An introduction to the Calculus, with an excellent balance between theory and technique. Integration is treated before d
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English Pages 689 Year 1991
Table of contents :
Preface
Contents
1. INTRODUCTION
Part 1. Historical Introduction
I 1.1 The two basic concepts of calculus
I 1.2 Historical background
I 1.3 'The method of exhaustion for the area of a parabolic segment
*I 1.4 Exercises
I 1.5 A critical analysis of Archimedes' method
I 1.6 The approach to calculus to be used in this book
Part 2. Some Basic Concepts of the Theory of Sets
I 2.1 Introduction to set theory
I 2.2 Notations for designating sets
I 2.3 Subsets
I 2.4 Unions, intersections, complements
I 2.5 Exercises
Part 3. A Set of Axioms for the RealNumber System
I 3.1 Introduction
I 3.2 The field axioms
*I 3.3 Exercises
I 3.4 The order axioms
*I 3.5 Exercises
I 3.6 Integers and rational numbers
I 3.7 Geometric interpretation of real numbers as points on a line
I 3.8 Upper bound of a set, maximum element, least upper bound (supremum)
I 3.9 The leastUpperbound axiom (completeness axiom)
I 3.10 The Archimedean property of the realnumber system
I 3.11 Fundamental properties of the supremum and infimum
*I 3.12 Exercises
*I 3.13 Existence of square roots of nonnegative real numbers
*I 3.14 Roots of higher order. Rational powers
*I 3.15 Representation of real numbers by decimals
Part 4. Mathematical Induction, Summation Notation, and Related Topics
I 4.1 An example of a proof by mathematical induction
I 4.2 The principle of mathematical induction
*I 4.3 The wellordering principle
I 4.4 Exercises
*I 4.5 Proof of the wellordering principle
I 4.6 The summation notation
I 4.7 Exercises
I 4.8 Absolute values and the triangle inequality
I 4.9 Exercises
*I 4.10 Miscellaneous exercises involving induction
1. THE CONCEPTS OF INTEGRAL CALCULUS
1.1 The basic ideas of Cartesian geometry
1.2 Functions. Informal description and examples
*1.3 Functions. Formal definition as a set of ordered pairs
1.4 More examples of real functions
1.5 Exercises
1.6 The concept of area as a set function
1.7 Exercises
1.8 Intervals and ordinate sets
1.9 Partitions and step functions
1.10 Sum and product of step functions
1.11 Exercises
1.12 The definition of the integral for step functions
1.13 Properties of the integral of a step function
1.14 Other notations for integrals
1.15 Exercises
1.16 The integral of more general functions
1.17 Upper and lower integrals
1.18 The area of an ordinate set expressed as an integral
1.19 Informal remarks on the theory and technique of integration
1.20 Monotonic and piecewise monotonic functions. Definitions and examples
1.21 Integrability of bounded monotonic functions
1.22 Calculation of the integral of a bounded monotonic function
1.23 Calculation of the integral $\int_0^b x^p dx$ when p is a positive integer
1.24 The basic properties of the integral
1.25 Integration of polynomials
1.26 Exercises
1.27 Proofs of the basic properties of the integral
2. SOME APPLICATIONS OF INTEGRATION
2.1 Introduction
2.2 The area of a region between two graphs expressed as an integral
2.3 Worked examples
2.4 Exercises
2.5 The trigonometric functions
2.6 Integration formulas for the sine and cosine
2.7 A geometric description of the sine and cosine functions
2.8 Exercises
2.9 Polar coordinates
2.10 The integral for area in polar coordinates
2.11 Exercises
2.12 Application of integration to the calculation of volume
2.13 Exercises
2.14 Application of integration to the concept of work
2.15 Exercises
2.16 Average value of a function
2.17 Exercises
2.18 The integral as a function of the upper limit. Indefinite integrals
2.19 Exercises
3. CONTINUOUS FUNCTIONS
3.1 Informal description of continuity
3.2 The definition of the limit of a function
3.3 The definition of continuity of a function
3.4 The basic limit theorems. More examples of continuous functions
3.5 Proofs of the basic limit theorems
3.6 Exercises
3.7 Composite functions and continuity
3.8 Exercises
3.9 Balzano's theorem for continuous functions
3.10 The intermediatevalue theorem for continuous functions
3.11 Exercises
3.12 The process of inversion
3.13 Properties of functions preserved by inversion
3.14 Inverses of piecewise monotonic functions
3.15 Exercises
3.16 The extremevalue theorem for continuous functions
3.17 The smallspan theorem for continuous functions (uniform continuity)
3.18 The integrability theorem for continuous functions
3.19 Meanvalue theorems for integrals of continuous functions
3.20 Exercises
4. DIFFERENTIAL CALCULUS
4.1 Historical introduction
4.2 A problem involving velocity
4.3 The derivative of a function
4.4 Examples of derivatives
4.5 The algebra of derivatives
4.6 Exercises
4.7 Geometric interpretation of the derivative as a slope
4.8 Other notations for derivatives
4.9 Exercises
4.10 The chain rule for differentiating composite fu nctions
4.11 Applications of the chain rule. Related rates and implicit differentiation
4.12 Exercises
4.13 Applications of differentiation to extreme values of functions
4.14 The meanvalue theorem for derivatives
4.15 Exercises
4.16 Applications of the meanvalue theorem to geometric properties of functions
4.17 Secondderivative test for extrema
4.18 Curve sketching
4.19 Exercises
4.20 Worked examples of extremum problems
4.21 Exercises
*4.22 Partial derivatives
*4.23 Exercises
5. THE RELATION BETWEEN INTEGRATION AND DIFFERENTIATION
5.1 The derivative of an indefinite integral. The first fundamental theorem of calculus
5.2 The zeroderivative theorem
5.3 Primitive functions and the second fundamental theorem of calculus
5.4 Properties of a function deduced from properties of its derivative
5.5 Exercises
5.6 The Leibniz notation for primitives
5.7 Integration by substitution
5.8 Exercises
5.9 Integration by parts
5.10 Exercises
*5.11 Miscellaneous review exercises
6. THE LOGARITHM, THE EXPONENTIAL, AND THE INVERSE TRIGONOMETRIC FUNCTIONS
6.1 Introduction
6.2 Motivation for the definition of the natural logarithm as an integral
6.3 The definition of the logarithm. Basic properties
6.4 The graph of the natural logarithm
6.5 Consequences of the functional equation L(ab) = L(a) + L(b)
6.6 Logarithms referred to any positive base b ≠ 1
6.7 Differentiation and integration formulas involving logarithms
6.8 Logarithmic differentiation
6.9 Exercises
6.10 Polynomial approximations to the logarithm
6.11 Exercises
6.12 The exponential function
6.13 Exponentials expressed as powers of e
6.14 The definition of e^x for arbitrary real x
6.15 The definition of a^x for a > 0 and x real
6.16 Differentiation and integration formulas involving exponentials
6.17 Exercises
6.18 The hyperbolic functions
6.19 Exercises
6.20 Derivatives of inverse functions
6.21 Inverses of the trigonometric functions
6.22 Exercises
6.23 Integration by partial fractions
6.24 Integrals which can be transformed into integrals of rational functions
6.25 Exercises
6.26 Miscellaneous review exercises
7. POLYNOMIAL APPROXIMATIONS TO FUNCTIONS
7.1 Introduction
7.2 The Taylor polynomials generated by a function
7.3 Calculus of Taylor polynomials
7.4 Exercises
7.5 Taylor's formula with remainder
7.6 Estimates for the error in Taylor' s formula
*7.7 Other forms of the remainder in Taylor' s formula
7.8 Exercises
7.9 Further remarks on the error in Taylor' s formula. The onotation
7.10 Applications to indeterminate forms
7.11 Exercises
7.12 L'Hôpital's rule for the indeterminate form 0/0
7.13 Exercises
7.14 The symbols +∞ and ∞. Extension of L'Hôpital's rule
7.15 Infinite limits
7.16 The behavior of log x and e^x for large x
7.17 Exercises
8. INTRODUCTION TO DIFFERENTIAL EQUATIONS
8.1 Introduction
8.2 Terminology and notation
8.3 A firstorder differential equation for the exponential function
8.4 Firstorder linear differential equations
8.5 Exercises
8.6 Some physical problems leading to firstorder linear differential equations
8.7 Exercises
8.8 Linear equations of second order with constant coefficients
8.9 Existence of solutions of the equation y" + by= 0
8.10 Reduction of the general equation to the special case y" + by = 0
8.11 Uniqueness theorem for the equation y" + by = 0
8.12 Complete solution of the equation y" + by = 0
8.13 Complete solution of the equation y" + ay' + by = 0
8.14 Exercises
8.15 Nonhomogeneous linear equations of second order with constant coefficients
8.16 Special methods for determining a particular solution of the nonhomogeneous equation y" + ay' + by = R
8.17 Exercises
8.18 Examples of physical problems leading to linear secondorder equations with constant coefficients
8.19 Exercises
8.20 Remarks concerning nonlinear differential equations
8.21 Integral curves and direction fields
8.22 Exercises
8.23 Firstorder separable equations
8.24 Exercises
8.25 Homogeneous firstorder equations
8.26 Exercises
8.27 Some geometrical and physical problems leading to firstorder equations
8.28 Miscellaneous review exercises
9. COMPLEX NUMBERS
9.1 Historical introduction
9.2 Definitions and field properties
9.3 The complex numbers as an extension of the real numbers
9.4 The imaginary unit i
9.5 Geometric interpretation. Modulus and argument
9.6 Exercises
9.7 Complex exponentials
9.8 Complexvalued functions
9.9 Examples of differentiation and integration formulas
9.10 Exercises
10. SEQUENCES, INFINITE SERIES, IMPROPER INTEGRALS
10.1 Zeno's paradox
10.2 Sequences
10.3 Monotonic sequences of real numbers
10.4 Exercises
10.5 Infinite series
10.6 The linearity property of convergent series
10.7 Telescoping series
10.8 The geometric series
10.9 Exercises
*10.10 Exercises on decimal expansions
10.11 Tests for convergence
10.12 Comparison tests for series of nonnegative terms
10.13 The integral test
10.14 Exercises
10.15 The root test and the ratio test for series of nonnegative terms
10.16 Exercises
10.17 Alternating series
10.18 Conditional and absolute convergence
10.19 The convergence tests of Dirichlet and Abel
10.20 Exercises
*10.21 Rearrangements of series
10.22 Miscellaneous review exercises
10.23 Improper integrals
10.24 Exercises
11. SEQUENCES AND SERIES OF FUNCTIONS
11.1 Pointwise convergence of sequences of functions
11.2 Uniform convergence of sequences of functions
11.3 Uniform convergence and continuity
11.4 Uniform convergence and integration
11.5 A sufficient condition for uniform convergence
11.6 Power series. Circle of convergence
11.7 Exercises
11.8 Properties of functions represented by real power series
11.9 The Taylor' s series generated by a function
11.10 A sufficient condition for convergence of a Taylor's series
11.11 Powerseries expansions for the exponential and trigonometric functions
*11.12 Bernstein's theorem
11.13 Exercises
11.14 Power series and differential equations
11.15 The binomial series
11.16 Exercises
12. VECTOR ALGEBRA
12.1 Historical introduction
12.2 The vector space of ntuples of real numbers
12.3 Geometric interpretation for n \leq 3
12.4 Exercises
12.5 The dot product
12.6 Length or norm of a vector
12.7 Orthogonality of vectors
12.8 Exercises
12.9 Projections. Angle between vectors in nspace
12.10 The unit coordinate vectors
12.11 Exercises
12.12 The linear span of a finite set of vectors
12.13 Linear independence
12.14 Bases
12.15 Exercises
12.16 The vector space V_n(C) of ntuples of complex numbers
12.17 Exercises
13. APPLICATIONS OF VECTOR ALGEBRA TO ANALYTIC GEOMETRY
13.1 Introduction
13.2 Lines in nspace
13.3 Some simple properties of straight lines
13.4 Lines and vectorvalued functions
13.5 Exercises
13.6 Planes in Euclidean nspace
13.7 Planes and vectorvalued functions
13.8 Exercises
13.9 The cross product
13.10 The cross product expressed as a determinant
13.11 Exercises
13.12 The scalar triple product
13.13 Cramer's rule for solving a system of three linear equations
13.14 Exercises
13.15 Normal vectors to planes
13.16 Linear Cartesian equations for planes
13.17 Exercises
13.18 The conic sections
13.19 Eccentricity of conic sections
13.20 Polar equations for conic sections
13.21 Exercises
13.22 Conic sections symmetric about the origin
13.23 Cartesian equations for the conic sections
13.24 Exercises
13.25 Miscellaneous exercises on conic sections
14. CALCULUS OF VECTORVALUED FUNCTIONS
14.1 Vectorvalued functions of a real variable
14.2 Algebraic operations. Components
14.3 Limits, derivatives, and integrals
14.4 Exercises
14.5 Applications to curves. Tangency
14.6 Applications to curvilinear motion. Velocity, speed, and acceleration
14.7 Exercises
14.8 The unit tangent, the principal normal, and the osculating plane of a curve
14.9 Exercises
14.10 The definition of arc length
14.11 Additivity of arc length
14.12 The arclength function
14.13 Exercises
14.14 Curvature of a curve
14.15 Exercises
14.16 Velocity and acceleration in polar coordinates
14.17 Plane motion with radial acceleration
14.18 Cylindrical coordinates
14.19 Exercises
14.20 Applications to planetary motion
14.21 Miscellaneous review exercises
15. LINEAR SPACES
15.1 Introduction
15.2 The definition of a linear space
15.3 Examples of linear spaces
15.4 Elementary consequence of the axioms
15.5 Exercises
15.6 Subspaces of a linear space
15.7 Dependent and independent sets in a linear space
15.8 Bases and dimension
15.9 Exercises
15.10 Inner products, Euclich planes, norms
15.11 Orthogonality in a Euclidean space
15.12 Exercises
15.13 Construction of orthogonal sets. The GramSchmidt process
15.14 Orthogonal complements. Projections
15.15 Best approximation of elements in a Euclidean space by elements in a finite dimensional subspace
15.16 Exercises
16. LINEAR TRANSFORMATIONS AND MATRICES
16.1 Linear transformations
16.2 Null space and range
16.3 Nullity and rank
16.4 Exercises
16.5 Algebraic operations on linear transformations
16.6 Inverses
16.7 Onetoone linear transformations
16.8 Exercises
16.9 Linear transformations with prescribed values
16.10 Matrix representations of linear transformations
16.11 Construction of a matrix representation in diagonal form
16.12 Exercises
16.13 Linear spaces of matrices
16.14 Isomorphism between linear transformations and matrices
16.15 Multiplication of matrices
16.16 Exercises
16.17 Systems of linear equations
16.18 Computation techniques
16.19 Inverses of square matrices
16.20 Exercises
16.21 Miscellaneous exercises on matrices
Answers to exercises
I1.4I4.7
I4.91.15
1.262.8
2.112.17
2.193.6
3.84.6
4.9
4.12
4.154.19
4.214.23
5.55.8
5.106.9
6.17
6.25
6.267.8
7.118.5
8.78.14
8.178.19
8.228.28
9.69.10
10.410.14
10.1610.22
10.2411.13
11.16
12.412.11
12.1513.5
13.813.17
13.2113.24
13.2514.4
14.714.13
14.1514.19
14.2115.9
16.1216.4
16.8
16.12
16.16
16.20
16.21
Index
Calculus)))
M.
Torn
Apostol)
VOLUME
OneVariable
I)
with
Calculus,
to Linear
Introduction
JOHN New
WILEY &
York.
Algebra)
EDITION)
SECOND
Chichester.
SONS) Brisbane
\302\267 Toronto.
an
Singapore)))
EDITOR)
CONSULTING
Springer, Indiana
George
Second First All
Edition Edition
rights
Copyrigh t @1967by copyright
@
1961
University)
John
Wiley & Sons,
by Xerox
Inc.
Corporation.
reserved.
Reproduction or translation
of any part of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyof the copyright owner is unlawright Act without the permission ful. for permission or further information should be Requests addressed to the Permissions Department, John Wiley & Sons. Inc.) ISBN Library
of Congress
Printed
in
0 471 00005
Catalog Card
the United
1
Number:
States of
20 19 18 17)))
America.)
6714605
Th
Jane
and
Stephen)))
PREFACE)
Excerpts from the Preface to
the
First
Edition)
in seems to be no general agreement as to what should constitute a first course and analytic geometry. Some peopleinsist that the only way to really understand and develop calculus is to start off with a thorough treatment of the realnumber system is the subject step by step in a logical and rigorous fashion. Others argue that calculus for believe course should stress a and the tool engineers applicaprimarily physicists;they and tions of the calculus by appeal to intuition develop by extensive drill on problems which is Calculus is sound in both these points of view. skills. There is much that manipulative of pure a deductive science and a branch mathematics. At the same time, it is very imporand that it derives has strong roots in physical tant to remember that calculus problems much of its power and beauty from the variety of its applications. It is possible to combine
There
calculus
a strong theoretical with development an attempt to strike a sensiblebalance
sound between
training the
in
two.
this book represents While treating the calculus as a
technique;
of the book does not neglect applications to physical problems. Proofs are of of mathematical as an essential the theorems growth important presented part discussion to give the the proofs are often ideas; preceded by a geometric or intuitive disintuitive a form. these student some insight into take Although why they particular cussions will satisfy readers who are not interested in detailed the complete proofs proofs, are also included for those who prefer a more rigorous presentation. and philosophical developThe in this book has been suggested the historical by approach before is treated of calculus and analytic geometry. For example, integration ment and is correct some seem it differentiation. to this unusual, historically Although may connection the true sound. it is the best way to make meaningful Moreover, pedagogically between the integral and the derivative. Since the integral of a step of the integral is defined first for step functions. The concept case is is merely a finite sum, integration theory in this function extremely simple. As the in the he gains of the integral for step functions, student learns the properties experience the notation with use of the summation notation and at the same time becomes familiar to more general from for integrals. This setsthe stage so that the transit\037on step functions deductive
science,
all the
functions
seems
easy
and natural.)
Vll)))
... VIlt)
Preface)
Preface edition
second
The incorporated,
at
an
earlier
on an
important
better
motivation
As in the
from the theorems
first
Second Edition) in
many
Linear algebra has been of calculusare introduced
respects.
and routine applications
have been and many new and easier exercises that the book has been divided into smaller
stage,
table of contents
differs
meanvalue
the
to the
reveals
first
Several sections have been rewritten concept. and to improve the flow of ideas. a historical introduction precedes edition,
and
added.
A glance at the
chapters, reorganized
each important
each centering to provide new
concept,
to its precise mathematical notion tracing its development from an early intuitive physical The of the past and of the triumphs formulation. student is told something of the struggles of the men who contributed most to the Thus the student becomes an active subject. in the evolution observer of results. of ideas rather than a passive participant of The secondedition, first two thirds like the first, is divided into two volumes. The V olume series and I deals with the calculus of functions of one variable, including infinite an introduction I introduces linear to differential The last third of Volume equations. leans to geometry and analysis. Much of this material algebra with applications heavily on the calculus for examples that the general illustrate theory. It provides a natural of algebra from oneand analysis and helps pave the blending way for the transition Further variable to multi variable calculus, discussed in Volume II. calculus development of linear algebra will occur as needed in the second edition of Volume II. Once again I acknowledge with my debt to Professors H. F. Bohnenblust, pleasure A. Erdelyi, influence F. B. Fuller, K. Hoffman, and H. S. Zuckerman. Their G. Springer, on the first edition continued into In preparing the second edition, I received the second. Thanks additional help from Professor Basil Gordon, who suggested many improvements. are also due GeorgeSpringer The staff William P. Ziemer, who read the final draft. and of the Blaisdell Publishing their has, as always, been helpful; I appreciate symCompany consideration and of wishes format pathetic concerning typography. my it gives me special pleasure to expressmy gratitude to my wife for the many ways Finally, she has contributed In grateful acknowledgment the of both editions. during preparation I happily dedicate this book to her.)
T. M. A.) Pasadena,
September
California
16,
1966)))
CONTENTS)
I.
The
I 1.2
Historical
I 1.3
The method
basic
two
Historical Introduction)
1.
Part
I 1.1
INTRODUCTION)
of calculus
concepts
I
2
background
of exhaustion
for
the
area
of a
*1 1.4 Exercises
I 1.5
A
I 1.6
The
critical
analysis
I 2.2
Notations
I 2.3
Subsets
I 2.4
Unions,
I 2.5
Exercises)
for
used in
this
book)
Concepts of the
10)
Theory of
Sets)
11
12
sets
12
I 3.2
The field
13
complements
15)
A Set
3.
Introduction
*1 3.5
designating
intersections,
I 3.1
of Axioms for the
Rea/Number
System)
17
17
axioms
Exercises The
8 8
method
to set theory
Introduction
I 3.4
Basic
Some
2.
I 2.1
Part
to be
to calculus
approach
Part
*1 3.3
of Archimedes'
3
parabolic segment
order
19
19
axioms
21
Exercises
I 3.6 Integersand
rational
numbers)
21)
ix)))
Contents)
x)
I 3.7
I 3.8 I 3.9
I 3.10
as points on a line Geometric interpretation of real numbers bound of a maximum least set, element, Upper upper bound
The leastupperboundaxiom The Archimedean property
I 3.11Fundamental
of the
properties
(supremum)
realnumber
and
supremum
25
system
26
infimum
28
Exercises
*1 3.12
*1 3.13 Existence
of
order. Rational of real numbers Representation
*1 3.15
29
30
powers
decimals
by
Mathematical
4.
Part
real numbers
of nonnegative
roots
square
Roots of higher
*1 3.14
23
25
axiom)
(completeness
of the
22
Induction,
and Related
30)
Summation Notation,
Topics)
of a proof by mathematical The principle of mathematical induction The wellordering principle
An example
I 4.1
I 4.2 *1 4.3
I 4.4
of the wellordering
Proof
I 4.6
The
I 4.7
Exercises Absolute
I 4.9
Exercises
*1 4.10
34
35 37 37
principle
39 the triangle
and
values
The
43
induction)
of Cartesian
ideas
basic
1.2 Functions.
Informal
geometry and
description
Functions. Formal definition of real
examples
as
a set
1.6 1.7
Exercises
1.8
Intervals and
1.9
Partitions
concept
and
of ordered
1.10 Sum and 1.11
set function
60 60
61
functions
product of step functions
63
63
definition
of
1.13 Propertiesof the 1.14
Other
53
56 57
Exercises
1.12 The
pairs
54
ordinate sets step
50
examples
functions
as a
of area
CALCULUS) 48
1.5 Exercises The
44)
INTEGRAL
OF
CONCEPTS
THE
More
41
inequality
Miscellaneousexercisesinvolving
1.
1.4
34
notation
summation
I 4.8
*1.3
32
Exercises
*1 4.5
1.1
induction
notations
the
integral
integral
for
integrals)
of a
for step functions step
function
64
66 69)))
. Contents)
1.15
Xl)
70
Exercises
1.16 The integral of more general functions 1.17 and lower integrals Upper
set expressed as an Informal remarks on the theory and technique functions. Monotonic and piecewisemonotonic
1.18 The 1.19
1.20
of an
area
ordinate
of bounded 1.21 Integrability 1.22 Calculation of the integral
1.23 Calculation 1.24
of
the
integral
1.25 Integration
of
integration
and examples
Definitions
when
dx
monotonic function p is a positive integer
of the integral
basic properties
The
bounded
P Sg x
of
functions
monotonic of a
integral
81
polynomials
1.26 Exercises
83
1.27 Proofs of the
2.
2.1 2.2
basic
of the
properties
84)
integral)
OF
INTEGRATION)
expressed
as an integral
APPLICATIONS
SOME
88
Introduction
The
72 74 75 75 76 77 79 79 80
area of
a region between
two
graphs
88
2.3 Worked examples
89
2.4 Exercises
94
2.5 2.6
The
94 97
functions
trigonometric
Integration formulas for the sine and cosine A geometric description of the sine and cosine functions
2.8 Exercises
102 104
2.9 Polar coordinates
108
2.7
2.10
The
2.11
Exercises
integral
for area
in
polar
2.12 Application of integration
to
109
coordinates
110 the
calculation
111
of volume
2.13 Exercises
2.14
114
of
Application
integration
to the
concept of work
115
2.15 Exercises 2.16
116
Average
value
of a function
117
2.17 Exercises
2.18
The
2.19
integral
119 as a
function
of the upper limit.
Indefinite
integrals
Exercises)
3. CONTINUOUS
3.1
Informal
3.2
The definition of the
description
of continuity limit
of a
function)
120 124)
FUNCTIONS) 126 127)))
.. XIl)
Contents)
3.3
The
definition
3.4
The
basic
3.5
Proofs of the
3.6
Exercises
of
More examples
basic limit
of continuous
functions
135
theorems
and
140
continuity
142
Exercises
3.9 Bolzano'stheorem
142
functions
continuous
for
3.10 The intermediatevalue
for
theorem
144
functions
continuous
145
3.11 Exercises
3.12
The
146
of inversion
process
3.13 Properties of
functions
147
by inversion
preserved
3.14 Inversesof piecewisemonotonic
148
functions
149 150
3.15 Exercises
3.16 The 3.17 The 3.18
for continuous
theorem
extremevalue
for continuous
theorem
smallspan
theorem for continuous theorems for integrals of
functions
functions
The integrability
3.19 Meanvalue
(uniform
continuity)
154
functions
155)
CALCULUS)
DIFFERENTIAL
4.
4.1 Historical introduction 4.2
A problem
4.3
The derivative Examples
4.5
The
156
157
velocity
involving
of a function
159
161
of derivatives
164
of derivatives
algebra
167
4.6 Exercises
4.7
Geometric
4.8
Other
for
derivative as
of the
interpretation notations
169
a slope
171
derivatives
4.9 Exercises
4.10 The chain 4.11 Applications 4.12
rule of
for differentiating the chain rule.
differentiation
of
The meanvalue
4.15
Exercises
Secondderivative Curve
sketching
Exercises
to extreme
differentiation
values of functions
theorem for derivatives
4.16 Applications of the
4.18
176
179
4.14
4.19
173 174
composite functions Related rates and implicit
Exercises
4.13 Applications
4.17
152
152
functions
continuous
3.20 Exercises
4.4
131
138
3.7 Compositefunctions 3.8
130
of a function
continuity
theorems.
limit
181 183
186 meanvalue
test for
theorem
extrema
to geometric
properties of functions
187
188 189 191)))
4.20
examples of extremum
Worked
Contents)
... XlII)
problems
191
4.21 Exercises *4.22
194
196
derivatives
Partial
*4.23 Exercises)
201)
DIFFERENTIATION)
AND
5.1
integral. The
an indefinite
of
derivative
The
INTEGRATION
BETWEEN
RELATION
THE
5.
fundamental
first
theorem
calculus
202
5.2
The
5.3
Primitive functions and the
5.4
Properties
5.5
Exercises
5.6
The
Integration
204
of calculus
theorem
fundamental
second
deduced from
a function
of
properties of
its
207
derivative
210
for primitives
notation
substitution
212
Exercises
216
Integration
parts
217
by
by
220
5.10 Exercises
*5.11Miscellaneous
6.
205 208
Leibniz
5.8
5.9
theorem
zeroderivative
5.7
of
THE
exercises
review
222)
THE
LOGARITHM,
EXPONENTIAL,
AND THE
INVERSE TRIGONOMETRIC
FUNCTIONS)
6.1
Introduction
6.2
Motivation
6.3
The
6.4
The
226
for the
of
the
natural
Basic logarithm. natural logarithm
of the
definition graph
definition
of the
6.5 Consequences of the
6.6
Logarithms
6.7
Differentiation
6.8
Logarithmic
functional
equation
230
to any
differentiation
approximations
to the
logarithm
6.11 Exercises
6.12 The 6.13
exponential
Exponentials
6.14
The
definition
6.15 The
definition
function
expressed as powersof e of eX for arbitrary real x of aX for a > 0 and x real)
= L(a)
L(ab)
positive base b \037 and integration formulas involving
referred
227
229
properties
6.9 Exercises
6.10 Polynomial
logarithm as an integral
+ L(b)
1 logarithms
230
232 233 235 236 238 242 242 244 244 245)))
. Contents)
XIV)
6.16
formulas
and integration
Differentiation
involving
245
exponentials
248
6.17 Exercises
6.18 The
251
functions
hyperbolic
251
6.19 Exercises
6.20 Derivatives
6.23 Integration
256 can
258
fractions
partial
by
into integrals
be transformed
Introduction
The
7.3
Calculus of
7.4
Exercises
APPROXIMATIONS
Taylor
generated
polynomials
Taylor's formula with remainder for the error in Taylor's
Other forms of the 7.8 Exercises
*7.7
7.10 Applicationsto 7.11
error
formula
formula
in
formula.
Taylor's
The onotation
forms
indeterminate
Exercises
7.12 L'Hopital's rule 7.13 Exercises
7.14 The
symbols
Infinite
7.15
FUNCTIONS)
a function
in Taylor's
remainder
on the
remarks
Further
by
TO
Taylor polynomials
Estimates
7.9
7.16 The
for the and
+00
form
indeterminate
00.
0/0
Extension ofL'Hopital's rule
limits
behavior of
log
x and
eX
for
large
x
7.17 Exercises)
8.1 8.2
8.3
TO
DIFFERENTIAL
EQUATIONS) 305
Introduction
Terminology A
firstorder
272 273 275 278 278 280 283 284 286 289 290 292 295 296 298 300 303)
INTRODUCTION
8.
264
268)
exercises
POLYNOMIAL
7.2
7.5
functions
267
7.1
7.6
of rational
Exercises
6.26 Miscellaneous review
7.
253
functions
trigonometric
6.24 Integrals which 6.25
252
functions
inverse
of
6.21 Inversesof the 6.22 Exercises
and notation differential
8.4 Firstorderlinear
differential
306
equation
for the exponential function
equations)
307 308)))
xv)
Contents
311
Exercises
8.5
8.6 Somephysical
leading
problems
linear differential equations
to firstorder
8.7 Exercises
8.8
Linear
8.9
Existence of solutions of the
equations
8.10 Reduction 8.11 Uniqueness 8.12 Complete 8.13 Complete
of
order
of second
the
theorem solution
= 0 y\" + by to the general equation special casey\" for the equation y\" + by = 0 of the equation y\" + by = 0 of the
solution
323 324 324 326
equation
equation
+
y\"
ay' +
=
by
+
= 0 by
326
0
8.14 Exercises 8.15
328
linear
Nonhomogeneous
equations
of second
order with
constant
coeffi
329
cients
8.16 Specialmethods for +
y\"
equation
a particular
determining
ay' +
=
by
solution of the
nonhomogeneous 332
R
8.17 Exercises 8.18
333
problems leading to
of physical
Examples
constant
linear
secondorder
equations
Remarks
339
nonlinear
concerning
8.21 Integral curves 8.22 Exercises Firstorder
8.23
direction
and
339
differential equations fields
341 344
345
equations
separable
8.24 Exercises 8.25
347
firstorder
Homogeneous
347
equations
8.26 Exercises
8.27
Some
8.28
350 and
geometrical
Miscellaneous
physical
9.2
Definitions
The complex numbers
COMPLEX
firstorder
and
field
an extension
9.4
The
Geometric interpretation.
9.6
Exercises
9.7
Complex exponentials
imaginary
Complexvalued
Examples
9.10 Exercises)
unit
NUMBERS)
358
of the real numbers
360
361
i
Modulusand
argument
362
365 366
functions
of differentiation
351 355)
properties as
equations
358
9.5
9.9
to
Historical introduction
9.3
9.8
problems leading
review exercises
9.
9.1
with
334
coefficients
8.19Exercises 8.20
319 322
coefficients
constant
with
313
368
and integration formulas
369 371)))
Contents)
XVI)
10.
INFINITE
SEQUENCES,
SERIES,
INTEGRALS)
IMPROPER
10.1
Zeno's
10.2
Sequences
378
374
paradox
10.3
Monotonic sequencesof real numbers
381
10.4
Exercises
382
10.5
Infinite series
383
10.6 The linearity property 10.7 Telescopingseries 10.8
The
series
geometric
10.9 Exercises *10.10Exercises on 10.11 Tests
385 386
series
of convergent
388
391
decimal
393
expansions
for convergence
394
10.12
Comparison
tests for series
10.13
The integral
test
of nonnegative
394
terms
397
10.14 Exercises
10.15
and the
test
root
The
398 ratio test for
of nonnegative
series
terms
10.16 Exercises
402
10.17Alternating 10.18
403
series
and absolute
Conditional
10.19The convergence 10.20 Exercises
tests
of
Rearrangements
10.22
10.24
406
convergence
of Dirichlet
and Abel
407
409
*10.21
Miscellaneous
10.23Improper
411
series
414
review exercises
416
integrals
Exercises)
11.
420)
AND
SEQUENCES
11.1
Pointwise
convergence
of sequences
SERIES
OF FUNCTIONS)
of functions
11.2 Uniform convergenceof sequencesof functions 11.3
Uniform
11.4
Uniform
11.5
A
11.6
Power
convergence
and
convergence
and integration for uniform convergence
sufficient
condition
series.
424
continuity
Circle
of convergence
The
11.10
A
Properties
Taylor's sufficient
of
functions
represented
series
422
423
425
427
428 430
11.7 Exercises
11.8 11.9
399
by real
power series
by a function
generated for convergence condition
of a
431 434
Taylor'sseries)
435)))
.. Contents)
11.11
Powerseries
*11.12
Bernstein's
expansions
XVll)
for the exponential
and trigonometric
functions
theorem
437
11.13 Exercises
11.14Power
438 differential
and
series
11.15 The binomial
439
equations
441 443)
series
Exercises)
11.16
12.
12.1
Historical
12.2
The vector
445 446 448 450
introduction
12.4
12.5
The
12.6
Length or norm
12.7
ALGEBRA)
VECTOR
space of ntuples Geometric interpretation for Exercises
12.3
dot
of
numbers.
real
3
n
x means
that
x
< y;
x
0 if
and
x is
say
nonnegative.
usually
that
if x is
only
A
briefly as < z, x
0,
Ifa
,
>,
O.
a
exercises.)))
indicate
how
the proofs
rational
and
Integers
numbers)
21)
 a. If x = 0, then b  a = a  b = 0, and hence, Axiom by 8 tells us that either x > 0 or x < 0, 9, we cannot have a > b or b > a. If x \037 0, Axiom but not that is, either a < b or b < a, but not both. both; Therefore, exactly one of the three a = b, a < b, b < a, holds.) relations,
1.16.Let
Proof of
add to
obtain
Proof of
a < b.
 a)
(b means
this
If a
If a
of 1.21.
(b
 a
b
then
 a)e
c. Then that x
>
 a)e
(b
7.
1.20
Theorem
Apply
0,
=
a . a > 0 by Axiom 7. In either case we have
> 0, then Axiom
with a
=
1. Prove Theorems1.22through 2 through In Exercises may use Axioms 1 through
is no real 3. The sum of two
2. There
4. If a
> 0, then
5. If 0
< a
6. If
< band
a
If a
using
10, prove the
given
9 and
negative
Ija
< b, then
real
1.25,
number
b < < band b
 a
b
But
may
e.)
a
0 since
7 we
Axiom
< 0,
multiply
> 0, and
a
then
may
be  ae > 0, and
ae. Hence
hence
o.)
1.)
a and
Theorems
statements
1.1 through
theorems and or establish
1 3.6
Integers
by Axiom
9.)
inequalities.
given
You
1.25.
a a
< C. = c,
b we have a 2
1 ja
< O.
then b = C. +
b 2 > O. If a
and b are
not
both
number
0, then
a2
h, then
+
b 2 > O.
x =
O.)
and rational numbers
There exist certain
introduce
the
1 through
that x 2 + 1 = O. numbers is negative.
subsets
of R
erties not shared by all real numbers. integers and the rational numbers. To
Axioms
x such
> 0; if a < 0, then 0 < b I < aI.
c, then c, and
the earlier
a such that x < a for all real X. 9. There is no real number 10. If x has the property that 0 < x < h for every positive real
anteed
O.
hence
y.)
e >
If
O.
But
O.
and
x = b

y
 b >
e
0,
Exercises)
*1 3.5
7.
=
y
e,
< be, as asserted.)
that ae
(a) . (a) > 0 by Proof
< b,
obtain
to
of 1.20.
Proof
b
1.18. Let x = a +  x > 0, and y
of 1.19.
Proof e by
Hence
=
17. If a < band b < e, then b  a > 0 and  a) + (e  b) > O. That is, e  a > (b
of I.
Proof
x
which are distinguished becausethey In this section we shall discuss two
the number the positive integers we begin with 1, whose 4. The number 1 + 1 is denoted by 2, the number 2 +
have such
special subsets,
propthe
existence is guar1 by 3, and so on.
in this way by repeated addition of 1 are all positive, The numbers 1, 2, 3, . . . , obtained and they are called the positive integers. Strictly this description of the positive speaking, is not because we have not in detail what we mean by entirely complete explained integers of I.\" Although or \"repeated addition the intuitive the expressions \"and so on,\" meaning)))
Introduction)
22)
expressions may seem clear, in to give a more precisedefinition necessary is to do this. One convenient method
of these
DEFINITION
OF
the
of
to
A set
SET.
INDUCTIVE
AN
of the realnumber system it is positive integers. There are many ways introduce first the notion of an inductive set.) treatment
a careful
of real numbers
an inductive
is called
set if it
has
two properties: 1 is in the set. number
the following
(a) The (b) For every
x in
the set,
example, R is an inductive integers to be those real numbers For
is the
So
set. which
OF POSITIVE INTEGERS.
DEFINITION
to every
1 is
x +
number
the
belong
A
real
also
in
the
set.)
set R+. Now we
shall
to every inductive
number
is called
a positive
define
the positive
set.)
integer
if it
belongs
set.)
inductive
Then P is itself an inductive set because(a) set of all positive integers. x. Since the members of P x + 1 whenever it contains (b) it contains ind uctive set. This property of we refer to P as the smallest ind uctive to set, every belong call proof by basis for a type of reasoning that the logical the set P forms mathematicians , in Part 4 of this Introduction. of which is given discussion a detailed induction, The positive integers, The negatives of the positive integersare calledthe negative integers. a Z form set which we call simply the and 0 with the (zero), integers negative together Let
it
P denote the 1, and
contains
set of
integers. it would be necessaryat this stage to In a thorough treatment of the realnumber system, For the about certain theorems sum, difference,or product of two example, integers. prove be an integer. of need not two but the an is However, we integers quotient integer, integers shall not enter into the details of such proofs. of integers afb (where b \037 0) are called rational numbers. The set of rational Quotients realize that all the field should denoted numbers, by Q, contains Z as a subset. The reader we say that the set of this For axioms and the order axioms are satisfied reason, by Q. that are not in Q are called irrational.) rational numbers is an orderedfield. Real numbers
I 3.7
Geometric
The by
means
reader
interpretation
is undoubtedly
of points
on a
straight
of real
familiar line.
numbers as points on with
a line
of real numbers representation geometric to the is selected to represent 0 and another, the scale. I. This choice determines 7. Figure the
A point
in right of 0, to represent 1, as illustrated for Euclidean If one adopts an appropriate set of axioms geometry, then each real number on the line correon line one this to and, conversely, each point point exactly corresponds called the real line line is often the n this reason real urn ber . For one and to one only sponds and point to use the words real number or the real axis, and it is customary interchangeably. x. Thus we often speak of the point x rather than the point corresponding to the real number a simple geometric interpretation. has relation among the real numbers The ordering If x < y, the point x lies to the left of the point y, as shown in Figure I. 7. Positive numbers)))
to the
lie
right of
inequalities a
of a
bound
Upper
0 and
set,
maximum
if and
to the left
numbers
negative
< x b  1. For this n we have n + 1 > b. Sincen + 1 is in fact that b is an upper bound for P.) 10 tells us
Axiom
 1, being less than
P,
above.
P is bounded
Assume
Proof
SinceP is
P of positive integers
The set
1.28.
THEOREM
be
integer
an
n such
that
for
bound
upper
consequences:)
that nx >
Proof
arbitrary
real
number,
there exists
a positive
integer
y.)
Theorem
Apply
ify is an
> 0 and
If x
1.30.
THEOREM
x.)
>
P, contradicting
Theorem 1.28.)
n such
n
1.29
x replaced
with
by yJx.)
1.30 is called the Archimedean property of the realThe property described in Theorem that any line segment, no matter how long, may system. Geometrically it means number of line segments of a given no matter how covered by a finite positive length,
number be
words, a
small. In other
often enough can measure arbitrarily small ruler used large realized that this was a fundamental property of the straight line In the 19th and 20th centuries, as one of the axioms of geometry. and stated it explicitly in which this axiom is rejected. nonArchimedeangeometrieshave been constructed From the Archimedean property, we can prove the following theorem, which will be calculus.) useful in our discussion of integral Archimedes
distances.
If three real
THEOREM 1.31.
(1.14
Proof
n(x
This we
shall
x >
If
 a) >
I 3.11
n >
y,
x
a, Theorem
=
discusses
use in our
with a supremum infimum contains
us
Hence
of the
properties
arbitrarily close
first
the
property
close to its to its
posItIve integer
n satisfying have x =
> a, so we
must
supremum
and infimum
a.)
infimum
properties of
development of calculus. points
inequalities)
a+\037)
supremum and
three fundamental
points arbitrarily
the
there is a that we cannot have x
The
contains
satisfy
a.)
1.30 tells
(1.14).
contradicting
Fundamental
section
1, then
x
is similar.)
(b)
A and
subsets
nonen1pty

B of R, let
C denote
set)
the
{a + b
C = If each
(a)
B has
and
A
of
a supremum,
If each
(b)
of
B has
and
A
an
and b
E
B has
and
A
c
sup
a +
since
these
b > a +
a supremum,
an
inf B
+
and)
.)
and)
infimum,
.)
a supremum. sup B; so sup A
then c = a + b, where B is an upper bound for C. sup
If
C
+
E C,
< sup A
sup B
+
1.32 (with
.)
Iln) there is an
h =
a
in
A and
that)
B such
Adding
A
.)
B}
sup B
+
C has
By Theorem
integer.
positive
any
inf
E
that)
and
sup C Now
A
then
=
A, b
C has
sup
infimum,
inf C
a E A
a E
I
then
C =
sup
b
\037t'ehave)
bound.
upper
.)
all x in
for
h
PROPERTY.
ADDITIVE
1.33.
THEOREM
< sup S 
If we had x
of (a).
h
S +
x < inf
Proof
vre hare)
S
in
27)
infimum)
S be a set of real numbers.
positive number and let
a given
be
h
supremum and
of the
properties
sup A b
')

1
n .)
obtain)
2
n ')
Therefore
sup C
sup
or)
we
O. If x2 = a, then x \037 0 and (_X)2 = a, then, (by Theorem 1.1I). Suppose, so both x and its negative are square roots. In other words, if a has a square root, then it has two square roots, one positive one negative. and Also, it has at most two because if x2 = a and y2 = a, then x 2 = y2 and (x  y)(x + y) = 0, and so, by Theorem 1.11, either x = y or x = yo Thus, if a has a square root, it has exactly two. an important The existence of at least one square root can be deducedfrom theorem in calculus known as the intermediatevalue theorem for continuous functions, but it root can be proved directly from may be instructive to see how the existence of a square was
It
Axiom
pointed
10.)
1.35.
THEOREM
Note: the
by
a 1 / 2 or
a)2 and hence a2 /(1 shall call b. Note that
we
2 2 a, b < a, or b = a. 2 Suppose b > a and let 2 = b2 +
+
a)2
>

square root. If a
by\037.
> 0,
\037.)
then 0 is the only square root. Assume, that a > then, 2 all positive x such that x < a. Since (I + a)2 > a, the number bound for S. Also, S is nonempty the number because al(l + a) is in
a(1 +
which
b2
or
root
square
nonnegative
a unique nonnegative
of
set
0,
square
negative
Proof the
a
If
nonnegative real
Every
Axiom
By
a) so b
>
O.
10,
S has
There
are
O. Let S be 1 + a is an S;
in fact,
a least upper bound only three possibilities:
Then 0 < c < band \037(b + alb). 2 2 2 2 (b  a)2j(4b ) > a. Therefore c > x This means that c is an upper bound for for each x in S, and hence c > x for each S. Sincec < b, we have a contradiction because b was the least upper bound for S. the inequality b 2 > a is impossible. Therefore 2 Suppose b < a. Since b > 0, we may choose a positive number c such that c < band
c
2
such
(b
c
b, 2 inequality b b +
+
this
0 has
yn
one
only
the
by
positive
symbols
positive
y such that yn = x. This y in (1.15). Since n is even, ( 
and
y
Yo
do not
We
root.
nth
later as consequencesof will be deduced they continuous functions (see Section 3.10). If r is a positive rational number, say r = m/n, where r
x
to be
r
*13.15
number of the
numbers
r=a a nonnegative
usually
o
is said
to be
1 = 5 Real
10
power
')
like these
a is
an integer.
decimal
lon')
Ion
1 = 50
2 2
an are
integers
0
satisfying
0,
x
if
= x =
g(x) = 0
g(x)
x >
if
+ 1. 2x.
=x
2 g(x) = x 2 g(x) = x .
g(x) 2
the limits
Calculate
22.
> 0, >
= V x + V\037
10. [(x)
15. lim
x x
if if
9. [(x) = 0
in each
each
In
[and g are defined by the formulas given. Unless otherLet h(x) = [[g(x)] whenever g consist of all real numbers. g(x) for case, describe the domain of h and give one or more formulas functions
g(x)
\037 \037
= x = sin = \037
8. [(x)
21.
of [and
the domain off
determining
13.
10, the
1 through domains
Exercises
In
wise noted, the
> 1
, g(x) ,)
2
= {
2)
 x2
if
Ix I
if
Ixl >
OF CONTINUOUS FUNCTIONS. Let f be conthere is an interval (c  \037,c + \037) about c in
O.
continuity,
By
for
every
0 there
\342\202\254 >
IS
a
that)
f(c)
take the

\342\202\254
0,fee) < 0, and
interval
S has
0, as shown = O. Our
>
f(b)
f(x)
=
f(x)
For
O.
[a, b] for which S is a nonempty
a supremum. Let
shall prove

\037, e +
\037), or
lie to the right e, and e is the
S can \037
0, there Therefore no bound for the set S. inequality fee) > 0
O.
an upper
\037is
S.
=
If fee)
f is positive.
which
in
the
Therefore
is an interval is impossible. If fee) < 0, there (e  \037, e + \037), or [e, e + \037) if e = a, in e is an 0 for some x which is Hence > e, contradicting the fact that < f negative. f(x) and 0 is also the bound for S. on'ly remaining possibility Thereforef(e)< impossible, upper Bolzano's is fee) = o. Also, a < e < b because < 0 and f(b) > O. This proves f(a)
theorem.)
3.10 The
immediate
An
continuous
for
theorem
intermediatevalue
of Bolzano's
consequence
illustrated
functions,
in
functions
continuous
theorem is the intermediatevaluetheorem
Figure
at each point Let f be continuous of a closed interval Xl < X 2 in [a, b] such that f(x l ) \302\245:f(x 2 ). Then f takes points (Xl' x 2 ).) f(x2) somea'here in the interval
THEOREM 3.8. arbitrary
f(x
t
) and
Proof function
Suppose defined
f(x 1) on
[Xl'
< f(x2) and X 2]
k be any
let
for
3.8.)
value between f(x
[a, on
l) and
Choose
b].
every
f(x
value
2 ).
t}vo
between
Let g
be the
as follows:)
g(x) =

f(x)
=
f{x) 
k
.)
k)
.
a:
I a)
FIGURE
X2)
Xl)
3.8
b)
the intermediatevalue theorem.)
Illustrating
b) I
f(a)') FIGURE
3.9
An example
theorem
is not
for
which
applicable.)))
Balzano's
145)
Exercises)
Then g is
continuous at each point =
g(X!)
means f(c) = k, and
this
Note: that
at both
that
We state
If n n = a.)
3.9.
THEOREM
b such
b
that
is a
Proof Choose c > 1 such interval [0, c] by the equation we have f(O) = endpoints
between the
one
= a for b such
positive on
increasing
values
function
we have f(x)
some b
that
[0, c].
have g(c) = 0 for
n
>
k
0
.)
c between
some
and
the
intermediatevalue
b], including the endpoints we refer to the necessary,
is
in [a, b] except at a. Although b] for which I(x) = o.
application of the has a positive
a formal
as
this

f(x2)
Xl
and
x 2.
But
positive integer
it is assumed To understand in Figure 3.9. Here negative and I(b) is
theorem, a and b. curve
is
I(a)
theorem
intermediatevalue
real number
positive
every
I 3.14.
Section
=
g(X 2 )
,)
have)
of [a,
with an
section
this
conclude
We
we prove
[a,
and we
is complete.)
endpoints
in
0
b.
which is Prove
that
g(x)
on
continuous
I has a
fixed
the point
closed interval in [a,
I(x) [a, b].
Assume
b]. (See Exercise5.))))
that
Continuous
146)
The process of
3.12 This
section
functions
from
a simple
function on the interval [0, 2] by f defined is the interval [1, 5]. Each point x in [0, 2] is
off in
y
=
x, we
y in
[1, 5],
x
as a function
defines x
equation
=
each
for
y in [1,
If we
of y.
5]. The function
and
[0,2],

!(y
g(y) =
each x in
will
the
equation
carried
by
it with
illustrate
f(x) f onto
new
= 2x
+ 1.
exactly one
2x + 1.)
there is exactly one x in solve Equation (3.21) to obtain)
Conversely, for every
This
used to construct we
5], namely)
[1,
(3.21 ))
this
is often in detail,
example.
range
point y
inversion
describes another important method that ones. Before we describe the method given
Considerthe
The
functions)
[0,
which y =
I(x). To find
1).)
denote
!(y 
function
this
by g,
we have)
1))
the inverse g is called = y for eachy in [1,5].
thatf[g(y)]
2] for
of f
Note
that
g[f(x)]
= x
for
more general functionfwith A and range B. For each x in A, domain = f(x). For each y in B, there is at least one x in A there is exactly one y in B such that)l such that f(x) = y. Supposethat there is exactly one such x. Then we can define a new function on B as follows:) g now a
Consider
g(y) = In other
words, the
of g
value
means
x)
at each point y
= f(x).)
y
in
that unique x in A such that f(x) = y. process by which g is obtained fromfis
B is
called the inverse off The = x for all x in A, and thatf[g(y)] = y for all y in B. that g[f(x)] The process of inversion can be applied to any function for f having the property that = each y in the range off, there is exactly one x in the domain such that off f(x) y. In that is continuous and strictly monotonic on an interval [a, b] has this particular, a function An example is shown in Figure 3.10. Let c = f(a), d = f(b). The intermediateproperty. value theorem for continuous functions tells us that in the interval [a, b], f takeson every value between c and d. Moreover,f cannot take on the same value twice becausef(x l) \302\245=g is
function
new
This
called inversion.
2)
f(x
Note
whenever
Xl
Therefore,
\302\245=X2'
every
continuous
strictly
monotonic
function
has an
Inverse.
The relation
in the g can also be simply explained we 1.3 describeda function concept. have the same first element. The inverse (x, y) no two of which x and y. the the elements taking pairs (x, y) infand interchanging if is then no two monotonic, (x, y) Ef Iff strictly only pairs in! a function
between
formulation of the
orderedpair a set of orderedpairs function g is formed by
f as
That is, (y, x) E have
Thus
the
g
same
g if
and
second
f and its inverse
element, and hence no
is, indeed, a function.)))
In Section
function
two
pairs
of g
have the same
first
element.
function. If
The nthroot
EXAMPLE.
preserved
of functions
Properties
n
a
is
Then f is strictly on every interval [a, b] with 0 < increasing is the nthroot function, definedfor y > 0 by the equation) g(y) = 3.13
of functions
Properties
possessed relationship
Many properties illustrates the
3.11
reflection
merely
by
and
only
if
preserved
through
if the point
(v, u)
yl/n
a
0 such that) (3.22))
of
process
Inversion.)
x 2 , which,
Yo
in
\037.)
generality
\342\202\254 in are
the open 0 there is
\342\202\254 >
[a,
if we
b].) Let
\037)))
Continuous
148)
of the
be the smaller
numbers)
two
f(x o) easy to check g is continuous
It is that
There is a
 f(x o
from the
f(x
+
o

E)
f(x
in (3.22). A slight modification the c, and continuous fronl
correspondingtheorem
decreasing continuous Theorem 3.1 0 to applying
and)
E))
at
right
function
strictly by

b works
this
that
functions)
 f)
o ) .)
of the
argument proves
at d.
left
inverse of continuous. This follows That is, the
functions.
for
decreasing
f is
strictly decreasing and
a
b)
+
g(yo)
f)
is the
()
smaller of these
distances)
two
d
+ f)
f(xo

g(yo)
\037f I I I I I I I I I I I
f) Yo
f( x 0
f))
)
c
a
FIGURE
EXAMPLE.
[c, d] Th\037s
\\vith
gives
of the
Continuity
c < d, since an alternate proof 0
in
rthpower
monotonic
to apply the process
this
of a
function,
Xo+
Xo
f
b)
function.)
function g, defined for interval and continuous on every
increasing
strictly
inverse
the
f
nthroot
The
function. is
inverse

increasing
strictly
continuous
function.
function, independent of the is continuous, we again = where r is a m/n positive yr,
nthroot
functions
=
hey)
o.)
of piecewise
Inverses
3.14
is
of the
continuity
of the continuity of the Since the product of continuous
of integration.
thebry
it
the
nthroot
equation g(y) = yl/n,
0 by the
y >
Proof of
3.12
Xo
functions
of inversion = x2
suppose
thatf(x)
interval
is carried
by
f
that is not monotonic on of the form [c, c] on the xaxis. 2 into exacdy one point y in the interval [0, c ],
on
to
a function
an interval
namely,)
y =
(3.23)) We
can
solve Equation
to each y
in
(0,
(3.23) for x in
terms
x2
of y,
.)
but there
are t}t1;'O values
c 2 ], namely,)
x=0)
and)
x
=
0)))
of x
corresponding
149)
Exercises) As we have
mentionedonce before, there was g in this case is a doublevalued
But since the
more modern in a
functions,
= vY)
gl(y)
(3.24))
2 y = x
as
defining
These
x 2)
=
f1(x)
=
g2(y)
x
0
in every
E.)
bisections. Assume that for some the theorem is false. I'hat is, assume [a, b] EO , the interval in each of which the span of f into a finite number of subintervals cannot be partitioned is false in of [a, b]. Then for the same than EO' Let c be the midpoint is less EO , the theorem were true in both intervals subintervals at least one of the two [a, c] or [c, b]. (If the theorem [a, c] and [c, b], it would also be true in the full interval [a, b].) Let [aI' b l ] be that half of [a, b] in which the theorem is false for EO' If it is false in both halves, let [aI' bl] be the left half, [a, c]. Now continue the bisection processrepeatedly, denoting by [a n / I , b n + l ] that that of [an, brJ in which the theoren1 is false for EO , with the understanding we choose half the left half if the theorem is false in both halves of [an, b n ]. Note that the span off in each is at least EO . subinterval [an, b n ] so constructed and let Let A denote the collection of leftmost a, a l , a 2 , . . . , so constructed, endpoints lJ.. be the least upper bound of A. Then lJ.. lies in [a, b]. By continuity off at lJ.., there is an = a, this interval is interval (lJ..  c5, lJ.. + c5) in which the span off' is less than EO' (If lJ.. the interval [an, bnl lies inside (lJ..  c5, c5, b].) However, [a, a + c5), and if lJ.. = b, it is (b n lJ.. + n is so large that c5) when (b a)j2 < 0, so the span off in [an, b n ] is also less than of f is at least EO in [an, b n ]. This contradiction the fact that the EO , contradicting span Proof
We
argue
by contradiction,
using
the
of successive
method
E, say for
completesthe 3.18 The
proof
[a, b] is also integrable
for
theorem
The smallspan theorem
can on
[a,
=
3.13.)
of Theorem
integrability
E
be b].)))
continuous used
functions
to prove
that
a
function
which
is continuous
on
The
3.14.
THEOREM
Theorem 3.11 shows lower and a integral, l(f). l(f), Choose an integer N > 1 and
that
offin
partition P =
is a
there
E
maximum and
than
1,2, . . .,
k =
each
Denote
E.
in the
off
an
has
continuous
integral,
upper

111k
tn
choice
we have)
Then
'\"k]'
span
the absolute
E)
defined on [a, b]
step functions
two
be
this
such that the
respectively,
(f),
[X k  1 ,
1.
for every that
proves
j\"
Therefore,
by
functions)
= l(f).
have l(f)
1.31, we must
Theorem
3.19 Meanvaluetheorems
for
of continuous
integrals
functions)
of a function f over an interval In Section 2.16 we defined the average value A(f) we can prove that this to be the quotient S\037f(x) dx/(b  a). Whenfis continuous, in value is equal to the value off at some [a, b].) point
MEANVALUE
3.15.
THEOREM
then for
somec in
[a,
b] we
m and
Let
b]. b
f(e)(b 
dx =
M denote, respectively,
for Then m < fCx) <  a, we find m < A(f) < M, mediatevalue theorem tells us that [a,
by
If f is
FOR INTEGRALS.
THEOREM
on
continuous
[a, b] average
[a, b],
have)
J: f(x)
Proof
This
on [a, b].)
is integrable
b].
=
these
fCc) for
some c
mean
values.)
in
now
But
This completes
b].
[a,
on
and dividing the inter
inequalities
 a).
dx/(b
S\037f(x)
values of f
maximum
and
Integrating
where A(f) =
A(f)
.)
minimum
the
all x in [a,
M
a)
the
proof.)
There is a correspondingresult 3.16.
THEOREM
on
continuous
WEIGHTED
[a, b].
weighted
MEANVALUE
If g never changessign
J: f(x)g(x)
(3.27))
FOR
THEOREM
in
[a,
dx = fee)J: g(x)
somec in
dx
Mg(x)
to
m dx
=
0, this
J:g(x)
is theorem as before to g.) weighted
to
complete
meanvalue
of a product of two functions, compute. Examplesare given
dx
for

0 becauseS\037; (sin t)/t dt > S\037; Isin tl/t dt. S\037; (sin 27T and 3.16, for some c between (b) The integral t)/ t dt = 0 because, by Theorem S\037; (sin we
47T
have) 41T
121T 5.
0,
have)
If n
is a positive
integer,
V If(c).]
of
f is continuous on is continuous on [a, b].
[a, b]. Prove
Assume
that
f(x)
on [a,
If f(c) also
b]. If S\037f(x)
> 0 at
that
= 0 for
a
point
J\037f(x)g(x) all
x in
[a,
dx
of continuity dx b].)))
= 0
that
f(x)
= 0
c, there is an
for every
function
g
4)
CALCULUS)
DIFFERENTIAL
4.1
Historical
introduction)
were for quite independently of one another, largely responsible calculus to the where hitherto insurmountable integral point problems or less routine methods. The successfulaccomplishments of these men were primarily due to the fact that were able to fuse the calculus they together integral with the second main branch of calculus, differential calculus. The central idea of differential calculus is the notion of derivative. Like the integral, the derivative from a of finding the tangent originated problem in geometrythe problem line at a point of a curve. Unlike the integral, however, the derivative evolved very late in the history of mathematics. The was not formulated in the 17th until concept early when the French mathematician Pierre to determine de the Fermat, century attempted maxima and minima of certain functions. special Fermat's idea, basically can be understood if we refer to the curve in very simple, 4.1. It is assumed that at each of has a its this curve definite direction that Figure points can be described line. Some of these tangents are indicated by broken lines by a tangent in the figure. Fermat noticed that at certain the where curve has a maximum or) points Newton
and
Leibniz,
developing the ideas of could be solved by more
Xl)
Xo)
FIGURE 156)))
4.1
The curve has
horizontal
tangents
above
the points
Xo
and
Xl
.)
A problem
mInImum, such as those must be horizontal. Thus on the solution of another
the
problem
of
that
problem,
157)
velocity) abscissae
figure with of locating
the
in
shown
involving
Xl , the
and
Xo
such extreme values is seen the horizontal locating tangents.
tangent line to depend
the direction of the tangent line general question of determining of the curve. It was the to solve this that arbitrary point attempt general problem to discover some of the rudimentary led Fermat ideas underlying the notion of derivative. At first between the problem of finding sight there seems to be no connection whatever the area of a region lying under a curve and the problem of finding the line at tangent a point of a curve. The first person to realize that two seemingly these remote ideas are, in rather related appears to have been Newton'steacher,Isaac Barrow fact, intimately and Leibniz were the first to understand Newton the real imporHowever, (16301677).
more
the
raises
This
at an
and
relation
this
of
tance
dented era
the
in
it to the fullest, exploited of mathematics.
they
development derivative was
thus
an
inaugurating
unprece
the of tangents, it originally formulated to study problem a to calculate more the provides way velocity and, generally, rate of change of a function. In the next section we shall consider a specialproblem inthe calculation of a The solution of this contains all the essential volving velocity. problem features of the derivative concept and to motivate the general definition of may help derivative which is given in Section 4.3.) the
Although
that
found
soon
was
A problem
4.2
also
it
velocity
involving
a projectile is fired
Suppose
friction,
that
back along
up and
moves
it
projectile attains
t
up
straight
per second. Neglect
assume
and
after
seconds
a straight firing.
the ground with initial of 144 feet velocity the projectile is influenced only so by gravity in feet that the line. Let f(t) denote the height If the force of gravity were not acting on it, the from
to move upward with a constant velocity, traveling a distance 144 feet every second, and at time t we would of have f(t) = 144t. In actual practice, causes the projectile to slow down until its velocity decreases to zero and then it gravity back to earth. Physical experiments that as long as the projectile is aloft, drops suggest its height f(t) is given by the formula) (4.1
continue
would
projectile
term
The
t =
when
16t 2 is due to the 9. This means that
be understood
is
meant
of average to
be
the
formula
that
The problem we each instant of its what
wish
influence of the
.)
returns
time
interval,
instant.
that
Note
the this
do
To
say from
= 0
f(t)
this, t to
time
when
t =
9 secondsand
to earth after < t < 9.
valid only for 0 is this: To determine Before we can understand
a
velocity during
 16t2
gravity.
projectile
velocity at each
by the
144t
(4.1) is to consider
motion.
velocity
it
0 and is to
of the projectile at must decide on
we
problem,
we introduce first the time t + h. This is
notion
defined
quotient)
change
in
distance
length
This
=
f(t)
))
quotient,
called
during of
time
a difference quotient,
time interval
interval) is a

J(t +
h)

J(t)
h)
number
which may
be calculated
whenever)))
158)
both
values of
keep
For example,considerthe
instant
and smaller absolute value. t = 2. The distance traveled =
288
covered
is)
f(2)
t =
time
At
2+
f(2 + the
Therefore
distance
the
h,
=
h)
+
J(2
h)
 J(2) =

80h
h
2 seconds
after
is)
224.)
224 + 80h
=
t =
from
interval
the
in
 64 =
 16(2+ h)2
144(2 + h)
velocity
average
be positive or negative, to the difference quotient as
may
and
fixed
t
smaller
with
h
[0, 9]. The number h see what happens
interval
the
in
We shall
zero.
we take
are
t + h
and
t
not
but
calculus)
Differential
16h 2
2 to =
=
t
80 _
 16h2 .)
2 +
h
is)
16h .
h)
smaller and smaller absolute value, this average As we take values of h with velocity closer and closer to 80. For example,if h =.0.1, we get an average velocityof 78.4; h = 0.001, we get 79.984; when h = 0.00001, we obtain the value 79.99984; and h =
0.00001, close
as
velocity
velocity
average
this limiting
we obtain 80.00016. The to 80 as we please by 80 as a limit approaches
value the
The same kind velocity for an J(t
+ h)
of
instantaneous
arbitrary
 J'(t) =
[144(t
at time t = 2. be carried out for any other from t to t + h is given by
interval
+
h)
 16(t+
h
this
and
limit
stantaneous

h)2]
[144t
 16t2 ]
The
instant. the
average
quotient)
= 144

32t

16h .
h)
h approaches
When
Ihl sufficiently h approaches
when
when
can make the average small. In other words, the to call zero. It seems natural
is that we
velocity
can
calculation time
thing
important taking
gets when
zero, the expression on the
is defined to
velocity
be the
instantaneous
by vet), we may
velocity
 32t as a limit, time t. If we denote the in
approaches
at
144
write)
vet)
(4.2))
right
= 144
distance f(t)

32t.)
a function f which tells us how high refer to We may f as the position function. projectile Its domain is the closed interval [0, 9] and its graph is shown in Figure [The scale 4.2(a). on the vertical axis is distorted in both Figures and The formula in (b).] 4.2(a) (4.2) for the velocity vet) defines a new function v which tells us how fast the projectile is moving at each instant of its motion. This is called the velocity function, and its graph is shown in = 144 to v(9) = As t from 0 to increases decreases from 9, 4.2(b). vet) v(O) Figure steadily 144. To find the time t for which vet) = 0, we solve the equation 144= 32t to obtain t = 9/2. Therefore, at the midpoint of the motion the influence of gravity reduces the to and the is at rest. The at this instant zero, velocity projectile momentarily height = 324. When t > 9/2, the velocity is negative, indicating that the is is f(9/2) height The
formula
the
decreasing.)))
in (4.1) for the
is at
each
instant
of
its motion.
defines
The derivative The
limit
159)
a function)
vet) is obtained from the
by which
process
of
differencequotient
written
is
sym
bolically as follows:)
v(t)
(4.3))
=
+ h)
lim f(t h+O
This
is used to
equation
for any particle
generally,
is such
difference
the
that
 f(t) .
h)
define velocity not only for this particular example but, the position function line, moving along a straight provided h tends to a definite limit as zero.) approaches quotient
f(t)
more
f
100
300
50
200
t
0
100

50)
t) o)

9)
\037)
100)
(b))
(a))
(a) Graph of the
FIGURE 4.2
velocity
derivative
The
4.3 The
the difference
in the
function:
16t
[(t)
the xaxis.
foregoing section points begin
with
Then we
choose
We
derivative.
(a, b) on
= 144/  32t.) = 144 v(t)
function
2
.
(b) Graph
of
the
function
described
example
the concept of interval
of a
position
a
function
a fixed
the
way
point x
in
to
the introduction
at least
f defined this
interval
of
on some open and introduce
quotient)
l(x
+ h)
 f(x)
h)
where
the
number
also lies in
(a, b).
h, The
which
may
numerator
be positive of this
or negative (but not zero), is such that x + h quotient measures the change in the function)))
160)
h
let
we
Now
approaches
zero
approaches
called the
OF
DEFINITION
(which
definition
+
= Iim f(x
f'(x)
(4.4))
The number
exists.
the limit
If the quotient
quotient.
(read as
f'ex)
symbol
as follows:) defined by

h)
h+O
provided
average rate of
that the limit is the same whether implies then this limit is values), through negative
f' (x) is
derivative
The
DERIVATIVE.)
this
to
happens
at x
off
Thus, the formal
limit
positive values or and is denoted by the of f'ex) may be stated
through
derivative
as a
value
definite
some
to as the
is referred
itself
quotient
x + h. and what zero see approach
in the
off
change
x to x + h. The interval joining x to
from
x changes
when
h
calculus)
Differential
f(x)
the
''/
of
prime
x\.
equation)
,
h)
called
is also
f'ex)
rate
the
of change
off at x.)
is we see that the concept of instantaneous (4.3), velocity By comparing (4.4) with derivative. the derivative the of The is to an of velocity vet) concept equal merely example which measures This is often described by saying position. f'(t), where f is the function of position with respect to time. In the example worked that velocity is the rate of change in Section the function is described the equation) out 4.2, by position f
f' is a
its derivative
and
new
function
(velocity)
f'(t) In
the limit
general,
a new
function
process
from
f'
called the first derivative
compute its
first
denoted
denoted by
= 144
f in
given
by)

.)
32t
from f(x) gives us a way f'ex) The process is called differentiation, is defined on an open interval,
produces
function
off. If f',
derivative,
the nth derivative off,
which
a given
 16t2 ,)
= 144t
f(t)
turn,
by f\" and called is defined to f
Derivative
of
difference
The
O.
(x +
sin
that
lh)
+
nthroot
the
u =
un
v
n
(x + , and
let
and
h)l/n
+
h)
 f(x)
v = xl/no
h)

the
obtain
Then we have
= xl/n
=
un
hand
x +
v
n =
x, so
h
=
1 V
n)
u
n
the nth:'root function shows on the right has the limit has the limit vlnln. quotient
difference
f(x)
becomes)
_
un
let
l/n h)l/n _ x
+
(x
of continuity denominator
so the
a positive integer,
uv)
f(x)
h)
The
we
(4.5),
h)
the difference quotient
f(x +
in
from
is)
h)

+ 0;
x as h
sin
If n is
function.
for f
quotient f(x
Let
163)
x.)
6.
EXAMPLE
for x
sine shows
the
of
Continuity
c'(x) =
of derivatives)
algebra
l
+
n 2
U
n
+
u + V
that v
V
. . .
l as
h
n 2
UV
+
V
n 1
h + O.
as
+ O.
v =
Since
+
There
xl/n,
this
Therefore are n terms
proves
.)
each term
altogether,
that)
= 1. x l / n  l .
f'ex)
n)
7.
EXAMPLE
a point
x,
+
f(x
f'
for h
is valid
which
(x) and, since this
the
right
approaches
\037
O.
quotient
that f is continuous at
If we
0 .f'ex)
h) =
let O.

+ h
(I(X
function f has the identity)
a derivative
at
we use
this,
f(X) ))
\037
the difference quotient on the right approaches tends to 0, the second term on by a factor which shows that f(x + h) + f(x) as h + 0, and hence
+ 0,
is multiplied
=
To prove
f(x) +
h
If a
derivatives.
Continuity of functions having it is also continuous at x.
then
This
X.)
that functions are continuous. Every provides a new way of showing the existence of a derivative we also establish, at the same time, f'ex), the continuity at X. It should be that the converse is not true. Connoted, however, off at x does not mean that the derivative tinuity necessarily f'ex) exists. For example, when = lxi, the of continuity f(x) point x = 0 is a point off [since f(x) + 0 as x + 0] but there is no derivative at O. (See Figure The difference 4.3.) quotient [f(O + h) f(O)]lh is This
time
example
we establish
y)
x) o)
FIGURE
4.3
The function is
continuous
at
0 but.f'(O)
does
not
exist.)))
164)
calculus)
Differential
equal to
Ihllh.
to a limit
as
The
4.5
quotient
product point
rules for computing
Let f and
THEOREM 4.1.
g
a
have
f' g, and in question.)
g)' =
(I.
(iii)
'
I
(iv)
( g)
. g

A
at
case
in
a
functions
above,
of
each point
At
f
difference g is not
 g,
zero at
the follo}1'ingformulas:)
0 .
:;f=
the
with
for every
the
but first we
occurs
pair of constants
property
linearity
the
C1 , . . . , Cn
derivative
I c 2g) =
of the
or as
an
function
involving
f +
If'
times
is the
sum of
(c
f'.
f)'
the derivative
off. derivatives [property
have)
c 2g I.)
+
and
derivative,
it
is
analogous
to the the
linearity
linearity
as follows:)
sums
' Ci
and fl , can be written
involving
equality
are equations
C
of its is constant, \302\267 = C. In some
mention
becomes
(iii) constant
we
C2
to
Using mathematical induction, we can extend
are constants formula
and
C1
want
the two functions
of
one
when
In this case, consideration. constant times f is the fact that the derivative of a sum
integral. to arbitrary finite
tive of the
g(x)
of a
( i\037
Every
are given by
functions
moment,
of (iii)
n
where
of
x under
derivative
that
find
is called
property
common interval.
sum f + g, the the extra proviso that
need
we
x where
points
( C If +
property
on a the
,
f'
I' g'
special all
other words,the Combiningthis
This
is
\302\267
g
this theorem
prove
consequences. say g(x) = c for
(0], we
a
with
g' ,
+
g'
f'
provides us
g2)
shall
We
=
sum, differ
the
of
+ g' ,
\302\267
f
defined true
functions
the same
(For fig quotient fig. The derivatives of these
f' 
g)' =

(f
tend
derivatives.)
g be t}1'O
derivative,
the
(i) (f + g)' = f' (ii)
of Section 3.4 tell us how to compute limits of two functions, so the next theorem
theorems
set of
corresponding
the
hence does not
0, and
h
* O.)
algebra
where f and
+ 1 if h
the value
has
This h
Ii
\037/i
)
. . . ,fn in
For
of the
are
\302\267 I\037 ,)
functions
two
example, functions
with derivatives
f\037
, . . . ,f\037 \302\267
as an equality between two of Theorem 4.1, as written property (i) states that the derivathese functions))) f' and g'. When
ways, either The properties
two
numbers.
functions.
g is the sum
n
\302\267 =
The algebra
at a point
evaluated
are
x,
formulas
obtain
we
of derivatives)
165)
Thus
numbers.
involving
formula
(i)
implies)
g)'(x) =
(f + We
of (i). for f
Proof quotient
Let x be a point + g is) g(x +
h) +
+
[f(x
proof of Theorem
now to the
proceed

h)]
4.1.)
derivatives
both
where
f'(x) + g'(x).)
f(x +
+ g(x)] =
[f(x)
h)
h h * 0
When
the
and hence the
first
sum

f(x)
+
g(x +
on
the right
approaches f'ex), the This proves (i), and
+ g'(x).
approachesf'(x)
difference
for the
quotient
+
+ h)g(x
f(x
product f' 
h)
difference
The

h)
g(x)
.)
h
h
quotient
Proof of (iii). The
exist.
and g'(x)
f'(x)
second the
g'(x),
approaches
of (ii) is
proof
similar.)
g is)
f(x)g(x)
(4.6)) h)
To
study
as h
this quotient
write (4.6) as a sum and subtracting g(x)/(x us to
f(x + h)g(x+

h)
two
of
+
we
h),
subtract
we add and
* 0,
f(x)g(x)
=gx(
)
j(x +
Proof of(iv). for
all
x and
A
term
first
*
I(x),
of (iv)
(iv) reduces to the
h
occurs whenf(x)
writingflg
(iii),
using
formula
1
[11 g( x
( g
g
to prove
+ h)] 
(4.8)) h)
h)
+
h)
 g(x) .
h)
x.
all
In this
at x,
(iii).)
case f'(x)
(4.7).
[11 x ) ] g(
(iv)
this
from
=0
special case
since)
I
it remains
g(x
enables
g. Adding
:\037)
deduce the general
can
and
= 1 for
' 1 1 f' f' g'  , == f'='f +f' ) ) ( Therefore
+ f( x+
which
I and

=
g
quotients of
formula)
G )' g(x) \037 O. We as a product
f(x)
a term
numerator
I
(4.7))
provided
the
on the right approaches g(x)f'(x). Since is continuous This proves so the second term f(x)g'(x). approaches case
special

h)
h
When h * 0 the we have f(x + h)
in
terms involving difference see that (4.6) becomes)
The
g2
g
difference
g(x
g . f'
+
quotient
h) h)
 f' g' . g2)
for I/g
 g(x) 1 ..
is)
1
g(x) g(x+
h))))
by
166)
calculus)
Differential
When
The
I/g(x).
g(x) as h
on the
the first quotient
+ 0,
h
continuity
+ O.
the quotient
Hence
in
(4.8)
Theorem
n is
where
a positive integer,
then
the
all
approaches
that
+
g(x
this
h) + (4.7).
proves small
sufficiently
out
worked
examples
Section
in
4.4,
formulas.)
= nx n  1. The
f'ex)
and
0 for \302\245:
of Section4.4 we
In Example 3
1. Polynomials.
EXAMPLE
thatg(x
with
using
+ h)
factor
third
the fact
g'(x)/g(X)2,
examplesof differentiation
us to derive new
enables
know
in conjunction
used
when
4.1,
are
approaches
In order to write (4.8) we need to Note: h. This follows from Theorem 3.7.
and the
g'(x)
approaches
right
required since we
g at x is
of
reader may
find
it
= xn ,
if f(x)
that
showed
to
instructive
consequence of the specialcasen = 1, using mathematical induction a product. with the formula for differentiating conjunction we can differentiate any polynomial the linearity Using this result along with property, and the derivatives. the derivative of each term Thus, if) adding computing
rederive in
by
as a
result
this
n
=
f(x)
!CkX
k
,
k=O)
term
by differentiating
then,
obtain)
we
term,
by
n
=
f'ex)
!
kC
kX
k 1
.
k=O)
the derivative
that
Note
For example, if EXA\037IPLE 2.
p(x)lq(x), Theorem
f(x)
the
4.1.
The
of
polynomial 2 
5x
that the function 1Ixm , where m is a
is written
exponents
positive integer
3.
We
Rational
have already
form
the
in
to negative
EXAMPLE
and
exponents of pou,'ers.
proved the
x
\037
r =
1In,
lOx


1.
7.)
X
where differentiating
n
is a
Let I(x)
=
m X)
1 , it for
for
.
m+l
an extension from nth powers.) differentiating provides
x >
0, where r is
a
positive
number.
rational
formula)
integer.
a product
m
= xr
differentiation
positive
2m
formula
the
find)
m 1
xm'0mx
f'(x)
formula for
we
0,
= mx
r'(x)
(4.9)) for
+
functions.
r' (x ) = this
a new polynomial of degreen
derivative
Note
If
n is
degree
8, thenf'(x) = 6x2
7x +
= If r is the quotient of two polynomials, say rex) be the formula (iv) in r'(x) may computed by quotient derivative r'(x) exists at every x for which the denominator q(x) \037 O. r' so defined In particular, when rex) = is itself a rational function.
Rational
then
of a 2x 3 +
=
shows
= rx r  1) Now we extend that Equation
it
to
all rational
(4.9) is also valid
powers. for
r =
The 21n)))
=
by induction, for r refers to m.) Therefore
and, for
Thus,
each case,we
2/3
X
m is any
argument positive integer. (The induction formula for all positive rational r. The r. is also valid for negative rational (4.9) = X 1/ 2 , then f'ex) = tx3/2. In If f(x)
x >
valid
now shows
that
= !X1/3 .
have f'ex)
, we
require
167)
Equation (4.9)is
a quotient
differentiating if f(x) =
where
m/n,
Exercises)
O.)
Exercises)
4.6
2. If
+ x  x2 ,
= 2
1. Iff(x)
= lx 3
f(x)
In Exercises 3 2 x) = x
+
4. f(x)
= x4
+
5. f(x)
= x4
3. f(
2x,
find
a formula for f'(x)
obtain
12,
through
f'(O),f'(t), f'(I),f'( 10). all x for which (a)f'(x)
compute

+ ix 2
3x + 2.)
sin
x.)
if
= 0; (b)f'(x) =
6. f(x)
=
7. f(x)
=
=
9. f(x)
=
x+
ground
x
'
1
1 2
1. \302\245:
+ x 5 cos
+ 1
that
the
with
an
height f(t) of initial velocity
of
Vo
ft/sec,
f(t) (a)
method described in
Use the
during
velocity
a time
interval
time
t is
at
(b) Compute (in
(c) What

Vo
Section
t +
1 to
from
h
to
(a)
area
The volume
16. f'(x) 1 7.
j>( x)
volume
321
required for the
x
of
sphere
+
sIn x
1 +x
2
.
directly
from
upward
the
formula)
the
that
velocity of
average
 16h ftlsec, and
the
23,
obtain
> O.
for the
that
to drop
velocity
to
to return
the
projectile
instantaneous
the
to zero. earth
of a
cube
the
radius
r is
with
respect to
r
\037
'V
' x)
x
>
O.)
to the
respect
its circumference
to
is equal 3 /3 and
the
after
1
sec?
after
if
f(x)
constant
accelera
length of eachedge? that the rate of Show
area is 471'r to
surface
the
=
19. f(x)
= X 3 / 2,)
3/2
,
X > O.
x
2
.
Show area.)
as indicated.
is defined
18. f(x)
X
271'r.
to a
circumference.
the
is equal
radius
a formula for .f'(x)
is
its surface
471'r
lead
will
which
height
1 1
x
acceleration.
constant
radius
with
volume
16 through
= \037, =
cosx .
the
projectile
r is 71'r 2 and
respect to
with
of a
rate of change of the In Exercises
16t 2.)
show
be for
The
change of the (b)
=
4.2 to
x
sin
being fired
by
is V o 
l
+)
2
after

+x 
2
'
2
T sec?
is the rate of change of the area of a circle of radius
What
=
2
3x +
+
earth?
moves with (e) Show that the projectile formula (f) Give an example of another tion of  20 ft/sec/sec.
15.
11. j>(x)
\302\267
cos x
x
is given
1. \302\245:
32t ft/sec. of v o ) the time
terms
velocity on return the initial velocity
is the
(d) What must 10 sec? after
14.
= x4
vol
indicated.)
l +
10. f(x)
a projectile,t seconds
10.
1
2
12.f(x) =
x.
x
'
x
1
X
13. Assume
x.)
=
(c)f'(x)
x
8. f(x)
2
sin
as
is described
f(x)
2;
>
O.)))
that the
]68)
= Xl/
20. [(x) 21.
24.
Let [1
x,
2
= xl/
[(x)
the
+ X1/3
2
+
3 +
+ xl/
>
x
Xl/4,
xl/
4
O.)
x >
,
where
23. f(x)
=
the entries in those x for
for
the
[(x)
= tan x
[(x)
= x
28. [(x)
>
O.)
'
x
>
O.)
1 + \037 x)
[\037(x)
g(x)
[1
.fn(x)
[(x)
=
30.
=
[(x)
(X)
short table
+
2
tan
x
sec
cot
x)
 csc2
compute
x
sec
x
X)
csc
x)
derivative
the
3 .
'
x
+x
1
x+x)
x
(b) 12x +
+ x2
+
1
=
33. [(x)
=
34. [(x)
=
35. [(x)
= ax.
formula
sin
x)
+ b
ax
cx +) d
.
cos x .)
sin
3
+
2X2
cosx,
=
'
.
x +
bx +
c
+ cos
x)
+ x
of the
values
. . \302\267x n +
1), determine, by differentiation, n 1 + . . . + IlX , 2 2 3 . . . 22X 3 x + + + n 2x n .))) ::;6
2x + 3x2
each
x)
2
1+ x
if
that
x)
32. [(x)
= (ax + b) sin x + (ex + d) cosx, determine = x cos x. 2 2 = If 37. g(x) (ax + bx + c) sin x + (dx + ex + f) 2 = such x that sin x. a, b, c, d, e,f g/(x) 38. Given the formula)
1 +
x
It is understood
.((x))
2
2.
f'ex)
(valid
x sec
tan
cot x csc x)
f'ex).
If f'(x)
that
.(/(x))
sin
31.
x)
1
that the formulas
is defined.)
2x
1 x)
understood
It is
[(x))
3
2
x2 2
of derivatives.
f'ex))
x.)
+
.)
is defined.)
sec x.)
tan
1 =x
29. [(x)
(a)
x
x
[;(x)
following
which
In Exercises 26 through 35, holds for those x for which [(x)
36.
o
1 +x) '
g/(x)
[(x))
27.
=
a rule for differentiating , . . . ,[n be n functions having derivatives [\037 , . . . , [\037. Develop = [1 . . .[n and prove it by mathematical induction. Show that for those points g of the function values flex), . . . ,fn(x) are zero, we have) none
Verify hold
26.
O.)
22. [(x)
product
=+...+
25.
calculus)
Differential
determine
xn+l
.
constants a, b, values
of
 1
xI)
formulas for
the
following
sums:
c, d such
the constants
Geon1etric
= xn ,
Let f(x)
39.
f(x +

h)
=
f(x)
h' on
the sum
Express
a positive
n is
where
the
Geometric
The
a
f is
used to the
way
shown
in
Figure
nx
n 1
+
of
 1)
n(n
xn
2

2
h
169)
slope)
+
theorem to expand (x
binomial
was
as a
the derivative
n+ . . . + nxh
Let h
notation.
using. (This result
you are
interpretation
procedure
natural
the
Use
integer.
in summation
right
State which limit theorems 3 of Section 4.4.))
4.7
derivative as a
h)n
the formula)
derive
and
of the
interpretation
0 and
+
derived
2
h n 1.)
+
= nx n  1.
deduce that f'(x) in another way in
Example
slope
which leads in to define the derivative has a geometric interpretation line to a curve. A portion of the graph of a function idea of a tangent P and Q are shown with coordinates) 4.4. Two of its points respective
/
Q / / / / /
/
Vertical
slope))
71 1 I I I I
If(x +
/
( no

h)
f(x)
_J
/\037
0
m =
I) \037I I
h)
Horizontal)
I I \037)
I f)
I I I I
x)
FIGURE
quotient
as the
(x,f(x))
and
+
and P.
(x
+ h)
f(x
of
interpretation
difference
altitude,
m indicates
x+h)
Geometric
4.4
m=)
of an
tangent
+ h)).
h,f(x
 f(x), represents
Therefore, the
difference
Lines
4.5
FIGURE
the
the slope)
of various
slopes.)
angle.)
Consider
the
right
with
triangle
of the
difference
the
hypotenuse
PQ;
two
points
ordinates of
the
its
Q
quotient)
f(x +
(4.1 0))
h)

f(x)
h)
represents
The real
the
trigonometric
number
tan
r:J..
tangent
is called
of the angle r:J.. that PQ of the line through
makes
For example,
if f is
the slope
way of measuring the \"steepness\"of = mx + b, the difference quotient f(x)
this
line.
(4.10)
has the value
P
m,
with
so
the horizontal.
Q and
and
a
linear
m is
it
a
provides
function,
the slope
say
of
the
line.
Some
examples
of lines
of various
slopes
are
shown
in Figure
4.5. For a horizontal
line,)))
170)
rJ..
calculus)
Differential 0 and the
=
from
to
left
we move from rJ..
of
slope
to
from
tan
rJ..
have no
lines
rJ..,
and the
left
increases
As
is also O. If rJ..liesbetween 0 and as we move !1T, the line is rising slope is positive. If rJ.. lies between !1T and 1T, the line is falling as right and the slope is negative. A line for which rJ.. = !1T has slope 1. 0 to !1T,tan rJ.. increases without bound, and the corresponding lines Since tan !1T is not defined, we say that vertical a vertical position.
slope, tan
right
approach
slope.
a derivative at x. This that means the difference quotient h as O. When this is f' (x) approaches interpreted geometrically it tells us that, as h gets nearer to 0, the point P remains fixed, Q moves along the curve its toward P, and the line through direction in such a that its slope PQ changes way the limit. number as a For this it seems reason to define the natural f'(x) approaches slope The line through P having this is called the of the curve at P to be the number f'(x). slope
Suppose now that approaches a certain
line
tangent
has
f
limit
at P.
tangent to a circle(and to a few other special curves) was a tangent Greeks. They defined line to a circleas a line having one of its points on the circle and all its other points outside the circle. From this definiof tangent lines to circlescan be derived. For example, we can prove tion, many properties that the tangent to the radius at that point. However, the at any point is perpendicular Greek definition of tangent line is not easily extended to more general curves. The method line is defined in terms described of a derivative, has proved to above, where the tangent be far more satisfactory. this we can definition, Using prove that for a circlethe tangent line has all the properties ascribed to it by the Greek geometers. Concepts such as per
The conceptof
Note:
consideredby
the
a line
ancient
pendicularity and parallelism can be explained rather of slopes of lines. For example, from the trigonometric

tan a
tan
it
that two
follows
nonvertical
 (3) = 1
(a
lines
with
+
in analytic terms identity)
simply
tan
{3
a tan
tan
use
,
{3)
slope are parallel.
the same
making
Also,
from
the
identity)
cot (a
we
find that
two nonverticallines

(3)
with
=
1 +
tan
tan a
slopes
rJ.

having
derivative of a function is a point
tan
tan
(3 , (3)
product
1 are perpendicular.)
information about the interval where the derivative immediate from left to vicinity of x as we move in A This derivative an interval means the negative right. at while a means a zero derivative at a horizontal is as shown Xl' point graph falling, tangent At a maximum or minimum, such as those shown at X2, Xs, and x 6, the slope must line. be Fermat was the first to notice that points like X2, Xs, and x 6' wheref has a maximum zero. or minimum, must occur among the roots of the equation f'(x) = O. It is important to realize thatf'(x) may also be zero at points where there is no maximum or minimum, such as above the point x 4. Note that this particular tangent line crosses the graph. This is an not covered of tangency.))) by the Greek definition example of a situation The
a1gebraic
behavior of its then is positive,
sign of the
For example, if x graph. the graph is rising in the occurs at X 3 in Figure 4.6.
gives
in
us useful
an open
notations for
Other
171)
derivatives)
!'(X s) = 0)
>
!'(X3)
= 0
!'(X2)
XI
I I I I I I I I
X2 FIGURE
I I I 1 I I I I 1 01)
Geometric
4.6
X4
X3
X 6)
Xs
significance of
the
of the
sign
derivative.)
The foregoing remarks concerning the significance of the algebraic sign of the derivative obvious when we seem them of these quite interpret may geometrically. Analytic proofs of will be in Section 4.16.) statements, based on general derivatives, properties given
4.8
has played an
Notation Some
mathematical
statements not
derivatives
for
notations
Other
symbols,
or formulas
only remind us
of
into the
extremely
such as a short process
role in the
important
xn
developmenI of
mathematics.
merely abbreviations that compress Others, like the integration symbol S\037f(x) but also help us in\037carrying represented
n!, are
or
space. being
long dx, out
computations.
Sometimes
several
different
notations
or another being dependent on This is especially true in differential
the
are
used
circumstances calculus
where
for the same idea, preference that surround the use of many
different
notations
for the
are
one
symbols.
use9 for
of a function derivatives. The derivative discussions f has been denoted in our previous introduced J. a notation L. in the 18th late This (17361813) by by f', Lagrange century. new fact that is a function the obtained from its value differentiation, f' emphasizes f by Each point (x, y) on the graph at x being denoted by f'(x). off has its coordinates x and the derivative by the equation y = f(x), and the symbol y' is also used to represent y related the derivatives .. For Similarly, y\",... ,y(n) represent higher f\"(x),. f'(x). ,f(n)(x). = = = sin if sin then cos etc. is notation not too x, x, x, y' example, y y\" Lagrange's used by Newton who wrote y and Y, instead far removed from that of y' and y\". Newton's dots are still used by some authors, especially to denote velocity and acceleration. introduced in 1800 Another was L. by Arbogast (17591803) who denoted the symbol of f by Df, a symbol that has derivative use today. The symbol D is called a))) widespread
172)
calculus)
Differential
obtained from f that Df is a new function Higher derivativesf\", fill, . . . , r
c
,)
if
Ixl
+
0 if
+
2(3x
1)(2x) =
(x
1)4
3x
\037. This
2
0, f'(x) > function increases over the negative maximum at x = O. Differentiating = (x
=
f(x)
f'(x)
Thus f\"(x) >
a, is called a vertical asymptote or from the left. In the foregoing
the right
from
for all x, and
positive
is given
j\"(x)
x =
line,
191)
a vertical asymptote.)
The graph
This is an even The first derivative
problems)
x
2 =
in
i, where the second
Exercises)
4.19
x such that j\"(x) = 0 ; (b) examine the sign of [' determine monotonic the and examine ; (c) [is sign of,f\" in which ,f' is monotonic ; (d) make In each case, the a sketch of the graph of f those intervals is defined for all x for which the given function formula for [(x) is meaningful.
In the
and
1. [(x)
= x2
those

2. [(x) = x3 3. j{x) = (x 4. j{x) = x3 
intervals
all
8. [(x)
4x. 1 )2(X
+ 2).
6x 2 + 9x
examples
points
which
+ 5.
= x + Ifx2 .)
Worked
4.20
in
3x + 2.)
5. [(x) = 2 + (x  1)4. 6. [(x) = 1 / x 2 . 7. [(x)
(a) find
exercises,
following
determine
9. [(x) 10. [(x)
1
(x  1)(x
. 3)
= xf(1 + x2 ). = (x2  4)f(x 2 
11.[(x) = sin
2
9).
x.
12. [(x)
= x

13..r(x)
=x
+cosx.
14. [(x) of extremum
=
sin
= !Jx2 +
x.
l2 cas
2x.)
problems
in both pure and applied mathematics can be attacked differential calculus. As a of of matter the rudiments fact, systematically calculus were first developed when Fermat tried to find general methods for differential maxima and minima. We shall solve a few examples in this section and give determining to solve others in the next set of exercises. the reader an opportunity we formulate two simple principles which First can be used to solve many extremum Many
extremum
with
pro
blems.)))
problems use of
the
192)
1.
EXAMPLE
Constantsum,
principle. x and y
maximumproduct
that among all choices of when x = y = is.) largest
Prove is
calculus)
Differential
numbers
positive
Given a with
x
+ Y
S.
posItIve
number
= S, the
product
xy
Proof If x + y = S, then y = S  x and the product xy is equal to xeS  x) = = xS  x2 . Let I(x) = xS  x2 . This quadratic polynomial has first derivative f'ex) of S 2x which is positive for x < ts and negative for x > tS. Hence the maximum use of x = ts, y = S  x = is. This can the also be proved without xy occurs when = !S2  (x  is)2and note that I(x) is largest when calculus. We simply write I(x) =
x
tS.) 2.
EXAMPLE
mlnlmumsum
Constantproduct,
that among all choices of when x = y = vi p .)
Prove smallest
numbers
positive
a
principle. Given x and y with xy
P.
number
posItIve
P, the
=
sum x
+y
is
= x + Pix for x > O. We must determine the minimum of the function I(x) first derivative is f'ex) = 1  Plx 2 . This is negative for x2 < P and positive for at x = vIP. P, so I(x) has its minimum Hence, the sum x + y is smallest when = vIP.) y
Proof
The
x2 x
> =
3.
EXAMPLE
Proof value
If the x
when
x
sum
mean of
< t(a +
0, b > 0, let is smallest when x =
Given
Proof
geometric
That is, VQb
mean.
metic
=
+ y
vIP +
vIP
occurs
if and
=
a >
2 vIP.
In particular,
only if a
=
P
two
positive
numbers
has the largest
square
the result of Example 1. Let x and y denote perimeter is fixed, then x + y is constant, so the y. Hence, the maximizing rectangle is a square.)
4. The
EXAMPLE
the
perimeter,
given
use
We
rectangle.
of
all rectangles
Among
does
the sides area
of a has its
xy
not exceed
area.)
general
largest
arith
their
b).)
=
ab.
y = vIP.
a+
b
>
Among all positive
In other 2vIP
x
words, if
= 2VQb,
so VQb
and
y
=
xy
Equality
b.)
EXAMPLE 5. A block of weight W is to be moved along a fiat table by a force inclined at an angle 8 with the line of motion, where0 < 8 < t1T, as shown in Figure 4.15. Assume the motion is resisted by a frictional force which is proportional to the normal force with which the block presses perpendicularly against the surface of the table. Find the angle 8 for which the propelling force needed to overcome friction will be as small as possible.)
Solution. Let F(8) denote F(8)
the force. It has an upward vertical component propelling so the net normal force pressingagainst the table is N = W  F(8) sin 8. The of force is flN, where fl (the Greek letter mu) is a constant called the coefficient is))) this The horizontal component of the propelling force is F(8) cos 8. When
sin 8,
frictional
friction.
Worked
equated to
frictional
the
we get F(O) cas 0
force,
O. In other words,if b < 2, the minimum does not occur hence b < 2, we see that at a critical point. In fact, when > 0 when y > 0, and f'(y) the for O. Therefore the absolute minimum occurs at is > endpoint increasing y strictly f fey)
Although minimum
y=
2.
minimum d is V b2 = Ibl. corresponding is a crItical > 2, legitimate point at y and th e abso lute hence derivative f' is increasing, O.
The
If b
the

only
there
d is
minimum
The
point.
minimum distanceis I bl if value referred to above.))
V 4(b 
b
2.
we
Thus
f\"(y)
2 for all y,
=
occurs at
shown
have
critical
this
that the
(The value b = 2 is the
special
4.21 Exercises)
1.
the square has the smallest that Prove area, perimeter. among all rectangles of a given a rectangular to enclose 2. A farmer has L feet of fencing pasture adjacent to a long stone wall. area of the pasture? What dimensions give the maximum of area A adjacent to a long stone wall. What 3. A farmer wishes to enclose a rectangular pasture the least amount of fencing? dimensions require sum S > O. Prove that among all positive numbers x and y with x + Y = S, the 4. Given x = y. x 2 + y2 is smallest when numbers x and y with x 2 + y2 = R, the sum 5. Given R > O. Prove that among all positive = when x x + y is largest y. 6. Each edge of a square has length L. Prove that among all squares inscribed in the given area has edges of length l L 0. square, the one of minimum area that can be 7. Each edgeof a square has length L, Find the size of the square of largest circumscribedabout the given square. all rectangles that can be inscribed in a given circle, the square has the 8. Prove that among
area.
largest
9. Prove
that among
all
circle.
10.Given
of radius R. Find lateral surface area 21Trh
a sphere
largest
11. Among
of a given
rectangles
all
right
circular
scribed sphere has radius
the that
cylinders of
0
times
area,
the
r and altitude be inscribed in
radius can given that
lateral of the
surface cylinder.)))
has
square h of the
the smallest
the right circular
circumscribed cylinder
with
sphere.
area, prove
that
the
smallest
circum
195)
Exercises)
12. Given
a right
right
cylinder
13. Find the dimensions
a
cone
circular
right
with radius of largest lateral
cone
circular
circular
of the
right
circular
of radius R and
a sphere of radius R. right circular cone of maximum 15. Find the rectangle of largest
14. Given
R
H. Find the radius and altitude of the that can be inscribed in the cone. cylinder of maximum volume that can be inscribed in and
altitude area
surface
altitude
H.
in terms of R, the radius r and the altitude h of the volume that can be inscribed in this sphere. that can be inscribed in a semicircle, the lower base being on
Compute, area
the diameter.
16.
the
Find the
trapezoid
of largest
area that
be inscribed
can
in a
semicircle,the
lower
base being
on
diameter.
open box is made from
a rectangular piece of material by removing equal squares at each the sides. Find the dimensions of the box of largest volume that can turning up be made in this manner if the material has sides (a) 10and 10; (b) 12 and 18. 18. If a and b are the legs ofa right triangle whose hypotenuse is 1, find the largest value of 2a + b. A truck is to be driven 300 miles on a freeway at a constant speed of x miles per hour. Speed laws require 30 < x < 60. Assume that fuel costs 30 cents per gallon and is consumed at the rate of 2 + x2 /600 gallons per hour. If the driver's D are dollars if he hour and wages per obeys all speed laws, find the most economical speed and the cost of the trip if (a) D = 0, (b) D = (c) D = 2, (d) D = 3, (e) D = 4. 20. A cylinder is obtained by revolving a rectangle about the xaxis, the base of the rectangle between the curve y = x/(x2 + 1) lying on the xaxisand the entire rectangle lying in the region and the xaxis. Find the maximum volume of the possible cylinder. 21. The lower righthand corner of a page is folded over so as to reach the leftmost edge. (See If the width of the page is six inches, find the minimum 4.17.) Figure length of the crease. What angle will this minimal crease make with the rightmost Assume the edge of the page? to prevent the crease reaching the top of the page.) page is long enough
17. An
cornerand
19.
1,
__________.1) 4.17
FIGURE
22. (a)
An
Exercise
is inscribed triangle the apex is restricted to lie
isosceles
angle
2a. at
value
of the
perimeter of the
FIGURE
21.)
triangle.
in
a circle
between
Give
4.18
Exercise 22.)
of radius r as shown in Figure 4.18. If the !1T, find the largest value and the smallest
0 and
full details
of your
reasoning.)))
196)
(b) A
the
is
What
of
radius
the
given perimeter is to be made in
window
circular
smallest
L? Give full
of a
triangle
23.
calculus)
Differential
the
details
disk large enough of your reasoning.
of a rectangle rectangular
form
surmounted
by
to
cover
a semicircle
every isosceles with
diameter
portion is to be of clear glass, and the equal rectangle. semicircular portion is to be of a coloredglass per square only half as much light admitting P. Find, foot as the clear glass. The total perimeter of the window frame is to be a fixed length in terms of P, the dimensions of the window which will admit the most light. cone with diameters 4 feet and 24. A log 12feet long has the shap\037 of a frustum of a right circular of h, the volume of the largest (4 + h) feet at its ends, where h > O. Determine, as a function that of the log. right circular cylinder that can be cut from the log, if its axis coincideswith 25. Given n real numbers at, . . . , an' Prove that the sum L\037=t (x  ak)2 is smallest when x is the arithmetic mean of at, . . . , an. = 5x 2 + Ax 5 , where A is a 26. If x > 0, let [(x) positive constant. Find the smallest A such that > 24 for all x > O. [(x) of [(x) over the 27. For each real t, let [(x) = !x 3 + t 2 x, and let met) denote the minimum 1 < t < 1. interval 0 < x < 1. Determine the value of met) for each t in the interval Remember that for some values of t the minimum of [(x) may occur at the endpoints of the 0 < x < 1. interval 28. A number x is known to lie in an interval a < x < b, where a > O. We wish to approximate x by another number t in [a, b] so that the relative error, It  xl/x, will be as small as possible. Let M(t) denote the maximum value of It  xlix as x varies from a to b. (a) Prove that this maximum occurs at one of the endpoints x = a or x = b. (b) Prove that M(t) is smallest when t is the harmonic mean of a and b, that is, when lit = !(lla + lib).) to the
This section notation
The
derivatives)
Partial
*4.22
base of the
explainsthe
else in In Chapter 1, a function object
of partial
concept
and terminology. We shall Volume I, so this material may
omitted or postponed without loss in continuity. to be a correspondencewhich with each associates in another set Y; the set X is referred to as the object
of the
of points on is not difficult
be
defined
X one and only one function. Up to now, we have dealt with functions a domain consisting having the xaxis. Such functions are usually called of one real variable. It functions to extend many of the ideas of calculus of two or more real to functions
a set
in
domain
was
derivative and introduces the reader to some the of this section anywhere results
not make use of
variables. X is a set of two real variables we mean one whose domain denotes such a value at a its function, (x, y) is a real points xyplane. point in a physical written f(x, y). It is easy to imagine how such a function arise number, might For a in a circular disk of radius fiat metal the of example, suppose problem. plate shape 4 centimeters is placed on the xyplane, with the center of the disk at the origin and with the disk heated in such a way that its temperature at each point (x, y) is 16  x 2  y2 the temperature at (x, y) by f(x, y), then f is a function degrees centigrade. If we denote
By a
realvaluedfunction
in
of
two
of
If f
the
variables
defined
by
the
( 4.27))
The domain
equation)
f(x,y)
of this
does not exceed4.
function The
theorem
is the
= 16 
set of all
of Pythagoras
x
2

y2.)
points (x,y) tells us
that
whose all
distance points
from the origin (x, y) at a distance)))
r from the origin
the
satisfy
equation)
x2
(4.28))
x2
y2 < 16.
+
16 
r
2
.
in this
domain
the
Therefore
Note is, the
That
function
One is by
variables.
a third
axis
coordinate
means
(called
all
r2.)
methods for obtaining of a
surface the zaxis);
in it
(x, y)
points
circle described by is constant on each f
two useful
describe
shall
We
=
the
Figure 4.19.) two
+ y2
case consists of
on
that
197)
derivatives)
Partial
which
the
(4.28),
circle with
the
inequality is f(x, y) = temperature center at the origin. (See satisfy
of picture of a function this surface, we introduce the origin and is perpendicular)
a geometric To construct
space.
through
passes
z) y)
(0,4))
z=
16
2
2
xy)
(4,0)) x)
y) x)
(X,y,O))
FIGURE 4.19 each
to
the
obtained
The temperature is constant with center at the origin.)
on
FIGURE
circle
xyplane. the
from
Above each point (x, y) z = f(x, y). equation
4.20
The
equation
we
plot
the
point
surface represented z = 16  x2  y2.)
(x, y, z)
by
the
whose zcoordinate is
deseribed above is shown in Figure 4.20. If we placed a on the plate, the top of the mercury column would just touch z = f(x, y) provided, of course,that the surface at the point (x, y, z) where unit distances on the zaxis are properly chosen. A kind of picture of a function of two variables in the different can be drawn entirely is of contour lines This the method that is used makers to xyplane. by map represent a threedimensionallandscapeby a twodimensional We the surface that drawing. imagine described above has been cut by various horizontal to the planes (parallel xyplane). They the surface at those points (x,y, z) whose z is constant. By projecting intersect elevation of contour these points on the xyplane, we get a family lines or level curves. Each level of those and only curve consists those whose coordinates satisfy the equation))) (x, y) points The
thermometer
for
surface
at
the example
a point
(x, y)
198)
calculus)
Differential z)
y)
y)
x)
z =
(a))
equation
curves
I(x, y) mentioned
= c,
constant
or
surface and
= xy.
The
they are close together, the next; this happens
apart
the
level
elevation
in the vicinity is changing
elevation
The corresponding
and
steep
We
for
shown
can
Another
in
When
100
ft
of
elevation.
from
one
contour
the
idea of
When
to lines are far contour
the
steepness
this
curve)
z)
/
Plane
where
z =
y = Yo)
f(x,yo) on Surface
whose
is z = f(x,y))
y)
x)
FIGURE
4.22
The curve
of intersection
of a
surface z
case
this
paraboloid.
every
a general
get
example
curves. of
weather map. The equation
rapidly as we move mountain.
the
represent
they
on a
drawn
be
might
is shown in Figure 4.21. is known as a hyperbolic
of a
level
particular curve. In
circles,
is changing
slowly .
xy. (b)
for that
topographic maps are often the
=
c)
constant.)
concentric
as
curves surface
\"saddleshaped\"
lines on
Contour
curves are
isothermals, its
is z
=
elevation
constant
the
the level
temperature,
of a
example
is z
where c is
above,
xy
=
Levelcurves:xy
(b)
(a) A surface whose
4.21
FIGURE
xy)
= f(x, y) and
a plane
y
=
Yo.)))
of
a)
derivatives)
Partial
landscape concerning
by considering the rate of
of its
the spacing
change of the
199)
curves.
level
to get precise information describe the surface in terms
However,
we must
elevation,
of a
calculus. apply the ideas of differential The rate at which is changing at a point (xo, Yo) depends the elevation on the direction in which we move away from this For the sake of we shall consider at point. simplicity, this time directions to the xand we examine the two just yaxes. Suppose special parallel a surface described by an equation of the form z = f(x, y); let us cut this surface with a as in to the shown 4.22. Such a consists of all plane perpendicular yaxis, Figure plane for which the ycoordinate is constant, say y = Yo. (The equation (x, y, z) in space points The intersection of this plane with the surface y = Yo is called an equation of this plane.) = z is a plane curve, all points of which satisfy the equation f(x, Yo)' On this curve the elevation f(x, Yo) is a function of x alone. Supposenow we move from a point (xo,Yo) to a point (xo + h, Yo)' The corresponding we form is f(x o + h, Yo)  f(x o , Yo)' This suggests that the difference change in elevation
function to
can
we
which
quotient)
f(x o + h,
( 4.29))

Yo)
f(x
Yo)
o,
h)
+ O. If this quotient as h + 0, we call a definite limit approaches to x at (xo, Yo)' There are various symbols partial derivative off tvith respect ones being) to denote partial derivatives, some of the most common let h
and
\037f(xo' Yo) f \037(xo, Yo) ,)
Yo)
fx(xo,
O'
fl(X
,)
Yo)
limit are
the used
o , Yo) .)
DJf(x
,)
this that
ox)
The
the last
1 in
subscript
allowed to change
the difference quotient
form
t J 1(xo , Yo
)
.
=
1
1m
f (x 0
+
h,
the partial
we define
I'
J2
( Xo
, Yo) =
.
1
notations
that
f( x0
1mf(xo,
Yo +
k)
first coordinate we have)
the
only
Thus
(4.29).
to y at
respect
k+O
alternative
in
, Yo)
h a'ith
derivative

Yo)
hO)
Similarly,
fact
to the
refers
notations
two
we
when
.
(xo , Yo)
 f(xo , Yo)
is
the
by
equation)
,
k)
being)
of(x
o
, Yo) f\037(xo
, Yo) ,)
fy(x
o , Yo) ,)
D2f(xo,
Yo)
.)
oy)
If
and ozjoy are also used to we write z = f(x, y), then ozjox Partial differentiation is not a new concept. If we introduce
variable,
defined
by the equation)
g(x) =
f(x, Yo)
,)))
denote
another
partial
function
derivatives.
g of
one
200)
calculus)
Differential
then the ordinary derivative g'(xo) is exactly the partial derivative f1(X, Yo) Geometrically,
as the partial
same
the
the
represents
derivative f1(XO , Yo)'
of the tangent line at a when x is constant, say
slope
typical point of the curve shown in Figure 4.22. In the same way, with of the surface z = f(x o , y) describes the curve of intersection x = Xo, the equation is x = Xo. The partial derivative f2(X O , y) gives the slope of the the plane whose equation line tangent to this curve. From theseremarks we see that to compute the partial derivative and use the ordinary it were constant off(x, y) with respect to x, we can treat y as though for example, if f(x, y) = 16  x 2  y2, we get calculus. rules of differential Thus, = 2x. we find f2(x, y) = 2y. y) Similarly, if we hold x fixed, fleX, Another
is the
example
function given
Its partial
derivatives
f2

of/oy from a
=
given
02f ox
that fl,2
Notice
we indicate the
sin
,)
f
=
oy
of derivatives
not always yield the
X
cos
y
02f
=

sin
xy2
+ 2y
xy
02f
=
/2,1 =/YX
offl
with
respect
=
/2,2
ox oy')
OX')
cos xy
.)
/YY
=
and
we of
02f .)
oy2 In the onotation,
to y.
by writing)
oy ox does
xy.)
which new functions f1 = of/ ox produces Since fl and f2 are also functions of two variables, derivatives These are called secondorderpartial
02f
This
y)
f2(x,
means (f1)2 , the partial derivative order
cos
y2
process
derivatives.
/1,2 =/XY
2 ')
xy
function
can consider their partial f, denoted as follows:)
fl,l = fxx
y3
is a
differentiation
Partial
siny +
are)
= sin y
y)
flex,
=
= x
f(x,y)
( 4.30))
by)
=
0
Of
oy ( ox )
same result as the
\302\267)
mixed
other
02j = 0 Of ox oy ax ( oy)
derivative,)
partial
.)
derivatives does hold under certain conditions of the two mixed partial in that occur most functions We shall discuss these satisfied by practice. usually in Volume II. further conditions we find that its secondorder partial derivatives are Referring to the example in (4.27), However,
that
given
equality
are
by the
formulas:)
following
fl,l(x,
y) =
2,)
fl,2(x,y) = J2,1(X,
y)
=
0,)
f2,2(x,
y)
=
2
.)))
201)
Exercises)
For the
exan1ple in

= !l,l(X, y) = !1,2(X, y) = !2,1(X, y)
y4
x
=
* 4.23
mare
cos
xy
cos y cos y 
f2,2(x,y) = x
A
we obtain)
(4.30),
detailed
sin
xy3 xy3
y
siny
, cos cos
 x2 y2
 x2y2
xy xy cas

cas
sin
3y2 y2
xy xy
sin

of partial derivatives
study
xy ,

xy
2y2
sin xy = !1,2(X, y) ,  2xy sin xy + 2 cosxy + 2 cas xy.)
2xy
sin
xy
4xy
sin
xy
will
be
undertaken
in Volume I I.)
Exercises)
derivatives. all first and secondorder In Exercises 1 through 8, compute partial are and derivatives mixed the that ,2(X, y) equal. y) 12.1(X, II partial verify
1. f(x, y) 2. f(x, y)
=
x 4 + y4
= x sin
3. f(x,y)
= xy
4.f(x,y)
=
9.
Show
(x
 4x2 y2.
5. f(x,
+ y).
6. f(x,
that
10. If f(x, y)
+

(y
x
2
+
xyj(X2
8. f(x,
+
= 2z if (a) z = (x ay) for (x, y) \037 (0, 0), show y2)2
y( azj
a2j
ax 2
+
a2j ay2
=
=
sin
=
sin [cas
2 (X y3).
(x
xy) y)

o.)))
3y)].
\037y).
x)
=
2y)2,
that)

(2x
= x+y
7. f(x,y)
\037 0).
y2.)
x( azj ax) +
=
y)
x
y Y
y)
In each
V x 2 +
(b)
z =
(x, y)
y2
(x
4
+ y4)1/2.
\037 (0,
0).)
case
5)
DIFFERENTIATION)
AND
5.1
the to We come now differentiation. The relationship
holds between
which
that
number and
then
back again.
Similarly,
new
function
original
defined
by
the
f
the
and
\"taking
positive
equation)
x
c is
a general
that exists between and integration two processes is somewhat analogous to the square root.\" If we square a positive
these
between
A(x) =
where
of calculus
connection
remarkable
\"squaring\"
theorem
fundamental
we get the original number square root of the result, if we operate on a continuous function we get a f by integration, of when indefinite leads back to the which, differentiated, integral f) For example, if f(x) = x2 , then an indefinite integral A of f may be
take
(an
function
The first
indefinite integral.
of an
derivative
The
INTEGRATION
BETWEEN
RELATION
THE
f
x
f(t) dt =
t
f
c
a constant. Differentiating, we find result, called the first fundamental
2
=
dt
X3 3
c
=
A'(x)
3
 c , 3)
x2 =
f(x).
of calculus,
theorem
This example illustrates which
be stated
may
as
follows:) THEOREM
integrable function
5.1.
FIRST FUNDAMENTAL
on [a, x] for A as follows:)
each
A(x)
Then the and
for
derivative
such x lve
A'(x)
in
x
=
exists at
r J(t)
a
0 such that)
IG(h)1
0. Assume mechanism propelsa
of the
particle
2t sin
t. The
occurs.
at time t
particle
 2a
an
+ (2  a2 ) sin
along initial point
perfectly
the interval
over
[0, a].
All
is)
[(a).
so that the
is designed
is given =
solid
a)
calculate It
> O.
\302\267
The volume of the
are squares.
on [0, a] and line. 0 on the line until time t
x
x
all
.
+ x)
(1
function./
nonnegative
cos a
x
2
dt =
.(t)
J0
a straight
works
nlechanism
a
interval
continuous
[is
from
dt =
la for
formu
given
X2(1+X)
dt = x 2 (1 + x) .
a solidis the
the
2
I: 11' when the particle returns to the initial point 0, or else prove that it never returns to O. line. Its position at time t is .(t). When 0 < t < 1, the 25. A particle moves along a straight is the position given by integral) then
From
on the
t
[(t)
not attempt to evaluate
(Do
acceleration (the ation f'(t)
26.
at

time
[(1)
t = when
acceleration
2; (b) t >
its
= r Jo) this
1
+ 2 sin
integral.)
cos
TTX
1 +x)
7TX
dx
2
For
t >
1,
\302\267
the
particle
it acquires at time t = 1). Compute the when t = 1; (c) its velocity when velocity
moves following:
t >
with constant (a) its acceler
1; (d) the
difference
1.
second derivative case, find a function f with a continuous [II which satisfies conditions or else such cannot an exist. explain why example given for every x, O. (a) fll(X) > 0 [1(0) = 1, for every x, (b) fll(X) > 0 .(1(0) = 1, for every x, for all x > O. > 0 (c) f\"(x) .(1(0) = 1, = 1 0 for 00) for all x < O.))) > 1, x, (d) f\"(x) every [/(0)
In
each
[1(1)= [1(1)= 3. [(x) < 100 [(x)
(o). = P\"(I) = O. P'(l)
the
and)
tn7T))
nth derivative
h(n)(x) =
g(n>(x)
of
h
= sin
tn7T).)
by the formula)
is given
! (:)1
(x +
,
(k)(x)g(nk)(x)
k=O)
5.
the binomial
where
(\037) denotes
Given
two functions
(5.30)) for
every
whenf(x)
f'(x) x
in
some
= sin
,( and g
= g(x) ,) open
coefficient. This is called Leibniz'sformula. derivatives
whose
g/(x)
interval J
x andg(x)
=
containing
= cosx.)))
f' and
g/
f(O) =
f(x),) o.
For
the
satisfy
0 ,)
example,
equations) g(O)
= 1
,)
these equations are
satisfied
(a) Provethatf2(x)
(5.30). Prove that F(x) = ,((x) and = J. Consider [Hint: hex) [F(x) ,f(x)]2 + [G(x)  g(X)]2.] every can you say about functions.f and g satisfying more (c) What (5.30)? 2 = x 3 for A function f, defined for all positive real numbers, satisfies the f(x ) equation every x > O. Determine f'(4). A function g, defined for all positive real numbers, satisfies the two conditions: following 2 = 3 = 1 x and for an x O. > ) g(l) g'(x Compute g(4). G(x) = g(x)for
7.
= 1 for every x in J. of functions satisfying pair
+ g2(X)
(b) Let F and G
6.
8. Show
be
another
x in
that)
x sin Io
C2
C l and
Let
9.
223)
exercises)
review
Miscellaneous
C is said to
be
\"bisect
two in
given
t +
> dt )
1
0
x >
for all
0 .)
curves passing through the origin as indicated in Figure the region between Cl and C2 if, for each point
area\"
regions A and B shown that the bisecting curve = ix 2 .)
shaded
t
5.2.
A curve
P of C, the
two
figure have equal areas. Determine the upper curve C 2, = x 2 and that the lower curve Cl has the C has the equation y in the
y
equation
y)
C 2)
C 1)
x) o)
5.2
FIGURE
10.
f is
function
A
defined for
all
x as
follows:) 2
X
f(x)
if
x is
rational,
if
x is
irrational.)
=
{0) if h \037 O. Let Q(h) = f(h)/h and conlpute.f'(O).
9.)
Exercise
(a) Prove
that Q(h) \037 0
as h
\037
O.
(b) Prove
thatfhas
a
derivative
at 0,
11. 12.
13.
14. 15.
11 through 20, of substitution
Exercises
In
using
the
r (2 \037
I
method
+ 3x)
xV 1
f2
X(X
sin 5x dx.
+ x2 dx.
2
x4(1 +
the
Try to whenever
integrals.
given
by
integration
l
parts
16. J0 x4(1

X)20
sin
17.

1)9 dx.
X
5
18.
dx.
19.
)5 dx.
20.
I I
I
sin
the
simplify
possible.
dx.
dx . \037
1\\2
(1 2x + 3 Jo (6x + 7)3 I
evaluate and/or
dx.
V\" xI
x sin
x2
VI +
3
cos
cos
x 2 dx. 2x
sin 2x
dx.)))
calculations
by
21. Show
23. Let
= In
12 ,
1 3, 14,
pair
x2)n dx.
S\037(1 and
15'
=
t
m
S\037
(1 + (m +
dt,
m >
25. Let/(n) = S\037/4 tan (a)
+ fen
(b) fen)
1
(c)
=
2)
if
n > 2.)
n _ 1

if
2 we
one
only
is defined
S\037Pn(x) dx
and)
=
P(x) such
polynomial as
inductively
= 0)
if
the
term
n >
follows:)
1.)
. . . , P 5 (x). in x
of degree
n,
of highest
degree
nx n 1 if n > 1.
have)
kl
L r=l)
225)
2.
n >
Pn(x)
one and
polynomials)
explicit formulas for P1(x), P2 (x), that P n(x) is a polynomial
by
(d) Prove that (e) Prove that
prove that there is
polynomial,
= Q(x).
 3P(x)
review exercises)
rn
=
(k Jo
Pn(x) dx
=
P n+1 (k )
Pn+1 (0)
.
n \037 1
Prove that Pn(1  x) = ( n (x) if n > 1. = 0 and P2n  1(!) = 0 if n > 1. P that Prove (0) 2n 1 + (g) on its largest 36. Assume that If\"(X) I < m for each x in the interval [0, a], and assume that [takes < am. You may assume Show that 1['(0)\\ + If'(a)\\ value at an interior point of this interval.
1)np
(f)
that [\"
is
continuous
in
[0,
a].)))
6)
THE
man
Whenever
focuses
his
on
attention
of a known function properties The function concept function. endless
variety
or
is so broad
in
in
and
so
general What
nature.
that
it is not surprising
is surprising
one who studies
chapterfirst and secondly, the inverses of the trigonometric mathematics, either as an abstract discipline or as
scientific field, will find erties is indispensable.
that
functions.
a tool for
knowledge of these functions
a good working
a few
that
to
find
rather
of natural We shall phenomena. of all, the logarithm and its inverse
this
function)
exponential
is
kinds
different
totally
many
he is either studying relationships, to discover the properties of an unknown
quantitative
trying
of functions occurring
special functions govern so study some of these functions (the
FUNCTIONS)
Introduction)
the
an
EXPONENTIAL,
THE INVERSE TRIGONOMETRIC
AND
6.1
THE
LOGARITHM,
and
Any
some
other
their
prop
The reader probably has had occasionto work with logarithms to the base 10 in an definition algebra or trigonometry course. The given in elementary elementary usually is If this: x the of x to the base denoted > 0, logarithm 10, x, is that loglo algebra by real number u such that IOu = x. If x = IOu and y = 10v , the law of exponents yields =
xy
(6.1
10 u + v . In terms
of logarithms,
))
this
becomes)
=
loglo(xy)
+ 10gloY')
loglox
makes logarithms particularly adaptable to computanumber 10 is useful The as a base because real numbers involving multiplication. in the decimal system, and certain important like 0.01, are commonly written numbers for their the 0.1, 1, 10, 100,1000,... have 2, 1,0, 1,2,3,..., logarithms integers
It is this
fundamental
property
that
tions
respectively.
It
is not
necessary to restrict
serve equally
well.
ourselvesto u =
( 6.2)) and
(6.3)
226)
the
fundamental
10.
base
Any other
positive base b
Thus)
property
in
10gb
(6.1)
10gb
(xy)
x =
means)
X)
becomes) =
10g b
x
+
10g b Y
.)))
bu
,)
\037
1 would
Motivation
If we
logical gaps.
of
First
of
natural
the
from a
in (6.2)
definition
the
examine
from several
the definition
for
critical
we
of view,
point
to understand
all,
227)
as an integral
logarithm
(6.2) we
find
know
must
that
it suffers
what
is meant
define when u is an integer or a rational number of two (the quotient by u a trivial matter to define b when u is irrational. For integers), example, how should Even if we we define 10\\l2? a satisfactory definition for bU , manage to obtain there are further difficulties to overcome before we can use (6.2)as a good definition of a number u such exists logarithms. I t must be shown that for every x > 0, there actually that x = bU. the law of exponents, bUb v = b U + v , must be established for all real Also, exponents u and v in order to derive (6.3) from (6.2). It is possible to overcome these difficulties and at a satisfactory definition of arrive is long and tedious. Fortunately, however, logarithms by this method, bur the process the of logarithms can proceed in an entirely different study way which is much simpler and the power and elegance of the methods of calculus. The idea is to which illustrates introduce logarithms and then to use define bU.) first, logarithms is easy to but it is not
This
bU.
for the
Motivation
6.2
The
an
is
logarithm
different
When
ways.
We shall
of the
by
itself and
want
this
from
which
how
this
An
properties of the
to have
leads
equation
xy
are
defined
used to
in many
a concept, such wants this concept of
propertieshe
may be
procedure
us.
If we
the property
in the
(6.4), which
of
think
the
expressedby = f(x)
arrive
or
process as
logical
definition
at the
the
Let
logarithm the
logarithm
us consider
as a
function
of a this
product property
f,
then
we
formula)
+ fey))
domain off
expresses a is called a functional points,
like
be
in the next
f(xy)
and
a number of
section. logarithms to have is that factors. logarithms of the individual
see where it
x, y,
as an integral
we want
(6.4)) whenever
logarithm
he is often led to a simple formula all the desired properties spring forth
properties,
is given
sum
function
has in mind
illustrate
which
logarithm
One of the should be the
natural
the
of a mathematical conceptthat can example a mathematician tries to formulate a definition
as the logarithm, he usually to have. By examining these that might serve as a definition deductions.
of
definition
relationship
between
the
values
of a
function
Many problems can a solution function which satisfies solving equation, being any the equation. Ordinarily an equation of this sort has many different and it is solutions, difficult to find them all. It is easier to seek only those solutions which have usually very such as or For the most these some additional continuity differentiability. property part, We shall adopt this are the only solutions we are interested in anyway. of view and point determine all differentiablesolutions of (6.4). But first let us try to deduce what information without any further restrictions on f we can from alone, (6.4) is zero everywhere on the real axis. In fact, of (6.4) is the function that One solution is defined for all real numbers. To prove let f this is the only solution of (6.4)that this, = of f, then we may be any function that satisfies (6.4). If 0 is in the domain 0 in put y = = and for in this 0 x the domain to (6.4) obtainf(O) f(x) + f(O), thatf(x) every implies In other words, if 0 is in the domain off, then f must be identically zero. Therefore, off of (6.4) that is not identically zero cannot be defined at O.))) a solution at
two
or more
be reducedto
a functional
equation.
mathematical
If f is a solution (6.4) to obtainf(l)
of
and the
the exponential,
The logarithm,
228
this
1 are
1 and
both
thatf(l) = 2f(I); hencef(I) = O. we may we
=
y
put
1
in
If
deduce
to
(6.4)
may take x = 1 x, x, 1, and 1 = f( 1) + f(x) x) f( now
other
1 in
and
y
are in
the
since
and,
to deduce
1
=
domain
off,
1) = 0,
f(
= f(x)
and
(6.4)
any solution of (6.4) is necessarilyan even function. at each x \037 O. If we hold y fixed now, we assume thatfhas a derivative f'(x) rule on the left), we find) differentiate with respect to x (using the chain =
yf'(xy)
x =
When
1,
this
= f'(I),
us yf'(y)
gives
every
for
second fundamental f(x)
x > 0, this Since f( 1) =
If
0,
the
choice
c =
f(x)
If x is
negative
then
x
These
two
positive
and
formulas negative
for f(x)
= 1'(1)
is monotonic and
f'
Also, f'
(x
c, and
dt)
may be
(x
if
x
x
if)
dt)
x
combined into
= f( x), if)
on
such interval,
write)
0, it holds
0 and
y
the
the
on
continuous
> 0,
the
real axis, has the
positive
of x
A(x)
integral
35. A
=
function
f,
functions)
for
that
property
dt)
/(t)
(and therefore dependsonly dt for
ff[(t)
Xy
= 2, compute
If [(2)
y).
real axis, has the
on the positive
continuous
on
dt =
Y
x J 1 [(t)
dt + x
that)
property
y
J 1 [(t)
dt)
= 3, compute[(x) for each x > O. set of a function over the is continuous [which [1, a]. All cross sections perpendicular to the interval [1, a] are squares. The volume solid is ta310g2a  fa 3 10g a + 2\037ia3  2\037ifor every a > 1. Compute [(a).)
6.10 In
> 0 and
all x
for
base of
this
To
y > O. If [(1) is the ordinate
show that can be used to
we will
section
which the
simplify
to
the logarithm
the
if x
is valid
< 1.
The
(1
log
first
x)
a simple
linear
From the algebraic identity
for
IX dt 
=
1
 x) =
x
io
lu real u
\037
1.
du 1

this
to the form)
defining
,

u
(1
converts
x < 1.)
for
valid
u)
u) the
by
To
logarithm.
0 to x, x +
2
= x
obtain
then
integrate
to
the method, we
the formula)
,
lu)
x 2
we
illustrate
2
u
 x) = x +
which
polynomials
=l+u+
this from
Integrating
Jog
( 6.19))
The graph of
integral
,
t = 1
of variable
change
1
any
the
t)
approximation to the integrand. 1  u 2 = (1  u)(1 + u), we
(6.18)) valid
replace
J
Now we approximate the integrand 1/(1 obtain corresponding approximations for
begin with
the
obtain)
Jog (1 which
of
approximated by certain degree of accuracy.
to any desired x by 1  x in
logarithms
we
can be
function
logarithm
compute
formulas,
resulting
interval
the logarithm
to
approximations
Polynomial
polynomials
all
a solid
of the
value
the
> O.
x
all
J 1 [(t)
36. The
of
all choices
integral)
J:Y is independent
inverse trigonometric
and the
exponential,
where
x
1, we
every
of degree
X3 +
X2 +
+
X
log
( 6.20))
 log (1  x)
y =
the polynomial
denote
Pn(x)
Then,
\"\",\"'\" \"\",'\"
polynomial approximation to
A quadratic
6.5
FIGURE
\"'\"
for
rewrite
u
\037
u
+
u
2
..+
+ .
this from
Integrating
u
n
0 to
l
+
un
log
(1
 x) =
Pn(x)
x, where
+
En(x),)
du
.
integral,)
x
En(x) =
fo 1
un

u)))
,
1u)
(6.20) in the form)
(6.21)) where En(x) is the
1.
= 1 +
x
0, a =fi
a f(x)
=
dx
af(x>.!,(x)
3x 2 dx, and we
=
du
eX3 (3x
1 J
u =
write
we
dx.)
Then
=
dx
,)
f
Sx
x 3.
=
u
c
+
tI(x)
derivative. If
become)
(6.39)
being valid for a
of these
second
the
=
ef(x\"j'(x) dx
247)
exponentials)
involving
a continuous
with
represents any function dx, the formulas in
f'(x)
formulas
integration
+ C .)))
arbitrary
=
The
248
the exponential, and
logarithm,
inverse
the
functions)
trigonometric
dx 5.
EXAMPLE
f
1 +
way to treat
One
Solution.
.
Integrate
eX)
this
is to
example
rewrite the integrand
ex)
1
Now
u =
put
eX f e
x
+
result
The
e X
X
f
1
(1 +
e X)
way
to treat
ways if we 1
=
log
=
Another
log
same
this
=
1+
1)
1+

1
eX +
dx
u =
where
1 +
eX.
we
Thus
eX
 log (1 +
.)
eX)
eX)
1 +
eX)
f 1
which
is one
6.17
Exercises)
of the forms
In Exercises 1 through for all real x for = e3x  1. 1.
xI f
+
+
= x eX)
 log (1 +
eX)
f(x)
2. f(x) 3. f(x)
8. f(x) 9. f'(x)
= e4x2 . = e x2 . = e Vx . f(x) = el/x . f(x) = 2x . f(x)
du

,)
;;
+
C.,
obtained above.)
12, find which
x
f
the derivative.f'(x). In given formula for f(x) 7. f(x)
defined
=
dx eX
find)
dx
6.
instance,)
write)
eX
=
f 1+
5.
C .
1)
1) = x
+

eX
) +
we have)
Then
4.
X
eX
= log
is to
example
e
log (1 +
manipulate the logarithm. For
e(eX)  log (eX x
C=
luJ +
log
u)
f
1)
we get)
and
dx,
du
=
other
in
written
log
dx
e x + 1
X
e
=
du
 e

dx =
can be
1. Then
+
e x +
eX
1 +
as follows:)
the
10. 11. 12.
each
case the
is meaningful. = 2x2 [which = esin x.
= = f(x) = f(x) = f(x)
function ,r is assumed to 2
means
2
definition
0 and
r is
TT (x

X)1/3
ai)b
.) X)2/3
i.
i=l)
The formula.f'(x) = rxT 1 was proved
any real number.
holds
also
aX = eX loga
for arbitrary
of
the result
conditions
to derive
the
real r. (a)
part
[Hint:
applies
properties
following
Write
aXa Y
= a X+ Y .
= (aY)X = a XY . a \037 1, then y = aX if and only if x = If Let f(x) = l(a X + aX) if a > O. Show that)
(d)
(aX)Y
(e) 37.
j'(x
+ y)
loga
+ f(x

y.
y)
=
2f(x)f(y)
x T = e T Jog x.]
for x < O. of general exponentials:
(a) log aX = x log a. (b) (ab)X = aXb x . (c)
x.
r.
Show that this formula (b) Discuss under what (a)
36. Usethe
=
34. f(x)
xxx.)

(cos x)sin
+
(1 _ x)(3 + n
f(x)
differenti
(log x)X
x)].
V I +
fis assumed to be
Logarithmic
(log x)x.
=
f(x)
\302\267
the function
case,
30. f(x) x
a
a
+ C
meaningful.
29. f(x) = xlog
.
ex
cos bx)
a 2 +) b 2
28.
c ,
+
 b
bx
bx)
2
derivative.f'(x). In each for f(x) is
 e X
eX
b
+)
given formula some cases.
the
= xx. = (1 + x)(1 + eX 2).)
+ aX + = log [log(log
f (x)
find the
which
the work
simplify
25. f(x)
24.
34,
all real
bx + b sin
a2
\302\267)))
The
250
38. Letf(x)
the exponential,
logarithm,
= eC\037,where
a constant.
c is
and the
Show
inverse
e CX 
the following
1)
= c.)
lim) x)
x\037o)
be a
Let f the
to deduce
relation:)
limit
39.
and use this
= c,
that,fl(O)
functions)
trigonometric
defined
function
on the real axis,
everywhere
fl which satisfies
a derivative
with
equation)
=
fl(X)
CX Let g(x) = f'(x)eand
[Hint:
be a
40. Let f
Prove that
a constant.
c is
where
functional
defined
function
cf(x)) is a
there
consider
for every x
constant K such
gl(X).] on the real
everywhere
,)
that
= Ke CX for
f(x)
axis. Supposealso
that
every
f satisfies
x.
the
equation)
f(x
+ y)
for all
x.
(i))
(a)
Using
f(O)
\037 0
only the then f(x)
functional \037 0
= f(x)f(y)) prove
equation,
and y
all x
for
that f(O) is
.)
0 or
either
1. Also,
prove
addition to (i), thatfl(x) exists for all x, and prove the following statements: = for all x and (b) ,fl(x)f(y) y. fl(y)f(x) (c) There is a constant c such that f'(x) = cf(x) for all x. CX See Exercise 39.] (d) f(x) = e if j'(O) \037 O. [Hint: 41. (a) Let f(x) = eX  1  x for all x. Prove that f'(x) > 0 if x > 0 and f'(x) < 0 if Use this fact to deduce the inequalities)
if
that
Assume, in
eX
for all
valid
x > these
Integrate
O. (When
(b)
eX
>
1 +
x +
(c)
eX
>
1 +
x +
(d)
Guess
x2 2! ') x2
42.
If n is
x
inequalities
2!
+
x3
3!
')
>
1 +
x
= 0, these to derive the
e X >
,)
1
 x
further
e x < 1
 x +x
e x > 1
 x +x
(
43.
By choosing Let f(x, y)
eX,)
and
2!
eX
O. Show that)
of = yxVl ax)
valid
.)

x3
3!
.)
result.
X
0, show that)
1 +
inequalities,
2
2
that
and)
n if
) ( 1\037)
2.5
O. 5 x = 13. 18. Find sinh x and cosh x if tanh 19. Find cosh (x + y) if sinh x = t and sinh 20. Find tanh 2x if tanh x = 1. 21 through = cosh x.
x
sinh
22. D coshx D tanh x
23.
= sinh
26, prove
differentiability
a
Assume
the inverse of f then the derivative g'(y) g be
f(x).
two
the
Moreover,
Ifwe
dxfdy
g'(y),
to
function
f is
which
x. x.)
on an interval [a, b], and and is nonzero at a point x in (a, b), the corresponding point y, where y = also existsand that is, we have) derivatives are reciprocals of each other,.
If
derivative
use the Leibniz then Equation
exists f'(x) is nonzero at
=
write y
(6.42) becomes)
for f(x),
dyfdx
x for
forf'(x),
g(y), and
1
=
,)
(
in Assume x is a point Proof We shall show that the difference
\302\267)
f';X)
and
notation
appearance of a trivial
continuous
and
increasing
strictly
the
dy
has the
= sech x tanh = csch x coth
its inverse.)
dx
which
csch 2 x.
construct the exponential function from the the trigonometric functions. It is convenient shows that the process of inversion transmits
to
invert
g'(y)
for
=
functions)
( 6.42))
Note:
x
D sech x D csch x
25. 26.
of inverse
6.7.
THEOREM
formulas.
24. D coth
x.
from
I.
differentiation
the
We have applied the process of inversion logarithm. In the next section, we shall at this point to discuss a generaltheorem
let
=
y
= sech2 x.
Derivatives
6.20
trigonometric functions)
x
In Exercises 21. D
the inverse
and
exponential,
= coshx + 1. sech 2 x = 1.
2 cosh 2 !x
13.
the
:) identity.)
algebraic
(a, b) where
f'(x)
exists and is nonzero, and
let
y
=
f(x).
quotient)
g(y +
k)

g(y)
k)
the
approaches
Let
g(y +
h = g(y
k).
limit
Iff' (x)
k) 
as k \037
O.
x = g(y), Since + g(y). = Therefore y + k f(x + h), and
this
implies
hence
h =
k = I(x
g(y + k)
+
h)

 x or x + h I(x).
Note
=
that)))
of the
Inverses
h
:;1=
in
0 if
k
:;1=
g is
0 because
trigonometric
if k
Therefore,
increasing.
strictly
253)
functions)
difference
the
0,
:;1=
quotient
is)
question
k)
g(y +

g(y)
h

( 6.43))
f(x +
k)
+ 0, the (b) of Theorem
As k


h)
1
[f(x +
f(x))

h)
f(x)]/h)
0 because of the continuity of g at y [property that h + 0 as k + O. But we know that the difference on the extreme right of (6.43) quotient in the denominator approaches f'ex) as h + 0 + k of (6.43) 0, the quotient on the extreme left [since f'ex) exists]. Therefore, when ' Theorem 6.7.) the limit 111(x). This proves approaches
6.21 The
begin
difference
+ k) means
g(y This
3.10].
Inverses of the trigonometric function.
sine
the
over some interval
it
where
functions
may be appliedto the trigonometric To determine a unique inverse, is monotonic. There are, of course,
of inversion
process with
+
g(y)
such
many
we
sine
intervals,
for)
y)
y)
I
functions. Suppose must consider the
we
x)
x)
I 71\",
2: I)
2)
2)
6.9
FIGURE
=
Y
SIn
FIGURE 6.10
x.)
i1T], [!1T, .\0371T], [!1T,
example
[i1T,
these we
choose. It is customary to
etc., and
!1T], select
[
 !1T,i1T]
f(x) = sin x) The
function
f so
defined is
2
y
arc sine, and
its
u
value
=
at y is
arCSIn
denoted
v)
by
v =
means
sIn
6.9.)
to each
function
y, or by
arcsin
graph
outside the
of the interval
arc sine is shown [1,
1].)))
in
Figure
one
of
follows:)
sin l y.
and)
6.1 O. Note
1
number y in [1, 1] that the inverse sine or
g is called Thus,)

x)
x
17/ 2
(x + 2) dx \302\267 I x 2  4x + 4
< 1).)
< a
\037 0).)
dx
32.
 6.
+ x
.
dx.
'2
+
2 I a
I
x2dx
13.
dx
I 1 + acosx
31.
x /x x'
(0
x
a cos
+
II
I(xX:\037)2'
I
dx.
x + 5
::os
x
sin
1)2
dx)
11. 12.
I2
2)2
 1
x +
+
27.
+ 1)2
I (x +
(x
I
dx
I x(x
5
.
2x +
4X5
25.
2
2
X2 dx
(x2 +
j
.
1
7
8. xX+2 dx. I +x 9.
\302\267
1
x/:
I
1
+ 5x2
dx
I x4
23.
dx.

3 X4
I x4
I X(\0372::/X.
x + 1
+
x
I
21.
dx.
1)(2x +
(x +
2 x3
'1l
8x3 + 5.
.
.
x 2 _ 2x
+
3)
,
x4
I
:x 2 )(X +
 3x +
x3
20.)
5)
X4 +2X6
4.
7.
dx.
I (x _ 2)(x+
X
integrals:
following
2X+3
dx.
dx.
dx.
5 dx.
dx.
x2+x+l)))
x
\037 0).)
The
268
and
the exponential,
logarithm,
inverse
the
dx 39.
f
2 V x
/
+x
[Hint:
Exercise
In
1. Let f(x) obtain
=
f(2)
a
2. Find
= t
+ f(l)
function
> O. Compute
if x
log2
for
3.
evaluate fex/x dx
Try to
/2
Integrate
5. A
log (e
S\037
functionfis
COS
defined
x
generating and
integral
6. A
under
Lo
(b) (c)
In a similar
Sf
of x is it et/(t + a) dt
at the point and above the
true =
xet f1
f1 In each
= eX.
dt
= 1
dt
=
(b)
S\037\037r(t)
(c)
S\037f(t)
8. If
f(x
prove
9.
dt
S\037f(t)
Given
 2x2 .
log x < F(x)? + a)  F(l + ea[F(x
integrals
')
t
f 1
f2(X)
X2
0
.)
a)].
means
2t)
of F:)
terms
in
xet
dt
[2
x >
that
case, give an example of a continuous else explain why there is no such
(a)
the
t)
x, or
real
1.
is rotated about the xaxis, thus of this solid. Compute this volume
integral:) if
dt
x =
.)
[1, 4]
for
indefinite
> 0
x
which
for interval
an integral (25/8).
x eat
7.
zero), such that)
.
+ 2))
I)(x
the following
express
way,
you should
t)
if
+
x(x
the following
by
what values that
dt
cos
4x +2
F(x) =
Prove
check,
by parts.
off
graph
the graph
function F is defined
For
a
the equation)
by
a solid of revolution. Write show that its value is 1T log
(a)
As
sin t
f(t) 2 +
integration
using
by
of the
the slope
Find
(b) The region
 x2 .])
X) dx.
[(x) = J (a)
f(x) + f(I/x).
not everywhere
x (and
all
f2(X) =
4.
 x
2.
continuous
f,
Y 2
by
exercises)
+ 1) dt
t)/(t
(log
Si
dx.
x2
f
numerator and denominator
40, multiply
review
Miscellaneous
6.26
 x2
 x
2
V
40.
\302\267
functions)
trigonometric
e 1/t
dt,
dt
\302\267)
1'\" function
f satisfying
the conditions stated for
all
function:
2 CX2}.]
 1.
= 1 + + y) = f(x)f(y) for all x and y and if f(x) xg(x), where = that exists for and (b) f(x) eX. (a),fl(x) every x, a function g which has a derivative gl(X) for every real x and which
+ 1 as
g(x)
x +
0,
the following
satisfies
equations:
g'(O) = (a)
Show
that
and
2
= 2eX g(x)
g(2x)
(b) Generalize
(a)
n. Prove your
result
g(x
by
finding
by
and
+ y) =
find a similar
a formula
induction.)))
eYg(x)
+
eXg(y)
formula
for
for
all
x and
y .
g(3x).
relating g(nx) to g(x), valid
for
every
positive
integer
= 0 and
269)
exercises)
review
Miscellaneous
the limit of g(h)/h as h \037 O. and that g'(x) = g(x) + Ce x for all x. Prove this statement Use the definition of the derivative [Hint: g'(x).] = all x in its domain. What can 10. A periodic function for with period a satisfies [(x + a) which and satisfies an equation of has a derivative everywhere you conclude about a function
(c)
that
Show
g(O)
(d) There is a constant find the value of C.
find
C such
[(x)
form)
the
where a
all x,
for
12.Let
=
A
differentiation
Express the
of the
values
_
t
LaI
differences.
following
dt. a _ 1
1)2
dt.
(d)
13.
I ot
(l
dt.
1
+
2tet2
1
et
i0
log
let f(x) = eXp(x). Co + CIX + C2X2 and of [at 0, is Co + nCI + n(n that [(n>(o), the nth derivative is a (b) Solve the problem when p polynomial of degree3. Let p(x) (a) Show
a polynomial Show that
to
= x
Let [(x)
+ t) dt.
=
(c) Generalize 14.
and
of A:
in terms
integrals
1 et (c) i 0 (t +
I (b)
of products
differentiation
for
formulas
e t
a
(a)
+ 1) dt.
/(t
S\037
b[(x))
formulas for sums and
the corresponding
from et
quotients
positive constants? to derive the
b are
and
11. Use logarithmic
=
+ a)
[(x
ax.
sin
15. Prove that)
(
k=O
1/(k
[Hint:
+
In
n
k
\037
L(I)
k
1) =
+
of degree nl. 2n [(2n>(x) = ( I)n(a x 1
_
k m t +
(
k=O)
1 )C2
'
k ) k+n+l
dt.]
a formula (or formulas) for computing 16. Let F(x) = S\037[(t) dt. Determine if [is defined as follows: = e 1tl . = (t + Itl)2. (c) [(t) (a) [(t) 2 I if t I tl < 1, of (d) [(t) = the maximum (b) [(t) = 1 _
solid of
if
It I
{
It I >
.
 2na2n  1 cos ax). 1
m
k
L.t(I)
) k+m+I
S\037
\037
sin ax

F(x) for
1
and
all
real
x
t2.
1.
function [ around + a, find the function f 18. Let [(x) = e2x for all x. Denote by S(t) the ordinate set of.f over the interval [0, t], where of the solid obtained by rotating t > O. Let A(t) be the area of S(t), V(t) the volume S(t) about the xaxis, and W(t) the volume of the solid obtained by rotating 5(t) about the yaxis. (a) A(t) ; (b) V(t) ; (c) W(t) ; (d) lim t _ O V(t)/ A(t). Compute the following: to compute such that sinh C =!. (Do not 19. Let C be the number c.) In each case attempt in terms of answers the given equation. find all those x (if any exist) satisfying Express your 2 3. and log log (a) log (eX + v e2X + 1) = c. (b) log (eX  Ve 2X  1) = c. is true or false. Prove eachtrue statement.) each of the following statements 20. Determine whether 17. A
the
interval
(a)
2 10g
revolution
on
[0, a]
is generated by xaxis. If, for
the
rotating
the
every a
a continuous
of
graph
> 0, the
volume
is a 2
n 5
=
5 10g
2.)
(c)
k 1/2
L
1.
k=l)
(b)
log2 5
=
log3 I
5
og2) 3
.
(d)
1
+
sinh x
< cosh x
for every
x.)))
The
270
21 through
In Exercises an appropriate 2
21.
22.
;;
1
each
establish
24,
< x
x
sin
inverse
the sign
examining
by
inequality
functions)
trigonometric
of the
of
derivative
function.
n
x
that)
Show
e 2X
=
['(x)
29.
) if
271)
exercises)
[has
2

3
x
+)
2
.
an inverse,
computingg(y)
and denote for each y
in
this
inverse
the
domain
by g. of g.
What
Sketch
to evaluate this integral.) dt if x > 0. (Do not attempt real axis. strictly increasing on the nonnegative to g2 the inverse the second derivative of g is proportional of f Show that 2 for in the domain of and find the constant of each proportionality.))) y g] cg (y)
7)
7.1
TO FUNCTIONS)
APPROXIMATIONS
POLYNOMIAL
Introduction
in analysis. occur are among the simplest functions that They are pleasant in numerical computations because their values may be found by performing In Chapter 6 weshowedthat the logarithm and additions. a finite number of multiplications that enable us to compute logarithms to any function can be approximated by polynomials Polynomials
to
with
work
desired
degree
In
of accuracy.
this
show
we will
chapter
that
other
many
such
functions,
can also be approximated by polynomials. as the exponentialand trigonometric functions, and its polynomial If the difference between a function small, approximation is sufficiently with the polynomial in place of the original then we can, for practical purposes, compute function. There are many ways to approximate a given function f by polynomials, depending on In we is to be made of the shall be use this interested in what chapter approximation. which with and some of at a a its derivatives agrees f given obtaining polynomial point. We begin our discussion with a simple example. the exponential function,f(x) = eX. At the point x = 0, the functionfand Supposefis linear all its derivatives have the value 1. The polynomial) g(x) = 1
also hasg(O) = 1andg'(O)= 1,soit the graph of g is the tangent this means If we approximate f by a quadratic at 0, we
derivatives
least near the
(0,
point
withfand line off at the
agrees
= Q' (0)
Q(O)
the
approximates
We can agree
improve
withfin
the
= 1
and
Q\" (0)
= y
eX
further
the
linear
two
g, at
closely
than
=
!X2
Figure
that
shows
7.1
the line y = 1 +
of the approximation
x near
using
by
derivatives as well. It is easy to verify
\037
Xk = 1
\037k! k=O)
272)))
= 1.
accuracy
and higher P(x)
x+
= f\" (0)
more
n
(7.1 ))
the
1). The polynomial
curve
third
f and its first function
with
Q which agrees approximation to f than
Q(x) = 1 + has
derivative at O. Geometrically, in Figure 7.1. point (0, 1),as shown its first
polynomial
a better
expect
might
+ x
+ x+
 + ...
x
2
2!
X +

n
n!
the graph the
point
of Q (0, 1).
which polynomials the polynomial) that
The
generated
polynomials
Taylor
273)
a function)
by
y)
2)
y = 1 + x
+
\037X2)
y = eX) x) 2)
o)
y=l+x
FIGURE 7.1
the exponential before we can use
with
agrees
course,
function,
we
than discuss
this
exponential Rather
to the
approximations
Polynomial
need
such
to
polynomials
some information example
particular
eX
near
(0,
1).)
at the point x = O. Of values for the approximate
first n derivatives
its
and
function
curve y =
compute
about the error made in in more detail, we turn
the now
approximation. to the general
theory.)
The
7.2
Suppose to
try
find
conditions
(7.2))
so
we try
polynomials
Taylor
derivatives up
f has
P
a polynomial
to be
generated
which
order
to
x = 0,
n at the point
with f
agrees
function
a
by
and
its
first
11
n >
where
1, and let us
at O. There
derivatives
are
n
+
satisfied,namely)
P(O)
= f(O)
a polynomial
P' (0)
,)
of degreen, P(x)
(7.3))
= f'
0 in then
(7.3)
= Co +
and
substitute
=
p(n>(o)
f(n>(o)
,)
say)
C1X
+
with n + 1 coefficients to be determined.We these coefficients in succession.
we put x = First, both sides of (7.3)and
. . . ,)
(0),)
C 2X
2
shall
+
\302\267 \302\267 \302\267
+
use
cnx
n ,)
the conditions
in
(7.2)
to determine
we find P(O) = Co,so Co = f(O). Next, we differentiate x = 0 oncemore to find P'(O) = Cl ; hence C 1 = f'(O).)))
1
274)
(7.3) again and put k times, we find
differentiate
we
If After
differentiating
to functions)
approximations
Polynomial
that
(7.4))
Ck
0, we
=
x
that
find
= k!
P(k)(O)
ck
p\" (0) = this , and
2c2
= C2 f\" (0)/2. us the formula) gives , so
j(k)(O)
=
k!)
0, 1, 2, . . . , n.
k = 0, we interpret 1(0)(0) to mean This argument f(O).] of which n exists satisfies then its coefficients < (7.2), degree proves polynomial will of P be if and are necessarily given by (7.4). to n \302\245= 0.) (The degree equal only if/(n)(o) P with coefficients that the polynomial it is easy to verify given by (7.4) satisfies Conversely, we have the following theorem.) (7.2), and therefore k =
for
one and
exists
This
one
only
=
P(O)
f(O)
is given
polynomial
a function
1 be
Let
7.1.
THEOREM
there
[When
if a
that
with
= f'
P' (0)
,)
the
by
of order
derivatives
degree
(o)
x =
the point
at
n
satisfies
O.
Then
conditions)
.)
formula)
=
P(x)
j(k)(O)
\037
L
Xk .
k!
k=O)
there same way, we may show that with f and its first n derivatives
the
In which
agrees
may
P
write
of 0, we
in place
 a and
of x
in powers
are led to
the
x = a.
as before.
proceed
one
only
of degree < n polynomial of (7.3), we instead fact,
In
If we evaluate the
derivativesat
a
polynomial) n
\037 P(x) = L
(7.5))
and at a point
is one
j
(k)
(a )
k!
_ a)7c .
(x
k=O)
is the
This
one and
Pea) = it is
and
Taylor
f(a) ,)
referred to as
a
P'ea)
Taylor
=
f,
a new
it produces
function
function at x is denoted by Tn/(x; dependence on a, we write of this
EXAMPLE
for
all k,
1.
When
f is the
so E(k>(O)=
eO
=
which
satisfies
. ,)
p(n)(a)
of the
in honor
polynomial
Tnf, Tn/(x)
a) instead
the or
the conditions) =
f(n>(a)
,)
Brook
mathematician
English
the polynomial
that
say
by f at the point polynomial of degreen generated that indicates It is convenient to have a notation P on f and n. We shall indicate this dependence of degree Tn is called the Taylor operator symbol function
n
..
.f'(a),)
More precisely, we
(16851731).
0 and
all x function
function
Define
T(x)
xy'
equation Prove
1,and
=
dt.
S\037set)
 y = x sin
x
Prove that the the interval
on
the differential this does
that
why
explain
equation has contradict
not
f,
X
l
= 1 +
x
real axis, such that)
l
1)
let) dt
by the equation)
[defined
> 0 has the
on the positive
continuous
function.
this
find
f(x) = for x
1.
on this interval. condition [(0) =
[(x)
for
limit
8.3.
Theorem
12. The
following
a finite
to
tend
+ 3.
differential
the
of the
each
on
2)
that all solutions
Prove
00).
let s(O) =
\302\245:0, and
x
[(x)
 1)(x
= (x
2y
as x
limit
finite
equations)
(i) it is
that
properties
2 xe{1X
)/2
 xe
t 2 e t2 / 2 dt)
x2f2
f:
on
continuous
the positive
real axis, and (ii) it
satisfies
the equation)
f(x) for all
x >
O. Find
The Bernoulli
with these
functions
all
 x f(t) J:
= 1
two
properties.
of the form
differential
A
equation. equation This equation 1, is called a Bernoulli equation. it can always be transformed The next exercise shows that  n. new unknown function v, where v = yk, k = 1 not
0 or
13.
b
v' +
is any
real
kP(x)v =
y = [(x),
number,
kQ(x) on is never
which
only
if
the
into
y' + P(x)y
where n is Q(x)yn, of the presence of yn.
=
because
a linear
firstorder
kth power
=
Q(x)yn)
of [is equal to g
on on
I,)
with
[(aYc
=
for
equation
and on an interval I. Q are continuous solution of the inititalvalue g(x) be the unique that 1 and k = 1  n, prove I, with g(a) = b. If n \302\245:of the initialvalue zero on I, is a solution problem)
v =
let
y' + P(x)y if and
is nonlinear
Assume P
a nonzero constant.
k be
Let
if
dt)
If a E
a
I and
problem
a function
b)
I.
In each of Exercises 14through interval. 17, solve the initialvalue problem on the specified 14. y'  4y = 2e Xy l/2 on (  00, + 00), with Y = 2 when x = O. 15.y'  Y = _y2(X2 + x + 1) on ( 00, + 00), with Y = 1 when x = O. 16. xy'  2y = 4x 3y l/2 on ( 00, + 00), with Y = 0 when x = 1. 17.xy' + Y = y 2x 2 log x on (0, + 00),with Y = t when x = 1. 18.2xyy' + (1 + X)y2 = eX on (0, + 00),with (a) y = yI; when x = 1; (b) Y = yI; when x = 1; (c) a finite limit as x + O. of the form y' + P(x)y 19. An equation + Q(X)y2 = R(x) is called a Riccati (There equation. method for solving the general Riccati equation.) Prove that if u is a known is no known = u + 1 Iv, where v there then are further solutions of the form solution of this satisfies
equation, linear
a firstorder
y
equation.)))
Some
((
+
00,
+
Some
8.6
equation
physical
problems leading to
section we will
In this
313)
equations
Exercise
use
 00,
linear differential
= 2 has two constant Start with each of these solutions. y' + Y + y2 19 to find further solutions as follows: (a) If  2 < b < 1, find a solution on  2,find a solution on the interval 00) for which y = b when x = O. (b) If b > 1 or b < = O.) = x b when 00) for which y
20. The Riccati and
leading to firstorder
problems
physical
matically as
firstorder linear
differential
various
discuss
physical problems that In each case, the differential and is called physical problem
equations
can
eq uations.
differential
idealized simplification of
the
a
eq uation
mathean represents
mathematical
model
be formulated
of
occurs The differential as a translation of some physical the problem. such law, equation a \"conservation\" as Newton's second law of motion, law, etc. Our purpose here is not to to deduce the choice of the mathematical model but rather logical consequences justify to reality, and its justification from it. Each model is only an approximation properly
to the science
belongs evidence
agrees
If not,
one.
useful
we
which
from
to
try
the
find
mathematically,
a more
If
emanates.
problem
deduced
results
the
with
or
intuition
we
then
feel
that
experimental
the model is
a
suitable model.)
various radioactive elements show marked decay. Although rates of decay, they all seem to share a common propertythe rate at is proportional to the amount present which a given substance decomposesat any instant If we denote by y = f(t) the amount present at time t, the derivative y' = instant. that at of decay\" the rate of change of y at time states that) t, and the \"law J'(t) represents 1.
EXAMPLE
differences
Radioactive
in their
y' = k is
where the
a positive
particular
as t
increases,
mathematical y = J(t) of this
constant (called the
ky
decay
,)
constant)
whose
actual value
depends on
comes in because y decreases element that is decomposing. The minus sign = ky is the and hence y' is always negative. The differential equation y' radioactive model used for problems concerning decay. Every solution differential
equation
has the J(t)
(8.13))
form)
= J(O)e
kt
.)
the amount present at time f, we need to know the initial amount k. constant the value of decay f(O) can be deduced from (8.13),without It is interesting to see what information the knowing there is no finite time t at which J(t) will exact value of f(O) or of k. First we observe that kt vanishes. it is not useful to study be zero because the exponential e never Therefore, it is possible to determine the the \"total lifetime\" of a radioactive substance. However, of a to The fraction time for \037is usually sample decay. any particular fraction required T at which f( T)I.f(O)= \037is called the halflife of the chosen for convenience and the time k l' = substance. This can be determined T. Taking \037for by solving the equation e= = T 2 or This kT relates the we halflife 2)/k. (log equation log get logarithms, to the decay constant. Sincewe have) Therefore,
to determine
and the
J(t
+ T)
J(t))
kU
+ 1')
J(O)ekt) J(O)e
=e
kT
1) 2')))
to differential
Introduction
314)
equations)
y)
j(O))
y =
j(O)e
1{(0))
kt)
, I I I I I I I
\037
tj(O))
I I I
\037j(O))
    T
I  r
 
I
 
 
:
I I)
I I I
T)
2T)
3T)
I
t) o)
8.1
FIGURE
see that
we the
general
2.
resistance) motion.
There
an
forces
acting
due to
gravity,
is
= f(t) denote The
are
upward
a
in
in the
height
which
velocity.
body
Palling
rest from a great and that the only the acceleration
its
decay
the halflife is the same for every sample of of a radioactive decay curve.) shape
EXAMPLE
Let s
Radioactive
proportional
assumption
Figure 8.1 illustrates
material.
a given
of mass
A body
m
is
from
dropped
that
has fallen at
the body it
rest means
from
falls
let
t and
time
that
v
=
f'(O)
= s'
= f' (t)
denote
O.
on the body, a downward force acting mg (due to its weight) and constant. Newton's (due to air resistance), wherek is somepositive the net sum of the forces acting on the body that at any instant is equal m and its acceleration. If we denote its mass the acceleration at time t = s\" and Newton's law us the equation) gives
forces
two
kv
force
law states to the product of = v' by a, then a second
ma =
This can be as a firstorder can be written
T.)
earth's atmosphere. Assume that it falls in a straight line on it are the earth's gravitational attraction (mg, where g is to be constant) and a resisting force assumed to air (due to its velocity. It is required to discuss the resulting
distance
the
medium.
resisting
halflife
with
considered equation in
the
as a for
secondorder

mg
differential
the velocity v. As
kv . the displacement for v, it is linear
for
equation
a firstorder equation
s or and
form) v
,
k +v=g. m)
This
equation
is the
mathematical model
of
the
problem.
Since
v
=
0 when
t
=
0,
the)))
Some
solution of the
unique
differential
(8.14))
e
kt
/m
v
\037
acceleration
du =
a =
instant
is
means
that
the air
m
+
t
k
s =
Since
0
t =
when
0, we
mg
S =
t +
If
the
initial
replaced
velocity is
/m
)
.)
Equation (8.14),we find that the a + 0 as t \037 + 00. Interpreted the force of gravity. for the displacement
out
to note as t increases
is interesting
velocity,
should convince EXAMPLE to
portional
for
that
c .
the equation
and
(e
/m
kt
 1) .
(8.14) for
the
velocity
on
3. A cooling the difference
\037
(1
at time
must
t
a number independent of that this seems reasonable.)
is mg/k,
grounds,
physical
problem. The between
(positive, negative,or zero),the
initial velocity
bound,
its
which
at
rate
and
be
laB'
temperature
of cooling.)
y' =
(8.15))

k[y
or)
M(t)])
a positive constant. This firstorder The unique cooling problems. = the b is formula) f(a) given by
k is use
condition
for
J(t) =
(8.16)) Consider
now
while immersed
a specific in
a medium
be kt
problem whose
in
+
which
e
linear
y'
J:
kM(u)e
temperature
limiting
The
reader
+ ky
the
kU
of the
temperature
medium,
surrounding
= kM(t)
,)
is the mathematical the equation satisfying
du
model initial
.)
cools from 200\302\260to is kept constant, say
a body
.
equation
solution of
kl
Vo
a body changes temperature is prothat of the surrounding medium.
If y = f(t) is the (unknown) if at time t and denotes the M(t) (known) temperature of the body Newton's law leads to the differential equation)
we
of motion becomes)
kt/m  ekt/m ) + voe.)
mg
=
every
without
himself,
(This is called Ne\037rton's
where
s, and
by)
V
It
/k
0, formula
t =
when
Vo
kt
2
2)
k
2
2 gm
gm
k
+
2)

e
equation
kt/m
C =
_
balance
g m2 e k
that
find
that
Note
.
a differential
g
s=
(1
resistance tends to
s', Equation (8.14) is itself be integrated directly to give)
it may
ktjm
ge
v =
Since
\037g
If we differentiate
+ 00.
\037
at every this
physically,
t
as
mg/k
315)
equations
by the formula)
is given
equation
itgeku/m
that
Note
=
v
linear differential
leading to firstorder
problems
physical
100\302\260 in
M(t)
=
40 minutes 10\302\260.If
we)))
to differential
Introduction
316)
minutes and f(t)
t in
measure
l(t) =
(8.17))
We
can 90
from the
k compute 190e 40\\
=
+
lOke
+
10(1
information
=
40k
so
kt
200e
J(O) =
we have
degrees,
200ekl
=
find
in
kl
equations)
eku
J;
 ekt )
=
10 + =
log (90/190),
=
190ekt
100.
lo(log 19
Putting
the time required for this same material compute 100\302\260 if the temperature of the medium is kept at 5\302\260.Then Equation same constant k but with M(u) = 5. Instead of (8.17),we get the
= 5+
J(t) the time
find
To
log (19/39),
which
for
t
!
t =
9 =
log
kt 95 = 195e,
2.1972 so,
 log
19) =
log
40
39 19
log
table of
a fourplace
39
(log
k
From
=
40 in (8.17), we
to cool from (8.16) is valid
200\302\260to with
the
formula)
so
kt
= log
(95/195)=
hence)
and
and
t
195ekt .)
= 100, we get
J(t)
.)
 log 9).
us
let
Next,
gives us)
(8.16)
Equation
du
thatJ(40) k
200 and
natural
we find
logarithms, sliderule
with
we
accuracy,
 log 19 .  log 9)
log 39 get t
= 3.6636,log 19= =
40(0.719)/(0.747)
2.9444,
= 38.5
minutes.
The differential
the rate of cooling decreases considerably that of the medium. to To temperature approach the temperature body begins 100\302\260 to 10\302\260 with illustrate, let us find the time required to cool the same substance from the medium leads to log (5/95) = kt, or) kept at 5\302\260.The calculation
as
of the
t =
! log 19 =
that
change
200\302\260 to
19
log
the temperature
from
drop
19
Jog
40
k
Note
tells us
in (8.15)
equation
the
 log 9
100\302\260 to
from
= 40(2.944)
= 158minutes.
0.747)
more
10\302\260 takes
than
four times
as long as the
100\302\260.)
100 gallons of brine whose concentration A dilution problem. A tank contains of Brine 2 of salt per gallon runs salt into the pounds per gallon. containing pounds tank at a rate of 5 gallons minute and the mixture uniform out runs per (kept by stirring) at the same rate. Find the amount of salt in the tank at every instant. Let y = J(t) denote the of pounds of salt in the tank at time t minutes number after which cause y to change, the incoming brine which mixing begins. There are two factors salt in at a rate of 10 pounds removes salt brings per minute and the outgoing mixture which at a rate of 5(y/l00)pounds minute. (The fraction y/l00 represents the concentration per EXAMPLE
4.
is 2.5
at time
t.) Hence the
y'
This linear
eq uation
differential
equation is
the
=
10

loY)
mathematical
is)
or)
model
y'
+
loY
for our
=
10.)
problem.
Since y = 250 when)))
Son1e
0, the unique solution is given
t =
(8.18))
y
=
t / 20
e
+
Hence,
the
This enablesus to 200
that
need
we
all
. a given
be
will
amount y, provided
a resistor, voltage
with
to know about the
denoted by
and the current,
of the
be solved
an electric circuit which Figure 8.2(a), page 318,shows and an inductor connected in series. The electrowhich in the circuit. causes an electric current to flow he should not be concerned.For our electric circuits,
circuits.
Electric
5.
has an electromotiveforce, motive force produces a If the reader is not familiar purposes,
also
content
salt
the
which
as t
200))

(y
.)
increases without bound. have been guessed from for t in terms of y to yield)
200
could
\037
50
20 log
50e f / 20
200 +
< 250.)
< y
EXAMPLE
time at
the
find
(8.18) can
Equation
problem.)
t =
let),
circuit is
denoted a differential by
the
that
of time
functions
are
voltage,
related
t
by
V(t),
equation
form)
R are assumed to be positive Land Here inductance and resistance of the circuit. as of a conservation law known ulation
Those readers unfamiliar
analogous
water
to
friction
in the which
influence
with
constants.
are called,
They
equation is a
The differential
voltage
Kirchhoff's
laH',
pipe, tends
to a which
pipe. The
pump
respectively, the form
mathematical
it serves
and
to oppose
water
the
the flow;
sudden changes in
the
of
think
force
electromotive
causes
which
tends
to oppose
may find it helpful to
circuits
in a
flowing
is analogous
generator)
V(t).)
as a
mathe
the circuit.
for
nl0del
matical
=
+ RI(t)
LI'(t)
(8.19))
the
=
du
equation
the statement of
to
20
lOe\"/
J:
shows that y > 200 for all t and that y the minimum salt content is 200 pounds. (This
This
317)
equations
formula)
the
by
t / 20
250e
linear differential
leading to firstorder
problems
physical
the
(usually
current
as being
a battery
or a
to flow; the resistor is analogous and the inductance is a stabilizing current due to sudden changes in
voltage. circuits is this: If a given current l(t)? Since we are solution is a routine matter. If
such usual type of question concerning the what is the on circuit, resulting impressed
The
firstorder linear differential initial current at time t
the
equation,
=
0, the
equation has the
An
special
important
for all t. In
this
case,
= J(O)e Rt
case occurs the
integration
J(t) =
/L
+
the
impressed
is easy to
+ \037
x) Vi
eRtlLi
when
perform
e Rx / L dx
voltage and
(J(O)
V(t)
dealing
with
1(0)
denotes
is
a the
solution)
t
J(t)
voltage
\037)eRt/L.)))
we
.)
is constant, say Vet) = E are led to the formula)
to
Introduction
318)
equations)
differential
when
Current
1(0))
E
I (0) >
Ii)
/
when 1(0)
Current
E Inductor)
= E R
\037)
R)
Electromotive force)
when 1(0)
Current
1(0))
E
E/ R, the coefficientof the exponential \037 R decreases to the value as t 00. If and the current is positive + 1(0) < E/R, E/ limiting value the steadystate the current increases to the limiting E/ R. The constant E/R is called current. Examand the exponential term [/(0)  E/ R]e Rt / L is called the transient current, that the nature and the quotient 1(0)
shows
This
current
ples are illustrated
in
8.2(b).)
Figure
examples illustrate the unifying They show how several different same type of differential equation.
The foregoing equations. the
exactly
The
differential
of attacking
a physical
a wide
equation variety
problem
leads
in
differential
a is a positive constant electric circuit with inductance
and
ay
analog
computer.)))
=
Q
practical
physical
because it means.
of the
equation
utility
problems
of
differential
may lead
to
the possibility For example, suppose
suggests
form)
,)
known function. We can try to construct an resistance R in the ratio R/ L = a and then try to the We would then have an electric circuit with exactly we to can numerical Thus, problem. hope get physical Q
is a
L and
impress a voltageLQ on the circuit. same mathematical model as the data about the solution of the physical problem the electric circuit. This idea has been used in the
types
special interest problems by electrical
y' + where
of
is of
(8.19)
of physical
to a
and
power
by making practice
and
in measurements of current of development
has led to the
319)
Exercises)
8.7 Exercises) In
use
exercises, following of the problem.
the
model
1. The halflife of
1600years. Find
what
of a given
percentage
quantity
100
in
disintegrates
equation as a mathematical
differential
firstorder
appropriate
is approximately
radium
for
radium
an
years. at a rate
of bacteria grows and if the population proportional to the amount present doubles in one hour, by how much will it increase at the end of two hours? of a substance 3. Denote by y = f'(t) the amount at a present at time t. Assume it disintegrates T for which If n is a positive integer, the number rate proportional to the amount present. = f'(O)/ n is called the 1/nth life of the substance. f(T) (a) Prove that the 1/ nth life is the same for every sample of a given material, and compute T in terms of n and the decay k. constant
2. If a strain
(b) If a and b are given,
can be expressedin
that f
prove
f'(t) = and nlean
determine of
4. A man chute
Refer
present
amount
the
that
t =
instants
two
at
t
is a
to
use
2 of
Example
open, assume the
while the parachute is
Thereafter,
pounds.
of gravity is 32 the approximation
ft/sec2
Use the
Section 8.6.
=
show that the
thus
differential
dv
ds dv
dv
dt
dt ds
ds)
where
with
c =
m/k
bv)
cv)
c =
and
vet)
write)
to express s in terms this equation gln/k. Integrate for v and s derived in the example. 8.6 by assuming 6. Modify Example 2 of Section the air resistance is proportional be in each that the differential can of the following forms:) equation put where b
pounds.
speed
example can be expressedas follows:)
in the
equation
dv)
=
to
rule
chain
is 12v(t)
resistance
explicit formulas for the 37/128 in your calculations.) find
and
e 5 / 4
ds
result
geometric
weighted
t = b.
and
==v
and
at time
present
a
a parachute The combined weight of man and parajumps from a great height. 192 pounds. Let vet) denote his speed (in feet per second)at time t seconds after the first 10 seconds, beforethe parachute During opens, assume the air resistance is
Assume the acceleration at time t. (You may
5.
This shows
f(a)w(t1(b)1W(t))
wearing
is
falling. ;fu(t)
wet). the amounts
form)
the
of v.
Check your
the formulas
vi mg/
k .
k)
m.
In
dv
k
Integrate
mg v2 = 
where b = V kg/
ds 
c
each of
(1 
Determine
v 2
e

the
v

) ;
limiting
v =
value
e bt _
c

2
the
obtain
e
bt
v 2 . Show
1
c
k
dv
these and
2ks / m
 m
dt 2 ')
to
v2
')
formulas
following
e
+e)
of v as t
bt
bt
\037
= c
tanh
+ 00.)))
bt
,
for
v:)
7.
A body in a
from
60\302\260 cools
120\302\260 in
200\302\260 to
after its temperature the time t required to
is 60
half an hour. + 140ekt , where
Show
that
Show
that
(c)
at which the temperature is 90\302\260. of the body at Find a formula for the temperature \302\260 constant but falls at a rate of 1 each ten minutes.
(d) kept when
[log 140the
Find
log
 60)]/k,
(T
where
minutes
t
reach a temperature 60 < T < 200.
k =
is
T degrees
of
(log 7 by
given

log 3)/30.
the formula
time
A thermometer outdoors
room temperature is
t if the
time
Assume
temperature is 200\302\260. has been storedin a room whose temperature is five minutes, it reads another it reads 65\302\260.After
the body
taken
9.
room at
equations)
(a) (b)
t =
8.
to differential
Introduction
320)
room
the
Five 75\302\260.
temperature minutes
is
not 60\302\260
after being outdoor
the
60\302\260.
Compute
temperature.
salt. Water runs into the of dissolved 50 pounds of brine containing a tank are 100 gallons is concentration the and tank at the rate of 3 gallons minute, by stirring. kept uniform per How much salt is in the tank at the end of one hour if the mixture runs out at a rate of 2 gallons In
minute?
per
of salt and ina mixture of the tank is covered with 10. Refer to Exercise9. Supposethe bottom to the difference between Assume that the salt dissolves at a rate proportional soluble material. of salt per gallon), of the solution and that of a saturated solution (3 pounds the concentration How much salt dissolve and that if the water were fresh 1 pound of salt would per minute. will be in solution at the end of one hour? the electromotive 5 of Section 8.6. Assume 11. Consideran electric circuit like that in Example = E sin wt, where E which a current force is an alternating Vet) voltage produces generator the current If 1(0) = 0, prove that constants and ware positive (w is the Greek letter on1ega). has the form) let) =
where
\037
depends
12.Refer
to
Example
=
Make
E (1 R)
sin
V
+
< b,
a
< t
by
the following
if
eR 0;
a
a < t
nature
(wt
2
L2
e Rt/L')
L =
O.
voltage is a step function t. If 1(0) = all other
defined
0 prove
as that
t < a;)
E eRt R)
2
/L
(e
Rb
/L
 eRa
/
L)
if
t > b
.
graph of 1.)
or bacanimal, (whether human, present at time t is necessarily a step on only integer values. Thereforethe true rate of growth dx/dt is zero (when t lies taking function x jumps in an dxjdt does not exist (when open interval where x is constant), or else the derivative from one integer to another). can often be obtained if we assume Nevertheless, useful information x is a continuous function of t with a continuous derivative that the population at each dxjdt for the population, instant. We then postulate various \"laws of growth\" depending on the factors in the environment which or hinder growth. may stimulate For example, if environment to assume that the rate has little or no effect, it seems reasonable of growth is proportional to the amount present. The simplest kind of growth law takes the form) Population
terial),
the
growth.
function
which
of a population
x of
dx (8.20))
individuals
 =kx '))) dt
321)
Exercises)
where k is a constant that on the depends which cause the factor k to change with
of population. Conditions may develop the growth law (8.20) can be generalized
kind
particular
and
time,
as
follows:)
dx (8.21))
for some
If, the
= k(t)x \302\267)
dt
reason,
food
supply to proportional
may
the population be exhausted),
both
x and
cannot
we
 x.
M
may
Thus
exceed a certain maximum M (for example,because is jointly that the rate of growth suppose reasonably we have a second type of growth law:)
dx
 = kx
(8.22))
as in
where,
(8.21), k
improvements
(8.22)even
further
x as a
13. Express both
by
) ')
or, more generally, or decrease the value allowing M to change with time.)
k may change with time. Technological of M slowly, and hence we can generalize
to increase
of t for each that the result
function
Show
constant).
X
be constant
may
tend
may

(M
dt
of the for
laws\"
\"growth
can be
(8.22)
in (8.20)
and
(8.22)
(with
k and
M
expressed as follows:)
M)
x=
(8.23))
1
rf..
where
14. Assume at
t
this
law
growth
in
16. The
at
x =
which
X 3 (X 2 X2

Xl) 2
X2

XI(X XI
States Use
Equation
1850,
and
(b)
3

x 2)
a census
suppose being
Xl'
X2 , x 3
is .
taken Show
.
X 3)
that
generalizes
constant.
necessarily Census
9.6,12.9,
92, 108,122,135,150. (a)
M/2.
law (8.22) when k is (8.23) of Exercise 13 for the growth in terms of the time to for which X = M /2. Express the result Bureau (in millions) for the United population figures reported the following at tenyear intervals from 1790 to 1950: 3.9, 5.3,7.2, 17, 23, 31, 39, 50, 63, 76,
a formula
Derive not
the time
formula
M =
(8.24))
15.
1 is
(8.23) of Exercise 13, and equally spaced times t I , t 2 , t 3 , the resulting numbers suffices to determine M and that, in fact, we have)
the
three
that
and
constant
is a
,)
ea(tfl)
+
(8.24) to
determine a value
of
M on
the basis of the
census
figures
for 1790,
1910.
Same as
(a) for the
(c) On the
basis
the
law (8.23)
years
1910,
1930, 1950.
to accept or reject in (a) and (b), would you be inclined of the United States? for the population 17. (a) of t, where x denotes the population figures quoted graph of log x as a function in Exercise law (8.20) was very nearly satisfied from 16. Use this graph to show that the growth 1790to 1910. Determine a reasonable average value of k for this period. from 1920 to 1950,assume that of k for the period value (b) Determine a reasonableaverage for the the United States population the growth law (8.20) will hold for this k, and predict years 2000 and 2050. of bacteria at a rate jointly 18. The presence of toxins in a certain medium destroys a strain proIf there were no))) the to amount of toxin. and to the of bacteria number portional present growth Plot a
of your
calculations
to differential
Introduction
322)
x)
equations)
x)
x)
t)
t)
(a))
t)
(b))
x)
(c))
x)
x)
t)
t)
(d))
t)
(e)) FIGURE
8.3
(f))
Exercise
18.)
to the amount grow at a rate proportional present. Let x bacteria present at time t. Assume that the amount of toxin is and that the production of toxin begins at time t = O. Set up a increasing at differential One of the curves shown in Figure equation for x. Solvethe differential equation. 8.3 best representsthe general behavior of x as a function of t. State your choice and explain
toxins present,
denotethe
your
bacteria
the
reasonIng.)
Linear equations
8.8
would
of living a constant rate
number
A differential
of second order with
equation of the
form)
y\" +
is said to
be a linear
unknown
function
coefficients
constant
P 2 (x)y
Pl(X)Y' +
=
R(x))
P 2 which of second order. The functions PI and multiply the of the equation. derivative y' are called the coefficients For firstorder linear we proved an existenceuniqueness theorem and deterequations, mined all solutions by an explicit formula. Although there is a correspondingexistencetheorem for the secondorder linear equation, there is no explicit uniqueness general all in formula which some A of cases. the general solutions, gives except special study linear equation of secondorder is undertaken in Volume II. Here we treat the case only P 2 are constants. When the righthand in which the coefficients PI and member R(x) is the is said to be identically zero, equation homogeneous. The homogeneous linear equation with constant coefficients was the first differential a A was first of to be solved. solution general type completely published by Euler equation in a in 1743. Apart from its historical interest, this arises of great variety equation applied so its study is of practical formulas Moreover, we can give explicit problems, importance. for all the solutions. equation
Y and its
Consider a homogeneous linear
equation
with
constant
follows:) y\"
+
ay' +
by
=
0
.)))
coefficients
which
we
write
as
of solutions
Existence
y
=
O.
and we begin
our study
= y\" + by
O.
The
1.
EXAMPLE
easily determine Then
of the equation
equation
=
y\"
y' is constant,
its derivative
say
=
0, and the
k >
treat
b \037 0 and
+
y\"
solutions
ekX ,
The
3.
EXAMPLE
>
the
0, and
in
equation
we
obtain
is y = sin kx.
= c 1ekX
all
C 1 and
solutions.)))
(
we
relation,
+
00,
(0).
find that y
differential
+
these
C2
are
arbitrary
C2 e

b >
0, we
further solutions
y = CI where
zero, = 0 on
formula.)
this
y\"
k 2),
=
another
and
y\"
Again
and we can
are
y\"
the form)
takes
constants.
arbitrary
are included
< O. Sinceb
linear combinations
by constructing
and C2 are
all solutions
the cases b
separately
equation
y
k
b
and
C 2 ,)
= 0, u'here b
by
differential
is y =
solution
obvious
CI
+
CIX
y\"
where
a
constants. Conversely,for any choice of constants C1 and C2 , the linear = 0, so we have found all solutions in this case.) + C 2 satisfies CIX J'\"
The equation
2.
EXAMPLE
further
0
form)
Next we assume that
One
solutions, can be found
solutions
and C2 are
polynomial y =
where
nontrivial
finding
satisfying y is any function = C1 . Integrating this y'
y'
CI
by =
+
y\"
both coefficients
O. Here
Assume
solutions.
all
necessarily has the
where
in
nontrivial
which
323)
0)
these cases, the coefficientof y' is zero, and the equation has the form is tantamount find that solving this to solving the special equation
solutions
of
Existence
=
b.y
solution is the constant function
+ (0). One are interested
00,
cases for
special
y\" +
equation
shall
We
general case.)
8.9
the
We
solution.
some
with
I n all
inspection.
by
real axis ( 
solutions on the entire This is called the trivial
We seek
of
forming
by
sin
Theorem 8.6
kx
will
is
y'
=
linear
cos kx, and another combinations,)
,)
show
that
this formula
includes
to differential
Introduction
324)
The problem of be reduced to
y,
u,
solving
equation
more general equations. The = uv. Differentiation gives express the combination y\" +
we
Now
u\"v.
by =
+
=
ay' +
for doing
consider three functions
uv'
and y\" = uv\" + of u and v. We
u'v,
+
terms
in
by
can
coefficients
method
is a
is to
idea
us y'
0
constant
with
cases just discussed. There
the special
applies to v such that y
and
2u'v' +
linear
secondorder
a
solving
of
that
also
that
this
general equation to the special case y\"
of the
Reduction
8.10
equations)
have)
y\" +
(8.25))
ay' +
=
by
= Next
we choose
may
choose
u
in
e ax
/ 2.
\" v
(8.25)
Equation
av
+
'
reduces
by
of u' zero. have
the
0 if and
=
function if
only
+b
the
This
vu\".)
2 /2 = a v/4,
av/2, so we of
v' =
that
requires
av'
2 2 av  aV + bv
v =
a
4b
=
424)
ay' +
=
by
(
is never
v
u satisfies
u\" +
the coefficient
and
2
v .
!( 4b
+
 a2
4b
v.)
u
4
)
so y satisfies the
zero, u\"
Let y and u be two functions (0), y satisfies the differential equation
 a2 )u
such
THEOREM 8.4. +
00,
=
v\"
+
av)u'
buv
=
O.
differential
y\" +
equation
we have
Thus,
proved the
theorem.)
following
(
e aX / 2 ,
v =
ay' +
+ (2v' +
+
to)
y\" +
Since
we
v
+ u'v)
a(uv'
becomes)
(8.25)
Thus,
this
For
+
u\"v
av' + bv)u
+
(v\"
the coefficient
make
to
v
v =
2u'v' +
uv\" +
that
=
y
+
y\" + ay'
ue=
by
ax
/ 2.
0 if
Then,
and
only
on the interval if u
satisfies
equation)
differential
u
\"
4b +
 a2
u =
0
.
4)
reduces the study of the equation + ay' + by = 0 to the special y\" = O. We have exhibited nontrivial solutions of this equation but, for y\" + by except case b = 0, we have not yet shown that we have found all solutions.) theorem
This
8.11
The problem of with
the
00,
+
(0).
the equation
Assume Assume
tH'O
also
uniqueness
functions that
f(O) Then
f(x)
= g(x)
for all x.)))
solutions
all
determining
help of the following
THEOREM 8.5.
on ( 
for
theorem
Uniqueness
f
y\"
+
by =
case the
0
of the equation
y\"
+
by
= 0
can be
solved
theorem.)
f and
g satisfy the and g satisfy the initial = g(O) ,)
f'(O)
=
differential
conditions)
g'(O)
.)
equation
y\" +
by
=
0
Uniqueness
= f(x)
Let hex)
Proof
this by expressing
do
First we
conditions the initial has derivatives equation
We
wish
g(x).
of the
differential
by
of odd order that all derivatives
those
0 has all
h at
are y 0, all If B An
=fi.
is shown
example
to 
00
as
solution
x
+
as x +
to 0
tend
solutions
0, each
will
sign
change
in Figure
sizes of
this
=
d=
(2C)2 2
c

and
k .
eCX(A +
> 0,
form)
Bx).)
This case is referred to as exactly once because of the linear
+ 00.
< 0, eachnontrivial
8. 6(a). If c
of
=
have the
all solutions
case,
2 2 4(c  k ), the nature The three cases d = 0, d
4k 2 2
solution
critical factor tends
damping. A + Ex. to + 00
or
+ 00.
(b) Positive discriminant: y =
c2
>
k 2. By hX
eCX(Ae
+
Be
Theorem 8.7 all hX
)
=
Ae(hc)X +
have
solutions
the form)
Be(h+c)x,)
2 2 2 2 2 2 2  c)(h+ c) < O. !Vd = V c  k . Since h = c  k , we have h  c < 0 so (h h + c is the numbers h c and h + c have opposite signs. If c > 0, then Therefore,  c is e(hc)X and e(h+c)x tend to zero negative, and hence both exponentials positive so h tend to 0 for all solutions as x + + 00. In this case, referred to as overcritical damping, can An is in Each solution shown x. 8.6(a). sign at most change Figure example large
where
h =
once. If c < 0, then
h
 c is
positive but
h +
c is
negative. Thus,
both
exponentials
e(hc)X)))
to differential
Introduction
336)
tend
e(h+c)X
and
to + 00 for
absolute values.
c2
discriminant:
Negative
(c)
large x, so again k 2.
0, the ordinate a solid of volume x:!(x) when out rotated about the xaxis, find the function f = O. 16. A nonnegative differentiable function [is defined on the closed interval [0, 1] with [( 1) For each a, 0 < a < 1,the line x = a cuts the ordinate set of [into two regions having areas A and B, respectively, A being the area of the If A  B = 2[(a) + 3a + b, leftmost region. where b is a constant of find the the function constant b. a, independent [and = (0, 1) and PI = (1,0). For 17. The graph of a function the two Po points [passes through every P = (x, y) on the graph, the curve lies above the chord of the PoP, and the area A(x) point 3 the function f region between the curve and the chord PPo is equal to x . Determine 18. A tank with vertical sides has a square crosssectionof area 4 square feet. Water is leaving the 2 feet above inches. If the water level is initially tank through an orifice of area 5/3 square the orifice, find the time required for the level to drop 1 foot. 19. Referto the preceding problem. If water also flows into the tank at the rate of 100 cubic inches per second,show that the water level approaches the value (25/24)2 feet above the orifice, of the initial water level. regardless 20. A tank has the shape of a right circular cone with its vertex up. Find the time required to the tank in its a from an orifice base. result in terms of the empty liquid through Express your dimensions of the cone and the area Ao of the orifice.))) A
in
point such
Q moves a way that
to differential
Introduction
356)
equations)
mx  x)y = 0 possesses a solution m of the form y = e , where y' + (1 equation xy\" is constant. Determine this solution explicitly. a suitable change of variable 22. Solve the differential (x + y3) + 6xy2y' = 0 by making equation which converts it into a linear equation. 2 2x a change of variable of 23. Solve the differential (1 + y e )y' + y = 0 by introducing equation mx = ue , where m is constant and u is a new unknown function. the form y satisfies the relations) 24. (a) Given a function I which
21.
The
=
let y
I(x) and
that
show
=fG)) a
y satisfies
>
(b)
a and b. a and b are constants. Determine Find a solution of the form I(x) = Cx n .
(a)
Let u be
where
a nonzero solution
the substitution y
that
Show
=
linear
a firstorder
(b) Obtain a nonzero and use the method of
for
v'.
of
the
(a) to find
y\" that
such
y
26. Scientists at
= 0 and y'
one
(a) Set up (b) Evaluate
27. In
the
when
x =
! gram remained.
solve
and the
preceding
the differential constant in
decay
problem, the
same.
equation
a
suppose
y\"

R(x))
2 4y' + x (y'

4y) =
0
by
inspection
of)
solution

4y) =
2xex3f3)
o.
isolated one gram of a new radioactive element called a rate proportional to the square of the amount present.
equation for
mass
the
of gm 1 yr
units
word
the
Show that
=
+ Q(x)y
2 4y' + x (y'
Ajax Atomics Works It was found to decay at
year,
data remaining a finite time,
=4

the
Deteriorum. After
the equation)
solution
,)
= 0 .)
converts
+ P(x)y'
,)
equation)
+ Q(x)y
equation
part
0
+ P(x)y'
uv
y\"
into
the secondorder
of
y\"
=
by
2
of the form)
equation
+
axy'
= I( 1)
0,)
differential
+
x2y\"
25.
x
if
2f'(x)
in
this
remaining at time t.
of Deteriorum
1.
square were replaced by case the substance would
square decay
the other within entirely
root,
find this time. 28. At the beginning of the Gold Rush, the population of Coyote Gulch, Arizona was 365. From then on, the population would have grown by a factor of e each year, exceptfor the high rate of \"accidental\" to one victim death, per day among every 100 citizens. By solving amounting an appropriate differential equation determine, as functions of time, (a) the actual population of from the the Rush Gulch t Gold and the cumulative of number (b) Coyote years day began, and
fatalities.
29.
With all
what forces
speed except
should a the earth's
rocket be fired gravitational
upward attraction.))))
so that
it
never
returns
to earth?
(Neglect
30. Lety
= [(x) be
that
of the
solution
357)
review exercises)
Miscellaneous
differential
equation)
+ x
2y2 I y = 3 y2
+
5)
= O. (Do not condition to solve this differential satisfies the initial [(0) attempt equation.) shows a relative maximum The differential whether (a) that.f/(O) = O. Discuss [has equation or minimum or neither at O. (b) Notice that .f'(x) > 0 for each x > 0 and that [' (x) > j for each x > l3Q . Exhibit two positive numbers a and b such that [(x) > ax  b for eachx > 3Q. \037 \037 0 as x Give full details of your +00. (c) Show that X/y2 reasoning. limit as x + + 00 and determine this limit. (d) Show that y/ x tends to a finite 31. Given a function [which satisfies the differential equation) which
1
Xf'/(X) + for
all real
(a)
If [has
(b) If [has (c) If [(0)
x. (Do not attempt an extremum at a an
extremum
= .f/(O)
=
c
this
\037 0,
differential
= 1
e
X)
equation.)
that this extremum is a minimum. or a minimum? Justify conclusion. your A such that [(x) < AX2 for all x >
show
is it a maximum the smallest constant
at 0, 0, find
to solve point
3x[f/(x)]2
O.)))
9)
NUMBERS)
COMPLEX
9.1
Historical
The
introduction
quadratic
x2
equation
+
0 has no
1 =
solution in
the
realnumber
system
because
New types of numbers,called complex numbers, square In this have been introduced to provide solutions to such brief equations. chapter we are in and show that discuss numbers they important solving algebraic equations complex and and calculus. that integral they have an impact on differential As early as the 16th century, a symbol of the solutions vi  1 was introduced to provide 2 = O. This symbol, later denoted by the letter i, was regarded quadratic equation x + 1 which could be manipulated or number as a fictitious like an imaginary algebraically that its square was 1. the quadratic Thus, for example, ordinary real number, except 2 x 2 + 1 = x 2  i 2 = (x  i)(x+ i), and the polynomial x + 1 was factored by writing 2 = of the equation x + 1 0 were exhibited as x = :1:;,without solutions concern any of such the meaning or validity formulas. Expressions such as 2 + 3i were regarding in were used a formal way for nearly 300 years called and numbers, they purely complex before they were described in a manner that would be considered satisfactory by presentday there is no real number
is
whose
1.
standards.
Early in the 19th century, Karl Friedrich Gauss Hamilton (18051865) independently and almost simultaneously as ordered pairs (a, b) of real numbers numbers complex defining
(17771855)
special properties. 9.2
Definitions
This
and
is widely
idea
field
accepted today and is
and
the
proposed
endowed
described in
Rowan
William
the
with next
the pair (a, b) is called a conzp/ex If a and b are real numbers, and that of pairs is definedasfollows: multiplication provided equality, addition, = d. = = a c b means and (c, d) (a, b) (a) Equality:
+ (c, d) = (a + Product: (a, b)(c,d) = (ac(c)
(b) Sum: (a, b)
definition
of equality
Thus, the complex 358)))
number
section.)
properties)
DEFINITION.
The
of certain
idea
c, b
+
d).
bd,
ad
+
tells us that (2,
3) is not
bc).)
(a, b) is to be regarded as an to the (3, 2). equal complex number the
number,
pair
ordered
pair.
The numbers
and field
Definitions
a and b are called components of the complex
Note that
the
vi 
i as fictitious
symbol
do this,
will
discuss
the basic
The
9.1.
THEOREM the
 1 by
V
the
properties
the
\037re have
x + y
faa's:
Associative faa's:
x+
Distributive
x(y
la\037t':
= y +
(y
+
laws are easily verified to prove the associative example, = z , , Z2) and note that) (Zl (YJ Y2),
y
x(yZ) = = = =
(Xl' X 2 )(Y
l Zl

(Xl(YlZl
\302\253XlYl


(XlYl
However, before we defined.) just
operations
multiplication That
The commutative and
Y2

is, if
of complex numbers satisfy x, y, and z are arbitrary complex

X2)'2)Zl X 2 Y2,
X 2 (Y

laws
distributive
+
the definition
from
(xy)z.
of
and
sum
x =
we write
multiplication,
X
X2 Yl)Z2,
X2Yl)(Zl ,
X l Y2 +
x(}'z)=
.)
for
l Z2 + Y2 Z l),
l Y2
(X
y) + z,
directly law
yx .
=
xy
product.
(Xl' x2 ),
+ Y2 Z l)
)'2Z 2 , Y l Z2
Z 2)
x,
+ z) = (x + z) = xy + xz
All these
Proof
the
la\037rs.
Presently properties
follo\037ring.
Commutative
For =
part
mathematicians.
early
of
and
operations of addition and distributive
associative
commutative,
numbers,
appear
a particular complex number
the
real
the
called the imaginary part. in this definition. anywhere which has all the algebraic
b, is
component,
1 doesnot
ascribedto we
component, a, is also called
The first
(a, b).
second
the
i =
symbol
introduce
shall
we
of
number;
359)
properties)
l (Y l Z2 +
Y2 Z l)
(X l Y2 +
X2Yl)Zl
=
Z2)
X 2 (Y
+
l Zl
(XlYl


Y2
Z 2))
X 2 Y2)Z2)
(xy)z.)
be similarly
may
+
proved.)
Theorem 9.1 shows that the set of all complex numbers satisfies the first three field axioms for the real number as in Section I 3.2. Now will we show that given system, Axioms 4, 5, and 6 are also satisfied. Since (0, 0) + (a, b) = (a, b) for all complex numbers (a, b), the complex number (0, 0) is an identity element for addition. It is called the zero complex number. Similarly, the (1, 0) is an
number
complex
for
identity
0) = (a, b))
(a, b)(I, all (a,
for
as the To
for
identity verify
negative Finally, identity
b). Thus, Axiom
of (a,
b).
we
show
element
(1,
Axiom
4 is
multiplication. 5, we simply We
write
satisfied
note
(a,
with (0,
that
b) for
each nonzero That is, if (a, b) 0). that
fact,
this equation
is equivalent ac

to bd
the
=
0) as
the
(a, b). complex number \037
(0,
has
is a
of equations) ad
+
be =
0 ,)))
for
addition
= (0, 0), so (a, a reciprocal
complex
d) = (1,0).)
pair 1,)
0), there
identity
+ (a, b)
b)
(a,
(a, b)(c, In
because)
multiplication
number
and (1, 0)
b)
is
the
relative to the (c,
d) such that)
360)
numbers)
Complex
which has the (9.1
solution)
unique
c =
))
condition
The
write
We
(a, b)
\037
(0,
(a, b)lor
1
b
+
b2
+
2 (a2 + b
shows
discussion
a
b
+
.
2)
b).
the reciprocal Thus, we have)
if
(a,
\037
(a,
2
0,
is well defined.
so
b
a
=
b
d=
,
that a 2
0) ensures
(a, b) foregoing
a
2)
I/(a, b) for the reciprocal of
(9.2))
The
a 2
'
a
2
+
)
set
the
that
2
b
b)
\037
(0,0).)
of all complex numbers satisfies the six all the laws of algebra deducible from
for the realnumber system. Therefore, 1.15 Theorems 1.1 through also hold for complex numbers. In particular, Theorem real numbers. as for well of Section I 3.2are all valid for complex numbers as numbers exist. That is, if (a, b) and (c, d) are two of complex 1.8 tells us that quotients with (a, b) \037 (0, 0), then there is exactly one complex number numbers (x, y) complex we have (x, y) = (c,d)(a,b)I.) In fact, such that (a, b)(x,y) = (c,d).
field
axioms
field
the
axioms
The
9.3
complex numbers
as
an
real numbers
of the
extension
Co of C consisting of complex numbers. Consider the subset with zero all numbers that form is, (a, 0), imaginary complex complex we in In have) of is of members or two The sum fact, Co. Co again product part.
Let C denote
set of all
the
of the
numbers
all
(a, 0)
(9.3))
+ (b,
0) = (a +
and)
0))
b,
(a, O)(b,
0) = (ab, 0) .)
in Co by adding or multiplying the two numbers shows that we can add or multiply real parts alone. Or, in other words, with respect to addition and multiplication, the were real numbers. The same is true for numbers in Co act exactly as though they and division, since (a, 0) = (a, 0) and (b, 0)1 = (b 1, 0) if b =;t:. O. For this subtraction make between the real number x and the complex no distinction reason, we ordinarily real we to whose is x and (x, 0), and we write x = (x, 0). number x; (x, 0) agree identify part 0 = (0, 0), 1 = (1, 0), 1 = (1, 0),and so on. In particular, we write Thus, we can of the real number system. number system as an extension think of the complex in a slightly relation between The Co and the realnumber system can be described which different way. Let R denote the set of all real numbers, and letf denote the function E if number That let) real number onto the x each x is, R, (x, 0). complex maps
This
=
f(x)
The
onto
function distinct
f so
defined has domain
elements
correspondencebetween preserved under
this
I(a these
equations
R
of Co. Because R and Co. correspondence.
+ b)
= f(a)
being merely
and
(x, 0) range
.)
Co, and
it
maps
of these properties, I is said The operations of addition That
is, we
+ f(b))
a restatement of
and)
(9.3).
distinct to establish and
of R a onetoone
elements
multiplication
are
have)
f(ab) Since
=
f(a)/(b)
,)
R satisfies the six field
axioms,)))
The
unit
imaginary
361)
i)
the function f fields R and Co are said to be isomorphic; As far as the algebraic above is called an isomorphism. between are concerned, we make no distinction and multiplication operations of addition with the number number the real x we That is fields. complex identify why isomorphic R of the realnumber an is called extension C The (x, 0). system system complexnumber
is true of
same
the
because
it
Co'
two
The
as described
them
relates
which
a subset
co\037tains
Co
to R.
is isomorphic
which
I 3.4 in such a way that the three order axioms of Section The field Cocan also be ordered define (x, 0) to be positive if and only if x > O. It is trivial are satisfied. In fact, we simply so Co is an ordered field. The that Axioms 7, 8, and 9 are satisfied, to verify isomorphism elements of R onto the it the positive ordersince also above described maps preserves f
The
9.4
of Co
elements
positive
.)
unit
imaginary
i
Complex numbers have some algebraic propertiesnot possessed has no x 2 + 1 = 0, which equation example, the quadratic numbers, can now be solved with the use of complexnumbers. since we have) number (0, 1) is a solution, 1)2 =
(0,
(0, 1)(0,
1) = (0 . 0 
1. 1,
0
1 . 0)
. 1 +
=
numbers. For the real
by real
solution In
among
the
fact,
complex
1
( 1,0) =
.)
It has the denoted by i and is called the imaginary unit. 2 = 1. The reader can that i 1, (i)2 = 1, easily verify square property so x =  i is another solution of the equation x2 + 1 = O. idea with the notation used by the early matheNow we can relate the orderedpair of complex numbers of multiplication First we note that the definition maticians. gives hence we have) us (b, 0)(0, 1) = (0, b), and
The
(0, 1) is
number
complex
is
its
that
(a, b)
other words,we
The
using
have
9.2.
THEOREM
involving
write a
if we
Therefore,
compute
bi)(c + di)
is in agreement of a nonzero reciprocal
a + formula
can be expressedin
a + hi. In
(a, b) =
get
(a, b)
the form
notation is that it aids us in algebraic manipulations For example, if we multiply a + hi multiplication. and associative laws, and replace i2 by 1, we find that)
1
This
(0, 1), we
and
a bi)
(a
+
is in agreement
bi)(a
with
= ac 
+
bd
(ad +
definition
the
with
course, the
+ (b,O)(O,1).)
of this
distributive
of
number (a, b)
complex
(a, 0)
and i =
= (b,O), the following.)
(a + which,
=
= (a, 0), b
proved
Every
advantage addition the
= (a, 0) + (0, b)
complex number a
that
bi))
given
a
2
+
b
2)
of multiplication. we may write) bi,
in (9.2).)))
a
a
2
+
+
bi.)
of formulas by
c +
di,
bc)i,)
+
 bi
a
bi)
= a
bi
b
2)
a
2
+
b
2
.)
Similarly, to
362)
Complex numbers) the
By
solve
to
equation
we may
introduction of complex numbers, we have gained much more than the ability the simple quadratic equation x2 + 1 = O. Consider, for example, the quadratic 2 ax the square, + bx + c = 0, where a, b, c are real and a \037 O. By completing write this equation in the form)
x+
(
b
2
2a)
 b2
4ac
+
=0.
4a2)
 b2 < 0, the equation has the real roots (b ::I::vi b2  4ac)j(2a). If 4ac  b 2 > 0, In this case, the left member is positive for every real x and the equation has no real roots. roots, however, there are two complex given by the formulas) If 4ac
b + .

r1 =
(9.4))
vi
Gauss proved that
every
a 1x
ao +
. . . , an
ao, a 1 ,
where
complex
exists
a solution
are
if n
numbers
the
in
arbitrary
+ a2 x 2
+
+
real numbers,
complexnumber
n =
\302\267 \302\267 \302\267
with
anx
an
coefficients
the
Geometric
9.5
Modulus and
interpretation.
0
form)
,)
0,
has
complex
among the
a solution
ao, a 1 ,
This fact is known system. is no need to construct
algebra.t It shows that there numbers to solve polynomialequations with complex
theorem of than
\037
2
2a)
of the
equation
polynomial
b
1
2a
Moreover, even if
> 1.
. vi 4ac 
b 

r2 =
and)
2a)
2a
In 1799,
 b2
4ac
1
. . . , an as
are
complex, the fundamental
numbers more
general
coefficients.)
argument
ordered pair of or by an arrow
it may be represented from the origin vector geometrically by the xyplane is often referred in Figure 9.1. In this context, to the point (x, y), as shown is the imaginary axis. The xaxis is called the real axis; the yaxis to as the complex plane. It is customary to use the words complex number and point interchangeably. Thus, we to the complex number z. the point refer to the point z rather than corresponding a simple geometric and of have of addition subtraction The complex numbers operations and are If numbers two Zl Z2 represented complex by arrows from the interpretation. is determined and then the sum to Z2 , respectively, Zl + Z2 Zl by the parallelogram origin of the from to is a the origin law. The arrow Zl + Z2 diagonal parallelogram determined in The 9.2. other diagonal is related 0, Zl , and Z2 , as illustrated Figure by the example by to of Zl and Z2' The arrow from to and equal in length to the difference Zl to Z2 is parallel the arrow from 0 to Z2  Zl ; the arrow in the opposite direction, from Z2 to Zl , is related a complex
Since
number (x, y) is an in the plane,
a point
in
the
same
t A proof
way to
of the
Zl

fundamental
For of a complex variable. 1945, or E. Hille, Analytic proof Publishing
is given
in
O. Schreier
Company,
Z2
real
numbers,
or geometric
.)
theorem
example, Function
and E.
New York,
of algebra can be found see K. Knopp, Theory Theory, Vol. I, Blaisdell
Sperner, 1951.)))
Introduction
in almost any book on the theory of functions New York, of Functions, Dover Publications, Co., 1959. A more elementary Publishing to Modern Algebra and Matrix Theory, Chelsea
Geometric
If (x,
y)
\037
x and
can express
0), we
(0,
and
Modulus
interpretation.
in polar
y
x = r cos e,)
363)
argument)
coordinates,)
y = r sin
e ,)
we obtain)
and
(9.5)) The
x
number
positive
the modulus or
r,
which
iy = r
+
of
x +
Ix +
iyl
e)
=
.)
of (x, y) from by Ix + iyl.
is denoted
and
iy
+ i sin
distance
the
represents
absolutevalue
(cos e
vi x 2 + y2
the
is called
origin,
Thus,
have)
we
.)
y)
2\\+2
x t)
I (x,y) I I I I I
.\037\\ \\J
\\). \037
.. Y = I
= x
r
. sIn
+
2)
iy
lJ u
I I) x) o)
2 1  22)
representation of x + iy.)
Geometric
9.1
FIGURE
complex
of and subtraction Addition numbers represented geometrically
9.2
FIGURE
the
number
complex
by
the
law.)
parallelogram
the rather than is called an argument of x + iy. We sayan argument for a given (x, y) the angle e is determined point only up to multiples it is desirable to assign a unique to a complex number. This of 27T. Sometimes argument e to lie in a 27T. The intervals interval of be done restricting halfopen length by may used for this purpose. We shall use the interval (7T, 7T] [0, 27T) and (7T, 7T] are commonly of x + iy; we denote this e by refer to the corresponding e as the principal and argument = 0 and r if x we define \037 + + + Thus, Ix (x iy iyl, arg (x + iy) to be the iy). arg
The polar
angle
argument
because
unique real
()
e satisfying
the
x = r the
For
Since is not
cos
e,)
y
=
7T < e
r sin e,)
number, we assign the
zero complex
used as an
conditions)
0 and
modulus
0)
if
z
z is
simply
the usual properties
\037
0,)
and)
IZ I
the
of
length
of absolute values 
z21
=
IZ 2
 zll
a line of
real
\302\267)))
segment, numbers.
it
364)
the absolute value
Geometrically, and
numbers)
Complex
in
Z2
the
IZ 1
z21
O.
and
the
pseudoordering is
axioms of
whether = x +
Section13.4are
the
we
IY,
say
satisfied
We say that z is positive
follows:
as
If z
numbers.
complex
the order
defined
to decide
try
if
and
if Izi > O.
only
10. Solve Exercise8 positive
11. Make
only
a sketch
(a)
(b)
Iz
+
11
12. Let w
=
(az
0, prove
If z
follows:
as
defined
= x +
iy,
we
say that
z is
if x
(d)
+ d), where w
 be
is
pseudoordering
1.
y. showing the set of all
and
if
12z +
If ad
of
of positive?
8 if the
Exercise
among Which
and
that

w
a, b, =
(ad
the imaginary
e, and d
 iI
Iz Iz I
obtain
T, we
by
manner
the
+
377)
described
time.
The general theory of infinite sums tend to a finite partial
whose partial
like (10.5)
those
and
between serieslike (10.1)whose
a distinction
makes
series limit,
have
sums
no finite
y) \037
, I
y=
x
Sum
of rectangles is
of areas
I
+!
.\037 i'
+ .,.+
+ l 3
2
!
n
;'; j(
of shaded
Area
is
region
x
I dx =
1\"+1 1
log (n
I ))
+
\\..1 I I I I
\037, \"
tJ ,'!,t,''!11,i;
,1 :Pin>
\037
1;' '\"'i'''

,,,cc\"
;l',\"\037l\",J\037;\037
FIGURE
10.2
Geometric

meaning of the
.... x)
..0
3
2
I
0
n
1 +
inequality
n +
1/2 +
1)
...
+
l/n
> log
(n +
1).)
former are called convergent, the latter divergent. Early investigators in the little or no attention to questions of convergence or divergence. They treated paid were infinite series as though finite to the usual laws of sums, they ordinary subject algebra, not extended that these laws cannot be universally to infinite series. Therefore, realizing that some of the results it is not were later shown to be incorrect. surprising they obtained of the unusual intuition and skill which Fortunately, many early pioneers possessed them from at too false even conclusions, prevented arriving many though they could not all their methods. men Foremost these was Leonard Euler who discovered among justify one beautiful formula after another and at the same time used infinite series as a unifying The
limit.
field
idea
to
bring of
quantity
together
Euler's
work
many branches that has survived
of the
test
instinct for what is mathematically correct. The widespread use of infinite series began late Euler was born, and coincided with before the early
Nicholas Mercator (16201687) and series
for
the logarithm
segment. Shortly
thereafter,
in
1668
Newton
William
hitherto
mathematics,
of history is in
the
a
The
unrelated. tribute
to
17th
century, nearly fifty years of the integral calculus. development Brouncker discovered an infinite (16201684)
to calculate the area of a attempting discovered the binomial series. This discovery
while
great
his remarkable
hyperbolic proved)))
378)
Sequences,
a
to be
binomial
integrals)
improper
history of mathematics. theorem which states that)
the
in
landmark
nowfamiliar
the
series,
infinite
A
special
case
of the
binomial series is
n
(1 +
x)n =
L
,
k=O)
(\037)Xk
n is a nonnegative integer, and (;:) is the binomial formula could be extended from values of integer real values of n by replacing the finite sum on the right the exponent n to arbitrary by a he gave no proof of this fact. a suitable infinite series, although careful treatment Actually, series raises some rather delicate questions of convergence of the binomial that could not x is
where
an
real
arbitrary
coefficient. Newton
found
number,
this
that
in Newton's have been answered time. in and the after Euler's death 1783, the flood of new discoveriesbegan to recede Shortly of series came to a close. A new and more critical period formal period in the history which for the first contained, began in 1812 when Gauss published a celebrated memoir of in history, a thorough and rigorous treatment of the a time convergence particular A few years later Cauchy introduced an definition of the limit series. infinite analytic in his treatise Cours d' analyse in 1821) and laid the foundaconcept algebrique (published and of that of tions of the modern theory convergence divergence. The rudiments theory are discussed in the sections that follow.)
10.2 Sequences
In
\"series\" are of the English language, the words \"sequence\"and usage or events are used a succession of in some and to suggest things arranged they synonyms, The word \"sequence\" In mathematics these words have technical order. meanings. special is employedas in the comn10n use of the term to conveythe idea of a set of things arranged different sense. The concept of a in order, but the word \"series\" is used in a somewhat sequencewill be discussed in this section, and series will be defined in Section 10.5. If for every positive integer n there is associated a real or complex number an, then the everyday
set)
ordered
al , to define an
said
is
the set term
infinite
sequence.
has beenlabeled with
a 2 , and,
there is no
\"last\"
an
in general, the
integer term
nth
a2
, a3 , . . . ,
an ,
. . .)
of The important thing here is that each member so that we may speak of the first term at , the second Each term an has a successor a n + l and hence an.
term.
if we give some rule or common examples of sequencescan be constructed the formula a for formula the nth term. Thus, for example, an = Ifn defines describing sequence whose first five terms are)
The
most
I,
Sometimes
two
or more
formulas
may
II
\"2,
be
a2nl =
3'
1 1 \302\267)
4'
\"5
as, for
employed
1,)
a2n
=
2n 2 ,)))
example,)
379)
Sequences)
the first
case
in this
terms
few
being)
after a given
to carryon
define a
way to
common
Another
a2 =
a 1 = This
rule is
particular
we may a n+ 1
1,)
1, 1, In
any
the essential
sequence
integers such that
an
terms
the
function
the
of
in order,
(that is, the
value
sequence
are)
terms
f defined n = 1, 2,
function
each
for
of
definition
set of all is called
term
nth
is usually
values)
3, . . ..
In
fact,
sequence.) 1, 2, 3, . . . is the sequence.) of
integers
positive the
on the positive
displayed
by
writing
thus:)
.
. . . ,fen),
{fen)} is used
the notation
brevity,
some
set of function
.)
a famous
defines
it
few
a technical
fen)
f(I),f(2),f(3), For
how
explains
.)
sequence
state
is the
domain
\037rhose
The function
sequence.
infinite
The range
f
and
The first
there be to
which
2
n >
for
1)
formula,
term of the
convenient way
A function
DEFINITION. called
fen)
a n
+
an
2, 3, 5, 8, 13,21,34 is that
thing
is the nth
the most
is probably
this
=
numbers.
Fibonaccit
of instructions
a set
by have)
a recursion
as
known
whose terms are called the
sequenceis
Thus
start.
1, 18, 1,32, 1 .)
1, 8,
1, 2,
to
the
denote
..
.)
sequence
whose
nth
term
is fen).
an , Sn , X n , Un , by using subscripts, and we write Very often the dependence on n is denoted Unless otherwise or something similar instead of fen). specified, all sequences in this are assumed to have real or complex terms. chapter we are concerned with here is to decide whether or not the terms The main question we must extend To treat this problem, fen) tend to a finite limit as n increases indefinitely. the limit concept to sequences. This is done as follows.)
A
DEFINITION.
is said to have
{fen)}
sequence
positive number N
is another
there
this
case,
'ri'e
say
all n
\342\202\254)
for
{fen)} converges to
the sequence
= L,)
limf(n)
N
L and
f( n) \037 L)
number
\342\202\254,
.)
write)
we
n
as
positive
\037
00 .)
n\037oo)
A
this
In
If f f = t
which does not
sequence
u
+
Fibonacci,
concerning
definition L
and
are
the
complex,
iv and
converge is calleddivergent.)
function
L = a
and
valuesf(n)
we may
+ lb.
also known as Leonardo the offspring of rabbits.)))
the limit
decompose them
Then
we
have
into
fen)
L may their
 L=
be
real
or complex
real and imaginary  a + u(n)
of Pisa (circa 11751250), encountered
i[v(n)
this
sequence
numbers. parts, say b]. The)
in a problem
380)
infinite series,
Sequences,
inequalities)
 al < If(n)
lu(n)
the relationf(n)
that
show
\037
the two
that
relations
complexvaluedsequencef part
imaginary
v
lim
that
is clear
sequence by
restricting
the
definition
between
The analogy the
+
was
as
=
\037
do
that
not
f(n) = The
basic
convergent himself.
f is
when
have infinite
,)
sequence\"is used only for said to diverge. There limits. Examples are defined =
f(n)
sin
,) \037'TT

by
the
1 +
(_l)n(
for dealing with limits of sums, The reader no should have sequences. Their proofs are somewhat similar to
,)
) \037
+ 00
f(n)
as
sequences
difficulty
those given
in
Section
7.)
Jim
1. = 0 na.
if
Iim x n = 0
if
(log n)a n)
lim n
lim n\03700
(
f(n)
=
e
1Tin / 2
.)
3.5.
many sequences may be determined by using properties a few important defined for all positive x. We mention some of whose limits may be found directly or by using
n\03700
lim
formulas:)
in formulating
a
>
O.
Ixl < 1
.
= 0
b
1/n =
1
all a
for
> 0, b
> 0 .)
.)
n\037oo)
(10.13))
write
etc., also hold for limits of these theorems
products,
of
Chapter
n\03700
(10.12))
00)
following
n\037oo)
(10.11))
a
a sequence whose limit is finite. A of are, course, divergent sequences
=
f(n)
rules
examples of realvalued the results derived in
(10.10)
to construct
7.14
If f is complex,we
realvalued.
limit is
The convergence or divergence are of familiar functions that
(10.9)
used
+ 00.
an infinite
(l)n
be
may
limf(n) =
If(n)1 The phrase \"convergent with
the
This
Section
and)
+ 00)
other
and
n\037oo)
7.15
in Section
done
00 if
+
sequence
for
real x
symbols)
limf(n)
In u
.)
v(n)
values.
in
n\037oo)
n
bl)
explains the strong analogy for more general functions. well, and we leave it for the reader to define
as
limits
infinite
00. Conversely,the
have)
lim
i
for all positive
take only integer given and the one
to
over
carries

LI)
n\037oo)
x to
just
n +
b as
\037

f(n) \037 L as n \037 00. only if both the real part
case we
n\037oo)
defined
function
any
If(n)
b imply
if and
= lim u(n)
n\037oo)
It
\037
v(n)
in which
f(n)
v(n)
N, as'
n
N)
that the sequence convergesto L, as asserted. in this case being the greatest proof is similar, the limit values.)))
that
f(n))
f(N)
must
means
M such
L)
increasing sequence convergesto
o
with because their
work
fact, we
unbounded
an
a bounded
that
for all
f(n)\037.
{f(n)} is called bounded A sequence that is not
A sequence
Note:
If(n)1 < M for
write
to
easy to determine. In
THEOREM
n
+ 1))
> fen
Monotonic sequencesare pleasant is particularly
for all
381)
If, on the other hand, we
by writingf(n),?1.
decreasing and decreasing.
the sequence or if it is increasing we
numbers)
if)
+ 1))
< fen
fen)
of real
numbers)
be increasing
fen)
We
sequences
lower bound of the
382)
Sequences,
infinite series,
improper
integrals)
Exercises)
10.4
22, a sequence {f(n)} is defined by the formula given. In each case,(a) and the sequence converges or diverges, of each convergent (b) find the limit the it some cases be to n by an arbitrary positive real x In replace may helpful integer sequence. of x by the methods of Chapter7. You to study the resulting function and (10.9) may use formulas listed the of Section 10.2. at end (10.13) through
In Exercises 1
through
whether
determine
n

1. fen)
=
2. fen)
=
3. fen)
= cos
1
+
= Vn

3n
2
15. fen)
.)
n .)
1 + (
=
17.fen)
(l)n.
1
)n
+
(_:)n
+;on
19.
=
21/n.
20.
10. fen)
=
n(l)n.)
21. fen)
E
assigned L
=
sin (n!) n+ l '
23. an
of E: =
{an}
in
Exercises
2\"
\302\267
+
(1 \037rn. = e 1Tin / 2 .
fen)
1 e1Tin / 2 .
=
n)
E
=
1, 0.1,
28 is convergent. Therefore,for every preon E) such that Ian  LI < E if n > N, where of N that is suitable for each of the following
0.01, 0.001, 0.0001.
26.
. n
n + 1
=
1
, . n.)
2n = n
. (_\037n+1
an =
27. an
.
= ne 1Tin / 2 .)
fen)
23 through
n)
24. an = 25. an
1 
sequences
22.
N (depending 0, there exists an integer In each case, determine a value an.
>
limn__oo
values
n7T
cos 1
2/ 3
of the
Each
n +
n
fen)
=
=n
\302\267) \037r
= 1+
1
< 1.)
2 '
1 +
18. fen)
9. fen)
11. fen)
+n
lal
a > 1.
,
(
n)
=
n
100,000n
=
\037.)
where
n
1

1
+
loga
=
16.f(n)
+ (_2)n+l
nan,)
=
2)n \302\267)
3n + 1
=
14. fen)
\302\267
2
= 1 +
8.f(n)
13. fen)
n 2
7. fen)
=
n7T
5n2
=
6. fen)
n)
(
3n +
12. fen)
n2 + 1
n2 +
=
5. fen)
n)
n2 n
4. fen)
1
+
n
1
n +
28.
an =
3
+
(on(
l
'
. \037o r
limits. 29. Prove that a sequence cannot converge to two different = the of limit to Use definition 30. Assume limn__oo O. an prove that lim n__ oo a\037 = O. = A and lim __ oo b n = B, use the we have of limit to prove that definition 31. If limn__oo an n = cA, where c is a constant. lim n __ oo (an + b n ) = A + B, and limn__oo(ca n) 32. From the results of Exercises 30 and 31, prove that if limn__oo an = A then limn__oo a; = A2. that limn__oo(anb 2a n b n = (an + bn )2  a\037  b\037 to prove Then use the identity n ) = AB if = = A and b B.))) 00 n limn__ limn__ 00 an
33. If
rJ. is a real number and the equation)
n
rJ.
( When
(a)
=
rJ.
n
show
t,
in teger,
a nonnegative

rJ.(rJ.
=
383)
series)
Infinite

1)(rJ.
. . .
2)
(\037)=
an =
Let
(b)
34. Let.f be a
)
1
(
State
(c)
35. UseExercise
,
( ;;)
n
(a)
(
n n
that 0
OO L
k=1
n
that
rJ.
16
(\037)=)
function
Sn =
(a)
(\037)
that)
( 1)n (}!2).Prove
Define
[0,1].
1)
n.),
(\037)=\037,
realvalued
 n +
(rJ.
5 \037,
coefficient
binomial
the
+ an
=
\037 ak k=l)))
\302\267
a new generate has the terms)
384)
infinite series,
Sequences,
The
{sn} of
sequence
denoted
the
by
partial sums is called an
integrals)
series,
infinite
a series, and is also
or simply
symbols:)
following
a 1 + a2
(10.15))
improper
. . . ,)
a3 +
+
...+
a2 +
+
a1
00 an
. . \302\267 ,)
+
\037ak . k=1)
For
the series
example,
{sn} for
the sequence
represents
Ilk
\037\0371
n
Sn =
1
L k
which)
.
k=1)
The
are intended to remind sequence {an} by addition number S such complex
in (10.15)
symbols
from the
is obtained
If there is a
or
real
Sn =
Iim
of partial sums {sn}
the sequence
that
us
terms.
of successive that)
S ,
n\037oo)
we
that the
say
series \037\0371
a k is
and has the
convergent
sum
S,
case we
in which
write)
00
a k =
\037 k=l)
If
we say
{sn} diverges,
the partial sums
Sn
the
of
series
the
THE HARMONIC
1.
EXAMPLE
that
S . has no sum.)
and
a k diverges
\037\0371
SERIES.In the discussion of Zeno's paradox, 1 I k satisfy the inequality) \037\0371
we
showed
that
series
n
Sn =
1
L k
> log (n
+
1) .
k=l)
Since
+ 1) \037 00 as n \037 00, the same is true of This seriesis called the harmonic series.)
log (n
diverges.
EXAMPLE 2.
of the
series
In the discussion
1 +
! +
! +
of Zeno's paradox,
\302\267 \302\267 \302\267
, given
by the
n
1
\0372
\037 k=1)
is easily
which
and
hence the
proved
by
The reader should of
has
sum
and
also
the series
hence
encountered
\037;:1
1 Ik
the partial sums
formula)
  1 1 ' 2n

\037
00,
these
2. We
partial
sums approach the
may indicate this
by
limit
2,
writing)
1 +t+!+\"'==2.)
(10.16))
sum
As n
induction.
series convergesand
2
k 1
we
Sn,
a convergent
realize
that
the word
\"sum\"
series is not obtained
by
is used ordinary
here in a very special sense. The addition but rather as the limit)))
The
of convergent series)
property
linearity
385)
should note that for a convergent series, of partial sums. Also, the reader the two are \037::1a k is used to denote both the series and its sum, even though a number and it is not capable of being conconceptually distinct. The sum represents or Once the distinction between a series and its sum has been realized, vergent divergent. the use of one symbol to represent both should cause no confusion. As in the case of finite summation the letter k used in the symbol notation, L::l a k is a index\" and be other convenient The letters n, m, replacedby any \"dummy may symbol. and r are commonly used for this purpose. Sometimes it is desirable to start the summation k = 2 or from some other from k = 0 or from value of k. Thus, for example, the series 1 /2 k. In general, if p > 0, we define as L::o in (10.16) could be written the sym bol L::p ak the same as L::l bk , where b k = a p + k  1 ' Thus b 1 = a p , b 2 = a p+ 1 , etc. When there to mean or when is no danger of confusion we write \037 a k instead the starting point is unimportant, the sequence
of
the
symbol
of L::p a k \302\267 I t is easy to prove Suppose t n+ 1
have
as n
\037
we let
= ao
Sn
Sn , so if Sn
Sn
\037
The same holds
 ao
T

omitted or addedat
The
linearity
Ordinary finite
the
\037
.
as n
often
p = 1, we
the following
Sn = t n
have
diverge
for p
, and
p =
when
> 1,
we
O.
have
\037ak
+
k=l
k=l
series) important properties:) n
n
b k) =
+
converge or both
both series
of convergent
have
\037(ak
00,
L::p
ap
that the sequences {sn} and {tn } both converge described number of terms may be by saying that a finite of a series without its or divergence.) affecting convergence
n
(10.17))
\037
a k both converge or both diverge. . . . a + + a p + n  1 ' If p = 0, we p+ 1 + then t n \037 a o + S and, conversely, if t n \037 T and
it follows
again
beginning
sums
an
S
tn =
and
1. For
property
ak
series L::l
Therefore,
p >
if
true
sn+pl Spl, and or both diverge. This is
10.6
..+
. al +
+
00, then
tn =
two
the
that
=
property))
(additive
\037bk k=l)
and) n
n
(10.18))
k=l
and
series
THEOREM rJ.
and
is given
extension of these properties to convergent infinite provides a natural series are justifies many algebraic manipulations in which convergent were finite sums. Both additivity and they homogeneity may be comcalled linearity which be described as follows:) property may
10.2.
Let
f3 be complex by the equation)
\037
and
an
constants.
\037
bn
Then
be convergent the
series
00
(10.19))
.)
thereby
treated as though bined into one
let
property)
(homogeneous
k=l)
theorem
next
The
k
L(cak)=cLa
\037
n=l
\037
(rJ.a
an
+
00
(exa n
+ f3brJ
=
ex
\037 n=l
terms and of complex also and its sum n) converges,
series
infinite
n +
f3b
00 f3 \037 b n=l)))
n .
386)
infinite series,
Sequences,
and (10.18), we
(10.17)
Using
Proof
When
n to
\037 f3
series
10.3.
THEOREM
If\037 an
Proof
Since bn
convergence
of L
converge if
If L
bn
\037
and
an
\037
example,
when
bn
for all
1
f3\037bk'
k=l
corollary
converges and
(an + b n )
=
(an + bn )
k=1)
second term proves that
, and
an
bn
if\037
then L
diverges,
since
\037
establish the divergence
+
(an
b n) diverges.)
Theorem 10.2 tells us that converges, . b + b n) cannot Therefore, n (an \037 \037
an
of
convergence
implies
used to
is often
which
(10.19).)
diverges.) series
The
EXAMPLE.
n +
(X\037ak
sum indicated by
{3b,J converges to the
(exa k +
10.2 has an interesting
Theorem
=
b k.
L::I
\037
k)
first term on the right of (10.20) tends to ex L:I a k and the Therefore the lefthand side tends to their sum, and this
the
00,
of a series.)
=
f3b
k=l
tends the
n +
\037(exak
integrals)
write)
may
n (10.20))
improper
bn
an n,
\037
+
(Ilk
are both divergent, b n = 1 for all n,
the seriesL
=
then
L (an
because
1/2k) diverges
+ bn )
then
\037
and
diverges
\037
1/2k
converges.)
may not converge. For But when an = 1 and
+ b n) mayor
(an
(an + b n )
L
Ilk
diverges.
converges.)
Telescoping series
10.7 Another
of finite sums is the
property
important
telescoping property which
n
(10.21))
\037(bk

 b k 1) = b i +
b
that)
states
n+1 .
k=1)
we
When
L
an
for
try to extend this each term an
which
may
be expressed
an = b n
(10.22)) are known
series
These
following
THEOREM
Let
10.4.
the series
as telescopingseries and
{an} and {bn } be t}1'O sequences an = b n
L
an
b n + 1)
if and only
converges 00
(10.24))
bn + 1
consider
series
those
form)
.)
behavior
their
is characterized
by
the
theorem.)
(10.23)) Then
series we are led to as a differenceof the
to infinite
property
L an n=l)
= bi
if
 L,
for)
the
of complex numbers n =
1, 2,
such
that)
3, . . . .)
sequence
{b n } converges,
where
L = lim b n . n to 00)))
in
which
case
we have)
Let
Proof
sum of L
nth partial
the
denote
Sn
n Sn =
of
because
Note: Every b1 to be arbitrary
\037
=
\037ak
then
\037
Sn
+
holds
(10.23)
b n
with
+
for
each
ges
+ l)(n + x
x)(n + x
+
n(n +
the
00
n=l)
the
series
3. Since
\037
Note: infinite
log
[nl(n
log
+ 1)]
We
verified
b 2)
by
+ (b 2 
merely
same operationson
leave b l
, cancelb2
every bn
+
the
have
cancels

+
obtain)
decomposition)
1
1)
(n
1)(n + x
+ x +
+
log (n +
1),and
\302\267 2)
log n
since
00 as
n
sums
and
\037
\037
, cancel
with the
b 3) +
+
(b 2

between
. . \302\267 + (b n  bn + l ) = bl
b 3) +
b 3 , and so on. exception of bl
(b3
finite
 bn + 1)
and canceling.
parentheses series)
removing infinite
the
 b2 )
difference it becomes)
form,

)
1
2(x + 1)(x+
2)
an important extended
2))
the following series conver
property,
= x
0, we

telescoping
log n
illustrate
series

=
= 1.
x)(n + x +
+ 1)(n+
1)] =
L
diverges.)
(bl
Thus
the
write (10.21)in
If we
can be
form the
+
[nl(n
Telescoping
series.
(bl which
2
al
1)
1 ( (n
choose + . . . + an .)
if we first =
Sn
,
n +
1
(n + x)(n + x
L
EXAMPLE
1)
1
n > 1. Therefore, by sum indicated:)
integer has
and
+ 2)
=
1, where
1
n
1)
a negative integer, we
If x is not 1
(n
1 
1
n=1)
diverge.
(10.24).)
(10.22)
satisfy
Since b1 = 1 and
= I/n.
or both
converge
proves
n >
for
Sn
=
L
2.
this
and
,
n+l
b
have)
1
00
EXAMPLE

Then we
n).
n(n
hence
 L,
bl
because we can always
an =
and

bk + l ) = b l
choose bn + 1 = b l
= I/(n 2
have)
we
Then
both sequences {sn} and {bn }
00,
then
and

\037(bk k=l)
k=1
is telescoping
series
1. Let an
EXAMPLE
n
b n \037 L as
an'
n
Therefore,
(10.21).
Moreover, if
387)
series)
Telescoping
b4) +
For each n . This leads
Supposenow
we per
. . \302\267) \302\267
> 1, at some stage we cancelbn . us to the conclusion that the sum)))
00,
388)
series is bl
of the
This
that
shows
parentheses
The geometric
10.8 The
as
a geometric
of a
real
fixed
understanding
Let Sn
denote
x =
1, each term + 00.
the
on
sums may
+ x2
+ x
last sum
the
1, we may
=
x)sn
(1
 x) L Xk
telescopes. Dividing  xn
1
>
1 the
proved the THEOREM and
has sum
that
term x n
the nth this
start
important example of the terms
addition
successive
is the nth power series with n = 0, with
this
Xk+l)
diverges since
series
the
case,
writing)
by
=
1
 xn
,
k=O)
1
 x, we

x
obtain the
n
if x
Ix) large
formula)
n depends
=/=
1 .)
behavior
the
on
entirely
of x n .
theorem.)
following
10.5.
1/(1 
If x
x).
1 + >
in a
 x). 00, and the series converges to the sum 1/(1 n + of {sn} implies x 0 as n + 00. Therefore, if n in this x not tend to 0 case. Thus we have since does sequence {sn} diverges
(10.25)
If I xl
for
a very
study
for Sn
nI k 2 (x
=
Ix
the behavior of Sn n + When x 0 as n + Ixl < 1, then n = Sincesn+1 x , sn convergence Ixl
by
1
=
Ix shows
= O.
x n I .)
+
=
Sn
nI

sn = This
they
can
\302\267 \302\267 \302\267
+
n. In the sum simplify
is 1 and
right
lc=O
since
n __ oo b n
series
If x =/=
(1
integrals)
10.9.))
be usedto seriesis generated by and has the form! xn , where progression or complex number x. It is convenient to that the initial term, xo, is equal to 1. the nth partial sum of this so that series,
00 as n
+ Sn
Section
in
Sn = 1 If
improper
Theorem 10.4, this conclusion is falseunless lim cannot always be removed in an infinite series as
of finite property the geometric series. This
telescoping
known
the
. Becauseof
(See alsoExercise24
finite sum.
in
infinite series,
Sequences,
1, the
series
is
complex,
is to
That
x + x2
+
. . .
with
say,
Ixl
2.
00
k
Therefore
k
l  , 
each
is bounded
cannot exceedM.)
We
necessary and
the following
obtain
to
10.1
Theorem
395)
terms)
nonnegative
convergence.)
10.7.
THEOREM
if and only If the
use
we may
increasing,
series of
tests for
Comparison
is therefore
Iln!
is e
series
 I, where
convergent and e is
the Euler
number.)
The convergence
of
the
example was
foregoing
the given series with those of a series a number of tests known as to yield
10.8.
THEOREM
exists a positive
COMPARISON constant
Assume
an
that
(cb
n
).
of!
convergence
said to
.)
an
be asymptotically
.
1
b n)
writing)
by
b n)
n +
as
00 .)
bn
an and!
terms
with
2
for s
is
In
ZETAFUNCTION.
section
Greek
since
convergent,
> 2, and that
this
series
s
use 2
+
n) as
for every s function
important
n
for every
converges
also converges
definesan
zetajunction:)
1/(n
Section 10.7, we proved this as a comparison series,
1 of
Example
l/n2\"\"\"
l/n
therefore!
letter zeta),
positive and asymptotically
that are
together.)
a convergent telescopingseries.If we
Ijn
to
asymptotically
diverge
they
RIEMANN
L 1/(n + n) is follows
comparison
together
converge
is
\"an
Two series!
10.10.
THEOREM
by
Theorem
equal to b n ,\" and it is intended the for large n. Using this same way essentially terminology, test in the following manner.)
b n is read b n behave in
\"\"\"
an
that
prove
\"\"\"
an
(10.40))
it
Therefore
I.
numbers are
of complex
indicated symbolically
is often
relation
that
1,
if)
.
equal
n
.
1
also
n+ 00
This
n
b n)
= c, holds if limn+oooan/b n provided 1 and we may compare! an with! 0, we conclude only that convergence of! b n implies
=
n
an =
N such that n > N implies t < > N, and the theorem follows by
lin1 n +ooo a n /(cb
DEFINITION.
equal
all
0 and b n > 0 for
>
b n converges.)
if!
only
10.9
Theorem
have
limn+oooan/b
all n
n for
that
then
a'n
is
(10.38)
that)
(10.39))
Then!
that
Assume
TEST.
COMPARISON
LIMIT
con
its
affect
Omitting
+
00.
real s >
Also,!
2.
> 1. Its sum,
in analysis
1/n2
shall
We
denoted
known
as
the
00 \037(s)
=
\037
1
\037ns
if
s >
1
.)
n=l)
Euler 1T
2
j6,
discovered a result
many beautiful formulas involving which is not easy to derive at this
\037(s).
stage.)))
In
particular,
he found
that
\037(2)
=
The integral
!
1 In 3. Since also to must Iin diverge. equal
EXAMPLE
series
every
diverges, For
397)
test)
example,
of the
1
sin
and)
2:
sin Iin
The
10.13
\037
To use comparison tests series
behavior.
purpose.
New
proved
by Cauchy
1 \302\267 n
that
x)lx
(sin
+ 1 as
x +O.)
test
integral
of known
asymptotically
series)
n=l)
the fact
from
follows
Iln
L
+ 10)
Vn(n
n=l)
relation
two
00
00
The
terms
positive
having
this is true
at our
have
must
we
effectively,
The geometric seriesand can be obtained very simply
are useful for
this
the integral test,
first
zetafunction
the
examples in 1837.)
disposal some examplesof
applying
by
y)
y)
n
f(k)
1,
Sn =
'Lf(k)
Let
1 be
a
n
2
0
n
FIGURE 10.4
all real x
x
x
2
o)
test.)
integral
decreasing
positive
function,
defined for
let)
n
tn =
and)
k=l)
Then
both
Proof
sequences
By
{sn} and {tn }
comparing
or both
converge
1 with appropriate
Li(x)
dx
\302\267)
diverge.)
step functions as
suggestedin
Figure
10.4,
we
obtain the inequalities)
J/(k)
(k))
Since both sequences {sn} and {tn } are monotonic or Sn  1(1) < t n < Snl' these inequalities show that both are bounded above or both are unbounded. as asserted.))) both sequences convergeor both diverge,
increasing, Therefore,
398)
Sequences, 1.
EXAMPLE
The
series,
infinite
to prove
test enables us
integral
integrals)
improper that)
00
1
2:
n
s > 1 .)
only if
if and
converges
S
n=1)
x s, we
=
Takingf(x)
have)
n t
= n
rn..!.dx J1
X
S

1s
1
=
n)
{ log
1 the
s >
When this
n
1 s +
1, then
s
1. and the series diverges. The
series)was discussedearlierin EXAMPLE
0 as
if
integral test,
of the
convergence
implies When
term
1 S
10.5.
case s = 1 (the harmonic special Its divergence was known to Leibniz.)
may be used to
that)
prove
00
1
2: n=2)
the sum
start
(We The
n =
with
in tegral
corresponding
n
tn =
i
2
2 to avoid n in this case
x)S
if and only {t n } converges for the series in question.)
true
if s
be zero.)
log n may
 (log2)1S
1 s
dx = { log
Thus
which
is)
(lOg n )1S
1 x(log
for
s > 1 .)
only if
if and
converges
n(log n)'
(log n)
> 1, and
 log (Jog
hence,
2))
the
by
if
s;j6. 1
,)
if
s =
.)
integral
1
test, the
same holds
10.14 Exercises) the
Test
following
series for
convergence or divergence.
\302\267
5.
case, give a
each
In
decision.) 00
1.
2: n=l
00
(4n

\037 V2n 2. L
n=1

3\0374n
1)
 Ilog(4n n(n
+
I sin
2:
nxl n
2
'
n=1
+ 1)
.
6.
+(l)n
\0372
1)
.
2n
\037
n=l)
00
3.
+ 1
\037n
L
\302\267
7.
2n
00
2: n=1)
\302\267
\037
n=1
4.
n!
\037
(n
n=1 00
2
;n
+ 2)!
\302\267
8.
2:
.n
n=2)))
log n V n
+ 1
.
reason for
your
test and
root
The
series of
test for
ratio
the
00
00
1
9.
2
n=l
A
14.
\302\267
/
+ 1)
'v n(n
399)
terms)
nonnegative
n cos 2 (n1Tf3)
\037
\302\267
L n=l
2n
00
10.
1+\037
\037 n=l
1
1.
(n + 1)3 
L
15.
2
n=3
00
2
n=2
\302\267
log
n)S
00
1
11.
log n (log
n
.
16.
(logn)S
ne n2 .
2
n=l) 00
12. \037Ianl L lon n=l
'
00
13.
f o
18.
\302\267
2
1000\037
1
+
2 dx.
n+ 1
\037
L
vxdx.
e
f
n
n=l)
a nonnegative increasing function the test to show that) by proof of the integral
Assumefis
for all
defined
n
N.
x so
an+l
an
X
X
 N
.)))
Then
there
must
be an N
test and
root
The
In other words, the we must have an/x n
or,
a\037v/xN,
cxn
an < Therefore
is dominated
L an
the
by
xn .
This
If the test
Warning. limit L
ratio
+
n/(n
divergence
it
terms
(n +
=
;:
that
>
(n
+
I)! 1)
example,
I,
not
does
it
harmonic
the
series,
cannot
an
series
the
the other hand, for all sufficiently large n
On
O.)
approach
convergence of
some N,
follow that necessarily which diverges, has test
1 but the limit L equals 1. ratio be greaterthan 1 for
an and
n
.
n+ 1
n
(n
n!
n
n
=
1
=
+
))
0
n
(1 + 1/ )
+ 1)
n
10.2. Sincel/e < 1,the of the series tends to 0; n!
( 10.44
less than
is always
N for
>
L n !/nn
by
as
n +
test.
ratio
the
is)
(10.13) of Section the general term
formula
implies
have a n + 1
of consecutive a n+ 1
by
n
1. For
may establish the
3. We
EXAMPLE
The ratio
than
1) which is always less than is sufficient that the test
such n we
for
because
ratio a n + 1/a
less
be
will
N,
(a).
proves
To prove (b),we simply observe that L > I implies a n + 1 > an for all n O. and hence an cannot approach 10.12.) is (c) proved by using the same examples as in Theorem Finally,
the
n >
N. x)
series L
convergent
particular, when
aN c =
where)
,)
401)
terms)
nonnegative
> N. In
for n is decreasing in other words,)
sequence {an/xn}
the
remaining
product of the first factors does not (10.44) also follows
2, the
Relation
both
the
root test and the when
we have
ratio test are, in case (a), convergence
reality,
special
is deduced
can be dominated geometric series by a suitable a is that a of com parison knowledge practice particular be deduced Further tests required. convergence may by using as Raabe's test and Two important examples known ways. Exercises 16 and 17 of Section 10.16.Theseare often helpful
of these
x n is not explicitly the comparison test in other Gauss' test are describedin when the ratio test fails.)))
series L
1)
n)
large n.
for
(10.41).
cases of the comparison test. In both tests in question the series from the fact that n Lx.
k +
even, and k = (n  1 )/2 if n is odd. does not exceed (i)k, and each of right 1. Since (i)k \037 0 as n + 00, this (10.44). proves
k =
nl2 if factors on the
where
(10.44)
n!
than
as follows:
tests in
402)
improper
integrals)
Exercises)
10.16
Test the each
infinite series,
Sequences,
or divergence and
for convergence
series
following
for your
a reason
give
decision in
case.) 00
1.
(n !)2
\037
.
8.
L (2n)! n=l
 1)n.
(nl/n
L n=l 00
2.
(n
\037
00
00
10.
.
nn
L
e n2 .
L
n=l)
2 n n!
n=l
4.
9.
\302\267
2n2
L n=l
3.
!)2
 en2
iG n=l)
).
00
3 n n!
\037 \037 nn
(1000)n
11.
.
n. \"
L n=l
n=l)
00
5.
n!
\037
3n
\037
nn+l/n
12.
\302\267
n!
\037
13.
\302\267
2 2n
\037
(n
\037
n 3 [v/2
n=l
n=l)
6.
L
1 / n)n
+
+ (I)n]n
.
3n
\037 n=l)
n=l)
.
00
00
1 7.
15.
L (log n=2 Let {an}
and {bn }
(a) If there is
(b) If Cn [Hint:
such
be two
Prove
[Hint:
a
sequences
positive
constant
Show that
ak
.LZ=N
for n > N
and
Show that .L
an
< 0
an be
with
y > O.
nx[,
>
0 for all n
>
let
and
N,
Cn
= bn
a seriesof
r >
>
0 for all
then .L an
1/ b n diverges,
dominates
.L l/b
n
n
>
N, then
.L
an
converges.
diverges.
.]
test: If there is an
Prove Raabe's
terms.
The series .L an

a n +l
1
r
n
n)
Use Exercise
r >
0 and
an
N
> 1
1
b n+1
=1+an)
< M for all
[Hint:
If
A
\037 1,
n,
=
for
all n
> N
.)
then
use Exercise
n.])
Gauss' test: If there is an
Prove
terms.
a n +l
If(n)1
,)
if)
diverges
> 1
> N
n)
15 with
an be a series of positive Let.L an M > 0 such that)
all n
for
an
[Hint:

< aNbN/r.]
if.L
positive
that Cn
that)
.L an converges.
where
and bn
> 0
an
r such
 2 and
if k
diverges
. 5 . . . (2n
2 .4 .6
(
n=l)
series)
the
that
. . .

k
1)
(2n)
For this
< 2.
403)
)
test
the ratio
example
fails.)
series)
Alternating
Up to now we have been concernedlargely our attention next to serieswhose terms
examplesoccur when
alternate
terms
the
or negative.
be positive
may
terms.
of nonnegative
series
with
to turn
These are called alternating
in sign.
We
wish
The simplest and
series
they have the form) 00
(10.45))
l
.2 (1)n
an
=
 a2
al
a3
+
 a4
la
+ (l)n
+...
n
+...,
n=l)
where
an >
each
Examplesof mentioned the
O.
+ x)
shall prove later < x < 1. For positive obtain the formula) we
As
1
which
Closely
(10.47))
x
series
this
x,
3
interest
us that the alternating in view of the fact
related
We have
investigators.
early
many
x
4
+
 1 4
+
+ (_1)nl
already
has the sum log (I + series. In particular, when
. . .
+
(_1)nl n
harmonic series has the
that
harmonic
to (10.46) is the interesting 7r
1
1
1
4
3
5
7
=1++\"'+
n
. . . .) ; +
x
and
converges
1
. . .
\"4
an alternating
it is
1
tells
special
on,
2
x 2:+ 3
= x
log 2 = 1  2 + 3
(10.46))
to
series)
logarithmic
log (1
were known
series
alternating
+ .
sum
the
series .2 1In
x)
whenever x
=
1 we
. . ')
log
2.
This result is of
diverges.
formula)
( _1)nl
2n

1)
+...
discovered Leibniz rediscovered this result in 1673 while by James Gregory in 1671. the area of a unit circular disk. computing and in (10.47) are alternating series of the form (10.45) in which the Both series in (10.46) Leibniz noticed, in 1705, that this simple to zero. sequence {an} decreases monotonically series.))) property of the an implies the convergence of any alternating
404)
Sequences,
the
then
series .L:=l
alternating
sum, lve also have
partial
the
The
(1 )n(s 
in (10.48) provide a sum Sn' The partial
inequalities
sum S
the
the sign
any
by
(I)n,
that
states
with limit
sequence
decreasing
denotes its sum
If S
O. the partial sums S2nlform + + Similarly, sequence because s2n+2 Therea decreasing Both sequences are boundedbelow sequence. by S2 and above by Sl' to a limit, fore, each sequence{S2n} and {S2nl}, being monotonic and bounded, converges say S2n + S', and S2nl + S\". But S' = S\" because) Proof
10.5.
S' 
= lim S2n
S\"

n+ 00) If
this common limit by the inequalities in (10.48)
we denote To derive
S2n
we have
Therefore
o< which,
S
taken
EXAMPLE
harmonic sum
< s2n+2
S2n
EXAMPLE
< S)

< s2n+l
= lim
S2nl)
series
S
< s2n+l
convergent series \037 an , where each an > O. Provethat \037 V an n converges = a counterexample Give for p !. or disprove the following statements: Prove (a) If \037 an converges absolutely, then so does\037 a\037/(l + a\037). = 1, then a n /(1 + an) convergesabsolutely.) (b) If \037 an converges absolutely, and if no an .2
51. Given a 52.
*10.21
of series
Rearrangements
The order of the terms the sum. In 1833 Cauchy For
series.
infinite
in a finite
1+
this
series,
taking
1+ l
this
1 3
.1 +

4

!
1. 5
.1 6 +

.)
\302\267 . .
log 2 was shown
log 2
+
! +

\037
+
! + i
 t
t1.f
10.17.
Section
in
value of true for
series)
two positive terms alternately can be designated as follows:)
series which
term, we get a new
(10.57))
f2
the affecting is not always
without that
series to the sum
of this
convergence
be rearranged
surprising discovery consider the alternating harmonic
1
the terms of
sum can
the
made
example,
(10.56)) The
\037.
followed by
 ..
+ +
If we rearrange one
negative
.)
harmonic series occurs exactly once in this occurs in the alternating this new series has a sum and vice versa. But we can easily prove that rearrangement, than log 2. We proceed as follows: greater of 3, say n = 3m, the Let t n denote the nth partial sum of (10.57). If n is a multiple 2m positive terms and m negative terms and is given by) partial sum t 3m contains
t 3m
1

=
?; In
which
term
Each
2k
\037
\037
of the last
each
=
2k
1
( 2: i

three sums, we use the n
2
1
=
log n +
2\037 \037
)
\037
\037
\037
\037
\037
\037
\037
\302\267) \037
\037
relation)
asymptotic
C+

= \037
0(1)
as
n
\037
00
,)
k
k=l)
to
obtain) 13m =
=
(log
!
log
4nl
+
2 +
C + 0(1)
0(1)) .)))
 !(log2m
+
C +
0(1))
 !(logm
+
C +
0(1))
i
412)
infinite series,
Sequences,
improper
integrals)
\037 00. But t 3m + 1 = t 3m + Ij(4m + 1) and t 3m  1 = t 3m  Ij(2m), log 2 as m so t 3rn + 1 and t 3m  1 have the same limit as t 3m when m \037 00. Therefore, every partial sum is ! log t n has the limit! log 2 n \037 00, so the sum of the series in (10.57) series of a convergent The shows that rearrangement of the terms foregoing example
t 3rn \037!
Thus
2.
as
We shall prove next that this can happen only if the given series is does series of an absolutely convergent That is, rearrangement conditionally convergent. more precisely what is meant by a not alter its sum. Before we prove this, we will explain sum.
its
alter
may
rearrangement.)
= {I, 2, 3, . . .}denote the set of positive integers. Let f be assume is and whose P, range f has the following property:)
Let P
DEFINITION.
is P
whose domain
and
m Such
a function
2
and
an
f is
2 bn
such
every
for
=
some
f,
permutation
2 bn
series
the
then
.)
\037 fen)
onetoone mapping
or a
P,
of that
bn
for
f(m)
implies
calleda permutation
two series
are
\037n)
n > 1 we
P onto
of
af(n))
to be
is said
a rearrangementof 2
If L an denotes the alternating harmonic series in (10.56) and if the permutation definedby series in (10.57), we have bn = af(n) , wherefis
f(3n
+ 1) =
Let L
10.20.
THEOREM
1
+
4n
an
,)
+
f(3n
an
also
Let L b n
be
a rearrangement, is a series of
Proof
absolutely
converges
absolutely because 21b n l above by 2 lanl. To prove that
has sum
b n also
2 n
En =
An =
,
An
\037
S and
A\037
2 ak
S, we
S*
IAN
as n
\037
 51
0 for all
Show
(a)
(b) If
0 implies
In Exercises 7
an
b
e n)
= an +
converges.
1
+n2 n).
(V I
9.
.
2
n=l)
n=2)
(log
n)logn
00
00
nS(V
2
n
+

1
v n
+
2V\037
1
 1).
10.
\302\267
2
n=l)
11.
n)
(bnla
00
2
8.
2
convergence.)
00
7.
that
show
converges,
given series for
test the
11,
through
2
have)
> O.
bn
if
n and
>
an
n we
each
for
415)
exercises)
review
Miscellaneous
nHl/n
n=l
an = Iln if n is odd, infinite series)
, where
2\0371 an
12. Show that
the
=
an
I1n
2
00
2
is even.
if n
 Vn a)
a (v n + 1
n=O)
> 2 and diverges for a = 2. for each n. For each of the following
converges
for a
Given
> 0
13.
an
counterexample. (a) If 2:=1 an diverges, 14.
give a proof
or exhibit
a
then
2:=1 a; diverges. I:=l anln converges. the series 2:=1 (n !)C 1(3n)
(b) If 2:=1 a\037 converges, Find all real c for which
15. Find
statenlents,
a > 1
then
for
series
! converges.
(n!)3j(an)! converges. 16. Let n1 < n 2 < n 3 < . . . denote those positive integers that do not involve the digit 0 in their decimal representations. Thus n1 = 1, n2 = 2, . . . , n 9 = 9, n10 = 11, . . . , n I 8 = 19, n19 = 21, etc. Show that the series of reciprocals 2\0371 link converges and has a sum less than 90. all
integers
17. If a is
the series by
Dominate
[Hint: an
real
arbitrary
the
which
9
2:=1
(9/10)n.]
2:=0
let sn(a) =
number,
1a
2a +
+
...
+ n a.
Determine
the following
limit:)
+ 1)
sn(a
. hm
noo both
(Consider
18.
If p
(a)
and q
positive and
are fixed
negative
integers,
a, as
p > q
well as
pn n 00
The
(b)
following
appear, alternately, 1 Show
[Hint:
2
1 = log
k=qn
\342\202\254 \302\267
k
q)
harmonic alternating two terms:) by negative
11+
the series
converges and has sum log 2 +
Consider
the
partial
sum
sSn and
0.)
that)
series is a rearrangement of the three terms followed positive
+l+it!+t+t+lI
that
a =
> 1, show
lim
.
nSn(a))
use
part
+
!
log
(a).])))

+
i.
'\
series in which there
416)
infinite series,
Sequences, (c)
the alternating harmonic series, writing terms. Then use part (a) to show that
Rearrange q negative
by
sum log 2 + 10.23
improper
t
integrals)
alternately this
rearranged
p positive terms series converges
followed and has
(p/q).)
log
integrals
Improper
in Chapter 1 under the restriction dx was introduced integral S\037f(x) on a interval is and bounded [a, b]. The scope of integration finite defined f these restrictions. extended be relaxing by theory may dx as b + + 00. This leads to the we may study the behavior of S\037f(x) To begin with, an of an infinite called notion (also integral improper integral of the first kind) denoted by is if we keep the interval Another extension obtained dx. the symbol S: f(x) [a, b] finite new or more The so unbounded at one obtained and allow f to become integrals points.
The concept of
an
function
the
that
(by a suitable limit process) the integrals of Chapter
called
are
1 from
improper
improper integrals of the second the former are integrals,
To distinguish
kind.
often called \"proper\"
integrals.
calculus.
advanced the
In
theory.
examples. It will be
fact,
evident
integral
proper
the
function
=
I defined
in
and it is
denoted
kind,
first
if the
definitions
the
this
direct
f f(x)
by
the
lim
I(b) = Iim b++oo
b++oo
a new function
I
.)
or an integral
f'
These
are similar to
definitions
play the role the
b
Ja)
The
improper integral of is said to converge
dx
f(x)
Otherwise, the integral S: f(x) dx equals A, the number A is called the
exists and
(10.61)
10.5.))))
define
integral,
is finite.
and
exists
that
a
of
many
limit)
(10.61))
when
may
that
integrals.
dx.
infinite
S: f(x)
symbol
we
b >
each
for
dx)
is called an
way
> a,
for every b
exists
it is not surprising for improper
analogs
integrals bear a
to improper
pertaining Therefore
series.
infinite
dx
S\037f(x)
I(b)
The
for
theorems on series have
the elementary
If the as follows:)
in
that
presently to those
resemblance
strong
as improper integrals of one kind or analysis appear undertaken in courses in of such functions is ordinarily the most elementary aspects of with We shall be concernedhere only we shall merely state some definitions and theorems and give some
functions important and a detailed study
Many another,
symbol
the
of the
\"partial
those and
dx = for
given
A
the with
infinite
integral the
of
the
integral,
If the
in
limit
and we
write)
\302\267)
series.
may be referred to
is used both for converges. (Compare
S: f(x) dx
integral
sums\"
f(x)
to diverge.
said
is value
as
The \"partial
function
integrals.\"
and for the value of near the end
remarks
values
the
I(b)
Note integral
of Section
The
1.
EXAMPLE
To prove
we
this,
integral
improper note that)
S\037
417)
integrals)
Improper
x
S
dx
S  1 1  s
b1 I(b) =
xs dx
J:
=
{ log b) Therefore
to a
tends
I(b)
if and
limit
finite
only if s

00
s
x
dx =
11
of this integral
behavior
The
S: sin x dx
The integral
2.
EXAMPLE
I(b) =
Infinite
to show
to be the
1
.)
the linlit
case
which
is)
for
of the serIes
the
zetafunction,
diverge
3.
EXAMPLE
that
the
The
integral
Hence
e
S:
a1xl
dx
e a1xl
dx
2ja. Note, in
the
simplest
of
e a1xl
however,
case these
ealx/
dx also that
similarly defined. integral
Also, if
S\037oo f{x)
dx and
S\037oo f(x)
dx is
convergent,
dx + f') f(x) dx .)
S\037oo f(x) unimportant.) The integral on the right of (10.62) is divergent.)
ax
if a
dx converges
dx
=
e

ab
1
+
a
0
dx =
o f b
eax
dx
integral
> 0, for 1
=

eS\037oo
dx is
if b
> 0, we
b
00 .)
as
+
said to
have)
a)
value 1ja. Also, if
converges and has the the
the
that
t:J(x)
and has the
converges o
S\037oo
ef
f b
Therefore
=
o
00.)
c, we say
b
e a / x /
,)
of c is
choice
S\037oo
+
dx are
dx =
b
f
+
 cosb
sum)
one of the integrals
least
at
if
b
S\037oo f(x)
L:f(x)
is easy
As
1)
= 1
x dx
sin
for some
convergent
is defined
value
its
(10.62))
(It
s =
because)
diverges
JOb
as
limit
form
of the
integrals
S: f(x) dx are both and
a
does not tend to
this
and
if
1 ,) \302\245=
.
1
that
to
analogous
s
1.
s
0
dt +
the
I100
s, by
S:
integral
eft
Example
s 1
dt
4.
1 e
l
t
s1
1/X
dt  I 1
el/uu sl d
t
t
S
1 dt
converges.
This
.)
To test the
that)
Ix
e
u.)))
first
integral
we put
420)
infinite series,
Sequences,
But
e
f\037
1!u
u S1
du
t S 1
0
s >
for
converges
improper
with
comparison
by
integrals)
dt converges for s > O. When s > 0, the sum integral S\037+ e t r so defined is called the gamma function, first r(s). The function 1729. It has the interesting property that n + 1) = n! when r(n Exercise 19 of Section 10.24 for an outline of the proof.)) The
convergence
analogs
for
tests given
these tests for
formulating
S
1 duo
(10.65) introduced
is any
x
1. (00
V x4 + 1
Jo
oo
3 \302\267
fo
dx.
j
a certain
real
Determine
a certain real
C, the
Cx 2 (x +
C and evaluate
the
1
Determine
a certain real
C, the
1
2x + 1)
dx)
integral.
integral)
2 ( 2X
1 converges.
x)8
0:;
C the integral)
00
13 . For
x \302\267)
x
1
J2 converges.
log
00
[00
12. For
V\037
10.
dx.
f 0+ 'vx
11. For
.
Jo+
evx A
dx. X
dx
[1
'vex oo
x
f 00 cosh
9.
dx.
x dx.
Jo+ :og
1
5.
x
(1
8.
convergence.
dx.
J o q\037\037 + 'V x
oo
3 'V x + 1
.. \037
fo
6. (
dx.
7.
j
for
integral
1 A
oo
4.
improper
dx.
ex2
J:
test the
10,
C and evaluate
: 2C the

x \037 1 )
dx)
integral.
integral) 00
1 converges.
14. Find
Determine
the values of a and
C and evaluate b such
1
\037
1)
integral.
that)
oo
f
the
dx) X
\037 2X2
( VI
2X2 + (
X (2
bx +
X + a))))
a_
1
)
dx
_
is denoted by Euler
integer >
have
himself.)
In each of Exercises1 through
Therefore
the by in
0. (See
10.25 have straightforward
through
Exercises)
10.24
2.
in
second kind. The readershould
of the
integrals
improper
10.23
Theorems
in
u
S\037
1.
no difficulty
in
421)
Exercises)
15. For
constants a and b will
of the
values
what
p
X
1)_+00 f 1)
Prove
(a)
h dx

Do the
+
x
(f 1
ho+

,h(,
Prove that
(b) Prove that (c) Does the 18. (a)
If
the integral lim x _ o+ x
integral
integral S\037[(x)
J\037+
dx
proof
(b) Give an
example
of a
=
Each of
S\037
t S 1e t dt,
sres). Then
Exercises20 > 1.
all x
defined
for
integral
S\037[(x)
21. If lim
In
if s
is
22.
If the
23.
If [is
sequence {In} positive and
\302\267)
dx converges. = 1.
dt
00,
the
that
prove
integral test.]
> O.
which
(Thegamma
the
series Use
function.)
to prove
that
r(n
+ 1)
converges and the
L fen)
integration
= 11!if
by
n is a
parts
in
to show
positive integer.
a statement, not necessarily true, about a function [ a positive integer, and the exercises, n denotes In denotes assumed to exist. For each statement a either proof or give
25 contains
always
and
if lim n _ oo In exists, In = A, then
then the
integral
Sf .((x) dx converges then the integral J'; .r(x) dx converges. converges, if lim n_oc1n = A, then J? [(x) dx converges and and
0.
of these
decreasing
= 0
dx =
or diverge?) sin x dx
induction
use
through each
dx, which
x _ oo [(x)
sin x
Jh)
L:
nonmonotonic [for
provide a counterexample.
20. If.{ismonotonic
00
2
of the
Sf [(x) dx diverges. 
res + 1) =
2
(h
lim
or diverge? dt converge decreasing for all x > 1 and if [(x) + 0 as x + + and the series L .((n)both converge or both diverge.
Recall the
res)
that)
converge
x)/x
(sin
t)/t
(cos
[Hint:
19. Let
.
h+
S; (cost)/t
f\037+
monotonic
[is
tegral
dx 1)
)
integrals
improper
following
and
=0)
X)
JI (a)
+
+ x
1 dX
fl dx ; X)
17.
be equal to 1?)
that)
lim
(b)
2
limit exist and
following
ax 2 + bx
X3 +
lim
16.
the
lim
n _ oo
24. Assume.f'(x)exists for each x > 1 and for all x > 1. If limn_x In = A, then 25. If S\037 [(x) dx converges, then limx_.oc
suppose the [(x)
there is a
integral.f? = O.)))
constant
[(x) dx
M
S? and
has
> 0
[(x) dx has the
converges. value A.
the value
such
converges and
that
has
A. If'(x)
I
the value
integer N such that
n >
N
implies)  f(t)1
Ifn(t)
Hence, if
x E
 g(X) I
Ign(x)
so gn
\037
[a, b] and if
a corollary,
THEOREM
11.4.
f on an
Assume interval
\037vhere
b],