TRENCH_REAL_ANALYSIS

INTRODUCTION TO REAL ANALYSIS William F. Trench Andrew G. Cowles Distinguished Professor Emeritus Department of Mathematics Trinity University San Antonio, Texas, USA [email protected] This book has been judged to meet the evaluation criteria set by the Editorial Board of the American Institute of Mathematic s in connection with the Institute’s Open Textbook Initiative . It may be copied, modified, redistributed, translated, and built u pon sub- ject to the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported Lice nse. FREE DOWNLOADABLE SUPPLEMENTS FUNCTIONS DEFINED BY IMPROPER INTEGRALS THE METHOD OF LAGRANGE MULTIPLIERS Library of Congress Cataloging-in-Publication Data Trench, William F. Introduction to real analysis / William F. Trench p. cm. ISBN 0-13-045786-8 1. Mathematical Analysis. I. Title. QA300.T667 2003 515-dc21 2002032369 Free Hyperlinked Edition 2.04 December 2013 This book was published previously by Pearson Education. 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TO BEVERLY Contents Preface vi Chapter 1 The Real Numbers 1 1.1 The Real Number System 1 1.2 Mathematical Induction 10 1.3 The Real Line 19 Chapter 2 Differential Calculus of Functions of One Variable 30 2.1 Functions and Limits 30 2.2 Continuity 53 2.3 Differentiable Functions of One Variable 73 2.4 L’Hospital’s Rule 88 2.5 Taylor’s Theorem 98 Chapter 3 Integral Calculus of Functions of One Variable 113 3.1 Definition of the Integral 113 3.2 Existence of the Integral 128 3.3 Properties of the Integral 135 3.4 Improper Integrals 151 3.5 A More Advanced Look at the Existence of the Proper Riemann Integral 171 Chapter 4 Infinite Sequences and Series 178 4.1 Sequences of Real Numbers 179 4.2 Earlier Topics Revisited With Sequences 195 4.3 Infinite Series of Constants 200 iv Contents v 4.4 Sequences and Series of Functions 234 4.5 Power Series 257 Chapter 5 Real-Valued Functions of Several Variables 281 5.1 Structure of RRRn281 5.2 Continuous Real-Valued Function of nVariables 302 5.3 Partial Derivatives and the Differential 316 5.4 The Chain Rule and Taylor’s Theorem 339 Chapter 6 Vector-Valued Functions of Several Variables 361 6.1 Linear Transformations and Matrices 361 6.2 Continuity and Differentiability of Transformations 37 8 6.3 The Inverse Function Theorem 394 6.4. The Implicit Function Theorem 417 Chapter 7 Integrals of Functions of Several Variables 435 7.1 Definition and Existence of the Multiple Integral 435 7.2 Iterated Integrals and Multiple Integrals 462 7.3 Change of Variables in Multiple Integrals 484 Chapter 8 Metric Spaces 518 8.1 Introduction to Metric Spaces 518 8.2 Compact Sets in a Metric Space 535 8.3 Continuous Functions on Metric Spaces 543 Answers to Selected Exercises 549 Index 563 Preface This is a text for a two-term course in introductory real anal ysis for junior or senior math- ematics majors and science students with a serious interest in mathematics. Prospective educators or mathematically gifted high school students ca n also benefit from the mathe- matical maturity that can be gained from an introductory rea l analysis course. The book is designed to fill the gaps left in the development of calculus as it is usually presented in an elementary course, and to provide the backgr ound required for insight into more advanced courses in pure and applied mathematics. The s tandard elementary calcu- lus sequence is the only specific prerequisite for Chapters 1 –5, which deal with real-valued functions. (However, other analysis oriented courses, suc h as elementary differential equa- tion, also provide useful preparatory experience.) Chapte rs 6 and 7 require a working knowledge of determinants, matrices and linear transforma tions, typically available from a first course in linear algebra. Chapter 8 is accessible after completion of Chapters 1–5. Without taking a position for or against the current reforms in mathematics teaching, I think it is fair to say that the transition from elementary co urses such as calculus, linear algebra, and differential equations to a rigorous real anal ysis course is a bigger step to- day than it was just a few years ago. To make this step today’s s tudents need more help than their predecessors did, and must be coached and encoura ged more. Therefore, while striving throughout to maintain a high level of rigor, I have tried to write as clearly and in- formally as possible. In this connection I find it useful to ad dress the student in the second person. I have included 295 completely worked out examples t o illustrate and clarify all major theorems and definitions. I have emphasized careful statements of definitions and theo rems and have tried to be complete and detailed in proofs, except for omissions left t o exercises. I give a thorough treatment of real-valued functions before considering vec tor-valued functions. In making the transition from one to several variables and from real-v alued to vector-valued functions, I have left to the student some proofs that are essentially re petitions of earlier theorems. I believe that working through the details of straightforwar d generalizations of more elemen- tary results is good practice for the student. Great care has gone into the preparation of the 761 numbered e xercises, many with multiple parts. They range from routine to very difficult. Hi nts are provided for the more difficult parts of the exercises. vi Preface vii Organization Chapter 1 is concerned with the real number system. Section 1 .1 begins with a brief dis- cussion of the axioms for a complete ordered field, but no atte mpt is made to develop the reals from them; rather, it is assumed that the student is fam iliar with the consequences of these axioms, except for one: completeness. Since the diffe rence between a rigorous and nonrigorous treatment of calculus can be described largely in terms of the attitude taken toward completeness, I have devoted considerable effort to developing its consequences. Section 1.2 is about induction. Although this may seem out of place in a real analysis course, I have found that the typical beginning real analysi s student simply cannot do an induction proof without reviewing the method. Section 1.3 i s devoted to elementary set the- ory and the topology of the real line, ending with the Heine-B orel and Bolzano-Weierstrass theorems. Chapter 2 covers the differential calculus of functions of o ne variable: limits, continu- ity, differentiablility, L’Hospital’s rule, and Taylor’s theorem. The emphasis is on rigorous presentation of principles; no attempt is made to develop th e properties of specific ele- mentary functions. Even though this may not be done rigorous ly in most contemporary calculus courses, I believe that the student’s time is bette r spent on principles rather than on reestablishing familiar formulas and relationships. Chapter 3 is to devoted to the Riemann integral of functions o f one variable. In Sec- tion 3.1 the integral is defined in the standard way in terms of Riemann sums. Upper and lower integrals are also defined there and used in Section 3.2 to study the existence of the integral. Section 3.3 is devoted to properties of the integr al. Improper integrals are studied in Section 3.4. I believe that my treatment of improper integ rals is more detailed than in most comparable textbooks. A more advanced look at the exist ence of the proper Riemann integral is given in Section 3.5, which concludes with Lebes gue’s existence criterion. This section can be omitted without compromising the student’s p reparedness for subsequent sections. Chapter 4 treats sequences and series. Sequences of constan t are discussed in Sec- tion 4.1. I have chosen to make the concepts of limit inferior and limit superior parts of this development, mainly because this permits greater fle xibility and generality, with little extra effort, in the study of infinite series. Section 4.2 provides a brief introduction to the way in which continuity and differentiability can be s tudied by means of sequences. Sections 4.3–4.5 treat infinite series of constant, sequenc es and infinite series of functions, and power series, again in greater detail than in most compar able textbooks. The instruc- tor who chooses not to cover these sections completely can om it the less standard topics without loss in subsequent sections. Chapter 5 is devoted to real-valued functions of several var iables. It begins with a dis- cussion of the toplogy of Rnin Section 5.1. Continuity and differentiability are discu ssed in Sections 5.2 and 5.3. The chain rule and Taylor’s theorem a re discussed in Section 5.4. viii Preface Chapter 6 covers the differential calculus of vector-value d functions of several variables. Section 6.1 reviews matrices, determinants, and linear tra nsformations, which are integral parts of the differential calculus as presented here. In Sec tion 6.2 the differential of a vector-valued function is defined as a linear transformatio n, and the chain rule is discussed in terms of composition of such functions. The inverse funct ion theorem is the subject of Section 6.3, where the notion of branches of an inverse is int roduced. In Section 6.4. the implicit function theorem is motivated by first considering linear transformations and then stated and proved in general. Chapter 7 covers the integral calculus of real-valued funct ions of several variables. Mul- tiple integrals are defined in Section 7.1, first over rectang ular parallelepipeds and then over more general sets. The discussion deals with the multip le integral of a function whose discontinuities form a set of Jordan content zero. Section 7 .2 deals with the evaluation by iterated integrals. Section 7.3 begins with the definition o f Jordan measurability, followed by a derivation of the rule for change of content under a linea r transformation, an intuitive formulation of the rule for change of variables in multiple i ntegrals, and finally a careful statement and proof of the rule. The proof is complicated, bu t this is unavoidable. Chapter 8 deals with metric spaces. The concept and properti es of a metric space are introduced in Section 8.1. Section 8.2 discusses compactne ss in a metric space, and Sec- tion 8.3 discusses continuous functions on metric spaces. Corrections–mathematical and typographical–are welcome and will be incorporated when received. William F. Trench [email protected] Home: 659 Hopkinton Road Hopkinton, NH 03229 CHAPTER 1 The Real Numbers IN THIS CHAPTER we begin the study of the real number system. T he concepts discussed here will be used throughout the book. SECTION 1.1 deals with the axioms that define the real numbers , definitions based on them, and some basic properties that follow from them. SECTION 1.2 emphasizes the principle of mathematical induc tion. SECTION 1.3 introduces basic ideas of set theory in the conte xt of sets of real num- bers. In this section we prove two fundamental theorems: the Heine–Borel and Bolzano– Weierstrass theorems. 1.1 THE REAL NUMBER SYSTEM Having taken calculus, you know a lot about the real number sy stem; however, you prob- ably do not know that all its properties follow from a few basi c ones. Although we will not carry out the development of the real number system from t hese basic properties, it is useful to state them as a starting point for the study of real a nalysis and also to focus on one property, completeness, that is probably new to you. Field Properties The real number system (which we will often call simply the reals ) is first of all a set fa;b;c;:::gon which the operations of addition and multiplication are d efined so that every pair of real numbers has a unique sum and product, both r eal numbers, with the following properties. (A)aCbDbCaandabDba(commutative laws). (B).aCb/CcDaC.bCc/and.ab/cDa.bc/ (associative laws). (C)a.bCc/DabCac(distributive law). (D) There are distinct real numbers 0and1such thataC0Daanda1Dafor alla. (E) For eachathere is a real number /NULasuch thataC./NULa/D0, and ifa¤0, there is a real number 1=asuch thata.1=a/D1. 1 2 Chapter 1 The Real Numbers The manipulative properties of the real numbers, such as the relations .aCb/2Da2C2abCb2; .3aC2b/.4cC2d/D12acC6adC8bcC4bd; ./NULa/D./NUL1/a; a./NULb/D./NULa/bD/NULab; and a bCc dDadCbc bd.b;d¤0/; all follow from (A) –(E). We assume that you are familiar with these properties. A set on which two operations are defined so as to have properti es(A) –(E) is called a field. The real number system is by no means the only field. The rational numbers (which are the real numbers that can be written as rDp=q, wherepandqare integers and q¤0) also form a field under addition and multiplication. The simp lest possible field consists of two elements, which we denote by 0and1, with addition defined by 0C0D1C1D0; 1C0D0C1D1; (1.1.1) and multiplication defined by 0/SOH0D0/SOH1D1/SOH0D0; 1/SOH1D1 (1.1.2) (Exercise 1.1.2 ). The Order Relation The real number system is ordered by the relation <, which has the following properties. (F) For each pair of real numbers aandb, exactly one of the following is true: aDb; a<b; orb<a: (G) Ifa<b andb<c , thena<c . (The relation <istransitive .) (H) Ifa<b , thenaCc<bCcfor anyc, and if0<c , thenac<bc . A field with an order relation satisfying (F)–(H) is an ordered field. Thus, the real numbers form an ordered field. The rational numbers also form an ordered field, but it is impossible to define an order on the field with two elements defi ned by ( 1.1.1 ) and ( 1.1.2 ) so as to make it into an ordered field (Exercise 1.1.2 ). We assume that you are familiar with other standard notation connected with the order relation: thus, a >b means thatb < a ;a/NAKbmeans that either aDbora >b ;a/DC4b means that either aDbora < b ; the absolute value of a, denoted byjaj, equalsaif a/NAK0or/NULaifa/DC40. (Sometimes we call jajthemagnitude ofa.) You probably know the following theorem from calculus, but w e include the proof for your convenience. Section 1.1 The Real Number System 3 Theorem 1.1.1 (The Triangle Inequality) Ifaandbare any two real numbers ; then jaCbj/DC4jajCjbj: (1.1.3) Proof There are four possibilities: (a) Ifa/NAK0andb/NAK0, thenaCb/NAK0, sojaCbjDaCbDjajCjbj. (b) Ifa/DC40andb/DC40, thenaCb/DC40, sojaCbjD/NULaC./NULb/DjajCjbj. (c) Ifa/NAK0andb/DC40, thenaCbDjaj/NULjbj. (d) Ifa/DC40andb/NAK0, thenaCbD/NULjajCjbj. Eq.1.1.3 holds in cases (c)and(d), since jaCbjD( jaj/NULjbjifjaj/NAKjbj; jbj/NULjajifjbj/NAKjaj: The triangle inequality appears in various forms in many con texts. It is the most impor- tant inequality in mathematics. We will use it often. Corollary 1.1.2 Ifaandbare any two real numbers ;then ja/NULbj/NAKˇˇjaj/NULjbjˇˇ (1.1.4) and jaCbj/NAKˇˇjaj/NULjbjˇˇ: (1.1.5) Proof Replacingabya/NULbin (1.1.3 ) yields jaj/DC4ja/NULbjCjbj; so ja/NULbj/NAKjaj/NULjbj: (1.1.6) Interchanging aandbhere yields jb/NULaj/NAKjbj/NULjaj; which is equivalent to ja/NULbj/NAKjbj/NULjaj; (1.1.7) sincejb/NULajDja/NULbj. Since ˇˇjaj/NULjbjˇˇD(jaj/NULjbjifjaj>jbj; jbj/NULjajifjbj>jaj; (1.1.6 ) and ( 1.1.7 ) imply ( 1.1.4 ). Replacingbby/NULbin (1.1.4 ) yields ( 1.1.5 ), sincej/NULbjD jbj. Supremum of a Set A setSof real numbers is bounded above if there is a real number bsuch thatx/DC4b wheneverx2S. In this case, bis an upper bound ofS. Ifbis an upper bound of S, then so is any larger number, because of property (G) . Ifˇis an upper bound of S, but no number less than ˇis, thenˇis asupremum ofS, and we write ˇDsupS: 4 Chapter 1 The Real Numbers With the real numbers associated in the usual way with the poi nts on a line, these defini- tions can be interpreted geometrically as follows: bis an upper bound of Sif no point of S is to the right of b;ˇDsupSif no point of Sis to the right of ˇ, but there is at least one point ofSto the right of any number less than ˇ(Figure 1.1.1 ). (S = dark line segments)β b Figure 1.1.1 Example 1.1.1 IfSis the set of negative numbers, then any nonnegative number i s an upper bound of S, and supSD0. IfS1is the set of negative integers, then any number a such thata/NAK/NUL1is an upper bound of S1, and supS1D/NUL1. This example shows that a supremum of a set may or may not be in t he set, since S1 contains its supremum, but Sdoes not. Anonempty set is a set that has at least one member. The empty set , denoted by;, is the set that has no members. Although it may seem foolish to speak of such a set, we will see that it is a useful idea. The Completeness Axiom It is one thing to define an object and another to show that ther e really is an object that satisfies the definition. (For example, does it make sense to d efine the smallest positive real number?) This observation is particularly appropriat e in connection with the definition of the supremum of a set. For example, the empty set is bounded above by every real number, so it has no supremum. (Think about this.) More impor tantly, we will see in Example 1.1.2 that properties (A) –(H) do not guarantee that every nonempty set that is bounded above has a supremum. Since this property is indis pensable to the rigorous development of calculus, we take it as an axiom for the real nu mbers. (I) If a nonempty set of real numbers is bounded above, then it has a supremum. Property (I)is called completeness , and we say that the real number system is a complete ordered field. It can be shown that the real number system is essentially the only complete ordered field; that is, if an alien from another planet were to construct a mathematical system with properties (A) –(I), the alien’s system would differ from the real number system only in that the alien might use different symbols for the real numbers and C,/SOH, and<. Theorem 1.1.3 If a nonempty set Sof real numbers is bounded above ;then supSis the unique real number ˇsuch that (a)x/DC4ˇfor allxinSI (b) if/SI>0. no matter how small /;there is anx0inSsuch thatx0>ˇ/NUL/SI: Section 1.1 The Real Number System 5 Proof We first show that ˇDsupShas properties (a) and(b). Sinceˇis an upper bound ofS, it must satisfy (a). Since any real number aless thanˇcan be written as ˇ/NUL/SI with/SIDˇ/NULa > 0 ,(b) is just another way of saying that no number less than ˇis an upper bound of S. Hence,ˇDsupSsatisfies (a)and(b). Now we show that there cannot be more than one real number with properties (a) and (b). Suppose that ˇ1< ˇ 2andˇ2has property (b); thus, if/SI > 0 , there is anx0inS such thatx0>ˇ 2/NUL/SI. Then, by taking /SIDˇ2/NULˇ1, we see that there is an x0inSsuch that x0>ˇ 2/NUL.ˇ2/NULˇ1/Dˇ1; soˇ1cannot have property (a). Therefore, there cannot be more than one real number that satisfies both (a)and(b). Some Notation We will often define a set Sby writingSD˚ xˇˇ/SOH/SOH/SOH/TAB , which means that Sconsists of all xthat satisfy the conditions to the right of the vertical bar; thus, in Example 1.1.1 , SD˚ xˇˇx<0/TAB (1.1.8) and S1D˚ xˇˇxis a negative integer/TAB : We will sometimes abbreviate “ xis a member of S” byx2S, and “xis not a member of S” byx…S. For example, if Sis defined by ( 1.1.8 ), then /NUL12Sbut0…S: The Archimedean Property The property of the real numbers described in the next theore m is called the Archimedean property . Intuitively, it states that it is possible to exceed any pos itive number, no matter how large, by adding an arbitrary positive number, no matter how small, to itself sufficiently many times. Theorem 1.1.4 ( Archimedean Property) If/SUBand/SIare positive;thenn/SI > /SUBfor some integer n: Proof The proof is by contradiction. If the statement is false, /SUBis an upper bound of the set SD˚ xˇˇxDn/SI;n is an integer/TAB : Therefore,Shas a supremum ˇ, by property (I). Therefore, n/SI/DC4ˇfor all integers n: (1.1.9) 6 Chapter 1 The Real Numbers SincenC1is an integer whenever nis, (1.1.9 ) implies that .nC1//SI/DC4ˇ and therefore n/SI/DC4ˇ/NUL/SI for all integers n. Hence,ˇ/NUL/SIis an upper bound of S. Sinceˇ/NUL/SI<ˇ , this contradicts the definition of ˇ. Density of the Rationals and Irrationals Definition 1.1.5 A setDisdense in the reals if every open interval .a;b/ contains a member ofD. Theorem 1.1.6 The rational numbers are dense in the reals Ithat is, ifaandbare real numbers with a<b; there is a rational number p=q such thata<p=q<b . Proof From Theorem 1.1.4 with/SUBD1and/SIDb/NULa, there is a positive integer qsuch thatq.b/NULa/>1 . There is also an integer jsuch thatj >qa . This is obvious if a/DC40, and it follows from Theorem 1.1.4 with/SID1and/SUBDqaifa>0 . Letpbe the smallest integer such that p>qa . Thenp/NUL1/DC4qa, so qa<p/DC4qaC1: Since1<q.b/NULa/, this implies that qa<p<qaCq.b/NULa/Dqb; soqa<p<qb . Therefore,a<p=q<b . Example 1.1.2 The rational number system is not complete; that is, a set of r ational numbers may be bounded above (by rationals), but not have a ra tional upper bound less than any other rational upper bound. To see this, let SD˚rˇˇris rational and r2<2/TAB: Ifr2S, thenr <p 2. Theorem 1.1.6 implies that if /SI>0 there is a rational number r0 such thatp 2/NUL/SI<r 0<p 2, so Theorem 1.1.3 implies thatp 2DsupS. However,p 2is irrational ; that is, it cannot be written as the ratio of integers (Exerc ise1.1.3 ). Therefore, ifr1is any rational upper bound of S, thenp 2<r 1. By Theorem 1.1.6 , there is a rational numberr2such thatp 2<r 2<r1. Sincer2is also a rational upper bound of S, this shows thatShas no rational supremum. Since the rational numbers have properties (A) –(H) , but not (I), this example shows that(I)does not follow from (A) –(H) . Theorem 1.1.7 The set of irrational numbers is dense in the reals Ithat is, ifaandb are real numbers with a<b; there is an irrational number tsuch thata<t <b: Section 1.1 The Real Number System 7 Proof From Theorem 1.1.6 , there are rational numbers r1andr2such that a<r 1<r2<b: (1.1.10) Let tDr1C1p 2.r2/NULr1/: Thentis irrational (why?) and r1<t <r 2, soa<t <b , from ( 1.1.10 ). Infimum of a Set A setSof real numbers is bounded below if there is a real number asuch thatx/NAKa wheneverx2S. In this case, ais alower bound ofS. Ifais a lower bound of S, so is any smaller number, because of property (G) . If˛is a lower bound of S, but no number greater than˛is, then˛is an infimum ofS, and we write ˛DinfS: Geometrically, this means that there are no points of Sto the left of˛, but there is at least one point ofSto the left of any number greater than ˛. Theorem 1.1.8 If a nonempty set Sof real numbers is bounded below ;then infSis the unique real number ˛such that (a)x/NAK˛for allxinSI (b) if/SI>0. no matter how small /, there is anx0inSsuch thatx0<˛C/SI: Proof (Exercise 1.1.6 ) A setSisbounded if there are numbers aandbsuch thata/DC4x/DC4bfor allxinS. A bounded nonempty set has a unique supremum and a unique infimu m, and infS/DC4supS (1.1.11) (Exercise 1.1.7 ). The Extended Real Number System A nonempty set Sof real numbers is unbounded above if it has no upper bound, or un- bounded below if it has no lower bound. It is convenient to adjoin to the real number system two fictitious points, C1 (which we usually write more simply as 1) and/NUL1, and to define the order relationships between them and any rea l numberxby /NUL1<x<1: (1.1.12) We call1and/NUL1 points at infinity . IfSis a nonempty set of reals, we write supSD1 (1.1.13) to indicate that Sis unbounded above, and infSD/NUL1 (1.1.14) to indicate that Sis unbounded below. 8 Chapter 1 The Real Numbers Example 1.1.3 If SD˚xˇˇx<2/TAB; then supSD2and infSD/NUL1 . If SD˚ xˇˇx/NAK/NUL2/TAB ; then supSD1 and infSD/NUL2. IfSis the set of all integers, then sup SD1 and infSD/NUL1 . The real number system with 1and/NUL1 adjoined is called the extended real number system , or simply the extended reals . A member of the extended reals differing from /NUL1 and1isfinite ; that is, an ordinary real number is finite. However, the word “finite” in “finite real number” is redundant and used only for emphasis, since we would never refer to1or/NUL1 as real numbers. The arithmetic relationships among 1,/NUL1, and the real numbers are defined as follows. (a) Ifais any real number, then aC1D 1C aD 1; a/NUL1D/NUL1C aD/NUL1; a 1Da /NUL1D0: (b) Ifa>0 , then a1 D 1aD 1; a./NUL1/D./NUL1/aD/NUL1: (c) Ifa<0 , then a1 D 1aD/NUL1; a./NUL1/D./NUL1/aD 1: We also define 1C1D11D ./NUL1/./NUL1/D1 and /NUL1/NUL1D1 ./NUL1/D./NUL1/1D/NUL1: Finally, we define j1jDj/NUL1jD1 : The introduction of 1and/NUL1, along with the arithmetic and order relationships defined above, leads to simplifications in the statements of theorem s. For example, the inequality (1.1.11 ), first stated only for bounded sets, holds for any nonempty s etSif it is interpreted properly in accordance with ( 1.1.12 ) and the definitions of ( 1.1.13 ) and ( 1.1.14 ). Exer- cises 1.1.10(b) and1.1.11(b) illustrate the convenience afforded by some of the arith- metic relationships with extended reals, and other example s will illustrate this further in subsequent sections. Section 1.1 The Real Number System 9 It is not useful to define 1/NUL1 ,0/SOH1,1=1, and0=0. They are called indeterminate forms , and left undefined. You probably studied indeterminate for ms in calculus; we will look at them more carefully in Section 2.4. 1.1 Exercises 1. Write the following expressions in equivalent forms not inv olving absolute values. (a)aCbCja/NULbj (b)aCb/NULja/NULbj (c)aCbC2cCja/NULbjCˇˇaCb/NUL2cCja/NULbjˇˇ (d)aCbC2c/NULja/NULbj/NULˇˇaCb/NUL2c/NULja/NULbjˇˇ 2. Verify that the set consisting of two members, 0and1, with operations defined by Eqns. ( 1.1.1 ) and ( 1.1.2 ), is a field. Then show that it is impossible to define an order <on this field that has properties (F),(G) , and(H) . 3. Show thatp 2is irrational. H INT:Show that ifp 2Dm=n; wheremandnare integers;then bothmandnmust be even:Obtain a contradiction from this : 4. Show thatppis irrational if pis prime. 5. Find the supremum and infimum of each S. State whether they are in S. (a)SD˚ xˇˇxD/NUL.1=n/CŒ1C./NUL1/n/c141n2;n/NAK1/TAB (b)SD˚ xˇˇx2<9/TAB (c)SD˚ xˇˇx2/DC47/TAB (d)SD˚xˇˇj2xC1j<5/TAB (e)SD˚xˇˇ.x2C1//NUL1>1 2/TAB (f)SD˚ xˇˇxDrational andx2/DC47/TAB 6. Prove Theorem 1.1.8 . HINT:The setTD˚xˇˇ/NULx2S/TABis bounded above if Sis bounded below :Apply property (I)and Theorem 1.1.3 toT: 7. (a) Show that infS/DC4supS . A/ for any nonempty set Sof real numbers, and give necessary and sufficient conditions for equality. (b) Show that if Sis unbounded then (A) holds if it is interpreted according to Eqn. ( 1.1.12 ) and the definitions of Eqns. ( 1.1.13 ) and ( 1.1.14 ). 8. LetSandTbe nonempty sets of real numbers such that every real number i s inS orTand ifs2Sandt2T, thens<t . Prove that there is a unique real number ˇ such that every real number less than ˇis inSand every real number greater than ˇis inT. (A decomposition of the reals into two sets with these prope rties is a Dedekind cut. This is known as Dedekind’s theorem .) 10 Chapter 1 The Real Numbers 9. Using properties (A) –(H) of the real numbers and taking Dedekind’s theorem (Exercise 1.1.8 ) as given, show that every nonempty set Uof real numbers that is bounded above has a supremum. H INT:LetTbe the set of upper bounds of Uand Sbe the set of real numbers that are not upper bounds of U: 10. LetSandTbe nonempty sets of real numbers and define SCTD˚sCtˇˇs2S;t2T/TAB: (a) Show that sup.SCT/DsupSCsupT . A/ ifSandTare bounded above and inf.SCT/DinfSCinfT . B/ ifSandTare bounded below. (b) Show that if they are properly interpreted in the extended re als, then (A) and (B) hold ifSandTare arbitrary nonempty sets of real numbers. 11. LetSandTbe nonempty sets of real numbers and define S/NULTD˚ s/NULtˇˇs2S;t2T/TAB : (a) Show that ifSandTare bounded, then sup.S/NULT/DsupS/NULinfT . A/ and inf.S/NULT/DinfS/NULsupT: . B/ (b) Show that if they are properly interpreted in the extended re als, then (A) and (B) hold ifSandTare arbitrary nonempty sets of real numbers. 12. LetSbe a bounded nonempty set of real numbers, and let aandbbe fixed real numbers. Define TD˚ asCbˇˇs2S/TAB . Find formulas for sup Tand infTin terms of supSand infS. Prove your formulas. 1.2 MATHEMATICAL INDUCTION If a flight of stairs is designed so that falling off any step in evitably leads to falling off the next, then falling off the first step is a sure way to end up at th e bottom. Crudely expressed, this is the essence of the principle of mathematical induction : If the truth of a statement depending on a given integer nimplies the truth of the corresponding statement with n replaced bynC1, then the statement is true for all positive integers nif it is true for nD1. Although you have probably studied this principle before, i t is so important that it merits careful review here. Peano’s Postulates and Induction The rigorous construction of the real number system starts w ith a set Nof undefined ele- ments called natural numbers , with the following properties. Section 1.2 Mathematical Induction 11 (A) Nis nonempty. (B) Associated with each natural number nthere is a unique natural number n0called thesuccessor of n. (C) There is a natural number nthat is not the successor of any natural number. (D) Distinct natural numbers have distinct successors; that is , ifn¤m, thenn0¤m0. (E) The only subset of Nthat contains nand the successors of all its elements is N itself. These axioms are known as Peano ’s postulates . The real numbers can be constructed from the natural numbers by definitions and arguments based o n them. This is a formidable task that we will not undertake. We mention it to show how litt le you need to start with to construct the reals and, more important, to draw attention t o postulate (E), which is the basis for the principle of mathematical induction. It can be shown that the positive integers form a subset of the reals that satisfies Peano’s postulates (with nD1andn0DnC1), and it is customary to regard the positive integers and the natural numbers as identical. From this point of view , the principle of mathematical induction is basically a restatement of postulate (E). Theorem 1.2.1 (Principle of Mathematical Induction) LetP1;P2;. . .; Pn;. . . be propositions ;one for each positive integer ;such that (a)P1is trueI (b) for each positive integer n;P nimpliesPnC1: ThenPnis true for each positive integer n: Proof Let MD˚nˇˇn2NandPnis true/TAB: From(a),12M, and from (b),nC12Mwhenevern2M. Therefore, MDN, by postulate (E). Example 1.2.1 LetPnbe the proposition that 1C2C/SOH/SOH/SOHCnDn.nC1/ 2: (1.2.1) ThenP1is the proposition that 1D1, which is certainly true. If Pnis true, then adding nC1to both sides of ( 1.2.1 ) yields .1C2C/SOH/SOH/SOHCn/C.nC1/Dn.nC1/ 2C.nC1/ D.nC1//DLEn 2C1/DC1 D.nC1/.nC2/ 2; or 1C2C/SOH/SOH/SOHC.nC1/D.nC1/.nC2/ 2; 12 Chapter 1 The Real Numbers which isPnC1, since it has the form of ( 1.2.1 ), withnreplaced bynC1. Hence,Pnimplies PnC1, so ( 1.2.1 ) is true for all n, by Theorem 1.2.1 . A proof based on Theorem 1.2.1 is an induction proof , or proof by induction . The assumption that Pnis true is the induction assumption . (Theorem 1.2.3 permits a kind of induction proof in which the induction assumption takes a di fferent form.) Induction, by definition, can be used only to verify results c onjectured by other means. Thus, in Example 1.2.1 we did not use induction to findthe sum snD1C2C/SOH/SOH/SOHCnI (1.2.2) rather, we verified that snDn.nC1/ 2: (1.2.3) How you guess what to prove by induction depends on the proble m and your approach to it. For example, ( 1.2.3 ) might be conjectured after observing that s1D1D1/SOH2 2; s 2D3D2/SOH3 2; s 3D6D4/SOH3 2: However, this requires sufficient insight to recognize that these results are of the form (1.2.3 ) fornD1,2, and3. Although it is easy to prove ( 1.2.3 ) by induction once it has been conjectured, induction is not the most efficient way to fi ndsn, which can be obtained quickly by rewriting ( 1.2.2 ) as snDnC.n/NUL1/C/SOH/SOH/SOHC1 and adding this to ( 1.2.2 ) to obtain 2snDŒnC1/c141CŒ.n/NUL1/C2/c141C/SOH/SOH/SOHCŒ1Cn/c141: There arenbracketed expressions on the right, and the terms in each add up tonC1; hence, 2snDn.nC1/; which yields ( 1.2.3 ). The next two examples deal with problems for which induction is a natural and efficient method of solution. Example 1.2.2 Leta1D1and anC1D1 nC1an; n/NAK1 (1.2.4) (we say thatanis defined inductively ), and suppose that we wish to find an explicit formula foran. By considering nD1,2, and3, we find that a1D1 1; a 2D1 1/SOH2;anda3D1 1/SOH2/SOH3; Section 1.2 Mathematical Induction 13 and therefore we conjecture that anD1 nŠ: (1.2.5) This is given for nD1. If we assume it is true for some n, substituting it into ( 1.2.4 ) yields anC1D1 nC11 nŠD1 .nC1/Š; which is ( 1.2.5 ) withnreplaced bynC1. Therefore, ( 1.2.5 ) is true for every positive integern, by Theorem 1.2.1 . Example 1.2.3 For each nonnegative integer n, letxnbe a real number and suppose that jxnC1/NULxnj/DC4rjxn/NULxn/NUL1j; n/NAK1; (1.2.6) whereris a fixed positive number. By considering ( 1.2.6 ) fornD1,2, and3, we find that jx2/NULx1j/DC4rjx1/NULx0j; jx3/NULx2j/DC4rjx2/NULx1j/DC4r2jx1/NULx0j; jx4/NULx3j/DC4rjx3/NULx2j/DC4r3jx1/NULx0j: Therefore, we conjecture that jxn/NULxn/NUL1j/DC4rn/NUL1jx1/NULx0jifn/NAK1: (1.2.7) This is trivial for nD1. If it is true for some n, then ( 1.2.6 ) and ( 1.2.7 ) imply that jxnC1/NULxnj/DC4r.rn/NUL1jx1/NULx0j/;sojxnC1/NULxnj/DC4rnjx1/NULx0j; which is proposition ( 1.2.7 ) withnreplaced bynC1. Hence, ( 1.2.7 ) is true for every positive integer n, by Theorem 1.2.1 . The major effort in an induction proof (after P1,P2, . . . ,Pn, . . . have been formulated) is usually directed toward showing that PnimpliesPnC1. However, it is important to verify P1, sincePnmay implyPnC1even if some or all of the propositions P1,P2, . . . ,Pn, . . . are false. Example 1.2.4 LetPnbe the proposition that 2n/NUL1is divisible by 2. IfPnis true thenPnC1is also, since 2nC1D.2n/NUL1/C2: However, we cannot conclude that Pnis true forn/NAK1. In fact,Pnis false for every n. The following formulation of the principle of mathematical induction permits us to start induction proofs with an arbitrary integer, rather than 1, a s required in Theorem 1.2.1 . 14 Chapter 1 The Real Numbers Theorem 1.2.2 Letn0be any integer .positive;negative;or zero/:LetPn0;Pn0C1; . . .;Pn;. . . be propositions ;one for each integer n/NAKn0;such that (a)Pn0is trueI (b) for each integer n/NAKn0;PnimpliesPnC1: ThenPnis true for every integer n/NAKn0: Proof Form/NAK1, letQmbe the proposition defined by QmDPmCn0/NUL1. ThenQ1D Pn0is true by (a). Ifm/NAK1andQmDPmCn0/NUL1is true, thenQmC1DPmCn0is true by (b) withnreplaced bymCn0/NUL1. Therefore,Qmis true for all m/NAK1by Theorem 1.2.1 withPreplaced byQandnreplaced bym. This is equivalent to the statement that Pnis true for alln/NAKn0. Example 1.2.5 Consider the proposition Pnthat 3nC16>0: IfPnis true, then so is PnC1, since 3.nC1/C16D3nC3C16 D.3nC16/C3>0C3(by the induction assumption) >0: The smallest n0for whichPn0is true isn0D/NUL5. Hence,Pnis true forn/NAK/NUL5, by Theorem 1.2.2 . Example 1.2.6 LetPnbe the proposition that nŠ/NUL3n>0: IfPnis true, then .nC1/Š/NUL3nC1DnŠ.nC1//NUL3nC1 >3n.nC1//NUL3nC1(by the induction assumption) D3n.n/NUL2/: Therefore,PnimpliesPnC1ifn > 2 . By trial and error, n0D7is the smallest integer such thatPn0is true; hence, Pnis true forn/NAK7, by Theorem 1.2.2 . The next theorem is a useful consequence of the principle of m athematical induction. Theorem 1.2.3 Letn0be any integer .positive;negative;or zero/:LetPn0;Pn0C1;. . .; Pn;. . . be propositions ;one for each integer n/NAKn0;such that (a)Pn0is trueI (b) forn/NAKn0;PnC1is true ifPn0;Pn0C1;. . .;Pnare all true. ThenPnis true forn/NAKn0: Section 1.2 Mathematical Induction 15 Proof Forn/NAKn0, letQnbe the proposition that Pn0,Pn0C1, . . . ,Pnare all true. Then Qn0is true by (a). SinceQnimpliesPnC1by(b), andQnC1is true ifQnandPnC1are both true, Theorem 1.2.2 implies thatQnis true for all n/NAKn0. Therefore,Pnis true for alln/NAKn0. Example 1.2.7 An integerp>1 is aprime if it cannot be factored as pDrswhere randsare integers and 1<r ,s<p . Thus, 2, 3, 5, 7, and 11 are primes, and, although 4, 6, 8, 9, and 10 are not, they are products of primes: 4D2/SOH2; 6D2/SOH3; 8D2/SOH2/SOH2; 9D3/SOH3; 10D2/SOH5: These observations suggest that each integer n/NAK2is a prime or a product of primes. Let this proposition be Pn. ThenP2is true, but neither Theorem 1.2.1 nor Theorem 1.2.2 apply, sincePndoes not imply PnC1in any obvious way. (For example, it is not evident from24D2/SOH2/SOH2/SOH3that 25 is a product of primes.) However, Theorem 1.2.3 yields the stated result, as follows. Suppose that n/NAK2andP2, . . . ,Pnare true. Either nC1is a prime or nC1Drs; (1.2.8) whererandsare integers and 1<r ,s<n , soPrandPsare true by assumption. Hence, r andsare primes or products of primes and ( 1.2.8 ) implies that nC1is a product of primes. We have now proved PnC1(thatnC1is a prime or a product of primes). Therefore, Pnis true for alln/NAK2, by Theorem 1.2.3 . 1.2 Exercises Prove the assertions in Exercises 1.2.1 –1.2.6 by induction. 1. The sum of the first nodd integers is n2. 2.12C22C/SOH/SOH/SOHCn2Dn.nC1/.2nC1/ 6: 3.12C32C/SOH/SOH/SOHC.2n/NUL1/2Dn.4n2/NUL1/ 3: 4. Ifa1,a2, . . . ,anare arbitrary real numbers, then ja1Ca2C/SOH/SOH/SOHCanj/DC4ja1jCja2jC/SOH/SOH/SOHCjanj: 5. Ifai/NAK0,i/NAK1, then .1Ca1/.1Ca2//SOH/SOH/SOH.1Can//NAK1Ca1Ca2C/SOH/SOH/SOHCan: 6. If0/DC4ai/DC41,i/NAK1, then .1/NULa1/.1/NULa2//SOH/SOH/SOH.1/NULan//NAK1/NULa1/NULa2/SOH/SOH/SOH/NULan: 16 Chapter 1 The Real Numbers 7. Suppose that s0>0andsnD1/NULe/NULsn/NUL1,n/NAK1. Show that0<s n<1,n/NAK1. 8. Suppose that R>0 ,x0>0, and xnC1D1 2/DC2R xnCxn/DC3 ; n/NAK0: Prove: Forn/NAK1,xn>x nC1>p Rand xn/NULp R/DC41 2n.x0/NULp R/2 x0: 9. Find and prove by induction an explicit formula for anifa1D1and, forn/NAK1, (a)anC1Dan .nC1/.2nC1/(b)anC1D3an .2nC2/.2nC3/ (c)anC1D2nC1 nC1an (d)anC1D/DC2 1C1 n/DC3n an 10. Leta1D0andanC1D.nC1/anforn/NAK1, and letPnbe the proposition that anDnŠ (a) Show thatPnimpliesPnC1. (b) Is there an integer nfor whichPnis true? 11. LetPnbe the proposition that 1C2C/SOH/SOH/SOHCnD.nC2/.n/NUL1/ 2: (a) Show thatPnimpliesPnC1. (b) Is there an integer nfor whichPnis true? 12. For what integers nis 1 nŠ>8n .2n/Š‹ Prove your answer by induction. 13. Letabe an integer/NAK2. (a) Show by induction that if nis a nonnegative integer, then nDaqCr, where q(quotient) and r(remainder) are integers and 0/DC4r <a . (b) Show that the result of (a)is true ifnis an arbitrary integer (not necessarily nonnegative). (c) Show that there is only one way to write a given integer nin the formnD aqCr, whereqandrare integers and 0/DC4r <a . 14. Take the following statement as given: If pis a prime and aandbare integers such thatpdivides the product ab, thenpdividesaorb. Section 1.2 Mathematical Induction 17 (a) Prove: Ifp,p1, . . . ,pkare positive primes and pdivides the product p1/SOH/SOH/SOHpk, thenpDpifor someiinf1;:::;kg. (b) Letnbe an integer > 1. Show that the prime factorization of nfound in Example 1.2.7 is unique in the following sense: If nDp1/SOH/SOH/SOHprandnDq1q2/SOH/SOH/SOHqs; wherep1, . . . ,pr,q1, . . . ,qsare positive primes, then rDsandfq1;:::;q rg is a permutation offp1;:::;p rg. 15. Leta1Da2D5and anC1DanC6an/NUL1; n/NAK2: Show by induction that anD3n/NUL./NUL2/nifn/NAK1. 16. Leta1D2,a2D0,a3D/NUL14, and anC1D9an/NUL23a n/NUL1C15a n/NUL2; n/NAK3: Show by induction that anD3n/NUL1/NUL5n/NUL1C2,n/NAK1. 17. TheFibonacci numbersfFng1 nD1are defined by F1DF2D1and FnC1DFnCFn/NUL1; n/NAK2: Prove by induction that FnD.1Cp 5/n/NUL.1/NULp 5/n 2np 5; n/NAK1: 18. Prove by induction that Z1 0yn.1/NULy/rdyDnŠ .rC1/.rC2//SOH/SOH/SOH.rCnC1/ ifnis a nonnegative integer and r >/NUL1. 19. Suppose that mandnare integers, with 0/DC4m/DC4n. The binomial coefficient n m! is the coefficient of tmin the expansion of .1Ct/n; that is, .1Ct/nDnX mD0 n m! tm: From this definition it follows immediately that n 0! D n n! D1; n/NAK0: For convenience we define n /NUL1! D n nC1! D0; n/NAK0: 18 Chapter 1 The Real Numbers (a) Show that nC1 m! D n m! C n m/NUL1! ; 0/DC4m/DC4n; and use this to show by induction on nthat n m! DnŠ mŠ.n/NULm/Š; 0/DC4m/DC4n: (b) Show that nX mD0./NUL1/m n m! D0andnX mD0 n m! D2n: (c) Show that .xCy/nDnX mD0 n m! xmyn/NULm: (This is the binomial theorem .) 20. Use induction to find an nth antiderivative of log x, the natural logarithm of x. 21. Letf1.x1/Dg1.x1/Dx1. Forn/NAK2, let fn.x1;x2;:::;x n/Dfn/NUL1.x1;x2;:::;x n/NUL1/C2n/NUL2xnC jfn/NUL1.x1;x2;:::;x n/NUL1//NUL2n/NUL2xnj and gn.x1;x2;:::;x n/Dgn/NUL1.x1;x2;:::;x n/NUL1/C2n/NUL2xn/NUL jgn/NUL1.x1;x2;:::;x n/NUL1//NUL2n/NUL2xnj: Find explicit formulas for fn.x1;x2;:::;x n/andgn.x1;x2;:::;x n/. 22. Prove by induction that sinxCsin3xC/SOH/SOH/SOHC sin.2n/NUL1/xD1/NULcos2nx 2sinx; n/NAK1: HINT:You will need trigonometric identities that you can derive f rom the identities cos.A/NULB/DcosAcosBCsinAsinB; cos.ACB/DcosAcosB/NULsinAsinB: Take these two identities as given : Section 1.3 The Real Line 19 23. Suppose that a1/DC4a2/DC4/SOH/SOH/SOH/DC4anandb1/DC4b2/DC4/SOH/SOH/SOH/DC4bn. Letf`1;`2;:::` ngbe a permutation off1;2;:::;ng, and define Q.` 1;`2;:::;` n/DnX iD1.ai/NULb`i/2: Show that Q.` 1;`2;:::;` n//NAKQ.1;2;:::;n/: 1.3 THE REAL LINE One of our objectives is to develop rigorously the concepts o f limit, continuity, differen- tiability, and integrability, which you have seen in calcul us. To do this requires a better understanding of the real numbers than is provided in calcul us. The purpose of this section is to develop this understanding. Since the utility of the co ncepts introduced here will not become apparent until we are well into the study of limits and continuity, you should re- serve judgment on their value until they are applied. As this occurs, you should reread the applicable parts of this section. This applies especially t o the concept of an open covering and to the Heine–Borel and Bolzano–Weierstrass theorems, w hich will seem mysterious at first. We assume that you are familiar with the geometric interpret ation of the real numbers as points on a line. We will not prove that this interpretation i s legitimate, for two reasons: (1) the proof requires an excursion into the foundations of Eucl idean geometry, which is not the purpose of this book; (2) although we will use geometric t erminology and intuition in discussing the reals, we will base all proofs on properties (A) –(I)(Section 1.1) and their consequences, not on geometric arguments. Henceforth, we will use the terms real number system andreal line synonymously and denote both by the symbol R; also, we will often refer to a real number as a point (on the real line). Some Set Theory In this section we are interested in sets of points on the real line; however, we will consider other kinds of sets in subsequent sections. The following de finition applies to arbitrary sets, with the understanding that the members of all sets und er consideration in any given context come from a specific collection of elements, called t heuniversal set . In this section the universal set is the real numbers. Definition 1.3.1 LetSandTbe sets. (a)ScontainsT, and we write S/ESCTorT/SUBS, if every member of Tis also inS. In this case,Tis asubset ofS. (b)S/NULTis the set of elements that are in Sbut not inT. (c)SequalsT, and we write SDT, ifScontainsTandTcontainsS; thus,SDTif and only ifSandThave the same members. 20 Chapter 1 The Real Numbers (d)Sstrictly contains TifScontainsTbutTdoes not contain S; that is, if every member ofTis also inS, but at least one member of Sis not inT(Figure 1.3.1 ). (e) Thecomplement ofS, denoted bySc, is the set of elements in the universal set that are not inS. (f) Theunion ofSandT, denoted byS[T, is the set of elements in at least one of S andT(Figure 1.3.1(b)). (g) Theintersection ofSandT, denoted byS\T, is the set of elements in both Sand T(Figure 1.3.1(c)). IfS\TD; (the empty set), then SandTaredisjoint sets (Figure 1.3.1(d)). (h) A set with only one member x0is asingleton set , denoted byfx0g. T S S T (a)S ∪ T = shaded region (b) (c) (d)S ∩ T = shaded region S ∩ T = ∅ T ST S T S Figure 1.3.1 Example 1.3.1 Let SD˚xˇˇ0<x<1/TAB; TD˚xˇˇ0<x<1 andxis rational/TAB; and UD˚ xˇˇ0<x<1 andxis irrational/TAB : ThenS/ESCTandS/ESCU, and the inclusion is strict in both cases. The unions of pair s of these sets are S[TDS; S[UDS; andT[UDS; and their intersections are S\TDT; S\UDU; andT\UD;: Section 1.3 The Real Line 21 Also, S/NULUDTandS/NULTDU: Every setScontains the empty set ;, for to say that;is not contained in Sis to say that some member of;is not inS, which is absurd since ;has no members. If Sis any set, then .Sc/cDSandS\ScD;: IfSis a set of real numbers, then S[ScDR. The definitions of union and intersection have generalizati ons: If Fis an arbitrary col- lection of sets, then [˚SˇˇS2F/TABis the set of all elements that are members of at least one of the sets in F, and\˚SˇˇS2F/TABis the set of all elements that are members of every set in F. The union and intersection of finitely many sets S1, . . . ,Snare also written asSn kD1SkandTn kD1Sk. The union and intersection of an infinite sequence fSkg1 kD1of sets are written asS1 kD1SkandT1 kD1Sk. Example 1.3.2 IfFis the collection of sets S/SUBD˚ xˇˇ/SUB<x/DC41C/SUB/TAB ; 0</SUB/DC41=2; then [˚S/SUBˇˇS/SUB2F/TABD˚xˇˇ0<x/DC43=2/TABand\˚S/SUBˇˇS/SUB2F/TABD˚xˇˇ1=2<x/DC41/TAB: Example 1.3.3 If, for each positive integer k, the setSkis the set of real numbers that can be written as xDm=k for some integer m, thenS1 kD1Skis the set of rational numbers andT1 kD1Skis the set of integers. Open and Closed Sets Ifaandbare in the extended reals and a<b , then the open interval .a;b/ is defined by .a;b/D˚xˇˇa<x<b/TAB: The open intervals .a;1/and./NUL1;b/aresemi-infinite ifaandbare finite, and ./NUL1;1/ is the entire real line. Definition 1.3.2 Ifx0is a real number and /SI>0 , then the open interval .x0/NUL/SI;x 0C/SI/ is an/SI-neighborhood ofx0. If a setScontains an/SI-neighborhood of x0, thenSis a neighborhood ofx0, andx0is an interior point ofS(Figure 1.3.2 ). The set of interior points ofSis the interior ofS, denoted byS0. If every point of Sis an interior point (that is,S0DS), thenSisopen . A setSisclosed ifScis open. 22 Chapter 1 The Real Numbers ( )x0 + x0 − x0 x0 = interior point of S S = four line segments Figure 1.3.2 The idea of neighborhood is fundamental and occurs in many ot her contexts, some of which we will see later in this book. Whatever the context, th e idea is the same: some defi- nition of “closeness” is given (for example, two real number s are “close” if their difference is “small”), and a neighborhood of a point x0is a set that contains all points sufficiently close tox0. Example 1.3.4 An open interval .a;b/ is an open set, because if x02.a;b/ and /SI/DC4minfx0/NULa;b/NULx0g, then .x0/NUL/SI;x 0C/SI//SUB.a;b/: The entire line RD./NUL1;1/is open, and therefore ;.DRc/is closed. However, ;is also open, for to deny this is to say that ;contains a point that is not an interior point, which is absurd because ;contains no points. Since ;is open, R.D;c/is closed. Thus, Rand;are both open and closed. They are the only subsets of Rwith this property (Exercise 1.3.18 ). Adeleted neighborhood of a pointx0is a set that contains every point of some neigh- borhood ofx0except forx0itself. For example, SD˚ xˇˇ0<jx/NULx0j</SI/TAB is a deleted neighborhood of x0. We also say that it is a deleted/SI-neighborhood ofx0. Theorem 1.3.3 (a) The union of open sets is open : (b) The intersection of closed sets is closed : These statements apply to arbitrary collections, finite or i nfinite, of open and closed sets : Proof (a) LetGbe a collection of open sets and SD[˚ GˇˇG2G/TAB : Ifx02S, thenx02G0for someG0inG, and sinceG0is open, it contains some /SI- neighborhood of x0. SinceG0/SUBS, this/SI-neighborhood is in S, which is consequently a neighborhood of x0. Thus,Sis a neighborhood of each of its points, and therefore open, by definition. (b) LetFbe a collection of closed sets and TD \˚FˇˇF2F/TAB. ThenTcD [˚FcˇˇF2F/TAB(Exercise 1.3.7 ) and, since each Fcis open,Tcis open, from (a). There- fore,Tis closed, by definition. Section 1.3 The Real Line 23 Example 1.3.5 If/NUL1<a<b<1, the set Œa;b/c141D˚ xˇˇa/DC4x/DC4b/TAB is closed, since its complement is the union of the open sets ./NUL1;a/and.b;1/. We say thatŒa;b/c141 is aclosed interval . The set Œa;b/D˚xˇˇa/DC4x<b/TAB is ahalf-closed orhalf-open interval if/NUL1<a<b<1, as is .a;b/c141D˚xˇˇa<x/DC4b/TABI however, neither of these sets is open or closed. (Why not?) Semi-infinite closed intervals are sets of the form Œa;1/D˚xˇˇa/DC4x/TABand./NUL1;a/c141D˚xˇˇx/DC4a/TAB; whereais finite. They are closed sets, since their complements are t he open intervals ./NUL1;a/and.a;1/, respectively. Example 1.3.4 shows that a set may be both open and closed, and Example 1.3.5 shows that a set may be neither. Thus, open and closed are not opposi tes in this context, as they are in everyday speech. Example 1.3.6 From Theorem 1.3.3 and Example 1.3.4 , the union of any collection of open intervals is an open set. (In fact, it can be shown that ev ery nonempty open subset of Ris the union of open intervals.) From Theorem 1.3.3 and Example 1.3.5 , the intersection of any collection of closed intervals is closed. It can be shown that the intersection of finitely many open set s is open, and that the union of finitely many closed sets is closed. However, the int ersection of infinitely many open sets need not be open, and the union of infinitely many clo sed sets need not be closed (Exercises 1.3.8 and1.3.9 ). Definition 1.3.4 LetSbe a subset of R. Then (a)x0is alimit point ofSif every deleted neighborhood of x0contains a point of S. (b)x0is aboundary point ofSif every neighborhood of x0contains at least one point inSand one not in S. The set of boundary points of Sis the boundary ofS, denoted by@S. The closure ofS, denoted byS, isSDS[@S. (c)x0is an isolated point ofSifx02Sand there is a neighborhood of x0that contains no other point of S. (d)x0isexterior toSifx0is in the interior of Sc. The collection of such points is the exterior ofS. Example 1.3.7 LetSD./NUL1;/NUL1/c141[.1;2/[f3g. Then 24 Chapter 1 The Real Numbers (a) The set of limit points of Sis./NUL1;/NUL1/c141[Œ1;2/c141 . (b)@SDf/NUL1;1;2;3gandSD./NUL1;/NUL1/c141[Œ1;2/c141[f3g. (c)3is the only isolated point of S. (d) The exterior of Sis./NUL1;1/[.2;3/[.3;1/. Example 1.3.8 Forn/NAK1, let InD/DC41 2nC1;1 2n/NAK andSD1[ nD1In: Then (a) The set of limit points of SisS[f0g. (b)@SD˚xˇˇxD0orxD1=n.n/NAK2//TABandSDS[f0g. (c)Shas no isolated points. (d) The exterior of Sis ./NUL1;0/["1[ nD1/DC21 2nC2;1 2nC1/DC3# [/DC21 2;1/DC3 : Example 1.3.9 LetSbe the set of rational numbers. Since every interval contain s a rational number (Theorem 1.1.6 ), every real number is a limit point of S; thus,SDR. Since every interval also contains an irrational number (Th eorem 1.1.7 ), every real number is a boundary point of S; thus@SDR. The interior and exterior of Sare both empty, and Shas no isolated points. Sis neither open nor closed. The next theorem says that Sis closed if and only if SDS(Exercise 1.3.14 ). Theorem 1.3.5 A setSis closed if and only if no point of Scis a limit point of S: Proof Suppose that Sis closed and x02Sc. SinceScis open, there is a neighborhood ofx0that is contained in Scand therefore contains no points of S. Hence,x0cannot be a limit point of S. For the converse, if no point of Scis a limit point of Sthen every point in Scmust have a neighborhood contained in Sc. Therefore,Scis open andSis closed. Theorem 1.3.5 is usually stated as follows. Corollary 1.3.6 A set is closed if and only if it contains all its limit points : Theorem 1.3.5 and Corollary 1.3.6 are equivalent. However, we stated the theorem as we did because students sometimes incorrectly conclude fro m the corollary that a closed set must have limit points. The corollary does not say this. I fShas no limit points, then the set of limit points is empty and therefore contained in S. Hence, a set with no limit points is closed according to the corollary, in agreement wi th Theorem 1.3.5 . For example, any finite set is closed. More generally, Sis closed if there is a ı>0 suchjx/NULyj/NAKıfor every pairfx;ygof distinct points in S. Section 1.3 The Real Line 25 Open Coverings A collection Hof open sets is an open covering of a setSif every point in Sis contained in a setHbelonging to H; that is, ifS/SUB[˚ HˇˇH2H/TAB . Example 1.3.10 The sets S1DŒ0;1/c141;S 2Df1;2;:::;n;:::g; S3D/SUB 1;1 2;:::;1 n;:::/ESC ;andS4D.0;1/ are covered by the families of open intervals H1D/SUB/DC2 x/NUL1 N;xC1 N/DC3ˇˇˇˇ0<x<1/ESC ;(NDpositive integer), H2D/SUB/DC2 n/NUL1 4;nC1 4/DC3ˇˇˇˇnD1;2;:::/ESC ; H3D( 1 nC1 2;1 n/NUL1 2!ˇˇˇˇnD1;2;:::) ; and H4Df.0;/SUB/j0</SUB<1g; respectively. Theorem 1.3.7 ( Heine –Borel Theorem) IfHis an open covering of a closed and bounded subset Sof the real line ;thenShas an open covering eHconsisting of finitely many open sets belonging to H: Proof SinceSis bounded, it has an infimum ˛and a supremum ˇ, and, sinceSis closed,˛andˇbelong toS(Exercise 1.3.17 ). Define StDS\Œ˛;t/c141 fort/NAK˛; and let FD˚ tˇˇ˛/DC4t/DC4ˇand finitely many sets from HcoverSt/TAB : SinceSˇDS, the theorem will be proved if we can show that ˇ2F. To do this, we use the completeness of the reals. Since˛2S,S˛is the singleton setf˛g, which is contained in some open set H˛from Hbecause HcoversS; therefore,˛2F. SinceFis nonempty and bounded above by ˇ, it has a supremum /CR. First, we wish to show that /CRDˇ. Since/CR/DC4ˇby definition of F, it suffices to rule out the possibility that /CR <ˇ . We consider two cases. 26 Chapter 1 The Real Numbers CASE 1. Suppose that /CR <ˇ and/CR62S. Then, since Sis closed,/CRis not a limit point ofS(Theorem 1.3.5 ). Consequently, there is an /SI>0 such that Œ/CR/NUL/SI;/CRC/SI/c141\SD;; soS/CR/NUL/SIDS/CRC/SI. However, the definition of /CRimplies thatS/CR/NUL/SIhas a finite subcovering from H, whileS/CRC/SIdoes not. This is a contradiction. CASE 2. Suppose that /CR < ˇ and/CR2S. Then there is an open set H/CRinHthat contains/CRand, along with /CR, an intervalŒ/CR/NUL/SI;/CRC/SI/c141for some positive /SI. SinceS/CR/NUL/SIhas a finite coveringfH1;:::;H ngof sets from H, it follows that S/CRC/SIhas the finite covering fH1;:::;H n;H/CRg. This contradicts the definition of /CR. Now we know that /CRDˇ, which is inS. Therefore, there is an open set HˇinHthat containsˇand along with ˇ, an interval of the form Œˇ/NUL/SI;ˇC/SI/c141, for some positive /SI. SinceSˇ/NUL/SIis covered by a finite collection of sets fH1;:::;H kg,Sˇis covered by the finite collectionfH1;:::;H k;Hˇg. SinceSˇDS, we are finished. Henceforth, we will say that a closed and bounded set is compact . The Heine–Borel theorem says that any open covering of a compact set Scontains a finite collection that also coversS. This theorem and its converse (Exercise 1.3.21 ) show that we could just as well define a set Sof reals to be compact if it has the Heine–Borel property; tha t is, if every open covering of Scontains a finite subcovering. The same is true of Rn, which we study in Section 5.1. This definition generalizes to more abs tract spaces (called topological spaces ) for which the concept of boundedness need not be defined. Example 1.3.11 SinceS1in Example 1.3.10 is compact, the Heine–Borel theorem implies thatS1can be covered by a finite number of intervals from H1. This is easily veri- fied, since, for example, the 2Nintervals from H1centered at the points xkDk=2N.0/DC4 k/DC42N/NUL1/coverS1. The Heine–Borel theorem does not apply to the other sets in Ex ample 1.3.10 since they are not compact: S2is unbounded and S3andS4are not closed, since they do not contain all their limit points (Corollary 1.3.6 ). The conclusion of the Heine–Borel theorem does not hold for these sets and the open coverings that we have giv en for them. Each point in S2is contained in exactly one set from H2, so removing even one of these sets leaves a point ofS2uncovered. If eH3is any finite collection of sets from H3, then 1 n62[˚ HˇˇH2eH3/TAB fornsufficiently large. Any finite collection f.0;/SUB 1/;:::;.0;/SUB n/gfrom H4covers only the interval.0;/SUB max/, where /SUBmaxDmaxf/SUB1;:::;/SUB ng<1: The Bolzano–Weierstrass Theorem As an application of the Heine–Borel theorem, we prove the fo llowing theorem of Bolzano and Weierstrass. Section 1.3 The Real Line 27 Theorem 1.3.8 ( Bolzano –Weierstrass Theorem) Every bounded infinite set of real numbers has at least one limit point : Proof We will show that a bounded nonempty set without a limit point can contain only a finite number of points. If Shas no limit points, then Sis closed (Theorem 1.3.5 ) and every pointxofShas an open neighborhood Nxthat contains no point of Sother thanx. The collection HD˚Nxˇˇx2S/TAB is an open covering for S. SinceSis also bounded, Theorem 1.3.7 implies thatScan be covered by a finite collection of sets from H, sayNx1, . . . ,Nxn. Since these sets contain onlyx1, . . . ,xnfromS, it follows that SDfx1;:::;x ng. 1.3 Exercises 1. FindS\T,.S\T/c,Sc\Tc,S[T,.S[T/c, andSc[Tc. (a)SD.0;1/ ,TD/STX1 2;3 2/ETX(b)SD˚xˇˇx2>4/TAB,TD˚xˇˇx2<9/TAB (c)SD./NUL1;1/,TD;(d)SD./NUL1;/NUL1/,TD.1;1/ 2. LetSkD.1/NUL1=k;2C1=k/c141 ,k/NAK1. Find (a)1[ kD1Sk(b)1\ kD1Sk(c)1[ kD1Sc k(d)1\ kD1Sc k 3. Prove: IfAandBare sets and there is a set Xsuch thatA[XDB[Xand A\XDB\X, thenADB. 4. Find the largest /SIsuch thatScontains an/SI-neighborhood of x0. (a)x0D3 4,SD/STX1 2;1/SOH(b)x0D2 3,SD/STX1 2;3 2/ETX (c)x0D5,SD./NUL1;1/(d)x0D1,SD.0;2/ 5. Describe the following sets as open, closed, or neither, and findS0,.Sc/0, and .S0/c. (a)SD./NUL1;2/[Œ3;1/(b)SD./NUL1;1/[.2;1/ (c)SDŒ/NUL3;/NUL2/c141[Œ7;8/c141 (d)SD˚xˇˇxDinteger/TAB 6. Prove that.S\T/cDSc[Tcand.S[T/cDSc\Tc. 7. LetFbe a collection of sets and define ID\˚FˇˇF2F/TABandUD[˚FˇˇF2F/TAB: Prove that (a)IcD[˚ FcˇˇF2F/TAB and(b)UcD˚ \FcˇˇF2F/TAB . 8. (a) Show that the intersection of finitely many open sets is open. 28 Chapter 1 The Real Numbers (b) Give an example showing that the intersection of infinitely m any open sets may fail to be open. 9. (a) Show that the union of finitely many closed sets is closed. (b) Give an example showing that the union of infinitely many clos ed sets may fail to be closed. 10. Prove: (a) IfUis a neighborhood of x0andU/SUBV, thenVis a neighborhood of x0. (b) IfU1, . . . ,Unare neighborhoods of x0, so isTn iD1Ui. 11. Find the set of limit points of S,@S,S, the set of isolated points of S, and the exterior ofS. (a)SD./NUL1;/NUL2/[.2;3/[f4g[.7;1/ (b)SDfall integersg (c)SD[˚ .n;nC1/ˇˇnDinteger/TAB (d)SD˚ xˇˇxD1=n;nD1;2;3;:::/TAB 12. Prove: A limit point of a set Sis either an interior point or a boundary point of S. 13. Prove: An isolated point of Sis a boundary point of Sc. 14. Prove: (a) A boundary point of a set Sis either a limit point or an isolated point of S. (b) A setSis closed if and only if SDS. 15. Prove or disprove: A set has no limit points if and only if each of its points is isolated. 16. (a) Prove: IfSis bounded above and ˇDsupS, thenˇ2@S. (b) State the analogous result for a set bounded below. 17. Prove: IfSis closed and bounded, then inf Sand supSare both inS. 18. If a nonempty subset SofRis both open and closed, then SDR. 19. LetSbe an arbitrary set. Prove: (a)@Sis closed. (b)S0is open. (c)The exterior ofSis open. (d) The limit points of Sform a closed set. (e)/NULS/SOHDS. 20. Give counterexamples to the following false statements. (a) The isolated points of a set form a closed set. (b) Every open set contains at least two points. (c) IfS1andS2are arbitrary sets, then @.S1[S2/D@S1[@S2. (d) IfS1andS2are arbitrary sets, then @.S1\S2/D@S1\@S2. (e) The supremum of a bounded nonempty set is the greatest of its l imit points. (f) IfSis any set, then @.@S/D@S. (g) IfSis any set, then @SD@S. (h) IfS1andS2are arbitrary sets, then .S1[S2/0DS0 1[S0 2. Section 1.3 The Real Line 29 21. LetSbe a nonempty subset of Rsuch that if His any open covering of S, thenS has an open covering eHcomprised of finitely many open sets from H. Show that Sis compact. 22. A setSis. in a setTifS/SUBT/SUBS. (a) Prove: IfSandTare sets of real numbers and S/SUBT, thenSis dense inT if and only if every neighborhood of each point in Tcontains a point from S. (b) State how (a) shows that the definition given here is consistent with the re - stricted definition of a dense subset of the reals given in Sec tion 1.1. 23. Prove: (a).S1\S2/0DS0 1\S0 2 (b)S0 1[S0 2/SUB.S1[S2/0 24. Prove: (a)@.S1[S2//SUB@S1[@S2 (b)@.S1\S2//SUB@S1[@S2 (c)@S/SUB@S (d)@SD@Sc (e)@.S/NULT//SUB@S[@T CHAPTER 2 Differential Calculus of Functions of One Variable IN THIS CHAPTER we study the differential calculus of functi ons of one variable. SECTION 2.1 introduces the concept of function and discusse s arithmetic operations on functions, limits, one-sided limits, limits at ˙1, and monotonic functions. SECTION 2.2 defines continuity and discusses removable disc ontinuities, composite func- tions, bounded functions, the intermediate value theorem, uniform continuity, and addi- tional properties of monotonic functions. SECTION 2.3 introduces the derivative and its geometric int erpretation. Topics covered in- clude the interchange of differentiation and arithmetic op erations, the chain rule, one-sided derivatives, extreme values of a differentiable function, Rolle’s theorem, the intermediate value theorem for derivatives, and the mean value theorem an d its consequences. SECTION 2.4 presents a comprehensive discussion of L’Hospi tal’s rule. SECTION 2.5 discusses the approximation of a function fby the Taylor polynomials of fand applies this result to locating local extrema of f. The section concludes with the extended mean value theorem, which implies Taylor’s theore m. 2.1 FUNCTIONS AND LIMITS In this section we study limits of real-valued functions of a real variable. You studied limits in calculus. However, we will look more carefully at t he definition of limit and prove theorems usually not proved in calculus. A rulefthat assigns to each member of a nonempty set Da unique member of a set Y is afunction from DtoY. We write the relationship between a member xofDand the memberyofYthatfassigns toxas yDf.x/: The setDis the domain off, denoted byDf. The members of Yare the possible values off. Ify02Yand there is an x0inDsuch thatf.x 0/Dy0then we say that fattains 30 Section 2.1 Functions and Limits 31 orassumes the valuey0. The set of values attained by fis the range off. Areal-valued function of a real variable is a function whose domain and range are both subsets of the reals. Although we are concerned only with real-valued func tions of a real variable in this section, our definitions are not restricted to this situatio n. In later sections we will consider situations where the range or domain, or both, are subsets of vector spaces. Example 2.1.1 The functions f,g, andhdefined on./NUL1;1/by f.x/Dx2; g.x/Dsinx; andh.x/Dex have rangesŒ0;1/,Œ/NUL1;1/c141, and.0;1/, respectively. Example 2.1.2 The equation Œf.x//c1412Dx (2.1.1) does not define a function except on the singleton set f0g. Ifx<0 , no real number satisfies (2.1.1 ), while ifx>0 , two real numbers satisfy ( 2.1.1 ). However, the conditions Œf.x//c1412Dxandf.x//NAK0 define a function fonDfDŒ0;1/with valuesf.x/Dpx. Similarly, the conditions Œg.x//c1412Dxandg.x//DC40 define a function gonDgDŒ0;1/with valuesg.x/D/NULpx. The ranges of fandgare Œ0;1/and./NUL1;0/c141, respectively. It is important to understand that the definition of a functio n includes the specification of its domain and that there is a difference between f, the name of the function, and f.x/ , thevalue offatx. However, strict observance of these points leads to annoyi ng verbosity, such as “the function fwith domain ./NUL1;1/and valuesf.x/Dx.” We will avoid this in two ways: (1) by agreeing that if a function fis introduced without explicitly defining Df, thenDfwill be understood to consist of all points xfor which the rule defining f.x/ makes sense, and (2) by bearing in mind the distinction betwe enfandf.x/ , but not emphasizing it when it would be a nuisance to do so. For exampl e, we will write “consider the functionf.x/Dp 1/NULx2,” rather than “consider the function fdefined onŒ/NUL1;1/c141 byf.x/Dp 1/NULx2,” or “consider the function g.x/D1=sinx,” rather than “consider the functiongdefined forx¤k/EM(kDinteger) byg.x/D1=sinx.” We will also write fDc(constant) to denote the function fdefined byf.x/Dcfor allx. Our definition of function is somewhat intuitive, but adequa te for our purposes. More- over, it is the working form of the definition, even if the idea is introduced more rigorously to begin with. For a more precise definition, we first define the Cartesian productX/STXY of two nonempty sets XandYto be the set of all ordered pairs .x;y/ such thatx2Xand y2Y; thus, X/STXYD˚ .x;y/ˇˇx2X;y2Y/TAB : 32 Chapter 2 Differential Calculus of Functions of One Variable A nonempty subset fofX/STXYis afunction if noxinXoccurs more than once as a first member among the elements of f. Put another way, if .x;y/ and.x;y 1/are inf, then yDy1. The set ofx’s that occur as first members of fis the off. Ifxis in the domain off, then the unique yinYsuch that.x;y/2fis the value offatx, and we write yDf.x/ . The set of all such values, a subset of Y, is the range off. Arithmetic Operations on Functions Definition 2.1.1 IfDf\Dg¤;;thenfCg;f/NULg;andfgare defined on Df\Dg by .fCg/.x/Df.x/Cg.x/; .f/NULg/.x/Df.x//NULg.x/; and .fg/.x/Df.x/g.x/: The quotient f=g is defined by /DC2f g/DC3 .x/Df.x/ g.x/ forxinDf\Dgsuch thatg.x/¤0: Example 2.1.3 Iff.x/Dp 4/NULx2andg.x/Dp x/NUL1;thenDfDŒ/NUL2;2/c141 and DgDŒ1;1/;sofCg;f/NULg;andfgare defined on Df\DgDŒ1;2/c141 by .fCg/.x/Dp 4/NULx2Cp x/NUL1; .f/NULg/.x/Dp 4/NULx2/NULp x/NUL1; and .fg/.x/D.p 4/NULx2/.p x/NUL1/Dp .4/NULx2/.x/NUL1/: (2.1.2) The quotient f=g is defined on .1;2/c141 by /DC2f g/DC3 .x/Dr 4/NULx2 x/NUL1: Although the last expression in ( 2.1.2 ) is also defined for /NUL1< x </NUL2;it does not representfgfor suchx;sincefandgare not defined on ./NUL1;/NUL2/c141. Example 2.1.4 Ifcis a real number, the function cfdefined by.cf/.x/Dcf.x/ can be regarded as the product of fand a constant function. Its domain is Df. The sum and product ofn./NAK2/functionsf1, . . . ,fnare defined by .f1Cf2C/SOH/SOH/SOHCfn/.x/Df1.x/Cf2.x/C/SOH/SOH/SOHCfn.x/ Section 2.1 Functions and Limits 33 and .f1f2/SOH/SOH/SOHfn/.x/Df1.x/f 2.x//SOH/SOH/SOHfn.x/ (2.1.3) onDDTn iD1Dfi, provided that Dis nonempty. If f1Df2D/SOH/SOH/SOHDfn, then ( 2.1.3 ) defines thenthpower off: .fn/.x/D.f.x//n: From these definitions, we can build the set of all polynomials p.x/Da0Ca1xC/SOH/SOH/SOHCanxn; starting from the constant functions and f.x/Dx. The quotient of two polynomials is a rational function r.x/Da0Ca1xC/SOH/SOH/SOHCanxn b0Cb1xC/SOH/SOH/SOHCbmxm.bm¤0/: The domain of ris the set of points where the denominator is nonzero. Limits The essence of the concept of limit for real-valued function s of a real variable is this: If L is a real number, then lim x!x0f.x/DLmeans that the value f.x/ can be made as close toLas we wish by taking xsufficiently close to x0. This is made precise in the following definition. y xL + L − Ly = f(x) x0 − δ x0 + δ x0 Figure 2.1.1 34 Chapter 2 Differential Calculus of Functions of One Variable Definition 2.1.2 We say thatf.x/ approaches the limit Lasxapproachesx0, and write lim x!x0f.x/DL; iffis defined on some deleted neighborhood of x0and, for every /SI >0 , there is aı >0 such that jf.x//NULLj</SI (2.1.4) if 0<jx/NULx0j<ı: (2.1.5) Figure 2.1.1 depicts the graph of a function for which lim x!x0f.x/ exists. Example 2.1.5 Ifcandxare arbitrary real numbers and f.x/Dcx, then lim x!x0f.x/Dcx0: To prove this, we write jf.x//NULcx0jDjcx/NULcx0jDjcjjx/NULx0j: Ifc¤0, this yields jf.x//NULcx0j</SI (2.1.6) if jx/NULx0j<ı; whereıis any number such that 0<ı/DC4/SI=jcj. IfcD0, thenf.x//NULcx0D0for allx, so (2.1.6 ) holds for all x. We emphasize that Definition 2.1.2 does not involve f.x 0/, or even require that it be defined, since ( 2.1.5 ) excludes the case where xDx0. Example 2.1.6 If f.x/Dxsin1 x; x¤0; then lim x!0f.x/D0 even thoughfis not defined at x0D0, because if 0<jxj<ıD/SI; then jf.x//NUL0jDˇˇˇˇxsin1 xˇˇˇˇ/DC4jxj</SI: On the other hand, the function g.x/Dsin1 x; x¤0; has no limit as xapproaches0, since it assumes all values between /NUL1and1in every neighborhood of the origin (Exercise 2.1.26 ). Section 2.1 Functions and Limits 35 The next theorem says that a function cannot have more than on e limit at a point. Theorem 2.1.3 Iflimx!x0f.x/ exists;then it is uniqueIthat is;if lim x!x0f.x/DL1and lim x!x0f.x/DL2; (2.1.7) thenL1DL2: Proof Suppose that ( 2.1.7 ) holds and let /SI>0 . From Definition 2.1.2 , there are positive numbersı1andı2such that jf.x//NULLij</SI if0<jx/NULx0j<ıi; iD1;2: IfıDmin.ı1;ı2/, then jL1/NULL2jDjL1/NULf.x/Cf.x//NULL2j /DC4jL1/NULf.x/jCjf.x//NULL2j<2/SI if0<jx/NULx0j<ı: We have now established an inequality that does not depend on x; that is, jL1/NULL2j<2/SI: Since this holds for any positive /SI,L1DL2. Definition 2.1.2 is not changed by replacing ( 2.1.4 ) with jf.x//NULLj<K/SI; (2.1.8) whereKis a positive constant, because if either of ( 2.1.4 ) or ( 2.1.8 ) can be made to hold for any/SI > 0 by makingjx/NULx0jsufficiently small and positive, then so can the other (Exercise 2.1.5 ). This may seem to be a minor point, but it is often convenient to work with (2.1.8 ) rather than ( 2.1.4 ), as we will see in the proof of the following theorem. A Useful Theorem about Limits Theorem 2.1.4 If lim x!x0f.x/DL1and lim x!x0g.x/DL2; (2.1.9) then lim x!x0.fCg/.x/DL1CL2; (2.1.10) lim x!x0.f/NULg/.x/DL1/NULL2; (2.1.11) lim x!x0.fg/.x/DL1L2; (2.1.12) and, ifL2¤0, (2.1.13) lim x!x0/DC2f g/DC3 .x/DL1 L2: (2.1.14) 36 Chapter 2 Differential Calculus of Functions of One Variable Proof From ( 2.1.9 ) and Definition 2.1.2 , if/SI>0 , there is aı1>0such that jf.x//NULL1j</SI (2.1.15) if0<jx/NULx0j<ı1, and aı2>0such that jg.x//NULL2j</SI (2.1.16) if0<jx/NULx0j<ı2. Suppose that 0<jx/NULx0j<ıDmin.ı1;ı2/; (2.1.17) so that ( 2.1.15 ) and ( 2.1.16 ) both hold. Then j.f˙g/.x//NUL.L1˙L2/jDj.f.x//NULL1/˙.g.x//NULL2/j /DC4jf.x//NULL1jCjg.x//NULL2j<2/SI; which proves ( 2.1.10 ) and ( 2.1.11 ). To prove ( 2.1.12 ), we assume ( 2.1.17 ) and write j.fg/.x//NULL1L2jDjf.x/g.x//NULL1L2j Djf.x/.g.x//NULL2/CL2.f.x//NULL1/j /DC4jf.x/jjg.x//NULL2jCjL2jjf.x//NULL1j /DC4.jf.x/jCjL2j//SI(from ( 2.1.15 ) and ( 2.1.16 )) /DC4.jf.x//NULL1jCjL1jCjL2j//SI /DC4./SICjL1jCjL2j//SIfrom ( 2.1.15 ) /DC4.1CjL1jCjL2j//SI if/SI<1 andxsatisfies ( 2.1.17 ). This proves ( 2.1.12 ). To prove ( 2.1.14 ), we first observe that if L2¤0, there is aı3>0such that jg.x//NULL2j<jL2j 2; so jg.x/j>jL2j 2(2.1.18) if 0<jx/NULx0j<ı3: To see this, let LDL2and/SIDjL2j=2in (2.1.4 ). Now suppose that 0<jx/NULx0j<min.ı1;ı2;ı3/; so that ( 2.1.15 ), (2.1.16 ), and ( 2.1.18 ) all hold. Then Section 2.1 Functions and Limits 37 ˇˇˇˇ/DC2f g/DC3 .x//NULL1 L2ˇˇˇˇDˇˇˇˇf.x/ g.x//NULL1 L2ˇˇˇˇ DjL2f.x//NULL1g.x/j jg.x/L 2j /DC42 jL2j2jL2f.x//NULL1g.x/j D2 jL2j2jL2Œf.x//NULL1/c141CL1ŒL2/NULg.x//c141j(from ( 2.1.18 )) /DC42 jL2j2ŒjL2jjf.x//NULL1jCjL1jjL2/NULg.x/j/c141 /DC42 jL2j2.jL2jCjL1j//SI(from ( 2.1.15 ) and ( 2.1.16 )): This proves ( 2.1.14 ). Successive applications of the various parts of Theorem 2.1.4 permit us to find limits without the/SI–ıarguments required by Definition 2.1.2 . Example 2.1.7 Use Theorem 2.1.4 to find lim x!29/NULx2 xC1and lim x!2.9/NULx2/.xC1/: Solution Ifcis a constant, then lim x!x0cDc, and, from Example 2.1.5 , lim x!x0xD x0. Therefore, from Theorem 2.1.4 , lim x!2.9/NULx2/Dlim x!29/NULlim x!2x2 Dlim x!29/NUL.lim x!2x/2 D9/NUL22D5; and lim x!2.xC1/Dlim x!2xClim x!21D2C1D3: Therefore, lim x!29/NULx2 xC1Dlim x!2.9/NULx2/ lim x!2.xC1/D5 3 and lim x!2.9/NULx2/.xC1/Dlim x!2.9/NULx2/lim x!2.xC1/D5/SOH3D15: One-Sided Limits The function f.x/D2xsinpx 38 Chapter 2 Differential Calculus of Functions of One Variable satisfies the inequality jf.x/j</SI if0 < x < ıD/SI=2. However, this does not mean that lim x!0f.x/D0, sincefis not defined for negative x, as it must be to satisfy the conditions of Definition 2.1.2 with x0D0andLD0. The function g.x/DxCjxj x; x¤0; can be rewritten as g.x/D/SUBxC1; x>0; x/NUL1; x<0I hence, every open interval containing x0D0also contains points x1andx2such that jg.x 1//NULg.x 2/jis as close to 2as we please. Therefore, lim x!x0g.x/ does not exist (Exercise 2.1.26 ). Althoughf.x/ andg.x/ do not approach limits as xapproaches zero, they each exhibit a definite sort of limiting behavior for small positive value s ofx, as doesg.x/ for small negative values of x. The kind of behavior we have in mind is defined precisely as fo llows. y xx0 x x0 − x x0 +f(x) = λy = f(x) f(x) = µ lim limµ λ Figure 2.1.2 Definition 2.1.5 (a) We say thatf.x/ approaches the left-hand limit Lasxapproachesx0from the left , and write lim x!x0/NULf.x/DL; iffis defined on some open interval .a;x 0/and, for each /SI > 0 , there is aı > 0 such that jf.x//NULLj</SI ifx0/NULı<x<x 0: Section 2.1 Functions and Limits 39 (b) We say thatf.x/ approaches the right-hand limit Lasxapproachesx0from the right , and write lim x!x0Cf.x/DL; iffis defined on some open interval .x0;b/and, for each /SI > 0 , there is aı > 0 such that jf.x//NULLj</SI ifx0<x<x 0Cı: Figure 2.1.2 shows the graph of a function that has distinct left- and righ t-hand limits at a pointx0. Example 2.1.8 Let f.x/Dx jxj; x¤0: Ifx<0 , thenf.x/D/NULx=xD/NUL1, so lim x!0/NULf.x/D/NUL1: Ifx>0 , thenf.x/Dx=xD1, so lim x!0Cf.x/D1: Example 2.1.9 Let g.x/DxCjxj.1Cx/ xsin1 x; x¤0: Ifx<0 , then g.x/D/NULxsin1 x; so lim x!0/NULg.x/D0; since jg.x//NUL0jDˇˇˇˇxsin1 xˇˇˇˇ/DC4jxj</SI if/NUL/SI<x<0 ; that is, Definition 2.1.5(a)is satisfied with ıD/SI. Ifx>0 , then g.x/D.2Cx/sin1 x; which takes on every value between /NUL2and2in every interval .0;ı/ . Hence,g.x/ does not approach a right-hand limit at xapproaches0from the right. This shows that a function may have a limit from one side at a point but fail to have a limit from the other side. 40 Chapter 2 Differential Calculus of Functions of One Variable Example 2.1.10 We leave it to you to verify that lim x!0C/DC2jxj xCx/DC3 D1; lim x!0/NUL/DC2jxj xCx/DC3 D/NUL1; lim x!0CxsinpxD0; and lim x!0/NULsinpxdoes not exist. Left- and right-hand limits are also called one-sided limits . We will often simplify the notation by writing lim x!x0/NULf.x/Df.x 0/NUL/and lim x!x0Cf.x/Df.x 0C/: The following theorem states the connection between limits and one-sided limits. We leave the proof to you (Exercise 2.1.12 ). Theorem 2.1.6 A functionfhas a limit at x0if and only if it has left- and right-hand limits atx0;and they are equal. More specifically ; lim x!x0f.x/DL if and only if f.x 0C/Df.x 0/NUL/DL: With only minor modifications of their proofs (replacing the inequality0<jx/NULx0j<ı byx0/NULı < x < x 0orx0< x < x 0Cı), it can be shown that the assertions of Theo- rems 2.1.3 and2.1.4 remain valid if “lim x!x0” is replaced by “lim x!x0/NUL” or “lim x!x0C” throughout (Exercise 2.1.13 ). Limits at ˙1 Limits and one-sided limits have to do with the behavior of a f unctionfnear a limit point ofDf. It is equally reasonable to study ffor large positive values of xifDfis unbounded above or for large negative values of xifDfis unbounded below. Definition 2.1.7 We say thatf.x/ approaches the limit Lasxapproaches1, and write lim x!1f.x/DL; iffis defined on an interval .a;1/and, for each /SI>0 , there is a number ˇsuch that jf.x//NULLj</SI ifx>ˇ: Section 2.1 Functions and Limits 41 Figure 2.1.3 provides an illustration of the situation described in Defin ition 2.1.7 . x ∞ lim f(x) = L βy L + L −L x Figure 2.1.3 We leave it to you to define the statement “lim x!/NUL1f.x/DL” (Exercise 2.1.14 ) and to show that Theorems 2.1.3 and2.1.4 remain valid if x0is replaced throughout by 1or /NUL1 (Exercise 2.1.16 ). Example 2.1.11 Let f.x/D1/NUL1 x2; g.x/D2jxj 1Cx;andh.x/Dsinx: Then lim x!1f.x/D1; since jf.x//NUL1jD1 x2</SI ifx>1p/SI; and lim x!1g.x/D2; since jg.x//NUL2jDˇˇˇˇ2x 1Cx/NUL2ˇˇˇˇD2 1Cx<2 x</SI ifx>2 /SI: However, lim x!1h.x/ does not exist, since hassumes all values between /NUL1and1in any semi-infinite interval ./FS;1/. We leave it to you to show that lim x!/NUL1f.x/D1, lim x!/NUL1g.x/D /NUL2, and limx!/NUL1h.x/ does not exist (Exercise 2.1.17 ). 42 Chapter 2 Differential Calculus of Functions of One Variable We will sometimes denote lim x!1f.x/ and lim x!/NUL1f.x/ byf.1/andf./NUL1/, respectively. Infinite Limits The functions f.x/D1 x; g.x/D1 x2; p.x/Dsin1 x; and q.x/D1 x2sin1 x do not have limits, or even one-sided limits, at x0D0. They fail to have limits in different ways: /SIf.x/ increases beyond bound as xapproaches0from the right and decreases beyond bound asxapproaches0from the left; /SIg.x/ increases beyond bound as xapproaches zero; /SIp.x/ oscillates with ever-increasing frequency as xapproaches zero; /SIq.x/ oscillates with ever-increasing amplitude and frequency a sxapproaches0. The kind of behavior exhibited by fandgnearx0D0is sufficiently common and simple to lead us to define infinite limits . Definition 2.1.8 We say thatf.x/ approaches1asxapproachesx0from the left , and write lim x!x0/NULf.x/D1 orf.x 0/NUL/D1; iffis defined on an interval .a;x 0/and, for each real number M, there is aı > 0 such that f.x/>M ifx0/NULı<x<x 0: Example 2.1.12 We leave it to you to define the other kinds of infinite limits (E xer- cises 2.1.19 and2.1.21 ) and show that lim x!0/NUL1 xD/NUL1; lim x!0C1 xD1I lim x!0/NUL1 x2Dlim x!0C1 x2Dlim x!01 x2D1I lim x!1x2Dlim x!/NUL1x2D1I and lim x!1x3D1; lim x!/NUL1x3D/NUL1: Section 2.1 Functions and Limits 43 Throughout this book, “lim x!x0f.x/ exists” will mean that lim x!x0f.x/DL; whereLisfinite . To leave open the possibility that LD˙1 , we will say that lim x!x0f.x/ exists in the extended reals. This convention also applies to one-sided limits and limits asxapproaches˙1. We mentioned earlier that Theorems 2.1.3 and2.1.4 remain valid if “lim x!x0” is re- placed by “lim x!x0/NUL” or “lim x!x0C.” They are also valid with x0replaced by˙1. Moreover, the counterparts of ( 2.1.10 ), (2.1.11 ), and ( 2.1.12 ) in all these versions of The- orem 2.1.4 remain valid if either or both of L1andL2are infinite, provided that their right sides are not indeterminate (Exercises 2.1.28 and2.1.29 ). Equation ( 2.1.14 ) and its counterparts remain valid if L1=L2is not indeterminate and L2¤0(Exercise 2.1.30 ). Example 2.1.13 Results like Theorem 2.1.4 yield lim x!1sinhxDlim x!1ex/NULe/NULx 2D1 2/DLE lim x!1ex/NULlim x!1e/NULx/DC1 D1 2.1/NUL0/D1; lim x!/NUL1sinhxDlim x!/NUL1ex/NULe/NULx 2D1 2/DLE lim x!/NUL1ex/NULlim x!/NUL1e/NULx/DC1 D1 2.0/NUL1/D/NUL1; and lim x!1e/NULx xDlim x!1e/NULx lim x!1xD0 1D0: Example 2.1.14 If f.x/De2x/NULex; we cannot obtain lim x!1f.x/ by writing lim x!1f.x/Dlim x!1e2x/NULlim x!1ex; because this produces the indeterminate form 1/NUL1 . However, by writing f.x/De2x.1/NULe/NULx/; we find that lim x!1f.x/D/DLE lim x!1e2x/DC1/DLE lim x!11/NULlim x!1e/NULx/DC1 D1.1/NUL0/D1: 44 Chapter 2 Differential Calculus of Functions of One Variable Example 2.1.15 Let g.x/D2x2/NULxC1 3x2C2x/NUL1: Trying to find lim x!1g.x/ by applying a version of Theorem 2.1.4 to this fraction as it is written leads to an indeterminate form (try it!). However, b y rewriting it as g.x/D2/NUL1=xC1=x2 3C2=x/NUL1=x2; x¤0; we find that lim x!1g.x/Dlim x!12/NULlim x!11=xClim x!11=x2 lim x!13Clim x!12=x/NULlim x!11=x2D2/NUL0C0 3C0/NUL0D2 3: Monotonic Function A functionfisnondecreasing on an interval Iif f.x 1//DC4f.x 2/wheneverx1andx2are inIandx1<x 2; (2.1.19) ornonincreasing onIif f.x 1//NAKf.x 2/wheneverx1andx2are inIandx1<x 2: (2.1.20) In either case, fis onI. If/DC4can be replaced by <in (2.1.19 ),fisincreasing onI. If/NAK can be replaced by >in (2.1.20 ),fisdecreasing onI. In either of these two cases, fis strictly monotonic onI. Example 2.1.16 The function f.x/D(x; 0/DC4x<1; 2; 1/DC4x/DC42; is nondecreasing on IDŒ0;2/c141 (Figure 2.1.4 ), and/NULfis nonincreasing on IDŒ0;2/c141 . 2 2 11y x Section 2.1 Functions and Limits 45 Figure 2.1.4 The function g.x/Dx2is increasing on Œ0;1/(Figure 2.1.5 ), y xy = x2 Figure 2.1.5 andh.x/D/NULx3is decreasing on ./NUL1;1/(Figure 2.1.6 ). y = − x3y x Figure 2.1.6 46 Chapter 2 Differential Calculus of Functions of One Variable In the proof of the following theorem, we assume that you have formulated the definitions called for in Exercise 2.1.19 . Theorem 2.1.9 Suppose that fis monotonic on .a;b/ and define ˛Dinf a<x<bf.x/ andˇDsup a<x<bf.x/: (a) Iffis nondecreasing ;thenf.aC/D˛andf.b/NUL/Dˇ: (b) Iffis nonincreasing ;thenf.aC/Dˇandf.b/NUL/D˛: .HereaCD/NUL1 ifaD/NUL1 andb/NULD1 ifbD1:/ (c) Ifa<x 0<b, thenf.x 0C/andf.x 0/NUL/exist and are finiteImoreover; f.x 0/NUL//DC4f.x 0//DC4f.x 0C/ iffis nondecreasing ;and f.x 0/NUL//NAKf.x 0//NAKf.x 0C/ iffis nonincreasing : Proof (a) We first show that f.aC/D˛. If M > ˛ , there is an x0in.a;b/ such thatf.x 0/ < M . Sincefis nondecreasing, f.x/<M ifa <x <x 0. Therefore, if ˛D/NUL1 , thenf.aC/D/NUL1 . If˛ >/NUL1, let MD˛C/SI, where/SI>0 . Then˛/DC4f.x/<˛C/SI, so jf.x//NUL˛j</SI ifa<x<x 0: (2.1.21) IfaD/NUL1 , this implies that f./NUL1/D˛. Ifa >/NUL1, letıDx0/NULa. Then ( 2.1.21 ) is equivalent to jf.x//NUL˛j</SI ifa<x<aCı; which implies that f.aC/D˛. We now show that f.b/NUL/Dˇ. IfM <ˇ , there is anx0in.a;b/ such thatf.x 0/>M . Sincefis nondecreasing, f.x/ > M ifx0< x < b . Therefore, if ˇD 1 , then f.b/NUL/D1 . Ifˇ<1, letMDˇ/NUL/SI, where/SI>0 . Thenˇ/NUL/SI<f.x//DC4ˇ, so jf.x//NULˇj</SI ifx0<x<b: (2.1.22) IfbD1 , this implies that f.1/Dˇ. Ifb <1, letıDb/NULx0. Then ( 2.1.22 ) is equivalent to jf.x//NULˇj</SI ifb/NULı<x<b; which implies that f.b/NUL/Dˇ. (b) The proof is similar to the proof of (a)(Exercise 2.1.34 ). (c)Suppose that fis nondecreasing. Applying (a) tofon.a;x 0/and.x0;b/sepa- rately shows that f.x 0/NUL/D sup a<x<x 0f.x/ andf.x 0C/D inf x0<x<bf.x/: Section 2.1 Functions and Limits 47 However, ifx1<x 0<x 2, then f.x 1//DC4f.x 0//DC4f.x 2/I hence, f.x 0/NUL//DC4f.x 0//DC4f.x 0C/: We leave the case where fis nonincreasing to you (Exercise 2.1.34 ). Limits Inferior and Superior We now introduce some concepts related to limits. We leave th e study of these concepts mainly to the exercises. We say thatfisbounded on a setSif there is a constant M <1such thatjf.x/j/DC4M for allxinS. Definition 2.1.10 Suppose that fis bounded on Œa;x 0/, wherex0may be finite or1. Fora/DC4x<x 0, define Sf.xIx0/Dsup x/DC4t<x 0f.t/ and If.xIx0/D inf x/DC4t<x 0f.t/: Then the left limit superior of fatx0is defined to be lim x!x0/NULf.x/Dlim x!x0/NULSf.xIx0/; and the left limit inferior of fatx0is defined to be lim x!x0/NULf.x/Dlim x!x0/NULIf.xIx0/: (Ifx0D1 , we definex0/NULD1 .) Theorem 2.1.11 Iffis bounded on Œa;x 0/;thenˇDlimx!x0/NULf.x/ exists and is the unique real number with the following properties W (a) If/SI>0 , there is ana1inŒa;x 0/such that f.x/<ˇC/SIifa1/DC4x<x 0: (2.1.23) (b) If/SI>0 anda1is inŒa;x 0/;then f.x/>ˇ/NUL/SIfor somex2Œa1;x0/: Proof Sincefis bounded on Œa;x 0/,Sf.xIx0/is nonincreasing and bounded on Œa;x 0/. By applying Theorem 2.1.9(b) toSf.xIx0/, we conclude that ˇexists (finite). Therefore, if /SI>0 , there is anainŒa;x 0/such that ˇ/NUL/SI=2<S f.xIx0/<ˇC/SI=2 ifa/DC4x<x 0: (2.1.24) 48 Chapter 2 Differential Calculus of Functions of One Variable SinceSf.xIx0/is an upper bound of˚ f.t/ˇˇx/DC4t <x 0/TAB ,f.x//DC4Sf.xIx0/. Therefore, the second inequality in ( 2.1.24 ) implies ( 2.1.23 ) witha1Da. This proves (a). To prove (b), leta1be given and define x1Dmax.a1;a/. Then the first inequality in ( 2.1.24 ) implies that Sf.x1Ix0/>ˇ/NUL/SI=2: (2.1.25) SinceSf.x1Ix0/is the supremum of˚ f.t/ˇˇx1<t <x 0/TAB , there is anxinŒx1;x0/such that f.x/>S f.x1Ix0//NUL/SI=2: This and ( 2.1.25 ) imply thatf.x/>ˇ/NUL/SI. Sincexis inŒa1;x0/, this proves (b). Now we show that there cannot be more than one real number with properties (a) and (b). Suppose that ˇ1<ˇ 2andˇ2has property (b); thus, if/SI > 0 anda1is inŒa;x 0/, there is anxinŒa1;x0/such thatf.x/>ˇ 2/NUL/SI. Letting/SIDˇ2/NULˇ1, we see that there is anxinŒa1;b/such that f.x/>ˇ 2/NUL.ˇ2/NULˇ1/Dˇ1; soˇ1cannot have property (a). Therefore, there cannot be more than one real number that satisfies both (a)and(b). The proof of the following theorem is similar to this (Exerci se2.1.35 ). Theorem 2.1.12 Iffis bounded on Œa;x 0/;then˛Dlimx!x0/NULf.x/ exists and is the unique real number with the following properties: (a) If/SI>0; there is ana1inŒa;x 0/such that f.x/>˛/NUL/SIifa1/DC4x<x 0: (b) If/SI>0 anda1is inŒa;x 0/;then f.x/<˛C/SIfor somex2Œa1;x0/: 2.1 Exercises 1. Each of the following conditions fails to define a function on any domain. State why. (a)sinf.x/Dx (b)ef .x/D/NULjxj (c)1Cx2CŒf.x//c1412D0 (d)f.x/Œf.x//NUL1/c141Dx2 2. If f.x/Dr .x/NUL3/.xC2/ x/NUL1andg.x/Dx2/NUL16 x/NUL7p x2/NUL9; findDf,Df˙g,Dfg, andDf =g. Section 2.1 Functions and Limits 49 3. FindDf. (a)f.x/Dtanx (b)f.x/D1p 1/NULjsinxj (c)f.x/D1 x.x/NUL1/(d)f.x/Dsinx x (e)eŒf .x//c1412Dx; f.x//NAK0 4. Find lim x!x0f.x/ , and justify your answers with an /SI–ıproof. (a)x2C2xC1; x 0D1 (b)x3/NUL8 x/NUL2; x 0D2 (c)1 x2/NUL1; x 0D0 (d)px; x 0D4 (e)x3/NUL1 .x/NUL1/.x/NUL2/Cx; x 0D1 5. Prove that Definition 2.1.2 is unchanged if Eqn. ( 2.1.4 ) is replaced by jf.x//NULLj<K/SI; whereKis any positive constant. (That is, lim x!x0f.x/DLaccording to Defini- tion2.1.2 if and only if lim x!x0f.x/DLaccording to the modified definition.) 6. Use Theorem 2.1.4 and the known limits lim x!x0xDx0, lim x!x0cDcto find the indicated limits. (a) lim x!2x2C2xC3 2x3C1(b) lim x!2/DC21 xC1/NUL1 x/NUL1/DC3 (c) lim x!1x/NUL1 x3Cx2/NUL2x(d) lim x!1x8/NUL1 x4/NUL1 7. Find lim x!x0/NULf.x/ and lim x!x0Cf.x/ , if they exist. Use /SI–ıproofs, where ap- plicable, to justify your answers. (a)xCjxj x; x 0D0 (b)xcos1 xCsin1 xCsin1 jxj; x 0D0 (c)jx/NUL1j x2Cx/NUL2; x 0D1(d)x2Cx/NUL2pxC2; x 0D/NUL2 8. Prove: Ifh.x//NAK0fora<x<x 0and lim x!x0/NULh.x/ exists, then lim x!x0/NULh.x/ /NAK0. Conclude from this that if f2.x//NAKf1.x/fora<x<x 0, then lim x!x0/NULf2.x//NAKlim x!x0/NULf1.x/ if both limits exist. 50 Chapter 2 Differential Calculus of Functions of One Variable 9. (a) Prove: If lim x!x0f.x/ exists, there is a constant Mand a/SUB > 0 such that jf.x/j /DC4Mif0 <jx/NULx0j< /SUB. (We say then that fisbounded on˚xˇˇ0<jx/NULx0j</SUB/TAB.) (b) State similar results with “lim x!x0” replaced by “lim x!x0/NUL.” (c) State similar results with “lim x!x0” replaced by “lim x!x0C.” 10. Suppose that lim x!x0f.x/DLandnis a positive integer. Prove that lim x!x0Œf.x//c141nD Ln(a) by using Theorem 2.1.4 and induction; (b) directly from Definition 2.1.2 . HINT:You will find Exercise 2.1.9 useful for.b/: 11. Prove: If lim x!x0f.x/DL>0 , then lim x!x0p f.x/Dp L. 12. Prove Theorem 2.1.6 . 13. (a) Using the hint stated after Theorem 2.1.6 , prove that Theorem 2.1.3 remains valid with “lim x!x0” replaced by “lim x!x0/NUL.” (b) Repeat(a)for Theorem 2.1.4 . 14. Define the statement “lim x!/NUL1f.x/DL.” 15. Find lim x!1f.x/ if it exists, and justify your answer directly from Definitio n2.1.7 . (a)1 x2C1(b)sinx jxj˛.˛>0/ (c)sinx jxj˛.˛/DC40/ (d)e/NULxsinx (e)tanx (f)e/NULx2e2x 16. Theorems 2.1.3 and2.1.4 remain valid with “lim x!x0” replaced throughout by “lim x!1” (“lim x!/NUL1 ”). How would their proofs have to be changed? 17. Using the definition you gave in Exercise 2.1.14 , show that (a) lim x!/NUL1/DC2 1/NUL1 x2/DC3 D1 (b) lim x!/NUL12jxj 1CxD/NUL2 (c) lim x!/NUL1sinxdoes not exist 18. Find lim x!/NUL1f.x/ , if it exists, for each function in Exercise 2.1.15 . Justify your answers directly from the definition you gave in Exercise 2.1.14 . 19. Define (a) lim x!x0/NULf.x/D/NUL1 (b) lim x!x0Cf.x/D1 (c) lim x!x0Cf.x/D/NUL1 20. Find (a) lim x!0C1 x3(b) lim x!0/NUL1 x3 (c) lim x!0C1 x6(d) lim x!0/NUL1 x6 (e) lim x!x0C1 .x/NULx0/2k(f) lim x!x0/NUL1 .x/NULx0/2kC1 (kDpositive integer) Section 2.1 Functions and Limits 51 21. Define (a) lim x!x0f.x/D1 (b) lim x!x0f.x/D/NUL1 22. Find (a) lim x!01 x3(b) lim x!01 x6 (c) lim x!x01 .x/NULx0/2k(d) lim x!x01 .x/NULx0/2kC1 (kDpositive integer) 23. Define (a) lim x!1f.x/D1 (b) lim x!/NUL1f.x/D/NUL1 24. Find (a) lim x!1x2k(b) lim x!/NUL1x2k (c) lim x!1x2kC1(d) lim x!/NUL1x2kC1 (k=positive integer) (e) lim x!1pxsinx (f) lim x!1ex 25. Suppose that fandgare defined on .a;1/and.c;1/respectively, and that g.x/ >a ifx >c . Suppose also that lim x!1f.x/DL, where/NUL1/DC4L/DC41 , and lim x!1g.x/D1 . Show that lim x!1f.g.x//DL. 26. (a) Prove: lim x!x0f.x/ does not exist (finite) if for some /SI0>0, every deleted neighborhood of x0contains points x1andx2such that jf.x 1//NULf.x 2/j/NAK/SI0: (b) Give analogous conditions for the nonexistence of lim x!x0Cf.x/; lim x!x0/NULf.x/; lim x!1f.x/; and lim x!/NUL1f.x/: 27. Prove: If/NUL1< x 0<1, then lim x!x0f.x/ exists in the extended reals if and only if lim x!x0/NULf.x/ and lim x!x0Cf.x/ both exist in the extended reals and are equal, in which case all three are equal. In Exercises 2.1.28 –2.1.30 consider only the case where at least one of L1andL2is˙1. 28. Prove: If lim x!x0f.x/DL1, lim x!x0g.x/DL2, andL1CL2is not indetermi- nate, then lim x!x0.fCg/.x/DL1CL2: 52 Chapter 2 Differential Calculus of Functions of One Variable 29. Prove: If lim x!1f.x/DL1, lim x!1g.x/DL2, andL1L2is not indeterminate, then lim x!1.fg/.x/DL1L2: 30. (a) Prove: If lim x!x0f.x/DL1, lim x!x0g.x/DL2¤0, andL1=L2is not indeterminate, then lim x!x0/DC2f g/DC3 .x/DL1 L2: (b) Show that it is necessary to assume that L2¤0in(a)by considering f.x/D sinx,g.x/Dcosx, andx0D/EM=2. 31. Find (a) lim x!0Cx3C2xC3 2x4C3x2C2(b) lim x!0/NULx3C2xC3 2x4C3x2C2 (c) lim x!12x4C3x2C2 x3C2xC3(d) lim x!/NUL12x4C3x2C2 x3C2xC3 (e)limx!1.ex2/NULex/ (f) lim x!1xCpxsinx 2xCe/NULx 32. Find lim x!1r.x/ and lim x!/NUL1r.x/ for the rational function r.x/Da0Ca1xC/SOH/SOH/SOHCanxn b0Cb1xC/SOH/SOH/SOHCbmxm; wherean¤0andbm¤0. 33. Suppose that lim x!x0f.x/ exists for every x0in.a;b/ andg.x/Df.x/ except on a setSwith no limit points in .a;b/ . What can be said about lim x!x0g.x/ for x0in.a;b/ ? Justify your answer. 34. Prove Theorem 2.1.9(b), and complete the proof of Theorem 2.1.9(b) in the case wherefis nonincreasing. 35. Prove Theorem 2.1.12 . 36. Show that iffis bounded on Œa;x 0/, then (a) lim x!x0/NULf.x//DC4lim x!x0/NULf.x/ . (b) lim x!x0/NUL./NULf/.x/D/NUL lim x!x0/NULf.x/ and lim x!x0/NUL./NULf/.x/D/NUL lim x!x0/NULf.x/ . (c) lim x!x0/NULf.x/D lim x!x0/NULf.x/ if and only if lim x!x0/NULf.x/ exists, in which case lim x!x0/NULf.x/Dlim x!x0/NULf.x/Dlim x!x0/NULf.x/: 37. Suppose that fandgare bounded on Œa;x 0/. Section 2.2 Continuity 53 (a) Show that lim x!x0/NUL.fCg/.x//DC4lim x!x0/NULf.x/Clim x!x0/NULg.x/: (b) Show that lim x!x0/NUL.fCg/.x//NAKlim x!x0/NULf.x/Clim x!x0/NULg.x/: (c) State inequalities analogous to those in (a)and(b) for lim x!x0/NUL.f/NULg/.x/ and lim x!x0/NUL.f/NULg/.x/: 38. Prove: lim x!x0/NULf.x/ exists (finite) if and only if for each /SI > 0 there is aı >0 such thatjf.x 1//NULf.x 2/j< /SI ifx0/NULı < x 1,x2< x 0. H INT:For sufficiency ; show thatfis bounded on some interval .a;x 0/and lim x!0/NULf.x/Dlim x!x0/NULf.x/: Then use Exercise 2.1.36.c/: 39. Suppose that fis bounded on an interval .x0;b/c141. Using Definition 2.1.10 as a guide, define limx!x0Cf.x/ (the right limit superior of fatx0) and limx!x0Cf.x/ (the right limit inferior of fatx0). Then prove that they exist. H INT:Use Theorem 2.1.9: 40. Suppose that fis bounded on an interval .x0;b/c141. Show that limx!x0Cf.x/D limx!x0Cf.x/ if and only if lim x!x0Cf.x/ exists, in which case lim x!x0Cf.x/Dlim x!x0Cf.x/Dlim x!x0Cf.x/: 41. Suppose thatfis bounded on an open interval containing x0. Show that lim x!x0f.x/ exists if and only if lim x!x0/NULf.x/Dlim x!x0Cf.x/Dlim x!x0/NULf.x/Dlim x!x0Cf.x/; in which case lim x!x0f.x/ is the common value of these four expressions. 2.2 CONTINUITY In this section we study continuous functions of a real varia ble. We will prove some impor- tant theorems about continuous functions that, although in tuitively plausible, are beyond the scope of the elementary calculus course. They are access ible now because of our better understanding of the real number system, especially of thos e properties that stem from the completeness axiom. 54 Chapter 2 Differential Calculus of Functions of One Variable The definitions of f.x 0/NUL/Dlim x!x0/NULf.x/; f.x 0C/Dlim x!x0Cf.x/; and lim x!x0f.x/ do not involve f.x 0/or even require that it be defined. However, the case where f.x 0/is defined and equal to one or more of these quantities is importa nt. Definition 2.2.1 (a) We say thatfiscontinuous at x0iffis defined on an open interval .a;b/ containing x0and lim x!x0f.x/Df.x 0/. (b) We say thatfiscontinuous from the left at x0iffis defined on an open interval .a;x 0/andf.x 0/NUL/Df.x 0/. (c) We say thatfiscontinuous from the right at x0iffis defined on an open interval .x0;b/andf.x 0C/Df.x 0/. The following theorem provides a method for determining whe ther these definitions are satisfied. The proof, which we leave to you (Exercise 2.2.1 ), rests on Definitions 2.1.2 , 2.1.5 , and 2.2.1 . Theorem 2.2.2 (a) A functionfis continuous at x0if and only iffis defined on an open interval .a;b/ containingx0and for each /SI>0 there is aı>0 such that jf.x//NULf.x 0/j</SI (2.2.1) wheneverjx/NULx0j<ı: (b) A functionfis continuous from the right at x0if and only if fis defined on an intervalŒx0;b/and for each /SI>0 there is aı >0 such that (2.2.1 )holds whenever x0/DC4x<x 0Cı: (c) A functionfis continuous from the left at x0if and only iffis defined on an interval .a;x 0/c141and for each /SI>0 there is aı>0 such that (2.2.1 )holds whenever x0/NULı<x/DC4x0: From Definition 2.2.1 and Theorem 2.2.2 ,fis continuous at x0if and only if f.x 0/NUL/Df.x 0C/Df.x 0/ or, equivalently, if and only if it is continuous from the rig ht and left atx0(Exercise 2.2.2 ). Example 2.2.1 Letfbe defined on Œ0;2/c141 by f.x/D/SUBx2; 0/DC4x<1; xC1; 1/DC4x/DC42 Section 2.2 Continuity 55 (Figure 2.2.1 ); then f.0C/D0Df.0/; f.1/NUL/D1¤f.1/D2; f.1C/D2Df.1/; f.2/NUL/D3Df.2/: Therefore,fis continuous from the right at 0and1and continuous from the left at 2, but not at1. If0<x ,x0<1, then jf.x//NULf.x 0/jDjx2/NULx2 0jDjx/NULx0jjxCx0j /DC42jx/NULx0j</SI ifjx/NULx0j</SI=2: Hence,fis continuous at each x0in.0;1/ . If1<x ,x0<2, then jf.x//NULf.x 0/jDj.xC1//NUL.x0C1/Djx/NULx0j </SI ifjx/NULx0j</SI: Hence,fis continous at each x0in.1;2/ . 23 2 11y xy = x + 1, 1 ≤ x ≤ 2 y = x2, 0 ≤ x < 1 Figure 2.2.1 Definition 2.2.3 A functionfiscontinuous on an open interval .a;b/ if it is continu- ous at every point in .a;b/ . If, in addition, f.b/NUL/Df.b/ (2.2.2) or f.aC/Df.a/ (2.2.3) 56 Chapter 2 Differential Calculus of Functions of One Variable thenfiscontinuous on .a;b/c141 orŒa;b/ , respectively. If fis continuous on .a;b/ and (2.2.2 ) and ( 2.2.3 ) both hold, then fis continuous on Œa;b/c141 . More generally, if Sis a subset ofDfconsisting of finitely or infinitely many disjoint intervals , thenfiscontinuous on S iffis continuous on every interval in S. (Henceforth, in connection with functions of one variable, whenever we say “ fis continuous on S” we mean that Sis a set of this kind.) Example 2.2.2 Letf.x/Dpx,0/DC4x<1. Then jf.x//NULf.0/jDpx</SI if0/DC4x</SI2; sof.0C/Df.0/ . Ifx0>0andx/NAK0, then jf.x//NULf.x 0/jDjpx/NULpx0jDjx/NULx0jpxCpx0 /DC4jx/NULx0jpx0</SI ifjx/NULx0j</SIpx0; so lim x!x0f.x/Df.x 0/. Hence,fis continuous on Œ0;1/. Example 2.2.3 The function g.x/D1 sin/EMx is continuous on SDS1 nD/NUL1.n;nC1/. However,gis not continuous at any x0Dn (integer), since it is not defined at such points. The function fdefined in Example 2.2.1 (see also Figure 2.2.1 ) is continuous on Œ0;1/ andŒ1;2/c141 , but not on any open interval containing 1. The discontinuit y offthere is of the simplest kind, described in the following definition. Definition 2.2.4 A functionfispiecewise continuous onŒa;b/c141 if (a)f.x 0C/exists for allx0inŒa;b/ ; (b)f.x 0/NUL/exists for allx0in.a;b/c141 ; (c)f.x 0C/Df.x 0/NUL/Df.x 0/for all but finitely many points x0in.a;b/ . If(c)fails to hold at some x0in.a;b/ ,fhas a jump discontinuity at x0. Also,fhas a jump discontinuity at aiff.aC/¤f.a/ oratbiff.b/NUL/¤f.b/ . Example 2.2.4 The function f.x/D8 ˆˆˆˆˆˆ< ˆˆˆˆˆˆ:1; xD0; x; 0<x<1; 2; xD1; x; 1<x/DC42; /NUL1; 2<x<3; 0; xD3; (Figure 2.2.2 ) is the graph of a piecewise continuous function on Œ0;3/c141 , with jump discon- tinuities atx0D0,1,2, and3. Section 2.2 Continuity 57 23 2 3 11 −1y x Figure 2.2.2 The reason for the adjective “jump” can be seen in Figures 2.2.1 and2.2.2 , where the graphs exhibit a definite jump at each point of discontinuity . The next example shows that not all discontinuities are of this kind. Example 2.2.5 The function f.x/D8 ˆ< ˆ:sin1 x; x¤0; 0; xD0; is continuous at all x0exceptx0D0. Asxapproaches0from either side, f.x/ oscillates between/NUL1and1with ever-increasing frequency, so neither f.0C/norf.0/NUL/exists. Therefore, the discontinuity of fat0is not a jump discontinuity, and if /SUB >0 , thenfis not piecewise continuous on any interval of the form Œ/NUL/SUB;0/c141,Œ/NUL/SUB;/SUB/c141, orŒ0;/SUB/c141 . Theorems 2.1.4 and2.2.2 imply the next theorem (Exercise 2.2.18 ). Theorem 2.2.5 Iffandgare continuous on a set S;then so arefCg;f/NULg;and fg:In addition;f=g is continuous at each x0inSsuch thatg.x 0/¤0: Example 2.2.6 Since the constant functions and the function f.x/Dxare continu- ous for allx, successive applications of the various parts of Theorem 2.2.5 imply that the function r.x/D9/NULx2 xC1 58 Chapter 2 Differential Calculus of Functions of One Variable is continuous for all xexceptxD/NUL1(see Example 2.1.7 ). More generally, by starting from Theorem 2.2.5 and using induction, it can be shown that if f1,f2, . . . ,fnare continuous on a set S, then so are f1Cf2C/SOH/SOH/SOHCfnandf1f2/SOH/SOH/SOHfn. Therefore, any rational function r.x/Da0Ca1xC/SOH/SOH/SOHCanxn b0Cb1xC/SOH/SOH/SOHCbmxm.bm¤0/ is continuous for all values of xexcept those for which its denominator vanishes. Removable Discontinuities Letfbe defined on a deleted neighborhood of x0and discontinuous (perhaps even unde- fined) atx0. We say that fhas a atx0if lim x!x0f.x/ exists. In this case, the function g.x/D8 < :f.x/ ifx2Dfandx¤x0; lim x!x0f.x/ ifxDx0; is continuous at x0. Example 2.2.7 The function f.x/Dxsin1 x is not defined at x0D0, and therefore certainly not continuous there, but lim x!0f.x/D0 (Example 2.1.6 ). Therefore, fhas a removable discontinuity at 0. The function f1.x/Dsin1 x is undefined at 0and its discontinuity there is not removable, since lim x!0f1.x/does not exist (Example 2.2.5 ). Composite Functions We have seen that the investigation of limits and continuity can be simplified by regarding a given function as the result of addition, subtraction, mult iplication, and division of simpler functions. Another operation useful in this connection is composition of functions; that is, substitution of one function into another. Definition 2.2.6 Suppose that fandgare functions with domains DfandDg. If Dghas a nonempty subset Tsuch thatg.x/2Dfwheneverx2T, then the composite functionfıgis defined on Tby .fıg/.x/Df.g.x//: Section 2.2 Continuity 59 Example 2.2.8 If f.x/Dlogxandg.x/D1 1/NULx2; then DfD.0;1/andDgD˚xˇˇx¤˙1/TAB: Sinceg.x/>0 ifx2TD./NUL1;1/, the composite function fıgis defined on ./NUL1;1/ by .fıg/.x/Dlog1 1/NULx2: We leave it to you to verify that gıfis defined on .0;1=e/[.1=e;e/[.e;1/by .gıf/.x/D1 1/NUL.logx/2: The next theorem says that the composition of continuous fun ctions is continuous. Theorem 2.2.7 Suppose that gis continuous at x0;g.x 0/is an interior point of Df; andfis continuous at g.x 0/:Thenfıgis continuous at x0: Proof Suppose that /SI>0 . Sinceg.x 0/is an interior point of Dfandfis continuous atg.x 0/, there is aı1>0such thatf.t/ is defined and jf.t//NULf.g.x 0//j</SI ifjt/NULg.x 0/j<ı1: (2.2.4) Sincegis continuous at x0, there is aı>0 such thatg.x/ is defined and jg.x//NULg.x 0/j<ı1ifjx/NULx0j<ı: (2.2.5) Now ( 2.2.4 ) and ( 2.2.5 ) imply that jf.g.x///NULf.g.x 0//j</SI ifjx/NULx0j<ı: Therefore,fıgis continuous at x0. See Exercise 2.2.22 for a related result concerning limits. Example 2.2.9 In Examples 2.2.2 and2.2.6 we saw that the function f.x/Dpx is continuous for x>0 , and the function g.x/D9/NULx2 xC1 is continuous for x¤/NUL1. Sinceg.x/ > 0 ifx </NUL3or/NUL1 < x < 3 , Theorem 2.2.7 implies that the function .fıg/.x/Ds 9/NULx2 xC1 is continuous on ./NUL1;/NUL3/[./NUL1;3/. It is also continuous from the left at /NUL3and3. 60 Chapter 2 Differential Calculus of Functions of One Variable Bounded Functions A functionfisbounded below on a setSif there is a real number msuch that f.x//NAKmfor allx2S: In this case, the set VD˚f.x/ˇˇx2S/TAB has an infimum ˛, and we write ˛Dinf x2Sf.x/: If there is a point x1inSsuch thatf.x 1/D˛, we say that ˛is the minimum offonS, and write ˛Dmin x2Sf.x/: Similarly,fisbounded above on Sif there is a real number Msuch thatf.x//DC4Mfor allxinS. In this case, Vhas a supremum ˇ, and we write ˇDsup x2Sf.x/: If there is a point x2inSsuch thatf.x 2/Dˇ, we say that ˇis the maximum offonS, and write ˇDmax x2Sf.x/: Iffis bounded above and below on a set S, we say thatfisbounded onS. Figure 2.2.3 illustrates the geometric meaning of these definitions for a functionf bounded on an interval SDŒa;b/c141 . The graph of flies in the strip bounded by the linesyDMandyDm, whereMis any upper bound and mis any lower bound forfonŒa;b/c141 . The narrowest strip containing the graph is the one bounded above by yDˇDsupa/DC4x/DC4bf.x/ and below by yD˛Dinfa/DC4x/DC4bf.x/ . y xy = αy = β y = my = M Figure 2.2.3 Section 2.2 Continuity 61 Example 2.2.10 The function g.x/D(1 2; xD0orxD1; 1/NULx; 0<x<1;C (Figure 2.2.4(a)) is bounded on Œ0;1/c141 , and sup 0/DC4x/DC41g.x/D1; inf 0/DC4x/DC41g.x/D0: Therefore,ghas no maximum or minimum on Œ0;1/c141 , since it does not assume either of the values0and1. The function h.x/D1/NULx; 0/DC4x/DC41; which differs from gonly at0and1(Figure 2.2.4(b)), has the same supremum and infi- mum asg, but it attains these values at xD0andxD1, respectively; therefore, max 0/DC4x/DC41h.x/D1and min 0/DC4x/DC41h.x/D0: 2 11 1y x 11y x (a) (b)y = g(x) y = 1 − x Figure 2.2.4 Example 2.2.11 The function f.x/Dex.x/NUL1/sin1 x.x/NUL1/; 0<x<1; oscillates between˙ex.x/NUL1/infinitely often in every interval of the form .0;/SUB/ or.1/NUL/SUB;1/ , where0</SUB<1 , and sup 0<x<1f.x/D1; inf 0<x<1f.x/D/NUL1: However,fdoes not assume these values, so fhas no maximum or minimum on .0;1/ . 62 Chapter 2 Differential Calculus of Functions of One Variable Theorem 2.2.8 Iffis continuous on a finite closed interval Œa;b/c141; thenfis bounded onŒa;b/c141: Proof Suppose that t2Œa;b/c141 . Sincefis continuous at t, there is an open interval It containingtsuch that jf.x//NULf.t/j<1 ifx2It\Œa;b/c141: (2.2.6) (To see this, set /SID1in (2.2.1 ), Theorem 2.2.2 .) The collection HD˚ Itˇˇa/DC4t/DC4b/TAB is an open covering of Œa;b/c141 . SinceŒa;b/c141 is compact, the Heine–Borel theorem implies that there are finitely many points t1,t2, . . . ,tnsuch that the intervals It1,It2, . . . ,Itncover Œa;b/c141 . According to ( 2.2.6 ) withtDti, jf.x//NULf.ti/j<1 ifx2Iti\Œa;b/c141: Therefore, jf.x/jDj.f.x//NULf.ti//Cf.ti/j/DC4jf.x//NULf.ti/jCjf.ti/j /DC41Cjf.ti/jifx2Iti\Œa;b/c141:(2.2.7) Let MD1Cmax 1/DC4i/DC4njf.ti/j: SinceŒa;b/c141/SUBSn iD1/NUL Iti\Œa;b/c141/SOH , (2.2.7 ) implies thatjf.x/j/DC4Mifx2Œa;b/c141 . This proof illustrates the utility of the Heine–Borel theor em, which allows us to choose Mas the largest of a finite set of numbers. Theorem 2.2.8 and the completeness of the reals imply that iffis continuous on a finite closed interval Œa;b/c141 , thenfhas an infimum and a supre- mum onŒa;b/c141 . The next theorem shows that factually assumes these values at some points inŒa;b/c141 . Theorem 2.2.9 Suppose that fis continuous on a finite closed interval Œa;b/c141: Let ˛Dinf a/DC4x/DC4bf.x/ andˇDsup a/DC4x/DC4bf.x/: Then˛andˇare respectively the minimum and maximum of fonŒa;b/c141Ithat is;there are pointsx1andx2inŒa;b/c141 such that f.x 1/D˛andf.x 2/Dˇ: Proof We show that x1exists and leave it to you to show that x2exists (Exercise 2.2.24 ). Suppose that there is no x1inŒa;b/c141 such thatf.x 1/D˛. Thenf.x/ > ˛ for all x2Œa;b/c141 . We will show that this leads to a contradiction. Suppose that t2Œa;b/c141 . Thenf.t/>˛ , so f.t/>f.t/C˛ 2>˛: Section 2.2 Continuity 63 Sincefis continuous at t, there is an open interval Itabouttsuch that f.x/>f.t/C˛ 2ifx2It\Œa;b/c141 (2.2.8) (Exercise 2.2.15 ). The collection HD˚ Itˇˇa/DC4t/DC4b/TAB is an open covering of Œa;b/c141 . Since Œa;b/c141 is compact, the Heine–Borel theorem implies that there are fi nitely many points t1, t2, . . . ,tnsuch that the intervals It1,It2, . . . ,ItncoverŒa;b/c141 . Define ˛1Dmin 1/DC4i/DC4nf.ti/C˛ 2: Then, sinceŒa;b/c141/SUBSn iD1.Iti\Œa;b/c141/ , (2.2.8 ) implies that f.t/>˛ 1; a/DC4t/DC4b: But˛1>˛, so this contradicts the definition of ˛. Therefore,f.x 1/D˛for somex1in Œa;b/c141 . Example 2.2.12 We used the compactness of Œa;b/c141 in the proof of Theorem 2.2.9 when we invoked the Heine–Borel theorem. To see that compact ness is essential to the proof, consider the function g.x/D1/NUL.1/NULx/sin1 x; which is continuous and has supremum 2on the noncompact interval .0;1/c141 , but does not assume its supremum on .0;1/c141 , since g.x//DC41C.1/NULx/ˇˇˇˇsin1 xˇˇˇˇ /DC41C.1/NULx/<2 if0<x/DC41: As another example, consider the function f.x/De/NULx; which is continuous and has infimum 0, which it does not attain, on the noncompact interval .0;1/. The next theorem shows that if fis continuous on a finite closed interval Œa;b/c141 , thenf assumes every value between f.a/ andf.b/ asxvaries fromatob(Figure 2.2.5 , page 64). Theorem 2.2.10 (Intermediate Value Theorem) Suppose that fis con- tinuous onŒa;b/c141;f.a/¤f.b/; and/SYNis betweenf.a/ andf.b/: Thenf.c/D/SYNfor somecin.a;b/: 64 Chapter 2 Differential Calculus of Functions of One Variable a b xxy y = f(x) y = µ Figure 2.2.5 Proof Suppose that f.a/</SYN<f.b/ . The set SD˚xˇˇa/DC4x/DC4bandf.x//DC4/SYN/TAB is bounded and nonempty. Let cDsupS. We will show that f.c/D/SYN. Iff.c/ > /SYN , thenc > a and, sincefis continuous at c, there is an /SI > 0 such thatf.x/ > /SYN if c/NUL/SI < x/DC4c(Exercise 2.2.15 ). Therefore, c/NUL/SIis an upper bound for S, which contradicts the definition of cas the supremum of S. Iff.c/</SYN , thenc<b and there is an/SI>0 such thatf.x/</SYN forc/DC4x<cC/SI, socis not an upper bound for S. This is also a contradiction. Therefore, f.c/D/SYN. The proof for the case where f.b/</SYN<f.a/ can be obtained by applying this result to/NULf. Uniform Continuity Theorem 2.2.2 and Definition 2.2.3 imply that a functionfis continuous on a subset Sof its domain if for each /SI>0 and eachx0inS, there is aı>0 ,which may depend upon x0as well as/SI, such that jf.x//NULf.x 0/j</SI ifjx/NULx0j<ı andx2Df: The next definition introduces another kind of continuity on a setS. Definition 2.2.11 A functionfisuniformly continuous on a subsetSof its domain if, for every/SI>0 , there is aı>0 such that jf.x//NULf.x0/j</SIwheneverjx/NULx0j<ıandx;x02S: We emphasize that in this definition ıdepends only on /SIandSand not on the particular choice ofxandx0, provided that they are both in S. Example 2.2.13 The function f.x/D2x Section 2.2 Continuity 65 is uniformly continuous on ./NUL1;1/, since jf.x//NULf.x0/jD2jx/NULx0j</SI ifjx/NULx0j</SI=2: Example 2.2.14 If0<r <1, then the function g.x/Dx2 is uniformly continuous on Œ/NULr;r/c141. To see this, note that jg.x//NULg.x0/Djx2/NUL.x0/2jDjx/NULx0jjxCx0j/DC42rjx/NULx0j; so jg.x//NULg.x0/j</SI ifjx/NULx0j<ıD/SI 2rand/NULr/DC4x;x0/DC4r: Often a concept is clarified by considering its negation: a fu nctionfisnotuniformly continuous on Sif there is an/SI0>0such that ifıis any positive number, there are points xandx0inSsuch that jx/NULx0j<ı butjf.x//NULf.x0/j/NAK/SI0: Example 2.2.15 The function g.x/Dx2is uniformly continuous on Œ/NULr;r/c141 for any finiter(Example 2.2.14 ), but not on ./NUL1;1/. To see this, we will show that if ı > 0 there are real numbers xandx0such that jx/NULx0jDı=2 andjg.x//NULg.x0/j/NAK1: To this end, we write jg.x//NULg.x0/jDjx2/NUL.x0/2jDjx/NULx0jjxCx0j: Ifjx/NULx0jDı=2andx;x0>1=ı , then jx/NULx0jjxCx0j>ı 2/DC21 ıC1 ı/DC3 D1: Example 2.2.16 The function f.x/Dcos1 x is continuous on .0;1/c141 (Exercise 2.2.23(i)). However,fis not uniformly continuous on .0;1/c141 , sinceˇˇˇˇf/DC21 n/EM/DC3 /NULf/DC21 .nC1//EM/DC3ˇˇˇˇD2; nD1;2;:::: Examples 2.2.15 and2.2.16 show that a function may be continuous but not uniformly continuous on an interval. The next theorem shows that this c annot happen if the interval is closed and bounded, and therefore compact. 66 Chapter 2 Differential Calculus of Functions of One Variable Theorem 2.2.12 Iffis continuous on a closed and bounded interval Œa;b/c141; thenf is uniformly continuous on Œa;b/c141: Proof Suppose that /SI >0 . Sincefis continuous on Œa;b/c141 , for eachtinŒa;b/c141 there is a positive number ıtsuch that jf.x//NULf.t/j</SI 2ifjx/NULtj<2ı tandx2Œa;b/c141: (2.2.9) IfItD.t/NULıt;tCıt/, the collection HD˚Itˇˇt2Œa;b/c141/TAB is an open covering of Œa;b/c141 . SinceŒa;b/c141 is compact, the Heine–Borel theorem implies that there are finitely many points t1,t2, . . . ,tninŒa;b/c141 such thatIt1,It2, . . . ,ItncoverŒa;b/c141 . Now define ıDminfıt1;ıt2;:::;ı tng: (2.2.10) We will show that if jx/NULx0j<ı andx;x02Œa;b/c141; (2.2.11) thenjf.x//NULf.x0/j</SI. From the triangle inequality, jf.x//NULf.x0/jDj.f.x//NULf.tr//C.f.t r//NULf.x0//j /DC4jf.x//NULf.tr/jCjf.tr//NULf.x0/j:(2.2.12) SinceIt1,It2, . . . ,ItncoverŒa;b/c141 ,xmust be in one of these intervals. Suppose that x2Itr; that is, jx/NULtrj<ıtr: (2.2.13) From ( 2.2.9 ) withtDtr, jf.x//NULf.tr/j</SI 2: (2.2.14) From ( 2.2.11 ), (2.2.13 ), and the triangle inquality, jx0/NULtrjDj.x0/NULx/C.x/NULtr/j/DC4jx0/NULxjCjx/NULtrj<ıCıtr/DC42ıtr: Therefore, ( 2.2.9 ) withtDtrandxreplaced byx0implies that jf.x0//NULf.tr/j</SI 2: This, ( 2.2.12 ), and ( 2.2.14 ) imply thatjf.x//NULf.x0/j</SI. This proof again shows the utility of the Heine–Borel theore m, which allowed us to defineıin (2.2.10 ) as the smallest of a finite set of positive numbers, so that ıis sure to be positive. (An infinite set of positive numbers may fail to hav e a smallest positive member; for example, consider the open interval .0;1/ .) Corollary 2.2.13 Iffis continuous on a set T;thenfis uniformly continuous on any finite closed interval contained in T: Section 2.2 Continuity 67 Applied to Example 2.2.16 , Corollary 2.2.13 implies that the function g.x/Dcos1=x is uniformly continuous on Œ/SUB;1/c141 if0</SUB<1 . More About Monotonic Functions Theorem 2.1.9 implies that if fis monotonic on an interval I, thenfis either continuous or has a jump discontinuity at each x0inI. This and Theorem 2.2.10 provide the key to the proof of the following theorem. Theorem 2.2.14 Iffis monotonic and nonconstant on Œa;b/c141; thenfis continuous on Œa;b/c141 if and only if its range RfD˚f.x/ˇˇx2Œa;b/c141/TABis the closed interval with endpoints f.a/ andf.b/: Proof We assume that fis nondecreasing, and leave the case where fis nonincreasing to you (Exercise 2.2.34 ). Theorem 2.1.9(a)implies that the set eRfD˚ f.x/ˇˇx2.a;b//TAB is a subset of the open interval .f.aC/;f.b/NUL//. Therefore, RfDff.a/g[eRf[ff.b/g/SUBff.a/g[.f.aC/;f.b/NUL//[ff.b/g: (2.2.15) Now suppose that fis continuous on Œa;b/c141 . Thenf.a/Df.aC/,f.b/NUL/Df.b/ , so (2.2.15 ) implies that Rf/SUBŒf.a/;f.b//c141 . Iff.a/ < /SYN < f.b/ , then Theorem 2.2.10 implies that/SYNDf.x/ for somexin.a;b/ . Hence,RfDŒf.a/;f.b//c141 . For the converse, suppose that RfDŒf.a/;f.b//c141 . Sincef.a//DC4f.aC/andf.b/NUL//DC4 f.b/ , (2.2.15 ) implies that f.a/Df.aC/andf.b/NUL/Df.b/ . We know from Theo- rem2.1.9(c)that iffis nondecreasing and a<x 0<b, then f.x 0/NUL//DC4f.x 0//DC4f.x 0C/: If either of these inequalities is strict, Rfcannot be an interval. Since this contradicts our assumption, f.x 0/NUL/Df.x 0/Df.x 0C/. Therefore, fis continuous at x0(Exer- cise2.2.2 ). We can now conclude that fis continuous on Œa;b/c141 . Theorem 2.2.14 implies the following theorem. Theorem 2.2.15 Suppose thatfis increasing and continuous on Œa;b/c141; and letf.a/D candf.b/Dd:Then there is a unique function gdefined onŒc;d/c141 such that g.f.x//Dx; a/DC4x/DC4b; (2.2.16) and f.g.y//Dy; c/DC4y/DC4d: (2.2.17) Moreover;gis continuous and increasing on Œc;d/c141: Proof We first show that there is a function gsatisfying ( 2.2.16 ) and ( 2.2.17 ). Sincef is continuous, Theorem 2.2.14 implies that for each y0inŒc;d/c141 there is anx0inŒa;b/c141 such that f.x 0/Dy0; (2.2.18) 68 Chapter 2 Differential Calculus of Functions of One Variable and, sincefis increasing, there is only one such x0. Define g.y 0/Dx0: (2.2.19) The definition of x0is illustrated in Figure 2.2.6 : withŒc;d/c141 drawn on the y-axis, find the intersection of the line yDy0with the curve yDf.x/ and drop a vertical from the intersection to the x-axis to findx0. y d c a bxy = f(x) x0 y0 Figure 2.2.6 Substituting ( 2.2.19 ) into ( 2.2.18 ) yields f.g.y 0//Dy0; and substituting ( 2.2.18 ) into ( 2.2.19 ) yields g.f.x 0//Dx0: Dropping the subscripts in these two equations yields ( 2.2.16 ) and ( 2.2.17 ). The uniqueness of gfollows from our assumption that fis increasing, and therefore only one value of x0can satisfy ( 2.2.18 ) for eachy0. To see thatgis increasing, suppose that y1<y 2and letx1andx2be the points in Œa;b/c141 such thatf.x 1/Dy1andf.x 2/Dy2. Sincefis increasing, x1<x 2. Therefore, g.y 1/Dx1<x 2Dg.y 2/; sogis increasing. Since RgD˚ g.y/ˇˇy2Œc;d/c141/TAB is the interval Œg.c/;g.d//c141DŒa;b/c141 , Theorem 2.2.14 withfandŒa;b/c141 replaced bygandŒc;d/c141 implies thatgis continuous on Œc;d/c141 . The function gof Theorem 2.2.15 is the inverse off, denoted by f/NUL1. Since ( 2.2.16 ) and ( 2.2.17 ) are symmetric in fandg, we can also regard fas the inverse of g, and denote it byg/NUL1. Section 2.2 Continuity 69 Example 2.2.17 If f.x/Dx2; 0/DC4x/DC4R; then f/NUL1.y/Dg.y/Dpy; 0/DC4y/DC4R2: Example 2.2.18 If f.x/D2xC4; 0/DC4x/DC42; then f/NUL1.y/Dg.y/Dy/NUL4 2; 4/DC4y/DC48: 2.2 Exercises 1. Prove Theorem 2.2.2 . 2. Prove that a function fis continuous at x0if and only if lim x!x0/NULf.x/Dlim x!x0Cf.x/Df.x 0/: 3. Determine whether fis continuous or discontinuous from the right or left at x0. (a)f.x/Dpx .x 0D0/(b)f.x/Dpx .x 0>0/ (c)f.x/D1 x.x0D0/(d)f.x/Dx2.x0arbitrary/ (e)f.x/D/SUBxsin1=x; x¤0; 1; xD0.x0D0/ (f)f.x/D/SUBxsin1=x; x¤0 0; xD0.x0D0/ (g)f.x/D8 < :xCjxj.1Cx/ xsin1 x; x¤0 1; x D0.x0D0/ 4. Letfbe defined on Œ0;2/c141 by f.x/D(x2; 0/DC4x<1; xC1; 1/DC4x/DC42: On which of the following intervals is fcontinuous according to Definition 2.2.3 : Œ0;1/ ,.0;1/ ,.0;1/c141 ,Œ0;1/c141 ,Œ1;2/ ,.1;2/ ,.1;2/c141 ,Œ1;2/c141 ? 5. Let g.x/Dpx x/NUL1: On which of the following intervals is gcontinuous according to Definition 2.2.3 : Œ0;1/ ,.0;1/ ,.0;1/c141 ,Œ1;1/,.1;1/? 70 Chapter 2 Differential Calculus of Functions of One Variable 6. Let f.x/D(-1 ifxis irrational; 1 ifxis rational: Show thatfis not continuous anywhere. 7. Letf.x/D0ifxis irrational and f.p=q/D1=q ifpandqare positive inte- gers with no common factors. Show that fis discontinuous at every rational and continuous at every irrational on .0;1/. 8. Prove: Iffassumes only finitely many values, then fis continuous at a point x0in D0 fif and only if fis constant on some interval .x0/NULı;x 0Cı/. 9. Thecharacteristic function Tof a setTis defined by T.x/D(1; x2T; 0; x62T: Show that Tis continuous at a point x0if and only if x02T0[.Tc/0. 10. Prove: Iffandgare continuous on .a;b/ andf.x/Dg.x/ for everyxin a dense subset (Definition 1.1.5 ) of.a;b/ , thenf.x/Dg.x/ for allxin.a;b/ . 11. Prove that the function g.x/Dlogxis continuous on .0;1/. Take the following properties as given. (a) limx!1g.x/D0. (b)g.x 1/Cg.x 2/Dg.x 1x2/ifx1;x2>0. 12. Prove that the function f.x/Deaxis continuous on ./NUL1;1/. Take the following properties as given. (a) limx!0f.x/D1. (b)f.x 1Cx2/Df.x 1/f.x 2/;/NUL1<x 1;x2<1. 13. (a) Prove that the functions sinh xand coshxare continuous for all x. (b) For what values of xare tanhxand cothxcontinuous? 14. Prove that the functions s.x/Dsinxandc.x/Dcosxare continuous on ./NUL1;1/. Take the following properties as given. (a) limx!0c.x/D1. (b)c.x1/NULx2/Dc.x1/c.x 2/Cs.x1/s.x 2/;/NUL1<x 1;x2<1. (c)s2.x/Cc2.x/D1;/NUL1<x<1. 15. (a) Prove: Iffis continuous at x0andf.x 0/> /SYN , thenf.x/ > /SYN for allxin some neighborhood of x0. (b) State a result analogous to (a)for the case where f.x 0/</SYN . (c) Prove: Iff.x//DC4/SYNfor allxinSandx0is a limit point of Sat whichfis continuous, then f.x 0//DC4/SYN. (d) State results analogous to (a),(b), and(c)for the case where fis contin- uous from the right or left at x0. Section 2.2 Continuity 71 16. Letjfjbe the function whose value at each xinDfisjf.x/j. Prove: Iffis continuous at x0, then so isjfj. Is the converse true? 17. Prove: Iffis monotonic on Œa;b/c141 , thenfis piecewise continuous on Œa;b/c141 if and only iffhas only finitely many discontinuities in Œa;b/c141 . 18. Prove Theorem 2.2.5 . 19. (a) Show that iff1,f2, . . . ,fnare continuous on a set Sthen so aref1Cf2C /SOH/SOH/SOHCfnandf1f2/SOH/SOH/SOHfn. (b) Use(a) to show that a rational function is continuous for all values ofx except the zeros of its denominator. 20. (a) Letf1andf2be continuous at x0and define F.x/Dmax.f1.x/;f 2.x//: Show thatFis continuous at x0. (b) Letf1,f2, . . . ,fnbe continuous at x0and define F.x/Dmax.f1.x/;f 2.x/;:::;f n.x//: Show thatFis continuous at x0. 21. Find the domains of fıgandgıf. (a)f.x/Dpx; g.x/D1/NULx2(b)f.x/Dlogx; g.x/Dsinx (c)f.x/D1 1/NULx2; g.x/Dcosx(d)f.x/Dpx; g.x/Dsin2x 22. (a) Suppose that y0Dlimx!x0g.x/ exists and is an interior point of Df, and thatfis continuous at y0. Show that lim x!x0.fıg/.x/Df.y 0/: (b) State an analogous result for limits from the right. (c) State an analogous result for limits from the left. 23. Use Theorem 2.2.7 to find all points x0at which the following functions are contin- uous. (a)p 1/NULx2 (b) sine/NULx2(c)log.1Csinx/ (d)e/NUL1=.1/NUL2x/(e)sin1 .x/NUL1/2(f)sin/DC21 cosx/DC3 (g).1/NULsin2x//NUL1=2(h) cot.1/NULe/NULx2/ (i)cos1 x 24. Complete the proof of Theorem 2.2.9 by showing that there is an x2such that f.x 2/Dˇ. 72 Chapter 2 Differential Calculus of Functions of One Variable 25. Prove: Iffis nonconstant and continuous on an interval I, then the set SD˚yˇˇyDf.x/;x2I/TABis an interval. Moreover, if Iis a finite closed interval, then so isS. 26. Suppose that fandgare defined on ./NUL1;1/,fis increasing, and fıgis con- tinuous on./NUL1;1/. Show thatgis continuous on ./NUL1;1/. 27. Letfbe continuous on Œa;b/ , and define F.x/Dmax a/DC4t/DC4xf.t/; a/DC4x<b: (How do we know that Fis well defined?) Show that Fis continuous on Œa;b/ . 28. Letfandgbe uniformly continuous on an interval S. (a) Show thatfCgandf/NULgare uniformly continuous on S. (b) Show thatfgis uniformly continuous on SifSis compact. (c) Show thatf=g is uniformly continuous on SifSis compact and ghas no zeros inS. (d) Give examples showing that the conclusion of (b) and(c)may fail to hold ifSis not compact. (e) State additional conditions on fandgwhich guarantee that fgis uniformly continuous on Seven ifSis not compact. Do the same for f=g. 29. Suppose that fis uniformly continuous on a set S,gis uniformly continuous on a setT, andg.x/2Sfor everyxinT. Show thatfıgis uniformly continuous on T. 30. (a) Prove: Iffis uniformly continuous on disjoint closed intervals I1,I2, . . . , In, thenfis uniformly continuous onSn jD1Ij. (b) Is(a)valid without the word “closed”? 31. (a) Prove: Iffis uniformly continuous on a bounded open interval .a;b/ , then f.aC/andf.b/NUL/exist and are finite. H INT:See Exercise 2.1.38: (b) Show that the conclusion in (a)does not follow if .a;b/ is unbounded. 32. Prove: Iffis continuous on Œa;1/andf.1/exists (finite), then fis uniformly continuous on Œa;1/. 33. Suppose that fis defined on ./NUL1;1/and has the following properties. (i) lim x!0f.x/D1and(ii)f.x 1Cx2/Df.x 1/f.x 2/;/NUL1<x 1;x2<1: Prove: (a)f.x/>0 for allx. (b)f.rx/DŒf.x//c141rifris rational. (c) Iff.1/D1thenfis constant. Section 2.3 Differentiable Functions of One Variable 73 (d) Iff.1/D/SUB>1 , thenfis increasing, lim x!1f.x/D1;and lim x!/NUL1f.x/D0: (Thus,f.x/Deaxhas these properties if a>0 .) HINT:See Exercises 2.2.10 and2.2.12: 34. Prove Theorem 2.2.14 in the case where fis nonincreasing. 2.3 DIFFERENTIABLE FUNCTIONS OF ONE VARIABLE In calculus you studied differentiation, emphasizing rule s for calculating derivatives. Here we consider the theoretical properties of differentiable f unctions. In doing this, we assume that you know how to differentiate elementary functions suc h asxn,ex, and sinx, and we will use such functions in examples. Definition of the Derivative Definition 2.3.1 A functionfisdifferentiable at an interior point x0of its domain if the difference quotient f.x//NULf.x 0/ x/NULx0; x¤x0; approaches a limit as xapproachesx0, in which case the limit is called the derivative off atx0, and is denoted by f0.x0/; thus, f0.x0/Dlim x!x0f.x//NULf.x 0/ x/NULx0: (2.3.1) It is sometimes convenient to let xDx0Chand write ( 2.3.1 ) as f0.x0/Dlim h!0f.x 0Ch//NULf.x 0/ h: Iffis defined on an open set S, we say thatfisdifferentiable on Siffis differentiable at every point of S. Iffis differentiable on S, thenf0is a function on S. We say that fiscontinuously differentiable onSiff0is continuous on S. Iffis differentiable on a neighborhood of x0, it is reasonable to ask if f0is differentiable at x0. If so, we denote the derivative of f0atx0byf00.x0/. This is the second derivative of fatx0, and it is also denoted byf.2/.x0/. Continuing inductively, if f.n/NUL1/is defined on a neighborhood of x0, then thenthderivative of fatx0, denoted byf.n/.x0/, is the derivative of f.n/NUL1/at x0. For convenience we define the zeroth derivative offto befitself; thus f.0/Df: We assume that you are familiar with the other standard notat ions for derivatives; for example, f.2/Df00; f.3/Df000; 74 Chapter 2 Differential Calculus of Functions of One Variable and so on, and dnf dxnDf.n/: Example 2.3.1 Ifnis a positive integer and f.x/Dxn; then f.x//NULf.x 0/ x/NULx0Dxn/NULxn 0 x/NULx0Dx/NULx0 x/NULx0n/NUL1X kD0xn/NULk/NUL1xk 0; so f0.x0/Dlim x!x0n/NUL1X kD0xn/NULk/NUL1xk 0Dnxn/NUL1 0: Since this holds for every x0, we drop the subscript and write f0.x/Dnxn/NUL1ord dx.xn/Dnxn/NUL1: To derive differentiation formulas for elementary functio ns such as sin x, cosx, andex directly from Definition 2.3.1 requires estimates based on the properties of these functio ns. Since this is done in calculus, we will not repeat it here. Interpretations of the Derivative Iff.x/ is the position of a particle at time x¤x0, the difference quotient f.x//NULf.x 0/ x/NULx0 is the average velocity of the particle between times x0andx. Asxapproachesx0, the average applies to shorter and shorter intervals. Therefor e, it makes sense to regard the limit (2.3.1 ), if it exists, as the particle’s instantaneous velocity at time x0. This interpretation may be useful even if xis not time, so we often regard f0.x0/as the instantaneous rate of change off.x/ atx0, regardless of the specific nature of the variable x. The derivative also has a geometric interpretation. The equation of the line thr ough two points .x0;f.x 0//and .x1;f.x 1//on the curveyDf.x/ (Figure 2.3.1 ) is yDf.x 0/Cf.x 1//NULf.x 0/ x1/NULx0.x/NULx0/: Varyingx1generates lines through .x0;f.x 0//that rotate into the line yDf.x 0/Cf0.x0/.x/NULx0/ (2.3.2) Section 2.3 Differentiable Functions of One Variable 75 asx1approachesx0. This is the tangent to the curveyDf.x/ at the point.x0;f.x 0//. Figure 2.3.2 depicts the situation for various values of x1. y xy = f(x) x0 x1 Figure 2.3.1 y xy = f(x) x0 x1x1x1''Tangent line Figure 2.3.2 Here is a less intuitive definition of the tangent line: If the function T.x/Df.x 0/Cm.x/NULx0/ approximates fso well nearx0that lim x!x0f.x//NULT.x/ x/NULx0D0; we say that the line yDT.x/ istangent to the curve yDf.x/ at.x0;f.x 0//. 76 Chapter 2 Differential Calculus of Functions of One Variable This tangent line exists if and only if f0.x0/exists, in which case mis uniquely determined bymDf0.x0/(Exercise 2.3.1 ). Thus, ( 2.3.2 ) is the equation of the tangent line. We will use the following lemma to study differentiable func tions. Lemma 2.3.2 Iffis differentiable at x0;then f.x/Df.x 0/CŒf0.x0/CE.x//c141.x/NULx0/; (2.3.3) whereEis defined on a neighborhood of x0and lim x!x0E.x/DE.x 0/D0: Proof Define E.x/D8 < :f.x//NULf.x 0/ x/NULx0/NULf0.x0/; x2Dfandx¤x0; 0; x Dx0:(2.3.4) Solving ( 2.3.4 ) forf.x/ yields ( 2.3.3 ) ifx¤x0, and ( 2.3.3 ) is obvious if xDx0. Defini- tion2.3.1 implies that lim x!x0E.x/D0. We defined E.x 0/D0to makeEcontinuous atx0. Since the right side of ( 2.3.3 ) is continuous at x0, so is the left. This yields the following theorem. Theorem 2.3.3 Iffis differentiable at x0;thenfis continuous at x0: The converse of this theorem is false, since a function may be continuous at a point without being differentiable at the point. Example 2.3.2 The function f.x/Djxj can be written as f.x/Dx; x>0; (2.3.5) or as f.x/D/NULx; x<0: (2.3.6) From ( 2.3.5 ), f0.x/D1; x>0; and from ( 2.3.6 ), f0.x/D/NUL1; x<0: Neither ( 2.3.5 ) nor ( 2.3.6 ) holds throughout any neighborhood of 0, so neither can be used alone to calculate f0.0/. In fact, since the one-sided limits lim x!0Cf.x//NULf.0/ x/NUL0Dlim x!0Cx x(2.3.7) and lim x!0/NULf.x//NULf.0/ x/NUL0Dlim x!0/NUL/NULx xD/NUL1 (2.3.8) Section 2.3 Differentiable Functions of One Variable 77 are different, lim x!0f.x//NULf.0/ x/NUL0 does not exist (Theorem 2.1.6 ); thus,fis not differentiable at 0, even though it is continu- ous at0. Interchanging Differentiation and Arithmetic Operations The following theorem should be familiar from calculus. Theorem 2.3.4 Iffandgare differentiable at x0;then so arefCg;f/NULg;andfg; with (a).fCg/0.x0/Df0.x0/Cg0.x0/I (b).f/NULg/0.x0/Df0.x0//NULg.x 0/I (c).fg/0.x0/Df0.x0/g.x 0/Cf.x 0/g0.x0/: The quotientf=g is differentiable at x0ifg.x 0/¤0;with (d)/DC2f g/DC30 .x0/Df0.x0/g.x 0//NULf.x 0/g0.x0/ Œg.x 0//c1412: Proof The proof is accomplished by forming the appropriate differ ence quotients and applying Definition 2.3.1 and Theorem 2.1.4 . We will prove (c)and leave the rest to you (Exercises 2.3.9 ,2.3.10 , and 2.3.11 ). The trick is to add and subtract the right quantity in the nume rator of the difference quotient for.fg/0.x0/; thus, f.x/g.x//NULf.x 0/g.x 0/ x/NULx0Df.x/g.x//NULf.x 0/g.x/Cf.x 0/g.x//NULf.x 0/g.x 0/ x/NULx0 Df.x//NULf.x 0/ x/NULx0g.x/Cf.x 0/g.x//NULg.x 0/ x/NULx0: The difference quotients on the right approach f0.x0/andg0.x0/asxapproachesx0, and limx!x0g.x/Dg.x 0/(Theorem 2.3.3 ). This proves (c). The Chain Rule Here is the rule for differentiating a composite function. Theorem 2.3.5 (The Chain Rule) Suppose that gis differentiable at x0andf is differentiable at g.x 0/:Then the composite function hDfıg;defined by h.x/Df.g.x//; is differentiable at x0;with h0.x0/Df0.g.x 0//g0.x0/: 78 Chapter 2 Differential Calculus of Functions of One Variable Proof Sincefis differentiable at g.x 0/, Lemma 2.3.2 implies that f.t//NULf.g.x 0//DŒf0.g.x 0//CE.t//c141Œt/NULg.x 0//c141; where lim t!g.x 0/E.t/DE.g.x 0//D0: (2.3.9) LettingtDg.x/ yields f.g.x///NULf.g.x 0//DŒf0.g.x 0//CE.g.x///c141Œg.x//NULg.x 0//c141: Sinceh.x/Df.g.x// , this implies that h.x//NULh.x0/ x/NULx0DŒf0.g.x 0//CE.g.x///c141g.x//NULg.x 0/ x/NULx0: (2.3.10) Sincegis continuous at x0(Theorem 2.3.3 ), (2.3.9 ) and Theorem 2.2.7 imply that lim x!x0E.g.x//DE.g.x 0//D0: Therefore, ( 2.3.10 ) implies that h0.x0/Dlim x!x0h.x//NULh.x0/ x/NULx0Df0.g.x 0//g0.x0/; as stated. Example 2.3.3 If f.x/Dsinxandg.x/D1 x; x¤0; then h.x/Df.g.x//Dsin1 x; x¤0; and h0.x/Df0.g.x//g.x/D/DC2 cos1 x/DC3/DC2 /NUL1 x2/DC3 ; x¤0: It may seem reasonable to justify the chain rule by writing h.x//NULh.x0/ x/NULx0Df.g.x///NULf.g.x 0// x/NULx0 Df.g.x///NULf.g.x 0// g.x//NULg.x 0/g.x//NULg.x 0/ x/NULx0 and arguing that lim x!x0f.g.x///NULf.g.x 0// g.x//NULg.x 0/Df0.g.x 0// Section 2.3 Differentiable Functions of One Variable 79 (because lim x!x0g.x/Dg.x 0//and lim x!x0g.x//NULg.x 0/ x/NULx0Dg0.x0/: However, this is not a valid proof (Exercise 2.3.13 ). One-Sided Derivatives One-sided limits of difference quotients such as ( 2.3.7 ) and ( 2.3.8 ) in Example 2.3.2 are called one-sided orright- and left-hand derivatives . That is, iffis defined on Œx0;b/, the right-hand derivative of fatx0is defined to be f0 C.x0/Dlim x!x0Cf.x//NULf.x 0/ x/NULx0 if the limit exists, while if fis defined on .a;x 0/c141, the left-hand derivative of fatx0is defined to be f0 /NUL.x0/Dlim x!x0/NULf.x//NULf.x 0/ x/NULx0 if the limit exists. Theorem 2.1.6 implies thatfis differentiable at x0if and only if f0 C.x0/ andf0 /NUL.x0/exist and are equal, in which case f0.x0/Df0 C.x0/Df0 /NUL.x0/: In Example 2.3.2 ,f0 C.0/D1andf0 /NUL.0/D/NUL1. Example 2.3.4 If f.x/D8 < :x3; x/DC40; x2sin1 x; x>0;(2.3.11) then f0.x/D8 < :3x2; x<0; 2xsin1 x/NULcos1 x; x>0:(2.3.12) Since neither formula in ( 2.3.11 ) holds for all xin any neighborhood of 0, we cannot simply differentiate either to obtain f0.0/; instead, we calculate f0 C.0/Dlim x!0Cx2sin1 x/NUL0 x/NUL0Dlim x!0Cxsin1 xD0; f0 /NUL.0/Dlim x!0/NULx3/NUL0 x/NUL0Dlim x!0/NULx2D0I hence,f0.0/Df0 C.0/Df0 /NUL.0/D0. 80 Chapter 2 Differential Calculus of Functions of One Variable This example shows that there is a difference between a one-s ided derivative and a one- sided limit of a derivative, since f0 C.0/D0, but, from ( 2.3.12 ),f0.0C/Dlimx!0Cf0.x/ does not exist. It also shows that a derivative may exist in a n eighborhood of a point x0 (D0in this case), but be discontinuous at x0. Exercise 2.3.4 justifies the method used in Example 2.3.4 to computef0.x/forx¤0. Definition 2.3.6 (a) We say thatfisdifferentiable on the closed interval Œa;b/c141 iffis differentiable on the open interval .a;b/ andf0 C.a/andf0 /NUL.b/both exist. (b) We say thatfiscontinuously differentiable on Œa;b/c141 iffis differentiable on Œa;b/c141 , f0is continuous on .a;b/ ,f0 C.a/Df0.aC/, andf0 /NUL.b/Df0.b/NUL/. Extreme Values We say thatf.x 0/is alocal extreme value offif there is aı>0 such thatf.x//NULf.x 0/ does not change sign on .x0/NULı;x 0Cı/\Df: (2.3.13) More specifically, f.x 0/is alocal maximum value offif f.x//DC4f.x 0/ (2.3.14) or alocal minimum value offif f.x//NAKf.x 0/ (2.3.15) for allxin the set ( 2.3.13 ). The point x0is called a local extreme point off, or, more specifically, a local maximum orlocal minimum point off. y x 1 2 23 4 −1 −1 21 Figure 2.3.3 Section 2.3 Differentiable Functions of One Variable 81 Example 2.3.5 If f.x/D8 ˆˆˆˆ< ˆˆˆˆ:1;/NUL1<x/DC4/NUL1 2 jxj;/NUL1 2<x/DC41 2; 1p 2sin/EMx 2;1 2<x/DC44 (Figure 2.3.3 ), then0,3, and everyxin./NUL1;/NUL1 2/are local minimum points of f, while1, 4, and everyxin./NUL1;/NUL1 2/c141are local maximum points. It is geometrically plausible that if the curve yDf.x/ has a tangent at a local extreme point off, then the tangent must be horizontal; that is, have zero slop e. (For example, in Figure 2.3.3 , seexD1,xD3, and everyxin./NUL1;/NUL1=2/ .) The following theorem shows that this must be so. Theorem 2.3.7 Iffis differentiable at a local extreme point x02D0 f;thenf0.x0/D0: Proof We will show that x0is not a local extreme point of fiff0.x0/¤0. From Lemma 2.3.2 , f.x//NULf.x 0/ x/NULx0Df0.x0/CE.x/; (2.3.16) where lim x!x0E.x/D0. Therefore, if f0.x0/¤0, there is aı>0 such that jE.x/j<jf0.x0/jifjx/NULx0j<ı; and the right side of ( 2.3.16 ) must have the same sign as f0.x0/forjx/NULx0j< ı. Since the same is true of the left side, f.x//NULf.x 0/must change sign in every neighborhood of x0(sincex/NULx0does). Therefore, neither ( 2.3.14 ) nor ( 2.3.15 ) can hold for all xin any interval about x0. Iff0.x0/D0, we say that x0is acritical point off. Theorem 2.3.7 says that every local extreme point of fat whichfis differentiable is a critical point of f. The converse is false. For example, 0is a critical point of f.x/Dx3, but not a local extreme point. Rolle’s Theorem The use of Theorem 2.3.7 for finding local extreme points is covered in calculus, so we will not pursue it here. However, we will use Theorem 2.3.7 to prove the following fundamental theorem, which says that if a curve yDf.x/ intersects a horizontal line at xDaand xDband has a tangent at .x;f.x// for everyxin.a;b/ , then there is a point cin.a;b/ such that the tangent to the curve at .c;f.c// is horizontal (Figure 2.3.4 ). 82 Chapter 2 Differential Calculus of Functions of One Variable y xb c a Figure 2.3.4 Theorem 2.3.8 ( Rolle’s Theorem) Suppose that fis continuous on the closed intervalŒa;b/c141 and differentiable on the open interval .a;b/; andf.a/Df.b/: Then f0.c/D0for somecin the open interval .a;b/: Proof Sincefis continuous on Œa;b/c141 ,fattains a maximum and a minimum value on Œa;b/c141 (Theorem 2.2.9 ). If these two extreme values are the same, then fis constant on .a;b/ , sof0.x/D0for allxin.a;b/ . If the extreme values differ, then at least one must be attained at some point cin the open interval .a;b/ , andf0.c/D0, by Theorem 2.3.7 . Intermediate Values of Derivatives A derivative may exist on an interval Œa;b/c141 without being continuous on Œa;b/c141 . Neverthe- less, an intermediate value theorem similar to Theorem 2.2.10 applies to derivatives. Theorem 2.3.9 (Intermediate Value Theorem for Derivatives )Suppose thatfis differentiable on Œa;b/c141;f0.a/¤f0.b/;and/SYNis betweenf0.a/andf0.b/:Then f0.c/D/SYNfor somecin.a;b/: Proof Suppose first that f0.a/</SYN<f0.b/ (2.3.17) and define g.x/Df.x//NUL/SYNx: Then g0.x/Df0.x//NUL/SYN; a/DC4x/DC4b; (2.3.18) and ( 2.3.17 ) implies that g0.a/<0 andg0.b/>0: (2.3.19) Sincegis continuous on Œa;b/c141 ,gattains a minimum at some point cinŒa;b/c141 . Lemma 2.3.2 and ( 2.3.19 ) imply that there is a ı>0 such that g.x/<g.a/; a<x<a Cı; andg.x/<g.b/; b/NULı<x<b Section 2.3 Differentiable Functions of One Variable 83 (Exercise 2.3.3 ), and therefore c¤aandc¤b. Hence,a < c < b , and therefore g0.c/D0, by Theorem 2.3.7 . From ( 2.3.18 ),f0.c/D/SYN. The proof for the case where f0.b/</SYN<f0.a/can be obtained by applying this result to/NULf. Mean Value Theorems Theorem 2.3.10 (Generalized Mean Value Theorem) Iffandgare con- tinuous on the closed interval Œa;b/c141 and differentiable on the open interval .a;b/; then Œg.b//NULg.a//c141f0.c/DŒf.b//NULf.a//c141g0.c/ (2.3.20) for somecin.a;b/: Proof The function h.x/DŒg.b//NULg.a//c141f.x//NULŒf.b//NULf.a//c141g.x/ is continuous on Œa;b/c141 and differentiable on .a;b/ , and h.a/Dh.b/Dg.b/f.a//NULf.b/g.a/: Therefore, Rolle’s theorem implies that h0.c/D0for somecin.a;b/ . Since h0.c/DŒg.b//NULg.a//c141f0.c//NULŒf.b//NULf.a//c141g0.c/; this implies ( 2.3.20 ). The following special case of Theorem 2.3.10 is important enough to be stated separately. Theorem 2.3.11 (Mean Value Theorem) Iffis continuous on the closed intervalŒa;b/c141 and differentiable on the open interval .a;b/; then f0.c/Df.b//NULf.a/ b/NULa for somecin.a;b/: Proof Apply Theorem 2.3.10 withg.x/Dx. Theorem 2.3.11 implies that the tangent to the curve yDf.x/ at.c;f.c// is parallel to the line connecting the points .a;f.a// and.b;f.b// on the curve (Figure 2.3.5 , page 84). Consequences of the Mean Value Theorem Iffis differentiable on .a;b/ andx1,x22.a;b/ thenfis continuous on the closed interval with endpoints x1andx2and differentiable on its interior. Hence, the mean value theorem implies that f.x 2//NULf.x 1/Df0.c/.x 2/NULx1/ for somecbetweenx1andx2. (This is true whether x1<x 2orx2<x 1.) The next three theorems follow from this. 84 Chapter 2 Differential Calculus of Functions of One Variable Theorem 2.3.12 Iff0.x/D0for allxin.a;b/; thenfis constant on .a;b/: Theorem 2.3.13 Iff0exists and does not change sign on .a;b/; thenfis monotonic on.a;b/Wincreasing;nondecreasing ;decreasing;or nonincreasing as f0.x/>0; f0.x//NAK0; f0.x/<0; orf0.x//DC40; respectively;for allxin.a;b/: Theorem 2.3.14 If jf0.x/j/DC4M; a<x<b; then jf.x//NULf.x0/j/DC4Mjx/NULx0j; x;x02.a;b/: (2.3.21) A function that satisfies an inequality like ( 2.3.21 ) for allxandx0in an interval is said to satisfy a Lipschitz condition on the interval. y xb c ay = f(x) f(b) f(c) f(a) Figure 2.3.5 2.3 Exercises 1. Prove that a function fis differentiable at x0if and only if lim x!x0f.x//NULf.x 0//NULm.x/NULx0/ x/NULx0D0 for some constant m. In this case, f0.x0/Dm. Section 2.3 Differentiable Functions of One Variable 85 2. Prove: Iffis defined on a neighborhood of x0, thenfis differentiable at x0if and only if the discontinuity of h.x/Df.x//NULf.x 0/ x/NULx0 atx0is removable. 3. Use Lemma 2.3.2 to prove that if f0.x0/>0 , there is aı>0 such that f.x/<f.x 0/ifx0/NULı<x<x 0andf.x/>f.x 0/ifx0<x<x 0Cı: 4. Suppose that pis continuous on .a;c/c141 and differentiable on .a;c/ , whileqis con- tinuous onŒc;b/ and differentiable on .c;b/ . Let f.x/D(p.x/; a<x/DC4c; q.x/; c<x<b: (a) Show that f0.x/D(p0.x/; a<x<c; q0.x/; c<x<b: (b) Under what additional conditions on pandqdoesf0.c/exist? Prove that your stated conditions are necessary and sufficient. 5. Find all derivatives of f.x/Dxn/NUL1jxj, wherenis a positive integer. 6. Suppose that f0.0/exists andf.xCy/Df.x/f.y/ for allxandy. Prove thatf0 exists for allx. 7. Suppose that c0.0/Daands0.0/Dbwherea2Cb2¤0, and c.xCy/Dc.x/c.y//NULs.x/s.y/ s.xCy/Ds.x/c.y/Cc.x/s.y/ for allxandy. (a) Show thatcandsare differentiable on ./NUL1;1/, and findc0ands0in terms ofc,s,a, andb. (b) (For those who have studied differential equations.) Find candsexplicitly. 8. (a) Suppose that fandgare differentiable at x0,f.x 0/Dg.x 0/D0, and g0.x0/¤0. Without using L’Hospital’s rule, show that lim x!x0f.x/ g.x/Df0.x0/ g0.x0/: (b) State the corresponding results for one-sided limits. 9. Prove Theorem 2.3.4(a). 86 Chapter 2 Differential Calculus of Functions of One Variable 10. Prove Theorem 2.3.4(b). 11. Prove Theorem 2.3.4(d). 12. Prove by induction: If n/NAK1andf.n/.x0/andg.n/.x0/exist, then so does .fg/.n/.x0/, and .fg/.n/.x0/DnX mD0 n m! f.m/.x0/g.n/NULm/.x0/: HINT:See Exercise 1.2.19:This is Leibniz ’s rule for differentiating a product. 13. What is wrong with the “proof” of the chain rule suggested aft er Example 2.3.3 ? Correct it. 14. Suppose that fis continuous and increasing on Œa;b/c141 . Letfbe differentiable at a pointx0in.a;b/ , withf0.x0/¤0. Ifgis the inverse of fTheorem 2.2.15 ), show thatg0.f.x 0//D1=f0.x0/. 15. (a) Show thatf0 C.a/Df0.aC/if both quantities exist. (b) Example 2.3.4 shows thatf0 C.a/may exist even if f0.aC/does not. Give an example where f0.aC/exists butf0 C.a/does not. (c) Complete the following statement so it becomes a theorem, an d prove the theorem: “Iff0.aC/exists andfis ata, thenf0 C.a/Df0.aC/.” 16. Show thatf.aC/andf.b/NUL/exist (finite) if f0is bounded on .a;b/ . H INT:See Exercise 2.1.38: 17. Suppose that fis continuous on Œa;b/c141 ,f0 C.a/exists, and/SYNis betweenf0 C.a/and .f.b//NULf.a//=.b/NULa/. Show thatf.c//NULf.a/D/SYN.c/NULa/for somecin.a;b/ . 18. Suppose that fis continuous on Œa;b/c141 ,f0 C.a/</SYN<f0 /NUL.b/, and .f.b//NULf.a//=.b/NULa/¤/SYN: Show that either f.c//NULf.a/D/SYN.c/NULa/orf.c//NULf.b/D/SYN.c/NULb/for somec in.a;b/ . 19. Let f.x/Dsinx x; x¤0: (a) Definef.0/ so thatfis continuous at xD0. HINT:Use Exercise 2.3.8: (b) Show that ifxis a local extreme point of f, then jf.x/jD.1Cx2//NUL1=2: HINT:Express sinxand cosxin terms off.x/ andf0.x/; and add their squares to obtain a useful identity : (c) Show thatjf.x/j/DC41for allx. For what value of xis equality attained? Section 2.3 Differentiable Functions of One Variable 87 20. Letnbe a positive integer and f.x/Dsinnx nsinx; x¤k/EM (kDinteger): (a) Definef.k/EM/ so thatfis continuous at k/EM. HINT:Use Exercise 2.3.8: (b) Show that ifxis a local extreme point of f, then jf.x/jD/STX1C.n2/NUL1/sin2x/ETX/NUL1=2: HINT:Express sinnxandcosnxin terms off.x/ andf0.x/; and add their squares to obtain a useful identity : (c) Show thatjf.x/j/DC41for allx. For what values of xis equality attained? 21. We say thatfhas at leastnzeros, counting multiplicities , on an interval Iif there are distinct points x1,x2, . . . ,xpinIsuch that f.j /.xi/D0; 0/DC4j/DC4ni/NUL1; 1/DC4i/DC4p; andn1C/SOH/SOH/SOHCnpDn. Prove: Iffis differentiable and has at least nzeros, counting multiplicities, on an interval I, thenf0has at leastn/NUL1zeros, counting multiplicities, on I. 22. Give an example of a function fsuch thatf0exists on an interval .a;b/ and has a jump discontinuity at a point x0in.a;b/ , or show that there is no such function. 23. Letx1,x2, . . . ,xnandy1,y2, . . . ,ynbe in.a;b/ andyi<x i,1/DC4i/DC4n. Show that iffis differentiable on .a;b/ , then nX iD1Œf.x i//NULf.y i//c141Df0.c/nX iD1.xi/NULyi/ for somecin.a;b/ . 24. Prove or give a counterexample: If fis differentiable on a neighborhood of x0, then fsatisfies a Lipschitz condition on some neighborhood of x0. 25. Let f00.x/Cp.x/f.x/D0andg00.x/Cp.x/g.x/D0; a<x<b: (a) Show thatWDf0g/NULfg0is constant on .a;b/ . (b) Prove: IfW¤0andf.x 1/Df.x 2/D0wherea < x 1< x 2< b, then g.c/D0for somecin.x1;x2/. HINT:Considerf=g: 88 Chapter 2 Differential Calculus of Functions of One Variable 26. Suppose that we extend the definition of differentiability b y saying that fis differ- entiable atx0if f0.x0/Dlim x!x0f.x//NULf.x 0/ x/NULx0 exists in the extended reals. Show that if f.x/D(px; x/NAK0; /NULp/NULx; x<0; thenf0.0/D1 . 27. Prove or give a counterexample: If fis differentiable at x0in the extended sense of Exercise 2.3.26 , thenfis continuous at x0. 28. Assume thatfis differentiable on ./NUL1;1/andx0is a critical point of f. (a) Leth.x/Df.x/g.x/ , wheregis differentiable on ./NUL1;1/and f.x 0/g0.x0/¤0: Show that the tangent line to the curve yDh.x/ at.x0;h.x 0//and the tangent line to the curve yDg.x/ at.x0;g.x 0/intersect on the x-axis. (b) Suppose that f.x 0/¤0. Leth.x/Df.x/.x/NULx1/, wherex1is arbitrary. Show that the tangent line to the curve yDh.x/ at.x0;h.x 0//intersects the x-axis atxDx1. (c) Suppose that f.x 0/¤0. Leth.x/Df.x/.x/NULx1/2, wherex1¤x0. Show that the tangent line to the curve yDh.x/ at.x0;h.x 0//intersects the x-axis at the midpoint of the interval with endpoints x0andx1. (d) Leth.x/D.ax2CbxCc/.x/NULx1/, wherea¤0andb2/NUL4ac¤0. Let x0D/NULb 2a. Show that the tangent line to the curve yDh.x/ at.x0;h.x 0// intersects the x-axis atxDx1. (e) Lethbe a cubic polynomial with zeros ˛,ˇ, and/CR, where˛andˇare distinct and/CRis real. Letx0D˛Cˇ 2. Show that the tangent line to the curve yDh.x/ at.x0;h.x 0//intersects the axis at xD/CR. 2.4 L’HOSPITAL’S RULE The method of Theorem 2.1.4 for finding limits of the sum, difference, product, and quo- tient of functions breaks down in connection with indetermi nate forms. The generalized mean value theorem (Theorem 2.3.10 ) leads to a method for evaluating limits of indetermi- nate forms. Theorem 2.4.1 ( L’Hospital ’s Rule) Suppose that fandgare differentiable andg0has no zeros on .a;b/: Let lim x!b/NULf.x/Dlim x!b/NULg.x/D0 (2.4.1) Section 2.4 L’Hospital’s Rule 89 or lim x!b/NULf.x/D˙1 and lim x!b/NULg.x/D˙1; (2.4.2) and suppose that lim x!b/NULf0.x/ g0.x/DL . finite or˙1/: (2.4.3) Then lim x!b/NULf.x/ g.x/DL: (2.4.4) Proof We prove the theorem for finite Land leave the case where LD˙1 to you (Exercise 2.4.1 ). Suppose that /SI>0 . From ( 2.4.3 ), there is anx0in.a;b/ such that ˇˇˇˇf0.c/ g0.c//NULLˇˇˇˇ</SI ifx0<c<b: (2.4.5) Theorem 2.3.10 implies that if xandtare inŒx0;b/, then there is a cbetween them, and therefore in.x0;b/, such that Œg.x//NULg.t//c141f0.c/DŒf.x//NULf.t//c141g0.c/: (2.4.6) Sinceg0has no zeros in .a;b/ , Theorem 2.3.11 implies that g.x//NULg.t/¤0ifx;t2.a;b/: This means that gcannot have more than one zero in .a;b/ . Therefore, we can choose x0 so that, in addition to ( 2.4.5 ),ghas no zeros in Œx0;b/. Then ( 2.4.6 ) can be rewritten as f.x//NULf.t/ g.x//NULg.t/Df0.c/ g0.c/; so (2.4.5 ) implies that ˇˇˇˇf.x//NULf.t/ g.x//NULg.t//NULLˇˇˇˇ</SI ifx;t2Œx0;b/: (2.4.7) If (2.4.1 ) holds, letxbe fixed inŒx0;b/, and consider the function G.t/Df.x//NULf.t/ g.x//NULg.t//NULL: From ( 2.4.1 ), lim t!b/NULf.t/Dlim t!b/NULg.t/D0; so lim t!b/NULG.t/Df.x/ g.x//NULL: (2.4.8) 90 Chapter 2 Differential Calculus of Functions of One Variable Since jG.t/j</SI ifx0<t <b; because of ( 2.4.7 ), (2.4.8 ) implies that ˇˇˇˇf.x/ g.x//NULLˇˇˇˇ/DC4/SI: This holds for all xin.x0;b/, which implies ( 2.4.4 ). The proof under assumption ( 2.4.2 ) is more complicated. Again choose x0so that ( 2.4.5 ) holds andghas no zeros in Œx0;b/. LettingtDx0in (2.4.7 ), we see that ˇˇˇˇf.x//NULf.x 0/ g.x//NULg.x 0//NULLˇˇˇˇ</SI ifx0/DC4x<b: (2.4.9) Since lim x!b/NULf.x/D˙1 , we can choose x1>x 0so thatf.x/¤0andf.x/¤f.x 0/ ifx1<x<b . Then the function u.x/D1/NULg.x 0/=g.x/ 1/NULf.x 0/=f.x/ is defined and nonzero if x1<x<b , and lim x!b/NULu.x/D1; (2.4.10) because of ( 2.4.2 ). Since f.x//NULf.x 0/ g.x//NULg.x 0/Df.x/ g.x/1/NULf.x 0/=f.x/ 1/NULg.x 0/=g.x/Df.x/ g.x/u.x/; (2.4.9 ) implies thatˇˇˇˇf.x/ g.x/u.x//NULLˇˇˇˇ</SI ifx1<x<b; which can be rewritten as ˇˇˇˇf.x/ g.x//NULLu.x/ˇˇˇˇ</SIju.x/jifx1<x<b: (2.4.11) From this and the triangle inequality, ˇˇˇˇf.x/ g.x//NULLˇˇˇˇ/DC4ˇˇˇˇf.x/ g.x//NULLu.x/ˇˇˇˇCjLu.x//NULLj/DC4/SIju.x/jCjLjju.x//NUL1j:(2.4.12) Because of ( 2.4.10 ), there is a point x2in.x1;b/such that ju.x//NUL1j</SI and thereforeju.x/j<1C/SIifx2<x<b: This, ( 2.4.11 ), and ( 2.4.12 ) imply that ˇˇˇˇf.x/ g.x//NULLˇˇˇˇ</SI.1C/SI/CjLj/SIifx2<x<b; Section 2.4 L’Hospital’s Rule 91 which proves ( 2.4.4 ) under assumption ( 2.4.2 ). Theorem 2.4.1 and the proof given here remain valid if bD 1 and “x!b/NUL” is replaced by “ x!1 ” throughout. Only minor changes in the proof are required to show that similar theorems are valid for limits from the right, li mits at/NUL1, and ordinary (two- sided) limits. We will take these as given. The Indeterminate Forms 0=0and1=1 We say thatf=g is of the form 0=0asx!b/NULif lim x!b/NULf.x/Dlim x!b/NULg.x/D0; orof the form1=1asx!b/NULif lim x!b/NULf.x/D˙1 and lim x!b/NULg.x/D˙1: The corresponding definitions for x!bCandx!˙1 are similar. If f=g is of one of these forms as x!b/NULand asx!bC, then we say that it is of that form as x!b. Example 2.4.1 The ratio sinx=x is of the form 0=0asx!0, and L’Hospital’s rule yields lim x!0sinx xDlim x!0cosx 1D1: Example 2.4.2 The ratioe/NULx=xis of the form1=1asx!/NUL1 , and L’Hospital’s rule yields lim x!/NUL1e/NULx xDlim x!/NUL1/NULe/NULx 1D/NUL1: Example 2.4.3 Using L’Hospital’s rule may lead to another indeterminate f orm; thus, lim x!1ex x2Dlim x!1ex 2x if the limit on the right exists in the extended reals. Applyi ng L’Hospital’s rule again yields lim x!1ex 2xDlim x!1ex 2D1: Therefore, lim x!1ex x2D1: More generally, lim x!1ex x˛D1 for any real number ˛(Exercise 2.4.33 ). 92 Chapter 2 Differential Calculus of Functions of One Variable Example 2.4.4 Sometimes it pays to combine L’Hospital’s rule with other ma nipula- tions. For example, lim x!04/NUL4cosx/NUL2sin2x x4Dlim x!04sinx/NUL4sinxcosx 4x3 D/DC2 lim x!0sinx x/DC3/DC2 lim x!01/NULcosx x2/DC3 D/DC2 lim x!0sinx x/DC3/DC2 lim x!0sinx 2x/DC3 D1 2/DC2 lim x!0sinx x/DC32 D1 2.1/2D1 2(Example 2.4.1 ): As another example, L’Hospital’s rule yields lim x!0e/NULx2log.1Cx/ xDlim x!0/NUL2xe/NULx2log.1Cx/Ce/NULx2.1Cx//NUL1 1D1: However, it is better to remove the “determinate” part of the ratio before using L’Hospital’s rule: lim x!0e/NULx2log.1Cx/ xD/DC2 lim x!0e/NULx2/DC3/DC2 lim x!0log.1Cx/ x/DC3 D.1/lim x!0log.1Cx/ x Dlim x!01=.1Cx/ 1D1: In using L’Hospital’s rule we usually write, for example, lim x!bf.x/ g.x/Dlim x!bf0.x/ g0.x/(2.4.13) and then try to find the limit on the right. This is convenient, but technically incorrect, since (2.4.13 ) is true only if the limit on the right exists in the extended r eals. It may happen that the limit on the left exists but the one on the right does not. I n this case, ( 2.4.13 ) is incorrect. Example 2.4.5 If f.x/Dx/NULx2sin1 xandg.x/Dsinx; then f0.x/D1/NUL2xsin1 xCcos1 xandg0.x/Dcosx: Section 2.4 L’Hospital’s Rule 93 Therefore, lim x!0f0.x/=g0.x/does not exist. However, lim x!0f.x/ g.x/Dlim x!01/NULxsin.1=x/ .sinx/=xD1 1D1: The Indeterminate Form 0/SOH1 We say that a product fgis of the form 0/SOH1asx!b/NULif one of the factors approaches 0and the other approaches ˙1 asx!b/NUL. In this case, it may be useful to apply L’Hospital’s rule after writing f.x/g.x/Df.x/ 1=g.x/orf.x/g.x/Dg.x/ 1=f.x/; since one of these ratios is of the form 0=0and the other is of the form 1=1asx!b/NUL. Similar statements apply to limits as x!bC,x!b, andx!˙1 . Example 2.4.6 The productxlogxis of the form 0/SOH1asx!0C. Converting it to an1=1form yields lim x!0CxlogxDlim x!0Clogx 1=x Dlim x!0C1=x /NUL1=x2 D/NUL lim x!0CxD0: Converting to a 0=0form leads to a more complicated problem: lim x!0CxlogxDlim x!0Cx 1=logx Dlim x!0C1 /NUL1=x. logx/2 D/NUL lim x!0Cx.logx/2D‹ Example 2.4.7 The productxlog.1C1=x/ is of the form 0/SOH1asx!1 . Converting it to a0=0form yields lim x!1xlog.1C1=x/Dlim x!1log.1C1=x/ 1=x Dlim x!1Œ1=.1C1=x//c141./NUL1=x2/ /NUL1=x2 Dlim x!11 1C1=xD1: 94 Chapter 2 Differential Calculus of Functions of One Variable In this case, converting to an 1=1form complicates the problem: lim x!1xlog.1C1=x/Dlim x!1x 1=log.1C1=x/ Dlim x!11/DC2/NUL1 Œlog.1C1=x//c1412/DC3/DC2/NUL1=x2 1C1=x/DC3 Dlim x!1x.xC1/Œlog.1C1=x//c1412D‹ The Indeterminate Form 1/NUL1 A differencef/NULgis of the form1/NUL1 asx!b/NULif lim x!b/NULf.x/Dlim x!b/NULg.x/D˙1: In this case, it may be possible to manipulate f/NULginto an expression that is no longer indeterminate, or is of the form 0=0or1=1asx!b/NUL. Similar remarks apply to limits asx!bC,x!b, orx!˙1 . Example 2.4.8 The difference sinx x2/NUL1 x is of the form1/NUL1 asx!0, but it can be rewritten as the 0=0form sinx/NULx x2: Hence, lim x!0/DC2sinx x2/NUL1 x/DC3 Dlim x!0sinx/NULx x2Dlim x!0cosx/NUL1 2x Dlim x!0/NULsinx 2D0: Example 2.4.9 The difference x2/NULx is of the form1/NUL1 asx!1 . Rewriting it as x2/DC2 1/NUL1 x/DC3 ; which is no longer indeterminate as x!1 , we find that lim x!1.x2/NULx/Dlim x!1x2/DC2 1/NUL1 x/DC3 D/DLE lim x!1x2/DC1 lim x!1/DC2 1/NUL1 x/DC3 D.1/.1/D1 Section 2.4 L’Hospital’s Rule 95 The Indeterminate Forms 00,11, and10 The function fgis defined by f.x/g.x/Deg.x/ logf .x/Dexp.g.x/ logf.x// for allxsuch thatf.x/>0 . Therefore, if fandgare defined and f.x/>0 on an interval .a;b/ , Exercise 2.2.22 implies that lim x!b/NULŒf.x//c141g.x/Dexp/DC2 lim x!b/NULg.x/ logf.x//DC3 (2.4.14) if lim x!b/NULg.x/ logf.x/ exists in the extended reals. (If this limit is ˙1 then ( 2.4.14 ) is valid if we define e/NUL1D0ande1D1 .) The product glogfcan be of the form 0/SOH1 in three ways as x!b/NUL: (a) If lim x!b/NULg.x/D0and lim x!b/NULf.x/D0. (b) If lim x!b/NULg.x/D˙1 and lim x!b/NULf.x/D1. (c) If lim x!b/NULg.x/D0and lim x!b/NULf.x/D1 . In these three cases, we say that fgis of the form 00,11, and10, respectively, as x! b/NUL. Similar definitions apply to limits as x!bC,x!b, andx!˙1 . Example 2.4.10 The function xxis of the form 00asx!0C. Since xxDexlogx and lim x!0CxlogxD0(Example 2.4.6 ), lim x!0CxxDe0D1: Example 2.4.11 The function x1=.x /NUL1/is of the form 11asx!1. Since x1=.x /NUL1/Dexp/DC2logx x/NUL1/DC3 and lim x!1logx x/NUL1Dlim x!11=x 1D1; it follows that lim x!1x1=.x /NUL1/De1De: Example 2.4.12 The function x1=xis of the form10asx!1 . Since x1=xDexp/DC2logx x/DC3 and lim x!1logx xDlim x!11=x 1D0; 96 Chapter 2 Differential Calculus of Functions of One Variable it follows that lim x!1x1=xDe0D1: 2.4 Exercises 1. Prove Theorem 2.4.1 for the case where lim x!b/NULf0.x/=g0.x/D˙1 . In Exercises 2.4.2 –2.4.40 , find the indicated limits. 2. lim x!0tan/NUL1x sin/NUL1x3. lim x!01/NULcosx log.1Cx2/4. lim x!0C1Ccosx ex/NUL1 5. lim x!/EMsinnx sinx6. lim x!0log.1Cx/ x7. lim x!1exsine/NULx2 8. lim x!1xsin.1=x/ 9. lim x!1px.e/NUL1=x/NUL1/10. lim x!0Ctanxlogx 11. lim x!/EMsinxlog.jtanxj/ 12. lim x!0C/DC41 xClog.tanx//NAK 13. lim x!1.p xC1/NULpx/ 14. lim x!0/DC21 ex/NUL1/NUL1 x/DC3 15. lim x!0.cotx/NULcscx/ 16. lim x!0/DC21 sinx/NUL1 x/DC3 17. lim x!/EMjsinxjtanx18. lim x!/EM=2jtanxjcosx 19. lim x!0jsinxjx 20. lim x!0.1Cx/1=x 21. lim x!1xsin.1=x/ 22. lim x!0/DC2x 1/NULcosx/NUL2 x/DC3 23. lim x!0Cx˛logx 24. lim x!elog.logx/ sin.x/NULe/ 25. lim x!1/DC2xC1 x/NUL1/DC3p x2/NUL1 26. lim x!1C/DC2xC1 x/NUL1/DC3p x2/NUL1 27. lim x!1.logx/ˇ x28. lim x!1.coshx/NULsinhx/ 29. lim x!1.x˛/NULlogx/30. lim x!/NUL1ex2sin.ex/ Section 2.4 L’Hospital’s Rule 97 31. lim x!1x.xC1/Œlog.1C1=x//c141232. lim x!0sinx/NULxCx3=6 x5 33. lim x!1ex x˛34. lim x!3/EM=2 /NULetanxcosx 35. lim x!1C.logx/˛log.logx/ 36. lim x!1xx xlogx 37. lim x!/EM=2.sinx/tanx 38. lim x!0ex/NULnX rD0xrrŠ xn.nDinteger/NAK1/ 39. lim x!0sinx/NULnX rD0./NUL1/rx2rC1 .2rC1/Š x2nC1.nDinteger/NAK0/ 40. lim x!0e/NUL1=x2 xnD0(nDinteger) 41. (a) Prove: Iffis continuous at x0and lim x!x0f0.x/exists, thenf0.x0/exists andf0is continuous at x0. (b) Give an example to show that it is necessary to assume in (a) thatfis con- tinuous atx0. 42. Theiterated logarithms are defined by L0.x/Dxand Ln.x/Dlog.Ln/NUL1.x//; x>a n; n/NAK1; wherea1D0andanDean/NUL1;n/NAK1. Show that (a)Ln.x/DLn/NUL1.logx/; x>a n; n/NAK1. (b)Ln/NUL1.anC/D0andLn.anC/D/NUL1 . (c) lim x!anC.Ln/NUL1.x//˛Ln.x/D0if˛>0 andn/NAK1. (d) lim x!1.Ln.x//˛=Ln/NUL1.x/D0if˛is arbitrary and n/NAK1. 43. Letfbe positive and differentiable on .0;1/, and suppose that lim x!1f0.x/ f.x/DL; where0<L/DC41: Definef0.x/Dxand fn.x/Df .f n/NUL1.x//; n/NAK1: Use L’Hospital’s rule to show that lim x!1.fn.x//˛ fn/NUL1.x/D1 if˛>0 andn/NAK1: 98 Chapter 2 Differential Calculus of Functions of One Variable 44. Letfbe differentiable on some deleted neighborhood Nofx0, and suppose that f andf0have no zeros in N. Find (a) lim x!x0jf.x/jf .x/if lim x!x0f.x/D0; (b) lim x!x0jf.x/j1=.f .x/ /NUL1/if lim x!x0f.x/D1; (c) lim x!x0jf.x/j1=f .x/if lim x!x0f.x/D1 . 45. Suppose that fandgare differentiable and g0has no zeros on .a;b/ . Suppose also that lim x!b/NULf0.x/=g0.x/DLand either lim x!b/NULf.x/Dlim x!b/NULg.x/D0 or lim x!b/NULf.x/D1 and lim x!b/NULg.x/D˙1: Find lim x!b/NUL.1Cf.x//1=g.x/. 46. We distinguish between 1/SOH1.D1/and./NUL1/1.D/NUL1/and between1C1 .D1/and/NUL1/NUL1.D/NUL1/. Why don’t we distinguish between 0/SOH1 and 0/SOH./NUL1/,1/NUL1 and/NUL1C1 ,1=1and/NUL1=1, and11and1/NUL1? 2.5 TAYLOR’S THEOREM Apolynomial is a function of the form p.x/Da0Ca1.x/NULx0/C/SOH/SOH/SOHCan.x/NULx0/n; (2.5.1) wherea0, . . . ,anandx0are constants. Since it is easy to calculate the values of a po lyno- mial, considerable effort has been devoted to using them to a pproximate more complicated functions. Taylor’s theorem is one of the oldest and most imp ortant results on this question. The polynomial ( 2.5.1 ) is said to be written in powers of x/NULx0, and is of degreenif an¤0. If we wish to leave open the possibility that anD0, we say that pis of degree /DC4n. In particular, a constant polynomial p.x/Da0is of degree zero if a0¤0. If a0D0, so thatpvanishes identically, then phas no degree according to our definition, which requires at least one coefficient to be nonzero. For con venience we say that the identically zero polynomial phas degree/NUL1. (Any negative number would do as well as /NUL1. The point is that with this convention, the statement that pis a polynomial of degree /DC4nincludes the possibility that pis identically zero.) Taylor Polynomials We saw in Lemma 2.3.2 that iffis differentiable at x0, then f.x/Df.x 0/Cf0.x0/.x/NULx0/CE.x/.x/NULx0/; Section 2.5 Taylor’s Theorem 99 where lim x!x0E.x/D0: To generalize this result, we first restate it: the polynomia l T1.x/Df.x 0/Cf0.x0/.x/NULx0/; which is of degree/DC41and satisfies T1.x0/Df.x 0/; T0 1.x0/Df0.x0/; approximates fso well nearx0that lim x!x0f.x//NULT1.x/ x/NULx0D0: (2.5.2) Now suppose that fhasnderivatives at x0andTnis the polynomial of degree /DC4n such that T.r/ n.x0/Df.r/.x0/; 0/DC4r/DC4n: (2.5.3) How well does Tnapproximatefnearx0? To answer this question, we must first find Tn. SinceTnis a polynomial of degree /DC4n, it can be written as Tn.x/Da0Ca1.x/NULx0/C/SOH/SOH/SOHCan.x/NULx0/n; (2.5.4) wherea0, . . . ,anare constants. Differentiating ( 2.5.4 ) yields T.r/ n.x0/DrŠar; 0/DC4r/DC4n; so (2.5.3 ) determinesaruniquely as arDf.r/.x0/ rŠ; 0/DC4r/DC4n: Therefore, Tn.x/Df.x 0/Cf0.x0/ 1Š.x/NULx0/C/SOH/SOH/SOHCf.n/.x0/ nŠ.x/NULx0/n DnX rD0f.r/.x0/ rŠ.x/NULx0/r: We callTnthenthTaylor polynomial of faboutx0. The following theorem describes how Tnapproximates fnearx0. Theorem 2.5.1 Iff.n/.x0/exists for some integer n/NAK1andTnis thenth Taylor polynomial of faboutx0;then lim x!x0f.x//NULTn.x/ .x/NULx0/nD0: (2.5.5) 100 Chapter 2 Differential Calculus of Functions of One Variable Proof The proof is by induction. Let Pnbe the assertion of the theorem. From ( 2.5.2 ) we know that ( 2.5.5 ) is true ifnD1; that is,P1is true. Now suppose that Pnis true for some integer n/NAK1, andf.nC1/exists. Since the ratio f.x//NULTnC1.x/ .x/NULx0/nC1 is indeterminate of the form 0=0asx!x0, L’Hospital’s rule implies that lim x!x0f.x//NULTnC1.x/ .x/NULx0/nC1D1 nC1lim x!x0f0.x//NULT0 nC1.x/ .x/NULx0/n(2.5.6) if the limit on the right exists. But f0has annth derivative at x0, and T0 nC1.x/DnX rD0f.rC1/.x0/ rŠ.x/NULx0/r is thenth Taylor polynomial of f0aboutx0. Therefore, the induction assumption, applied tof0, implies that lim x!x0f0.x//NULT0 nC1.x/ .x/NULx0/nD0: This and ( 2.5.6 ) imply that lim x!x0f.x//NULTnC1.x/ .x/NULx0/nC1D0; which completes the induction. It can be shown (Exercise 2.5.8 ) that if pnDa0Ca1.x/NULx0/C/SOH/SOH/SOHCan.x/NULx0/n is a polynomial of degree /DC4nsuch that lim x!x0f.x//NULpn.x/ .x/NULx0/nD0; then arDf.r/.x0/ rŠI that is,pnDTn. Thus,Tnis the only polynomial of degree /DC4nthat approximates fnear x0in the manner indicated in ( 2.5.5 ). Theorem 2.5.1 can be restated as a generalization of Lemma 2.3.2 . Lemma 2.5.2 Iff.n/.x0/exists;then f.x/DnX rD0f.r/.x0/ rŠ.x/NULx0/rCEn.x/.x/NULx0/n; (2.5.7) where lim x!x0En.x/DEn.x0/D0: Section 2.5 Taylor’s Theorem 101 Proof Define En.x/D8 < :f.x//NULTn.x/ .x/NULx0/n; x2Df/NULfx0g; 0; x Dx0: Then ( 2.5.5 ) implies that lim x!x0En.x/DEn.x0/D0, and it is straightforward to verify (2.5.7 ). Example 2.5.1 Iff.x/Dex, thenf.n/.x/Dex. Therefore,f.n/.0/D1forn/NAK0, so thenth Taylor polynomial of faboutx0D0is Tn.x/DnX rD0xr rŠD1Cx 1ŠCx2 2ŠC/SOH/SOH/SOHCxn nŠ: (2.5.8) Theorem 2.5.1 implies that lim x!0ex/NULnX rD0xr rŠ xnD0: (See also Exercise 2.4.38 .) Example 2.5.2 Iff.x/Dlogx, thenf.1/D0and f.r/.x/D./NUL1/.r/NUL1/.r/NUL1/Š xr; r/NAK1; so thenth Taylor polynomial of faboutx0D1is Tn.x/DnX rD1./NUL1/r/NUL1 r.x/NUL1/r ifn/NAK1. (T0D0.) Theorem 2.5.1 implies that lim x!1logx/NULnX rD1./NUL1/r/NUL1r.x/NUL1/r .x/NUL1/nD0; n/NAK1: Example 2.5.3 Iff.x/D.1Cx/q, then f0.x/Dq.1Cx/q/NUL1 f00.x/Dq.q/NUL1/.1Cx/q/NUL2 ::: f.n/.x/Dq.q/NUL1//SOH/SOH/SOH.q/NULnC1/.1Cx/q/NULn: 102 Chapter 2 Differential Calculus of Functions of One Variable If we define q 0! D1and q n! Dq.q/NUL1//SOH/SOH/SOH.q/NULnC1/ nŠ; n/NAK1; then f.n/.0/ nŠD q n! ; and thenth Taylor polynomial of fabout0can be written as Tn.x/DnX rD0 q r! xr: (2.5.9) Theorem 2.5.1 implies that lim x!0.1Cx/q/NULnX rD0 q r! xr xnD0; n/NAK0: Ifqis a nonnegative integer, then q n! is the binomial coefficient defined in Exer- cise1.2.19 . In this case, we see from ( 2.5.9 ) that Tn.x/D.1Cx/qDf.x/; n/NAKq: Applications to Finding Local Extrema Lemma 2.5.2 yields the following theorem. Theorem 2.5.3 Suppose that fhasnderivatives at x0andnis the smallest positive integer such that f.n/.x0/¤0: (a) Ifnis odd;x0is not a local extreme point of f: (b) Ifnis even;x0is a local maximum of fiff.n/.x0/<0; or a local mininum of fif f.n/.x0/>0: Proof Sincef.r/.x0/D0for1/DC4r/DC4n/NUL1, (2.5.7 ) implies that f.x//NULf.x 0/D" f.n/.x0/ nŠCEn.x/# .x/NULx0/n(2.5.10) in some interval containing x0. Since lim x!x0En.x/D0andf.n/.x0/¤0, there is a ı>0 such that jEn.x/j<ˇˇˇˇˇf.n/.x0/ nŠˇˇˇˇˇifjx/NULx0j<ı: Section 2.5 Taylor’s Theorem 103 This and ( 2.5.10 ) imply that f.x//NULf.x 0/ .x/NULx0/n(2.5.11) has the same sign as f.n/.x0/if0<jx/NULx0j<ı. Ifnis odd the denominator of ( 2.5.11 ) changes sign in every neighborhood of x0, and therefore so must the numerator (since the ratio has constant sign for 0 <jx/NULx0j< ı). Consequently, f.x 0/cannot be a local extreme value of f. This proves (a). Ifnis even, the denominator of ( 2.5.11 ) is positive forx¤x0, sof.x//NULf.x 0/must have the same sign as f.n/.x0/for0<jx/NULx0j<ı. This proves (b). FornD2,(b) is called the second derivative test for local extreme points. Example 2.5.4 Iff.x/Dex3, thenf0.x/D3x2ex3, and0is the only critical point off. Since f00.x/D.6xC9x4/ex3 and f000.x/D.6C54x3C27x6/ex3; f00.0/D0andf000.0/¤0. Therefore, Theorem 2.5.3 implies that0is not a local extreme point off. Sincefis differentiable everywhere, it has no local maxima or mini ma. Example 2.5.5 Iff.x/Dsinx2, thenf0.x/D2xcosx2, so the critical points of f are0and˙p .kC1=2//EM ,kD0;1;2;::: . Since f00.x/D2cosx2/NUL4x2sinx2; f00.0/D2andf00/DLE ˙p .kC1=2//EM//DC1 D./NUL1/kC1.4kC2//EM: Therefore, Theorem 2.5.3 implies thatfattains local minima at 0and˙p .kC1=2//EM for odd integersk, and local maxima at ˙p .kC1=2//EM for even integers k. Taylor’s theorem Theorem 2.5.1 implies that the error in approximating f.x/ byTn.x/approaches zero faster than.x/NULx0/nasxapproachesx0; however, it gives no estimate of the error in approximating f.x/ byTn.x/for a fixedx. For instance, it provides no estimate of the error in the approximation e0:1/EMT2.0:1/D1C0:1 1ŠC.0:1/2 2ŠD1:105 (2.5.12) obtained by setting nD2andxD0:1in (2.5.8 ). The following theorem provides a way of estimating errors of this kind under the additional assum ption thatf.nC1/exists in a neighborhood of x0. 104 Chapter 2 Differential Calculus of Functions of One Variable Theorem 2.5.4 (Taylor’s Theorem) Suppose that f.nC1/exists on an open in- tervalIaboutx0;and letxbe inI:Then the remainder Rn.x/Df.x//NULTn.x/ can be written as Rn.x/Df.nC1/.c/ .nC1/Š.x/NULx0/nC1; wherecdepends upon xand is between xandx0: This theorem follows from an extension of the mean value theo rem that we will prove below. For now, let us assume that Theorem 2.5.4 is correct, and apply it. Example 2.5.6 Iff.x/Dex, thenf000.x/Dex, and Theorem 2.5.4 withnD2 implies that exD1CxCx2 2ŠCecx3 3Š; wherecis between0andx. Hence, from ( 2.5.12 ), e0:1D1:105Cec.0:1/3 6; where0<c<0:1 . Since0<ec<e0:1, we know from this that 1:105<e0:1<1:105Ce0:1.0:1/3 6: The second inequality implies that e0:1/DC4 1/NUL.0:1/3 6/NAK <1:105; so e0:1<1:1052: Therefore, 1:105<e0:1<1:1052; and the error in ( 2.5.12 ) is less than0:0002 . Example 2.5.7 In numerical analysis, forward differences are used to approximate derivatives. If h > 0 , the first and second forward differences with spacing hare defined by /c129f.x/Df.xCh//NULf.x/ and /c1292f.x/D/c129Œ/c129f.x//c141D/c129f.xCh//NUL/c129f.x/ Df.xC2h//NUL2f.xCh/Cf.x/:(2.5.13) Higher forward differences are defined inductively (Exerci se2.5.18 ). Section 2.5 Taylor’s Theorem 105 We will find upper bounds for the magnitudes of the errors in th e approximations f0.x0//EM/c129f.x 0/ h(2.5.14) and f00.x0//EM/c1292f.x 0/ h2: (2.5.15) Iff00exists on an open interval containing x0andx0Ch, we can use Theorem 2.5.4 to estimate the error in ( 2.5.14 ) by writing f.x 0Ch/Df.x 0/Cf0.x0/hCf00.c/h2 2; (2.5.16) wherex0<c<x 0Ch. We can rewrite ( 2.5.16 ) as f.x 0Ch//NULf.x 0/ h/NULf0.x0/Df0.c/h 2; which is equivalent to /c129f.x 0/ h/NULf0.x0/Df00.c/h 2: Therefore, ˇˇˇˇ/c129f.x 0/ h/NULf0.x0/ˇˇˇˇ/DC4M2h 2; whereM2is an upper bound for jf00jon.x0;x0Ch/. Iff000exists on an open interval containing x0andx0C2h, we can use Theorem 2.5.4 to estimate the error in ( 2.5.15 ) by writing f.x 0Ch/Df.x 0/Chf0.x0/Ch2 2f00.x0/Ch3 6f000.c0/ and f.x 0C2h/Df.x 0/C2hf0.x0/C2h2f00.x0/C4h3 3f000.c1/; wherex0<c 0<x 0Chandx0<c 1<x 0C2h. These two equations imply that f.x 0C2h//NUL2f.x 0Ch/Cf.x 0/Dh2f00.x0/C/DC44 3f000.c1//NUL1 3f000.c0//NAK h3; which can be rewritten as /c1292f.x 0/ h2/NULf00.x0/D/DC44 3f000.c1//NUL1 3f000.c0//NAK h; because of ( 2.5.13 ). Therefore, ˇˇˇˇ/c1292f.x 0/ h2/NULf00.x0/ˇˇˇˇ/DC45M3h 3; whereM3is an upper bound for jf000jon.x0;x0C2h/. 106 Chapter 2 Differential Calculus of Functions of One Variable The Extended Mean Value Theorem We now consider the extended mean value theorem, which impli es Theorem 2.5.4 (Exer- cise2.5.24 ). In the following theorem, aandbare the endpoints of an interval, but we do not assume that a<b . Theorem 2.5.5 (Extended Mean Value Theorem) Suppose thatfis con- tinuous on a finite closed interval Iwith endpoints aandb.that is, either ID.a;b/ or ID.b;a//;f.nC1/exists on the open interval I0;and;ifn> 0; thatf0, . . . ,f.n/exist and are continuous at a:Then f.b//NULnX rD0f.r/.a/ rŠ.b/NULa/rDf.nC1/.c/ .nC1/Š.b/NULa/nC1(2.5.17) for somecinI0: Proof The proof is by induction. The mean value theorem (Theorem 2.3.11 ) implies the conclusion for nD0. Now suppose that n/NAK1, and assume that the assertion of the theorem is true with nreplaced byn/NUL1. The left side of ( 2.5.17 ) can be written as f.b//NULnX rD0f.r/.a/ rŠ.b/NULa/rDK.b/NULa/nC1 .nC1/Š(2.5.18) for some number K. We must prove that KDf.nC1/.c/for somecinI0. To this end, consider the auxiliary function h.x/Df.x//NULnX rD0f.r/.a/ rŠ.x/NULa/r/NULK.x/NULa/nC1 .nC1/Š; which satisfies h.a/D0; h.b/D0; (the latter because of ( 2.5.18 )) and is continuous on the closed interval Iand differentiable onI0, with h0.x/Df0.x//NULn/NUL1X rD0f.rC1/.a/ rŠ.x/NULa/r/NULK.x/NULa/n nŠ: (2.5.19) Therefore, Rolle’s theorem (Theorem 2.3.8 ) implies that h0.b1/D0for someb1inI0; thus, from ( 2.5.19 ), f0.b1//NULn/NUL1X rD0f.rC1/.a/ rŠ.b1/NULa/r/NULK.b1/NULa/n nŠD0: If we temporarily write f0Dg, this becomes g.b1//NULn/NUL1X rD0g.r/.a/ r.b1/NULa/r/NULK.b1/NULa/n nŠD0: (2.5.20) Section 2.5 Taylor’s Theorem 107 Sinceb12I0, the hypotheses on fimply thatgis continuous on the closed interval J with endpoints aandb1,g.n/exists onJ0, and, ifn/NAK1,g0, . . . ,g.n/NUL1/exist and are continuous at a(also atb1, but this is not important). The induction hypothesis, appl ied to gon the interval J, implies that g.b1//NULn/NUL1X rD0g.r/.a/ rŠ.b1/NULa/rDg.n/.c/ nŠ.b1/NULa/n for somecinJ0. Comparing this with ( 2.5.20 ) and recalling that gDf0yields KDg.n/.c/Df.nC1/.c/: Sincecis inI0, this completes the induction. 2.5 Exercises 1. Let f.x/D/SUB e/NUL1=x2; x¤0; 0; xD0: Show thatfhas derivatives of all orders on ./NUL1;1/and every Taylor polynomial offabout0is identically zero. H INT:SeeExercise 2.4.40: 2. Suppose that f.nC1/.x0/exists, and let Tnbe thenth Taylor polynomial of fabout x0. Show that the function En.x/D8 < :f.x//NULTn.x/ .x/NULx0/n; x2Df/NULfx0g; 0; x Dx0; is differentiable at x0, and findE0 n.x0/. 3. (a) Prove: Iffis continuous at x0and there are constants a0anda1such that lim x!x0f.x//NULa0/NULa1.x/NULx0/ x/NULx0D0; thena0Df.x 0/,f0is differentiable at x0, andf0.x0/Da1. (b) Give a counterexample to the following statement: If fandf0are continuous atx0and there are constants a0,a1, anda2such that lim x!x0f.x//NULa0/NULa1.x/NULx0//NULa2.x/NULx0/2 .x/NULx0/2D0; thenf00.x0/exists. 4. (a) Prove: iff00.x0/exists, then lim h!0f.x 0Ch//NUL2f.x 0/Cf.x 0/NULh/ h2Df00.x0/: 108 Chapter 2 Differential Calculus of Functions of One Variable (b) Prove or give a counterexample: If the limit in (a) exists, then so does f00.x0/, and they are equal. 5. A functionfhas a simple zero (or a zero of multiplicity 1) atx0iffis differentiable in a neighborhood of x0andf.x 0/D0, whilef0.x0/¤0. (a) Prove thatfhas a simple zero at x0if and only if f.x/Dg.x/.x/NULx0/; wheregis continuous at x0and differentiable on a deleted neighborhood of x0, andg.x 0/¤0. (b) Give an example showing that gin(a)need not be differentiable at x0. 6. A functionfhas a double zero (or a zero of multiplicity 2) atx0iffis twice dif- ferentiable on a neighborhood of x0andf.x 0/Df0.x0/D0, whilef00.x0/¤0. (a) Prove thatfhas a double zero at x0if and only if f.x/Dg.x/.x/NULx0/2; wheregis continuous at x0and twice differentiable on a deleted neighborhood ofx0,g.x 0/¤0, and lim x!x0.x/NULx0/g0.x/D0: (b) Give an example showing that gin(a)need not be differentiable at x0. 7. Letnbe a positive integer. A function fhas a zero of multiplicity natx0iff isntimes differentiable on a neighborhood of x0,f.x 0/Df0.x0/D /SOH/SOH/SOH D f.n/NUL1/.x0/D0andf.n/.x0/¤0. Prove thatfhas a zero of multiplicity nat x0if and only if f.x/Dg.x/.x/NULx0/n; wheregis continuous at x0andntimes differentiable on a deleted neighborhood of x0,g.x 0/¤0, and lim x!x0.x/NULx0/jg.j /.x/D0; 1/DC4j/DC4n/NUL1: HINT:Use Exercise 2.5.6 and induction : 8. (a) Let Q.x/D˛0C˛1.x/NULx0/C/SOH/SOH/SOHC˛n.x/NULx0/n be a polynomial of degree /DC4nsuch that lim x!x0Q.x/ .x/NULx0/nD0: Show that˛0D˛1D/SOH/SOH/SOHD˛nD0. Section 2.5 Taylor’s Theorem 109 (b) Suppose that fisntimes differentiable at x0andpis a polynomial p.x/Da0Ca1.x/NULx0/C/SOH/SOH/SOHCan.x/NULx0/n of degree/DC4nsuch that lim x!x0f.x//NULp.x/ .x/NULx0/nD0: Show that arDf.r/.x0/ rŠif0/DC4r/DC4nI that is,pDTn, thenth Taylor polynomial of faboutx0. 9. Show that iff.n/.x0/andg.n/.x0/exist and lim x!x0f.x//NULg.x/ .x/NULx0/nD0; thenf.r/.x0/Dg.r/.x0/,0/DC4r/DC4n. 10. (a) LetFn,Gn, andHnbe thenth Taylor polynomials about x0off,g, and their product hDfg. Show that Hncan be obtained by multiplying Fn byGnand retaining only the powers of x/NULx0through thenth. H INT:Use Exercise 2.5.8.b/: (b) Use the method suggested by (a)to computeh.r/.x0/,rD1;2;3;4 . (i)h.x/Dexsinx; x 0D0 (ii)h.x/D.cos/EMx=2/. logx/; x 0D1 (iii)h.x/Dx2cosx; x 0D/EM=2 (iv)h.x/D.1Cx//NUL1e/NULx; x 0D0 11. (a) It can be shown that if gisntimes differentiable at xandfisntimes dif- ferentiable at g.x/ , then the composite function h.x/Df.g.x// isntimes differentiable at xand h.n/.x/DnX rD1f.r/.g.x//X rrŠ r1Š/SOH/SOH/SOHrnŠ/DC2g0.x/ 1Š/DC3r1/DC2g00.x/ 2Š/DC3r2 /SOH/SOH/SOH g.n/.x/ nŠ!rn whereP ris over alln-tuples.r1;r2;:::;r n/of nonnegative integers such that r1Cr2C/SOH/SOH/SOHCrnDr and r1C2r2C/SOH/SOH/SOHCnrnDn: (This is Faa di Bruno ’s formula ). However, this formula is quite complicated. Justify the following alternative method for computing the derivatives of a composite function at a point x0: 110 Chapter 2 Differential Calculus of Functions of One Variable LetFnbe thenth Taylor polynomial of fabouty0Dg.x 0/, and letGnand Hnbe thenth Taylor polynomials of gandhaboutx0. Show thatHncan be obtained by substituting GnintoFnand retaining only powers of x/NULx0 through thenth. H INT:See Exercise 2.5.8.b/: (b) Compute the first four derivatives of h.x/Dcos.sinx/atx0D0, using the method suggested by (a). 12. (a) Ifg.x 0/¤0andg.n/.x0/exists, then the reciprocal hD1=gis alsontimes differentiable at x0, by Exercise 2.5.11(a), withf.x/D1=x. LetGnandHn be thenth Taylor polynomials of gandhaboutx0. Use Exercise 2.5.11(a)to prove that ifg.x 0/D1, thenHncan be obtained by expanding the polynomial nX rD1Œ1/NULGn.x//c141r in powers ofx/NULx0and retaining only powers through the nth. (b) Use the method of (a) to compute the first four derivatives of the following functions atx0. (i)h.x/Dcscx; x 0D/EM=2 (ii)h.x/D.1CxCx2//NUL1; x 0D0 (iii)h.x/Dsecx; x 0D/EM=4 (iv)h.x/DŒ1Clog.1Cx//c141/NUL1; x 0D0 (c) Use Exercise 2.5.10 to justify the following alternative procedure for obtaini ng Hn, again assuming that g.x 0/D1: If Gn.x/D1Ca1.x/NULx0/C/SOH/SOH/SOHCan.x/NULx0/n (where, of course, arDg.r/.x0/=rŠ/ and Hn.x/Db0Cb1.x/NULx0/C/SOH/SOH/SOHCbn.x/NULx0/n; then b0D1; b kD/NULkX rD1arbk/NULr; 1/DC4k/DC4n: 13. Determine whether x0D0is a local maximum, local minimum, or neither. (a)f.x/Dx2ex3(b)f.x/Dx3ex2 (c)f.x/D1Cx2 1Cx3(d)f.x/D1Cx3 1Cx2 (e)f.x/Dx2sin3xCx2cosx (f)f.x/Dex2sinx (g)f.x/Dexsinx2(h)f.x/Dex2cosx 14. Give an example of a function that has zero derivatives of all orders at a local mini- mum point. Section 2.5 Taylor’s Theorem 111 15. Find the critical points of f.x/Dx3 3Cbx2 2CcxCd and identify them as local maxima, local minima, or neither. 16. Find an upper bound for the magnitude of the error in the appro ximation. (a) sinx/EMx;jxj</EM 20 (b)p 1Cx/EM1Cx 2;jxj<1 8 (c) cosx/EM1p 2/STX 1/NUL/NUL x/NUL/EM 4/SOH/ETX ;/EM 4<x<5/EM 16 (d) logx/EM.x/NUL1//NUL.x/NUL1/2 2C.x/NUL1/3 3;jx/NUL1j<1 64 17. Prove: If Tn.x/DnX rD0xr rŠ; then Tn.x/<T nC1.x/<ex</DC4 1/NULxnC1 .nC1/Š/NAK/NUL1 Tn.x/ if0<x<Œ.nC1/Š/c1411=.n C1/. 18. The forward difference operators with spacing h>0 are defined by /c1290f.x/Df.x/; /c129f.x/Df.xCh//NULf.x/; /c129nC1f.x/D/c129Œ/c129nf.x//c141; n/NAK1: (a) Prove by induction on n: Ifk/NAK2,c1, . . . ,ckare constants, and n/NAK1, then /c129nŒc1f1.x/C/SOH/SOH/SOHCckfk.x//c141Dc1/c129nf1.x/C/SOH/SOH/SOHCck/c129nfk.x/: (b) Prove by induction: If n/NAK1, then /c129nf.x/DnX mD0./NUL1/n/NULm n m! f.xCmh/: HINT:See Exercise 1.2.19: In Exercises 2.5.19 –2.5.22 ,/c129is the forward difference operator with spacing h>0 . 112 Chapter 2 Differential Calculus of Functions of One Variable 19. Letmandnbe nonnegative integers, and let x0be any real number. Prove by induction onnthat /c129n.x/NULx0/mD/SUB0 if0/DC4m/DC4n; nŠhnifmDn: Does this suggest an analogy between “differencing" and dif ferentiation? 20. Find an upper bound for the magnitude of the error in the appro ximation f00.x0//EM/c1292f.x 0/NULh/ h2; (a) assuming that f000is bounded on .x0/NULh;x 0Ch/; (b) assuming that f.4/is bounded on .x0/NULh;x 0Ch/. 21. Letf000be bounded on an open interval containing x0andx0C2h. Find a constant ksuch that the magnitude of the error in the approximation f0.x0//EM/c129f.x 0/ hCk/c1292f.x 0/ h2 is not greater than Mh2, whereMDsup˚ jf000.c/jˇˇjx0<c<x 0/TAB . 22. Prove: Iff.nC1/is bounded on an open interval containing x0andx0Cnh, then ˇˇˇˇ/c129nf.x 0/ hn/NULf.n/.x0/ˇˇˇˇ/DC4AnMnC1h; whereAnis a constant independent of fand MnC1D sup x0<c<x 0Cnhjf.nC1/.c/j: HINT:See Exercises 2.5.18 and2.5.19: 23. Suppose that f.nC1/exists on.a;b/ ,x0, . . . ,xnare in.a;b/ , andpis the polyno- mial of degree/DC4nsuch thatp.x i/Df.x i/,0/DC4i/DC4n. Prove: Ifx2.a;b/ , then f.x/Dp.x/Cf.nC1/.c/ .nC1/Š.x/NULx0/.x/NULx1//SOH/SOH/SOH.x/NULxn/; wherec, which depends on x, is in.a;b/ . HINT:Letxbe fixed;distinct from x0; x1;. . . ,xn;and consider the function g.y/Df.y//NULp.y//NULK .nC1/Š.y/NULx0/.y/NULx1//SOH/SOH/SOH.y/NULxn/; whereKis chosen so that g.x/D0:Use Rolle’s theorem to show that KD f.nC1/.c/for somecin.a;b/: 24. Deduce Theorem 2.5.4 from Theorem 2.5.5 . CHAPTER 3 Integral Calculus of Functions of One Variable IN THIS CHAPTER we discuss the Riemann on a finite interval Œa;b/c141 , and improper inte- grals in which either the function or the interval of integra tion is unbounded. SECTION 3.1 begins with the definition of the Riemann integra l and presents the geo- metrical interpretation of the Riemann integral as the area under a curve. We show that an unbounded function cannot be Riemann integrable. Then we define upper and lower sums and upper and lower integrals of a bounded function. The section concludes with the definition of the Riemann–Stieltjes integral. SECTION 3.2 presents necessary and sufficient conditions fo r the existence of the Riemann integral in terms of upper and lower sums and upper and lower i ntegrals. We show that continuous functions and bounded monotonic functions are R iemann integrable. SECTION 3.3 begins with proofs that the sum and product of Rie mann integrable functions are integrable, and that jfjis Riemann integrable if fis Riemann integrable. Other topics covered include the first mean value theorem for integrals, a ntiderivatives, the fundamental theorem of calculus, change of variables, integration by pa rts, and the second mean value theorem for integrals. SECTION 3.4 presents a comprehensive discussion of imprope r integrals. Concepts de- fined and considered include absolute and conditional conve rgence of an improper integral, Dirichlet’s test, and change of variable in an improper inte gral. SECTION 3.5 defines the notion of a set with Lebesgue measure z ero, and presents a necessary and sufficient condition for a bounded function fto be Riemann integrable on an intervalŒa;b/c141 ; namely, that the discontinuities of fform a set with Lebesgue masure zero. 3.1 DEFINITION OF THE INTEGRAL The integral that you studied in calculus is the Riemann integral , named after the German mathematician Bernhard Riemann , who provided a rigorous formulation of the integral to 113 114 Chapter 3 Integral Calculus of Functions of One Variable replace the intuitive notion due to Newton andLeibniz . Since Riemann’s time, other kinds of integrals have been defined and studied; however, they are all generalizations of the Riemann integral, and it is hardly possible to understand th em or appreciate the reasons for developing them without a thorough understanding of the Rie mann integral. In this section we deal with functions defined on a finite interval Œa;b/c141 . Apartition ofŒa;b/c141 is a set of subintervals Œx0;x1/c141; Œx 1;x2/c141;:::;Œx n/NUL1;xn/c141; (3.1.1) where aDx0<x 1/SOH/SOH/SOH<x nDb: (3.1.2) Thus, any set of nC1points satisfying ( 3.1.2 ) defines a partition PofŒa;b/c141 , which we denote by PDfx0;x1;:::;x ng: The pointsx0,x1, . . . ,xnare the partition points ofP. The largest of the lengths of the subintervals ( 3.1.1 ) is the norm ofP, written askPk; thus, kPkD max 1/DC4i/DC4n.xi/NULxi/NUL1/: IfPandP0are partitions of Œa;b/c141 , thenP0is arefinement of Pif every partition point ofPis also a partition point of P0; that is, ifP0is obtained by inserting additional points between those of P. Iffis defined on Œa;b/c141 , then a sum /ESCDnX jD1f.c j/.xj/NULxj/NUL1/; where xj/NUL1/DC4cj/DC4xj; 1/DC4j/DC4n; is aRiemann sum of fover the partition PDfx0;x1;:::;x ng. (Occasionally we will say more simply that /ESCis a Riemann sum of foverŒa;b/c141 .) Sincecjcan be chosen arbitrarily inŒxj;xj/NUL1/c141, there are infinitely many Riemann sums for a given function fover a given partitionP. Definition 3.1.1 Letfbe defined on Œa;b/c141 . We say that fisRiemann integrable on Œa;b/c141 if there is a number Lwith the following property: For every /SI>0 , there is aı>0 such that j/ESC/NULLj</SI if/ESCis any Riemann sum of fover a partition PofŒa;b/c141 such thatkPk<ı. In this case, we say thatListhe Riemann integral of foverŒa;b/c141 , and write Zb af.x/dxDL: Section 3.1 Definition of the Integral 115 We leave it to you (Exercise 3.1.1 ) to show thatRb af.x/dx is unique, if it exists; that is, there cannot be more than one number Lthat satisfies Definition 3.1.1 . For brevity we will say “integrable” and “integral” when we m ean “Riemann integrable” and “Riemann integral.” Saying thatRb af.x/dx exists is equivalent to saying that fis integrable on Œa;b/c141 . Example 3.1.1 If f.x/D1; a/DC4x/DC4b; thennX jD1f.c j/.xj/NULxj/NUL1/DnX jD1.xj/NULxj/NUL1/: Most of the terms in the sum on the right cancel in pairs; that i s, nX jD1.xj/NULxj/NUL1/D.x1/NULx0/C.x2/NULx1/C/SOH/SOH/SOHC.xn/NULxn/NUL1/ D/NULx0C.x1/NULx1/C.x2/NULx2/C/SOH/SOH/SOHC.xn/NUL1/NULxn/NUL1/Cxn Dxn/NULx0 Db/NULa: Thus, every Riemann sum of fover any partition of Œa;b/c141 equalsb/NULa, so Zb adxDb/NULa: Example 3.1.2 Riemann sums for the function f.x/Dx; a/DC4x/DC4b; are of the form /ESCDnX jD1cj.xj/NULxj/NUL1/: (3.1.3) Sincexj/NUL1/DC4cj/DC4xjand.xjCxj/NUL1/=2is the midpoint of Œxj/NUL1;xj/c141, we can write cjDxjCxj/NUL1 2Cdj; (3.1.4) where jdjj/DC4xj/NULxj/NUL1 2/DC4kPk 2: (3.1.5) Substituting ( 3.1.4 ) into ( 3.1.3 ) yields /ESCDnX jD1xjCxj/NUL1 2.xj/NULxj/NUL1/CnX jD1dj.xj/NULxj/NUL1/ D1 2nX jD1.x2 j/NULx2 j/NUL1/CnX jD1dj.xj/NULxj/NUL1/:(3.1.6) 116 Chapter 3 Integral Calculus of Functions of One Variable Because of cancellations like those in Example 3.1.1 , nX jD1.x2 j/NULx2 j/NUL1/Db2/NULa2; so (3.1.6 ) can be rewritten as /ESCDb2/NULa2 2CnX jD1dj.xj/NULxj/NUL1/: Hence, ˇˇˇˇ/ESC/NULb2/NULa2 2ˇˇˇˇ/DC4nX jD1jdjj.xj/NULxj/NUL1//DC4kPk 2nX jD1.xj/NULxj/NUL1/(see ( 3.1.5 )) DkPk 2.b/NULa/: Therefore, every Riemann sum of fover a partition PofŒa;b/c141 satisfies ˇˇˇˇ/ESC/NULb2/NULa2 2ˇˇˇˇ</SI ifkPk<ıD2/SI b/NULa: Hence,Zb axdxDb2/NULa2 2: The Integral as the Area Under a Curve An important application of the integral, indeed, the one in variably used to motivate its definition, is the computation of the area bounded by a curve yDf.x/ , thex-axis, and the linesxDaandxDb(“the area under the curve”), as in Figure 3.1.1 . y x b ay = f(x) Figure 3.1.1 Section 3.1 Definition of the Integral 117 For simplicity, suppose that f.x/>0 . Thenf.c j/.xj/NULxj/NUL1/is the area of a rectangle with basexj/NULxj/NUL1and heightf.c j/, so the Riemann sum nX jD1f.c j/.xj/NULxj/NUL1/ can be interpreted as the sum of the areas of rectangles relat ed to the curve yDf.x/ , as shown in Figure 3.1.2 . y xac1x1x2c2x3 c3c4by = f(x) Figure 3.1.2 An apparently plausible argument, that the Riemann sums app roximate the area under the curve more and more closely as the number of rectangles in creases and the largest of their widths is made smaller, seems to support the assertion thatRb af.x/dx equals the area under the curve. This argument is useful as a motivation for Definition 3.1.1 , which without it would seem mysterious. Nevertheless, the logic i s incorrect, since it is based on the assumption that the area under the curve has been previ ously defined in some other way. Although this is true for certain curves such as, for exa mple, those consisting of line segments or circular arcs, it is not true in general. In fact, the area under a more complicated curve is defined to be equal to the integral, if the integral exists. That this new definition is consistent with the old one, where the latter applies, is evi dence that the integral provides a useful generalization of the definition of area. Example 3.1.3 Letf.x/Dx,1/DC4x/DC42(Figure 3.1.3 , page 118). The region under the curve consists of a square of unit area, surmounted by a tr iangle of area 1=2; thus, the area of the region is 3=2. From Example 3.1.2 , Z2 1xdxD1 2.22/NUL12/D3 2; so the integral equals the area under the curve. 118 Chapter 3 Integral Calculus of Functions of One Variable y x2 1y = x Figure 3.1.3 y xy = x2 2 1 Figure 3.1.4 Example 3.1.4 If f.x/Dx2; 1/DC4x/DC42 (Figure 3.1.4 ), thenZ2 1f.x/dxD1 3.23/NUL13/D7 3 (Exercise 3.1.4 ), so we say that the area under the curve is 7=3. However, this is the defini- tionof the area rather than a confirmation of a previously known fa ct, as in Example 3.1.3 . Section 3.1 Definition of the Integral 119 Theorem 3.1.2 Iffis unbounded on Œa;b/c141; thenfis not integrable on Œa;b/c141: Proof We will show that if fis unbounded on Œa;b/c141 ,Pis any partition of Œa;b/c141 , and M >0 , then there are Riemann sums /ESCand/ESC0offoverPsuch that j/ESC/NUL/ESC0j/NAKM: (3.1.7) We leave it to you (Exercise 3.1.2 ) to complete the proof by showing from this that f cannot satisfy Definition 3.1.1 . Let /ESCDnX jD1f.c j/.xj/NULxj/NUL1/ be a Riemann sum of fover a partition PofŒa;b/c141 . There must be an integer iin f1;2;:::;ngsuch that jf.c//NULf.c i/j/NAKM xi/NULxi/NUL1(3.1.8) for somecinŒxi/NUL1xi/c141, because if there were not so, we would have jf.x//NULf.c j/j<M xj/NULxj/NUL1; x j/NUL1/DC4x/DC4xj; 1/DC4j/DC4n: Then jf.x/jDjf.c j/Cf.x//NULf.c j/j/DC4jf.c j/jCjf.x//NULf.c j/j /DC4jf.c j/jCM xj/NULxj/NUL1; x j/NUL1/DC4x/DC4xj; 1/DC4j/DC4n: which implies that jf.x/j/DC4 max 1/DC4j/DC4njf.c j/jCM xj/NULxj/NUL1; a/DC4x/DC4b; contradicting the assumption that fis unbounded on Œa;b/c141 . Now suppose that csatisfies ( 3.1.8 ), and consider the Riemann sum /ESC0DnX jD1f.c0 j/.xj/NULxj/NUL1/ over the same partition P, where c0 jD/SUBcj; j¤i; c; jDi: 120 Chapter 3 Integral Calculus of Functions of One Variable Since j/ESC/NUL/ESC0jDjf.c//NULf.c i/j.xi/NULxi/NUL1/; (3.1.8 ) implies ( 3.1.7 ). Upper and Lower Integrals Because of Theorem 3.1.2 , we consider only bounded functions throughout the rest of t his section. To prove directly from Definition 3.1.1 thatRb af.x/dx exists, it is necessary to discover its valueLin one way or another and to show that Lhas the properties required by the definition. For a specific function it may happen that this can be done by straightforward calculation, as in Examples 3.1.1 and3.1.2 . However, this is not so if the objective is to find general conditions which imply thatRb af.x/dx exists. The following approach avoids the difficulty of having to discover Lin advance, without knowing whether it exists in the first place, and requires only that we compare two numbers that mus t exist iffis bounded on Œa;b/c141 . We will see thatRb af.x/dx exists if and only if these two numbers are equal. Definition 3.1.3 Iffis bounded on Œa;b/c141 andPDfx0;x1;:::;x ngis a partition of Œa;b/c141 , let MjD sup xj/NUL1/DC4x/DC4xjf.x/ and mjD inf xj/NUL1/DC4x/DC4xjf.x/: Theupper sum of foverPis S.P/DnX jD1Mj.xj/NULxj/NUL1/; and the upper integral of fover,Œa;b/c141 , denoted by Zb af.x/dx; is the infimum of all upper sums. The lower sum of foverPis s.P/DnX jD1mj.xj/NULxj/NUL1/; and the lower integral of foverŒa;b/c141 , denoted by Zb af.x/dx; is the supremum of all lower sums. Section 3.1 Definition of the Integral 121 Ifm/DC4f.x//DC4Mfor allxinŒa;b/c141 , then m.b/NULa//DC4s.P//DC4S.P//DC4M.b/NULa/ for every partition P; thus, the set of upper sums of fover all partitions PofŒa;b/c141 is bounded, as is the set of lower sums. Therefore, Theorems 1.1.3 and1.1.8 imply thatRb af.x/dx andRb af.x/dx exist, are unique, and satisfy the inequalities m.b/NULa//DC4Zb af.x/dx/DC4M.b/NULa/ and m.b/NULa//DC4Zb af.x/dx/DC4M.b/NULa/: Theorem 3.1.4 Letfbe bounded on Œa;b/c141 , and letPbe a partition of Œa;b/c141: Then (a) The upper sum S.P/ offoverPis the supremum of the set of all Riemann sums of foverP: (b) The lower sum s.P/ offoverPis the infimum of the set of all Riemann sums of f overP: Proof (a) IfPDfx0;x1;:::;x ng, then S.P/DnX jD1Mj.xj/NULxj/NUL1/; where MjD sup xj/NUL1/DC4x/DC4xjf.x/: An arbitrary Riemann sum of foverPis of the form /ESCDnX jD1f.c j/.xj/NULxj/NUL1/; wherexj/NUL1/DC4cj/DC4xj. Sincef.c j//DC4Mj, it follows that /ESC/DC4S.P/ . Now let/SI>0 and choosecjinŒxj/NUL1;xj/c141so that f.cj/>M j/NUL/SI n.xj/NULxj/NUL1/; 1/DC4j/DC4n: The Riemann sum produced in this way is /ESCDnX jD1f.cj/.xj/NULxj/NUL1/>nX jD1/DC4 Mj/NUL/SI n.xj/NULxj/NUL1///NAK .xj/NULxj/NUL1/DS.P//NUL/SI: Now Theorem 1.1.3 implies thatS.P/ is the supremum of the set of Riemann sums of f overP. (b) Exercise 3.1.7 . 122 Chapter 3 Integral Calculus of Functions of One Variable Example 3.1.5 Let f.x/D/SUB0ifxis irrational; 1ifxis rational; andPDfx0;x1;:::;x ngbe a partition of Œa;b/c141 . Since every interval contains both ratio- nal and irrational numbers (Theorems 1.1.6 and1.1.7 ), mjD0andMjD1; 1/DC4j/DC4n: Hence, S.P/DnX jD11/SOH.xj/NULxj/NUL1/Db/NULa and s.P/DnX jD10/SOH.xj/NULxj/NUL1/D0: Since all upper sums equal b/NULaand all lower sums equal 0, Definition 3.1.3 implies that Zb af.x/dxDb/NULaandZb af.x/dxD0: Example 3.1.6 Letfbe defined on Œ1;2/c141 byf.x/D0ifxis irrational and f.p=q/D 1=q ifpandqare positive integers with no common factors (Exercise 2.2.7 ). IfPD fx0;x1;:::;x ngis any partition of Œ1;2/c141 , thenmjD0,1/DC4j/DC4n, sos.P/D0; hence, Z2 1f.x/dxD0: We now show thatZ2 1f.x/dxD0 (3.1.9) also. SinceS.P/>0 for everyP, Definition 3.1.3 implies that Z2 1f.x/dx/NAK0; so we need only show thatZ2 1f.x/dx/DC40; which will follow if we show that no positive number is less th an every upper sum. To this end, we observe that if 0</SI <2 , thenf.x//NAK/SI=2for only finitely many values of xin Œ1;2/c141 . Letkbe the number of such points and let P0be a partition of Œ1;2/c141 such that kP0k</SI 2k: (3.1.10) Section 3.1 Definition of the Integral 123 Consider the upper sum S.P 0/DnX jD1Mj.xj/NULxj/NUL1/: There are at most kvalues ofjin this sum for which Mj/NAK/SI=2, andMj/DC41even for these. The contribution of these terms to the sum is less than k./SI=2k/D/SI=2, because of (3.1.10 ). SinceMj</SI=2 for all other values of j, the sum of the other terms is less than /SI 2nX jD1.xj/NULxj/NUL1/D/SI 2.xn/NULx0/D/SI 2.2/NUL1/D/SI 2: Therefore,S.P 0/</SI and, since/SIcan be chosen as small as we wish, no positive number is less than all upper sums. This proves ( 3.1.9 ). The motivation for Definition 3.1.3 can be seen by again considering the idea of area under a curve. Figure 3.1.5 shows the graph of a positive function yDf.x/ ,a/DC4x/DC4b, withŒa;b/c141 partitioned into four subintervals. a x1x2x3 by = f(x)y x Figure 3.1.5 The upper and lower sums of fover this partition can be interpreted as the sums of the area s of the rectangles surmounted by the solid and dashed lines, r espectively. This indicates that a sensible definition of area Aunder the curve must admit the inequalities s.P//DC4A/DC4S.P/ for every partition PofŒa;b/c141 . Thus,Amust be an upper bound for all lower sums and a lower bound for all upper sums of fover partitions of Œa;b/c141 . If Zb af.x/dxDZb af.x/dx; (3.1.11) 124 Chapter 3 Integral Calculus of Functions of One Variable there is only one number, the common value of the upper and low er integrals, with this property, and we define Ato be that number; if ( 3.1.11 ) does not hold, then Ais not defined. We will see below that this definition of area is consistent wi th the definition stated earlier in terms of Riemann sums. Example 3.1.7 Returning to Example 3.1.3 , consider the function f.x/Dx; 1/DC4x/DC42: IfPDfx0;x1;:::;x ngis a partition of Œ1;2/c141 , then, sincefis increasing, MjDf.x j/DxjandmjDf.x j/NUL1/Dxj/NUL1: Hence, S.P/DnX jD1xj.xj/NULxj/NUL1/ (3.1.12) and s.P/DnX jD1xj/NUL1.xj/NULxj/NUL1/: (3.1.13) By writing xjDxjCxj/NUL1 2Cxj/NULxj/NUL1 2; we see from ( 3.1.12 ) that S.P/D1 2nX jD1.x2 j/NULx2 j/NUL1/C1 2nX jD1.xj/NULxj/NUL1/2 D1 2.22/NUL12/C1 2nX jD1.xj/NULxj/NUL1/2:(3.1.14) Since 0<nX jD1.xj/NULxj/NUL1/2/DC4kPknX jD1.xj/NULxj/NUL1/DkPk.2/NUL1/; (3.1.14 ) implies that 3 2<S.P//DC43 2CkPk 2: SincekPkcan be made as small as we please, Definition 3.1.3 implies that Zb af.x/dxD3 2: A similar argument starting from ( 3.1.13 ) shows that 3 2/NULkPk 2/DC4s.P/<3 2; Section 3.1 Definition of the Integral 125 soZb af.x/dxD3 2: Since the upper and lower integrals both equal 3=2, the area under the curve is 3=2accord- ing to our new definition. This is consistent with the result i n Example 3.1.3 . The Riemann–Stieltjes Integral TheRiemann–Stieltjes integral is an important generalization of the Riemann integral. We define it here, but confine our study of it to the exercises in th is and other sections of this chapter. Definition 3.1.5 Letfandgbe defined on Œa;b/c141 . We say that fisRiemann –Stieltjes integrable with respect to gonŒa;b/c141 if there is a number Lwith the following property: For every/SI>0 , there is aı>0 such that ˇˇˇˇˇˇnX jD1f.c j//STX g.x j//NULg.x j/NUL1//ETX /NULLˇˇˇˇˇˇ</SI; (3.1.15) provided only that PDfx0;x1;:::;x ngis a partition of Œa;b/c141 such thatkPk<ıand xj/NUL1/DC4cj/DC4xj; jD1;2;:::;n: In this case, we say that Listhe Riemann–Stieltjes integral of fwith respect to gover Œa;b/c141 , and writeZb af.x/dg.x/DL: The sumnX jD1f.c j//STX g.x j//NULg.x j/NUL1//ETX in (3.1.15 ) isa Riemann–Stieltjes sum of fwith respect to gover the partition P. 3.1 Exercises 1. Show that there cannot be more than one number Lthat satisfies Definition 3.1.1 . 2. (a) Prove: IfRb af.x/dx exists, then for every /SI > 0 , there is aı > 0 such that j/ESC1/NUL/ESC2j</SIif/ESC1and/ESC2are Riemann sums of fover partitions P1andP2 ofŒa;b/c141 with norms less than ı. 126 Chapter 3 Integral Calculus of Functions of One Variable (b) Suppose that there is an M >0 such that, for every ı>0 , there are Riemann sums/ESC1and/ESC2over a partition PofŒa;b/c141 withkPk<ısuch thatj/ESC1/NUL/ESC2j/NAK M. Use(a)to prove thatfis not integrable over Œa;b/c141 . 3. Suppose thatRb af.x/dx exists and there is a number Asuch that, for every /SI >0 andı>0 , there is a partition PofŒa;b/c141 withkPk<ıand a Riemann sum /ESCoff overPthat satisfies the inequality j/ESC/NULAj</SI. Show thatRb af.x/dxDA. 4. Prove directly from Definition 3.1.1 that Zb ax2dxDb3/NULa3 3: Do not assume in advance that the integral exists. The proof o f this is part of the problem. H INT:LetPDfx0;x2;:::;x ngbe an arbitrary partition of Œa;b/c141: Use the mean value theorem to show that b3/NULa3 3DnX jD1d2 j.xj/NULxj/NUL1/ for some points d1;. . . ,dn;wherexj/NUL1< d j< x j. Then relate this sum to arbitrary Riemann sums for f.x/Dx2overP: 5. Generalize the proof of Exercise 3.1.4 to show directly from Definition 3.1.1 that Zb axmdxDbmC1/NULamC1 mC1 ifmis an integer/NAK0. 6. Prove directly from Definition 3.1.1 thatf.x/ is integrable on Œa;b/c141 if and only if f./NULx/is integrable on Œ/NULb;/NULa/c141, and, in this case, Zb af.x/dxDZ/NULa /NULbf./NULx/dx: 7. Letfbe bounded on Œa;b/c141 and letPbe a partition of Œa;b/c141 . Prove: The lower sum s.P/ offoverPis the infimum of the set of all Riemann sums of foverP. 8. Letfbe defined on Œa;b/c141 and letPDfx0;x1;:::;x ngbe a partition of Œa;b/c141 . (a) Prove: Iffis continuous on Œa;b/c141 , thens.P/ andS.P/ are Riemann sums of foverP. (b) Name another class of functions for which the conclusion of (a)is valid. (c) Give an example where s.P/ andS.P/ are not Riemann sums of foverP. Section 3.1 Definition of the Integral 127 9. FindR1 0f.x/dx andR1 0f.x/dx if (a)f.x/D/SUBxifxis rational; /NULxifxis irrational:(b)f.x/D/SUB1ifxis rational; xifxis irrational: 10. Given thatRb aexdxexists, evaluate it by using the formula 1CrCr2C/SOH/SOH/SOHCrnD1/NULrnC1 1/NULr.r¤1/ to calculate certain Riemann sums. H INT:See Exercise 3.1.3: 11. Given thatRb 0sinxdx exists, evaluate it by using the identity cos.j/NUL1//DC2/NULcos.jC1//DC2D2sin/DC2sinj/DC2 to calculate certain Riemann sums. H INT:See Exercise 3.1.3: 12. Given thatRb 0cosxdx exists, evaluate it by using the identity sin.jC1//DC2/NULsin.j/NUL1//DC2D2sin/DC2cosj/DC2 to calculate certain Riemann sums. H INT:See Exercise 3.1.3: 13. Show that ifg.x/DxCc(c=constant), thenRb af.x/dg.x/ exists if and only ifRb af.x/dx exists, in which case Zb af.x/dg.x/DZb af.x/dx: 14. Suppose that/NUL1<a<d <c<1and g.x/D/SUBg1; a<x<d; g2; d <x<b;(g1;g2Dconstants), and letg.a/,g.b/, andg.d/ be arbitrary. Suppose that fis defined on Œa;b/c141 , continuous from the right at aand from the left at b, and continuous at d. Show thatRb af.x/dg.x/ exists, and find its value. 15. Suppose that/NUL1< aDa0< a 1</SOH/SOH/SOH< a pDb <1, letg.x/Dgm (constant) on .am/NUL1;am/,1/DC4m/DC4p, and letg.a0/,g.a1/, . . . ,g.ap/be arbitrary. Suppose that fis defined on Œa;b/c141 , continuous from the right at aand from the left atb, and continuous at a1,a2, . . . ,ap/NUL1. EvaluateRb af.x/dg.x/ . HINT:See Exercise 3.1.14: 16. (a) Give an example whereRb af.x/dg.x/ exists even though fis unbounded onŒa;b/c141 . (Thus, the analog of Theorem 3.1.2 does not hold for the Riemann– Stieltjes integral.) (b) State and prove an analog of Theorem 3.1.2 for the case where gis increasing. 128 Chapter 3 Integral Calculus of Functions of One Variable 17. For the case where gis nondecreasing and fis bounded on Œa;b/c141 , define upper and lower Riemann–Stieltjes integrals in a way analogous to Defi nition 3.1.3 . 3.2 EXISTENCE OF THE INTEGRAL The following lemma is the starting point for our study of the integrability of a bounded functionfon a closed interval Œa;b/c141 . Lemma 3.2.1 Suppose that jf.x/j/DC4M; a/DC4x/DC4b; (3.2.1) and letP0be a partition of Œa;b/c141 obtained by adding rpoints to a partition PDfx0;x1;:::;x ng ofŒa;b/c141: Then S.P//NAKS.P0//NAKS.P//NUL2MrkPk (3.2.2) and s.P//DC4s.P0//DC4s.P/C2MrkPk: (3.2.3) Proof We will prove ( 3.2.2 ) and leave the proof of ( 3.2.3 ) to you (Exercise 3.2.1 ). First suppose that rD1, soP0is obtained by adding one point cto the partition PD fx0;x1;:::;x ng; thenxi/NUL1< c < x ifor someiinf1;2;:::;ng. Ifj¤i, the prod- uctMj.xj/NULxj/NUL1/appears in both S.P/ andS.P0/and cancels out of the difference S.P//NULS.P0/. Therefore, if Mi1D sup xi/NUL1/DC4x/DC4cf.x/ andMi2Dsup c/DC4x/DC4xif.x/; then S.P//NULS.P0/DMi.xi/NULxi/NUL1//NULMi1.c/NULxi/NUL1//NULMi2.xi/NULc/ D.Mi/NULMi1/.c/NULxi/NUL1/C.Mi/NULMi2/.xi/NULc/:(3.2.4) Since ( 3.2.1 ) implies that 0/DC4Mi/NULMir/DC42M; rD1;2; (3.2.4 ) implies that 0/DC4S.P//NULS.P0//DC42M.x i/NULxi/NUL1//DC42MkPk: This proves ( 3.2.2 ) forrD1. Now suppose that r > 1 andP0is obtained by adding points c1,c2, . . . ,crtoP. Let P.0/DPand, forj/NAK1, letP.j /be the partition of Œa;b/c141 obtained by adding cjto P.j/NUL1/. Then the result just proved implies that 0/DC4S.P.j/NUL1///NULS.P.j ///DC42MkP.j/NUL1/k; 1/DC4j/DC4r: Section 3.2 Existence of the Integral 129 Adding these inequalities and taking account of cancellati ons yields 0/DC4S.P.0///NULS.P.r///DC42M.kP.0/kCkP.1/kC/SOH/SOH/SOHCkP.r/NUL1/k/: (3.2.5) SinceP.0/DP,P.r/DP0, andkP.k/k/DC4kP.k/NUL1/kfor1/DC4k/DC4r/NUL1, (3.2.5 ) implies that 0/DC4S.P//NULS.P0//DC42MrkPk; which is equivalent to ( 3.2.2 ). Theorem 3.2.2 Iffis bounded on Œa;b/c141; then Zb af.x/dx/DC4Zb af.x/dx: (3.2.6) Proof Suppose that P1andP2are partitions of Œa;b/c141 andP0is a refinement of both. LettingPDP1in (3.2.3 ) andPDP2in (3.2.2 ) shows that s.P 1//DC4s.P0/andS.P0//DC4S.P 2/: Sinces.P0//DC4S.P0/, this implies that s.P 1//DC4S.P 2/. Thus, every lower sum is a lower bound for the set of all upper sums. SinceRb af.x/dx is the infimum of this set, it follows that s.P 1//DC4Zb af.x/dx for every partition P1ofŒa;b/c141 . This means thatRb af.x/dx is an upper bound for the set of all lower sums. SinceRb af.x/dx is the supremum of this set, this implies ( 3.2.6 ). Theorem 3.2.3 Iffis integrable on Œa;b/c141; then Zb af.x/dxDZb af.x/dxDZb af.x/dx: Proof We prove thatRb af.x/dxDRb af.x/dx and leave it to you to show thatRb af.x/dxD Rb af.x/dx (Exercise 3.2.2 ). Suppose that Pis a partition of Œa;b/c141 and/ESCis a Riemann sum of foverP. Since Zb af.x/dx/NULZb af.x/dxD Zb af.x/dx/NULS.P/! C.S.P//NUL/ESC/ C /ESC/NULZb af.x/dx! ; 130 Chapter 3 Integral Calculus of Functions of One Variable the triangle inequality implies that ˇˇˇˇˇZb af.x/dx/NULZb af.x/dxˇˇˇˇˇ/DC4ˇˇˇˇˇZb af.x/dx/NULS.P/ˇˇˇˇˇCjS.P//NUL/ESCj Cˇˇˇˇˇ/ESC/NULZb af.x/dxˇˇˇˇˇ:(3.2.7) Now suppose that /SI>0 . From Definition 3.1.3 , there is a partition P0ofŒa;b/c141 such that Zb af.x/dx/DC4S.P 0/<Zb af.x/dxC/SI 3: (3.2.8) From Definition 3.1.1 , there is aı>0 such that ˇˇˇˇˇ/ESC/NULZb af.x/dxˇˇˇˇˇ</SI 3(3.2.9) ifkPk<ı. Now suppose that kPk<ıandPis a refinement of P0. SinceS.P//DC4S.P 0/ by Lemma 3.2.1 , (3.2.8 ) implies that Zb af.x/dx/DC4S.P/<Zb af.x/dxC/SI 3; so ˇˇˇˇˇS.P//NULZb af.x/dxˇˇˇˇˇ</SI 3(3.2.10) in addition to ( 3.2.9 ). Now ( 3.2.7 ), (3.2.9 ), and ( 3.2.10 ) imply that ˇˇˇˇˇZb af.x/dx/NULZb af.x/dxˇˇˇˇˇ<2/SI 3CjS.P//NUL/ESCj (3.2.11) for every Riemann sum /ESCoffoverP. SinceS.P/ is the supremum of these Riemann sums (Theorem 3.1.4 ), we may choose /ESCso that jS.P//NUL/ESCj</SI 3: Now ( 3.2.11 ) implies that ˇˇˇˇˇZb af.x/dx/NULZb af.x/dxˇˇˇˇˇ</SI: Since/SIis an arbitrary positive number, it follows that Zb af.x/dxDZb af.x/dx: Section 3.2 Existence of the Integral 131 Lemma 3.2.4 Iffis bounded on Œa;b/c141 and/SI>0; there is aı>0 such that Zb af.x/dx/DC4S.P/<Zb af.x/dxC/SI (3.2.12) andZb af.x/dx/NAKs.P/>Zb af.x/dx/NUL/SI ifkPk<ı. Proof We show that ( 3.2.12 ) holds ifkPkis sufficiently small, and leave the rest of the proof to you (Exercise 3.2.3 ). The first inequality in ( 3.2.12 ) follows immediately from Definition 3.1.3 . To establish the second inequality, suppose that jf.x/j/DC4Kifa/DC4x/DC4b. From Definition 3.1.3 , there is a partitionP0Dfx0;x1;:::;x rC1gofŒa;b/c141 such that S.P 0/<Zb af.x/dxC/SI 2: (3.2.13) IfPis any partition of Œa;b/c141 , letP0be constructed from the partition points of P0andP. Then S.P0//DC4S.P 0/; (3.2.14) by Lemma 3.2.1 . SinceP0is obtained by adding at most rpoints toP, Lemma 3.2.1 implies that S.P0//NAKS.P//NUL2KrkPk: (3.2.15) Now ( 3.2.13 ), (3.2.14 ), and ( 3.2.15 ) imply that S.P//DC4S.P0/C2KrkPk /DC4S.P 0/C2KrkPk <Zb af.x/dxC/SI 2C2KrkPk: Therefore, ( 3.2.12 ) holds if kPk<ıD/SI 4Kr: 132 Chapter 3 Integral Calculus of Functions of One Variable Theorem 3.2.5 Iffis bounded on Œa;b/c141 and Zb af.x/dxDZb af.x/dxDL; (3.2.16) thenfis integrable on Œa;b/c141 and Zb af.x/dxDL: (3.2.17) Section 3.2 Existence of the Integral 133 Proof If/SI>0 , there is aı>0 such that Zb af.x/dx/NUL/SI<s.P//DC4S.P/<Zb af.x/dxC/SI (3.2.18) ifkPk<ı(Lemma 3.2.4 ). If/ESCis a Riemann sum of foverP, then s.P//DC4/ESC/DC4S.P/; so (3.2.16 ) and ( 3.2.18 ) imply that L/NUL/SI</ESC <LC/SI ifkPk<ı. Now Definition 3.1.1 implies ( 3.2.17 ). Theorems 3.2.3 and3.2.5 imply the following theorem. Theorem 3.2.6 A bounded function fis integrable on Œa;b/c141 if and only if Zb af.x/dxDZb af.x/dx: The next theorem translates this into a test that can be conve niently applied. Theorem 3.2.7 Iffis bounded on Œa;b/c141; thenfis integrable on Œa;b/c141 if and only if for each/SI>0 there is a partition PofŒa;b/c141 for which S.P//NULs.P/</SI: (3.2.19) Proof We leave it to you (Exercise 3.2.4 ) to show that ifRb af.x/dx exists, then ( 3.2.19 ) holds forkPksufficiently small. This implies that the stated condition i s necessary for in- tegrability. To show that it is sufficient, we observe that si nce s.P//DC4Zb af.x/dx/DC4Zb af.x/dx/DC4S.P/ for allP, (3.2.19 ) implies that 0/DC4Zb af.x/dx/NULZb af.x/dx</SI: Since/SIcan be any positive number, this implies that Zb af.x/dxDZb af.x/dx: Therefore,Rb af.x/dx exists, by Theorem 3.2.5 . The next two theorems are important applications of Theorem 3.2.7 . 134 Chapter 3 Integral Calculus of Functions of One Variable Theorem 3.2.8 Iffis continuous on Œa;b/c141; thenfis integrable on Œa;b/c141 . Proof LetPDfx0;x1;:::;x ngbe a partition of Œa;b/c141 . Sincefis continuous on Œa;b/c141 , there are points cjandc0 jinŒxj/NUL1;xj/c141such that f.c j/DMjD sup xj/NUL1/DC4x/DC4xjf.x/ and f.c0 j/DmjD inf xj/NUL1/DC4x/DC4xjf.x/ (Theorem 2.2.9 ). Therefore, S.P//NULs.P/DnX jD1/STXf.c j//NULf.c0 j//ETX.xj/NULxj/NUL1/: (3.2.20) Sincefis uniformly continuous on Œa;b/c141 (Theorem 2.2.12 ), there is for each /SI>0 aı>0 such that jf.x0//NULf.x/j</SI b/NULa ifxandx0are inŒa;b/c141 andjx/NULx0j<ı. IfkPk<ı, thenjcj/NULc0 jj<ıand, from ( 3.2.20 ), S.P//NULs.P/</SI b/NULanX jD1.xj/NULxj/NUL1/D/SI: Hence,fis integrable on Œa;b/c141 , by Theorem 3.2.7 . Theorem 3.2.9 Iffis monotonic on Œa;b/c141; thenfis integrable on Œa;b/c141 . Proof LetPDfx0;x1;:::;x ngbe a partition of Œa;b/c141 . Sincefis nondecreasing, f.x j/DMjD sup xj/NUL1/DC4x/DC4xjf.x/ and f.x j/NUL1/DmjD inf xj/NUL1/DC4x/DC4xjf.x/: Hence, S.P//NULs.P/DnX jD1.f.x j//NULf.x j/NUL1//.x j/NULxj/NUL1/: Since0<x j/NULxj/NUL1/DC4kPkandf.x j//NULf.x j/NUL1//NAK0, S.P//NULs.P//DC4kPknX jD1.f.x j//NULf.x j/NUL1// DkPk.f.b//NULf.a//: Section 3.2 Existence of the Integral 135 Therefore, S.P//NULs.P/</SI ifkPk.f.b//NULf.a//</SI; sofis integrable on Œa;b/c141 , by Theorem 3.2.7 . The proof for nonincreasing fis similar. We will also use Theorem 3.2.7 in the next section to establish properties of the integral. In Section 3.5 we will study more general conditions for inte grability. 3.2 Exercises 1. Complete the proof of Lemma 3.2.1 by verifying Eqn. ( 3.2.3 ). 2. Show that iffis integrable on Œa;b/c141 , then Zb af.x/dxDZb af.x/dx: 3. Prove: Iffis bounded on Œa;b/c141 , there is for each /SI>0 aı>0 such that Zb af.x/dx/NAKZb af.x/dx/NUL/SI<s.P/ ifkPk<ı. 4. Prove: Iffis integrable on Œa;b/c141 and/SI > 0 , thenS.P//NULs.P/ < /SI ifkPkis sufficiently small. H INT:Use Theorem 3.1.4: 5. Suppose that fis integrable and gis bounded on Œa;b/c141 , andgdiffers fromfonly at points in a set Hwith the following property: For each /SI>0 ,Hcan be covered by a finite number of closed subintervals of Œa;b/c141 , the sum of whose lengths is less than/SI. Show thatgis integrable on Œa;b/c141 and that Zb ag.x/dxDZb af.x/dx: HINT:Use Exercise 3.1.3: 6. Suppose that gis bounded on Œ˛;ˇ/c141 , and letQW˛Dv0<v 1</SOH/SOH/SOH<v LDˇbe a fixed partition of Œ˛;ˇ/c141 . Prove: (a)Zˇ ˛g.u/duDLX `D1Zv` v`/NUL1g.u/duI(b)Zˇ ˛g.u/duDLX `D1Zv` v`/NUL1g.u/du: 7. A functionfisof bounded variation on Œa;b/c141 if there is a number Ksuch that nX jD1ˇˇf.a j//NULf.a j/NUL1/ˇˇ/DC4K wheneveraDa0<a 1</SOH/SOH/SOH<a nDb. (The smallest number with this property is the total variation of fonŒa;b/c141 .) 136 Chapter 3 Integral Calculus of Functions of One Variable (a) Prove: Iffis of bounded variation on Œa;b/c141 , thenfis bounded on Œa;b/c141 . (b) Prove: Iffis of bounded variation on Œa;b/c141 , thenfis integrable on Œa;b/c141 . HINT:Use Theorems 3.1.4 and3.2.7: 8. LetPDfx0;x1;:::;x ngbe a partition of Œa;b/c141 ,c0Dx0Da,cnC1DxnDb, andxj/NUL1/DC4cj/DC4xj,jD1,2, . . . ,n. Verify that nX jD1g.cj/Œf.x j//NULf.x j/NUL1//c141Dg.b/f.b//NULg.a/f.a//NULnX jD0f.x j/Œg.c jC1//NULg.cj//c141: Use this to prove that ifRb af.x/dg.x/ exists, then so doesRb ag.x/df.x/ , and Zb ag.x/df.x/Df.b/g.b//NULf.a/g.a//NULZb af.x/dg.x/: (This is the integration by parts formula for Riemann–Stieltjes integrals.) 9. Letfbe continuous and gbe of bounded variation (Exercise 3.2.7 ) onŒa;b/c141 . (a) Show that if /SI > 0 , there is aı > 0 such thatj/ESC/NUL/ESC0j< /SI=2 if/ESCand/ESC0 are Riemann–Stieltjes sums of fwith respect to gover partitions PandP0 ofŒa;b/c141 , whereP0is a refinement of PandkPk< ı. H INT:Use Theo- rem2.2.12: (b) Letıbe as chosen in (a). Suppose that /ESC1and/ESC2are Riemann–Stieltjes sums offwith respect to gover any partitions P1andP2ofŒa;b/c141 with norm less thanı. Show thatj/ESC1/NUL/ESC2j</SI. (c) Ifı >0 , letL.ı/ be the supremum of all Riemann–Stieltjes sums of fwith respect togover partitions of Œa;b/c141 with norms less than ı. Show thatL.ı/ is finite. Then show that LDlimı!0CL.ı/ exists. H INT:Use Theorem 2.1.9: (d) Show thatRb af.x/dg.x/DL. 10. Show thatRb af.x/dg.x/ exists iffis of bounded variation and gis continuous on Œa;b/c141 . HINT:See Exercises 3.2.8 and3.2.9: 3.3 PROPERTIES OF THE INTEGRAL We now use the results of Sections 3.1 and 3.2 to establish the properties of the integral. You are probably familiar with most of these properties, but not with their proofs. Theorem 3.3.1 Iffandgare integrable on Œa;b/c141; then so isfCg;and Zb a.fCg/.x/dxDZb af.x/dxCZb ag.x/dx: Section 3.3 Properties of the Integral 137 Proof Any Riemann sum of fCgover a partition PDfx0;x1;:::;x ngofŒa;b/c141 can be written as /ESCfCgDnX jD1Œf.c j/Cg.cj//c141.x j/NULxj/NUL1/ DnX jD1f.c j/.xj/NULxj/NUL1/CnX jD1g.cj/.xj/NULxj/NUL1/ D/ESCfC/ESCg; where/ESCfand/ESCgare Riemann sums for fandg. Definition 3.1.1 implies that if /SI > 0 there are positive numbers ı1andı2such that ˇˇˇˇˇ/ESCf/NULZb af.x/dxˇˇˇˇˇ</SI 2ifkPk<ı1 andˇˇˇˇˇ/ESCg/NULZb ag.x/dxˇˇˇˇˇ</SI 2ifkPk<ı2: IfkPk<ıDmin.ı1;ı2/, then ˇˇˇˇˇ/ESCfCg/NULZb af.x/dx/NULZb ag.x/dxˇˇˇˇˇDˇˇˇˇˇ /ESCf/NULZb af.x/dx! C /ESCg/NULZb ag.x/dx!ˇˇˇˇˇ /DC4ˇˇˇˇˇ/ESCf/NULZb af.x/dxˇˇˇˇˇCˇˇˇˇˇ/ESCg/NULZb ag.x/dxˇˇˇˇˇ </SI 2C/SI 2D/SI; so the conclusion follows from Definition 3.1.1 . The next theorem also follows from Definition 3.1.1 (Exercise 3.3.1 ). Theorem 3.3.2 Iffis integrable on Œa;b/c141 andcis a constant;thencfis integrable onŒa;b/c141 andZb acf.x/dxDcZb af.x/dx: Theorems 3.3.1 and3.3.2 and induction yield the following result (Exercise 3.3.2 ). Theorem 3.3.3 Iff1; f2;. . .; fnare integrable on Œa;b/c141 andc1; c2;. . .; cnare constants;thenc1f1Cc2f2C/SOH/SOH/SOHCcnfnis integrable on Œa;b/c141 and Zb a.c1f1Cc2f2C/SOH/SOH/SOHCcnfn/.x/dxDc1Zb af1.x/dxCc2Zb af2.x/dx C/SOH/SOH/SOHCcnZb afn.x/dx: 138 Chapter 3 Integral Calculus of Functions of One Variable Theorem 3.3.4 Iffandgare integrable on Œa;b/c141 andf.x//DC4g.x/ fora/DC4x/DC4b; thenZb af.x/dx/DC4Zb ag.x/dx: (3.3.1) Proof Sinceg.x//NULf.x//NAK0, every lower sum of g/NULfover any partition of Œa;b/c141 is nonnegative. Therefore,Zb a.g.x//NULf.x//dx/NAK0: Hence,Zb ag.x/dx/NULZb af.x/dxDZb a.g.x//NULf.x//dx DZb a.g.x//NULf.x//dx/NAK0;(3.3.2) which yields ( 3.3.1 ). (The first equality in ( 3.3.2 ) follows from Theorems 3.3.1 and3.3.2 ; the second, from Theorem 3.2.3 .) Theorem 3.3.5 Iffis integrable on Œa;b/c141; then so isjfj, and ˇˇˇˇˇZb af.x/dxˇˇˇˇˇ/DC4Zb ajf.x/jdx: (3.3.3) Proof LetPbe a partition of Œa;b/c141 and define MjDsup˚ f.x/ˇˇxj/NUL1/DC4x/DC4xj/TAB ; mjDinf˚f.x/ˇˇxj/NUL1/DC4x/DC4xj/TAB; MjDsup˚ jf.x/jˇˇxj/NUL1/DC4x/DC4xj/TAB ; mjDinf˚jf.x/jˇˇxj/NUL1/DC4x/DC4xj/TAB: Then Mj/NULmjDsup˚ jf.x/j/NULjf.x0/jˇˇxj/NUL1/DC4x;x0/DC4xj/TAB /DC4sup˚ jf.x//NULf.x0/jˇˇxj/NUL1/DC4x;x0/DC4xj/TAB DMj/NULmj:(3.3.4) Therefore, S.P//NULs.P//DC4S.P//NULs.P/; where the upper and lower sums on the left are associated with jfjand those on the right are associated with f. Now suppose that /SI>0 . Sincefis integrable on Œa;b/c141 , Theorem 3.2.7 implies that there is a partition PofŒa;b/c141 such thatS.P//NULs.P/ < /SI . This inequality and ( 3.3.4 ) imply thatS.P//NULs.P/ < /SI . Therefore,jfjis integrable on Œa;b/c141 , again by Theorem 3.2.7 . Since f.x//DC4jf.x/jand/NULf.x//DC4jf.x/j; a/DC4x/DC4b; Section 3.3 Properties of the Integral 139 Theorems 3.3.2 and3.3.4 imply that Zb af.x/dx/DC4Zb ajf.x/jdx and/NULZb af.x/dx/DC4Zb ajf.x/jdx; which implies ( 3.3.3 ). Theorem 3.3.6 Iffandgare integrable on Œa;b/c141; then so is the product fg: Proof We consider the case where fandgare nonnegative, and leave the rest of the proof to you (Exercise 3.3.4 ). The subscripts f,g, andfgin the following argument identify the functions with which the various quantities ar e associated. We assume that neitherfnorgis identically zero on Œa;b/c141 , since the conclusion is obvious if one of them is. IfPDfx0;x1;:::;x ngis a partition of Œa;b/c141 , then Sfg.P//NULsfg.p/DnX jD1.Mfg;j/NULmfg;j/.xj/NULxj/NUL1/: (3.3.5) Sincefandgare nonnegative, Mfg;j/DC4Mf;jMg;jandmfg;j/NAKmf;jmg;j. Hence, Mfg;j/NULmfg;j/DC4Mf;jMg;j/NULmf;jmg;j D.Mf;j/NULmf;j/Mg;jCmf;j.Mg;j/NULmg;j/ /DC4Mg.Mf;j/NULmf;j/CMf.Mg;j/NULmg;j/; whereMfandMgare upper bounds for fandgonŒa;b/c141 . From ( 3.3.5 ) and the last inequality, Sfg.P//NULsfg.P//DC4MgŒSf.P//NULsf.P//c141CMfŒSg.P//NULsg.P//c141: (3.3.6) Now suppose that /SI > 0 . Theorem 3.2.7 implies that there are partitions P1andP2of Œa;b/c141 such that Sf.P1//NULsf.P1/</SI 2MgandSg.P2//NULsg.P2/</SI 2Mf: (3.3.7) IfPis a refinement of both P1andP2, then ( 3.3.7 ) and Lemma 3.2.1 imply that Sf.P//NULsf.P/</SI 2MgandSg.P//NULsg.P/</SI 2Mf: This and ( 3.3.6 ) yield Sfg.P//NULsfg.P/</SI 2C/SI 2D/SI: Therefore,fgis integrable on Œa;b/c141 , by Theorem 3.2.7 . 140 Chapter 3 Integral Calculus of Functions of One Variable Theorem 3.3.7 (First Mean Value Theorem for Integrals) Suppose that uis continuous and vis integrable and nonnegative on Œa;b/c141: Then Zb au.x/v.x/dxDu.c/Zb av.x/dx (3.3.8) for somecinŒa;b/c141 . Proof From Theorem 3.2.8 ,uis integrable on Œa;b/c141 . Therefore, Theorem 3.3.6 implies that the integral on the left exists. If mDmin˚u.x/ˇˇa/DC4x/DC4b/TABandMDmax˚u.x/ˇˇa/DC4x/DC4b/TAB (recall Theorem 2.2.9 ), then m/DC4u.x//DC4M and, sincev.x//NAK0, mv.x//DC4u.x/v.x//DC4Mv.x/: Therefore, Theorems 3.3.2 and3.3.4 imply that mZb av.x/dx/DC4Zb au.x/v.x/dx/DC4MZb av.x/dx: (3.3.9) This implies that ( 3.3.8 ) holds for any cinŒa;b/c141 ifRb av.x/dxD0. IfRb av.x/dx¤0, let uDZb au.x/v.x/dx Zb av.x/dx(3.3.10) SinceRb av.x/dx > 0 in this case (why?), ( 3.3.9 ) implies that m/DC4u/DC4M, and the intermediate value theorem (Theorem 2.2.10 ) implies that uDu.c/ for somecinŒa;b/c141 . This implies ( 3.3.8 ). Ifv.x//DC11, then ( 3.3.10 ) reduces to uD1 b/NULaZb au.x/dx; souis the average of u.x/ overŒa;b/c141 . More generally, if vis any nonnegative integrable function such thatRb av.x/dx¤0, thenuin (3.3.10 ) is the weighted average of u.x/ over Œa;b/c141 with respect to v. Theorem 3.3.7 says that a continuous function assumes any such weighted average at some point in Œa;b/c141 . Theorem 3.3.8 Iffis integrable on Œa;b/c141 anda/DC4a1<b 1/DC4b;thenfis integrable onŒa1;b1/c141: Section 3.3 Properties of the Integral 141 Proof Suppose that/SI>0 . From Theorem 3.2.7 , there is a partition PDfx0;x1;:::;x ng ofŒa;b/c141 such that S.P//NULs.P/DnX jD1.Mj/NULmj/.xj/NULxj/NUL1/</SI: (3.3.11) We may assume that a1andb1are partition points of P, because if not they can be inserted to obtain a refinement P0such thatS.P0//NULs.P0//DC4S.P//NULs.P/ (Lemma 3.2.1 ). Let a1Dxrandb1Dxs. Since every term in ( 3.3.11 ) is nonnegative, sX jDrC1.Mj/NULmj/.xj/NULxj/NUL1/</SI: Thus,PDfxr;xrC1;:::;x sgis a partition of Œa1;b1/c141over which the upper and lower sums offsatisfy S.P//NULs.P/</SI: Therefore,fis integrable on Œa1;b1/c141, by Theorem 3.2.7 . We leave the proof of the next theorem to you (Exercise 3.3.8 ). Theorem 3.3.9 Iffis integrable on Œa;b/c141 andŒb;c/c141; thenfis integrable on Œa;c/c141; andZc af.x/dxDZb af.x/dxCZc bf.x/dx: (3.3.12) So far we have definedRˇ ˛f.x/dx only for the case where ˛<ˇ . Now we define Z˛ ˇf.x/dxD/NULZˇ ˛f.x/dx if˛<ˇ , andZ˛ ˛f.x/dxD0: With these conventions, ( 3.3.12 ) holds no matter what the relative order of a,b, andc, provided that fis integrable on some closed interval containing them (Exer cise3.3.9 ). Theorem 3.3.8 and these definitions enable us to define a function F.x/DRx cf.t/dt , wherecis an arbitrary, but fixed, point in Œa;b/c141 . Theorem 3.3.10 Iffis integrable on Œa;b/c141 anda/DC4c/DC4b;then the function F defined by F.x/DZx cf.t/dt satisfies a Lipschitz condition on Œa;b/c141; and is therefore continuous on Œa;b/c141: 142 Chapter 3 Integral Calculus of Functions of One Variable Proof Ifxandx0are inŒa;b/c141 , then F.x//NULF.x0/DZx cf.t/dt/NULZx0 cf.t/dtDZx x0f.t/dt; by Theorem 3.3.9 and the conventions just adopted. Since jf.t/j/DC4K .a/DC4t/DC4b/for some constant K,ˇˇˇˇZx x0f.t/dtˇˇˇˇ/DC4Kjx/NULx0j; a/DC4x;x0/DC4b (Theorem 3.3.5 ), so jF.x//NULF.x0/j/DC4Kjx/NULx0j; a/DC4x;x0/DC4b: Theorem 3.3.11 Iffis integrable on Œa;b/c141 anda/DC4c/DC4b;thenF.x/DRx cf.t/dt is differentiable at any point x0in.a;b/ wherefis continuous ;withF0.x0/Df.x 0/:If fis continuous from the right at a;thenF0 C.a/Df.a/ . Iffis continuous from the left atb;thenF0 /NUL.b/Df.b/: Proof We consider the case where a < x 0< b and leave the rest to you (Exer- cise3.3.14 ). Since 1 x/NULx0Zx x0f.x 0/dtDf.x 0/; we can write F.x//NULF.x 0/ x/NULx0/NULf.x 0/D1 x/NULx0Zx x0Œf.t//NULf.x 0//c141dt: From this and Theorem 3.3.5 , ˇˇˇˇF.x//NULF.x 0/ x/NULx0/NULf.x 0/ˇˇˇˇ/DC41 jx/NULx0jˇˇˇˇZx x0jf.t//NULf.x 0/jdtˇˇˇˇ: (3.3.13) (Why do we need the absolute value bars outside the integral? ) Sincefis continuous at x0, there is for each /SI>0 aı>0 such that jf.t//NULf.x 0/j</SI ifjx/NULx0j<ı andtis betweenxandx0. Therefore, from ( 3.3.13 ), ˇˇˇˇF.x//NULF.x 0/ x/NULx0/NULf.x 0/ˇˇˇˇ</SIjx/NULx0j jx/NULx0jD/SIif0<jx/NULx0j<ı: Hence,F0.x0/Df.x 0/. Section 3.3 Properties of the Integral 143 Example 3.3.1 If f.x/D(x; 0/DC4x/DC41; xC1; 1<x/DC42; then the function F.x/DZx 0f.t/dtD8 ˆˆ< ˆˆ:x2 2; 0<x/DC41; x2 2Cx/NUL1; 1<x/DC42; is continuous on Œ0;2/c141 . As implied by Theorem 3.3.11 , F0.x/D8 < :xDf.x/; 0<x<1; xC1Df.x/; 1<x<2; F0 C.0/Dlim x!0CF.x//NULF.0/ xDlim x!0C.x2=2//NUL0 xD0Df.0/; F0 /NUL.2/Dlim x!2/NULF.x//NULF.2/ x/NUL2Dlim x!2/NUL.x2=2/Cx/NUL1/NUL3 x/NUL2 Dlim x!2/NULxC4 2D3Df.2/: Fdoes not have a derivative at xD1, wherefis discontinuous, since F0 /NUL.1/D1andF0 C.1/D2: The next theorem relates integration and differentiation i n another way. Theorem 3.3.12 Suppose that Fis continuous on the closed interval Œa;b/c141 and dif- ferentiable on the open interval .a;b/; andfis integrable on Œa;b/c141: Suppose also that F0.x/Df.x/; a<x<b: ThenZb af.x/dxDF.b//NULF.a/: (3.3.14) Proof IfPDfx0;x1;:::;x ngis a partition of Œa;b/c141 , then F.b//NULF.a/DnX jD1.F.x j//NULF.x j/NUL1//: (3.3.15) From Theorem 2.3.11 , there is in each open interval .xj/NUL1;xj/a pointcjsuch that F.x j//NULF.x j/NUL1/Df.c j/.xj/NULxj/NUL1/: 144 Chapter 3 Integral Calculus of Functions of One Variable Hence, ( 3.3.15 ) can be written as F.b//NULF.a/DnX jD1f.c j/.xj/NULxj/NUL1/D/ESC; where/ESCis a Riemann sum for foverP. Sincefis integrable on Œa;b/c141 , there is for each /SI>0 aı>0 such that ˇˇˇˇˇ/ESC/NULZb af.x/dxˇˇˇˇˇ</SI ifkPk<ı: Therefore, ˇˇˇˇˇF.b//NULF.a//NULZb af.x/dxˇˇˇˇˇ</SI for every/SI>0 , which implies ( 3.3.14 ). Corollary 3.3.13 Iff0is integrable on Œa;b/c141; then Zb af0.x/dxDf.b//NULf.a/: Proof Apply Theorem 3.3.12 withFandfreplaced byfandf0, respectively. A functionFis an antiderivative offonŒa;b/c141 ifFis continuous on Œa;b/c141 and differ- entiable on.a;b/ , with F0.x/Df.x/; a<x<b: IfFis an antiderivative of fonŒa;b/c141 , then so isFCcfor any constant c. Conversely, ifF1andF2are antiderivatives of fonŒa;b/c141 , thenF1/NULF2is constant on Œa;b/c141 (Theo- rem2.3.12 ). Theorem 3.3.12 shows that antiderivatives can be used to evaluate integral s. Theorem 3.3.14 (Fundamental Theorem of Calculus) Iffis continu- ous onŒa;b/c141; thenfhas an antiderivative on Œa;b/c141: Moreover;ifFis any antiderivative offonŒa;b/c141; thenZb af.x/dxDF.b//NULF.a/: Proof The function F0.x/DRx af.t/dt is continuous on Œa;b/c141 by Theorem 3.3.10 , andF0 0.x/Df.x/ on.a;b/ by Theorem 3.3.11 . Therefore,F0is an antiderivative of f onŒa;b/c141 . Now letFDF0Cc(cDconstant) be an arbitrary antiderivative of fonŒa;b/c141 . Then F.b//NULF.a/DZb af.x/dxCc/NULZa af.x/dx/NULcDZb af.x/dx: Section 3.3 Properties of the Integral 145 When applying this theorem, we will use the familiar notatio n F.b//NULF.a/DF.x/ˇˇˇˇb a: Theorem 3.3.15 (Integration by Parts) Ifu0andv0are integrable on Œa;b/c141; thenZb au.x/v0.x/dxDu.x/v.x/ˇˇˇˇb a/NULZb av.x/u0.x/dx: (3.3.16) Proof Sinceuandvare continuous on Œa;b/c141 (Theorem 2.3.3 ), they are integrable on Œa;b/c141 . Therefore, Theorems 3.3.1 and3.3.6 imply that the function .uv/0Du0vCuv0 is integrable on Œa;b/c141 , and Theorem 3.3.12 implies that Zb aŒu.x/v0.x/Cu0.x/v.x//c141dxDu.x/v.x/ˇˇˇˇb a; which implies ( 3.3.16 ). We will use Theorem 3.3.15 here and in the next section to obtain other results. Theorem 3.3.16 (Second Mean Value Theorem for Integrals) Suppose thatf0is nonnegative and integrable and gis continuous on Œa;b/c141: Then Zb af.x/g.x/dxDf.a/Zc ag.x/dxCf.b/Zb cg.x/dx (3.3.17) for somecinŒa;b/c141: Proof Sincefis differentiable on Œa;b/c141 , it is continuous on Œa;b/c141 (Theorem 2.3.3 ). Sincegis continuous on Œa;b/c141 , so isfg(Theorem 2.2.5 ). Therefore, Theorem 3.2.8 implies that the integrals in ( 3.3.17 ) exist. If G.x/DZx ag.t/dt; (3.3.18) thenG0.x/Dg.x/; a<x<b (Theorem 3.3.11 ). Therefore, Theorem 3.3.15 withuDf andvDGyields Zb af.x/g.x/dxDf.x/G.x/ˇˇˇˇb a/NULZb af0.x/G.x/dx: (3.3.19) Sincef0is nonnegative and Gis continuous, Theorem 3.3.7 implies that Zb af0.x/G.x/dxDG.c/Zb af0.x/dx (3.3.20) 146 Chapter 3 Integral Calculus of Functions of One Variable for somecinŒa;b/c141 . From Corollary 3.3.12 , Zb af0.x/dxDf.b//NULf.a/: From this and ( 3.3.18 ), (3.3.20 ) can be rewritten as Zb af0.x/G.x/dxD.f.b//NULf.a//Zc ag.x/dx: Substituting this into ( 3.3.19 ) and noting that G.a/D0yields Zb af.x/g.x/dxDf.b/Zb ag.x/dx/NUL.f.b//NULf.a//Zc ag.x/dx; Df.a/Zc ag.x/dxCf.b/ Zb ag.x/dx/NULZa cg.x/dx! Df.a/Zc ag.x/dxCf.b/Zb cg.x/dx: Change of Variable The following theorem on change of variable is useful for eva luating integrals. Theorem 3.3.17 Suppose that the transformation xD/RS.t/ maps the interval c/DC4 t/DC4dinto the interval a/DC4x/DC4b;with/RS.c/D˛and/RS.d/Dˇ;and letfbe continuous onŒa;b/c141: Let/RS0be integrable on Œc;d/c141: Then Zˇ ˛f.x/dxDZd cf./RS.t///RS0.t/dt: (3.3.21) Proof Both integrals in ( 3.3.21 ) exist: the one on the left by Theorem 3.2.8 , the one on the right by Theorems 3.2.8 and3.3.6 and the continuity of f./RS.t// . By Theorem 3.3.11 , the function F.x/DZx af.y/dy is an antiderivative of fonŒa;b/c141 and, therefore, also on the closed interval with endpoints ˛andˇ. Hence, by Theorem 3.3.14 , Zˇ ˛f.x/dxDF.ˇ//NULF.˛/: (3.3.22) By the chain rule, the function G.t/DF./RS.t// Section 3.3 Properties of the Integral 147 is an antiderivative of f./RS.t///RS0.t/onŒc;d/c141 , and Theorem 3.3.12 implies that Zd cf./RS.t///RS0.t/dtDG.d//NULG.c/DF./RS.d///NULF./RS.c// DF.ˇ//NULF.˛/: Comparing this with ( 3.3.22 ) yields ( 3.3.21 ). Example 3.3.2 To evaluate the integral IDZ1=p 2 /NUL1=p 2.1/NUL2x2/.1/NULx2//NUL1=2dx we let f.x/D.1/NUL2x2/.1/NULx2//NUL1=2;/NUL1=p 2/DC4x/DC41=p 2; and xD/RS.t/Dsint;/NUL/EM=4/DC4t/DC4/EM=4: Then/RS0.t/Dcostand IDZ1=p 2 /NUL1=p 2f.x/dxDZ/EM=4 /NUL/EM=4f.sint/costdt DZ/EM=4 /NUL/EM=4.1/NUL2sin2t/.1/NULsin2t//NUL1=2costdt:(3.3.23) .1/NULsin2t/1=2Dcost;/NUL/EM=4/DC4t/DC4/EM=4 and 1/NUL2sin2tDcos2t; (3.3.23 ) yields IDZ/EM=4 /NUL/EM=4cos2tdtDsin2t 2ˇˇˇˇ/EM=4 /NUL/EM=4D1: Example 3.3.3 To evaluate the integral IDZ5/EM 0sint 2Ccostdt; we take/RS.t/Dcost. Then/RS0.t/D/NUL sintand ID/NULZ5/EM 0/RS0.t/ 2C/RS.t/dtD/NULZ5/EM 0f./RS.t///RS0.t/dt; where f.x/D1 2Cx: 148 Chapter 3 Integral Calculus of Functions of One Variable Therefore, since /RS.0/D1and/RS.5/EM/D/NUL1, ID/NULZ/NUL1 1dx 2CxD/NUL log.2Cx/ˇˇˇˇ/NUL1 1Dlog3: These examples illustrate two ways to use Theorem 3.3.17 . In Example 3.3.2 we evalu- ated the left side of ( 3.3.21 ) by transforming it to the right side with a suitable substit ution xD/RS.t/, while in Example 3.3.3 we evaluated the right side of ( 3.3.21 ) by recognizing that it could be obtained from the left side by a suitable subs titution. The following theorem shows that the rule for change of varia ble remains valid under weaker assumptions on fif/RSis monotonic. Theorem 3.3.18 Suppose that /RS0is integrable and /RSis monotonic on Œc;d/c141; and the transformation xD/RS.t/ mapsŒc;d/c141 ontoŒa;b/c141: Letfbe bounded on Œa;b/c141: Then g.t/Df./RS.t///RS0.t/ is integrable on Œc;d/c141 if and only if fis integrable over Œa;b/c141; and in this case Zb af.x/dxDZd cf./RS.t//j/RS0.t/jdt: Proof We consider the case where fis nonnegative and /RSis nondecreasing, and leave the the rest of the proof to you (Exercises 3.3.20 and3.3.21 ). First assume that /RSis increasing. We show first that Zb af.x/dxDZd cf./RS.t///RS0.t/dt: (3.3.24) LetPDft0;t1;:::;t ngbe a partition of Œc;d/c141 andPDfx0;x1;:::;x ngwithxjD/RS.tj/ be the corresponding partition of Œa;b/c141 . Define UjDsup˚/RS0.t/ˇˇtj/NUL1/DC4t/DC4tj/TAB; ujDinf˚ /RS0.t/ˇˇtj/NUL1/DC4t/DC4tj/TAB ; MjDsup˚f.x/ˇˇxj/NUL1/DC4x/DC4xj/TAB; and MjDsup˚f./RS.t///RS0.t/ˇˇtj/NUL1/DC4t/DC4tj/TAB: Since/RSis increasing, uj/NAK0. Therefore, 0/DC4uj/DC4/RS0.t//DC4Uj; t j/NUL1/DC4t/DC4tj: Sincefis nonnegative, this implies that 0/DC4f./RS.t//u j/DC4f./RS.t///RS0.t//DC4f./RS.t//U j; t j/NUL1/DC4t/DC4tj: Therefore, Mjuj/DC4Mj/DC4MjUj; Section 3.3 Properties of the Integral 149 which implies that MjDMj/SUBj; (3.3.25) where uj/DC4/SUBj/DC4Uj: (3.3.26) Now consider the upper sums S.P/DnX jD1Mj.tj/NULtj/NUL1/andS.P/DnX jD1Mj.xj/NULxj/NUL1/: (3.3.27) From the mean value theorem, xj/NULxj/NUL1D/RS.tj//NUL/RS.tj/NUL1/D/RS0./FSj/.tj/NULtj/NUL1/; (3.3.28) wheretj/NUL1</FSj<tj, so uj/DC4/RS0./FSj//DC4Uj: (3.3.29) From ( 3.3.25 ), (3.3.27 ), and ( 3.3.28 ), S.P//NULS.P/DnX jD1Mj./SUBj/NUL/RS0./FSj//.tj/NULtj/NUL1/: (3.3.30) Now suppose thatjf.x/j/DC4M,a/DC4x/DC4b. Then ( 3.3.26 ), (3.3.29 ), and ( 3.3.30 ) imply that ˇˇS.P//NULS.P/ˇˇ/DC4MnX jD1.Uj/NULuj/.tj/NULtj/NUL1/: The sum on the right is the difference between the upper and lo wer sums of /RS0overP. Since/RS0is integrable on Œc;d/c141 , this can be made as small as we please by choosing kPk sufficiently small (Exercise 3.2.4 ). From ( 3.3.28 ),kPk/DC4KkPkifj/RS0.t/j/DC4K,c/DC4t/DC4d. Hence, Lemma 3.2.4 implies thatˇˇˇˇˇS.P//NULZb af.x/dxˇˇˇˇˇ</SI 3andˇˇˇˇˇS.P//NULZd cf./RS.t///RS0.t/dtˇˇˇˇˇ</SI 3(3.3.31) ifkPkis sufficiently small. Now ˇˇˇˇˇZb af.x/dx/NULZd cf ./RS.t///RS0.t/dtˇˇˇˇˇ/DC4ˇˇˇˇˇZb af.x/dx/NULS.P/ˇˇˇˇˇCjS.P//NULS.P/j CˇˇˇˇˇS.P//NULZd cf./RS.t///RS0.t/dtˇˇˇˇˇ: ChoosingPso thatjS.P//NULS.Pj</SI=3 in addition to ( 3.3.31 ) yields ˇˇˇˇˇZb af.x/dx/NULZd cf./RS.t///RS0.t/dtˇˇˇˇˇ</SI: Since/SIis an arbitrary positive number, this implies ( 3.3.24 ). 150 Chapter 3 Integral Calculus of Functions of One Variable If/RSis nondecreasing (rather than increasing), it may happen th atxj/NUL1Dxjfor some values ofj; however, this is no real complication, since it simply mean s that some terms in S.P/ vanish. By applying ( 3.3.24 ) to/NULf, we infer that Zb af.x/dxDZd cf./RS.t///RS0.t/dt; (3.3.32) since Zb a./NULf/.x/dxD/NULZb af.x/dx and Zd c./NULf./RS.t//RS0.t//dtD/NULZd cf./RS.t///RS0.t/dt: Now suppose that fis integrable on Œa;b/c141 . Then Zb af.x/dxDZb af.x/dxDZb af.x/dx; by Theorem 3.2.3 . From this, ( 3.3.24 ), and ( 3.3.32 ), Zd cf./RS.t///RS0.t/dtDZd cf./RS.t///RS0.t/dtDZb af.x/dx: This and Theorem 3.2.5 (applied tof./RS.t///RS0.t/) imply thatf./RS.t///RS0.t/is integrable on Œc;d/c141 andZb af.x/dxDZd cf./RS.t///RS0.t/dt: (3.3.33) A similar argument shows that if f./RS.t///RS0.t/is integrable on Œc;d/c141 , thenfis integrable onŒa;b/c141 , and ( 3.3.33 ) holds. 3.3 Exercises 1. Prove Theorem 3.3.2 . 2. Prove Theorem 3.3.3 . 3. Canjfjbe integrable on Œa;b/c141 iffis not? 4. Complete the proof of Theorem 3.3.6 . HINT:The partial proof given above implies that ifm1andm2are lower bounds for fandgrespectively on Œa;b/c141; then .f/NULm1/.g/NULm2/is integrable on Œa;b/c141: 5. Prove: Iffis integrable on Œa;b/c141 andjf.x/j/NAK/SUB>0 fora/DC4x/DC4b, then1=f is integrable on Œa;b/c141 Section 3.3 Properties of the Integral 151 6. Suppose that fis integrable on Œa;b/c141 and define fC.x/D(f.x/ iff.x//NAK0; 0 iff.x/<0 ,andf/NUL.x/D(0 iff.x//NAK0; f.x/ iff.x/<0 . Show thatfCandf/NULare integrable on Œa;b/c141 , and Zb af.x/dxDZb afC.x/dxCZb af/NUL.x/dx: 7. Find the weighted average uofu.x/ overŒa;b/c141 with respect to v, and find a point c inŒa;b/c141 such thatu.c/Du. (a)u.x/Dx,v.x/Dx,Œa;b/c141DŒ0;1/c141 (b)u.x/Dsinx,v.x/Dx2,Œa;b/c141DŒ/NUL1;1/c141 (c)u.x/Dx2,v.x/Dex,Œa;b/c141DŒ0;1/c141 8. Prove Theorem 3.3.9 . 9. Show thatZc af.x/dxDZb af.x/dxCZc bf.x/dx for all possible relative orderings of a,b, andc, provided that fis integrable on a closed interval containing them. 10. Prove: Iffis integrable on Œa;b/c141 andaDa0<a 1</SOH/SOH/SOH<a nDb, then Zb af.x/dxDZa1 a0f.x/dxCZa2 a1f.x/dxC/SOH/SOH/SOHCZan an/NUL1f.x/dx: 11. Suppose that fis continuous on Œa;b/c141 andPDfx0;x1;:::;x ngis a partition of Œa;b/c141 . Show that there is a Riemann sum of foverPthatequalsRb af.x/dx . 12. Suppose that f0exists andjf0.x/j/DC4MonŒa;b/c141 . Show that any Riemann sum /ESC offover any partition PofŒa;b/c141 satisfies ˇˇˇˇˇ/ESC/NULZb af.x/dxˇˇˇˇˇ/DC4M.b/NULa/kPk: HINT:See Exercise 3.3.11: 13. Prove: Iffis integrable and f.x//NAK0onŒa;b/c141 , thenRb af.x/dx/NAK0, with strict inequality iffis continuous and positive at some point in Œa;b/c141 . 14. Complete the proof of Theorem 3.3.11 . 15. State theorems analogous to Theorems 3.3.10 and3.3.11 for the function G.x/DZc xf.t/dt; and show how your theorems can be obtained from them. 152 Chapter 3 Integral Calculus of Functions of One Variable 16. The symbolR f.x/dx denotes an antiderivative of f. A plausible analog of The- orem 3.3.1 would state that if fandghave antiderivatives on Œa;b/c141 , then so does fCg, which is true, and Z .fCg/.x/dxDZ f.x/dxCZ g.x/dx: . A/ However, this is not true in the usual sense. (a) Why not? (b) State a correct interpretation of (A). 17. (See Exercise 3.3.16 .) Formulate a valid interpretation of the relation Z .cf/.x/dxDcZ f.x/dx .c¤0/: Is your interpretation valid if cD0? 18. (a) Letf.nC1/be integrable on Œa;b/c141 . Show that f.b/DnX rD0f.r/.a/ rŠ.b/NULa/rC1 nŠZb af.nC1/.t/.b/NULt/ndt: HINT:Integrate by parts and use induction : (b) What is the connection between (a)and Theorem 2.5.5 ? 19. In addition to the assumptions of Theorem 3.3.16 , suppose that f.a/D0,f6/DC10, andg.x/ > 0.a < x < b/ . Show that there is only one point cinŒa;b/c141 with the property stated in Theorem 3.3.16 . HINT:Use Exercise 3.3.13: 20. Assuming that Theorem 3.3.18 is true under the additional assumption that fis nonnegative on Œa;b/c141 , show that it is true without this assumption. 21. Assuming that the conclusion of Theorem 3.3.18 is true if/RSis nondecreasing, show that it is true if /RSis nonincreasing. H INT:Use Exercise 3.1.6: 22. Supposeg0is integrable and fis continuous on Œa;b/c141 . Show thatRb af.x/dg.x/ exists and equalsRb af.x/g0.x/dx . 23. Supposefandg00are bounded and fg0is integrable on Œa;b/c141 . Show thatRb af.x/dg.x/ exists and equalsRb af.x/g0.x/dx . HINT:Use Theorem 2.5.4: 3.4 IMPROPER INTEGRALS So far we have confined our study of the integral to bounded fun ctions on finite closed intervals. This was for good reasons: /SIFrom Theorem 3.1.2 , an unbounded function cannot be integrable on a finite close d interval. Section 3.4 Improper Integrals 153 /SIAttempting to formulate Definition 3.1.1 for a function defined on an infinite or semi- infinite interval would introduce questions concerning con vergence of the resulting Riemann sums, which would be infinite series. In this section we extend the definition of integral to includ e cases where fis unbounded or the interval is unbounded, or both. We sayfislocally integrable on an interval Iiffis integrable on every finite closed subinterval of I. For example, f.x/Dsinx is locally integrable on ./NUL1;1/; g.x/D1 x.x/NUL1/ is locally integrable on ./NUL1;0/,.0;1/ , and.1;1/; and h.x/Dpx is locally integrable on Œ0;1/. Definition 3.4.1 Iffis locally integrable on Œa;b/ , we define Zb af.x/dxDlim c!b/NULZc af.x/dx (3.4.1) if the limit exists (finite). To include the case where bD1 , we adopt the convention that 1/NULD1 . The limit in ( 3.4.1 ) always exists if Œa;b/ is finite andfis locally integrable and bounded onŒa;b/ . In this case, Definitions 3.1.1 and3.4.1 assign the same value toRb af.x/dx no matter howf.b/ is defined (Exercise 3.4.1 ). However, the limit may also exist in cases wherebD1 orb <1andfis unbounded as xapproachesbfrom the left. In these cases, Definition 3.4.1 assigns a value to an integral that does not exist in the sense of Def- inition 3.1.1 , andRb af.x/dx is said to be an improper integral thatconverges to the limit in (3.4.1 ). We also say in this case that fisintegrable on Œa;b/ and thatRb af.x/dx exists . If the limit in ( 3.4.1 ) does not exist (finite), we say that the improper integralRb af.x/dx diverges , andfisnonintegrable on Œa;b/ . In particular, if lim c!b/NULRc af.x/dxD˙1 , we say thatRb af.x/dx diverges to˙1, and we write Zb af.x/dxD1 orZb af.x/dxD/NUL1; whichever the case may be. Similar comments apply to the next two definitions. 154 Chapter 3 Integral Calculus of Functions of One Variable Definition 3.4.2 Iffis locally integrable on .a;b/c141 , we define Zb af.x/dxDlim c!aCZb cf.x/dx provided that the limit exists (finite). To include the case w hereaD/NUL1 , we adopt the convention that/NUL1CD/NUL1 . Definition 3.4.3 Iffis locally integrable on .a;b/; we define Zb af.x/dxDZ˛ af.x/dxCZb ˛f.x/dx; wherea<˛<b , provided that both improper integrals on the right exist (fi nite). The existence and value ofRb af.x/dx according to Definition 3.4.3 do not depend on the particular choice of ˛in.a;b/ (Exercise 3.4.2 ). When we wish to distinguish between improper integrals and i ntegrals in the sense of Definition 3.1.1 , we will call the latter proper integrals . In stating and proving theorems on improper integrals, we wi ll consider integrals of the kind introduced in Definition 3.4.1 . Similar results apply to the integrals of Defini- tions 3.4.2 and3.4.3 . We leave it to you to formulate and use them in the examples an d exercises as the need arises. Example 3.4.1 The function f.x/D2xsin1 x/NULcos1 x is locally integrable and the derivative of F.x/Dx2sin1 x onŒ/NUL2=/EM;0/ . Hence, Zc /NUL2=/EMf.x/dxDx2sin1 xˇˇˇˇc /NUL2=/EMDc2sin1 cC4 /EM2 andZ0 /NUL2=/EMf.x/dxDlim c!0/NUL/DC2 c2sin1 cC4 /EM2/DC3 D4 /EM2; according to Definition 3.4.1 . However, this is not an improper integral, even though f.0/ is not defined and cannot be defined so as to make fcontinuous at 0. If we define f.0/ arbitrarily (say f.0/D10), thenfis bounded on the closed interval Œ/NUL2=/EM;0/c141 and con- tinuous except at 0. Therefore,R0 /NUL2=/EMf.x/dx exists and equals 4=/EM2as a proper integral (Exercise 3.4.1 ), in the sense of Definition 3.1.1 . Section 3.4 Improper Integrals 155 Example 3.4.2 The function f.x/D.1/NULx//NULp is locally integrable on Œ0;1/ and, ifp¤1and0<c<1 , Zc 0.1/NULx//NULpdxD.1/NULx//NULpC1 p/NUL1ˇˇˇˇc 0D.1/NULc//NULpC1/NUL1 p/NUL1: Hence, lim c!1/NULZc 0.1/NULx//NULpdxD/SUB.1/NULp//NUL1; p<1; 1; p>1: ForpD1, lim c!1/NULZc 0.1/NULx//NUL1dxD/NUL lim c!1/NULlog.1/NULc/D1: Hence,Z1 0.1/NULx//NULpdxD/SUB.1/NULp//NUL1; p<1; 1; p/NAK1: Example 3.4.3 The function f.x/Dx/NULp is locally integrable on Œ1;1/and, ifp¤1andc>1 , Zc 1x/NULpdxDx/NULpC1 /NULpC1ˇˇˇˇc 1Dc/NULpC1/NUL1 /NULpC1: Hence, lim c!1Zc 1x/NULpdxD/SUB.p/NUL1//NUL1; p>1; 1; p<1: ForpD1, lim c!1Zc 1x/NUL1dxDlim c!1logcD1: Hence,Z1 1x/NULpdxD/SUB.p/NUL1//NUL1; p>1; 1; p/DC41: Example 3.4.4 If1<c<1, then Zc 11 xlog1 xdxD/NULZc 11 xlogxdxD/NUL1 2.logx/2ˇˇˇˇc 1D/NUL1 2.logc/2: Hence, lim c!1Zc 11 xlog1 xdxD/NUL1; so Z1 11 xlog1 xdxD/NUL1: 156 Chapter 3 Integral Calculus of Functions of One Variable Example 3.4.5 The function f.x/Dcosxis locally integrable on Œ0;1/and lim c!1Zc 0cosxdxDlim c!1sinc does not exist; thus,R1 0cosxdx diverges, but not to ˙1. Example 3.4.6 The function f.x/Dlogxis locally integrable on .0;1/c141 , but un- bounded asx!0C. Since lim c!0CZ1 clogxdxDlim c!0C.xlogx/NULx/ˇˇˇˇ1 cD/NUL1/NULlim c!0C.clogc/NULc/D/NUL1; Definition 3.4.2 yieldsZ1 0logxdxD/NUL1: Example 3.4.7 In connection with Definition 3.4.3 , it is important to recognize that the improper integralsR˛ af.x/dx andRb ˛f.x/dx must converge separately forRb af.x/dx to converge. For example, the existence of the symmetric lim it lim R!1ZR /NULRf.x/dx; which is called the principal value ofR1 /NUL1f.x/dx , does not imply thatR1 /NUL1f.x/dx converges; thus, lim R!1ZR /NULRxdxDlim R!10D0; butR1 0xdx andR0 /NUL1xdx diverge and therefore so doesR1 /NUL1xdx . Theorem 3.4.4 Suppose that f1;f2;. . .;fnare locally integrable on Œa;b/ and thatRb af1.x/dx;Rb af2.x/dx; . . .;Rb afn.x/dx converge:Letc1; c2;. . .; cnbe constants: ThenRb a.c1fCc2f1C/SOH/SOH/SOHCcnfn/.x/dx converges and Zb a.c1f1Cc2f2C/SOH/SOH/SOHCcnfn/.x/dxDc1Zb af1.x/dxCc2Zb af2.x/dx C/SOH/SOH/SOHCcnZb afn.x/dx: Proof Ifa<c<b , then Zc a.c1f1Cc2f2C/SOH/SOH/SOHCcnfn/.x/dxDc1Zc af1.x/dxCc2Zc af2.x/dx C/SOH/SOH/SOHCcnZc afn.x/dx; by Theorem 3.3.3 . Lettingc!b/NULyields the stated result. Section 3.4 Improper Integrals 157 Improper Integrals of Nonnegative Functions The theory of improper integrals of nonnegative functions i s particularly simple. Theorem 3.4.5 Iffis nonnegative and locally integrable on Œa;b/; thenRb af.x/dx converges if the function F.x/DZx af.t/dt is bounded on Œa;b/ , andRb af.x/dxD1 if it is not. These are the only possibilities, and Zb af.t/dtDsup a/DC4x<bF.x/ in either case : Proof SinceFis nondecreasing on Œa;b/ , Theorem 2.1.9(a)implies the conclusion. We often write Zb af.x/dx<1 to indicate that an improper integral of a nonnegative funct ion converges. Theorem 3.4.5 justifies this convention, since it asserts that a divergent integral of this kind can only di- verge to1. Similarly, if fis nonpositive andRb af.x/dx converges, we write Zb af.x/dx>/NUL1 because a divergent integral of this kind can only diverge to /NUL1. (To see this, apply Theorem 3.4.5 to/NULf.) These conventions do not apply to improper integrals of fu nctions that assume both positive and negative values in .a;b/ , since they may diverge without diverging to˙1. Theorem 3.4.6 (Comparison Test) Iffandgare locally integrable on Œa;b/ and 0/DC4f.x//DC4g.x/; a/DC4x<b; (3.4.2) then (a)Zb af.x/dx<1 ifZb ag.x/dx <1 and (b)Zb ag.x/dxD1 ifZb af.x/dxD1 . 158 Chapter 3 Integral Calculus of Functions of One Variable Proof (a) Assumption ( 3.4.2 ) implies that Zx af.t/dt/DC4Zx ag.t/dt; a/DC4x<b (Theorem 3.3.4 ), so sup a/DC4x<bZx af.t/dt/DC4sup a/DC4x/DC4bZx ag.t/dt: IfRb ag.x/dx <1, the right side of this inequality is finite by Theorem 3.4.5 , so the left side is also. This implies thatRb af.x/dx<1, again by Theorem 3.4.5 . (b)The proof is by contradiction. IfRb ag.x/dx<1, then(a)implies thatRb af.x/dx< 1, contradicting the assumption thatRb af.x/dxD1 . The comparison test is particularly useful if the integrand of the improper integral is complicated but can be compared with a function that is easy t o integrate. Example 3.4.8 The improper integral IDZ1 02Csin/EMx .1/NULx/pdx converges ifp<1 , since 0<2Csin/EMx .1/NULx/p/DC43 .1/NULx/p; 0/DC4x<1; and, from Example 3.4.2 , Z1 03dx .1/NULx/p<1; p<1: However,Idiverges ifp/NAK1, since 0<1 .1/NULx/p/DC42Csin/EMx .1/NULx/p; 0/DC4x<1; andZ1 0dx .1/NULx/pD1; p/NAK1: Iffis any function (not necessarily nonnegative) locally inte grable onŒa;b/ , then Zc af.x/dxDZa1 af.x/dxCZc a1f.x/dx ifa1andcare inŒa;b/ . SinceRa1 af.x/dx is a proper integral, on letting c!b/NULwe conclude that if either of the improper integralsRb af.x/dx andRb a1f.x/dx converges then so does the other, and in this case Zb af.x/dxDZa1 af.x/dxCZb a1f.x/dx: Section 3.4 Improper Integrals 159 This means that any theorem implying convergence or diverge nce of an improper integralRb af.x/dx in the sense of Definition 3.4.1 remains valid if its hypotheses are satisfied on a subinterval Œa1;b/ofŒa;b/ rather than on all of Œa;b/ . For example, Theorem 3.4.6 remains valid if ( 3.4.2 ) is replaced by 0/DC4f.x//DC4g.x/; a 1/DC4x<b; wherea1is any point in Œa;b/ . From this, you can see that if f.x//NAK0on some subinterval Œa1;b/ofŒa;b/ , but not necessarily for all xinŒa;b/ , we can still use the convention introduced earlier for posi tive functions; that is, we can writeRb af.x/dx <1if the improper integral converges orRb af.x/dxD1 if it diverges. Example 3.4.9 Ifp/NAK0, then x/NULp 2/DC4.x/NUL1/p.2Csinx/ .x/NUL1=3/2p/DC44x/NULp forxsufficiently large. Therefore, Theorem 3.4.6 and Example 3.4.3 imply that Z1 1.x/NUL1/p.2Csinx/ .x/NUL1=3/2pdx converges ifp>1 or diverges if p/DC41. Theorem 3.4.7 Suppose that fandgare locally integrable on Œa;b/;g.x/>0 and f.x//NAK0on some subinterval Œa1;b/ofŒa;b/; and lim x!b/NULf.x/ g.x/DM: (3.4.3) (a) If0<M <1;thenRb af.x/dx andRb ag.x/dx converge or diverge together. (b) IfMD1 andRb ag.x/dxD1;thenRb af.x/dxD1 . (c) IfMD0andRb ag.x/dx<1;thenRb af.x/dx<1. Proof (a) From ( 3.4.3 ), there is a point a2inŒa1;b/such that 0<M 2<f.x/ g.x/<3M 2; a 2/DC4x<b; and thereforeM 2g.x/<f.x/<3M 2g.x/; a 2/DC4x<b: (3.4.4) Theorem 3.4.6 and the first inequality in ( 3.4.4 ) imply that Zb a2g.x/dx <1 ifZb a2f.x/dx<1: 160 Chapter 3 Integral Calculus of Functions of One Variable Theorem 3.4.6 and the second inequality in ( 3.4.4 ) imply that Zb a2f.x/dx<1 ifZb a2g.x/dx<1: Therefore,Rb a2f.x/dx andRb a2g.x/dx converge or diverge together, and in the latter case they must diverge to 1, since their integrands are nonnegative (Theorem 3.4.5 ). (b) IfMD1 , there is a point a2inŒa1;b/such that f.x//NAKg.x/; a 2/DC4x/DC4b; so Theorem 3.4.6(b) implies thatRb af.x/dxD1 . (c)IfMD0, there is a point a2inŒa1;b/such that f.x//DC4g.x/; a 2/DC4x/DC4b; so Theorem 3.4.6(a)implies thatRb af.x/dx<1. The hypotheses of Theorem 3.4.7(b) and(c)do not imply thatRb af.x/dx andRb ag.x/dx necessarily converge or diverge together. For example, if bD 1 , thenf.x/D1=x andg.x/D1=x2satisfy the hypotheses of Theorem 3.4.7(b), whilef.x/D1=x2and g.x/D1=x satisfy the hypotheses of Theorem 3.4.7(c). However,R1 11=xdxD1 , whileR1 11=x2dx<1. Example 3.4.10 Letf.x/D.1Cx//NULpandg.x/Dx/NULp. Since lim x!1f.x/ g.x/D1 andR1 1x/NULpdxconverges ifp >1 or diverges if p/DC41(Example 3.4.3 ), Theorem 3.4.7 implies that the same is true of Z1 1.1Cx//NULpdx: Example 3.4.11 The function f.x/Dx/NULp.1Cx//NULq is locally integrable on .0;1/. To see whether IDZ1 0x/NULp.1Cx//NULqdx converges according to Definition 3.4.3 , we consider the improper integrals I1DZ1 0x/NULp.1Cx//NULqdx andI2DZ1 1x/NULp.1Cx//NULqdx Section 3.4 Improper Integrals 161 separately. (The choice of 1as the upper limit of I1and the lower limit of I2is completely arbitrary; any other positive number would do just as well.) Since lim x!0Cf.x/ x/NULpDlim x!0C.1Cx//NULqD1 andZ1 0x/NULpdxD/SUB.1/NULp//NUL1; p<1; 1; p/NAK1; Theorem 3.4.7 implies thatI1converges if and only if p<1 . Since lim x!1f.x/ x/NULp/NULqDlim x!1.1Cx//NULqxqD1 and Z1 1x/NULp/NULqdxD/SUB.pCq/NUL1//NUL1; pCq>1; 1; pCq/DC41; Theorem 3.4.7 implies thatI2converges if and only if pCq > 1 . Combining these results, we conclude that Iconverges according to Definition 3.4.3 if and only if p < 1 andpCq>1 . Absolute Integrability Definition 3.4.8 We say thatfisabsolutely integrable on Œa;b/ iffis locally inte- grable onŒa;b/ andRb ajf.x/jdx<1. In this case we also say thatRb af.x/dx converges absolutely oris absolutely convergent . Example 3.4.12 Iffis nonnegative and integrable on Œa;b/ , thenfis absolutely integrable on Œa;b/ , sincejfjDf. Example 3.4.13 Sinceˇˇˇˇsinx xpˇˇˇˇ/DC41 xp andR1 1x/NULpdx<1ifp>1 (Example 3.4.3 ), Theorem 3.4.6 implies that Z1 1jsinxj xpdx<1; p>1I that is, the function f.x/Dsinx xp is absolutely integrable on Œ1;1/ifp > 1 . It is not absolutely integrable on Œ1;1/if p/DC41. To see this, we first consider the case where pD1. Letkbe an integer greater than3. Then 162 Chapter 3 Integral Calculus of Functions of One Variable Zk/EM 1jsinxj xdx >Zk/EM /EMjsinxj xdx Dk/NUL1X jD1Z.jC1//EM j/EMjsinxj xdx >k/NUL1X jD11 .jC1//EMZ.jC1//EM j/EMjsinxjdx:(3.4.5) ButZ.jC1//EM j/EMjsinxjdxDZ/EM 0sinxdxD2; so (3.4.5 ) implies that Zk/EM 1jsinxj xdx>2 /EMk/NUL1X jD11 jC1: (3.4.6) However, 1 jC1/NAKZjC2 jC1dx x; jD1;2;:::; so (3.4.6 ) implies that Zk/EM 1jsinxj x>2 /EMk/NUL1X jD1ZjC2 jC1dx x D2 /EMZkC1 2dx xD2 /EMlogkC1 2: Since lim k!1logŒ.kC1/=2/c141D1 , Theorem 3.4.5 implies that Z1 1jsinxj xdxD1: Now Theorem 3.4.6(b) implies that Z1 1jsinxj xpdxD1; p/DC41: (3.4.7) Theorem 3.4.9 Iffis locally integrable on Œa;b/ andRb ajf.x/jdx <1;thenRb af.x/dx convergesIthat is;an absolutely convergent integral is convergent : Proof If g.x/Djf.x/j/NULf.x/; Section 3.4 Improper Integrals 163 then 0/DC4g.x//DC42jf.x/j andRb ag.x/dx<1, because of Theorem 3.4.6 and the absolute integrability of f. Since fDjfj/NULg; Theorem 3.4.4 implies thatRb af.x/dx converges. Conditional Convergence We say thatfisnonoscillatory atb/NUL.D1 ifbD1/iffis defined on Œa;b/ and does not change sign on some subinterval Œa1;b/ofŒa;b/ . Iffchanges sign on every such subinterval, fisoscillatory atb/NUL. For a function that is locally integrable on Œa;b/ and nonoscillatory at b/NUL, convergence and absolute convergence ofRb af.x/dx amount to the same thing (Exercise 3.4.16 ), so absolute convergence is not an interesting concept in connection with such functions. However, an oscillatory function may be integrable, but not absolutely integrable, on Œa;b/ , as the next example shows. We then say that fis conditionally integrable on Œa;b/ , and thatRb af.x/dx converges conditionally . Example 3.4.14 We saw in Example 3.4.13 that the integral I.p/DZ1 1sinx xpdx is not absolutely convergent if 0<p/DC41. We will show that it converges conditionally for these values of p. Integration by parts yields Zc 1sinx xpdxD/NULcosc cpCcos1/NULpZc 1cosx xpC1dx: (3.4.8) Sinceˇˇˇcosx xpC1ˇˇˇ/DC41 xpC1 andR1 1x/NULp/NUL1dx <1ifp > 0 , Theorem 3.4.6 implies thatx/NULp/NUL1cosxis absolutely integrableŒ1;1/ifp>0 . Therefore, Theorem 3.4.9 implies thatx/NULp/NUL1cosxis integrable Œ1;1/ifp>0 . Lettingc!1 in (3.4.8 ), we find that I.p/ converges, and I.p/Dcos1/NULpZ1 1cosx xpC1dx ifp>0: This and ( 3.4.7 ) imply thatI.p/ converges conditionally if 0<p/DC41. The method used in Example 3.4.14 is a special case of the following test for convergence of improper integrals. 164 Chapter 3 Integral Calculus of Functions of One Variable Theorem 3.4.10 ( Dirichlet ’s Test) Suppose that fis continuous and its an- tiderivativeF.x/DRx af.t/dt is bounded on Œa;b/: Letg0be absolutely integrable on Œa;b/; and suppose that lim x!b/NULg.x/D0: (3.4.9) ThenRb af.x/g.x/dx converges: Proof The continuous function fgis locally integrable on Œa;b/ . Integration by parts yields Zc af.x/g.x/dxDF.c/g.c//NULZc aF.x/g0.x/dx; a/DC4c<b: (3.4.10) Theorem 3.4.6 implies that the integral on the right converges absolutely asc!b/NUL, sinceRb ajg0.x/jdx<1by assumption, and jF.x/g0.x/j/DC4Mjg0.x/j; whereMis an upper bound for jFjonŒa;b/ . Moreover, ( 3.4.9 ) and the boundedness of F imply that lim c!b/NULF.c/g.c/D0. Lettingc!b/NULin (3.4.10 ) yields Zb af.x/g.x/dxD/NULZb aF.x/g0.x/dx; where the integral on the right converges absolutely. Dirichlet’s test is useful only if fis oscillatory at b/NUL, since it can be shown that if fis nonoscillatory at b/NULandFis bounded on Œa;b/ , thenRb ajf.x/g.x/jdx <1if onlygis locally integrable and bounded on Œa;b/ (Exercise 3.4.14 ). Example 3.4.15 Dirichlet’s test can also be used to show that certain integr als di- verge. For example,Z1 1xqsinxdx diverges ifq > 0 , but none of the other tests that we have studied so far implie s this. It is not enough to argue that the integrand does not approach ze ro asx!1 (a common mistake), since this does not imply divergence (Exercise 4.4.31 ). To see that the integral diverges, we observe that if it converged for some q > 0 , thenF.x/DRx 1xqsinxdx would be bounded on Œ1;1/, and we could let f.x/Dxqsinxandg.x/Dx/NULq in Theorem 3.4.10 and conclude that Z1 1sinxdx also converges. This is false. Section 3.4 Improper Integrals 165 The method used in Example 3.4.15 is a special case of the following test for divergence of improper integrals. Theorem 3.4.11 Suppose that uis continuous on Œa;b/ andRb au.x/dx diverges:Let vbe positive and differentiable on Œa;b/; and suppose that limx!b/NULv.x/D1 andv0=v2 is absolutely integrable on Œa;b/: ThenRb au.x/v.x/dx diverges: Proof The proof is by contradiction. Let fDuvandgD1=v, and suppose thatRb au.x/v.x/dx converges. Then fhas the bounded antiderivative F.x/DRx au.t/v.t/dt onŒa;b/ , lim x!1g.x/D0andg0D/NULv0=v2is absolutely integrable on Œa;b/ . Therefore, Theorem 3.4.10 implies thatRb au.x/dx converges, a contradiction. If Dirichlet’s test shows thatRb af.x/g.x/dx converges, there remains the question of whether it converges absolutely or conditionally. The next theorem sometimes answers this question. Its proof can be modeled after the method of Exampl e3.4.13 (Exercise 3.4.17 ). The idea of an infinite sequence, which we will discuss in Sect ion 4.1, enters into the statement of this theorem. We assume that you recall the conc ept sufficiently well from calculus to understand the meaning of the theorem. Theorem 3.4.12 Suppose that gis monotonic on Œa;b/ andRb ag.x/dxD1:Letf be locally integrable on Œa;b/ and ZxjC1 xjjf.x/jdx/NAK/SUB; j/NAK0; for some positive /SUB;wherefxjgis an increasing infinite sequence of points in Œa;b/ such thatlimj!1xjDbandxjC1/NULxj/DC4M;j/NAK0;for someM:Then Zb ajf.x/g.x/jdxD1: Change of Variable in an Improper Integral The next theorem enables us to investigate an improper integ ral by transforming it into another whose convergence or divergence is known. It follow s from Theorem 3.3.18 and Definitions 3.4.1 ,3.4.2 , and 3.4.3 . We omit the proof. Theorem 3.4.13 Suppose that /RSis monotonic and /RS0is locally integrable on either of the half-open intervals IDŒc;d/ or.c;d/c141; and letxD/RS.t/ mapIonto either of the half-open intervals JDŒa;b/ orJD.a;b/c141: Letfbe locally integrable on J:Then the improper integrals Zb af.x/dx andZd cf ./RS.t//j/RS0.t/jdt 166 Chapter 3 Integral Calculus of Functions of One Variable diverge or converge together ;in the latter case to the same value. The same conclusion holds if/RSand/RS0have the stated properties only on the open interval .a;b/; the transfor- mationxD/RS.t/ maps.c;d/ onto.a;b/; andfis locally integrable on .a;b/: Example 3.4.16 To apply Theorem 3.4.13 to Z1 0sinx2dx; we use the change of variable xD/RS.t/Dpt, which takes Œc;d/DŒ0;1/intoŒa;b/D Œ0;1/, with/RS0.t/D1=.2pt/. Theorem 3.4.13 implies that Z1 0sinx2dxD1 2Z1 0sintptdt: Since the integral on the right converges (Example 3.4.14 ), so does the one on the left. Example 3.4.17 The integral Z1 1x/NULpdx converges if and only if p > 1 (Example 3.4.3 ). Defining/RS.t/D1=tand applying Theorem 3.4.13 yields Z1 1x/NULpdxDZ1 0tpj/NULt/NUL2jdtDZ1 0tp/NUL2dt; which implies thatR1 0tqdtconverges if and only if q>/NUL1. 3.4 Exercises 1. (a) Letfbe locally integrable and bounded on Œa;b/ , and letf.b/ be defined arbitrarily. Show that fis properly integrable on Œa;b/c141 , thatRb af.x/dx does not depend on f.b/ , and that Zb af.x/dxDlim c!b/NULZc af.x/dx: (b) State a result analogous to (a)which ends with the conclusion that Zb af.x/dxDlim c!aCZb cf.x/dx: 2. Show that neither the existence nor the value of the improper integral of Defini- tion3.4.3 depends on the choice of the intermediate point ˛. Section 3.4 Improper Integrals 167 3. Prove: IfRb af.x/dx exists according to Definition 3.4.1 or3.4.2 , thenRb af.x/dx also exists according to Definition 3.4.3 . 4. Find all values of pfor which the following integrals exist (i)as proper integrals (perhaps after defining fat the endpoints of the interval) or (ii) as improper inte- grals.(iii) Evaluate the integrals for the values of pfor which they converge. (a)Z1=/EM 0/DC2 pxp/NUL1sin1 x/NULxp/NUL2cos1 x/DC3 dx (b)Z2=/EM 0/DC2 pxp/NUL1cos 1xCxp/NUL2sin1 x/DC3 dx (c)Z1 0e/NULpxdx(d)Z1 0x/NULpdx(e)Z1 0x/NULpdx. 5. Evaluate (a)Z1 0e/NULxxndx .nD0;1;:::/ (b)Z1 0e/NULxsinxdx (c)Z1 /NUL1xdx x2C1(d)Z1 0xdxp 1/NULx2 (e)Z/EM 0/DC2cosx x/NULsinx x2/DC3 dx (f)Z1 /EM=2/DC2sinx xCcosx x2/DC3 dx 6. Prove: IfRb af.x/dx exists as a proper or improper integral, then lim x!b/NULZb xf.t/dtD0: 7. Prove: Iffis locally integrable on Œa;b/ , thenRb af.x/dx exists if and only if for each/SI>0 there is a number rin.a;b/ such that ˇˇˇˇZx2 x1f.t/dtˇˇˇˇ</SI wheneverr/DC4x1,x2<b. HINT:See Exercise 2.1.38 . 8. Determine whether the integral converges or diverges. (a)Z1 1logxCsinxpxdx (b)Z1 /NUL1.x2C3/3=2 .x4C1/3=2sin2xdx (c)Z1 01Ccos2xp 1Cx2dx (d)Z1 04Ccosx .1Cx/pxdx (e)Z1 0.x27Csinx/e/NULxdx (f)Z1 0x/NULp.2Csinx/dx 168 Chapter 3 Integral Calculus of Functions of One Variable 9. Find all values of pfor which the integral converges. (a)Z/EM=2 0sinx xpdx (b)Z/EM=2 0cosx xpdx (c)Z1 0xpe/NULxdx (d)Z/EM=2 0sinx .tanx/pdx(e)Z1 1dx x.logx/p(f)Z1 0dx x.jlogxj/p (g)Z/EM 0xdx .sinx/p 10. LetLn.x/be the iterated logarithm defined in Exercise 2.4.42 . Show that Z1 adx L0.x/L 1.x//SOH/SOH/SOHLk.x/ŒL kC1.x//c141p converges if and only if p > 1 . Hereais any number such that LkC1.x/> 0 for x/NAKa. 11. Find conditions on pandqsuch that the integral converges. (a)Z1 /NUL1.cos/EMx=2/q .1/NULx2/pdx (b)Z1 /NUL1.1/NULx/p.1Cx/qdx (c)Z1 0xpdx .1Cx2/q(d)Z1 1Œlog.1Cx//c141p.logx/q xpCqdx (e)Z1 1.log.1Cx//NULlogx/q xpdx (f)Z1 0.x/NULsinx/q xpdx 12. Letfandgbe polynomials and suppose that ghas no real zeros. Find necessary and sufficient conditions for convergence of Z1 /NUL1f.x/ g.x/dx: 13. Prove: Iffandgare locally integrable on Œa;b/ and the improper integralsRb af2.x/dx andRb ag2.x/dx converge, thenRb af.x/g.x/dx converges absolutely. H INT:.f˙ g/2/NAK0: 14. Suppose that fis locally integrable and F.x/DRx af.t/dt is bounded on Œa;b/ , and letfbe nonoscillatory at b/NUL. Letgbe locally integrable and bounded on Œa;b/ . Show thatZb ajf.x/g.x/jdx<1: 15. Suppose that gis positive and nonincreasing on Œa;b/ andRb af.x/dx exists as a proper or absolutely convergent improper integral. Show t hatRb af.x/g.x/dx exists and Section 3.4 Improper Integrals 169 lim x!b/NUL1 g.x/Zb xf.t/g.t/dtD0: HINT:Use Exercise 3.4.6: 16. Show that iffis locally integrable on Œa;b/ and nonoscillatory at b/NUL, thenRb af.x/dx exists if and only ifRb ajf.x/jdx<1. 17. (a) Prove Theorem 3.4.12 . HINT:See Example 3.4.13: (b) Show thatgsatisfies the assumptions of Theorem 3.4.10 ifg0is locally inte- grable,gis monotonic on Œa;b/ , and lim x!b/NULg.x/D0. 18. Find all values of pfor which the integral converges (i)absolutely; (ii) condition- ally. (a)Z1 1cosx xpdx (b)Z1 2sinx x.logx/pdx(c)Z1 2sinx xplogxdx (d)Z1 1sin1=x xpdx (e)Z1 0sin2xsin2x xpdx(f)Z1 /NUL1sinx .1Cx2/pdx 19. Suppose thatg00is absolutely integrable on Œ0;1/, lim x!1g0.x/D0, and lim x!1g.x/D L(finite or infinite). Show thatR1 0g.x/ sinxdx converges if and only if LD0. HINT:Integrate by parts : 20. Lethbe continuous on Œ0;1/. Prove: (a) IfR1 0e/NULs0xh.x/dx converges absolutely, thenR1 0e/NULsxh.x/dx converges absolutely ifs>s 0. (b) IfR1 0e/NULs0xh.x/dx converges, thenR1 0e/NULsxh.x/dx converges ifs>s 0. 21. Suppose that fis locally integrable on Œ0;1/, lim x!1f.x/DA, and˛ >/NUL1. Find lim x!1x/NUL˛/NUL1Rx 0f.t/t˛dt, and prove your answer. 22. Suppose thatfis continuous and F.x/DRx af.t/dt is bounded on Œa;b/ . Suppose also thatg>0 ,g0is nonnegative and locally integrable on Œa;b/ , and lim x!b/NULg.x/D 1. Show that lim x!b/NUL1 Œg.x//c141/SUBZx af.t/g.t/dtD0; /SUB>1: HINT:Integrate by parts : 23. In addition to the assumptions of Exercise 3.4.22 , assume thatRb af.t/dt converges. Show that lim x!b/NUL1 g.x/Zx af.t/g.t/dtD0: HINT:LetF.x/DRb xf.t/dt; integrate by parts ;and use Exercise 3.4.6: 170 Chapter 3 Integral Calculus of Functions of One Variable 24. Suppose that fis continuous, g0.x//DC40, andg.x/>0 onŒa;b/ . Show that if g0is integrable on Œa;b/ andRb af.x/dx exists, thenRb af.x/g.x/dx exists and lim x!b/NUL1 g.x/Zb xf.t/g.t/dtD0: HINT:LetF.x/DRb xf.t/dt; integrate by parts ;and use Exercise 3.4.6: 25. Find all values of pfor which the integral converges (i)absolutely; (ii) condition- ally. (a)Z1 0xpsin1=xdx (b)Z1 0jlogxjpdx (c)Z1 1xpcos.logx/dx (d)Z1 1.logx/pdx (e)Z1 0sinxpdx 26. Letu1be positive and satisfy the differential equation u00Cp.x/uD0; 0/DC4x<1: . A/ (a) Prove: IfZ1 0dx u2 1.x/<1; then the function u2.x/Du1.x/Z1 xdt u2 1.t/ also satisfies (A), while if Z1 0dx u2 1.x/D1; then the function u2.x/Du1.x/Zx 0dt u2 1.t/ also satisfies (A). (b) Prove: If (A) has a solution that is positive on Œ0;1/, then (A) has solutions y1andy2that are positive on .0;1/and have the following properties: y1.x/y0 2.x//NULy0 1.x/y 2.x/D1; x>0; /DC4y1.x/ y2.x//NAK0 < 0; x>0; and lim x!1y1.x/ y2.x/D0: Section 3.4 Improper Integrals 171 27. (a) Prove: Ifhis continuous on Œ0;1/, then the function u.x/Dc1e/NULxCc2exCZx 0h.t/sinh.x/NULt/dt satisfies the differential equation u00/NULuDh.x/; x>0: (b) Rewriteuin the form u.x/Da.x/e/NULxCb.x/ex and show that u0.x/D/NULa.x/e/NULxCb.x/ex: (c) Show that if lim x!1a.x/DA(finite), then lim x!1e2xŒb.x//NULB/c141D0 for some constant B. HINT:Use Exercise 3.4.24:Show also that lim x!1exŒu.x//NULAe/NULx/NULBex/c141D0: (d) Prove: If lim x!1b.x/DB(finite), then lim x!1u.x/e/NULxDlim x!1u0.x/e/NULxDB: HINT:Use Exercise 3.4.23: 28. Suppose that the differential equation u00Cp.x/uD0 . A/ has a positive solution on Œ0;1/, and therefore has two solutions y1andy2with the properties given in Exercise 3.4.26(b). (a) Prove: Ifhis continuous on Œ0;1/andc1andc2are constants, then u.x/Dc1y1.x/Cc2y2.x/CZx 0h.t/Œy 1.t/y 2.x//NULy1.x/y 2.t//c141 dt . B/ satisfies the differential equation u00Cp.x/uDh.x/: For convenience in (b) and(c), rewrite (B) as u.x/Da.x/y 1.x/Cb.x/y 2.x/: 172 Chapter 3 Integral Calculus of Functions of One Variable (b) Prove: IfR1 0h.t/y 2.t/dt converges, thenR1 0h.t/y 1.t/dt converges, and lim x!1u.x//NULAy1.x//NULBy2.x/ y1.x/D0 for some constants AandB. H INT:Use Exercise 3.4.24 withfDhy2and gDy1=y2: (c) Prove: IfR1 0h.t/y 1.t/dt converges, then lim x!1u.x/ y2.x/DB for some constant B. H INT:Use Exercise 3.4.23 withfDhy1andgD y2=y1: 29. Suppose that f,f1, andgare continuous, f > 0 , and.f1=f/0is absolutely inte- grable onŒa;b/ . Show thatRb af1.x/g.x/dx converges ifRb af.x/g.x/dx does. 30. Letgbe locally integrable and fcontinuous, with f.x//NAK/SUB > 0 onŒa;b/ . Sup- pose that for some positive Mand for every rinŒa;b/ there are points x1andx2 such that (a)r < x 1< x 2< b;(b)gdoes not change sign in Œx1;x2/c141; and (c)Rx2 x1jg.x/jdx/NAKM. Show thatRb af.x/g.x/dx diverges. H INT:Use Exer- cise3.4.7 and Theorem 3.3.7: 3.5 A MORE ADVANCED LOOK AT THE EXISTENCE OF THE PROPER RIEMANN INTEGRAL In Section 3.2 we found necessary and sufficient conditions f or existence of the proper Riemann integral, and in Section 3.3 we used them to study the properties of the integral. However, it is awkward to apply these conditions to a specific function and determine whether it is integrable, since they require computations o f upper and lower sums and upper and lower integrals, which may be difficult. The main re sult of this section is an integrability criterion due to Lebesgue that does not require computation, but has to do with how badly discontinuous a function may be and still be in tegrable. We emphasize that we are again considering proper integrals of bounded functions on finite intervals. Definition 3.5.1 Iffis bounded on Œa;b/c141 , the oscillation of fonŒa;b/c141 is defined by WfŒa;b/c141D sup a/DC4x;x0/DC4bjf.x//NULf.x0/j; which can also be written as WfŒa;b/c141Dsup a/DC4x/DC4bf.x//NULinf a/DC4x/DC4bf.x/ Section 3.5 Advanced Look at the Existence of the Proper Riemann Integra l173 ( Exercise 3.5.1 ). Ifa<x<b , the oscillation of fatxis defined by wf.x/Dlim h!0CWf.x/NULh;xCh/: The corresponding definitions for xDaandxDbare wf.a/Dlim h!0CWf.a;aCh/ andwf.b/Dlim h!0CWf.b/NULh;b/: For a fixedxin.a;b/ ,Wf.x/NULh;xCh/is a nonnegative and nondecreasing function ofhfor0 < h < min.x/NULa;b/NULx/; therefore,wf.x/exists and is nonnegative, by Theorem 2.1.9 . Similar arguments apply to wf.a/andwf.b/. Theorem 3.5.2 Letfbe defined on Œa;b/c141: Thenfis continuous at x0inŒa;b/c141 if and only ifwf.x0/D0:.Continuity at aorbmeans continuity from the right or left, respectively./ Proof Suppose that a<x 0<b. First, suppose that wf.x0/D0and/SI>0 . Then WfŒx0/NULh;x 0Ch/c141</SI for someh>0 , so jf.x//NULf.x0/j</SI ifx0/NULh/DC4x;x0/DC4x0Ch: Lettingx0Dx0, we conclude that jf.x//NULf.x 0/j</SI ifjx/NULx0j<h: Therefore,fis continuous at x0. Conversely, if fis continuous at x0and/SI>0 , there is aı>0 such that jf.x//NULf.x 0/j</SI 2andjf.x0//NULf.x 0/j</SI 2 ifx0/NULı/DC4x,x0/DC4x0Cı. From the triangle inequality, jf.x//NULf.x0/j/DC4jf.x//NULf.x 0/jCjf.x0//NULf.x 0/j</SI; so WfŒx0/NULh;x 0Ch/c141/DC4/SIifh<ıI therefore,wf.x0/D0. Similar arguments apply if x0Daorx0Db. Lemma 3.5.3 Ifwf.x/ < /SI fora/DC4x/DC4b;then there is a ı > 0 such that WfŒa1;b1/c141/DC4/SI;provided that Œa1;b1/c141/SUBŒa;b/c141 andb1/NULa1<ı: Proof We use the Heine–Borel theorem (Theorem 1.3.7 ). Ifwf.x/ < /SI , there is an hx>0such that jf.x0//NULf.x00/j</SI (3.5.1) 174 Chapter 3 Integral Calculus of Functions of One Variable if x/NUL2hx<x0;x00<xC2hxandx0;x002Œa;b/c141: (3.5.2) IfIxD.x/NULhx;xChx/, then the collection HD˚Ixˇˇa/DC4x/DC4b/TAB is an open covering of Œa;b/c141 , so the Heine–Borel theorem implies that there are finitely many pointsx1,x2, . . . ,xninŒa;b/c141 such thatIx1,Ix2, . . . ,IxncoverŒa;b/c141 . Let hDmin 1/DC4i/DC4nhxi and suppose that Œa1;b1/c141/SUBŒa;b/c141 andb1/NULa1< h. Ifx0andx00are inŒa1;b1/c141, then x02Ixrfor somer.1/DC4r/DC4n/, so jx0/NULxrj<h xr: Therefore, jx00/NULxrj/DC4jx00/NULx0jCjx0/NULxrj<b 1/NULa1Chxr < hChxr/DC42hxr: Thus, any two points x0andx00inŒa1;b1/c141satisfy ( 3.5.2 ) withxDxr, so they also satisfy (3.5.1 ). Therefore, /SIis an upper bound for the set ˚ jf.x0//NULf.x00/jˇˇx0;x002Œa1;b1/c141/TAB ; which has the supremum WfŒa1;b1/c141. Hence,WfŒa1;b1/c141/DC4/SI. In the following, L.I/ is the length of the interval I. Lemma 3.5.4 Letfbe bounded on Œa;b/c141 and define E/SUBD˚x2Œa;b/c141ˇˇwf.x//NAK/SUB/TAB: ThenE/SUBis closed;andfis integrable on Œa;b/c141 if and only if for every pair of positive numbers/SUBandı;E /SUBcan be covered by finitely many open intervals I1;I2;. . .;Ipsuch thatpX jD1L.I j/<ı: (3.5.3) Proof We first show that E/SUBis closed. Suppose that x0is a limit point of E/SUB. Ifh>0 , there is anxfromE/SUBin.x0/NULh;x 0Ch/. SinceŒx/NULh1;xCh1/c141/SUBŒx0/NULh;x 0Ch/c141for sufficiently small h1andWfŒx/NULh1;xCh1/c141/NAK/SUB, it follows that WfŒx0/NULh;x 0Ch/c141/NAK/SUB for allh>0 . This implies that x02E/SUB, soE/SUBis closed (Corollary 1.3.6 ). Now we will show that the stated condition in necessary for in tegrability. Suppose that the condition is not satisfied; that is, there is a /SUB>0 and aı>0 such that pX jD1L.I j//NAKı Section 3.5 Advanced Look at the Existence of the Proper Riemann Integra l175 for every finite setfI1;I2;:::;I pgof open intervals covering E/SUB. IfPDfx0;x1;:::;x ng is a partition of Œa;b/c141 , then S.P//NULs.P/DX j2A.Mj/NULmj/.xj/NULxj/NUL1/CX j2B.Mj/NULmj/.xj/NULxj/NUL1/; (3.5.4) where AD˚jˇˇŒxj/NUL1;xj/c141\E/SUB¤;/TABandBD˚jˇˇŒxj/NUL1;xj/c141\E/SUBD;/TAB: SinceS j2A.xj/NUL1;xj/contains all points of E/SUBexcept any of x0,x1, . . . ,xnthat may be inE/SUB, and each of these finitely many possible exceptions can be co vered by an open interval of length as small as we please, our assumption on E/SUBimplies that X j2A.xj/NULxj/NUL1//NAKı: Moreover, ifj2A, then Mj/NULmj/NAK/SUB; so (3.5.4 ) implies that S.P//NULs.P//NAK/SUBX j2A.xj/NULxj/NUL1//NAK/SUBı: Since this holds for every partition of Œa;b/c141 ,fis not integrable on Œa;b/c141 , by Theorem 3.2.7 . This proves that the stated condition is necessary for integ rability. For sufficiency, let /SUBandıbe positive numbers and let I1,I2, . . . ,Ipbe open intervals that coverE/SUBand satisfy ( 3.5.3 ). Let eIjDŒa;b/c141\Ij: (IjDclosure ofI.) After combining any of eI1,eI2, . . . ,eIpthat overlap, we obtain a set of pairwise disjoint closed subintervals CjDŒ˛j;ˇj/c141; 1/DC4j/DC4q./DC4p/; ofŒa;b/c141 such that a/DC4˛1<ˇ 1<˛ 2<ˇ 2/SOH/SOH/SOH<˛ q/NUL1<ˇ q/NUL1<˛ q<ˇ q/DC4b; (3.5.5) qX iD1.ˇi/NUL˛i/<ı (3.5.6) and wf.x/</SUB; ˇ j/DC4x/DC4˛jC1; 1/DC4j/DC4q/NUL1: Also,wf.x/</SUB fora/DC4x/DC4˛1ifa<˛ 1and forˇq/DC4x/DC4bifˇq<b. 176 Chapter 3 Integral Calculus of Functions of One Variable LetP0be the partition of Œa;b/c141 with the partition points indicated in ( 3.5.5 ), and refine P0by partitioning each subinterval Œˇj;˛jC1/c141(as well asŒa;˛ 1/c141ifa < ˛ 1andŒˇq;b/c141 ifˇq< b) into subintervals on which the oscillation of fis not greater than /SUB. This is possible by Lemma 3.5.3 . In this way, after renaming the entire collection of partit ion points, we obtain a partition PDfx0;x1;:::;x ngofŒa;b/c141 for whichS.P//NULs.P/ can be written as in ( 3.5.4 ), with X j2A.xj/NULxj/NUL1/DqX iD1.ˇi/NUL˛i/<ı (see ( 3.5.6 )) and Mj/NULmj/DC4/SUB; j2B: For this partition, X j2A.Mj/NULmj/.xj/NULxj/NUL1//DC42KX j2A.xj/NULxj/NUL1/<2Kı; whereKis an upper bound for jfjonŒa;b/c141 and X j2B.Mj/NULmj/.xj/NULxj/NUL1//DC4/SUB.b/NULa/: We have now shown that if /SUBandıare arbitrary positive numbers, there is a partition Pof Œa;b/c141 such that S.P//NULs.P/<2KıC/SUB.b/NULa/: (3.5.7) If/SI>0 , let ıD/SI 4Kand/SUBD/SI 2.b/NULa/: Then ( 3.5.7 ) yields S.P//NULs.P/</SI; and Theorem 3.2.7 implies thatfis integrable on Œa;b/c141 . We need the next definition to state Lebesgue’s integrabilit y condition. Definition 3.5.5 A subsetSof the real line is of Lebesgue measure zero if for every /SI>0 there is a finite or infinite sequence of open intervals I1,I2, . . . such that S/SUB[ jIj (3.5.8) and nX jD1L.I j/</SI; n/NAK1: (3.5.9) Section 3.5 Advanced Look at the Existence of the Proper Riemann Integra l177 Note that any subset of a set of Lebesgue measure zero is also o f Lebesgue measure zero. (Why?) Example 3.5.1 The empty set is of Lebesgue measure zero, since it is contain ed in any open interval. Example 3.5.2 Any finite set SD fx1;x2;:::;x ngis of Lebesgue measure zero, since we can choose open intervals I1,I2, . . . ,Insuch thatxj2IjandL.I j/ < /SI=n , 1/DC4j/DC4n. Example 3.5.3 An infinite set is denumerable if its members can be listed in a se- quence (that is, in a one-to-one correspondence with the pos itive integers); thus, SDfx1;x2;:::;x n;:::g: (3.5.10) An infinite set that does not have this property is nondenumerable . Any denumerable set (3.5.10 ) is of Lebesgue measure zero, since if /SI>0 , it is possible to choose open intervals I1,I2, . . . , so thatxj2IjandL.I j/<2/NULj/SI,j/NAK1. Then ( 3.5.9 ) holds because 1 2C1 22C1 23C/SOH/SOH/SOHC1 2nD1/NUL1 2n<1: (3.5.11) There are also nondenumerable sets of Lebesgue measure zero , but it is beyond the scope of this book to discuss examples. The next theorem is the main result of this section. Theorem 3.5.6 A bounded function fis integrable on a finite interval Œa;b/c141 if and only if the set Sof discontinuities of finŒa;b/c141 is of Lebesgue measure zero : Proof From Theorem 3.5.2 , SD˚x2Œa;b/c141ˇˇwf.x/>0/TAB: Sincewf.x/>0 if and only if wf.x//NAK1=ifor some positive integer i, we can write SD1[ iD1Si; (3.5.12) where SiD˚ x2Œa;b/c141ˇˇwf.x//NAK1=i/TAB : Now suppose that fis integrable on Œa;b/c141 and/SI >0 . From Lemma 3.5.4 , eachSican be covered by a finite number of open intervals Ii1,Ii2, . . . ,Iinof total length less than /SI=2i. We simply renumber these intervals consecutively; thus, I1;I2;/SOH/SOH/SOHDI11;:::;I 1n1;I21;:::;I 2n2;:::;I i1;:::;I ini;:::: Now ( 3.5.8 ) and ( 3.5.9 ) hold because of ( 3.5.11 ) and ( 3.5.12 ), and we have shown that the stated condition is necessary for integrability. 178 Chapter 3 Integral Calculus of Functions of One Variable For sufficiency, suppose that the stated condition holds and /SI > 0 . ThenScan be covered by open intervals I1;I2;::: that satisfy ( 3.5.9 ). If/SUB>0 , then the set E/SUBD˚ x2Œa;b/c141ˇˇwf.x//NAK/SUB/TAB of Lemma 3.5.4 is contained in S(Theorem 3.5.2 ), and therefore E/SUBis covered by I1;I2;:::. SinceE/SUBis closed (Lemma 3.5.4 ) and bounded, the Heine–Borel theorem implies that E/SUB is covered by a finite number of intervals from I1;I2;:::. The sum of the lengths of the latter is less than /SI, so Lemma 3.5.4 implies thatfis integrable on Œa;b/c141 . 3.5 Exercises 1. In connection with Definition 3.5.1 , show that sup x;x02Œa;b/c141jf.x//NULf.x0/jD sup a/DC4x/DC4bf.x//NULinf a/DC4x/DC4bf.x/: 2. Use Theorem 3.5.6 to show that if fis integrable on Œa;b/c141 , then so isjfjand, if f.x//NAK/SUB>0.a/DC4x/DC4b/, so is1=f. 3. Prove: The union of two sets of Lebesgue measure zero is of Leb esgue measure zero. 4. Use Theorem 3.5.6 and Exercise 3.5.3 to show that if fandgare integrable on Œa;b/c141 , then so arefCgandfg. 5. Supposefis integrable on Œa;b/c141 ,˛Dinfa/DC4x/DC4bf.x/ , andˇDsupa/DC4x/DC4bf.x/ . Letgbe continuous on Œ˛;ˇ/c141 . Show that the composition hDgıfis integrable onŒa;b/c141 . 6. Letfbe integrable on Œa;b/c141 , let˛Dinfa/DC4x/DC4bf.x/ andˇDsupa/DC4x/DC4bf.x/ , and suppose thatGis continuous on Œ˛;ˇ/c141 . For eachn/NAK1, let aC.j/NUL1/.b/NULa/ n/DC4uj n;vj n/DC4aCj.b/NULa/ n; 1/DC4j/DC4n: Show that lim n!11 nnX jD1jG.f.u j n///NULG.f.v j n//jD0: 7. Leth.x/D0for allxinŒa;b/c141 except forxin a set of Lebesgue measure zero. Show that ifRb ah.x/dx exists, it equals zero. H INT:Any subset of a set of measure zero is also of measure zero : 8. Suppose that fandgare integrable on Œa;b/c141 andf.x/Dg.x/ except forxin a set of Lebesgue measure zero. Show that Zb af.x/dxDZb ag.x/dx: CHAPTER 4 Infinite Sequences and Series IN THIS CHAPTER we consider infinite sequences and series of c onstants and functions of a real variable. SECTION 4.1 introduces infinite sequences of real numbers. T he concept of a limit of a sequence is defined, as is the concept of divergence of a seque nce to˙1. We discuss bounded sequences and monotonic sequences. The limit infer ior and limit superior of a sequence are defined. We prove the Cauchy convergence criter ion for sequences of real numbers. SECTION 4.2 defines a subsequence of an infinite sequence. We s how that if a sequence converges to a limit or diverges to ˙1, then so do all subsequences of the sequence. Limit points and boundedness of a set of real numbers are discussed in terms of sequences of members of the set. Continuity and boundedness of a function are discussed in terms of the values of the function at sequences of points in its domain. SECTION 4.3 introduces concepts of convergence and diverge nce to˙1 for infinite series of constants. We prove Cauchy’s convergence criterion for a series of constants. In con- nection with series of positive terms, we consider the compa rison test, the integral test, the ratio test, and Raabe’s test. For general series, we conside r absolute and conditional con- vergence, Dirichlet’s test, rearrangement of terms, and mu ltiplication of one infinite series by another. SECTION 4.4 deals with pointwise and uniform convergence of sequences and series of functions. Cauchy’s uniform convergence criteria for sequ ences and series are proved, as is Dirichlet’s test for uniform convergence of a series. We g ive sufficient conditions for the limit of a sequence of functions or the sum of an infinite se ries of functions to be continuous, integrable, or differentiable. SECTION 4.5 considers power series. It is shown that a power s eries that converges on an open interval defines an infinitely differentiable functi on on that interval. We define the Taylor series of an infinitely differentiable function, and give sufficient conditions for the Taylor series to converge to the function on some interva l. Arithmetic operations with power series are discussed. 178 Section 4.1 Sequences of Real Numbers 179 4.1 SEQUENCES OF REAL NUMBERS Aninfinite sequence (more briefly, a sequence ) of real numbers is a real-valued function defined on a set of integers˚nˇˇn/NAKk/TAB. We call the values of the function the terms of the sequence. We denote a sequence by listing its terms in order; thus, fsng1 kDfsk;skC1;:::g: (4.1.1) For example, /SUB1 n2C1/ESC1 0D/SUB 1;1 2;1 5;:::;1 n2C1;:::/ESC ; f./NUL1/ng1 0Df1;/NUL1;1;:::;./NUL1/n;:::g; and/SUB1 n/NUL2/ESC1 3D/SUB 1;1 2;1 3;:::;1 n/NUL2;:::/ESC : The real number snis thenthterm of the sequence. Usually we are interested only in the terms of a sequence and the order in which they appear, but not in the particular value of k in (4.1.1 ). Therefore, we regard the sequences /SUB1 n/NUL2/ESC1 3and/SUB1 n/ESC1 1 as identical. We will usually write fsngrather thanfsng1 k. In the absence of any indication to the contrary, we take kD0unlesssnis given by a rule that is invalid for some nonnegative integer, in which case kis understood to be the smallest positive integer such that snis defined for all n/NAKk. For example, if snD1 .n/NUL1/.n/NUL5/; thenkD6. The interesting questions about a sequence fsngconcern the behavior of snfor largen. Limit of a Sequence Definition 4.1.1 A sequencefsngconverges to a limit sif for every/SI >0 there is an integerNsuch that jsn/NULsj</SI ifn/NAKN: (4.1.2) In this case we say that fsngisconvergent and write lim n!1snDs: A sequence that does not converge diverges , or is divergent 180 Chapter 4 Infinite Sequences and Series As we saw in Section 2.1 when discussing limits of functions, Definition 4.1.1 is not changed by replacing ( 4.1.2 ) with jsn/NULsj<K/SI ifn/NAKN; whereKis a positive constant. Example 4.1.1 IfsnDcforn/NAKk, thenjsn/NULcjD0forn/NAKk, and lim n!1snDc. Example 4.1.2 If snD/SUB2nC1 nC1/ESC ; then lim n!1snD2, since jsn/NUL2jDˇˇˇˇ2nC1 nC1/NUL2nC2 nC1ˇˇˇˇD1 nC1I hence, if/SI>0 , then ( 4.1.2 ) holds withsD2ifN/NAK1=/SI. Definition 4.1.1 does not require that there be an integer Nsuch that ( 4.1.2 ) holds for all/SI; rather, it requires that for each positive /SIthere be an integer Nthat satisfies ( 4.1.2 ) for that particular /SI. Usually,Ndepends on/SIand must be increased if /SIis decreased. The constant sequences (Example 4.1.1 ) are essentially the only ones for which Ndoes not depend on/SI(Exercise 4.1.5 ). We say that the terms of a sequence fsng1 ksatisfy a given condition for allnifsnsatisfies the condition for all n/NAKk, orfor largenif there is an integer N >k such thatsnsatisfies the condition whenever n/NAKN. For example, the terms of f1=ng1 1are positive for all n, while those off1/NUL7=ng1 1are positive for large n(takeND8). Uniqueness of the Limit Theorem 4.1.2 The limit of a convergent sequence is unique : Proof Suppose that lim n!1snDsand lim n!1snDs0: We must show that sDs0. Let/SI>0 . From Definition 4.1.1 , there are integers N1andN2 such that jsn/NULsj</SI ifn/NAKN1 (because lim n!1snDs), and jsn/NULs0j</SI ifn/NAKN2 Section 4.1 Sequences of Real Numbers 181 (because lim n!1snDs0). These inequalities both hold if n/NAKNDmax.N1;N2/, which implies that js/NULs0jDj.s/NULsN/C.sN/NULs0/j /DC4js/NULsNjCjsN/NULs0j</SIC/SID2/SI: Since this inequality holds for every /SI > 0 andjs/NULs0jis independent of /SI, we conclude thatjs/NULs0jD0; that is,sDs0. Sequences Diverging to ˙1 We say that lim n!1snD1 if for any real number a,sn>afor largen. Similarly, lim n!1snD/NUL1 if for any real number a,sn<afor largen. However, we do not regard fsngas convergent unless lim n!1snis finite, as required by Definition 4.1.1 . To emphasize this distinction, we say thatfsngdiverges to1./NUL1/if lim n!1snD1./NUL1/. Example 4.1.3 The sequencefn=2C1=ngdiverges to1, since, ifais any real num- ber, then n 2C1 n>a ifn/NAK2a: The sequencefn/NULn2gdiverges to/NUL1, since, ifais any real number, then /NULn2CnD/NULn.n/NUL1/<a ifn>1Cp jaj: Therefore, we write lim n!1/DC2n 2C1 n/DC3 D1 and lim n!1./NULn2Cn/D/NUL1: The sequencef./NUL1/nn3gdiverges, but not to /NUL1 or1. Bounded Sequences Definition 4.1.3 A sequencefsngisbounded above if there is a real number bsuch that sn/DC4bfor alln; bounded below if there is a real number asuch that sn/NAKafor alln; orbounded if there is a real number rsuch that jsnj/DC4rfor alln: 182 Chapter 4 Infinite Sequences and Series Example 4.1.4 IfsnDŒ1C./NUL1/n/c141n, thenfsngis bounded below .sn/NAK0/but unbounded above, and f/NULsngis bounded above ./NULsn/DC40/but unbounded below. If snD ./NUL1/n, thenfsngis bounded. If snD./NUL1/nn, thenfsngis not bounded above or below. Theorem 4.1.4 A convergent sequence is bounded : Proof By taking/SID1in (4.1.2 ), we see that if lim n!1snDs, then there is an integer Nsuch that jsn/NULsj<1 ifn/NAKN: Therefore, jsnjDj.sn/NULs/Csj/DC4jsn/NULsjCjsj<1Cjsjifn/NAKN; and jsnj/DC4maxfjs0j;js1j;:::;jsN/NUL1j;1Cjsjg for alln, sofsngis bounded. Monotonic Sequences Definition 4.1.5 A sequencefsngisnondecreasing ifsn/NAKsn/NUL1for alln, ornonin- creasing ifsn/DC4sn/NUL1for alln:Amonotonic sequence is a sequence that is either nonin- creasing or nondecreasing. If sn>sn/NUL1for alln, thenfsngisincreasing , while ifsn<sn/NUL1 for alln,fsngisdecreasing . Theorem 4.1.6 (a) Iffsngis nondecreasing ;then limn!1snDsupfsng: (b) Iffsngis nonincreasing ;then limn!1snDinffsng: Proof (a) . LetˇDsupfsng. Ifˇ<1, Theorem 1.1.3 implies that if /SI>0 then ˇ/NUL/SI<s N/DC4ˇ for some integer N. SincesN/DC4sn/DC4ˇifn/NAKN, it follows that ˇ/NUL/SI<s n/DC4ˇifn/NAKN: This implies thatjsn/NULˇj</SIifn/NAKN, so lim n!1snDˇ, by Definition 4.1.1 . IfˇD1 andbis any real number, then sN> b for some integer N. Thensn> b forn/NAKN, so limn!1snD1 . We leave the proof of (b) to you (Exercise 4.1.8 ) Example 4.1.5 Ifs0D1andsnD1/NULe/NULsn/NUL1, then0<s n/DC41for alln, by induction. Since snC1/NULsnD/NUL.e/NULsn/NULesn/NUL1/ifn/NAK1; Section 4.1 Sequences of Real Numbers 183 the mean value theorem (Theorem 2.3.11 ) implies that snC1/NULsnDe/NULtn.sn/NULsn/NUL1/ifn/NAK1; (4.1.3) wheretnis betweensn/NUL1andsn. Sinces1/NULs0D/NUL1=e<0 , it follows by induction from (4.1.3 ) thatsnC1/NULsn<0for alln. Hence,fsngis bounded and decreasing, and therefore convergent. Sequences of Functional Values The next theorem enables us to apply the theory of limits deve loped in Section 2.1 to some sequences. We leave the proof to you (Exercise 4.1.13 ). Theorem 4.1.7 Letlimx!1f.x/DL;whereLis in the extended reals ;and suppose thatsnDf.n/ for largen:Then lim n!1snDL: Example 4.1.6 Let snDlogn nandf.x/Dlogx x: By L’Hospital’s rule, lim x!1logx xDlim x!11=x 1D0: Hence, lim n!1logn=nD0. Example 4.1.7 LetsnD.1C1=n/nand f.x/D/DC2 1C1 x/DC3x Dexlog.1C1=x/: By L’Hospital’s rule, lim x!1xlog/DC2 1C1 x/DC3 Dlim x!1log.1C1=x/ 1=x Dlim x!1/NUL1 x21 1C1=x /NUL1=x2D1I hence, lim x!1/DC2 1C1 x/DC3x De1Deand lim n!1/DC2 1C1 n/DC3n De: The last equation is sometimes used to define e. 184 Chapter 4 Infinite Sequences and Series Example 4.1.8 Suppose that snD/SUBnwith/SUB>0 , and letf.x/D/SUBxDexlog/SUB. Since lim x!1exlog/SUBD8 ˆ< ˆ:0; if log/SUB<0 .0</SUB<1/; 1; if log/SUBD0 ./SUBD1/; 1;if log/SUB>0 ./SUB>1/; it follows that lim n!1/SUBnD8 < :0; 0</SUB<1; 1; /SUBD1; 1; /SUB>1: Therefore, lim n!1rnD8 < :0;/NUL1<r <1; 1; rD1; 1; r >1; a result that we will use often. A Useful Limit Theorem The next theorem enables us to investigate convergence of se quences by examining simpler sequences. It is analogous to Theorem 2.1.4 . Theorem 4.1.8 Let lim n!1snDsand lim n!1tnDt; (4.1.4) wheresandtare finite:Then lim n!1.csn/Dcs (4.1.5) ifcis a constantI lim n!1.snCtn/DsCt; (4.1.6) lim n!1.sn/NULtn/Ds/NULt; (4.1.7) lim n!1.sntn/Dst; (4.1.8) and lim n!1sn tnDs t(4.1.9) iftnis nonzero for all nandt¤0. Proof We prove ( 4.1.8 ) and ( 4.1.9 ) and leave the rest to you (Exercises 4.1.15 and 4.1.17 ). For ( 4.1.8 ), we write sntn/NULstDsntn/NULstnCstn/NULstD.sn/NULs/tnCs.tn/NULt/I Section 4.1 Sequences of Real Numbers 185 hence, jsntn/NULstj/DC4jsn/NULsjjtnjCjsjjtn/NULtj: (4.1.10) Sinceftngconverges, it is bounded (Theorem 4.1.4 ). Therefore, there is a number Rsuch thatjtnj/DC4Rfor alln, and ( 4.1.10 ) implies that jsntn/NULstj/DC4Rjsn/NULsjCjsjjtn/NULtj: (4.1.11) From ( 4.1.4 ), if/SI>0 there are integers N1andN2such that jsn/NULsj</SI ifn/NAKN1 (4.1.12) and jtn/NULtj</SI ifn/NAKN2: (4.1.13) IfNDmax.N1;N2/, then ( 4.1.12 ) and ( 4.1.13 ) both hold when n/NAKN, and ( 4.1.11 ) implies that jsntn/NULstj/DC4.RCjsj//SIifn/NAKN: This proves ( 4.1.8 ). Now consider ( 4.1.9 ) in the special case where snD1for allnandt¤0; thus, we want to show that lim n!11 tnD1 t: First, observe that since lim n!1tnDt¤0, there is an integer Msuch thatjtnj/NAKjtj=2 ifn/NAKM. To see this, we apply Definition 4.1.1 with/SIDjtj=2; thus, there is an integer Msuch thatjtn/NULtj<jt=2jifn/NAKM. Therefore, jtnjDjtC.tn/NULt/j/NAKjjtj/NULjtn/NULtjj/NAKjtj 2ifn/NAKM: If/SI>0 , chooseN0so thatjtn/NULtj</SIifn/NAKN0, and letNDmax.N0;M/ . Then ˇˇˇˇ1 tn/NUL1 tˇˇˇˇDjt/NULtnj jtnjjtj/DC42/SI jtj2ifn/NAKNI hence, lim n!11=tnD1=t. Now we obtain ( 4.1.9 ) in the general case from ( 4.1.8 ) with ftngreplaced byf1=tng. Example 4.1.9 To determine the limit of the sequence defined by snD1 nsinn/EM 4C2.1C3=n/ 1C1=n; we apply the applicable parts of Theorem 4.1.8 as follows: lim n!1snDlim n!11 nsinn/EM 4C2h lim n!11C3lim n!1.1=n/i lim n!11Clim n!1.1=n/ D0C2.1C3/SOH0/ 1C0D2: 186 Chapter 4 Infinite Sequences and Series Example 4.1.10 Sometimes preliminary manipulations are necessary before applying Theorem 4.1.8 . For example, lim n!1.n=2/Clogn 3nC4pnDlim n!11=2C.logn/=n 3C4n/NUL1=2 Dlim n!11=2Clim n!1.logn/=n lim n!13C4lim n!1n/NUL1=2 D1=2C0 3C0(see Example 4.1.6 ) D1 6: Example 4.1.11 Suppose that/NUL1<r <1 and s0D1; s 1D1Cr; s 2D1CrCr2;:::; s nD1CrC/SOH/SOH/SOHCrn: Since sn/NULrsnD.1CrC/SOH/SOH/SOHCrn//NUL.rCr2C/SOH/SOH/SOHCrnC1/D1/NULrnC1; it follows that snD1/NULrnC1 1/NULr: (4.1.14) From Example 4.1.8 , lim n!1rnC1D0, so ( 4.1.14 ) and Theorem 4.1.8 yield lim n!1.1CrC/SOH/SOH/SOHCrn/D1 1/NULrif/NUL1<r <1: Equations ( 4.1.5 )–(4.1.8 ) are valid even if sandtare arbitrary extended reals, provided that their right sides are defined in the extended reals (Exer cises 4.1.16 ,4.1.18 , and 4.1.21 ); (4.1.9 ) is valid ifs=tis defined in the extended reals and t¤0(Exercise 4.1.22 ). Example 4.1.12 If/NUL1<r <1 , then lim n!1rn nŠDlim n!1rn lim n!1nŠD0 1D0; from ( 4.1.9 ) and Example 4.1.8 . However, if r >1 , (4.1.9 ) and Example 4.1.8 yield lim n!1rn nŠDlim n!1rn lim n!1nŠD1 1; an indeterminate form. If r/DC4/NUL1, then lim n!1rndoes not exist in the extended reals, so (4.1.9 ) is not applicable. Theorem 4.1.7 does not help either, since there is no elementary functionfsuch thatf.n/Drn=nŠ. However, the following argument shows that Section 4.1 Sequences of Real Numbers 187 lim n!1rn nŠD0;/NUL1<r <1: (4.1.15) There is an integer Msuch that jrj n<1 2ifn/NAKM: LetKDrm=MŠ . Then jrjn nŠ/DC4Kjrj MC1jrj MC2/SOH/SOH/SOHjrj n<K/DC21 2/DC3n/NULM ; n>M: Given/SI >0 , chooseN/NAKMso thatK=2N/NULM</SI. Thenjrjn=nŠ</SI ifn/NAKN, which verifies ( 4.1.15 ). Limits Superior and Inferior Requiring a sequence to converge may be unnecessarily restr ictive in some situations. Of- ten, useful results can be obtained from assumptions on the limit superior andlimit inferior of a sequence, which we consider next. Theorem 4.1.9 (a) Iffsngis bounded above and does not diverge to /NUL1;then there is a unique real numberssuch that;if/SI>0; sn<sC/SIfor largen (4.1.16) and sn>s/NUL/SIfor infinitely many n: (4.1.17) (b) Iffsngis bounded below and does not diverge to 1;then there is a unique real numberssuch that;if/SI>0; sn>s/NUL/SIfor largen (4.1.18) and sn<sC/SIfor infinitely many n: (4.1.19) Proof We will prove (a) and leave the proof of (b) to you (Exercise 4.1.23 ). Since fsngis bounded above, there is a number ˇsuch thatsn<ˇ for alln. Sincefsngdoes not diverge to/NUL1, there is a number ˛such thatsn>˛ for infinitely many n. If we define MkDsupfsk;skC1;:::;s kCr;:::g; 188 Chapter 4 Infinite Sequences and Series then˛/DC4Mk/DC4ˇ, sofMkgis bounded. SincefMkgis nonincreasing (why?), it converges, by Theorem 4.1.6 . Let sDlim k!1Mk: (4.1.20) If/SI>0 , thenMk<sC/SIfor largek, and sincesn/DC4Mkforn/NAKk,ssatisfies ( 4.1.16 ). If (4.1.17 ) were false for some positive /SI, there would be an integer Ksuch that sn/DC4s/NUL/SIifn/NAKK: However, this implies that Mk/DC4s/NUL/SIifk/NAKK; which contradicts ( 4.1.20 ). Therefore, shas the stated properties. Now we must show that sis the only real number with the stated properties. If t <s, the inequality sn<tCs/NULt 2Ds/NULs/NULt 2 cannot hold for all large n, because this would contradict ( 4.1.17 ) with/SID.s/NULt/=2. If s<t , the inequality sn>t/NULt/NULs 2DsCt/NULs 2 cannot hold for infinitely many n, because this would contradict ( 4.1.16 ) with/SID.t/NULs/=2. Therefore,sis the only real number with the stated properties. Definition 4.1.10 The numbers sandsdefined in Theorem 4.1.9 are called the limit superior andlimit inferior , respectively, offsng, and denoted by sDlim n!1snandsDlim n!1sn: We also define lim n!1snD 1 iffsngis not bounded above ; lim n!1snD/NUL1 if lim n!1snD/NUL1; lim n!1snD/NUL1 iffsngis not bounded below ; andlim n!1snD 1 if lim n!1snD1: Theorem 4.1.11 Every sequencefsngof real numbers has a unique limit superior ;s; and a unique limit inferior ;s, in the extended reals ;and s/DC4s: (4.1.21) Section 4.1 Sequences of Real Numbers 189 Proof The existence and uniqueness of sandsfollow from Theorem 4.1.9 and Defini- tion4.1.10 . Ifsandsare both finite, then ( 4.1.16 ) and ( 4.1.18 ) imply that s/NUL/SI<sC/SI for every/SI>0 , which implies ( 4.1.21 ). IfsD/NUL1 orsD1 , then ( 4.1.21 ) is obvious. If sD1 orsD/NUL1 , then ( 4.1.21 ) follows immediately from Definition 4.1.10 . Example 4.1.13 lim n!1rnD8 < :1;jrj>1; 1;jrjD1; 0;jrj<1I and lim n!1rnD8 ˆˆˆˆ< ˆˆˆˆ:1; r >1; 1; rD1; 0;jrj<1; /NUL1; rD/NUL1; /NUL1; r </NUL1: Also, lim n!1n2Dlim n!1n2D1; lim n!1./NUL1/n/DC2 1/NUL1 n/DC3 D1; lim n!1./NUL1/n/DC2 n/NUL1 n/DC3 D/NUL1; and lim n!1Œ1C./NUL1/n/c141n2D1;lim n!1Œ1C./NUL1/n/c141n2D0: Theorem 4.1.12 Iffsngis a sequence of real numbers, then lim n!1snDs (4.1.22) if and only if lim n!1snDlim n!1snDs: (4.1.23) Proof IfsD˙1 , the equivalence of ( 4.1.22 ) and ( 4.1.23 ) follows immediately from their definitions. If lim n!1snDs(finite), then Definition 4.1.1 implies that ( 4.1.16 )– (4.1.19 ) hold withsandsreplaced bys. Hence, ( 4.1.23 ) follows from the uniqueness of sands. For the converse, suppose that sDsand letsdenote their common value. Then (4.1.16 ) and ( 4.1.18 ) imply that s/NUL/SI<s n<sC/SI for largen, and ( 4.1.22 ) follows from Definition 4.1.1 and the uniqueness of lim n!1sn (Theorem 4.1.2 ). 190 Chapter 4 Infinite Sequences and Series Cauchy’s Convergence Criterion To determine from Definition 4.1.1 whether a sequence has a limit, it is necessary to guess what the limit is. (This is particularly difficult if the sequ ence diverges!) To use Theo- rem4.1.12 for this purpose requires finding sands. The following convergence criterion has neither of these defects. Theorem 4.1.13 ( Cauchy ’s Convergence Criterion) A sequencefsngof real numbers converges if and only if ;for every/SI>0; there is an integer Nsuch that jsn/NULsmj</SI ifm;n/NAKN: (4.1.24) Proof Suppose that lim n!1snDsand/SI>0 . By Definition 4.1.1 , there is an integer Nsuch that jsr/NULsj</SI 2ifr/NAKN: Therefore, jsn/NULsmjDj.sn/NULs/C.s/NULsm/j/DC4jsn/NULsjCjs/NULsmj</SI ifn;m/NAKN: Therefore, the stated condition is necessary for convergen ce offsng. To see that it is suffi- cient, we first observe that it implies that fsngis bounded (Exercise 4.1.27 ), sosandsare finite (Theorem 4.1.9 ). Now suppose that /SI > 0 andNsatisfies ( 4.1.24 ). From ( 4.1.16 ) and ( 4.1.17 ), jsn/NULsj</SI; (4.1.25) for some integer n>N and, from ( 4.1.18 ) and ( 4.1.19 ), jsm/NULsj</SI (4.1.26) for some integer m>N . Since js/NULsjDj.s/NULsn/C.sn/NULsm/C.sm/NULs/j /DC4js/NULsnjCjsn/NULsmjCjsm/NULsj; (4.1.24 )–(4.1.26 ) imply that js/NULsj<3/SI: Since/SIis an arbitrary positive number, this implies that sDs, sofsngconverges, by Theorem 4.1.12 . Example 4.1.14 Suppose that jf0.x/j/DC4r <1;/NUL1<x<1: (4.1.27) Show that the equation xDf.x/ (4.1.28) has a unique solution. Section 4.1 Sequences of Real Numbers 191 Solution To see that ( 4.1.28 ) cannot have more than one solution, suppose that xD f.x/ andx0Df.x0/. From ( 4.1.27 ) and the mean value theorem (Theorem 2.3.11 ), x/NULx0Df0.c/.x/NULx0/ for somecbetweenxandx0. This and ( 4.1.27 ) imply that jx/NULx0j/DC4rjx/NULx0j: Sincer <1 ,xDx0. We will now show that ( 4.1.28 ) has a solution. With x0arbitrary, define xnDf.x n/NUL1/; n/NAK1: (4.1.29) We will show thatfxngconverges. From ( 4.1.29 ) and the mean value theorem, xnC1/NULxnDf.x n//NULf.x n/NUL1/Df0.cn/.xn/NULxn/NUL1/; wherecnis betweenxn/NUL1andxn. This and ( 4.1.27 ) imply that jxnC1/NULxnj/DC4rjxn/NULxn/NUL1jifn/NAK1: (4.1.30) The inequality jxnC1/NULxnj/DC4rnjx1/NULx0jifn/NAK0; (4.1.31) follows by induction from ( 4.1.30 ). Now, ifn>m , jxn/NULxmjDj.xn/NULxn/NUL1/C.xn/NUL1/NULxn/NUL2/C/SOH/SOH/SOHC.xmC1/NULxm/j /DC4jxn/NULxn/NUL1jCjxn/NUL1/NULxn/NUL2jC/SOH/SOH/SOHCjxmC1/NULxmj; and ( 4.1.31 ) yields jxn/NULxmj/DC4jx1/NULx0jrm.1CrC/SOH/SOH/SOHCrn/NULm/NUL1/: (4.1.32) In Example 4.1.11 we saw that the sequence fskgdefined by skD1CrC/SOH/SOH/SOHCrk converges to 1=.1/NULr/ifjrj<1; moreover, since we have assumed here that 0<r <1 , fskgis nondecreasing, and therefore sk<1=.1/NULr/for allk. Therefore, ( 4.1.32 ) yields jxn/NULxmj<jx1/NULx0j 1/NULrrmifn>m: Now it follows that jxn/NULxmj<jx1/NULx0j 1/NULrrNifn;m>N; and, since lim N!1rND0,fxngconverges, by Theorem 4.1.13 . IfbxDlimn!1xn, then (4.1.29 ) and the continuity of fimply thatbxDf.bx/. 192 Chapter 4 Infinite Sequences and Series 4.1 Exercises 1. Prove: Ifsn/NAK0forn/NAKkand lim n!1snDs, thens/NAK0. 2. (a) Show that lim n!1snDs(finite) if and only if lim n!1jsn/NULsjD0. (b) Suppose thatjsn/NULsj /DC4tnfor largenand lim n!1tnD0. Show that limn!1snDs. 3. Find lim n!1sn. Justify your answers from Definition 4.1.1 . (a)snD2C1 nC1(b)snD˛Cn ˇCn(c)snD1 nsinn/EM 4 4. Find lim n!1sn. Justify your answers from Definition 4.1.1 . (a)snDn 2nCpnC1(b)snDn2C2nC2 n2Cn (c)snDsinnpn(d)snDp n2Cn/NULn 5. State necessary and sufficient conditions on a convergent se quencefsngsuch that the integerNin Definition 4.1.1 does not depend upon /SI. 6. Prove: If lim n!1snDsthen lim n!1jsnjDjsj. 7. Suppose that lim n!1snDs(finite) and, for each /SI>0 ,jsn/NULtnj</SIfor largen. Show that lim n!1tnDs. 8. Complete the proof of Theorem 4.1.6 . 9. Use Theorem 4.1.6 to show thatfsngconverges. (a)snD˛Cn ˇCn.ˇ>0/ (b)snDnŠ nn (c)snDrn 1Crn.r >0/ (d)snD.2n/Š 22n.nŠ/2 10. LetyDTan/NUL1xbe the solution of xDtanysuch that/NUL/EM=2<y </EM=2 . Prove: Ifx0>0andxnC1DTan/NUL1xn.n/NAK0/, thenfxngconverges. 11. Suppose that s0andAare positive numbers. Let snC1D1 2/DC2 snCA sn/DC3 ; n/NAK0: (a) Show thatsnC1/NAKp Aifn/NAK0. (b) Show thatsnC1/DC4snifn/NAK1. (c) Show thatsDlimn!1snexists. (d) Finds. 12. Prove: Iffsngis unbounded and monotonic, then either lim n!1snD1 or lim n!1snD /NUL1. 13. Prove Theorem 4.1.7 . Section 4.1 Sequences of Real Numbers 193 14. Use Theorem 4.1.7 to find lim n!1sn. (a)snD˛Cn ˇCn.ˇ>0/ (b)snDcos1 n (c)snDnsin1 n(d)snDlogn/NULn (e)snDlog.nC1//NULlog.n/NUL1/ 15. Suppose that lim n!1snDs(finite). Show that if cis a constant, then lim n!1.csn/D cs. 16. Suppose that lim n!1snDswheresD˙1 . Show that if cis a nonzero constant, then lim n!1.csn/Dcs. 17. Prove: If lim n!1snDsand lim n!1tnDt, wheresandtare finite, then lim n!1.snCtn/DsCtand lim n!1.sn/NULtn/Ds/NULt: 18. Prove: If lim n!1snDsand lim n!1tnDt, wheresandtare in the extended reals, then lim n!1.snCtn/DsCt ifsCtis defined. 19. Suppose that lim n!1tnDt, where0 <jtj<1, and let0 < /SUB < 1 . Show that there is an integer Nsuch thattn>/SUBt forn/NAKNift >0 , ortn</SUBt forn/NAKNif t <0 . In either case,jtnj>/SUBjtjifn/NAKN. 20. Prove: If lim n!1sn/NULs snCsD0; then lim n!1snDs: HINT:DefinetnD.sn/NULs/=.s nCs/and solve for sn: 21. Prove: if lim n!1snDsand lim n!1tnDt, wheresandtare in the extended reals, then lim n!1sntnDst provided that stis defined in the extended reals. 22. Prove: If lim n!1snDsand lim n!1tnDt, then lim n!1sn tnDs t.A/ ifs=tis defined in the extended reals and t¤0. Give an example where s=tis defined in the extended plane, but (A) does not hold. 23. Prove Theorem 4.1.9(b). 24. Findsands. (a)snDŒ./NUL1/nC1/c141n2(b)snD.1/NULrn/sinn/EM 2 194 Chapter 4 Infinite Sequences and Series (c)snDr2n 1Crn.r¤/NUL1/ (d)snDn2/NULn (e)snD./NUL1/ntnwhere lim n!1tnDt 25. Findsands. (a)snD./NUL1/n(b)snD./NUL1/n/DC2 2C3 n/DC3 (c)snDnC./NUL1/n.2nC1/ n(d)snDsinn/EM 3 26. Suppose that lim n!1jsnjD/CR(finite). Show thatfsngdiverges unless /CRD0or the terms infsnghave the same sign for large n. HINT:Use Exercise 4.1.19: 27. Prove: The sequence fsngis bounded if, for some positive /SI, there is an integer N such thatjsn/NULsmj</SIwhenevern,m/NAKN. In Exercises 4.1.28 –4.1.31 , assume that s,s.ors/,t, andtare in the extended reals, and show that the given inequalities or equations hold whenever their right sides are defined .not indeterminate /. 28. (a) lim n!1./NULsn/D/NULs (b) lim n!1./NULsn/D/NULs 29. (a) lim n!1.snCtn//DC4sCt (b) lim n!1.snCtn//NAKsCt 30. (a) Ifsn/NAK0,tn/NAK0, then(i) lim n!1sntn/DC4stand(ii) lim n!1sntn/NAKst. (b) Ifsn/DC40,tn/NAK0, then(i) lim n!1sntn/DC4stand(ii) lim n!1sntn/NAKst. 31. (a) If lim n!1snDs>0 andtn/NAK0, then(i)lim n!1sntnDstand(ii) lim n!1sntnDst. (b) If lim n!1snDs<0 andtn/NAK0, then(i) lim n!1sntnDstand(ii) lim n!1sntnDst. 32. Suppose thatfsngconverges and has only finitely many distinct terms. Show tha tsn is constant for large n. 33. Lets0ands1be arbitrary, and snC1DsnCsn/NUL1 2; n/NAK1: Use Cauchy’s convergence criterion to show that fsngconverges. 34. LettnDs1Cs2C/SOH/SOH/SOHCsn n,n/NAK1. (a) Prove: If lim n!1snDsthen lim n!1tnDs. (b) Give an example to show that ftngmay converge even though fsngdoes not. Section 4.2 Earlier Topics Revisited with Sequences 195 35. (a) Show that lim n!1/DLE 1/NUL˛ 1/DC1/DLE 1/NUL˛ 2/DC1 /SOH/SOH/SOH/DLE 1/NUL˛ n/DC1 D0; if˛>0: HINT:Look at the logarithm of the absolute value of the product : (b) Conclude from (a)that lim n!1 q n! D0ifq>/NUL1; where q n! is the generalized binomial coefficient of Example 2.5.3 . 4.2 EARLIER TOPICS REVISITED WITH SEQUENCES In Chapter 2.3 we used /SI–ıdefinitions and arguments to develop the theory of limits, continuity, and differentiability; for example, fis continuous at x0if for each/SI>0 there is aı > 0 such thatjf.x//NULf.x 0/j< /SI whenjx/NULx0j< ı. The same theory can be developed by methods based on sequences. Although we will no t carry this out in detail, we will develop it enough to give some examples. First, we nee d another definition about sequences. Definition 4.2.1 A sequenceftkgis asubsequence of a sequencefsngif tkDsnk; k/NAK0; wherefnkgis an increasing infinite sequence of integers in the domain o ffsng. We denote the subsequenceftkgbyfsnkg. Note thatfsngis a subsequence of itself, as can be seen by taking nkDk. All other subsequences offsngare obtained by deleting terms from fsngand leaving those remaining in their original relative order. Example 4.2.1 If fsngD/SUB1 n/ESC D/SUB 1;1 2;1 3;:::;1 n;:::/ESC ; then lettingnkD2kyields the subsequence fs2kgD/SUB1 2k/ESC D/SUB1 2;1 4;:::;1 2k;:::/ESC ; and lettingnkD2kC1yields the subsequence fs2kC1gD/SUB1 2kC1/ESC D/SUB 1;1 3;:::;1 2kC1;:::/ESC : 196 Chapter 4 Infinite Sequences and Series Since a subsequence fsnkgis again a sequence (with respect to k), we may ask whether fsnkgconverges. Example 4.2.2 The sequencefsngdefined by snD./NUL1/n/DC2 1C1 n/DC3 does not converge, but fsnghas subsequences that do. For example, fs2kgD/SUB 1C1 2k/ESC and lim k!1s2kD1; while fs2kC1gD/SUB /NUL1/NUL1 2kC1/ESC and lim k!1s2kC1D/NUL1: It can be shown (Exercise 4.2.1 ) that a subsequence fsnkgoffsngconverges to 1if and only ifnkis even forksufficiently large, or to /NUL1if and only if nkis odd forksufficiently large. Otherwise,fsnkgdiverges. The sequence in this example has subsequences that converge to different limits. The next theorem shows that if a sequence converges to a finite lim it or diverges to˙1, then all its subsequences do also. Theorem 4.2.2 If lim n!1snDs ./NUL1/DC4s/DC41/; (4.2.1) then lim k!1snkDs (4.2.2) for every subsequence fsnkgoffsng: Proof We consider the case where sis finite and leave the rest to you (Exercise 4.2.4 ). If (4.2.1 ) holds and/SI>0 , there is an integer Nsuch that jsn/NULsj</SI ifn/NAKN: Sincefnkgis an increasing sequence, there is an integer Ksuch thatnk/NAKNifk/NAKK. Therefore, jsnk/NULLj</SI ifk/NAKK; which implies ( 4.2.2 ). Theorem 4.2.3 Iffsngis monotonic and has a subsequence fsnkgsuch that lim k!1snkDs ./NUL1/DC4s/DC41/; then lim n!1snDs: Section 4.2 Earlier Topics Revisited with Sequences 197 Proof We consider the case where fsngis nondecreasing and leave the rest to you (Ex- ercise 4.2.6 ). Sincefsnkgis also nondecreasing in this case, it suffices to show that supfsnkgDsupfsng (4.2.3) and then apply Theorem 4.1.6(a). Since the set of terms of fsnkgis contained in the set of terms offsng, supfsng/NAKsupfsnkg: (4.2.4) Sincefsngis nondecreasing, there is for every nan integernksuch thatsn/DC4snk. This implies that supfsng/DC4 supfsnkg: This and ( 4.2.4 ) imply ( 4.2.3 ). Limit Points in Terms of Sequences In Section 1.3 we defined limit point in terms of neighborhoods: xis a limit point of a set Sif every neighborhood of xcontains points of Sdistinct from x. The next theorem shows that an equivalent definition can be stated in terms of sequen ces. Theorem 4.2.4 A pointxis a limit point of a set Sif and only if there is a sequence fxngof points inSsuch thatxn¤xforn/NAK1;and lim n!1xnDx: Proof For sufficiency, suppose that the stated condition holds. Th en, for each/SI > 0 , there is an integer Nsuch that0<jxn/NULxj</SIifn/NAKN. Therefore, every /SI-neighborhood ofxcontains infinitely many points of S. This means that xis a limit point of S. For necessity, let xbe a limit point of S. Then, for every integer n/NAK1, the interval .x/NUL1=n;xC1=n/ contains a point xn.¤x/inS. Sincejxm/NULxj/DC41=n ifm/NAKn, limn!1xnDx. We will use the next theorem to show that continuity can be defi ned in terms of se- quences. Theorem 4.2.5 (a) Iffxngis bounded;thenfxnghas a convergent subsequence : (b) Iffxngis unbounded above ;thenfxnghas a subsequencefxnkgsuch that lim k!1xnkD1: (c) Iffxngis unbounded below ;thenfxnghas a subsequencefxnkgsuch that lim k!1xnkD/NUL1: 198 Chapter 4 Infinite Sequences and Series Proof We prove (a) and leave (b) and(c) to you (Exercise 4.2.7 ). LetSbe the set of distinct numbers that occur as terms of fxng. (For example, if fxngDf./NUL1/ng, SDf1;/NUL1g; iffxngDf1;1 2;1;1 3;:::;1;1=n;:::g,SDf1;1 2;:::;1=n;:::g.) IfS contains only finitely many points, then some xinSoccurs infinitely often in fxng; that is, fxnghas a subsequencefxnkgsuch thatxnkDxfor allk. Then lim k!1xnkDx, and we are finished in this case. IfSis infinite, then, since Sis bounded (by assumption), the Bolzano–Weierstrass the- orem (Theorem 1.3.8 ) implies that Shas a limit point x. From Theorem 4.2.4 , there is a sequence of pointsfyjginS, distinct from x, such that lim j!1yjDx: (4.2.5) Although each yjoccurs as a term of fxng,fyjgis not necessarily a subsequence of fxng, because if we write yjDxnj; there is no reason to expect that fnjgis an increasing sequence as required in Defini- tion 4.2.1 . However, it is always possible to pick a subsequence fnjkgoffnjgthat is increasing, and then the sequence fyjkgDfsnjkgis a subsequence of both fyjgandfxng. Because of ( 4.2.5 ) and Theorem 4.2.2 this subsequence converges to x. Continuity in Terms of Sequences We now show that continuity can be defined and studied in terms of sequences. Theorem 4.2.6 Letfbe defined on a closed interval Œa;b/c141 containingx:Thenfis continuous at x.from the right if xDa;from the left if xDb/if and only if lim n!1f.x n/Df.x/ (4.2.6) wheneverfxngis a sequence of points in Œa;b/c141 such that lim n!1xnDx: (4.2.7) Proof Assume thata<x<b ; only minor changes in the proof are needed if xDaor xDb. First, suppose that fis continuous at xandfxngis a sequence of points in Œa;b/c141 satisfying ( 4.2.7 ). If/SI>0 , there is aı>0 such that jf.x//NULf.x/j</SI ifjx/NULxj<ı: (4.2.8) From ( 4.2.7 ), there is an integer Nsuch thatjxn/NULxj< ı ifn/NAKN. This and ( 4.2.8 ) imply thatjf.x n//NULf.x/j</SIifn/NAKN. This implies ( 4.2.6 ), which shows that the stated condition is necessary. For sufficiency, suppose that fis discontinuous at x. Then there is an /SI0>0such that, for each positive integer n, there is a point xnthat satisfies the inequality jxn/NULxj<1 n Section 4.2 Earlier Topics Revisited with Sequences 199 while jf.x n//NULf.x/j/NAK/SI0: The sequencefxngtherefore satisfies ( 4.2.7 ), but not ( 4.2.6 ). Hence, the stated condition cannot hold if fis discontinuous at x. This proves sufficiency. Armed with the theorems we have proved so far in this section, we could develop the theory of continuous functions by means of definitions and pr oofs based on sequences and subsequences. We give one example, a new proof of Theorem 2.2.8 , and leave others for exercises. Theorem 4.2.7 Iffis continuous on a closed interval Œa;b/c141; thenfis bounded on Œa;b/c141: Proof The proof is by contradiction. If fis not bounded on Œa;b/c141 , there is for each positive integer na pointxninŒa;b/c141 such thatjf.x n/j>n. This implies that lim n!1jf.x n/jD1: (4.2.9) Sincefxngis bounded,fxnghas a convergent subsequence fxnkg(Theorem 4.2.5(a)). If xDlim k!1xnk; thenxis a limit point of Œa;b/c141 , sox2Œa;b/c141 . Iffis continuous on Œa;b/c141 , then lim k!1f.x nk/Df.x/ by Theorem 4.2.6 , so lim k!1jf.x nk/jDjf.x/j (Exercise 4.1.6 ), which contradicts ( 4.2.9 ). Therefore, fcannot be both continuous and unbounded on Œa;b/c141 4.2 Exercises 1. LetsnD./NUL1/n.1C1=n/ . Show that lim k!1snkD1if and only if nkis even for largek, lim k!1snkD/NUL1if and only if nkis odd for large k, andfsnkgdiverges otherwise. 2. Find all numbers Lin the extended reals that are limits of some subsequence of fsng and, for each such L, choose a subsequence fsnkgsuch that lim k!1snkDL. (a)snD./NUL1/nn (b)snD/DC2 1C1 n/DC3 cosn/EM 2 (c)snD/DC2 1/NUL1 n2/DC3 sinn/EM 2(d)snD1 n (e)snDŒ./NUL1/nC1/c141n2(f)snDnC1 nC2/DLE sinn/EM 4Ccosn/EM 4/DC1 200 Chapter 4 Infinite Sequences and Series 3. Construct a sequence fsngwith the following property, or show that none exists: for each positive integer m,fsnghas a subsequence converging to m. 4. Complete the proof of Theorem 4.2.2 . 5. Prove: If lim n!1snDsandfsnghas a subsequencefsnkgsuch that./NUL1/ksnk/NAK0, thensD0. 6. Complete the proof of Theorem 4.2.3 . 7. Prove Theorem 4.2.5(b) and(c). 8. Suppose thatfsngis bounded and all convergent subsequences of fsngconverge to the same limit. Show that fsngis convergent. Give an example showing that the conclusion need not hold if fsngis unbounded. 9. (a) Letfbe defined on a deleted neighborhood Nofx. Show that lim x!xf.x/DL if and only if lim n!1f.x n/DLwheneverfxngis a sequence of points in N such that lim n!1xnDx. HINT:See the proof of Theorem 4.2.6: (b) State a result like (a)for one-sided limits. 10. Give a proof based on sequences for Theorem 2.2.9 . H INT:Use Theorems 4.1.6; 4.2.2;4.2.5;and4.2.6: 11. Give a proof based on sequences for Theorem 2.2.12 . 12. Suppose that fis defined on a deleted neighborhood Nofxandff.x n/gap- proaches a limit whenever fxngis a sequence of points in Nand lim n!1xnD x. Show that iffxngandfyngare two such sequences, then lim n!1f.x n/D limn!1f.y n/. Infer from this and Exercise 4.2.9 that lim x!xf.x/ exists. 13. Prove: Iffis defined on a neighborhood Nofx, thenfis differentiable at xif and only if lim n!1f.x n//NULf.x/ xn/NULx exists wheneverfxngis a sequence of points in Nsuch thatxn¤xand lim n!1xnD x. HINT:Use Exercise 4.2.12: 4.3 INFINITE SERIES OF CONSTANTS The theory of sequences developed in the last two sections ca n be combined with the fa- miliar notion of a finite sum to produce the theory of infinite s eries. We begin the study of infinite series in this section. Definition 4.3.1 Iffang1 kis an infinite sequence of real numbers, the symbol 1X nDkan Section 4.3 Infinite Series of Constants 201 is an infinite series , andanis thenth term of the series. We say thatP1 nDkanconverges to the sumA, and write 1X nDkanDA; if the sequencefAng1 kdefined by AnDakCakC1C/SOH/SOH/SOHCan; n/NAKk; converges toA. The finite sum Anis thenth partial sum ofP1 nDkan. IffAng1 kdiverges, we say thatP1 nDkandiverges ; in particular, if lim n!1AnD1 or/NUL1, we say thatP1 nDkandiverges to1or/NUL1, and write 1X nDkanD1 or1X nDkanD/NUL1: A divergent infinite series that does not diverge to ˙1 is said to oscillate , orbe oscillatory . We will usually refer to infinite series more briefly as series . Example 4.3.1 Consider the series 1X nD0rn;/NUL1<r <1: HereanDrn.n/NAK0/and AnD1CrCr2C/SOH/SOH/SOHCrnD1/NULrnC1 1/NULr; (4.3.1) which converges to 1=.1/NULr/asn!1 (Example 4.1.11 ); thus, we write 1X nD0rnD1 1/NULr;/NUL1<r <1: Ifjrj>1, then ( 4.3.1 ) is still valid, butP1 nD0rndiverges; ifr >1 , then 1X nD0rnD1; (4.3.2) while ifr </NUL1,P1 nD0rnoscillates, since its partial sums alternate in sign and the ir magnitudes become arbitrarily large for large n. IfrD/NUL1, thenA2mC1D0andA2mD1 form/NAK0, while ifrD1,AnDnC1; in both cases the series diverges, and ( 4.3.2 ) holds ifrD1. 202 Chapter 4 Infinite Sequences and Series The seriesP1 nD0rnis called the geometric series with ratio r. It occurs in many appli- cations. An infinite series can be viewed as a generalization of a finite sum ADNX nDkanDakCakC1C/SOH/SOH/SOHCaN by thinking of the finite sequence fak;akC1;:::;a Ngas being extended to an infinite se- quencefang1 kwithanD0forn>N . Then the partial sums ofP1 nDkanare AnDakCakC1C/SOH/SOH/SOHCan; k/DC4n<N; and AnDA; n/NAKNI that is, the terms of fAng1 kequal the finite sum Aforn/NAKk. Therefore, lim n!1An DA. The next two theorems can be proved by applying Theorems 4.1.2 and4.1.8 to the partial sums of the series in question (Exercises 4.3.1 and4.3.2 ). Theorem 4.3.2 The sum of a convergent series is unique : Theorem 4.3.3 Let 1X nDkanDAand1X nDkbnDB; whereAandBare finite:Then 1X nDk.can/DcA ifcis a constant; 1X nDk.anCbn/DACB; and1X nDk.an/NULbn/DA/NULB: These relations also hold if one or both of AandBis infinite, provided that the right sides are not indeterminate : Dropping finitely many terms from a series does not alter conv ergence or divergence, although it does change the sum of a convergent series if the t erms dropped have a nonzero sum. For example, suppose that we drop the first kterms of a seriesP1 nD0an, and consider the new seriesP1 nDkan. Denote the partial sums of the two series by AnDa0Ca1C/SOH/SOH/SOHCan; n/NAK0; and A0 nDakCakC1C/SOH/SOH/SOHCan; n/NAKk: Section 4.3 Infinite Series of Constants 203 Since AnD.a0Ca1C/SOH/SOH/SOHCak/NUL1/CA0 n; n/NAKk; it follows that ADlimn!1Anexists (in the extended reals) if and only if A0Dlimn!1A0 n does, and in this case AD.a0Ca1C/SOH/SOH/SOHCak/NUL1/CA0: An important principle follows from this. Lemma 4.3.4 Suppose that for nsufficiently large .that is;forn/NAKsome integer N/ the terms ofP1 nDkansatisfy some condition that implies convergence of an infini te series: ThenP1 nDkanconverges:Similarly, suppose that for nsufficiently large the termsP1 nDkan satisfy some condition that implies divergence of an infinit e series:ThenP1 nDkandiverges: Example 4.3.2 Consider the alternating series test, which we will establi sh later as a special case of a more general test: The seriesP1 kanconverges if./NUL1/nan>0;janC1j<janj;andlimn!1anD0: The terms of1X nD116C./NUL2/n n2n do not satisfy these conditions for all n/NAK1, but they do satisfy them for sufficiently large n. Hence, the series converges, by Lemma 4.3.4 . We will soon give several conditions concerning convergenc e of a seriesP1 nDkanwith nonnegative terms. According to Lemma 4.3.4 , these results apply to series that have at most finitely many negative terms, as long as anis nonnegative and satisfies the conditions fornsufficiently large. When we are interested only in whetherP1 nDkanconverges or diverges and not in its sum, we will simply say “Panconverges” or “Pandiverges.” Lemma 4.3.4 justifies this convention, subject to the understanding thatPanstands forP1 nDkan, wherekis an integer such that anis defined for n/NAKk. (For example, X1 .n/NUL6/2stands for1X nDk1 .n/NUL6/2; wherek/NAK7.) We writePanD1./NUL1/ifPandiverges to1./NUL1/. Finally, let us agree that 1X nDkanand1X nDk/NULjanCj (where we obtain the second expression by shifting the index in the first) both represent the same series. 204 Chapter 4 Infinite Sequences and Series Cauchy’s Convergence Criterion for Series The Cauchy convergence criterion for sequences (Theorem 4.1.13 ) yields a useful criterion for convergence of series. Theorem 4.3.5 (Cauchy’s Convergence Criterion for Series) A seriesPanconverges if and only if for every /SI>0 there is an integer Nsuch that janCanC1C/SOH/SOH/SOHCamj</SI ifm/NAKn/NAKN: (4.3.3) Proof In terms of the partial sums fAngofPan, anCanC1C/SOH/SOH/SOHCamDAm/NULAn/NUL1: Therefore, ( 4.3.3 ) can be written as jAm/NULAn/NUL1j</SI ifm/NAKn/NAKN: SincePanconverges if and only if fAngconverges, Theorem 4.1.13 implies the conclu- sion. Intuitively, Theorem 4.3.5 means thatPanconverges if and only if arbitrarily long sums anCanC1C/SOH/SOH/SOHCam; m/NAKn; can be made as small as we please by picking nlarge enough. Example 4.3.3 Consider the geometric seriesPrnof Example 4.3.1 . Ifjrj/NAK1, then frngdoes not converge to zero. ThereforePrndiverges, as we saw in Example 4.3.1 . If jrj<1andm/NAKn, then jAm/NULAnjDjrnC1CrnC2C/SOH/SOH/SOHCrmj /DC4jrjnC1.1CjrjC/SOH/SOH/SOHCjrjm/NULn/NUL1/ DjrjnC11/NULjrjm/NULn 1/NULjrj<jrjnC1 1/NULjrj:(4.3.4) If/SI>0 , chooseNso that jrjNC1 1/NULjrj</SI: Then ( 4.3.4 ) implies that jAm/NULAnj</SI ifm/NAKn/NAKN: Now Theorem 4.3.5 implies thatPrnconverges ifjrj<1, as in Example 4.3.1 . LettingmDnin (4.3.3 ) yields the following important corollary of Theorem 4.3.5 . Corollary 4.3.6 IfPanconverges;then limn!1anD0: Section 4.3 Infinite Series of Constants 205 It must be emphasized that Corollary 4.3.6 gives a necessary condition for convergence; that is,Pancannot converge unless lim n!1anD0. The condition is not sufficient ;Pan may diverge even if lim n!1anD0. We will see examples below. We leave the proof of the following corollary of Theorem 4.3.5 to you (Exercise 4.3.5 ). Corollary 4.3.7 IfPanconverges;then for each /SI > 0 there is an integer Ksuch that ˇˇˇˇˇ1X nDkanˇˇˇˇˇ</SI ifk/NAKKI that is; lim k!11X nDkanD0: Example 4.3.4 Ifjrj<1, then ˇˇˇˇˇ1X nDkrnˇˇˇˇˇDˇˇˇˇˇrk1X nDkrn/NULkˇˇˇˇˇDˇˇˇˇˇrk1X nD0rnˇˇˇˇˇDjrjk 1/NULr: Therefore, if jrjK 1/NULr</SI; then ˇˇˇˇˇ1X nDkrnˇˇˇˇˇ</SI ifk/NAKK; which implies that lim k!1P1 nDkrnD0. Series of Nonnegative Terms The theory of seriesPanwith terms that are nonnegative for sufficiently large nis simpler than the general theory, since such a series either converge s to a finite limit or diverges to 1, as the next theorem shows. Theorem 4.3.8 Ifan/NAK0forn/NAKk;thenPanconverges if its partial sums are bounded;or diverges to1if they are not :These are the only possibilities and ;in either case; 1X nDkanDsup˚ Anˇˇn/NAKk/TAB ; where AnDakCakC1C/SOH/SOH/SOHCan; n/NAKk: 206 Chapter 4 Infinite Sequences and Series Proof SinceAnDAn/NUL1Canandan/NAK0.n/NAKk/, the sequencefAngis nondecreasing, so the conclusion follows from Theorem 4.1.6(a)and Definition 4.3.1 . Ifan/NAK0for sufficiently large n, we will writePan<1ifPanconverges. This con- vention is based on Theorem 4.3.8 , which says that such a series diverges only ifPanD 1. The convention does not apply to series with infinitely many negative terms, because such series may diverge without diverging to 1; for example, the seriesP1 nD0./NUL1/nos- cillates, since its partial sums are alternately 1and0. Theorem 4.3.9 (The Comparison Test) Suppose that 0/DC4an/DC4bn; n/NAKk: (4.3.5) Then (a)Pan<1ifPbn<1: (b)PbnD1 ifPanD1: Proof (a) If AnDakCakC1C/SOH/SOH/SOHCanandBnDbkCbkC1C/SOH/SOH/SOHCbn; n/NAKk; then, from ( 4.3.5 ), An/DC4Bn: (4.3.6) Now we use Theorem 4.3.8 . IfPbn<1, thenfBngis bounded above and ( 4.3.6 ) implies thatfAngis also; therefore,Pan<1. On the other hand, ifPanD1 , thenfAngis unbounded above and ( 4.3.6 ) implies thatfBngis also; therefore,PbnD1 . We leave it to you to show that (a)implies (b). Example 4.3.5 Since rn n<rn; n/NAK1; andPrn<1if0<r <1 , the seriesPrn=nconverges if0<r <1 , by the comparison test. Comparing these two series is inconclusive if r > 1 , since it does not help to know that the terms ofPrn=nare smaller than those of the divergent seriesPrn. Ifr <0 , the comparison test does not apply, since the series then have in finitely many negative terms. Example 4.3.6 Since rn<nrn andPrnD1 ifr/NAK1, the comparison test implies thatPnrnD1 ifr/NAK1. Compar- ing these two series is inconclusive if 0 < r < 1 , since it does not help to know that the terms ofPnrnare larger than those of the convergent seriesPrn. Section 4.3 Infinite Series of Constants 207 The comparison test is useful if we have a collection of serie s with nonnegative terms and known convergence properties. We will now use the compar ison test to build such a collection. Theorem 4.3.10 (The Integral Test) Let cnDf.n/; n/NAKk; (4.3.7) wherefis positive;nonincreasing ;and locally integrable on Œk;1/:Then X cn<1 (4.3.8) if and only ifZ1 kf.x/dx<1: (4.3.9) Proof We first observe that ( 4.3.9 ) holds if and only if 1X nDkZnC1 nf.x/dx<1 (4.3.10) (Exercise 4.3.9 ), so it is enough to show that ( 4.3.8 ) holds if and only if ( 4.3.10 ) does. From (4.3.7 ) and the assumption that fis nonincreasing, cnC1Df.nC1//DC4f.x//DC4f.n/Dcn; n/DC4x/DC4nC1; n/NAKk: Therefore, cnC1DZnC1 ncnC1dx/DC4ZnC1 nf.x/dx/DC4ZnC1 ncndxDcn; n/NAKk (Theorem 3.3.4 ). From the first inequality and Theorem 4.3.9(a) withanDcnC1and bnDRnC1 nf.x/dx , (4.3.10 ) implies thatPcnC1<1, which is equivalent to ( 4.3.8 ). From the second inequality and Theorem 4.3.9(a)withanDRnC1 nf.x/dx andbnDcn, (4.3.8 ) implies ( 4.3.10 ). Example 4.3.7 The integral test implies that the series X1 np;X1 n.logn/p;andX 1 nlognŒlog.logn//c141p converge ifp>1 and diverge if 0<p/DC41, because the same is true of the integrals Z1 adx xp;Z1 adx x.logx/p;andZ1 adx xlogxŒlog.logx//c141p ifais sufficiently large. (See Example 3.4.3 and Exercise 3.4.10 .) The three series di- verge ifp/DC40: the first by Corollary 4.3.6 , the second by comparison with the divergent seriesP1=n, and the third by comparison with the divergent seriesP1=.n logn/. (The 208 Chapter 4 Infinite Sequences and Series divergence of the last two series for p/DC40also follows from the integral test, but the divergence of the first does not. Why not?) These results can b e generalized: If L0.x/DxandLk.x/DlogŒLk/NUL1.x//c141; k/NAK1; thenX 1 L0.n/L 1.n//SOH/SOH/SOHLk.n/ŒL kC1.n//c141p converges if and only if p>1 (Exercise 4.3.11 ). This example provides an infinite family of series with known convergence properties that can be used as standards for the comparison test. Except for the series of Example 4.3.7 , the integral test is of limited practical value, since convergence or divergence of most of the series to whic h it can be applied can be determined by simpler tests that do not require integration . However, the method used to prove the integral test is often useful for estimating the ra te of convergence or divergence of a series. This idea is developed in Exercises 4.3.13 and4.3.14 . Example 4.3.8 The series 1X1 .n2Cn/q(4.3.11) converges ifq>1=2 , by comparison with the convergent seriesP1=n2q, since 1 .n2Cn/q<1 n2q; n/NAK1: This comparison is inconclusive if q/DC41=2, since then X1 n2qD1; and it does not help to know that the terms of ( 4.3.11 ) are smaller than those of a divergent series. However, we can use the comparison test here, after a little trickery. We observe that1X nDk/NUL11 .nC1/2qD1X nDk1 n2qD1; q/DC41=2; and1 .nC1/2q<1 .n2Cn/q: Therefore, the comparison test implies that X1 .n2Cn/qD1; q/DC41=2: Section 4.3 Infinite Series of Constants 209 The next theorem is often applicable where the integral test is not. It does not require the kind of trickery that we used in Example 4.3.8 . Theorem 4.3.11 Suppose that an/NAK0andbn>0forn/NAKk:Then (a)X an<1 ifX bn<1 and lim n!1an=bn<1: (b)X anD1 ifX bnD1 and lim n!1an=bn>0: Proof (a) Iflimn!1an=bn<1, thenfan=bngis bounded, so there is a constant M and an integer ksuch that an/DC4Mb n; n/NAKk: SincePbn<1, Theorem 4.3.3 implies thatP.Mb n/ <1. NowPan<1, by the comparison test. (b) If limn!1an=bn>0, there is a constant mand an integer ksuch that an/NAKmbn; n/NAKk: SincePbnD1 , Theorem 4.3.3 implies thatP.mb n/D1 . NowPanD1 , by the comparison test. Example 4.3.9 Let X bnDX1 npCqandX anDX2Csinn/EM=6 .nC1/p.n/NUL1/q: Then an bnD2Csinn/EM=6 .1C1=n/p.1/NUL1=n/q; so lim n!1an bnD3and lim n!1an bnD1: SincePbn<1if and only if pCq >1 , the same is true ofPan, by Theorem 4.3.11 . The following corollary of Theorem 4.3.11 is often useful, although it does not apply to the series of Example 4.3.9 . Corollary 4.3.12 Suppose that an/NAK0andbn>0forn/NAKk;and lim n!1an bnDL; where0<L<1:ThenPanandPbnconverge or diverge together : 210 Chapter 4 Infinite Sequences and Series Example 4.3.10 With this corollary we can avoid the kind of trickery used in t he second part of Example 4.3.8 , since lim n!11 .n2Cn/q/RS1 n2qDlim n!11 .1C1=n/qD1; soX1 .n2Cn/qandX1 n2q converge or diverge together. The Ratio Test It is sometimes possible to determine whether a series with p ositive terms converges by comparing the ratios of successive terms with the correspon ding ratios of a series known to converge or diverge. Theorem 4.3.13 Suppose that an>0;b n>0; and anC1 an/DC4bnC1 bn: (4.3.12) Then (a)Pan<1ifPbn<1: (b)PbnD1 ifPanD1: Proof Rewriting ( 4.3.12 ) asanC1 bnC1/DC4an bn; we see thatfan=bngis nonincreasing. Therefore, limn!1an=bn<1, and Theorem 4.3.11(a) implies (a). To prove (b), suppose thatPanD1 . Sincefan=bngis nonincreasing, there is a number/SUBsuch thatbn/NAK/SUBanfor largen. SinceP./SUBa n/D1 ifPanD1 , Theo- rem4.3.9(b) (withanreplaced by/SUBan) implies thatPbnD1 . We will use this theorem to obtain two other widely applicabl e tests: the ratio test and Raabe’s test. Theorem 4.3.14 (The Ratio Test) Suppose that an>0forn/NAKk:Then (a)Pan<1iflimn!1anC1=an<1: (b)PanD1 iflimn!1anC1=an>1: If lim n!1anC1 an/DC41/DC4lim n!1anC1 an; (4.3.13) then the test is inconclusive Ithat is;Panmay converge or diverge : Section 4.3 Infinite Series of Constants 211 Proof (a) If lim n!1anC1 an<1; there is a number rsuch that0<r <1 and anC1 an<r fornsufficiently large. This can be rewritten as anC1 an<rnC1 rn: SincePrn<1, Theorem 4.3.13(a)withbnDrnimplies thatPan<1. (b) If lim n!1anC1 an>1; there is a number rsuch thatr >1 and anC1 an>r fornsufficiently large. This can be rewritten as anC1 an>rnC1 rn: SincePrnD1 , Theorem 4.3.13(b) withanDrnimplies thatPbnD1 . To see that no conclusion can be drawn if ( 4.3.13 ) holds, consider X anDX1 np: This series converges if p>1 or diverges if p/DC41; however, lim n!1anC1 anDlim n!1anC1 anD1 for everyp. Example 4.3.11 If X anDX/DLE 2Csinn/EM 2/DC1 rn; then anC1 anDr2Csin.nC1//EM 2 2Csinn/EM 2 which assumes the values 3r=2 ,2r=3 ,r=2, and2r, each infinitely many times; hence, lim n!1anC1 anD2r and lim n!1anC1 anDr 2: Therefore,Panconverges if 0 < r < 1=2 and diverges if r > 2 . The ratio test is inconclusive if 1=2/DC4r/DC42. 212 Chapter 4 Infinite Sequences and Series The following corollary of the ratio test is the familiar rat io rest from calculus. Corollary 4.3.15 Suppose that an>0.n/NAKk/and lim n!1anC1 anDL: Then (a)Pan<1ifL<1: (b)PanD1 ifL>1: The test is inconclusive if LD1: Example 4.3.12 The seriesPanDPnrn/NUL1converges if0<r <1 or diverges if r >1 , since anC1 anD.nC1/rn nrn/NUL1D/DC2 1C1 n/DC3 r; so lim n!1anC1 anDr: Corollary 4.3.15 is inconclusive if rD1, but then Corollary 4.3.6 implies that the series diverges. The ratio test does not imply thatPan<1if merely anC1 an<1 (4.3.14) for largen, since this could occur with lim n!1anC1=anD1, in which case the test is inconclusive. However, the next theorem shows thatPan<1if (4.3.14 ) is replaced by the stronger condition thatanC1 an/DC41/NULp n for somep>1 and largen. It also shows thatPanD1 if anC1 an/NAK1/NULq n for someq<1 and largen. Theorem 4.3.16 ( Raabe ’s Test) Suppose that an>0for largen:Let MDlim n!1n/DC2anC1 an/NUL1/DC3 andmDlim n!1n/DC2anC1 an/NUL1/DC3 : Then (a)Pan<1ifM </NUL1: (b)PanD1 ifm>/NUL1: The test is inconclusive if m/DC4/NUL1/DC4M: Section 4.3 Infinite Series of Constants 213 Proof (a) We need the inequality 1 .1Cx/p>1/NULpx; x>0; p>0: (4.3.15) This follows from Taylor’s theorem (Theorem 2.5.4 ), which implies that 1 .1Cx/pD1/NULpxC1 2p.pC1/ .1Cc/pC2x2; where0<c<x . (Verify.) Since the last term is positive if p>0 , this implies ( 4.3.15 ). Now suppose that M </NULp</NUL1. Then there is an integer ksuch that n/DC2anC1 an/NUL1/DC3 </NULp; n/NAKk; soanC1 an<1/NULp n; n/NAKk: Hence, anC1 an<1 .1C1=n/p; n/NAKk; as can be seen by letting xD1=nin (4.3.15 ). From this, anC1 an<1 .nC1/p/RS1 np; n/NAKk: SinceP1=np<1ifp>1 , Theorem 4.3.13(a)implies thatPan<1. (b) Here we need the inequality .1/NULx/q<1/NULqx; 0<x<1; 0<q<1: (4.3.16) This also follows from Taylor’s theorem, which implies that .1/NULx/qD1/NULqxCq.q/NUL1/.1/NULc/q/NUL2x2 2; where0<c<x . Now suppose that/NUL1</NULq<m . Then there is an integer ksuch that n/DC2anC1 an/NUL1/DC3 >/NULq; n/NAKk; soanC1 an/NAK1/NULq n; n/NAKk: Ifq/DC40, thenPanD1 , by Corollary 4.3.6 . Hence, we may assume that 0<q<1 , so the last inequality implies that anC1 an>/DC2 1/NUL1 n/DC3q ; n/NAKk; 214 Chapter 4 Infinite Sequences and Series as can be seen by setting xD1=n in (4.3.16 ). Hence, anC1 an>1 nq/RS1 .n/NUL1/q; n/NAKk: SinceP1=nqD1 ifq<1 , Theorem 4.3.13(b) implies thatPanD1 . Example 4.3.13 If X anDX nŠ ˛.˛C1/.˛C2//SOH/SOH/SOH.˛Cn/NUL1/; ˛>0; then lim n!1anC1 anDlim n!1nC1 ˛CnD1; so the ratio test is inconclusive. However, lim n!1n/DC2anC1 an/NUL1/DC3 Dlim n!1n/DC2nC1 ˛Cn/NUL1/DC3 Dlim n!1n.1/NUL˛/ ˛CnD1/NUL˛; so Raabe’s test implies thatPan<1if˛ > 2 andPanD1 if0 <˛ < 2 . Raabe’s test is inconclusive if ˛D2, but then the series becomes XnŠ .nC1/ŠDX1 nC1; which we know is divergent. Example 4.3.14 Consider the seriesPan, where a2mD.mŠ/2 ˛.˛C1//SOH/SOH/SOH.˛Cm/ˇ.ˇC1//SOH/SOH/SOH.ˇCm/ and a2mC1D.mŠ/2.mC1/ ˛.˛C1//SOH/SOH/SOH.˛Cm/ˇ.ˇC1//SOH/SOH/SOH.ˇCmC1/; with0<˛<ˇ . Since 2m/DC2a2mC1 a2m/NUL1/DC3 D2m/DC2mC1 ˇCmC1/NUL1/DC3 D/NUL2mˇ ˇCmC1 and .2mC1//DC2a2mC2 a2mC1/NUL1/DC3 D.2mC1//DC2mC1 ˛CmC1/NUL1/DC3 D/NUL.2mC1/˛ ˛CmC1; we have lim n!1n/DC2anC1 an/NUL1/DC3 D/NUL2˛ and lim n!1n/DC2anC1 an/NUL1/DC3 D/NUL2ˇ: Raabe’s test implies thatPan<1if˛ >1=2 andPanD1 ifˇ < 1=2 . The test is inconclusive if 0<˛/DC41=2/DC4ˇ. Section 4.3 Infinite Series of Constants 215 The next theorem, which will be useful when we study power ser ies (Section 4.5), con- cludes our discussion of series with nonnegative terms. Theorem 4.3.17 (Cauchy’s Root Test) Ifan/NAK0forn/NAKk;then (a)Pan<1iflimn!1a1=n n<1: (b)PanD1 iflimn!1a1=n n>1: The test is inconclusive if limn!1a1=n nD1: Proof (a) Iflimn!1a1=n n< 1, there is anrsuch that0 < r < 1 anda1=n n< r for largen. Therefore,an<rnfor largen. SincePrn<1, the comparison test implies thatPan<1. (b) Iflimn!1a1=n n>1, thena1=n n>1for infinitely many values of n, soPanD1 , by Corollary 4.3.6 . Example 4.3.15 Cauchy’s root test is inconclusive if X anDX1 np; because then lim n!1a1=n nDlim n!1/DC21 np/DC31=n Dlim n!1exp/DLE /NULp nlogn/DC1 D1 for allp. However, we know from the integral test thatP1=np<1ifp > 1 andP1=npD1 ifp/DC41. Example 4.3.16 If X anDX/DLE 2Csinn/EM 4/DC1n rn; then lim n!1a1=n nDlim n!1/DLE 2Csinn/EM 4/DC1 rD3r; and soPan<1ifr < 1=3 andPanD1 ifr > 1=3 . The test is inconclusive if rD1=3, but thenja8mC2jD1form/NAK0, soPanD1 , by Corollary 4.3.6 . Absolute and Conditional Convergence We now drop the assumption that the terms ofPanare nonnegative for large n. In this case,Panmay converge in two quite different ways. The first is defined a s follows. Definition 4.3.18 A seriesPanconverges absolutely , or is absolutely convergent ;ifPjanj<1: 216 Chapter 4 Infinite Sequences and Series Example 4.3.17 A convergent seriesPanof nonnegative terms is absolutely conver- gent, sincePanandPjanjare the same. More generally, any convergent series whose terms are of the same sign for sufficiently large nconverges absolutely (Exercise 4.3.22 ). Example 4.3.18 Consider the series Xsinn/DC2 np; (4.3.17) where/DC2is arbitrary and p>1 . Since ˇˇˇˇsinn/DC2 npˇˇˇˇ/DC41 np andP1=np<1ifp>1 , the comparison test implies that Xˇˇˇˇsinn/DC2 npˇˇˇˇ<1; p>1: Therefore, ( 4.3.17 ) converges absolutely if p>1 . Example 4.3.19 If0<p<1 , then the series X./NUL1/n np does not converge absolutely, since Xˇˇˇˇ./NUL1/n npˇˇˇˇDX1 npD1: However, the series converges, by the alternating series te st, which we prove below. Any test for convergence of a series with nonnegative terms c an be used to test an arbi- trary seriesPanfor absolute convergence by applying it toPjanj. We used the compar- ison test this way in Examples 4.3.18 and4.3.19 . Example 4.3.20 To test the series X anDX ./NUL1/nnŠ ˛.˛C1//SOH/SOH/SOH.˛Cn/NUL1/; ˛>0; for absolute convergence, we apply Raabe’s test to X anDX nŠ ˛.˛C1//SOH/SOH/SOH.˛Cn/NUL1/: From Example 4.3.13 ,Pjanj<1if˛>2 andPjanjD1 if˛<2 . Therefore,Pan converges absolutely if ˛>2 , but not if˛<2 . Notice that this does not imply thatPan diverges if˛<2 . Section 4.3 Infinite Series of Constants 217 The proof of the next theorem is analogous to the proof of Theo rem3.4.9 . We leave it to you (Exercise 4.3.24 ). Theorem 4.3.19 IfPanconverges absolutely ;thenPanconverges: For example, Theorem 4.3.19 implies that Xsinn/DC2 np converges ifp>1 , since it then converges absolutely (Example 4.3.18 ). The converse of Theorem 4.3.19 is false; a series may converge without converging abso- lutely. We say then that the series converges conditionally , or is conditionally convergent ; thus,P./NUL1/n=npconverges conditionally if 0 < p/DC41. Dirichlet’s Test for Series Except for Theorem 4.3.5 and Corollary 4.3.6 , the convergence tests we have studied so far apply only to series whose terms have the same sign for lar gen. The following theo- rem does not require this. It is analogous to Dirichlet’s tes t for improper integrals (Theo- rem3.4.10 ). Theorem 4.3.20 (Dirichlet’s Test for Series) The seriesP1 nDkanbncon- verges if limn!1anD0;X janC1/NULanj<1; (4.3.18) and jbkCbkC1C/SOH/SOH/SOHCbnj/DC4M; n/NAKk; (4.3.19) for some constant M: Proof The proof is similar to the proof of Dirichlet’s test for inte grals. Define BnDbkCbkC1C/SOH/SOH/SOHCbn; n/NAKk and consider the partial sums ofP1 nDkanbn: SnDakbkCakC1bkC1C/SOH/SOH/SOHCanbn; n/NAKk: (4.3.20) By substituting bkDBkandbnDBn/NULBn/NUL1; n/NAKkC1; into ( 4.3.20 ), we obtain SnDakBkCakC1.BkC1/NULBk/C/SOH/SOH/SOHCan.Bn/NULBn/NUL1/; which we rewrite as SnD.ak/NULakC1/BkC.akC1/NULakC2/BkC1C/SOH/SOH/SOH C.an/NUL1/NULan/Bn/NUL1CanBn:(4.3.21) 218 Chapter 4 Infinite Sequences and Series (The procedure that led from ( 4.3.20 ) to ( 4.3.21 ) is called summation by parts . It is analo- gous to integration by parts.) Now ( 4.3.21 ) can be viewed as SnDTn/NUL1CanBn; (4.3.22) where Tn/NUL1D.ak/NULakC1/BkC.akC1/NULakC2/BkC1C/SOH/SOH/SOHC.an/NUL1/NULan/Bn/NUL1I that is,fTngis the sequence of partial sums of the series 1X jDk.aj/NULajC1/Bj: (4.3.23) Since j.aj/NULajC1/Bjj/DC4Mjaj/NULajC1j from ( 4.3.19 ), the comparison test and ( 4.3.18 ) imply that the series ( 4.3.23 ) converges absolutely. Theorem 4.3.19 now implies thatfTngconverges. Let TDlimn!1Tn. Since fBngis bounded and lim n!1anD0, we infer from ( 4.3.22 ) that lim n!1SnDlim n!1Tn/NUL1Clim n!1anBnDTC0DT: Therefore,Panbnconverges. Example 4.3.21 To apply Dirichlet’s test to 1X nD2sinn/DC2 nC./NUL1/n; /DC2¤k/EM (kDinteger); we take anD1 nC./NUL1/nandbnDsinn/DC2: Then lim n!1anD0, and janC1/NULanj<3 n.n/NUL1/ (verify), soX janC1/NULanj<1: Now BnDsin2/DC2Csin3/DC2C/SOH/SOH/SOHC sinn/DC2: To show thatfBngis bounded, we use the trigonometric identity sinr/DC2Dcos/NULr/NUL1 2/SOH/DC2/NULcos/NULrC1 2/SOH/DC2 2sin./DC2=2/; /DC2¤2k/EM; Section 4.3 Infinite Series of Constants 219 to write BnD.cos3 2/DC2/NULcos5 2/DC2/C.cos5 2/DC2/NULcos7 2/DC2/C/SOH/SOH/SOHC/NUL cos/NUL n/NUL1 2/SOH /DC2/NULcos.nC1 2//DC2/SOH 2sin./DC2=2/ Dcos3 2/DC2/NULcos.nC1 2//DC2 2sin./DC2=2/; which implies that jBnj/DC4ˇˇˇˇ1 sin./DC2=2/ˇˇˇˇ; n/NAK2: Sincefangandfbngsatisfy the hypotheses of Dirichlet’s theorem,Panbnconverges. Dirichlet’s test takes a simpler form if fangis nonincreasing, as follows. Corollary 4.3.21 ( Abel’s Test) The seriesPanbnconverges ifanC1/DC4anfor n/NAKk;limn!1anD0;and jbkCbkC1C/SOH/SOH/SOHCbnj/DC4M; n/NAKk; for some constant M: Proof IfanC1/DC4an, then mX nDkjanC1/NULanjDmX nDk.an/NULanC1/Dak/NULamC1: Since lim m!1amC1D0, it follows that 1X nDkjanC1/NULanjDak<1: Therefore, the hypotheses of Dirichlet’s test are satisfied , soPanbnconverges. Example 4.3.22 The seriesXsinn/DC2 np; which we know is convergent if p > 1 (Example 4.3.18 ), also converges if 0 < p/DC41. This follows from Abel’s test, with anD1=npandbnDsinn/DC2(see Example 4.3.21 ). The alternating series test from calculus follows easily fr om Abel’s test. Corollary 4.3.22 (Alternating Series Test) The seriesP./NUL1/nanconverges if0/DC4anC1/DC4anandlimn!1anD0: 220 Chapter 4 Infinite Sequences and Series Proof LetbnD./NUL1/n; thenfjBnjgis a sequence of zeros and ones and therefore bounded. The conclusion now follows from Abel’s test. Grouping Terms in a Series The terms of a finite sum can be grouped by inserting parenthes es arbitrarily. For example, .1C7/C.6C5/C4D.1C7C6/C.5C4/D.1C7/C.6C5C4/: According to the next theorem, the same is true of an infinite s eries that converges or diverges to˙1. Theorem 4.3.23 Suppose thatP1 nDkanDA;where/NUL1/DC4A/DC41:Letfnjg1 1be an increasing sequence of integers, with n1/NAKk. Define b1DakC/SOH/SOH/SOHCan1; b2Dan1C1C/SOH/SOH/SOHCan2; ::: brDanr/NUL1C1C/SOH/SOH/SOHCanr: Then1X jD1bnjDA: Proof IfTris therth partial sum ofP1 jD1bnjandfAngis thenth partial sum ofP1 sDkas, then TrDb1Cb2C/SOH/SOH/SOHCbr D.a1C/SOH/SOH/SOHCan1/C.an1C1C/SOH/SOH/SOHCan2/C/SOH/SOH/SOHC.anr/NUL1C1C/SOH/SOH/SOHCanr/ DAnr: Thus,fTrgis a subsequence offAng, so lim r!1TrDlimn!1AnDAby Theorem 4.2.2 . Example 4.3.23 IfP1 nD0./NUL1/nansatisfies the hypotheses of the alternating series test and converges to the sum S, Theorem 4.3.23 enables us to write SDkX nD0./NUL1/nanC./NUL1/kC11X jD1.akC2j/NUL1/NULakC2j/ and SDkX nD0./NUL1/nanC./NUL1/kC12 4akC1/NUL1X jD1.akC2j/NULakC2j/NUL1/3 5: Since0/DC4anC1/DC4an, these two equations imply that S/NULSkis between0and./NUL1/k/NUL1akC1. Section 4.3 Infinite Series of Constants 221 Example 4.3.24 Introducing parentheses in some divergent series can yield seem- ingly contradictory results. For example, it is tempting to write 1X nD1./NUL1/nC1D.1/NUL1/C.1/NUL1/C/SOH/SOH/SOHD0C0C/SOH/SOH/SOH and conclude thatP1 nD1./NUL1/nD0, but equally tempting to write 1X nD1./NUL1/nC1D1/NUL.1/NUL1//NUL.1/NUL1//NUL/SOH/SOH/SOH D1/NUL0/NUL0/NUL/SOH/SOH/SOH and conclude thatP1 nD1./NUL1/nC1D1. Of course, there is no contradiction here, since Theorem 4.3.23 does not apply to this series, and neither of these operation s is legitimate. Rearrangement of Series A finite sum is not changed by rearranging its terms; thus, 1C3C7D1C7C3D3C1C7D3C7C1D7C1C3D7C3C1: This is not true of all infinite series. Let us say thatPbnis arearrangement ofPanif the two series have the same terms, written in possibly diffe rent orders. Since the partial sums of the two series may form entirely different sequences , there is no apparent reason to expect them to exhibit the same convergence properties, a nd in general they do not. We are interested in what happens if we rearrange the terms of a convergent series. We will see that every rearrangement of an absolutely converge nt series has the same sum, but that conditionally convergent series fail, spectacularly , to have this property. Theorem 4.3.24 IfP1 nD1bnis a rearrangement of an absolutely convergent seriesP1 nD1an;thenP1 nD1bnalso converges absolutely ;and to the same sum : Proof Let AnDja1jCja2jC/SOH/SOH/SOHCjanjandBnDjb1jCjb2jC/SOH/SOH/SOHCjbnj: For eachn/NAK1, there is an integer knsuch thatb1,b2, . . . ,bnare included among a1,a2, . . . ,akn, soBn/DC4Akn. SincefAngis bounded, so isfBng, and thereforePjbnj<1 (Theorem 4.3.8 ). Now let AnDa1Ca2C/SOH/SOH/SOHCan; B nDb1Cb2C/SOH/SOH/SOHCbn; AD1X nD1an;andBD1X nD1bn: 222 Chapter 4 Infinite Sequences and Series We must show that ADB. Suppose that /SI>0 . From Cauchy’s convergence criterion for series and the absolute convergence ofPan, there is an integer Nsuch that jaNC1jCjaNC2jC/SOH/SOH/SOHCjaNCkj</SI; k/NAK1: ChooseN1so thata1,a2, . . . ,aNare included among b1,b2, . . . ,bN1. Ifn/NAKN1, thenAnandBnboth include the terms a1,a2, . . . ,aN, which cancel on subtraction; thus, jAn/NULBnjis dominated by the sum of the absolute values of finitely many terms fromPan with subscripts greater than N. Since every such sum is less than /SI, jAn/NULBnj</SI ifn/NAKN1: Therefore, lim n!1.An/NULBn/D0andADB. To investigate the consequences of rearranging a condition ally convergent series, we need the next theorem, which is itself important. Theorem 4.3.25 IfPDfanig1 1andQDfamjg1 1are respectively the subsequences of all positive and negative terms in a conditionally conver gent seriesPan;then 1X iD1aniD1 and1X jD1amjD/NUL1: (4.3.24) Proof If both series in ( 4.3.24 ) converge, thenPanconverges absolutely, while if one converges and the other diverges, thenPandiverges to1or/NUL1. Hence, both must diverge. The next theorem implies that a conditionally convergent se ries can be rearranged to produce a series that converges to any given number, diverge s to˙1, or oscillates. Theorem 4.3.26 Suppose thatP1 nD1anis conditionally convergent and /SYNand/ETBare arbitrarily given in the extended reals ;with/SYN/DC4/ETB:Then the terms ofP1 nD1ancan be rearranged to form a seriesP1 nD1bnwith partial sums BnDb1Cb2C/SOH/SOH/SOHCbn; n/NAK1; such that lim n!1BnD/ETBand lim n!1BnD/SYN: (4.3.25) Proof We consider the case where /SYNand/ETBare finite and leave the other cases to you (Exercise 4.3.36 ). We may ignore any zero terms that occur inP1 nD1an. For convenience, we denote the positive terms by PDf˛ig1 1and and the negative terms by QDf/NULˇjg1 1. We construct the sequence fbng1 1Df˛1;:::;˛ m1;/NULˇ1;:::;/NULˇn1;˛m1C1;:::;˛ m2;/NULˇn1C1;:::;/NULˇn2;:::g; (4.3.26) Section 4.3 Infinite Series of Constants 223 with segments chosen alternately from PandQ. Letm0Dn0D0. Ifk/NAK1, letmkand nkbe the smallest integers such that mk>m k/NUL1,nk>n k/NUL1, mkX iD1˛i/NULnk/NUL1X jD1ˇj/NAK/ETB; andmkX iD1˛i/NULnkX jD1ˇj/DC4/SYN: Theorem 4.3.25 implies that this construction is possible: sinceP˛iDPˇjD1 , we can choosemkandnkso that mkX iDmk/NUL1˛iandnkX jDnk/NUL1ˇj are as large as we please, no matter how large mk/NUL1andnk/NUL1are (Exercise 4.3.23 ). Since mkandnkare the smallest integers with the specified properties, /ETB/DC4BmkCnk/NUL1</ETBC˛mk; k/NAK2; (4.3.27) and /SYN/NULˇnk<B mkCnk/DC4/SYN; k/NAK2: (4.3.28) From ( 4.3.26 ),bn<0ifmkCnk/NUL1<n/DC4mkCnk, so BmkCnk/DC4Bn/DC4BmkCnk/NUL1; m kCnk/NUL1/DC4n/DC4mkCnk; (4.3.29) whilebn>0ifmkCnk<n/DC4mkC1Cnk, so BmkCnk/DC4Bn/DC4BmkC1Cnk; m kCnk/DC4n/DC4mkC1Cnk: (4.3.30) Because of ( 4.3.27 ) and ( 4.3.28 ), (4.3.29 ) and ( 4.3.30 ) imply that /SYN/NULˇnk<B n</ETBC˛mk; m kCnk/NUL1/DC4n/DC4mkCnk; (4.3.31) and /SYN/NULˇnk<B n</ETBC˛mkC1; m kCnk/DC4n/DC4mkC1Cnk: (4.3.32) From the first inequality of ( 4.3.27 ),Bn/NAK/ETBfor infinitely many values of n. However, since lim i!1˛iD0, the second inequalities in ( 4.3.31 ) and ( 4.3.32 ) imply that if /SI >0 thenBn> /ETBC/SIfor only finitely many values of n. Therefore, limn!1BnD/ETB. From the second inequality in ( 4.3.28 ),Bn/DC4/SYNfor infinitely many values of n. However, since limj!1ˇjD0, the first inequalities in ( 4.3.31 ) and ( 4.3.32 ) imply that if /SI > 0 then Bn</SYN/NUL/SIfor only finitely many values of n. Therefore, limn!1BnD/SYN. Multiplication of Series The product of two finite sums can be written as another finite s um: for example, .a0Ca1Ca2/.b0Cb1Cb2/Da0b0Ca0b1Ca0b2 Ca1b0Ca1b1Ca1b2 Ca2b0Ca2b1Ca2b2; 224 Chapter 4 Infinite Sequences and Series where the sum on the right contains each product aibj.i;jD0;1;2/ exactly once. These products can be rearranged arbitrarily without changing th eir sum. The corresponding situation for series is more complicated. Given two series 1X nD0anand1X nD0bn (because of applications in Section 4.5, it is convenient he re to start the summation index at zero), we can arrange all possible products aibj.i;j/NAK0/in a two-dimensional array: a0b0a0b1a0b2a0b3/SOH/SOH/SOH a1b0a1b1a1b2a1b3/SOH/SOH/SOH a2b0a2b1a2b2a2b3/SOH/SOH/SOH a3b0a3b1a3b2a3b3/SOH/SOH/SOH ::::::::::::(4.3.33) where the subscript on ais constant in each row and the subscript on bis constant in each column. Any sensible definition of the product 1X nD0an! 1X nD0bn! clearly must involve every product in this array exactly onc e; thus, we might define the product of the two series to be the seriesP1 nD0pn, wherefpngis a sequence obtained by ordering the products in ( 4.3.33 ) according to some method that chooses every product exactly once. One way to do this is indicated by a0b0!a0b1a0b2!a0b3/SOH/SOH/SOH # " # a1b0 a1b1a1b2a1b3/SOH/SOH/SOH # " # a2b0!a2b1!a2b2a2b3/SOH/SOH/SOH # a3b0 a3b1 a3b2 a3b3/SOH/SOH/SOH # ::::::::::::(4.3.34) Section 4.3 Infinite Series of Constants 225 and another by a0b0!a0b1a0b2!a0b3a0b4/SOH/SOH/SOH . % . % a1b0a1b1a1b2a1b3/SOH/SOH/SOH # % . % a2b0a2b1a2b2a2b3/SOH/SOH/SOH . % a3b0a3b1a3b2a3b3/SOH/SOH/SOH # % a4b0:::::::::(4.3.35) There are infinitely many others, and to each corresponds a se ries that we might consider to be the product of the given series. This raises a question: If 1X nD0anDAand1X nD0bnDB whereAandBare finite, does every product seriesP1 nD0pnconstructed by ordering the products in ( 4.3.33 ) converge to AB? The next theorem tells us when the answer is yes. Theorem 4.3.27 Let 1X nD0anDAand1X nD0bnDB; whereAandBare finite, and at least one term of each series is nonzero. The nP1 nD0pnD ABfor every sequence fpngobtained by ordering the products in (4.3.33 )if and only ifPanandPbnconverge absolutely :Moreover;in this case,Ppnconverges absolutely : Proof First, letfpngbe the sequence obtained by arranging the products faibjgaccord- ing to the scheme indicated in ( 4.3.34 ), and define AnDa0Ca1C/SOH/SOH/SOHCan;AnDja0jCja1jC/SOH/SOH/SOHCjanj; BnDb0Cb1C/SOH/SOH/SOHCbn;BnDjb0jCjb1jC/SOH/SOH/SOHCjbnj; PnDp0Cp1C/SOH/SOH/SOHCpn;PnDjp0jCjp1jC/SOH/SOH/SOHCjpnj: From ( 4.3.34 ), we see that P0DA0B0; P 3DA1B1; P 8DA2B2; and, in general, P.mC1/2/NUL1DAmBm: (4.3.36) 226 Chapter 4 Infinite Sequences and Series Similarly, P.mC1/2/NUL1DAmBm: (4.3.37) IfPjanj<1andPjbnj<1, thenfAmBmgis bounded and, since Pm/DC4P.mC1/2/NUL1, (4.3.37 ) implies thatfPmgis bounded. Therefore,Pjpnj<1, soPpnconverges. Now 1X nD0pnDlim n!1Pn (by definition) Dlim m!1P.mC1/2/NUL1 (by Theorem 4.2.2 ) Dlim m!1AmBm (from ( 4.3.36 )) D/DLE lim m!1Am/DC1/DLE lim m!1Bm/DC1 (by Theorem 4.1.8 ) DAB: Since any other ordering of the products in ( 4.3.33 ) produces a a rearrangement of the absolutely convergent seriesP1 nD0pn, Theorem 4.3.24 implies thatPjqnj<1for every such ordering and thatP1 nD0qnDAB. This shows that the stated condition is sufficient. For necessity, again letP1 nD0pnbe obtained from the ordering indicated in ( 4.3.34 ), and suppose thatP1 nD0pnand all its rearrangements converge to AB. ThenPpnmust converge absolutely, by Theorem 4.3.26 . Therefore,fPm2/NUL1gis bounded, and ( 4.3.37 ) implies thatfAmgandfBmgare bounded. (Here we need the assumption that neitherPan norPbnconsists entirely of zeros. Why?) Therefore,Pjanj<1andPjbnj<1. The following definition of the product of two series is due to Cauchy. We will see the importance of this definition in Section 4.5. Definition 4.3.28 TheCauchy product ofP1 nD0anandP1 nD0bnisP1 nD0cn, where cnDa0bnCa1bn/NUL1C/SOH/SOH/SOHCan/NUL1b1Canb0: (4.3.38) Thus,cnis the sum of all products aibj, wherei/NAK0,j/NAK0, andiCjDn; thus, cnDnX rD0arbn/NULrDnX rD0bran/NULr: (4.3.39) Henceforth,/NULP1 nD0an/SOH/NULP1 nD0bn/SOHshould be interpreted as the Cauchy product. Notice that 1X nD0an! 1X nD0bn! D 1X nD0bn! 1X nD0an! ; and that the Cauchy product of two series is defined even if one or both diverge. In the case where both converge, it is natural to inquire about the relat ionship between the product of their sums and the sum of the Cauchy product. Theorem 4.3.27 yields a partial answer to this question, as follows. Section 4.3 Infinite Series of Constants 227 Theorem 4.3.29 IfP1 nD0anandP1 nD0bnconverge absolutely to sums AandB; then the Cauchy product ofP1 nD0anandP1 nD0bnconverges absolutely to AB: Proof LetCnbe thenth partial sum of the Cauchy product; that is, CnDc0Cc1C/SOH/SOH/SOHCcn (see ( 4.3.38 )). LetP1 nD0pnbe the series obtained by ordering the products fai;bjgac- cording to the scheme indicated in ( 4.3.35 ), and definePnto be itsnth partial sum; thus, PnDp0Cp1C/SOH/SOH/SOHCpn: Inspection of ( 4.3.35 ) shows thatcnis the sum of the nC1terms connected by the diagonal arrows. Therefore, CnDPmn, where mnD1C2C/SOH/SOH/SOHC.nC1//NUL1Dn.nC3/ 2: From Theorem 4.3.27 , lim n!1PmnDAB, so lim n!1CnDAB. To see thatPjcnj< 1, we observe that nX rD0jcrj/DC4mnX sD0jpsj and recall thatPjpsj<1, from Theorem 4.3.27 . Example 4.3.25 Consider the Cauchy product ofP1 nD0rnwith itself. Here anD bnDrnand ( 4.3.39 ) yields cnDr0rnCr1rn/NUL1C/SOH/SOH/SOHCrn/NUL1r1Crnr0D.nC1/rn; so 1X nD0rn!2 D1X nD0.nC1/rn: Since1X nD0rnD1 1/NULr;jrj<1; and the convergence is absolute, Theorem 4.3.29 implies that 1X nD0.nC1/rnD1 .1/NULr/2;jrj<1: Example 4.3.26 If 1X nD0anD1X nD0˛n nŠand1X nD0bnD1X nD0ˇn nŠ; 228 Chapter 4 Infinite Sequences and Series then ( 4.3.39 ) yields cnDnX mD0˛n/NULmˇm .n/NULm/ŠmŠD1 nŠnX mD0 n m! ˛n/NULmˇmD.˛Cˇ/n nŠI thus, 1X nD0˛n nŠ! 1X nD0ˇn nŠ! D1X nD0.˛Cˇ/n nŠ: (4.3.40) You probably know from calculus thatP1 nD0xn=nŠconverges absolutely for all xtoex. Thus, ( 4.3.40 ) implies that e˛eˇDe˛Cˇ; a familiar result. The Cauchy product of two series may converge under conditio ns weaker than those of Theorem 4.3.29 . If one series converges absolutely and the other converges condi- tionally, the Cauchy product of the two series converges to t he product of the two sums (Exercise 4.3.40 ). If two series and their Cauchy product all converge, then t he sum of the Cauchy product equals the product of the sums of the two se ries (Exercise 4.5.32 ). However, the next example shows that the Cauchy product of tw o conditionally convergent series may diverge. Example 4.3.27 If anDbnD./NUL1/nC1 p nC1; thenP1 nD0anandP1 nD0bnconverge conditionally. From ( 4.3.39 ), the general term of their Cauchy product is cnDnX rD0./NUL1/rC1./NUL1/n/NULrC1 prC1pn/NULrC1D./NUL1/nnX rD01prC11pn/NULrC1; so jcnj/NAKnX rD01pnC11pnC1DnC1 nC1D1: Therefore, the Cauchy product diverges, by Corollary 4.3.6 . 4.3 Exercises 1. Prove Theorem 4.3.2 . 2. Prove Theorem 4.3.3 . 3. (a) Prove: IfanDbnexcept for finitely many values of n, thenPanandPbn converge or diverge together. Section 4.3 Infinite Series of Constants 229 (b) LetbnkDakfor some increasing sequence fnkg1 1of positive integers, and bnD0ifnis any other positive integer. Show that 1X nD1bnand1X nD1an diverge or converge together, and that in the latter case the y have the same sum. (Thus, the convergence properties of a series are not change d by inserting zeros between its terms.) 4. (a) Prove: IfPanconverges, then lim n!1.anCanC1C/SOH/SOH/SOHCanCr/D0; r/NAK0: (b) Does(a)imply thatPanconverges? Give a reason for your answer. 5. Prove Corollary 4.3.7 . 6. (a) Verify Corollary 4.3.7 for the convergent seriesP1=np.p >1/ . HINT:See the proof of Theorem 4.3.10: (b) Verify Corollary 4.3.7 for the convergent seriesP./NUL1/n=n. 7. Prove: If0/DC4bn/DC4an/DC4bnC1, thenPanandPbnconverge or diverge together. 8. Determine convergence or divergence. (a)Xp n2/NUL1p n5C1(b)X 1 n2/STX1C1 2sin.n/EM=4//ETX (c)X1/NULe/NULnlogn n(d)X cos/EM n2 (e)X sin/EM n2(f)X1 ntan/EM n (g)X1 ncot/EM n(h)Xlogn n2 9. Suppose that f.x//NAK0forx/NAKk. Prove thatR1 kf.x/dx<1if and only if 1X nDkZnC1 nf.x/dx<1: HINT:Use Theorems 3.4.5 and4.3.8 . 10. Use the integral test to find all values of pfor which the series converges. (a)Xn .n2/NUL1/p(b)Xn2 .n3C4/p(c)X sinhn .coshn/p 230 Chapter 4 Infinite Sequences and Series 11. LetLnbe thenth iterated logarithm. Show that X 1 L0.n/L 1.n//SOH/SOH/SOHLk.n/ŒL kC1.n//c141p converges if and only if p>1 . HINT:See Exercise 3.4.10 . 12. Suppose that g,g0, and.g0/2/NULgg00are all positive on ŒR;1/. Show that Xg0.n/ g.n/<1 if and only if lim x!1g.x/<1. 13. Let S.p/D1X nD11 np; p>1: Show that 1 .p/NUL1/.NC1/p/NUL1<S.p//NULNX nD11 np<1 .p/NUL1/Np/NUL1: HINT:See the proof of Theorem 4.3.10 . 14. Suppose that fis positive, decreasing, and locally integrable on Œ1;1/c141, and let anDnX kD1f.k//NULZn 1f.x/dx: (a) Show thatfangis nonincreasing and nonnegative, and 0< lim n!1an<f.1/: (b) Deduce from (a)that /CRDlim n!1/DC2 1C1 2C1 3C/SOH/SOH/SOHC1 n/NULlogn/DC3 exists, and0</CR <1 . (/CRisEuler ’s constant;/CR/EM0:577 .) 15. Determine convergence or divergence. (a)X2Csinn/DC2 n2Csinn/DC2(b)XnC1 nrn.r >0/ (c)X e/NULn/SUBcoshn/SUB./SUB>0/ (d)XnClogn n2.logn/2 (e)XnClogn n2logn(f)X.1C1=n/n 2n Section 4.3 Infinite Series of Constants 231 16. LetLnbe thenth iterated logarithm. Prove that X 1 ŒL0.n//c141q0C1ŒL1.n//c141q1C1/SOH/SOH/SOHŒLm.n//c141qmC1 converges if and only if there is at least one nonzero number i nfq0;q1;:::;q mgand the first such is positive. H INT:See Exercises 4.3.11 and2.4.42.b/: 17. Determine convergence or divergence. (a)X2Csin2.n/EM=4/ 3n(b)Xn.nC1/ 4n (c)X3/NULsin.n/EM=2/ n.nC1/(d)XnC./NUL1/n n.nC1/ 18. Determine convergence or divergence, with r >0 . (a)XnŠ rn(b)X nprn(c)Xrn nŠ (d)Xr2nC1 .2nC1/Š(e)Xr2n .2n/Š 19. Determine convergence or divergence. (a)X.2n/Š 22n.nŠ/2(b)X.3n/Š 33nnŠ.nC1/Š.nC3/Š (c)X2nnŠ 5/SOH/SOH/SOH7/SOH.2nC3/(d)X˛.˛C1//SOH/SOH/SOH.˛Cn/NUL1/ ˇ.ˇC1//SOH/SOH/SOH.ˇCn/NUL1/.˛;ˇ>0/ 20. Determine convergence or divergence. (a)Xnn.2C./NUL1/n/ 2n(b)X/DC21Csin3n/DC2 3/DC3n (c)X .nC1//DC21Csin.n/EM=6/ 3/DC3n (d)X/DC2 1/NUL1 n/DC3n2 21. Give counterexamples showing that the following statement s are false unless it is assumed that the terms of the series have the same sign for nsufficiently large. (a)Panconverges if its partial sums are bounded. (b) Ifbn¤0forn/NAKkand lim n!1an=bnDL, where0<L<1, thenPan andPbnconverge or diverge together. (c) Ifan¤0andlimn!1anC1=an<1, thenPanconverges. (d) Ifan¤0andlimn!1nŒ.a nC1=an//NUL1/c141</NUL1, thenPanconverges. 22. Prove: If the terms of a convergent seriesPanhave the same sign for n/NAKk, thenPanconverges absolutely. 23. Suppose that an/NAK0forn/NAKmandPanD1 . Prove: IfNis an arbitrary integer /NAKmandJis an arbitrary positive number, thenPNCk nDNan> J for some positive integerk. 24. Prove Theorem 4.3.19 . 232 Chapter 4 Infinite Sequences and Series 25. Show that the series converges absolutely. (a)X ./NUL1/n1 n.logn/2(b)Xsinn/DC2 2n (c)X ./NUL1/n1pnsin/EM n(d)X cosn/DC2p n3/NUL1 26. Show that the series converges. (a)Xnsinn/DC2 n2C./NUL1/n./NUL1</DC2 <1/(b)Xcosn/DC2 n./DC2¤2k/EM;kDinteger/ 27. Determine whether the series is absolutely convergent, con ditionally convergent, or divergent. (a)Xbnpn.b4mDb4mC1D1; b 4mC2Db4mC3D/NUL1/ (b)X1 nsinn/EM 6(c)X1 n2cosn/EM 7 (d)X1/SOH3/SOH5/SOH/SOH/SOH.2nC1/ 4/SOH6/SOH8/SOH/SOH/SOH.2nC4/sinn/DC2 28. Letgbe a rational function (ratio of two polynomials). Show thatPg.n/rncon- verges absolutely if jrj< 1 or diverges ifjrj> 1. Discuss the possibilities for jrjD1. 29. Prove: IfPa2 n<1andPb2 n<1, thenPanbnconverges absolutely. 30. (a) Prove: IfPanconverges andPa2 nD1 , thenPanconverges condition- ally. (b) Give an example of a series with the properties described in (a). 31. Suppose that 0/DC4anC1<a nand lim n!1b1Cb2C/SOH/SOH/SOHCbn wn>0; wherefwngis a sequence of positive numbers such that X wn.an/NULanC1/D1: Show thatPanbnD1 . HINT:Use summation by parts. 32. (a) Prove: If0<2/SI</DC2 </EM/NUL2/SI, then lim n!1jsin/DC2jCj sin2/DC2jC/SOH/SOH/SOHCj sinn/DC2j n/NAKsin/SI 2: HINT:Show thatjsinn/DC2j>sin/SIat least “half the time”; more precisely, show that ifjsinm/DC2j/DC4sin/SIfor some integer mthenjsin.mC1//DC2j>sin/SI. Section 4.3 Infinite Series of Constants 233 (b) Show that Xsinn/DC2 np converges conditionally if 0<p/DC41and/DC2¤k/EM(kDinteger). H INT:Use Exercise 4.3.31 and see Example 4.3.22 . 33. Show that1X nD1./NUL1/nC1 nD1 21X nD11 n.2n/NUL1/: 34. Letb3mC1,b3mC2D/NUL2, andb3mC3D1form/NAK0. Show that 1X nD1bn nD2 31X mD01 .mC1/.3mC1/.3mC2/: 35. LetPbnbe obtained by rearranging finitely many terms of a convergen t seriesPan. Show that the two series have the same sum. 36. Prove Theorem 4.3.26 for the case where (a)/SYNis finite and/ETBD1 ;(b)/SYND/NUL1 and/ETBD1 ;(c)/SYND/ETBD1 . 37. Give necessary and sufficient conditions for a divergent ser ies to have a convergent rearrangement. 38. A series diverges unconditionally to1if every rearrangement of the series diverges to1. State necessary and sufficient conditions for a series to ha ve this property. 39. Suppose that fandghave derivatives of all orders at 0, and lethDfg. Show formally that 1X nD0f.n/.0/ nŠxn! 1X nD0g.n/.0/ nŠxn! D1X nD0h.n/.0/ nŠxn in the sense of the Cauchy product. H INT:See Exercise 2.3.12 . 40. Prove: IfPjanj<1andPbnconverges (perhaps conditionally), withP1 nD0anD AandP1 nD0bnDB, then the Cauchy product 1X nD0cnD 1X nD0an! 1X nD0bn! converges toAB. HINT:LetfAng,fBng, andfCngbe the partial sums of the series. Show that Cn/NULAnBDnX rD0ar.Bn/NULr/NULB/ and apply Theorem 4.3.5 toPjanj. 234 Chapter 4 Infinite Sequences and Series 41. Suppose that ar/NAK0for allr/NAK0and andP1 0arDA<1. Show that lim n!11 nn/NUL1X r;sD0arCsD0and lim n!11 nn/NUL1X r;sD0ar/NULsD2A/NULa0: 42. Prove: If lim i!1a.i/ jDaj(j/NAK1) andja.i/ jj/DC4/ESCj(i;j/NAK1), whereP1 jD1/ESCj< 1, then lim i!1P1 jD1a.i/ jDP1 jD1aj. 43. Prove: Ifan>0,n/NAK1, andP1 nD1anD1 , thenP1 nD1an=.1Can/D1 . 4.4 SEQUENCES AND SERIES OF FUNCTIONS Until now we have considered sequences and series of constan ts. Now we turn our attention to sequences and series of real-valued functions defined on s ubsets of the reals. Throughout this section, “subset” means “nonempty subset.” IfFk,FkC1, . . . ,Fn;::: are real-valued functions defined on a subset Dof the reals, we say thatfFngis an infinite sequence or (simply a sequence )of functions on D. If the sequence of valuesfFn.x/gconverges for each xin some subset SofD, thenfFngdefines a limit function on S. The formal definition is as follows. Definition 4.4.1 Suppose thatfFngis a sequence of functions on Dand the sequence of valuesfFn.x/gconverges for each xin some subset SofD. Then we say that fFng converges pointwise on Sto the limit function F, defined by F.x/Dlim n!1Fn.x/; x2S: Example 4.4.1 The functions Fn.x/D/DC2 1/NULnx nC1/DC3n=2 ; n/NAK1; define a sequence on DD./NUL1;1/c141, and lim n!1Fn.x/D8 < :1; x<0; 1; xD0; 0; 0<x/DC41: Therefore,fFngconverges pointwise on SDŒ0;1/c141 to the limit function Fdefined by F.x/D/SUB1; xD0; 0; 0<x/DC41: Example 4.4.2 Consider the functions Fn.x/Dxne/NULnx; x/NAK0; n/NAK1; (Figure 4.4.1 ). Section 4.4 Sequences and Series of Functions 235 y xy =Fn(x)=xne−nxy = e−n Figure 4.4.1 Equating the derivative F0 n.x/Dnxn/NUL1e/NULnx.1/NULx/ to zero shows that the maximum value of Fn.x/onŒ0;1/ise/NULn, attained at xD1. Therefore, jFn.x/j/DC4e/NULn; x/NAK0; so lim n!1Fn.x/D0for allx/NAK0. The limit function in this case is identically zero on Œ0;1/. Example 4.4.3 Forn/NAK1, letFnbe defined on ./NUL1;1/by Fn.x/D8 ˆˆˆˆˆˆˆˆˆˆ< ˆˆˆˆˆˆˆˆˆˆ:0; x< /NUL2 n; /NULn.2Cnx/;/NUL2 n/DC4x</NUL1 n; n2x;/NUL1 n/DC4x<1 n; n.2/NULnx/;1 n/DC4x<2 n; 0; x/NAK2 n (Figure 4.4.2 , page 236), SinceFn.0/D0for alln, lim n!1Fn.0/D0. Ifx¤0, thenFn.x/D0ifn/NAK2=jxj. Therefore, lim n!1Fn.x/D0;/NUL1<x<1; so the limit function is identically zero on ./NUL1;1/. Example 4.4.4 For each positive integer n, letSnbe the set of numbers of the form xDp=q, wherepandqare integers with no common factors and 1/DC4q/DC4n. Define Fn.x/D/SUB1; x2Sn; 0; x62Sn: 236 Chapter 4 Infinite Sequences and Series Ifxis irrational, then x62Snfor anyn, soFn.x/D0,n/NAK1. Ifxis rational, then x2Sn andFn.x/D1for all sufficiently large n. Therefore, lim n!1Fn.x/DF.x/D/SUB1ifxis rational; 0ifxis irrational: y x y = −ny = n n1n1 n2n2y =Fn(x) − − Figure 4.4.2 Uniform Convergence The pointwise limit of a sequence of functions may differ rad ically from the functions in the sequence. In Example 4.4.1 , eachFnis continuous on ./NUL1;1/c141, butFis not. In Example 4.4.3 , the graph of each Fnhas two triangular spikes with heights that tend to 1asn!1 , while the graph of F(thex-axis) has none. In Example 4.4.4 , eachFn is integrable, while Fis nonintegrable on every finite interval. (Exercise 4.4.3 ). There is nothing in Definition 4.4.1 to preclude these apparent anomalies; although the definiti on implies that for each x0inS,Fn.x0/approximates F.x 0/ifnis sufficiently large, it does not imply that any particular Fnapproximates Fwell over allofS. To formulate a definition that does, it is convenient to introduce the notat ion kgkSDsup x2Sjg.x/j and to state the following lemma. We leave the proof to you (Ex ercise 4.4.4 ). Lemma 4.4.2 Ifgandhare defined on S;then kgChkS/DC4kgkSCkhkS and kghkS/DC4kgkSkhkS: Moroever;if eithergorhis bounded on S;then kg/NULhkS/NAKjkgkS/NULkhkSkj: Section 4.4 Sequences and Series of Functions 237 Definition 4.4.3 A sequencefFngof functions defined on a set Sconverges uniformly to the limit function FonSif lim n!1jjFn/NULFkSD0: Thus,fFngconverges uniformly to FonSif for each/SI>0 there is an integer Nsuch that kFn/NULFkS</SI ifn/NAKN: (4.4.1) IfSDŒa;b/c141 andFis the function with graph shown in Figure 4.4.3 , then ( 4.4.1 ) implies that the graph of yDFn.x/; a/DC4x/DC4b; lies in the shaded band F.x//NUL/SI<y<F.x/C/SI; a/DC4x/DC4b; ifn/NAKN. From Definition 4.4.3 , iffFngconverges uniformly on S, thenfFngconverges uniformly on any subset of S(Exercise 4.4.6 ). y xa by =F(x) −y =F(x) + y =F(x) Figure 4.4.3 Example 4.4.5 The sequencefFngdefined by Fn.x/Dxne/NULnx; n/NAK1; converges uniformly to F/DC10(that is, to the identically zero function) on SDŒ0;1/, since we saw in Example 4.4.2 that kFn/NULFkSDkFnkSDe/NULn; 238 Chapter 4 Infinite Sequences and Series so kFn/NULFkS</SI ifn>/NULlog/SI. For these values of n, the graph of yDFn.x/; 0/DC4x<1; lies in the strip /NUL/SI/DC4y/DC4/SI; x/NAK0 (Figure 4.4.4 ). The next theorem provides alternative definitions of pointw ise and uniform convergence. It follows immediately from Definitions 4.4.1 and4.4.3 . Theorem 4.4.4 LetfFngbe defined on S:Then (a)fFngconverges pointwise to FonSif and only if there is, for each /SI>0 andx2S, an integerN .which may depend on xas well as/SI/such that jFn.x//NULF.x/j</SI ifn/NAKN: (b)fFngconverges uniformly to FonSif and only if there is for each /SI>0 an integer N .which depends only on /SIand not on any particular xinS/such that jFn.x//NULF.x/j</SI for allxinSifn/NAKN: y xy = e−ny = e y = −ey =xne−nx Figure 4.4.4 The next theorem follows immediately from Theorem 4.4.4 and Example 4.4.6 . Section 4.4 Sequences and Series of Functions 239 Theorem 4.4.5 IffFngconverges uniformly to FonS;thenfFngconverges pointwise toFonS:The converse is false Ithat is;pointwise convergence does not imply uniform convergence. Example 4.4.6 The sequencefFngof Example 4.4.3 converges pointwise to F/DC10 on./NUL1;1/, but not uniformly, since kFn/NULFk./NUL1;1/DFn/DC21 n/DC3 DˇˇˇˇFn/DC2/NUL1 n/DC3ˇˇˇˇDn; so lim n!1kFn/NULFk./NUL1;1/D1: However, the convergence is uniform on S/SUBD./NUL1;/SUB/c141[Œ/SUB;1/ for any/SUB>0 , since kFn/NULFkS/SUBD0ifn>2 /SUB: Example 4.4.7 IfFn.x/Dxn,n/NAK1, thenfFngconverges pointwise on SDŒ0;1/c141 to F.x/D/SUB1; xD1; 0; 0/DC4x<1: The convergence is not uniform on S. To see this, suppose that 0</SI<1 . Then jFn.x//NULF.x/j>1/NUL/SIif.1/NUL/SI/1=n<x<1: Therefore, 1/NUL/SI/DC4kFn/NULFkS/DC41 for alln/NAK1. Since/SIcan be arbitrarily small, it follows that kFn/NULFkSD1 for alln/NAK1. However, the convergence is uniform on Œ0;/SUB/c141 if0</SUB<1 , since then kFn/NULFkŒ0;/SUB/c141D/SUBn and lim n!1/SUBnD0. Another way to say the same thing: fFngconverges uniformly on every closed subset of Œ0;1/ . The next theorem enables us to test a sequence for uniform con vergence without guessing what the limit function might be. It is analogous to Cauchy’s convergence criterion for sequences of constants (Theorem 4.1.13 ). 240 Chapter 4 Infinite Sequences and Series Theorem 4.4.6 (Cauchy’s Uniform Convergence Criterion) A sequence of functionsfFngconverges uniformly on a set Sif and only if for each /SI >0 there is an integerNsuch that kFn/NULFmkS</SI ifn;m/NAKN: (4.4.2) Proof For necessity, suppose that fFngconverges uniformly to FonS. Then, if/SI>0 , there is an integer Nsuch that kFk/NULFkS</SI 2ifk/NAKN: Therefore, kFn/NULFmkSDk.Fn/NULF/C.F/NULFm/kS /DC4kFn/NULFkSCkF/NULFmkS(Lemma 4.4.2 ) </SI 2C/SI 2D/SIifm;n/NAKN: For sufficiency, we first observe that ( 4.4.2 ) implies that jFn.x//NULFm.x/j</SI ifn;m/NAKN; for any fixed xinS. Therefore, Cauchy’s convergence criterion for sequences of constants (Theorem 4.1.13 ) implies thatfFn.x/gconverges for each xinS; that is,fFngconverges pointwise to a limit function FonS. To see that the convergence is uniform, we write jFm.x//NULF.x/jDjŒFm.x//NULFn.x//c141CŒFn.x//NULF.x//c141j /DC4jFm.x//NULFn.x/jCjFn.x//NULF.x/j /DC4kFm/NULFnkSCjFn.x//NULF.x/j: This and ( 4.4.2 ) imply that jFm.x//NULF.x/j</SICjFn.x//NULF.x/jifn;m/NAKN: (4.4.3) Since lim n!1Fn.x/DF.x/ , jFn.x//NULF.x/j</SI for somen/NAKN, so ( 4.4.3 ) implies that jFm.x//NULF.x/j<2/SI ifm/NAKN: But this inequality holds for all xinS, so kFm/NULFkS/DC42/SI ifm/NAKN: Since/SIis an arbitrary positive number, this implies that fFngconverges uniformly to F onS. The next example is similar to Example 4.1.14 . Section 4.4 Sequences and Series of Functions 241 Example 4.4.8 Suppose that gis differentiable on SD./NUL1;1/and jg0.x/j/DC4r <1;/NUL1<x<1: (4.4.4) LetF0be bounded on Sand define Fn.x/Dg.F n/NUL1.x//; n/NAK1: (4.4.5) We will show thatfFngconverges uniformly on S. We first note that if uandvare any two real numbers, then ( 4.4.4 ) and the mean value theorem imply that jg.u//NULg.v/j/DC4rju/NULvj: (4.4.6) Recalling ( 4.4.5 ) and applying this inequality with uDFn/NUL1.x/andvD0shows that jFn.x/jDjg.0/C.g.F n/NUL1.x///NULg.0//j/DC4jg.0/jCjg.F n/NUL1.x///NULg.0/j /DC4jg.0/jCrjFn/NUL1.x/jI therefore, since F0is bounded on S, it follows by induction that Fnis bounded on Sfor n/NAK1. Moreover, if n/NAK1, then ( 4.4.5 ) and ( 4.4.6 ) withuDFn.x/andvDFn/NUL1.x/ imply that jFnC1.x//NULFn.x/jDjg.F n.x///NULg.F n/NUL1.x//j/DC4rjFn.x//NULFn/NUL1.x/j;/NUL1<x<1; so kFnC1/NULFnkS/DC4rkFn/NULFn/NUL1kS: By induction, this implies that kFnC1/NULFnkS/DC4rnkF1/NULF0kS: (4.4.7) Ifn>m , then kFn/NULFmkSDk.Fn/NULFn/NUL1/C.Fn/NUL1/NULFn/NUL2/C/SOH/SOH/SOHC.FmC1/NULFm/kS /DC4kFn/NULFn/NUL1kSCkFn/NUL1/NULFn/NUL2kSC/SOH/SOH/SOHCkFmC1/NULFmkS; from Lemma 4.4.2 . Now ( 4.4.7 ) implies that kFn/NULFmkS/DC4kF1/NULF0kS.1CrCr2C/SOH/SOH/SOHCrn/NULm/NUL1/rm <kF1/NULF0kSrm 1/NULr: Therefore, if kF1/NULF0kSrN 1/NULr</SI; thenkFn/NULFmkS< /SI ifn,m/NAKN. Therefore,fFngconverges uniformly on S, by Theorem 4.4.6 . 242 Chapter 4 Infinite Sequences and Series Properties Preserved by Uniform Convergence We now study properties of the functions of a uniformly conve rgent sequence that are inherited by the limit function. We first consider continuit y. Theorem 4.4.7 IffFngconverges uniformly to FonSand eachFnis continuous at a pointx0inS;then so isF. Similar statements hold for continuity from the right and l eft: Proof Suppose that each Fnis continuous at x0. Ifx2Sandn/NAK1, then jF.x//NULF.x 0/j/DC4jF.x//NULFn.x/jCjFn.x//NULFn.x0/jCjFn.x0//NULF.x 0/j /DC4jFn.x//NULFn.x0/jC2kFn/NULFkS:(4.4.8) Suppose that /SI >0 . SincefFngconverges uniformly to FonS, we can choose nso that kFn/NULFkS</SI. For this fixed n, (4.4.8 ) implies that jF.x//NULF.x 0/j<jFn.x//NULFn.x0/jC2/SI; x2S: (4.4.9) SinceFnis continuous at x0, there is aı>0 such that jFn.x//NULFn.x0/j</SI ifjx/NULx0j<ı; so, from ( 4.4.9 ), jF.x//NULF.x 0/j<3/SI; ifjx/NULx0j<ı: Therefore,Fis continuous at x0. Similar arguments apply to the assertions on continuity from the right and left. Corollary 4.4.8 IffFngconverges uniformly to FonSand eachFnis continuous on S;then so isFIthat is;a uniform limit of continuous functions is continuous. Now we consider the question of integrability of the uniform limit of integrable func- tions. Theorem 4.4.9 Suppose thatfFngconverges uniformly to FonSDŒa;b/c141 . Assume thatFand allFnare integrable on Œa;b/c141: Then Zb aF.x/dxDlim n!1Zb aFn.x/dx: (4.4.10) Proof Since ˇˇˇˇˇZb aFn.x/dx/NULZb aF.x/dxˇˇˇˇˇ/DC4Zb ajFn.x//NULF.x/jdx /DC4.b/NULa/kFn/NULFkS and lim n!1kFn/NULFkSD0, the conclusion follows. Section 4.4 Sequences and Series of Functions 243 In particular, this theorem implies that ( 4.4.10 ) holds if each Fnis continuous on Œa;b/c141 , because then Fis continuous (Corollary 4.4.8 ) and therefore integrable on Œa;b/c141 . The hypotheses of Theorem 4.4.9 are stronger than necessary. We state the next theorem so that you will be better informed on this subject. We omit th e proof, which is inaccessible if you skipped Section 3.5, and quite involved in any case. Theorem 4.4.10 Suppose thatfFngconverges pointwise to Fand eachFnis inte- grable onŒa;b/c141: (a) If the convergence is uniform ;thenFis integrable on Œa;b/c141 and(4.4.10 )holds. (b) If the sequencefkFnkŒa;b/c141gis bounded and Fis integrable on Œa;b/c141; then (4.4.10 ) holds. Part(a)of this theorem shows that it is not necessary to assume in The orem 4.4.9 thatF is integrable on Œa;b/c141 , since this follows from the uniform convergence. Part (b) is known as the bounded convergence theorem . Neither of the assumptions of (b) can be omitted. Thus, in Example 4.4.3 , wherefkFnkŒ0;1/c141gis unbounded while Fis integrable on Œ0;1/c141 , Z1 0Fn.x/dxD1; n/NAK1; butZ1 0F.x/dxD0: In Example 4.4.4 , wherekFnkŒa;b/c141D1for every finite interval Œa;b/c141 ,Fnis integrable for alln/NAK1, andFis nonintegrable on every interval (Exercise 4.4.3 ). After Theorems 4.4.7 and4.4.9 , it may seem reasonable to expect that if a sequence fFng of differentiable functions converges uniformly to FonS, thenF0Dlimn!1F0 nonS. The next example shows that this is not true in general. Example 4.4.9 The sequencefFngdefined by Fn.x/Dxnsin1 xn/NUL1 converges uniformly to F/DC10onŒr1;r2/c141if0 < r 1< r 2< 1 (or, equivalently, on every compact subset of .0;1/ ). However, F0 n.x/Dnxn/NUL1sin1 xn/NUL1/NUL.n/NUL1/cos1 xn/NUL1; sofF0 n.x/gdoes not converge for any xin.0;1/ . Theorem 4.4.11 Suppose that F0 nis continuous on Œa;b/c141 for alln/NAK1andfF0 ng converges uniformly on Œa;b/c141: Suppose also thatfFn.x0/gconverges for some x0inŒa;b/c141: ThenfFngconverges uniformly on Œa;b/c141 to a differentiable limit function F;and F0.x/Dlim n!1F0 n.x/; a<x<b; (4.4.11) while F0 C.a/Dlim n!1F0 n.aC/andF0 /NUL.b/Dlim n!1F0 n.b/NUL/: (4.4.12) 244 Chapter 4 Infinite Sequences and Series Proof SinceF0 nis continuous on Œa;b/c141 , we can write Fn.x/DFn.x0/CZx x0F0 n.t/dt; a/DC4x/DC4b (4.4.13) (Theorem 3.3.12 ). Now let LDlim n!1Fn.x0/ and G.x/Dlim n!1F0 n.x/: (4.4.14) SinceF0 nis continuous andfF0 ngconverges uniformly to GonŒa;b/c141 ,Gis continuous on Œa;b/c141 (Corollary 4.4.8 ); therefore, ( 4.4.13 ) and Theorem 4.4.9 (withFandFnreplaced by GandF0 n) imply thatfFngconverges pointwise on Œa;b/c141 to the limit function F.x/DLCZx x0G.t/dt: (4.4.15) The convergence is actually uniform on Œa;b/c141 , since subtracting ( 4.4.13 ) from ( 4.4.15 ) yields jF.x//NULFn.x/j/DC4jL/NULFn.x0/jCˇˇˇˇZx x0jG.t//NULF0 n.t/jdtˇˇˇˇ /DC4jL/NULFn.x0/jCjx/NULx0jkG/NULF0 nkŒa;b/c141; so kF/NULFnkŒa;b/c141/DC4jL/NULFn.x0/jC.b/NULa/kG/NULF0 nkŒa;b/c141; where the right side approaches zero as n!1 . SinceGis continuous on Œa;b/c141 , (4.4.14 ), (4.4.15 ), Definition 2.3.6 , and Theorem 3.3.11 imply ( 4.4.11 ) and ( 4.4.12 ). Infinite Series of Functions In Section 4.3 we defined the sum of an infinite series of consta nts as the limit of the sequence of partial sums. The same definition can be applied t o series of functions, as follows. Definition 4.4.12 Ifffjg1 kis a sequence of real-valued functions defined on a set D of reals, thenP1 jDkfjis an infinite series (or simply a series ) of functions on D. The partial sums of ,P1 jDkfjare defined by FnDnX jDkfj; n/NAKk: IffFng1 kconverges pointwise to a function Fon a subsetSofD, we say thatP1 jDkfj converges pointwise to the sum FonS, and write FD1X jDkfj; x2S: Section 4.4 Sequences and Series of Functions 245 IffFngconverges uniformly to FonS, we say thatP1 jDkfjconverges uniformly to F onS. Example 4.4.10 The functions fj.x/Dxj; j/NAK0; define the infinite series1X jD0xj onDD./NUL1;1/. Thenth partial sum of the series is Fn.x/D1CxCx2C/SOH/SOH/SOHCxn; or, in closed form, Fn.x/D8 < :1/NULxnC1 1/NULx; x¤1; nC1; xD1 (Example 4.1.11 ). We have seen earlier that fFngconverges pointwise to F.x/D1 1/NULx ifjxj<1and diverges ifjxj/NAK1; hence, we write 1X jD0xjD1 1/NULx;/NUL1<x<1: Since the difference F.x//NULFn.x/DxnC1 1/NULx can be made arbitrarily large by taking xclose to1, kF/NULFnk./NUL1;1/D1; so the convergence is not uniform on ./NUL1;1/. Neither is it uniform on any interval ./NUL1;r/c141 with/NUL1<r <1 , since kF/NULFnk./NUL1;r//NAK1 2 for everynon every such interval. (Why?) The series does converge unif ormly on any intervalŒ/NULr;r/c141with0<r <1 , since kF/NULFnkŒ/NULr;r/c141DrnC1 1/NULr and lim n!1rnD0. Put another way, the series converges uniformly on closed s ubsets of ./NUL1;1/. 246 Chapter 4 Infinite Sequences and Series As for series of constants, the convergence, pointwise or un iform, of a series of functions is not changed by altering or omitting finitely many terms. Th is justifies adopting the convention that we used for series of constants: when we are i nterested only in whether a series of functions converges, and not in its sum, we will omi t the limits on the summation sign and write simplyPfn. Tests for Uniform Convergence of Series Theorem 4.4.6 is easily converted to a theorem on uniform convergence of se ries, as fol- lows. Theorem 4.4.13 (Cauchy’s Uniform Convergence Criterion) A seriesPfnconverges uniformly on a set Sif and only if for each /SI > 0 there is an integer N such that kfnCfnC1C/SOH/SOH/SOHCfmkS</SI ifm/NAKn/NAKN: (4.4.16) Proof Apply Theorem 4.4.6 to the partial sums ofPfn, observing that fnCfnC1C/SOH/SOH/SOHCfmDFm/NULFn/NUL1: SettingmDnin (4.4.16 ) yields the following necessary, but not sufficient, condit ion for uniform convergence of series. It is analogous to Coroll ary4.3.6 . Corollary 4.4.14 IfPfnconverges uniformly on S;then limn!1kfnkSD0: Theorem 4.4.13 leads immediately to the following important test for unifo rm conver- gence of series. Theorem 4.4.15 (Weierstrass’s Test) The seriesPfnconverges uniformly onSif kfnkS/DC4Mn; n/NAKk; (4.4.17) wherePMn<1: Proof From Cauchy’s convergence criterion for series of constant s, there is for each /SI>0 an integerNsuch that MnCMnC1C/SOH/SOH/SOHCMm</SI ifm/NAKn/NAKN; which, because of ( 4.4.17 ), implies that kfnkSCkfnC1kSC/SOH/SOH/SOHCkfmkS</SI ifm;n/NAKN: Lemma 4.4.2 and Theorem 4.4.13 imply thatPfnconverges uniformly on S. Section 4.4 Sequences and Series of Functions 247 Example 4.4.11 TakingMnD1=n2and recalling that X1 n2<1; we see thatX1 x2Cn2andXsinnx n2 converge uniformly on ./NUL1;1/. Example 4.4.12 The series X fn.x/DX/DC2x 1Cx/DC3n converges uniformly on any set Ssuch that ˇˇˇˇx 1Cxˇˇˇˇ/DC4r <1; x2S; (4.4.18) because ifSis such a set, then kfnkS/DC4rn and Weierstrass’s test applies, with X MnDX rn<1: Since ( 4.4.18 ) is equivalent to /NULr 1Cr/DC4x/DC4r 1/NULr; x2S; this means that the series converges uniformly on any compac t subset of./NUL1=2;1/. (Why?) From Corollary 4.4.14 , the series does not converge uniformly on SD./NUL1=2;b/ withb <1or onSDŒa;1/witha>/NUL1=2, because in these cases kfnkSD1for all n. Weierstrass’s test is very important, but applicable only t o series that actually exhibit a stronger kind of convergence than we have considered so far. We say thatPfnconverges absolutely on SifPjfnjconverges pointwise on S, and absolutely uniformly onSifPjfnjconverges uniformly on S. We leave it to you (Exercise 4.4.21 ) to verify that our proof of Weierstrass’s test actually shows thatPfnconverges absolutely uniformly on S. We also leave it to you to show that if a series converges absol utely uniformly on S, then it converges uniformly on S(Exercise 4.4.20 ). The next theorem applies to series that converge uniformly, but perhaps not absolutely uniformly, on a set S. 248 Chapter 4 Infinite Sequences and Series Theorem 4.4.16 (Dirichlet’s Test for Uniform Convergence) The se- ries1X nDkfngn converges uniformly on Sifffngconverges uniformly to zero on S;P.fnC1/NULfn/con- verges absolutely uniformly on S;and kgkCgkC1C/SOH/SOH/SOHCgnkS/DC4M; n/NAKk; (4.4.19) for some constant M: Proof The proof is similar to the proof of Theorem 4.3.20 . Let GnDgkCgkC1C/SOH/SOH/SOHCgn; and consider the partial sums ofP1 nDkfngn: HnDfkgkCfkC1gkC1C/SOH/SOH/SOHCfngn: (4.4.20) By substituting gkDGkandgnDGn/NULGn/NUL1; n/NAKkC1; into ( 4.4.20 ), we obtain HnDfkGkCfkC1.GkC1/NULGk/C/SOH/SOH/SOHCfn.Gn/NULGn/NUL1/; which we rewrite as HnD.fk/NULfkC1/GkC.fkC1/NULfkC2/GkC1C/SOH/SOH/SOHC.fn/NUL1/NULfn/Gn/NUL1CfnGn; or HnDJn/NUL1CfnGn; (4.4.21) where Jn/NUL1D.fk/NULfkC1/GkC.fkC1/NULfkC2/GkC1C/SOH/SOH/SOHC.fn/NUL1/NULfn/Gn/NUL1:(4.4.22) That is,fJngis the sequence of partial sums of the series 1X jDk.fj/NULfjC1/Gj: (4.4.23) From ( 4.4.19 ) and the definition of Gj, ˇˇˇˇˇˇmX jDnŒfj.x//NULfjC1.x//c141G j.x/ˇˇˇˇˇˇ/DC4MmX jDnjfj.x//NULfjC1.x/j; x2S; Section 4.4 Sequences and Series of Functions 249 so /CR/CR/CR/CR/CR/CRmX jDn.fj/NULfjC1/Gj/CR/CR/CR/CR/CR/CR S/DC4M/CR/CR/CR/CR/CR/CRmX jDnjfj/NULfjC1j/CR/CR/CR/CR/CR/CR S: Now suppose that /SI>0 . SinceP.fj/NULfjC1/converges absolutely uniformly on S, The- orem 4.4.13 implies that there is an integer Nsuch that the right side of the last inequality is less than/SIifm/NAKn/NAKN. The same is then true of the left side, so Theorem 4.4.13 implies that ( 4.4.23 ) converges uniformly on S. We have now shown that fJngas defined in ( 4.4.22 ) converges uniformly to a limit functionJonS. Returning to ( 4.4.21 ), we see that Hn/NULJDJn/NUL1/NULJCfnGn: Hence, from Lemma 4.4.2 and ( 4.4.19 ), kHn/NULJkS/DC4kJn/NUL1/NULJkSCkfnkSkGnkS /DC4kJn/NUL1/NULJkSCMkfnkS: SincefJn/NUL1/NULJgandffngconverge uniformly to zero on S, it now follows that lim n!1kHn/NUL JkSD0. Therefore,fHngconverges uniformly on S. Corollary 4.4.17 The seriesP1 nDkfngnconverges uniformly on Sif fnC1.x//DC4fn.x/; x2S; n/NAKk; ffngconverges uniformly to zero on S;and kgkCgkC1C/SOH/SOH/SOHCgnkS/DC4M; n/NAKk; for some constant M: The proof is similar to that of Corollary 4.3.21 . We leave it to you (Exercise 4.4.22 ). Example 4.4.13 Consider the series 1X nD1sinnx n withfnD1=n(constant),gn.x/Dsinnx, and Gn.x/DsinxCsin2xC/SOH/SOH/SOHC sinnx: We saw in Example 4.3.21 that jGn.x/j/DC41 jsin.x=2/j; n/NAK1; n¤2k/EM (kDinteger): 250 Chapter 4 Infinite Sequences and Series Therefore,fkGnkSgis bounded, and the series converges uniformly on any set Son which sinx=2 is bounded away from zero. For example, if 0<ı</EM , then ˇˇˇsinx 2ˇˇˇ/NAKsinı 2 ifxis at leastıaway from any multiple of 2/EM; hence, the series converges uniformly on SD1[ kD/NUL1Œ2k/EMCı;2.kC1//EM/NULı/c141: SinceXˇˇˇˇsinnx nˇˇˇˇD1; x¤k/EM (Exercise 4.3.32(b)), this result cannot be obtained from Weierstrass’s test. Example 4.4.14 The series 1X nD1./NUL1/n nCx2 satisfies the hypotheses of Corollary 4.4.17 on./NUL1;1/, with fn.x/D1 nCx2; g nD./NUL1/n; G 2mD0; andG2mC1D/NUL1: Therefore, the series converges uniformly on ./NUL1;1/. This result cannot be obtained by Weierstrass’s test, sinceX1 nCx2D1 for allx. Continuity, Differentiability, and Integrability of Serie s We can obtain results on the continuity, differentiability , and integrability of infinite series by applying Theorems 4.4.7 ,4.4.9 , and 4.4.11 to their partial sums. We will state the theorems and give some examples, leaving the proofs to you. Theorem 4.4.7 implies the following theorem (Exercise 4.4.23 ). Theorem 4.4.18 IfP1 nDkfnconverges uniformly to FonSand eachfnis contin- uous at a point x0inS;then so isF:Similar statements hold for continuity from the right and left: Example 4.4.15 In Example 4.4.12 we saw that the series F.x/D1X nD0/DC2x 1Cx/DC3n Section 4.4 Sequences and Series of Functions 251 converges uniformly on every compact subset of ./NUL1=2;1/. Since the terms of the series are continuous on every such subset, Theorem 4.4.4 implies thatFis also. In fact, we can state a stronger result: Fis continuous on ./NUL1=2;1/, since every point in ./NUL1=2;1/lies in a compact subinterval of ./NUL1=2;1/. The same argument and the results of Example 4.4.13 show that the function G.x/D1X nD1sinnx n is continuous except perhaps at xkD2k/EM (kDinteger). From Example 4.4.14 , the function H.x/D1X nD1./NUL1/n1 nCx2 is continuous for all x. The next theorem gives conditions that permit the interchan ge of summation and inte- gration of infinite series. It follows from Theorem 4.4.9 (Exercise 4.4.25 ). We leave it to you to formulate an analog of Theorem 4.4.10 for series (Exercise 4.4.26 ). Theorem 4.4.19 Suppose thatP1 nDkfnconverges uniformly to FonSDŒa;b/c141: Assume thatFandfn;n/NAKk;are integrable on Œa;b/c141: Then Zb aF.x/dxD1X nDkZb afn.x/dx: We say in this case thatP1 nDkfncan be integrated term by term overŒa;b/c141 . Example 4.4.16 From Example 4.4.10 , 1 1/NULxD1X nD0xn;/NUL1<x<1: The series converges uniformly, and the limit function is in tegrable on any closed subinter- valŒa;b/c141 of./NUL1;1/; hence, Zb adx 1/NULxD1X nD0Zb axndx; so log.1/NULa//NULlog.1/NULb/D1X nD0bnC1/NULanC1 nC1: LettingaD0andbDxyields log.1/NULx/D/NUL1X nD0xnC1 nC1;/NUL1<x<1: 252 Chapter 4 Infinite Sequences and Series The next theorem gives conditions that permit the interchan ge of summation and differ- entiation of infinite series. It follows from Theorem 4.4.11 (Exercise 4.4.28 ). Theorem 4.4.20 Suppose that fnis continuously differentiable on Œa;b/c141 for eachn/NAK k;P1 nDkfn.x0/converges for some x0inŒa;b/c141; andP1 nDkf0 nconverges uniformly on Œa;b/c141: ThenP1 nDkfnconverges uniformly on Œa;b/c141 to a differentiable function F;and F0.x/D1X nDkf0 n.x/; a<x<b; while F0.aC/D1X nDkf0 n.aC/andF0.b/NUL/D1X nDkf0 n.b/NUL/: We say in this case thatP1 nDkfncan be differentiated term by term onŒa;b/c141 . To apply Theorem 4.4.20 , we first verify thatP1 nDkfn.x0/converges for some x0inŒa;b/c141 and then differentiateP1 nDkfnterm by term. If the resulting series converges uniformly, t hen term by term differentiation was legitimate. Example 4.4.17 The series 1X nD1./NUL1/n1 ncosx n(4.4.24) converges atx0D0. Differentiating term by term yields the series 1X nD1./NUL1/nC11 n2sinx n(4.4.25) of continuous functions. This series converges uniformly o n./NUL1;1/, by Weierstrass’s test. By Theorem 4.4.20 , the series ( 4.4.24 ) converges uniformly on every finite interval to the differentiable function F.x/D1X nD1./NUL1/n1 ncosx n;/NUL1<x<1; and F0.x/D1X nD1./NUL1/nC11 n2sinx n;/NUL1<x<1: Example 4.4.18 The series E.x/D1X nD0xn nŠD1CxCx2 2ŠCx3 3ŠC/SOH/SOH/SOH (4.4.26) Section 4.4 Sequences and Series of Functions 253 converges uniformly on every interval Œ/NULr;r/c141by Weierstrass’s test, because jxjn nŠ/DC4rn nŠ;jxj/DC4r; and Xrn nŠ<1 for allr, by the ratio test. Differentiating the right side of ( 4.4.26 ) term by term yields the series1X nD1xn/NUL1 .n/NUL1/ŠD1X nD0xn nŠ; which is the same as ( 4.4.26 ). Therefore, the differentiated series is also uniformly c onver- gent onŒ/NULr;r/c141for everyr, so the term by term differentiation is legitimate and E0.x/DE.x/;/NUL1<x<1: This is not surprising if you recognize that E.x/Dex. Example 4.4.19 Failure to verify that the given series converges at some poi nt can lead to erroneous conclusions. For example, differentiati ng 1X nD1cosx n(4.4.27) term by term yields /NUL1X nD11 nsinx n; which converges uniformly on Œ/NULr;r/c141for everyr, since ˇˇˇˇ1 nsinx nˇˇˇˇ/DC4jxj n2(Exercise 2.3.19 ) /DC4r n2ifjxj/DC4r; andP1=n2<1. We cannot conclude from this that ( 4.4.27 ) converges uniformly on Œ/NULr;r/c141. In fact, it diverges for every x. (Why?) 4.4 Exercises 1. Find the setSon whichfFngconverges pointwise, and find the limit function. (a)Fn.x/Dxn.1/NULx2/ (b)Fn.x/Dnxn.1/NULx2/ 254 Chapter 4 Infinite Sequences and Series (c)Fn.x/Dxn.1/NULxn/ (d)Fn.x/Dsin/DC2 1C1 n/DC3 x (e)Fn.x/D1Cxn 1Cx2n(f)Fn.x/Dnsinx n (g)Fn.x/Dn2/DLE 1/NULcosx n/DC1 (h)Fn.x/Dnxe/NULnx2 (i)Fn.x/D.xCn/2 x2Cn2 2. Prove: IffFngconverges toFonŒa;b/c141 andFnis nondecreasing for each n, thenF is nondecreasing. 3. Show that the functions fFngof Example 4.4.4 are integrable and FDlimn!1Fn.x/ is nonintegrable on every finite interval. 4. Prove Lemma 4.4.2 . 5. FindF.x/Dlimn!1Fn.x/onS. Show thatfFngconverges uniformly to Fon closed subsets of S, but not onS. (a)Fn.x/Dxnsinnx,SD./NUL1;1/ (b)Fn.x/D1 1Cx2n,SDfxjx¤˙1g (c)Fn.x/Dn2sinx 1Cn2x,SD.0;1/HINT:See Exercise 2.3.19: 6. (a) Show that iffFngconverges uniformly on S, thenfFngconverges uniformly on every subset of S. (b) Show that iffFngconverges uniformly on S1,S2, . . . ,Sm, thenfFngcon- verges uniformly onSm kD1Sk. (c) Give an example where fFngconverges uniformly on each of an infinite se- quence of sets S1,S2, . . . , but not onS1 kD1Sk. 7. Describe the sets on which the sequences of Exercise 4.4.1 converge uniformly. Re- strict your attention to sets that are the union of finitely ma ny intervals and singleton sets. 8. Suppose thatfFngconverges pointwise on Œa;b/c141 and, for each xinŒa;b/c141 , there is an open interval Ixcontainingxsuch thatfFngconverges uniformly on Ix\Œa;b/c141 . Show thatfFngconverges uniformly on Œa;b/c141 . 9. Prove: IffFngconverges uniformly to FonS, then lim n!1kFnkSDkFkS. 10. Prove: IffFngconverges uniformly to FonS, thenFis bounded on Sif and only iflimn!1fkFnkSg<1. 11. Prove: IffFngandfGngconverge uniformly to FandGonS, thenfFnCGng converges uniformly to FCGonS. 12. (a) Prove: IffFngandfGngconverge uniformly to bounded functions FandG onS, thenfFnGngconverges uniformly to FG onS. Section 4.4 Sequences and Series of Functions 255 (b) Give an example showing that the conclusion of (a)may fail to hold if For Gis unbounded on S. 13. (a) Suppose thatfFngconverges uniformly to Fon.a;b/ . Prove: Ifx0<a<b andLnDlimx!x0Fn.x/exists (finite) for every n, thenLDlimn!1Ln exists (finite) and lim x!x0F.x/DL: (b) State similar results for limits from the right and left. 14. Find the limits. (a) lim n!1Z4 1n xsinx ndx (b) lim n!1Z2 0dx 1Cx2n (c) lim n!1Z1 0nxe/NULnx2dx (d) lim n!1Z1 0/DLE 1Cx n/DC1n dx 15. Prove (without using Theorem 4.4.10 ): If eachFnis integrable andfFngconverges uniformly on Œa;b/c141 , then lim n!1Rb aFn.x/dx exists. 16. Prove (without using Theorem 4.4.10 ): If eachFnis nondecreasing and fFngcon- verges uniformly to FonŒa;b/c141 , then lim n!1Zb aFn.x/dxDZb aF.x/dx: 17. Use Weierstrass’s test to determine sets on which the series converges absolutely uniformly. (a)X1 n1=2/DC2x 1Cx/DC3n (b)X1 n3=2/DC2x 1Cx/DC3n (c)X nxn.1/NULx/n(d)X1 n.x2Cn/ (e)X1 nx(f)X.1/NULx2/n .1Cx2/nsinnx 18. Show that ifPjanj<1, thenPancosnxandPansinnxdefine continuous functions on./NUL1;1/. 19. (a) Give an example showing that the following “comparison test ” is invalid: IfPfnconverges uniformly on SandkgnkS/DC4kfnkS, thenPgnconverges uniformly on S. (b) This “comparison test” can be corrected by adding one word to its hypothesis and conclusion. What is the word? 20. (a) Explain the difference between the following statements: (i)Pfnconverges absolutely and uniformly on S;(ii)Pfnconverges absolutely uniformly onS. 256 Chapter 4 Infinite Sequences and Series (b) Show that ifPfnconverges absolutely uniformly on S, thenPfnconverges uniformly on S. 21. Show that the hypotheses of Weierstrass’s test imply thatPfnconverges absolutely uniformly on S. 22. Prove Corollary 4.4.17 . 23. Prove Theorem 4.4.18 . 24. Suppose thatfang1 1is monotonic and lim n!1anD0. Show that 1X nD1ansinnx and1X nD1ancosnx define functions continuous for all x¤2k/EM (kDinteger). 25. Prove Theorem 4.4.19 . 26. Formulate an analog of Theorem 4.4.10 for series. 27. In Section 4.5 we will see that e/NULx2D1X nD0./NUL1/nx2n nŠand sinxD1X nD0./NUL1/nx2nC1 .2nC1/Š for allx, and in both cases the convergence is uniform on every finite i nterval. Find series that converge to (a)F.x/DZx 0e/NULt2dt and(b)G.x/DZx 0sint tdt for allx. 28. Prove Theorem 4.4.20 . 29. Show from Example 4.4.17 thatP1 nD1./NUL1/nsin.x=n/ converges uniformly on any finite interval. 30. Prove: If0 < a nC1< a nandPak n<1for some positive integer k, thenP./NUL1/nsinanxconverges uniformly on any finite interval. 31. Forn/NAK2, define fn.x/D8 ˆˆ< ˆˆ:n4.x/NULnC1=n3/; n/NUL1=n3/DC4x/DC4n; /NULn4.x/NULn/NUL1=n3/; n/DC4x/DC4nC1=n3; 0; jx/NULnj>1=n3; and letF.x/DP1 nD2fn.x/. Show thatR1 0F.x/dx <1, and conclude that ab- solute convergence of an improper integralR1 0F.x/dx does not imply that lim n!1F.x/D 0, even ifFis continuous on Œ0;1/. Section 4.5 Power Series 257 4.5 POWER SERIES We now consider a class of series sufficiently general to be in teresting, but sufficiently specialized to be easily understood. Definition 4.5.1 An infinite series of the form 1X nD0an.x/NULx0/n; (4.5.1) wherex0anda0,a1, . . . , are constants, is called a power series in x/NULx0. The following theorem summarizes the convergence properti es of power series. Theorem 4.5.2 In connection with the power series (4.5.1 );defineRin the extended reals by 1 RDlim n!1janj1=n: (4.5.2) In particular;RD0iflimn!1janj1=nD1 , andRD1 iflimn!1janj1=nD0:Then the power series converges (a) only forxDx0ifRD0I (b) for allxifRD1;and absolutely uniformly in every bounded set I (c) forxin.x0/NULR;x 0CR/if0<R <1;and absolutely uniformly in every closed subset of this interval. The series diverges if jx/NULx0j>R: No general statement can be made concerning conver- gence at the endpoints xDx0CRandxDx0/NULRWthe series may converge absolutely or conditionally at both ;converge conditionally at one and diverge at the other ;or diverge at both: Proof In any case, the series ( 4.5.1 ) converges to a0ifxDx0. If X janjrn<1 (4.5.3) for somer > 0 , thenPan.x/NULx0/nconverges absolutely uniformly in Œx0/NULr;x0C r/c141, by Weierstrass’s test (Theorem 4.4.15 ) and Exercise 4.4.21 . From Cauchy’s root test (Theorem 4.3.17 ), (4.5.3 ) holds if lim n!1.janjrn/1=n<1; which is equivalent to rlim n!1janj1=n<1 (Exercise 4.1.30(a)). From ( 4.5.2 ), this can be rewritten as r < R , which proves the assertions concerning convergence in (b) and(c). If0/DC4R<1andjx/NULx0j>R, then 258 Chapter 4 Infinite Sequences and Series 1 R>1 jx/NULx0j; so (4.5.2 ) implies that janj1=n/NAK1 jx/NULx0jand thereforejan.x/NULx0/nj/NAK1 for infinitely many values of n. Therefore,Pan.x/NULx0/ndiverges (Corollary 4.3.6 ) if jx/NULx0j>R. In particular, the series diverges for all x¤x0ifRD0. To prove the assertions concerning the possibilities at xDx0CRandxDx0/NULR requires examples, which follow. (Also, see Exercise 4.5.1 .) The number Rdefined by ( 4.5.2 ) is the radius of convergence ofPan.x/NULx0/n. If R > 0 , the open interval.x0/NULR;x 0CR/, or./NUL1;1/ifRD1 , is the interval of convergence of the series. Theorem 4.5.2 says that a power series with a nonzero radius of convergence converges absolutely uniformly in every com pact subset of its interval of convergence and diverges at every point in the exterior of th is interval. On this last we can make a stronger statement: Not only doesPan.x/NULx0/ndiverge ifjx/NULx0j>R, but the sequencefan.x/NULx0/ngis unbounded in this case (Exercise 4.5.3(b)). Example 4.5.1 For the series Xsinn/EM=6 2n.x/NUL1/n; we have lim n!1janj1=nDlim n!1/DC2jsinn/EM=6 2n/DC31=n D1 2lim n!1.jsinn/EM=6j/1=n(Exercise 4.1.30(a)) D1 2.1/D1 2: Therefore,RD2and Theorem 4.5.2 implies that the series converges absolutely uniformly in closed subintervals of ./NUL1;3/ and diverges if x</NUL1orx>3 . Theorem 4.5.2 does not tell us what happens when xD/NUL1orxD3, but we can see that the series diverges in both these cases since its general term does not approach zero. Example 4.5.2 For the seriesXxn n; lim n!1janj1=nDlim n!1/DC21 n/DC31=n Dlim n!1exp/DC21 nlog1 n/DC3 De0D1: Therefore,RD1and the series converges absolutely uniformly in closed sub intervals of./NUL1;1/ and diverges ifjxj> 1. ForxD/NUL1the series becomesP./NUL1/n=n, which converges conditionally, and at xD1the series becomesP1=n, which diverges. Section 4.5 Power Series 259 The next theorem provides an expression for Rthat, if applicable, is usually easier to use than ( 4.5.2 ). Theorem 4.5.3 The radius of convergence ofPan.x/NULx0/nis given by 1 RDlim n!1ˇˇˇˇanC1 anˇˇˇˇ if the limit exists in the extended reals : Proof From Theorem 4.5.2 , it suffices to show that if LDlim n!1ˇˇˇˇanC1 anˇˇˇˇ(4.5.4) exists in the extended reals, then LDlim n!1janj1=n: (4.5.5) We will show that this is so if 0<L<1and leave the cases where LD0orLD1 to you (Exercise 4.5.7 ). If (4.5.4 ) holds with0<L<1and0</SI<L , there is an integer Nsuch that L/NUL/SI<ˇˇˇˇamC1 amˇˇˇˇ<LC/SIifm/NAKN; so jamj.L/NUL/SI/<jamC1j<jamj.LC/SI/ifm/NAKN: By induction, jaNj.L/NUL/SI/n/NULN<janj<jaNj.LC/SI/n/NULNifn>N: Therefore, if K1DjaNj.L/NUL/SI//NULNandK2DjaNj.LC/SI//NULN; then K1=n 1.L/NUL/SI/<janj1=n<K1=n 2.LC/SI/: (4.5.6) Since lim n!1K1=nD1ifKis any positive number, ( 4.5.6 ) implies that L/NUL/SI/DC4lim n!1janj1=n/DC4lim n!1janj1=n/DC4LC/SI: Since/SIis an arbitrary positive number, it follows that lim n!1janj1=nDL; which implies ( 4.5.5 ). 260 Chapter 4 Infinite Sequences and Series Example 4.5.3 For the power series Xxn nŠ; lim n!1ˇˇˇˇanC1 anˇˇˇˇDlim n!1nŠ .nC1/ŠDlim n!11 nC1D0: Therefore,RD1 ; that is, the series converges for all x, and absolutely uniformly in every bounded set. Example 4.5.4 For the power series X nŠxn; lim n!1ˇˇˇˇanC1 anˇˇˇˇDlim n!1.nC1/Š nŠDlim n!1.nC1/D1: Therefore,RD0, and the series converges only if xD0. Example 4.5.5 Theorem 4.5.3 does not apply directly to X./NUL1/n 4nnpx2n(pDconstant); (4.5.7) which has infinitely many zero coefficients (of odd powers of x). However, by setting yDx2, we obtain the seriesX./NUL1/n 4nnpyn; (4.5.8) which has nonzero coefficients for which lim n!1ˇˇˇˇanC1 anˇˇˇˇDlim n!14nnp 4nC1.nC1/pD1 4lim n!1/DC2 1C1 n/DC3/NULp D1 4: Therefore, ( 4.5.8 ) converges ifjyj< 4 and diverges ifjyj> 4. SettingyDx2, we conclude that ( 4.5.7 ) converges ifjxj< 2 and diverges ifjxj> 2. AtxD˙2, (4.5.7 ) becomesP./NUL1/n=np, which diverges if p/DC40, converges conditionally if 0<p/DC41, and converges absolutely if p>1 . Properties of Functions Defined by Power Series We now study the properties of functions defined by power seri es. Henceforth, we consider only power series with nonzero radii of convergence. Theorem 4.5.4 A power series f.x/D1X nD0an.x/NULx0/n Section 4.5 Power Series 261 with positive radius of convergence Ris continuous and differentiable in its interval of convergence;and its derivative can be obtained by differentiating term b y termIthat is; f0.x/D1X nD1nan.x/NULx0/n/NUL1; (4.5.9) which can also be written as f0.x/D1X nD0.nC1/anC1.x/NULx0/n: (4.5.10) This series also has radius of convergence R: Proof First, the series in ( 4.5.9 ) and ( 4.5.10 ) are the same, since the latter is obtained by shifting the index of summation in the former. Since lim n!1..nC1/janj/1=nDlim n!1.nC1/1=njanj1=n D/DLE lim n!1.nC1/1=n/DC1/DLE lim n!1janj1=n/DC1 (Exercise 4.1.30(a)/ D/DC4 lim n!1exp/DC2log.nC1/ n/DC3/NAK/DLE lim n!1janj1=n/DC1 De0 RD1 R; the radius of convergence of the power series in ( 4.5.10 ) isR(Theorem 4.5.2 ). Therefore, the power series in ( 4.5.10 ) converges uniformly in every interval Œx0/NULr;x0Cr/c141such that 0<r <R , and Theorem 4.4.20 now implies ( 4.5.10 ) for allxin.x0/NULR;x 0CR/. Theorem 4.5.4 can be strengthened as follows. Theorem 4.5.5 A power series f.x/D1X nD0an.x/NULx0/n with positive radius of convergence Rhas derivatives of all orders in its interval of convergence ; which can be obtained by repeated term by term differentiati onIthus; f.k/.x/D1X nDkn.n/NUL1//SOH/SOH/SOH.n/NULkC1/an.x/NULx0/n/NULk: (4.5.11) The radius of convergence of each of these series is R: Proof The proof is by induction. The assertion is true for kD1, by Theorem 4.5.4 . Suppose that it is true for some k/NAK1. By shifting the index of summation, we can rewrite (4.5.11 ) as f.k/.x/D1X nD0.nCk/.nCk/NUL1//SOH/SOH/SOH.nC1/anCk.x/NULx0/n;jx/NULx0j<R: 262 Chapter 4 Infinite Sequences and Series Defining bnD.nCk/.nCk/NUL1//SOH/SOH/SOH.nC1/anCk; (4.5.12) we rewrite this as f.k/.x/D1X nD0bn.x/NULx0/n;jx/NULx0j<R: By Theorem 4.5.4 , we can differentiate this series term by term to obtain f.kC1/.x/D1X nD1nbn.x/NULx0/n/NUL1;jx/NULx0j<R: Substituting from ( 4.5.12 ) forbnyields f.kC1/.x/D1X nD1.nCk/.nCk/NUL1//SOH/SOH/SOH.nC1/na nCk.x/NULx0/n/NUL1;jx/NULx0j<R: Shifting the summation index yields f.kC1/.x/D1X nDkC1n.n/NUL1//SOH/SOH/SOH.n/NULk/an.x/NULx0/n/NULk/NUL1;jx/NULx0j<R; which is ( 4.5.11 ) withkreplaced bykC1. This completes the induction. Example 4.5.6 In Example 4.4.10 we saw that 1 1/NULxD1X nD0xn;jxj<1: Repeated differentiation yields kŠ .1/NULx/kC1D1X nDkn.n/NUL1//SOH/SOH/SOH.n/NULkC1/xn/NULk D1X nD0.nCk/.nCk/NUL1//SOH/SOH/SOH.nC1/xn;jxj<1; so 1 .1/NULx/kC1D1X nD0 nCk k! xn;jxj<1: Example 4.5.7 By the method of Example 4.5.5 , it can be shown that the series S.x/D1X nD0./NUL1/nx2nC1 .2nC1/ŠandC.x/D1X nD0./NUL1/nx2n .2n/Š Section 4.5 Power Series 263 converge for all x. Differentiating yields S0.x/D1X nD0./NUL1/nxn .2n/ŠDC.x/ and C0.x/D1X nD1./NUL1/nx2n/NUL1 .2n/NUL1/ŠD/NUL1X nD0./NUL1/nx2nC1 .2nC1/ŠD/NULS.x/: These results should not surprise you if you recall that S.x/DsinxandC.x/Dcosx: (We will soon prove this.) Theorem 4.5.5 has two important corollaries. Corollary 4.5.6 If f.x/D1X nD0an.x/NULx0/n;jx/NULx0j<R; then anDf.n/.x0/ nŠ: Proof SettingxDx0in (4.5.11 ) yields f.k/.x0/DkŠak: Corollary 4.5.7 (Uniqueness of Power Series) If 1X nD0an.x/NULx0/nD1X nD0bn.x/NULx0/n(4.5.13) for allxin some interval .x0/NULr;x0Cr/;then anDbn; n/NAK0: (4.5.14) Proof Let f.x/D1X nD0an.x/NULx0/nandg.x/D1X nD0bn.x/NULx0/n: From Corollary 4.5.6 , anDf.n/.x0/ nŠandbnDg.n/.x0/ nŠ: (4.5.15) 264 Chapter 4 Infinite Sequences and Series From ( 4.5.13 ),fDgin.x0/NULr;x0Cr/. Therefore, f.n/.x0/Dg.n/.x0/; n/NAK0: This and ( 4.5.15 ) imply ( 4.5.14 ). Theorems 4.4.19 and4.5.2 imply the following theorem. We leave the proof to you (Exercise 4.5.15 ). Theorem 4.5.8 Ifx1andx2are in the interval of convergence of f.x/D1X nD0an.x/NULx0/n; thenZx2 x1f.x/dxD1X nD0an nC1/STX.x2/NULx0/nC1/NUL.x1/NULx0/nC1/ETXI that is;a power series may be integrated term by term between any two p oints in its interval of convergence : Example 4.5.16 presents an application of this theorem. Taylor’s Series So far we have asked for what values of xa given power series converges, and what are the properties of its sum. Now we ask a related question: What properties guarantee that a given function fcan be represented as the sum of a convergent power series in x/NULx0? A partial answer to this question is provided by what we alread y know: Theorem 4.5.5 tells us thatfmust have derivatives of all orders in some neighborhood of x0, and Corollary 4.5.6 tells us that the only power series in x/NULx0that can possibly converge to fin such a neighborhood is 1X nD0f.n/.x0/ nŠ.x/NULx0/n: (4.5.16) This is called the Taylor series offaboutx0(also, the Maclaurin series off, ifx0D0). Themth partial sum of ( 4.5.16 ) is the Taylor polynomial Tm.x/DmX nD0f.n/.x0/ nŠ.x/NULx0/n; defined in Section 2.5. The Taylor series of an infinitely differentiable function fmay converge to a sum dif- ferent fromf. For example, the function f.x/D/SUB e/NUL1=x2; x¤0; 0; xD0; Section 4.5 Power Series 265 is infinitely differentiable on ./NUL1;1/andf.n/.0/D0forn/NAK0(Exercise 2.5.1 ), so its Maclaurin series is identically zero. The answer to our question is provided by Taylor’s theorem (T heorem 2.5.4 ), which says that iffis infinitely differentiable on .a;b/ andxandx0are in.a;b/ then, for every integern/NAK0, f.x//NULTn.x/Df.nC1/.cn/ .nC1/Š.x/NULx0/n/NUL1; (4.5.17) wherecnis betweenxandx0. Therefore, f.x/D1X nD0f.n/.x0/ nŠ.x/NULx0/n for anxin.a;b/ if and only if lim n!1f.nC1/.cn/ .nC1/Š.x/NULx0/nC1D0: It is not always easy to check this condition, because the seq uencefcngis usually not pre- cisely known, or even uniquely defined; however, the next the orem is sufficiently general to be useful. Theorem 4.5.9 Suppose that fis infinitely differentiable on an interval Iand lim n!1rn nŠkf.n/kID0: (4.5.18) Then;ifx02I0;the Taylor series 1X nD0f.n/.x0/ nŠ.x/NULx0/n converges uniformly to fon IrDI\Œx0/NULr;x0Cr/c141: Proof From ( 4.5.17 ), kf/NULTnkIr/DC4rnC1 .nC1/Škf.nC1/kIr/DC4rnC1 .nC1/Škf.nC1/kI; so (4.5.18 ) implies the conclusion. Example 4.5.8 Iff.x/Dsinx, thenkf.k/k./NUL1;1/D1; k/NAK0. Since lim n!1rn nŠD0; 0<r <1 266 Chapter 4 Infinite Sequences and Series (Example 4.1.12 ), (4.5.18 ) holds for all r. Since f.2m/.0/D0andf.2mC1/.0/D./NUL1/m; m/NAK0; we see from Theorem 4.5.9 , withID./NUL1;1/,x0D0, andrarbitrary, that sinxD1X nD0./NUL1/nx2nC1 .2nC1/Š;/NUL1<x<1; and the convergence is uniform on bounded sets. A similar argument shows that cosxD1X nD0./NUL1/nx2n .2n/Š;/NUL1<x<1; with uniform convergence on bounded sets. Example 4.5.9 Iff.x/Dex, thenf.k/.x/Dexandkf.k/kIDer,k/NAK0, if IDŒ/NULr;r/c141. Since lim n!1rn nŠerD0; we conclude as in Example 4.5.8 that exD1X nD0xn nŠ;/NUL1<x<1; with uniform convergence on bounded sets. Example 4.5.10 Iff.x/D.1Cx/q, then f.n/.x/ nŠD q n! .1Cx/q/NULn;sof.n/.0/ nŠD q n! (4.5.19) (Example 2.5.3 ). The Maclaurin series 1X nD0 q n! xn is called the binomial series . We saw in Example 2.5.3 that this series equals .1Cx/qfor allxifqis a nonnegative integer. We will now show that if qis an arbitrary real number, then 1X nD0 q n! xnDf.x/D.1Cx/q; 0/DC4x<1: (4.5.20) Since Section 4.5 Power Series 267 lim n!1ˇˇˇˇˇ q nC1!/RS q n!ˇˇˇˇˇDlim n!1ˇˇˇˇq/NULn nC1ˇˇˇˇD1; the radius of convergence of the series in ( 4.5.20 ) is1. From ( 4.5.19 ), kf.n/kŒ0;1/c141 nŠ/DC4Œmax.1;2q//c141ˇˇˇˇˇ q n!ˇˇˇˇˇ; n/NAK0: Therefore, if 0<r <1 , lim n!1rn nŠkf.n/kŒ0;1/c141/DC4Œmax.1;2q//c141lim n!1ˇˇˇˇˇ q n!ˇˇˇˇˇrnD0; where the last equality follows from the absolute convergen ce of the series in ( 4.5.20 ) on ./NUL1;1/. Now Theorem 4.5.9 implies ( 4.5.20 ). We cannot prove in this way that the binomial series converge s to.1Cx/qon./NUL1;0/. This requires a form of the remainder in Taylor’s theorem tha t we have not considered, or a different kind of proof altogether (Exercise 4.5.20 ). The complete result is that .1Cx/qD1X nD0 q n! xn;/NUL1<x<1; (4.5.21) for allq, and, as we said earlier, the identity holds for all xifqis a nonnegative integer. Arithmetic Operations with Power Series We now consider addition and multiplication of power series , and division of one by an- other. We leave the proof of the next theorem to you (Exercise 4.5.21 ). Theorem 4.5.10 If f.x/D1X nD0an.x/NULx0/n;jx/NULx0j<R 1; (4.5.22) g.x/D1X nD0bn.x/NULx0/n;jx/NULx0j<R 2; (4.5.23) and˛andˇare constants ;then ˛f.x/Cˇg.x/D1X nD0.˛a nCˇbn/.x/NULx0/n;jx/NULx0j<R; whereR/NAKminfR1;R2g: 268 Chapter 4 Infinite Sequences and Series Theorem 4.5.11 Iffandgare given by (4.5.22 )and(4.5.23 );then f.x/g.x/D1X nD0cn.x/NULx0/n;jx/NULx0j<R; (4.5.24) wherecnDnX rD0arbn/NULrDnX rD0an/NULrbr andR/NAKminfR1;R2g: Proof Suppose that R1/DC4R2. Since the series ( 4.5.22 ) and ( 4.5.23 ) converge abso- lutely tof.x/ andg.x/ ifjx/NULx0j<R 1, their Cauchy product converges to f.x/g.x/ if jx/NULx0j<R 1, by Theorem 4.3.29 . Thenth term of this product is nX rD0ar.x/NULx0/rbn/NULr.x/NULx0/n/NULrD nX rD0arbn/NULr! .x/NULx0/nDcn.x/NULx0/n: Example 4.5.11 If f.x/D1 1/NULxD1X nD0xn;jxj<1; and g.x/D1X nD0bnxn;jxj<R; then g.x/ 1/NULxD1X nD0snxn;jxj<minf1;Rg; where snD.1/b 0C.1/b 1C/SOH/SOH/SOHC.1/b n Db0Cb1C/SOH/SOH/SOHCbn: Example 4.5.12 From the paragraph following Example 4.5.10 , .1Cx/pD1X nD0 p n! xn;jxj<1; and .1Cx/qD1X nD0 q n! xn;jxj<1: Section 4.5 Power Series 269 Since .1Cx/p.1Cx/qD.1Cx/pCqD1X nD0 pCq n! xn; while the Cauchy product isP1 nD0cnxn, with cnDnX rD0 p r! q n/NULr! ; Corollary 4.5.7 implies that cnD pCq n! : This yields the identity pCq n! DnX rD0 p r! q n/NULr! ; valid for allpandq. The quotient f.x/Dh.x/ g.x/(4.5.25) of two power series h.x/D1X nD0cn.x/NULx0/n;jx/NULx0j<R 1; and g.x/D1X nD0bn.x/NULx0/n;jx/NULx0j<R 2; can be represented as a power series f.x/D1X nD0an.x/NULx0/n(4.5.26) with a positive radius of convergence, provided that b0Dg.x 0/¤0: This is surely plausible. Since g.x 0/¤0andgis continuous near x0, the denominator of (4.5.25 ) differs from zero on an interval about x0. Therefore,fhas derivatives of all orders on this interval, because gandhdo. However, the proof that the Taylor series of f aboutx0converges to fnearx0requires the use of the theory of functions of a complex variable. Therefore, we omit it. However, it is straightfor ward to compute the coefficients in (4.5.26 ) if we accept the validity of the expansion. Since f.x/g.x/Dh.x/; 270 Chapter 4 Infinite Sequences and Series Theorem 4.5.11 implies that nX rD0arbn/NULrDcn; n/NAK0: Solving these equations successively yields a0Dc0 b0; anD1 b0 cn/NULn/NUL1X rD0bn/NULrar! ; n/NAK1: It is not worthwhile to memorize these formulas. Rather, it i s usually better to view the procedure as follows: Multiply the series f(with unknown coefficients) and gaccording to the procedure of Theorem 4.5.11 , equate the resulting coefficients with those of h, and solve the resulting equations successively for a0,a1, . . . . Example 4.5.13 Suppose that we wish to find the coefficients in the Maclaurin s eries tanxDa0Ca1xCa2x2C/SOH/SOH/SOH: We first observe that since tan xis an odd function, its derivatives of even order vanish at x0D0, soa2mD0,m/NAK0. Therefore, tanxDa1xCa3x3Ca5x5C/SOH/SOH/SOH: Since tanxDsinx cosx; it follows from Example 4.5.8 that a1xCa3x3Ca5x5C/SOH/SOH/SOHDx/NULx3 6Cx5 120C/SOH/SOH/SOH 1/NULx2 2Cx4 24C/SOH/SOH/SOH so .a1xCa3x3Ca5x5C/SOH/SOH/SOH//DC2 1/NULx2 2Cx4 24C/SOH/SOH/SOH/DC3 Dx/NULx3 6Cx5 120C/SOH/SOH/SOH; or, according to Theorem 4.5.11 , a1xC/DLE a3/NULa1 2/DC1 x3C/DLE a5/NULa3 2Ca1 24/DC1 x5C/SOH/SOH/SOHDx/NULx3 6Cx5 120C/SOH/SOH/SOH: From Corollary 4.5.7 , coefficients of like powers of xon the two sides of this equation must be equal; hence, a1D1; a 3/NULa1 2D/NUL1 6; a 5/NULa3 2Ca1 24D1 120; so a1D1; a 3D/NUL1 6C1 2.1/D1 3; a 5D1 120C1 2/DC21 3/DC3 /NUL1 24.1/D2 15: Section 4.5 Power Series 271 Therefore, tanxDxCx3 3C2 15x5C/SOH/SOH/SOH: Example 4.5.14 To find the reciprocal of the power series g.x/D1CexD2C1X nD1xn nŠ; we lethD1in (4.5.25 ). If 1 g.x/D1X nD0anxn; then 1D.a0Ca1xCa2x2Ca3x3C/SOH/SOH/SOH//DC2 2CxCx2 2Cx3 6C/SOH/SOH/SOH/DC3 D2a0C.a0C2a1/xC/DLEa0 2Ca1C2a2/DC1 x2 C/DLEa0 6Ca1 2Ca2C2a3/DC1 x3C/SOH/SOH/SOH: From Corollary 4.5.7 , 2a0D1; a0C2a1D0; a0 2Ca1C2a2D0; a0 6Ca1 2Ca2C2a3D0: Solving these equations successively yields a0D1 2; a1D/NULa0 2D/NUL1 4; a2D/NUL1 2/DLEa0 2Ca1/DC1 D/NUL1 2/DC21 4/NUL1 4/DC3 D0; a3D/NUL1 2/DLEa0 6Ca1 2Ca2/DC1 D/NUL1 2/DC21 12/NUL1 8C0/DC3 D1 48; so 1 1CexD1 2/NULx 4Cx3 48C/SOH/SOH/SOH: 272 Chapter 4 Infinite Sequences and Series Example 4.5.15 To find the reciprocal of g.x/DexD1X nD0xn nŠ; (4.5.27) we again lethD1in (4.5.25 ). If .ex//NUL1D1X nD0anxn; then 1D 1X nD0anxn! 1X nD0xn nŠ! D1X nD0cnxn; where cnDnX rD0ar .n/NULr/Š: From Corollary 4.5.7 ,c0Da0D1andcnD0ifn/NAK1; hence, anD/NULn/NUL1X rD0ar .n/NULr/Š; n/NAK1: (4.5.28) Solving these equations successively for a0,a1, . . . yields a1D/NUL1 1Š(4.5.1 )D/NUL1; a2D/NUL/DC41 2Š.1/C1 1Š./NUL1//NAK D1 2; a3D/NUL/DC41 3Š.1/C1 2Š./NUL1/C1 1Š/DC21 2/DC3/NAK D/NUL1 6 a4D/NUL/DC41 4Š.1/C1 3Š./NUL1/C1 2Š/DC21 2/DC3 C1 1Š/DC2 /NUL1 6/DC3/NAK D1 24: From this, we see that akD./NUL1/k kŠ for0/DC4k/DC44and are led to conjecture that this holds for all k. To prove this by induction, we assume that it is so for 0/DC4k/DC4n/NUL1and compute from ( 4.5.28 ): anD/NULn/NUL1X rD01 .n/NULr/Š./NUL1/r rŠ D/NUL1 nŠn/NUL1X rD0./NUL1/r n r! (Exercise 1.2.19(a)) D./NUL1/n nŠ(Exercise 1.2.19(b)): Section 4.5 Power Series 273 Thus, we have shown that .ex//NUL1D1X nD0./NUL1/nxn nŠ: Since this is precisely the series that results if xis replaced by/NULxin (4.5.27 ), we have verified a fundamental property of the exponential function : that .ex//NUL1De/NULx: This also follows from Example 4.3.26 . Abel’s Theorem From Theorem 4.5.4 , we know that a function fdefined by a convergent power series f.x/D1X nD0an.x/NULx0/n;jx/NULx0j<R; (4.5.29) is continuous in the open interval .x0/NULR;x 0CR/. The next theorem concerns the behavior offasxapproaches an endpoint of the interval of convergence. Theorem 4.5.12 (Abel’s Theorem) Letfbe defined by a power series (4.5.29 ) with finite radius of convergence R: (a) IfP1 nD0anRnconverges;then lim x!.x0CR//NULf.x/D1X nD0anRn: (b) IfP1 nD0./NUL1/nanRnconverges;then lim x!.x0/NULR/Cf.x/D1X nD0./NUL1/nanRn: Proof We consider a simpler problem first. Let g.y/D1X nD0bnyn and 1X nD0bnDs(finite): We will show that lim y!1/NULg.y/Ds: (4.5.30) 274 Chapter 4 Infinite Sequences and Series From Example 4.5.11 , g.y/D.1/NULy/1X nD0snyn; (4.5.31) where snDb0Cb1C/SOH/SOH/SOHCbn: Since 1 1/NULyD1X nD0ynand therefore 1D.1/NULy/1X nD0yn;jyj<1; (4.5.32) we can multiply through by sand write sD.1/NULy/1X nD0syn;jyj<1: Subtracting this from ( 4.5.31 ) yields g.y//NULsD.1/NULy/1X nD0.sn/NULs/yn;jyj<1: If/SI>0 , chooseNso that jsn/NULsj</SI ifn/NAKNC1: Then, if0<y<1 , jg.y//NULsj/DC4.1/NULy/NX nD0jsn/NULsjynC.1/NULy/1X nDNC1jsn/NULsjyn <.1/NULy/NX nD0jsn/NULsjynC.1/NULy//SIyNC11X nD0yn <.1/NULy/NX nD0jsn/NULsjC/SI; because of the second equality in ( 4.5.32 ). Therefore, jg.y//NULsj<2/SI if .1/NULy/NX nD0jsn/NULsj</SI: This proves ( 4.5.30 ). To obtain (a) from this, let bnDanRnandg.y/Df.x 0CRy/; to obtain (b), let bnD./NUL1/nanRnandg.y/Df.x 0/NULRy/. Section 4.5 Power Series 275 Example 4.5.16 The series f.x/D1 1CxD1X nD0./NUL1/nxn diverges atxD1, while lim x!1/NULf.x/D1=2. This shows that the converse of Abel’s theorem is false. Integrating the series term by term yields log.1Cx/D1X nD0./NUL1/nxnC1 nC1;jxj<1; where the power series converges at xD1, and Abel’s theorem implies that log2D1X nD0./NUL1/nC1 nC1: Example 4.5.17 Ifq/NAK0, the binomial series 1X nD0 q n! xn converges absolutely for xD˙1. This is obvious if qis a nonnegative integer, and it follows from Raabe’s test for other positive values of q, since ˇˇˇˇanC1 anˇˇˇˇDˇˇˇˇˇ q nC1!/RS q n!ˇˇˇˇˇDn/NULq nC1; n>q; and lim n!1n/DC2ˇˇˇˇanC1 anˇˇˇˇ/NUL1/DC3 Dlim n!1n/DC2n/NULq nC1/NUL1/DC3 Dlim n!1n nC1./NULq/NUL1/D/NULq/NUL1: Therefore, Abel’s theorem and ( 4.5.21 ) imply that 1X nD0 q n! D2qand1X nD0./NUL1/n q n! D0; q/NAK0: 4.5 Exercises 1. The possibilities listed in Theorem 4.5.2(c) for behavior of a power series at the endpoints of its interval of convergence do not include abso lute convergence at one endpoint and conditional convergence or divergence at the o ther. Why can’t these occur? 276 Chapter 4 Infinite Sequences and Series 2. Find the radius of convergence. (a)X/DC2nC1 n/DC3n2 Œ2C./NUL1/n/c141nxn(b)P2pn.x/NUL1/n (c)X/DLE 2Csinn/EM 6/DC1n .xC2/n(d)Pnpnxn (e)X/DLEx n/DC1n 3. (a) Prove: Iffanrngis bounded andjx1/NULx0j< r, thenPan.x1/NULx0/ncon- verges. (b) Prove: IfPan.x/NULx0/nhas radius of convergence Randjx1/NULx0j> R , thenfan.x1/NULx0/ngis unbounded. 4. Prove: Ifgis a rational function defined for all nonnegative integers, thenPanxn andPang.n/xnhave the same radius of convergence. H INT:Use Exercise 4.1.30.a/: 5. Suppose that f.x/DPan.x/NULx0/nhas radius of convergence Rand0 < r < R1<R. Show that there is an integer ksuch that ˇˇˇˇˇf.x//NULkX nD0an.x/NULx0/nˇˇˇˇˇ/DC4/DC2r R1/DC3kC1R1 R1/NULr ifjx/NULx0j/DC4randk/NAKk. 6. Suppose that kis a positive integer and f.x/D1X nD0anxn has radius of convergence R. Show that the series g.x/Df.xk/D1X nD0anxkn has radius of convergence R1=k. 7. Complete the proof of Theorem 4.5.3 by showing that (a)RD0if lim n!1janC1jıjanjD1 ; (b)RD1 if lim n!1janC1jıjanjD0. 8. Find the radius of convergence. (a)P.logn/xn(b)P2nnp.xC1/n (c)X ./NUL1/n 2n n! xn(d)X ./NUL1/nn2C1 n4n.x/NUL1/n (e)Xnn nŠ.xC2/n(f)X˛.˛C1//SOH/SOH/SOH.˛Cn/NUL1/ ˇ.ˇC1//SOH/SOH/SOH.ˇCn/NUL1/xn (˛,ˇ¤negative integer) Section 4.5 Power Series 277 9. Suppose that an¤0fornsufficiently large. Show that (a) lim n!1ˇˇˇˇanC1 anˇˇˇˇ/DC4lim n!1janj1=nand(b) lim n!1janj1=n/DC4lim n!1ˇˇˇˇanC1 anˇˇˇˇ: Show that this implies Theorem 4.5.3 . 10. Given that 1 1/NULxD1X nD0xn;jxj<1; use Theorem 4.5.4 to expressP1 nD0n2xnin closed form. 11. The function Jp.x/D1X nD0./NUL1/n nŠ.nCp/Š/DLEx 2/DC12nCp .pDinteger/NAK0/ is the Bessel function of order p. Show that (a)J0 0D/NULJ1. (b)J0 pD1 2.Jp/NUL1/NULJpC1/; p/NAK1. (c)x2J00 pCxJ0 pC.x2/NULp2/JpD0. 12. Given that the power series f.x/DP1 nD0anxnsatisfies f0.x/D/NUL2xf.x/; f.0/D1; findfang. Do you recognize f? 13. Let f.x/D1X nD0anxn;jxj<R; andg.x/Df.xk/, wherekis a positive integer. Show that g.r/.0/D0ifr¤kn andg.kn/.0/D.kn/Š nŠf.n/.0/; n/NAK0: 14. Let f.x/D1X nD0an.x/NULx0/n;jx/NULx0j<R; andf.tn/D0, wheretn¤x0and lim n!1tnDx0. Show that f.x//DC10 .jx/NULx0j<R/ . HINT:Rolle’s theorem helps here : 15. Prove Theorem 4.5.8 . 16. ExpressZx 1logt t/NUL1dt as a power series in x/NUL1and find the radius of convergence of the series. 278 Chapter 4 Infinite Sequences and Series 17. By substituting/NULx2forxin the geometric series, we obtain 1 1Cx2D1X nD0./NUL1/nx2n;jxj<1: Use this to express f.x/DTan/NUL1x .f.0/D0/as a power series in x. Then evaluate all derivatives of fatx0D0, and find a series of constants that converges to/EM=6. 18. Prove: If f.x/D1X nD0an.x/NULx0/n;jx/NULx0j<R; andFis an antiderivative of fon.x0/NULR;x 0CR/, then F.x/DCC1X nD0an nC1.x/NULx0/nC1;jx/NULx0j<R; whereCis a constant. 19. Suppose that some derivative of fcan be represented by a power series in x/NULx0 in an interval about x0. Show thatfand all its derivatives can also. 20. Verify Eqn. ( 4.5.21 ) by showing that .1Cx//NULq1X nD0 q n! xnD1;jxj<1; HINT:Differentiate: 21. Prove Theorem 4.5.10 . 22. Find the Maclaurin series of cosh xand sinhxfrom the definition in Eqn. ( 4.5.16 ), and also by applying Theorem 4.5.10 to the Maclaurin series for exande/NULx. 23. Give an example where the radius of convergence of the produc t of two power series is greater than the smaller of the radii of convergence of the factors. 24. Use Theorem 4.5.11 to find the first four nonzero terms in the Maclaurin. (a)exsinx(b)e/NULx 1Cx2(c)cosx 1Cx6(d).sinx/log.1Cx/ 25. Derive the identity 2sinxcosxDsin2x from the Maclaurin series for sin x, cosx, and sin2x. 26. (a) Given that .1/NUL2xtCx2//NUL1=2D1X nD0Pn.t/xn;jxj<1; . A/ Section 4.5 Power Series 279 if/NUL1<t <1 , show thatP0.t/D1,P1.t/Dt, and PnC1.t/D2nC1 nC1tPn.t//NULn nC1Pn/NUL1.t/; n/NAK1: HINT:First differentiate (A)with respect to x: (b) Show from (a) thatPnis a polynomial of degree n. It is thenthLegendre polynomial , and.1/NUL2xtCx2//NUL1=2is the generating function of the sequence fPng. 27. Define (if necessary) the given function so as to be continuou s atx0D0, and find the first four nonzero terms of its Maclaurin series. (a)xex sinx(b)cosx 1CxCx2(c)secx (d)xcscx (e)sin2x sinx 28. Leta0Da1D5andanC1Dan/NUL6an/NUL1; n/NAK1. (a) ExpressF.x/DP1 nD0anxnin closed form. (b) WriteFas the difference of two geometric series, and find an explici t formula foran. 29. Starting from the Maclaurin series log.1/NULx/D/NUL1X nD0xnC1 nC1;jxj<1; use Abel’s theorem to evaluate 1X nD01 .nC1/.nC2/: 30. In Example 4.5.17 we saw that 1X nD0 q n! D2q; q/NAK0: Show that this also holds for /NUL1 < q < 0 , but not forq/DC4/NUL1. H INT:See Exer- cise4.1.35: 31. (a) Prove: IfP1 nD0bnconverges, then the series g.x/DP1 nD0bnxnconverges uniformly on Œ0;1/c141 . HINT:If/SI>0 , there is an integer Nsuch that jbnCbnC1C/SOH/SOH/SOHCbmj</SI ifn;m/NAKN: Use summation by parts to show that then jbnxnCbn/NUL1xn/NUL1C/SOH/SOH/SOHCbmxmj<2/SI if0/DC4x<1; n;m/NAKN: This is also known as Abel’s theorem : 280 Chapter 4 Infinite Sequences and Series (b) Show that (a) implies the restricted form of Theorem 4.5.12 (concerningg) proved in the text. 32. Use Exercise 4.5.31 to show that ifP1 nD0an,P1 nD0bn, and their Cauchy productP1 nD0cnall converge, then 1X nD0an! 1X nD0bn! D1X nD0cn: 33. Prove: If g.x/D1X nD0bnxn;jxj<1; andbn/NAK0, then 1X nD0bnDlim x!1/NULg.x/ (finite or infinite) : 34. Use the binomial series and the relation d dx.sin/NUL1x/D.1/NULx2//NUL1=2 to obtain the Maclaurin series for sin/NUL1x .sin/NUL10D0/. Deduce from this series and Exercise 4.5.33 that 1X nD0 2n n! 1 22n.2nC1/D/EM 2: CHAPTER 5 Real-Valued Functions of Several Variables IN THIS CHAPTER we consider real-valued function of nvariables, where n>1 . SECTION 5.1 deals with the structure of Rn, the space of ordered n-tuples of real numbers, which we call vectors . We define the sum of two vectors, the product of a vector and a real number, the length of a vector, and the inner product of t wo vectors. We study the arithmetic properties of Rn, including Schwarz’s inequality and the triangle inequali ty. We define neighborhoods and open sets in Rn, define convergence of a sequence of points in Rn, and extend the Heine–Borel theorem to Rn. The section concludes with a discussion of connected subsets of Rn. SECTION 5.2 deals with boundedness, limits, continuity, an d uniform continuity of a func- tion ofnvariables; that is, a function defined on a subset of Rn. SECTION 5.3 defines directional and partial derivatives of a real-valued function of n variables. This is followed by the definition of differentia blity of such functions. We define the differential of such a function and give a geometric inte rpretation of differentiablity. SECTION 5.4 deals with the chain rule and Taylor’s theorem fo r a real-valued function of nvariables. 5.1 STRUCTURE OF RRRn In this chapter we study functions defined on subsets of the re aln-dimensional space Rn, which consists of all ordered n-tuples XD.x1;x2;:::;x n/of real numbers, called the coordinates orcomponents ofX. This space is sometimes called Euclideann-space . In this section we introduce an algebraic structure for Rn. We also consider its topologi- calproperties; that is, properties that can be described in ter ms of a special class of subsets, the neighborhoods in Rn. In Section 1.3 we studied the topological properties of R1, which we will continue to denote simply as R. Most of the definitions and proofs in Section 1.3 were stated in terms of neighborhoods in R. We will see that they carry over to Rnif the concept of neighborhood in Rnis suitably defined. 281 282 Chapter 5 Real-Valued Functions of nVariables Members of Rhave dual interpretations: geometric, as points on the real line, and alge- braic, as real numbers. We assume that you are familiar with t he geometric interpretation of members of R2andR3as the rectangular coordinates of points in a plane and three - dimensional space, respectively. Although Rncannot be visualized geometrically if n/NAK4, geometric ideas from R,R2, and R3often help us to interpret the properties of Rnfor arbitraryn. As we said in Section 1.3, the idea of neighborhood is always a ssociated with some definition of “closeness” of points. The following definitio n imposes an algebraic structure onRn, in terms of which the distance between two points can be defin ed in a natural way. In addition, this algebraic structure will be useful later f or other purposes. Definition 5.1.1 Thevector sum of XD.x1;x2;:::;x n/and YD.y1;y2;:::;y n/ is XCYD.x1Cy1;x2Cy2;:::;x nCyn/: (5.1.1) Ifais a real number, the scalar multiple of Xbyais aXD.ax1;ax 2;:::;ax n/: (5.1.2) Note that “C” has two distinct meanings in ( 5.1.1 ): on the left, “C” stands for the newly defined addition of members of Rnand, on the right, for addition of real numbers. However, this can never lead to confusion, since the meaning of “ C” can always be deduced from the symbols on either side of it. A similar comment applies to the use of juxtaposition to indicate scalar multiplication on the left of ( 5.1.2 ) and multiplication of real numbers on the right. Example 5.1.1 InR4, let XD.1;/NUL2;6;5/ and YD/NUL 3;/NUL5;4;1 2/SOH : Then XCYD/NUL 4;/NUL7;10;11 2/SOH and 6XD.6;/NUL12;36;30/: We leave the proof of the following theorem to you (Exercise 5.1.2 ). Section 5.1 Structure of Rn283 Theorem 5.1.2 IfX;Y;andZare in Rnandaandbare real numbers ;then (a) XCYDYCX.vector addition is commutative /: (b).XCY/CZDXC.YCZ/.vector addition is associative /: (c) There is a unique vector 0;called the zero vector ;such that XC0DXfor all Xin Rn: (d) For each XinRnthere is a unique vector /NULXsuch that XC./NULX/D0: (e)a.bX/D.ab/X: (f).aCb/XDaXCbX: (g)a.XCY/DaXCaY: (h)1XDX: Clearly, 0D.0;0;:::;0/ and, if XD.x1;x2;:::;x n/, then /NULXD./NULx1;/NULx2;:::;/NULxn/: We write XC./NULY/asX/NULY. The point 0is called the origin . A nonempty set VDfX;Y;Z;:::g, together with rules such as ( 5.1.1 ), associating a unique member of Vwith every ordered pair of its members, and ( 5.1.2 ), associating a unique member of Vwith every real number and member of V, is said to be a vector space if it has the properties listed in Theorem 5.1.2 . The members of a vector space are called vectors . When we wish to emphasize that we are regarding a member of Rnas part of this algebraic structure, we will speak of it as a vector; otherwi se, we will speak of it as a point. Length, Distance, and Inner Product Definition 5.1.3 Thelength of the vector XD.x1;x2;:::;x n/is jXjD.x2 1Cx2 2C/SOH/SOH/SOHCx2 n/1=2: Thedistance between points XandYisjX/NULYj; in particular,jXjis the distance between Xand the origin. IfjXjD1, then Xis aunit vector . IfnD1, this definition of length reduces to the familiar absolute v alue, and the distance between two points is the length of the interval having them a s endpoints; for nD2and nD3, the length and distance of Definition 5.1.3 reduce to the familiar definitions for the plane and three-dimensional space. Example 5.1.2 The lengths of the vectors XD.1;/NUL2;6;5/ and YD/NUL3;/NUL5;4;1 2/SOH are jXjD.12C./NUL2/2C62C52/1=2Dp 66 284 Chapter 5 Real-Valued Functions of nVariables and jYjD.32C./NUL5/2C42C.1 2/2/1=2Dp 201 2: The distance between XandYis jX/NULYjD..1/NUL3/2C./NUL2C5/2C.6/NUL4/2C.5/NUL1 2/2/1=2Dp 149 2: Definition 5.1.4 Theinner product X/SOHYofXD.x1;x2;:::;x n/andYD.y1;y2;:::;y n/ is X/SOHYDx1y1Cx2y2C/SOH/SOH/SOHCxnyn: Lemma 5.1.5 ( Schwarz ’s Inequality) IfXandYare any two vectors in Rn; then jX/SOHYj/DC4j XjjYj; (5.1.3) with equality if and only if one of the vectors is a scalar mult iple of the other : Proof IfYD0, then both sides of ( 5.1.3 ) are 0, so ( 5.1.3 ) holds, with equality. In this case, YD0X. Now suppose that Y¤0andtis any real number. Then 0/DC4nX iD1.xi/NULtyi/2 DnX iD1x2 i/NUL2tnX iD1xiyiCt2nX iD1y2 i DjXj2/NUL2.X/SOHY/tCt2jYj2:(5.1.4) The last expression is a second-degree polynomial pint. From the quadratic formula, the zeros ofpare tD.X/SOHY/˙p .X/SOHY/2/NULjXj2jYj2 jYj2: Hence, .X/SOHY/2/DC4jXj2jYj2; (5.1.5) because if not, then pwould have two distinct real zeros and therefore be negative between them (Figure 5.1.1 ), contradicting the inequality ( 5.1.4 ). Taking square roots in ( 5.1.5 ) yields ( 5.1.3 ) ifY¤0. IfXDtY, thenjX/SOHYj D j XjjYjD jtjjYj2(verify), so equality holds in ( 5.1.3 ). Conversely, if equality holds in ( 5.1.3 ), thenphas the real zero t0D.X/SOHY/=jYk2, and nX iD1.xi/NULt0yi/2D0 from ( 5.1.4 ); therefore, XDt0Y. Section 5.1 Structure of Rn285 y ty = p(t) r1 r2 Figure 5.1.1 Theorem 5.1.6 (Triangle Inequality) IfXandYare in Rn;then jXCYj/DC4j XjCjYj; (5.1.6) with equality if and only if one of the vectors is a nonnegativ e multiple of the other : Proof By definition, jXCYj2DnX iD1.xiCyi/2DnX iD1x2 iC2nX iD1xiyiCnX iD1y2 i DjXj2C2.X/SOHY/CjYj2 /DC4jXj2C2jXjjYjCjYj2(by Schwarz’s inequality) D.jXjCjYj/2:(5.1.7) Hence, jXCYj2/DC4.jXjCj Yj/2: Taking square roots yields ( 5.1.6 ). From the third line of ( 5.1.7 ), equality holds in ( 5.1.6 ) if and only if X/SOHYDjXjjYj, which is true if and only if one of the vectors XandYis a nonnegative scalar multiple of the other (Lemma 5.1.5 ). Corollary 5.1.7 IfX;Y;andZare in Rn;then jX/NULZj/DC4j X/NULYjCjY/NULZj: Proof Write X/NULZD.X/NULY/C.Y/NULZ/; and apply Theorem 5.1.6 with XandYreplaced by X/NULYandY/NULZ. 286 Chapter 5 Real-Valued Functions of nVariables Corollary 5.1.8 IfXandYare in Rn;then jX/NULYj/NAKjj Xj/NULjYjj: Proof Since XDYC.X/NULY/; Theorem 5.1.6 implies that jXj/DC4j YjCjX/NULYj; which is equivalent to jXj/NULjYj/DC4j X/NULYj: Interchanging XandYyields jYj/NULjXj/DC4j Y/NULXj: SincejX/NULYjDj Y/NULXj, the last two inequalities imply the stated conclusion. Example 5.1.3 The angle between two nonzero vectors XD.x1;x2;x3/andYD .y1;y2;y3/inR3is the angle between the directed line segments from the orig in to the points XandY(Figure 5.1.2 ). X 0 YYX X−Yθ Figure 5.1.2 Applying the law of cosines to the triangle in Figure 5.1.2 yields jX/NULYj2DjXj2CjYj2/NUL2jXjjYjcos/DC2: (5.1.8) However, jX/NULYj2D.x1/NULy1/2C.x2/NULy2/2C.x3/NULy3/2 D.x2 1Cx2 2Cx2 3/C.y2 1Cy2 2Cy2 3//NUL2.x1y1Cx2y2Cx3y3/ DjXj2CjYj2/NUL2X/SOHY: Section 5.1 Structure of Rn287 Comparing this with ( 5.1.8 ) yields X/SOHYDjXjjYjcos/DC2: Sincejcos/DC2j/DC41, this verifies Schwarz’s inequality in R3. Example 5.1.4 Connecting the points 0,X,Y, and XCYinR2orR3(Figure 5.1.3 ) produces a parallelogram with sides of length jXjandjYjand a diagonal of length jXCYj. 0X YYY X XX+YX+Y Figure 5.1.3 Thus, there is a triangle with sides jXj,jYj, andjXCYj. From this, we see geometrically that jXCYj/DC4j XjCjYj inR2orR3, since the length of one side of a triangle cannot exceed the s um of the lengths of the other two. This verifies ( 5.1.6 ) forR2andR3and indicates why ( 5.1.6 ) is called the triangle inequality. The next theorem lists properties of length, distance, and i nner product that follow di- rectly from Definitions 5.1.3 and5.1.4 . We leave the proof to you (Exercise 5.1.6 ). Theorem 5.1.9 IfX;Y;andZare members of Rnandais a scalar, then (a)jaXjDjajjXj: (b)jXj/NAK0;with equality if and only if XD0: (c)jX/NULYj/NAK0;with equality if and only if XDY: (d) X/SOHYDY/SOHX: (e) X/SOH.YCZ/DX/SOHYCX/SOHZ: (f).cX//SOHYDX/SOH.cY/Dc.X/SOHY/: 288 Chapter 5 Real-Valued Functions of nVariables Line Segments in RRRn The equation of a line through a point X0D.x0;y0;´0/inR3can be written parametri- cally as xDx0Cu1t; yDy0Cu2t; ´D´0Cu3t;/NUL1<t <1; whereu1,u2, andu3are not all zero. We write this in vector form as XDX0CtU;/NUL1<t <1; (5.1.9) with UD.u1;u2;u3/, and we say that the line is through X0in the direction of U. There are many ways to represent a given line parametrically . For example, XDX0CsV;/NUL1<s<1; (5.1.10) represents the same line as ( 5.1.9 ) if and only if VDaUfor some nonzero real number a. Then the line is traversed in the same direction as sandtvary from/NUL1 to1ifa>0 , or in opposite directions if a<0 . To write the parametric equation of a line through two points X0andX1inR3, we take UDX1/NUL0in (5.1.9 ), which yields XDX0Ct.X1/NULX0/DtX1C.1/NULt/X0;/NUL1<t <1: The line segment from X0toX1consists of those points for which 0/DC4t/DC41. Example 5.1.5 The lineLdefined by xD/NUL1C2t; yD3/NUL4t; ´D/NUL1;/NUL1<t <1; which can be rewritten as XD./NUL1;3;/NUL1/Ct.2;/NUL4;0/;/NUL1<t <1; (5.1.11) is through X0D./NUL1;3;/NUL1/in the direction of UD.2;/NUL4;0/ . The same line can be represented by XD./NUL1;3;/NUL1/Cs.1;/NUL2;0/;/NUL1<s<1; (5.1.12) or by XD./NUL1;3;/NUL1/C/FS./NUL4;8;0/;/NUL1</FS <1: (5.1.13) Since .1;/NUL2;0/D1 2.2;/NUL4;0/; Lis traversed in the same direction as tandsvary from/NUL1 to1in (5.1.11 ) and ( 5.1.12 ). However, since ./NUL4;8;0/D/NUL2.2;/NUL4;0/; Section 5.1 Structure of Rn289 Lis traversed in opposite directions as tand/FSvary from/NUL1 to1in (5.1.11 ) and ( 5.1.13 ). SettingtD1in (5.1.11 ), we see that X1D.1;/NUL1;/NUL1/is also onL. The line segment from X0toX1consists of all points of the form XDt.1;/NUL1;/NUL1/C.1/NULt/./NUL1;3;/NUL1/; 0/DC4t/DC41: These familiar notions can be generalized to Rn, as follows: Definition 5.1.10 Suppose that X0andUare in RnandU¤0. Then the line through X0in the direction of Uis the set of all points in Rnof the form XDX0CtU;/NUL1<t <1: A set of points of the form XDX0CtU; t 1/DC4t/DC4t2; is called a line segment . In particular, the line segment from X0toX1is the set of points of the form XDX0Ct.X1/NULX0/DtX1C.1/NULt/X0; 0/DC4t/DC41: Neighborhoods and Open Sets in RRRn Having defined distance in Rn, we are now able to say what we mean by a neighborhood of a point in Rn. Definition 5.1.11 If/SI>0 , the/SI-neighborhood of a point X0inRnis the set N/SI.X0/jD˚XˇˇjX/NULX0j</SI/TAB: An/SI-neighborhood of a point X0inR2is the inside, but not the circumference, of the circle of radius /SIabout X0. InR3it is the inside, but not the surface, of the sphere of radius /SIabout X0. In Section 1.3 we stated several other definitions in terms of /SI-neighborhoods: neigh- borhood ,interior point ,interior of a set ,open set ,closed set ,limit point ,boundary point , boundary of a set ,closure of a set ,isolated point ,exterior point , and exterior of a set . Since these definitions are the same for Rnas for R, we will not repeat them. We advise you to read them again in Section 1.3, substituting RnforRandX0forx0. Example 5.1.6 LetSbe the set of points in R2in the square bounded by the lines xD˙1,yD˙1, except for the origin and the points on the vertical lines xD˙1 (Figure 5.1.4 , page 290); thus, SD˚ .x;y/ˇˇ.x;y/¤.0;0/;/NUL1<x<1;/NUL1/DC4y/DC41/TAB : 290 Chapter 5 Real-Valued Functions of nVariables Every point of Snot on the lines yD˙1is an interior point, so S0D˚ .x;y/ˇˇ.x;y/¤.0;0/;/NUL1<x;y<1/TAB : Sis a deleted neighborhood of .0;0/ and is neither open nor closed. The closure of Sis SD˚.x;y/ˇˇ/NUL1/DC4x;y/DC41/TAB; and every point of Sis a limit point of S. The origin and the perimeter of Sform@S, the boundary ofS. The exterior of Sconsists of all points .x;y/ such thatjxj>1orjyj>1. The origin is an isolated point of Sc. y x(1, 1) (−1, 1) (1, −1) (−1, −1)x Figure 5.1.4 Example 5.1.7 IfX0is a point in Rnandris a positive number, the openn-ball of radiusrabout X0is the setBr.X0/D˚ XˇˇjX/NULX0j<r/TAB . (Thus,/SI-neighborhoods are openn-balls.) If X1is inSr.X0/and jX/NULX1j</SIDr/NULjX/NULX0j; then Xis inSr.X0/. (The situation is depicted in Figure 5.1.5 fornD2.) Thus,Sr.X0/contains an/SI-neighborhood of each of its points, and is therefore open. We leave it to you (Exercise 5.1.13 ) to show that the closure of Br.X0/is the closedn-ball of radiusrabout X0, defined by Section 5.1 Structure of Rn291 Sr.X0/D˚XˇˇjX/NULX0j/DC4r/TAB: X0 X1 X r r− X1−X0 Figure 5.1.5 Open and closed n-balls are generalizations to Rnof open and closed intervals. The following lemma will be useful later in this section, whe n we consider connected sets. Lemma 5.1.12 IfX1andX2are inSr.X0/for somer >0 , then so is every point on the line segment from X1toX2: Proof The line segment is given by XDtX2C.1/NULt/X1; 0<t <1: Suppose that r >0 . If jX1/NULX0j<r;jX2/NULX0j<r; and0<t <1 , then jX/NULX0jDjtX2C.1/NULt/X1/NULtX0/NUL.1/NULt/X0j Djt.X2/NULX0/C.1/NULt/X1/NULX0/j /DC4tjX2/NULX0jC.1/NULt/jX1/NULX0j <trC.1/NULt/rDr: The proofs in Section 1.3 of Theorem 1.3.3 (the union of open sets is open, the intersec- tion of closed sets is closed) and Theorem 1.3.5 and its Corollary 1.3.6 (a set is closed if and only if it contains all its limit points) are also valid in Rn. You should reread them now. 292 Chapter 5 Real-Valued Functions of nVariables The Heine–Borel theorem (Theorem 1.3.7 ) also holds in Rn, but the proof in Section 1.3 is valid only for nD1. To prove the Heine–Borel theorem for general n, we need some preliminary definitions and results that are of interest in t heir own right. Definition 5.1.13 A sequence of points fXrginRnconverges to the limit Xif lim r!1jXr/NULXjD0: In this case we write lim r!1XrDX: The next two theorems follow from this, the definition of dist ance in Rn, and what we already know about convergence in R. We leave the proofs to you (Exercises 5.1.16 and 5.1.17 ). Theorem 5.1.14 Let XD.x1;x2;:::;xn/and XrD.x1r;x2r;:::;x nr/; r/NAK1: Then limr!1XrDXif and only if lim r!1xirDxi; 1/DC4i/DC4nI that is;a sequencefXrgof points in Rnconverges to a limit Xif and only if the sequences of components offXrgconverge to the respective components of X: Theorem 5.1.15 (Cauchy’s Convergence Criterion) A sequencefXrgin Rnconverges if and only if for each /SI>0 there is an integer Ksuch that jXr/NULXsj</SI ifr;s/NAKK: The next definition generalizes the definition of the diamete r of a circle or sphere. Definition 5.1.16 IfSis a nonempty subset of Rn, then d.S/Dsup˚jX/NULYjˇˇX;Y2S/TAB is the diameter ofS. Ifd.S/<1;SisboundedIifd.S/D1 ,Sisunbounded . Theorem 5.1.17 (Principle of Nested Sets) IfS1;S2;. . . are closed nonempty subsets of Rnsuch that S1/ESCS2/ESC/SOH/SOH/SOH/ESCSr/ESC/SOH/SOH/SOH (5.1.14) and lim r!1d.S r/D0; (5.1.15) then the intersection ID1\ rD1Sr contains exactly one point : Section 5.1 Structure of Rn293 Proof LetfXrgbe a sequence such that Xr2Sr.r/NAK1/. Because of ( 5.1.14 ),Xr2Sk ifr/NAKk, so jXr/NULXsj<d.S k/ifr;s/NAKk: From ( 5.1.15 ) and Theorem 5.1.15 ,Xrconverges to a limit X. Since Xis a limit point of everySkand everySkis closed, Xis in everySk(Corollary 1.3.6 ). Therefore, X2I, so I¤;. Moreover, Xis the only point in I, since if Y2I, then jX/NULYj/DC4d.S k/; k/NAK1; and ( 5.1.15 ) implies that YDX. We can now prove the Heine–Borel theorem for Rn. This theorem concerns compact sets. As in R, a compact set in Rnis a closed and bounded set. Recall that a collection Hof open sets is an open covering of a set Sif S/SUB[˚ HˇˇH2H/TAB : Theorem 5.1.18 (Heine–Borel Theorem) IfHis an open covering of a com- pact subsetS;thenScan be covered by finitely many sets from H: Proof The proof is by contradiction. We first consider the case wher enD2, so that you can visualize the method. Suppose that there is a coverin gHforSfrom which it is impossible to select a finite subcovering. Since Sis bounded,Sis contained in a closed square TDf.x;y/ja1/DC4x/DC4a1CL;a 2/DC4x/DC4a2CLg with sides of length L(Figure 5.1.6 ). T(1) S(1)S(2) S(3)S(4)T(2) T(3)T(4) Figure 5.1.6 294 Chapter 5 Real-Valued Functions of nVariables Bisecting the sides of Tas shown by the dashed lines in Figure 5.1.6 leads to four closed squares,T.1/;T.2/,T.3/, andT.4/, with sides of length L=2. Let S.i/DS\T.i/; 1/DC4i/DC44: EachS.i/, being the intersection of closed sets, is closed, and SD4[ iD1S.i/: Moreover, Hcovers eachS.i/, but at least one S.i/cannot be covered by any finite sub- collection of H, since if all the S.i/could be, then so could S. LetS1be a set with this property, chosen from S.1/,S.2/,S.3/, andS.4/. We are now back to the situation we started from: a compact set S1covered by H, but not by any finite subcollection of H. However,S1is contained in a square T1with sides of length L=2 instead ofL. Bisecting the sides ofT1and repeating the argument, we obtain a subset S2ofS1that has the same properties as S, except that it is contained in a square with sides of length L=4. Continuing in this way produces a sequence of nonempty closed sets S0.DS/,S1,S2, . . . , such that Sk/ESCSkC1andd.S k//DC4L=2k/NUL1=2.k/NAK0/. From Theorem 5.1.17 , there is a point XinT1 kD1Sk. Since X2S, there is an open set HinHthat contains X, and thisHmust also contain some /SI-neighborhood of X. Since every XinSksatisfies the inequality jX/NULXj/DC42/NULkC1=2L; it follows that Sk/SUBHforksufficiently large. This contradicts our assumption on H, which led us to believe that no Skcould be covered by a finite number of sets from H. Consequently, this assumption must be false: Hmust have a finite subcollection that covers S. This completes the proof for nD2. The idea of the proof is the same for n > 2 . The counterpart of the square Tis the hypercube with sides of length L: TD˚.x1;x2;:::;x n/ˇˇai/DC4xi/DC4aiCL;iD1;2;:::;n/TAB: Halving the intervals of variation of the ncoordinatesx1,x2, . . . ,xndividesTinto2n closed hypercubes with sides of length L=2: T.i/D˚.x1;x2;:::;x n/ˇˇbi/DC4xi/DC4biCL=2;1/DC4i/DC4n/TAB; wherebiDaiorbiDaiCL=2. If no finite subcollection of HcoversS, then at least one of these smaller hypercubes must contain a subset of Sthat is not covered by any finite subcollection of S. Now the proof proceeds as for nD2. The Bolzano–Weierstrass theorem is valid in Rn; its proof is the same as in R. Connected Sets and Regions Although it is legitimate to consider functions defined on ar bitrary domains, we restricted Section 5.1 Structure of Rn295 our study of functions of one variable mainly to functions de fined on intervals. There are good reasons for this. If we wish to raise questions of contin uity and differentiability at every point of the domain Dof a function f, then every point of Dmust be a limit point ofD0. Intervals have this property. Moreover, the definition ofRb af.x/dx is obviously applicable only if fis defined on Œa;b/c141 . It is not productive to consider questions of continuity and differentiability of functions defined on the union of disjoint intervals, since many import ant results simply do not hold for such domains. For example, the intermediate value theor em (Theorem 2.2.10 ; see also Exercise 2.2.25 ) says that if fis continuous on an interval Iandf.x 1/ < /SYN < f.x 2/ for somex1andx2inI, thenf.x/D/SYNfor somexinI. Theorem 2.3.12 says thatfis constant on an interval Iiff0/DC10onI. Neither of these results holds if Iis the union of disjoint intervals rather than a single interval; thus, if fis defined on ID.0;1/[.2;3/ by f.x/D/SUB1; 0<x<1; 0; 2<x<3; thenfis continuous on I, but does not assume any value between 0and1, andf0/DC10on I, butfis not constant. It is not difficult to see why these results fail to hold for thi s function: the domain of f consists of two disconnected pieces. It would be more sensib le to regardfas two entirely different functions, one defined on .0;1/ and the other on .2;3/ . The two results mentioned are valid for each of these functions. As we will see when we study functions defined on subsets of Rn, considerations like those just cited as making it natural to consider functions d efined on intervals in Rlead us to single out a preferred class of subsets as domains of fun ctions ofnvariables. These subsets are called regions . To define this term, we first need the following definition. Definition 5.1.19 A subsetSofRnisconnected if it is impossible to represent Sas the union of two disjoint nonempty sets such that neither con tains a limit point of the other; that is, ifScannot be expressed as SDA[B, where A¤;; B¤;;A\BD;;andA\BD;: (5.1.16) IfScan be expressed in this way, then Sisdisconnected . Example 5.1.8 The empty set and singleton sets are connected, because they cannot be represented as the union of two disjoint nonempty sets. Example 5.1.9 The space Rnis connected, because if RnDA[BwithA\BD; andA\BD;, thenA/SUBAandB/SUBB; that is,AandBare both closed and therefore are both open. Since the only nonempty subset of Rnthat is both open and closed is Rn itself (Exercise 5.1.21 ), one ofAandBisRnand the other is empty. 296 Chapter 5 Real-Valued Functions of nVariables y x(3, 3) (3, 2) (1, 1)(1, 2) Figure 5.1.7 IfX1;X2;:::; Xkare points in RnandLiis the line segment from XitoXiC1,1/DC4i/DC4 k/NUL1, we say that L1,L2, . . . ,Lk/NUL1form a polygonal path from X1toXk, and that X1 andXkareconnected by the polygonal path. For example, Figure 5.1.7 shows a polygonal path in R2connecting.0;0/ to.3;3/ . A setSispolygonally connected if every pair of points inScan be connected by a polygonal path lying entirely in S. Theorem 5.1.20 An open setSinRnis connected if and only if it is polygonally connected: Proof For sufficiency, we will show that if Sis disconnected, then Sis not polygonally connected. Let SDA[B, whereAandBsatisfy ( 5.1.16 ). Suppose that X12Aand X22B, and assume that there is a polygonal path in Sconnecting X1toX2. Then some line segment Lin this path must contain a point Y1inAand a point Y2inB. The line segment XDtY2C.1/NULt/Y1; 0/DC4t/DC41; is part ofLand therefore in S. Now define /SUBDsup˚ /FSˇˇtY2C.1/NULt/Y12A; 0/DC4t/DC4/FS/DC41/TAB ; and let X/SUBD/SUBY2C.1/NUL/SUB/Y1: Then X/SUB2A\B. However, since X/SUB2A[BandA\BDA\BD;, this is impossible. Therefore, the assumption that there is a polygonal path in Sfrom X1toX2must be false. Section 5.1 Structure of Rn297 For necessity, suppose that Sis a connected open set and X02S. LetAbe the set consisting of X0and the points in Scan be connected to X0by polygonal paths in S. Let Bbe set of points in Sthat cannot be connected to X0by polygonal paths. If Y02S, then Scontains an/SI-neighborhood N/SI.Y0/ofY0, sinceSis open. Any point Y1inN/SI.Y0can be connected to Y0by the line segment XDtY1C.1/NULt/Y0; 0/DC4t/DC41; which lies in N/SI.Y0/(Lemma 5.1.12 ) and therefore in S. This implies that Y0can be connected to X0by a polygonal path in Sif and only if every member of N/SI.Y0/can also. Thus,N/SI.Y0//SUBAifY02A, andN/SI.Y0/2BifY02B. Therefore,AandBare open. SinceA\BD;, this implies that A\BDA\BD; (Exercise 5.1.14 ). SinceAis nonempty.X02A/, it now follows that BD;, since ifB¤;,Swould be disconnected (Definition 5.1.19 ). Therefore, ADS, which completes the proof of necessity. We did not use the assumption that Sis open in the proof of sufficiency. In fact, we actu- ally proved that any polygonally connected set, open or not, is connected. The converse is false. A set (not open) may be connected but not polygonally c onnected (Exercise 5.1.29 ). Our study of functions on Rnwill deal mostly with functions whose domains are regions, defined next. Definition 5.1.21 AregionSinRnis the union of an open connected set with some, all, or none of its boundary; thus, S0is connected, and every point of Sis a limit point of S0. Example 5.1.10 Intervals are the only regions in R(Exercise 5.1.31 ). Then-ball Br.X0/(Example 5.1.7 ) is a region in Rn, as is its closure Sr.X0/. The set SD˚ .x;y/ˇˇx2Cy2/DC41orx2Cy2/NAK4/TAB (Figure 5.1.8(a), page 298) is not a region in R2, since it is not connected. The set S1 obtained by adding the line segment L1WXDt.0;2/C.1/NULt/.0;1/; 0<t <1; toS(Figure 5.1.8(b)) is connected but is not a region, since points on the line seg ment are not limit points of S0 1. The setS2obtained by adding to S1the points in the first quadrant bounded by the circles x2Cy2D1andx2Cy2D4and the line segments L1and L2WXDt.2;0/C.1/NULt/.1;0/; 0<t <1 (Figure 5.1.8(c)), is a region. More about Sequences in RRRn From Definition 5.1.13 , a sequencefXrgof points in Rnconverges to a limit Xif and only if for every/SI>0 there is an integer Ksuch that jXr/NULXj</SI ifr/NAKK: 298 Chapter 5 Real-Valued Functions of nVariables TheRndefinitions of divergence, boundedness, subsequence, and s ums, differences, and constant multiples of sequences are analogous to those give n in Sections 4.1 and 4.2 for the case where nD1. Since Rnis not ordered for n>1 , monotonicity, limits inferior and superior of sequences in Rn, and divergence to˙1 are undefined for n>1 . Products and quotients of members of Rnare also undefined if n>1 . L2L1 (c)(a)L1 (b)y xy x y x Figure 5.1.8 Several theorems from Sections 4.1 and 4.2 remain valid for s equences in Rn, with proofs unchanged, provided that “ j j" is interpreted as distance in Rn. (A trivial change is re- quired: the subscript n, used in Sections 4.1 and 4.2 to identify the terms of the sequ ence, must be replaced, since nhere stands for the dimension of the space.) These include Th e- orems 4.1.2 (uniqueness of the limit), 4.1.4 (boundedness of a convergent sequence), parts of4.1.8 (concerning limits of sums, differences, and constant mult iples of convergent se- quences), and 4.2.2 (every subsequence of a convergent sequence converges to th e limit of the sequence). Section 5.1 Structure of Rn299 5.1 Exercises WithRreplaced by Rn, the following exercises from Section 1:3are also suitable for this section: 1.3.7 -1.3.10;1.3.12 -1.3.15;1.3.19;1.3.20.except(e)/;and1.3.21: 1. FindaXCbY. (a) XD.1;2;/NUL3;1/,YD.0;/NUL1;2;0/ ,aD3,bD6 (b) XD.1;/NUL1;2/,YD.0;/NUL1;3/,aD/NUL1,bD2 (c) XD.1 2;3 2;1 4;1 6/,YD./NUL1 2;1;5;1 3/,aD1 2,bD1 6 2. Prove Theorem 5.1.2 . 3. FindjXj. (a).1;2;/NUL3;1/ (b)/NUL1 2;1 3;1 4;1 6/SOH (c).1;2;/NUL1;3;4/ (d).0;1;0;/NUL1;0;/NUL1/ 4. FindjX/NULYj. (a) XD.3;4;5;/NUL4/,YD.2;0;/NUL1;2/ (b) XD./NUL1 2;1 2;1 4;/NUL1 4/,YD.1 3;/NUL1 6;1 6;/NUL1 3/ (c) XD.0;0;0/ ,YD.2;/NUL1;2/ (d) XD.3;/NUL1;4;0;/NUL1/,YD.2;0;1;/NUL4;1/ 5. Find X/SOHY. (a) XD.3;4;5;/NUL4/,YD.3;0;3;3/ (b) XD.1 6;11 12;9 8;5 2/,YD./NUL1 2;1 2;1 4;/NUL1 4/ (c) XD.1;2;/NUL3;1;4/ ,YD.1;2;/NUL1;3;4/ 6. Prove Theorem 5.1.9 . 7. Find a parametric equation of the line through X0in the direction of U. (a) X0D.1;2;/NUL3;1/,UD.3;4;5;/NUL4/ (b) X0D.2;0;/NUL1;2;4/ ,UD./NUL1;0;1;3;2/ (c) X0D./NUL1 2;1 2;1 4;/NUL1 4/,UD.1 3;/NUL1 6;1 6;/NUL1 3/ 8. Suppose that U¤0andV¤0. Complete the sentence: The equations XDX0CtU;/NUL1<t <1; and XDX1CsV;/NUL1<s<1; represent the same line in Rnif and only if ... 9. Find the equation of the line segment from X0toX1. (a) X0D.1;/NUL3;4;2/ ,X1D.2;0;/NUL1;5/ (b) X0D.3;1/NUL2;1;4/ ,X1D.2;0;/NUL1;4;/NUL3/ (c) X0D.1;2;/NUL1/,X1D.0;/NUL1;/NUL1/ 300 Chapter 5 Real-Valued Functions of nVariables 10. Find sup˚ /SIˇˇN/SI.X0//SUBS/TAB . (a) X0D.1;2;/NUL1;3/;SDthe open 4-ball of radius 7 about .0;3;/NUL2;2/ (b) X0D.1;2;/NUL1;3/;SD˚.x1;x2;x3;x4/ˇˇjxij/DC45;1/DC4i/DC44/TAB (c) X0D.3;5 2/;SDthe closed triangle with vertices .2;0/ ,.2;2/ , and.4;4/ 11. Find(i)@S;(ii)S;(iii)S0;(iv) exterior ofS. (a)SD˚ .x1;x2;x3;x4/ˇˇjxij<3;iD1;2;3/TAB (b)SD˚.x;y;1/ˇˇx2Cy2/DC41/TAB 12. Describe the following sets as open, closed, or neither. (a)SD˚ .x1;x2;x3;x4/ˇˇjx1j>0;x 2<1;x 3¤/NUL2/TAB (b)SD˚.x1;x2;x3;x4/ˇˇx1D1;x 3¤/NUL4/TAB (c)SD˚.x1;x2;x3;x4/ˇˇx1D1;/NUL3/DC4x2/DC41;x 4D/NUL5/TAB 13. Show that the closure of the open n-ball Br.X0/D˚ XˇˇjX/NULX0j<r/TAB is the closedn-ball Br.X0/D˚XˇˇjX/NULX0j/DC4r/TAB: 14. Prove: IfAandBare open and A\BD;, thenA\BDA\BD;. 15. Show that if lim r!1Xrexists, then it is unique. 16. Prove Theorem 5.1.14 . 17. Prove Theorem 5.1.15 . 18. Find lim r!1Xr. (a) XrD/DLE rsin/EM r;cos/EM r;e/NULr/DC1 (b) XrD/DC2 1/NUL1 r2;logrC1 rC2;/DC2 1C1 r/DC3r/DC3 19. Findd.S/ . (a)SD˚.x;y;x/ˇˇjxj/DC42;jyj/DC41;j´/NUL2j/DC42/TAB (b)SD/SUB .x;y/ˇˇ.x/NUL1/2 9C.y/NUL2/2 4D1/ESC (c)SDthe triangle in R2with vertices .2;0/ ,.2;2/ , and.4;4/ (d)SD˚.x1;x2;:::;x n/ˇˇjxij/DC4L;iD1;2;:::;n/TAB (e)SD˚.x;y;´/ˇˇx¤0;jyj/DC41;´>2/TAB 20. Prove thatd.S/Dd.S/for any setSinRn. 21. Prove: If a nonempty subset SofRnis both open and closed, then SDRn. Section 5.1 Structure of Rn301 22. Use the Bolzano–Weierstrass theorem to show that if S1,S2, . . . ,Sm, . . . is an infinite sequence of nonempty compact sets and S1/ESCS2/ESC/SOH/SOH/SOH/ESCSm/ESC/SOH/SOH/SOH , thenT1 mD1Smis nonempty. Show that the conclusion does not follow if the s ets are assumed to be closed rather than compact. 23. Suppose that a sequence U1,U2, . . . of open sets covers a compact set S. Without using the Heine–Borel theorem, show that S/SUBSN mD1Umfor someN. H INT: Apply Exercise 5.1.22 to the setsSnDS\/NULSn mD1Um/SOHc: (This is a seemingly restricted version of the Heine–Borel t heorem, valid for the case where the covering collection His denumerable. However, it can be shown that there is no loss of generality in assuming this.) 24. Thedistance from a point X0to a nonempty set Sis defined by dist.X0;S/Dinf˚ jX/NULX0jˇˇX2S/TAB : (a) Prove: IfSis closed and X02Rn, there is a point XinSsuch that jX/NULX0jDdist.X0;S/: HINT:Apply Exercise 5.1.22 to the sets CmD˚ XˇˇX2SandjX/NULX0j/DC4dist.X0;S/C1=m/TAB ; m/NAK1: (b) Show that ifSis closed and X062S, then dist.X0;S/>0 . (c) Show that the conclusions of (a)and(b) may fail to hold if Sis not closed. 25. Thedistance between two nonempty sets SandTis defined by dist.S;T/Dinf˚jX/NULYjˇˇX2S;Y2T/TAB: (a) Prove: IfSis closed and Tis compact, there are points XinSandYinT such that jX/NULYjDdist.S;T/: HINT:Use Exercises 5.1.22 and5.1.24: (b) Under the assumptions of (a), show that dist .S;T/>0 ifS\TD;. (c) Show that the conclusions of (a) and(b) may fail to hold if SorTis not closed orTis unbounded. 26. (a) Prove: If a compact set Sis contained in an open set U, there is a positive numberrsuch that the set SrD˚ Xˇˇdist.X;S//DC4r/TAB is contained in U. (You will need Exercise 5.1.24 here.) (b) Show thatSris compact. 302 Chapter 5 Real-Valued Functions of Several Variables 27. LetD1andD2be compact subsets of Rn. Show that DD˚.X;Y/ˇˇX2D1;Y2D2/TAB is a compact subset of R2n. 28. Prove: IfSis open andSDA[BwhereA\BDA\BD;, thenAandBare open. 29. Give an example of a connected set in Rnthat is not polygonally connected. 30. Prove that a region is connected. 31. Show that the intervals are the only regions in R. 32. Prove: A bounded sequence in Rnhas a convergent subsequence. H INT:Use Theo- rems 5.1.14;4.2.2;and4.2.5.a/: 33. Define “lim r!1XrD1 ” iffXrgis a sequence in Rn,n/NAK2. 5.2 CONTINUOUS REAL-VALUED FUNCTIONS OF nVARI- ABLES We now study real-valued functions of nvariables. We denote the domain of a function f byDfand the value of fat a point XD.x1;x2;:::;x n/byf.X/orf.x 1;x2;:::;x n/. We continue the convention adopted in Section 2.1 for functi ons of one variable: If a func- tion is defined by a formula such as f.X/D/NUL1/NULx2 1/NULx2 2/NUL/SOH/SOH/SOH/NULx2 n/SOH1=2(5.2.1) or g.X/D/NUL1/NULx2 1/NULx2 2/NUL/SOH/SOH/SOH/NULx2 n/SOH/NUL1(5.2.2) without specification of its domain, it is to be understood th at its domain is the largest subset of Rnfor which the formula defines a unique real number. Thus, in th e absence of any other stipulation, the domain of fin (5.2.1 ) is the closed n-ball˚ XˇˇjXj/DC41/TAB , while the domain of gin (5.2.2 ) is the set˚XˇˇjXj¤1/TAB. The main objective of this section is to study limits and cont inuity of functions of n variables. The proofs of many of the theorems here are simila r to the proofs of their coun- terparts in Sections 2.1 and . We leave most of them to you. Definition 5.2.1 We say thatf.X/approaches the limit LasXapproaches X0and write lim X!X0f.X/DL ifX0is a limit point of Dfand, for every /SI>0 , there is aı>0 such that jf.X//NULLj</SI for all XinDfsuch that 0<jX/NULX0j<ı: Section 5.2 Continuous Real-Valued Functions of nVariables 303 Example 5.2.1 If g.x;y/D1/NULx2/NUL2y2; then lim .x;y/ !.x0;y0/g.x;y/D1/NULx2 0/NUL2y2 0 (5.2.3) for every.x0;y0/. To see this, we write jg.x;y//NUL.1/NULx2 0/NUL2y2 0/jDj.1/NULx2/NUL2y2//NUL.1/NULx2 0/NUL2y2 0/j /DC4jx2/NULx2 0jC2jy2/NULy2 0j Dj.xCx0/.x/NULx0/jC2j.yCy0/.y/NULy0/j /DC4jX/NULX0j.jxCx0jC2jyCy0/j/;(5.2.4) since jx/NULx0j/DC4j X/NULX0jandjy/NULy0j/DC4j X/NULX0j: IfjX/NULX0j<1, thenjxj<jx0jC1andjyj<jy0jC1. This and ( 5.2.4 ) imply that jg.x;y//NUL.1/NULx2 0/NUL2y2 0/j<KjX/NULX0jifjX/NULX0j<1; where KD.2jx0jC1/C2.2jy0jC1/: Therefore, if /SI>0 and jX/NULX0j<ıDminf1;/SI=Kg; thenˇˇg.x;y//NUL.1/NULx2 0/NUL2y2 0/ˇˇ</SI: This proves ( 5.2.3 ). Definition 5.2.1 does not require that fbe defined at X0, or even on a deleted neighbor- hood of X0. Example 5.2.2 The function h.x;y/Dsinp 1/NULx2/NUL2y2 p 1/NULx2/NUL2y2 is defined only on the interior of the region bounded by the ell ipse x2C2y2D1 (Figure 5.2.1(a), page 304). It is not defined at any point of the ellipse itself or on any deleted neighborhood of such a point. Nevertheless, lim .x;y/ !.x0;y0/h.x;y/D1 (5.2.5) 304 Chapter 5 Real-Valued Functions of Several Variables if x2 0C2y2 0D1: (5.2.6) To see this, let u.x;y/Dp 1/NULx2/NUL2y2: Then h.x;y/Dsinu.x;y/ u.x;y/: (5.2.7) Recall that lim r!0sinr rD1I therefore, if/SI>0 , there is aı1>0such that ˇˇˇˇsinu u/NUL1ˇˇˇˇ</SI if0<juj<ı1: (5.2.8) From ( 5.2.3 ), lim .x;y/ !.x0;y0/.1/NULx2/NUL2y2/D0 if (5.2.6 ) holds, so there is a ı>0 such that 0<u2.x;y/D.1/NULx2/NUL2y2/<ı2 1 ifXD.x;y/ is in the interior of the ellipse and jX/NULX0j<ı; that is, if Xis in the shaded region of Figure 5.2.1(b). Therefore, 0<uDp 1/NULx2/NUL2y2<ı1 (5.2.9) ifXis in the interior of the ellipse and jX/NULX0j<ı; that is, if Xis in the shaded region of Figure 5.2.1(b). This, ( 5.2.7 ), and ( 5.2.8 ) imply that jh.x;y//NUL1j</SI for such X, which implies ( 5.2.5 ). (a)y x x2+ 2y2 = 1 (b)y x x2+ 2y2 = 1 X−X0 = δX0 Figure 5.2.1 Section 5.2 Continuous Real-Valued Functions of nVariables 305 The following theorem is analogous to Theorem 2.1.3. We leav e its proof to you (Exer- cise5.2.2 ). Theorem 5.2.2 Iflim X!X0f.X/exists;then it is unique. When investigating whether a function has a limit at a point X0, no restriction can be made on the way in which Xapproaches X0, except that Xmust be inDf. The next example shows that incorrect restrictions can lead to incor rect conclusions. Example 5.2.3 The function f.x;y/Dxy x2Cy2 is defined everywhere in R2except at.0;0/ . Does lim .x;y/ !.0;0/f.x;y/ exist? If we try to answer this question by letting .x;y/ approach.0;0/ along the line yDx, we see the functional values f.x;x/Dx2 2x2D1 2 and conclude that the limit is 1=2. However, if we let .x;y/ approach.0;0/ along the line yD/NULx, we see the functional values f.x;/NULx/D/NULx2 2x2D/NUL1 2 and conclude that the limit equals /NUL1=2. From Theorem 5.2.2 , these two conclusions cannot both be correct. In fact, they are both incorrect. Wha t we have shown is that lim x!0f.x;x/D1 2and lim x!0f.x;/NULx/D/NUL1 2: Since lim x!0f.x;x/ and lim x!0f.x;/NULx/must both equal lim .x;y/ !.0;0/f.x;y/ if the latter exists (Exercise 5.2.3(a)), we conclude that the latter does not exist. The sum, difference, and product of functions of nvariables are defined in the same way as they are for functions of one variable (Definition 2.1.1 ), and the proof of the next theorem is the same as the proof of Theorem 2.1.4 . Theorem 5.2.3 Suppose that fandgare defined on a set D;X0is a limit point of D;and lim X!X0f.X/DL1; lim X!X0g.X/DL2: Then lim X!X0.fCg/.X/DL1CL2; (5.2.10) lim X!X0.f/NULg/.X/DL1/NULL2; (5.2.11) lim X!X0.fg/. X/DL1L2; (5.2.12) and;ifL2¤0; lim X!X0/DC2f g/DC3 .X/DL1 L2: (5.2.13) 306 Chapter 5 Real-Valued Functions of Several Variables Infinite Limits and Limits as jXj!1 Definition 5.2.4 We say thatf.X/approaches1asXapproaches X0and write lim X!X0f.X/D1 ifX0is a limit point of Dfand, for every real number M, there is aı>0 such that f.X/>M whenever0<jX/NULX0j<ı and X2Df: We say that lim X!X0f.X/D/NUL1 if lim X!X0./NULf/.X/D1: Example 5.2.4 If f.X/D.1/NULx2 1/NULx2 2/NUL/SOH/SOH/SOH/NULx2 n//NUL1=2; then lim X!X0f.X/D1 ifjX0jD1, because f.X/D1 jX/NULX0j; so f.X/>M if0<jX/NULX0j<ıD1 M: Example 5.2.5 If f.x;y/D1 xC2yC1; then lim .x;y/ !.1;/NUL1/f.x;y/ does not exist (why not?), but lim .x;y/ !.1;/NUL1/jf.x;y/jD1: To see this, we observe that jxC2yC1jDj.x/NUL1/C2.yC1/j /DC4p 5jX/NULX0j(by Schwarz’s inequality), where X0D.1;/NUL1/, so jf.x;y/jD1 jxC2yC1j/NAK1p 5jX/NULX0j: Section 5.2 Continuous Real-Valued Functions of nVariables 307 Therefore, jf.x;y/j>M if0<jX/NULX0j<1 Mp 5: Example 5.2.6 The function f.x;y;´/Dˇˇˇˇsin/DC21 x2Cy2C´2/DC3ˇˇˇˇ x2Cy2C´2 assumes arbitrarily large values in every neighborhood of .0;0;0/ . For example, if XkD .xk;yk;´k/, where xkDykD´kD1q 3/NUL kC1 2/SOH /EM; then f.Xk/D/DC2 kC1 2/DC3 /EM: However, this does not imply that lim X!0f.X/D1 , since, for example, every neighbor- hood of.0;0;0/ also contains points XkD/DC21p 3k/EM;1p 3k/EM;1p 3k/EM/DC3 for whichf.Xk/D0. Definition 5.2.5 IfDfis unbounded ;we say that lim jXj!1f.X/DL(finite) if for every/SI>0 , there is a number Rsuch that jf.X//NULLj</SI wheneverjXj/NAKRand X2Df: Example 5.2.7 If f.x;y;´/Dcos/DC21 x2C2y2C´2/DC3 ; then lim jXj!1f.X/D1: (5.2.14) To see this, we recall that the continuity of cos uatuD0implies that for each /SI>0 there is aı>0 such that jcosu/NUL1j</SI ifjuj<ı: 308 Chapter 5 Real-Valued Functions of Several Variables Since1 x2C2y2C´2/DC41 jXj2; it follows that ifjXj>1=p ı, then 1 x2C2y2C´2<ı: Therefore, jf.X//NUL1j</SI: This proves ( 5.2.14 ). Example 5.2.8 Consider the function defined only on the domain DD˚.x;y/ˇˇ0<y/DC4ax/TAB; 0<a<1 (Figure 5.2.2 ), by f.x;y/D1 x/NULy: We will show that lim jXj!1f.x;y/D0: (5.2.15) It is important to keep in mind that we need only consider .x;y/ inD, sincefis not defined elsewhere. InD, x/NULy/NAKx.1/NULa/ (5.2.16) and jXj2Dx2Cy2/DC4x2.1Ca2/; so x/NAKjXjp 1Ca2: This and ( 5.2.16 ) imply that x/NULy/NAK1/NULap 1Ca2jXj;X2D; so jf.x;y/j/DC4p 1Ca2 1/NULa1 jXj;X2D: Therefore, jf.x;y/j</SI ifX2Dand jXj>p 1Ca2 1/NULa1 /SI: This implies ( 5.2.15 ). Section 5.2 Continuous Real-Valued Functions of nVariables 309 y xy = ax Figure 5.2.2 We leave it to you to define lim jXj!1f.X/D1 and lim jXj!1f.X/D/NUL1 (Exer- cise5.2.6 ). We will continue the convention adopted in Section 2.1: “lim X!X0f.X/exists” means that lim X!X0f.X/DL, whereLis finite; to leave open the possibility that LD˙1 , we will say that “lim X!X0f.X/exists in the extended reals.” A similar convention applies to limits asjXj!1 . Theorem 5.2.3 remains valid if “lim X!X0” is replaced by “lim jXj!1,” provided that Dis unbounded. Moreover, ( 5.2.10 ), (5.2.11 ), and ( 5.2.12 ) are valid in either version of Theorem 5.2.3 if either or both of L1andL2is infinite, provided that their right sides are not indeterminate, and ( 5.2.13 ) remains valid if L2¤0andL1=L2is not indeterminate. Continuity We now define continuity for functions of nvariables. The definition is quite similar to the definition for functions of one variable. Definition 5.2.6 IfX0is inDfand is a limit point of Df, then we say that fis continuous at X0if lim X!X0f.X/Df.X0/: The next theorem follows from this and Definition 5.2.1 . 310 Chapter 5 Real-Valued Functions of Several Variables Theorem 5.2.7 Suppose that X0is inDfand is a limit point of Df:Thenfis con- tinuous at X0if and only if for each /SI>0 there is aı>0 such that jf.X//NULf.X0/j</SI whenever jX/NULX0j<ı and X2Df: In applying this theorem when X02D0 f, we will usually omit “and X2Df,” it being understood that Sı.X0//SUBDf. We will say that fiscontinuous on Siffis continuous at every point of S. Example 5.2.9 From Example 5.2.1 , we now see that the function f.x;y/D1/NULx2/NUL2y2 is continuous on R2. Example 5.2.10 If we extend the definition of hin Example 5.2.2 so that h.x;y/D8 ˆ< ˆ:sinp 1/NULx2/NUL2y2 p 1/NULx2/NUL2y2; x2C2y2<1; 1; x2C2y2D1; then it follows from Example 5.2.2 thathis continuous on the ellipse x2C2y2D1: We will see in Example 5.2.13 thathis also continuous on the interior of the ellipse. Example 5.2.11 It is impossible to define the function f.x;y/Dxy x2Cy2 at the origin to make it continuous there, since we saw in Exam ple5.2.3 that lim .x;y/ !.0;0/f.x;y/ does not exist. Theorem 5.2.3 implies the next theorem, which is analogous to Theorem 2.2.5 and, like the latter, permits us to investigate continuity of a given f unction by regarding the function as the result of addition, subtraction, multiplication, an d division of simpler functions. Section 5.2 Continuous Real-Valued Functions of nVariables 311 Theorem 5.2.8 Iffandgare continuous on a set SinRn;then so arefCg;f/NULg; andfg:Also;f=g is continuous at each X0inSsuch thatg.X0/¤0: Vector-Valued Functions and Composite Functions Suppose that g1,g2, . . . ,gnare real-valued functions defined on a subset TofRm, and define the vector-valued function GonTby G.U/D.g1.U/;g2.U/;:::;g n.U//; U2T: Theng1,g2, . . . ,gnare the component functions ofGD.g1;g2;:::;g n/. We say that lim U!U0G.U/DLD.L1;L2;:::;L n/ if lim U!U0gi.U/DLi; 1/DC4i/DC4n; and that Giscontinuous atU0ifg1,g2, . . . ,gnare each continuous at U0. The next theorem follows from Theorem 5.1.14 and Definitions 5.2.1 and5.2.6 . We omit the proof. Theorem 5.2.9 For a vector-valued function G; lim U!U0G.U/DL if and only if for each /SI>0 there is aı>0 such that jG.U//NULLj</SI whenever0<jU/NULU0j<ı and U2DG: Similarly, Gis continuous at U0if and only if for each /SI>0 there is aı>0 such that jG.U//NULG.U0/j</SI wheneverjU/NULU0j<ı and U2DG: The following theorem on the continuity of a composite funct ion is analogous to Theo- rem2.2.7 . Theorem 5.2.10 Letfbe a real-valued function defined on a subset of Rn;and let the vector-valued function GD.g1;g2;:::;g n/be defined on a domain DGinRm:Let the set TD˚ UˇˇU2DGand G.U/2Df/TAB .Figure 5.2.3/, be nonempty ;and define the real-valued composite function hDfıG onTby h.U/Df.G.U//; U2T: Now suppose that U0is inTand is a limit point of T;Gis continuous at U0;andfis continuous at X0DG.U0/:Thenhis continuous at U0: 312 Chapter 5 Real-Valued Functions of Several Variables mnR(G) = range of G G DG Df Figure 5.2.3 Proof Suppose that /SI > 0 . Sincefis continuous at X0DG.U0/, there is an/SI1>0 such that jf.X//NULf.G.U0//j</SI (5.2.17) if jX/NULG.U0/j</SI1and X2Df: (5.2.18) Since Gis continuous at U0, there is aı>0 such that jG.U//NULG.U0/j</SI1ifjU/NULU0j<ı and U2DG: By taking XDG.U/in (5.2.17 ) and ( 5.2.18 ), we see that jh.U//NULh.U0/jDjf.G.U//NULf.G.U0//j</SI if jU/NULU0j<ı and U2T: Example 5.2.12 If f.s/Dps and g.x;y/D1/NULx2/NUL2y2; thenDfDŒ0;1/c141,DgDR2, and TD˚.x;y/ˇˇx2C2y2/DC41/TAB: From Theorem 5.2.7 and Example 5.2.1 ,gis continuous on R2. (We can obtain the same conclusion by observing that the functions p1.x;y/Dxandp2.x;y/Dyare continuous onR2and applying Theorem 5.2.8 .) Sincefis continuous on Df, the function h.x;y/Df .g.x;y//Dp 1/NULx2/NUL2y2 is continuous on T. Section 5.2 Continuous Real-Valued Functions of nVariables 313 Example 5.2.13 If g.x;y/Dp 1/NULx2/NUL2y2 and f.s/D8 < :sins s; s¤0; 1; sD0; thenDfD./NUL1;1/and DgDTD˚ .x;y/ˇˇx2C2y2/DC41/TAB : In Example 5.2.12 we saw thatg(we called it hthere) is continuous on T. Sincefis continuous on Df, the composite function hDfıgdefined by h.x;y/D8 ˆ< ˆ:sinp 1/NULx2/NUL2y2 p 1/NULx2/NUL2y2; x2C2y2<1; 1; x2C2y2D1; is continuous on T. This implies the result of Example 5.2.2 . Bounded Functions The definitions of bounded above, bounded below , and bounded on a setSare the same for functions ofnvariables as for functions of one variable, as are the definit ions of supremum andinfimum of a function on a set S(Section 2.2). The proofs of the next two theorems are similar to those of Theorems 2.2.8 and2.2.9 (Exercises 5.2.12 and5.2.13 ). Theorem 5.2.11 Iffis continuous on a compact set SinRn;thenfis bounded onS: Theorem 5.2.12 Letfbe continuous on a compact set SinRnand ˛Dinf X2Sf.X/; ˇDsup X2Sf.X/: Then f.X1/D˛andf.X2/Dˇ for some X1andX2inS: The next theorem is analogous to Theorem 2.2.10 . Theorem 5.2.13 (Intermediate Value Theorem) Letfbe continuous on a regionSinRn:Suppose that AandBare inSand f.A/<u<f. B/: Thenf.C/Dufor some CinS: 314 Chapter 5 Real-Valued Functions of Several Variables Proof If there is no such C, thenSDR[T, where RD˚XˇˇX2Sandf.X/<u/TAB and TD˚XˇˇX2Sandf.X/>u/TAB: IfX02R, the continuity of fimplies that there is a ı>0 such thatf.X/<u ifjX/NULX0j< ıandX2S. This means that X062T. Therefore,R\TD;. Similarly,R\TD;. Therefore,Sis disconnected (Definition 5.1.19 ), which contradicts the assumption that S is a region (Exercise 5.1.30 ). Hence, we conclude that f.C/Dufor some CinS. Uniform Continuity The definition of uniform continuity for functions of nvariables is the same as for functions of one variable; fis uniformly continuous on a subset Sof its domain in Rnif for every /SI>0 there is aı>0 such that jf.X//NULf.X0/j</SI wheneverjX/NULX0j<ıandX;X02S. We emphasize again that ımust depend only on /SI andS, and not on the particular points XandX0. The proof of the next theorem is analogous to that of Theorem 2.2.12 . We leave it to you (Exercise 5.2.14 ). Theorem 5.2.14 Iffis continuous on a compact set SinRn;thenfis uniformly continuous on S: 5.2 Exercises WithRreplaced by Rn;the following exercises from Sections 2:1and2:2have analogs for this section: 2.1.5 ,2.1.8 –2.1.11 ,2.1.26 ,2.1.28 ,2.1.29 ,2.1.33 ,2.2.8;2.2.9;2.2.10 ,2.2.15 , 2.2.16 ,2.2.20 ,2.2.29 ,2.2.30 . 1. Find lim X!X0f.X/and justify your answer with an /SI–ıargument, as required by Definition 5.2.1 . HINT:See Examples 5.2.1 and5.2.2: (a)f.X/D3xC4yC´/NUL2,X0D.1;2;1/ (b)f.X/Dx3/NULy3 x/NULy,X0D.1;1/ (c)f.X/Dsin.xC4yC2´/ xC4yC2´,X0D./NUL2;1;/NUL1/ Section 5.2 Continuous Real-Valued Functions of nVariables 315 (d)f.X/D.x2Cy2/log.x2Cy2/1=2,X0D.0;0/ (e)f.X/Dsin.x/NULy/px/NULy,X0D.2;2/ (f)f.X/D1 jXje/NUL1=jXj,X0D0 2. Prove Theorem 5.2.2 . 3. If lim x!x0y.x/Dy0and lim x!x0f .x;y.x//DL, we say that f.x;y/ ap- proachesLas.x;y/ approaches.x0;y0/along the curve yDy.x/ . (a) Prove: If lim .x;y/ !.x0;y0/f.x;y/DL, thenf.x;y/ approachesLas.x;y/ approaches.x0;y0/along any curve yDy.x/ through.x0;y0/. (b) We saw in Example 5.2.3 that if f.x;y/Dxy x2Cy2; then lim .x;y/ !.0;0/f.x;y/ does not exist. Show, however, that f.x;y/ ap- proaches a value Laas.x;y/ approaches.0;0/ along any curve yDy.x/ that passes through .0;0/ with slopea. FindLa. (c) Show that the function g.x;y/Dx3y4 .x2Cy6/3 approaches0as.x;y/ approaches.0;0/ along a curve as described in (b), but that lim .x;y/ !.0;0/f.x;y/ does not exist. 4. Determine whether lim X!X0f.X/D˙1 . (a)f.X/Djsin.xC2yC4´/j .xC2yC4´/2,X0D.2;/NUL1;0/ (b)f.X/D1px/NULy,X0D.0;0/ (c)f.X/Dsin1=xpx/NULy,X0D.0;0/ (d)f.X/D4y2/NULx2 .x/NUL2y/3,X0D.2;1/ (e)f.X/Dsin.xC2yC4´/ .xC2yC4´/2,X0D.2;/NUL1;0/ 5. Find lim jXj!1f.X/, if it exists. (a)f.X/Dlog.x2C2y2C4´2/ x2Cy2C´2(b)f.X/Dsin.x2Cy2/p x2Cy2 (c)f.X/De/NUL.xCy/2(d)f.X/De/NULx2/NULy2 316 Chapter 5 Real-Valued Functions of Several Variables (e)f.X/D8 < :sin.x2/NULy2/ x2/NULy2; x¤˙y; 1; xD˙y 6. Define(a)limjXj!1f.X/D1 and(b) limjXj!1f.X/D/NUL1 . 7. Let f.X/Djx1ja1jx2ja2/SOH/SOH/SOHjxnjan Xjb: For what nonnegative values of a1,a2, . . . ,an,bdoes lim X!0f.X/exist in the extended reals? 8. Let g.X/D.x2Cy4/3 1Cx6y4: Show that lim jxj!1g.x;ax/D1 for any real number a. Does lim jXj!1g.X/D1‹ 9. For eachfin Exercise 5.2.1 , find the largest set Son whichfis continuous or can be defined so as to be continuous. 10. Repeat Exercise 5.2.9 for the functions in Exercise 5.2.5 . 11. Give an example of a function fonR2such thatfis not continuous at .0;0/ , butf.0;y/ is a continuous function of yon./NUL1;1/andf.x;0/ is a continuous function ofxon./NUL1;1/. 12. Prove Theorem 5.2.11 . HINT:See the proof of Theorem 2.2.8: 13. Prove Theorem 5.2.12 . HINT:See the proof of Theorem 2.2.9: 14. Prove Theorem 5.2.14 . HINT:See the proof of Theorem 2.2.12: 15. Suppose that X2Df/SUBRnandXis a limit point of Df. Show thatfis continuous atXif and only if lim k!1f.Xk/Df.X/wheneverfXkgis a sequence of points inDfsuch that lim k!1XkDX. HINT:See the proof of Theorem 4.2.6: 5.3 PARTIAL DERIVATIVES AND THE DIFFERENTIAL To say that a function of one variable has a derivative at x0is the same as to say that it is differentiable at x0. The situation is not so simple for a function fof more than one variable. First, there is no specific number that can be calle dthederivative offat a point X0inRn. In fact, there are infinitely many numbers, called the directional derivatives of fatX0(defined below), that are analogous to the derivative of a fun ction of one variable. Second, we will see that the existence of directional deriva tives at X0does not imply that f is differentiable at X0, if differentiability at X0is to imply (as it does for functions of one variable) that f.X//NULf.X0/can be approximated well near X0by a simple linear function, or even thatfis continuous at X0. Section 5.3 Partial Derivatives and the Differential 317 We will now define directional derivatives and partial deriv atives of functions of several variables. However, we will still have occasion to refer to d erivatives of functions of one variable. We will call them ordinary derivatives when we wish to distinguish between them and the partial derivatives that we are about to define. Definition 5.3.1 Letˆbe a unit vector and Xa point in Rn.The directional derivative offatXin the direction of ˆis defined by @f.X/ @ˆDlim t!0f.XCtˆ//NULf.X/ t if the limit exists. That is, @f.X/=@ˆis the ordinary derivative of the function h.t/Df.XCtˆ/ attD0, ifh0.0/exists. Example 5.3.1 LetˆD./RS1;/RS2;/RS3/and f.x;y;´/D3xy´C2x2C´2: Then h.t/Df.xCt/RS1;yCt/RS2;´Ct/RS3/; D3.xCt/RS1/.yCt/RS2/.´Ct/RS3/C2.xCt/RS1/2C.´Ct/RS3/2 and h0.t/D3/RS1.yCt/RS2/.´Ct/RS3/C3/RS2.xCt/RS1/.´Ct/RS3/ C3/RS3.xCt/RS1/.yCt/RS2/C4/RS1.xCt/RS1/C2/RS3.´Ct/RS3/: Therefore, @f.X/ @ˆDh0.0/D.3y´C4x//RS 1C3x´/RS 2C.3xyC2´//RS 3: (5.3.1) The directional derivatives that we are most interested in a re those in the directions of the unit vectors E1D.1;0;:::;0/; E2D.0;1;0;:::;0/;:::; EnD.0;:::;0;1/: (All components of Eiare zero except for the ith, which is1.) Since XandXCtEidiffer only in theith coordinate, @f.X/=@Eiis called the partial derivative of fwith respect to xiatX. It is also denoted by @f.X/=@x iorfxi.X/; thus, @f.X/ @x1Dfx1.X/Dlim t!0f.x 1Ct;x2;:::;x n//NULf.x 1;x2;:::;x n/ t; 318 Chapter 5 Real-Valued Functions of Several Variables @f.X/ @xiDfxi.X/Dlim t!0f.x 1;:::;x i/NUL1;xiCt;xiC1;:::;x n//NULf.x 1;x2;:::;x n/ t if2/DC4i/DC4n, and @f.X/ @xnDfxn.X/Dlim t!0f.x 1;:::;x n/NUL1;xnCt//NULf.x 1;:::;x n/NUL1;xn/ t; if the limits exist. If we write XD.x;y/ , then we denote the partial derivatives accordingly; thus, @f.x;y/ @xDfx.x;y/Dlim h!0f.xCh;y//NULf.x;y/ h and @f.x;y/ @yDfy.x;y/Dlim h!0f.x;yCh//NULf.x;y/ h: It can be seen from these definitions that to compute fxi.X/we simply differentiate f with respect to xiaccording to the rules for ordinary differentiation, while treating the other variables as constants. Example 5.3.2 Let f.x;y;´/D3xy´C2x2C´2(5.3.2) as in Example 5.3.1 . Taking ˆDE1(that is, setting /RS1D1and/RS2D/RS3D0) in ( 5.3.1 ), we find that @f.X/ @[email protected]/ @E1D3y´C4x; which is the result obtained by regarding yand´as constants in ( 5.3.2 ) and taking the ordinary derivative with respect to x. Similarly, @f.X/ @[email protected]/ @E2D3x´ and @f.X/ @´[email protected]/ @E3D3xyC2´: The next theorem follows from the rule just given for calcula ting partial derivatives. Theorem 5.3.2 Iffxi.X/andgxi.X/exist;then @.fCg/.X/ @xiDfxi.X/Cgxi.X/; @.fg/. X/ @xiDfxi.X/g.X/Cf.X/gxi.X/; Section 5.3 Partial Derivatives and the Differential 319 and;ifg.X/¤0; @.f=g/. X/ @xiDg.X/fxi.X//NULf.X/gxi.X/ Œg.X//c1412: Iffxi.X/exists at every point of a set D, then it defines a function fxionD. If this function has a partial derivative with respect to xjon a subset of D, we denote the partial derivative by @ @xj/DC2@f @xi/DC3 D@2f @xj@xiDfxixj: Similarly, @ @xk/DC2@2f @xj@xi/DC3 D@3f @xk@xj@xiDfxixjxk: The function obtained by differentiating fsuccessively with respect to xi1;xi2;:::;x iris denoted by @rf @xir@xir/NUL1/SOH/SOH/SOH@xi1Dfxi1/SOH/SOH/SOHxir/NUL1xirI it is anrth-order partial derivative of f. Example 5.3.3 The function f.x;y/D3x2y3Cxy has partial derivatives everywhere. Its first-order partia l derivatives are fx.x;y/D6xy3Cy; f y.x;y/D9x2y2Cx: Its second-order partial derivatives are fxx.x;y/D6y3; f yy.x;y/D18x2y; fxy.x;y/D18xy2C1; f yx.x;y/D18xy2C1: There are eight third-order partial derivatives. Some exam ples are fxxy.x;y/D18y2; f xyx.x;y/D18y2; f yxx.x;y/D18y2: Example 5.3.4 Computefxx.0;0/ ,fyy.0;0/ ,fxy.0;0/ , andfyx.0;0/ if f.x;y/D8 < :.x2yCxy2/sin.x/NULy/ x2Cy2; .x;y/¤.0;0/; 0; .x;y/ D.0;0/: Solution If.x;y/¤.0;0/ , the ordinary rules for differentiation, applied separate ly to xandy, yield fx.x;y/D.2xyCy2/sin.x/NULy/C.x2yCxy2/cos.x/NULy/ x2Cy2 /NUL2x.x2yCxy2/sin.x/NULy/ .x2Cy2/2; .x;y/¤.0;0/;(5.3.3) 320 Chapter 5 Real-Valued Functions of Several Variables and fy.x;y/D.x2C2xy/ sin.x/NULy//NUL.x2yCxy2/cos.x/NULy/ x2Cy2 /NUL2y.x2yCxy2/sin.x/NULy/ .x2Cy2/2; .x;y/¤.0;0/:(5.3.4) These formulas do not apply if .x;y/D.0;0/ , so we findfx.0;0/ andfy.0;0/ from their definitions as difference quotients: fx.0;0/Dlim x!0f.x;0//NULf.0;0/ xDlim x!00/NUL0 xD0; fy.0;0/Dlim y!0f.0;y//NULf.0;0/ yDlim y!00/NUL0 yD0: SettingyD0in (5.3.3 ) and ( 5.3.4 ) yields fx.x;0/D0; f y.x;0/Dsinx; x¤0; so fxx.0;0/Dlim x!0fx.x;0//NULfx.0;0/ xDlim x!00/NUL0 xD0; fyx.0;0/Dlim x!0fy.x;0//NULfy.0;0/ xDlim x!0sinx/NUL0 xD1: SettingxD0in (5.3.3 ) and ( 5.3.4 ) yields fx.0;y/D/NUL siny; f y.0;y/D0; y¤0; so fxy.0;0/Dlim y!0fx.0;y//NULfx.0;0/ yDlim y!0/NULsiny/NUL0 yD/NUL1; fyy.0;0/Dlim y!0fy.0;y//NULfy.0;0/ yDlim y!00/NUL0 yD0: This example shows that fxy.X0/andfyx.X0/may differ. However, the next theorem shows that they are equal if fsatisfies a fairly mild condition. Theorem 5.3.3 Suppose thatf;f x;fy;andfxyexist on a neighborhood Nof.x0;y0/; andfxyis continuous at .x0;y0/:Thenfyx.x0;y0/exists, and fyx.x0;y0/Dfxy.x0;y0/: (5.3.5) Proof Suppose that /SI>0 . Chooseı>0 so that the open square Section 5.3 Partial Derivatives and the Differential 321 SıD˚.x;y/ˇˇjx/NULx0j<ı;jy/NULy0j<ı/TAB is inNand jfxy.bx;by//NULfxy.x0;y0/j</SI if.bx;by/2Sı: (5.3.6) This is possible because of the continuity of fxyat.x0;y0/. The function A.h;k/Df.x 0Ch;y 0Ck//NULf.x 0Ch;y 0//NULf.x 0;y0Ck/Cf.x 0;y0/(5.3.7) is defined if/NULı<h ,k<ı ; moreover, A.h;k/D/RS.x 0Ch//NUL/RS.x 0/; (5.3.8) where /RS.x/Df.x;y 0Ck//NULf.x;y 0/: Since /RS0.x/Dfx.x;y 0Ck//NULfx.x;y 0/;jx/NULx0j<ı; (5.3.8 ) and the mean value theorem imply that A.h;k/DŒfx.bx;y 0Ck//NULfx.bx;y 0//c141h; (5.3.9) wherebxis betweenx0andx0Ch. The mean value theorem, applied to fx.bx;y/ (wherebx is regarded as constant), also implies that fx.bx;y 0Ck//NULfx.bx;y 0/Dfxy.bx;by/k; wherebyis betweeny0andy0Ck. From this and ( 5.3.9 ), A.h;k/Dfxy.bx;by/hk: Now ( 5.3.6 ) implies that ˇˇˇˇA.h;k/ hk/NULfxy.x0;y0/ˇˇˇˇDˇˇfxy.bx;by//NULfxy.x0;y0/ˇˇ</SI if0<jhj;jkj<ı: (5.3.10) Since ( 5.3.7 ) implies that lim k!0A.h;k/ hkDlim k!0f.x 0Ch;y 0Ck//NULf.x 0Ch;y 0/ hk /NULlim k!0f.x 0;y0Ck//NULf.x 0;y0/ hk Dfy.x0Ch;y 0//NULfy.x0;y0/ h; it follows from ( 5.3.10 ) that ˇˇˇˇfy.x0Ch;y 0//NULfy.x0;y0/ h/NULfxy.x0;y0/ˇˇˇˇ/DC4/SIif0<jhj<ı: 322 Chapter 5 Real-Valued Functions of Several Variables Taking the limit as h!0yields jfyx.x0;y0//NULfxy.x0;y0/j/DC4/SI: Since/SIis an arbitrary positive number, this proves ( 5.3.5 ). Theorem 5.3.3 implies the following theorem. We leave the proof to you (Exe rcises 5.3.10 and5.3.11 ). Theorem 5.3.4 Suppose that fand all its partial derivatives of order /DC4rare contin- uous on an open subset SofRn:Then fxi1xi2;:::;x ir.X/Dfxj1xj2;:::;x jr.X/;X2S; (5.3.11) if each of the variables x1;x2;. . .;xnappears the same number of times in fxi1;xi2;:::;x irgandfxj1;xj2;:::;x jrg: If this number is rk;we denote the common value of the two sides of (5.3.11 )by @rf.X/ @xr1 1@xr2 2/SOH/SOH/SOH@xrnn; (5.3.12) it being understood that 0/DC4rk/DC4r; 1/DC4k/DC4n; (5.3.13) r1Cr2C/SOH/SOH/SOHCrnDr; (5.3.14) and;ifrkD0;we omit the symbol @x0 kfrom the “denominator” of (5.3.12 ): For example, if fsatisfies the hypotheses of Theorem 5.3.4 withkD4at a point X0in Rn(n/NAK2), then fxxyy.X0/Dfxyxy.X0/Dfxyyx.X0/Dfyyxx.X0/Dfyxyx.X0/Dfyxxy.X0/; and their common value is denoted by @4f.X0/ @x2@y2: It can be shown (Exercise 5.3.12 ) that iffis a function of .x1;x2;:::;x n/and.r1;r2;:::;r n/ is a fixed ordered n-tuple that satisfies ( 5.3.13 ) and ( 5.3.14 ), then the number of partial derivativesfxi1xi2/SOH/SOH/SOHxirthat involve differentiation ritimes with respect to xi,1/DC4i/DC4n, equals the multinomial coefficient rŠ r1Šr2Š/SOH/SOH/SOHrnŠ: Section 5.3 Partial Derivatives and the Differential 323 Differentiable Functions of Several Variables A function of several variables may have first-order partial derivatives at a point X0but fail to be continuous at X0. For example, if f.x;y/D(xy x2Cy2; .x;y/¤.0;0/; 0; .x;y/D.0;0/;(5.3.15) then fx.0;0/Dlim h!0f.h;0//NULf.0;0/ hDlim h!00/NUL0 hD0 and fy.0;0/Dlim k!0f.0;k//NULf.0;0/ kDlim k!00/NUL0 kD0; butfis not continous at .0;0/ . (See Examples 5.2.3 and5.2.11 .) Therefore, if differentia- bility of a function of several variables is to be a stronger p roperty than continuity, as it is for functions of one variable, the definition of differentia bility must require more than the existence of first partial derivatives. Exercise 2.3.1 characterizes differentiability of a func- tionfof one variable in a way that suggests the proper generalizat ion:fis differentiable atx0if and only if lim x!x0f.x//NULf.x 0//NULm.x/NULx0/ x/NULx0D0 for some constant m, in which case mDf0.x0/. The generalization to functions of nvariables is as follows. Definition 5.3.5 A functionfisdifferentiable at X0D.x10;x20;:::;x n0// ifX02D0 fand there are constants m1,m2, . . .;mnsuch that lim X!X0f.X//NULf.X0//NULnX iD1mi.xi/NULxi0/ jX/NULX0jD0: (5.3.16) Example 5.3.5 Let f.x;y/Dx2C2xy: We will show that fis differentiable at any point .x0;y0/, as follows: 324 Chapter 5 Real-Valued Functions of Several Variables f.x;y//NULf.x 0;y0/Dx2C2xy/NULx2 0/NUL2x0y0 Dx2/NULx2 0C2.xy/NULx0y0/ D.x/NULx0/.xCx0/C2.xy/NULx0y/C2.x0y/NULx0y0/ D.xCx0C2y/.x/NULx0/C2x0.y/NULy0/ D2.x0Cy0/.x/NULx0/C2x0.y/NULy0/ C.x/NULx0/.x/NULx0C2y/NUL2y0/ Dm1.x/NULx0/Cm2.y/NULy0/C.x/NULx0/.x/NULx0C2y/NUL2y0/; where m1D2.x0Cy0/Dfx.x0;y0/andm2D2x0Dfy.x0;y0/: (5.3.17) Therefore, jf.x;y//NULf.x 0;y0//NULm1.x/NULx0//NULm2.y/NULy0/j jX/NULX0jDjx/NULx0jj.x/NULx0/C2.y/NULy0/j jX/NULX0j /DC4p 5jX/NULX0j; by Schwarz’s inequality. This implies that lim X!X0f.x;y//NULf.x 0;y0//NULm1.x/NULx0//NULm2.y/NULy0/ jX/NULX0jD0; sofis differentiable at .x0;y0/. From ( 5.3.17 ),m1Dfx.x0;y0/andm2Dfy.x0;y0/in Example 5.3.5 . The next theorem shows that this is not a coincidence. Theorem 5.3.6 Iffis differentiable at X0D.x10;x20;:::;x n0/;thenfx1.X0/; fx2.X0/;. . .;fxn.X0/exist and the constants m1;m2;. . .;mnin(5.3.16 )are given by miDfxi.X0/; 1/DC4i/DC4nI (5.3.18) that is; lim X!X0f.X//NULf.X0//NULnX iD1fxi.X0/.xi/NULxi0/ jX/NULX0jD0: Proof Letibe a given integer in f1;2;:::;ng. Let XDX0CtEi, so thatxiDxi0Ct, xjDxj 0ifj¤i, andjX/NULX0jDjtj. Then ( 5.3.16 ) and the differentiability of fatX0 imply that lim t!0f.X0CtEi//NULf.X0//NULmit tD0: Section 5.3 Partial Derivatives and the Differential 325 Hence, lim t!0f.X0CtEi//NULf.X0/ tDmi: This proves ( 5.3.18 ), since the limit on the left is fxi.X0/, by definition. Alinear function is a function of the form L.X/Dm1x1Cm2x2C/SOH/SOH/SOHCmnxn; (5.3.19) wherem1,m2, . . .;mnare constants. From Definition 5.3.5 ,fis differentiable at X0if and only if there is a linear function Lsuch thatf.X//NULf.X0/can be approximated so well near X0by L.X//NULL.X0/DL.X/NULX0/ that f.X//NULf.X0/DL.X/NULX0/CE.X/.jX/NULX0j/; (5.3.20) where lim X!X0E.X/D0: (5.3.21) Theorem 5.3.7 Iffis differentiable at X0;thenfis continuous at X0. Proof From ( 5.3.19 ) and Schwarz’s inequality, jL.X/NULX0/j/DC4MjX/NULX0j; where MD.m2 1Cm2 2C/SOH/SOH/SOHCm2 n/1=2: This and ( 5.3.20 ) imply that jf.X//NULf.X0/j/DC4.MCjE.X/j/jX/NULX0j; which, with ( 5.3.21 ), implies that fis continuous at X0. Theorem 5.3.7 implies that the function fdefined by ( 5.3.15 ) is not differentiable at .0;0/ , since it is not continuous at .0;0/ . However,fx.0;0/ andfy.0;0/ exist, so the converse of Theorem 5.3.7 is false; that is, a function may have partial derivatives at a point without being differentiable at the point. The Differential Theorem 5.3.7 implies that if fis differentiable at X0, then there is exactly one linear functionLthat satisfies ( 5.3.20 ) and ( 5.3.21 ): L.X/Dfx1.X0/x1Cfx2.X0/x2C/SOH/SOH/SOHCfxn.X0/xn: 326 Chapter 5 Real-Valued Functions of Several Variables This function is called the differential of fatX0. We will denote it by dX0fand its value by.dX0f/.X/; thus, .dX0f/.X/Dfx1.X0/x1Cfx2.X0/x2C/SOH/SOH/SOHCfxn.X0/xn: (5.3.22) In terms of the differential, ( 5.3.16 ) can be rewritten as lim X!X0f.X//NULf.X0//NUL.dX0f/.X/NULX0/ jX/NULX0jD0: For convenience in writing dX0f, and to conform with standard notation, we introduce the functiondxi, defined by dxi.X/DxiI that is,dxiis the function whose value at a point in Rnis theith coordinate of the point. It is the differential of the function gi.X/Dxi. From ( 5.3.22 ), dX0fDfx1.X0/dx 1Cfx2.X0dx2C/SOH/SOH/SOHCfxn.X0/dx n: (5.3.23) If we write XD.x;y;:::;/ , then we write dX0fDfx.X0/dxCfy.X0/dyC/SOH/SOH/SOH; wheredx,dy, . . . are the functions defined by dx.X/Dx; dy. X/Dy;::: When it is not necessary to emphasize the specific point X0, (5.3.23 ) can be written more simply as dfDfx1dx1Cfx2dx2C/SOH/SOH/SOHCfxndxn: When dealing with a specific function at an arbitrary point of its domain, we may use the hybrid notation dfDfx1.X/dx 1Cfx2.X/dx 2C/SOH/SOH/SOHCfxn.X/dx n: Example 5.3.6 We saw in Example 5.3.5 that the function f.x;y/Dx2C2xy is differentiable at every XinRn, with differential dfD.2xC2y/dxC2xdy: To finddX0fwith X0D.1;2/ , we setx0D1andy0D2; thus, dX0fD6dxC2dy and .dX0f/.X/NULX0/D6.x/NUL1/C2.y/NUL2/: Section 5.3 Partial Derivatives and the Differential 327 Sincef.1;2/D5, the differentiability of fat.1;2/ implies that lim .x;y/ !.1;2/f.x;y//NUL5/NUL6.x/NUL1//NUL2.y/NUL2/p .x/NUL1/2C.y/NUL2/2D0: Example 5.3.7 The differential of a function fDf.x/ of one variable is given by dx0fDf0.x0/dx; wheredxis the identity function; that is, dx.t/Dt: For example, if f.x/D3x2C5x3; then dfD.6xC15x2/dx: Ifx0D/NUL1, then dx0fD9dx; .d x0f/.x/NULx0/D9.xC1/; and, sincef./NUL1/D/NUL2, lim x!/NUL1f.x/C2/NUL9.xC1/ xC1D0: Unfortunately, the notation for the differential is so comp licated that it obscures the simplicity of the concept. The peculiar symbols df,dx,dy, etc., were introduced in the early stages of the development of calculus to represent very small (“infinitesimal”) increments in the variables. However, in modern usage they a re not quantities at all, but linear functions. This meaning of the symbol dxdiffers from its meaning inRb af.x/dx , where it serves merely to identify the variable of integrati on; indeed, some authors omit it in the latter context and write simplyRb af. Theorem 5.3.7 implies the following lemma, which is analogous to Lemma 2.3.2 . We leave the proof to you (Exercise 5.3.13 ). Lemma 5.3.8 Iffis differentiable at X0;then f.X//NULf.X0/D.dX0f/.X/NULX0/CE.X/jX/NULX0j; whereEis defined in a neighborhood of X0and lim X!X0E.X/DE.X0/D0: Theorems 5.3.2 and5.3.7 and the definition of the differential imply the following theorem. 328 Chapter 5 Real-Valued Functions of Several Variables Theorem 5.3.9 Iffandgare differentiable at X0;then so arefCgandfg. The same is true of f=g ifg.X0/¤0. The differentials are given by dX0.fCg/DdX0fCdX0g; dX0.fg/Df.X0/dX0gCg.X0/dX0f; and dX0/DC2f g/DC3 Dg.X0/dX0f/NULf.X0/dX0g Œg.X0//c1412: The next theorem provides a widely applicable sufficient con dition for differentiability. Theorem 5.3.10 Iffx1;fx2;. . .;fxnexist on a neighborhood of X0and are contin- uous at X0;thenfis differentiable at X0: Proof LetX0D.x10;x20;:::;x n0/and suppose that /SI > 0 . Our assumptions imply that there is a ı>0 such thatfx1;fx2;:::;f xnare defined in the n-ball Sı.X0/D˚XˇˇjX/NULX0j<ı/TAB and jfxj.X//NULfxj.X0/j</SI ifjX/NULX0j<ı; 1/DC4j/DC4n: (5.3.24) LetXD.x1;x;:::;x n/be inSı.X0/. Define XjD.x1;:::;x j;xjC1;0;:::;x n0/; 1/DC4j/DC4n/NUL1; andXnDX. Thus, for1/DC4j/DC4n,Xjdiffers from Xj/NUL1in thejth component only, and the line segment from Xj/NUL1toXjis inSı.X0/. Now write f.X//NULf.X0/Df.Xn//NULf.X0/DnX jD1Œf.Xj//NULf.Xj/NUL1//c141; (5.3.25) and consider the auxiliary functions g1.t/Df.t;x 20;:::;x n0/; gj.t/Df.x 1;:::;x j/NUL1;t;x jC1;0;:::;x n0/; 2/DC4j/DC4n/NUL1; gn.t/Df.x 1;:::;x n/NUL1;t/;(5.3.26) where, in each case, all variables except tare temporarily regarded as constants. Since f.Xj//NULf.Xj/NUL1/Dgj.xj//NULgj.xj 0/; the mean value theorem implies that f.Xj//NULf.Xj/NUL1/Dg0 j./FSj/.xj/NULxj 0/; Section 5.3 Partial Derivatives and the Differential 329 where/FSjis betweenxjandxj 0. From ( 5.3.26 ), g0 j./FSj/Dfxj.bXj/; wherebXjis on the line segment from Xj/NUL1toXj. Therefore, f.Xj//NULf.Xj/NUL1/Dfxj.bXj/.xj/NULxj 0/; and ( 5.3.25 ) implies that f.X//NULf.X0/DnX jD1fxj.bXj/.xj/NULxj 0/ DnX jD1fxj.X0/.xj/NULxj 0/CnX jD1Œfxj.bXj//NULfxj.X0//c141.x j/NULxj 0/: From this and ( 5.3.24 ), ˇˇˇˇˇˇf.X//NULf.X0//NULnX jD1fxj.X0/.xj/NULxj 0/ˇˇˇˇˇˇ/DC4/SInX jD1jxj/NULxj 0j/DC4n/SIjX/NULX0j; which implies that fis differentiable at X0. We say thatfiscontinuously differentiable on a subsetSofRnifSis contained in an open set on which fx1,fx2, . . .;fxnare continuous. Theorem 5.3.10 implies that such a function is differentiable at each X0inS. Example 5.3.8 If f.x;y/Dx2Cy2 x/NULy; then fx.x;y/D2x x/NULy/NULx2Cy2 .x/NULy/2andfy.x;y/D2y x/NULyCx2Cy2 .x/NULy/2: Sincefxandfyare continuous on SD˚ .x;y/ˇˇx¤y/TAB ; fis continuously differentiable on S. Example 5.3.9 The conditions of Theorem 5.3.10 are not necessary for differentiabil- ity; that is, a function may be differentiable at a point X0even if its first partial derivatives are not continuous at X0. For example, let f.x;y/D8 < :.x/NULy/2sin1 x/NULy; x¤y; 0; x Dy: 330 Chapter 5 Real-Valued Functions of Several Variables Then fx.x;y/D2.x/NULy/sin1 x/NULy/NULcos1 x/NULy; x¤y; and fx.x;x/Dlim h!0f.xCh;x//NULf.x;x/ hDlim h!0h2sin.1=h//NUL0 hD0; sofxexists for all.x;y/ , but is not continuous on the line yDx. The same is true of fy, since fy.x;y/D/NUL2.x/NULy/sin1 x/NULyCcos1 x/NULy; x¤y; and fy.x;x/Dlim k!0f.x;xCk//NULf.x;x/ kDlim k!0k2sin./NUL1=k//NUL0 kD0: Now, f.x;y//NULf.0;0//NULfx.0;0/x/NULfy.0;0/yp x2Cy2D8 < :.x/NULy/2 p x2Cy2sin1 x/NULy; x¤y; 0; x Dy; and Schwarz’s inequality implies that ˇˇˇˇˇ.x/NULy/2 p x2Cy2sin1 x/NULyˇˇˇˇˇ/DC42.x2Cy2/p x2Cy2D2p x2Cy2; x¤y: Therefore, lim .x;y/ !.0;0/f.x;y//NULf.0;0//NULfx.0;0/x/NULfy.0;0/yp x2Cy2D0; sofis differentiable at .0;0/ , butfxandfyare not continuous at .0;0/ . Geometric Interpretation of Differentiability In Section 2.3 we saw that if a function fof one variable is differentiable at x0, then the curveyDf.x/ has a tangent line yDT.x/Df.x 0/Cf0.x0/.x/NULx0/ that approximates it so well near x0that lim x!x0f.x//NULT.x/ x/NULx0D0: Moreover, the tangent line is the “limit” of the secant line t hrough the points .x1;f.x 0// and.x0;f.x 0//asx1approachesx0. Section 5.3 Partial Derivatives and the Differential 331 Dyz xz = f(x, y) Figure 5.3.1 Differentiability of a function of nvariables has an analogous geometric interpretation. We will illustrate it for nD2. Iffis defined in a region DinR2, then the set of points .x;y;´/ such that ´Df.x;y/; .x;y/2D; (5.3.27) is asurface inR3(Figure 5.3.1 ). yz xz = f(x,y) (x0, y0) Tangent plane Figure 5.3.2 Iffis differentiable at X0D.x0;y0/, then the plane ´DT.x;y/Df.X0/Cfx.X0/.x/NULx0/Cfy.X0/.y/NULy0/ (5.3.28) intersects the surface ( 5.3.27 ) at.x0;y0;f.x 0;y0//and approximates the surface so well near.x0;y0/that 332 Chapter 5 Real-Valued Functions of Several Variables lim .x;y/ !.x0;y0/f.x;y//NULT.x;y/p .x/NULx0/2C.y/NULy0/2D0 (Figure 5.3.2 ). Moreover, ( 5.3.28 ) is the only plane in R3with these properties (Exer- cise 5.3.25 ). We say that this plane is tangent to the surface ´Df.x;y/ at the point .x0;y0;f.x 0;y0//. We will now show that it is the “limit” of “secant planes” ass ociated with the surface ´Df.x;y/ , just as a tangent line to a curve yDf.x/ inR3is the limit of secant lines to the curve (Section 2.3). LetXiD.xi;yi/.iD1;2;3/ . The equation of the “secant plane” through the points .xi;yi;f.x i;yi//.iD1;2;3/ on the surface ´Df.x;y/ (Figure 5.3.3 ) is of the form ´Df.X0/CA.x/NULx0/CB.y/NULy0/; (5.3.29) whereAandBsatisfy the system f.X1/Df.X0/CA.x 1/NULx0/CB.y 1/NULy0/; f.X2/Df.X0/CA.x 2/NULx0/CB.y 2/NULy0/: Solving forAandByields AD.f.X1//NULf.X0//.y 2/NULy0//NUL.f.X2//NULf.X0//.y 1/NULy0/ .x1/NULx0/.y2/NULy0//NUL.x2/NULx0/.y1/NULy0/(5.3.30) and BD.f.X2//NULf.X0//.x 1/NULx0//NUL.f.X1//NULf.X0//.x 2/NULx0/ .x1/NULx0/.y2/NULy0//NUL.x2/NULx0/.y1/NULy0/(5.3.31) if .x1/NULx0/.y2/NULy0//NUL.x2/NULx0/.y1/NULy0/¤0; (5.3.32) which is equivalent to the requirement that X0,X1, and X2do not lie on a line (Exer- cise5.3.23 ). If we write X1DX0CtUand X2DX0CtV; where UD.u1;u2/andVD.v1;v2/are fixed nonzero vectors (Figure 5.3.3 ), then (5.3.30 ), (5.3.31 ), and ( 5.3.32 ) take the more convenient forms ADf.X0CtU//NULf.X0/ tv2/NULf.X0CtV//NULf.X0/ tu2 u1v2/NULu2v1; (5.3.33) BDf.X0CtV//NULf.X0/ tu1/NULf.X0CtU//NULf.X0/ tv1 u1v2/NULu2v1; (5.3.34) and u1v2/NULu2v1¤0: Section 5.3 Partial Derivatives and the Differential 333 yz xX0 X2 X1 V U Figure 5.3.3 Iffis differentiable at X0, then f.X//NULf.X0/Dfx.X0/.x/NULx0/Cfy.X0/.y/NULy0/C/SI.X/jX/NULX0j; (5.3.35) where lim X!X0/SI.X/D0: (5.3.36) Substituting first XDX0CtUand then XDX0CtVin (5.3.35 ) and dividing by tyields f.X0CtU//NULf.X0/ tDfx.X0/u1Cfy.X0/u2CE1.t/jUj (5.3.37) and f.X0CtV//NULf.X0/ tDfx.X0/v1Cfy.X0/v2CE2.t/jVj; (5.3.38) where E1.t/D/SI.X0CtU/jtj=t andE2.t/D/SI.X0CtV/jtj=t; so lim t!0Ei.t/D0; iD1;2; (5.3.39) because of ( 5.3.36 ). Substituting ( 5.3.37 ) and ( 5.3.38 ) into ( 5.3.33 ) and ( 5.3.34 ) yields ADfx.X0/C/c1291.t/; BDfy.X0/C/c1292.t/; (5.3.40) where 334 Chapter 5 Real-Valued Functions of Several Variables /c1291.t/Dv2jUjE1.t//NULu2jVjE2.t/ u1v2/NULu2v1 and /c1292.t/Du1jVjE2.t//NULv1jUjE1.t/ u1v2/NULu2v1; so lim t!0/c129i.t/D0; iD1;2; (5.3.41) because of ( 5.3.39 ). From ( 5.3.29 ) and ( 5.3.40 ), the equation of the secant plane is ´Df.X0/CŒfx.X0/C/c1291.t//c141.x/NULx0/CŒfy.X0/C/c1292.t//c141.y/NULy0/: Therefore, because of ( 5.3.41 ), the secant plane “approaches” the tangent plane ( 5.3.28 ) as tapproaches zero. Maxima and Minima We say that X0is alocal extreme point offif there is aı>0 such that f.X//NULf.X0/ does not change sign in Sı.X0/\Df. More specifically, X0is alocal maximum point if f.X//DC4f.X0/ or alocal minimum point if f.X//NAKf.X0/ for all XinSı.X0/\Df. The next theorem is analogous to Theorem 2.3.7 . Theorem 5.3.11 Suppose thatfis defined in a neighborhood of X0inRnandfx1.X0/; fx2.X0/;. . .;fxn.X0/exist:LetX0be a local extreme point of f:Then fxi.X0/D0; 1/DC4i/DC4n: (5.3.42) Proof Let E1D.1;0;:::;0/; E2D.0;1;0;:::;0/;:::; EnD.0;0;:::;1/; and gi.t/Df.X0CtEi/; 1/DC4i/DC4n: Thengiis differentiable at tD0, with g0 i.0/Dfxi.X0/ Section 5.3 Partial Derivatives and the Differential 335 (Definition 5.3.1 ). Since X0is a local extreme point of f,t0D0is a local extreme point ofgi. Now Theorem 2.3.7 implies thatg0 i.0/D0, and this implies ( 5.3.42 ). The converse of Theorem 5.3.11 is false, since ( 5.3.42 ) may hold at a point X0that is not a local extreme point of f. For example, let X0D.0;0/ and f.x;y/Dx3Cy3: We say that a point X0where ( 5.3.42 ) holds is a critical point off. Thus, iffis defined in a neighborhood of a local extreme point X0, then X0is a critical point of f; however, a critical point need not be a local extreme point of f. The use of Theorem 5.3.11 for finding local extreme points is covered in calculus, so we will not pursue it here. 5.3 Exercises 1. [email protected]/=@ˆ. (a)f.x;y/Dx2C2xycosx,ˆD 1p 3;/NULr 2 3! (b)f.x;y;´/De/NULxCy2C2´,ˆD/DC21p 3;/NUL1p 3;1p 3/DC3 (c)f.X/DjXj2,ˆD/DC21pn;1pn;/SOH/SOH/SOH;1pn/DC3 (d)f.x;y;´/Dlog.1CxCyC´/,ˆD.0;1;0/ 2. Let f.x;y/D8 < :xysinx x2Cy2; .x;y/¤.0;0/; 0; .x;y/D.0;0/; and let ˆD./RS1;/RS2/be a unit vector. Find @f.0;0/=@ ˆ. 3. [email protected]/=@ˆ, where ˆis the unit vector in the direction of X1/NULX/. (a)f.x;y;´/Dsin/EMxy´ ;X0D.1;1;/NUL2/,X1D.3;2;/NUL1/ (b)f.x;y;´/De/NUL.x2Cy2C2´/;X0D.1;0;/NUL1/,X1D.2;0;/NUL1/ (c)f.x;y;´/Dlog.1CxCyC´/;X0D.1;0;1/ ,X1D.3;0;/NUL1/ (d)f.X/DjXj4;X0D0,X1D.1;1;:::;1/ 4. Give a geometrical interpretation of the directional deriv [email protected] 0;y0/=@ˆof a function of two variables. 5. Find all first-order partial derivatives. (a)f.x;y;´/Dlog.xCyC2´/(b)f.x;y;´/Dx2C3xy´C2xy (c)f.x;y;´/Dxey´(d)f.x;y;´/D´Csinx2y 6. Find all second-order partial derivatives of the functions in Exercise 5.3.5 . 336 Chapter 5 Real-Valued Functions of Several Variables 7. Find all second-order partial derivatives of the following functions at.0;0/ . (a)f.x;y/D8 < :xy.x2/NULy2 x2Cy2; .x;y/¤.0;0/; 0; .x;y/ D.0;0/ (b)f.x;y/D( x2tan/NUL1y x/NULy2tan/NUL1x y; x¤0; y¤0; 0; x D0oryD0 (Herejtan/NUL1uj</EM=2 .) 8. Find a function fDf.x;y/ such thatfxyexists for all .x;y/ , butfyexists nowhere. 9. Letuandvbe functions of two variables with continuous second-order partial derivatives in a region S. Suppose that uxDvyanduyD/NULvxinS. Show that uxxCuyyDvxxCvyyD0 inS. 10. Letfbe a function of .x1;x2;:::;x n/.n/NAK2/such thatfxi,fxj, andfxixj.i¤ j/exist on a neighborhood of X0andfxixjis continuous at X0. Use Theorem 5.3.3 to prove thatfxjxi.X0/exists and equals fxixj.X0/. 11. Use Exercise 5.3.10 and induction on rto prove Theorem 5.3.4 . 12. Letr1;r2;:::;r nbe nonnegative integers such that r1Cr2C/SOH/SOH/SOHCrnDr/NAK0: (a) Show that .´1C´2C/SOH/SOH/SOHC´n/rDX rrŠ r1Šr2Š/SOH/SOH/SOHrnŠ´r1 1´r2 2/SOH/SOH/SOH´rn n; whereP rdenotes summation over all n-tuples.r1;r2;:::;r n/that satisfy the stated conditions. H INT:This is obvious if nD1;and it follows from Exercise 1.2.19 ifnD2:Use induction on n: (b) Show that there are rŠ r1Šr2Š/SOH/SOH/SOHrnŠ orderedn-tuples of integers .i1;i2;:::;i n/that containr1ones,r2twos, . . . , andrnn’s. (c) Letfbe a function of .x1;x2;:::;x n/. Show that there are rŠ r1Šr2Š/SOH/SOH/SOHrnŠ partial derivatives fxi1xi2/SOH/SOH/SOHxirthat involve differentiation ritimes with respect toxi, foriD1;2;:::;n . 13. Prove Lemma 5.3.8 . Section 5.3 Partial Derivatives and the Differential 337 14. Show that the function f.x;y/D8 < :x2y x6C2y2; .x;y/¤.0;0/; 0; .x;y/ D.0;0/; has a directional derivative in the direction of an arbitrar y unit vectorˆat.0;0/ , but fis not continuous at .0;0/ . 15. Prove: Iffxandfyare bounded in a neighborhood of .x0;y0/, thenfis continuous at.x0;y0/. 16. Show directly from Definition 5.3.5 thatfis differentiable at X0. (a)f.x;y/D2x2C3xyCy2,X0D.1;2/ (b)f.x;y;´/D2x2C3xC4y´,X0D.1;1;1/ (c)f.X/DjXj2,X0arbitrary 17. Suppose that fxexists on a neighborhood of .x0;y0/and is continuous at .x0;y0/, whilefymerely exists at .x0;y0/. Show thatfis differentiable at .x0;y0/. 18. FinddfanddX0f, and write.dX0f/.X/NULX0/. (a)f.x;y/Dx3C4xy2C2xysinx,X0D.0;/NUL2/ (b)f.x;y;´/De/NUL.xCyC´/,X0D.0;0;0/ (c)f.X/Dlog.1Cx1C2x2C3x3C/SOH/SOH/SOHCnxn/,X0D0 (d)f.X/DjXj2r,X0D.1;1;1;:::;1/ 19. (a) Suppose that fis differentiable at X0andˆD./RS1;/RS2;:::;/RS n/is a unit vector. Show that @f.X0/ @ˆDfx1.X0//RS1Cfx2.X0//RS2C/SOH/SOH/SOHCfxn.X0//RSn: (b) For what unit vector ˆ[email protected]/=@ˆattain its maximum value? 20. Letfbe defined on Rnby f.X/Dg.x 1/Cg.x 2/C/SOH/SOH/SOHCg.x n/; where g.u/D( u2sin1 u; u¤0; 0; uD0: Show thatfis differentiable at .0;0;:::;0/ , butfx1,fx2, . . . ,fxnare all discon- tinuous at.0;0;:::;0/ . 21. The purpose of this exercise is to show that if f,fxandfyexist on a neighborhood Nof.x0;y0/andfxandfyare differentiable at .x0;y0/, thenfxy.x0;y0/D fyx.x0;y0/. Suppose that the open square ˚.x;y/ˇˇjx/NULx0j<jhj;jy/NULy0j<jhj/TAB 338 Chapter 5 Real-Valued Functions of Several Variables is inN. Consider B.h/Df.x 0Ch;y 0Ch//NULf.x 0Ch;y 0//NULf.x 0;y0Ch/Cf.x 0;y0/: (a) Use the mean value theorem as we did in the proof of Theorem 5.3.3 to write B.h/DŒfx.bx;y 0Ck//NULfx.bx;y 0//c141h; wherebxis betweenx0andx0Ch. Then use the differentiability of fxat .x0;y0/to infer that B.h/Dh2fxy.x0;y0/ChE1.h/; where lim h!0E1.h/ hD0: (b) Use the mean value theorem to write B.h/D/STX fy.x0Ch;by//NULfy.x0;by//ETX h; wherebyis betweeny0andy0Ch. Then use the differentiability of fyat .x0;y0/to infer that B.h/Dh2fyx.x0;y0/ChE2.h/; where lim h!0E2.h/ hD0: (c) Infer from (a)and(b) thatfxy.x0;y0/Dfyx.x0;y0/. 22. (a) Letfxiandfxjbe differentiable at a point X0inRn. Show from Exer- cise5.3.21 that fxixj.X0/Dfxjxi.X0/: (b) Use(a)and induction on rto show that all .r/NUL1/-st order partial derivatives offare differentiable on an open subset SofRn, thenfxi1xi2/SOH/SOH/SOHxir.X/(X2S) depends only on the number of differentiations with respect to each variable, and not on the order in which they are performed. 23. Prove that.x0;y0/,.x1;y1/, and.x2;y2/lie on a line if and only if .x1/NULx0/.y2/NULy0//NUL.x2/NULx0/.y1/NULy0/D0: 24. Find the equation of the tangent plane to the surface ´Df.x;y/ at.x0;y0;´0/D.x0;y0;f.x 0;y0//: (a)f.x;y/Dx2Cy2/NUL1; .x 0;y0/D.1;2/ (b)f.x;y/D2xC3yC1; .x 0;y0/D.1;/NUL1/ (c)f.x;y/Dxysinxy; .x 0;y0/D.1;/EM=2/ (d)f.x;y/Dx2/NUL2y2C3xy; .x 0;y0/D.2;/NUL1/ Section 5.4 The Chain Rule and Taylor’s Theorem 339 25. Prove: Iffis differentiable at .x0;y0/and lim .x;y/ !.x0;y0/f.x;y//NULa/NULb.x/NULx0//NULc.y/NULy0/p .x/NULx0/2C.y/NULy0/2D0; thenaDf.x 0;y0/,bDfx.x0;y0/, andcDfy.x0;y0/. 5.4 THE CHAIN RULE AND TAYLOR’S THEOREM We now consider the problem of differentiating a composite f unction h.U/Df.G.U//; where GD.g1;g2;:::;g n/is a vector-valued function, as defined in Section 5.2. We begin with the following definition. Definition 5.4.1 A vector-valued function GD.g1;g2;:::;g n/isdifferentiable at U0D.u10;u20;:::;u m0/ if its component functions g1,g2, . . . ,gnare differentiable at U0. We need the following lemma to prove the main result of the sec tion. Lemma 5.4.2 Suppose that GD.g1;g2;:::;g n/is differentiable at U0D.u10;u20;:::;u m0/; and define MD0 @nX iD1mX jD1/[email protected] @uj/DC321 A1=2 : Then;if/SI>0; there is aı>0 such that jG.U//NULG.U0/j jU/NULU0j<MC/SIif0<jU/NULU0j<ı: Proof Sinceg1,g2, . . . ,gnare differentiable at U0, applying Lemma 5.3.8 togishows that gi.U//NULgi.U0/D.dU0gi/.U/NULU0/CEi.U/j.U/NULU0j DmX [email protected]/ @uj.uj/NULuj 0/CEi.U/j.U/NULU0j;(5.4.1) 340 Chapter 5 Real-Valued Functions of Several Variables where lim U!U0Ei.U/D0; 1/DC4i/DC4n: (5.4.2) From Schwarz’s inequality, jgi.U//NULgi.U0/j/DC4.MiCjEi.U/j/jU/NULU0j; where MiD0 @mX jD1/[email protected]/ @uj/DC321 A1=2 : Therefore, jG.U//NULG.U0/j jU/NULU0j/DC4 nX iD1.MiCjEi.U/j/2!1=2 : From ( 5.4.2 ), lim U!U0 nX iD1.MiCjEi.U/j/2!1=2 D nX iD1M2 i!1=2 DM; which implies the conclusion. The following theorem is analogous to Theorem 2.3.5 . Theorem 5.4.3 (The Chain Rule) Suppose that the real-valued function fis differentiable at X0inRn;the vector-valued function GD.g1;g2;:::;g n/is differentiable atU0inRm;andX0DG.U0/:Then the real-valued composite function hDfıGdefined by h.U/Df.G.U// (5.4.3) is differentiable at U0;and dU0hDfx1.X0/dU0g1Cfx2.X0/dU0g2C/SOH/SOH/SOHCfxn.X0/dU0gn: (5.4.4) Proof We leave it to you to show that U0is an interior point of the domain of h(Exer- cise5.4.1 ), so it is legitimate to ask if his differentiable at U0. LetX0D.x10;x20;:::;x n0/. Note that xi0Dgi.U0/; 1/DC4i/DC4n; by assumption. Since fis differentiable at X0, Lemma 5.3.8 implies that f.X//NULf.X0/DnX iD1fxi.X0/.xi/NULxi0/CE.X/jX/NULX0j; (5.4.5) where lim X!X0E.X/D0: Section 5.4 The Chain Rule and Taylor’s Theorem 341 Substituting XDG.U/andX0DG.U0/in (5.4.5 ) and recalling ( 5.4.3 ) yields h.U//NULh.U0/DnX iD1fxi.X0/.gi.U//NULgi.U0//CE.G.U//jG.U//NULG.U0/j:(5.4.6) Substituting ( 5.4.1 ) into ( 5.4.6 ) yields h.U//NULh.U0/DnX iD1fxi.X0/.dU0gi/.U/NULU0/C nX iD1fxi.X0/Ei.U/! jU/NULU0j CE.G.U//jG.U//NULG.U0j: Since lim U!U0E.G.U//Dlim X!X0E.X/D0; (5.4.2 ) and Lemma 5.4.2 imply that h.U//NULh.U0//NULnX iD1fxi.X0dU0gi.U/NULU0/ jU/NULU0jD0: Therefore,his differentiable at U0, anddU0his given by ( 5.4.4 ). Example 5.4.1 Let f.x;y;´/D2x2C4xyC3y´; g1.u;v/Du2Cv2; g 2.u;v/Du2/NUL2v2; g 3.u;v/Duv; and h.u;v/Df.g 1.u;v/;g 2.u;v/;g 3.u;v//: LetU0D.1;/NUL1/and X0D.g1.U0/;g2.U0/;g3.U0//D.2;/NUL1;/NUL1/: Then fx.X0/D4; f y.X0/D5; f ´.X0/D/NUL3; @g1.U0/ @uD2;@g1.U0/ @vD/NUL2; @g2.U0/ @uD2;@g2.U0/ @vD4; @g3.U0/ @uD/NUL1;@g3.U0/ @vD1: Therefore, dU0g1D2du/NUL2dv; d U0g2D2duC4dv; d U0g3D/NULduCdv; 342 Chapter 5 Real-Valued Functions of Several Variables and, from ( 5.4.4 ), dU0hDfx.X0/dU0g1Cfy.X0/dU0g2Cf´.X0/dU0g3 D4.2du/NUL2dv/C5.2duC4dv//NUL3./NULduCdv/ D21duC9dv: Since dU0hDhu.U0/duChv.U0/dv we conclude that hu.U0/D21 andhv.U0/D9: (5.4.7) This can also be obtained by writing hexplicitly in terms of .u;v/ and differentiating; thus, h.u;v/D2Œg1.u;v//c1412C4g1.u;v/g 2.u;v/C3g2.u;v/g 3.u;v/ D2.u2Cv2/2C4.u2Cv2/.u2/NUL2v2/C3.u2/NUL2v2/uv D6u4C3u3v/NUL6uv3/NUL6v4: Hence, hu.u;v/D24u3C9u2v/NUL6v3andhv.u;v/D3u3/NUL18uv2/NUL24v3; sohu.1;/NUL1/D21andhv.1;/NUL1/D9, consistent with ( 5.4.7 ). Corollary 5.4.4 Under the assumptions of Theorem 5.4.3; @h.U0/ @uiDnX [email protected]/ @[email protected]/ @ui; 1/DC4i/DC4m: (5.4.8) Proof Substituting [email protected]/ @[email protected]/ @u2du2C/SOH/SOH/[email protected]/ @umdum; 1/DC4i/DC4n; into ( 5.4.4 ) and collecting multipliers of du1,du2, . . . ,dumyields dU0hDmX iD10 @nX [email protected]/ @[email protected]/ @ui1 Adui: However, from Theorem 5.3.6 , dU0hDmX [email protected]/ @uidui: Comparing the last two equations yields ( 5.4.8 ). Section 5.4 The Chain Rule and Taylor’s Theorem 343 When it is not important to emphasize the particular point X0, we write ( 5.4.8 ) less formally as @h @uiDnX jD1@f @xj@gj @ui; 1/DC4i/DC4m; (5.4.9) with the understanding that in calculating @h.U0/=@u i,@gj=@u iis evaluated at U0and @f=@x jatX0DG.U0/. The formulas ( 5.4.8 ) and ( 5.4.9 ) can also be simplified by replacing the symbol Gwith XDX.U/; then we write h.U/Df.X.U// and @h.U0/ @uiDnX [email protected]/ @[email protected]/ @ui; or simply @h @uiDnX jD1@f @xj@xj @ui: (5.4.10) Example 5.4.2 Let.r;/DC2/ be polar coordinates in the xy-plane; that is, xDrcos/DC2; yDrsin/DC2: Suppose that fDf.x;y/ is differentiable on a set S, and let h.r;/DC2/Df.rcos/DC2;rsin/DC2/: If.rcos/DC2;rsin/DC2/2S, (5.4.10 ) implies that @h @rD@f @x@x @rC@f @y@y @rDcos/DC2@f @xCsin/DC2@f @y(5.4.11) and @h @/DC2D@f @x@x @/DC2C@f @y@y @/DC2D/NULrsin/DC2@f @xCrcos/DC2@f @y; wherefxandfyare evaluated at .x;y/D.rcos/DC2;rsin/DC2/. The proof of Corollary 5.4.4 suggests a straightforward way to calculate the partial derivatives of a composite function without using ( 5.4.10 ) explicitly. If h.U/Df.X.U//, then Theorem 5.4.3 , in the more casual notation introduced before Example 5.4.2 , implies that dhDfx1dx1Cfx2dx2C/SOH/SOH/SOHCfxndxn; (5.4.12) wheredx1,dx2, . . . ,dxnmust be written in terms of the differentials du1,du2, . . . ,dum of the independent variables; thus, 344 Chapter 5 Real-Valued Functions of Several Variables dxiD@xi @u1du1C@xi @u2du2C/SOH/SOH/SOHC@xi @umdum: Substituting this into ( 5.4.12 ) and collecting the multipliers of du1,du2, . . . ,dumyields ( 5.4.10 ). Example 5.4.3 If h.r;/DC2;´/Df.x.r;/DC2/;y.r;/DC2/;´/; then dhDfxdxCfydyCf´d´: But dxD@x @rdrC@x @/DC2d/DC2 anddyD@y @rdrC@y @/DC2d/DC2I hence, dhDfx/DC2@x @rdrC@x @/DC2d/DC2/DC3 Cfy/DC2@y @rdrC@y @/DC2d/DC2/DC3 Cf´d´ D/DC2 fx@x @rCfy@y @r/DC3 drC/DC2 fx@x @/DC2Cfy@y @/DC2/DC3 d/DC2Cf´d´; so hrDfx@x @rCfy@y @r; h /DC2Dfx@x @/DC2Cfy@y @/DC2; h ´Df´: Example 5.4.4 Let h.x/Df.x;y.x;´.x//;´.x//: Then dhDfxdxCfydyCf´d´; (5.4.13) dyDyxdxCy´d´; (5.4.14) and d´D´0dx; (5.4.15) where the prime indicates differentiation with respect to x. Substituting ( 5.4.15 ) into (5.4.14 ) yields dyD.yxCy´´0/dx and substituting this and ( 5.4.15 ) into ( 5.4.13 ) yields dhDŒfxCfy.yxCy´´0/Cf´´0/c141dxI hence, h0DfxCfy.yxCy´´0/Cf´´0: Herefx,fy, andf´are evaluated at .x;y.x;´.x//;´.x// ,yxandy´are evaluated at .x;´.x// , and´0is evaluated at x. Section 5.4 The Chain Rule and Taylor’s Theorem 345 Higher Derivatives of Composite Functions Higher derivatives of composite functions can be computed b y repeatedly applying the chain rule. For example, differentiating ( 5.4.10 ) with respect to ukyields @2h @uk@uiDnX jD1@ @uk/DC2@f @xj@xj @ui/DC3 DnX jD1@f @xj@2xj @uk@uiCnX jD1@xj @ui@ @uk/DC2@f @xj/DC3 :(5.4.16) We must be careful finding @ @uk/DC2@f @xj/DC3 ; which really stands here for @ @uk/[email protected]// @xj/DC3 : (5.4.17) The safest procedure is to write temporarily g.X/[email protected]/ @xjI then ( 5.4.17 ) becomes @g.X.U// @ukDnX [email protected]// @[email protected]/ @uk: Since @g @xsD@2f @xs@xj; this yields @ @uk/DC2@f @xk/DC3 DnX sD1@2f @xs@xj@xs @uk: Substituting this into ( 5.4.16 ) yields @2h @uk@uiDnX jD1@f @xj@2xj @uk@uiCnX jD1@xj @uinX sD1@2f @xs@xj@xs@uk: (5.4.18) To computehuiuk.U0/from this formula, we evaluate the partial derivatives of x1,x2, . . . ,xnatU0and those offatX0DX.U0/. The formula is valid if x1,x2, . . . ,xnand their first partial derivatives are differentiable at U0andf,fxi,fx2, . . . ,fxnand their first partial derivatives are differentiable at X0. Instead of memorizing ( 5.4.18 ), you should understand how it is derived and use the method, rather than the formula, when calculating second pa rtial derivatives of composite functions. The same method applies to the calculation of hig her derivatives. 346 Chapter 5 Real-Valued Functions of Several Variables Example 5.4.5 Suppose that fxandfyin Example 5.4.2 are differentiable on an open setSinR2. Differentiating ( 5.4.11 ) with respect to ryields @2h @r2Dcos/DC2@ @r/DC2@f @x/DC3 Csin/DC2@ @r/DC2@f @y/DC3 Dcos/DC2/DC2@2f @x2@x @rC@2f @y@x@y @r/DC3 Csin/DC2/DC2@2f @x@y@x @rC@2f @y2@y @r/DC3(5.4.19) if.x;y/2S. Since @x @rDcos/DC2;@y @rDsin/DC2; and@2f @x@yD@2f @y@x if.x;y/2S(Exercise 5.3.21 ), (5.4.19 ) yields @2h @r2Dcos2/DC2@2f @x2C2sin/DC2cos/DC2@2f @x@yCsin2/DC2@2f @y2: Differentiating ( 5.4.11 ) with respect to /DC2yields @2h @/DC2@rD/NUL sin/DC2@f @xCcos/DC2@f @yCcos/DC2@ @/DC2/DC2@f @x/DC3 Csin/DC2@ @/DC2/DC2@f @y/DC3 D/NUL sin/DC2@f @xCcos/DC2@f @yCcos/DC2/DC2@2f @x2@x @/DC2C@2f @y@x@y @/DC2/DC3 Csin/DC2/DC2@2f @x@y@x @/DC2C@2f @y2@y @/DC2/DC3 : Since @x @/DC2D/NULrsin/DC2and@y @/DC2Drcos/DC2; it follows that @2h @/DC2@rD/NUL sin/DC2@f @xCcos/DC2@f @y/NULrsin/DC2cos/DC2/DC2@2f @x2/NUL@2f @y2/DC3 Cr.cos2/DC2/NULsin2/DC2/@2f @x@y: The Mean Value Theorem For a composite function of the form h.t/Df.x 1.t/;x 2.t/;:::;x n.t// wheretis a real variable, x1,x2, . . . ,xnare differentiable at t0, andfis differentiable at X0DX.t0/, (5.4.8 ) takes the form h0.t0/DnX jD1fxj.X.t0//x0 j.t0/: (5.4.20) This will be useful in the proof of the following theorem. Section 5.4 The Chain Rule and Taylor’s Theorem 347 Theorem 5.4.5 (Mean Value Theorem for Functions of nVariables) Letfbe continuous at X1D.x11;x21;:::;x n1/andX2D.x12;x22;:::;x n2/and dif- ferentiable on the line segment Lfrom X1toX2:Then f.X2//NULf.X1/DnX iD1fxi.X0/.xi2/NULxi1/D.dX0f/.X2/NULX1/ (5.4.21) for some X0onLdistinct from X1andX2. Proof An equation of Lis XDX.t/DtX2C.1/NULt/X1; 0/DC4t/DC41: Our hypotheses imply that the function h.t/Df.X.t// is continuous on Œ0;1/c141 and differentiable on .0;1/ . Since xi.t/Dtxi2C.1/NULt/xi1; (5.4.20 ) implies that h0.t/DnX iD1fxi.X.t//.x i2/NULxi1/; 0<t <1: From the mean value theorem for functions of one variable (Th eorem 2.3.11 ), h.1//NULh.0/Dh0.t0/ for somet02.0;1/ . Sinceh.1/Df.X2/andh.0/Df.X1/, this implies ( 5.4.21 ) with X0DX.t0/. Corollary 5.4.6 Iffx1;fx2;. . .;fxnare identically zero in an open region SofRn; thenfis constant in S: Proof We will show that if X0andXare inS, thenf.X/Df.X0/. SinceSis an open region,Sis polygonally connected (Theorem 5.1.20 ). Therefore, there are points X0;X1;:::; XnDX such that the line segment Lifrom Xi/NUL1toXiis inS,1/DC4i/DC4n. From Theorem 5.4.5 , f.Xi//NULf.Xi/NUL1/DnX iD1.deXif/.Xi/NULXi/NUL1/; whereeXis onLiand therefore in S. Therefore, fxi.eXi/Dfx2.eXi/D/SOH/SOH/SOHDfxn.eXi/D0; 348 Chapter 5 Real-Valued Functions of Several Variables which means that deXif/DC10. Hence, f.X0/Df.X1/D/SOH/SOH/SOHDf.Xn/I that is,f.X/Df.X0/for every XinS. Higher Differentials and Taylor’s Theorem Suppose that fis defined in an n-ballB/SUB.X0/, with/SUB>0 . IfX2B/SUB.X0/, then X.t/DX0Ct.X/NULX0/2B/SUB.X/; 0/DC4t/DC41; so the function h.t/Df.X.t// is defined for 0/DC4t/DC41. From Theorem 5.4.3 (see also ( 5.4.20 )), h0.t/DnX iD1fxi.X.t/.x i/NULxi0/ iffis differentiable in B/SUB.X0/, and h00.t/DnX jD1@ @xj nX [email protected]// @xi.xi/NULxi0/! .xj/NULxj 0/ DnX i;[email protected]// @[email protected]/NULxi0/.xj/NULxj 0/ iffx1,fx2, . . . ,fxnare differentiable in B/SUB.X0/. Continuing in this way, we see that h.r/.t/DnX i1;i2;:::;i [email protected]// @xir@xir/NUL1/SOH/SOH/[email protected]/NULxi1;0/.xi2/NULxi2;0//SOH/SOH/SOH.xir/NULxir;0/(5.4.22) if all partial derivatives of fof order/DC4r/NUL1are differentiable in B/SUB.X0/. This motivates the following definition. Definition 5.4.7 Suppose that r/NAK1and all partial derivatives of fof order/DC4r/NUL1 are differentiable in a neighborhood of X0. Then therthdifferential of fatX0, denoted byd.r/ X0f, is defined by d.r/ X0fDnX i1;i2;:::;i [email protected]/ @xir@xir/NUL1/SOH/SOH/SOH@xi1dxi1dxi2/SOH/SOH/SOHdxir; (5.4.23) wheredx1,dx2, . . . ,dxnare the differentials introduced in Section 5.3; that is, dxiis the function whose value at a point in Rnis theith coordinate of the point. For convenience, we define .d.0/ X0f/Df.X0/: Notice thatd.1/ X0fDdX0f. Section 5.4 The Chain Rule and Taylor’s Theorem 349 Under the assumptions of Definition 5.4.7 , the value of @rf.X0/ @xir@xir/NUL1/SOH/SOH/SOH@xi1 depends only on the number of times fis differentiated with respect to each variable, and not on the order in which the differentiations are perfor med (Exercise 5.3.22 ). Hence, Exercise 5.3.12 implies that ( 5.4.23 ) can be rewritten as d.r/ X0fDX rrŠ r1Šr2Š/SOH/SOH/SOHrnŠ@rf.X0/ @xr1 1@xr2 2/SOH/SOH/[email protected] 1/r1.dx 2/r2/SOH/SOH/SOH.dx n/rn; (5.4.24) whereP rindicates summation over all ordered n-tuples.r1;r2;:::;r n/of nonnegative integers such that r1Cr2C/SOH/SOH/SOHCrnDr and@xri iis omitted from the “denominators” of all terms in ( 5.4.24 ) for whichriD0. In particular, ifnD2, d.r/ X0fDrX jD0 r j! @rf.x 0;y0/ @xj@yr/NULj.dx/j.dy/r/NULj: Example 5.4.6 Let f.x;y/D1 1CaxCby; whereaandbare constants. Then @rf.x;y/ @xj@yr/NULjD./NUL1/rrŠajbr/NULj .1CaxCby/rC1; so d.r/ X0fD./NUL1/rrŠ .1Cax0Cby0/rC1rX jD0 r j! ajbr/NULj.dx/j.dy/r/NULj D./NUL1/rrŠ .1Cax0Cby0/rC1.adxCbdy/r if1Cax0Cby0¤0. Example 5.4.7 Let f.X/Dexp0 @/NULnX jD1ajxj1 A; wherea1,a2, . . . ,anare constants. Then @rf.X/ @xr1 1@xr2 2/SOH/SOH/SOH@xrnnD./NUL1/rar1 1ar2 2/SOH/SOH/SOHarn nexp0 @/NULnX jD1ajxj1 A: 350 Chapter 5 Real-Valued Functions of Several Variables Therefore, .d.r/ X0f/.ˆ/D./NUL1/r X rrŠ r1Šr2Š/SOH/SOH/SOHrnŠar1 1ar2 2/SOH/SOH/SOHarn n.dx 1/r1.dx 2/r2/SOH/SOH/SOH.dx n/rn! /STXexp0 @/NULnX jD1ajxj 01 A D./NUL1/r.a1dx1Ca2dx2C/SOH/SOH/SOHCandxn/rexp0 @/NULnX jD1ajxj 01 A (Exercise 5.3.12 ). The next theorem is analogous to Taylor’s theorem for functi ons of one variable (Theo- rem2.5.4 ). Theorem 5.4.8 (Taylor’s Theorem for Functions of nVariables) Suppose thatfand its partial derivatives of order /DC4kare differentiable at X0andXinRnand on the line segment Lconnecting them :Then f.X/DkX rD01 rŠ.d.r/ X0f/.X/NULX/C1 .kC1/Š.d.kC1/ eXf/.X/NULX0/ (5.4.25) for someeXonLdistinct from X0andX. Proof Define h.t/Df.X0Ct.X/NULX0//: (5.4.26) With ˆDX/NULX0, our assumptions and the discussion preceding Definition 5.4.7 imply thath,h0, . . . ,h.kC1/exist onŒ0;1/c141 . From Taylor’s theorem for functions of one variable, h.1/DkX rD0h.r/.0/ rŠCh.kC1/./FS/ .kC1/Š; (5.4.27) for some/FS2.0;1/ . From ( 5.4.26 ), h.0/Df.X0/andh.1/Df.X/: (5.4.28) From ( 5.4.22 ) and ( 5.4.23 ) with ˆDX/NULX0, h.r/.0/D.d.r/ X0f/.X/NULX0/; 1/DC4r/DC4k; (5.4.29) and h.kC1/./FS/D/DLE dkC1 eXf/DC1 .X/NULX0/ (5.4.30) Section 5.4 The Chain Rule and Taylor’s Theorem 351 where eXDX0C/FS.X/NULX0/ is onLand distinct from X0andX. Substituting ( 5.4.28 ), (5.4.29 ), and ( 5.4.30 ) into (5.4.27 ) yields ( 5.4.25 ). Example 5.4.8 Theorem 5.4.8 and the results of Example 5.4.6 with X0D.0;0/ and ˆD.x;y/ imply that if1CaxCby>0 , then 1 1CaxCbyDkX rD0./NUL1/r.axCby/rC./NUL1/kC1.axCby/kC1 .1Ca/FSxCb/FSy/kC2 for some/FS2.0;1/ . (Note that/FSdepends onkas well as.x;y/ .) Example 5.4.9 Theorem 5.4.8 and the results of Example 5.4.7 with X0D0and ˆDXimply that exp0 @/NULnX jD1ajxj1 ADkX rD0./NUL1/r rŠ.a1x1Ca2x2C/SOH/SOH/SOHCanxn/r C./NUL1/kC1 .kC1/Š.a1x1Ca2x2C/SOH/SOH/SOHCanxn/kC1 /STXexp2 4/NUL/FS0 @nX jD1ajxj1 A3 5; for some/FS2.0;1/ . By analogy with the situation for functions of one variable, we define the kthTaylor polynomial of fabout X0by Tk.X/DkX rD01 rŠ.d.r/ X0f/.X/NULX0/ if the differentials exist; then ( 5.4.25 ) can be rewritten as f.X/DTk.X/C1 .kC1/Š.d.kC1/ eXf/.X/NULX0/: A Sufficient Condition for Relative Extreme Values The next theorem leads to a useful sufficient condition for lo cal maxima and minima. It is related to Theorem 2.5.1 . Strictly speaking, however, it is not a generalization of T heo- rem2.5.1 (Exercise 5.4.18 ). 352 Chapter 5 Real-Valued Functions of Several Variables Theorem 5.4.9 Suppose that fand its partial derivatives of order /DC4k/NUL1are differ- entiable in a neighborhood Nof a point X0inRnand allkth-order partial derivatives of fare continuous at X0:Then lim X!X0f.X//NULTk.X/ jX/NULX0jkD0: (5.4.31) Proof If/SI > 0 , there is aı > 0 such thatBı.X0//SUBNand allkth-order partial derivatives of fsatisfy the inequality ˇˇˇˇˇ@kf.eX/ @xik@xik/NUL1/SOH/SOH/SOH@xi1/[email protected]/ @xik@xik/NUL1/SOH/SOH/SOH@xi1ˇˇˇˇˇ</SI;eX2Bı.X0/: (5.4.32) Now suppose that X2Bı.X0/. From Theorem 5.4.8 withkreplaced byk/NUL1, f.X/DTk/NUL1.X/C1 kŠ.d.k/ eXf/.X/NULX0/; (5.4.33) whereeXis some point on the line segment from X0toXand is therefore in Bı.X0/. We can rewrite ( 5.4.33 ) as f.X/DTk.X/C1 kŠh .d.k/ eXf/.X/NULX0//NUL.d.k/ X0f/.X/NULX0/i : (5.4.34) But ( 5.4.23 ) and ( 5.4.32 ) imply that ˇˇˇ.d.k/ eXf/.X/NULX0//NUL.d.k/ X0f/.X/NULX0/ˇˇˇ<nk/SIjX/NULX0jk(5.4.35) (Exercise 5.4.17 ), which implies that jf.X//NULTk.X/j jX/NULX0jk<nk/SI kŠ;X2Bı.X0/; from ( 5.4.34 ). This implies ( 5.4.31 ). Letrbe a positive integer and X0D.x10;x20;:::;x n0/. A function of the form p.X/DX rar1r2:::rn.x1/NULx10/r1.x2/NULx20/r2/SOH/SOH/SOH.xn/NULxn0/rn; (5.4.36) where the coefficients far1r2:::rngare constants and the summation is over all n-tuples of nonnegative integers .r1;r2;:::;r n/such that r1Cr2C/SOH/SOH/SOHCrnDr; is a homogeneous polynomial of degree rinX/NULX0, provided that at least one of the coefficients is nonzero. For example, if fsatisfies the conditions of Definition 5.4.7 , then the function p.X/D.d.r/ X0f/.X/NULX0/ Section 5.4 The Chain Rule and Taylor’s Theorem 353 is such a polynomial if at least one of the rth-order mixed partial derivatives of fatX0is nonzero. Clearly,p.X0/D0ifpis a homogeneous polynomial of degree r/NAK1inX/NULX0. Ifp.X//NAK0for all X, we say that pispositive semidefinite ; ifp.X/ > 0 except when XDX0,pispositive definite . Similarly,pisnegative semidefinite ifp.X//DC40ornegative definite ifp.X/<0 for all X¤X0. In all these cases, pissemidefinite . Withpas in ( 5.4.36 ), p./NULXC2X0/D./NUL1/rp.X/; sopcannot be semidefinite if ris odd. Example 5.4.10 The polynomial p.x;y;´/Dx2Cy2C´2CxyCx´Cy´ is homogeneous of degree 2inXD.x;y;´/ . We can rewrite pas p.x;y;´/D1 2/STX.xCy/2C.yC´/2C.´Cx/2/ETX; sopis nonnegative, and p.x;y;´/D0if and only if xCyDyC´D´CxD0; which is equivalent to .x;y;´/D.0;0;0/ . Therefore, pis positive definite and /NULpis negative definite. The polynomial p1.x;y;´/Dx2Cy2C´2C2xy can be rewritten as p1.x;y;´/D.xCy/2C´2; sop1is nonnegative. Since p1.1;/NUL1;0/D0,p1is positive semidefinite and /NULp1is negative semidefinite. The polynomial p2.x;y;´/Dx2/NULy2C´2 is not semidefinite, since, for example, p2.1;0;0/D1andp2.0;1;0/D1: From Theorem 5.3.11 , iffis differentiable and attains a local extreme value at X0, then dX0fD0; (5.4.37) sincefx1.X0/Dfx2.X0/D/SOH/SOH/SOHDfxn.X0/D0. However, the converse is false. The next theorem provides a method for deciding whether a point satis fying ( 5.4.37 ) is an extreme point. It is related to Theorem 2.5.3 . 354 Chapter 5 Real-Valued Functions of Several Variables Theorem 5.4.10 Suppose that fsatisfies the hypotheses of Theorem 5.4.9 withk/NAK 2;and d.r/ X0f/DC10 .1/DC4r/DC4k/NUL1/; d.k/ X0f6/DC10: (5.4.38) Then (a) X0is not a local extreme point of funlessd.k/ X0fis semidefinite as a polynomial in X/NULX0:In particular;X0is not a local extreme point of fifkis odd: (b) X0is a local minimum point of fifd.k/ X0fis positive definite ;or a local maximum point ifd.k/ X0fis negative definite : (c) Ifd.k/ X0fis semidefinite ;then X0may be a local extreme point of f;but it need not be: Proof From ( 5.4.38 ) and Theorem 5.4.9 , lim X!X0f.X//NULf.X0//NUL1 kŠ.d.k/ X0/.X/NULX0/ jX/NULX0jkD0: (5.4.39) IfXDX0CtU, where Uis a constant vector, then .d.k/ X0f/.X/NULX0/Dtk.d.k/ X0f/.U/; so (5.4.39 ) implies that lim t!0f.X0CtU//NULf.X0//NULtk kŠ.d.k/ X0f/.U/ tkD0; or, equivalently, lim t!0f.X0CtU//NULf.X0/ tkD1 kŠ.d.k/ X0f/.U/ (5.4.40) for any constant vector U. To prove (a), suppose that d.k/ X0fis not semidefinite. Then there are vectors U1andU2 such that .d.k/ X0f/.U1/>0 and.d.k/ X0f/.U2/<0: This and ( 5.4.40 ) imply that f.X0CtU1/>f. X0/andf.X0CtU2/<f. X0/ fortsufficiently small. Hence, X0is not a local extreme point of f. To prove (b), first assume that d.k/ X0fis positive definite. Then it can be shown that there is a/SUB>0 such that .d.k/ X0f/.X/NULX0/ kŠ/NAK/SUBjX/NULX0jk(5.4.41) Section 5.4 The Chain Rule and Taylor’s Theorem 355 for all X(Exercise 5.4.19 ). From ( 5.4.39 ), there is aı>0 such that f.X//NULf.X0//NUL1 kŠ.d.k/ X0f/.X/NULX0/ jX/NULX0jk>/NUL/SUB 2ifjX/NULX0j<ı: Therefore, f.X//NULf.X0/>1 kŠ.d.k/ X0/.X/NULX0//NUL/SUB 2jX/NULX0jkifjX/NULX0j<ı: This and ( 5.4.41 ) imply that f.X//NULf.X0/>/SUB 2jX/NULX0jkifjX/NULX0j<ı; which implies that X0is a local minimum point of f. This proves half of (b). We leave the other half to you (Exercise 5.4.20 ). To prove (c)merely requires examples; see Exercise 5.4.21 . Corollary 5.4.11 Suppose that f;f x;andfyare differentiable in a neigborhood of a critical point X0D.x0;y0/offandfxx;fyy;andfxyare continuous at .x0;y0/:Let DDfxx.x0;y0/fxy.x0;y0//NULf2 xy.x0;y0/: Then (a).x0;y0/is a local extreme point of fifD >0I.x0;y0/is a local minimum point if fxx.x0;y0/>0 , or a local maximum point if fxx.x0;y0/<0: (b).x0;y0/is not a local extreme point of fifD<0: Proof Write.x/NULx0;y/NULy0/D.u;v/ and p.u;v/D.d.2/ X0f/.u;v/DAu2C2BuvCCv2; whereADfxx.x0;y0/,BDfxy.x0;y0/, andCDfyy.x0;y0/, so DDAC/NULB2: IfD>0 , thenA¤0, and we can write p.u;v/DA/DC2 u2C2B AuvCB2 A2v2/DC3 C/DC2 C/NULB2 A/DC3 v2 DA/DC2 uCB Av/DC32 CD Av2: This cannot vanish unless uDvD0. Hence,d.2/ X0fis positive definite if A > 0 or negative definite if A<0 , and Theorem 5.4.10(b) implies (a). IfD<0 , there are three possibilities: 356 Chapter 5 Real-Valued Functions of Several Variables 1.A¤0; thenp.1;0/DAandp/DC2 /NULB A;1/DC3 DD A. 2.C¤0; thenp.0;1/DCandp/DC2 1;/NULB C/DC3 DD C. 3.ADCD0; thenB¤0andp.1;1/D2Bandp.1;/NUL1/D/NUL2B. In each case the two given values of pdiffer in sign, so X0is not a local extreme point off, from Theorem 5.4.10(a). Example 5.4.11 If f.x;y/Deax2Cby2; then fx.x;y/D2axf.x;y/; f y.x;y/D2byf.x;y/; so fx.0;0/Dfy.0;0/D0; and.0;0/ is a critical point of f. To apply Corollary 5.4.11 , we calculate fxx.x;y/D.2aC4a2x2/f.x;y/; fyy.x;y/D.2bC4b2y2/f.x;y/; fxy.x;y/D4abxyf.x;y/: Therefore, DDfxx.0;0/f yy.0;0//NULf2 xy.0;0/D.2a/.2b//NUL.0/.0/D4ab: Corollary 5.4.11 implies that.0;0/ is a local minimum point if aandbare positive, a local maximum ifaandbare negative, and neither if one is positive and the other is n egative. Corollary 5.4.11 does not apply if aorbis zero. 5.4 Exercises In the exercises on the use of the chain rule, assume that the f unctions satisfy appropriate differentiability conditions. 1. Under the assumptions of Theorem 5.4.3 , show that U0is an interior point of the domain ofh. Section 5.4 The Chain Rule and Taylor’s Theorem 357 2. Leth.U/Df.G.U//and finddU0hby Theorem 5.4.3 , and then by writing h explicitly as a function of U. (a)f.x;y/D3x2C4xy2C3x, g1.u;v/DveuCv/NUL1, g2.u;v/De/NULuCv/NUL1,.u0;v0/D.0;1/ (b)f.x;y;´/De/NUL.xCyC´/, g1.u;v;w/Dlogu/NULlogvClogw, g2.u;v;w/D/NUL2logu/NUL3logw, g3.u;v;w/DloguClogvC2logw,.u0;v0;w0/D.1;1;1/ (c)f.x;y/D.xCy/2, g1.u;v/Ducosv, g2.u;v/Dusinv,.u0;v0/D.3;/EM=2/ (d)f.x;y;´/Dx2Cy2C´2, g1.u;v;w/Ducosvsinw, g2.u;v;w/Ducosvcosw, g3.u;v;w/Dusinv;.u0;v0;w0/D.4;/EM=3;/EM=6/ 3. Leth.r;/DC2;´/Df.x;y;´/ , wherexDrcos/DC2andyDrsin/DC2. Findhr,h/DC2, and h´in terms offx,fy, andf´. 4. Leth.r;/DC2;/RS/Df.x;y;´/ , wherexDrsin/RScos/DC2,yDrsin/RSsin/DC2, and´D rcos/RS. Findhr,h/DC2, andh/RSin terms offx,fy, andf´. 5. Prove: (a) Ifh.u;v/Df.u2Cv2/, thenvhu/NULuhvD0. (b) Ifh.u;v/Df.sinuCcosv/, thenhusinvChvcosuD0. (c) Ifh.u;v/Df.u=v/ , thenuhuCvhvD0. (d) Ifh.u;v/Df.g.u;v/;/NULg.u;v// , thendhD.fx/NULfy/dg. 6. Findhyandh´if h.y;´/Dg.x.y;´/;y;´;w.y;´//: 7. Suppose that u,v, andfare defined on ./NUL1;1/. Letuandvbe differentiable andfbe continuous for all x. Show that d dxZv.x/ u.x/f.t/dtDf.v.x//v0.x//NULf.u.x//u0.x/: 8. We say thatfDf.x 1;x2;:::;x n/ishomogeneous of degree rifDfis open and there is a constant rsuch that f.tx 1;tx2;:::;tx n/Dtrf.x 1;x2;:::;x n/ 358 Chapter 5 Real-Valued Functions of Several Variables whenevert > 0 and.x1;x2;:::;x n/and.tx1;tx2;:::;tx n/are inDf. Prove: If fis differentiable and homogeneous of degree r, then nX iD1xifxi.x1;x2;:::;x n/Drf.x 1;x2;:::;x n/: (This is Euler’s theorem for homogeneous functions .) 9. Ifh.r;/DC2/Df.rcos/DC2;rsin/DC2/, show that fxxCfyyDhrrC1 rhrC1 r2h/DC2/DC2: HINT:Rewrite the defining equation as f.x;y/Dh.r.x;y/;/DC2.x;y//; withr.x;y/Dp x2Cy2and/DC2.x;y/Dtan/NUL1.y=x/; and differentiate with respect to xandy: 10. Leth.u;v/Df.a.u;v/;b.u;v// , whereauDbvandavD/NULbu. Show that huuChvvD.fxxCfyy/.a2 uCa2 v/: 11. Prove: If u.x;t/Df.x/NULct/Cg.xCct/; thenuttDc2uxx. 12. Leth.u;v/Df.uCv;u/NULv/. Show that (a)fxx/NULfyyDhuv(b)fxxCfyyD1 2.huuChvv/ 13. Returning to Exercise 5.4.4 , findhrrandhr/DC2in terms of the partial derivatives of f. 14. LethuvD0for all.u;v/ . Show thathis of the form h.u;v/DU.u/CV.v/: Use this and Exercise 5.4.12(a)to show that if fxx/NULfyyD0for all.x;y/ , then f.x;y/DU.xCy/CV.x/NULy/: 15. Prove or give a counterexample: If fis differentiable and fxD0in a regionD, thenf.x 1;y/Df.x 2;y/whenever.x1;y/and.x2;y/are inD; that isf.x;y/ depends only on y. 16. FindT3.X/. (a)f.x;y/Dexcosy,X0D.0;0/ (b)f.x;y/De/NULx/NULy,X0D.0;0/ (c)f.x;y;´/D.xCyC´/NUL3/5,X0D.1;1;1/ (d)f.x;y;´/Dsinxsinysin´,X0D.0;0;0/ 17. Use Eqns. ( 5.4.23 ) and ( 5.4.32 ) to prove Eqn. ( 5.4.35 ). Section 5.4 The Chain Rule and Taylor’s Theorem 359 18. Carefully explain why Theorem 5.4.9 is not a generalization of Theorem 2.5.1 . 19. Suppose that pis a homogeneous polynomial of degree rinYandp.Y/>0 for all nonzero YinRn. Show that there is a /SUB > 0 such thatp.Y//NAK/SUBjYjrfor all Yin Rn. HINT:passumes a minimum on the set˚YˇˇjYjD1/TAB:Use this to establish the inequality in Eqn. ( 5.4.41 ): 20. Complete the proof of Theorem 5.4.10(b). 21. (a) Show that.0;0/ is a critical point of each of the following functions, and th at they have positive semidefinite second differentials at .0;0/ . p.x;y/Dx2/NUL2xyCy2Cx4Cy4I q.x;y/Dx2/NUL2xyCy2/NULx4/NULy4: (b) Show thatDas defined in Corollary 5.4.11 is zero for both pandq. (c) Show that.0;0/ is a local minimum point of pbut not a local extreme point ofq. 22. Suppose that pDp.x 1;x2;:::;x n/is a homogeneous polynomial of degree r (Exercise 5.4.8 ). Leti1,i2, . . . ,inbe nonnegative integers such that i1Ci2C/SOH/SOH/SOHCinDk; and let q.x1;x2;:::;x n/[email protected] 1;x2;:::;x n/ @xi1 1@xi2 2/SOH/SOH/SOH@xinn: Show thatqis homogeneous of degree /DC4r/NULk, subject to the convention that a homogeneous polynomial of negative degree is identically z ero. 23. Suppose that fDf.x 1;x2;:::;x n/is a homogeous function of degree r(Exer- cise 8), with mixed partial derivative of all orders. Show th at nX i;[email protected] 1;x2;:::;x n/ @[email protected]/NUL1/f.x 1;x2;:::;x n/ and nX i;j;k [email protected];x2;:::;x n/ @xi@[email protected]/NUL1/.r/NUL2/f.x 1;x2;:::;x n/: Can you generalize these results? 24. Obtain the result in Example 5.4.7 by writing F.X/De/NULa1x1e/NULa2x2/SOH/SOH/SOHe/NULanxn; formally multiplying the series e/NULaixiD1X riD0./NUL1/ri.aixi/ri riŠ; 1/DC4i/DC4n together, and collecting the resulting products appropria tely. 360 Chapter 5 Real-Valued Functions of Several Variables 25. Let f.x;y/DexCy: By writing f.x;y/D1X rD0.xCy/r rŠ; and expanding .xCy/rby means of the binomial theorem, verify that d.r/ (0;0/fDrX jD0 r j! @rf.0;0/ @xj@yr/NULj.dx/j.dy/r/NULj: CHAPTER 6 Vector-Valued Functions of Several Variables IN THIS CHAPTER we study the differential calculus of vector -valued functions of several variables. SECTION 6.1 reviews matrices, determinants, and linear tra nsformations, which are inte- gral parts of the differential calculus as presented here. SECTION 6.2 defines continuity and differentiability of vec tor-valued functions of several variables. The differential of a vector-valued function Fis defined as a certain linear trans- formation. The matrix of this linear transformation is call ed the differential matrix of F, denoted by F0. The chain rule is extended to compositions of differentiab le vector-valued functions. SECTION 6.3 presents a complete proof of the inverse functio n theorem. SECTION 6.4. uses the inverse function theorem to prove the i mplicit function theorem. 6.1 LINEAR TRANSFORMATIONS AND MATRICES In this and subsequent sections it will often be convenient t o write vectors vertically; thus, instead of XD.x1;x2;:::;x n/we will write XD2 6664x1 x2 ::: xn3 7775 when dealing with matrix operations. Although we assume tha t you have completed a course in linear algebra, we will review the pertinent matri x operations. We have defined vector-valued functions as ordered n-tuples of real-valued functions, in connection with composite functions hDfıG, wherefis real-valued and Gis vector- valued. We now consider vector-valued functions as objects of interest on their own. 361 362 Chapter 6 Vector-Valued Functions of Several Variables Iff1,f2, . . . ,fmare real-valued functions defined on a set DinRn, then FD2 6664f1 f2 ::: fm3 7775 assigns to every XinDanm-vector F.X/D2 6664f1.X/ f2.X/ ::: fm.X/3 7775: Recall thatf1,f2, . . . ,fmare the component functions , or simply components , ofF. We write FWRn!Rm to indicate that the domain of Fis inRnand the range of Fis inRm. We also say that Fis a transformation from RntoRm. IfmD1, we identify Fwith its single component function f1and regard it as a real-valued function. Example 6.1.1 The transformation FWR2!R3defined by F.x;y/D2 42xC3y /NULxC4y x/NULy3 5 has component functions f1.x;y/D2xC3y; f 2.x;y/D/NULxC4y; f 3.x;y/Dx/NULy: Linear Transformations The simplest interesting transformations from RntoRmare the linear transformations , defined as follows Definition 6.1.1 A transformation LWRn!Rmdefined on all of Rnislinear if L.XCY/DL.X/CL.Y/ for all XandYinRnand L.aX/DaL.X/ for all XinRnand real numbers a. Section 6.1 Linear Transformations and Matrices 363 Theorem 6.1.2 A transformation LWRn!Rmdefined on all of Rnis linear if and only if L.X/D2 6664a11x1Ca12x2C/SOH/SOH/SOHCa1nxn a21x1Ca22x2C/SOH/SOH/SOHCa2nxn ::: am1x1Cam2x2C/SOH/SOH/SOHCamnxn3 7775; (6.1.1) where theaij’s are constants : Proof If can be seen by induction (Exercise 6.1.1 ) that if Lis linear, then L.a1X1Ca2X2C/SOH/SOH/SOHCakXk/Da1L.X1/Ca2L.X2/C/SOH/SOH/SOHCakL.Xk/ (6.1.2) for any vectors X1,X2, . . . , Xkand real numbers a1,a2, . . . ,ak. Any XinRncan be written as XD2 6664x1 x2 ::: xn3 7775Dx12 66641 0 ::: 03 7775Cx22 66640 1 ::: 03 7775C/SOH/SOH/SOHCxn2 66640 0 ::: 13 7775 Dx1E1Cx2E2C/SOH/SOH/SOHCxnEn: Applying ( 6.1.2 ) withkDn,XiDEi, andaiDxiyields L.X/Dx1L.E1/Cx2L.E2/C/SOH/SOH/SOHCxnL.En/: (6.1.3) Now denote L.Ej/D2 6664a1j a2j ::: amj3 7775; so (6.1.3 ) becomes L.X/Dx12 6664a11 a21 ::: am13 7775Cx22 6664a12 a22 ::: am23 7775C/SOH/SOH/SOHCxn2 6664a1n a2n ::: amn3 7775; which is equivalent to ( 6.1.1 ). This proves that if Lis linear, then Lhas the form ( 6.1.1 ). We leave the proof of the converse to you (Exercise 6.1.2 ). We call the rectangular array AD2 6664a11a12/SOH/SOH/SOHa1n a21a21/SOH/SOH/SOHa2n :::::::::::: am1am2/SOH/SOH/SOHamn3 7775(6.1.4) 364 Chapter 6 Vector-Valued Functions of Several Variables thematrix of the linear transformation ( 6.1.1 ). The number aijin theith row andjth column of Ais called the.i;j/ th entry of A. We say that Ais anm/STXnmatrix, since A hasmrows andncolumns. We will sometimes abbreviate ( 6.1.4 ) as ADŒaij/c141: Example 6.1.2 The transformation Fof Example 6.1.1 is linear. The matrix of Fis 2 42 3 /NUL1 4 1/NUL13 5: We will now recall the matrix operations that we need to study the differential calculus of transformations. Definition 6.1.3 (a) Ifcis a real number and ADŒaij/c141is anm/STXnmatrix, thencAis them/STXnmatrix defined by cADŒcaij/c141I that is,cAis obtained by multiplying every entry of Abyc. (b) IfADŒaij/c141andBDŒbij/c141arem/STXnmatrices, then the sum ACBis them/STXn matrix ACBDŒaijCbij/c141I that is, the sum of two m/STXnmatrices is obtained by adding corresponding entries. The sum of two matrices is not defined unless they have the same number of rows and the same number of columns. (c) IfADŒaij/c141is anm/STXpmatrix and BDŒbij/c141is ap/STXnmatrix, then the product CDABis them/STXnmatrix with cijDai1b1jCai2b2jC/SOH/SOH/SOHCaipbpjDpX kD1aikbkj; 1/DC4i/DC4m; 1/DC4j/DC4n: Thus, the.i;j/ th entry of ABis obtained by multiplying each entry in the ith row of Aby the corresponding entry in the jth column of Band adding the products. This definition requires that Ahave the same number of columns as Bhas rows. Otherwise, ABis undefined. Example 6.1.3 Let AD2 42 1 2 /NUL1 0 3 0 1 03 5;BD2 40 1 1 /NUL1 0 2 3 0 13 5; and CD2 45 0 1 2 3 0/NUL3 1 1 0/NUL1 13 5: Section 6.1 Linear Transformations and Matrices 365 Then 2AD2 42.2/ 2.1/ 2.2/ 2./NUL1/ 2.0/ 2.3/ 2.0/ 2.1/ 2.0/3 5D2 44 2 4 /NUL2 0 6 0 2 03 5 and ACBD2 42C0 1C1 2C1 /NUL1/NUL1 0C0 3C2 0C3 1C0 0C13 5D2 42 2 3 /NUL2 0 5 3 1 13 5: The (2, 3) entry in the product ACis obtained by multiplying the entries of the second row of Aby those of the third column of Cand adding the products: thus, the (2, 3) entry ofACis ./NUL1/.1/C.0/./NUL3/C.3/./NUL1/D/NUL4: The full product ACis 2 42 1 2 /NUL1 0 3 0 1 03 52 45 0 1 2 3 0/NUL3 1 1 0/NUL1 13 5D2 415 0/NUL3 7 /NUL2 0/NUL4 1 3 0/NUL3 13 5: Notice that ACC,BCC,CA, and CBare undefined. We leave the proofs of next three theorems to you (Exercises 6.1.7 –6.1.9 ) Theorem 6.1.4 IfA;B;andCarem/STXnmatrices;then .ACB/CCDAC.BCC/: Theorem 6.1.5 IfAandBarem/STXnmatrices and randsare real numbers ;then(a) r.sA/D.rs/AI(b).rCs/ADrACsAI(c)r.ACB/DrACrB: Theorem 6.1.6 IfA;B;andCarem/STXp;p/STXq;andq/STXnmatrices;respectively; then.AB/CDA.BC/: The next theorem shows why Definition 6.1.3 is appropriate. We leave the proof to you (Exercise 6.1.11 ). Theorem 6.1.7 (a) If we regard the vector XD2 6664x1 x2 ::: xn3 7775 as ann/STX1matrix;then the linear transformation (6.1.1 )can be written as L.X/DAX: 366 Chapter 6 Vector-Valued Functions of Several Variables (b) IfL1andL2are linear transformations from RntoRmwith matrices A1andA2 respectively;thenc1L1Cc2L2is the linear transformation from RntoRmwith matrixc1A1Cc2A2: (c) IfL1WRn!RpandL2WRp!Rmare linear transformations with matrices A1 andA2;respectively;then the composite function L3DL2ıL1;defined by L3.X/DL2.L1.X//; is the linear transformation from RntoRmwith matrix A2A1: Example 6.1.4 If L1.X/D2 42xC3y 3xC2y /NULxCy3 5 and L2.X/D2 4/NULx/NULy 4xCy x3 5; then A1D2 42 3 3 2 /NUL1 13 5 and A2D2 4/NUL1/NUL1 4 1 1 03 5: The linear transformation LD2L1CL2 is defined by L.X/D2L1.X/CL2.X/ D22 42xC3y 3xC2y /NULxCy3 5C2 4/NULx/NULy 4xCy x3 5 D2 43xC5y 10xC5y /NULxC2y3 5: The matrix of Lis AD2 43 5 10 5 /NUL1 23 5D2A1CA2: Example 6.1.5 Let L1.X/D/DC4xC2y 3xC4y/NAK WR2!R2; and L2.U/D2 4uCv /NULu/NUL2v 3uCv3 5WR2!R3: Section 6.1 Linear Transformations and Matrices 367 Then L3DL2ıL1WR2!R3is given by L3.X/DL2..L1.X//D2 4.xC2y/C.3xC4y/ /NUL.xC2y//NUL2.3xC4y/ 3.xC2y/C.3xC4y/3 5D2 44xC6y /NUL7x/NUL10y 6xC10y3 5: The matrices of L1andL2are A1D/DC41 2 3 4/NAK and A2D2 41 1 /NUL1/NUL2 3 13 5; respectively. The matrix of L3is CD2 44 6 /NUL7/NUL10 6 103 5DA2A1: Example 6.1.6 The linear transformations of Example 6.1.5 can be written as L1.X/D/DC41 2 3 4/NAK/DC4x y/NAK ;L2.U/D2 41 1 /NUL1/NUL2 3 13 5/DC4u v/NAK ; and L3.X/D2 44 6 /NUL7/NUL10 6 103 5/DC4x y/NAK : A New Notation for the Differential If a real-valued function fWRn!Ris differentiable at X0, then dX0fDfx1.X0/dx 1Cfx2.X0/dx 2C/SOH/SOH/SOHCfxn.X0/dx n: This can be written as a matrix product dX0fDŒfx1.X0/ f x2.X0//SOH/SOH/SOHfxn.X0//c1412 6664dx1 dx2 ::: dxn3 7775: (6.1.5) We define the differential matrix of fatX0by f0.X0/DŒfx1.X0/ f x2.X0//SOH/SOH/SOHfxn.X0//c141 (6.1.6) and the differential linear transformation by dXD2 6664dx1 dx2 ::: dxn3 7775: 368 Chapter 6 Vector-Valued Functions of Several Variables Then ( 6.1.5 ) can be rewritten as dX0fDf0.X0/dX: (6.1.7) This is analogous to the corresponding formula for function s of one variable (Exam- ple5.3.7 ), and shows that the differential matrix f0.X0/is a natural generalization of the derivative. With this new notation we can express the definin g property of the differential in a way similar to the form that applies for nD1: lim X!X0f.X//NULf.X0//NULf0.X0/.X/NULX0/ jX/NULX0jD0; where X0D.x10;x20;:::;x n0/andf0.X0/.X/NULX0/is the matrix product Œfx1.X0/ f x2.X0//SOH/SOH/SOHfxn.X0//c1412 6664x1/NULx10 x2/NULx20 ::: xn/NULxn03 7775: As before, we omit the X0in (6.1.6 ) and ( 6.1.7 ) when it is not necessary to emphasize the specific point; thus, we write f0D/STX fx1fx2/SOH/SOH/SOHfxn/ETX anddfDf0dX: Example 6.1.7 If f.x;y;´/D4x2y´3; then f0.x;y;´/DŒ8xy´34x2´312x2y´2/c141: In particular, if X0D.1;/NUL1;2/, then f0.X0/DŒ/NUL64 32/NUL48/c141; so dX0fDf0.X0/dXDŒ/NUL64 32/NUL48/c1412 4dx dy d´3 5 D/NUL64dxC32dy/NUL48d´: The Norm of a Matrix We will need the following definition in the next section. Definition 6.1.8 Thenorm;kAk;of anm/STXnmatrix ADŒaij/c141is the smallest number such that jAXj/DC4k AkjXj for all XinRn: Section 6.1 Linear Transformations and Matrices 369 To justify this definition, we must show that kAkexists. The components of YDAX are yiDai1x1Cai2x2C/SOH/SOH/SOHCainxn; 1/DC4i/DC4m: By Schwarz’s inequality, y2 i/DC4.a2 i1Ca2 i2C/SOH/SOH/SOHCa2 in/jXj2: Summing this over 1/DC4i/DC4myields jYj2/DC40 @mX iD1nX jD1a2 ij1 AjXj2: Therefore, the set BD˚ KˇˇjAXj/DC4KjXjfor all XinRn/TAB is nonempty. Since Bis bounded below by zero, Bhas an infimum ˛. If/SI>0 , then˛C/SI is inBbecause if not, then no number less than ˛C/SIcould be inB. Then˛C/SIwould be a lower bound for B, contradicting the definition of ˛. Hence, jAXj/DC4.˛C/SI/jXj;X2Rn: Since/SIis an arbitrary positive number, this implies that jAXj/DC4˛jXj;X2Rn; so˛2B. Since no smaller number is in B, we conclude thatkAkD˛. In our applications we will not have to actually compute the n orm of a matrix A; rather, it will be sufficient to know that the norm exists (finite). Square Matrices Linear transformations from RntoRnwill be important when we discuss the inverse func- tion theorem in Section 6.3 and change of variables in multip le integrals in Section 7.3. The matrix of such a transformation is square ; that is, it has the same number of rows and columns. We assume that you know the definition of the determinant det.A/Dˇˇˇˇˇˇˇˇˇa11a12/SOH/SOH/SOHa1n a21a22/SOH/SOH/SOHa2n :::::::::::: an1an2/SOH/SOH/SOHannˇˇˇˇˇˇˇˇˇ of ann/STXnmatrix AD2 6664a11a12/SOH/SOH/SOHa1n a21a22/SOH/SOH/SOHa2n :::::::::::: an1an2/SOH/SOH/SOHann3 7775: 370 Chapter 6 Vector-Valued Functions of Several Variables Thetranspose ,At, of a matrix A(square or not) is the matrix obtained by interchanging the rows and columns of A; thus, if AD2 41 2 3 3 1 4 0 1/NUL23 5;then AtD2 41 3 0 2 1 1 3 4/NUL23 5: A square matrix and its transpose have the same determinant; thus, det.At/Ddet.A/: We take the next theorem from linear algebra as given. Theorem 6.1.9 IfAandBaren/STXnmatrices;then det.AB/Ddet.A/det.B/: The entriesaii,1/DC4i/DC4n, of ann/STXnmatrix Aare on the main diagonal ofA. Then/STXn matrix with ones on the main diagonal and zeros elsewhere is c alled the identity matrix and is denoted by I; thus, ifnD3, ID2 41 0 0 0 1 0 0 0 13 5: We call Ithe identity matrix because AIDAandIADAifAis anyn/STXnmatrix. We say that ann/STXnmatrix Aisnonsingular if there is ann/STXnmatrix A/NUL1, the inverse of A, such that AA/NUL1DA/NUL1ADI. Otherwise, we say that Aissingular Our main objective is to show that an n/STXnmatrix Ais nonsingular if and only if det.A/¤0. We will also find a formula for the inverse. Definition 6.1.10 LetADŒaij/c141be ann/STXnmatrix;withn/NAK2:Thecofactor of an entryaijis cijD./NUL1/iCjdet.Aij/; where Aijis the.n/NUL1//STX.n/NUL1/matrix obtained by deleting the ith row andjth column ofA:Theadjoint ofA;denoted by adj .A/;is then/STXnmatrix whose .i;j/ th entry iscj i: Example 6.1.8 The cofactors of AD2 44 2 1 3/NUL1 2 0 1 23 5 Section 6.1 Linear Transformations and Matrices 371 are c11Dˇˇˇˇ/NUL1 2 1 2ˇˇˇˇD/NUL4; c 12D/NULˇˇˇˇ3 2 0 2ˇˇˇˇD/NUL6; c 13Dˇˇˇˇ3/NUL1 0 1ˇˇˇˇD3; c21D/NULˇˇˇˇ2 1 1 2ˇˇˇˇD/NUL3; c 22Dˇˇˇˇ4 1 0 2ˇˇˇˇD8; c 23D /NULˇˇˇˇ4 2 0 1ˇˇˇˇD /NUL4; c31Dˇˇˇˇ2 1 /NUL1 2ˇˇˇˇD5; c 32D/NULˇˇˇˇ4 1 3 2ˇˇˇˇD/NUL5; c 33Dˇˇˇˇ4 2 3/NUL1ˇˇˇˇD/NUL10; so adj.A/D2 4/NUL4/NUL3 5 /NUL6 8/NUL5 3/NUL4/NUL103 5: Notice that adj .A/is the transpose of the matrix 2 4/NUL4/NUL6 3 /NUL3 8/NUL4 5/NUL5/NUL103 5 obtained by replacing each entry of Aby its cofactor. For a proof of the following theorem, see any elementary line ar algebra text. Theorem 6.1.11 LetAbe ann/STXnmatrix: (a) The sum of the products of the entries of a row of Aand their cofactors equals det.A/; while the sum of the products of the entries of a row of Aand the cofactors of the entries of a different row equals zero Ithat is; nX kD1aikcjkD/SUBdet.A/; iDj; 0; i¤j:(6.1.8) (b) The sum of the products of the entries of a column of Aand their cofactors equals det.A/;while the sum of the products of the entries of a column of Aand the cofactors of the entries of a different column equals zero Ithat is; nX kD1ckiakjD/SUBdet.A/; iDj; 0; i¤j:(6.1.9) If we compute det .A/from the formula det.A/DnX kD1aikcik; 372 Chapter 6 Vector-Valued Functions of Several Variables we say that we are expanding the determinant in cofactors of its ith row . Since we can chooseiarbitrarily fromf1;:::;ng, there arenways to do this. If we compute det .A/ from the formula det.A/DnX kD1akjckj; we say that we are expanding the determinant in cofactors of its jth column . There are also nways to do this. In particular, we note that det .I/D1for alln/NAK1. Theorem 6.1.12 LetAbe ann/STXnmatrix:Ifdet.A/D0;then Ais singular:If det.A/¤0;then Ais nonsingular ;andAhas the unique inverse A/NUL1D1 det.A/adj.A/: (6.1.10) Proof If det.A/D0, then det.AB/D0for anyn/STXnmatrix, by Theorem 6.1.9 . Therefore, since det .I/D1, there is no matrix n/STXnmatrix Bsuch that ABDI; that is, A is singular if det .A/D0. Now suppose that det .A/¤0. Since ( 6.1.8 ) implies that Aadj.A/Ddet.A/I and ( 6.1.9 ) implies that adj.A/ADdet.A/I; dividing both sides of these two equations by det .A/shows that if A/NUL1is as defined in (6.1.10 ), then AA/NUL1DA/NUL1ADI. Therefore, A/NUL1is an inverse of A. To see that it is the only inverse, suppose that Bis ann/STXnmatrix such that ABDI. Then A/NUL1.AB/DA/NUL1, so.A/NUL1A/BDA/NUL1. Since AA/NUL1DIandIBDB, it follows that BDA/NUL1. Example 6.1.9 In Example 6.1.8 we found that the adjoint of AD2 44 2 1 3/NUL1 2 0 1 23 5 is adj.A/D2 4/NUL4/NUL3 5 /NUL6 8/NUL5 3/NUL4/NUL103 5: We can compute det .A/by finding any diagonal entry of Aadj.A/. (Why?) This yields det.A/D/NUL25. (Verify.) Therefore, A/NUL1D/NUL1 252 4/NUL4/NUL3 5 /NUL6 8/NUL5 3/NUL4/NUL103 5: Section 6.1 Linear Transformations and Matrices 373 Now consider the equation AXDY (6.1.11) with AD2 6664a11a12/SOH/SOH/SOHa1n a21a22/SOH/SOH/SOHa2n :::::::::::: an1an2/SOH/SOH/SOHann3 7775;XD2 6664x1 x2 ::: xn3 7775;and YD2 6664y1 y2 ::: yn3 7775: Here AandYare given, and the problem is to find X. Theorem 6.1.13 The system (6.1.11 )has a solution Xfor any given Yif and only if Ais nonsingular :In this case;the solution is unique and is given by XDA/NUL1Y. Proof Suppose that Ais nonsingular, and let XDA/NUL1Y. Then AXDA.A/NUL1Y/D.AA/NUL1/YDIYDYI that is, Xis a solution of ( 6.1.11 ). To see that Xis the only solution of ( 6.1.11 ), suppose thatAX1DY. Then AX1DAX, so A/NUL1.AX/DA/NUL1.AX1/ and .A/NUL1A/XD.A/NUL1A/X1; which is equivalent to IXDIX1, orXDX1. Conversely, suppose that ( 6.1.11 ) has a solution for every Y, and let Xisatisfy AXiD Ei,1/DC4i/DC4n. Let BDŒX1X2/SOH/SOH/SOHXn/c141I that is, X1,X2, . . . , Xnare the columns of B. Then ABDŒAX1AX2/SOH/SOH/SOHAXn/c141DŒE1E2/SOH/SOH/SOHEn/c141DI: To show that BDA/NUL1, we must still show that BADI. We first note that, since ABDI and det.BA/Ddet.AB/D1(Theorem 6.1.9 ),BAis nonsingular (Theorem 6.1.12 ). Now note that .BA/.BA/DB.AB/A/DBIAI that is, .BA/.BA/D.BA/: Multiplying both sides of this equation on the left by BA//NUL1yields BADI. The following theorem gives a useful formula for the compone nts of the solution of (6.1.11 ). 374 Chapter 6 Vector-Valued Functions of Several Variables Theorem 6.1.14 ( Cramer ’s Rule) IfADŒaij/c141is nonsingular ;then the solu- tion of the system a11x1Ca12x2C/SOH/SOH/SOHCa1nxnDy1 a21x1Ca22x2C/SOH/SOH/SOHCa2nxnDy2 ::: an1x1Can2x2C/SOH/SOH/SOHCannxnDyn .or;in matrix form ;AXDY/is given by xiDDi det.A/; 1/DC4i/DC4n; whereDiis the determinant of the matrix obtained by replacing the ith column of Awith YIthus; D1Dˇˇˇˇˇˇˇˇˇy1a12/SOH/SOH/SOHa1n y2a22::: a 2n :::::::::::: ynan2/SOH/SOH/SOHannˇˇˇˇˇˇˇˇˇ; D 2Dˇˇˇˇˇˇˇˇˇa11y1a13/SOH/SOH/SOHa1n a21y2a23/SOH/SOH/SOHa2n ::::::::::::::: an1ynan3/SOH/SOH/SOHannˇˇˇˇˇˇˇˇˇ;/SOH/SOH/SOH; DnDˇˇˇˇˇˇˇˇˇa11/SOH/SOH/SOHa1;n/NUL1y1 a21/SOH/SOH/SOHa2;n/NUL1y2 :::::::::::: an1/SOH/SOH/SOHan;n/NUL1ynˇˇˇˇˇˇˇˇˇ: Proof From Theorems 6.1.12 and6.1.13 , the solution of AXDYis 2 6664x1 x2 ::: xn3 7775DA/NUL1YD1 det.A/2 6664c11c21/SOH/SOH/SOHcn1 c12c22/SOH/SOH/SOHcn2 /SOH/SOH/SOH /SOH/SOH/SOH:::/SOH/SOH/SOH c1nc2n/SOH/SOH/SOHcnn3 77752 6664y1 y2 ::: yn3 7775 D2 6664c11y1Cc21y2C/SOH/SOH/SOHCcn1yn c12y1Cc22y2C/SOH/SOH/SOHCcn2yn ::: c1ny1Cc2ny2C/SOH/SOH/SOHCcnnyn3 7775: But c11y1Cc21y2C/SOH/SOH/SOHCcn1ynDˇˇˇˇˇˇˇˇˇy1a12/SOH/SOH/SOHa1n y2a22::: a 2n :::::::::::: ynan2/SOH/SOH/SOHannˇˇˇˇˇˇˇˇˇ; Section 6.1 Linear Transformations and Matrices 375 as can be seen by expanding the determinant on the right in cof actors of its first column. Similarly, c12y1Cc22y2C/SOH/SOH/SOHCcn2ynDˇˇˇˇˇˇˇˇˇa11y1a13/SOH/SOH/SOHa1n a21y2a23/SOH/SOH/SOHa2n ::::::::::::::: an1ynan3/SOH/SOH/SOHannˇˇˇˇˇˇˇˇˇ; as can be seen by expanding the determinant on the right in cof actors of its second column. Continuing in this way completes the proof. Example 6.1.10 The matrix of the system 4xC2yC´D1 3x/NULyC2´D2 yC2´D0 is AD2 44 2 1 3/NUL1 2 0 1 23 5: Expanding det .A/in cofactors of its first row yields det.A/D4ˇˇˇˇ/NUL1 2 1 2ˇˇˇˇ/NUL2ˇˇˇˇ3 2 0 2ˇˇˇˇC1ˇˇˇˇ3/NUL1 0 1ˇˇˇˇ D4./NUL4//NUL2.6/C1.3/D/NUL25: Using Cramer’s rule to solve the system yields xD/NUL1 25ˇˇˇˇˇˇ1 2 1 2/NUL1 2 0 1 2ˇˇˇˇˇˇD2 5; yD/NUL1 25ˇˇˇˇˇˇ4 1 1 3 2 2 0 0 2ˇˇˇˇˇˇD/NUL2 5; ´D/NUL1 25ˇˇˇˇˇˇ4 2 1 3/NUL1 2 0 1 0ˇˇˇˇˇˇD1 5: A system ofnequations innunknowns a11x1Ca12x2C/SOH/SOH/SOHCa1nxnD0 a21x1Ca22x2C/SOH/SOH/SOHCa2nxnD0 ::: an1x1Can2x2C/SOH/SOH/SOHCannxnD0(6.1.12) (or, in matrix form, AXD0) ishomogeneous . It is obvious that X0D0satisfies this system. We call this the trivial solution of (6.1.12 ). Any other solutions of ( 6.1.12 ), if they exist, are nontrivial . 376 Chapter 6 Vector-Valued Functions of Several Variables We will need the following theorems. The proofs may be found i n any linear algebra text. Theorem 6.1.15 The homogeneous system (6.1.12 )ofnequations innunknowns has a nontrivial solution if and only if det.A/D0: Theorem 6.1.16 IfA1;A2;. . .;Akare nonsingular n/STXnmatrices;then so isA1A2/SOH/SOH/SOHAk; and .A1A2/SOH/SOH/SOHAk//NUL1DA/NUL1 kA/NUL1 k/NUL1/SOH/SOH/SOHA/NUL1 1: 6.1 Exercises 1. Prove: If LWRn!Rmis a linear transformation, then L.a1X1Ca2X2C/SOH/SOH/SOHCakXk/Da1L.X1/Ca2L.X2/C/SOH/SOH/SOHCakL.Xk/ ifX1;X2;:::; Xkare in Rnanda1,a2, . . . ,akare real numbers. 2. Prove that the transformation Ldefined by Eqn. ( 6.1.1 ) is linear. 3. Find the matrix of L. (a)L.X/D2 43xC4yC6´ 2x/NUL47C2´ 7xC2yC3´3 5 (b) L.X/D2 6642x1C4x2 3x1/NUL2x2 7x1/NUL4x2 6x1Cx23 775 4. FindcA. (a)cD4;AD2 42 2 4 6 0 0 1 3 3 4 7 113 5(b)cD/NUL2;AD2 41 3 0 0 1 2 1/NUL1 33 5 5. Find ACB. (a)AD2 4/NUL1 2 3 1 1 4 0/NUL1 43 5;BD2 4/NUL1 0 3 5 6/NUL7 0/NUL1 23 5 (b) AD2 40 5 3 2 1 73 5;BD2 4/NUL1 2 0 3 4 73 5 6. Find AB. (a)AD2 4/NUL1 2 3 0 1 4 0/NUL1 43 5;BD2 4/NUL1 2 0 3 4 73 5 (b) AD/DC45 3 2 1 6 7 4 1/NAK ;BD2 6641 3 4 73 775 Section 6.1 Linear Transformations and Matrices 377 7. Prove Theorem 6.1.4 . 8. Prove Theorem 6.1.5 . 9. Prove Theorem 6.1.6 . 10. Suppose that ACBandABare both defined. What can be said about AandB? 11. Prove Theorem 6.1.7 . 12. Find the matrix of aL1CbL2. (a) L1.x;y;´/D2 43xC2yC´ xC4yC2´ 3x/NUL4yC´3 5, L2.x;y;´/D2 4/NULxCy/NUL´ /NUL2xCyC3´ yC´3 5; aD2; bD/NUL1 (b) L1.x;y/D2 42xC3y x/NULy 4xCy3 5;L2.x;y/D2 43x/NULy xCy /NULx/NULy3 5; aD4; bD 2 13. Find the matrices of L1ıL2andL2ıL1, where L1andL2are as in Exercise 6.1.12(a). 14. Write the transformations of Exercise 6.1.12 in the form L.X/DAX. 15. Findf0andf0.X0/. (a)f.x;y;´/D3x2y´,X0D.1;/NUL1;1/ (b)f.x;y/Dsin.xCy/,X0D./EM=4;/EM=4/ (c)f.x;y;´/Dxye/NULx´,X0D.1;2;0/ (d)f.x;y;´/Dtan.xC2yC´/,X0D./EM=4;/NUL/EM=8;/EM=4/ (e)f.X/DjXjWRn!R,X0D.1=pn;1=pn;:::;1=pn/ 16. LetADŒaij/c141be anm/STXnmatrix and /NAKDmax˚ jaijjˇˇ1/DC4i/DC4m;1/DC4i/DC4n/TAB : Show thatkAk/DC4/NAKpmn. 17. Prove: If Ahas at least one nonzero entry, then kAk¤0. 18. Prove:kACBk/DC4k AkCk Bk. 19. Prove:kABk/DC4k AkkBk. 20. Solve by Cramer’s rule. (a)xCyC2´D1 2x/NULyC´D/NUL1 x/NUL2y/NUL3´D2(b)xCy/NUL´D5 3x/NUL2yC2´D0 4xC2y/NUL3´D14 378 Chapter 6 Vector-Valued Functions of Several Variables (c)xC2yC3´D/NUL5 x/NUL´D/NUL1 xCyC2´D/NUL4(d)x/NULyC´/NUL2wD1 2xCy/NUL3´C3wD4 3xC2yCwD13 2xCy/NUL´D4 21. Find A/NUL1by the method of Theorem 6.1.12 . (a)/DC41/NUL2 3 4/NAK (b)2 41 2 3 1 0/NUL1 1 1 23 5 (c)2 44 2 1 3/NUL1 2 0 1 23 5 (d)2 41 0 1 0 1 1 1 1 03 5 (e)2 6641 2 0 0 /NUL2 3 0 0 0 0 2 3 0 0/NUL1 23 775(f)2 6641 1 2/NUL1 2 2/NUL1 3 /NUL1 4 1 2 3 1 0 13 775 22. For1/DC4i;j/DC4m, letaijDaij.X/be a real-valued function continuous on a compact setKinRn. Suppose that the m/STXmmatrix A.X/DŒaij.X//c141 is nonsingular for each XinK, and define the m/STXmmatrix B.X;Y/DŒbij.X;Y//c141 by B.X;Y/DA/NUL1.X/A.Y//NULI: Show that for each /SI>0 there is aı>0 such that jbij.X;Y/j</SI; 1/DC4i;j/DC4m; ifX;Y2KandjX/NULYj<ı. HINT:Show thatbijis continuous on the set ˚ .X;Y/ˇˇX2K;Y2K/TAB : Then assume that the conclusion is false and use Exercise 5.1.32 to obtain a contradiction : 6.2 CONTINUITY AND DIFFERENTIABILITY OF TRANS- FORMATIONS Throughout the rest of this chapter, transformations Fand points Xshould be considered as written in vertical form when they occur in connection with m atrix operations. However, we will write XD.x1;x2;:::;x n/when Xis the argument of a function. Section 6.2 Continuity and Differentiability of Transformations 379 Continuous Transformations In Section 5.2 we defined a vector-valued function (transfor mation) to be continuous at X0 if each of its component functions is continuous at X0. We leave it to you to show that this implies the following theorem (Exercise 1). Theorem 6.2.1 Suppose that X0is in;and a limit point of ;the domain of FWRn! Rm:Then Fis continuous at X0if and only if for each /SI>0 there is aı>0 such that jF.X//NULF.X0/j</SI ifjX/NULX0j<ı and X2DF: (6.2.1) This theorem is the same as Theorem 5.2.7 except that the “absolute value” in ( 6.2.1 ) now stands for distance in Rmrather than R. IfCis a constant vector, then “lim X!X0F.X/DC” means that lim X!X0jF.X//NULCjD0: Theorem 6.2.1 implies that Fis continuous at X0if and only if lim X!X0F.X/DF.X0/: Example 6.2.1 The linear transformation L.X/D2 4xCyC´ 2x/NUL3yC´ 2xCy/NUL´3 5 is continuous at every X0inR3, since L.X//NULL.X0/DL.X/NULX0/D2 4.x/NULx0/C.y/NULy0/C.´/NUL´0/ 2.x/NULx0//NUL3.y/NULy0/C.´/NUL´0/ 2.x/NULx0/C.y/NULy0//NUL.´/NUL´0/3 5; and applying Schwarz’s inequality to each component yields jL.X//NULL.X0/j2/DC4.3C14C6/jX/NULX0j2D23jX/NULX0j2: Therefore, jL.X//NULL.X0/j</SI ifjX/NULX0j</SIp 23: Differentiable Transformations In Section 5.4 we defined a vector-valued function (transfor mation) to be differentiable at X0if each of its components is differentiable at X0(Definition 5.4.1 ). The next theorem characterizes this property in a useful way. 380 Chapter 6 Vector-Valued Functions of Several Variables Theorem 6.2.2 A transformation FD.f1;f2;:::;f m/defined in a neighborhood of X02Rnis differentiable at X0if and only if there is a constant m/STXnmatrix Asuch that lim X!X0F.X//NULF.X0//NULA.X/NULX0/ jX/NULX0jD0: (6.2.2) If(6.2.2 )holds;then Ais given uniquely by AD/[email protected]/ @xj/NAK D2 [email protected]/ @[email protected]/ @x2/SOH/SOH/[email protected]/ @xn @f2.X0/ @[email protected]/ @x2/SOH/SOH/[email protected]/ @xn:::::::::::: @fm.X0/ @[email protected]/ @x2/SOH/SOH/[email protected]/ @xn3 7777777775: (6.2.3) Proof LetX0D.x10;x20;:::;x n0/. IfFis differentiable at X0, then so are f1,f2, . . . ,fm(Definition 5.4.1 ). Hence, lim X!X0fi.X//NULfi.X0//NULnX [email protected]/ @xj.xj/NULxj 0/ jX/NULX0jD0; 1/DC4i/DC4m; which implies ( 6.2.2 ) with Aas in ( 6.2.3 ). Now suppose that ( 6.2.2 ) holds with ADŒaij/c141. Since each component of the vector in (6.2.2 ) approaches zero as Xapproaches X0, it follows that lim X!X0fi.X//NULfi.X0//NULnX jD1aij.xj/NULxj 0/ jX/NULX0jD0; 1/DC4i/DC4m; so eachfiis differentiable at X0, and therefore so is F(Definition 5.4.1 ). By Theo- rem5.3.6 , [email protected]/ @xj; 1/DC4i/DC4m; 1/DC4j/DC4n; which implies ( 6.2.3 ). A transformation TWRn!Rmof the form T.X/DUCA.X/NULX0/; where Uis a constant vector in Rm,X0is a constant vector in Rn, and Ais a constantm/STXn matrix, is said to be affine . Theorem 6.2.2 says that if Fis differentiable at X0, then Fcan be well approximated by an affine transformation. Section 6.2 Continuity and Differentiability of Transformations 381 Example 6.2.2 The components of the transformation F.X/D2 4x2C2xyC´ xC2x´Cy x2Cy2C´23 5 are differentiable at X0D.1;0;2/ . Evaluating the partial derivatives of the components there yields AD2 42 2 1 5 1 2 2 0 43 5: (Verify). Therefore, Theorem 6.2.2 implies that the affine transformation T.X/DF.X0/CA.X/NULX0/ D2 43 5 53 5C2 42 2 1 5 1 2 2 0 43 52 4x/NUL1 y ´/NUL23 5 satisfies lim X!X0F.X//NULT.X/ jX/NULX0jD0: Differential of a Transformation IfFD.f1;f2;:::;f m/is differentiable at X0, we define the differential of FatX0to be the linear transformation dX0FD2 6664dX0f1 dX0f2 ::: dX0fm3 7775: (6.2.4) We call the matrix Ain (6.2.3 )the differential matrix of FatX0and denote it by F0.X0/; thus, F0.X0/D2 [email protected]/ @[email protected]/ @x2/SOH/SOH/[email protected]/ @xn @f2.X0/ @[email protected]/ @x2/SOH/SOH/[email protected]/ @xn :::::::::::: @fm.X0/ @[email protected]/ @x2/SOH/SOH/[email protected]/ @xn3 777777777775: (6.2.5) 382 Chapter 6 Vector-Valued Functions of Several Variables (It is important to bear in mind that while Fis a function from RntoRm,F0is not such a function; F0is anm/STXnmatrix.) From Theorem 6.2.2 , the differential can be written in terms of the differential matrix as dX0FDF0.X0/2 6664dx1 dx2 ::: dxn3 7775(6.2.6) or, more succinctly, as dX0FDF0.X0/dX; where dXD2 6664dx1 dx2 ::: dxn3 7775; as defined earlier. When it is not necessary to emphasize the particular point X0, we write ( 6.2.4 ) as dFD2 6664df1 df2 ::: dfm3 7775; (6.2.5 ) as F0D2 666666666664@f1 @x1@f1 @x2/SOH/SOH/SOH@f1 @xn @f2 @x1@f2 @x2/SOH/SOH/SOH@f2 @xn :::::::::::: @fm @x1@fm @x2/SOH/SOH/SOH@fm @xn3 777777777775; and ( 6.2.6 ) as dFDF0dX: With the differential notation we can rewrite ( 6.2.2 ) as lim X!X0F.X//NULF.X0//NULF0.X0/.X/NULX0/ jX/NULX0jD0: Section 6.2 Continuity and Differentiability of Transformations 383 Example 6.2.3 The linear transformation F.X/D2 6664a11x1Ca12x2C/SOH/SOH/SOHCa1nxn a21x1Ca22x2C/SOH/SOH/SOHCa2nxn ::: am1x1Cam2x2C/SOH/SOH/SOHCamnxn3 7775 can be written as F.X/DAX, where ADŒaij/c141. Then F0DAI that is, the differential matrix of a linear transformation is independent of Xand is the matrix of the transformation. For example, the differentia l matrix of F.x1;x2;x3/D/DC41 2 3 2 1 0/NAK2 4x1 x2 x33 5 is F0D/DC41 2 3 2 1 0/NAK : IfF.X/DX(the identity transformation), then F0DI(the identity matrix). Example 6.2.4 The transformation F.x;y/D2 66664x x2Cy2 y x2Cy2 2xy3 77775 is differentiable at every point of R2except.0;0/ , and F0.x;y/D2 666664y2/NULx2 .x2Cy2/2/NUL2xy .x2Cy2/2 /NUL2xy .x2Cy2/2x2/NULy2 .x2Cy2/2 2y 2x3 777775: In particular, F0.1;1/D2 66640/NUL1 2 /NUL1 20 2 23 7775; 384 Chapter 6 Vector-Valued Functions of Several Variables so lim .x;y/ !.1;1/1p .x/NUL1/2C.y/NUL1/20 [email protected];y//NUL2 66641 2 1 2 23 7775/NUL2 66640/NUL1 2 /NUL1 20 2 23 7775/DC4x/NUL1 y/NUL1/NAK1 CCCA D2 40 0 03 5: IfmDn, the differential matrix is square and its determinant is ca lled the Jacobian of F. The standard notation for this determinant is @.f1;f2;:::;f n/ @.x1;x2;:::;x n/Dˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇ@f1 @x1@f1 @x2/SOH/SOH/SOH@f1 @xn @f2 @x1@f2 @x2/SOH/SOH/SOH@f2 @xn :::::::::::: @fn @x1@fn @x2/SOH/SOH/SOH@fn @xnˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇ: We will often write the Jacobian of Fmore simply as J.F/, and its value at X0asJF.X0/. Since ann/STXnmatrix is nonsingular if and only if its determinant is nonze ro, it follows that if FWRn!Rnis differentiable at X0, then F0.X0/is nonsingular if and only if JF.X0/¤0. We will soon use this important fact. Example 6.2.5 If F.x;y;´/D2 6664x2/NUL2xC´ xC2xyC´2 xCyC´3 7775; then @.f1;f2;f3/ @.x1;x2;x3/DJF.X/Dˇˇˇˇˇˇ2x/NUL2 0 1 1C2y 2x 2´ 1 1 1ˇˇˇˇˇˇ D.2x/NUL2/ˇˇˇˇ2x 2´ 1 1ˇˇˇˇCˇˇˇˇ1C2y 2x 1 1ˇˇˇˇ D.2x/NUL2/.2x/NUL2´/C.1C2y/NUL2x/: Section 6.2 Continuity and Differentiability of Transformations 385 In particular, JF.1;/NUL1;1/D/NUL3, so the differential matrix F0.1;/NUL1;1/D2 40 0 1 /NUL1 2 2 1 1 13 5 is nonsingular. Properties of Differentiable Transformations We leave the proof of the following theorem to you (Exercise 6.2.16 ). Theorem 6.2.3 IfFWRn!Rmis differentiable at X0;then Fis continuous at X0: Theorem 5.3.10 and Definition 5.4.1 imply the following theorem. Theorem 6.2.4 LetFD.f1;f2;:::;f m/WRn!Rm;and suppose that the partial derivatives @fi @xj; 1/DC4i/DC4m; 1/DC4j/DC4n; (6.2.7) exist on a neighborhood of X0and are continuous at X0:Then Fis differentiable at X0: We say that Fiscontinuously differentiable on a setSifSis contained in an open set on which the partial derivatives in ( 6.2.7 ) are continuous. The next three lemmas give properties of continuously differentiable transformatio ns that we will need later. Lemma 6.2.5 Suppose that FWRn!Rmis continuously differentiable on a neigh- borhoodNofX0:Then;for every/SI>0; there is aı>0 such that jF.X//NULF.Y/j<.kF0.X0/kC/SI/jX/NULYjifA;Y2Bı.X0/: (6.2.8) Proof Consider the auxiliary function G.X/DF.X//NULF0.X0/X: (6.2.9) The components of Gare gi.X/Dfi.X//NULnX [email protected]/@xj x j; so @gi.X/ @[email protected]/ @xj/[email protected]/ @xj: 386 Chapter 6 Vector-Valued Functions of Several Variables Thus,@gi=@x jis continuous on Nand zero at X0. Therefore, there is a ı>0 such that ˇˇˇˇ@gi.X/ @xjˇˇˇˇ</SIpmnfor1/DC4i/DC4m; 1/DC4j/DC4n; ifjX/NULX0j<ı: (6.2.10) Now suppose that X,Y2Bı.X0/. By Theorem 5.4.5 , gi.X//NULgi.Y/DnX [email protected]/ @xj.xj/NULyj/; (6.2.11) where Xiis on the line segment from XtoY, soXi2Bı.X0/. From ( 6.2.10 ), (6.2.11 ), and Schwarz’s inequality, .gi.X//NULgi.Y//2/DC40 @nX jD1/[email protected]/ @xj/NAK21 AjX/NULYj2</SI2 mjX/NULYj2: Summing this from iD1toiDmand taking square roots yields jG.X//NULG.Y/j</SIjX/NULYjifX;Y2Bı.X0/: (6.2.12) To complete the proof, we note that F.X//NULF.Y/DG.X//NULG.Y/CF0.X0/.X/NULY/; (6.2.13) so (6.2.12 ) and the triangle inequality imply ( 6.2.8 ). Lemma 6.2.6 Suppose that FWRn!Rnis continuously differentiable on a neigh- borhood of X0andF0.X0/is nonsingular :Let rD1 k.F0.X0///NUL1k: (6.2.14) Then;for every/SI>0; there is aı>0 such that jF.X//NULF.Y/j/NAK.r/NUL/SI/jX/NULYjifX;Y2Bı.X0/: (6.2.15) Proof LetXandYbe arbitrary points in DFand let Gbe as in ( 6.2.9 ). From ( 6.2.13 ), jF.X//NULF.Y/j/NAKˇˇjF0.X0/.X/NULY/j/NULjG.X//NULG.Y/jˇˇ; (6.2.16) Since X/NULYDŒF0.X0//c141/NUL1F0.X0/.X/NULY/; (6.2.14 ) implies that jX/NULYj/DC41 rjF0.X0/.X/NULYj; so jF0.X0/.X/NULY/j/NAKrjX/NULYj: (6.2.17) Now chooseı>0 so that ( 6.2.12 ) holds. Then ( 6.2.16 ) and ( 6.2.17 ) imply ( 6.2.15 ). See Exercise 6.2.19 for a stronger conclusion in the case where Fis linear. Section 6.2 Continuity and Differentiability of Transformations 387 Lemma 6.2.7 IfFWRn!Rmis continuously differentiable on an open set containing a compact set D;then there is a constant Msuch that jF.Y//NULF.X/j/DC4MjY/NULXjifX;Y2D: (6.2.18) Proof On SD˚ .X;Y/ˇˇX;Y2D/TAB /SUBR2n define g.X;Y/D8 < :jF.Y//NULF.X//NULF0.X/.Y/NULX/j jY/NULXj;Y¤X; 0; YDX: Thengis continuous for all .X;Y/inSsuch that X¤Y. We now show that if X02D, then lim .X;Y/!.X0;X0/g.X;Y/D0Dg.X0;X0/I (6.2.19) that is,gis also continuous at points .X0;X0/inS. Suppose that /SI > 0 andX02D. Since the partial derivatives of f1,f2, . . . ,fmare continuous on an open set containing D, there is aı>0 such that ˇˇˇˇ@fi.Y/ @xj/[email protected]/ @xjˇˇˇˇ</SIpmnifX;Y2Bı.X0/; 1/DC4i/DC4m; 1/DC4j/DC4n: (6.2.20) (Note that@fi=@x jis uniformly continuous on Bı.X0/forısufficiently small, from The- orem 5.2.14 .) Applying Theorem 5.4.5 tof1,f2, . . . ,fm, we find that if X,Y2Bı.X0/, then fi.Y//NULfi.X/DnX [email protected]/ @xj.yj/NULxj/; where Xiis on the line segment from XtoY. From this, 2 4fi.Y//NULfi.X//NULnX [email protected]/ @xj.yj/NULxj/3 52 D2 4nX jD1/[email protected]/ @xj/[email protected]/ @xj/NAK .yj/NULxj/3 52 /DC4jY/NULXj2nX jD1/[email protected]/ @xj/[email protected]/ @xj/NAK2 (by Schwarz’s inequality) </SI2 mjY/NULXj2(by ( 6.2.20 )): Summing from iD1toiDmand taking square roots yields jF.Y//NULF.X//NULF0.X/.Y/NULX/j</SIjY/NULXjifX;Y2Bı.X0/: This implies ( 6.2.19 ) and completes the proof that gis continuous on S. 388 Chapter 6 Vector-Valued Functions of Several Variables SinceDis compact, so is S(Exercise 5.1.27 ). Therefore, gis bounded on S(Theo- rem5.2.12 ); thus, for some M1, jF.Y//NULF.X//NULF0.X/.Y/NULX/j/DC4M1jX/NULYjifX;Y2D: But jF.Y//NULF.X/j/DC4jF.Y//NULF.X//NULF0.X/.Y/NULX/jCjF0.X/.Y/NULX/j /DC4.M1CkF0.X/k/j.Y/NULXj:(6.2.21) Since kF0.X/k/DC40 @mX iD1nX jD1/[email protected]/ @xj/NAK21 A1=2 and the partial derivatives f@fi=@x jgare bounded on D, it follows thatkF0.X/kis bounded onD; that is, there is a constant M2such that kF0.X/k/DC4M2;X2D: Now ( 6.2.21 ) implies ( 6.2.18 ) withMDM1CM2. The Chain Rule for Transformations By using differential matrices, we can write the chain rule f or transformations in a form analogous to the form of the chain rule for real-valued funct ions of one variable (Theo- rem2.3.5 ). Theorem 6.2.8 Suppose that FWRn!Rmis differentiable at X0;GWRk!Rnis differentiable at U0;andX0DG.U0/:Then the composite function HDFıGWRk! Rm;defined by H.U/DF.G.U//; is differentiable at U0:Moreover; H0.U0/DF0.G.U0//G0.U0/ (6.2.22) and dU0HDdX0FıdU0G; (6.2.23) whereıdenotes composition : Proof The components of Hareh1,h2, . . . ,hm, where hi.U/Dfi.G.U//: Applying Theorem 5.4.3 tohiyields dU0hiDnX [email protected]/ @xjdU0gj; 1/DC4i/DC4m: (6.2.24) Section 6.2 Continuity and Differentiability of Transformations 389 Since dU0HD2 6664dU0h1 dU0h2 ::: dU0hm3 7775anddU0GD2 6664dU0g1 dU0g2 ::: dU0gn3 7775; themequations in ( 6.2.24 ) can be written in matrix form as dU0HDF0.X0/dU0GDF0.G.U0//dU0G: (6.2.25) But dU0GDG0.U0/dU; where dUD2 6664du1 du2 ::: duk3 7775; so (6.2.25 ) can be rewritten as dU0HDF0.G.U0//G0.U0/dU: On the other hand, dU0HDH0.U0/dU: Comparing the last two equations yields ( 6.2.22 ). Since G0.U0/is the matrix of dU0Gand F0.G.U0//DF0.X0/is the matrix of dX0F, Theorem 6.1.7(c)and ( 6.2.22 ) imply ( 6.2.23 ). Example 6.2.6 LetU0D.1;/NUL1/, G.U/DG.u;v/D2 6664pu p u2C3v2 pvC23 7775;F.X/DF.x;y;´/D"x2Cy2C2´2 x2/NULy2# ; and H.U/DF.G.U//: Since Gis differentiable at U0D.1;/NUL1/andFis differentiable at X0DG.U0/D.1;2;1/; Theorem 6.2.8 implies that His differentiable at .1;/NUL1/. To find H0.1;/NUL1/from ( 6.2.22 ), we first find that 390 Chapter 6 Vector-Valued Functions of Several Variables G0.U/D2 66666641 2pu0 up u2C3v23vp u2C3v2 01 2pvC23 7777775 and F0.X/D/DC42x 2y 4´ 2x/NUL2y 0/NAK : Then, from ( 6.2.22 ), H0.1;/NUL1/DF0.1;2;1/ G0.1;/NUL1/ D/DC42 4 4 2/NUL4 0/NAK2 66641 20 1 2/NUL3 2 01 23 7775D/DC43/NUL4 /NUL1 6/NAK : We can check this by expressing Hdirectly in terms of .u;v/ as H.u;v/D2 64/NULpu/SOH2C/DLEp u2C3v2/DC12 C2/NULpvC2/SOH2 /NULpu/SOH2/NUL/DLEp u2C3v2/DC123 75 D/DC4uCu2C3v2C2vC4 u/NULu2/NUL3v2/NAK and differentiating to obtain H0.u;v/D/DC41C2u 6vC2 1/NUL2u/NUL6v/NAK ; which yields H0.1;/NUL1/D/DC43/NUL4 /NUL1 6/NAK ; as we saw before. 6.2 Exercises 1. Show that the following definitions are equivalent. (a) FD.f1;f2;:::;f m/is continuous at X0iff1,f2, . . . ,fmare continuous atX0. Section 6.2 Continuity and Differentiability of Transformations 391 (b) Fis continuous at X0if for every/SI > 0 there is aı > 0 such thatjF.X//NUL F.X0/j</SIifjX/NULX0j<ıandX2DF. 2. Verify that lim X!X0F.X//NULF.X0//NULF0.X0/.X/NULX0/ jX/NULX0jD0: (a) F.X/D2 43xC4y 2x/NULy xCy3 5;X0D.x0;y0;´0/ (b) F.X/D2 42x2CxyC1 xy x2Cy23 5;X0D.1;/NUL1/ (c) F.X/D2 4sin.xCy/ sin.yC´/ sin.xC´/3 5;X0D./EM=4;0;/EM=4/ 3. Suppose that FWRn!RmandhWRn!Rhave the same domain and are continuous at X0. Show that the product hFD.hf1;hf2;:::;hf m/is continuous at X0. 4. Suppose that FandGare transformations from RntoRmwith common domain D. Show that if FandGare continuous at X02D, then so are FCGandF/NULG. 5. Suppose that FWRn!Rmis defined in a neighborhood of X0and continuous at X0,GWRk!Rnis defined in a neighborhood of U0and continuous at U0, and X0DG.U0/. Prove that the composite function HDFıGis continuous at U0. 6. Prove: If FWRn!Rmis continuous on a set S, thenjFjis continuous on S. 7. Prove: If FWRn!Rmis continuous on a compact set S, thenjFjis bounded on S, and there are points X0andX1inSsuch that jF.X0/j/DC4j F.X/j/DC4j F.X1/j;X2SI that is,jFjattains its infimum and supremum on S. HINT:Use Exercise 6.2.6: 8. Prove that a linear transformation LWRn!Rmis continuous on Rn. Do not use Theorem 6.2.8 . 9. LetAbe anm/STXnmatrix. (a) Use Exercises 6.2.7 and6.2.8 to show that the quantitites M.A/Dmax/SUBjAXj jXjˇˇX¤0/ESC andm.A/Dmin/SUBjAXj jXjˇˇX¤0/ESC exist. H INT:Consider the function L.Y/DAYonSD˚YˇˇjYjD1/TAB: 392 Chapter 6 Vector-Valued Functions of Several Variables (b) Show thatM.A/DkAk. (c) Prove: Ifn>m ornDmandAis singular, then m.A/D0. (This requires a result from linear algebra on the existence of nontrivial so lutions of AXD0.) (d) Prove: IfnDmandAis nonsingular, then m.A/M.A/NUL1/Dm.A/NUL1/M.A/D1: 10. We say that FWRn!Rmisuniformly continuous on Sif each of its components is uniformly continuous on S. Prove: If Fis uniformly continuous on S, then for each/SI>0 there is aı>0 such that jF.X//NULF.Y/j</SI ifjX/NULYj<ı and X;Y2S: 11. Show that if Fis continuous on RnandF.XCY/DF.X/CF.Y/for all XandY inRn, then Ais linear. H INT:The rational numbers are dense in the reals : 12. Find F0andJF. Then find an affine transformation Gsuch that lim X!X0F.X//NULG.Y/ X/NULX0D0: (a) F.x;y;´/D2 4x2CyC2´ cos.xCyC´/ exy´3 5;X0D.1;/NUL1;0/ (b) F.x;y/D/DC4excosy exsiny/NAK ;X0D.0;/EM=2/ (c) F.x;y;´/D2 4x2/NULy2 y2/NUL´2 ´2/NULx23 5;X0D.1;1;1/ 13. Find F0. (a)F.x;y;´/D/DC4.xCyC´/ex .x2Cy2/e/NULx/NAK (b) F.x/D2 6664g1.x/ g2.x/ ::: gn.x/3 7775 (c)F.x;y;´/D2 4exsiny´ eysinx´ e´sinxy3 5 14. Find F0andJF. (a)F.r;/DC2/D/DC4rcos/DC2 rsin/DC2/NAK (b) F.r;/DC2;/RS/D2 4rcos/DC2cos/RS rsin/DC2cos/RS rsin/RS3 5 (c)F.r;/DC2;´/D2 4rcos/DC2 rsin/DC2 ´3 5 Section 6.2 Continuity and Differentiability of Transformations 393 15. Prove: If G1andG2are affine transformations and lim X!X0G1.X//NULG2.Y/ jX/NULX0jD0; then G1DG2. 16. Prove Theorem 6.2.3 . 17. Show that if FWRn!Rmis differentiable at X0and/SI >0 , there is aı >0 such that jF.X//NULF.X0/j/DC4.kF0.X0/kC/SI/jX/NULX0jifjX/NULX0j<ı: Compare this with Lemma 6.2.5 . 18. Suppose that FWRn!Rnis differentiable at X0andF0.X0/is nonsingular. Let rD1 kŒF0.X0//c141/NUL1k and suppose that /SI>0 . Show that there is a ı>0 such that jF.X//NULF.X0/j/NAK.r/NUL/SI/jX/NULX0jifjX/NULX0j<ı: Compare this with Lemma 6.2.6 . 19. Prove: If LWRn!Rmis defined by L.X/DA.X/, where Ais nonsingular, then jL.X//NULL.Y/j/NAK1 kA/NUL1kjX/NULYj for all XandYinRn. 20. Use Theorem 6.2.8 to find H0.U0/, where H.U/DF.G.U/. Check your results by expressing Hdirectly in terms of Uand differentiating. (a) F.x;y;´/D2 4x2Cy2 ´ x2Cy23 5;G.u;v;w/D2 664wcosusinv wsinusinv wcosv3 775,U0D ./EM=2;/EM=2;2/ (b) F.x;y/D2 4x2/NULy2 y x3 5;G.u;v/D"vcosu vsinu# ;U0D./EM=4;3/ (c) F.x;y;´/D2 43xC4yC2´C6 4x/NUL2yC´/NUL1 /NULxCyC´/NUL23 5;G.u;v/D2 4u/NULv uCv u/NUL2v3 5, U0arbitrary 394 Chapter 6 Vector-Valued Functions of Several Variables (d) F.x;y/D/DC4xCy x/NULy/NAK ;G.u;v;w/D/DC42u/NULvCw eu2/NULv2/NAK ;U0D.1;1;/NUL2/ (e) F.x;y/D/DC4x2Cy2 x2/NULy2/NAK ;G.u;v/D/DC4eucosv eusinv/NAK ;U0D.0;0/ (f) F.x;y/D2 4xC2y x/NULy2 x2Cy3 5;G.u;v/D/DC4uC2v 2u/NULv2/NAK ;U0D.1;/NUL2/ 21. Suppose that FandGare continuously differentiable on Rn, with values in Rn, and letHDFıG. Show that @.h1;h2;:::;h n/ @.u1;u2;:::;u n/[email protected];f2;:::;f n/ @.x1;x2;:::;x n/@.g1;g2;:::;g n/ @.u1;u2;:::;u n/: Where should these Jacobians be evaluated? 22. Suppose that FWRn!RmandXis a limit point of DFcontained inDF. Show thatFis continuous at Xif and only if lim k!1F.Xk/DF.X/wheneverfXkgis a sequence of points in DFsuch that lim k!1XkDX. HINT:See Exercise 5.2.15: 23. Suppose that FWRn!Rmis continuous on a compact subset SofRn. Show that F.S/is a compact subset of Rm. 6.3 THE INVERSE FUNCTION THEOREM So far our discussion of transformations has dealt mainly wi th properties that could just as well be defined and studied by considering the component func tions individually. Now we turn to questions involving a transformation as a whole, tha t cannot be studied by regarding it as a collection of independent component functions. In this section we restrict our attention to transformation s from Rnto itself. It is useful to interpret such transformations geometrically. If FD.f1;f2;:::;f n/, we can think of the components of F.X/D.f1.X/;f2.X/;:::;f n.X// as the coordinates of a point UDF.X/in another “copy” of Rn. Thus, UD.u1;u2;:::;u n/, with u1Df1.X/; u 2Df2.X/; :::; u nDfn.X/: We say that Fmaps XtoU, and that Uis the image of Xunder F. Occasionally we will also write@ui=@x jto mean@fi=@x j. IfS/SUBDF, then the set F.S/D˚ UˇˇUDF.X/;X2S/TAB is the image ofSunder F. We will often denote the components of Xbyx,y, . . . , and the components of Ubyu, v, . . . . Section 6.3 The Inverse Function Theorem 395 Example 6.3.1 If /DC4u v/NAK DF.x;y/D/DC4x2Cy2 x2/NULy2/NAK ; then uDf1.x;y/Dx2Cy2; vDf2.x;y/Dx2/NULy2; and ux.x;y/[email protected];y/ @xD2x; u y.x;y/[email protected];y/ @yD2y; vx.x;y/[email protected];y/ @xD2x; v y.x;y/[email protected];y/ @yD/NUL2y: To find F.R2/, we observe that uCvD2x2; u/NULvD2y2; so F.R2//SUBTD˚.u;v/ˇˇuCv/NAK0;u/NULv/NAK0/TAB; which is the part of the uv-plane shaded in Figure 6.3.1 . If.u;v/2T, then F/DC2puCv 2;pu/NULv 2/DC3 D/DC4u v/NAK ; soF.R2/DT. v u u + v = 0u − v = 0 Figure 6.3.1 396 Chapter 6 Vector-Valued Functions of Several Variables Invertible Transformations A transformation Fisone-to-one , orinvertible , ifF.X1/andF.X2/are distinct whenever X1andX2are distinct points of DF. In this case, we can define a function Gon the range R.F/D˚ UˇˇUDF.X/for some X2DF/TAB ofFby defining G.U/to be the unique point in DFsuch that F.U/DU. Then DGDR.F/andR.G/DDF: Moreover, Gis one-to-one, G.F.X//DX;X2DF; and F.G.U//DU;U2DG: We say that Gis the inverse ofF, and write GDF/NUL1. The relation between FandGis symmetric; that is, Fis also the inverse of G, and we write FDG/NUL1. Example 6.3.2 The linear transformation /DC4u v/NAK DL.x;y/D/DC4x/NULy xCy/NAK (6.3.1) maps.x;y/ to.u;v/ , where uDx/NULy; vDxCy:(6.3.2) Lis one-to-one and R.L/DR2, since for each .u;v/ inR2there is exactly one .x;y/ such that L.x;y/D.u;v/ . This is so because the system ( 6.3.2 ) can be solved uniquely for.x;y/ in terms of.u;v/ : xD1 2.uCv/; yD1 2./NULuCv/:(6.3.3) Thus, L/NUL1.u;v/D1 2/DC4uCv /NULuCv/NAK : Example 6.3.3 The linear transformation /DC4u v/NAK DL1.x;y/D/DC4xCy 2xC2y/NAK maps.x;y/ onto.u;v/ , where uDxCy; vD2xC2y:(6.3.4) Section 6.3 The Inverse Function Theorem 397 L1is not one-to-one, since every point on the line xCyDc(constant) is mapped onto the single point .c;2c/ . Hence, L1does not have an inverse. The crucial difference between the transformations of Exam ples 6.3.2 and6.3.3 is that the matrix of Lis nonsingular while the matrix of L1is singular. Thus, L(see ( 6.3.1 )) can be written as /DC4u v/NAK D/DC41/NUL1 1 1/NAK/DC4x y/NAK ; (6.3.5) where the matrix has the inverse 2 41 21 2 /NUL1 21 23 5: (Verify.) Multiplying both sides of ( 6.3.5 ) by this matrix yields 2 41 21 2 /NUL1 21 23 5/DC4u v/NAK D/DC4x y/NAK ; which is equivalent to ( 6.3.3 ). Since the matrix /DC41 1 2 2/NAK ofL1is singular, ( 6.3.4 ) cannot be solved uniquely for .x;y/ in terms of.u;v/ . In fact, it cannot be solved at all unless vD2u. The following theorem settles the question of invertibilit y of linear transformations from RntoRn. We leave the proof to you (Exercise 6.3.2 ). Theorem 6.3.1 The linear transformation UDL.X/DAX.Rn!Rn/ is invertible if and only if Ais nonsingular ;in which case R.L/DRnand L/NUL1.U/DA/NUL1U: Polar Coordinates We will now briefly review polar coordinates, which we will us e in some of the following examples. The coordinates of any point .x;y/ can be written in infinitely many ways as xDrcos/DC2; yDrsin/DC2; (6.3.6) 398 Chapter 6 Vector-Valued Functions of Several Variables where r2Dx2Cy2 and, ifr > 0 ,/DC2is the angle from the x-axis to the line segment from .0;0/ to.x;y/ , measured counterclockwise (Figure 6.3.2 ). y xx2 + y2(x, y) θ Figure 6.3.2 For each.x;y/¤.0;0/ there are infinitely many values of /DC2, differing by integral multiples of2/EM, that satisfy ( 6.3.6 ). If/DC2is any of these values, we say that /DC2is an argument of.x;y/ , and write /DC2Darg.x;y/: By itself, this does not define a function. However, if /RSis an arbitrary fixed number, then /DC2Darg.x;y/; /RS/DC4/DC2 </RSC2/EM; does define a function, since every half-open interval Œ/RS;/RSC2/EM/ contains exactly one argument of.x;y/ . We do not define arg .0;0/ , since ( 6.3.6 ) places no restriction on /DC2if.x;y/D.0;0/ and thereforerD0. The transformation /DC4r /DC2/NAK DG.x;y/D2 4p x2Cy2 arg.x;y/3 5; /RS/DC4arg.x;y/</RSC2/EM; is defined and one-to-one on DGD˚ .x;y/ˇˇ.x;y/¤.0;0//TAB ; and its range is R.G/D˚.r;/DC2/ˇˇr >0;/RS/DC4/DC2 </RSC2/EM/TAB: Section 6.3 The Inverse Function Theorem 399 For example, if /RSD0, then G.1;1/D2 64p 2 /EM 43 75; since/EM=4 is the unique argument of .1;1/ inŒ0;2/EM/ . If/RSD/EM, then G.1;1/D2 64p 2 9/EM 43 75; since9/EM=4 is the unique argument of .1;1/ inŒ/EM;3/EM/ . If arg.x0;y0/D/RS, then.x0;y0/is on the half-line shown in Figure 6.3.3 andGis not continuous at .x0;y0/, since every neighborhood of .x0;y0/contains points .x;y/ for which the second component of G.x;y/ is arbitrarily close to /RSC2/EM, while the second component of G.x0;y0/is/RS. We will show later, however, that Gis continuous, in fact, continuously differentiable, on the plane with this half-l ine deleted. y x(x0, y0) φ Figure 6.3.3 Local Invertibility A transformation Fmay fail to be one-to-one, but be one-to-one on a subset SofDF. By this we mean that F.X1/andF.X2/are distinct whenever X1andX2are distinct points of S. In this case, Fis not invertible, but if FSis defined on Sby FS.X/DF.X/;X2S; and left undefined for X62S, then FSis invertible. We say that FSis the restriction of F toS, and that F/NUL1 Sis the inverse of Frestricted toS. The domain of F/NUL1 SisF.S/. 400 Chapter 6 Vector-Valued Functions of Several Variables IfFis one-to-one on a neighborhood of X0, we say that Fislocally invertible at X0. If this is true for every X0in a setS, then Fislocally invertible on S. Example 6.3.4 The transformation /DC4u v/NAK DF.x;y/D/DC4x2/NULy2 2xy/NAK (6.3.7) is not one-to-one, since F./NULx;/NULy/DF.x;y/: (6.3.8) It is one-to-one on Sif and only if Sdoes not contain any pair of distinct points of the form .x0;y0/and./NULx0;/NULy0/; (6.3.8 ) implies the necessity of this condition, and its sufficienc y follows from the fact that if F.x1;y1/DF.x0;y0/; (6.3.9) then .x1;y1/D.x0;y0/or.x1;y1/D./NULx0;/NULy0/: (6.3.10) To see this, suppose that ( 6.3.9 ) holds; then x2 1/NULy2 1Dx2 0/NULy2 0 (6.3.11) and x1y1Dx0y0: (6.3.12) Squaring both sides of ( 6.3.11 ) yields x4 1/NUL2x2 1y2 1Cy4 1Dx4 0/NUL2x2 0y2 0Cy4 0: This and ( 6.3.12 ) imply that x4 1/NULx4 0Dy4 0/NULy4 1: (6.3.13) From ( 6.3.11 ), x2 1/NULx2 0Dy2 1/NULy2 0: (6.3.14) Factoring ( 6.3.13 ) yields .x2 1/NULx2 0/.x2 1Cx2 0/D.y2 0/NULy2 1/.y2 0Cy2 1/: If either side of ( 6.3.14 ) is nonzero, we can cancel to obtain x2 1Cx2 0D/NULy2 0/NULy2 1; which implies that x0Dx1Dy0Dy1D0, so ( 6.3.10 ) holds in this case. On the other hand, if both sides of ( 6.3.14 ) are zero, then x1D˙x0; y 1D˙y0: From ( 6.3.12 ), the same sign must be chosen in these equalities, which pro ves that ( 6.3.8 ) implies ( 6.3.10 ) in this case also. Section 6.3 The Inverse Function Theorem 401 We now see, for example, that Fis one-to-one on every set Sof the form SD˚.x;y/ˇˇaxCby >0/TAB; whereaandbare constants, not both zero. Geometrically, Sis an open half-plane; that is, the set of points on one side of, but not on, the line axCbyD0 (Figure 6.3.4 ). Therefore, Fis locally invertible at every X0¤.0;0/ , since every such point lies in a half-plane of this form. However, Fis not locally invertible at .0;0/ . (Why not?) Thus, Fis locally invertible on the entire plane with .0;0/ removed. y xax + by = 0 (a, b) ax + by > 0 Figure 6.3.4 It is instructive to find F/NUL1 Sfor a specific choice of S. Suppose that Sis the open right half-plane: SD˚ .x;y/ˇˇx>0/TAB : (6.3.15) Then F.S/is the entireuv-plane except for the nonpositive uaxis. To see this, note that every point in Scan be written in polar coordinates as xDrcos/DC2; yDrsin/DC2; r >0;/NUL/EM 2</DC2 </EM 2: Therefore, from ( 6.3.7 ),F.x;y/ has coordinates .u;v/ , where uDx2/NULy2Dr2.cos2/DC2/NULsin2/DC2/Dr2cos2/DC2; vD2xyD2r2cos/DC2sin/DC2Dr2sin2/DC2: 402 Chapter 6 Vector-Valued Functions of Several Variables Every point in the uv-plane can be written in polar coordinates as uD/SUBcos˛; vD/SUBsin˛; where either/SUBD0or /SUBDp u2Cv2>0;/NUL/EM/DC4˛</EM; and the points for which /SUBD0or˛D/NUL/EMare of the form .u;0/ , withu/DC40(Figure 6.3.5 ). If.u;v/DF.x;y/ for some.x;y/ inS, then ( 6.3.15 ) implies that /SUB>0 and/NUL/EM <˛ < /EM. Conversely, any point in the uv-plane with polar coordinates ./SUB;˛/ satisfying these conditions is the image under Fof the point .x;y/D./SUB1=2cos˛=2;/SUB1=2sin˛=2/2S: Thus, F/NUL1 S.u;v/D2 4.u2Cv2/1=4cos.arg.u;v/=2/ .u2Cv2/1=4sin.arg.u;v/=23 5;/NUL/EM < arg.u;v/</EM: v u(u,v) α α = −πu2 + v2 Figure 6.3.5 Because of ( 6.3.8 ),Falso maps the open left half-plane S1D˚.x;y/ˇˇx<0/TAB onto F.S/, and F/NUL1 S1.u;v/D2 4.u2Cv2/1=4cos.arg.u;v/=2/ .u2Cv2/1=4sin.arg.u;v/=2/3 5; /EM < arg.u;v/<3/EM; D/NULF/NUL1 S.u;v/: Section 6.3 The Inverse Function Theorem 403 Example 6.3.5 The transformation /DC4u v/NAK DF.x;y/D/DC4excosy exsiny/NAK (6.3.16) is not one-to-one, since F.x;yC2k/EM/DF.x;y/ (6.3.17) ifkis any integer. This transformation is one-to-one on a set Sif and only if Sdoes not contain any pair of points .x0;y0/and.x0;y0C2k/EM/ , wherekis a nonzero integer. This condition is necessary because of ( 6.3.17 ); we leave it to you to show that it is sufficient (Exercise 6.3.8 ). Therefore, for example, Fis one-to-one on S/RSD˚.x;y/ˇˇ/NUL1<x<1;/RS/DC4y</RSC2/EM/TAB(6.3.18) where/RSis arbitrary. Geometrically, S/RSis the infinite strip bounded by the lines yD/RSand yD/RSC2/EM. The lower boundary is in S/RS, but the upper is not (Figure 6.3.6 ). Since every point is in the interior of some such strip, Fis locally invertible on the entire plane. y xy = φy = φ + 2π Figure 6.3.6 The range of FS/RSis the entireuv-plane except the origin, since if .u;v/¤.0;0/ , then .u;v/ can be written uniquely as /DC4u v/NAK D/DC4/SUBcos˛ /SUBsin˛/NAK ; where /SUB>0; /RS/DC4˛</RSC2/EM; so.u;v/ is the image under Fof .x;y/D.log/SUB;˛/2S: The origin is not in R.F/, since jF.x;y/j2D.excosy/2C.exsiny/2De2x¤0: 404 Chapter 6 Vector-Valued Functions of Several Variables Finally, F/NUL1 S/RS.u;v/D2 4log.u2Cv2/1=2 arg.u;v/3 5; /RS/DC4arg.u;v/</RSC2/EM: The domain of F/NUL1 S/RSis the entireuv-plane except for .0;0/ . Regular Transformations The question of invertibility of an arbitrary transformati onFWRn!Rnis too general to have a useful answer. However, there is a useful and easily ap plicable sufficient condition which implies that one-to-one restrictions of continuousl y differentiable transformations have continuously differentiable inverses. To motivate our study of this question, let us first consider t he linear transformation F.X/DAXD2 6664a11a12/SOH/SOH/SOHa1n a21a22/SOH/SOH/SOHa2n :::::::::::: an1an2/SOH/SOH/SOHann3 77752 6664x1 x2 ::: xn3 7775: From Theorem 6.3.1 ,Fis invertible if and only if Ais nonsingular, in which case R.F/D Rnand F/NUL1.U/DA/NUL1U: Since AandA/NUL1are the differential matrices of FandF/NUL1, respectively, we can say that a linear transformation is invertible if and only if its diffe rential matrix F0is nonsingular, in which case the differential matrix of F/NUL1is given by .F/NUL1/0D.F0//NUL1: Because of this, it is tempting to conjecture that if FWRn!Rnis continuously differen- tiable and A0.X/is nonsingular, or, equivalently, JF.X/¤0, for Xin a setS, then Fis one-to-one on S. However, this is false. For example, if F.x;y/D/DC4excosy exsiny/NAK ; then JF.x;y/Dˇˇˇˇexcosy/NULexsiny exsiny excosyˇˇˇˇDe2x¤0; (6.3.19) butFis not one-to-one on R2(Example 6.3.5 ). The best that can be said in general is that if Fis continuously differentiable and JF.X/¤0in an open set S, then Fis locally invertible on S, and the local inverses are continuously differentiable. T his is part of the inverse function theorem, which we will prove presently. Fi rst, we need the following definition. Section 6.3 The Inverse Function Theorem 405 Definition 6.3.2 A transformation FWRn!Rnisregular on an open set SifFis one-to-one and continuously differentiable on S, andJF.X/¤0ifX2S. We will also say that Fis regular on an arbitrary set SifFis regular on an open set containing S. Example 6.3.6 If F.x;y/D/DC4x/NULy xCy/NAK (Example 6.3.2 ), then JF.x;y/Dˇˇˇˇ1/NUL1 1 1ˇˇˇˇD2; soFis one-to-one on R2. Hence, Fis regular on R2. If F.x;y/D/DC4xCy 2xC2y/NAK (Example 6.3.3 ), then JF.x;y/Dˇˇˇˇ1 1 2 2ˇˇˇˇD0; soFis not regular on any subset of R2. If F.x;y/D/DC4x2/NULy2 2xy/NAK (Example 6.3.4 ), then JF.x;y/Dˇˇˇˇ2x/NUL2y 2y 2xˇˇˇˇD2.x2Cy2/; soFis regular on any open set Son which Fis one-to-one, provided that .0;0/62S. For ex- ample, Fis regular on the open half-plane˚ .x;y/ˇˇx>0/TAB , since we saw in Example 6.3.4 thatFis one-to-one on this half-plane. If F.x;y/D/DC4excosy excosy/NAK (Example 6.3.5 ), thenJF.x;y/De2x(see ( 6.3.19 )), so Fis regular on any open set on which it is one-to-one. The interior of S/RSin (6.3.18 ) is an example of such a set. Theorem 6.3.3 Suppose that FWRn!Rnis regular on an open set S;and let GDF/NUL1 S:Then F.S/is open;Gis continuously differentiable on F.S/; and G0.U/D.F0.X///NUL1;where UDF.X/: Moreover;since Gis one-to-one on F.S/; Gis regular on F.S/: 406 Chapter 6 Vector-Valued Functions of Several Variables Proof We first show that if X02S, then a neighborhood of F.X0/is in F.S/. This implies that F.S/is open. SinceSis open, there is a /SUB > 0 such thatB/SUB.X0//SUBS. LetBbe the boundary of B/SUB.X0/; thus, BD˚ˇˇX/TABjX/NULX0jD/SUB: (6.3.20) The function /ESC.X/DjF.X//NULF.X0/j is continuous on Sand therefore on B, which is compact. Hence, by Theorem 5.2.12 , there is a point X1inBwhere/ESC.X/attains its minimum value, say m, onB. Moreover,m>0 , since X1¤X0andFis one-to-one on S. Therefore, jF.X//NULF.X0/j/NAKm>0 ifjX/NULX0jD/SUB: (6.3.21) The set˚ UˇˇjU/NULF.X0/j<m=2/TAB is a neighborhood of F.X0/. We will show that it is a subset of F.S/. To see this, let Ube a fixed point in this set; thus, jU/NULF.X0/j<m=2: (6.3.22) Consider the function /ESC1.X/DjU/NULF.X/j2; which is continuous on S. Note that /ESC1.X//NAKm2 4ifjX/NULX0jD/SUB; (6.3.23) since ifjX/NULX0jD/SUB, then jU/NULF.X/jDj.U/NULF.X0//C.F.X0//NULF.X//j /NAKˇˇjF.X0//NULF.X/j/NULjU/NULF.X0/jˇˇ /NAKm/NULm 2Dm 2; from ( 6.3.21 ) and ( 6.3.22 ). Since/ESC1is continuous on S,/ESC1attains a minimum value /SYNon the compact set B/SUB.X0/ (Theorem 5.2.12 ); that is, there is an XinB/SUB.X0/such that /ESC1.X//NAK/ESC1.X/D/SYN; X2B/SUB.X0/: Setting XDX0, we conclude from this and ( 6.3.22 ) that /ESC1.X/D/SYN/DC4/ESC1.X0/<m2 4: Because of ( 6.3.20 ) and ( 6.3.23 ), this rules out the possibility that X2B, soX2B/SUB.X0/. Section 6.3 The Inverse Function Theorem 407 Now we want to show that /SYND0; that is, UDF.X/. To this end, we note that /ESC1.X/ can be written as /ESC1.X/DnX jD1.uj/NULfj.X//2; so/ESC1is differentiable on Bp.X0/. Therefore, the first partial derivatives of /ESC1are all zero at the local minimum point X(Theorem 5.3.11 ), so nX [email protected]/ @xi.uj/NULfj.X//D0; 1/DC4i/DC4n; or, in matrix form, F0.X/.U/NULF.X//D0: Since F0.X/is nonsingular this implies that UDF.X/(Theorem 6.1.13 ). Thus, we have shown that every Uthat satisfies ( 6.3.22 ) is in F.S/. Therefore, since X0is an arbitrary point ofS,F.S/is open. Next, we show that Gis continuous on F.S/. Suppose that U02F.S/andX0is the unique point in Ssuch that F.X0/DU0. Since F0.X0/is invertible, Lemma 6.2.6 implies that there is a /NAK>0 and an open neighborhood NofX0such thatN/SUBSand jF.X//NULF.X0/j/NAK/NAKjX/NULX0jifX2N: (6.3.24) (Exercise 6.2.18 also implies this.) Since Fsatisfies the hypotheses of the present theorem onN, the first part of this proof shows that F.N/ is an open set containing U0DF.X0/. Therefore, there is a ı>0 such that XDG.U/is inNifU2Bı.U0/. Setting XDG.U/ andX0DG.U0/in (6.3.24 ) yields jF.G.U///NULF.G.U0//j/NAK/NAKjG.U//NULG.U0/jif U2Bı.U0/: Since F.G.U//DU, this can be rewritten as jG.U//NULG.U0/j/DC41 /NAKjU/NULU0jifU2Bı.U0/; (6.3.25) which means that Gis continuous at U0. Since U0is an arbitrary point in F.S/, it follows thatGis continous on F.S/. We will now show that Gis differentiable at U0. Since G.F.X//DX;X2S; the chain rule (Theorem 6.2.8 ) implies that ifGis differentiable at U0, then G0.U0/F0.X0/DI 408 Chapter 6 Vector-Valued Functions of Several Variables (Example 6.2.3 ). Therefore, if Gis differentiable at U0, the differential matrix of Gmust be G0.U0/DŒF0.X0//c141/NUL1; so to show that Gis differentiable at U0, we must show that if H.U/DG.U//NULG.U0//NULŒF0.X0//c141/NUL1.U/NULU0/ jU/NULU0j.U¤U0/; (6.3.26) then lim U!U0H.U/D0: (6.3.27) Since Fis one-to-one on SandF.G.U//DU, it follows that if U¤U0, then G.U/¤ G.U0/. Therefore, we can multiply the numerator and denominator o f (6.3.26 ) byjG.U//NUL G.U0/jto obtain H.U/DjG.U//NULG.U0j jU/NULU0j G.U//NULG.U0//NULŒF0.X0//c141/NUL1.U/NULU0/ jG.U//NULG.U0/j! D/NULjG.U//NULG.U0/j jU/NULU0j/STX F0.X0//ETX/NUL1/DC2U/NULU0/NULF0.X0/.G.U//NULG.U0// jG.U//NULG.U0/j/DC3 if0<jU/NULU0j<ı. Because of ( 6.3.25 ), this implies that jH.U/j/DC41 /NAKkŒF0.X0//c141/NUL1kˇˇˇˇU/NULU0/NULF0.X0/.G.U//NULG.U0// jG.U//NULG.U0/jˇˇˇˇ if0<jU/NULU0j<ı. Now let H1.U/DU/NULU0/NULF0.X0/.G.U//NULG.U0// jG.U//NULG.U0/j To complete the proof of ( 6.3.27 ), we must show that lim U!U0H1.U/D0: (6.3.28) Since Fis differentiable at X0, we know that if H2.X/Dlim X!X0F.X//NULF.X0//NULF0.X0/.X/NULX0/ jX/NULX0j; then lim X!X0H2.X/D0: (6.3.29) Since F.G.U//DUandX0DG.U0/, H1.U/DH2.G.U//: Section 6.3 The Inverse Function Theorem 409 Now suppose that /SI>0 . From ( 6.3.29 ), there is aı1>0such that jH2.X/j</SI if0<jX/NULX0jDj X/NULG.U0/j<ı1: (6.3.30) Since Gis continuous at U0, there is aı22.0;ı/ such that jG.U//NULG.U0/j<ı1if0<jU/NULU0j<ı2: This and ( 6.3.30 ) imply that jH1.U/jDj H2.G.U//j</SI if0<jU/NULU0j<ı2: Since this implies ( 6.3.28 ),Gis differentiable at X0. Since U0is an arbitrary member of F.N/, we can now drop the zero subscript and conclude that Gis continuous and differentiable on F.N/, and G0.U/DŒF0.X//c141/NUL1;U2F.N/: To see that Giscontinuously differentiable onF.N/, we observe that by Theorem 6.1.14 , each entry of G0.U/(that is, each partial derivative @gi.U/=@u j,1/DC4i;j/DC4n) can be written as the ratio, with nonzero denominator, of determin ants with entries of the form @fr.G.U// @xs: (6.3.31) Since@fr=@x sis continuous on NandGis continuous on F.N/, Theorem 5.2.10 implies that ( 6.3.31 ) is continuous on F.N/. Since a determinant is a continuous function of its entries, it now follows that the entries of G0.U/are continuous on F.N/. Branches of the Inverse IfFis regular on an open set S, we say that F/NUL1 Sis abranch of F/NUL1. (This is a convenient terminology but is not meant to imply that Factually has an inverse.) From this definition, it is possible to define a branch of F/NUL1on a setT/SUBR.F/if and only if TDF.S/, where Fis regular on S. There may be open subsets of R.F/that do not have this property, and therefore no branch of F/NUL1can be defined on them. It is also possible that TDF.S1/D F.S2/, whereS1andS2are distinct subsets of DF. In this case, more than one branch of F/NUL1is defined on T. Thus, we saw in Example 6.3.4 that two branches of F/NUL1may be defined on a set T. In Example 6.3.5 infinitely many branches of F/NUL1are defined on the same set. It is useful to define branches of the argument To do this, we th ink of the relationship between polar and rectangular coordinates in terms of the tr ansformation /DC4x y/NAK DF.r;/DC2/D/DC4rcos/DC2 rsin/DC2/NAK ; (6.3.32) where for the moment we regard rand/DC2as rectangular coordinates of a point in an r/DC2- plane. LetSbe an open subset of the right half of this plane (that is, S/SUB˚.r;/DC2/ˇˇr >0/TAB) 410 Chapter 6 Vector-Valued Functions of Several Variables that does not contain any pair of points .r;/DC2/ and.r;/DC2C2k/EM/ , wherekis a nonzero integer. Then Fis one-to-one and continuously differentiable on S, with F0.r;/DC2/D/DC4cos/DC2/NULrsin/DC2 sin/DC2 r cos/DC2/NAK (6.3.33) and JF.r;/DC2/Dr >0; .r;/DC2/2S: (6.3.34) Hence, Fis regular on S. Now letTDF.S/, the set of points in the xy-plane with polar coordinates in S. Theorem 6.3.3 states thatTis open and FShas a continuously differentiable inverse (which we denote by G, rather than F/NUL1 S, for typographical reasons) /DC4r /DC2/NAK DG.x;y/D2 4p x2Cy2 argS.x;y/3 5; .x;y/2T; where argS.x;y/ is the unique value of arg .x;y/ such that .r;/DC2/D/DLEp x2Cy2;argS.x;y//DC1 2S: We say that argS.x;y/ is abranch of the argument defined on T. Theorem 6.3.3 also implies that G0.x;y/D/STXF0.r;/DC2//ETX/NUL1D"cos/DC2 sin/DC2 /NULsin/DC2 rcos/DC2 r# (see ( 6.3.33 )) D2 64xp x2Cy2yp x2Cy2 /NULy x2Cy2x x2Cy23 75 (see ( 6.3.32 )): Therefore, @argS.x;y/ @xD/NULy x2Cy2;@argS.x;y/ @yDx x2Cy2: (6.3.35) A branch of arg .x;y/ can be defined on an open set Tof thexy-plane if and only if the polar coordinates of the points in Tform an open subset of the r/DC2-plane that does not intersect the/DC2-axis or contain any two points of the form .r;/DC2/ and.r;/DC2C2k/EM/ , where kis a nonzero integer. No subset containing the origin .x;y/D.0;0/ has this property, nor does any deleted neighborhood of the origin (Exercise 6.3.14 ), so there are open sets on which no branch of the argument can be defined. However, if o ne branch can be defined onT, then so can infinitely many others. (Why?) All branches of ar g.x;y/ have the same partial derivatives, given in ( 6.3.35 ). Section 6.3 The Inverse Function Theorem 411 Example 6.3.7 The set TD˚ .x;y/ˇˇ.x;y/¤.x;0/ withx/NAK0/TAB ; which is the entire xy-plane with the nonnegative x-axis deleted, can be written as TD F.Sk/, where Fis as in ( 6.3.32 ),kis an integer, and SkD˚.r;/DC2/ˇˇr >0;2k/EM </DC2 <2.k C1//EM/TAB: For each integer k, we can define a branch argSk.x;y/ of the argument in Skby taking argSk.x;y/ to be the value of arg .x;y/ that satisfies 2k/EM< argSk.x;y/<2.kC1//EM: Each of these branches is continuously differentiable in T, with derivatives as given in (6.3.35 ), and argSk.x;y//NULargSj.x;y/D2.k/NULj//EM; .x;y/2T: Example 6.3.8 Returning to the transformation /DC4u v/NAK DF.x;y/D/DC4x2/NULy2 2xy/NAK ; we now see from Example 6.3.4 that a branch GofF/NUL1can be defined on any subset Tof theuv-plane on which a branch of arg .u;v/ can be defined, and Ghas the form /DC4x y/NAK DG.u;v/D2 4.u2Cv2/1=4cos.arg.u;v/=2/ .u2Cv2/1=4sin.arg.u;v/=2/3 5; .u;v/2T; (6.3.36) where arg.u;v/ is a branch of the argument defined on T. If G1andG2are different branches of F/NUL1defined on the same set T, then G1D˙ G2. (Why?) From Theorem 6.3.3 , G0.u;v/D/STXF0.x;y//ETX/NUL1D/DC42x/NUL2y 2y 2x/NAK/NUL1 D1 2.x2Cy2//DC4x y /NULy x/NAK : Substituting for xandyin terms ofuandvfrom ( 6.3.36 ), we find that @x @uD@y @vDx 2.x2Cy2/D1 2.u2Cv2/1=4cos.arg.u;v/=2/ (6.3.37) and @x @vD/NUL@y @uDy 2.x2Cy2/D1 2.u2Cv2/1=4sin.arg.u;v/=2/: (6.3.38) It is essential that the same branch of the argument be used he re and in ( 6.3.36 ). 412 Chapter 6 Vector-Valued Functions of Several Variables We leave it to you (Exercise 6.3.16 ) to verify that ( 6.3.37 ) and ( 6.3.38 ) can also be obtained by differentiating ( 6.3.36 ) directly. Example 6.3.9 If/DC4u v/NAK DF.x;y/D/DC4excosy exsiny/NAK (Example 6.3.5 ), we can also define a branch GofF/NUL1on any subset Tof theuv-plane on which a branch of arg .u;v/ can be defined, and Ghas the form /DC4x y/NAK DG.u;v/D/DC4 log.u2Cv2/1=2 arg.u;v//NAK : (6.3.39) Since the branches of the argument differ by integral multip les of2/EM, (6.3.39 ) implies that ifG1andG2are branches of F/NUL1, both defined on T, then G1.u;v//NULG2.u;v/D/DC40 2k/EM/NAK (kDinteger): From Theorem 6.3.3 , G0.u;v/D/STXF0.x;y//ETX/NUL1D/DC4excosy/NULexsiny exsiny excosy/NAK/NUL1 D/DC4e/NULxcosy e/NULxsiny /NULe/NULxsiny e/NULxcosy/NAK : Substituting for xandyin terms ofuandvfrom ( 6.3.39 ), we find that @x @uD@y @vDe/NULxcosyDe/NUL2xuDu u2Cv2 and @x @vD/NUL@y @uDe/NULxsinyDe/NUL2xvDv u2Cv2: The Inverse Function Theorem Examples 6.3.4 and6.3.5 show that a continuously differentiable function Fmay fail to have an inverse on a set Seven ifJF.X/¤0onS. However, the next theorem shows that in this case Fis locally invertible on S. Theorem 6.3.4 (The Inverse Function Theorem) LetFWRn!Rnbe continuously differentiable on an open set S;and suppose that JF.X/¤0onS:Then;if X02S;there is an open neighborhood NofX0on which Fis regular:Moreover;F.N/ is open and GDF/NUL1 Nis continuously differentiable on F.N/; with G0.U/D/STX F0.X//ETX/NUL1.where UDF.X//; U2F.N/: Section 6.3 The Inverse Function Theorem 413 Proof Lemma 6.2.6 implies that there is an open neighborhood NofX0on which Fis one-to-one. The rest of the conclusions then follow from app lying Theorem 6.3.3 toFon N. Corollary 6.3.5 IfFis continuously differentiable on a neighborhood of X0andJF.X0/¤ 0;then there is an open neighborhood NofX0on which the conclusions of Theorem 6.3.4 hold: Proof By continuity, since JF0.X0/¤0,JF0.X/is nonzero for all Xin some open neighborhood SofX0. Now apply Theorem 6.3.4 . Example 6.3.10 LetX0D.1;2;1/ and 2 4u v w3 5DF.x;y;´/D2 4xCyC.´/NUL1/2C1 yC´C.x/NUL1/2/NUL1 ´CxC.y/NUL2/2C33 5: Then F0.x;y;´/D2 41 1 2´/NUL2 2x/NUL2 1 1 1 2y/NUL4 13 5; so JF.X0/Dˇˇˇˇˇˇ1 1 0 0 1 1 1 0 1ˇˇˇˇˇˇD2: In this case, it is difficult to describe Nor find GDF/NUL1 Nexplicitly; however, we know that F.N/ is a neighborhood of U0DF.X0/D.4;2;5/ , that G.U0/DX0D.1;2;1/ , and that G0.U0/D/STX F0.X0//ETX/NUL1D2 41 1 0 0 1 1 1 0 13 5/NUL1 D1 22 41/NUL1 1 1 1/NUL1 /NUL1 1 13 5: Therefore, G.U/D2 41 2 13 5C1 22 41/NUL1 1 1 1/NUL1 /NUL1 1 13 52 4u/NUL4 v/NUL2 w/NUL53 5CE.U/; where lim U!.4;2;5/E.U/p .u/NUL4/2C.v/NUL2/2C.w/NUL5/2D0I thus we have approximated Gnear U0D.4;2;5/ by an affine transformation. Theorem 6.3.4 and ( 6.3.34 ) imply that the transformation ( 6.3.32 ) is locally invertible onSD˚ .r;/DC2/ˇˇr >0/TAB , which means that it is possible to define a branch of arg .x;y/ in a neighborhood of any point .x0;y0/¤.0;0/ . It also implies, as we have already seen, that 414 Chapter 6 Vector-Valued Functions of Several Variables the transformation ( 6.3.7 ) of Example 6.3.4 is locally invertible everywhere except at .0;0/ , where its Jacobian equals zero, and the transformation ( 6.3.16 ) of Example 6.3.5 is locally invertible everywhere. 6.3 Exercises 1. Prove: If Fis invertible, then F/NUL1is unique. 2. Prove Theorem 6.3.1 . 3. Prove: The linear transformation L.X/DAXcannot be one-to-one on any open set ifAis singular. H INT:Use Theorem 6.1.15: 4. Let G.x;y/D"p x2Cy2 arg.x;y/# ; /EM=2/DC4arg.x;y/<5/EM=2: Find (a)G.0;1/ (b) G.1;0/ (c)G./NUL1;0/ (d) G.2;2/ (e)G./NUL1;1/ 5. Same as Exercise 6.3.4 , except that/NUL2/EM/DC4arg.x;y/<0 . 6. (a) Prove: IffWR!Ris continuous and locally invertible on .a;b/ , thenfis invertible on.a;b/ . (b) Give an example showing that the continuity assumption is ne eded in (a). 7. Let F.x;y/D/DC4x2/NULy2 2xy/NAK (Example 6.3.4 ) and SD˚ .x;y/ˇˇaxCby>0/TAB .a2Cb2¤0/: Find F.S/andF/NUL1 S. If S1D˚.x;y/ˇˇaxCby<0/TAB; show that F.S1/DF.S/andF/NUL1 S1D/NULF/NUL1 S. 8. Show that the transformation /DC4u v/NAK DF.x;y/D/DC4excosy exsiny/NAK (Example 6.3.5 ) is one-to-one on any set Sthat does not contain any pair of points .x0;y0/and.x0;y0C2k/EM/ , wherekis a nonzero integer. Section 6.3 The Inverse Function Theorem 415 9. Suppose that FWRn!Rnis continuous and invertible on a compact set S. Show thatF/NUL1 Sis continuous. H INT:IfF/NUL1 Sis not continuous at UinF.S/; then there is an/SI0>0and a sequencefUkginF.S/such that limk!1UkDUwhile jF/NUL1 S.Uk//NULF/NUL1 S.U/j/NAK/SI0; k/NAK1: Use Exercise 5.1.32 to obtain a contradiction : 10. Find F/NUL1and.F/NUL1/0: (a)/DC4u v/NAK DF.x;y/D/DC44xC2y /NUL3xCy/NAK (b)2 4u v w3 5DF.x;y;´/D2 4/NULxCyC2´ 3xCy/NUL4´ /NULx/NULyC2´3 5 11. In addition to the assumptions of Theorem 6.3.3 , suppose that all qth-order.q>1/ partial derivatives of the components of Fare continuous on S. Show that all qth- order partial derivatives of F/NUL1 Sare continuous on F.S/. 12. If /DC4u v/NAK DF.x;y/D/DC4x2Cy2 x2/NULy2/NAK (Example 6.3.1 ), find four branches G1,G2,G3, and G4ofF/NUL1defined on T1D˚.u;v/ˇˇuCv>0;u/NULv>0/TAB; and verify that G0 i.u;v/D.F0.x.u;v/;y.u;v////NUL1,1/DC4i/DC44. 13. Suppose that Ais a nonsingular n/STXnmatrix and UDF.X/DA2 6664x2 1 x2 2::: x2 n3 7775: (a) Show that Fis regular on the set SD˚Xˇˇeixi>0; 1/DC4i/DC4n/TAB; whereeiD˙1,1/DC4i/DC4n. (b) Find F/NUL1 S.U/.(c)Find.F/NUL1 S/0.U/. 14. Let/DC2.x;y/ be a branch of arg .x;y/ defined on an open set S. (a) Show that/DC2.x;y/ cannot assume a local extreme value at any point of S. (b) Prove: Ifa¤0and the line segment from .x0;y0/to.ax0;ay 0/is inS, then /DC2.ax 0;ay 0/D/DC2.x0;y0/. (c) Show thatScannot contain a subset of the form ADn .x;y/ˇˇ0<r 1/DC4p x2Cy2/DC4r2o : 416 Chapter 6 Vector-Valued Functions of Several Variables (d) Show that no branch of arg .x;y/ can be defined on a deleted neighborhood of the origin. 15. Obtain Eqn. ( 6.3.35 ) formally by differentiating: (a)arg.x;y/Dcos/NUL1xp x2Cy2(b) arg.x;y/Dsin/NUL1yp x2Cy2 (c)arg.x;y/Dtan/NUL1y x Where do these formulas come from? What is the disadvantage o f using any one of them to define arg .x;y/ ? 16. For the transformation /DC4u v/NAK DF.x;y/D/DC4x2/NULy2 2xy/NAK (Example 6.3.4 ), find a branch GofF/NUL1defined onTD˚ .u;v/ˇˇauCbv>0/TAB . Find G0by means of the formula G0.U/DŒF0.X//c141/NUL1of Theorem 6.3.3 , and also by direct differentiation with respect to uandv. 17. A transformation F.x;y/D/DC4u.x;y/ v.x;y//NAK isanalytic on a setSif it is continuously differentiable and uxDvy; u yD/NULvx onS. Prove: If Fis analytic and regular on S, then F/NUL1 Sis analytic on F.S/; that is, xuDuvandxvD/NULuu. 18. Prove: If UDF.X/andXDG.U/are inverse functions, then @.u1;u2;:::;u n/ @.x1;x2;:::;x [email protected];x2;:::;x n/ @.u1;u2;:::;u n/D1: Where should the Jacobians be evaluated? 19. Give an example of a transformation FWRn!Rnthat is invertible but not regular onRn. 20. Find an affine transformation Athat so well approximates the branch GofF/NUL1 defined near U0DF.X0/that lim U!U0G.U//NULA.U/ jU/NULU0jD0: (a)/DC4u v/NAK DF.x;y/D/DC4x4y5/NUL4x x3y2/NUL3y/NAK ;X0D.1;/NUL1/ Section 6.4 The Implicit Function Theorem 417 (b)/DC4u v/NAK DF.x;y/D/DC4x2yCxy 2xyCxy2/NAK ;X0D.1;1/ (c)2 4u v w3 5DF.x;y;´/D2 42x2yCx3C´ x3Cy´ xCyC´3 5;XD.0;1;1/ (d)2 4u v w3 5DF.x;y;´/D2 4xcosycos´ xsinycos´ xsin´3 5;X0D.1;/EM=2;/EM/ 21. IfFis defined by 2 4x y ´3 5DF.r;/DC2;/RS/D2 4rcos/DC2cos/RS rsin/DC2cos/RS rsin/RS3 5 and Gis a branch of F/NUL1, find G0in terms ofr,/DC2, and/RS. H INT:See Exer- cise6.2.14.b/: 22. IfFis defined by2 4x y ´3 5DF.r;/DC2;´/D2 4rcos/DC2 rsin/DC2 ´3 5 and Gis a branch of F/NUL1, find G0in terms ofr,/DC2, and´. H INT:See Exer- cise6.2.14.c/: 23. Suppose that FWRn!Rnis regular on a compact set T. Show that F.@T/D @F.T/; that is, boundary points map to boundary points. H INT:Use Exercise 6.2.23 and Theorem 6.3.3 to show [email protected]//SUBF.@T/: Then apply this result with Fand Treplaced by F/NUL1andF.T/to show that F.@T//[email protected]/: 6.4 THE IMPLICIT FUNCTION THEOREM In this section we consider transformations from RnCmtoRm. It will be convenient to denote points in RnCmby .X;U/D.x1;x2;:::;x n;u1;u2;:::;u m/: We will often denote the components of Xbyx,y, . . . , and the components of Ubyu,v, . . . . To motivate the problem we are interested in, we first ask whet her the linear system of mequations inmCnvariables a11x1Ca12x2C/SOH/SOH/SOHCa1nxnCb11u1Cb12u2C/SOH/SOH/SOHCb1mumD0 a21x1Ca22x2C/SOH/SOH/SOHCa2nxnCb21u1Cb22uxC/SOH/SOH/SOHCb2mumD0 ::: am1x1Cam2x2C/SOH/SOH/SOHCamnxnCbm1u1Cbm2u2C/SOH/SOH/SOHCbmmumD0(6.4.1) 418 Chapter 6 Vector-Valued Functions of Several Variables determinesu1,u2, . . . ,umuniquely in terms of x1,x2, . . . ,xn. By rewriting the system in matrix form as AXCBUD0; where AD2 6664a11a12/SOH/SOH/SOHa1n a21a22/SOH/SOH/SOHa2n :::::::::::: am1am2/SOH/SOH/SOHamn3 7775;BD2 6664b11b12/SOH/SOH/SOHb1m b21b22/SOH/SOH/SOHb2m :::::::::::: bm1bm2/SOH/SOH/SOHbmm3 7775; XD2 6664x1 x2 ::: xn3 7775;and UD2 6664u1 u2 ::: um3 7775; we see that ( 6.4.1 ) can be solved uniquely for Uin terms of Xif the square matrix Bis nonsingular. In this case the solution is UD/NULB/NUL1AX: For our purposes it is convenient to restate this: If F.X;U/DAXCBU; (6.4.2) where Bis nonsingular, then the system F.X;U/D0 determines Uas a function of X, for all XinRn. Notice that Fin (6.4.2 ) is a linear transformation. If Fis a more general transformation fromRnCmtoRm, we can still ask whether the system F.X;U/D0; or, in terms of components, f1.x1;x2;:::;x n;u1;u2;:::;u m/D0 f2.x1;x2;:::;x n;u1;u2;:::;u m/D0 ::: fm.x1;x2;:::;x n;u1;u2;:::;u m/D0; can be solved for Uin terms of X. However, the situation is now more complicated, even ifmD1. For example, suppose that mD1and f.x;y;u/D1/NULx2/NULy2/NULu2: Section 6.4 The Implicit Function Theorem 419 Ifx2Cy2>1, then no value of usatisfies f.x;y;u/D0: (6.4.3) However, infinitely many functions uDu.x;y/ satisfy ( 6.4.3 ) on the set SD˚.x;y/ˇˇx2Cy2/DC41/TAB: They are of the form u.x;y/D/SI.x;y/p 1/NULx2/NULy2; where/SI.x;y/ can be chosen arbitrarily, for each .x;y/ inS, to be1or/NUL1. We can narrow the choice of functions to two by requiring that ube continuous on S; then u.x;y/Dp 1/NULx2/NULy2 (6.4.4) or u.x;y/D/NULp 1/NULx2/NULy2: We can define a unique continuous solution uof (6.4.3 ) by specifying its value at a single interior point of S. For example, if we require that u/DC21p 3;1p 3/DC3 D1p 3; thenumust be as defined by ( 6.4.4 ). The question of whether an arbitrary system F.X;U/D0 determines Uas a function of Xis too general to have a useful answer. However, there is a theorem, the implicit function theorem, that answers th is question affirmatively in an important special case. To facilitate the statement of th is theorem, we partition the differential matrix of FWRnCm!Rm: F0D2 6666666664@f1 @x1@f1 @x2/SOH/SOH/SOH@f1 @xnj@f1 @u1@f1 @u2/SOH/SOH/SOH@f1 @um @f2 @x1@f2 @x2/SOH/SOH/SOH@f2 @xnj@f2 @u1@f2 @u2/SOH/SOH/SOH@f2 @um ::::::::::::j:::::::::::: @fm @x1@fm @x2/SOH/SOH/SOH@fm @xnj@fm @u1@fm @u2/SOH/SOH/SOH@fm @um3 7777777775(6.4.5) or F0DŒFX;FU/c141; where FXis the submatrix to the left of the dashed line in ( 6.4.5 ) and FUis to the right. For the linear transformation ( 6.4.2 ),FXDAandFUDB, and we have seen that the system F.X;U/D0defines Uas a function of Xfor all XinRnifFUis nonsingular. The next theorem shows that a related result holds for more gener al transformations. 420 Chapter 6 Vector-Valued Functions of Several Variables Theorem 6.4.1 (The Implicit Function Theorem) Suppose that FWRnCm! Rmis continuously differentiable on an open set SofRnCmcontaining.X0;U0/:Let F.X0;U0/D0;and suppose that FU.X0;U0/is nonsingular :Then there is a neighborhood Mof.X0;U0/;contained inS;on which FU.X;U/is nonsingular and a neighborhood N ofX0inRnon which a unique continuously differentiable transformat ionGWRn!Rm is defined;such that G.X0/DU0and .X;G.X//2M and F.X;G.X//D0ifX2N: (6.4.6) Moreover; G0.X/D/NULŒFU.X;G.X///c141/NUL1FX.X;G.X//; X2N: (6.4.7) Proof Define ˆWRnCm!RnCmby ˆ.X;U/D2 66666666666664x1 x2 ::: xn f1.X;U/ f2.X;U/ ::: fm.X;U/3 77777777777775(6.4.8) or, in “horizontal”notation by ˆ.X;U/D.X;F.X;U//: (6.4.9) Then ˆis continuously differentiable on Sand, since F.X0;U0/D0, ˆ.X0;U0/D.X0;0/: (6.4.10) The differential matrix of ˆis ˆ0D2 6666666666666666666666641 0/SOH/SOH/SOH0 0 0 /SOH/SOH/SOH0 0 1/SOH/SOH/SOH0 0 0 /SOH/SOH/SOH0 :::::::::::::::::::::::: 0 0/SOH/SOH/SOH1 0 0 /SOH/SOH/SOH0 @f1 @x1@f1 @x2/SOH/SOH/SOH@f1 @xn@f1 @u1@f1 @u2/SOH/SOH/SOH@f1 @um @f2 @x1@f2 @x2/SOH/SOH/SOH@f2 @xn@f2 @u1@f2 @u2/SOH/SOH/SOH@f2 @um :::::::::::::::::::::::: @fm @x1@fm @x2/SOH/SOH/SOH@fm @xn@fm @u1@fm @u2/SOH/SOH/SOH@fm @um3 777777777777777777777775D/DC4I 0 FXFU/NAK ; Section 6.4 The Implicit Function Theorem 421 where Iis then/STXnidentity matrix, 0is then/STXmmatrix with all zero entries, and FX andFUare as in ( 6.4.5 ). By expanding det .ˆ0/and the determinants that evolve from it in terms of the cofactors of their first rows, it can be shown in nsteps that JˆDdet.ˆ0/Dˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇ@f1 @u1@f1 @u2/SOH/SOH/SOH@f1 @um @f2 @u1@f2 @u2/SOH/SOH/SOH@f2 @um :::::::::::: @fm @u1@fm @u2/SOH/SOH/SOH@fm @umˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇDdet.FU/: In particular, Jˆ.X0;U0/Ddet.FU.X0;U0/¤0: Since ˆis continuously differentiable on S, Corollary 6.3.5 implies that ˆis regular on some open neighborhood Mof.X0;U0/and thatcMDˆ.M/ is open. Because of the form of ˆ(see ( 6.4.8 ) or ( 6.4.9 )), we can write points of cMas.X;V/, where V2Rm. Corollary 6.3.5 also implies that ˆhas a a continuously differentiable inverse /c128.X;V/defined oncMwith values in M. Since ˆleaves the “ Xpart" of.X;U/ fixed, a local inverse of ˆmust also have this property. Therefore, /c128must have the form /c128.X;V/D2 66666666666666664x1 x2 ::: xn h1.X;V/ h2.X;V/ ::: hm.X;V/3 77777777777777775 or, in “horizontal” notation, /c128.X;V/D.X;H.X;V//; where HWRnCm!Rmis continuously differentiable on cM. We will show that G.X/D H.X;0/has the stated properties. From ( 6.4.10 ),.X0;0/2cMand, sincecMis open, there is a neighborhood NofX0in Rnsuch that.X;0/2cMifX2N(Exercise 6.4.2 ). Therefore,.X;G.X//D/c128.X;0/2M ifX2N. Since /c128Dˆ/NUL1,.X;0/Dˆ.X;G.X//. Setting XDX0and recalling ( 6.4.10 ) shows that G.X0/DU0, since ˆis one-to-one on M. 422 Chapter 6 Vector-Valued Functions of Several Variables Henceforth we assume that X2N. Now, .X;0/Dˆ./c128.X;0// (since ˆD/c128/NUL1/ Dˆ.X;G.X// (since /c128.X;0/D.X;G.X//) D.X;F.X;G.X/// (since ˆ.X;U/D.X;F.X;U//): Therefore, F.X;G.X//D0; that is, Gsatisfies ( 6.4.6 ). To see that Gis unique, suppose thatG1WRn!Rmalso satisfies ( 6.4.6 ). Then ˆ.X;G.X//D.X;F.X;G.X///D.X;0/ and ˆ.X;G1.X//D.X;F.X;G1.X///D.X;0/ for all XinN. Since ˆis one-to-one on M, this implies that G.X/DG1.X/. Since the partial derivatives @hi @xj; 1/DC4i/DC4m; 1/DC4j/DC4n; are continuous functions of .X;V/oncM, they are continuous with respect to Xon the subset˚.X;0/ˇˇX2N/TABofcM. Therefore, Gis continuously differentiable on N. To verify (6.4.7 ), we write F.X;G.X//D0in terms of components; thus, fi.x1;x2;:::;x n;g1.X/;g2.X/;:::;g m.X//D0; 1/DC4i/DC4m; X2N: Sincefiandg1,g2, . . . ,gmare continuously differentiable on their respective domai ns, the chain rule (Theorem 5.4.3 ) implies that @fi.X;G.X// @xjCmX [email protected];G.X// @[email protected]/ @xjD0; 1/DC4i/DC4m; 1/DC4j/DC4n; (6.4.11) or, in matrix form, FX.X;G.X//CFU.X;G.X//G0.X/D0: (6.4.12) Since.X;G.X//2Mfor all XinNandFU.X;U/is nonsingular when .X;U/2M, we can multiply ( 6.4.12 ) on the left by F/NUL1 U.X;G.X//to obtain ( 6.4.7 ). This completes the proof. In Theorem 6.4.1 we denoted the implicitly defined transformation by Gfor reasons of clarity in the proof. However, in applying the theorem it i s convenient to denote the transformation more informally by UDU.X/; thus, U.X0/DU0, and we replace ( 6.4.6 ) and ( 6.4.7 ) by .X;U.X//2M and X.X;U.X//D0ifX2N; and U0.X/D/NULŒFU.X;U.X///c141/NUL1FX.X;U.X//; X2N; Section 6.4 The Implicit Function Theorem 423 while ( 6.4.11 ) becomes @fi @xjCmX rD1@fi @ur@ur @xjD0; 1/DC4i/DC4m; 1/DC4j/DC4n; (6.4.13) it being understood that the partial derivatives of urandfiare evaluated at Xand.X;U.X//, respectively. The following corollary is the implicit function theorem fo rmD1. Corollary 6.4.2 Suppose that fWRnC1!Ris continuously differentiable on an open set containing .X0;u0/;withf.X0;u0/D0andfu.X0;u0/¤0. Then there is a neighborhood Mof.X0;u0/;contained inS;and a neighborhood NofX0inRnon which is defined a unique continuously differentiable function uDu.X/WRn!Rsuch that .X;u.X//2M andfu.X;u.X//¤0; X2N; u.X0/Du0;andf.X;u.X//D0; X2N: The partial derivatives of uare given by uxi.X/D/NULfxi.X;u.X// fu.X;u.X//; 1/DC4i/DC4n: Example 6.4.1 Let f.x;y;u/D1/NULx2/NULy2/NULu2 and.x0;y0;u0/D.1 2;/NUL1 2;1p 2/. Thenf.x 0;y0;´0/D0and fx.x;y;u/D/NUL2x; f y.x;y;u/D/NUL2y; f u.x;y;u/D/NUL2u: Sincefis continuously differentiable everywhere and fu.x0;y0;u0/D/NULp 2¤0, Corol- lary6.4.2 implies that the conditions 1/NULx2/NULy2/NULu2D0; u.1=2;/NUL1=2/D1p 2; determineuDu.x;y/ near.x0;y0/D.1 2;/NUL1 2/so that ux.x;y/D/NULfx.x;y;u.x;y// fu.x;y;u.x;y//D/NULx u.x;y/; (6.4.14) and uy.x;y/D/NULfy.x;y;u.x;y// fu.x;y;u.x;y//D/NULy u.x;y/: (6.4.15) It is not necessary to memorize formulas like ( 6.4.14 ) and ( 6.4.15 ). Since we know that fanduare differentiable, we can obtain ( 6.4.14 ) and ( 6.4.15 ) by applying the chain rule to the identity f.x;y;u.x;y//D0: 424 Chapter 6 Vector-Valued Functions of Several Variables Example 6.4.2 Let f.x;y;u/Dx3y2u2C3xy4u4/NUL3x6y6u7C12x/NUL13 (6.4.16) and.x0;y0;u0/D.1;/NUL1;1/, sof.x 0;y0;u0/D0. Then fx.x;y;u/D3x2y2u2C3y4u4/NUL18x5y6u7C12; fy.x;y;u/D2x3yu2C12xy3u4/NUL18x6y5u7; fu.x;y;u/D2x3y2uC12xy4u3/NUL21x6y6u6: Sincefu.1;/NUL1;1/D/NUL7¤0, Corollary 6.4.2 implies that the conditions f.x;y;u/D0; u.1;/NUL1/D1 (6.4.17) determineuas a continuously differentiable function of .x;y/ near.1;/NUL1/. If we try to solve ( 6.4.16 ) foru, we see very clearly that Theorem 6.4.1 and Corol- lary6.4.2 areexistence theorems; that is, they tell us that there is a function uDu.x;y/ that satisfies ( 6.4.17 ), but not how to find it. In this case there is no convenient for mula for the function, although its partial derivatives can be expre ssed conveniently in terms of x, y, andu.x;y/ : ux.x;y/D/NULfx.x;y;u.x;y// fu.x;y;u.x;y//; u y.x;y/D/NULfy.x;y;u.x;y// fu.x;y;u.x;y//: In particular, since u.1;/NUL1/D1, ux.1;/NUL1/D/NUL0 /NUL7D0; u y.1;/NUL1/D/NUL4 /NUL7D4 7: Example 6.4.3 Let XD2 4x y ´3 5 and UD/DC4u v/NAK ; and F.X;U/D/DC42x2Cy2C´2Cu2/NULv2 x2C´2C2u/NULv/NAK : IfX0D.1;/NUL1;1/ andU0D.0;2/ , then F.X0;U0/D0. Moreover, FU.X;U/D/DC42u/NUL2v 2/NUL1/NAK and FXD/DC44x 2y 2´ 2x 0 2´/NAK ; so det.FU.X0;U0//Dˇˇˇˇ0/NUL4 2/NUL1ˇˇˇˇD8¤0: Section 6.4 The Implicit Function Theorem 425 Hence, the conditions F.X;U/D0;U.1;/NUL1;1/D.0;2/ determine UDU.X/near X0. Although it is difficult to find U.X/explicitly, we can approximate U.X/near X0by an affine transformation. Thus, from ( 6.4.7 ), U0.X0/D/NULŒFU.X0;U.X0///c141/NUL1FX.X0;U.X0// (6.4.18) D/NUL/DC40/NUL4 2/NUL1/NAK/NUL1/DC44/NUL2 2 2 0 2/NAK D/NUL1 8/DC4/NUL1 4 /NUL2 0/NAK/DC44/NUL2 2 2 0 2/NAK D/NUL1 8/DC44 2 6 /NUL8 4/NUL4/NAK : Therefore, lim X!.1;/NUL1;1//DC4u.x;y/ v.x;y//NAK /NUL/DC40 2/NAK C1 8/DC44 2 6 /NUL8 4/NUL4/NAK2 4x/NUL1 yC1 ´/NUL13 5 Œ.x/NUL1/2C.yC1/2C.´/NUL1/2/c1411=2D/DC40 0/NAK : Again, it is not necessary to memorize ( 6.4.18 ), since the partial derivatives of an implic- itly defined function can be obtained from the chain rule and C ramer’s rule, as in the next example. Example 6.4.4 LetuDu.x;y/ andvDv.x;y/ be differentiable and satisfy x2C2y2C3´2Cu2CvD6 2x3C4y2C2´2CuCv2D9(6.4.19) and u.1;/NUL1;0/D/NUL1; v.1;/NUL1;0/D2: (6.4.20) To finduxandvx, we differentiate ( 6.4.19 ) with respect to xto obtain 2xC2uu xCvxD0 6x2CuxC2vv xD0: Therefore, /DC42u 1 1 2v/NAK/DC4ux vx/NAK D/NUL/DC42x 6x2/NAK ; 426 Chapter 6 Vector-Valued Functions of Several Variables and Cramer’s rule yields uxD/NULˇˇˇˇ2x 1 6x22vˇˇˇˇ ˇˇˇˇ2u 1 1 2vˇˇˇˇD6x2/NUL4xv 4uv/NUL1 and vxD/NULˇˇˇˇ2u 2x 1 6x2ˇˇˇˇ ˇˇˇˇ2u 1 1 2vˇˇˇˇD2x/NUL12x2u 4uv/NUL1 if4uv¤1. In particular, from ( 6.4.20 ), ux.1;/NUL1;0/D/NUL2 /NUL9D2 9; v x.1;/NUL1;0/D14 /NUL9D/NUL14 9: Jacobians It is convenient to extend the notation introduced in Sectio n 6.2 for the Jacobian of a trans- formation FWRm!Rm. Iff1,f2, . . . ,fmare real-valued functions of kvariables, k/NAKm, and/CAN1,/CAN2, . . . ,/CANmare anymof the variables, then we call the determinant ˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇ@f1 @/CAN1@f1 @/CAN2/SOH/SOH/SOH@f1 @/CANm @f2 @/CAN1@f2 @/CAN2/SOH/SOH/SOH@f2 @/CANm :::::::::::: @fm @/CAN1@fm @/CAN2/SOH/SOH/SOH@fm @/CANmˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇˇ; theJacobian off1,f2, . . . ,fmwith respect to /CAN1,/CAN2, . . . ,/CANm. We denote this Jacobian by @.f1;f2;:::;f m/ @./CAN1;/CAN2;:::;/CAN m/; and we denote the value of the Jacobian at a point Pby @.f1;f2;:::;f m/ @./CAN1;/CAN2;:::;/CAN m/ˇˇˇˇˇ P: Example 6.4.5 If F.x;y;´/D/DC43x2C2xyC´2 4x2C2xy2C´3/NAK ; Section 6.4 The Implicit Function Theorem 427 then @.f1;f2/ @.x;y/Dˇˇˇˇ6xC2y 2x 8xC2y24xyˇˇˇˇ;@.f1;f2/ @.y;´/Dˇˇˇˇ2x 2´ 4xy 3´2ˇˇˇˇ; and @.f1;f2/ @.´;x/Dˇˇˇˇ2´ 6xC2y 3´28xC2y2ˇˇˇˇ: The values of these Jacobians at X0D./NUL1;1;0/ are @.f1;f2/ @.x;y/ˇˇˇˇˇ X0Dˇˇˇˇ/NUL4/NUL2 /NUL6/NUL4ˇˇˇˇD4;@.f1;f2/ @.y;´/ˇˇˇˇˇ X0Dˇˇˇˇ/NUL2 0 /NUL4 0ˇˇˇˇD0; and @.f1;f2/ @.´;x/ˇˇˇˇˇ X0Dˇˇˇˇ0/NUL4 0/NUL6ˇˇˇˇD0: The requirement in Theorem 6.4.1 thatFU.X0;U0/be nonsingular is equivalent to @.f1;f2;:::;f m/ @.u1;u2;:::;u m/ˇˇˇˇˇ .X0;U0/¤0: If this is so then, for a fixed j, Cramer’s rule allows us to write the solution of ( 6.4.13 ) as @ui @xjD/[email protected];f2;:::;f i;:::;f m/ @.u1;u2;:::;x j;:::;u m/ @.f1;f2;:::;f i;:::;f m/ @.u1;u2;:::;u i;:::;u m/; 1/DC4i/DC4m; Notice that the determinant in the numerator on the right is o btained by replacing the ith column of the determinant in the denominator, which is 2 6666666664@f1 @ui @f2 @ui::: @fm @ui3 7777777775;by2 6666666664@f1 @xj @f2 @xj ::: @fm @xj3 7777777775: So far we have considered only the problem of solving a contin uously differentiable system F.X;U/D0.FWRnCm!Rm/ (6.4.21) for the lastmvariables,u1,u2, . . . ,um, in terms of the first n,x1,x2, . . . ,xn. This was merely for convenience; ( 6.4.21 ) can be solved near .X0;U0/for anymof the variables in terms of the other n, provided only that the Jacobian of f1,f2, . . . ,fmwith respect to the chosenmvariables is nonzero at .X0;U0/. This can be seen by renaming the variables and applying Theorem 6.4.1 . 428 Chapter 6 Vector-Valued Functions of Several Variables Example 6.4.6 Let F.x;y;´/D/DC4f.x;y;´/ g.x;y;´//NAK be continuously differentiable in a neighborhood of .x0;y0;´0/. Suppose that F.x0;y0;´0/D0 and @.f;g/ @.x;´/ˇˇˇˇˇ .x0;y0;´0/¤0: (6.4.22) Then Theorem 6.4.1 with XD.y/andUD.x;´/ implies that the conditions f.x;y;´/D0; g.x;y;´/D0; x.y 0/Dx0; ´.y 0/D´0; (6.4.23) determinexand´as continuously differentiable functions of yneary0. Differentiating (6.4.23 ) with respect to yand regarding xand´as functions of yyields fxx0CfyCf´´0D0 gxx0CgyCg´´0D0: Rewriting this as fxx0Cf´´0D/NULfy gxx0Cg´´0D/NULgy; and solving for x0and´0by Cramer’s rule yields x0Dˇˇˇˇ/NULfyf´ /NULgyg´ˇˇˇˇ ˇˇˇˇfxf´ gxg´ˇˇˇˇD/[email protected];g/ @.y;´/ @.f;g/ @.x;´/(6.4.24) and ´0Dˇˇˇˇfx/NULfy gx/NULgyˇˇˇˇ ˇˇˇˇfxf´ gxg´ˇˇˇˇD/[email protected];g/ @.x;y/ @.f;g/ @.x;´/: (6.4.25) Equation ( 6.4.22 ) implies that @.f;g/[email protected];´/ is nonzero ifyis sufficiently close to y0. Example 6.4.7 LetX0D.1;1;2/ and F.x;y;´/D/DC4f.x;y;´/ g.x;y;´//NAK D/DC46xC6yC4´3/NUL44 /NULx2/NULy2C8´/NUL14/NAK : Section 6.4 The Implicit Function Theorem 429 Then F.X0/D0, @.f;g/ @.x;´/Dˇˇˇˇ6 12´2 /NUL2x 8ˇˇˇˇ; and @.f;g/ @.x;´/ˇˇˇˇˇ .1;1;2/Dˇˇˇˇ6 48 /NUL2 8ˇˇˇˇD144¤0: Therefore, Theorem 6.4.1 with XD.y/andUD.x;´/ implies that the conditions f.x;y;´/D0; g.x;y;´/D0; and x.1/D1; ´.1/D2; (6.4.26) determinexand´as continuously differentiable functions of yneary0D1. From ( 6.4.24 ) and ( 6.4.25 ), x0D/[email protected];g/ @.y;´/ @.f;g/ @.x;´/D/NULˇˇˇˇ6 12´2 /NUL2y 8ˇˇˇˇ ˇˇˇˇ6 12´2 /NUL2x 8ˇˇˇˇD/NUL2Cy´2 2Cx´2 and ´0D/[email protected];g/ @.x;y/ @.f;g/ @.x;´/D/NULˇˇˇˇ6 6 /NUL2x/NUL2yˇˇˇˇ ˇˇˇˇ6 12´2 /NUL2x 8ˇˇˇˇDy/NULx 4C2x´2: These equations hold near yD1. Together with ( 6.4.26 ) they imply that x0.1/D/NUL1; ´0.1/D0: Example 6.4.8 Continuing with Example 6.4.7 , Theorem 6.4.1 implies that the con- ditions f.x;y;´/D0; g.x;y;´/D0; y.1/D1; ´.1/D2 determineyand´as functions of xnearx0D1, since @.f;g/ @.y;´/Dˇˇˇˇ6 12´2 /NUL2y 8ˇˇˇˇ and @.f;g/ @.y;´/ˇˇˇˇˇ .1;1;2/Dˇˇˇˇ6 48 /NUL2 8ˇˇˇˇD144¤0: However, Theorem 6.4.1 does not imply that the conditions f.x;y;´/D0; g.x;y;´/D0; x.2/D1; y.2/D1 430 Chapter 6 Vector-Valued Functions of Several Variables definexandyas functions of ´near´0D2, since @.f;g/ @.x;y/Dˇˇˇˇ6 6 /NUL2x/NUL2yˇˇˇˇ and @.f;g/ @.x;y/ˇˇˇˇˇ .1;1;2/Dˇˇˇˇ6 6 /NUL2/NUL2ˇˇˇˇD0: We close this section by observing that the functions u1,u2, . . . ,umdefined in Theo- rem6.4.1 have higher derivatives if f1;f2;:::;f mdo, and they may be obtained by differ- entiating ( 6.4.13 ), using the chain rule. (Exercise 6.4.17 ). Example 6.4.9 Suppose that uandvare functions of .x;y/ that satisfy f.x;y;u;v/Dx/NULu2/NULv2C9D0 g.x;y;u;v/Dy/NULu2Cv2/NUL10D0: Then @.f;g/ @.u;v/Dˇˇˇˇ/NUL2u/NUL2v /NUL2u 2vˇˇˇˇD/NUL8uv: From Theorem 6.4.1 , ifuv¤0, then uxD1 [email protected];g/ @.x;v/D1 8uvˇˇˇˇ1/NUL2v 0 2vˇˇˇˇD1 4u; uyD1 [email protected];g/ @.y;v/D1 8uvˇˇˇˇ0/NUL2v 1 2vˇˇˇˇD1 4u; vxD1 [email protected];g/ @.u;x/D1 8uvˇˇˇˇ/NUL2u 1 /NUL2u 0ˇˇˇˇD1 4v; vyD1 [email protected];g/ @.u;y/D1 8uvˇˇˇˇ/NUL2u 0 /NUL2u 1ˇˇˇˇD/NUL1 4v: These can be differentiated as many times as we wish. For exam ple, uxxD/NULux 4u2D/NUL1 16u3; uxyD/NULuy 4u2D/NUL1 16u3; and vyxDvx 4v2D1 16v2: Section 6.4 The Implicit Function Theorem 431 6.4 Exercises 1. Solve for UD.u;:::/ as a function of XD.x;:::/ . (a)/DC41 1 1/NUL1/NAK/DC4u v/NAK C/DC41/NUL1 2/NUL3/NAK/DC4x y/NAK D/DC40 0/NAK (b)u/NULvCwC3xC2yD0 /NULuCvCw/NULxCyD0 uCv/NULwCyD0 (c)3uCvCyDsinx uC2vCxDsiny (d)2uC2vCwC2xC2yC´D0 u/NULvC2wCx/NULyC2´D0 3uC2v/NULwC3xC2y/NUL´D0 2. Suppose that X02RnandU02Rm. Prove: IfN1is a neighborhood of .X0;U0/ inRnCm, there is a neighborhood NofX0inRnsuch that.X;U0/2N1ifX2N. 3. Let.X0;U0/be an arbitrary point in RnCm. Give an example of a function FW RnCm!Rmsuch that Fis continuously differentiable on RnCm,F.X0;U0/D0, FU.X0;U0/is singular, and the conditions F.X;U/D0andU.X0/DY0 (a) determine Uas a continuously differentiable function of Xfor all X; (b) determine Uas a continuous function of Xfor all X, but Uis not differentiable atX0; (c) do not determine Uas a function of X. 4. LetuDu.x;y/ be determined near .1;1/ by x2yuC2xy2u3/NUL3x3y3u5D0; u.1;1/D1: Findux.1;1/ anduy.1;1/ . 5. LetuDu.x;y;´/ be determined near .1;1;1/ by x2y5´2u5C2xy2u3/NUL3x3´2uD0; u.1;1;1/D1: Findux.1;1;1/ ,uy.1;1;1/ , andu´.1;1;1/ . 6. Findu.x 0;y0/,ux.x0;y0/, anduy.x0;y0/. (a)2x2Cy2CueuD6; .x 0;y0/D.1;2/ (b)u.xC1/Cx.yC2/Cy.u/NUL2/D0; .x 0;y0/D./NUL1;/NUL2/ (c)1/NULeusin.xCy/D0; .x 0;y0/D./EM=4;/EM=4/ (d)xloguCylogxCulogyD0; .x 0;y0/D.1;1/ 432 Chapter 6 Vector-Valued Functions of Several Variables 7. Findu.x 0;y0/,ux.x0;y0/, anduy.x0;y0/for all continuously differentiable func- tionsuthat satisfy the given equation near .x0;y0/. (a)2x2y4/NUL3uxy3Cu2x4y3D0;.x0;y0/D.1;1/ (b) cosucosxCsinusinyD0;.x0;y0/D.0;/EM/ 8. Suppose that UD.u;v/ is continuously differentiable with respect to .x;y;´/ and satisfies x2C4y2C´2/NUL2u2Cv2D/NUL4 .xC´/2Cu/NULvD/NUL3 and u.1;1 2;/NUL1/D/NUL2; v.1;1 2;/NUL1/D1: Find U0.1;1 2;/NUL1/. 9. Letuandvbe continuously differentiable with respect to xand satisfy uC2u2Cv2Cx2C2v/NULxD0 xuvCeusin.vCx/D0 andu.0/Dv.0/D0. Findu0.0/andv0.0/. 10. LetUD.u;v;w/ be continuously differentiable with respect to .x;y/ and satisfy x2yCxy2Cu2/NUL.vCw/2D/NUL3 exCy/NULu/NULv/NULwD/NUL2 .xCy/2CuCvCw2D3 andU.1;/NUL1/D.1;2;0/ . Find U0.1;/NUL1/. 11. Two continuously differentiable transformations UD.u;v/ of.x;y/ satisfy the system xyu/NUL4yuC9xvD0 2xy/NUL3y2Cv2D0 near.x0;y0/D.1;1/ . Find the value of each transformation and its differential matrix at.1;1/ . 12. Suppose that u,v, andware continuously differentiable functions of .x;y;´/ that satisfy the system excosyCe´cosuCevcoswCxD3 exsinyCe´sinuCevcoswD1 extanyCe´tanuCevtanwC´D0 near.x0;y0;´0/D.0;0;0/ , andu.0;0;0/Dv.0;0;0/Dw.0;0;0/D0. Find ux.0;0;0/ ,vx.0;0;0/ , andwx.0;0;0/ . Section 6.4 The Implicit Function Theorem 433 13. LetFD.f;g;h/ be continuously differentiable in a neighborhood of P0D.x0;y0;´0;u0;v0/, F.P0/D0, and @.f;g;h/ @.y;´;u/ˇˇˇˇ P0¤0: Then Theorem 6.4.1 implies that the conditions F.x;y;´;u;v/D0; y.x 0;v0/Du0; ´.x 0;v0/D´0; u.x 0;v0/Du0 determiney,´, anduas continuously differentiable functions of .x;v/ near.x0;v0/. Use Cramer’s rule to express their first partial derivatives as ratios of Jacobians. 14. Decide which pairs of the variables x,y,´,u, andvare determined as functions of the others by the system xC2yC3´CuC6vD0 2xC4yC´C2uC2vD0; and solve for them. 15. Letyandvbe continuously differentiable functions of .x;´;u/ that satisfy x2C4y2C´2/NUL2u2Cv2D/NUL4 .xC´/2Cu/NULvD/NUL3 near.x0;´0;u0/D.1;/NUL1;/NUL2/, and suppose that y.1;/NUL1;/NUL2/D1 2; v.1;/NUL1;/NUL2/D1: Findyx.1;/NUL1;/NUL2/andvu.1;/NUL1;/NUL2/. 16. Letu,v, andxbe continuously differentiable functions of .w;y/ that satisfy x2yCxy2Cu2/NUL.vCw/2D/NUL3 exCy/NULu/NULv/NULwD/NUL2 .xCy/2CuCvCw2D3 near.w0;y0/D.0;/NUL1/, and suppose that u.0;/NUL1/D1; v.0;/NUL1/D2; x.0;/NUL1/D1: Find the first partial derivatives of u,v, andxwith respect to yandwat.0;/NUL1/. 17. In addition to the assumptions of Theorem 6.4.1 , suppose that Fhas all partial derivatives of order /DC4qinS. Show that UDU.X/has all partial derivatives of order/DC4qinN. 434 Chapter 6 Vector-Valued Functions of Several Variables 18. Calculate all first and second partial derivatives at .x0;y0/D.1;1/ of the functions uandvthat satisfy x2Cy2Cu2Cv2D3 xCyCuCvD3;u.1;1/D0; v.1;1/D1: 19. Calculate all first and second partial derivatives at .x0;y0/D.1;/NUL1/of the func- tionsuandvthat satisfy u2/NULv2Dx/NULy/NUL2 2uvDxCy/NUL2;u.1;/NUL1/D/NUL1; v.1;/NUL1/D1: 20. Suppose that f1,f2, . . . ,fnare continuously differentiable functions of Xin a regionSinRn,/RSis continuously differentiable function of Uin a regionTofRn, .f1.X/;f2.X/;:::;f n.X//2T; X2S; /RS.f 1.X/;f2.X/;:::;f n.X//D0; X2S; and nX jD1/RS2 uj.U/>0; U2T: Show that @.f1;f2;:::;f n/ @.x1;x2;:::;x n/D0; X2S: CHAPTER 7 Integrals of Functions of Several Variables IN THIS CHAPTER we study the integral calculus of real-value d functions of several variables. SECTION 7.1 defines multiple integrals, first over rectangul ar parallelepipeds in Rnand then over more general sets. The discussion deals with the mu ltiple integral of a function whose discontinuities form a set of Jordan content zero, ove r a set whose boundary has Jordan content zero. SECTION 7.2 deals with evaluation of multiple integrals by m eans of iterated integrals. SECTION 7.3 begins with the definition of Jordan measurabili ty, followed by a derivation of the rule for change of content under a linear transformati on, an intuitive formulation of the rule for change of variables in multiple integrals, and fi nally a careful statement and proof of the rule. This is a complicated proof. 7.1 DEFINITION AND EXISTENCE OF THE MULTIPLE IN- TEGRAL We now consider the Riemann integral of a real-valued functi onfdefined on a subset of Rn, wheren/NAK2. Much of this development will be analogous to the developme nt in Sections 3.1–3 for nD1, but there is an important difference: for nD1, we considered integrals over closed intervals only, but for n > 1 we must consider more complicated regions of integration. To defer complications due to geome try, we first consider integrals over rectangles in Rn, which we now define. Integrals over Rectangles The S1/STXS2/STX/SOH/SOH/SOH/STXSn of subsetsS1,S2, . . . ,SnofRis the set of points .x1;x2;:::;x n/inRnsuch thatx12 S1;x22S2;:::;x n2Sn. For example, the Cartesian product of the two closed interv als 435 436 Chapter 7 Integrals of Functions of Several Variables Œa1;b1/c141/STXŒa2;b2/c141D˚.x;y/ˇˇa1/DC4x/DC4b1; a2/DC4y/DC4b2/TAB is a rectangle in R2with sides parallel to the x- andy-axes (Figure 7.1.1 ). y xa1 b1a2b2 Figure 7.1.1 The Cartesian product of three closed intervals Œa1;b1/c141/STXŒa2;b2/c141/STXŒa3;b3/c141D˚.x;y;´/ˇˇa1/DC4x/DC4b1; a2/DC4y/DC4b2; a3/DC4´/DC4b3/TAB is a rectangular parallelepiped in R3with faces parallel to the coordinate axes (Figure 7.1.2 ). z y x Figure 7.1.2 Section 7.1 Definition and Existence of the Multiple Integral 437 Definition 7.1.1 Acoordinate rectangle RinRnis the Cartesian product of nclosed intervals; that is, RDŒa1;b1/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒan;bn/c141: Thecontent ofRis V.R/D.b1/NULa1/.b2/NULa2//SOH/SOH/SOH.bn/NULan/: The numbers b1/NULa1,b2/NULa2, . . . ,bn/NULanare the edge lengths ofR. If they are equal, thenRis acoordinate cube . IfarDbrfor somer, thenV.R/D0and we say that Ris degenerate ; otherwise,Risnondegenerate . IfnD1,2, or3, thenV.R/ is, respectively, the length of an interval, the area of a rectangle, or the volume of a rectangular parallelepiped. H enceforth, “rectangle” or “cube” will always mean “coordinate rectangle” or “coordinate cub e” unless it is stated otherwise. If RDŒa1;b1/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒan;bn/c141 and PrWarDar0<a r1</SOH/SOH/SOH<a rmrDbr is a partition of Œar;br/c141,1/DC4r/DC4n, then the set of all rectangles in Rnthat can be written as Œa1;j1/NUL1;a1j1/c141/STXŒa2;j2/NUL1;a2j2/c141/STX/SOH/SOH/SOH/STXŒan;jn/NUL1;anjn/c141; 1/DC4jr/DC4mr; 1/DC4r/DC4n; is apartition ofR. We denote this partition by PDP1/STXP2/STX/SOH/SOH/SOH/STXPn (7.1.1) and define its norm to be the maximum of the norms of P1,P2, . . . ,Pn, as defined in Section 3.1; thus, kPkD maxfkP1k;kP2k;:::;kPnkg: Put another way,kPkis the largest of the edge lengths of all the subrectangles in P. Geometrically, a rectangle in R2is partitioned by drawing horizontal and vertical lines through it (Figure 7.1.3 ); inR3, by drawing planes through it parallel to the coordinate axe s. Partitioning divides a rectangle Rinto finitely many subrectangles that we can number in arbitrary order as R1,R2, . . . ,Rk. Sometimes it is convenient to write PDfR1;R2;:::;R kg rather than ( 7.1.1 ). 438 Chapter 7 Integrals of Functions of Several Variables y xa1 b1a2b2 Figure 7.1.3 IfPDP1/STXP2/STX/SOH/SOH/SOH/STXPnandP0DP0 1/STXP0 2/STX/SOH/SOH/SOH/STXP0 nare partitions of the same rectangle, then P0is arefinement ofPifP0 iis a refinement of Pi,1/DC4i/DC4n, as defined in Section 3.1. Suppose thatfis a real-valued function defined on a rectangle RinRn,PDfR1;R2;:::;R kg is a partition of R, and Xjis an arbitrary point in Rj,1/DC4j/DC4k. Then /ESCDkX jD1f.Xj/V.R j/ is aRiemann sum of foverP. Since Xjcan be chosen arbitrarily in Rj, there are infinitely many Riemann sums for a given function fover any partition PofR. The following definition is similar to Definition 3.1.1 . Definition 7.1.2 Letfbe a real-valued function defined on a rectangle RinRn. We say thatfisRiemann integrable on Rif there is a number Lwith the following property: For every/SI>0 , there is aı>0 such that j/ESC/NULLj</SI if/ESCis any Riemann sum of fover a partition PofRsuch thatkPk<ı. In this case, we say thatLis the Riemann integral of foverR, and write Z Rf.X/dXDL: IfRis degenerate, then Definition 7.1.2 implies thatR Rf.X/dXD0for any function f defined onR(Exercise 7.1.1 ). Therefore, it should be understood henceforth that whene ver we speak of a rectangle in Rnwe mean a nondegenerate rectangle, unless it is stated to the contrary. Section 7.1 Definition and Existence of the Multiple Integral 439 The integralR Rf.X/dXis also written as Z Rf.x;y/d.x;y/ .n D2/;Z Rf.x;y;´/d.x;y;´/ .n D3/; or Z Rf.x 1;x2;:::;x n/d.x 1;x2;:::;x n/(narbitrary): HeredXdoes not stand for the differential of X, as defined in Section 6.2. It merely identifiesx1,x2, . . . ,xn, the components of X, as the variables of integration. To avoid this minor inconsistency, some authors write simplyR Rfrather thanR Rf.X/dX. As in the case where nD1, we will say simply “integrable” or “integral” when we mean “Riemann integrable” or “Riemann integral.” If n/NAK2, we call the integral of Defi- nition 7.1.2 amultiple integral ; fornD2andnD3we also call them double andtriple integrals , respectively. When we wish to distinguish between multipl e integrals and the integral we studied in Chapter .nD1/, we will call the latter an ordinary integral. Example 7.1.1 FindR Rf.x;y/d.x;y/ , where RDŒa;b/c141/STXŒc;d/c141 and f.x;y/DxCy: Solution LetP1andP2be partitions of Œa;b/c141 andŒc;d/c141 ; thus, P1WaDx0<x 1</SOH/SOH/SOH<x rDbandP2WcDy0<y 1</SOH/SOH/SOH<y sDd: A typical Riemann sum of foverPDP1/STXP2is given by /ESCDrX iD1sX jD1./CANijC/DC1ij/.xi/NULxi/NUL1/.yj/NULyj/NUL1/; (7.1.2) where xi/NUL1/DC4/CANij/DC4xiandyj/NUL1/DC4/DC1ij/DC4yj: (7.1.3) The midpoints of Œxi/NUL1;xi/c141andŒyj/NUL1;yj/c141are xiDxiCxi/NUL1 2andyjDyjCyj/NUL1 2; (7.1.4) and ( 7.1.3 ) implies that j/CANij/NULxij/DC4xi/NULxi/NUL1 2/DC4kP1k 2/DC4kPk 2(7.1.5) and j/DC1ij/NULyjj/DC4yj/NULyj/NUL1 2/DC4kP2k 2/DC4kPk 2: (7.1.6) 440 Chapter 7 Integrals of Functions of Several Variables Now we rewrite ( 7.1.2 ) as /ESCDrX iD1sX jD1.xiCyj/.xi/NULxi/NUL1/.yj/NULyj/NUL1/ CrX iD1sX jD1/STX./CANij/NULxi/C./DC1ij/NULyj//ETX.xi/NULxi/NUL1/.yj/NULyj/NUL1/:(7.1.7) To findR Rf.x;y/d.x;y/ from ( 7.1.7 ), we recall that rX iD1.xi/NULxi/NUL1/Db/NULa;sX jD1.yj/NULyj/NUL1/Dd/NULc (7.1.8) (Example 3.1.1 ), and rX iD1.x2 i/NULx2 i/NUL1/Db2/NULa2;sX jD1.y2 j/NULy2 j/NUL1/Dd2/NULc2(7.1.9) (Example 3.1.2 ). Because of ( 7.1.5 ) and ( 7.1.6 ) the absolute value of the second sum in ( 7.1.7 ) does not exceed kPkrX jD1sX jD1.xi/NULxi/NUL1/.yj/NULyj/NUL1/DkPk"rX iD1.xi/NULxi/NUL1/#2 4sX jD1.yj/NULyj/NUL1/3 5 DkPk.b/NULa/.d/NULc/ (see ( 7.1.8 )), so ( 7.1.7 ) implies that ˇˇˇˇˇˇ/ESC/NULrX iD1sX jD1.xiCyj/.xi/NULxi/NUL1/.yj/NULyj/NUL1/ˇˇˇˇˇˇ/DC4kPk.b/NULa/.d/NULc/: (7.1.10) It now follows that rX iD1sX jD1xi.xi/NULxi/NUL1/.yj/NULyj/NUL1/D"rX iD1xi.xi/NULxi/NUL1/#2 4sX jD1.yj/NULyj/NUL1/3 5 D.d/NULc/rX iD1xi.xi/NULxi/NUL1/(from ( 7.1.8 )) Dd/NULc 2rX iD1.x2 i/NULx2 i/NUL1/ (from ( 7.1.4 )) Dd/NULc 2.b2/NULa2/ (from ( 7.1.9 )): Similarly, rX iD1sX jD1yj.xi/NULxi/NUL1/.yj/NULyj/NUL1/Db/NULa 2.d2/NULc2/: Section 7.1 Definition and Existence of the Multiple Integral 441 Therefore, ( 7.1.10 ) can be written as ˇˇˇˇ/ESC/NULd/NULc 2.b2/NULa2//NULb/NULa 2.d2/NULc2/ˇˇˇˇ/DC4kPk.b/NULa/.d/NULc/: Since the right side can be made as small as we wish by choosing kPksufficiently small, Z R.xCy/d.x;y/D1 2/STX.d/NULc/.b2/NULa2/C.b/NULa/.d2/NULc2//ETX: Upper and Lower Integrals The following theorem is analogous to Theorem 3.1.2 . Theorem 7.1.3 Iffis unbounded on the nondegenerate rectangle RinRn;thenfis not integrable on R: Proof We will show that if fis unbounded on R,PDfR1;R2;:::;R kgis any parti- tion ofR, andM >0 , then there are Riemann sums /ESCand/ESC0offoverPsuch that j/ESC/NUL/ESC0j/NAKM: (7.1.11) This implies that fcannot satisfy Definition 7.1.2 . (Why?) Let /ESCDkX jD1f.Xj/V.R j/ be a Riemann sum of foverP. There must be an integer iinf1;2;:::;kgsuch that jf.X//NULf.Xi/j/NAKM V.R i/(7.1.12) for some XinRi, because if this were not so, we would have jf.X//NULf.Xj/j<M V.R j/;X2Rj; 1/DC4j/DC4k: If this is so, then jf.X/jDjf.Xj/Cf.X//NULf.Xj/j/DC4jf.Xj/jCjf.X//NULf.Xj/j /DC4jf.Xj/jCM V.R j/;X2Rj; 1/DC4j/DC4k: However, this implies that jf.X/j/DC4max/SUB jf.Xj/jCM V.R j/ˇˇ1/DC4j/DC4k/ESC ;X2R; which contradicts the assumption that fis unbounded on R. 442 Chapter 7 Integrals of Functions of Several Variables Now suppose that Xsatisfies ( 7.1.12 ), and consider the Riemann sum /ESC0DnX jD1f.X0 j/V.R j/ over the same partition P, where X0 jD/SUBXj; j¤i; X; jDi: Since j/ESC/NUL/ESC0jDjf.X//NULf.Xi/jV.R i/; (7.1.12 ) implies ( 7.1.11 ). Because of Theorem 7.1.3 , we need consider only bounded functions in connection with Definition 7.1.2 . As in the case where nD1, it is now convenient to define the upper and lower integrals of a bounded function over a rectangle. T he following definition is analogous to Definition 3.1.3 . Definition 7.1.4 Iffis bounded on a rectangle RinRnandPDfR1;R2;:::;R kg is a partition of R, let MjDsup X2Rjf.X/; m jDinf X2Rjf.X/: Theupper sum offoverPis S.P/DkX jD1MjV.R j/; and the upper integral of foverR, denoted by Z Rf.X/dX; is the infimum of all upper sums. The lower sum of foverPis s.P/DkX jD1mjV.R j/; and the lower integral of foverR, denoted by Z Rf.X/dX; is the supremum of all lower sums. The following theorem is analogous to Theorem 3.1.4 . Section 7.1 Definition and Existence of the Multiple Integral 443 Theorem 7.1.5 Letfbe bounded on a rectangle Rand let Pbe a partition of R: Then (a) The upper sum S.P/offover Pis the supremum of the set of all Riemann sums of fover P: (b) The lower sum s.P/offover Pis the infimum of the set of all Riemann sums of f over P: Proof Exercise 7.1.5 . If m/DC4f.X//DC4M forXinR; then mV.R//DC4s.P//DC4S.P//DC4MV.R/I therefore,R Rf.X/dXandR Rf.X/dXexist, are unique, and satisfy the inequalities mV.R//DC4Z Rf.X/dX/DC4MV.R/ and mV.R//DC4Z Rf.X/dX/DC4MV.R/: The upper and lower integrals are also written as Z Rf.x;y/d.x;y/ andZ Rf.x;y/d.x;y/ .n D2/; Z Rf.x;y;´/d.x;y;´/ andZ Rf.x;y;´/d.x;y;´/ .n D3/; orZ Rf.x 1;x2;:::;x n/d.x 1;x2;:::;x n/ and Z Rf.x 1;x2;:::;x n/d.x 1;x2;:::;x n/ (narbitrary): Example 7.1.2 FindR Rf.x;y/d.x;y/ andR Rf.x;y/d.x;y/ , withRDŒa;b/c141/STX Œc;d/c141 and f.x;y/DxCy; as in Example 7.1.1 . Solution LetP1andP2be partitions of Œa;b/c141 andŒc;d/c141 ; thus, P1WaDx0<x 1</SOH/SOH/SOH<x rDbandP2WcDy0<y 1</SOH/SOH/SOH<y sDd: 444 Chapter 7 Integrals of Functions of Several Variables The maximum and minimum values of fon the rectangle Œxi/NUL1;xi/c141/STXŒyj/NUL1;yj/c141arexiCyj andxi/NUL1Cyj/NUL1, respectively. Therefore, S.P/DrX iD1sX jD1.xiCyj/.xi/NULxi/NUL1/.yj/NULyj/NUL1/ (7.1.13) and s.P/DrX iD1sX jD1.xi/NUL1Cyj/NUL1/.xi/NULxi/NUL1/.yj/NULyj/NUL1/: (7.1.14) By substituting xiCyjD1 2Œ.xiCxi/NUL1/C.yjCyj/NUL1/C.xi/NULxi/NUL1/C.yj/NULyj/NUL1//c141 into ( 7.1.13 ), we find that S.P/D1 2.†1C†2C†3C†4/; (7.1.15) where †1DrX iD1.x2 i/NULx2 i/NUL1/sX jD1.yj/NULyj/NUL1/D.b2/NULa2/.d/NULc/; †2DrX iD1.xi/NULxi/NUL1/sX jD1.y2 j/NULy2 j/NUL1/D.b/NULa/.d2/NULc2/; †3DrX iD1.xi/NULxi/NUL1/2sX jD1.yj/NULyj/NUL1//DC4kPk.b/NULa/.d/NULc/; †4DrX iD1.xi/NULxi/NUL1/sX jD1.yj/NULyj/NUL1/2/DC4kPk.b/NULa/.d/NULc/: Substituting these four results into ( 7.1.15 ) shows that I <S.P/<ICkPk.b/NULa/.d/NULc/; where ID.d/NULc/.b2/NULa2/C.b/NULa/.d2/NULc2/ 2: From this, we see thatZ R.xCy/d.x;y/DI: After substituting xi/NUL1Cyj/NUL1D1 2Œ.xiCxi/NUL1/C.yjCyj/NUL1//NUL.xi/NULxi/NUL1//NUL.yj/NULyj/NUL1//c141 into ( 7.1.14 ), a similar argument shows that I/NULkPk.b/NULa/.d/NULc/<s.P/<I; Section 7.1 Definition and Existence of the Multiple Integral 445 so Z R.xCy/d.x;y/DI: We now prove an analog of Lemma 3.2.1 . Lemma 7.1.6 Suppose thatjf.X/j/DC4MifXis in the rectangle RDŒa1;b1/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒan;bn/c141: LetPDP1/STXP2/STX/SOH/SOH/SOH/STXPnandP0DP0 1/STXP0 2/STX/SOH/SOH/SOH/STXP0 nbe partitions of R;whereP0 j is obtained by adding rjpartition points to Pj;1/DC4j/DC4n:Then S.P//NAKS.P0//NAKS.P//NUL2MV.R/0 @nX jD1rj bj/NULaj1 AkPk (7.1.16) and s.P//DC4s.P0//DC4s.P/C2MV.R/0 @nX jD1rj bj/NULaj1 AkPk: (7.1.17) Proof We will prove ( 7.1.16 ) and leave the proof of ( 7.1.17 ) to you (Exercise 7.1.7 ). First suppose that P0 1is obtained by adding one point to P1, andP0 jDPjfor2/DC4j/DC4n. IfPris defined by PrWarDar0<a r1</SOH/SOH/SOH<a rmrDbr; 1/DC4r/DC4n; then a typical subrectangle of Pis of the form Rj1j2/SOH/SOH/SOHjnDŒa1;j1/NUL1;a1j1/c141/STXŒa2;j2/NUL1;a2j2/c141/STX/SOH/SOH/SOH/STXŒan;jn/NUL1;anjn/c141: Letcbe the additional point introduced into P1to obtainP0 1, and suppose that a1;k/NUL1<c<a 1k: Ifj1¤k, thenRj1j2/SOH/SOH/SOHjnis common to PandP0, so the terms associated with it in S.P0/ andS.P/ cancel in the difference S.P//NULS.P0/. To analyze the terms that do not cancel, define R.1/ kj2/SOH/SOH/SOHjnDŒa1;k/NUL1;c/c141/STXŒa2;j2/NUL1;a2j2/c141/STX/SOH/SOH/SOH/STXŒan;jn/NUL1;anjn/c141; R.2/ kj2/SOH/SOH/SOHjnDŒc;a 1k/c141/STXŒa2;j2/NUL1;a2j2/c141/STX/SOH/SOH/SOH/STXŒan;jn/NUL1;anjn/c141; Mkj2/SOH/SOH/SOHjnDsup˚f.X/ˇˇX2Rkj2/SOH/SOH/SOHjn/TAB(7.1.18) and M.i/ kj2/SOH/SOH/SOHjnDsupn f.X/ˇˇX2R.i/ kj2/SOH/SOH/SOHjno ; iD1;2: (7.1.19) 446 Chapter 7 Integrals of Functions of Several Variables ThenS.P//NULS.P0/is the sum of terms of the form h Mkj2/SOH/SOH/SOHjn.a1k/NULa1;k/NUL1//NULM.1/ kj2/SOH/SOH/SOHjn.c/NULa1;k/NUL1//NULM.2/ kj2/SOH/SOH/SOHjn.a1k/NULc/i /STX.a2j2/NULa2;j2/NUL1//SOH/SOH/SOH.anjn/NULan;jn/NUL1/:(7.1.20) The terms within the brackets can be rewritten as .Mkj2/SOH/SOH/SOHjn/NULM.1/ kj2/SOH/SOH/SOHjn/.c/NULa1;k/NUL1/C.Mkj2/SOH/SOH/SOHjn/NULM.2/ kj2/SOH/SOH/SOHjn/.a1k/NULc/; (7.1.21) which is nonnegative, because of ( 7.1.18 ) and ( 7.1.19 ). Therefore, S.P0//DC4S.P/: (7.1.22) Moreover, the quantity in ( 7.1.21 ) is not greater than 2M.a 1k/NULa1;k/NUL1/, so ( 7.1.20 ) implies that the general surviving term in S.P//NULS.P0/is not greater than 2MkPk.a2j2/NULa2;j2/NUL1//SOH/SOH/SOH.anjn/NULan;jn/NUL1/: The sum of these terms as j2, . . . ,jnassume all possible values 1/DC4ji/DC4mi,2/DC4i/DC4n, is 2MkPk.b2/NULa2//SOH/SOH/SOH.bn/NULan/D2MkPkV.R/ b1/NULa1: This implies that S.P//DC4S.P0/C2MkPkV.R/ b1/NULa1: This and ( 7.1.22 ) imply ( 7.1.16 ) forr1D1andr2D/SOH/SOH/SOHDrnD0. Similarly, ifriD1for someiinf1;:::;ngandrjD0ifj¤i, then S.P//DC4S.P0/C2MkPkV.R/ bi/NULai: To obtain ( 7.1.16 ) in the general case, repeat this argument r1Cr2C/SOH/SOH/SOHCrntimes, as in the proof of Lemma 3.2.1 . Lemma 7.1.6 implies the following theorems and lemma, with proofs analo gous to the proofs of their counterparts in Section 3.2. Theorem 7.1.7 Iffis bounded on a rectangle R;then Z Rf.X/dX/DC4Z Rf.X/dX: Proof Exercise 7.1.8 . The next theorem is analogous to Theorem 3.2.3. Theorem 7.1.8 Iffis integrable on a rectangle R;then Z Rf.X/dXDZ Rf.X/dXDZ Rf.X/dX: Proof Exercise 7.1.9 . Section 7.1 Definition and Existence of the Multiple Integral 447 Lemma 7.1.9 Iffis bounded on a rectangle Rand/SI>0; there is aı>0 such that Z Rf.X/dX/DC4S.P/<Z Rf.X/dXC/SI and Z Rf.X/dX/NAKs.P/>Z Rf.X/dX/NUL/SI ifkPk<ı: Proof Exercise 7.1.10 . The next theorem is analogous to Theorem 3.2.5. Theorem 7.1.10 Iffis bounded on a rectangle Rand Z Rf.X/dXDZ Rf.X/dXDL; thenfis integrable on R;and Z Rf.X/dXDL: Proof Exercise 7.1.11 . Theorems 7.1.8 and7.1.10 imply the following theorem, which is analogous to Theo- rem3.2.6 . Theorem 7.1.11 A bounded function fis integrable on a rectangle Rif and only if Z Rf.X/dXDZ Rf.X/dX: The next theorem translates this into a test that can be conve niently applied. It is analo- gous to Theorem 3.2.7 . Theorem 7.1.12 Iffis bounded on a rectangle R;thenfis integrable on Rif and only if for every /SI>0 there is a partition PofRsuch that S.P//NULs.P/</SI: Proof Exercise 7.1.12 . Theorem 7.1.12 provides a useful criterion for integrability. The next the orem is an important application. It is analogous to Theorem 3.2.8 . Theorem 7.1.13 Iffis continuous on a rectangle RinRn;thenfis integrable on R: 448 Chapter 7 Integrals of Functions of Several Variables Proof Let/SI > 0 . Sincefis uniformly continuous on R(Theorem 5.2.14 ), there is a ı>0 such that jf.X//NULf.X0/j</SI V.R/(7.1.23) ifXandX0are inRandjX/NULX0j<ı. LetPDfR1;R2;:::;R kgbe a partition of Rwith kPk<ı=pn. Sincefis continuous on R, there are points XjandX0 jinRjsuch that f.Xj/DMjDsup X2Rjf.X/andf.X0 j/DmjDinf X2Rjf.X/ (Theorem 5.2.12 ). Therefore, S.P//NULs.P/DnX jD1.f.Xj//NULf.X0 j//V.R j/: SincekPk<ı=pn,jXj/NULX0 jj<ı, and, from ( 7.1.23 ) with XDXjandX0DX0 j, S.P//NULs.P/</SI V.R/kX jD1V.R j/D/SI: Hence,fis integrable on R, by Theorem 7.1.12 . Sets with Zero Content The next definition will enable us to establish the existence ofR Rf.X/dXin cases where fis bounded on the rectangle R, but is not necessarily continuous for all XinR. Definition 7.1.14 A subsetEofRnhas zero content if for each /SI>0 there is a finite set of rectangles T1,T2, . . . ,Tmsuch that E/SUBm[ jD1Tj (7.1.24) and mX jD1V.T j/</SI: (7.1.25) Example 7.1.3 Since the empty set is contained in every rectangle, the empt y set has zero content. If Econsists of finitely many points X1,X2, . . . , Xm, then Xjcan be enclosed in a rectangle Tjsuch that V.T j/</SI m; 1/DC4j/DC4m: Then ( 7.1.24 ) and ( 7.1.25 ) hold, soEhas zero content. Section 7.1 Definition and Existence of the Multiple Integral 449 Example 7.1.4 Any bounded set Ewith only finitely many limit points has zero con- tent. To see this, we first observe that if Ehas no limit points, then it must be finite, by the Bolzano–Weierstrass theorem (Theorem 1.3.8 ), and therefore must have zero content, by Example 7.1.3 . Now suppose that the limit points of EareX1,X2, . . . , Xm. LetR1,R2, . . . ,Rmbe rectangles such that Xi2R0 iand V.R i/</SI 2m; 1/DC4i/DC4m: (7.1.26) The set of points of Ethat are not in[m jD1Rjhas no limit points (why?) and, being bounded, must be finite (again by the Bolzano–Weierstrass th eorem). If this set contains p points, then it can be covered by rectangles R0 1,R0 2, . . . ,R0 pwith V.R0 j/</SI 2p; 1/DC4j/DC4p: (7.1.27) Now, E/SUB m[ iD1Ri![0 @p[ jD1R0 j1 A and, from ( 7.1.26 ) and ( 7.1.27 ), mX iD1V.R i/CpX jD1V.R0 j/</SI: Example 7.1.5 Iffis continuous on Œa;b/c141 , then the curve yDf.x/; a/DC4x/DC4b (7.1.28) (that is, the set˚ .x;y/ˇˇyDf.x/; a/DC4x/DC4b/TAB /, has zero content in R2. To see this, suppose that/SI>0 , and chooseı>0 such that jf.x//NULf.x0/j</SI ifx;x02Œa;b/c141 andjx/NULx0j<ı: (7.1.29) This is possible because fis uniformly continuous on Œa;b/c141 (Theorem 2.2.12 ). Let PWaDx0<x 1</SOH/SOH/SOH<x nDb be a partition of Œa;b/c141 withkPk<ı, and choose/CAN1,/CAN2, . . . ,/CANnso that xi/NUL1/DC4/CANi/DC4xi; 1/DC4i/DC4n: Then, from ( 7.1.29 ), jf.x//NULf./CAN i/j</SI ifxi/NUL1/DC4x/DC4xi: This means that every point on the curve ( 7.1.28 ) above the interval Œxi/NUL1;xi/c141is in a rect- angle with area 2/SI.x i/NULxi/NUL1/(Figure 7.1.4 ). Since the total area of these rectangles is 2/SI.b/NULa/, the curve has zero content. 450 Chapter 7 Integrals of Functions of Several Variables y xy = f(ξi) + y = f(ξi) y = f(ξi) − a b xi−1xi ξi Figure 7.1.4 The next lemma follows immediately from Definition 7.1.14 . Lemma 7.1.15 The union of finitely many sets with zero content has zero cont ent: The following theorem will enable us to define multiple integ rals over more general subsets of Rn. Theorem 7.1.16 Suppose that fis bounded on a rectangle RDŒa1;b1/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒan;bn/c141 (7.1.30) and continuous except on a subset EofRwith zero content :Thenfis integrable on R: Proof Suppose that /SI > 0 . SinceEhas zero content, there are rectangles T1,T2, . . . , Tmsuch that E/SUBm[ jD1Tj (7.1.31) and mX jD1V.T j/</SI: (7.1.32) We may assume that T1,T2, . . . ,Tmare contained in R, since, if not, their intersections withRwould be contained in R, and still satisfy ( 7.1.31 ) and ( 7.1.32 ). We may also assume that ifTis any rectangle such that T\0 @m[ jD1T0 j1 AD;;thenT\ED; (7.1.33) Section 7.1 Definition and Existence of the Multiple Integral 451 since if this were not so, we could make it so by enlarging T1,T2, . . . ,Tmslightly while maintaining ( 7.1.32 ). Now suppose that TjDŒa1j;b1j/c141/STXŒa2j;b2j/c141/STX/SOH/SOH/SOH/STXŒanj;bnj/c141; 1/DC4j/DC4m; letPi0be the partition of Œai;bi/c141(see ( 7.1.30 )) with partition points ai;bi;ai1;bi1;ai2;bi2;:::;a im;bim (these are not in increasing order), 1/DC4i/DC4n, and let P0DP10/STXP20/STX/SOH/SOH/SOH/STXPn0: ThenP0consists of rectangles whose union equals [m jD1Tjand other rectangles T0 1,T0 2, . . . ,T0 kthat do not intersect E. (We need ( 7.1.33 ) to be sure that T0 i\ED;;1/DC4i/DC4k:/ If we let BDm[ jD1TjandCDk[ iD1T0 i; thenRDB[Candfis continuous on the compact set C. IfPDfR1;R2;:::;R kgis a refinement of P0, then every subrectangle RjofPis contained entirely in Bor entirely inC. Therefore, we can write S.P//NULs.P/D†1.Mj/NULmj/V.R j/C†2.Mj/NULmj/V.R j/; (7.1.34) where†1and†2are summations over values of jfor whichRj/SUBBandRj/SUBC, respectively. Now suppose that jf.X/j/DC4M forXinR: Then †1.Mj/NULmj/V.R j//DC42M† 1V.R j/D2MmX jD1V.T j/<2M/SI; (7.1.35) from ( 7.1.32 ). Sincefis uniformly continuous on the compact set C(Theorem 5.2.14 ), there is aı>0 such thatMj/NULmj</SIifkPk<ıandRj/SUBC; hence, †2.Mj/NULmj/V.R j/</SI† 2V.R j//DC4/SIV.R/: This, ( 7.1.34 ), and ( 7.1.35 ) imply that S.P//NULs.P/<Œ2MCV.R//c141/SI ifkPk< ı andPis a refinement of P0. Therefore, Theorem 7.1.12 implies thatfis integrable on R. 452 Chapter 7 Integrals of Functions of Several Variables Example 7.1.6 The function f.x;y/D(xCy; 0/DC4x<y/DC41; 5; 0/DC4y/DC4x/DC41; is continuous on RDŒ0;1/c141/STXŒ0;1/c141 except on the line segment yDx; 0/DC4x/DC41 (Figure 7.1.5 ). Since the line segment has zero content (Example 7.1.5 ),fis integrable on R. y xf(x, y) = x + y f(x, y) = 5y = x 1 1 Figure 7.1.5 Integrals over More General Subsets of Rn We can now define the integral of a bounded function over more g eneral subsets of Rn. Definition 7.1.17 Suppose that fis bounded on a bounded subset of SofRn, and let fS.X/D(f.X/;X2S; 0; X62S:(7.1.36) LetRbe a rectangle containing S. Then the integral of foverSis defined to be Z Sf.X/dXDZ RfS.X/dX ifR RfS.X/dXexists. Section 7.1 Definition and Existence of the Multiple Integral 453 To see that this definition makes sense, we must show that if R1andR2are two rect- angles containing SandR R1fS.X/d Xexists, then so doesR R2fS.X/dX , and the two integrals are equal. The proof of this is sketched in Exercis e7.1.27 . Definition 7.1.18 IfSis a bounded subset of Rnand the integralR SdX(with inte- grandf/DC11) exists, we callR SdXthecontent (also, area ifnD2orvolume ifnD3) ofS, and denote it by V.S/ ; thus, V.S/DZ SdX: Theorem 7.1.19 Suppose that fis bounded on a bounded set Sand continuous ex- cept on a subset EofSwith zero content. Suppose also that @Shas zero content :Thenf is integrable on S: Proof LetfSbe as in ( 7.1.36 ). Since a discontinuity of fSis either a discontinuity of f or a point of@S, the set of discontinuities of fSis the union of two sets of zero content and therefore is of zero content (Lemma 7.1.15 ). Therefore, fSis integrable on any rectangle containingS(from Theorem 7.1.16 ), and consequently on S(Definition 7.1.17 ). Differentiable Surfaces Differentiable surfaces , defined as follows, form an important class of sets of zero co ntent inRn. Definition 7.1.20 Adifferentiable surface SinRn.n>1/ is the image of a compact subsetDofRm, wherem < n , under a continuously differentiable transformation GW Rm!Rn. IfmD1,Sis also called a differentiable curve . Example 7.1.7 The circle ˚.x;y/ˇˇx2Cy2D9/TAB is a differentiable curve in R2, since it is the image of DDŒ0;2/EM/c141 under the continuously differentiable transformation GWR!R2defined by XDG./DC2/D/DC43cos/DC2 3sin/DC2/NAK : Example 7.1.8 The sphere ˚.x;y;´/ˇˇx2Cy2C´2D4/TAB is a differentiable surface in R3, since it is the image of DD˚./DC2;/RS/ˇˇ0/DC4/DC2/DC42/EM;/NUL/EM=2/DC4/RS/DC4/EM=2/TAB under the continuously differentiable transformation GWR2!R3defined by XDG./DC2;/RS/D2 42cos/DC2cos/RS 2sin/DC2cos/RS 2sin/RS3 5: 454 Chapter 7 Integrals of Functions of Several Variables Example 7.1.9 The set ˚ .x1;x2;x3;x4/ˇˇxi/NAK0.iD1;2;3;4/; x 1Cx2D1; x 3Cx4D1/TAB is a differentiable surface in R4, since it is the image of DDŒ0;1/c141/STXŒ0;1/c141 under the continuously differentiable transformation GWR2!R4defined by XDG.u;v/D2 664u 1/NULu v 1/NULv3 775: Theorem 7.1.21 A differentiable surface in Rnhas zero content : Proof LetS,D, and Gbe as in Definition 7.1.20 . From Lemma 6.2.7 , there is a constantMsuch that jG.X//NULG.Y/j/DC4MjX/NULYjifX;Y2D: (7.1.37) SinceDis bounded,Dis contained in a cube CDŒa1;b1/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒam;bm/c141; where bi/NULaiDL; 1/DC4i/DC4m: Suppose that we partition CintoNmsmaller cubes by partitioning each of the intervals Œai;bi/c141intoNequal subintervals. Let R1,R2, . . . ,Rkbe the smaller cubes so produced that contain points of D, and select points X1,X2, . . . , Xksuch that Xi2D\Ri,1/DC4i/DC4k. IfY2D\Ri, then ( 7.1.37 ) implies that jG.Xi//NULG.Y/j/DC4MjXi/NULYj: (7.1.38) Since XiandYare both in the cube Riwith edge length L=N , jXi/NULYj/DC4Lpm N: This and ( 7.1.38 ) imply that jG.Xi//NULG.Y/j/DC4MLpm N; which in turn implies that G.Y/lies in a cube eRiinRncentered at G.Xi/, with sides of length2MLpm=N . Now kX iD1V.eRi/Dk/DC22MLpm N/DC3n /DC4Nm/DC22MLpm N/DC3n D.2MLpm/nNm/NULn: Sincen > m , we can make the sum on the left arbitrarily small by taking Nsufficiently large. Therefore, Shas zero content. Theorems 7.1.19 and7.1.21 imply the following theorem. Section 7.1 Definition and Existence of the Multiple Integral 455 Theorem 7.1.22 Suppose that Sis a bounded set in Rn;with boundary consisting of a finite number of differentiable surfaces :Letfbe bounded on Sand continuous except on a set of zero content. Then fis integrable on S: Example 7.1.10 Let SD˚.x;y/ˇˇx2Cy2D1; x/NAK0/TABI thus,Sis bounded by a semicircle and a line segment (Figure 7.1.6 ), both differentiable curves in R2. Let f.x;y/D(.1/NULx2/NULy2/1=2; .x;y/2S; y/NAK0; /NUL.1/NULx2/NULy2/1=2; .x;y/2S; y<0: Thenfis continous on Sexcept on the line segment yD0; 0/DC4x<1; which has zero content, from Example 7.1.5 . Hence, Theorem 7.1.22 implies thatfis integrable on S. y xx2 + y2 = 1, x ≥ 0 Figure 7.1.6 Properties of Multiple Integrals We now list some theorems on properties of multiple integral s. The proofs are similar to those of the analogous theorems in Section 3.3. Note: Because of Definition 7.1.17 , if we say that a function fis integrable on a set S, thenSis necessarily bounded. 456 Chapter 7 Integrals of Functions of Several Variables Theorem 7.1.23 Iffandgare integrable on S;then so isfCg;and Z S.fCg/.X/dXDZ Sf.X/dXCZ Sg.X/dX: Proof Exercise 7.1.20 . Theorem 7.1.24 Iffis integrable on Sandcis a constant;thencfis integrable on S;and Z S.cf/. X/dXDcZ Sf.X/dX: Proof Exercise 7.1.21 . Theorem 7.1.25 Iffandgare integrable on Sandf.X//DC4g.X/forXinS;then Z Sf.X/dX/DC4Z Sg.X/dX: Proof Exercise 7.1.22 . Theorem 7.1.26 Iffis integrable on S;then so isjfj;and ˇˇˇˇZ Sf.X/dXˇˇˇˇ/DC4Z Sjf.X/jdX: Proof Exercise 7.1.23 . Theorem 7.1.27 Iffandgare integrable on S;then so is the product fg: Proof Exercise 7.1.24 . Theorem 7.1.28 Suppose that uis continuous and vis integrable and nonnegative on a rectangleR:ThenZ Ru.X/v.X/dXDu.X0/Z Rv.X/dX for some X0inR: Proof Exercise 7.1.25 . Lemma 7.1.29 Suppose that Sis contained in a bounded set Tandfis integrable onS:ThenfS.see(7.1.36 )/is integrable on T;and Z TfS.X/dXDZ Sf.X/dX: Proof From Definition 7.1.17 withfandSreplaced byfSandT, Section 7.1 Definition and Existence of the Multiple Integral 457 .fS/T.X/D/SUBfS.X/;X2T; 0; X62T: SinceS/SUBT,.fS/TDfS. (Verify.) Now suppose that Ris a rectangle containing T. ThenRalso contains S(Figure 7.1.7 ), R T Figure 7.1.7 soZ Sf.X/dXDZ RfS.X/dX (Definition 7.1.17 , applied tofandS/ DZ R.fS/T.X/dX(since.fS/TDfS) DZ TfS.X/dX (Definition 7.1.17 , applied tofSandT/; which completes the proof. Theorem 7.1.30 Iffis integrable on disjoint sets S1andS2;thenfis integrable on S1[S2;andZ S1[S2f.X/dXDZ S1f.X/dXCZ S2f.X/dX: (7.1.39) Proof ForiD1,2, let fSi.X/D(f.X/;X2Si; 0; X62Si: From Lemma 7.1.29 withSDSiandTDS1[S2,fSiis integrable on S1[S2, and Z S1[S2fSi.X/dXDZ Sif.X/dX; iD1;2: Theorem 7.1.23 now implies that fS1CfS2is integrable on S1[S2and Z S1[S2.fS1CfS2/.X/dXDZ S1f.X/dXCZ S2f.X/dX: (7.1.40) 458 Chapter 7 Integrals of Functions of Several Variables SinceS1\S2D;, /NUL fS1CfS2/SOH .X/DfS1.X/CfS2.X/Df.X/;X2S1[S2: Therefore, ( 7.1.40 ) implies ( 7.1.39 ). We leave it to you to prove the following extension of Theorem 7.1.30 . (Exercise 7.1.31(b)). Corollary 7.1.31 Suppose that fis integrable on sets S1andS2such thatS1\S2 has zero content :Thenfis integrable on S1[S2;and Z S1[S2f.X/dXDZ S1f.X/dXCZ S2f.X/dX: Example 7.1.11 Let S1D˚ .x;y/ˇˇ0/DC4x/DC41; 0/DC4y/DC41Cx/TAB and S2D˚ .x;y/ˇˇ/NUL1/DC4x/DC40; 0/DC4y/DC41/NULx/TAB (Figure 7.1.8 ). Sy xy = 1 − x y = 1 + x 1 −1 Figure 7.1.8 Then S1\S2D˚ .0;y/ˇˇ0/DC4y/DC41/TAB has zero content. Hence, Corollary 7.1.31 implies that if fis integrable on S1andS2, then fis also integrable over SDS1[S2D˚.x;y/ˇˇ/NUL1/DC4x/DC41; 0/DC4y/DC41Cjxj/TAB (Figure 7.1.9 ), and Z S1[S2f.X/dXDZ S1f.X/dXCZ S2f.X/dX: Section 7.1 Definition and Existence of the Multiple Integral 459 y y x xy = 1 − x y = 1 + x S1S2 Figure 7.1.9 We will discuss this example further in the next section. 7.1 Exercises 1. Prove: IfRis degenerate, then Definition 7.1.2 implies thatR Rf.X/dXD0iff is bounded on R. 2. Evaluate directly from Definition 7.1.2 . (a)R R.3xC2y/d.x;y/ ;RDŒ0;2/c141/STXŒ1;3/c141 (b)R Rxyd.x;y/ ;RDŒ0;1/c141/STXŒ0;1/c141 3. Suppose thatRb af.x/dx andRd cg.y/dy exist, and letRDŒa;b/c141/STXŒc;d/c141 . Criticize the following “proof” thatR Rf.x/g.y/d.x;y/ exists and equals Zb af.x/dx! Zd cg.y/dy! : (See Exercise 7.1.30 for a correct proof of this assertion.) “Proof.” Let P1WaDx0<x 1</SOH/SOH/SOH<x rDbandP2WcDy0<y 1</SOH/SOH/SOH<y sDd be partitions of Œa;b/c141 andŒc;d/c141 , andPDP1/STXP2. Then a typical Riemann sum of fgoverPis of the form /ESCDrX iD1sX jD1f./CAN i/g./DC1 j/.xi/NULxi/NUL1/.yj/NULyj/NUL1/D/ESC1/ESC2; where /ESC1DrX iD1f./CAN i/.xi/NULxi/NUL1/and/ESC2DsX jD1g./DC1j/.yj/NULyj/NUL1/ 460 Chapter 7 Integrals of Functions of Several Variables are typical Riemann sums of foverŒa;b/c141 andgoverŒc;d/c141 . Sincefandgare integrable on these intervals, ˇˇˇˇˇ/ESC1/NULZb af.x/dxˇˇˇˇˇandˇˇˇˇˇ/ESC2/NULZd cg.y/dyˇˇˇˇˇ can be made arbitrarily small by taking kP1kandkP2ksufficiently small. From this, it is straightforward to show that ˇˇˇˇˇ/ESC/NUL Zb af.x/dx! Zd cg.y/dy!ˇˇˇˇˇ can be made arbitrarily small by taking kPksufficiently small. This implies the stated result. 4. Suppose that f.x;y//NAK0onRDŒa;b/c141/STXŒc;d/c141 . Justify the interpretation ofR Rf.x;y/d.x;y/ , if it exists, as the volume of the region in R3bounded by the surfaces´Df.x;y/ and the planes ´D0,xDa,xDb,yDc, andyDd. 5. Prove Theorem 7.1.5 . HINT:See the proof of Theorem 3.1.4: 6. Suppose that f.x;y/D8 ˆˆ< ˆˆ:0 ifxandyare rational, 1 ifxis rational and yis irrational, 2 ifxis irrational and yis rational, 3 ifxandyare irrational. Find Z Rf.x;y/d.x;y/ andZ Rf.x;y/d.x;y/ ifRDŒa;b/c141/STXŒc;d/c141: 7. Prove Eqn. ( 7.1.17 ) of Lemma 7.1.6 . 8. Prove Theorem 7.1.7 HINT:See the proof of Theorem 3.2.2: 9. Prove Theorem 7.1.8 HINT:See the proof of Theorem 3.2.3: 10. Prove Lemma 7.1.9 HINT:See the proof of Lemma 3.2.4: 11. Prove Theorem 7.1.10 HINT:See the proof of Theorem 3.2.5: 12. Prove Theorem 7.1.12 HINT:See the proof of Theorem 3.2.7: 13. Give an example of a denumerable set in R2that does not have zero content. 14. Prove: (a) IfS1andS2have zero content, then S1[S2has zero content. (b) IfS1has zero content and S2/SUBS1, thenS2has zero content. (c) IfShas zero content, then Shas zero content. 15. Show that a degenerate rectangle has zero content. Section 7.1 Definition and Existence of the Multiple Integral 461 16. Suppose that fis continuous on a compact set SinRn. Show that the surface ´Df.X/,X2S, has zero content in RnC1. HINT:See Example 7.1.5: 17. LetSbe a bounded set such that S\@Sdoes not have zero content. (a) Suppose that fis defined on Sandf.X//NAK/SUB>0 on a subsetTofS\@S that does not have zero content. Show that fis not integrable on S. (b) Conclude that V.S/ is undefined. 18. (a) Suppose that his bounded and h.X/D0except on a set of zero content. Show thatR Sh.X/dXD0for any bounded set S. (b) Suppose thatR Sf.X/dXexists,gis bounded on S, andf.X/Dg.X/except forXin a set of zero content. Show that gis integrable on Sand Z Sg.X/dXDZ Sf.X/dX: 19. Suppose that fis integrable on a set SandS0is a subset of Ssuch that@S0has zero content. Show that fis integrable on S0. 20. Prove Theorem 7.1.23 HINT:See the proof of Theorem 3.3.1: 21. Prove Theorem 7.1.24 . 22. Prove Theorem 7.1.25 HINT:See the proof of Theorem 3.3.4: 23. Prove Theorem 7.1.26 HINT:See the proof of Theorem 3.3.5: 24. Prove Theorem 7.1.27 HINT:See the proof of Theorem 3.3.6: 25. Prove Theorem 7.1.28 HINT:See the proof of Theorem 3.3.7: 26. Prove: Iffis integrable on a rectangle R, thenfis integrable on any subrectangle ofR. HINT:Use Theorem 7.1.12Isee the proof of Theorem 3.3.8: 27. Suppose that RandeRare rectangles, R/SUBeR,gis bounded on eR, andg.X/D0if X62R. (a) Show thatR eRg.X/dXexists if and only ifR Rg.X/dXexists and, in this case, Z eRg.X/dXDZ Rg.X/dX: HINT:Use Exercise 7.1.26: (b) Use(a)to show that Definition 7.1.17 is legitimate; that is, the existence and value ofR Sf.X/dXdoes not depend on the particular rectangle chosen to containS. 28. (a) Suppose that fis integrable on a rectangle RandPDfR1;R2;:::;R kgis a partition of R. Show that Z Rf.X/dXDkX jD1Z Rjf.X/dX: HINT:Use Exercise 7.1.26: 462 Chapter 7 Integrals of Functions of Several Variables (b) Use(a)to show that if fis continuous on RandPis a partition of R, then there is a Riemann sum of foverPthat equalsR Rf.X/dX. 29. Suppose that fis continuously differentiable on a rectangle R. Show that there is a constantMsuch that ˇˇˇˇ/ESC/NULZ Rf.X/dXˇˇˇˇ/DC4MkPk if/ESCis any Riemann sum of fover a partition PofR. HINT:Use Exercise 7.1.28.b/ and Theorem 5.4.5: 30. Suppose thatRb af.x/dx andRd cg.y/dy exist, and let RDŒa;b/c141/STXŒc;d/c141 . (a) Use Theorems 3.2.7 and7.1.12 to show that Z Rf.x/d.x;y/ andZ Rg.y/d.x;y/ both exist. (b) Use Theorem 7.1.27 to prove thatR Rf.x/g.y/d.x;y/ exists. (c) Justify using the argument given in Exercise 7.1.3 to show that Z Rf.x/g.y/d.x;y/D Zb af.x/dx! Zd cg.y/dy! : 31. (a) Suppose that fis integrable on SandS0is obtained by removing a set of zero content from S. Show thatfis integrable on S0andR S0f.X/dXDR Sf.X/dX. (b) Prove Corollary 7.1.31 . 7.2 ITERATED INTEGRALS AND MULTIPLE INTEGRALS Except for very simple examples, it is impractical to evalua te multiple integrals directly from Definitions 7.1.2 and7.1.17 . Fortunately, this can usually be accomplished by evalu- atingnsuccessive ordinary integrals. To motivate the method, let us first assume that fis continuous on RDŒa;b/c141/STXŒc;d/c141 . Then, for each yinŒc;d/c141 ,f.x;y/ is continuous with respect toxonŒa;b/c141 , so the integral F.y/DZb af.x;y/dx exists. Moreover, the uniform continuity of fonRimplies thatFis continuous (Exer- cise7.2.3 ) and therefore integrable on Œc;d/c141 . We say that I1DZd cF.y/dyDZd c Zb af.x;y/dx! dy Section 7.2 Iterated Integrals and Multiple Integrals 463 is an iterated integral offoverR. We will usually write it as I1DZd cdyZb af.x;y/dx: Another iterated integral can be defined by writing G.x/DZd cf.x;y/dy; a/DC4x/DC4b; and defining I2DZb aG.x/dxDZb a Zd cf.x;y/dy! dx; which we usually write as I2DZb adxZd cf.x;y/dy: Example 7.2.1 Let f.x;y/DxCy andRDŒ0;1/c141/STXŒ1;2/c141 . Then F.y/DZ1 0f.x;y/dxDZ1 0.xCy/dxD/DC2x2 2Cxy/DC3ˇˇˇˇ1 xD0D1 2Cy and I1DZ2 1F.y/dyDZ2 1/DC21 2Cy/DC3 dyD/DC2y 2Cy2 2/DC3ˇˇˇˇ2 1D2: Also, G.x/DZ2 1.xCy/dyD/DC2 xyCy2 2/DC3ˇˇˇˇ2 yD1D.2xC2//NUL/DC2 xC1 2/DC3 DxC3 2; and I2DZ1 0G.x/dxDZ1 0/DC2 xC3 2/DC3 dxD/DC2x2 2C3x 2/DC3ˇˇˇˇ1 0D2: In this example, I1DI2; moreover, on setting aD0,bD1,cD1, anddD2in Example 7.1.1 , we see thatZ R.xCy/d.x;y/D2; so the common value of the iterated integrals equals the mult iple integral. The following theorem shows that this is not an accident. 464 Chapter 7 Integrals of Functions of Several Variables Theorem 7.2.1 Suppose that fis integrable on RDŒa;b/c141/STXŒc;d/c141 and F.y/DZb af.x;y/dx exists for each yinŒc;d/c141: ThenFis integrable on Œc;d/c141; and Zd cF.y/dyDZ Rf.x;y/d.x;y/I (7.2.1) that is;Zd cdyZb af.x;y/dxDZ Rf.x;y/d.x;y/: (7.2.2) Proof Let P1WaDx0<x 1</SOH/SOH/SOH<x rDbandP2WcDy0<y 1</SOH/SOH/SOH<y sDd be partitions of Œa;b/c141 andŒc;d/c141 , and PDP1/STXP2. Suppose that yj/NUL1/DC4/DC1j/DC4yj; 1/DC4j/DC4s; (7.2.3) so /ESCDsX jD1F./DC1 j/.yj/NULyj/NUL1/ (7.2.4) is a typical Riemann sum of FoverP2. Since F./DC1 j/DZb af.x;/DC1 j/dxDrX iD1Zx xi/NUL1f.x;/DC1 j/dx; (7.2.3 ) implies that if mijDinf˚f.x;y/ˇˇxi/NUL1/DC4x/DC4xi;yj/NUL1/DC4y/DC4yj/TAB and MijDsup˚f.x;y/ˇˇxi/NUL1/DC4x/DC4xi;yj/NUL1/DC4y/DC4yj/TAB; thenrX iD1mij.xi/NULxi/NUL1//DC4F./DC1 j//DC4rX iD1Mij.xi/NULxi/NUL1/: Multiplying this by yj/NULyj/NUL1and summing from jD1tojDsyields sX jD1rX iD1mij.xi/NULxi/NUL1/.yj/NULyj/NUL1//DC4sX jD1F./DC1 j/.yj/NULyj/NUL1/ /DC4sX jD1rX iD1Mij.xi/NULxi/NUL1/.yj/NULyj/NUL1/; Section 7.2 Iterated Integrals and Multiple Integrals 465 which, from ( 7.2.4 ), can be rewritten as sf.P//DC4/ESC/DC4Sf.P/; (7.2.5) wheresf.P/andSf.P/are the lower and upper sums of fover P. Now letsF.P2/and SF.P2/be the lower and upper sums of FoverP2; since they are respectively the infimum and supremum of the Riemann sums of FoverP2(Theorem 3.1.4 ), (7.2.5 ) implies that sf.P//DC4sF.P2//DC4SF.P2//DC4Sf.P/: (7.2.6) Sincefis integrable on R, there is for each /SI > 0 a partition PofRsuch thatSf.P//NUL sf.P/ < /SI , from Theorem 7.1.12 . Consequently, from ( 7.2.6 ), there is a partition P2of Œc;d/c141 such thatSF.P2//NULsF.P2/</SI , soFis integrable on Œc;d/c141 , from Theorem 3.2.7 . It remains to verify ( 7.2.1 ). From ( 7.2.4 ) and the definition ofRd cF.y/dy , there is for each/SI>0 aı>0 such that ˇˇˇˇˇZd cF.y/dy/NUL/ESCˇˇˇˇˇ</SI ifkP2k<ıI that is, /ESC/NUL/SI<Zd cF.y/dy </ESCC/SIifkP2k<ı: This and ( 7.2.5 ) imply that sf.P//NUL/SI<Zd cF.y/dy <S f.P/C/SIifkPk<ı; and this implies that Z Rf.x;y/d.x;y//NUL/SI/DC4Zd cF.y/dy/DC4Z Rf.x;y/d.x;y/C/SI (7.2.7) (Definition 7.1.4 ). Since Z Rf.x;y/d.x;y/DZ Rf.x;y/d.x;y/ (Theorem 7.1.8 ) and/SIcan be made arbitrarily small, ( 7.2.7 ) implies ( 7.2.1 ). Iffis continuous on R, thenfsatisfies the hypotheses of Theorem 7.2.1 (Exercise 7.2.3 ), so (7.2.2 ) is valid in this case. IfR Rf.x;y/d.x;y/ and Zd cf.x;y/dy; a/DC4x/DC4b; 466 Chapter 7 Integrals of Functions of Several Variables exist, then by interchanging xandyin Theorem 7.2.1 , we see that Zb adxZd cf.x;y/dyDZ Rf.x;y/d.x;y/: This and ( 7.2.2 ) yield the following corollary of Theorem 7.2.1 . Corollary 7.2.2 Iffis integrable on Œa;b/c141/STXŒc;d/c141; then Zb adxZd cf.x;y/dyDZd cdyZb af.x;y/dx; provided thatRd cf.x;y/dy exists fora/DC4x/DC4bandRb af.x;y/dx exists forc/DC4y/DC4d: In particular;these hypotheses hold if fis continuous on Œa;b/c141/STXŒc;d/c141: Example 7.2.2 The function f.x;y/DxCy is continuous everywhere, so ( 7.2.2 ) holds for every rectangle R. For example, let RD Œ0;1/c141/STXŒ1;2/c141 . Then ( 7.2.2 ) yields Z R.xCy/d.x;y/DZ2 1dyZ1 0.xCy/dxDZ2 1"/DC2x2 2Cxy/DC3ˇˇˇˇ1 xD0# dy DZ2 1/DC21 2Cy/DC3 dyD/DC2y 2Cy2 2/DC3ˇˇˇˇ2 1D2: Sincefalso satisfies the hypotheses of Theorem 7.2.1 withxandyinterchanged, we can calculate the double integral from the iterated integra l in which the integrations are performed in the opposite order; thus, Z R.xCy/d.x;y/DZ1 0dxZ2 1.xCy/dyDZ1 0"/DC2 xyCy2 2/DC3ˇˇˇˇ2 yD1# dx DZ1 0/DC2 xC3 2/DC3 dxD/DC2x2 2C3x 2/DC3ˇˇˇˇ1 0D2: A plausible partial converse of Theorem 7.2.1 would be that ifRd cdyRb af.x;y/dx exists then so doesR Rf.x;y/d.x;y/ ; however, the next example shows that this need not be so. Example 7.2.3 Iffis defined on RDŒ0;1/c141/STXŒ0;1/c141 by f.x;y/D/SUB2xy ifyis rational; y ifyis irrational; Section 7.2 Iterated Integrals and Multiple Integrals 467 thenZ1 0f.x;y/dxDy; 0/DC4y/DC41; andZ1 0dyZ1 0f.x;y/dxDZ1 0ydyD1 2: However,fis not integrable on R(Exercise 7.2.7 ). The next theorem generalizes Theorem 7.2.1 toRn. Theorem 7.2.3 LetI1;I2;. . .;Inbe closed intervals and suppose that fis integrable onRDI1/STXI2/STX/SOH/SOH/SOH/STXIn:Suppose that there is an integer pinf1;2;:::;n/NUL1gsuch that Fp.xpC1;xpC2;:::;x n/DZ I1/STXI2/STX/SOH/SOH/SOH/STX Ipf.x 1;x2;:::;x n/d.x 1;x2;:::;x p/ exists for each .xpC1;xpC2;:::;x n/inIpC1/STXIpC2/STX/SOH/SOH/SOH/STXIn:Then Z IpC1/STXIpC2/STX/SOH/SOH/SOH/STX InFp.xpC1;xpC2;:::;x n/d.x pC1;xpC2;:::;x n/ exists and equalsR Rf.X/dX. Proof For convenience, denote .xpC1;xpC2;:::;x n/byY. DenotebRDI1/STXI2/STX/SOH/SOH/SOH/STX IpandTDIpC1/STXIpC2/STX/SOH/SOH/SOH/STXIn. LetbPDfbR1;bR2;:::;bRkgandQDfT1;T2;:::;T sg be partitions of bRandT, respectively. Then the collection of rectangles of the for mbRi/STXTj (1/DC4i/DC4k,1/DC4j/DC4s) is a partition PofR; moreover, every partition PofRis of this form. Suppose that Yj2Tj; 1/DC4j/DC4s; (7.2.8) so /ESCDsX jD1Fp.Yj/V.T j/ (7.2.9) is a typical Riemann sum of Fpover Q. Since Fp.Yj/DZ bRf.x 1;x2;:::;x p;Yj/d.x 1;x2;:::;x p/ DkX jD1Z bRjf.x 1;x2;:::;x p;Yj/d.x 1;x2;:::;x p/; (7.2.8 ) implies that if mijDinfn f.x 1;x2;:::;x p;Y/ˇˇ.x1;x2;:::;x p/2bRi;Y2Tjo and MijDsupn f.x 1;x2;:::;x p;Y/ˇˇ.x1;x2;:::;x p/2bRi;Y2Tjo ; 468 Chapter 7 Integrals of Functions of Several Variables then kX iD1mijV.bRi//DC4Fp.Yj//DC4kX iD1MijV.bRi/: Multiplying this by V.T j/and summing from jD1tojDsyields sX jD1kX iD1mijV.bRi/V.T j//DC4sX jD1Fp.Yj/V.T j//DC4sX jD1kX iD1MijV.bRi/V.T j/; which, from ( 7.2.9 ), can be rewritten as sf.P//DC4/ESC/DC4Sf.P/; (7.2.10) wheresf.P/andSf.P/are the lower and upper sums of fover P. Now letsFp.Q/and SFp.Q/be the lower and upper sums of Fpover Q; since they are respectively the infimum and supremum of the Riemann sums of Fpover Q(Theorem 7.1.5 ), (7.2.10 ) implies that sf.P//DC4sFp.Q//DC4SFp.Q//DC4Sf.P/: (7.2.11) Sincefis integrable on R, there is for each /SI > 0 a partition PofRsuch thatSf.P//NUL sf.P/</SI , from Theorem 7.1.12 . Consequently, from ( 7.2.11 ), there is a partition QofT such thatSFp.Q//NULsFp.Q/</SI , soFpis integrable on T, from Theorem 7.1.12 . It remains to verify that Z Rf.X/dXDZ TFp.Y/dY: (7.2.12) From ( 7.2.9 ) and the definition ofR TFp.Y/dY, there is for each /SI>0 aı>0 such that ˇˇˇˇZ TFp.Y/dY/NUL/ESCˇˇˇˇ</SI ifkQk<ıI that is, /ESC/NUL/SI<Z TFp.Y/dY</ESCC/SIifkQk<ı: This and ( 7.2.10 ) imply that sf.P//NUL/SI<Z TFp.Y/dY<S f.P/C/SIifkPk<ı; and this implies that Z Rf.X/dX/NUL/SI/DC4Z TFp.Y/dY/DC4Z Rf.X/dXC/SI: (7.2.13) SinceZ Rf.X/dXDZ Rf.X/dX(Theorem 7.1.8 ) and/SIcan be made arbitrarily small, (7.2.13 ) implies ( 7.2.12 ). Section 7.2 Iterated Integrals and Multiple Integrals 469 Theorem 7.2.4 LetIjDŒaj;bj/c141;1/DC4j/DC4n, and suppose that fis integrable on RDI1/STXI2/STX/SOH/SOH/SOH/STXIn:Suppose also that the integrals Fp.xpC1;:::;x n/DZ I1/STXI2/SOH/SOH/SOH/STXIpf.X/d.x 1;x2;:::;x p/; 1/DC4p/DC4n/NUL1; exist for all .xpC1;:::;x n/inIpC1/STX/SOH/SOH/SOH/STXIn: Then the iterated integral Zbn andxnZbn/NUL1 an/NUL1dxn/NUL1/SOH/SOH/SOHZb2 a2dx2Zb1 a1f.X/dx 1 exists and equalsR Rf.X/dX: Proof The proof is by induction. From Theorem 7.2.1 , the proposition is true for nD2. Now assume n>2 and the proposition is true with nreplaced byn/NUL1. Holdingxnfixed and applying this assumption yields Fn.xn/DZbn/NUL1 an/NUL1dxn/NUL1Zbn/NUL2 an/NUL2dxn/NUL2/SOH/SOH/SOHZb2 a2dx2Zb1 a1f.X/dx 1: Now Theorem 7.2.3 withpDn/NUL1completes the induction. Example 7.2.4 LetRDŒ0;1/c141/STXŒ1;2/c141/STXŒ0;1/c141 and f.x;y;´/DxCyC´: Then F1.y;´/DZ1 0.xCyC´/dxD/DC2x2 2CxyCx´/DC3ˇˇˇˇ1 xD0D1 2CyC´; F2.´/DZ2 1F1.y;´/dyDZ2 1/DC21 2CyC´/DC3 dy D/DC2y 2Cy2 2Cy´/DC3ˇˇˇˇ2 yD1D2C´; and Z Rf.x;y;´/d.x;y;´/ DZ1 0F2.´/d´DZ1 0.2C´/d´D/DC2 2´C´2 2/DC3ˇˇˇˇ1 0D5 2: The hypotheses of Theorems 7.2.3 and 7.2.4 are stated so as to justify successive in- tegrations with respect to x1, thenx2, thenx3, and so forth. It is legitimate to use other orders of integration if the hypotheses are adjusted accord ingly. For example, suppose that 470 Chapter 7 Integrals of Functions of Several Variables fi1;i2;:::;i ngis a permutation off1;2;:::;ngandR Rf.X/dXexists, along with Z Ii1/STXIi2/STX/SOH/SOH/SOH/STX Iijf.X/d.x i1;xi2;:::;x ij/; 1/DC4j/DC4n/NUL1; (7.2.14) for each .xijC1;xijC2;:::;x in/inIijC1/STXIijC2/STX/SOH/SOH/SOH/STXIin: (7.2.15) Then, by renaming the variables, we infer from Theorem 7.2.4 that Z Rf.X/dXDZbin aindxinZbin/NUL1 ain/NUL1dxin/NUL1/SOH/SOH/SOHZbi2 ai2dxi2Zbi1 ai1f.X/dx i1: (7.2.16) Since there are nŠpermutations off1;2;:::;ng, there arenŠways of evaluating a mul- tiple integral over a rectangle in Rn, provided that the integrand satisfies appropriate hy- potheses. In particular, if fis continuous on Randfi1;i2;:::;i ngis any permutation of f1;2;:::;ng, thenfis continuous with respect to .xi1;xi2;:::;x ij/onIi1/STXIi2/STX/SOH/SOH/SOH/STXIij for each fixed .xijC1;xijC2;:::;x in/satisfying ( 7.2.15 ). Therefore, the integrals ( 7.2.14 ) exist for every permutation of f1;2;:::;ng(Theorem 7.1.13 ). We summarize this in the next theorem, which now follows from Theorem 7.2.4 . Theorem 7.2.5 Iffis continuous on RDŒa1;b1/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒan;bn/c141; thenR Rf.X/dXcan be evaluated by iterated integrals in any of the nŠways indicated in (7.2.16 ): Example 7.2.5 Iffis continuous on RDŒa1;b1/c141/STXŒa2;b2/c141/STXŒa3;b3/c141, then Z Rf.x;y;´/d.x;y;´/ DZb3 a3d´Zb2 a2dyZb1 a1f.x;y;´/dx DZb2 a2dyZb3 a3d´Zb1 a1f.x;y;´/dx DZb3 a3d´Zb1 a1dxZb2 a2f.x;y;´/dy DZb1 a1dxZb3 a3d´Zb2 a2f.x;y;´/dy DZb2 a2dyZb1 a1dxZb3 a3f.x;y;´/d´ DZb1 a1dxZb2 a2dyZb3 a3f.x;y;´/d´: Section 7.2 Iterated Integrals and Multiple Integrals 471 Integrals over More General Sets We now consider the problem of evaluating multiple integral s over more general sets. First, suppose thatfis integrable on a set of the form SD˚.x;y/ˇˇu.y//DC4x/DC4v.y/; c/DC4y/DC4d/TAB(7.2.17) (Figure 7.2.1 ). Ifu.y//NAKaandv.y//DC4bforc/DC4y/DC4d, and fS.x;y/D(f.x;y/; .x;y/2S; 0; .x;y/62S;(7.2.18) then Z Sf.x;y/d.x;y/DZ RfS.x;y/d.x;y/; whereRDŒa;b/c141/STXŒc;d/c141 .. From Theorem 7.2.1 , Z RfS.x;y/d.x;y/DZd cdyZb afS.x;y/dx provided thatRb afS.x;y/dx exists for each yinŒc;d/c141 . From ( 7.2.17 ) and ( 7.2.18 ), this integral can be written asZv.y/ u.y/f.x;y/dx: (7.2.19) Thus, we have proved the following theorem. y xb ax = v(y) y = cy = d x = u(y) Figure 7.2.1 Theorem 7.2.6 Iffis integrable on the set Sin(7.2.17 )and the integral (7.2.19 ) exists forc/DC4y/DC4d;then Z Sf.x;y/d.x;y/DZd cdyZv.y/ u.y/f.x;y/dx: (7.2.20) 472 Chapter 7 Integrals of Functions of Several Variables From Theorem 7.1.22 , the assumptions of Theorem 7.2.6 are satisfied if fis continuous onSanduandvare continuously differentiable on Œc;d/c141 . Interchanging xandyin Theorem 7.2.6 shows that if fis integrable on SD˚ .x;y/ˇˇu.x//DC4y/DC4v.x/; a/DC4x/DC4b/TAB (7.2.21) (Figure 7.2.2 ) andZv.x/ u.x/f.x;y/dy exists fora/DC4x/DC4b, then Z Sf.x;y/d.x;y/DZb adxZv.x/ u.x/f.x;y/dy: (7.2.22) y x a by = v(x) y = u(x)S Figure 7.2.2 Example 7.2.6 Suppose that f.x;y/Dxy andSis the region bounded by the curves xDy2andxDy(Figure 7.2.3 ). SinceScan be represented in the form ( 7.2.17 ) as SD˚.x;y/ˇˇy2/DC4x/DC4y; 0/DC4y/DC41/TAB; (7.2.20 ) yieldsZ Sxyd.x;y/DZ1 0dyZy y2xydx; which, incidentally, can be written as Z Sxyd.x;y/DZ1 0ydyZy y2xdx; Section 7.2 Iterated Integrals and Multiple Integrals 473 sinceyis independent of x. Evaluating the iterated integral yields Z Sxyd.x;y/DZ1 0 x2 2ˇˇˇˇy y2! ydyD1 2Z1 0.y3/NULy5/dy D1 2/DC2y4 4/NULy6 6/DC3ˇˇˇˇ1 0D1 24: y xx = y2x = y (1, 1) S Figure 7.2.3 In this case we can also represent Sin the form ( 7.2.21 ) as SD˚ .x;y/ˇˇx/DC4y/DC4px; 0/DC4x/DC41/TAB I hence, from ( 7.2.22 ), Z Sxyd.x;y/DZ1 0xdxZpx xydyDZ1 0 y2 2ˇˇˇˇpx yDx! xdx D1 2Z1 0.x2/NULx3/dxD1 2/DC2x3 3/NULx4 4/DC3ˇˇˇˇ1 0D1 24: Example 7.2.7 To evaluate Z S.xCy/d.x;y/; where SD˚ .x;y/ˇˇ/NUL1/DC4x/DC41; 0/DC4y/DC41Cjxj/TAB 474 Chapter 7 Integrals of Functions of Several Variables (see Example 7.1.11 and Figure 7.2.4 ), Sy xy = 1 − x y = 1 + x 1 −1 Figure 7.2.4 we invoke Corollary 7.1.31 and write Z S.xCy/d.x;y/DZ S1.xCy/d.x;y/CZ S2.xCy/d.x;y/; where S1D˚.x;y/ˇˇ0/DC4x/DC41; 0/DC4y/DC41Cx/TAB and S2D˚.x;y/ˇˇ/NUL1/DC4x/DC40; 0/DC4y/DC41/NULx/TAB (Figure 7.2.5 ). From Theorem 7.2.6 , Z S1.xCy/d.x;y/DZ1 0dxZ1Cx 0.xCy/dyDZ1 0" .xCy/2 2ˇˇˇˇ1Cx yD0# dx D1 2Z1 0/STX.2xC1/2/NULx2/ETXdx D1 2/DC4.2xC1/3 6/NULx3 3/NAKˇˇˇˇ1 0D2 andZ S2.xCy/d.x;y/DZ0 /NUL1dxZ1/NULx 0.xCy/dyDZ0 /NUL1" .xCy/2 2ˇˇˇˇ1/NULx yD0# dx D1 2Z0 /NUL1.1/NULx2/dxD1 2/DC2 x/NULx3 3/DC3ˇˇˇˇ0 /NUL1D1 3: Therefore, Z S.xCy/d.x;y/D2C1 3D7 3: Section 7.2 Iterated Integrals and Multiple Integrals 475 y y x xy = 1 − x y = 1 + x S1S2 Figure 7.2.5 Example 7.2.8 To find the area Aof the region bounded by the curves yDx2C1andyD9/NULx2 (Figure 7.2.6 ), we evaluate ADZ Sd.x;y/; where SD˚.x;y/ˇˇx2C1/DC4y/DC49/NULx2;/NUL2/DC4x/DC42/TAB: According to Theorem 7.2.6 , ADZ2 /NUL2dxZ9/NULx2 x2C1dyDZ2 /NUL2/STX.9/NULx2//NUL.x2C1//ETXdx DZ2 /NUL2.8/NUL2x2/dxD/DC2 8x/NUL2x3 3/DC3ˇˇˇˇ2 /NUL2D64 3: y xy = x2 + 1 y = 9 − x2(2, 5) (−2, 5) S Figure 7.2.6 476 Chapter 7 Integrals of Functions of Several Variables Theorem 7.2.6 has an analog for n>2 . Suppose that fis integrable on a set Sof points XD.x1;x2;:::;x n/satisfying the inequalities uj.xjC1;:::;x n//DC4xj/DC4vj.xjC1;:::;x n/; 1/DC4j/DC4n/NUL1; and an/DC4xn/DC4bn: Then, under appropriate additional assumptions, it can be s hown by an argument analogous to the one that led to Theorem 7.2.6 that Z Sf.X/dXDZbn andxnZvn.xn/ un.xn/dxn/NUL1/SOH/SOH/SOHZv2.x3;:::;x n/ u2.x3;:::;x n/dx2Zv1.x2;:::;x n/ u1.x2;:::;x n/f.X/dx 1: These additional assumptions are tedious to state for gener aln. The following theorem contains a complete statement for nD3. Theorem 7.2.7 Suppose that fis integrable on SD˚.x;y;´/ˇˇu1.y;´//DC4x/DC4v1.y;´/; u 2.´//DC4y/DC4v2.´/; c/DC4´/DC4d/TAB; and let S.´/D˚ .x;y/ˇˇu1.y;´//DC4x/DC4v1.y;´/; u 2.´//DC4y/DC4v2.´//TAB for each´inŒc;d/c141: Then Z Sf.x;y;´/d.x;y;´/ DZd cd´Zv2.´/ u2.´/dyZv1.y;´/ u1.y;´/f.x;y;´/dx; provided thatZv1.y;´/ u1.y;´/f.x;y;´/dx exists for all.y;´/ such that c/DC4´/DC4dandu2.´//DC4y/DC4v2.´/; and Z S.´/f.x;y;´/d.x;y/ exists for all´inŒc;d/c141: Example 7.2.9 Suppose that fis continuous on the region SinR3bounded by the coordinate planes and the plane xCyC2´D2 (Figure 7.2.7 ); thus, Section 7.2 Iterated Integrals and Multiple Integrals 477 y xz x + y + 2z = 1 Figure 7.2.7 SD˚ .x;y;´/ˇˇ0/DC4x/DC42/NULy/NUL2´; 0/DC4y/DC42/NUL2´; 0/DC4´/DC41/TAB : From Theorem 7.2.7 , Z Sf.x;y;´/d.x;y;´/ DZ1 0d´Z2/NUL2´ 0dyZ2/NULy/NUL2´ 0f.x;y;´/dx: There are five other iterated integrals that equal the multip le integral. We leave it to you to verify that Z Sf.x;y;´/d.x;y;´/ DZ2 0dyZ1/NULy=2 0d´Z2/NULy/NUL2´ 0f.x;y;´/dx DZ1 0d´Z2/NUL2´ 0dxZ2/NULx/NUL2´ 0f.x;y;´/dy DZ2 0dxZ1/NULx=2 0d´Z2/NULx/NUL2´ 0f.x;y;´/dy DZ2 0dxZ2/NULx 0dyZ1/NULx=2/NULy=2 0f.x;y;´/d´ DZ2 0dyZ2/NULy 0dxZ1/NULx=2/NULy=2 0f.x;y;´/d´ (Exercise 7.2.15 ). Thus far we have viewed the iterated integral as a tool for eva luating multiple integrals. In some problems the iterated integral is itself the object o f interest. In this case a result 478 Chapter 7 Integrals of Functions of Several Variables like Theorem 7.2.6 can be used to evaluate the iterated integral. The procedure is as follows. (a) Express the given iterated integral as a multiple integral, and check to see that the multiple integral exists. (b) Look for another iterated integral that equals the multiple integral and is easier to evaluate than the given one. The two iterated integrals must be equal, by Theo- rem7.2.6 . This procedure is called changing the order of integration of an iterated integral. Example 7.2.10 The iterated integral IDZ1 0dyZy 0e/NUL.x/NUL1/2dx is hard to evaluate because e/NUL.x/NUL1/2has no elementary antiderivative. The set of points .x;y/ that enter into the integration, which we call the region of integration , is SD˚ .x;y/ˇˇ0/DC4x/DC4y; 0/DC4y/DC41/TAB (Figure 7.2.8 ). y xy = x 1 1 Figure 7.2.8 Therefore, IDZ Se/NUL.x/NUL1/2d.x;y/; (7.2.23) and this multiple integral exists because its integrand is c ontinuous. Since Scan also be written as SD˚.x;y/ˇˇx/DC4y/DC41; 0/DC4x/DC41/TAB; Section 7.2 Iterated Integrals and Multiple Integrals 479 Theorem 7.2.6 implies that Z Se/NUL.x/NUL1/2d.x;y/DZ1 0e/NUL.x/NUL1/2dxZ1 xdyD/NULZ1 0.x/NUL1/e/NUL.x/NUL1/2dx D1 2e/NUL.x/NUL1/2ˇˇˇˇ1 0D1 2.1/NULe/NUL1/: This and ( 7.2.23 ) imply that ID1 2.1/NULe/NUL1/: Example 7.2.11 Suppose that fis continuous on Œa;1/andysatisfies the differen- tial equation y00.x/Df.x/; x>a; (7.2.24) with initial conditions y.a/Dy0.a/D0: Integrating ( 7.2.24 ) yields y0.x/DZx af.t/dt; sincey0.a/D0. Integrating this yields y.x/DZx adsZs af.t/dt; sincey.a/D0. This can be reduced to a single integral as follows. Since th e function g.s;t/Df.t/ is continuous for all .s;t/ such thatt/NAKa,gis integrable on SD˚.s;t/ˇˇa/DC4t/DC4s; a/DC4s/DC4x/TAB (Figure 7.2.9 ), and Theorem 7.2.6 implies that Z Sf.t/d.s;t/DZx adsZs af.t/dtDy.x/: (7.2.25) However,Scan also be described as SD˚.s;t/ˇˇt/DC4s/DC4x; a/DC4t/DC4x/TAB so Theorem 7.2.6 implies that Z Sf.t/d.s;t/DZx af.t/dtZx tdsDZx a.x/NULt/f.t/dt: Comparing this with ( 7.2.25 ) yields y.x/DZx a.x/NULt/f.t/dt: 480 Chapter 7 Integrals of Functions of Several Variables t xsaas = t S Figure 7.2.9 7.2 Exercises 1. Evaluate (a)Z2 0dyZ1 /NUL1.xC3y/dx (b)Z2 1dxZ1 0.x3Cy4/dy (c)Z2/EM /EM=2xdxZ2 1sinxydy (d)Zlog2 0ydyZ1 0xex2ydx 2. LetIjDŒaj;bj/c141,1/DC4j/DC43, and suppose that fis integrable on RDI1/STXI2/STXI3. Prove: (a) If the integral G.y;´/DZb1 a1f.x;y;´/dx exists for.y;´/2I2/STXI3, thenGis integrable on I2/STXI3and Z Rf.x;y;´/d.x;y;´/ DZ I2/STXI3G.y;´/d.y;´/: (b) If the integral H.´/DZ I1/STXI2f.x;y;´/d.x;y/ Section 7.2 Iterated Integrals and Multiple Integrals 481 exists for´2I3, thenHis integrable on I3and Z Rf.x;y;´/d.x;y;´/ DZb3 a3H.´/d´: HINT:For both parts ;see the proof of Theorem 7.2.1: 3. Prove: Iffis continuous on Œa;b/c141/STXŒc;d/c141 , then the function F.y/DZb af.x;y/dx is continuous on Œc;d/c141 . HINT:Use Theorem 5.2.14: 4. Suppose that f.x0;y0//NAKf.x;y/ ifa/DC4x/DC4x0/DC4b; c/DC4y/DC4y0/DC4d: Show thatfsatisfies the hypotheses of Theorem 7.2.1 onRDŒa;b/c141/STXŒc;d/c141 . HINT: See the proof of Theorem 3.2.9: 5. Evaluate by means of iterated integrals: (a)Z R.xyC1/d.x;y/ ;RDŒ0;1/c141/STXŒ1;2/c141 (b)Z R.2xC3y/d.x;y/ ;RDŒ1;3/c141/STXŒ1;2/c141 (c)Z Rxyp x2Cy2d.x;y/ ;RDŒ0;1/c141/STXŒ0;1/c141 (d)R Rxcosxycos2/EMxd.x;y/ ;RDŒ0;1 4/c141/STXŒ0;2/EM/c141 6. LetAbe the set of points of the form .2/NULmp;2/NULmq/, wherepandqare odd integers andmis a nonnegative integer. Let f.x;y/D( 1; .x;y/62A; 0; .x;y/2A: Show thatfis not integrable on any rectangle RDŒa;b/c141/STXŒc;d/c141 , but Zb adxZd cf.x;y/dyDZd cdyZb af.x;y/dxD.b/NULa/.d/NULc/: . A/ HINT:For(A);use Theorem 3.5.6 and Exercise 3.5.6: 7. Let f.x;y/D/SUB2xy ifyis rational; y ifyis irrational; andRDŒ0;1/c141/STXŒ0;1/c141 (Example 7.2.3 ). 482 Chapter 7 Integrals of Functions of Several Variables (a) CalculateR Rf.x;y/d.x;y/ andR Rf.x;y/d.x;y/ , and show that fis not integrable on R. (b) CalculateR1 0/DLER1 0f.x;y/dy/DC1 dxandR1 0/DLER1 0f.x;y/dy/DC1 dx. 8. LetRDŒ0;1/c141/STXŒ0;1/c141/STXŒ0;1/c141 ,eRDŒ0;1/c141/STXŒ0;1/c141 , and f.x;y;´/D8 ˆˆ< ˆˆ:2xyC2x´ ifyand´are rational; yC2x´ ifyis irrational and ´is rational; 2xyC´ ifyis rational and ´is irrational; yC´ ifyand´are irrational: Calculate (a)Z Rf.x;y;´/d.x;y;´/ andZ Rf.x;y;´/d.x;y;´/ (b)Z eRf.x;y;´/d.x;y/ andZ eRf.x;y;´/d.x;y/ (c)Z1 0dyZ1 0f.x;y;´/dx andZ1 0d´Z1 0dyZ1 0f.x;y;´/dx . 9. Suppose that fis bounded on RDŒa;b/c141/STXŒc;d/c141 . Prove: (a)Z Rf.x;y/d.x;y//DC4Zb a Zd cf.x;y/dy! dx. HINT:Use Exercise 3.2.6(a): (b)Z Rf.x;y/d.x;y//NAKZb a Zd cf.x;y/dy! dx. HINT:Use Exercise 3.2.6(b): 10. Use Exercise 7.2.9 to prove the following generalization of Theorem 7.2.1 : Iffis integrable on RDŒa;b/c141/STXŒc;d/c141 , then Zb af.x;y/dy andZd cf.x;y/dy are integrable on Œa;b/c141 , and Zb a Zd cf.x;y/dy! dxDZb a Zd cf.x;y/dy! dxDZ Rf.x;y/d.x;y/: 11. Evaluate (a)Z R.x/NUL2yC3´/d.x;y;´/ ;RDŒ/NUL2;0/c141/STXŒ2;5/c141/STXŒ/NUL3;2/c141 (b)Z Re/NULx2/NULy2sinxsin´d.x;y;´/ ;RDŒ/NUL1;1/c141/STXŒ0;2/c141/STXŒ0;/EM=2/c141 (c)Z R.xyC2x´Cy´/d.x;y;´/ ;RDŒ/NUL1;1/c141/STXŒ0;1/c141/STXŒ/NUL1;1/c141 Section 7.2 Iterated Integrals and Multiple Integrals 483 (d)Z Rx2y3´exy2´2d.x;y;´/ ;RDŒ0;1/c141/STXŒ0;1/c141/STXŒ0;1/c141 12. Evaluate (a)Z S.2xCy2/d.x;y/ ;SD˚ .x;y/ˇˇ0/DC4x/DC49/NULy2;/NUL3/DC4y/DC43/TAB (b)Z S2xyd.x;y/ ;Sis bounded by yDx2andxDy2 (c)Z Sexsiny yd.x;y/ ;SD˚.x;y/ˇˇlogy/DC4x/DC4log2y; /EM=2/DC4y/DC4/EM/TAB 13. EvaluateR S.xCy/d.x;y/ , whereSis bounded by yDx2andyD2x, using iterated integrals of both possible types. 14. Find the area of the set bounded by the given curves. (a)yDx2C9,yDx2/NUL9,xD/NUL1,xD1 (b)yDxC2,yD4/NULx,xD0 (c)xDy2/NUL4,xD4/NULy2 (d)yDe2x,yD/NUL2x,xD3 15. In Example 7.2.9 , verify the last five representations ofR Sf.x;y;´/d.x;y;´/ as iterated integrals. 16. LetSbe the region in R3bounded by the coordinate planes and the plane xC 2yC3´D1. Letfbe continuous on S. Set up six iterated integrals that equalR Sf.x;y;´/d.x;y;´/ . 17. Evaluate (a)Z Sxd.x;y;´/ ;Sis bounded by the coordinate planes and the plane 3xCyC´D2. (b)Z Sye´d.x;y;´/ ;SD˚ .x;y;´/ˇˇ0/DC4x/DC41;0/DC4y/DC4px;0/DC4´/DC4y2/TAB (c)Z Sxy´d.x;y;´/ ; SDn .x;y;´/ˇˇ0/DC4y/DC41; 0/DC4x/DC4p 1/NULy2; 0/DC4´/DC4p x2Cy2o (d)Z Sy´d.x;y;´/ ;SD˚ .x;y;´/ˇˇ´2/DC4x/DC4p´; 0/DC4y/DC4´; 0/DC4´/DC41/TAB 18. Find the volume of S. (a)Sis bounded by the surfaces ´Dx2Cy2and´D8/NULx2/NULy2. (b)SDf.x;y;´/j0/DC4´/DC4x2Cy2; .x;y;0/ is in the triangle with vertices .0;1;0/ ,.0;0;0/ , and.1;0;0/ } (c)SD˚.x;y;´/ˇˇ0/DC4y/DC4x2; 0/DC4x/DC42; 0/DC4´/DC4y2/TAB (d)SD˚.x;y;´/ˇˇx/NAK0; y/NAK0; 0/DC4´/DC44/NUL4x2/NUL4y2/TAB 484 Chapter 7 Integrals of Functions of Several Variables 19. LetRDŒa1;b2/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒan;bn/c141. Evaluate (a)R R.x1Cx2C/SOH/SOH/SOHCxn/dX (b)R R.x2 1Cx2 2C/SOH/SOH/SOHCx2 n/dX (c)R Rx1x2;/SOH/SOH/SOHxndX 20. Assuming that fis continuous, express Z1 1=2dyZp 1/NULy2 /NULp 1/NULy2f.x;y/dx as an iterated integral with the order of integration revers ed. 21. EvaluateR S.xCy/d.x;y/ of Example 7.2.7 by means of iterated integrals in which the first integration is with respect to x. 22. EvaluateZ1 0xdxZp 1/NULx2 0dyp x2Cy2: 23. Suppose that fis continuous on Œa;1/, y.n/.x/Df.x/; t/NAKa; andy.a/Dy0.a/D/SOH/SOH/SOHDy.n/NUL1/.a/D0. (a) Integrate repeatedly to show that y.x/DZx adtnZtn adtn/NUL1/SOH/SOH/SOHZt3 adt2Zt2 af.t1/dt1: . A/ (b) By successive reversals of orders of integration as in Examp le7.2.11 , deduce from (A) that y.x/D1 .n/NUL1/ŠZx a.x/NULt/n/NUL1f.t/dt: 24. LetT/SUBDŒ0;/SUB/c141/STXŒ0;/SUB/c141;/SUB>0 . By calculating I.a/Dlim /SUB!1Z T/SUBe/NULxysinaxd.x;y/ in two different ways, show that Z1 0sinax xdxD/EM 2ifa>0: 7.3 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS In Section 3.3 we saw that a change of variables may simplify t he evaluation of an ordinary integral. We now consider change of variables in multiple in tegrals. Section 7.3 Change of Variables in Multiple Integrals 485 Prior to formulating the rule for change of variables, we mus t deal with some rather involved preliminary considerations. Jordan Measurable Sets In Section we defined the content of a set Sto be V.S/DZ SdX (7.3.1) if the integral exists. If Ris a rectangle containing S, then ( 7.3.1 ) can be rewritten as V.S/DZ R S.X/dX; where Sis the characteristic function of S, defined by S.X/D/SUB1;X2S; 0;X62S: From Exercise 7.1.27 , the existence and value of V.S/ do not depend on the particular choice of the enclosing rectangle R. We say that SisJordan measurable ifV.S/ exists. ThenV.S/ is the Jordan content of S. We leave it to you (Exercise 7.3.2 ) to show that Shas zero content according to Defini- tion7.1.14 if and only if Shas Jordan content zero. Theorem 7.3.1 A bounded set Sis Jordan measurable if and only if the boundary of Shas zero content : Proof LetRbe a rectangle containing S. Suppose that V.@S/D0. Since Sis bounded onRand discontinuous only on @S(Exercise 2.2.9 ), Theorem 7.1.19 implies thatR R S.X/dXexists. For the converse, suppose that @Sdoes not have zero content and letPDfR1;R2;:::;R kgbe a partition of R. For eachjinf1;2;:::;kgthere are three possibilities: 1.Rj/SUBS; then min˚ S.X/ˇˇX2Rj/TABDmax˚ S.X/ˇˇX2Rj/TABD1: 2.Rj\S¤; andRj\Sc¤;; then min˚ S.X/ˇˇX2Rj/TAB D0and max˚ S.X/ˇˇX2Rj/TAB D1: 3.Rj/SUBSc; then min˚ S.X/ˇˇX2Rj/TAB Dmax˚ S.X/ˇˇX2Rj/TAB D0: 486 Chapter 7 Integrals of Functions of Several Variables Let U1D˚ jˇˇRj/SUBS/TAB and U2D˚ jˇˇRj\S¤; andRj\Sc¤;/TAB : (7.3.2) Then the upper and lower sums of SoverPare S.P/DX j2U1V.R j/CX j2U2V.R j/ Dtotal content of the subrectangles in Pthat intersect S(7.3.3) and s.P/DX j2U1V.R j/ Dtotal content of the subrectangles in Pcontained inS:(7.3.4) Therefore, S.P//NULs.P/DX j2U2V.R j/; which is the total content of the subrectangles in Pthat intersect both SandSc. Since these subrectangles contain @S, which does not have zero content, there is an /SI0>0such that S.P//NULs.P//NAK/SI0 for every partition PofR. By Theorem 7.1.12 , this implies that Sis not integrable on R, soSis not Jordan measurable. Theorems 7.1.19 and7.3.1 imply the following corollary. Corollary 7.3.2 Iffis bounded and continuous on a bounded Jordan measurable set S;thenfis integrable on S: Lemma 7.3.3 Suppose that Kis a bounded set with zero content and /SI;/SUB >0: Then there are cubes C1;C2;. . .;Crwith edge lengths </SUBsuch thatCj\K¤;;1/DC4j/DC4r; K/SUBr[ jD1Cj; (7.3.5) andrX jD1V.C j/</SI: Proof SinceV.K/D0,Z C K.X/dXD0 ifCis any cube containing K. From this and the definition of the integral, there is a ı>0 such that ifPis any partition of CwithkPk/DC4ıand/ESCis any Riemann sum of Kover P, then 0/DC4/ESC/DC4/SI: (7.3.6) Section 7.3 Change of Variables in Multiple Integrals 487 Now suppose that PDfC1;C2;:::;C kgis a partition of Cinto cubes with kPk<min./SUB;ı/; (7.3.7) and letC1,C2, . . . ,Ckbe numbered so that Cj\K¤; if1/DC4j/DC4randCj\KD; ifrC1/DC4j/DC4k. Then ( 7.3.5 ) holds, and a typical Riemann sum of KoverPis of the form /ESCDrX jD1 K.Xj/V.C j/ with Xj2Cj,1/DC4j/DC4r. In particular, we can choose XjfromK, so that K.Xj/D1, and /ESCDrX jD1V.C j/: Now ( 7.3.6 ) and ( 7.3.7 ) imply thatC1,C2, . . . ,Crhave the required properties. Transformations of Jordan-Measurable Sets To formulate the theorem on change of variables in multiple i ntegrals, we must first con- sider the question of preservation of Jordan measurability under a regular transformation. Lemma 7.3.4 Suppose that GWRn!Rnis continuously differentiable on a bounded open setS;and letKbe a closed subset of Swith zero content :Then G.K/ has zero content. Proof SinceKis a compact subset of the open set S, there is a/SUB1> 0 such that the compact set K/SUB1D˚ Xˇˇdist.X;K//DC4/SUB1/TAB is contained in S(Exercise 5.1.26). From Lemma 6.2.7 , there is a constant Msuch that jG.Y//NULG.X/j/DC4MjY/NULXjifX;Y2K/SUB1: (7.3.8) Now suppose that /SI > 0 . SinceV.K/D0, there are cubes C1,C2, . . . ,Crwith edge lengthss1,s2, . . . ,sr</SUB 1=pnsuch thatCj\K¤;,1/DC4j/DC4r, K/SUBr[ jD1Cj; andrX jD1V.C j/</SI (7.3.9) (Lemma 7.3.3 ). For1/DC4j/DC4r, letXj2Cj\K. IfX2Cj, then jX/NULXjj/DC4sjpn</SUB 1; 488 Chapter 7 Integrals of Functions of Several Variables soX2KandjG.X//NULG.Xj/j/DC4MjX/NULXjj/DC4Mpnsj, from ( 7.3.8 ). Therefore, G.Cj/ is contained in a cube eCjwith edge length 2Mpnsj, centered at G.Xj/. Since V.eCj/D.2Mpn/nsn jD.2Mpn/nV.C j/; we now see that G.K//SUBr[ jD1eCj and rX jD1V.eCj//DC4.2Mpn/nrX jD1V.C j/<.2Mpn/n/SI; where the last inequality follows from ( 7.3.9 ). Since.2Mpn/ndoes not depend on /SI, it follows thatV.G.K//D0. Theorem 7.3.5 Suppose that GWRn!Rnis regular on a compact Jordan measur- able setS:Then G.S/is compact and Jordan measurable : Proof We leave it to you to prove that G.S/is compact (Exercise 6.2.23). Since S is Jordan measurable, V.@S/D0, by Theorem 7.3.1 . Therefore, V.G.@S//D0, by Lemma 7.3.4 . But G.@S/[email protected]// (Exercise 6.3.23 ), soV.@. G.S///D0, which implies that G.S/is Jordan measurable, again by Theorem 7.3.1 . Change of Content Under a Linear Transformation To motivate and prove the rule for change of variables in mult iple integrals, we must know howV.L.S// is related toV.S/ ifSis a compact Jordan measurable set and Lis a nonsin- gular linear transformation. (From Theorem 7.3.5 ,L.S/is compact and Jordan measurable in this case.) The next lemma from linear algebra will help to establish this relationship. We omit the proof. Lemma 7.3.6 A nonsingular n/STXnmatrix Acan be written as ADEkEk/NUL1/SOH/SOH/SOHE1; (7.3.10) where each Eiis a matrix that can be obtained from the n/STXnidentity matrix Iby one of the following operations W (a) interchanging two rows of II (b) multiplying a row of Iby a nonzero constant I (c) adding a multiple of one row of Ito another: Matrices of the kind described in this lemma are called elementary matrices. The key to the proof of the lemma is that if Eis an elementary n/STXnmatrix and Ais anyn/STXnmatrix, then EAis the matrix obtained by applying to Athe same operation that must be applied toIto produce E(Exercise 7.3.6 ). Also, the inverse of an elementary matrix of type (a), (b), or(c)is an elementary matrix of the same type (Exercise 7.3.7 ). The next example illustrates the procedure for finding the fa ctorization ( 7.3.10 ). Section 7.3 Change of Variables in Multiple Integrals 489 Example 7.3.1 The matrix AD2 40 1 1 1 0 1 2 2 03 5 is nonsingular, since det .A/D4. Interchanging the first two rows of Ayields A1D2 41 0 1 0 1 1 2 2 03 5DbE1A; where bE1D2 40 1 0 1 0 0 0 0 13 5: Subtracting twice the first row of A1from the third yields A2D2 41 0 1 0 1 1 0 2/NUL23 5DbE2bE1A; where bE2D2 41 0 0 0 1 0 /NUL2 0 13 5: Subtracting twice the second row of A2from the third yields A3D2 41 0 1 0 1 1 0 0/NUL43 5DbE3bE2bE1A; where bE3D2 41 0 0 0 1 0 0/NUL2 13 5: Multiplying the third row of A3by/NUL1 4yields A4D2 41 0 1 0 1 1 0 0 13 5DbE4bA3bE2bE1A; where bE4D2 41 0 0 0 1 0 0 0/NUL1 43 5: 490 Chapter 7 Integrals of Functions of Several Variables Subtracting the third row of A4from the first yields A5D2 41 0 0 0 1 1 0 0 13 5DbE5bA4bE3bE2bE1A; where bE5D2 41 0/NUL1 0 1 0 0 0 13 5: Finally, subtracting the third row of A5from the second yields IDbE6bE5bE4bE3bE2bE1A; (7.3.11) where bE6D2 41 0 0 0 1/NUL1 0 0 13 5: From ( 7.3.11 ) and Theorem 6.1.16 , AD.bE6bE5bE4bE3bE2bE1//NUL1DbE/NUL1 1bE/NUL1 2bE/NUL1 3bE/NUL1 4bE/NUL1 5bE/NUL1 6: Therefore, ADE6E5E4E3E2E1; where E1DbE/NUL1 6D2 41 0 0 0 1 1 0 0 13 5, E2DbE/NUL1 5D2 41 0 1 0 1 0 0 0 13 5, E3DbA/NUL1 4D2 41 0 0 0 1 0 0 0/NUL43 5,E4DbE/NUL1 3D2 41 0 0 0 1 0 0 2 13 5, E5DbE/NUL1 2D2 41 0 0 0 1 0 2 0 13 5, E6DbE/NUL1 1D2 40 1 0 1 0 0 0 0 13 5 (Exercise 7.3.7(c)). Lemma 7.3.6 and Theorem 6.1.7(c)imply that an arbitrary invertible linear transforma- tionLWRn!Rn, defined by XDL.Y/DAY; (7.3.12) can be written as a composition LDLkıLk/NUL1ı/SOH/SOH/SOHı L1; (7.3.13) where Li.Y/DEiY; 1/DC4i/DC4k: Section 7.3 Change of Variables in Multiple Integrals 491 Theorem 7.3.7 IfSis a compact Jordan measurable subset of RnandLWRn!Rn is the invertible linear transformation XDL.Y/DAY;then V.L.S//Djdet.A/jV.S/: (7.3.14) Proof Theorem 7.3.5 implies that L.S/is Jordan measurable. If V.L.R//Djdet.A/jV.R/ (7.3.15) wheneverRis a rectangle, then ( 7.3.14 ) holds ifSis any compact Jordan measurable set. To see this, suppose that /SI > 0 , letRbe a rectangle containing S, and letPD fR1;R2;:::;R kgbe a partition of Rsuch that the upper and lower sums of SoverP satisfy the inequality S.P//NULs.P/</SI: (7.3.16) LetU1andU2be as in ( 7.3.2 ). From ( 7.3.3 ) and ( 7.3.4 ), s.P/DX j2U1V.R j//DC4V.S//DC4X j2U1V.R j/CX j2U2V.R j/DS.P/: (7.3.17) Theorem 7.3.7 implies that L.R1/,L.R2/, . . . , L.Rk/andL.S/are all Jordan measurable. Since [ j2U1Rj/SUBS/SUB[ j2S1[S2Rj; it follows that L0 @[ j2U1Rj1 A/SUBL.S//SUBL0 @[ j2S1[S2Rj1 A: SinceLis one-to-one on Rn, this implies that X j2U1V.L.Rj///DC4V.L.S///DC4X j2U1V.L.Rj//CX j2U2V.L.Rj//: (7.3.18) If we assume that ( 7.3.15 ) holds whenever Ris a rectangle, then V.L.Rj//Djdet.A/jV.R j/; 1/DC4j/DC4k; so (7.3.18 ) implies that s.P//DC4V.L.S// jdet.A/j/DC4S.P/: This, ( 7.3.16 ) and ( 7.3.17 ) imply that ˇˇˇˇV.S//NULV.L.S// jdet.A/jˇˇˇˇ</SII hence, since/SIcan be made arbitrarily small, ( 7.3.14 ) follows for any Jordan measurable set. 492 Chapter 7 Integrals of Functions of Several Variables To complete the proof, we must verify ( 7.3.15 ) for every rectangle RDŒa1;b1/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒan;bn/c141DI1/STXI2/STX/SOH/SOH/SOH/STXIn: Suppose that Ain (7.3.12 ) is an elementary matrix; that is, let XDL.Y/DEY: CASE 1. If Eis obtained by interchanging the ith andjth rows of I, then xrD8 < :yrifr¤iandr¤jI yjifrDiI yiifrDj: Then L.R/is the Cartesian product of I1,I2, . . . ,InwithIiandIjinterchanged, so V.L.R//DV.R/Djdet.E/jV.R/ since det.E/D/NUL1in this case (Exercise 7.3.7(a)). CASE 2. If Eis obtained by multiplying the rth row of Ibya, then xrD/SUByrifr¤i; ayiifrDi: Then L.R/DI1/STX/SOH/SOH/SOH/STXIi/NUL1/STXI0 i/STXIiC1/STX/SOH/SOH/SOH/STXIn; whereI0 iis an interval with length equal to jajtimes the length of Ii, so V.L.R//DjajV.R/Djdet.E/jV.R/ since det.E/Dain this case (Exercise 7.3.7(a)). CASE 3. If Eis obtained by adding atimes thejth row of Ito itsith row (j¤i), then xrD/SUByr ifr¤iI yiCayjifrDi: Then L.R/D˚ .x1;x2;:::;x n/ˇˇaiCaxj/DC4xi/DC4biCaxjandar/DC4xr/DC4brifr¤i/TAB ; which is a parallelogram if nD2and a parallelepiped if nD3(Figure 7.3.1 ). Now V.L.R//DZ L.R/dX; which we can evaluate as an iterated integral in which the firs t integration is with respect toxi. For example, if iD1, then V.L.R//DZbn andxnZbn/NUL1 an/NUL1dxn/NUL1/SOH/SOH/SOHZb2 a2dx2Zb1Caxj a1Caxjdx1: (7.3.19) Section 7.3 Change of Variables in Multiple Integrals 493 SinceZb1Caxj a1Caxjdy1DZb1 a1dy1; (7.3.19 ) can be rewritten as V.L.R//DZbn andxnZbn/NUL1 an/NUL1dxn/NUL1/SOH/SOH/SOHZb2 a2dx2Zb1 a1dx1 D.bn/NULan/.bn/NUL1/NULan/NUL1//SOH/SOH/SOH.b1/NULa1/DV.R/: Hence,V.L.R//Djdet.E/jV.R/ , since det.E/D1in this case (Exercise 7.3.7(a)). a1b1y1 y1y2y3b2a2y2 i = 1, j = 2, a > 0 i = 2, j = 3, a > 0 Figure 7.3.1 From what we have shown so far, ( 7.3.14 ) holds if Ais an elementary matrix and Sis any compact Jordan measurable set. If Ais an arbitrary nonsingular matrix, 494 Chapter 7 Integrals of Functions of Several Variables then we can write Aas a product of elementary matrices ( 7.3.10 ) and apply our known result successively to L1,L2, . . . , Lk(see ( 7.3.13 )). This yields V.L.S//Djdet.Ek/jjdet.Ek/NUL1/j/SOH/SOH/SOHj detE1jV.S/Djdet.A/jV.S/; by Theorem 6.1.9 and induction. Formulation of the Rule for Change of Variables We now formulate the rule for change of variables in a multipl e integral. Since we are for the present interested only in “discovering” the rule, we wi ll make any assumptions that ease this task, deferring questions of rigor until the proof . Throughout the rest of this section it will be convenient to t hink of the range and domain of a transformation GWRn!Rnas subsets of distinct copies of Rn. We will denote the copy containing DGasEn, and write GWEn!RnandXDG.Y/, reversing the usual roles of XandY. IfGis regular on a subset SofEn, then each XinG.S/can be identified by specifying the unique point YinSsuch that XDG.Y/. Suppose that we wish to evaluateR Tf.X/dX, whereTis the image of a compact Jordan measurable set Sunder the regular transformation XDG.Y/. For simplicity, we take Sto be a rectangle and assume that fis continuous on TDG.S/. Now suppose that PDfR1;R2;:::;R kgis a partition of SandTjDG.Rj/(Fig- ure7.3.2 ). Tj TRj Sy x uv X = G(U) Figure 7.3.2 ThenZ Tf.X/dXDkX jD1Z Tjf.X/dX (7.3.20) (Corollary 7.1.31 and induction). Since fis continuous, there is a point XjinTjsuch that Z Tjf.X/dXDf.Xj/Z TjdXDf.Xj/V.T j/ Section 7.3 Change of Variables in Multiple Integrals 495 (Theorem 7.1.28 ), so ( 7.3.20 ) can be rewritten as Z Tf.X/dXDkX jD1f.Xj/V.T j/: (7.3.21) Now we approximate V.T j/. If XjDG.Yj/; (7.3.22) then Yj2Rjand, since Gis differentiable at Yj, G.Y//EMG.Yj/CG0.Yj/.Y/NULYj/: (7.3.23) Here GandY/NULYjare written as column matrices, G0is a differential matrix, and “ /EM” means “approximately equal” in a sense that we could make pre cise if we wished (Theo- rem6.2.2 ). It is reasonable to expect that the Jordan content of G.Rj/is approximately equal to the Jordan content of A.Rj/, where Ais the affine transformation A.Y/DG.Yj/CG0.Yj/.Y/NULYj/ on the right side of ( 7.3.23 ); that is, V.G.Rj///EMV.A.Rj//: (7.3.24) We can think of the affine transformation Aas a composition ADA3ıA2ıA1, where A1.Y/DY/NULYj; A2.Y/DG0.Yj/Y; and A3.Y/DG.Yj/CY: LetR0 jDA1.Rj/. Since A1merely shifts Rjto a different location, R0 jis also a rectangle, and V.R0 j/DV.R j/: (7.3.25) Now letR00 jDA2.R0 j/. (In general, R00 jis not a rectangle.) Since A2is the linear transfor- mation with nonsingular matrix G0.Yj/, Theorem 7.3.7 implies that V.R00 j//DjdetG0.Yj/jV.R0 j/DjJG.Yj/jV.R j/; (7.3.26) whereJGis the Jacobian of G. Now letR000 jDA3.R00 j/. Since A3merely shifts all points in the same way, V.R000 j/DV.R00 j/: (7.3.27) Now ( 7.3.24 )–(7.3.27 ) suggest that V.T j//EMjJG.Yj/jV.R j/: 496 Chapter 7 Integrals of Functions of Several Variables (Recall thatTjDG.Rj/.) Substituting this and ( 7.3.22 ) into ( 7.3.21 ) yields Z Tf.X/dX/EMkX jD1f.G.Yj//jJG.Yj/jV.R j/: But the sum on the right is a Riemann sum for the integral Z Sf.G.Y//jJG.Y/jdY; which suggests that Z Tf.X/dXDZ Sf.G.Y//jJG.Y/jdY: We will prove this by an argument that was published in the American Mathematical Monthly [Vol. 61 (1954), pp. 81-85] by J. Schwartz. The Main Theorem We now prove the following form of the rule for change of varia ble in a multiple integral. Theorem 7.3.8 Suppose that GWEn!Rnis regular on a compact Jordan measur- able setSandfis continuous on G.S/: Then Z G.S/f.X/dXDZ Sf.G.Y//jJG.Y/jdY: (7.3.28) Since the proof is complicated, we break it down to a series of lemmas. We first observe that both integrals in ( 7.3.28 ) exist, by Corollary 7.3.2 , since their integrands are continu- ous. (Note that Sis compact and Jordan measurable by assumption, and G.S/is compact and Jordan measurable by Theorem 7.3.5 .) Also, the result is trivial if V.S/D0, since then V.G.S//D0by Lemma 7.3.4 , and both integrals in ( 7.3.28 ) vanish. Hence, we assume thatV.S/>0 . We need the following definition. Definition 7.3.9 IfADŒaij/c141is ann/STXnmatrix;then max8 < :nX jD1jaijjˇˇ1/DC4i/DC4n9 = ; is the infinity norm of A;denoted bykAk1. Lemma 7.3.10 Suppose that GWEn!Rnis regular on a cube CinEn;and let Abe a nonsingular n/STXnmatrix:Then V.G.C///DC4jdet.A/j/STXmax˚kA/NUL1G0.Y/k1ˇˇY2C/TAB/ETXnV.C/: (7.3.29) Section 7.3 Change of Variables in Multiple Integrals 497 Proof Letsbe the edge length of C. Let Y0D.c1;c2;:::;c n/be the center of C, and suppose that HD.y1;y2;:::;y n/2C. IfHD.h1;h2;:::;h n/is continuously differen- tiable onC, then applying the mean value theorem (Theorem 5.4.5 ) to the components of Hyields hi.Y//NULhi.Y0/DnX [email protected]/ @yj.yj/NULcj/; 1/DC4i/DC4n; where Yi2C. Hence, recalling that H0.Y/D/DC4@hi @yj/NAKn i;jD1; applying Definition 7.3.9 , and noting thatjyj/NULcjj/DC4s=2,1/DC4j/DC4n, we infer that jhi.Y//NULhi.Y0/j/DC4s 2max˚ kH0.Y/k1ˇˇY2C/TAB ; 1/DC4i/DC4n: This means that H.C/is contained in a cube with center X0DH.Y0/and edge length smax˚kH0.Y/k1ˇˇY2C/TAB: Therefore, V.H.C///DC4/STXmaxfkH0.Y/k1/c141nˇˇY2C/TABsn D/STX maxfkH0.Y/k1/c141nˇˇY2C/TAB V.C/:(7.3.30) Now let L.X/DA/NUL1X and set HDLıG; then H.C/DL.G.C// and H0DA/NUL1G0; so (7.3.30 ) implies that V.L.G.C////DC4/STX max˚ kA/NUL1G0.Y/k1ˇˇY2C/TAB/ETXnV.C/: (7.3.31) Since Lis linear, Theorem 7.3.7 with Areplaced by A/NUL1implies that V.L.G.C///Djdet.A//NUL1jV.G.C//: This and ( 7.3.31 ) imply that jdet.A/NUL1/jV.G.C///DC4/STX max˚ kA/NUL1G0.Y/k1ˇˇY2C/TAB/ETXnV.C/: Since det.A/NUL1/D1=det.A/, this implies ( 7.3.29 ). Lemma 7.3.11 IfGWEn!Rnis regular on a cube CinRn;then V.G.C///DC4Z CjJG.Y/jdY: (7.3.32) 498 Chapter 7 Integrals of Functions of Several Variables Proof LetPbe a partition of Cinto subcubes C1,C2, . . . ,Ckwith centers Y1,Y2, . . . , Yk. Then V.G.C//DkX jD1V.G.Cj//: (7.3.33) Applying Lemma 7.3.10 toCjwith ADG0.Aj/yields V.G.Cj///DC4jJG.Yj/j/STXmax˚k.G0.Yj///NUL1G0.Y/k1ˇˇY2Cj/TAB/ETXnV.C j/: (7.3.34) Exercise 6.1.22 implies that if /SI>0 , there is aı>0 such that max˚k.G0.Yj///NUL1G0.Y/k1ˇˇY2Cj/TAB<1C/SI; 1/DC4j/DC4k; ifkPk<ı: Therefore, from ( 7.3.34 ), V.G.Cj///DC4.1C/SI/njJG.Yj/jV.C j/; so (7.3.33 ) implies that V.G.C///DC4.1C/SI/nkX jD1jJG.Yj/jV.C j/ifkPk<ı: Since the sum on the right is a Riemann sum forR CjJG.Y/jdYand/SIcan be taken arbi- trarily small, this implies ( 7.3.32 ). Lemma 7.3.12 Suppose that Sis Jordan measurable and /SI;/SUB > 0: Then there are cubesC1; C2;. . .; CrinSwith edge lengths < /SUB; such thatCj/SUBS; 1/DC4j/DC4r; C0 i\C0 jD; ifi¤j;and V.S//DC4rX jD1V.C j/C/SI: (7.3.35) Proof SinceSis Jordan measurable, Z C S.X/dXDV.S/ ifCis any cube containing S. From this and the definition of the integral, there is a ı>0 such that ifPis any partition of CwithkPk< ıand/ESCis any Riemann sum of Sover P, then/ESC >V.S//NUL/SI=2. Therefore, if s.P/ is the lower sum of Sover P, then s.P/>V.S//NUL/SIifkPk<ı: (7.3.36) Now suppose that PD fC1;C2;:::;C kgis a partition of Cinto cubes withkPk< min./SUB;ı/ , and letC1,C2, . . . ,Ckbe numbered so that Cj/SUBSif1/DC4j/DC4rand Cj\Sc¤; ifj > r . From ( 7.3.4 ),s.P/DPr jD1V.C k/. This and ( 7.3.36 ) imply (7.3.35 ). Clearly,C0 i\C0 jD; ifi¤j. Section 7.3 Change of Variables in Multiple Integrals 499 Lemma 7.3.13 Suppose that GWEn!Rnis regular on a compact Jordan measur- able setSandfis continuous and nonnegative on G.S/: Let Q.S/DZ G.S/f.X/dX/NULZ Sf.G.Y//jJG.Y/jdY: (7.3.37) ThenQ.S//DC40: Proof From the continuity of JGandfon the compact sets SandG.S/, there are constantsM1andM2such that jJG.Y/j/DC4M1ifY2S (7.3.38) and jf.X/j/DC4M2ifX2G.S/ (7.3.39) (Theorem 5.2.11 ). Now suppose that /SI > 0 . SincefıGis uniformly continuous on S (Theorem 5.2.14 ), there is aı>0 such that jf.G.Y///NULf.G.Y0//j</SI ifjY/NULY0j<ıandY;Y02S: (7.3.40) Now letC1,C2, . . . ,Crbe chosen as described in Lemma 7.3.12 , with/SUBDı=pn. Let S1D8 < :Y2SˇˇY…r[ jD1Cj9 = ;: ThenV.S 1/</SI and SD0 @r[ jD1Cj1 A[S1: (7.3.41) Suppose that Y1,Y2, . . . , Yrare points inC1,C2, . . . ,CrandXjDG.Yj/,1/DC4j/DC4r. From ( 7.3.41 ) and Theorem 7.1.30 , Q.S/DZ G.S1/f.X/dX/NULZ S1f.G.Y//jJG.Y/jdY CrX jD1Z G.Cj/f.X/dX/NULrX jD1Z Cjf.G.Y//jJG.Y/jdY DZ G.S1/f.X/dX/NULZ S1f.G.Y//jJG.Y/jdY CrX jD1Z G.Cj/.f.X//NULf.Aj//dX CrX jD1Z Cj..f.G.Yj///NULf.G.Y///jJ.G.Y/jdY CrX jD1f.Xj/ V.G.Cj///NULZ CjjJG.Y/jdY! : 500 Chapter 7 Integrals of Functions of Several Variables Sincef.X//NAK0,Z S1f.G.Y//jJG.Y/jdY/NAK0; and Lemma 7.3.11 implies that the last sum is nonpositive. Therefore, Q.S//DC4I1CI2CI3; (7.3.42) where I1DZ G.S1/f.X/dX; I 2DrX jD1Z G.Cj/jf.X//NULf.Xj/jdX; and I3DrX jD1Z Cjjf.G/.Yj///NULf.G.Y//jjJG.Y/jdY: We will now estimate these three terms. Suppose that /SI>0 . To estimateI1, we first remind you that since Gis regular on the compact set S,Gis also regular on some open set OcontainingS(Definition 6.3.2 ). Therefore, since S1/SUBS andV.S 1/</SI ,S1can be covered by cubes T1,T2, . . . ,Tmsuch that rX jD1V.T j/</SI (7.3.43) andGis regular onSm jD1Tj. Now, I1/DC4M2V.G.S1// (from ( 7.3.39 )) /DC4M2mX jD1V.G.Tj// . sinceS1/SUB[m jD1Tj/ /DC4M2mX jD1Z TjjJG.Y/jdY(from Lemma 7.3.11 ) /DC4M2M1/SI (from ( 7.3.38 ) and ( 7.3.43 )): To estimateI2, we note that if XandXjare in G.Cj/then XDG.Y/andXjDG.Yj/ for some YandYjinCj. Since the edge length of Cjis less thanı=pn, it follows that jY/NULYjj<ı, sojf.X//NULf.Xj/j</SI, by ( 7.3.40 ). Therefore, I2</SIrX jD1V.G.Cj// /DC4/SIrX jD1Z CjjJG.Y/jdY(from Lemma 7.3.11 ) /DC4/SIM1rX jD1V.C j/ (from ( 7.3.38 )/ /DC4/SIM1V.S/ . since[r jD1Cj/SUBS/: Section 7.3 Change of Variables in Multiple Integrals 501 To estimateI3, we note again from ( 7.3.40 ) thatjf.G.Yj///NULf.G.Y//j< /SI ifYand Yjare inCj. Hence, I3</SIrX jD1Z CjjJG.Y/jdY /DC4M1/SIrX jD1V.C j/(from ( 7.3.38 ) /DC4M1V.S//SI becauseSr jD1Cj/SUBSandC0 i\C0 jD; ifi¤j. From these inequalities on I1,I2, andI3, (7.3.42 ) now implies that Q.S/<M 1.M2C2V.S///SI: Since/SIis an arbitrary positive number, it now follows that Q.S//DC40. Lemma 7.3.14 Under the assumptions of Lemma 7.3.13;Q.S//NAK0: Proof Let G1DG/NUL1; S 1DG.S/; f 1D.jJGj/fıG; (7.3.44) and Q1.S1/DZ G1.S1/f1.Y/dY/NULZ S1f1.G1.X//jJG1.X/jdX: (7.3.45) Since G1is regular on S1(Theorem 6.3.3 ) andf1is continuous and nonnegative on G1.S1/DS, Lemma 7.3.13 implies thatQ1.S1//DC40. However, substituting from ( 7.3.44 ) into ( 7.3.45 ) and again noting that G1.S1/DSyields Q1.S1/DZ Sf.G.Y//jJG.Y/jdY /NULZ G.S/f.G.G/NUL1.X///jJG.G/NUL1.X//jjJG/NUL1.X/jdX:(7.3.46) Since G.G/NUL1.X//DX,f.G.G/NUL1.X///Df.X/. However, it is important to interpret the symbolJG.G/NUL1.X//properly. We are not substituting G/NUL1.X/intoGhere; rather, we are evaluating the determinant of the differential matrix of Gat the point YDG/NUL1.X/. From Theorems 6.1.9 and 6.3.3 , jJG.G/NUL1.X//jjJG/NUL1.X/jD1; so (7.3.46 ) can be rewritten as Q1.S1/DZ Sf.G.Y//jJG.Y/jdY/NULZ G.S/f.X/dXD/NULQ.S/: SinceQ1.S1//DC40, it now follows that Q.S//NAK0. 502 Chapter 7 Integrals of Functions of Several Variables We can now complete the proof of Theorem 7.3.8 . Lemmas 7.3.13 and7.3.14 imply (7.3.28 ) iffis nonnegative on S. Now suppose that mDmin˚ f.X/ˇˇX2G.S//TAB <0: Thenf/NULmis nonnegative on G.S/, so ( 7.3.28 ) withfreplaced byf/NULmimplies that Z G.S/.f.X//NULm/d XDZ S.f.G.Y//NULm/jJG.Y/jdY: (7.3.47) However, setting fD1in (7.3.28 ) yields Z G.S/dXDZ SjJG.Y/jdY; so (7.3.47 ) implies ( 7.3.28 ). The assumptions of Theorem 7.3.8 are too stringent for many applications. For example, to find the area of the disc˚ .x;y/ˇˇx2Cy2/DC41/TAB ; it is convenient to use polar coordinates and regard the circ le as G.S/, where G.r;/DC2/D/DC4rcos/DC2 rsin/DC2/NAK (7.3.48) andSis the compact set SD˚.r;/DC2/ˇˇ0/DC4r/DC41; 0/DC4/DC2/DC42/EM/TAB(7.3.49) (Figure 7.3.3 ). SX = G(r, θ)2π 1θy rxx2 + y2 = 1 G(S) Figure 7.3.3 Section 7.3 Change of Variables in Multiple Integrals 503 Since G0.r;/DC2/D/DC4cos/DC2/NULrsin/DC2 sin/DC2 r cos/DC2/NAK ; it follows that JG.r;/DC2/Dr. Therefore, formally applying Theorem 7.3.8 withf/DC11 yields ADZ G.S/dXDZ Srd.r;/DC2/DZ1 0rdrZ2/EM 0d/DC2D/EM: Although this is a familiar result, Theorem 7.3.8 does not really apply here, since G.r;0/D G.r;2/EM/ ,0/DC4r/DC41, soGis not one-to-one on S, and therefore not regular on S. The next theorem shows that the assumptions of Theorem 7.3.8 can be relaxed so as to include this example. Theorem 7.3.15 Suppose that GWEn!Rnis continuously differentiable on a bounded open set Ncontaining the compact Jordan measurable set S;and regular on S0:Suppose also that G.S/is Jordan measurable ;fis continuous on G.S/; andG.C/ is Jordan measurable for every cube C/SUBN. Then Z G.S/f.X/dXDZ Sf.G.Y//jJG.Y/jdY: (7.3.50) Proof Sincefis continuous on G.S/and.jJGj/fıGis continuous on S, the integrals in (7.3.50 ) both exist, by Corollary 7.3.2 . Now let /SUBDdist.@S;Nc/ (Exercise 5.1.25), and PD˚Yˇˇdist.Y;@S//TAB/DC4/SUB 2: ThenPis a compact subset of N(Exercise 5.1.26) and @S/SUBP0(Figure 7.3.4 ). SinceSis Jordan measurable, V.@S/D0, by Theorem 7.3.1 . Therefore, if /SI > 0 , we can choose cubes C1,C2, . . . ,CkinP0such that @S/SUBk[ jD1C0 j (7.3.51) and kX jD1V.C j/</SI (7.3.52) Now letS1be the closure of the set of points in Sthat are not in any of the cubes C1, C2, . . . ,Ck; thus, S1DS\/DLE [k jD1Cj/DC1c : 504 Chapter 7 Integrals of Functions of Several Variables Because of ( 7.3.51 ),S1\@SD;, soS1is a compact Jordan measurable subset of S0. Therefore, Gis regular on S1, andfis continuous on G.S1/. Consequently, if Qis as defined in ( 7.3.37 ), thenQ.S 1/D0by Theorem 7.3.8 . N = open set bounded by outer curve S = closed set bounded by inner curve∂S D ρ Figure 7.3.4 Now Q.S/DQ.S 1/CQ.S\Sc 1/DQ.S\Sc 1/ (7.3.53) (Exercise 7.3.11 ) and jQ.S\Sc 1/j/DC4ˇˇˇˇˇZ G.S\Sc 1/f.X/dXˇˇˇˇˇCˇˇˇˇˇZ S\Sc 1f.G.Y//jJG.Y/jdYˇˇˇˇˇ: But ˇˇˇˇˇZ S\Sc 1f.G.Y//jJG.Y/jdYˇˇˇˇˇ/DC4M1M2V.S\Sc 1/; (7.3.54) whereM1andM2are as defined in ( 7.3.38 ) and ( 7.3.39 ). SinceS\Sc 1/SUB[k jD1Cj, (7.3.52 ) implies that V.S\Sk 1/</SI ; therefore, ˇˇˇˇˇZ S\Sc 1f.G.Y//jJG.Y/jdYˇˇˇˇˇ/DC4M1M2/SI; (7.3.55) from ( 7.3.54 ). Also ˇˇˇˇˇZ G.S\Sc 1/f.X/dXˇˇˇˇˇ/DC4M2V.G.S\Sc 1///DC4M2kX jD1V.G.Cj//: (7.3.56) Section 7.3 Change of Variables in Multiple Integrals 505 By the argument that led to ( 7.3.30 ) withHDGandCDCj, V.G.Cj///DC4/STXmax˚kG0.Y/k1ˇˇY2Cj/TAB/ETXnV.C j/; so (7.3.56 ) can be rewritten as ˇˇˇˇˇZ G.S\Sc 1/f.X/dXˇˇˇˇˇ/DC4M2/STXmax˚kG0.Y/k1ˇˇY2P/TAB/ETXn/SI; because of ( 7.3.52 ). Since/SIcan be made arbitrarily small, this and ( 7.3.55 ) imply that Q.S\Sc 1/D0. NowQ.S/D0, from ( 7.3.53 ). The transformation to polar coordinates to compute the area of the disc is now justi- fied, since GandSas defined by ( 7.3.48 ) and ( 7.3.49 ) satisfy the assumptions of Theo- rem7.3.15 . Polar Coordinates IfGis the transformation from polar to rectangle coordinates /DC4x y/NAK DG.r;/DC2/D/DC4rcos/DC2 rsin/DC2/NAK ; (7.3.57) thenJG.r;/DC2/Drand ( 7.3.50 ) becomes Z G.S/f.x;y/d.x;y/DZ Sf.rcos/DC2;rsin/DC2/rd.r;/DC2/ if we assume, as is conventional, that Sis in the closed right half of the r/DC2-plane. This transformation is especially useful when the boundaries of Scan be expressed conveniently in terms of polar coordinates, as in the example preceding Th eorem 7.3.15 . Two more examples follow. Example 7.3.2 Evaluate IDZ T.x2Cy/d.x;y/; whereTis the annulus TD˚ .x;y/ˇˇ1/DC4x2Cy2/DC44/TAB (Figure 7.3.5(b)). Solution We writeTDG.S/, with Gas in ( 7.3.57 ) and SD˚.r;/DC2/ˇˇ1/DC4r/DC42; 0/DC4/DC2/DC42/EM/TAB 506 Chapter 7 Integrals of Functions of Several Variables (Figure 7.3.5(a)). Theorem 7.3.15 implies that IDZ S.r2cos2/DC2Crsin/DC2/rd.r;/DC2/; which we evaluate as an iterated integral: IDZ2 1r2drZ2/EM 0.rcos2/DC2Csin/DC2/d/DC2 DZ2 1r2drZ2/EM 0/DLEr 2Cr 2cos2/DC2Csin/DC2/DC1 d/DC2/DC4 since cos2/DC2D1 2.1Ccos2/DC2//NAK DZ2 1r2/DC4r/DC2 2Cr 4sin2/DC2/NULcos/DC2/NAKˇˇˇˇ2/EM /DC2D0drD/EMZ2 1r3drD/EMr4 4ˇˇˇˇ2 1D15/EM 4: T Sy rx2π (a) (b)θ 2 1 Figure 7.3.5 Example 7.3.3 Evaluate IDZ Tyd.x;y/; whereTis the region in the xy-plane bounded by the curve whose points have polar coor- dinates satisfying rD1/NULcos/DC2; 0/DC4/DC2/DC4/EM (Figure 7.3.6(b)). Solution We writeTDG.S/, with Gas in ( 7.3.57 ) andSthe shaded region in Figure 7.3.6(a). From ( 7.3.50 ), IDZ S.rsin/DC2/rd.r;/DC2/; Section 7.3 Change of Variables in Multiple Integrals 507 which we evaluate as an iterated integral: IDZ/EM 0sin/DC2d/DC2Z1/NULcos/DC2 0r2drD1 3Z/EM 0.1/NULcos/DC2/3sin/DC2d/DC2 D1 12.1/NULcos/DC2/4ˇˇˇˇ/EM 0D4 3: TSr y x θπ (b) (a) Figure 7.3.6 Spherical Coordinates IfGis the transformation from spherical to rectangular coordi nates, 2 4x y ´3 5DG.r;/DC2;/RS/D2 4rcos/DC2cos/RS rsin/DC2cos/RS rsin/RS3 5; (7.3.58) then G0.r;/DC2;/RS/D2 4cos/DC2cos/RS/NULrsin/DC2cos/RS/NULrcos/DC2sin/RS sin/DC2cos/RS r cos/DC2cos/RS/NULrsin/DC2sin/RS sin/RS 0 r cos/RS3 5 andJG.r;/DC2;/RS/Dr2cos/RS, so ( 7.3.50 ) becomes Z G.S/f.x;y;´/d.x;y;´/ DZ Sf.rcos/DC2cos/RS;rsin/DC2cos/RS;rsin/RS/r2cos/RSd.r;/DC2;/RS/(7.3.59) if we make the conventional assumption that j/RSj/DC4/EM=2 andr/NAK0. 508 Chapter 7 Integrals of Functions of Several Variables Example 7.3.4 Leta>0 . Find the volume of TD˚.x;y;´/ˇˇx2Cy2C´2/DC4a2; x/NAK0; y/NAK0; ´/NAK0/TAB; which is one eighth of a sphere (Figure 7.3.7(b)). (a) (b)yzφ θ r x2π 2π aa aa Figure 7.3.7 Solution We writeTDG.S/with Gas in ( 7.3.58 ) and SD˚.r;/DC2;/RS/ˇˇ0/DC4r/DC4a; 0/DC4/DC2/DC4/EM=2; 0/DC4/RS/DC4/EM=2/TAB Section 7.3 Change of Variables in Multiple Integrals 509 (Figure 7.3.7(a)), and letf/DC11in (7.3.59 ). Theorem 7.3.15 implies that V.T/DZ G.S/dXDZ Sr2cos/RSd.r;/DC2;/RS/ DZa 0r2drZ/EM=2 0d/DC2Z/EM=2 0cos/RSd/RSD/DC2a3 3/DC3/DLE/EM 2/DC1 (7.3.1 )D/EMa3 6: Example 7.3.5 Evaluate the iterated integral IDZa 0xdxZp a2/NULx2 0dyZp a2/NULx2/NULy2 0´d´ .a>0/: Solution We first rewrite Ias a multiple integral IDZ G.S/x´d.x;y;´/ where GandSare as in Example 7.3.4 . From Theorem 7.3.15 , IDZ S.rcos/DC2cos/RS/.r sin/RS/.r2cos/RS/d.r;/DC2;/RS/ DZa 0r4drZ/EM=2 0cos/DC2d/DC2Z/EM=2 0cos2/RSsin/RSd/RSD/DC2a5 5/DC3 (7.3.1 )/DC21 3/DC3 Da5 15: Other Examples We now consider other applications of Theorem 7.3.15 . Example 7.3.6 Evaluate IDZ T.xC4y/d.x;y/; whereTis the parallelogram bounded by the lines xCyD1; xCyD2; x/NUL2yD0; andx/NUL2yD3 (Figure 7.3.8(b)). Solution We define new variables uandvby /DC4u v/NAK DF.x;y/D/DC4xCy x/NUL2y/NAK : 510 Chapter 7 Integrals of Functions of Several Variables Sv u 23 1 (a)x y= F−1(u,v) Ty x (b)x − 2y = 0 x − 2y = 3 x + y = 2 x + y = 1 Figure 7.3.8 Then /DC4x y/NAK DF/NUL1.u;v/D2 642uCv 3u/NULv 33 75; JF/NUL1.u;v/Dˇˇˇˇˇ2 31 3 1 3/NUL1 3ˇˇˇˇˇD/NUL1 3; andTDF/NUL1.S/, where SD˚ .u;v/ˇˇ1/DC4u/DC42; 0/DC4v/DC43/TAB Section 7.3 Change of Variables in Multiple Integrals 511 (Figure 7.3.8(a)). Applying Theorem 7.3.15 with GDF/NUL1yields IDZ S/DC22uCv 3C4u/NUL4v 3/DC3/DC21 3/DC3 d.u;v/D1 3Z S.2u/NULv/d.u;v/ D1 3Z3 0dvZ2 1.2u/NULv/duD1 3Z3 0.u2/NULuv/ˇˇˇˇ2 uD1dv D1 3Z3 0.3/NULv/dvD1 3/DC2 3v/NULv2 2/DC3ˇˇˇˇ3 0D3 2: Example 7.3.7 Evaluate IDZ Te.x2/NULy2/2e4x2y2.x2Cy2/d.x;y/; whereTis the annulus TD˚ .x;y/ˇˇa2/DC4x2Cy2/DC4b2/TAB witha > 0 andb > 0 (Fig- ure7.3.9(a)). y xy xa bT a bT1T2 T3T4 (a) (b) Figure 7.3.9 Solution The forms of the arguments of the exponential functions sugg est that we introduce new variables uandvdefined by /DC4u v/NAK DF.x;y/D/DC4x2/NULy2 2xy/NAK and apply Theorem 7.3.15 toGDF/NUL1. However, Fis not one-to-one on T0and therefore has no inverse on T0(Example 6.3.4 ). To remove this difficulty, we regard Tas the union of the quarter-annuli T1,T2,T3, andT4in the four quadrants (Figure 7.3.9 )(b)), and let IjDZ Tje.x2/NULy2/2e4x2y2.x2Cy2/d.x;y/: 512 Chapter 7 Integrals of Functions of Several Variables Since the pairwise intersections of T1,T2,T3, andT4all have zero content, IDI1C I2CI3CI4(Corollary 7.1.31 ). Theorem 7.3.8 implies thatI1DI2DI3DI4(Exer- cise7.3.12 ), soID4I1. SinceI1does not contain any pairs of distinct points of the form .x0;y0/and./NULx0;/NULy0/,Fis one-to-one on T1(Example 6.3.4 ), F.T1/DS1D˚ .u;v/ˇˇa4/DC4u2Cv2/DC4b4;v/NAK0/TAB (Figure 7.3.10(b)), S1s1v u ρα π a2b2a2b2 (a) (b) Figure 7.3.10 and a branch GofF/NUL1can be defined on S1(Example 6.3.8 ). Now Theorem 7.3.15 implies that I1DZ S1e.x2/NULy2/2e4x2y2.x2Cy2/jJG.u;v/jd.u;v/; wherexandymust still be written in terms of uandv. Since it is easy to verify that JF.x;y/D4.x2Cy2/ and therefore JG.u;v/D1 4.x2Cy2/; doing this yields I1D1 4Z S1eu2Cv2d.u;v/: (7.3.60) To evaluate this integral, we let /SUBand˛be polar coordinates in the uv-plane (Figure 7.3.11 ) and define Hby/DC4u v/NAK DH./SUB;˛/D/DC4/SUBcos˛ /SUBsin˛/NAK I thenS1DH.eS1/, where eS1D˚./SUB;˛/ˇˇa2/DC4/SUB/DC4b2; 0/DC4˛/DC4/EM/TAB Section 7.3 Change of Variables in Multiple Integrals 513 (Figure 7.3.10(a)); hence, applying Theorem 7.3.15 to (7.3.60 ) yields I1D1 4Z eS1e/SUB2jJH./SUB;˛/jd./SUB;˛/D1 4Z eS1/SUBe/SUB2d./SUB;˛/ D1 4Z/EM 0d˛Zb2 a2/SUBe/SUB2d/SUBD/EM.eb4/NULea4/ 8I hence, ID4I1D/EM 2.eb4/NULea4/: v uρ α(u, v) Figure 7.3.11 Example 7.3.8 Evaluate IDZ Tex1Cx2C/SOH/SOH/SOHC xnd.x 1;x2;:::;x n/; whereTis the region defined by ai/DC4x1Cx2C/SOH/SOH/SOHCxi/DC4bi; 1/DC4i/DC4n: Solution We define the new variables y1,y2, . . . ,ynbyYDF.X/, where fi.X/Dx1Cx2C/SOH/SOH/SOHCxi; 1/DC4i/DC4n: IfGDF/NUL1thenTDG.S/, where SDŒa1;b1/c141/STXŒa2;b2/c141/STX/SOH/SOH/SOH/STXŒan;bn/c141; andJG.Y/D1, sinceJF.X/D1(verify); hence, Theorem 7.3.8 implies that 514 Chapter 7 Integrals of Functions of Several Variables IDZ Seynd.y 1;y2;:::;y n/ DZb1 a1dy1Zb2 a2dy2/SOH/SOH/SOHZbn/NUL1 an/NUL1dyn/NUL1Zbn aneyndyn D.b1/NULa1/.b2/NULa2//SOH/SOH/SOH.bn/NUL1/NULan/NUL1/.ebn/NULean/: 7.3 Exercises 1. Give a counterexample to the following statement: If S1andS2are disjoint subsets of a rectangle R, then either Z R S1.X/dXCZ R S2.X/dXDZ R S1[S2.X/dX orZ R S1.X/dXCZ R S2.X/dXDZ R S1[S2.X/dX: 2. Show that a set Ehas content zero according to Definition 7.1.14 if and only if E has Jordan content zero. 3. Show that ifS1andS2are Jordan measurable, then so are S1[S2andS1\S2. 4. Prove: (a) IfSis Jordan measurable then so is S, andV.S/DV.S/ . MustSbe Jordan measurable if Sis? (b) IfTis a Jordan measurable subset of a Jordan measurable set S, thenS/NULT is Jordan measurable. 5. Suppose that His a subset of a compact Jordan measurable set Ssuch that the inter- section ofHwith any compact subset of S0has zero content. Show that V.H/D0. 6. Suppose that Eis ann/STXnelementary matrix and Ais an arbitrary n/STXpmatrix. Show that EAis the matrix obtained by applying to Athe operation by which Eis obtained from the n/STXnidentity matrix. 7. (a) Calculate the determinants of elementary matrices of types (a),(b), and(c) of Lemma 7.3.6 . (b) Show that the inverse of an elementary matrix of type (a),(b), or(c)is an elementary matrix of the same type. (c) Verify the inverses given for bE1;:::;bE6in Example 7.3.1 . Section 7.3 Change of Variables in Multiple Integrals 515 8. Write as a product of elementary matrices. (a)2 41 0 1 1 1 0 0 1 13 5(b)2 42 3/NUL2 0/NUL1 5 0/NUL2 43 5 9. Suppose that ad/NULbc¤0,u1<u 2, andv1<v 2. Find the area of the parallelogram bounded by the lines axCbyDu1; axCbyDu2; cxCdyDv1; cxCdyDv2: 10. Find the volume of the parallelepiped defined by 1/DC42xC3y/NUL2´/DC42; 5/DC4/NULxC5y/DC47; 1/DC4/NUL2xC4y/DC46: 11. In writing Eqn. ( 7.3.53 ) we assumed that Z G.S/f.X/dXDZ G.S1/f.X/dXCZ G.S\Sc 1/f.X/dX: Justify this. H INT:Show that G.S1/\G.S\Sc 1/has zero content : 12. Use Theorem 7.3.8 to show thatI1DI2DI3DI4in Example 7.3.7 . 13. LeteiD˙1,0/DC4i/DC4n. LetTbe a bounded subset of Rnand bTD˚ .e1x1;e2x2;:::;e nxn/ˇˇ.x1;x2;:::;x n/2T/TAB : Suppose that fis defined on Tand definegonbTby g.e1x1;e2x2;:::;e nxn/De0f.x 1;x2;:::;x n/: (a) Prove directly from Definitions 7.1.2 and7.1.17 thatfis integrable on Tif and only ifgis integrable on bT, and in this case Z bTg.Y/dYDe0Z Tf.X/dX: (b) Suppose that bTDT, f.e 1x1;e2x2;:::;e nxn/D/NULf.x 1;x2;:::;x n/; andfis integrable on T. Show that Z Tf.X/dXD0: 14. Find the area of (a)˚.x;y/ˇˇy/DC4x/DC44y; 1/DC4xC2y/DC43/TAB; 516 Chapter 7 Integrals of Functions of Several Variables (b)˚ .x;y/ˇˇ2/DC4xy/DC44; 2x/DC4y/DC45x/TAB . 15. Evaluate Z T.3x2C2yC´/d.x;y;´/; where TD˚.x;y;´/ˇˇjx/NULyj/DC41;jy/NUL´j/DC41;j´Cxj/DC41/TAB: 16. Evaluate Z T.y2Cx2y/NUL2x4/d.x;y/; whereTis the region bounded by the curves xyD1; xyD2; yDx2; yDx2C1: 17. Evaluate Z T.x4/NULy4/exyd.x;y/; whereTis the region in the first quadrant bounded by the hyperbolas xyD1; xyD2; x2/NULy2D2; x2/NULy2D3: 18. Find the volume of the ellipsoid x2 a2Cy2 b2C´2 c2D1 .a;b;c>0/: 19. EvaluateZ Tex2Cy2C´2 p x2Cy2C´2d.x;y;´/; where TD˚.x;y;´/ˇˇ9/DC4x2Cy2C´2/DC425/TAB: 20. Find the volume of the set Tbounded by the surfaces ´D0,´Dp x2Cy2, and x2Cy2D4. 21. Evaluate Z Txy´.x4/NULy4/d.x;y;´/; where TD˚ .x;y;´/ˇˇ1/DC4x2/NULy2/DC42; 3/DC4x2Cy2/DC44; 0/DC4´/DC41/TAB : 22. Evaluate (a)Zp 2 0dyZp 4/NULy2 ydx 1Cx2Cy2(b)Z2 0dxZp 4/NULx2 0ex2Cy2dy (c)Z1 /NUL1dxZp 1/NULx2 /NULp 1/NULx2dyZp 1/NULx2/NULy2 0´2d´ Section 7.3 Change of Variables in Multiple Integrals 517 23. Use the change of variables 2 664x1 x2 x3 x43 775DG.r;/DC2 1;/DC22;/DC23/D2 664rcos/DC21cos/DC22cos/DC23 rsin/DC21cos/DC22cos/DC23 rsin/DC22cos/DC23 rsin/DC233 775 to compute the content of the 4-ball TD˚ .x1;x2;x3;x4/ˇˇx2 1Cx2 2Cx2 3Cx2 4/DC4a2/TAB : 24. Suppose that ADŒaij/c141is a nonsingular n/STXnmatrix andTis the region in Rn defined by ˛1/DC4ai1x1Cai2x2C/SOH/SOH/SOHCainxn/DC4ˇi; 1/DC4i/DC4n: (a) FindV.T/ . (b) Show that ifc1,c2, . . . ,cnare constants, then Z T0 @nX jD1cjxj1 AdXDV.T/ 2nX iD1di.˛iCˇi/; where 2 6664d1 d2 ::: dn3 7775D.At//NUL12 6664c1 c2 ::: cn3 7775: 25. IfVnis the content of the n-ballTD˚XˇˇjXj/DC41/TAB, find the content of the n- dimensional ellipsoid defined by nX jD1x2 j a2 j/DC41: Leave the answer in terms of Vn. CHAPTER 8 Metric Spaces IN THIS CHAPTER we study metric spaces. SECTION 8.1 defines the concept and basic properties of a metr ic space. Several examples of metric spaces are considered. SECTION 8.2 defines and discusses compactness in a metric spa ce. SECTION 8.3 deals with continuous functions on metric space s. 8.1 INTRODUCTION TO METRIC SPACES Definition 8.1.1 Ametric space is a nonempty set Atogether with a real-valued func- tion/SUBdefined onA/STXAsuch that ifu,v, andware arbitrary members of A, then (a)/SUB.u;v//NAK0, with equality if and only if uDv; (b)/SUB.u;v/D/SUB.v;u/ ; (c)/SUB.u;v//DC4/SUB.u;w/C/SUB.w;v/ . We say that/SUBis ametric onA. Ifn/NAK2andu1,u2, . . . ,unare arbitrary members of A, then(c)and induction yield the inequality /SUB.u 1;un//DC4n/NUL1X iD1/SUB.u i;uiC1/: Example 8.1.1 The set Rof real numbers with /SUB.u;v/Dju/NULvjis a metric space. Definition 8.1.1(c)is the familiar triangle inequality: ju/NULvj/DC4ju/NULwjCjw/NULuj: Motivated by this example, in an arbitrary metric space we ca ll/SUB.u;v/ the distance from utov, and we call Definition 8.1.1(c)the triangle inequality . 518 Section 8.1 Introduction to Metric Spaces 519 Example 8.1.2 IfAis an arbitrary nonempty set, then /SUB.u;v/D/SUB0ifuDv; 1ifu¤v is a metric on A(Exercise 8.1.5 ). We call it the discrete metric. Example 8.1.2 shows that it is possible to define a metric on any nonempty set A. In fact, it is possible to define infinitely many metrics on any se t with more than one member (Exercise 8.1.3 ). Therefore, to specify a metric space completely, we must s pecify the couple.A;/SUB/ , whereAis the set and /SUBis the metric. (In some cases we will not be so precise; for example, we will always refer to the real number s with the metric /SUB.u;v/D ju/NULvjsimply as R.) There is an important kind of metric space that arises when a d efinition of length is imposed on a vector space. Although we assume that you are fam iliar with the definition of a vector space, we restate it here for convenience. We confi ne the definition to vector spaces over the real numbers. Definition 8.1.2 Avector space Ais a nonempty set of elements called vectors on which two operations, vector addition and scalar multiplic ation (multiplication by real numbers) are defined, such that the following assertions are true for all U,V, and Win Aand all real numbers rands: 1.UCV2A; 2.UCVDVCU; 3.UC.VCW/D.UCV/CW; 4. There is a vector 0inAsuch that UC0DU; 5. There is a vector /NULUinAsuch that UC./NULU/D0; 6.rU2A; 7.r.UCV/DrUCrV; 8..rCs/UDrUCsU; 9.r.sU/D.rs/U; 10.1UDU. We say thatAisclosed under vector addition if (1) is true, and that Aisclosed under scalar multiplication if (6) is true. It can be shown that if Bis any nonempty subset of A that is closed under vector addition and scalar multiplicat ion, thenBtogether with these operations is itself a vector space. (See any linear algebra text for the proof.) We say that Bis asubspace ofA. Definition 8.1.3 Anormed vector space is a vector space Atogether with a real-valued functionNdefined onA, such that if uandvare arbitrary vectors in Aandais a real number, then (a)N.u//NAK0with equality if and only if uD0; (b)N.au/DjajN.u/ ; (c)N.uCv//DC4N.u/CN.v/ . We say thatNis anorm onA, and.A;N/ is anormed vector space . 520 Chapter 8 Metric Spaces Theorem 8.1.4 If.A;N/ is a normed vector space ;then /SUB.x;y/DN.x/NULy/ (8.1.1) is a metric on A: Proof From(a) withuDx/NULy,/SUB.x;y/DN.x/NULy//NAK0, with equality if and only ifxDy. From(b) withuDx/NULyandaD/NUL1, /SUB.y;x/DN.y/NULx/DN./NUL.x/NULy//DN.x/NULy/D/SUB.x;y/: From(c)withuDx/NUL´andvD´/NULy, /SUB.x;y/DN.x/NULy//DC4N.x/NUL´/CN.´/NULy/D/SUB.x;´/C/SUB.´;y/: We will say that the metric in ( 8.1.1 ) isinduced by the norm N. Whenever we speak of a normed vector space .A;N/ , it is to be understood that we are regarding it as a metric space.A;/SUB/ , where/SUBis the metric induced by N. We will often write N.u/ askuk. In this case we will denote the normed vector space as .A;k/SOHk/. Theorem 8.1.5 Ifxandyare vectors in a normed vector space .A;N/; then jN.x//NULN.y/j/DC4N.x/NULy/: (8.1.2) Proof Since xDyC.x/NULy/; Definition 8.1.3(c)withuDyandvDx/NULyimplies that N.x//DC4N.y/CN.x/NULy/; or N.x//NULN.y//DC4N.x/NULy/: Interchanging xandyyields N.y//NULN.x//DC4N.y/NULx/: SinceN.x/NULy/DN.y/NULx/(Definition 8.1.3(b) withuDx/NULyandaD/NUL1), the last two inequalities imply ( 8.1.2 ). Metrics for RRRn In Section 5.1 we defined the norm of a vector XD.x1;x2;:::;x n/inRnas kXkD nX iD1x2 i!1=2 : Section 8.1 Introduction to Metric Spaces 521 The metric induced by this norm is /SUB.X;Y/D nX iD1.xi/NULyi/2!1=2 : Whenever we write Rnwithout identifying the norm or metric specifically, we are r eferring toRnwith this norm and this induced metric. The following definition provides infinitely many norms and m etrics on Rn. Definition 8.1.6 Ifp/NAK1andXD.x1;x2;:::;x n/, let kXkpD nX iD1jxijp!1=p : (8.1.3) The metric induced on Rnby this norm is /SUBp.X;Y/D nX iD1jxi/NULyijp!1=p : To justify this definition, we must verify that ( 8.1.3 ) actually defines a norm. Since it is clear thatkXkp/NAK0with equality if and only if XD0, andkaXkpDjajkXkpifais any real number and X2Rn, this reduces to showing that kXCYkp/DC4kXkpCkYkp (8.1.4) for every XandYinRn. Since jxiCyij/DC4jxijCjyij; summing both sides of this equation from iD1tonyields ( 8.1.4 ) withpD1. To handle the case where p > 1 , we need the following lemmas. The inequality established i n the first lemma is known as Hölder ’s inequality . Lemma 8.1.7 Suppose that/SYN1;/SYN2;. . .;/SYNnand/ETB1;/ETB2;. . .;/ETBnare nonnegative numbers : Letp>1 andqDp=.p/NUL1/Ithus; 1 pC1 qD1: (8.1.5) Then nX iD1/SYNi/ETBi/DC4 nX iD1/SYNp i!1=p nX iD1/ETBq i!1=q : (8.1.6) Proof Let˛andˇbe any two positive numbers, and consider the function f.ˇ/D˛p pCˇq q/NUL˛ˇ; 522 Chapter 8 Metric Spaces where we regard ˛as a constant. Since f0.ˇ/Dˇq/NUL1/NUL˛andf00.ˇ/D.q/NUL1/ˇq/NUL2>0 forˇ>0 ,fassumes its minimum value on Œ0;1/atˇD˛1=.q /NUL1/D˛p/NUL1. But f.˛p/NUL1/D˛p pC˛.p/NUL1/q q/NUL˛pD˛p/DC21 pC1 q/NUL1/DC3 D0: Therefore, ˛ˇ/DC4˛p pCˇq qif˛;ˇ/NAK0: (8.1.7) Now let ˛iD/SYNi0 @nX jD1/SYNp j1 A/NUL1=p andˇiD/ETBi0 @nX jD1/ETBq j1 A/NUL1=q : From ( 8.1.7 ), ˛iˇi/DC4/SYNp i p0 @nX jD1/SYNp j1 A/NUL1 C/ETBq i q0 @nX jD1/ETBq j1 A/NUL1 : From ( 8.1.5 ), summing this from iD1tonyieldsPn iD1˛iˇi/DC41, which implies ( 8.1.6 ). Lemma 8.1.8 ( Minkowski ’s Inequality) Suppose that u1;u2;. . .;unandv1; v2;. . .;vnare nonnegative numbers and p>1: Then nX iD1.uiCvi/p!1=p /DC4 nX iD1up i!1=p C nX iD1vp i!1=p : (8.1.8) Proof Again, letqDp=.p/NUL1/. We write nX iD1.uiCvi/pDnX iD1ui.uiCvi/p/NUL1CnX iD1vi.uiCvi/p/NUL1: (8.1.9) From Hölder’s inequality with /SYNiDuiand/ETBiD.uiCvi/p/NUL1, nX iD1ui.uiCvi/p/NUL1/DC4 nX iD1up i!1=p nX iD1.uiCvi/p!1=q ; (8.1.10) sinceq.p/NUL1/Dp. Similarly, nX iD1vi.uiCvi/p/NUL1/DC4 nX iD1vp i!1=p nX iD1.uiCvi/p!1=q : This, ( 8.1.9 ), and ( 8.1.10 ) imply that nX iD1.uiCvi/p/DC42 4 nX iD1up i!1=p C nX iD1vp i!1=p3 5 nX iD1.uiCvi/p!1=q : Section 8.1 Introduction to Metric Spaces 523 Since1/NUL1=qD1=p, this implies ( 8.1.8 ), which is known as Minkowski’s inequality . We leave it to you to verify that Minkowski’s inequality impl ies (8.1.4 ) ifp>1 . We now define the1-norm onRnby kXk1Dmax˚ jxijˇˇ1/DC4i/DC4n/TAB : (8.1.11) We leave it to you to verify (Exercise 8.1.15 ) thatk/SOHk 1is a norm on Rn. The associated metric is /SUB1.X;Y/Dmax˚jxi/NULyijˇˇ1/DC4i/DC4n/TAB: The following theorem justifies the notation in ( 8.1.11 ). Theorem 8.1.9 IfX2Rnandp2>p 1/NAK1;then kXkp2/DC4kXkp1I (8.1.12) moreover, lim p!1kXkpDmax˚ jxijˇˇ1/DC4i/DC4n/TAB : (8.1.13) Proof Letu1,u2, . . . ,unbe nonnegative and MDmax˚uiˇˇ1/DC4i/DC4n/TAB. Define /ESC.p/D nX iD1up i!1=p : Sinceui=/ESC.p//DC41andp2>p 1, /DC2ui /ESC.p 2//DC3p1 /NAK/DC2ui /ESC.p 2//DC3p2 I therefore, /ESC.p 1/ /ESC.p 2/D nX iD1/DC2ui /ESC.p 2//DC3p1!1=p 1 /NAK nX iD1/DC2ui /ESC.p 2//DC3p2!1=p 1 D1; so/ESC.p 1//NAK/ESC.p 2/. SinceM/DC4/ESC.p//DC4Mn1=p, lim p!1/ESC.p/DM. LettinguiDjxij yields ( 8.1.12 ) and ( 8.1.13 ). Since Minkowski’s inequality is false if p<1 (Exercise 8.1.19 ), (8.1.3 ) is not a norm in this case. However, if 0<p<1 , then kXkpDnX iD1jxijp is a norm on Rn(Exercise 8.1.20 ). Vector Spaces of Sequences of Real Numbers In this section and in the exercises we will consider subsets of the vector space R1con- sisting of sequences XDfxig1 iD1, with vector addition and scalar multiplication defined by XCYDfxiCyig1 iD1andrXDfrxig1 iD1: 524 Chapter 8 Metric Spaces Example 8.1.3 Suppose that 1<p<1and let `pD( X2R1ˇˇ1X iD1jxijp<1) : Let kXkpD 1X iD1jxijp!1=p : Show that.`p;k/SOHk p/is a normed vector space. Solution Suppose that X,Y2`p. From Minkowski’s inequality, nX iD1jxiCyijp!1=p /DC4 nX iD1jxijp!1=p C nX iD1jyijp!1=p for eachn. Since the right side remains bounded as n!1 , so does the left, and 1X iD1jxiCyijp!1=p /DC4 1X iD1jxijp!1=p C 1X iD1jyijp!1=p ; (8.1.14) soXCY2`p. Therefore,`pis closed under vector addition. Since `pis obviously closed under scalar multiplication, `pis a vector space, and ( 8.1.14 ) implies thatk/SOHk pis a norm on`p. The metric induced by k/SOHk pis /SUBp.X;Y/D 1X iD1jxi/NULyijp!1=p : Henceforth, we will denote .`p;k/SOHk p/simply by`p. Example 8.1.4 Let `1D˚X2R1ˇˇfxig1 iD1is bounded/TAB: Let kXk1Dsup˚jxijˇˇi/NAK1/TAB: We leave it to you (Exercise 8.1.26 ) to show that .`1;k/SOHk 1/is a normed vector space. The metric induced by k/SOHk 1is /SUB1.X;Y/Dsup˚ jxi/NULyijˇˇi/NAK1/TAB : Henceforth, we will denote .`1;k/SOHk 1/simply by`1. Section 8.1 Introduction to Metric Spaces 525 Familiar Definitions and Theorems At this point you may want to review Definition 1.3.1 and Exercises 1.3.6 and1.3.7 , which apply equally well to subsets of a metric space .A;/SUB/ . We will now state some definitions and theorems for a general m etric space.A;/SUB/ that are analogous to definitions and theorems presented in Secti on 1.3 for the real numbers. To avoid repetition, it is to be understood in all these definiti ons that we are discussing a given metric space.A;/SUB/ . Definition 8.1.10 Ifu02Aand/SI>0 , the set N/SI.u0/D˚u2Aˇˇ/SUB.u 0;u/</SI/TAB is called an/SI-neighborhood ofu0. (Sometimes we call S/SItheopen ball of radius /SIcentered atu0.) If a subset SofAcontains an/SI-neighborhood of u0, thenSis aneighborhood of u0, andu0is an interior point ofS. The set of interior points of Sis the interior ofS, denoted byS0. If every point of Sis an interior point (that is, S0DS), thenSisopen . A setSisclosed ifScis open. Example 8.1.5 Show that ifr >0 , then the open ball Sr.u0/D˚u2Aˇˇ/SUB.u 0;u/<r/TAB is an open set. Solution We must show that if u12Sr.u0/, then there is an /SI>0 such that S/SI.u1//SUBSr.u0/: (8.1.15) Ifu12Sr.u0/, then/SUB.u 1;u0/<r . Since /SUB.u;u 0//DC4/SUB.u;u 1/C/SUB.u 1;u0/ for anyuinA,/SUB.u;u 0/ < r if/SUB.u;u 1/ < r/NUL/SUB.u 1;u0/. Therefore, ( 8.1.15 ) holds if /SI<r/NUL/SUB.u 1;u0/. The entire space Ais open and therefore ;.DAc/is closed. However, ;is also open, for to deny this is to say that it contains a point that is not an interior point, which is absurd because;contains no points. Since ;is open,A.D;c/is closed. IfADR, these are the only sets that are both open and closed, but this is not so in al l metric spaces. For example, if/SUBis the discrete metric, then every subset of Ais both open and closed. (Verify!) Adeleted neighborhood of a pointu0is a set that contains every point of some neigh- borhood ofu0exceptu0itself. (If/SUBis the discrete metric then the empty set is a deleted neighborhood of every member of A!) The proof of the following theorem is identical to the proof T heorem 1.3.3 . 526 Chapter 8 Metric Spaces Theorem 8.1.11 (a) The union of open sets is open. (b) The intersection of closed sets is closed. Definition 8.1.12 LetSbe a subset of A. Then (a)u0is alimit point ofSif every deleted neighborhood of u0contains a point of S. (b)u0is aboundary point ofSif every neighborhood of u0contains at least one point inSand one not in S. The set of boundary points of Sis the boundary ofS, denoted by@S. The closure ofS, denoted byS, is defined by SDS[@S. (c)u0is an isolated point ofSifu02Sand there is a neighborhood of u0that contains no other point of S. (d)u0isexterior toSifu0is in the interior of Sc. The collection of such points is the exterior ofS. Although this definition is identical to Definition 1.3.4 , you should not assume that con- clusions valid for the real numbers are necessarily valid in all metric spaces. For example, ifADRand/SUB.u;v/Dju/NULvj, then Sr.u0/D˚ uˇˇ/SUB.u;u 0//DC4r/TAB : This is not true in every metric space (Exercise 8.1.6 ). For the proof of the following theorem, see the proofs of Theo rem 1.3.5 and Corol- lary1.3.6 . Theorem 8.1.13 A set is closed if and only if it contains all its limit points : Completeness Since metric spaces are not ordered, concepts and results co ncerning the real numbers that depend on order for their definitions must be redefined and ree xamined in the context of metric spaces. The first example of this kind is completeness . To discuss this concept, we begin by defining an infinite sequence (more briefly, a sequence ) in a metric space .A;/SUB/ as a function defined on the integers n/NAKkwith values in A. As we did for real sequences, we denote a sequence in Aby, for example,fungDfung1 nDk. A subsequence of a sequence inAis defined in exactly the same way as a subsequence of a sequenc e of real numbers (Definition 4.2.1 ). Definition 8.1.14 A sequencefungin a metric space .A;/SUB/ converges tou2Aif lim n!1/SUB.u n;u/D0: (8.1.16) In this case we say that lim n!1unDu. We leave the proof of the following theorem to you. (See the pr oofs of Theorems 4.1.2 and4.2.2 .) Section 8.1 Introduction to Metric Spaces 527 Theorem 8.1.15 (a) The limit of a convergent sequence is unique : (b) Iflimn!1unDu;then every subsequence of fungconverges tou: Definition 8.1.16 A sequencefungin a metric space .A;/SUB/ is aCauchy sequence if for every/SI>0 there is an integer Nsuch that /SUB.u n;um/</SI andm;n>N: (8.1.17) We note that if /SUBis the metric induced by a norm k/SOHk onA, then ( 8.1.16 ) and ( 8.1.17 ) can be replaced by lim n!1kun/NULukD0 and kun/NULumk</SI andm;n>N; respectively. Theorem 8.1.17 If a sequencefungin a metric space .A;/SUB/ is convergent;then it is a Cauchy sequence. Proof Suppose that lim n!1unDu. If/SI > 0 , there is an integer Nsuch that /SUB.u n;u/</SI=2 ifn>N . Therefore, if m,n>N , then /SUB.u n;um//DC4/SUB.u n;u/C/SUB.u;u m/</SI: Definition 8.1.18 A metric space .A;/SUB/ iscomplete if every Cauchy sequence in A has a limit. Example 8.1.6 Theorem 4.1.13 implies that the set Rof real numbers with /SUB.u;v/ Dju/NULvjis a complete metric space. This example raises a question that we should resolve before going further. In Section 1.1 we defined completeness to mean that the real numbers have the following property: Axiom(I). Every nonempty set of real numbers that is bounded above has a supremum. Here we are saying that the real numbers are complete because every Cauchy sequence of real numbers has a limit. We will now show that these two usa ges of “complete” are consistent. 528 Chapter 8 Metric Spaces The proof of Theorem 4.1.13 requires the existence of the (finite) limits inferior and superior of a bounded sequence of real numbers, a consequenc e of Axiom (I). However, the assertion in Axiom (I)can be deduced as a theorem if Axiom (I)is replaced by the assumption that every Cauchy sequence of real numbers has a l imit. To see this, let Tbe a nonempty set of real numbers that is bounded above. We first sh ow that there are sequences fuig1 iD1andfvig1 iD1with the following properties for all i/NAK1: Section 8.1 Introduction to Metric Spaces 529 (a)ui/DC4tfor somet2Tandvi/NAKtfor allt2T; (b).vi/NULui//DC42i/NUL1.v1/NULu1/. (c)ui/DC4uiC1/DC4viC1/DC4vi SinceTis nonempty and bounded above, u1andv1can be chosen to satisfy (a) with iD1. Clearly, (b) holds withiD1. Letw1D.u1Cv1/=2, and let .u2;v2/D/SUB.w1;v1/ifw1/DC4tfor somet2T; .u1;w1/ifw1/NAKtfor allt2T: In either case, (a)and(b) hold withiD2and(c)holds withiD1. Now suppose that n >1 andfu1;:::;u ngandfv1;:::;v nghave been chosen so that (a) and(b) hold for 1/DC4i/DC4nand(c)holds for1/DC4i/DC4n/NUL1. LetwnD.unCvn/=2and let .unC1;vnC1/D/SUB.wn;vn/ifwn/DC4tfor somet2T; .un;wn/ifwn/NAKtfor allt2T: Then(a)and(b) hold for1/DC4i/DC4nC1and(c)holds for1/DC4i/DC4n. This completes the induction. Now(b) and(c)imply that 0/DC4uiC1/NULui/DC42i/NUL1.v1/NULu1/and0/DC4vi/NULviC1/DC42i/NUL1.v1/NULu1/; i/NAK1: By an argument similar to the one used in Example 4.1.14 , this implies thatfuig1 iD1and fvig1 iD1are Cauchy sequences. Therefore the sequences both converg e (because of our assumption), and (b) implies that they have the same limit. Let lim i!1uiDlim i!1viDˇ: Ift2T, thenvi/NAKtfor alli, soˇDlimi!1vi/NAKt; therefore,ˇis an upper bound of T. Now suppose that /SI >0 . Then there is an integer Nsuch thatuN>ˇ/NUL/SI. From the definition ofuN, there is atNinTsuch thattN/NAKuN>ˇ/NUL/SI. Therefore,ˇDsupT. Example 8.1.7 (The Metric Space CŒa;b/c141)LetCŒa;b/c141 denote the set of all real-valued functions fcontinuous on the finite closed interval Œa;b/c141 . From Theorem 2.2.9 , the quantity kfkD max˚ jf.x/jˇˇa/DC4x/DC4b/TAB is well defined. We leave it to you to verify that it is a norm on CŒa;b/c141 . The metric induced by this norm is /SUB.f;g/Dkf/NULgkD max˚ jf.x//NULg.x/jˇˇa/DC4x/DC4b/TAB : Whenever we refer to CŒa;b/c141 , we mean this metric space or, equivalently, this normed linear space. From Theorem 4.4.6 , a Cauchy sequence ffnginCŒa;b/c141 converges uniformly to a func- tionfonŒa;b/c141 , and Corollary 4.4.8 implies thatfis inCŒa;b/c141 ; hence,CŒa;b/c141 is complete. 530 Chapter 8 Metric Spaces The Principle of Nested Sets We say that a sequence fTngof sets is nested ifTnC1/SUBTnfor alln. Theorem 8.1.19 (The Principle of Nested Sets) A metric space .A;/SUB/ is complete if and only if every nested sequence fTngof nonempty closed subsets of Asuch thatlimn!1d.T n/D0has a nonempty intersection : Proof Suppose that .A;/SUB/ is complete andfTngis a nested sequence of nonempty closed subsets of Asuch that lim n!1d.T n/D0. For eachn, choosetn2Tn. Ifm/NAKn, thentm,tn2Tn, so/SUB.tn;tm/ < d.T n/. Since lim n!1d.T n/D0,ftngis a Cauchy se- quence. Therefore, lim n!1tnDtexists. Since tis a limit point of TnandTnis closed for alln,t2Tnfor alln. Therefore,t2\1 nD1Tn; in fact,\1 nD1TnDftg. (Why?) Now suppose that .A;/SUB/ is not complete, and let ftngbe a Cauchy sequence in Athat does not have a limit. Choose n1so that/SUB.tn;tn1/ < 1=2 ifn/NAKn1, and letT1D˚ tˇˇ/SUB.t;t n1//DC41/TAB . Now suppose that j > 1 and we have specified n1,n2, . . . ,nj/NUL1 andT1,T2, . . . ,Tj/NUL1. Choosenj> n j/NUL1so that/SUB.tn;tnj/ < 2/NULjifn/NAKnj, and let TjD˚ tˇˇ/SUB.t;t nj//DC42/NULjC1/TAB . ThenTjis closed and nonempty, TjC1/SUBTjfor allj, and lim j!1d.T j/D0. Moreover, tn2Tjifn/NAKnj. Therefore, if t2\1 jD1Tj, then/SUB.tn;t/ < 2/NULj,n/NAKnj, so lim n!1tnDt, contrary to our assumption. Hence, \1 jD1TjD;. Equivalent Metrics When considering more than one metric on a given set Awe must be careful, for example, in saying that a set is open, or that a sequence converges, etc ., since the truth or falsity of the statement will in general depend on the metric as well as t he set on which it is imposed. In this situation we will alway refer to the metric space by it s “full name;" that is, .A;/SUB/ rather than just A. Definition 8.1.20 If/SUBand/ESCare both metrics on a set A, then/SUBand/ESCareequivalent if there are positive constants ˛andˇsuch that ˛/DC4/SUB.x;y/ /ESC.x;y//DC4ˇfor allx;y2Asuch thatx¤y: (8.1.18) Theorem 8.1.21 If/SUBand/ESCare equivalent metrics on a set A;then.A;/SUB/ and.A;/ESC/ have the same open sets. Proof Suppose that ( 8.1.18 ) holds. LetSbe an open set in .A;/SUB/ and letx02S. Then there is an/SI > 0 such thatx2Sif/SUB.x;x 0/ < /SI , so the second inequality in ( 8.1.18 ) implies thatx02Sif/ESC.x;x 0//DC4/SI=ˇ. Therefore,Sis open in.A;/ESC/ . Conversely, suppose that Sis open in.A;/ESC/ and letx02S. Then there is an /SI > 0 such thatx2Sif/ESC.x;x 0/ < /SI , so the first inequality in ( 8.1.18 ) implies that x02Sif /SUB.x;x 0//DC4/SI˛. Therefore,Sis open in.A;/SUB/ . Section 8.1 Introduction to Metric Spaces 531 Theorem 8.1.22 Any two norms N1andN2onRninduce equivalent metrics on Rn: Proof It suffices to show that there are positive constants ˛andˇsuch ˛/DC4N1.X/ N2.X//DC4ˇif X¤0: (8.1.19) We will show that if Nis any norm on Rn, there are positive constants aNandbNsuch that aNkXk2/DC4N.X//DC4bNkXk2ifX¤0 (8.1.20) and leave it to you to verify that this implies ( 8.1.19 ) with˛DaN1=bN2andˇDbN1=aN2. We write X/NULYD.x1;x2;:::;x n/as X/NULYDnX iD1.xi/NULyi/Ei; where Eiis the vector with ith component equal to 1and all other components equal to 0. From Definition 8.1.3(b),(c), and induction, N.X/NULY//DC4nX iD1jxi/NULyijN.Ei/I therefore, by Schwarz’s inequality, N.X/NULY//DC4KkX/NULYk2; (8.1.21) where KD nX iD1N2.Ei/!1=2 : From ( 8.1.21 ) and Theorem 8.1.5 , jN.X//NULN.Y/j/DC4KkX/NULYk2; soNis continuous on Rn 2DRn. By Theorem 5.2.12 , there are vectors U1andU2such thatkU1k2DkU2k2D1, N.U1/Dmin˚ N.U/ˇˇkUk2D1/TAB ;andN.U2/Dmax˚ N.U/ˇˇkUk2D1/TAB : IfaNDN.U1/andbNDN.U2/, thenaNandbNare positive (Definition 8.1.3(a)), and aN/DC4N/DC2X kXk2/DC3 /DC4bNif X¤0: This and Definition 8.1.3(b) imply ( 8.1.20 ). We leave the proof of the following theorem to you. 532 Chapter 8 Metric Spaces Theorem 8.1.23 Suppose that /SUBand/ESCare equivalent metrics on A:Then (a) A sequencefungconverges touin.A;/SUB/ if and only if it converges to uin.A;/ESC/: (b) A sequencefungis a Cauchy sequence in .A;/SUB/ if and only if it is a Cauchy sequence in.A;/ESC/: (c).A;/SUB/ is complete if and only if .A;/ESC/ is complete: 8.1 Exercises 1. Show that (a),(b), and(c)of Definition 8.1.1 are equivalent to (i)/SUB.u;v/D0if and only if uDv; (ii)/SUB.u;v//DC4/SUB.w;u/C/SUB.w;v/ . 2. Prove: Ifx,y,u, andvare arbitrary members of a metric space .A;/SUB/ , then j/SUB.x;y//NUL/SUB.u;v/j/DC4/SUB.x;u/C/SUB.v;y/: 3. (a) Suppose that .A;/SUB/ is a metric space, and define /SUB1.u;v/D/SUB.u;v/ 1C/SUB.u;v/: Show that.A;/SUB 1/is a metric space. (b) Show that infinitely many metrics can be defined on any set Awith more than one member. 4. Let.A;/SUB/ be a metric space, and let /ESC.u;v/D/SUB.u;v/ 1C/SUB.u;v/: Show that a subset of Ais open in.A;/SUB/ if and only if it is open in .A;/ESC/ . 5. Show that ifAis an arbitrary nonempty set, then /SUB.u;v/D/SUB0ifvDu; 1ifv¤u; is a metric on A. 6. Suppose that .A;/SUB/ is a metric space, u02A, andr >0 . (a) Show thatSr.u0//SUB˚uˇˇ/SUB.u;u 0//DC4r/TABifAcontains more than one point. (b) Verify that if /SUBis the discrete metric, then S1.u0/¤˚ uˇˇ/SUB.u;u 0//DC41/TAB . Section 8.1 Introduction to Metric Spaces 533 7. Prove: (a) The intersection of finitely many open sets is open. (b) The union of finitely many closed sets is closed. 8. Prove: (a) IfUis a neighborhood of u0andU/SUBV, thenVis a neighborhood of u0. (b) IfU1,U2, . . . ,Unare neighborhoods of u0, so is\n iD1Ui. 9. Prove: A limit point of a set Sis either an interior point or a boundary point of S. 10. Prove: An isolated point of Sis a boundary point of Sc. 11. Prove: (a) A boundary point of a set Sis either a limit point or an isolated point of S. (b) A setSis closed if and only if SDS. 12. LetSbe an arbitrary set. Prove: (a)@Sis closed. (b)S0is open. (c)The exterior ofSis open. (d) The limit points of Sform a closed set. (e)/NULS/SOHDS. 13. Prove: (a).S1\S2/0DS0 1\S0 2 (b)S0 1[S0 2/SUB.S1[S2/0 14. Prove: (a)@.S1[S2//SUB@S1[@S2 (b)@.S1\S2//SUB@S1[@S2 (c)@S/SUB@S (d)@SD@Sc (e)@.S/NULT//SUB@S[@T 15. Show that kXkD maxfjx1j;jx2j;:::;jxnjg is a norm on Rn. 16. Suppose that .Ai;/SUBi/,1/DC4i/DC4k, are metric spaces. Let ADA1/STXA2/STX/SOH/SOH/SOH/STXAkD˚ XD.x1;x2;:::;x k/ˇˇxi2Ai;1/DC4i/DC4k/TAB : IfXandYare inA, let /SUB.X;Y/DkX iD1/SUB.xi;yi/: (a) Show that/SUBis a metric on A. 534 Chapter 8 Metric Spaces (b) LetfXrg1 rD1Df.x1r;x2r;:::;x kr/g1 rD1be a sequence in A. Show that lim r!1XrDbXD.bx1;bx2;:::;bxk/ if and only if lim r!1xirDbxi; 1/DC4i/DC4k: (c) Show thatfXrg1 rD1is a Cauchy sequence in .A;/SUB/ if and only iffxirg1 rD1is a Cauchy sequence in .Ai;/SUBi/,1/DC4i/DC4k. (d) Show that.A;/SUB/ is complete if and only if .Ai;/SUBi/is complete,1/DC4i/DC4k. 17. For each positive integer i, let.Ai;/SUBi/be a metric space. Let Abe the set of all objects of the form XD.x1;x2;:::;x n;:::/ , wherexi2Ai,i/NAK1. (For example, ifAiDR,i/NAK1, thenADR1.) Letf˛ig1 iD1be any sequence of positive numbers such thatP1 iD1˛i<1. (a) Show that /SUB.X;Y/D1X iD1˛i/SUBi.xi;yi/ 1C/SUBi.xi;yi/ is a metric on A. (b) LetfXrg1 rD1Df.x1r;x2r;:::;x nr;:::/g1 rD1be a sequence in A. Show that lim r!1XrDbXD.bx1;bx2;:::;bxn;:::/ if and only if lim r!1xirDbxi; i/NAK1: (c) Show thatfXrg1 rD1is a Cauchy sequence in .A;/SUB/ if and only iffxirg1 rD1is a Cauchy sequence in .Ai;/SUBi/for alli/NAK1. (d) Show that.A;/SUB/ is complete if and only if .Ai;/SUBi/is complete for all i/NAK1. 18. LetCŒ0;1/be the set of all real-valued functions continuous on Œ0;1/. For each nonnegative integer n, let kfknDmax˚ jf.x/jˇˇ0/DC4x/DC4n/TAB and /SUBn.f;g/Dkf/NULgkn 1Ckf/NULgkn: Define /SUB.f;g/D1X nD11 2n/NUL1/SUBn.f;g/: (a) Show that/SUBis a metric on CŒ0;1/. Section 8.1 Introduction to Metric Spaces 535 (b) Letffkg1 kD1be a sequence of functions in CŒ0;1/. Show that lim k!1fkDf in the sense of Definition 8.1.14 if and only if lim k!1fk.x/Df.x/ uniformly on every finite subinterval of Œ0;1/. (c) Show that.CŒ0;1/;/SUB/ is complete. 19. Show that Minkowski’s inequality is false if 0<p<1 . 20. Suppose that 0<p<1 . Show that if uandvare nonnegative, then .uCv/p/DC4upCvp: Use this to show that if X,Y2Rn, /SUB.X/DnX iD1jxijp;and/SUB.Y/DnX iD1jyijp; then /SUB.XCY//DC4/SUB.X/C/SUB.Y/: Is/SUBa norm on Rn? 21. Suppose that XDfxig1 iD1is in`p, wherep>1 . Show that (a) X2`rfor allr >p ; (b) Ifr >p , thenkXkr/DC4kXkp; (c) limr!1kXkrDkXk1. 22. Let.A;/SUB/ be a metric space. (a) Suppose thatfungandfvngare sequences in A, lim n!1unDu, and lim n!1vnD v. Show that lim n!1/SUB.u n;vn/D/SUB.u;v/ . (b) Conclude from (b) that if lim n!1unDuandvis arbitrary in A, then limn!1/SUB.u n;v/D/SUB.u;v/ . 23. Prove: Iffurg1 rD1is a Cauchy sequence in a normed vector space .A;k/SOHk/, then fkurkg1 rD1is bounded. 24. Let AD( X2R1ˇˇthe partial sums1X iD1xi;n/NAK1;are bounded) : (a) Show that kXkD sup n/NAK1ˇˇˇˇˇnX iD1xiˇˇˇˇˇ is a norm onA. (b) Let/SUB.X;Y/DkX/NULYk. Show that.A;/SUB/ is complete. 536 Chapter 8 Metric Spaces 25. (a) Show that kfkDZb ajf.x/jdx is a norm onCŒa;b/c141 , (b) Show that the sequence ffngdefined by fn.x/D/DLEx/NULa b/NULa/DC1n is a Cauchy sequence in .CŒa;b/c141;k/SOHk/. (c) Show that.CŒa;b/c141;k/SOHk/is not complete. 26. (a) Verify that`1is a normed vector space. (b) Show that`1is complete. 27. LetAbe the subset of R1consisting of convergent sequences XDfxig1 iD1. Define kXkD supi/NAK1jxij. Show that.A;k/SOHk/is a complete normed vector space. 28. LetAbe the subset of R1consisting of sequences XDfxig1 iD1such that lim i!1xiD 0. DefinekXkD max˚ jxijˇˇi/NAK1/TAB . Show that.A;k/SOHk/is a complete normed vector space. 29. (a) Show that Rn pis complete if p/NAK1. (b) Show that`pis complete if p/NAK1. 30. Show that if XDfxig1 iD12`pandYDfyig1 iD12`q, where1=pC1=qD1, then ZDfxiyig2`1. 8.2 COMPACT SETS IN A METRIC SPACE Throughout this section it is to be understood that .A;/SUB/ is a metric space and that the sets under consideration are subsets of A. We say that a collection Hof open subsets of Ais an open covering ofTifT/SUB [˚ HˇˇH2H/TAB . We say that Thas the Heine–Borel property if every open covering H ofTcontains a finite collection bHsuch that T/SUB[n HˇˇH2bHo : From Theorem 1.3.7 , every nonempty closed and bounded subset of the real number s has the Heine–Borel property. Moreover, from Exercise 1.3.21 , any nonempty set of reals that has the Heine–Borel property is closed and bounded. Giv en these results, we defined a compact set of reals to be a closed and bounded set, and we now draw the following conclusion: A nonempty set of real numbers has the Heine–Borel property i f and only if it is compact . Section 8.2 Compact Sets in a Metric Space 537 The definition of boundedness of a set of real numbers is based on the ordering of the real numbers: if aandbare distinct real numbers then either a<b orb <a . Since there is no such ordering in a general metric space, we introduce th e following definition. Definition 8.2.1 Thediameter of a nonempty subset SofAis d.S/Dsup˚ /SUB.u;v/ˇˇu;v2T/TAB : Ifd.S/<1thenSisbounded . As we will see below, a closed and bounded subset of a general m etric space may fail to have the Heine–Borel property. Since we want “compact" an d “has the Heine–Borel property" to be synonymous in connection with a general metr ic space, we simply make the following definition. Definition 8.2.2 A setTiscompact if it has the Heine–Borel property. Theorem 8.2.3 An infinite subset TofAis compact if and only if every infinite subset ofThas a limit point in T: Proof Suppose that Thas an infinite subset Ewith no limit point in T. Then, ift2T, there is an open set Htsuch thatt2HtandHtcontains at most one member of E. Then HD[˚ Htˇˇt2T/TAB is an open covering of T, but no finite collection fHt1;Ht2;:::;H tkg of sets from Hcan coverE, sinceEis infinite. Therefore, no such collection can cover T; that is,Tis not compact. Now suppose that every infinite subset of Thas a limit point in T, and let Hbe an open covering ofT. We first show that there is a sequence fHig1 iD1of sets from Hthat covers T. If/SI > 0 , thenTcan be covered by /SI-neighborhoods of finitely many points of T. We prove this by contradiction. Let t12T. IfN/SI.t1/does not cover T, there is at22Tsuch that/SUB.t1;t2//NAK/SI. Now suppose that n/NAK2and we have chosen t1,t2, . . . ,tnsuch that /SUB.ti;tj//NAK/SI,1/DC4i < j/DC4n. If[n iD1N/SI.ti/does not cover T, there is atnC12Tsuch that/SUB.ti;tnC1//NAK/SI,1/DC4i/DC4n. Therefore,/SUB.ti;tj//NAK/SI,1/DC4i < j/DC4nC1. Hence, by induction, if no finite collection of /SI-neighborhoods of points in TcoversT, there is an infinite sequenceftng1 nD1inTsuch that/SUB.ti;tj//NAK/SI,i¤j. Such a sequence could not have a limit point, contrary to our assumption. By taking/SIsuccessively equal to 1,1=2, . . . ,1=n, . . . , we can now conclude that, for eachn, there are points t1n,t2n, . . . ,tkn;nsuch that T/SUBkn[ iD1N1=n.tin/: DenoteBinDN1=n.tin/,1/DC4i/DC4n,n/NAK1, and define fG1;G2;G3;:::gDfB11;:::;B k1;1;B12;:::;B k2;2;B13;:::;B k3;3;:::g: 538 Chapter 8 Metric Spaces Ift2T, there is anHinHsuch thatt2H. SinceHis open, there is an /SI >0 such thatN/SI.t//SUBH. Sincet2Gjfor infinitely many values of jand lim j!1d.G j/D0, Gj/SUBN/SI.t//SUBH for somej. Therefore, iffGjig1 iD1is the subsequence of fGjgsuch thatGjiis a subset of someHiinH(thefHigare not necessarily distinct), then T/SUB1[ iD1Hi: (8.2.1) We will now show that T/SUBN[ iD1Hi: (8.2.2) for some integer N. If this is not so, there is an infinite sequence ftng1 nD1inTsuch that tn…n[ iD1Hi; n/NAK1: (8.2.3) From our assumption, ftng1 nD1has a limittinT. From ( 8.2.1 ),t2Hkfor somek, so N/SI.t//SUBHkfor some/SI>0 . Since lim n!1tnDt, there is an integer Nsuch that tn2N/SI.t//SUBHk/SUBn[ iD1Hi; n>k; which contradicts ( 8.2.3 ). This verifies ( 8.2.2 ), soTis compact. Any finite subset of a metric space obviously has the Heine–Bo rel property and is there- fore compact. Since Theorem 8.2.3 does not deal with finite sets, it is often more convenient to work with the following criterion for compactness, which is also applicable to finite sets. Theorem 8.2.4 A subsetTof a metricAis compact if and only if every infinite se- quenceftngof members of Thas a subsequence that converges to a member of T: Proof Suppose that Tis compact andftng/SUBT. Ifftnghas only finitely many distinct terms, there is a tinTsuch thattnDtfor infinitely many values of n; if this is so for n1<n 2</SOH/SOH/SOH, then lim j!1tnjDt. Ifftnghas infinitely many distinct terms, then ftng has a limit point tinT, so there are integers n1< n 2</SOH/SOH/SOHsuch that/SUB.tnj;t/ < 1=j ; therefore, lim j!1tnjDt. Conversely, suppose that every sequence in Thas a subsequence that converges to a limit inT. IfSis an infinite subset of T, we can choose a sequence ftngof distinct points in S. By assumption,ftnghas a subsequence that converges to a member tofT. Sincetis a limit point offtng, and therefore of T,Tis compact. Theorem 8.2.5 IfTis compact;then every Cauchy sequence ftng1 nD1inTconverges to a limit inT: Section 8.2 Compact Sets in a Metric Space 539 Proof By Theorem 8.2.4 ,ftnghas a subsequenceftnjgsuch that lim j!1tnjDt2T: (8.2.4) We will show that lim n!1tnDt. Suppose that /SI > 0 . Sinceftngis a Cauchy sequence, there is an integer Nsuch that /SUB.tn;tm/</SI ,n>m/NAKN. From ( 8.2.4 ), there is anmDnj/NAKNsuch that/SUB.tm;t/</SI . Therefore, /SUB.tn;t//DC4/SUB.tn;tm/C/SUB.tm;t/<2/SI; n/NAKm: Theorem 8.2.6 IfTis compact;thenTis closed and bounded. Proof Suppose that tis a limit point of T. For eachn, choosetn¤t2B1=n.t/\T. Then lim n!1tnDt. Since every subsequence of ftngalso converges to t,t2T, by Theorem 8.2.3 . Therefore,Tis closed. The family of unit open balls HD˚ B1.t/ˇˇt2T/TAB is an open covering of T. SinceTis compact, there are finitely many members t1,t2, . . . ,tnofTsuch thatS/SUB[n jD1B1.tj/. Ifuandvare arbitrary members of T, thenu2B1.tr/andv2B1.ts/for somerandsin f1;2;:::;ng, so /SUB.u;v//DC4/SUB.u;t r/C/SUB.tr;ts/C/SUB.ts;v/ /DC42C/SUB.tr;ts//DC42Cmax˚ /SUB.ti;tj/ˇˇ1/DC4i <j/DC4n/TAB : Therefore,Tis bounded. The converse of Theorem 8.2.6 is false; for example, if Ais any infinite set equipped with the discrete metric (Example 8.1.2 .), then every subset of Ais bounded and closed. However, ifTis an infinite subset of A, then HD˚ftgˇˇt2T/TABis an open covering of T, but no finite subfamily of HcoversT. Definition 8.2.7 A setTistotally bounded if for every/SI > 0 there is a finite set T/SI with the following property: if t2T, there is ans2T/SIsuch that/SUB.s;t/</SI . We say that T/SIis afinite/SI-net forT. We leave it to you (Exercise 8.2.4 ) to show that every totally bounded set is bounded and that the converse is false. 540 Chapter 8 Metric Spaces Theorem 8.2.8 IfTis compact;thenTis totally bounded. Proof We will prove that if Tis not totally bounded, then Tis not compact. If Tis not totally bounded, there is an /SI>0 such that there is no finite /SI-net forT. Lett12T. Then there must be a t2inTsuch that/SUB.t1;t2/ > /SI . (If not, the singleton set ft1gwould be a finite/SI-net forT.) Now suppose that n/NAK2and we have chosen t1,t2, . . . ,tnsuch that /SUB.ti;tj//NAK/SI,1/DC4i < j/DC4n. Then there must be a tnC12Tsuch that/SUB.ti;tnC1//NAK/SI, 1/DC4i/DC4n. (If not,ft1;t2;:::;t ngwould be a finite /SI-net forT.) Therefore, /SUB.ti;tj//NAK/SI, 1/DC4i <j/DC4nC1. Hence, by induction, there is an infinite sequence ftng1 nD1inTsuch that/SUB.ti;tj//NAK/SI,i¤j. Since such a sequence has no limit point, Tis not compact, by Theorem 8.2.4 . Section 8.2 Compact Sets in a Metric Space 541 Theorem 8.2.9 If.A;/SUB/ is complete and Tis closed and totally bounded ;thenTis compact. Proof LetSbe an infinite subset of T, and letfsig1 iD1be a sequence of distinct members ofS. We will show that fsig1 iD1has a convergent subsequence. Since Tis closed, the limit of this subsequence is in T, which implies that Tis compact, by Theorem 8.2.4 . Forn/NAK1, letT1=nbe a finite1=n-net forT. Letfsi0g1 iD1Dfsig1 iD1. SinceT1is finite andfsi0g1 iD1is infinite, there must be a member t1ofT1such that/SUB.si0;t1//DC41 for infinitely many values of i. Letfsi1g1 iD1be the subsequence of fsi0g1 iD1such that /SUB.si1;t1//DC41. We continue by induction. Suppose that n > 1 and we have chosen an infinite subse- quencefsi;n/NUL1g1 iD1offsi;n/NUL2g1 iD1. SinceT1=nis finite andfsi;n/NUL1g1 iD1is infinite, there must be member tnofT1=nsuch that/SUB.si;n/NUL1;tn//DC41=n for infinitely many values of i. Letfsing1 iD1be the subsequence of fsi;n/NUL1g1 iD1such that/SUB.sin;tn//DC41=n. From the triangle inequality, /SUB.sin;sj n//DC42=n; i;j/NAK1; n/NAK1: (8.2.5) Now letbsiDsii,i/NAK1. Thenfbsig1 iD1is an infinite sequence of members of T. Mo- roever, ifi;j/NAKn, thenbsiandbsjare both included in fsing1 iD1, so ( 8.2.5 ) implies that /SUB.bsi;bsj//DC42=n; that is,fbsig1 iD1is a Cauchy sequence and therefore has a limit, since .A;/SUB/ is complete. Example 8.2.1 LetTbe the subset of `1such thatjxij/DC4/SYNi,i/NAK1, where lim i!1/SYNiD 0. Show thatTis compact. Solution We will show that Tis totally bounded in `1. Since`1is complete (Exer- cise8.1.26 ), Theorem 8.2.9 will then imply that Tis compact. Let/SI>0 . ChooseNso that/SYNi/DC4/SIifi >N . Let/SYNDmax˚/SYNiˇˇ1/DC4i/DC4n/TABand letp be an integer such that p/SI>/SYN . LetQ/SID˚ri/SIˇˇriDinteger inŒ/NULp;p/c141/TAB. Then the subset of`1such thatxi2Q/SI,1/DC4i/DC4N, andxiD0,i >N , is a finite/SI-net forT. Compact Subsets of CŒa;b/c141 In Example 8.1.7 we showed that CŒa;b/c141 is a complete metric space under the metric /SUB.f;g/Dkf/NULgkD max˚jf.x//NULg.x/jˇˇa/DC4x/DC4b/TAB: We will now give necessary and sufficient conditions for a sub set ofCŒa;b/c141 to be compact. Definition 8.2.10 A subsetTofCŒa;b/c141 isuniformly bounded if there is a constant M such that jf.x/j/DC4M ifa/DC4x/DC4bandf2T: (8.2.6) A subsetTofCŒa;b/c141 isequicontinuous if for each/SI>0 there is aı>0 such that jf.x 1//NULf.x 2/j/DC4/SIifx1;x22Œa;b/c141;jx1/NULx2j<ı; andf2T: (8.2.7) 542 Chapter 8 Metric Spaces Theorem 2.2.8 implies that for each finCŒa;b/c141 there is a constant Mfwhich depends onf, such that jf.x/j/DC4Mfifa/DC4x/DC4b; and Theorem 2.2.12 implies that there is a constant ıfwhich depends on fand/SIsuch that jf.x 1//NULf.x 2/j/DC4/SIifx1;x22Œa;b/c141 andjx1/NULx2j<ıf: The difference in Definition 8.2.11 is that the sameMandıapply to allfinT. Theorem 8.2.11 A nonempty subset TofCŒa;b/c141 is compact if and only if it is closed ; uniformly bounded ;and equicontinuous. Proof For necessity, suppose that Tis compact. Then Tis closed (Theorem 8.2.6 ) and totally bounded (Theorem 8.2.8 ). Therefore, if /SI > 0 , there is a finite subset T/SID fg1;g2;:::;g kgofCŒa;b/c141 such that iff2T, thenkf/NULgik/DC4/SIfor someiinf1;2;:::;kg. If we temporarily let /SID1, this implies that kfkDk.f/NULgi/Cgik/DC4kf/NULgikCkgik/DC41Ckgik; which implies ( 8.2.6 ) with MD1Cmax˚ kgikˇˇ1/DC4i/DC4k/TAB : For ( 8.2.7 ), we again let /SIbe arbitary, and write jf.x 1//NULf.x 2/j/DC4jf.x 1//NULgi.x1/jCjgi.x1//NULgi.x2/jCjgi.x2//NULf.x 2/j /DC4jgi.x1//NULgi.x2/jC2kf/NULgik <jgi.x1//NULgi.x2/jC2/SI:(8.2.8) Since each of the finitely many functions g1,g2, . . . ,gkis uniformly continuous on Œa;b/c141 (Theorem 2.2.12 ), there is aı>0 such that jgi.x1//NULgi.x2/j</SI ifjx1/NULx2j<ı; 1/DC4i/DC4k: This and ( 8.2.8 ) imply ( 8.2.7 ) with/SIreplaced by3/SI. Since this replacement is of no consequence, this proves necessity. For sufficiency, we will show that Tis totally bounded. Since Tis closed by assumption andCŒa;b/c141 is complete, Theorem 8.2.9 will then imply that Tis compact. Letmandnbe positive integers and let /CANrDaCr m.b/NULa/; 0/DC4r/DC4m; and/DC1sDsM n;/NULn/DC4s/DC4nI that is,aD/CAN0< /CAN 1</SOH/SOH/SOH< /CAN mDbis a partition of Œa;b/c141 into subintervals of length .b/NULa/=m , and/NULMD/DC1/NULn< /DC1 /NULnC1</SOH/SOH/SOH< /DC1 n/NUL1< /DC1 nDMis a partition of the Section 8.2 Compact Sets in a Metric Space 543 segment of the y-axis between yD/NULMandyDMinto subsegments of length M=n . LetSmnbe the subset of CŒa;b/c141 consisting of functions gsuch that fg./CAN0/;g./CAN 1/;:::;g./CAN m/g/SUBf/DC1/NULn;/DC1/NULnC1:::;/DC1 n/NUL1;/DC1ng andgis linear onŒ/CANi/NUL1;/CANi/c141,1/DC4i/DC4m. Since there are only .mC1/.2nC1/points of the form./CANr;/DC1s/,Smnis a finite subset of CŒa;b/c141 . Now suppose that /SI > 0 , and choose ı > 0 to satisfy ( 8.2.7 ). Choosemandnso that .b/NULa/=m<ı and2M=n</SI . Iffis an arbitrary member of T, there is aginSmnsuch that jg./CANi//NULf./CAN i/j</SI; 0/DC4i/DC4m: (8.2.9) If0/DC4i/DC4m/NUL1, jg./CANi//NULg./CANiC1/jDjg./CANi//NULf./CAN i/jCjf./CAN i//NULf./CAN iC1/jCjf./CAN iC1//NULg./CANiC1/j:(8.2.10) Since/CANiC1/NUL/CANi<ı, (8.2.7 ), (8.2.9 ), and ( 8.2.10 ) imply that jg./CANi//NULg./CANiC1/j<3/SI: Therefore, jg./CANi//NULg.x/j<3/SI; /CAN i/DC4x/DC4/CANiC1; (8.2.11) sincegis linear onŒ/CANi;/CANiC1/c141. Now letxbe an arbitrary point in Œa;b/c141 , and chooseiso thatx2Œ/CANi;/CANiC1/c141. Then jf.x//NULg.x/j/DC4jf.x//NULf./CAN i/jCjf./CAN i//NULg./CANi/jCjg./CANi//NULg.x/j; so (8.2.7 ), (8.2.9 ), and ( 8.2.11 ) imply thatjf.x//NULg.x/j<5/SI ,a/DC4x/DC4b. Therefore,Smn is a finite5/SI-net forT, soTis totally bounded. Theorem 8.2.12 ( Ascoli –Arzela Theorem) Suppose that Fis an infinite uni- formly bounded and equicontinuous family of functions on Œa;b/c141: Then there is a sequence ffnginFthat converges uniformly to a continuous function on Œa;b/c141: Proof LetTbe the closure of F; that is,f2Tif and only if either f2Torf is the uniform limit of a sequence of members of F. ThenTis also uniformly bounded and equicontinuous (verify), and Tis closed. Hence, Tis compact, by Theorem 8.2.12 . Therefore, Fhas a limit point in T. (In this context, the limit point is a function fin T.) Sincefis a limit point of F, there is for each integer na functionfninFsuch that kfn/NULfk<1=n ; that isffngconverges uniformly to fonŒa;b/c141 . 8.2 Exercises 1. Suppose that T1,T2, . . . ,Tkare compact sets in a metric space .A;/SUB/ . Show that [k jD1Tjis compact. 544 Chapter 8 Metric Spaces 2. (a) Show that a closed subset of a compact set is compact. (b) Suppose that Tis any collection of closed subsets of a metric space .A;/SUB/ , and somebTinTis compact. Show that \˚TˇˇT2T/TABis compact. (c) Show that if Tis a collection of compact subsets of a metric space .A;/SUB/ , then\˚ TˇˇT2T/TAB is compact. 3. IfSandTare nonempty subsets of a metric space .A;/SUB/ , we define the distance fromStoTby dist.S;T/Dinf˚ /SUB.s;t/ˇˇs2S;t2T/TAB : Show that if SandTare compact, then dist .S;T/D/SUB.s;t/ for somesinSand sometinT. 4. (a) Show that every totally bounded set is bounded. (b) Let ıirD/SUB1ifiDr; 0ifi¤r; and letTbe the subset of `1consisting of the sequences XrDfıirg1 iD1, r/NAK1. Show thatTis bounded, but not totally bounded. 5. LetTbe a compact subset of a metric space .A;/SUB/ . Show that there are members s andtofTsuch thatd.s;t/Dd.T/ . 6. LetTbe the subset of `1such thatjxij/DC4/SYNi,i/NAK1, whereP1 iD1/SYNi<1. Show thatTis compact. 7. LetTbe the subset of `2such thatjxij/DC4/SYNi,i/NAK1, whereP1 i/SYN2 i<1. Show thatTis compact. 8. LetSbe a nonempty subset of a metric space .A;/SUB/ and letu0be an arbitrary member ofA. Show thatSis bounded if and only if DD˚/SUB.u;u 0/ˇˇu2S/TABis bounded. 9. Let.A;/SUB/ be a metric space. (a) Prove: IfSis a bounded subset of A, thenS(closure ofS) is bounded. Find d.S/. (b) Prove: If every bounded closed subset of Ais compact, then .A;/SUB/ is com- plete. 10. Let.A;/SUB/ be the metric space defined in Exercise 8.1.16 Let TDT1/STXT2/STX/SOH/SOH/SOH/STXTk; whereTi/SUBAiandTi¤;,1/DC4i/DC4k. Show thatTis compact if and only Tiis compact for1/DC4i/DC4k. 11. Let.A;/SUB/ be the metric space defined in Exercise 8.1.17 . Let TDT1/STXT2/STX/SOH/SOH/SOH/STXTn/STX/SOH/SOH/SOH; Section 8.3 Continuous Functions on Metric Spaces 545 whereTi/SUBAiandTi¤;,i/NAK1. Show that if Tis compact, then Tiis compact for alli/NAK1. 12. LetfTng1 nD1be a sequence of nonempty closed sets of a metric space such th at(a) T1is compact; (b)TnC1/SUBTn,n/NAK1; and(c)limn!1d.T n/D0. Show that \1 nD1Tncontains exactly one member. 8.3 CONTINUOUS FUNCTIONS ON METRIC SPACES In Chapter we studied real-valued functions defined on subse ts ofRn, and in Chapter 6.4. we studied functions defined on subsets of Rnwith values in Rm. These are examples of functions defined on one metric space with values in another m etric space.(Of course, the two spaces are the same if nDm.) In this section we briefly consider functions defined on subse ts of a metric space .A;/SUB/ with values in a metric space .B;/ESC/ . We indicate that fis such a function by writing fW.A;/SUB/!.B;/ESC/: Thedomain andrange offare the sets DfD˚ u2Aˇˇf.u/ is defined/TAB and RfD˚v2BˇˇvDf.u/ for someuinDf/TAB: Definition 8.3.1 We say that lim u!buf.u/Dbv ifbu2Dfand for each/SI>0 there is aı>0 such that /ESC.f.u/;bv/</SI ifu2Dfand0</SUB.u;bu/<ı: (8.3.1) Definition 8.3.2 We say thatfiscontinuous atbuifbu2Dfand for each/SI>0 there is aı>0 such that /ESC.f.u/;f.bu//</SI ifu2Df\Nı.bu/: (8.3.2) Iffis continuous at every point of a set S, thenfiscontinuous on S. Note that ( 8.3.2 ) can be written as f.D f\Nı.bu///SUBN/SI.f.bu//: Also,fis automatically continuous at every isolated point of Df. (Why?) 546 Chapter 8 Metric Spaces Example 8.3.1 If.A;k/SOHk/is a normed vector space, then Theorem 8.3.5 implies that f.u/Dkukis a continuous function from .A;/SUB/ toR, since jkuk/NULkbukj/DC4ku/NULbuk: Here we are applying Definition 8.3.2 with/SUB.u;bu/Dku/NULbukand/ESC.v;bv/Djv/NULbvj. Theorem 8.3.3 Suppose that bu2Df:Then lim u!buf.u/Dbv (8.3.3) if and only if lim n!1f.u n/Dbv (8.3.4) for every sequencefunginDfsuch that lim n!1unDbu: (8.3.5) Proof Suppose that ( 8.3.3 ) is true, and letfungbe a sequence in Dfthat satisfies (8.3.5 ). Let/SI > 0 and chooseı > 0 to satisfy ( 8.3.1 ). From ( 8.3.5 ), there is an inte- gerNsuch that/SUB.u n;bu/ < ı ifn/NAKN. Therefore, /ESC.f.u n/;bv/ < /SI ifn/NAKN, which implies ( 8.3.4 ). For the converse, suppose that ( 8.3.3 ) is false. Then there is an /SI0> 0 and a sequence funginDfsuch that/SUB.u n;bu/<1=n and/ESC.f.u n/;bv//NAK/SI0, so ( 8.3.4 ) is false. We leave the proof of the next two theorems to you. Theorem 8.3.4 A functionfis continuous at buif and only if lim u!buf.u/Df.bu/: Theorem 8.3.5 A functionfis continuous at buif and only if lim n!1f.u n/Df.bu/ wheneverfungis a sequence in Dfthat converges to bu. Theorem 8.3.6 Iffis continuous on a compact set T;thenf.T/ is compact. Proof Letfvngbe an infinite sequence in f.T/ . For eachn,vnDf.u n/for someun2 T. SinceTis compact,funghas a subsequencefunjgsuch that lim j!1unjDbu2T (Theorem 8.2.4 ). From Theorem 8.3.5 , lim j!1f.u nj/Df.bu/; that is, lim j!1vnjD f.bu/. Therefore,f.T/ is compact, again by Theorem 8.2.4 . Definition 8.3.7 A functionfisuniformly continuous on a subsetSofDfif for each /SI>0 there is aı>0 such that /ESC.f.u/;f.v//</SI whenever/SUB.u;v/<ı andu;v2S: Section 8.3 Continuous Functions on Metric Spaces 547 Theorem 8.3.8 Iffis continuous on a compact set T;thenfis uniformly continuous onT. Proof Iffis not uniformly continuous on T, then for some /SI0>0there are sequences fungandfvnginTsuch that/SUB.u n;vn/<1=n and /ESC.f.u n/;f.v n///NAK/SI0: (8.3.6) SinceTis compact,funghas a subsequencefunkgthat converges to a limit buinT(Theo- rem8.2.4 ). Since/SUB.u nk;vnk/<1=n k, lim k!1vnkDbualso. Then lim k!1f.u nk/Dlim k!1f.v nk/Df.bu/ (Theorem 8.3.5 ), which contradicts ( 8.3.6 ). Definition 8.3.9 IffW.A;/SUB/!.A;/SUB/ is defined on all of Aand there is a constant ˛ in.0;1/ such that /SUB.f.u/;f.v///DC4˛/SUB.u;v/ for all.u;v/2A/STXA; (8.3.7) thenfis acontraction of.A;/SUB/ . We note that a contraction of .A;/SUB/ is uniformly continuous on A. Theorem 8.3.10 (Contraction Mapping Theorem) Iffis a contraction of a complete metric space .A;/SUB/; then the equation f.u/Du (8.3.8) has a unique solution : Proof To see that ( 8.3.8 ) cannot have more than one solution, suppose that uDf.u/ andvDf.v/ . Then /SUB.u;v/D/SUB.f.u/;f.v//: (8.3.9) However, ( 8.3.7 ) implies that /SUB.f.u/;f.v///DC4˛/SUB.u;v/: (8.3.10) Since ( 8.3.9 ) and ( 8.3.10 ) imply that /SUB.u;v//DC4˛/SUB.u;v/ and˛<1 , it follows that /SUB.u;v/D0. HenceuDv. We will now show that ( 8.3.8 ) has a solution. With u0arbitrary, define unDf.u n/NUL1/; n/NAK1: (8.3.11) We will show thatfungconverges. From ( 8.3.7 ) and ( 8.3.11 ), /SUB.u nC1;un/D/SUB.f.u n/;f.u n/NUL1///DC4˛/SUB.u n;un/NUL1/: (8.3.12) 548 Chapter 8 Metric Spaces The inequality /SUB.u nC1;un//DC4˛n/SUB.u 1;u0/; n/NAK0; (8.3.13) follows by induction from ( 8.3.12 ). Ifn>m , repeated application of the triangle inequality yields /SUB.u n;um//DC4/SUB.u n;un/NUL1/C/SUB.u n/NUL1;un/NUL2/C/SOH/SOH/SOHC/SUB.u mC1;um/; and ( 8.3.13 ) yields /SUB.u n;um//DC4/SUB.u 1;u0/˛m.1C˛C/SOH/SOH/SOHC˛n/NULm/NUL1/<˛m 1/NUL˛: Now it follows that /SUB.u n;um/</SUB.u 1;u0/ 1/NUL˛˛Nifn;m>N; and, since lim N!1˛ND0,fungis a Cauchy sequence. Since Ais complete,funghas a limitbu. Sincefis continuous at bu, f.bu/Dlim n!1f.u n/NUL1/Dlim n!1unDbu; where Theorem 8.3.5 implies the first equality and ( 8.3.11 ) implies the second. Example 8.3.2 Suppose that hDh.x/ is continuous on Œa;b/c141 ,KDK.x;y/ is con- tinuous onŒa;b/c141/STXŒa;b/c141 , andjK.x;y/j /DC4Mifa/DC4x;y/DC4b. Show that ifj/NAKj< 1=M.b/NULa/there is a unique uinCŒa;b/c141 such that u.x/Dh.x/C/NAKZb aK.x;y/u.y/dy; a /DC4x/DC4b: (8.3.14) (This is Fredholm ’s integral equation .) Solution LetAbeCŒa;b/c141 , which is complete. If u2CŒa;b/c141 , letf.u/Dv, where v.x/Dh.x/C/NAKZb aK.x;y/u.y/dy; a /DC4x/DC4b: Sincev2CŒa;b/c141 ,fWCŒa;b/c141!CŒa;b/c141 . Ifu1,u22CŒa;b/c141 , then jv1.x//NULv2.x/j/DC4j/NAKjZb ajK.x;y/jju1.y//NULv1.y/jdy; so kv1/NULv2k/DC4j/NAKjM.b/NULa/ku1/NULu2k: Sincej/NAKjM.b/NULa/<1 ,fis a contraction. Hence, there is a unique uinCŒa;b/c141 such that f.u/Du. Thisusatisfies ( 8.3.14 ). Section 8.3 Continuous Functions on Metric Spaces 549 8.3 Exercises 1. Suppose that fW.A;/SUB/!.B;/ESC/ andDfDA. Show that the following state- ments are equivalent. (a)fis continuous on A. (b) IfVis any open set in .B;/ESC/ , thenf/NUL1.V/is open in.A;/SUB/ . (c) IfVis any closed set in .B;/ESC/ , thenf/NUL1.V/is closed in.A;/SUB/ . 2. A metric space .A;/SUB/ isconnected ifAcannot be written as ADA1[A2, where A1andA2are nonempty disjoint open sets. Suppose that .A;/SUB/ is connected and fW.A;/SUB/!.B;/ESC/ , whereDfDA,RfDB, andfis continuous on A. Show that.B;/ESC/ is connected. 3. Letfbe a continuous real-valued function on a compact subset Sof a metric space .A;/SUB/ . Let/ESCbe the usual metric on R; that is,/ESC.x;y/Djx/NULyj. (a) Show thatfis bounded on S. (b) Let˛Dinfu2Sf.u/ andˇDsupu2Sf.u/ . Show that there are points u1 andu2inŒa;b/c141 such thatf.u 1/D˛andf.u 2/Dˇ. 4. LetfW.A;/SUB/!.B;/ESC/ be continuous on a subset UofA. Letube inUand define the real-valued function gW.A;/SUB/!Rby g.u/D/ESC.f.u/;f.u//; u2U: (a) Show thatgis continuous on U. (b) Show that ifUis compact, then gis uniformly continuous on U. (c) Show that if Uis compact, then there is a bu2Usuch thatg.u//DC4g.bu/, u2U. 5. Suppose that .A;/SUB/ ,.B;/ESC/ , and.C;/CR/ are metric spaces, and let fW.A;/SUB/!.B;/ESC/ andgW.B;/ESC/!.C;/CR/; whereDfDA,RfDDgDB, andfandgare continuous. Define hW.A;/SUB/! .C;/CR/ byh.u/Dg.f.u// . Show thathis continuous on A. 6. Let.A;/SUB/ be the set of all bounded real-valued functions on a nonempty setS, with/SUB.u;v/Dsups2Sju.s//NULv.s/j. Lets1,s2, . . . ,skbe members of S, and f.u/Dg.u.s 1/;u.s 2/;:::;u.s k//, wheregis real-valued and continuous on Rk. Show thatfis a continuous function from .A;/SUB/ toR. 7. Let.A;/SUB/ be the set of all bounded real-valued functions on a nonempty setS, with/SUB.u;v/Dsups2Sju.s//NULv.s/j. Show thatf.u/Dinfs2Su.s/ andg.u/D sups2Su.s/ are uniformly continuous functions from .A;/SUB/ toR. 8. LetIŒa;b/c141 be the set of all real-valued functions that are Riemann inte grable on Œa;b/c141 , with/SUB.u;v/Dsupa/DC4x/DC4bju.x//NULv.x/j. Show thatf.u/DZb au.x/dx is a uniformly continuous function from IŒa;b/c141 toR. 550 Answers to Selected Exercises Answers to Selected Exercises Section 1.1 pp. 9–10 1:1:1(p.9)(a)2max.a;b/ (b)2min.a;b/ (c)4max.a;b;c/ (d)4min.a;b;c/ 1:1:5(p.9) (a)1(no);/NUL1(yes)(b)3(no);/NUL3(no)(c)p 7(yes);/NULp 7(yes) (d)2(no);/NUL3(no)(e)1(no);/NUL1(no)(f)p 7(no);/NULp 7(no) Section 1.2 pp. 15–19 1:2:9(p.16) (a)2n=.2n/Š (b)2/SOH3n=.2nC1/Š(c)2/NULn.2n/Š=.nŠ/2(d)nn=nŠ 1:2:10(p.16) (b) no 1:2:11(p.16) (b) no 1:2:20(p.18)AnDxn nŠ0 @lnx/NULnX jD11 j1 A 1:2:21(p.18)fn.x1;x2;:::;x n/D2n/NUL1max.x1;x2;:::;x n/,gn.x1;x2;:::;x n/D 2n/NUL1min.x1;x2;:::;x n/ Section 1.3 pp. 27–29 1:3:1(p.27)(a)Œ1 2;1/;./NUL1;1 2/[Œ1;1/;./NUL1;0/c141[.3 2;1/;.0;3 2/c141;./NUL1;0/c141[.3 2;1/; ./NUL1;1 2/c141[Œ1;1/(b)./NUL3;/NUL2/[.2;3/ ;./NUL1;/NUL3/c141[Œ/NUL2;2/c141[Œ3;1/;;;./NUL1;1/;;; ./NUL1;/NUL3/c141[Œ/NUL2;2/c141[Œ3;1/(c);;./NUL1;1/;;;./NUL1;1/;;;./NUL1;1/ (d);;./NUL1;1/;Œ/NUL1;1/c141;./NUL1;/NUL1/[.1;1/;Œ/NUL1;1/c141;./NUL1;1/ 1:3:2(p.27) (a).0;3/c141(b)Œ0;2/c141(c)./NUL1;1/[.2;1/(d)./NUL1;0/c141[.3;1/ 1:3:4(p.27) (a)1 4(b)1 6(c)6(d)1 Answers to Selected Exercises 551 1:3:5(p.27) (a) neither;./NUL1;2/[.3;1/;./NUL1;/NUL1/[.2;3/ ;./NUL1;/NUL1/c141[.2;3/ ; ./NUL1;/NUL1/c141[Œ2;3/c141(b) open;S;.1;2/ ;Œ1;2/c141(c)closed;./NUL3;/NUL2/[.7;8/ ;./NUL1;/NUL3/[ ./NUL2;7/[.8;1/;./NUL1/NUL3/c141[Œ/NUL2;7/c141[Œ8;1/(d) closed;;;S˚.n;nC1/ˇˇnDinteger/TAB; ./NUL1;1/ 1:3:20(p.28) (a)˚xˇˇxD1=n; nD1;2;:::/TAB;(b);(c),(d)S1Drationals, S2Dirrationals (e)any set whose supremum is an isolated point of the set (f),(g) the rationals (h)S1Drationals,S2Dirrationals Section 2.1 pp. 48–53 2:1:2(p.48)DfDŒ/NUL2;1/[Œ3;1/,DgD./NUL1;/NUL3/c141[Œ3;7/[.7;1/,Df˙gD DfgDŒ3;7/[.7;1/,Df =gD.3;4/[.4;7/[.7;1/ 2:1:3(p.48) (a) ,(b)˚xˇˇx¤.2kC1//EM=2 wherekDinteger/TAB(c)˚xˇˇx¤0;1/TAB (d)˚xˇˇx¤0/TAB(e)Œ1;1/ 2:1:4(p.49) (a)4(b)12(c)/NUL1(d)2(e)/NUL2 2:1:6(p.49) (a)11 17(b)/NUL2 3(c)1 3(d)2 2:1:7(p.49) (a)0;2(b)0, none (c)/NUL1 3;1 3(d) none,0 2:1:15(p.50) (a)0(b)0(c)none (d)0(e)none(f)0 2:1:18(p.50) (a)0(b)0(c)none (d) none (e)none(f)0 2:1:20(p.50) (a)1(b)/NUL1(c)1(d)1(e)1(f)/NUL1 2:1:22(p.51) (a) none(b)1(c)1(d) none 2:1:24(p.51) (a)1(b)1(c)1(d)/NUL1(e)none(f)1 2:1:31(p.52) (a)3 2(b)3 2(c)1(d)/NUL1(e)1(f)1 2 2:1:32(p.52)limx!1r.x/D1 ifn > m andan=bm> 0;D/NUL1 ifn > m and an=bm<0;Dan=bmifnDm;D0ifn<m . lim x!/NUL1r.x/D./NUL1/n/NULmlimx!1r.x/ 2:1:33(p.52)limx!x0f.x/Dlimx!x0g.x/ 2:1:37(p.52)(c) limx!x0/NUL.f/NULg/.x//DC4limx!x0/NULf.x//NULlimx!x0/NULg.x/ ; limx!x0/NUL.f/NUL g/.x//NAKlimx!x0/NULf.x//NULlimx!x0/NULg.x/ Section 2.2 pp. 69–73 2:2:3(p.69) (a) from the right (b) continuous (c)none (d) continuous (e) none (f)continuous (g)from the left 2:2:4(p.69)Œ0;1/ ,.0;1/ ,Œ1;2/ ,.1;2/ ,.1;2/c141 ,Œ1;2/c141 2:2:5(p.69)Œ0;1/ ,.0;1/ , .1;1/2:2:13(p.70) (b) tanhxis continuous for all x, cothxfor allx¤0 552 Answers to Selected Exercises 2:2:16(p.70)No 2:2:21(p.71)(a)Œ/NUL1;1/c141,Œ0;1/(b)S1 nD/NUL1.2n/EM;.2nC1//EM/ , .0;1/(c)S1 nD/NUL1.n/EM;.nC1//EM/ ,./NUL1;/NUL1/[./NUL1;1/[.1;1/(d)S1 nD/NUL1Œn/EM;.nC 1 2//EM/c141,Œ0;1/ 2:2:23(p.71) (a)./NUL1;1/(b)./NUL1;1/(c)x0¤.2kC3 2/EM/; kDinteger (d) x¤1 2(e)x¤1(f)x¤.kC1 2/EM/; kDinteger (g)x¤.kC1 2/EM/; kD integer(h)x¤0(i)x¤0 Section 2.3 pp. 84–88 2:3:4(p.85) (b)p.c/Dq.c/ andp0 /NUL.c/Dq0 C.c/ 2:3:5(p.85)f.k/.x/Dn.n/NUL1//SOH/SOH/SOH.n/NULk/NUL1/xn/NULk/NUL1jxjif1/DC4k/DC4n/NUL1;f.n/.x/DnŠ ifx >0 ;f.n/.x/D/NULnŠifx < 0 ;f.k/.x/D0ifk >n andx¤0;f.k/.0/does not exist ifk/NAKn. 2:3:7(p.85) (a)c0Dac/NULbs,s0DbcCas(b)c.x/Deaxcosbx,s.x/D eaxsinbx 2:3:15(p.86) (b)f.x/D/NUL1ifx/DC40,f.x/D1ifx > 0 ; thenf0.0C/D0, but f0 C.0/does not exist. (c)continuous from the right 2:3:22(p.87)There is no such function (Theorem 2.3.9). 2:3:24(p.87)Counterexample: Let x0D0,f.x/Djxj3=2sin.1=x/ ifx¤0, and f.0/D0. 2:3:27(p.88)Counterexample: Let x0D0,f.x/Dx=jxjifx¤0,f.0/D0. Section 2.4 pp. 96–98 2:4:2(p.96)12:4:3(p.96)1 22:4:4(p.96)1 2:4:5(p.96)./NUL1/n/NUL1n 2:4:6(p.96)12:4:7(p.96)02:4:8(p.96)1 2:4:9(p.96)0 2:4:10(p.96)0 2:4:11(p.96)02:4:12(p.96)/NUL1 2:4:13(p.96)0 2:4:14(p.96)/NUL1 22:4:15(p.96)02:4:16(p.96)0 2:4:17(p.96)1 2:4:18(p.96)1 2:4:19(p.96)12:4:20(p.96)e 2:4:21(p.96)1 2:4:24(p.96)1=e 2:4:22(p.96)0 2:4:23(p.96)/NUL1 if˛/DC40,0if˛>0 2:4:25(p.96)e22:4:26(p.96)12:4:27(p.96)0 2:4:28(p.96)0 2:4:29(p.96)1if˛>0 ,/NUL1 if˛/DC40 2:4:30(p.96)1 2:4:31(p.97)12:4:32(p.97)1=120 2:4:33(p.97)1 2:4:34(p.97)/NUL1 2:4:35(p.97)/NUL1 if˛/DC40,0if˛>0 2:4:36(p.97)1 2:4:37(p.97)12:4:38(p.97)02:4:39(p.97)0 2:4:40(p.97)02:4:41(p.97) (b) Suppose that g0is continuous at x0andf.x/D g.x/ ifx/DC4x0,f.x/D1Cg.x/ ifx>x 0. 2:4:44(p.97) (a)1(b)e(c)12:4:45(p.98)eL Answers to Selected Exercises 553 Section 2.5 pp. 107–112 2:5:2(p.107)f.nC1/.x0/=.nC1/Š.2:5:4(p.107) (b) Counterexample: Let x0D0 andf.x/Dxjxj. 2:5:5(p.108) (b) Letg.x/D1Cjx/NULx0j, sof.x/D.x/NULx0/.1Cjx/NULx0j/. 2:5:6(p.108) (b) Letg.x/D1Cjx/NULx0j, sof.x/D.x/NULx0/2.1Cjx/NULx0j/. 2:5:10(p.109) (b) (i)1,2,2,0(ii)0,/NUL/EM,3/EM=2 ,/NUL4/EMC/EM3=2 (iii)/NUL/EM2=4,/NUL2/EM,/NUL6C/EM2=4,4/EM(iv)/NUL2,5,/NUL16,65 2:5:11(p.109) (b)0,/NUL1,0,5 2:5:12(p.110) (b) (i) 0,1,0,5(ii)/NUL1,0,6,/NUL24(iii)p 2,3p 2,11p 2, 57p 2 (iv)/NUL1,3,/NUL14,88(a) min(b) neither (c) min(d) max(e) min(f)neither (g) min(h) min 2:5:14(p.110)f.x/De/NUL1=x2ifx¤0,f.0/D0(Exercise 2:5:1(p.107)) 2:5:15(p.111)None ifb2/NUL4c<0 ; local min at x1D./NULbCp b2/NUL4c/=2 and local max atx1D./NULb/NULp b2/NUL4c/=2 ifb2/NUL4c>0 ; ifb2D4cthenxD/NULb=2is a critical point, but not a local extreme point. 2:5:16(p.111) (a)1 6/DLE/EM 20/DC13 (b)1 83(c)/EM2 512p 2(d)1 4.63/4 2:5:20(p.112) (a)M3h=3, whereM3Dsupjx/NULcj/DC4hjf.3/.c/j (b)M4h2=12whereM4Dsupjx/NULcj/DC4hjf.4/.c/j 2:5:21(p.112)kD/NULh=2 Section 3.1 pp. 125–128 3:1:8(p.126) (b) monotonic functions (c) LetŒa;b/c141DŒ0;1/c141 andPDf0;1g. Let f.0/Df.1/D1 2andf.x/Dxif0<x<1 . Thens.P/D0andS.P/D1, but neither is a Riemann sum of foverP. 3:1:9(p.127) (a)1 2,/NUL1 2(b)1 2,13:1:10(p.127)eb/NULea3:1:11(p.127) 1/NULcosb3:1:12(p.127)sinb 3:1:14(p.127)f.a/Œg 1/NULg.a//c141Cf.d/.g 2/NULg1/Cf.b/Œg.b//NULg2/c141 3:1:15(p.127)f.a/Œg 1/NULg.a//c141Cf.b/Œg.b//NULgp/c141CPp/NUL1 mD1f.a m/.gmC1/NULgm/ 3:1:16(p.127) (a) Ifg/DC11andfis arbitrary, thenRb af.x/dg.x/D0. Section 3.3 pp. 149–151 3:3:7(p.150) (a)uDcD2 3(b)uDcD0(c)uD.e/NUL2/=.e/NUL1/; cDp u 554 Answers to Selected Exercises Section 3.4 pp. 165–171 3:4:4(p.166) (a) (i)p/NAK2(ii)p>0 (iii)0 (b) (i)p/NAK2(ii)p>0 (iii)0 (c) (i) none (ii)p>0 (iii)1=p (d) (i)p/DC40(ii)0<p<1 (iii)1=.1/NULp/ (e) (i) none (ii) none 3:4:5(p.166) (a)nŠ(b)1 2(c)divergent (d)1(e)/NUL1(f)0 3:4:8(p.166) (a) divergent (b) convergent (c)divergent (d) convergent (e) convergent (f)divergent 3:4:9(p.166) (a)p <2 (b)p <1 (c)p >/NUL1(d)/NUL1<p <2 (e)none (f)none (g)p<1 3:4:11(p.167) (a)p/NULq<1 (b)p;q<1 (c)/NUL1<p<2q/NUL1(d)q>/NUL1, pCq>1 (e)pCq>1 (f)qC1<p<3qC1 3:4:12(p.167)degg/NULdegf/NAK2 3:4:18(p.168) (a) (i)p>1 (ii)0<p/DC41 (b) (i)p>1 (ii)p/DC41 (c) (i)p>1 (ii)0/DC4p/DC41 (d) (i)p>0 (ii) none (e) (i)1<p<4 (ii)0<p/DC41 (f) (i)p>1 2(ii)0<p/DC41 2 3:4:25(p.169) (a) (i)p>/NUL1(ii)/NUL2<p/DC4/NUL1 (b) (i)p>/NUL1(ii) none (c) (i)p</NUL1(ii) none (d) (i) none (ii) none (e) (i)p</NUL1(ii)p>1 Section 4.1 pp. 192–195 4:1:3(p.192) (a)2(b)1(c)04:1:4(p.192)(a)1=2(b)1=2(c) 1=2(d)1=2 4:1:11(p.192) (d)p A 4:1:14(p.193) (a)1(b)1(c)1(d)/NUL1 (e)0 4:1:22(p.193)IfsnD1andtnD/NUL1=n, then.limn!1sn/=.limn!1tn/D1=0D1 , but lim n!1sn=tnD/NUL1 . 4:1:24(p.193) (a)1,0(b)1,/NUL1 ifjrj>1;2,/NUL2ifrD/NUL1;0,0ifrD1;1, /NUL1ifjrj<1(c)1,/NUL1 ifr </NUL1;0,0ifjrj<1;1 2,1 2ifrD1;1,1ifr >1 (d)1,1(e)jtj,/NULjtj Answers to Selected Exercises 555 4:1:25(p.194) (a)1,/NUL1(b)2,/NUL2(c)3,/NUL1(c)p 3=2,/NULp 3=2 4:1:34(p.194) (b) IffsngDf1;0;1;0;:::g, then lim n!1tnD1 2 Section 4.2 pp. 199–200 4:2:2(p.199) (a) limm!1s2mD1 , lim m!1s2mC1D/NUL1 (b) limm!1s4mD1, lim m!1s4mC2D/NUL1, lim m!1s2mC1D0 (c)limm!1s2mD0, lim m!1s4mC1D1, lim m!1s4mC3D/NUL1 (d) limn!1snD0(e)limm!1s2mD1 , lim m!1s2mC1D0 (f)limm!1s8mDlimm!1s8mC2D1, lim m!1s8mC1Dp 2, limm!1s8mC3Dlimm!1s8mC7D0, lim m!1s8mC5D/NULp 2, limm!1s8mC4Dlimm!1s8mC6D/NUL1 4:2:3(p.199)f1;2;1;2;3;1;2;3;4;1;2;3;4;5;::: g 4:2:8(p.200)Letftngbe any convergent sequence and fsngDft1;1;t 2;2;:::;t n;n;:::g. Section 4.3 pp. 228–234 4:3:4(p.229) (b) No; considerP1=n 4:3:8(p.229) (a) convergent (b) convergent (c)divergent (d) divergent (e)convergent (f)convergent (g) divergent (h) convergent 4:3:10(p.229) (a)p>1 (b)p>1 (c)p>1 4:3:15(p.230) (a) convergent (b) convergent if 0<r <1 , divergent if r/NAK1 (c)divergent (d) convergent (e)divergent (f)convergent 4:3:17(p.231) (a) convergent (b) convergent (c)convergent (d) convergent 4:3:18(p.231) (a) divergent (b) convergent if and only if 0<r <1 orrD1and p</NUL1(c)convergent (d) convergent (e)convergent 4:3:19(p.231) (a) divergent (b) convergent (c) convergent (d) convergent if ˛<ˇ/NUL1, divergent if ˛/NAKˇ/NUL1 4:3:20(p.231) (a) divergent (b) convergent (c)convergent (d) convergent 4:3:21(p.231) (a)P./NUL1/n(b)P./NUL1/n=n,P/DC4./NUL1/n nC1 nlogn/NAK (c)P./NUL1/n2n(d)P./NUL1/n 4:3:27(p.232) (a) conditionally convergent (b) conditionally convergent (c)abso- lutely convergent (d) absolutely convergent 4:3:28(p.232)Letkandsbe the degrees of the numerator and denominator, respec- tively. IfjrjD1, the series converges absolutely if and only if s/NAKkC2. The series converges conditionally if sDkC1andrD/NUL1, and diverges in all other cases, where s/NAKkC1andjrjD1. 4:3:30(p.232) (b)P./NUL1/n=pn4:3:41(p.233) (a)0(b)2A/NULa0 556 Answers to Selected Exercises Section 4.4 pp. 253–256 4:4:1(p.253) (a)F.x/D0;jxj/DC41(b)F.x/D0;jxj/DC41 (c)F.x/D0;/NUL1<x/DC41(d)F.x/Dsinx;/NUL1<x<1 (e)F.x/D1;/NUL1<x/DC41;F.x/D0;jxj>1(f)F.x/Dx;/NUL1<x<1 (g)F.x/Dx2=2;/NUL1<x<1(h)F.x/D0;/NUL1<x<1 (i)F.x/D1;/NUL1<x<1 4:4:5(p.254) (a)F.x/D0(b)F.x/D1;jxj<1;F.x/D0;jxj>1 (c)F.x/Dsinx=x 4:4:6(p.254) (c)Fn.x/Dxn;SkDŒ/NULk=.kC1/;k=.kC1//c141 4:4:7(p.254) (a)Œ/NUL1;1/c141(b)Œ/NULr;r/c141[f1g[f/NUL1g; 0<r <1 (c)Œ/NULr;r/c141[f1g; 0< r <1 (d)Œ/NULr;r/c141; r >0 (e)./NUL1;/NUL1=r/c141[Œ/NULr;r/c141[Œ1=r;1/[f1g; 0<r <1 (f)Œ/NULr;r/c141; r >0 (g)Œ/NULr;r/c141; r >0 (h)./NUL1;/NULr/c141[Œr;1/[f0g; r >0 (i)Œ/NULr;r/c141; r >0 4:4:12(p.254) (b) LetSD.0;1/c141 ,Fn.x/Dsin.x=n/ ,Gn.x/D1=x2; thenFD0, GD1=x2, and the convergence is uniform, but kFnGnkSD1 . 4:4:14(p.255) (a)3(b)1(c)1 2(d)e/NUL1 4:4:17(p.255) (a) compact subsets of ./NUL1 2;1/(b)Œ/NUL1 2;1/(c) closed sub- sets of 1/NULp 5 2;1Cp 5 2! (d)./NUL1;1/(e)Œr;1/; r > 1 (f)compact subsets of./NUL1;0/[.0;1/ 4:4:19(p.255) (a) LetSD./NUL1;1/,fnDan(constant), wherePanconverges conditionally, and gnDjanj.(b) “absolutely" 4:4:20(p.255) (a) (i) means thatPjfn.x/jconverges pointwise andPfn.x/con- verges uniformly on S, while(ii) means thatPjfn.x/jconverges uniformly on S. 4:4:27(p.256) (a)1X nD0./NUL1/nx2nC1 nŠ.2nC1/(b)1X nD0./NUL1/nx2nC1 .2nC1/.2nC1/Š Section 4.5 pp. 275–280 4:5:2(p.276) (a)1=3e (b)1(c)1 3(d)1(e)1 4:5:8(p.276) (a)1(b)1 2(c)1 4(d)4(e)1=e(f)1 4:5:10(p.277)x.1Cx/=.1/NULx/34:5:12(p.277)e/NULx2 4:5:16(p.277)1X nD1./NUL1/n/NUL1 n2.x/NUL1/nIRD1 4:5:17(p.277)Tan/NUL1xD1X nD0./NUL1/nx2nC1 .2nC1/If.2n/.0/D0If.2nC1/.0/D./NUL1/2.2n/Š ; Answers to Selected Exercises 557 /EM 6DTan/NUL11p 3D1X nD0./NUL1/n .2nC1/3nC1=2 4:5:22(p.278)coshxD1X nD0x2n .2n/Š, sinhxD1X nD0x2nC1 .2nC1/Š 4:5:23(p.278).1/NULx/P1 nD0xnD1converges for all x 4:5:24(p.278) (a)xCx2Cx3 3/NUL3x5 40C/SOH/SOH/SOH(b)1/NULx/NULx2 2C5x3 6C/SOH/SOH/SOH(c) 1/NULx2 2Cx4 24/NUL721x6 720C/SOH/SOH/SOH (d)x2/NULx3 2Cx4 6/NULx5 6C/SOH/SOH/SOH 4:5:27(p.279)(a)1CxC2x2 3Cx3 3C/SOH/SOH/SOH(b)1/NULx/NULx2 2C3x3 2C/SOH/SOH/SOH(c)1Cx2 2C5x4 24C61x6 720C/SOH/SOH/SOH (d)1Cx2 6C7x4 360C31x6 15120C/SOH/SOH/SOH(e)2/NULx2Cx4 12/NULx6 360C/SOH/SOH/SOH 4:5:28(p.279)F.x/D5 .1/NUL3x/.1C2x/D3 1/NUL3xC2 1C2xD1X nD0Œ3nC1/NUL./NUL2/nC1/c141xn 4:5:29(p.279)1 Section 5.1 pp. 299–302 5:1:1(p.299) (a).3;0;3;3/ (b)./NUL1;/NUL1;4/(c).1 6;11 12;23 24;5 36/ 5:1:3(p.299) (a)p 15(b)p 65=12 (c)p 31(d)p 3 5:1:4(p.299) (a)p 89(b)p 166=12 (c)3(d)p 31 5:1:5(p.299) (a)12(b)1 32(c)27 5:1:7(p.299)XDX0CtU./NUL1<t <1/in all cases. 5:1:8(p.299):::UandX1/NULX0are scalar multiples of V. 5:1:9(p.299) (a) XD.1;/NUL3;4;2/Ct.1;3;/NUL5;3/ (b) XD.3;1;/NUL2;1;4;/Ct./NUL1;/NUL1;1;3;/NUL7/ (c)XD.1;2;/NUL1/Ct./NUL1;/NUL3;0/ 5:1:10(p.300) (a)5(b)2(c)1=2p 5 5:1:11(p.300)(a)(i)˚.x1;x2;x3;x4/ˇˇjxij/DC43.iD1;2;3/ with at least one equality/TAB (ii)˚ .x1;x2;x3;x4/ˇˇjxij/DC43.iD1;2;3//TAB(iii)S (iv)˚.x1;x2;x3;x4/ˇˇjxij>3for at least one of iD1;2;3/TAB (b) (i)S(ii)S(iii);(iv)˚.x;y;´/ˇˇ´¤1orx2Cy2>1/TAB 5:1:12(p.300) (a) open(b) neither(c)closed 5:1:18(p.300) (a)./EM;1;0/ (b).1;0;e/ 5:1:19(p.300) (a)6(b)6(c)2p 5(d)2Lpn(e)1 5:1:29(p.302)˚ .x;y/ˇˇx2Cy2D1/TAB 558 Answers to Selected Exercises 5:1:33(p.302):::if forAthere is an integer Rsuch thatjXrj>A ifr/NAKR. Section 5.2 pp. 314–316 5:2:1(p.314) (a)10(b)3(c)1(d)0(e)0(f)0 5:2:3(p.315) (b)a=.1Ca2/ 5:2:4(p.315) (a)1(b)1(c)no(d)/NUL1(e)no 5:2:5(p.315) (a)0(b)0(c)none (d)0(e)none 5:2:6(p.316) (a) . . . ifDfis unbounded and for each Mthere is anRsuch that f.X/>M ifX2DfandjXj>R.(b) Replace “>M ” by “<M ” in(a). 5:2:7(p.316)lim X!0f.X/D0ifa1Ca2C/SOH/SOH/SOHCan>b; no limit ifa1Ca2C/SOH/SOH/SOHCan/DC4 banda2 1Ca2 2C/SOH/SOH/SOHCa2 n¤0; lim X!0f.X/D1 ifa1Da2D/SOH/SOH/SOHDanD0andb>0 . 5:2:8(p.316)No; for example, lim x!1g.x;px/D0. 5:2:9(p.316) (a) R3(b)R2(c)R3(d)R2(e)˚.x;y/ˇˇx/NAKy/TAB(f)Rn 5:2:10(p.316) (a) R3/NULf.0;0;0/g(b)R2(c)R2(d)R2(e)R2 5:2:11(p.316)f.x;y/Dxy=.x2Cy2/if.x;y/¤.0;0/ andf.0;0/D0 Section 5.3 pp. 335–339 5:3:1(p.335)(a)2p 3.xCycosx/NULxysinx//NUL2r 2 3.xcosx/(b)1/NUL2yp 3e/NULxCy2C2´ (c)2pn.x1Cx2C/SOH/SOH/SOHCxn/(d)1=.1CxCyC´/ 5:3:2(p.335)/RS2 1/RS25:3:3(p.335) (a)/NUL5/EM=p 6(b)/NUL2e(c)0(d)0 5:3:5(p.335) (a)fxDfyD1=.xCyC2´/,f´D2=.xCyC2´/ (b)fxD2xC3y´C2y,fyD3x´C2x,f´D3xy(c)fxDey´,fyDx´ey´, f´Dxyey´(d)fxD2xycosx2y,fyDx2cosx2y,f´D1 5:3:6(p.335) (a)fxxDfyyDfxyDfyxD/NUL1=.xCyC2´/2,fx´Df´xD fy´Df´yD/NUL2=.xCyC2´/2,f´´D/NUL4=.xCyC2´/2 (b)fxxD2,fyyDf´´D0,fxyDfyxD3´C2,fx´Df´xD3y,fy´Df´yD3x (c)fxxD0,fyyDx´2ey´,f´´Dxy2ey´,fxyDfyxD´ey´,fx´Df´xDyey´, fy´Df´yDxey´ (d)fxxD2ycosx2y/NUL4x2y2sinx2y,fyyD/NULx4sinx2y,f´´D0,fxyDfyxD 2xcosx2y/NUL2x3ysinx2y,fx´Df´xDfy´Df´yD0 5:3:7(p.336) (a)fxx.0;0/Dfyy.0;0/D0,fxy.0;0/D/NUL1,fyx.0;0/D1 (b)fxx.0;0/Dfyy.0;0/D0,fxy.0;0/D/NUL1,fyx.0;0/D1 5:3:8(p.336)f.x;y/Dg.x;y/Ch.y/ , wheregxyexists everywhere and his nowhere differentiable. Answers to Selected Exercises 559 5:3:18(p.337)(a)dfD.3x2C4y2C2ysinxC2xycosx/dxC.8xyC2xsinx/dy , dX0fD16dx ,.dX0f/.X/NULX0/D16x (b)dfD/NULe/NULx/NULy/NUL´.dxCdyCd´/,dX0fD/NULdx/NULdy/NULd´, .dX0f/.X/NULX0/D/NULx/NULy/NUL´ (c)dfD.1Cx1C2x2C/SOH/SOH/SOHCnxn//NUL1Pn jD1jdx j,dX0fDPn jD1jdx j, .dX0f/.X/NULX0/DPn jD1jxj, (d)dfD2rjXj2r/NUL2Pn jD1xjdxj,dX0fD2rnr/NUL1Pn jD1dxj, .dX0f/.X/NULX0/D2rnr/NUL1Pn jD1.xj/NUL1/, 5:3:19(p.337)(b) The unit vector in the direction of .fx1.X0/;fx2.X0/;:::;f xn.X0// provided that this is not 0; if it is 0, [email protected]/=@ˆD0for every ˆ. 5:3:24(p.338)(a)´D2xC4y/NUL6(b)´D2xC3yC1(c)´D./EMx/=2Cy/NUL/EM=2 (d)´DxC10yC4 Section 5.4 pp. 356–360 5:4:2(p.357) (a)5duC34dv (b)0(c)6du/NUL18dv (d)8du 5:4:3(p.357)hrDfxcos/DC2Cfysin/DC2,h/DC2Dr./NULfxsin/DC2Cfycos/DC2/,h´Df´ 5:4:4(p.357)hrDfxsin/RScos/DC2Cfysin/RSsin/DC2Cf´cos/RS,h/DC2Drsin/RS./NULfxsin/DC2C fycos/DC2/,h/RSDr.fxcos/RScos/DC2Cfycos/RSsin/DC2/NULf´sin/RS/ 5:4:6(p.357)hyDgxxyCgyCgwwy,h´Dgxx´Cg´Cgww´ 5:4:13(p.358)hrrDfxxsin2/RScos2/DC2Cfyysin2/RSsin2/DC2Cf´´cos2/RSCfxysin2/RSsin2/DC2C fy´sin2/RSsin/DC2Cfx´sin2/RScos/DC2, hr/DC2D./NULfxsin/DC2Cfycos/DC2/sin/RSCr 2.fyy/NULfxx/sin2/RSsin2/DC2Crfxysin2/RScos2/DC2C r 2.f´ycos/DC2/NULf´xsin/DC2/sin2/RS 5:4:16(p.358) (a)1CxCx2 2/NULy2 2Cx3 6/NULxy2 2 (b)1/NULx/NULyCx2 2CxyCy2 2/NULx3 6/NULx2y 2/NULxy2 2/NULy3 6 (c)0(d)xy´ 5:4:21(p.359) (a).d2 .0;0/p/.x;y/D.d2 .0;0/q/.x;y/D2.x/NULy/2 Section 6.1 pp. 376–378 6:1:3(p.376) (a)2 43 4 6 2/NUL4 2 7 2 33 5(b)2 6642 4 3/NUL2 7/NUL4 6 13 775 6:1:4(p.376) (a)2 48 8 16 24 0 0 4 12 12 16 28 443 5(b)2 4/NUL2/NUL6 0 0/NUL2/NUL4 /NUL2 2/NUL63 5 560 Answers to Selected Exercises 6:1:5(p.376) (a)2 4/NUL2 2 6 6 7/NUL3 0/NUL2 63 5(b)2 4/NUL1 7 3 5 5 143 5 6:1:6(p.376) (a)2 413 25 16 31 16 253 5(b)/DC429 50/NAK 6:1:10(p.377)AandBare square of the same order. 6:1:12(p.377) (a)2 47 3 3 4 7 7 6/NUL9 13 5(b)2 414 10 6/NUL2 14 23 5 6:1:13(p.377)2 4/NUL7 6 4 /NUL9 7 13 5 0/NUL143 5,2 4/NUL5 6 0 4/NUL12 3 4 0 33 5 6:1:15(p.377) (a)/STX 6xy´ 3x´23x2y/ETX ;/STX /NUL6 3/NUL3/ETX (b) cos.xCy//STX1 1/ETX;/STX0 0/ETX (c)/STX.1/NULx´/ye/NULx´xe/NULx´/NULx2ye/NULx´/ETX;/STX2 1/NUL2/ETX (d) sec2.xC2yC´//STX 1 2 1/ETX ;/STX 2 4 2/ETX (e)jXj/NUL1/STX x1x2/SOH/SOH/SOHxn/ETX ;1pn/STX 1 1/SOH/SOH/SOH1/ETX 6:1:20(p.377) (a).2;3;/NUL2/(b).2;3;0/ (c)./NUL2;0;/NUL1/(d).3;1;3;2/ 6:1:21(p.378) (a)1 10/DC44 2 /NUL3 1/NAK (b)1 22 4/NUL1 1 2 3 1/NUL4 /NUL1/NUL1 23 5 (c)1 252 44 3/NUL5 6/NUL8 5 /NUL3 4 103 5(d)1 22 41/NUL1 1 /NUL1 1 1 1 1/NUL13 5 (e)1 72 6643/NUL2 0 0 2 1 0 0 0 0 2/NUL3 0 0 1 23 775(f)1 102 664/NUL1/NUL2 0 5 /NUL14/NUL18 10 20 21 22/NUL10/NUL25 17 24/NUL10/NUL253 775 Section 6.2 pp. 390–394 6:2:12(p.392)(a) F0.X/D2 642x 1 2 /NULsin.xCyC´//NULsin.xCyC´//NULsin.xCyC´/ y´exy´x´exy´xyexy´3 75; JF.X/Dexy´sin.xCyC´/Œx.1/NUL2x/.y/NUL´//NUL´.x/NULy//c141; Answers to Selected Exercises 561 G.X/D2 40 1 13 5C2 42 1 2 0 0 0 0 0/NUL13 52 4x/NUL1 yC1 ´3 5 (b) F0.X/D/DC4excosy/NULexsiny exsiny excosy/NAK ;JF.X/De2x; G.X/D/DC40 1/NAK C/DC40/NUL1 1 0/NAK/DC4x y/NUL/EM=2/NAK (c)F0.X/D2 42x/NUL2y 0 0 2y/NUL2´ /NUL2x 0 2´3 5;JFD0; G.X/D2 42/NUL2 0 0 2/NUL2 /NUL2 0 23 52 4x/NUL1 y/NUL1 ´/NUL13 5 6:2:13(p.392) (a) F0.X/D/DC4.xCyC´C1/exexex .2x/NULx2/NULy2/e/NULx2ye/NULx0/NAK (b) F0.X/D2 6664g0 1.x/ g0 2.x/ ::: g0 n.x/3 7775 (c)F0.r;/DC2/D2 4exsiny´ ´excosy´ yexcosy´ ´eycosx´ eysinx´ xeycosx´ ye´cosxy xe´cosxy e´sinxy3 5 6:2:14(p.392) (a) F0.r;/DC2/D/DC4cos/DC2/NULrsin/DC2 sin/DC2 r cos/DC2/NAK ;JF.r;/DC2/Dr (b) F0.r;/DC2;/RS/D2 4cos/DC2cos/RS/NULrsin/DC2cos/RS/NULrcos/DC2sin/RS sin/DC2cos/RS r cos/DC2cos/RS/NULrsin/DC2sin/RS sin/RS 0 r cos/RS3 5; JF.r;/DC2;/RS/Dr2cos/RS (c)F0.r;/DC2;´/D2 4cos/DC2/NULrsin/DC2 0 sin/DC2 r cos/DC2 0 0 0 13 5;JF.r;/DC2;´/Dr 6:2:20(p.393) (a)/DC40 0 4 0/NUL1 20/NAK (b)/DC4/NUL18 0 2 0/NAK (c)2 49/NUL3 3/NUL8 1 03 5 (d)/DC44/NUL3 1 0 1 1/NAK (e)/DC42 0 2 0/NAK (f)2 45 10 9 18 /NUL4/NUL83 5 562 Answers to Selected Exercises Section 6.3 pp. 414–417 6:3:4(p.414) (a)Œ1;/EM=2/c141 (b)Œ1;2/EM/c141 (c)Œ1;/EM/c141 (d)Œ2p 2;9/EM=4/c141 (e) Œp 2;3/EM=4/c141 6:3:5(p.414) (a)Œ1;/NUL3/EM=2/c141 (b)Œ1;/NUL2/EM/c141(c)Œ1;/NUL/EM/c141(d)Œ2p 2;/NUL7/EM=4/c141 (e)Œp 2;/NUL5/EM=4/c141 6:3:6(p.414) (b) Letf.x/Dx.0/DC4x/DC41 2/,f.x/Dx/NUL1 2.1 2<x/DC41/; thenfis locally invertible but not invertible on Œ0;1/c141 . 6:3:7(p.414)F.S/D˚.u;v/ˇˇ/NUL/EMC2/RS< arg.u;v/</EMC2/RS/TAB, where/RSis an argu- ment of.a;b/ ; F/NUL1 S.u;v/D.u2Cv2/1=4"cos.arg.u;v/=2/ sin.arg.u;v/=2/# ; 2/RS/NUL/EM < arg.u;v/<2/RSC/EM 6:3:10(p.415) (a)/DC4x y/NAK D1 10/DC4u/NUL2v 3uC4v/NAK ;.F/NUL1/0D1 10/DC41/NUL2 3 4/NAK (b)2 4x y ´3 5D1 22 4uC2vC3w u/NULw uCvC2w3 5;.F/NUL1/0D1 22 41 2 3 1 0/NUL1 1 1 23 5 6:3:12(p.415)G1.u;v/D1p 2/DC4puCvpu/NULv/NAK ,G0 1.u;v/D1 2p 2/DC41=puCv 1=puCv 1=pu/NULv/NUL1=pu/NULv/NAK G2.u;v/D1p 2/DC4/NULpuCvpu/NULv/NAK ,G0 2.u;v/D1 2p 2/DC4/NUL1=puCv/NUL1=puCv 1=pu/NULv/NUL1=pu/NULv/NAK G3.u;v/D1p 2/DC4puCv /NULpu/NULv/NAK ,G0 3.u;v/D1 2p 2/DC41=puCv 1=puCv /NUL1=pu/NULv 1=pu/NULv/NAK G4.u;v/D1p 2/DC4/NULpuCv /NULpu/NULv/NAK ,G0 4.u;v/D1 2p 2/DC4/NUL1=puCv/NUL1=puCv /NUL1=pu/NULv 1=pu/NULv/NAK 6:3:15(p.416)From solving xDrcos/DC2,yDrsin/DC2for/DC2Darg.x;y/ . Each equation is satisfied by angles that are not arguments of .x;y/ , since none of the formulas identifies the quadrant of .x;y/ uniquely. Moreover, (c)does not hold if xD0. 6:3:16(p.416)/DC4x y/NAK DG.u;v/D.u2Cv2/1=4" cosŒ1 2arg.u;v//c141 sin.arg.u;v/=2/# , whereˇ/NUL/EM=2< arg.u;v/<ˇC/EM=2 andˇis an argument of .a;b/ ; G0.u;v/D1 2.x2Cy2//DC4x y /NULy x/NAK 6:3:19(p.416)IfF.x1;x2;:::;x n/D.x3 1;x3 2;:::;x3 n/, then Fis invertible, but JF.0/D0. 6:3:20(p.416) (a) A.U/D/DC41 /NUL1/NAK /NUL1 25/DC45 5 3 8/NAK/DC4uC5 v/NUL4/NAK Answers to Selected Exercises 563 (b) A.U/D/DC41 1/NAK C1 6/DC44/NUL2 /NUL3 3/NAK/DC4u/NUL2 v/NUL3/NAK (c)A.U/D2 40 1 13 5C2 40/NUL1 1 /NUL1 1 0 1 0 03 52 4u/NUL1 v/NUL1 w/NUL23 5 (d) A.U/D2 41 /EM=2 /EM3 5C2 40/NUL1 0 1 0 0 0 0/NUL13 52 4u vC1 w3 5 6:3:21(p.417)G0.x;y;´/D2 66664cos/DC2cos/RS sin/DC2cos/RS sin/RS /NULsin/DC2 rcos/RScos/DC2 rcos/RS0 /NUL1 rcos/DC2sin/RS/NUL1 rsin/DC2sin/RS1 rcos/RS3 77775 6:3:22(p.417)G0.x;y;´/D2 6664cos/DC2 sin/DC2 0 /NUL1 rsin/DC21 rcos/DC2 0 0 0 13 7775 Section 6.4 pp. 431–434 6:4:1(p.431) (a)/DC4u v/NAK D1 2/DC4/NUL3 4 1/NUL2/NAK/DC4x y/NAK (b)2 4u v w3 5D/NUL1 22 43 3 /NUL1 2 2 33 5/DC4x y/NAK (c)/DC4u v/NAK D1 5/DC42/NUL1 /NUL1 3/NAK/DC4/NULyCsinx /NULxCsiny/NAK (d)uD/NULx,vD/NULy,´D/NULw 6:4:3(p.431)fi.X;U/D0 @nX jD1aij.xj/NULxj 0/1 Ar /NUL.ui/NULui0/s,1/DC4i/DC4m, wherer andsare positive integers and not all aijD0.(a)rDsD3;(b)rD1,sD3;(c) rDsD2 6:4:4(p.431)ux.1;1/D/NUL5 8,uy.1;1/D/NUL1 2 6:4:5(p.431)ux.1;1;1/D5 8,uy.1;1;1/D/NUL9 8,u´.1;1;1/D1 2 6:4:6(p.431) (a)u.1;2/D0,ux.1;2/Duy.1;2/D/NUL4 (b)u./NUL1;/NUL2/D2,ux./NUL1;/NUL2/D1,uy./NUL1;/NUL2/D/NUL1 2 (c)u./EM=2;/EM=2/Dux./EM=2;/EM=2/Duy./EM=2;/EM=2/D0 (d)u.1;1/D1,ux.1;1/Duy.1;1/D/NUL1 6:4:7(p.431) (a)u1.1;1/D1,@u1.1;1/ @xD5,@u1.1;1/ @yD2 564 Answers to Selected Exercises u2.1;1/D2,@u2.1;1/ @xD/NUL14;@u2.1;1/ @yD/NUL2 (b)uk.0;/EM/D.2kC1//EM=2 ,@uk.0;/EM/ @xD0,@uk.0;/EM/ @yD/NUL1,kDinteger 6:4:8(p.432)1 5/DC4/NUL1/NUL2 1 /NUL1/NUL2 1/NAK 6:4:9(p.432)u0.0/D3,v0.0/D/NUL1 6:4:10(p.432)1 62 45 5 /NUL5/NUL5 6 63 5 6:4:11(p.432)U1.1;1/D/DC43 1/NAK ,U0 1.1;1/D/DC41 3 /NUL1 2/NAK ; U2.1;1/D/NUL/DC43 1/NAK ,U0 2.1;1/D/NUL/DC41 3 /NUL1 2/NAK 6:4:12(p.432)ux.0;0;0/D2,vx.0;0;0/Dwx.0;0;0/D/NUL2 6:4:13(p.433)yxD/[email protected];g;h/ @.x;´;u/ @.f;g;h/ @.y;´;u/,yvD/[email protected];g;h/ @.v;´;u/ @.f;g;h/ @.y;´;u/,´xD/[email protected];g;h/ @.y;x;u/ @.f;g;h/ @.y;´;u/, ´vD/[email protected];g;h/ @.y;v;u/ @.f;g;h/ @.y;´;u/,uxD/[email protected];g;h/ @.y;´;x/ @.f;g;h/ @.y;´;u/,uvD/[email protected];g;h/ @.y;´;v/ @.f;g;h/ @.y;´;u/ 6:4:14(p.433)xD/NUL2y/NULu,´D/NUL2v;xD/NUL2y/NULu,vD/NUL´ 2;yD/NULx 2/NULu 2, ´D/NUL2v;yD/NULx 2/NULu 2,vD/NUL´ 2;´D/NUL2v,uD/NULx/NUL2y;uD/NULx/NUL2y,vD/NUL´ 2 6:4:15(p.433)yx.1;/NUL1;/NUL2/D/NUL1 2,vu.1;/NUL1;/NUL2/D1 6:4:16(p.433)uw.0;/NUL1/D5 6,uy.0;/NUL1/D0,vw.0;/NUL1/D/NUL5 6,vy.0;/NUL1/D0, xw.0;/NUL1/D1,xy.0;/NUL1/D/NUL1 6:4:18(p.434)ux.1;1/D0,uy.1;1/D0,vx.1;1/D/NUL1,vy.1;1/D/NUL1,uxx.1;1/D 2, uxy.1;1/D1,uyy.1;1/D2,vxx.1;1/D/NUL2,vxy.1;1/D/NUL1,vyy.1;1/D/NUL2 6:4:19(p.434)ux.1;/NUL1/D0,uy.1;/NUL1/D1 2,vx.1;/NUL1/D/NUL1 2,vy.1;/NUL1/D0, uxx.1;/NUL1/D/NUL1 8,uxy.1;/NUL1/D1 8,uyy.1;/NUL1/D1 8,vxx.1;/NUL1/D/NUL1 8, vxy.1;/NUL1/D/NUL1 8,vyy.1;/NUL1/D1 8 Index 565 Section 7.1 pp. 459–462 7:1:2(p.459)(a)28(b)1 47:1:6(p.460)3.b/NULa/.d/NULc/,07:1:13(p.460)˚.m;n/ˇˇm;nDintegers/TAB Section 7.2 pp. 480–484 7:2:1(p.480) (a)12(b)79 20(c)/NUL1(d).1/NULlog2/=2 7:2:5(p.481) (a)7 4(b)17(c)2 3.p 2/NUL1/(d)1=4/EM 7:2:7(p.481) (a)3 8,5 8(b)3 8,5 87:2:8(p.482) (a)3 4,5 4(b)3 4/NUL´C1 2/SOH, 5 4/NUL´C1 2/SOH(c)´C1 2,1 7:2:11(p.482) (a)/NUL285(b)0(c)0(d)1 4.e/NUL5 2/ 7:2:12(p.483) (a)324(b)1 6(c)17:2:13(p.483)52 15 7:2:14(p.483) (a)36(b)1(c)64 3(d).e6C17/=2 7:2:17(p.483) (a)2 27(b)1 2.e/NUL5 2/(c)1 24(d)1 36 7:2:18(p.483) (a)16/EM(b)1 6(c)128 21(d)/EM 2 7:2:19(p.484) (a)1 2.b1/NULa1//SOH/SOH/SOH.bn/NULan/Pn jD1.ajCbj/ (b)1 3.b1/NULa1//SOH/SOH/SOH.bn/NULan/Pn jD1.a2 jCajbjCb2 j/ (c)2/NULn.b2 1/NULa2 1//SOH/SOH/SOH.b2 n/NULa2 n/ 7:2:20(p.484)Rp 3=2 /NULp 3=2dxRp 1/NULx2 1=2f.x;y/dy 7:2:22(p.484)1 2 Section 7.3 pp. 514–517 7:3:1(p.514)LetS1andS2be dense subsets of Rsuch thatS1[S2DR. 7:3:7(p.514)(a)/NUL1;c(constant);17:3:9(p.515).u2/NULu1/.v2/NULv1/=jad/NULbcj 7:3:10(p.515)5 67:3:14(p.515) (a)4 9(b) log5 27:3:15(p.516)3 7:3:16(p.516)1 27:3:17(p.516)5 4e.e/NUL1/ 7:3:18(p.516)4 3/EMabc 7:3:19(p.516)2/EM.e25/NULe9/7:3:20(p.516)16/EM=3 7:3:21(p.516)21=64 7:3:22(p.516) (a)./EM=8/ log5(b)./EM=4/.e4/NUL1/(c)2/EM=15 7:3:23(p.517)/EM2a4=2 7:3:24(p.517)(a).ˇ1/NUL˛1//SOH/SOH/SOH.ˇn/NUL˛n/=jdet.A/j7:3:25(p.517)ja1a2/SOH/SOH/SOHanjVn Index A Abel’s test, 219 Abel’s theorem, 273,279 Absolute convergence, 215 of an improper integral, 160 of a series of constants, 215 of a series of functions, 247 Absolute integrability, 160 Absolute uniform convergence, 247,255 (Exercises 4.4.17 and4.4.20 ), 256(Exercise 4.4.21 ) of a power series, 257 Absolute value, 2 Addition of power series, 267 Adjoint matrix, 370 Affine transformation, 380 Alternating series, 203 test, 203,219 Analytic transformation, 416(Exercise 6.3.17 ) Angle between two vectors, 286 Antiderivative, 143,150(Exercise 3.3.16 ) Archimedean property, 5 Area under a curve, 116 Argument, 398 branch of, 409,410,415(Exercise 6.3.14 ) Ascoli–Arzela theorem, 543 Associative laws for the real numbers, 2(see p. 1) for vector addition, 283 B Bessel function, 277(Exercise 4.5.11 )Binomial coefficient, 17(Exercise 1.2.19 ), 102,194(Exercise 4.1.35 ) Binomial series, 266 Binomial theorem, 17(Exercise 1.2.19 ) Bolzano–Weierstrass theorem, 27, 294, 301(Exercise 5.1.22 ) Bound lower, 7 upper, 3 Boundary, 526 point, 289,526 of a set, 23,289 Bounded convergence theorem, 243 Bounded function, 47,60,313 Boundedness of a continuous function on a closed interval, 62,199 on a compact set, 313 Boundedness of an integrable function, 119 on a metric space, 537 Bounded sequence, 181,197,292 Bounded set above, 3,313 below 7,313 Bounded variation, 134–135(Exercises 3.2.7 , 3.2.9 ,3.2.10 ) Branch of an argument, 409,415 of an inverse, 409 C C[a,b], 521 equicontinuous subset of, 541 566 Index 567 uniformly bounded subset of, 541 Cartesian product, 31, 435 Cauchy product of series, 226,233(Ex- ercise 4.3.40 ),280(Exercise 4.5.32 ) Cauchy sequence, 527 Cauchy’s convergence criterion for sequences of real numbers, 190 for sequences of vectors, 292 for series of real numbers, 204 Cauchy’s root test, 215 Cauchy’s uniform convergence criterion for sequences, 239 for series, 246 Chain rule, 77,340,388 Change of variable, 145,147 in an improper integral, 164 in a multiple integral, 496 formulation of the rule for, 494 in an ordinary integral, 145,147 Changing the order of integration, 478 Characteristic function, 70(Exercise 2.2.9 ), 485 Closed under scalar multiplication, 519 under vector addition, 519 Closed interval, 23 Closedn-ball, 291 Closed set, 21,289,525 Closure of a set, 23,289 Cofactor, 370 expanding a determinant in, 371–372 Commutative laws for the reals, 2(See p. 1) for vector addition, 283 Compact set, 20,293,537 Comparison test for improper integrals, 156 for series, 206 Complement of a set, 20 Complete metric space, 527 Completeness axiom, 4 Complete ordered field, 4 Component function, 311 Components, 284(see p. 281) of a vector-valued function, 311,362Composite function, 58,311 continuity of, 59,311 differentiability of, 77,340 higher derivatives of, 345 Taylor polynomial of, 109–110 (Exercise 2.5.11 ) Composition of functions, 58 Conditional convergence of an improper integral, 162 of a series, 217 Conditionally integrable, 162 Connected metric space, 549(Exercise 8.3.2 ) Connected set, 295 polygonally, 296 Containment of a set, 19 Content, 453 of a coordinate rectangle, 437 of a set, 485 zero, 448,514(Exercise refexer:7.3.2) Continuity, 54,302 of a composite function, 59,311 of a differentiable function, 76,325 of a function of nvariables, 309 of a function of one variable, 54 on an interval, 55 from the left, 54 of a monotonic function, 67 piecewise, 56 from the right, 54 on a set, 56,311 of a sum, difference, product, and quotient, 57,311 in terms of sequences, 198 of a transformation, 379 uniform, 64,66,314,392(Exercise 6.2.10 ) of a uniform limit, 242 of a uniformly convergent series, 250 Continuous function 54,309 boundedness of, 62,313 extreme values of on a closed inter- val,62 integrability of, 133 intermediate values of, 63,313 on a metric space, 545 Continuous transformation, 379 568 Index Continuously differentiable, 73, 80, 329, 385, 409 Contraction mapping theorem, 547 Convergence absolute of an improper integral, 160 of a series of constants, 215 absolute uniform, 247 conditional of a series, 217 of an improper integral, 162 of an improper integral, 152 of an infinite series, 201 interval of, 258 pointwise of a sequence of functions, 234, 238 of a series of functions, 244 of a power series, 257 radius of, 258 of a sequence in a metric space, 526 of a sequence in Rn,292 of a sequence of real numbers, 179 of a series of constants, 200 of a sum, difference, or product of sequences, 184 of a Taylor series, 264 uniform, 246 of a sequence, 237 of a series, 246 Coordinate cube, 437 degenerate, 437 nondegenerate, 437 Coordinate rectangle, 437 Coordinates, polar, 397,502,505 spherical, 507 Covering, open, 25,293,536 Cramer’s rule, 373 Critical point, 81,335 Curve, differentiable, 453 D Decreasing sequence, 182 Dedekind cut, 9(Exercise 1.1.8 )Dedekind’s theorem, 9(Exercise 1.1.8 ) Defined inductively, 12 Degree of a homogeneous polynomial, 352 of a polynomial, 98 Deleted/SI-neighborhood, 22 Deleted neighborhood, 525 Dense set, 6,29(Exercise 1.3.22 ),70(Ex- ercise 2.2.10 ) Density of the rationals, 6,392(Esercise 6.2.11 ) Density of the irrationals, 6 Denumerable set, 176 Derivative, 73 of a composite function, 77 directional, 317 infinite, 88(Exercise 2.3.26 ) of an inverse function, 86(Exercise 2.3.14 ) left-hand, 79 nth,73 one-sided, 79 ordinary, 317 partial, 317 of a power series, 261–262 right-hand, 79 rth order, 319 second, 73 of a sum, difference, product, and quotient, 77 zeroth, 73 Determinant, 368(see p. 369) expanding in cofactors, 371–372 of a product of square matrices, 370 Diameter of a set, 292, 586 Difference quotient, 73 Differentiability of a composite function, 340 continuous, 329 of a function of one variable, 73 of a function of several variables, 323 of the limit of a sequence, 243 of a power series, 260–262 of a series, 252 Differentiable 73,323 continuously, 73,80,409 curve, 453 Index 569 function, continuity of, 76,325,385 on an interval, 80 on a set, 73 surface, 453 transformation, 380 vector-valued function, 339 Differential, 326 higher, 348 of a linear transformation, 367 matrix, 367,381 of a real-valued function, 326 of a sum, difference, product, and quotient, 328 of a transformation, 381 Differential equation, 170–171 (Exercises 3.4.27 –3.4.29 ) Directional derivative, 317 Dirichlet’s test for improper integrals, 163 for series of constants, 217 for uniform convergence of series, 248 Disconnected set, 295 Discontinuity jump, 56 removable, 58 Discrete metric, 519 Disjoint sets, 20 Distance in a metric space, 518 from a point to a set, 301 (Exercise 5.1.24 ) between subsets of a metric space, 549(Exercise 8.3.3 ) between two sets, 301 (Exercise 5.1.25 ) between two vectors, 283 Distributive law, 2(see p. 1) Divergence, unconditional, 233 (Exercise 4.3.38 ) Divergent improper integral, 152 Divergent sequence, 179 Divergent series, 201 Domain of a function, 31(see p. 30), 545 Double integral, 438E Edge lengths of a coordinate rectangle, 437 Elementary matrix, 488 Empty set, 4 Entries of a matrix, 364 /SI-neighborhood, 21,289,525 /SI-net, 539 Equicontinuous subset of CŒa;b/c141 ,541 Equivalent metrics, 530 Error in approximating derivatives, 112 (Exercises 112–112) Euclideann-space, 282(see p. 281) Euler’s constant, 230(Exercise 4.3.14 ) Euler’s theorem, 357–358(Exercise 2.4.8 ) Existence of an improper integral, 152 Existence theorem, 420 Expanding a determinant, 362–372 Exponential function, 70(Exercise 2.2.12 ), 72(Exercise 2.2.33 ),228,273 Extended mean value theorem, 106 Extended reals, 7, Exterior point, 289,526 Exterior of a set, 23,289,526 F Faa di Bruno’s formula, 109 (Exercise 2.5.11 ) Fibonnacci numbers, 17(Exercise 1.2.17 ) Field complete ordered, 4 ordered, 2 properties, 2(see p. 1) Finite real, 7 First mean value theorem for integrals, 139 Forward differences, 104,71(Example 2.2.18 ), 112(Exercises 2.5.19 –2.5.22 ) Fredholm’s integral equation, 548 Function 31,32 absolutely integrable, 160 Bessel, 277(Exercise 277) bounded, 47,60,313 above, 60,313 below, 60,313 570 Index of bounded variation, 134(Exercise 3.2.7 ) characteristic, 70(Exercise 2.2.9 ),485 composite, 58,311 decreasing, 44 differentiable at a point, 73,323 domain of, 31,32 exponential, 70(Exercise 2.2.12 ),72 (Exercise 2.2.33 ),227,273 generating, 278(Exercise 4.5.26 ) homogeneous, 357(Exercise 5.4.8 ) increasing, 44 infimum of, 55,313 inverse of, 68 linear, 325 locally integrable, 152 maximum of, 60 monotonic, 44,67 nondecreasing, 44 nonincreasing, 44 nonoscillatory at a point, 162 nth power of, 33 oscillation of, 171 piecewise continuous, 56 range of, 31,32 rational, 33,232, (Exercise 4.3.28 ), 276(Exercise 4.5.4 ) real-valued, 302 restriction of, 399 Riemann integrable, 114,438 Riemann–Stieltjes integrable, 125 strictly monotonic, 44 supremum of, 313 value of, 31,32 vector-valued, 311 Functions, composition of, 58,311 difference of, 32 product of, 32 quotient of, 32 sum of, 32 Fundamental theorem of calculus, 143 G Generalized mean value theorem, 83Generating function, 278(Exercise 4.5.26 ) Geometric series, 202 Grouping terms of series, 220 H Heine–Borel property, Heine–Borel theorem, 172,66,172,293 Higher derivatives of a composite func- tion, 345 Higher differential, 348 Homogeneous function, 357(Exercise 5.4.8 ), 359(Exercise 5.4.23 ) Homogeneous polynomial, 359(Exercise 5.4.22 ), Homogeneous system, 375 Hypercube, 295(see p. 294) Hölder’s inequality, 521 I Identity matrix, 370 Image, 394 Implicit function theorem, 420,423 Improper integrability, 146 Improper integral, 152 absolutely convergent, 160 change of variable in, 164 conditionally convergent, 162 convergence of, 152 divergence of, 152 existence of, 152 of a nonnegative function, 156 Incompleteness of the rationals, 6 Increasing sequence, 182 Indeterminate forms, 91,93–95 Induction assumption, 12 Induction proof, 12 Inequality, Hölder, 521 Minkowski, 522 Schwarz, 284 triangle, 2,285 Infimum of a function, 60,313 of a set, 7 existence and uniqueness of, 7,9 (Exercise 1.1.6 ) Index 571 Infinite derivative, 88(Exercise 2.3.26 ) Infinite limits, 42,306,317,316(Exercise 5.2.6 ) Infinite sequence, 179 in a metric space, 526 Infinite series, 210,244 convergence of, 201 integrability of, 251 oscillatory, 201 Infinity norm, 496,523,524 Inner product, 284 Instantaneous rate of change, 74 velocity, 74 Integrability conditional, 162 of a continuous function, 133 of a function of bounded variation, 134(Exercise 3.2.7 ) improper, 152 of an infinite series, 251 local, 152 of a monotonic function, 133 of a power series, 264 Integrable Riemann, 114,438 Riemann–Stieltjes, 125 Integral over an arbitrary set in Rn,452 of a constant times a function, 136, 456 double, 439 improper, 151 iterated, 462 lower for Riemann integral, 120,442 for Riemann–Stieltjes integral 128 (Exercise 3.1.17 ) multiple, 439 ordinary, 439 of a product, 138,456 proper, 153 over a rectangle in Rn,436(See p. 435) Riemann, 114,438Riemann–Stieltjes, 125,127(Exer- cise3.1.16 ),135(Exercises 3.2.8 – 3.2.10 ),151(Exercise 3.3.23 ) over subsets of Rn,436(See p. 435), 450,452,471–472 of a sum, 136,456 test, 207 triple, 439 Integration by parts, 144 for Riemann–Stieltjes integrals, 135 (Exercise 3.2.8 ) Interior of a set, 21,289 Interior point, 21,289,525 Intermediate value theorem for continuous functions, 63,313 for derivatives 82 Intersection of sets, 20 Interval closed, 23 half closed, 23 half open, 23 open, 21 semi-infinite, 21,23 Interval of convergence, 258 for derivatives, 82 Inverse function, 68 branch of, 409 derivative of, 86(Exercise. 2.3.14 ) of a function restricted to a set, 399 of a matrix, 370 of a transformation, 396 Inverse function theorem, 412 Invertible, locally, 400 Invertible transformation, 396 Irrational number, 6 Isolated point, 23,289,526 Iterated integral, 462 Iterated logarithm, 97(Example 2.4.42 ), 167(Exercise 3.4.10 ),208 230 (Exercise 4.3.11 ),230(Exercise 4.3.16 ) J Jacobian, 384,426 Jordan content, 485 changed by linear transformation, 488 572 Index Jordan measurable set, 485,488 Jump discontinuity, 56 L Lebesgue measure zero, 175,177(Exer- cises 3.5.7 ,3.5.8 ) Lebesgue’s existence criterion, 176 Left limit inferior, 47 Left limit superior, 47 Left-hand derivative, 79 Left-hand limit, 38 Legendre polynomial, 278(Exercise 4.5.27 ) Leibniz’s rule, 86, (Exercise. 2.3.12 ) Length of a vector, 283 l’Hospital’s rule, 88 Limit of a real-valued function, 302 Limit along a curve, 315(Exercise 5.2.3 ) in the extended reals, 43 inferior of a sequence, 188 left,47 infinite, 42,306,316(Exercise 5.2.6 ) at infinity, 307,316(Exercise 5.2.6 ) left-hand, 38 one-sided, 37,40 point, 23,289,526 pointwise, 234,238,244 at˙1,40 of a real-valued function asxapproachesx0,34 asxapproaches1,40 asxapproaches/NUL1,50(Exer- cise2.1.14 ) right-hand, 39 of a sequence, 179,292 uniqueness of, 35,305 of a sum, product, or quotient, 35, 305 superior, left, 47 superior of a sequence, 188 uniform, 237 uniqueness of, 35,305 Line segments in Rn,288 Line, parametric representation of, 288– 289Linear function, 325 Linear transformation, 362 change of content under, 490 differential of, 367 matrix of, 363 Lipschitz condition, 84,87(Exercise 2.3.24 ), 140 Local extreme point, 80,334 Local extreme value, 80 Local integrability, 152 Local maximum point, 80,334 Local minimum point, 80,334 Locally invertible, 400 Lower bound, 7 Lower integral, 120,442 Lower sum, 120,442 M Maclaurin’s series, 264 Magnitude, 2 Main diagonal of a matrix, 370 Mathematical induction, 10,13 Matrices product of, 364 sum of, 364 Matrix adjoint, 370 of a composition of linear transfor- mations, 366 differential, 367,381 elementary, 488 identity, 370 inverse, 370 of a linear transformation, 363 main diagonal of, 370 nonsingular, 370 norm of, 368 scalar multiple of, 364 singular, 370 square, 368(See p. 369) transpose of, 370 Maximum value, local, 80 Maximum of a function, 60 Mean value theorem, 83,347 extended, 106 Index 573 generalized, 83 for integrals, 138,144 Metric, 518 discrete, 519 induced by a norm, 520 Metrics, equivalent, 530 Metric space, 518 complete, 527 connected, 549(Exercise 8.3.2 ) Minimum of a function, 60 Minimum value, local, 80 Minkowski’s inequality, 522 Monotonic function, 44,67,84 integrability of, 133 Monotonic sequence, 182 Multinomial coefficient, 322,336(Exer- cise5.3.12 ) Multiple integral, 439 Multiplication of matrices, 364 of series, 223 scalar, 519 Multiplicityof a zero, 87(Exercise. 2.3.21 ), 108(Exercises 2.5.5 –2.5.7 ) N Natural numbers, 10 n-ball, 290–291 Negative definite polynomial, 353 Negative semidefinite polynomial, 353 Neighborhood, 21,289,525 deleted, 22,525 deleted/SI,22 /SI,21 Nested sets, 292,530 principle of, 292,530 Nondecreasing sequence, 182 Nondegenerate coordinate cube, 437 Nondenumerable set, 176 Nonempty set, 4 Nonincreasing sequence, 182 Nonoscillatory at a point, 162 Nonsingular matrix, 370 Nontrivial solution, 375 Norminfinity, 496,523,524 of a matrix, 368 metric induced by, 520 of a partition, 114,437 on a vector space, 519 Normed vector space, 519 nth derivative, 73 nth partial sum of a series, 201 nth term of a series, 201 Number, natural, 10 Number, prime, 15 O One-sided derivative, 79 One-sided limit, 37 One-to-one transformation, 396 Open ball, 525 Open covering, 25,293,536 Open interval, 21 Openn-ball, 290 Open set, 21,289,525 Ordered field, 2 complete, 4 Order relation, 2 Ordinary derivative, 317 Ordinary integral, 439 Origin of Rn,283 Oscillation of a function, 171 at a point, 172 Oscillatory infinite series, 201 P Parametric representation of a line, 288, 289 Partial derivative, 317 rth order, 319 Partial sums, 244 Partition, 114,437 norm of, 114,437 points, 114 refinement of, 114,438 Path, polygonal, 296 Peano’s postulates, 10–11 Piecewise continuous function, 56 Point, 19 574 Index boundary, 23,289,526 critical, 81,335 exterior, 23,289,526 at infinity, 7 interior, 21,289 isolated, 23,289,526 limit, 23,289,526 local extreme, 80,334 local maximum, 80,334 local minimum, 80,334 in terms of sequences, 197 Pointwise convergence of a sequence of functions, 234,238 of a series, 244 Pointwise limit, 234,238,244 Polar coordinates, 397,502,505 Polygonal path, 296 Polygonally connected, 296 Polynomial, 33,98 homogeneous, 352 negative definite, 353 negative semidefinite, 353 positive definite, 353 positive semidefinite, 353 semidefinite, 353 Taylor, 99,351 Power series, 257 arithmetic operations with, 267 continuity of, 260–261 convergence of, 257 differentiability of, 260–261 integration of, 264 of a product, 268 of a reciprocal, 271 of a quotient, 269 uniqueness of, 263 Prime, 15 Principal value, 155 Principle of mathematical induction, 11, 14 Principle of nested sets, 530 Product Cartesian, 31,436(see p. 435) Cauchy, 226,233(Example 4.3.40 ) inner, 284of matrices, 364 of power series, 268 of series, 223 Proper integral, 153 R Rn,282(see p. 281) rth order partial derivative, 319 Raabe’s test, 212 Radius of convergence, 258 Range of a function, 31,32,545 Ratio of a geometric series, 202 Ratio test, 210 Rational function, 33,232(Exercise 4.3.28 ), 276(Exercise 4.5.4 ) Rational numbers, 2 density of, 6 incompleteness of, 6 Real line, 19 Real number system, 19 Real-valued function, ofnvariables, 302 of a real variable, 31 Reals, extended, 7 Rearrangement of series, 221 Rectangle, coordinate, 437 Refinement of a partition, 114,438 Region, 295,297 Region of integration, 476 Regular transformation, 405 Remainder in Taylor’s formula, 405 Removable discontinuity, 58 Restriction of a function, 399 Riemann integrable, 114,438 Riemann integral 114(see p. 113), 438 uniqueness of, 125(Exercise 3.1.1 ) Riemann sum, 114,438 Riemann–Stieltjes integral, 125 integration by parts for, 135(Exer- cise3.2.8 ) Riemann–Stieltjes sum, 125 Right limit inferior, 53(Exercise 2.1.39 ) Right limit superior, 53(Exercise 2.1.39 ) Right-hand derivative, 79 Right-hand limit, 39 Index 575 Rolle’s theorem, 82 S Scalar multiple, 282 Scalar multiplication, 519 Schwarz’s inequality, 284 Secant plane, 332–333 Second derivative, 73 Second derivative test, 103 Second mean value theorem for integrals, 144 Sequence, 179,526 bounded, 181,292 bounded above, 181 bounded below, 181 Cauchy, 527 convergence of, 179,292,526 decreasing, 182 divergent, 179 to˙1,181 of functional values, 183 of functions, pointwise, 234 increasing, 182 limit of, 179,292 uniform, 237 limit inferior of, 188 limit superior of, 188 monotonic, 182 nondecreasing, 182 nonincreasing, 182 nth term of, 179 terms of, 179 unbounded, 292 uniformly convergent, 237 Series alternating, 203 binomial, 266 Cauchy product of, 226,233(Exer- cise4.3.40 ),280(Exercise 4.5.32 ) differentiability of, 252 divergent, 201 geometric, 202 grouping terms in, 220 Maclaurin, 264multiplication of, 223 of nonnegative terms, 205 partial sums of, 244 power, 257 product of, 218 rearrangement of, 221 Taylor, 223 term by term differentiation of, 252 term by term integration of, 251 uniformly convergent, 246 Set boundary of, 23,289,526 bounded, 7,537 above, 3 below, 7 closed, 21,289,525 closure of, 23,289,526 compact, 26,293,537 complement of, 20 connected, 295 containment of, 19 content of, 485 dense, 6,29(Example 1.3.22 ),70 (Exercise 2.2.10 ) denumerable, 176 diameter of, 292,537 disconnected, 295 empty, 4 exterior of, 23,289,526 interior of, 21,289,525 nondenumerable, 176 nonempty, 4 open, 21,289,525 singleton, 20 strict containment of, 20 subset of, 19 totally bounded, 539 unbounded below, 7 uniformly bounded, 541 universal, 19 Sets disjoint, 20 equality of, 19 intersection of, 20 nested, 530 576 Index union of, 20 Simple zero, 108(Exercise 2.5.5 ) Singleton set, 20 Singular matrix, 370 Solution of a system of linear equations nontrivial, 375 trivial, 375 Space metric, 518 vector, 519 Spherical coordinates, 507 Square matrix, 368(see p. 369) Subsequence, 195 of a convergent sequence, 196,527 Subset, 19 Subspace of a vector space, 519 Successor, 11 Sum of matrices, 364 Riemann, 114,438 lower, 120,442 upper, 120,442 Riemann–Stieltjes, 125 of vectors, 282 Summation by parts, 218 Supremum of a function, 60,313 of a set, 3 existence and uniqueness of, 4 Surface, 331 differentiable, 453 T Tangent to a curve, 75 line, 75 plane, 332 Taylor polynomial, 99,351 of a composite function, 109(Exer- cise2.5.11 ) of a product, 109(Exercise 2.5.10 ) of a reciprocal, 110(Exercise 2.5.12 ) Taylor series, 264 convergence of, 264 Taylor’s theoremfor functions of nvariables, 350 for a function of one variable, 104 Terms of a sequence, 179 Term by term differentiation, 252 Term by term integration, 251 Test Cauchy’s root, 215 comparison for improper integrals, 156 for series, 206 integral, 207 Raabe, 212 ratio, 210 second derivative, 103 Topological properties of Rn,282 (See p. 281) Topological space, 26 Total variation, 134(Exercise 3.2.7 ) Totally bounded, 539 Transformation, 362 affine, 380 analytic, 416(Exercise 6.3.17 ) continuous, 379 differentiable, 339,379–380 differential of, 381 inverse of, 396 invertible, 396396 linear, 362 one-to-one, 396 regular, 405 Transitivity of <,31 Transpose of a matrix, 370 Triangle inequality, 2,285 in a metric space, 518 Triple integral, 439 Trivial solution, 375 U Unbounded above, 7 below, 7 sequence, 292 Unconditional divergence, 233(Exercise 4.3.38 ) Uniform continuity, 64,72(Exercises 2.2.30 – 2.2.32 ),546 Index 577 for functions of nvariables, 314,392 (Exercise 6.2.10 ) Uniform convergence properties preserved by continuity, 242 differentiability, 243 integrability, 242 of a sequence, 236 of a series, 246 Uniformly bounded set in CŒa;b/c141 ,541 Union of sets, 20 Uniqueness of infimum, 7 of limit, 35,305,527 of power series, 263 of prime factorization, 16(Exercise 1.2.14 ) of Riemann integral, 125(Exercise 3.1.1 ) of supremum, 4 Uniform continuity, 64,66,72(Exercises 2.2.30 – 2.2.32 ) Unit vector, 283 Universal set, 19 Upper bound, 3 Upper integral, 120,442 Upper sum, 120,442 V Value of a function, 31,32 local maximum, 80 local minimum, 80 principal, 155 Variation, total, 134(Exercise 3.2.7 ) Vector, 283,519 Vector space, 283,519 normed, 519 subspace of, 519 Vector sum, 282 Vector, unit, 283 Vector-valued function, 362(see p. 361) continuous, 379 differentiable, 379–380 W Weighted average, 139Weierstrass’s test, 246 Z Zero content, 448,460(Exercises 7.1.14 , 7.1.15 ),461(Exercises 7.1.16 – 7.1.19 ),487,514(Exercise 7.3.2 ), 515(Exercise. 7.3.11 ) Zero multiplicityof, 108(Exercises 2.5.5 – 2.5.7 ) simple 108(Exercise 2.5.5 ) Zeroth derivative, 73