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Real Analysis
PURE AND APPLIED MATHEMATICS A Wiley Series of Texts, Monographs, and Tracts Founded by RICHARD COURANT Editors Emeriti: MYRON B. ALLEN III, DAVID A. COX, PETER HILTON, HARRY HOCHSTADT, PETER LAX, JOHN TOLAND A complete list of the titles in this series appears at the end of this volume.
Real Analysis: A Historical Approach Second Edition
Saul Stahl
The University of Kansas Department of Mathematics Lawrence, KS
)WILEY A JOHN WILEY & SONS, INC., PUBLICATION
Copyright © 2011 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representation or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print, however, may not be available in electronic formats. For more information about Wiley products, visit our web site at www.wiley.com. Library of Congress Cataloging-in-Publication Data: Stahl, Saul. Real analysis : a historical approach / Saul Stahl. — 2nd ed. p. cm. Includes index. ISBN 978-0-470-87890-3 (hardback) 1. Mathematical analysis. 2. Functions of real variables. I. Title. QA300.S882 2011 515'.8—dc22 2011010976 Printed in the United States of America. oBook ISBN: 978118096864 ePDF ISBN: 978118096840 ePub ISBN: 978118096857 10 9 8 7 6 5 4 3 2 1
This book is dedicated to the memory of my parents and my brother, Finkla, Moses, and Dan Stahl
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Contents
Preface to the Second Edition
xi
Acknowledgments
xv
1
Archimedes and the Parabola 1.1 The Area of the Parabolic Segment 1.2* The Geometry of the Parabola
1 1 8
2
Fermat, Differentiation, and Integration 2.1 Fermat's Calculus
13 13
3
Newton's Calculus (Part 1) 3.1 The Fractional Binomial Theorem 3.2 Areas and Infinite Series 3.3 Newton's Proofs
19 19 23 30
4
Newton's Calculus (Part 2) 4-1 The Solution of Differential Equations 4-2* The Solution of Algebraic Equations Chapter Appendix: Mathematica Implementations of Newton's Algorithm
35 35 40 48 VII
νιϊι
CONTENTS
5 Euler 5.1 Trigonometrie Series
51 51
6
The Real Numbers 6.1 An Informal Introduction 6.2 Ordered Fields 6.3 Completeness and Irrational Numbers 6.4* The Euclidean Process 6.5 Functions
61 61 64 71 76 80
7
Sequences and Their Limits 7.1 The Definitions 7.2 Limit Theorems
85 85 92
8
The Cauchy Property 8.1 Limits of Monotone Sequences 8.2 The Cauchy Property
103 103 111
9
The 9.1 9.2 9.3 9.4*
115 115 120 124 132
Convergence of Infinite Series Stock Series Series of Positive Terms Series of Arbitrary Terms The Most Celebrated Problem
10 Series of Functions 10.1 Power Series 10.2 Trigonometric Series
139 139 145
11 Continuity 11.1 An Informal Introduction 11.2 The Limit of a Function 11.3 Continuity 11.4 Properties of Continuous Functions
149 149 150 155 163
12 Differentiability 12.1 An Informal Introduction to Differentiation 12.2 The Derivative 12.3 The Consequences of Differentiability
169 169 171 177
CONTENTS
12.4 Integrability
ιχ
182
13 Uniform Convergence 13.1 Uniform and Nonuniform, Convergence 13.2 Consequences of Uniform Convergence
187 187 197
14 The Vindication 14.1 Trigonometric Series 14.2 Power Señes
207 207 212
15 The Riemann Integral 15.1 Continuity Revisited 15.2 Lower and Upper Sums 15.3 Integrability
217 217 221 227
Appendix A: Excerpts from "Quadrature of the Parabola" by Archimedes
237
Appendix B: On a Method for the Evaluation of Maxima and Minima by Pierre de Fermat
245
Appendix C: From a Letter to Henry Oldenburg on the Binomial Series (June 13, 1676) by Isaac Newton
247
Appendix D: From a Letter to Henry Oldenburg on the Binomial Series (October 24, 1676) by Isaac Newton
249
Appendix E: Excerpts from "Of Analysis by Equations of an Infinite Number of Terms" by Isaac Newton
253
Appendix F: Excerpts from "Subsiduum Calculi Sinuum" by Leonhard Euler
265
Solutions to Selected Exercises
267
Bibliography
287
Index
291
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Preface to the Second Edition
A Focused and Historical Approach
The need for rigor in analysis is often presented as an end in itself. Historically, however, the excitement and impatience that characterized the first century and a half of calculus and its applications induced mathematicians to place this issue on the proverbial back burner while they explored new territories. Ironically, it was developments in physics, especially the study of sound and heat, that brought the pathological behavior of trigonometric series into the foreground and forced the mathematical world to pay closer attention to foundational issues. It is the purpose of this text to provide a picture of analysis that reflects this evolution. Rigor is therefore introduced as an explanation of the convergence of series in general and of the puzzling behavior of trigonometric series in particular. The first third of this book describes the utility of infinite, power, and trigonometric series in both pure and applied mathematics through several snapshots from the works of Archimedes, Fermat, Newton, and Euler offering glimpses of the Greeks' method of exhaustion, pre-Newtonian calculus, Newton's concerns, and Euler's miraculously effective, though often logically unsound, mathematical wizardry. The infinite geometric progression is the scarlet thread that unifies Chapters 1 to 5 wherein the nondifferentiability of xi
xii
PREFACE TO THE SECOND EDITION
Euler's trigonometric series provides an example that clarifies the need for this careful examination of the foundations of calculus. Chapters 6 to 10 consist of a fairly conventional discussion of various aspects of the completeness of the real number system. These culminate in Cauchy's criterion for the convergence of infinite series which is in turn applied to both power and trigonometric series. Sequential continuity and differentiability are discussed in Chapters 11 and 12 as is the maximum principle for continuous functions and the mean value theorem for differentiable ones. Chapter 13 covers a discussion of uniform convergence proving the basic theorems on the continuity, integrability, and differentiability of uniformly convergent series and applying them to both power and trigonometric series. While the exposition here does not follow the historical method, a considerable amount of discussion and quoted material is included to shed some light on the concerns of the mathematicians who developed the key concepts and on the difficulties they faced. Chapter 14, The Vindication, uses the tools developed in Chapters 6 to 13 to prove the validity of most of the methods and results of Newton and Euler that were described in the motivational chapters.
Intended Audience for this Book
This text and its approach have been used in a junior/senior-level college introductory analysis course with a class size between 20 and 50 students. Roughly 40% of these students were mathematics majors, of whom about one in four planned to go on to graduate work in mathematics; another 40% were prospective high school teachers. The balance of the students came from a variety of disciplines, including business, economics, and biology.
Pedagogy
Experience indicates that about three quarters of the material in this text can be covered by the above described audience in a one-semester collegelevel course. In the context of this fairly standard time constraint, a trade-off between the historical background material and the more theoretical last two chapters needs to be made. One strategy favors covering uniform convergence [through Cauchy's theorem about the continuity of uniformly convergent series of continuous functions (13.2.1-3)] instead of the more traditional Maximum Principle and Mean Value Theorem. Each section is followed by its own set of exercises that vary from the routine to the challenging. Hints or solutions are given for most of the oddnumbered exercises. Each chapter concludes with a summary. The excerpts in the appendices can be used as independent reading or as an in-class lineby-line reading with commentary by the professor.
PREFACE TO THE SECOND EDITION
xiii
This book contains sections which can be considered optional material for a college course; these sections are indicated by an asterisk. The least optional of the optional sections is 4.2, which describes the rudiments of Newton's polygon method. While the details look daunting, the material can be covered in two 50-minute-long lectures. The purpose of Section 6.4 is to provide an elementary and self-contained proof of the existence of irrational numbers. Section 9.4 is included as an explanation of some concerns of higher mathematics. All of these optional sections reinforce the importance of infinite series and infinite processes in mathematics. The Riemann integral proved to be a thorny problem. Within the limitations of one semester it is impossible to do justice to historical motivation, the traditional contents of the course, uniform convergence, and the Riemann integral. Therefore I have decided to state and make use of the integrability of continuous functions without proof. This is applied for the first time in the proof of Theorem 13.2.4, a place to which a typical class is unlikely to get in one semester. For the sake of completeness, a new Chapter 15 has been included in the second edition of this book. Its purpose is to establish the rigorous foundations of the Riemann integral. Riemann's definition of this integral, appeared in an (unpublished) manuscript on trigonometric series. However, the only example he gave is too technical for the purposes of this text. For that reason, the development in this new chapter is the traditional one in terms of partitions and the least upper and greatest lower bounds. SAUL STAHL Lawrence, Kansas
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Acknowledgments
Above all I wish to acknowledge the help given to me by Fred Galvin, who carefully read and corrected major portions of the first edition. Fred van Vleck and Seungly Oh proofread parts of the manuscript. I am indebted to them. I am also indebted to my colleague Bill Paschke and the late Pawel Szeptycki for their patience when they were subjected to my ramblings and half-baked ideas. My thanks are also due to Susanne Steitz-Filler, Christine Punzo, Jacqueline Palmieri and all the folks at John Wiley & Sons, as well as Larisa Martin, for their help in preparing this edition. Thanks also to Judy Roitman for making some of her class notes available to me. Your comments are welcome at
[email protected] S.S.
XV
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ί_
Archimedes and the Parabola How did Archimedes evaluate the area of the parabolic segment almost 2,000 years before the birth of calculus? 1.1
THE AREA OF THE PARABOLIC SEGMENT
One of the most important issues considered in the context of elementary geometry is that of the area of polygonal regions. The only curvilinear figure commonly studied at this elementary level is the circle, and once again its area is a major concern, as are its circumference and the construction of tangents. It is only natural to extend these notions to other curvilinear figures. Analytic geometry and calculus were developed, at least in part, in order to facilitate investigations of this advanced level. The ancient Greeks, in whose culture modern science and mathematics are rooted, developed their own version of calculus in order to compute areas and volumes. This Greek calculus, which they named the method of exhaustion, was invented by Eudoxus (408?-355 B.C.) and perfected by Archimedes (287212 B.C.). The latter actually had several different techniques, not all of which bear a resemblance to the methods used in today's calculus texts. Only one of Archimedes's accomplishments is discussed here, and that in summary form— the determination of the area of the region bounded by a parabolic arc and its chord. Such details as Archimedes supplied appear in Appendix A and are given some justification in Section 1.2. The present section is limited to
1
2
ARCHIMEDES AND THE PARABOLA
a discussion of those aspects of Archimedes's arguments that are germane to the subsequent development of calculus. The parabola is one of several curves, collectively known as the Conic Sections, that were first studied in depth by Menaechmus (circa 350 B.C.) of the Platonic school. The book Conic Sections written by Archimedes's contemporary Apollonius of Perga (circa 262-circa 190 B.C.) contained over 4Ö0 propositions about these curves. The parabola was originally defined as the section of a right circular cone made by a plane that is parallel to one of the cone's generating lines (Fig. 1.1).
Fig. 1.1 The parabola as a conic section.
Figure 1.2 illustrates definitions needed to describe Archimedes's argument. Given any two points Q and Q' on a parabola, the portion of the parabola
Fig. 1.2 A typical parabolic segment.
bounded by them is an arc, and the straight line segment QQ' is its chord. The region bounded by an arc of the parabola and its chord is called a segment (of the parabola) and the chord is said to be the base of this segment. Given any segment of a parabola, that point P on the arc at which the tangent is parallel to the base is called the vertex of the segment (see Exercise 1.2.3). The crux of Archimedes's argument is the following geometrical proposition (see Fig. 1.3). Here and elsewhere we adopt Euclid's convention that the equality (or inequality) of geometrical figures refers to their areas only. Thus, the triangle's median divides it into two equal (though in general not congruent) portions.
THE AREA OF THE PARABOLIC SEGMENT
3
R
Fig. 1.3 Triangle's parabolic segments. Proposition 1.1.1 Let QQ' be the base of a parabolic segment with vertex P. If R is the vertex of the parabolic segment with base PQ, then APQQ' =
8APRQ.
Archimedes's proof of this fact appears both in the next section and in Appendix A. Here we are concerned only with how this proposition can be applied to the evaluation of the area of the parabolic segment. Archimedes's idea was to use this proposition to obtain successive approximations of the area of the parabolic segment by polygons that consist of conglomerations of inscribed triangles. For the parabolic arc QQ' with vertex P of Figure 1.4, the first approximating polygon Pi is APQQ'. If R and R! are the vertices
Fig. 1.4 Approximating a parabolic arc. of the parabolic arcs with bases PQ and PQ' respectively, then the second approximating polygon P 2 is the pentagon QRPR'Q' obtained by augmenting Pi with APQR and APQ'R'. Since each of these augmenting triangles has area one-eighth that of APQQ', it follows that P2 = Pi + 7P1. 4 If Si, S2, S3, S4 denote the vertices of the segments with bases QR, RP, PR', R'Q' respectively, then the third approximating polygon P3 is obtained
4
ARCHIMEDES AND THE PARABOLA
by augmenting P 2 with AQSiR, has area
ARS2P,
APS3R',
AR'S^Q',
each of which
\APQR = \APQ'R' = i · \APQQ'. Thus, P 3 = P 2 + 4 · —Pi = Pi + I ? ! + ¿ P i . 64 4 16 In general, each iteration of this process adjoins to each triangle Δ of the previous iteration two new triangles whose total area equals one-fourth that of Δ. Hence the total area of the chain of triangles adjoined in one iteration is also one-fourth the total area of the triangles adjoined in the previous iteration. Consequently,
P f c + 1 = P1 + ip1 + ... + - L p 1 + ¿p 1 = (i + i + ... + ¿)p 1 . Archimedes next argued that the parabolic segment is indeed the limit of the polygon Ρ^. This he justified by pointing out that in Fig 1.5, where P is
Fig. 1.5 A triangle inside a parabolic segment. the vertex of the parabolic arc QQ', and PS is an abbreviation for parabolic segment PQQ'', PS < parallelogram QSTQ' = 2APQQ' .·.
Í P S < APQQ'
.·.
PS - APQQ' < i p S .
Thus, each iteration of the approximating process reduces the uncovered portion of the parabolic segment by more than half. Formally, 0*>* Equation (3) is the principle that underlies the array commonly known as Pascal's triangle. Exercises 6 and 7 outline a proof of the Binomial Theorem. As Newton explained (see Appendix D), he was led to a fractional version of the Binomial Theorem in deriving the area under the curves of the form {l±x'2)m/n.
y=
He was aware that for integral values of the exponent, i.e., when n = 1, the Binomial Theorem yielded m —0 m —l m —2 m —3 ~ 2 3 4
m— (fc — 1) jfc
^'
as the coefficient of x2k in the expansion of (1 +x2)m. He also noted that each of these coefficients could be obtained from the previous one by multiplying it with an appropriate fraction. Thus, m — 0 m— 1 m —2 m —3 1 2 and, in general,
m \ k + l)
=
/ m —0 m — 1 m — 2 \
ι/ιτι\ \kj
m —k k +1
m —3
(5) .
iJ U ; · ^
(see Exercise 3). By analogy, Newton then reasoned that this recursion would also remain valid for fractional values of m. In other words, he guessed that the first coefficient is 1 and that the subsequent coefficients in the expansion of (1 ± x 2 ) m / / " satisfy the relation m/n\ ^k + lj
fm/n\ \ k )
m/n — k fc + 1
Accordingly, the first few coefficients of (1 + x 2 ) 1 / 2 are 1/2 0 l/2\ 1 ) l/2\
= 1, _ / l / 2 \ 1/2-0 1_ 1 ~ \ 0 ) ' 1 2 ~2' _ Λ/2Ν 1 / 2 - 1 _ 1 ( V
2/
vi y
l/2\ 3
/l/2\ \ 2
=
2 1/2-2 3
"2"u =
/ 1\ V 8/
(_1\=J_ V 2/ 16'
THE FRACTIONAL BINOMIAL THEOREM
(ΤΗΤ)^άΗ)
21
5 128'
In short,
The reason Newton examined ( 1 + x 2 ) 1 / 2 rather than the simpler expression ( 1 + x ) 1 / 2 was that the latter was too simple for his purposes. He was looking to integrate new functions and the substitution u = 1 + x converts (1 + x ) 1 / 2 to u 1//2 , which is easily integrated by means of the formula undu =
un+1
+ C, n+ 1 which, as will be seen below, Newton already knew. The extension of the Binomial Theorem to fractional exponents has many uses besides facilitating integration, and so the expansion of (1 + x ) 1 / 2 is of interest for other reasons. When x 2 is replaced by x, Equation (6) becomes /
(i
+
.>""-i + | . - i . · ^ - ^ - .
(T)
The replacement of x by — x2 transforms Equation (7) to
(i - χψ2 = i + \(-χ2) - \(-χΎ + ¿(-X2)3 - ¿ ( - z 2 ) 4 · · · =
1_IX2
2
1 4_
1
8
16
6_
5
128
8
w
which is one of the expansions that Newton wrote out explicitly (see Appendix D). Newton's fractional version of the Binomial Theorem is stated in full detail. The complete proof appears in Section 14.2. Theorem 3.1.1 (The Fractional Binomial Theorem) For every rational number r and nonnegative integer k, let the symbol (£) be defined recursively by
Then (1 + x)r = 1 + Qx
+ Q x
2
+ Q x
3
+ ■ ··
for |x| < 1.
It is clear from the above definition that if r is a positive integer, then (r+i) = 0 a n d hence (£) = 0 for all k > r, so that the binomial expansion is in fact finite. However, for fractional values of r, such as the above m = 1/2 (Equation 4), zero will not be encountered, and the expansion will continue ad
22
NEWTONS CALCULUS (PART 1)
infinitum. Newton was fully aware that his extension of the Binomial Theorem from integral to fractional values of m was lacking in logical grounding and so he verified Equation (8) by multiplying 1 _ 1 2 _ 1χ4 _ 1 6 _ 5 2 8 16 128
8
by itself and observing that the result was indeed 1 — x2. Note 1 x
_1 a 2
_1 4 8
1 6 16
_ 5 128
-~-x° .
16
_IX2 2
128
1 4 _ 1 β _ 5 8 8 16 128
+
-V
___
5
1 2o -=-x* 1 Λ -~x16 6 1 ---x X
1
8
r 4 +vex6 +kx* 1
"8*
4
1
6
+ x
ü
1
x
6
-Te
+
1
mx X
8 8
+
T2X -AX8 128
1
-x2.
Similarly, having derived the expression
(1 - χψ3 K
'
= 1 - \x2 - i.x4 3
5
6
oi. ■
81-X
9
Newton multiplied it twice by itself to obtain the value 1 — x2. The extended version of the Binomial Theorem turned out to be valid for negative exponents as well. Thus, according to this theorem,
(-l)(-2)(-3)(-4) +
4
+ 1-2-3-4 ' " 2 3 4 = 1- x +x - x + x + ··· .
Replacing x by — x in this equation results in the well-known (and tested) infinite geometric progression 1—x
= 1 + x + x2 + x3 + x4 + ■ ■ ■ .
AREAS AND INFINITE SERIES
23
Exercises 3.1 1. Find the first five terms of the expansion of a)
(1+x)1/3
b)
(1-x)1/3
c)
(1+x2)1/3
d)
( 1 _ 2x3)1/3
e)
(1+x)2/3
f)
(1 - x ) 2 / 3
g)
(1 + x 2 ) 2 / 3
h)
(1 - 2x 3 ) 2 / 3
i)
(1 + x ) 5 / 3
j)
(1 - x ) 5 / 3
k)
(1 + x 2 ) 5 / 3
3 5 3 1) (1 - 2x ) /
m)
(1+X)-2
n)
(1-x)-2
0)
(1+x2)-2
P)
(1 - 2 x 3 ) - 2
q)
(I+X)-4/3
r)
(I-X)-4/3
s)
(l+x2)~4/3
t)
(1 - 2x 3 )- 4 /3
u)
(1-x)-3.
2. Let the binomial coefficient (£) be defined by means of Equation (2). Prove that it satisfies the recurrence of Equation (3). 3. Let the binomial coefficient (£) be defined by means of Equation (2). Prove that it satisfies the recurrence of Equation (5). 4. Explain the relevance of the sequence 11°, l l 1 , l l 2 , l l 3 , l l 4 to the Binomial Theorem (see Appendix D). 5. Explain the statement of the Binomial Theorem in Appendix C. 6. Prove that if n is a positive integer then (1 + x ) n = Q + (")x + (tyx2 + l· (")x n · [Hint: It is clear that (1 + x ) n = (1 + x)(l + x) · · · (1 + x) = d0 + dix+d2x2 H \-dnxn for some integers d0> di> ^2, · · · An- That dk = (£) follows by differentiating this equation k times and setting x = 0.] 7. Use Exercise 6 to prove the Binomial Theorem [Equation (1)]. 3.2
AREAS AND INFINITE SERIES
Newton developed his version of the calculus in the years 1665-1666. Two or three years later he summarized this work in a tract On Analysis by Infinite Equations (De Analyst...) that he sent to Isaac Barrow (1630-1677), the Lucasian mathematics professor at Cambridge University. Translated excerpts from this tract appear in Appendix E. In this summary Newton stated his general principles as rules which are quoted and discussed here. The first rule refers to Figure 3.1 where A denotes the origin, AB = x, and BD = y. Rule I. Ifaxm/n
= y; it shall be _^_x(m+n)/n
m +n
=
area
ΑβΓ)
NEWTONS CALCULUS (PART 1)
A
B
Fig. 3.1 A typical curve. As Newton's comments toward the end of the tract indicate, he was fully aware of the relationship between the area under a curve and the antiderivative and of the difference between them. This statement about areas should be interpreted as one about antiderivatives, or indefinite integrals, whose modern equivalent is /
xrdx=-^—xr+1+C, r + 1
where r is any real number different from — 1. It is clear from Newton's examples that he knew exactly how to handle situations where the curve dips below the x-axis, or where A is not the origin. R u l e II. If the Value of y be made up of several such Terms, the Area likewise shall be made up of the Areas which result from every one of the Terms. This is the linearity property of integrals, which is written today as / \af(x) + bg(x)\ dx = a
f(x) dx + b / g(x) dx,
where a and b are any two real numbers. These two rules were applied (see Appendix E) in a series of carefully chosen examples to the evaluation of proper and improper definite integrals. Two of these are
Γ{Γ2+Γ*'2)άί
and
Γ Í2t3-3t5-Y-4
+
r3Adt.
The examples for the third of Newton's rules constitute the main portion of the tract. R u l e I I I . But if the value of y, or any of it's Terms be more compounded than the foregoing, it must be reduced into more simple Terms; by performing the Operation in Letters, after the same Manner as Arithmeticians divide in Decimal Numbers, extract the Square Root, or resolve affected Equations . . . . For pedagogical reasons the discussion of the first of Newton's examples for Rule III is preceded by the consideration of the special case 1+ x
AREAS AND INFINITE SERIES
25
Having noted that Rules I and II are of no avail in finding the area under this curve, we turn to Newton's suggestion for . . . performing the Operation in letters after the same Manner as Arithmeticians In particular, Rule III suggests that the same process of long division that is used to get rid of the troublesome denominators of fractions and converts 2345 to 75.64516129... 31 should be used to convert the function in Equation (9) to a more manageable form. Accordingly, to obtain the quotient when dividing the numerator of 1 by the denominator of 1 + x, focus on the leading terms of both expressions and obtain a quotient of 1 + 1 = 1. The process of long division then yields 1
1+x Γ~ϊ
1+ x -x + 0. To divide 1 + x into —x + 0, focus on the respective leading terms 1 and — x whose quotient is — x + 1 = —x and the process of long division now yields 1-x
1+x
fl
1+ x -x + 0 —x — x'5 P+0.
At the next stage, the quotient of the leading terms is x 2 + 1 = x 2 and two more repetitions are now easily seen to yield 1 - x + x2 - x3 + x4
1 +x [ I
1+ x -x + 0 — X — X¿
x2 + 0
X2 + X 3
-x3 + 0 - x 3 - x4 x4+0 x4 + x5 - x 5 + 0. The above process is summarized as the equation
1+x
= 1 - x + x2 - x3 + x4
.
(10)
26
NEWTONS CALCULUS (PART 1)
Of course, if x is replaced by —x, this equation is transformed into the infinite geometric progression = 1 + x + x2 + x3 + x4 H
. 1—x that was used by both Archimedes and Fermat in their evaluations of areas and that also occurred as a special case of Newton's Fractional Binomial Theorem. The first example to Rule III actually given by Newton deals with the hyperbola 0+X 2
The leading term of the quotient, a /b, is obtained by dividing the numerator a2 by the first term b of the denominator. The product of this quotient with the whole divisor is a2(b + x)/b = a2 + a2x/b, which is then subtracted from the dividend o 2 , yielding a remainder of —a2x/b: a2/b
b + x I a2
a2 + a2x/b -a2x/b + 0.
The dividend's second term —a2x/b2 is obtained by dividing the remainder —a2x/b by the same 6 of the denominator, leading to a new remainder of a2x2/b2 and so on. Thus, b+x
a2/b - a2x/b2 + a2x2/b3 | a2 a2 + a2x/b -a2x/b + 0 —a2x/b - a2x2/b2
- a2x3/b4 + ■■■
a2x2/b2 + 0 a2x2/b2 + a2x3/b3 -a2x3/b3 -a2x3/b3
+0 - a2x4/b4 a2x4/b4
and so
■ ■■
a2 a2 a2x a2x2 a2x3 = + b+ x T~~&2~ &3 ^4 · Newton then used Rules I and II to conclude that the area under the hyperbola [Equation (11)] was given by the expression a2x
a2x2
a2x3 3
~b
W^Hb
a2x4
W" '
AREAS AND INFINITE SERIES
27
As Newton noted, a similar long division yields the series expansion x2 + x 4 - x 6 + x 8 ·
1+x2
Actually, the same series may be obtained simply by replacing the x of Equation (10) with x 2 . Consequently, it follows from Rules I and II that the area under the graph of y = (1 + x 2 ) - 1 , from the origin to the point x on the x-axis, is 1
3
1 5
1 7
However, this same area is recognized as dx
/
1+x
tan
_i
_
x + C.
Since the area in question is 0 when x = 0, it follows that C = 0 so that we get tan_1x = x - - x 3 + - x 5 - - x 7 · · · , (12) 3 5 7 a series that was first derived by James Gregory (1638-1675). If we now set x = 1 in Gregory's series and recall that t a n - 1 1 = π/4, we obtain the series of Leibniz, π 1 1 1 Λ 4 = 1 - 3 + 5 - 7 - · This last series is intriguing and will be justified in Section 14.1. While it looks as though it could be used for evaluating the decimal expansion of π, this is not the case in practice. Estimates derived by calculating the partial sums of this series are not very accurate. Nevertheless, in combination with some other information, Gregory's series [Equation (12)] does indeed provide a practical method for evaluating π with great accuracy (Exercises 2, 3). Newton took the trouble to point out that in converting 1/(1 + x 2 ) to an infinite series by means of long division, the decision to consider 1 as the leading term of the denominator 1 + x 2 was subjective. If x 2 were taken as the leading term of this fraction, the long-division process would yield
1 + x2
x2 + 1
an expression, he noted, that is useful when the values of x are large. Newton also mentioned, without going into details, that the process of long division can be used to convert V
to
2x*/2-x3/2
= l + xV2-3x
2x 1 / 2 - 2x + 7x 3 / 2 - 13x2 + 34x 5 / 2 · · · .
n_.
(13)
(14)
28
NEWTONS CALCULUS (PART 1)
The following are the details omitted by Newton. The required computations can be simplified by setting u = x1/2 so that Equation (13) becomes u2 1+ u—
y
3M 2
To apply long division to the fraction (2 - M 2 ) / ( 1 + u - 3M 2 ), focus on the leading terms of both the numerator and the denominator and obtain 2-^-1 = 2 as the first term of the quotient. The required long division is 2 1+ u-
-
2 + 2 +
3M2
7M2
2M + OM
_
M2 6M 2
+
5M 2
2M
-2u
-
-2M
13M 3 +
-
2M 2
+
6M 3
7M2 7M2
—
6M 3 7M3
+
-13M3 -13M3
34u4
-
21M 4
+
21M 4 13M 4
+
39M 5
34M 5
-
39M 5
—
Consequently 1+ M-
2 - 2M + 7M2 - 13M 3 + 34M 4
3M2
from which expression (14) is easily obtained. The same expression could also have been obtained in a different manner. It follows from Equation (10) and several applications of the Binomial Theorem that 1 1+ M-
3M 2
1+
1 -
(M
3M 2 )
= 1 - (M - 3M 2 ) + (M - 3 M 2 ) 2 - (M - 3 M 2 ) 3 4- (« - 3 M 2 ) 4 · ■ ·
= 1 - (u -
3M 2 )
= 1 - u + 4M 2
+
(M2
7M
3
+
-
6M 3 19M
+
9M 4 )
-
(M 3
-
9M 4
+ ···) +
(W4
)·· ·
4
where the omitted portions contain terms of degree 5 or higher. Consequently, 2 - M2 T-5 = (2 1 + M - 3M^
M 2 )(1
-
M
+
4M 2
-
7M 3
+
19M 4
■ · ·)
= 2 - 2M + 7M2 - 13M 3 + 34M 4
Newton's Rule III also refers to analogs of decimal methods for the extraction of roots and the solution of equations. There was at that time a
AREAS AND INFINITE SERIES
29
well-known procedure for extracting square (and higher order) roots that was commonly taught in high schools until very recently and whose details were so onerous as to make the long-division process seem like child's play. Nowadays, however, subsequent to the advent of the hand-held calculators, few are familiar with it, and so there is no further discussion beyond its inclusion in Appendix E. On the other hand, the utility of infinite series in the solution of equations, a topic to which Newton dedicates a major portion of the De Analyst tract, is the subject matter of the next chapter. Exercises 3.2
1. Find the first five nonzero terms of the infinite series expansion of a) '
1 - 2x
b) '
2+ x
.x
1
v
1
;
e; 1-x-x2 . x +1 . g) l-9x2 ' .. l+2x ,, k J) 7 ) 1 +7x 3 3 2. Use the trigonometric
c) ' n
4x + 3 1
1-x2 ' l+x-2x2 x+ 1 .. 1+x 1} x2 + x + 1 ϊ ^ 2 1+x+x 2 ,. 2 + 3x 2 1 1 +, „x2 +■ x-44 *) 1 + x - x 2 + x 3 formula
t a n ( 4 4- B) = (tan A + tan B)/{1 - tan A tan B) to prove that π/4 = t a n _ 1 ( l / 2 ) + t a n _ 1 ( l / 3 ) . Use Equation (12) through the x 1 1 term and a calculator (with eight decimal places at least) to estimate 4[tan _ 1 (l/2) + t a n _ 1 ( l / 3 ) ] . Compare the resulting estimate of π with its actual value. 3. Use the trigonometric formula t&n(A + B) =
tan A + tan B 1 — tan A tan B
to prove that ττ/4 = 4 t a n _ 1 ( l / 5 ) - t a n _ 1 ( l / 2 3 9 ) . Use Equation (12) through the x 7 term and a calculator (with eight decimal places at least) to estimate 4[4tan _ 1 (l/5) — tan _ 1 (l/239)]. Compare the resulting estimate of π with its actual value. 4. Assuming that s i n - 1 x = / ( l — x 2 ) _ 1 / / 2 dx, obtain a power series expansion for s i n _ 1 x . 5. Assuming that ln(l +X) = J(l + x ) - 1 dx, obtain a power series expansion for ln(l + x ) . 6. Describe the main mathematical achievements of a) Isaac Barrow b) James Gregory
30
NEWTON 'S CAL CUL US (PART 1)
3.3
NEWTON'S PROOFS
De Analyst, Newton's first written description of the calculus, contains very little by way of formal proofs. Instead, he relied on lengthy computational verifications of specific examples to check on the validity of those conclusions to which his reasoning by analogy led. As his opening sentence states: "The General Method, which I had devised some considerable Time ago, for measuring the Quantities of Curves, by Means of Series, infinite in the Number of their Terms, is rather shortly explained, than accurately demonstrated in what follows." Nevertheless, in the concluding paragraphs, he did offer some arguments to support his methods. In subsequent works he formulated a fairly rigorous theory of differentiation but he never returned to the issue of the convergence of series. The following paragraphs are lightly edited extracts from these arguments. A more faithful rendition of the original text appears in Appendix E. Newton's argument for Rule I is quite reminiscent of Fermat's max/min method. However after discussing the specific example of the area under the curve y = x1/2, Newton gives a more general argument for y = xmln. Figure 3.2 illustrates this proof.
Fig. 3.2 The area as an antiderivative.
Let then ADS be any Curve whose Base AB = x, the perpendicular Ordinate BD = y and the Area ABD = z, as at the Beginning. Likewise put Bß = dx, BK = v; and the rectangle BßHK ((dx)v) = Space BßSD. Therefore it is Aß = x + dx and Αδβ = z + (dx)v. Which things being premised, assume any Relation betwixt x and z that you please and seek for y in the following Manner. z; or | x 3 Take, at Pleasure, | x 3 ' 2 z 2 . Then x + dx(Aß) being substituted for x and z + (dx)v(Aöß) for z, there arises ~(x3 + 3x2dx + 3x(dx)2 + (dx)3) = z2 + 2zv{dx) + (dxfv2
NEWTON'S PROOFS
31
(from the Nature of the Curve). And taking away Equals ( | x 3 and z2) and dividing the Remainders by dx, there arises -(3x 2 + 3x(dx) + 3(dx)2) = 2zv + (dx)v2. Now if we suppose Bß to be diminished infinitely and to vanish, or dx to be nothing, v and y in that Case will be equal, and the Terms which are multiplied by dx will vanish: So that there will remain Í ■ 3x2 = 2zv or
\x2 (= zy) = | x 3 / 2 y ;
or x 1 / 2 (= x2/xa/2)
= y.
Wherefore conversely, if it be x1//2 = y, it shall be | x 3 / 2 = z. Or, universally, if [n/(m+n)]ax^m+n''n = z; or putting na/(m + n) = c and m + n = p, if cxP/'n = z; or cnxp = zn: Then by substituting x + dx for x and z + (dx)v (or which is the same z + {dx)y) for z, there arises c n (x p + p(dx)x p_1 · · ·) = zn + n{dx)yzn-1 the other Terms, which would at length vanish being neglected. Now taking away c"xp and z" which are equal and dividing the Remainders by dx, there remains / \ / cTI pxp— 1 = nyz n—1 /(= nyz n ¡z) = nyc n xyp/cx'
p/n
.
n p
or, by dividing by c x , it shall be or pcx^p~n''n = nj/;
px~ = ny/cx
or by restoring na/(m + n) for c and m + n for p, that is m for p — n and na for pc, it becomes axm'n = y. Wherefore conversely, if axm'n = y, it shall be [n/(m + n)]ax ( m + n ) / n = z. Q.E.D. The above essentially valid derivation of the standard differentiation formula (xr)' = r x r _ 1 is followed in the conclusion of De Analyst by a considerably less satisfactory discussion of the convergence of power series. Here Newton seems to be trying to mimic the method used by Archimedes to argue that the successive polygons inscribed in the parabolic segment converge to it. This argument, as was mentioned in Section 1.1, was based on the principle that if one repeatedly removes more than half of a quantity the remainder converges to zero. Newton's discussion makes use of the easily verified (see Exercise la) fact that for any positive integer n
+
+
+
+
G)'-í[G)" Gr Gr Gr Here are Newton's own words: Thus, if it be x = 1/2, you have x the half of all these x + x 2 + x 3 + x 4 · · · and x 2 the half of all these x 2 + x 3 + x4 + x 5 · · -. Therefore if x < 1/2
32
NEWTON'S CALCULUS (PART 1)
x shall be greater than half of all these x + x2 + x3 ■ ■ ■ and x2 greater than half of all these x2 + x3 + x4 ■ · ·. Thus if x/b < 1/2, you shall have x more than half of all these x + x2 /b + x3/b2 · · ·, And the same Way of others. And as to the numerical Coefficients, for most part they perpetually decrease: Or if at any Time they increase, you need only suppose x some few Times less. If the readers feel that this argument is weak, they are right. After all, the infinite geometric progression also converges for values of x that are greater than 1/2, to which values Newton's argument does not apply. Moreover, modern standards of rigor cannot be satisfied with a proof that uses the phrase "...for the most part... ." Chapters 9, 10, 13, and 14 will supply the details missing from Newton's tract. In defense of Newton's arguments it should be pointed out that several of these modern proofs regarding the behavior of the general power series are indeed based on comparisons with the infinite geometric progression. Exercise 3.3
Use the infinite geometrical progression to verify the following assertions made by Newton. 1. For every positive integer n,
■ » i - a G)**G) b) xn > Uxn + xn+1 + xn+2
+
+ xn+3
c) anxn > - (anxn + an+1xn+l
+
G) G)
+
+ ■ ■ ■) if 0 < x < |
+ an+2xn+2
+ an+3xn+3
+ ■■■) if 0 < x
03 > 04 > · · · > 0 d) anxn > - (anxn + an+lxn+1
+ an+2xn+2
— and 0 < Ofc+i < 2α& for each k = 1,2,3,... e) anxn
> - (anxn + an+1xn+1
+ an+2xn+2
+ an+3xn+3
+ · · ·) if b > 0,
0 < x < —, and 0 < ak+i < ba^ for each k = 1,2,3,... 2b Chapter Summary
Isaac Newton is credited with consolidating the disparate methods of several of his predecessors and contemporaries into a unified branch of mathematics
NEWTONS PROOFS
33
that became known as calculus. This corpus related to both differentiation (and its reverse, integration) and the topic of (infinite) power series. He was led to many of his major contributions to this topic by reasoning by analogy. In particular, he figured that the same binomial theorem that was known to hold for positive-integer exponents should also hold for arbitrary rational exponents. Similarly, the four arithmetical operations that hold for real numbers should also hold for variables. Both of these assumptions led him to infinite power series. Newton did offer some arguments to support his new mathematics, but these remain ultimately unsatisfactory.
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L-
Newton's Calculus (Part 2) As Newton recognized and stressed, power series provide a powerful technique for solving a variety of algebraic and differential equations. 4.1
THE SOLUTION OF DIFFERENTIAL EQUATIONS
An ordinary differential equation is an equation that relates x, y, y' and possibly higher derivatives. Such equations are y' = x + y, y' = 1 - 3x 4- y + x2 + xy, (1 + x2)y" + 2xy' + Ax2y = 0. Many important mathematical and scientific problems can be reduced to the solution of such equations. Newton knew this and he devoted several pages of "On the Method of Fluxions..." to their study. In these he presented a variety of techniques for solving such differential equations. The most general of these calls for replacing y and its derivatives by power series with unknown coefficients and then using the given differential equation to. evaluate these coefficients. This is the only one of Newton's techniques that will be discussed in this section by means of several examples that do not involve second or higher order derivatives. The general outline of his method is the same for all of the examples and seems eminently reasonable. In fact, however, several of the steps need justifications that will only be provided in Chapters 13 and 14. The variable y 35
36
NEWTON'S CALCULUS (PART 2)
in the proposed differential equation is assumed to have an (unknown) power series expansion in terms of x. The power series of y' is then obtained by differentiating this power series term by term and both y and y' are replaced by their series expansions. The given differential equation is then broken down into an infinite number of equations, one for each power xn and, if possible, these derived equations are used to evaluate the coefficients of the power series of y. Example 4.1.1 Consider the differential equation y' = x + y. Suppose the solution of this equation has the infinite series expansion oo
y = Co + c\x + C2X2 + C3X3 + C4X4 -\
= ¿2 °ηχη·
(!)
n=0
Term-by-term differentiation yields 00
y' = ci + 2c2x + 3c3x2 + 4c 4 x 3 H
= J ^ nCnXn~l 71=1
00
= £ > + l)c+iiB.
(2)
n=0
The substitution of Equations (1) and (2) into the given equation transforms the latter into 00
00
n=0
n=0
^T(n 4- l)c„+ix n = x + Σ
v"-
With Newton we take it for granted that for each power xn the corresponding coefficients on the two sides of the equation must balance out. This results in the following infinite number of level equations: Level
Equation
Recxj
x°
C\ = C 0
c\ =
x1
2c2 = 1 + ci
C2 =
xn (n > 2)
(n+ l)cn+i
Hence, if Co is replaced by c, we have c\ = c , 1+c C2
:
2 ' _ 1+c
°'S ~ 2 - 3 '
=cn
Cn+1
1+ci 2 n+1
THE SOLUTION OF DIFFERENTIAL EQUATIONS
37
1+C
C4
2-3-4' 1 + c
ttor n >^ 2. o
cn = —— n!
In other words, the general solution to the given differential equation is the function (x2 x3 x4 \
Example 4.1.2 Consider the differential equation y' = 1 — 3x + y + x2 + xy. Once again it is assumed that y and y' satisfy Equations (1) and (2), respectively, so that the given equation is transformed into oo
oo
^ ( n + l ) c „ + i x n = 1 - 3x + x2 + ^
oo
c„x n + x ^
n=0 oo
= 1 - 3x + X2 + Σ
n=0 oo
CnXn + Σ CnXnA
n=0 op
= 1 - 3x + X2 + Σ
cnxn
n=0 oo
CnXn + Σ
n=0
On-lXn.
n=l
This translates to the table: Level
Equation
Recurrence
x°
ci = 1 + Co
ci = 1 + co - 3 + ci +co C 2 = 2 1
X
2c2 = - 3 + cx + Co
X2
3c 3 = 1 + c 2 + c\
1
η
τ
(n. > '.X\
InA-Wr.
— r. A-r
,i
Hence, if we replace Co by c, we get ci = 1 + c, C2 = — 1 + C, C3 = C4 =
05
l + 2c
—g—,
- 2 +5c
_
12~' _ 2 +13c
~—m~'
r
1 1 —
n +1
38
NEWTON'S CALCULUS (PART 2)
The particular case c = 0 yields the solution derived by Newton, „2
,.3
^5
„.4
x
x
x
E x a m p l e 4.1.3 This section's method can be used to provide Newton's Fractional Binomial Theorem with a somewhat firmer grounding than the reasoning by analogy given in Section 3.1. Observe that the function y = (1 + x)r has the derivative y' = r ( l + x ) r _ 1 and hence it is a solution of the differential equation r-y = (l+x)-y'. (3) The substitution of Equations (1) and (2) into this equation yields oo
oo
χΤΙ
r ^2 °η n=0
x
oo
η
χΎΙ Χ
= (1 + ) 5Z °η ~
= 5Z η ~
n=\
n=\ oo
oo
= Σ(η
+ ι
)°η+ιχη
oo
ηα χη 1
+ X]
n=0
+ ^Ζ nc « x ™ n=l
ncnxn.
n=0
The level-fc equation is rck = (fe + l)c fc+ i +kck,
k = 0,1,2, ■· ·
or cii+i = T—r-Cfc,
A; = 0,1,2, ··■ .
The choice of the particular value co = 1 then yields * = (!)·
^0'1'2····
Thus, both the binomial series and the function (1 + x)r are solutions of the same differential equation (3) that have the same initial value of 1 at x = 0. It therefore seems reasonable to conclude that they are equal and that we have here a proof of Theorem 3.1.1. There are several problems with this purported proof of the Fractional Binomial Theorem. In the first place, unless r is a positive integer, the binomial series never converges for x > 1 and at this point it is quite conceivable that for some values of r it may never converge at all. Thus, it is not clear that the binomial series is in fact a function in any sense of the word. This problem will be resolved in Example 10.1.5, where it will be demonstrated that, regardless of the value of the exponent r, the binomial series always converges for - 1 < x < 1. Another deficiency of this argument is that even after the binomial series is proven to actually define a function, it is not immediately clear that this
THE SOLUTION OF DIFFERENTIAL EQUATIONS
39
function is in fact differentiable. That this is indeed the case is proven in Corollary 13.2.10. Finally, the overall logic of this argument is that once the initial value at x = 0 has been specified, there is only one solution to any given first-order differential equation. This need not be the case, as is demonstrated by the two functions /(x)=0
x2 / ( « ) = —,
and
both of which are solutions of the initial-value problem y' = y1/2,
y(o) = o.
This last difficulty will be resolved by Corollary 14.2.7. In conclusion it is necessary to point out that the method presented in this section is not infallible. Exercise 6 exhibits an instance of the failure of this method or, rather, indicates that it sometimes requires further refinements. Exercises 4.1 1. Use the method of infinite series to solve the following differential equations: a)
y' = x - y
b)
y' = 1 + x + y
c)
y' = 2 - x + xy
d)
1+2x-2y+
e)
y' = 1 + xy
g)
y' = y +
T
^
3y' = xy 2
f)
y' = 2 - x y
h)
y' = y +
Y^
y' = y +
tan-1x
i)
y' = V + v 7 ! + x
j)
k)
xy' = 1 + y
1) x2y' =
\-y
m)
3-x2
a)
y" - 2xy' + y = 0
b)
y" = xy
c)
y" + y' + y = x
d)
ν" + ν' + ν =
e)
y"-Zy'
f)
y"-3y'-10y
+ 2xy = y' + n) y' = ——¿, j/(0) = 0 1—x 1 + x¿ o) y'= ΓΛ y(0)=0 vT x 2 2. Use the method of infinite series to solve the following differential equations:
+ 2y = 0
γ ^ = x
40
NEWTON'S CALCULUS (PART 2)
3. Use the fact that y = sinrr satisfies the differential equation y" = — y to show that 9 x3 x5 x7 _x . . . . sinx = a : _ _ + _ _ _ + 4. Use the fact that y = cosx satisfies the differential equation y" = —y to show that 2 A 8 G _ _x _x _ _X + _x . . . . cosx = 1
+
5. Use the fact that y = ex satisfies the differential equation y' = y to show that 7
ΧΔ
■}
X°
4
X*
Ϊ ! + 2 ! + 3! + 4 Γ · · · 6. Show that y = 1/x is a solution of the differential equation xy' + y = 0 that cannot be obtained directly by this section's method. Explain how the substitution u = 1 + x resolves this difficulty. β
4.2*
= 1 +
X
THE SOLUTION OF ALGEBRAIC EQUATIONS
By the time Newton was developing his version of calculus, mathematicians had devised a variety of numerical methods for solving algebraic equations such as x7 - 5x4 + 3x2 - 2 = 0. These methods employed recursive procedures in which each estimate to the solution was used to obtain a better estimate, much like the Newton-Raphson method that is commonly taught in first-year calculus. In his tract De Analyst Newton explained one such technique that is actually different from the aforementioned Newton-Raphson method. He then went on to show how the same technique could be used to solve an algebraic equation such as y3 + y - 2 + xy - z 3 = 0
(4)
for y. The solution here is no longer a number, of course, but a power series expansion of y in terms of x. The overall strategy is the same for these algebraic equations as for the differential equations of the previous section. It is first assumed that a power series expansion of the form of Equation (1) actually exists. This series is then substituted into the given equation and, by focusing on the successive powers of x, the coefficients of the power series are evaluated one at a time. Unfortunately, the given equation may include powers of y, leading to powers of Equation (1) and a situation where the resultant level equations are difficult to describe. This complication calls for a further refinement of the method. Rather than write down all of the resultant level equations and then solve for as many coefficients as needed, these level equations are derived, and solved, one at a time. As each of the level equations is solved, the solution is used
THE SOLUTION OF ALGEBRAIC EQUATIONS
41
to modify the given equation and prepare it for the next iteration of the process. The formal description of this process necessitates two definitions. First, assuming Equation (1), we define the tail series tn = cnxn + Cn+ixn+1 + c „ + 2 x n + 2 + · ■ · , Clearly, to = y
and
tn = c„x n 4- t n +i
for
n = 0,1,2,....
(5)
all n.
Second, the given algebraic equation will be assumed to be F(x,y)
= 0.
As the given equation will be modified in each iteration of the solution process, we begin by setting F0(x,y) = F(x,y). The kth iteration of the solution process consists of the execution of the following three steps: Step 1: Extract the xfc-level equation out of Fk(x,tk)
= 0.
fc
Step 2: Solve the x -level equation for ck. Step 3: Substitute tfc = CkXk + ife+l into Fk(x, tk) = 0 and use the Binomial Theorem to obtain the new equation Fk+1(x,tk+i) = 0. The logic and limitations of this iterative procedure will be discussed following an example. E x a m p l e 4.2.1 Consider the equation y3 - x - 1 = 0. Here we have F 0 (x,to):
ig - x - 1 = 0 .
Iteration # 0 Step 1: Since io = co+cix+C2X2-\ , and since the x°-level equation ignores all nonconstant terms in Fo(x,to), it follows that x° level:
eg - 1 = 0.
Step 2: The only real solution of this equation is Co = 1. Step 3: In Fo(x,to) above replace io with 1 +t\ to obtain (l+ii)3-x-l=0,
42
NEWTON'S CALCULUS (PART 2)
which simplifies to Fi(a;,ii):
3*i + 3if + if - x = 0.
Iteration # 1 Step 1: Since t\ = cyx + C2X2 + C3X3 + · · ■ and since the i 1 -level equation ignores all terms of degree >1 in Fx(x,ti), it follows that x1 level:
3ci — 1 = 0 .
Step 2: ci = - . Step 3: In Fi(x,t\)
replace ¿i with x / 3 4- t 2 to obtain
3
(Ι + ί 2 ) + 3 (Ι + ί 2 ) 2 + (Ι + ί 2 ) 3 -^= 0 '
which simplifies to F2(x, t 2 ):
3i 2 + 3*1 + i | + 2t2x + t\x + y + ^ - + | ^ = 0.
Iteration # 2 Step 1: Since t2 = c 2 x 2 + C3X3 + C4X4 + ■ ■ ■ and since the x2-level equation ignores all terms of degree >2 in F2(x, t2), it follows that x? level:
3c2 + \ = 0.
Step 2: c 2 = — - . Step 3: Stop. Thus, xz
2: y
3 9 is an approximate solution to the given equation. The Fractional Binomial Theorem affords us with another way to solve this example's equation. Accordingly, # 1 + x == ( l + x ) 1/3 1/3 1+ X —
+m.
=
1 +
1/3- - 1 2—x
( 2 ■
1
M K 2
1+ -
■ f
X2
9 "
We now restate and comment on the three steps in each iteration of Newton's procedure.
THE SOL UTION OF AL GEBRAIC EQUATIONS
43
Step 1: Extract the xfc-level equation out of Fk(x,tk) = 0. If the term tk in the equation Fk(x, tk) =0 were replaced by its power series expansion of Equation (5), each summand would have xk as a factor. The reason for this is that the Step 3 substitutions of iterations 0 , 1 , 2 , . . . , k — 1 have eliminated all the terms of degrees 0 , 1 , 2 , . . . , k — 1. Consequently, the only summands of Fk(x,tk) that make a contribution to the xfe-level equation have forms Atk and Bxk for some numbers A, B. In other words, the xfe-level equation has the form = 0.
Ack+B
Step 2: Solve the Xfc-level equation for Cfc. If the number A above is not zero, we get B Ck =
-
T
If A is zero, the method fails, and a refinement is needed. This refinement is known as Newton's Polygon Method and falls outside the scope of this text. Step 3: Substitute tk = CkXk + ifc+i into Fk{x,tk) = 0 and use the Binomial Theorem to obtain the new equation Fk+i(x, tfc+i). By the time we get to iteration # 3 this step is of course quite messy, and it was this complexity that dictated our decision to stop after iteration # 2 . Computerized symbolic manipulators can be quite helpful here. All of this section's examples are redone and extended at the end with the aid of the application Mathematica. E x a m p l e 4.2.2 Consider the equation y3 + y — 2 + xy — y3 = 0. Replacing y with i 0 yields F0{x, t0):
t^+t0-2
+ t0x-x3
= 0.
Iteration # 0 Step 1: Ignoring all the nonconstant terms in Fo(x,to), we get x° level:
c | + CQ - 2 = 0.
Step 2: The only real solution of this equation is Co = 1. Step 3: In FQ(X, to) above replace to with 1 + ii to obtain (1 + i i ) 3 + (i+ti)-2
+ x(l + h) - x3 = 0,
which simplifies to Fi{x,ti):
4t1+3tj+tl
+ x + tix-x3
= 0.
44
NEWTON'S CALCULUS (PART 2)
Iteration # 1 Step 1: Ignoring all xn with n > 1 in F\(x,ti) xl level: Step 2:
Cl
yields
Ac\ + 1 = 0 .
= -- .
Step 3: In Fi(x,t\)
replace t\ with — x/A + t2 to obtain
4 ( - | + ί 2 ) + 3 ( - | + ί 2 ) 2 + ( - | + ί 2 ) 3 + χ + ( - | + ί 2 ) χ - χ 3 = 0, which simplifies to
F2(x,t2):
2
2
„ o t2x 3Üx At2 + *tl +Qtl----f-x _ +3t2_x -
65x 3
—
=0.
Iteration # 2 Step 1: Ignoring all a:n with n > 2 in
, t2) yields
x2 level: Step
2: C2 =
Step 3: Stop.
4c2 -
έ·
The above computations lead to the conclusion that υ = 1
X
ΧΔ
1 4 64 is an approximate solution to the given equation. More terms can be obtained by further iterations of this procedure. y
As is the case for equations with a single unknown, solutions of algebraic equations need not be unique. Example 4.2.3 Suppose F(x, y) = 3z 2 + 2xy + y2 + x - y - 2 = 0.
(6)
This gives F0(x,t0):
3x2 + 2xt0 + t% +x-t0-2
= 0.
Iteration # 0 Step 1: Ignoring all the nonconstant powers of x in FQ(X, to), we get
THE SOLUTION OF ALGEBRAIC EQUATIONS
45
c2, - CQ - 2 = 0,
x° level:
which has solutions Co = — 1, 2. (Accordingly, there will be two solutions for y. We continue here by deriving the solution that corresponds to y = 1 and will pursue the other solution below.) Step 2: Co = — 1 (see comment above). Step 3: In F0(x, to) replace io with —1 + ίχ to obtain 3x 2 + 2 x ( - l + ii) + ( - 1 + i i ) 2 + x - ( - 1 + ii) - 2 = 0, which simplifies to - 3 t i + tf - x + 2iix + 3x 2 = 0.
Fi(x,ii): Iteration # 1
Step 1: Ignoring all xn with n > 1 in Fi(x,ti), x 1 level: Step 2:
Cl
we get
—3ci — 1 = 0 .
= --.
Step 3: In Fi(x,ti) replace *i with —x/3 + Í2 to obtain - x + 2x ( - | + 1 2 ) + 3x 2 = 0,
- 3 ( - | + t2) + ( - | + i2) which simplifies to
4i x 22x 2 „ - 3 i 2 + i i9 + ^ 2- + - r - = 0 .
F2(x.,,ta): Iteration # 2 Step 1: Igiíoring
22 -3c 2 + — = Λ0.
x 2 level: Step 2: c 2
22 ~ 27
Step 3: Stop. The above procedure yields y
„
=
x
22x 2
-1+3+^7-
as a partial solution to the given equation.
46
NEWTON 'S CAL OIL US (PART 2)
On the other hand, if we begin with CQ = 2, the recursive procedure yields: Iteration # 0 Step 1: Same as above. Step 2: c 0 = 2. Step 3: In Fo(x, to) set to = 2 + ti to obtain 3x 2 + 2x(2 + ti) + (2 + i i ) 2 + x - (2 + ii) - 2 = 0, which simplifies to 3ii + t\ + 5x + 2*ix + 3x 2 = 0.
Fi(x,ti): Iteration # 1
Step 1: Ignoring all x" with n > 1 in ί \ ( χ , * ι ) , we get x 1 level:
3c^ + 5 = 0.
Step 2: cx = - - . Step 3: In F\{x,t\)
set fi = - ( 5 x / 3 ) + *2 to obtain \
OX
3(-y+i
2
/
OX
) + {-γ+ί
2
)
\
. „
.
n
dX
I
\
which simplifies to „,
*
F 2 (x, t 2 ):
9
4* 2 x
22x 2
3i2 + t\ - -f- + —
Iteration # 2 Step 1: Ignoring all x" with n > 2 in F2(x,*2)> we get 22 3c2 + — = 0.
x 2 level: 22 Step 2: c2 = - —. Step 3: Stop. Starting with CQ = 2, we obtained „
5x
. r» 2
+ 5 x + 2l - y + ί 2 ) χ + 3 x ^ = 0 ,
22x2
as a partial solution to the given equation.
=
THE SOLUTION OF ALGEBRAIC EQUATIONS
47
Because Equation (6) is of the second degree in y there is another way to derive the infinite series expansion of y. This equation can be solved for y by applying the quadratic formula and then going on to evaluate the radical in the solution by means of the Fractional Binomial Theorem (Exercise 3). As was the case for the differential equations of the previous section, the method described here needs further amplification (see Exercise 6). The details again fall outside the scope of this text. Exercises 4.2
1. Find the first three terms of the infinite series expansion of y in terms of x in the equations below in two different ways: using the method described in this section and using the Fractional Binomial Theorem. a) c) e)
b) y2 — x — 1 = 0 d) y3 - (1 + x)2 = 0 f) y — x2y = 1
y — xy = 1 y2 - (1 + x)3 = 0 y — Ixy + x2y = 1
2. Without the use of a calculating device, find the first three terms of the expansion of y in terms of x where the two variables are related by:
a) c) e) g) i)
y-xy =0 b) y3 y3 + xy = 1 d) y2 2 x - 2y + y = 0 f) y2 3 3 y + 2y - 3 + xy - x = 0 h) y3 2 2 3 x y - 2y + x - y + 18 = 0 j) x2
+y +x =0 +y+x = 0 - y = x + 3y - 4 - x2y + 2x = 0 + 2xy - 2y2 + x - y + 1 = 0
3. Solve Equation (6) for y by means of the quadratic equation and then use the fractional version of the Binomial Theorem to find the first three terms of the power series expansion of both solutions. 4. Use a symbol manipulator to derive the first five terms of the expansion of y in terms of x for the equations listed in Exercise 2. 5. Use a symbol manipulator to derive the first five terms of the expansion of y in terms of x where they are related by: a) c)
x2 - 3xy3 + y4 - 16 = 0 x2 - Zxy3 + y6 - 1 = 0
b) d)
x2 - 3xy3 + y5 - 1 = 0 x2 - 3xy3 + y7 - 1 = 0
6. Show that this section's method fails to solve the equation xy = 1. Explain how this is remedied by the substitution u = 1 + x. 7. Find the first five terms of the infinite series expansion of the y in terms of x in the equations below. a
) b) c)
2/3 + V = 2 cos a; y3 — y = sinx y = tanx
(Hint: Use Exercise 4.1.4) (Hint: Use Exercise 4.1.3) [Hint: Use Equation (12) of Chapter 3]
48
NEWTONS CALCULUS (PART 2)
d) e)
siny = x ey = 1 + x
(Hint: Use Exercise 4.1.3) (Hint: Use Exercise 4.1.5)
8. Let P(x) be a polynomial in x such that P{0) is positive. Prove that this section's method will produce a solution to the equation y2 = P(x).
Chapter Appendix: Mathematica Implementations of Newton's Algorithm Example 4.2.1 y3 - x - 1 = 0 In[31]: = Original = y"3 - 1 - x Out[31] = -l-x + y3 In[32]: = FO = Original / / . y -> tO Out[32] = - 1 + t0 3 -x In[33]: = Solve[(FO//.x -> 0) = = 0,t0] Out[33] = {{tO -> 1}, {tO - - ( - l ) 1 ' 3 } , {tO - . (-1) 2 / 3 }} In[34]: = F l = Expand[FO//.tO -> 1 + t l ] Out[34] = 3tl + 3tl 2 + t l 3 - x In[35]: = Solve[3tl-x = = 0, t l ] Out{35] = {{tl — §}} In[36]: = F2 = E x p a n d [ F l / / . t l -> x / 3 + t 2 ] Out[36] = 3i2 + 3t22 + Í2 3 + 2t2x + í22x + ¿ + ^ + £ In[37j: = Solve[3t2+x"2/3 = = 0,t2] Out[37]= { { t 2 - - ^ } } In[38]: = F3 = Expand[F2//.t2 -> -x"2/9+t3] Out[38]= 3 t 3 + 3 t 3 2 + t 3 3 + 2 t 3 i + t 3 2 x - í 3 f ^ - í 3 4 ^ - ^ - - 2 i | ^ + i ^ + f ^ - ^ In[39]: = Solve[3t3-5x"3/27 = = 0,t3] Out[39]= { { t 3 - ^ } } In[40]: = F4 = Expand[F3//.t3 -> 5x*3/81+t4] Out[40j = 3t4 + 3t42 + Í43 + 2i4x + t 4 2 x - *£■ - *Af- + *$- + ^ ^ - + ^ + 1 3 t 4 j 4 _ 2 ¿ _ 10t4i 5 _ 8 x 6 , 25t4x 6 ,r 4 0 l 7 81 243 243 2187 "·" 2187 " 6561
In[41]: = Solve[3t4+10x~4/8 = = 0,t4] Out[411 = { { t 4 ^ - ^ } } Hence x x2 5x3 ICte4 3
9
81
243
Example 4.2.2 y3 + y - 2 + xy - x3 = 0 In[15]: = Original = y~3+y-2+x*y-*~3 Out[15] = -2 - x3 + y + xy + y3 ln[16j: = FO = Original / / . y -> tO Out[16] = - 2 + tO + tO3 + tOi - i 3 In[22]: = Solve[(F0//.x -> 0) = = 0,t0] Out[22] = { {Í0 — 1}, { tO — \{-\ - Iy/7)} , { Í0 - | ( - 1 + Is/7)} } In[23j: = F l = Expand[FO//.tO -> 1 + t l ] Out[23] = 4tl + 3tl 2 + i l 3 + x + tlx - x 3 In[24]: = Solve[4tl+x = = 0, t l ] Out[24] = { { U - . - Í } } In[25]: = F2 = E x p a n d [ F l / / . t l -> (-x/4)+t2]
APPENDIX
49
Out[25] = 4t2 + 3t2 2 + Í2 3 - ψ - ^ í p - f§ + ^ - -131x"3/128+t4] 2 Out[29] = 4Í4 + 3Í4 + Í 4 3 4- ψ - &&*■ - &$£+ 35í £ ¿ - ®&έ. _ Z8f§f¿1 4 5 6 6 I 2111x , 6291t4x 4719x Λί\αη ~*~ i Λί\αα ΙΛΟΟΛ 128 "'1" 4096 4096 16384 412257a: 7 , 5 1 4 8 3 x 8 2248091x9 5 2 4 2 8 8 "*" 1048576 2097152
393t4x 4Anna 096
. 830017x , 51483t4x ' 262144 οβθΐΛ/t "*'" 1£Q9/| "·" 16384
In[30]: = Solve[4t4+2111x~4/409e = = 0,t4] Out[30]={{t4->-*-}££}} Hence
-ι_Ξ
y
χ2
131χ3
2111χ4
+
~ ~ 4 64 ~ 512 ~~ 16384 Example 4.2.3 3x + 2xy + y2 + x-y-2 =0 2
In/42/: = Original = 3x"2+2x*y+y"2-Hx-y-2 Out[42] = - 2 + x + 3z 2 - y + 2xy + y2 In[43]: = F0 = Original / / . y - > tO Out[43] = - 2 - tO + tO2 + a: + 2t0x + 3x2 In[44j: = Solve[(F0//.x -> 0) = = 0,t0] Out[44] = {{tO -* 1}, {tO -> 2}} (* First branch *) In/45]: = F l = Expand[F0//.t0 - > - l + t l ] Out[45] = - 3 t l + t l 2 - x + 2tlx + 3x2 In[46]: = Solve[-3tl-x = = 0, t l ] Out/46] = {{il — - f } } In[47]: = F2 = E x p a n d [ F l / / . t l -> - x / 3 + t 2 ] Out[47] = -3Í2 + Í2 2 + *ψ. + 22sl In/49]: = Solve[-3t2+22x"2/9 = = 0,t2]
Out/49] = { { t 2 ^ 2 | ¿ } }
In[51]: = F3 = Expand[F2//.t2 -> 22x~2/27+t3] Out/51] = -3Í3 + t3 2 + * § * + ^ ^ - + fiff^ + ^ In/57/: = Solve[-3t3+88x~3/81 = = 0,t3] Out/57] = { { í 3 - f f ¿ } }
Hence
,
x
22x2
88a:3
3
27
243
(* Second branch *) In[52]: = F l = Expand[F0//.t0 - > 2 + t l ] Out/52] = 3tl + t l 2 + 5a: + 2tla: + 3a:2 In/53]: = Solve[3tl+5x = = 0, t l ] Out/53] = {{α^-ψ}} In/54]: = F2 = E x p a n d [ F l / / . t l -> -5x/3+t2] Out/54] = 3t2 + t2 2 - *ψ + &j£
50
NEWTON'S CALCULUS (PART 2)
In/557: = Solve[3t2+22x"2/9 = = 0,t2]
Outf55/={{t2--22¿}}
In/567: = F3 = Expand[F2//.t2 -> 22x"2/27+t3]
Out/567 = 3t3 + Í32 - *ψ - ^ ^ + ^
In/587: = Solve[3t3+88x"3/81 = = 0,t3] Out/587 = { { « - - f f ¿ } } Hence
„
+ iHfj'
5x
22x2
88x 3
3
27
243
Chapter Summary
Mathematics progresses by posing problems and finding their solutions. Newton's power series above and beyond providing methods for evaluating complicated functions also make possible the effective solution of interesting differential and algebraic equations.
5
Euler
The most dominant figures on the mathematical landscape of the eighteenth century were Leonhard Euler (1707-1783) and Joseph Louis Lagrange (17361813). One of their many contributions to calculus is their recognition of the importance of infinite series of trigonometric functions to both pure and applied mathematics. The discussion below, extracted from an article of Euler's, has several interesting aspects: complex numbers are used to obtain new information about real numbers, and manifestly false equations lead to correct and interesting results.
5.1
TRIGONOMETRIC SERIES
In his article Subsiduum Calculi Sinnum, Euler put the infinite geometric progression 1+x + x2 +x3 + ■■■ l-x to a very surprising use. He replaced the variable x by the complex number z = a(cos φ + i sin φ), where i = y/—l to obtain the series . . = 1 + [a(cos φ + i sin φ)] 1 — a(cos φ + ι sin φ) + [a(cos φ + i sin φ)]2 + [a(cos φ + i sin φ)]3 · · ·
(1) 51
52
EULER
Euler discussed the case a = — 1 in some detail and we shall restrict attention to it immediately. Applying standard procedures for rationalizing denominators, the left side of Equation (1) is transformed as follows: 1 1 + (cos φ + ¿sin φ)
1 1 + cos φ — i sin 0 14- cos φ 4- i sin φ 1 + cos φ — i sin φ 1 4- cos φ — i sin φ 1 + cos φ — i sin φ 2 2 ~~ ( l 4 - c o s 0 ) 4 - s i n 0 ~ 2(1+ eosφ) 1 i sin φ (2) ~ 2 2(1 + eos φ)'
On the right side of Equation (1), De Moivre's Theorem (cos φ + i sin 0) = cosfc04- i sin Κ,φ can be used to obtain 1 — (cos φ 4- i sin φ) 4- (cos φ 4- i sin φ)2 — (cos φ + i sin φ)3 ■ ■ ■ = 1 — (cos φ + i sin φ) 4- (cos 2φ + i sin 20) — (cos 30 4- i sin 30) · · · = (1 — eos0 4- eos 20 — eos 30 4- · · ·) — i(sin 0 — sin20 4- sin30 · · ·). (3) Equating the real parts of Equations (2) and (3), Euler obtained the equation — = 1 — cos 0 4- cos 20 — cos 30 4- cos 40 — cos 50 ■ · ■ or cos 0 — cos 20 4- cos 30 — cos 40 4- cos 50 · · · = —.
(4)
When this equation is integrated, with integration constant C\, we get sin 0 — - sin 20 4- - sin 30 — - sin 40 4- - sin 50 · · · = — 4- C\. 2 3 4 5 2
(5)
Since sin 0 = 0, the substitution 0 = 0 yields C\ = 0, so that another integration results in cos0
7; cos 20 -I—r-cos 30 — —r cos 40 4- -rrcos50· ■· = CI —. 2Z 3¿ \¿ bz 4 Since cosO = 1, the substitution 0 = 0 yields _, . 1 1 1 1 ^ 2 - 1 - ^ + ^ - ^ 4 - ^ · · · .
(6)
._. (7)
The right side of Equation (7) had been studied by Euler (and others) before, and he already knew that it actually equals 7r2/12, and in the article under discussion Euler merely referred to this fact in order to evaluate the integration
TRIGONOMETRIC SERIES
53
constant Ci in Equation (6). Here, instead, this interesting fact will be proved by means of an argument that, while not rigorous by today's standards, would certainly have been accepted by Euler and his contemporaries.
Proposition 5.1.1 *- = 1 - 1 + ¿ - ¿ + ¿ · · ■ · Proof. Recall that, according to Equation (7), 2
2 2 + 32
4 2 + 5 2 '"' '
On the other hand, bearing in mind that cos(2n + a) = cosa, the substitution of φ = π / 2 into Equation (6) yields (ΤΓ/2) 2
C2
7Γ
1
1
3π
1
„
1
π
— = cos - - ^ cos π + ^ cos - ^ - ^ cos 2π + ^ cos - · · ·
or
Ξ! = 0 - 1 ( - Ι ) + Ο - 1 + Ο - 1 ( - Ι ) _ l /
1
1
1
+
Ο - 1
\ _ 1
Solving for C2 now yields π 2 / 1 6 _ 7Γ2
3/4
_
D
12
The above proposition begs the question of evaluating — — — + 2 2 + 32 + 42 + " " · This problem had also been considered by several of Euler contemporaries and was eventually solved by Euler in a manner different from that presented here. This particular series appears in several surprising contexts. For example, it yields the probability that two integers chosen at random are relatively prime (see Proposition 9.4.2 below). 1
Proposition 5.1.2 -5- = 1 + ^ + ^ + τ?Η 6 2Z 3¿ 4 J Proof. By Proposition 5.1.1,
ÜÍ - 1 12~
J_
~¥
+
l
1
_L
J
l
J_
3? ~ i 2 + δ2 ~ β2 + Y1 ~ δ2 ' "
_/J_J_J_J_J_J_J^ ~ V
2 2 + 3 2 + 4 2 + 5 2 + 62 + 72 + 8 2
'")
\
54
EULER
2
(
^22
I-I
+
1 2
4 2 + 62 + 8 2 ' " J 1
2
—2 (λ 2 I
-i/Ί
~ 2 V
It follows that
1 3
2
1
1
1
4
2
1
1 5
2
1
1 6
2
1
1 7
2
1
82
.
—2 + i -2 + —2 2 3 4 "
I I i i i 22
π2 6
+
32
+
42
1
+
52
1
+
62
1
+
I
72
2 + -^ 2 + ···. = 1 + 2-^ 3 2 + 4-^
+
I
82
D
It may be felt that the content of Propositions 5.1.1 and 5.1.2, while perhaps interesting, hardly deserve the status of a proposition, since they seem to be lacking in generality. It is, however, common practice among mathematicians to label interesting observations as propositions, or theorems. In this particular case, the significance of these formulas is enhanced by the fact that the complex function, which is defined indirectly by
and which for s = 2 yields the sum evaluated in Proposition 5.1.2, contains much information regarding the distribution of prime numbers. So much so that the nature of the solutions of the equation
C(s) = 0 is considered by many to be the most important unresolved problem of pure mathematics (see Section 9.4 for some more details). Of course, such solutions have to be imaginary numbers, but that does not detract from their importance. Euler's aforementioned algebraic manipulations are interesting because they also bring to the foreground a new type of infinite series. To explain this it is best to reexamine the power series of the previous chapters. The assumption that underlies Newton's work is that every function can be expressed as a power series of the type Co + c\x + c^x2 + C3X3 + ■ ■ ■ and that these series by and large behave like the polynomials they resemble. This assumption was justified solely on the grounds of its utility. It made possible the effective solution of otherwise unsolvable problems, and since these solutions did not seem to lead to any outrageous contradictions, the assumption might as well be taken for granted. Just like polynomials, trigonometric functions also have properties that make them particularly useful under certain circumstances. The most prominent of these are the periodicity / ( χ + 2π) = f(x) enjoyed by all trigonometric
TRIGONOMETRIC SERIES
55
functions and the periodicities of the derivatives of the sin x and cos x functions, namely that these two functions satisfy the differential equation y" = -y
(8)
and, consequently, y (iv) = y. It so happens that the periodicity exhibited in Equation (8) turns the trigonometric functions into the building blocks of the solutions to the two partial differential equations k
dx-2
=
-äi
and
fc
^
=
(9)
ä^'
where / — f(x,t) and A; is a positive constant. Equations (9) are known as the heat equation and the wave equation, respectively. They are of fundamental importance in physics since, as their names imply, they govern the propagation of heat, sound, and light. A detailed exposition of the genesis of these equations and the role they played in the evolution of the concept of a mathematical function can be found in the references. At this point, suffice it to say that it is easily seen that for each positive integer n the function / ( x , t) = sin nx cos nkt constitutes a solution of the wave equation of Equation (9), as is every combination of the form oi sin x cos kt + a r, then for every positive integer m, b — (b/m) is not an upper bound of S and hence there exists an x e S such that x > b — (b/m). If we now choose an integer m > nbn/(bn — r), then, by Lemma 6.2.2.12,
r>x->ffc-AV = \
mj
\
ft»fl-iy>ft»(l-IL) mj
V
m/
COMPLETENESS AND IRRATIONAL NUMBERS
75
^ ( • - ^ ) which is impossible. Hence, r = bn, or b = \/r.
□ m n
r
It was Newton who proposed the definitions a l = i/a™ and a~ = \/ar for every positive real number a, positive rational number r, and positive integers m and n (see Appendix C). The resultant Fractional Binomial Theorem shows this to be much more than a mere notational convenience. Every real number that is not a rational number is said to be irrational. Exercise 6.1.5 asserts that if p is a prime and n > 1 is an integer, then tfp is irrational. Each of the several parts of the following proposition is proved by a straighforward reductio ad absurdum (Exercise 6). Proposition 6.3.5 If r is a nonzero rational number and a is an irrational number, then the numbers a±r, ar, a/r, r/a are all irrational. The following proposition says that both rational and irrational numbers can be found everywhere on the real line. It will later facilitate the construction of some interesting functions. P r o p o s i t i o n 6.3.6 Ifa R is a rule that assigns a number / ( x ) to every element x of D. For example, the rule / ( x ) = y/l — x 2 defines a function / : [—1,1] —> R. On the other hand, the rule / ( x ) = (x 2 - I ) - 1 / 2 defines a function / : D -»· R, where D = ( - o o , - l ) U (l,oo). The set D is called the domain of the function / : D —» R and, in this text, consists of an interval or a ray or the union of some intervals and/or rays. The domain will quite frequently be left implicit, unless its nature turns out to be significant. Correspondingly, the notation / : D —* R will usually be abbreviated to / .
82
THE REAL NUMBERS
The four arithmetical operations on the real numbers induce analogous operations on functions. Thus, given two functions / : D - » R and g : E —> R, the functions / + g, f — g, f x g, f/g are defined by the following rules:
(/ +
