set of 0. It maximum member. + log(« 1) — log/z < n If show = + l This number, is not even Suppose /(l) Now is = y it H { a 2 , J decreasing, is lim I + 1 • • / is • + ••• + /(/!- 1)< / = on increasing that each a n log > log n H [1, oo). and show < Show f f(x)dx < /(2) quite refractory; that + • • • + /(«). that < nl 1)" +1 + follows that Vn! ,. lim n->-oo This result shows that V/i! /; is ratio of these two quantities conclude that n\ estimate for n\ is is 1 = e approximately n/e, is close to 1 in the sense that the for large n . But we cannot close to (n/e)" in this sense; in fact, this is false. An very desirable, even for concrete computations, because cannot be calculated easily The standard right information will be found even with logarithm difficult) theorem which provides the Problem 27-19. (and in and number e n-\ re! /; ;? (n it log 1 known as Euler's number, has proved to be known whether y is rational. that choose l l + r that the sequence {«„} follows that there (b) always bounded. 1 1 (a) is that Prove that fl« 13. and 0, n—>oo a„ actually has a n (b) = tables. . . 464 Infinite Sequences and Infinite Series 14. (a) Show that the tangent line to the graph of / at (x\, f(x\)) intersects the horizontal axis at (X2, 0), where X2= — X\ f'(xi) may This intersection point be regarded as a rough approximation to where the graph of / intersects the horizontal axis. If we now start at X2 and repeat the process to get #3, then use X3 to get X4, etc., we have a sequence {x„ defined inductively by the point } f(Xn) x n+l Xn f'(Xn) Figure 7 suggests that this is we of this problem FIGURE method works 7 facts (b) Suppose (c) Let (Figures 8 8k that /,' > = > xo Xk — some conditions under which Newton's and 9 show two cases where it doesn't). A few will establish about convexity that x\ converge to a number c with f(c) — 0; In the remainder will {x,,} called Newton's method for finding a zero of /. may /" > and 0, > > Then that useful; see we choose Chapter 1 1 , Appendix. 0. Show [c, x\]. Show x\ with f(x\) > c. xt, c. be found f(Xk) 8k for some £* in (c, Xk). Show that <$*+i = /'(&) X\ X3 X5 X 2 X4 Conclude f'(Xk) that f(x k ) f"{m)(xk-$k) FIGURE 8 for some rjk in (c, Xk), and then (*) s k +i that < 8k l . f'iXk) (d) Let that (e) m = = on [c,A|] and Newton's method works if x\ — c f'(c) inf /' x3 x4 which is = = = = . 1 RE 9 x2 — A? 1.4142857 1.4142136 1.4142136, already correct to 7 decimals! Notice that the inequality (*) when M/in < 1 on 1.4 decimals at least doubled each time. ( = sup/" < m/M \ x2 I M What is the formula for xn+ when f(x) = If we take A = 2 and x\ = 1.4 we get X] I let This is essentially number of correct guaranteed by the 22. 15. Use Newton's method f{x) = — (iii) f(x) = x +x - (iv) /U)=x fix) (i) (ii) 16. cos* 3 Prove that 3 near 0. near 0. on on +l = lim a n if x 1 -3.r 2 465 to estimate the zeros of the following functions. — cos — x2 tanx Sequences Infinite [0, 1]. [0,1]. then /, n—»co — lim = w—>-oo problem Problem 13-40. Hint: This 17. (a) Prove that is very similar to lim a n+ n—>oo if \ /. n — = a„ (in fact, /, then actually a special case of) it is lim a n /n h—>oo = See the Hint: /. previous problem. (b) Suppose that / lim f(x)/x that A + [n,n *18. Suppose lim x (a) — > 20. that = {a,,} is is 1 for a , — f(x) = I. be the inf and sup of not in Prove / on sort of I. argument Prove that that works in > 0. all in [0, 1]. Prove also in [0, 1]. {a,,} of points all in (0, 1) such that lim a n (0, 1). / that is continuous and that the sequence x, converges to same = lim a n+ \/a n n—>co a convergent sequence of points Find a convergent sequence Suppose b„ 1) except using multiplication instead of addition, together with 16, Suppose is and each n and that for Hint: This requires the I. that lim a n n—>oo (b) Hint: Let a„ I. + 1]. the fact that lim \fa n—>oo 19. = lim f(x x—*-cx> CO that a„ '{fa^ Problem — continuous and is /. fix), fif(x)), fififix))), Prove that / is ... a "fixed point" for /, i.e., /(/) = /. Hint: Two special cases have occurred already. 21. (a) < fix) < 1 for all x in Suppose that / is continuous on [0, 1] and that [0, 1]. Problem 7-11 shows that / has a fixed point (in the terminology of Problem 20). If / is increasing, a much stronger statement can be made: For any x in [0,1], the sequence x, fix), fifix)),... is necessarily a fixed point, by Problem 20). Prove this by examining the behavior of the sequence for / (.v > x and fix) < x, or by looking at Figure 10. A diagram of this sort is used in Littlewood's Mathematician's Miscetlany to preach the value of drawing ." pictures: "For the professional the only proof needed is [this Figure] has a limit (which assertion, riiil'RE 10 ) 466 Infinite Sequences and Infinite Series *(b) / and g are two continuous functions on [0, I], with < < g(x) < for all x in [0, 1], which satisfy f o g = f(x) < 1 and o g /. Suppose, moreover, that / is increasing. Show that / and g have a common fixed point; in other words, there is a number / such that Suppose that 1 = = /(/) g(l). Hint: / Begin by choosing a fixed point for g. For a long time mathematicians amused themselves by asking whether the conclusion of part holds without the assumption that (b) two independent announcements cal Society, a pretty Volume Number 14, problem silly / is increasing, but American Mathematicounterexamples, so it was probably in the Notices of the 2 give along. all Problem 20 is really much more valuable than Problem 20 might suggest, and some of the most important "fixed point theorems" depend upon looking at sequences of the form x, f(x), f(f(x)), .... A special, but representa- The tive, is trick in theorem case of one such is Problem 23 treated in (for which the next problem preparation). 22. (a) Use Problem 2-5 c show m to that if c ^ 1, then +c m+\ + ... +c n = 1 (b) Suppose that < \c\ 1. Prove that lim c m m,n—* oo (c) Suppose that Prove that 23. Suppose that / c < 1 . Prove that (b) Prove that (c) By / / — • • + c n = 0. \x„ — xn+ i\ < where c", < c < 1. R such c\x is — that for all y\, x and y, called a contraction.) continuous. is has at most one fixed point. considering the sequence for any general (a) < f(y)\ x, 24. • a sequence with (Such a function (a) + Cauchy sequence. a a function on is \f(x) (*) where {x,,} is {x,,} is -C f(x), /(/(*)) / does have a fixed point. (This known as the "contraction lemma.") x, prove that setting, Prove that is / if is differentiable and \f'\ < 1, then / result, in a more has at most one fixed point. (b) Prove that (c) Give an example to 25. if \f'(x)\ ensure that This problem is / lo <c<\ show for all .v, then / has a fixed point. not sufficient problem. Let b„ be a = lim b„ exists 1 s has a fixed point. a sort of converse to the previous sequence defined by < i- that the hypothesis |/'0O| b\ = a, b„ + \ = f(h„). Prove that if h 22. and /' continuous at is = have b n b for some an interval around / on consider 26. Hint: If \f'{b)\ n). and b n b, be will the interval [b, b = (a) Note lim b„ exist? n—<-oo If b exists, > Suppose that — (c) e < we < a < e more is if b b n+ for all n. x in Now is < /y for < a < that that {b„} \ = a ", when does by Problem 20. form y (and also that b 1 1 enough some e y. Describe l/e increasing . and also b„ < e. e). difficult. b then e~ exists, < b < l e. Then show ]/e . From now on we Show then a in the Show . — a, =b define b\ exists, Using Problem 25, show that e 1, don't already the symbol and conclude ^e x exists analysis for a that e~ (d) b if 1/v y <a < 1 This proves that b The if then a can be written the graph of g(y) (b) that > 467 b,,]. In other words, sense. we then |/'(jc)I > (provided that 1 in this interval for large This problem investigates for which a makes < then \f'(b)\ b, Sequences Infinite will < suppose that e~ e a < 1. that the function fix) ax = logjc is (e) decreasing on the interval Using part / (f) (g) 27. = (e), lim «2»+l exists and that a° n—KX Using part Finally, (e) show show again, that that lim «2„ + 2 Let {x„} be a sequence which yn (a) (0, 1). number such that a b = b. Show that a < b < 1. show that if < x < b, then jc < a a < b. Conclude that Let b be the unique = is = I = = b. b, so that {y,,} lim b n bounded, and sup{x„,x„ +I Prove that the sequence /. ,.v„ + 2 converges. = b. let }. The limit lim yn is denoted by II—>00 lim xn or lim sup x n of the sequence (b) , and called the limit {x,,}. Find lim x„ for each of the following: n—>oo (i) xn = -. n (ii) x„ = (-l)"I. n superior, or upper limit, 468 Infinite Sequences and Infinite Series *„ (iii) 1 = (-!)" +- 1 n x„ (ivj = s/n. Define lim x„ (or liminf xn ) and prove that (c) < lim x„ lim xn —*oo . II (d) Prove that lim x„ = case lim x„ — lim x n if lim x n = n—>oo Recall the definition, in Problem 8-18, of lim (e) Prove that the if lim x n and that in this lim x„. n—*oo n-+oo and only exists if numbers x„ are distinct, then A for a lim x„ bounded = A. set lim A, where II—>oo A = 28. {*„ n in N}. : In the Appendix on an interval. If Chapter 8 we defined uniform continuity of a function f(x) is defined only for rational x, this concept still makes sense: we / there and say that some is \x to 5 - y\ < > 8, is uniformly continuous on an interval such that, then \f(x) if - if for every e x and y are rational numbers < f(y)\ in the interval e. Let x be any (rational or irrational) point in the interval, and (a) a sequence of rational > points in the interval such that lim xn let {x n = } be Show x. that the sequence {f(x n )} converges. (b) Prove that the limit of the sequence {f(x n )} doesn't depend on the choice of the sequence {x„}. We whole (c) denote will this limit by f(x), so that / is an extension of / to the interval. / Prove that the extended function is uniformly continuous on the inter- val. 29. Let a > x f(x) = a as defined in the usual elementary This problem shows directly that / can be extended to a and for 0, algebraic way. rational / on continuous function A' let , the whole show that a line. Problem 28 provides the necessary machinery. (a) (b) For rational x Using Problem rational (c) y, 10, numbers x show close Using the equation a interval (d) < Show / is x — x < any that for enough ay = < 1 and and a x > a y we have > \a x — for a 1| < < 1. s for {a x ~ y -- 1 ), prove that on any closed in the sense of Problem 28. / of Problem 28 is increasing for a > satisfies f(x + y) = f(.\)f(y). that the extended function for a e 1 to 0. ay uniformly continuous, and decreasing > a y for a 1 . I 22. "30. The Theorem Bolzano-Weierstrass differently than in the text A points. point x point a in (a) Find A all (i) with - a limit point of the set |jc — a\ < also proved, quite x s but / A > for every s if there a is a. of the following sets. N n in : and 469 Sequences the classical statement uses the notion of limit is limit points I — usually stated, is Infinite . I" — 1+1 (-!>" iii) m and n : m ;/ : N in N n in n (b) (iv) Z (v) Q is a limit point of A if and only if for every e > — a\ < e. many points a of A satisfying Prove that lim A is the largest limit point of A and lim A the Prove that * infinitely (c) an usual form of the Bolzano-Weierstrass infinite set point of a limit point of A. Prove is Using the form already proved numbers (a) this in x\, X2, *3, • • • one must contain A is then some [a, b], A is infinite, there in A. If [a, b] many infinitely is divided into two A points of 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 [a b] with , (b) states that if two ways: in the text. Hint: Since Using the Nested Intervals Theorem. Hint: intervals, at least 31. Theorem of numbers contained in a closed interval [a, b] are distinct (e) smallest. , The (d) there are |jc Theorem Also use the Bolzano-Weierstrass uous on [a, b], / then / (x„ > to ) prove that uniformly continuous on is n if / [a, b] (see is contin- Chapter 8, Appendix). **32. (a) Let {a n } be the sequence 112123123412 2 ' 3 ' Suppose that j < 3 4 ' ' 4 ' 4 < a < b < n such that aj is 5 • 1. ' 5 ' 5 5 ' ' 6 Let N{n; a, b) be the (Thus N(2\ in {a, b). 6 ' \, |) number of integers = 2, and N(4; A, |) = 3.) Prove that N(n;a,b) lim n->oo (b) A sequence {«„} —b — a. n of numbers in [0, 1] is called in [0, 1] if .. lim N(n;a,b) = b — a uniformly distributed 470 Infinite Sequences and Infinite Series for all < a and b with defined on and [0, 1], a {a,,} is l ,. s / Jo Prove that on [0, 1], if {a n } is < b < = hm 1 . Prove that uniformly distributed s(a\)-\ a step function if 5 is in [0, 1], then \-s(a„) uniformly distributed in [0, 1] and / is integrable then , / .. — hm /(«!) + + /(0.) /• **33. (a) Let / be a function defined on [0, 1] such that lim /(y) ' exists for all a y—*u in [0, 1]. For any in [0, 1] with | e > prove that there are only lim f(y) — y—>a f(a)\ > s. Hint: finitely Show many points a that the set of such points cannot have a limit point x, by showing that lim f(y) could not exist. (b) Prove / is that, in the terminology of Problem 21-5, the discontinuous Problem 6-17: uous except If at a / is countable. This finally set of points where answers the question of has only removable discontinuities, then countable discontinuous everywhere. set of points, and in particular, / is contin/ cannot be CHAPTER mm ^/ Infinite [NFINITE SERIES sequences were introduced + ai + «3 + a\ This the "partial sum = a\ \-a n -\ sum must presumably be infinite in the a\ — H fl 3 proximations as n be lim to sum of What can be defined are , defined in terms of these partial sums. this definition has already been devel- is to be any hope of computing the infinite sums sn should represent closer and closer apand larger. This last assertion amounts to little previous chapter. If there + 02 + more than • completely undefined. as yet is Fortunately the mechanism for formulating oped • 8111118" sn and the • not an entirely straightforward matter, for the is many numbers infinitely previous chapter with the specific inten- "sums" tion of considering their in this chapter. in the is • ? the partial chosen larger a sloppy definition of limits: the "infinite This approach s„. will necessarily leave undefined, since the sequence {s n } may easily fail sum" a\ +0.2 + 013 H ought the "sum" of many sequences to have a limit. For example, the sequence -1, 1, with a„ — (— 1)" +1 yields the si S2 st, 54 for which lim sn does not 1, -1, new sequence = a\ = = = = Cl[ a\ Cl\ exist. 1, + G2 = 0, + 02 + a 3 = 1, +CL2 + «3 + CI4 = unavoidable that some sequences sum of 0, Although there happen n—>oo tensions of the definition suggested here (see definition of the ... will to be some clever ex- Problems 12 and 24-20) have no sum. For this it seems reason, an acceptable a sequence should contain, as an essential component, terminology which distinguishes sequences for which sums can be defined from less 471 fortunate sequences. . 472 Infinite. Sequences and Infinite Series DEFINITION The sequence {a n summable is } the sequence {s„ } converges, =flH s„ is denoted by an (or, less In this case, lim sn n—>oo if \-a n where . oo y formally, + ai + «3 + ci\ n=\ and is sum of the called the The terminology sequence introduced in expressions; indeed the title {a u ). this definition of this chapter is usually replaced by less precise is derived from such everyday language. 00 An infinite sum >^ an is usually called an infinite series, the word emphasiz- "series" n=\ ing the connection with the infinite sequence The {a,,}. statement that {a,,} is, or oo is not, summable is conventionally replaced by the statement that the series does, or does not, converge. This terminology somewhat is ^ an n=\ peculiar, because at 00 best the symbol \_. a n denotes a number n= note anything at is (so it can't "converge"), and it doesn't de- \ all unless {a n } is summable. Nevertheless, this informal language convenient, standard, and unlikely to yield to attacks on logical grounds. Certain elementary arithmetical operations on quences of the definition. It is infinite series are direct a simple exercise to show that if {a n } and conse- {b„} are summable, then + b„) y^c an y^(a« = 2J a„ + 22 bn, oo =c } a„ n=\ n=\ As yet these equations are not very interesting, since we have no examples of summable sequences (except for the trivial examples in which the terms are eventually all 0). Before we actually exhibit a summable sequence, some general conditions for There is summability will be recorded. one necessary and stated immediately. sufficient condition for The sequence {a„} is converges, which happens, according to 1 s„ = 0; this ' summability which can be summable if and only Theorem 22-3, if and if the sequence only if lim m./i->00 {s,,} sm — condition can be rephrased in terms of the original sequence as follows. 23. THE CAUCHY CRITERION The sequence {a n } summable is a n+ lim m.n—>oo Although the Cauchy criterion if \ and only + • THE VANISHING CONDITION is it is not very useful However, one simple any particular sequence. consequence of the Cauchy criterion provides a which 0. of theoretical importance, is for deciding the summability of 473 if +a m — • Infinite Series condition for summability necessary too important not to be mentioned explicitly. If [a n ] summable, then is = lim a n 0. n—>oo This condition follows from the Cauchy criterion by taking be proved directly as lim a, = lim — (s„ n—»oo = 1-1 = Unfortunately, this condition is it can also —I, then follows. If lim s n «—>00 m = n + 1; s n -\) = from sufficient. lim n—>oo s„ — lim s n -\ n—*oo 0. far For example, lim \jn = n—>oo but the sequence {l/n} numbers l/n shows 1 + ^ is + 3^4 + 1 ^ 5 6 ^ 7 the following grouping of the fact, not bounded: is + + 9 + •• + * + •• 1 1 8 1 1 1 ^2 ^2 ^2 terms, (4 terms, each >|) each > g) (2 The method not summable; in that the sequence {s n } of proof used in this 0, terms, (8 each>^) example, a clever trick which one might never see, reveals the need for some more standard methods for attacking these problems. These methods shall be developed soon (one of them will give an alternate proof oo that y_\ l/ n does not converge) but examples of convergent The most important it be necessary to will first procure a few series. of all infinite series are the l+r + r 2 "geometric series" + r3 + «=o Only the cases if \r\ > 1. \r\ These < 1 series can be managed because the partial sn can be evaluated do not approach sums are interesting, since the individual terms = 1+r+ in simple terms. s„ rs„ = = 1 The two + r + r2 + 2 r + r + • • • + r" equations • + r" + r" +r" +1 . . 474 Infinite Sequences and Infinite Series lead to s„(\-r)= l-r" +1 or 1 -r" +1 -r 1 by (division lim r" = — 1 r 0, since is \r\ we valid since < It 1. Y* r" = are not considering the case r = Now 1). follows that 1 -r" +] lim 1 «=0 1 |r| -r < 1. = 1, In particular, E(iy--V2 «=0 EG) „=1 that X 7 is, 2 an 1-2 + 4^8^ 16 — ^ L > sum which can always be remembered from infinite the picture in Figure 1 I FIGURE 1 Special as they are, geometric series are standard examples from which important summability tests for we For a while will be derived. shall consider only sequences sequences are called nonnegative. If [a n quence {s n } is A all itself, sn is sn is theorem i (THE COMPARISON TEST) other Theorem 22-2, is summable if and only if the set of partial bounded. this criterion bounded to obtain such a nonnegative sequence, then the se- is is just is not very helpful —deciding whether or not what we are unable to do. On convergent series are already available for comparison, all 0; summability: nonnegative sequence {a„} sums By test for > with each an clearly nondecreasing. This remark, combined with provides a simple-minded THE BOUNDEDNESS CRITERION ] {a,,} a result whose simplicity belies its importance tests). Suppose that < an < bn for all // the set of the other hand, this criterion (it is if some can be used the basis for almost ' . Then if \^ b„ converges, so does 23. Infinite Series {s,,} is 475 >^ a„ n=l PROOF If sn = a\ t„ =b\ +--- an , + bn , -\ (- then < Now {t n } is bounded, sn < /.^n since for all n. tn Therefore converges. bounded; COnse- oo quently, by the boundedness criterion yj. a converges. >\ | n=\ Quite frequently the comparison test can be used to analyze very complicated looking series in which most of the complication A2 + 3 + sin (n 2" n=\ + is irrelevant. For example, 1) n2 converges because 2 + sin J (n + 2" + n 2 l) 3_ * 2" and ~ OO 00 . n=\ is a convergent (geometric) Similarly, we would series. expect the series 00 £ n=\ to converge, since the nth > - 2" term of the 2 1 +sin n 3 series is practically 1/2" for large n, and we would expect the series £ n=\ to diverge, since (n + l)/(n + 1) is + n2 + n 1 \ practically \/n for large n. These facts can be derived immediately from the following theorem, another kind of "comparison test." THEOREM 2 (THE LIMIT COMPARISON TEST) If a„, b„ > and lim a n /b n n—>oo converges. — c / 0, then 7 a n converges *—^ n=\ if and only if 7 *—' n=\ bn 476 Infinite Sequences and Infinite Series PROOF Suppose / b„ converges. Since lim a n /b n *—^ — there c, N some is n—»oc such that n=\ < 2cb n an > N. for n 00 Then Theorem But the sequence 2c >^ b„ certainly converges. n=N a„ converges, y and The finitely shows that implies convergence of the whole series a 2_] n n=l additional terms. this n=N has only 1 many we converse follows immediately, since also have lim b n /a„ = 1/c , which ^ /I-+00 The comparison test yields other important when we tests 0. | use previously an- 00 Choosing the geometric alyzed series as catalysts. we series par excellence, THEOREM 3 (THE ratio TEST) Let a„ > for all n , obtain the most important of all and suppose a n+\ ,- converges if r < 1 . tests for , the convergent summability. that hm Then >^ a n /_V W series On = r. the other hand, if r > 1 , then the terms a n 00 are unbounded, so >_, a n diverges. (Notice that it is therefore essential to compute n=\ PROOF lim a u+ \/a n n—>oo and not lim a„/a n+ \l) Suppose that r first n—»-oo < Choose any number 1. = lim implies that there is some N r with r s < 1 such that < lor n s > N. This can be written a n+ \ a N+ \ < sa n for n > N. Thus tf/v+2 < sa N < sa N+ < c'N+k < , \ s k aN . s aN , < s < 1. The hypothesis 23. = Since 2_. a NS &N /_, Infinite Series ^.11 converges, the comparison test shows that s A=0 fc=0 00 00 ~N+k N n= k=0 converges. This implies the convergence of the whole series 2_. a n 77=1 The case r > 1 is even < easier. If 1 a »+l > < s r, then there c lor n s is a number TV such that xr > N, an which means that (*N+k so that the terms are > QNS k unbounded. As a simple application of = k 0, 1 , . . . , | the ratio consider the series test, 2, ^/n\. Letting 77=1 an = we \/n\ obtain 1 (n «77+l + an n\ 1)! (n 1 + 1 n 1)1 + 1 n\ Thus *" +1 lim 77-*OC which shows that the -o, Qr series 2_. \/n\ converges. If we consider instead the series n=\ n 2_\f /n\ , where r is some number, then fixed positive 77=1 r lim /I—>-oo n+l ^L + f n M= , _J_ = o. im n—t-oo yi -\- n\ so Y^r"/n! converges. It follows that 77=1 r" lim — 77—>O0 n\ = 0, 1 478 Infinite Sequences and Infinite Series a result already proved in Chapter 16 (the proof given there was based on the same 00 ideas as those used in the ratio we Finally, if test). consider the series we nr" } n=\ have (n + + lim (// n—>oo since \)/n = \)r" +x 1 n—»oo nr' This proves that 1. + // lim r lim n—»oo < if < r 1, Ynr ^-^ then converges, n=\ and consequently lim nr" = 0. n—foo (This result clearly holds for direct proof of this Although the tool it A more exist. ratio test will 0, also.) It a useful exercise to provide a is test as an intermediary. be of the utmost theoretical importance, as a practical may be quite One drawback might equal difficult to of the ratio determine, and which appears with maddening serious deficiency, fact that the limit < be found disappointing. lim a n+ \/a n the fact that r 1 without using the ratio limit, will frequently — < 1. The — case lim an+ i/an 1 is may test is not even regularity, is precisely the the one n—>oc which inconclusive: is {a,,} might not be summable (for example, if = a„ \/n), 00 but then again might be. In it our very next fact, test will show that } {\/n)~~ n=\ converges, even though n lim This THEOREM 4 (THE INTEGRAL TEST) provides a quite different test gence of infinite series comparison but Suppose test, that / is — + 1 method like the ratio test, the series positive determining convergence or diver- for it is an immediate consequence of the chosen for comparison and decreasing on [1, oo), is quite novel. and that /(//) = 00 Then >_, a n converges if and only if the limit n=\ /»oo rA exists. A l'K< )( )l I he existence of lim f / /is equivalent to convergence jf'*jf '£'••• ol the series a n for all //. ' . 23. Now, / since is we have (Figure 2) /(n+ < decreasing I) /< / Infinite Series 479 /(«)• oo The half of this double inequality shows that the series first oo pared to the series 2_\ n=\ i . lim if Ja n= i (and hence 2_, a n) converges n=[ i i \ exists. / A^oo J ^^a„ + /? proving that I °° °° pn+\ x 00 n n + The second half of the inequality shows that the series 1 pn+\ 2, i pared to the FIGURE m ay be COm- y, a »+\ series > c/ /; , proving that lim / must / f maY I De com- Jn a n converges. > exist if | n=\ 2 Only one example using the integral test will be given here, but question of convergence for infinitely many series at once. If p > it settles 0, the the conver- oo gence of ^, \/n p is by the equivalent, integral test, to the existence of n=\ r°° l dx. xP J\ Now 1 1 f XP dx I ap- Cp-1) log A + 1 p#1 p-i' /? , = 1 °° r-A This shows that lim \/x p dx exists / if p > 1, but not if p < 1. A-^ooJ, Thus ) 1/h p ^— ll=\ oo converges precisely for p > In particular, 1. 2^ I/' 2 diverges. n=\ The tests sequences considered so far apply only to nonnegative sequences, but nonpositive may be handled in precisely the OO , OO \=l n=\ all same way. In fact, since v ; considerations about nonpositive sequences can be reduced to questions invok- ing nonnegative sequences. terms are quite another Sequences which contain both positive and negative story. 00 If 2_j a" *s a se q uence w tn i both positive and negative terms, one can con- n=\ oo sider instead the sequence 2, \ a n\, all of whose terms are nonnegative. Cheerfully 480 Infinite Sequences and Infinite Series we may have thrown away all the interesting informawe proceed to eulogize those sequences which ignoring the possibility that tion about the original sequence, are converted by this procedure into convergent sequences. The DEFINITION (In series /_, a " n=\ *s absolutely convergent more formal language, sequence the sequence THEOREM 5 Every absolutely convergent if formed from If 2_. \ 2, absolutely \ a n\ convergent. is summable if the The be of any interest, it turns following theorem shows that the definition not entirely useless. convergent PROOF is right to expect this definition to out to be exceedingly important. at least {a,,} the series summable.) {|fl„|} is Although we have no is if and only its if series convergent. Moreover, a series is the series formed from its positive is absolutely terms and the series negative terms both converge. Cauchy a n\ converges, then, by the criterion, n=\ lim m,n—>oo \a n+ h \a m -I | \ = 0. \ Since it < \-a,„\ \a„+i H \a„+\\ + follows that lim m,n—*oo which shows that To prove 2, a » an+ \ + am = H converges. n=\ the second part of the theorem, a„ + _ — f a„, 0, on , 0, so that 2_. a n^ is 0, the series let > ifa„<0, if a„ S ifa„>0, if #71 formed from the positive terms of / n=\ n=\ is the series formed from oo oo If the negative terms. a + and 2_[ n y an ~ both «=i n=\ l converge, then oo J2 k'»l = ]O + fl " ~ ( a» )] = ^L, a» + ' ^2 a » an and } , an , 23. also converges, so >^ Infinite Series 481 a„ converges absolutely. n=\ OO On the other hand, if 00 >_. n = \ an converges, then, as I we have just shown, 2, a n n=\ \ Therefore also converges. + o(Z> = Cl " Jl n=\ + J2 n=\ n=\ 00 OO n=\ n=\ la " and oo , 1 n=\ both converge. It | follows from be used Theorem 5 that every convergent series with positive terms can many to obtain infinitely other convergent series, simply by putting in 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 stateminus signs at — ment we need a test for convergence which applies specifically to series with positive and negative terms. THEOREM 6 (LEIBNIZ'S THEOREM) Suppose that ^ a\ and > a^ > lim a„ — ^2 • > 0, that 0. tt—>oo Then the series n=\ converges. PROOF Figure 3 illustrates relationships (1) s2 (2) Si (3) Sk Sl S4 FIGURE < > < s4 s3 sb s5 < > if si S(, 3 < > S8 -SlO k S]2 between the , is even and .V 1 1 J9 .¥7 S5 / is S3 odd. Sl partial sums which we will establish: 482 Infinite Sequences and Infinite Series To prove the To prove two first inequalities, observe that (1) S2n+2 = s 2n > s 2n (2) S2n+3 = S2n+\ < S2„+i, + ain+] ~ since a 2 „+\ , the third inequality, notice $2n S2n-1 < S2n-\ is easy: if k is even and > - «2n+3- that tt2n > since a2„ in 0. conjunction with (1) and (2) the general odd, choose n such that / is In a 2n+ 2 since a 2ll+ 2 first = > - ain+2 + «2«+3 This proves only a special case of (3), but case «2«+2 > k Sk < and In - < S2n-\ < > 1 /; then which proves Now, S2n the sequence {s2 n } converges, because above (by si Sl, (3). any odd for it is nondecreasing and is bounded Let /). = a sup{s2 n } = inf{5 2 7+ i} = lim S2„n-*oo Similarly, let = ft It follows from (3) that a < S2n+\ it is actually the case that The standard example ft; — a , since = S2„ = lim s 2n +i- n—»oo and ci2n+\ This proves that a ft. derived from Theorem 6 is = ft = = lim s„. | the series 4^5 2^3 1 lim a n oo which is convergent, but not absolutely convergent (since 2_, verge). If the sum of this series is 1 /" does not con- denoted by x, the following manipulations lead to quite a paradoxical result: r-1_I + I_ I4^5 + I-I + ^ ... — 2^3 * J — 6 I 4 I 1 J " 2 "'" (the pattern I I 1 3 6 8 here is one t _i_ W 1 J_ J_ _u I -L 5 10 12 14 positive I± 16 "•" ' term followed by two negative ones) = (i ~ - i + (5 - i> - i + <i " to) " A + (7 " A) - n + — 2 4^6 8 ^ 10 12 ^ 14 16 ^ — 2 U 2^3 4^5 6^7 8 ^ 2-) ; — — J, x, 2 • • . 23. = so x = x/2, implying that x sum the partial On 0. the other hand, equals i, and the proof of Leibniz's si it is 483 Infinite Series / easy to see that x Theorem shows that x > 0: S2- 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 — \Vnl contains In fact, bn — that = \, {b,,} is a 1 i 2' 3 3' 8' 10' 5 12' sequence in the 1 '4' number = 3m + 6' 1 1 1 1 1 J_ JL 5 6' 7' 8' 9' 10' 11 of 12' each in the following precise sense: {a,,} a certain function which "permutes" the natural numbers, is every natural 1) 4' ' rearrangement where / /(2m + 2 numbers [a n ) fl/(„) is, the all i, 1 m f{n) for precisely one is terms (the 1, n. In go into the 5, 5 our example 1st, 4th, 7th, . . places), = 3m /(4m) (the . /(4m + 2) = 3m + 2 is . (the . Nevertheless, there . . . — 4, — g, — -rj' terms • • g° i nto tne 3rd, 6th, 9th, • places), terms -\, -\, -^ go into the 2nd, 5th, 8th, places). no reason to assume that /_^^ n should equal n=\ sums are, by definition, lim b\ n—>oo + b„ + and lim a\ n—xx + +a„, ^ an : these n=\ so the particular 00 order of the terms can quite conceivably matter. special in this regard; indeed, — solutely convergent 7 If Yj «» behavior is how bad series /_\— 1)" typical of series the following result (really than a theorem) shows THEOREM its The + /n is not which are not ab- more of a grand counterexample conditionally convergent series are. converges, but does not converge absolutely, then for any number a there 00 is a rearrangement {b„} of {a,,} such that /.&„ = a - n=\ PROOF Let 2_. Pn denote the series formed from the positive terms of series let /_/7„ one had bounded partial follows from Theorem 5 that at least one of As a matter of fact, both must fail to converge, for sums, and the other had unbounded partial sums, then of negative terms. these series does not converge. if and n=\ n=\ denote the {a,,} It 484 Infinite Sequences and Infinite Series >^ a„ would the original series the assumption Now also have a be any number. Assume, let unbounded partial sums, contradicting n=\ that it converges. for simplicity, that a > (the proof for oo a < be a simple modification). will Since the series Y^ Pn is not convergent, „=] there a is N number such that N ^2 Pn > a. ,1=1 We will choose N\ to be the N smallest with this property. This means that N\-\ J2 p " - a (i) - n=\ but ^T p n > (2) a. n=\ Then if = ^Pn, Si n=\ we have S\ -a < pi PNi p\ H FIGURE V pn a -\ 1 P\-\ x p N] -i + Pn, 4 This relation, which is clear from Figure 4, follows immediately from equation (1): /V,-l Si -a < Si - ^jT p„ = p Ni . n=\ To the which sum is less we now add on just enough negative terms to obtain a new sum T\ than a. In other words, we choose the smallest integer M\ for which S\ T\ As before, = S\ + ^2 Q» < a - we have a -T\ < -q M} We now continue this procedure indefinitely, obtaining sums alternately larger and smaller than a, each time choosing the smallest Nk or M^ possible. The . 23. 485 Infinite Series sequence pi, ..., pN\, q\, is a rearrangement of The {a,,}. . . , Pni + \, <?a*i, •• Pm, . sums of this rearrangement increase partial to S\ , then decrease to T\, then increase to 52, then decrease to 72, etc. To complete the proof we simply note that S* — a and 7* — a are less than or equal to p^k or —qM | | respectively, and 1 | that these terms, being k members of , the original sequence {a n }, 00 must decrease to 0, since y_. a n converges. | Together with Theorem tion THEOREM 8 If 7, the next theorem establishes conclusively the distinc- between conditionally convergent and absolutely convergent \] a n converges absolutely, and {b n } is any rearrangement of series. {a n then }, \]b n n=\ also converges (absolutely), and CO 00 PROOF Let us denote the partial sums of {«„} by sn , and the sums of partial {b n by } tn . 00 Suppose that e > 0. Since converges, there 2, a n some is TV such that n=\ 00 ' - an / < sN £. Moreover, since >. a n converges, we can also choose \ TV so that I n=l 7, i.e., ~ \a,A (l«ll H h I«nI) < e, so that \<*N+\\ + + l«/V+2l Now choose M so large that each of a\, whenever m > M, the difference m — s^ t are definitely excluded. \ON+3\ H . is . , < £• a# appear among the sum of certain a,-, Consequently, \t,n — Sn\ < \Qn+\ I + k'/V + 2l + |«/V + 3 1 + • • • b\ • , . . . , Then but- «>/^/? aj , . . . , a^ 486 Infinite Sequences and Infinite Series Thus, if m > M, then 2_, - an - SN < 22 a n ~ SN y^<3» m t - (t m -Sn) n=\ n=\ oo + \t,n - SN \ n=] < E + £. oo Since this is true for every e > 0, the series 2. b n converges to 2, a n oo To show V, b„ that converges absolutely, note that { \b n \ } is a rearrangement oo of { \a„\ }; since V", a »\ converges absolutely, \ n=\ V, \bn\ converges by the first part of n=\ the theorem. | Absolute convergence series. is also Unlike the situation for addition, oo oo y^Qii isn't oo + ^2 b „ = y^dn+bn), n=\ there when we want to multiply two infinite where we have the simple formula important n=\ n=\ quite so obvious a candidate for the product \Y2 a ") ' (£M =(fl i + a2 + •••)• (b\+b 2 + ••)• It would seem that we ought to sum all the products ajbj. The trouble form a two-dimensional array, rather than a sequence: Nevertheless, all a\b\ a\bj a\b?, aib\ ^2^2 #2^3 dT,b\ a^bi fl 3 Z?3 that these the elements of this array can be arranged in a sequence. picture below shows one way of doing many is this, and of course, there are other ways. a\b\ ciT,b\ a\b2 a\b-), 02^2 #2^3 03/^2 03 bj, The (infinitely) 23. Suppose that {c,,} is some sequence of Then we might just once. Infinite Series 487 containing each product a/bj this sort, naively expect to have OO OO £ E = c» n=\ 00 a »-£^n=\ n=\ But this isn't true (see Problem 10), nor is this really so surprising, since we've said nothing about the specific arrangement of the terms. The next theorem shows that the result does hold when the arrangement of terms is irrelevant. THEOREM 9 If E an<^ fl " products conver §e absolutely, and E^" a- bj for x each pair OO 00 E £ = c« n=\ PROOF Notice first «£^- n=\ n=l sequence that the L = ^\a,\-^\bj\ pL and converges, since {a„} is > 0, if and since the limit of a Cauchy sequence, which means {b„} are absolutely convergent, So the product of the limits. that for any s any sequence containing the 00 fl L product is } then j), (/, {c n a {/?/.} is V are large enough, L and V L L' then L < It follows that £ (1) or i Now suppose that every term djbj for TV i, j \ai\-\bj\< e < S. j>L any number so large that the terms is < L. Then c„ for n < N the difference N L L J2 c * - J2 a '-J2 b J n=\ consists of terms cijbj with / > L E 2) > or j N ( L, so L L c" j=\ i=\ -E"'-E^ n=\ ^ E <e But since the limit of a product is L OO E «E^ - E a >=1 by(l). the product of the limits, OO (3) tai-w iovj>L 7=1 i=\ ./ = ! i=\ we L a '-£^ 7=1 < E also have include — 488 Infinite Sequences and Infinite Series for large will enough L. Consequently, if we choose and then N, L, large enough, we have 00 J2 a >-J2 b J2 c » -- J j=\ ,=1 oo 00 00 J2 a '-Jl b - Jl a '-J2 b - J ,=1 i=l J N L L a bJ + Jl >-Jl which proves the theorem. by 2s - I2 C » n=\ 7=1 i=\ < j=\ i=\ J=\ (2) and (3), | Unlike our previous theorems, which were merely concerned with summability, this result says no reason to something about the actual sums. presume terms. However, Taylor's sum can be that a given infinite many simple Generally speaking, there "evaluated" in any simpler expressions can be equated to infinite sums by using Theorem. Chapter 20 provides many examples of functions (i) f where lim R n , a (x) = 0. This is f(x) = hm for (a) precisely equivalent to — - > n—>oo *—-^ - (x a) , 1 i i=0 which means, in turn, that /(*) = £ f n (a) (x-a) 1 , V. i=0 As particular examples we have 4 2 cos* = arctan x log(l+*) =' + v + = 6 X X x l--+-~+ X e X3 x — = ,~ is T" X2 + x*2 v. X 14 + xJ + + v. 5 X X4 X3 x* + ••• , \x\ < I, 5 + X' T+T T T+ .- - v 7 =- "^ ..., - 1 <X< 1 which 23. in addition, = — 1, when x + x) and log(l (Notice that the series for arctanx do not even converge log(l + x) becomes the series for _1 1 Infinite Series i_I_ I 2 4 3 for \x\ 489 > 1; which does not converge.) Some pretty impressive results are obtained with particular values of x ' =1 + + + II 2i 111 3i : + -- _ = ,__ + ___ + ..., 7T , „ log2 More sinjc significant and cos* a have obtained if = l-- + 1 , developments i - 5 J may be 1 + .... anticipated if we compare the series for little more carefully. The series for cosjc is just the one we would we had enthusiastically differentiated both sides of the equation sill X =X— x x — 1 3! 5! we have never proved anything about the derivatives of infinite sums. Likewise, if we differentiate both sides of the formula for cosx formally (i.e., without justification) we obtain the formula cos'(x) = — sinx, and if we differentiate the formula for e x we obtain exp'(x) = exp(x). In the next chapter we shall see that such term-by-term differentiation of infinite term-by-term, ignoring the fact that sums is indeed valid in certain important cases. PROBLEMS 1 . Decide whether each of the following gent. The tools ison, ratio, and which you integral infinite series tests. aforethought; two series which look quite similar (and then again, they used. oo „ . E SU1H0 n2 l i-i+i00 iv n=\ Jogn may is convergent or diver- need are Leibniz's Theorem and the comparA few examples have been picked with malice will not). The hint may require different below indicates which tests tests may be : 490 Infinite Sequences and Infinite Series i V^ v) ( . - n=2 ifn* OO 1 (The summation begins with n the meaningless term obtained i Ei VI -) oo (v.) j E^ oo (viiT logn E n= n \ 00 IX , ^— E log/? ' oo i ^ A *-*> (lo (log/?)* S n) oo 1 ^ ^ - (log/7)' (logn)" n=2 00 M E- V-1V1) (log "oo^r ( oo i ? EH" oo (xiv) , E sin oo (xv) V- ^" i oo (xvi)7 7 ^— i ' — on(log n ) oo (xvii) 7 ^— i ~ ' -^7; 2 (log/?) /? n=2 ( XV111 • *-^ n log n n=1 )Z.^7- n=l =2 for n simply to avoid = 1). . 23. > (xx) Hint: Use the comparison test for (vii), (xviii), (xix), (v), (vi), (ix), (x), (xi), (xiii), (i), the integral test for (xx); The next two problems examine, with hints, some more delicate analysis than those in Problem 1 (a) If 491 . the ratio test for *2. Infinite Series you have examples successfully solved (viii), (xiv), (xvii); (xv), (xvi). infinite series that (xix) and require (xx) from Problem < and diverges 1, 00 it n should be clear that a n\/n >_. n converges for a e for oo a > = For a e. diverges, e the ratio test fails; show that " " 2_, e n}-/ n actually by using Problem 22-13. 00 (b) Decide when 2_. n " " converges, again resorting to Problem 22-13 / a n} - n- when the ratio test fails. oo oo *3. Problem the first 1 presented the two series >_,(log«) n=2 diverges while the second converges. _A ' and V\log«)~", of which n=2 The series oo V ^ n=2 which lies between these two, is 1 - (log«) lo §"' analyzed in parts (a) and (b). oo (a) Show that f^° (b) Show that ey y /y dy exists, by considering the 00 series \]( e / n )" . V ^(logrc) ! 11=1 u converges, by using the integral tion (c) and part Show lo g« test. Hint: Use an appropriate substitu- (a). that oo v ^— ' diverges, . ! (log«) lo s (1 °g") by using the integral in part (b), and show test. Hint: Use the same substitution as directly that the resulting integral diverges. 492 Infinite Sequences and Infinite Series i 4. —r— Decide whether or not > —. converges. n=\ oo 5. (a) Prove that oo n=\ *(b) 6. Let Show / be a,, . n=\ that this does not hold for conditional convergence. a continuous function on an interval around 0, and (for large 7, /_]a„ converges absolutely, then so does / if enough = let a„ /(1/n) n). oo (a) Prove that } an if = converges, then /(0) "=1 0. oo (b) Prove that if f'(0) exists and (c) Prove that if /"(0) and /(0) \] a" converges, then f'(0) n=] exists /'(0) = a converges. 0, then 2_] " »=1 Suppose 2_] a " converges. Must f'(0) "=1 (e) 0. oo = oo (d) = exist? 00 = Suppose /(O) /'(O) = 0. Must /. a " converge? n=\ oo 7. (a) Let < be a sequence of integers with {a,,} < an 9. Prove that 2, a n 10~" n=\ (and exists we (b) and between lies usually denote by 0. a xaia^aA, Suppose < x < that 1. (This, of course, 1). . . . is the number which .) Prove that there a sequence of integers is {a,,} 00 < with < 9 and /Ja„ 10~" = a„ .v. »=1 (where [j] denotes the greatest integer (c) Show that if {a,,} repeating, is i.e., Hint: For example, aj which is < = y). of the form a\, ai, is [10x] • • • ,dk, 00 a i , ai , . . • , cik:,<* l ^2 • , , then Y^ «„ 1 0~" a rational is number (and find n=] it). The same result naturally holds the sequence {a^+k} is repeating for if {a n } eventually repeating, is i.e., if some N. 00 (d) Prove that if x — 0~" \_] a n 1 is rational, then {a„ } is eventually repeat- n=\ ing. (Just look at the process of finding the decimal expansion of dividing q into 8. Suppose that {a n } p by long the hypothesis of Leibniz's satisfies proof of Leibniz's Theorem +1 Jj-1)" n=\ «„ p/q— division.) Theorem. to obtain the following estimate: -[ a] -a 2 + ---±a N ] < a N+ \, Use the ' . 23. 9. Prove that (a) > an if ~ and lim ?Ja n = then > r, ^— n—»oo 493 Infinite Series a„ converges < if r 1, ,;=1 and diverges if r > 1 (The proof is very similar to that of the . ratio test.) oo This result known is More as the "root test." converges 2, a n generally, n=] there if is some < s such that 1 all but many finitely < ^/a^ are and s, 00 a„ diverges y many if infinitely aja^ are > 1 . This result known as is the 71=1 "delicate root test" (there a similar delicate ratio is follows, using It test). oo the notation of and diverges Prove that (b) Problem 22-27, test 1 — chapter. easy to construct series for which the ratio works. For example, the root i < n—>00 ; problem from the previous It is lim ^/a^ if 1 lim 'ifa^ > 1 no conclusion is possible if lim t/a^ n—*oo n—>oo the ratio test works, the root test will also. Hint: Use a if if > a converges *—^ n that + test shows that the + a) 2 +(i) 2 +(i) 3 +(l) 3 + i test fails, while the root series --- converges, even though the ratios of successive terms do not approach a limit. is 10. Most examples are of this rather artificial nature, but the root test nevertheless quite an important theoretical tool. For two sequences {a n ) and {b n }, let c n = a kbn+\-k- (Then c„ 22. is The of the terms on the nth diagonal in the picture on page 486.) c„ y is called the Cauchy product of /. a » ancl n=l >i=\ {—\) n /y/n, show that \cn \ > 1, sum series oo 00 00 the /^ bn If • <*n = bn = n=\ so that the Cauchy product does not con- verge. 11. (a) A Consider the collection of natural numbers that do their usual (base 10) representation. A converges. Hint: how many between 10 and of the numbers in 9 are (b) If B in is 9 the A sequence in B then the do not have sum all 1 digits of the reciprocals of all ten digits representation.) {a n } is called Cesaro summable, si +s„ -\ lim 7!->-0O Sk that converges. (So "most" integers must have .. (where contain a 9 in 99?; etc. numbers in their usual representation, numbers in their 12. A?; the collection of natural not Show that the sum of the reciprocals How many numbers between 1 and = a\ H \- a*). = with Cesaro sum /, if / n Problem 22-16 shows that a summable sequence l 494 Infinite Sequences and Infinite Series is 13. sum automatically Cesaro summable, with a sequence which Suppose > that a n equal to summable, but which not is and {a,,} its Cesaro sum. Find Cesaro summable. is Cesaro summable. Suppose is also that the oo El sequence {na„} bounded. Prove that the is n / 14. n=\ n . n v^ and an = — > s,, *-^ a, prove that sn n = /_. a » converges. Hint: If series 1 '= + n -an bounded. is 1 1 This problem outlines an alternative proof of Theorem 8 which does not rely on the Cauchy (a) criterion. Suppose that a n > and let s„ — a\ + • there is m some (b) Show (d) that Let {b n } be a rearrangement of n. and < tm s„ = t„ b\ + • • • + bn Show that for . {a,,}, each n . 00 — /. that 2_. a " n=\ Show + a„ 00 ^« /, bn if exists. n=\ n=\ 00 (c) • with 00 each for • 00 2, a n = /_. ^" n=\ n=\ • Now replace the condition a n > by the hypothesis that >^ a n converges absolutely using the second part of Theorem 5. 15. (a) Prove that if /_, a " converges absolutely and {b n } is any subsequence of n=\ 00 {</„}, then 2_j^" converges (absolutely). OO (b) Show that this is false if Y_] o-n does not converge absolutely. n=\ 00 *(c) Prove that 2_. a » converges absolutely then if n=\ 00 an } = (a\ + a-j + as H ) + («2 + «4 + «6 + • )• n=\ 00 00 16. Prove that if V an is absolutely convergent, then /, Problem 19-43 shows Prove that *18. / |(sin that the x)/x\dx Find a continuous function exists, i_, l a "l" improper integral f 00, (sinx)/xdx converges. diverges. / but lim f(x) does not Jt-»-00 — n=\ «=] *17. 00 a" with f(x) exist. > for all x such that / f(x)dx ) . 23. 19. — Let f(x) x < x < l/x for sin not bounded (thus is 1] 21. €(/, P P) for a 2 2 we 1 1)tt shown Figure in In this problem 2 . the function shaded region all has "infinite length"). Hint: Try / 0, (2/?+ / be that the set of 2 P= Let Recall the definition 0. form partitions of the 20. = /(0) let Show of i(f, P) from Problem 13-25. partition of [0, and 1, 495 Infinite Series Find in Figure 5. and /, fQ also the area of the 5. "binomial series" will establish the a = £; (!+*)« < \x\ 1, k=0 for any a, by showing that lim R n o(x) = and uses the Cauchy and Lagrange forms (a) Use the for ratio test to < |r| follows in particular that lim FIGURE as found that the series 2_^ not to say that (this is 1 show The proof is 0. it in Problem 20-21 r does indeed converge + necessarily converges to (1 \r" I \ I in several steps, = for < |r| r) a ). It 1. 5 (b) Suppose < x < that first Show 1. lim R„q(x) that = H—>-00 < Lagrange's form of the remainder, noticing that (1 + > a. Now suppose Cauchy 's form of n (c) + by using 0, a~n ~ t) l for 1 \ remainder +t) a |.v(l —1 < x < 0; the number / — < x < t < 0. Show that that satisfies ~ < ] where \x\M, \ in the 1 M = max + xf (1, (1 '), and x - t 1 TT7 - t/x < x \+t Using Cauchy 's form of the remainder, and the + (n a 1 Ji show that lim n-+oo 22. (a) Rn Suppose that the o(x) = partial + = a a- fact that 1 1 0. sums of the sequence {a n } are bounded and that oo {/?„} is a sequence with b n > b n+ and lim \ b„ = 0. Prove that II—>0O Y^ a n b„ ' < n=\ converges. This is known as Dirichlet's (Problem 19-36) to check the Cauchy (b) Derive Leibniz's Theorem from test. Hint: criterion. this result. Use Abel's Lemma 496 Infinite Sequences and Infinite Series Prove, using Problem 15-33, that the series 2~]( cosnx )/ n converges (c) x if n=\ not of the form Ikn for any integer k is (in which case clearly diverges). it oo Prove (d) Abel's If >_, test: a converges and {b n } a sequence which is >i is either bounded, then j a n bn n=\ nondecreasing or nonincreasing and which is n=\ — converges. Hint: Consider b„ where b b, — lim b„. n—>oo oo *23. Suppose {a n ) decreasing and lim a n is — Prove that 0. if n—KX) / t—^1 n=\ a„ converges, oo 2,^ a then a l so converges (the "Cauchy Condensation Theorem"). No- 2n OO tice that the 00 divergence of >^ \ jn is a special case, for if n=\ 2, l/ n converged, n=\ oo 2; 2" (1/2") would then remark may serve also converge; this as a hint. n=\ |: 24. Prove that (a) and 2_,^ n " converge, then 2_, a nb„ converges. 2_, a n if »=1 n=\ /;=1 OO Prove that (b) 00 /_, a " if converges, then 2_, a n/n a converges for any a. n=\ ^25. Suppose {a„} then lim na n decreasing and each a n is — 0. Hint: Write down > 0. Prove that Cauchy the j an if criterion converges, and be sure to — 0. n^-oo use the fact that [a n ] is decreasing. 00 ^26. If a„ converges, then the partial 7 sums s„ and lim are bounded, a„ n=\ It is tempting to conjecture that boundedness of the partial sums, together oo with the condition lim a„ = 0, implies convergence of the series H->-00 > ^-^ a„. n=\ Find a counterexample to show that subsequence of the partial this to happen, without sums will this have letting the is not true. Hint: to converge; Prove that if a n > and 7 ^ n=\ Compare the partial sums. you must somehow allow sequence of partial sums oo 27. Notice that some oo a n diverges, then 7 <-" - —+a„— 1 11=1 Does the converse hold:' - itself converge. also diverges. Hint: . 23. 28. we For b n > say that the infinite product 1 b„ 1 497 Infinite Series converges if the sequence n=l P" = l I converges, and also lim p„ hi =£ 0. i=\ (a) Prove that if I b n converges, then b n approaches 1 n=\ OO (b) Prove that 00 1 b„ converges 1 and only if Y^ log b„ if converges. «=] (c) > For a n 0, prove that + a„) ( 1 1 J converges if and only Y~, an con- if n=\ Use Problem 27 verges. Hint: + a) for log(l n=\ for one implication, and a simple estimate for the reverse implication. The remaining Problem show parts of this that the hypothesis a n > is needed. (d) Use the Taylor series for log(l + x) show to that for sufficiently small x we have \x 2 <x -log(l an > —1 and + x) < \x 2 . oo Conclude that if all 2_. a " converges, then the series n=\ 00 oo + 2_^log(l a„) converges if and only /_]a« if n=l 2 converges. )i=l 00 if all > —1 and an oo + an converges, then V^log(l 2_. an ) converges oo and only y, a if >\ Use the Cauchy converges. Hint: criterion. n=\ {€) Show that (-1)" ^ v^ =2 n n=2 converges, but nc + n=l diverges. (f) Consider the sequence In \ I""/ Similarly, — — 1 1 ' 1 -Li 2 3 ' pair -L ' 4 L ' 3 ' -II 4 3 3 pairs ' -L ' -L L 4; 5 ' 6 L ' .V .1 6 1 ' 5 pairs 5 1 ' 6' if ' 498 Infinite Sequences and Infinite Series (compare Problem Show 26). that diverges, but 2_, a " n=\ oo Y](\+a„)=--\. n= B=l 29. (a) Compute f\(\ --A n=2 oo (b) Compute J~](l + .v 2 ") for |.v| < 1. 00 30. The divergence of \] \/ n related to the following remarkable fact: s i- Any n=\ positive rational of the form \/n. number x can be written as definite sum of distinct numbers The idea of the proof is shown by the following calculation for j[: Since 27 _ i 31 2 23 62 3 1 _ — _ — 23 62 7 186 186 ^ 4' 7 J_ _J_ 186 27 _ — • • • 26 ' 1674 we have 27_i_ — 2 i_i,J__ "^ "^ 31 3 27 | Notice that the numerators 23, (a) Prove that sort if \/{n + 1) < jc 1_ | " 7, < 1 1674 of the differences are decreasing. 1/rc for some n, then the numerator in as a finite sum of distinct numbers \/k. 00 (b) this of calculation must always decrease; conclude that x can be written Now prove the result for all x by using the divergence of V, 1 /«. UNIFORM CONVERGENCE AND POWER SERIES CHAPTER The considerations at the end of the previous chapter suggest an entirely of looking Our at infinite series. equations to attention will shift from particular new way infinite sums like 2 X X =!+_ + _ + ... «• depend on x. defined by equations of the form which concern sums of quantities interested injunctions / (*) (in the previous = /l(*) = example f„(x) In other words, that + /2OO + /3 (*) + •• l x"~ /(n — we are • In such a situation {/„} will be 1)!). some sequence of functions; for each x we obtain a sequence of numbers {/„(.*)}, and fix) is the sum of this sequence. In order to analyze such functions it will certainly be necessary to remember that each sum /1W + /2W+/3W + is, by sequence definition, the limit of the /!(*), If we define a f l (x) new sequence + f2 (x), of functions sn then we can express this fact = more we shall therefore FIGURE on functions defined such functions can be true actually first is — we have of these shows that even Contrary to if = \- fn, by writing lim s n (x). concentrate on functions defined as = The body of results about nothing one would hope to be very easily: total a splendid collection of counterexamples. each /„ is continuous, the function f„ will shows the graphs of the functions < = 1, 499 limits, lim f„ix), n—*oo what you may expect, the functions fnix) .... by as infinite sums. summed up instead } succinctly f(x) rather than {s n + f2 (x) + Mx), f\ H f(x) For some time fi(x) X > A" 1. < 1 / may The not be! be very simple. Figure 1 , 500 Infinite . Sequences and Infinite Series 1- These functions are — continuous, but the function f(x) all lim f„(x) is not n—>oo continuous; in fact, < x < 0, lim fn (x) Another example of tions /„ are defined same phenomenon this > x 1 1 1 illustrated in Figure 2; the func- is by 1 x< -1, n FIGURE 2 /«(*)= 1 n n - < x < - nx, ' 1 1 - < 1, x. n In this case, and if x > if 0, x < 0, then then f„{x) fn (x) eventually is eventually is lim f„(x) so, once again, the function f(x) By rounding FIGURE 3 fn (x) is while /„(0) -1, A<0 jc=0 1, x>0: lim f„(x) One enough for all n. n) equal to — 1, Thus not continuous. is examples it is even possible to functions {/„} for which the function f(x) dijferentiable not continuous. = 0, off the corners in the previous produce a sequence of lim = = 1, for large (i.e., such sequence easy to define explicitly: is 1 < x = n /H7TX\ fnix) 1 1 < 1, These functions are <x < sin differentiable (Figure 3), but lim fn {x) = x. we still -1, x<0 0, x = 1, x > have 0. Continuity and differentiability are, moreover, not the only properties for which problems arise. Another Figure 4; on the interval altitude n, while f„(x) I I I , I l< 1 . 4 as follows: = by the sequence {/„} shown in the graph of /„ forms an isosceles triangle of difficulty [0. 1//;] for is illustrated x > \/n. These functions may be defined explicitly 24. Uniform Convergence and Power Series 2n 2 x, = fn(x) 2n — < _ < a — ~ In 1 1 2n 2 x, 501 to**' n I <x < 0, Because instincts limit In this sequence varies so erratically near 0, 1. our primitive mathematical might suggest that lim f„(x) does not always does fact, if n—*oo x, and the function f(x) exist for all x > 0, then for all n, so that we fn (x) is = lim f„{x) = lim f„(0) 0. n—*oo — lim f,,(x) for all On x. even continuous. is n—»oo eventually 0, so lim f„(x) certainly have Nevertheless, this exist. = 0; = moreover, /„(0) In other words, f(x) = the other hand, the integral quickly reveals the n-+oo strange behavior of this sequence; we have f„(x)dx = j, /' /(*)</* = 0. Jo FKiURE 5 but Jo Thus, lim / -*°°Jo f,,(x)dx y£ n I lim f„(x)dx. JO "^°° This particular sequence of functions behaves in a imagined when we first considered functions defined by way that limits. we really never Although it is true that f(x) = the graphs of the functions /„ lying close to it — if, for each x in [0, lim f„(x) 1], n—foo do not "approach" the graph of / in the sense of we draw a strip around / of total width 2s (al- as in Figure 5, lowing a width of e above and below), then the graphs of /„ do not lie completely within this strip, no matter how large an n we choose. Of course, for each jc there is some TV such that the point (x, fn (x)) lies in this amounts to the fact that lim f„(x) = f(x). But just strip for n it is > N; this assertion necessary to choose larger n—>oo and all x larger /V's 6 x is chosen closer and closer to situation actually occurs, examples given previously. Figure 6 though 0< 1. A strip < of total less blatantly, for illustrates this MX) If e and no one 0, Af will work for at once. The same FIGURK as X > each of the other point for the sequence x < 1 1. width 2s has been drawn around the graph of f(x) 2? this strip consists = lim f„(x). n—>-oo of two pieces, which contain no points with second . 502 Infinite Sequences and Infinite Series coordinate equal to |; since each function f„ takes on the value j, the graph of each /„ fails to lie within this strip. Once again, for each point x there is some N such that It is fn (x)) (x, which works for > N; but it is not possible to pick one N at once. easy to check that precisely the same situation occurs for each of the other we have examples. In each case some defined on all the strip for n lies in x all a function /, and a sequence of functions {/„}, A, such that set /(*)= lim f„(x) for all x A in n—>oo This means that FIGURE 7 for s all > 0, and \f(x)-f But in for all x A, there in N some is such that n if > N, then <£. n (x)\ each case different TV's must be chosen for different x's, and in each case n > N, it not true that is for all |/(x) Although > e there -/B (x)| this phrase "for some is N such that for x if then <e. condition differs from the all x in A, all in A," it first only by a minor displacement of the has a totally different significance. If a sequence {/„} second condition, then the graphs of /„ eventually lie close to the graph of /, as illustrated in Figure 7. This condition turns out to be just the one which makes the study of limit functions feasible. satisfies this DEFINITION Let {/„} be a sequence of functions defined on A, and let / be a function which Then / is called the uniform limit of /„ } on A if for is also defined on A . > every e there { some is if We f also say that {/„} n TV such that for > yv\ then all x in A, - fn (x)\ < |/(jc) s. converges uniformly to/ on^4, or that /„ approaches uniformly on A. As a contrast to this definition, f(x)= then we say that {/„} if we know lim f„(x) only that for each x in A, n—>oo converges pointwise to f on A Clearly, uniform conver- gence implies pointwise convergence (but not conversely!). Evidence for the usefulness of uniform convergence Integrals represent a particularly easy topic; Figure 7 if to is not at all difficult to makes it amass. almost obvious that converges uniformly to /, then the integral of /„ can be made as close the integral of / as desired. Expressed more precisely, we have the following {/„} theorem. 503 24. Uniform Convergence and Power Series THEOREM 1 Suppose that /„ { [a b] , that {/„} a sequence of functions which are integrable on [tf,&], and is converges uniformly on [a ,b] to a function } / e on pb / = lim n Jit Let integrable is Then . pb PROOF / which > Thus, 0. if n There N > is some N \f(x) — fn (x)\ < /„. / ^°° J a such that for N > n all we have for all x in [a, b]. £ we have pb pb f{ X I )dx- f [f(x)~ fn (x)]dx fn (x)dx I Ja Ja Ja 4 \f(x)-fn (x)\dx edx = Since this is true for any e > 0, pb /= / lim Ja treatment of continuity a). follows that it pb The — s(b /„.| / Ja is only a bit little "e/3-argument," a three-step estimate of \f(x) more involving an difficult, — f(x + /?)!• If {/„} of continuous functions which converges uniformly to /, then there is is a sequence some n such that Moreover, since /„ (1) \f(x)-fn (x)\ < (2) \f {x+h )- |, fn {x + < h)\ |. we have continuous, for sufficiently small h is ~ \f„(x) (3) + h)\ < fn(x s 3 It will follow from however, we must (1), (2), that THEOREM 2 (2) true, it is \f(x)- f(x+h)\ < that of \h\ in a way that s. In order to obtain cannot be predicted therefore quite essential that there be no matter how small \h\ may be — it is some (3), until n fixed n precisely at this point uniform convergence enters the proof. Suppose that {/„} is a sequence of functions which are continuous on that {/„} converges uniformly PROOF (3) restrict the size has already been chosen; which makes and For each x in [a b] , with x in (a, /?); on [a, b] to /. we must prove the cases x — that a and x — / is Then / is [a, b], also continuous continuous at x . We on and [a, b]. will deal only b require the usual simple modifications. 504 Infinite Sequences and Infinite Series Let £ > / on Since {/„} converges uniformly to 0. [a, b], there is some n such that 1/00 In particular, for Now /„ is -/»(.v)l h such that x all \f(x < + h)- 8, h is (2) |/(.v+//)-./; is some 8 all in [a, b], |/(*) -/„(*)! continuous, so there if \h\ + for (1) < ) > v in [a,b]. we have |. U+ /z)| < |. such that for |/j| < <5 we have \fn(x)-Mx+h)\ <-. (3) Thus, <\ then f(x)\ + fn (x) - f(x)\ - fn (x)\ + \f„(x) - f (x)\ = \f(x +h)- fn (x +h) + fn (x +h)- f < \f( X +h) - /„(X+A)| + n SEE \fn(-X+h) (x) "3+3+3 = This proves that £. / is continuous at x. | f(x) FIGURE = \x\ S After the two noteworthy successes provided by the situation for differentiability turns out to / is 1 and Theorem 2, each /„ is If uniformly to /, it is still not necessarily true differentiable. For example, Figure 8 shows that there is a sequence of differentiable, that and Theorem be very disappointing. if {/„} converges differentiable functions { f„ } which converges uniformly to the function f(x) = \x\. 505 24. Uniform Convergence and Power Series Even / if is differentiable, it may not be true that fix) this is not at all by very rapidly surprising if we = lim n-»oo fn '{x); a smooth function can be approximated example (Figure 9), if reflect that oscillating functions. For 1 2 fn (x) = -sin(n x), n then {/„} converges uniformly to the function f(x) f„'{x) and lim n cos(h 2 *) does not always — = 0, but n cos(n x), exist (for example, it does not exist if x = 0). fn FIGURE -19 Despite such examples, the Fundamental antees that rem 1; some sort Theorem of Calculus of theorem about derivatives the crucial hypothesis will practically guar- be a consequence of Theo- that {/,/} converges uniformly (to some continuous is function). THEOREM 3 Suppose that {/„} is a sequence of functions which are differentiable with integrable derivatives moreover, that Then / is {/„'} /„', and Applying Theorem 1 [a, b] to some continuous function and f'{x) PROOF = lim /„'(*) to the interval [a, x], / g = = lim / we see that for each x /„' lim [/„(*)- f„(a)] n-»-oo = [a,£], that {/„} converges (pointwise) to /. Suppose, converges uniformly on differentiable on f(x)-f(a). we have g. . 506 Infinite Sequences and Infinite Series Since g is continuous, = follows that f'(x) it = g(x) lim f„'(x) for all x in the n—>oo interval [a, b\. | Now how that the basic facts about uniform limits have been established, it is clear to treat functions defined as infinite sums, /(x) = /l(*) + /2(*) + /3 (*) + •• • This equation means that f(x)= lim + /iOO ••• + /„(*); our previous theorems apply when the new sequence /l, + /2, /1+/2 + /3, /l converges uniformly to /. Since in, DEFINITION we The single series it this the only case is we shall ever be interested out with a definition. / fn converges uniformly (more uniformly summable) to/ on A, h, converges uniformly to may be + /2, /1 formally: the sequence {/„} is the sequence if /1 + /2 + /3, ... / on A We can now apply each the results ••• of Theorems stated in one 1, 2, common and 3 to uniformly convergent series; corollary. 00 COROLLARY Let 2. fn converge uniformly to / on [a, b\. n=\ (1) If (2) If each /„ is continuous on [a, b], then / is continuous on / and each /„ is integrable on [a, b], then Ja 1 [a, b]. Ja 00 Moreover, \] fn if converges (pointwise) to / on [a b] , , each /„ has an integrable n=\ derivative /„' and 2_. fn' converges uniformly on n=\ tion, then 00 (3) f'{x) = ^2f ,'(x) t n=\ for all x in [a.b\. [a, b] to some continuous func- ) 507 24. Uniform Convergence and Power Series PROOF each /„ is continuous, then so is each of the sequence f\ f\ + fj, f\ + fi + /3, (1) If , (2) Since f\, f\ Theorem 1 + fn, + fi + f3, f\ • . . . , / so is and / is the uniform limit continuous by converges uniformly to /, • it Theorem 2. follows from that f= f f (/,+•• + lim = lim rt—>00 tf E/ (3) • \-fn , f\-\ Each function /i + • • • 4- /« /i+ /„) -+ r/ " /» differentiable, with derivative is and f\ ', f\' + f2, f\ + fi + yV converges uniformly by hypothesis. It follows from Theorem 3 that ' , • /'(*)= lim [fi'(x) to a /V + • • • + /,/, continuous function, + /„'(*)] + 00 n=l At the moment this corollary is not very useful, since it seems quite difficult to when the sequence f\ f\ +/2, /1+/2+/3, will converge uniformly The most important condition which ensures such uniform convergence is provided by the following theorem; the proof is almost a triviality because of the cleverness predict • , • • with which the very simple hypotheses have been chosen. THEOREM 4 (THE WEIERSTRASS M-TEST) Let {/„} be a sequence of functions defined on A, and suppose that {M„} a is sequence of numbers such that \fn(x)\ < M„ x for all in A. 00 00 Suppose moreover that \_. M„ converges. Then for each A' in A the series Y_] fn U" 00 converges (in fact, it converges absolutely), and 2_. fn converges uniformly on n=\ to the function 00 /(*) = 2 n=\ /»(*) A 508 Infinite Sequences and Infinite Series PROOF For each x A in the series /J|/n (jc)| converges, by the comparison test; COnse- oo quently /_/«(*) converges (absolutely). Moreover, for all x in A we have n=\ /U)-[/lOO + ••• + = /*(*)]! oo n = /V+l oo £ < M- n=W+l Since 2_. M„ number converges, the = A/V f x (x) T r \{2x) FIGURE M n can be made as small as desired, n=N+\ n=\ by choosing 1 \_. N sufficiendy large. | 10 T 1 The following sequence {/„} illustrates a simple application of the Weierstrass M-test. (a) f{x) = Let {x} denote the distance from x to the nearest integer (the graph of {x} is illustrated in Figure 10). Now define f2 (x) = \{Ax) The functions f\ 10" has and fj are been replaced by shown 2"). in Figure 1 1 (but to the Weierstrass M-test automatically applies: clearly (b) l/«(*)| FIG UK I. I I make the drawings simpler, This sequence of functions has been defined so that < for all x , . 24. Uniform Convergence and Power Series and y^ 1/10" converges. Thus /_,f" converges uniformly; since each /„ is 509 con- tinuous, the corollary implies that the function 00 oo i /(*) = £/»<*> = is also continuous. • • • + /„ As n . E lo^lO"*} n=\ n=\ Figure 12 shows the graph of the few partial sums first become harder and harder increases, the graphs to draw, f\ + and oo the infinite sum N /„ is shown by quite undrawable, as the following theorem n=\ (included mainly as an interesting sidelight, to be skipped if you find the going too rough). THEOREM 5 The function /^» = ET7^l 10" | 0".v| n=\ is PROOF continuous everywhere and differentiable nowhere! We have just shown that / uses uniform convergence. is continuous; We this is the only part of the proof which prove that will / not differentiable at a, for is by the straightforward method of exhibiting a particular sequence approaching for which any a, + h m )-f(a) f(a .. lim m-KX> does not 0<a< exist. hm obviously suffices to consider only those numbers a satisfying It 1. Suppose that the decimal expansion of a = a Let h m {h,„} = 10 - '" if / am 4 or 9, but {).a\a2CiT,a$ .... let ~ km ^ = - 10"'" hm these two exceptions will appear soon). f(a+h m )-f(a) is am if - 4 or 9 )} - (the reason for Then ' ±10"'" 10" n=l oo = This infinite series is ^±10m - M [{10"(a +h m really a finite sum, because if n > m, then {10"«}], 10"/?,,, is so {\0"(a+h„,)}-{\0"a} On the other hand, for n 10"<7 \0"(a + hm ) = = < m we = can write + 0.a n+ ian+ 2a n+ 2 integer + 0.a n+ \a n+ 2an +3 integer 0. . . . am . (a,„ . ± 1) • • • an integer, 510 Infinite Sequences and Infinite Series (in order for the second equation to be true — 10 -m when = a,„ Now 9). 0.a n+ \a n+ 2a n+ 3 Then we . . hm m = the special case =— 10~ m essential that is we choose when a m = n 4). {10 .a h for n = si- n + iflm ± 1) • • • < the second equation 1 i m we we chose true because is This means that (a+h m )}-{\0 n a} = ±l(y m i , and exactly the same equation can be derived when 0.a n+ [Ci tl+ 2a n+ < hm have also 0.a„ + \a n+2 a n +3 (in it suppose that T, • • • > \- Thus, have m 10 -"[{10"(fl + //„,)} -{10"«}] = ±1. In other words, f(a+h m )-f(a) sum of m — 1 numbers, each of which is ± Now adding +1 or — 1 to a number changes it from odd to even, and vice versa. The sum of m — 1 numbers each ±1 is therefore an even integer if m is odd, and an odd integer if m is even. is fi + f2 +h the 1 . Consequently the sequence of ratios + h m )-f{a) f(a hm cannot possibly converge, since odd and even. a sequence of integers which are alternately | In addition to its role in the previous tool for analyzing functions A + fi + h + h is it attention to functions of the theorem, the Weierstrass M-test which are very form well behaved. We is an ideal will give special oo = J2 a n(x-a)'\ f(x) n=0 which can also be described by the equation 00 «=0 FIGURE 12 for f„(x) = a„(.x on powers of simplicity, we (x — a)". —a), is will usually Such an called a infinite power sum, of functions which depend only series centered at a. For the sake of concentrate on power series centered at 0, 00 f(x) = y\/„.v". n=0 — • 511 24. Uniform. Convergence and Power Series One group of power especially important where / of the form some function which has derivatives of all orders at a; Taylor series for/ at a. Of course, it is not necessarily this series is called the equation holds only when is true that " ;/< Y ) = ft , f(x) this series are those ,(fl) r (x - -a)V ; the remainder terms satisfy lim R„ M (x) n—>oo = 0. 00 We all already x. For know power that a example, the power series \x\ < 1 , while the power series — 1 < jc < converges only for x — 0. x1 x5 X2 converges only for a n x" does not necessarily converge for series x3 converges only for >_^ 1 It is . X4 X3 X5 even possible to produce a power series which For example, the power series Ix" E» n=0 does not converge for x / 0; indeed, the ratios (n+ l)!(.t" +1 ) = (n n xY n + l)x I ! are unbounded for any x ^ 0. If a power series V,a„x" does converge for 00 some xq \x\ theorem 6 < |x Suppose 7^ however, then a great deal can be said about the series 2, a n* n for 1 that the series oo /(.to) = ^2a„x " «=0 converges, and let < a be any number with 00 n=0 a < \xq\. Then on [—«,«] the series 1 512 Infinite Sequences and Infinite Series converges uniformly (and absolutely). Moreover, the same is true for the series oo g(x) Finally, / is = }na n x n ~ l . and differentiable oo n=\ for all x with PROOF Since < |.v| \xq\. Yja„*o" converges, the terms a n XQ ,i=0 bounded: there is some number n |<z„jco Now if x is | [—a, a], then in \a n x n \ = \a n < \a n . \ approach M such that \xq"\ < M = \a„\ \x\ < \x n 0. for Hence they are surely all n. so \a\, n \ • \a"\ • 1*0 \ a KM < M the clever step) (this is XQ a XQ But < |«/jcq| 1, so the (geometric) series EM f «=o converges. Choosing M a ME = *o n=0 number M„ \a/xo\" as the in the Weierstrass M-test, it 00 follows that j>a„x" converges uniformly on [—a, a]. n=0 To prove the same assertion for g(x) \na n x n | — >/2a„jc" = n\an < n\a„\ M \ \x" \a"- n \x \ 1 a n XQ \a\ < ' Since \q/xq\ < 1, Mn — a — \a\ notice that ,1 *0 the series Am a M XQ rr / a y-^ n XQ n=\ converges (this fact was proved in Chapter 23 as an application of the ratio test) 513 24. Uniform Convergence and Power Series Another appeal " 2_, na n x to the Weierstrass M-test proves that converges uni- n=\ formly on [—a, Finally, a]. our corollary proves, that g first continuous, and then that is oo = /"(a) g(x) — 2_, na nX n~ for [—a, in A" a]. n=\ Since all we could have chosen any number a < |ao|. | x with a | We are < with < a holds for \xo\, this result | now manipulate power in a position to manipulations are fairly series with ease. Most algebraic straightforward consequences of general theorems about OO For example, suppose that f(x) infinite series. where the two power series — 00 x" 2_. a » „=0 both converge for some = y^b„x n and g(x) , n=0 Then ao. for |a| < |ao| we have OO OO 00 00 n " )xn J2 "nx + J2 b » x " = J2 {a " x + b » x ") = J2 {a " + b " n=0 n=0 n=0 - ,1=0 oo So the series for these a The h(x) = + b n )x" /,(«« |a| < |ao|, and h = f+g < then . treatment of products is just a that the series more little \^a„x n and 2_,b n x n converge |a| |ao|, we So absolutely. it follows from n=0 m=0 OO Theorem 23-9 involved. If OO oo know also converges for 00 V^ a n x" \\ b n x " that the product given by /;=0 /i=0 oo 1S oo 1=0 7=0 where the elements djX'bjXJ are arranged choose the arrangement aobo + (aob\ which can be written + a\bo)x + (ao^2 \cn x / the In particular, + a\b\ + a2bo)x L we can -\ n n for cn n=0 is any order. as oo This in = ^Y\a k b„- k . k=0 "Cauchy product" that was introduced in Problem 23-10. Thus, the = fg for 00 Cauchy product h(x) = / j »=o these a. Cn x n also converges for |a| < |aq| and // a 514 Infinite Sequences and Infinite Series Finally, — y^ suppose that f{x) n nx where , so that /(0) ¥" 0, ciq = ^ ao 0. „=0 oo Then we can try to find a power >^ b n x" which series represents \/f. This means „=0 that we want have to oo oo J2 " J2 b " X = anX* left + 1 2 + .V • • A" + • • • . „=0 ,i=0 Since the = 1 be given by the Cauchy product, we want side of this equation will have to a b 1 aob\ «0^2 Since «o 7^ 0, we can solve the = + a\bo = + a\b\ + aibQ = Then we can of these equations for bo. first solve oo the second for b\ , etc. Of course, we have still prove that the to new >^ b„x" series „=o ^ does converge for some x 0. This Theorem 6 gives us all when we apply Theorem 6 to the infinite For derivatives, J = 1 cosjr get precisely the results A v x + + we need. In particular, series 5 X + 1 -a ={+ e we x the information 18). --- + X--..., 3 X :-- sin*= an exercise (Problem as is left 4 6 8 .V 4!-& ? A 9 7 X A3 + v. .V ^-"-' + + 4 A + ---< v. y. which are expected. Each of these converges hence the conclusions of Theorem 6 apply 3a 2 5a —+— sin (a) = , 1 cos(A) = 2a 4a -+ ^ exp (x) =1+—+ for = —+ For the functions arctan and f(x) = : 3a~ = H log(l ... +a) = -smA, exp(A). the situation complicated. Since the series arctan a =a —— + — A 3 ao, cos a, 6a 5 3 -^- any 4 r 2a any a for A 5 A7 — -\ is only slightly more .. . 24. Uniform Convergence and Power Series = converges for xq 1 , = arctan'(jc) also converges for it 1 |jc —x +x —x + • < | • 1 and , = • x + xz 1 = In this case, the series happens to converge forx for the derivative is = not correct for x =— or x 1 l-x 2 +x 4 -x 6 + 515 -1 1 also. for \x < 1 However, the formula indeed the ; \ series ... diverges for x — 1 and x = — 1 Notice that this does not contradict Theorem 6, which proves that the derivative is given by the expected formula only for \x\ < |*o|. . Since the series "•345 X X X X log(l+ , )=x __ + ___+__... converges for xq 1+* — = 1 , it log'( also converges for \x 1 +x = ) 1 — x < \ 1 , and + x — x' + • • • differentiated series does not converge for x All the considerations its which apply derivative, at the points where the | = — 1; < 1 moreover, the 1 power to a jc | In this case, the original series does not converge for x = for derivative series will automatically is apply to represented by a power series. If 00 converges for all x in some interval ( /"(a) — R, = ^na nu n for all x in ( — R. R). R), then = x nx Theorem 6 ' ni-i \ Applying Theorem 6 once again we find f"{x) implies that = ^n{n- \)a n x"- that 2 , n=2 and proceeding by induction we find that oo Thus, a function defined by a power ( — R, R) is series which converges in automatically infinitely differentiable in that interval. previous equation implies that f«\0)=k\ak so that ai w = / (0) . , some interval Moreover, the 516 Infinite Sequences and Infinite Series In other words, a convergent power function which On happy note we could this A series. series centered at always is the Taylor series at of the it defines. end our study of power easily series and Taylor some unexplained careful assessment of our situation will reveal facts, however. The Taylor series they converge for of x, all and exp are sin, cos, and can be — differentiated term-by-term for — Taylor series of the function f(x) we could as satisfactory as + x) log(l all slightly less pleasing, is desire; x. The because it < x < 1 but this deficiency is a necessary consequence of power series. If the Taylor series for / converged for any xo with |xo > 1? then it would converge on the interval (— |jto|, |*ol), 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 (— ), where it equals converges only for 1 , the basic nature of I 1 , 1 log(l+*). The Taylor series for arctan more is be no possible excuse for the refusal of this mysterious behavior + 1/(1 x an ), exemplified even is f(x) = > If \x = = 1 series + x2 - 1 of 2 .v strikingly which / + is 4 - .v answer: always dangerous, since is it there does |jc| > 1. This by the function f(x) = the next best thing to a 8 .r . Why? What unseen and —1? Asking 1 we may have + to settle for obstacle this sort of an unsympathetic — intelligently Taylor when to happens because it happens that's the way things are! In this case happen to be an explanation, but this explanation is impossible to give at the present time; to devote 6 .v all. prevents the Taylor series from extending past question is — there seems given by the Taylor series does not converge at 1 I converge series to more infinitely differentiable function polynomial function. The Taylor comprehend difficult to only although the question when placed two chapters series in in to quite is about real numbers, a broader context. new It will it can be answered therefore be necessary material before completing our discussion of Chapter 27. PROBLEMS 1. For each of the following sequences {/„} (if it exists) uniformly to (i) (ii) on the indicated — determine the pointwise and decide whether this function. fn (x)=yx~, fn (x) {,/*„}, interval, I \ x on — n, [0.1]. x > on ii. \a,b], and on R. limit of {/„} converges 24. Uniform Convergence and Power Series fn (x) = — x (iv) fn {x) = e~ nx \ (v) /„(*) (iii) 2. 517 on (l,oo). , n on [-1,1]. = ^, onR. This problem asks for the same information as in Problem 1 but the functions are not so easy to analyze. Some hints are given at the end. , 2n fn (x) = x"-x on (i) [0,1]. nx (ii) on [0, oo). fn (x) = Jx 2 + -ron 2 [a, oo), /„(*)= +n + x 1 (iii) a > 0. n = J x2 +~2 onR (iv) /„(*) (v) fn (x) = Jx H /„(x) = Jx -\ on [a, oo), yfx on [0, oo). a > 0. n V (vii) /„(x) = « Jx H v^ pn[a, oo), a (viii) fn (x) - nlJx A Jx oo) Hints: (i) on maximum - fn (x)\ for x large, separate estimate of \f(x) — fn (x)\ Find the Taylor for (i) /(*)= x series at — = log(* (iii) f(x) = -== = (iv) — - a), x fix) ' Vl-x 2 v) fix) = and on (0, oo). (iii) - f„\ on [0, 1]. Mean Value Theorem, of \f for small (ii) |jc|. (vii) Use (v). each of the following functions. a^0. , f(x) 1 0. a (ii) V [0, > J For each n, find the consider \f(x) 3. y/x h V (vi) - arcsinx. (1 a < - x)~ 0. x '2 . (Use Problem 20-21.) For each n, (iv) Give a 518 Infinite Sequences and Infinite Series 4. Find each of the following (i) l -' + (ii) 1 - Hint: i- i + i-- What X2 ..... x + x 6 -x 9 + x3 X J3 y-r^ (111) sums. infinite is + 1 - --- x for \x | + x2 - x3 X*4 XJ < 1. ? H 5 4T3-5^ + "' lor|x!<1 - Hint: Differentiate. 5. Evaluate the following is infinite sums. (In most cases they are f(a) where a some obvious number and f(x) is given by some power series. To evalupower series, manipulate them until some well-known power ate the various series emerge.) A ^ ,.x (-1)"2 2 "tt 2 " (2n)! oo -, y y, ^2,, + /l\ 2 " +l 1 in lUJ OO iv n=0 oo , ^3»(»+l) A ^ VI If f{x) power 2n = + 1 2»n! (smx)/x for x ^ and /(0) = 1, find / w (0). series for /. In this problem without all in part (a) the we deduce the binomial series (1 work of Problem 23-2 of that problem — the 1 , series although f(x) +x) a = / we = Vj < 1. ( l.v". \x\ < l will use a fact established x" I ) ^"' «=0 \x\ Hint: Find the does converge for 519 24. Uniform Convergence and Power Series (a) (b) Prove that (1 +x)f'(x) = oef(x) for |x| < 1. Now show that any function / satisfying part (a) is of the form f(x) = a c(l + x) for some constant c, and use this fact to establish the binomial a series. Hint: Consider g(x) = /(jc)/(1 + x) . 8. Suppose that /„ are nonnegative bounded functions on A and OO on A, does sup/,,. If 2_jf" conver Ses uniformly verges 9. (a let M„ = 00 it follow that \] M„ con- converse to the Weierstrass M-test)? Prove that the series 00 y ^ x n(\ +nx 2 ) converges uniformly on R. 10. (a) Prove that the series 00 , ^ 3nx n=0 converges uniformly on (b) By considering series 11. (a) the [a, oo) for a N sum from > Hint: 0. x to oo for does not converge uniformly on = \im(smh)/h = 1. 2/(n3 N ), show that the (0, oo). Prove that the series oo nx > 4 + rrx- , , 1 converges uniformly on of nx/(l (b) Show +n 4 x 2 ) on [a, oo) for a > 0. L «3« Hint: First find the maximum [0, oo). that ' \NJ ~ 2 n>VN and by using an integral to estimate the sum, show that f(\/N~) > Conclude that the series does not converge uniformly on R. (c) What about the series oo E n=U nx ;> 1/4. 520 Infinite Sequences and Infinite Series 12. (a) Use Problem 15-33 and Abel's Lemma (Problem 19-36) form Cauchy condition", showing that for any e > 0, to obtain a "uni- sin/cx E k=m can be ing m made on the whole interval [e, In - e] by chooslarge enough. Conclude that the series arbitrarily small (and n) 00 E converges uniformly on (b) For jc = nx sin [e, 27T — e] for s show 7r/N, with TV large, > 0. that N IN Vj sin kx > yv 7T k=N Conclude fc=0 that 2N sin A:x E > 27r' k=N and that the series does not converge uniformly on [0, 2n\. oo 13. (a) (b) Suppose that /(*) = ^a x n converges for n all x in some interval (-R, R) and that f(x) = for all x in (-/?, /?). Prove that each a„ = 0. (If you remember the formula for a„ this is easy.) for some sequence {xn } with Suppose we know only that /(*„) = show that lim x„ = 0. Prove again that each a„ = 0. Hint: First n—>oo /(0) = = fl 0; then that ; / (0) = a\ This result shows that if /(*) = e~ possibly be written as a power series. =0, l/x2 etc. l/x for x sin / 0, then / cannot shows that a function defined thus a but nonzero for x > for x < by a power series cannot be remained has which power series cannot describe the motion of a particle at rest until time 0, and then begins to move! It also oo oo (c) Suppose in some that /(*) = £>„*" m } converging to 0. Show and g{x) and interval containing {t — = that /(/,„) that a„ = ^b x" n = converge for g(tm ) for b n for each all x some sequence n. 00 14. Prove that and if / is if f(x) - ^a nx n is an even function, then a„ an odd function, then a„ = for n even. = for n odd, — 521 24. Uniform Convergence and Power Series 15. Show x < power series for fix) — log(l — x) converges only for —1 < and that the power series for g(x) = log[(l + x)/(\ — x)] converges that the 1, only for x in (— 1, *16. 1). Recall that the Fibonacci sequence {a n } (a) Show (b) Let that a n+ \/a n f(x) < = J2 a n is — defined by a\ = ai \, a n+ \ = 2. x"- = ] + 1 + a- 2 2.x- + 3.v 3 + n=\ Use the (c) ratio test to prove that Prove that < if \x\ f(x) converges partial fraction series for \/(x (e) —a), +x — x2 1/2. f(x)—xf(x) — xf(x)= decomposition for i/(x to obtain series another power +x — 1. and the power 1 ), series for /. / must be the same (they are of the function), conclude that — an < 1 Since the two power series obtained for both the Taylor | -1 = Hint: This equation can be written Use the |jc 1/2, then /(*) (d) if . V5 17. Let f(x) = 2_, a nx" = / and g(x) — } ]cn x n Suppose we merely knew . that n=0 n={) oo f(x)g(x) bn x n for some cn , but we didn't know how to multiply series n=0 series for Use Leibniz's formula (Problem 10-20) to show directly that this fg must indeed be the Cauchy product of the series for / and g. Suppose that f(x) in general. 18. = 2_.°nx" converges for some A'o, and that «o 7^ 0; n=0 for simplicity, we'll recursively assume that ciq = 1. Let {b„} be the sequence defined by b = 1 n-\ b„ = -}b a k k=0 n -k- ' 522 Infinite . Sequences and Infinite Series The aim of problem this show to is y^b n x" that some also converges for «=0 xc 7^ 0, so that represents 1\/f fc small // for it < M, show If all |a„.vo"| a) enough \x that n-\ |/WI<M^|W|k=0 Choose (b) M so that |c/„.v () "| < M, and also so that M/(M 2 - 1) < Show 1. that |/WI<M 2 ". oo Conclude (c) \]b„x n that converges for |x| sufficiently small. n=0 oo s 19. Suppose that /_, n=0 oo a " conver ges - We know that the series f(x) = >_^a„jc" n=0 must converge uniformly on [—«,«] for < a < 1, but it may not converge uniformly on [—1,1]; in fact, it may not even converge at the point —1 (for example, f(x) if shows that the = log(l + x)). However, a beautiful theorem of Abel converge uniformly on series does / is oo m Theorem and, in particular, /_, an ~ ^ n=0 by noticing that if \am + -+a„\ < s, then by Abel's Lemma continuous on Consequently, [0, 1]. oo [0, 1] /. a » x n=0 \a m x m '" + Prove Abel's + a„x"\ < s, (Problem 19-36). 00 20. A sequence {a,,} is called Abel summable if lim \] a n x " exists; Prob- n=0 lem 19 shows that a summable sequence is necessarily Abel summable. Find a sequence which is Abel summable, but which is not summable. Hint: Look over the list of Taylor series until you find one which does not converge at 1, even though the function it represents is continuous at 1 21. (a) 1111 Using Problem W 2 " • 1 19, find the following infinite sums. 3 • 2 + 4 3 5 • 4 +" ' 00 OO (b) Let 2_. c n be the Cauchy product of two convergent power series 2_^ a n=0 oo oo and y_\ b>n ancl suppose merely that 00 converges to the product y. c n „=o n=0 it >, «=0 00 2, a n /^ b n ' n=0 «=0 . converges. Prove that, in fact, . 24. Uniform Convergence and Power Series 22. (a) Show that the series x E + In converges uniformly to A 1 converges to log it (Problem 23. (a) Suppose x n+\ 2n+\ 71=0 at + log(A' (Why 2. +2 In 1 1) on [—a, / (b) doesn't this contradict Abel's that {/„} Find a sequence of continuous functions on unbounded / that function / on [a, b] [a, b]. Prove that which converge point- [a b] , Prove that the function /' differentiable. is on to ples of discontinuous derivatives, this provides another pointwise limit of continuous functions 26. Theorem is the pointwise know examexample where the sequence of continuous functions. (Since we already limit of a 25. but that a sequence of bounded (not necessarily continuous) is on [a, b] which converge uniformly bounded on [a, b]. Suppose 1, 19)?) wise to an 24. < a < a] for functions is 523 is not continuous.) Find a sequence of integrable functions {/„} which converges to the (nonin- on the tegrable) function / Each except at a few points. (a) /„ will be Prove that integrable / if on that is is 1 rationals and the uniform limit of {/„} then so [a, b], is on the irrationals. Hint: on [a,b] and each /. (So one of the hypotheses in /„ is Theorem 1 was unnecessary.) (b) In Theorem we assumed only 3 that {/„} converges pointwise to /. Show- that the remaining hypotheses ensure that {/„} actually converges uniformly to /. (c) Suppose that in Theorem (d) Show we do not assume assume only tion /, but instead [a, b]. 3 {/„} converges to a func- that f„(xo) converges for that /„ does converge (uniformly) to Prove that the some / some xo in = g). (with /' series E (-D" +n , converges uniformly on 27. Suppose to /. x [0, oo). that /„ are continuous functions l//i Km [0, 1] that true if / the convergence /•! ./"„ "^°° Jo Is this on Prove that = / Jo isn't uniform? /. converge uniformly 524 Infinite Sequences and Infinite Series 28. (a) Suppose that {/„} pointwise. > and for all n uniformly on x all Suppose moreover that we have fn (x) > fn+ \(x) Prove that {/„} actually approaches in [a,/?]. Hint: Suppose not, choose an appropriate sequence [a, b]. of points x„ in (b) a sequence of continuous functions on [a, b] which is approaches [a, b], and apply the Bolzano-Weierstrass theorem. Prove Dini's Theorem: If {/„} is a nonincreasing sequence of continuous on [a,b] which approaches the continuous function / point- functions approaches / uniformly on a nondecreasing sequence.) wise, then {/„} also (c) holds if Does Dini's is {/„} Theorem hold replaced by the open interval 29. (a) Suppose that {/„} How continuous? if /' isn't (The same [a, b]. about result b] if [a, is (a, b)? a sequence of continuous functions on [a,b] that is converges uniformly to /. Prove that if x n approaches x, then fn (x n ) approaches f{x). (b) (c) Is this statement true without assuming that the /„ are continuous? / is continuous on [a,b] and {/„} is a sequence with the property that fn (x n ) approaches f(x) whenever x n approaches x, then /„ converges uniformly to / on [a,/?]. Hint: If not, and a sequence x n with \fn (x n — f(x„)\ > s for infinitely there is an s > many distinct x n Then use the Bolzano-Weierstrass theorem. Prove the converse of part If (a): ) . 30. This problem outlines a completely different approach to the integral; consequently, (a) Let s unfair to use any facts about integrals learned previously. it is be a step function on [a,fr], so that s is constant on (/,_i,r,) for " pb some partition {to, . . . , tn } of [a b] , Define . / V_, s s as i= Si the (constant) value of s is not depend on the partition (b) A function limit {to, (f,-_i, . . Show i ) where l that this definition does regulated function on [a b] if it is the uniform {s n } on [a,b]. Show that in this every s > 0, some /V such that for m n > jV we have called a is — ?,•_ tn ). , . U). (U , of a sequence of step functions case there |^m (jc) / on i is, for — sm (x)\ < (c) Show that (d) Suppose , £ for all .v in [a, b]. the sequence of numbers that {t„} is N we > Conclude that s„\ will be a I Cauchy sequence. another sequence of step functions on [a,b] which converges uniformly to /. that for n \ have lim / n—*oo JI a \s n s„ Show that for every e (x) — = lim t„(x)\ oo /;— / < tn . s for x > there N such we can k'hne is an in [a, b}. This means that J/ a . b f I f to be lim s„ for any sequence of step functions {s,,} converging J (I uniformly to /. regulated? The only remaining question is: Which functions are . 24. Uniform Convergence and Power Series *(f) Prove that a continuous function is regulated. Hint: To 525 find a step func- on [a, b] with \f(x) — s(x)\ < e for all x in [a, b], consider for which there is such a step function on [a v] Every step function s has the property that lim s(x) and lim s(x) tion s all v , (g) for all and x—>a + a. Conclude that every regulated function has the find an integrable function that that, conversely, every function lim f(x) exist for x—>a~ *31. all a is / is not regulated. with the property that exist x—*-a~ same property, (It also true is lim f(x) and x—>a + regulated.) Find a sequence {/„} approaching / uniformly on lim (length of /„ on [0, 1]) / length of / on [0, [0, 1] for 1]. which we have (Length is defined in n—»oo Problem 13-25, but the simplest example whose graphs will be obvious.) will involve functions the length of b CHAPTER #VV COMPLEX NUMBERS With the exception of the few paragraphs of the previous chapter, last has presented unremitting propaganda for the real numbers. root. numbers do have a great deficiency not every polynomial function has a The simplest and most notable example is the fact that no number x can satisfy x2 + =0. 1 This deficiency number x satisfying + the family of especially / = 0." = 1 0, nonsensical to say "let 1 numbers seemed to simplify greatly a + bi many =0. For a since there no is be the number / Nevertheless, admission of the "imaginary" when "complex numbers" valid. + /" was quite mysterious: i is it so severe that long ago mathematicians with the property that number to i algebraic computations, a and b in R) were allowed, and (for enumerated the laws of arithmetical computation be to 1 + x satisfying 2 i is number "number" the need to "invent" a long time the status of the all Nevertheless, the — real felt book this in Chapter 1 were assumed For example, every quadratic equation ax + bx + c = 2 (a / 0) can be solved formally to give x = -b + \Jb - 4ac 2 -b - v' = x or b 2 — 4ac > 0, these 4a c . formulas give correct solutions; allowed the formulas seem to - La La If 2 make x 2 sense in all cases. +x + = 1 when complex numbers are For example, the equation has no real root, since x But the formula 2 +x + = 1 (.v + 2 + \) \ > for 0, for the roots of a quadratic equation suggest the "solutions" x -l — + V-3 , and x = -l-V-3 2 if all .v. we understand y — 3 to mean ; 2 y/3 ( — 1) = v 3-y — = v3 i, 1 then these numbers would be 73 -- + —-/ 1 2 It is — and 2 1 V3 2 2 i. not hard to check that these, as yet purely formal, numbers do indeed the equation x 526 — 2 +x+ 1 =0. satisfy 527 25. Complex Numbers It is even possible to "solve" quadratic equations whose coefficients are themselves complex numbers. For example, the equation x +x + + =0 2 1 / ought to have the solutions ±71-4(1+/) 1 where the symbol v — 3 — —3 — 4/. In ±7-3-4/ -1 means a complex number a 4/ + fti whose square is order to have (a + pi) 2 = a2 - p2 + lafii 2 2 a - p = = -4. = -3 - 4/ we need 2oeP These two equations can -3, be solved for real a and easily fi\ in fact, there are two possible solutions: = a and = -2 Thus - —3 — the two "square roots" of reasonable way to v— 3 — 4/; a 1 4/ are 1 =— = 2. — 1 and — 1 2/ decide which one of these should be called which case y/x denotes the nonnegative root. For (real) 1 = x this There 2/. V —3 — makes sense only the conventional usage of y/x + 4/, is no and which for real x > 0, in reason, the solution ± 7-3 - 4/ 2 must be understood x With this = an abbreviation as -1 +r — — - where , understanding we r is for: one of the square roots of —3 — 4/. arrive at the solutions -1 + -2/ 1 2 -1-1+2/ x= as you can easily check, these ^ = - 1+ , . ' ; numbers do provide formal x 2 +x + 1 + = i solutions for the equation 0. For cubic equations complex numbers are equally useful. Every cubic equation ax with real coefficients a, ax~ 3 + b, c, bx and + bx + ex + d by — a we the other roots of .v 2 + ex + d = d, has, as (a we know, # 0) a real root a, and if we divide obtain a second-degree polynomial whose roots are ax +bx -\-cx+d = 0; the roots of this second-degree polynomial 528 Infinite Sequences and Infinite Series may be complex numbers. Thus a cubic equation or one real root and 2 complex roots. The will have either three real roots 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 equation actually find a formula for all The the roots. we can, with sufficient cleverness, following derivation is presented not only as an interesting illustration of the ingenuity of early mathematicians, but as numbers (whatever they may further evidence for the importance of complex To solve the most general cubic equation, it be). obviously suffices to consider only equations of the form x It is 3 + bx 2 + ex + d = 0. even possible to eliminate the term involving x manipulation. If we , by a fairly straight-forward let b then o , b2 v 2 b3 SO = x 3 + bx 2 + T. Jv 3 + 1 \ The for y (b 2 ~T + d 2b 2 - 3 i right-hand side we can cx now find x\ this contains no term with y shows that it suffices to . If we can solve the equation consider in the first place only equations of the form x + px +q = 0. = we obtain the equation x 3 = —q. We shall see later on has three, so that this that every complex number does have a cube root, in fact equation has three solutions. The case p ^ 0, on the other hand, requires quite In the special case p it an ingenious step. Let (*) x =w- — 25. Complex Numbers 529 Then = x 3 + px ( a 2 3wp~ 5w 2 p -2 3 = — +q = w- — + 3w + pw - — + + p (w - ) p~3 ) q ,2 u; 4 3 P 3 This equation can be written 27(u; which is 3 2 ) a quadratic equation in + 27^(w 3 )-/? 3 = w 0, (!!). Thus 3 u; = -21q ± y(27)V 2-27 V 4 2 Remember that this really + 4-27/i 3 27 means: 2 w 3 We 9 =—— + r, , where r • r is a square root ol P q — +— 3 can therefore write If = 3 N this equation means that root of q~/4 + substituted into x jll. p some cube This allows root of —q/2 + r, where six possibilities for 3 N 2 V 4 + r is some square when these are 27 3- 3 \ it w, but yielding (*), = if is ?*y /«!4 + £ 27 / turns out that only 3 different values for x will be obtained! surprising feature of this solution arises when we whose roots are real; the formula derived above may in an essential way. For example, the roots of 15jc-4 = An even more consider a cubic equation still involve all of complex numbers 530 Infinite Sequences and Infinite Series are 4, (with —2 + V3, and —2 — V3. On the other p — — 15, q = —4) gives as one solution = ^/2 hand, the formula derived above 15 + V4" 125 V2 3-y/2 V4 + V4- 125 V/2TT17 + 3- ^2 + 11/ Now, (2 + /) 3 = 2 3 + 3-2 2 + 3-2-/ 2 +/ 3 = 8+ 12/ -6-i / -2+ one of the cube roots of so we 2+ 11/ is 11/, 2+ Thus, for one solution of the equation r. obtain x =2+ + 15 i =2+ + 6 i = 2 + /+ 6 + 3/ 15 6- -3/ + 3/ 6 - -3/ - 90 45/ + 36 9 = 4(!). The The other roots can also be found fact that even one of these real roots depends on complex numbers complex numbers cannot be is + 11/ are known. obtained from an expression which is impressive enough to suggest that the use of As a matter of fact, entirely nonsense. for the solutions of the quadratic in the other cube roots of 2 if the formulas and cubic equations can be interpreted entirely terms of real numbers. Suppose we agree, writing the real moment, to number a as a + 0/ and for the ordinary arithmetic and the relation (a + bi) + + bi) (a Thus, an equation (c + di) = + di) = = (a all complex numbers the number / —1 show that + (ac as + bd) + d)i + (ad + bc)i. = +7/ + c) - (b like (1 may be (c 2 i write +2/) .(3 + 1/) 1 regarded simply as an abbreviation for the two equations 1-3-2-1 = 1-1+2-3 = 1. 7. 1 / . as a The + bi, laws of 531 25. Complex Numbers The + ax solution of the quadratic equation bx + c = with real coefficients could be paraphrased as follows: — v = b — 4ac, uv = 0, = b — 4a c) i.e., if (u + vi) u II u 2a then -b + + c = 0, +b V J 2a 2a u v L2^J 2a then a 0, —b +- u + vi\ , i.e., = , / +b\ | [ —b + 2a down + vi . +c= 2a not very hard to check this assertion about real It is u numbers without a single "/," but the complications of the statement itself writing- should convince you that equations about complex numbers are worthwhile as abbreviations for pairs of equations about real numbers. (If you are still not convinced, try paraphrasing the solution of the cubic equation.) If we really intend to use complex numbers consistently, however, it is going to be necessary to present some reasonable definition. One 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 A complex number is an ordered pair of real numbers; if z = is called the real part of z, and b is called part of z. The set of all complex numbers is denoted by C. If are two complex numbers we define plex number, then a (a,b) (The + + and and • • + (c,d) (a (a,b)-(c,d) (a appearing on the left (a, b) is the a com- imaginary (a, b) and (c, d) + c,b + d) c — side are b d,a d + b-c). new symbols being defined, while the appearing on the right side are the familiar addition and multiplication for real numbers.) When complex numbers were first introduced, bers were, in particular, complex numbers; is not true a real number is if it was understood our definition is not a pair of real numbers, after that real num- taken seriously all. This this difficulty 532 Infinite Sequences and Infinite Series is only a minor annoyance, however. Notice that + (a, 0) (a, 0) = + b,0 + 0) = (a + b, 0), = (a b - 0, a + b) = (b, 0) (b, 0) • (a • • shows that the complex numbers of the form this {a • b, 0); behave precisely the same {a 0) , with respect to addition and multiplication of complex numbers as real numbers do with own their The complex numbers can now be recovered if one more be denoted simply by a will , for DEFINITION = i Notice that 2 i we addition and multiplication. For this reason convention that (a 0) = (0, 1) • definition will + is adopt the bi notation made. (0,1). (—1,0) (0, 1) familiar a . = —1 (the last equality sign depends on our convention). Moreover (a,b) = {a,0) + = (a,0) + — a bi. (0,b) (b,0)-(0, 1) -\- You may it is our definition was merely an elaborate device for defining feel that complex numbers as "expressions of the a firmly established prejudice of be defined as something specific, modern modern mathematics not as "expressions." esting to note that mathematicians numbers form a +/?/." That is essentially correct; that new objects Nevertheless, it must inter- is were sincerely worried about using complex 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. until the definition For example, the solution of the cubic equation required writing x so we want to know that l/w makes sense. Moreover, w =w— p/3w, was found by solving a quadratic equation, which requires numerous other algebraic manipulations. In short, real we are likely to use, at numbers. We Fortunately this real is certainly some time or do not want not necessary. Since numbers can be justified other, to stop any manipulations performed on each time and is quite easy, by the properties and these every step. algebraic manipulations performed all facts will on Chapter 1, it is only neccomplex numbers. In most listed in essary to check that these properties are also true for cases this justify not be listed as formal theorems. For example, the proof of PI, | (a, b) + (c, d)] + (e, f) = (a, b) + [ (c, d) + {e, /)] requires only the application of the definition of addition for complex numbers. The left side becomes (\a +c] +e, [b + d\ + ./'). J , 533 25. Complex Numbers and the right side becomes {a + two are equal because PI these + e\,b+\d + f\)\ [c true for real numbers. is It is a good idea to check P2-P6 and P8 and P9. Notice that the complex numbers playing the role and of out what P2 and P6 are in 1 — (a,b) trickier: if is, ^ (a,b) and (0, 0) (1, 0), respectively. It but the multiplicative inverse for (0, 0), + b ^0 2 then a 2 not hard to figure is a little and ^—+ — ,^—b- a- + b- (a,fc). is required in P7 (a, b) =(1,0). \aThis fact = (a,b)- (x,y) it is The = solutions are x \/(a + bi) — bx + ay + b~), a/(a~ means ax a Once y by — (1,0) + it a 1 a bi + bi a the existence of inverses has actually by some method), it = = 1 0. —bj(cr anything, then 1 + b~). the inverse of a precisely this trick — — a bi + bi + b2 been proved is (after 6 + 90 really valid; is no set numbers. In P of complex numbers such that PI fact, if —1 = it is easy to prove that satisfied for all complex ) not have disastrous consequences, but ), and this it P would have to contain 1 (since 1 = 1 and would contradict P10. The absence of P10-P12 will < w PI 2 are there were, then —1 i one — +9 also (since the easiest 6 3/ Unlike P1-P9, the rules P10 PI 2 do not have analogues: is it is actually being sought - 45/ 36 there guessing the inverse 6-3/ 15 3/ also possible to reason to evaluate 15 6 2 a bi — complex number which we used It is should be true that follows that this manipulation remember when was find (x, y) with only necessary to solve the equations that if to To could have been guessed in two ways. it does mean that we cannot define complex z and w. Also, you may remember that for the real numbers, z P10 PI 2 were used to prove that 1 + ^ 0. Fortunately, the corresponding fact for complex numbers can be reduced to this one: clearly (1, 0) + (1 0) ^ (0, 0). for 1 , Although we remembering will usually write that the set of all numbers. Long ago complex numbers complex numbers in the C is form a + bi, it is worth just the collection of all 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 real axis, and the pairs of real this collection 534 Infinite Sequences and Infinite Series vertical axis If z = x of is defined as z Two important definitions are also related geometric picture. to this DEFINITION called the imaginary axis. is + iy is complex number a = z — x modulus and the absolute value or and y (with x \z\ iy real), then the conjugate , of z is defined as length z\^»z = (x,y) = x + iy +y > (Notice that x + y is x^ + y \x 0, so that the nonnegative real square root of •z FIGURE = x — ly Geometrically, z it denotes .) simply the reflection of z in the real is distance from z to (0,0) (Figure 1 defined unambiguously; axis, while |z| is the Notice that the absolute value notation for 1). 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 w and z (1) l (2) z be complex numbers. Then = z. = z number (5) (6) Fy = (8) \z (9) \z (4) z real is (i.e., is + of the form a 0/', for some real z • w\ • + Assertions (=r = w\ (1) 1 < \z\ and \w\. • \z\ c#o. ,if + (2) \w\. are obvious. Equations forward calculations and 1 Equations (7) difficult and = (8) (4) = = only if a). + w — z + w. =£=-*. w = z w. z (3) PRCJOF and only if 1 and + may (6) (-*) = = + . may also (3) and (5) may be checked by straight- then be proved by a so -z, 7^\ —c = = so trick: z, (z)" be proved by a straightforward calculation. The part of the theorem is This inequality has, (9). in fact, curred (Problem 4-9), but the proof will be repeated here, using already oc- slightly different terminology. It is is clear that equality holds in true X < 0). if z = ^-W for any real (9) if z = number X Suppose, on the other hand, that or w= 0. It is also easy to see that (9) (consider separately the cases X :. ^ Xw for any real number k, > and and that 535 25. Complex Numbers w ^ Then, 0. for numbers real all k, = = ,2 < kw\ — kw) — kw) (z (z _ + - zz = Notice that wz + zw Thus + zw = the right side of (*) real solutions; its follows, since is + zw this inequality it and \w\ + + = = w\ + Two complex numbers 2). segment from geometric The a a FIGURE 2 +c z + zw. Thus < 0; and fact (z w = \z\. + w) (z + w) + \w\ 2 + (wz + zw) + \w\~ + 2\w\ \z\ (\z\ + w\ < \w\r, + \w\. I is its z — The numbers both have picture for addition w = (a,b) and sides the line (c, left to you [compare Appendix (a one-line — may our attention more to is z (0, 0) to z, +w Chapter involved. very simple (Fig- (a and the proof of been shown If z to follow nonzero complex numbers. this 4]). = or w = computational proof can be given, but even the assertion has already to is 1 is d) determine a paral- segment from (0, 0) to w; the vertex opposite (0, 0) unnecessary restrict 0, that 2 interpretation of multiplication • > |it;| operations of addition and multiplication of complex lelogram having for two of then 2 \z\ zw) < 2\w\ important geometric interpretations. line wz that \z +w + zw = are real numbers, \z\ = ure wz + zw) 2 - 4\w\ 2 < The \z\ follows that \z which implies + zw) k(wz + zw). discriminant must therefore be negative. wz _ _ k(wz a quadratic equation in k with real coefficients and no (wz From \w\ + zw = wz (wz it (z _ ww — + 9 k real, since is wz k — kw) — kw) (z this in this expression, number |z| is ;^0we can write a positive real number, while is from P1-P9), so we We begin by putting every nonzero complex number into a special form (compare Appendix 3 to Chapter For any complex 0, 4). . 536 Infinite Sequences and Infinite Series so that z/\z\ a = x + iy is number of absolute a complex with = 1 =x +y \a\ a for angle of 9. = (cos 9, sin 9) + cos 9 Now any 1. i sin = Z + r(cos9 complex number form in the 9 Thus every nonzero complex number radians z can be written sin#) i > and some number 9. The number r is unique (it equals |z|), but 9 any 9q is one possibility, then the others are 9q + 2&?r for k in Z one of these numbers is called an argument of z. Figure 3 shows z in terms of r and 9. (To find an argument 9 for z = x + iy we may note that the equation for is FIGURE some number = value can be written 3 some r not unique; — if x means + =z= iy sin 9) i that x v So, for example, 9 + |z|(cos 9 if = 7r/2 when y > Now the product x > and # = = |z| cos#, |z| s\n9. = we can take 9 = when y < 37r/2 arctany/jt; if x = 0, we can take 0.) of two nonzero complex numbers = w = Z + s{cos(p + r(cos# i sin^), i sin0), is z w— = = rs (cos 9 + i — cos rs [cos(6> + 0) + sin / 9 sin sin(<9 Thus, the absolute value of a product factors, while the sum now an known = as De \z\" This formula describes 2 )F (sin cos + cos sin 0)] + 0)] is the product of the absolute values of the = r(cos9 + sin 9) i Moivre's Theorem): (cosn9 + i sin«#), for any argument 9 of z. z" so explicitly that precisely, for it is any complex number there are precisely n different )( </>) /' easy to decide just Every nonzero complex number has exactly n complex nth More I'R( + easy matter to prove by induction the following very important formula z" THEOREM sin nonzero complex number Z (sometimes i </>) of any argument for each of the factors will be an argument for the product. For a it is + sin 9) (cos rs [(cos w / 0, w = 0) + i sin — w: roots. z satisfying z" ,9(cos0 z" and any natural number complex numbers Let when = w. n, . — . 537 25. Complex Numbers = for s and some number \w\ — Z =w satisfies z" if and only r" (cos which happens if Then a complex number 0. + nO and only first equation it + / nO sin some + s(cos0 i sin0), = = s, + cos sin 0. i follows that t/s denotes the positive real follows that for = nO) sin i — r where sinO) if cos nO the i if r" From + r(cosO integer k V?, nth root of From s. the second equation it we have Ikn n Conversely, + r(cos8 is it if we choose r = i — and ^/s = sin^) will satisfy z" n 9k for To determine w. some k, then the number z = the number of nth. roots of w, Now therefore only necessary to determine which such z are distinct. any integer k can be written — k some for integer q , and some integer + cos Ok This shows that every Z Moreover, k = 0, . . . it is n , — = >/s (cos Ok 4 + easy to see that these 1 by differ less k' =w than i -\- k' and n — between = sin Ok z satisfying z" In the course of proving FIGURE i nq cos + 0k> i sin 1 . Then Ot can be written k sin Ok) numbers = are 0, all . . . , — n 1 different, since any two Ok for 2tt. | Theorem 2, we have actually developed a method complex number. For example, to find the cube |/| = 1 and that tt/2 is an argument for i. The cube for finding the «th roots of a roots of i (Figure 4) note that roots of i are therefore r 1 [cos n -+ n • i sin -^ j [ (n 2tt\ It ( tc 2tt = COS 57T — —h : . / sin 6 it An 4tt jt + 6 T = cos 3tt — —h 2 57T —— 6 . / sin 3jt — 2 — 538 Infinite , Sequences and Infinite Series Since x/3 71 — cos 6 — V3 6 2 5jt ' Sm "6- sin —— — = 0, 2 the cube roots of -s/3 / - we cannot expect the cube roots of 2 is =— 1 are / \/3 arctan -y 2' 2 + —— In general, 1 = 37T 3tt cos 1 6=2' 2 5tt cos IT Sm ? + for i —i. , to obtain such simple results. For Hi, note that an argument + , + |2 1 1/| = y/2 2+11/. One of the cube (arctan U- \ 2 + 2 ll example, = ^125 + 1/ roots of 2 is 1 to find and that therefore I arctan -y VT25 (arctan cos Previously V2 + 2 l 2 = we noted V5, and that 2 + / is also a since arctan ^ is ' ^\ / arctan arctan + 11/. Since 2 + /, we can cube root of 2 an argument of -y- |2 + /| write this cube root as 2 + = v 5 (cos arctan / These two cube roots are actually the arctan A + / sin arctan 5 ) • same number, because — = 1 arctan 2 3 (you can check this by using the formula in Problem 15-9), but this sort of thing The one might fact that every complex order to provide a solution for number has an «th root for all n is just a special The number was originally introduced in =0. The Fundamental Theorem the equation x + / 1 one addition automatically provides of Algebra states the remarkable solutions for other polynomial equations: every equation z" hardly the notice! case of a very important theorem. all is +an -iz n ~ fact that this ] H hflo = ao fl„_i in C has a complex root! we an almost complete proof of the Fundamental Theorem of Algebra; the slight gap left in the text can be filled in as an exercise (Problem 26-5). The proof of the theorem will rely on several new concepts which In the next chapter come up shall give quite naturally in a more thorough investigation of complex numbers. . 25. Complex Numbers 539 PROBLEMS 1 Find the absolute value and argument(s) of each of the following. (iv) + 4/. (3 + 4/r d+/) 5 ^3 + 4/. (v) |3+4/|. (i) (ii) (iii) 2. 3 1 . . Solve the following equations. (i) (ii) (iii) x2 x 4 + ix + = I 0. +x +\=0. 2 + 2ix - = 0. ix - (1 + i)y = 3, (2 + i)x+iy=4 x 3 - x 2 - x - 2 = 0. x 2 1 1 (v) 3. Describe the }) ii) "i) ;iv) y) 4. = z = z~ of all complex numbers < is (z such that l . \z—a\ = \z — b\. \z — a\ + \z — b\ = \z\ z -z. Prove that part 5. z set 1 |z| — Prove that — c. real part of z = \z\, and that the real part of z + — is (z+z)/2, while the imaginary z)/2i. \z w| 2 + \z w\ 2 = 2{\z\ 2 + \w\ 2 ), and interpret this statement geometrically. 6. What is the pictorial relation between z and V zv—i ? (Note that there may be more than one answer, because v/ and V— 7 both have two different possible values.) Hint: Which line goes into the real axis under multiplic <ui< ui / • by 7. (a) Prove that if ao, the equation z" . . . , a„_i are + a„_iz" _1 + • real • • and a + ao = + bi 0, (for then a a and — £> real satisfies bi also satisfies this equation. (Thus the nonreal roots of such an equation always occur in and the number of such Conclude that z" +a n _\z"~ pairs, roots is even.) l (b) -\ (whose '8. (a) hflo is divisible by z 2 — 2az + (a 2 + b 2 ) coefficients are real). Let c be an integer which is not the square of another integer. If a and b we define the conjugate of a + b^/c, denoted by a + by/c, as a — b<J~c. Show that the conjugate is well defined In showing thai a number can be written a + b*Jc, for integers a and b, in onh one way. are integers 540 Infinite Sequences and Infinite Series (b) Show that for and only a rr (c) = *10. is (a)~ a and an if if ao, + z" integer; . . . + a = /3 a„_iz" a„_i are , _1 a • /3; /3; + • • and integers + • gq = = z = z then 0, a + £>\/c satisfies the a — b-Jc also satisfies equation. Find all form that does not involve the 4th roots of (a) Prove that (b) A number a> is + b-Jc, we have a = a; a = a if and + —a — —a; a ft = a of the form a /3 a ^0. } Prove that this 9. a if equation all if co is the set of is there for n any trigonometric functions. an nth root of roots of 3, 4, 5, 1, then so is or. primitive nth root of called a all /7th = express the one having smallest argument in a i; How many 1. 1 if [l,eo,co , ..., co"~ primitive nth roots of 1 l } are 9? n-l (c) Let co be an «th root of 1, with w^ Prove that /^oo 1. = 0. k=0 *11. (a) Prove that then if z\, ... + z\ + Zk / he on one side of some straight line through Zk , Hint: This 0. is obvious from the geometric pretation of addition, but an analytic proof clear if the line is the real axis, and a is 0, inter- also easy: the assertion is reduce the general case trick will to this one. (b) Show through *12. z\~\ further that Prove that 0, so that if |zi | = z\~ \zi\ — ... Zk~ , x H \zs\ h and { all lie Zk~ zi l + # 0. Z2 + the vertices of an equilateral triangle. Hint: real, and this can be done with no loss on one Z3 = It will 0, side of a straight line then help to of generality. and Z3 are assume that z\ is zi, Zi, Why? CHAPTER mm ^^ COMPLEX FUNCTIONS You will probably not be surprised to learn that a deeper investigation of 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 formal a function definition): does not contain two real numbers in to new case of the order to not even accord will a collection of pairs of is distinct pairs with the same first it the complex numbers which we will sometimes which a function clarify the context in number we element. Since be certain complex numbers, the old definition one. Nevertheless, honors of a full is really a special resort to special terminology being considered. is consider A function domain of / f(z) /, and complex-valued to emphasize that it is not necessarily real-valued. Similarly, we will usually state explicitly that a function / is defined on [a subset of] R in those cases 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 is called complex real-valued if is a real for all z in the z as well as real z. Among on C, certain ones are important. Foremost among these are the functions of the form the multitude of functions defined f(z)=a n z" +a„_ where ao, ... , l z"~ +--- l + ao. These functions are a n are complex numbers. particularly called, as in the polynomial functions; they include the function f(z) — z (the "identity function") and functions of the form f(z) —a for some complex number a ("conreal case, stant functions"). Another important generalization of a familiar function — "absolute value function" f(z) Two and Im complex numbers are Re Re(.r "conjugate function" + />) — x, + iy) = y, . lm(x is x and y = l = - Re(z) Familiar real-valued functions defined on i 541 real. Im(z). R may be combined in produce new complex-valued functions defined on + . , tor . defined by f(z ) fix (the "real "imaginary part function"), defined by (the T the for all z in C. functions of particular importance for part function") The \z\ is iy) = e y s'm(x — y) C — an example + ix cos v. many ways is to the function 542 Infinite Sequences and Infinite Series The formula for this particular function illustrates a decomposition which possible. Any complex-valued function / can be written in the form some and v(z) and real-valued functions u as the v — simply define u{z) as the real part of f(z), imaginary part. This decomposition always; for example, it always = u + iv f for is would be inconvenient is often very useful, but not to describe a polynomial function in this way. One other function will play an important role in argument of a nonzero complex number Z There are In. If we infinitely call this = |z|(cos# many arguments unique argument "argument function") on [z in C z : z for is a + (real) i sin such that (9). but just one which z, Recall that an this chapter. number 0(z), then is < satisfies < a (real-valued) function (the ^ 0}. "Graphs" of complex-valued functions defined on C, since they lie in 4-dimenpresumably not very useful for visualization. The alternative picture of a function mentioned in Chapter 4 can be used instead: we draw two sional space, are and arrows from copies of C, FIGURE z in one copy, to f(z) in the other (Figure 1). 1 The most common pictorial representation of a complex-valued function is pro- duced by labeling a point in the plane with the value f(z), instead of with z (which can be estimated from the position of the point in the picture). Figure 2 shows this Certain features of the function are sort of picture for several different functions. by such a "graph." For example, the absolute value function is constant on concentric circles around 0, the functions Re and Im are constant on the vertical and horizontal lines, respectively, and the function f(z) = z wraps illustrated very clearly the circle of radius r twice around the circle of radius r . Despite the problems involved in visualizing complex-valued functions in general, it is still possible to define analogues of important properties previously defined lor real-valued functions to visualize in on R, and in some cases these properties may be easier the complex case. For example, the notion of limit can be defined as follows: lim f(z) = I means that for every (real) number s > there is a (real) z—*a 8 >() such that, lor all z, if < \z - u\ < 8, then \f(z) -l\ <S. number * 1 26. Complex Functions -i' It f |.. -i. 1" -i' §.. -i. i" -" -" i- -i- I" 3 1" i ' ' li §1 -i. -i- , , 3 3 3 3 3 2 2 2 2 7 1 1 1 1 1 1 2 2 7 ? 2 2 : 1 3 > 3 3 3 3 3 i 2 2 2 2 1 1 1 1 1 1 i i i 2 2 7. 1 1 -1 -1 -1 -1 -1 -1 3 3 3 3 3 2 2 2 2 2 (c) 3 2 f(z) 1 1 i 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 : 2 2 2 2 2 2 7T 7T TZ It = „o .1(1 U "I < ,1 1 .in ni < ,1 1 -III 2 _In i i < . I r i , : (e) 1 2 liy > « ,1 _1|, , _i,i 1 „0 ol < ,,o ni < 1 1 I —i — i — i < ' * i -i" "0 "i _In ,,n U ni _^„ „ I.' -il lo 4 2 • , -i. I < »1 1 < 2 2 ,,3 2 2 ,,3 2 ,,3 2 ni 2 .3 1 , 2 , 1 ^ 2 ..I 2 ni 2 2 ,11 ol il1 li 2 71 ff il 2 2 2 Im(z) TT i,l , ,1 -1 -1 -1 -1 -1 -1 3 ,1 1 Hi 2 2 »§ „i () 1 1 „ a 2 ( _i„ f(z) 3 ,1 1 ,,o _i„ i 1 Pi « _!,, . T1 < 2 2 , I 1 ol i>n (b) (a) P T° 2 — -h -- !-. §< "2T —In > 2 |. , f(z)=6(z) f(z) = Re(z) . 2 543 544 Infinite Sequences and Infinite Series Although the definition reads precisely — w\ equation \imf(z) = ferent. Since \z means / as before, the interpretation is slightly dif- z and w, the between the complex numbers the distance is made that the values of f(z) can be to lie inside z—>a any given around circle small circle around a /, provided that This assertion . copy" picture of a function (Figure is z restricted to is lie inside a sufficiently particularly easy to visualize using the "two 3). © FIGURE 3 Certain facts about can be proved exactly as limits lim c = c, lim z — a, in the real case. In particular, z.^>-a + g(z)] lim f(z) g(z) lim[/(z) = = 1 1 g(z) lim g(z) lim a The < limg(z), + £— >a lim f(z) lim g(z), if ' essential property of absolute values + lim f(z) z—>a Z—>a limg(z) 7^0. upon which these results are based is the + \w\, and this inequality holds for complex numbers as These facts already provide quite a few limits, but many more can be obtained from the following theorem. inequality \z w\ \z\ well as for real numbers. THEOREM 1 Let f(z) real = u(z) + iv(z) numbers a and )3. for real-valued functions u Then = and I if and only lim u{z) = a, = (3. lim f(z) v, and let / = a + if z->« Z—>C1 lim v(z) Z->a PROOF Suppose first that lim /(-) = /. If e ' > 8 > then \f(z) — 0, there is such that, for z->a if The second < \z — a\ < 8. I\ inequality can be written \[u(z)-a\ +i[v(z)~P]\ <£, < £ all z, ifi for 545 26. Complex Functions or [u(z) Since u(z) —a. and v(z) — /3 - a] 2 + - [v(z) 2 <e 2 P] . are both real numbers, their squares are positive; this inequality therefore implies that < [m(z) —or] £~ and [u(z) — {5\~ and \v(z) — >3 < s , which implies that — \u(z) Since this is true for all s > < a\ 0, it 8 s. follows that \imu(z)=a and limu(z) Z—>a = /J. z-*a Now suppose that these two equations hold. < \z — a\ < 8, then for all z, if \u(z) < 1 — If e £ < — a\ and > — \v{z) 0, there a\ S < — is a 8 > such that, , which implies that \f(z)-l\ = \[u(z)-a] < |«( z )- a £ This proves that lim f(z) In order to apply limit lim z = a, we = I. Theorem + i[v(z)-P\\ + |i|. \V(Z)-P\ | £ | 1 fruitfully, notice that since we already know the can conclude that lim Re (z) = Re(a). = Irn(a). = sin(Re(a)) Z—*a lim Im(z) Z-*a A limit like lim sin (Re (z)) Z—*a follows easily, using continuity of sin. Many applications of these principles prove such limits as the following: lim z =a , Z->tf lim |z| = \a\, z-*a lim e y sin x + cos y ix = e sin a + ia~ cos b. (x+iy)—*a+bi Now that the notion of limit has of continuity can also be extended: been extended / is to complex continuous at a if functions, the notion lim f(z) = /(«), and ' 546 Infinite Sequences and Infinite Series f is continuous if / is continuous at a for all a in the domain of /. The previous work on limits shows that all the following functions are continuous: f(z) f(z) f(z) fix = = = +an -iz an z l ^ h«0, z, \z\, v ix cosy. +,-3 e y sinx + iy) • Examples of discontinuous functions are easy to produce, and certain ones come up very naturally One particularly frustrating example is the "argument function" 9, which is discontinuous at all nonnegative real numbers (see the "graph" Figure in for ities; 2). By suitably redefining 9 example (Figure 4), it is possible to change the discontinu- 9'(z) denotes the unique argument of if z with < 0'(z) < 5tt/2, then is discontinuous at ai for every nonnegative real number a. But, no matter how 9 is redefined, some discontinuities will always 9' tt/2 occur. 2^ TT ' 2 2&* n • .l\n 2 ' TT 2ft» 2 ' 2 2 * * n 2 TT &*2\n i* 2 5* •2\n ' 2 ' 2 ' *2±n TT TT 2h *2i» ' 2 4 ?l 2 3 TT 4 2 2' TT TT 71 71 71 71 71 71 •2i;r • TT •2j» 27r 27r 27r 2;r 2;r 2;r 2;r 27r 2;t TT ~2 in T ( 3jT , 3tt T T ( ' 3rr , 3tt T T ( , 3* 2 3tt T FIGURE The 4 f(z) < = 0'{Z) discontinuity of 9 has an important bearing on the problem of defining a "square-root function," that is, a function / such that (f(z))~ = z for all -. For real numbers the function \/~ had as domain only the nonnegative real numbers. If complex numbers arc allowed, then every number has two square roots (except 0, which has only one). Although this situation may seem better, it is in 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. — . 547 26. Complex Functions One way / to define is the following. = j(z) Clearly (f(z))~ = Z, As a matter of fact, In all z. prove fact, it is / but the function it is set 9& — h is even impossible / by — /). Then we argument). But for this claim that for is left to + / you is + even though cos # + d— functions [a, b] x [c, / —> sin ^ 1 i for we set —& / is discontinuous. = Z for — To such that (/(z)) — all z with 1 |z| . always a standard type of least upper bound that f{\) = 9 cos — (since 1 < 6 with 9 < sin — Itc 9 + = cos n + = -l sin 9) any n > — as A similar argument shows that ' # we could we have sin 9) (it for z i implies that (*) lim /(cos 9 1 sin be defined for all 0-^2tt tion. / and 0, discontinuous, since for f(z) to /(cos 9 (*) The argument = /(0) impossible to find a continuous by contradiction, we can assume this replace /TiY \/\z\[ cos We it is 2^ . i sin n Thus, we have our contradic- impossible to define continuous "nth-root 2. d] 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 < b and c < y < d (Figure 5). This set is called a closed rectangle, and is denoted by [a, b] x [c, d]. figure If 5 then That / it seems reasonable, and is, there is some does not [a,b] x make is < / is in [a, b] x indeed true, that number real |/(z)| It domain is bounded on a continuous complex-valued function whose is M sense to say that [a, b] x [c, d], [a, b] x [c, d]. M such that for / all z has a [c, d]. maximum and a minimum value on complex numbers. If / is a make sense, and is true. In if / is any complex-valued continuous function on [a, b] x [c, d], then continuous, so there is some zo in [a, b] x [c, d] such that [c, d], since there is no notion of order for real-valued function, however, then this assertion does particular, |/| is also |/(zo)l a similar statement "/ attains The its is <l/(z)l for all -in [a,b] x [c,d]; true with the inequality reversed. maximum and minimum modulus on It is [a, b\ x sometimes said that [c, d~\." various facts listed in the previous paragraph will not be proved here, though proofs are outlined in Problem 5. Assuming these facts, however, al- we can ' 548 Infinite Sequences and Infinite Series now Fundamental Theorem of Algebra, which give a proof of the surprising, since we have much not yet said is really quite polynomial functions to distinguish from other continuous functions. THEOREM 2 (THE fundamental THEOREM OF ALGEBRA) Let ao, . . . , fl„_i be any complex numbers. Then there number z such that n Z PROOF a complex is + an - lZ n - + ] a„_2Z n-2 + • + = fl 0. Let "+a n _ lZ,«-! +---+O0. 1 f(z)=z" Then / is continuous, and so = \f\(z) the function \f\ defined is by = \z"+a n ^z"- +---+a l \f{z)\ \. Our proof is based on the observation that a point zo with f(zo) = would clearly minimum point for / To prove the theorem we will first show that / does indeed have a smallest value on the whole complex plane. The proof will be almost be a . 1 1 | identical to the proof, in | Chapter on (with real coefficients) has a smallest value fact that if We |z| is large, then |/(z)| begin by writing, for of R; both proofs depend on the all large. is ^0, z f(z) polynomial function of even degree 7, that a = zn + 1 + ••• + a so that l/(z)l = 1 |z| ^1 + + Let M = max(l, 2«|a„_i Then for all z with k > M, we have |z| \a„-k\ 7 k\ \z \a„-k\ < — \ > |, . |z| . . , 2n\ao\). and \a n -k\ < J_ 2n 2n\an . so a n -\ «0 0-n-\ z" Z which implies < + ••• + z 1 2' that fl i + »-' +... + "" 7 > ~n 1 - a„-\ clq Z z" This means that 1 l/(z)l In particular, < -11 if |z| > M and also \z\ \.f ~l" ^ > > (z)l for |z| > M. (/2|/(0)|, then >l/ [0)\. 1 > — - 2 549 26. Complex Functions Now let [a, b] x max(M, y2|/(0)| [c, )}, d] be a closed rectangle (Figure and suppose that the which contains 6) minimum of on |/| [a, b] {z x : [c, |z| < d] is attained at zo, so that (1) It |/(zo)l < l/(z)l for z in [a,b] < follows, in particular, that |/(zo)l (2) if \z\ x [c,d]. Thus 1/(0)1- > max(M, ^21/(0)1), then |/(z )| and (2) we see that |/(zo)l S \f(z)\ minimum value on the whole complex plane at zq- Combining (1) > |/(0)| for all z, > |/Uo)l- so that |/| attains its max(M, ^21/(0)1) l/(z)l > 1/(0)1 for z here FIGURE 6 To complete the proof of the theorem we now show that /(zo) = 0. It is convenient to introduce the function g defined by = g(z) Then g is occurs at 0. + zo). f(z a polynomial function of degree n, We want to show that g(0) Suppose instead that g(0) = a which occurs in the expression for = whose minimum absolute value 0. m / 0. g, we can If is the smallest positive ft ^ 0. Now, according to Theorem 25-2 such that y m __ ~ z write m+l + g(z)=a + Pzm +cm+l z where power of « fi' --- + there f„; is a complex number y 550 Infinite Sequences and Infinite Series Then, setting dk = Cky k we have , = + Py m z m + dm+[Z = \a-az'"+dm+] z"1+l '" \g(yz)\ +l \c* a[ l-,'" + + dn z"\ +] d„ + ^±l -'"+'+. a a = - 1 I z m m+\ l + zm + a dm A a 1 + Z i a This expression, so tortuously arrived diction. Notice first that if \z\ is l m+ a If we choose, from among all z enable us to reach a quick contra- at, will chosen small enough, we \ z have this inequality holds, some < + which for will 1. z which is real and positive, then dm+ a Consequently, 1 - < if z m + z < 1 \ z+ :." -z a + 'II- < is \g(yz)\ the original assumption /(zo) 1 1 - the desired contradiction: for such a contradicting the fact that - _'" i we have = This jn < |a| is must be the < z"' m z m z d„ + 1 + + z a dm +\ + z a + + zm number z we have |a|, minimum incorrect, I of \g\ and g(0) on the whole = 0. plane. Hence, This implies, finally, that 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 step is further the analogues of properties of real functions. to define derivatives: a function .. / differentiable at a f(a+z)-f(a) lim z-»0 is exists, z The if next obvious ; 551 26. Complex Functions which case the in limit denoted by f'{a). is f(a) = if/(z)=c, f(a)=\ (f+g)\a) {f - g)'(a) 1\' J the proofs of all = = i£f(z)=Z, + g'(a), f\a)gia) + f{a)g'(a), f{a) -g'(a) = («)-t4^ (fl) if*(«)^0. Mtf = x if differentiability of rational functions however. , / is to . + iy) = be differentiable at 0, - x iy f{z) (i.e., f(x+iy)-f(0) x (x+iy)-*-0 exist. other obvious candidates are = z). then the limit lim must Many Suppose, for example, that fix If same as before. It follows, in These formulas only prove the these formulas are exactly the ~ nz n f(z) = z" then f'(z) particular, that not differentiate. easy to prove that It is = + iy lim (x+iy)^0 x-iy x + iy Notice however, that if j = ~ x — iy = + iy x — x + iy x , then 0, 1, and if x = then 0, iy =— 1 therefore this limit cannot possibly exist, since the quotient has both the values and — 1 for x + 1 iy arbitrarily close to 0. where other differentiable functions are to come from. If you recall the definitions of sin and exp, you will sec 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 solved. In view of this example, it is not at all clear PROBLEMS 1. (a) a(x) = valued function defined on R). Show that immediately from a theorem in this chapter.) For any real x (b) + iy is Let / be f(x + number (a) rectangle + iy a is (so that or is a complex- continuous. (This follows Show a continuous function defined on C. iy). Suppose x v) = g(x) = similarly that fi( continuous. Show that g similarly that hiy) 2. v, define that / [a, b] is x = is fix For fixed v. let a continuous function (defined on R). + iy) is continuous. Hint: Use part Show (a). a continuous real-valued function defined on a closed [c, d]. Prove that if / takes on the values f(z) and f(w) 552 Infinite Sequences and Infinite Series and w in [a, /?] x [c, d], then / also takes all values between f(z) and f(w). Hint: Consider g(t) = /(rz + (1 - r)w) for t in [0, 1]. If / is a continuous complex-valued function defined on [a, b] x [c, o"], the assertion in part (a) no longer makes any sense, since 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 f(z) and f{w), but even this is false. Find an example which shows this. for z "(b) Prove that 3. if «o, • complex numbers Z n (not necessarily distinct) + a,^ ]Z n z" complex numbers, then there are are any 1 - ] +...+ao = Y\(z ~ such that zi). i=\ (b) Prove that if oo, . . . on _i are , real, then z+a and written as a product of linear factors all of whose coefficients are + a n _\z n z" ~ + + «o can be quadratic factors z 2 +az+b { • • • (Use Problem 25-7.) real. In this problem we will consider only polynomials with real coefficients. Such a polynomial called a is for polynomials (a) Prove that if (b) Prove that if (c) Suppose that sum of squares if it can be written as /?i~ + - +^, • 2 ; with real coefficients. //, / is a sum of squares, then f(x) > for all x. / and g are sums of squares, then so is f g. for all x. Show that / is a sum of squares. f(x) > • Hint: k First write f(x) = \(x I — where g(x) > a{) g(x), for all x. Then use i=] Problem 5. (a) (a) 3(b). A be Let limit point of the real case, a a point a in A with \z — < a\ [a, b] x [c, d], divide [a, b] x is FIGURE infinite, at least this in half (b) [c, then A if e but z ^ number z by a a. title > 0, there If A is an infinite subset [c, d]. Hint: First vertical line as in Figure 7(a). Since infinitely line, as in Figure is Prove the two-dimensional many 7(b). and horizontal A points of A. Divide Continue in this way, lines. (The two-dimensional bisection argument outlined method of called, as in the for every (real) s Theorem: one half contains by a horizontal dard that the is has a limit point in [a, b] x d] in half alternately dividing by vertical 7 A set version of the Bolzano-Weierstrass of A a set of complex numbers. in this hint is so stan- "Bolzano-Weierstrass" often serves to describe the proof, in addition to the H. Petard, "A Contribution to the theorem See, for example, itself. Mathematical Theory of Big Game Hunting," Amer. Math. Monthly, 45 (1938), 446-447.) (b) Prove that a continuous (complex-valued) function on [a,b] x \c\d] bounded on (c) Prove that if [a,b\ x / is on a / can use the same then takes [c, d\. (Imitate a real-valued continuous function on \a,b\ x maximum and minimum trick that works for is Problem 22-31.) value on Theorem 7-3.) [a, b\ x [c, d], [c, d\. (You 26. Complex Functions *6. The proof of Theorem 2 cannot be considered 25-2, interest to tion z" (a) — c a convex subset of the plane explicit be found for n 1 " 0. Make an why Explain (b) (a) — computation show to any complex number — c = can be reduced to the case where odd. is / absolute value. If zo rem 2 is equal to 1; show 0, — z" — that the integer m in the we can since remainder of the proof works minimum certainly find for /. It its minimum proof of Theo1 y with y therefore suffices to / does not occur absolute value of c has = — ar/)6, show at 0. small 8 at least one of these points has smaller absolute value than 8 — c, thereby obtaining a contradiction. a noneonvex subset of the plane FIGURE the that the Suppose instead that / has its minimum absolute value at 0. Since n is odd, the points ±<5, ±8i go under / into — c±<5", — c±5"i. Show that for (d) (b) can c. —c = the solution of z" that solutions of z Let zo be the point where the function f(z) (c) be completely elementary to = —a/fi depends on Theoand thus on the trigonometric functions. It is therefore of some provide an elementary proof that there is a solution for the equa- because the possibility of choosing y with y rem 553 7. Let/(z) = (z-zi) mj C for :)'"< mk > mi, 0. k Show that /'(z) = (z - zi)"" • . . • . (z - m Zk) J2 nia{z ~ ~l Za) - a=\ Let g(z) (b) cannot — }^m a (z — all lie za ) on the same . Show that if g(z) side of a straight line = 0, then through z. z\, ..., Zk Hint: Use Problem 25-11. A subset K of the plane any two points set FIGURE 9 containing in it, it is convex (Figure which is 8). if K For any called the contains the line segment joining set A, there is a smallest convex convex hull of A (Figure 9); if a 554 Infinite Sequences and Infinite Series point P side of is not in the convex hull of A, then some the roots of f'(z) = all through P. Using straight line lie of A is contained on one information, prove that this within the convex hull of the set Further information on convex sets will {z\, . . . , Zk}- be found in reference [18] of the Suggested Reading. 8. *9. Prove that Suppose (a) if that / is / = For fixed yo differentiable at u let + = u(x + / is continuous = + iyo). Show that if g'(xo) = a and h'(xo) = fi. = u(xo + iy) and /(y) = v(xo + iy). and h(x) a and /3, then iyo) (c) (a) Using the expression ifi 1/1 1 1 (b) find / w (Jc) Use this result to find + jc z at z. v are real-valued functions. for real f'{xQ + /yo) = a + On the other hand, suppose that k(y) Show that /'(yo) = a and k'(yo) = —fi. Suppose that f'(z) — for all z. Show that / (b) 10. then where u and iv g(x) z, 1 Zi for all*. arctan (A) (0) for all k. v(x is a constant function. . COMPLEX POWER CHAPTER If to come from, the title of complex functions are going differentiable chapter should give the secret away: this we intend to means of infinite series. This will necessitate a discussion of infinite sequences of complex numbers, and sums of such sequences, but (as was the case with limits and continuity) the basic definitions are almost exacdy the same as for real sequences and series. An infinite sequence of complex numbers is, formally, a complex-valued function whose domain is N; the convenient subscript notation for sequences of real numbers will also be used for sequences of complex numbers. A sequence {a„ of complex numbers is most conveniently pictured by labeling the points a„ in the define functions by «4 a\ «6 aj* as «3 a5 'a 7 an FIGURE you have not already guessed where SERIES *«12 } plane (Figure 1). 1 The sequence shown Figure in converges to 1 0, "convergence" of complex sequences being defined precisely as for real sequences: converges to /, lim a„ d\. ,°2 if aN .03 for every e > sequence {a n } the symbols in there is a natural if number > N, then n = I, N such — \a„ that, for all n < l\ , e. «6 This condition means that any drawn around / will contain a n for all suffiexpressed more colloquially, the sequence is eventually ciently large n (Figure 2); circle drawn around /. Convergence of complex sequences any inside FIGURE circle sequences, but can even be reduced to 2 THEOREM 1 is not only defined precisely as for real this familiar case. Let and a„ = b>i I = fi + for real b„ icn and cn , let Then lim a„ = / if + iy and only if lim b n = for real and y /3 n—>oc and /3 lim c„ n—>oo n—»oo PROOF The proof is left as consult the similar The an easy Theorem sum of a sequence exercise. ) is s„ 555 is of Chapter 26. 1 \a n If there = y. any doubt a i H h to | defined, once again, as lim sn n—>oo = how as to an . , where proceed, 556 Infinite Sequences and Infinite Series Sequences for which are this limit exists summable; alternatively, we may say that 00 the infinite series 2_. an converges limit exists, if this and diverges otherwise. It »=i is unnecessary to develop any new convergence of infinite tests for series, because of the following theorem. THEOREM 2 Let an = b„ + ic„ oo 00 Then 2, a n converges if and only and for real b n 2, c n n=\ n=\ . oo b n and 2, if cn both converge, and in this n=\ case oo OO J2 a 'i = J2 + \Jl c " h" i n=\ PROOF This an immediate consequence of Theorem is sums of {a,,}. | There is also a notion of absolute convergence for complex an 2> converges absolutely if the series following theorem is 2^ \an \ converges (this is a series of real which our to may be earlier tests applied). The not quite so easy as the preceding two. Let a„ = b„ + ic„ cn . oo Then >, a n converges absolutely if and only oo 2, bn and 2, c n if n=\ absolutely. Suppose and for real b„ 00 PROOF the series n=[ n=\ numbers, and consequently one 3 series: oo oo THEOREM applied to the sequence of partial 1 first that /_]b n and /£„ ,;=1 z,= both converge n=\ 71=1 both converge absolutely, i.e., that 2_. \b,i\ and |<7„| and ;;=1 l oo oo 2_] \cn \ both converge. It follows that 2, + \bn\ \cn \ converges. Now, n=\ n=\ \a n \ = \b„ +icn < \b n \ \ + \c„\. 00 It follows from the comparison test that 2_. a n\ converges (the \ numbers ,;=1 oo \b„\ + \c n \ are real and nonnegative). Thus /^an converges absolutely. a 557 27. Complex Power Series Now suppose that 2, \ a »\ converges. Since n=\ |0b it is = I \/^ 2 Cn 2 + , clear that \b n < \ \a n and \ < \c„\ \a n \. oo Once oo again, the comparison test shows that 2_. n=\ \b„\ and \] \ cn\ converge. | n=\ oo Two consequences of Theorem 3 are particularly noteworthy. If \_. a " con " n=\ OO 00 verges absolutely, then /_]£« and /Jc„ n=\ also converge absolutely; consequently n=\ OO 00 00 2_\b„ and /_/'„ converge, by Theorem 23-5, so n=\ n=\ 2_, it a " converges by shows that any rearrangement of an absolutely convergent facts can also be proved orems for real numbers, by (see Problem 13). first With these preliminaries power series, that is, 2. =1 In other words, absolute convergence implies convergence. sum. These Theorem Similar reasoning series has the same directly, without using the corresponding the- establishing an analogue of the Cauchy criterion we can now safely disposed of, functions of the consider complex form oo f(z) — ^2a„(z - a) n — a + a\(z - a) + a2(z - a) 2 -\ . n=0 Here the numbers a and a„ are allowed to be complex, interested in the behavior of consider power series / for complex z. As and we are naturally in the real case, we shall usually centered at 0, oo f( Z ) = J2 a n >i? ' n=0 f(zo) converges, then f(z) will also converge for \z\ < \zo\. The fact will be similar to the proof of Theorem 24-6, but, for reasons in this case, if proof of this that will soon become clear, we will not use all the paraphernalia of uniform con- vergence and the Weierstrass M-test, even though they have complex analogues. Our theorem 4 next theorem consequently generalizes only a small part of Theorem 24-6. Suppose that oo /, n ZQ n = ao + a\zo + a 2 zo 2 H . 558 Infinite Sequences and Infinite Series converges for some zo Then 7^ 0. if |z| < the two series \zo\, 00 ^ an z n = + a\ z + aiz 2 a H n=0 oo — a\ + 2a2Z + 3«3Z 2 + na,,z" } j n=\ both converge absolutely. PROOF As in the a n zo n is proof of Theorem 24-6, we bounded: there is a need only the fact that the set of numbers M such that < M for number all n. \a n Zo"\ We will then have \a„z and, for I — \a„zo < M z^O, \na n z I = —-n\a„zo \z\ < M — n Since the series \_. \z/zo\" and /_] n l-/"ol" converge, n=0 n=\ 00 and \_. na nZ"~ n converge absolutely (the argument for 2_\ na nZ ~ assumed that n=\ n=\ 7^0, but this series certainly converges for Theorem 4 \] a nz" n=0 OO ; shows that both this z = also). | evidently restricts greatly the possibilities for the set | z '. 2_. a "Z" converges | oo For example, the shaded set A in Figure 3 cannot be the set of all z where / a u z" n=0 ikuri; i converges, since It it contains z, but not the seems quite unlikely that the set number w satisfying \w\ < \z\. 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 complication which we converges at will all points), then this assertion is true (with one soon mention); the proof requires only Theorem 4 and a knack for good organization. . 559 27. Complex Power Series theorem 5 For any power series 00 ^ = ao+a\z + aiz 2 + a^z 3 + a„z" n=0 one of the following three must be possibilities true: 00 2_. a nZ (1) n converges only for z — 0. oo 2_. a nZ" converges absolutely for «=o (2) There (3) a is number R > all z in C. such that 2_, a nZ n converges absolutely if |z| < R ,i=0 and diverges when PROOF Id if = R.) x in > \z\ R. (Notice that we do not mention what happens Let = S Suppose I first that R 2, a wn n : S converges for some unbounded. Then is for w =x with \w\ \ any complex number z, there is oo a number x in S such that < |z| By x. definition of S, this means /^ a„ that w" n=0 00 converges for some w with X > w\ converges absolutely. Thus, in Now R — on for then 2_\ a nZ n is number x in this case possibility bounded, and converges only for n=0 the other hand, that a IS 0, suppose that S R > S with some w with < \z\ \w\ 0. < Then x. Once > R, then 2_, a nZ n R be let z = if z is a complex again, this occurs in case that 2, anZ n (2) is true. the least upper 0, so possibility (1) is bound of means 2_, a nW n=0 that < |z| n If Suppose, true. number with S. R, there converges Moreover, if = does not converge, since The number R which from Theorem 4 so that 2_, a n z " converges absolutely. 7! |z| follows It (3) is |z| is not in S. called the | radius of convergence of oo V is a„z" . and and (2) it is oo, respectively. When In cases (1) customary < R < to say that the radius oo, the circle {z : \z\ of convergence = R} is called v 560 Infinite Sequences and Infinite Series Eri—Kj /]an z n of convergence of the circle If Z . outside the circle, then, of course, is n=0 much a n z' does not converge, but actually a stronger statement can be made: n=0 the terms a n z" are not even bounded. > |z| > \w\ show R; if 2_, a nW that n converges, which ^— 11=0 _ convergence the the terms a„z" are not of convergence circle bounded To prove this, let w be any number with were bounded, then the proof of Theorem 4 would the terms a n z" y. a nZ series n Thus false. is (Figure 4), inside converges in the best possible n=0 outside the circle the series diverges in the worst possible way way the circle of and (absolutely) terms a„z" are (the not bounded). What happens We will series on the circle of convergence not consider that question at all, which converge everywhere on the converge nowhere on the anything a much more difficult question. except to mention that there are power circle of convergence, power series which of convergence, and power series that do just about circle between. (See Problem in is 5.) Algebraic manipulations on complex power series can be justified just as in the oo oo Thus, real case. FIGURE if — f(z) = " 2_, a " z an<^ S(z) n=0 4 2-,^" z " k° tri nave racnus °f n=0 oo convergence > R, then h(z) > R and h = f+g = + /](fl« b n )z" also has radius inside the circle of radius R. oo h(z) = Similarly, the of convergence Cauchy product n — / for c„ c 2_, nz!\ «=0 > R and akbn -k, has radius of convergence h = fg k=0 oo inside the circle of radius R. And if f(z) — 2~]a„z" has radius of convergence > n=0 oo and «o ¥" 0, then we can find a power series /_\b„z n with radius of convergence n=0 > which represents But our real goal side // inside its circle is of convergence. to produce differentiable functions. power to generalize the result Chapter 24, even if we were willing proved for to introduce real (which could also have been used in Chapter 24). notice that at least there produced by term-by-term is series in uniform convergence, because no analogue of Theorem 24-3 seems available. Instead we we We Chapter 24, a function defined by a power series can be differentiated term-by-term inthe circle of convergence. At this point we can no longer imitate the proof of therefore that want 1 in this chapter will use a direct argument Before beginning the proof, no problem about the convergence of the differentiation. If the series /^ctnz" has radius n=0 vergence R, then converges for Theorem 4 immediately < R. Moreover, if |z| > implies that the series series of con- oo / na n z n ~ also R, so that the terms a„z n arc unbounded, 27. Complex Power Series then the terms na n z" l are surely unbounded, so y. nanZ n 561 does not converge. oo n This shows that the radius of convergence of 2_, na nl ~ also exactly R. is n=\ theorem 6 If the power series = J^a n z n f(z) n=0 has radius of convergence R > / then 0, differentiable at z for all z with is \z\ < R, and f {z) = Y j na n z n- n=\ PROOF We The another "e/3 argument." will use theorem fact that the is clearly true for polynomial functions suggests writing f(z + h)-f(z) - }na nz A 2 n~ h dz A / na + hr-z") an n=0 A < A + h)"-z") ((z l^ a » n «z 1 „=0 n=0 A A 2^na + hy-z") dz + l^ a » 7 n= n=0 + / nan z n l j We will show that for any N s > nz \ - }jia n z n n=\ < e/3 by choosing nz \ + h)"-z") l^ a " 1 = ~ n=\ each absolute value on the right side can be 0, and h sufficiently large This sufficiently small. made will clearly prove the theorem. Only the term first + < convenient way with \z h\ | < |z| I- -n _ ] < ((z v" — = x "} x-y Applying \zo\ The expression if we remember that co of (*) in the right side with, choose some zo with + xn ~2 + v h) n — we difficulties. will + h) n z")/ h can be written in a + x"" 3 v 2 + • + • y"" 1 . + h)" (z + h)-z (z h ' obtain <- _i_ iiV ^-^ 1 _ -" = + hf- + 1 (z z(z + To begin consider only h this to (z we present any will R; henceforth h)"- 2 + + n z ~\ more . 562 Infinite Sequences and Infinite Series Since + h)"- + +h)'- 2 1 \(z z(z + + l z"- \ < ] n\zo\"~ , we have + h) n -z n ((z ] < a„ But the series Y_]n|a„| converges, so \zq\" • ,/j-l n\a„\ N if \zo sufficiently large, is then n=\ n a n\ 2 < \Z0\"~ \ , n=N+1 This means that A {{z A + hT-z") ((z + h)"-z n /; ) // n=0 71=0 + ((z an J2 n -z") h)" E * h = N+\ «z + h) n -z") h n = N+\ s < — 3 In short, if N A A 1; is then sufficiently large, «z a» + h) n -zn A L ) 1 n=0 ((z + h)"-z "- n £ ) h n=0 for all h with It is fl + \z h\ < \zq\. easy to deal with the third term on the right side of (*): Since converges, it follows that N if sufficiently large, Y,nan z n~ l - J2 na " z "~ < n=\ n=\ choosing an .. N such that A hm ) Hz an (1) and (2) + hr-z") - are true, = n=0 n ((z + h)" A > we na n z note that „_! n=\ since the polynomial function g(z) ^a then N oo Finally, is y. na nZ' n=\ -z") = 2_\ a nZ" n=0 ^ certainly differentiable. Therefore N -1 (3) n=0 is < £ 3' n=\ for sufficientlv small h. As we have already indicated, (1), (2), and (3) prove the theorem. | 27. Complex Power Series Theorem 6 has an obvious corollary: a function represented by a power series of convergence, and the power infinitely differentiable inside the circle Taylor series / at 0. It follows, in particular, that convergence, since a function differentiable at The 563 continuity of a power series inside z at z is its circle of (Problem 26-8). of convergence helps explain the circle its continuous is series continuous inside the is is behavior of certain Taylor series obtained for real functions, and gives the promised answers to the questions raised end of Chapter 24. at the — that the Taylor series for the function f(z) 4 2 l-z + 5 -z + < z 2 ), We have already seen namely, ---, and consequently has radius of converof convergence contains the two points i and —i at If this power series converged in a circle of radius greater than 1, then (Figure 5) it would represent a function which was continuous in that circle, in particular at 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 s approaches / or —i from inside the unit circle. The use of complex numbers also sheds some light on the strange behavior of converges for real gence FIGURE when 6 z + 1/(1 1 . z only \z\ no accident that the which / is undefined. It is 1, circle the Taylor series for the function -l/.x e fix) 2 Although we have not yet defined if y is real and unequal e z for = 0. x 0, complex z, it will 2 f (iy) = e -Wy) =e The interesting fact about Thus / so hardly surprising that will expression this small. it is it is equal to x x3 =x— = 1 — V x2 when x = 1 X + x5 + x + (as — 4 " 4! x 2 + 2! V. therefore e z 5! + define 5 3 sinz —z- + 3! = 1 - - 2 exp(z) = e z = 1 — ~5\ 74 2 cose yi + + 4! + + large as y ^— becomes defined for complex numbers, know x we 2! e becomes Taylor series only for its clear. For real cosx we that 2 it will actually define e sin z l/.v that is not even be continuous at The method by which we complex z should by now be For complex presumably be true to 0, then + well as sin z that z = 0. and cos z) for 564 Infinite Sequences and Infinite Series Then = sin'(z) Moreover, if of the terms we = — sinz, cosz, cos'(z) and replace z by iz in the series for e (justified z , = exp(z) by Theorem 6. and make a rearrangement exp'(z) by absolute convergence), something particularly interesting happens: (iz) 3 (iz) 4 . (iz) 5 , so = z e' It is clear from the definitions cosz the (i.e., also have u e From series) that = — sin z, = cosz, cos(— z) we sinz. i power sin(— z) so + the equations for e' z and — ' e cosz — we can c i sin derive the formulas sin 2/ cosz 2 The development of complex power series thus places the exponential function the very core of the development of the elementary functions — it at reveals a con- nection between the trigonometric and exponential functions which was never imagined when these functions were 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 it: if in the formula z e' we take z = tt, we = generally, e^'^" is With these remarks we tions. And cosz + i sinz obtain the remarkable result e (More defined, first yet there are been mentioned. Thus in =-1. an «th root of will power series which have not we have seldom considered power series centered at a. still far, 1.) bring to a close our investigation of complex func- several basic facts about f(z) = J^a n (z-a) n , 27. Complex Power Series — except for a 0. This omission was adopted partly to simplify the exposition. For power series centered at a there are obvious versions of this chapter (the proofs require only (possibly with \z — all the theorems in there trivial modifications): oo is number R a n or "oo") such that the series 2_, a n(z — a) converges absolutely for z n=0 a\ < R, and has unbounded terms for z with \z — a\ > R; moreover, for with all z 565 — a\ < R \z the function 00 f(z)=Y an (z-a) n / has derivative = J^na n (z-a) n - f\z) It is less FIGURE 6 . straightforward to investigate the possibility of representing a function power as a 1 series centered at b, if power already written as a it is series centered at a. If oo f(z) = J2a„(z-a)" ,<=o has radius of convergence is and b /?, a point with is power true that f(z) can also be written as a \b series f iz) = J2b„(: — a\ < R (Figure centered at 6), then it b, bf n=0 numbers b n are (the convergence We at least necessarily R— prove the will not \b — a\ facts several other important facts f [it (n) {b)/n\)\ moreover, this series has radius of may be larger). mentioned we shall f{z)=Y, aniz-a) in the previous paragraph, and there are not prove. For example, n and if <Xz)=YJ b n(Z-b)\ n=0 and g(b) = centered at new and at a, then b. we would expect that All such facts could be ideas, but the proofs would not be The reciprocals of power series. a into one centered at b fog more of computations, we will requires still proved can be written as a power now about sums, products of changing a power series centered more Rather than end skill. series without introducing any basic as easy as the proofs possibility quite a bit is fog involved, and the treatment of this section with a tour deforce instead give a preview of "complex analysis," one of most beautiful branches of mathematics, where all these facts are derived as straightforward consequences of some fundamental results. the Power tions series were introduced which are entiable, it is differentiable. in this chapter in order to provide complex func- Since these functions are actually infinitely differ- natural to suppose that we have therefore selected only a very special 566 Infinite Sequences and Infinite Series complex collection of differentiable show analysis that this If a complex function then it is not at is basic theorems of complex all true: defined in is The functions. some A region of the plane and A automatically infinitely differentiable in . is differentiable in A, A the Moreover, for each point a in a\ Taylor series for These to give a-) facts are f at a among will converge the first f to in any be proved to any idea of the proofs themselves from anything in elementary calculus. — contained in A (Figure 7). complex analysis. It is impossible methods used are quite different in the If these facts are granted, mentioned before can be proved very the facts circle however, then easily. Suppose, for example, that / and g are functions which can be written as power series. Then, as we have shown, / and g are differentiable it then follows from — f + easy general theorems that FIGURE 7 Appealing as power We to the results f g, from complex 1/g and g, • analysis, it fog are also differentiable. follows that they can be written series. know how to compute the power series for f + g, f g and \/g from and g. It is also easy to guess how one would compute an expression / fog as a power series in (z — b) when we are given the power series expansions already those for for oo f(z) = J2 an(z-a) n n=0 = g(z) Y J b k(z-b) k , k=0 with a = g{b) = bo, so that Y h(z-b) g(z)-a = k . J k=\ First of all, we know how to compute the power (g(z)-a) = l r£ b f and this power series will begin with (z — b) k 1 . series (z-b) k \ . Consequently, the coefficient of z" in = f(g(z)) J2cii(g(z)-a) 1 1=0 can be calculated as a powers of g(z) Similarly, — finite sum, involving only coefficients arising from the first n a. if oo f(z) has radius of convergence R). Thus, if b is in A, it /?, is then = J^an (z-a) n / is differentiable in the region possible to write / as a power A = series {z : \z—a\ centered < at b, 567 27. Complex Power Series which will converge in the be f (,1) {b)/n\ will . This circle series R — of radius may — \b The a\. coefficient of z n actually converge in a larger circle, because 00 2^ a n(z — may be a)" the series for a function differentiable in a larger region than A. For example, suppose that f(z) — 1/(1 + Z ). Then / is differentiable, except at i and —i, where it is not defined. Thus f(z) can be written as a power 00 2_, a nZ series ain — n with radius of convergence (—1)" and cik = if k is odd). It is a matter of (as 1 fact, we know that also possible to write oo f {z) = Y^b (Z-\)\ lr n=0 where b n FIGURE series: = f it is 8 {n) VI + As an added will (^)/n\ 2 (^) is z the not at easily predict the radius the distance from ^ to be mentioned, which For real can lies or — i of convergence of (Figure 8). analysis further, quite near the surface, this and which one more will result be found in subject. values of sin z always all true. i complex incentive to investigate any treatment of the this , We . In fact, if z = gi(iy) between lie iy, for _ real, y — 1 and 1 , but for complex z then g—i(iy) sin zy 2/ If y is large, then sin iy is 2/ also large in absolute value. This behavior of sin is typical of functions 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: Liouville's Theorem: The only bounded entire functions are the constant junctions. As a simple application of Liouville's Theorem, consider a polynomial function = z"+a n _ lZ "~ +---+a l f(z) where n > large z, 1, so that so Liouville's , / is not a constant. We already know that Theorem tells us nothing interesting about f(z) is large for /. But consider the function g(z) 1 = f(z) If f(z) were never 0, then g would be entire; since f(z) becomes large for large z, would also be bounded, contradicting Liouville's Theorem. Thus for some z, and we have proved the Fundamental Theorem of Algebra. the function g f(z) = PROBLEMS 1 . Decide whether each of the following verges absolutely. series converges, and whether it con- * 568 Infinite Sequences and Infinite Series n=\ y, + 2/ 1 E oo iv ) E ( + 5 2'')"' — Alogw v 2^ + .„log« n n n=2 Use the ratio test to show that the radius of convergence of each of the power following approach a < limit > to a limit series 1 is if 1 |z| (In . < 1, each case the ratios of successive terms but for |z| > 1 will the ratios will tend to oo or 1.) n=\ n=\ oo in n=\ oo iv oo (v) 3. E 2 "— Use the root test power the following problems to 2 + z y+ oo n Y-z n . n=\ in Z^ nn n=\ oo iv 9 T-z n »=] oo E 2V n=l series. (In Chapter 2 z (i) (Problem 23-9) to find the radius of convergence of each of ' ! - 3 4 2^ + z some cases, 22.) z 5 ? 23 + z 6 3^ + "- you will need limits derived in the 27. Complex Power Series The root test can always be used, in theory at convergence of a power to a least, to find 569 the radius of a close analysis of the situation leads series; in fact, formula for the radius of convergence, known as the "Cauchy-Hadamard formula." Suppose that the set of numbers \f\a n first bounded. is \ 00 (a) Use Problem 23-9 show to that lim ^/\a„\ H—>0O if \z\ < 1, then } a n z" con*-^ «=0 verges. OO (b) Also show that lim >/|a n if | n—*oo \z\ > /a„z" then 1, has *—~' unbounded terms. n=0 00 (c) Parts and (a) show (b) that the radius of convergence of >_,a„z" is #1=0 1/ lim vjflj (where "1/0" means To complete "oo"). the formula, de- II—>oo lim ^/\a„ fine \ = oo the set of if ^/\a„\ all unbounded. Prove is that in n—>-oo oo this case, (which is 5. 2_, a nZ n diverges for z may be 00 e Show E-"- n=\ n=\ 9. converges everywhere on the unit circle; that the nowhere on the unit circle; and that the second series one point on the unit circle and diverges for at least circle. e w — e w . + that sin(z w— w) = w for sin z sin sin z cos all w + complex z cos z sin w and some real number y. x+iy x e for real |e = and (b) Prove that (a) Prove that exp takes on every complex value except (b) Prove that sin takes on every complex value. | jc (i) f(z) (ii) f(z) = = tan z. z(\ -z)- ]/2 . + w) = 1 can be written e iy y. For each of the following functions, compute the the Taylor series centered at cos(z and w. Prove that every complex number of absolute value for 8. 00 „ z+w for all complex numbers z and w by showing z+w is the Cauchy product that the infinite series for e of the series for e z cos z cos (a) 00 „ 2: E-. least Prove that e z and 7. from Problem £%• one point on the unit (b) of convergence first series converges for at (a) series n=l third series converges 6. 0, so that the radius considered as "l/oo"). Consider the following three Prove that the ^ first 0. three nonzero terms of by manipulating power series. 570 Infinite Sequences and Infinite Series ,sin z = iii) f(z) iv) /(z)=log(l f(z) = sin v) f(z) = sin(r) vi) f(z) = vii) ). y~ z z cos z 4 = /(7) 3' -2z 2 + : viii) -z 2 l[^<V^I-l)_ !j complex function / as / = u + /u, where u and u are real-valued. Let u and D denote the restrictions of u and v to the real numbers. In other words, u{x) = u{x) for real numbers x (but it is not defined for other x). Using Problem 26-9, show Suppose 10. that we write a differentiable we have that for real x f'(x) where = complex /' denotes the + u'(x) iv'ix), and derivative, while u' v' denote the ordinary derivatives of these real-valued functions on R. (b) Show, more generally, that (k f \x) (c) Suppose that / satisfies / (*) w u {k) (x) + iv a) (x). the equation +fl»-l/ w| - 1, + -" + flO/ = I where the a, are real numbers, and where the f [k) denote higher-order complex derivatives. Show that u and v satisfy the same equation, where (k) u and v (k) now denote higher-order derivatives of real -valued functions (d) onR. Show that ...-(- aq if — 0, solutions of 11. (a) (b) Show that is l a \ (*). exp w^ Given complex root of the equation z" + an - z."~ + then f(x) = e bx sin ex and f(x) = e bx coscx are both a =b-\- ci 0, is not show one-one on C. that e z =w if and only if z (here log denotes the real logarithm function), '(c) Show = x + iy = with x log | w \ and y an argument of w. that there does not exist a continuous function log defined for nonzero complex numbers, such that exp(log(z)) (Show that log cannot even Since there is no way be defined continuously to define a = Z for all j for |z| = 1 / 0. .) continuous logarithm function we can- not 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. And , . 571 27. Complex Power Series complex numbers a and b we define a b to be a set of complex numbl ° sa bers, namely the set of all numbers e or, more precisely, the set of bz all numbers e where z is a logarithm for a for (d) If m is an then a m consists of only one number, the one given by integer, the usual elementary definition of a'" (e) (f) (g) (h) . and n are integers, then the set a m n coincides with the set of values given by the usual elementary definition, namely the set of all b m where b is an nth root of a. b If a and b are real and b is irrational, then a contains infinitely many members, even for a > 0. Find all logarithms of /, and find all values of /'. By (a b ) c we mean the set of all numbers of the form z c for some number z If m ' in the set a b Show . many that (1')' has infinitely 1' values, while ' has only one. 12. (i) Show that all values of a bc (a) For real x show that (It (b) will The are also values of (a we can choose log(.t + / ) = log( v logU' — i) = log(\/l 1 + i) log(x + x 2 + I— — + x2 — help to note that tt/2 : — I i ) (a and log(x — /) to > 0.) ) Is . arctan x ) — — arctan x 1 i ) a bc = b c = arctan x b c ) n(a c ) b ? be . arctan \/x for x expression 1/1 1 +x 1 2 \x — 2/ 1 x i + i yields, formally, / Use part 13. (a) A sequence lim real. are (b) (c) (a) — \a„, + x2 1 to = ^IJogC* check that {a„} a„\ -0 - log(* +i)J. 2/ this answer agrees with the usual one. of complex numbers = 0. Prove that {a„} Suppose a is that a n called a bn + if Cauchy sequence icn , where b n and and only if {b n } if c„ are and {c„} Cauchy sequences. Prove that every Cauchy sequence of complex numbers converges. Give direct proofs, without using theorems about real has the same sum. is that an permitted, and in fact advisable, to use the proofs (It is series.) Prove that " 1 X^ e ikx = J*1 A=1 series, convergent and that any rearrangement of the corresponding theorems for real (a) — Cauchy sequence absolutely convergent series 14. is - e inx - - e ix sin -a V2 / ( ) x sin 2 J(n+l)x/2 * 572 Infinite . Sequences and Infinite Series (b) Deduce \ the formulas for and 2_\ cos kx k=\ m kx that are given in k=\ Problem 15-33. 15. = a2 = Let {a„} be the Fibonacci sequence, a\ = If r„ (b) Show that (c) Prove that the limit does - r„ , = if r < 1 = an+ i/an show that rn +\ (a) and lim < r„ r„ +2 + l/r„. r = + 1 then r„ exists, 1, 1 1 + a n+ a„ /r, so that r < Hint: If r„ exist. — a„+2 + v5 (1 = \. (l + v5 )/2. )/2, then — 2 r,, . oo (d) Show has radius of convergence 2/(1 2_, a nZ" that + V5). (Using n=\ oo the unproved theorems in this chapter and the = " 2_, a n z fact that n=\ — l/(z +z — from Problem 24-16 we could have predicted that the 1) radius of convergence — Z 1 = + + 16. Since (e z which v5)/2. (—1 ± v5)/2, the radius of convergence Notice that number this is indeed equal to 75).) — \)/z can be written as the nonzero is + the smallest absolute value of the roots of z 0; since the roots are should be (— 1 2/(1 is at 0, it power follows that there is a series 1 power + z/2! + z /3! + • • • series n=0 with nonzero radius of convergence. Using the unproved theorems in we can even chapter, predict the radius of convergence; the smallest absolute value of the non-zero e z — =0. The 1 numbers numbers z it is = this 2tt, since this 2kni b„ appearing here are called the is for which Bernoulli numbers.* (a) Clearly bo — 1 . Now show that z e z - 2 1 +\ e -z-\ b | z 2 e : e z + - 1 1 + e z-\ e~ z and deduce + ez 1 that = —^ b„ , = if n is odd and n > 1 _ if n *Sometimes the numbers Bn = ( — 1)" »2« arc called the Bernoulli numbers, because />„ = is odd and > (see part (a)) and because the numbers bin alternate in sign, although we will not 1 prove this. ( )ihei modifications of this nomenclature are also in use. 27. Complex Power Series By (b) 573 finding the coefficient of z" in the right side of the equation H £ £-•% oo show ,3 ,1 + £ + fr + that This formula allows us to compute any bk in terms of previous ones, and shows that each i>2 Part *(c) (a) = rational. Calculate is *4 \, b 2n Show "2 Show — = -^ z e" 2 + e~^ 2 2 e*/2 - "" ^/ 2 oo , £^<-'>" 22 "; 2 "- n=0 = cot z — 2 cot 2z. that tanz = , Y a- '"( Bernoulli 1)- 2 2 "(2 2 " 1 l); 2 "" 1 . \ (This series converges for The ^8 43. that oo 17. = by 2/z and show that z tan z *(e) e'+l ^-1 z „ 2n 2-* (2n)!* *(d) ^6 ~^J. shows that yReplace = two or three of the following: \z\ < tc/2.) numbers play an important role in a theorem which is best D to denote the "differThen D k f will mean f and introduced by some notational nonsense. Let us use entiation operator," so that Df denotes /'. (k) oo e D f but will it mean /_\f ( ' n=0 will make sense !l /n\ (of course this series if / is makes no sense in general, a polynomial function, for example). Finally, A denote the "difference operator" for which Af(x) = f(x + 1) — f(x). Now Taylor's Theorem implies, disregarding questions of convergence, that let f(x+\) = J2 f w (x) n=0 or (*) f(x+\)-f(x) = J2 n=\ f W (x) , 574 Infinite Sequences and Infinite Series we may Af = write this symbolically as D Even more symbolically "identity operator." which suggests (e — 1) where /', stands for the 1 can be written this A= e D — 1 that D Thus we obviously ought have to oo D= , Y^D A. k fc=0 i.e. oo . bi k=0 The (a) beautiful thine" about Prove that (**) the infinite a formula for f Deduce from \x + 1 ) of —f (x); f ^{x) that it works! a polynomial function By applying (in (*) which case to f (k \ find then use the formula in Problem 16(b) in the right side of (**). { (**) that 00 /'(O) is is really a finite sum). Hint: is (k to find the coefficient (b) / literally true if is sum nonsense all this '.-> + • • u + f{n) = ]T -£U (k) (n + 1 - / w (0)]. ) k=0 (c) Show g(0) (d) + any polynomial function g we have that for ... Apply + g(n)= [' g(x) this to = Tk' = £-*> Using the The first Y ^[g [ - ]) —+ ^r^(k\k-\ p i + \ ' fact that b\ (n + \) - (k g - ]) (0)\. = — i, p-k+] v k=\ show that ten instances of this formula which offered {k x p to show that P k=\ + g (t)dt were written out in Problem as a challenge the discovery of the general pattern. may now seem to this way! After writing out these 10 formulas, Bernoulli claims (in his posthumously printed Ars Conjectandi, 1713): may be This be a preposterous suggestion, but the Bernoulli num- bers were actually discovered in precisely larity 2-7, "Whoever will examine the able to continue the table." He work series as to their regu- then writes down the above . . 575 27. Complex Power Series formula, offering no proof at merely noting that the all, (which he denoted simply by A, B, C, lem power the *18. The 16(b). The formula Problem some new (a) The coefficients bk equation in Prob- sum must be where g to the case replaced by a an expression coefficients in Euler. finite for the remainder, not a is sum plus a it is useful functions. Bernoulli polynomials are defined by <p n =^\V) bn-kX fc=0 first ) can be generalized to find <Pn(x) The satisfy the was discovered by 1) infinite remainder term. In order to introduce — 17(c) polynomial function; the . between these numbers and the relation series for z/(e z in . k . ' ^ three are 1 cpi(x) =x - (p~>(x) = x~2 — x + -, ! -, o 3x , (p } Show (x) x =x --rT+y: that (p„(0) <p n (\) (pn '{x) <pn (x) =bn = , bn if // > 1 =n<p„-i(x), = (-l) n <pn -x). Hint: Prove the last equation by induction (b) Ru Let k be the remainder term {x) interval [x x , (*) + 1 ] , f«\x + on n. in Taylor's Theorem for (k f \ on so that i) - k f \x) = Y l—r^ + R » k (- X) - n=\ Prove that k=0 Hint: Imitate (c) Use k=0 ' Problem the integral 17(a). Notice the subscript form of the remainder ££*»VM = /' to show N—k that +1 ^±f^/<-'>«*. on R. the ) 576 Infinite Sequences and Infinite Series Deduce (d) + g(x+ g(x) = Summation Formula": the "Euler-Maclaurin \) + f --- + g(x+n) g(t)dt Jx + J2^[g lk - l) (x+n + - 1) {k - ]] g (x)] + SN (x,n), ' k=\ where + ; +1 <p . Let (e) \jjn = since <p„(l) Show 1. iN) g (ndt. X *x+j A J (Part and ^,,(0), (a) implies that also that \(/ n is n if even which 1, > 1, if n then satisfies \[/„(t) %//„ is even and odd is = continuous, if n is that r.x+n- SN (x,n) -{) " • be the periodic function, with period ^(f)forO<? < odd.) +j+\ ';,' N (X — =- N\ {N) g {t)dt c+fl+l + " +l ( = (-l) N+l r "^~-g^ N) {t)dt if x is an integer! Unlike the remainder in Taylor's Theorem, the remainder Sn(x,h) usually does not become satisfy lim Sn(x,h) = 0, because the Bernoulli numbers and functions large very rapidly (although the first few examples do not suggest Nevertheless, important information can often be obtained from the formula. The general situation is this). summation best discussed within the context of a specialized study ("asymptotic series"), but the next problem shows one particularly important example. **19. (a) log 1 N= Use the Euler-Maclaurin Formula, with + • • • + log(n — 2, to show that 1 /*" 1/1 \ 2 (t) f\ = / U «""-2 l08 " + 12U- 1 j + y. "2^ 1 (b) Show that i0 (c) \lr Sy n n+l/2 e -n+l/a2n)J- Explain why the improper integral that if a = exp(fi + 1 ft + = It 2 J ] dt xj/2(t)/2t I 1/12), then n\ ttts \2 \ -rrr^r-r a n n+l/2 e -n+l/a2n) J =- r / J init) _ 2t , at . exists, and show . 27. Complex Power Series (d) Problem 19-4 1(d) shows 577 that 2 {n\) 2 2n V^= n^-oc (ln)\y/n i Use part (c) show to that a 2 n 2n+\ e -2n22n = s/tt lim a(2n) 2n+l / 2 e- 2n s/n n ^°° and conclude (e) Show = a that \J1ti ' . that 1/2 <f>2(t)dt / = (You can do the computations mediately from Problem f \J/(x) tfr(n) — attention to i/2(t)dt of (f) <p2 (Figure Noting that for its 0. Conclude that > forO <0 for Graph Hint: all n. but the result also follows im- explicitly, 18(a).) Jo with = (p2(t)dt / Jo Jo values at xq, I, and x\, <x < 1/2 1/2<jc< 1, \js on [0, 1], paying particular where xq an d x\ are the roots 9). i}/{x) = f(x) —\j/{\ = — x), show that f(f)dt > / on [0, 1], Jo and hence everywhere, with (g) Finally, use this \[r(n) = f 2t 2 Jn Using the all n information and integration by parts to show that t2it) (h) for fact that the dt maximum > 0. value of |<^2U)I for x in conclude that o<rm«< / It Jn (i) Finally, 2 ' \2n conclude that ^n" +l/2 e-" < n\ < >/2^/i" +,/ V" + 1/,l2 " ) . [0, 1] is g, 578 Infinite Sequences and Infinite Series The n\ is of Problem 19, a strong form of Stirling's Formula, shows that approximately s/lix n" + / e~", in the sense that this expression differs from n\ final result to n when n is large. For example, for we obtain 3598696 instead of 3628800, with an error < 1%. A more general form of Stirling's Formula illustrates the "asymptotic" nature of the summation formula. The same argument which was used in Problem 19 can now be used to show that for N > 2 we have by an amount which n = is small compared 10 Y b +r )=r^k(k-\)n J -' io g k- \ 1 n Since i/f/v is bounded, we can obtain estimates of the form rr dt Nt N , If A^ 1_/v 77 is large, the constant will make Mm will also /2 e- be large; but for very large n the factor expf a very bad approximation for depends on N) ~ nN~v the product very small. Thus, the expression ^n" +l may be Nt N it will n\ V when ^-— ) n is small, but for large n (how large be an extremely good one (how good depends on N). PART EPILOGUE 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 Foundation. JONATHAN SWIFT . CHAPTER AiV Throughout FIELDS book a conscientious attempt has been made this portant concepts, even terms like "function," for often considered sufficient. But which an to define all Q and R, the two main protagonists of have only been named, never defined. What im- intuitive definition is this story, has never been defined can never be analyzed thoroughly, and "properties" P1-P13 must be considered assumptions, not theorems, about numbers. Nevertheless, the term "axiom" has been purposely avoided, and in this chapter the logical status of P1-P13 will be scrutinized more carefully. Q and Like talk about from a all R, the sets N Z and have also remained undefined. True, some four was inserted in Chapter 2, but those rough descriptions are far definition. To say, for N example, that consists of 1, merely 2, 3, etc., names some elements of N without identifying them (and the "etc." is useless). natural numbers can be defined, but the procedure is involved and not quite The pertinent to the rest of the book. to this The Suggested Reading list contains references problem, as well as to the other steps that are required develop calculus from its The basic logical starting point. one wishes if further to development program would proceed with the definition of Z, in terms of N, and the in terms of Z. This program results in a certain well-defined set Q_, certain explicitly defined operations + and •, and properties P1-P12 as theorems. The final step in this program is the construction of R, in terms of It is this last construction which concerns us. Assuming that Q_ has been defined, and that PI -PI 2 have been proved for Q, we shall ultimately define R and prove all of this definition of Q Q of P1-P13 for R. Our means that we must define not only and multiplication of real numbers. Indeed, the intention of proving PI -PI 3 bers, but also addition bers are of interest only as a set together with these operations: real real how numnum- 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 number systems of this book. This modern mathematics incorporates the as special cases the three important extraordi- narily important abstraction of properties P1-P9 common to Q_, soever), together with rules which associate in F) for A field is + b) + c — a + (a (2) There (i) (ii) a set F + and two "binary operations" and b to elements a in which the following conditions are (1) 581 R, and C. (b + c) is some element a + =a in for F all (of objects of • defined on F, other elements a any F sort what- is, two and a • b (that +b satisfied: a, b, and c in F. such that for all a in F, for every a in F, there is some element b in F such that a + b — 0. 582 Epilogue +b=b+a a (3) (4) (a b) • There (5) =a c • a* (i) =a 1 a.fc= (6) a * (7) a • The + b and why in F for a • all a and b b +a examples of 1^0 such that / with a 0, there and some element b is in F such that in F. for c • all and a, b, fields are, as mean are simply restatements of PI c in F. Q, R, and already indicated, + and • these are fields, but the explanation are understood to • F c in F. 1. a — 1 in being the familiar operations of • and » + c) (/? familiar explain + = b and for all a in F, For every a (ii) in F. for all a, b, c) • some element is and b for all a (b • + the ordinary probably unnecessary to It is . is, at and any , • C, with rate, quite brief. the rules When (1), (3), (4), (6), (7) P4, P5, P8, P9; the elements which play the role of , and 1 are the numbers and 1 (which accounts for the choice of the symbols 0, 1); and the number b in (2) or (5) is —a or a respectively. (For this reason, in an arbitrary field F we denote by —a the element such that a + (—a) — 0, and by , the element such that a a In addition to Q_, R, a • -1 = ^ for a 1, and C, there are 0.) several other fields which can be described numbers a + b\2 for a, b in Q. The operations + and will, once again, be the usual + and for real numbers. It is necessary to point out that these operations really do produce new elements One example easily. is the collection F\ of all • • of Fi: + bV2) + (a + bVl)- (a Conditions all real 1 = 1 (c (1), (3), (4), (6), (7) + c) + (b + d)\fl, which in F\\ which in (ac + 2bd) + (bc + ad)V2, (a the (2) for a field are for all is fi in F\, difficult point. If a F\. obvious for F\: since these hold for real numbers of + b\2. = a + b\2 form a the number = 0+0v 2 is in F\ and, for a = (—a) + (—b)w2 in F\ satisfies a + ji = 0. Similarly, so (5i) is satisfied. The verification of (5ii) is the only slighdy holds because the number +0v2 is is numbers, they certainly hold Condition in F\ + dV2) = + dy/2) = (c + b\f2 / 0, a then —== X + bV2 1; a+b^/2 it is therefore necessary to _ 1 a+bV2 (The division by a only a if a = + by/l^ The b — that \/{a + bv2 a-bs/2 — b\2 (since is ) _ is in F\. is irrational) which -2b is is true because (-b) 2 valid because the relation a v2 This a a2 (a-bV2)(a+bV2) a — 2 r- -2b by/2 = 2 could be true ruled out by the hypothesis 0.) next example of a tains only show field, Fj, is considerably simpler in one respect: two elements, which we might as well denote by and 1. The it con- operations 583 28. Fields + and • are described by the following tables. 01 + .01 1 1 1 The verification of conditions (1) (7) are straightforward, case-by-case checks. For example, condition be written Our — or = — 1. 1 may (1) setting a, b, c in 1 1 be proved by checking the 8 equations obtained by Notice that in 1. example of a field is rather R, and + and • are defined by final (a, a) + (a, a) + (The for R.) and The = (b, b) = (b, b) may also consists of all pairs (a, a) for a Ft, silly: equation 0; this (a + b,a + b), (a • b,a b). • appearing on the right side are ordinary addition and multiplication • verification that F3 A detailed • +1= this field 1 is a field you to is left investigation of the properties of fields is as a simple exercise. a study in but for our itself, purposes, fields provide an ideal framework in which to discuss the properties of numbers most economical way. For example, the consequences of P1-P9 in the which were derived for "numbers" in particular, they are true for the fields Notice that certain For example, C — the assertion Q fields Q, common properties of Q, — b b —a + 1/0 1 and R, however, any actually hold for 1 field; in R, and C. possible for the equation it is consequently a Chapter R, and + 1 1 = C do not hold for to hold in some = does not necessarily imply that a all fields. fields, For the b. and field was derived from the explicit description of C; for the was derived from further properties which the conditions for a field. There is a related concept this assertion do not have analogues in which does use these properties. An ordered field and •) together with a certain subset P of F (the is a field F (with operations + "positive" elements) with the following properties: (8) For all (i) a = 0, (ii) a is in P, —a (iii) (9) If The is and b are in P, then a have already seen that the field F2, implies that 1 consisting of true: +b • b is field is in P. in P. G cannot be made + all 1 = (8), is applied to in numbers a 1 — — 1, P, contradicting + shows that (8). /?v2 with a,b in into made with only two elements, likewise cannot be condition field: in fact, is in P. a and b are in P, then a (10) If a We a in F, one and only one of the following 1 an ordered into must be field. an ordered in P; then (9) On the other hand, the field F\ Q, certainly can be made , into . 584 . Epilogue an ordered (in field: let P be the the ordinary sense). description of P The set of + b\/7. a all field Ft, which are positive real can also be made into an ordered numbers field; the to you. is left natural to introduce notation for an arbitrary ordered field which corre- It is sponds to that used for Q and R: > < < > a a a a we define b if a —b b if b b if a b if a > < > is a, b or a b or a Using these definitions we can reproduce, definitions of A Chapter in P, for = = b, b. an arbitrary ordered field F, the 7: A of elements of F is bounded above if there is some x in F such that x > a for all a in A Any such x is called an upper bound for A An element x of F is a least upper bound for A if x is an upper bound for A and x < y for every y in F which is an upper bound for A set . Finally, last it is . possible to state an analogue of property PI 3 for R; this leads to the abstraction of this chapter: A complete which The is ordered field is an ordered field in which every nonempty bounded above has a least upper bound. consideration of fields may seem constructing the real numbers. However, means of formulating Is 1 2. Is Our now provided with an There are two questions which will intelligible be answered there a complete ordered field? there only one complete ordered field? starting point for these considerations will field, containing N and Z as certain subsets. necessary to assume another fact about Let x be an element of in have taken us far from the goal of are remaining two chapters: in the dered this goal. to we set N such that nx > Q with x Q, assumed be At one to be an or- crucial point it will be Q: > 0. Then for any y in Q_ there is some n y. This assumption, which asserts that the rational numbers have the Archimedean property of the real numbers, does not follow from the other properties of an ordered field (for the example of the Suggested Reading). that demonstrates this conclusively see reference [14] The important point for us constructed, properties PI -PI 2 appear as theorems, is that and when Q is explicitly so does this additional 28. Fields assumption; if we really 585 began from the beginning, no assumptions about Q_ would be necessary. PROBLEMS Let 1. F be the set {0, 1, 2} (The rule tables. in the usual way, = 4 = 3+1, 2-2 and define operations for constructing these tables + and so 2 2 1 F is a add or multiply 1 2 that by the following 2-2=1.) 1 Show F possible multiple of 3; thus 1 1 on as follows: is and then subtract the highest • field, and prove that it 1 2 2 1 made cannot be into an ordered field. Suppose now 2. 2, we that