534 95 16MB
English Pages 330 Year 1956
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INTERMEDIATE ANALYSIS
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THE APPLETONCENTURY MATHEMATICS SERIES Edited by Raymond W. Brink
A First Year of College Mathematics, Second Edition
by Raymond W. Brink College Al,gebra, Second Edition
by Raymond W. Brink Al,gebraCollege Course, Second Edition
by Raymond W. Brink IntermediafR Al,gebra, Second Edition
by Raymond W. Brink Calculus
by Lloyd L. Smail Analytic Geometry and Calculus
by Lloyd L. Smail Solid Analytic Geometry by John M . H. Olmsted Intermediate Analysis by John M . H. Olmsted The Mathematics of Finance
by Franklin C. Smith Plane Trigonometry, Revised Edition
by Raymond W. Brink Spherical Trigonometry
by Raymond W. Brink Analytic Geometry, Revised Edition
by Raymond W. Brink Essentials of Analytic Geometry
by Raymond W. Brink
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Intermediate Analysis AN INTRODUCTION TO THE THEORY OF FUNCTIONS OF ONE REAL VARIABLE
:by .
~·, ..

John M. H: Olmsted PROFESSOR OF MATHEMATICS UNIVERSITY OF MINNESOTA
NEW YORK: APPLETONCENTURYCROFTS, INC.
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Copyright,
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©
1956 by
APPLETONCENTURYCROFTS, INC. AU rights reserved. Thia book, or parts thereof, must not be reproduced in any form without permiatsion of the publiahera. 5261
Library of Congress Card Number: 565844
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 intermediate 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 prove 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 term before studying uniform 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 affecting 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 fortyfive meetings. Inclusion of most of the singly starred material pennits 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 nondifferentiable 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) = x 2 + 5." Examples of subjunctive uses are "let x = n," and "let E 1," 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 parentheses 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.
M inneapoli8 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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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
DIFFERENTIATION
./
401. Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402. The derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403. Onesided derivatives . .. . ... . . .... . .. . . .. . .. ... . . ... . ... ..... 404. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405. Rolle's Theorem and the Law of the Mean.. . . ... . . . .. ... .... . . 406. Consequences of the Law of the Mean. . . . . . . . . . . . . . . . . . . . . . . . . 407. The Extended Law of the Mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408. Exercises . .. .. . . .. .. ... . . . . ... . . . . . . . . . . . . . : . . . ...... . . .... 409. Maxima and minima. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410. Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ...... ....... 411. Approximations by differentials.. . .. . . .. . . . . . . . . . . . . . . . . . . . . . . 412. Exercises... .. ... . .... .. . .. .. . . ..... . . . ... . .... . . .. .. . . . . . . 413. L'Hospital's Rule. Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414. The indeterminate form 0/0.. .. ..... . . . . . . . . . . . . . . . . . . . . . . . . . 415. The indeterminate form~;~.. . . .. . . .. . . .. . . . . ... .. . . .. . .. . . 416. Other indeterminate forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417. Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418. Curve tracing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419. Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . *420. Without loss of generality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . *421. Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 5
85 85 88 91 93 97 98 100 105 107 109 111 115 116 117 120 121 123 127 129 129
V/
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.
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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 RiemannStieltjes integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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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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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
*Chapter 9 UNIFORM CONVERGENCE
240 243 244 246 248 249 250 253 255 259 261 264 264 266 267 268
v
*901. *902. *903. *904. *905. *906. *907. *908.
Uniform convergence of sequences...... .. . .. ............... .. Uniform convergence of series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dominance and the Weierstrass Mtest . . . . . . . . . . . . . . . . . . . . . . . . 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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UNIVERSITY OF MICHIGAN
j
INTERMEDIATE ANALYSIS
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UNIVERSITY OF MICHIGAN
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UNIVERSITY OF MICHIGAN
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 trueprobably by people who know such things simply because they have been told. He knows pretty well why some of these familiar faets (like 2 · 3 = 3 · 2) are true, but is entitled to be unsure of the reasons for others (like y2 · y3 = y3 ·y2) . 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 (18581932), 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, 1'heory of Functions (Washington, Harren Press, 1950). 1
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UNIVERSITY OF MICHIGAN
THE REAL NUMBER SYSTEM
2
[§ 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 comp"lete 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 comp"leteness. 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 multiplication. As 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.
t For a proof of this fact and a discll88ion 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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UNIVERSITY OF MICHIGAN
l
§ 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
= 0.
That is, if x and y are any numbers,
+ y = y + X.
The number 0 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 17, § 103. II. 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 x1 of axiom (iv) is called the reciprocal of the number x. The quotient of x and y (y ¢ 0) is defined: X  s
x·,r1•
y
The resulting operation is called division. Some of the properties that can be derived from Axioms II alone are given in Exercises 811, § 103.
m. Addition and Multiplication (i) The distributive law holds.
x(y
That is, if x, y, and z are any numbers,
+ z)
= xy + xz.
The distributive law, together with Axioms I and II, yields further familiar relations, some of which are given in Exercises 1233, § 103.
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UNIVERSITY OF MICHIGAN
THE REAL NUMBER SYSTEM
4
[§ 103
103. EXERCISES
In Exercises 133, prove the given statement or establish the given equation. 1. There is only one number having the property of the number O of 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
+
+
( x)
+ (x + y)
+
=
(x)
+
+ (x + z).
Use the associative law. 3. The negative of a number is unique. Hint: If y has the property of x in Axiom I (iv), § 102, x + y = x + (x) = 0. • Use the law of cancellation, given in Exercise 2. 4. 0 = 0. 6. (x) = x. 6. 0  X = x. 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) + [( x) + (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 X "F 0. 10. The reciprocal of a number ( "F 0) is unique. 11. 11 = 1. 12. x•0 = 0. Hint: x•0 + 0 = x•0 = x(O + 0) = x·0 + x•0. Use the law of cancellation on the extreme members. 13. Zero has no reciprocal. 14. If x "F 0 and y "F 0, then xy "F 0. Equivalently, if xy = 0, then either x = 0 or y = 0. Hint: Assume x "F 0, y "F 0, and xy = 0. Then x 1 (xy) = x1.0 = 0. Use the associative law to infer y = 0. 16. If x "F 0, then x1 "F 0 and (x1 )1 = x. 16. :!: y
17.
= 0 (y
.! =
"F 0) if and only if x
= 0.
x1 (x "F 0).
X
18. If x "F 0 and y "F 0, then (xy)1 = x1y1 , or 19.
If b "F
0 and d "F 0, then
..!.. Xy
=
! • !. X y
a
b = ad bi
= (ad)(d1b1 ) = a[(dd1)b1 ] = ab·1• 20. If b "F 0 and d "F 0 then ~ • £ = ac_
Hint: (ad)(bd) 1
'
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b
d
bd
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UNIVERSITY OF MICHIGAN
§ 104]
AXIOMS OF ORDER
21. If b ~ 0 and d ~ 0, then
5
a
b + dc = ad+bc bd ·
Hint: (bd)1 (ad + be) = (b1d1)(ad) + (b 1d1)(bc). 22. (1)( 1) = I. 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 ~ 0). y y y 27. x(y  z) = xy  xz. *28. (x  y) + (y  z) = x  z. *29. (a  b)  (c  d) = (a + d)  (b + c). *30. (a + b)(c + d) = (ac + bd) + (ad + be). *31. (a  b)(c  d) = (ac + bd)  (ad+ be). *32. a  b = c  d if and only if a + d = b + c. 33. The general linear equation ax + b = 0, a ~ 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 >
 x is positive; y if and only if x  y is positive.
The number x is negative if and only if  x is 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 1J and only (f either x
~ ?J
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< y or x = y; > y or x = y.
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UNIVERSITY OF MICHIGAN
THE REAL NUMBER SYSTEM
6
[§ 105
NoTE 1. The two statements x < y and y > x are equivalent. The two statements x ;:ii y and y si:; x are equivalent. NOTE 2. The sense of an inequality of the form x < y or x ;:ii y is said to be the reverse of that of an inequality of the form x > y or x si:; y. NOTE 3. The simultaneous inequalities x < y, y < z are usually written x < y < z, and the simultaneous inequalities x > y, y > z are usually written x > y > a. Similar interpretations are given the compound inequalities x ;:ii y ;:ii z and x si:; y ~ z.
105. EXERCISES
In Exercises 124, prove the given statement or establish the given equation. 1. x > 0 if and only if x is positive. 2. The transitive law holds for < and for >: x < y, y < z imply x < z; x > y, y > z imply x > z. (Cf. Ex. 19.) 3. The law of trichotomy holds: for any x and y, exactly one of the following holds: x
y.
4. Addition of any number to both members of an inequality preserves the order relation: x < y implies x + z < y + z. A similar fact holds for subtraction. Hint: (y + z)  (x + z) = y  x. 6. x < 0 if and only if x is negative. 6. The sum of two negative numbers is negative. 7. The product of two negative numbers is positive. Hint: xy = (x)(y). (Cf. Ex. 24, § 103.) 8. The square of any nonzero real number is positive. 9. 1 > 0. 10. The equation x2 + 1 = 0 has no real root. 11. The product of a positive number and a negative number is negative. 12. The reciprocal of a positive number is positive. The reciprocal of a negative number is negative. 13. 0
< x < y imply O 0 imply xz < yz and x/z < y/z. 16. Multiplication or division of both members of an inequality by a negative number reverses the order relation: x < y, z < 0 imply xz > yz and x/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 x and y are nonnegative numbers, then x < y if and only if x2 < y 2 (cf. Ex. 10, § 107). 19. The transitive law holds for ;:ii (also for ~): x ;:ii y, y ;:ii z imply x :ii z. (Cf. Ex. 2.) 20. x ;:ii y, y ;i x imply x = y. 21. x + x = 0 implies x = 0, x + x + x = 0 implies x = 0. 22. x2 + y 2 e:; O; x2 + y 2 > 0 unless x = y = 0.
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§ 106]
7
POSITIVE INTEGERS
23. If x is a fixed number satisfying the inequality x < E for every positive number E, then x ~ 0. Hint: If x were positive one could choose E = x. 24. There is no largest number, and therefore the real number system is infinite. (Cf. Ex. 11, § 109.) Hint: x + l > x. 26. If x < a < y or if x > a > y , then a is said to be between x and y. Prove that if x and y are distinct numbers, their arithmetic mean ½(x + y) is between them. 26. If x2 = y, then x is called a square root of y. By Exercise 8, if such a number x exists, y must be nonnegative, and if y = 0, x = 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 x1 = z2 = y. Then x'  z' = (x  z)(x + z) = 0. Therefore x = z or x = z.
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 posi,tive int,egers 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): The "positive int,egers" are positive. (Cf. Ex. 9, § 105.) If n is a positive integer, n s;;; 1. 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. (vi) If m and n are positive int,egers, and m < n, then n  mis a positive integer. (vii) If n is a positive integer, there is no positive integer m such that n I and n > I such that p = mn. A positive integer pis prime if and only if p > 1 and pis not composite. Prove that if m,, i = I, 2, · · · , n, are integers > I, then m1m1 · · · m,. > n. Hence prove that any integer > I 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. •26. 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 + I 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 cl •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 a1, 02, • • • , 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 I 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,,. = q1q2 • • • q,.. A prime number remains which is a factor of both members but which violates Ex. 28. (Cf. Ex. 22.) •30. 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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UNIVERSITY OF MICHIGAN
THE REAL NUMBER SYSTEM
12
[§ 107
34. If n is a positive integer, n factorial, written n ! , is defined: n ! = 1 •2 · 3 · ~ 1. If r is a positive integer or zero
• • • · n. Zero factorial is defined: 0 !
and if 0;:;; r;:;; n, the binomial coefficient(;) (also written ,.C,) is defined:
(;) = (n  nr; ! r t" Prove that (;) is a positive integer.
Hint: Establish the law of Pascal's
Triangle (cf. any College Algebra text): ( n ;"
1
)
=
(r ~ 1 ) + (;}
36. Prove the Binomial Theorem for positive integral exponents (cf. V, § 807): If x and y are any numbers and n is a positive integer,
+ y)"
(x
+ (~) x"1y + ... + (;) X",Y' + ... + (:) y".
= (~) X"
(Cf. Ex. 34.) 36. The sigma summation notation is defined: n
L J(k) = f(m) + J(m + 1) + · · · + f(n), km (i) k is a dummy variable:
n :E
k=m n
:E
(ii) Lis additive:
[f(k)
k=m
(iii)
+ g(k)]
n
:E
:E
is homogeneous:
cf(k)
k=m
n
(iv)
f(k) =
L
k=m
1
=n  m
=
~
L"
where n ~ m.
Prove:
f(i).
i=m n
n
L f(k) + k=m L g(k). k=m n
c
L f(k). k=m
+ 1.
•37. A useful summation formula is (1)
t,
kl
[f(k)  f(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 f(n) = n 2• Then (l) becomes n
L [k k=l
n
2

(k  1) 2] =
,.
2
•39. By means Let f(n) = n 3 in •40. By means •41. By means
L
k=l
k = n2
E (2k k=l
 1) = n2,
or
+ n.
of Exercises 3638, derive the formula of Exercise 13.
Hint:
(1 ).
of Exercises 3639, derive the formula of Exercise 14. of Exercises 3640, derive the formula of Exercise 15. n
•42. Use mathematical induction to prove that
E
k1
k"' is a polynomial in n of
degree m + 1 whose leading coefficient (coefficient of the term of highest degree) is 1/(m + 1). (Cf. Exs. 3641.)
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UNIVERSITY OF MICHIGAN
§ 108]
INTEGERS AND RATIONAl NUMBERS
13
•43. If a1, a2, · · · , a,. and b1, b2, · · · , b,. are real numbers, show that 2 b, 2 )  ( a,b,) = L (a,b;  a;b,) 2.
( ta/) ( t i=l
t
i1
i1
l~i 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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UNIVERSITY OF MICHIGAN
EXERCISES
§ 203)
33
¾, i,
Example 3. Find the limit of each sequence: (a) ½,
'
1 ... . 1  , 2n
(b) 3, 3, 3, · · · , 3, · · · ; (c) 1, ½, 1, ¾, 1, l, · · · . Solution. (a) The expression Ian  11 is equal to ;n' which is less than any
preassigned positive number whenever n is sufficiently large, as shown in Exl\mple 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 oddnumbered terms form a subsequence identically 1, while the evennumbered terms form a subsequence which is the sequence of part (a). The limit is 1.
Example 4. Show that each of the following sequences diverges : (a) 1, 2, 1, 2, 1, 2, · · · ; (c) 1, 2, 1, 3, 1, 4, · · · ;
(b) 1, 2, 4, 8, 16, · · · ;
(d) 1, 2, 4, 8, 16, · · · 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 +00 . (c) The comment of part (b) applies to this sequence, since there is a subsequence tending toward +00 . This sequence has no limit, finite or infinite. (d) The subsequence of the oddnumbered terms tends toward +00, and that of the evennumbered terms tends toward 00. The sequence of absolute values tends toward +00, so that the sequence itself has the limit 00.
203. EXERCISES
In Exercises 110, 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. 510, § 216.) 1. y = Vx2
3. Y

2. y = ±¥25  x2 •
9.
= v=;_
4. Y
9. y
=
=
±vixi.
6. y = v1x 2  161. 8. y = (x  [x]) 2 •
6. y = v'4x  x 2 • 7. y = x  [x].
10. y
v'x [x].
=
[x]
+
v'x=[x=].
In Exercises 1118, give a rule for finding the general term of the sequence.
n,
12. ½, ¼, h, 11\.r, •••• 11. 2, !, ! , l, · · · . 14. 1, 2, 24, 720, 40320, · · · . 13. 1, 1, ½, t, h, r½1r, · · · · 16. 1·3, 1 · 3 · 5, 1·3·5 •7, 1· 3 · 5 · 7 •9, · · · 16. 1, 2, 3, 2, 1, 2, 3, 2, l, · · · .
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UNIVERSITY OF MICHIGAN
FUNCTIONS AND SEQUENCES
34
[§ 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 1924, find the limit of the sequence and justify your contention (cf. Exe. 2530). 2n + 1 20. J,l,t, ... • ~ • .... 19. 2, 2, 2, 2, 2, · · · . 21. 23.
l, 9, t, !r, 1., *' ··· . 9, 9, ¥, ¥, · • • .
22. 1, 4, 9, 16, · · · , n', · · · . 2'. 9, 16, 21, 24, • • • , 10n  n 1,
• • • •
In Exercises 2530, give a simple explicit function N(E) or N(B), in accord with Definition I or II, for the sequence of the indicated Exercise.
•25. For Ex. 19. •28. For Ex. 22.
•26. For Ex. 20. •29. For Ex. 23.
•27. For Ex. 21. *30. For Ex. 24.
In Exercises 3134, 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, · · · .
32. 1, 2, 3, 1, 2, 3, 1, 2, 3, · · · . 3' 21, 21, 21, 2•, 21, 21, ....
204. LIMIT THEOREMS FOR SEQUENCES
Theorem I. The alteration 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 sequences and if M and N are two positive integers such that aM+ .. = 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 infinite 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 some 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 subsequence of a convergent sequence converges, and its limit is the limit of the original sequence. (Cf. Ex. 12, § 205.)
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Proof. Assume a" _. a. Since every neighborhood of a contains all but a finite number of terms of {a,.} 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 {a,.} is bounded if and only if there exists a positive number P such that la,.I ~ P for all n. Theorem V. Any ronvergent sequence is bounded. (Cf. Ex. 2, § 205.) Proof. Assume a,. _. 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 {a,.}, a suitable enlargement will contain these missing terms as well. Definition II. If {a,.} and {bn} are two sequences, the sequences {a,. + b,.}, {a,.  bn}, 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 {a,./b,.} is called their quotient. The definitions of sum and product extend to any finite number of sequences. Theorem VI. The sum of two ronvergent sequences is a ronvergent sequence, and the limit of the sum is the sum of the limits:
lim (a,.+ b,.) = lim a,.+ lim b,,. n..... +eo
n...... +•
n .....
+•
(Cf. Ex. 4, § 205.) This rule extends to the sum of any finite number of sequences. Proof. Assume a,, _. a and b,. _. b, and let E > 0 be given. Choose N so large that the following two inequalities hold simuli,aneously for n > N :
la,.  al < ½E, lb,.  bl < ½E. Then, by the triangle inequality, for n > N l(an + b,.)  (a+ b)I = l(a,,  a) + (b,.  b)I ~
la,. 
al
+ lb,. 
bl
< ½t: + 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 ronvergent sequences is a convergent sequence, and the limit of the difference is the difference of the limits:
lim (a,.  bn) = lim a,, 
n ..... +co
n...... +t0
lim b,,. n++oo
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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lim (a,.b,.)
=
[§ 204
lim a,.· lim b,..
n.+• This rule extends to the product of any finite ti,umber of n~+•
n++oo
(Cf. Ex. 5, § 205.) 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,.} ronverges to O and {d,.} ronverges, then {c,.d,.} ronverges too. Proof of lemma. By Theorem V the sequence {d,.} is bounded, and there exists a positive number P such that ld,.I ~ P for all n. If E > 0 is given, choose N so large that lc..l < E/P for n > N. Then for n > N, lc,.d,.I
= lc.. l ·
ld,.I
< (E/P)·P = E.
This inequality completes the proof of the lemma, and hence of the theorem. The extension to the product of an arbitrary number of sequences is provided by mathematical induction. (Cf. Ex. 3, § 205.)
Theorem IX. The quotient of two ronvergent sequences, where the denominators and their limit are nonzero, is a ronvergent sequence and the limit of the quotient is the quotient of the limits: lim a,. lim _a,.  ,..... +.. • n+• b,.  lim b,. n++oo
Proof. Assume a,. + a, b,. + b, and that b and b,. are nonzero for all n. Inasmuch as a,./b,. = (a,.)· (1/b,.), Theorem VIII permits the reduction of this proof to that of showing that 1/b,. + 1/b or, equivalently, that
_!_  l  b  b,. . .!.  0 b,. bb b,. . Let c,. = (b  b,.)/b and d,. = 1/b,. and observe that the conclusion of the Lemma of Theorem VIII is valid (with no change in the proof) when the sequence {d,.} is assumed to be merely bounded (instead of convergent). Sincethesequence {c,.} = {(b  b,.)·(1/b)} convergestozero(bythissame lemma), we have only to show that the sequence {d,.} = {1/b,.} is bounded. We proceed now to prove this fact. Since b ;= 0, we can choose neighborhoods of 0 and b which have no points in common. Since b,. + b, the
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LIMIT THEOREMS FOR SEQUENCES
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neighborhood of b contains all but a finite number of the terms of {b,.}, so that only a finite number of these terms can lie in the neighborhood of 0. Since b,. is nonzero for all n, there is a (smaller) neighborhood of O that excludes all terms of the sequence {b,.}. If this neighborhood is the open interval (  E, E), where E > 0, then for all n, jb,.j ~ E, or ld,.j = 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 constant 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: lim (k a,.) = k • lim a,.. n~+•
Proof.
n➔+•
This is a consequence of Theorems III and VIII.
Theorem XI. If {a,.} is 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..l  +oo and if E > 0 is given, there exists a number N such that for n > N, la.. l > 1/E, and therefore ll/a,.I < E. Conversely, if 1/a,.+ 0 and Bis any given positive number, there exists a number N such that for n > N, ll/a,.I < 1/B, and therefore la.. l > B. Theorem m.
If a> 1, lim a"= +oo. n~+• 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)
p2
+ .. · ~ 1 + np.
Therefore, if Bis a given positive number and if n
a" Theorem XIII.
Proof.
~ 1
If lrl < 1,
> B/p, then
+ np > 1 + B > B. lim r" = 0.
n+• 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,. ~ a,.+1 (a,. ~ a11t1) 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,. ~ 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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[§ 205
*
Proof. We give the details only for the case a,. j (cf. Ex. 7, § 205). Since the set A of points consisting of the terms of the sequence {an} is bounded above, it has a least upper bound a(§ 114), and since Pis an upper bound of A, the following inequalities must hold for all n: a,. ~ 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:
a  E < aN Consequently a  E< a,. < a +
< a + E. E, or la,.  al < E, and the proof is com
~
a,.
~
a
plete. As a consequence of Theorem XIV we can say in general that any +oo, or oo ), 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,.
~
b,.for all n, and if lim a,. and lim b,. exist (finite, n~+oo
+00,
or
Proof.
00 ), then
lim a,.
n~+oo
~
lim b,.. n~+oo n~+• If the two limits are finite we can form the difference
c,. = b,.  a,. and, by appealing to Theorem VII, reduce the problem to the special case: if e,. ~ 0 for all n and if C = lim e,. exists and is finit,e then C ~ 0. n~+• By the definition of a limit, for any positive E 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,. + El < E, and hence e,. + E < E. This contradicts the nonnegativeness of e,.. On the other hand, if it is assumed that a,.+ +00 and b,.+ B (finite), we may take E = 1 and find first an N1 such that n > N1 implies a,.> B + 1, and then an N2 such that n > N2 implies b,. < B + 1. Again the inequality a,. ~ b,. is contradicted (for n greater than both N1 and N2). The student should complete the proof for the cases a,.+ A (finite), b,.+ 00 and a,.+ +00, b,.+ oo. 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,. + +oo and either {b,.} converges or b,. + +oo, then a,.+ b,.+ +oo.
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§ 205)
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,.!, 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,. ~ b) and a,.+ a, then a ~ b (a ~ b). Show by an example that from the strict inequality a,. < b (a,. > b) we cannot infer the strict inequality a < b (a > b) . H. 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 {rn} 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 {s,.} is a given sequence, define,;,. =
.!n (s1 + s2 + · · · + s,.).
that if {s,.} converges to O then {,;,.} also converges to 0. Let m be a positive integer < n, and write
,;,. = .! (s1 n
(Cf. Ex. 17.)
+ ;} Prove
Hint:
+ · · · + s..) + .!n (s.. +i + · · · + s,.).
If E > 0, first choose m so large that whenever k > m, !ski < ½E. Holding m fixed, choose N > m and so large that ls1 + · · · + s,,.1/N < ½E. Then choose n > N, and consider separately the two groups of terms of,;,., given above. With the notation of Exercise 16, prove that if {s,.} converges, then {,;,.} also converges and has the same limit. Show by the example 0, 1, 0, 1, · · · that the convergence of {,;,.} does not imply that of {s,.}. Can you find a divergent sequence {s,.} such that,;,.+ O? Hint: Assumes,.+ l, let t,. = s,.  l, and use the result of Ex. 16. ••18. With the notation of Exercise 16, prove that if lim s,. = +00, then
••17.
Jim ,,.,. =
n++oo
+00.
n++oo
Show by the example 0, 1, 0, 2, 0, 3, · · · that the reverse
implication is not valid. Hint: If B is a given positive number, first choose m so large that whenever k > m, Bk > 3B. Then choose N > 3m and so large that ls1 + · · · + s.. l/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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of Exercise 16,
lim s,.
n++.,
an example where Jim
...... +..
= oo does not imply Jim n++ oo 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 L is 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.
:i:+a
t In the terminology of the next chapter (§ 309), a is a limit point of the domain of definition D of /(x). It can be shown that every neighborhood of a contains infinitely many points of D.
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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 twosided limit may be used to distinguish the first of these three limits from the other two in case of ambiguity arising 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 Las x ap'/J7'oaches a, written
limf(x) = L, or f(x) +Las x+ a, if and only if corre8p0nding to an arbitrary 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 L as x approaches a from the right (left),t written
lim f(x)
= L,
or f(x)
+
L as 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 < E, for values of x for which f(x) is defined. These onesided 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 lim f(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 < lz  al < &, then, define a deleted neighborhood of the point a. t Parentheses are 118ed 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. 1314, § 208), lim f(x) may fail to exist either by lim f(x) and lim f(x) being unequal or by either or xoa+
zoa 
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) has the limit L as x becomes positively (negatively) infinite, written
f(+00) = lim f(x) :e++oo
(f(00)
= z+,0 lim
f(x)
=
L,
or f(x) 
L g,s x 
+00
= L,
or f(x) 
Las x 
00 ),
if and only if corresponding to an arbitrary positive number E there exists a number N = N(E) such that x > N (x < N) implies lf(x)  LI < E1 for values of x for which f(x) is 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) has the limit +00 (00) as x approaches
a, written limf(x)
=
+00 (00 ), or f(x) 
+00 (00) as x  a,
if and only if corresponding to an arbitrary number B there exists a positive numbero = o(B)suchthatO < Ix  al< oimpliesf(x) > B(f(x) < B),for values of x for which f(x) is defined.
As with limits of sequences it is often convenient to use an unsigned inWhen we say that a variable, dependent Or independent, tends toward 00, we shall mean that its absolute value approaches +00. Thus limf(x) = Lis defined as in Definition II, with the inequality x > N re
finity, 00.
.....
.,
> N,
placed by lxl
and lim f(x)
= 00
is equivalent to lim lf(x)I :,:+a
=
+00 .
Discuss the limits of each of the following functions as x apxt + X proaches 0, o+, and 0, and in each case sketch the graph: (a) /(x) =  x if x '¢ 0, undefined for x = O; (b) /(x) = lxl if x '¢ 0, /(0) 3; (c) the signum function, J(x) sgn x I if x > 0, J(x) sgn x 1 if x < 0, /(0) sgn 0 Example 1.
=
=
= O; (d)t J(x) = sin! if X
x = O; (J) J(x)
x ¢ 0, J(O)
= ~ if x ¢ X
a
=
=
=
O; (e) J(x) = ! if x X
=
¢
0, undefined if
0, undefined if x = 0.
t For illll8trative examples and exercises the familiar properties of the trigonometric funetions will be assumed. An analytic treatment is given in §§ 603604.
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43
Solution. The graphs are given in Figure 203. In part (a) if x "F 0, f(x) is identically equal to the function x + l, and its graph is therefore the straight line y = x + l with the single point (0, 1) deleted; lim f(x) = f(0+) = f(0) = 1. z....0
In part (b) limf(x) = f(0+) = /(0) = 0. The fact that f(0) = 3 has no z....o bearing on the statement of the preceding sentence. For the signum function (c), f(O+) = 1, /(0) = 1, and lim f(x) does not exist. In part (d) all z....o three limits fail to exist. In part (e) f(0+) = +co, /(0) = co , and lim f(x) = co (unsigned infinity) (cf. Exs. 3132, § 208). In part (!), z....o f(0+) = f(0) = lim f(x) = +co (cf. Ex. 32, § 208). z....0
y
'II
11
11X
1 (b)
11
X
1
(c)
\_ j "\_ 1
\
(d)
1
X
1
X
(f)
(e) FIG. 203
•Example 2.
Show that Jim xi; z2
x
X 
+ 18 = 4. Find an explicit function o(E) 1
as demanded by Definition I. Solution. Form the absolute value of the difference: (1)
1x2  x + 18  41
= 1x2  13:c + 221 =
Ix  21 · Ix  111,
3x1 The first factor , 1 Ix  21, is certainly small if xis near 2; and the second factor, l:x_ !1• is not 3xl
3xl
We wish to show that this expression is small if xis near 2.
dangerously large if x is near 2 and at the same time not too near ¼ 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 Ix  11 I < 10 and l3x  l I > 2. Thus the second factor is less than Jl = 5. Now let a positive number E be given. Since th e
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expression (1) will be less than E if simultaneously Ix we have only to take o(E)
= min ( 1,
t}
o=
 21 < ~ and
[§ 207
l:x_1!I < 5,
o(E) to be the smaller of the two numbers 1 and
t,
The graph of this function o(E} is shown in Figure 204.
0 I f
5 FIG. 204
207. LIMIT THEOREMS FOR FUNCTIONS
Many of the theorems of § 204 are special cases of more general limit theorems which apply to real valued functions, whether the independent variable tends toward a (finite) number, toward a number from one side only, or toward +oo, oo or oo. Some of these are stated below. Since in each case the statement is essentially the same regardless of the manner in which the independent variable approaches its limit, t this latter behavior is unspecified. Furthermore, since the proofs are mere reformulations of those given in § 204, only one sample is given here. Others are requested in the Exercises of § 208.
Theorem I. If lim f(x) exists it is unique. Theorem II. If the function f(x) is equal to a constant k, then limf(x) exists and is equal to k . Theorem m. If lim f(x) and lim g(x) exist and are finite, then lim [f(x) g(x)] exist~ and is finite, and
+
lim [f(x)
+ g(x)] = limf(x) + lim g(x).
In short, the limit of the sum is the sum of the limits. the sum of any finite number offunctions.
This rule extends to
Theorem IV. Under the hypotheses of Theorem III, the limit of the difference is the difference of the limits: lim [f(x)  g(x)]
= limf(x)
 lim g(x) .
t The word limit has been given specif!' mPaning in this chapter only for functions, but it is convenient to extend its use to apply to both dependent and independent variables. The reader should recognize, however, that such an isolated statement as "the limit of the variable v is l" is meaningless, and takes on meaning only if v is aeeociated with another variable, thus: .,...... Jim v(x)  l, or lim /(v) .. k .
....,
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45
Theorem V. Under the hypotheses of Theorem Ill, the limit of the product is the product of the limits:
lim [f(x)g(x)]
= limf(x)·limg(x) .
This rule extends to the product of any finite number of functions. Proof for the case x+ a. Assume f(x)+ Land g(x)+ M as x+ a. We wish to show that f(x) g(x) + LM or, equivalently, that f(x) g(x)  LM
+
0, as x
+
a.
By addition and subtraction of the quantity L · g(x) and by appeal to Theorem III, we can use the relation
f(x) g(x)  LM = [f(x)  L]g(x)
+ L[g(x)
 M]
to reduce the problem to that of showing that [J(x)  L]g(x) + 0 and L[g(x)  M]+ 0 as x+ a. The fact that they do is a consequence of the following lemma: Lemma. If 0, f(x)
=
a
6.
;;:i; 0.
x  1 if x
< 0, f(0) = 0.
11. Prove the equivalence of the two definitions of continuity, § 209, under the assumptions of the first paragraph of that section. 12. Prove the Theorem of§ 209 (a) for the case t  c; (b) for the case t  c. 13. Give an example to show that the Theorem of § 209 is false if the continuity assumption is omitted. 14. Prove that the negation of continuity of a function at a point of its domain can be expressed: f(x) is discontinuous at x = a if and only if there is a positive number E having the property that corresponding to an arbitrary positive number 8 (however small), there exists a number x such that Ix  al < 8 and 1/(x)  J(a)I ~
E.
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[§ 213
15. Show that the function defined for all real numbers, /(x) s 1 if x is rational, f(x) = 0 if x is irrational, is everywhere discontinuous. (Cf. Example 6, § 201.) 16. Prove that Ix! is everywhere continuous. •17. Prove that lf(x)I is continuous wherever f(x) is. Give an example of a function defined for all real numbers which is never continuous but whose absolute value is always continuous. Hint: Consider a function like that of Ex. 15, with values ± 1. 18. Prove that if /(x) is continuous and positive at x = a, then there is a neighborhood of a in which /(x) is positive. Prove that, in fact, there exist a positive number E and a neighborhood of a such that in this neighborhood /(x) > E. State and prove corresponding facts if /(a) is negative. Hint: Cf. Ex. 9, § 208. •19. Prove that if f(x) is continuous at x = a, and if E > 0 is given, then there exists a neighborhood of a such that for any two points in this neighborhood the values of /(x) differ by less than E. **20. Show that the function with domain the closed interval [0, 1], /(x) x if x is rational, f(x) = 1  x if x is irrational, is continuous only for x = ½**21. Show that the following example is a counterexample to the following False Theorem, which strives to generalize the Theorem of § 209, and Theorem VI, § 211: If Jim q,(t) = a, where a is finite and t approaches some limit, and if lim/(.x) = b, where bis finite, then lim/(q,(t)) = b, as t approaches its limit.
=
=
=
Counterexample: q,(t) aia 0 for all t, J(x) 0 if x ~ 0, /(0) 1; t+ 0, x+ 0. Prove that this False Theorem becomes a True Theorem with the additional hypothesis that q,(t) is never equal to a, except possibly for the limiting value oft.
213. MORE THEOREMS ON CONTINUOUS FUNCTIONS
We list below a few important theorems on continuous functions, whose proofs depend on certain rather sophisticated ideas discussed in Chapter 3, which is starred for possible omission or postponement. The proofs of these particular theorems are given in § 306.
Theorem I. A function continuous on a closed interval is bounded there. That is, if f(x) is continuous on [a, b], there exists a number B such that a ~ x ~ b implies lf(x) I ~ B. Theorem II. A function continuous on a closed interval has a maximum and a minimum there. That is, if f(x) is continuous on [a, b], there exist points Xi and X2 in [a, b] such that a ~ x ~ b implies f(x1) ~ f(x) ~ f(x2). Theorem m. If f(x) is continuous on the closed interval [a, b] and if f (a) and f(b) have opposite signs, there is a point x 0 between a and bfor which f(xo) = 0.
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\/2
AND OTHER ROOTS
55
Theorem IV. A function continuous on an interval assumes (as a value) every number between any two of its values. (Cf. Ex. 47, § 408.) NOTE . For other properties of a function continuous on a closed interval, see § 307 and § 501.
214. EXISTENCE OF
\12 AND
OTHER ROOTS
In Chapter 1 (Ex. 8, § 109) y'2 was mentioned as an example of an irrational number, but proof of its existence was deferred. We are able now, with the aid of the last theorem of the preceding section, to give a simple proof that there exists a positive number whose square is 2. The idea is to consider the function f(x) = x 2 , which is continuous everywhere (Ex. 1, § 212) and, in particular, on the closed interval from x = 0 to x = 2. Since the values of the function at the endpoints of this interval are 02 = 0 and 22 = 4, and since the number 2 is between these two extreme values, there must be a (positive) number between these endpoints for which the value of the function is 2. That is, there is a positive number whose square is 2. The following theorem generalizes this result, and establishes the uniqueness of positive nth roots : Theorem. If p is a positive number and n is a positive integer, there exists a unique positive number x such that x" = p. This number is called the nth root of and is written X = Olp. Proof. We establish uniqueness first. If x" = y" = p, where x and y are positive, then (Ex. 9, § 107) x"  y" = (x  y)(xn1 xn2y y"1) = 0. Since the second factor is positive (Ex. 8, § 107) the first factor must vanish (Ex. 14, § 103): x  y = 0, or .x = y. For the proof of existence, we note first that since x" ~ x for x ;?; 1, lim x" = +oo . Therefore there exists a number b such that b" > p.
p
+
+ ••• +
z+oo
The number p is thus between the extreme values assumed by x" on the closed interval [O, b], and therefore, since x" is continuous, this function must assume the value pat some point x between O and b: x" = p. 215. MONOTONIC FUNCTIONS AND THEIR INVERSES
Definition. A function f(x) is monotonically increasing (decreasing), written f(x) j ( L), on a set A if and only if whenever a and b are elements of A and a < b, then f(a) ~ f(b) (f(a) ~ f(b) ). In either case it is called monotonic. Whenever a < b implies f(a) < f(b) (f(a) > f(b) ), f(x) is called strictly increasing (decreasing), and in either case strictly monotonic.
(Cf. Fig. 208.)
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Theorem XIV, § 204, states facts about monotonic sequences. If we permit the inclusion of infinite limits we can drop the assumption of boundedness and state that any monotonic sequence has a limit (cf. the Note, § 204). In Exercise 27, § 216, the student is asked to generalize this fact and prove that a monotonic function f(x) always has onesided limits. Consequently the only type of discontinuity that a monotonic function can have (at a point where it is defined) is a finite jump (Ex. 28, § 216). 11
•
11

• •
Monotonically increasing
Strictly decreasing FIG. 208
Consider now the case of a function that is continuous and strictly monotonic on a closed interval [a, b]. For definiteness assume that f(x) is strictly increasing there, and let c = f(a) and d = f(b). Then c < d, and (Theorem IV, § 213) f(x) assumes in the interval [a, b] every value between c and d. Furthermore, it cannot assume the same value twice, for if a < fJ, then f(a) < f(/J). (Cf. Fig. 209.) The function f(x), thered 11
C
X
a
a
fJ
b
FIG. 209
fore, establishes a onetoone correspondence between the points of the two closed intervals [a, b] and [c, d]. Thus to each pointy of the closed interval [c, d] corresponds a unique point x of [a, b] such that y = f(x). Since y determines x uniquely, x can be considered as a singlevalued furn:
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57
tion of y, x = q,(y), and is called the inverse function of y = f(x), the latter being referred to, then, as the direct function. The nth roots obtained in the preceding section are inverse functions: if y = f(x) = x", then x = f/,(y) = ~ We now state an important fact regarding the continuity of inverse functions. The proof is given in § 306. Theorem. If y = f(x) is continuous and stricay monotonic on a closed interval [a, b], its inverse function x = q,(y) is continuous and stricay monotonic on the corresponding cwsed interval [c, d]. Corollary. The Junction y = O'x, where n is a psitive integer, is a continuous and stricUy increas·ing Junction of x for x ?; 0. (Cf. Ex. 18, § 216.)
216. EXERCISES 1. Show that the equation x 4 + 2x  11 = 0 has a real root. Prove that if f(x) is a real polynomial (real coefficients) of odd degree, the equation/(x) = 0 has a real root. 2. Prove that the maximum (minimum) value of a function on a closed interval is unique, but show by examples that the point ~ at which f(x) is a maximum (minimum) may or may not be unique. 3. Prove that Yf(x) is continuous wherever f(x) is continuous and positive, and that ~ is continuous wherever f(x) is continuous. 4. Assuming continuity and other standard properties of sin x and cos x (cf. §§ 603604), discuss continuity of each function. (a) Vl
+ sin x;
(b)
1
· Vl  sinx'
(d)
(c) vlcos x2 ;
1
vii
+ cos2 x
.
 sin2 x
In Exercises 510, discuss the discontinuities of each function, and draw a graph in each case. The bracket function [x] is defined in Example 5, § 201. (Also cf. Exs. 710, § 203.)
+ [x].
6. [ x].
6. [x]
7. [ ✓x"].
8. [x2].
9. [x]
+ Vx 
10. [x]
[x].
+
(x  [x)) 2 •
11. Prove the two laws of radicals: (i)
Vab = r
(ii)
~a ~b, ~
~i = ~i• b ~ 0,
where n is a positive integer, and in case n is even a and bare nonnegative. 12. Prove that lim ~; = +ao, where n is a positive integer . .,....+ ..
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*13. Prove that
lim v'x (v'x x++oo
+ a  Vx) = ½a.
Hint:
[§ 216
Multiply
by
( ~ + Vx)/(V + a + Vx). X
*14. Let /(x) be a polynomial of degree m and leading coefficient ao > 0, and let g(x) be a polynomial of degree n and leading coefficient b0 > 0, and let k be a positive integer. Prove that the limit lim ¢'/(x) / ~
z+oo
is equal to O if m < n, to ¢'ao/bo if m = n, and to +oo if m > n. (Cf. Ex. 35, § 208.) *16. Give an example of a function which is continuous and bounded on the open interval (0, 1), but which has neither maximum nor minimum there. *16. Give an example of a function which is defined and singlevalued on the closed interval [0, 1), but which is not bounded there. *17. Give an example of a function which is defined, singlevalued, and bounded on the closed interval [0, 1), but which has neither maximum nor minimum there. *18. Prove that is an increasing function of n for any fixed x between 0 and 1, and a decreasing function of n for any fixed x > l. Hint: Show that the desired order relation between {1/x and n+yx follows from Ex. 10, § 107, by taking n(n + l)th powers.
erx
*19. Prove that if x > 0 then lim
n++oo
that O Jim n++oo
0, 1v;  va1 = Iv;  va . v; + val = Ix  31 < Ix  3J. 1 vx + v3 ¼ + v3
•33. Vx2  4, a= 2. •36. ~;, a = 5. Hint: If x
I~;  ~51 = •36.
L, a=
v'x
•3' V3x2
x, a= 1.

> 0,
1~; ~5 · ~~ ++ ~s; ++ ~25 < Ix  51. 1
~x 2
~ 5x
1
6.
•37. ~;• a
j
~25
=
·
l.
**38. If a is a fixed positi\'e rational number prove that the set of all numbers
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(§ 216
of the form r + sv;;, where r and s 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.
**'°·
Prove that lim ( 1 + .! )" exists and is between 2 and 3.
n
n++oo
definition of e.
!r
For another treatment see §§ 601602.)
the binomial expansion of a,.
= (1+
and thereby show that a,.
a,.
N and n > N together imply lam  a,.I < E. A sequence {a,.} satisfying the condition lim (am  a,.) = 0 m,n+oo
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.+oo
be a number such that n > N implies Ja,.  al < ½E. both greater than N, we have simultaneously
Jam  al < ½E and Ja,. so that by the triangle inequality (§ 110)
Ja,,.  a,.J = J(a,,.  a) · (a,.  a)I
~
Then if m and n are
 aJ < ½E, lam  al + Ja,.  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 Ja,.  aN+1I < 1. (This is the Cauchy Criterion with E = 1 and m = N + 1.) Then, by the triangle inequality, for n > N,
Ja,.I = l(a,.  aN+i) + aN+1I
~
Ja,.  aN+1I + JaN+1I < JaN+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,.J < 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 for k, 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 V2 (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 16 and 1522. *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) w be continuous at x = a is that whenever {x,.} is a sequence of numbers which converges w a (and for which f(x) is defined), then {f(x,.)} is a sequence of numbers converging w 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 def,ned). Since x,. + a, there exists a number N such that, for n > N, Ix,.  al < o. Therefore, for n > N, lf(x,.)  f(a)I < E. This establishes the desired convergence. Next we prove sufficiency, by assuming that f(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 lf(x)  f(a)I ~ E. We construct the sequence {x,.} by requiring that x,. satisfy the two inequalities Ix,.  al < 1/n and lf(x,.)  f(a)I ~ E. The former guarantees the con
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vergence of {x,.} to a, while the latter forbids the convergence of {J(x,.)} to f(a) . This completes the proof of Theorem I. 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 {J(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 {J(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 nei,ghborhood 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 f: 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 lim/(x) =Land let E > 0. Then there ex:r+a
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 If(x')  f (x") I = I(f(x')  L)  (f(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
[/(xm)  f(x,.)]
m,n++oo
= 0.
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 If(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 < lxm  al < 8 and 'O < Jx,.  al < 8, so that If(xm)  f (x,.) I < E. Thus the sequence {J(x,.)} is a Cauchy sequence, and the proof is complete.
Theorem II. The limit lim f(x) exists and is finite if and only if correx++«I
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) co~verges 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 ja,. _: 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. •4. Prove that the condition lim (a,.+P  a,.) = 0 for every positive inn++ 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, 3½, 3¼, 4, 4¼, 4½,
4¾, 5, 5¼, · · · .
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67
•6. Find the error in the "theorem" and "proof": If a,. + a, then a,. = a for 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 , lam  aN+nl < E. But this means that Jim aN+" = a,,., which in turn implies that Jim a,. = a = am for all m > N. n++ao
n++ao
•6. If {b,.} is a convergent sequence and if {a,.} is a sequence such that lam  a,.I ;:;;; 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 x.....O+
· is finite, and if lf(b)  f(a)I ;:;;; lg(b)  g(a)I for all positive numbers a and b, prove that Jim f(x) exists and is finite .
x .... 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 x1, x1', x2, x2', xa, xa', · · · . For the casef(x,.)+ +oo, assume that lim f(x) ~ +oo, and show that there must exist a constant Band
a sequence {x,.} such that O < Ix.  al of Jim f(x,.), proceed similarly.
c; (ii) all x such that f(x) < c. Prove that the following sets are closed: (iii) all x such that f(x) iii::; c; (iv) all x such that f(x) ~ c; (v) all x such that /(x) = c. If /(x) 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 x, 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..+t• This property is symbolized A,.!. (A constant sequence, where A,. = A, for all n, is an extreme example.) Prove the theorem: If {A,.} is a monotonically decreasing
&(A, B)
sequence of nonempty compact sets there exists a point x common to every set of the sequence. The nested intenals 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 a,.. + x. For any N, a... belongs to AN for sufficiently large k, so that x 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, ~)} 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 HeineBorel 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 F is 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 I 2,
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[§ 312
and relabel I = Ii. Let Ia be a closed half of /2 that is not covered by any finite collection of sets of F, and repeat the process to obtain a decreasing sequence {/,.} of closed intervals whose lengths tend toward zero. If x is a point common to every interval !Ji (Ex. 26), let B be an open set of the family F which contains x. Then B contains a neighborhood (x  E, x + t) 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. 1617, § 305.) Extend these results to unbounded sequences. **30. Let A be a set contained in the domain of definition of a bounded function /(x). The oscillation of f(x) 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 (f(x))  inf (f(x)). x in A
x in A
Prove that if A is contained in B, then w(A) ~ w(B). Hence show that if 8 > 0, and if A, (xo  8, xo + 8), then w(A,) is a monotonically increasing function of 8, so that lim w(Aa) exists (cf. § 215). Make appropriate modifica&o0+ tions to take care of the possibility that A or A, may be only partly contained in the domain of definition of f(x). **31. Let f(x) be defined on a closed interval I. The oscillation w(xo) of f(x) at a point x 0 of I is defined to be the limit of the function w(A,) of Exercise 30: w(xo) lim w(A,). &o0+ Prove that w(xo) ;;; 0 and that f(x) is continuous at x = Xo if and only if w(xo) = 0. **32. Let f(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(x0 ) < E, let 7J = E  w(x0 ) and show that there exists a neighborhood N of x 0 for which w(l) < w(xo) + 11, so that for x in N, w(x) < w(xo) + 1J = E. **33. Let /(x) be defined on a closed interval I, and consider the sequence of closed sets D1, D½, D , • • • , D¼, • • • (cf. Ex. 32). Prove that each of these sets 1 is contained in its successor, and that the set of points x such that x is a member of some D! is precisely the set D of points of discontinuity of f(x) •
=
=
.
/
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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 assumed to be familiar are repeated here for the sake of completeness, without the full discussion and motivation which they deserve when fi.r st 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 singlevalued realvalued 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 following limit exists and is finite; thefunctionf'(x) defined by the limit is caUed its derivative:
(I)
'El_ = f'(x) = lim f(x dx
hO
+ h)
 f(x),
h
85
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86
DIFFERENTIATION
[§ E(f(x) > f(U for x < Eandf(x) < f(E) for x > E).
Hint: If lim f(t Ax.....0
+ A:)X 
f(t)
> 0, then by Ex. 9,
§ 208, there exists a deleted
neighborhood of 0 such that for any Ax within this deleted neighborhood
t o.
8. Derive the formula! (x")
= nx" 1, if n is a positive integer.
•9. Derive the formula :x (x") = nx"•, if n is a positive rational number and :e is positive (without assuming in the process that x" is differentiable) . Hint: With standard notation, 'II = zP19, y4 = zP, and (y Ay) 4 = (x Ax)P, so that
+
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DIFFERENTIATION
+
+
[§ 404
zP + pxpt ll.x + ••• . Cancel first terms, divide by Since ll.y+ 0 (Ex. 21, § 216), the formula follows by algebraic simplification (cf. Ex. 22, § 216).
y9 qy 9 1 ll.y ll.x, and let ll.x+ 0.
=
*10. Derive the formula :x (xn)
=
nx"1 , if n is any rational number and xis
positive. Hint: Use Ex. 9 to show that x" is differentiable. let m n and differentiate the quotient 1/x"'.
=
NOTE
1.
=
The formula :x (x")
If n is negative,
nxn 1 is shown in Exercise 7, § 602, to be
valid for all real values of n for x > 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 ll.x + 0 does not imply ll.y > 0. Give an example of a continuous function for which ll.y + 0 does not imply ll.x+ 0. 13. By mathematical induction extend the chain rule for differentiating a composite function to the case of n functions: y = J1(f2(/a • • • (f,.(x)) • • • ) ). 14. Prove that the derivative of a monotonically increasing (decreasing) differentiable function f(x) satisfies the inequality f'(x) ~ 0 ( ~ 0). Does strict increase (decrease) imply a strict inequality?
In Exercises 1520, discuss differentiability of the given function f(x) (where f(0) = 0). Is the derivative continuous wherever it is defined? 16. x2 sgn x (cf. Example 1, § 206).
16.
1 17. x2 cos·
18. x2 sin~
X COS.!_, X
x
X
t . 1 *19.xsm•
*20. xi cos ];·
X
*21. Discuss differentiability of the function f(x) = 0 if x ~ 0, f(x) = x" if > 0. For what values of n does f'(x) exist for all values of x? For what values of n isf'(x) continuous for all values of x? *22. If f(x) is the function of Exercise 21, for what values of n does the kth derivative of f(x) exist for all values of x? For what values of n is the kth derivative continuous for all values of x? x
*23. If f(x)
= x" sin.!. for x > 0 andf(0) = 0, findf'(x). X
n does f'(x) exist for all nonnegative values of x? continuous for all nonnegative values of x?
•24. If f(x)
For what values of
For what values of n isf'(x)
= xn sin.!. for x > 0 andf(0) = 0, findf"(x). X
n does f"(x) exist for all nonnegative values of x? f"(x) continuous for all nonnegative values of x?
For what values of
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
93
•25. By mathematical induction establish Leibnitz's Rule: If u and v are functions of x, each of which possesses derivatives of order n, then the product also does and
(n)1 d"u dx"1
d" d"u dx" (uv) = dx" · v
+
dv dx
1 •
(n)2 d"u dx"2
+
2 •
d2v dx 2
d"v
+ · · · + dx"'
where the coefficients are the binomial coefficients (Ex. 35, § 107). *26. If y is a function of u, and u is a function of x, each possessing derivatives of as high an order as desired, establish the following formulas for higher order derivatives of y with respect to x:
(du) + ~ d2u, dx du dx 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 t between a and x (t = a if x = a) such that (7) f(x) = f(a) + !'m(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 Bsuch that 0 < B < 1 and
(8)
f(a
+ h) = f(a) + f'(a + Bh)h.
(Ex. 13, § 408.) 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 + Bh) ·h becomes sin x = cos (Bx) •x. If 0 < x ~ 1, then 0 < Bx < l and cos (Bx) < 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
l
>
1 and h
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+h h < In (l + h) < h,
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UNIVERSITY OF MICHIGAN
§ 406)
Solution. f(a)
LAW OF THE MEAN
If a
= 1, and
f(x)
97
= In x, the equation
+ f'(a + 8h) ·h becomes In (1 + h)
=
1
~ eh·
If h
j(a
+ 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 O < 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)
~
0,
in contradiction to the basic assumption. Theorem II. Two differentiable functions whose derivatives are equal througlwut 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 ( ;;;i 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 x 1 and x2 be points of the interval such that X1 < X2. Then, by the Law of the Mean, there is a number xa between x1 and x2 such that f(x2)  f(x1) = 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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DIFFERENTIATION
[§ 407
tonic (over the interval) , and its derivative is of constant 8ign (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 t of [a, b] and a minimum value at some point fl of [a, b], where t ¢ fl (why must t and fl be distinct?) . By Theorem I, § 405, neither t nor fl can be an interior point of [a, b]. Therefore either t = a and fl = b or t = band fl = a. However, if t = a and fl = b, f'(a) ~ 0 and f'(b) ~ 0 (why?), whereas if t = band fl = a, f'(a) ~ 0 and f'(b) ~ 0. Either conclusion is a contradiction to the assumption that f'(a) and f'(b) h,ve opposite signs. ·
Corollary. If a function has a nonzero derivative over an interval, the inverse function exists and is differootiable, and consequenlly the formula
: = 1/~ is valid whenever its righthand 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 approximate f(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 (1)
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) = pn>(a). We first express p(x) by means of the substitution of a+ (x  a) for x, and subsequent expansion, in the form (2)
p(x)
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 a)
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a)"
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§ 407]
EXTENDED LAW OF THE MEAN
99
Successive differentiation and substitution in the equations p(a) = f(a), p'(a) = f'(a), · · · , p(a) lead to an evaluation of the coefficients in (2) in terms of the function f(x), and hence to the following expression for p(x) (check the details) : (3)
p(x) = f(a)
+ f'(a)(x 
11
a)
1 /n\ +2T (x 
a) 2
+ · · · + f(x) are continuous on the clo8ed interval [a, b], and if J(x) exists at every point of an interval I (open, closed, or halfopen) that contains the point x = a, then for any x belonging to I there exists a point t between a and x (t = a if x = a) such that (11)
/(x)
= /(a) + j'(a)(x
 a)
+ i}~)
(x  a) 2
+ ,(nJ(a + Oh) h".
(n 
n I
(Ex. 17, § 408.) 408. EXERCISES In Exercises 12, find a value fort as prescribed by Rolle's Theorem. a figure.
1. J(x) 2. J(x)
= cos x, for ½r ;a x ;a t,r. = x 3  6x2 + 6x  1, for ½(5  V21) ;a
x
;a
Draw
1.
In Exercises 34, find a value fort as prescribed by the Law of the Mean, (2), § 405. Draw a figure.
3. J(x)
= In x, for 1 ;a
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~
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EXERCISES
§ 408)
101
4. f(x) =1 px2 + qx + r, for a, ~ x ~ b. In Exercises 56, find a value for 8 as prescribed by the Law of the Mean, (8), § 405. Draw a figure. 6. J(x)
6. f(x)
= In x, for a =
=
px2
e, h
= 1 
e.
+ qx + r, a and h arbitrary.
In Exercises 78, find a value for~ as prescribed by the Generalized Law of the Mean, (6), § 405. Draw a figure. 7. f(x) 8. f(x)
= 2x + 5, g(x) = x for O < b ~ x = x g(x) = x for 1 ~ x ~ 3. 2
3,
,
~
a.
2,
In Exercises 910, find a value for~ as prescribed by the Extended Law of the Mean, (5), § 407. 9. f(x) 10. f(x)
1 = 1 x , a = 0, arbitrary n, and b < l.
=
lnx, a= 1, b = 3, n = .3.
11. Prove Note 1, § 405. Explain why the function xi on the closed interval [ 1, 1) is excluded. 12. Prove Note 2, § 405. 13. Prove Notes 3 and 4, § 405 . . 14. Prove Theorem IV, § 405. Hint: First show that g(b) '¢ g(a), by using Rolle's Theorem with the function h(x) = g(x)  g(a). Then let K = (f(b) f(a))/(g(b)  g(a)) and define the function q,(x) = f(b)  J(a)  K[g(b)  g(a)]. Proceed as with the proof of the Law of the Mean. 16. Prove Note 1, § 407. 16. Prove Note 2, § 407. 17. Prove Notes 3 and 4, § 407. 18. The functions f(x)
= !X and g(x) =!X + sgn x
(Example 1, § 206) have
identical derivatives, but do not differ by a constant. possible in the presence of Theorem II, § 406.
Explain how this is
In Exercises 1930, use the Law of the Mean to establish the given inequalities. (Assume the standard properties of the transcendental functions. Cf. §§ 428431.) 19. tan x > x for 0 < x < ½11'. 20. lsin a  sin bl ~ la  bl, (Cf. Ex. 44.)
< ln Q < b
 a, for 0 < a < b. a a 22. Vl + h < l + ½h, for 1 < h < 0 or h > 0. More generally, for these values of hand 0 < p < l, (1 + h) 11 < l + ph. (Cf. Exs. 3841.) 23. (1 + h) 11 > 1 + ph, for 1 < h < 0 or h > 0, and p > l or p < 0. (Cf. Exs. 3841.)
21. b b a
24.
1
+h h
2
< Arctan h < h, for
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[§ 408
DIFFERENTIATION
102
< Arcsin x
1('
27. lcos ax
~ cos bxl :ii la  bl, for z
28. sin pz
< p, for
X
1. ;,4
0.
> 0 and x > 0.
p
29. e0 (b  a) < eh  e0 < eb(b  a), for a 1  eu 30. aeu < ' < a, for a > 0 and x X
< b. > 0.
In Exercises 3134, use the Extended Law of the Mean to establish the given inequalities. 31. cos x e;; 1  x' ,
2
32. cos x 33. z 
>
1  x' , for x
f < sin
0.
2
;,4
x
< x, for x > 0.
1
34. 1
1
+ X + 2x < ez < 1 + z + 2x ez, for X > o.
•36. Use the trigonometric identity cos u  cos v = 2 sin ½(u + v) sin ½(u  v) to establish the inequality cos a  cos b < b2  2a', for O :ii a < b. Prove that cos ax x'  cos bx < b'  a 2, for O ~ _ a < b, x ;,4 o. 2
2 •36. Prove that r
sin z
< X < 1, for 0 < x < r2. Hint: For the first inequal
ity, show that sin x/x is a decreasing function. 4
•37. Prove that 
1"
~nz > > 1, for 0 < x < r4, (Cf. Ex. 36.) X
38. Use the Law of the Mean to establish the following inequalities, for the designated ranges of p, assuming in each case that z > 0 and z ;,4 1: p(x  l)zP1 < zP  1 < p(x  1), for 0 < p < 1, p(x  l)zP1 > zP  1 > p(x  1), for p < 0 or p > 1. •39. By solving the lefthand inequalities of Exercise 38 for xP, establish the following inequalities for the designated ranges of p and h = x  1· l
~ : ~ ph < (I + h)P < 1 + ph, for 0
I
I+ h
+h
_ ph
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+ h)
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I
< p < I and either
1
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+ ph,
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"§ 408]
EXERCISES
103
< h < 0 or 0 < h < p1 1, < 0 and either h > 0 or   l < h < 0.
for p > land either 1 or for p
1 p Conclude that for any real number p, the expression (1 + h)P, 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. 2223.) •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  ! < h < 0, then
>0
or
(Cf. Exs. 3940.) •42. Let/(x) be differentiable, withf'(x) e:; 0 (f'(x) ~ 0), on an interval, and assume that on no subinterval does f'(x) vanish identically. Prove that /(x) is strictly increasing (decreasing) on the interval. •43. Establish the inequality sin x < x, for x > 0 (Example 1, § 405) by applying Exercise 42 to the function x  sin x. Similarly, establish the inequality tan x > x for 0
< x < ~ (Ex.
19).
*"·
Let /(x) be differentiable, with f'(x) e:; k (f'(x) ~ k), on an interval and assume that on no subinterval does f'(x) equal k identically. Prove that for the graph of y = f(x), on the given interval, every secant line has slope > k ( < k). As a consequence, show that if a "F b, !sin a  sin bl < la  bl, !cosa  cosbl < la  bl. Hint: Let g(x) = f(x)  kx, and use Exercise 42. 46. Prove that a function differentiable at every point of an interval is monotonic there if and only if its derivative does not change sign there. 46. Let x = g(t) and y = f(t) be continuous over a closed interval a ~ t ~ b and differentiable in the interior, and assume that g'(t) does not vanish there. Prove that y, as a function of x, is continuous over a corresponding interval and differentiable in the interior, and that for any interior point
!bl.= tm.. dx
g'(t)
Interpret the results geometrically. •47. It was shown in Example 3, § 403, that although a function may be differentiable for all values of the independent variable, its deriiative may not be continuous. In spite of this fact, a derivative shares with continuous functions
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[§ 408
DIFFERENTIATION
104
the intermediate value property of Theorem IV, § 213.
Prove the theorem :
If f(x) is the derivative of some function g(x), on an interval, then f(x) assumes (as a value) every number between any two of its values. Hint: Let c and d be any two distinct values of f(x), let r be any number between c and d, and apply Theorem IV, § 406, to the function f(x) = g(x)  rx.
*48. Show by the example f(x) = x2 sin\ (f(0) = 0) that derivatives do not X
always share with continuous functions the property of being bounded on closed intervals (Theorem I, § 213). That is, exhibit a closed interval at every point of which f(x) is differentiable but on which f'(x) is unbounded. *49. Prove that among the discontinuities discussed in § 210 for functions in general, derivatives can have discontinuities only of type (iii), where not both onesided limits exist. In other words, show that the kinds of discontinuities exhibited by the derivatives of the functions of Example 3, § 403, and Exercise 48 are the rule and not the exception. Hint: If lim f'(x) exists and
,,......,+
+
;,t!f'(c), there exist positive numbers E and 6 such that c < x < c 6 implies lf'(x)  f'(c)I ~ E. Use Ex. 47 . *60. Show by the example f(x) x 2x2 sin (1/x)(f(0) 0) that the hypotheses (i) f'(x) exists in a neighborhood of the point x = a, and (ii) f'(a) > 0,
= +
=
do not imply that there exists some neighborhood of x = a throughout which f(x) is increasing.
*61. The Wronskian determinant of two differentiable functions, f(x) and g(x), is defined: W(f, g)
= ,~~~~)
:~t2)1·
Prove that if the Wronskian off and g never vanishes over an interval, then between any two roots of the equation f(x) = 0 (g(x) = 0) there must exist at least one root of the equation g(x) = 0 (f(x) = 0). (Example: f(x) = sin x, g(x) = cos x.) Hint: Observe first that f(x) and g(x) never vanish simultaneously. Let a and b (a < b) be two roots of f(x) and assume that gfx) does not vanish for a < x < b. Apply Rolle's Theorem to the quotient f(x)/g(x), to obtain a contradiction. *62. Prove that if a function has a bounded derivative in an open interval it is uniformly continuous there. *63. Prove that if f(x) has a bounded derivative on an open interval (a, b) , then f(a+) and f(b) exist and are finite. Hint: Cf. Ex. 52, and Ex. 15, § 308. *64. Prove that if the righthand (lefthand) limit of the derivative of a function (Definition III, § 403) exists and is finite, then the righthand (lefthand) derivative (Definition II, § 403) exists and is equal to it. Hint: For the case x+ a+, show that there exists an interval of the form (a, a + h), where h > 0, in which f'(x) is bounded (cf. Ex. 9, § 208), and use Exercise 53 to infer that f(a+) lim f(x) exists. Redefine f(a) f(a+) and use the Law of the z.a+ Mean in the form [f(a + h)  f(a+ )]/h = f'(a + Oh).
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§ 409]
MAXIMA AND MINIMA
105
409. MAXIMA AND MINIMA
We shall assume that the reader is familiar with the standard routine of finding maximum and minimum values of a function y = f(x): (i) differentiate; (ii) set the derivative equal to zero; (iii) solve the equation f'(x) = 0 for x; (iv) test the values of x thus obtained, by using either the first or second derivative of the function; and (v) substitute inf(x) the appropriate values of x to find the maximum and minimum values of y = f(x). Inasmuch as this routine gives only a partial answer to the story of maxima and minima, we present in this section a more complete summary of the pertinent facts and tests. Let f(x) be a function defined over a set A, and let ~ be a point of A. If the inequality fm ~ f(x) um ~ f(x)) holds for every X in A, we say thatf(x) has a maximum (minimum) value on A equal tofm. According to Theorem II, § 213, such maximum and minimum values exist if A is a closed interval and f(x) is continuous on A . If a function has a maximum (minimum) value on its domain of definition, this value is called the absolute maximum (minimum) value of the function. If ~ is a point where f(x) is defined, and if within some neighborhood of ~ the inequality f(~) ~ f(x) Um ~ f(x)) holds whenever f(x) is defined, we say that fm is a relative maximum (minimum) value of f(x). By a critical value of x for a function f(x) we mean any point c of the domain of definition of f(x) such that either (i) f'(c) does not exist (as a finite quantity) or (ii) f'(c) = 0. Theorem I. If a function has a maximum or minimum value on an interval at a point ~ of the interval, then ~ is either an endpoint of the interval or a critical value for the function. Proof. This theorem extends Theorem I, § 405, to an arbitrary (not
necessarily closed) interval and drops continuity and differentiability assumptions. The details of the proof, however, are identical. Some different kinds of maxima for a function continuous on a closed interval are illustrated in Figure 408.
~
(a)
(b) FIG. 408
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DIFFERENTIATION
106
[§ 409
Theorem II. First Derivative Test. If f(x) is continuous at a point x = Eand differentiable in a deleted neighborhood of E, and if in this deleted neighborhood r(x) > 0 for x < E and f'(x) < 0 for x > E (f'(x) < 0 for x < Eandf'(x) > Oforx > E),thenf(x)hasarelativemaximum(minimum) value at x = E. If, on the other hand, r(x) is of constant sign throughout the deleted neighborhood, f(x) has neither a relative maximum nor a relative minimum value at x = E. Proof. Let x be an arbitrary point in the deleted neighborhood· of E. By the Law of the Mean, § 405, there exists a point f between E and x such that f(x)  fm = f'(fl(x  E). Examination of each individual case leads to an appropriate inequality of the fonnf(x) > fm or f(x) < f(E). (Check the details.) NOTE I. The conditions assumed in Theorem II are sufficient but not necessary, even if j'(x) is continuous and f'W = 0 (cf. Exs. 2122, § 412).
Theorem II. Second Derivative Test. If f(x) is differentiable in a neighborhood of a critical value E, and if f"m exists and is negative (positive), then f(x) has a relative maximum (minimum) value at x = E. Proof. Assume f"W < 0. Then, by Exercise 7, § 404, within some neighborhood of E, r rm = 0 for x < E and r E, By the First Derivative Test, f(x) has a relative maximum value at x = E. (Supply the corresponding details for the casef"m > 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 valuethe 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 r (x), f" (x), • · · , f (x), of order ~n  1, where n > 1. If = f"m = · · · = pnll(E) = 0, and if f(E) exists and is different from zero, then (i) if n is even, f(x) has a relative maximum value or a relative minimum value at x = Ea,ecording as 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 = E. Proof. Owing to the vanishing of the derivatives of order 0.9, and the third member of (7) is between h2/6 and 0. Therefore
(9)
1
+ ½h 
¼h2
(Illustration: 0.9694
N1, the generalized mean value theorem guarantees the relation f(x)  [(N1) g(x)  g(N,)
(1)
and therefore,
= &). g(x)
. 1  f(N,)/f(x) 1  g(N1)/g(x)
= Lill, g'W
f'm
&). _ 1  g(N1)/g(x) g(x)  g'W 1  f(N1)/f(x)'
(2)
for x > N1 and a suitable ~ between x and N,. First choose N1 so large that if ~ > Ni, then f' W / g' W 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 = +00, let N2 be so z+• z+• large that if x > N2, then the fraction [1  g(N,)/g(x)]/[1  f(N,)/f(x)] is within a specified degree of approximation of the number I. 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 number. We shall show first that there exists a positive number o such that IY  LI < ½E and lz  II < o, imply lyz  LI < E. To do this we use the
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UNIVERSITY OF MICHIGAN
§415]
T H E I N D ET E RM I N A TE F O RM «J /«J
119
triangle inequality to write
lyz  LI ;;; lyz  YI + IY  LI = IYI · lz  11 + IY  LI. If IY  LI < ½E, IYI inequalities imply
< ILi + ½E, so that if 8 = ½E(ILI + ½E)i the assumed
lyz  LI < (ILi + ½E) 2(ILI E+ ½E) + ~ = E, which is the desired result. The rest is simple. First choose Ni such that
~> and second choose N,
x
Ni implies
1~:m  LI < ½E,
> Ni such that
. . 11 _ g(Ni)/g(x) > N 2 1mphes 1 f(Ni)/f(x)
I< 8.
 1
Then, by (2), X
Example 1.
Solution.
Show tha.t Jim X"
+• es
If a X"
Then Jim 
+• es
> N2 implies
=
~
1rc:~  LI
0.
•1
Jim ~ . 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+ +oo, faster than any power of x, and therefore faster than any polynomial. In x Example 2. Show that Jim = 0 for any a > 0 . .,..... +.. x• 1 Solution.
Jim In x = Jim
+..
X"
+ ..
x _1 = Jim _!_ = 0.
+• az
aX"
(Cf. Ex. 7, § 602.)
In other words, In x increases, as x+ +oo, more slowly than any positive power of x.
Show that Jim 1e" = 0. ,...... + .. n . Solution. This is an indeterminate form to which !'Hospital's Rule does not apply, since n ! (unless the Gamma function is used to define n ! for all positive real numbers) cannot be differentiated. We can establish the limit as follows: Let n be greater than 3. Then Example 3.
: .., =
G·;n(r·· !) < f ·(1ra·
As n + +oo, the last factor approaches zero. Example'
1·1m sinx . Cr1·t·1c1ze:
.,.... +•
X
=
1·1m cos x an d th ere fore does no t  ,
+•
1
exist!
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UNIVERSITY OF MICHIGAN
[§ 416
DIFFERENTIATION
120
Solution. L'Hospital's Rule does not apply, since the given expression is not indeterminate . . Since Jsin x/xl ~ 1/x, the limit is 0. 416. OTHER INDETERMINATE FORMS
In the sense discussed in § 414, the forms 0 • co,
co 
co,
00,
00°, and l'°
are indeterminate (Ex. 42, § 417.) The first type can often be evaluated by writing the product f(x) g(x) as a quotient and then using !'Hospital's Rule (Example 1). The second type sometimes lends itself to rearrangement, use of identities, or judicious multiplication by unity (Examples 23). The remaining three forms are handled by first taking a logarithm: if y = f(x)uttl, then In y = g(x) In (f(x) ), and an indeterminacy of the first type above results. Then, by continuity of the exponential function (cf. § 602), limy= Iim (e 1n u) = elim Un u>. (Examples 46). Finally, other devices, including separation of determinate from indeterminate expressions and substitution of a reciprocal variable, are possible (Examples 79). Find lim x• In x. zo+ Sofotion. If a ~ 0, the expression is not indeterminate (cf. Ex. 7, § 602), and the limit is  co . If a > 0, the limit can be written . x . . In x = ,m _ 1/ _,___ = 1,m x• = 0 . 1 I1m zo+ x• ro+  ax• 1 rO+  a In other words, whenever the expression x4 In x is indeterminate, the algebraic factor "dominates" the logarithmic factor (cf. Example 2, § 415).
Example 1.
Example 2.
lim (sec x  tan x)
=
r
COS X
r
z+2
.1:2
Alternatively, . hm (sec x  tan x) r z+2
Example 3.
Jim 1  sin x
=
lim [Vx 2
x..

. sec2 x  tan2 x hm + t an x r sec x z+2 a2

=
JxJ] = Jim v'x z,o
=
=
Jim c?s x r sin X :r.+2
Jim r sec x zt2
2 
1
= o.
+ tan x
= 0.
2 a  Ix! . v ' ~ + !xi 1 v'x2  a2 + Jxl
at Jim ,=== = 0. z,o v'x2  a2 + JxJ •
Example 4.
Find Jim X"'. zo+ Solution. Let y = x". Then In y = x In x and, by Example 1, In y> 0. Therefore, by continuity of the function ez, y = e 10 M+ e0 = 1.
!
Example 6.
Find
lim z++GO
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(1
+ ax)"', a > 0.
Original from
UNIVERSITY OF MICHIGAN
EXERCISES
§ 417]
Therefore y
+ ax)~,
If y = (1
Solution.
121
+ ax)
In y = In (1
+
0.
X
= e10 v+ e0 =
Example 6.
l.
Show that lim (1 ~
. I1m 1n y =
1
+ ax);, . In (I + ax) 11m =
If a ~ 0 and if y
Solution.
!
+ ax)r = e'.
= (l
x (Cf. Ex. 10, § 602.)
x....O
and y = e1n v+ e•.
z0
. a 11m +1 = a, ax
.r....O
! Find Jim x e".
Example 7.
z....O+
! e" If this is written Jim  , differentiation leads to the answer. z....O+ 1 
Solution.
X
However, the limit can be written Jim e_'  +oo.
t++• t
Jim (csc2 x  x csc 3 x)
Example 8.
z....o
x  x x3 = Jim sin  •  .33 z0
Sill X
X
!) .(~~
=(
si:
xr =  !·
by Example 2, § 414, and the continuity of the function x (the limit of the cube is the cube of the limit). 3
~
12 sin; In x
Example 9.
Jim
(x3
z+t
x
+ 5)(x 
z+1
.I()
~ ~...,
{
1)
. Iim
,t ';,
),
/; 
12 sin r
2X
X
a
+5
.
~~
/s 1[1=, )(
~ 41
12 · 1 • )"Im X = 2 . =6 X+1 1
IIm  .,..... x  l
417. EXERCISES
In Exercises 130, evaluate the limit. l. Jim 3x: z+2
. 3 • I1m
z+1 X
X
+x 
14.

X 
2
In x
2
+X
x+3
1

2
z+1
x4
•
x 3 + X + 30 · 4x3 + 1lx 2 + 9
4:. Jim cos ,rx.
•
2 ½x• . eos 5. I1m  x  l '+ 
z+0
. 2• I1m
2
l
6. Jim In (I + x)  x. z....O cosx1 sin (1/x) • 8 • Jim .,....,. Arc tan (1/x)
7. J i m2 ~,,....,, tan 4x
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UNIVERSITY OF MICHIGAN
[§ 417
DIFFERENTIATION
122 • a"'  1 9. h m· z,0,,,. 1
z,0
. 8x5 5x2 +1
11. hm
:c•
13.
3X 5
tanxx Arc sin x  x . tan x  6 12. I1m     · ztr sec X + 5 1._ lim In ~in x . zrlnsm2x . In (1  2x) 16• I1 m    ~.. :cl tan rx a• 18. lim ;::;; a > 1, b
10. lim·
+X ·
In sin 2x lim :clr In cosx
lim cosh x (cf. § 607). +• ~ 17. lim (In x)"'n > 0 . X +• 15.
+•
> 0.
;i;
19.
lim x(ln x)", n zo0+
> 0.
20. lim (x  a) tan n.
21.
~Ifr [xtanx 
~secxJ
22.
,..._
lim
z1+
2a
[x1  1...LJ. n X 
1
23.
lim (1 + x1 );, +•
2' lim (1 z,0
1
lim xln "'· zo0+ 27. Jim x"''(x""' zoO+
X
+ 2 sin x)oe>t z, 1
28. lim (x
26.
z,0
e
x).
.! 29. lim (cos 2x)zl. z,0
28.
lim zo0+
30. ~+
+ et.);, [In {l + x]"'.
L/~xx)' 
In 1
~ :z]
31. Prove Case 2 of !'Hospital's Rule, § 414. 32. Prove Case 3 of !'Hospital's Rule, § 414. Hint: Make sensible use of Cases 1 and 2. 33. Prove Case 5 of !'Hospital's Rule, § 414. 3'. Prove Case 6 of !'Hospital's Rule, § 414. 36. Prove Case 1 of !'Hospital's Rule, § 415. Hint: Apply Case 4 (cf. proof of Case 4, § 414). •36. Prove Case 2 of !'Hospital's Rule, § 415. •37. Prove Case 3 of !'Hospital's Rule, § 415. •38. Prove Case 5 of !'Hospital's Rule, § 415. •39. Prove Case 6 of !'Hospital's Rule, § 415. For Case 4 of !'Hospital's Rule, § 415, supply the ''epsilon" details for the case L = +oo. 41. Prove that the forms ( +0)+•, ( +oo )+•, and a+• (where a> 0 and a "F 1) are determinate. What can you say about (+0)•? (+oo)•? (+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 x > a, there exist ~1 and~, between a and x such that J(x)  J(a+) = f'{~ 1) and g(x)  g(a+) = g'(~s).
*'°·
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UNIVERSITY OF MICHIGAN
§ 418]
CURVE TRACING
123
Therefore
&1 = f(x) g(x)
 [(a+)= Bfil_Bill g(x)  g(a+) g'(~s) g'(a+)
= 1·
f1=J.
~+ g'(x) ·
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 ordinat,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
'11cosx+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
xis large (Fig. 413). (iii) Vertical and horizontal asymptoks. A function represented as a quotient /(x)/g(x) of continuous functions has a vertical asymptote at a point a where g(a) = 0 and /(a) ¢ 0, If lim/(x)/g(x), as x becomes infinite ( +oo,  00, or oo), exists and is a finite number b, then y = b is a horizontal asymptote. (Fig. 414.) (iv) Other asymptot,es. If /(x)  mx  b approaches zero as x becomes infinite ( +oo, 00, or 00 ), then the line y = mx + b is an asymptote for
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UNIVERSITY OF MICHIGAN
DIFFERENTIATION
124
[§ 418
fl
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 facurrs. Certain principles used for functions represented as sums have their applications to functions represented as products. The functions eaz: sin bx and ea" cos bx, useful in electrical theory, a.re 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 l/ I
I I I
I I
I
~~~~=~~=~~,I 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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UNIVERSITY OF MICHIGAN
§ 418]
CURVE TRACING
125
FIG. 415
= x2 (2 
y2
x) is approximated by that of y 2 4(2  x) for x near 2. (Fig. 417.)
= 2x 2 for x near Oand by that
of y = 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: I ~  [JD. . (1) y  dx  f'(t)' 2
(2)
y
II 
~
13J.: 
..2:J_•
 d,x2  f'(t)
The folium of Descartes,
.... __
y



  FIG. 416
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UNIVERSITY OF MICHIGAN
126
DIFFERENTIATION
[§ 418
11
:,;
2
11' • 4(2z) FIG. 417
3at
(3)
X
= t' + 1'
3at2
y
= t'
+ 1'
is illustrated in Figure 419. Since ~
(4)
_ t(t'  2)
dx 2t'l'
horizontal tangents correspond to the values of t for which the numerator of (4) vanishes: t = 0 (the point (0, 0)) and t = ~ (the point (a~, a¢4) ). , Vertical tangents correspond to the values oft for which (4)" becomes infinite: t = oo (the point (0, 0)) and t = ~ (the point (a¼, a~)).
11
fl
z1 (2z)
FIG. 418
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UNIVERSITY OF MICHIGAN
EXERCISES
§ 419]
Since x + y = 3at/((1.  t + 1), the line x (let t  1).
127
+y +a =
0 is an asymptote
11
2a
2a
FIG. 419
419. EXERCISES
In Exercises 150, graph the equation, showing essential shape and asymptotic behavior. 1. y = x 2 (x1 3. y
x1

= x' 

9).
4 9'
6.y=x+t X 7. y
2
2. y
= x' + 1·
4:. Y
= x' 
6. y
= x1 + ~x
x'  9
4'
= X + ~•
 4 9• y' = x' x'  9'
11. y' = (x  l)(x  2)1 (x  3) 8• 13. xi + xy + y' = 4. 16. y = ez•. 17. y = ~19. y = xezt.
10. y1
x'  9· =x2  4
12. y'
=
H. x'
+ 4xy + y1 = 4.
18, 18. 20. 22.
l)•(x  2)(x  3) 2•
(x 
Y
= e1l",
y
= x'e,,.
y
=
y
= x In x.
e" 
x.
21. y
= In lxl.
23. y
In x =· X
24. y
= 2.
25. y
1 =· x In x
28. y
= x' In x.
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In x
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UNIVERSITY OF MICHIGAN
DIFFERENTIATION
128
[§ 419
28. y
= In xe"'.
29. y
= x + sin x. = In (1 + .x2).
30. y
= In sin .x.
31. y
= ez cos 2.x.
32. y
=
27. y
f (1 
ekz).
33. The Witch of Agnesi, x 2y = 4a2(2a  y).
34. The Cissoid of Diocles, y2(2a  .x) = ,x3.
35. The Catenary,
36. The Folium of Descartes,
y
=
+ ez/a).
½a(ez/a
.x3
t2
X

= t' +
I I, y
=
3axy.
xi+ yt = at.
= a½.
40. The Ovals of Cassini, (.x2 + y' + a2)2 _ 4a2,x2
39. The Lemniscate of Bernoulli, (x' + y')' = a2(,x2  y'). 41.
3
38. The Hypocycloid,
37. The Parabola,
±.x½ ± y½
+y
=
t2
2t
+ I.
42. x = t
+ In t, y
= t
+e
= c4. 1 •
44. The Cycloid, x = a(t  sin t), y = a(l  cos t).
43. The Ellipse,
x = a cos t, y = b sin t.
46. The Cardioid,
45. The Hypocycloid, x = a cos3 t, y = a sin 3 t.
x
= a(2 cost
y
=
 cot 2t),
a(2 sin t  sin 2t).
47. The Serpentine, x = a cot t, y = b sin t cos t.
48. The Witch of Agnesi, x = a cot t, y = a sin2 t.
49. The Hypocycloid of Three Cusps, x = 2a COB t a COB 2t y = 2a sin t  a sin 2t.
50. The Hyperbolic Spiral,
+
51. Graph /(.x)
= f'(0) = 0. 52. Graph /(.x)
=
I
1
+ e ,., 1
,
X
= ta
and show that /(0+)
= e /a:', .x ;,6. 0, /(0) = 0. 1
t y = ta Bill . t•
COS ,
=
1 and /(0)
= f'(0+)
Prove that Jim x"J(x) r~
= 0 for
every positive integer n, and hence show that /(.x) has everywhere continuous derivatives of all orders, all of which vanish at x = 0. (Cf. Example 2, § 805, for a use of this function as a counterexample.)
•53. Show by a graph that a function f(x) exists having the following properties: (i) f(x) is strictly increasing, (ii) f'(.x) exists for every real x, (iii) f(x) is bounded above, (iv) the statement lim f'(x) = 0 is false. :r+oo
=
=
Show that the function f(x) x/(l + e 1'"'), x ;r6 0, /(0) 0, is everywhere continuous, and has unequal righthand and lefthand derivatives at X = 0.
•M.
•55. Graph (y  .x2) 2 = .x 6, with particular attention to a neighborhood of the origin.
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§420]
WITHOUT LOSS OF GENERALITY
129
*420. WITHOUT LOSS OF GENERALITY
One of the standard techniques of an analytic proof is to reduce a general proposition to a special case "without loss of generality." This means that it is possible to construct a proof of the general theorem on the assumption that the special form of that theorem is true. Following the establishment of this inference, proof of the special case implies proof of the general proposition. This device was used in two proofs given in § 204: in the proof that the limit of the product of two sequences is the product of their limits (Theorem VIII) we saw that it could be assumed without loss of generality that one of the limits is zero; in the proof that the limit of the quotient of two sequences is the quotient of their limits (Theorem IX) we assumed without loss of generality that the numerators were identically equal to unity. Another instance of the same principle is the proof of the Law of the Mean(§ 405) by reducing it to the special case called Rolle's Theorem. In the following Exercises are a few problems in showing that the special implies the general. "421. EXERCISES
In Exercises 110, prove the stated proposition on the basis of the assumption given in the braces { } . *1· If m and n are relatively prime positive integers, there exist positive integers x and y such that mx  ny = 1. {If m and n are relatively prime positive integers, there exist positive integers x and y such that lmx  nyl = 1.} *2. If m and n are nonzero integers and if d is their greatest common divisor, then there exist integers x and y such that mx + ny = d. {If m and n are relatively prime positive integers, there exist integers x and y such that
+ ny
= 1.} If Ii, /2, · · · I,. are open intervals and if I is the set of all x such that x is a member of every Ik, k = 1, 2, · · · , n, then I is either empty or an open interval. {n = 2.} *4. The Schwarz inequality (Ex. 43, § 107). {All of the numbers ai, a,., b,, • • • , b,. are positive.} mx
*3.
*6. sinXpx < p, for *6. lim zP In x z,()+
=
> 0 and x > 0. {p = 0, p > 0. {p = 1.}
p
I.l .
*7. Any bounded sequence {a,.} that does not converge to a contains a subsequence {an.} converging to some b F a. {la,.  al ~ E > 0 for all n.} •S. The trigonometric functions are continuous where they are defined. {sin x and cos x are continuous at x = O.} •9. If f'(x) ~ g'(x) for a ~ x ~ b, then f(x)  f(a) ~ g(x)  g(a) . {f(x) = 0 for a ~ x ~ b.} *10. Theorem III, § 213. {a = 0, b = 1.}
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130
[§ 421
DIFFERENTIATION
*11. Show that in proving that if f(x) and f'(x) are defined in a neighborhood · of ;r; = a and if f"(a) exists, then (2)
f"(a)
=
lira f(a ~
+ h) + /(:1 . h)
 2/(a),
it may be assumed without loss of generality, that a = 0, f(O) = 0, f'(O) = 0, and f"(O) = 0. Hence prove (2). Hints: To show that one may assume f(O) = 0, define a new function g(;r;) • f(x)  f(O). To show that one may assume f'(O) = 0, define a new function g(x) • f(x)  /'(O) •z. (Cf. Ex. 36, § 412.)
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5 Integration
501. THE DEFINITE INTEGRAL
It will be assumed that the reader is already familiar with some of the properties and many of the applications of the definite integral. It is our purpose in this section to give a precise definition and a few of the simpler properties of the integral, with analytical proofs that do not depend on the persuasion of a geometrical picture. We shall be dealing in the main with a fixed closed interval [a, b] . On such an interval a finite set of points a = ao < a1 < 0,
then J(x) is integrable on [ a, a] and faf(x) dx = 2 {af(x) dx. a
Jo
Prove that if f(x) is even on (00, +00) and a and bare any real numbers, then f
af(x) dx b
= (bf(x) dx, whenever the integrals exist. )a
8. A functionf(x) is said to be odd if and only if the equality J(x) = f(x) holds for all x in the domain of definition of the function. (Examples: lOx, x2n+ 1, sin x, the signum function of Example 1, § 206.) Prove that if f(x) is odd on [ a, a], and integrable on [O, a], where a > 0, then f(x) is integrable
on [ a, a] and j~f(x) dx = 0.
Prove that if f(x) is odd on ( 00, +00) and
a and b are any real numbers, then f~bf(x) dx
= .£bf(x) dx,
whenever the
integrals exist. 9. Prove that any sum of even functions is even and any sum of odd functions is odd. Prove that the product of two even functions and the product of two odd functions are even, and that the product of an even function and an odd function is odd. (Cf. Exs. 78.) 10. Prove that the only function that is both even and odd is identically zero. (Cf. Exs. 79.) 11. Prove that any function whose domain contains the negative of every one of its members is uniquely representable as the sum of an even function and an odd function. (Cf. Exs. 710.) Hint:
+
+
J(x) = ½[f(x) J( x)] ½[f(x)  J( x)]. 12. Prove that if J(x) is even (odd) and differentiable, then f'(x) is odd
(even) . (Cf. Exs. 711.) *13. Let f(x) be defined to be x + 1 for all real x, and let g(x) be defined to be x2 + 1 for x > 0, x2 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 functionf(x) is an open interval of the form (a, a) or ( 00, +00) and if f'(x) is odd, then f(x) is even; similarly, that if f'(x) is even, then f(x) plus a suitable constant is odd. (Cf. Exs. 712.) Hint: J(x) and/( x) ha,ve the same derivative, and must therefore differ by a constant. 14. If f(x) is the bracket function of Example 5, § 201, J(x) = [x], evaluate
.fo J(x) dx and .£ /(x) dx. 3
1
More generally, if n is a positive integer, evaluate
.fonf(x) dx.
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UNIVERSITY OF MICHIGAN
EXERCISES
§ 503]
16. Prove that _£\:2 dx
=
145
i(b3

a 3).
(Cf. the Example, § 501, and Ex. 13,
= ¼(b 4

a 4).
(Cf. the Example, § 501, and Ex. 14,
= !W 
a 6).
(Cf. the Example, § 501, and Ex. 15,
§ 107.)
16. Prove that .£bx3 dx § 107 .)
17. Prove that .£bx 4 dx
§ 107.) •18. Prove that if mis a positive integer, then 1
=
(bx"' dx
)a
(b"'+I  a"'+ 1).
+ l
m
(Cf. the Example, § 501, and Ex. 42, § 107 .)
= cos a
•19. Prove that .£bsin x dx § 501, let b
>
0 and a
fo
= 0, and
Hint : As in the Example,
write
. ( sm . b 11m
bsm . x d x .=
Multiply each term by 2 sin
 cos b.
n.... +..
n
+ · · · + sm . nb) · ·b n
n
:n' and use the identity
2 sin A sin B
= cos (A
 B)  cos (A
+ B)
to obtain
fobsin x dx =
. [( cos b  cos 3b) hm 2n 2n
n .... +..
+( cos •20. Prove that _£\os x dx
•21. Prove that .£bez dx
 cos 5b) 2n
+ ···
(2nl)b_ (2n+l)b)]· cos 2n 2n
= sin b
= ,!>
3b + ( cos 2n
 sin a.
b
. b 2n sm n 2
(Cf. Ex. 19.)
 e".
•22. Prove the Trapezoidal Rule for approximating a definite integral: If f(x) is integrable on [a, b], if [a, b] is subdivided into n equal subintervals of length ~x, and if the values of f(x) at then 1 points of subdivision, xo, x1, Xt, • • • , x,., are Yo, Y1, Y2, · · • , y .. , respectively, then
+
(bf(x) dx
)a
= lim (½Yo+ Y1 + Y2 + · · · n++o:.
+ Yn1 + ½Y .. ) .£lx.
Also prove the following estimate for the error in the trapezoidal formula, assuming existence off" (x) on [a, b]: ba )af(x) dx  [½Yo+ Yi+··· + ½y,.] ~x = J"m .£lz2,
r
12
where a < E < b. Hints: For the first part, write the expression in brackets in the form (y1 + · · · + y,.) + (½Yo  ½y .. ). For the second part, reduce the problem to
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[§ 503
INTEGRATION
146
that of approximating .£bf(x) dx 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), · · · of open intervals covering Sm, of lengths L1 + ••• + L,. < E/2"'. The entire collection of intervals In(x) that is continuous on [a, b] such that 4>'(x) exists and is equal to f(x) there with at most a finite number of exceptions must differ from F(x) by at most a constant. If 4>(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
sin 10 x dx, we might proceed: I =
= so that (1)
I
J
sin9 x d(cos x) = sin9 x cos x
sin9 xcosx
+
9J
J
+9
sins x cos2 x dx
sinsxdx  91,
= lo sin9 x cos x +hfsins 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 formul,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.
Express
f
sin"' x cos" x dx, where m
+n
¢.
0, in terms of an
integral with reduced exponent on sin x (cf. Exs. 12, § 510).
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[§ 510
INTEGRATION
162
The derivative of the product sinP x cost x is p sinpt x cost+• x  q sinP+l x cosr1 x ... p sinP1 x cosr1 x (1  sin' x)  q sinP+l x cosr1 x = p sinP1 x cosr1 x  (p + q) sinP+l x cosr1 x. In other words, Solution.
pfsinP
1
(2)
Letting p with the (3)
f
+ q) f sinP+l x cosr1 x dx = sinP x cost x +C. 1 and q = n + 1, and absorbing the constant of integration
x cosr1 x dx  (p
m 
!!!
J,
we have the formula sought: 1
sin• x
COB"
x dx
1
x cos + x + m  1 =  sin"'====m+n m+n 11
f
sin•t x
COB"
x dx.
Equation (1) 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 ""F 1) (cf. Exs. 12, § 510) : x dx cos +1 x n  m +2 COB" x dx 4 ( ) sin• x =  (m  1) sin•1 x m  1 sin•t x • Solution. In (2), let p == m + l and q • n + 1.
f
Eumple 3.
f
11
COB"
Establish the reduction formula (m # 1):
f
f
dz X + 2n  3 d,x (a' + x')" = 2(n  l)a2 {a2 + x 2) 1 2(n  l)a2 (a'+ x 2) Solution. The derivative of the product x(a' + x2)"' is (a2 + x 2)"' + 2mx2(a2 + x 2)"'1 = (a' + x 2)"' + 2m[{a2 + x')  a2 ](a2 + x')"'1 • (1 + 2m)(a2 + x 2) ..  2ma'(a2 + x') .. 1• In other words, (5)
(6)
(1
11
+ 2m)
Now let m
f
e

(a'+ x')"' dx  2ma'f (a'+ x')"'1 dx
n
11

1•
= x(a2 + x 2)• + C.
+ 1, from which we obtain (5).
Useful reduction formulas are given in nearly every Table of Integrals. *510. EXERCISES In Exercises 112, establish the reduction formula.
f.
srn • x cos" x dx
f *3. f *2.
= sin"'+' x cos

1
x
m+n
+ n__l_
f
sin.. x cos"t x dx (m + n ,;4 0). m+n sin•+lx _ m  n +2 sin"' x dx (n # 1). (n  1) cos11  1 x n  1 cos11 t x
sin"' x dx cos" x
=
tan" x dx
= tan"n _ x 1
1
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='....;;;.....:..;.;;;...._...;;;
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f
f
tan"t x dx (n # 1).
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UNIVERSITY OF MICHIGAN
EXERCISES
§ 510]
*'
cot"1 x cot" x ch =  "n1
6•
,. x ch= sec"➔ se e nxtan l x
f * f f *1. f f *6.
*8.
*10.
f
+n n  2 l
n _x cot x 1
x" sin x ch = x" cos x x" cos x ch = x" sin x
x"'(ax
cot.., x ch (n "6 1).
CBC"_,
csc"xch = 
*9• fx•(ax
f
+ nn _
f
X"
ch
sec
2 1
+ nx"1 cos x 
r:
1
x
ch (n "6 1) .
csc",xch (n
J J
 n(n  1) n(n  1)
J
x•1 (ax
x
,6. 1) . 11

2
sin x ch.
x"' cos x dx.
+ b)"dx
(m
+ n + 1 ,6. 0) .
dx (m
+ n + 1 ,6. 0).
~ br
+ n +~ + 1
*11•
f ,., f
+ n:r:"1 sin x
+ b)" ch = x•(ax + b)"+1 a(m + n + 1)  a(m +m! + l) + b)" ch = x:+
163
J
x"'(ax
+ b)
11 
1
xn1
Vax'+ bx+ c = a(n  1) 1 bf x" dx a Vax 2 +bx+c
2 cf x" ch (n a Vax'+bx+c
,6. l).
1(2ax  x2 )i x""'='"'n +2 + a(2n + l) fx 1V2ax  x2 ch (n ,6. 2). n+2 In'Exercises 1320, perform the integration, using reduction formulas above.
*12.
f
. 1 2ax  x2 ch x"v
=
11 
*13. *16. *17. *19.
f f ~ f f Vx +
sin• x ch.
*1'
cot 6 ch.
*16.
x 4 sin 2x dx.
*18.
2
xa ch x
+1
f f f
cos6 x ch. sec7 x ch.
xi(x
+ 4)½ dx.
*20. fx 3V6x  x2 dx.
**21. Use mathematical induction to verify the formula:
i
b(x
 a)"'(b  x)" ch = (b  a)"'+n+i
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1 m! n (m and n positive integers). (m+n+l)I
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UNIVERSITY OF MICHIGAN
[§ 511
INTEGRATION
164
511. IMPROPER INTEGRALS, INTRODUCTION
The definite integral [f(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 .lbf(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. Let f (x) be ( Riemann) integrab/,e on the interval [a, b  t] ( [a E, b]) for every number E such that O < E < b  a, but not integrab/,e on the interval [a, b], and assume that
+
lim (1>f(x) dx ....o+ Ja exists.
( Jim (1' f(x) dx) ....o+ Ja+•
Under these conditions the improper integral lf(x) dx is defined w
be this limit:
(1)
rbf(x) dx
Ja
= Jim
....o+
.r:f(x) dx
a
( lim (1' f(x) dx). ....o+ Ja+•
If the limit in (1) is finite the integral .lbf(x) dx is convergent to this limit and the function f(x) is said to be improperly integrable on the halfopen interval [a, b) ( (a, b]) ; if the limit in ( 1) is infinite or does not exist, the integral is divergent. NoTE I. A function improperly integrable on a halfopen interval is necessarily unbounded there. In fact, it is unbounded in every neighborhood of the endpoint 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, l.et a
< c < b,
and
l.et 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 lf(x) dx
is convergent and defined to be: [f(x) dx
(2)
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UNIVERSITY OF MICHIGAN
IMPROPER INTEGRALS, FINITE INTERVAL
§ 512]
165
If either integral on the righthand side of (2) diverges, so does i~f(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 /(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) l, Jim [ x'
1
11 ]
....o+ 1  p •
if and only if p
0, r rational.
(Cf. § 214 and Exs. 2122, § 216.) (v) holds for any real r. Hints: (i):
(%11 ~ = rx ~ )1 t )1 t
(ii) : use (i) with x
+
=
It follows from Exercise 6, below, that
rx
rx11 ~ = ~ Jx t )1 t
+ (XII~; )x
t/µ;
y • ~; y
= (t1/:;)". *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 all real x; (ii) e0 = 1, Jim e" = +oo, Jim e" = 0; x~+• ~co (iii) e" is continuous and, in fact, differentiable, with derivative e".
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UNIVERSITY OF MICHIGAN
§ 603)
191
THE tRIGONOMETRIC FUNCTION$
•7. The function x", for x cise 6: ·
>0
and arbitrary real a, is defined as in Exerx4 e
e" In"'•
Prove the following properties of x": (i) If a > 0 (a < 0), x" is strictly increasing (decreasing); (ii) if a > 0 (a < 0), .lim x" = +oo (0) and lim x" = 0 ( +oo); z+oo zo+ (iii) x" is continuous and, in fact, differentiable, with derivative ax 11  1• •8. If a > 0, prove that the function x", defined as in Exercise 7 for x > 0, and defined to be O when x = 0, has the three properties stated in Exercise 7, except that if 0 < a < 1, the derivative of x4 at x = 0 is +oo. 1
*9. Prove that e
.!!.1n x dx
=
lim (1 Z0
+ x);.
= lim In (1 + h)
h
Ji,()
Hint: At x
=
l, l
 In (1)
= lim In (1 + h)h =
1.
h0
1
•10. Prove that lim (1 Z0
+ ax); = e".
Define log., x, where a > 0 and a ;,4 1, x > 0, as the inverse of the function a"'. That is, log., xis that unique number y such that aw = x. Prove the standard laws of logarithms, and the change of base formulas: log., x = log., b lo~ x = 1°S6 x. 1o~a Prove that a 1oa:.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.
•11.
*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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UNIVERSITY OF MICHIGAN
SOME ELEMENTARY FUNCTIONS
192
[§ 604
*604. EXERCISES
•1.
Prove that the function defined by the equation
. .£" . ~ •t
Arcs1n x =
dt
0
Vl 
2
1
< X < 1,
is strictly increasing, continuous, and, in fact, differentiable with derivative (1  x2)½. (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 ¾. f/
:z:
FIG. 603
If the number 1r is defined : 1r = 4 Arcsin ½V2, and if the number b is defined: b = Arcsin i, prove that over the interval (b, b), containing r, sin xis strictly
increasing, continuous, and, in fact, differentiable with derivative (1  sin2 x)! [l  (sin x) 2]½. (Fig. 604.) Also, sin 0 = 0, sin = ½V2.
vr
=
11
:z:
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 ¼r = ½V2. Hint: If y = cos x ~ 0, y2 = 1  sin2 x, 2 cos x dy/dx = 2 sin x cos x.
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§ 604]
EXERCISES
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:
+ (c(x) ) 2 = 1,
(1)
(s(x) ) 2
(2)
d dx (s(x))
= c(x),
(3)
!
=
(c(x))
s(x),
c(O) = 1. Prove that if a and f3 are any two numbers which, together with their sum a + /3, belong to the interval ( a, a), then the following two identities hold:
(4)
(5)
s(a
(6)
c(a
+ {3) + {3)
+ c(a) s(/3),
=
s(a) c(/3)
=
c(a) c(/3)  s(a) s(/3).
Hence prove that if a and 2a belong to ( a, a), then (7) s(2a) = 2s(a) c(a), (8)
c(2a)
=
(c(a) ) 2

(s(a) ) 2•
+
+
Hint: For (5), let 'Y = a {3, 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 t,wo 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. Prove that sin x and cos x, as defined for all real numbers x, are odd and even functions, respectively. (Cf. Exs. 78, § 503.) •S. Prove that sin ½,r = 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 ~ ½1r, and if sin 2k = 1, then cos 2k = cos2 k  sin2 k = 0, and cos k = sink = ½V2, and
=
•7.
k
= ¼1r. •9. Prove that sin 1r = 0 and cos ,r = 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 2,r is the smallest positive number whose cosine is 1.
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UNIVERSITY OF MICHIGAN
[§ 605
SOME ELEMENTARY FUNCTIONS
194
•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>. and µ. are any two real numbers such that >.2 + µ.2 = 1, then there is a unique value of x in the half open interval (0, 2r) such that sin x = >. and cos x = µ.. Hint: Let Xo be the unique number in the closed interval [0, ½r] whose sine is l>1, and choose for x the appropriate number according to Ex. 12 : Xo, 1r  Xo, r + Xo, 2r  Xo. •H. Prove that 1 ~ sin x ~ 1 and 1 ~ COB x ~ 1 for all x. •16. Prove that sin x and cos x have, everywhere, continuous derivatives of all orders. . sin x •16. Prove that hm   = 1. Z0
Hint: Let f(x)
E!!
X
sin x.
Then f'(0) • lim f(O
+ hk 
f(O)_
0, f(O}
s
h....O
••17. Prove that the function f(x)
sin x, x
e
;,4
X
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 (ii..) If
(i) If
J ~ = ln x +
C, and if In x
> 0, how does one evaluate the simple integral f3
J22=2n+ 1 I + a x a ax a
dx
x
C an d
f x a 2
dx 2=2n+ 1 I x  a a x a
2
dx? X
+ C' '
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
(a+
a)
1 x x 1 2 a In ~ · x + a + C + C' or 2a In ( 1)
+C+
C'
which does not even exist! The answers to these questions lie most simply in the use of absolute values. We know that if x
> 0, then :fx (ln (x))
then dd (ln (x)) = _!_ (1) = X
X
!. X
=
~
and that if x
< 0,
That is, ln x and In (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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UNIVERSITY OF MICHIGAN
§ 605]
SOME INTEGRATION FORMULAS
195
tion In lxl encompasses both and is defined for any nonzero x. single integration formula
f~
(1)
= ln lxl +
Thus the
c
is applicable under all possible circumstances. The integration of question (i) is therefore simple:
J 2d:,; 3

=
X
[ln lxl ]2 = In l21 3
In l31 = In
J.
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 +  = 2a la+ a x ax f1xa1 2a +a f
(2)
2
dx
1 ln 
2
C·
'
dx 2 l1n   + C 2 
(2')
x

a
x

·
The seeming paradox is resolved by the fact that I 11 = 1, whose logarithm certainly exists (although it vanishes). More generally, any of the standard integration formulas that involve logarithms become universally applicable when absolute values are inserted. The student should establish this fact in detail for the following formulas (Ex. 12, § 606) . (In formulas (7)(14), a represents a positive constant.) (3) (4) (5) (6) (7) (8)
f tan =  In Icos xi + = ln 1sec xi + f cot = ln lsin xi + .C = In Iese xi + C; f sec = ln !sec + tan xi + f csc x = In Iese x  cot xi + f v'xt ± = ln Ix+ v'x ± a 1+ lxl > for the  case; f v'x ± a = ½[xv'x ± a ± a In Ix + v'x ± a I] + lxl > x dx
C
C;
x dx
x dx
C;
x
dx
C;
2
2
dx
C,
a2
2
2
2
dx
2
2
a
2
2
C,
for the  case; (9)
f :ry'
2
dx 02
±
x2
2
=!In Iv' a ± x a x
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a
SOME ELEMENTARY FUNCTIONS
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[§ 606
Since the derivative of sec x tan x is sec• x
+ sec x tan x = sec• x + sec x (sec2 x 2
and since by (5), above, the derivative of have, by addition,
f sec• x
(10)
dx
= 2 sec• x  sec x, In 1sec x + tan xi is sec x, we 1)
= ½sec x tan x + ½ln 1sec x + 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 ...;at _ xi = Arcsin~a + f a + = a Arctan a + dx
(11)
dx
(12)
1
J...; a
2
(13)
1
f xv'x
2
'
11"
C 
'
2
< a'
_! ~ Arcsin~ ~ !: . 2 a  2'
r < Arctan Xa < 2' ·
x2 dx = ½xVa2  x2 + ½a2 Arcsin ~ + C, lxl < a;

=
dx
(14)
X
x2
C a< x
{¼
2

a
+ C, x > a, 1 Arcsm . (a) a ;; + C, X < a. Arcsin (;)
606. EXERCISES
In Exercises 110, perform the integration, and specify any limitations on the variable x.
f f dx f dx f ~==· 'f + 1.
tan 5x dx.
3 •
v'2  x1 •
6
dx v'4x2 4x+5·
•
7.
9.
xv'5  2x1
dx '• 3x1 5x  7
f ' f f f ee:?dx. f + +
2.
sec 4x dx.
v'x1 + 4x dx.
6.
v'x  3x1 dx.
8.
lO.
3x1
~x
1·
11. Show that in formula (2), § 605, the absolute value signs can be removed if lxl < 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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§ 607]
HYPERBOLIC FUNCTIONS
197
differentiation of the righthand 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"  ez 1 , smhx= cothx =   , 2 tanhx e"' + ez 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):
= 1;
(2)
cosh2 x  sinh2 x
(3) (5)
= sech2 x; coth2 x  1 = csch2 x; sinh (x ± y) = sinh x cosh y ±
(6)
cosh (x ± y) = cosh x cosh y ± sinh x sinh y;
(7)
tanh (x ± )
(8)
sinh 2x = 2 sinh x cosh x;
(9)
cosh 2x = cosh2 x
(4)
2
1  tanh x
y
=
cosh x sinh y;
tanh x ± tanh y , 1 ± tanh x tanh y'
+ 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; = sinh x; d(tanh x)/dx = sech2 :r; d(coth x)/dx = csch x;
(14)
d(sech x)/dx = sech x tanh x;
(15)
d(csch x)/dx
(10) (11) (12)
d(sinh x)/dx
d(cosh x)/dx
2
= csch x coth x .
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[§ 607
SOME ELEMENTARY FUNCTIONS
198
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 J J
tanh x dx = In cosh x
(16)
coth
(17)
x dx = In jsinh xi + C;
sech x dx
(18) (19)
+ C;
= Arctan (sinh x) +
csch x dx = In ltanh
~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 x 2 + y2 = a 2 can be written x = a cos fJ, y = a sin fJ. In analogy with this, the hyperbolic functions are related to a hyperbola. For example, the parametric equations of the rectangular hyperbola x 2  y 2 = a2 can be written x = a cosh 0, y = a sinh fJ. In this latter case, however, the parameter fJ does not represent the polar coordinate angle for the point (x, y). y
1 y=sinhx
y=coshx
y
1 X
'l=tanhx
y
=
FIG. 605
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coth x
§ 608]
INVERSE HYPERBOLIC FUNCTIONS
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 a.re the most useful in this connection a.re the first four, sinh x, cosh x, ta.oh x, and coth x. The graphs and notation a.re 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 cosh1 x, and write for these principal values Cosh1 x. y
y
y
y
I
I I
I
,1
1
I I
I
I I I I
1/
= sinh1 X all real z
I
y Coeh•x z ~1
'J
= tanh1 X lxl < 1
l I
X
I
I
l
X
11
I I I
I
I
I I
I
I
y
= coth1 x lzl> 1
FIG. 606
The inverse hyperbolic functions can be expressed in terms of functions discussed previously:
= In (x + v'x2 +
(1)
sinh1 x
(2)
Cosh1 x
= In (x + ~ ) , x
(3)
tanh1 x
= ½In
!: :; lxl
x 1
1.
1), a.II real x;
~ 1;
We prove (2) and leave the rest for the student (Ex. 25, § 609) . If
x
11 41 = e +2 e , let
the positive quantity e11 be denoted by p.
Then
1 = p + , or p 2  2xp + 1 = 0. Solving this equation we p p = x ± ~  Since y is to be chosen nonnegative, p = e" ~ 1.
2x
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get If
200
SOME ELEMENTARY FUNCTIONS
[§ 609
1, x + ~ > 1, and since the product of this and x  v'7"=I is equal to 1, x  v'x 2  1 < 1. We therefore reject the minus sign and have p = x + v'x 2  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  x 2 > 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: x
>
+
(5)
d sinh 1 x = ~ '2 all real x; dx 1+x
(6)
Cosh1 x =. ~ •x dx vx2  l
(7)
1:_ tanh1 x =   ,2
lxl
1.
1
d
1
dx
(8)
1x
>
1;
'
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 righthand members (Ex. 28, § 609). The letter a represents a positive number throughout.
dx
sinh 1 ~
,=dx==
Cosh1 ~
f v'a2 + x2 = f v'x2 _ a2 = f a2___x_2 = a f x2 _ a2 = a
(9) (10)
dx
(11)
dx
(12)
1
a
tanh1 1
+C
a
'
all real x·
'
+ C, x > a;
a+ C, lxl < a; X
X
coth1 ~
+ C, lxl > a.
609. EXERCISES
In Exercises 110, differentiate the given function. 1. cosh 3x. 2. sinh2 x. 3. tanh (2  x). 4. x coth x2 • 6. In sinh 2x. 6. e"" cosh bx. 7. sinh1 4x. 8. Cosh1 es. 9. tanh1 x2 • 10. coth1 (sec x).
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NUMBERS AND FUNCTIONS
§ 610]
201
In Exercises 1120, perform the indicated integration, expressing your answer in terms of hyperbolic functions or their inverses. 11. 13. 16.
17.
f f f f f
tanh 6x dx.
12.
cosh3 x dx.
14.
sinh2 x dx.
16.
f f f f f
e" coth e" dx.
tanh2 lOx dx. x• tanh1 x dx.
dx . 18. dx . ..lx2  2 V4x 2 4x+5 dx 19 20 1  5!x  3x2' • • 3x2 + 5x  7 2 (7  5x  3x2 > 0) . (3x + 5x  7 > 0). 21. Prove that cosh x and sech x are even functions and that the other fout hyperbolic functions are odd (cf. Exs. 78, § 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_ _!E£.._ _ coshx  1 _ 2sinh2 ½x. . h h +  2 , where u  cosh x, and h .1 Sill X U2  1 COS X 1 COS 2 ,X 26. Establish formulas (1), (3), (4), § 608. 26. Establish formulas (5)(8), § 608, by use of (10)(13), § 607. Hint for (5): Let y = sinh1 x . Then x = sinh y, dx/dy = cosh y, and dy/dx = 1/Vl + sinh' 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
esch x dx
= coth1 (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) aoxn + a1xnl + ... + a,. = 0, where the coefficients ao, ai, • • • , a,. are integers. Examples of algebraic numbers are ¾ (a root of the equation 4x  3 = 0) (in fact,. every rational number is algebraic),  ~ (a root of the equation x' + 5 = 0), and~+ V3 (a root of the equation x'  9x4  10x8 + 27~2  90x  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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202
[§ 610
quotient of algebraic numbers is algebraic, and that any (real) root of an algebraic number is an algebraic number. Therefore any number (like
V 3  v1s/9 / 4V'61) 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, Elementary Mathematics from 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 transcrendental 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 /(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 function is any function that is not algebraic. Examples of polynomials are 2x  V3 and rx' + 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 Vx and 1/V x' + r. Examples of transcendental functions are ez, In x, sin x, and cos x. (Cf. Exs. 1113, § 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 Theory (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 el~ mentary calculus, like sin 2x, e~, and ½Arctan ½x, are examples of what are called ekmentary functwns. In order to define this concept, we describe first the elementary operations on functions. The elementary operations on functions f(x) and g(x) are those that yield any of the following: f(x) ± g(x), f(x) g(x), f(x)/g(x), {/(x)} 0 , a1 ½ Finally, since :En• converges for every r < 1, we have only to chooser between ! 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.....,+oo
= p,
where O ~ p ~ +oo . Then (i) if O ~ p < 1, :Ea,. converges; (ii) if 1 < p ~ +oo, :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 Since lim r,. n+oo
< 1. = p < r, we may choose a neighborhood of p that excludes r.
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[§ 709
INFINITE SERIES OF CONSTANTS
216
Since every rn, for n greater than or equal to some N, lies in this neighborhood of p, we have: n ~ N implies r,. < r. This gives the following sequence of inequalities:
aN+a aN+2
< r, or aN+a < raN+s < r'aN, ••••
•, , , •
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 Lan 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 rn ~ 1 whenever n ~ N. In other words, (3)
n
~ N implies
az
1
~ 1, or an+l ~
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 i:a,. diverges.
Proof of (iii). For any pseries the test ratio is r,. and since the function x" is continuous at x lim rn = lim (
n ....
+..
n ....
+.. n
= a~:1 =
= 1 (Ex. 7,
(n ~ 1)1',
§ 602),
+n 1)" = [ lim +n 1]" = l" ...... + .. n
= 1.
If p > 1 the pseries converges and if p ~ 1 the pseries diverges, but in either case p = 1. 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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UNIVERSITY OF MICHIGAN
§ 709)
THE RATIO TEST
217
a,.+i ;:ii r < 1 holds for all 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 fact, if r,. i1:;; 1 for all 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 fail, 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 r,,._1 = 2"/311 + 0 and r2,. = 3"/211 +1+ +oo. NOTE 4. The ratio test provides a simple proof that certain limits are zero. That is, if lim a,.+ifa,., where a,. > 0 for all n, exists and is less than 1, then a,.+ 0.
..+•
This is true because a,. is the general term of a convergent series. X" = 0, for every real number x . ..+• 1n.
Prove that lim
Example 1.
Without loss of generality we can assume x
Solution.
xn+I
values).
Then
Example 2.
. Sol ution.
(n
X"
Use the ratio test to establish convergence of the series 1 1 1 1 1 + 11 + 2 ! + ai + ... + (n  1) I + ... '
s·mce a,. = (n =
Eltample 3.
(take absolute
+ l) ! + ;i = n + 1 + 0.
r,.
Therefore p
>0
X
lim r,.
n+•
_1 l) , a,.+1
1
1 and = nl
= a,.+1 = (n a,.
=0
N, does guarantee convergence. NoTE 2. For divergence it is sufficient to have an inequality of the form {1/a,. ~ 1, for n > N, since this precludes lim a,. = 0. n++ oo
NoTE 3. The ratio test is usually easier to apply than the root test, but the latter is more powerful. (Cf. Exs. 3435, § 711.) Example.
Use the root test to establish convergence of the series
1+;+~+i+ .... Solution.
Since {1/a,. =
n /
\J 2
::,_ 1 =
~~1 ,
the problem of finding
2 n
can be reduced to that of finding the two limits
lim {1/;;: and
n++oo
lim {1/;;:
n++•
1.! Jim 2 "·
n++oo
The second of these is not indeterminate, owing to the continuity of the func1
tion 2" at x = l, and has the value 21 = 2. let y
=
To evaluate lim {1/;;:, or lim x;, n++ co x++ co ! In x xx, and take logarithms: (cf. § 416): In y =  · Then, by !'Hospital's
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UNIVERSITY OF MICHIGAN
§ 711]
Rule
EXERCISES (§
219
lim In y = lim 1/ x = 0 and y x+oo ..+ .. 1 ' = ½ < 1, and La,. converges.
415),
lim {1/ a,.
n.+oo
+
e0 = 1.
Therefore
711. EXERCISES
120,
In ExerciseE'
+ v2 + va 1' 1 3 5 7
establish convergence or divergence of the given series.
+ vi+ ... 9
.
2.
1 + 1 + 1 + 1 + .... v1 2 v2.3 v'3.4 v'4.5
3•
2·4 + 4·6 + 6·8 + 8· 10 + ....
va
V5
' v2 + 1 + 33  1 1 1·2 6• 3 + 3.5 + 6• 7 • 8•
9. 11.
16.
V9
v'a + 1 + .vi + 1 + vs + 1 + .... 3 1 1
4  1 5  1 6 1·2·3 1·2·3·4 3.5.7 + 3.5.7.9 + ....
1!
2I
3!
4!
2I
3I
4I
5I

1
25+25+27+ 28 + ··· · 1 1 1 1 In 2 + In 3 + In 4 + In 5 + · · · · 41+51+6i+71+ .... + .. Vn
L '+ 4·
10.
n1n
+..
L
+"'
L
n•l
11 '.
tv~v;;_ n"'
n•l
+•
L
Inn _,1
+1
32111
12. ,..1n L ~+1· +.. 1 l' L   · n•2 (Inn)"
1
e
+•
n1nvn
+•
4
.!!....
n1n
13.
V7
1
18.
17• ..~2 (In n)Q. +oo
1s. n•l L (\1/n 
16.
+..
!!..
+ ..
1
L ,a> n1l+a" +oo
1) 11 •
20.
I
L
n1n"
L
n•l
r" jsin n al, a
1.
> 0, r > 0.
21. Prove the comparison test (Theorem I, § 708) for divergence. 22. Prove Theorem II, § 708. 23. Prove the root test, § 710. 2' Prove that if a,. ~ 0 and there exists a number k > 1 such that lim nta,.
exists and is finite, then
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INFINITE SERIES OF CONSTANTS
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25. Prove that if an
~
0 and
[§ 711
lim nan exists and is positive, then
n~+oo
La.
diverges. •26. Prove the Schwarz (or Cauchy) inequality for nonnegative series: If a,. ~ 0 and b,. ~ 0, n = 1, 2, · · · , then
+., L anb,. ( n1
(I)
)2 ~ +ao +ao L a,. L b,. 2
n1
1,
n1
with the following interpretations: (i) if both series on the righthand side of (l) converge, then the series on the lefthand side also converges and (1) holds; (ii) if either series on the righthand 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 l:a,.2 is a convergent series, then L la..1/n is also convergent. (Cf. Ex. 26.) +.. d •28. Prove that any series of the form L 1111, where d,. "" 0, 1, 2, • • • , 9, n•l 10
converges. Hence show that any decimal expansion O.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 O.d,dt · · · 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 onetoone correspondence between the points of the halfopen interval (0, l] and all nonterminating decimals O.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 nonterininating decimals O.d,dt · · · is nondenumerable. Finally, prove that the set Sis nondenumerable by assuming the contrary and obtaining a contradiction, as follows: Assume Sis made up of the sequence {z,.}: z1: .du d12 du du z,: .dt1 du du du z,: .du dn daa du Construct a new sequence {a,.}, where a,. is one of the digits 1, 2, · · · , 9, and ;,I: d,.,., for n = I, 2, • • • . Finally, show that the decimal O.a,a,a, · • · simultaneously must and cannot belong to S. •31. Establish the following form of the ratio test: The positive series l:a..
a,.
converges if
ITiii !'.!.!!±! < 1, and diverges if lim !!!!±! >
n+.. a,.
n+• a,.
1.
(Cf. Ex. 16,
§ 305.)
•32. Establish the following form of the root test: If l:a.. is a nonnegative series and if u = ITiii ~. then l:a,. converges if u < 1 and diverges if u
>
n+ao
1.
(Cf. Ex. 31.)
•33. Apply Exercises 31 and 32 to the series~+!+~+~+ · · · of Ex
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UNIVERSITY OF MICHIGAN
§
712]
MORE REFINED TESTS
221
ample 2, § 708, and show that ITiii ~ = +co, lim ~ = 0, ITiii C'a,. = 1/V2 a,. an . ' and lim C'a,. = l/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 ~ exists, then lim +.. a,. n++ ..
.e;;,: also exists and is equal
Hints: For the case lim a,.+i = L, where O < L < +co, let a and fJ be n++oo a,. arbitrary numbers such that O < a < L < fJ. Then for n ~ some N, a a,. < a,.+1 < fJ a,.. Hence to it.
a 2
a aN
Thus, for n
aN
< a aN+I
< aN +1 < fJ aN < aNH < fJ aN+I < fJ2 aN
> N,
and
~ ~ {·:=~} ~ fJ.
n++oo *35. The example 1, 2, 1, 2, • • • shows that
lim .e;;,: may exist when n++• Find an example of a convergent positive series :Ea,. for
lim ~ does not. n++• a,., which this situation is also true.
*36. Prove that lim " ~ = e. n++• '\j;j
(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 ~ n++• a,. Theorem I. Kummer's Test.
= 1.
Let :Ea,. be a positive series, and
be a sequence of positive constants such that (1)
......+..
[p . a..+1 a,. 
r. 1mp1esp,.a,.+i sequence of inequalities :
For any positive integer m, then, we have the
PNaN  PN+1aN+1
> raN+i,
PN+1111aN+•  PN+maN+111 > raN+m• Adding on both sides we have, because of cancellations by pairs on the left: PNaN  PN+maN+... > r(aN+i + · · · + aN+m). Using the notation S,. for the partial sum a1 + · · · + a,., we can write the sum in parentheses of (2) as SN+...  SN, and obtain by rearrangement of terms : rSN+111 < rSN + PNaN  PN+maN+m < 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 l:a,. are bounded and hence (§ 705) the series converges. · (2)
To prove the second part we infer from the inequality p,. ..!!!!....  p,.+1 ~ 0, an+I which holds for n ~ 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
> N,
L ..!., that l:a,. also diverges. p,.
Theorem II. Raabe's Test. Let l:a,. be a posi,tive series and assume that lim n ( a,. ,....+.. a,.+1
1) = L
exi8t8 (finite or infinite) . Then (i) if 1 < L ~ +oo, l:a,. converges; (ii) if oo ~ L < 1, l:a,. diverges. (iii) if L = 1, l:a,. may either converge or diverge, and the test fail8.
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UNIVERSITY OF MICHIGAN
§ 713]
Proof. p,.
223
EXERCISES
= n.
The first two parts are a consequence of Kummer's Test with (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 I:a,. be a positive series and assume that lim
n+•
lnn[n(~  1) 1] L a..+1
exists (finite or infinite). Then (i) if l < L ~ +oo, I:a.. converges; (ii) if oo ~ L < l, I:a,. diverges; (iii) if L  l, I:a,. may either converge or diverge, and the test fail8. Proof. The first two parts are a consequence of Kummer's Test with p,. = n n. (Cf. Exs. 910, § 713, for further suggestions.)
ln
Example 1. Test for convergence or divergence:
(!)P 2 +
(U)p (!..:.!:,.Q)P ... . 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,.  t )  n [ ( 2n ),.l] 2n .O+z)P1 a11+1 2n  1 2n  1 2z
1
where z a (2n  1)1• The limit of this expression, as n+ +co, is (by !'Hospital's Rule) : lim (1 + z)P  1 = lim p(l + z)1>1 = P. ~ 2z ~ 2 2 Therefore the given series converges for p > 2 and diverges for p < 2. For ~he case p = 2, see Example 2.
EJ:am.ple 2.
Test for convergence or divergence
!.:l )' + ( 2•4·6 !..:.!:.§ )' + (!)' 2 + (2·4
•.. .
Solution. Both the ratio test and Raabe's test fail (cf. Example 1). paring to use Theorem III, we simplify the expression
(...!!!!.. _ 1 )
_ 1 =n 4n  l _ 1 = 3n  1 . a,.+1 4n'  4n + 1 (2n  1)1 1 . 1·1m In n ( 3n  l) . . d"1verges. Smce = 0 < 1, th e given series 2 n+• n  2
n
*713. EXERCISES
In Exercises 14, test for convergence or divergence. ~
2· 4·6 • •· 2n
•1. ,.':"1 l · 3. 5
... (2n
Digitized by
+ I).
(Cf. Ex. 4.)
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UNIVERSITY OF MICHIGAN
Pre
[§ 713
INFINITE SERIES OF CONSTANTS
224
*2· E 13 ... c2n  o. _1__ n 1
2 · 4 · · · 2n
2n + 1
E 1·3 • • • (2n  1). 4n + 3_ 2 · 4 · · · 2n 2n + 2 *4• E [ 2 · 4·6 • • • 2n ],,· 1·3·5 · · · (2n + 1) •a.
n 1
n1
*6. Prove Theorem II, § 712. terms of
+~
l: a,. n1
Hint: For examples for part (iii), define the
inductively, with a1
= 1, and a,.+1 .l!!!.... = 1 + ! + n
k
 n 1n n
Then
use Theorem III, § 712. *6. Prove that the hypergeometric series 1 + ~ + a(a + 1)(3@ + 1) l•y 1·2•y·('Y+l)
+ ...
+ a(a +
1) ···(a+ n  O·t3@ + 1) · · · (8 + n  1) + ... n ! 'Y('Y + 1) · · · (y + n  1) converges if and only if 'Y  a  fJ > 0 (n not O or a negative integer). •7. Prove Gauss's Test: If the positive series l:a,. is such that
(!.. ),
~ = 1 + !! + 0 2 a,.+1 n n then l:a,. converges if h > 1, and diverges if h ~ 1. Show that 0(1/n2) could be replaced by the weaker condition 0(1/n°), where a > 1. *8. State and prove limit superior and limit inferior forms for the Theorems of § 712. (Cf. Exs. 3132, § 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 Inn) that Inn [n (.i!!!.... a,.+1
1) 1]  Sn Inn .l!!!....  (n + 1) In (n 1 a.+1
+ l)t+ 1, f
or (n + 1) In (n + 1) 
[n Inn
+ Inn] = (n +
1) In ( n
~ 1 )+ 1.
**10. Prove the following refinement of Theorem III, § 712: Let positive series such that
Ea.
be a
Iim In Inn S1nn[n(.i!!!....1)1] 1l = L 1 a,.+1 f exists (finite or infinite). Then (i) if 1 < L ~ +ao, l:a., converges, and (ii) if ao ~ L < 1, l:a,. diverges. Hint: The problem is to show (cf. Ex. 9) that (1) (n + 1) In (n + 1) In In (n + 1)  {n Inn In Inn +Inn In Inn + In Inn} + 1. To establish this limit, show first that if E,.+ 0, then In (1 + E,.) = E,. + O(E,.2) (cf. Ex. 21, § 408), and, more generally, if {a,.} is a sequence of positive numn++~
bers and if E,./ a,.+ 0, then In (a,. + E,.)
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= In
(a,.) +
~+
a"
2 0 (E" 2)•
Original from
UNIVERSITY OF MICHIGAN
a:"
Thus
225
SERIES OF ARBITRARY TERMS
§ 714]
In (n + 1) =Inn+~+ 0 (~) and In In n = In [ In n +
~+
0 ( ~2 ) ] = In In n + lnl n [ ; + 0 ( ~2 ) ]
+ 0
([!n + 0 (!...)] ~
2 )
= In Inn + 
1
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 bas terms of one sign, as we have seen for nonnegative series, there is only one kind of divergenceto infinityand 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 negativeor 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)
Ci 
where Cn
C2
+ Ca 
c,
+ ··· ,
> 0 for every n.
Theorem. An alternating series (1) whose terms satisfy the two conditions (i) Cn+1 < Cn for every n, (ii)
Cn+
0 as
n+
+oo,
converges. If S and Sn denote the sum, and the partial sum of the first n terms, respectively, of the series (1),
is ..  si < cn11
(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 S2,._1 (consisting of an odd number of terms) form a decreasing sequence. C. For every m and every n, S2m < S2,._1. D. Sexists. E. For every m and every n, S2m < S < S2,._1. F. The inequality (2) holds.
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INFINITE SERIES OF CONSTANTS
226
A: Since S1,....1
= S2,. + (c, ....., 
B: Since S2,.+1
= S2,.1 
s,....., > s,,.. Stn+l
< S2,._,.
c,,....2), and since es,....,
(c2,.  c2,....,), and since C2n+1
[§ 715
< c,,....,, < es,.,
C: If 2m < 2n  1, S,,._1  s,,,. = (c2w1+1  c2...+2) + · · · + (c2,._,  es,._,) + es,._, > o, and S2m < S2,._,. If 2m > 2n  1, S2.,.  S2nl = (Ctn  Ctn+!)  • • •  (Ct...2  C21)  C2.. < 0, and S2m < S2nl•
s, s. . . . s.,.
S
S1n1
• • •
S,
s,
FIG. 702
D: From A, B, and C it follows that {S2,.} and {S2,._1} are bounded monotonic sequences and therefore converge. We need only show that their limits are equal. But this is true, since
lim S2..+1  lim S2,. = lim (S2n+1  S2,.) = lim es,....1 = 0. n~+• n➔+• n++• n++oo E: By the fundamental theorem on convergence of monotonic sequences (Theorem XIV, § 204), for an arbitrary fixed n and variable m, Si.. < S2,....1 implies S ~ S2n+1 < S2,._1. Similarly, for an arbitrary fixed m and variable n, S2,,a+2 < S2,....1 implies St... < S2,n+2 ~ S. F: On the one hand, 0
< St,._!
S

< S2,._1 
S2n
= C2,.,
while, on the other hand 0
m >N
*18. Prove that an infinite series ing to Ja,.
>
E
>
+•
L
n1
+ ..
•19. Prove that an infinite series E
implies la..
L
n1
E
a,. converges if and only if correspond
0 there exists a number N such that n (Cf. Ex. 17.)
+ a,.+1 + · · · + a,.+PI < E.
ing to
> 0 there exists a + a ..+1 + · · · + a,.I < E. (Cf.
a,. converges if and only if corresponding to
>
N and p
>
0 imply
a,. converges if and only if correspond
0 there exists a positive integer N such that n
>
N implies
+ aN+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 : Jim (a,. + ••• + a,.+P) = 0 for every p > 0. (Cf. Ex. 4, n+oo iaN
§ 305, Ex. 43, § 904.) •21. · Use the Cauchy criterion of Exercise 17 to prove that an absolutely convergent series is convergent.
+oo ••22. Prove the following Abel test: If the partial sums of a series L a,. are n1
bounded and if {b,.} is a monotonically decreasing sequence of nonnegative numbers whose limit is 0, then La,.b,. converges. Use this fact to establish the convergence in the alternating series test as phrased in Exercise 11. Hint: Let S,. = a1 a,. , and assume IS,.I < K for all n. Then
+ ·· · +
,m (S; l,.tma;b;I = l.t
S;1)b;I
=
ln:i:l ,m S;(b; 
b;+1)
+ S,.b,.
 s,,._,b .. 1
**23. Prove the following Abel test: If La,. converges and if {b,.} is a bounded monotonic sequence, then La,.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  · · · ; (ii ) in groups of 1, 2, 3, 4, • • •: 1  ½  ½+ + +     · · · ; (iii) in groups of 1, 2, 4, 8, • • • : 1  ½  l + + + + · · · .
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UNIVERSITY OF MICHIGAN
§ 718]
GROUPINGS AND REARRANGEMENTS
Show that {i) and (ii) converge and (iii) diverges.
In Abel's test, Ex. 22, let a,. = ±
(Cf. Ex. 22.)
231
Hint for (ii):
);i and b,. = );;·
**25. If A ·is an arbitrary finite or denumerable set of real numbers, a1, a 2, 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:
n1
p,. be a conver
gent series of positive numbers with sum p. Define f(x) to be O if x is any lower l?ound of A, and otherwise equal to the sum of all terms p,.. of I:p,. such that a,.. < x. Prove the following five properties of /(x): (i) /(x) is monotonically increasing on (oo, +oo); (ii) Jim /(x) = O; (iii) Jim f(x) = p; ~  ao
(iv) f(a,.+)  /(a,.) not in A.
=
z++oo
p,. for every n; (v) f(x) is continuous at every point
718. GROUPINGS AND REARRANGEMENTS
A series I:b,. is said to arise from a given series I:a,. by grouping of terms (or by the introduction of parentheses) if every b,. is the sum of a finite number of consecutive terms of I:a,., 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 ;;; q. For example, the grouping (a1 + at) + (aa) + (a. +
= a1 + · · · + a,., S = L
(The sum is
.fo ex• dx. 1
Cf. Example 4, § 810.J
The sum S, is between 1.4625 and 1.4626. Since r,.+ p = 0, we
i°'~o~~~l
< 0.00013. 1 ~ 0 = 0.000107, and 1 ~ r 7 = 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
+ 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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INFINITE SERIES OF CONSTANTS
236
[§720
Theorem II. If f(x) is a positive monotonically decreasing function, + .. for x ~ a, if ~ an is a convergent positive series, with f(n) = an for n > a, n 1
if S
=
+ · · · + an, then for n > a, +.. i+ .. f(x) dx < S < S,. + f(x) dx. S,. + i +1 n
~On, and if Sn= a1
(6)
A much sharper estimate is provided by the following theorem (hints for a proof are given in Ex. 21, § 721) :
m.
**Theorem Under the hypotheses of The0rem I I and the additional assumption that f"(x) is a positive monotonically decreasing function for x ~ a, then for n > a, (7)
Sn+ !'.!!:I:! +i+ ..f(x) dx  f'(n
2
n+l
+ 2)
12
< S < S,. + an+l +i+"'f(x) dx 2
 f'(n)_
n+l
12
Example 2. Estimate the sum of the pseries with p = 2, using IO terms. Solution. With the aid of a table of reciprocals we find that S10 = 1.549768.
With f(x)
=
x2,
(+'°f(x) dx =
)11
n: =
0.090909 and
(+'°f(x) dx =
)10
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. ••Further computations give! au= 0.004132,  / f'(l2) = _~23 = 0.000096, 2 6 and  / !'(10) = _~ = 0.000167. ThustheestimateofTheoremlilplaces 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 = l +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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UNIVERSITY OF MICHIGAN
§ 721]
237
EXERCISES
slowly converging alternating series: Example 3.
Evaluate the alternating harmonic series, S=l½+½l+··· 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: X=n:n+nrn+ .... If we double, remove parentheses (the student should justify this step), and introduce parentheses, we find 2x = (h + tr)  (h + h) + (h + ts) 
··• =tr+trhh+nr+nr ··· = tr+ (tr  n)  (h  nr) + (nr  h)  ..• 1 1 1 1  11 + 11·12  12·13 + 1314  ....
Again doubling, and removing and introducing parentheses, we have 2 1 2 2 4 x = 11 + 11·12 + 11·12·13 12·13•14 + ·· ·' or 1 1 2 x = 25 + 12·13•14+···· 11·24 11·12·13 Once more: 3 + 11·12·13 1 + 11·12·13·14 3 12_·_1_3_·14_·_1_5 + .... 1112 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.) 4x
= __1Q_
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 +00, (ii) tend toward 00, (iii) tend toward 00 but neither +00 nor 00, (iv) are bounded and have no limit. 3. Prove Theorem I, § 719. •4. Prove that if :Ea.. and :Eb.. are nonnegative series with sums A and B, respectively, and if :Ee,. is their product series, with sum C, then C = AB under all circumstances of convergence or divergence, with the usual conventions about infinity ( ( +00) · ( +00) = +00, (positive number)· ( +00) = +00) and the additional convention O· ( +00) = 0.
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UNIVERSITY OF MICHIGAN
238
INFINITE SERIES OF CONSTANTS
+.,
•5. Prove that if
[§ 721
+.,
L a,. and n=O L b,. are both the series n=O 1
1 + = 
1  =
v2
+.,
v'3
1
=
v'4
+ ...
L c,. is their product series, then l:a,. and l:b,. converge, while I:c. n o diverges. Hint: Show that ic,.I ii;:; 1. •6. Prove Theorem I. § 720. •7. Prove Theorem II, § 720. •8. Prove the commutative law for product series : The product series of l:a,. and l:b,. is the same as the product series of l:b,. and l:a,.. •9. Prove the associative law for product series: Let l:d,. be the product series of l:a,. and l:b .., and let l:e .. be the product series of l:b,. and LC•. Then the product series of l:a.. and l:e,. is the same as the product series of l:d,. and l:c,.. •10. Prove the distributive law for multiplying and adding series: The product series of l:a,. and l:(b,. + c,.) is the sum of the product series of I:a. and l:b,. and the product series of l:a,. and l:c,.. •11. Using the evaluation of the series l:n2 given in the Note, § 720, show that
and if
r2
1
1
1
8 = l +32+52+72+ . .. .
•12. Using the evaluation of the series l:n2 given in the Note, § 720, show that r2 12
=
1
1
1
22 + 3'  42 + .. . .
1 
In Exercises 1318, compute to four significant digits. •13. (e =) 1
* 14.
(In 2 =)
1 + l! + 21! + 31! + ••• . 1
1
•17. (~
1
4
3
•15. (!(3) =) 1 5 •16. 22.32
1
2 + 2 · 22 + 3 •2 + 4 · 2 + · · · ·
+ 231 + 331 + 431 + .. . .
9
13
17
+ 42 . 52 + 62.72 + s2 . 92 + .. . .
=)
*18• ( In 32 =· )
1 
! + ½ t + · ·· .
1
1
2  2 · 22
1
+ 3·2
1
3 
4 · 24
+ ....
••19. Prove that the sequence {C,.}
l
= 1 + ½+ ••• + ~  In n f
is decreasing and bounded below.
Hence C =
Jim C,. exists. n++oo
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UNIVERSITY OF MICHIGAN
(The number
§
EXERCISES
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 E a,. converges absolutely to A and no +oo +oo
if E
n0
b,. converges to B, then the product series
E
n•O
c,. converges to AB.
Hint:
By virtue of Exs. 810, it may be assumed without loss of generality that Ea,. is nonnegative, A > 0, and B = 0. Under these assumptions prove that
C,. For a given therefore
E
n
n
n
E ak, B,. = k0 E bk, C,. = k0 Eck, Then k0 = aJ>o + (aJ>1 + a1bo) + · · · + (aJ>,. + · · · + a,.bo) = aoB,. + a1B,._1 + · · · + a,.Bo.
Ee,. converges to 0 : Define .4,. =
> 0, first choose N such that m > N implies IB,.I < E/2A, and
laoB,. +
Then choose N'
>
· · · + a,._N1BN+1I < ½E.
N such that n
la,._NI
N' implies
(N E+ l) · max
**21. Prove Theorem III, § 720.
(IBol, IB1I, · · · , IBNI, = a,.+1 ½a,.+ T,.+1
Hints: Let R,.
+ ½(a,.+1 + a,.+2) + ½(a,.+2 + a,.+3) + · •· =
lJ.
+ a,.+2 + •••
Interpret T,.+t as a sum of areas of trapezoids, and use the trapezoidal formula error estimate rm+l (Ex. 22, § 503) to write ½(a .. + a .. +1) = ]m J(x) dx + rd"(~ .. ), where ½a,.+1
m
h
< ~ ..
R.
al_ Therefore, by the ra

al
< R and diverges if
After the radius of convergence has been found, the endpoints of the interval of convergence should be tested. For such points the ratio test cannot give any information, for if a,.±1 R had a limit for R > 0, the limit a,. (5) would exist, and the absolute value of the test ratio must have the limit 1. To test the endpoints one is forced to use some type of comparison test, or a refined test of the type discussed in § 712. (The root test also fails at the endpointscf. Ex. 16, § 802.) *NOTE 4. A universally valid formula for the radius of convergence of a power series "E,a,.(x  a)" makes use of limit superior (Ex. 16, § 305):
R =
1
'
um {/1a..1
n++oo
with the conventions 1/0 = +oo, 1/ +oo § 802, and cf. Ex. 16, § 802.) Example 1.
= 0. (Give the proof in Ex. 18,
Determine the interval of convergence for the series x• :z;6 l + _x2 + 2! + 3! + · · · .
(Cf. § 807.) Solution. This should be treated as a power series in powers of x2 • The radius of convergence is lim (n + l) ! = lim (n + 1) = +oo. n! ,..... + .. Therefore the interval of convergence is (00, +oo ). R =
Example 2.
,..... +..
Determine the interval of convergence for the series (x 
(x1)' l)   2 
+ (x1)3  3  
(x1)•  4 
+ · ·· ·
(Cf. § 807 J
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§ 802]
Solution.
The radius of convergence is R
=
243 1
Jim n +
n
n++oo
= l, and the mid
point of the interval of convergence is x = l. We test for convergence at the endpoints of the interval, x = 2 and x = 0. The value x = 2 gives the convergent alternating harmonic series and x = 0 gives minus the divergent harmonic series. The interval of convergence is (0, 2]. Determine the interval of convergence for the series l x x 2 x3 x• x 6 x• x1 2+a+22+ai+:i+ai+24+34+ ··· ·
Example 3.
(Cf. Example 2, § 708.) Solution. The limit (5) does not exist. gence of the two series
!
2
However, the intervals of conver
+ ~2 + ~3 + • • • and ! + £2 + 3 + ... 2 2 3 3 3
t
are (Y2, V2) and (v'a, V3}, respectively. Therefore the given series converges absolutely for !xi < V2 and diverges for lxl ~ V2. The interval of convergence is (V2, V2}. Determine the values of x for which the series xe" + 2x2et.: + 3x3e3z + 4x 4ev + ...
Example 4.
converges. Solution. Either the ratio test or the root tei,t shows that the series converges absolutely for Ix e"I < l (otherwise it diverges). The inequality Ix e"I < l is equivalent to lxl < e", and is satisfied for all nonnegative x. To determine the negative values of x which satisfy this inequality, we let a = x, and solve the equation a = e« (a = 0.567, approximately). Then the given series converges if and only if x > a, and the convergence is absolute. 802. EXERCISES
In Exercises 110, determine the interval oi convergence, and specify the nature of any convergence at each endpoint of the interval of convergence.
1' l _ l2x!
+ (2x)
2
_
2!
(3x) 3 3!
+ .. .
.
7
x x' x 2. x  3 + 5  7 + · · · · 3
x2 x4 x6 3· 1 +27+4!+6!+ · · ·. 4. 1 + x + 2!x2 + 3!xx + 4!.r 4 + • • • . 6• (X + l)  (x +4 1 )2 + (x +9 1)3  (.r +16 1)• + .. • •
 2) 3 + 6• (X _ 2 ) + (x 3!
(x 
5!
2)• +
(.r 
7!
2)7 + ...
•
3 7_ (ln2)(~  5) + (ln3Hr  !i)2 + (ln4\(.r  !i) + ....
V2
~3
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[§ 803
8. (x  1) + (x ~ 1)3 + (x ~ I)•+ (x ~ 1)7 + ....
*9• 1 *10•
X
1
2x +
1 ·3 2 2·4 x

1 ·3·5 1 ·3·5·7 2 · 4 •6 x3 + 2 4 6 · 8 x'  · • • •
x1 1 · 3 x• 1 •3 · 5 x7 + 6 + 2•4. 5 + 2•4 •6 7 + . • • •
11. Determine the values of x for which the series
_1_+ 1 + 1 + 1 +.;; 3 2(.x  3) 2 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 . sin• x  sin x + smx  sin 3x + 57..• • X 
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 Jim tia,. exists (replacing
um by Jim). •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 ~ c,. ~ c,.+1 < d,.+1 ~ 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. (CC..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. 3133, § 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)
f(x)
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 a)
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TAYLOR SERIES
We now differentiate termbyterm, as if the infinite series were simply a finite sum: f'(x) = a1 + 2at(x  a) + 3aa(x  a) 2 + • • • • Again : f"(x) = 202 + 2·3aa(x  a)+ 3 ·4a.c(x  a) 2 + • • •. And so forth : f"'(x)
= 3 I aa +
2 ·3 •4a4(X  a)+ 3 ·4·5(a) vergence) equality in (3), we define the series I: (x  a)"t on the n•O n 1 righthand side of (3) as the Taylor series for the function /(x) at x = a. +• J(z) • /(z), and recall that Ol  l. Although for z  a and n  0 an indeterminancy 00 develops, let us agree that in this instance 00 shall be defined to be 1.
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[§ 804
POWER SERIES
The answer to this question is a major concern of the remaining sections of this chapter. (ii) For a given power series l;a,.(x  a)", converging to a function f(x) in an interval with positive radius of convergence, is the Taylor series equation (3) true? We answer this second question affirmatively, now, but defer the proof to the next chapter (Ex. 13, § 911): Theorem I.
If
+ ..
L
n•O
a,.(x  a)" has a positive radius of convergence, and
if f(x) = l;a,.(x  a)" in the interval of convergence of the series, then throughout the interior of that interval f(x) is continuous and has (continuous) derivatives of all orders, and relations (2) hold:a,. = J(0) = 1, and the Maclaurin series is 11
= 0, 1, 2, · · · ,
11
x2
(1)
1
xa
x"
+ X + 2! + 3! + •· • + n ! + ••• .
The test ratio is~. whose limit is 0. Therefore the series (1) converges abn aolutely for all x, and the radius of convergence is infinite: R = +ao . To show that the series (1) represents the function e"' for all real x, we choose an arbitrary x ¢ 0, and use the Lagrange form of the remainder: R,.(x)
(2)
et•
= n ! x",
where~ .. is between O and x. We observe first that for a fixed x, ~.. (although it depends on n and is not constant) satisfies the inequality~.. < lxl, and therefore eE· is bounded above by the constant e1"' 1• Thus the problem has been reduced to showing that lim x", = 0. But x", is the general n++oo n . n. term of (1), and since (1) has already been shown to converge for all x, the general term must tend toward 0. II. The sine function, f(x) = sin x. The sequence {f 1). Assume 1 ~ x < 0. Then the Cauchy form of the remainder is (14)
where x
R,.(x)
= nx (x

< ~n < 0. We rewrite
~..) .. 1 ( ~ ) (1
+ ~..)'"",
(14):
)n1 R,.(x) = n ( ~) ;r"(l + ~ .. )'"1 (: : ~ . ~
(15)
~X < 1 +~. (check this), so that the last factor of (15) is less than 1 for n > 1. Also, if m > 1, the inequality 1 + tn < 1 implies (1 + ~,.)mI < 1, while if m < 1 the inequality 1 + ~.. > 1 + x implies (1 + ~..)m 1 < (1 + x)m 1• ln any case, then, the For 1
~
x
< ~.. < 0, 0
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 +oo
E b,.x" be two power Berie8 repre8enting the function8 f1(x)
no
+•
E
n0
a,.x" and
and f2(x), reapec
lively, within their interval8 of convergence, and let 'Y be an arbitrary comtant. Then (i) the p'ower BerieB E"(a..x" repreBents the function 'Yf1 (x) throughout the interval of convergence of Ea..x"; (ii1 the power Berie8 E(a,. ± b,.) x" repreBents the function f1(x) ± f2(x) for all points common to the interval8 of conn
vergence of the two given power BerieB; and (iii1 if c.. = E a,.b..J:, n = 0, 1, 2, · · · , then the power Berie8
k0
+•
E c,.x" repreaents the function f1(x) f2(x) for
no
all points interior to both interval8 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. Example 1. Since the series for
2
e"
is I
+ x + ; ! + •· · , the series for e"
is found by substituting x for x: e" = I  x + x2  x' 2!
3!
+ ••• .
This series expansion is valid for all real x, and is therefore the Maclaurin series fore". E:s:ample 2. The Maclaurin series for cos 2x is cos2x = I  ~ + 0 can be found by writing In a+ In [ 1 + x ~ a]
The Taylor series for In x at x = a
In x = In [a+ (x  a)] =
x  a  (x  a)' (x  a)' =Ina+ +~    .... a 2a' 3a3 This is valid for O < x :a 2a. Example 8.
The Taylor series for x• at x = a
x• =[a+ (x  a)]•= a•[1 + (x
~
> 0 can
be found by writing
a)]"'
l
= a• [ l +me~ a)+ m(~ ~ 1) (X ~ ay + .. · This is valid for O < x < 2a; also at O for m > 0 and at 2a for m > 1. Example 9. The Maclaurin series for es sin ax is found by multiplying the series: [ 1 + x + ;' I + ; ! + · · ·] [ ax If we wish the terms of degree
( 1+ x + = ax +
x'
2
+
x1
6
ax' +
~

~~• + ~~• 
···
J.
5, we have
x' ) ( a3x3 a1x 1 ) + 24 + · · · ax  6 + 120 + · · ·
~ (3  a 2)x' + ~ (1  a 2)x' +
6
6
.J!_
·~
(5  10a2 + a')x1 + • • •
'
809. SUBSTITUTION OF POWER SERIES
Sometimes it is important to obtain a power series for a composite function, where each of the constituent functions has a known power series.
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256
The most useful special case of a general theorem for such substitutions is the principal theorem of thi 0, there exists a finite sequence of terms of (1) whose sum exceeds C  E. If M is the largest index of the rows from which these terms are selected, then
M
I: Cm must also exceed
C 
E.
Therefore
m1
+oo
I: c,,.
~
C
E
and, since E is arbitrarily small,
 n,1
tion with a preceding inequality this gives
+oo
I: c,,.
~
C.
In combina
m1
+ ..
I: c,,. = C.
m1
We now remove the assumption that Cmn ~ 0, and (by splitting the entire array into nonnegative and nonpositive parts in the manner of§ 716) immediately draw every conclusion stated in the lemma, except the equality +oo
I: Cm =
C.
But this equality follows from the fact that for any
m1
E
>0
there exists a number M such that the sum of the absolute values of all terms of (1) appearing below the Mth row is less than E (check the details of this carefully in Ex. 28, § Sll). This completes the proof. Theorem I. Substitution.
y
(1)
(2)
Let
= f(u) = ao + a1u + a2u2 + · · · , and u = g(x) = b1x + b2X2 + · · · ,
where both power series have positive radii of convergence. Then the composite function h(x) = f(g(x)) is represeni,ed by a power series having a positive radius of convergence, obtained by BUbstituting the entire serie.'! (2) for the quantity u in (1), expanding, and collecting terms:
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§ 809)
(3)
y
SUBS TIT UT ION OF POWER SER IE S
257
= h(x) = ao + a1(b1x + b,x2 + · · ·) + aa(b1x + •••)2 + •••
+ a1b1x + (a1b2 + Otb1 x + (a1ba + 202b1b2 + aab11) x3 + (a1b• + 202b1ba + Otb2 + 3aab1 b2) x4 + · ·· ·· · = ao
2
2
)
2
2
*Proof. We first exploit the eontinuity of g(x) at x = 0 (Theorem I, § 803) and observe that if x belongs to a sufficiently small neighborhood of x = 0 and if u is defined in terms of x by (2), then u belongs to the interior of the interval of convergence of (1), and the expansions indicated by (1) and the first line of (3) are valid. It now remains to justify the removal of parentheses and the subsequent rearrangement in (3) . For this purpose we shall insist on restricting x to so small an interval about x = 0 that + .. v = I: lb..x"I is inside the interval of convergence of (1). Then, as a conn1
sequence of the absolute convergence of all series concerned, we can apply the preceding lemma to the double array 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. INDOERMINATE 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 O" or "order of magnitude" concepts. (Cf. Ex. 10, § 713.) . d 1·1m 1  cos x · Fm 2 ~ X
E xample 1.
Solution. Since cos x
= 1  ~ + £_ 21
Example 2.
Solution.
4!
Find lim n{ln (n n.+t0
+ 1)
4)
1  cos x x2
'
=!
2
+ O(x
2)
+ !. 2
 Inn} .
We write the expression In (n
Therefore n{ln (n Example 3.
Solution.
t2 + O(x
=1
 •••
+ 1) 
+ 1)
Inn
 Inn}
Find Iim In ~l ~
= In { 1 + ~) = ~ + 0 (~}
=
1
+ 0 {!) 1.
+ t>.
e"
In (1 + x) _ x + O(x 2) eis  1  2x + O(x 2 )
=
l + O(x) 2 + O(x) [1 + O(x)J[½
+ O(x)] = 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 [ x++++ x8 x6 x7 (1) ln=2 ··· ] 1x 3 5 7 (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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265
We illustrate some of these techniques in the following examples: Eumple 1.
Compute the sine of one radian to five decimal places.
sin i . 0.000003.
Solution 1/9 I
0 belong to I. Then :E la,.! {!xi converges. Dirichlet's rearrangement theorem (Theorem II) of § 718 shows that {laol + la1IE + laM + · · ·} + {la1I + 2la2lt + 3la,IE2 + · · ·}x + · · · converges. Hence {ao + •••} + {a1 + •••}x + ••• converges. Show that it converges to f(x) by use of the Lemma, § 809. *6. Prove Theorem IV, § 815. **1· A theorem regarding inverse .functions, easily proved in the theory of analytic functions of a complex variable, but whose proof by use of techniques of real variable theory only is much more difficult (cf. K. Knopp, Theory and Application of Infinite Series (London, Blackie, 1928)) follows: If y = f(x) is analytic and has a nonzero derivative at x = a, then in some neighborhood of b · = f(a), a singlevalued inverse function x = q,(y) is defined and is analytic. This means that if y = b + a1(x  a) + a,(x  a) 2 + · · · in a neighborhood of a, and a 1 ~ 0, then it is possible to solve this equation for x in terms of a power series in powers of (y  b). Prove that in a proof of this theorem it may be assumed without loss of generality that a = b = 0 and a1 = 1. (Do not attempt to prove the theorem.) **8. Obtain a set of recursion formulas for the coefficients b,., where X = q,(y) = b1y + b2y2 + bay• + · · · , if q,(y) is the inverse function of Y = f(x) = x + a2x2 + a,x 3 + ••• . (Cf. Ex. 7.) Use this method to obtain the first few terms of the Maclaurin !x' series for x = tan y, the inverse function of y = Arctan x = x  ½x 3 tx7 (Cf. Example 3, § 810, Ex. 3, § 811.) Hint: Substitute the known series for yin the unknown series (or conversely) and equate coefficients in x = q,(f(x)) (or y = f(q,(y)) ).
+
+
+ · •• .
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9
* Uniform Convergence *901. UNIFORM CONVERGENCE OF SEQUENCES
Let (1)
be a sequence of functions defined on a set A. We say that this sequence of functions converges on A in case, for every fixed x of A, the sequence of constants {Sn(x)} converges. Assume that (1) converges on A, and define
=
lim S,.(x). n~+• Then the rapidity with which Sn(x) approaches S(x) can be expected to depend (rather heavily) on the value of x. Let us write down explicitly the analytic formulation of (2) : S(x)
(2)
N
Corresponding to any point x in A and any E > 0, there exists a number = N(x, E) (dependent on both x and · E) such that n > N implies
1s..(x)
 S(x)I
< E.
FIG. 901
270
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§901]
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UNIFORM CONVERGENCE OF SEQUENCES
A = (oo, +oo ), then Jim S,.(x) exists n n+~ and is equal to 0 for every x in A. (Cf. Fig. 901.) If E > 0, the inequality IS,.(x)  S(x)I < E is equivalent to n > lxl/E. Therefore N = N(x, E) can be defined to be N(x, E) = lxl/E. We can see how N must depend on x. For instance, if we were asked, "How large must n be in order that IS,.(x)  S(x)I =
Example 1.
If S,.(x)
= !, and if
l!1 < 0.001 ?" we should be entitled to reply, "Tell us first how large xis." If n = 0.001, the second term of the sequence provides the desired degree of approx
x
imation, if x = 1, we must proceed at least past the 1000th term, and if x we must choose n > 1,000,000.
= 1000,
If it is possible to find N, in the definition of (2), as a function of E alone, independent of x, then a particularly powerful type of convergence occurs. This is prescribed in the definition: Definition.
A sequence of functions {S,. (x)} , defined on a set A, conto a function S(x) defined on A, this being written
verges uniformly on A
S,.(x) =; S(x), if and only if corresponding to E > 0 there exists a number N = N(E), dependent on E alone and not on the point x, such that n > N implies
1s.. (x) for every x in A .
 S(x)I·
0, and (iii) the number N, which depends on both x and E. In the case of uniform convergence we have (i) the number E > 0, (ii) the number N, which depends only on E, and (iii) the point x.
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UNIFORM CONVERGENCE
272 Eumple 2.
If S,.(x)
[§ 901
= !, and if A is the interval (  1000, 1000), then the
n sequence {S,.(x)} converges uniformly to 0 on the set A . The function N(E) can be chosen: N(E) = 1000/E. Then n > N(E) and lxl ~ 1000 imply l!l 1000 IS,.(x)  S(x) I = n < lOOO/E = E.
A quick and superficial reading of Example 1 might lead one to believe, since N seems to depend so thoroughly on x, that even when the set A is restricted as it is in Example 2 the convergence could not be uniform. We have seen, however, that it is. Are we really sure, now, that the convergence in Example 1 fails to be uniform? In order to show this, we formulate the negation of uniform convergence (give the proof in Ex. 25, § 904): Negation of Uniform Convergence. A sequence of functions {S,.(x)}, defined on a set A, fails to converge uniformly on A to a function S(x) defined on A if and only if there exists a positive number E having the property that for any number N there exist a positive integer n > Nanda point x of A such that IS,.(x)  S(x)I ~ E. Example 3. Show that the convergence of Example 1 is not uniform on the interval (oo, +oo ). Solution. We can take E a 1. If N is an arbitrary number, let us choose any n > N, and hold this n fixed . Next we .pick x > n. Then (for this pair n and x), S,.(x)  S(x)
= ~ > !! = 1 = n
n
E.
NOTE. Any convergent sequence of constants (constant functions) converges uniformly on any set. Example '
Show that the sequence
{e!.,} converges uniformly on [O, +oo ).
Solution. For a fixed value of n, the nonnegative function f(x) • xe has a maximum value on [0, +oo) given by x = 1/n, and equal to 1/ne. Therefore the sequence approaches 0 uniformly on [0, +oo ).
Example 6.
Show that the sequence
{:!} converges uniformly on [a, +oo ),
for any a > 0, but not uniformly on (0, +oo ). Solution. For x ~ a, and a fixed n > 1/E, the function f(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 f(x) = nxenz has a maximum value given by x = 1/n and equal to 1/e. The convergence is therefore not uniform.
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902]
UNIFORM CONVERGENCE OF SERIES
273
*902. UNIFORM CONVERGENCE OF SERIES
Let (1)
be a series of functions defined on a set A, and let
=
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 :Ea,., we have:
Theorem. If the series :Eu,.(x) C01Werges uniformly on a set A, then the general term u,.(x) converges to O uniformly on A. Proof. By the triangle inequality, if S,.(x) = u1(x) + · · · + u,.(x) and +•
S(x)
= I:
u,.(x),
n1
lu,.(x)I = IS,.(x)  S,._1(x)I = l[S,.(x)  S(x)] ~ IS,.(x)  S(x)I
Let
E
> 0 be given.
+ IS,._1(x)
 s,._1(x)]I
 S(x)I.
If N is chosen such that n IS,.(x)  S(x)I
for all x in A, then n
+ [S(x)
> N implies lu,.(x)I
>N
 I implies
< ½E < E for all x in A.
Example. Show that the Maclaurin series for e" 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,
IS,.(x)  S(x)I
= =, IR,.(x)I
ef•
n.
ff%
lxl" ;:ii ,a".
n.
Since lim ff%! a" = 0 and ff%' a" is independent of x, the uniform convergence n+• n n. on A is established.
'
+•
If the set A is unbounded, we can show that I: x", fails to converge uniformly no n. on A by showing that the general term does not approach 0 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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UNIVERSITY OF MICHIGAN
274
UNIFORM CONVERGENCE
[§ 903
*903. DOMINANCE AND THE WEIERSTRASS MTEST 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 Evn(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 p0sitive integer n. Theorem I. Comparison Test. Any series of functions Eu,.(x) dominated on a set A lYy 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 ~ V11+1(x) + Vn+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
nt
First re
§ 904]
EXERCISES
277
•42. Ex. 19. •43. Prove that the Cauchy condition for convergence of an infinite series + a,.+1 + · · · + a,.+P =: 0, uniformly in p > 0. (Cf. Exs. 18, 20, § 717.) Explain what you would mean by saying that a series of functions is uniformly summable (C, 1) on a given set. Prove that a uniformly convergent series is uniformly summable, but not conversely. Generalize to uniform summability (C, r). Give examples. (Cf. Exs. 1516, § 717 .) •46. Prove the Cauchy criterion for uniform convergence of a sequence of functions: A sequence {S,.(x)} converges uniformly on a set A if and only if corresponding to E > 0 there exists N = N(E) such that m > N and n > N imply ISm(x)  S,.(x)I < E for every x in A. Hint: By § 302, S(x) = lim S,.(x) n+co exists. Assume that the convergence is not uniform, use the Negation of § 901, N as prescribed above for ½E, and the triangle inequality, for fixed n > N and arbitrary m > N: IS,.(x)  S(x)I ~ IS,.(x)  Sm(x)I + ISm(X)  S(x)I. •46. State and prove the Cauchy criterion for uniform convergence of a series of functions . (Cf. Ex. 45, above, Exs. 1719, § 717.) +oo **47. Show that :E +x 2 2 converges for all x, that it converges uniformly 11  1 1 nx for x ~ 11 > 0, and that it does not converge uniformly in any neighborhood of x = 0. Hint: Use the Cauchy criterion (Ex. 46), together with the ideas used in establishing the integral test (§ 706) to show that uniform convergence of the given series is equivalent to showing that (" xdt . Jm 1 + x 2t2 = Arctan mx  Arctan nx is uniformly small for x in a given set, and large m and n. +oo ••48. Show that :E +x 2 2 converges uniformly for all x. Hint: See Ex. 47, n=Jn nx and writexln [(mnx 2 + m)/(mnx 2 + n)] = x ln [1 + 1/ nx 2]  x ln [1 + 1/ mx 2] . Show that x ln [1 + 1/ nx 2] < ½In 5n½. **49. Prove the Abel test for uniform convergence: If the partial sums of a +oo series L u,.(x) of functions are uniformly bounded on a set A (Ex. 31), and if
l:;a,. (Ex. 17, § 717) can be expressed in the form: a,.
**'"·
n1
{v,.(x)} is a monotonically decreasing sequence of nonnegative functions converging +oo uniformly to O on A, then the series L u,.(x) v,.(x) converges uniformly on A. na)
(Cf. Ex. 22, § 717.) **60. Adapt and prove the Abel test of Exercise 23, § 717, for uniform convergence. **61, Prove the alternating series test for uniform convergence: If {v,.(x)} is a monotonically decreasing sequence of nonnegative functions converging uni+oo formly to Oona set A, then :E (1)"1v,.(x) converges uniformly on A . (Cf. n1
Ex. 49.)
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UNIVERSITY OF MICHIGAN
[§ 905
UNIFORM CONVERGENCE
278 ••62. Show that
+
00
l)n1
:E  + X n• l n (
converges uniformly for x ~ 0, but does not con
verge absolutely for any x. (Cf. Ex. 51.) +oo (l)"x2" ••63. Show that L + 2,. converges uniformly on [ 1 n•l
1
X
+ 1/ , 1 
11], 1/
> 0.
(Cf. Ex. 51.)
+'°(1)" :E n" converges uniformly for x ~ 1/ > 0, but not uniformly for x > 0. (Cf. Ex. 51.) +• x1 + nx ••66. Show that :E (1)" converges uniformly on every bounded set. n•l n2 Where does it converge absolutely? (Cf. Ex. 51.) +oo .,, **66. Show that :E (1)" ~ e ;;; converges uniformly on every bounded set, n•l n but absolutely only for x = 0. (Cf. Ex. 51.)
**M. Show that
n• l
*905. UNIFORM CONVERGENCE AND CONTINUITY _1_
The example S,.(x) = ;t2n 1, 1 ;:;; x;:;; I (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 I, § 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
FIG. 903
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UNIVERSITY OF MICHIGAN
§ 906]
279
CONVERGENCE AND INTEGRATION l
l
ISn(x)  S(x)I < E, which is equivalent to 1  x ~ < ½, or x ~ fails. In case of uniform 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 < tE for every x of A. Holding N fixed, and using the continuity of SN(x) at x = a, we can find a 8 > 0 such that Ix  al < 8 implies ISN(X)  SN(a)I < tE, (The number 8 apparently depends on both N and E, but since N is determined by E, 8 is a function of E alonefor the fixed value x = a.) Now we use the triangle inequality: IS(x)  S(a)I ~ IS(x)  SN(x)I
+ ISN(x)
 SN(a)I
+ ISN(a)
 S(a)I,
Then Ix  al < 8 implies IS(x)  S(a)I
< ½E
+ ½E + ½E = 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 wntinuous on A, then f(x) is continuous
on A.
(Ex. 19, § 908.)
Example.
.
+•
Show that the function f(x) defined by the series :Ee"" cos nx n0
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. Thereforef(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 ~ l/2n, 2n(l  nx) for l/2n ~ x ~ 1/n, and 0 for 1/n ~ x ~ I. The limit function S(x) is identically 0 for 0 ~ x ~ I. For every n,
.£
1
s,.(x) dx
= ½ (the integral is the area of a triangle of altitude n and
base 1/n), but
.£
1
s(x) dx
= 0. 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 Sn(x) are defined by single analytic formulas, see Ex. 25, § 908.)
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280
[§ 906
UNIFORM CONVERGENCE
In case of uniform convergence we have the theorem : Theorem.
If S,.(x) is integrable on [a, b] for n = I, 2, la, b] , then S(x) is integrabl,e on [a, b] and
, and if
S,.(x) :::; S(x) on
lim ibS,.(x) dx =
(I)
a
n+ao
r" Jim S,.(x) dx Ja n+oo
=i~S(x) dx.
a
y
n 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 stepfunctions (§ 502), and finally to construct two new stepfunctions that squeeze S(x). Accordingly, for a given E > 0, 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 stepfunctions u1(x) and r 1(x) such that u1(x)
~
S.v(x)
~
r1(x) on [a, b], and ib[r1(x)  u1(x)] dx
fine the new stepfunctions: u(x)
Then, for a
= u1(x) 
~ x ~
= r1(x) + 4(b E_
b,
u(x) r(x)
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4 (b _ a)' r(x)
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< ui(x) > r1(x)
+ [S(x) + [S(x)
 SN(x)] ~ S(x),  SN(x)] ~ S(x),
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UNIVERSITY OF MICHIGAN
a)°
+•
(1)
S(x)
=
i'"
T(t) dt  S(.xo).
Since T(x) is continuous on [a, b] (§ 905), the functionS(x) is differentiable with derivative S'(x) ' = T(x) (Theorem I, § 504). Finally, since
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EXERCISES
§ 908]
ISn(X)  S(x)I
283
= li"[sn'(t)
 S'(t)] dt
+ [Sn(Xo)
 S(Xo)]I
~ lf:\Sn'(t)
 S'(t)I dtl
+ 1s..(xo)
 S(Xo)I,
the convergence of {Sn(x)} is uniform (cf. Ex. 23, § 908).
Corollary. If :Eun(x) is a series of differentiable functions on [a, b], convergent at one point of [a, b], and if the derived series :Eu,.'(x) converges uniformly on [a, b], then the original series converges uniformly on [a, b] to a differentiable function whose derivative is represented on [a, b] by the derived series. (Ex. 21, § 908.)
+•
Show that if J(x) is the function defined by the series L enz cos nx
Example.
n0
of Example 2, § 903, then +oo
= L
f'(x)
for every x
n•O
> 0.
nenz (cos nx
+ sin nx),
Let x be a given positive number, and choose a and b such that 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 Mtest, with M,. = 2ne.. (check the details). Solution.
0
< a < x < b. 0
*908. EXERCISES
In Exercises i6, show that the convergence fails to be uniform by showing that the limit function is not continuous.
*1.
lim sin" x, for O ~ x n~+• lim
n++oo
*'*6·
X" + ,.• for O ~ x X
lim enzt, for n++•
. hm S,.(x), where S,.(x)
~
1.
sin nx = for O < x ~ r, and S,.(0) = 1. ~
+ x(l  x) + x (1  x) + ••• , for 0 x1 x2 •6. x' + 1 + x2 + (1 + x2)1 + ... 'for lxl ~ 1. 2
(1  x)
!xi
~ 2.
1
+..
*2.
~ r.
~ x ~
l.
In Exercises 712, show that the equation is true.
*7•
. IIm n++•
*9.
.J:
J;" sin nx d
X
'" nx +• sin nx L  n1dx n1
= 0.
+•
*8. 2
= n1 L (2n
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Jim (2 e=• dx = 0. n++oo J1
1)3 •
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UNIVERSITY OF MICHIGAN
UNIFORM CONVERGENCE
284
*10.
rf 2
In ~x dx
J1 n=l
*1l. (•
f
n
= ~ 2
e"
l
t
2+00
J: L
E In 4\1. n
n1
n sin nx dx
)0 n = 1
*12.
=
[§ 908
nen"' dx
=
_e_,
e2
n= l

1
In Exercises 1318, show that the equation is true. d
[+'°
*13. dx n1n L (n *14. d
dx
+oo
X"
+ l) ]
X"
= n=on L +2 , for \x\
l, is dominated by a convergent series of constants, then the series l:u,.(x) can be integrated term by term, from a to +co: ••33. Show that the series +oo L 1 ( n=ln
(+'° )a
___£_ )"
X
i: u,.(x) dx = f
n1
(+"' u,.(x) dx.
n=l)a
Hint: Use Ex. 46, § 515.
The hint given in that Exercise, together with the Weierstrass Mtest 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 Sn.(Xk) is:; E > 0. Assume without loss of generality that {xk} converges: Xk + x. Show that for every n, S,.(x) is:; E. **37. Prove the MooreOsgood Theorem: If S,.(x) is a sequence of functions
=
defined for (i) S,.
=
X
in a deleted neighborhood J of X
=
c, and if
= lim S,.(x) exists and is finite for every n = 1, 2, · · · ; X+C
(ii) f(x)
=
lim S,.(x) exists and is finite for et•ery x in J(x
~
c);
n++oo
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286
[§ 908
UNIFORM CONVERGENCE
(iii) the convergence in (ii) i8 uniform in J; then (iv) lim S,. exists and is finite;
+•
(v) lim f(x) exists and i8 finite; (vi) the limits in (iv) and (v) are equal.
Similar results hold for xc+, c, +oo, 00, and oo. Hints: For (iv), write IS..  S,.I :ii IS..  S .. (x)I + IS.. (x)  S,.(x)I + JS,.(x)  S,.J, let N be such that m > N and n > N imply JS.,(x)  S,.(x)I < e/3, for all x in J , and let x  c. For (v), write lf(x')  f(x")I ~ IJ(x')  S,.(x')I + IS,.(x')  S,.(x")I + IS,.(x")  f(x")I, let N be such that n > N implies IS,.(x)  f(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 .. (x)  S,.(x) = {[S .. (x)  S,.(x)]  [S.. (xo)  S,.(xo)]} + {S.,(Xo)  S,.(Xo)} = [S.. '(~)  S,.'W] (x  Xo) + {S.. (xo)  S,.(Xo)}, to establish S,.(x) ~ S(x). Then use the MooreOsgood Theorem (Ex. 37) to obtain lim Jim S,.(x)  S,.(c) = Jim Jim S,.(x)  S,.(c), z..: n++ •
X 
C
X 
n++ • 
C
by writing
~W~M_&W&M_~W&W_~M&M xc
X 
= S.,'(~)
(Cf. Ex. 45, § 904.)
xc
xc
C
 S,.'(E).
**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 IS,.'(x)I :ii K on [a, b]. First use JS,.(x')  S,.(x")I :ii K · Ix'  x"I and IS(x')  S(x")I :ii IS(x')  S,.(x')I + JS,.(x')  S,.(x")I + IS,.(x")  S(x")I to show that S(x) is continuous. Assume the convergence is not uniform, and let Xt  x such that ISn.CXt)  S(xt)I ~ E > 0. But JS,,.(xt)  S(Xt)I ~ ISn.(Xt)  S,..(z)I + 1s... (z)  S(x)I + IS(z)  S(x.)I. **40. Let f,.(x) and f(x) be RiemannStieltjes integrable with respect to g(x) on [a, b], and assume f,.(x) ~ f(x) on [a, b]. Prove that
j). (x) dg(x) ibf(x) dg(x). **4.1. Let f1(x), fi(x), • • • be a sequence of "sawtooth" functions (Fig. 906), defined for all real x, where the graph of f,.(x) is made up of line segments of slope ±1, such thatf,.(x) = 0 forx = ±m•4.., m = 0, 1, 2, • ·· , andf,.(x) =
= ½·4•
+ m4•, m = 0, 1, 2,
=
+•
~ f,.(x). Prove no that f(x) is everywhere continuous and nowhere differentiable.t Hint for non
½· 4• for x
••• .
Letf(x)
t This example is modeled after one due to Van der Waerden. Cf. E. C. Titchmarsh, Th ~ory of Funclions (Oxford, Oxford University Press, 1932).
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§ 908]
EXERCISES
287
differentiability: If a is any fixed point, show that for any n = 1, 2, • • • a number h,. can be chosen as one of the numbers 4..i or 4n 1 such thatf,.(a + h,.)  f,.(a) = ±h,.. Then f ,,.(a + h,.)  f ,,.(a) has the value ±h,. for m ~ n, and otherwise vanishes. Hence the difference quotient [f(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+ +oo cannot exist as a finite quantity.
L\L\v f.,.(x)
,~AAAA/ J,.+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 positivevalued function defined for t > 0 such that lim p(t) = 0. Then there exists afunctionf(x), defined and continuous t.....O+
for all real x, having the property that corresponding to any real number a there is a sequence of numbers {h,.} such that h,.+ 0 and lf(a + h,.)  f(a)J / p(lh,.J)+ +oo. 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 o:,. defined inductively to be a sufficiently high integral power of ¼ to ensure the following inequality, where Ak = 4kak 1 is the absolute value of the slopes of the segments in the graph of fk(x), k = 1, 2, • • • , n  1, (0:1 being defined to be i) : A,. > A, + A2 + · · · + A,._, + n[p(o:,./4)]/o:,., n > 1. Then choose h,. = ±o:,./4 such that Jf,.(a + h,.)  f,.(a)I/Jh,.J = A,.. Define
+oo
f(x)
= I:
f,.(x).
n•l
**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 lf(x,)  f(x2)l, formed for all x, and X2 of I such that Jx,  x2l < 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 f(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. 358360.
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288
UNIFORM CONVERGENCE
*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. 1013, § 911) of the following basic theorem:
Theorem I. A power series converges uniformly on any interval whose endpoints 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,.x" (cf. Ex. 9, § 911). Let R > 0 be the radius of converno gence of :Ea..x", and let I be an interval with endpoints a and /3, where max (Jal, J/31) = r < R. Choose a fixed 'Y such that r < 'Y < R, and define M,. = Ja,.'Y"I, n = 0, 1, · · · . Since 'Y is interior to the interval of convergence of :Ea,.x", :Ea,.'Y" converges absolutely. Therefore the convergent series of constants :EM,. dominates the series :Ea,.x" throughout I, and by the Weierstrass Mtest, 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 endpoints 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 endpoints 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. +oo
Theorem II. Abel's Theorem. If L a,. is a convergent series of conno +..
slants, then the power series
L
n=O
a,.x" converges uniformly for O ~ x
~
1.
Corollary I. If a power series converges at an endpoint P of its interval of convergence I, it converges uniformly on any closed interval that has P as one of its endpoints and any interior point of I as its other endpoint. If a power series converges at both endpoints 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 endpoint of that interval at which the series converges.
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UNIVERSITY OF MICHIGAN
§ 910]
PROOF OF ABEL'S THEOREM
Example. Show that (1) In (1 for 1 < x
~
+ x) = x
x x 2 + 3  ••• , 2

289
3
+oo l, by integrating :E (l)"x". n0
+oo
:E (  l)"x" converges for !xi < 1. Thereno fore the relation (1) is valid for !xi < 1 (Theorem II, § 810). Since In (1 + x) is continuous at x = l, the relation (1) is also true for x = l, by Corollary II. Solution.
The geometric series
**910. PROOF OF ABEL'S THEOREM
Notation.
If ¼>, u., "'2, · · • , u,., • • · is a sequence of real numbers, we
denote by (1)
the maximum of the n  m luml, lum
+ 1 numbers
+ Umtil,
· · · , lum
+ Um+1 + · · · + u,.1.
Lemma. If {a,.} is any sequence of real numbers, and if {b,.} is a monotonically decreasirig sequence of nonnegative numbers (b,. L and b,. ~ 0), then (2)
and
Ik'f. a.1,bkl ~ bm·max ;jam+ Om+1 + · ·· + akl•
(3)
Proof. The two statements (2) and (3) are identical, except for notation . For convenience we shall prove (2), and then apply (3) in the proof of Abel's Theorem. Letting A,.= ao a1 a,., n = 0, 1, 2, · · · , we can write the lefthand member of (2) in the form
+ + ·· · +
IAobo
+ (A1 
+ (A2  A1)b2 + · · · + (A,.  A,._1)b,.I = IAo(bo  b1) + A1(b1  b2) + •' • + An1(b,._1 Ao)b1
b,.)
+ Anbnl •
By the triangle inequality and the assumptions on {b,.}, this quantity is less than or equal to IAol(bo  b1)
+ IA1l(b1 
b2)
+ · · · + !Anil(bn1 
~ {maximum of IAol, IA1I, · · · , IA,.!}· {(bo  b1)
= bo·max ~ao +
+ IA,.lbn
b,.)
+ · · · + b,.}
· · · + akl•
Proof of Abel's Theorem. By the Lemma, if b,.
= x",
Iktm akXkl ~ 1 ·max ;jam+ ••• + a,.I.
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290
[§ 911
UNIFORM CONVERGENCE
The convergence of La,. implies that for any E > 0 there exists a number N such that n ~ m > N implies ja,,. + ••• + a,.j < E. Therefore, by the Cauchy criterion for uniform convergence (Ex. 45, § 904), the proof is complete. *911. EXERCISES In Exercises 16, obtain the given expansions by integration, and justify the inclusion of the endpoints specified. •1. In (1  x) = x  xi  x2  · · · ; 1 :i x < 1.
2
3
x•
3 + 5  · · · ; lxl ~ 1. 1 x2 1 · 3 x• 1 •3 · 5 x7 = x + 2 3 + 2·4 5 + 2·4 · 6 7 + ... ; lxl
•2. Arctan x = x .
•3. Arcsm x
*'**6.
x2
+ ~+ x) =x  ! ~+~~ 23 2 ·4 5 ½[xv'l  x• + Arcsin x] In (x
=x
**6.
½[xv'l

lx1
23 
l •lx• 2·4 5
5 l ·3 · ~ 2·4·6 7

:i 1.
+ · · · ' lxl
1· 1•3x7 246 7
:S 1

.
 · · · ; lxl ~
+ x + In (x + Vl + x')] = x + ! ~ _ l:! ~ + l · l · 3 £  · · · ; lxl 23 2·4 5 246 7
l.
2
•7. Show that
., Io
In (1
+ t) dt
x2
x•
= 1·2  2 ·3
~
1.
+ 3x'. 4  · •· .
1 1 1 Hence evaluate   +   .. · . 1· 2 2 ·3 3.4 •8. Show that ("' x' x' x• Jo Arctan t dt = ~  . 4 + 5 _6  .. • . 3 Hence evaluate 1  ½  i + l + l  ¼ t + +   · · · , •9. Prove Theorem I, § 909, for La,.(x  a)". •10. Prove that a function defined by a power series is continuous throughout the interval of convergence. •11. Prove Theorem II, § 810. *12. Prove Theorem III, § 810. •13. Prove Theorem I, § 803. •14. Prove Corollary I of Theorem II, § 909. •16. Prove Corollary II of Theorem II, § 909. *16. Show that
E (
fl Arctan x dx = (1 J!!!...!L dx _ 1)" . 1 x Jo 1 + x2  ... o (2n + 1)
Jo
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UNIVERSITY OF MICHIGAN
§ 911]
EXERCISES
*17. Show that 1
f
1
Arcain x 0, then there exists a polynomial P(x) such that lf(x)  P(x)I < E for all x in [a, b] .t Suggested outline: (i) The binomial series for v'1 + x converges uniformly to v'I+x for 1 ~ x ~ 0. (ii) The corresponding series for v'1 + (x'  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 ~ x ~ c (c ~ x ~ b) and linear for c ~ x ~ b (a ~ x ~ b). (vi) The theorem is true for any continuous function with a polygonal (brokenline) graph. (vii) Any continuous function on [a, b] can be uniformly approximated on [a, b] by a continuous function with a polygonal graph.
**23.
Prove that ifibx"f(x) ... min
69.
(1,l)·
!· 17
67. 3; N(e) •
e
62. l; li(e) • min (1, 20e).
(!, .!.)·
6'• 6·' li(e) • min 8 42
66. 0; N(e) • max { 1, ~)•
+00 ; if B > 0, li(B)
70. 00; if B
00.
60. 9; li{e) • min { 1, ~)
61. 28; li(e) • min { 1, ~)
611. 0; N(e) •
·u.
< 0, li(B)
68.
t;
N(e) • max { 1, ~)
• min { 1, ~) ; otherwi~e li(B) • 392. • min { 1,
~B); otherwise li(B) ""1. 1216, page 117
82. li(e) • min (3, e).
SS. li{e) • min { 1, ~)
3'. li(e) • min { 1, ~)
811. li(e) • min (5, e).
36. li{e) • min (5, &).
37. li(e) • min
G• ~)
18011, page 66 20. +00; 00. 1;0. 80. 1; 1. l'2. 1; 1. .K. 1; 1.
19. 1; 0. 21. 1; 0. 29. +00; 00. 81. 1; 1. ss. 1; 1.
n.
§ aoa, page 78 8. li(e)  ~
' li(e) • !.. 4
Ii. li(e) • 2e.
6. li{e) • el.
7. li{e) • e.
8. li(e) •
13. li(e) • min
(f,~e)
el
4.
1'. li(e) • min
(1, 1.;. x)· 2
1'°4, page 91 2. 3x1_
1. 2x  4. 2
22 '  (5x  4)1.
3.  ;;· Ii _1__
1 6. 3vx· ·~·
• 2VX
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Original from
UNIVERSITY OF MICHIGAN
ANSWERS 11. Yes. 18. No. 21. n > 1; n
16. Yee. 19. No.
> J. 
x•➔ cos!, x > 0;f'(0) X
l)x•➔
24. f"(x)  [n(n 
/"(0)  0, n
> 3.
n

> k; n > k.  0, n > J. n > l.
:i;•4) sin~  2(n  I)x11a cos~• x
> 3.
n
n
> 2.
> 0;
> 4. § '°8, page 100 2. 2 
1. 1r, 21r, or 31r.
a. e 
17. No. 20. Yee.
22. n
= nx•1 sin!X
23. f'(x)
295
V2.
,. ½(a+ b).
l.
I. _1_.
6.
e 1
½
b. 1• a+ 2
9. 1 
"+Vf"="'o. § ,12, page 111
1. Relativemaximum 9;relativeminimum  5;increasingon(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 [l, +co). , . Relative maximum  2V3/9; absolute minimum  0; increasing on [0, ½] and [l, +co); decreasing on
[i, 1).
I. Maximum  ½;minimum"" J. 6. Maximum  4; minimum  109/16. 7. Maximum  2; minimum  9/8. 8. No maximum; minimum  1/e. 11. (a) x  20; (b) x .. 25; (c) x • 21; (d) no profit possible. 16. (a) x • 20; (b) x • 37.4; (c) full speed, x • 60. 19. If t ~ ,, x = b; if t > ,, x • min (b, aa/V/1 at). 20. r ~ 1 : no minimum; 1 < r ~ ½ : minimum• a1 /4(r + l); ½ ;:ii r ;:ii 0 : minimum .. ra1 ; r ii:; 0: minimum• 0. as. e(~x)  (6:i;I  5)~ + 4x~ + ~
M. e(~) = ~/x'(x + ~). 11. e(~)  v'x ~/2(:e + v:,;t + ~)1 • 87. 10.48810. 88. h.
89. 0.06.
'°· x.
'2.
'1. x. h
+;· 1 + x.
'8. x.
" 1
46. h.
,1.
½ ~ (x i}
u.o. '8. 0.
§ f17, page 121
a.
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i
Original from
UNIVERSITY OF MICHIGAN
296
ANSWERS
7. ft•
6. h
6. 1.
8. 1.
9
t.
Ina. 'In b
10. 2. 13. 1. 16. 0.
14. Meaningless. 17. 0.
18.
19. 0.
20.  2a_
21. 1.
22. ½, 211. e. 28. 1.
26. e1 . 29. et.
11.
12. 1.
16. , .
,r
23. 1.
+oo.
24. e'. 27. 0. so. 0.
§ li03, page 143
14. 0; 3; ½(n2

n).
211. ~
3. Ii. 7. 8.
26.
i
27. 1~2.
§ 1106, page 1116 0. 4. sin x2• sin x2• 6. 3x2 sin x8• 4x 3 sin x 8  3x2 sin x8 • 2x cos x 2 @in (sin• x2)  sin 2x sin (sin' x). § li08, page 160
3. No.
No. § 1110, page 162
13. 14.
l sin6x cos x  ,h sin1x cos x  T•1I sin x cos x + r°7Ix + C. ¼sin X cos'x + n sin X cos2x +/.,;sin X + C.
111. 
5
X
5
X
4cot' 5 + 2cot2 5 +
.X
5 In sm
5+
C.
16. l sec6x tan x + ,h sec'x tan x + I,; sec x tan x + /,r In 1sec x + tan xi + C. 17. ¼[(4x3  6x) sin 2x + (2x' + 6x2  3) cos 2x] + C. 18. ¼(x2 + 4x)i  (x + 2)Vx2+4x + 4 In Ix+ 2 + Vx2+4xl + C. 19. ½x2  Hx + ½) v'x2 + x + 1 + i In Ix+½+ v'x• + x + lj + C. 20. ~ (4x2 + 21x + 105)(6x  x2)i + 1.p. (x  3)V6x  x2 1701 . x   3 + C. + A re sm 8 3 § 1116, page 171
1. ~
2.
o.
3. Divergent.
4. 2.
Ii. 2.
6. Divergent.
8. (k _ l)~ln 2 )H' k 16. Convergent.
7. ~
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>
1; divergent, k
~
1.
9. 2.
16. Divergent
Original from
UNIVERSITY OF MICHIGAN
10.
'K
i
ANSWERS 17. Convergent. 19. Divergent. 21. Convergent. 23. Divergent.
18. 20. 22. 24.
297
Convergent. Convergent. Convergent. Convergent.
§ 618, page 184
1.
J.
2. 3.
4.  (1 + e + et). 6. e + e1  2. 36. v(x) = x + [x]; p(x) = x; n(x) = [x] . 36. v(x) = 3 + 2x  x1, 1 ~ x ~ l; 5  2x + x1, 1 p(x) = 3 + 2x  x1 , 1 ~ x ~ 1; 4, 1 ~ x ~ 2; n(x) = 0, 1 ~ x ~ 1; 1  2x + xt, 1 ~ x ~ 2.
6. r  2. ~ x ~
2;
§ 606, page 196 1. 2. 3. 4. 6.
¼In 1sec 5xl + C. ¼In !sec 4x + tan 4xl + C. Arc sin (x/V2) + C, xi < 2. ½(x + 2}Vx1 + 4x  21n Ix+ 2 + Vx1 + 4xl + C, x ~ 0 or x ~ 4. ½In l2x  1 + v' 4x1  4x + 51 + C.
6x1~ 6. ~ X  3xl +
7.
Js
8. sec 9. 10.
In l(v'5  2x1
Vx tan Vx
_l_lnl
V3 Arc BID . 7Z

(6x  1) + C, 0
V5}/xl + C, lxl
< X < }.
< VS/2.
+ In 1sec Vx + tan Vxl + C, x
> 0.
+ 5  V1091 + C. + 5 + v'109 2 6x + 5 _ ~ Arc tan _~ + C.
vi09
6x
6x
v59
v59
§ 609, page 200 2. sinh 2x. 4. coth xi  2xl csch1xl. 6. e=(a cosh bx + b sinh bx).
1. 3 sinh 3x. 3. sech1 (2  x). 6. 2 coth 2x.
ea
7 4 . • VI+ J6xl 2x
9. 1,,
x
8.  ~ • x > 0. Ve..  1
\xi < 1.
10. csc x, x F' nr.
11. ½In cosh 6x + C. 13. sinh x + ½sinh1 x + C. 16. ¼sinh 2x  ½x + C.
12. In (sinh ea) + C. 14. x tanh lOx + C. 16. ¼[(x•  I) tanh1x + ½xi + x] + C.
17. Cosh1 ~ + C.
1 . h, 2x  1 18. 2am 2 
+ 5 + C. V109
2 6x 19.  Vl0 coth1 9
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*
20. 
+ C.
2  tanh 1 5x + 5 + C.
vi09
vi09
Original from
UNIVERSITY OF MICHIGAN
298
ANSWERS § 707, page 211
n 1·n+l;l.
2· ~
3. 1  m• ; 1. 6. 2 + ½ +
= n(n
l;
r,(71
•·
l + t~; a, = 2, a.
2n:+ 2
~ o' n >
½.
I).
'
+oo.
l.
6. l + 2  2 + 2; a, = 1, a. = 2(1)•, n > l. 7. 1.3  0.39 + 0.117  0.0351; 1.3 (0.3)••. 8. 2  0.7  0.21  0.063; a,  2, a.  0.7(0.3)•t, n > l. 9. ¥10. Divergent. 11. 0.3. . 1..•• 10,201 201 13. •· 16. ffi. 18. Divergent. 21. Convergent.
16. 19. 22. 26.
2'. Convergent.
1. ,. 7. 10. 13. 16. 16. 18. 19. 20.
Divergent. Convergent. Divergent. Convergent. Convergent. Convergent for a Convergent. Convergent for a Convergent. Convergent for 0
,.
17. 20. 23. 26.
Convergent. Divergent. Convergent.
> i;
§ 711, page 219 2. Divergent. 6. Convergent. 8. Convergent. 11. Convergent. lf. Convergent. divergent for a ;:;ii ½.
> 1;
divergent for a ;:;ii I.
S. Convergent. 6. Divergent. 9. Convergent. 12. Divergent.
17. Divergent.
< r < l; divergent for r
~
1 unless a ,. k1r.
§ 713, page 223 2. Convergent.
1. Divergent.
,. Convergent for p
Divergent. Convergent. Divergent. Convergent.
> 2;
a.
Divergent.
divergent for p :iii 2.
1. S. Ii. 7. 8. 9. 10.
§ 717, Pllge 229 Conditionally convergent. 2. Divergent. Conditionally convergent. 4. Absolutely convergent. Absolutely convergent. 6. Absolutely convergent. Absolutely convergent for x > 0; divergent for x ;::ii 0. Absolutely convergent for lrl < 1; divergent for lrl ~ I. Absolutely convergent for p > l; conditionally convergent for p ;::ii l. Absolutely convergent for p > 2; conditionally convergent for 0 < p ;:;ii 2; divergi>nt for p ;::ii 0.
13. 16. 17. 22.
2.718. l.202. 0.7854. 0.5770 < C
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< 0.5774.
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Original from
UNIVERSITY OF MICHIGAN
299
ANSWERS
§ 802, 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 ;:ii 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 Ix! ;:Ii 1. Absolutely convergent for x < 2 or x > 4; conditionally convergent for x "" 2. 2 1 2 1 12. Absolutely convergent if !xi 'F n 1r; conditionally convergent if !xi • n ir.
1. 2. 8. 4. 6. 6. 7. 8. 9. 10. 11.
i
i
§ 811, page 261
2.1x'+x•x'+ · · · ·
9. 2 [ X + x'
zl
6• In 2
3x 27x' + 1·2  9x' 2 ·4 + 3·8  · · · .
+ 2!·4 + 3! •8 + . . . .
lO.
x' x1 + 2! •5  31•7 + • • • •
x' x1 xii x" 12• 3  31·7 + 51·11  71·15 + ....
X 
3
18•
2  4 + 48 
1
X
lll.
X 
•
zl
if.2
f + f + f + . . J
l1.
17
X
,. l +
x'
1
x'
1' 1
480°
17
2
3 x' + 15 x'  315 X7• ~£. _±... 6 180 2835
19. cosa [ 1
(x  a)' 2!
X 
+ (x
x' x' x1 3 •31 + 5•51  7 •7! + .. . •
+x
16• l +
X
a  x4if 
x' 
+
x'
2
+
x'
2
+
x'
30· 3x4
8
37x' + 120 .
x' x4 x' 17x' 18•  2  12  45  2520·
 a)• 41
· ·· ]
sin a [