Intermediate analysis: an introduction to the theory of functions of one real variable 0891977961, 9780891977964

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Intermediate analysis: an introduction to the theory of functions of one real variable
 0891977961, 9780891977964

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INTERMEDIATE ANALYSIS

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THE APPLETON-CENTURY MATHEMATICS SERIES Edited by Raymond W. Brink

A First Year of College Mathematus, Second Edition

by Raymond W. Brink College Algebra, Second Edition

by Raymond W. Brink Algebra-College Course, Second Edition

by Raymond W. Brink InrermediafR Algebra, Second Edition

by Raymond W. Brink Calculus

by Lloyd L. Smail Analyti,c Geometry and Calculus

by Lloyd L. Smail Solul A nalyti,c Geometry

by John M. H. Olmsted Inrermediare Analysis

by John M. H. Olmsted The M athemati,cs of Finance

by Franklin C. Smith Plane Trigonometry, Revised Edition

by Raymond W. Brink Spherual Trigonometry

by Raymond W. Brink Analyti,c Geometry, Revised Edition

by Raymond W. Brink Essentials of Analyti,c Geometry

by Raymond W. Brink

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Intermediate Analysis AN INTRODUCTION TO THE THEORY OF FUNCTIONS OF ONE REAL VARIABLE

:by . ~.--~ J

'

~

-

John M. H: Olmsted PROFESSOR OF MATHEMATICS UNIVERSITY OF MINNESOTA

NEW YORK: APPLETON-CENTURY-CROFTS, INC.

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© 1956 by

APPLETON-CENTURY-CROFTS, INC. AU right8 reserved. Thia book, or parts thereof, must not be reproduced in any form without permiatsion of the publiahera.

526-1 Library of Congress Card Number: 56-5844

PRINTED IN THE UNITED STATES OF AMERICA

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PREFACE

When should the student of mathematics first begin to develop the techniques of a precise analytic proof, complete with "deltas and epsilons"? Such techniques are quite distinct from the manipulative skills normally acquired in the standard elementary and intennediate courses of the mathematics curriculum, and are often not developed until the student is in the graduate school with a course in Advanced Calculus, and possibly several other graduate courses, behind him. It is sometimes difficult for a student to adjust himself to what seems like an entirely new way of thinking, and particularly disturbing to him to realize that what he had considered to be a "proof" is no longer acceptable. Although, under these circumstances, he is not actually faced with unlearning things he has acquired, he must at least relearn in certain areas. The author of the present book holds that a student of mathematics should properly begin to make his acquaintance with the tools of analysis as soon as possible after the completion of the first course in calculus. These early overtures may be scattered and limited, but they should be thorough and precise as far as they go. In particular, the student should be encouraged to prove (in full detail) statements which previously he has been persuaded to accept because of their immediate obviousness. Who, for instance, could doubt that if a,. -+ + 00 , then 2a,. -+ + 00? On the other hand, how many students fresh from calculus can really 'ffT'OVe this? The purpose of this book is to present the basic ideas and techniques of analysis, for functions of a single real variable, in such a way that students who have studied calculus can proceed at whatever pace and degree of intensity are considered suitable. The organization of the book is designed for maximum flexibility. Topics which are of a more difficult or theoretical nature than most of the material normally studied at the level of Intermediate Calculus are starred for possible omission or postponement. For example, the student can learn to differentiate and integrate power series term by tenn before studying unifonn convergence. Frequently, if a theorem is easy to understand, whereas its proof is more difficult, only the V

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proof is starred. Occasionally such a proof is presented in a later section or chapter. An instance is the boundedness of a function continuous on a closed interval. Some topics which can be omitted from the starred portions without aft'ecting the continuity of what remains receive double stars. This system of starring and double starring applies to the exercises as well as the text. The resultant arrangement of the contents makes the book unusually adaptable for courses of varying lengths and levels. The unstarred portions of the book are immediately suitable for any student who has completed a first course in calculus. They contain extremely careful and thorough discussion of elementary ideas and can be reasonably covered in a course with approximately forty-five meetings. Inclusion of most of the singly starred material permits a richer and more substantial course of some sixty meetings. The doubly starred portions of the book, particularly in the exercises, permit the more advanced student to explore well into the graduate curriculum. Instances are the criterion for Riemann integrability in terms of continuity almost everywhere, and the Lebesgue dominated convergence theorem for Riemann and improper integrals; to these the student is led by suitable sequences of exercises. Special attention should be called to the abundant sets of exercises. These include routine drills for practice, intermediate exercises that extend the material of the text while retaining its character, and advanced exercises that go beyond the standard textual subject matter. Whenever guidance seems desirable, generous hints are included. In this manner the student is led to such items of interest as Weierstrass's theorem on uniform approximation of continuous functions by polynomials, and the construction of a continuous nqndifferentiable function. The analytic treatment of the logarithmic, exponential, and trigonometric functions is presented entirely in the exercises, where sufficient hints are given to make these topics available to all. Answers to all problems are given at the end of the book. Illustrative examples abound throughout. · A few words regarding notation should be given. The equal sign = is used for equations, both conditional and identical, and the triple bar sign = is reserved for definitions. For simplicity, if the meaning is clear from the context, neither symbol is restricted to the indicative mood as in "(a+b) 2 = a2 + 2ab + b2," or "where f(x) = x2 + 5." Examples of subjunctive uses are "let x = n," and "let E = l," which would be read "let x be equal ton," and "let Ebe defined to be 1," respectively. A similar freedom is granted the inequality symbols. For instance, the symbol > in the following constructions "if E > 0, then • · •," "let E > O," and "let E > 0 be given," could be translated "is greater than," "be greater than," and "greater than," respectively. In a few places parentheses are used to indicate alternatives. The principal instances of such uses are heralded by announcements or footnotes in the text. Here again it is hoped that the context will prevent any

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ambiguity. Such a sentence as "The function f(x) is integrable from a to b (a < b)" would mean that "f(x) is integrable from a to b, where it is assumed that a < b," whereas a sentence like "A function having a positive (negative) derivative over an interval is strictly increasing (decreasing) there" is a compression of two statements into one, the P.arentheses indicating an alternative formulation. The author wishes to express his deep appreciation of the aid and suggestions given by Professor R. W. Brink of the University of Minnesota in the preparation of the manuscript. He is also indebted to others, including in particular Professor W. D. Munro of the University of Minnesota, for their friendly and helpful counsel. Minneapoli8

J.M.H.O.

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CONTENTS

PAGE

PREFACE.....................................................

V

Chapter 1

THE REAL NUMBER SYSTEM SECTION

101. 102. 103. 104. 105. 106. 107. 108. 109. 110. 111. 112. 113. •114. 115. •116.

*

Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Axioms of the basic operations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Axioms of order.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Positive integers and mathematical induction. . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Integers and rational numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Geometrical representation and absolute value. . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some further properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Axiom of completeness...................................... Further remarks on mathematical induction . . . . . . . . . . . . . . . . . . . Exercises..................................................

1 2 4 5 6 7 10 13 13 16 18 19 20 22 24 25

Chapter 2

FUNCTIONS, SEQUENCES, LIMITS, CONTINUITY 201. 202. 203. 204. 205. 206. 207. 208. 209. 210. 211.

Functions and sequences. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limit of a sequence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limit theorems for sequences................ . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limits of functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limit theorems for functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Types of discontinuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continuity theorems........................................ ix

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CONTENTS

X

212. 213. 214. 215. 216.

Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . More theorems on continuous functions........................ Existence of V2 and other roots.............................. Monotonic functions and their inverses. . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

53 54 55 55 57

*Chapter 3

SOME THEORETICAL CONSIDERATIONS *301. *302. *303. *304. *305. *306. *307. *308. **309. **310. **311. **312.

A fundamental theorem on bounded sequences.................. The Cauchy Criterion for convergence of a sequence. . . . . . . . . . . . Sequential criteria for continuity and existence of limits. . . . . . . . . . The Cauchy Criterion for functions. . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proofs of some theorems on continuous functions. . . . . . . . . . . . . . . Uniform continuity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercise.'!.................................................. Point sets: open, closed, compact, connected sets. . . . . . . . . . . . . . . Point sets and sequences......... . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some general theorems...................................... Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

61 62 64 65 66 69 71 73 74 78 79 81

Chapter 4

I/

DIFFERENTIATION 401. 402. 403. 404. 405. 406. 407. 408. 409. 410. 411. 412. 413. 414. 415. 416. 417. 418. 419. *420. *421.

Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The derivative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . One-sided derivatives............................ . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Rolle's Theorem and the Law of the Mean. . . . . . . . . . . . . . . . . . . . . Consequences of the Law of the Mean......................... The Extended Law of the Mean. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises .................................. : : .............. Maxima and minima. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Differentials... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Approximations by differentials...... . ..... . .................. Exercises.................................................. L'Hospital's Rule. Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . The indeterminate form 0/0.................................. The indeterminate form co /co . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Other indeterminate forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises.................................................. Curve tracing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Without loss of generality... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5

85 85 88 91 93 97 98 100

105 107 109 111 115 116 117 120 121 123 127 129 129



INTEGRATION 501. The definite integral......................................... *502. More integration theorems...................................

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CONTENTS

503. 504. 505. 506. *507. *508. *509. *510. 511. 512. 513. 514. 515. **516. *517. *518.

xi

Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Fundamental Theorem of Integral Calculus. . . . . . . . . . . . . . . . Integration by substitution................................... Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sectional continuity and smoothness. . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reduction formulas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Improper integrals, introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Improper integrals, finite interval. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Improper integrals, infinite interval. . . . . . . . . . . . . . . . . . . . . . . . . . . Comparison tests. Dominance............................... Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bounded variation....................................... . .. The Riemann-Stieltjes integral. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises..................................................

v

Chapter 6

SOME ELEMENTARY FUNCTIONS *601. *602. *603. *604. 605. 606. 607. 608. 609, *610. *611. *612.

143 154 156 156 159 160 161 162 164 164 167 168 171 175 179 184

The exponential and logarithmic functions. . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The trigonometric functions.................................. Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some integration formulas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Inverse hyperbolic functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Classification of numbers and functions........................ The elementary functions.................................... Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

188 188 191 192 194 196 197 199 200 201 203 204

Chapter 7

INFINITE SERIES OF CONSTANTS 701. 702. 703. 704. 705. 706. 707. 708. 709. 710. 711. *712. *713.

Basic definitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Three elementary theorems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A necessary condition for convergence. . . . . . . . . . . . . . . . . . . . . . . . . The geometric series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Positive series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The integral test............................. . .. . ........... Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Comparison tests. Dominance................... . .. . ........ The ratio test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The root test.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises......................................... . ........ More refined tests. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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206 207 207 208 208 209 211 212 215 218 219 221 223

CONTENTS

xii

714. 715. 716. 717. 718. 719. *720. 721.

Series of arbitrary terms..................................... Alternating series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Absolute and conditional convergence. . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Groupings and rearrangements................................ Addition, subtraction, and multiplication of series............... Some aids to computation.................................... Exercises ............. : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

225 225 227 229 231 233 235 237

Chapter 8



POWER SERIES

801. 802. 803. 804. 805. 806. 807. 808. 809. 810. 811. 812. 813. 814. *815. *816.

Interval of convergence...................................... Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Taylor series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Taylor's Formula with a Remainder. . . . . . . . . . . . . . . . . . . . . . . . . . . Expansions of functions...................................... Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Some Maclaurin series.. . .. .. .. . . .. .. .. . . .. .. . . .. .. . . . . . . . . . . Elementary operations with power series....................... Substitution of power series.. . . .. .. . . .. .. .. . . .. . . . . .. . . . . . . . . Integration and differentiation of power series. . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Indeterminate expressions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Computations ........ '. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Analytic functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

240 243 244 246 248 249 250 253 255 259 261 264 264 266 267 268

*Chapter 9 UNIFORM CONVERGENCE

*901. *902. ~3. *904. *905. *906. *907. *908.

Uniform convergence of sequences. . . . . . . . . . . . . . . . . . . . . . . . . . . . Uniform convergence of series................................ Dominance and the Weierstrass M-test. . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Uniform convergence and continuity. . . . . . . . . . . . . . . . . . . . . . . . . . Uniform convergence and integration.......................... Uniform convergence and differentiation....................... Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . *909. Power series. Abel's Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . **910. Proof of Abel's Theorem..................................... •911. Exercises..................................................

270 273 27 4 275 278 279 281 283 288 289 290

ANSWERS TO ALL PROBLEMS. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

293

INDEX......................................................

301

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INTERMEDIATE ANALYSIS

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I The Real Number System

101. INTRODUCTION

The reader is already familiar with many of the properties of real numbers. He knows, for example, that 2 + 2 = 4, that the product of two negative numbers is a positive number, and that if x, y, and z are any real numbers, then x(y + z) = xy + xz. The average reader at the level of Calculus usually knows these things because he has been told that they are true-probably by people who know such things simply because they have been told. He knows pretty well why some of these familiar facts (like 2 · 3 = 3 · 2) are true, but is entitled to be unsure of the reasons for others (like V2 ·v3 = y3 •y12). It is important to know that the properties of real numbers which we use almost daily are true, not by fiat or decree, but by rigorous mathematical proof. The situation is not unlike that of Euclidean Plane Geometry. In either case (numbers or geometry), all properties within one particular mathematical system follow by logical inference from a few basic assumptions, or hypotheses, or axioms, which are properties that are "true" only in the relative sense that they are assumed as a working basis for that particular mathematical system. What is true in one system may be false or meaningless in another, but within one logical system "true" means "implied by the axioms." For the system of real numbers the axioms, as usually given, t are five in number, and are concerned only with the natural numbers, 1, 2, 3, · · · . These five axioms were given in 1889 by the Italian mathematician G. Peano (1858-1932), and are called the Peano axioms for the natural numbers. In the book by Landau, referred to in the footnote, the reader is led by a sequence of steps (consisting of 73 definitions and 301 theorems, most of which are very easy) through the entire construction of the real and complex number systems. The natural numbers are used to build the

t For a detailed discussion of the real number system see E. Landau, Foundations of Analysis (New York, Chelsea Publishing Company, 1951) or E. W. Hobson, 7'heory of Functions (Washington, Harren Press, 1950). 1

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THE REAL NUMBER SYSTEM

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[§ 102

larger system of positive rational numbers (fractions); these in turn form the basis for constructing the positive real numbers; the positive real numhers then lead to the system of all real numbers; finally, the complex numbers are constructed from the real numbers. In each successive class the basic operations of addition and multiplication and (through the class of real numbers) the relation of order are defined, and shown to be consistent with those of the preceding class considered as a subclass. The result is the number system as we know it. In this book we shall begin our discussion of the number system at a point far along the route just outlined. We shall, in fact, assume that we already have the entire real number system, and describe this system by means of a few fundamental properties which we shall accept without further question. These fundamental properties are so chosen that they give a complete description of the real number system, so that all properties of the real numbers are deducible from them. t For this reason, and to distinguish them from other properties, we call these fundamental properties axioms, and refer to them as such in the text. These axioms are arranged according to three categories: basic operations, order, and completeness. The present chapter is devoted exclusively to the real number system. The first part of the chapter is ele~entary in nature, and includes the axioms for the basic operations and order, a review of mathematical induction, a brief treatment of the integers and rational numbers, and a. discussion of geometrical representation and absolute value. The last part, of a more advanced character, is starred for possible omission or postponement, and includes the axiom of completeness and a further treatment of topics introduced earlier. Except for passing mention, complex numbers are not used in this book. Unless otherwise specified, the word number should be interpreted to mean real number.

.

The reader should be advised that many of the properties of real numbers that are given in the Exercises of this chapter are used throughout the book without specific reference. In many cases (like the laws of cancellation and the unique factorization theorem) such properties, when desired, can be located by means of the index. 102. AXIOMS OF THE BASIC OPERATIONS

The basic operations of the number system are addition and multiplicaAs shown below, subtraction and division can be defined in terms of these two. The axioms of the basic operations are further subdivided into three subcategories: addition, multiplication; and addition and multiplication.

tion.

t For a proof of this fact and a discl188ion of the axioms of the real number system, see G. Birkhoff and S. MacLane, A Survey of Modern Algebra (New York, The Macmillan Company, 1944). Also cf. Ex. 13, § 116.

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§ 102]

3

BASIC OPERATIONS

I. Addition (i) Any two numbers, x and y, have a unique sum, x + y. (ii) The associative law holds. That is, if x, y, and z are any numbers, x

+ (y + z) = (x + y) + z.

(iii) There exists a number O such that for any number x, x + 0 = x. (iv) Corresponding to any number x there exists a number -x such that

+ (-x)

X

(v) The commutative law holds. X

+y

= 0.

That is, if x and y are any numbers,

=

y

+ X.

The number O of axiem (iii) is called zero. The number -x of axiom (iv) is called the negative of the number x. The difference between x and y is defined:

y

X -

= X + (-y).

The resulting operation is called subtraction. Some of the properties that can be derived from Axioms I alone are given in Exercises 1-7, § 103.

n.

Multiplication (i) Any two numbers, x and y, have a unique product, xy. (ii) The associative law holds. That is, if x, y, and z are any numbers,

x(yz) = (xy)z. (iii) There exists a number 1 ¢ 0 such that for any number x, x · 1 = x. (iv) Corresponding to any number x ¢ 0 there exists a number x-1 such that x-x-1 = 1. (v) The commutative law holds. That is, if x and y are any numbers,

xy

= yx.

The number 1 of axiom (iii) is called one or unity. The number x-1 of axiom (iv) is called the reciprocal of the number x. The quotient of x and y (y ¢ O) is defined: X

- = x•7r1.

y

The resulting operation is called division. Some of the properties that can be derived from Axioms II alone are given in Exercises 8-11, § 103.

m. Addition and Multiplication (i) The distributive law holds.

That is, if x, y, and z are any numbers,

x(y + z) = xy + xz. The distributive law, together with Axioms I and II, yields further familiar relations, some of which are given in Exercises 12-33, § 103.

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THE REAL NUMBER SYSTEM

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[§ 103

103. EXERCISES

In Exercises 1-33, prove the given statement or establish the given equation. 1. There is only one number having the property of the number Oof Axiom I (iii). Hint: Assume that the numbers O and O' both have the property. Then simultaneously 0' 0 = 0' and 0 0' = 0.

+

+

2. The law of cancellation for addition holds: x + y = x + z implies y = z. Hint: Let ( -x) be a number satisfying Axiom I (iv). Then

(-z) + (z + y) = (-z) + (x + a). Use the associative law. 3. The negative of a number is unique. Hint: If y has the property of -z in Axiom I (iv), § 102, z + y = x + (-z) = 0. • Use the law of cancellation, given in Exercise 2. 4. -0 = 0.

6. -(-z) = x. 6. 0 - X = -z. 7. -(x y) = -x - y; -(x ....:. y)

+ = y - x. Hint: By the uniqueness of the negative it is sufficient for the first part to prove that (x + y) + [(-z) + (-y)] = 0. Use the commutative and associative laws. 8. There is only one number having the property of the number 1 of Axiom II (iii). 9. The law of cancellation for multiplication holds: xy = xz implies y = z if z ,;,t, 0. 10. The reciprocal of a number (,;,t, 0) is unique. 11. 1-1 = 1. 12. x•0 = 0. Hint: x•0 + 0 = x•0 = x(0 + 0) = x·0 + x•0. Use the law of cancellation on the extreme members. 13. Zero has no reciprocal. 14. If x ,;,t, 0 and y ,;,t, 0, then xy ,;,t, 0. Equivalently, if xy = 0, then either x = 0 or y = 0. Hint: Assume x ,;,t, 0, y ,;,t, 0, and xy = 0. Then x-1 (xy) = x-1.0 = 0. Use the associative law to infer y = 0. 16. If x ,;,t, 0, then z-1 ,;,t, 0 and (x-1) -1 = x. 16. ~ y

17.

= 0 (y

,;,t, 0) if and only if x

! = z-1 (x X

= 0.

,;,t, 0).

18. If x ,;,t, 0 and y ,;,t, 0, then {xy)-1 = x-1y-1, or ..!_ = ! xy X 19. If b ,;,t, 0 and d ,;,t, 0, then

• !. y

i = ::-

Hint: (ad)(bd)-1 = (ad)(d-1b-1) = a[(dd-1 )b-1] = ab·-1 • 20. If b ,;,t, 0 and d ,;,t, 0 then ~ · £ = !!Q. '

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b

d

bd

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§ 104]

AXIOMS OF ORDER

21. If b F- 0 and d F- 0, then

5

a

b + de = ad+be bd ·

Hint: (bd)-1 (ad + be) = (b- 1d-1 )(ad) + (b- 1d-1 )(be). 22. (-1)(-1) = 1. Hint: (-1)(1 + (-1)) = 0. The distributive law gives ( -1) + (-1)( -1) = 0. Add 1 to each member. 23. (-l)x = -x. Hint: Multiply each member of the equation 1 + (-1) = 0 by x. 24. (-x)(-y) = xy. Hint: Write -x = (-l)x and -y = {-l)y. 25. -(xy) = (-x)y = x{-y).

=-=-

26. - ~ = =~ (y F- 0). y y -y 27. x(y - z) = xy - xz. *28. (x - y) (y - z) = x - z. *29. (a - b) - (e - d) = (a+ d) - (b + e). *30. (a + b)(e + d) = (ac + bd) + (ad + be). *31. (a - b)(e - d) = (ae + bd) - (ad+ be). *32. a - b = e - d if and only if a + d = b + e. 33. The general linear equation ax + b = 0, a F- 0, has a unique solution x = -b/a.

+

104. AXIOMS OF ORDER

In addition to the basic operations, the real numbers have an order relation subject to certain axioms. One form of these axioms expresses order in terms of the primitive concept of positiveness: (i) Some numbers have the property of being positive. (ii) For any number x exactly one of the following three statements is true: x = 0; x is positive; - x is positive. (iii) The sum of two positive numbers is positive. (iv) The product of two positive numbers is positive. Definition I. The symbols< and> (read "less than" and "greater than," respectively) are defined by the statements

x x

Definition II.

< >

y if and only if y - xis positive; y if and only if x - y is positive.

The number xis negative if and only if-xis positive.

Definition m. The symbols ~ and ~ (read "less than or equal to" and "greater than or equal to," respectively) are defined by the statements

x x

~ ~

y if and only if either x y 1J and only (f either x

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< >

y or x = y; y or x = y.

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THE REAL NUMBER SYSTEM

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[§ 105

NOTE 1. The two statements :i; < y and y > :i; are equivalent. The two statements :i; ~ y and y e:; :i; are equivalent. NOTE 2. The &eme of an inequality of the form :i; < y or :i; ~ y is said to be the reverse of that of an inequality of the form :i; > y or :i; e:; y. NOTE 3. The simultaneous inequalities :i; < y, y < z are usually written :i; < y < z1 and the simultaneous inequalities :i; > y, y > z are usually written :i; > y > z. Similar interpretations are given the compound inequalities :i; ~ y ~ z and :i; e:; y e;;; z.

105. EXERCISES

In Exercises 1-24, prove the given statement or establish the given equation. 1. :i; > 0 if and only if :i; is positive. 2. The transitive law holds for < and for >: :i; < y, y < z imply :i: < z; :i: > y, y > z imply :i: > z. (Cf. Ex. 19.) 3. The law of trichotomy holds: for any :i; and y, e:i:actly one of the following holds: :i: < y, :i: = y, :i: > y. 4. Addition of any number to both members of an inequality preserves the order relation: :i; < y implies :i; z < y z. A similar fact holds for subz) - (:i; z) = y - :i:. traction. Hint: (y 5. :i; < 0 if and only if :i; is negative. 6. The sum of two negative numbers i_s negative. 7. The product of two negative numbers is positive. Hint: :i:y = (-:i;)(-y). (Cf. Ex. 24, § 103.) 8. The square of any nonzero real number is positive.

+

+ +

+

1 > 0. The equation :i:2 + 1 = 0 has no real root. The product of a positive number and a negative number is negative. The reciprocal of a positive number is positive. The reciprocal of a negative number is negative. 9. 10. 11. 12.

13. 0

< :i: < y imply 0

0 imply :i:z < yz and :i;/z < y/a. 15. Multiplication or division of both members of an inequality by a negative number reverses the order relation: :i; < y, z < 0 imply :i:z > yz and :i;/z > y/z. 16. a < b, c < d imply a + c < b + d. 17. 0 < a < b, 0 < c < d imply ac < bd and a/d < b/c. 18. If :i; and y are nonnegative numbers, then :i; < y if and only if :i:2 < y2 (cf. Ex. 10, § 107). 19. The transitive law holds for ~ (also for e:;): :i; ~ y, y ~ z imply :i; ~ z. (Cf. Ex. 2.) 20. :i; ~ y, y ~ :i: imply :i; = y. 21. :i; + :i; = 0 implies :i; = 0, :i; + :i; + x = 0 implies :i; = 0. 22. :i:2 + y 2 e:; 0; :i:2 + y2 > 0 unless :i; = y = 0.

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§ 106]

7

POSITIVE INTEGERS

23. If :,; is a fixed number satisfying the inequality :,; < E for every positive number e, then :,; ~ 0. Hint: If :,; were positive one could choose E = :,;. 24. There is no largest number, and therefore the real number system is infinite. (Cf. Ex. 11, § 109.) Hint: :,; + 1 > :,;. 26. If :,; < a < y or if :,; > a > y, then a is said to be between z and y. Prove that if z and y are distinct numbers, their arithmetic mean ½(z + y) is between them. 26. If x2 = y, then:,; is called a square root of y. By Exercise 8, if such a number :,; exists, y must be nonnegative, and if y = 0, :,; = 0 is the only square root of y. Show that if a positive number y has square roots, it has exactly two square roots, one positive and one negative. The unique positive square root is called the square root and is written Vy. It is shown in § 214 that such square roots do exist. Hint: Let :,;1 = z2 = y. Then z 1 - z1 = (:,; - z)(:,; + z) = 0. Therefore z = z or z = -s. 106. POSITIVE INTEGERS AND MATHEMATICAL INDUCTION

In the development of the real number system, as discussed briefly in the Introduction, the natural numbers become absorbed in successively larger number systems, losing their identity but not their properties, and finally emerge in the real number system as positive integers, 1, 2, 3, · · · . The positive integers have certain elementary properties that follow directly from the role of the natural numbers in the construction of the number system (cf. § 115 for further remarks): (i) (ii) (iii) (iv) (v) (vi)

The "positive int,egers" are positive. If n i8 a positive integer, n ~ 1.

(Cf. Ex. 9, § 105.)

2 = 1 + 1, 3 = 2 + 1, 4 = 3 + 1, 5 = 4 + 1, • · · . 0 < 1 < 2 < 3 < 4 < •· • . The sum and product of two positive integers are positive integers. If m and n are positive inl,egers, and m < n, then n - m i8 a positive integer. (vii) If n i8 a positive integer, there i8 no positive integer m sueh that n n, where n is a positive integer. 12. Establish the formula 1 2 n = ½n(n 1). (Cf. Ex. 38.) 13. Establish the formula (cf. Ex. 39) l2 22 n2 = ¼n(n 1)(2n 1). 14. Establish the formula (cf. Ex. 40) l8 21 n 3 = tn'(n 1)2 = (1 2 n) 2• 15. Establish the formula (cf. Ex. 41) 1' 2' n' = io-n(n 1)(2n 1)(3n2 3n - 1). *16. Let A be a nonempty set of real numbers with the property that whenever z is a member of A and y is a member of A then z - y is a member of A. Prove that whenever z1, z2, · · · , z,. are members of A then z1 z2 z,. is a member of A. 17. Establish the law of exponents: a"'a" = a..+.., where a is any number and m and n are positive integers. Hint: Hold m fixed and use induction onn. 18. Establish the law of exponents: a.,_,. if m < n a"' ' 1 a" = { - - if n < m

+

+

+ ••• +

+ + ••• +

+ + •, • +

+

+

+ + · ·· +

+

+ + ••• +

+

+

+

+ + ••• +

+

+

+ + ·· · +

a"-•

'

where a is any nonzero number and m and n are positive integers. 19. Establish the law of exponents: (a"')" = a..", where a is any number and m and n are positive integers. 20. Establish the law of exponents: (ab)" = a"b", where a and b are any numbers and n is a positive integer. Generalize to m factors. 21. Establish the law of exponents (

r

1

= ::, where a and bare any numbers

(b ;,t: 0) and n is a positive integer.

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EXERCISES

§ 107]

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22. A positive integer m is a factor of a positive integer p if and only if there exists a positive integer n such that p = mn. A positive integer p is called composite if and only if there exist integers m > 1 and n > 1 such that p = mn. A positive integer p is prime if and only if p > 1 and p is not composite. Prove that if mi, i = 1, 2, • • • , n, are integers > 1, then m1m1 • • • m,. > n. Hence prove that any integer > 1 is either a prime or a product of primes. (Cf. Ex. 29.) 23. Two positive integers are relatively prime if and only if they have no common integral factor greater than 1. A fraction p/q, where p and q are positive integers, is in lowest terms if and only if p and q are relatively prime. Prove that any quotient of positive integers is equal to such a fraction in lowest terms. (Cf.. Ex. 33.) *24. Prove that the positive integers are Archimedean (cf. § 114): If a and b are positive integers, then there exists a positive integer n such that na > b. Hint: ba ~ b. •25. Prove the Fundamental Theorem of Euclid: If a and b are positive integers, there exist unique numbers n and r, each of which is either O or a positive integer and where r < a, such that b = na + r. Hint: For existence, let n + 1 be the smallest positive integer such that (n + l)a > b (cf. Ex. 24). •26. Prove that a prime number p cannot be a factor of the product of two positive integers, a and b, each of which is less than p. Hint: Assuming that p is a factor of ab, let c be the smallest positive integer such that p is a factor of ac, and let n be a positive integer such that nc < p < (n + l)c (cf. Ex. 24). Then p - nc is a positive integer less than c having the property assumed for c l •27. Prove that if a prime number p is a factor of the product of two positive integers, a and b, then pis a factor of either a orb (or both). Hint: Use Exs. 25 and 26. •28. Prove that if a prime number p is a factor of the product of n positive integers a,, at, · · • , a,., then p is a factor of at least one of these numbers. (Cf. Ex. 27.) •29. Prove the Unique Factorization Theorem: Every positive integer greater than 1 can be represented in one and only one way as a product of primes. Hint: Assume P1P2 • • • p,.. and q1q2 • • • q,. are two nonidentical factorizations of a positive integer. Cancel all identical prime factors from both members of the equation P1P2 • • • p,,. = q,q2 • • • q,.. A prime number remains which is a factor of both members but which violates Ex. 28. (Cf. Ex. 22.) •SO. Prove that if a, b, and c are positive integers such that a and c are relatively prime and band c are relatively prime, then ab and care relatively prime. (Cf. Ex. 23.) •31. Prove that if a, b, and c are positive integers such that a and c are relatively prime and c is a factor of ab, then c is a factor of b. (Cf. Ex. 23.) •32. Prove that two positive integers each of which is a factor of the other are equal. *33. Prove that if a fraction p/q, where p and q are positive integers, is in lowest terms, then p and q are uniquely determined. (Cf. Ex. 23.)

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THE REAL NUMBER SYSTEM

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[§ 107

=

34. If n is a positive integer, n factorial, written n ! , is defined: n ! 1 ·2·3· • • • •n. Zero factorial is defined: 0 ! ~ 1. If r is a positive integer or zero

and if O ~ r

~

n, the binomial coefficient(;) (also written ,.C,) is defined:

(;) = (n - nr; ! r I Prove that (;) is a positive integer.

Hint: Establish the law of Pascal's 1

)

= (r ~

+ (;}

) 1 35. Prove the Binomial Theorem for positive integral exponents (cf. V, § 807): If x and y are any numbers and n is a positive integer,

Triangle (cf. any College Algebra text): (n;

(:z;

+ y)" = (~) x" + (~) :z; -1y + ... + (;) :z;"-Ty• + ... + (:) y". 11

(Cf. Ex. 34.) 36. The sigma summation notation is defined: n

E J(k) = J(m) + J(m + 1) + · · · + f(n), k-m (i) k is a dummy variable:

En

E

is additive:

E

(iii)

E

is homogeneous:

(iv)

E 1=n k=m

11

[J(k)

k=m

"

i=m

+ g(k)]

~

- m

E "

J(k)

k=m n

n

n

Prove:

= E J(i).

J(k)

k=m

(ii)

where n ~ m.

+ k=m Ln g(k).

E cf(k) = c k=m E J(k). k=1n

+ 1.

•37. A useful summation formula is (1)

t,

[J(k) - J(k - 1)]

k-1

= J(n) -

J(0).

Establish this by mathematical

induction. •38. By means of Exercises 36 and 37 derive the formula of Exercise 12. Hint: Let J(n) = n 2• Then (]) becomes n

n

E

k-1

[k1

-

(k - 1) 2]

= E

k=l

(2k - 1)

= n2, or

n

2

•39. By Let J(n) •40. By •41. By

means 3 in means means

=n

Ek= n + n. k=l 2

of Exercises 36-38, derive the formula of Exercise 13. (1). of Exercises 36-39, derive the formula of Exercise 14. of Exercises 36-40, derive the formula of Exercise 15. n

Hint:

> km is a polynomial in n of k-1 degree m + 1 whose leading coefficient (coefficient of the term of highest degree) is 1/(m + 1). (Cf. Exs. 36-41.) •42. Use mathematical induction to prove that

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§ 108]

•43. If

INTEGERS AND RATIONAl NUMBERS

a,,

a2, · · • ,

a,. and b,,

(t a; t b; 2

i-1

)

(

2

)

-

13

b2, · · · , b,. are real numbers, show that

t a,b;)

(

i-1

i-1

2

L

=

l:iii -n > 0, use property (vi), § 106. 2. Prove that the product of two integers is an integer. 3. Prove that the difference between two integers is an integer. 4. Prove that the quotient of two integers need not be an integer. 6. Prove that the integers are rational numbers. 6. Prove that the sum, difference, and product of two rational numbers are rational numbers, and that the quotient of two rational numbers the second l•f which is nonzero is a rational number.

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THE REAL NUMBER SYSTEM

14

[§ 109

7. Define integral powers an, where a "'F O and n is any integer, and establish the laws of exponents of Exercises 17-21, § 107, for integral exponents. 8. The square root of a positive number was defined in Exercise 26, § 105. Existence is proved in § 214. Prove that v'2 is irrational. That is, prove that there is no positive rational number whose square is 2. Hint: Assume V2 = p/q, where p and q are relatively prime positive integers (cf. Ex. 23, § 107). Then p2 = 2q', and pis even and of the form 2k. Repeat the argument to show that q is also even! *9. Assume that f(x) = aoZ" + a1z11- 1 + a,._1x + a,. is a polynomial with integral coefficients ao, • • • , a,., of which the leading coefficient ao and the constant term an are nonzero. Prove that if p/q is a rational root of the equation f(x) = O, where p and q are relatively prime integers (that is, they are nonzero and they have no common integral factor greater than 1; cf. Ex. 23, § 107), then p is a factor of a,. and q is a factor of ao. Hint: By assumption aopn + a1p"-1q + ••• + a,.q" = 0. Use Ex. 31, § 107, and the fact that each of p and q is a factor of n of these terms, and therefore also of the remaining term. *10. Prove that a positive integer m cannot have a rational nth root (n is a positive integer), unless m is a perfect nth power (of a positive integer), and hence generalize Exercise 8. Hint: Use Ex. 9. *11. Prove that a positive rational number p/q, where p and q are relatively prime positive integers, cannot have a rational nth root (n is a positive integer) unless both p and q are perfect nth powers. (Cf. Ex. 23, § 107; Ex. 10 above.) *12. Let a and b be any two nonzero integers, and consider the set

+ •••

+ nb} consisting of all numbers of the form ma + nb, where m G(a, b)

e

{ma

and n are arbitrary integers. Prove that G(a, b) consists precisely of all integral multiples of a positive integer k. Hint: Show first that whenever c and d belong to G(a, b), so doc+ d and c - d, and that whenever c belongs to G(a, b) and his an arbitrary integer, then ch belongs to G(a, b). Then let k be the smallest positive integer in G(a, b). If there were a member of G(a, b) not a multiple of k, use the Fundamental Theorem of Euclid (Ex. 25, § 107) to obtain a contradiction. •13. Prove that the number k of Exercise 12 is unique. Pi;ove that k is a factor of both a and b, and that any common factor of a and b is a factor of k. (In other words, k is the highest common factor of a and b.) •14. Prove that if a and bare relatively prime integers, there exist integers m and n such that ma+ nb

= 1.

(Cf. Exs. 12-13.) NoTE. In the following problems, nonzero polynomial means any polynomial different from the zero polynomial all of whose coefficients are zero. •16. Prove the Fundamental Theorem of Euclid for polynomials: If f(x) and g(x) are nonzero polynomials, there exist unique polynomials Q(x) and R(x), where R(x) either is O or has degree less than that of g(x), such that f(x)

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= Q(x)g(x) + R(x).

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UNIVERSITY OF MICHIGAN

EXERCISES

§ 109]

15

The polynomials Q(z) and R(z) are called the quotient and remainder, respectively, when /(z) is divided by g(z). (Cf. Ex. 25, § 107 .) •16. Let /(z) and g(z) be any two nonzero polynomials, and consider the set

+ ~(z)g(z)} consisting of all polynomials of the form J + ~g, where and ~ are arbitrary G(J, g)

!!!!

{(z)/(z)

polynomials. Prove that G(J, g) consists precisely of all polynomial multiples of a nonzero polynomial P(z). Hint: See Ex. 12, observing that P(z) is not uniquely determined, but is any nonzero polynomial in G(J, g) of lowest degree. •17. State and prove the analogue of Exercise 13 for Exercise 16. Hint: P(z) is unique except for a nonzero constant factor. •18. Prove that if /(z) and g(z) are relatively prime polynomials (that is, they are nonzero and have only constants as common polynomial factors), there exist polynomials q,(z) and ~(z) such that q,(z)/(z)

+ ~(z)g(z)

= 1.

(Cf. Exs. 16-17.) •19. Prove that any fraction of the form N(z)//(z)g(z), where N(z) is a polynomial and /(z) and g(z) are relatively prime polynomials, can be written in the form

Jl.J&... = /(z)g(z)

A(z) /(z)

+ B(z), g(z)

where A(z) and B(z) are polynomials. Hint: Take the equation of Ex. 18, divide by /(z)g(z), and multiply by N(z). •20. Prove that if /(z) and g(z) are nonconstant polynomials and if the degree of N(z) is less than that of /(z)g(z), then the polynomials A(z) and B(z) of Exercise 19 can be chosen so that their degrees are less than those of /(z) and g(z), respectively. Hint: Write

Jl.J&... = cl>(z) + "IJl(z),

f(z)g(z)

/(z)

g(z)

where cl>(z) and "IJl(z) are polynomials, and write cl>(z) = Q(z) /(z) + A(z), where the degree of A(z) is less than that of /(z), and define B(z) "IJl(z) + Q(z) g(z). Under the assumption that the degree of B(z) is at least that of g(z), obtain a contradiction from the equation N(z) = A(z)g(z) + B(z)/(z). •21. Prove that any fraction of the form A(z)/[p(z)]•, where A(z) and p(z) are nonconstant polynomials and n is a positive integer, and where the degree of the numerator A(z) is less than that of the denominator [p(z)]", can be written in the form

=

B(z) [p(z) ]"-1

+

C(z) , [p(z) ]•

where B(z) and C(z) are polynomials such that B(z) either is O or has degree less than that of [p(z) J•-1, and C(z) either is O or has degree less than that of p(z). Hint: Use Ex. 15 to write A(z) = B(z)p(z) + C(z). •22. Recall the fact from College Algebra that any real polynomial (any polynomial with real coefficients) of positive degree can be factored into real linear and quadratic factors; more precisely, as a constant times a product of factors of the form (z + a) and (z1 + bx + c), where the discriminant b2 - 4c is negative. Use this fact and those contained in Exercises 19-21 to prove the

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THE REAL NUMBER SYSTEM

16

[§ 110

Fundamental Theorem on Partial Fractions (used in Integral Calculus): Any quotient of real polynomials, where the degree of the numerator is less than that of the denominator, can be expressed as a sum of fractions of the form

A

(z

and

+ a)"'

Bx+ C ------'---, 2 (z

+ bz + c)"

where A, B, and Care constants and m and n are positive integers. 110. GEOMETRICAL REPRESENTATION AND ABSOLUTE VALUE

The reader has doubtless made use of the standard representation of real numbers by means of points on a straight line. It is conventional, when considering this line to lie horizontally as in Figure 101, to adopt a · · · -2

-1

0

1

2

•••

FIG. 101

uniform scale, with numbers increasing to the right and decreasing to the left. With an appropriate axiomatic system for Euclidean geometryt there is a one-to-one correspondence between real numbers and points on a line. That is, to any real number there corresponds precisely one point of the line, and to any point of the line, there corresponds precisely one real number. For this reason it is often immaterial whether one speaks of numbers or points. In this book we shall frequently use these two words interchangeably, and feel free, for example, to speak of the point 3. In this sense, in Figure 101, positive numbers lie to the right of the point 0, and x < y if and only if the point x is to the left of the pointy. Again, if x < z, then the number y satisfies the simultaneous inequalities x < y < z if and only if the pointy is between the points x and z (cf. Ex. 25, § 105). Properties of the real numbers, axiomatized and obtained in this chapter, lend strength to our intuitive conviction that a straight line with a number scale furnishes a reliable picture of the real number system. Definition I. If a and b are any two real numbers such that a < b, the open interval from a to b, written (a,b), is the set of all numbers x between a and b, a < x < b. The closed interval from a to b, written [a,b], includes the points a. and b and is the set of all x such that a ~ x ~ b. The halfopen intervals (a,b] and [a,b) are defined by the inequalities a < x ~ b and a

~

x

< b,

respectively.

In any of these cases the interval is called a

finite interval and the points a and bare called end-points.

Infinite intervals

are denoted and defined as fallows, the point a, where it appears, being the

t Cf. D. Hilbert, The Foundations of Geometry (La Salle, Ill., The Open Court Publishing Company, 1938).

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GEOMETRICAL REPRESENTATION

§ 110]

17

end-point of the interval: (a, +oo), x > a; [a, +oo), x ~ a; (-00, a), x < a; (-oo, a], x ~ a; (-oo, +oo ), all x.t Any point of an interval that is not an end-point is called an interior point of the interval. Definition Il.

The absolute value of a number x, written lxl

= { x if~ ~ -X

Ix!,

is defined:

0,

if X < 0.

The absolute value of a number can be thought of as its distance from the origin O in Figure 101. Similarly, the absolute value of the difference between two numbers, Ix - YI, is the distance between the two points x and y. Some of the more useful properties of the absolute value are given below. For hints for some of the proofs, see § 111. Properties of Absolute Value I. lxl ~ O; lxl = 0 if and only if x := 0. II. lxyl = lxl · IYI-

III.

ltl = ~

(y

¢

O).

IV. If E > 0, then (i) the--inequal,ity lxl < E is equivalent to the simultaneous inequalities -E < X < E; (ii) the inequality lxl ~ Eis equivalent to the simultaneous inequalities - E ~ X ~ E. V. The triangle inequalityt holds: Ix + YI ~ lxl + IYIVI. I-xi = !xi; Ix - YI = IY - xi. VII. !xl 2 = x2 ; lxl = Vx2• (Cf. Ex. 26, § 105.) VIII. Ix - YI ~ lxl + IYIIX. I lxl - IYI I~ Ix - YI• Definition m. A neighborhood or epsilon-neighborhood of a point a is an open interval of the form (a - E, a+ E), where Eis a positive number.

By Property IV, the neighborhood (a - E, a + E) consists of all x satisfying the inequalities a - E < x < a + E, or - E < x - a < E, or Ix - al < E. It consists, therefore, of all points whose distance from a is less than E. The point a is the midpoint of each of its neighborhoods.

t These infinite intervals are also sometimes designated directly by means of inequalities: a < x < +00, a ~ x < +00, -00 < x < a, -00 < x ~ a, and -00 < x < +00, respectively. · t Property V is called the triangle inequality because the corresponding inequality for complex numbers states that any side of a triangle is less than or equal to the sum of the other two.

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THE REAL NUMBER SYSTEM

18

[§ 111

111. EXERCISES 1. Prove 2. Prove 3. Prove that x = yz. 4. Prove 5. Prove

Property I, § 110. Property II, § 110. Property Ill, § 110.

Hint: Use Property II with z

=

x/y, so

Property IV, § 110. Property V, § 110. Hint: For the case x x + y < x + 0 < x - y = lxl -(x

+

y) = -x - y

0 and y < 0, + IYI, lxl + IYI•

Use Property IV. Consider all possible cases of sign. 6. Prove Property VI, § 110. 7. Prove Property VII, § 110. 8. Prove Property VIII, § 110. 9. Prove Property IX, § 110. Hint: The inequality lxl follows by Property V from l(x - y) + YI ~ Ix - YI + IYl10. Prove that lx1 ·x2 · · · · x,.I = lx1I · lx,I · · · · · Ix..!. 11. Prove the general triangle inequality: lx1

+

X2

+ ··· +

x,.I ~

- IYI

~ Ix -

YI

lx1I + lx2l + · · · + lxnl•

12. Replace by an equivalent single inequality: x > a + b, x > a - b. In Exercises 13-24, find the values of x that satisfy the given inequality or inequalities. Express your answer without absolute values.

13. 15. •16. •18. •20. •21. •23.

< 3. < Ix + 1I. > x - 2. > x - 4. x 2x - 15 < 0. x2 + 10 < 6x. x < x2 - 12 < 4x. 21 51 Ix - 41 Ix - 21

H.

Ix Ix -

2

-

Ix+ 31

~

Hint: Square both members.

2. (Cf. Ex. 18, § 105.)

*17. Ix - 41 ~ 2 - x. *19. lx2 - 21 ~ 1. Hint: Factor and graph the left-hand member. •22. Ix + 5J < 214 •24. Ix - 71 < 5 < l5x - 251.

In Exercises 25-28, solve for x, and express your answer in a simple form by using absolute value signs. x-a

•25. -+ X a

>0.

*27.

>

X -

1

x-3

X

X ~ 0. a+x-

•26. a -

+ 3.

x+l

In Exercises 29-38, sketch the graph. Geometry.)

•29. y

Digitized by

(These proble ns presuppose Analytic X

*30• y = lxl·

= Ix!.

•31. y = x · Ix!. •33. y = 1 lxl •35. IYI < Ix!.

*28_ x - a :;: x + a. x-b x+b

11.

Google

•32. Y = vf ~. •34. IYI = lz . •36. !xi + I~ = I.

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UNIVERSITY OF MICHIGAN

§112)

•37.

19

SOME FURTHER PROPERTIES

lxl + IYI < 1.

•38.

lxl - IYI

~ 1.

112. SOME FURTHER PROPERTIES

In this section we give a few relations which exist between certain real numbers (integers and rational numbers, to be precise) and real numbers in general. In the constructive development of the real number system, discussed in the introduction, these properties follow easily and naturally from the definitions. From the point of view of a set of axioms descriptive of the real numbers, it is of interest that the properties of this section are implied by the properties listed as axioms in other sections of this chapter, in particular the axiom of completeness (cf.§ 114 for statements and proofs not given in this section). (i) If xis any real number there exist8 a positive integer n such that n > x. (ii) If x is any real number there exist integers m and n such that m < x < n. (iii) If x is any real number there exist8 a unique integer n such that

n~x x, let p be a positive integer such that p > -x, and let m -p. Proof of (iii). Existence: By (ii) there exist integers r and s such that r < x < s. If x is an integer, let n = x. If x is not an integer, x must lie between two consecutive integers of the finite set r, r + 1, r + 2, · · · , s. Uniqueness: If m and n are distinct integers such that m ~ x < m + 1 and n ~ x < n + 1, assume n < m. Then n < m ~ x < n + 1, in contradiction to property (vii), § 106. Proof of (iv). Let n > 1/E, by (i). Proof of (v). Letaandb be two real numbers, where a< b, orb - a> 0.

=

Let q be a positive integer such that! < b - a, by (iv). q

We now seek an

integer p so that l!. shall satisfy the relation q

a< E :::=a+!< b. q q

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20

[§ 113

THE REAL NUMBER SYSTEM

This will hold if p is chosen so that b. Proof. If the theorem were false, the inequality na ~ b would hold for all positive integers n. That is, the set a, 2a, 3a, • • • would be bounded above. Let c be the least upper bound of this set. Then na ~ c for all n, and hence (n l)a ~ c for all n. Therefore na a ~ c, or na ~ c - a, for all n. Thus c - a is an upper bound that is less than the least upper bound c. This is the desired contradiction.

+

.

+

Corollary ( (i), § 112). If x is any real number there exists a positive integer n such that n > x. Proof. If x ~ O, let n = 1. If x > 0, use Theorem II with a = 1 and b = x. *115. FURTHER REMARKS ON MATHEMATICAL INDUCTION

If the real number system is considered as defined by the descriptive axioms of this chapter rather than by the constructive development outlined in the Introduction, the simple properties of positive integers given in § 106 are no longer obvious. However, they are still true if we superimpose on the axiomatized real number system the structure of the natural numhers given by the Peano axioms. The positive integers greater than 1 are then defined inductively, 2 1 + 1, 3 2 + 1, · · · , and the remaining properties of § 106 are proved by mathematical induction. We

=

t Cf.

=

D. Hilbert, op. cit.

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§ 116]

25

EXERCISES

omit further discussion and refer the reader to the book by Birkhoff and MacLa.ne cited in the footnote on page 2. We present now the formal detailed proofs by mathematical induction of the general associative and commutative laws stated in § 106. For simplicity of notation we restrict ourselves to the multiplicative form. Proof of the general associative law. Let P(n) be the proposition: "Any product of the m numbers xi, x2, · · • , x,,., in that order, is equal to the special product xi(x2(xa( · · · (x,,._ix,,.) • • ·) ) ), whenever m ~ n." ,For n = 1 and n = 2 the proposition is trivia.I, and for n = 3 it follows from the associative law, (xix2)xa = xi(xtXa). Assume now the truth of P(n), for a fixed n, and consider any possible form for the product of the n + 1 numbers xi, x2, · · · , Xn+i, in that order. Such a product must have the form ab, where a and b are products of at most n of the x's. By the induction assumption that P(n) is true, ea.ch of these two factors can be rewritten, if necessary, in the form a = XiY and b = Xk+iz, where y is either 1 or a product of the factors x2, • • • , x.,, and z is either 1 or a product .of the factors XH2, • • • , X11-ri• By the associative law of § 102, ab = (.xtY)(.x1:+1Z) = xi(y(xk+iz) ). Again using the induction hypothesis, we can write the product y(xk+iz) in the special form x2(xa( • • • (x..x,.+1) · • ·) ). This fact, with the aid of the Fundamental Theorem of Mathematical Induction, § 106, establishes the truth of P(n) for every positive integer n. Finally, since any two products of n numbers in a given order are equal to the same special product, they must be equal to each other. Proof of the general commutative law. Let P(n) he the proposition: "Any two products of n numbers are equal regardless of the order of the factors." For n = 1 the proposition is trivial and for n = 2 it follows from the commutative law of § 102: XtXi = xix2. Assume now the truth of P(n) for a particular n and consider any possible product of then 1 numbers Xi, X2, • • • 1 Xn-ti• This product must have the form XXiY, where x is either 1 or the product of some of the x's and y is either 1 or the product of some of the x's. By the commutative and associative laws, xx 1y = (xxi)Y = (xix)y = xi(xy). The product xy contains the n factors x2, x3, • • • , x,.+i which, by the induction assumption that P(n) is true, can be rearranged according to the order of the subscripts. Therefore P(n 1) follows from P(n), and application of the Fundamental Theorem completes the proof.

+

+

*116. EXERCISES

*1· Prove that the system of integers satisfies the axiom of completeness. (Cf. Ex. 2, below.) *2, Prove that the system of rational numbers does not satisfy the axiom of completeness. Hint: Consider the set S of all rational numbers less than V2. Then S has an upper bound in the system of rational numbers (the

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THE REAL NUMBER SYSTEM

26

[§ 116

rational number 2 is one such). Assume that S has a least upper bound r. Use the density of the rational numbers to show that if r < v'2 then r is not even an upper bound of S, and that if r > v'2 then r is not the least upper bound of S. •3. Let x be a real number and let S be the set of all rational numbers less than x. Show that x is the least upper bound of S. Prove that if S is a bounded nonempty set there is a smallest closed interval I containing S. That is, I has the property that if J is any closed interval containing S, then J contains I. •6. Prove by counterexample that the statement of Exercise 4 is false if the word closed is replaced by the word open. •6. Prove that if S is a set of numbers dense in the system of real numbers, and if any finite number of points are deleted from S1 the remaining set is still dense. •7. Let S be a nonempty set of numbers bounded above,' and let x be the least upper bound of S. Prove that x has the two properties corresponding to an arbitrary positive number E: (i) every elements of S satisfies the inequality s < x + Ei (ii) at least one elements of S satisfies the inequality s > x - E. •8. Prove that the two properties of Exercise 7 characterize the least upper bound. That is, prove that a number' x subject to these two properties is the least upper bound of S. **9. Prove the analogue of Exercise 7 for greatest lower bounds. **10. Prove the analogue of Exercise 8 for greatest lower bounds. **11. Prove Dedekind's Theorem: Let the real numbers be divided into two nonempty sets A and B such that (i) if xis an arbitrary member of A and if y is an arbitrary member of B then x < y and (ii) if x is an arbitrary real number then either xis a member of A or xis a member of B. Then there exists a number c (which may belong to either A or B) such that any number less than c belongs to A and any number greater than c belongs to B. **12. Prove that the real number system is both minimal and maximal in the sense that it is impossible to have two systems R and S both satisfying the axioms of this chapter, where every member of R belongs to S but not every member of S belongs to R, and where the members of R are combined algebraically and related by order in the same way whether they are thought of as members of R or as members of S. Hint: The multiplicative unit 1 of S must be the multiplicative unit of R, and therefore R must contain all rational members of S. Since R is dense in S and complete it must be the same as S. · **13. Discuss the essential uniqueness of the real number system in the following sense: Prove that if Rand Sare any two "real number systems" subject to the axioms of this chapter, then it is possible to establish a one-to-one correspondence between their members which preserves algebraic operations and order. Hint: Let correspond, first, the additive units O and O', then the multiplicative units 1 and 1', then the positive integers n and n', then the integers n and n', and then the rational numbers p/q and p' /q', and show that these correspondences preserve operations and order. Finally, since any real num-

*'·

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§ 116]

27

EXERCISES

her is the least upper bound of all rational numbers less than it (Ex. 3), the correspondence can be extended to all elements of R and S. NOTE. By virtue of Exercise 13 it is possible to describe completely the real number system by the definition: The real numbera are a complete ordered fi,eul. (Cf. Ex. 11, § 113.)

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Functions, Sequences, Limits, Continuity

201. FUNCTIONS AND SEQUENCES

Whenever one says that y is a function of x, one has in mind some mechanism that assigns values to y corresponding to given values of x. The most familiar examples are real-valued functions of a real variable given by formulas, like y = 3x2 - 12.x or y = ±V.x2 - 4. These and other examples (where the variables x and y need not be related by formula, or even be real numbers) are given below. Definition I. Let D and R be two sets of objects. Then y = f(x) is called a function with domain (of definition) D and range (of values) R if and only if to each member x of D there corresponds at least one member y of R, and for each member y of R there is at least one member x of D to. which y corresponds. The general member of D and the general member of R are called the independent and the dependent variable, respectively. In case no two members of R correspond to the same member of D, J(x) is called single-valued. In case R consists of just one object, J(x) is called a constant function. In case D consists of real numbers, f(x) is called a Junction of a real variable. In case R consists of real numbers, J(x) is called real-valued. The symbol f(.xo) denotes the members of R that correspond to the member Xe of D. Example 1. The function y = 3x1 - 12x is defined for all real numbers. If we take D to be the set of all real numbers, R consists of all real numbers ~ -12, since the function has an absolute minimum (cf. § 409) when x = 2.

The function is a single-valued real-valued function of a real variable. Example 2. The function of Example 1 restricted to the domain D == (1, 5) (the open interval from 1 to 5) has range R equal to the half-open interval [ -12, 15). ·This function is not the same as that of Example 1, since it has a different domain. It is also, however, a single-valued real-valued function of a real variable.

v

The function y = x1 - 4 with domain D consisting of all real numbers x such that lxl ~ 2 is a single-valued real-valued function of a Example 3.

28

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§ 201]

FUNCTIONS AND SEQUENCES

real variable, with range R consisting of all nonnegative real numbers.

29 (Cf.

§ 214.)

Example 4. The function y = ±'Vz 1 - 4 with the same domain as the function of Example 3 is real-valued, but is single-valued only for z = ±2. Otherwise it is double-valued. Its range is the set of all real numbers. Example 6. The bracket function or greatest integer function, /(z) = [z], is defined to be the largest integer less than or equal to z, with domain all real numbers (Fig. 201). It is a single-valued real-valued function of a real variable. Its range is the set of all integers. y



2

••





1

z -1

••

1

2

3

• FIG. 201

Enmple 6. Let D be the closed interval [O, 1], and define /(z), for z in D, to be 1 if z is rational and O if z is irrational. (See Fig. 202.) Then R consists of the two numbers O and 1. Enmple 7. Let D be the contestants in· a radio quiz show and define /(z) as follows: If z is a contestant who has correctly answered all questions, then y

1 - t - - - - -...............

··----+----1

- - - - ! - •...........

FIG. 202

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30

[§ 201

FUNCTIONS AND SEQUENCES

!!!! n where n is the number of dollars of prize money; if z is a contestant who has incorrectly answered a question, then /(z) is a box of breakfast food. In this example /(z) is a single-valued function. In some cases its value is a real number, and in some cases its value is a box of breakfast food.

/(z)

A type of function of particular importance in mathematics is specified in the following definition. Definition II. An infinite sequence is a single-valued function whose domain of definition is the positive integers.

This means that corresponding to any positive integer there is a unique value or term determined. In particular, there is a first term a1 coiresponding to the number 1, a second term 02 corresponding to the number 2, etc. An infinite sequence can thus be represented: a1, ezi, • • • , a,., · · · , or {a,.}. The nth term, a,., is sometimes called the general term of the infinite sequence. Since it is a function of n, (a,. = f(n) ), it must be prescribed by some rule. If the terms are numbers, this rule may sometimes be expressed as a simple algebraic formula. Such a formula may be impractical, but a definite rule must exist. NoTE 1. Frequently an infinite sequence is indicated by an explicit listing of only the first few terms, in case the general rule for procedure is clear beyond reasonable doubt. For instance, in part (c) of the following Example 8 the rule that is clearly implied by alternating 1's and O's for the first six terms is alternating 1's and O's for all terms, although the ingenious artificer could construct any number of infinite sequences that start with alternating 1's and O's (the terms could continue by being identically O, or with alternating 6's and 7's, for example). Such interpretations, we hold, are not only unnatural, but deliberately mischievous. NoTE 2. For convenience, if the meaning is clear, the single word sequence will be used henceforth to mean infinite sequence. Example 8. Give a rule for obtaining the general term for each of the following sequences: (b) 1, t, ½, ¼, · · · ; (a) ½, -l, 1-, --h, · · · ; (c) 1, 0, 1, O, 1, O, • • • ; (d) 1, 2, 3, 1, 2, 3, 1, 2, 3, · · · (e)

½, ¼, ¼, l, l, n,

Solution.

· ·· ·

(a) The factor (-1)" or (-1) 11+1 is a standard device to take care

The general term is (-1) 11+1 n ~ - (b) If n = 1, 1 3 a,.= 1; if n > 1, a,.= _!_ , (c) First formulation: if n is odd, a. = 1; if n n- 1 is even, a,. = 0. Second formulation: a2,._1 = 1; a2,. = 0. Third formulation:

of alternating signs.

a11 = ½[( -1) 11+1 + 1].

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(d) a1n-2

= 1;

a1n-1

= 2; aa.. = 3.

(e) 0211-1

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UNIVERSITY OF MICHIGAN

= ~;

§ 202]

LIMIT OF A SEQUENCE

31

202. LIMIT OF A SEQUENCE

A sequence is said to tend toward, or converge to, a number if and only if the absolute value of the difference between the general term of the sequence and this number is less than any preassigned positive number (however small) whenever the subscript n of the general term is sufficiently large. Symbolically, this is written lim n .....

+•

On

=a

Or lim

On

n~•

=a

Or a,. -+ a,

where an is the nth term of the sequence and a is the number to which it converges. If {a,.} converges to a, a is called the limit of the sequence. A more concise form of the definition given above is the following:

= a, if 11--0+• and only if corresponding w an ar'bitrary positive number E there exists a number N = N(E) 8UCh that la,. - al < E whenever n > N. Definition I. The sequence {a,.} has the limit a, written lim a,.

NoTE 1. In conformity with the discussion following Definition III, § 110, the statement that {a.} converges to a is equivalent to the statement that every neighborhood of a contains all of the terms of {a.} from aome point on, and is also equivalent to the statement that every neighborhood of a contains all but a finite number of the terms of {a,.} (that is, all of the terms except for a finite number of the subscripts).

If a sequence converges to some number, the sequence is said to be

convergent; otherwise it is divergent. The concept of an infinite limit is important, and will be formulated in precise symbolic form. AB an exercise, the student should reformulate the following definition in his own words, without the use of mathematical symbols.

Definition II. The sequence {an} has the limit +00, written lim a,.=

-+•

+00,

0T

an-+

+00,

if and onl11 if corresponding wan arbitrary number B (however large) there exists a number N = N(B) 8UCh that an> B whenever n > N; the sequence {a,.} has the limit - 00 , written

lim a,.= -00,oran-+

,...... +.

-00,

if and only if corresponding w an arbitrary number B (however large its nega,.. tive) there exists a number N = N(B) 8UCh that a,. < B whenever n > N; the sequence has the limit 00 (unsigned infinity), written

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32

[§ 202

FUNCTIONS AND SEQUENCES

lim a,.= n-++ao

oo, or a,.-+ 00 1

if and only if lim la,.I = +oo. ,.__,+ ..

NOTE 2. Although the word limit is applied to both the finite and infinite cases, the word converge is used only for finite limits. Thus, a sequence tending toward +00 diverges. NOTE 3. In any extensive treatment of limits there are numerous statements which can be interpreted to apply to both finite and infinite cases, and which are of such a nature that the proofs for the finite and infinite particularizations are in essence identical. In such instances these proofs can be combined into a single proof by appropriate extensions of the word neighborhood. We define "neighborhoods of infinity" as follows: (i) a neighborhood of +00 is an open interval of the form (a, +00); (ii) a neighborhood of - 00 is an open interval of the form ( -00, b); (iii) a neighborhood of 00 is the set of all z satisfying an inequality of the form lzl > a. With these conventions, for example, all cases of Definitions I and II can be included in the following single formulation for lim a,.= a (where a may be a number, or +00, -00, or 00): Corre-

n-++•

.

sponding to every neighborhood Na of a there ezists a neighborhood N +a, of +00 such that whenever n belongs to N +a,, a,. belongs to N 0 • In the sequel we formulate theorems and proofs separately for the finite and infinite forms, but suggest that the student interested in exploring the simplifying techniques available with general neighborhoods try his hand at combining the separate formulations into unified ones. A word of warning is in order, however: Do not confuse infinite symbols with numbers, and write such nonsense as la,. - oo I < E when dealing with an infinite limit! It is to avoid such possible confusion of ideas that we have adopted the policy of maintaining (in the main) the separation of the finite an°d infinite.

Definition m. A subsequence of a sequence is a sequence whose terms are terms of the original sequence arranged in the same order. That is, a subsequence of a sequence {a,.} has the form a,.., a,.., a,.,, · · • , where

n1 It is denokd by {a,..} .

< ni < na < · · ·.

Example 1. The sequence ½, ¼, i, • •• is a subsequence of the sequence ½, i, ¼, t, i, · · · of Example 8, § 201. The sequence 0, 1, 0, 1, 0, 1, · · · is a

subsequence of the sequence 1, 0, 1, 0, 0, 1, 0, 0, 0, I, • • • . Show that the following sequences converge to 0: 1 1 (-1)"+1 · · • , ;• • · • ; (b) ½, ¼, i, · · ·, 2,., · · · ; (c) ½, -¼, i, · · · , 2,. 1

Example 2. (a) 1,

½, ½,

Solution.

(a) Since

la,. - al = 11n - OI = !, n

and since

!n < E

• • •

whenever

n > 1/E, we can choose as the function N(E) of Definition I the expression 1/E. (Cf. § 112.) (b) By Ex. 11, § 107, 2" > n for all positive integers, so that we can choose N(E) = 1/E. (Cf. Ex. 14, § 205.) (c) This reduces immediately to (b).

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EXERCISES

Example 3.

Find the limit of each sequence: (a)

(b) 3, 3, 3, · · · , 3, · · · ; (c) 1,

Solution.

33

½,

(a) The expression

1, ¾, 1,

la,. - 11

½, ¾, i, •• • , 1 - ~• • • • ·

l, · · · . is equal to ;,., which is less than any

preassigned positive number whenever n is sufficiently large, as shown in Ex1Lmple 2, (b). Therefore the limit is 1. (b) The absolute value of the difference between the general term and 3 is identically zero, which is less than any preassigned positive number for any n, and certainly for n sufficiently large. Therefore the limit is 3. (c) By combining the reasoning in parts (a) and (b) we see that the general term differs numerically from 1 by less than any preassigned positive number if n is sufficiently large. The odd-numbered terms form a subsequence identically 1, while the even-numbered terms form a subsequence which is the sequence of part (a). The limit is 1. Show that each of the following sequences diverges: (b) 1, 2, 4, 8, 16, · · · ; (c) 1, 2, 1, 3, 1, 4, · · · ; (d) 1, -2, 4, -8, l6, · · · Solution. (a) If {a,.} converges to a, every neighborhood of a must contain all terms from some point on, and therefore must contain both numbers 1 and 2. On the other hand, no matter what value a may have, a neighborhood of a of length less than 1 cannot contain both of these points! (b) No finite interval about any point can contain all terms of this sequence, from some point on. The limit is +oo. (c) The comment of part (b) applies to this sequence, since there is a subsequence tending toward +oo. This sequence has no limit, finite or infinite. (d) The subsequence of the odd-numbered terms tends toward +ao, and that of the even-numbered terms tends toward -oo. The sequence of absolute values tends toward +ao, so that the sequence itself has the limit oo. Example 4.

(a) 1, 2, 1, 2, 1, 2, · · · ;

203. EXERCISES

In Exercises 1-10, draw the graph of the given function, assuming the domain of definition to be as large as possible. Give in each case the domain and the range of values. The bracket function [x] is defined in Example 5, § 201, and square roots are discussed in § 214. (Also cf. Exs. 5-10, § 216.) 1. y 3. y

6. y 7. y 9. y

= v'x• - 9. = -vC;, = v'4z - z1 • = z - [z]. = v'z - [z].

2. y

,. y 6. y 8. y

= = =

±v'25 -

.:i:2 •

±v'ixi.

v'l.:r: 2 - 161. = (z - [z]) 1 •

10. y = [z]

+ v'z~-- [x].

In Exercises 11-18, give a rule for finding the general term of the sequence.

11. 13. 16. 16.

*· .••.

2, t, t, t, . . . . 12. 1, --1, rr, --h, 1, -1, ½, -¼, -h, -Th, · · · . 14. 1, 2, 24,720, 40320, · · · . 1·3, 1·3·5, 1·3·5·7, 1·3·5·7•9, .. ·. 1, 2, 3, 2, 1, 2, 3, 2, 1, · · · .

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[§ 204

17. -1, 1, 1, -2, 2, 2, -3, 3, 3, -4, 4, 4, · · · . 18. 1, 2·4, 1·3·5, 2·4·6·8, 1·3·5·7·9, · · · . In Exercises 19-24, find the limit of the sequence and justify your contention (cf. Exs. 25-30). 2n 1 19. 2, 2, 2, 2, 2, · · · . 20. J, ¼, t, . . . ' 2n ... .

+

21. 23.

1-, .... I, fr, n, n, · · · . ♦, -f, ¥, -l,/1, · · · .

22. 1, 4, 9, 16, · · · , n1, • • • • 2'. 9, 16, 21, 24, • • • , 10n - n 1,

• • • •

In Exercises 25-30, give a simple explicit function N(E) or N(B), in accord with Definition I or II, for the sequence of the indicated Exercise. *26. For Ex. 19. *26. For Ex. 20. *27. For Ex. 21. *28. For Ex. 22. d9. For Ex. 23. *30. For Ex. 24. In Exercises 31-34, prove that the given sequence has no limit, finite or infinite.

31. 1, 5, 1, 5, 1, 5, • · · . 33. 1, 2, 1, 4, 1, 8, 1, 16, · · · .

82. 1, 2, 3, 1, 2, 3, 1, 2, 3, · · · . 8'. 21, 2-1, 21, 2-•, 21, 2-1, ....

204. LIMIT THEOREMS FOR SEQUENCES

Theorem I. The aUeration of a finite number of terms of a sequence has no effect on convergence or divergence or limit. In other words, if {a,.} and {b,.} are two 8equences and if M and N are two positive integer8 such that aM+n = bN+" for all positive integers n, then the two sequences {a,.} and {b,.} must either both converge to the same limit or both diverge; in case of divergence either both have the same infinit,e limit or neither has an infinite limit. Proof. If {a,.} converges to a, then every neighborhood of a contains all but a finite number of the terms of {a,.}, and therefore all but a finite number of the terms of {b,.}. Proof for the case of an infinite limit is similar. Theorem II. If a sequence converges, its limit is unique. Proof. Assume a,. - a and a,. - a', where a¢ a'. Take neighborhoods of a and a' so small that they have no points in common. Then each must contain all but a finite number of the terms of {a,.}. This is clearly impossible.

Theorem m. If all terms of a sequence, from 80me point on, are equal to a constant, the sequence converges to this constant. Proof. Any neighborhood of the constant contains the constant and therefore all but a finite number of the terms of the sequence.

Theorem IV. Any 8Ub8e(J'IJ,e11,Ce of a convergent sequence converges, and its limit is the limit of the original sequence. (Cf. Ex. 12, § 205.)

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LIMIT THEOREMS FOR SEQUENCES

35

Proof. Assume an -+ a. Since every neighborhood of a contains all but a finite number of terms of {an} it must contain all but a finite number of terms of any subsequence. Definition I. A sequence is bounded if and only if all of its terms are conf,ained in some interval. Equivalently, the sequence {an} is bounded if and only if there exists a positive number P such that lanl ~ P for all n. Theorem V. Any convergent sequence is bounded. (Cf. Ex. 2, § 205.) Proof. Assume an -+ a, and choose a definite neighborhood of a, say the open interval (a - 1, a + 1). Since this neighborhood contains all but a finite number of terms of {an}, a suitable enlargement will contain these missing terms as well. Definition II. If {a,.} and {b,.} are two sequences, the sequences {a,.+ b,.}, {a,. - b,.}, and {a,.b,.} are called their sum, difference, and product, respectively. If {a,.} and {b,.} are two sequences, where b,. is never zero, the sequence {an/b,.} is called their quotient. The definitions of sum and product extend to any finite number of sequences. Theorem VI. The sum of two convergent sequences is a convergent sequence, and the limit of the sum is the sum of the limits: lim (a,.+ b,.) = lim a,.+ lim b,..

-+•

......+..

......+..

(Cf. Ex. 4, § 205.) This rule extends to the sum of any finite number of sequences. Proof. Assume a.. -+ a and bn -+ b, and let E > 0 be given. Choose N so large that the following two inequalities hold simult,aneously for n > N:

la,. - al < ½E, lb,. - bl Then, by the triangle inequality, for n > N l(a..

+ b,.) -

(a+ b)I

= l(a,. - a) + (bn ~

- b)I la,. - al

< ½E.

+

lb,. -

bl < ½E + ½E =

E.

The extension to the sum of an arbitrary number of sequences is provided by mathematical induction. (Cf. Ex. 3, § 205.) Theorem VII. The difference of two convergent sequences is a convergent sequence, and the limit of the difference is the difference of the limits: lim (a,. - bn) = lim a.. n-++ oo

n-++ t0

lim b,.. n-++ co

Proof. The details are almost identical with those of the preceding proof. (Cf. Ex. 6, § 205.) Theorem vm. The product of two convergent sequences is a convergent sequence and the limit of the product is the product of the limits:

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36

=

lim a,.· lim b,.. n-+oo n-+• This rule extends to the Foduct of any jinit.e ~umber of lim (a,.b,.)

n-+oo

(Cf. Ex. 5, § 205.)

[§ 204

sequences. Proof. Assume a,. -+ a and b,. -+ b. We wish to show that a,.b" -+ ab or, equivalently, that a,.b,. - ab -+ 0. By addition and subtraction of the quantity ab,. and by appeal to Theorem VI, we can use the relation a,.b,. - ab= (a,. - a)b,. + a(b,. - b) to reduce the problem to that of showing that both sequences {(a,. - a)b,.} and {a(b,. - b)} converge to zero. The fact that they do is a consequence of the following lemma:

Lemma. If {c,.} converges to O and {d,.} converges, then {c,.d,.} converges

too. Proof of lemma. By Theorem V the sequence {d,.} is bounded, and there exists a positive number P such that ld..l ~ P for all n. If E > 0 is given, choose N so large that lcnl < E/P for n > N. Then for n > N,

lc,.d,.I

=

lc,.I · ld,.I
0, then for all n, lb,.I i6;; E, or ld,.I = I1/b,.I ~ 1/E. The sequence {d,.} is therefore bounded, and the proof is complete. Theorem X. Multiplication of the terms of a sequence by a nonzero conswnt k does not affect convergence of divergence. If the original sequence converges, the new sequence converges to k times the limit of the original, for any conswnt k:

Proof.

lim (k a,.) = k • lim a,.. n-+• n-+,. This is a consequence of Theorems III and VIII.

Theorem XI. If {a,.} iB a sequence of nonzero numbers, then a,.--+ oo if and only if 1/a,.--+ 0; equivalenUy, a,.--+ 0 if and only if 1/a,.--+ oo. Proof. If la,.I --+ +oo and if E > 0 is given, there exists a number N such that for n > N, la..l > 1/E, and therefore 11/a,.I < E. Conversely, if 1/a,.--+ 0 and Bis any given positive number, there exists a number N such that for n > N, 11/a,.I < 1/B, and therefore la..l > B. Theorem m.

If a

>

1, lim a" =

n-+•

+oo.

+

Proof. Let p = a - 1 > 0. Then a = 1 p, and by the Binomial Theorem (cf. Ex. 35, § 107), if n is a positive integer,

a"

= (1 + p)" == 1 + np + n(n 2-

1)

-pl

+ ••• ~ 1 + np.

Therefore, if Bis a given positive number and if n

> B/p, then

a" i6;; 1 + np Theorem XIII.

Proof.

If lrl

< 1,

lim

-+•

> 1 + B > B. r" == 0.

This is a consequence of the two preceding theorems.

Definition m. A sequence {a,.} is monotonically increasing (decreasing),t written a,. j (a,.!), if and only if a,. ~ an+1 (a,. i6;; an+1) for every n. A sequence is monotonic if and only if it is monotonically increasing or, monotonically decreasing. Theorem XIV. Any bounded monotonic sequence converges. If a,. j (a,.!) and if a,. ~ P (a,. i6;; P) for all n, then {a,.} converges; moreover, if a,. --+ a, then a,. ~ a ~ P (a,. ~ a ~ P) for all n.

t Parentheses are used here to indicate an alternative statement. of the use of parentheses for alternatives, see the Preface.

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FU NC TI O NS AND SEQUENCES

[§ 205

*

Proof. We give the details only for the case an j (cf. Ex. 7, § 205). Since the set A of points consisting of the terms of the sequence {a,.} is bounded above, it has a least upper bound a ( § 114), and since P is an upper bound of A, the following inequalities must hold for all n: an ~ a ~ P. To prove that a,. ---+ a we let E be a given positive number and observe that there must exist a positive integer N such that aN > a - t (cf. Ex. 7, § 116). Therefore, for n > N, the following inequalities hold: Consequently a plete.

E

a - E < aN ~ a,. ~ a < a + E. < a,. < a + E, or Ian - al < r:, and the proof is com-

As a consequence of Theorem XIV we can say in general that any +co, or -co), and that the limit is finite if and only if the sequence is bounded. (The student should give the details in Ex. 7, § 204.) NOTE.

monotonic sequence has a limit (finite,

Theorem XV.

If a,. :i bnfor all n, and if lim an and 1im b,. exi8t (finite,

+oo, or -oo ), then lim an :i lim b,.. ......+..

-+..

-+ ..

,.....+ ..

Proof. If the two limits are finite we can form the difference e,. = b,. - a,. and, by appealing to Theorem VII, reduce the problem to the special case: ~ 0 for all n and if C = lim e,. exists and is finite then C ~ 0 . ...... + .. By the definition of a limit, for any positive r: we can find values of n (arbitrarily large) such that le,. - Cl < E. Now if C < 0, let us choose E = -C > 0. We can then find arbitrarily large values of n such that le,.· - Cl = le,. + r:I < r:, and hence e,. + E < E. This contradicts the nonnegativeness of e,.. On the other hand, if it is assumed that an---+ +oo and b,.---+ B (finite), we may take r: = 1 and find first an N1 such that n > N1 implies an> B + 1, and then an N2 such that n > N2 implies b,. < B + 1. Again the inequality a,. :i b,. is contradicted (for n greater than both N 1 and N 2). The student should complete the proof for the cases an---+ A (finite), bn ---+ -oo and an---+ +oo, b,.---+ -oo.

if e,.

205. EXERCISES

1. Prove that if two subsequences of a given sequence converge to distinct limits, the sequence diverges. 2. Show by a counterexample that the converse of Theorem V, § 204, is false. That is, a bounded sequence need not converge. 3. Prove the extensions of Theorems VI and VIII, § 204, to an arbitrary finite number of sequences. 4. Prove that if a,. - +co and either {b,.} converges or b,. - +ao, then

a,.+b,.-+co.

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EXERCISES

39

6. Prove that if a,. - +00 and either b,. - b > 0 or b,. - +00, then a,.b,. - +00. Prove that if a,. - 00 and b,. - b ~ 0, then a,.b,. - 00. 6. Prove Theorem VII, § 204. 7. Prove Theorem XIV, § 204, for the case a,. L, and the statement of the Note that follows. Hint: Let b,. a -a,. and use Theorem XIV, § 204, for the case b,. j . Cf. Ex. 23.

8. Show by counterexamples that the sum (difference, product, quotient) of two divergent sequences need not diverge. 9. Prove that if the sum and the difference of two sequences converge, then both of the sequences converge. 10. Prove that a,. - 0 if and only if la.I - 0. 11. Prove that a,. - a implies la..1 - lal. Is the converse true? Prove, or give a counterexample. Hint: Use Property IX, § 110. 12. Prove that if a sequence has the limit +00 (-00, 00) then any subsequence has the limit +00 (-00, 00 ). 13. Prove that if a,. ~ b (a,. i1:; b) and a,. - a, then a ~ b (a i1:; b). Show by an example that from the strict inequality a,. < b (a,. > b) we cannot infer the strict inequality a < b (a > b). 14. Prove that if O ~a,.~ b,. and b,. - 0, then a,. - 0. More generally, prove that if a,. ~ b,. ~ c,. and {a,.} and {c,.} converge to the same limit, then {b,.} also converges to this same limit. 16. Prove that if x is an arbitrary real number, there is a sequence {r,.} of rational numbers converging to x. Hint: By the density of the rationals ( (v), § 112), there is a rational number r,. in the open interval ( x - ~• x

**16. If {8,.} is a given sequence, define u,.

a

!n (81 + 82 + ••· + 8,.).

that if {8,.} converges to O then {u,.} also converges to 0. Let m be a positive integer < n, and write

(Cf. Ex. 17.)

+

!).

Prove Hint:

!n (81 + ••• + 8•) + !n (8,,.+1 + ••• + 8,.). If E > 0, first choose m so large that whenever k > m, 18.1:I < ½E. Holding m fixed, choose N > m and so large that ls1 + ·· · + 8.. I/N < ½E. Then choose n > N, and consider separately the two groups of terms of u,., given above. u,. =

**17. With the notation of Exercise 16, prove that if {s,.} converges, then {u,.} also converges and has the same limit. Show by the example O, 1, O, 1, · · · that the convergence of {u,.} does not imply that of {8,.}. Can you find a divergent sequence {8,.} such that u,. - O? Hint: Assume 8,. - l, let t,. = 8,. - l, and use the result of Ex. 16. ••18. With the notation of Exercise 16, prove that if lim s,. = +00, then n--.+oo

lim u,. = +00. Show by the example 0, 1, 0, 2, 0, 3, • • · that the reverse n~+• implication is not valid. Hint: If B is a given positive number, first choose m so large that whenever k > m, B.t > 3B. Then choose N > 3m and so large that 181 + · · · + 8.. I/N < B. Then follow the hint of Ex. 16. **19. Show by the example 1, -1, 2, -2, 3, -3, · • • that with the notation

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FUNCTIONS AND SEQUENCES

of Exercise 16,

lim s,.

= oo

does not imply

..-+• an example where lim s,. = oo and lim ..-+• ..-+•

N. A function f(x) is said to tend toward or approach or have a limit L as x approaches a number a if and only if the absolute value of the difference between f(x) and Lis less than any preassigned positive number (however small) whenever the point x belonging to the domain of definition of f(x) is sufficiently near a but not equal to a. This is expressed symbolically: limf(x)

= L.

%->a

t In the terminology of the next chapter (§ 309), a is a limit point of the domain of definition D of J(x). It can be shown that every neighborhood of a contains infinitely many points of D.

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41

If in this definition the independent variable x is restricted to values greater than a, we say that x approaches a from the right or from above and write lim f(x) = L.

-+

Again, if x is restricted to values less than a, we say that x approaches a from the left or from below and write lim f(x)

= L.

The terms undirected limit or two-sided limit may be used to distinguish the first of these three limits from the other two in case of ambiguity a.rising from use of the single word limit. A more concise formulation for these limits is given in the following definition : Definition I.

The function f(x) has the limit L as x awroaches a, written

limf(x)

= L, or f(x)-+ Las x-+ a,

if and only if corruponding to an arbi,trary positive number E there exists a positive number 8 = 8(E) such that O < Ix - al < 8 implies lf(x) - LI < E, for values of x for which f(x) is defined;t f(x) has the limit Las x approaches a from the right (left),t written

lim f(x)

= L,

or f(x)-+ Las x-+ a+

( lim f(x)

= L,

or f(x) -+ L as x -+a-),

-+

if and only if corresponding to an arbitrary positive number E there exists a positive number 8 = 8(E) such that a < x < a 8 (a - 8 < x < a) implies . lf(x) - LI < E1 for values of x for which f(x) is defined. These one-sided limits (if they exist) are also denoted:

+

f(a+)

= lim f(x), -+

f(a·-)

= lim f(x). --

Since the definition of limit employs only values of x different from a, it is completely immaterial what the value of the function is at x = a or, indeed, whether it is defined there at all. Thus a function can fail to have a limit as x approaches a only by its misbehavior for values of x near a but not equal to a. Since limf(x) exists if and only if lim f(x) and

-

-+

t An open interval with the midpoint removed is called a deleted neighborhood of the miBBing point. The inequalities O < iz - al < &, then, define a deleted neighborhood of the point a. t Parentheses are used here and in the following two definitions to indicate an alternative statement. For a discussion of the use of parentheses for alternatives, see the Preface.

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[§ 206

lim f(x) both exist and are equal (cf. Exs. 13-14, § 208), lim f(x) may

.,....... fail to exist either by lim f(x) and lim f(x) being unequal or by either or

-+

--

both of the latter failing to exist in one way or another. These possibilities are illustrated in Example 1 below. Limits as the independent variable becomes infinite have a similar formulation:

Definition II. The function f(x) haa the limit L aa x become8 positively (negatively) infinite, written f( +oo) = lim f(x) .,....+ .. (f(-00)

= L, or f(x) - L

= .,...._,. lim f(x) = L,

fJ,8

x-

+oo

or f(x) - Laa x -

-oo ),

if and only if corre8ponding to an arbitrary positive number E there exists a number N = N(E) Buch that x > N (x < N) implie8 lf(x) - LI < E, for value8 of x for which f(x) i8 defined. In an analogous fashion, infinite limits can be defined. Only a sample definition is given here, others being requested in the Exercises of § 208.

Definition m. The function f(x) haa the limit +oo ( - oo) aa x approache8 a, wriUen limf(x) = +oo (-00), or f(x)-+oo (-00) aax-a,

-

if and only if corresponding to an arbitrary number B there exists a positive number o = o(B) such that O < Ix - al < oimplie8 f(x) > B (f(x) < B), for value8 of x for which f(x) is defined. As with limits of sequences it is often convenient to use an unsigned infinity, oo. When we say that a variable, dependent Or independent, tends toward 00, we shall mean that its absolute value approaches +oo. Thus limf(x) = Lis defined as in Definition II, with the inequality x > N re-

.,....

.

> N,

-

is equivalent to lim 1/(x)I = +oo. .,....... Example 1. Discuss the limits of each of the following functions as x ap-

placed by !xi

proaches 0,

a;nd lim/(x)

= 00

xi + X o+, and 0-, and in each case sketch the graph: (a) f(x) = x

if x ¢ 0, undefined for x = 0; (b) f(x) = function, J(x) = sgn x 1 if x > 0, J(x)

lxl if x ¢ 0, /(0) = 3; (c) the signum = sgn x = -1 if x < 0, /(0) = sgn 0

= 0; (d)t f(x) = sin! if x

0; (e) f(x) = ! if x

=

X

x

=

0; (/) f(x)

= ~ if x ¢ X

¢

0, /(0)

a

X

0, undefined if x

=

¢

0, undefined if

0.

t For illll8trative examples and exercises the familiar properties of the trigonometric functions will be assumed. An analytic treatment is given in H 603-604.

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Solution. The graphs are given in Figure 203. In part (a) if x ~ 0, /(x) is identically equal to the function x + 1, and its graph is therefore the straight line y = x + 1 with the single point (0, l} deleted; lim /(x) -= /(0+) ... /(0-) = 1. z---.0

In part (b) lim /(x) Z---.0

= /(0+) = /(0-) = 0.

The fact that /(0)

=3

has no

bearing on the statement of the preceding sentence. For the signum function (c), /(0+) = 1, /(0-) = -1, and lim /(x) does not exist. In part (d) all Z---.0

three limits fail to exist. In part (e) /(0+) = +oo, /(0-) "" -oo, and lim /(x) = 00 (unsigned infinity) (cf. Exs. 31-32, § 208). In part (/}, Z---.0

/(0+)

= /(0-) = Jim /(x) ... +oo (cf. Ex. 32, § 208). Z---.0

y

1J

11

1i-----

x 1 (b)

11

X

1

(cl)

1

(c)

\_ _) '\_ 1

\

X

1

X

(/)

(e) FIG. 203

•Eumple 2.

Show that Jim _.2

zl - X + 18 = 4. 3X - 1

Find an explicit function o(E)

as demanded by Definition I. Solution. Form the absolute value of the difference: 2

1x

-

z + 18

1

13x + 221 = lz _ 21 · Ix - 111· 3x-1 3x-1 3x-1 We wish to show that this expression is small if xis near 2. The first factor, 1 Ix - 21, is certainly small if xis near 2; and the second factor, !1• is not (1)

_ 41 • lx

l:z-_

dangerously large if x is near 2 and at the same time not too near t. Let us make this precise by first requiring that o ~ 1. If x is within a distance less than o of 2, then 1 < x < 3, and hence also -10 < x - 11 < -8 and 2 < 3x - 1 < 8, so that lz - lll < 10 and !3x - 11 > 2. Thus the second factor is less than J2Q. = 5. Now let a positive number E be given. Since the

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[§ 207

i and l::r:--l !I < 5, we have only to take 6 = 6(E) to be the smaller of the two numbers 1 and i• 6(E) = min ( 1, l} The graph of this function 6(E} is shown in Figure 204.

expression (1) will be less than E if simultaneously iz

- 21
0, as a monotonic function of r. Show that for O < x < 1, h(r) ! as r i and that for x > 1, h(r) j as r j . Find lim h(r) and lim h(r). (Cf .

......+..

,....._,.

Ex. 18.) **26. Prove that the function h(r) of Exercise 25 is continuous, if only rational values of r are considered. Hint: Write x• - x"' = x10 (x•-ro - 1), and use Exs. 19 and 25. *27. Prove that a monotonic function always has one-sided limits (finite or infinite). *28. Prove that the only discontinuities that a monotonic function can have at points where it is defined are finite jumps. (Cf. Ex. 27.) *29. Prove that the set of points of discontinuity of a monotonic function is either finite or denumerable. (Cf. Ex. 12, § 113. Also cf. Ex. 25, § 717.) Hint: There are at most two values of x at which y = J(x) can become infinite. Any other point of discontinuity, on the x-axis, corresponds to a finite interval of jump on the y-axis, and to distinct points of jump discontinuity correspond open intervals having no point in common. In each of these intervals of jump lies a rational number ((v), § 112), and the rational numbers are denumerable (Ex. 14, § 113). *30. Prove that any monotonic function can be redefined at the points of discontinuity at which it is defined so that it becomes everywhere continuous from the right (or from the left). *31. Prove the following converse of the statement of § 215 that a continuous strictly monotonic function has a single-valued inverse: If /(x) is continuous on a closed interval [a, b] and if J(x) does not assume there any value twice (that is, at distinct points of [a, b] J(x) has distinct values), then/(x) is strictly monotonic there. In Exercises 32-37, find a specific function 8 = 8(E) as specified by Definition II, § 209, for the given function at the prescribed point. (Cf. Exs. 59-64, § 208.)

*32.

v;, a

= 3. Hint: If x > 0,

1vx - va1 = Iv; - va . v; + val = Ix - 3! < Ix - a1. 1 vx + va v; + va *33. Vx2

-

*35. ~. a = 5.

I~x

-

*as. L, a=

vx

*3'- V3x2

4, a= 2.

Hint: If x

~51 =

-

x, a= -1.

> 0,

1~; -~5 · ~~ + ~~ + ~25, < Ix - 51. 1

6.

~x2 +~5x+~25

*37.

·

1

~x• a = 1.

**38. If a is a fixed positive rational number prove that the set of all numbers

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(§ 216

of the form r + av';;, where r and a are arbitrary rational numbers, form a field (cf. Ex. 8, § 113). **39. Prove that each example demanded by Exercises 57 and 58, § 208, must be discontinuous and that, in fact, it must have infinitely many points of discontinuity.

••iO. Prove that -+• lim (1 + .! )• exists and is between 2 and 3. n definition of e.

r

For another treatment see §§ 601-602.)

the binomial expansion of a,.

= (1 +¼

Hint: First express

in the form

!) 1-(

1 + 1 + J_ ( 1 + 3! 1 - nl) ( 1 - n2 ) 2! n and thereby show that an j and, furthermore, that 1 1 1 1 a. < 1 + 1 + 2 + ~ + ~ + ••• + 2•-l •

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*301. A FUNDAMENTAL THEOREM ON BOUNDED SEQUENCES

t

One of the properties of the real number system that was assumed as an axiom in Chapter 1 is that of completeness. A useful consequence (which is actually one of several alternative formulations of the concept of completeness) is stated in the following theorem:

Theorem. Fundamental Theorem on Bounded Sequences. Every bounded sequence (of real numbers) contains a convergent subsequence.

In order to prove this theorem we observe first that we already know (Theorem XIV, § 204) that every bounded monotonic sequence converges. Therefore the proof will be complete as soon as we establish the following lemma: Lemma I.

Every sequence (of real numbers) contains a monotonic sub-

sequence. Before proceeding with the details of the proof of this lemma we discuss briefly the idea of largest term of a sequence, and give a few illustrative examples. By a largest term of a sequence a 1, a,, a8, • • • we mean any term a,. with the property a1c ~ an for every positive integer n. The examples below show that a sequence may not have a largest term, and that if it does have a largest term it may have several. In case a sequence has a largest term, there must be a first largest term (first in the order of the terms according to the subscripts). This particular largest term will be called for convenience the largest term. Example 1. The sequence 0, I, 0, 1, • • • contains the two monotonic constant subsequences 0, 0, 0, • • • and 1, 1, I, • • • • It has a largest term a,. = 1,

t Unless otherwise qualified the word sequence should be interpreted in this chapter to mean infinite aequence of real numbera. 61

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where k may be any even positive integer. is the second.

[§ 302

The largest term of the sequence

Example 2. The sequence 1, -1, 2, -2, 3, -3, , • • contains the monotonic subsequences 1, 2, 3, • • • and -1, -2, -3, • • • . It has no largest term, but the subsequence -1, - 2, -3, • • • has the largest term -1. Example 3. The sequence 1, -1, ½, - ½, ½, -½, • • • contains the monotonic su.b,.seq uences 1, ½, ½, • • • and -1, - ½, - ½, • • • , and has the largest term·i'. The subsequence -1, - ½, -½, · · · has no largest term. · Example 4. The sequence 1, ½, i, ¾, j, • • • contains the monotonic subsequence ½, ¾, ¾, • · • • The sequence has the largest term 1, but no monotonic subsequence has a largest term.

As an aid in the proof of Lemma I we establish a simple auxiliary lemma:

Lemma II. If a sequence S has a largest term equal to x, and if a subsequence T of S has a largest term equal toy, then x ~ y.

Proof of Lemma II. Let S = {a,.}. Since Tis a subsequence of S, y is equal to some term a,. of S. If ak is the largest term of S, then ak ~ a,., or x ~ y. We are now ready to prove Lemma I: Proof of Lemma I. Let S = a1, '12, aa, ·· · · be a given sequence of real numbers, and let certain subsequences of S be denoted as follows: So = S, Si '12, as, a., •· · , S2 as, a., a6 , • • , etc. There are two cases to consider: Case I. Every sequence S0 , S1, 82, · • · contains a largest term. Denote by a,., the largest term of S 0 , by a,., the largest term of Sn., by an, the largest term of Sn., etc. By construction, n1 < n2 < na · · · , so that a,.,, a,.,, a,.., · · • is a subsequence of S, and by Lemma II it is a monotonically decreasing subsequence. Case II. There exists a sequence SN containing no largest term. Then every term of SN is followed ultimately by some larger term (for otherwise there would be some term of SN which could be exceeded only by its predecessors, of which there are at most a finite number, so that SN would have a largest term). We can therefore obtain inductively an increasing subsequence of S by letting a,., = aN+i, a,., = the first term following ani that is greater than an,, a,., = the first term following an, that is greater than a,.,, etc. This completes the proof of Lemma I, and therefore that of the Fundamental Theorem.

=

=

*302. THE CAUCHY CRITERION FOR CONVERGENCE OF A SEQUENCE

Convergence of a sequence means that there is some number (the limit of the sequence) that has a particular property, formulated in terms of t and N. In order to test the convergence of a given sequence, then, it

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might seem that one is forced to obtain its limit first.

63

That this is not

always the case was seen in Theorem XIV, § 204, where the convergence of a monotonic sequence is guaranteed by the simple condition of boundedness. A natural question to ask now is whether there is some way of testing an arbitrary sequence for convergence without knowing in advance what its limit is. It is the purpose of this section to answer this question by means of the celebrated Cauchy Criterion which, in a crude way, says that the terms of a sequence get arbitrarily close to something fixed, for sufficiently large subscripts, if and only if they get arbitrarily close to each other, for sufficiently large subscripts. Definition. If {a,.} is a sequence (of real numbers), the notation lim (am - a,.) = 0 means that corresponding to an arbitrary positive

m,n-++ao

number E there exists a number N such that m > N and n > N together imply la,,. - a,.I < E. A sequence {a,.} satisfying the condition lim (am - a,.) = 0 m,n--++ao

is called a Cauchy sequence. Theorem. Cauchy Criterion. A sequence (of real numbers) converges if and only if it is a Cauchy sequence. Proof. The ''only if" part of the proof is simple. Assume that the sequence {a,.} converges, and let its limit be a: a,.---+ a. We wish to show that lim (am - a,.) = 0. Corresponding to a preassigned E ~ 0, let N m,n-++ao

be a number such that n > N implies la,. - al both greater than N, we have simultaneously

la,,. - al

< ½E

< ½E.

and la,. - al so that by the triangle inequality (§ 110)

la,,. -

a,.I =

l(a,,. - a) -· (a,. - a)I

~

Then if m and n are

< ½E,

la,,. - al

+

la,. - al

< E.

For the "if" half of the proof we assume that a sequence {a,.} satisfies the Cauchy condition, and prove that it converges. The first step is to show that it is bounded. To establish boundedness we choose N so that n > N implies la,. - aN+1I < 1. (This is the Cauchy Criterion with E = 1 and m = N + 1.) Then, by the triangle inequality, for n > N,

la,.1 = l(a,. - aN+1)

+ aN+1I

~ la,. - aN+1I

+ laN+1I < laN+1I + 1.

Since the terms of the sequence are bounded for all n > N, and since there are only a finite number of terms with subscripts less than N, the entire sequence is bounded. According to the Fundamental Theorem (§ 301), then, there must be a convergent subsequence {a,..}. Let the limit of this subsequence be a. We shall show that a,. ---+ a. Let E be an arbitrary positive number. Then, by the Cauchy condition, there exists a numl:er N such that for m > N and n > N; lam - a,.I < ½E. Let n be an arl!itrary

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SOME THEORETICAL CONSIDERATIONS

positive integer greater than N. the triangle inequality,

We shall show that la,. - al


N and (ii) la... - al < ½E. If we choose any such value fork, the inequality above implies la,. - al < ½E + ½E = E, and the proof is complete. NOTE. Without the Axiom of Completeness, the Cauchy condition is not a criterion for convergence. For example, in the system of rational numbers, where the Axiom of Completeness fails (Ex. 2, § 116), a sequence of rational numbers converging to the irrational number v'2 (Ex. 15, § 205) satisfies the Cauchy condition, but does not converge in the system of rational numbers.

At this point, the student may wish to do a few of the Exercises of § 305. Some that are suitable for performance now are Exercises 1-6 and 15-22. *303. SEQUENTIAL CRITERIA FOR CONTINUITY AND EXISTENCE OF LIMITS

For future purposes it will be convenient to have a necessary and sufficient condition for continuity of a function at a point, in a form that involves only sequences of numbers, and a similar condition for existence of limits. (Cf. the Theorem, § 209.) Theorem I. A necessary and sufficient condition for a function f(x) ta be continuous at x = a is that whenever {x,.} is a sequence of numbers which converges to a (and for which f(x) is defined), then {f(x,.)} is a sequence of numbers converging to f(a); in short, that x,.-+ a implies f(x,.) -+ f(a). Proof. We first establish necessity. Assume that f(x) is continuous at x = a, and let x,.-+ a. We wish to show that f(x,.)-+ f(a). Let E > 0 be given. Then there exists a positive number o such that Ix - ·al < o implies lf(x) - f(a)I < E (for values of x for which f(x) is def1ned). Since x,. -+ a, there exists a number N such that, for n > N, Ix.. - al < o. Therefore, for n > N, IJ(x,.) - f(a)I < E. This establishes the desired convergence. Next we prove sufficiency, by assuming that J(x) is discontinuous at x = a, and showing that we can then obtain a sequence {x,.} converging to a such that the sequence {f(x,.)} does not converge tof(a). By Exercise 14, § 212, the discontinuity of f(x) at x = a means that there exists a positive number E such that however small the positive number o may be, there exists a number x such that Ix - al < o and 1/(x) - /(a)I E: E. We construct the sequence {x,.} by requiring that x,. satisfy the two inequalities Ix,. - al < I/n and IJ(x,.) - f(a)I E: E. The former guarantees the con-

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65

vergence of {x,.} to a, while the latter forbids the convergence of {f(x,.)} This completes the proof of Theorem I.

to f(a).

The formulation and the proof of a sequential criterion for the existence of a limit of a function are similar to those for continuity. The statement in the following theorem does not specify the manner in which the independent variable approaches its limit, since the result is independent of the behavior of the independent variable, subject to the restrictions of the first paragrap4 of § 207. The details of the proof, with hints, are left to the student in Exercises 8 and 9, § 305.

Theorem II. The limit, lim f(x), of the function f(x) exists (finite or infinite) if and only if, for every sequence {x,.} of numbers having the same limit as x but never equal to this limit (and for which f(x) is defined), thesequence {f(x,.)} has a limit (finite or infinite); in short, if and only if x,. -lim x, x,. ¢ lim x imply f(x,.) - limit. As might be expected, the Cauchy Criterion for convergence of a sequence has its application to the question of the existence of a limit of a function. With the same understanding regarding the behavior of the independent variable as was assumed for Theorem II, we have as an immediate corollary of that theorem the following:

Theorem m. The limit, lim f(x), of the function f(x) exists and is finite if and only if, for every sequence {x,.} of numbers approaching the same limit as x but never equal to this limit (and for which f(x) is defined), the sequence {f(x,.)} is a Cauchy sequence; in short, if and only if x,. - lim x, x,. ¢ lim x imply {f(x,.)} is a Cauchy sequence. *304. THE CAUCHY CRITERION FOR FUNCTIONS

The Cauchy Criterion for sequences gives a test for the convergence of a sequence that does not involve explicit evaluation of the limit of the sequence. Similar tests can be formulated for the existence of finite limits for more general functions, where explicit evaluation of the limits is not a part of the test. Such criteria are of great theoretical importance and practical utility whenever direct evaluation of a limit is difficult. Owing to the latitude in the behavior granted the independent variable, we have selected for specific formulation in this section only two particular cases, and the proof of one. The remaining proof and other special cases are treated in the exercises of § 305.

Theorem I. Assume that every deleted neighborhood of the point a contains one point of the domain of definition of the function f(x). Then the limit lim f(x) exists and is finite if and only if corresponding to an arbitrary at least

-

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[§ 305

positive number t= there exists a positive number 8 such that O < Ix' - al < 8 and O < Ix" - al < 8 imply lf(x') L f(x")I < E, for values of x' and x" for which f(x) is defined. Proof. "Only if": Assume limf(x) =Land let E > 0. Then there ex.,....,. ists 8 > 0 such that O < Ix - al < 8 implies lf(x) - LI < ½E. If x' and x" are any two numbers such that O < Ix' - al < 8 and O < Ix" - al < 8, the triangle inequality gives

lf(x') - f(x")I = l(J(x') - L) - (J(x") - L)I ~ lf(x') - LI + lf(x") - LI < ½E + ½E = E. "If": Let {x,.} be an arbitrary sequence of numbers (for which f(x) is defined) such that x,. - a and x,. ¢. a. By Theorem III, § 303, we need only show that the sequence {f(x,.)} is a Cauchy sequence: lim

[f(xm) - f(x,.)] = 0.

m,n--++oo

If E is a preassigned positive number, let 8 be a positive number having the assumed property that O < Ix' - al < 8 and O < Ix" - al < 8 imply lf(x') - f(x") I < E. Since x,. - a and x,. ¢. a, there exists a number N such that n > N implies O < Ix,. - al < 8. Accordingly, if m > N and n > N, we have simultaneously O < Ix,,. - al < 8 and ' O < Ix,. - al, < 8, so that If(xm) - f(x,.) I < E. Thus the sequence {f(x,.)} is a Cauchy sequence, and the proof is complete.

Theorem II. The limit Jim f(x) exists and is finite if and only if corre.,._.+,.

sponding to an arbitrary positive number E there exists a number N such that x' > N and x" > N imply lf(x') - .f(x") I < E (for values of x' and x" for which f(x) is defined). *305. EXERCISES

*1· Prove that a sequence {a,.} (of real numbers) converges if and only if corresponding to an arbitrary positive number E there exists a positive integer N such that for all positive integers p and q, laN+p - aN+ql < E. *2. Prove that a sequence {a,.} (of real numbers) converges if and only if corresponding to an arbitrary positive number E there exists a number N such that n > N implies la,. _: aNI < E. *3. Prove that a sequence {a,.} (of real numbers) converges if and only if corresponding to an arbitrary positive number E there exists a positive integer N such that for all positive integers p, laN+p - aNI < E. Prove that the condition lim (a,.+,, - a,.) = 0 for every positive in-

*'·

n-++ co

teger p is necessary but not sufficient for the convergence of the sequence {a,.}. Hint: Consider a sequence suggested by the terms 1, 2, 2½, 3, 3¼, 3¾, 4, 4¼, 4½,

4¾, 5, 5!-, · · ·.

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67

•5. Find the error in the "theorem" and "proof": If a,, --. a, then a,. = a fur sufficiently large n. Proof. Any convergent sequence is a Cauchy sequence. Therefore, if E > 0 there exists· a positive integer N such that for all positive integers m and n, with m > N, la .. - aN+nl < E. But this means that Jim aN+• = a,,., which in turn implies that Jim a,. = a = a., for all m > N.

n-+ao n ...... +oo •6. If {b,.} is a convergent sequence and if {a,,} is a sequence such that ja., - a,.j ~ lb .. - b,.I for all positive integers m and n, prove that {a,.} converges. •7. If f(x) and g(x) are functions defined for x > 0, if Jim g(x) exists and z....o+ · is finite, and if lf(b) - f(a)I ~ lg(b) - g(a)I for all positive numbers a and b, prove that lim f(x) exists and is finite . .,.....o+ •S. Prove Theorem II, § 303, for the case x --. a. Hint: First show that if x,,--. a, x,. ¢ a imply that the sequence {f(x,.)} has a limit then this limit is unique, by considering, for any two sequences {x,.} and {x,.'} which converge to a, the compound sequence x,, x,', x2, x2', xa, xa', • • · . For the case f(x,,) --. +oo, assume that Jim f(x) ¢ +oo, and show that there must exist a constant Band a sequence {x,.} such that O < Ix,, of lim f(x,.), proceed similarly.

al < l/n and f(x,,)

~

B. For other cases

%'-0(1

•9. Prove Theorem II, § 303, for the case x --. a+; x --. a-; x--. +oo; x--. -oo; x--. oo. (Cf. Ex. 8.) •10. Assuming only Theorem I, § 303, and the limit theorems for sequences (§ 204), prove the continuity theorems of § 211. Hint for Theorem IV: Let {x,.} be an arbitrary sequence converging to a. Then f(x,.) --. J(a) and g(x,.) --. g(a); and therefore f(x,.)g(x,.) --. f(a)g(a). *11. Assuming only Theorem ~I, § 303, and the limit theorems for sequences (§ 204), prove the limit theorems of § 207. (Cf. Ex. 10.) *12. Reformulate and prove Theorem I, § 304, for the case x --. a+; for the " case x--. a-. •13. Prove Theorem II, § 304. *14. Reformu_late and prove Theorem II, § 304, for the case x --. - oo ; for the case x --. oo • *16. Let {a,.} be a sequence of real numbers, and let A,. be the least upper bound of the set {a,., a,.+1, a,.+2, • • •} (A,. = +oo if this set is not bounded above). Prove that either A,. = +oo for every n, or A,, is a monotonically decreasing sequence of real numbers, and that therefore Jim A,. exists ( +00, . , n-+CIO finite, or - oo ) • Prove a similar result for the sequence {B,.} , where B,. is the greatest lower bound of the set {a,., a,.+1, a,,+2, • · ·}. *16. The limit superior and the limit inferior of a sequence {a,,}, denoted ,·· Tiiii a,., or Jim sup a,., and lim a,., or lim inf a,., respectively, are defined n-+• n--++oo n.-+oo n--++oo as the limits of the sequences {A,.} and {B,.}, respectively, of Exercise 15. Justify the following formulations: ·

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. A .. = mf um a,.= ......Jim + ..

.......+.. Jim

....... +..

a,.=

Jim B,.

...... + ..

[§ 305

+• [ sup +• (a.,) ] , m•n

n- 1

+•1 [·inf m•n +• (a.. )] . = sup n•

(Cf. § 114.) •17. Prove that a number L is the limit superior of a sequence {a..} if and only if it has the following two properties, where E is an arbitrary preassigned positive number: (i) The inequality a,. < L + E holds for all but a finite number of terms. (ii) The inequality a,. > L - E holds for infinitely many terms. State and prove a similar result for the limit inferior. Prove that the limit superior and limit inferior of a bounded sequence are limit points of that sequence (cf. Exs. 24-26, § 205), and that any other limit point of the sequence is between these two. Extend this result to unbounded sequences. (Cf. Exs. 15-16, and Ex. 29, § 312.) •18. Prove that for any sequence {a,.}, bounded or not, lim a,. ~ Tiiii a•. n_.+• n-++• Prove that a sequence converges if and only if its limit superior and limit inferior are finite and equal, and that in the case of convergence, Jim a,. = Jim a. = "~+oo n-++• Tiiii a,.. Extend these results to include infinite cases. (Cf. Exs. 15-17,)

......+.

In Exercises 19-22, find the limit superior and the limit inferior. 15-18.)

(Cf. Exs.

•19. 0, 1, 0, 1, 0, 1, · · · . •20. 1, -2, 3, -4, • • • , ( -l)n+l n, • • • .

•21. i, i, ¾, ¼, t, !-, t, t, · · · · •22. ¾, -½, t, -½, ¾, -¼, !, -¼, · · · . •23. Let /(:r,) be a real-valued function, defined for at least some values of x neighboring the point :r, = a (except possibly at a itself). If 6 is an arbitrary positive number, let q,(6) and "1(6) be the least upper bound and the greatest lower bound, respectively, of the values of /(:r,) for all :r, such that 0 < Ix - al < o (and for which J(:r,) is defined). Prove that, in a sense that includes infinite values (cf. Ex. 15), q,(6) and "1(6) are monotonic functions and therefore have limits as 6-+ o+. (Cf. § 215.) •2~ The limit superior and the limit inferior of a function /(:r,) at a point :r, = a, denoted um /(:r,), or lim sup J(:r,), and lim /(:r,), or lim inf J(:r,), respecx--+a

X--+a

;:;;

X--+a

tively, are defined as the limits of the functions q,(6) and "1(6), respectively, of Exercise 23. Justify the following formulations:

um J(:r,) = lim q,(6) = inf [sup J(x>]· z---a a--.o+ 4>0 O 0 and points x,.' and x,." of I such that 1 Ix,.' - x,."I < and lf(x,.') - f(x,.")I ~ E. The bounded sequence {z,.'}

n

contains a convergent subsequence {x,..'}, converging to some point xo of I (cf. proof of Theorem I, § 213, given in § 306). We show that the corresponding sequence {x,..''} also converges to xo by use of the triangle inequality:

lxn." - Xol = I(x,.." - x...') + (x,..' - Xo) I ~ Ix,.." - x...'I + Ix...' - Xol < l/n1 + Ix...' - 2:ol. As k becomes infinite each of the last two terms approaches zero, and therefore so does the quantity Ix..." - xol- By continuity of f(x) at Xo, lim f(x,..') = lim f(x,..") = f(Xo), a.nd therefore lim [f(z...') - f(x,..")] k-++• k->+• lo--t+co = 0. This last statement is inconsistent with the inequality lf(x,..') - f(z,..") I ~ E, which must hold for all values of k, and the proof is complete. *308. EXERCISES

*1, Prove Theorem I, § 307. *2. Establish the Negation of Uniform Continuity, § 307.

In Exercises 3-8, find an explicit function definition of uniform continuity.

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a = 8(E)

in conformity with the

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•3. y = x2, for O ~ x ~ 1. Hint: lx"2 - x'2I = Ix" - x'I · Ix" + x'I ~ 2lx" - x'!. 2 y = z , for O ~ z ~ 2. _,_, _,z" - z' y = vx, for 1 ~ x ~ 2. Hint: vz" - vx' = --,,=-------,-=

•6.

y

v'x"

= v';-, for 0 ~ z ~ v'z"

•7.

y

= !, for z ~ 1.

•8.

y

= Vl

+

1. Hint: v''? ~ v'~iz-,,-_-x-'I

+ v'x'

(cf. Ex. 5).

X

- z 2, for lzl ~ 1.

v'1 _ x'2 _ Vl _

Hint: For 0 ~ x'

< x"

~ 1,

( - x x"2 = x + z')z vT+? v'1 - x' + v'1 + x'' ~ (

II

II

')

~ 2(x" - _ x') ,.,. 2(x" - x') ~ ,;---; .a:. _ - v 1 - z' + v 1 1 - x" - v 1 (1 - x') - (1 - x")

(cf. Ex. 6).

In Exercises 9-12, use the Negation of § 307 to show that the given function is not uniformly continuous on the given interval. (Cf. Exs. 13-14.)

•9. !, 0 < X

•10,

X 1, X

X

< 1.

e:;; 1.

•11. sin !, 0

0. X

•12.

1.

In Exercises 13-14, find an explicit function o = o(E, zo), in conformity with the O-E definition of continuity, for the given function at a given point xo of the specified interval. Observe that o depends essentially on ZQ. (Cf. Exs. 9-10; also Exs. 59-64, § 208.)

•13.

!,

•14.

X2, X

X

0

< z < 1. e:;; 1.

•15. If f(x) is uniformly continuous on an open interval (a, b), prove that the two limits f(a+) and f(b-) exist and are finite. Hint: Use the Cauchy Criterion for functions. •16. Prove the following converse to Exercise 15: If J(x) is continuous on an open interval (a, b) and if the two limits J(a+) and f(b-) exist and are finite, then f(x) is uniformly continuous on (a, b). Hint: Extend f(z) to a function continuous on the closed interval [a, b].

**309. POINT SETS: OPEN, CLOSED, COMPACT, CONNECTED SETS Many of the results obtained in the preceding portions of this chapter are special cases of more general theorems which can be formulated for Euclidean spaces of any number of dimensions. In the remaining sections of this chapter we shall establish some of these general theorems for the particular one-dimensional space of real numbers. The statements of the

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theorems and the concepts and techniques involved are of such a nature that their analogues in higher dimensional spaces are immediately available, with only a few minor adjustments in basic definitions. In this chapter the single word set means set of real numbers, and the word point means real number. · Definition I. If A is a set, the complement of A, written A', is the set of all real numbers that are not members of A.

=

=

Examples. If A (-oo, 23), then A' = (23, +oo ). If A [l, 3), then A' consists of all real numbers less than 1, or greater than or equal to 3.

Definition II. A set A is open if and only if every member of A has some neighborhood contained entirely in A.

=

Examples. The set A (2, 7) is open, because if 2 < x < 7 and E is a positive number less than both 7 - x and x - 2, then the open interval (x - E, x + E) is a neighborhood of x lying within A. More generally, any open interval, finite or infinite, is open (Ex. 1, § 312). The set B = ( -1, 8) is not open, since every neighborhood of the point 8 extends to the right, outside the set B.

Definition m. A point p is a limit point of a set A if and only if every deleted neighborhood of p contains at least one point of A. NOTE 1. If p is a limit point of a set A we say that the set A has p as a limit point, whether p is a member of A or not. For example, the two sets B = [2, 5] and C (2, 5) both have 5 as a limit point, but only B contains 5 as a member. NoTE 2. If pis a limit point of a set A, every neighborhood of p contains infinitely many points of A (Ex. 2, § 312).

=

Definition IV. limit points.

A set A is closed if and only if it contains all of its

=

Examples. The set A [1, 6) is closed. More generally, any finite closed interval is closed, and the infinite intervals [a, +oo) and (-oo, b] are closed (Ex. 5, § 312). The set B = (3, 5) is not closed, since the point 3 is a limit point of B but not a member of B. It is neither open nor closed. The set C of all integers is a closed set since it has no limit points and therefore contains all of its limit points. For the same reason any finite set is closed. The set D of all rational numbers is not closed since every real number, rational or irrational, is a limit point of D. It is neither open nor closed. NoTE 3. The empty set, denoted 9), is both open and closed by double default. Since it has no points, every member of 9) has every property, including that of having a neighborhood contained in 9), so that 9) is open; and since j;) has no points it has no limit points and is therefore closed. The entire space S of real numbers is also both open and closed. The student is asked

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to show in Exercise 11, § 312, that fi) and Sare the only sets of real numbers that are both open and closed. Any set with at least one member is called nonempty.

Definition V.

A set A is compact if and only if it is closed and bounded. t

Examples. Any finite closed interval is compact. Any finite set is compact. The set consisting of the reciprocals of the positive integers together with the number O is compact. The set of all integers is not compact because it is unbounded. The open interval (0, 1) is not compact because it is not closed.

The preceding examples show that a set may be neither open nor closed (a half-open interval and the set of rational numbers are two instances). On the other hand, as explained in the Note above, a set may be both open and closed. One might wonder whether ther~ is any relation between these two concepts of openness and closedness. The following theorem gives the answer.

Theorem I. A set is open if and only if its compkment is closed. Equivalently, a set is closed if and only if its complement is open. Proof. "Only if": Let A be an open set. We wish to show that the complement A' contains all of its limit points. Assume that x is a limit point of A' that does not belong to A'. Then x belongs to A. · Since A is open, x has a neighborhood lying entirely in A and therefore containing no points of A'. Therefore x is not a limit point of A'. Contradiction. "If": Let A' be a closed set. We wish to prove that A is open. Let x be any point of A. Then xis not a limit point of A'. Therefore x has a neighborhood that contains no points of A' and thus lies entirely in A.

Therefore A is open, and the proof is complete. A useful relation between least upper bounds (or greatest lower bounds) and limit points is the following, whose proof is requested in Exercise 8, § 312:

Theorem Il. If p is the least upper bound (or the greatest lower bound) of a set A and if p is not a member of A, then p is a limit point of A. An immediate consequence (Ex. 9, § 312) is the following:

Theorem m. Every compact set of real numbers has a greatest member and a least member. Definition VI.

Two sets are disjoint if and only if they have no point

in common.

t This formulation of compactneBB is suitable for finite dimensional Euclidean spaces, but not for general topological or metric spaces. For a treatment of metric spaces the reader is referred to M. H. A. Newman, Elements of the Topowgy of Pl,ane Sets of Points (Cambridge, Cambridge l"niversity PreBB, 1939).

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ED1Dples. The intervals (0, l} and (1, 2) are disjoint. The intervals (0, 1] and (1, 2] are disjoint. The intervals [O, 1] and [I, 2] are not disjoint, since they have the point 1 in common.

Definition VII. Two sets are separated if and only if they are disjoint and neither contains a limit point of the other. ED1Dples. The intervals (O; l} and (1, 2) are separated. The intervals (0, 1] and (1, 2] are not separated, since the point 1 belongs to (0, 1] and is a limit point of (1, 2]. The intervals (0, 1] and (1, 2] are not separated, since they are not disjoint. The set of rational numbers and the set of irrational numbers are disjoint, but are about as far from being separated as two disjoint sets of real numbers can be!

Definition VIII. A set A is said to be split into two part, B and C if and only if Band Care disjoint and every point of A belongs either to B or to C. Definition IX. A set A is connected if and only if it cannot be split into two separated nonempty parts. The following theorem shows that for real numbers, connectedness is not a sufficiently rich concept to excite anyone. Our reason for presenting the subject here is that connectedness, along with openness, closedness, and compactness, is a very significant idea in the theory of spaces of more than one dimension. In the study of functions of several variables it is important to know the substance of the theorem that we now state and prove. Theorem IV. A nonempty set of real numbers is connected if and only if

it is an interoal or consists of one point. Proof. The "only if" half of this proof is fairly easy. Let A be any connected set consisting of more than one point. We divide the proof that A is an interval into two parts, the second of which is left to the student, with hints (Ex. 10, § 312): (i) if a and b are any two points belonging to A, then any point between a and b must also belong to A; (ii) any set having the property just specified in (i) is an interval. To prove (i), we seek a contradiction to the assumption that there exist two points a and b (a < b) which are members of A and a third point c between a and b (a < c < b) which is not a member of A. The point c provides a splitting of A into two parts (one consisting of all points of A less than c and the other consisting of all points of A greater than c) neither of which contains a limit point of the other. Since the set A is assumed to be connected, the desired contradiction has been obtained. The "if" half is more difficult. We wish to show that every interval is connected. Let us assume the contrary, and let A be an interval which is not connected, and letA consist of the two nonempty parts B and C neither of which contains a limit point of the other. Let b be any point

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in B and c any point in C and assume for definiteness that b < c. Since A is an interval, every point between band c must belong either to R or to C. Denote by D the set of points of the closed interval [b, c] which belong to B, and let d = sup (D). There are two cases: (i) d belongs to B. In this case d < c since c belongs to C, and every point of the half-open interval (d, c] belongs to C (by the definition of d). But d, being a limit point of (d, c] must thereby be a limit point of C. But a member of B cannot be a limit point of C! (ii) d belongs to C. In this cased must be a limit point of D (Theorem Ii), and therefore of B. But a member of C cannot be a limit point of BI In either case we obtain a contradiction, and the theorem is proved. **310. POINT SETS AND SEQUENCES

Much of our earlier work has been based on sequential arguments resting ultimately on the Fundamental Theorem for sequences. In order to exploit the techniques already used, we obtain in this section some useful facts about certain types of sets, phrased in terms of sequences. It turns out that the key concept that permits the useful extension of the Fundamental Theorem for sequences to more general sets than intervals is compactness. This is shown in Theorem IV of this section. Definition. A sequence {an} of points is called a sequence of distinct points if and only if no two terms are equal; that is, if and only if m ~ n implies a,,. F an. · Theorem I. If a sequence {a,.} (of real numbers) has a finite limit a, then either all but a finite number of the terms are equal to a or there exisis a subsequence of distinct terms converging to a. Proof. Assume that a,. -+ a and that for every N there exists an n > N such that a,. F a. The subsequence sought can be obtained inductively. Let an, be the first term different from a. Take E = la.., - al > 0 and let a,,, be the first term a,. satisfying the inequalities O < la,. - al < E = la.., - al, let a,., be the first term a,. satisfying the inequalities

< Ian - al < Ian, - al, etc. By construction, n 1 < n2 < n 3 < · · · , so that {an.} 0

is a subsequence, and

no two terms are equal. Theorem Il. A point p is a limit point of a set A if and only if there exists a sequence {a,.} of distinct points of A converging top. Proof. If {a,.} is a sequence of distinct points of A converging top, then every neighborhood of p contains all points of the sequence from some index on, and therefore infinitely many points of A, so that pis a limit point

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of A. On the other hand, if pis a limit point of A, we can find, for each positive integer n, a point p,. of A such that O < Ip,. - Pl < 1/n. The sequence {p,.} therefore converges to p, and since none of the terms are equal to p it contains (by Theorem I) a subsequence {p,..} no two terms of p,.•. which are equal, such that p,.. --. p. Let a,.

=

Theorem m. A set A of real numbers is closed if and only if the limit of every convergent sequence {a,.} of points of A is a point of A. Proof. "If": Assume that the limit of every convergent sequence of points of A is a point of A and let p be a limit point of A. We wish to show that pis a point of A. Since pis a limit point of A, Theorem II guarantees the existence of a sequence of points of A converging top, so that p must belong to A. "Only if": Let A be a closed set, and let {a,.} be a sequence of points of A converging to the point a. We wish to show that a belongs to A. According to Theorem I there are two possibilities. Either a,. = a for all but a finite number of n (in which case a must belong to A) or the sequence {a,.} contains a subsequence of distinct terms (in which case, by Theorem II, a must be a limit point of A). In either case, since A contains all of its limit points, a must belong to A. Theorem IV. A set A of real numbers is compact if and only if every sequence {a,.} of points of A contains a BUbsequence convergi,ng to a point of A. Proof. "If": Assume that every sequence {a,.} of points of A contains a convergent subsequence whose limit belongs to A. We wish to show that A is bounded and closed. If A were unbounded there would exist a sequence {a,.} of points of A such that la,.I > n, so that no subsequence could converge to any point. If A were not closed, there would exist (by Theorem III) a sequence {a,.} of points of A converging to a point p not a member of A. Since every subsequence would also converge top, no subsequence could converge to a point of A. These contradictions show that A must be both bounded and closed, and therefore compact. "Only if": Assume that A is compact and let {a,.} be an arbitrary sequence of points of A. Since A is bounded, {a,.} contains a convergent subsequence {a,..}, and since A is closed, by Theorem III the limit of this subsequence must belong to A. tt311. SOME GENERAL THEOREMS

With the aid of the theoren,.s on sets and sequences given in the preceding section, we can now establish four of the most important general theorems on continuous functions. We remind the reader that in this chapter only real-valued functions of a real variable are considered, but the theorems

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are stated in general terms, so that they may be taken without change for more general use. Joining compactneas as a key concept, now, is connect.edness.

Theorem I. A function continuous on a compact domain has a compact range. Proof. Let the function bey = f(x), with compact domain A and range B. If B is not compact, there is a sequence {bn} of points of B such that {b,.} contains no subsequence converging to a point of B (Theorem IV, § 310). For each n, let an be a point of A such that bn = f(an). Since A is assumed to be compact, the sequence {an} contains a subsequence {a,,.} converging to some point a of A. But from the continuity of f(x) at x = a we can infer that an. -+ a implies /(an.) -+ /(a) (Theorem I, § 303). In other words, the subsequence {bn.} of the sequence {bn} converges to the point b = f(a} of B. This contradiction completes the proof.

Since any compact set is bounded and since any compact set of real numbers has a greatest member and a least member (Theorem III, § 309), we have immediately the following two corollaries, of which the first two theorems of § 213 are special cases where the domain is a closed interval:

Corollary I. A function continuous on a compact domain is bounded there. Corollary II. A real-valued function continuous on a compact domain has a maximum value and a minimum value there. Theorem II. A Junction continuous on a connected domain has a connected range. Proof. Let the function be y = f(x), with connected domain A and range B. If B is not connected, then B can be split into two disjoint nonempty parts, B1 and B2, neither of which contains a limit point of the other. Denote by Ai the points x of A such that f(x) is a point of B1, and by A2 the points x of A such that f(x) is a point of B2. Then A is split into the two disjoint nonempty subsets, A1 and A2. Since A is connected, one of these sets must contain a limit point of the other. For definiteness, assume that p belongs to A1 and is a limit point of A2, and let {an} be a sequence of points of A2 such that an-+ p. Since f(x) is assumed to be continuous at x = p, an-+ p implies f(a,.)-+ f(p}. But this means that a sequence of points of B2 converges to a point of B1, in contradiction to the assumption that no·point of B1 is a limit point of B2, and the proof is complete. Since any connected set of real numbers is an interval or a one-point set, we have the following corollary, of which Theorems III and IV, § 213, are special cases where the domain is an interval:

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Corollary. A r~valued funct,ion contintwus on a connected domain assumes (as a value) every number between any two of its va/.ues. The Theorem of § 215 on the continuity of the inverse function of a strictly monotonic function finds its generalization again based on compactness, with the monotonic property replaced by the mere existence of a single-valued inverse:

Theorem m. If a function y = f(x) is continiwus on a compact domain A and never assumes the same value at distinct points of A, then the inverse function x = q,(y) is continiwus on the range B of f(x). Proof. We observe first that since f(x) always has distinct values at distinct points of A, f(x) establishes a one-to-one correspondence between the points of A and the points of B, so that the inverse function x = q,(y) exists on B. To establish continuity of q,(y) we wish to show that b,.--+ b (where b,. and b are points of B) implies a,. --+ a, where a,. = q,(b,.) and a= q,(b) (a,. and a are points of A). Let us assume that a,.-++ a, so that there exists a neighborhood of a outside of which lie infinitely many terms of the sequence {a,.}. Since these infinitely many terms form a subsequence of {a,.}, and since A is assumed to be compact, there must be a subsequence of this subsequence which converges to some point of A (Theorem IV, § 310) different from a. Denote by {a...} this new convergent subsequence, and denote by a' its limit: a'= lim a..., where A,--,+oo

a'¢. a.

Since f(x) is assumed to be continuous at a', a...--+ a' implies f(a...) = b,..--+ f(a'). On the other hand, b,.--+ b implies b,..--+ b = f(a). By the uniqueness of the limit of a sequence (Theorem II, § 204), as applied to the subsequence {b,..), we infer thatf(a) = f(a'), in contradiction to our assumption that f(x) never assumes the same value at distinct points. Finally, the proof given in § 307 that a function continuous on a closed interval is uniformly continuous there generalizes with only trivial notational changes (Ex. 23, § 312) to a function continuous on any compact set:

Theorem IV. A function continiwus on a compact domain is uniformly continiwus there. tt312. EXERCISES

••1.

Prove that any open interval, finite or infinite, is open. Give some examples of open sets (of real numbers) that are not intervals. •*2. Prove the statement of Note 2, § 309. ••3. Prove that every nonempty open set (of real numbers) contains both rational and irrational numbers. Show, in fact, that it contains infinitely many of each type. **'- Find an example of a pair of nonempty open sets (of real numbers) such that every member of each set is exceeded by some member of the other.

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**5. Prove that any finite closed interval and the infinite intervals [a, +co) and ( - oo, b] are closed. **6. Give some more examples of closed sets. **7· Give some more examples of sets (of real numbers) that are neither open nor closed. **8. Prove Theorem II, § 309. **9. Prove Theorem Ill, § 309. **10. Prove that if A is a set of real numbers containing more than one point, with the property that whenever two points belong to A every point between these two also belongs to A, then A is an interval (finite or infinite). Hint: If A is bounded, show that any point c between inf (A) and sup (A) is flanked by two members, a and b, of A; a < c < b. If A is unbounded proceed similarly. **11. Prove that the only sets of real numbers both open and closed are the empty set and the entire space S. Hint: Assume that A is a nonempty, open, and closed set of real numbers not containing all real numbers. Then its complement B = A' is also nonempty, open, and closed. But this means that the connected set Sis split into two nonempty separated parts, A and B. **12. Prove that if A and B are two nonempty disjoint open sets (of real numbers) there exist numbers a, b, and c, where c is between a and b, such that a belongs to A, b belongs to B, and c belongs to neither A nor B. Hint: Show that the contrary assumption implies that the set C made up of all of the points of A together with all of the points of Bis connected (cf. Exs. 10-11). **13. Prove the Bolzano-Weierstrass Theorem: Any infinite bounded set of real numbers has a limit point. Hint: If A is an infinite bounded set, let {an} be a sequence of distinct points of A, and let {an.} be a convergent subsequence of {an} converging to a point p. Show that pis a limit point of A. **14. Give some examples of bounded sets (of real numbers) that have no limit points. **15. Give some examples of infinite sets that have no limit points. **16. Give some examples of sets having the property that each point of the set is a limit point of the set. ••17. If A is an arbitrary set of real numbers, prove that the set B of all limit points of A is closed. **18. The closure of a set A, denoted A, is the set made up by adjoining to A all limit points of A. Prove that A is always closed. **19. The distance between a point panda nonempty set A, written 6(p, A), is defined to be the greatest lower bound of the set of distances IP - al between p and arbitrary points a of A. Prove that the distance between p and A is 0 if and only if p is either a point of A or a limit point of A. Prove that if p is not a point of a nonempty closed set A, then 6(p, A) > 0. **20. Prove that if p is a point that is not a member of a nonempty closed set A, then there is a point a of A such that 6(p, A) = IP - al, (Cf. Ex. 19.) Hint: Let {a,.} be a sequence of points of A such that

IP - a,.I
0. (Cf. Exe. 20-21.) Hint: ABBume that A is compact, and choose points a,. of A and b,. of B such that 1 la. - b,.I < 8(A, B) + -· Let a,.. -+ a, and choose a convergent subsequence

n of {b,.1}. **23. Prove Theorem IV,§ 311. **M. Let f(z) be continuous for all real numbers z, and let c be a constant. Prove that the following sets are open: {i) all z such that f(z) > c; (ii) all z such that f(z) < c. Prove that the following sets are closed: (iii) all z such that f(z) ~ c; (iv) all z such that f(z) :ii c; (I,) all z such that f(z) = c. If /(z) is bounded, must any of these sets be bounded? (Prove or give a counterexample.) **26. Give an example of a function, defined for all real numbers z, such that the set of all points of continuity is (i) open but not closed; (ii) closed but not open; (iii) neither open nor closed. **26. A sequence {A,.} of sets is called monotonically decreasing if and only if every set A,. of the sequence contains its successor A,.+1• This property is symbolized A,. t . (A constant sequence, where A,. = A, for all n, is an extreme example.) Prove the theorem: If {A,.} is a monotonically decreasing sequence of nonempty compact sets there ezists a poinl z common to every set of the sequence. The nested intervals theorem is the special case of this theorem where every compact set A,. is a (finite) closed interval. Hint: For every positive integer n let a,. be a point of A,., and let an• -+ z. For any N, an. belongs to AN for sufficiently large k, so that z also belongs to AN, **27. Show that the theorem of Exercise 26 is false if the assumption of compactness is replaced by either boundedness or closedness alone. Hint: Consider the sequences {(0,

n} and {[n, +ao)}.

**28. A collection of open sets is said to cover a set A if and only if every point of A is a member of some open set of the collection. Such a collection of open sets is called an open covering of A. An open covering of a set A is said to be reducible to a finite covering if and only if there exists some finite subcollection of the open sets of the covering which also covers A. Prove the Heine-Borel Theorem: Any open covering of a compact set is reducible to a finite covering. Hint: First prove the theorem for the special case where the compact set is a (finite) closed interval I, as follows: Assume that Fis a collection of open sets (each member of F is an open set) which covers I and which is not reducible to a finite covering of I. Consider the two closed intervals into which I is divided by its midpoint. At least one of these two subintervals cannot be covered by any finite collection of sets of F. Call this closed subinterval h

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and relabel I = Ii. Let Ia be a closed half of 12 that is not covered by any finite collection of sets of F, and repeat the process to obtain a decreasing sequence {J,.} of closed intervals whose lengths tend toward zero. If x is a point common to every interval !.a (Ex. 26), let B be an open set of the family F which contains x. Then B contains a neighborhood (z - E, x + f) of x which, in turn, must contain one of the intervals I,.. But this means that I,. is covered by a finite collection of eets of F, namely, the single set B. This contradiction establishes the special case. Now let A be an arbitrary compact set and let F be an arbitrary open covering of A. Let I be a closed interval containing A, and adjoin to the family F the open set A' (A' is the complement of A). This larger collection is an open covering of I, and accordingly is reducible to a finite covering of I, by the first part of the proof. Those sets of F which belong to this finite covering of I cover A. **29. A limit point of a sequence {a,.} was defined in Exercise 20, § 205, to be a number x to which some subsequence converges. Prove that the set of all limit points of a given sequence is closed, and that the set of all limit points of a bounded sequence is compact. Hence show that any bounded sequence has a largest limit point and a smallest limit point. Prove that these are the limit superior and limit inferior, respectively, of the sequence. (Cf. Exs. 16-17, § 305.) Extend these results to unbounded sequences. **30. Let A be a set contained in the domain of definition of a bounded function /(z). The oscillation of /(z) on the set A, w(A), is defined to be the difference between the least upper bound of its values there and the greatest lower bound of its values there, w(A) sup (J(z)) - inf (/(z)).

= zinA

:rinA

Prove that if A is contained in B, then w(A) ~ w(B). Hence show that if 6 > 0, and if Aa s (xo - 6, x 0 + 6), then w(Aa) is a monotonically increasing function of 6, so that Jim w(A1) exists (cf. § 215). Make appropriate modifica-

1--.o+

tions to take care of the possibility that A or A, may be only partly contained in the domain of definition of /(x). **31. Let /(z) be defined on a closed interval I. The oscillation w(xo) of /(x) at a point xo of I is defined to be the limit of the function w(A1) of Exercise 30: w(xo) = lim w(A,).

1--.o+

Prove that w(xo) ~ 0 and that J(x) is continuous at x = xo if and only if w(xo) = 0. **32. Let /(x) be defined on a closed interval I. If E > 0, let D, be the set of all points x such that w(x) ~ E. Prove that D, is closed. (Cf. Ex. 31.) Hint : If w(xo) < E, let 71 E - w(zo) and show that there exists a neighborhood N of zo for which w(J) < w(zo) + '7, so that for x in N, w(x) < w(xo) + '1 = E. **33. Let /(x) be defined on a closed interval I, and consider the sequence of closed sets D1, Dt' D¼, • • • (cf. Ex. 32). Prove that each of these sets is contained in its successor, and that the set of points x such that xis a member of some D!II is precisely the set D of points of discontinuity of J(x).

=

n., •••,

/

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4 Differentiation

401. INTRODUCTION

This and the following chapter contain a review and an amplification of certain topics from a first course in Calculus. Some of the theorems that are usually stated without proof in a first introduction are established here. Other results are extended beyond the scope of a first course. On the other hand, many definitions and theorems with which the student can be asswned to be familiar are repeated here for the sake of completeness, without the full discussion and motivation which they deserve when first encountered. For illustrative material we have felt free to use calculus formulas which either are assumed to be well known or are established in later sections. For example, the trigonometric, exponential, and logarithmic functions provide useful examples and exercises for these chapters, but their analytic treatment is deferred to Chapter 6. The only inverse trigonometric functions used in this book are the inverse sine and inverse tangent, denoted Arcsin x and Arctan x, respectively, with values restricted to the principal value ranges -~ ~ Arcsin x ~~and -~

< Arctan x < ~

(the

upper case A indicates principal values). 402. THE DERIVATIVE

We shall consider only single-valued real-valued functions of a real variable defined in a neighborhood of the particular value of the independent variable concerned (or possibly just for values of the independent variable neighboring the particular value on one side). Definition. A function y = f(x) is said to have a derivative or be differentiable at a point x if and only if the followi1UJ limit exists and is finite; the function f'(x) defined by the limit is called its derivative: (1)

0. 11. Prove that any rational function (cf. Ex. 20, § 208) of a single variable is differentiable wherever it is defined. 12. Give an example of a function for which Ax - 0 does not imply Ay -+ 0. Give an example of a continuous function for which Ay - 0 does not imply

Ax-o. 13. By mathematical induction extend the chain rule for differentiating a composite function to the case of n functions: y = J1(h(la • • • (J,.(x)) • • • ) ). 14. Prove that the derivative of a monotonically increasing (decreasing) differentiable function j(x) satisfies the inequality f'(x) ~ 0 ( ~ O) . Does strict increase (decrease) imply a strict inequality?

In Exercises 15-20, discuss differentiability of the given function j(x) (where = O). Is t~e derivative continuous wherever it is defined?

j(O)

15. x2 sgn x (cf. Example 1, § 206).

16.

17. x2 cos!.

18. x2 sin~-

X COS!. X

x

X

*19. xt sin !.

*20.

X

x• cos h·

*21. Discuss differentiability of the function J(x) = 0 if x ~ 0, J(x) s x• if x > 0. For what values of n does f'(x) exist for all values of x? For what values of n is j'(x) continuous for all values of x? *22. If J(x) is the function of Exercise 21, for what values of n does the kth derivative of j(x) exist for all values of x? For what values of n is the kth derivative continuous for all values of x? *23. If j(x) • x" sin! for x X

> 0 and J(O)

ii!!

0, find j'(x).

n does j'(x) exist for all nonnegative values of x? continuous for all nonnegative values of x? *24. If J(x) .

;a;;

x" sin! for x X

For what values of

For what values of n isj'(x)

> 0 andj(O) "" 0, findf"(x). For what values of

n does J"(x) exist for all nonnegative values of x? J"(x) continuous for all nonnegative values of x?

For what values of n is

1

2. For the function e-;;, which behaves in a curious fashion near the origin, see Exercise 52, § 419. NOTE

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ROLLE'S THEOREM

§ 405]

93

•26. By mathematical induction establish Leibnitz's Rule: If u and v are Junctions of x, each of which possesses derivatives of order n, then the product also does and dn dxn (uv)

=

d"u dxn .

dn-lu dv (n) d"- U d2v d"v + (n) 1 d:i;n-1 . dx + 2 d:i;n----1 . dx2 + ... + b. (Ex. 12, § 408.) NoTE 3. If f(x) is differentiable in a neighborhood of x = a, then for any x in this neighborhood there exists a point~ between a and x (~ = a if x = a) such that (7)

f(x)

= J(a) + f'(~)(x

- a).

(Ex. 13, § 408.) ·

NoTE 4. If J(x) is differentiable in a neighborhood of x = a, and if h is sufficiently small numerically, then there exists a number 8 such that 0 < 8 < 1 and

(8) (Ex. 13, § 408.)

J(a

+ h) = J(a) + f'(a + 8h)h.

Example 1. Use the Law ot the Mean to establish the inequality sin x < x, for X > 0. Solution. If a = 0, h = x, andf(x) = sin x, the equation/(a + h) = J(a) + f'(a + 8h) ·h becomes sin x = cos (8x) ·x. If 0 < x ~ 1, then 0 < 8x < l and cos (8x) < 1, so that sin x < x. If x > 1, sin x ~ 1 < x. · Example 2.

Use the Law of the Mean to establish the inequalities

(9)

if h

1

>

-1 and h

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~

h + h < ln (1 + h) < h,

0.

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UNIVERSITY OF MICHIGAN

§ 406]

Solution.

f(a)

LAW OF THE MEAN

If a = 1, and /(x) = In x,

+ f'(a + 8h) ·h becomes In (1 + h)

=

97

the equation J(a 1

: eh·

If h

>

+ h)

=

0, the inequalities

< 8 < 1 imply 1 < 1 + 8h < 1 + h, and hence 1 ~ h < 1 ~ Bh < 1, whence. (9) follows. If -1 < h < 0, the inequalities 0 < 8 < 1 imply 1 > 1 + 8h > 1 1 1 + h > 0, and hence + h > + Bh > 1, whence (9) follows. 1 1

0

406. CONSEQUENCES OF THE LAW OF THE MEAN

· It is a trivial fact that a constant function has a derivative that is identically zero. The converse, though less trivial, is also true. Theorem I. A function with an identically vanishing derivative throughout an interval must be constant in that interval. Proof. If f(x) is differentiable and nonconstant in an interval, there are two points, a and b, of that interval wheref(a) ¢ f(b). By the Law of the Mean there must be a point ~ between a and b such that f'W

=

[f(b) - f(a)]/(b - a)

¢

o,

in contradiction to the basic assumption.

Theorem II. Two different-iable functions whose derivatives are equal throughout an interval must differ by a constant in that interval. Proof. This is an immediate consequence of Theorem I, since the difference of the two functions has an identically vanishing derivative and must therefore be a constant. A direct consequence of the definition of a derivative is that monotonic differentiable functions have derivatives of an appropriate sign (Ex. 14, § 404). A converse is stated in the following theorem:

Theorem m. If f(x) is continuous over an interval and differentiable in the interior, and if f(x) ~ 0 ( ~ 0) there, then f(x) is monotonically increasing (decreasing) on the interval. If furthermore f'(x) > 0 ( < 0), then f(x) is strictly increasing (decreasing). Proof. Let xi and x2 be points of the interval such that xi < x2. Then, by the Law of the Mean, there is a number xa between xi and x2 such that f(x2) - f(xi) = f'(xa)(x2 - xi). The resulting inequalities give the desired conclusions. An important relation between monotonicity of a function and the nonvanishing of its derivative is contained in the following theorem:

Theorem IV. A function continuous over an interval and having a nonzero derivative throughout at least the interior of that interval is strictly mono-

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98

DIFFERENTIATION

[§ 407

-tonic (over the interval), and its derivative is of constant sign (wherever it is defined in the interval). Proof. By Theorem III, it is sufficient to show that the derivative is of constant sign. Accordingly, we seek a contradiction to the assumption that there exist two points, a and b (a < b), of an interval throughout which the function f(x) has a nonzero derivative, and such that f'(a) and f'(b) have opposite signs. On the closed interval [a, b] the continuous function f(x) has a maximum value at some point Eof [a, b] and a minimum value at some point 11 of [a, b], where E¢ 11 (why must Eand,,, be distinct?). By Theorem I, § 405, neither Enor 11 can be an interior point of [a, b]. Therefore either E = a and 11 = b or E = band 11 = a. However, if E = a and 11 = b, f'(a) ~ 0 and f'(b) ~ 0 (why?), whereas if E = band 11 = a, f'(a) ~ 0 and f'(b) ~ 0. Either conclusion is a contradiction to the assumption thatf'(a) andf'(b) ~ve opposite signs. ·

Corollary. If a function has a nonzero derivative over an interval, the inverse function exists and is differentiable, and consequenay the formula

: = 1/~ is valid whenever its right-hand member exists over an interval. 407. THE EXTENDED LAW OF THE MEAN

In order to motivate an extension of the Law of the Mean to include higher order derivatives, we consider heuristically the problem of trying to approximate a given function f(x) in a neighborhood of a point x = a by means of a polynomial p(x). The higher the degree of p(x), the better we should expect to be able to approximatef(x) (assuming, as we shall for this introductory discussion, that f(x) not only is continuous but has derivatives of as high an order as we wish to consider, for x in the neighborhood of a). If p(x) is a constant (degree zero), a reasonable value for this constant, if it is to approximate f(x) for x near a, is clearly f(a). If p(x) is a polynomial of (at most) the first degree, it should certainly approximate f(x) if its graph is the line tangent to the graph of y = f(x) at the point (a, f(a) ), that is, if p(a) = f(a) and p'(a) = f'(a). In this case, p(x) has the form p(x) = f(a) + f'(a)(x - a). More generally, let us approximate f(x) in the neighborhood of x = a by a polynomial p(x) of degree ~ n with the property that p(a) = f(a), p'(a) = f'(a), · · · , p(a). We first express p(x) by means of the substitution of a + (x - a) for x, and subsequent expansion, in the form (1)

(2)

p(x)

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- a)

+ ··· + p..(x -

a)"

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UNIVERSITY OF MICHIGAN

§ 407]

EXTENDED LAW OF THE MEAN

99

Successive differentiation and substitution in the equations p(a) = f(a), = f'(a), · • · , p 0. More generally, for these values of h and O < p < l, (1 + h)P < 1 + ph. (Cf. Exs. 38-41.) 23. (1 + h)P > 1 + ph, for -1 < h < 0 or h > 0, and p > l or p < 0, (Cf. Exs. 38-41.)

21. b -b a

24.

< In~ < b -

·h

1

+ ht < Arctan h < h, for h > O.

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[§ 408

DIFFERENTIATION

102

26. z < Arcsin z < ~ • for 0 < z < 1. 1 - x1

+ h) ;:ii!+~• for h > -1. 27. . l°°s ax ~ cos bzl ;:ii la - bl, for z F 0. 26. Arctan (1

> 0 and x > 0.

28. sin pz < p, for p X

29. e4 (b - a) < e'> - ea < e'>(b - a), for a < b. 1 - e-as 30. ae-as < .;;;......_;;;_ < a, for a > 0 and x > 0. X

In Exercises 31-34, use the Extended Law of the Mean to establish the given inequalities. zl 31. COS X i1= 1 ,

2

32. cos x

zl >1- 2 , for x

33. x - : 34. 1

F 0.

< sin x < z, for x

+ X + 2zl
0.

*36. Use the trigonometric identity cos u - cos 11 = -2 sin ½(u to establish the inequality 1 cos a - cos b < bl - - -a, 2

+ 11) sin ½(u -

11)

for O ;:ii a < b.

Prove that cosaz-cosbx - - > 1, for 0 < x < - . (Cf. Ex. 36.)

11" X 4 38. Use the Law of the Mean to establish the following inequalities, for the designated ranges of p, assuming in each case that x > 0 and z F 1: p(z - l)zP-1 < zP - 1 < p(z - 1), for 0 < p < 1, p(x - l)zP-1 > zP - 1 > p(x - 1), for p < 0 or p > 1.

*39. By solving the left-hand inequalities of Exercise 38 for zP, establish the following inequalities for the designated ranges of p and h = x - 1 · l

~ : !:_ ph < (1 + h)P < 1 + ph,

l

+ h+_h ph > ( 1 + h) > 1 + ph,

for 0 < p < 1 and either -1 < h < 0 or h

1

p

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> 0.

"§ 408]

EXERCISES

103

> land either -1 < h < 0 or O < h < -p-1- 1, or for p < 0 and either h > 0 or - l < h < 0. for p

1 -p Conclude that for any real number p, the expression (1 + h)", for sufficiently small lhl, is between the two numbers (1 + h)/(1 + h - ph) and 1 + ph (being equal to them if they are equal). (Cf. Exs. 22-23.) •40. Show that if n is an integer greater than 1, and if either h > 0 or -1 < h < 0, then

(Cf. Ex. 39.) •41. Show that if n is an integer greater than 1, and if either h - l < h < O, then

>0

or

(Cf. Exs. 39-40.) •42. Let/{z) be differentiable, withj'(z) ~ 0 (f'{z) ~ 0), on an interval, and assume that on no subinterval does f'(z) vanish identically. Prove that /(z) is strictly increasing (decreasing) on the interval. •43. Establish the inequality sin z < z, for z > 0 (Example 1, § 405) by applying Exercise 42 to the function z - sin z. Similarly, establish the inequality tan z

> z for O < z
0.) NOTE 2. No conclusion regarding maximum or minimum of a function can be drawn from the vanishing of the second derivative at a critical value-the function may have a maximum value or a minimum value or neither at such a point (cf. Ex. 9, § 412).

A useful extension of the Second Derivative Test is the following: Theorem IV. Let f(x) be a function which, in some neighborhood of the point x = E, is defined and has derivatives f' (x), f" (x), · · · , f 1. If f'W = f"(E) = · · · = J(E) is negative or positive, and (ii) if n is odd, f(x) has neither a relative maximum value nor a relative minimum value at x = EProof. Owing to the vanishing of the derivatives of order E, according to the various possibilities for the sign of J(E) < 0, and suggest that the student furnish the corresponding details for the casej(E) > 0. By Exercise 7, § 404 (applied to the function J E, J E, whence f(x) < f(E) for x in the deleted neighborhood of E, and f(E) is a relative maximum value of f(x). If n is odd, the right-hand member of (1) is positive for x < E and negative for x > E, so that /(E) is neither a relative maximum value nor a relative minimum value of f(x) at x = E. x

+

Example.

Examine the function J(x)

x2

a

~8

for critical values of x and

relative and absolute maxima and minima. Find its maximum and minimum values (when they exist) for the intervals [l, 3), ( -1, 2), and [l, +00 ). 11

-3 -2 -1

0

1

2

3

FIG. 409

Solution. The graph of /(x) (Fig. 409) is symmetrical with respect to the y-axis, lim /(x) = 0, and from the definition of one-sided derivatives, the derivative from the right at x is - 00 • For x :p O,

= 0 is +00

and the derivative from the left at x 4(4 -

f'(x)

=0

X1 )

= 3-?'i (x2 + 8)2

The critical values of x are x = 0, 2, and -2. At x = 0 the function has both a relative and an absolute minimum -value of 0. At x = ±2 the function has a relative and an absolute maximum value oft~¼.= 0.1326 (approximately, to four decimal places). On the interval [1, 3) the function has a maximum of /(2) and a minimum of /(1) = t = 0.1111 (!(3) = 0.1224). On the interval ( -1, 2) J(x) has a minimum of /(0) = O, but no maximum. On the interval [1, +00) /(x) has a maximum of /(2), but no minimum.

At this point, the student may wish to do a few of the Exercises of § 412. Some that are suitable for performance now are Exercises 1-22. 410. DIFFERENTIALS

The student of Calculus becomes familiar with the differential notation, and learns to appreciate its convenience in the treatment of composite,

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108

[§ 410

DIFFERENTIATION

inverse, and implicit functions and functions defined parametrically, and in the technique of integration by substitution. Differentials also lend themselves simply and naturally to such procedures as the solving of differential equations. Such techniques and their legitimacy will not be discussed here. In the present section we restrict ourselves to basic definitions and theoretical facts. If y = f(x) is a differentiable function of x, we introduce two symbols, dx and dy, devised for the purpose of permitting the derivative symbol to be regarded and manipulated as a fraction. To this end we let dx denote an arbitrary real number, and define dy = d(f(x)) to be a function of the two independent variables x and dx, prescribed by the equation (1)

dy

=f'(x) dx.

The differentials dx and dy are interpreted geometrically in Figure 410. Although Calculus was the common invention of Sir Isaac Newton (1642-1727) and Gottfried Wilhelm von Leibnitz (1646-1716), the differential notation is due to Leibnitz. Its principal importance lies ultimately y

:r: :r: FIG. 410

in the fact that formula (1), initially true under the hypothesis that y is a remains true under any possible function of the independent variable reinterpretation of the dependence or independence of the variables x and y. This fact is given explicit formulation in the two theorems that follow. Before proceeding to these theorems we point out a further justification of the differential notation which is of so elementary a character that it is frequently overlooked, but which follows directly from the definition (1): If two variables are related by the identity relation, y x, their differdx (Ex. 23, § 412). entials are also related by the identity relation, dy

x,

=

=

Theorem I. If x = q,(y) is a differentiable Junction of the independent variable yin a certain interval, if q,(y) ¢ 0 in this interval, and if dy and dx denote the differentials of the independent variable y and the dependent variable x, respectively, related by definition by the equation dx = q,'(y) dy, then if

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§ 411]

APPROXIMATIONS BY DIFFERENTIALS

109

y = f(x) denotes the inverse function of x = ct,(y), these differential,s are also related by equation (1): dy = f'(x) dx.

Proof. By the Corollary to Theorem IV, § 406, the derivatives dx/dy and dy/dx are reciprocals, so that ct,'(y) = 1/f'(x). Therefore

and dy

dx

= f'(x) dx.

= (1/f'(x)) dy,

Theorem II. If y = f(x) is a differentiable function of the variable x, and if x is a differentiable function of the variable t, then if x and y are both regarded as dependent variables, depending on the independent variable t, their differentials are related by equation (1): dy = f'(x) dx. Proof. Let x = ct,(t) and y = i[t(t) = f(ct,(t)) denote the dependence of x and yon the independent variable t. Then by definition, dx = ct,'(t) dt and dy = i/t'(t) dt. By the Chain Rule (Theorem I, § 402), dy/dt

=

(dy/dx)(dx/dt),

or i/t'(t) = f'(x)ct,'(t) = f'(ct,(t) )ct,'(t), so that dy

= i/t'(t) dt = f'(ct,(t) )(ct,'(t) dt) = f'(x)

dx.

This completes the proof and shows that Theorem II is, in essence, simply a reformulation of the Chain Rule. 411. APPROXIMATIONS BY DIFFERENTIALS

In order to compare the differentials dx and dy, on the one hand, and the increments ax and t:,.y, on the other, we observe that whereas the differentials can be associated with the tangent to the curve y = f(x) (Fig. 410), the increments are associated with the curve itself. It is often convenient to regard the real number dx as an increment in the variable x : dx = ax. In this case the quantities just discussed find their geometric interpretation in Figure 411. Throughout this section we shall identify dx and ax. It y

FIG. 411

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[§ 411

DIFFERENTIATION

110

shonld be noted, however, that under this assumption dy and t:.y are not in general the same. Although in theory the increment t:.y is a simpler concept than the differential dy, in practice the differential is usually easier to compute than the increment, and dy is often a useful approximation to t:.y. The statement that for a numerically small increment dx = llz, the quantity dy is a good approximation to the quantity t:.y, has a precise formulation, and means much more than the statement that dy and t:.y are approximately equal (which is trivially true since they are both approximately equal to zero). The precise relation (under the assumption that the given function is differentiable at the point in question) is given by equation (4), § 402, rewritten in the form (1)

where, for a fixed x, E = E{llz) is an infinitesimal function of /lz (E-+ 0 as llz-+ 0). Using equation (1) we can formulate the statement that (for small ldxl = lllzl) dy is a good approximation to t:.y by means of the equation . t:.y - dy -- 0, l1m

(2)

.v-0

llz

or, in case dy is not identically zero, by the equation lim ~d = 1.

(3)

.v-0

E:1:ample 1. If y

y

(Ex. 24, § 412).

= J(:r:) = zl, thent:.y = J(:r: + t:.:r:)

+ t:.x1, and dy = 3:r:' dx. 3x t:.:r: + t:.:r:1, equation (2) 1 tion (3) becomes lim 3z

.v-0

- J(:r:)

= 3:r:'llz + 3:r:t:.x2

Therefore, in equation (1), the function E(t:.x) is takes the form lim {3x t:.z + llz') = 0, and equa.v-0

+ 3z /lz, + ilz' = 1. 3:r;

E:1:ample 2. Prove that for numerically small h the quantity closely approximated by 1 + ½h.

'V1+k is

Let y = J(:r:) = ~. and let z change from 1 to 1 + h. Then dz h dx = t:.x = h, and dy = v; = - Therefore the value of y changes from 2 2 Vl = 1 by an amount approximately equal to dy = ½h, The new value of y (which is Vl + h) is therefore approximately equal to 1 + ½h. (Cf. Ex. 41, Solution.

§ 408.)

The Extended Law of the Mean provides a measure of the accuracy of approximation of dy for t:.y. Under the assumption that J"(x) exists in the neighborhood of x = a, the Extended Law of the Mean ((12), § 407, for n = 2) ensures the existence of a number 8 between O and 1 such that (-1)

f(a

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§ 412]

EXERCISES

111

for numerically small h. Expressed in terms of differentials, with Ax 11y = f(a + h) - f(a) and dy = f'(a)h, (4) becomes

Ay - dy

(5)

=

½f"(a

+ fJAx)Ax

2

= h,



If Bis a bound for the absolute value of f"(x) (that is, lf"(x)J ~ B) for x in a certain neighborhood of x = a, and if Ax is so restricted that a Ax (and therefore a+ fJAx) is in this neighborhood, then (5) gives the inequality

+

(6)

Enmple 3. Find an estimate for the accuracy of the approximation established in Example 2, if lhl ;:;i 0.1. 1 Solution. If/(,;) = v';, then J'(x) = \_, f"(x) = , ' and formula 2v,;

4,;v,;

(5) becomes Ay - dy

(7)

=

(VI

+h -

1) - (½h)

= -

ht

.

8(1 + fJh)i If h is positive (whether h ;:;i 0.1 or h > 0.1), 1 + (Jh > 1 and therefore the third member of (7) is between -h1/8 and 0. We thus have the inequalities

< ~ < 1 + ½h, h > 0. 1.058 < v'1.12 < 1.06.) h < 0, 1 + (Jh > 0.9, and the third member of

(8)

1

(Illustration: If -0.1 ;:;i -h2/6 and 0. (9)

+ ½h -

th1

(7) is between

Therefore

1

+ ½h -

¼ht 1.

0), x near 0.

In Exercises 49-60, verify the given inequalities, which give estimates of the errors in the approximations of Exercises 37-48.

d9. 10.48808 < Vll0 < 10.48810. *60. h - ½h' < In (1 + h) < h, h > 0, h - th' < In (1 + h) < h, -0.1 *61. -0.0622 < In 0.94 < -0.06.

f < sin z < z, z > 0.

*62. z -

!:1y - dy =

*63. z + ½z3 < tan z 4y - dy = i(l *54.

½- ½V3 (z

h

< 0.

Hint: Show that

-¼ cos (8x) z',

0

< 8 < 1.

< x + ½zl, 0 < z < 0.1.

Hint: Show that + 3t )(1 + t )z', where t = tan (8x), 0 < 8 < 1. 2

1

!11")1 ~ cos z ~ ½ -

- ir) - ½(x -

0 :i! X :i! tr. z, 0 < x < ½r.

½V3 (z - ir),

< Arctan z < + !! - (n - 2l)ht < ~1 + h < 1 + !!, h > 0·

*66. z - iz' •66. 1

~

n

1

n

2n

+ !! n

(n - 1 l)hl n


O+ hm -g '(l/t),• 2 (multiplying and dividing by - t-2) d

=

,

dt f(l/t)

lim - - t--->O+ 1._ g(l/t) dt ·

=

.

f(l/t)

hm t---t0+ g(l/t)

=

JU,,.\)

lim ~ -

..-+ .. g(x)

The next-to-the-last equality is true by Case 1. The sequence of equalities implies that the limit under consideration exists and is equal to L. Example 1.

~ si: x = ~ co: x =

1.

(Cf. Ex. 16, § 604.)

Example 2.

. sinx - x IIm - - 8 z-0

:r:

. -cosx . cosx -1 . -sinx = 1Im .c....;;..;_--'--- = 11m - - = 1Im - - = 2 z-0

3:r:

z-0

6:r:

z-0

6

1

6

In this case l'Hospital's Rule is iterated. The existence of each limit implies that of the preceding and their equality. e-"' -e-"' e-• Example 3. lim - = lim - - = lim - ! 1 1 ..-+ .. ..-+.. z-+• -2 ;i;

;i;2

;i;3

Things are getting worse! See Example 1, § 415. It is important before applying l'Hospital's Rule to check on the indeterminacy of the expression being treated. The following example illustrates this. Example 4. A routine and thoughtless application of }'Hospital's Rule may yield an incorrect result as follows: . 2:r:2-- -;i;IIm -1 = l'1m 4z ---1 = l'1m -4 = 2 . z-1 z2 - x z-1 2z - 1 z-+1 2 The first equality is correct, and the answer is obtained by direct substitution of 1 for z in the continuous function (4z - 1)/(2:r: - 1), to give the correct value of 3. 415. THE INDETERMINATE FORM r:,:,

/r:o

The symbol 00 /00 indicates that a limit is being sought for the quotient of two functions, e,ach of which is becoming infinite (the absolute value

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118

DIFFERENTIATION

[§ 415

approaches +co) as the independent variable approaches some limit. L'Hospital's Rule is again applicable, but the proof is more difficult.

Theorem. L'Hospital's Rule. If f(x) and g(x) are differentiable Junctions and g'(x) ~ 0, for values of x concerned, if limf(x) = lim g(x) = oo, and if lim ~:~=~

=L

. &} 11m g(x)

= .

(finite,

+co,

-co, or co),

then

Case 1. Case 2. Case 3. Case 4. Case 5. Case 6.

L

x--+ a+. (Ex. 35, § 417.) x--+ a-. (Ex. 36, § 417.) x--+ a. (Ex. 37, § 417.) x --+ +co. (Proof below.) x--+ -co. (Ex. 38, § 417.) x--+ co. (Ex. 39. § 417.)

* Proof of Case 4.

Observe that whenever xis sufficiently large to prevent the vanishing of f(x) and g(x), and Ni is sufficiently large to prevent the vanishing of g'(E) for E > Ni, the generalized mean value theorem guarantees the relation (1)

and therefore, (2)

for x > Ni and a suitable E between x and Ni. First choose Ni so large that if E > Ni, thenf'(E)/g'(E) is within a specified degree of approximation of L. (If L is infinite, the term approximate should be interpreted liberally, in accordance with the definitions of infinite limits, § 206). Second, using the hypotheses that lim lf(x)I = lim lg(x)I = +co, let N2 be so

-+•

-+•

large that if x > N2, then the fraction [1 - g(Ni)/g(x)]/[1 - f(Ni)/f(x)] is within a specified degree of approximation of the number 1. In combination, by equation (2), these two approximations guarantee that f(x)/g(x) approximates L. This completes the outline of the proof, but for more complete rigor, we present the "epsilon" details for the case where L is finite. (Cf. Ex. 40, § 417.) Let L be an arbitrary real number and E an arbitrary positive munber. We shall show first that there exists a positive number a such that IY - LI < ½E and lz - 11 < a, imply lyz - LI < E. To do this we use the

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§415]

T H E I N D ET ERM I N A TE FO RM rs, /rs,

119

triangle inequality to write

lyz - LI

;;;i

If IY - LI < ½E, IYI inequalities imply

lyz - YI + IY - LI = IYI ·lz

- 11 + IY - LI,

< ILi + ½E, so that if 8 = ½E(ILI + ½E)-1 the assumed

i=

lyz - LI < (ILi + ½E) 2(ILI E+ ½E) +

The rest is simple. First choose N1 such that

which is the desired result.

~>Ni implies and second choose N,

E,

1~:m - LI < ½E,

> Ni such that

I

. . 11 - g(N1)/g(x) x > N 2 imp11es 1 _ f(Ni)/f(x) - 1 < 8. Then, by (2), X

E,

= 0 for any real a.

Example 1. Show thll,t Jim X"

-+• es

Solution.

1rc:~ - LI
N2 implies

If a :ii O the expression is not indeterminate. X"

Then Jim · -

-+• es

=

Assume a

>

0.

•-1

lim ~ . and if this process is continued, an exponent for

-+• es

x is ultimately found that is zero or negative. This example shows that es increases, as z-+ +co, faster than any power of x, and therefore faster than any polynomial. Example 2.

Show that Jim In :,;

-+• :,;•

= 0 for any a > 0.

1

Solution.

Jim In:,;

-+•

X"

=

Jim _x_

-+•

=

a:,;a-1

Jim _!_

-+•

=

0.

(Cf. Ex. 7, § 602.)

ax 0.

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EXERCISES

§ 417] 1

If y = (1

Solution.

Therefore y

= e10 11 -

+ ax);,, =

eo

In y

- 0.

X

1.

Show that Jim (1

Example 6.

+ ax)

In (1

=

121

~

!

+ ax)r = e-. 1

+ ax);, . In (1 + ax) 1Im =

If a F 0 and if y

Solution.

= (1

. . a 1Im - + IIm 1n y = = a, .r.....o ~ x z.....o 1 ax 10 and y = e 11 - ea. (Cf. Ex. 10, § 602.) ! Example 7. Find Jim x e"'. ~+ ! e"' Solution. If this is written Jim - , differentiation leads to the answer. :P--tO+ 1 X 1

+oo.

However, the limit can be written Jim ~t =

1-+•

3 Jim (csc2 x - x csc 3 x) = Jim sinx-x --'--• -.x- 3 :P--tO x3 sm x

Example 8.

z.....O

!) ·(~ si: xr = - !·

=(-

by Example 2, § 414, and the continuity of the function x3 (the limit of the cube is the cube of the limit). 12 sin; In x (x3 + 5 )(: _ l)

~

Example 9.

~

12 sin;

xi+/

417. EXERCISES

In Exercises 1-30, evaluate the limit. l. lim 3x1 + x - 14 ,,_2 x 2 - x - 2

2 •

3. lim

4:. lim cos p:z.

z-1 X

. 5. I1m

2

lnx

+X

2

1+

z-1

½z

2

COS X .;;..;;_;..;..;__..;;......,_.o..;.;_.

x'

z.....o



2

X -

1

G. lim In (1 + x) - x, z.....O cosx-1

sin (1/x) . 8 lim • ,,_.. Arc tan (1/x)

7. J i m ~ ,,_,, tan2 4x

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DIFFERENTIATION

122 9. Jim a• - 1. - 0 ~ -1

[§ 417

10. lim

tan z - z .

-o Arcsinz - z

. 8x1 -5z1 +1 +X • 3X 1 z-•

11. hm

l2. lim tan z - 6 ..-,.. sec z + 5

13_ Jim ln sin 2z ..-1.-- ln cos x

l'- lim In ~in z .

16. 17. 19. 21.

-Insm2z

Jim coeh x (cf. § 607).

16. lim ln (1 - 2z) OP-+6- tan rx

(ln z)•, n Jim

> 0.

18.

> 0.

20. lim (z - a) tan rz. .,...... 2a 1 22. lim -]• -i+ z - 1 lnx

..-+•

es

z-+•

X

lim z(ln z)•, n

-o+

!.If.. [ztanz -

~secx}

lim ~ a

-+• ;,;-

lim (1

-+•

+ X1);,

27.

+ 2 sin z)

2" Jim (1 -0

1

26.

00

t "'.

1

+ eta);, [In (1 + z]•.

lim xii";,

28. lim (z

lim z"'•(z.,. • x).

28.

-o+

-o+

> 0.

1, b

[-z- - -

1

23.

>

-0

1

29. lim (cos 2z);._ -0

lim

-o+

30. 2!:+ [c1 l~zz)' -

In

1~

xl

Prove Case 2 of l'Hospital's Rule, § 414. Prove Case 3 of l'Hospital's Rule, § 414. Hint: Make sensible use of 1 and 2. Prove Case 5 of l'Hoepital's Rule, § 414. 3'- Prove Case 6 of l'Hospital'e Rule, § 414. 36. Prove Case 1 of l'Hoepital'e Rule, § 415. Hint: Apply Case 4 (cf. proof of Case 4, § 414). •36. Prove Case 2 of l'Hospital's Rule, § 415. •37. Prove Case 3 of l'Hospital's Rule, § 415. •38. Prove Case 5 of l'Hospital'e Rule, § 415. •39. Prove Case 6 of l'Hospital's Rule, § 415. For Case 4 of }'Hospital's Rule, § 415, supply the ''epsilon" details for the case L = +ao . 41. Prove that the forms (+0)+•, (+ao)+•, and a+• (where a> 0 and a ~ 1) are determinate. What can you say about ( +o)- •? ( +ao )- •? (+0)•? (+oo)•? 42. Show by examples that all of the forms of § 416 are indeterminate, as stated. 43. Criticize the following alleged "proof" of !'Hospital's Rule for the form 0/0 as x -+ a+ : By the Law of the Mean, for any z > a, there exist f1 and f1 between a and x such that /(z) - /(a+) = f'{f1) and g(x) - g(a+) ""' g'(f1). 31. 32. Cases 33.

*'°·

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§ 418]

CURVE TRACING

123

Therefore

&1 = [(x> - [(a+>= Bhl_tlill = r g(:c)

g(:c) - g(a+)

g'(~s)

g'(a+)

fJ.:2

~+ {/(:c) •

418. CURVE TRACING

It is not our purpose in this section to give an extensive treatment of curve tracing. Rather, we wish to give the reader an opportunity to review in practice such topics from differential calculus as increasing and decreasing functions, maximum and minimum points, symmetry, concavity, and points of inflection. Certain basic principles we do wish to recall explicitly, however. (i) Composition of ordinaf.es. The graph of a function represented as the sum of terms can often be obtained most simply by graphing the separate terms, and adding the ordinates visually, as indicated in Figure 412. 11

11-cosx+sin~ FIG. 412

(ii) Dominant terms. If different terms dominate an expression for different values of the independent variable, the general shape of the curve can often be inferred. For example, for positive values of x, the function

x

+ !X is dominated by the second term if x is small and by the first term if

x is large (Fig. 413). (iii) Vertical and horizontal asymptof.es. A function represented as a quotient f(x)/g(x) of continuous functions has a vertical asymptote at a

point a where g(a) = 0 and f(a) ¢ 0, If limf(x)/g(x), as x becomes infinite ( +00, - 00, or oo), exists and is a finite number b, then y = b is a horizontal asymptote. (Fig. 414.) (iv) Other asymptof.es. If f(x) - mx - b approaches zero as x becomes infinite ( +oo, -00, or oo ), then the line y = mx +bis an asymptote for

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DIFFERENTIATION

124

[§ 418

11

FIG. 413

the graph of f(:x). For the function x + e"', for example, the line y = x is an asymptote as x -+ - oo • (Fig. 415.) (v) Two facfms. Certain principles used for functions represented as sums have their applications to functions represented as products. The functions e-a:,; sin bx and e-a:,; cos bx, useful in electrical theory, are good examples. (Fig. 416.) Vanishing factors often determine the general shape of a curve in neighborhoods of points where they vanish. For example, the graph of

I

I I I

I I

I

--~---=---~-~-==-~---~--~---1 ----r-------------I 1

-2,

X

12

I I I I

I I I I

I I I

I

I I I

I FIG. 414

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§ 418]

125

CURVE TRACING

y

t I I I

I I

/

I I I I

I

// /

/

/

/

,,"

/

//

'/

/

/

FIG. 415

y2 = x2 (2 - x) is approximated by that of y2 = 2x 2 for x near Oand by that of y2 = 4(2 - x) for x near 2. (Fig. 417.) A further aid in graphing an equation like that of Figure 417, of the form y2 = f(x), is graphing the function f(x) itself to determine the values of x for whichf(x) is positive, zero, or negative, and hence for which y is doublevalued, zero, or imaginary. (Fig. 418.) (tn) Parametric equatiom. If x = f(t) and y = g(t), we recall two formulas:

' - rJ:u - [JD_ .

(1)

y - dx - f'(t)'

0 there exists a > 0 BUCh that for any net m. of norm less than 8 and any choice of points x; BUCh that a.-1 ~ Xi ~ ai, i = 1, 2, · · · , n, the inequality _'Ef(Xi) &J;i - JI < I ,-1

(3)

holds.

E

NoTE 1. Closely though Definition I may resemble the definition of the limit of a single-valued function of a real variable as the independent variable approaches 0, the type of limit just introduced should be recognized as a new concept. Although, for a given net m. and points x1, x2, • · • , x,., the sum (1) is uniquely determined, the limit (2) is formed with respect to the norm alone as the independent variable. For a given positive number p < b - a there are infinitely many nets m. of norm lffi'.I equal top, and for each such net there are infinitely many choices of the points x1, x2, • • · , x,.. In other words, as a function of the independent variable lffi'.I, the sum (1) is an infinitely many valued function, t and it is the limit of such a function that is prescribed in Definition I. It should not be forgotten, however, that each sum (1) appearing in the inequality (3) is simply a number obtained by adding together a finite number of terms.

Definition II. A function f(x), defined on [a, b], is integrablet there if and only if the limit (2) exists (and is finite). In case the limit exists it is called the definite integralt of the function, and denoted

l

(4)

Definition m.

If b

b

f(x) dx = lim

\\ll\.....O i=l

< a, ibf(x) dx = -

(5)

" E f(x;) &xi.

£f(x)

dx,

in case the latter integral exists. Furthermore, (6)

Certain questions naturally come to mind.

Does the limit (2) always

t The expression "infinitely many valued" should be interpreted here to mean "poSBibly infinitely many valued" since for any constant function/(x), the sums (1) can have only one value (cf. Theorem VII). It can be shown that for any function f(x) that is not constant on [a, b], and for any positive number 8 less than half the length of the interval [a, b], the sum (1), as a function of Im.I, is strictly infinitely many valued for the particular value Im.I = 6. · t The terms Riemann integrable and Riemann integral are also used, especially if it is important to distinguish the integral defined in this section from some other type, such as the Riemann-Stieltjes integral (§ 517) or the Lebesgue integral (not discussed in this book). When a definite integral is to be distinguished from an improper integral (§§ 511-515), it is customary to call it a 'fYTOper integral.

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§ 501]

133

THE DEFINITE· INTEGRAL

exist? If it does not always exist, under what circumstances does it exist'? When it does exist is it unique? Let us remark first that the limit (2) never exists for unbounded functions. In other words, every integrable function is bounded. To see this, let m, be an arbitrary net and letf(x) be unbounded in the kth subinterval [a1:-1, a1:], Then whatever may be the choice of points x, for i ¢ k, the point Xk can be picked so that the sum (1) is numerically larger than any preassigned quantity. On the other hand, not all bounded fur1ctions are integrable. For example, the function of Example 6, § 201, which is 1 on the rational numbers from 0 to 1 and 0 on the irrational numbers from 0 to 1, is not integrable on [0, 1]. For, no matter how small the norm of a net m, may be, every subinterval contains both rational and irrational points ((v), § 112, and Ex. 2, § 113) and the sums (1) can be made to have either the extreme value 1 (if every point x, is chosen to be rational) or the extreme value 0 (if every point x, is chosen to be irrational). The limit (2), then, cannot exist for this function (let E ½). One answer to the question of integrability lies in the concept of continuity. The function just considered, which is not integrable, is nowhere continuous. At the opposite extreme is a function which is everywhere continuous (on a closed interval). We shall see that such a function is al'l!)ays integrable (Theorem VIII). Between these two extremes are bounded functions which are somewhere but not everywhere continuous. It will be seen that such functions are certainly integrable if they have only finitely many discontinuities (Theorem IX), that they may be integrable even with infinitely many discontinuities (Example 2, § 502), and that a criterion for integrability lies in the intriguing concept of continuity almost everywhere (Ex. 54, § 503). We proceed now to the establishment of some of the simpler properties of the definite integral.

=

Theorem I.

E" f(x,) t:.x, exists, the limit is unique.

If lim

l!nl--oO • - 1

Proof. Assume that

" " lim Ef(x,).t:.x, = I and lim Ef(x,) /lx; = J,. l!nl--oO •- 1

l!nl--oO •- 1

where I> J, and let E = ½(I - J). Then there exists a positive number 8 so small that for any net m, of norm less than 8, and for any choice of points X1, Xi, • • • , x,., the following inequalities hold simultaneously:

I But this implies I -

E

Digitized by

E

" f(x,) t:.x, < J + E. ½(I

- J).

(Contradiction.)

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134

[§ 501

INTEGRATION

Theorem II. If f(x) and g(x) are integrable on [a,b], and if f(x) there, then [f(x) dx Proof.

Let I

=

~

[f(x) dx and J

~

g(x)

£

g(x) dx.

= J: g(x) dx, assume I > J, and let

E = ½(/ J). Then there exists a positive number 8 so small that for any net~ of norm less than 8 and any choice of points x1, X!, • • • , x,., the following inequalities hold simultaneously:

" I:

i-1

g(x,) llx,

0 be such that l~I

< 8 implies

I'i:, f(x,) •- I

;a!!

0, .and if e > 0 is given,

llx, -

JI < e/lkl, and there-

fore

Theorem IV. If f(x) and g(x) are integrable on [a, b], then so are their sum and difference, and (8)

j)J(x)

Proof. 8

Let l

± g(x)] dx = J:!(x) dx ± ibg(x) dx.

=lbf(x) dx and J =.['g(x) dx, and if e > 0 is given, let

> 0 be such that l~I < 8 implies simultaneously the inequalities

I'f:. f(x~) llx; - 1\ < ½e, It g(x,) ~, - JI < ½e, 1•1

,-1

and hence, by the triangle inequaiity,.

Iit1 [/(x,) =':= g(x,)] llx; -

(l

± J)

I

~ j•=.E1f(x,) llx; -

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11 + 1.E g(x;) Ax; ,-1

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UNIVERSITY OF MICHIGAN

JI


(9)

Let lf(x)I

Proof.

dx - J:f dx

< K for a~ I a

+ltf dx.

x ~ c, let

_ff(x) dx and J a J:f(x) dx,

and let E > 0 be given. Let abe a positive number less than E/4K and so small that for any net on [a, b] or [b, c] with norm less than aany sum of the form (1) differs from I or J, respectively, by less than ¼E. We shall

I

J)I

< aimplies 'f:, /(xi) Axi - (/ + < E. Accord•-1 net m. let a1:-1 < b ~ a1: (the kth subinterval is the first

show that for [a, c], Im.I

ingly, for such a containing the point b), and write the sum (1) in the form

The following sum,

[1:£ f(xi) Axi + J(b)(b i-1 1

S'

a

a1:-1)]

+ [!(b)(a1: -

b)

+ i-k+l E /(xi) Axi],

can be considered as made up of two parts, which approximate I and J, each by less than ¼E. Thus IS' - (/ + J) I < ½E. On the other hand, IS - S'I - l/(x1:) - f(b)I Ax1: < 2K(E/4K) = ½E. Therefore, IS - (I+ J)I ~ IS - S'I

+ IS' -

(I+ J)I < ½E

+ ½E -

E.

NoTE 2. By virtue of Definition III, the relation (9) is universally true whenever the three integrals exist, whatever may be the order relation between the numbers a, b, and c. For example, if c < a < b, then

J:b/(z) dz

0

= J: /(z) dz +.£b/(z) dz .

. Hence

i

c/(z) ch

0

= - f /(z) dz = fb/(z)

Jc

J,. .

dz - fb/(z) dz

Jc

= fb/(z) dz +., fc/(z) dz.

J,..

Jb

By mathematical induction, the relation (9) can be extended to an arbitrary number of terms: where ao, a1, • • • , a,. are any n + 1 real numbers, and where every integral of (10) is assumed to exist. (The student should satisfy himself regarding (9), by considering other order relations between a, b, and c, including poBSible equalities of some of these letters, and he should give the details of the proof of (10). (Cf.

Ex. 1, § 503.)

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[§ 501

INTEGRATION

136

Theorem VI. If the values of a function defined on a closed interval, are changed at a finite number of points of the interval, neither the integrability nor the value of the integral is affected. Proof. Thanks to mathematical induction, the proof can (and will) be reduced to showing the following: If f(x) is integrable on [a, b], with integral I, and if g(x) is defined on [a, b] and equal to f(x) at every point of [a, b] except for one point c, then g(x) is integrable on [a, b] with integral I. For any net

m.,

n

the terms of the sum

E g(xi) tui must be identical with

i=l n

the terms of the sum

E f(x;) tu; with the exception of at most two terms

i=l

(in case Xi-1 =

Xi

Thus, for a given

= c

E

for some i). Therefore

> 0, let o be a positive number less than E/4(\f(c)i

and so small that

+ lu(c)I)

\m.\ < o implies \;tJ(x;) tu;

-

II< ½E.

Then

\m.\ < o

implies

Iit tui - I~ I;ti g(xi)

I

tu;I

g(x;) ll.x; - f(xi)

I

+ ,-1 .E f(xi) tu, -

Ij < ½E + ½E = E.

NOTE 3. Theorem VI makes it possible to define integrability and integral for a function which is defined on a closed interval except for a finite number of points. This is done by assigning values to the function at the exceptional points in any manner whatsoever. Theorem VI assures us that the result of applying Definition II is independent of the values assigned. Since the assignment of values does not affect the value of the integral, where it exists, we shall assume that the definition is extended to include such functions even though they remain undefined at the exceptional points.

Theorem VII. If f(x) is conswnt, f(x) f (x) is integrab'le there and

=

k, on the interval [a, b], then

ibf(x) dx = k(b - a). Proof.

For any net m.,

fl

E f(x,)

i-1

fl

tu,

=k L

i-1

x.

= k(b

- a).

For the sake of convenience and accessibility we state now the three best-known sufficient conditions for integrability (the first being a, special case of the second). The proofs are given in § 502 and Exercise 33, § 503.

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§

T H E D·E F I N I TE I N T EG RA L

501]

Theorem there.

vm.

137

A function continuous on a closed interval is inregrable

Theorem IX. A function defined and bounded on a closed interval and continuous there except for a finite number of points is infR,grable there. Theorem X. A function defined and monotonic on a closed interval is inregrable there. Prove that

Example.

fo\~

dx

= ½lr

if b

i\x dx = ½(b

2

> 0,

-

and, more generally, that

a1 ),

where a and b are any real numbers. Solution. Let us first observe that since the function /(x) = x is everywhere continuous the integrals exist. The problem is one of evaluation. To evaluate

fob x dx, where b > 0, we form a particular simple net by means of the points ao

= 0,

and choose x,

a1

= b/n, · · · , a,

a,, i =

=

a

ib/n, · · · , a,.

a

nb/n

= b,

n

1, 2, · · · , n.

The sum

L f(x,) .1x, becomes ,-1

. ~ = ~ [l + 2 + ... + n] = ~ . n(n + 1), n2 2 n n2 the last equality having been obtained in Exercise 12, § 107. Therefore dx = lim ½b2 n + 1 = ½b2, as stated. It is left to the student to Jo ......+.. n

+ ... + nb] [ n£ + 2b n n

(bx

show, first, that

.£bx dx = ½lr if b ;a; 0

(cf. Exs. 7-8, § 503) and then, by

using formula (9) of Theorem V and formula (5) of Definition III, that

.£bx dx = ½(b2 -

a1 ).

Other evaluations of this type are given in Exercises 15-21, § 503.

In conclusion we present a useful basic theorem (give the proof in Ex. 5, § 503). Theorem XI. First Mean Value Theorem for Integrals. If f(x) is continuous on [a, b] there exists a point~ such that a < ~ O, the existence of step-functions a(x) and T(x) satisfying (1) and (2). Let us observe initially that for any step-functions satisfying (1), [ a(x) dx

~ i\cx) dx

(Theorem II, § 501), so that the leaat upper bound I of the integrals i"a(x) dx, for all a(x)

~ f(x), and the grootest lower bound J of the integrals

ibT(x) dx, for all T(x)

~ f(x),

both exist and are finite, and I

~J

(supply

the details). Because of (2) and the arbitrariness of E > 0, integrals of the form [ a(x) dx and [ T(x) dx can be found arbitrarily close to each other, so that I cannot be 'less than J, and therefore I = J. Now let E > 0 be given, and choose step-functions a(x) and T(x) satisfying (1) and (2) and such that [ a(x) dx

> I - ½E and [

Then choose a > O so that

T(x) dx

< J + ½E.

1ml < a implies

I: a{x,) ~. > J.f" a(x) dx -

i-1

itl T(Xi)

~i

0. Since there are only a finite number of positive integers ;:ii 1/E, there are only a finite number of rational numbers p/q of the interval [O, i] for which J(p/q) ~ E. If 8 > 0 is chosen so small that. the interval (x0 - 8, :eo + 8) excludes all of these finitely many rational points, then Ix - xol < 8 implies J(x) < E, whether x is rational or irrational, and continuity at x0 is established. If :eo is rational and if J(x.) is redefined to be 0, continuity at xo is established by the same argument. Finally, if E > 0, since J(x) ~ 0 all that remains to be shown is the existence of a step-

function T(x)

~ J(x)

such that fo\(x) dx

< E,

Let

X1, X2, • • • ,

x,. be the ra

tional points of [O, 1] such that J(x,) ~ ½E, i = 1, 2, • • · , n, and let I, be a neighborhood of x, of length 0, xi for x < 0, undefined for x = 0. By means of these examples show that the derivative of a function which is neither even nor odd may be either even or odd. On the other hand, show that if the domain of a differentiable function/(x} is an open interval of the form ( -a, a) or ( -00, +00) and if f'(x) is odd, then /(x) is even; similarly, that if f'(:z;) is even, then /(x) plus a suitable constant is odd. (Cf. Exs. 7-12.) Hint: J(x) and/( -x) h~ve the same derivative, and must therefore differ by a constant. 14. If /(x) is the bracket function of Example 5, § 201, f(x) = [x], evaluate

.fo J(x) dx and .fo /(x) dx. 3

1

More generally, if n is a positive integer, evaluate

.fo"!(x) dx.

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EXERCISES

§ 503]

15. Prove that .£bx2 dx

=

l(b3 - a1 ).

145

(Cf. the Example, § 501, and Ex. 13,

§ 107.)

16. Prove that .£\;a dx = ¼(b4

-

a 4}.

(Cf. the Example, § 501, and Ex. 14,

-

a 5 ).

(Cf. the Example, § 501, and Ex. 15,

§ 107.)

17. Prove that .£bx4 dx

= !(b5

§ 107.) •18. Prove that if mis a positive integer, then (bx"' dx = - 1- (b•+I - a"'+i). )a m +1 (Cf. the Example, § 501, and Ex. 42, § 107.)

= cos a

•19. Prove that .£bsin x dx § 501, let b

- cos b.

Hint: As in the Example,

> 0 and a = 0, and write

fo sm x dx =.

. ( sm . -b 11m n

b •

Multiply each term by 2 sin

n--++oo

+ · · • + sm . -nb) · -·b n

n

:n' and use the identity

2 sin A sin B

= cos (A

- B) - cos (A + B)

to obtain .(bsin x dx =

lim [(cos.!_ - cos ab) + (cos ab - cos 5b) + ... n--++.. 2n 2n 2n 2n +(cos(2n-l)b_cos(2n+l)b)]· b. 2n 2n 2n sin

:n

•20. Prove that .£bcos x dx

•21. Prove that .£be" dx

= sin b - sin a. (Cf. Ex. 19.)

= fl>

-

e-.

•22. Prove the Trapezoidal Rule for approximating a definite integral: If /(x) is integrable on [a, b], if [a, b] is subdivided into n equal subintervals of length 1 points of subdivision, xo, x1, x,, · · · , x,., llx, and if the values of f(x) at then are Yo, Y1, Y2, • • • , y,., respectively, then

+

(bf(x) dx =

)a

lim (½Yo+ Y1 + Y2 + · · · + Yn-1 + ½Y..) llx.

n~+•

Also prove the following estimate for the error in the trapezoidal formula, assuming existence of f"(x) on [a, b]: ~ b-a )a f(x) dx - [½Yo+ Y1 + · · · + !y,.] llx = - l 2 J"(E) llx2, where a < t < b. Hints: For the first part, write the expression in brackets in the form (y1 + · · · + y,.) + ( ½Yo - !y,.). For the second part, reduce the problem to

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INTEGRATION

146

that of approximating .£"/(z) dz by a single trapezoid, and assume without loss The problem reduces to that

of generality that the interval [a, b] is [ -h, h]. of evaluating K, defined by the equation

J) 0 on [a, b], then 1//(x) is integrable on [a, b]. Hint: Use Ex. 34 and

I: l,c1x,·) - f( x,1.,)1 Ax, = ,-1 t 17 O? *37. Prove that the quotient of two integrable functions is integrable if the second (denominator) function is bounded from zero (cf. Ex. 36). *38. Prove that if J(x) is integrable on [a, b], then so is I/(x)I. Construct an example to show that the converse implication is false. Hints:

L

II/(x,)I - l/(x,')II Ax; ~

L

lf(x,) - /(x/)I Ax;

(cf. Ex. 34). Consider a function like that of Example 6, § 201, with values ±1. *39. If f(x) and g(x) are defined on a set A, the functions M(x) = max [f(x), g(x)] and m(x) = min [/(x), g(x)] are defined, for each x in A, to be the larger and smaller, respectively, of the two numbers f(x) and g(x) {equal to them if they are equal). (See Fig. 507.) Prove that if J(x) and g(x) are integrable on [a, b], then so are M(x) and m(x). I/int: M(x) = ~ [f(x) + g(x)] + ½lf(x) - g(x)I, m(x) = ½[/(x) + g(x)] - ½l/(x) - g(x)I. (Cf. Ex. 38.)

FIG. 507

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EXERCISES

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*40. If /(z) is defined on a set A, the nonnegative functions j+(z) and J-(x) are defined: j+(z) max [J(z), OJ, J-(z) max [ -/(z), OJ. (See Fig. 508.) Prove that if /(z) is integrable on [a, bJ then so are j+(z) and J-(z). Prove that the integrability of any two of the following four functions implies that of all: J(x), j+(x), J-(z), IJ(z)I. Hint: J(x) = J+(z) - J-(x), IJ(z)I = j+(z) + J-(x). (Cf. Ex. 39.)

=

=

IJ I

J(x) FIG. 508

*41. Prove Bliss's Theorem:t If J(x) and g(x) are integrable on [a, bJ, then n

the limit lim

i

= 1, 2,

I\Jll-0

I:1/(z;) g(z;') Ax;,

where

a;-1

;:ii z; ;:ii a, and

a;-1

;:ii z;' ;:ii a,,

I-

· · · , n, the limit being interpreted in the sense of Definition I, § 501,

exists and is equal to J:"J(z)g(z) dz. "E,J(x;) g(x;') Ax;

Hint: Use the identity

= "E,J(x;) g(x;) Ax; + "E,J(x;)

[g(x.') - g(x;)J Ax;,

the inequality 1/(z;) [g(z;') - g(x;)] Ax;I ;:ii K · "E. lu(x;') - g(x;) I Ax;,

and Ex. 34. *42. Bliss's Theorem (Ex. 41) is used in such applied problems as work performed by a variable force and total fluid force on a submerged plate. In the latter case, for example, assume that a vertical plane area is submerged in a liquid of constant density p, and that the width at depth z is w(x) (cf. Fig. 509). Then (with appropriate continuity assumptions) the element of area, between

a

b FIG. 509

t Cf. G. A. Bliss, "A Substitute for Duhamel's Theorem," Annals of Mathematica, Vol. 16 (1914-15), pp. 45-49.

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[§ 503

x, and x, + .1x, is neither w(x,) .1x, nor w(x, + .1x,) .1x,, but w(,.,,) .1:i;, for some appropriate intermediate m, (cf. the Mean Value Theorem for Integrals, and Ex. 5). Furthermore, the total force on this element of area is greater than the area times the pressure at the top of the strip, and less than the area times the pressure at the bottom of the strip. It is therefore equal to product of the area and an intermediate pressure, p~;. Therefore the total force on the plate n

is the sum of the individual elements of force: F =

:E p~;w(,.,,) .:ix,.

i-1

According

to Bliss's Theorem, the limit of this sum can be expressed as a definite integral, and we have the standard formula, F

= pibxw(x) dx.

Discuss the applica-

tion of Bliss's Theorem to the work done in pumping the fluid contents of a tank to a certain height above the tank. *4:3. It can be proved (cf. Ex. 55) that a bounded continuous function of an integrable function is integrable. The reverse happens to be false: an integrable function of a bounded continuous function may not be integrable. (A counterexample can be constructed with the aid of the "Cantor set," well known in Lebesgue Theory.) By means of the following two functions show that an integrable function of an integrable function may not be integrable: Let g(x) be the function of Example 2, § 502, and let/(:,;) be the signum function (Example 1, § 206). Then J(g(x)) is the function of Example 6, § 201, discussed in a paragraph following Definition III, § 501. The numbers I and J defined in the first part of the proof of Theorem II, § 502, are called the lower and upper integral, respectively, of f(x) on [a, b].,

*"·

and are written Lbf(x) dx

= I, rf(x) dx = J. Prove that for a bounded func-

tionf(x) the lower and upper integrals always exist, that Lbf(x) dx

~ rf(x) dx,

and that J(x) is integrable if and only if its lower and upper integrals are equal and, in the case of integrability, its integral is equal to their common value.

I '

j



14 ►

' '

FIG. 510

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§ 503]

EXERCISES

151

*45. Prove that if u(z) is any step-function on the interval [a, b], and if E > 0, then there exist continuous functions q,{z) and 1/t(z) on [a, b] such that q,(z) ~ u{z) :ii 1/t(z) and _[b[l/t(z) - q,(z)] dz

< E.

bounded function on [a, b], .£.bf(z) dz

= sup .[bq,(z) dz,

~ f(z),

ous functions q,{z) such that q,(z)

Hence prove that if f{z) is any taken for all continu-

and rf{x) dz = inf .[bl/t(x) dx,

taken for all continuous functions 1/t(x) such that 1/t(z) ~ f(x). Prove that Theorem II, § 502, remains true if the word step-functions is replaced by the words continuo'UB functions. (Cf. Ex. 44.) Hint: See Fig. 510, for q,(x). **46. Define a step-function of positive compact type to be either the function identically zero or a nonnegative step-function whose nonzero values are assumed as constants on disjoint closed intervals (of positive length). (See Fig. 511.) Prove that if f(x) is a bounded nonnegative function on [a, b], then fbf(z) dz = sup fbu(z) dx taken for all step-functions u(x) of positive com~ )a pact type such that (0 ~) u(x) ~ f(z).

FIG. 511

~ f(z) ~ g(z) ~

**47. Assume K

~

such that f(z)

u(z)

~

0 on [a, b], and let u(x) be a step-function

0 and .[bu(z) dz

> .£.bf(x) dx

-

E,

where

E

> 0.

Es-

tablish the existence of a step-function T(x) of positive compact type (Ex. 46)

~

such that g(x)

T(z) (~ 0), u(z)

~

T(z), and _(\(z) dx

f/F(z)

Hint: Abbreviating _[bF(x) dx and .•

tively, define 11 that O ~ p(z) 1/t(x)

=J

~

a- dx - i f dx

g(x) and

dx by

f

> .£.bg(z) dx

-

E.

F dx and LF dx, respec-

+ E > 0, and let p(z) be a .step-function such

J >i p dz

g dz - 11· If q,(z)

= min (u(z), p(z)) a~d

= max (u(z), p(z) ), then q,(z) and 1/t(z) are step-functions, J1/t dz ~ .f!dz,

J

+ p(z) = q,(z) + 1/t(x). ThenJq,dx + 1/tdx = Ja-dx +Jpdx > f dx + g dx - E, and J q, dx > .[g dx - E. Finally, use Ex. 46.

and u(z)

L i

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**48. Assume K s;; fi(x) s;; f2(x) s;; • · · s;; f ,.(x) lim f,.(x) = 0

[§ 503 ~ f ,.+h)

s;; • · · and

tl-++oo

for every x on [a, b]. of Ex. 47, assume

Prove that .Lhf,.(x) dx-+ 0.

ff,.

~

dx-++ O.'

Then, since

ff,.

Hint: With the notation dx

~

!,

lim n-++ao

ff.

dx exists

and is equal to a positive number 3E. By Ex. 47, there exists a sequenl'e {un(x)} of step-functions of positive compact type such that (0 ;;i) u,.(z) ;;i f,.(.r).

f

> 2E, for n = 1, 2, • • · . the set of points such that un(x) s;; E/(b - a), then {A,.}

un(x)

s;;

u,.+1(x), and

u,. dx

If A,. is defined to be

is a compact nonempty set, for n = 1, 2, • • • , and A,. is a decreasing sequence (Ex. 26, § 312). If xo is a point common to every set of this sequence (Ex. 26, § 312), f,.(Xo) ~ u,.(xo) s;; E/b - a), andf,.(xo)-++ 0. (Contradiction.) **49. Prove the Lebesgue Theorem on Bounded Convergence for Riemann lntegrals:t If {f,.(x)} is a seqmnce of Riemann-integrable functions converging on [a, b] to a Riemann-integrablefunctionf(x), and if lf,.(x)I ;;i K for a ;;i x ~ b and n

= 1, 2, · · · ,

then

=

Jim n--++oo

(bf,.(x) dx

)a

=

=

(bf(x) dx

Ja

=

(b lim f,.(x) dx.

)a n-++oo

Hint: Define g,.(x) Jf,.(x) - f(x)I, h,.(x) sup (g,.(x), g,.+h), · · · ). Then (Exs. 16-18, § 305), h,.(x) ! , h,.(x) -+ 0, and the result of Ex. 48 shows that Lh,. dz-+ 0.

-+ff dx.

Therefore, sinceO ;;i gn(z) ;;i h,.(x),

f

g,. dz-+ 0, whenceJf,. d.r

(Cf. Ex. 46, § 515.)

**50. Let r1, r2, ra, · · · be the rational numbers on the interval [O, 1], and define the function f,.(x) to be 1 if x = r1, r2, · · · , or r,., and O otherwise, 0 ;;i x ;;i 1. Show that Jim f,.(x) = f(x) exists and that the convergence is bounded and n-++• monotonic (Ex. 49), but that f(x) is not Riemann-integrable. (Even if all of the functions f,.(x) are continuous on [O, l] and the convergence is bounded and monotonic, the limit function need not be Riemann-integrable.) **51. Let f(x) be defined on a closed interval [a,~] and assume that there exists a positive number 11 such that the oscillation w(x) (cf. Ex. 31, § 312) of f(x) at each point x of [a,~] is less than 11· Prove that there exists a net ~ for [a,~] such that on each closed subinterval of ~ the difference between the least upper bound of f(x) and the greatest lower bound of f(x) is less than 11 (that is, the oscillation of f(x) < 11), Hint: Each point of [a,~] has a neighborhood I in which the oscillation of f(x) is less than.,,. Use the Heine-Borel theorem (Ex. 28, § 312) to reduce this covering of [a,~] to a finite covering. Finally, obtain the desired net from this finite covering.

t This theorem was stated by ArzelA in 1885, before the invention of the Lebesgue Integral, and given a simplifed proof (whose essential ideas we have outlined here) by Hausdorff in 1927. If the integrals concerned are Lebesgue integrals, the integrability of the limit function f(x) is a consequence of the other hypotheses and need not be specifically assumed. (( f. Ex. 50.) However, the theorem as stated above has useful applications in cases where the limit function is known to be Riemann-integrable.

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153

**62. A set S of real numbers is said to be of (Lebesgue) measure zero if and only if corresponding to E > 0 there exists a (finite or infinite) sequence of open intervals 11, /2, · · · of lengths Li, L2, · · · , which cover S (cf. Ex. 28, § 312) and such that for every n, L1 + L2 + ••• + L,. < E. Prove that any finite set of points is of measure zero. Prove that any denumerable set (cf. Ex. 12, § 113) is of measure zero. (The rational numbers form a set of measure zero.) Prove that any subset of a set of measure zero is of measure zero. Hint: For a denumerable set xi, x2, • • • , x,., • • • , if E > 0, let I,. be a neighborhood of x,. of length L,. ~ E/2", n = 1, 2, · · · . **63. Let 81, S2, · · · be a (finite or infinite) sequence of sets of measure zero and let S be the set of points x such that x is a number of at least one set S,. of the sequence (Sis the "union" of the sets of the sequence). Prove that S is of measure zero. Hint: If E > 0, define for each m the sequence /1, L2 + ••• + L,. can be arranged as a sequence (cf. Ex. 13, § 113). ••64. A property is said to hold almost everywhere if and only if the set of points where it fails is of measure zero. Prove that a functionf.(x), defined on a closed interval, is integrable there if and only if it is bounded and almost everywhere continuous. Hints: (i) If J(x) is integrable on [a, b] and if E > 0 and f/ > 0, there exist step-functions u(x) and T(x), over the same net mt, such that u(x)

~ J(x) ~ T(x)

and J)T(x) - u(x)] dx < er,.

Then the sum of the

lengths of the subintervals of mt for which Ti - O"i ii;;; E is less than f/. By appropriateiy enclosing the points of mt in small neighborhoods, show that the set DE of points x where w(x) ii;;; E (cf. Ex. 32, § 312) is contained in a finite set of open intervals the sum of whose lengths < 11, and is therefore a set of measure zero. Conclude by referring to Ex. 33, § 312, and Ex. 53 above. (ii) With the notation of Ex. 33, § 312, assume that D is of measure zero. Then (Ex. 52) D., for any E > 0, is of measure zero. If IJ(x)I 0, let Ii, h · · · be a sequence of open intervals covering D., where fl = E/2(b - a), of such small lengths L1, L2, • • • that for every n, L1 + L2 + · · · + L,. < E/4K. By the Heine-Borel Theorem (Ex. 28, § 312) there exists a finite sequence /1, 12, · · · , I,. covering D 1 • On each of the closed subintervals [a, 13] of [a, b] remaining after removal of / 1, 1 2, • • • , I,., use Ex. 51 to construct ste,v-functions u(x) and T(x) such that u(x) ~ J(x) ~ T(x) and such that the sum of the integrals

f

[T(x) - u(x)] dx over these intervals < 11(b - a) = ½E.

Extend u(x)

and T(x) to the entire interval [a, b] by defining them equal to -Kand K, respectively on the intervals Ii, 12, • • • , I,..

Then .£b[T(x) - u(x)] dx < E.

**66. Prove that a bounded continuous function of an integrable function is integrable. In particular, if J(x) is integrable on [a, b] and p > 0, prove that IJ(x)IP is integrable on [a, b]. Hint: The composite function is continuous almost everywhere (cf. Ex. 54). ••66. Prove that if f(x) is integrable and positive on a closed interval [a, b],

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154

then lbf(x) dx

> 0. Hint:

By Ex. 54 J(x) is continuous at some point of

[a, b].

**57. Prove that in Exercise 56 the word positive may be replaced by the expression positive almost everywhere. In fact, prove that if J(x) is a nonnegative integrable function and if lbf(x) dx

=

0, then J(x)

=

0 almost everywhere.

504. THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS

Evaluation of a definite integral by actually taking the limit of a sum, as in the Example of § 501 and Exercises 15-21, § 503, is usually extremely arduous. Historically, many of the basic concepts of the definite integral as the limit of a sum were appreciated by the ancient Greeks long before the invention of the Differential Calculus by Newton and Leibnitz. Before the invention of derivatives, an area or a volume could be computed only by such a limiting process as is involved in the definition of the definite integral. The introduction of the notion of a derivative provided a spectacular impetus to the development of mathematics, not only by its immediate application as a rate of change, but also by permitting the evaluation of a definite integral by the process of reversing differention, known as antidifferention or integration. It is the purpose of this section to study certain basic relations between the two fundamental concepts of Calculus, the derivative and the definite integral.

Theorem I. Let f(x) be defined and continoous on a c'losed interval [a, b] or [b, a], and define the function F(x) on this interval: (1)

F(x)

= f) 0, J(x) = x + l if x < 0. On [ -1, 1] the following functions are sectionally continuous but not sectionally smooth: (d) Vl - x1 ; (e) ~x1 ; (/) x2 sin (1/z) (Example 3, § 403).

•sos. EXERCISES. •1. Show that the function

[x] (Example 5, § 201) is sectionally smooth on any finite closed interval. Draw the graph of its derivative. •2. Show that !sin xi is sectionally smooth on any finite interval. Draw the graphs of the function and its derivative. •3. Prove that a function that is sectionally continuous on a closed interval is bounded there. Must it have a maximum value and a minimum value there? Must it take on all values between any two of its values? •4. Prove the Theorem of § 507. Hint: For the interval [a,-1, a,], if /(z) is defined or redefined: f(a,_1) = /(a,-1+) and f(a,) = f(a,-), then /(x) becomes continuous there. Apply Theorem V, § 501, and induction. •6. Let F(x) be sectionally smooth on [a, b], with exceptional points a;, i = 1, 2, · · · , n - 1, and possibly ao = a and an = b, and denote by J, the jump of F(x) at a,: J, = F(a,+) - F(a,- ), i = 1, 2, · · · , n - l. Prove that (1)

(bF'(x) dx

Ja

= F(b-)

- F(a+) -

nt,1 J;. i-1

(Cf. Ex. 20, § 506.) •6. Prove that if J(x) and g(x) are sectionally continuous (smooth) then so are /(x) + g(x) and /(x) g(x). Extend by induction ton terms and n factors. •7. Generalize Exercise 5 to the following integration by parts formula: If F(x) is sectionally smooth on [a, b] and if G(x) has a continuous derivatfre there, then (with the notation of Ex. 5)

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§ 509]

REDUCTION FORMULAS

(2)

=

.£bF'(x) G(x) dx

161

F(b-) G(b) - F(a+) G(a) -

(bF(x) G'(x) dx -

)a

1

"'t, ,-1 J; G(a;).

(Cf. Ex. 21, § 506.) *8. Prove that if f'(x) is sectionally continuous on [a, b], then J(x) is sectionally smooth there. Furthermore, prove that J(x) has a right-hand and a lefthand derivative at each point between a and b, and a right-hand derivative at a and a left-hand derivative at b. (Cf. Exs. 53-54, § 408.) *9. If J(x) is sectionally continuous on the closed interval [a, b], then F(x)

=.["'J(t) dt is continuous there and F'(x) exists and is equal to J(x) there

with at most a finite number of exceptions (Exs. 22-23, § 506). Prove that any function ~(x) that is continuous on [a, b] such that ~'(x) exists and is equal to /(x) there with at most a finite number of exceptions must differ from F(x) by at most a constant. If ~(x) is not everywhere continuous, what can you say? *509. REDUCTION FORMULAS

It often happens that the routine evaluation of an integral involves repeated applications of integration by parts, all such integrations by parts being of the same tedious type. For example, in evaluating I

=

J

sin10 x dx, we might proceed:

I=

=

f

sin9 xd(-cosx) = -sin9 xcosx

-sin9 x cos x

J

+9

so that (1)

I

= --trr sin9 x cos x +

+

9f

sinsxcos2 xdx

sins x dx - 91,

to-f sins x dx.

We have succeeded in reducing the exponent from 10 to 8. We could repeat this labor to reduce the new exponent from 8 to 6; then from 6 to 4; etc. A more satisfactory method is to establish a single f ormul,a to handle all integrals of a single type. We present a few derivations of such reduction formulas in Examples, and ask for more in the Exercises 6f the following section. Since differentiation is basically a simpler process than integration, we perform our integrations by parts by means of differentiating certain products. Example 1. E~press

f

sinm x cos" x dx, where m

+ n ~ 0,

in terms of an

integral with reduced exponent on sin x (cf. Exs. 1-2, § 510).

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162

Solution. The derivative of the product sin" x cos• x is p sin,,-1 x cost+1 x - q sin,,+1 x cosr-1 x ... p sin,,-1 x cosr-1 x (1 - sin1 x) - q sin,,+1 x cosr-1 x = p sin,,-1 x cosr-1 x - (p + q) sin,,+1 x cosr-1 x. In other words, (2) pfsin,,-1 x cosr-1 x dz - (p q) sin,,+1 x cosr-1 x dz = sin" x cost x

f

Letting p with the (3)

f

+ +C. 1 and q = n + 1, and absorbing the constant of integration

m -

E

J,

we have the formula sought:

sin• X

C08" X

dz

=-

1 x cos11 +1 x sin"'=;..._....;.;;...;;.;;.;;;.__ _

m+n

m - 1 + --m+n

J

sin•-t X

C08" X

dz.

Equation {l) is a special case of (3), with m = 10, n = 0.

Often it is desirable to increase a negative exponent. Eumple 2. Establish the reduction formula (m ¢ 1) (cf. Exe. 1-2, § 510): cos"xdz = _ cos"+ix _ n - m + 2 f cos"xdz 4

f

sin• x (m - 1) sin•-1 x m - 1 Solution. In (2), let p = -m + 1 and q • n + 1.

( )

Eumple 3. Establish the reduction formula (m (5)

f

(at

(6)

(1

+ 2m)

sin•-t x

1):

¢

f

X + 2(n2n- - l)at 3 dz l)a2(a1 + zl)11-1 (al + zt) ..-1· Solution. The derivative of the product x{a1 + x2)"' is (a2 + x2)"' + 2mx1(a1 + x1) • -1 = (a1 + x1)• + 2m[(a1 + x1) - a1]{a1 + x1)•-I • (1 + 2m){a1 + x2)• - 2ma2(a2 + x1) ..- 1. In other words,

dz + zl)" = 2(n -

Now let m

f

e

+ x dz - 2ma•f (a + x -n + 1, from which we obtain (5). (a1

1

1

)•

1

) ..- 1

dx = x(a1

+ x + C. 2)"'

Useful reduction formulas are given in nearly every Table of Integrals. *510. EXERCISES In Exercises 1-12, establish the reduction formula. sin•+I X C0811 - 1 X sm., x cos" x dx = ='--....:;:..='---=

f f *3. f

*2.



m+n n - 1 + --m+n

sin• x dz C08" X

tan" x dz

Digitized by

=

sin•+1z

(n - 1) cos11- 1 X 1

= tan•n _ x 1

Google

-

f

J.

SID"' X

_ m - n

+2

n - l

tan"-t x dz (n

¢

.,_

cos"-t X tu (m + n ,,_ 0).

J

sin"' x dx (n cos"-t X

1).

Original from

UNIVERSITY OF MICHIGAN

¢

1).

EXERCISES

§ 510]

*'*Ii•

f

f *1. f f f

1

cot-- x=- n-1

secx dz

xl tan x = sec"-1 n-

csc"xdz

=-

J ,.

*6.

11

cot• x dz

f

cot"-1 x dz (n 7': 1).

J.

+ n -_ 2 sec -1 x dz (n 7': 1) • 11 1 csc"-1 _x cot x + n -_ 2 csc -lzdz (n 7': 1). 11 11 1 1

J

= Z" sin:,; + nz11- 1 COB:,;

:,;"cos:,; dz

*9·

:,;•(az + b)"+i :,;•(a:,;+ b)" dz= a(m + n + 1) - a(m +m! + l)

*10. Jz.. (a:,; + b)" dz

*12.

f f

11

J:,; J

x" sin x dz = -:,;"cos:,; + nz•-1 sin:,; - n(n - 1)

*8.

*11.

163

-

n(n - 1)

d:,;.

z•-1 (az + b)" dz (m + n + 1 7': 0).

= zws+l(az +

b)" m+n+l

+ n +

:,;"

':!' + 1 Jz•(a:,; + b)11-

bf

a

I

:,;•-I COB:,;

J

dz :,;11-1 v'azt+bz+c = a(n -1)

_

11-1 sin:,; dz.

z"v 2az - :,;2 dz

=

cf

:,;11-1 dz Vaz 2 +bz+c a 1(2az - :,;2)f - z"--------n +2

+ a(!n++2 l)

f

1

dz (m + n + 1 7': 0).

:,;11-1 dz (n 7': 1). 'Vaz2 +bz+c

:,;11-1v'2az -

:,;2

dz (n 7': -2).

ln'Exercises 13-20, perform the integration, using reduction formulas above.

*13. *15. *17. *19.

f f f :,; f

sin•:,; dz.

*1'-

cot'~ dz.

*16.

4

*18.

sin 2x dz.

f f f

cos1 :,; dz. sec7 :,; dz. :,;I-(:,; + 4)1- dz.

:,;a dz *20. fz 1 V6z - :,;2 dz. v'z 2 +z+l **21. Use mathematical induction to verify the formula:

i

b (:,; -

a)•(b - :,;)" dz

Digitized by

= (b -

a) ..+11+1

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m ! 11 ! (m and n positive integers). (m+n+l) I

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[§ 511

INTEGRATION

164

511. IMPROPER INTEGRALS, INTRODUCTION

The definite integral lf(x) dx, as defined in § 501, has meaning only if

a and b are finite and f(x) is bounded on the interval [a, b]. In the following two sections we shall extend the definition so that under certain circumstances the symbol ibf(x) dx shall be meaningful even when the interval of integration is infinite or the function f(x) is unbounded. For the sake of conciseness, parentheses will be used in some of the definitions to indicate alternative statements. For a discussion of the use of parentheses for alternatives, see the Preface. 512. IMPROPER INTEGRALS, FINITE INTERVAL

Definition I. Letf(x) be (Riemann-) integrable on the interval [a, b - E] ( [a + E, b]) for every number E such that O < E < b - a, but not integrabl,e on the interval [a, b], and assume that lim r,-f(x) dx ....o+ Ja

exists.

( lim

....o+

(6 f(x) dx)

Ja+•

Under these conditions the improper integral lf(x) dx is defined to

be this limit: (1)

("f(x) dx

Ja'

=

lim r,-f(x) dx ....o+ Ja

( lim (6 f(x) dx). ....o+ Ja+•

If the limit in (1) is finite the integral lf(x) dx is convergent to this limit and the function f(x) is said to be improperly integrable on the half-open interval [a, b) ( (a, b]); if the limit in (1) is infinite or does not exist, the integral, is divergent. NOTE 1. A function improperly integrable on a half-open interval is necessarily unbounded there. In fact, it is unbounded in every neighborhood of the end-point of the interval that is not included. (Cf. Theorem III, § 502, and Ex. 28, § 503.)

Definition Il.

Let [a, b] be a given finite interval, let a

< c < b,

and

let both integrals .[f(x) dx and J.b f(x) dx be convergent improper integrals in the sense of Definition I.

Then the improper integral

is convergent and defined to be: lf(x) dx

(2)

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= .[f(x) dx

+J.bf(x) dx.

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J:

f(x) dx

§ 512)

IMPROPER INTEGRALS, FINITE INTERVAL

165

If either integral on the right-hand side of (2) diverges, so does .[bf(x) dx. NOTE 2. There are four ways in which the integrals (2) may be improper, corresponding to the points in the neighborhoods of which f(x) is unbounded: (i) c- and c+, (ii) a+ and c+, (iii) c- and b-, and (iv) a+ and b-. In this last case (iv), the definition (2) is meaningful only if the value of

J) 1, Jim [ :,;l-_p ] = +co. Therefore the given integral converges ....o+ 1 - p • if and only if p < 1.

and if p

Example 4. Evaluate

fb

4 -

-2

:,;I

Solution. The integrand becomes infinite at both end-points of the interval [ -2, 2], and we therefore evaluate according to Definition II, choosing some number c in the interior of this interval. The simplest value of c is 0. We find, then, that

fO dz J-2,V4 and

=

:,;I

fo b 2

Jim [Arcsin

....O+

= !a~ [Arcsin

~Jo

2 -2+•

=

-Arcsin ( -1)

= !, 2

i]:-• = Arcsin (1) = ~•

and the value of the given integral is r. Notice that any other value of c between -2 and 2 could have been used (Note 2, above):

J

c

-2

dz V4 -

:,;I

+f c

dz V4 -

:,;I

= (Arcsin E

2

+ ~) + (!2 - Arcsin E.) = 2 2

r.

NoTE 4. Under certain circumstances a student may evaluate an improper integral on a finite interval, with correct result, without forming a limit, or even without recognizing that the given integral is improper. This would be the case with Example 1, above-but not with Example 2. The following theorem justifies such a method and simplifies many evaluations:

Theorem. If f(x) is continuous in the open interval (a, b), and if there exists a function F(x) which is continuous over the closed interval [a, b] and such thatF'(x)

= f(x) in the open interval (a, b), then the integral [f(x) dx,

whether proper or improper, converges and ibf(x) dx

= F(b)

- F(a).

(Cf. Ex. 15, § 515.) Proof. By the Fundamental Theorem of Integral Calculus, for any c between a and b, and sufficiently small positive E and .,,, . [+•f(x) dx

+Jct'-i(x) dx = F(b -

11) - F(a

+ E),

and the result follows from the continuity of F(x) at a and b. Example 6. In Example 1, above, .let F(x) = 2-Vx - 4. Then

is v/~ 4 = [ 2✓x-::4]: = 2.

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§513]

IMPROPER INTEGRALS, INFINITE INTERVAL

167

513. IMPROPER INTEGRALS, INFINITE INTERVAL

Definition I.

Let J(x) be (Riemann-) integrable on the interval [a, u]

for every number u

>

a, and assume that

lim [!(x) dx

-+• ..

exists.

Under these conditions the improper integral J..+•J(x) dx is defined to be this limit: (I)

If the limit in (1) is finite the improper integral is convergent to this limit and the Junction f(x) is said to be improperly integrable on the interval [a, +oo); if the limit in (1) is infinite or does not exist, the integral is diver-

gent. A similar definition holds for the improper integral[. J(x) dx.

Definition II.

Let J(x) be (Riemann-) integrable on every finite cwsed

interval, and assume that both improper integrals

l.J(x) dx converge.

f/•J(x)

dx and

Then the improper integral J_:•f(x) dx is con-

vergent and defined to be: (2)

J_+: J(x) dx • l.J(x) dx +j/•J(x) dx.

If either integral on the right-hand side of (2) diverges, so does J_:•J(x) dx. NoTE 1. The improper integral (2) could have been defined unambigu-

ously: ber.

f_+,.•J(:z:) d:z: = f~_J(:z:)

th+ i+•J(:z;) th, where c is an arbitrary num-

(Cf. Ex. 12, § 515.)

Improper integrals on finite and infinite intervals are often combined: Definition m. Let J(x) be improperly integrable on the interval (a, c] and on the interval [c, +oo ), where c is any constant greater than a.

Then the improper integral (3)

. defined to be: f.. +•f(x) dx is convergent and

• J(x) dx. f.. +•J(x) dx = J,.reJ(x) dx +5.+•

If either integral on the right-hand side of (3) diverges, so does

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f.. +•J(x) dx.

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168

INTEGRATION

[§ 514

Similar statements hold for a similarly improper integral [j 0. 2

=

f

> 0.

-• a

X

Solution.

f

[!. Arctan a~],. + .-+.. lim [!. Arc tan ~J• x ,....._,. a a a lim [ - !. Arctan ~J + lim [!. Arctan !] !.. !: + !. •!: = !. --• a a .-+.. a a a 2 a 2 a

+ • 2 ck+

-• a

=

= lim

2

o

=

E:z:ample 3.

Solution.

0

For what values of p does

If p

J:+

00

-dx converge? X"

F1, Ji"x-,,dx = [/:"PJ:, and the integral converges if

> 1 and diverges if p < 1. Similarly, f+'° dx

=

fore the given integral converges if and only if p is 1/(p - 1).

>

p

Ji

E:z:ample 4.

Solution.

For what values of p does

J:+•

lim In u = u-++oo

X

There-

1, and its value, for such p,

dx - converge? X"

For this improper integral to converge, both

J: 5i+• 1

and

converge. But they never converge for the same value of p. §§ 512, 513.) Answer: none.

514. COMPARISON TESTS.

+ao.

must

(Examples 3,

DOMINANCE

· For a nonnegative function, convergence of an improper integral means (in a sense determined by the definition of the improper integral concerned) that the function is not too big: on a finite interval the function does not become infinite too fast, and on an infinite interval the function does not approach O too slowly. This means that whenever a nonnegative function g(x) has a convergent improper integral, any well-behaved nonnegative function f(x) less than or equal to g(x) also has a convergent improper integral. We make these ideas precise: Definition I. The statement that a function g(x) dominates a function f(x) on a set A means that both functions are defined for every member x of A and that for every such x, lf(x)I ~ g(x).

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§514]

COMPARISON TESTS.

DOMINANCE

169

NOTE 1. Any dominating function is automatically nonnegative, although a dominated function may have negative values.

Theorem I. Comparison Test. If a function 'g(x) dominates a nonnegative function f(x) on the interval [a, b) ( [a, +oo) ), if both functions are integrable on the _interval [a, c] for every c such that a and if the improper integralibg(x) dx

-1 and

th, for 0

q-

< c < 1, is equivalent to 1

1 > -1. Therefore .fo /(x) dx

converges if and only if both p and q are positive.

Enmple 9.

(The Gamma Function.) r(a)

converges. Solution.

x--1 as

x-

5o+ .. x-- e-• th 1

If /(x) is the integrand, /(x) is of the same order of magnitude as

o+,

Rule in case a

5.+• ';:

a

Determine the values of a for which

>

and /(x)

=0

-1). Since

(~) as x -

+oo

(this is true by l'Hospital's

1

.fo x--1 dx converges if and only if a > 0, and

converges, the given integral converges if and only if a is positive.

515. EXERCISES

In Exercises 1-10, evaluate every convergent improper integral and specify those that diverge.

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INTEGRATION

172

2.

[§ 515

ldJ:

J

_.,-·

-lv:,; r

4.

2

.fo vsin x tan x dx.

6. (+'° sin x dx.

Jo-

8 •

(+00

dx

J2

x(ln x)k.

+oo

lO.

dx

J

1

-oo

+ 4x

2•

11. Prove that the existence and the value of the improper integral (2), § 512, does not depend on the value of c in case (iv) of Note 2, § 512. Hint: Let a+ E < c < d < b - 1/· Then



Ja+•

+J.b-.,



= (•

Ja+•

+J."• +Jtl

+ rb-.,_ Ja+• Jtl

= ("

(b-.,

12. Prove the independence of c of the improper integrals referred to in the two Notes of § 513. (Cf. Ex. 11.) Illustrate each by an example. 13. Let f(v) be continuous for c < v < d, let v(x) have values between c and d, and a continuous derivative v'(x), for a < x < b, and assume that v(x) --+ c as x--+ a+ and v(x) --+ d as x--+ b-. Prove that if the integral

J."!(x) dv converges (whether proper or improper) then the left-hand member

in the following integration by substitution formula also converges and ibf(v(x)) v'(x) dx = J.!(v) dv.

(1)

Evaluate the integral of Example 4, § 512, by use of this formula and the substitution x = 2 sin"· Illustrate with other examples. Hint: If F(v) is a primitive of f(v), c < " < d, (b-J(v(x)) v'(x) dx

Ja+•

=

+ E) ).

F(v(b - 11)) - F(v(a

14. Discuss the integration by substitution formula (1) for cases involving infinite intervals. Illustrate with examples.

In Exercises 15-24, merely establish convergence or divergence. try to evaluate. 16.

(+'°

Jo

dx

Vl

+x

17.

J_+.,.. e-x• dx.

19.

f1 ~. Jo -v'x In x

21•

J+'° _.,

Digitized by

. 4

2x dx . ~

- e-s

Google

16.

+,.

.l

x dx _1 4 vx 1 -

:t·

18. 5o11n 20. 22•

f-.. ~

In

lxl dx.

fl In x dx. Jo 1 -x

Original from

UNIVERSITY OF MICHIGAN

Do not

EXERCISES

§ 515]

23.

f

l -1

173

fo ~ 1

~dx.

24.

x ln (1/x) dx.

26. Prove Theorem II, § 514. 26. Prove that if p is an arbitrary positive number, then (2) ln x = o(x"), x = o(eJ>Z), as x-+ +ao;

1

In x = o(x-"),

(3)

'/!

- = o(e"'),

as

X

x-+

o+.

27. Show that the functions in any "big 0" relationship can be multiplied or divided by any function whose values are positive: If f(x) = O(g(x)) and h(x) > 0, thenf(x) h(x) = O(g(x) h(x)); conversely, if f(x) h(x) = O(g(x) h(x) ), then f(x) = O(g(x) ). Is the same thing true for the "little o" relation? 28. Prove that (+"° )2

X

(/x ) converges if and only if p nX P

>

1.

Ia+"° x(ln x)t~ In x)" converges if and only if p > 1.

29. Prove that

generally, prove that



+..

dx

.

)(l (I ) ) , where a 1s sufnx n 1nx .. · n 1n · · · 1nx" ficiently large, converges if and only if p > 1. x

(l

More

*30. The Cauchy principal value of the improper integral

J_+

00

f(x) dx is

00

denoted and defined (P)

provided the integral improper integral

J_+

00

00

f(x) dx

= ..~~..

.J-J(x) dx,

.J-,. f(x) dx and its limit exist.

Prove that whenever the

J_+....f(x) dx converges, its Cauchy principal value exists and

is equal to it. Give examples to show that the converse is false. (Cf. Ex. 31.) *31. For each of the cases (i) and (ii), below, define a Cauchy principal value of the improper integral J~J(x) dx assuming f(x) is integrable on [u, v] for every u and v such that (i) -a < u < v < a; (ii) -a ;:a u < v < 0 and O < u < v ;:a a. Give a sufficient condition for the existence of the integral in each case. (Cf. Ex. 30.)

In Exercises 32-35, establish the given relation. (1 xv-1 (+"° x-v dx _

*32· Jo x + 1 dx - J1 x + 1' p > O. (+ x"- dx _ (+ x-,, dx, *33• Jo x + 1 - Jo x + 1 O < P < 1. 1

00

.fo\,,+ *36. .fo (

1(1

*34.

00

- x)q dx

dx

00

X

=;

Jo

1

x"(l - x)q-t dx, p

1r

+ p. ) v-X = ---=• p Vp

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> 0, q >

0.

> 0.

Original from

UNIVERSITY OF MICHIGAN

INTEGRATION

174

[§ 515

In Exercises 36 and 37, use integration by parts and mathematical induction to verify the formulas.

*36. Wallis's Formulas.

fo ~ =.foi 1

=

sin" X dx

=

foi

COS" X

dx

2·4·6 .. • (n _ _ ... 3 5 7 (n { 1 ·3·5_ •_• • ... 2 4 6

- 1) . • • n ' 1f n 1s an odd mteger

- 1)

n

·

>

1;

1r .

• • 2, 1f n 18 an even mteger > 0.

5o+ •x"e_,, dx = ;1. (n a positive integer, a > 0).

*37.

In Exercise 38-41, state and prove the analogue for improper integrals of the specified theorem of § 501.

*38. Theorem II. *40. Theorem IV.

*39. Theorem III. *41. Theorem V.

*42. Show by an example that a continuous function improperly integrable on [O, +ao) need not have a zero limit at +ao, in contrast to the fact that the general term of a convergent infinite series must tend toward zero (§ 703). (For another example see Ex. 46.) Hint: See Figure 513. 11

1 FIG. 513

*43. Prove the Cauchy Criterion for convergence of improper integrals: If f(x) is integrable on [a, c] for every c such that a < c < b (a < c), then the 00

(.£+ /(x) dx) converges if and only if corresponding to E > 0 there exists a number c = c(E) such that a < c < b (a < c) and such that c < u < v < b (c < u < v) implies IJ) dg(x) =.f,cx) dg(x) +J:1cx~~x),

(16)

'J}1'wided the three integrals exist. Proof. We shall assume a < b < c (cf. Exs. 16-17, § 518). For E > 0, let a > 0 be such that 1ml < a implies that any sum corresponding to any of the three integrals approximates that integral within ½E. Then the sum of two such approximating sums for the two integrals on the right-hand side of (16) must approximate the left-hand side within ½E. Therefore the two sides of (16) are constants differing by less than E, and must consequently be equal. Finally, a simple application of Theorems II-IV and the fact (Theorem VIII, § 516) that any function of bounded variation can be represented as the difference between two monotonically increasing functions yields the following generalization of Theorem III:

**

**Theorem VI. If, on the interval [a, b], one of the functions f(x) and g(x) is continuous and the other of bounded variation, the integral lf(x) dg(x) exista. ?:

f/x d sin x.

Evaluate the Riemann-Stieltjes integral

E%8Dlple 3.

J/

:!

Solution.

By Theorem I, this integral is equal to

x cos x dx.

However,

it is simpler to use the integration by parts formula (Theorem II) directly: T

T

r\ d sin x = !2 sin !2 -

Jo Example 4.

0 sin O - r\in x dx = ! - 1. Jo 2

Evaluate the Riemann-Stieltjes integral

.fo e2z d [x], where xis 3

the bracket function of § 201. Solution. The integration by parts formula gives J:e2z d [x]

= 3e9

which, by Theorem I, is equal to 3e6

-

2

f

Digitized by



3

2

e2"' dx

0 ell - Jo[x] d e2z,

-

- 4

Google

et:r dx

= e2 + e4 + e9•

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INTEGRATION

184

[§ 518

The result could be obtained directly by using the fact that the function [x] in the original form of the integral makes contributions only at its jumps, of amounts determined by the values of etz and the size of the jumps (cf. Example 1, and Exs. 8-9, § 518.) *518. EXERCISES

The notation [x] indicates the bracket function of § 201.

In Exercises 1-6, evaluate the Riemann-Stieltjes integral.

•1.

Jo xdx 1

1•

•2. .r:xd [x].

*'-

.J::es d{x -

[x]}.

•6. 5o•x d !cos xi. •7. Prove that if g(x) is defined on [a, b], then J:bdg(x) exists and is equal to g(b) - g(a).

•8. Prove that if J(x) is continuous on [0, n], then /(2)

J;

J(x) d [x]

= J(l)

+

+ ••• + J(n).

•9. For a net m.: a = ao < a1 < • •• < a,. = b, let g(x) be a step-function constant for a;-1 < x < a;, i = 1, 2, • • • , n, and having jumps Jo = g(a+) - g(a), J; g(a;+) - g(a;-), i = 1, 2, • • •, n - 1, and J,. g(b) - g(b-). · Iff(x) is continuous on [a, b], prove that

=

=

£1cx) dg(x)

(1)

=

it

J;J(a;).

Hint: Since the integral exists (Theorem VI, § 517) choose a net of arbitrarily small norm such that each a;, i = 1, 2, • • • , n - 1, is interior to some subinterval. Then let each a;, i = 0, 1, • • • , n, be a point chosen for evaluatingf(x). •10. Prove that if J(x) and g(x) are both discontinuous at a point c, where a ;;; c ;;; b, then .£bf(x) dg(x) does not exist.

Hint: For a

< c < b, consider let ak-1 < c < ak,

any net m., of arbitrarily small norm, not containing c and Then IAutl can always be made ii;;; .,,, a fixed positive number. Also, for two suitable numbers, xk and x/, in [at-1, ak], IJ(xk) - J(xt')I ii;;; ~. a fixed positive number. Hence, for any 6 > 0, there is a net m. of norm < 6, and two sums :E J(x;) Ag; which differ numerically by at least.,,~ (for i F- k, let X; = x;'). •11. Prove Theorem II, § 517.

Hint: Expansion of

:E" g(x;)[f(a;)

- J(a;-1)]

i-1

and rearrangement of terms leads to the following identical formulation, where Xo a and X11+1 b:

=

=

J(b) g(b) - f(a) g(a) -

L" J(a;)

(g(.r;+1) - g(x,) ).

i-0

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185

•12. Prove (8), § 517. Hint: If mt and m. are arbitrary nets on [a, b], let CP he the net consisting of all points appearing in either mt or m. (or both). Then, since CP contains both mt and m., L(mt)

~

L(CP)

~

U( 0, y > 0; (ii) In(;) = In x - In y, x > 0, y > 0; (iii) In (x") = n In x, x > 0, n an integer; (iv) In (~x = ! In x, X > o, n a positive integer; n

(v) In (x•)

= r In x,

x

> 0, r rational.

(Cf. § 214 and Exs. 21-22, § 216.) (v) holds for any real r. Hints: (i):

rx

(Z11 ~ = ~ )1 t )1 t

(ii): use (i) with x =

It follows from Exercise 6, below, that

+ {XII~ = rx dt + rXll ~ j Jx

t

)1 t

Jx

t/~

y • ~; y

= (~)". •3. Let the function e" = exp (x) be defined as the inverse of the logarithmic (iv): use (iii) with x

function: y = e" = exp (x) if and only if x = In y. Prove that e" has the following three properties: (i) e" is defined, positive, and strictly increasing for a.II real x; (ii) e0 = 1, lim e" = +oo, lim e"' = O;

.,.....+.

.,....._,.

(iii) e"' is continuous and, in fa.ct, differentiable, with derivative e".

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§ 603)

THE tRIGONOMETRIC FUNCTION$

•7. The function x•, for x cise 6: ·

>0

191

and arbitrary real a, is defined as in Exerza

e

e4 In"'•

Prove the following properties of za: (i) If a > 0 (a < 0), x• is strictly increasing (decreasing); (ii) if a > 0 (a < 0), Jim x• = +ao (0) and lim x• = 0 ( +ao ); --+• ,.....o+ (iii) x• is continuous and, in fact, differentiable, with derivative ax•- 1• •8. If a > 0, prove that the function x•, defined as in Exercise 7 for x > 0, and defined to be 0 when x = 0, has the three properties stated in Exercise 7, except that if 0 < a < 1, the derivative of za at x = 0 is +ao. 1

•9.

Prove that e

= lim (1 ,.....0

!!_ In x

dx

=

lim In (1

+ x);, + h)

h

h-0

Hint: At x

= 1, 1

- In (l)

= lim In (1 + h)h = 1. h....O

1

*10. Prove that lim (1 ,.....0

+ ax); = e4.

*11. Define log. x, where a > 0 and a ~ 1, x > 0, as the inverse of the function a". That is, log. x is that unique number y such that au = x. Prove the standard laws of logarithms, and the change of base formulas: log. x = log.. b lo~ x = 1°gb x. 1o~a

Prove that a1oa:.z = log.. (a"') = x. **12. Show that the function log.. x defined as In x/ln a is the inverse of the function a"' and hence identical with the function of Exercise 11. Derive its other properties from this definition alone. *603. THE TRIGONOMETRIC FUNCTIONS

In a first course in Trigonometry the six basic trigonometric functions are defined. Their definitions there and their subsequent treatment in Calculus, however, are usually based on geometric arguments and intuitive appeal unfortified by a rigorous analytic background. It is the purpose of this section and the following section of exercises to present purely analytic definitions and discussion of the trigonometric functions. That these definitions correspond to those of the reader's previous experience is easily shown after the concept of arc length is available. The development here is restricted to the sine and cosine functions, since the remaining four trigonometric functions are readily defined in terms of those two. Furthermore, the calculus properties of the other four are immediately obtainable, once they are established for sin x and cos x. These properties, as well as those of the inverse trigonometric functions will be assumed without specific formulation here.

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[§ 604

*604. EXERCISES

•1. Prove that the function defined by the equation

. .£"

Arcsmx =

dt

_r:----7,.'

-1

< X < 1,

o v l - t1 is strictly increasing, continuous, and, in fact, differentiable with derivative (1 - x2)-¼. (Fig. 603.) •2. Let the function sin x be defined as the inverse of the function prescribed in Exercise 1 : y = sin x if and only if x = Arcsin y, for -Arcsin i < x < Arcsin f. fl

:,:

FIG. 603

If the number 1r is defined: 1r = 4 Arcsin ½¥2, and if the number bis defined: b = Arcsin i, prove that over the interval (-b, b), containing 1r, sin xis strictly

increasing, continuous, and, in fact, differentiable with derivative (1 - sin' x)¼ [1 - (sin x) 2]½. (Fig. 604.) Also, sin 0 = 0, sin = ½¥2.

vr

=

'II

FIG. 604

*3. Let the function cos x be defined, on the interval (-b, b) of Exercise 2, as the positive square root of 1 - sin2 x. Prove that cos x is differentiable and that the derivatives of sin x and cos x are cos x and -sin x, respectively. Also, cos 0 = 1 and cos ¼,r = ½¥2. Hint: If y = cos x ~ 0, y2 = 1 - sin2 x, 2 cos x dy/dx = -2 sin x cos x.

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193

•4. Let s(x) and c(x) be any two functions defined and differentiable on an open interval (-a, a), a > 0, and possessing there the following four properties: (1) (s(x) )2 + (c(x) )2 = 1, d

(2)

dx (s(x))

= c(x),

(3)

!

=

(c(x))

-s(x),

c(O) = 1. Prove that if a and fJ are any two numbers which, together with their sum a + {J, belong to the interval (-a, a), then the following two identities hold: (5) s(a + fJ) = s(a) c(fJ) + c·(a) s(fJ), (6) c(a + fJ) = c(a) c(fJ) - s(a) s(fJ). Hence prove that if a and 2a belong to (-a, a), then (7) s(2a) = 2s(a) c(a), (8) c(2a) = (c(a) )1 - (s(a) ) 2• Hint: For (5), let 'Y a + {J, and prove that the function s(x) c(-y - x) + c(x) s(-y - x) has an identically vanishing derivative and is therefore a constant. Then let x = a and x = 0, in turn. •5. Let s(x) and c(x) be any two functions satisfying (1)-(4) of Exercise· 4, on the open interval (-a, a), a > 0. Define two new functions in the open interval ( -2a, 2a) by means of the formulas (9) S(x) = 2s(½x) c(½x), (10) C(x) (c( ½x) )2 - (s(½x)) 2 • Prove that S(x) = s(x) and C(x) = c(x) on ( -a, a), and that properties (1)-(4), and therefore also (5)-(8), hold for S(x) and C(x) on (-2a, 2a) (where sand care replaced by Sand C, respectively). •6. Prove that a repeated application of the extension definitions of Exercise 5 provide two functions, sin x and cos x, defined and differentiable for all real numbers, possessing properties (1)-(8) (where s(x) and c(x) are replaced by sin x and cos x, respectively), and agreeing with the originally defined sin x and cos x on the interval (-b, b) of Exercise 2. •7. Prove that sin x and cos x, as defined for all real numbers x, are odd and even functions, respectively. (Cf. Exs. 7-8, § 503.) •S. Prove that sin ½1r = 1, and cos ½r = 0, and that ½r is the smallest positive number whose sine is 1 and also the smallest positive number whose cosine is 0. Hence show that in the closed interval [O, ½r], sin x and cos x are strictly increasing and decreasing, respectively. Hint: If 2k ~ ½r, and if sin 2k = 1, then cos 2k = cos2 k - sin2 k = 0, and cos k = sink = ½V2, and

(4)

=

=

k

= ¼r. •9. Prove that sin 1r = 0 and cos 1r = -1, and that 1r is the smallest positive

number whose sine is O and also the smallest positive number whose cosine is -1. •10. Prove that sin 2r = 0 and cos 2r = 1, and that 21r is the smallest positive number whose cosine is 1.

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•11. Prove that sin x and cos x are periodic with period 2r. That is, sin (x + 2r) = sin x and cos (x + 2r) = cos x for all x, and if either sin (x + k) = sin x or cos (x + k) = cos x for all x, then k = 2nr, for some integer n. •12; Prove that sin x and cos x have the familiar signs in the appropriate "quadrants"; namely, +, +, -, -, and +, -, -, +, respectively, for x in the ranges (0, ½r), (½r, r), (r, Jr), (Jr, 2r). •13. Prove that if X and µ are any two real numbers such that X2 + µ 2 = 1, then there is a unique value of x in the half open interval [0, 2r) such that sin x = X and cos x = µ. Hint: Let :co be the unique number in the closed interval [0, ½1r] whose sine is IXI, and choose for x the appropriate number according to Ex. 12: Xo, r - Xo, r + Xo, 2r - Xo. •H. Prove that -1 ~ sin x ~ 1 and -1 ~ cos x :i 1 for all x. •15. Prove that sin x and cos x have, everywhere, continuous derivatives of all orders. . sin x •16. Prove that bm - - = 1. z-,()

Hint: Let J(x)

X

sin x.

a

Then J'(0) • lim J(O h....O

••17. Prove that the function J(x)

e

sin x, x



X

+ hi - f(O)_ 0, J(0)

5i

1, (i) is everywhere

continuous; (ii) is everywhere differentiable; (iii) has everywhere a continuous nth derivative for all positive integers n. 605. SOME INTEGRATION FORMULAS

We start by asking two questions. is defined only for x

f a -x

(i) If

f ~ = ln x +

C, and if ln x

> 0, how does one evaluate the simple integral

f-

-3

2

dx? X

f

+ x + C and x2 dx-a2 = 21a ln xx +- aa + C'' a a-x where a > 0, why cannot each of these integration formulas be obtained from the other by a mere change in sign? In other words, why does their sum, which should be a constant, give formally the result (ii) If

2

dx

.!. ln ~

2

= 21 ln a

(aa-x + x•x+a x- a) + C + C' or .!_ ln ( -1) + C + C' ~

which does not even exist l The answers to these questions lie most simply in the use of absolute values. We know that if x then dd (ln ( -x)) = X

> O, then fx (ln (x)) = ~

...!... (-1) -X

=

!. X

and that if x

< 0,

That is, ln x and ln ( -x) both have

the same derivative, formally, but have completely distinct domains of definition, ln x being defined for x > 0 and ln (-x) for x < 0. The func-

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195

tion In lxl encompasses both and is defined for any nonzero x. single integration formula

Thus the

(1)

is applicable under all possible circumstances. The integration of question (i) is therefore simple: -2dx - = [ ln lxl = ln 1-21 - ln l-31 = ln j.

J

-3

]-2

X

-3

For similar reasons the formulas of question (ii) are only apparently incompatible, since they apply to different domains of definition. The first is applicable if x2 < a2 and the second if x2 > a2 • (Ex. 11, § 606.) However, again ·a single integration formula is available which is universally applicable, and can be written alternatively in the two forms:

xi a-x f a -x 2a1 la+ f ~ = ;a 1n 1: ~ :1 +

dx=2 - l n - - + C · -2 '

(2)

a for the

a2 dx = ½[xv' x2 ± a 2 ± a2 ln Ix + v' x2 ± a 2 I]

- case;

+ C, lxl > a

for the - case; (9)

f

dx --=== = xv'a2 ± x2

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al

a ±x -1 ln Iv' -----"'---+ C, lxl a x

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Since the derivative of sec x tan x is sec• x

+ sec x tan x = sec• x + sec x (sec2 x 2

= 2 secl x

- sec x, and since by (5), above, the derivative of In 1sec x + tan xi is sec z, we have, by addition,

f

(10)

sec• x dx

1)

= ½sec z tan z + ½ln 1sec z + tan xi + C.

Finally, we give four more integration formulas, which involve inverse trigonometric functions. These are valid for the principal value ranges specified. The student should draw the graphs of the functions involved and verify the statements just made, as well as show that the formulas that follow are not all valid if the range of the inverse function is unrestricted (Ex. 13, § 606).

f va2-z2 = f - -+- = a f =

Arcsin ~

dx

(11)

dx

(12)

a2

dx

f xvx• - a

(14)

+ C'

-a< x

1 X 11" -Arctan - + C - a ' 2

xt

Vat - z 2 dx

(13)

a

=

½xv'at - z 2

{-¾

2

< a' 2 _! ~ -

Arcsin ~ ~ '.!. a-2'

1r < Arctan -Xa < -· 2'

+ ½a

2

Arcsin ~

+ C, lzl < a;

+ C, x > a, 1 . (a) a Arcsm X + C, X < -a. Arcsin (;)

606. EXERCISES

In Exercises 1-10, perform the integration, and specify any limitations on the variable x.

f . f v2 - x'· f f

1.

tan 5x dx.

3

dx

6

dx V4x2 -4x+5

1 •

dx xV5 - 2x2

·

f '- f + f 'V:,; f f +

2.

sec 4x dx.

Vx 2

3x' dx.

6.

8.

lO.

4x dx.

se~dx. 3x2

~x

+ 7•

11. Show that in formula (2), § 605, the absolute value signs can be removed if lzl < a, and similarly for formula (2'), § 605, if lxl > a. Hints: The quotient of a + x and a - x is positive if and only if their product is positive. 12. Establish formulae (3)-(9), § 605, by direct evaluation rather than mere

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197

differentiation of the right-hand members. Hints: For (3)-(6) express the functions in terms of sines and cosines. For (7)-(9) make trigonometric substitutions and, if neceBBary, use (10), § 605. 13. Establish formulas (11)-(14), § 605, by direct evaluation. (Cf. Ex. 12.)

607. HYPERBOLIC FUNCTIONS

The hyperbolic functions, called hyperbolic sine, hyperbolic cosine, etc., are defined: . e" - e-z 1 smhx= cothx = -- , , 2 tanhx e"' + e-z 1 (1) cosh x = , sech x = - - , 2 coshx 1 csch x = -.-- . smhx

tanh x = sinh x, coshx

These six functions bear a close resemblance to the trigonometric functions. · For example, sinh x, tanh x, coth x, and csch x are odd functions and cosh x and sech x are even functions. (Ex. 21, § 609.) Furthermore, the hyperbolic functions satisfy identities that are similar to the basic trigonometric identities (verify the details in Ex. 22, § 609): (2)

cosh2 x - sinh2 x 2

= 1;

= sech2 x; 1 = csch2 x; y) = sinh x cosh y ±

(3)

1 - tanh x

(4)

coth2 x -

(5)

sinh (x ±

(6)

cosh (x ± y) = cosh x cosh y ± sinh x sinh y;

(7)

tanh (x ± ) y

(8)

sinh 2x = 2 sinh x cosh x;

=

cosh x sinh y;

tanh x ± tanh y : 1 ± tanh x tanh y'

(9) cosh 2x = cosh2 x + sinh2 x = 2 cosh2 :r - 1 = 2 sinh2 x + 1. The differentiation formulas (and therefore the corresponding integration formulas, which are omitted here) also have a familiar appearance (Ex. 23, § 609):

(13)

= cosh x; d(cosh x)/dx = sinh x; d(tanh x)/dx = sech2 x; d(coth x)/dx = -csch2 x;

(14)

d(sech x)/dx = -sech x tanh x;

(15)

d(csch x)/dx = -csch x coth x.

(10) (11) (12)

d(sinh x)/dx

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The graphs of the first four hyperbolic functions are given in Figure 605. A set of integration formulas (omitting those that are mere reformulations of the differentiation formulas (10)-(15) follows (Ex. 24, § 609):

f f f f

(16) (17) (18)

(19)

tanh x dx coth

= ln cosh x + C;

x dx = ln lsinh xi

+ C;

sech x dx = Arctan (sinh x) csch x dx

= lnltanh

~I+

+ C;

C.

NOTE. ·The trigonometric functions are sometimes called the circular functions because of their relation to a circle. For example, the parametric equations of the circle x1 + y1 = a1 can be written x = a cos 8, y = a sin 8. In analogy with this, the hyperbolic functions are related to a hyperbola. For example, the parametric equations of the rectangular hyperbola x2 - y1 = a' can be written x "" a cosh 8, y = a sinh 8. In this latter case, however, the parameter 8 does not represent the polar coordinate angle for the point (x, y).

y

1 y

= sinh X

y=coshx

_ _ _ _ _ :L_ __

y

1

'I= tanh

x

y

= coth x

FIG. 605

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199

608. INVERSE HYPERBOLIC FUNCTIONS

The integrals of certain algebraic functions a.re expressed in terms of inverse trigonometric functions(§ 605). In a similar fashion, the integrals of certain other algebraic functions can be expressed in terms of inverse hyperbolic fqnctions. The four hyperbolic functions whose inverses are the most useful in this connection are the first four, sinh x, cosh x, tanh x, and coth x. The graphs and notation are given in Figure 606. For simplicity, by analogy with the principal value ranges for the inverse trigonometric functions, we choose only the nonnegative values of cosh-1 x, and write for these principal values Cosh-1 x. 'Y

'Y

'Y

'Y I

I I

I

1

,1 I I

I

I I I I

'IJ

= sinh-1 :e all real z

I

y - Cosh-1 :e z ~1

'Y

= tanh-1 :e lxl < 1

l I

X

I

I

l

:e

11

I I I

I I

I

I

I

I

y

= coth-1 :e lzl> 1

FIG. 606

The inverse hyperbolic functions can be expressed in terms of functions discussed previously:

= In (x + v'x2 + 1), all real x;

(1)

sinh-1 x

(2)

Cosh-1 x

= In (x + ~ ) , x

(3)

tanh-1 x

1 = ½In 1+ xx'· lxl < 1 ·'

(4)

coth-1 x

= ½In xx+ 11 ; lxl >

~ 1;

1.

We prove (2) and leave the rest for the student (Ex. 25, § 609). If

x

11 = e +2 e-11, let

the positive quantity e11 be denoted by p. Then

= p + !, or p2 - 2xp + 1 = 0. Solving this equation p p = x ± ~ - Since y is to be chosen nonnegative, p = e

2x

11

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SOME ELEMENTARY FUNCTIONS

[§ 609

x > 1, x + ~ > 1, and since the product of this and x - ~ is equal to 1, x - v'x2- 1 < 1. We therefore reject the minus sign and have p = x + Vx2 - 1, or equation (2). Notice that the expressions in (3) and (4) exist for the specified values of x. For example, in (3), (1 + x) and (1 - x) have a positive quotient if and only if they have a positive product, and 1 - x2 > 0 if and only if lxl < 1. The derivatives of the four inverse hyperbolic functions considered here can either be obtained from the derivatives of the corresponding hyperbolic functions (Ex. 26, § 609) or from formulas (1)-(4) (Ex. 27, § 609). They are:

k, 1+

(5)

d sinh-1 x = dx

(6)

d - Cosh-1 x dx

(7)

1 1:.._ tanh-1 x = - - , dx 1-x2 d - coth-1 x dx

(8)

x2

all real x;

1 Vx - l

= --===, x > 1; 2

lxl < 1 · '

1

=- , lxl > 1. 1 - x2

The corresponding integration formulas are mere reformulations of formulas (2) and (7) of § 605, with appropriate ranges specified. They can be established independently by differentiations of the right-hand members (Ex. 28, § 609). The letter a represents a positive number throughout. dx

sinh-1 ~

--;=dx== 2 2

Cosh-1 ~

f v'a2+x2 = f Vx -a = f -a2___x_2 = a f -x2___a_2 = -a

(9)

dx

(11)

dx

(12)

1

a

tanh-1

1

+C

a

'

all real x·

'

+ C, x > a;

a+ C, lxl X

coth-1

< a;

a+ C, lxl > a. X

609. EXERCISES

In Exercises 1-10, differentiate the given function. 1. cosh 3x. 2. sinh2 x. 3. 6. 7. 9.

tanh (2 - x). In sinh 2x. sinh-1 4x. tanh-1 x2•

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4. x coth x2 • 6. e"" cosh bx. 8. Cosh-1 es. 10. coth-1 (sec x).

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§ 610]

201

In Exercises 11-20, perform the indicated integration, expressing your answer in terms of hyperbolic functions or their inverses.

f f f J • Vx f ++

11.

tanh 6x dx.

12.

13.

cosh8 x dx.

14.

16.

sinh2 x dx.

16.

17

dx

19 •

2 -

1

es coth es dx. tanh2 !Ox dx. x8 tanh-1 x dx.

1s.J

2

dx

3x1

f f f

20 •

5x - 7 5x - 7 > 0).

dx • v'4x2 -4x+5

f

1 - 5!x_ 3x1 ' (7 - 5x - 3x2 > 0).

(3x 21. Prove that cosh x and sech x are even functions and that the other fom hyperbolic functions are odd (cf. Exs. 7-8, § 503). 22. Establish the identities (2)-(9), § 607. 23. Establish the differentiation formulas (10)-(15), § 607. 24. Establish the integration formulas (16)-(19), § 607. Hint for (19): _k_ du _ h cosh x - 1 _ 2 sinh2 ½x. . h ~ l where u, - cos x, and COSh X Sill X U, 1 - 2 COS h2 .1irX 26. Establish formulas (1), (3), (4), § 608. 26. Establish formulas (5)-(8), § 608, by use of (10)-(13), § 607. Hint for (5): Let y = sinh-1 x. Then x = sinh y, dx/dy = cosh y, and dy/dx = 1/v'l + sinh1 y. 27. Establish formulas (5)-(8), § 608, by use of (1)-(4), § 608. 28. Establish formulas (9)-(12), § 608.

f

-f

+

*29. Show that

f

csch x dx

= -coth-1 (cosh x)

+ C.

*30. Establish formulas (7)-(9), § 605, by means of hyperbolic substitutions. *610. CLASSIFICATION OF NUMBERS AND FUNCTIONS

Certain classes of numbers (integers, rational numbers, and irrational numbers) were defined in Chapter 1. Another important class of numbers is the algebraic numbers, an algebraic number being defined to be a root of a polynomial equation (1) aox" + a1x•-1 + ··· + a. = O, where the coefficients ao, a1, · • • , a,. are integers. Examples of algebraic numbers are ¾(a root of the equation 4x - 3 ·= 0) (in fact,. every rational number is algebraic), - -¢'5 (a root of the equation :,;I+ 5 = 0), and-¢'5 + V3 (a root of the equation x" - 9x' - 10x3 + 27:,;2 - 90:,; - 2 = 0). It can be shownt that any number that is the sum, difference, product, or

t Cf. Birkhoff and MacLane, A Survey of Modern Algebra (New York, The Macmillan Company, 1944).

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[§ 610

quotient of algebraic numbers is algebraic, and that any (real) root of an algebraic number is an algebraic number. Therefore any number (like

V3 - v's/9 / 4-v'fil) that can be obtained from the integers by a finite sequence of sums, differences, products, quotients, powers, and roots is algebraic. However, it should be appreciated that not every algebraic number is of the type just described. In fact, it is shown in Galois Theory (cf. the Birkhoff and MacLane book just referred to) that the general equation of degree 5 or higher cannot be solved in terms of radicals. In particular, the real root of the equation x• - x - 1 = 0 (this number is algebraic by definition) cannot be expressed in the finite form described above. A transcendental number is any number that is not algebraic. The most familiar transcendental numbers are e and r. The transcendental character of e was established by Hermite in 1873, and that of r by Lindemann in 1882. t For a discussion of the transcendance of these two numbers see Felix Klein, Ekmenl,ary Mathematics frqm an Advanced Standpoint (New York, Dover Publications, 1945). The first number to be proved transcendental was neither e nor r, but a number constructed artificially for the purpose by Liouville in 1844. (Cf. the Birkhoff and MacLane reference, page 413.) The existence of a vast infinite supply of transiiendental numbers was provided by Cantor in 1874 in his theory of transfinite cardinal numbers. This amounts to showing (Ex. 10, § 612) that the algebraic numbers are denumerable (Ex. 12, § 113) and that the real numbers are not (Ex. 30, § 711). This method, however, merely establishes existence without specifically producing examples. Functions are classified in a manner similar to that just outlined for numbers. The role played by integers above is now played by polynomials. A rational function is any function that can be expressed as the quotient of two polynomials, an algebraic function is any function f(x) that satisfies (identically in x) a polynomial equation (2)

ao(x)(f(x) )"

+ a1(x)(f(x) )"-1 + · ·· + a,.(x) = 0,

where ao(x), a1(x), • • • , a,.(x) are polynomials, and a transcendental func-

tion is any function that is not algebraic.

Examples of polynomials are 2x - V3 and rr' + Ve. Examples of rational functions that are not polynomials are 1/x and (x + V3)/(5x - 6). Examples of algebraic functi~ns that are not rational are Vz and 1/v'xi + r. Examples of transcendental functions are e"', In x, sin x, and cos x. (Cf. Exs. 11-13, § 612.)

t It is easy to show that e is irrational (Ex. 34, § 811). A proof of the irrationality of r was given in 1761 by Lambert. For an elementary proof, given in 1947 by Niven, see Trygve Nagell, Introduction to Number Th«rry (New York, John Wiley and Sons, Inc., 1951).

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§

611]

THE ELEMENTARY FUNCTIONS

203

*611. THE ELEMENTARY FUNCTIONS

Most of the familiar functions which one encounters at the level of ele-. mentary calculus, like sin 2z, e-z1, and ½Arctan ½x, are examples of what are called elementary function8. In order to define this concept, we describe first the elementary operations on functions. The elementary operations on functions f(z) and g(x) are those that yield any of the following: f(x) ± g(x), f(z) g(x), f(z)/g(x), {J(z)}•, al, log..f(x), and T(f(z) ), where T is any trigonometric or inverse trigonometric function. The elementary functions are those generated by constants and the independent variable by means of a finite sequence of elementary operations. Thus (zt

+ 17r)• Arcsin [(loga cos x)(e"°';)]

is an elementary function. Its derivative is also an elementary function. In fact (cf. Ex. 9, § 612), the derivative of any elementary function is an elementary function. Reasonable questions to be asked at this point are, "Is there ever any occasion to study functions that are not elementary? What are some examples?" The answer is that nonelementary functions arise in connection with infinite series, differential equations, integral equations, and equations defining functions implicitly. For example, the differential equation ~ -

si:

x ( -1 if x - 0) has a nonelementary solution

J:

{l)

si: t dt

whose power series expansion (cf. Chapter 7) is

zl

zl

x7

x - 3.3 I+ 5·51- 7.7 I+ ....

Other examples of nonelementary functions which are integrals of elementary functions are the probability integral,

J:

(2)

e-P dt,

the Fresnel sine inf,egral and cosine int,egral,

J:

(3)

sin P dt and

J:

cos P dt,

and the elliptic integrals of the first and second kind, (4)

r v' 1 - dtk sin t

Jo

2

2

and

r Vl -

Jo

k2 sin2 t dt.

Extensive tables for these and other nonelementary functions exist. t

t Cf. Eugen Jahnke and Fritz Emde, Tablu of Functions (New York, Dover Publications, 1943).

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204

L§ 612

Proofs that these functions are not elementary are difficult and will not be given in this book. t The inverse of an elementary function need not be elementary. For example, the inverse of the function x - a sin x (a r6 0), the function defined implicitly by Kep'ler's equation y - a sin y - x = O,

(5)

is not elementary.t Again, the proof is omitted. An important lesson to be extracted from the above considerations is that inability to integrate an elementary function in terms of elementary functions does not mean that the integral fails to exist. In fact, many important nonelementary functions owe their existence to the (Riemann) integration process. This is somewhat similar to the position of the logarithmic, exponential, and trigonometric functions as developed in this chapter, which were defined, ultimately, in terms of Riemann integrals of algebraic functions. One more final remark. The reader might ask, "Apart from the importance of the functions defined above, if all we seek is an example of a nonelementary function, isn't the bracket function [x] of§ 201 one such?" Perhaps the most satisfactory reply is that whereas [x] is elementary in intervals, the functions presented above might be described as "nowhere elementary." A behind-the-scenes principle here is that of analytic continuation, important in the theory of analytic functions of a complex variable. (Cf. § 815.) *612. EXERCISES

•1. Prove that v'5 - V2 is algebraic by finding a polynomial equation with integral coefficients of which it is a root. •2. Prove that 1/x is not a polynomial. Hint: It is not sufficient to remark that it does not look like one. Show that 1/x has some property that no polynomial can have, such as a certain kind of limit as x - oo. (Cf. Ex. 4.) *3• Prove that x :

1

is not a polynomial (cf. Ex. 2).

•4. Prove that v' x2 + 1 is not a polynomial. tion a large number of times.

Hint: Differentiate the func-

•5. Prove that v' x2 + 1 is not rational. (Cf. Ex. 4.) •6. Prove that the hyperbolic functions and their inverses are elementary. •7. Prove that x" is elementary.

*8. Prove that fo"'t,. sin t dt, where n is a positive integer, is elementary. •9. Prove that the derivative of any elementary function is elementary.

t Cf. J. F. Ritt,

Integration in Finite Terms (New York, Columbia University Press,

1948).

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EXERCISES

§ 612]

205

•10. Prove that the algebraic numbers are denumerable (cf. Ex. 12, § 113). Hint: A useful device of Cantor was to let p{x) = aox" + · · · + a,. = 0 be the (unique!) polynomial equation of lowest degree and smallest positive leading coefficient ao (ao, • • • , a,. being integers) satisfied by a given algebraic number a and to define the height of a to be the positive integer n - 1 + laol + • • • la.. l. Show that there are on,ly finitely many algebraic numbers having a given height. •11. Prove that ez is transcendental. Hint: Let ao(x) e"" + a,(x) e• + ••• + a,._1(x) ez + a,.(x) = 0, where ao(x), • • • , a,.(x) are polynomials and a,.(x) is not identically 0. Form the limit as x-+ -oo. Then Iim a,.(x) = 0 (!). •12. Prove that In xis transcendental. Hint: Let y a In x, and assume that y satisfies identically a polynomial equation, arranged in the form bo(y) x"

+ b (y) x 1 + ••• + b,._ (y) x + b.. (y) = 0, 1

11 -

1

where bo(y) is not identically 0. Divide every term by x•, and let x-+ +oo. •13. Prove that sin x and cos x are transcendental. Hint for sin x: Let ao(x) sin" x + a,(x) sin•-1 x + ••• + a,......,(x) sin x + a,.(x) = O, where ao(x), • • • , a,.(x) are polynomials and a,.(x) is not identically 0. Let x = kr, where k is so large that a,.(x) -;,6, 0.

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7 Infinite Series of Constants

701. BASIC DEFINITIONS If

a1, tit,

••• ,

a,., ••• is a sequence of numbers, the expression

(1)

(also written

Ea,.

if there is no possible misinterpretation) is called an

infinite series or, in brief, a series. The numbers ai, a,, • • • , a,., • • • are called terms, and the number a,. is called the nth term or _general term. The sequence {8,.}, where

81 = a1, 82 = a1 + a,, · · · , 8,. E a1 + a, + · · · + a,., is called the sequence of partial sums of the series (1). An infinite series converges or diverges according as its sequence of partial sums converges or diverges. In case of convergence, the series (1) is said to have a sum equal to the limit of the sequence of partial sums, and if 8 = lim S,., ......+..

we write (2)

+•

Ea,. = n-1 E a,. = a1 + · · · + a,. + · · · = 8,

and say that the series converges to S. In case lim 8,. exists and is infinite the series

-+•

Ea,., although divergent,

is said to have an infinite sum:

+•

Ea,. = n-1 E a,. = a1 + ·· • · + a,. + • • • = + 00, - 00 , or 00 ,

(3)

and the series is said to diverge to +00, - 00 , or 00 . Example 1.

! + ¼+ ••• + :.. + ••• =

1, since the sum of the first n

terms is S,.

1

1

1

= 2 + 4 + ... + 2" = 1 -

1 211'

/

and S,.-+ 1 (cf. Example 3, § 202). 206

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§702]

THREE ELEMENTARY THEOREMS

Ezample 2. The series whose general term is 2n diverges to

207

+ao:

2 + 4 + 6 + ••• + 2n + ••• = +ao. Eumple 3. The series 1 - 1 + 1 - 1 + ... + (-1) 11+1 + ... diverges since its sequence of partial sums is 1, 0, 1, 0, 1, 0, 1, · · · . This sequence diverges since, if it converged to S, every subsequence would also converge to S, and S would equal both 0 and 1, in contradiction to the .uniqueness of the limit of a sequence. (Cf. § 204; also Example 4, § 202.)

702. THREE ELEMENTARY THEOREMS

Direct consequences of Theorems I, II, and X, § 204, are the following:

Theorem I. The alteration of a finit,e number of terms of a series ~s no effect on corwergence or divergence (aUhough it will in general change the sum in ca8e of corwergence). Theorem D. If a aeries corwerges or has an injinit,e sum, this sum is unique.

m.

Theorem Multiplication of the terms of a series b-y a nonzero constant k does not affect corwergence or divergence. If the original series corwerges, the new series converges to k times the sum of the original, Jor any constant k: t!' +• 2.,;ka,.=kI:a,..

(1)

,a-1

n-1

(Cf. Ex. 27, § 707.) 703. A NECESSARY CONDITION FOR CONVERGENCE

Theorem. If a series corwerges, its general term tends toward zero as n becomes infinire. Equivalently, if the general term of a series does not rend toward zero as n becomes infinire, the series diverges.

+ + · ·· ,

Proof. Assume the series, a1 at converges to S. Then S,. = S,._1 + a,., or a,. = S,. - S,._1. Since the sequence of partial sums converges to S, both S,. and S,._1 tend toward S as n becomes infinite. Hence (Theorem VII, § 204), lim a,. ,.. 1im (8,. - S,._1) = S - S = 0.

-+•

-+•

Eumple. Show that the series 2 + 4 + ••· + 2n + · · · and 0 + 1 + 0 + 1 +••·diverge. Solution. In neither case does the general term tend toward zero, although the second series contains a subsequence converp;inp; to zero.

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208

[§ 704

NOTE. The condition of the theorem is necessary but not sufficient for convergence. It will be shown later (Theorem IV, § 708) that the harmonic series

1 + ½+ ½+

· · · diverges, although the general term, !, converges to zero. n

704. THE GEOMETRIC SERIES

The series

a + ar

(1)

+

ar2

+ ••• +

arn-1

+ ...

is known as a geometric series. Theorem. with

1

The geometric series (1), where a

¢

0, converges if

lrl
1. 3 9 . an dll r < 1. Th e sum 1s _ ( _ j) = . 1 5 (b)

705. POSITIVE SERIES

For a positive series (a series whose terms are positive) or, more generally, a nonnegative series (a series whose terms are nonnegative), there is a simple criterion for convergence.

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§ 706]

THE INTEGRAL TEST

Tile series

Theorem. .

+•

E

a,., where a,.

~

209

0, oorwerges if and only if tile

11-1

sequence of partial sums is bounded; that is, if and only if there is a number A such that a1 + · ·· + a,. ~ A for all values of n. If this oondition is satisfied, tile sum of tile series is l.ess than or equal to A. A series of nonnegative terms either oorwerges, or diverges to + oo. Proof. Since B..+1 = S,. + a,., the condition a,. ~ 0 is equivalent to the condition S,. f , that is, that the sequence {S,.} is monotonically increasing. The Theorem of this section is therefore a direct consequence of the Note following Theorem XIV, § 204. 706. THE INTEGRAL TEST

A convenient test for establishing convergence or divergence for certain series of positive te~s makes use of the technique of integration.

Theorem. Integral Test. If Ea,. is a positive series and if f(x) is a positive monotonically decreasing function, defined for x ~ a, wllere a is some real number, and if (for integral values of n) f(n) = a,. for n > a, tlten tile im'J)Toper integral

Jar+• f(x) dx

(1)

and tile infinite series (2)

eitlter both oorwerge, or both diverge to +oo. Proof. Let us observe that both (1) and (2) always exist, in the finite or + 00 sense. (Why? Supply details in Ex. 28, § 707.) We wish to show that the finiteness of either implies the finiteness of the other. With the aid of results given in §§ 511, 702, and 705, we have only to show that if K is a fixed positive integer ~ a and if N is an arbitrary integer > K, then the boundedness (as a function of N) of either (3)

or N

(4)

J(N)

= E a,. n-K

implies that of the other. (Cf. Ex. 29, § 707). Let us look at the geometric representations of (3) and (4) given in Figure 701. Each term of (4) is the area of a rectangle of width 1 and height a,.. This suggests that if (3) is written as the sum

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INFINITE SERIES OF CONSTANTS

210

(5)

l(N) =



N

K

f(x) dx =

N-1E f+l

n-K

n

[§ 706

f(x) dx,

then each term of (5) is trapped between successive terms of (4). Indeed, for the interval n ~ x ~ n + 1, since f(x) is monotonically decreasing,

a,. so that

a..

=

[ +I

= f(n)

a,. dx i5;

i5; f(x) i5; f(n

f+l n

+ 1) = a..+1,

f(x) dx i5;

f+l 11

ao,+1

dx

= ao,+1-

FIG. 701

Therefore

Consequently (in the notation of (3) and (4) ), since

J(N - 1) i5; l(N) i5; J(N) -

aK,

These two inequalities show that the boundedness of either J(N) or l(N) implies that of the other (Ex. 30, § 707) and the proof is complete. Example 1. Show that the po.series

+• 1

En:P

n-1

converges if p > l and diverges if p ;:a; I. Solution. This follows immediately from Example 3, § 51', if p p < 0, the conclusion is trivial. ~ 13

Example 2. Show that Solution.

+• E (I 1 ) converges if and only if p 11-2n nn:P

The integral (+•~converges if and only if p J2 z(ln z):1>

28, § 513.)

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~

0.

If

l.

>

l.

(Ex.

§ 707]

EXERCISES

211

707. EXERCISES

In Exercises 1-4, find an expression for the sum of the first n terms of the series. Thus find the sum of the series, if it exists. = ! __ 1_, 1 1 1 . 1 1 n n + 1 • f:2 + 2·3 + 3·4 + · · · · Hint: n(n + l) 1 1 1 2• i::'a + 3.5 + 5.7 + .... 3.

½+i+1'r+lr+ ···.

4. l +2+3+4+ ....

In Exercises 5-8, write down the first four terms and the general term of the series whose sequence of partial sums is given. 6. 2, 2½, 21, 2¾, · · · . 6. 1, 3, 1, 3, 1, 3, · · · 7. 1.3, 0.91, 1.027, 0.9919, · · · , 1 - ( - .3)", · · · . 8. 2, 1.3, 1.09, 1.027, ... ' 1 + (.3) 11- 1, ••••

In Exercises 9-12, determine whether the given geometric series is convergent or not, and find its sum in· case of convergence.

+ ···.

9. 12 - 8 11. 0.27 + 0.027 + · · · •

+ ···.

10. 8 - 12 12. 101 - 100 + .. · .

In Exercises 13-16, find as a quotient of integers the rational number represented by the repeating decimal, by means of geometric series. 13. 0.5555 · · · . Hint: 0.5555 · • • means 0.5 + 0.05 + 0.005 + · · · • 14. 3.285555 • • •· Hint: Write the number 3.28 0.005555 • • • . 16. 6.1727272 .. · . 16. 0.428571428571428571 , · · .

+

In Exercises 17-26, establish convergence or divergence by the integral test of§ 506. . 1 18. L 3' 11. :E 2n + i·

nz:

19. :E n 2

1

+ 4'

20. :E!!:.,

e" +..

+..

1 21. :E --;--• n-3 n - 4 23.

L lnn. n

1

1

26. :E (n

1

22. :E--===n-4v'n2 - 9 +.. 1 24 • n~3 n(ln n)(ln In n)2°

+ l)I°

26.:En1+1·

27. Establish formula (1) of Theorem III, § 702, for the case of an infinite sum (+co, -co, or co). 28. Explain why both (1) and (2), § 706 must exist, as either finite or infinite limits. (Cf. Theorem X, § 501.) 29. For the Theorem of § 706 prove that it is sufficient to show that the boundedness of either (3) or (4) implies that of the other. In particular, explain why we may use K in place of a in (3) and Kin place of 1 in (4).

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INFINITE SERIES OF CONSTANTS

212

30. Supply the details needed in the last sentence of the proof of § 706. +oo 1 31. Prove that I: (l )(I ) converges if and only if p > 1. More n-3 n nn n 1nn P +oo 1 • ) ) , where N is sufgenerally, prove that I: (l )(l (l 1 1 1 n-N n n n n n n · · · n n · · · n n P ficiently large for (In In · · · Inn) to be defined, converges if and only if p > 1. (Cf. Ex. 29, § 513.) *32. Indicate by a graph how you could define a function f(x) having the

~

properties: f(x) is positive, continuous, defined for x verges, and

+..

I:

n-1

/(n) diverges.

j/ /(x) dx con00

0,

(Cf. Ex. 40, § 513.)

*33. Indicate by a graph how you could define a function /(x) having the properties:/(x) is positive, continuous, defined for x

~

0,

f/

00

/(x) dx diverges,

+oo and

I:

n-1

/(n) converges.

708. COMPARISON TESTS.

DOMINANCE

In establishing the integral test of § 706 we have already made use of the fact that a series of positive terms converges if and only if its sequence of partial sums is bounded. In order to prove convergence for a series of positive terms it is necessary to show only that the terms approach zero fast enoUfJh to keep the partial sums bounded. One of the simplest and most useful ways of doing this is to compare the terms of the given series with those of a series whose behavior is known. The ideas are similar to those relevant to comparison tests for improper integrals (§ 512). · Definition I. means that ja,.1

The statement that a series I:b,. dominates a series Ea,. b,. for every positive integer n.

~

NOTE 1. Any dominating series consists automatically of nonnegative terms, although a dominated series may have negative terms.

Theorem I. Comparison Test. Any nonnegative series dominated by a convergent series converges. Equivalently, any series that dominates a divergent nonnegative series diverges. Proof. We shall prove only the first form of the statement (cf. Ex. 21, § 711). Assume O ~ a,. ~ b,. and that I:b,. converges. The convergence N

of I:b,. implies the boundedness of the partial sums

I: b,., which implies

n•l

N

the boundedness of the partial sums

I: a,.. By the Theorem of § 705, n=l

the proof is complete.

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§ 708]

COMPARISON TESTS.

Eumple 1. Prove that the series 1 1 1 + 2' + 3'

DOMINANCE

213

+ ... n"1 + ...

converges. Solution.

Since

l. ~

n"

b for n > 1, the given series is dominated by a con~--

vergent geometric series (except for the first term).

Eumple 2. Prove that the series 1 1 1 1

1

1

This series is dominated by 1 1 1 1 1

1

2+3+21+3,+23+31+ ...

converges. Solution.

2 + 2 + 21 + 21 + 21 + 21 + .. ·,

.

whose partial sums are bounded by twice the sum of the convergent geometric . ~ 1

senes

L-

,.. 2

Let us adapt the "big 0" notation of§ 512 to the present topic:t

Definition II. If {a.} and {b,.} are two 8equences, ~ notation a,. = O(b,.) (read "a sub n is big O of b 8ub n"), means that there exi8t a positive integer N and a positive number K 8UCh that n > N implie8 la.I ~ K b,.. If simuUaneously a,. = O(b,.) and b,. = O(a,.), the two 8equence8 (and also the two 8erie8 Ea,. and Eb.) are 8aid to be of the same order of magnitude at +ao. NoTE 2. As in§ 512, the "big O" notation is also used in the following sense: a,. = b,. O(c..) means a,. - b,. = O(c,.).

+

Big O and order of magnitude relationships are usually established by taking limits:

Theorem II. Let {a,.} and {b,.} be 8equence8 of positive terms and as8ume that lim ab,. = L exists, in a finite or infinite 8ense. Then: n-++• " (i) 0 ~ L < +ao implie8 a,. = O(b,.), (ii) 0 < L ~ +ao implie8 b,. = O(a,.), (iii) 0 < L < +ao implies that {a..} and {b,.} are of the 8ame order of magnitude at +ao. (Give the details of the proof in Ex. 22, § 711. Cf. Theorem II, § 512.)

t The "little o" notation of § 512 takes the form for sequences: a,. - o(b,.) means that there exist a positive integer N and a sequence of nonnegative terms K,. such that K,.-+ 0 and such that n > N implies Ja,.J ~ K,.b,.. Obviously a,. - o(b,.) implies a,. - O(b,.).

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INFINITE SERIES OF CONSTANTS

[§ 708

Theorem m. If I:a,. and I:b,. are series of nonnegative terms and if a,. = O(b,.), then the convergence of I:b,. implies that of I:a,.. Equivalenlly, the divergence of I:a,. implies that I:b,.. +• Proof. The convergence of L Kb,. implies that of n-N+l

Theorem I.

Corollary. Two positive series that have the same order of magnitude either both converge or both diverge. Because of the usefulness of the p-series,

I:~,,,

given in Example 1,

§ 706, as a "test series," we restate here the facts established in that Example.

1:' .!.n" converges if p > 1 and diverges if As a special case, the harmonic series, 1:' .!, diverges. t n

Theorem IV. The p-series

n-1

~

p

1.

n-1

(Cf. Ex. 29, § 707, for a sequence of test series similar to the p-series, each converging more slowly than the preceding.) The technique of the comparison test for positive series, particularly in the form of Theorem III, consists usually of four steps: (i) get a "feeling" for how rapidly the terms of the given series are approaching zero; (ii) construct a new test-series of positive terms which dominates, is dominated by, or is of the same order of magnitude as the original series; (iii) establish the necessary inequalities, or the limit of the quotient of the general terms; (iv) infer convergence if the dominating series converges, and divergence if the dominated series diverges. In practice, one frequently postpones applying a comparison test to a given series until after one has tried one of the more automatic tests (like the ratio test) given in subsequent sections. Example 3.

Test for convergence or divergence: 1 2 3 4 i-:'a + 5.7 + 9·11 + 13·15 +

....

t The function defined by the convergent p-series is called the Riemann Zeta-function: +• 1

rczl = n-1 I: -· n•

c1>

This function is defined for complex values of z, by (1) if the real part of z > 1, and by other means (infinite products, improper integrals) for other values of z. Its values have been tabulated (cf. Eugen Jahnke and Fritz Emde, Tables of Functiom (New York, Dover Publications, 1943).

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§ 709]

THE RATIO TEST

215

Solution. The general term is ( n _ ~ n _ l) which evidently is of the 4 3 4 same order of magnitude at +oo as! (the limit of the quotient is -h-). Since n the test series

:E !n diverges, so does the given series.

Example 4. Prove that if a,. = O(c,.) and b,. = O(d,.), then a,.+ b,. = O(c,. + d,.) and a,.b,. = O(c,.d,.). Prove that if a,. = O(c,.) and b,. and d,. are sequences of positive numbers which are of the same order of magnitude at

+oo, then:: = 0

(t}

Solution. For the first part, if la.I ~ Kie,. and lb,.I ~ Kt 0 (by l'Hospital's Rule), the numerator = O(n9 ) for every q > ½ (cf. Example 4). Since the denominator is of the same order of magnitude as n1 at +oo, the general term = O(n•) for every r > -J. Finally, since :En• converges for every r < -1, we have only to chooser between -½ and -1 to establish the convergence of the given series.

709. THE RATIO TEST

One of the most practical routine tests for convergence of a positive series makes use of the ratio of consecutive terms.

Theorem. Ratio test. Let :Ea,. be a positive series, and define the test ratio

Assume that the limit of this test ratio exists: lim r,. n--++•

= p,

where O ~ p ~ +co. Then (i) if O ~ p < 1, :Ea,. converges; (ii) if 1 < p ~ +00, :Ea,. diverges; (iii) if p = 1, :Ea,. may either converge or diverge, and the test fails. Proof of (i). Assume p < ~. and let r be any number such that p < r < 1. Since lim r,. = p < r, we may choose a neighborhood of p that excludes r. n--++co

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[§ 709

INFINITE SERIES OF CONSTANTS

216

Since every r,., for n greater than or equal to some N, lies in this neighborhood of p, we have: n i5;; N implies r,. < r. This gives the following sequence of inequalities:

aN+a aN+2

< r, or aN+a < raN+s < rJaN, • • ••

• ••• •

Thus, each term of the series (1)

aN+i

+ aN+t + aN+a + ···

is less than the corresponding term of the series (2) But the series (2) is a geometric series with common ratio r < 1, and therefore converges. Since (2) dominates (1), the latter series also converges. Therefore (Theorem I, § 702), the original series :Ea,. converges. Proof of (ii). By reasoning analogous to that employed above, we see that since lim r,. > 1, whether the limit is finite or infinite, there Dlllst

n-+•

be a number N such that r,. i5;; 1 whenever n i5;; N.

n

(3)

In other words,

1

i5;; N implies az- i5:; 1, or a..+1 i5:;

a,..

The inequalities (3) state that beyond the first N terms, each term is at least as large as the preceding term. Since these terms are positive, the limit of the general term cannot be 0 (take E aN > 0), and therefore (§ 703) the series :Ea,. diverges.

=

Proof of (iii).

For any p-series the test ratio is r,.

and since the function zP is continuous at x

)P

= 1 (Ex. 7,

a;: = (n ~ 1y, 1

§ 602),

JP lP

lim ( +n = [ lim +n 1 = = 1. 1 n-+• n n-+• n If p > 1 the p-series converges and if p ~ 1 the p-series diverges, but in either case p = 1. lim

n-+•

r,.

=

=

NoTE 1. For convergence it is important that the limit of the test ratio be less than 1. It is not sufficient that the test ratio itself be always less than 1. This is shown by the harmonic series, which diverges, whereas the test ratio n/(n + 1) is always less than 1. However, if an inequality of the form

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THE RATIO TEST

709]

§

217

a,.+i ~ r

< 1 holds for a.II sufficiently large n (whether the limit p exists or not), a,. the series Ea,. converges. NOTE 2. For divergence it is sufficient that the test ratio itself be greater than 1. In fa.ct, if r,. ~ 1 for a.II sufficiently large n, the series Ea,. diverges, since this inequality is the inequality (3) upon which the proof of (ii) rests. NOTE 3. The ratio test may fa.ii, not only by the equality of p and 1, but by the failure of the limit p to exist, finitely or infinitely. For example, the series of Example 2, § 708, converges, although the test ratio r,. = a,.+ifa,. has no limit, since r2n-1 = 2"/3 11 -+ 0 and r2,. = 3"/2 11+1 -+ +ao. NOTE 4. The ratio test provides a. simple proof that certain limits a.re zero. That is, if lim a11+1/a.., where a,. > 0 for a.II n, exists and is less than 1, then a,.

-+

0.

..-+•

This is true because a,. is the genera.I term of a. convergent series.

Example 1. Prove that lim -X" = 0, for every real number x . ..-+• n 1 Solution. Without loss of generality we can assume x > 0 (take absolute zn+l

values).

X"

Then (n + l) ! + n 1 = n

Example 2.

Use the ratio test to establish convergence of the series 1+

. Sol ution.

X

+ 1 -+ 0.

1

1

1

11 + 21 + ai +

s·mce a,. =

1

••• + (n - 1) I + ••• '

(n _1 l) , a11+1

1

r,.

= a..+1 = (n a11

Therefore p =

Example 3.

1 and = nl

lim r,. = 0

n-+•

- 1) I = !. n ! n

< 1.

Use the ratio test to establish convergence of the series

1 ~ • ~ ~ 2+22+23+24+ •.• +2,.+ ... • Solution.

. _ n1 _ (n + 1) 1 Smee a11 - 211, a11+1 211+1 and

(n +n 1) 2k11 . = l(n + 1) 2 n 1< 1. lim r11 = ! lim (1 + !) 2 ..-+• 2 ..-+• n 1

, 11 = a11+1 = a,.

1

..



+1

2

Therefore p =

= -

E%8Dlple 4. Test for convergence or divergence: 1 2I 3I nI

a+ 3; + 3a' + ... + 3" + ....

Solution.

. _ n! _ (n + 1) ! Smee a11 - ,., a11+1 311+1 and 3 r

Digitized by

_ (n

" -

+ 1) ! . ~ nI

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3..+1

_ n -

+ 1. 3

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218

[§ 710

INFINITE SERIES OF CONSTANTS

Therefore

p =

lim n

n-.+ao

+3 1 = +ao, and the series diverges.

NOTE 5. Experience teaches us that if the general term of a series involves the index n either exponentially or factorially (as in the preceding Examples) the ratio test can be expected to answer the question of convergence or divergence, while if the index n is involved only algebraically or logarithmically (as in the p-series and Examples 3 and 5, § 708), the ratio test can be expected to fail.

710. THE ROOT TEST

A test somewhat similar to the ratio test is the following, whose proof is requested in Exercise 23, § 711:

Theorem. Root Test. Let :Ea,. be a nonnegative series, and assume the existence of the f oUowing limit: lim n--++ao

.zya: = u,

where O ~ u ~ +co. Then (i) if O ~ u < 1, :Ea,. converges; (ii) if 1 < u ~ +co, :Ea,. diverges; (iii) if u = 1, :Ea,. may either converge or diverge, and the te8t fail,s. NOTE 1. The inequality -e-'a,. < 1, for all n, is not sufficient for convergence, although an inequality of the form -e-'a,. ~ r < 1, for n > N, does guarantee convergence. NoTE 2. For divergence it is sufficient to have an inequality of the form -e-'a,. ~ 1, for n > N, since this precludes Jim a,. = 0. n--++ao

3. The ratio test is usually easier to apply than the root test, but the latter is more powerful. (Cf. Exs. 34-35, § 711.) NOTE

Example.

Use the root test to establish convergence of the series 1

Solution.

Since

-e-'a,.

+i+~+i+····

= n / :,_1 = 2

'\J

~5

1

2

,

the problem of finding

n

can be reduced to that of finding the two limits

lim

n--++•

-e'n

and

Jim

n--++•

.e,,;,;

1-.! Jim 2 "·

n--++o0

The second of these is not indeterminate, owing to the continuity of the function 2"' at x = 1, and has the value 21 = 2. let y

To evaluate lim

n--++ oo

In x

!

= xx.. and take logarithms: (cf. § 416): In y = -X ·

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.e,,;;:, or

1

lim x;,

x--++ oo

Then, by l'Hospital's

Original from

UNIVERSITY OF MICHIGAN

§ 711]

EXERCISES

Rule (§ 415),

lim In y

-+•

lim -e'a,. = ½ < 1, and n-++•

=

lim l/x

-+•

1

=

0

219

and y-+ e0

,

= 1.

Therefore

I:a,. converges.

711. EXERCISES

In Exercisei;, 1-20, establish convergence or divergence of the given series. + va + vi+ ... 1'31 + v2 5 7 9 . 2.

1 + 1 + 1 + 1 + .... v1,2 v2.3 v3.4 v4.5

va

V5

V7

V9

3• 2·4 + 4·6 + 6·8 + 8·10 + ....

va

'- v2 + 1 + + 1 +vi+ 1 + v'5 + 1 + ..•• 33 - 1 43 - 1 51 - 1 63 - 1 1 1·2 1·2·3 1·2·3•4 6• 3 + 3.5 + 3.5.7 + 3.5.7.9 + ... ' 1! 2I 3I 4! 6• 25+29+27+29+ ... ·

1 1 1 1 7 • 1n 2 + In 3 + In 4 + In 5 + · · · · 21

31

41

51

8• 41+51+61+71+ ... ·

+• Vn

9. n-1n I: 2

11.

13. 16.

+•

+ 4·

Inn

I: _./ + l

4

I: ~I

+•

I: e-"

1



n-1

"i:v'n+i-Vn. n"'

+• 1 17 • ..~2 (In n)'"

21. 22. 23. 2'9

+•

n-1 nvn

n-1n

n-1

19.

10.

+•

I: (-e'n -

n-1

18.

0

1) 11 •

20.

+•

1

I: +. ,. a> a

n-1 1

+•

I:

n-1

r"

lsin n al, a

-1.

> 0, r > 0.

Prove the comparison test (Theorem I, § 708) for divergence. Prove Theorem II, § 708. Prove the root test, § 710. Prove that if a,. e:; 0 and there exists a number k > 1 such that Jim nka,.

exists and is finite, then

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n~+•

I:a,. converges.

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220

INFINITE SERIES OF CONSTANTS

25. Prove that if a,.

~

0 and

[§ 711

lim n a,. exists and is positive, then

......+..

Ea•

diverges. *26. Prove the Schwarz (or Cauchy) inequality for nonnegative series: If a,.~ 0 and b,. ~ 0, n = 1, 2, · · · , then

(E a,.b,.) ~ E a,. E b,. 1

1

(1)

n.-1

n.-1

1,

n-1

with the following interpretations: (i) if both series on the right-hand side of (1) converge, then the series on the left-hand side also converges and (1) holds; (ii) if either series on the right-hand side of (1) has a zero sum, then so does the series on the left. (Cf. Ex. 43, § 107, Ex. 29, § 503, Ex. 14, § 717.) *27. Prove that if Ea,.2 is a convergent series, then E la..1/n is also convergent. (Cf. Ex. 26.) +• d •28. Prove that any series of the form E ",., where d,. '""' 0, 1, 2, • • • , 9, .. -1 10 converges. Hence show that any decimal expansion 0.d1dt • • • represents some real number r, where O ~ r ~ 1. *29. Prove the converse of Exercise 28: If O ~ r ~ 1, then there exists a decimal expansion 0.d1dt • • • representing r. Show that this decimal expansion is unique unless r is positive and representable by a (unique) terminating decimal (cf. Ex. 5, § 113), in which case r is also representable by a (unique) decimal composed, from some point on, of repeating 9's. •30. Show that Exercise 29 establishes a one-to-one· correspondence between the points of the half-open interval (0, 1] and all nonterminating decimals 0.d1d2 · · • • Prove that any subset of a denumerable set (cf. Ex. 12, § 113) is either finite or denumerable, and hence in order to establish that the real numbers are nondenumerable (neither finite nor denumerable) it is sufficient to show that the set S of nonterminating decimals 0.d1dt · • • is nondenumerable. Finally, prove that the set S is nondenumerable by assuming the contrary and obtaining a contradiction, as follows: Assume Sis made up of the sequence {z,.}: z1: .du d12 du du Z2: .dt1 dn du du Za: .da1 dn du dM Construct a new sequence {a,.}, where a,. is one of the digits 1, 2, · · • , 9, and a,.¢ d,.,., for n = 1, 2, • • • . Finally, show that the decimal 0.a,a,aa • • · simultaneously must and cannot belong to S. *31. Establish the following form of the ratio test: The positive series Ea. converges if

um

On+I

,..... +.. a,.

< 1,

and diverges if

lim ~

...... +•

a,.

>

1.

(Cf. Ex. 16,

§ 305.) *32. Establish the following form of the root test: If Ea. is a nonnegative then Ea,. converges if " < 1 and diverges if series and if " =

um ~.

...... + ..

u

>

1.

(Cf. Ex. 31.)

*33. Apply Exercises 31 and 32 to the series

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½+ ½+ ~ + ~ + · · · of Ex-

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UNIVERSITY OF MICHIGAN

§ 712]

MORE REFINED TESTS

221

a a ,.,ample 2, § 708, and show that IIiii ~ = +oo, lim ~ = 0, IIiii v a,. = 1/V2, a,. a,. and lim -e-'a,. = 1/V3. Thus show that the ratio test of § 709 and that of Exercise 31 both fail, that the root test of § 710 fails, and that the root test of Exercise 32 succeeds in establishing convergence of the given series.

•34. Prove that if lim a,.+, exists, then lim

-+• a.

n-+•

.e,-;: also exists and is equal

to it. Hints: For the case lim ~ = L, where O < L < +oo , let a and {3 be n-+• a,. arbitrary numbers such that O < a < L < {3. Then for n ~ some N, a a,. < a.+1 < {3 a.. Hence a aN < aN+1 < /3 aN a 2 aN < a aN+1 < aN+2 < /3 aN+l < /32 aN Thus, for n

> N,

a~{-:•::}

and

•H. The example 1, 2, Jim !!!ti! does not.

..-+• a,.

;a {3 •

..-+•

1, 2, • • • shows that

lim

..-+•

.e,-;:

may exist when

Find an example of a convergent positive series :Ea,. for

which this situation is also true.

-+- '\J;-j

" ~ - e.

•36. Prove that lim

(Cf. Ex. 34, above; also Ex. 48, § 515.)

*712. MORE REFINED TESTS The tests discussed in preceding sections are those most commonly used in practice. There are occasions, however, when such a useful test as the ratio test fails, and it is extremely difficult to devise an appropriate test series for the comparison test. We give now some sharper criteria. which may sometimes be used in the event that lim ~ ..-+• a,.

= 1.

Theorem I. X:umm.er's Test. Let :Ea,. be a positive 8erie8, and kt {p,.} be a sequence of poaitive constants such that

..-+•

[p.. a..+1 a,. -

r. For any positive integer m, then, we have the a,.+1 sequence of inequalities:

> raN+1, PN+2"N+2 > raN+2,

PNaN - PN+1aN+1 PN+1aN+1 -

PN+m-laN+-1 - PN+.,.aN+"' > raN+m• Adding on both sides we have, because of cancellations by pairs on the left: (2)

Using the notation S,. for the partial sum a 1 + •· · + an, we can write the sum in parentheses of (2) as SN+... - SN, and obtain by rearrangement of terms: rSN+m < rSN + PNaN - PN+maN+... < rSN + PNaN, Letting B denote the constant (rSN + pNaN)/r, we infer that S,. < B for n > N. In other words, the partial sums of Ea,. are bounded and hence (§ 705) the series converges. · To prove the second part we infer from the inequality p,. J!!!.... a,.+1

which holds for n

~

Pn+i

~ 0,

N, the sequence of inequalities

PNaN ~ PN+1aN+1 ~ · · • ~ p,.a,., for any n > N. Denoting by A the positive constant pNaN, we conclude from the comparison test, the inequality 1 a,. ~ A · p,.' for n

and the divergence of E

> N,

..!., that Ean also diverges. Pn

Theorem II. Rube's Test. Let Ea,. be a positive series and assume that lim n ( a,. -

11-+•

a,.+1

1) = L

exists (finite or infinite). Then (i) if 1 < L ~ +oo, Ean converges; (ii) if -oo ~ L < 1, Ea,. diverges. (iii) if L = 1, Ean may either converge or diverge, and the test fail8.

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§ 713]

p ..

223

EXERCISES

Proof. The first two parts are a consequence of Kummer's Test with n. (Cf. Ex. 5, § 713, for further suggestions.)

=

In case the limit L of Raabe's Test exists and is equal to 1, a refinement is possible:

Theorem

m.

Let :Ea. be a posi,tive series and assume that lim ln

-+•

n[n (..!!!!.. - 1) - 1] - L a..+1

exists (finite or infinite). Then (i) if 1 < L ~ +co, :Ea. convergeB; (ii) if -co ~ L < 1, I:;a,. divergeB; (iii) if L = 1, I:;a,. may either converge or diverge, and the test failB. Proof. The first two parts are a consequence of Kummer's Test with p,. = n ln n. (Cf. Exs. 9-10, § 713, for further suggestions.) Example 1.

Test for convergence or divergence:

(~)p (~)p •••' (!)P 2 + 2·4 + 2·4·6 +

Solution. Since lim !!!!±! n-+• a,. test. We find

= 1, the ratio test fails, and we turn to Raabe's

n ( a. _ 1 ) _ n [( 2n

2n-1

aa+1

)P _ 1] _

2n . (1 2n-1

+ z)P -

1,

2z

where z a (2n - 1)-1• The limit of this expression, as n--+ +co, is (by !'Hospital's Rule): lim (I + z)P - 1 = lim p(I + z)P-1 = P. -o 2x -o 2 2 Therefore the given series converges for p > 2 and diverges for p < 2. For ~he case p = 2, see Example 2. Example 2. Test for convergence or divergence

(!)' 2 + (~)' 2·4 + (~)' 2•4•6 + ... . Solution. Both the ratio test and Raabe's test fail (cf. Example l}. Preparing to use Theorem III, we simplify the expression n ( a,. _ 1 ) _ 1 = n 4n - 1 _ 1 = 3n - 1 a..+1 4n1 - 4n + 1 (2n - 1)1 1 1·1m In n ( an - l)' = 0 < 1, th e given . . d'1verges. . Smce series 2n n-+• *713. EXERCISES

In Exercises 1-4, test for convergence or divergence. ~ 2·4·6 •·· 2n •1...~11 ·3·5 ... (2n + 0·

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(Cf. Ex. 4.)

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224

INFINITE SERIES OF CONSTANTS

*2.

[§ 713

"'i: 1-32·4··•2n ... c2n - o . _1__ 2n+l

n-1

*3• E 1 ·3 .. · (2n -

2·4···2n

n-1

1) . 4n + 3. 2n+2

*'· E [ 1·3·52·4·6· · · •(2n • • 2n JP· + 1) n-1

•5. Prove Theorem II, terms of

§ 712.

Hint: For examples for part (iii), define the

Ea" inductively, with a, = 1, and an+l ..!!!..

n-1

5

l

1+

n

+ - k • Then 1 nnn

use Theorem III, § 712. •6. Prove that the hypergeometric series 1 + ~ + a(a + l)t3(p + 1) + ... 1•-y 1·2•-y•('Y + 1) + a(a + 1) · · · (a+ n - 1) ·t3@ + 1) • • • (tl + n - 1) + ... n ! -y(-y + 1) · · • (-y + n - 1) converges if and only if 'Y - a - fJ > 0 (n not 0 or a negative integer). Prove Gauss's Test: If the positive series Ea.. is such that

•7.

a.

a,.+1

= 1 + !! + 0 n

(!.n ), 1

then Ea,. converges if h > 1, and diverges if h ~ 1. Show that O(1/n1 ) could be replaced by the weaker condition 0(1/n°), where a > 1. •S. State and prove limit superior and limit inferior forms for the Theorems of § 712. (Cf. Exs. 31-32, § 711.) •9. Prove the first two parts of Theorem III, § 712. (Cf. Ex. 11.) Hint: This is equivalent to showing (using Kummer's test with p. = n In n) that

Inn [ n

(a::, - 1) - 1] - ln Inn a::_, - (n + 1) In (n + l)f 1, -t

or (n + 1) In (n

+ 1) -

[n In n

+ In n] = (n +

1) In ( n

~ 1 ) - 1.

**10. Prove the following refinement of Theorem III, § 712: Let Ea,. be a positive series such that lim In Inn jln

n-+•

l

n[n (..!!!.. - 1) - 1] a..+1

1t

f

=L

exists (finite or infinite). Then (i) if 1 < L ~ +oo, Ea.. converges, and (ii) if -00 ~ L < 1, Ea,. diverges. Hint: The problem is to show (cf. Ex. 9) that (1) (n + 1) In (n + 1) In In (n + 1) - {n Inn Jn Inn+ Inn In Inn + In Inn} - t 1. To establish this limit, show first that if E,. -+ 0, then In (1 + E..) = E,. + O(E,.1) (cf. Ex. 21, § 408), and, more generally, if {a,.} is a sequence of positive numbers and if E,./a,. -+ 0, then In (a,. + E,.)

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= In

(a,.) + E,. a,.

+ 0 {E"a,.11)•

Original from

UNIVERSITY OF MICHIGAN

Thus

225

SERIES OF ARBITRARY TERMS

§ 714]

In (n

+

= In n + ¼+

1)

0

G 2)

and In Inn

Gz)] = In Inn + l: n [; + 0 G + 0 ([! + 0 (!...)] = In In n + - 1

= In [inn + ¼+ 0

2) ]

2

~

n

)

n~n

+0

(!...). ~

Now form the product (n + 1) In (n + 1) In In (n + 1), subtract the terms in the braces of (1), and take a limit. **11, Prove the third part of Theorem III, § 712. Hint: Cf. Exs. 5 and 10. 714. SERIES OF ARBITRARY TERMS

If a series has terms of one sign, as we have seen for nonnegative series, there is only one kind of divergence-to infinity-and convergence of the series is equivalent to the boundedness of the partial sums. We wish to turn our attention now to series whose terms may be either positive or negative-or zero. The behavior of such series is markedly different from that of nonnegative series, but we shall find that we can make good use of the latter to clarify the former. 715. ALTERNATING SERIES

An alternating series is a series of the form (1)

C1 -

where Cn

C2

+ Ca -

C4

+ •••

1

> 0 for every n.

Theorem. An alternating series (1) whose i,erms satisfy the two conditions (i) C..+1 < Cn for every n, (ii)

Cn-+

0 as

n-+

+co,

converges. If S and S. denote the sum, and the partial sum of the first n terms, respectively, of the series (1),

1s. - s1 < cn-1-1-

(2)

Proof. We break the proof into six parts (cf. Fig. 702.): A. The partial sums S2n (consisting of an even number of terms) form an increasing sequence. B. The partial sums 82-1 (consisting of an odd number of terms) form a decreasing sequence. C. For every m and every n, S2m < S2n-1D. Sexists. E. For every m and every n, S2m < S < S2n-1• F. The inequality (2) holds.

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226

INFINITE SERIES OF CONSTANTS

A: Since 81,.,...1 8s......1

>

= 82,. + (cs......1 -

81,.. B: Since 81,.,...1 = 82,._1 81,.,...1 < 82n-l• C: If 2m < 2n - 1, 82,._1 - 81... = (Ctw.+1 and 82,,. < 81,._1. If 2m > 82m - 82n-1 = -(c1,. and S2m < 82n-l•

Cltt,f-1), and since es,.,...2

(c2,. - c1,.,...1), and since es,.,...1

[§ 715

< Cltt,f-1, < es..,

C21n+1) + · · · + (c211-1 - es,._1) + es,._1 > o, 2n - 1, c2,.,...1) - · • · - (Ctm--2 - c.,,._1) - c2,,. < 0,

s

811,-1 ••• 8.

8,

FIG. 702

D: From A, B, and C it follows that {82,.} and {82,._1} are bounded monotonic sequences and therefore converge. We need only show that their limits are equal. But this is true, since lim 82,. = lim (81,.,...1 - 82,.) = lim es,.,...1 = 0. n-+• n-+• E: By the fundame~tal theorem on convergence of monotonic sequences (Theorem XIV, § 204), for an arbitrary fixed n and variable m, 82m < 81,.,...1 implies 8 ~ 81,.,...1 < 82,._1. Similarly, for an arbitrary fixed m and variable n, 82-+2 < 82,.,...1 implies 8s... < 82m+2 ~ 8. F: On the one hand, lim 81,.,...1 -

-+•

..-+•

0

< 8111-1

- 8

< 8111-1

- 82,.

= C1,.,

while, on the other hand 0 Example 1.

< 8 - 8111 < 8s......1 - 8111 = es,.,...1.

The alternating harmonic series

1-½+i-¼+-··· converges, since the conditions of the Theorem above are satisfied. Example S.

Solution.

Prove that the series

+•

I: (-1)" Inn converges. n

n-1

All of the conditions of the alternating series test are obvious ex-

cept for the inequality In

!"'-t+l 1) < Innn.

The simplest way of establishing

this is to show that /(z) • In z is strictly decreasing since its derivative is X

f'(z)

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=1-

X

!n z

< 0, for z > e.

Original from

UNIVERSITY OF MICHIGAN

§716]

ABSOlUTE, CONDITIONAL CONVERGENCE

227

716. ABSOLUTE AND CONDITIONAL CONVERGENCE

We introduce some notation.

Let

+•

:E

a,. be a given series of arbitrary

.. -1

terms. For every positive integer n we define p,. to be the larger of the two numbers a,. and 0, and q,. to be the larger of the two numbers - a,. and 0 (in case a,.= 0, p,. = q,. = O): (1)

p,.

=max (a,., 0), q,. =max (-a,., 0).

Then p,. and q,. are nonnegative numbers (at least one of them being zero) satisfying the two equations (2) p,. - q,. = a,., p,. + q,. = la,.1, and the two inequalities

la..l, o ~ q,. ~ la..l. The two nonnegative series :Ep,. and :Eq,. are called the nonnegative and nonpositive parts, respectively, of the series :Ea,.. A third nonnegative series related to the original is :E la,.I, called the series of absolute values of :Ea,.. o~

(3)

p,. ~

We now assign labels to the parti.al sums of the series considered:

+ · · · + a,., P,. = Pi + · · · + p,.,

S,. = ai

A,. =

lail + · · · + la,.I,

Q,. = qi

+ · · · + q,..

From (1) and (2) we deduce: P,. - Q,. = S,.,

(4)

P,.

+ Q,. =

A,.,

and are ready to draw some conclusions about convergence. From the first equation of (4) we can solve for either P,. or Q,. (P,.

= Q,. + S,.

and Q,.

= P,. -

S,.),

and conclude (with the aid of a limit theorem) that the convergence. of any two of the three series :Ea,., :Ep,., and :Eq,. implies the convergence of the third. This means that if both the nonnegative and nonpositive parts of a series converge the series itself must also converge. It also means that if a series converges, then the nonnegative and nonpositive parts must either both converge or both diverge. (Why?) The alternating harmonic series of § 715 is an example of a convergent series whose nonnegative and nonpositive parts both diverge. The inequalities (3) and the second equation of (4) imply (by the comparison test and a limit theorem) that the series of absolute values, :Ela..! converges if and only if both the nonnegative and nonpositive parts of the series converge. Furthermore, in case the series :Ela..l converges, if we define P = lim P "' Q = lim Q,., S = lim S,., and A = lim A,., we n-+oo

n-+•

have from (4), P - Q = S, P

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-+•

,._+,.

A.

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228

INFINITE SERIES OF CONSTANTS

[§ 716

We give a definition and a theorem embodying some of the results just obtained: Definition. A series :Ea,. converges absolutely (or is absolutely convergent) if and only if the series of absolute values, :E la,.! converges. A series converges conditionally (or is conditionally convergent) if and

only if it converges and does not converge absolutely. Theorem I. An absolutely convergent series is convergent. absolute convergence, l:Ea,.I ;;;; :E la,.!.

In case of

NOTE 1. For an absolutely convergent series, the nonnegative and nonpositive parts both converge. For a conditionally convergent series, the nonnegative and nonpositive parts both diverge. NOTE 2. An alternative proof of Theorem I is provided by the Cauchy Criterion (cf. Exs. 17-21, § 717 for a discussion). NoTE 3. The tests established for convergence of nonnegative series are of course immediately available for absolute convergence of arbitrary series. We state here the ratio test for arbitrary series, of nonzero terms, both because of its practicality and because the conclusion for p > 1 is not simply that the series fails to converge absolutely but that it diverges (cf. Ex. 12, § 717).

Theorem II. Ratio Test. Let :Ea,. be a series of nonzero terms, and a..+-1/a,.. Assume that the limit of the absolute value

define the test ratio r,. = of this test ratio exiats:

Iim a,....11 =p, .......+., a,.

l

where O ~ p ~ +co. Then (i) if O ~ p < 1, :Ea,. converges absolutely; (ii) if 1 < p ~ +co, :Ea.. diverges; (iii) if p = 1, :Ea,. may converge absolutely, or converge conditionally, or diverge, and the test fails. NOTE 4. The ratio test never establishes convergence in the case of a conditionally convergent series. The alternating p-series, 1 1 1 1 - 2P + 3P - 4P

Example 1.

converges absolutely if p verges if p ~ 0. *Example 2.

> 1,

+ •••

I

converges conditionally if O < p

.

Show that the series

1

2-

1·3

_ 2 4

+ 21·3·5 _ _ 4 6

~

1·3•5·7

_ _ _ 2 4 6 8

1, and di-

+ ···

converges conditionally. Solution. By Example 1, § 712, the series fails to converge absolutely. In order to prove that the series converges, we can use the alternating series test to reduce the problem to showing that the general term tends toward O (clearly c,. ! ). Letting p = 3 in Example 1, § 712, we know that c,.1 - 0. Therefore

c,.-o.

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§ 717]

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EXERCISES

717. EXERCISES In Exercises 1-10, test for absolute convergence, conditional convergence, or divergence. (-l)n-1 1. +• E ...____.____ n-1

2n

2.

+3

+ + +

~ ( - l)nv'nt 5 3. '- ""-:":.;::::::::======,._1 v'n3 n 1

5.

t (-l)nn. +

n-1

4.

n

2

+• ,.~/-l)n-1 (n

n4

+ 1) t'

+•

l E (- l)n !'!....!!..!!. en

n-1

7. e-• cos x + e-t• cos 2x + e-a.. cos 3x + ••• . 8. 1 +rcos8+r1 cos28+r'cos38+ •··. 1 1 9 • 2(ln 2)P - 3(ln 3)P

*10.

1

+ 4(ln 4)P -

··· ·

(!)P - (U)p + (~)p - .... 2 2·4 2•4•6

11. Prove that if the condition (i) of § 715 is replaced by Cn+1 ~ c,., the conclusion is altered only by replacing (2), § 715, by IS,. - SI ~ c,.+1• 12. Prove the ratio test (Theorem II) of § 716. 13. Show by three counterexamples that each of the three conditions of the alternating series test, § 715, is needed in the statement of that test (that is, the alternating of signs, the decreasing nature of c,., and the limit of Cn being 0). 14. Prove the Schwarz (or Cauchy) and the Minkowski inequalities for series:

*

+•

+•

+•

n-1

n-1

n-1

If E a,.2 and E b,.2 converge, then so do

E

+•

anb,. and E (a,. n-1

+ b,.)

1,

and

[ E a,.b,.J ~ E a,. E b,. 2

n-1

[ E (a,.+ b,.)

n-1

2

]

1

fl-1

n-1

2 ,

~ [f a.. ]½ + [f b,. n-1 n-1

2

2

]

1 .

(Cf. Exs. 43-44, § 107, Exs. 29-30, § 503, Ex. 26, § 711.)

**15. Let

+•

E

n-1

a,. be a given series, with partial sums S,.

=

a1

+ ·· · + an,

Define the sequence of arithmetic means tT,.

- 81

=

+ ... + s,. n

The series Ea. is said to be summable by Cesaro's method of arithmetic means of order 1 (for short, summable (C, 1)) if and only if lim tT,. exists and is finite. n--++• Show that Exercises 16-19, § 205, prove that sum1IU1,bility (C, 1) is a generalization of convergence: that any convergent series is summable (C, 1) with lim tT,. = lim S,., that for nonnegative series summability (C, 1) is identical

-+•

-+•

.

with convergence, and that there are divergent series (whose terms are not of one sign) which are summable (C, 1). Hint: Consider 1 - 1 + 1 - 1 + · · · .

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[§ 717

**16. A series Ea,. is summable by Cesaro's method of arithmetic means of order 2 (for short, summable (C, 2)) if and only if lim a-,

+ · · · + a-,.

n-+•

n (in the notation of Exercise 15) exists and is finite. Show that summability (C, 1) implies summability (C, 2), but not conversely. Generalize to summability (C, r). *17. Prove the Cauchy criterion for convergence of an infinite series: An infinite series

+•

E

n-1

a,. converges if and only if corresponding to

number N such that n § 302.)

>m >N

*18. Prove that an infinite series

implies

+•

E

n-1

>

0 there exists a

la.. + a..+1 + •· · + a,.I < E.

(Cf.

a,. converges if and only if correspond-

E > 0 there exists a number N such Ja,. + a,.+1 + ••• + a,.+PI < E. (Cf. Ex. 17.)

ing to

•19. Prove that an infinite series

E

that n

>

N and p

>

0 imply

+•

E a,. converges if and only if correspond-

n-1

ing to E > 0 there exists a positive integer N such that n > N implies JaN + UN+1 + · •· + a,.I < E. (Cf. Ex. 2, § 305.) *20. Show by an example that the condition of Exercise 18 is not equivalent to the following: lim (a,. + ••• + a,.+J>) = 0 for every p > 0. (Cf. Ex. 4, n--++oo

-

§ 305, Ex. 43, § 904.) •21. ·Use the Cauchy criterion of Exercise 17 to prove that an absolutely convergent series is convergent.

**22. Prove the following Abel test: If the partial sums of a series

+•

E

n-1

a,. are

bounded and if {b,.} is a monotonically decreasing sequence of nonnegative numbers whose limit is 0, then Ea,.b,. converges. Use this fact to establish the convergence in the alternating series test as phrased in Exercise 11. Hint: Let S,. = a, + · · · + a,., and assume JS,.I < K for all n. Then

I.E

i-m

a,b,,

=

l.f:.

,-m

l):

1

(S, - s,_,)b;I

=

,-m

S;(b, - b1+1>

+ S,.b,. -

s,._,b,.I

**23. Prove the following Abel test: If Ea,. converges and if {b,.} is a bounded monotonic sequence, then Ea,.b,. converges. Hint: Assume for definiteness that b,. ! , let b,. - b, write a,.b,. = a,.(b,. - b) + a,.b, and use Ex. 22. **24. Start with the harmonic series, and introduce + and - signs according to the following patterns: (i) in pairs: 1 + ½ - ½- ¼+ ¼+ ¼- - • • · ; (ii) in groups of 1, 2, 3, 4, · · ·: 1 - ½ - ½+ + + - - - - · · · ; (iii) in groups of 1, 2, 4, 8, • • • : 1 - ½ - ½+ + + + · · · .

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§ 718]

GROUPINGS AND REARRANGEMENTS

Show that (i) and (ii) converge and (iii) diverges.

In Abel's test, Ex. 22, let a,. = ±

(Cf. Ex. 22.)

J; and b,. = );t

231

Hint for (ii):

**25. If A ·is an arbitrary finite or denumerable set of real numbers, a 1, a2, aa, · · · (which may be discrete like the integers, or dense like the rational numbers, or anywhere in between), construct a bounded monotonic function whose set of points of discontinuity is precisely A, as follows: Let

+•

I:

11-1

p,. be a conver-

gent series of positive numbers with sum p. Define f(x) to be O if x is any lower Qound of A, and otherwise equal to the sum of all terms Pm of :Ep,. such that a., < x. Prove the following five properties of f(x): (i) f(x) is monotonically increasing on (-oo, +oo); (ii) Jim /(x) = O; (iii) Jim /(x) = p; (iv) /(a,.+) - /(a,.-)

not in

=

--•

-+•

p,. for every n; (v) f(x) is continuous at every point

A.

718. GROUPINGS AND REARRANGEMENTS

A series :Eb,. is said to arise from a given series :Ea,. by grouping of terms (or by the introduction of parentheses) if every b,. is the sum of a finite number of consecutive terms of :Ea,., and every pair of terms a.. and a,., where m < n, appear as terms in a unique pair of terms bp and bq, respectively, where p ;;i q. For example, the grouping (a1 + ll2) + (aa) + (a. + tJ&) + (tJ&) + · · · gives rise to the series :Eb,., where b1 = a 1 + ll2, b2 = aa, b3 = a4 b, = ae, · · · .

+ '2&,

Theorem I. Any series arising from a convergent series by grouping of terms is convergent, and has the same sum as the original series. Proof. The partial sums of the new series form a subsequence of the partial sums of the original. The example (2 - 1½) + (1½ - 1¼) + (ll- - 1¼) + · · · shows that grouping of terms may convert a divergent series into a convergent series. NOTE.

A series :Eb,. is said to arise from a given series :Ea,. by rearrangement (of terms) if there exists a one-to-one correspondence between the terms of :Ea,. and those of :Eb,. such that whenever a.. and b,. correspond, a.. = b,.. For example, the series 1 1 1 1 1 4 + l + 16 + 9 + 36 + 25 + ... is a rearrangement of the p-series, with p = 2.

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[§ 718

Theorem II. Dirichlet's Theorem. Any series arising from an absolutely convergent series by rearrangement of terms is absolutely convergent, and has the same sum as the original series. Proof. We prove first that the theorem is true for nonnegative series. Let Ea,. be a given nonnegative series, convergent with sum A and let Eb,. be any rearrangement. If B,. is any partial sum of Eb,., the terms of B,. consist of a finite number of terms of Ea.., and therefore form a part of some partial sum A.,. of Ea,.. Since the terms are assumed to be nonnegative, B,. ~ Am, and hence, B,. ~ A. Therefore the partial sums of the nonnegative series Eb,. are bounded, and Eb,. converges to a sum B ~ A. Since Ea.. is a rearrangement of Eb,., the symmetric relation A ~ B also holds, and A = B. If Ea,. is absolutely convergent, and if Eb,. is any rearrangement, the nonnegative and nonpositive parts of Eb,. are rearrangements of the non-

negative and nonpositive parts of Ea.., respectively. Since these latter both converge, say with sums P and Q, respectively, their rearrangements will also both converge, with sums P and Q, respectively, by the preceding paragraph. Finally, Ea,.= P - Q = Eb,., and the proof is complete. *Theorem m. The terms of any conditionally convergent series can be rearranged to give either a divergent series or a conditionally convergent series whose sum is an arbitrary preassigned number. *Proof. We prove one case, and leave the rest as an exercise. (Ex. 2, § 721.) Let Ea,. be a conditionally convergent series, with divergent nonnegative and nonpositive parts Ep,. and Eq,., respectively, and let c

be an arbitrary real number. Let the rearrangement be determined as follows: first put down terms pi+ P2 + · · · + Pm, until the partial sum first exceeds c. Then attach terins -qi - q2 - q3 - • • · - q,., until the total partial sum first falls short of c. Then attach terms Pm,+i + · · · + Pm, until the total partial sum first exceeds c. Then terms -q,.,+, - · · · - q,.., etc. Each of these steps is possible because of the divergence of Ep,. and Eq,.. The resulting rearrangement of Ea,. converges to c since p,. --+ 0 and q,.--+ 0. Example. Rearrange the terms of the series obtained by doubling all terms of the alternating harmonic series so that the resulting series is the alternating harmonic series, thus convincing the unwary that 2 = I. Solution. Write the terms

2[1 - ½+ ½- ¼+ 1- - ¼+ t - t + l - · · · J =2-l+i-½+t-!+t-¼+i- ··· = (2 - 1) - ½+ (¾ - l) - ¼+ (! _, ¼) - ¼+ ::1-½+½-i-+¼-¼+t-t+l-····

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§ 719]

233

OPERATIONS WITH SERIES

719. ADDITION, SUBTRACTION, AND MULTIPLICATION OF SERIES

Definition I. If Ea,. and Eb,. are two series, their sum Ee,. and difference Ed,. are series defined by the equations (1)

Theorem I. The sum and difference of two C01Wergent series, Ea,. = A and Eb,.= B, corwerge to A+ Band A - B, respectively. The sum and difference of two absolutely C01Wergent series are absolutely corwergent. The proof is left as an exercise (Ex. 3, § 7~1). The product of two series is a more difficult matter. The definition of a product series is motivated by the form of the product of polynomials or, more generally, power series (treated in Chapter 8): (ao

+ a1x + aa + · · ·)(bo + b1x + b,p:2 + •••) = aobo + (aob1 + a1bo)x + (aob2 + a1b1 + atbo)x2 + · · · . 2

For convenience we revise slightly our notation for an infinite series, letting the terms have subscripts 0, 1, 2, · • • , and write

+• +• +• Definition II. If E a,. and E b,. are two series, their product E c,. n-o

n-o

n-o

is defined: (2)

Co

= aobo, c1 = aobi + a1bo, n

c,. =

E

1:-0

··· ,

a1;b,._11; = aob.. + aibn-1

+ ••• + a..bo,

••• ,

The basic questions are these: If Ea,. and Eb,. converge, with sums A and B, and if Ee,. is their product series, does Ee,. converge? If Ee,. converges to C is C = AB? If Ee,. does not necessarily converge, what conditions on Ea,. and Eb,. guarantee convergence of Ee,.? The answers, in brief, are: The convergence of Ea,. and Eb,. does not guarantee convergence of Ee,. (Ex. 5, § 721). If Ee,. does converge, then C = AB. (This result is due to Abel. Cf. Ex. 20, § 911.) If both Ea,. and Eb,. converge, and if one of them converges absolutely, then Ee,. converges (to AB). (This result is due to Mertens. Cf. Ex. 19, § 721.) If both Ea and Eb,. converge absolutely, then Ee,. converges absolutely (to AB). (This is our next theorem.) 11

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INFINITE SERIES OF CONSTANTS

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[§ 719

Theorem II. The product series of two absolutely corwergent series is absolutely corwergent. I ts sum is the product of their sums. +• +• Proof. Let I: a,. and I: b,. be the given absolutely convergent series, n-0 n-o +• +• let I: c,. be their product series, and define the series I: d,. to be n-o

aobo

n-o

+ aob1 + a1bo + aob2 + a1b1 + aibo + aoba + a1b2 + aib1 + aabo + aob, + · · · ,

the terms following along the diagonal lines suggested in Figure 703. • • •

...

.. . •



••

FIG. 703

Furthermore, let

= ao + a1 + · · · + a,., C,.= Co+ c1 + · · · + c,.,

A,.

= bo + b1 + · · · + b,., D,. =do+ d1 + · · · + d,..

B,.

Observe that every c,. is obtained by the grouping of (n + 1) terms of the series I:d,., that every C,. is a partial sum of the series I:d,. with terms occupying a triangle in the upper left-hand corner of Figure 703, and t~t every A,.B,. is a sum (not a strict "partial sum") of certain terms of the series I:d,. occupying a square in the upper left-hand corner of Figure 703. We prove the theorem first for the case of nonnegative series I:a,. and I:b,.. In this caS'e, since every finite set of terms from I:d,. is located in some square in the upper left-hand corner of Figure 703, every D,,. is less than or equal to some A,.B,.. That is, if A lim A,., and B lim B., n-+• n-+• the inequality D,,. ~ AB holds for every m. Therefore the series I:d,. converges, the limit D = lim D,. is finite, and D ~ AB. On the other hand, n-+• since A,.B,. is a sum of terms of the convergent series I:d,., the inequality A..Bn ~ D holds for every n, and hence AB ~ D. Thus D = AB. Finally, since {C,.} is a subsequence of {D,.} (resulting from introducing parentheses in the series I:d,.), C = lim C,. = D = AB. n-+•

=

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§ 720]

SOME AIDS TO COMPUTATION

235

If the given series, :Ea,. and :Eb,., have terms of arbitrary sign, the conclusion sought is a consequence of Dirichlet's Theorem (§ 718): Since, by the preceding paragraph, the series :Ed,. converges absolutely, any rearrangement converges absolutely to the same sum. The sequence {C,.} is a subsequence of the sequence of partial sums of Ed,., and the sequence {A,.B,.} is a subsequence of the sequence of partial sums of an appropriate rearrangement of :Ed,.. Therefore lim C,. = lim A,.B,. = lim D,..

n-++ •

-+ •

n-++ •

*720. SOME AIDS TO COMPUTATION

The only techniques available from the preceding sections of this chapter for evaluating series are (i) the sum a/(1 - r) of a geometric series and (ii) the estimate IS,. - SI < c,.+1 for an alternating series. We give in this section some further means of estimating the sum of a convergent series, with illustrative examples. These will be available for computation work presented in the next chapter. If a nonnegative series is dominated by a convergent geometric series, the formula a/(1 - r) provides an estimate for the sum. A useful formulation of this method makes use of the test ratio (give the proof in Ex. 6, § 721):

Theorem I. r,. ! , and r,.-+

If p

a,. > 0,

=

a1

+ · · · + a,.,

+oo

S

= E a,., n-1

_ an+l

r,. = a,. '

< 1, then for any n for which r,. < 1,

+~ < S < S,. + 1 -a,.+ir,.+1 • 1- p

S,.

(1)

S,.

Compute the sum of the series 1 1 1 1 1 + 3 + 2 ! • 5 + 3 ! • 7 + · · · + n ! (2n + 1)

Example 1.

to three decimal places.

Solution. estimate:

(The sum is

fo

1

&" dx.

+ · ·· '

Cf. Example 4, § 810.J

The sum Se is between 1.4625 and 1.4626. Since r,.-+ p

= 0, we

i°'~ ~\o;

~ = 0.000107, and 1 ~ r 7 = 0 1 < 0.00013. 1 0 Therefore, from (1), S must lie between 1.4626 and 1.4628. Its value to three decimal places is therefore 1.463. If a nonnegative series is not dominated by a convergent geometric

series ( for example, if a;:•-+ 1), Theorem I cannot be used.

However,

in this case the process of integration (cf. the integral test, § 706) can sometimes be used, as expressed in the following theorem (give the proof in Ex. 7, § 721):

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236

Theorem II. for x ~ a, if if S (6)

+•

:E

n-1

[§ 720

If f(x) is a positive monotonically decreasing function, an is a convergent positive series, with f(n) = an for n

> a,

= :Ea.., and if Sn = a1 + · · · + a,., then for n > a, S,. + +•f(x) dx < S < S,. +i+•f(x) dx. i +l n

A much sharper estimate is provided by the following theorem (hints for a proof are given in Ex. 21, § 721):

**Theorem m. Under the hypotheses of Theorem I I and the additional assumption that f"(x) is a positive monotonically decreasing function for x ~ a, then for n > a, (7)

S,.

+ !!!!±! +i+•f(x) dx - f'(n + 2) 2

n+l

12

< S < S,. + a,.±1 +i+"'f(x) dx 2

n+l

-

B!!l. 12

Example 2. Estimate the sum of the p-series with p = 2, using 10 terms. Solution. With the aid of a table of reciprocals we find that S10 = 1.549768.

With /(x) = x-t,

(+•f(x) dx J11

= -h = 0.090909

and

(+•f(x) dx

Jw

= -h =

0.100000. Thus the estimate of Theorem II places the sum S between 1.640 and 1.650, with an accuracy of one digit in the second decimal place. 1 ••Further computations give½ au= 0.004132, !'(12) = . ~23 = 0.000096, 12 6 and - / /'(10) = _~ = -0.000167. Thus the estimate of Theorem III places 2 6 03 the sum S between 1.6449 and 1.6450, with an accuracy of one digit in the fourth decimal place. NOTE. The sum in Example 2 can be specifically evaluated by means of the techniques of Fourier series (cf. H. S. Carslaw, Fourier's Series and Integrals, 3rd ed. (New York, Dover Publications, Inc., 1930), p. 234): r6 = 1 +21+32+42+ 1 1 1 ....

If an alternating series converges slowly, the estimate given in the alternating series test(§ 715) is very crude unless an excessively large number of terms is used. For example, to compute the value of the alternating harmonic series to four decimal places would require at least ten thousand terms! This particular series happens to converge to In 2 (cf. § 807), and this fortuitous circumstance permits a simpler and more speedy evaluation (cf. Example 4, § 813). However, this series will be used in the followiQg example to illustrate a technique frequently useful in evaluating

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§ 721]

237

EXERCISES

slowly converging alternating series: Evaluate the alternating harmonic series, S=l-½+i-i+··· to four decimal places. Solution. We start by evaluating the sum of the first 10 terms: S 10 = 0.645635. We wish to estimate the remainder: Example 3.

x-rr-n-+n-n+···· If we double, remove parentheses (the student should justify this step), and introduce parentheses, we find 2x

= (-tr + -tr) - (n- + n-) + (n + -h-) - • · • =rr+-tr-n--n-+n+n- •·· = rr + - + - ••• 1 1 1 1 = 11 + 11·12 - 12·13 + 13·14 - ....

Again doubling, and removing and introducing parentheses, we have 2 1 2 2 4 x = U + 11·12 + 11·12·13 12·13 •14 + · · ·' or 25 1 1 2 12·13•14 + ... • x = 11·24 + 11·12·13 Once more: 3 4x=~+ 1 + 11·12·13·14 3 -12_·_1_3_·1_4_·_1_5 + .... 11·12 11·12·13 The sum of the first two terms of this series is 0.189977, and the remainder is less than the term 3/11 · 12· 13· 14 < 0.000126. Therefore xis between 0.04749, and 0.04753, and Sis between 0.69312 and 0.69317. An estimate to four places is 0.6931 +. (The actual value to five places is 0.69315.)

721. EXERCISES

1. Prove that any series aris~ng from a divergent nonnegative series by grouping of terms is divergent. Equivalently, if the introduction of parentheses into a nonnegative series produces a convergent series, the original series is convergent. 2. Prove that the terms of a conditionally convergent series can be rearranged to give a series whose partial sums (i) tend toward +oo, (ii) tend toward -oo, (iii) tend toward oo but neither +oo nor -oo, (iv) are bounded and have no limit. 3. Prove Theorem I, § 719. *'- Prove that if l:a.. and l:b,. are nonnegative series with sums A and B, respectively, and if I:c. is their product series, with sum C, then C = AB under all circumstances of convergence or divergence, with the usual conventions about infinity (( +oo) · (+co) = +oo, (positive number)· ( +oo) = +oo) and the additional convention O· ( +oo) = 0.

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238

INFINITE SERIES OF CONSTANTS +oo

•5.

Prove that if

I:

[§ 721

+oo a,. and

n=O

I:

b,. are both the series

n=O

1

1

1

1----=+---=----=+ ... v'2 v'3 v'4

+oo

I: e,. is their product series, then :Ea,. and :Eb,. converge, while l:c. n-o diverges. Hint: Show that le.I e:; 1. •6. Prove Theorem I, § 720. *7. Prove Theorem II, § 720. *8. Prove the commutative law for product series: The product series of :Ea,. and :Eb,. is the same as the product series of :Eb,. and :Ea,.. *9. Prove the associative law for product series: Let :Ed,. be the product series of :Ea,. and :Eb,., and let :Ee,. be the product series of :Eb,. and I:c... Then the product series of :Ea.. and :Ee,. is the same as the product series of :Ed.. and :Ee,.. *10. Prove the distributive law for multiplying and adding series: The product series of :Ea,. and l:(b,. + e,.) is the sum of the product series of l:a. and :Eb,. and the product series of :Ea,. and :Ee,.. *11. Using the evaluation of the series :En-2 given in the Note, § 720, show that r2 1 1 1 8 = 1 +32+52+71+ .... and if

•12. Using the evaluation of the series :En-2 given in the Note, § 720, show that r2 1 1 1 12 = 1 - 22 + 32 - 42 + ....

In Exercises 13-18, compute to four significant digits. 1 1 ! + ! + ••• . 2 3 1 1 1 1 * 14. (In 2 =) 2 + 2·22 + 3·23 + 4•2 4 + ••••

*13. (e =) 1 +

1 I! +

1 *15. csC3> = > 1 + 23 + 331 + 431 5

*16· 22.32 + •17.

(! =)

+ ....

9 13 17 42.52 + 62,72 + g2.92 + ..•• 1-!+½-t+ ....

*18• ( In23 =')

1 1 1 2 - 2·22 + 3·23

-

1 4·2 4 + ...•

**19. Prove that the sequence {C,.}

=

ll + ½+ · · · + ~ -

is decreasing and bounded below.

Hence C =

Inn~

lim C,. exists. n-++oo

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§ 721]

239

It is believed to be transcendental, but its transcendence has never been established.) (Cf. Ex. 22.)

C is known as Euler's constant.

**20. Prove the theorem of Mertens: If if

+•

:E b,.

n-o

+ ..

L

n-o

a,. converges absolutely to A and

converges to B, then the product series

+•

L

n-o

c,. converges to AB.

Hint:

By virtue of Exs. 8-10, it may be assumed without loss of generality that

:Ea,. is nonnegative, A > 0, and B = 0. Under these assumptions prove that n n n :Ee,. converges to 0: Define .4.,. = :E a1t, B,. = L b1r, C,. = L C1t. Then k-0 k-0 k-0 = aJ>o + (aob, + a,bo) + · · · + (aob,. + •· · + a,.bo) = aoB.. + a,B,._, + · · · + a.Bo, given E > 0, first choose N such that m > N implies IB,,.I < E/2A, C,.

For a therefore

Then choose N'

>N

and

laoB,. + · · · + a•-N-1BN+1I < ½E. > N' implies

such that n

la•-NI < 2 (N E+

l) · max

**21. Prove Theorem Ill, § 720.

(JBol, JB,I, · · · , IBNI,

Hints: Let R,.

+ ½(a,.+1 + a,.+z) + ½(a,.+1 + a,.+a) + · · · =

lJ.

= a,.+1 + a,.+2 + •••

+ T •+'·

Interpret T ..+1 as a sum of areas of trapezoids, and use the trapezoidal formula error estimate ½a,.+1

(Ex. 22, § 503) to write ½(a.,

m

-h

< E..
-a, and the convergence is absolute. 802. EXERCISES

In Exercises 1-10, determine the interval of convergence, and specify the nature of any convergence at each end-point of the interval of convergence. 2 3 1 I _ 2x + (2x) _ (3x) ' 1! 2! 3! x3 x• x1

+ ...

'

2.x-3+5-7+····

3· 1

4. 6•

x2

x4

x6

+2,+41+6,+···· 1 + x + 2!x2 + 3 !x3 + 4!x 4 + · • • . ( + 1) - (x + 1)2 + (x + ])3 - (.r + 1)• + .. • X 4 9 16 • ( _ 2) + (x - 2) + (x - 2)• + (.r. - 2)7 + ... 3

6•

3!

X

7• (In

2)(x -

5)

5!

+ (In 3H.r -

V2

.'i)2

~3

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+ (In 4) (:r



-

/i)

3

+ ....

V4

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S. (x _ l) + (x ~ 1) + (x ~ 1) + (x ~ 1)7 + ...• 3

6

1 1·3 1·3·5 1·3·5·7 *9• l-2x+2·4x -2·4·6x3+2-4-6·8x 1·3 x 1 • 3 •5 x *10• + x6 + 2•4 . 5 + 2•4•6 7 + • • • • 1

1

8

4 -

···•

7

X

11. Determine the values of x for which the series

_1_ +

1 + 1 + .;; 3(x - 3) 3 4(x - 3) 4 converges, and specify the type of convergence. 12. Determine the values of x for which the series 3 7 . sin1 x - sin x + smx - sin -3-x + --5-7.•• • x - 3

1

2(x - 3) 2

+

converges, and specify the type of convergence. 13. Prove Note 2, § 801. 14. Prove Note 3, § 801. , 16. Prove Theorem III, § 801, for the cases R = 0 and R = +oo. 16. Prove Note 4, § 801, for the case where lim ~;,;: exists (replacing

um by lim). •17. Give a proof of Theorem I, § 801, based on convergent sequences, as follows: Assume that neither condition (i) nor (ii) of that theorem holds. Show first that there exist positive points of convergence and of divergence which are arbitrarily close. Then construct two sequences {c,.} and {d,.} of points of convergence and divergence, respectively, where 0 :ii c,. :ii c,.+1 < d,.+1 :ii d,., for n = 1, 2, · · · , and define R to be their common limit. Prove that R is the radius of convergence of the given series. •18. Prove Note 4, § 801. (Cf. Ex. 32, § 711.) •19. Apply Note 4, § 801 to Example 3, § 801. (Cf..Ex. 33, § 711.) •20. Show by examples that Theorem III, § 801, cannot be generalized in the manner of Note 4, § 801, by the use of limits superior or inferior. (Cf. Exs. 31-33, § 711.) 803. TAYLOR SERIES

We propose to discuss in this section some formal procedures, nearly all of which need justification and will be discussed in future sections. The purpose of this discussion is to motivate an important formula, and raise some questions. Let us suppose that a power series :Ea,.(x - a)" has a positive radius of convergence (0 < R ~ +oo ), and let/(x) be the function defined by this series wherever it converges. That is, (1)

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We now differentiate term-by-term, a.s if the infinite series were simply a finite sum: f'(x) = a1 + 2at(x - a) + 3aa(x - a) 2 + ••• . Aga.in: J"(x)

=

202

+ 2·3aa(x -

a)

+ 3·4a.c(x -

a) 2

+ ••• .

And so forth: J"'(x)

= 3 I aa + 2·3·~(x

- a)+ 3·4·5a,;(x - a) 2

+ ... ,

= a, we have: = a1, f"(a) = 2 ! 02, !"'(a) = 3 ! aa,

Upon substitution of x

f(a) = ao, f'(a) or, if we solve for the coefficients a,.:

ao = f(a), a1 = f'(a),

(2)

··•,

/(R)/a\

/If( n\

= ~ ' .· · · , a,. = ~ ' · · · ·

at

This suggests that if a power series l:a,.(x - a)" converges to a function J(x), then the coefficients of the power series should be determined by the

values of that function and its successive derivatives according to equa.tions (2). In other words, we.should expect: (3)

J(x)

= f(a) + f'(a)(x

- a)

+ ~ (x -

a) 2

+ •••

+~ n I (x and, in particular, for a

- a)"

+ ... '

= 0: /(R)fn\

(4)

J(x)

= /(0) + f'(0)x + .. · + ~ x" + .. ·.

Now suppose /(x) has derivatives of all orders, at least in a neighborhood of the point x = a. Thenj< >(a) is defined for every n, and the series (3) exists. Regardless of any question of convergence or (in case of con+• J (x) exist.s at every point of an interval I containing the point x = a, then the Taylor's Formula with a Remainder for the function f(x), for any point x of I, is (1)

f(x) = J(a)

+ f'(a)(x

- a)

f (n-ll(a) (x + (n _.__....__,_ - 1) I

+ f"(a) ! 2 - a) 11-

1

(x - a) 2

+ ···

+ R" (x) •

The quantity R,.(x) is called the remainder after n terms. The principal substance of Note 3, § 407, is an explicit evaluation of the remainder R,.(x):

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Theorem I. Lagrange Form of the Remainder. Let n be a fixed positive integer. If f(x) exists at every point of an interval I (open, c1.osed, or halfopen) containing x = a, and if xis any point of I, then there exists a point ~" between a and x (~,. = a if x = a) such that the remainder after n terms, in Taylor's Formula with a Remainder, is

R,.(x) = J (~n) (x - a)".

(2)

n.

The principal purpose of this section is to obtain two other forms of the remainder ·R,.(x). We first establish an integral form of the remainder (Theorem II), from which both the Lagrange form and the Cauchy form (Theorem Ill) can be derived immediately (cf. Ex. 1, § 806). Assuming continuity of all of the derivatives involved, we start with the obvious identity L"-af'(x - t) dt = f(x) - f(a), and integrate by parts, repeatedly: f(x) - f(a) = Lx-af'(x - t) dt = [tf'(x - t)

= f'(a)(x - a)

= f'(a)(x

+Jor,:-af"(x

]:-a +L"-atf"(x -

- t) d

t) dt

(21t2)

]:-a +L"-a:,f'"(x - t) dt rz-af"'(x - t) d (t8) a) + Jo 31

- a)+ [:,f"(x - t)

= f'(a)(x - a)

+ EJQl (x 21

-

2

= ..... Iteration of this process an appropriate number of times leads to the formula written out in the following theorem: Theorem II. Integral Form of the Remainder. Let n be a fixed positive integer. If /(~ ..) •(x

- a).

On the other hand, if x ~ a, we obtain the same result by reversing the inequalities or replacing them by equalities. We have the conclusion:

Theorem m. Cauchy Form of the Remainder. Let n be a fixed posiiive inf.eger. If f (x) exists and is continuous throughout an interval I containing the point x = a, and if xis any point of I, then there exists a point~.. between a and x (~.. = a if x = a) such that the remainder aff.er n terms, in Taylm's F ormul,a with a Remainder is (4)

805. EXPANSIONS OF FUNCTIONS

Definition. A series of functions Eu,.(x) represents a function f(x) on a certain set A if and only if for every point x of the set A the series Eu,.(x) converges f,o the value of the function f(x) at that point. Immediately after the question of convergence of the Taylor series of a function comes the question, "Does the Taylor series of a given function represent the function throughout the interval of convergence?" The clue to the answer lies in Taylor's Formula with a Remainder. To clarify the situation we introduce the notation of partial sums (which now depend on x): f 0.

A tech-

nique for doing this is suggested in Exercise 37, § 811. We conclude that the binomial series represents the binomial function throughout the interval of convergence. 808. ELEMENTARY OPERATIONS WITH POWER SERIES

From results obtained in Chapter 7 for series of constants (§§ 702, 719) we have the theorem for power series (expressed here for simplicity in terms of powers of x, although similar formulations are valid for power series in powers of (x - a)):

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[§ 808

Theorem. Addition, Subtraction, and Multiplication. Let

+•

E b,.xn be two power 8erie8 repre8enting the functiom f1(x)

+•

E

n-0

anx" and

and h(x), respec-

n-o

tively, within their intervals of convergence, and let 'Y be an arbitrary comtant. Then (i) the p'ower BerieB E-ra,.x" repreBents thefunction-rf1(x) throughout the interval of convergence of Eanx"; (ii) the power Berie8 E(a,. ± b,.) x" represents the function f1(x) ± f2(x) for all points common to the intervals of conn

vergence of the two given power series; and (iii) if

Cn

=E

a,.blHO, n = O,

1,-0

1, 2, · · · , then the power Beries

+•

E c,.x" represents the function fi(x) h(x) for

n-o

all points interior to both intervals of convergence of the two given power Beries (cf. § 909). In finding the Maclaurin or Taylor series for a given function, it is well to bear in mind the import of the uniqueness theorem (Theorem II, § 803) for power series. This means that the Maclaurin or Taylor series need not be obtained by direct substitution in the formulas defining those series. Any means that produces an appropriate power series representing the function automatically produces the Maclaurin or Taylor series. Since the series for

Example 1.

1

e"

is 1 + x + ; ! + • • • , the series for e~

is found by substituting -x for x: e-s..,1-x+:,;I _xi+··· 21

31

.

This series expansion is valid for all real x, and is therefore the Maclaurin series fore-". The Maclaurin series for cos 2x is 0. Hint: Use the method sugn-+• n gested in the hints of Ex. 36, k being a positive integer > 1/m. **38. Prove Theorem I, § 810, by using limits superior and the formula of Note 4, § 801.

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[§ 812

812. INDETERMINATE EXPRESSIONS

It is frequently possible to find a simple evaluation of an indeterminate expression by means of Maclaurin or Taylor series. This can often be expedited by the "big 0" or "order of magnitude" concepts. (Cf. Ex. 10, § 713.) . d 1·1m 1 - cos x · Fm 2 Z-0 X

E:umpIe 1. Solution.

Since

cos x = 1 - ~ 21 Example 2.

Solution.

+ 4! ~ - ••• =

1 - ~ 2

Find lim n{In (n

..-+•

+ O(x

+ 1)

4)

1 - cos x = ! x2 2

'

+ O(x

1)

-+ !.

2

- Inn} .

We write the expression

¼) = !+ 0 (~} Therefore n{In + 1) - Inn} = 1 + 0 (!)-+ 1. Example 3. Find lim In ~I + t>. z-o e"In (n

+ 1) -

Inn= In ( 1 +

(n

Solution.

In (1 e1" -

+ x) 1

_ x - 2x

+ O(x1) + O(x

1)

=

+ O(x) + O(x) [1 + O(x)][½ + O(x)] 1 2

= ½

+ O(x)-+ ½-

813. COMPUTATIONS

The principal techniques most commonly used for computations by means of power series have already been established. In any such computation one wishes to obtain some sort of definite range within which the sum of a series must lie. The usual tools for this are (i) an estimate provided by a dominating series (Theorem I, § 720), (ii) an estimate provided by the integral test (Theorems II and III, § 720), (iii) the alternating series estimate (Theorem, § 715), and (iv) some form of the remainder in Taylor's Formula (§ 804). Another device is to seek a different series to represent a given quantity. For instance, some logarithms can be computed more efficiently with the series l+x [ xa x6 :,;7 ] (1) ln --=2 1-x 3 5 7

x+-+-+-+ ···

+

(cf. Ex. 9, § 811) than with the Maclaurin series for In (1 x). Finally, one must not forget that infinite series are not the only means for computation. We include in Example 5 one illustration of the use of an approximation (Simpson's Rule) to a definite integral.

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§ 813)

COMPUTATIONS

265

We illustrate some of these techniques in the following examples: Eumple 1. Compute the sine of one radian to five decimal places. Solution sin i = 1 - 1. + 1- - .l. = 0.84147 with an error of less than • 31 5! 71 1/9 I


0, but not uniformly on (0, +oo ). Solution. For x ~ a, and a fixed n > 1/E, the function J(x) = nxe-- is monotonically decreasing, with a maximum value of nae--. This quantity is independent of x and approaches Oas n-. +oo. On the entire interval (0, +oo ), the function J(x) nxe-"" has a maximum value given by x = 1/n and equal to 1/e. The convergence is therefore not uniform.

=

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273

*902. UNIFORM CONVERGENCE OF SERIES

Let

u1(.x) + ttt(x) + · · · + Un(x) + · · · be a series of functions defined on a set A, and let (1)

=

S,.(x)

u1(x)

+ · ·· + u,.(x).

We say that this series of functions converges on A in case the sequence {S,.(x)} converges on A. The series (1) converges uniformly on A if and only if the sequence {S,.(x)} converges uniformly on A. NOTE. Any convergent series of constants (constant functions) converges uniformly on any set.

Corresponding to the condition a,. -+ 0 which is necessary for the convergence of the series of constants Ean, we have:

Theorem. If the series Eun(x) converges uniformly on a set A, then the general term u,.(x) converges to O uniformly on A. Proof. S(x)

By the triangle inequality, if S,.(x)

+•

= E

= u1(x)

+ · · · + u,.(x) and

Un(x),

.. -1

lu..(x)I

1s..(x)

=

- S-1(.x)I = l[Sn(X) - S(x)]

;:;;; ISn(x) - S(x)I

Let

E

> 0 be given.

+ ISn-1(.x) -

- Sn-1(x)]I

S(x)I,

If N is chosen such that n ISn(x) - S(x)I

+ [S(x)

>N

- 1 implies

< ½E

for all x in A, then n > N implies lu,.(x)I < E for all x in A. Example. Show that the Maclaurin series fore" converges uniformly on a set A if and only if A is bounded. Solution. If the set A is bounded, it is contained in some interval of the form [ -a, a]. Using the Lagrange form of the remainder in Taylor's formula fore" at x = a = 0, we have (with the standard notation) for any x, et. ff" IS,.(x) - S(x)I = IR,.(x)I = 1 lxl" ;:;i 1 a". n. n. Since lim

ff"! a"

..-+• n

= 0 and

ff"! a" is independent of x, the uniform convergence

n.

on A is established.

'

+•

If the set A is unbounded, we can show that E x" fails to converge uniformly n-o n 1 on A by'showing that the general term does not approach O uniformly on A. This we do with the aid of the Negation of Uniform Convergence formulated in § 901: letting e be 1 and n be any fixed positive integer, we can find an x in A such that lxl" > n I

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[§ 903

UNIFORM CONVERGENCE

*903. DOMINANCE AND THE WEIERSTRASS M-TEST The role of dominance in uniform convergence of series of functions is similar to that of dominance in convergence of series of constants (§ 708). Definition. The statement that a series of functions Ev,.(x) dominates a series of functions Eu,.(x) on a set A means that all terms are defined on A and that for any x in A lu..(x)I ~ v,.(x) for every positive integer n. Theorem I. Comparison Test. Any series of functions Eu..(x) dominated on a set A by a series of functions Ev,.(x) which is uniformly convergent , on A is uniformly convergent on A. Proof. From previous results for series of constants, we know that the series Eu,.(x) converges for every x in A. If u(x) Eu,.(x) and v(x} = Ev,.(x), we have (cf. Theorem I, § 716):

=

+ U2(x) + · · · + u,.(x)] - u(x)I = lu..+1(x) + U..+2(x) + ···I ~ Vn+1(x) + V..+2(x) + · · · = I[v1(x) + v2(x) + · · · + v,.(x)] - v(x)I.

l[u1(x)

If E > 0 and if N

= N(E)

is such that for n

> N,

+ · · · + v,.(x)] for all x in A, then l[u1(x} + · · · + u,.(x)] I[v1(x)

v(x)I < E u(x)I < E for all x in A.

Corollary. A series of functions converges uniformly on a set whenever its series of absolute values converges uniformly on that set. Example 1. Since the Maclaurin series for sin x and cos x are dominated on any set by the series obtained by substituting lxl for x in the Maclaurin series for ez, these series for sin x and cos x converge uniformly on any bounded set. Neither converges uniformly on an unbounded set. (Cf. the Example, § 902.)

Since any convergent series of constants converges uniformly on any set, we have as a special case of Theorem I the extremely useful test for uniform convergence due to Weierstrass: Theorem II. Weierstrass M-Test. If Eu,.(x) is a series of functions defined on a set A, if EM,. is a convergent series of nonnegative constants, and if for every x of A,

lu,.(x)I ~ M,., n then Eu,.(x) converges uniformly on A. (1)

=

1, 2, · · ·,

+

Example 2. Prove that the series 1 + e-" cos x + e-2" cos 2x converges uniformly on any set that is bounded below by a positive constant. Solution. If a > 0 is a lower bound of the set A, then for any x in A, le-,,., cos nxJ ~ e-"" ~ e--. By the Weierstrass M-test, with M,. s e--..«, the given series converges uniformly on A (the series Ee-- is a geometric series with common ratio e- < 1). (Cf. Ex. 26, § 904.)

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EXERCISES

*904. EXERCISES

In Exercises 1-10, use the Weierstrass M-test to show that the given series · converges uniformly on the given set.

x211-1 _ l) ; (-1000, 2000). 1

•3. I: (2n •6.

~

sinnx

+ 1i (-ao,

~ n'

+ao).

•7.

I: ne--; [a, +ao ), a > 0.

•9.

In X)" ; [1, L ( -;:

+ao).

•2.

L X" ,; n

*'-

~ -· x ~n"'

•6.

~ ~

•8.

L

•10.

[-1, 1].

X"

= 1, 2'

· · · ' K.

sinnx

- - ; ( -ao, +ao ).

e" e"" ,.; (-ao, a], a 5

I: (x In x)";

< In 5.

(0, 1].

In Exercises 11-20, show that the sequence whose general term is given converges uniformly on the first of the two given intervals, and that it converges but not uniformly on the second of the two given intervals. The letter 11 denotes an arbitrarily small positive number.

•11. sin" x; [0, ½r - 11]; [0, ½r). x *13. -+; [0, b]; [0, +ao). x n nx •16. + nx; [11, +ao); (0, +ao).

*12. {1/sin x; [11, ½r]; (0, Jr]. •H. x+n - - ; [a, b]; (-ao, +ao). n nx •16. 1 + n'x2 ; [11, +ao); (0, +ao ).

•17. In (1 + nx); [0, b]; [0, +ao ).

•18. n2x2e--; [11, +ao); (0, +ao).

1

n

*19. l

+X" x"; [0, 1 -

11]; [0, 1).

•20.

sin nx -+; nx

1

[11, +ao); (0, +ao),

·

In Exercises 21-24, show that the given series converges uniformly on the first of the two given intervals, and that it converges but not uniformly on the second of the two given intervals. The letter 11 denotes an arbitrarily small positive number.

•21.

L

x"; [0, 1 - 11]; [0, 1).

1 •23. I:-;

n"

[1 +11, +ao); (1, +ao).

•22.

•24.

L ~

~

X"; [0, 1 - 11]; [0, 1). n nx - ; (71, +ao); (0, +ao }. e""

•26. Prove the Negation of Uniform Convergence, § 901. •26. Prove that the convergence of the series in Example 2, § 903, is not uniform on (0, +ao ). •27. Prove that if a sequence of functions converges uniformly on a set, then any subsequence converges uniformly on that set. Show by an example that the converse is false; that is, show that a sequence which converges nonuniformly on a set may have a uniformly convergent subsequence. (Cf. Ex. 28.) •28. Prove that if a monotonic sequence of functions (B..+1Cx) ~ S,.(x) for

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[§ 904

UNIFORM CONVERGENCE

each x) converges on a set and contains a uniformly convergent subsequence, then the convergence of the original sequence is uniform. (Cf. Ex. 27.) *29. Prove that if a sequence converges uniformly on a set A, then it converges uniformly on any set contained in A. *30. Prove that if a sequence converges uniformly on a set A and on a set B, then it converges uniformly on the combined set made up of points that belong either to A or to B (or to both). Show that any convergent sequence converges uniformly on every finite set. Show that if a sequence converges uniformly on an open interval (a, b) and converges at each endpoint, then the sequence converges uniformly on the closed interval [a, b]. •31. Prove that if S,.(x) :::; S(x) on a set A, and if S(x) is bounded on A, then the functions S,.(x) are uniformly bounded on A: there exists a constant K such that IS,.(x)I ~ K for all n and all x in A. •32. Prove that if S,.(x):::; S(x) on a set A, and if each S,.(x) is bounded on A (there exists a sequence of constants {K,.} such that IS,.(x)I ~ K,. for each n and all x in A), then the functions S,.(x) are uniformly bounded on A. (Cf. Ex. 31.) •33. If f(x) is defined on [0, 1] and continuous at 1, prove that {x"f(x)} converges for every x of [0, 1], and that this convergence is uniform if and only if j(x) is bounded and J(l) = 0. •34. If two sequences {f,.(x)} and {g.,(x)} converge uniformly on a set A, prove that their sum {f,.(x) + g,.(x)} also converges uniformly on A. Show by an example that their product {f,.(x)g,.(x)} need not converge ·uniformly on A. Prove, however, that if both original sequences are uniformly bounded (cf. Ex. 31), then the sequence {f,.(x)g,.(x)} converges uniformly. What happens if only one of the sequences is uniformly bounded? *35. Construct a series :Eu,.(x) such that (i) :Eu,.(x) converges on a set A, (ii) u,.(x)-➔ 0 on A, and (iii) :Eu,.(x) does not converge uniformly on A. •36. Prove the ratio test for uniform convergence: If :Eu,.(x) is a series of nonvanishing functions on a set A, and if there exist constants N and p, where p < 1, such that ,u~:(;~),

~

p for all n

>

N and all x in A, then :Eu,.(x) con-

verges uniformly on A. Use this test to prove that for any fixed x such that lxl < 1, the binomial series ((11), § 807), considered as a series of functions of m, converges uniformly for 1ml ~ M, where M is fixed. In Exercises 37-42, find an appropriate function N(E), as prescribed in the Definition of uniform convergence, § 901, for the sequence of the given Exercise (and the set specified in that exercise). *38. Ex. 13.

•37. Ex. 11.

•39. Ex. 15.

*40. Ex. 17. Hint: The inequality In (1 + nb) 0,

11], 1/

> 0.

but not uni-

(Cf. Ex. 51.)

+ nz converges uniformly on every bounded set.

+• (-1)" x' **55. Show that :E

n2

n-1

Where does it converge absolutely?

+..

**56. Show that

+ 1/, 1 -

:E (-1)" n~ e-

n-1

but absolutely only for x

= 0.

.,.

;;i

(Cf. Ex. 51.) converges uniformly on every bounded set,

(Cf. Ex. 51.)

*905. UNIFORM CONVERGENCE AND CONTINUITY 1

r"-

1 The example S,.(x) = , -1 ;:ia x ;:ia 1 (Fig. 903), shows that the limit of a sequence of continuous functions need not be continuous. The limit in this case is the signum function (Example 1, § 206), which is discontinuous at x = 0. As we shall see, this is possible because the convergence is not uniform. For example, let us set E = ½, Then however large n may be there will exist a positive number x so small that the inequality

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CONVERGENCE AND INTEGRATION 1

1

IS,.(x) - S(x)I < E, which is equivalent to 1 - x ~ < ½, or x ~ fails. In case of unifonn convergence we have the basic theorem:

> ½,

Theorem. If S,.(x) is continuous at every point of a set A, n = 1, 2, • • • , and if S,.(x) =t S(x) on A, then S(x) is continuous at every point of A. Proof. Let a be any point of A, and let E > 0. We first choose a positive integer N such that ISN(x) - S(x)I < ½E for every x of A. Holding N fixed, and using the continuity of SN(x) at x = a, we can find a a> 0 such that Ix - al < a implies ISN(X) - SN(a)I < ½E. (The number a apparently depends on both N and E, but since N is determined by E, ais a function

of E alone-for the fixed value x

= a.)

IS(x) - S(a)I ;;;; IS(x) - SN(x)I

Then Ix - al

Now we use the triangle inequality:

+ ISN(x)

- SN(a)I

+ ISN(a)

- S(a)I,

< a implies IS(x) - S(a)I

< ½E + ½E + }E = E,

and the proof is complete. Corollary. If f(x) = :Eu..(x), where the series converges uniformly on a set A, and if every term of the series is continuous on A, thenf(x) is continuous

on A. (Ex. 19, § 908.) Example.

.

Show that the function f(x) defined by the series

+•

:E e-n" cos nx

n-0

of Example 2, § 903, is continuous on the set (0, +oo )-that is, for positive x. Solution. Let x = a > 0 be given, and let a = ½a, Then the given series converges uniformly on [½a, +oo ), by Example 2, § 903. Therefore/(x) is continuous on [ ½a, +oo) and, in particular, at x = a. *906. UNIFORM CONVERGENCE AND INTEGRATION

The example illustrated in Figure 904 shows that the limit of the integral (of the general term of a convergent sequence of functions) need not equal the integral of the limit (function). The function S,. (x) is defined to be 2n2x for O ;;;; x ~ 1/2n, 2n(l - nx) for 1/2n ;;;; x ;;;; 1/n, and 0 for 1/n ;;;; x ;;;; 1. The limit function S(x) is identically 0 for 0 ;;;; x ;;;; 1. For every n,



1

s.(x) dx = ½ (the integral is the area of a triangle of altitude n and

.fo s(x) dx = 0. 1

base· 1/n), but

Again, the reason that this kind of mis-

behavior is possible, is that the convergence is not uniform (cf. Ex. 22, § 908). (For another example of the same characte~, where the functions S.(x) are defined by single analytic formulas, see Ex. 25, § 908.)

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UNIFORM CONVERGENCE

In case of uniform convergence we have the theorem: Theorem. S,.(x) =i S(x)

, and if

If S,.(x) is integrable on [a, b] for n = I, 2, on La, b], then S(x) is integrable on [a, b] and

Jim ibS,.(x) dx = . [ lim S,.(x) dx =ihS(x) dx.

(1)

n-+oo

a

an-+•

a

y

S,.(x)

3 2

1

z 1

1..

a

n

1

1

2

FIG. 904

/

Proof. We shall first prove that S(x) is integrable on [a, b(,The idea is to approximate S(x) by a particular SN(x), then to squeeze SN(x) between two step-functions (§ 502), and finally to construct two new stepfunctions that squeeze S(x). Accordingly, for a given E > O, we find an index N such that ISN(x) - S(x)I

< 4(b

~

a) for a

~

x

~ b. Since SN(x)

is integrable on [a, b], there must exist step-functions u1(x) and T1(x) such that u1(x)

~

S.v(x)

~

T1(x) on [a, b], and i\n(x) - ui(x)] dx

fine the new step-functions: u(x)

Then, for a

Digitized by

= u1(x) -

~ x ~

E

4 (b _ a)' T(x)

= T1(x)

+ 4 (b E_

b,

u(x)

< ui(x) + [S(x)

- S.v(x)] ~ S(x),

T(x)

>

+ [S(x)

- S.v(x)] ;;; S(x),

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UNIVERSITY OF MICHIGAN

a)°


0. Solution. Let x be a given positive number, and choose a and b such that 0 < a < x < b. Since the original series has already been shown to converge (uniformly) on [a, b], it remains only to show that the derived series converges uniformly there. This is easily done by the Weierstrass M-test, with M,. = 2ne-.. (check the details). 11

*908. EXERCISES

In Exercises i-6, show that the convergence fails to be uniform by showing that the limit function is not continuous.

•1.

lim sin" x, for O :iii x :iii r.

"-+•

X" •3. lim 1-+, for O ~ ..-+• X"

•2.

lim e-ue, for

11-+•

lxl

~

1.

x ;::ii 2 •

. sin nx hm S,.(x), where S,.(x) • - - for O < x ~ r, and 8,.(0) = 1. nx •IS. (1 - x) + x{l - x) + x1{1 - x) + ••• , for 0 :iii x ~ 1. x2 x1 1 •6. x + 1 + x' + (l + x')' + · · · , for lxl ~ 1.

*'-

..-+•

In Exercises 7-12, show that the equation is true.

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284

*1l. ( ..

f

Jon= t

*12.

i

2+•

I:

n=l

=~ 2

n sin nx dx e"

ne--- dx

[§ 908

e -l

= _e __ 2 e -l

In Exercises 13-18, show that the equation is true.

[+• + l} = +•L -+2, for \x\ < 1. n=on d [+• *14. dL .!!. ] = - +• L n+i, for \x\ > 1. d

*13. d-x n-1n L (n

X"

]

X"

2

X

n= 1 X"

n- 1 X"

d [~ sin nx] cos nx *15. dL., - =~ L.. - - , for any x. 32 X

n=l

n

n-1

n

+• . ] = +• . nx, for x "F- 0. *16. -d [ L ~ L cos2 nx - sm 3 3 2 dx

n=l

nx

n=l

nx

nx

[+• +1 nx ] = -2x n-1n +•L (l +1 nx for any x. d [+• . knx ] = +• . knx], for x > 0. *18. dL e_,.., sm L ne-""[k cos knx - sm d

*17. d-x n=ln L 1 (l X

n-1

2)

2) 2

2

n=l

*19. *20. *21. *22.

Prove the Corollary to the Theorem of § 905. Prove the Corollary to the Theorem of § 906. Prove the Corollary to the Theorem of § 907. Prove that the convergence of the sequence illustrated in Figure 904, § 906, is not uniform. *23. Complete the final details of the proof of § 907 .•

= n! e-,..,, S(x) = n-+• lim S,.(x). Show that lim S,.'(O) "F- S'(0). n-+oo Let S,.(x) = nxe-nzl, S(x) = lim S,.(x). Show that n-+oo

*24. Let S,.(x) *25.

1

1

lim Cs,.(x) dx "F- C s(x) dx. n-+oo Jo Jo

*26. Construct an example to show that it is possible to have a sequence of continuous functions converge nonuniformly to a continuous function, on a closed interval [a, b]. *27. Construct an example to show that it is possible to have a sequence {S,.(x)} of continuous functions converging to a continuous function S(x} nonuniformly on a closed interval [a, b], but still have

.£b

S,.(x) dx

-Lb

S(x) dx.

Hint: Construct functions like those illustrated in Figure 904, § 906, but having

uniform maximum values. *28. Construct an example to show that it is possible to have S,.(x) ~ S(x) and S,.'(x) --. S'(x) nonuniformly on an interval. Hint: Work backwards from

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EXERCISES

§ 908]

the result of Ex. 27, letting {S,.'(x)} be the sequence of that exercise, and S,.(x)

=IazS,.'(t) dt.

*29. Show that S,.(x)

=1 + n x n2x2

2 2

furnishes an example of the type requested

in Exercise 27, for the interval [ -1, 1]. sin nx •30. Show that S,.(x) ~ 0 for all x, but that {S,.'{x)} converges only n for integral multiples of 2r. ••31. Prove that 1 _ m + m(m - I) _ m(m - l)(m - 2) + ... 2! 3! eonverges, but not uniformly, for 0 ~ m ~ I. Hint: The limit function is discontinuous. ••32. Show that although nx{l - x)"-+ 0 nonuniformly for 0 ~ x ~ 1,

=--

.J:/nx(l - x)" -+ 0. § 904).

Hint: The functions are uniformly bounded (Ex. 31,

Cf. Ex. 49, § 503, or proceed from first principles.

••33. Show that the series +oo E -1 ( ___..!_l )" converges uniformly for a ~ x ~ ½ n=l n X and that the derived series converges uniformly for a ~ x ~ b < ½. ••34. Prove that if u,.(x) is improperly integrable on [a, +oo ), for n = 1, 2, · · · , and if Exku,.(x), where k > 1, is dominated by a convergent series of constants, then the series Eu..(x) can be integrated term by term, from a to +oo:

r+oo n-1 E u,.(x) dx = n=l)a f r+oo u,.(x) dx.

)a

Hint: Use Ex. 46, § 515. The hint given in that Exercise, together with the Weierstrass M-test provides a proof of the present proposition without the necessity of using Ex. 49, § 503. **36. Show by an example that uniform convergence of a sequence of functions on an infinite interval is not sufficient to guarantee that the integral of the limit is the limit of the integral. (Cf. Ex. 46, §1515.) Hint: Consider S,.(x) 1/n for 0 ~ x ~ n, and S,.(x) 0 for x > n. **36. Prove that any monotonic convergence of continuous functions on a closed interval (more generally, on any compact set) to a continuous function there is uniform. Hint: Assume without loss of generality that for each x belonging to the closed interval A, S,.(x) ! and S,.(x) -+ 0. If the convergence were not uniform there would exist a sequence {xk} of points of A such that S.,,.(xk) ~ E > 0. Assume without loss of generality that {xk} converges: Xk -+ x. Show that for every n, S,.(x) ~ E. **37. Prove the Moore-Osgood Theorem: If S,.(x) is a sequence of Junctions defined for X in a deleted neighborhood J of X = c, and if (i) S,. lim S,.(x) exists and is finite for every n = 1, 2, · · · ;

=

=

=

x--+C

(ii) J(x) = lim S,.(x) exists and is finite for et'ery x in J(x

;,E,

c);

n--++ao

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[§908

(iii) the convergence in (ii) i8 uniform in J; then (iv) lim S,. exists and i8 finite;

-+•

(v) lim /(x) exists and i8 finite; (vi) the limits in (iv) and (v) are equal. Similar results hold for x - c+, c-, +co, -co, and co. Hints: For (iv), write 1s.. - s..1 ~ 1s.. - S .. (x)I 1s.. (x) - S,.(x)I jS,.(x) - s.. ,, let N be such that m > N and n > N imply IS.. (x) - S,.(x)I < e/3, for all x in J, and let x -+ c. For (v), write 1/(x') - /(x")I ~ IJ(x') - S,.(x')I + jS,.(x') - S,.(x")I + jS,.(x") - J(x")I, let N be such that n > N implies jS,.(x) - /(x)I < e/3 for all x in J, and use (i). **38. Prove the Theorem of § 907, without the assumption of continuity for the derivatives. Hints: First write {S.,(Xo) - S,.(:ro)} S.,(x) - S,.(x) = {(S,.(x) - S,.(x)] - [S.,(xo) - S,.(xo)]} = [S,,.'W - S,.'W] (x - Xo) {S.. (Xo) - S,.(Xo)}, to establish S,.(x) ~ S(x). Then use the Moore-Osgood Theorem (Ex. 37) to obtain lim lim S,.(x) - S,.(c) = lim lim S,.(x) - S,.(c),

+

+

+

+

.,....., n-+•

X -

n-+• .,_....,

C

X -

C

by writing

~W-~M_&W-Mtj_~W-&W_~W-Mtj x-c

x-c

x-c

x-c

= s..'(U - S,.'(U.

(Cf. Ex. 45, § 904.)

**39. Prove that if a sequence of differentiable functions with uniformly bounded derivatives (Ex. 31, § 904) converges on a closed interval, then the convergence is uniform. Hints: Assume S,.(x)-+ S(x) and jS,.'(x)I ~ K on [a, b]. First use IS,.(x') - S,.(x")I ~ K · Ix' - x"I and IS(x') - S(x")I ~ IS(x') - S,.(x')I jS,.(x') - S,.(x")I jS,.(x") - S(x")I to show that S(x) is continuous. Assume the convergence is not uniform, and let z1o-+ x such that jS..,.(x1o) - S(x1o)I i1:; e > 0. But 1s...(z1o) - S(x1o)I :a 1s...(x1,) - S..,.(z)I IS..,.(z) - S(z)I IS(z) - S(x1o)I. Let /,.(x) and /(x) be Riemann-Stieltjes integrable with respect to g(x) on [a, b], and assume /,.(x) ~ /(x) on [a, b]. Prove that

+

+

+

**'°·

+

J:"J,.(x) dg(x) -J:"!(x) dg(x). **4.1. Let /,(x), /2(x), • • • be a sequence of "sawtooth" functions (Fig. 906), defined for all real x, where the graph of /,.(x) is made up of line segments of slope ±1, such that/,.(x) = 0 for x = ±m•4-.., m = 0, 1, 2, · · ·, and/,.(x) = ½·4-,, for x

=

½·4-,.

+ m-4-

11

,

m

= O, 1, 2,

••• .

Let/(x)

+• = I: /,.(x). Prove n-o

that /(x) is everywhere continuous and nowhere differentiable.t

t This example is modeled after one due to Van der Waerden. 1'h~ory of F1mctions (Ol(ford, Oxford University Press, 1932).

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Cf. E. C. Titchmarsh,

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§ 908]

EXERCISES

287

differentiability: If a is any fixed point, show that for any n = 1, 2, · · · a numh,.) ber h,. can be chosen as one of the numbers 4-..-i or _4--..- 1 such thatf,.(a - f,.(a) = ±h,.. Then J,,.(a h,.) - J,,.(a) has the value ±h,. for m ~ n, and otherwise vanishes. Hence the difference quotient [J(a h,.) - f(a)]/h,. is an integer of the same parity as n (even if n is even and odd if n is odd).

+

+

+

Therefore its limit as n - +ao cannot exist as a finite quantity.

'A./VV'../\.LVVV f.+i

(x)

FIG. 906

••42. Modify the construction of Exercise 41 to prove the following generalization (cf. the Note below): Let p(t) be a positive-valued Junction defined fort > 0 such that lim p(t) = 0. Then there exists afunctionf(x), defined and continuous 1-0+

for all real x, having the property that corresponding to any real number a there h,.) - J(a)IIP(lh,.I) is a sequence of numbers {h.} such that h. - 0 and IJ(a +ao. Hint: Let f,.(x) be a sawtooth function, somewhat like that of Ex. 41, having maximum value ½•4--, and minimum value O occurring for x = all integral multiples of a number a,. defined inductively to be a sufficiently high integral power of l to ensure the following inequality, where AA: == 4-A:ak-1 is the absolute value of the slopes of the segments in the graph of fk(x), k = I, 2, • • • , n - 1, (a1 being defined to be i): A. > A1 + A, + · · · + A,._1 + n[p(a,./4)]/a,., n > 1. Then choose h. == ±a./4 such that lf.(a h,.) - f,.(a)l/lh,.I = A,.. Define

+

J(x)

==

+

+..

L

n-1

f,.(x).

**NOTE. The statement of Exercise 42 is of interest in connection with the concept of modulus of continuity of a function f(x) on an interval I, which is a function of a positive independent variable 8, denoted w(8) and defined: w(8) == sup IJ(x1) - f(x,)I, formed for all x1 and x, of I such that lx1 - xii < 8. The function w(8) is monotonic, and hence approaches a limit w, as 8 - o+. The statement w = 0 is clearly equivalent to the statement that J(x) is uniformly continuous on I. The example of Exercise 42 shows that for functions continuous (and hence uniformly continuous) over a closed interval, there is no bound to the slowness with which the modulus of continuity can approach O as 8 - o+. (With the notation of Ex. 42, w(8,.) > p(8,.) for a sequence {8,.} approaching 0.) The method of proof suggested in Exercise 42 is due to F. Koehler. Also cf. W. S. Loud, "Functions with Prescribed Lipschitz condition," Proc. A. M. S., Vol. 2, No. 3 (June 1951), pp. 358-360.

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*909. POWER SERIES.

[§ 909

ABEL'S THEOREM

Chapter 8 contains three important theorems on power series (Theorem I, § 803, Theorems II and III, § 810), whose proofs have been deferred. These theorems (having to do with continuity, integration, and differentiation of power series within the interval of convergence) are simple corollaries (cf. Exs. 10-13, § 911) of the following basic theorem:

Theorem I. A power series converges uniformly on any interval whose end-points lie in the interior of its interval of convergence. Proof. For simplicity of notation we shall assume that the series has +oo the form E a,.xn (cf. Ex. 9, § 911). Let R > 0 be the radius of conver-

n-o

gence of Ea,.xn, and let I be an interval with endpoints a and fJ, where max (Jal, lfJI) = r < R. Choose a fixed 'Y such that r < 'Y < R, and define Mn = lan-rnl, n = O, 1, · · · . Since 'Y is interior to the interval of convergence of Ea..x", Ean'Y" converges absolutely. Therefore the convergent series of constants EM n dominates the series Ea..x" throughout I, and by the Weierstrass M-test, the uniform convergence desired is established. Theorem I and its corollaries in Chapter 8 have to do only with points in the interior of the interval of convergence of a power series. Behavior at the end-points of the interval usually involves more subtle and delicate questions. One of the most useful and elegant tools for treating convergence at and near the end-points is due to Abel. In this section we present the statement of Abel's Theorem, together with two corollaries (whose proofs are requested in Exercises 14 and 15, § 911). The proof of Abel's Theorem is given in the following section.

Theorem II. Abel's Theorem. If st ants, then the power series

+oo

E

n=O

+oo

E

n-o

an is a convergent series of con-

anxn converges uniformly for O ~ x

~

1.

Corollary I. If a power series converges at an end-point P of its interval of convergence I, it converges uniformly on any closed interval that has P as one of its end-points and any interior point of I as its other end-point. If a power series converges at both end-points of its interval of convergence it converges uniformly throughout that interval. Corollary II. If a function continuous throughout the interval of convergence of a power series is represented by that power series in the interior of that interval, then it is represented by that power series at any end-point of that interval at which the series converges.

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§ 910]

PROOF OF ABEL'S THEOREM

Example. Show that (1) In (1 for -1

-1. •20. Prove that if I:a.. and I:b,. are convergent series, with sums A and B, respectively, if I:c.. is their product series, and if I:c,. converges, with sum C, then C = AB. Hint: Define the three functionsf(x) I:a..x", g(x) I:b..x", and k(x) = LC..X", for O ~ x ;:ia 1. Examine the continuity of these functions for O ~ x ~ 1, and the equation k(x) = f(x) g(x) for O ~ x < 1. **21. Verify the expansion of the following elliptic integral (cf. § 611):

f

=

!: 2

fo Vl - ~ for lkl

~

1

sin1 t =

=

i[ 1 + (½)'k' + G::rk + G::::rk• + ... ], 4

Hint: Expand by the binomial series: 11 . 1 1·3k. ( 1 - k 2 sm1 t)1 -..- = 1 k sm t _4 4 sm 4 t

1.

.

+2

+2

+ 1·3·5. . 4. k sm• t + •· · , 2 6 8

and use Wallis's formulas (Ex. 36, § 515). **22. Prove the Weierstrass Uniform Approximation Theorem: If f(x) is continuous on [a, b], and if E > 0, then there exists a polynomial P(x) suck that lf(x) - P(x)I < E for all x in [a, b].t Suggested outline: (i) The binomial series for Vl + x converges uniformly to vi'"+z for -1 ~ x ~ 0. (ii) The corresponding series for Vl + (x2 - 1) converges uniformly to lxl for lxl ~ 1. (iii) lxl can be uniformly approximated by polynomials for lxl ~ 1. (iv) The theorem is true for any function of the form mix - cl, (v) The theorem is true for any continuous function identically O for a ;i x ;:ia c (c ~ x ~ b) and linear for c ~ x ;:ia b (a ~ x ;:ia b). (vi) The theorem is true for any continuous function with a polygonal (broken-line) graph. (vii) Any continuous function on [a, b] can be uniformly approximated on [a, b] by a continuous function with a polygonal graph.

**23.

L"

Prove that if

x"f(x) dx

= 0 for n = 0, 1, 2,

ous on [a, b], then f(x) is identically O on [a, b].

• • • , and if f(x) is continuHint: Use the Weierstrass

Uniform Approximation Theorem (Ex. 22) to show that since

= 0 for any polynomial P(x),

L"

[/(x)] 1 dx

L"

P(x) f(x) dx

= 0.

t Several proofs of this theorem have been given, the original by Weierstrass in 1885. The proof outlined here was given by Lebesgue in 1898. For references and additional discussion see Hobson, Thwry of Functions (Washington, D. C., Harren Press, 1950).

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Answers

§ 111, page 18

x > a + lbl, -1 < X < 5. X > 2. No values. - VJ ~ X ~ -1 or I 20. -3 < X < 5. 22. x < -torx > 5. 2'. 2 < x < 4 or 6 < x < 26. lxl < !al or x = a(a ~ 28. If a - b, no values; if

12. 13. 16, 17. 19.

X

< -lbl,

-5orx ;i;; -1. 16, X < 3. 18. All values.

14,, x ~

~

X

~

VJ. 21. No values. 23.4 2. -lbl < x < 0 or x > lbl; if a> b: 0 < x < lbl or 26. 27.

§ 113, page 20 17. ad= be.

§ 203, page 33 (-})t1+l 12, nl + 2 •

2n 11 • 2n - ( (-l)n-1 13. (n - 1) i'

14. (2n - 2) I.

16,

0211

17. aan

+

(2n 1) I ,.nl • 2 3, 0411-I = ] ,

16.1·3·5···(2n+l)=

!4n-l = = a,._, = n, a,.._1 = -n.

= 2,

18. n even, 2"n '.; nodd , (2n) ,. n !( 2 20. 1. 19. 2. 23. 00 , 22. +oo. 26. N(,) • 1. 27. N(E) - 1/E. 29. N(B) - ¼(7B + 2).

21.

H. 26. N(,) • 1/,. 28. N(B) • B.

SO. N(B) •

IBI

½,

-oo.

+ 10.

§ 208, page 46

21. 4. 24. -¼, 293

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Original from

UNIVERSITY OF MICHIGAN

294

ANSWERS (c) oo; (d) +oo; (e) -oo; (J) oo. (J) 0; (a), (b), (d): there ill no limit. 18. +oo. at. -oo, '1. 0. ·'2. -oo.

(b) +oo; (e) l;

83. (a) -oo;

3'. (c) O; 8'1 • •.

'°·

+oo.

i

19. 6; 6(e) •

60. 9; 6(e) • min ( 1, ~)·

62. -t; &(e) • min (1, 20e).

61. 28; 6(e) • min ( 1, ~)63.

f;

6(e) • min ( 1,

1)·

6'. 6; &(e) • min



81. 0; N(e) •

G• ~)-

86. 0; N(e) • -max ( 1, ~)·

87. 3; N(e) •

1e7•

69. +oo; if B

> O, &(B)

• min ( 1,

70. - oo ; if B

< 0, 6(B)

• min ( 1, _ ;B); otherwise 6(B) .., 1.

63.

j; N(e) •

k); otherwise &(B) •

max

(1, ~)·

392.

I 218, page 11 82. &(e) • min (3, e).

83. cl(e) • min ( 1, ~)-

3'. 6(e) • min ( 1, ~)-

SI. &(e) • min (5, e).

86. &(e) • min (5, 6e).

81. &(e) • min

(f ;)-

I IOI, page 86 19. l; O. 21. l; 0. 29. +oo; -oo.

20. +oo; -oo.

22. SO. 2'2. .16.

11. l; -1. 83. l; -1.

l; 0. l; -1.

l; -1. l; -1.

I aoe, page 71 S. ll(e) • ;.

'- &(e) • !.. 4

I. 6(e) • 2e.

6. 6(e) • ~-

7. &(e) • e.

8. 6(e) •

18. 6(e) • min

(r, ~)-

(1, 1 .;. 2xJ

page 91 2. azt. 22

a. - ;;·

,. - (5:c - 4)1

I _1__

6

. 2-v;

Digitized by

4.

1'. &(e) • min

' '°'•

1. 2x - 4. 2

~

Google

1

' 3~i

Original from

UNIVERSITY OF MICHIGAN

ANSWERS 11. Yes. 18. No. 21. n > l; n

> l.

22. n

1. r,

:,;11➔ cos!, :,; > 0; /'(0) :,;

l):,;11➔

24. f"(:r;) - [n(n -

/"(0) - 0, n

1'1. No.

16. Yes. 19. No.

28. f'(:r;) == n:,;•-1 sin !:,; -

> 3.

n

-

:r;A--4]

> 3.

n

295

>

20. Yes. k; n > k.

- 0, n

> 1. n > 1. n > 2.

sin~ - 2(n - l)r--• cos~• :,;

> O;

> 4. §'°8, page 100 2. 2 - V2. ,. ½(a+ b).

2..-, or 3..-.

8. e - 1. 1.-1__

6.

e- 1

½-

'1• a+b. 2 9. 1 -

•+\fr=o.

§ ,12, page 111 1. Relative maximum -9; relative minimum - 5; increasing on (-co, l] and [3, +co);

decreasing on [l, 3]. 2. Absolute maximum - l; absolute minimum=- -1; increasing on [-1, l]; de-

creasing on (-co, -1] and [l, +co). 8. Relative maximum - 3~/25; relative minimum - 0; increasing on [0, t]; decreasing on ( - co, 0] and [t, +co). ,. Relative maximum - 2V3/9; absolute minimum - 0; increasing on [0, ½] and [l, +co); decreasing on [½, l]. I. Maximum - ½;minimum.,. -1. 6. Maximum - -4; minimum - -109/16. 7. Maximum - 2; minimum - -9/8. 8. No maximum; minimum - -1/e. 11. (a) :,; - 20; (b) x - 25; (c) x - 21; (d) no profit possible. 16. (a) :,; - 20; (b) x - 37.4; (c) full speed, x ,. 60. 19. If t :;; ,, :,; = b; if t > ,, x - min {b, a,/'\/t' -al). 20. r ;:ii -1: no minimum; -1 < r ~ -½: minimum - -a1 /4(r + l); -½ ;:ii r ~ 0: minimum ... ra•; r ~ 0: minimum,. 0. 88. e(~) - (6:,;1 - 5)~ + 4:r;~ + ~8'. e(~) = ~/zl(x + ~). 11. e(~) - -Vi ~/2(:i: + V zl + ~)•. 8'1. 10.48810. 88. h. 89. -0.06.

,1. x.

'°· x.

+;· 1 +z.

"- 1 ,.,.

'2. h

½-

~ (z - i)-

U. O.

§ ,1'1, page 121

1.

2. f.

¥-

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a.

½-

Original from

UNIVERSITY OF MICHIGAN

296

7.

ANSWERS

fi•

6. -h,

6. 1.

8. 1.

9

Ina

0

10. 2. 13. 1. 16. o.

11. J. 14. Meaningless. 17. o.

· In b 12. 1. 16. ½, 18. +co.

19. 0.

!.10. - 2a_

21. -1.

11"

22.

½,

!.IS. 1. 26. e'. 29. e-t.

16. e. 28. 1.

24. e'. 27. o. 30. o.

§ 603, page 143

14. 0; 3; ½(n1

-

n).

26. ~-

3. 6. 7. 8.

26.

i

§ 606, page 166 0. 4. sin x1 • -sin x1 . 6. 3x1 sin x8. 4x1 sin :r8 - 3x2 sin x8. 2x cos x1 P.in (sin1 x1) - sin 2x sin (sin' x). § 608, page 160

3. No.

No. § 610, page 162

-¼ ein6x cos x - ,h sin•x cos x - •A sin x cos x + H. ¼sin x cos•x + !J-sinx cos2x + /g-sin x + C. 13.

16. -

5

X

5

X

4cot• 5 + 2cot' 5 +

.X

5 In sm

5+

y\X

+ C.

C.

16. ¼sec'x tan x + ,h sec1x tan x + -A sec x tan x + lw ln 1sec x + tan xi + C. 17. H(4xl - 6x) sin 2x + (-2x' + 6x2 - 3) cos 2x] + C. 18. l(x' + 4x)i - (x + 2 ) ~ + 4 ln Ix + 2 + ~ I + C. 19. ½xi - ½(x + ½)v'x2 + x + 1 + i lo Ix+½+ v'x2 + x + 11 + C. 20. --h (4x1 + 21x

+ 105)(6x -

x2)l + .i.p (x - 3)v'6x - x2 . -x - A re sm + -1701 8 § 616, page 171

1. ~-

2. 0.

3. Divergent.

4. 2.

6. 2.

6. Divergent.

8. (k _ l)~ln 2 )t-1' k 16. Convergent. 7. ~-

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> I; divergent, k ~

1.

9. 2.

16. Divergent

Original from

UNIVERSITY OF MICHIGAN

- -3

3

+C.

ANSWERS 17. 19. 21. 23.

Convergent. Divergent. Convergent. Divergent.

18. 20. 22. 24.

297

Convergent. Convergent. Convergent. Convergent.

§ 618, page 184

1. f.

2. 3.

,. -(1 + e + e'). 36. v(x) = x + [x]; p(x) 36. v(x) = 3 + 2x - xi, p(x) = 3 + 2x - xi, n(x) = O, -1 ;:;; x ;:;;

6. e + e-1

-

2.

6.

2.

-r -

.. x; n(x) - [x]. -1 ~ x;:;; l; 5 - 2x x1 , 1 ;:;; x;:;; 2; -1 ~ x :ii l; 4, 1 ;:;; x ~ 2; l; 1 - 2x x1 , 1 ;:;; x ~ 2.

+

+

§ 606, page 196

1. t In 1sec 5xl + C. 2. ¼In 1sec 4x tan 4xl C. 3. Arc sin (x/V2) + C, xi < 2.

+

+

4. ½(x + 2) v' x1 + 4x - 2 In Ix + 2 + v' xi + 4xl + C, x !;;; 0 or x ~ -4. 6. ½In l2x - 1 + v' 4xt - 4x + 5I + C. 6. 7.

6x - 1 . r--;::-: v x - 3xt

~

Js

8. sec

V3 Arc sm . (6x + 72

In l(v'5 - 2xl - V5)/xl

Vx tan vi +

+ C, !xi < VS/2.

Vx +

tan

Vxl +

C, x

> 0.

I + 5 - 'V1091 + C.

9. _l_ln 6x 6x

'V109

In !sec

+ C, 0 < x < i.

1)

+ 5 + Vl09 + C.

2 6x +s 10. . r::= Arc tan . r::= v59 v59

§ 809, page 200 2. sinh 2x. 4. coth xi - 2xl csch1xl. 6. e-z(a cosh bx b sinh bx).

1. 3 einh 3x. 8. -eech1 (2 - x). 6. 2 coth 2x.

7

+

e"

4

8. . r=---=-' ve"' - 1

. VI+ 16xl 2x

9. 1 _ v' !xi < 1. 11. ¼In coeh 6x + C. 13. einh x + l einh1 x + C. 16. ¼einh 2x - ½x + C. 17. Coeh-1 ~

-CBC X, X

> 0.

F n-r.

+

12. In (sinh e") C. 14. x - -h tanh lOx C. 16. H(v - 1) tanh-1:z; + ix' + x] + C. 18"28lll ! · h-1 2x 2- l C .

+

+

+ C.

2 coth-16x+S+C. 19 . ./109 v'i09

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2

20. . r.=tanh v 109

-16x+5 .r.= vl09

+C.

Original from

UNIVERSITY OF MICHIGAN

298

ANSWERS § 707, page 211

2· 2n:l; ½. 3. 1 -

m·:

~

1.

7,{71

•·

+ J + ¼+ -h; a, = 2, an =- n(n 1__ 1), n > 1 + 2 - 2 + 2; a, = 1, a. = 2( - 1)•, n > l. 1.3 - 0.39 + 0.lJ 7 - 0.0351; J.3 (-0.3)•-I.

6. 2

6. 7.

+ o.

2

'

+oo.

l.

8. 2 - 0.7 - 0.21 - 0.063; a, - 2, a. - -0.7(0.3)•-t, n > 1. 9. y.. 10. Divergent. 11. 0.3.

1..Ao 10,201. 201

13. •·

H. Wf.

15. ffi. 18. Divergent. 21. Convergent. 114. Convergent.

16. 19. 22. 26.

17. Divergent. 20. Convergent. 23. Divergent. 26. Convergent.

1. Divergent. 4. Convergent. 7. Divergent. 10. Convergent. 13. Convergent. 16. Convergent for a 16. Convergent. 18. Convergent for a 19. Convergent. 20. Convergent for 0

1. Divergent. 4. Convergent for p

•. Convergent. Divergent. Convergent.

> i;

§ 711, page 219 2. Divergent. 6. Convergent. 8. Convergent. 11. Convergent. 14. Convergent. divergent for a ~ i-

> 1;

divergent for a

S. Convergent. 6. Divergent. 9. Convergent. 12. Divergent.

17. Divergent.

< r < l;

>

~

1.

divergent for r

~

1 unless a =- kr.

§ 713, page 223 2. Convergent. 2; divergent for p ~ 2.

S. Divergent.

§ 717, Pllge 229 2. Divergent. 4. Absolutely convergent. 6. Absolutely convergent. Absolutely convergent for x > 0; divergent for x ~ 0. Absolutely convergent for lrl < 1; divergent for lrl ~ 1. Absolutely convergent for p > 1; conditionally convergent for p ;:ii 1. Absolutely convergent for p > 2; conditionally convergent for 0 < p ~ 2; divergent for p ;:ii 0.

1. Conditionally convergent. 3. Conditionally convergent. 6. Absolutely convergent. 7. 8. 9. 10.

13. 2.718. 16. 1.202. 17. 0.7854. 21. 0.5770 < C

Digitized by

§ 721, page 237 H. 0.6931. 16. 0.1775. 18. 0.4055.

< 0.5774.

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Original from

UNIVERSITY OF MICHIGAN

299

ANSWERS 1. 2. 8. 4. I. 6. 7.

8. 9. 10.

11. 12.

§SOI, page HS Absolutely convergent for -oo < x < +oo. Absolutely convergent for !xi < 1; conditionally convergent for x - ± 1. Absolutely convergent for -oo < x < +oo. Absolutely convergent for x - 0. Absolutely convergent for -2 ~ x ~ 0. Absolutely convergent for -oo < x < +oo. Absolutely convergent for 4 < x < 6; conditionally convergent for x = 4. Absolutely convergent for 0 < x < 2. Absolutely convergent for !xi < 1; conditionally convergent for x - 1. Absolutely convergent for !xi :ii 1. Absolutely convergent for x < 2 or x > 4; conditionally convergent for x = 2. 2 1 1 2 Absolutely convergent if !xi F n ... ; conditionally convergent if !xi ... n r.

i

i

§ 811, page 261

xi

x4

x'

z1

zS

~

z4

zl

xii

1. 1+ 21 + 8•

X -

41 + 61 +-···

3+5- 7

S.1-xl+x4-x'+···· X xi z1 '• l + 11•2 + 21•4 + 31•8 + ••• •

+ ••• •

3z 9xl 27zl 6• In 2 + -1·2 - -2·4 + -3·8 - .•• •

I. 1 - -2! + -41 - 6-1 + . • . •

! z4 +

l •3 zS + 1 •3•5 xii+ • • • • 2·4 2·4·6 2 2 2·1 2·1·3 2·1·3·5 8• 2 + sx - 8·16 xi+ 8·16•24 zl - 8·16·24·32x4 + ....

7. 1 +

9. 2 [ X +

f + f + f + ••J

zl

x8 x7 + 2!•5 - 31•7 + • • • •

ll.

X -

3

18•

2 - 4+

1

X

15.

X -

3 z1 +

zl zS 48 - 480°

1

17 - £ • 6

zl x7 x11 x16 12• 3 - 31·7 + 51·11 - 71·15 + ....

2 17 15 zS - 315 x7.

- ±_ 180

a - 6x4 -

1'- 1 + x -

zl

l&. l +

2

X

+

zl

+

zl

2

+

x8 30· 3x4

8

37x8 + 120 .

zl x4 x' 17x' 18• - 2 - 12 - 45 - 2520°

±... 2835

19 cos a [1 - (x - a)•+ (x - a)' - • • •] " 2! 41 -sin a

[ex - a) -

(x - a)•+ (x - a)• - . ··]·

31

V2[

51

/,.._?:\I {x-?:)' ] 4 20• ~ 1 - { x4- ? : ) +--··· · 2 21 ~ + 3! - e)1 - e) 1 u. 1 + -x f-J -e - (x 2fJI + (x &' - ··· . 22. 1 + m(x - 1) +

m(m - 1)

-

- (x - 1)1 + • • • •

21

28. (1 - x)--1. U. H(zl + 1) Arc tan x - x].

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ANSWERS

+

26, (1 X 1)~. 26. 3ln½ - 1. 36. Bo= 1, B1

= -½, B1 = ¼, B, = -irs, B, = -b, Ba= -irs, Bio=- A· § 814, page 266

1. &.

9. 11. 18. 16. 17. 19.

l-

2. 1

-ff

6. 1.

2el. 7.38906. 1.0986. 2.15443. 0.4931. 0.4920.

8. b1

=

-½-

s.

,. t-

¾-

7. t-

8. -2.

10. 0.

12. 14. 16. 18. 20.

1.64872. 0.8776. 0.10033. 0.0976. 0.7468.

§ 816, page 268 1, b2 + a2 = 0, ba + ~ + a, = 0, b, + 3asba

+ (a;, + 2aa)bs + a. = 0.

§ 9CK, page ff&

87

In E • · In sin {½r - 71)

t

E

2b

n.L

40. -;.·

E7J

41.

38.

max(;•~)-

42

lnE • In (1 - 71)

§ 911, page 290

7. In 4 - 1.

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8. ¼r - ½In 2.

Original from

UNIVERSITY OF MICHIGAN

INDEX (The numbers refer to pages)

Abel test, 230 (Exs. 22, 23) for uniform convergence, 277 (Exs. 49, 50)

Abel's theorem, 288 Absolute convergence of improper integrals, 174 (Ex. 44) Absolute convergence of series, 228 Absolute maximum (minimum), 105 Absolute value, 17 Addition of numbers, 3 Addition of series, 233 Algebraic function, 202 Algebraic number, 201 Almost everywhere, 153 (Ex. 54) Alternating harmonic series, 226 Alternating series, 225 Analytic functions, 267-269 Antiderivative, 155 Approximations by differentials, 109-111 Archimedean property, 11 (Ex. 24), 24 Associative law, 2, 3, 8, 25 Asymptotes, 123 Axiom of completeness, 22 Axioms of order, 5 Axioms of the basic operations, 2, 3 Axioms, Peano, 1 Beta function, 171 Between, 7 Big O notation, 169, 213 Binary number, 20 (Ex. 4) Binomial series, 252 Binomial theorem, 12 (Ex. 35), 252 Bliss's theorem, 149 (Ex. 42) Bolzano-Weierstrass theorem, 82 (Ex. 13) Bonnet form of the second mean value theorem, 186 (Ex. 29) Bound, lower, 23 upper, 22 Bounded convergence, 152 (Ex. 49) Bounded function, 54

Bounded sequence, 35 Bounded set, 23 Bounded variation, 175-179 Bracket function, 29, 51 Cancellation law, 4 (Exs. 2, 9) Cauchy criterion, for functions, 65 for improper integrals, 174 (Ex. 43) for sequences, 62-63, 277 (Ex. 45) for series, 230 (Ex. 17), 277 (Ex. 46) for uniform convergence, 277 (Exs. 45, 46) Cauchy form of the remainder, 248 Cauchy inequality, 13 (Ex. 43), 147 (Ex. 29), 220 (F..x. 26), 229 (Ex. 14) Cauchy principal value, 173 Cauchy sequence, 63 CesAro summability, 229 (Ex. 15), 230 (Ex. 16), 277 (Ex. 44) Chain rule, 87 Classification of numbers and functions, 201-205 Closed interval, 16 Closed set, 75 Closure of a set, 82 (Ex. 18) Commutative law, 3, 9, 25 Compact set, 76 Comparison tests, improper integrals, 168171 series, 212-215, 274 Complement of a set, 75 Complete ordered field, 27 Completeness, axiom of, 22 Composite number, 11 (Ex. 22) Composition of ordinates, 123 Computation of series, 235-239 Computations with series, 264-267 Conditional convergence, improper integrals, 174 (Ex. 45) series, 228 Connected set, 77

301

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UNIVERSITY OF MICHIGAN

INDEX

302

Constant function, 28 Constant of integration, 156 Continued fraction, 21 (Ex. 15) Continuity, 49 and integrability, 133, 137, 153 (Ex. 54) and uniform convergence, 278 from the right or left, 50 modulus of, 287 negation of, 53 (Ex. 14) of inverse function, 57, 81 sectional, 159 Continuity theorems, 52-55, 57, 80, 81, 137 Continuous nondifferentiable function, 86, 286 (Ex. 41) Convergence, bounded, 152 (Ex. 49) dominated, 175 (Ex. 47) interval of, 240-241 of improper integrals, 164-175 of sequences, 31 of series, 206 uniform, of sequences, 270 uniform, of series, 273 Covering of a set, 83 (Ex. 28) Criterion, Cauchy, 63, 65 Criterion for continuity, sequential, 64 Critical value, 105 Curve tracing, 123-128 Decimal expansions, 20 (Ex. 5), 220 (Exs. 28-30) Dedekind's theorem, 26 (Ex. 11) Definite integral, 131-154 Deleted neighborhood, 41 Dense, 19 Denumerable set, 21 (Ex. 12) Dependent variable, 28 Derivative, 85 one-sided, 88-91 Differentiation, 85-130 and uniform convergence, 281-283 of power series, 259 Dirichlet's theorem, 232 Discontinuity, jump, 52 removable, 51 Discontinuous derivative, 90 Disjoint sets, 76 Distance between a point and a set, 82 (Ex. 19) Distance between two sets, 82 (Ex. 21) Distributive law, 3, 9 Division of numbers, 3 Domain of definition, 28 Dominance, 168,212,274 Dominant terms, 123 Dominated convergence, 175 (Ex. 47)

e, 60 (Ex. 40), 190 (Ex. 5) Elementary functions, 203

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Empty set, 75 Enumerable set, 21 (Ex. 12) Euclid, fundamental theorem of, 11 (Ex. 25), 14 (Ex. 15) Euler's constant, 239 (Exs. 19, 22) Even function, 144 (Exs. 7-13) Exponential functions, 188-191 Factor, 11 (Ex. 22) highest common, 14 (Ex. 13) Factorial, 12 (Ex. 34) Field, 20 (Ex. 8) complete ordered, 27 ordered, 21 (Ex. 11) First derivative test, 106 First mean value theorem for integrals, 137, 143 (Exs. 5, 6), 186 (Ex. 25) Function, 28 algebraic, 202 analytic, 267-269 binomial, 252 bounded, 54 bracket, 29, 51 constant, 28 continuous, 49 differentiable, 85 direct, 57 elementary, 203 even, 144 (Exs. 7-13) greatest integer, 29, 51 inverse, 57 monotonic, 55 nondifferentiable, 86, 286 (Ex. 41) odd, 144 (Exs. 8-13) rational, 202 real-valued, 28 signum, 42, 52 single-valued, 28 transcendental, 202 Functions, hyperbolic, 197-201 Fundamental theorem of integral calculus, 154-155, 158 (Exs. 19, 20) Gamma function, 171 Gauss's test, 224 (Ex. 7) Geometric series, 208 Greatest integer function, 29, 51 Greatest lower bound, 23 Group, 20 (Ex. 7) Grouping of terms, 231 Half-open interval, 16 Harmonic serie!!, 210 Heine-Borel theorem, 83 (Ex. 28) Highest common factor, 14 (Ex. 13)

Original from

UNIVERSITY OF MICHIGAN

INDEX l'Hospital's rule, 115-121 Hyperbolic functions, 197-201 Hypergeometric series, 224 (Ex. 6)

Improper integrals, 164-175 Indefinite integral, 155 Independent variable, 28 . Indeterminate forms, 115-121, 264 Inequality, Minkowski, 13 (Ex. 44), 147 (Ex. 30), 229 (Ex. 14) Schwarz (Cauchy), 13 (Ex. 43), 147 (Ex. 29), 220 (Ex. 26), 229 (Ex. 14) triangle, 17 Infinite derivative, 86 Infinite interval, 16 Infinite limit, of a function, 42 of a sequence, 31 Infinite sequence, 30 Infinite series (see Series, infinite), 206-299 Integer, negative, 13 positive, 7, 13 Integral, definite, 131-154 improper, 164-175 indefinite, 155 lower, 150 (Ex. 44) proper, 132, 165 Riemann, 131-154 Riemann-Stieltjes, 179-187 upper, 150 (Ex. 44) Integral form of the remainder, 247 Integral test, 209 Integration, 131-187 and uniform convergence, 279-281 by parts, 157 (Exs. 10-13), 160 (Ex. 7), 181 by substitution, 156, 185 (Ex. 18) of power series, 259 Interior point, 17 Intermediate value property, 55, 103 (Ex. 47) Interval, 16 Interval of convergence, 240-241 Inverse function, 57 continuity of, 57, 81 differentiability of, 88, 98 Inversion of power series, 269 (Exs. 7, 8) Irrational number, 13 Irrationality of e, 202, 263 (Ex. 34) Irrationality of r, 202

Jump discontinuity, 52

Kummer's test, 221

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303

Lagrange form of the remainder, 247 Landau, E., 1 Law, associative, 2, 3, 8, 25 cancellation, 4 (Exs. 2, 9) commutative, 3, 9, 25 distributive, 3, 9 transitive, 6 (Exs. 2, 19) trichotomy, 6 (Ex. 3) Law of the mean, 94 extended, 99 generalized, 95 Least upper bound, 23 Lebesgue measure zero, 153 (Exs. 52-54) Lebesgue theorem on bounded convergence, 152 (Ex. 49) Lebesgue theorem on dominated convergence, 175 (Ex. 47) Leibnitz's rule, 93 (Ex. 25) l'HRpital's rule, 115-121 Limit inferior, 67 (Ex. 16), 68 (Ex. 24) Limit, of a function, 40 of a sequence, 31 of a variable, 44 Limit point, of a sequence, 40 (Ex. 20) of a set, 75 Limit superior, 67 (Ex. 16), 68 (Ex. 24) Limit theorems, for functions, 44 for sequences, 34 Little o notation, 169, 213 Logarithm, 188-191 Lower bound, 23 Lower integral, 150 (Ex. 44) Lower semicontinuity, 69 (Ex. 35) Lowest terms, 11 (Ex. 23) Maclaurin series, 245-253 for cos x, 251 fore", 250 for ln(l x), 251 for sin x, 250 for (1 x)•, 252 Mathematical induction, 7, 24 Maxima and minima, 105-107 Maximum of a function, 54, 80 Mean value theorem for derivatives, 94 extended, 99 generalized, 95 Mean value theorem for integrals, first, 137, 143 (Exs. 5, 6), 186 (Ex. 25) second, 157 (Ex. 14), 186 (Exs. 27-29) Measure zero, 153 (Ex. 52) Minimum of a function, 54, 80 Minkowski inequality, 13 (Ex. 44), 147 (Ex. 30), 229 (Ex. 14) Modulus of continuity, 287 Monotonic function, 55 Monotonic ,u&nce, 37

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INDEX

Monotonically decreasing sequence of sets, 83 (Ex. 26) Moore-Osgood theorem, 285 (Ex. 37) Multiplication of numbers, 3 Multiplication of series, 233 Natural numbers, 1 Negation of continuity, 53 (Ex. 14) Negation of uniform continuity, 73 Negation of uniform convergence, 272 Negative integer, 13 Negative number, 5 Negative of a number, 3 Neighborhood, 17 deleted, 41 of infinity, 32 Nested intervals theorem, 83 (Ex. 26) Net, 131 Nondenumerability of the reals, 220 (Ex. 30) Nondifferentiable function, 86, 286 (Ex. 41) Norm of a net, 131 Number, algebraic, 201 binary, 20 (Ex. 4) composite, 11 (Ex. 22) integral, 13 irrational, 13 natural, 1 negative, 5 positive, 5 prime, 11 (Ex. 22) rational, 13 real, 1-27 transcendental, 202 Odd function, 144 (Exs. 8-13) One, 3 One-sided derivatives, 88-91 Open covering of a set, 83 (Ex. 28) Open interval, 16 Open set, 75 Order of magnitude, 169, 213 Order relation, 5 Ordered field, 21 (Ex. 11) Oscillation of a function, 84 (Ex. 30) Parametric equations, 125 Partial fractions, 16 (Ex. 22) Parts, integration by, 157 (Exs. 10-13), 160 (Ex. 7), 181 Peano axioms, 1 Point sets, 74-34 Polynomial, 46 (Ex. 19) Polynomial equation, 201 .

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Positive integer, 7, 13 Positive number, 5 Power series, 240-269 differentiation of, 259 integration of, 259 inversion of, 269 (Exs. 7, 8) Prime numbers, 11 (Ex. 22) relat.ively, 11 (E.,:. 23) Primitive, 155 Principal value, Cauchy, 173 (Ex. 30) Proper integral, 132, 165 p-series, 210 Quotient of numbers, 3 Quotient of polynomials, 14 (Ex. 15) Raabe's test, 222 Radius of convergence, 241, 242 Range of values, 28 Ratio test, 215, 220 (Ex. 31), 228 Rational function, 47 (Ex. 20), 48 (Ex. 35), 202 Rational number, 13 Real number, 1-27 Real-valued function, 28 Rearrangement of terms, 231 Reciprocal of a number, 3 Reducible covering, 83 (Ex. 28) Reduction formulas, 161-163 Relative maximum (minimum), 105 Remainder, 14 (E.,:. 15) Removable discontinuity, 51 Riemann integral, 131-154 Riemann-Stieltjes integral, 179-187 Riemann Zeta-function, 214 Rolle's theorem, 93 Root test, 218, 220 (Ex. 32) Roots of numbers, 55 Schwarz inequality, 13 (Ex. 43), 147 (Ex. 29), 220 (Ex. 26), 229 (Ex. 14) Second derivative test, 106 Second mean value theorem for integrals, 157 (Ex. 14), 186 (Exs. 27-29) Sectional continuity, 159, Sectional smoothness, 159 Semicontinuity, 69 (Ex. 35) Separated sets, 77 Sequence, 30 Sequence of distinct points, 78 Sequential criterion for continuity, 64 Sequential criterion for limits, 64 Series, infinite, 206-299 · addition of, 233 alternating, 225

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UNIVERSITY OF MICHIGAN

INDEX alternating harmonic, 226 basic definitions, 206 computation of, 235-239 computations with, 264-267 geometric, 208 harmonic, 214 Maclaurin, 245-253 multiplication of, 233 p-series, 210, 214 power, 240-269 subtraction of, 233 Taylor, 244-253 uniform convergence of, 273 Set, bounded, 23 closed, 75 denumerable (enumerable), 21 (Ex. 12) open, 75 Sets of points, 74-84 Signum function, 42, 52 Simple continued fraction, 21 (Ex. 15) Simpson's rule, 146 (Ex. 23) Single-valued function, 28 Smoothness, sectional, 159 Square root, 7, 55 Step-function, 138 Subsequence, 32 Substitution of power series, 255-259 Subtraction of numbers, 3 Subtraction of series, 233 Summability of series, 229 (Ex. \5), 230 (Ex. 16) uniform, 277 (Ex. 44) Summation notation, 12 (Ex. 36) Taylor series, 244-253 Taylor's formula with a remainder, 246248 Cauchy form, 248 integral form, 247 Lagrange form, 247 Terminating decimal, 20 (Ex. 5) Total variation, 175-177 Transcendental function, 202 Transcendental number, 202

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Transitive law, 6 (Exs. 2, 19) Trapezoidal rule, 145 (Ex. 22) Triangle inequality, 17 Trichotomy, 6 (Ex. 3) Trigonometric functions, 191-194

Uniform approximation theorem of Weierstrass, 291 (Ex. 22) Uniform continuity, 71 Uniform convergence, 270-299 and continuity, 278 and differentiation, 281-283 and integration, 279-281 Uniform summability, 277 (Ex. 44) Unique factorization theorem, 11 (Ex. 29) Unity, 3 Upperbound, 22 Upper integral, 150 (Ex. 44) Upper semicontinuity, 69 (Ex. 35) Variable, dependent, 28 independent, 28 real, 28 Variation, bounded, 175-179 negative, 177 positive, 177 total, 175-177 Wallis's formulas, 174 (Ex. 36) Weierstrass continuous nondifferentiable function, 86 Weierstrass M-test, 274 Weierstrass uniform approximation theorem, 291 (Ex. 22) Well-ordering principle, 8 Without loss of generality, 129 Wronskian determinant, 104 (Ex. 51) Zero, 3 Zeta-function of Riemann, 214

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