Calculus, Vol. 1: One-Variable Calculus, with an Introduction to Linear Algebra [2 ed.] 9780471000051

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Calculus)))

M.

Torn

Apostol)

VOLUME

One-Variable

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.)

67-14605

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 real-number 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

mean-value

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 least-upper-boundaxiom The Archimedean property

I 3.11Fundamental

of the

properties

(supremum)

real-number

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 well-ordering principle

An example

I 4.1

I 4.2 *1 4.3

I 4.4

of the well-ordering

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 intermediate-value

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

extreme-value

for continuous

theorem

small-span

theorem for continuous theorems for integrals of

functions

functions

The integrability

3.19 Mean-value

(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 mean-value

4.15

Exercises

Second-derivative 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 mean-value

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

zero-derivative

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 o-notation

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

first-order

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 First-orderlinear

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 first-order

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

second-order

equations

Remarks

339

nonlinear

concerning

8.21 Integral curves 8.22 Exercises First-order

8.23

direction

and

339

differential equations fields

341 344

345

equations

separable

8.24 Exercises 8.25

347

first-order

Homogeneous

347

equations

8.26 Exercises

8.27

Some

8.28

350 and

geometrical

Miscellaneous

physical

9.2

Definitions

The complex numbers

COMPLEX

first-order

and

field

an extension

9.4

The

Geometric interpretation.

9.6

Exercises

9.7

Complex exponentials

imaginary

Complex-valued

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

Power-series

*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 n-tuples 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 < a-I.

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 real-number 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 real-number 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. non-Archimedeangeometrieshave 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 intermediate-value 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

intermediate-value

of Bolzano's

consequence

illustrated

functions,

in

functions

continuous

theorem is the intermediate-valuetheorem

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

intermediate-value

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

intermediate-value

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

ordered-pair 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 nth-root

EXAMPLE.

preserved

of functions

Properties

n

a

is

Then f is strictly on every interval [a, b] with 0 < increasing is the nth-root 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

rth-power

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

nth-root

The

function. is

inverse

-

increasing

strictly

continuous

function.

function, independent of the is continuous, we again = where r is a m/n positive yr,

nth-root

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

nth-root

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 x-axis. 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 double-valued

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 small-span 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 Mean-valuetheorems

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

MEAN-VALUE

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, mediate-value 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

MEAN-VALUE

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

mean-value

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 geometry-the 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, (1630-1677).

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 x-axis.

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)

---+

nth-root

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 vl-nln. quotient

difference

f(x)

becomes)

_

un

let

l/n h)l/n _ x

+

(x

of continuity denominator

so the

a positive integer,

u-v)

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'0-mx

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) = -tx-3/2. In If f(x)

x >

valid

now shows

that

= !X-1/3 .

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

= x-l/

[(x)

the

+ X1/3

2

+

3 +

+ x-l/

>

x

Xl/4,

x-l/

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

x-I)

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 (1736-1813) 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 (1759-1803) 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.

l-2 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

Constant-sum,

principle. x and y

maximum-product

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

mlnlmum-sum

Constant-product,

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 x-axis, the base of the rectangle between the curve y = x/(x2 + 1) lying on the x-axisand the entire rectangle lying in the region and the x-axis. Find the maximum volume of the possible cylinder. 21. The lower right-hand 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 x-axis. 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 xy-plane. 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 xy-plane, 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

real-valuedfunction

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 z-axis);

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

-x-y)

(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

xy-plane. 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 z-coordinate 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 z-axis 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 xy-plane. by map represent a three-dimensionallandscapeby a two-dimensional We the surface that drawing. imagine described above has been cut by various horizontal to the planes (parallel xy-plane). They the surface at those points (x,y, z) whose z is constant. By projecting intersect elevation of contour these points on the xy-plane, 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

\"saddle-shaped\"

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 y-axes. 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 y-axis, Figure plane for which the y-coordinate 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)

h--O)

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 o-notation,

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 second-orderpartial

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 second-order 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 second-order 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

x-y) 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(n-k)(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\" x-I

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)

k-l

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

chapter-first 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

I-X dt -

=

1

- x) =

x

io

l-u 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)

,

l-u)

-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)))

,

1-u)

(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

e-x)

1

Now

u =

put

e-X 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.

x-I 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 + a-X) 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

.

e-x

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 = yxV-l 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 +2X-6

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 + e-a[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 x-axis, 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

La-I

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) = e-2x 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 x-axis, and W(t) the volume of the solid obtained by rotating 5(t) about the y-axis. (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 x-axis. 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

(1685-1731).


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{1-X

)/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

first-order

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 initital-value g(x) be the unique that 1 and k = 1 - n, prove I, with g(a) = b. If n \302\245:of the initial-value 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 initial-value 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 first-order

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 first-order

problems

physical

matically as

first-order 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 property-the 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 half-life 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 half-life 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 half-life 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 first-order 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

half-life

with

considered equation in

the

as a for

second-order

-

mg

differential

the velocity v. As

kv . the displacement for v, it is linear

for

equation

a first-order 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 first-order 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 - e-kt/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 first-order

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,

200e-kl

=

find

in

kl

equations)

eku

J;

- e-kt )

=

10 + =

log (90/190),

=

190e-kt

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 four-place

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

195e-kt .)

= 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, slide-rule

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'

+

-l-oY

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 first-order

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

first-order 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

e-RtlLi

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)e-Rt/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 steady-state 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 half-life of

1600years. Find

what

of a given

percentage

quantity

100

in

disintegrates

equation as a mathematical

differential

first-order

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)

c-v)

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)1-W(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. + 140e-kt , 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

e-R 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 e-Rt 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 ten-year 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

,)

e-a(t-fl)

+

(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 first-order linear we proved an existence-uniqueness theorem and deterequations, mined all solutions by an explicit formula. Although there is a correspondingexistencetheorem for the second-order 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 right-hand 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

second-order

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 .

e-CX(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

e-CX(Ae

+

Be-

Theorem 8.7 all hX

)

=

Ae(h-c)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(h-c)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(h-c)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 x-axis, 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 cross-sectionof 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 first-order

(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) =

2xe-x3f3)

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 second-order

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 de-ath, 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

real-number

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 present-day there is no real number

is

whose

-1.

standards.

Early in the 19th century, Karl Friedrich Gauss Hamilton (1805-1865) independently and almost simultaneously as ordered pairs (a, b) of real numbers numbers complex defining

(1777-1855)

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 con-zp/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 real-number 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 real-number 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 one-to-one

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 real-number an is called extension C The (x, 0). system system complex-number

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 ordered-pair 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,

complex-number

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 xy-plane is often referred in Figure 9.1. In this context, to the point (x, y), as shown is the imaginary axis. The x-axis is called the real axis; the y-axis 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 half-open 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

pseudo-ordering 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

pseudo-ordering

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 (1620-1687) 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 (1620-1684)

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

now-familiar

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

a2n-l =

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 1175-1250), 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

complex-valuedsequencef 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 real-valued the results derived in

(10.10)

to construct

7.14

If f is complex,we

real-valued.

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 e-1Tin / 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,

real-valued

- 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

\037--2-

\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+p-l Sp-l, 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 left-hand 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

I-x) 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

n-I k 2 (x

=

I-x

the behavior of Sn n -+ When x 0 as n -+ Ixl < 1, then n = Sincesn+1 x , sn convergence Ixl

by

1

=

I-x shows

= O.

x n I .)

+

=

Sn

n-I

-

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

ZETA-FUNCTION.

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

zeta-function

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 < Sn-l' 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

-

1-s

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)1-S

1- s

dx = { log

Thus

which

is)

(lOg n )1-S

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

v-xdx.

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

- e-n2

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)n-l

already

has the sum log (I + series. In particular, when

. . .

+

(_1)n-l 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)n-l

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 S2n-lform + + 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 {S2n-l}, being monotonic and bounded, converges say S2n ---+ S', and S2n-l ---+ 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

S2n-l)

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)

one-to-one 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

n--oo 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

-1-1+

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+i--t-!+t+t+l-I

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) =

x-s 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

zeta-function,

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

e-alx/

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

e-ft

Example

s- 1

dt

4.

1 e

-l

t

s-1

1/X

dt -- I 1

e-l/uu -s-l 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 -S-1

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+

e-vx 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.

e-x2

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

h-o+

-

,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

J-h)

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)

J-I (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],


0 since

B

(B \302\267 B)(A which

is (12.2).

= O. But one of the vectors

if C

. B)2(B. B) + 0,

\302\267

A)

This proof also shows that C

so we

- (A the

\302\267

B)2(B

(A

\302\267

B)2

equality

>

B)

0, we

may divide

0 if and only if xA = yB. This is a scalar multiple of the other.)))

=

\302\267 2xy(A

C \302\267 C >

inequality

7\"=

-

A)

\302\267 B) .)

get)

\302\267

B)

by

+ y2(B

(B

0

> .

.)

B) to

obtain)

0,)

sign holds

in

equation holds,in

(12.2) turn,

if and if and

only only if

of a

or norm

Length

453)

vector)

applications inequality has important Cauchy-Schwarz next.) we discuss norm of a vector, a conceptwhich

The length

or

12.6

Length or

of the

properties

a vector

of

norm

the

to

vector Figure 12.7 shows the geometric of the theorem From Pythagoras, plane.

from we

the origin to a point find that the length

=

A

(a l ,

A is

of

a2)

given

in the the

by

formula)

of

length

A

=

V

a\037 +

a\037 .)

A)

G3) A) \"

....... \"\" \"

\" \"

....... \" \"

G'l.)

o)

A

In

12.7

of

, the l ength

+

twice,

product In

in either

that of

with

A

In

12.8

FIGURE

of A

length

Note

....... \" \"

\" \"

\" \"

\

V 3

, the

length of

A

is V

ai +

a\037 +

.) a\037

.) a\037

12.8. picture in 3-space is shown in Figure of a geometric vector we find that the length

corresponding

Pythagoras

V2

is Vai

A

....... ....... \"

G2)

G))

FIGURE

\"

case the

A is

of

length

This formula

itself.

= V a\037 + given

(A

way

in

is given

3-space

of

by)

a\037 .

a\037 +

by

suggests a

the theorem

Applying A

\302\267

A)1/2,

to

the

introduce

square root the concept

of the dot of length

n-space.)

If A is

DEFINITION.

the

a vector in

Vn ,

its length

or

norm

is denoted

by

IIA

II

and

is defined

by

equation)

IIA II

The

properties of the

fundamental

(a) (b) (c)

If A is

12.4.

THEOREM IIA

II

>

=

II A II IlcA

II

=

0

if

0

if IclllA

II

a vector in A

7\"=

0

A

=

0,

=

(A

\302\267

dot product Vn

and if

A)l/2.)

lead

to corresponding

c is a scalar,}-,ve

(positivity), (homogeneity).)))

have

properties the folloa'ing

of norms.) properties:

Vector algebra)

454)

Properties (a) and (b) follow at once from properties (d) and (e) prove (c), we use the homogeneity property of dot productsto obtain)

Proof 12.2. To

IlcA

The

=

II

\302\267

(cA

the

in

inequality

equivalent

\302\267

BI

IA

use the Cauchy-Schwarz

we shall

TRIANGLE

12.5.

THEOREM

the

Icl

II .)

IIA

It states

of norms.

terms

that)

IIBI12.)

also write the

can

we

IIA

Cauchy-Schwarz

BII


then

(12.6)

(12.5),

then

IIAI/IIB\" o.)))

=

The triangle length

illustrated

is

inequality

side of a

of one

triangle

12.9. It states

in Figure

geometrically exceed the

not

does

455)

of vectors)

Orthogonality

sum of

the

of the

lengths

the

that

other two

sides.)

12.7

of

Orthogonality

In the course formula)

of

the

vectors of the

proof

(12.7))

IIA

+

triangle

BI12 =

2 11

IIA

+

B)

B)

---)--x

---

, I I I I I , I

--

the

\302\267

2A

+

IIBI12

we obtained

12.5),

(Theorem

inequality

A+B)

A+B)

II BII)

,'IIBII

I

\037\037

, I , , I I

\037\037)

A)

IIAII)

Geometric

12.9

FIG URE

the

of

meaning

vectors

inequality:

triangle

BII


II

12.9 Projections.

Angle

The dot product of

12.11(a)shows two example,

IIA II

two

nonzero

we have

V n , prove that there exist vectors C and D 23 and express C and D in terms of A and B.

B in

A and

vectors

two nonperpendicular the three in Vn satisfying

this

quadrilateral exceeds the length of the line segment

of any of the

sides square

y the vectors of 2A + length

satisfying

In

three vectors A, B, provides a proof of the

involving V n . This

length 6. A vector Bin Vn has the property that for every pair of scalars xA + yB and 4yA - 9xB are orthogonal. Compute the length of Band 3B. vectors A = (1, 2, 3,4,5) and B = (1, i, \037,!, !) in V s . Find two vectors C and D three conditions: C is parallelto A, D is orthogonal to A, and B = the following

A in

x and 23.

457)

n-space)

suggests a vector identity that it holds for vectors in

in geometry and prove

by four

diagonals

in

methods.

sum

\"The

22.

identity

by vector of the

theorem

the

theorem

following

and C.

vectors

between

Angle

Projections.

Exercise

statements

concerning

B, then IIA + xB\" > IIA II for all real A is orthogonal for all real x, then to

vectors

in

0


is

zero volume.)

B, C

parallele-

the product

and

negative

are on a plane through product is zero. In this case,

If A,

their scalar triple

and

=

cos

BII (IICII

by A, B, C. When negative of the volume.

dependent

linearly

degenerates

algebra to

base times the altitude.

of the

area

the 4\302\273,

IIA

are

of vector

Applications

the origin, they the parallelepiped

AxB)

=

Altitude

II

ell

Volume =

cos (jJ-)

=

Geometric interpretation

FIGURE 13.6

of

the scalar

triple

as

product

II

A

x B . e)

A xB11)

the

volume

of

a parallelepiped.)

This

geometric

properties the

leaves

scalar

product unchanged.

(13.9))

A

An algebraic proof of this that the dot and implies combined

the

of

commutativity

with the

C =

B.

X

property

cross are

B

X

A =

C'

C

X

vectors

algebraic

A, B,

A.

B.)

in Exercise 7 of Section 13.14.This in a scalar triple product. In (B X C) and when implies (B X C). A = A \302\267

property

fact, the

we

find

C

that)

is outlined

in (13.9),

equation

we mean

this

By

the

of

interchangeable

dot product

first

suggests certain

For example, a cyclic permutation

product.

triple

scalar triple product

the

of

interpretation

of this

this

is

that)

AxB'C=A.BxC.)

(13.10))

A \302\267 B X C is often denoted by The scalar triple the [ABC] without indiproduct symbol dot or cross. Because of Equation in this notationcating the (13.10), there is no ambiguity the product depends only on the order of the factors A, B, C and not on the positions of the

and cross.)

dot

13.13 Cramer's The

scalar

equations

(13.11))

rule

for

solving

triple product

in three

unknowns

may

x,

of three

a system be

y, z.

linear

equations

a system of Suppose the system is written

used

to solve

a1x + b1y +

clZ

a 2x +

b 2y +

c 2Z = d 2 ,

a3 x

b 3y

+

+ c3 Z

=

=

d 1 ,)

d 3 .)))

simultaneous

three

in

the

form

linear

491)

Exercises)

Let A be the three

vector

equations

in

with

are equivalent xA +

(13.12))

D

Then

similarly.

the

equation)

.)

B x

with

equation

C,

for

[ABC]

writing

A .

B x

C,

[BBC]

= 0, the

= [CBC]

+ z[CBC]= [DBC].)

+ y[BBC]

x[ABC] Since

zC =

yB +

D

B, C, and

that)

find

we

of this

multiply both sides

we dot

If

, a 3 and define to the single vector

a I , a2

components I 1)

(13.

of y

and z

if

[ABC]

coefficients

x= [DBC]

(13.13))

drop =;f

we obtain)

and

out

0 .)

[ABC])

similar

A

(13.14))

[ADC]

=

Y

In

case,

if

[ABC]

=;f

that the three vectors A, every vector D in 3-space

(13.12)

B,

C are

is spanned

determined x, y, z are uniquely by the formulas in (13.13) multipliers occur in these formulas are written the scalar triple products that rule for solving the system (13.11):) as Cramer's result is known

If

= [ABC] unless

infinitely

so there

0,

+

dl

CI

al

bi

dl

d2

b2

C2

a

2

d2

C2

a

b2

d2

d 3

b3

C3

a 3

d3

C3

a3

b3

d3

a l bi

CI

al

bi

CI

al

bI

CI

a

Y=

z=

b2

C2

a

2

b2

C2

a

2

b2

C2

a3

b3

C3

a 3

b3

C3

a 3

b3

C 3)

then

D lies

(y +

so

same

the

of the M,'

tv)B

+

In

plane.

the triple (z + xA

this

In fact,

system.

all zero

not

does

on a plane

C lie

B,

A, in

such

(x +

that tu,

(b) A (c)

A

the

scalar

(3, 0,0),

=(2,3,

= (2,1,3),

-1),

triple

product

latter

origin

B =

and

it is

case,

t}v)C + yB

B = (0,4, B =(3,

the

the system has no to show that there are easy the vectors A, B, C are linearly dependent uA + vB + }vC = O. If the (x, y, z) triple y + tv, z + tH') for all real t, since we have) the

through

+ zC

+

t(uA

+

vB +

Exercises)

1. Compute (a) A =

2

2

= 13.14

determinants,

al

many solutions scalars exist u, v,

tu)A +

as

CI

satisfies (13.12),then (x

linearly independent. the by A, B, C and and (13.14). When

dl bi

x=

solution

0 .)

[ABC])

[A BC] =;f 0 means shows that

condition this

we have)

Thus

z.

and

z= [ABD]

and)

[ABC]) The

for y

yields analogous formulas

argument

A .B 0),

-7,5),

(-3,0,6),

x C in

each

case.

C =

(0, 0, 8). C =(1, -5,2).

C = (4,5,-1).)))

}vC) =

xA

+ yB

+ zC

.)

492)

Find all real 3. Compute the 4. Prove that A 2.

5. Prove 6. (a)

= A (A x i)

vectors

all

of the

x B

geometry)

vectors (1, t, 1), (t, 1, 0), (0, 1, t) are linearly parallelepiped determined by the vectors i + j, j x i)i + A . (B x j)j + A . (B x k) k. x (A x j) + k x (A x k) = 2A.

the three

which

volume

i x

that Find

for

t

algebra to analytic

of vector

Applications

.

(B

+.i

ai + bj

+ ck which

(ai + bj

the relation)

satisfy

+ ck) . k

dependent. k + i.

+ k,

+ 4k) =

+ 3j

x (6;

3 .)

in (a). (b) Find that vector a; + bj + ck of shortest length which satisfies the relation of of the dot and cross products to derive the 7. Use algebraicproperties properties following the scalar triple product. . (a) (A + B) (A + B) x C = o. . B x C = - B . A x C. This shows that A the first two vectors reversesthe (b) switching [Hint:

sign.

(c) A reverses

(d)

.B x .

A

the sign. B x C =

the sign.

(a) and

Use part

C = -A . C x B.

[Hint:

[Hin

laws.]

that

shows

This

Use skew-symmetry.] x A. This shows that

t :

- C.B

Use (b) and

the first

switching

of (b), (c),and

we find

(d),

xB

that a cyclic permutation of A, B, C leaves their a proof of the vector identity)

shows

and

third

reverses

vectors

third

vectors

that)

A'BxC=B.CxA=C'A which

the second and

switching

(c).]

the right members

Equating

distributive

,)

scalar triple

product

unchanged.

exercise outlines

9. This

A x

(13.15)) sometimes

referred

and prove

that)

to as

the

= (C . A)B

(B x C) minus

\"cab

i x

-

(B

Let B = (bl , b2

bac\" formula.

(B x

. A)C ,)

C) = clB -

, b 3),

C =

(CI, c2

, C3)

biC.)

for A = j and (13.15) in the special case A =;. Prove correspondingfornlulas then combine them to obtain (13.15). 10. Use the \"cab minus bac\" formula of Exercise9 to derive the following vector identities. (a) (A x B) x (C x D) = (A x B . D)C - (A x B. C)D. (b) A x (B x C) + B x (C x A) + C x (A x B) = O. (c) A x (B x C) = (A x B) x C if and only if B x (C x A) = O. . . . . . D). (d) (A x B) (C x D) = (B D)(A C) - (B C)(A 11. Four vectors A, B, C, D in V 3 satisfy the relations A xC, B = 5, A x D . B = 3, C + D = of i, j, k. ; + 2j + k, C - D = i - k. Compute (A x B) x (C x D) in terms 12. Prove that (A x B) . (B x C) x (C x A) = (A . B x C)2. 13. Prove or disprove the formula A x [A x (A x B)] . C = - IIA 112 A . B x C. are A, B, C, D is) whose 14. (a) Prove that the volume of the tetrahedron vertices This A

proves k, and

=

! I(B (b)

15. (a)

this volume Compute B \037 C, prove that

If

Compute

this distance

A)

.

(C

-

A)

x

(D

-

.)

A)I

D = (4, 0, 0). A = (1, 1, 1), B = (0,0, 2), C = (0,3,0), and when the perpendicular distance from A to the line through Band C is)

- B) x

II(A

(b)

-

when

A

= (1,

(C

-

B)II/IIB

-2, -5), B

-

CII

.)

= (-1,1,1),

and

C

= (4,5,1).)))

493)

vectors to planes)

Normal

16. Heron'sform

ula for computing the area 5 of a triangle whose sides have lengths 0, b, estates a exercise outlines 5 = \\/ s(s - a)(s - b)(s - c), where s = (a + b + c)/2. This vectorial proof of this formula. = a, IIBII = b, liB - A II = c. Assume the triangle has v\037rtices at 0, A, and B, with IIA \" two identities) Combine the (a)

that

45 2

a 2b 2 -

=

!(c

the formula

Rewrite

(b)

52 = and

2

a2

-

in

part

- (A

. B)2,)

rule to

Cramer's

- b2 )2

= !(2ab

(a) to

obtain)

-2A

. B

- c2

+

-

IIA

18.

solve the

of equations

system

satisfies

the

of linear

system

to

vectors

Normal

+ a

b)(c

each

in

in

addition,

N is

equations x

- y +z

3z =

+

5y

each

point 1, x +

=

z) on the = 5, 3x 3z + y

and

B is

normal

to a

THEOREM

= N .B

(b) M is the

plane,so is tN

of all X

for

proves

M is

,a) since A

x

{P + fA}

+ 7z

+ y

= 11.)

sA

tB}, where A be described

+

V 3 can

real

f

so a

linear

the

in

vector

span of

A

and

A

by

A and

B.

and in

B.

If,

B an

A in

to perpendicular and B. Also, if

\037 O.)

+

tB} through

P spanned by

and

A

B.

to M. in

the equation)

V 3 satisfying

a plane,

B is

= 0,

+ tB)

(sA

= {P+ sA

(X Since

19.

the following:

(13.16)) Proof

N.

every

plane M

\037vehave

vector

set

= 0, then

perpendicular to every vector

Then

.)

planes

Given a

13.15.

Let N = A X B. (a) N is a normal

b 2)

line

(x, y,

P spanned {P + sA + tB} be the plane through be perpendicularto M if N is perpendicular to both nonzero, then N is called a norlnal vector to the plane. If N. A

-

3.

said to

V 3 is

Note:

- a2

17, 18, and

M =

Let

DEFINITION.

vector N

N is

IIBI12)

- b) ,)

of Exercises

A plane was defined in Section 13.6 as a set of the form {P + are linearly independent vectors. Now we show that planes in a vector.) different the of normal way, using entirely concept

A

-

IIAI12

+ c2

a 2 + b 2 )(2ab

-y+z=3.

= 4, x + 2x + 3x Y z = 2, y + 2z z = 5. 19. x + Y = 5, x + z = 2, Y + 20. If P = (1,1,1)and A = (2,1, -1), prove that

both

-

BI12

formula.)

2x-y+4z=II, - -

17.x+2y+3z=5,

13.15

=

- c)(c- a +

c)(a + b

b +

+

l-6(a

deduce Heron's

thereby Use

11211BI12

IIA

the formula)

obtain

to

=

X BI12

IIA

A

and

orthogonal

P)

B are

to both

. N

= 0

.)

linearly independent, A

and

B.)))

so

A

x

B

=;f

O.

This

494)

To

which B,

A,

(b), let

prove

- P is

X

M' be the

e M'.

V 3 satisfying

in

- P is

X

B, so

and

A

proves

that M

linearly independent (Theorem 13.13),they

we

some

scalars

P=

(13.16). If X EM, then Therefore X E M' Then X satisfies (13.16).Since to N.

in

vector

every

span

V 3

so,

in

have)

sA

s,

t,

+

tB.

u.

=

- P

X

so X -

M'.

XE

suppose

Conversely,

geometry)

Equation orthogonal

N are

particular,

for

X

of all

set

span of

linear

the

in

algebra to analytic

of vector

Applications

uN)

product of each member X E M. Hence, M' c

the dot

Taking

This shows

tB +

+

sA

that

with

M,

N,

we find u =

which

0, the

completes

of (b).

proof

The geometric are on the plane

13.15 is shown

of Theorem

meaning

and the

in

normal vector N is orthogonal

The points

13.7.

Figure

- P.

X

to

This

P and X the

suggests

figure

following theorem.) 13.16.

THEOREM

to M,

M

a plane

Given

P, and given

a point

through

a nonzerovector

N normal)

let)

d =

(13.17))

IP . NI

.

IINII)

Then

X on

every

M has length

projection of P along

= tN

x

Proof as

way

the

By

on M the

same

13.16

argument

Cartesian

Linear

write

N

=

a Cartesian

points (x, y, z) each we multiply We may

(13.19))

vector

=

II XII

d if and

only

if X

is the

the

same

which

of a,

lie

P'N

N.N)

.

is the distance

-

to the

origin

13.16 can also be expressedin and X = (x, y, z), Equation

bey

equation for

the

Xl)

the plane.

b, c by

the

from

plane.)

of components. becomes) (13.16)

terms

, Zl),

+

on

on M, then among all points X - Q is the projection of P - Q along and is called the distancefrom Q to the

for planes and

13.15

exactly

in V2 .)

lines

X

NI/IINII

Q)

in

inequality

for

is a point not when

occurs

c), P = (Xl'YI a(x

is called

of normal

-

(13.17)

Theorems

(a, b,

if Q

that

find

equations

(13.18)) This

we II X

d in

t=

where)

,)

Q II is length I(P -

number

The results of If we

Theorem

smallest length

The

plane.

have

M,'e

from the Cauchy-Schwarz 13.6, the corresponding result

follows

proof

minimum

This

N.

The

we proved

d. Moreover,

>

II XII

N:)

-

YI)

+

scalar

-

Zl)

0

.)

satisfying

points

t. This

=

satisfied

and it is

plane,

The set of

a nonzero

c(z

simply

amounts

and only those is not altered if (13.18) to a different choice

by

those

in (13.16).

transpose the

terms

not

x, y, and z, and

involving

ax +

by

+

cz =

d l ,)))

write

(13.18)

in the

form)

Cartesian

Linear

where dl =

aX

bYl +

l +

We have just shown (13.19) in which not

that

CZ

of

An

l .

equation point (x, y,

every

this

495)

for planes)

equations

to be linear

is said

type

z) on a plane satisfies

in

and z.

x, y,

Cartesian

a linear

equation

of a, b, C are zero. Conversely, every linear equation with this a plane. (The readermay verify this as an exercise.) property represents d of the plane to the distance The number d l in Equation (13.19) bears a simple relation In particular Idll = d from the origin. Since d l = p. N, we have Idll = IP' NI = diINII. if d l = O.) if and only 1. The plane passes through if the normal N has length the origin all three

z)

N)

(0,

1.

0)) y)

x)

X

Cartesian

The

N =

2i +

P and

through

vector

normal

with

EXAMPLE.

vector

A plane

13.7

FIGURE

FIGURE

A plane with 3, 1, 2.)

13.8

N.)

2x + 6y

equation

6j + 3k. We

=

3z

Cartesian

the

rewrite

+

a plane

6 represents

equation

in

intercepts

with

normal

points

(3, 0,

form)

the

\037+[+\037=1

312)

it is

which

from

(0, 1, 0), and interceptsof quickly.

d=

A

two

intersects the coordinate axes at 3, 1, 2 are called,respectively, knowledge of the intercepts makes it possible

only

plane

plane is shown

in

13.8.

Figure

Its distance

the the

to

0),

and zsketch the plane X-, y-,

d from the origin is

6/7.)

planes

parallel

have

will

difference

the perpendicular

being

a common

can be

planes

ax +

the

the

that

The numbers

A plane. of the portion

Two parallel

tions of

0, 2).

the

=

6/IINII

apparent

(0,

in

by

the

distance between

+

as

written

cz =

normal N.

d l ,)

right-hand the two

If N = (a, b,

c),

the

Cartesian

equa-

follows:)

ax

+

members. planes,

a

by +

cz =

d2 ,)

The number Idl definition

suggested

- d2 1/ II Nil by Theorem

is called

13.16.)))

496)

vector

of

Applications

Two planes are called perpendicularif More generally, if the normals of that

13.17

Exercises

is an

()

1. Given vectorsA

(a)

=

(b) Give a Cartesian

(c) Give

two

two

each

with

()

of the

other,

then

planes.)

- 4k and

3j

for the equation x +

a Cartesian

perpendicular to a normal an angle

make

planes

B = j + k. vector N perpendicular to both for the plane through equation

2; +

a nonzero

Find

between the

angle

of one is

a normal

other.

we say

to analytic geometry)

algebra

equation

through

plane

B.

A and the

spanned

origin

(1, 2,

3) spanned by

by A and B. A and B.

- 2z + 7 = O. Find has Cartesian the following: plane 2y (a) a normal vector of unit length; (b) the intercepts of the plane; from the origin; (c) the distance of the plane (d) the point Q on the plane nearest the origin. 3. Find a Cartesian equation of the plane which passes through (1, 2, -3) and is parallel to = 4. What is the 3x 2z distance between the two planes? + plane y given by 4. Four planes have Cartesian equations x + 2y - 2z = 5, 3x - 6y + 3z = 2, 2x + y + 2z + z = 7. -1, and x

2.

A

the

=

2y

(a) (b)

three

from

the

(1,

points

the

to

normal

1, -1),

(b) a

plane;

7. Determine an line

(1, 2,

angle

parallel

(3, 3, 2), and (3, Cartesianequation

other

two are

perpendicular.

planes. - 2) determine -1, for the plane;

3) and

the plane determined the planes with Cartesian

(2,

perpendicular

10. A point moves (1 - t); + (2

equation

(1,

2, 3),

equations

the plane given

plane this

through

(2, 3,

(2, 3, 4), and

x + y = 1 and to a plane M if

- 7), given

point

( -1, y + N is

the line

that

the line which contains the point - 3 + z = 5. by Y a way that at time t its position is given by

(2, 1,

7, -2). z = 2. normal

through

- 3) and is

4x

X(I) =

vector

the

l)k.

2z + 1 = O? part

(c) which

contains

X(3).

(e) Find a

11. Find

of the

plane.

in space in such - 3t)j + (2t that the point moves along a line. Prove it L.) (a) (Call (b) Find a vector N parallel to L. (c) At what time does the point strike the plane given by 2x + 3y + a Cartesian equation for that to the one in (d) Find plane parallel the

(a) a vector plane

a plane. Find the distance

(c)

for

equation

parametric

to

by

vector N is said to be perpendicular

for the 4, 12) is perpendicular to

a Cartesian

Find

a vector

9. Find

the

for

between

to a nonzero

parallel

to M.

two

origin.

a Cartesian equation

6. Find

8. A

them are parallel and between the

distance

the

Find

5. The

two of

that

Show

Cartesian equation

for

plane perpendicular to L which (1, 1, 1) if a normal plane through

that

contains

the point

X(2).

makes angles equation . \" h 1,], k , I \"37T, 1f7T, \"37T, respectIvey. 12. Compute the volume of the tetrahedron whose vertices are at the origin and at the points where the coordinate axes intersect the plane given by x + 2Y + 3z = 6. 1 perpendicular 13. Find a vector A of length to the plane with to ; + 2j - 3k and parallel = 1. Cartesian x 5z + y equation 14. Find a Cartesian equation of the plane which is parallel to both vectors; + j and j + k and intersects the x-axis at (2, 0, 0). = 5, 15. Find all points which lie on the intersection of the three planes given by 3x + y + z 3x + y + 5z = 7, x - y + 3z = 3.

111

16. Prove one

for

a Cartesian

. WIt

that

point.)))

three

planes

the

whose normals

are

linearly

independent

vector N

intersect

in one

and

only

17. A

(1,

containing

4z =

5.

18. Given

a

one

vector

direction

with

line

each

parametric equation for this parallel to a plane M, prove that

a vector

Find line

to be parallel to a plane of the planes given

is said

A

2, 3) is parallelto

L not

M by

if A

x +

2Y

is parallel to M. A line + 3z = 4, 2x + 3 Y +

line.

intersection

the

L n

M contains exactly

with

Cartesian

point.

19. (a) Prove that the distance ax + by + cz + d = 0 is)

from the

(x o , Yo , zo)

point

lax o + byo

2 2 (a + b (b) point

20. Find

497)

sections)

conic

The

Find the point Q = ( -2,

15, -7).

a Cartesian

equation

point (3, 2,

-1)

if the

P

21. (a) If three points A, plane is I(Q A)' (B

on the

is

plane

for

the

equidistant

given

plane

from

by

+

cZ o + dl

+

2 C )1/2)

5x

parallel

to the plane

-

14y +

2z + 9

to the plane

given

=

0 which

by

2x

-

equation

is nearest to y +

2z + 4

the

=

0

both planes.

that the distance from a point determine a plane,prove Q to this x x A) A) (C A)I/II(B (C A)II. this distance if Q = (1, 0, 0), A = (0, 1, 1), B = (1, -1, 1), and C = (2, 3, 4). (b) Compute 22. Prove that if two planes M and M' are not parallel, their intersection M n M' is a line. 23. Find a Cartesian equation which for the plane which is parallel to j and passes through the intersection x + 2Y + 3z = 4, and 2x + y + z = 2. of the planes described by the equations 24. Find a Cartesian equation for the plane parallel to the vector 3; - j + 2k if it contains every = 3 and 2y + 3z = 4.) on the line the with of intersection of point planes equations x + y

13.18

The

conic

B, C -

sections)

A at a given point P, making G which intersects a fixed line a constant called a right circular A, where 0 < () < !7T, generates a surface in 3-space cone. The line G is called a generator of the cone, A is its axis, and P its vertex. Each of the cones shown in Figure 13.9 has a vertical axis. The upper and lower portions of the cone at are The curves obtained by slicing the vertex called of the cone. the meeting nappes cone with a plane not passing through the vertex are called conic sections,or simply conics. If the cutting plane is paraIlel to a line of the cone through the vertex, the conic is called a) A

angle

moving

line

() with

\\ Hyperbola

Parabola

/)

/)

FIGURE

13.9

The

conic sections.)))

498)

Applications of

Otherwise the intersection is called an just one or both nappes. (See Figure

parabola. cuts

plane

vector algebra to

\"branches,\" one on

each

geometry)

analytic

ellipse 13.9.)

or

a hyperbola, The

hyperbola

according as the consists of two

nappe.

and applied mathematics have been related pure of conicsas early as the 3rd century B.C.\",'as treatment Appolonius' Greek one of the most profound achievementsof classical Nearly 2000 years geometry. that a projectile fired horizontally from the Galileo discovered later, top of a tower falls to earth along a parabolic path (if air resistance is neglected and if the motion takes place above of the earth that can be regarded as a flat plane). One of the turning a part in points 1600 when the history of astronomy occurred around that all Kepler suggested planets was able to demonstrate that Newton move in elliptical orbits. Some 80 years later, an law of an attraction. This led gravitational elliptical path implies inverse-square planetary Newton to formulate his famous theory of universal which has often been gravitation not only as referred to as the greatest scientific discovery ever made. Conicsections appear orbits of planets and but also as trajectories of elementary atomic satellites particles. They are used in the design of lenses and mirrors, and in architecture. These examples and many of the conic sectionscan hardly be overestimated.) others show that the importance Many

to

important

conic

the

discoveries

in

both

sections.

of the conic sections. One of these refers to special are other equivalent definitions An be as the set of all points in a known defined may ellipse points asfoci (singular:focus). d d from two and fixed the sum of whose distances 2 l points F 1 and F2 (the foci) is) plane

There

I Directrix I I I)

d 2)

d 1 + d 2 = constant

Id 1

- d 2 = constant 1

(hyperbola))

(ellipse))

d) = d2 (parabola))

FIGURE 13.1 0

Focal definitions

of the

conic sections.)

constant. (See Figure If the foci coincide, the ellipse reducesto a circle. A hyper13.10.) bola is the set of all points for which the difference Id1 - d2 1 is constant. A parabola is the set of all points in a plane for which the distance to a fixed point F (called the focus) is to the distance to a line the directrix). given (called equal There is a very simple and elegant argument which shows that the focal property of an a is of its of a definition as a section cone. This which we may consequence ellipse proof, in 1822 a refer to as the \"ice-cream-cone was discovered mathematician, Belgian by proof,\" and makes use of the two G. P. Dandelin are drawn SI and S2 which (1794-1847), spheres in so as to be tangent to the cutting the as and illustrated 13.11. These cone, Figure plane the shall touch cone two circles C and C . We that the 2 along 1 spheres prove parallel points F I and F 2 , where the spheres contact the plane, can serve as foci of the ellipse.)))

The

499)

sections)

conic

Sphere 5.)

Circle

C2 52)

Sphere

The ice-cream-coneproof.)

FIGURE 13.11

P be

Let

an

is constant,

that is,

cone

the

and

-+

liP

from

C2,

\037

IIPF 2 11

\037

+

2

this

its

be

IIPF 2 11 =

-+

liP Alii

+

\037

prove

purpose,

intersections

two tangents to Sl therefore we have)

-+

\037

\037

A

are

PAl

IIPA 2 11, and

IIPFll1

For

of P. and

\037

Then PFI and respectively. --+-+ =

The problem is to

ellipse.

independent of the choice 0 to P and let Al

vertex

Similarly

Alii.

of the

point

arbitrary

\037

that

IIPF111 +

draw

that line

with

from P, and

IIPF 2 11

on the

the circles -+

hence

IIPFll1

Cl -

-+ IIPA 2 11 \302\267)

\037

C l and circles lA211, which is the distance between the parallel the surface of the cone. This C2 measured that Fl and F 2 can serve as foci of along proves the ellipse, as asserted. Modifications work also for the hyperbola of this proof and the parabola. In the case of the hyperbola, the proof of the cone. F or the))) employs one sphere in each portion But

liP Alii

+

liP

A 2 11 =

II A

500)

tangent to the

parabola

one

the cone

along a circle which

sphere

directrix of the

properties of a cone.) 13.19

lies

With

parabola. the

of

cutting

focus F is used.

at the

plane

geometry)

plane whose intersectionwith these hints the reader should be able in a

may be deducedfrom

and parabola

hyperbola

This

touches

sphere

is the

the

cutting plane to show that the

focal

as sections

definitions

their

sections

of conic

Eccentricity

algebra to analytic

of vector

Applications

characteristic of conic sections involves a conceptcalled property eccentricity. section can be defined as a curve traced out by a point moving in a plane in such a way that the ratio of its distances from a fixed point and a fixed line is constant. This constant ratio is called the eccentricity of the curve and is denoted bye. (This should not be confused with the Euler number is The curve an ellipse if 0 < e < 1, a parabola if e.) e = 1, and a hyperbola if e > 1. The fixed point is called a focus and the fixed line a Another

A conic

directrix.

definition as the basis for treatment of all three types it is understood discussion

this adopt a simultaneous

shall

We

permits

methods.

vector

In

this

our of

conic sections since it

and lends itself to the and lines are in the points

all

that

of the

study conics

of

use

same

plane.)

distance

the

a point X

from

N is

L is given

eccentricity if e > 1.)

L

by

=

FII

L,

and a positive all X satisfying

e d(X,

Let

e.

number

d(X, L)

relation)

the

L))

conicis called

is any point on

if P

and

on

set of

e. The

with

section

hyperbola

a vector normal to

X to

point

-

II X

is calleda conic if e = 1,and a

not

The

L.

to

(13.20))

If

F

L, a point

a line

Given

DEFINITION.

denote

an

ellipse

< 1, a parabola

if e

L the distance d(X,L) from

any

formula

the

d(X , L)

= I(X

. P)

NI

.)

II Nil

N has

When

(13.20) for

length 1,

the

conic

this

L separates the

positive half-plane, and vector

L itself

line

N dictates

left are in

the

that

negative

plane to the

according

\"negative\"

On the

-

II X

line

The

if

(X

we have points

= I(X -

=

- P). NI.)

P) .

NI,

and

basic

the

equation

becomes)

sections

(13.21))

and

d(X, L)

to

simplifies

FII

two

into

parts

choice of

- P)

(X -

to the

\302\267

P)

N

e I(X


0, we say that is in the negative

13.12 the choice of

positive

half-plane

half-plane. the

normai

and those to the

half-plane.

as indicated in Now we place the focus F in the negative half-plane, P - F = dN, where choose P to be that point on L nearest to F. Then

13.12,

Figure Idl

=

liP

-

and FII

is)))

Polar

conic sections)

for

equations

501)

Directrix

L)

x \",

IIX-FII/// /

N unit

/1 I I

I

(X

- F)

---

---

\",\037_________L______

F

L)

:

/

/'\" \", Focus

normal to

.N

d

-

- F)

(X

+

P = F

+

dN)

an

X

.N

d)

FIGURE

A conic

13.12

section with IIX

from

distance

the

(F -

P).

N

=

the focus to the

< 0, so d is

-d

following theorem,

which

-

=

FII

e is

eccentricity

- F)' N

e I(X

directrix. SinceF is

satisfying

dl.)

the

in

negative

+ dN

in

we have the

half-plane, (13.21),

we obtain

13.12.)

in Figure

is illustrated

the set of

P by F

Replacing

positive.

-

13.17. Let C be a conic section }fith L e, focus F, and directrix eccentricity distance d from F. If N is a unit normal to L and if F is in the negative half-plane deterthe equation) by N, then C consists of all points X satisfying

THEOREM

at a mined

II X

( 13.22))

Polar

13.20

The

equation

position.

equations in

for conic

Theorem

For example, if the

(13.23))

-

=

FII

- F) \302\267 N -

e I(X

13.17 can

be

simplified

focus is at

the

origin

II XII

if we place the focus the equation becomes)

and

r = e

to the left of the directrix, we becomes r = e(d - r cos()), (13.24)

X lies

(13.25))

r=

in

a

special

=eIX'N-dl.)

especiallyuseful if we wish to express X in the directrix L to be vertical, as shown in Figure 13.13, rand (), we have II XII = r, X. N = r cos (), and ordinates

If

.)

sections)

This form is

(13.24))

dl

Ir

()

cos

dl

solving

ed

e cos

()

+

of polar

1)))

coordinates.

= i. If

Equation

X has

polar

Take co-

(13.23) becomes)

.)

< d, so Ir cos for r, we obtain)

r cos ()

have or,

-

terms

and let N

()

- dl =

d - r cos()

502)

lies to the

If X

the

of

right

algebra to

of vector

Applications

cos (j

we have r

directrix,

r =

e(r cos (j

-

geometry)

analytic

d)

(13.24) becomes)

d, so

>

,)

. . us)

gIvIng

ed

r=

(13.26))

r >

Since

0,

the

is

illustrated

last

this

directrix

of

-

(j

1)

1. In other Thus, we have

e >

implies

equation

e cos

only for the hyperbola. in Figure 13.13.)

the following

proved

F)

d)

(a) r

cos

(j

left

and

FIGURE

13.18.

THEOREM

a vertical

with

sections

Conic

Let

the

with

directrix L at

curve

is a

the

right

on C

lies to

the

left

Polar

with

right

equations

set of

corresponding

exercises.)))

e,

eccentricity the

of F. of L

}-vith

The focus F

F at

afocus

If 0 < e < 1, the

and satisfies

the

origin,

C is

conic

the

polar

on

the left

equation)

ed

e cos (j

+

1)

on each

a branch

branch

r=

(13.28))

with

of

branch

right

r = e Ir cos () - dl. of the directrix.)

d to

point

hyperbola

satisfy (13.27) and points on

left

the

a distance

r= the

equation

polar

a conicsection

(13.27))

1,

(b) r

and lies to

origin

C be

an ellipse or a parabola;every

e >

cos (j > d on the the hyperbola)

d on the ellipse,parabola, branch of the hyperbola)


1, we

let

b =

(c, 0) and lalVe

2

(-c,

-

< 1)and

the

(ae, 0) and

(e >

hyperbola

( -ae,

x2

y2

a

b

0); the

I) and

directrices

write

the

standard

form)

2)

0), where

1 and

of the ellipse in

the equation

-+-=1. 2

Its

=1.

-ale. write

(13.36))

If e

_ e2 ))

represents both the ellipse (e form. The foci are at the points

equation in standard

the vertical

If e

2 a (1

a

Cartesian

This

y2

-+2

(13.35))

c = ae

the

= Va2

-

of the

equation

b2 .

An

example

hyperbola

in

the

is shown standard

form)

x2 - -

(13.37))

Its

foci

shown

a2 are at the in

Figure

points (c, 0) and 13.14(b).)))

(-c,

y2 --1 2

.)

b

0), where

c=

lal

e =

V a 2 + b 2 . An

example

is

506)

of vector

Applications

Note:

of x

y -in terms

for

Solving

in

algebra to analytic geometry) obtain

, we

(13.37)

two

solutions)

b ( 13.38))

Y x2-a2.)

y= ::I:\037

For large positive x, the (13.38) is nearly :l:bxllal.

- a2 /1al

2 by x

=

Y2

-

=

Y2

It is

approaches

b

-

Yl

(x

-

A / -V

x2

- a2 is nearly easy to prove that the 0 as x -+ + 00. This 2 Y x

number

- a 2)

Yl

-

Y2

-+ 0

The

hyperbola.

as x -+ + 00. Therefore, line y = -bxllal is another

- (x2

the

between

difference

is)

a 2)

2

difference

-a

Ia I

=

2

line y =

bxllal

The

asymptote.

member

right =

Yl

b

V x2-a

x+

The asymptotes are shown

lines asymptotically.

these

proach

x2

lalx+yx

lal

so

= -b

so the

equal to x,

2


0 the

=

+ y2

C)2

4cx.)

Y

: I)

C

(13.20)

Equation

equation (13.20) focus at (c, 0).

the

directrix (the origin in Figure 13.15) is cal1ed the the vertex and focus is the axis of the passing through is symmetric about its axis. If c > 0, the parabola lies to the right 13.15. When c < 0, the curve lies to the left of the y-axis.)

parabola

y-axis, as in

- c and

x =

form)

y2 =

The point midway between vertex of the parabola, and of the

and

c,y),

( 13.40))

line

507)

to the basic

we return

parabola, vertical

to the standard

simplifies

the conic sections)

for

equations

opens

parabola

=

The parabolax2

(0, c) and

Cartesian

=

4cy.)

horizontal

if the

becomes)

equation

4cy .

as shown

upward

= - c)

y

in

13. 16.

Figure

When c

< 0, it

opens

downward.

If the

parabola

in

Figure

equation

corresponding

13.15

(y focus

The

axis of

the

Similarly,

focus

-

YO)2

=

is now at the point (xo + c, Yo) is the line y = Yo . parabola a translation of the parabola in

(x with

at (xo , Yo + c).

The reader

may

find

so

is translated

that

is at the point

vertex

its

(xo , Yo),

the

becomes)

-

XO)2

4c(x and

-

4c(y

.)

directrix

the

-

Yo)

is the line x

leads

13.16

Figure

=

xo)

=

-

c. The

to the equation)

,)

The line y = Yo - c is its directrix, the line x = Xo to prove that a parabola does not have any

it amusing

Xo

its

axis.

asymptotes.)))

508

Each of the the

x2

y2

+

100

36

+

(x -

and

2)2

9x

= 1.

5.

4y2

+ 3)2

(y

+

9

=

Find the coordinates ellipse. determine the eccentricity.

and

vertices,

4.

36

16

the

= 1.

x2

y2

100

3.

geometry)

analytic

1 through 6 represents an sketch each curve. Also

in Exercises

equations

the foci,

center,

2.

algebra to

Exercises)

13.24

1.

of vector

Applications

6.

1.)

2

+

25y2

+

3x

+ 1)2

(x

16

2

= 25.) =

1.)

+ 2)2 =

+ (y

1.)

25

of Exercises 7 through standard 12, find a Cartesian equation (in the appropriate that satisfies the conditions given. Sketch each curve. ellipse 7. Center at (0, 0), one focus at (!, 0), one vertex at (1, 0). 8. Center at ( -3, 4), semiaxes of lengths 4 and 3, major axis parallel to the x-axis. 9. Same as Exercise 8, exceptwith axis parallel to the y-axis. major 10. Vertices at ( -1, 2),( -7, 2),minor axis of length 2. In each

for

12.

at (3,

Center

at (2,

and (2,

3).

Each of the

of the

center,

Also,

compute x2

14. 15.

100 y2

100

the

and

the

at (4,

- 2), (12,- 2). x-axis, the curve

to the

13 through Sketch

vertices.

x2

-

show

and

1.

16. 9x 2

-

16y2 =

=

1.

17.4x2

-

5y2 +

64

-

(y

3)2 = 1.)

18. 23,

which satisfies

asymptotes.

19. Center

at (0,

(x

- 1)2 4)

of Exercises19through

hyperbola

the

points

(6,

1)

the coordinates the

asymptotes.

eccentricity.)

=

-

through

passing

a hyperbola. Find the positions of

18 represents

each curve

64

4

of the

in Exercises

equations

y2

(x + 3)2

the

- 2), (13, - 2), foci 1), major axis parallel

the foci,

-

In each for

fornl)

the

11. Vertices

13.

of

0), one

focus at

the

fi nd

a Cartesian

conditions

(4, 0),

given.

one vertex

at

(y

144.)

20 =

O.)

+ 2)2 9

= 1.)

standard form) (in the appropriate the positions Sketch each curve and show

equation

(2, 0).

::!:1). 21. Vertices at (::!:2, 0), asymptotes y = ::!:2x. 22. Center at ( -1, 4), one focus at ( -1, 2), one vertex at ( -1, 3). 23. Center at (2, -3), transverse axis parallel to one of the coordinate axes, the curve passing and ( -1, 0). (3, -1) through 24. For what value (or values) of C will the line 3x - 2y = C be tangent to the hyperbola x 2 - 3y2 = 1 ? 2x + y = O. Find 25. The asymptotes of a hyperbola are the lines 2x - y = 0 and a Cartesian if for the curve it the 5). equation passes through point (3, in Exercises Each of the equations 26 through 31 represents a parabola. Find the coordinates of the vertex, an equation for the directrix, and an equation for the axis. Sketch eachof the curves. 26. y2 = -8x. 29. x2 = 6y. 27. y2 = 3x. 30. x2 + 8y = O. 28. (y - 1)2 = 12x - 6. 31. (x + 2)2 = 4 Y + 9.)))

20. Foci

at

(0,

::!:

V2),

vertices

at (0,

the

sections)

509)

of Exercises 32 through 37, find a Cartesian equation (in appropriate the curve. and sketch that satisfies the conditions given

each

In

for

on conic

exercises

Miscellaneous

32. Focusat

(0,

33.

(0, 0);

at

Vertex

34. Vertex at 35. Focus at

-t);

of

equation

directrix,

y

equation of directrix, focus at (-4,1).

(-4,3);

x

- 2.

-1);

(3,

s is

i

13.25 Miscellaneousexerciseson 1. Show

2

+ y2/b

is the

What

3. Find

all

the

its major

about

result

5. 6.

By2

=

arch has a

bounded

the

(b)

(c) 8. 9.

of the

2

=

1 is ab

times the

without

integral,

revolution generated the volume of a unit

by

times

be proved from

general

properties

the

rotating

ellipse

sphere.

of the

without

integral,

ellipse

A and

B, A

B)x2

+

(A

-

the

B)y2 = 3 .)

base of length b and altitude h. Determine the area of the region and the base. by = 8x and the line x = 2 is rotated about the The bounded x-axis. region by the parabola y2 Find the volume of the solid of revolution so generated. - 2) enclosea - 1)and = = R. Two 2(x 4(x plane parabolas y2 region having the equations y2 A parabolic

(a)

7.

of

solid

axis is ab 2

(A + 4.

properties

+ y2/b

is rotated about its minor axis? of the region enclosed by area > B, such that the 3 is equal to the area of the region enclosed by the ellipse) the

if

numbers

positive

AX2 +

ellipse

general

x 2/a 2

ellipse

any integrations.

performing

(b)

1

of

volume

the

2=

This statement can

Note:

whose

parabola

any integrations.

that

Show

the

by

This statement can be proved from

Note:

the

sections

of the region bounded radius 1.

of

circle

performing

x 2 /a

conic

area

the

that

area of a

2. (a)

form)

= t.

=

x = !. equation of directrix, the to 36. y-axis; passes through (0, 1), (1,0), and (2, 0). parallel vertex at (1, 3); passesthrough 37. Axis is parallel to the x-axis; ( -1, -1). the focal definition, find a Cartesian from 38. Proceeding directly equation for focus is the origin and whose directrix is the line 2x + y = 10.) Ax

standard

parabola

arch

Compute the area of R by integration. Find the volume of the solid of revolution generated the y-axis. as (b), but revolve R about Same

by

revolving

R about

the x-axis.

(x, y) whose distance equation for the conic section consisting of all points = is half the distance the line 8. from (0, 2) point y whose focus is at the origin and whose directrix Find a Cartesian equation for the parabola is the line x + y + 1 = O. the origin, given that its asymptotes Find a Cartesian equation for a hyperbola passing through = 2x + 1 and y = - 2x + 3. are the lines Find

a Cartesian

from

the

y

10. (a)

For each p

> 0,

the

equation

(in terms of p) the eccentricity (b) Find a Cartesian equation part

(a)

and

which has

eccentricity

and for

= pX2 + (p + 2)y2 p2 + 2p the coordinates of the foci. the hyperbola which has the

represents same

foci

an

ellipse.

as the

Find

ellipse of

V3.

Section 13.22 we proved that a conic symmetric about the origin satisfies the equation - Fi! = leX' N IIX ai, where a = ed + eF' N. Usethis relation to prove that II X - Fi! + = from any X 2a if the sum of the distances the conic is an ellipse. In other + Fir II words, to its foci is on an constant.))) point ellipse

11. In

510)

12.

+ F

II X

difference

the

hyperbola

IIX

-

-

PII

(replacing x by tx and y by ty) carries an ellipse words, ellipse with the same eccentricity. In other

transformation

a similarity

at

center

with

of a

branch

each

on

geometry)

II is constant.

Prove that

13. (a)

11. Prove that

to Exercise

Refer

algebra to analytic

of vector

Applications

the origin

another

into

similar ellipses have the same eccentricity. Prove also the converse. That is, if two concentric ellipses have the same eccentricity and major axes on the sameline, then they are related by a similarity transformation. to and for (c) Prove results (a) (b) corresponding hyperbolas. 14. Usethe Cartesian which represents all conics of eccentricity e and center at the equation = (e 2 to prove that these conics are integral curves of the differential origin l)xly. equation yl (b)

Note: Since this is a homogeneous differential equation (Section 8.25),the such conics of eccentricity e is invariant under a transformation. similarity with Exercise 13.))

15. (a)

set

of all

(Compare

that the collection of all parabolas is invariant under a similarity transformation. transformation carries a parabola into a parabola. is, a similarity 2 to y = x . (b) Find all the parabolas similar = 16. The line x y + 4 0 is tangent to the parabola y2 = 16x. Find the point of contact. 17. (a) Given a \037 O. If the two parabolas y2 = 4p(x - a) and x2 = 4qy are tangent to each other, show that the x-coordinate of the point of contact is determined by a alone. on a, p, and q which the fact that the two parabolas are tangent (b) Find a condition expresses Prove

That

to each other.

18.Consider

of the

P in the plane for which the distance of P from the point distances of P from the two coordinate axes. (a) Show that the part of this locus which lies in the first quadrant is part of a hyperbola. Locate the asymptotes and make a sketch. (b) Sketch the graph of the locus in the other quadrants. 19. Two parabolashave the same point as focus and the same line as axis, but their vertices lie on opposite sides of the focus. Prove that the parabolas intersect (i.e., their orthogonally lines are the at of intersection). perpendicular tangent points

(2, 3)

the

locus

is equal to

20. (a) Prove that

sum

the

the

points

of the

Cartesian

equation)

x2

y2

a2 +

represents

all conics

(b) Keep c fixed

numbers

\037C2.

and Prove

d-y

22.

Show that the and

are

locus of the tangent

to a

2

( dx )

(c) Prove that S is self-orthogonal; S is S itself. [Hint: Replace yl point

- c2)

= 1

the origin with foci at (c, 0) and ( -c, 0). symmetric about let S denote the set of all such conics obtained asa2 varies over all positive that every curve in S satisfies the differential equation)

xy

21.

a2

+ (x2

the locus of the centers Show that to a given circle and also to a given to be a specialcase.))))

y2

that is, -ll y

by

of a

centers given

-

line,

straight

the

dx)

xy

=

0 .

of all orthogonal differential equation

set

in the

family

is a

of a

l

- - c2 ) dy

of

circles,

all of

trajectories of curves in

which

in

(b).] pass

through

a given

parabola.

family

of circles, all of line, is a parabola.

which

are

tangent

(Exercise21 can

(externally) be considered

23. (a) P

A chord and

of length 8

Q be the

is perpendicular

(b) The chord

to

that

is drawn the

chord

from 0 to

perpendicular meets the

to

the

sections) axis

parabola.

of the

Show

that

511) parabola

y2

the vector

=

4cx.

Let

from 0 to P

Q.

and parallel to the directrix is called the distance latus rectum is twice from the the and then show that the tangents to the parabola at both ends of the directrix, latus rectum intersect the axis of the parabola on the directrix. Two points P and Q are said to be symmetric with to a circle if P and Q are collinear respect with the center, if the center is not between of their distances from the them, and if the product center is equal to the square of the radius. Given that the straight line Q describes x + 2 Y - 5 = 0, find the P symmetric to Q with locus of the point to the circle respect x 2 + y2 = 4.)))

the latus focus to

24.

points

lei

where

on conic

exercises

Miscellaneous

rectum.

of

a parabola drawn Show first that the

through length

the

of the

focus

14)

14.1

This chapter

some describes algebra with the methods of calculus and of a The concept of curves and to some problems in mechanics. in this study.) fundamental vector

combines

to the

applications

study

vector-valuedfunction

variable

a real

of

functions

Vector-valued

is

is a set A function whose dOl11ain is called a vector-valuedfunction

DEFINITION.

of real

subset of n-space V n

We have encountered such point P parallel to a nonzero

in

functions

vector

Vector-valued functions small bold-face italic In the

endpoints

letters.t: g, etc.

The usual operations of or to combine a functions are vector-valued new

(F + The

sum

real

valued.

example, the

+

is a

range

line

a

through

X given

function

by)

tA)

+

may be

be

will

an interval

Y,

or by

etc.,

as usual,

which may

by

contain

infinite.)

Components vector

be applied to combine two vector-valued If F and G with a real-valued function. a common real-valued function, all having domain, can

algebra

vector-valued

function

G(t)

,)

(uF)(t)

G and the product uF are and G(t) are in 3-space,

If F(t)

of F

domain

the

G by =

vector

the

u(t)F(t)

equations)

,)

(F x G)(t) =

(F

.

G)(t)

whereas

valued,

we can

also define the

the formula)

512)))

ued

vector-val

value

The

functions, and if u is a F + G, uF, and F. functions

G)(t) = F(t)

F +

study,

or which

operations.

Algebraic

we define

13. For

by capital letters such as F, G, X, of a function Fat t is denoted,

denoted

be

will

examples we shall

one or both 14.2

variable.)

range of the

= P

X(t)

F(t).

Chapter

A is the

whose

and

numbers

of a real

t.

all real

for

FUNCTIONS)

VECTOR-VALUED

OF

CALCULUS

F(t) x

G(t).)

= F(t)

the dot

. G(t) .)

product F. G is

cross product F

X

G by

The operation

of

For example, if

of a

cludes the range

valued function

may be

composition

functions.

real-valued

by the

defined

applied to combine vector-valued F

is a

F0

u is

in-

domain

a new

vector-

F[u(t)])

domain of u.

t in the

each

If a function

G=

composition

with

functions whose

function

equation)

G(t) = for

vector-valued

u, the

function

real-valued

513)

and integrals)

derivatives,

Limits,

F has

its values

then each

Vn ,

in

vector F(t) has n

and

components,

we can

write)

=

F(t)

Thus,

14.3

of

rise to

integral

. . ,fn)

I: the

components

(

We also say

these definitions,

view of continuity,

THEOREM

F . G,

(F+

it

differentiation,

14.1.

and

H'e

If F,

We

state

G, and

u

Iimfl(t),

. . . , l imfn(t)

)

t\037p)

t\037p

...

=

. . . , fn(t) I: U: fl(t) dt, are

we define

function,

(f{(t),

right

F is continuous, the corresponding

vector-valued functions.

and

the

vector-valued

=

derivative,

limit,

,

dt)

,)

meaningful.)

property on the

or integrableon

an

if each

interval

F' +

G',)

com-

interval.

to find that many of the theorems on is not surprising are also valid for and integration of real-valuedfunctions we use in this chapter.) of the theorems that some

are

differentiable

on an

interval,

then

so are

F +

have)

G)' =

its

,f\037(t)),)

differentiable,

that

has

F

F(t) dt

on

a

is

=

F(t)

F'(t)

In

we

of calculus,

t\037p

limits,

at

and

and integrals

lim

ponent

values

F.)

= (fl , . If F the by equations

DEFINITION.

of

relation

such as limit, derivative, and integral, can alsobe extended function in terms of the vector-valued We simply express on the components.) and perform the operations of calculus

components

\"\"henever

this

. ,fn whose (fl , . . . ,fn)'

fl , . . by writing F =

functions

n real-valued

indicate

We

.)

functions.

vector-valued

and

of F(t).

derivatives,

Limits,

The basicconcepts to

F gives

vector-valued

each

t are the components call he the kth component

. . ,fn(t))

(fl(t),f2(t)\",

(uF)'

=

u'F + uF',)

(F' G)' =

F'

\302\267

G

+

F.

G'.)))

G, uF,

G have values in

F and

If

of vector-valued

Calculus

514)

V 3, H'e

also

To indicate

Proof

The proofs of

F =

Writing

others

the

(fl , . . . ,fn)'

, . . . , ufn)

kth

=

(uF)'

=

G'.)

X

\302\253ufl)\"

=

for (uF)'.

formula

the

reader.

us

uf; ,

+

in

\302\267)

so we

have)

+ uF'.)

u'F

formulas

one must

F. G gives

=

or product

a sum

. . . , (ufn)')

U'fk

+ u(f{,. . . ,f\037)

differentiation

the

that

for differentiating

formula

F

is (ufk)'

of uF

. . ,fn)

U'(fl,'

The reader should note

The

+

(uF)'

,)

component

to the usual formulas for differentiating the cross product is not commutative, the formula for (F X G)'. .

G

X

have)

we

(Ufl

of the

derivative

the

= F'

the routine nature of the proofs we discuss are similar and are left as exercises for the

uF = But

have)

X G)'

(F

functions)

are analogous

14.1

Theorem

of real-valued functions. to the order of the

pay

attention

the

following

theorem

Since factors

in

shall

use

we

which

frequently.) 14.2.

THEOREM

Let get)

Proof

= 0 on I. But g' F . F' = we have next

The

3.5 and

4.2

THEOREM

u'(t) and

a vector-valued F. F' = 0 on

=

II F(t)

=

is

function

.

F(t). dot product, we F(t)

lfords, F'(f) is perpendicular

g is

hypothesis,

By

have

and has constant

differentiable

In other

I.

= F'. g'

F +

constant

on I,

on

length

to

and

F(t)

for

hence

F. F' = 2F.F'. Therefore

O.)

deals

theorem

with

the

contain

which

G = F 0 u, F if is continuous then G'(t)

exist,

Its proof follows easilyfrom results for real-valued functions.)

functions.

composite

corresponding

Let

14.3.

F'[u(t)]

2 11

g is a

since

at t and

continuous

then

If

open interval I, each t in I.)

an

is vector

F

\037vhere

at

u(t),

continuous

is given

by the

also exists and G'(t)

valued and

then G is

= F'[u(t)]u'(t)

u

is at

chain

real t.

Theorems

valued.

{f u is the derivatives If

rule,)

.)

function F is continuous on a closed interval [a, b], then comeach is continuous and on hence so F is on The next [a, b], integrable integrable [a, b]. ponent I n each of the integral of vector-val ued functions. three theorems give basic properties the results for integrals of real-valued case, the proofs follow at once from corresponding If

a vector-valued

functions.)

THEOREM

are integrable

14.4. on

LINEARITY

[a, b],

f:

AND

so is clF

(cIF(t)

+

ADDITIVITY. C2G

for

+ C2G (t)) dt

all

=

Cl

If the vector-valued c 2, and }1'ehave)

functions F and G

and

dt c clf: F(t) + 2 I:

G(t) dt

\302\267)))

A

Iso,

each c

for

in

b], H'e have)

[a,

dt I: F(t) 14.5.

THEOREM valued

function

on [a,

given

function

A'(x)

and

exists,

J:

=

have

MJe

b]. If

E [a,

J: F(t)

next by

is an extension

theorem

c replaced

scalar

the

14.7.

THEOREM

(c1, . . . , c n )

If

the dot

J \" we

Now

THEOREM

(14.1

0 \302\245=

(14.2))

14.7

and

in

Let

C =

S\037F(t)

dt.

of

If C

Theorem 14.7to 1IC112 =

on

[a,

I: C

.

C .

F(t)

dt

we

have)




Compute the

can divide

derivatives

F'(I)


O. 1. r(t) = a(1 - cost)i + aCt - sin t)j, 0 < t < 2. 2. r(t) = et cos t i + et sin tj, 0 < t < 217, a > O. 3. r(t) = a(cost + t sin t)i + a(sin t - t cos t)j, 2 c2 ( 3 c 0 < t < 217, c 2 = a 2 - b 2 , 0 < b < a. 4. r(t) = cos t i + sin 3 t j, -;;

5. r(t)

=

b

a(sinh

6. r(t) = 7. r(t) =

sin t

ti

-

+

t)i

a(cosh

+ (1

+ tj

3t 2j

+ 6t 3 k + log (sec t)j

i +

8. r(t)

=

t i

9.

=

a cos

r(t)

t

wt

t

- l)j,

- cost)k (0 < t


0). Show of this curve is equal to the area of the region bounded by that c times the length (b) = c cosh (xlc), the x-axis, the y-axis, and the line x = a. y a = 2. this integral and find the length of the curve when (c) Evaluate 15. Show that the length of the curve y = coshx joining the (0, 1) and (x, coshx) is points sinh x if x > O. function has 16. A nonnegative that its ordinate set over an arbitrary interval f has the property an area proportional to the arc length of the graph the interval. Find f above 17. Use the vector equation r(t) = a sin t i + b cos t j, where 0 < b < a, to show that the cir-

cumference L of

an

ellipse

is given

by

L =

where

of an

the

integral)

J:12Y

4a

v a2 - b 2 /a. (The number e is the integral of the form) e =

E(k)

called

tabulated

an elliptic

integral of the

for various

second

values of k.)))

=

f:12 kind,

l-

e

2

sin

2 t

k

where 0

2

,)

of the

eccentricity

Y1-

dt

sin 2 t

\037 k

dt

ellipse.) This is a special

case

,)

< 1.

The numbers E(k)

have

been

18.

of vector-valued

Calculus

536)

< b
< 1T, between = r and v cos = curve. If the curve is expressed in polar coordinates, 4> prove that v sin 1> the v where is dr/de, speed. Let

(b)

out

15.

A

is designed

missile

tion

actual

in

flight if it

to move directly

makes a

fixed

toward angle

its target. Due with the line

ex \037 0

to mechanical failure, its direcfrom the missile to the target.

Discuss how the path varies with c(. Does the takes placein a plane.) (Assume a ground crew has lost control of a missile It is fired. recently at a constant speed on a straight course of unknown the missile will proceed that known it is sighted for an instant the missile is 4 miles away, and lost again. Immedirection. When missile is fired with a constant speed three times that of the first missile. an anti-missile diately in order for it to overtake missile the first one? be the course of the second What should (Assume both missiles move in the same plane.) of the form y' = ,(x, y) is re17.Prove that if a homogeneous first-order differential equation it reduces to a Use this method to solve written in polar coordinates, separableequation. = + x). x)/(y y' (y

is fired the path Find missile ever reachthe target? 16. Due to a mechanical failure,

18. A

and

f is

speed

19. A

the

w.

particle with circle

a fixed

target.

the motion

w is a v = wk x f, where vector positive constant the with that moves a circle constant particle along position angular to be Id8/dtl, where 8 is the polar (The speed is defined angle at time t.) angular The motion takes place along a moves in a plane perpendicular to the z-axis. center on this axis.)))

(moving

particle

at

in space) has vector.

Prove

velocity

(a) Show that

is a

there

vector

wet)

z-axis such

to the

parallel

x r(t)

= wet)

vet)

545)

motion)

to planetary

Applications

that)

,)

and velocity vectors at time t. The vectorwet) is called r(t) and vet) denote the position = Ilw(t) II is called the angular velocity vector and its magnitude wet) the angular speed. vector. Show that the accelera(b) The vector aCt) = w/(t) is called the angular acceleration tion vector aCt) [= v'et)] is given by the formula) where

aCt) =

.

[wet)

-

r(t)]w(t)

w

2

+ aCt)

(t)r(t)

x r(t)

.)

particle lies in the xy-plane and if the angular speed wet) is constant, say wet) = w, that the acceleration vector aCt) is centripetal and that, in fact, aCt) = - w 2 r(t). prove A body is said to undergo a rigid motion if, for every pair of particlesp and q in the body, - r q(t) II is r r the distance of where and denote the IIr llt) t, pet) q(t) independent position vectors of p and q at time t. Prove that for a rigid motion in which each particle p rotates about the z-axis we have v p(t) = wet) x r pet), where and wet) is the same for each particle, vp(t) is the velocity of particle p.) (c)

20.

14.20

By

If the

to planetary

Applications

data on

the voluminous

analyzing

German

motion)

Johannes

astronomer

up to

accumulated

motion

planetary

tried to discover the

(1571-1630)

Kepler

the motions of the planets. There were six known to their the orbits were Copernican theory, according thought sun. the the spherical shells about Kepler attempted to show that were linked up with the five regular solids of geometry. He proposed that the solar system was designed something like a Chinese puzzle.

laws

at that time to lie on concentric radii of these shells

planets

governing

and,

system he placed the that can

sun.

innermost

The

path.

correspondedto and inside orbit,

being

within

five

much

smaller

After

the

sphere,

much

corresponded more After

Kepler's

first

which

la\037v:

as

rate.)))

occurred

elliptical

years

of

paths

unceasing

explained

all

the

at

time

that

finally

realized

to him

that

were

spheres

Mercury's

sphere, theory

accurate

Jupiter's

containing

correct

to to a percentageerror seemed

he had to modify this theory. data concerning the orbits

that the

concentric

idea of the

solids-the octahedron, order (from inside out).

correspondedto

this

Although

ingenious center

observed

than the circular paths of the three effort, Kepler set forth the astronomical phenomena

system.

Copernican

famous known

laws, at

that

empiritime.

follows:)

Planets

Kepler's secondla\037v: constant

it

study

to

several more

cally discovered, They may be stated

observations

and Kepler

on

lay

the

and inscribed the icosahedron, sphere around the icosahedron

octahedron

on, the outermost

around the cube.

astronomical this,

the

so

and

dodecahedron,

further

regular octahedron,

an At

six

the

the five regular cube, in respective

Earth's orbit

of Venus.

circumscribed

percent, than

in the

circumscribed

which

orbit

the

and

tetrahedron,

inscribed

sphere,

The next

around

circumscribed

icosahedron,dodecahedron,

he arranged

in succession,

Then,

be inscribed and

1600, the

mathematical

The

move

in ellipses

position

vector

with

the

from the

sun sun

at one to

focus.

a planet

sweeps out area at

a

of vector-valuedfunctions)

Calculus

546)

the

from

elliptical

planet is proportional

of a

period

to the

cube

of its

sun.)

By the period of a planet orbit. The mean distance

Note:

of the

square of the

la'tv: The

third

Kepler's

mean distance

one

sun is

the

the go once around the axis of length major

the time required to

is meant fronl

half

the

ellipse.

of these laws from a study of astronomical tables was a remarkable all of Kepler's laws are 50 that three Newton Nearly years later, proved of gravilaw of universal law of his own second motion and his celebrated consequences laws may be how Kepler's tation. In this section we shall use vector methods to show The

formulation

achievement.

deducedfrom

Newton's.)

Orbit)

Sun)

of

have

we

Assume

the

sun

a force

by

states

motion

the acceleration

from the

sun

the same

direction as r,

to

sun to

a

planet.)

that)

F=

a is

the

to of mass nl attracted a fixed sun of mass M and a moving planet of all other forces.) Newton's secondlaw F. (We neglect the influence

( 14.28)) where

vector from

The position

14.19

FIGURE

the

planet

vector of (as

so that

in

the

Figure r = ru

ma

moving 14.19),

,)

Denote by r the position vector planet. with vector let r = \\Ir 1\\, and let Ur be a unit

The universal law of gravitation

r .

-

F =

states

that)

mM G

Ur ,

\037 r)

where

G is

a constant.

Combining this

a=

( 14.29))

which

lies in

tells

we obtain)

(14.28),

-

GM

2

Ur

,

r)

us that the

a plane.

the position

with

acceleration is radial. In a

Once we know

vector sweeps

out

this,

it follows

area

at a

moment

at once

constant

rate.)))

we

shall

prove

from the results of

that the orbit

Section

14.17

that

To prove that the

introduce

the

r X

Since

a =

= 0,

If e

0,

shows

dv

-

X

r

position

is not

r

-dv + -dr dt dt

X

parallel to

is

acceleration

-d (r

v =

X

have e

0, so the position vector lies in a radial, r sweeps out area at a constant

.

v)

v =

X

e.

a

is along

motion

the

a straight line, we must

X

dt)

vector, say r

and

v

r . C =

that

parallel. If we

a are

r and

that

fact

a constant

v is

X

r is

547)

have)

v =

X

v

that

of a planet

we

drfdt,

+

dt

means

this

=

v

vector

the

the path

Since

vector

a= r

r x

plane we use the

lies in a

path

velocity

motion)

to planetary

Applications

line.

straight

r x

relation

The

\037 O.

v

=

e

Since the perpendicular This proves Kepler's second to e.

plane

rate.

law.

to prove that

is easy

It

fact,

we

if

(14.25), we

e =

hence the

which

r

X

e,

a

e =

a =

( dt

dB dt U(J )

r

Ur +

X

e. In

in Equation

as u(J

dO

\"

=

r-

Ur

dt

x

U(J ,)

equal to 2IA'(t)l, where

A'(t)

is the

rate at

vector

- GM U r \037 )

dvfdt and

u(J

in Figure in equal time

illustrated

is

position

X

=

(

r

2

dB Ur

dt

X

=

U(J

the foregoing

durfdB,

dB

-GM

Ur

dt

)

equation

X (u r

for

a

X

u(J)

X e

=

GM

can also

dB de U().)

be

written

follows:)

d

dt (v

X

v X

where

h is

another constant

vector.

(14.31))

v

= h.

where

GMe

for r.

For this

r.

Equating

=

e)

d dt (GMu

.)

r)

us)

gives

Integration

by

of Ur and

which are 14.20. The two shaded regions, intervals, have equal areas. next that the path is an ellipse. First of all, we form the cross product prove (14.29) and (14.30), and we find that) the

(

Since as

by

using

X

dr

(ru r )

2 lie II = Ir dBfdtl. By (14.27) this is radius vector sweeps out area.

shall

We

a

v =

X

Kepler's secondlaw swept out

vector

the

of

length

that)

find

(14.30))

and

rate is exactly half the and express the velocityin terms

constant

this

coordinates

use polar

X

purpose

the two

GMu r

e =

GM(u

with

(14.30) scalar

,)

this as +

r

multiply both sides

expressions for the

h

+

rewrite

can

We

We shall combine this we dot

e =

e)

follows:)

,)

to eliminate

of (14.30)by

triple

product

v

and

c and

r

\302\267

v X

an equation sides of (14.31) e, we are led to the obtain

both

equation)

GMr(1 +

( 14.32)) where

e =

II ell,

c =

II ell,

and

1>

represents

= e cos 1\302\273 the

angle

c 2 ,)

between the constant

vector e and the)))

of vector-valued

Calculus

548)

Figure 14.21.) If we

r. (See

vector

radius

ed (14.33))

e cos 1>

By Theorem

13.18,

this

the

is

d =

let

c 2 f(GMe),

r =

or)

+

functions)

e( d

becomes)

(14.32)

Equation

- r cos1\302\273 .)

1)

equation of a conic section with eccentricity the directrix drawn perpendicular to e at to the directrix is d - r cos1>, and planet

polar

focus at the sun. Figure 14.21 shows d from the sun. The distance from the

v

v)

d -

d)

14.20.

Kepler's

regions,

rf(d

- r cos

and a

swept

have

intervals,

is 1\302\273

out

equal

the

hyperbola if

second

e

time

e. The eccentricity > 1. Since planets

r cos

a

ratio)

I I I I)

e) I I ljJ I 1 I I --I I)

The ratio rl(d - r coscp) = Ileli.) eccentricity e

FIGURE 14.21.

The

law.

in equal areas.)

.

DIrectrIX

r cos ljJ)

FIGURE

the

.

Orbit)

two shaded

e and a distance

is the

conic is an are known

if e

ellipse

move

to

< 1, a parabolaif e = 1, on closed paths, the orbit

under consideration must be an ellipse. This proves Kepler's first law. we deduce Kepler's third law. the ellipse has major axis of length 2a Finally, Suppose and minor axis of length 2b. Then the area of the ellipse is 7Tab. T be the time it takes Let for the planet to go once around the Since the position vector sweepsout area at ellipse. the rate ic, we have icT = 7Tab, or T = 27Tab/c. We wish to prove that T2 is proportional to a3 . From Section 13.22 we have b2 = a 2(1 - e2 ), ed = a(l - e2 ), so) c2 = and

hence

we

2 =

47T

2

a

c T2 is

GMa(1-

e

- e2 ) - e2 )

=

2

),)

have)

T

Since

GMed =

a constant times

a3 ,

this

2 2

b

=

47T

2

4 a (1

2

GMa(l

proves

Kepler's

third

47T

2

a

GM

law.)))

3

.)

14.21 Miscellaneousreview

549)

review exercises)

Miscellaneous

exercises)

r denote the vector from the origin to an arbitrary point on the parabola angle that r makes with the tangent line, 0 < rJ.. < 7T, and let e be the angle rJ.. in terms of e. with the positive x-axis, 0 < e < 7T. Express = 4cx at the to the parabola 2. Show that the vector T = yi + 2cj is tangent y2 to T. and that the vector N = 2ci - yj is perpendicular 1. Let

y2

be the

[Hint:

3. Prove

that

a vector

Write

for

equation

the

using

parabola,

y as

=

that

point

x, let rJ.. r makes (x, y),

a parameter.]

= 4cx can of the line of slope /11 that is tangent to the parabola y2 form y = Inx + c/ /11. What are the coordinates of the point of contact? 3 for the parabola (y - Yo)2 = 4c(x - x o )' x 2 = 4cy and, more generally, for the parabola Exercise 3 for the parabola an equation

in the be written 4. (a) SolveExercise

(b)

(x

Solve

-

xo)2

=

4c(y

-

Yo)'

= 4cx at the that an equation of the line that is tangent to the parabola y2 point = in be written the form can + 2c(x Xl)' (Xl' Yl) YlY 4. 6. Solve Exercise5 for each of the parabolas describedin Exercise of intersection on the parabola y = X2. Let Q be the point of the normal 7. (a) Let P bea point is the limiting position of Q as P tends to the y-axis? line at P with the y-axis. What (b) Solve the same problem for the curve y = [(x), where ['(0) = O. = x 2 at two Find the radius of 8. Given that the line y = c intersects the parabola points. y two points and through the vertex of the parabola. The radius the circle passing through these to this radius as c \037 O? depends on c. What happens you determine as the ellipse x2/a 2 + y2/b 2 = 1 according 9. Prove that a point (x o , Yo) is inside, on, or outside or is less than 1. than, equal to, greater x5/a2 + Y5/b2 10. Given an ellipse x 2/a 2 + y2/b 2 = 1. Show that the vectors T and N given by) 5.

Prove

T

=

-

Y. b

are,

tangent

respectively,

eccentric angle of (xo

, Yo)

and

norl11al

is 00 , show

2

I +

to that

X

- cos a

x.

J

N

x.

=-I+-J a2

Y.

a2

,)

the

(x, y). If the ellipse when placed at the point line at (x o , Yo) has the Cartesian equation)

b 2)

the tangent

e0 + Y- sin b

e0 =

1 .)

line to the ellipse x 2 /a 2 + y2/b 2 = 1 at the point (x o , Yo) has the 2 = 1. + yoy/b xox/a equation from the foci of an ellipse to any tangent 12. Prove that the product of the perpendicular distances line is constant, this constant being of half the minor axis. the square of the length 13. Two tangent lines are drawn to the ellipse x 2 + 4y2 = 8, each parallel to the line X + 2 Y = 7. Find the points of tangency. 14. A circle passes through both foci of an ellipse and is tangent to the ellipse at two points. Find the eccentricity of the ellipse. 15. Let V be one of the two vertices of a hyperbola whose transverse axis has length 2a and whose P A is be on the as of the Let a same branch V. Denote the area 2. point by eccentricity r and be the of bounded the and the line let VP. VP, region hyperbola length by segment for the hyperbola. Place the coordinate axes in a convenient (a) position and write an equation to evaluate this integral, show (b) Expressthe area A as an integral and, without attempting as the point P tends to V. Find this limit.))) that Ar- 3 tends to a limit

11.

Show

that

the

2

tangent

16.

that the

Show

by

the

to the

2

)i

tangent 2 xox/a equation

+ (x/a x 2/a 2 -

hyperbola

the

that

Show

vectors T = (y/b

normal

and

gent

17.

of vector-valued

Calculus

550)

line to

the

-

2

hyperbola =

yoy/b

2

functions)

)j and N = 2 y2/b = 1 if

2 (y/b )j

(x/a2 )i

-

placed

at the

x 2 /a 2 - y2/b 2

= 1

at

are, respectively, tan-

point (x, y) the

point(x

on o

the curve.

, Yo)

is given

1.

and the line from that to the origin form an point of a curve point isoscelestriangle whose base is on the x-axis. Show that the curve is a hyperbola. 19.The normal line at a point P of a curve intersects the x-axis at X and the y-axis at Y. Find the curve if each P is the mid-point of the corresponding line segment X Y and if the point

18. The normal

20.

(4, 5) is on the

curve.

Prove that the

product

to 21.

at

line

A

points of

the

the difference

formed by 22. If a curve terval

(a, b).

[a, b],

by a

has arc

of

radial

the

arc and

the

prove this

that result

the

r =

equation

radii

to

is described

3-space

Interpret

perpendicular

distances from

an

arbitrary

point

on a hyperbola

two distinct f if an arbitrary arc joining to the subtended at the (a) ; (b) length proportional angle origin from the origin to its endpoints; distances (c) the area of the sector

polar

curve

the in

of the

is constant.

its

asymptotes curve is given

each

scalar

f( 0).

Find

its endpoints. a vector-valued

by

triple

geometrically.)))

function

. product r'(t) r(a)

r defined on a parametric inis zero for at least one t in

x r(b)

15)

SPACES)

LINEAR

15.1

Introduction)

added

be

can

infinite

examples and addition and

do not

space is

a set of by

multiplication

linear

space,

which

which

certain

take

The

15.2

Let

V

definition

can

numbers)

kind on be performed.

of any

elements

I

In

a linear

(called

operations

space, we are to be peroperations certain properties which a linear

defining

nature

of

all these

includes

cases.

them.

on

the complex we chapter

this

of the elements nor do we tell how the we require that the have Instead, operations as axioms for a linear space. We turn now to a detailed descriptionof the

specify

formed

space

concept, called a

as special

others

many

a linear

Briefly,

objects First of all, the real

numbers.

by real

vectors

series,

a general mathematical

discuss

mathematical

of

examples

many

functions, objects. Other examplesare real-valued in n-space, and vector-valued functions. In

are such

themselves

numbers

numbers,

we

book we have encountered to each other and multiplied

this

Throughout that

these

space)

a nonempty The set V is called set of objects, called elements. satisfies the following ten axioms which we list in three groups.)

denote

if it

axioms.)

a

linear

Closureaxioms) AXIOM

1.

corresponds a AXIOM

2.

CLOSURE unique

UNDER

CLOSURE

UNDER

every real number a there by

For

ADDITION.

in V

element

called

the

sum

corresponds

For

BY REAL NUMBERS.

MULTIPLICATION an

every pair of elements x and of x and y, denoted by x +

element

in V

called

the

product

y

in

V there

y.)

every x in V and of a and x, denoted

ax.)

Axioms

for addition) LAW.

AXIOM

3.

COMMUTATIVE

AXIOM

4.

ASSOCIATIVELAW.

For all

x and y

For all x,y, and

z in

in

V,

}ve have

V, we have (x

x + y

+ y)

+

= y+

z =

x.)

x + (y

+ z). 551)))

Linear

552)

EXISTENCE OF ZERO

5.

AXIOM

spaces)

for all

x+O=x)

AXIOM

6.

For ever.y x

OF NEGATIVES.

EXISTENCE

x +

Axioms

DISTRIBUTIVE

8.

that)

V .)

in

element

V, the

has

(-I)x

the

property)

.)

LAW

x

IN

ADDITION

FOR

all real

V and

in

numbers a and b,

}-ve

have)

= (ab)x.)

a(bx)

AXIOM

= 0

(-I)x

For every

LAW.

ASSOCIATIVE

7.

AXIOM

in

x

by 0, such

V, denoted

by numbers)

multiplication

for

element in

is an

There

ELEMENT.

For all

V.

x and y

in

all real

V and

a,

H'ehave) + y)

a(x

9.

AXIOM

a and b,

}-ve

LAW

DISTRIBUTIVE

ay

.)

OF NUMBERS.

ADDITION

FOR

+

For

all x

in

V and

all real

have)

(a + 10.

AXIOM

= ax

OF

EXISTENCE

b)x =

IDENTITY.

For

ax + bx .)

every x

in

V,

we have

Ix =

x.)

to emphasize above, are sometimescalled real linear spaces If real number the elements of V by real numbers. the fact that we are multiplying is restructure is called a com2, 7, 8, and 9, the resulting placed by complex number in Axioms a linear space is referred to as a linear vector space or simply plex linear space. Sometimes are also called scalars. A real linear a vector space; the numbers used as multipliers space linear has complex numbers as scalars. has real numbers as scalars; a complex space of real linear spaces, all the theorems are we shall deal primarily with examples Although we use the term linear space without valid linear spaces as well. When for complex further can be real or complex.) the space it is to be understood that designation, Linear

15.3

If

spaces,

Examples we

specify

numbers,

of the

as defined

of linear the

we get a

following

set

V

spaces and

tell

concreteexample

examples

satisfies

how to add its elements of a linear space. The all the

EXAMPLE 1. Let V = R, the set of all addition and multiplication of real numbers.)

axioms for a

real

numbers,

real

and

linear

and let

to

how

reader can

easily

them

multiply

verify

by

that each

space.)

x + y

and

ax

be ordinary

define x + y to be ordinary 2. Let V = C, the set of all complex numbers, of the complex number of complex numbers, and defineax to be multiplication x V are the of Even elements a. real number this is the a real numbers, though complex by the scalars are real.))) linear because space EXAMPLE

addition

of linear

Examples

V =

Let

3.

EXAMPLE

and multiplication

it

is a

by

space of all

vector

n , the

in the usual

defined

plane

of

domain

reader

EXAMPLE

6.

EXAMPLE

7.

consider this

set of all

way:)

it is

understood

of degreeequal

satisfied. For example, the 8. The

EXAMPLE

polynomials of degree
1. Dividing .} where !Pn = u n / II Un

1T

, !P2' . .

1f2n-l(X) =

,

orthogonal basis. The relative

dx = 1T,

= V

Ilunll

obtain an orthonormal set {!Po,!PI

a

the zero element, S is independent. We have (u o , u o ) = S\0371Tdx = 21T

S is

1, we have) (U 2n - l

In

= 0,

dx

un(x)um(x)

fornlulas)

Linear

566)

Taking the inner product of

Proof

spaces) of (15.7)

member

each

obtain)

e j , we

with

n

e j) =

(x,

= 0 if

(e i , e j )

since

If {e1, . . . , en}

This

i \037 j.

implies

orthonormal

is an

! i=

ci(e 1)

i

, e j)

(15.8), and

= cj(e j ,

= 1, we

(e j , e j)

when

basis, Equation

e j) obtain (15.9).)

(15.7) can be written

in

form)

the

n ei)e

i=l)

theorem shows that

The next

of

basis the inner product

{e 1 , . . . , en}

is an

Euclidean

finite-dimensional

be computed

can

elements

i .

a finite-dimensionalEuclidean orthonormal basisfor V.

in

space

of

space Then

an

with

of

terms

their

ortho

components.)

n, and of elefnents

dimension

every pair

for

normal

}t'e have)

in V,

y

in a two

V be

Let

15.12.

THEOREM

assume that

x and

! (x,

X =

(15.10))

n

(x, Y) =

(15.11))

In particular,

=

x

when

y,

! i=l)

l-ve

(x,

ei )

n

2 /lx/l =

Taking the

Proof

of

property

linearity

(15.11) reduces to Note:

product

the inner

!

I(x,

i=l)

e i )1

2

.

members of Equation (15.10)with we obtain (15.11). When x = uct, prod

of both

y and

using

y, Eq uation

(15.12).) who (15.11) is named in honor of M. A. Parseval (circa1776-1836), in a special function space.) of formula

Equation this type

obtained

inner

formula).)

(Parseval's

have)

(15.12))

the

ei)( y,

15.12 Exercises)

1. Let

x

= (Xl'

whether

(x, y)

. . . , xn ) and y = (Y1 is an inner product

(x, y)

is

inner

an

not

product,

be arbitrary vectors in V n . In each case, determine for V n if (x, y) is defined by the formula given. In tell which axioms are not satisfied.) , . . . , Yn)

n (a)

(x, y) =

L

Xi

i=l)

( i\037l

n

(b) (x, y)

(c) (x,y) 2.

Suppose

=

\302\267)

x;y; )

n

n

.

\037XiYi

i=l) n

1/2

n

=

(d) (x, y)

IYil.

(e) (x,y)

case

= !(Xi i=l

+ Yi)2

-

!x;

i=l

-

n !y\037.

i=l)

n

=!Xi\037Yj' i=l j=l) we retain

the

first

three

axioms

for a

real inner

product

(symmetry,

linearity,

and)))

567)

Exercises)

but homogeneity) x = O. Prove that Assume

[Hint:

space spanned Euclidean

3. (x, y)

=

4. (x, y) = 5. (x, y) =

x) > 0 for

(x,

(x, x)

> 0 for

0 if 0 if

and and

if

Ilx +

only

if

IIx

only

if Ilx + cy

=

yll

+ yll2 II

>

8. In

linear space

real

the

=

yll2

Ilx II =

C(I, e), define

If [(x) Find a

f'(x)

= 1.

(a)

9.

In U1

real

the

, U2 , u3

= V\037, linear

III

compute

=

g(x)

C( -I, = 1

ul(t)

two of

that

Prove

an angle

make

10. In the

linear

are

them

1T/6with

space

P n of

real

all

+

y

if

and

only

if

\037 O.

\037 O.

In the

elements x and y

in

a

2

Ilyli . for all real +

c.

ilyll.

-

IIyll2

each other,

0 with

angle

inner

cos ().)

2 Ilxlillyll

product

by

then)

the

equation)

dx \302\267) f: (log x)f(x)g(x)

= a

I), let ,)

+ bx g)

(f,

f\037l

= 1 + t

an

of degree

Consider

dt.

U 3 (t)

,)

to the

orthogonal

I(t)g(t)

two make

polynomials

(f,g)

is

that

=

= t

U2(t)

orthogonal, other.

each

some

x

0.])

valid for all

7 is

through

all

II.

polynomial

linear space given by)

2

an

(f,g)

(b)

3

an

IIxl1

z) =

\037 0

= 0

(x, x)

0 for

yll. 2

IIxl1

6. (x + y,x - y) = 0 if and only if Ilxll 7. If x and yare nonzero elements making -

-

Ilx

=

< 0 for

and (y, y) with (z,

\037 0

in Exercises

statements

Ilx

x

some

space. 0 if and only

x

all

find an element z

{x, y},

the

/

by a new axiom (4 ): \037 0 or else (x, x)


n

of linear

then

transformation

R(S The

U,

W), and let

(S

(b)

= S[T(ax

for

composition

all scalars

U and

in

be combined

ff(

16.7.

THEOREM

are in

by)

can

Composition

tion of scalarsin

=

linear

Ware

V,

TTn-l)

linear transformations,

-+ Ware

Proof

V into

maps

The reader may verify Tm+n for all nonnegative

transformation.

identity

16.6.

THEOREM

=

Tn

,)

implies the law of exponents TmTn the The next theorem shows that

and

\037vhich

function

as follo\037's:)

TO= Here

a

V be

-+

V

585)

definition of

.)

composition and

is left

as

an exerCIse.)

16.6

Inverses)

of real-valued functions we learned how functions. Now we wish to extend functions. class of more general if possible, another Given a function T, our goal is to find, In

our

with

study

of monotonic

inversion

T

is the

identity transformation.

we have to distinguish which we call left and

DEFINITION. is called

Given

a left inverse

ST

between right

tM:'o

and

new

construct

to

function S

Since composition is in general TS. Therefore we introduce two

functions

by

of inversion to a

the process

whose composition not kinds

commutative, of inverses

inverses.)

sets

V and

Wand a function

of T if S[T(x)]= x

ST

x in V,

all

for

=

T:

Iv

,)))

V -+

W.

that is,

if)

A

function

S:

T( V) -+

V

Linear

586)

where

Iv

is the identity

=

ifT[R(y)]

ofT

transformation all y

for

Y

in

that

T(V),

is the

IT(V)

is,

A function

EXAMPLE.

no

\037vith

and let W = {O}. Define R : W two right inverses

=

have a

cannot

=

1 =

This

be

shows

example

simple

I

inverse

right

by)

R' (0)

= 2

Let

inverses.

right

T(I) = T(2) =

=

V

{I, 2} has

function

This

O.

.)

require)

2 =

and)

need not

inverses

left

that

two

\037vith

V given

,)

= S(O))

S[T(I)]

but

this would

S since

inverse

left

called a

,)

follows:

as

W

R(O)

It

IT(v)

R' : W \037

V and

\037

V is

if)

inverse

left

V \037

T:

T(V) \037

on T(V).)

transforn1ation

identity

R :

A function

V.

on

TR

where

and matrices)

transformations

= S(O) .)

S[T(2)]

exist and

that

right

need not

inverses

unique.)

T:

function

Every

form y = T(x) for T[R(y)] = T(x) = because there may (in prove presently inverses then right First we prove

at

V \037

W has

at least

least

one x

in

each

for

y

more

be

16.9) that if

that

unique. if a left inverse

each y

exists

such x and

select one

If we

V.

In fact,

inverse.

right

y in T(V), so R is a than one x in V which

Theorem are

one

it

T( V)

of exactly

has the = x, then

T( V)

R(y)

Nonuniqueness y in T(

may occur V). We shall one

at the same time, is

and,

unique

in

define

onto a given is the image

maps

in

is

inverse.

right

each y

x in V,

a

right

Inverse.)

16.8.

THEOREM

inverse S,

S is

then

Assume

Proof

y in T(V).

A function also a

right

T has

two

T:

V

\037

W

have at

can

most one left

inverse.

If T

has a left

inverse.)

left

We shall prove that

=

and)

x)

V and

Now y = T(x) for

S'(y).

S(y)

S[T(x)]

S: T( V) \037

inverses, =

S': T( V) \037 V. Choose any x in V, so we have) some

= x

S'[T(x)]

,)

= x and S'(y) = x, so S(y) = S'(y) that left inverses are unique. for all y in T( V). Therefore S = S' which proves inverse. Choose we prove that every left inverse S is also a right Now any element y in = = E T(V), Since we have for some x in We shall prove that T(x) T(V). y y. y T[S(y)] so) But S is a left inverse, V. x = S[T(x)] = S(y). since

both

T,

Applying

S' are

Sand

we

get T(x)

left

Therefore

inverses.

= T[S(y)].

But

Y

=

S(y)

T(x),

so

Y

=

T[S(y)],

proof.) The

next

theorem

characterizes

all functions

having

left

inverses.)))

which completes the

linear

One-to-one

onto

of V

elements

T:

A function

16.9.

THEOREM

V

elements

distinct

(16.5))

W has

of

W; that

is, if and

implies)

T(x)

(16.5) is equivalent

Condition

Note:

T(x) =

( 16.6))

= y.

this

implies

domain.

if and

inverse

left

\037

only if

if, for all

only

T maps distinct

x and y

in

T(y).)

the statement)

to

x =

implies)

T(y))

(16.6)for

all

x and

y

y.)

V is

in

said to be one-to-one on

a left inverse S, and aSSUIne that T(x) = T(y). We wish to prove = x and S[T(y)] = y, we find S[T(x)] = S[T(y)]. SinceS[T(x)] Applying x = y. This proves that a function with a left inverse is one-to-one on its

the converse. Assume T is one-to-oneon V. We shall exhibit prove -+ V which is a left inverse of T. If Y E T( V), then y = T(x) for some T( V) there is (16.6), exactly one x in V for which y = T(x). Define S(y) to be this we define S on T( V) as follows:)

a

S(y) = we

S[T(x)] =

have

defined is a left

know

T-I

the

a inverse of

transformations.)

16.7

One-to-one this

In

section, a linear

zero

Therefore,

the

S so

function

The

V.

unique

We say that

left inverse invertible,

of T

T is

and

(a'hich we

call

to

functions.

arbitrary

Now

we apply these

same

scalars,

ideas to

transformations)

several

equivalent

T: V -+ W

Let

the

of

T enables

and T: V \037 W us to express the

forms.)

be a

linear

transformation

in

2(V,

W).

Then

the

V.

on and

x in

element

Iv.

are equivalent.

invertible

(c) For all the

section refer

in

16.10.

T is

one-to-one on by T-l.

is denoted

V and W denote linear spaceswith transformation in 2( V, W). The linearity

(a) T is one-to-one (b)

W be

inverse)

linear

statements

following

ST =

V, so

x in

each

= }' .)

T.)

one-to-one property THEOREM

for

T( x)

By is,

That

x.

T.)

-+

V

right

of this

linear

denotes

T:

Let

is also

results

The

of

inverse

DEFINITION. we

x

that)

means

x)

function

x in V.

S:

Then

V.)

S,

we

Now

V,)

T has

Assume

Proof

that x

(16.5) or

T satisfying

function

A

a

-+

x\037y)

587)

transformations)

of

V,

its inverse

T(x)

V.)))

= 0

T-I: T( V)

-+

implies x = O.

V is That

linear. is,

the

null space

N(T) contains only

Linear

588)

and matrices)

transformations

and (c) implies (a). First (c), Proof We shall prove that (a) implies (b), (b) implies assume (a) holds. Then T has an inverse (by Theorem 16.9),and we must show that T-l is linear. Take any two elements u and v in T(V). Then u = T(x) and v = T(y) for some x and y in V. For any scalars a and b, we have)

au

linear.

T is

since

Hence,

bv =

+

Next that

find

we have

-

T(u

=

,)

v)

T(u)

=

T(x)

O.

Applying

T-\\ we

(b) implies (c). u and v in V with T(u)

elements =

T(v)

which

Therefore,

two

any

-

V for

in

0, so

proof of the theorem is complete. is finite-dimensional, the one-to-one

V

and dimensionality,

independence

T:

Let

16.11.

THEOREM

V is

+ bT-l(v)

aT-leu)

-

u

v

=

= T(v). By Therefore, Tis one-to-one

O.

the

and

When

Take

(c) holds.

assume

linearity, V,

=

by

Therefore (a) implies (b). assume that (b) holds. Take any x x = T-l(O) = 0, since T-l is linear.

Finally,

on

ax +

is linear.

T-l

so

=

bv)

,)

by)

we have)

T-\\

applying

T-l(au +

bT(y) = T(ax +

aT(x) +

finite-dimensional,

on (a) T is one-to-one . . . e are e , p (b) If 1, elements in T( V). (c) dim T(V) = n.

(d) If {e1, . . . , en} We shall

Proof

V

W be

-+

say dim

as indicated

V

=

n.

property can be the

by

in

formulated

of

terms

next theorem.)

a linear tran\037rormation Then the folloli'ing

in

V, W) and assume are equivalent.

2(

statements

that

V.

elements

independent

is a

prove

(a). Assume (a) holds. elements T(e1), . . . , T(e p )

basis for that

(a)

V,

in T(

{T(e 1),

. . . , T(

T( e 1),

V,

then

..

. , T(en )}

is a

(b), (b) implies (c), (c) implies e p be independent elements of

V). Suppose

e p)

basis for

implies

el , . . . ,

Let

then

in

are independent

T(V).)

and

(d),

V and

(d) implies consider the

that)

p

:2 ciT(e\037)

=

0

i=l)

scalars

certain

for

C1 ,

. . . , C P'

T

(

since

T is

(a) implies Now

one-to-one.

But

(b).

assume

T(el), . . . , T(e 16.3,

f Cie;

we have

holds. in T( V)

(b) n)

)

\037=l)

we

By linearity,

p

= 0,

el , . . . ,

obtain)

and

Therefore

\037

= 0

i=l)

e p are

independent, so

Let {el ,. . . , en} be a are independent. Therefore,

dim T(V) < n.

, c.e. \037z

hence)

dim

T(V)

for

basis dim

= n,

C1

=

\302\267 \302\267 \302\267 = C

V.

T( V)

By

p

(b),

=

Therefore n elements

the

> n. But,

so (b) implies (c).)))

O.

by

Theorem

589)

Exercises)

Next,

T( V).

assume (c) holds and Then y = T(x)for some

. . . , en}

{e l , x in V,

let

so we

for

a basis

be

and hence

=

y

{T(e l ),...,

{T(el), . . . ,

T(e

will

= 0,

T(x)

then

T(x) = L cZT(e z=l)

and hence

,

= n,

{e l

,. . .,

so

O.

Let

en}

i ).

. . .,

. . . =

=

C1

T( V)

n

L czei

Cn = 0, since the elements T(el), on V. Thus, (d) implies (a) and one-to-one

Therefore x = 0, soT is

dim

write)

i=l) If

y in

ciT(ei ).

But we are assuming (c) implies (d). that T(x) = 0 impliesx =

n X =

element

Therefore

prove

we may

x E V,

If

i=l)

T( V).

spans

(d) holds. We

Finally, assume a basis for V.

be

T(en )}

basis for T(V).

n )} is a

= L

T(x)

i=l)

Therefore

any

n

n \"\" c.e. zZ' \302\243..,

X =

Take

V.

have)

are independent. is complete.)

n)

T(e

the

proof

16.8 Exercises)

1. Let

V = {O,

Tl , T2 ,

2. Let

=

V

of each

position In

Describe

2}.

Label

altogether.

where

as

them

3

. If it is, describe its formulas for determining give

on

3.

4.

V2

T(x,

y) y)

In

T(x,

y)

T(x, y, z), one on

3.

T(x, y,

14. T(x,

y,

formulas

T(x,

13 through z) is an

20, arbitrary

is, describe its and give formulas for

range

determining

3z). x + y +

(x,y,

\037

V be a

TO = law

(Tn)-l

I,

function Tn

V

3'

(u, v)

T( V2 ), let

in

=

of exponents:

= (T-l)n.)))

20.

which maps

TTn-l

T is

(x, y)

given for one-to-one

= T-l(U,v)

and

(x

-

= (2x

y)

V 3 \037

1).

V 3 is

In each

+ 1). y,x + y). - y, x + y).)

+ l,y

=(x =

e Y) .

defined

given for

formula

the

by

case, determine whether

Tis

each

T(x, y,

z) = (x

+ 1,Y

18. T(x,y,z)=(x + 19. T(x,y,z) =(x,x z).)

formula

the

by

inverses.

one-to-

point (u, v, w) in T( V3), let (x, y, x, y, and z in terms of u, v, and w.) for

17.

(x, 2y,

=

in

point T( V3);

z) = (z, y, x). z) = (x, y, 0). =

T:

a function

give their

are six the com-

v.)

10. T(x,y) 11. T(x,y)

If it

implies the that

9.

and

V,

There

showing

case determine whether

y) = (eX, T(x, y) = (x,

12.

(x, y,

on

8 . T (x,

= (x 2, y2).)

15. T(x,y, z) 16. T(x,y, z) 21. Let T: V and

point of u and

V.

table

defined

V 2 is

=

T( V)

multiplication

V 2 \037

each

= (x,x).

where

V, w)

T-l(U,

13.

V

for

terms

0).

of Exercises

each

V2); y in

- y).

= (x, = (x,

6. T(x,y)

7.

T( range x and

T:

V2 . In each

= (y, x).

T(x, y)

5. T(x,

in

point

which

V for

are one-to-one

a function

12,

through

inverses.

\037

make a

functions

which

the

showing

T: V

altogether. Label them as of each pair. Indicate composition

are four

There

V.

their

give

T6 and

Tl'.'\"

is an arbitrary

(x, y)

Vand

\037

table

functions

all

Indicate

pair.

of Exercises

each

on

are one-to-one

{O, 1,

T: V

functions

all

a multiplication

make

T4 and

functions

which

T(x, y),

I}. Describe

T3 ,

= (x T(x,y, z)

+

I,y +

y,x

+ y,y

1, z

+2,z + +

z) =

- 1). +3).

Y

z, x

+z).

+ z).)

V into itself. Powers are defined inductively by the for 11 > 1. Prove that the associative law for composition Tm Tn = Tm+n. Tn is also invertible If T is invertible, prove that

Linear

590)

matrices)

and

transformations

T denote functions with V and values in 22 through domain 25, Sand = TS, we say that Sand T If ST commute. TS. ST \037 general, n > O. that (ST)n = snTn for all integers 22. If Sand T commute, prove 1 that that ST is also invertible and 23. If Sand T are invertible, (ST)-l = T-1S- . In prove of inverses, taken in reverse order. words, the inverse of ST is the composition and commute, 24. If Sand T are invertible prove that their inverses also commute. If Sand T commute, prove that) 25. Let V be a linear space. In Exercises

(S

+ T)2 =

2ST +

S2 +

T)3 =

(S +

and)

T2)

S3 + 3S2 T

2 + 3ST

if ST \037 TS. be altered how these formulas must of V 3 into V 3 defined by the formulas Sand T be the linear transformations = (x, x + y, x + y + z), where (x, y, z) is an arbitrary y, x) and T(x, y, z) each of the following Determine the image of (x,y, z) under transformations:

+ T3

In

V.

other

.)

Indicate

26. Let (z,

(a)

ST

- TS,S2, T2,

(d)

Determine

Sex, y, z) = of V 3 . point TS,

ST,

- TS)2. (b) Prove that Sand T are one-to-one on V 3 and find the image of (u, v, w) under each of the transformations: S-1, T-1, (ST)-l, (TS)-l. following (c) Find the image of (x, y, z) under (T - I)n for each n > 1. 27. Let V be the linear space of all real polynomials p(x). Let D denotethe differentiation operator which maps each polynomial and let T denotethe integration operator p onto the polynomial q given by q(x) = S\037pet) dt. Prove that DT = I but that TD \037 I. Describe the null space and range of TD. 28. Let V be the linear space of all real polynomials p(x). Let D denQtethe differentiation operator that maps p(x) onto xp'(x). and let T be the linear transformation = 2 + 3x - x2 + 4x 3 and determine the image of p under each of the Let p(x) (a) following 2 2 transformations: D, T, DT, TD, DT - TD, T D - D2T2. (b) Determine thosep in V for which T(p) = p. in V for which (DT - 2D)(p) = O. those (c) Determine (ST

(TS)2,

(ST)2,

p

29. Let

V

and

those D be

xp(x). Prove

30. Let for

as

that

Sand T be in > 1.

p in

DT

V for

in

Exercise

- TD

Y(V,

which (DT - TD)n(p) 28 but let T be the

= I and

V) and

assume

-

DTn

that

Dn(p). transformation D = nTn-l for

Tn

- TS

ST

that

=

that maps

linear

= 1.

Prove

n

that

>

p(x) onto

2.

STn

- Tns

=

nTn-l

all n

V be the linear space of all real polynomials = Co + C1x + . an arbitrary polynomial p(x)

31. Let

..

Let R, S, Tbe the functions n in V onto the polynomials

p(x).

+ cnx

which rex),

map sex),

and t(x), respectively,where) n rex)

=

p(O)

,)

sex)

=

!

Ck Xk

k=l)

-1

n

,

t(x)

=

!

CkXk+1.

k=O)

= 2 + 3x - x2 + x 3 and determine the image of p under each of the following R, S, T, ST, TS,(TS)2, T2S2, S2T2, TRS, RST. (b) Provethat R, S, and T are linear and determine the null space and range of each. its inverse. (c) Prove that T is one-to-one on Vand determine > I and n and in terms of R. If snTn 1, express (Ts)n (d) T is one-to-one on V. If it is, describe whether 32. Refer to Exercise 28 of Section 16.4. Determine Let p(x) transformations: (a)

its

16.9

inverse.)

Linear

If V is with

transformations

with

prescribed

values

construct finite-dimensional, we can always values at the basis elements of V, as

prescribed

a linear transformation T: in the next theorem.))) described

V \037

W

Matrix

,.

Let e1, . . . , en elements in

16.12.

THEOREM ul

. . , un

T:

transformation

an

T maps

This

=

T(ek )

(16.7))

k =

for

Uk)

x in

element

arbitrary

space

. . .,n

1, 2,

then

k,

!

=

T(x)

x

Every

Proof

xn

Xl\"'\"

multipliers

(el , . . . , en)' linear. If x

some k, then all components = (16.8) gives T(ek ) Uk' as required. To prove that there is only one linear transformation and

=

e k for

= T(x) for

i = (1,0)andj =

( k\037lxkek

the linear

(0,1)

as

If x =

T(x)

Now,

, . . . , lV m be

has values in basis elements

is

V

shows

a

k\037/kT'(ek)

)

T' =

we have

transformation

= i +

+ x2 j

is an

x 2 T(j)

=

j

that a

,)

arbitrary

xl(i +

j) +

of linear

representations

16.12

Theorem

linear space el , . . . , en' WI

= xli

x1T(i) +

Matrix

16.10

let T'

1, so

be another

=

T(x).)

k\037/kllk

T,

which

completes

T:

V 2 \037

V 2 which

the proof.) maps the

basis elements

follows:)

T(i)

Solution.

is

which

n

=

=

x in V,

all

Determine

EXAMPLE.

the kth, (16.7),

satisfying

n

n

= T

T(x)

Since T'(x)

to

matter

0 except

X are

of el , . . . , en , ordered basis verify that T is

We find that)

T'(x).

compute

of

the

to

relative

X

components it is a straightforward

(16.8),

by

combination

a linear

as

uniquely

of

the

being

T

define

we

If

be expressed

V can

in

k .

XkU

k=l)

k=l)

the

.)

n

x =!xke

If

Let

linear

one

as folloH's:)

V

n

(16.8))

space V. only

that)

W such

V \037

a linear

591)

transformations)

for an n-dimensional linear W. Then there is one and

a basis

be

n arbitrary

be

of linear

representations

linear

=

T(j)

element of

x2 (2i

- j)

- j

2i

.)

T(x) is given

, then

V 2

=

(Xl

+

T:

V

\037

2x 2)i

+

(Xl

-

by)

x 2 )j.)

transformations)

transformation

W

of

a finite-dimensional

a given set of basis elements completely the W is also finite-dimensional,say dim W = m, and let suppose space basis for W. (The dimensionsnand m mayor may not be equal.) SinceT determined

W, each

element T(ek )

can

by its action on

be expressed

uniquely

as

a linear

combination

of the

WI , . . . , W m , say) m

T(e k) =

where

t lk ,

. . . , t mk

are

the components

!

tikw

i

,

i=l)

of T(ek )

relative

to the

ordered basis (WI'

. . . , W m)')))

Linear

592)

shall

We

(t ue

the m-tuple

display

and matrices)

transformations , . . . , t mk)

as follows:)

vertically, t

lk

t 2k) (16.9))

t mk) This

one

a

is called

array

each of

T(e l ),

of brackets to

pair

column matrix.

or a

vector

column

n elements

the

. . . , T(e

n ).

We place

We

have

them side by

vector for

a column

such

and

side

enclose

them

in

obtain the following rectangular array:)

is called a matrix

t il

t 12

tIn

t 21

t 22

t 2n)

t ml

t m2

t mn)

We call it an m by n matrix, of m rows and n columns. The m x 1 first row is the 1 x n matrix (t II , t I2 , . . . , tIn). matrix displayed in (16.9) is the kth column. The scalarst ik are indexed so the first subin which t ik i indicates the row, and the second script subscript k indicates the column or the ik-element of the matrix. The more compact notation) occurs. We call t ik the ik-entry

This

array or an m

X

n

consisting

The

matrix.

or)

(t ik ) ,)

used to

is also

denote the

whose

matrix

')

(tik)\037\037:l

ik-entry

is t ik

.

Thus, space V into an m-dimensional every linear transformation T of an n-dimensional space W gives rise to an m X n matrix (t ik ) whose columns consist of the components of to the basis (\037VI, . . . , Vt'm)' We call this the matrix T(e l ), . . . , T( en) relative representation of T relative to the given choice of ordered bases (el , . . en) for V and (WI' . . . , W m) for to the of any element w. Once we know the matrix (t ik ), the components T(x) relative basis (\037Vl, . . . , W m) can be determined as describedin the next theorem.)

.,

W

=

Let

16.13.

THEOREM

dim

m. Let

and let (t ik )

be the

T be

(el , . . . , en) m

X

a linear and

transformation

. . . ,

(WI'

W

whose entries

n matrix

in 2(V, W), be ordered bases for m)

are determinedby

m

(16.10)) Then

an arbitrary

T(e k )

=

!

tikw

i

for

,

i=l)

k =

element) 'Y1

(16.11))

X =

!xlce k k=l)))

1, 2, . . . , n

the

.)

Vt,here V

and

dim

V =

nand

W, respectively,

equations)

Matrix

in V

with

593)

transformations)

. . . , en)

to (e l ,

. . . , x n ) relative

(Xl'

components

of linear

representations

by T onto

is mapped

the

element)

m

=

T(x)

(16.12))

in W

with

components of

by

YiWi

. . . , W m)'

to (WI'

(YI, . . . , Y m) relative the linear equations)

components X

\037

i=l)

n

(16.13))

=

Yi

\037 tikx

each member of

T to

Applying

Proof

n

n

T(x) =

...,m

1, 2,

each

Y i

a pair

chosen

Having

every linear

m

m

k\037lXk

\"'!:/ikWi

V

T:

i\037

\037

W

(16.10), we

and

to the

obtain)

m

( k\037/ikXk

)

= Wi i\037/iWi')

proof.)

(\037i'l ,

a matrix

has

related

are

Yi

.)

n

This completes the

of bases(el , . . . , en)

transformation

using

=

by (16.13).

is given

and

(16.11)

= k\037lxkT(ek)

where

i =

for

k

k=l)

The

. . . , \037vm) for

Vand

representation (t ik ).

W, respectively, if Conversely,

as a rectangular matrix (t ik ) and choose a pair of easy to prove that there is exactly one linear transT at the basis this matrix representation. We simply define T: V \037 W having formation Theorem there is one and only elements of V by the equations in (16.10). Then, 16.12, by of an T: V \037 W with these prescribed values. The image one linear transformation T(x) x V and in is then (16.13).) (16.12) given by Equations arbitrary point

we start

with

3

arranged

it is

then

W,

of a linear

Construction 2 x

the

with

scalars

mn

for Vand

1.

EXAMPLE

start

any

bases

ordered

[\037

Choose

the usual bases

represents a in

linear

V 3 onto the

of

vector

(YI

, Y2)

V be

2.

in

Construction

the linear

V 2

of a

space of all sion 4, and we choose the basis maps each polynomial p(x) in transformation

of

degree

nlatrix.

a given

Suppose we

of V into < 2. In W we

-:l)

for

vectors

T: V 3 \037

Y2

EXAMPLE

\037

coordinate

unit

transformation

YI

Let

from

transformation

matrix)

to

according

V 2

.

Then

maps an arbitrary linear the equations)

=

3x I +

X 2

=

Xl +

OX 2

matrix

V 3 and

V 2 which

-

+

representation

the given matrix (Xl' X2 , x 3 )

vector

2X3)

4x 3 .) of

a given linear

3.

transforn1ation.

has dimenThis polynomials p(x) of degree < space which (1, x, X2, X3). Let D be the differentiation operator V onto D as a linear its derivative p'(x). We can regard

real

W, where W is the 3-dimensional space of all real polynomials choose the basis (1, X, X2). To find the matrix representation of

D)))

Linear

594)

relative to express

2

D(x

The

coefficients

D.

Therefore,

= 0

= 0

D(I)

) =

+

Ox 2 ,)

+

Ox

0 + 2x +

2x =

of these the

of the basis elements

combination

a linear

(differentiate)each basis element

we transform

of bases,

choice

this

as

it

and matrices)

transformations

1 =

) =

3x 2

3

D(x

,)

polynomials determine the

Ox +

1 + =

the

we

0 +

Ox

V and

of that)

find

2 ,)

Ox + 3x2

.)

matrix representation

following

3

X

of

4 matrix:)

0

0

I

o

by

Thus,

of the

columns

is given

representation

required

=

D(x)

2

Ox

W.

of

o 0 2 0 .

000 To also

3)

on the basis elements but in Wand use, instead, the transformed into the same poly-

that the matrix representation depends not only emphasize on their order, let us reverse the order of the basis elements 2 (x ,

basis

ordered

obtained

nomials

basis (x2 ,

x, 1).

x, 1)

appear

order.

reversed

in

V

are

of these

the components

but

above,

of

elements

basis

the

Then

to

relative

polynomials

the matrix representation

Therefore,

the new

of D now

becomes)

000

3

o 0 2 0

o I 0 Let 1 +

us compute a third x + x 2 + x 3) for

formed as

matrix V,

and

representation

matrix

representation

in

x) = I

+

D(1

D( 1 +

the

x, x

0)

for D, using the basis (1, 1 + x, for W. The basis elements of )

x + this

x2

= 1+

I

I

0 0 0

Since it is tion

by

matrix all

the

x + x2 ,

V are

trans-

will

have

0

=

I +

2x

,)

2x + 3x2 ,)

1

.

3)

matrix representation in

possible to obtain different choices of bases, it

different

entries

of a

+ x2 )

is

case

0 0 2 2

Construction

+ x

D( I

,)

x 3)

+

0

16.11

1+

follows:)

D(1)= 0 ,)

so

basis (1,

the

2

.

diagonal

form

of a given linear transformachoose the bases so that the resulting shows a particularly that we can make simple form. The next theorem from the upper left-hand corner))) starting except possibly along the diagonal matrix

is natural

representations

to try to

matrix.

the

of

entries

t

=

ik

this

Along

ones being

number of

the rank

to

be a

said to

k is \302\245=

in

matrix

matrix.)

diagonal

16.14. Let V and W be finite-dimensional linear spaces, with m. Assume T E 2( V, W) and let r = dim dim W T( V) denote the rank W such that) V and a basis (\"\"1' . . . , \",'m) for exists a basis (e1, . . en) for =

.,

T(e i) =

Wi)

for

0)

for

by zeros, the (t ik) with all

followed A

THEOREM

(16.14))

595)

form)

diagonal

will be a string of ones of the transformation.

there

diagonal

equal

0 when i

matrix representation

of a

Construction

i =

1, 2,

dim

V =

nand

of T.

Then

there

. . . , r ,)

and)

T(e

(16.15)) the

Therefore,

diagonal

matrix

of T

(t ik )

=

i)

+ 1, . . . , n

.)

bases has all entries

relative to these

zero

except

for

the r

entries)

t 22 =

t 11 =

First

Proof

r, the space T( form

elements

i = r

\302\267 \302\267 \302\267 =

t rr = 1 .)

we construct a basis for W. Since T( V) is a , . V) has a basis of r elementsin W, say \0371!1

of some

a subset

basis for

we can

Therefore

W.

subspaceof

W with

dim T( V)

=

15.7, these

By Theorem

. . , W r'

adjoin elements \",'r+

1, . . . ,

w m so that)

( \"\"1 ,

(16.16)) is a

basis for W. we construct a

N ow

one

least

i =

basis for

1, 2, . . . , r so 16.3 we have

elements, the

that

of

in

k elements

n

= V

a basis

k +

in

Now let k r. Since dim we

which

be

in (16.16)

V.

dim

Since

independent. Suppose

that

V = some

n

=

as er+l

Therefore, to

. . . , er =

linear

image of at

space N(T) has a basis . For each of these the proof, we must show

k, the

, . . . , e r+k

complete

, e r+ 1 ,

. . . , er+

k))

r + k, we need only show that these combination of them is zero, say) r+k

! cie =

(16.18))

O.

i

i=1)

Applying

is the

call it e i . Then T(e i ) = Wi for the dimension of the null space N(T). Vand

N(T)

designate

\",'i

set)

( e1, for

r elements

first

such element

(16.15) is satisfied.

Equation ordered

(16.17)) is

of the

Each

V.

in V. Choose one is satisfied. (16.14)

element

By Theorem consisting

. . . , W r , W r+ 1 , . . . , W m))

T and

using Equations

(16.14)and

r

r+k \037 c.T \302\243..1, ( 'l

e. )

i=l

we find that)

(16.15),

=

\037 c.w. \302\243..1,'l i=l)))

=

0

\302\267

elements are

Linear

transformations

independent,

and hence

596)

But

in (16.18)

terms

are

. . , \\Vr

H'l,'

are zero, so (16.18) reduces

and matrices) Cl

. . . =

=

Cr =

O. Therefore,

first

the

r

to)

r+k = o. .L cle i i=r+l)

e r+ l

But

. . , er + k

,.

are

all the Ci

c r+ k = O. Therefore,

basis for

This

V.

and they form a basis for N(T), are so in the elements zero, (16.18)

since

independent

\302\267 \302\267 \302\267 =

in

cr + l = form a

hence (16.17)

the proof.)

completes

2 of Section 16.10,where D is the differentiation operator of into W of polynomials of 3 the space < polynomials maps space degree degree< 2. J n this example, the range T(V) = W, so T has rank 3. Applying the method we choose any basis for W, for example the basis (1, X, x 2). used to prove Theorem 16.14, 3 in V which onto these elements is given We A set of polynomials map by (x, !X2, tx ). V the constant polynomial 1, which is a basis extend this set to get a basisfor by adjoining for the null space of D. Therefore, if we use the basis (x, ix2 , !x 3, 1) for V and the basis 2 for D has the diagonal form) (1, x, x ) for W, the corresponding matrix representation We

EXAMPLE.

refer

to Example

the

which

of

V

[\037

\037l)

vector space V l1 , the usual basis of unit coordinate vectors is to be In exercises matrix mentioned. concerned with of the specifically \037 W where V = W, we take the same basis in both Vand W unless

the involving basis is

exercises

all

chosenunless

another

a linear

transformation T:

another

choice

V

is indicated.

1. Determine the

matrix of each of the identity transformation,

(a)

(b)

the

zero

transformation, a fixed by

(c) multiplication

2. Deternline the T: (b) T: (a) (c)

3. A

V 3 \037

V2 ,

V 3 \037

V2 ,

(a)

of

T(3i

Compute

(c) Solve A

the

5. Let

the

- 4j) and

(b) if the transformation

= i

+j

- 4j) in and of T2.

T2(3i

of T

matrix

part

T(k)

:

T(j)

,)

terms

basis (i,j) is replacedby

of i

- j

= 2i

as follows:)

.)

andj.

(e l , e2),

where el

=i

as follows:. Each vector to yield T(x, y). Determinethe y-axis and then doubled in length T: V 3 \037 V 3 be a linear transformation such that)

linear

Vn

into

projections.

following

T(x l , X 2 , X 3 ) = (Xl' x 2 ). where T(XI, X2 , X3) = (X2, X3)' where T(x l , X2 , X 3 , X 4 , x 5 ) = (x 2 , x 3 , X4)' T: V2 \037 V 2 maps the basis vectors i and j

transformation

(b) Determine the

Vn

where

T: V5 \037 V 3 , linear

of

transformations

linear

following

scalar c.

for each

matrix

the

T(i)

4.

\037

Exercises)

16.12

In

r

= 2i

T:

+ 3j

+

V2

5k

\037

,)

V 2 is

T(j

defined

+ k)

= i

,)

T(i

+ j

+ k)

- j, e2

=

3; +

j.

(x, y) is reflected in matrix of T and

of T2.

=j -

k

.)))

Linear

(a) Compute T(i + 2j + (b) Determine the matrix 6. For

=

in Exercise

0), e3

(1, 0,

597) of T.

rank

and

nullity

of T.

transformation

linear

the

(2, 3, 5), e2

the

determine

and

3k)

of matrices)

spaces

=

5, choose both

to be

bases

-1), and determine

(0, 1,

of

matrix

the

(el , e2 , e3 ), where e l = T relative to the new

bases.

7. A

(1,1),T(k)

=

(a) Compute

(d) Find in

(e l form.

linear

Compute

T(2i

(b)

the

(c)

Find

- 3j) and

linear

for

such

Compute

(b)

Determine

linear

the basis

11.

space of

14.

(sin x,

W2

(WI'

follows:

T(i)

(1, 2).

T will

of

matrix

=

W2

= (1,

be

0, 1),

, w 3)

for

= (1,

T(j)

for

matrix

the

of

matrix

be

T will

1, 1).

V and of T will

V.

T(3e

the nullity

to

T relative

real-valued

operator.

to which

relative

2 and

dimension

determine

subspace In each

V 3

each

basis

with

(e l

, e2)'

Let T: V

\037

W

that)

the

given

a new

find

be

2e2) =

l +

and

of

rank

7eI + 23e2') T.

basis.

basis of

form

the

(e l

+ ae2,

2eI + be2)

for

W,

form.

in diagonal

each of

functions, Use the

case,

cos x).

given set find the matrix

15.

(-cos

16. (sin

( 1, 1 + x, X (eX, xe ).)

the

as a of

following for basis

D and

is independent and Vand let D: V \037 V be

sets

of D2 relative to

this

choice

1+x

+ eX).

x, sin x). x, x sin x, x cos x). x, eX cos x). 2x 3x, e cos 3x).)

x, cos

17. (eX sin 18. (e2x sin

x 2 , x 3) in the linear V of all real polynomials of degree < 3. space D denote Let the differentiation operator and let T: V \037 V be the linear transformation which to the given basis, determine the matrix of each of the maps p(x) onto xp'(x). Relative TD transformations: (a) T; (b) DT; (c) TD; (d) DT; (e) T2; (f)T2D2- D2T2. following Refer to Exercise 19. Let W be the image of V under Find bases for V and for W TD. the matrix of TD is in diagonal relative to which form.)

19. Choosethe

20.

, e2 ) all

the

(1, 1),

basis.)

12. (1,x, eX). 13.

(e l

finite-dimensional

the differentiation

of

matrix

the

which

=

of T.

and rank

nullity

WI

to which

vectors as

the basis

e2) = 3el + ge2,)

- el) and

T(e2

(a)

(c) Use the relative to

of

and

V2

transformation

T(el +

spans a

the

determine

V2 , where

V2 relative

for

W2)

V 3 maps

= (1, 0, 1) and T(i) linear spaces,eachwith

W be

Vand

the

\037

(wI'

in

of T.

matrix

(el , e 2) form.

and

Exercise8 if

9. Solve

In

V 3

T: V2

bases

diagonal

be a

for

1,0,1).

Determine

10. Let

, e 3)

in

transformation

(a)

in

=

0), T(j)

of T.

rank

and

nullity

V 3 and the basis (WI' w2) relative to these bases.

k)

, e2

the

determine

and

of T

(i,j,

bases

T(j) = ( -

= (0,

of T.

matrix

matrix

the

diagonal

8. A

T(i)

(1, -1).

T(4i - j + k)

(b) Determine the (c) Use the basis Determine

vectors as follows:

the basis

T: V3 \037 V 2 maps

transformation

linear

(1, x,

Linear spaces

16.13 We

basis

have

formations. necessarily

of

matrices)

seen how matrices arise in a natural Matrices can also be consideredas being connected

to linear

way as objects

transformations.

representations of

existing As

such,

linear

in their own right, they form another

transwithout

class

of)))

Linear

598)

and matrices)

transformations

on which can be defined. The connection algebraic operations transformations serves as motivation for these but this connection definitions, will be ignored for the moment. Let m and n be two positive integers, and let Im,n be the set of all pairs of integers (i,}) such that 1 < i < m, 1 < j < n. Any function A whose domain is Im,n is called an m X n matrix. The function value A(i,j) is called the ij-entry or ij-element of the matrix and will be denoted also by a ij . It is customary to display all the function values in a rectangular

mathematical objects linear

with

of m

array consisting

The

a ij

elements

numbers,

may

sometimes

but

for example,

objects,

ana

rows

as follows:)

n columns,

an

a12

a 1n

a 21

a 22

a 2n)

amI

a m2

a mn)

be arbitrary objects it is convenient

functions.

We

also

of any kind. Usually they will be real or complex to consider matrices whose elements are other denote matrices the more notation) by compact

= ( aij ) rn,n i ,j=1)

A

A =

or)

(aij)

.)

A 1 x n matrix is called a row n, the matrix is said to be a square matrix. Inatrix; 1 matrix is called a column matrix. Two functions are equal if and only if they have the same domain and take the same function value at each element in the domain. Since matrices are functions, two matrices A = (a ij ) and B = (bij ) are equal if and only if they have the same number of rows, the same number of columns, and equal entries aij = b ij for each pair (i,)). Now we assume the entries are numbers (real or complex) and of we define addition matrices and multiplication scalars the method used for realor same by any complexby If

m =

an m

X

valued

functions.)

we define

= (aij) and B = (bij) are cA as follows:) + Band

If A

DEFINITION.

matrices

A

+

A

The

sun1 is

defined

EXAMPLE.

only

have

A +

B=

.) b.1.}') the

cA =

same

matrices and if

fn X n

(ca ij )

c is any

scalar,

.)

size.)

If

0

1 2 and

[ -1 we

(a.. 1.} + B have

A and

when

A = then

B =

two

0

B=G

-:] 4

2 2A [\037

-2

-2

-\037l

= [-\037

0 -:l

(-l)B

:l =

0

-1 [-5

2 -I} -3)))

linear

between

Isomorphism

transformations

and

599)

matrices)

whose elements are O. With that the collection of all m X n If the entries are real this linear space by Mm. n' matrices is a linear space. We denote M m, n is a complex are complex, the space Mm,n is a real linear space. If the entries numbers, In fact, a basis for linear space. It is also easy to prove that this space has dimension Inn. define

We

these

of the

consists

Mm,n

For

0 to be the m x the zero matrix it is a straightforward exercise

n

:

[:

:l

a basis

form

verify

one entry

having

equal to 1 and all others equal

to

O.

the six matrices)

example,

[\037

matrices

Inn

all of

matrix

to

definitions,

\037

for the set

:l

of all

2

[:

:

X 3

matrices.)

16.14 Isomorphism between linear

\037l

[\037

:

and

transformations

[:

:l

\037

[:

:l

:

\037l)

matrices)

Let V now to the connection between matrices and linear transformations. V = n and dim W = m. Choose a dim finite-dimensional linear spaces with basis (el , . . . , en) for V and a basis (H'l , . . . , H'm) for W. I n this discussion, these basesare transformations of V into the linear space of all linear fixed. Let 2( V, W) denote kept of T relative to the given bases. We recall W. If T E fe( V, W), let In(T) denote the matrix that meT) is defined as follows. as a linear combination of the basis of each basis element ek is expressed The image

We

return

W be

and

in W:)

elements

m

(16.19))

T(e

=

k)

L

tikw

k =

for

i

1, 2, . . . , n

.)

i=l)

scalar

The

t ik

multipliers

the

are

n1(T) =

( 16.20))

(16.20)

Equation

are matrices in the range of m M m,n is

e,

defines a new function Since every m X

M m,n'

16.15.

ISOMORPHISM

(tik)\037\037:l

Thus, we .)

domain is \037(V, whose values W) and is the matrix meT) for some T in \037(V, W), shows that the transformation m: \037(V, W)-+

m whose

THEOREM.

For

all Sand

T

in

\037(V,

W)

have)

m(S

+ T)

= m(S) +

meT))

and)

meeT)

Moreover,)

m(S)

so

have)

n matrix

is M m,n' The next theorem linear and one-to-one on \037(V, W).)

THEOREM we

of meT).

ik-entries

m is

one-to-one

on

\037(V,

W).)))

= meT))

implies S

=

T,)

= em(T)

.)

and all

scalars

Linear

600)

Proof

matrix

The

matrix m(S) is

from the

is formed

m(T)

formed from

the

and matrices)

transformations

=

Seek)

L

for

SikWi

(16.19).

the

Similarly,

the equations)

m

(16.21))

in

tik

multipliers

Sik in

multipliers

k =

1, 2, . . . , n

and)

(cT)(

i=l)

.)

we have)

Since

m

+

(S

m(S + T) =

we obtain

k)

=

m

L

+

(Silc

i=l)

(Sik

tik) =

+

tik)W i

m(S) +

m(T)

suppose

that

and

e k) =

L

(ctik)W

i

= (ct ik ) = cm(T).

m(cT)

,

i=l)

This proves

is linear.

m

that

T)(e

To prove

is one-to-one,

m

that

(t ik ).

(16.19)

Equations so Sex) = T(x)

for all

x

and (16.21) show that in V, and hence S =

m(T), where S = (Silc) and T(e k) for each basis element

=

m(S)

=

Seek)

T = ek ,

T.)

function m is called an isomorphism. For a given choice of bases, m one-to-one between the set of linear transformations correspondence of addition and multipli2( V, W) and the set of m x n matrices M m,n' The operations cation by scalars are preserved under this correspondence. The linear spaces 2( V, W) and M m,n are said to be isomorphic. Incidentally, Theorem 16.11 shows that the domain of a one-to-one transformation linear has the same dimension as its range. Therefore, dim 2(V, W) = dim M m ,n = mn. The

Note:

a

establishes

If

V

=

to the

corresponds

choose the same basisin identity transformation I: V to 1 and all others eq ual to

if we

Wand

diagonal entry equal and is denoted by I 16.15

of

Multiplication

O.

matrices

transformations can be multiplied in such a way that of matrices to the composition of the linear transformations We recall that if T: U \037 Vand S: V \037 Ware ST: U \037 W is a linear transformation given by) Some linear

= S[T(x)])

ST(x)

U, V, and

that

dim

for

by

U,

U = n

Choose

bases

the matrix

T is a p

following

definition

of matrix

This

extends

V,

of composition.

means

the

of

product

matrices

two

they represent. linear transformations,

for all x

Now we shall corresponds

their composition

U .)

in

Ware finite-dimensional,say)

matrix,

m(S)m(T).

V is

or by In.)

definemultiplication

Suppose

W, then the matrix m(l) which an n X n diagonal matrix with each This is called the identity or unit matrix Vand

both \037

W.

and

the

x

n

V =

dim

,)

Relative matrix,

and

multiplication isomorphism

P

,)

dim

W =

to these bases, the matrix of will

enable

property to

us to products.)))

the

m

.)

matrix

ST is an

m

is an m m(S) X n matrix.

deduce the relation

X

p

The

nz(ST) =

A be any m

Let

DEFINITION.

=

A

and let

matrix,

p

the m

be

X n

p

(cij )

matrix,

say)

.)

(bij)\037,'jn=l

C =

n matrix

X

whose

is given

i)-entry

by)

'P \"'\" . a. \037 l k b k 3'

c.,l3 =

(16.22))

B be any B=

and)

(aij)2;:1)

AB is definedto

the product

Then

X

601)

of matrices)

Multiplication

k=l)

The product AB is of rows of B.)

Note: number

the

of columns

the number

unless

defined

not

of

If we write Ai for the ith row of A, and Bj for the jth column of B, and p-dimensional vectors, then the sum in (16.22) is simply the dot product of AB is the dot product of the ith row of A with the words, the ij-entry =

AB

Thus,

.

(Ai

is the

AB

1

0]

2

A

=

[A of AB

. BI

.4

A

I

= 3

A 2 \302\267 BI =

(-1)

.4

dot product.)

of

the

-1

5

. SinceA

is 2

o

X 3 and

B is 3

X

2,

2)

2

+ 1. 5

I

\302\267

BI

\302\267

BI

2

AI

\302\267

A

' B2 ]

2

B2

=

17 21 [ 1

-7)

.

]

as follows:)

computed

5 + + 1 \302\267

column of B:)

jth

\302\267 =

1 7,)

0 .

0=

0

+

B2 = A I \302\267 A 2 . B2

1,)

3 .

= (-

6 + 1 . (-

1)+ 2

1). 6 + 1

\302\267 =

2

- 1) +

\302\267

(

21

,)

0 .2 =

-

. 1+

. 2

7

Let)

2.

EXAMPLE

are

as

In other

2 matrix)

X

AB

The entries

=

B

and

[ -1) the product

2

A =

Let

1.

EXAMPLE

1

Bj.

6

4 3

of these

think Ai'

. Bj)\037\037\037 l,3-) l

can be regarded as a generalization

matrix multiplication

equal to

A is

-2 A =

1 .

B=)

and) \037

[\037

-:]) 2)

Here

A is

2

X

3 and

B is 3

X

AB is the

1, so

AB

=

AI [A

since 8.)))

Al

\302\267 =

Bl

2

\302\267

(-2)

+

1

\302\267

1 +

(-3)

2

-9

Bl

= Bl

\302\267 =

2

1 matrix

\302\267

\302\267 2

X

]

-9

[ and

8)

]

A 2

given

by)

, \302\267 =

BI

1 . (-2)

+ 2

4

=

.)

Linear

602)

3. If A and B are both defined. For example, if)

EXAMPLE

BA are

and matrices)

transformations

same size, then

of the

matrices

square

and

AB

both

1 =

A

AB= This

BIp = B for

in

is the p

If I p

4.

EXAMPLE

that

shows

example

and

m X

every

-28]

[I3 2

AB

general

xp

,

[-I \037

If AB

BA.

For

p matrix B.

we prove

the

that

matrix

A

for

A

every

\037 \037J[\037

of a

= BA, we say

and

B commute.)

p x

n

matrix

A,

example,)

[\037\037

Now

12) lOJ.

3

I pA =

then

matrix,

identity

=

BA

\037][\037]=[\037l

[\037\037

and

:l

[\037

\037]

that

find

we

B =

and

-1

[

\037

\037l)

\037]=[\037

composition ST is the

of the

product

matrices m(S)

m(T).)

Let T: U

16.16.

THEOREM

\037

V and

S:

the

by

dim

Assume

U, (VI'

. ..,

V p) a basis

W be

linear

a fixed choice

where U,

transforn1ations, of

the matrices

bases,

W

V,

of S,

T,

equation)

m(ST) Proof

V \037

linear spaces. Then, for

are finite-dimensional ST are related and

= m(S)m(T)

U = n, dim V = p, dim for V, and (WI' . . . , W m)

W

=

a basis

.)

Let (ul , . . . , un) for W. Relative to

m.

be

a basis

for

these bases,we

have)

m

m(S) =

where

(Sij):;:I'

S(v

k)

=

I

SikWi

for

k =

1, 2, . . . , p ,)

for

j =

1, 2, . . . , n

i=1)

and) p

meT)

=

(tij)\037j\0371

,

where

T(u j )

I tkjV k=I)

k

.)

we have)

Therefore,

ST(u

j)

= S[T(u j)]

p

p

=

m

m

= k\037/kjS(Vk)

so we

=

p

= k\037/kj

i\037/ikWi

i\037

( '1-1Siktkj

)

Wi'

that)

find

p

m(ST)

We

mutative laws.)))

\\m,n

= (

\"\037lSiktkj

=

matrix multiplication already noted that law. The next theorem shows that it does

have

m(S)m(T)

\302\267)

Ji,j\037l

does satisfy

not always satisfy the associative and

the distributive

com-

Exercises)

ASSOCIATIVEAND

16.17.

THEOREM

(a) If the

B are of the

A and

Assume

and CB

if CA

are meaningful,

These

Proof

plication, linear

the

prefer

A =

composition,

we obtain

equation,

proof of (b) can be

meR),)

B=

=

of matrix multi-

definition

\037

finite-dimensional

V \037

S:

V,

m(T)

W

R:

W,

\037

X

.)

From the associativelaw

= BC.

16.16

Theorem

= m(RS)m(T)or A(BC)= of argument.)

.)

Introduce

T: U

C

m(S),)

lalv)

the

from

type of argument.

bases,

which

(AB)C,

more to

once

proves

(a).

for this

The

is a

square matrix,

\037re

pOH'ers of

integral

define

A

inductively

as

:)

AO =

= AAn-l)

An

I,)

>

n

for

1.)

Exercises)

16.16

1. If A =

BA,

2.

,)

by a similar type

given

If A

DEFINITION. follo\037vs

deduced directly we have)

m(R)m(ST)

H'e have)

laa')

(left distributive

16.16, we have m(RS) = AB and m(ST) we find that R(ST) = (RS)T. Applying

Theorem

By

distributive

linear transformations

that, for a fixed choice of

such

and BC are meaningful,

(right

CB)

following

U, V, W, X and

spaces

BC)

+

CA

can be

properties we

but

If AC

.)

H'e have)

B) =

C(A +

(associativelaw)

size.

same

B)C = AC +

(A + whereas

MULTIPLICATION.

MATRIX

}1'e have)

meaningful,

= (AB)C)

A(BC)

(b)

are

and (AB)C

A(BC)

products

FOR

LAWS

DISTRIBUTIVE

C.

A, B,

matrices

Given

603)

Let

AC, A

1 -4 4 [ -1 CA, A(2B

= [\037

3. In each

1 0

1 0 0 0 (a)

[

- 3C). Find

an 2

a, b,

c, d to

-

c=

=

[: -3

-2;l

-;

B +

compute \037],

x 2 matrices B such that

AB

(a)

= 0;

(b) BA

=

O.

\037J.

case find

0 0

B -\037l

a b

0 1 0 0 c 0 0 0 1

d

the

satisfy

given

equation.

1 0 2 0

1

-

9

6 5

b

c

4

9

0

0

1

1

0

1

0

0

0

0

1

0)))

(b)

[:

\037J

= [\037

0

6

9

8 :l

C, AB,

Linear

604)

4. Calculate

- BA

AB

A =

(a)

B =

[-;

\037;l

[\037

(b)

\037

=

6. Let A

[ sin =

A

[

An and

[ -1

10. Find

11. (a)

O

. Prove that

A

\037 \037 \037

.

of the

four

compute

A3 and

Compute

O.

A

4

An.

Guess

.

a general

1]

2 =

- I and

2A

AIOO.

compute

matrices A such that A2 = O. 2 matrix A commutes with every 2 x

2 matrix

if and

only

if

A commutes

matrices)

\037l)

[\037

all such matrices A. A 2 = I is satisfied

Find

(b)

=

[

prove

and

>

by induction.

it

[\037

12.

2

cos 28]

[ sin 28

a 2 x

that

with each

A

0, n

28

-sin

26

=

A2

1J

x 2

2

all

Prove

that

. Verify

m >

An,

compute COS

that

. Verify

00100]

=

A

and

all integers

\037J

6

cos 6 ]

\037 \037

1

9. Let

for

A2 =

that

-sin

6

\037

for

formula

\037;l

Am+n

[\037

6

COS

8. Let

-;

AnAm =

\037J,

=

A

that

prove

Verify [\037

7. Let

l

[_:

;l

is a square matrix,

If A

;

\037

B =

=

A

[ _: 5.

case.

each

in

and matrices)

transformations

by each

The equation

[\037

\037l)

of the 2 x

\037l)

\037J.)

[\037

2 matrices) bl ')

13. If

c are

band

where

-1

2 A

=

[ and

14. (a)

DA

-2

3]

not

that

the algebraic + B)2

=

for the 2 x

hold

(b) Amend matrices A (c) For

B

=

Find all 2 x

[

7

6

9

8]

'

x 2

2

find

2 matrices A

that

such

matrices C and

D

A2 = 1.) that

such

AC

=

B)

= B.

Verify

(A

do

and

-IJ

[\037)

-\037l)

numbers.

real

arbitrary

[:

\037l)

[\037

the and

which

right-hand B. matrices

A2

identities) + 2AB

and)

1

2 matrices

A

2]

and

identities

B to

valid

- B) =

1 0

-1

of these identities

B are the

+ B)(A

(A

= [0

members A and

+ B2)

=

[1 obtain

as stated

2

-

B 2)

.

2] formulas in

A

(a)?)))

valid for all

square

of linear

/)ystems

16.17 Systems of Let A =

numbers.

(a ij ) set

A

linear

605)

equations)

equations

x

a given m of m equations be

n

of

matrix

of the

n =

2 aikx k

(16.23))

and let cl ,

numbers,

. . .,

further

be m

Cm

form)

1, 2, . . . , m

i =

for

Ci

k=l)

,)

are regarded as . . . , Xu) for which all the A is called the coefficient-matrix of the system. equations are satisfied. The matrix the help of linear transformations. Choose the usual Linear systems can be studied with a bases of unit coordinate vectors in V n and in V m' The coefficient-matrixA determines X = (Xl' . . . , xn ) in Vn vector T: V n -* Vm , which maps an arbitrary linear transformation, the vector y = (Yl , . . . , Ym) in V m given by the m linear equations) onto a system

is called

of

linear

m

in n unknowns. Here Xl of numbers (Xl' n-tuple

equations

of the

A solution

unknown.

system is any

n

=

Yi

2

i =

for

QikXk

1, 2,

, . . . , Xn

. ..,m

.)

k=l)

(ci , .

c =

Let

(16.23).

system

. . , cm)

vector

the

be

in

V

m whose

can be written

This system

=

T(x)

The

are the numbers

components

more

simply

c

.)

if c is in the range of T. If exactly has a solution if and only If more than one X maps solution. has one the c, exactly system

has more

than

1.

EXAMPLE

tion.

sum

The

2.

EXAMPLE

one

3.

of one equation is

sum

With

1 gives

each

V n

c, the

onto

maps

system

solution.)

A systenl of two

\037i'ith

The system

solution.

no

be both

numbers cannot

A system

H'ith

one

exactly

exactly one solution: (x,Y) EXAMPLE

x in

one

system

onto

appearing in

as)

=

x +

1 and

Y

The system

solution.

1, x

=

+

Y

=

2 has no

solu-

2.)

+ y

X

= 1,

X

-

y'

=

0 has

(1, 1).)

A system 'rvith nlore in two unknowns,

than

one solution.

has more

one

than

The system x

+y

= 1, consisting two numbers whose Any

solution.

a solution.) linear

system

(16.23), we can

associate another

system)

n

2aikxk

= 0

i =

for

1, 2,

...,m

,)

k=l)

obtained

by replacing

each Ci

in

(16.23)

to (16.23). If c \037 0, system sponding x in V n will satisfy the homogeneous

correby O. This is called the homogeneoussystem A a vector is called (16.23) nonhomogeneous system.

system if and T(x)

=

0,)))

only

if)

Linear

606)

where T is the

ous system

of the

solutions

between

relation

by the coefficient-matrix. The homogenenamely x = 0, but it may have others. The set of homogeneous system is the null space of T. The next theorem describesthe solutions of the homogeneous system and those of the nonhomogeneous determined

transformation

linear

always

and matrices)

transformations

one solution,

has

system.)

16.18.

(a) If

a vector x

is a

a

(b) If

solution

of v is

vector

Let T: V n as described

matrix,

T(b) = c.

the

x and

Let

T(x)

\037

system splits system,

that

=

v

+

x

-

b is

b

a

system.)

linear transformation determined by the coefficientis a solution of the nonhomogeneous system we have two vectors in V n such that v = x-b. Then we have)

v be

shows

=

T(x if

only

- b) = T(x)= O.

T(v)

of

that the problem

=

T(b)

- c .) and

(a)

all solutions (2)

finding

(b).)

of a

solutions

all

finding

(1) Finding parts: the null space of T; and

determining

T(x)

This proves both

two

into

naturally

is,

vector x

then the

system,

homogeneous

v =

be the

Vm

= c if and

theorem

This

the

b.

say

vector

the

above. Sinceb

T(v)

Therefore

of

solution,

then

system.

homogeneous

corresponding

a solution

of the nonhomogeneous

solution

Proof

of the nonhomogeneous system,

a solution

is

system (16.23)has a

the nonhomogeneous

Assume

THEOREM

v

of

nonhomogeneous the

homogeneous

one particular the null space of

solution b of

T, we thereby nonhomogeneous system. By adding b to each vector v in obtain all solutions x = v + b of the nonhomogeneous system. k independent of T). If we can find of N(T) (the nullity the dimension Let k denote will a basis for . V of form solutions . . the , k N(T), and we VI , homogeneous system, they in all combinations) v linear can obtain N(T) by forming possible every the

v

=

. \302\267 \302\267

t 1v l +

+

fkv

k ,)

is called the general solution scalars. This linear combination t 1 , . . . , t k are arbitrary solution of the nonhomogeneous the homogeneous system, system. If b is one particular all solutions x are given by)

where of then

x

This

linear

combination

Theorem 16.18can now THEOREM

x =

Let

16.19.

. . , x n)' (XI' \302\267

Y

=

be

=

b +

tlV

l +

\302\267 \302\267 \302\267

+

is called the general as follows.) restated

T: Vn (y 1 ,

\037

Vm

be the linear

tkVk

solution

of

the

nonhomogeneous

such

transformation

y m) and)

. , \302\267 \302\267

n

Yi

.)

=

! k=l)

aikx

k

for

i =

1, 2,

. . .,m

.)))

that

system.

T(x) = y, }vhere

the

k denote

Let

the

then

of

nullity

T.

If

solution

general

of

=

x

. . , tic

,.

tl

}t'here

This theorem

are

b, nor

system.

It does tell us

The

EXAMPLE.

x + }' = where

with fore

O.

to

what

Therefore,

t is

= (0,2)

(x, y)

16.18

t is

if a nonhomogeneous system has a particular determine solutions Vi , . . . , V k of the homogeneous when the nonhomogeneous system has a solution. expect the theorem.) very simple, ill ustrates

x + y = 2 has for its associated the null space consists of all

system

Since (t, arbitrary. basis (1, -1). A particular solution of the the general

where

k ,)

to decide

us how

tell

although

example,

following

tkv

does it tell us how to

solution

The

...+

tlv! +

scalars.)

arbitrary

does not

b +

c,

is)

system

nonhomogeneous

of the homogeneous nonhomogeneous system T(x) =

of the

solution

particular

the

k independent solutions

, . . . , V k are

Vi

0, and if b is one

T(x) =

system

607)

techniques)

Computation

-t) = t(l, -1),

this

system the equation homogeneous in V2 of the form (t, -t), a one-dimensional subspace of V 2

vectors is

nonhomogeneous system is (0, 2). nonhomogeneous system is given by) of the

solution

+ t(I, -1))

x

or)

=

There-

y=2-t,)

t,)

arbitrary.)

techniques

Computation

problem of actually computing the solutions of a nonhomogeneous methods have been developed for attacking this problem, system. Although many all of them require considerablecomputation if the system is large. For example, to solve a system of ten equations in as many unknowns can require several hours of hand comwith even the aid of a desk calculator. putation, shall discuss a widely-used method, known We as the Gauss-Jordan elimination method, which is relatively and can be for electronic simple easily programmed high-speed computing We

turn

to the

now

linear

machines. The method of a linear system: (1)

(2) (3)

Interchanging all

Multiplying

to

Adding

Each time we exactly

one

two equations; the terms of an equation a multiple

perform one of

the same

operations over

solutions. and

of applying

consists

over

these Two

again

three basic types of

scalar;

system we obtain a new system having systems By performing these a systematic fashion we finally arrive at an equivalent on the

operations

are called equivalent.

be solved

system by inspection. We shall illustrate the method with some specific examples. It method is to be appliedin general.) which

EXAMPLE

can

equations

of another.

such

in

a nonzero

by

equation

on the

operations

1. A system

\037i'ith

a unique

2x

solution.

-

xx

-

5y

+

2y

+

4y

+

will

Consider the system) 4z =

-3

z = 5 6z =

10 .)))

then

be clear

how the

Linear

608)

This particular

= 124,Y

solution, x

a unique

has

system

and matrices)

transformations

by the Gauss-Jordan elimination the letters x, y, z and the equals sign obtain

over again,

and

over

save

To

process.

= 31, which we labor we do not bother to =

but

75, z

instead

work

with the

shall copy

aug-

mented matrix) 4

-5 (16.24))

[:) obtained

the

by adjoining and

matrix

x, y, z back again ultimate

1

-4

6)

10)

]

system to the coefficient matrix. The on the rows of the augmented process we can put the letters

of the

members

right-hand

-\037

types of operations mentioned aboveare performed At any stage of the are called rOlV operations.

basic

three

-2

goal

insert equals signs along the vertical line to obtain equations. Our arrive at the augmented matrix)

and

is to

(16.25))

[

124

0

0

1

010

75

001)

31)

]

of row operations. The corresponding of equations is x = 124, system = 75, z 31, which gives the desired solution. Y The first step is to obtain a 1 in the upper left-hand corner of the matrix. We can do this either the second or third by interchanging the first row of the given matrix (16.24) with a succession

after =

row.

we can

Or,

row by!.

first

the

multiply

\037

[ 1 The next step is to the first row intact.

second row.

To do this we

Then

these two

After

make all the

=

\037

-4

and

first

rows, we

second

get)

.

-\037 10)

]

first column equal to zero, leaving - 2 and add the result to the by and add the result to the third row.

in the

entries

the first row

multiply

the first row

multiply

\037 6)

remaining

we

the

Interchanging

-1

by

operations, we obtain)

( 16.26)) \037

[\037

Now

we

repeat

adjacent to the the

second

two

row of

the zeros.

We

-I:})

on the smaller

process

(16.26)by

\037

a 1 in

can obtain -1.

-1 its

This gives us the 1

o [

- 2

matrix)

1 .))) 5)

-13 which

5 [ -2 left-hand upper

1-2

o -2

2

matrix

I:]

5] corner

by

appears multiplying

the second row

Multiplying

by

2 and

the result to the

adding

stage, the corresponding system -

31)]

is given

eq uations

of

x

get)

.

l\037

001)

[ At this

1-2

o

we

third,

1

1 -2 ( 16.27))

609)

techniques)

Computation

=

2y + z - 2z Y

by)

5

= 13

z = 31 .) These

z =

to give

31 ,) we

Or,

=

2 and

adding

succession,

62 =

2z = 13 +

13 +

the Gauss-Jordan

in the

elements

in

second and

the result to the

third

we row

EXAMPLE equations

the

multiply

one

third

by 2 and

2Y - z = 5

5 +

all

making

columns.

Multiplying

and

working

+ 150-

the entries zero the second row

H,ith

than

2x -

5y

+

2y

+

4y

+

above the of (16.27)

.

13

31)

] and in

then multiply (16.25).)

Consider the following system of

one solution.

more

124.)

31

third row by 3 and add the result to the first row, add the result to the second row to get the matrix

2. A system

31=

we obtain)

row,

[ 001) Finally,

=

process by

third

first

x

75,)

1 0 -3 o 1 -2

the

the

with

starting

us)

can continue

diagonal by

y

be solved

can

equations

backwards,

3

in 5 unknowns:)

( 16.28))

The corresponding

x

-

x

-

augmented matrix

of x, y, z and

u

-

v

-

u

+

v =

-

v

z

6z + 2u

2

-5

4

1

-2

1 -1

the

=

=

-3 5 10.)

is)

1 -4 The coefficients

4z +

6)

right-hand

1

-3

-1

1

2 -1) members

.

5

]

10)

are the same

as those in

Example

1.)))

Linear

610)

If we

used in

row operations

the same

perform

and matrices)

transformations

1, we

Example

arrive

finally

at the

augmented

terms

of u

matrix)

0

-16

19

o 1 0

-9

11

75

4)

31)

0

1

-3 [o 0 1 The

of equations

system

corresponding us) gIvIng

u

solution.

This

=

t2 ,

= (124

v)

=

9u-l1v

z=

31 +

3u

where

(x, y, z, u,

+ 16t1 - 19t2 , the parts

are

-

t 2 are

1 and

t

involving

+

(124,75,31,0,0)

t

gives the general solution of of the nonhomogeneous

the

of -4, 0, 1) are solutions a basis for independent, they form

real

Vs given

9t 1 -

111,31

in

t 1 and

t 2 , we

and

numbers,

+

3t 1

- 4t2

can rewrite

1,0) + 12 ( -19, The vector

system (16.28).

v,

determine

The

two

-11,

, t1 ,

this

-4,0,

as

t 2)) follows:)

1) .)

(124,75,31,0,0) vectors

corresponding homogeneous space of all solutions of the

the

and

by)

2

system.

the

-11,

and z

x, y,

arbitrary

9, 3,

1(16,

]

4v.)

v) in

75 +

.

19v

75+

solution

(-19,

for

y=

the vector

By separating

equation

particular

they

v

equations,

z, u,

(x, y,

and

t 1 and

z, u, v)

(x,y, is a

=

these

can be solved 124 + 16u -

x =

If we let x, y, z by

124

is a

(16, 9, 3, 1,0) system. Since

homogeneous

system.)

3. A

EXAMPLE

system

with

no

2x xx -

( 16.29))

This system third

identical

has been

the same row

Applying arrive

is almost

equation

at

the

augmented

to

5y

+

4z =

2y

+

z =

4y

+

5z =

of

that

changed from

the system)

Consider

solution.

6

-3 5

10.)

1 except that the coefficient of z in The corresponding augmented matrix

Example to 5.

2

-

1

-2

1

5

[ 1

-4

5)

10) ]

operations used in

5

Example

\302\267

1 to

transform

l\037

\302\267

1

-2 [\037

-\037)

is)

-3

4

matrix)

( 16.30))

the

o)

31)))

]

(16.24) into (16.27), we

Inverses

row is

bottom

the

When

has no

system

original

expressed as

an

that 0 = 31. Therefore (16.29) and (16.30) are equivalent.)

it states

equation,

solution since the

two

611)

matrices)

of square

systems

the

of equations did not exceed the number foregoing examples, the number are more equations than there the Gauss-Jordan unknowns, process is still For we consider the of which has the applicable. system Example 1, example, suppose solution x = 124,Y = 75, z = 31. If we adjoin a new eq uation to this which is also system satisfied by the same triple, for example, the equation 2x - 3y + z = 54, then the elimination process leads to the agumented matrix)

of the

each

In

of

If

unknowns.

124

0

0

1

o 1 0

the

by

of zeros

a row

with

a

O. \302\245=

The

has no

the system

0

1

31

o

0 0

0)

of

the

form)

1

0

0

matrix

z =

is not satisfied

which

eq uation

+

1, then

elimination

the

124

o 1 0

75

o

0

1

31

o

0 0

a)

gives a

last row now

adjoin a new equation x + y

if we

But

bottom.

0 =

equation

contradictory

a

that

shows

which

solution.)

of square

Inverses

16.19

o

for example the

leads to an augmented

process

where

along the

(124,75,31),

triple

75

matrices

= (a be a square n X n matrix. If there is another n X n matrix B such that ij ) I A is called nonsingular and B is called a where is the n x n I, identity matrix, then

Let A =

BA

of A.

inverse

left

Choose the usual basis

transformation

then B =

meT)

A

is

if and

nonsingular

let

and

Then we have the

T: Vn

the linear

V n be

-+

following.)

if

only

T is

invertible.

If

=

I,

=

0

BA

m(T-I).)

Assume

Proof

that A is nonsingular and that T(x) = 0,

x such implies x = O. Given the components of x. Since consisting of zeros, so B(AX) IX = X, so every component

implies that m(T-I)

= A.

in Vn

vectors

coordinate

The matrix

16.20.

THEOREM

of unit

matrix

with

=

m(T-l)m(T)

m(T)m(T-I)

B.

Conversely, is the identity

T(x) = is

of x

0, the

also

that let

matrix

a column

BA X be

= I. the n

product

matrix of

Therefore, Tis invertible, Am(T-l) = I. Multiplying

We shall prove X

AX

is O.

I or if T is invertible, then T-IT is the A is nonsingular matrix. Therefore

=

is an

zeros. and on

But

that

matrix

1 column n

X

T(x)

formed from =

B(AX)

the equation the left by

identity

matrix

1 column

= TT-I = I B, we find (BA)X

transformation

and m(T-l)A

=

I.)))

so

Linear

612)

the

All singular

In

inverse is also a =

AB

I. We

linear transformations have left inverses (if they exist)

of invertible

properties matrices.

particular,

call B the

for nonand unique, every left and BA = I, then

their

counterparts

are

other words, if A is nonsingular and denote it by A-I. The inverse A-I is also nonsingular

In

inverse.

right

and matrices)

transformations

of A

inverse

its inverse is A. of a of actually determining the entries of the inverse Now we show that the problem linear matrix is n to solving separate systems. equivalent nonhomogeneous nonsingular and let A-I = (bij ) be its inverse. The entries of A and Let A = (a ij ) be nonsingular A-I are related by the n 2 equations) and

n \"\" a. l.k b k '}, \302\243..,

(16.31))

=

D.. 'l.'} ,

k=l)

= 0 if i \302\245= For each fixed choice of oij = 1 if i = j, and Oii j. j, we can regard this in n unknowns of n linear equations as a nonhomogeneoussystem blj , b 2j , . . . , b nj . Since the A is nonsingular, each of these systems has a unique solution, jth column of B. All A and in differ their right members. have the same coefficient-matrix these only systems which can be expressed there are 9 equations in (16.31) if A is a 3 x 3 matrix, For example, linear as 3 separate systems having the following augmented matrices:)

where

[ If we

apply

all

al2

a l3

1

a\037n

a 22

a 23

0

a 31

a 32

a 33

0)

0

]

0

In and

actual solve

0

a 21

a 22

a 23

1

l_ a 31

a 32

a 33

0

process, we

o

1

0 b 22

[o

0

1 b 32)]

,

1 b 31)]

]

a l3

a 21

a 22

a 23

[ a 31

a 32

a 33)

process then

on the

of

right

of the line is the 3 x

3

the

. \037]

1

0

0 b 13

o

1

0 b 23

[o

0

1 b 33)]

matrices)

.

coefficient-matrix

0

a l3

1

a 2l

a 22

a 23

0

1

a 31 [all

a 32

a 33

0

0

. \037]

to)

leads

identity

O

respective augmented

,

a l2

0

0 bil

b12

b 13

o

1

0 b2l

b 22

b 23

[o

0

1 b 31

b 32

b 33)]

1

matrix

0 b 12

al2

all three systems have the same practice we exploit the fact that with the enlarged matrix) all three systerns at once by working

The elimination

The

0

all ')

at the

arrive

1

0 bll

o 1 0 b21 [o

a12 a 13

,

the Gauss-Jordan 1

all

vertical matrix.)))

line is the required

.

inverse.

The

matrix

on

the left

613)

Exercises)

It is not necessary to we can still

one of the to

the

A is

whether

advance

Gauss-Jordan

the

apply

elements

diagonal

in

know

nonsingular),

zero, and

will become

A

is singular

in the

somewhere

but

be possible to

not

will

it

If

nonsingular.

method,

(not

process

transform

A

matrix.)

identity

16.20 Exercises) exists, 1. x + y 2x - y

the

determine

-y

+ 3z +

general

=

5

4z =

11

+ z =

process to each

elimination solution.

Gauss-Jordan

the

Apply

5x + 3y

+

z = 1 3z = 2

7x +

+

5z =

+

+

2y

4y

4. 3x +

+

2y

+ 3z

7x + x +

+

4y Y

9. Prove

that

solution

3.

= 2 3 z = O.

the

if a

Find

x + y + 2z all solutions of the

=

of the

2y

-

-

tells how

y +

[:

;J

is nonsingular

y

+

3y

+

5

=

2

z

-

2z +

5x -

u

=

u = y

+

5

3. az =

6, has a

unique

-1

u =

-

z

=

2u

-

6z +

-2.)

z

=

-1

u =

-2

2u

-

-

d -b

b

a)

if and

only

J if

d

1

ad -

be

[ -

6.)

2 x

nonsingular

[ e dJ [ -e

that

2u

7.

=

system)

to determine all a

Deduce

-

z

3z = 2,

+

y

6z +

x - y + x+y+z exercise

u

= 8.

a

5x + 2Y

This

3z +

2

system)

x

solutions

-

2, 2x

when

5x +

11.

=

-

5x -

10. (a) Determine all solutions

(b) Determine all

=

3u

+

y

4x

system

\037 8.

2u

+ Y 7z + 3u = 3. + y + 2z + 3u + 4v = 0 + 2y + 7z + 11u + 14v = 0 + 3y + 6z + IOu + 15v = O. - 2y + z + 2u = -2 2x + 3 Y - z - 5u = 9

5z =

-

-

2x 7x 7. x 2x 3x 8. x

z = 1

5x + 3 Y

Y

6x + Y 6. x + Y -

3.

solution

u = 1

5z + 3z + - 4z +

-

If a

systems.

following

+

2y

x +

2. 3x + 2y + z = 1 5x + 3Y + 3z = 2 x + Y - z = 1. 3. 3x

-

5. 3x

of the

=

(ad

ad

2 matrices. Prove that)

-

- be

-b

ea'))) J

be)I .

\302\2450, in

which case

its

inverse

is)

Linear transformations

614)

the inverse of each of

Determine

:

\037 \037

I

[ -1

.

o 1 2

3

001

2

000

1)

15.

2)]

-0

[\037-;

\037l

o 0

\037:

[-\037

1. If

the

(a) If AB + BA = (b) If A and Bare (c) If A and Bare (d) If A, B, and A = 0, then (e) If A3 If the

statements

3. If A =

matrix

A2B3 =

nonsingular,

+ Bare

product of k

matrices

B is is nons

A +

then

AB

nonsingular.

in gular.

- B is

A

then

nonsingular.

nonsingular.

. . .

Al

2 4J

Ak is

matrix Ai is

each

then

nonsingular,

A =

matrix P such

a nonsingular

a i [ j b'J

j2 =

where

a =!(l

-1,

= A. Describe completely erty that A2

2 x

all

=

P-IAP

that

--

+ V5),and

2 matrices A

[ 0 -1 J b =

with

nonsingular.

O

6

, find

counter

B3A2.

then

nonsingular,

- I is

[5 The

0, then nonsingular,

A

1

or a column of zeros,prove it is singular. that about n x n matrices, give a proof or exhibit a

of zeros

row

following

example.

(f)

on matrices)

exercises

matrix has a

a square

2. For each of

020

1

0)

-:l)

Miscellaneous

16.21

0

0

1

00030 1 o 0 0 0 2

14.

4.

000

030

16.

0 0-

0 0

1

202

13.

16.)

234

I 12.

12 through

in Exercises

matrices

the

matrices)

and

-W

.

- V5), has

complex

entries

the

prop-

such

that

= A.

A2

5. If A2 6. The

=

A, prove

specialtheory

that of

(A

+ I)k

relativity -

= z, t' = aCt

= I

-

+ (2k

makes use 2 Here v

I)A.

of a set of

form x' a moving of velocity The linear transformation

equations

of the

the vxfc .) represents y' y, z' a = cfv c2 - v 2 , where and the speed of light, Ivl < c. (x', t') is called a Lorentz maps the two-dimensional vector (x, t) onto matrix relative to the usual bases is denoted by L(v) and is given by) =

L(v)

that L(v) is nonsingular 2 2 v)c f(uv + c ). In other Lorentz transformation.)))

Note

(u +

and

words,

= a

2)

I

c object, which

transformation.

Its

. ]

L(O) = I. Prove that the product of two Lorentz

that

vt),

_V

1 [ -vc-

= a(x -

L(v)L(u)

= L(w),

transformations

where w = is another

7. If

we

615)

on matrices)

exercises

Miscellaneous

the rows and columns of a rectangular matrix A, the of A and is denoted by At. For exanlple, if we transpose

interchange

is called the

1 A

_

[4

2

3

5

6

At =

then

')

]

so obtained

Inatrix

new have)

:})

[\037

Prove that transposes have the following properties: (At)t = A. (b) (A + B)t = At + Bt. (c) (cA)t = cAt. = = BtAt. if A is (d) (AB)t (e) (At)-I (A-I)t nonsingular. 8. A square matrix A is called an orthogonal matrix if AAt = I. Verify that the 2 x 2 (a)

. [

that its rows,

9. For

counter

(c) If A 10. Hadamard the

III.

entry

is 1 or

row,

considered

dot

The

Hadamard matrices arise in

give a proof

n matrices,

matrix, prove

set.

an orthonormal x

or elseexhibit

a

is orthogonal.

is orthogonal. B is orthogonal.

Jacques

as a vectpr in of any two distinct

product

x /1 orthogonal

n

any

AB

Hadamard

properties: -1.

following

I. Each II. Each

for

named

matrices,

+ B

A

is

If A

O.

in Vn , form considered as vectors statements about n following

example. and B are orthogonal, then and B are orthogonal, then and AB are orthogonal, then

A

(b) If A

with

real

each

for

of the

each

(a) If

is orthogonal

cos 0 ]

0

sIn

matrix

0

-sin

0

COS

(1865-1963), are

_ has length VIl

Vn ,

those

Il x

/l matrices

.

rows is O. in geometry and problems

of numbers, and the theory to in the construction of code comwords optimum applied recently space they In spite of their apparent munication. unsolved The problems. simplicity, they present many main unsolved problem at this time is to determine all n for which an /l x n Hadamard matrix certain

been

have

a partial solution. exists. This exerciseoutlines (a) Determine all 2 x 2 Hadamard matrices (there are exactly outlines a simple proof of the (b) This part of the exercise n

x n Hada/1zard simple

very

them to

the

1.

LEMMA

matrix,

lemmas rows

If

X,

n

where

concerning

of Hadamard

Y, Z

ponent

Xi

Write X = (Xl' Yi , Zi is either 1 or

2. ,

n

vectors

Y = (YI the

IIXI1

,.

product

lemmas

and

apply

then H'e have)

2

y).(X+Z)=

. . . , X n)' -1, then

If A is an on two

is based

theorem.)

Vn ,

in

theorem:

multiple 0.( 4. The proof Prove each of these

is a

vectors in n-space. matrix to prove the

are orthogonal

(X+

LEMMA

> 2, then

8). following

. . , Y n)' (Xi +

.)

Z = Yi)(X

. . . , Z n)' If each comZi) is either 0 or 4.)))

(Z1 , i

+

EXERCISES)

TO

ANSWERS

Introduc Exercises (page

*1 1.4

1. (a)

3 4

2.

(c)

iab

3.

(b)

Sn


n

> 0

if each ak 4)

1)

Chapter 1.5

(page 56)

Exercises

= 3,f( -2) = -1, -f(2) = -3\"f(\037) =\037, l/f(2) = l,f(a + b) = a + b + = a + b + 2\"f(a)f(b) = ab + a + b + 1 ,f(a) + f(b) = = = = -3\"f[g(2)] = 0, + ,f(2) g(2) 2,f(2) g(2) 4,f(2)g(2) -3,f(2)/g(2) = = = 2 + 2a,f(l)g(-I) (1 + t)2 + g( -a) g[f(2)] -2\"f(a)

1. f(2) 2. 3.

4. 5.

cp(

0)

(a) (f)

= 4,

All

x

All

a

1) =

cp(

2,

(b)

(a)

Ixl

< 2

(f)

Ix I

< 2,

2) = 2, cp( 3) = 2, All x and y (c)

cp(

Iyl < 1

(b)

x

\037 0

(c) (b) {x 10 < x < I} 7. Intersect when x = 0, 1, -1 when 8. Intersect x = -1, -3 6.

10. (a) p(x) = 1 11.

(d)

p(x)

(a)

p(x)

(c)

p(x)

(b)

= ax(x

= ax(1 = ax,

a

p(x)

- 1) +

- x) +

(c)

cp(

/tl

{x12


t

< 4}

= !x(x - 1) b, a and b arbitrary b, a and b arbitrary

arbitrary

p(x)

= c,

2) =

8,

(d)

All

(d)

0
0;

x

\037V);

x

< 0

if I x

>

I

1;

decreases if x if x < 1 or

> 0;

>

if x

increases

\037;

if Ixl >

decreasesif

increases

(b).f

[' increases if


3;

11, 12, and [increases (n

if

< 0, or if

x

13, if

for

2; decreasesif

if

decreasesif x

-3 or

if

Ixl

n denotes

> V); -3 <
2 1/ 3 ; decreases if 0 < x < 21/ 3 7. (a) 21/ 3 (b) [increases

4.

all

X < 2n1T)))

+ !)1T

to exercises)

Answers

626)

13. (a) (2n + !)17 (c) [/ increases 14.

0

(a)

for all x [increases + !)17 < x < (2n + i)l7;

(b) if (2n

(b) [increases

decreases

> 0;

x

if

if

decreases x < 0

if

< x

-l)l7

(2n




_ 8xy 2.'

= 4x3

ox

.fx

t

Exercises(page

!xx= 2

3.

if

\037V2

= 4r

min

2(\037l-)7/2

= 0

= arctan

angle

max = 3V3 r;

=

02f. oy

=

-2(x

ox

= -3xy2(X2 !xx = .fyx = y(2x [xy

+ y)(x

+ y2)-5/2;[yy

2

-

2 y2)(X +

+

- y)-3; ---:= 4y(x ox 2 02f

=

-x(x

y2)-5/2)))

2

-

2y2)(X

6

sin

-

2

(2x

-

- = 02[

y)-3;

oy2 2

3y) sin

+ y2)-5/2;

4x(x

[cos

-

(2x

y)-3

-

3y)]

y-2

Chapter 5.5

a 4) - a5 ) -

4.

+ !(b

- a2) -

2

(b

5.

7. 8. 9.

3/ 2

16.

f(t)

=

18. p(x)

=

2X 15

=

3

= 2;

.f\" ( 1)

(I +

(a)

f

25.

(a)

26.

(a)

x

8.

9.

10. 11.

-! 2

ix = 5

(I)

(b)

)-3

2x(1 +

X

4

2x(1 +

(c)

)-3

X

4

- 3x 2 (1 + X6 )-3

)-3

x 12

(b) 1 + \037V2 -

(c)

cos a)1/2 1 - TT (b)

(c)

(36)1/3

(d)

(d)

-TT2

0

* (e)

(TT-

3TT/2

- I) (d) \037(t (b) \037 (c) \037)(I t + 2 = None (c) (b) One exampleisf(x) x + x One exampleisf(x) = 1 + x + x 2 for x > O,f(x) = 1/(1rx

+ 1)3/2

C -

+

3X)5/2

1)7/2

2

(b)

\037;

inlplies

(c)

(X;

rx

implies

-

o

2x

arcsin

if

- x2

Ix I VI

4

18.)

29.

x

sin

V sin 2x)

23.

V\037

17.)

cos x + x

cosx

X=F(k+!)7T

x 2)

2( 1 +

Icosxl 16.)

if x)

1

_

-

+ cos4

sin 4 x

arcsin 2

X )5/2)

x

x

> 1

33.

2x-l

2

V7

V7

34. ![(1 + x2 )

x

9

2

36.

\037(1

37.

(1 + x) arctan

+ x )(arctan

(arctan

39.

!(arcsin x + 2v 1 + x

2

/

A

+

! log (1

+ x2 )

2'v 1 + x

arctan

+ C

44.

lo -\037

+ C

45.

2)

Ib

- al (b

-

a) arcsin

47.

t Ib

6.25

1.

Exercises (page 267) Ix 21 + log Ix + log

2.

_1 2

(x +

4.

2 .1 2x

-

+

2 log Ix

9. 10.

log

Ixl

9x 2

+ +

4(x

3

1

x+

x

-

-

-

\037arctan

! log

(x

+ 68

2)(x

+ 3)2

2

+ 1) +

+ .1 8 log

+ log -

Ix

11

+

- log Ix

x+!loglx-21-!loglx+31

14.

log

1

15.

2

-x)))

-

-

21

-

arctan

4 x_ (x

+ C

C 1

+ C

2(x

2

(x

+ l)(x (x

+

1)

+ 2)16

+ 3)17

C

11 +

13.

Ix

C

+

(xf2)

log Ix + 11 +

50x

12. t log Ix 2

3

+

1

11.

+ C

(2x + 1)2 2x + 1 - 11 + log (x 2 + x + 1) + C 11

7. x + 1arctan 8. 2 log Ixl

+ 2)

x-I

-

a)(b

+ C

2

x +

x3 (x

x + lo g

5. log Ix 6.

\037log

1)

-

C

+

x-I +

_

(x

tV

C

51 +

3

+

3)

1 3(x

: =;

J

2)4

(x + 1)(x +

-

eX

+

I

+

C

+C +

C

2

- 2)

+ C

g (1 +

a arcsin 2(b

3.

-

(

43.

+ C

x

arctan

2

46.

lo g

+ C

1 _

42.

X

+ 1)earctan

(x 41.

x

X

C

+C

-

e- 2X ) -

arcsin

al

x) [2x

arccot

eX

+ C

eX)

-x - Va 2 a)

- a) _

+c)

1 + x2 )

X

/

A

C

C

+

V\037

xV 1 - x2 )

- 1)earctan

(x

-

arctan

+

C

+

V\037)2

- x2

VI

- x

X)2 V\037

38.

40.

x2

2 +

-

x

arccos

3

+c

- x]

x

arctan

3

35.

c

+

arctan

631)

to exercises)

Answers

-

x2 +

C

x- a + J b_ a

- (a + b)]

+

C

C

16. 4log Ix

17. ! 18.

19.

+

-

11

(x

1

x2

-

1)2

+X + 2

x +

1 +

4x

4x2 Ixl

23. 24.

log

A / 'V

! log

x2

/-

4v 2 2

(x

log

-

25. -x/(x 1

5

1+3

28.

(x/2 )

'V a 2

casx +

1

log

(V2

-

35. Y 3

36. Y x2 37.

38. Y x2

39.

- x2

log

40. _

-x

2

+ x

x

ix v

x2

-

-Y 3

2

a cas x

tan x)

+ C

x

sin

+C)

+ C

x)

! log 2

33. txY 3 34.

1

1 +

x)

(\037

(-TT/4)

2 V a -

+ c)

tan

arctan

+ C

2)

a

1 +

a + -

\037

tan

(J

casx 31. - a a SIn. x b cas + ( 32.

- a

l

- a2

1

/

:b

+ C

+ c

arctan

1 A

x2

C

tan

2 Y

1-

+ C

+ 1)

(x

xV2

arctan

Y5

29. x - !Y2 arctan 30.

2'v/-2 A

arctan

Y5 27.

1 +

arctan

1) +

x +

+

C

x +

\037arctan

+ 1

xv 2

C

x +

+ 1

/-

2)-1 +

2x +

+

26.

x

A

+C

+ 1)1-

+ xV2

2

2

arctan

+

X

- 1)/(x

I(x

1 A

-

1)

+ C

1

x2

1 +

+ C

_

1 x log 8 x

+

C

+ C

1

1 +

Ixl

log

1

22.

2(x2

21 +

Ix +

\037log

x

x-I

'3 log

-

Ixl

\037log

x+1 -

log

20.

21.

10 exercises)

Answers

632)

-

+ 5

+

+ x

+! arcsin ( :3 +

C

Y3!og

-

x

! 2

+ x

+

1) +

4

+

+ 1) +

5) +

C

C

2Y x 2

+ 1)

+ x

+ C

C

V2 -x !og (

+ c

)

+ 1 +

Y2 +

'\\/3

+ 2x

V x2

(2 x

log 2

+

x

(

+ 2Y x

Y2 -x -x

- x2

V3

! log (2Y x 2 + ! log (x +

+ 1

(2 x + 1

c

+ )

x

-x

2

_

V2

4

2X

)

-arcsin

(

+ 3)))

1

)

+ C

6.26

Miscellaneousreview =

1.

f(x)

+ f(l/x)

2.

.f(x)

= log V3/(2

4.

exercises

(page

+ cosx)

1) 7

(a))

V =

(b)

-1-i)

f 6.

10.

,f(x)

12.

(a)

=

(b)

a

+

+ nC l

Co

(b)

16.

(c)

Ifp(x)

(a)

ix

(b)

x

(c)

1

2

+

(x

if

= V(2x

18.

(a)

i(1

20. (a) 25.

x >

+

- e-2t )

19. (a) log 3

-

2

True

Ixl


log2

where

v/2

IRI IRI




\037(k I)!

L k=l

Ixl > I

0

7)

(page 284)

55v/2

8.

2

-

Chapter

7.8

e log

(d)

?\037::)

if

1


v mg/k)

19.51b

10.

2.

(b

55\302\260

9.

1.

- t)/(b - a) 16 + 166e20 2t if

=

0

a2

\302\245n1T/k

- I

8.17 Exercises(page 1.

to exercises)

Answers

636)

= C1eX Y

2. y = clex

333)

+

C 2e-X

+

c2 -

- x

- x2

2x

-

!x

3

X

3

c l e- + C 2 + !x 3. Y 4. y = eX(c l cos ,l2x + C2 sin V2x) - 287 + ix + ix 2 + !x 3 5. Y = clex + c 2 e 4x + 392 + i-x + !X2 6. Y = c l e2X + c 2 e- 3X - -/-2 + !x - x 2 - lx 3 7. Y = (c l + tx)e 2X + c 2 e- 2X = c1 COS 2x + C 2 sin 2x + !e- 2X 8. Y = cl e- 2X + (c 2 + !x)e X 9. Y = c1e- 2X + c 2 e x + te 2x 10. Y = cl e- 2X + (c 2 + -lx)e X + t e2x 11. Y 12. Y = (c l + c2 x + !x 3)eX + x + 2 13. Y = (c i + c2 x - log Ixl)e-X - 2 = clsinx + (cl + 14. + cotxl)cosx Icscx Y log x X x = c 15. Y cle + 2 e- + (eX e- ) log (1 + eX) xe X - 1 16. Y = (c i + !x)e X + !e- X + c 2 e- 2X - t - !(eX + e- 2X ) log (1 + =

17.

18. 19.

Y Y

23.

c 1e = (cl = (c i

Y

21.

24. 25.

=

=

Y Y Y

3X

-

< 1 or x

if

.x

3X

x < 2

1
2,

3X

\037-x) cos

+ ixe 3x + (c 2 -

\037x) cos

x + C2 sin

l:!g)

sin

3x

x

+ (c 2 + tx) sin x + C 2 sin 2x + x cos x + \037sin x = c COS2x + C sin 2x + x sin x l 2 i cos x 2X = c + c e 3X i 2 -!e (3 sin x + cos x) C I COS x

CI

2x

+

sin x

+ 4,lOe2X (3 sin 3x

COS x

C2

- cos 3x))

(page 339)

Exercises

8.19

c 2 e-

+

= cl COS

=

Y

3X

+ bx)e-3X

{(a

Y

20. 22.

=

Y

c2 x)e-

(C1 +

1. 2V2

2. ::I: 14017 3.

A =

4.

m =

C,

5.

= 3 COS417X Y C = +

6.

Y

= !V6,

7.

Y

= -A

let)

9.

10.

12.

=

y\"

=.

17X

r(t) =

t +

{

1

.

- cos t t >

If

t

SIn

-

ct +

-

c

ct +

(;

t

w

-

0 < t

k t)))


1;

divergent

for

s
1

< 0;

15.

Absolutely

convergent

Absolutely

convergent

Absolutely

convergent

16.

17.

18.

Divergent

10.

11. 12.

13. 14.

25.

1;

Divergent

Conditionally

convergent

Absolutely

Divergent

convergent for s < 0;

Absolutely Divergent

convergent

for

convergent

Absolutely

convergent

convergent

Conditionally

convergent

Conditionally

convergent)

s


y,

= 3 - 4x + 36x 2 ; (DT)p(x) 2 = 24x 2x 12x 3 + + ; ; (DT (TD)p(x) TD)p(x) - 192x (b) p(x) = ax, a an arbitrary scalar (T2D2 D2T2)p(X) = 8 2 All P in V scalars (c) p(x) = ax + b, a and b arbitrary (d) = 2; Sp(x)= 3 - x + x 2 ; Tp(x) = 2x + 3x 2 - x 3 + x 4 ; (a) Rp(x) = 2 + 3x - x2 + x 3 ; = 3x - x2 + x 3 ; (TS)2 p (X) = 3x - x 2 + x 3 ; (ST)p(x) (TS)p(x) 2 3 = -x + x ; (S2T2)p(X) = 2 + 3x - x 2 + x 3 ; (TRS)p(x) = 3x; (T2S2)p(X) =

31.

-

= (0, - 2x

-2x

(RST)p(x) =

2

(b)

{p

N(R)

=

Tp(x)

= 3x

-

R(V) =

{p I p

is constant};

2X2

+

12x 3 ;

2

I

p(O)

S(V) = V; N(T)={O}; (d) (Ts)n = I - R; on 32. T is not one-to-one

= O}; snTn

T(V)={plp(O)=O} = I

V because

it

maps

all constant

N(S) = (c) T-l=S

{p I p

sequences onto

the

is constant};

same

sequence)))

to exercises)

Answers

1. (a)

2.

The

The zero

(c)

The

(a)

0

matrix

matrix

(c(5 jk ),

0

0

1 (a))

matrix

identity

(b)

[ 3.

(page 596)

Exercises

16.12

(c)

-\037].

4.

2 0]

5.

-t]\"47

-;

[\037

' [\037

4j +

\037]

\037]

4k;

-1 0, rank

nullity

3

-:

-3

[ -1

-5

(b)

0

:]

-1

6.

7.

]

'

3; +

(a)

[

\037]

[\037

[:

1

(a)

T(4; -

-\037]

j

+

- 2);

= (0,

k)

1 1, rank

nullity

2

(b) 1) [\037)

(c))

0

1

0 [

o

3

e2

(d) el =j,

-2)]

=

e 3 = i,

k,

8. (a)

(5,0, -1);

2

0, rank

nullity

[0

e2

=;,

(a)

(-1, -3,

(c)

e l = i,

(a)

e1 -

=;

+j,

-1);

WI

=

(1,0,1),

0, rank

nullity

2

10.

0 -1

11.

[

1

0

e2 ;

e2 =j -;,

0, rank

nullity

-

,

] [

WI

2

(0,0,2),

0

1

0-1

]

\037

;],

[\037

\037

;]

13. [\037

\037

-:l

[\037

\037-:J)))

w 3

=

(0,1,0)

=

(0,0,1)

5,

b =

(b) W2

(b)

=

:]

(0,1,0),

(c) [\037\037]

12. [\037

(1,0,1),

W2

1]

w2 =

[\037

=

(1,1),

0

(b) 1

(c) el

=

WI

-:])

_ 1

1

9.

\037 k

0

0

1

[ 00010)

- 12j

9;

(b)

[;

0 0

(c)

\037 \037])

[\037

+ 7j,

-5;

= 1 if = k, and (jjk = 0 if j (5 I = \302\2535 jk jle ), where j = 0 = (ajk) where each entry ajle matrix of part (a) where \302\2535 jk ) is the identity 0 1 0 0 0

(b)

010) ]

[-2 o

653)

W3

a =

4

=

(1, -1))

to exercises)

Answers

654)

14. [\037

0

[\037

\037J

-1

,

15.

0

1

\037J

-

0

1

0-1)

0 -1

[ 1

1

0

0

0

0

0 -1

0

0

1

[\037

-\037l

[

16.

]

]

-1

0

0

0

2

-1

0

1

0 -2

0

0

0 -1

0

0

0 -1

0

17. -5

18. [\037

19.

-\037]

[\037

12 -12] -5

[

-\037l

o 0 0 0 0 1 o 0 (a)

0 2

0

o 0 0

0

0

0 -2

0

0

0

2 0

9

0

0

0

0

0 0 0 0

(c)

0 10 0 (e)

-3 0 0

(f)

0 0 4 0 0 0 0 9

basis for

2

(x , x) as a

and

V,

6

0

0 -8

0

0

0 -48

0

0

0

0

0

0

0

0

basis for W. Then

the

0

matrix

200) 0] (page 603)

Exercises

16.16

0 0 0 0

0 0 0 0

0

1) as a

0

0

0

0

0 0 3 20. Choose(x , x 2 , X, 0 0 [6o

0 0 0

3

0

0

(b)

0

0 -1 (d)

0 1 0 0 0 0 4 0

1

1. B + C =

:

\037

[6

-5

AC =

] =

\037l

a and b

(a) [\037

3.

(a)

a =

4.

(a))

-: [ -7

\037

[4

- 2a (b)

1

5

A(2B _

-I\037

-5)]

= 1,

d

=

(b)

5)

-\037

[

12

a

3C) =

,a

[ -2b

b]

(b) a

= 1,

\037

-27

\037: 0) ]

- 28

7

[

8]

-16

\037

\037\037

,

I:

=4

-\037:l

\037

-\037

arbitrary

9, b = 6, c

=

BA

[ _\037\037

CA [\037

2.

, AB =

[ -30

,

14] -28

28]

b arbitrary

and

b

30

=\037

=

6,

c =

0, d =

-2)))

of T D

is

Answers

6.

=

An

:

[\037

7.

8.

655)

to exercises)

]

cos nO

-sin

[ sin nO)

cos

An =

no nO)

,

]

+ 1)

n(n

1

n

o

1

n

o 0

1

2)

=)

An

9. \037]

[-10\037

a

b

e

-a

, where

10.

[

where a is

[ \037

-1

14. (b)

ll-

l2\037

7

8

(A

the

b

a [

e-a

, where

e are

band

arbitrary

and

:4:- \037: -4-]

[4 A2 +

AB

+

BA +

B2;

+

(A

B)(A

=

- B)

A2

BA

+

those which commute) 613) \037)

3. (x,y,z)=(I,-1,0)+I(-3,4,1) = (1, -1,0) + t( -3,4,1) + 1(1,14,5,0) u) = (1,1,0,0) 3,0) u) = (1,8, 0, -4) + 1(2,7, 7. (x,y, z, u, v) = t 1( -1,1,0,0,0) + 12 ( -1,0,3, 8. (x, y, z, u) = (1, 1, 1, - 1) + 11( - 1, 3, 7,0)

4. (x,y, z) 5. (x,y, z, 6. (x,y, z,

-3,1)

+ 12( 4,

9, 0,

9. (x,y,z) = (t, \037,O) + 1(5,1, -3) + 11(4,11,7,0) + 12 (0,0,0,1) 10. (a) (x,y, z, u) = (1,6,3,0) 3 - 11,7, 22) , 0) + 1(4, ( b) (x, y, z, u) = (1 1' 4, { \037

7)

-

\037

[ -3

13.

[=!)

-

_:

_!

\037

5

4]

2 -3)

\037

o)))

1l

14.

I;

:

\037

[ 3 2 1]

1 -2

-

-iJ

a is

-be

any solution

]

No solution

12.

a2 =

equation

- be

]

16.20 Exercises(page 1. (x,y, z) = (\037, -t, 2.

1

D =

,

B)2 =

+

For

(c)

a2 =

equation

[

of

any solution

arbitrary

0 , and o 1] [ 0 -1 ] [

13. C =

a is

],

,

the

and

\037

0

1

arbitrary,

]

11. (b) 12.

e are

band

15.

o

1

0

1

0

0

1

o

0)

o)

-2

1

-2 1)

- AB

-

B2

of

16.)

0

\"2

1

0 -1

0

1

1

0

0

0

0

0

0

0

0

1

0 -1

-3 0

0 0

1 0

0 0

0 0

0 1

0

1

0)

-3

0

9

exercises

Miscellaneous

16.21

3. p = 0 4.

[

G 0

o 0]

,

[-\037

-

1 0

, [_Ole]

on matrices (page 614)

[ -1

=\037l

1

1 1

]

'

1- ] ,

l

a2

[1

[=\037

b

a

and

equation

1 (a)

\"2

_\037]

quadratic

10.

to exercises

Answers

656)

arbitrary and a is any

e are

band

where

a

- a

+ be

-1 1

1 -1' ]

-\037])))

= 0

[

1

'

] [

1 1

1

-1

-1

l']

[

1

-1

-1 -1

, -1 ] [ -1

l']

solution

of the

INDEX)

ABEL,

HENRIK, 407

NIELS

Abel's partial Abel's test for Abscissa, 48 Absolute

convergence,

Arc

407

formula,

408

as

convergence maximum

Absolute

values,

of series, 406 and minimum,

in

41, 363

Arc

polar

normal Addition Additive

and

tangential for

formulas

of area, of

components the sine and

of, 527

cosine, 96

length, 532 119 (Exercise 13) series,

convergent

of the

66,

integral,

of work, Alternating

67, 80, 514 and infimum,

403

Analytic

geometry, 48, 471

Analytic

model

in

a Euclidean

in

n-space,

Angular Angular

speed,

19)

522, 545 (Exercise 19) 545 (Exercise 19) velocity, 205 (primitive),

Arc cosine, Archimedean

function,

117-119

rate of change,

160

157

118

weighted,

575

field,

575

197 :

for

of real

numbers,

(continuity),

area,

space, 551, 552

58, 59

for the real-number for volume, 112 26)

25

17

for a linear

254

396

46, 149

Axes, 48,

171

property

506 equal,

completeness 272-304,

LOUIS,

190

Asymptote,

Axiom(s)

by trigonometric polynomials, in a Euclidean space, 574 constant, 211, 307 Arbitrary ARBOGAST,

for composition of functions, 141, 584 for multiplication of numbers, 18, 359 for union and intersection of sets, 14 in a linear space, 551, 552

velocity,

APPOLONIUS, 498 polynomials,

of numbers, 18, 359 of vectors, 447

addition

of a

545 (Exercise

Approximations: by

law:

Average,

measurement of, 102

88 363

117

mean, 46,

Asymptotically

458

graphs,

complexnumber,

of a hyperbola,

space, 564

acceleration,

Angular Antiderivative

geometry, 471

of Euclidean

Angles: radian

for

two

between

for addition

27

110

set,

Arithmetic

115 series,

radial

Associative

of the supremun1 112 of volume,

92

transformations,

similarity

axiomatic definition of, 57-59 in polar coordinates, 110 of an ordinate set, 75 Argument of a

385

40

sums,

4)

255

of a region

of derivatives, 164

of finite

and

of a

59

of averages,

544 (Exercise

coordinates, 253

polar sine,

Area:

property:

of arc

530, 531

of,

Arc tangent, 541

coordinates,

534

an integral,

function, 533 150

Acceleration, 160, 521 in

length:

definition

Absolute

26

2-9,

ARCHIMEDES,

summation

least-upper-bound, order, 20)

system,

17-25

25

657)))

Index)

658)

Axiomatic

development:

of inner

561

products,

of

17-25

Centrifugal,

522

Centripetal,

522

Chain

174,

rule,

Circle,49, 521 Base of logarithms, Basis, 327, 466

232

312

equation,

46 (Exercise 14)

inequality,

BERNSTEIN,

35) (Exercise JOHANN, 235, 292, 305, SERGEI, 437

Bernstein's

theorem, 437

BERNOULLI,

Besselfunctions,

443

Binary scale, Binomial

331

44, 383,

442

BOLZANO,

44 (Exercise 474 143 BERNARD,

Bolzano's

theorem, 143

4), 378, 442

theorem,

551

axioms,

matrix (column

dot

for

products,

inner

products,

Bounded

Boundedness

of

functions,

377,

150

281

of

TT,

285

10)

of square roots, 444(Exercises 20, 21) fundamental theorems of, 202, 205,

Calculus, 515

17

CARDANO,

HIERONIMO,

3

Cartesian

equation, 49, 475,494

Cartesian

geometry,

AUGUSTIN-LOUIS,

186,

284,368,378,397,399,411,452 42, 452, 563 inequality, Cauchy-Schwarz 186 formula, Cauchy's mean-value in remainder 284 formula, Taylor's Cauchy's BONAVENTURA, 3, 111 CAVALIERI,

Cavalieri solid, Cavalieri's

111

principle,

of, 130, 369, of, 153

theorems Contour

127, 172,

111, 112)

364

54

functions:

integrability

3,42,

497-507

function,

definition

48

58

sets,

Conjugate complexnumber, Continuous

CANTOR, GEORG, 11,

CAUCHY,

convergence, 406

Conic sections, Constant

584

transformations, 122, 189

function,

Congruenceof

240-242

(Exercise

of, 174, 514

of

Conditional

of logarithms,

141

of,

differentiation

Concave

468

space,

function, 140, 584

continuity

Calculation:

552

space,

numbers, 358-373

Composite

390)

562

space,

Conlplexvector

23

513

on, 132, 141-154 197

lines,

Convergence:

of improper integrals, of sequences, 379

416,

of series,384-425

pointwise,

422

tests for,

394-408

uniform,

424

Convex

67, 81

14

function, 368

Composition

of e,

a set,

of

Complex

numbers,

continuous

418

for integrals,

theorem

Complexlinear

381

WILLIAM,

BROUNCKER,

561

integrals,

improper

Complex

23-25

sequence,

451

COl)lplexEuclidean

73

Bounded set of real

359

for multiplication of numbers, 18, 359 for union and intersection of sets, 14 in a linear space, 551

Complement

24

and lower,

of numbers, 18, of vectors,447

of series, 394-396

greatest lower, 25 upper

vector), 592,598

law:

Comparison

Bounded function,

605

matrix,

for

of

11

GEORGE,

60

Closure

Comparison tests for convergence:

Bound:

least upper,

521

Closedinterval,

for addition for addition

JOHANN,

BOOLE,

motion,

523

Commutative

10)

(Exercise

Binomial series, 377,441 BOLYAI,

Circular

Column

393

coefficient,

Binomial

helix,

Coefficient

225

polynomials,

Circular

Class of sets, 14

Bernoulli: differential

428

of convergence,

157

ISAAC,

BARROW,

514

equation, 327

Characteristic

111-112)

volume,

mass, 118

Center of

of the real-number system, of vector algebra, 551,552

446

ARTHUR,

CAYLEY,

of area, 57-59

function,

122,

189)))

418

659)

Index)

112

Convex set,

Difference

Coordinates: 108, 540

polar,

Copernican

Cosinefunction:

second-orderlinear,

of, 134, equation

Dimension of a linear Direction field, 343

139 for,

323

Directrix of conic

differentiation of, 162 100, 207

of,

integration

power series for, 436

Dirichlet's

test for

483

product),

Curvature, 537

131

131

Distance:

between of,

length of,

529-535 536 (Exercise

530,

nonrectifiable,

two

22)

530

law:

Distributive

Cycloid,

536 (Exercise

Cylindrical

coordinates,

products, 483

for cross

20)

for

543)

inner

Divergent

16 (Exercise 10) space, 552 416, 418 inlproper integral,

Divergent

series,

set

for in

P., 498

GERMINAL

Decimal expansion of real Decreasing function, 76

Decreasingsequence,

30,

numbers,

393

operations,

a linear

Divergent sequence, 379 of

Deductive systems,

of numbers,

8

integral: definition of, 73 81

De Moivre's

theorem, 371 (Exercise5)

Dependence,

linear,

functions notations

of complex-valued of higher

functions,

functions,

DESCARTES,

RENE,

48, 446

functions,

of vectors,447)

545 of conic

Eccentricity

Electric circui 513

282

of,

irrationality

Earth,

ts,

592,

matrix,

of a set,

Elliptic

18

sections,

500

317, 336

Elemen t :

Elliptic

function,

Endpoints

282

500, 506 535 (Exercise 519

integral, reflector,

Empty set,

486 598

11

Ellipse, 498, 132

281

of, 231

Elementary

Difference:

of real numbers, of sets, 14

451, 469,

logarithm):

of,

definition

of a

on, 164

Determinant, 486 of

of natural

computation

of a determinant,

199-201

theorems

369

order, 160,200

of vector-valued partial,

196, 512, 578

50, 53,

function,

Dot product (inner product), Duodecimalscale,393) e (base

163

of one variable, 160 of several variables, 199-201 160, 171, 172, 199,200 for,

functions

18, 360

463, 557

Derivatives: continuity,

55

functions,

Domain of a

Definite

80,

384

Division:

381

DEDEKIND, RICHARD, 17

and

561

451,

products,

for numbers, 18, 359

vibrations, 335

propertiesof,

495

planes,

between two points, 364, 462 from a point to a line, 476, 477 from a point to a plane, 494

517

definition

DANDELIN,

14

sets,

Disjoint

Curve:

Damped

407

convergence,407

removable, 131

(vector

rectifiable,

500 LEJEUNE,

PETER

jump,

188

point,

sections,

GUSTAV

DIRICHLET,

infinite,

491

Cross product

559

space,

Discontinuity:

function, 103

Cramer'srule, Critical

96

of,

properties

322

345

separable,

differential

347-350

of, 439-443

solutions

power-series

theory, 545

continuity

Cotangent

linear, 308

homogeneousfirst-order, 197

48,

rectangular,

305-357

equations,

first-order

543

cylindrical,

517

157, 159,

quotient,

Differential

13 of

an interval,

60)))

17)

562

Index)

660)

342

Envelope,

factorial, 52

Equality:

of complex

358

numbers,

of functions,

gamma, 419, 421 (Exercise

of vectors,447,

hyperbolic, 251

468

EULER,

278, 280

472, 561 231, 377,

Euler's constant, function,

Even

integer,

informal description of, 50-52 inverse

396,405,420

logarithmic, 229-235

of, 246 power seriesfor,

95 77

55

polynomial,

power, 54

range of, 53

436

Riemann

for

theorem

51

real-valued,

tests for, 182,188,189 continuous

func-

151)

258-266

166,

rational,

of, 182

tions,

123

linear,

piecewise

Extremum:

Extreme-value

25)

variables, 196

monotonic,

integral

definition

196, 512

(Exercise

periodic,

243

of,

for, 50,

piecewise

definition of, 242 derivative

76

notation

of several

308, 323

367

complex-valued,

monotonic,

odd, 84

function:

Exponential

253-256

trigonometric,

54

linear,

84 (Exercise25)

theorems,

146, 252

inverse,

property of area, 59 Existence

73

integrable,

28 (Exercise 10) method of, 2-8

Exhaustion,

76

increasing,

405

Even

51

identity,

geometry, 9, 471

space, LEONARD,

19)

63

greatest-integer,

lines, 351 Equipotential Error in Taylor's formula, EUCLID, 9, 471 Euclidean

of, 53

definition

formal

54

12

of sets,

Euclidean

367

242,

exponential,

396

zeta,

step, 52 95-107

trigonometric,

73

unbounded,

vector-valued, 512 44, 52

Factorials, Family FERMAT,

of

curves,

PIERRE

341, 351 DE, 3, 156

LODOYICO,

FERRARI, FIBONACCI

Function

3 of Pisa),

(Leonardo

46 (Exercise Fibonacci numbers, Field axioms, 17 Fixed point, 145 (Exercise 5) Focus of a conic section,498

of simple :

Function(s)

73

bounded,

368

complex-valued,

122 130

continuous,

convex, 122

decreasin g, defined

by

76 an

integral,

domain of, 53 elementary, 282 even,

logarithm,

Fundamental

theorem of

Fundamental

theorems

243

algebra,362

of calculus,

202, 205,515)

498

GALILEO,

motion, 339

function, 227

function, 419, 421 (Exercise KARL

FRIEDRICH,

19)

362, 378, 607 process,

358,

473

Gauss-Jordan elimination Gauss' test for convergence, 402 (Exercise

17)

interpretation: of derivative as a slope, 169 of integral as area, 65, 75, 89 Geometricmean, 47 (Exercise 20)

Geometric series, 388-390

54

constant,

exponential

the

Geometric

characteristic, 64 (Exercise8) concave,

the

for

GAUSS,

harmonic

equation, 227

for

Gamma

Fourier coefficients, 575 Frequency

16), 379

575

JOSEPH, 127,

FOURIER,

379

553

space,

Functional

84 (Exercise

120

GIBBS,

JOSIAH

GRAM,

J0RGEN

WILLARD, 445 568 PEDERSON,

Gram-Schmidt

process, 568

Graph of a GRASSMANN,

function, HERMANN,

Gravitational 25))

Greatest-integer

51 446

attraction, Newton's function, 63)))

law

of,

546

661)

Index)

GREGORY,

403

390,

JAMES,

Gregory's Growth

320,

laws,

25

Infimum,

series, 403

Infinite

321)

297

Infinity,

Inflection Initial

Hadamard matrices, 615(Exercise

10)

313

Half-life,

WILLIAM

HAMILTON,

ROWAN, 358,445

Harmonic

series, 384

HEA

OLIVER, 445

VISIDE,

160, 200

derivatives,

equation, 347-350

differential

Homogeneous

Homogeneousproperty: of integrals,

66 of equations,

50

ROBERT,

Hooke's

116

law, 50,

Hyperbola, 498,500,

for

18

differentiation,

function,

179

Improper

function,

functions,

forms,

259

289-302

of

144

functions,

187 (Exercise 10) 14

sets,

146, 252 585, 146,

Inversion,

586

functions, 253-256 253

transformation, 585-588 numbers, 17, 22, 28, 31, 282

Invertible Irrational

Isoclines,344,

348

Isomorphism,

361, 600

Isothermals, 198,351)

Induction:

definition

by,

39

by, 32-37

Inductive set, 22 Inequality,

Bernoulli,

100, 207,

theorem:

trigonometric

120, 134

integral,

Indeterminate

258-264

functions,

transformation,

418

Increasing sequence,381

proof

trigonometric

function,

function, 76

Indefinite

rational

of

Inverse:

416

kind,

rational

Increasing

of

matrix, 612

first kind, second

79

79, 81

Intervals, 60, 310

179

Improper integral:

of the

212-216 functions,

Intersection

579

Implicit of the

fractions, 258-264 217-220

parts,

for continuous for derivatives,

600

Implicit

513

74

Integration: by partial

Intermediate-value

18

transformation,

369

function,

of polynomials,

198)

function, 51

matrix,

605

73

Intercepts, 190,495

multiplication,

Identity,

vector-valued

of monotonic

element:

for addition,

74

function,

by substitution,

506

Hyperbolic paraboloid,

Identity

bounded

of a complex-valuedfunction, of a step function, 65

by

251

function,

Hyperbolic

120

and upper,

Integrand,

Homogeneoussystem HOOKE,

indefinite,

test, 397

385

series,

416-420

of a

sums, 40

of infinite

211

73,

of a

7

SHERLOCK,

77

function,

improper, lower

471

DAVID,

of finite

monotonic

Integral: curve, 341

493

Higher-order

function, 153

a continuous

definite,

Heron'sformula, HOLMES,

product,

Integer, 22

of a

Helix, 523

HILBERT,

Inner

of

334

motion,

307

problem, 307 451, 469, 561

Integrability:

Harmonic mean, 46 Harmonic

191

point,

condition,

Initial-value

JACQUES, 615

HADAMARD,

300

limits, 299,

20

Jump

discontinuity,

Jupiter,

545)

KEPLER,

JOHANNES,

131

46

Cauchy-Schwarz, 42, 452,563 for the sine and cosine, 95 42, 364, 454, 563) triangle,

Kepler's

laws,

545,

498, 546)))

545

264

Index)

662)

JOSEPH

LAGRANGE,

Laplace's Least

of, 196

axiom, Least-upper-bound

continuity,

126

Left-hand

coordinate

Left-hand

system,

limit,

Left

(Exercise 25)

GOTTFRIED

LEIBNIZ,

210,222,

10, 157,

172,

for

nth

the

derivative

for

series,

alternating

404

of a curve, 530, 531 other

OF PISA

Level curve,

GUILLAUME

FRAN