Schaum's outline of theory and problems of advanced calculus

SCHAUM’S OUTLINE OF

THEORY AND PROBLEMS of

ADVANCED CALCULUS

MURRAY R. SPIEGEL, Ph.D. Former Professor and Chairman, Mathematics Department Rensselaer Polytechnic Institute Hartford Graduate Center

SCHAUM’S OUTLINE SERIES McGraw-Hill

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37th printing, 1998 Copyright 0 1963 by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. No part of this publication may be reproduced, stored i n a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.

ISBN 07-060229-8

38 39 40 BAWBAW 9

Preface The subject commonly called “Advanced Calculus” means different things to different people. To some it essentially represents elementary calculus from an advanced viewpoint, i.e. with rigorous statements and proofs of theorems. To others it represents a variety of special advanced topics which are considered important but which cannot be covered in an elementary course. In this book an effort has been made to adopt a reasonable compromise between these extreme approaches which, it is believed, will serve a variety of individuals. The early chapters of the book serve in general to review and extend fundamental concepts already presented in elementary calculus. This should be valuable to those who have forgotten some of the calculus studied previously and who need “a bit of refreshing”. It may also serve to provide a common background for students who have been given different types of courses in elementary calculus. Later chapters serve to present special advanced topics which are fundamental to the scientist, engineer and mathematician if he is to become proficient in his intended field. This book has been designed for use either as a supplement to all current standard textbooks or as a textbook for a formal course in advanced calculus. It should also prove useful to students taking courses in physics, engineering or any of the numerous other fields in which advanced mathematical methods are employed. Each chapter begins with a clear statement of pertinent definitions, principles and theorems together with illustrative and other descriptive material. This is followed by graded sets of solved and supplementary problems. The solved problems serve to illustrate and amplify the theory, bring into sharp focus those fine points without which the student continually feels himself on unsafe ground, and provide the repetition of basic principles so vital to effective learning. Numerous proofs of theorems and derivations of basic results are included among the solved problems. The large number of supplementary problems with answers serve as a complete review of the material of each chapter. Topics covered include the differential and integral calculus of functions of one or more variables and their applications. Vector methods, which lend themselves so readily to concise notation and to geometric and physical interpretations, are introduced early and used whenever they can contribute to motivation and understanding. Special topics include line and surface integrals and integral theorems, infinite series, improper integrals, gamma and beta functions, and Fourier series. Added features are the chapters on Fourier integrals, elliptic integrals and functions of a complex variable which should prove extremely useful in the study of advanced engineering, physics and mathematics. Considerably more material has been included here than can be covered in most courses. This has been done to make the book more flexible, to provide a more useful book of reference and to stimulate further interest in the topics. I wish to take this opportunity to thank the staff of the Schaum Publishing Company for their splendid cooperation in meeting the seemingly endless attempts at perfection by the author.

M.R. SPIEGEL

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CONTENTS

Chapter

1

NUMBERS

....................................................

Page

1

Sets. Real numbers. Decimal representation of real numbers. Geometric representation of real numbers. Operations with real numbers. Inequalities. Absolute value of real numbers. Exponents and roots. Logarithms. Axiomatic foundations of the real number system. Point sets. Intervals. Countability. Neighborhoods. Limit points. Bounds. Weierstrass-Bolzano theorem. Algebraic and transcendental numbers. The complex number system. Polar form of complex numbers. Mathematical induotion.

Chapter

Z

FUNCTIONS, LIMITS AND CONTINUITY.. . . . . . . . . . . . . . . . . .

20

Functions. Graph of a function. Bounded functions. Monotonic functions. Inverse functions. Principal values. Maxima and minima. Types of functions. Special transcendental functions. Limits of functions. Right and left hand limits. Theorems on limits. Infinity. Special limits. Continuity. Right and left hand continuity. Continuity in an interval. Theorems on continuity. Sectional continuity. Uniform continuity.

Chapter

3

SEQUENCES ..................................................

41

Definition of a sequence. Limit of a sequence. Theorems on limits of sequences. Infinity. Bounded, monotonic sequences. Least upper bound and greatest lower bound of a sequence. Limit superior. Limit inferior. Nested intervals. Cauchy’s convergence criterion. Infinite series.

Chapter

4

DERIVATIVES

................................................

57

Definition of a derivative. Right and left hand derivatives. Differentiability in an interval. Sectional differentiability. Graphical interpretation of the derivative. Differentials. Rules for differentiation. Derivatives of special functions. Higher order derivatives. Mean value theorems. Rolle’s theorem. The theorem of the mean. Cauchy’s generalized theorem of the mean. Taylor’s theorem of the mean. Special expansions. L’Hospital’s rules. Applications.

Chapter

5

INTEGRALS

..................................................

Definition of a definite integral. Measure zero. Properties of definite integrals. Mean value theorems for integrals. Indefinite integrals. Fundamental theorem of integral calculus. Definite integrals with variable limits of integration. Change of variable of integration. Integrals of special functions. Special methods of integration. Improper integrals. Numerical methods for evaluating definite integrals. Applications.

80

CONTENTS

Chapter

6

PARTIAL DERIVATIVES

Page

.....................................

101

Functions of two or more variables. Dependent and independent variables. Domain of a function. Three dimensional rectangular coordinate systems. Neighborhoods. Regions. Limits. Iterated limits. Continuity. Uniform continuity. Partial derivatives. Higher order partial derivatives, Differentials. Theorems on differentials. Differentiation of composite functions. Euler’s theorem on homogeneous functions, Implicit functions. Jacobians. Partial derivatives using Jacobians. Theorems on Jacobians. Transformations. Curvilinear coordinates. Mean value theorems. ~

Chapter

7

VECTORS .........................

~~~

. * ..........................

134

Vectors and scalars. Vector algebra. Laws of vector algebra. Unit vectors. Rectangular unit vectors. Components of a vector. Dot or scalar product. Cross or vector product. Triple products. Axiomatic approach to vector analysis. Vector functions. Limits, continuity and derivatives of vector functions. Geometric interpretation of a vector derivative. Gradient, divergence and curl. Formulas involving V . Vector interpretation of Jacobians. Orthogonal curvilinear coordinates. Gradient, divergence, curl and Laplacian in orthogonal curvilinear coordinates. Special curvilinear coordinates. ~~

Chapter

~~~

8

APPLICATIONS OF PARTIAL DERIVATIVES.

~

.............. 161

Applications to geometry. Tangent plane to a surface. Normal line to a surface. Tangent line to a curve. Normal plane to a curve. Envelopes. Directional derivatives. Differentiation under the integral sign. Maxima and minima. Method of Lagrange multipliers for maxima and minima. Applications to errors.

Chapter

9

MULTIPLE INTEGRALS .....................................

180

Double integrals. Iterated integrals. Triple integrals. Transformations of multiple integrals.

Chapter

IO

LINE INTEGRALS, SURFACE INTEGRALS AND INTEGRAL THEOREMS .....................................

195

Line integrals. Vector notation for line integrals. Evaluation of line integrals. Properties of line integrals. Simple closed curves. Simply and multiply-connected regions. Green’s theorem in the plane. Conditions for a line integral to be independent of the path. Surface integrals. The divergence theorem. Stokes’ theorem.

Chapter

ZI

INFINITE SERIES

...........................................

Convergence and divergence of infinite series. Fundamental facts concerning infinite series. Special series. Geometric series. The p series. Tests for convergence and divergence of series of constants. Comparison test. Quotient test. Integral test. Alternating series test. Absolute and conditional convergence. Ratio test. The nth root test. Raabe’s test. Gauss’ test. Theorems on absolutely convergent series. Infinite sequences and series of functions. Uniform convergence. Special tests for uniform convergence of series. Weierstrass M test. Dirichlet’s test. Theorems on uniformly convergent series. Power series. Theorems on

224

CONTENTS

Page power series. Operations with power series. Expansion of functions in power series. Some important power series. Special topics. Functions defined by series. Bessel and hypergeometric functions. Infinite series of complex terms. Infinite series of functions of two (or more) variables. Double series. Infinite products. Summability. Asymptotic series.

Chapter

IZ

IMPROPER INTEGRALS

.....................................

260

Definition of a n improper integral. Improper integrals of the first kind. Special improper integrals of the first kind. Geometric o r exponential integral. The p integral of the first kind. Convergence tests for improper integrals of the first kind. Comparison test. Quotient test. Series test. Absolute and conditional convergence. Improper integrals of the second kind. Cauchy principal value. Special improper integrals of the second kind. Convergence tests f o r improper integrals of the second kind. Improper integrals of the third kind. Improper integrals containing a parameter. Uniform convergence. Special tests for uniform convergence of integrals. Weierstrass M test. Dirichlet’s test. Theorems on uniformly convergent integrals. Evaluation of definite integrals. Laplace transforms. Improper multiple integrals.

Chapter

13

GAMMA AND BETA FUNCTIONS. ..........................

285

Gamma function. Table of values and graph of the gamma function. Asymptotic formula f o r r(n). Miscellaneous results involving the gamma function. Beta function. Dirichlet integrals.

Chapter

14

FOURIER SERIES

...........................................

298

Periodic functions. Fourier series. Dirichlet conditions. Odd and even functions. Half range Fourier sine or cosine series. Parseval’s identity. Differentiation and integration of Fourier series. Complex notation for Fourier series. Boundary-value problems. Orthogonal functions.

Chapter

15

FOURIER INTEGRALS ......................................

321

The Fourier integral. Equivalent forms of Fourier’s integral theorem. Fourier transforms. Parseval’s identities f o r Fourier integrals. The convolution theorem.

Chapter

16

ELLIPTIC INTEGRALS ......................................

331

The incomplete elliptic integral of the first kind. The incomplete elliptic integral of the second kind. The incomplete elliptic integral of the third kind. Jacobi’s forms f o r the elliptic integrals. Integrals reducible to elliptic type. Jacobi’s elliptic functions. Landen’s transformation.

Chapter

17

FUNCTIONS OF A COMPLEX VARIABLE.. . . . . . . . . . . . . . . . . . 345 Functions. Limits and continuity. Derivatives. Cauchy-Riemann equations. Integrals. Cauchy’s theorem. Cauchy’s integral formulas. Taylor’s series. Singular points. Poles. Laurent’s series. Residues. Residue theorem. Evaluation of definite integrals.

INDEX

........................................................

373

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Chapter 1 Numbers

SETS Fundamental in mathematics is the concept of a set, class or collection of objects having specified characteristics. For example we speak of the set of all university professors, the set of all letters A, B, C, D , . . .,Z of the English alphabet, etc. The individual objects of the set are called members or elements. Any part of a set is called a subset of the given set, e.g. A, B, C is a subset of A, B, C , D,. . .,Z. The set consisting of no elements is called the empty set or null set. REAL NUMBERS . The following types of numbers are already familiar to the student.

1. Natural numbers 1,2,3,4,. . ., also called positive integers, are used in counting members of a set. The symbols varied with the times, e.g. the Romans used I, 11,111, IV, . . .. The sum a + b and product a * b or ab of any two natural numbers a and b is also a natural number. This is often expressed by saying that the set of natural numbers is closed under the operations of addition and multiplication, or satisfies the closure property with respect to these operations.

2. Negative integers and zero denoted by -1, -2, -3, . . . and 0 respectively, arose to permit solutions of equations such as x b = a where a and b are any natural

+

numbers. This leads to the operation of subtraction, or inverse o f addition, and we write x = a - b. The set of positive and negative integers and zero is called the set of integers.

3. Rational numbers or fractions such as 8, --$ . . . arose to permit solutions of equations such as b x = a for all integers a and b where b # 0 . This leads to the operation of division, or inverse o f multiplication, and we write x = a/b or a + b where a is the numerator and b the denominator. The set of integers is a subset of the rational numbers, since integers correspond to rational numbers where b = 1.

fi and x are numbers which are not rational, i.e. cannot be expressed as (called the quotient of a and b ) where a and b are integers b and b # 0 . The set of rational and irrational numbers is called the set of real numbers.

4. Irrational numbers such as

DECIMAL REPRESENTATION of REAL NUMBERS Any real number can be expressed in decimal form, e.g. 17/10=1.7, 9/100=0.09, 1/6= 0.16666.. . In the case of a rational number the decimal expansion either terminates or, if it does not terminate, one or a group of digits in the expansion will ultimately repeat as, for example, in 4 = 0.142857 142857142. . In the case of an irrational number such as fi= 1.41423.. . or T = 3.14159. . no such repetition can occur. We can always consider a decimal expansion as unending, e.g. 1.375 is the same as 1.37500000.. . or 1.3749999.. .. To indicate recurring decimals we sometimes place dots over the repeating cycle of digits, e.g. 3 = 0.142857,7 = 3 . k The decimal system uses the ten digits 0, 1,2,. . .,9. It is possible to design number systems with fewer or more digits, e.g. the binary system uses only two digits 0 and 1 (see Problems 32 and 33).

.

.

......

1

..

2

NUMBERS

[CHAP. 1

GEOMETRIC REPRESENTATION of REAL NUMBERS The geometric representation of real numbers as points on a line called the r e d axis, as in the figure below, is also well known to the student. For each real number there corresponds one and only one point on the line and conversely, i.e. there is a one to one (1-1)correspondence between the set of real numbers and the set of points on the line. Because of this we often use point and number interchangeably. -_

-D

4

3

,

L

-s

-4

-3

-2

1

-1

I

0

I

1

2

I

3

I

4

I

5

Fig. 1-1

The set of real numbers to the right of 0 is called the set of positive numbers; the set to the left of 0 is-the set of negative numbers, while 0 itself is neither positive nor negative. Between any two rational numbers (or irrational numbers) on the line there are infinitely many rational (and irrational) numbers. This leads us to call the set of rational (or irrational) numbers an everywhere dense set.

OPERATIONS with REAL NUMBERS If a, b, c belong to the set R of real numbers, then: Closure law 1. a b and ab belong to R 2. a + b = b + a Commutative law of addition Associative law of addition 3. a + ( b + c ) = ( a + b ) + c Commutative law of multiplication 4. ab = ba Associative law of multiplication 5. a(bc) = (ab)c Distributive law 6. a ( b + c ) = a b + a c 7. a + O = O + a = a, l * a = a * l = a 0 is called the identity w i t h respect to addition, 1 is called the identity w i t h respect to mu1t iplica t ion. 8. For any a there is a number x in R such that x a = 0. x is called the inverse of a w i t h respect to addition and is denoted by -a. 9. For any a + 0 there is a number x in R such that ax = 1. x is called the inverse o f a w i t h respect to multiplication and is denoted by a-' or lla. These enable us to operate according to the usual rules of algebra. In general any set, such as R, whose members satisfy the above is called a field.

+

+

INEQUALITIES If a - b is a nonnegative number we say that a isgreater than or equal to b or b is less than or equal to a,and write respectively a h b or b d a. If there is no possibility that a = b, we write a > b or b < a. Geometrically, a > b if the point on the real axis corresponding to a lies to the right of the point corresponding to b. Examples: 3 < 6 or 5 > 3 ; - 2 < -1 or -1 > -2; z S 3 means that z is a real number which m a y be 3 or less than 3.

If a, b and c are any given real numbers, then: 1. Either a > b, a = b or a < b 2. If a > b and b > c, then a > c 3. If a > b, then a + c > b + c 4. If a > b and c > 0, then ac > bc 5 . If a > b and c < 0, then ac < bc

Law of trichotomy Law of transitivity

3

NUMBERS

CHAP. 11

ABSOLUTE VALUE of REAL NUMBERS The absolute value of a real number a, denoted by lal, is defined as a if a > 0 , -a if a < 0 , and 0 if a = 0 . Examples: 1-61 = 6, 1+21 = 2,

1. Jab1 = Ja(Jb(

+

2. la+b( 5 lal Ibl 3. la- bl 2 lal - Ibl

or or

1-21

= Q,

= fi, 101 = 0.

l-fil

labc.. .ml = lal Ibl Icl . . . Iml l a + b + c + ... +ml d lal+lbl+lcI+ ... +lml

The distance between any two points (real numbers) a and b on the real axis is

la - bl = J b- al.

EXPONENTS and ROOTS The product a-a.. .a of a real number a by itself p times is denoted by up where p is called the exponent and a is called the base. The following rules hold.

These and extensions to any real numbers are possible so long as division by zero is excluded. In particular by using 2, with p = q and p = O respectively, we are led to the definitions ao= 1, a-q = l/a? There If ap = N, where p is a positive integer, we call a a pth root of N, written For example since Z2 = 4 and (-2)2 = 4, there may be more than one real pth root of N . It is customary to denote the positive are two real square roots of 4, namely 2 and -2. square root by fi= 2 and the negative one by -fi= -2. If p and q are positive integers, we define aPlq= @.

m.

L

LOGARITHMS

If aP=N, p is called the Logarithm of N to the base a, written p = logaN. If a and N are positive and a # 1, there is only one real value for p . The following rules hold. I

+

M

2. logo N = log, M 1. logs MN = logs M log, N 3. log, M‘ = r logo M

- log, N

In practice two bases are used, the Briggsian sgstem uses base a = 10, the Napierian syst e m uses the natural base a = e = 2.71828.

. ..

AXIOMATIC FOUNDATIONS of the REAL NUMBER SYSTEM The number system can be built up logically, starting from a basic set of axioms or “self evident’’ truths, usually taken from experience, such as statements 1-9, Page 2. If we assume as given the natural numbers and the operations of addition and multiplication (although it is possible to start even further back with the concept of sets), we find that statements 1-6, Page 2, with R as the set of natural numbers, hold while 7-9 do not hold. Taking 7 and 8 as additional requirements, we introduce the numbers -1, -2, -3, . . . and 0. Then by taking 9 we introduce the rational numbers.

4

NUMBERS

[CHAP.1

Operations with these newly obtained numbers can be defined by adopting axioms 1-6, where R is now the set of integers. These lead to proofs of statements such as (-2)(-3) = 6, -(-4) = 4, (0)(5)= 0, etc., which are usually taken for granted in elementary mathematics. We can also introduce the concept of order or inequality for integers, and from these inequalities for rational numbers. For example if a, b, c , d are positive integers we define a/b > c/d if and only if ad > bc, with similar extensions to negative integers. Once we have the set of rational numbers and the rules of inequality concerning them, we can order them geometrically as points on the real axis, as already indicated. We can then show that there are points on the line which do not represent rational numbers (such as \/2,~,etc.). These irrational numbers can be defined in various ways one of which uses the idea of Dedekind cuts (see Problem 34). From this we can show that the usual rules of algebra apply to irrational numbers and that no further real numbers are possible.

POINT SETS, INTERVALS A set of points (real numbers) located on the real axis is called a one-dimensional point set. The set of points x such that a 5 x 5 b is called a closed interval and is denoted by [a,b ] . The set a < x < b is called an open interval, denoted by (a,b ) . The sets a < x S b and a S x < b, denoted by (a, b] and [a, b ) respectively, are called half open or half closed intervals. The symbol x, which can represent any number or point of a set, is called a variable. The given numbers a or b are called constants. Example:

The set of all z such that 1x1 < 4, i.e. -4 interval.

< z < 4, is represented by (-4,4), an open

The set x > a can also be represented by a < x < 00. Such a set is called an infinite or unbounded interval. Similarly --oo < x < 00 represents all real numbers x.

COUNTABILITY A set is called countable or denumerabze if its elements can be placed in 1-1 correspondence with the natural numbers. Example:

The even natural numbers 2 , 4 , 6 , 8 , . . . is a countable set because of the 1-1 correspondence shown. Given set

2

4

6

8

Natural numbers

1

2

3

4

$ $ $ $

A set is infinite if it can be placed in 1-1 correspondence with a subset of itself. An infinite set which is countable is called countably infinite. The set of rational numbers is countably infinite while the set of irrational numbers or all real numbers is non-countably infinite (see Problems 17-20). The number of elements in a set is called its cardinal number. A set which is countably infinite is assigned the cardinal number X, (the Hebrew letter aleph-null). The set of real numbers (or any sets which can be placed into 1-1 correspondence with this set) is given the cardinal number C, called the cardiwdity of the continuum.

CHAP. 11

5

NUMBERS

NEIGHBORHOODS The set of all points x such that 1x - a1 < 8 where 8 > 0, is called a 8 neighborhood of the point a. The set of all points x such that 0 < Ix -a1 < 8 in which x = a is excluded, is called a deleted 8 neighborhood of a. LIMIT POINTS A limit point, point o f accumulation or cluster point of a set of numbers is a number l such that every deleted 8 neighborhood of l contains members of the set. In other words for any 8 > 0, however small, we can always find a member x of the set which is not equal to l but which is such that 1x - lI < 8 . By considering smaller and smaller values of 6 we see that there must be infinitely many such values of x. A finite set cannot have a limit point. An infinite set may or may not have a limit point. Thus the natural numbers have no limit point while the set of rational numbers has infinitely many limit points. A set containing all its limit points is called a closed set. The set of rational numbers is not a closed set since, for example, the limit point fi is not a member of the set (Problem 5). However, the set 0 I x 5 1 is a closed set. BOUNDS If for all numbers x of a set there is a number M such that x 5 M , the set is bounded above and M is called an upper bound. Similarly if x z m , the set is bounded below and m is called a lower bound. If for all x we have m S x S M , the set is called bounded. If M is a number such that no member of the set is greater than &f but there is at least one member which exceeds - c for every c > 0, then &f is called the least upper bound (1.u.b.) of the set. Similarly if no member of the set is smaller than 6 but a t least one member is smaller than G + c for every ~ > 0 ,then f i is called the greatest lower bound (g.1.b.) of the set. WEIERSTRASS-BOLZANO THEOREM The Weierstrass-Bolzano theorem states that every bounded infinite set has at least one limit point. A proof of this is given in Problem 23, Chapter 3. ALGEBRAIC and TRANSCENDENTAL NUMBERS A number x which is a solution to the polynomial equation aox" + u ~ V - + ' ~ 1 2 + ~ ~...- + ~ a,-lx + an = 0 (4 where a0 + 0, U I , a2, . . . , U , , are integers and n is a positive integer, called the degree of the equation, is called an algebraic number. A number which cannot be expressed as a solution of any polynomial equation with integer coefficients is called a transcendental number. Examples:

5 and fiwhich are solutions of 3%- 2 = 0 and numbers.

X*

- 2 = 0 respectively, are algebraic

The numbers x and e can be shown to be transcendental numbers. We still cannot determine whether some numbers such as er or e x are algebraic or not. The set of algebraic numbers is a countably infinite set (see Problem 23) but the set of transcendental numbers is non-countably infinite.

+

6

NUMBERS

[CHAP. 1

The COMPLEX NUMBER SYSTEM Since there is no real number x which satisfies the polynomial equation x2 + 1 = 0 or similar equations, the set of complex numbers is introduced. We can consider a complex number as having the form a + bi where a and b are real numbers called the real and imaginary parts, and i = 6 1 is called the imaginary unit. Two complex numbers a bi and c d i are equal if and only if a = c and b = d. We can consider real numbers as a subset of the set of complex numbers with b = 0 . The complex number O + O i corresponds to the real number 0. The absolute value or modulus of a bi is defined as [a bi[ = (m.The complex conjugate of a bi is defined as a - bi. The complex conjugate of the complex number x is often indicated by 2 or x*. The set of complex numbers obeys rules 1-9 of Page 2, and thus constitutes a field. In performing operations with complex numbers we can operate as in the algebra of real numbers, replacing ? by -1 when it occurs. Inequalities for complex numbers are not defined. From the point of view of an axiomatic foundation of complex numbers, i t is desirable to treat a complex number as an ordered pair (a, b ) of real numbers a and b subject to certain operational rules which turn out to be equivalent to those above. For example, we define (a, b ) + ( c , d ) = ( a + c, b d ) , (a, b ) ( c ,d ) = (ac - bd, ad + bc), m(a, b ) = (mu,mb), etc. We then find that (a, b ) = a(1,O)+ b ( 0 , l ) and we associate this with a + bi, where i is the symbol for ( 0 , l ) .

+

+

+

+

+

+

POLAR FORM of COMPLEX NUMBERS If real scales are chosen on two mutually perpendicular axes X'OX and Y'OY (the x and y axes) as in Fig. 1-2 below, we can locate any point in the plane determined by these lines by the ordered pair of numbers ( x ,y) called rectangular coordinates of the point. Examples of the location of such points are indicated by P , & , R , S and T in Fig. 1-2.

t' I

X'

-'4

-'a

8

o

* f -'2

-'I

i

i

IY

P(3,4)

T(2.6,O)

i

i

x

-1

R(-2.6, -1.6)

-P

Y'

'S(2,-2)

-a

Fig. 1-2

Y' Fig. 1-3

+

Since a complex number x iy can be considered as an ordered pair (2,y), we can represent such numbers by points in an x y plane called the complex plane or Argand diagram. Referring to Fig. 1-3 above we see that x = pcos+, y = p s i n + where p = = Jx iyl and +, called the amplitude or argument, is the angle which line OP makes with the positive x axis O X . It follows that

d

w

+

x = x

+ iy

=

+OS+

+ isin+)

(2)

called the polar form of the complex number, where p and + are called polar coordinates. It is sometimes convenient to write cis + instead of cos + i sin +.

+

7

NUMBERS

CHAP. 13

zn =

{&OS

+ + i sin +)}"

= pn(cosn+

+ i sin%+)

where n is any real number. Equation ( 5 ) is sometimes called De Moivre's theorem. We can use this to determine roots of complex numbers. For example if n is a positive integer, xl/n

+ + i sin +)}'In P l / n { c o s ( ~ ) + isin(?)}

= {&OS

=

(6)

k=0,1,2,3,. . . , n - l

from which it follows that there are in general n different values for xl/". Later (Chap. 11) we will show that ei@ = cos i sin + where e = 2.71828. . .. This is called EuZer's f ormuza.

++

MATHEMATICAL INDUCTION The principle of mathematicd induction is an important property of the positive integers. It is especially useful in proving statements involving all positive integers when it is known for example that the statements are valid for n = 1,2,3 but it is suspected or conjectured that they hold for all positive integers. The method of proof consists of the following steps. 1. Prove the statement for n = 1 (or some other positive integer). 2. Assume the statement true for n = k where k is any positive integer. 3. From the assumption in 2 prove that the statement must be true for n = k 1. This is the part of the proof establishing the induction and may be difficult or impossible. 4. Since the statement is true for n = 1 [from step 11 it must [from step 31 be true for n = 1 1 = 2 and from this for n = 2 1 = 3, etc., and so must be true for all positive integers.

+

+

+

Solved Problems OPERATIONS with NUMBERS 1. If x = 4, y = 15, x = -3, p = Q , q = -9, and r = 2, evaluate (a) x (c) (d) (pq)r, ( e ) 4 P + d .

m-), +

+ (y + x ) ,

+ + x,

( b ) (x y)

+ + +

x ( 2 / + z ) = 4 [16 (-3)] = 4 + 12 = 16 ( b ) ( x + Y ) z = ( 4 + 16) (-3) = 19-3 = 16 The fact that (a) and ( b ) are equal illustrates the associative law of addition. (U)

+

(4 P(qr) = Q < < - & > < ~ > = > (Q)(--& = (Q> = -& = -j& (4 (PZ.)r= { ( Q > < - Q > > < ~=> (-A)($)= = --& = -&

The fact that ( c ) and ( d ) are equal illustrates the associative law of multiplication.

+ q) = 4(3 - Q) = A($-&) Another method: x ( p + q ) =

( e ) x(p

distributive law.

= 4(g) = xp

+ xq

=2

= (4)(3)

+ (4)(-9)

=

t -$

=

-Q =

= 2 using the

8 2.

NUMBERS

Explain why we do not consider (a)

0

(b)

[CHAP. 1

1

6 as numbers.

(a) If we define u/b as that number (if it exists) such that b z = a, then 0/0 is that number x such that Oz = 0. However, this is true for all numbers. Since there is no unique number which 0/0 can represent, we consider it undefined. (b)

As in (a), if we define 1/0 as that number x (if it exists) such that Ox = 1, we conclude that there is no such number. Because of these facts we must look upon division by zero as meaningless.

3. Simplify

+

s2- 6% 6 52 - 2s - 3 .

z ' - 5 6 2 + 6 - (x-3)(2-2) - -2 - 2 provided that the cancelled factor ( z - 3 ) 2' - 2%- 3 (2 - 3)(% 1) 2 1

+

+

is not zero, i.e.

x # 3. For z = 3 the given fraction is undefined.

RATIONAL and IRRATIONAL NUMBERS 4. Prove that the square of any odd integer is odd.

+

Any odd integer has the form 2m 1. Since (2m integer 4m' 4m = 2(2mL 2m), the result follows.

+

+

5.

+ 1)' = 4ma + 4m + 1 is 1 more than the even

Prove that there is no rational number whose square is 2. Let p / q be a rational number whose square is 2, where we assume that p / q is in lowest terms, i.e. p and q have no common integer factors except '1 (we sometimes call such integers relatively p r i m e ) . Then (p/q)' = 2, p' = 2q' and p' is even. From Problem 4, p is even since if p were odd, p* would be odd. Thus p = 2 m . Substituting p = 2m in p' = 2q' yields q' = 2mL, so that q' is even and q is even. Thus p and q have the common factor 2, contradicting the original assumption that they had no common factors other than al. By virtue of this contradiction there can be no rational number whose square is 2.

6.

Show how to find rational numbers whose squares can be made arbitrarily close to 2. We restrict ourselves to positive rational numbers. Since (1)' = 1 and (2)' = 4, we are led to choose .,1.9. rational numbers between 1 and 2, e.g. 1.1,1.2,1.3, Since (1.4)' = 1.96 and (1.5)' = 2.26, we consider rational numbers between 1.4 and 1.6, e.g. 1.41,

..

1.42,

. . .,1.49.

Continuing in this manner we can obtain closer and closer rational approximations, e.g. (1.414213662)' is less than 2 while (1.414213663)' is greater than 2.

+

+

+

.

. . . U,, = 0 where UO,UI,. .,a n are integers 7. Given the equation aoxn a1xn-l and a. and a,#O. Show that if the equation is to have a rational root p / q , then p must divide a, and q must divide a0 exactly. Since p / q is a root we have, on substituting in the given equation and multiplying by q", the result aop"

or dividing by p ,

+ alp"-'q + aap"-'q* + ,p"-1

+ an-1pqn-l + a,qn

+ a l p * - * q + ... + a , - l q " - l

= 0

= -a n q" P

Since the left side of (a) is an integer the right side must also be an integer. Then since p and q are relatively prime, p does not divide q" exactly and so must divide an. In a similar manner, by transposing the first term of (1) and dividing by q , we can show that q must divide ao.

8. Prove that

fl+ fi cannot be a rational

number.

If 2 = fi+fithen x* = 6+2&, x ' - 6 = 2 6 and squaring, x ' - 1 0 x 2 + 1 = 0. The only possible rational roots of this equation are -C1 by Problem 7, and these do not satisfy the equation. It follows that fi+ which satisfies the equation, cannot be a rational number.

6,

CHAP.11 9.

9

NUMBERS

Prove that between any two rational numbers there is another rational number. If a and b are rational numbers, then

-is

a rational number between a and b.

To prove this assume a < b. Then by adding a to both sides, 2a Similarly adding b to both sides, a Thus a < a+b < b.

+ b < 2b

a+b < b. 2

and

2 ~

Then

= :(E+:)

a+b 2

=

2 9

INEQUALITIES 10. For what values of x is x x+3(2-x) I 4-x

2 18

r

P

+ 5)

'(E

a+b

a = - and b = - where p , q, r, 8 are integers and

a+b is a rational number, let 2

To prove that Q f O , 8#0.

< a + b and a < 2 -

9

=

is a rational number.

+ 3(2 - x) I 4 - x ?

when x + 6 - 3 ~ h 4-x, 6-293 2 4-x, 6 - 4 2 2 x - x ,

< 10 - 2x

11. For what values of x is x2 - 3x - 2

2 2 x , i.e. z S 2 .

?

The required inequality holds when

xp-3x-2-10+2x


0, i.e. x < 4 and x > -3. This is possible when -3 Thus the inequality holds for the set of all z such that -3 < z < 4.

Case 2:

12. If a 2 0 and b 2 0, prove that +(a+ b) 2

< x < 4.

fi.

A method of proof is often arrived a t by assuming the required result to be true and performing valid operations until a result is obtained which is known to be true. By reversing the steps (assuming this possible) the proof follows. In this problem we start with the required result to obtain successively a b 2 2 m , ( a b)' I 4ab or a Z - 2 a b b' B 0, i.e. (a- b)' 1 0, which is known to be true. Retracing the steps, the result follows.

+

+

Another method:

Since

(6fi)2 10

we have a - 2

This result can be generalized to

as+

n

* ' *

+

an

6

+6 h 0

B

or J(a

7-

+

+ b) 1 6.

where a1,

. . .,a,

are non-

negative. The left and right sides are called respectively the arithmetic mean and geometric mean of the numbers al, ,U,,.

. ..

. . .,b n are any real numbers, prove Schwards + ... + a,b,)2 5 .(a; + a; + ... + a:)@; + b,2 + . * .+ 6:)

13. If a1,a2,. . .,a, and b ~b2, , (Ulbl

+

a2b2

For all real numbers h, we have (UIA

+ b~)' +

(USA+

Expanding and collecting terms yields where

A'h'

bz)'

+

*.*

+ (anh +

+ 2Ch + B'

A' = a : + a X + - . . + u ~ , B ' = b : + b i + * . - + b : ,

bn)'

Z

inequality

O

h 0

(1)

C = aibi+azbp+...+anbn

(2)

Now (1) can be written

But this last inequality is true for all real h if and only if gives the required inequality upon using (2).

B' - A C' 2 0 or C' 5 A'B'

which

10

NUMBERS

14. Prove that

1

1 1 +4+ + Sn =

Let Then

+Sn=

Subtracting,

&Sn

&

.. .

+ 2"1-'

+ +8+

. * *


1.

+ F1

1 & + & + * + *2"+1 Y +2" -

= Q -

2".

Thus S, = 1 -

1

< 1 for all n.

EXPONENTS, ROOTS and LOGARITHMS 15. Evaluate each of the following.

=

dZGiii=

(3)-3 or

(ur)g

= 5 lO-' or 0.00005

z = -3.

> 0 and a, b # 1.

y assuming a, b

Since

q%iij=

=

= a" = bu = U we have afy= a' or z y = 1 the required value.

COUNTABILITY 17. Prove that the set of all rational numbers between 0 and 1 inclusive is countable.

...

considering equivalent fractions such as Write all fractions with denominator 2, then 3, no more than once. Then the 1-1 correspondence with the natural numbers can be accomplished as follows.

Q,.:,

5, . . .

Rational numbers Natural numbers

O l & Q 3 * 2 i Q . . .

$ $ $ $ $ $ $ U 1 2 3 4 5 . 6 7 8 9 ...

Thus the set of all rational numbers between 0 and 1 inclusive is countable and has cardinal number (see Page 4).

x,

18. If A and B are two countable sets, prove that the set consisting of all elements from A or B (or both) is also countable. Since A is colintable, there is a 1-1 correspondence between elements of A and the natural numbers so t h a t we can denote these elements by a', a,as, Similarly we can denote the elements of B by b l , bz, b3,

. ...

. ...

Case 1: Suppose elements of A a r e all distinct from elements of B . Then the set consisting of elements from A o r B is countable since we can establish the following 1-1 correspondence.

A or B

ar

bi

Natural numbers

1

2

az

bz

a3

3

4

5

bs

$ $ $ $ $ $

6

... ...

Case 2: If some elements of A and B a r e the same, we count them only once as in Problem 17. Then the set of elements belonging to A or B (or both) is countable.

CHAP. 13

11

NUMBERS‘

The set consisting of all elements which belong to A or B (or both) is often called the union of A and B, denoted by A U B or A B. The set consisting of all elements which a r e contained in both A and B is called the intersection of A and B, denoted by A n B or AB. If A and B a r e countable, so is A n B . The set consisting of all elements in A but not in B is written A - B . If we let B be the set of elements which a r e not in B, we can also write A - B = AB. If A and B a r e countable, so is A - B .

+

19. Prove that the set of all positive rational numbers is countable. Consider all rational numbers x > 1. With each such rational number we can associate one and only one rational number l / x in ( O , l ) , i.e. there is a one to one correspondence between all rational numbers > 1 and all rational numbers in ( 0 , l ) . Since these last a r e countable by Problem 17, i t follows t h a t the set of all rational numbers > 1 a r e also countable. From Problem 18 i t then follows tha t the set consisting of all positive rational numbers is countable, since this is composed of the two countable sets of rationals between 0 and 1 and those greater than or equal to 1. From this we can show t h a t the set of all rational numbers is countable (see Problem 59).

20. Prove that the set of all real numbers in [0,1]is non-countable.

.

.

Every real number in [0,1] has a decimal expansion .a1 a2 a3 . . where al, apt,. . a r e any of the digits 0, 1 , 2 , .,9. W e assume t h a t numbers whose decimal expansions terminate such as 0.7324 a r e written 0.73240000.. . and t h a t this is the same as 0.73239999.. .. If all real numbers in [0,1] a r e countable we can place them in 1-1 correspondence with t he natural numbers as in the following list.

..

1 2 3

f,

O.all a 1 2 a13 a14

f)

0.a21 a 2 2 a 2 3 a 2 4

f)

O.Usi

W e now form a number

O.bi

.

b2 b3 bq

a 3 2 a33 a 3 4

. .. .. ...

...

where b l # all, b2 # a 2 2 , b3 # a 3 3 , b4 # a44, . . and where all b’s beyond some position a r e not all 9’s. This number, which is in [0,1], is different from all numbers in the above list and is thus not in the list, contradicting the assumption t h a t all numbers in [0,1] were included. Because of this contradiction i t follows t h a t the real numbers in [0,1] cannot be placed in 1-1 correspondence with the natural numbers, i.e. the set of real numbers in [0,1] is non-countable.

LIMIT POINTS, BOUNDS, WEIERSTRASS-BOLZANO THEOREM 21. (a) Prove that the infinite set of numbers 1, Q, Q, 4, . . . is bounded. ( b ) Determine the least upper bound (1.u.b.) and greatest lower bound (g.1.b.) of the set. ( c ) Prove that 0 is a limit point of the set. (d) Is the set a closed set? ( e ) How does this set illustrate the Weierstrass-Bolzano theorem ? ( a ) Since all members of the set a r e less than 2 and greater than -1 (for example), the set is bounded; 2 is a n upper bound, -1 is a lower bound. W e can find smaller upper bounds (e.g.%) a nd larger lower bounds (e.g. -4). ( b ) Since no member of the set is greater tha n 1 and since there is at least one.member of the set (namely 1) which exceeds 1 - e for every positive number e, we see t h a t 1 is the 1.u.b. of the set. Since no member of the set is less than 0 and since there is at least one member of the set which is less than O + C fo r every positive e (we can always choose for this purpose the number l/n where n is a positive integer greater tha n U€),we see t h a t 0 is the g.1.b. of the set.

12

[CHAP. 1

NUMBERS (c)

Let x be any member of the set. Since we can always find a number z such that 0 < 1x1 < 8 for any positive number 6 (e.g. we can always pick x to be the number l l n where n is a positive integer greater than UQ),we see that 0 is a limit point of the set. To put this another way, we see that any deleted 6 neighborhood of 0 always includes members of the set, no matter how small we take Q > 0.

( d ) The set is not a closed set since the limit point 0 does not belong to the given set. (e)

Since the set is bounded and infinite i t must, by the Weierstrass-Bolzano theorem, have a t least one limit point. We have found this to be the case, so t h a t the theorem is illustrated.

ALGEBRAIC and TRANSCENDENTAL NUMBERS

fi + fi is an algebraic

22. Prove that

+

number.

+

Let x = ;/z fi. Then z - fi = @. Cubing both sides and simplifying, we find x3 9%- 2 1). Then squaring both sides and simplifying we find x" - 92' - 42' 27x2 36%- 23 = 0. Since this is a polynomial equation with integral coefficients it follows that fi,which is a solution, is a n algebraic number.

= 3fi(x*

+

+

+

fi+

23. Prove that the set of all algebraic numbers is a countable set.

+

+. +

Algebraic numbers are solutions to polynomial equations of the form UGZ" alx"-' .. as = 0 where uo,al,. . .,a, are integers. lull . . . la,l n. For any given value of P there are only a finite number Let P = luol of possible polynomial equations and thus only a finite number of possible algebraic numbers. Write all algebraic numbers corresponding to P = 1 , 2 , 3 , 4 , . avoiding repetitions. Thus all algebraic numbers can be placed into 1-1correspondence with the natural numbers and so are countable.

+

+

+

+

..

COMPLEX NUMBERS 24. Perform the indicated operations. (U)

(4-2i)

+ (-6+Si)

=

( b ) (-7+ 3i) - ( 2 - 4 i ) (c)

(3-2i)(l

+ 3i)

+ 6i = 4 - 6 + ( - 2 + S ) i - 7 + 3i - 2 + 4i = - 9 + 7i

= 3(1 + 3 i ) - %(I + 3 i ) = 3 4- 9i - 2i - 6ta

+

+

6 6i4 3i -6+6i (d) - 4 - 3i 4-3i 4+3i

(4

i+aq+t3+i'+i5 l+i

-

i-1

- (-6

+ 6~')(4+ 323 16 - 9%-

-

+ (a")(i)+ (a")* + (ia))"i-l+i

-20

lx1zal

z1

=

= =

2 1

+ i y ~ , zz =

x2+

1 (Xl + iZ/l)(ZZ + iy2) I

&1X2

- 2/1y2y

+

Then

i92.

= j

+

(X12/2

XlX2

- 2/1Y2

x2?d1)2

=

3

i - 1-i

+ i(x1yz +

+ 9i - 2i + 6

- 16i + 2Oi + 16tq 16

25. If x1 and x2 are two complex numbers, prove that Let

= -2+3i

= 4 - 2i - 6

+9

+1 +i

l+i

I21221

22Y1)

I

=

1x11 I x 2 l .

dx:x2 + YfY2 + %:?I: + XIY:

= 9

+ 7i

CHAP. 11

13

NUMBERS

26. Solve x3 - 2x

- 4 = 0.

By trial we find x = 2 is a root. The possible rational roots using Problem 7 are *1,*2,*4. Then the given equation can be written ( z - 2 ) ( x P + 2 x + 2 ) = 0. The solutions to the quadratic equa-

.-

tion axP+b x + c = 0 are x = If: 2a 2 4 2 2 % --2*&i=?i - -1ki. 2 2 2 The set of solutions is 2, -1 i, -1 - i.

+

For a = l , b = 2 ,

c=2

this gives

POLAR FORM of COMPLEX NUMBERS 27. Express in polar form (a)3 + 3i, (b) - 1 + f i i , ( c ) - 1, ( d ) - 2 - 2fii.

Fig. 1-4 (a) Amplitude

3

+

+ = 46O

fm

= a/4 radians. Modulus p = = 3 f i Then 3i = p(cos # i sin #) = 3fi(cos u/4 4- i sin ~ 1 4 )= 3 f i cis a/4 = 3fieTit4

+

d

M

= fi = 2. ( b ) Amplitude # = 120° = 2 ~ / 3radians. Modulus p = -1 f l i = COS 2n13 i sin 2 ~ 1 3 )= 2 cis 2aI3 = 2etUits

+

(c) Amplitude 9 = 180' =

+

P

Modulus p = d(-l)2

radians.

+ i sin a) =

-1 = l(cosa

+ (O)*

= 1. Then

cisa = eTi

+

Modulus p = d(-2)z (-2fi)2 = 4. 3 (d) Amplitude 9 = 240° = 4 ~ / radians. -2 - 2 6 = COS 4 ~ / 3 i sin 4 ~ 1 3 )= 4 cis 4n13 = 4e'"/'

+

28. Evaluate (a) (-1

+ fli)lo,

(b) (-1

+

+

+

=

1024(-&

i = f i ( c o s 136' ( b ) -1 Then

+

+4 6 4

+

= -612

+

+

+ 612fii

+ i sin 136O) = fi[cos (136O + k

The results for k = 0, 1 , 2 are

Then

i)ll3.

(a) By Problem 27(b) and De Moivre's theorem, (-1 fl~]'O = COS 2a/3 i sin 2a/3)]1° = 210(c0s20r/3 i sin 20a/3) = 1024[cos (2a/3 6a) i sin (2a/3 6u)] = 1024(cos 2a/3

+

Then

+ i sin (,,,. +:*3600

360')

+ i sin 2n/3)

+ i sin (135' + k

360°)]

>I

+ i sin 4 6 O ) , f i ( c o s 166O + i sin 166O), *(cos 285' + i sin 286O) f i ( c o s 46O

The results for k = 3,4,6,6,7, . . . give repetitions of these. These complex roots are represented geometrically in the complex PJ on the circle of Fig. 1-6. plane by points PI,PP, Fig. 1-5

x

=

14

NUMBERS

MATHEMATICAL INDUCTION

+ + ... + n2 =

29. Prove that l2+ Z2 + 32 42

The statement is true for n = 1 since

[CHAP. 1

Assume the statement true for n = k .

+

Qn(n 1)(2n+ 1).

1' = Q(l)(l

+ 1)(2 1 + 1) =

1.

Then

+ + + + + + + + = Qk(k + 1)(2k + 1) + ( k + 1)' = ( k + 1)[Qk(2k+ 1) + k + 11 = Q(k + 1)(2k' + 7 k + 6 ) = &(k + l)(k + 2)(2k + 3 ) which shows that the statement is true for n = k + 1 i f it is true for n = k . But since it is true for n = 1, it follows that it is true for n = 1 + 1 = 2 and for n = 2 + 1 = 3, . . ., i.e. it is true for all Adding ( k 1' 2' 3'

1'

+ 1)'

+ 2' + 3' +

* * *

+ k'

= Qk(k 1)(2k 1)

to both sides, .. - k' ( k 1)'

positive integers n.

30. Prove that x n - g n has x-g

as a factor for a11 positive integers n.

The statement is true for n = 1 since xi- y' = z - -y. Assume the statement true for n = k , i.e. assume that z k - y k has x - - y xk+'

-

yk+1

-

= = xk(z-y) gk+1

zk-y

+

+

- yk+'

s k y

as a factor.

y(zk--yk)

The first term on the right has x - - y as a factor, and the second term on the right also a factor because of the above assumption. Thus zk+'- yk+' has x - - y as a factor if x k - yk does. Then since x' --yi has x - - y as factor, it follows that x'--y' x - y as a factor, etc.

+

31. Prove Bemoulli's inequality (1 x)"

Consider

has x - -y as

has x - - y as a factor, x'-ys

has

> 1+ nx for n = 2,3, . . . if x > -1, x # 0.

The statement is true for n = 2 since ( l + x ) ' = 1 + 2 x + x s > 14-2x. Assume the statement true for n = k , i.e., (1 x)lr > 1 kx. Multiply both sides by 1 x (which is positive since x > -1). Then we have (1 x ) k + ' > (1 x)(l kx) = 1 ( k 1)z kx' > 1 (k 1)z

+

+

Thus the statement is true for

+

+ n = k + 1 if

+

+ +

+

+

it is true for n = k.

+ + +

.

But since the statement is true for n = 2, it must be true for n = 2 1 = 3, . . and is thus true for all integers greater than or equal to 2. Note that the result is not true for n = 1. However, the modified result (1 z)" 2 1 nx is true for n = 1 , 2 , 3,....

+

+

MISCELLANEOUS PROBLEMS 32. Prove that every positive integer P can be expressed uniquely in the form P = a02" a 1 2 ~ - l ~ 2 2 " - ~ . . . a, where the a's are 0's or 1's.

+

+

+

+

+

+

+

+ + +

UI 2"-' ... a,-' d2. Dividing P by 2, we have P / 2 = a0 2"-' Then an is the remainder, 0 or 1, obtained when P is divided by 2 and is unique. a12"-' Let PI be the integer part of P / 2 . Then PI = a02""' &-I. Dividing PI by 2 we see that a,-1 is the remainder, 0 or 1, obtained when PIis divided by 2 and is unique. By continuing in this manner, all the a's can be determined as 0's or 1's and are unique.

+

33. Express the number 23 in the form of Problem 32. The determination of the coefficients can be arranged as follows. 2 23 2 Remainder 1

0

Remainder 1

CHAP. 11

NUMBERS

15

+

+

+

+

The coefficients a r e 1 0 1 1 1. Check: 23 = 1 2* 0 z3 1 22 1 2 1. The number 10111 is said to represent 23 in the scale of two or binary scale.

34. Dedekind defined a cut, section or purtition in the rational number system as a separa-

tion of all rational numbers into two classes or sets called L (the left hand class) and R (the right hand class) having the following properties: I. The classes are non-empty (i.e. a t least one number belongs to each class). 11. Every rational number is in one class or the other. 111. Every number in L is less than every number in R. Prove each of the following statements: (a) There cannot be a largest number in L and a smallest number in R.

( b ) It is possible for L to have a largest number and for R to have no smallest number. What type of number does the cut define in this case? ( c ) It is possible for L to have no largest number and for R to have a smallest number.

What type of number does the cut define in this case?

( d ) It is possible for L to have no largest number and for R to have no smallest number. What type of number does the cut define in this case? (a) Let a be the largest rational number in L, and b the smallest rational number in R . Then either a = b or a < b . We cannot have a = b since by definition of the cut every number in L is less than every number in R. We cannot have a < b since by Problem 9, &(a+ b) is a rational number which would be greater than a (and so would have to be in R) but less than b (and so would have to be in L), and by definition a rational number cannot belong to both L and R.

( b ) As a n indication of the possibility let L contain the number 8 and all rational numbers less than 8, while R contains all rational numbers greater than 3. In this case the cut defines the rational number 3. A similar argument replacing 8 by any other rational number shows t h a t in such case the cut defines a rational number. (c)

As a n indication of the possibility let L contain all rational numbers less than Q while R contains all rational numbers greater than 3. This c u t also defines the rational number 3. A similar argument shows t h a t this cut always defines a rational number.

( d ) A s a n indication of the possibility let L consist of all negative rational numbers and all positive rational numbers whose squares a r e less than 2, while R consists of all positive numbers whose squares a r e greater than 2. We can show t h a t if a is any number of the L class there is always a larger number of the L class, while if b is any number of the R class there is always a smaller number of the R class (see Problem 106). A cut of this type defines a n irrational number. From ( b ) , ( c ) , ( d )i t follows t h a t every cut in the rational number system, called a Dedekind cut, defines either a rational or a n irrational number. By use of Dedekind cuts we can define

operations (such as addition, multiplication, etc.) with irrational numbers.

NUMBERS

16

[CHAP. 1

Supplementary Problems OPERATIONS with NUMBERS 35. Given z = -3, y = 2, z = 5 , a = $ and b = -*, evaluate: 3a'b ab' xy - 22' (a) (2z - y)(3y + z)(5z - "), (b) (1' 2albL + 1

+

J

Am.

(U) 2200,

( b ) 32, (c) -S1/41,

+

(U% by)' ( d ) (ay + bz)l

9

+ ( a y - bx)' - by)'

+

*

(d) 1

36. Find the set of values of z fo r which the following equations are true. Justify all steps in each case. ~ 1)} 2 ( 2 ~ 1) = 12(x 2) - 2 (a) 4{(2 - 2) 3 ( 2 (c) ~ / x * + 8 ~ + 7 - ~ G = E~ + 1

+

+

37.

+

+

Prove th at

giving restrictions if any.

RATIONAL and IRRATIONAL NUMBERS 38. Find decimal expansions fo r ( a ) +, (b)

fi.

.eeamm

A m . (a) 0.428571, ( b ) 2.2360679.

..

..

.,16 ha s 16 digits in the repeating portion of its decimal expansion. Is there a ny relation between the orders of the digits in these expansions?

39. Show t h a t a fraction with denominator 17 and with numerator 1,2,3,

fi a r e irrational numbers. - fi,(b) fi+ fi+ 6 a r e irrational numbers.

40.

Prove t h a t (a)fi, (b)

41.

Prove t h a t (a)

42.

Determine a positive rational number whose square differs from 7 by less than .000001.

43.

Prove th a t every rational number can be expressed as a repeating decimal.

44.

Find the values of z fo r which (a)2 z 3 - S s z - 9 z + 18 = 0, ( b ) 3 ~ ~ + 4 2 ~ - 3 5 2 +=80, (c) ~ ' - 2 1 2 * + 4 = 0. A m . (a) 3, -2,3/2 ( b ) 8/3, -2 2 (c) i ( S f 3(-5 zt

m),

6

45. If a, 6, c, d are rational and rn is not a perfect square, prove that a 6 =d. l+fi+fi - 12fi-Zfi+l46-7 46. Prove t h a t 11 1-fi+fi-

+ 6 6= c + d fiif and only if a = c and

INEQUALITIES 47. Find the set of values of x fo r which each of the following inequalities holds. 1 3 2 x+3 ( U ) 1: 2s 2 6 , ( b ) z ( x i - 2 ) 5 24, (c) Iz+21 < 1z-61, ( d ) 2+2 >

m*

+

Ans. (a) 0 < z 5 4,

( b ) -6 5 % 5 4 , (c) x < 3 / 2 ,

+ IyI,

48.

Prove (a) Ix+yl I1 . 1

49.

Prove t h a t fo r all real x,y, z, z* y*

50. If a' 51. If

+ b'

= 1 and

( b ) I z + y + zl 5 1x1 4- 1

+ + z'

~ 4-1

+ yz + xz. ac + bd 5 1.

1x1,

-9, or

x < -2

(c) 1x-111 2 1x1 - 12/1.

1 zy

+ ds = 1, prove t h a t 1 z"+l + ~1n + > l z" + 2

' C

x > 0, prove t h a t

( d ) z > 3 , -1 < s
x o or x < x o respectively. We have lirn f ( x ) = l if and only if lirn f ( x ) = lirn f ( x ) = 1. 2420

2-+20+

2420-

24

[CHAP. 2

FUNCTIONS, LIMITS AND CONTINUITY

THEOREMS on LIMITS If lim f ( s ) = A and lim g(x) = B, then 2- 20

2-20

I. lim ( f ( x ) 2-20

+ g(x))

= lim f ( x ) 2-20

+

lim g(x) = A

2-xo

+B

lim ( f ( x ) - g(x)) = lim f ( x ) - lim g(x) = A - B

2.

2-20

2-20

2-20

2-20

Similar results hold for right and left hand limits.

INFINITY It sometimes happens that as x+ xo, f ( z ) increases or decreases without bound. In such case it is customary to write lim f ( x ) = +m or lirn f ( x ) = --oo respectively. The 2-xo

5-20

symbols +a (also written a) and -00 are read plus infinity (or infinity)and minus infinity respectively, but it must be emphasized that they are not numbers. In precise language, we say that lim f ( z ) = 00 if for each positive number M we 2-20 can find a positive number 6 (depending on M in general) such that f ( z ) > M whenever 0 N . A similar definition can be formulated for lim f(z).

If(%) - I1 < c

whenever

z-+-OO

SPECIAL LIMITS sinx 1. lim- 1,

x

2-0

+;I

(1

2. lim x-00

e2- 1

3. lim2-0

x

= e,

= 1,

lim 1- cosx x-bo

x

= o

+

lim (1 z ) " ~ = e

2-o+

x-1 limlnx

= 1

CONTINUITY Let f ( x ) be defined and single-valued for all values of x near x = xo as well as at x = x o (i.e. in a 6 neighborhood of XO). The function f ( x ) is called continuous at X = X O if lirn f ( z ) = f ( x 0 ) . Note that this implies three conditions which must be met in order that 2-20

f ( x ) be continuous a t x = x o .

CHAP. 21

25

FUNCTIONS, LIMITS AND CONTINUITY

1. lirn f ( x ) = l x-20

must exist.

2. ~ ( z omust ) exist, i.e. f ( x ) is defined at XO.

3. l = f ( x o )

Equivalently if f ( x ) is continuous a t XO, we can write this in the suggestive form Iim f ( x ) = f ( Iim x).

z-xo

x+xg

Examples: 1. If f(x) =

1

o,

%”

# =

then from the example on Page 23, lirn f ( z ) = 4. But f ( 2 ) = 0. t-, I

Hence lim f ( x ) # f ( 2 ) and the function is not continuous at z = 2. #-*S

2. If f(z) = xe for all z,then lirn f ( z ) CI, I

= f ( 2 ) = 4 and f ( x ) is continuous at z = 2.

Points where f ( x ) fails to be continuous are called discontinuities of f(s)and f ( x ) is said to be discontinuous a t these points. In constructing a graph of a continuous function the pencil need never leave the paper, while for a discontinuous function this is not true since there is generally a jump taking place. This is of course merely a characteristic property and not a definition of continuity or discontinuity. Alternative to the above definition of continuity, we can define f ( x ) as continuous at z = xo if for any E > 0 we can find 6 > 0 such that If($) - f ( x 0 ) l < c whenever Iz - x01 < S. Note that this is simply the definition of limit with l = f ( x o ) and removal of the restriction that x # XO.

RIGHT and LEFT HAND CONTINUITY If f ( x ) is defined only for x 2x0, the above definition does not apply. In such case

we call f ( x ) continuous (on the right) a t x = x o if

lim f ( x ) = ~ ( x o )i.e. , if f(xo+) = ~ ( z o ) .

x-xo+

Similarly, f ( x ) is continuous (on the left) a t x = z o if Definitions in terms of E and 6 can be given.

lirn f(x) = f(zo), i.e. f(xo-) = ~ ( x o ) .

x-20-

CONTINUITY in an INTERVAL A function f ( x ) is said to be continuous in an interval if it is continuous at all points of the interval. In particular, if f ( s ) is defined in the closed interval a S x 5 b or [a, b ] , then f ( x ) is continuous in the interval if and only if lirn f ( x ) = ~ ( x o )for a < xo < b, 2-+20 lirn f ( x ) = f(a) and lirn f ( x ) = f ( b ) .

xda+

zdb-

THEOREMS on CONTINUITY Theorem 1. If f ( x ) and g(x) are continuous at f ( x ) + g(4, f ( x ) - g(4, f ( x ) g ( x ) and (’)

go’

X=XO,

so also are the functions

the last only if g(x0) # 0. Similar results hold for continuity in an interval. Theorem 2. The following functions are continuous in every finite interval: (a) all polynomials; (b) sin x and cos x; ( c ) a2, a > 0. Theorem 3. If y = f ( x ) is continuous at X = ~ Oand z = g ( y ) is continuous at y=yo and if yo=f(xo), then the function z = g [ f ( z ) ] ,called a function of a function or composite function, is continuous at x = x o . This is sometimes briefly stated as: A continuous function of a continuous function i s continuous.

26

FUNCTIONS, LIMITS AND CONTINUITY

[CHAP. 2

Theorem 4. If f ( x ) is continuous in a closed interval, it is bounded in the interval. Theorem 5. If f ( x ) is continuous at x = xo and ~ ( x o>) 0 [or f ( x 0 )< 01, there exists an interval about x = 20 in which f ( x )> 0 [or f ( x )< 01. Theorem 6. If a function f ( x ) is continuous in an interval and either strictly increasing or strictly decreasing, the inverse function f-l(x) is single-valued, continuous and either strictly increasing or strictly decreasing. Theorem 7. If f ( x ) is continuous in [a, b] and if f(a) = A and f ( b )= B, then corresponding to any number C between A and B there exists at least one number c in [a,b] such that f ( c ) = C . This is sometimes called the intermediate value theorem. Theorem 8. If f ( x ) is continuous in [a,b] and if f(a) and f ( b ) have opposite signs, there is at least one number c for which f ( c ) = O where a < c < b . This is related to Theorem 7. Theorem 9. If f ( x ) is continuous in a closed interval, then f ( x ) has a maximum value M for at least one value of x in the interval and a minimum value m for at least one value of x in the interval. Furthermore, f ( x ) assumes all values between m and M for one o r more values of x in the interval. Theorem 10. If f ( x ) is continuous in a closed interval and if M and m are respectively the least upper bound (1.u.b.) and greatest lower bound (g.1.b.) of f ( x ) , there exists at least one value of x in the interval for which f ( x )= M or f ( x )= m. This is related to Theorem 9. SECTIONAL CONTINUITY A function is called sectionally continuous or piecewise continuous in an interval a 5 x 5 b if the interval can be subdivided into a finite number of intervals in each of which the function is continuous and has finite right and left hand limits. Such a function has only a finite number of discontinuities. An example of a function which is sectionally continuous in a 5 IC 5 b is shown graphically in Fig. 2-1 below. This function has discontinuities at XI, x2, x3 and x4. f (4 I I I

Fig.2-1

UNIFORM CONTINUITY Let f ( x ) be continuous in an interval. Then by definition a t each point xo of the interval and for any e > 0, we can find 8 > 0 (which will in general depend on both and - XO[< 8. If we can find the particular point XO) such that If(%) - f(x0)l < E whenever 8 for each c which holds for all points of the interval (i.e. if 8 depends only on E and not on XO), we say that f ( x ) is uniformly continuous in the interval. Alternatively, f ( x ) is uniformly continuous in an interval if for any E > 0 we can find 8 > 0 such that If(xl)-f(x2)l < whenever 1x1-n1 < 8 where $1 and x2 are any two points in the interval. Theorem. If f ( x ) is continuous in a closed interval, it is uniformly continuous in the interval.

IX

27

FUNCTIONS, LIMITS AND CONTINUITY

CHAP. 21

Solved Problems

2.

z

2

f(z)

0

3

4

5

5 8

6 9

7

8

2

8

6

0

.

5

7.5

2.75

2.75

.

Let g(x) = ( z - 2 ) ( 8 - x ) for 2 < x < 8 . (a)Discuss the difference between the graph of g(x) and that of f ( x ) in Problem 1. (b) What is the 1.u.b. and g.1.b. of g(x)? ( c ) Does g(x) attain its 1.u.b. and g.1.b. for any value of x in the domain of definition? (d) Answer parts (b) and ( c ) for the function f ( x ) of Problem 1. (a) The graph of g ( x ) is the same as that in Problem 1 except that the two points (2,O) and (8,O) are missing, since g ( x ) is not defined a t x = 2 and x = 8. ( b ) The 1.u.b. of g(z) is 9. The g.1.b. of g(z) is 0. (c) The 1.u.b. of g ( x ) is attained for the value x = 5. The g.1.b. of g(x) is not attained, since there is no value of x in the domain of definition such that g ( x ) = 0. ( d ) As in ( b ) , the 1.u.b. of f(x) is 9 and the g.1.b. of f(x) is 0. The 1.u.b. of f ( z )is attained for the value x = 5 and the g.1.b. of f ( x ) is attained at x = 2 and x = 8.

Note that a function, such as f ( x ) , which is continuous in a closed interval attains its 1.u.b. and g.1.b. at some point of the interval. However a function, such a s g(x), which is not continuous in a closed interval need not attain its 1.u.b. and g.1.b. See Problem 34.

3. Let

f(x) =

f(-5) =1 f(1.41423) = 1

1, if x is a rational number x is an irrational number

0, if

(a)Find f(Q),f(-5), f(1.414231, f ( f i ) ,

since -5 is a rational number since 1.41423 is a rational number f(fi) = o since fi is a n irrational number ( b ) The graph is shown in the adjoining Fig. 2-3. From its appearance i t would seem that there are two functional values 0 and 1 corresponding to each value of x, i.e. that f ( z )

f (2) 1

0

X

28 4.

FUNCTIONS, LIMITS AND CONTINUITY

[CHAP. 2

Referring to Problem 1, (a) construct the graph of f - ' ( x ) , (b) find an expression for and show that f-'(x) is not single-valued. Y =f-'(z)

f-l(x)

The graph of y = f ( z ) or x = f-l(y) is shown in Fig. 2-2 of Problem l ( e ) . To obtain the graph of y=f-'(x), we have only to interchange the z and y axes. We obtain the graph shown in the adjoining Fig. 2-4 after orienting the axes in the usual manner.

We have y = ( x - 2 ) ( 8 - x ) or x ' - l O x + 1 6 + y Using the quadratic formula, 2

= f-'(y) = l0 *

Then, y = f - ' ( z ) = 6

4 ' ' ' - 4(16+Y) 2

2

= 5

2

= 0.

fGy. Fig.2 4

d G .

In the graph, A P represents y = 6 + d G , BP represents y = 5 - d G . Thus for each value of x in 0 Sx g(xa) whenever x l < x ~ . If Z I < X S then 9--1 > 9 - x ~ , d G i > d G i , 6 + d G i > 6 + d G showing that g(x) is strictly decreasing. Yes, any strictly decreasing function is also monotonic decreasing, since if g(z1) > g(x2) i t is also true t h a t g(xl)h g ( x ~ ) . However if g ( x ) is monotonic decreasing, it is not necessarily strictly decreasing.

+

+

If y = 6 d G then y - 6 = d G or squaring, z = -16 1Oy - y' = (y - 2)(8 - y) and x is a single-valued function of y, i.e. the inverse function is single-valued. In general, any strictly decreasing (or increasing) function has a single-valued inverse (see Theorem 6, Page 26). The results of this problem can be interpreted graphically using the figure of Problem 4.

6.

Construct graphs for the functions greatest integer 5 x.

(a) j ( x )

= {;,sin 1/x, x > 0

x=o '

( b ) f ( x ) = [XI =

(a) The required graph is shown in Fig. 2-6 below. Since Ix sin l/s] 5 1x1, the graph is included between y = x and y = -x. Note that f ( z )= 0 when sin 1/x = 0 or 1/z = mr, m = 1,2,3,4, . ., i.e. where x = l/a,1/2a,1/3s, . . .. The curve oscillates infinitely often between x = l/lr and z = 0.

.

I

\

Fig. 2-5

Fig. 2-6

= 1, ( b ) The required graph is shown in Fig. 2-6 above. If 1 d x < 2, then [x]= 1. Thus [1.8] = 1, [fi] [1.99999]= 1. However, [2] = 2. Similarly for 2 5 x < 3, [z]= 2, etc. Thus there a r e jumps at the integers. The function is sometimes called the staircase function or step function.

CHAP. 21

7.

29

FUNCTIONS, LIMITS AND CONTINUITY

(a) Construct the graph of f ( x ) = tan x. (b) Construct the graph of tan-' x. ( c ) Show graphically why tan-' x is a multiple-valued function. (cl) Indicate possible principal values for tan-'2. (e) Using your choice, evaluate tan-l(-l). (a) The graph of f ( s )= tan x appears in Fig. 2-7 below.

Fig. 2-8

Fig. 2-7

(b)

If y = f ( x ) = tan x, then x = f-'(y) = tan-'^. Then the graph of f-'(x) = tan" x is obtained by interchanging the x and y axes in the graph of (a). The result, with axes oriented as usual, appears in Fig. 2-8 above.

(c)

In Fig. 2-8 of (b), any vertical line meets the graph in infinitely many points. Thus tan-'% is a multiple-valued function with infinitely many branches.

( d ) To define tan-lx as a single-valued function, it is clear from the graph that we can only do so by restricting its value to any of the following: -a/2 < tan-' R: < ~ / 2 , a/2 < tan-' z < 3 d 2 , etc. We shall agree to take the first as defining the principal value. Note that on any of these branches, tan-'% is a strictly increasing function with a singlevalued inverse. (e)

tan-'(-l) = -u/4 is the only value lying between -a/2 and a/2, i.e. it is the principal value according to our choice in (4.

fi+l 8. Show that f ( x ) = - x # -1, is an irrational algebraic function. x+l'

-

+

Vx+l

+

+

+

then (x 1)y - 1 = or squaring, (z 1)'~'- 2 ( x 1)y 1 - x = 0, a x+l polynomial equation in y whose coefficients are polynomials in x. Thus f ( z ) is an algebraic function. However, it is not the quotient of two polynomials, so that it is an irrational algebraic function. I f y = -

9.

6

If f(x) = coshx = i(ez+e-z), prove that we can choose as the principal value of the inverse function, cosh-' x = In (x d F l ) , x 2 1. 22/+q=iIf y = *(er + e -3 , eel - 2yez+ 1 = 0. Then using the quadratic formula, er = 2

+

j,*dm.

Thus z = l n ( y * d F i ) .

Since y - d

F i x

=

(U

-

Y+-1-

+

= --+ln(y =/I)

Y+dF=i or

)

-

Y

+

1

m

'

cosh-'y = * l n ( y

we can also write

+d

m )

+ sign as defining the principal value and replacing y by x, we have + d r l ) . The choice x 2 1 is made so that the inverse function is real.

Choosing the In (x

cosh-'x

=

LIMITS (a) We must show that given any E > 0 we can find 6 Ixl-41 < E when 0 < 1%-2) < 6 .

>0

(depending on

E

in general) such that

30

[CHAP. 2

FUNCTIONS, LIMITS AND CONTINUITY

Choose 6 S 1 so t h a t 0 < 12 - 2) < 1 o r 1 < x < 3, z # 2. Then 1a9-41 = \ ( ~ - 2 ) ( s + 2 ) 1 = 1%-21 ] x + 2 1 < 6 ] x + 2 1 < 56. Take 6 as 1 or 4 6 , whichever is smaller. Then we have 1 9 - 4) < e whenever 0 < ]x- 21 < 6 and the required result is proved. It is of interest to consider some numerical values. If for example we wish to make 1x* - 41 < .OS, we can choose 6 = 4 5 = .05/5 = .01. To see t h a t this is actually the case, note that if 0 < 12-21 < .01 then 1.99 < x < 2.01 ( z Z 2 ) and so 3.9601 < x* < 4.0401, -.0399 < x* - 4 < .0401 and certainly 1%' - 41 < .06 (x*# 4). The fact t h a t these inequalities also happen to hold at x = 2 is merely coincidental. If we wish to make Is*- 41 < 6, we can choose 6 = 1 and this will be satisfied. There is no difference between the proof for this case and the proof in (a),since in both cases we exclude x = 2.

11. Prove that lim

+ +

2x4- 6x3 x2 3

x-l

2 41

We must show t h a t for any

< 1% - 11 < 6 .

e

>0

= -8. we can find 6 > 0 such that

+ +

Since z# 1, we can write 2%' - 6xS x'

I

- 6 z a + + ' + 3 - (-8)l < e - (2xa- 42' - 3%- 3)(s - 1) 22'

3 x-1 x-1 2xs - 4s* - 3%- 3 on cancelling the common factor x - 1 # 0. Then we must show t h a t for any e > 0, we can find 6 > 0 such t h a t ]2xa- 4x' - 32 51 < e when 0 < 1% - 11 < 6. Choosing 6 i1, we have 0 < z < 2, z # 1. Now 12%'- 4%'- 3s 61 = 1% - 11 12%'- 2%- 51 < 6 12%'- 2%- 5 ) < 6 ( ] 2 ~ ' 1 12x1 6 ) < (8 4 6 ) s = 176. Taking 6 as the smaller of 1 and d17, the required result follows.

when 0

+

+

+ +

12. Let f(x) = Iim f(x).

z+S-

(a) Graph the function.

(b) Find lim f ( x ) .

(c) Find

2+3+

( d ) Find l i m f ( ~ ) . 2eS

Then the graph, shown in the adjoining Fig. 2-9, consists of the lines y = 1, x > 3; y = -1, x < 3 and the point (3,O).

As x+

+

+

t 1 I

3 from the right, f ( x )-P 1, i.e. lirn f ( x )= 1, ++a+

as seems clear from the graph. To prove this we must show t h a t given any e > 0, we can find 6 > 0 such t h a t If(%) 11 < e whenever 0 < x - 1 < 6 . Now since z > 1, f ( z )= 1 and so the proof consists in the triviality that 11 - 11 < e whenever 0 < x - 1 < 6.

-

As x + 3 from the left, f ( x )-P -1, i.e. lim f ( x ) = -1. ++a-

Fig. 2-9

A proof can be formulated as in (b).

Since lim f(x) # lim f ( x ) , lirn f ( x ) does not exist. rea+

=+a-

r-, 8

13. Prove that lim x sin l l x = 0. zeo

0


0, we can find 6 > 0 such t h a t 1% sin l/s - 01 < e when Jx-01 < 6. If 0 < 1x1 < 6, then Iz sin l / x J = 1x1 ]sin1/~1S 1x1 < 8 since lsinl/z] S 1 for all x # O . Making the choice 6 = r, we see that Ix sin l/xl < e when 0 < 1x1 < 6, completing the proof.

CHAP. 21

14. Evaluate

31

FUNCTIONS, LIMITS AND CONTINUITY

lirn 1

z-+o+

2

+ e-'/"'

As x+ O+ we suspect t h a t 1/x increases indefinitely, e"" increases indefinitely, e-''= approaches 0,

1

+ e-'IS approaches 1; th us the required limit is 2.

To prove this conjecture we must show that, given


0, we can

when

P

0

find 6

<x
0 such t h a t

6

Now 0 will work when

Since th e function on t h e right is smaller t h a n 1 for all s > 0, a ny 6 0I
--l, 2 c

E 2

1. If

1

In ( 2 / ~- 1)

15. Explain exactly what is meant by the statement

validity of this statement.

1 lirn -= z + l (X - 1 ) 4

00

and prove the

The statement means t h a t fo r each positive number M ,we can find a positive number 6 (depending on M in general) such t h a t (x

To prove this note t h a t Choosing 6 =

l/G,the

> M

- 1)'

when

O < Iz-11 < 6

-> M when O < ( z - l ) ' < - 1 or M ( x - 1)'

O < 1cc-11

O

.

(a) Construct a graph of f ( z ) .

Evaluate (b) lirn f ( z ) , (c) z+lima lim f@), (e) r lirn f ( z ) , (f) lim f ( z ) , justifying your answer - f ( z ) , (d) *+O+ 40r-e0 =+a in each case. A m . (b) 9, ( c ) -10, (d) 6 , (e) -1, ( f ) does not exist 53.

A m . (a) 2, ( b ) 3 54.

(a)If f ( z ) = z' cos Us, evaluate lirn f ( z ) , justifying your answer. (b) Does your answer to (a) etill =-+ 0 ' cos l/z, z # 0, f ( 0 ) = 2? Explain. remain the same if we consider f ( z ) = z

55. Prove that lirn 10-l'(z-s)a = 0 using the definition. =-,a 1 10-'/* Evaluate (a) lim f ( z ) , ( b ) lim f ( x ) , (c) lim 56. Let f ( z ) = - lo-l,a, z#O, f ( 0 ) 2+0+ 0-0r-b0

+

=t.

A m . (a)

fying answers in all cases. 57.

Find (a) Iim =+o+

w, x

( b ) lirn 040-

t.

4, ( b ) -1,

f(4, justi-

(c) does not exist.

Illustrate your answers graphically.

A m ; (a) 1, ( b ) -1

58. If f ( z ) is the function defined in Problem 66, does limf(ls1) exist? Explain. 2 40

59. Explain ezuctly what is meant when one writes:

(a) lirn

2--2 -

t + a (2:- 3 )'

60. Prove that

Explain why

=+O+

2x+6 - 2 -r4003X-2 3'

(c) lim

-00,

cosz ( b ) .liyOD5+p - 0. (a) lim s i n x does not exist, (b) lirn e'" s i n x does exist.

(a) lim 10-" = 0, t+

61.

(b) lirn (1- ellr) =

-00,

OD

0-b

m

r-b

m

62.

A m . (a) 2, ( b ) 1/6, (c) 2, (d)1/6, (e) does not exist 63. If

[XI

= largest integer S z, evaluate (a) =+1+ lim { x - [s]}, (b) lim (z- [z]}. =+a-

A m . (a) 0, ( b ) 1

= G. What generalizations of these do you suspect are true? Can you prove them?

64. If

65. If

lirn f ( z ) = A, prove that (a) r-, limro { f ( z ) } * = A',

r+ ' 0

lim f ( z ) = A and lirn g(z) = B, prove t h a t

r-e ro

2 4

(a) z+ limro { f ( z ) - g(z) 1 = A

fO

-B,

(b) lirn { af ( z ) r-,

Zo

(b) lirn

13 00

+ b g(z)} = aA + bB where a, b = any constants.

CHAP. 21

FUNCTIONS, LIMITS AND CONTINUITY

39

66. If the limits of f(x), g(x) and h(x) are A, B and C respectively, prove t h a t lim f(x) g(x) h(x) = A B C . Generalize these results. (a)2lim { f(x) g ( x ) h(x)} = A B C, (b) z+ 3 z0 fo

+

67.

+

+ +

Evaluate each of the following using the theorems on limits. (31:

+2 x2 8) (- 1k - 3) - z' - 6s + 3

Am.

(32 - 1)(2x + 3) Ji? (Sx - 3)(4x + 6)

4/21

(U)

( b ) 3/10

3x

(41

z4-w

(d) 1/32

p~~m-2 .

68. Evaluate

(Hint: Let 8

69. If lim f(x) = A and lim g(x) r 4 zo

70. Given

r4z0

Ans. 1/12

2'1).

0, prove directly that xlirn fO = AB 4 x 0 g(x)

x

x

1 - cos5

(b)

#

=

sin x lim -= 1, evaluate:

s+o

sin 3x (a)lim r 4 0

=B

+h

(c) lim

1 - cosx ~

r 3 O

(d)lirn (x - 3) csc PX z-b 8

A m . (a)3, ( b ) 0, (c) 1/2, ( d ) - l h ,

(f) lirn z+o

(e) 2/7,

6 x - sin 22 3 sin 4 2 cos ax - cos bx 2%

(e)

x*

1 - 2 cos x

+

(g)

(h) lirn

2'

(fl +(ba - a'),

(g)

ef- 1 = 1, prove t h a t 1; e-a.z - e-b+ ax- bx U = b - a; (b) lirn -= In T;, a,b > 0; (a)lim s40 x 2 4 0 x

0 4

-1,

+ cos 22

X1

3 sin T X

- sin 3nx XS

1

(h) 4 2

71. If lirn Z-0

72. Prove that lim f ( s )= 2

if and only if

CONTINUITY 73. Prove that f ( s ) = x ' - 32

+2

"310

(c) lim

230

tanh ax

= a. 1:

lim f(x) = + 4lirn f(x) = 1. 20-

S3Z0+

is continuous at z = 4.

74. Prove that f(x) = l / x is continuous

(a)at x = 2, (b) in 1 S x S 3.

75. Investigate the continuity of each of the following functions at the indicated points:

sin x , X Z O , f(0)=o; (4f ( 4= 7. ( b ) f(x) = x - 1x1 ; x = 0

x=o

(c)

m;

f(x) = xs - 8 x # 2, f(2) = 3; x = 2

A m . (a)discontinuous, (b) continuous, (c) continuous, (d)discontinuous

= greatest integer d z, investigate the continuity of f(z) = x (b) 1 5 x 5 2 .

76. If

[XI

- [XI

in the interval (a) 1 < x < 2,

77. Prove t h a t f(x) = xs is continuous in every finite interval. 78. If f ( x ) / g ( z ) and g(z) are continuous at

x = XO, prove that f(x) must be continuous at x = XO.

79. Prove that f ( z ) = (tan-'x)lx, f(0)= 1 is continuous at 80.

x = 0.

Prove that a polynomial is continuous in every finite interval.

81. If f ( x ) and g ( x ) are polynomials, prove that f(x)/g(x) is continuous at each point g(x0) z 0.

X=XO

for which

40

FUNCTIONS, LIMITS AND CONTINUITY

[CHAP. 2

82. Give the points of discontinuity of each of the following functions.

Am.

(U)z =

2,4, (b) none, (c) none, (d) z = 7a/6 f 2 m ~ ,l l d 6 -C 2ma, m = 0,1,2,

...

UNIFORM CONTINUITY 83. Prove that f ( z ) = z*is uniformly continuous in (a)0 < z < 2, ( b ) 0 5 z 5 2, (c) any finite interval. 84. Prove that f ( x ) = 5’ is not uniformly continuous in 0 < z
0, ( c ) not uniformly continuous in 0 < x < 1. 86. If f ( x ) and g(z) are uniformly continuous in the same interval, prove that (a) f ( z ) f g(z) and (b) f ( z ) g ( x ) are uniformly continuous in the interval. State and prove a n analogous theorem for ’ f(zVg(4.

MISCELLANEOUS PROBLEMS 87. Give an “c, 8” proof of the theorem of Problem 31. 88. (a)Prove that the equation tan z = x has a real positive root in each of the intervals ~ / < 2 z < 3d2, 3a/2 < z < 6a/2, 6a/2 < x < 7a/2, . (b) Illustrate the result in (a)graphically by constructing the graphs of y = tan z and y = z and locating their points of intersection. (c) Determine the value of the smallest positive root of t a n s = x. Ans. (c) 4.49 approximately

.. .

89.

Prove that the only real solution of sin z = z is z = 0.

+

1 = 0 has infinitely many real roots. 90. (a)Prove that cos z coshz (b) Prove that for large values of x the roots approximate those of cosx

= 0.

z’ sin (Us) = 0. 91. Prove that lim sins 92. Suppose f ( z ) is continuous at z = zo and assume f(z0) > 0. Prove that there exists an interval (20- h, zo h), where h > 0, in which f ( s ) > 0. (See Theorem 6, Page 26.) [Hint: Show that we can make If(z) -f(xo)l < +f(zo). Then show that f ( z ) 2 f(z0) - If(z)-f(zo)l > i f ( z 0 ) > 0.1

+

93. (a) Prove Theorem 10, Page 26, for the greatest lower bound m (see Problem 34). (b) Prove Theorem 9,

Page 26, and explain its relationship to Theorem 10.

Chapter

3

Sequences DEFINITION of a SEQUENCE A function of a positive integral variable, designated by f ( n ) o r Un, where n = 1,2,3, . . ., is called a sequence. Thus a sequence is a set of numbers U I , U S , U S , . . . in a definite order of arrangement (i.e. a correspondence with the natural numbers) and formed according to a definite rule. Each number in the sequence is called a term; Un is called the nth term. The sequence is called finite or infinite according as there are or are not a finite number of terms. The sequence U I , U ~ , U ~. ., . is also designated briefly by {Un}. Examples: 1. The set of numbers 2,7,12,17, . . ., 3 2 is a finite sequence; the nth term is given by U,, = f ( n ) = 2 S(n-1) = Sn-3, n = 1 , 2 , ..., 7.

+

2. The set of numbers 1, 1/3, 1/6, 1/7, U,,= 1/(2n - l ) , n = 1,2,3, . ..

.

...

is an infinite sequence with nth term

Unless otherwise specified, we shall consider infinite sequences only.

LIMIT of a SEQUENCE A number l is called the limit of an infinite sequence u ~ , u P , u . .~. , if for any positive number E we can find a positive number N depending on E such that I U n - l I < E for all integers n > N . In such case we write lim = 1.

+

Example: If U n = 3 l/n = (372 lirn un = 3. I+

+l)h,

n+oo

the sequence is 4,7/2,10/3,

m

.. .

and we can show that

If the limit of a sequence exists, the sequence is called convergent; otherwise it is called divergent. A sequence can converge to only one limit, i.e. if a limit exists i t is unique. See Problem 8. A more intuitive but unrigorous way of expressing this concept of limit is to say that a sequence U I , U ~ , U S ., . . has a limit 1 if the successive terms get “closer and closer” to 1. This is often used to provide a “guess” as to the value of the limit, after which the definition is applied to see if the guess is really correct. One should observe the similarities and differences between limits of functions and sequences. In defining Iim f(x) = I , the limit l is attained for all possible approaches to X+oo infinity. In defining lim f ( n ) = I, the limit Z need exist only along a certain approach to n+m infinity, namely along the positive integers. Other possibilities present themselves. For example, in some cases it may be important to consider the limit of f(x) as x approaches 00 (or in fact any number $0) along a sequence of rational numbers.

THEOREMS on LIMITS of SEQUENCES If lirn a n = A and lirn bn = B, then n+

n+m

1. lirn ( a n +

bn)

n+ m

2. lim n+ao

(an-bn)

50

= lim a n n+ca

+ Iim n-oo

bn

= A

+B

= lim a n - lim bn = A - B n+m

n+m

41

42

[CHAP. 3

SEQUENCES

3. lirn (an*b,) n-+m

= AB

= (lim U,,,)(lim b n ) n4oo

n-oo

If B = 0 and A # 0, lim 5 does not exist. n-oo

bn

If B = 0 and A = 0, lim 5 may or may not exist. n - + ab n

5. lim a,P = (lim n-, a

6.

n-,

an)'

lim pan = p?Ean

n 4 a

= AP,

for p = any real number if Ap exists.

= pA,

for p = any real number if p A exists.

03

INFINITY We write lim a, = -oo if for each positive number M we can find a positive numn-+a ber N (depending on M ) such that an > M for all n > N . Similarly we write lim an = --oo n-ta

if for each positive number M we can find a positive number N such that a n < -M for all n > N . It should be emphasized that -oo and --oo are not numbers and the sequences are not convergent. The terminology employed merely indicates that the sequences diverge in a certain manner.

BOUNDED, MONOTONIC SEQUENCES If U,, 5 M for n = 1,2,3, . . ., where M is a constant (independent of n), we say that the sequence {U,} is bounded above and M is called an upper bound. If U n Z m , the sequence is bounded below and m is called a lower bound. If m S U , 5 M the sequence is called bounded, often indicated by l U n l 5 P . Every convergent sequence is bounded, but the converse is not necessarily true. If U,+ 1 2 U n the sequence is called monotonic increasing; if U n +1 > U , it is called

strict1y increasing.

Similarly if Un+1 5 Un the sequence is called monotonic decreasing, while if it is strictly decvweasing.

U n +1

< Un

Examples: 1. The sequence l , l . l ,1.11,1.111, . . . is bounded and monotonic increasing. It is also strictly increasing. 2. The sequence 1, -1, 1, -1, 1, . . . is bounded but not monotonic increasing or decreasing. 3. The sequence -1, -1.5, -2, -2.5, -3, . . . is monotonic decreasing and not bounded. However, it is bounded above.

The following theorem is fundamental and is related to the Weierstrass-Bolzano theorem (Chapter l,Page 5) which is proved in Problem 23. Theorem. Every bounded monotonic (increasing or decreasing) sequence has a limit.

LEAST UPPER BOUND and GREATEST LOWER BOUND of a SEQUENCE A number M is called the least upper bound (1.u.b.) of the sequence {U,} if u n S M , for any E > 0. n = 1,2,3, . . . while a t least one term is greater than & - I A number 6 is called the greatest lower bound (g.1.b.) of the sequence {U,} if U , 2 nZ, n= 1,2,3, . . . while a t least one term is less than iii E for any C > 0.

+

CHAP. 31

SEQUENCES

43

Compare with the definition of 1.u.b. and g.1.b. for sets of numbers in general (see Page 5 ) .

LIMIT SUPERIOR, LIMIT INFERIOR A number t is called the limit superior, greatest limit or upper limit (lim sup or fi) of the sequence {U,} if infinitely many terms of the sequence are greater than !- E while only a finite number of terms are greater than z + t , where t is any positive number. A number _I is called the limit inferior, least limit or lower limit (lim inf or h) of the sequence {U,} if infinitely many terms of the sequence a r e less than _I + E while only a finite number of terms are less than l - E, where E is any positive number. These correspond to least and greatest limiting points of general sets of numbers. If infinitely many terms of {U,} exceed any positive number M , we define lim sup {U ,} = W . If infinitely many terms are less than - M , where M is any positive number, we define lim inf {U,} = --oo. If lirn U , = 00, we define lim sup {U,} = lim inf {U,} = W. n 4 00

If limu, = -00, we define lim sup {U,} = lim inf { u ? ~=} --oo. n-+m Although every bounded sequence is not necessarily convergent, i t always has a finite lim sup and lim inf. A sequence {U,} converges if and only if lim sup U , = lim inf Un is finite.

NESTED INTERVALS Consider a set of intervals [a,, b,], n = 1,2,3, . . ., where each interval is contained in the preceding one and lim (a,- b,) = 0. Such intervals a r e called nested intervals. n+m

We can prove that to every set of nested intervals there corresponds one and only one real number. This can be used to establish the Weierstrass-Bolzano theorem of Chap. 1. (See Problems 22 and 23.)

CAUCHY’S CONVERGENCE CRITERION Cauchy’s convergence criterion states that a sequence {U,} converges if and only if for each E > 0 we can find a number N such that lu,-u,l < t for all p , q > N . This criterion has the advantage that one need not know the limit l in order to demonstrate convergence. INFINITE SERIES Let U I , U Z , U ~ ,. . . be a given sequence.

Form a new sequence

S&,S3,.

..

where

S~=UI+U S 3~=, ~ 1 + ~ 2 + ~ 3 ..., , S n = u 1 + ~ 2 + ~ 3 + * . . + ~ n ,. . . where S,, called the nth partial sum, is the sum of the first n terms of the sequence {U,}. S1=u1,

The sequence SI,SZ,S~, . . . is symbolized by u l + u 2 + u 3 +

which is called an infinite series. If lim

new

Sn

a . .

=

2 U n

n=l

= S exists, the series is called convergent

and S is its sum, otherwise the series is called divergent. Further discussion of infinite series and other topics related to sequences is given in Chapter 11.

[CHAP. 3

SEQUENCES

44

Solved Problems SEQUENCES 1. Write the first five terms of each of the following sequences.

x@ -29 x= -x7 x - - Ans. l ! ' 3! ' 6 ! ' 7 ! ' 9 !

(-Un-l 2-l (2n - 1) !

Note that n ! = 1 * 2 * 3 * 4 . . . n . Thus l ! = l , 3! = 1 . 2 . 3 = 6, 6 ! = 1 . 2 . 3 . 4 . 6 = 120, etc. We define O! = 1.

Two students were asked to write an nth term for the sequence 1,16,81,256, . . . and to write the 5th term of the sequence. One student gave the nth term as u,,=n4. The other student, who did not recognize this simple law of formation, wrote Un = IonS- 35n2 50%- 24. Which student gave the correct 5th term? If U,,= n4, then u1= 1' = 1, us = 2' = 16, U S= 3' = 81, ur = 4' = 266 which agrees with the first

+

' = 626. four terms of the sequence. Hence the first student gave the 6th term as ua = 6 If U, = 10nS- 36nZ 6On - 24, then U I= 1, uz= 16, U S= 81, u4 = 266 which also agrees with the first four terms given. Hence the second student gave the 6th term as US= 601. Both students were correct. Merely giving a finite number of terms of a sequence does not define a unique nth term. In fact an infinite number of nth terms is poHible.

+

LIMIT of a SEQUENCE 3. A sequence has its nth term given by

Un

=

3n-1 (a)Write the Ist, Sth, loth, IOOth, m.

lOOOth, 10,000th and 100,000th terms of the sequence in decimal form. Make a guess as to the limit of this sequence as n+ CO. (b) Using the definition of limit verify that the guess in (a)is actually correct. (a)

n =1 n=6 n = 10 n = 100 n = 1000 n = 10,000 n = 100,000 ,22222. .66000. . . .64444. . . .73827. . . .74881. . .74988. . . ,74998.. . A good guess is that the limit is .76000.. . = %.Note that it is only for large enough values of n that a possible limit may become apparent.

..

.

( b ) We must shcw that for any given c > 0 (no matter how small) there is a number on e) such that Iun- $1 < c for all n > N.

Now

3n-1

I--al 4n+6

3

4'4:9+6)

1-1

-19

+

= 4(4n 6)
:,

e

4n+6

when

+

l9 4(4n 6)

~

> 19, 4e

n

tl


a(:-5)

Choosing N = $(19/4c - 6), we see that 1%< c for all n > N, so that Ulim un= 9 -bW and the proof is complete. Note that if e = .OOl (for example), N = &(19000/4- 6) = 1186t. This means that all terms of the sequence beyond the 1186th term differ from 9 in absolute value by less than .001.

CHAP. 31

4.

Prove that lim-

C

n+wnP

= 0 where C Z O and p > O are constants (independent of n).

We must show that for any Now on

5.

E),

1;1



>

or n

N such that Iclnp - 01 C

(F)

1l P

.

N =

Choosing

we see that Ic/n*l < e for all n > N, proving that lim (clnp) = 0 .

(v) e

for all n > N. 1fP

(depending

14 NI

2 lirn 1 + 2 . 1 o n - n+w5+3*10n - 3 '

Prove that

We must show that for any

n> N.

1 (6

Now

$46

45

SEQUENCES

+ 3.10")

+ 2.10" 2 + 3.10" - sl >

l/e,

3.10"

e

> 0 there

= 13(6

>

l1

is a number N such that

-7

+ 3.10")

7/3e-6,

1

when

c

3(6

9(7/3e- 6)

10"

+

.

+

lon

+

7 3.10n) C

or

n

>

- 21 < 3

e,

e

for all

i.e. when

loglo{&(7/3c- 6)) = N,

proving the existence of N and thus establishing the required result. Note that the above value of N is real only if 7/3e - 6 1 2.10" - 2 < e for all n > ~ . that 16 3.10"

+

>

0, i.e. 0 < e < 7/16. If

c

2 7/15, we see

+

6.

Explain exactly what is meant by the statements (a) lirn 32n--1 = 00, (b) lim (1- 272) = - W . n+ w

n-, w

(a) If for each positive number M we can find a positive number a , , > M for all n > N. then we write lirn a, = 0 0 . w

U+

In this case, 3*"-l > M when (2n- 1) log 3

> log M,

i.e. n

( b ) If for each positive number M we can find a positive number an < -M for all n > N, then we write lirn a,,= - W . In this case, 1 - 2n

< -M

N (depending on M ) such that

914

w

when 2n- 1

>M

or n

>&

N (depending on M ) such that

> J(M+ 1) = N .

It should be emphasized that the use of the notations w and -00 for limits does not in any way imply convergence of the given sequences, since 00 and --m are not numbers. Instead, these are notations used to describe that the sequences diverge in specific ways.

7. Prove that lirn x n = 0 if n+w

< 1.

Method 1: We can restrict ourselves to x Z 0 since if x = 0 the result is clearly true. Given e > 0, we must show that there exists N such that 1z"I < e for n > N. Now Ix"I = 1x1" C c when n log10 121 < log10 e. log10 e Dividing by log10 151, which is negative, yields n > -= N, proving the required result. log10 1x1 Method 2: Let 1x1 = l / ( l + p ) , where p > O . By Bernoulli's inequality (Prob. 31, Chap. 1), we have

=

121"

= 1/(1+ p)" < 1/(1+ np)
N.

THEOREMS on LIMITS of SEQUENCES 8. Prove that if limu, exists, it must be unique. 11.300

We must show that if lirn

n+ w

U,,= 11

and lim U,,= 18, then 11 = IS. U+

w

Thus

lim x" = 0.

n+ w

By hypothesis, given any lun-

Then 111

111

e>O

we can find

< Qe

when n

+

- 121 = 111 -Un

i.e. \Il - ZP1 is less than any positive

9.

[CHAP. 3

SEQUENCES

46

N such that

> N,

~ n ZsI -

d

IUn

-

+

unl

(11-


0, we can find N n > N. From inequality 2, Page 3, we have \(U,+

bn)

By hypothesis, given

- (A + B)I = I(u,-A) E

> 0 we can


0

c

ZI = IS.

+ bn) = A + B .

such that

+ (bn-B)[

+ bn) - (A + B)I
N S

i e

Then from (I),(2) and ( J ) , I(an+ b,)

111

(however small) and so must be zero. Thus

E

If lim a, = A and lirn b n = B, prove that lirn (a, n+w

when n > N

Qe

- (A +B)j
N

e

Ns. Thus the required result follows.

where N is chosen as the larger of NI and

10. Prove that a convergent sequence is bounded. Given

lim a,, = A, we must show that there exists a positive number 0 4

lunl

= lan-A+Al

Iun-Al

5

<e

But by hypothesis we can find N such that la,-Al

< It follows that [AI.

E

+

11. If lim b n = B n+m

P such that lu,l < P

a

for all n. Now

1ct.l

#



N, i.e.,

for all n > N

for all n if we choose P as the largest one of the numbers

UI,

a,. . ., UN,

N such that lb.1 > 31B1 for all n > N . IB - bnl + Ibn[.

0, prove there exists a number

+

Since B = R - b n bn, we have: (I) IBI S Now we can choose N so that IB - bn[ = Ibn - B [ hypothesis. lan[ or l b n l > Hence from (I), 1Bl
N, since lirn bn = B by n+ m

for all n > N.

12. If lim a, = A and lirn b n = B, prove that lim anbn= AB. n4aD

n4m

n+m

We have, using Problem 10, lanbn-

But since

AB[ = [ h ( b n - B )

+ B(un-A)I

5

lim a n = A and lim b, = B, given any

n+m

Hence from (I), la,b, - AB1 Nt. Thus the result is proved.

n+aa


O

+ [BI Ian-Al + (IBI + l ) l a , - A J

(11

we can find NI and NS such that

for all n > N, where

N is the larger of NI and

CHAP. 31

47

SEQUENCES

1 1 an = A 13. If l i m a , = A and l i m b n = B P O , prove (a) lim- = E , (b) nlim -mbn E' n-

n 4 m

00

n 4 m

(a) We must show that for any given

E

> 0, we

bn

can find N such that

By hypothesis, given any E > 0, we can find NI such that Ibn-BI < -&Bee for all n > NI. Also, since lim b n = B f 0, we can find Nz such that Ibnl > +IBI for all n > Nz (see Problem 11). n+w Then if N is the larger of NI and Nz, we can write (1) as

and the proof is complete. ( b ) From part (a)and Problem 12, we have

A B

This can also be proved directly (see Problem 41).

14. Evaluate each of the following, using theorems on limits. (a)

:!

3na- Sn 6na 2n - 6

+

=

lim

5

11-00

3-Sln

+ 2ln - 6In'

= u+m lim

{

+0 - - 3S+O+O -

}

n~+na+2n = (n l ) ( n a 1 )

+

+

-

lim n-m

3na+4n 2n-1

lim

n-,

6

{

+ l / n + 2/na + l/n)(l+ l/nz)

1 (1

1+0+0 = (1 0 ) (1 0 ) -

( d ) lim - -

3 -

+

+

I-

+

3 4ln 2/n - l / n a

Since the limits of the numerator and denominator are 3 and 0 respectively, the limit does not exist. 3na 3n Since 3ne+4n > =can be made larger than any positive number M by choosing ~

271-1

2n

2

n > N, we can write, if desired,

1

(g)

? m !

+ 2.10"

5 + 3-10;

= lim xdm

10-"

lim

n+m

+2 +3

6 10-"

3n2+4n -2n-1

- 2

- 3

BOUNDED MONOTONIC SEQUENCES 15. Prove that the sequence with nth term

W.

(Compare with Prob. 6.)

Un

=

2n-7

(a) is monotonic increasing,

(b) is bounded above, ( c ) is bounded below, ( d ) is bounded, (e) has a limit. (a)

{U,}

is monotonic increasing if

Un+l

I Un, n =1,2,3,

... .

Now

48

S E Q U E N C ES

+

2(n 1) - 7 2n - 7 3(n + 1) + 2

+

[CHAP. 3 2n-5 2n-7 1 3n+2 3n+5

if and only if

+

o r (272 - 5)(3n 2) 2 (2n - 7)(3n 5), 6n2- l l n - 10 2 6n2- l l n - 35, i.e. -10 h -35, which is true. Thus by reversal of steps in the inequalities, we see t h a t {U,} is monotonic increasing. Actually, since -10 > -35, the sequence is strictly increasing. By writing some terms of the sequence, we may guess t h a t a n upper bound is 2 (for example). To prove this we must show th at Un S 2. If (2n-7)/(3n+2) 5 2 then 2n-7 5 6 n + 4 or -4n < 11, which is true. Reversal of steps proves t h a t 2 is a n upper bound. Since this particular sequence is monotonic increasing, the first term -1 is a lower bound, i.e. Un 2 -1, n = 1,2,3, . . . . Any number less than -1 is also a lower bound. Since the sequence has a n upper and lower bound, i t is bounded. Thus for example we can write lUnl S 2 fo r all n. Since every bounded monotonic (increasing o r decreasing) sequence has a limit, the given sequence 2n-7 2-7/n 2 - h a s a limit. In fact, lim - - lim 3n+2 n e m 3 2/98 3' ~

+

16. A sequence {U,} is defined by the recursion formula Un+1 = (a) Prove that limu, exists. ( b ) Find the limit in (a). n 4 m

6, u1 = 1.

(a) The terms of the sequence a r e u1= 1, uz = = 3"*, u3 = = 31/4+1/4, .... n-1 The nth term is given by Un = 31/2+1/4+. . . + 1 / 2 as can be proved by mathematical induction (Chapter 1). Clearly, Un+l Z U,,. Then the sequence is monotone increasing. By Problem 14, Chapter 1, Un 5 3l = 3, i.e. Un is bounded above. Hence U n is bounded (since a lower bound is zero). Thus a limit exists, since the sequence is bounded and monotonic increasing.

6

U n + l = lim ( b ) Let x = required limit. Since nlirn -b m n-b m possibility, x = 0, is excluded since U n 2 1).

1/2+1/4+

Another method: lim 3

' . * t 1,2n-1

m

6, we have

1-1/2n - nlirn 3 -bm

-

x = fi and z = 3. (The other

lim (1-1/2") 3n-CO

= 31 =

17. Verify the validity of the entries in the following table. Sequence

. . . , 2 - (n-1)llO . . . 1, -1, 1, -1, . . . , ... a, -9, ), -&, . . ., (-l)m-l/(n+l), . . . .6, .66, .666, . . ., Q(l - l/lOn), .. . -1, +2, -3, +4, -6, . . ., (-l)"n, . . . 2, 1.9, 1.8, 1.7,

(-1)-1,

18. Prove that

Bounded

Monotonic Increasing

Monotonic Decreasing

Limit Exists

No

No

Yes

No

Yes

No

No

No

Yes

No

No

Yes (0) ~~

Yes

Yes

No

No

No

No

-

~

Yes

No

lim

n-, m

By the binomial theorem, if n is a positive integer (see Problem 96, Chapter l), n(n - l ) . . . ( n- n n(n - l)(n - 2) n(n - 1) xs (1+x)" = 1 nx -x22 ! n! 3! Letting x = l/n, 1 n(n-1) ... n ( n - l ) . . . ( n - n + l ) U,, = = 1 n-n n! 2! n2

+

+

+

+

+

+

I

+

+

+

+ 1)X" L

n"

(8)

CHAP. 31

49

SEQUENCES

Since each term beyond the first two terms in the last expression is an increasing function of n, it follows that the sequence U, is a monotonic increasing sequence. It is also clear that < l + l + -1 - +1- + . . . + L< 1 + 1 + 1- + -1+ . . . + 2 . - ‘ < 3 21 31 n! 2 2’ by Problem 14, Chapter 1. Thus U,,is bounded and monotonic increasing, and so has a limit which we denote by e. The value of e = 2.71828,.

..

19. Prove that lirn = e, where x+ 00 in any manner whatsoever (i.e. not necessarily along the positive integers, as in Problem 18). 2*0O

I f n = l a r g e s t i n t e g e r d x , t h e n n S x ~ n + l a n d (1 + L n +yl S

(1 + k > . S

+i)n

(1

Since and

it follows that

lirn (1

=400

+ $)‘= e,

LEAST UPPER BOUND, GREATEST LOWER BOUND, LIMIT SUPERIOR, LIMIT INFERIOR 20. Find the (a) I.u.b., (b) g.l.b., ( c ) lirn sup (E), and (d) lirn inf (]im) for the sequence 2, -2,1, -1,l, -l,l, -1, , , , . 1.u.b. = 2, since all terms are less than or equal to 2 while a t least one term (the 1st) is greater than 2 - o for any E > 0. g.1.b. = -2, since all terms are greater than or equal to -2 is less than - 2 + 0 for any E > O .

while at least one term (the 2nd)

lim sup or lim = 1, since infinitely many terms of the sequence are greater than 1 - E for any E > 0 (namely all 1’s in the sequence) while only a finite number of terms are greater than 1 E for any e > O (namely the 1st term).

+

+

lirn inf or lim = -1, since infinitely many terms of the sequence are less than -1 c for any E > 0 (namely all -1’s in the sequence) while only a finite number of terms are less than -1 - E for any E > 0 (namely the 2nd term).

I

Sequence

. . ., 2 - (n-l)/lO . .. 1, -1, 1, -1, . . ., (-1y-1, . . . 4, -9, *, -&, . . ., (-l)n-l/(n+l), .. . .6,.66,.666, . . ., 3(1- l / l O n ) , . . . -1, +2, -3, +4, -6, . . ., (-l).n, . . . 2, 1.9, 1.8, 1.7,

~

1.u.b.

g.1.b.

2

none

1 ~~

lim sup or lim -CO

-1 ~~

~

~~

lim inf or& -a

1

-1

~

4

-&

0

0

Q

6

Q

3

none

none

+a

-a

50

SEQUENCES

[CHAP. 3

NESTED INTERVALS 22. Prove that to every set of nested intervals [an,b,], n = 1 , 2 , 3 , . . ., there corresponds one and only one real number. a,+l 2

By definition of nested intervals,

an,

bn+l

S bn, n = 1,2,3, .

..

and lim ( a , - b,) = 0. ,-+a

Then al 5 a n S b,, 5 b ~ ,and the sequences { a n } and { b n } a r e bounded and respectively monotonic increasing and decreasing sequences and so converge to a and b.

To show t h a t a = b and thus prove the required result, we note t h a t

Now given any

e

> 0, we can find N such t h a t for all n > N lb-bnl

so t h a t from (Z), [ b - a\


1, prove

= P , ( c ) lim uf:= 1P

(1300

43.

If r

44.

If [r[> 1, prove that lim r" does not exist.

QO,

(2)" = 0.

I-+00

carefully explaining the significance of this statement.

*-boa

n+oo

45. Evaluate each of the following, using theorems on limits.

(a) lim 4 - 2n - 3n' 2d+n

BOUNDED MONOTONIC SEQUENCES 46. Prove that the sequence with nth term U, = G / ( n below, (c) is bounded above, (d) has a limit. 47.

If

Un

48. If un+l = 49.

1

= l+n

1 1 + 2+n +3+n +

d z ,

'*.

+-

12

1 +

n,

+ 1)

(a)is monotonic decreasing, (b) is bounded

prove that lirn U*

U,, exists

and lies between 0 and 1.

00

u 1 = l , prove that lim un = g ( l + f i ) . n-w

If % + I = +(U,,,+piu,,) where p > 0 and used to determine

> 0,

UI

prove that lirn n-+w

U,,

Show how this can. be = 6.

50. If U,,is monotonic increasing (or monotonic decreasing) prove that Sdn, where Sn = ui

is also monotonic increasing (or monotonic decreasing).

+ uo+ ... +

LEAST UPPER BOUND, GREATEST LOWER BOUND, LIMIT SUPERIOR, LIMIT INFERIOR 51. Find the l.u.b., g.l.b., lirn sup (G), lirn inf (b) for each sequence: (c) 1,-3,6, -7, . .,(-l),,-l (2n - l ) , . . . (a) -1, *, -9, .,(-1)"/(2n - 11,

a, .. ( b ) 3, -t,+, -9, .. ., (-l)m+l (n + l)/(n + 2), . . .

A m . (a) Q , - l , O , O

(b) 1,-l,l,-1

.

(d) 1,4,1,16,1,36,

( c ) none,none, +a, --oo

52. Prove that a bounded sequence {U,} is convergent if and only if

. . .,nl+(-I)", . . .

(d) none, 1,+a, 1

un = @ Un.

Un,

55

SE&U EN CES

CHAP. 31

INFINITE SERIES 53. Find the sum of the series

x(3).. 00

n-1

2 (-l)n-1/5n. n=1 00

54.

Evaluate

55.

Prove that

6

Am.

1.2 + 2.3 + 3.4 + 4.5 1

1

Am. 2

1

1

+

1 = 1. n(n + 1)

=

- . a

1 - 1 1 [Hint: n(n+l) - n --] n + l

56. Prove that multiplication of each term of an infinite series by a constant (not zero) does not affect the

convergence or divergence.

57.

Prove that the series 1 Then prove that

1 + ;Z1 + 51 + ... + ;+. ..

diverges.

= lirn

n-+ m

bn

n 4m

1 3 00

n 4 00

1 + -+ 2

1 34-

4-

1 -. n

- S n I > +, giving a contradiction with Cauchy's convergence criterion.]

15'2,

MISCELLANEOUS PROBLEMS 58. If a,, S U,,5 b n for all n > N, and lirn an = lim 59. If lim a,

[Hint: Let S n = 1

= I, prove that lirn

n-+ 00

Un

= 0, and e is independent of n, prove that lim (ancos ne

bn

n 4 m'

= I.

+

bn

sin ne) = 0. IS the

result true when e depends on n? 60. Let 61.

Un

= + { l + (-l)n}J n = 1,2,3,.

Prove that (a) lim U 4

62. If lirn

IUn+l/Unl

=

m

n*ln

lal

- 4 a0

63. If lal

< 1,


d x

+

+

+

60

[CHAP. 4

DERIVATIVES

DERIVATIVES of SPECIAL FUNCTIONS In the following we assume that U is a differentiable function of x; if U = x, d u l d x = 1. The inverse functions are defined according to the principal values given in Chapter 2. d 1. - ( C ) dx

2.

d dx

-Un

= 0 =

16.

d dx cosu

dx

d IS. -csc-’U = dx

du = - sinu dx

d

7

dx du

dz sinhu

du = coshudx

d 20. - coshu dx

du = sinhudx

= sech2u-d u dx

19.

du sec2U dx

5. ~ t a n u=

u = -- 1 dl + u 2 dz

d 17. dx

nun-’- du

d du 3. - s i n u = cos U dx dx

4.

d

dx cot-’u

d

rifu>l + ifu 0 , a#l = -U dx

d 9. -1Ogau dx d 10, &log,u

d = -1nu dx

u = -1dUdx

d du 11. z a U = a u I n a dx

12.

d eu 5

-

d 13. Z s i n - l u d dx d 15. -tan-’u dx

14.

-COS-’U

eu- d u

dx

= =

1

du

d w dx 1 du --

di=2

1 du = -1+u2 d x

dx

24.

d

du = - csch U coth U dx

dz each U

25. d sinh-’u

&x d

26.

dx cosh-’u

27.

&x tanh-l

28.

29. 30.

d

=

=

1

du

(TTiFZ

1 du -

d r i d~

u

d

dx d

dx d

dx

HIGHER ORDER DERIVATIVES If f ( x ) is differentiable in an interval, its derivative is given by ,”(x), y’ or d y l d x , where g = f ( x ) . If f’(x) is also differentiable in the interval, its derivative is denoted by f”(x), y” Or d& & - 9 dX2‘ Similarly the nth derivative of f ( x ) , if i t exists, is denoted

0 2,

where n is called the order of the derivative. Thus derivatives of by f(n)(x),gCn)or the first, second, third, . . . orders are given by f’(x), f”(x), f”’(x), Computation of higher order derivatives follows by repeated application of the differentiation rules given above.

. .. .

CHAP. 41

61

DERIVATIVES

MEAN VALUE THEOREMS 1. Rolle's theorem. If f(x) is continuous in [a,b] and differentiable in (a,b ) and if f(a)= f(b) = 0, then there exists a point 5 in (U, b ) such that f'(5) = 0. 2. The theorem of the mean. If f ( x ) is continuous in [a,b] and differentiable in (a$), then there exists a point 5 in (a,b) such that a O.

+ C.

Ans. d z a + 2 z + 6 - l n I x + 1 + d z 2 + 2 z + 6 1

57. Establish the validity of the method of integration by parts. 58. Evaluate

(a)

i= S z cos 32 dx, (b)

x' e-'" dx.

-

xa tan-'% dx =

59.

60. (a) If

U

( b ) -&e-lr(4z3

+ 65' + 62 + 3)+ c

+ Q In 2

Q

= f(z) and v = g ( z ) have continuous nth derivatives, prove that

called generalized integration b y parts. z4 sin z dx.

61. Show that

*

(b) What simplifications occur if

+

= O? Discuss.

(c) Use

a-2 - 8 '

zdx (z+l)a(zs+l)

+

- -

Prove that

U(")

A m . (c) 'P - 1212 48

[Hint: Use partial fractions, i.e. assume (z A , B, c, D.1 62.

A m . ( a ) -219,

3r

,

&Fi

l){z2

+ 1)

- - A - (z+ 1)s

+

z+l Fc z +F Di +

and find

a>l.

NUMERICAL METHODS for EVALUATING DEFINITE INTEGRALS 63. Evaluate

i& '

approximately, using ( a ) the trapezoidal rule, (b) Simpson's rule, taking n= 4.

Compare with the exact value, In 2 = 0.6931. 64. Using ( a ) the trapezoidal rule, ( 6 ) Simpson's rule evaluate

IT''

sin' x d z by obtaining the values of

sins z at z = O o , loo, . . .,90° and compare with the exact value u/4. 65.

Prove the (a)rectangular rule, (b) trapezoidal rule, i.e. (16) and (17)of Page 86.

66. Prove Simpson's rule.

67. Evaluate to 3 decimal places using numerical integration:

Ans. ( a ) 0.322, ( b ) 1.106

(a)

l* m, dz

(b)

J'cosh x2 d z . 0

APPLICATIONS 68. Find the ( a ) area and ( b ) moment of inertia about the y axis of the region in the zy plane bounded A m . (a)2, ( b ) U' - 4 by y = sin x, 0 S x 5 a and the x axis, assuming unit density. 69. Find the moment of inertia about the z axis of the region bounded by y = x'

is proportional to the distance from the x axis.

and y = x , if the density A m . +M,where M = mass of the region.

70. Show that the arc length of the catenary y = cosh z from

x = 0 to z = In 2 is

8.

CHAP. 51 71.

99

INTEGRALS

Show that the length of one arch of the cycloid s = a(e - sin e), y = a(1 - cos e), (0 f

72. Prove that the area bounded by the ellipse xa/as

+ ya/bs =

8

S 27) is 8a.

1 is rab.

73. Find the volume of the region obtained by revolving the curve g = sin x, 0 S x 5 a, about the x axis.

Am. a'l2

74. Prove that the centroid of the region bounded by y = @%?, -a 5 x d a and the x axis is located

at (0,4a/3~).

75. (a)If p = f(+) is the equation of a curve in polar coordinates, show that the area bounded by this

curve and the lines and # = # a is lemnkcate p' = ascos2#. Ans. ( b ) as

&:

p'd+.

(b) Find the area bounded by one loop of the

i:dps + (dp/d+)'

76. (a)Prove that the arc length of the curve in Problem 76(a) is A m . ( b ) 8a length of arc of the ccwdioid p = a(l - cos#).

d+.

( b ) Find the

MISCELLANEOUS PROBLEMS 77.

Establish the theorem of the mean for derivatives from the first mean value theorem for integrals. [Hint: Let f ( x ) = F'(x) in (4),Page 81.1

78.

and give a geometric interpretation of the results. [These limits, denoted usually by

dx

respectively, a r e called

improper integrals of the second kind (see Problem 33) since the integrands are not bounded in the range of integration. For further discussion of improper integrals, see Chapter 12.1

Jl

79. Prove that

(a)

lM

x5 e-' d x = 41 = 24,

x(2 - x )

P -

80.

Am.

(U)

2a 3J3

81. Evaluate

82.

lim

(b) 3

( c ) does not exist

ex'/=

- eal4 + Jzu" erin * d t . 1

z-+u/2

+ cosex

Am. e l 2 ~

Prove: (a) ZS,rJ ( t s + t + l ) d t = 3 1 ; * + s 6 - 2 x a + 3 x s - 2 x ,

83. Prove the result (IS) on Page 83.

84. 85.

Prove that

(a) xTd-1

Explain the fallacy:

+ sin x d x

I =

J:ldm x -

Hence I = 0. But I = tan-'(l) 86. Prove that 87.

cos P X

dx

= 4,

S

- tan'l(-l) 1 1 -tan-l%. 4

(b)

z i

( b ) d " " c o s f d t = 22 cosx'- cosx'.

17'9sin dx cos +

= -I,

= n/4 - (--a/4)

= filn



~-ox'+Y'

= lim re0

'X

= 1 and lim y+o

{ q} =

-1

lim OX'+

a r e not equal, lim f(s,y) r-0 U-0

cannot exist. Hence f(x,y) cannot be continuous at (0,O).

PARTIAL DERIVATIVES 8. If f(x,y) = 2x2-xz/+y2, find (a) afldx, and (b) af/az/ at ( x o , ~ o )directly from the definition. (a)

=

fr(z0, go)

~ l < r o , ~ o ~

m e + h, yo) - fb, go) = hlim -0

h

= lim

h

h-+O

= lim h-0

= lim

+ + ( v o + k)'] - [ 2 d - xovo + &]

[ 2 d - ~o(yo k)

k-0

=

- kxo + 2kyo + k' lim

k-0

k

k = lim k-+O

(-zo

+ 2y0+ k)

=

- zo + 2yo

Since the limits exist for all points (20,yo), we can write f&, which are themselves functions of x and y.

fu = - z + 2 y

y)

= f s = 4s - y, f&,

y)

=

Note that f o m a l l y f=(zo,yO) is obtained from f(x,y) by differentiating with respect to z, keeping y constant and then putting x = xo, y = yo. Similarly fU(zo,yo) is obtained by differentiating f with respect to y, keeping x constant. This procedure, while often lucrative in practice, need not always yield correct results (see Problem 9). It will work if the partial derivativee a r e continuous.

PARTIAL DERIVATIVES

CHAP.61

113

9.

exist but that (b) f ( x ,y) is discontinuous at (0'0).

(b)

Let x

+0

and y

+

0 along the line y = mx in the xy plane. Then

f ( ~ y) , =

!%

mxg xg

+

-

mgxg -

Y e 0

so that the limit depends on the approach and therefore does not exist. Hence f ( ~ , y )is

i+m9 not continuous at (0,O). Note that unlike the situation for functions of one variable, the existence of the first partial derivatives at a point does not imply continuity at the point. ya - x 'y - -xa- xy' and f r ( O J O ) J f # ( O , O) Note also that if (x,y) f (O,O), fr = cannot be computed from them by merely letting x = O and y=O. Problem S(b), Chapter 4.

See remark at the end of

10.

Note that 9, = in this case. This is because the second partial derivatives exist and are continuous for all (x,y) in a region q. When this is not true we may have 9- f $t (see Problem 43, for example).

+

11. Show that U($, y, X ) = (x2+ y2 x2)-lI2 satisfies Laplace's partial differential equation

a2u

a2u

=+a1/2+=

a2u - 0. -

We assume here that (5,y, z ) # (0, 0,O). Then au = -&(%a + y9 + z g ) - s / S 22 = - ~ ( 5 9 + y'+ ax

zg)-W'

114

[CHAP. 6

PARTIAL DERIVATIVES

a22

12. If x = x2

find - at (1'1). a 3 ay

X'

+ Y')(3X') - ( W 2 4 -

+

2 3

(x2 yS)*

-

2'

1 2 = 1

at (1, 1).

The result can be written xzlf(l, 1) = 1. Note: In this calculation we are using the fact that xty is continuous at (1,l) (see remark at the end of Problem 9).

and if 13. If f ( x , y ) is defined in a region of q,prove that f m = fus a t this point. Let

(XO,YO)

and

fux

exist and are continuous a t a point

be the point of T . Consider

G Define

fxu

(1)

+

+

= f ( x o + h, YO k) - f ( ~ o YO , k) - f ( x o

+ h, YO) + f ( z o , 2 / 0 )

+(x,Y) = f ( s+ h, Y) - f ( z , ~ ) (2) $.(z,Y)= f ( z ,Y + k) -

f(x,v)

Applying the theorem of the mean for functions of one variable (see Page 61) to (3) and (4), we have elk) - fy(xO,YO elk)} 0 < 8, < 1 ( 5 ) G = k+,(xo, go elk) = k{fy(xo h, YO (6) G = h$,(zo e,h, yo) = h { f z ( z o e,h,yo k) - fZ(xo e,h, yo)} 0 < 8, < 1

+

+

+ +

+

+

+

+

Applying the theorem of the mean again to ( 5 ) and ( 6 ) , we have (7) (8)

From (7) and (8) we have

Letting h

+

0 and k

+

f v z ( s o e,h, po

(9) --*

O<e,

From (.!?), u z F r = 1 and urFy

But by hypothesis

A

Y)

+ uu = 0

or

= g { F ( u , y ) , Y>

(2)

(5) F ,

=

-U&.

uZvy- u,v,

=

Using this, (4)becomes

= 0 identically,

so t h a t ( 6 ) becomes

d v = v Z F udu. This means essentially that referring to (2), av/dg = 0 which means that v is not dependent on y but depends only on U, i.e. v is a function of U , which is the same as saying that the functional relation +(U, U ) = 0 exists.

'+'

36. (a) If U = - and v = t a n - ' x + 1-xw (b) Are U and v functionally related? If so, find the relationship.

( b ) By Problem 36, since the Jacobian is identically zero in a region, there must be a functional relationship between U and U . This is seen to be tan v = U , i.e. +(u,v) = U - tan v = 0. We can show this directly by solving for x (say) in one of the equations and then substituting in the t a n - l y we find t a n - l x = v - tan-'y and so other. Thus, for example, from v = tan-'$

+

Then substituting this in

U

+

= ( x y)/(l - x y ) and simplifying, we find

U

= tan U .

CHAP. 61

123

PARTIAL DERIVATIVES

+

+

37. (a) If x = U - v w, y = u2- v2 - w 2 and x = u3 v, evaluate the Jacobian a($, Y,2) a(% v,w) and ( b ) explain the significance of the non-vanishing of this Jacobian..

( b ) The given equations can be solved simultaneously for the Jacobian is not zero in 54(.

U,v , w

in terms of x, y, z in a region 54( if

TRANSFORMATIONS, CURVILINEAR COORDINATES 38. A region T in the xy plane is bounded by x y = 6, x - y = 2 and y = 0. (a) Determine the region T’in the uv plane into which T is mapped under the transforma-

+

+

tion x = U v, y = U - v. ( b ) Compute a(xyy) ( c ) Compare the result of ( b ) with a(u,v) ’ the ratio of the areas of % and T’. ( a ) The region “Ip shown shaded in Fig. 6-6(a) below is a triangle bounded by the lines x + y = 6, x - y = 2 and y = 0 which for distinguishing purposes are shown dotted, dashed and heavy respectively.

Fig. 6-6

+

+ +

Under the given transformation the line x y = 6 is transformed into (U v ) i.e. 2 u = 6 or U = 3, which is a line (shown dotted) in the uv plane of Fig. 6-6(b) above.

(U

-v) = 6,

+

Similarly, x -y = 2 becomes (U v ) - (U- v ) = 2 or v = 1, which is a line (shown dashed) in the uv plane. In like manner, y = 0 becomes U - v = 0 or U = v , which is a line shown heavy in the uv plane. Then the required region is bounded by u = 3 , v = 1 and u = v , and is shown shaded in Fig. 6-6(b).

(c)

The area of triangular region % is 4, whereas the area of triangular region 54(’ is 2. Hence the ratio is 4 / 2 = 2, agreeing with the value of the Jacobian in ( b ) . Since the Jacobian is constant in this case, the areas of any regions % in the xy plane are twice the areas of corresponding mapped regions %’ in the uv plane.

124

PARTIAL DERIVATIVES

[CHAP. 6

+

+ +

39. A region % in the xy plane is bounded by x2 y2 = u2, x2 y2 = b2, x = 0 and y = 0, where 0 < U < b. (U) Determine the region %’ into which % is mapped under the transformation x = p cos 4, y = p sin 4, where p > 0, 0 5 < 2 ~ . ( b ) Discuss what

happens when u=O.

4% 4) (c) Compute - (d) Compute -

a(x, Y)

a(f 9 4)

Fig. 6-7 (a) The region l( [shaded in Fig. 6-7(a) above] is bounded by x= 0 (dotted), g = 0 (dotted and dashed), x 2 + y 2 = a2 (dashed), x 2 + y 2 = b2 (heavy). Under the given transformation, x 2 ya = aa and zz ya = b2 become p’ = a’ and p’ = b2 or p = a and p = b respectively. Also, x = O , a S y 5 b becomes $=.rr/2, a s p 5 b; y=O, a 5 x 5 b becomes + = 0, a Ip 5 b. The required region l(’ is shown shaded in Fig. 6-7(b) above. Another method: Using the fact that p is the distance from the origin 0 of the xy plane and 9 is the angle measured from the positive x axis, it is clear that the required region is given by a S p Ib, 0 S 9 5 ~ / 2as indicated in Fig. 6-7(b).

+

+

( b ) If a = 0, the region 9( becomes one-fourth of a circular region of radius b (bounded by 3 sides) while l(’ remains a rectangle. The reason for this is that the point x = 0, y = 0 is mapped into p = 0, + = an indeterminate and the transformation is not one to one at this point which is sometimes called a singular point.

=

p(cos2+

+ sin2+)

=

P

(d) From Problem 45(b) we have, letting U = p, v = 9,

This can also be obtained by direct differentiation. Note that from the Jacobians of these transformations it is clear why p = 0 (i.e. x = 0, y = 0) is a singular point.

MEAN VALUE THEOREMS, TAYLOR’S THEOREM 40. Prove the first mean value theorem for functions of two variables. Let F ( t ) = f ( x o

+ ht, yo + kt). By the mean value theorem for functions of one variable, F(1) - F(0) = F’(e)

0

<e O, then AIA is a unit vector, denoted by a, having the same direction as A. Then A = Aa.

RECTANGULAR UNIT VECTORS The rectangular unit vectors i, j and k are unit vectors having the direction of the positive x, y and z axes of a rectangular coordinate system [see Fig. 7-51. We use right-handed rectangular coordinate systems unless otherwise specified. Such systems derive their name from the fact that a right threaded screw rotated through 90" from Ox to Oy will advance in the positive x direction. In general,

Fig. 7-5

VECTORS

136

[CHAP. 7

three vectors A, B and C which have coincident initial points and are not coplanar are said to form a right-handed system o r dextral system if a right threaded screw rotated through an angle less than 180" from A to B will advance in the direction C [see Fig. 7-6 below].

COMPONENTS of a VECTOR Any vector A in 3 dimensions can be represented with initial point at the origin 0 of a rectangular coordinate system [see Fig. 7-7 above]. Let (Al,Az,A3) be the rectangular coordinates of the terminal point of vector A with initial point a t 0. The vectors Ali, A2j and Ask are called the rectangular component vectors, or simply component vectors, of A in the x, y and x directions respectively. AI, A2 and A3 are called the rectangular components, or simply components, of A in the x, y and x directions respectively. The sum or resultant of Ali, A2 j and Ask is the vector A, so that we can write A

= Ali

+ A2j + Ask

(1)

The magnitude of A is

A

= IAI

= dA:+A,2+Ai

In particular, the position vector o r radius vector r from 0 to the point (x,y,x) is written r = xi yj xk (8)

+

+

and has magnitude r = Irl = d x 2 + y 2 + z 2 .

DOT or SCALAR PRODUCT The dot or scalar product of two vectors A and B, denoted by A . B (read A dot B) is defined as the product of the magnitudes of A and B and the cosine of the angle between them. In symbols, A ~ B = ABCOS~, Note that A B is a scalar and not a vector.

oses

The following laws are valid:

1. A * B = B * A 2. A * ( B + C ) = A * B + A * C 3. m(A*B) = (mA) B = A (mB) 4. i * i = j e j = k * k = 1, i * j =

Commutative Law for Dot Products Distributive Law where m is a scalar. = (AB)m, j*k = k*i= 0

(4)

CHAP. 71

5. If A = Ali

137

VECTORS

+ A2j + A s k

+ B2j + Bsk, then AiBi + AzB2 + A3B3 A2 = A; + A: + A: B2 = B; + B,2 +

and B = Bli

A*B = A*A = B*B =

6. If A * B = 0 and A and B are not null vectors, then A and B are perpendicular. CROSS or VECTOR PRODUCT The cross or vector product of A and B is a vector C = A x B (read A cross B). The magnitude of A x B is defined as the product of the magnitudes of A and B and the sine of the angle between them. The direction of the vector C = A x B is perpendicular to the plane of A and B and such that A, B and C form a right-handed system. In symbols, A x B = ABsinh, OS8ST (5) where U is a unit vector indicating the direction of A x B. If A = B or if A is parallel to B, then sin8 = 0 and we define A x B = 0. The following laws are valid: 1. A X B = - B X A (Commutative Law for Cross Products Fails) 2. A X ( B + C ) = A x B + A X C Distributive Law 3. m(A X B) = (mA) x B = A X (mB) = (A x B)m, where m is a scalar. 4. i X i = j x j = k x k = O , ixj=k, jxk=i, k x i = j 5. If A = Ali A2j A3k and B = Bli B2j Bsk, then

+

+

AXB

=

1

+

+

j A1 A2 A3 I:

B2

z 3

1

6. IAxBI = the area of a parallelogram with sides A and B. 7. If A x B = 0 and A and B are not null vectors, then A and B are parallel.

TRIPLE PRODUCTS Dot and cross multiplication of three vectors A, B and C may produce meaningful products of the form (A*B)C, A *(B x C) and A x (B x C). The following laws are valid: 1. (A*B)C # A(B*C) in general 2. A * (BX C) = B . (CXA) = C * ( A x B ) = volume of a parallelepiped having A, B, and C as edges, or the negative of this volume according as A, B and C do or do not form a right-handed system. If A = Ali A2 j +Ask, B = Bli B2 j B3k and C = Cli+CPj+Cak, then

+

A*(BxC) =

+

+

A1 A2 A3

3. A x (B X C) # ( A x B) X C (Associative Law for Cross Products Fails) 4. A X (BXC) = (A*C)B- (A*B)C (A X B) X C = (A*C)B- (B*C)A

The product A (B x C) is sometimes called the scalar triple product or box product and may be denoted by [ABC]. The product A x (B x C) is called the vector triple product.

138

VECTORS

[CHAP. 7

In A (B x C) parentheses are sometimes omitted and we write A. B x C. However, parentheses must be used in A x (B x C) (see Problem 29). Note that A (B x C) = (A x B) C. This is often expressed by stating that in a scalar triple product the dot and the cross can be interchanged without affecting the result (see Problem 26).

AXIOMATIC APPROACH to VECTOR ANALYSIS From the above remarks it is seen that a vector 1: = xi yj xk is determined when its 3 components (x,y,x) relative to some coordinate system are known. In adopting an axiomatic approach i t is thus quite natural for us to make the following

+ +

Definition. A 3 dimensional vector is an ordered tripZet of real numbers (A1,A2,A3). With this as starting point we can define equality, vector addition and subtraction, etc. Thus if A = ( A I , A ~ , A s )and B = (BI, B2,B3), we define 1. A = B if and only if A I = & , AB=&, A s = & 2. A + B = ( A I + B I , A ~ + B ~ , A s + B ~ ) 3. A - B = (A1 -B1, Az-Bz, As-Bs) 4. 0 = (O,O,O) 5. mA = m(Al,At,A3) = (mA1,mAs,mA3) 6. A * B = AiBi A2B2 A3B3 7. Length or magnitude of A = IAI = d F A = d A : + A ; + A ; From these we obtain other properties of vectors, such as A B = B A, A (A B) C, A (B C) = A . B A . C, etc. By defining the unit vectors

+

+ +

+

+

+

+

i = (l,O,O),

j = (O,l,O),

we can then show that A

= AIi

k = (0,0,1)

+

+ (B + C) = (7)

+ A2j + Ask

In like manner we can define A x B = (Ad33 - A3B2, AlB3, A& - A2B1). After this axiomatic approach has been developed we can interpret the results geometrically or physically. For example, we can show that A *B = AB cos 8, IA x BI = AB sine, etc. In the above we have considered three dimensional vectors. It is easy to extend the idea of a vector to higher dimensions. For example, a f o u r dimensional vector is defined as an ordered quudrupZe (AI,AB,As, Al).

VECTOR FUNCTIONS If corresponding to each value of a scalar u we associate a vector A, then A is called a function of u denoted by A@). In three dimensions we can write A(u) = Al(u)i 4 u ) j + &(u)k. The function concept is easily extended. Thus if to each point (x,y,x) there corresponds a vector A, then A is a function of (x,y, x), indicated by A(x, y, x ) = AI(x,y, x)i A2 (x,y, x)j + A3 (x,y, x)k. We sometimes say that a vector function A(x,y,x) defines a vector field since i t associates a vector with each point of a region. Similarly +(x,y,z) defines a scalar field since it associates a scalar with each point of a region.

+

+

139

VECTORS

CHAP. '71

LIMITS, CONTINUITY and DERIVATIVES of VECTOR FUNCTIONS Limits, continuity and derivatives of vector functions follow. rules similar to those for scalar functions already considered. The following statements show the analogy which exists. 1. The vector function A(u) is said to be continuous a t uo i f given any positive number E, we can find some positive number 6 such that ] A @ )- A ( U Ol

find

=

3 sin as

- 2 sin as a

dx (2 - cos x)2'

(See Problem 62, Chapter 5.)

APPLICATIONS O F PARTIAL DERIVATIVES

CHAP. 81

171

INTEGRATION UNDER the INTEGRAL SIGN 18. Prove the result (18),Page 163, for integration under the integral sign. Consider

(1)

+(a)

=

I:{la

1

f ( x ,a ) d a d x

By Leibnitz's rule,

Then by integration,

(2)

$(a)

=

J

a

@(a)da

+c

xa{J-r

Since +(a)= 0 from (I),we have c = 0 in ( 8 ) . Thus from (I) and (2)with c = 0, we find

l:{l

f(s, a) d x } d z

Putting a = b, the required result follows.

=

From Problem 62, Chapter 5,

f ( z a) , dz}da

ff>1.

Integrating the left side with respect to a from a to b yields

Integrating the right side with respect to a from a to b yields

and the required result follows.

MAXIMA and MINIMA 20. Prove that a necessary condition for f(x,y) to have a relative extremum (maximum or = 0, fv(x0,yo) = 0. minimum) at (XO,YO)is that fX(x~,y~) Xf f ( z 0 , yo) is to be an extreme value for f ( x , y), then it must be an extreme value for both f ( s yo) , and ~ ( x oy)., But a necessary condition that these have extreme values at x = xo and y = yo respectively is f.(xo, yo) = 0 , fU(xo,yo) = 0 (using results for functions of one variable).

21. Let f(x,y) be continuous and have continuous partial derivatives, of order two at least, in some region including the point (z0,yo). Prove that a sufficient condition that , > 0 and f ( x 0 , yo) is a relative maximum is that A = f&~,y~) fvr(xo,yo) - ~ & ( x oYO) fxx(~o,z/o)< 0.

By Taylor's theorem of the mean (see Page log), using f*(xo,go) = 0, fi ( X O , go) = 0, we have

+ h, YO+ k ) - ~ ( x oVO),

= i(h'ffuc + 2hkfa3 + k*fyy) where the second derivatives on the right are evaluated at xo + eh, vo ek where f(xo

pleting the square on the right of (1) we find

+

(1)

0 < e C 1. On com-

172

[CHAP. 8

APPLICATIONS OF PARTIAL DERIVATIVES

Now by hypothesis there is a neighborhood of (ZO,go) such that f,, < 0. Also the sum of the terms in braces must be positive, since f,,f,, - f & > 0 by hypothesis. Thus i t follows t h a t

+

f ( ~ o h, yo

+ k)

5 f ( x o ,yo)

for all sufficiently small h and k. But this states that f(x0,yo) is a relative maximum. Similarly we can establish sufficient conditions for a relative minimum.

+

+

22. Find the relative maxima and minima of f ( x ,y) = x3 y3 - 3x - 12y 20. f , = 3 x 2 - 3 = 0 when x = '1, f y = 3y2- 12 = 0 when y = k 2 . Then critical points are P(1,2), Q(-1,2), R(1, - 2 ) s

s(-1,-2).

f , = 6x, f , , = 6y, f q = 0. Then A =

fzxfvv

- f & = 36xy.

A t P(l,2), A > 0 and f,, (or f,,)

> 0; hence P is a relative minimum point. < 0 and Q is neither a relative maximum or minimum point. R(1, -2), A < 0 and R is neither a relative maximum or minimum point. S(-1, -2), A > 0 and (or f,,)< 0 so S is a relative maximum point.

At Q(-1, 2), A At At

fzZ

Thus the relative minimum value of f(x,y) occurring at P is 2, while the relative maximum value occurring a t S is 38. Points Q and R are saddle points.

23. A rectangular box, open at the top, is to have a volume of 32 cubic feet. What must be the dimensions so that the total surface is a minimum? If x, y and x are the edges (see Fig. 8-4), then (I) (2)

Volume of box = V = zyx = 32 Surface area of box = S = xy 2yx

+

+ 2x2

or, since x = 32/xy from ( I ) ,

s a sax -

y

64

64

X

Y

= xy+-+-

64

-? = 0 when

Fig. 8-4

(3) x2y = 64,

-

ay

64 = 0 when (4) xyz= 64 = x --

Y2

Dividing equations (3) and ( 4 ) , we find y = x so that x3 = 64 or x = y = 4 and x = 2. 128 For x = y = 4, A = S,,S,, - Sty = -1 > 0 and S,, = x3 > 0. Hence it fol-

(T)c+)

lows that the dimensions 4 f t X 4 f t X 2 f t give the minimum surface.

LAGRANGE MULTIPLIERS for MAXIMA and MINIMA 24. Consider F(x, y, x ) subject to the constraint condition G(x,y, x ) = 0. Prove that a necessary condition that F ( x ,y, x ) have an extreme value is that F , G, - FUG, = 0. Since G ( s ,y,x) = 0 , we can consider x a s a function of x and y, say x = f ( x , y). A necessary condition that F[x,y, f(x,y)] have an extreme value is t h a t the partial derivatives with respect t o x and y be zero. This gives ( I ) F,

Since G ( x , y , x ) = 0, we also have (3) G,

+ Fzxz

= 0

+ Gzx, =

0

(2) F ,

+ FA,

= 0

( 4 ) Gg

+ Gzx,

= 0

From ( 1 ) and (3)we have ( 5 ) F,G,-F,G, = 0, and from (2)and (4) we have ( 6 ) F y G z - F z G u = 0 . Then from ( 5 ) and ( 6 ) we find F+G, - F y G , = 0. The above results hold only if F , # 0, G, # 0.

APPLICATIONS OF PARTIAL DERIVATIVES

CHAP. 81

173

25. Referring to the preceding problem, show that the stated condition is equivalent to the conditions +%=O, + r = O where = F+XG and x is a constant. If #,=O, Fa+AGa = 0. If # # = O , F,,+AG, = 0. Elimination of A between these equations

+

yields F,G,

- FUGz= 0.

The multiplier A is the Lagrange multiplier. If desired we can consider equivalently where +a = 0, ## = 0.

26. Find the shortest distance from the origin to the hyperbola x2

+

= AF

+G

+ 8x16 + 7g2 = 225,

x = 0. We must find the minimum value of x a + y a (the square of the distance from the origin to any point in the xy plane) subject to the constraint z L + 8 x y + 7y' = 226. According to the method of Lagrange multipliers, we consider # = xa 8xy + 7ya- 226 x(za+ya).

+

Then

$=

= 22

= 8x

+ 8y + 2x2 + 14y + 2Ay

From (1) and (a), since (x,y) # (O,O),

I

1

7:A

= 0 = 0

or or

+

+ 4y + (h+7)y

(1)

( A + 1)x

(2)

42

= 0 = 0

we must have i.e.

= 0,

Case 1: A = 1. From (1) or (a),x = - 2 y f o r which no real solution exists.

Aa+8A-9

= 0

or

A = 1,-9

and substitution in x ' + 8 x y + 7y' = 226 yields -6y2=226,

+

+

Case 2: X = -9. From (2) or (a),y = 2 s and substitution in xa 8xy 7ys = 226 yields 452' = 226. Then x' = 6 , yz = 42' = 20 and so xs yL = 26. Thus the required shortest distance is 6 6 = 6.

+

27. (a) Find the maximum and minimum values of x 2 + g 2 + z 2 subject to the constraint conditions x2/4 g2/5 x2/25 = 1 and x = x y. (b) Give a geometric interpretation of the result in (a). 2' (a) We must find the extrema of F = z'+y'+ z' subject to the constraint conditions #, = a +

+

$ +&-1

+

+

= 0 and #% = x + y - z

= 0. In this case w e use two Lagrange multipliers A,, A, and

consider the function

G = F

+ A ~ $ , + A,$,

= x9

+ ya + zs + h l ( $ + vsz + ~za- l

) + (x + y - - z ) A,

Taking the partial derivatives of G with respect to x , y , x and setting them equal to zero, we find

G= = 2~

x +7 + A, A1

0,

Gv = 2y

2x1 + A, +5 2/

= 0, G, = 22

+--2x126

2

A,

= 0

(1)

Solving these equations for x, y, z, we find

From the second constraint condition, x + y - z = 0, we obtain on division by A,, assumed different from zero (this is justified since otherwise we would have x = 0, y = 0, z = 0 which would not satisfy the first constraint condition), the result

+ 4)(X, + 6)(A, + 26) and simplifying yields 17X: + 246h, + 760 = 0 or (A, + 10)(17A1+ 76)

Multiplying both sides by 2(X,

from which A. = -10 or -76/17.

= 0

(CHAP. 8

APPLICATIONS O F PARTIAL DERIVATIVES

174

Case 1: xl=-lO. = &AL, z = #A,, Substituting in the first constraint condition, x2/4 From (Z), x = 2 / 2 5 = 1, yields hi = 180/19 or A, = 2 6 m 9 . This gives the two critical points

ix,,

( Z r n 9 , 3 W 9 , SJmiiT),

The value of x'

+ y2/5 +

(-2rn, - 3 m , - 5 m )

+ pL + 2 corresponding to these critical points

is (20

+ 45 + 125)/19

= 10.

Case 2: A, = -75/17. Substituting in the first constraint condition, x'/4 From ( 2 ) , x =?A,, y = z =#A2. y2/5 z*/26 = 1, yields A, = 2 1 4 0 / ( 1 7 m ) which gives the critical points

-yA2,

+

( 4 0 / a 6 , - 3 5 / m , 6/m),( - 4 O / m , 35/*,

+ +

The value of z* y a

corresponding to these is (1600

Z'

+

-5/~6C)

+ 1226 + 26)/646 = 75/17.

Thus the required maximum value is 10 and the minimum value is 75/17.

+ +

(b) Since x 2 y* z' represents the square of the distance of (x,y, z) from the origin (O,O, 0), the problem is equivalent to determining the largest and smallest distances from the origin to the curve of intersection of the ellipsoid x5/4+y215+2'/25 = 1 and the plane z = x + y . Since this curve is a n ellipse, we have the interpretation t h a t C O and d m a r e the lengths of the semi-major and semi-minor axes of this ellipse. The fact th at the maximum and minimum values happen to be given by -A, in both Case 1 and Case 2 is more than a coincidence. It follows, in fact, on multiplying equations (1) by 5, y and z in succession and adding, fo r we then obtain

x' +2 + x , x + 2y' + XI

22'

y+

+

X2y

225

+2x1zz 25 - x,z

= 0

i.e. Then using the constraint conditions, we find x ' + y 2 + z 2 = --Al.

For a generalization of this problem, see Problem 76.

APPLICATIONS to ERRORS 28. The period T of a simple pendulum of length I is given by T = 2 ~ 4 5 .Find the (a)error and ( b )percent error made in computing T by using I = 2 m and g = 9.75m/sec2,if the true values are I = 1.95m and g = 9.81m/sec2. ( a ) T = 2a11"g-'"'. Then

+

dT = (2ag-1/S)(42-1/*dZ) ( 2 ~ 1 ~ / ~ ) ( - 3 g - ~ / * d=g ) Error i n g = Ag

= dg =

fi dZ

-

dg

(1)

+0.06; error in I = A1 = dl = -0.05

The error in T is actually AT, which-is in this case approximately equal to dT. Thus we have from (I), ErrorinT

=

dT

dmm (-0.05) 7r

=

-

7~

/&

The valce of T for 1 = 2, g = 9.75 is T = 27r

-( +0.06)

$&

=

=

- 0.0444 sec (approx.)

2.846 sec (approx.)

dT -0.0444 = - 1.56%. ( b ) Percent error (or relative error) in T = T =2.846

+ 4 In I - 4 lng, L d l - Ldg = $+E)

Another method: Since In T = In 27r

dT T

-

-

2 1

2 g

-

L(?) 2

975

=

-1.5a

as before. Note that (2) can be written

Percent error in T

=

4 Percent error in 1 - 4 Percent error in g

(2)

APPLICATIONS O F PARTIAL DERIVATIVES

CHAP. 81

175

MISCELLANEOUS PROBLEMS 29. Evaluate

1'

dx.

In order to evaluate this integral, we resort to the following device. Define

Then by Leibnitz's rule

+ +

Integrating with respect to a, +(a) = In (a 1) c. But since $(O) = 0, c = 0 and so $(a) = In (a 1). Then the value of the required integral is + (l) = In 2.

+

The applicability of Leibnitz's rule can be justified here, since if we define F ( x , a) = (xa - l)/ln x , 0 < x < 1, F(0, a) = 0, F(1, a) = a, then F ( z , a) is continuous in both z and a for 0 S x 5 1 and all finite a > 0.

30. Find constants a and b for which ~ ( ab ),

ir

=

{sin x

- (az2

+b x ) ) ~ x

is a minimum. The necessary conditions for a minimum are dF/da = 0, dF/db = 0. Performing these differentiations, we obtain aa

~{sinx - (axe {sin x

db From these we find

+ bx)}2dx

=

xe{sinx - (ax2+ bx)} dx

=

0

- (ax2+ bx)I2dx =

- 2 J r x {sin x - (ux2+ bx)} dx

=

0

=x4dx

+

blr

x3dx

=

lTx2sinxdx

x3dx

+

b i n x2dx

=

fxsinxdx

or

='U

7r4b

= IT2-4

Solving for a and b, me find a

= 2o- -320 r3

77'

-0.40065,

b = 7 240

--p= l2

1.24798

We can show that for these values, F(a, b) is indeed a minimum using the sufficiency conditions on Page 164. The polynomial ax2 bx is said to be a least square approximation of sin x over the interval (0, r). The ideas involved here are of importance in many branches of mathematics and their applications.

+

APPLICATIONS OF PARTIAL DERIVATIVES

176

[CHAP. 8

Supplementary Problems TANGENT PLANE and NORMAL LINE to a SURFACE 31. Find the equations of the (a)tangent plane and (b) normal line to the surface s a + y a = 42 at X - 2 - y+4 - 2-6

Am.

(2,-4,6).

(U) ~

- 2 y - z = 6, ( b ) 7

- -2 -1

32. If z = f ( s , y ) , prove that the equations for the tangent plane and normal line at point P ( s o , ~ o , z o ) are given respectively by x - x o - y--0 2-20 (a) z - 20 = f.IP(x- 20) f v l p ( y - y 0 ) and (b) -=f 4 P ful, -1

+

-

33. Prove that the acute angle y between the z axis and the normal to the surface F(x,y,z) = 0 at any point is given by sec y = dF: J'i Fj//F,I.

+ +

34.

The equation of a surface is given in cylindrical coordinates by F ( p , # , x ) = 0 , where F is continuously differentiable, Prove that the equations of (a)the tangent plane and (b) the normal line at the point P(po,+o, 2 0 ) are given respectively by

+

+ C(z-ZO) =

A ( x - x ~ ) B(y-y~)

where

A 35.

= po sin +o and 1 = Fp[pcos+d- -J'41psin#o,

xo = po cos $o,

0

and

yo

P

B = F,I,sin@,

x--0 = A

y--0

B

1 + -F4(,cos#,, P

- 2-20 C

C = FsIp

Use Problem 34 to find the equation of the tangent plane to the surface u z = p # at the point where p = 2 , # = d 2 , z = 1. To check your answer work the problem using rectangular coordinates. Ans. 2 x - ~ y + 2 a z = 0

TANGENT LINE and NORMAL PLANE to a CURVE 36.

Find the equations of the (a) tangent line and (b) normal plane to the space curve x = 6 sin t, = 4 cos 3 4 z = 2 sin €it at the point where t = d4.

y

37.

The surfaces x + y + z = 3 and x * - y * + 2 2 * = 2 intersect in a space curve. Find the equations of the (a)tangent line (b) normal plane to this space curve at the point (1,1,1).

ENVELOPES 38. Find the envelope of each of the following families of curves in the xy plane. In each case construct xa y' a graph. (a)y = a2 - a', ( b ) y l-a = 1.

+

Am.

(U) d = 4 y ;

( b ) Z + Y = 21, 2 - - y = 2 1

39. Find the envelope of a family of lines having the property that the length intercepted between the

x and

40.

y

axes is a constant a.

Am. ~ ' / ~ + y= ~ /aaIs '

Find the envelope of the family of circles having centers on the parabola y = d and passing through its vertex. [Hint Let (a,(U') be any point on the parabola.] An& xa = -g/(2y 1)

+

41. Find the envelope of the normals (called an evolute) to the parabola y=Jxa and construct a graph.

A m . 8(y - 1)' = 27%'

42.

177

APPLICATIONS O F PARTIAL DERIVATIVES

CHAP. 81

Find the envelope of the following families of surfaces:

-

( b ) (x - a)2

(a)& - y ) a'z = 1, Ans. (a) 42 = ( ~ - y ) ~ (b) , y2 = x * + 2 x x

+ y'

= 2ax

43.

Prove t h a t the envelope of the two parameter family of surfaces F ( z , y , z , a , p ) = 0, if it exists, is obtained by eliminating a and p in the equations F = 0, Fa = 0, Fs = 0.

44.

Find the envelope of the two parameter families (a)x = ax py - 'a x cosy = a where cos2a cos*p cos2y = 1 and a is a constant. Ans. (a)42 = x s + y p , (b) x 2 + y 2 + z 2 = u2

+

+

+

-

and (b) z cos a

+ y cos /3 4

DIRECTIONAL DERIVATIVES 45.

(a) Find the directional derivative of U = 2xy-zz" at ( 2 , - 1 , l ) in a direction toward (3,1,-1). (b) In what direction is the directional derivative a maximum? (c) What is the value of this maximum?

A m . (a)10/3, ( b ) -2i

+ 4 j - 2k,

(c) 2fi

T = lOOxy/(z*+y*). (a) Find the directional derivative at the point (2,l) in a direction making a n angle of 60° with the positive z axis. (b) In what direction from ( 2 , l ) would the derivative be a maximum? (c) What is the value of this maximum? Ans. (a)1 2 f i - 6; ( b ) in a direction making an angle of P - tan-l2 with the positive x axis, or in the 2j; (c) 1 2 6 direction -i

46. The temperature at any point (x,y) in the xy plane is given by

+

47.

Prove that if F(p, 9 , x ) is continuously differentiable, the maximum directional derivative of F at any point is given by

s-:

DIFFERENTIATION UNDER the INTEGRALI SIGN 1 1 1 4-b cosax2dx, find Ans. x 2 sin ax2 dx - 2 cos- - -cos ap 48. If # ( a ) = da ' a 2 6 49. (a) If F(a) = i a p t a n - l t d x , find da dF 1 by Leibnitz's rule. (b) Check the result in (a)by direct

;S

Ans.

integration. 50.

1 Given l l x P d x = 1, p +

In (1

52. Prove that 53.

(U) 2a tan-' a

Show that

I* (5

+ a cos x) dx: dx

- 3 COS x)'

(-l)"'m!

> -1. Prove that

In (1 - 2a cos x

*

- +In (a' + 1)

=

P

+ a2)dx

In ( 1

=

+

{OS

( p + l ) ? n + l ,m

= 1,2,3,

m), 2 lal < 1. In

Iff'
1'

Discuss the case \a\= 1.

- -69n -

2048

INTEGRATION UNDER the INTEGRAL SIGN

55. Starting with the result

(a

- sinx) dx = 2 ~ a ,prove that for all constants a and b,

*

178

APPLICATIONS OF PARTIAL DERIVATIVES O=

56. Use the result

dx = a+sinx

caJ2

..

(b) Show that

1

812

2T

a>1

to prove that

,/='

cos - *

dx

sec x In (1

[CHAP. 8

+ +cos x) dx

5n2 72

= -

MAXIMA and MINIMA. LAGRANGE MULTIPLIERS 58. Find the maxima and minima of F ( s , y, z ) = xy2z3 subject to the conditions x z > 0. Ans. maximuin value = 108 a t x = 1, g = 2, z = 3

+ y + z = 6, x > 0, y > 0,

59. What is the volume of the largest rectangular parallelepiped which can be inscribed in the ellipsoid x2/9 y2/16 + z 2 / 3 6 = I ? Ans. 6 4 f i

+

+

60. (a) Find the maximum and minimum values of x2 y2 subject to the condition 32' (b) Give a geometrical interpretation of the results in (a). Ans. maximum value = 70, minimum value = 20

+ 4sy + 6y'

= 140.

61. Solve Problem 23 using Lagrange multipliers. 62. Prove t h a t in any triangle ABC there is a point that P is the intersection of the medians.

P such that

PAS + PB* + ?P

is a minimum and

63. (a) Prove that the maximum and minimum values of f(x,y) = x a + x y + y 3 in the unit square 0 S x 5 1, 0 5 y 5 1 are 3 and 0 respectively. (b) Can the result of (a) be obtained by setting the partial derivatives of f(x,y) with respect to x and y equal to zero. Explain. 64. Find the extreme values of z on the surface 2x2

A m . maximum = 5, minimum = -5

+ 3y2 + z2 - 12xy + 4 x 2 = 36.

65. Establish the method of Lagrange multipliers in the case where we wish to find the extreme values of F ( x ,y, z ) subject to the two constraint conditions G(x, y, z ) = 0, H ( x , y, z ) = 0.

66. Prove that the shortest distance from the origin to the curve of intersection of the surfaces x y z = a and y = bx where a > 0, b > 0, is 3da(bz 1)/2b.

+

+

67. Find the volume of the ellipsoid 11x2 9y2

+ 16xz - 4xy + 1Oyz - 20x2 = 80.

A m . 64nfi/3

APPLICATIONS to ERRORS 68. The diameter of a right circular cylinder is measured a s 6.0 +- 0.03 inches, while its height is measured a s 4.0 0.02 inches. What is the largest possible (a) error and (b) percent error made in computing Ans. (a) 1.70 ins, (b) 1.5% the volume? 69. The sides of a triangle are measured to be 12.0 and 15.0 feet, and the included angle 60.0°. If the lengths can be measured to within 1% accuracy while the angle can be measured to within 2% accuracy, find the maximum error and percent error in determining the (a) area and (b) opposite side Ans. (a) 2.501 ft2, 3.21%; (b) 0.287 ft, 2.08% of the triangle.

MISCELLANEOUS PROBLEMS 70.

If p and + a r e cylindrical coordinates, a and b a r e any positive constants and n is a positive integer, prove that the surfaces pmsin n$ = a and p" cos n+ = b are mutually perpendicular along their curves of intersection.

CHAP. 81

179

APPLICATIONS O F PARTIAL DERIVATIVES

71. Find an equation f or the ( a ) tangent plane and ( 6 ) normal line to the surface 8 r e + = n 2 at the point where r = 1, e = n/4, + = a/2, (r,8, +) being~- spherical coordinates. x y-fi/2 - z-*/2 Ans. ( U ) 4~ - (t2+47i)2/ (47-77')~ = - n ' f i , ( b ) --4 r2- 47 T2 4ii

+

+

72. (a) Prove t h a t the shortest distance from the point (a,b, c ) to the plane A x

I Aa + Bb + C c + D I

I

1

dA2+Be+C

(b) Find the shortest distance from (1,2,-3) to the plane 2 s

+ B y + Cx + D

- 3 y + 62 = 20.

= 0 is

Ans. ( 6 ) 6

73. The potential V due to a charge distribution is given in spherical coordinates (r,e,q,) by

t ' = - p COS e

r2 where p is a constant. Prove t h a t the maximum directional derivative at any point is pdsins e

pisI

74.

Prove t h a t

+ 4 cos2e

r3

dx

if

7n

> 0 , n > 0 . Can you extend the result to the case

wi > -1, n > -l? 75. ( a ) If

2r/d-1 ax2

76.

+

+

+

Prove t h a t the maximum and minimum distances from the origin to the curve of intersection defined by x P / a 2 + y 2 1 b 2 + ~ * I c=* 1 and A x + B y + C z = 0 can be obtained by solving for d the equation A B2b* c2c2

+ -+

77.

+

b2 - 4ac < 0 and a > 0, c > 0 , prove t h a t the area of the ellipse a t 2 b x y cy2 = 1 is [Hint: Find the maximum and minimum values of x 2 + v2 subject to the constraint b x y cy2 = 1.1

=

Prove t h a t the last equation in the preceding problem always has two real solutions d: and $ for any real non-zero constants a, b, c and any real constants A , B, C (not all zero). Discuss the geometrical significance of this.

78. (a) Prove th a t

IM

=

iM+ dx ( x 2 a')'

M +

- -tan-'1 22

a

M 2a* (a'+ M')

(b) Find lim IM. This can be denoted by E4-m

79.

Find the point on the paraboloid x = x 2 Ans. (1,-2,s)

+ 9'

80. Investigate the maxima and minima of f ( x , y)

Ans. minimum value = 0

81. ( a ) Prove th at

=I2

-

C O S d ~x

( b ) Use ( a ) to prove t h a t

812

which is closest to the point (3, -6,4).

In a

+

- -a72(a* 1)

cos2x d x

+ 4y' - 8 ~ ) ~ .

= ( x 2- 2x

-

a*

30

+1

*

+ 5 - 81n2 60

82. ( a ) Find sufficient conditions fo r a relative maximum o r minimum of w = f ( x , y,2 ) . ( b ) Examine w = x2 y2 2' - 6 x y 8x2 - 1Oyz for maxima and minima.

+ +

+

[Hint: F o r ( a ) use the fact th at the qzcadratic form (i.e. is positive definite) if

Aa2+BP2+Cy2+2DaP+2Eay+2Fpy

I

>

0

Chapter 9 Multiple Integrals DOUBLE INTEGRALS Let F(x,y) be defined in a closed region % of the xy plane (see Fig. 9-1). Subdivide % into n subregions A T , of area AA,, k = 1,2, . . .,n. Let (tk,vk) be some point of A%,. Form the sum

Consider where the limit is taken so that the number n of subdivisions increases without limit and such that the largest linear dimension of each A?(, approaches zero. If this limit exists it is denoted by

J-JF ( x ,Y) d A

Fig.9-1

(3)

9

and is called the double integral of F ( x , y) over the region %. It can be proved that the limit does exist if F(x,y) is continuous (or sectionally continuous) in %.

ITERATED INTEGRALS If % is such that any lines parallel to the y axis meet the boundary of % in at most two points (as is true in Fig. 9-l),then we can write the equations of the curves ACB and ADB bounding % as y = f&c) and y = f 2 ( x ) respectively, where fl(x) and fz(x) are single-valued and continuous in a S x 5 b. In this case we can evaluate the double integral (3) by choosing the regions A%, as rectangles formed by constructing a grid of lines parallel to the x and y axes and AAk as the corresponding areas. Then (3) can be written

=

s.l.(sfz(5) Y=fl(x)

F(x,?4 dll}dx

where the integral in braces is to be evaluated first (keeping x constant) and finally integrating with respect to x from a to b. The result (4) indicates how a double integral can be evaluated by expressing it in terms of two single integrals called iterated integrals. 180

181

MULTIPLE INTEGRALS

CHAP. 91

If % is such that any lines parallel to the x axis meet the boundary of % in a t most two points (as in Fig. 9-1),then the equations of curves CAD and CBD can be written x = g&) and x = g2@) respectively and we find similarly

=

f { s”(g)wa)

dx}dll

%=U1 ( 9 )

If the double integral exists, (4) and ( 5 ) yield the same value. (See, however, Problem 17.) In writing a double integral, either of the forms (8) or (5), whichever is appropriate, may be used. We call one form an interchange of the order of integration with respect to the other form. In case l( is not of the type shown in the above figure, it can generally be subdivided . . which are of this type. Then the double integral over % is found into regions . .. by taking the sum of the double integrals over ql,q2,

.

.

TRIPLE INTEGRALS The above results are easily generalized to closed regions in three dimensions. For example, consider a function F(x, y, x ) defined in a closed three dimensional region %. Subdivide the region into n subregions of volume A v k , k = 1,2, . . . n. Letting (&, vk, b,) be some point in each subregion, we form

where the number n of subdivisions approaches infinity in such a way that the largest linear dimension of each subregion approaches zero. If this limit exists we denote it by

9

called the triple integral of F ( x , y , x ) over q. The limit does exist if F ( x , y , x ) is continuous (or sectionally continuous) in T. If we construct a grid consisting of planes parallel to the xy, yz and xx planes, the region % is subdivided into subregions which are rectangular parallelepipeds. In such case we can express the triple integral over given by (7) as an iterated integral of the form f’ (z,ar) J b z=a

Jg’(z) g = g l (I)

F ( x , y , z )d x d y d x

~ = f (lz , ~ )

=

{z);;;;Ja -J [;

s”(”” X = f l (I,Y)

11

F(x,Y,4 dz dY dx (8)

(where the innermost integral is to be evaluated first) or the sum of such integrals. The integration can also be performed in any other order to give an equivalent result. Extensions to higher dimensions are also possible.

TRANSFORMATIONS of MULTIPLE INTEGRALS In evaluating a multiple integral over a region %, it is often convenient to use coordinates other than rectangular, such as the curvilinear coordinates considered in Chapters 6 and 7.

182

[CHAP. 9

MULTIPLE INTEGRALS

If we let (u,v) be curvilinear coordinates of points in a plane, there will be a set of transformation equations x = f ( u ,U), g = g(u,v ) mapping points ( x , y) of the x y plane into points (u,v) of the uv plane. In such case the region of the xg plane is mapped into a region !I(‘of the zcv plane. We then have

where

G(u,U ) = F { f ( & v, ) ,g ( u , v ) ) and

I ax ax I

is the Jacobian of x and y with respect to U and v (see Chapter 6). Similarly if (u,v,w) are curvilinear coordinates in three dimensions, there will be a set of transformation equations x = f(u,U,w),y = g(u,v, w), x = h(w,v,w) and we can write

is the Jacobian of x, y and x with respect to U , v and w . The results ( 9 ) and (11) correspond to change of variables for double and triple integrals. Generalizations to higher dimensions are easily made.

Solved Problems DOUBLE INTEGRALS 1. (a) Sketch the region

ss

in the xg plane bounded by y =x 2 , x = 2 , y = 1.

(b) Give a physical interpretation to

(x2+Y2)dXdZ/*

9

( c ) Evaluate the double integral in (b). (a) The required region “Ip is shown shaded in Fig. 9-2 below. ( b ) Since x* yz is the square of the distance from any point (z,y) to (0, 0), we can consider the double integral as representing the polar moment of inertia (i.e. moment of inertia with respect to the origin) of the region (assuming unit density).

+

We can also consider the double integral as representing the a density varying as x’ 8’.

+

712488

of the region “Ip assuming

183

MULTIPLE INTEGRALS

CHAP. 91

Fig. 9-2 (c)

Method 1:

Fig. 9-3

s'

The double integral can be expressed as the iterated integral ( x z + y2) dy}dx

=

x2y

z=1

+

$1"

dx y=l

The integration with respect to y (keeping x constant) from y = 1 to y = x 2 corresponds formally to summing in a vertical column (see Fig. 9-2). The subsequent integration with respect to x from x = 1 to x = 2 corresponds to addition of contributions from all such vertical columns between x = 1 and x = 2.

Method 2:

The double integral can also be expressed as the iterated integral ( x 2 + y2)dz}dy

=

(! + 2y2 -% - y5I2)dy

Jll$ +

=

3/2

#=I

=

xy'r

dy *=fi

1006 105

in Fig. 9-2 above is replaced by a horizontal In this case the vertical column of region column as in Fig. 9-3 above. Then the integration with respect to x (keeping y constant) from x = f i to x = 2 corresponds to summing in this horizontal column. Subsequent integration with respect to y from y = l to y = 4 corresponds to addition of contributions for all such horizontal columns between y = 1 and y = 4 .

2. Find the volume of the region common to the intersecting cylinders x 2 + 2 2 = a2.

x2+'y2

= a2 and

Required volume = 8 times volume of region shown in Fig. 9-4

= 8

J

T

z dy dx

As an aid in setting up this integral note that z d y d x corresponds to the volume of a column such a s shown darkly shaded in the figure. Keeping x constant and integrating with respect to y from y = O to y = d= corresponds to adding the volumes of all such columns in a slab parallel to the yz plane, thus giving the volume of this slab. Finally, integrating with respect to 2 from x = 0 to x = a corresponds to adding the volumes of all such slabs in the region, thus giving the required volume.

184

MULTIPLE INTEGRALS

[CHAP. 9

3. Find the volume of the region bounded by x = x+y, ~ = 6 x, = O , y=O, x = O Required volume

i;o+-)

= volume of region shown in Fig. 9-5 -

- (x+y)Wydx

y o ( 6 - x ) U - +Y2)

6-x

dx

y=o

-

s1,

+(6- ~ ) ~ d=x 36

In this case the volume of a typical column (shown darkly shaded) corresponds to (6 (x+y)}dydx. The limits of integration are then obtained by integrating over the region % of the (obtained figure. Keeping x constant and integrating with respect to y from y = O to ~t/ = 6 - x from x = 6 and x = x + y ) corresponds to summing all columns in a slab parallel to the yx plane. Finally, integrating with respect to x from x = O to x = 6 corresponds to adding the volumes of all such slabs and gives the required volume.

TRANSFORMATION of DOUBLE INTEGRALS 4. Justify equation (9), Page 182, for changing variables in a double integral. In rectangular coordinates, the double integral of F ( x , y ) over the region “Ip (shaded in Fig. 9-6) is

ss

F(x,y) dx dy. We can also evaluate this double 9 integral by considering a grid formed by a family of U and v curvilinear coordinate curves constructed on the region “Ip as shown in the figure. Let P be any point with coordinates (x,y) or (u,v), where x = f(u,v) and y = g(u,v). Then the vector r from 0 to P is given by r = xi y j = f(u, v)i g(u,v)j. The tangent vectors to the coordinate curves u = c 1 and V = C Z , where CI and c2 are constants, are drldv and arldu respectively. Then the area of region AT

+

+

I* El

Fig. 9-6

of Fig. 9-6 is given approximately by au X av Au AV. But

so that The double integral is the limit of the sum

taken over the entire region “Ip. An investigation reveals that this limit is

where 5‘(’ is the region in the uv plane into which the region % is mapped under the transformation

x = f(u, v), y = d u , 4.

Another method of justifying the above method of change of variables makes use of line integrals and Green’s theorem in the plane (see Chapter 10, Problem 32).

CHAP. 91

5.

If

U

MULTIPLE INTEGRALS

185

= x2 - y2 and v = 2xy, find a($, y)/d(u,v) in terms of U and v.

+

+ ( 2 ~ y )we~ have and x2 + ye

From the identity ( x 2 y2)' = (x2 - y2)2 ( x z + g 2 ) 2 = u2

+ v2

=

d

w

Then by Problem 45, Chapter 6, d(x,y)

a(% v) Another method: directly.

6.

=

1

-

1

a(% v)/a(x, ?I) 4 ( X 2 + Y 2 )

-

1

4 d m

Solve the given equations for x and. y in terms of

U

and v and find the Jacobian

Find the polar moment of inertia of the region in the xy plane bounded by x2 - y2 = 1, x2 - y2 = 9, xy = 2, xy = 4 assuming unit density.

Under the transformation x z - y 2 = U, 2 x y = v the required region % in the xy plane [shaded in Fig. 9-7(a)] is mapped into region of the uv plane [shaded in Fig. 9-7(b)]. Then: Required polar moment of inertia

=

ss 9

+

(x2 U') d x d y

=

ss 9'

( x 2+

1-

dudv

where we have used the results of Problem 5. Note that the limits of integration for the region %' can be constructed directly from the region % in the x y plane without actually constructing the region T'. I n such case we use a grid a s in Problem 4. The coordinates (U,v) are curvilinear coordinates, in this case called hyperbolic coordinates.

7. Evaluate

ss

d m d x dy, where

9

x 2 + y 2 = 4 and

x2+2j2

+

is the region in the xy plane bounded by

= 9.

The presence of x 2 y 2 suggests the use of polar coordinates (p, +), where x = p cos I$,y = p sin I$ (see Problem 38, Chapter 6). Under this transformation the region % [Fig. 9-8(a) below] is mapped into the region %' [Fig. 9-8(b) below].

186

MULTIPLE INTEGRALS

[CHAP. 9

Since d ( x ’ y ) = p, it follows that a(P,

+)

We can also write the integration limits for ‘3(’ immediately on observing the region ‘3(, since for fixed +, p varies from p = 2 to p = 3 within the sector shown dashed in Fig. 9-8(a). An integration with respect to + from = O to $ = 2 ~then gives the contribution from all sectors. Geometrically p d p d + represents the area d A a s shown in Fig. 9-8(a).

8. Find the area of the region in the xy plane bounded by the lemniscate p2 = u2 cos Z+. Here the curve is given directly in polar coordinates (p, 46). By assigning various values to + and finding corresponding values of p we obtain the graph shown in Fig. 9-9. The required area (making use of symmetry) is r/4

4 S @r =’ o4 Spa= om p d p d +

=

-a

4 i = o $ l p=o

2s,=o

d+ Iil 4

Ii/4

=

a2cos 2$ d+

=

a2 sin 2+

-

a2

TRIPLE INTEGRALS 9. (a) Sketch the 3 dimensional region (?! bounded by x + y + x = a (u>O), x = O , y=o,

x=o.

( b ) Give a physical interpretation to

JJJ (x2+ y2+ x2) dx dy dx 9

( c ) Evaluate the triple integral in (b). ( a ) The required region ‘3( is shown in Fig. 9-10.

+

Fig. 9-10

( b ) Since x 2 4-g2 x2 is the square of the distance from any point (x,y, x ) to (0, O,O), we can consider the triple integral a s representing the polar moment of inertia, (i.e. moment of inertia with

CHAP. 91

MULTIPLE INTEGRALS

187

respect to the origin) of the region % (assuming unit density). We can also consider the triple integral as representing the rnms of the region if the density varies a s x2 y2 2.

+ +

(c)

The triple integral can be expressed as the iterated integral

=

sa

+ y2z +

a-2

X ~ Z

2=0

=

j-l0iL2{

$1

x2(a- x) - x2y

a-r-y

dydx Z=O

+ ( a - x)y2 - y3 + ( a - x3 - y)3}dY

dX

The integration with respect to x (keeping x and y constant) from x = O to x = a- x - y corresponds to summing the polar moments of inertia (or masses) corresponding to each cube in a vertical column. The subsequent integration with respect to y from y = O to g = a-% (keeping x constant) corresponds to addition of contributions from all vertical columns contained in a slab parallel to the yx plane. Finally, integration with respect to z from x = 0 to x = a adds up contributions from all slabs parallel to the yx plane. Although the above integration has been accomplished in the order x , y, x, any other order is clearly possible and the final answer should be the same.

10. Find the (a) volume and ( b ) centroid of the region bounded by the parabolic cylinder x = 4 - x 2 and the planes x = O , y = O , y = 6 , x = O assuming the density to be a constant U. The region % is shown in Fig. 9-11. (a) Required volume

= 4

Fig. 9-11

Lo Lo 4 - 9

( b ) Total mass g

=

=

adxdyclx

Total moment about yx plane Total mass

=

= 32a by part (a), since

I:,L.L,'

U

ux dz dy dx

Total mass

is constant.

-

24a 32u

Then -

3 4

[CHAP. 9

MULTIPLE INTEGRALS

188

l:,-(1:,L0 4 - 9

@

i

=

=

Total moment about xx plane Total mass

=

Total moment about xy plane Total mass

=

oydxdydx Total mass

Total mass

-

960 - -

-

256016 -

3

320

320

-

8 -

5

Thus the centroid has coordinates (3/4,3,8/5). Note that the value f o r 3 could have been predicted because of symmetry.

TRANSFORMATION of TRIPLE INTEGRALS 11. Justify equation ( I I ) , Page 182, f o r changing variables in a triple integral.

Fig. 9-12 By analogy with Problem 4, we construct a grid of curvilinear coordinate surfaces which subdivide the region % into subregions, a typical one of which is A% (see Fig. 9-12). The vector r from the origin 0 to point P is r = xi

+ y j + xk

= f(u, U , w)i

+ g(u,

U , w)j

+ h(u,

U , w)k

x = f ( u ,U , w),y = g ( u , U , w) and x = h(u,?I,w ) . Tangent vectors to the coordinate curves corresponding to the intersection of pairs of coordinate surfaces are given by arlau, arlav, arlaw. Then the volume of the region AT of Fig, 9-12 is given approximately by assuming that the transformation equations are

The triple integral of F ( x , y, x ) over the region is the limit of the sum

An investigation reveals that this limit is

where

q’is the region in the uvw space into which the region % is mapped under the transformation.

Another method for justifying the above change of variables in triple integrals makes use of Stokes’ theorem (see Problem 84, Chapter 10).

CHAP. 91

12. Express

189

MULTIPLE INTEGRALS

JJJ F ( x ,y, x ) dx dy dx

in (a) cylindrical and ( b ) spherical coordinates.

%

(a) The transformation equations in cylindrical coordinates are x = p cos +, y = p sin $, x = x . As in Problem 39, Chapter 6, d(x, y, z ) / d ( p , +, x ) = p. Then by Problem 11 the triple integral becomes

JJJ

where T’ is the region in the sin 9, x ) .

G(P,

$9

4 P dP d+ dz

%’

p,

+, x

space corresponding to % and where

G(p, $ , x )

= F ( p cos +,

p

( b ) The transformation equations in spherical coordinates a r e x = r sin e cos +, y = r sin e sin $, x = r cose. By Problem 103, Chapter 6, d(x, y,x)/a(r, e,$) = r2 sin 8. Then by Problem 11 the triple integral becomes H ( r , 8, +) r 2sin e dr de d+

sss %I

where T’ is the region in the r , e , $ F ( r sin e cos +, r sin e sin +, T cos e).

space corresponding to

T, and

where

H(r,e,$)

E

+

13. Find the volume of the region above the xy plane bounded by the paraboloid x = x2 y2 and the cylinder x2 y2 = a2.

+

The volume is most easily found by using cylindrical coordinates. In these coordinates the equations for the paraboloid and cylinder a r e respectively x = p2 and p = a. Then Required volume = 4 times volume shown in Fig. 9-13 = 4

s””sa @=o

p=D

= 4sTJ2f @=o

p dz dp

d+

p3dpd+

p=o

Fig. 9-13 The integration with respect to x (keeping p and + constant) from x = 0 to x = p2 corresponds to summing the cubical volumes (indicated by d V ) in a vertical column extending from the xy plane t o the paraboloid. The subsequent integration with respect to p (keeping + constant) from p = 0 to p = a corresponds to addition of volumes of all columns in the wedge shaped region. Finally, integration with respect to + corresponds to adding volumes of all such wedge shaped regions. The integration can also be performed in other orders to yield the same result. We can also set up the integral by determining the region mapped by the cylindrical coordinate transformation.

T’ in

p,+,x

space into which

is

14. (a) Find the moment of inertia about the x axis of the region in Problem 13, assuming that the density is the constant U. ( b ) Find the radius of gyration. (a) The moment of inertia about the x axis is

MULTIPLE INTEGRALS

190

[CHAP. 9

The result can be expressed in terms of the mass M of the region, since by Problem 13,

M = volume

X

2 -na% - -nas 2M density = z a 4 ~so that I , = - - - - - - M u 2 3 aa4 3 2 3

Note that in setting up the integral for I , we can think of u p d x d p d $ as being the mass of the cubical volume element, p2 up dx dp d$, as the moment of inertia of this mass with respect to the x axis and

JJJ p2

-*

up

dx dp d$ as the total moment of inertia about the x axis.

The limits of integration are determined a s in Problem 13. ( b ) The radius of gyration is the value K such that MK2 = $Mu2, i.e. K 2 = #a2 or K = am.

The physical significance of K is that if all the mass M were concentrated in a thin cylindrical shell of radius K , then the moment of inertia of this shell about the axis of the cylinder would be I,.

15. (a) Find the volume of the region bounded above by the sphere x 2 + y 2 + x 2 = u2 and below by the cone x2 sin2(Y = (x2 y2)cos2 where (Y is a constant such that 0 5 S T. ( b ) From the result in (a),find the volume of a sphere of radius a. In spherical coordinates the equation of the sphere is r = a and t h a t of the cone is e = a. This can be seen directly or by u+ng the transformation equations x = r sin e cos $, y = r sin e sin 9, x = r cos 8. For example, x2 sin2 = ( x 2 y2) cos2 becomes, on using these equations,

+

(Y,

+

(Y

(Y

+

= (r2sin2e cos2@ r 2 sin2e sin24) cos2 a ? cos2e sin2(Y = r2 sin2e cos2 (Y

r2 cos2e sin2(Y

i.e., from which t a n e = say e = (Y.

-t-

t a n a and so

e=a

or e = r - a .

It is sufficient to consider one of these,

( a ) Required volume = 4 times volume (shaded) in Fig. 9-14 = 4$

-

-

2pa3 -(1 3

~

i:o ~

r2 ~ sin e ~dr de d@ o

- cosa)

The integration with respect to r (keeping e and $ constant) from r=O to r = a corresponds to summing the volumes of all cubical elements (such as indicated by dV) in a column extending from r = 0 to r = a. The subsequent integration with respect to e (keeping $ constant) from e = 0 to e = ~ / 4corresponds to summing the volumes of all columns in the wedge shaped region. Finally, integration with respect to $ corresponds to adding volumes of all such wedge shaped regions. ( b ) Letting a = -n, the volume of the sphere thus obtained is

2-na3 -(1 3

- COST)

4 3

= --nu3

CHAP. 91

MULTIPLE INTEGRALS

191

16. (a) Find the centroid of the region in Problem 16. ( b ) Use the result in (a) to find the centroid of a hemisphere. ) due to symmetry, given by 2 = P = O (a) The centroid ( # , # , iis, Total moment about zy plane I = Total mass

Since z = r cos e and 40.

U

is constant the numerator is

T

cos e

f"Sa KO 4=0

e=o

=

r' sin e d r de d+

and

- SJJ z U dV

sss @dV

4u f ' * $ a &=o

e=o

=

o=o

&=O

Then

$1

sin e cos e de d+

t=O

sin e cos e de d+

The denominator, obtained by multiplying the result of Pmb. 16(a) by #

)nua4 sin*a

=

+d(l

- COS a)

3 = -a(l 8

U,

is #ruaa(l- cos a).

+ cosa).

( b ) Letting a = ~ / 2 ,i = #U.

MISCELLANEOUS PROBLEMS

=

It&

- 5 1 ' - 1 - z+lo - 2

This follows a t once on formally interchanging z and y in and then multiplying both sides by -1.

(U)to

obtain

ll{J =5 f i d z ) d g

1

This example shows that interchange in order of integration may not always produce equal results. A sufficient condition under which the order may be interchanged is that the double integral over the corresponding region exists. In this case

ss

e

d

z dy, where

is the

s

region 0 6 %d 1, 0 S'yd 1 fails to exist because of the discontinuity of the integrand at the origin. The integral is actually an improper double integral (see Chapter 12).

18. Prove that Let I(%) =

f {f

F(u)du}dt, J(x) =

I'(z) =

A=

f F(u)du,

( x - u ) F ( u ) d u . Then

J'(z) =

F(u)du

using Leibnitz's rule, Page 163. Thus I'(z) = J ' ( z ) , and so I(%)- J ( z ) = c where c is a constant. Since Z(0) = J ( 0 ) = 0, c = 0 and so I(%) = J ( z ) .

MULTIPLE INTEGRALS

192

[CHAP. 9

The result is sometimes written in the form

The result can be generalized to give (see Problem 64)

Supplementary Problems DOUBLE INTEGRALS 19. (U)Sketch the region % in the x y plane bounded by y3 = 2% and y = z. (b) Find the area of 'Ip. (c) Find the polar moment of inertia of "Ip assuming constant density U. A m . (b) 3; (c) 48d36 = 72M/36, where M is the mass of "Ip.

Am. $=#, # = l

20. Find the centroid of the region in the preceding problem. 21.

i T ( z+

Given

s'

y) dx d y.

(a) Sketch the region and give a possible physical interpretation of

the double integral. ( b ) Interchange the order of integration. (c) Evaluate the double integral.

A m . (b) 22.

r=l

J4-s(% y-0

+ y) d y dx,

( c ) 241/60

Show that

dydx

23. Find the volume of the tetrahedron bounded by x / a

A m . abcI6

+ y/b + x/c

=

4(u + 2) 52

= 1 and the coordinate planes.

24.

Find the volume of the region bounded by x = x a + y 3 , z = O , x = --a, A m . 8a4/3

25.

Find (a)the moment of inertia about the z axis and (b) the centroid of the region in Problem 24 assuming a constant density U. Ans. ( a )%a% = #Mu3, where M = mass; (b) 3 = p = 0, Z = &U'

ss

X=U,

y=-a,

TRANSFORMATION of DOUBLE INTEGRALS 26. Evaluate

d w d x dy , where % is the region x'

s

27. If

is the region of Prob. 26, evaluate

28. By using the transformation x

+y =

ss

+ y'

e - @+ y 3)dx d y .

s

U,

y = uv, show that

5 ua.

An& Quas

Ana. ~ (- le-.')

y=u.

CHAP. 91

MULTIPLE INTEGRALS

193

29. Find the area of the region bounded by xy = 4, zy = 8, sys= 6, zy' = 15, [Hint: Let zy = U, sy' Ans. 2 In 3 30.

v.]

Show that the volume generated by revolving the region in the first quadrant bounded by the parabolas = x, ye = 8x, xz = y, xz = 8y about the x axis is 279a/2. [Hint: Let y' = ux, 'x = vy.]

y' 31.

ss

Find the area of the region in the first quadrant bounded by y = xs, y = 4x', x = ya, x = 4ya.

+

be the region bounded by x y = 1, x = 0, y = 0. Show t h a t [Hint: Let x - y = U, x + y = v.] Q

32. Let

x:,i:,

TRIPLE INTEGRALS 33.

34.

(a)Evaluate Ans. ( a ) Q

J'= x

zyx dx dy d

Z

cos( z ) d x dy

& =sin 1 Am. 2

.

dx. (b) Give a physical interpretation t o the integral in (a).

W

Find the (a)volume and (b) centroid of the region in the first octant bounded by x/u where a,b, c are positive. AnS. (a) abc/6; ( b ) 3 = a/4,g = b/4, 3 = c/4

+ y / b + z/c

= 1,

35. Find the (a)moment of inertia and (b) radius of gyration about the z axis of the region in Prob. 34.

Ans. (a)M ( d 36.

+ bS)/lO,

(b)

d(a*+ b*)/lO

Find the mass of the region corresponding to equal to xyz. Ans. 4/3

37. Find the volume of the region bounded by z

+ y' + z'

5%

5 4, z h 0, y 1 0 , z 2 0, if the density is

= x'+y2 and z = 2 ~

TRANSFORMATION of TRIPLE INTEGRALS Find the volume of the region bounded by z = 4 - x 4 - y a

38.

Am. r/2

and the xy plane.

39.

Find the centroid of the region in Problem 38, assuming constant density Ans. Z = g = O , Z = $

40.

(a)Evaluate

JJJ d

m dx dy dz,

where

Ans. 8a

U.

is the region bounded by the plane z = 3 and the

%

cone z = q m . (b) Give a physical interpretation of the integral in (a). [ H i n t Perform the integration in cylindrical coordinates in the order p, z, #.] A m . 2 7 ~ ( 2 f i- 1)/2

d

m

41.

Show that the volume of the region bounded by the cone z = is af6.

42.

Find the moment of inertia of a right circular cylinder of radius U and height b, about its axis if the density is proportional to the distance from the axis. An8. #Mu'

43.

(a)Evaluate

+

JJJ

dx dy dz ( x 2+

yp + 22)s,3

, where

and the paraboloid z = xL+ y'

is the region bounded by the spheres 'x

+ y* + z'

= 'a

Q

and xp y'+ z4 = b2 where a > b > 0. (b) Give a physical interpretation of the integral in (a). An8. (a)4n In (alb) 44.

(a)Find the volume of the region bounded above by the sphere r = 2a cos 8 , and below by the cone = (Y where 0 < (Y < a/2. ( b ) Discuss the case a = a/2. Ans. $aas(l- cos' a)

$I

45.

Find the centroid of a hemispherical shell having outer radius a and inner radius b if the density (a)is constant, (b) varies a s the square of the distance from the base. Discuss the case a = b. A m . Taking the z axis as axis of symmetry: (a) 2 = # = O , t = #(a4- b')/(as- b3); ( b ) 2 = # = 0 , E = #(a6 - b8)/(a5- b5)

[CHAP. 9

MULTIPLE INTEGRALS

194

MISCELLANEOUS PROBLEMS 46. Find the mass of a right circular cylinder of radius a and height b, if the density varies as the square of the distance from a point on the circumference of the base. Ans. &aa2bk(9a2 2b2), where k = constant of proportionality.

+

47.

Find the (a)volume and (b) centroid of the region bounded above by the sphere z 2 + y L + z 2= a2 and below by the plane z = b where a > b > 0, assuming constant density. Ans. ( a ) &r(2a3- 3a2b b3); ( 6 ) 2 = @ = 0, Z = i ( a b)2/(2a b)

+

48.

+

+

A sphere of radius a has a cylindrical hole of radius b bored from it, the axis of the cylinder coinciding with a diameter of the sphere. Show t h a t the voluine of the sphere which remains is #r[a3- (a2- b*)3/2].

49. A simple closed curve in a plane is revolved about a n axis in the plane which does not intersect the

curve. Prove t h a t the volume generated is equal to the a re a bounded by the curve multiplied by the distance traveled by the centroid of the area (Pappus' theorem).

50.

51.

Use Problem 49 to find the volume generated by revolving the circle x 2 about the x axis. Ans. 2r2a2b

+ (y - b)2 = as,

b

>a >0

Find the volume of the region bounded by the hyperbolic cylinders q = 1 , x y = 9 , x z = 4 , s z = 3 6 , Ans. 64

yz = 25, gz = 49. [Hint: Let xy = U , xz = U , yz = w.]

52. Evaluate

+

[fJ dl - (x2/u2+ y*/b2+ z2/c2) dx dy dz,

where

is the region interior t o the ellipsoid

%

+

x*/u2 V 2 / b Z zz/c2 = 1. [Hint: Let x = a24, y = bv, z = cw. Then use spherical coordinates.] Ans. ia2abc 53.

If

+ + y* 5

is the region so xy

Jf

1, prove t h a t

+

e-(=*+a+y2)

dr: dy

2a

=e6 ( e - 1).

Q

[Hint: Let x = U cos a - v sin a, y = U sin a v cosa and choose a so as t o eliminate the sy term in the integrand. Then let U = up cos @, v = bp sin @ where a and b a r e appropriately chosen.] 54.

Prove th at

ix xx.. f .

2

lz

F ( z )dz" = (n l)!

(z - u)n-lF(u)du f o r n = 1 , 2 , 3 ,

. . . (see Prob. 18).

Chapter

IO

Line Integrals, Surface Integrals and Integral Theorems LINE INTEGRALS Let C be a curve in the xy plane which connects points A(a1, b ~ )and 4 x 2 , b2), (see Fig. 10-1). Let P ( x , y ) and Q(x,y) be singlevalued functions defined at all points of C. Subdivide C into n parts by choosing ( n - 1 ) points on it given by ( x I , ~ I )(,x z , Y ~ )., . ., ( X n - 1 , yn-1). Call A X k = X k - X k - 1 and A y k = y k - y k - 1 , k = 1,2, . . . , n where (al,b ~ =) ( x o , ~ o )(a2, , b2) = (Xn, y,J and suppose that points ( t k , v k ) are chosen so that they are situated on C between points ( x k - 1 , yk-I) and ( X k , y k ) . Form the sum

I

I I_

a1

a4

2

Fig. 10-1

The limit of this sum as n + 00 in such a way that all the quantities A x k , A y k approach zero, if such limit exists, is called a line integral along C and is denoted by X P ( z , y ) d ~+ & ( ~ , y ) d z ~ or

s

( a 2 ,b z )

Pdx

(al.bl)

+ Qdy

(2)

The limit does exist if P and Q are continuous (or sectionally continuous) at all points of C. The value of the integral depends in general on P, Q , the particular curve C, and on the limits (al,b ~ and ) (a2, b2). In an exactly analogous manner one may define a line integral along a curve C in three dimensional space as lim

n

n-+aok=l

{ A 1 ( t k , v k ? ck) A X k

=

+ A2(tk9

L A l d z

+

v k 9 c k ) A?/k

A 2 d y

+

+

A3(5k9 vk9 0 at the point ( x o , ~ o ) . By hypothesis dP/dy and dQ/dx are continuous in %, so that there must be some region r containing ( x o , ~ oas ) a n interior point for which d P / d g - dQ/dx > 0. If r is the boundary of T, then by Green’s theorem $Pdx

contradicting the hypothesis that cannot be positive.

+ Qdy

= l f ( z - g ) d x d y

$P d x + Q d y

>

0

7

= 0 for all closed curves in %. Thus dQ/dx - dP/dy

Similarly we can show that dQ/dx-dP/dg cannot be negative, and it follows t h a t i t must be identically zero, i.e. dP/dy = dQ/& identically in q.

12, Let P and Q be defined as in Problem 11. Prove that a necessary and sufficient con-

dition that

s,”a&/& + Pdx

is that aP/ay =

Q d y be independent of the path in % joining points A and B

identically in T .

Sufficiency. If dP/dg = dQ/dx, then by Problem 11, S P d z + Q d y

=

0

A

ADBEA

(see Fig. 10-8). From this, omitting for brevity the integrand Pdx Q d y , we have

+

J+J=o, ADB

BEA

s

ADB

i.e. the integral is independent of the path.

=-J=J BEA

AEB

Fig. 10-8

206

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS

[CHAP. 10

Necessity. If the integral is independent of the path, then for all paths CI and CIin % we have

From this i t follows t h a t the line integral around any closed path in 'Ip ie zero, and hence by Problem 11 that W/dy = dQ/ax.

13. Let P and

Q be as in Problem 11.

(a) Prove that a necessary and sufficient condition that P d x differential of a function +(x,y) is that aP/ay = aQ/ax.

( b ) Show that in such case

iB

+ Qdy

Pdx

=

+ Qdz/

be an exact

lB

d+ = +(B) - +(A)

where A

and B are any two points. (a) Necessity.

If P d x

+ Qdy

= d+ = * d z az

11

+zdy,

a n exact differential, then (1) d+/ax = P , (2) d+/dy = Q. Thus by differentiating (1) and (2) with respect to y and x respectively, aP/dy = d Q / d x since we are assuming continuity of the partial derivatives.

Sufficiency. By Prob. 12, if Pdx

+ Q dy

aPlay = aQ/dx, then

is independent of the path

joining two points. In particular, let the two points be (a, b) and (5, y) and define +(x,y)

Then

=

I

Fig. 10-9

+ Qdy

frrar,Pdx

r / (a. b)

d z + A z , Y)

- +(wd

=

r+Ar. y

Pdx

s

(a, b)

=

+ Q dy

s

r)

(I,

-

Pdx

(a. b)

+ Qdy

( r + A r , U)

Pdx+Qdy

(2,U)

+

Since the last integral is independent of the path joining (x,y) and (x Ax, y), we can choose the path to be a straight line joining these points (see Fig. 10-9) so t h a t d y = O . Then by the mean value theorem for integrals,

Taking the limit as A x

+ 0,

we have d#/dx = P.

Similarly we can show t h a t d+/dy = Q. Thus it follows that

P dx

+ Q dy

=

a+ d x t -dy a+ ay

= a$.

CHAP. 101

207

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS (3.4)

14. (U) Prove that

(6xy2- y3)dx

(1.2)

+ (6x2y- 3xy2)dy

is independent of the path

joining (1,2) and (3,4). ( b ) Evaluate the integral in (a). (a) P = 6xy2-y3, Q = 6x2y-33xy2. Then aP/ay = 12xy-3y2 = aQ/ax and by Problem 12 the line integral is independent of the path.

( b ) Method 1: Since the line integral is independent of the path, choose any path joining (1,2) and (3,4), for example that consisting of lines from (1,2) to (3,2) [along which y = 2, dy = 01 and then (3,2) to (3,4) [along which x = 3, dx = 01. Then the required integral equals f

(24~--WdX

l

I=,

+ '

(64y-9y2)dy = 80

Method 2: Since

ay

= aQ we must have ax

= 236

ay

asb =

(I)

+ 166

6xy2 - y3, (2) asb = 6xey - 3 2 9 .

From ( I ) , 9 = 3xey2-zy3+f(y). From (2), 9 = 3x2yz-xy3+g(x). The only way in which these two expressions for d are equal is if f(y) = g ( x ) = c, a constant. Hence @ = 3x2y2-xy3+c. Then by Problem 13,

S

(3.4)

( 6 %-~us) ~ dx

+ (6x22/- 3~2/*)dy

s

(3,4)

1

(1,2)

d(3x2y2- xy3

+ c)

(1,2)

= 3x59 - xy3+ cl:::::

= 236

Note t h a t in this evaluation the arbitrary constant c can be omitted. See also Prob. 16, Page 115. We could also have noted by inspection that ( 6 ~ -9 y3) ~ dx

+ (6x2y - 3xy') dy

from which i t is clear that

f

15. Evaluate

hypocycloid

+

(x2ycosx

x213

P = x2y cos x

@

+

+

= (62y2dx 6 % ' dy) ~ - (usdz 3 4 d y ) = d(3x2y2)- d(2y3) = d(3x2y2- %y3)

= 3x2yZ-xy3+ c.

2xy sinx

+ y213 = u213.

+ 2xy sin x - y2es, +

-

y2ez)dx

Q = x2 sin x

+ (x2sinx -

2ye")dy

around the

- 2ye".

so that by Problem 11 the line integral Then aP/ay = x2 cos x 2x sin x - 2ye" = aQ/ax, around any closed path, in particular x213 yeI8 = a213, is zero.

+

SURFACE INTEGRALS 16. If y is the angle between the normal line t o any point (x,y,x) of a surface S and the positive x axis, prove that

according as the equation for S is x = f ( x ,y) or F(x,y, x ) = 0. If the equation of S is F(x,y,z) = 0, a normal to S at (x,y,z) is V F = F,i+F,j +F,k. Then

V F * k = (VFI lkl c o s y from which

lsecyl =

F'

'f

IF4

+

+

E

or

F, = dF!

+ F; + Ftcosy

a s required.

In case the equation is z = f(x,y), we can write F ( x , y , z ) = z -f(x,y) F,= -z,, F , = -zy, F , = 1 and we find lsecyl = dl z: 4- g.

+

= 0,

from which

208

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS

ss

17. Evaluate

[CHAP. 10

+

U ( x , y , x ) d S where S is the surface of the paraboloid x = 2 - (x2 y2)

S

above the xy plane and U(x,y,x) is equal to (a) 1, ( b ) x2+y2, ( c ) 32. Give a physical interpretation in each case.

ss

The required integral is equal to U(x, y, x )

d1 + x:

+ xE dx dy

(1)

%

where is the projection of S on the xa plane given by x2 y2 = 2, z = 0.

+

Since x x = -2x, xy = -2y,

(1) can be written

ff U(x, y, x ) d1+ 42' + 4y2dx dy

ss

(2)

(a) If U ( x ,y,x) = 1, (2) becomes

dl+4x2 + 4y2dx dy

%

To evaluate this, transform to polar coordinates ( p , (6). Then the integral becomes

s2= @=o

S p$= o d m p d p d $

~

=

If.

+

( 41~ ~ ) "d$ ~

137 = 3

Physically this could represent the surface area of S, or the mass of S assuming unit density. (b)

If

V(x,y,z) = x2

+ y2, (2) becomes

fs

+

+

(x2 y2)dl 4 8

+ 4y2 dx dy

or in polar coordinates

%

where the integration with respect to p is accomplished by the substitution

q l +4p2 = U.

Physically this could represent the moment of inertia of S about the x axis assuming unit density, or the mass of S assuming a density = x2 y2.

+

(c)

ss

If U(x, y, x ) = 32, ( 2 ) becomes

3xdl -I- 4x2 f 4y2 dx dy

SJ

=

+

3{2 - (x2 y2)} dl

%

%

+ 4x2 + 4y2 dx dy

or in polar coordinates,

Physically this could represent the mass of S assuming a density = 32, or three times the first moment of S about the xy plane.

18. Find the surface area of a hemisphere of radius a cut off by a cylinder having this radius as diameter. Equations f o r the hemisphere and cylinder (see Fig. 10-11) are given respectively by x2 y2 x2 = u2 (or x = d u 2 - x2 - y 2 ) and ( x - U / Z ) ~ y2 = u2/4 (or x2 y2 = ax).

+

+

+ +

Since 2s

we have

=

-X

a2-x2-

Y2

and x, =

-Y

?

CHAP. 101

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS Required surface area

=

2

fs

d -

dx dy

9 Two methods of evaluation are possible.

=

2

ss% d

a2

U

- x2

- y2

209

dx dy

Method 1: Using polar coordinates. Since x2 y2 = a x in polar coordinates is p = U cos +, the integral becomes

+

dp d#

= =

Method 2:

812

2 u L z o,RI2

2u21

d w l a cosmd$ p=o

(1 - sin$) d$

=

(7-2)u2

The integral is equal to

=

2uJa

sin-' d s d x

Letting x = a tan2e, this integral becomes

Note that the above integrals a r e actually improper and should be treated by appropriate limiting procedures (see Problem 78, Chapter 5, and also Chapter 12).

19. Find the centroid of the surface in Problem 17.

The numerator and denominator can be obtained from the results of Prob. 17(c) and 17(a) respec377110 - 111 tively, and we thus have X = -- 13713 130'

sJ

20. Evaluate

A * ndS,

where A = xy i

- x2 j

S

+ (x+ x ) k,

S is that portion of the

plane 2 x + 2 y + x = 6 included in the first octant, and n is a unit normal to S. A normal to S is V(2x and so n =

2i+2j+k

+ 2y + x - 6) = 2i + 2j + k,

-

2i+2j+k.

Then

3 + 22 + l2 {xyi - x2j + (x + x)k} (2i + + k, 2xy - 2x2 + ( x + x ) 3 2xy - 2x2 + ( ~ + 6 - 2 ~ - 2 2 2 / )

d22

A*n

-

3

2~2/-2~~-~-22/+6 3

The required surface integral is therefore

210

[CHAP. 10

L INE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS

JJ (

~ X Y- 2%'-3 x - 2 y

=

+ 6 ),

{J

-

S S ( ~ X Y - ~ X ~ -+ X+- ~ ~ + ~ 3

%

&

)j/l

z$ d x d y

+ + 22 d x d y

( 2 ~ ~ - - 23 2 ~ - ~ - 2 ~ + 6 , )d12 22

11,Lo 3-2

-

=

( ~ X Y- 2~~ - x - 2y

s,lo

(xy2- 2 x 2 y - XY

- y2

+ 6 ) dg dx

+ 62/)1:-=d~ =

27/4

21. In dealing with surface integrals we have restricted ourselves to surfaces which are two-sided. Give an example of a surface which is not two-sided. Take a strip of paper such as ABCD as shown in the adjoining Fig. 10-13. Twist the strip so t h a t points A and B fall on D and C respectively, as in the adjoining figure. If n is the positive normal at point P of the surface, we find th a t a s n moves around the surface i t reverses its original direction when i t reaches P again. If we tried to color only one side of the surface we would find th e whole thing colored. This surface, called a Moebius s t r i p , is a n example of a one-sided surface. This is sometimes called a non-orientable surface. A two-sided surface is orientable.

The DIVERGENCE THEOREM 22. Prove the divergence theorem.

Fig. 10-14 Let S be a closed surface which is such t h a t any line parallel to the coordinate axes cuts S in at most two points. Assume the equations of the lower and upper portions, SIand SZ, to be x = f~(x,y) and x = f z ( x ,y) respectively. Denote the projection of the surface on the xy plane by T. Consider

For the upper portion Sa, dy dx = cos y2 dS2 = k angle y2 with k.

n2

dS2 since the normal

n2

to

S 2

makes a n acute

CHAP. 101

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS

For the lower portion S1, d y d x = -cosyldS1 = - k * n l d S l an obtuse angle y1 with k.

211

since the normal nl to SI makes

Then

so that

Similarly, by projecting S on the other coordinate planes,

Adding (I), (2) and (3),

or The theorem can be extended to surfaces which are such that lines parallel to the coordinate axes meet them in more than two points. To establish this extension, subdivide the region bounded by S into subregions whose surfaces do satisfy this condition. The procedure is analogous to t h a t used in Green's theorem for the plane.

23. Verify the divergence theorem for A = (2x - x)i bounded by x = O , x = l , y=O, y = l , x = O , x = l .

ss

We first evaluate

+ x2y j - xx2 k

taken over the region

A n d S where S is the

S

surface of the cube in Fig. 10-15. Face DEFG: n = i , x = l . Then

=

s,'

i'(2-x)dydz

= 312

Face ABCO: n = -i, x = 0. Then

Jf A ABCO

n dS =

i1 s,' (-xi)

(-i) dy dx

= ~ ' S , ' x d y d z = 112

Fig. 10-15

212

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS Face ABEF: n = j, y = 1. Then

[f A

n dS =

ABEF

l11'

((2x - x)i

Face OGDC: n = -j, y = 0.

Jf A

Then

+ x*j - xx2k} j dx dz

=

l'l'

= 113

x'dxdx

l'l'

n dS =

((22 - z)i - xz*k} (-j) dx dz

OCDC

[CHAP. 10

= 0

Face BCDE: n = k, z = 1. Then

Jf

=

A * n dS

BCDE

l'i'

((22 - l)i

Face AFGO: n = -k, z = 0.

=

JfA*ndS

JJ A - n dS

- xk} * k dxdy =

- xdxdy

= -112

Then

AFCO

Adding,

+ x'yj

=

# +Q + Q + 0 - 9 + 0

S

=

j fV s V * A d V

= 0

~'J'(Zxi-x*yj~.(-k)didy

=

l's,' 1' +

#.

Since

(2 xe - 2x2) dx dy dz

=

11 6

the divergence theorem is verified in this case.

24. Evaluate

ss

r n dS, where S is a closed surface.

S

By the divergence theorem,

= JJJ($i V =

+G a j

$S(E+$+$)dV

*(xi+yj+zk)dV

=

3fJsdV

=

3V

V

V

where V is the volume enclosed by S .

25. Evaluate

ss

xz2dydx

+ (x2y-x3)dxdz + (2xy+y22)dxdy

where S is the entire

S

surface of the hemispherical region bounded by x = Va2-x2-y2 and x = O divergence theorem (Green's theorem in space), (b) directly. ( a ) Since

dy dz = d S cos a, dz dx = d S cos p, dx dy = d S cosy,

+

where A = xz'i (x'y - z3)j drawn unit normal.

+ (2x9 + y%)k

and

n = cos a i

(a) by the

the integral can be written

+ cos /3 j + cos y k,

Then by the divergence theorem the integral equals

where V is the region bounded by the hemisphere and the zy plane.

the outward

CHAP. 101

213

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS By use of spherical coordinates, as in Problem 15, Chapter 9, this integral is equal to xi2

4 &=o

i=o xi2

r2 r2 sin 8 d r de d+

2pa5

=

5

( b ) If S1 is the convex surface of the hemispherical region and S2 is the base ( z = 0 ) , then

ss ss S2

ss

x z 2 d y d z = 0, (2xy -t y2.z)dx dy

=

d z d x = 0,

(x'Y-z')

ii + (2xy

y2(0)}dx dy

=

2xydydx

=

s2

S2

By addition of the above, we obtain

4sas" y=o

22

a2-212

x2j/a2- x2 - x2 dz dx

- z2 dzdy -I4

+4sas"

a=O

r=O

y 2 d a 2- x2 - y2 dy dx

y=O

Since by symmetry all these integrals a r e equal, the result is, on using polar coordinates,

12sas" r=O

y 2 d a 2- x2 - y2 dy dx

g=O

=

12

sxJ2 s' @=o

p=o

p2

sin2+ d

m

dp d+

STOKES' THEOREM 26. Prove Stokes' theorem. Let S be a surface which is such that its projections on the xy, yz and xz planes are regions bounded by simple closed curves, a s indicated in Fig. 10-16. Assume S to have representation z = f ( x , y) or x = g ( y , z ) or y = h ( x , z ) , where f , g , h a r e single-valued, continuous and differentiable functions. We must show that

ss

( V X A ) * nd S

S

=

ff

[V

X

( A ~ i + A z j + A a k ) ]* n d S

S

where C is the boundary of S. Consider first

ss S

[ V X (Ali)] * n dS.

Fig. 10-16

=

2pa5 5

0

214

[CHAP. 10

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS

V x (Ali) =

Since

- *n*k)dS

[V X ( A l i ) ] - n dS =

aY

If z = f(x,y) is taken as the equation of S, then the position vector to any point of S is r = z i 4az ar y j + z k = x i + y j + f ( x , y ) k s o t h a t dr = j + - k = j + d f k . But isavectortangenttos and thus perpendicular to n, so that

ay

ay

ay

ay

Substitute in (1) to obtain

or

[V X ( A l i ) ] * ndS =

Then

ff

-($ + * * ) n * k

[V X (Ali)] * n dS

a2

ay

dS

= Js-gdxdy

s

8

where CIp is the projection of S on the xy plane. By Green's theorem for the plane the last integral equals

9;,

F d x where

r

is the boundary of T . Since at each point ( x , ~of ) I' the value of

F is the

same as the value of A1 at each point (x,y,z) of C, and since dx is the same for both curves, we must have $Fdx = $Aldx or

JJ[VX(Ali)]*ndS

= IAldx

8

Similarly, by projections on the other coordinate planes,

Thus by addition,

s s ( V X A ) - n d S = $A*& 8

c

The theorem is also valid for surfaces S which may not satisfy the restrictions imposed above. Sa, . . .,s k with boundaries CI,Ca, . .,Ck which For assume that S can be subdivided into surfaces SI, do satisfy the restrictions. Then Stokes' theorem holds for each such surface. Adding these surface integrals, the total surface integral over S is obtained. Adding the corresponding line integrals over cl,C,, , . .,Ck, the h e integral over c is obtained.

.

CHAP. 101

L INE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS

215

+

27. Verify Stokes’ theorem f o r A = 3y i - xx j yx2k, where S is the surface of the paraboloid 22 = x 2 + g 2 bounded by x = 2 and C is its

boundary.

+

The boundary C of S is a circle with equations x2 y2 = 4, x = 2 cost, y = 2 sin t, x = 2, where 0 5 t < 2n. Then x = 2 and parametric equations

$A*dr

= i 3 y d x - xxdy

+ yx’ddz

= 1 1 3 ( 2 sin t)(- 2 sin t) d t - (2 cos t)(2)(2 cos t) d t = 1 2 T ( 1 2sin2t

+ 8 cos2 t) dt

i Also,

VXA

=

j

k

ax ay 3y -xx

= ( x 2 + x)i - ( x + 3)k

ax

Then

ss

( V X A ) * n dS

=

S

Fig. 10-17

U22

V(x2fy2-22)

n =

and

= 20n

I V(X2+ Y2 - 22) I

ss T

=

xi + y j - k @-qjq3 ’

ss

( V x A ) - n - dx dy In*kl

+ + x + 3) dx dy

(xx2 x2

%

I n polar coordinates this becomes

Aodr = 0

28. Prove that a necessary and sufficient condition that curve C is that v x A = 0 identically. Sufficiency. Suppose V X A = 0.

f o r every closed

Then by Stokes’ theorem

fA*dr

= JJ(VXA)*ndS

= 0

S

Necessity. Suppose

$

A d r = 0 around every closed path C, and assume V X A # 0 a t some point P.

Then assuming V X A is continuous there will be a region with P as a n interior point, where V X A # 0. Let S be a surface contained in this region whose normal n at each point ha s the same direction a s V X A, i.e. V X A = an where (Y is a positive constant. Let C be the boundary of S. Then by Stokes’ theorem $A-dr

= Js(VXA)*ndS

which contradicts the hypothesis t h a t

It follows t h a t

VXA = 0

= aJln*ndS

4;,

0

A d r = 0 and shows t h a t V X A = 0.

is also a necessary and sufficient condition for a line integral

, f p z A * d r to be independent of the path joining points PI and P2. p1

>

S

S

216

LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS

29. Prove that a necessary and sufficient condition that Sufficiency. If A = V+,then V Necessity. If V

X

X

vXA

= 0 is that A = v+.

A = V X V + = 0 by Prob. 80, Chap. 7, Page 168.

$ A *dr = 0

A = 0, then by Prob. 28,

s

around every closed path

s

is independent of the path joining two points which we take as (a, b, c) and (r,m,

+(X,Y,Z)

=

Then

[CHAP. 10

(r.II, x)

L)

A*dr =

Aldx

(a. b, e)

(a. b, c )

( r + A r , U,

A dr

2). Let us define

+ Asdy + A ~ d z

I)

Aids

(r. v. x )

(2, y,

and

+ Azdy + Asdz

Since the last integral is independent of the path joining (2,y, x) and (xi-As,y, z), we can choose the path to be a straight line joining these points so that dy and dz are zero. Then

+(x + As,Y, 4

- +@,v , 4

=

Ax

S' As

(=+A',

U. x )

=

Aldx

O<e 0 but is not an integer, the series converges (absolutely) for -1 5 x I1. (c) If -1 < p < 0, the series converges for -1 < x S I. ( d ) If p S -1, the series converges for -1 < x < 1.

For all p the series certainly converges if -1 < x < l .

SPECIAL TOPICS 1. Functions defined by series are often useful in applioations and frequently arise as solutions of differential equations. For example, the function defined by Jp(z)

= -

&il 2

- 2 ( 2X2 p+2)

+

2 * 4 ( 2 p +x42 ) ( 2 p + 4 )

- ...}

(-1)s (x/2)p+2n

n=O

n!(n+p)!

+

+

is a solution of Bessel’s differential equation x2yr/ xy’ ( x 2 - p 2 ) y = 0 and is thus called a Bessel function of order p . See Problems 46, 106-109. Similarly, the hypergeometric function is a solution of Gauss’ differential equation x(1- x)y” aby = 0. These functions have many important properties.

+ { c - (a + b + 1)x)y’ -

2. Infinite series of complex terms, in particular power series of the form

2 a&, QI

n=O

where x = x + i y and a, may be complex, can be handled in a manner similar to real series. Such power series converge for Izl< R, i.e. interior to a circle of convergence x 2 + y 2 = R2, where R is the radius of convergence (if the series converges only for x = 0, we say that the radius of convergence R is zero; if it converges for all x, we say that the radius of convergence is infinite). On the boundary of this circle, i.e. IxI=R, the series may or may not converge, depending on the particular x . Note that for y = O the circle of convergence reduces to the interval of convergence for real power series. Greater insight into the behavior of power series is obtained by use of the theory of functions of a ccmplex variable (see Chapter 17).

3. Infinite series of functions of two (or more) variables, such as be treated in a manner analogous to series in one variable. can discuss power series in x and y having the form a00

+ (arox + ao1y) + (u20x2 + allxy + aony2) + -

X

U&, n=l

y) can

In particular, we (18)

using double subscripts for the constants. As for one variable, we can expand suitable functions of x and y in such power series using results of Chapter 6, Page 109 and showing that the remainder R,+ 0 as n+ QO. In general, such power series converge, inside a rectangular region 1x1 < A , Iy)< B and possibly on the boundary.

4.

Double Series.

Consider the array of numbers (or functions)

.. . .. . . ..

U11 U12 U13

UP1 U22 U23 U31 U32 U33

Let

233

INFINITE SERIES

CHAP. 111

Smn

=

9 2 %,

p=l q=l

\

.

. . . . . . . .

be the sum of the numbers in the first m rows and first n

columns of this array. If there exists a number S such that lim say that the double series

c m

m-r

m

Smn

x

= S, we

n 4m

p=1 q=1

upq converges to the sum S; otherwise it diverges.

Definitions and theorems for double series are very similar to those for series already considered.

+

+

Let P , = (1 ul)(l+ u2)(1+u3) .. . (1 un) denoted by

5. Infinite Products.

fi(1 +

uk)

k=l

where we suppose that uk # -1, k = 1,2,3, . . . . If there exists a number P Z 0

+

such that lim P n = P , we say that the infinite product (1 ul)(l+u2)(1+ us) . . . = n e w

n(1 + uk),or briefly n ( l + W

uk),

k=l

converges to P;otherwise it diverges.

+

If n ( l + I u k l ) converges, we call the infinite product n(l u k ) absohtely convergent. I t can be shown that an absolutely convergent infinite product con-

verges and that factors can in such cases be rearranged without affecting the result. Theorems about infinite products can (by taking logarithms), often be made to depend on theorems for infinite series. Thus, for example, we have the

+

Theorem. A necessary and sufficient condition that n(l uk) converge absolutely is that k k converge absolutely. 6.

Summability.

Let

S1, &S3,

Sl+S2

. ..

be the partial sums of a divergent series Xun.

, . . . (formed by taking arithmetic means If the sequence SI,7, 3 Sl+S2+S3

..

of the first n terms of S1, S2, S3, .) converges to S, we say that the series Xu, is summable in the C6saro sense, or C-1 summable to S (see Problem 51). If 2% converges to S, the C6saro method also yields the result S. For this reason the C6saro method is said to be a regular method of summability. In case the C6saro limit does not exist, we can apply the same technique to

, . . . . If the C-1 limit for this sequence 3 exists and equals S, we say that Zuk converges to S in the C-2 sense. The process can be continued indefinitely. the sequence SI, Sl+S2

2

Sl+S2+s3

'

7. Asymptotic series. Consider the series

S(x) =

and suppose that Sn(x) = uo

U0

+ -x + 2 x + a1

+ -x + 3 x + a1

a2

are the partial sums of this series.

U2

...

'..

+ -an-$ +

+X" an

* * .

n = 0 , 1 , 2 ...

(19)

(20)

234

[CHAP. 11

INFINITE SERIES

If Rn(x) = f ( x ) - S,(x), where f ( z ) is given, is such that for every n lirn xnRn(x) = 0 IZI

-

-

CO

then S ( x ) is called an asgmptotic expansion of f ( z ) and we denote this by writing

f @ ) S@).

In practice, the series (19) diverges. However, by taking the sum of successive terms of the series, stopping just before the terms begin t o increase, we may obtain a useful approximation for f ( x ) . Various operations with asymptotic series are permissible. For example, asymptotic series may be multiplied or integrated term by term to yield another asymptotic series.

Solved Problems CONVERGENCE and DIVERGENCE of SERIES of CONSTANTS 1 -+-+-+3.5

1

1 1. (a) Prove that 1.3 its sum.

(

Since lim Sn = lim - 1 -+:2 m+ w

n+ m

... =

5-7

1>

=

2

1 n=l(2n- 1)(2n 1)

i,

+

converges and ( b ) find

the series converges and its sum is

4.

The series is sometimes called a telescoping series since the terms of S,, other than the first and last, cancel out in pairs.

2.

(a) Prove that Q

+ (p)2 + (Q)3 + . . . = s, = QSm = &Sn =

Subtract:

9 (8)"

n= 1

.E + (8)s + ( # ) a + (#)*

.-*

converges and ( b ) find its sum.

+

($)U

+ (#)' + ..* + (3). + (Q)*"

Q - (3)""

sn = 2{1 - (Q)"}

or

Since lim S m = lim 2{1 - (#)"} = 2, the series converges and its sum is 2. n-

w

Another method:

,-00

Let a = Q ,

3. Prove that the series

T=#

in Prob. 25 of Chap. 3; then the sum is a / ( l - T ) = Q/(l-Q) = 2.

4 + 3 + 2 + 6 + . .. =

2- n+ 1 diverges.

nZln

n lim un = lim - = 1. Hence by Problem 26, Chapter 3, the series is divergent. newn+l

n+ 00

CHAP.111 4.

235

INFINITE SERIES U,, =

Show that the series whose nth term is The fact that lim

U,,

U+"

d a - fi diverges although

= 0 follows from Problem 14(c), Chapter 3.

lim un = 0.

n+m

Sn = u l + u a + . . - + u , , = ( ~ - ~ ) + ( ~ - ~ ) + . . . + =( 6~F ~ i - f- i -~ ) Then S, increases without bound and the series diverges. This problem shows that lim U,, = 0 is a necessary but not m&knt condition for the conNOW

%+

oo

vergence of Zun. See also Problem 6.

COMPARISON TEST and QUOTIENT TEST 5. If 0 Suns Vn, n = 1,2,3,. . . and if B V n converges, prove that Bun also converges (i.e. establish the comparison test for convergence).

+ + +

+ +

va Let S n = ui+ UI Un, Tn = ~i Since BV, converges, lim Tn existEl and C n+

Then

Sn

m

1..

Vn.

X ~ I ~ T, S say.

= ur+%+...+un S vl+vr+..-+vn

Also, since

Vn I0,

T,, S T.

T or O S S , , S T .

4

Thus S,,is a bounded monotonic increasing sequence and must have a limit (see Chap. 3), i.e. converges.

6.

Using the comparison test prove that 1 1 2 4

We have

*+* *+*+*++

2 *+$ =

* *

+ + + + + + +=

h

+ + & + & + . . - + &2 etc. Thus to any desired number of

" 1 + + + . .. = ,=In - diverges.

&+&+&+...+&(8terms)

tell118,

1+(*++)+(*+*+*++)+

.*.

2

=

4

& + * + * + ...

Since the right hand side can be made larger than any positive number by choosing enough terms, the given series diverges. " 1 By methods analogous to that used here, we can show that 3 -, where p is a constant, diverges np

n=t

if p S 1 and converges if p > 1. This can also be shown in other ways [see Problem 13(a)].

7. Test for convergence or divergence Since I n n

' (14w

3n+3

...

+(

1*4*7-..(3n-2) 3.609...(3n)

+

* . W .

= 1. However, by Raabe's test,

and so the series converges.

24. Test for convergence

(;J+(=J+(=J+ 2.4 204.6

The ratio test fails since

However, using long division,

80

that the series diverges by Gauss' test.

...

+(

1.3.5.-.(2n- 1) 2-4.6...(2%)

+

. a * .

= 1. Also, Raabe's test fails since

242

INFINITE SERIES

[CHAP.11

SERIES of FUNCTIONS 25. For what values of x do the following series converge?

x" - 1

=n.3"* Assuming x # 0

(a)

Then the series converges if

(if x = 0 the series converges), we have

If x = 3 the series becomes If x = - 3

1x1 > 1. If -

M3 < 1, and diverges if

3

3

= 1, i.e.

x

= 2 3 , the test fails.

" 1 1 " l 2= 5 8 - which diverges. n=13n ,=in

the series becomes

"

(-1)n-1

2

1

7

-

A5

which converges.

- 3 n=l

Then the interval of convergence is -3 d x < 3. The series diverges outside this interval. Note that the series converges absolutely for - 3 < x < 3 . conditionally.

(b)

Proceed as in part (a) with

U,,=

= lim 1-00

(-I)*-1 xlr-l (2n - l ) !

(2n

.

(2%- I)!

A t z=-3

the series converges

Then

+ 1)(2n)(2n- l ) !

%L

=

lim

1-00

+ 1)(2n) = o 5%

(2n

Then the series converges (absolutely) for all x, i.e. the, interval of (absolute) convergence is -CQ

In 1x1

since division by In 1x1 (which is negative since 1x1 < i)reverses the sense of the inequality. lne In But if 1x1 -> -2- = In14 In(+> of x, the series is uniformly convergent in the interval. In 1x1 d In(+) and n In this case 1x1 5 i, uniformly convergent in -3 d x d

4.

Reasoning similar to the above, with vergent in -.99 S x 5.99.

4

N. Thus since N is independent

In6 In >L 2= N, In(+) In1 . 1

so that the series is also

replaced by .99, shows that the series is uniformly con-

The arguments used above break down in this case, since

can be made larger than any In 1x1 positive number by choosing 1x1 sufficiently close to 1. Thus no N exists and it follows that the series is not uniformly convergent in -1 < x < 1. Since the series does not even converge a t all points in this interval, it cannot converge uniformly in the interval.

29. Discuss the continuity of the sum function S ( x ) = limS,(x) of Problem 27 for the n-oc interval 0 5 x 5 1. If 0 5 %< 1, S(x) = lim Sn(z) = lim (1 - 2 " ) = 1. *Moo

n em

If x = l , Sn(Z) = 0 and S(z) = 0. Thus

S(z) =

points in 0 5 x < 1.

i

1 ifOdz N, since Z M , converges. Since N is clearly independent of x, we have IRn(X)j < c for n > N, and the series is uniformly convergent. The absolute convergence follows at once from the comparison test.

32. Test for uniform convergence:

(a)

I

cos nx 1 1 7S 2 =

Mn. Then since EM, converges (p series with p = 4 > l), the series is uniformly (and absolutely) convergent for all x by the M test.

( b ) By the ratio test, the series converges in the interval -1 5 x 5 1, i.e. 1x1 5 1. F o r all x in this interval,

converges. (c)

(-1

151= $ d n~/". 1 Choosing

M n

= ~J/L, 1 we see that HMn

Thus the given series converges uniformly for -1 d x d l by the

-. n

sinnx d 1

M test.

However, BM,, where Mn =--, 1 does not converge. The M test cannot be used

in this case and we cannot conclude anything about the uniform convergence by this test (see, however, Problem 121).

IAI 2 ,

(4 n + x

S

1

1 and S 7 converges. Then by the M test the given series converges uniformly n

f o r all x.

33. If a power series Xanxn converges for x = $0, prove that i t converges (a) absolutely in the interval 1x1 < 1x01, (b) uniformly in the interval 131S lxll, where 1x11< 1 ~ 0 1 . (a) Since Band converges, lim Unxg = 0 and so we can make I&xo"l < 1 by choosing n large n-

enough, i.e. 1al.

00

1 N.

Ixol"

Then

246

[CHAP. 11

INFINITE SERIES

Since the last series in (1) converges for lzl < 1 ~ 0 1 , it follows by the comparison test that the first series converges, i.e. the given series is absolutely convergent.

( b ) Let

M n

IzI 5

='zlln Ixap *

Then 9 M n converges since 1x11< Iso~.

so that by the Weierstrass

As in part

(a),

1kz.l < Mn

for

M test, Ha,zn is uniformly convergent.

It follows that a power series is uniformly convergent in any interval within its interval of convergence.

THEOREMS on UNIFORM CONVERGENCE 34. Prove Theorem 6, Page 228. We must show that S(z) is continuous in [a, b ] . NOW S(Z) =

+ R n (z), that S(Z + h) - S(Z) = SO

Sn(Z)

where we choose h so that both z and z

S(Z

+ h) = Sn + h) + Rn(z + h) (Z

Sn(s

+ h ) - S~(Z) +

Rn(Z+

and thus

h) - Rn(z)

(1)

+ h lie in [a,b] (if z = b, for example, this will require h < 0).

Since Sn(z) is a sum of a finite number of continuous functions, it must also be continuous. Then given c > 0, we can find 6 so that

+ h) - S,(z)

I S.(z

I
N we have from ( I ) ,

e>O

(1)

we can choose N such that

Also, for x = 1, lRu(x)l = JRnI< c for n > N. Thus IR,(x)) < c for all n > N,where N is independent of the value of x in 0 S x 5 1, and the required result follows. Extensions to other power series are easily made..

43. Prove Abel's limit theorem (see Page 230). As in Problem 42, assume the power series to be Then we must show that

lim 2 4

1-

00

3 akxk

k=O

=

2 Q3

k=O

akxk, convergent for O S x S 1.

00

3 ak.

k=O

This follows at once from Problem 42 which shows that sakxk is uniformly convergent for z = 1.

0 S z 5 1, and from Problem 34 which shows that 9 a k X k is continuous at

Extensions to other power series are easily made.

+

xs x 5 2 7 tan-lx = x - -- - + . . . where the series is uniformly con44. (a) Prove that 3 5 7 vergent in -1 S x S 1. 7T 1 1 1 (b) Provethat - = I - - + 3 --- 7 f - . 4 (a) By Problem 26 of Chapter 3, with T = -x* and a = 1, we have -1 1-2' - 1 - x* + x4 S o + ... - l < x < l Integrating from 0 to x, where -1 < x < 1, yields

+

-

using Problems 33 and 36. Since the series on the right of (2) converges for x = + 1 , it follows by Problem 42 that the series is uniformly convergent in -1 S z S 1 and represents tan-'$ in this interval. ( b ) By Problem 43 and part (a), we have

250

INFINITE SERIES

45. Evaluate

[CHAP. 11

dx to 3 decimal place accuracy.

Then if u = - x ' ,

e-$ = 1 - x* + x4 U - -x6+ - - 3!

Thus ~ - e -- ~ 1--+-x* X' 2!

x4

3!

4!

5!

+ ..., -* < z < m.

2(1+ 28- ... . 4!

5!

Since the series converges for all x and so, in particular, converges uniformly for 0 d z 5 1, we can integrate term by term to obtain

=

dx

x--+--- x 3

x5 5*3!

3*2!

9*5!

...It

0

1 +---1 +--... 1 3*2! 5 * 3 ! 7 * 4 ! 9 * 5 ! 1 - 0.16666 0.03333 - 0.00595 0.00092

- I-=

x7 + x@ -

7*4!

+

+

-

* * -

=

0.862

Note that the error made in adding the first four terms of the alternating series is less than the fifth term, i.e. less than 0.001 (see Problem 15).

MISCELLANEOUS PROBLEMS Page 232, satisfies Bessel's differential equation 46. Prove that y = &(x) defined by (M), x2g" xy' (x2 - p2)g = 0

+

+

The series for Jp(x) converges for all x [see Problem 106(a)], Since a power series can be differentiated term by term within its interval of convergence, we have for all x,

v

=

y'

=

y"

=

Then, (x'-p')y

Adding, z'y"

=

xy'

=

xly"

=

+ xy' + (z'-p')y

=

(-1)s n=O

(p

+ 2n)(p + 2n - 1) xP+*"-' 2p+'r n! (n+ p ) !

INFINITE SERIES

CHAP. 111

47. Test for convergence the complex power series

For 1x1 = 3, the series of absolute values is convergent and thus convergent for 1x1 = 3.

zl

251

.3n- .

p - 1

m

n&3

lZp-1

p=

O0

84n

n=1

so that the series is absolutely

Thus the series converges within and on the circle 1x1 = 3.

48. Assuming the power series for e2 holds for complex numbers, show that eiz = cosx + i s i n x Letting z = iz in &

ex = 1

z' za +x +++ - - -we , have 2! 3!

= l+&+j-j-+llc+... ,pxi = cosz

Similarly,

e-cf

+ i sinz =

+ +- + + . . . +:

Letting

00

f(s) =

XS +-... 6!

cos z - i sin s. The results are called Euter's identities.

1 1 1 49. Prove that lim ( 1 2- 3 4n-

)+ i(s-5

=

*a 3!s

- In n

l/x in ( I ) , Problem 11, we find

-21+ -31+ - +14 . . . + - M1

1. . . + 1 I + -1+ - 1 +-+ 2 3 4 M-1

p 1nM

from which we have on replacing M by n, 1 1 1 1 1 S l+-+-+-+...+--lnn S 1 n 2 3 4 n 1 1 1 Thus the sequence Sn = 1 ... n1 I n n is bounded by 0 and 1. 3 4

++, -+ -+

Consider S,+I- S, =

+

- In n+l

+--

(9).

1 By integrating the inequality -5 5 L. with n+l-x-n

respect to x from n to n 1, we have Sl*(--)Sn+l 1 or --1 n+l n n+l i.e. S n + l - Sn 5 0, so that S , is monotonic decreasing.

1 n

- -11 n ( y ) n + l n+l

0

Since S , is bounded and monotonic decreasing, it has a limit. This limit, denoted by y , is equal y is rational or not.

to 0.677216.. . and is called Euler's conatant. It is not yet known whether

50. Prove that the infinite product

m

k=

+

(1 ur), where uk > 0, converges if

m

k=l

uk converges.

By equation (I) of Problem 28, Chapter 4, l + z d d for z>O, so that

n (1+uk) -- (1+ ut)(l+ a)... (1 + U,,) S d i e"s ... e'n = e'l+''+' u1 + UI + - -. converges, it follows that P, is a bounded monotonic increasing sequence and (I

Pn =

Since

k=l

so has a limit, thus proving the required result.

*

+mm

262

INFINITE SERIES

51. Prove that the series 1- 1

+ 1 - 1 + 1 - 1 + ...

The sequence of partial sums is 1,0,1,0,1,0,. .

Then

..

Sl+S¶- 1 + 0 - 1 S1+SnfSs S I =1, -- -2 ' 2 2 3

[CHAP. 11

is C - 1 summable to 1/2.

2 --1 + 0 + 1 -3

3'

' * * '

...,

the nth term being Continuing in this manner, we obtain the sequence 1, 4, Q, +,9, *, if n is even Thus lim T, = & and the required result follows. Tn = $2n - 1) if n is odd ' 00

{

(1-

52. (a) Prove that if x

> 0 and p > 0,

(b) Use (a) to prove that

\

i.e. the series on the right is an asymptotic expansion of the function on the left. (a) Integrating by parts, we have

e-'

Similarly, Zp+l = xp+l - ( p

+ 1) ZP+¶ so that

e

By continuing in this manner the required result follows.

Then

R,(x) =

f(z)

- Sn(z) = (-l)n+lp(p+l)...(p+n) sO-$&du. t

.s since i"e-"du

5 i"e-"du

= 1. Thus

P(P + W . ( P %p+r+l

+4

NOW

CHAP. 111

253

INFINITE SERIES

P(P+ 1 ) . . * ( p + n ) = 0 lzlP and it follows that lirn xuR,(z) = 0. Hence the required result is proved. lirn IsnRn(X)I S

I=l-,-

lirn

PI-,

aJ

I=l-+aJ

lim n-00

I*I un

=

Q)

and the series diverges for all x by the ratio test.

Supplementary Problems CONVERGENCE and DIVERGENCE of SERIES of CONSTANTS 1 53. (a)Prove that the series 3.7 7.11 11 16 + ... ( b ) find its sum. Ans. (b) 1/12

- 5

+-

+

n=l

1 (4n - 1)(4n 3)

+

converges and

54. Prove that the convergence or divergence of a series is not affected by (U) multiplying each term by the same non-zero constant, (b) removing (or adding) a finite number of terms.

55. If Bun and Bv, converge to A and B respectively, prove that C(un 56.

Prove that the series

# + ( f ) 2 + (#)3

57. Find the fallacy: Let S

S = (1 - 1)

+

-

Vn)

converges to A

+ B.

= 2(#)" diverges.

= 1-1+1-1+1-1+

+ (1 - 1) + (1 - 1) +

+

... .

= 0. Hence 1 = 0.

Then S = 1 - ( 1 - 1 ) - ( 1 - 1 ) -

... =

1 and

COMPARISON TEST and QUOTIENT TEST 58. Test for convergence:

Ans. 59.

(U)conv.,

(b) div., (c) div., (d) conv., (e) div., (f) conv.

Investigate the convergence of

(a)

O0

ncl

+

5 4n'+lOns

4nP+6n-2 (b) n(n2 l)'/*' n=l

*

Ans. (a) conv., (b) div.

60. Establish the comparison test for divergence (see Page 226). 61.

Use the comparison test to prove that " 1 converges if p > 1 and diverges if p S 1, ( b ) 2 " t7 an-'n (a) 2 diverges, (c) n=l np n=l

62. Establish the results (b) and ( c ) of the quotient test, Page 226. 63. Test for convergence:

Am. (a)conv., (b) div., (c) div., (d) div.

5 !2"f converges.

r=1

254

[CHAP. 11

INFINITE SERIES

64. If BIG, converges, where un 2 0 f or .n

> N, and if lim nun exists, prove that lirn nu,, = n-

65. (a)Test for convergence

"

1

?+=.

0.

m

( b ) Does your answer to (a.) contradict the statement about the

p series made on Page 225 t h a t B l/nP converges for p > l?

67. Prove t h a t

n-,

m

2where p is a constant, n(ln n)' '

Ans. (a) div.

(a)converges if p

Ot

n=z

> 1 and

(b) diverges if p 5 1.

1 5 59 < 2" ,s < 4.

68. Prove t h a t

r=l

69. Investigate the convergence of

A m . conv.

+

QnS/' Q S

70. (a) Prove t h a t

m rrl

v. n +1

fi + fi+ 6+ + 6 I 3n3/s4- nils - 8fi + fi + fi + . - -+ G,giving the maxirnum 6 . .

(b) Use (a) to estimate the value of

+

error.

+

Show how the accuracy in (b) can be improved by estimating, f o r example, 6 0 6 1 -.. + CO5 and adding on the value of fi fi fi computed to some desired degree of accuracy. A m . ( b ) 671.6 rt 4.5 (c)

+ + +

ALTERNATING SERIES 71. (e)

5 (-l)nfi Inn

n=z

72.

An8. (a) conv., ( 6 ) conv., (c) div., (d) conv., (e) div.

'

a (-1)" (a)What is the largest absolute error made in approximating the sum of the series ]E 2"(# 1- 1) by the sum of the first 6 terms? An8. U192

( b ) What is the least number of terms which must be taken in order t h a t 3 decimal place accuracy will result? Ans. 8 terms 73.

( b ) How many terms of the series on the right are needed in order to calculate S to six decimal place An8. ( h ) at least 100 terms accuracy?

ABSOLUTE and CONDITIONAL CONVERGENCE 74.

Test for absolute o r conditional convergence:

AnS. (a) abs. conv., (h) cond. conv., ( c ) cond. conv., ( d ) div., (e) abs. conv., (f) abs. conv. 75.

Prove t h a t

2 cosn-m --r--+ converges absolutely f o r all a! + g,

n=1

')1

real z and a.

CHAP. 111

255

INFINITE SERIES

+ + - + - - - converges to S, prove that the rearranged series 1+ Q - + + Q + 3- 4 + Q + + -.- = QS. Explain. [Hint Take 1/2 of the first series and write it as 0 + + + 0 -$ + 0 + Q + -..; then add term by

76. If 1-

Q

term to the first series. Note that S = In 2, as shown in Problem 96.1

77. Prove that the terms of an absolutely convergent series can always be rearranged without altering

the sum.

RATIO TEST 78. Test for convergence:

Am. (a) conv. (abs.); (b) conv., (c) div., (d)conv. (abs.), (e) div. 79. Show that the ratio test cannot be used to establish the conditional convergence of a series. 80. Prove that (a)

"2 nl 3 converges n"

,,=I

and (b) lim n-oo

nf = 0. n"

MISCELLANEOUS TESTS 81. Establish the validity of the nth root test on Page 226. 82. Apply the nth root test to work Problems 78(a),(c), (d) and (e). 83. Prove that

Q + (3)' +

+ (3)' + 1

84. Test for convergence: (a)5 Am. (a) div., (b) conv.

+ 1.4

+ (Q)"4-

(Q)5

1.4.7 6

converges.

0 . .

+ -..,

2 2.5 2.5.8 (b) -+ 7 9 9 12 9.12.16 +

> a, prove that a a(a+d) a ( a + d ) ( a + 2d) b b(b d ) b(b d)(b 2d) converges if b - a > d, and diverges if b - a 5 d.

85. If a, b and d are positive numbers and b

-+- +

+

+

+

+

...

SERIES of FUNCTIONS 86. Find the domain of convergence of the series:

Am. (a) -1 5 z S 1, (b) -1 87. Prove that

O0

< z S 3,

(e)

all x f 0, ( d ) x

> 0,

(e)

z d 0.

1 3 6.-.(2n- 1)zn converges for -1 S z < 1. 2 4 6.. .(2n)

UNIFORM CONVERGENCE 88. By use of the definition, investigate the uniform convergence of the series

5 [l+ (n - l)x][l + nz] X

n-1

[Hint: Resolve the nth term into partial fractions and show that the nth partial sum is S,(z) = 1 l-1+nz*l A m . Not uniformly convergent in any interval which includes z=O; uniformly convergent in any other interval.

INFINITE SERIES

266

89. Work Problem 30 directly by first obtaining

[CHAP. 11

S,(x).

90. Investigate by any method the convergence and uniform convergence of the series:

Ans. (a)conv. for 1x1 < 3; unit. conv. for 1x1 _ I T not unif. conv. for x L 0, but unif. conv. for

91. If

F(d =

sin nx

SJj-

< 3.

(b) unif. conv. for all

x 2 T > 0.

i'(n cos22

93. Prove that F(x) =

(c) conv. for X L 0;

, prove t h a t

(a)F ( x ) is continuous for all x, (b) lim F(x) = 0, (c) F'(x) = 92. Prove that

1c.

+

r'

cos42 3 5

sinnx

,

8 4 0

cw6x 5 7

+ ..)dx

OD

,,=I

na

= 0.

has derivatives of all orders for any real

94. Examine the sequence un(x) =

is continuous everywhere.

1c.

x,U, n = 1,2,3, . . ., for uniform convergence. 1

+

95.

POWER SERIES 96. (a)Prove that In (1 +x) = x

--+--+ .... (b) Prove that In2 = 1 - + Q - a + ..+. 1 = 1 - x + - + -.[Hint: Use the fact that and integrate.] l+x lxa 1 . 3 ~ ' 1.3*6x7 Prove that sin-lx = x + 2 3 + -- 4- -- + ..., - 1 s x s 1 . 2.4 6 2.406 7 2'

XI

2'

2'

97.

98. Evaluate (a)l " ' e - 9

dz, ( b )

1'

-:OS

XI

dx to 3 decimal places, justifying all steps.

A m . (a)0.461, ( b ) 0.486 99. Evaluate (a)sin 40°, ( b ) cos 66O, (c) tan 12O correct to 3 decimal places.

Ant?. (a)0.643, ( b ) 0.423, (c) 0.213

100. Verify the expansions 4, 6 and 6 on Page 231. 101. By multiplying the series for sin x and cos 2, verify that 2 sin x 102. Show that

.*.),

eecnO =

103. Obtain the expansions

(a) tanh-lx

f(x) =

I

e-"L'

X P O

= x + -x3+ 3

--oo

<x
0,

prove that a, = b n for n = 0,1,2, exists, the expansion is unique. 112. Suppose that lim

... .

Jp(x)tP.

n=

hx" =

= 1,2,3,

fl = L.

. .. .

( b ) Use (a)to show that if the Taylor expansion of a function

Prove that Bun converges or diverges according as L < 1 or L > 1.

If L = 1 the test fails.

113. Prove that the radius of convergence of the series BGX' can be determined by the following limits,

-

m'

when they exist, and give examples: (a) lirn 1 5 1 ,(b) lim a-+w & + I

n-)"

(c)

- 1 lim -

r-+w

qzj*

114. Use Problem 113 to find the radius of convergence of the series in Problem 22. 115. (a) Prove that a necessary and sufficient condition that the series Su,,converge is that, given any c > 0, we can find N > 0 depending on e such that ISp-SqI< e whenever p > N and q > N, where S k = ul+ul+ *.* + u k . " ' n (b) Use (a) to prove that the series converges. (c)

1 How could you use (a)to prove that the series " diverges? r = n~

[Hint: Use the Cauchy convergence criterion, Page 43.1 116. Prove that the hypergeometric series (Page 232) (a)is absolutely convergent for 121< 1, (b) is divergent for 1x1 > 1, (c) is absolutely convergent for l l ~ l = 1 if U b - c < 0, (d) satisfies the differential equation x(1- x)y" {c (a b 1)x)y' - a b y = 0.

+

+ - + +

F(a,b;c;x) is the hypergeometric function defined by the series on Page 232, prove that (a) F(--p,l; 1; -x) = (1 x)p, (b) zF(1,l;2; -2) = ln(1 x), (c) F(&,+;#; z*)= (sin-' x ) / x .

117, If

+

+

118. Find the sum of the series S ( x )

[Hint: Show that S'(x) = 1

= z

+ xS(z)

x5 +1.3 + m+ Xa

.*.*

and solve.]

Ans.

&Ir

l'e-s/adx

[CHAP. 11

INFINITE SERIES

258 119. Prove that

120. Establish the Dirichlet test on Page 228.

121. Prove that

&E !

r=i

is uniformly convergent in any interval which does not include 0, f a , -)-2n,

n

. . ..

[Hint Use the Dirichlet test, Page 228, and Problem 94, Chapter 1.1 122. Establish the results on Page 232 concerning the binomial series.

[Hint: Examine the Lagrange and Cauchy forms of the remainder in Taylor's theorem.] 123. Prove that 124. Prove that

OD

s=l

(-1)n-1

converges n + xi

1 1 -4

1 + ... + -71 - 10

125. If x=yc', prove that y =

-

5 (-1)n-tn !

a 1 3 6 + -3I n 2

8-1

126. Prove that the equation e-*

127. Let

uniformly for all x, but not absolutely.

xn for - l / e < x S l / e .

= A-1 has only one real root and show that it is given by

Bix' 2 = 1 + B l x + -+ ef- 1 2!

Bsx' -+ 3!

.... (a) Show that the numbers B,, called the Bernoulli

+

numbers, satisfy the recursion formula ( B 1)"- Bn = 0 where expanding. ( b ) Using (a) or otherwise, determine B1, ,B6. An8. ( b ) B1=-*, B s = & , Bs=O, B l = - & , & = 0 , B e = & 128. (a)Prove that

& = :coth%

if k = l , 2 , 3 ,....

129. Derive the series expansions:

xa + + -x3 - 46

- 1).

...

(b) Use Problem 127 and part (a)ta show that

(2~)'" + Bm(2n)! x

(U)

1 coth~ = -

(b)

cotx

1 -x- xs + ...(- 1)"B'"(24'" = x 3 46 (2n)!x

(c)

tan%

= x

(d)

cscx

= X

x

1

Bk is formally replaced by Br after

+

&k+1=

0

.. . +

...

Bt, (2~)'"-' + ... ++ 2x5 + ...(- 1)"-' 2(2*"- 1)(2n) ! 3 2'

7 + sx + -xa + ..*(-1)*-' 2(2'a-' 360

- 1) B1n X i"'~

+

...

(2n) ! [Hint: For (a)use Problem 128; for ( b ) replace x by iz in (a);for (c) use tan x = cotx - 2 cot 22; for (d) use csc 2 = cot x tan d 2 . l

+

130.

n=l

131. Use the definition to prove that

n (1 - where 0 for x L 2 and lnx x diverges.

Am

g ( z ) d x con-

f ( z ) d x also converges.

lm

Then if f ( x )2 g ( z ) for all x 2 a,

1-

also converges.

J.

f ( z )dx also diverges.

diverges ( p integral with

g ( x ) dx diverges.

262

IMPROPER INTEGRALS

[CHAP. 12

2. Quotient test for integrals with non-negative integrands.

(a) If f ( x ) Z O and g ( x ) Z O , and if l i m m = A Sh 0 or

3

/.m

a

9(x) g ( x ) d x either both converge or both diverge.

(b) If A = 0 in (a) and (c) If

00,

=-to3

A = a in (a) and

then l m f ( z ) d x and

im lW iw lw g ( z ) d x converges, then

f ( x ) d z converges.

g ( s ) dx diverges, then

f ( z ) dx diverges.

This test is related to the comparison test and is often a very useful alternative to it. In particular, taking g(x) = 1/xp, we have from known facts about the p integral, the

Theorem 1. Let lim x p f ( x ) = A. X+Q

Then

(i) J o 3 f ( x ) dx converges if p > 1 and A is finite a

(ii) J w f ( z ) d x diverges if p S l and A+O (A may be infinite). a

Example 1: Example 2:

xa d x

2' 1 - lim a9 7 4 2 $25 4'

converges since

Jm-

o+w

im q7-TFTT x dx

diverges since

lim x t-, 00

X

= 1.

@TzTi

Similar tests can be devised using g(x) = e-% 3. Series test for integrals with non-negative integrands.

diverges according as Bun, where

U,, = f ( n ) ,converges

4. Absolute and conditional convergence.

if then

Ja

or diverges.

I*

f ( z )dx is called absolute& convergent

If(x)l dx converges. If l m f ( z ) d x converges but d a

Theorem 2. If

f ( z ) d x converges or

lw

If(z)I dx diverges,

f ( z ) dx is called conditionuUy convergent.

im

[ f ( x ) dx [ converges, then l w f ( x ) d x converges. In words, an abso-

lutely convergent integral converges. Example 1:

iw2.+1 cos x d x is and

Example 2:

absolutely

JW&

convergent and thus convergent since converges.

l mdxTconverges (see Prob. l l ) , but l"IyId x does not converge

1

xa+ 1

(see Prob. 12).

m

Thus

%dsc

is conditionally convergent.

Any of the tests used for integrals with non-negative integrands can be used to test for absolute convergence.

IMPROPER INTEGRALS

CHAP. 121

263

IMPROPER INTEGRALS of the SECOND KIND If f ( x ) becomes unbounded only at the end point x = a of the interval a 5 x S b, then we define (4) - o t I:€f(z)cix x b f ( x ) d x = rlim If the limit on the right of (4) exists, we call the integral on the left convergent; otherwise it is divergent, Similarly if f ( z ) becomes unbounded only at the end point x = b of the interval a 5 x 5 b, then we define Jb-' f(x)dx (5) f ( x ) d x = E lim -bO+ In such case the integral on the left of ( 5 ) is called convergent or divergent according as the limit on the right exists or does not exist. If f ( x ) becomes unbounded only a t an interior point x = xo of the interval a S x S b, then we define

The integral on the left of (6) converges or diverges according as the limits on the right exist or do not exist. Extensions of these definitions can be made in case f ( x ) becomes unbounded at two or more points of the interval a 5 x 5 b.

CAUCHY PRINCIPAL VALUE It may happen that the limits on the right of (6) do not exist when cl and c2 approach zero independently. In such case it is possible that by choosing c1 = c 2 = c in (6),i.e. writing Jbf0dX

=

JlZ {Jxo-cf(x)dx +

f

6

f(z)dx}

(7)

the limit does exist. If the limit on the right of (7) does exist, we call this limiting value the Cauchy principal value of the integral on the left. See Problem 14.

SPECIAL IMPROPER INTEGRALS of the SECOND KIND 1. 2.

ib&

converges if p

Jb&

< 1 and diverges if

p 2 1.

converges if p < 1 and diverges if p 2 1.

These can be called p integrals of the second kind. Note that when p S 0 the integrals are proper.

CONVERGENCE TESTS for IMPROPER INTEGRALS of the SECOND KIND The following tests are given for the case where f ( x ) is unbounded only at x = a in the interval a 5 x 5 b. Similar tests are available if f ( x ) is unbounded a t x = b or at x = 50 where a < xo < b.

264

IMPROPER INTEGRALS

[CHAP. 12

1. Comparison test for integrals with non-negative integrands. (a) Convergence. Let g ( x ) 2 0 for a < x l b , and suppose that verges. Example:

I'

Then if O S f ( x ) S g ( z ) for a < x S b ,

1 < -for dF1 d a p=i),

J'&

converges ( p integral with a = 1,

also converges.

( b ) Divergence. Let g ( s ) 2 0 for a < x Ib, and suppose that

Then if f(x) l g ( z ) for a < x 5 b, Example:

g ( x ) d x con-

f ( x ) d x also converges.

dz

z> 1. Then since

s'

In z for >(2 - 3)' ( 2 - 3)' In z

z>3.

Jb

g ( x ) dx diverges.

f ( x )d z also diverges.

Then since

16m dz

diverges ( p integral with a = 3,

dz also diverges.

2. Quotient test for integrals with non-negative integrands.

= A # 0 or g(x) g ( x ) dx either both converge or both diverge.

(a) If f ( x )2 0 and g ( x ) I 0 for a < x S b, and if lim

ib

f(x) dx and

ib

( b ) If A = O in (a),and

then

f ( x ) d x converges.

g ( z ) d x converges, then

1 b

(c) If A = 00 in (a), and

00,

2-0

f ( 3 ) dx

g(x)dx diverges, then

diverges.

This test is related to the comparison test and is a very useful alternative to it. In particular taking g ( x ) = l/(x - a)P we have from known facts about the p integral the

Theorem 3. Let xlim (x- ~ ) ~ f (=x A. ) Then +o+ (i)

Jb a

f ( x )dx converges if p < 1 and A is finite

(ii) J b f ( x ) dx diverges if p 5 1 and A # 0 (A may be infinite). a

If f ( x ) becomes unbounded only at the upper limit these conditions are replaced by those in

Theorem 4. (i)

3

P b

Let z lim ( b - z ) P f ( x ) = B. Then db-

f ( x )dx converges if p < 1 and B is finite

0

3

/rb

(ii)

a

f ( x ) d x diverges if p l l and B#O (B may be infinite).

Example 1: Example 2:

f

x'

converges since

dx

(3

dz

-x) d 2 T i

lim (z- I)'/*

r+l+

diverges since

1

(2'

lim (3 - z)

++S-

- 1)"'

- r-,lim1 + 1

(3-x)d-i

4 s

=

- -1 -

+.

0'

CHAP. 121

I* ib ib

then

lf(z)l dx converges.

If

lb

f ( z )dx is called absolutely convergent

3. Absolute and conditional convergence.

if

265

IMPROPER INTEGRALS

lb

lb

If(z)l dx diverges,

f ( x )dx converges but

f ( x )d x is called conditionaZZy convergent.

Theorem 5. If

f ( x )dx converges. In words, an absolutely con-

lf(z)l dx converges, then

vergent integral converges. Example: Since

Iq z I

- dsinx

it follows that lutely).

vrr

dx

1

s,” I

converges ( p integral with a = ‘R, p = i), sin x

dx converges and thus

dx

converges (abso-

Any of the tests used for integrals with non-negative integrands can be used to test for absolute convergence.

IMPROPER INTEGRALS of the THIRD KIND Improper integrals of the third kind can be expressed in terms of improper integrals of the first and second kinds, and hence the question of their convergence or divergence is answered by using results already established.

IMPROPER INTEGRALS CONTAINING a PARAMETER. UNIFORM CONVERGENCE Let (8)

This integral is analogous to an infinite series of functions. In seeking conditions under which we may differentiate or integrate +(a) with respect to a, it is convenient to introduce the concept of uniform convergence for integrals by analogy with infinite series. We shall suppose that the integral (8) converges for a 1 S a 5 a 2 , or briefly [ a l , a 2 ] .

Definition. The integral (8) is said to be uniformly convergent in find a number N depending on c but not on a such that

I+(.) 1’ -

f ( x ,a)dzl

This can be restated by noting that


0 we can

for all U > N and all a in

I+(*) -

J ’ f ( z , ~ ~ ) d z /=

[ a l ,a,]

lim

f ( z , a ) d x l which is

analogous in an infinite series to the absolute value of the remainder after N terms. The above definition and the properties of uniform convergence to be developed are formulated in terms of improper integrals of the first kind. However, analogous results can be given for improper integrals of the second and third kinds.

IMPROPER INTEGRALS

266

[CHAP. 12

SPECIAL TESTS for UNIFORM CONVERGENCE of INTEGRALS 1. Weierstrass M test. If we can find a function M ( x ) 2 0 such that (U) If(x,a)l S M ( x ) a 1 S a S a 2 ,x > U

Jrn M ( x ) d x converges,

(b)

a

then

FWf ( x ,

a) d x

xrn-$&

is uniformly and absolutely convergent in al S a S

Ja

a2

E 5 - and converges, it follows I x ~ i - l l- X L + l is uniformly and absolutely convergent for all real values of a.

Example: Since

As in the case of infinite series, it is possible for integrals to be uniformly convergent without being absolutely convergent, and conversely. 2. Dirichlet’s test. Suppose that (a) $(x) is a positive monotonic decreasing function which approaches zero as X”W

all u > a and

alSaSa2.

Then the integral is uniformly convergent for

a15

(Y

5 a2.

THEOREMS on UNIFORMLY CONVERGENT INTEGRALS Theorem 6. If f ( x , a ) is continuous for x Z a and a l S a S a 2 , and if convergent for a l S a S a 2 , then particular, if

a.

is any point of lim

adao

If

a.

+(a)

=

+(a)

=

a IS & a z ,

f ( x , dx

im dx f ( x ,a)

is uniformly

is continuous in a l S a S a 2 . I n

J r na )

we can write

lim J r n f ( x , a ) d x =

a+ao

Jrn;;-no f ( x ,a ) d x

(9)

is one of the end points, we use right or left hand limits.

Theorem 7. Under the conditions of Theorem 6, we can integrate al to a2 to obtain +(a)da =

s,y’{~rnf(x,a)dx}da =

+(a)

with respect to

a

xrn{1;

f(z,4da}dz

from (10)

which corresponds to a change of the order of integration.

Theorem 8. If f ( x , a ) is continuous and has a continuous partial derivative with respect to

x S a and a l S a 5 a 2 , and if not depend on CY,

lrn 2 dx &!!da

If a depends on

converges uniformly in =

a1S

2 a2, then if

lrngd5

this result is easily modified (see Leibnitz’s rule, Page 163).

for

a does (11)

.

CHAP. 121

267

IMPROPER INTEGRALS

EVALUATION of DEFINITE INTEGRALS Evaluation of definite integrals which are improper can be achieved by a variety of techniques. One useful device consists of introducing an appropriately placed parameter in the integral and then differentiating or integrating with respect to the parameter, employing the above properties of uniform convergence.

LAPLACE TRANSFORMS The Laplace transform of a function F(x) is defined as f ( s ) = x{F(x)} =

I s

a

e-szF(x) dx (1%)

1

1a-e

eQ+

Jrn

I

I

8>a

1

I I

and is analogous to power series as seen by = t? Many replacing cSby t so that cSz properties of power series also apply to Laplace transforms. The adjacent short table of Laplace transforms is useful. In each case a is a real constant. One useful application of Laplace transforms is to the solution of differential equations (see Problems 34-36).

8>0

sin ax

8>0

cos ax

8>0

x" n = 1,2,3,.

..

I

n! p + 1

8>0

IMPROPER MULTIPLE INTEGRALS The definitions and results for improper single integrals can be extended to improper multiple integrals.

Solved Problems IMPROPER INTEGRALS 1. Classify according to the type of improper integral. 10

1

dx

+ tanx

(d)

1

x dx X2dx

* ~ 4 + ~ 2 + 1

Second kind (integrgnd is unbounded at x = 0 and x = -1). Third kind (integration limit is infinite and integrand is unbounded where tans = -1). This is a proper integral (integrand becomes unbounded at x = 2 , but this is outside the range of integration 3 5 x S 10). First kind (integration limits are infinite but integrand is bounded). This is a proper integral (since

jly+ -2'

=

L2

by applying L'Hospital's rule).

268 2.

IMPROPER INTEGRALS

[CHAP. 12

dx

Show how to transform the improper integral of the second kind,

into

(a)an improper integral of the first kind, ( b ) a proper integral. (a) Consider

fE

1 d-) ¶-c

vq&,

*

sideration of

dz

As

where 0 < c < 1, say.

c+O+,

sideration of 2

1 = -. Then the integral becomes

v

we see that consideration of the given integral is equivalent to conwhich is an improper integral of the first kind.

-lWvd&

( b ) Letting 2 - 2 = v'

Let 2 --z

in the integral of (a), it becomes 2

J'm dv

which is a proper integral.

dv

Jk7Z

.

We are thus led to con-

From the above we see that an improper integral of the first kind may be transformed into an improper integral of the second kind, and conversely (actually this can always be done). We also see that an improper integral may be transformed into a proper integral (this can only sometimes be done).

IMPROPER INTEGRALS of the FIRST KIND 3. Prove the comparison test (Page 261) for convergence of improper integrals of the first kind. Since 0 5 f ( z )S g(z) for z L a, we have using Property 7, Page 81,

But by hypothesis the last integral exists. Thus f ( x )dz converges

lim l b f ( z )dz exists, and hence

b+m

4.

Prove the quotient test (a)on Page 262. f (4= A > 0. Then given any c > 0, we can find N such that By hypothesis, lim 1-00

g(4

g(z)

- AI


d

+ ~)g(z)

P

lb

f ( z ) d z converges

f ( z )d x exists, and so

g(z) d z diverges, then by the inequality on the left of ( I ) ,

limx

b-

lb

f ( z )dz =

00

and so

iw

f ( z )dx diverges

For the cases where A = 0 and A = Q),see Problem 41.

As seen in this and the preceding problem, there is in general a marked similarity between proofs for infinite series and improper integrals.

CHAP. 121 5.

269

IMPROPER INTEGRALS

1

xdx

O0

Test for convergence:

(a)

3x4+5x2+1’

(b)

sm

x2-1

2

di7TiG

dx,

(a) Method 1: For large x, the integrand is approximately x/3z4 = 1/3xa. 1 X $ and converges (p integral with p = 3 ) , it follows Since 3x4+ 6xz 1 32”

$lm$

+

by the comparison test that

+z d z + 1

3x4 6x2

Jm

also converges.

Note that the purpose of examining ‘the integrand for large x is to obtain a suitable comparison integral.

verges,

J,

f(x) dx also converges by the quotient test.

Note that in the comparison function g(x), we have discarded the factor ever, just as well have been included.

!;\

Method 3: converges. (b)

For x 2 2, Method 2:

Let n-J

J

9

Method 3: Page 262.

X +

6xz l) = +

1 5.

Hence by Theorem 1, Page 262, the required integral

= l/x.

Foi large x, the integrand is approximately x’/@

Method 1:

diverges,

xa (3x4

x2- 1 > 1.1. Since

@Ti-2x

+. It could, how-

iIm $ i@7 dx

x2- 1 , g(x) =-. 1 f ( 4= X

d-

xa- 1

diverges,

Then since

x +16

lim

=4m

fo = g(x)

1,

also diverges.

and

lWg(x)dx

f(x)dz also diverges. Since

!itx

l/m (

-

)

= 1,

the required integral diverges by Theorem 1,

Note that Method 1 may (and often does) require one to obtain a suitable inequality factor (in this case +, or any positive constant less than 4) before the comparison test can be applied. Methods 2 and 3, however, do not require this.

6.

Prove that

I*

lim z ’ e - 3

=-+ 00

e-z2 dx converges.

= 0 (by L’Hospital’s rule or otherwise). Then by Theorem 1, with A = 0, p = 2,

the given integral converges. Compare Problem 10(a), Chapter 11.

7. Examine for convergence:

(a)

1

* lnx

Zadx,

where a is a positive constant; (b)

In x lim x -= (a) =+Oo x+a diverges.

QQ.

1 - cosx dx.

Jrn

0

x2

Hence by Theorem 1, Page 262, with A = w, p = 1, the given integral

The first integral on the right converges [see Problem l(e)]. 8 4 Oo

= 0, the second integral on the right converges by Theorem 1,

Page 262, with A = 0 and p = 3/2. Thus the given integral converges.

270

8.

[CHAP. 12

IMPROPER INTEGRALS

Test for convergence: (a)

s-'

c d x , (b)

-al

(a) Let x = - y .

x

Then the integral becomes e-'

-S

Method 1:

Y

Lal= *

x3+x2

dx.

- l w dy. y

e-r for y 21. Then since

e-y dy converges,

Im 5

dy converges; hence

the given integral converges. lim

Method 2:

u 4 m

ya(y)=

lim ye-# = 0.

V-w

Then the given integral converges by Theorem 1,

Page 262, with A = 0 and p = 2. Z

( b ) Write the given integral as

X

&

+

I*-

x'-txa dx. X ' f jA

Letting x = -y in the first integral,

= 1, this integral converges. = 1, the second integral converges.

0-b 00

Thus the given integral converges.

ABSOLUTE and CONDITIONAL CONVERGENCE for IMPROPER INTEGRALS of the FIRST KIND

Am

f(x) dx converges if

9. Prove that

iw ]f(x)ldx

converges, i.e. an absolutely con-

vergent integral is convergent. We have

lm

If(x)l dz, which converges, we see that

10. Prove that Method 1: cosx

1-7 dx

I

1 1 7d 39

Jw

+

I

=-+ 00

[f(x)

+ If(x)l]dx

xa/a

i.e.

converges.

Hence by subtract-

f(z)dx converges.

converges.

for x Z 1. Then by the comparison test, since

ICY dx converges,

Method 2: Since lim

Then

d f ( x ) d If(x)l, i.e. 0 S f ( s ) If(x)l d 2If(x)(.

lf(x)l dx converges, it follows that

If

ing

- If(x)l

rm

J1

a-

f

converges, it follows that

dx converges absolutely, and so converges by Problem 9.

=

0, it follows from Theorem 1, Page 262, with A

and hence p = 3/2, that ~ w l c ~ l converges, d x

l* x:c

dx converges (absolutely).

= 0 and

CHAP. 121

271

IMPROPER INTEGRALS

11. Prove that

lw 9

dz converges.

Since x l eX x

is continuous in O < x S 1 and

X

we need only show that

lim -x

r*o+

d x converges.

lMKX&

Method 1: Integration by parts yields sin x d x = -+

cosxI:

'X

or on taking the limit on both sides of (1) as

M+ =

i w sin T xd x

cos x

cos1 - cos M

=

X

M

and using the fact that

00

cos1

+l

cosM lim - - 0,

~4~

M

w y d x

Since the integral on the right of (2) converges by Problem 10, the required result follows. The technique of integration by parts to establish convergence is often useful in practice.

iv

Method 2:

-

sin x lm,dx

sin x d E

Ja2T

+

X

sin x n=O

Letting x = v

sin x

dx

+

...

+

la

(ntl)a

sin x 7 dx +

* -

dx

+ nn, the summation becomes

i ( - l ) n J T & vd v+ n u

n=O

This is an alternating series.

=

IT!&!!

dv - l v z d v

Since l s v+nn

v

+ ( 1n + l ) s

+

ia&dv

-

...

and sinv 2 0 in [O,n],it follows that

Also, Thus each term of the alternating series is in absolute value less than or equal to the preceding term, and the nth term approaches zero as n + W . Hence by the alternating series test (Page 226) the series and thus the integral converges.

12. Prove that

1

dx converges conditionally.

Since by Problem 11 the given integral converges, we must show that it is not absolutely convergent, i.e.

lobI

d x diverges.

As in Problem 11, Method 2, we have

1 1 NOW -z -(n l,.rr +

+

for 0 S v 5 n. sin v d v

Since Hence

2(n+l)a O0

iw 1 5 1 n=O

Hence

2

lr

( n ,ll)n

sin v d v

=

(n

2

+ 1)s

diverges, the series on the right of (1) diverges by the comparison test.

d x diverges and the required result follows.

272

IMPROPER INTEGRALS

[CHAP. 12

IMPROPER INTEGRALS of the SECOND KIND. CAUCHY PRINCIPAL VALUE 13. (a) Prove that

7Zi

J7 -1

converges and (b) find its value. Then we define the integral as

The integrand is unbounded a t x=-l.

This shows that the integral converges to 6.

dx

14. Determine whether

converges (a) in the usual sense, (b) in the Cauchy

principal value sense. (a) By definition,

and since the limits do not exist, the integral does not converge in the usual sense. (b)

Since

the integral exists in the Cauchy principal value sense. The principal value is 3/32.

s

15. Investigate the convergence of:

(4

dx x2(x3- 8)'13

s/2

dx TJ(5 - x ) ( x - I)

1

dx

( b ) S s0 % d x (a)

verges by Theorem (b) (c)

-

1 xz(xJ - 8 ) 2 / 3

)ly+ (x - 2)"s

lim+( x

-z+2+

1

'X

3(9, Page 264.

+ 22 + 4

)

213

=

z.

Hence the integral con-

sin x lim x2 - 1. Hence the integral diverges by Theorem 3(G] on Page 264.

r+O+

XS

Write the integral as Since

1

lim (z-

+-,

1+

dx

\ / ( S - x)(x

- 1) 4-

1

d(5 - x)(x

- 1)

I

Since

lim (6-x)"'

z-5-

I

{(S - X)(X - 1)

dx d(6- x)(x - 1)

=

t , the first integral converges.

=

$, I

the second integral converges.

Thus the given integral converges. Hence the integral diverges.

IMPROPER INTEGRALS

CHAP. 121 Another method: 2'1"-1

I

1-x (e)

lim I-,

%n-

ll& - z J ~ - ( ~ > I ' n

2 - niz 21-x'

(r/2 - x)""

and

1 (cosx)i/,, -

273

diverges.

Hence the given integral diverges.

r/2-x

= 1.

16. If m and n are real numbers, prove that

Hence the integral converges.

l1

xm-l(l- z)n-l dx (a) converges if m > 0

and n > 0 simultaneously and (b) diverges otherwise. For m 1 1 and n L 1 simultaneously, the integral converges since the integrand is continuous in O d x d 1. Write the integral as '/2

xm-l(l-s)n-lds

l,*

+

1

xm-' (1-

dx

(11

):+ XI-"' zm-'(l If 0 < m < 1 and 0 < n < 1, the first integral converges since ? = 1, using Theorem 3(i), Page 264, with p = 1 - m and a = 0. lim (1- x)l-* x * - l ( l - x)"--l = 1, using Similarly, the second integral converges since Z-1Theorem 4(4, Page 264, with p = 1 - n and b = 1. Thus the given integral converges if m > 0 and n > 0 simultaneously. If m S 0, lim x x"-'(l2+0+

x)"-l =

W.

Hence the first integral in (1) diverges, regardless of the

value of n, by Theorem 3(ii), Page 264, with p = l and a = 0 . Similarly, the second integral diverges if n d O regardless of the value of m, and the required result follows. Some interesting properties of the given integral, called the beta integral or beta function, are considered in Chapter 13.

17. Prove that

J,

sin 1 dx

converges conditionally.

Letting z = 1/11,the integral becomes

dy and the required result follows from Prob. 12.

IMPROPER INTEGRALS of the THIRD KIND 18. If n is a real number, prove that if n 5 0.

1-

e-z

dx (a) converges if n > 0 and (b) diverges

Write the integral as x*-'e-*dx

+

l*zn-le-zdz

(1)

(a) If n 1 1 , the first integral in (1) converges since the integrand is continuous in 0 5 x S 1. If O < n < l , the first integral in (1) is an improper integral of the second kind at z=O. Since lim d-" x"-'e'= = 1, the integral converges by Theorem 3 ( 4 , Page 264, with p = 1 - n z+o+

and a = 0. Thus the first integral converges for n > 0 . If n > 0 , the second integral in (1) is an improper integral of the first kind. Since lim x' ~ " - ' e -=~ 0 (by L'Hospital's rule o r otherwise), this integral converges by Theorem l(9,

2-

m

Page 262, with p = 2 . Thus the second integral also converges for n > 0 , and so the given integral converges for n > 0.

274

[CHAP. 12

IMPROPER INTEGRALS

z zn-le-f = ( b ) If n 5 0, the first integral of (1) diverges since r +lim O+

[Theorem 3(i4, Page 2641.

00

If n 5 0, th e second integral of ( I ) converges since lim z * x n - l e - r = 0 [Theorem l(i), t+m Page 2621. Since the first integral in (1) diverges while the second integral converges, their sum also diverges, i.e. the given integral diverges if n 5 0. Some interesting properties of the given integral, called the gamma function,a r e considered in Chapter 13.

UNIFORM CONVERGENCE of IMPROPER INTEGRALS Evaluate

+(a)

=

f

ae-*"dx for a > 0 .

Prove that the integral in (a) converges uniformly to 1 for a 2 a1 > 0. Explain why the integral does not converge uniformly to 1 for a > 0. xbae-azdx =

=

$(a)

lim - e - * * b-t

m

L0

-

lim 1 - e - a b = 1 if a > 0 . b+*

Thus the integral converges t o 1 for all a > 0. Method 1, using definition.

I lu

The integral converges uniformly to 1 in a 2 a l > 0 if for each r > O we can find N, depending on

but not on a, such t h a t

z

11

Since

-

lU

ae+Zdz

I

l -

ae-azdxl

N.

= 11 - (1 - e - f f u )I = e-*" 5 e-alu


1 In 1 = N, a1

the result follows. Method 2, using the Weierstrass M test. Since lim xp ae-"" = 0 for a 2 a, > 0, we can choose lae-azl t+ 00

say

2: 2x0.

1

Taking M ( z ) = - and noting t h a t XP

integral is uniformly convergent to 1 for a 2 a, > 0.

1;$

1 <XL

E

f o r sufficiently large x,

it follows t h a t the given

- converges,

As a1 + 0, the number N in the first method of (b) increases without limit, so t h a t the integral cannot be uniformly convergent for a > 0.

+(.)

= l w f ( z , a ) d z is uniformly convergent for a l I a 5 a p , prove that + ( a ) is

continuous in this interval. Let Then

Thus

$(a) =

xU

+ R(u, where R(u,a) = f(x, a + k ) dx + R(u,a + h) and so

f(x, a) d x

I

+ h) = $(a + h) - +(a)

=

Ju

I $(a + h) - $44I

5

fI

$(a

f

U),

~ ( za,

f(s,a) dx.

+ h ) - f ( z ,a)>d x + ~ ( ua ,+ h) - ~ ( ua),

f(x, a + h ) - f ( x ,a) I dx

+ I R(u, a + 4 I + I R(% a) I

Since the integral is uniformly convergent in a, d a Ia%,we can, f o r each ent of a such th a t fo r U > N, I R(u,a h) I < d3, R(u, a ) I < ~ / 3

+

Since f ( x , a) is continuous, we can find S /.U

c

> 0,

find

I

>0

corresponding to each

e

>0

(1)

N independ(a)

such t h a t

CHAP. 121

275

IMPROPER INTEGRALS

+

I

I

Using (2)and (8) in ( I ) , we see that +(a h) - +(a) < e for lhl < 6, so that +(a) is continuous. Note that in this proof we assume that a and a h are both in the interval a, S a 5 as. Thus if a = a,,for example, h > 0 and right hand continuity is assumed. Also note the analogy of this proof with that for infinite series.

+

Other properties of uniformly convergent integrals can be proved similarly.

21. (a) Show that (a)

lim xOOae-azdx Z. IOO(,&rp+ ~ ~ e - ~ ~ ) (b) d z Explain . the result in (a).

a+O+

:?.+ 1 ° 0 a e - a r dx

=

1(ii+y+ 1°0 ae-ar)

(b)

lim 1 = 1 by Problem 19(a).

a*O+

Since #(a) =

dx = i O O O d x = 0.

ae-ar

Thus the required result follows.

da: is not uniformly convergent for a 2 0 (see Problem 19), there is no

guarantee that +(a) will be continuous for a 2 0. Thus

e-ax cosrx dx =

22. (a) Prove that

d! for a2 + r2

lim +(a) may not be equal to +(O).

awO+

0 and any real value of r .

a>

(b) Prove that the integral in (a) converges uniformly and absolutely for a 5 a 5 b, where 0 < a < b and any r . (a) From integration formula 34, Page 84, we have

1°0

1 e-az

( b ) This follows at once from the Weierstrass M test for integrals, by noting that and

e-ardx

cos r x I 5 e-ar

converges.

EVALUATION of DEFINITE INTEGRALS

23. Prove that

6'"

'7r

I n s i n x d x = --ln2. 2

The given integral converges [Problem 42(f)].

Then 21 =

=

l 1

(In sinx

T/¶

U/'

In sin22 dx

Letting 2x = v , i H " l n sin 2 s dx

+ In cosx)dx - Jl''ln2

=

ii"

=

*(Z

In sin v dv

+ z)

=

Z

Letting

X

= 1""ln dx

(y) ds

f" In sin 2x dx -

=

t(

=

= d 2 - Y,

l r " l n sin v dv

+

In 2

s,:

In sin v dv

(letting v = P - U in the last integral)

Hence ( I ) becomes 21 = Z - E l n 2 or Z = --In 2 2 U

2.

[CHAP. 12

IMPROPER INTEGRALS

276

x2

Jyx In sinx dx = --ln2.

24. Prove that Let x =

2

U

J

Then, using the results in the preceding problem,

- y.

=

l T x In s i n x d x

=

iR

=

4

(P

-

- U) In sin U d u

=

lT (U

- x) In sin x d x

-2ln2 - J

or J = - - ln2. U '

2

25. (a) Prove that +(a) =

is uniformly convergent for

Jm&

.

+(a) = lr

(b) Show that

\

2 6

( c ) Evaluate

dx

( d ) Prove that

R/2

aZ

1.

1m(x,d."1,2.

cos2"e do =

1 3 5 - - - (2n- 1)x

204.6...(2n)

2'

1 1 (a) The result follows from the Weierstrass test, since 7p for a 2 1 and x+l x + a converges.

(c)

From (b),

lws =

2

c

Differentiating both sides with respect to a, we have

a

2 1 (because

1 and 1 (x' + a)' - (2' + 1)'

&

Jrn

ia uniformly convergent f o r

dz

the result being justified by Theorem 8, Page 266, since

converges).

Taking the limit a s a + 1t , using Theorem 6, Page 266, we find

(d) Differentiating both sides of

ice= dx

-

-- lrn(2'+ dx

U

1)'

4

n times, we find

where justification proceeds as in part (c). Letting a + 1+, we find

JW

dx

(2'

+ U"+,

Substituting x = tan

26. Prove that

S,

8,

-

1 * 3 * 6 . . . ( 2 n - l)U 2" n ! 2

-

1 3 6. .(2n- 1) 2 4 6...(2n) 2 +

8 d e and the required result is obtained. the integral becomes ~ R " c o s ' n

- e-bx 1 b2+r2 where a, b dx = -1nxsecrx 2 a2+r2

OOe--Ox

From Problem 22 and Theorem 7, Page 266, we have

> 0.

277

IMPROPER INTEGRALS

CHAP.121

or

e - ~

i.e.

e-b~

z secrz

e-Qz 1 - cosx

27. Prove that

x2

1 b'+r' = -1n2 a'+@

dz

1

a

d~ = tan+-a -zln(a2+1), a > 0 .

By Problem 22 and Theorem 7, Page 266, we have

{Im lm{lr } I* =

e-uzcosrz dz}dr

=

e-az s i n r z dz

or

e-Qzcosrz d r dx

=

i r f i d r

tan-1 E

z

a

Integrating again with respect to

T

from 0 to

T

yields

using integration by parts. The required result follows on letting r = 1.

x

dx = -

28. Prove that

2'

Since e'a*-

1

- cos z S 1 - cosz for a 2 0,s 2 0 and X' 2'

7 ( b ) ] , it follows by the Weirstrass test that

t3-az

*

dz converges [see Problem

dz is uniformly convergent and repre-

z'

sents a continuous function of a for a B 0 (Theorem 6, Page 266). Then letting a + 0+, using Prob. 27, we have

Integrating by parts, we have dz

= (-;)(l

Taking the limit as

E +

-cosz)lr

O+ and

M

+

Q)

+

l ' y d z

shows that

=

1 - cosc - 1 - COS M

z

278

IMPROPER INTEGRALS

-a(

= Then

eaiz

- - SLO 2i

[CHAP. 12

1 + -3s i n % ) +: ( = - -r sin32 ) -iz

,iZ

4

4

MISCELLANEOUS PROBLEMS 31. Prove that

iw dx

1/

= 6/2.

e-zx

By Problem 6, the integral converges. ZW =

U

e - f ' dx =

L'

6-9

Let

d y and let lim ZV = Z, M+a

(lMdz)(

the required value of the integral. Then 1;

= = =

e-$

JM

lM lM

ss

e-(s+@)

e-fl d>

d x dy

e-($+fl)dxdy

2

S R d

where qZy is the square OACE of side M (see Fig. 12-1). Since the integrand is positive, we have

ss

e-(z'+va)dZdy

I:

s

ss

Fig. 12-1 (11

e-($ 9)d x d y +

9%

91

where qZ1 and q, are the regions in the first quadrant bounded by the circles having radii M and ~ respectively. f i

Then taking the limit as M

32. Evaluate

da

in (3),we And

d)lm :Z = P = n/4

and Z = &/2.

e-zt cos ax dx.

Let I(a) =

cedures,

+ 00

e-$ cos az ds. - z e - f ' sin a x dx

Then using integration by parts and appropriate limiting pro-

=

*e-$ sin

-

+a

e-3 cos a2 d z

=

-%I 2

The differentiation under the integral sign is justified by Theorem 8, Page 266, and the fact that x e - 9 sin a% dz is uniformly convergent for all a (since by the Weierstrass test, ~ e - ~and *

lw

zed dx converges).

I ze-$

sin az I 5

CHAP. 121

279

IMPROPER INTEGRALS

From Problem 31 and the uniform convergence, and thus continuity, of the given integral (since

I e'ls

cos ax

Z(O)

= lim 0-0

I

S

eeZ1 and

~ ( a= )

Solving

36.

dcu =

- EZ 2

subject to I(0) =

33. (a) Prove that I ( a ) = (U)

We have

converges, so that the Weierstrass test applies), we have

Jwe-$dx

I'(a)

e-tz-alx)~

T, we find

I(a)

dx = -

*

= 2 e-ar/r.

(b) Evaluate s,"e

2 .

-(x2+z-2)

dx,

= 2 i r n e - ( z - a ' z ) * (1 - n / ~ *dz. )

The differentiation is proved valid by observing that the integrand remains bounded as

x + O+ and that for sufficiently large x, = e--rd+'Q--OL'/r*( 1 -a/x') e-(r-a/+)*( 1 -

5

e'ae-9

so that Z'(a) converges uniformly for a 2 0 by the Weierstrass test, since

Now

It(&)

=

2

Jae-(x-a/r)' dx

-

dx

=

Jrn

e-12 dz converges.

0

as seen by letting a / x = y in the second integral. Thus Z ( a ) = c , a constant. To determine c, let a + O+ in the required integral and use Problem 31 to obtain c = 6 / 2 .

1 S 34. Verify the results: (a) < { e a x } = s - a ' s > a ; ( b ) o(l{cosax} = $2 + (-&2 s > 0. 9

= (b) a

by Problem 22 with a = s, r = a.

Another method, using complex numbers. From part (a), 4 { e a 2 } = 1 . Replace a by ui. Then 8-a

x{eai2}

=

cos a x + i sin a x }

Equating real and imaginary parts:

=

COS a x }

+

i = 4{z>, 4{Y”(z)} s { Y ( z ) >= 1/8’ 8’ .(‘{Y(Z)}- 8 Y(0)- Y’(0) -k 4{Y(X)}.= 1/8’

80

Solving for x { Y ( z ) }using the given conditions, we find

by methods of partial fractions. Since

1

7

= s { x } and -- x{sinz), it follows that 8’ -k 1

+

1 1 a.+ 7 = x{z + sinz}. +1 8

Hence from (I), x { Y ( z ) }= x { x sinz}, from which we can conclude that Y ( z ) = z which is, in fact, found to be a solution.

+ sins

Another method:

If S{F(x)}= f ( 8 ) , we call By Problem 78,

4-l ( f ( 8 )

f(8)

the inverse Laplace transform of F ( z ) and write {f(s)} 4-l {g(8)}. Then from (I),

+ g ( 8 ) } = 4‘’

+

f(8)

= 4-l {F(x)}.

Inverse Laplace transforms can be read from the Table on Page 267.

Supplementary Problems IMPROPER INTEGRALS of the FIRST KIND 37. Test for convergence:

An8. (a)conv., ( b ) div., (c) conv., ( d ) conv., ( e ) conv., ( f ) div., (0) conv., (h) div.,

(9 conv.

CHAP. 121

IMPROPER INTEGRALS

38. Prove that 39.

s-:xa+2ax+ba dx

Test for convergence: ( a )

-

if b > lal.

7T

iOO e-*

281

In x dx, (b)

e‘’

In (1

+

6”)dx,

(c)

Ans. (a) conv., (b) conv., (c) div. 40.

Test for convergence, indicating absolute or conditional convergence where possible: (a) (b)

JQ

e-aZ

(d)

cos bx d x , where a, b are positive constants;

(e)

iaxx

41.

cos x

sin 22

dx;

sme&; O

-00

Ir m 1 @G

A m . ( a ) abs. conv., (b) abs. conv., ( c ) cond. conv., ( d ) div., ( e ) abs. conv.

dx.

Prove the quotient tests (b) and (c) on Page 262.

IMPROPER INTEGRALS of the SECOND KIND 42. Test for convergence:

. . ( f ) JT’’ln

dx 1

sin x d x

x

Ans. (a)conv., (b) div., (c) div., (d) conv., (e) conv., ( f ) conv., (g) dive, (h) dive, 43.

(a) Prove that

xs4;

(9conv., (i) conv.

diverges in the usual sense but converges in the Cauchy principal value

dx

sense. (b) Find the Cauchy principal value of the integral in (a)and give a geometric interpretation. A m . (b) In 4 44.

Test for convergence, indicating absolute or conditional convergence where possible: (a)

cos(

J1i cos( i) dx, 1’$ cos( i) dx. 14’r (32‘ sin 1. - x cos $)dx = pt. 32fi

$)d x ,

(b)

(c)

Ans. ( a ) abs. conv., (b) cond. conv., (c) div. 45.

Prove that

IMPROPER INTEGRALS of the THIRD KIND 46. Test for convergence: (a)

1% 6-O

Ans, (a) conv., (b) div., (c) conv.

In x dx, (b)

q-’

e-* d x

6-*dx

JOO

0

47. Test for convergence: (a)

sinh (ax)

A m . (a)conv., (b) conv. if a > 2, div. if 0 C a 5 2. 48.

Prove that

I*!?:

dx converges if 0 S [a1 < a and diverges if

lal

h U.

49. Test for convergence, indicating absolute or conditional convergence where possible: Ans. (a) cond. conv., (b) abs. conv.

282

[CHAP. 12

IMPROPER INTEGRALS

UNIFORM CONVERGENCE of IMPROPER INTEGRALS 50. (a) Prove that

+(a)

(b) Prove that 51. Let

i.e.

+(U)

=

=

+ ( a ) is

dx

continuous for all a. ( c ) Find lirn

lm lrn

a-a

+(U).

Ans.

F(x, a ) dx, where F ( x , 0 ) = a * 2 ~ 3 - ~(U) ~ . Show that lim F ( z , a ) d x .

lim J = F ( x , a ) d x #

a-ro

is uniformly convergent for all a. (c) a/2.

$(a)

is not continuous at

a

= 0,

(b) Explain the result in ( a ) .

Q M O

52. Work Problem 61 if F ( x , a ) = a'xe-"". 53. If F ( x ) is bounded and continuous for

< z < 00 and

--oo

prove that lirn V( s,y ) = F(x). Y-rO

54. Prove (a) Theorem 7 and (b) Theorem 8 on Page 266. 55. Prove the Weierstrass 56.

Prove that if

57.

Prove that

x*

F ( x ) ds converges, then

(a)

=

+(a)

dx =

(c)

r

M test for uniform convergence of integrals.

lrn e-ar

e-as

F ( x ) d x converges uniformly for

si; x ds converges uniformly for

a

2 0,

(b)

a h 0.

+(a)

= - tan-* a,

(compare Problems 27-29).

2

58. State the definition of uniform convergence for improper integrals of the. second kind. 59. State and prove a theorem corresponding to Theorem 8, Page 266, if a is a differentiable function of a.

EVALUATION of DEFINITE INTEGRALS Establish each of the following results. Justify all steps in each case. 60.

JQe-a=

61.

Jrne;;;:ibX dx =

62.

e-b=

sin rx dx

= In (bla), a , b > 0

ds

tan-'(blr) - tan-' ( u / r ) , u , b , r > 0

= i(l-e-r),

rB0

63. 64.

65. (a) Prove that

iwe-ax('"" ; ux

( b ) Use (a)to prove that

COS

Jrnc0s ux

b z )dx

;

COS

+

= ; l n ( 'am ) b2 ,

bx d x

a20.

- In(:). -

[The results of (b) and ProbIem 60 are speciaI cases of Frulluni's integral, F ( 0 ) In 66. Given

(t),

where F ( t ) is mntinuous fort > 0, F'(0) existe and

fwe-addz d 0

=

+a, Prove t h a t for a>0.

p

= 1,2,3,...,

CHAP. 121

283

IMPROPER INTEGRALS

(e-afEL - e-b/d)dx =

67. If a > 0, b > 0, prove that

68. Prove that

xw

tan-' ('Ib) X

tan-1 dx

69. Prove that

dx

- -4n

j 6

IIn( !)

=

-

6.

where a > 0, b > 0.

[Hint: Use Problem 38.1

MISCELLANEOUS PROBLEMS 70.

Prove that

2

Hint: Consider

dx

72. Prove that

=

U

+

In (1 a), a 2 0.

l T " l n sine de = - 3Ul n 2 .

(b) Use ( a ) to show that

dx = 3' R

74.

Prove that

75.

Evaluate (a) lal

(c) tan"

76. (a)If < { F ( x ) } = f ( 8 ) , prove that U

bB'

x { F ( x ) } = f ( 8 ) , prove that x { x " F ( x ) } = ( - 1 ) " f ( " ) ( 8 ) , giving suitable restrictions on F ( z ) .

(b) Evaluate ~ { x c o s x } . 78. Prove that

-1.

+

#-a

Letting x = e-y, the integral becomes (-l)n J0 yne-("'+')Bdy. If (m becomes

+ l)y = U, this last integral

Compare with Problem 50, Chapter 8, Page 177.

9.

A particle is attracted toward a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If the particle is released from rest, find the time for it to reach 0. At time t = 0 let the particle be located on the x axis at x = a > 0 and let 0 be the origin. Then

by Newton's law

m-d% dt'

-

k

--X

(11

where m is the mass of the particle and k > 0 is a constant of proportionality. dx d% dv d v . 3 Let - = v , the velocity of the particle. Then -- dt' dt dx d t dt mv-dv = - or dx X 2 = -klnx+c upon integrating. Since v = 0 at x = a, we find c = k In a. Then

where the negative sign is chosen since x is decreasing as t increases. taken for the particle to go from x = a to x = O is given by

=

-- '.dx dv

and ( 1 ) becomes

We thus find that the time T

Glm dx

Letting In a/x = U or x = ae-y, this becomes

The BETA FUNCTION 10. Prove that (a)B(m,n) = B(n,m), ( b ) B(m,n) = 2 (a) Using the transformation x = 1 - y, we have

(b)

Using the transformation x = sin' B(m, n) =

1'

8,

J

r/2

sin2m-1 8 cos2n-1 e de.

we have

xm-' (1 - x)"-I dx =

rY/a

(9)

(sina ep--l (cosae p - l 2 sin e cos e de

= 2 i r ' k P m - l e cossn-' e de

290

GAMMA AND BETA FUNCTIONS

11. Prove that

r(m) B(m,n)= r(m + n )

[CHAP. 13

m, > 0,

using the results of Problem 10. Hence the required result follows. The above argument can be made rigorous by using a limiting procedure as in h o b . 31, Chap. 12.

12. Evaluate each of the following integrals.

(b)

(a)

s**

olh=ii

.

Letting z = 2w, the infegral becomes

f y4d m d y .

1

0/2

13. Show that

Letting y' = U% or y = a f i , the integral becomes

sinzm-' 6 cos2"-l 6 d6 = r(m) 2 r ( m n)

+

> 0.

m,

This follows at once from Problems 10 and 11.

14. Evaluate (U)

f" sin66 d6,

(a)

0

(b)

s0I2 0

sin46 cos5 6 de, (c)

s" 0

Let 2m- 1 = 6, 2n- 1 = 0, i.e. m = 7/2, n = 1/2, in Problem 13. Then the required integral has the value r(7'2)r(1'2) 2 r(4)

-

- 32'

cos48 d6.

291

GAMMA AND BETA FUNCTIONS

CHAP. 131

( b ) Letting 2m - 1= 4, 2n - 1 = 5, the required integral has the value

r(6/2)r(8) - A. 2r(11/2) 316

H/a

(c)

The given integral = 2

i

de.

COS'B

Thus letting 2m - 1 = 0, 2n - 1= 4 in Problem 13, the value is 2 r(1/2) r(6/2) - 3p -7 2 r(3)

15.

From Problem 13 with 2 m - 1 = p, 2 n - 1 = 0, we have

(a) If p = 2r, the integral equals

(b)

If p = 2r+ 1, the integral equals

1

H/L

In both cases

f

16, Evaluate (a)

H/2

sinPe ds =

cos60 do,

0

(b)

cospe de, as seen by letting e = d 2 - 9.

SHi2 0

(a) From Problem 15 the integral equals (b)

o/a

sinS8 cos28 do, (c)

2-4-62

=

32

J2H

sins 8 do.

[compare Problem 14(a)].

The integral equals

The method of Problem 14(b) can also be used.

17. Given

"OxP-1

1+2dx

Letting

7T =sin p~ ' show that r(p)r(1- p) =

X =y l+x

i1

yp-1(1-

or

5

7r

where 0 < p

< 1.

'

=the given integral becomes 1--y'

y)-p dy

= B(p, 1- p ) = r(p)r(1-p)

18.

The result can also be obtained by letting ya = tan e.

and the result follows.

292

[CHAP. 13

GAMMA AND BETA FUNCTIONS

1 6 ~

19. Show that

Letting x3 = 8y or x = 2y1l3, the integral becomes

STIRLING'S FORMULA 20. Show that f o r large n, n ! = G We have

n nne - n approximately. (msne-xdx =

=

r(n+1)

Jw

enlnx-x

dx

JO

, is easily shown by elementary The function n l n x - x has a relative maximum for ~ = n as calculus. This leads us t o the substitution x = n + y. Then ( I ) becomes r ( n + 1) =

e-n

1:

en l n ( n + y )

nn e - n

1;

- y dy

enln(l+Yln)

e-nJ-:

=

enlnn

+ n l n ( l + y / n ) - dY

(2)

- Y d-Y

Up to now the analysis is rigorous. The following procedures in which we proceed formally can be made rigorous by suitable limiting procedures, but the proofs become involved and we shall omit them. I n (2) use the result x2 xs In(l+x) = x - - + - - . . 2 3 with x = y/n. Then on letting y = fiv, we find

r(,,,,+ 1) =

Sp.

%n e - n

e--y2/2n

+ u3/3n2 -

* * *

dzl

=

nne-n,&

J>e-v2/2+v3/&-

- - dv *

(4)

When n is large a close approximation is

It is of interest that from ( 4 ) we can also obtain the result (13) on Page 286. See Problem 74.

DIRICHLET INTEGRALS 21. Evaluate

Z = sssxaw1

x7-l

dx dy dx

V

where V is the region in the first octant bounded by the sphere x2 y2 x2 = I and the coordinate planes. Let 2 = U, y2 = w, 2 = w. Then

+ +

I JJJ

- 8

u(a/2)-i v ( B / 2 j - l

w~y/z)-l

d u d w dw

(1)

x

where Z'( is the region in the uvuw space bounded by the plane U v w = 1 and the UV, vw and uw planes as in Fig. 13-2. Thus

+ +

293

GAMMA AND BETA FUNCTIONS

CHAP. 131

(3)

Letting v = (1 - u)t, we have

80

that (8) becomes

where we have used ( y / 2 ) r ( y / 2 ) = r ( y / 2

+ 1).

The integral evaluated here is a special case of the Dirichlet integral (20), Page 287. The general case can be evaluated similarly.

+ +

22. Find the mass of the region bounded by x2 y2 x2 = a2 if the density is The required mass = 8 by the sphere

JJJz*y*zXdxdydz,

where

U

V is the region in the first octant bounded

V

= us and the coordinate planes.

In the Dirichlet integral (PO), Page 287, let b = c = a, p = q = r = 2 and required result is as us as r ( 3 / 2 ) 1?(3/2)r ( 3 / 2 )

+ + 3/2 + 3 / 2 )

2 2 2 r(1 3/2

-

-

a

= p = y = 3. Then the

4rae 945

MISCELLANEOUS PROBLEMS

Let z4= y. Then the integral becomes

From Problem 17 with p = 1/4, r ( 1 / 4 ) r(3/4) =

24. Prove the duplication formula Let

I =

= x2y2x2.

IT''

sin2Px dx,

xTla

Z2*-l

J =

~ fso ithat the required result follows.

r(p) r ( p

+ +) = Gr(2p).

singp2x dx.

294

GAMMA AND BETA FUNCTIONS

[CHAP. 13

Letting 22 = U, we And J

But

=

=

+x‘sin*Pu du

lufs

sin’pu du

=

(2 sin x COB s ) * p dx

J

2”

= I

iU/*

sinapx cosspz dx

Then since I = J,

and the required result follows.

Consider

as in Problem 23.

Letting fi sin e/2 = sin + in this last integral, it becomes the result follows.

7r

26. Prove that 2p 1

=

i&

xWup-’e-*du. ~

*

~

PI2

from which

O 0,

MISCELLANEOUS PROBLEMS 56. Prove that

lb-

(z a)P (b - z)q dz = (b - a)P+*+lB(p

+ 1, q + 1 )

where p

> -1, q > -1

[Hint: Let z - a = (b-u)y.]

68. Prove that

1' + xm-l

B(m,n) =

[Hint Let y = z/(l +z).]

(1

dz

2

-

60. Prove that

+

[Hint Let z = (r l)y/(r

='' sin'"-'

Z)rn+*

where m,n

> 0.

n p a tanpede = -8w-

59. If O < p < 1 prove that

61. Prove that

+ %m-1

8

+ y).] cos*"-'

8

2 '

B(m9n, r"(1 r ) m + "

+

d8

where m, n and v are positive constants.

- B(m, n) 2a"b"

where m,

[Hint Let z = sin'@ in Problem 60 and choose r appropriately.] 62.

Prove that

> o.

and b > a.

297

GAMMA AND BETA FUNCTIONS

CHAP. 131 63. Prove that for m = 2,3,4,

...

?h ... sin- ( m m- l ) a - 2"-1 zm- 1 = (z- 1)(z - al)(z- a,). - .(z- as-l),

sin-7r sin-27r sin-37r m

m

m

[Hint Use the factored form and consider the limit as z+ 1.1

xu'*

In sin x d z = - ~ / 2In 2

64. Prove that

divide both sides by z - 1,

using Problem 63.

[Hint Take logarithms of the result in Prob. 63 and write the limit as m+

00

as a definite integral.]

65.

[Hint Square the left hand side and use Problem 63 and equation ( 8 ) , Page 286.1

1'

In r(x) dx =

66. Prove that

+ In ( 2 ~ ) .

[Hint: Take logarithms of the result in Problem 6b and let m+ 67.

(U)

s i n s dz

Prove that

-

U

O

L{-

=

cos (n

+

cos (n- 1)z n-1

+ cosnu - 1 + cosnn

;{

1 1

=

+ 1)z

n+l

U

n+l

n-1

1.

= -. 4

= g(-cosz)lT

P

0

12. Expand f ( s ) = x, 0 < x < 2, in a half range (a) sine series, ( b ) cosine series. (a) Extend the definition of the given function to that of the odd function of period 4 shown in Fig. 14-11 below, This is sometimes called the odd e z t e b of f ( x ) . Then 2 L = 4 , L = 2 .

f (4 f

/

/

-: /

/ /'4

/

0

/

/ 0

I

/

-2

/

/

I

/

/

/

/

/

/

/

0 X

I

4

0

0

Fig. 14-11 Thus ccr= 0 and

b,

=

i x ' f ( z ) sin-nuz dz

=

L = { ( x ) (nus c o s y ) Then

fb) =

-

tl'

(I)($

nax 3 -4 cos nu sin nn 2

z s i nn m2 dz

sin?)}[

0

=

-4

nu cos nu

309

FOURIER SERIES

CHAP. 141

( b ) Extend the definition of f(x) to that of the even function of period 4 shown in Fig. 14-12 below. This is the even extension of f(z). Then 2 L = 4, L = 2.

f (4 0'

4

\

\

/ I

I

A

\

\ \

I

-4

-6

/

-2

\ 0

1

2

\

\

/ I 4

0 "\ I

2

6

Fig. 14-12

4

If n = 0,

=

T ( c o s n n - 1)

l'

nn

if n#O

x dx = 2.

Then

It should be noted that the given function f(x) = x, 0 < x < 2, is represented equally wetl by the two diferent series in ( a ) and (b).

PARSEVAL'S IDENTITY 13. Assuming that the Fourier series corresponding to f ( x ) converges uniformly to f ( z ) in (-L, L), prove Parseval's identity

where the integral is assumed to exist. If

f(x) =

by term from -L to

+ 5 (an n=1

nnx

cos-

L

+ bn sin-nEx),

then multiplying by f(x) and integrating term

L (which is justified since the series

is uniformly convergent) we obtain

where we have used the results La0

(2)

obtained from the Fourier coefficients. The required result follows on dividing both sides of ( I ) by L. Parseval's identity is valid under less restrictive conditions than that imposed here.

310 14.

FOURIER SERIES

[CHAP. 14

Write Parseval's identity corresponding to the Fourier series of Problem 12(b). 1 1 1 1 Determine from (a) the sum S of the series 14 9 34 . . . 2 . . . .

+ + +

4

(cosnr Here L = 2, a6 = 2, a, = n'r'

- l ) , n # 0,

bn = 0.

Then Parseval's identity becomes

f or

l:

s-+

=

{f(x)}' d x

-

xa d x

8 = 2 + p6 ( 41 L 4 + 31 + g 1+ . . ) , 3

15. Prove that for all positive integers

(W 2

+

* 16 ~ ( c o s n -r 1)' nrln~

=. -. P4 F 1 + 1+ +1g + - 96

i.e.

u4 s - 96 + 16,

+ +

from which S = 90 P4

M,

where U,, and b, are the Fourier coefficients corresponding to f ( x ) , and f ( x ) is assumed L). sectionally continuous in (4, For M = 1,2,3, We have

...

this is the sequence of partial sums of the Fourier series corresponding to f ( x ) .

Jl

{ f ( z )- SM(Z)}* dx 2

0

(9)

since the integrand is non-negative. Expanding the integrand, we obtain 21;

f ( 4SE&)

dx

-

ll

S;(x) d x

5

ll

{f(x)}' d x

(8)

Multiplying both sides of (1) by 2 f ( x ) and integrating from -L to L, using equations (2) of Problem 13, gives 2 f ( z )S M ( Z d) x = 2L n = 1 (a: bi)} (4)

1;

6+ 5

+

Also, squaring ( I ) and integrating from -L to L,using Problem 3, we And

Substitution of (4)and ( 5 ) into (8) and dividing by L yields the required result. Taking the limit as M

-+

OQ,

we obtain Bessel's inequality

If the equality holds, we have Parseval's identity (Problem 13). We can think of SM(Z) as representing an approximation to f ( x ) , while the left hand side of (a), divided by 2L,represents the mean square error of the approximation. Parseval's identity indicates that as 1111- OQ the mean square error approaches zero, while Bessel's inequality indicates the possibility that this mean square error does not approach zero. The results are connected with the idea of completeness of an orthonormal set. If, for example, we were to leave out one or more terms in a Fourier series (say cos 4axlL, for example) we could never get the mean square error to approach zero no matter how many terms we took. For an analogy with 3 dimensional vectors, see Problem 60.

CHAP. 141

311

FOURIER SERIES

DIFFERENTIATION and INTEGRATION of FOURIER SERIES 16. (U)Find a Fourier series for f ( x ) = x2, 0 < x < 2, by integrating the series of Problem 12(a). (b) Use (U)to evaluate the series (a) From Problem 12(a), x

=

(-I)%--1

n=l n2 ' m

A ( sin-ax - -1s i n2ax 2 + U 2 2

3sin1 3ax 2

-

...)

Integrating both sides from 0 to x (applying the theorem of Page 300) and multiplying by 2, we find 16 1 2ax 1 3ax x' = c -P '(c0sy -cos2' 2 p s 2 -

'..)

+

-

( b ) To determine C in another way, note that (a) represents the Fourier cosine aeries for x' in 0 < x < 2. Then since L = 2 in this case,

Then from the value of C in (a), we have

17. Show that term by term differentiation of the series in Problem 12(a) is not valid. Term by term differentiation yields

- cos=2

2

+ cos&2

-

-0.)

.

Since the nth term of this series does not approach 0, the series does not converge for any value of x.

CONVERGENCE of FOURIER SERIES 18. Prove that

We have

(a)

1 + cost + cos2t + . .. + cosMt

cos n t sin@

+

sin ( M #t 2 sin i t

= *{sin ( n + &)t - sin (n - 3 ) t ) .

Then summing from n = 1 to M, singt{cost

=

+ cos2t + ... + cos M t }

=

(sinit

+

- s i n i t ) + (sinjt - s i n j t ) ... + sin(M+*)t - s i n ( M - 4 1 9

(

= Q{sin(M+i)t - sinit} On dividing by s i n i t and adding

4,

the required result follows.

Integrate the result in (a)from -T to 0 and 0 to U respectively. This gives the required results, since the integrals of all the cosine terms are zero.

19. Prove that

lim

n-oo

1:

f ( x ) sinnx dx =

!zl: f ( x )

cosnx dx = 0

tionally continuous. This follows a t once from Problem 16, since if the series lim ccr = b, = 0. 00

!iy

The result is sometimes called Riemann'a theorem.

E!

2

if

+ 5 (at+ &) n=1

f(2)

is sec-

is convergent,

312

FOURIER S E R I E S

20. Prove that

l:

[CHAP. 14

f ( x )sin (M + Q)x dx = 0 if f(x) is sectionally continuous.

We have

J-: f (x)sin (M + Q)x d x

=

1;

+

{ f (5) sin Qx} cos Mx dz

{f(x) cos +E} sin Mz dz

Then the required result follows at once by using the result of Problem 19, with f ( z ) replaced by f ( x ) sin Qx and f ( x ) cos 8 x respectively which a r e sectionally continuous if f ( z ) is. The result can also be proved when the integration limits are a and b instead of -U and U.

21. Assuming that L = 7, i.e. that the Fourier series corresponding to f ( x ) has period 2L = 27, show that

Using the formulas for the Fourier coefficients with L = P, we have a, cos nx

+ b, sin nx

=

(: l: f ( u )cos nu du) cos nx

+

( k J - : f ( u ) sin nu

f ( u )cos n(u - x) du 2

Also, Then

SM(x) =

5 2

+

=

27T

1;

f ( u )du

Y

n=I

(a,cosnx

'

+ b,

nn=l

7T

using Problem 18. Letting

U

21

sinnx)

s" -R

f ( u )cos n(u - x) du

+ ' 2 cosn(u-x) n-1

- x = t, we have

Since the integrand has period 2a, we can replace the interval -P - 2, 'R - z by any other interval of length Z r , in particular -n,n. Thus we obtain the required result.

Subtracting ( 9 ) from ( I ) yields the required result.

FOURIER SERIES

CHAP. 141

23. If f ( z ) and

313

P(z) are sectionally continuous in lim S M ( X )=

(-T,T), prove that f ( x + 0) + f ( x - 0)

2

M-w

The function f ( t + $)sin - f('i t + O) is sectionally continuous in o < t s -a because ~ z is) sectionally continuous. - lim f ( t+ 4 - f(z + 0) Also, lim f ( t +z) - f('+ O) - lim f ( t +z) - f(z+ t t - ~ + 2 sinit t*O+ t 2 s i n i t - t+o+

.

exists, since by hypothesis f'(a) is sectionally continuous so that the right hand derivative of f(z) at each z exists.

Thus

f(t

+

z, - f ( z + 2 sinit

Similarly, f ( t

+

o s t s p.

is sectionally continuous in

$-

s t s 0.

O) is sectionally continuous in --a

Then from Problems 20 and 22, we have

BOUNDARY-VALUE PROBLEMS 24. Find a solution U(x,t) of the boundary-value problem t>0, o < x < 2

U(0,t ) = 0, U(2,t ) = 0 U ( z ,0 ) = x

t >0 o<x 0, while U ( x ,0) indicates the initial temperature a t any point x of the rod. In this problem the length of the rod is L = 2 units, while the diffusivity is k = 3 units.

ORTHOGONAL FUNCTIONS 25. (a) Show that the set of functions

7rx . 27rx 2XX 37rx 37rx sin-, L COST, s i n L , cos- L ' * ' . L' COST, forms an orthogonal set in the interval (-L, L).

XX

1, sin-

FOURIER SERIES

CHAP.141

316

( b ) Determine the corresponding normalizing constants for the set in (a) so that the

set is orthonormal in (4, L).

(a) This follows at once from the resulte of Problems 2 and 3. (b) By Problem 3, llLpX

1;cos'^ rnpX

= L,

sin'xdx

dx = L

Also, Thus the required orthonormal set is given by

MISCELLANEOUS 'PROBLEMS 26. Find a Fourier series for f ( x )= cos ax,

-X

S x 5 X , where a # 0, tl,*2, k3,

.. . .

We shall take the period as 2a so that 2L = 2a, L = U. Since the function is even, bn = 0 and an

4

L

=

;{

1 sin (a - n)r

=

Then cosas

Let

X=U

a

=

sin (a

tl'

- ~a'- c1'

cos ax cos nx dz

1-

+ n)a

a+n

-

2a sin ap u(2

COS

- n')

np

O0 cosnu cos 122 n=la'-n'

o

s

2a +x (II,cocos22

l-*)(l-*) (W

sinx = x ( l - $ (

(37+

2a --

....

in the Fourier series obtained in Problem 26. Then cosm

or

+

a-n

sinan - + -2a2sina7r U - @=(A a

27. Prove that

f ( ~ cos ) nx dx

-

T(; sinap

1

+d-l'+ 2a 2a + 'a - 2'

2a + d - 3'

..j

This result is of interest since it represents an expansion of the cotangent into partial fractions.

FOURIER S E R I E S

316

[CHAP.14

By the Weierstrass M test, the series on the right of (1)converges uniformly for 0 5 lal d 1x1 < 1 and the left hand side of ( 1 ) approaches zero as a+0, as is seen by using L'Hospital's rule. Thus we can integrate both sides of ( 1 ) from 0 to x to obtain

S

( r c o t a r - :)da

=

fm 2a da

=

*+a0

=

a+

0

+

+ ... is&& 2'

a'

or In(+)

i.e.,

!CEEE ux

=

lim

0400

Replacing z by z/r,we obtain

+ In (1

-

{ !% (1-

$)(I

-

m t '

+ In (1

-5)

g) (1- g)}

6-s)c-g) ...@-$)

=

sin% =

s) +

6-5))

lim00 ln{Q-g)(l-:)-.-

= In so that

( );

lim In 1 - -

x(l--~)(~-&).-,

(1-$)(1-~)

...

(3)

(8)

called the infinite product for sin r , which can be shown valid for all x. The result is of interest since it corresponds to a factorization of sin z in a manner analogous to factorization of a polynomial.

28. Prove that

* -2

Let x = 1/2 in equation (a) of Problem 27. Then,

Taking reciprocals of both sides, we obtain the required result which is often called W d W product.

FOURIER SERIES

CHAP. 141

317

Supplementary Problems FOURIER SERIES 29. Graph each of the following functions and And their corresponding Fourier series using properties of even and odd functions wherever applicable.

(0)

f(x)

AM.

=

(U)

22

4x, O C x C 10, Period 10

0

16 5 (1 - cos nu) sin n x n 2 U

(b) 2

n=t

40

OD

7zt 8

nrx

1

- 7s - t n-sin-

(c) 20

-

5

n=l

OD

(1

OS1F +7 sin32 sin6x 38 + 6a + cos42

**>

36. Use the preceding problem to show that

37. Showthat

1 1 F +F - F1 - F1 + F1 + F1 -

=

3 2 fi 16

cos

7

FOURIER SERIES

318

[CHAP.14

DIFFERENTIATION and INTEGRATION of FOURIER SERIES Show that for -U < x < U, sinx sin22 + sin32 x = 2(T2 3 By integrating the result of (a),show that f o r

--a

S x S -a,

--a

S x S T,

..>

3 By integrating the result of (b), show that for x(a - X ) ( P Show that for -r

+ x)

<x 0. (b) Interpret physically.

49. (a) Solve the boundary-value problem a'Y

Y(z, 0) = O.O6z(2 - 2), Yt(2, 0) = 0, A m . (a)Y(z,t) =

1.6 2 5

1 (2n - l ) u ~ 3(2n - 1)nt (2n- lysin 2 cos 2

50. Solve the boundary-value problem

+

au = aau at axis

U(0,t) = 1, U(u, t) = 3, U(%, 0) = 2.

[Hint Let U(%, t) = V ( z ,t) F ( z ) and choose F ( z ) so as to simplify the differential equation and boundary conditions for V ( z ,t).]

A m . U ( z ,t ) = 1

22 +7 +

4 cosma e"'''t mu

sinmz

51. Give a physical interpretation to Problem 60.

52. Solve Problem 49 with the boundary conditions for

Yt (2, 0) = O.O6x(2 - z), and give

Y(z,0) and Yt (2, 0) interchanged, i.e. Y ( z ,0) = 0,

a physical interpretation,

53. Verify that the boundary-value problem of Problem 24 actually has the solution (14),Page 314.

MISCELLANEOUS PROBLEMS 54. If - u C x C u and a # 0,*1,*2,

...,

prove that

sinaz -- -sin-z2 sin au 1'- a'

2 sin2x 2' -a'

U

55. If - P < x < u ,

3 sin 32 3'

- 'a

- ...

prove that U sinh a s s i-n z-- 2 sinhau a9+12

56. Prove that

+

sinhz

12a

2 sin22

3 sin 32

a * + ~ + ~ C O S X+

a'+

1'

a c 0 ~ 2-~... a'+ 2'

( ;:)( 1+ &!)(l+ &)*-

= z 1+ -

[Hint cos z = (sin 2 4 4 2 sin z).]

-

...

320

FOURIER SERIES

59. (a) Prove that if

a+

. . ., then

0,*l, ,2'

U

sin an (b) Prove that if

O 0, t > 0. We proceed 88 in Problem 24, Chapter 14. A solution satisfying the partial differential equation and the first boundary condition is given by Be-A't sin Ax. Unlike Problem 24, Chapter 14,the boundary conditions do not prescribe specific values for A, so we must assume that all values of A are possible. By analogy with that problem we sum over all possible values of A, which corresponds to an integration in this case, and are led to the possible solution

U ( x ,t )

=

i*

B(x) cALt sin Ax dx

where B(A) is undetermined. By the second condition, we have

from which we have by Fourier's integral formula

B(A)

=

:im 21* =

f ( x ) sin Ax dx

U

sin Xx dx

=

-aX

so that, at least formally, the solution is given by

See Problem 26.

14. Show that e-%'I2 is its own Fourier transform. Since e-='/' is even, its Fourier transform is given by

&

lm e-$Jg cos xa

dx.

Letting x = f l u and using Problem 32, Chapter 12, the integral becomes

which proves the required result.

15. Solve the integral equation

Y(4

= g(x)

+

l*2/(u)r(z--u)du

where g(z) and r ( z ) are given. Suppose that the Fourier transforms of ~ ( x )g(x) , and r(x) exist, and denote them by Y(a),G(a) and R(a) reepectively. Then taking the Fourier transform of both sides of the given integral equation, we have by the convolution theorem

Y(a) =

G(a)

+ 6Y(a)R(a)

or

Y(a) =

Gb)

1 - &R(a)

CHAP. 161

329

FOURIER INTEGRALS

Then assuming this integral exists.

Supplementary Problems The FOURIER INTEGRAL and FOURIER TRANSFORMS

{ '62'

f(x) =

16. (a) Find the Fourier transform of

(b) Determine the limit of this transform as

17. (a) Find the Fourier transform of

x cos x - sin x

(b) Evaluate

I

e -*

O+ and discuss the result.

1-x*

f(x) =

) cos:

$ 1x1

0

by using the result in (a).

(c) Explain from the viewpoint of Fourier's integral theorem why the result in (b) does not hold for m = 0.

Ans. (a)

[ a / ( l4- a*)]

20. Solve for Y(x) the integral equation

lm

Y(x) sin xt dx

= {

1 2

OSt 0.1

A m . (a) d 4 , ( b ) d 4

330

FOURIER INTEGRALS

22. Use Problem 18 to show that

= :,

)'dz

(x cos x - sin x ) ~ dx =

23. Show that

X6

[CHAP. 15

(b) f m F d s =

g.

6.

MISCELLANEOUS PROBLEMS 24.

(a)Solve

au = 2 s, at U( 0,t ) = 0, a2u

U(x,0) = e-*, x > 0, U(x, t) is bounded where z > 0, t > 0.

(b) Give a physical interpretation. An8. U(%,t ) =

26.

2P

lm A'+

sin " dh 1

(a)Show that the solution to Problem 13 can be written

U ( z ,t )

=

$l

+/a#

e-"' dv

(b) Prove directly that the function in (a)satisfies 27. Verify the convolution theorem for the functions 28. Establish equation

-

e-$ dv

at =

and the conditions of Problem 13.

f ( z ) = g(z)

=

I'

lx' 0 at time t = 0 and is released from rest, i.e. then c = - ( g l l ) cos 8,. We thus have

p = de/dt = 0 when e = e,,

deldt

=

Ifr

m d c o s e - cos 8,

When the pendulum goes froni e = e o to e = O (which corresponds to one quarter of a period, or T/4), del& is negative; hence choose the - sign. Then

Thus

de sin*e,/2

- sin'

e/2

Letting sin e/2 = sin e,/2 sin U, this integral becomes 7

T

=

4

~

~

~

du

*

'

'

dl - k1 sin'u

,

k = sin e,/2

If k = 0 , T = 2 a a which is the approximate period for small oscillations.

26. Prove that

sn (U + v) =

+

s n u c n v dnv cnu s n v d n u 1 - k2 sn2u sn2v

Let U + v = a, a constant. Then dvldu = -1.

Let us define U = s n u , V = sn U . It follows

that where dots denote differentiation with respect to

@

= ( 1 - US)(l-kgU*)

Then

U.

@ =

and

( l - V * ) ( l - k*V')

Differentiating and simplifying, we find (I)

If

= 2k*U8 - ( 1 + k * ) U

(2)

= 2k*V3 - (1

+ k')V

Multiplying (I) by V , ( 2 ) by U , and subtracting, we have

EV - U? = 2k*UV(UZ- V * )

It can be verified that (see Problem 68) 6 * V Z- U** = (1 - k1ULV2)(V* - U*) or

fiv-u3

(4)

= ( 1 - k*U'V)(V' - U*)

irv + u t

Dividing equations (3) and ( 5 ) we have - U'it.

irvBut 6 V

U+

= - 2k'UV( fiV + U+) 1 - L'UV

- Uv = ( f i V - U+) and - Bk'UV(fiV + Ut> = -(1 - k*LJIVL), so that du du d(fiV - Ut) - d ( l - k'LJLV') irv- U P 1 - k'VV' d

d

( 6 ) becomes

342

ELLIPTIC INTEGRALS

fiv - k-'xUv p -

An integration yields

[CHAP. 16

constant), i.e.,

c (a

+

cnusnvdnu = snucnvdnv 1 - kx snxu sn'v is a solution of the differential equation. It is also clear that u + v = solutions must be related as follows:

+

cnusnvdnu anucnvdnv 1 - k' s n x u an*v

Q

is a solution. The two

= 3 sin +,

= 2 cos + be the para-

= f(u+V)

Putting v = 0, we 8ee that f ( u ) = sn U. Then f(u

+

= an (tc + v ) and the required result follows.

U)

Supplementary Problems ELLIPTIC INTEGRALS 27. (a) Use the binomial theorem to show that if 1x1 < 1,

(b) If lkl

< 1, prove that =

6 ( k ,d 2 )

iU" \/1

kx sin' e de

28. Evaluate (a)E ( f i / 2 , d2), ( b ) F ( f i / 2 , d 2 ) , (c) E(O.6), (d)K ( 6 / 2 ) .

Ans. (a)1.3606, ( b ) 1.8641, (c) 1.4676, (d) 2.1666

a. 30. Find the perimeter of the ellipse a?/9 Am. 16.865 metric equations.] 31. Evaluate

32. Express

lu" Itd-

A w . LF(L

dsin'x

+ 2 cosas

dx

4

fi f i J

33. Show that

dx

[Hint: Let

in terms of elliptic integrals.

1 d-. +

#

d# 2 sine

3c

$K(+)

AW.

in terms of elliptic integrals.

, where 1

+ yx/4 = 1.

= sin"

(6sin t )

- F($, I

sin-*

4).

CHAP. 161 34.

ELLIPTIC INTEGRALS

Evaluate

l1

dX d x ( 1 - x)(1+2) dx

35.

.

Ans. fiK(L)

fi

(e,5>

Ans. 6 '{F

Im&

-F

Show that

37.

Express each of the following in terms of elliptic integrals.

(b)

i2

s' 3

(G,t)}

=

36.

(a)

343

s'

dx d ( 1 6 - x8)(25 - z ')

(c)

dx d ( x 2 - 9)(26 - x*)

( d ) Jm

d=dz 1-x2 dx d ( x 4 16)(x44- 25)

+

o

38.

39.

i m + + + + + dx

40. Show t h a t

[Hint: x 4 + x'

1 = (x'+ x

1)(x2- x

1

'3

3

1

XZ

Jx4

l).]

10

41.

Evaluate

(a)

CEX

4 ( x 2- 2 ~ , +4)(x2- 42

dx

+ 8)

1

d(XP

- 4x + 5)(x2- 4x + 10)

*

ELLIPTIC FUNCTIONS 42.

Show t h a t (a) snO = 0, (b) cnO = 1, (c) dnO = 1, ( d ) tnO = 0, (e) a m 0 = 0.

43.

Prove that (a)sn'u

44. Prove t h a t dn' 45. 46.

47.

Prove that

U

- k2 cn2U

(a) s n 2 u =

(b) dn'u

+ Ic'

s n 2 u = 1.

= kt2 where k' = J/-. 1 - cn 2u - sn U dn U 1 + c n 2 u - cn U

1 - cn 2u 1 dn 2 u '

+

d d Find (a) -(sn U cn U), (b) - (dn 14)~. du du Ans. (a) (cn' U - sn' U)dn U, (b) - 3k2 dn2U sn U cn U d Find (a) -(tn du

U),

48. Verify the results

49. Show t h a t

50.

+ cn2u = 1,

d - sech-l ( k sn U ) . du

(a)

cn U du =

sn2x dx =

1

1

dn u

cn U

Ans. (a) cn2u' ( b ) K

cos-' (dn U)

+ c,

(b)

du

= In(

snu cnu+dnu

)+

c.

{U - E ( k , am U)}.

cn U k' sn U (a) s n ( u $ - K ) =(b) c n ( u + K ) = -=, dnu' is the complementary modulus.

Prove that J-/

lu

(b)

( c ) d n ( u + K ) =- k'

dn u

where k' =

344 51.

ELLIPTIC INTEGRALS

f

dv

[CHAP. 16

= tn-'(x,k) = F(k, tan''%) where k' = \r(l+ vL)(l k ' V ) write tn-I ( x , k) briefly a s tn-I x , the modulus k being understood. Prove that

+

4 G .

We often

MISCELLANEOUS PROBLEMS 53. Prove that 54.

l*sin-lz'dx

= 5 2

-

2fiE($,t)

lr

+

fiE'(1,;).

fi

A pendulum of length 2 f t is released at a position in which it makes a n angle of 60° with the vertical. Determine its period of oscillation assuming the acceleration due to gravity = g = 32 ft/sec'. A m . 1.686 seconds

55. Show that a t any time t the angle e for the pendulum of Problem 26 is given by

sin e/2

= sin e,/2 sn (mt)

where t is measured from the instant where the pendulum rod is vertical.

[Hint sin e

de d s i n e cos8

Ttl

56. Show that

+

+ cos e

=

fi sin (e + n/4).]

57. Obtain the expansion

snu

=

U

U ' - (1 + kL)g

+ (1 + 14k'+ k') $ - (1 + 136 + 1 3 L ' + k') fu7i + :'

58. Verify equation (4)of Problem 26. 59. Prove that

-

(a) c n ( u + v )

=

cnucnv

(b)

=

dnudnv - k'snusnvcnucnv 1 - k' sn'u sn'v

1

-

snusnvdnudnv k' sn'u sn'v

60. Evaluate (a)F(@2, a/3),(b) F(0.6,a/4) by using Landen's transformation. A m . (a)1.1424, ( b ) 0.8044 61. Verify the result ( I J ) , Page 332, using Landen's transformation.

62. Prove that if k, =

2 d G , then 1 + kn-1

lim kn = 1. Hence verify (15), Page 333. a-,

00

* *

Chapter 17 Functions of a Complex Variable FUNCTIONS If to each of a set of complex numbers which a variable x may assume there corresponds one or more values of a variable w, then w is called a function of the complex variable x , written w = f ( x ) . The fundamental operations with complex numbers have already been considered in Chapter 1. A function is single-valued if for each value of x there corresponds only one value of w; otherwise it is multiple-valued or many-valued. In general we can write w = f ( x ) = u(x,y) i v ( x ,y), where U and v are real functions of x and y.

+

Example: w = za = (z+iy)) = za-ya+2ixy = u + i v so that u ( z , y ) = z L - y L , v(z,y) = 2x21. These are called the real and imaginary parts of w = zL respectively.

Unless otherwise specified we shall assume that f ( z ) is single-valued. A function which is multiple-valued can be considered as a collection of single-valued functions.

LIMITS and CONTINUITY Definitions of limits and continuity for functions of a complex variable are analogous to those for a real variable. Thus f ( x ) is said to have the limit l as x approaches xo if, given any c > 0, there exists a 6 > 0 such that lf(z) - lI < c whenever 0 < Iz - 2 0 1 < 6. Similarly, f ( x ) is said to be continuous at xo if, given any c > 0, there exists a 6 > 0 such that If@) - f(zo)l < whenever Ix - 201 < 6. Alternatively, f ( x ) is continuous a t xo if lim f ( x ) = f ( x 0 ) . z e xo

DERIVATIVES If f ( x ) is single-valued in some region of the x plane the derivative of f ( x ) , denoted by f ' ( x ) , is defined as

provided the limit exists independent of the manner in which A X + O . If the limit ( I ) exists for x = 20, then f ( x ) is called analytic at 20. If the limit exists for all x in a region T , then f ( x ) is called analytic in T . In order to be analytic, f ( x ) must be single-valued and continuous. The converse, however, is not necessarily true. We define elementary functions of a complex variable by a natural extension of the corresponding functions of a real variable. Where series expansions for real functions f ( x ) exist, we can use as definition the series with x replaced by x . The convergence of such complex series has already been considered in Chapter 11. z' - -z6 + _ 4! 6!

...

zs + x + -+ 2!

+

+

+

z3 z5 z7 ZL - - - '", cos2 = 1 - 3! 5 ! 7! 2! From these we can show that eL = eZti* = er (cosy i sin y), as well

Example 1: We define eL = 1

z3 %+ .-., sinz = z

as numerous other relations.

345

--

+

FUNCTIONS O F A COMPLEX VARIABLE

346 Example 2:

[CHAP. 17

We define ab as eblno even when a and b are complex numbers. Since eSkd= 1, i t follows that e'b = e'cb+ekn)and we define In x = In (peQ) = In p i(@ 2k~). Thus In z is a many-valued function. The various single-valued functions of which this manyvalued function is composed are called its branches.

+ +

Rules for differentiating functions of a complex variable are much the same as for d d those of real variables. Thus dx (x") = nxt'-l, d z (sin x ) = cos x , etc.

CAUCHY-RIEMANN EQUATIONS A necessary condition that w = f ( z ) = u ( x , g ) + i v ( x ,y) be analytic in a region 5l( is that U and v satisfy the Cauchy-Riemann equations

(see Problem 7). If the partial derivatives in ( 2 ) are continuous in q,the equations are sufficient conditions that f ( x ) be analytic in q. If the second derivatives of U and v with respect to x and y exist and are continuous, we find by differentiating ( 2 ) that

Thus the real and imaginary parts satisfy Laplace's equation in two dimensions. tions satisfying Laplace's equation are called harmonic functions.

Func-

INTEGRALS If f ( x ) is defined, single-valued and continuous in a region q,we define the integral of f ( z ) along some path C in from point x1 to point 22, where 21 = X I i y ~ 22 , = x2 iy2, as

s,

f(x)dx

s

(Z¶,V¶)

=

(u+iv)(dx+idy)

s

(X2.Y¶)

=

(Z,,Yl)

udx-vdy

(Z1,trl)

+ +

is

+

(XZ'Yt)

vdx+udy

(XI. Y1)

with this definition the integral of a function of a complex variable can be made to depend on line integrals for real functions already considered in Chapter 10. An alternative definition based on the limit of a sum, as for functions of a real variable, can also be formulated and turns out to be equivalent to the one above. The rules for complex integration are similar to those for real integrals. An important result is

where M is an upper bound of If(z)l on C, i.e. If(x)l 5 M, and L is the length of the path C.

CAUCHY'S THEOREM Let C be a simple closed curve. If f ( x ) is analytic within the region bounded by C as well as on C, then we have Cauchg's theorem that if(z)dz =

9;,

f W 2

=

0

where the second integral emphasizes the fact that C is a simple closed curve.

347

FUNCTIONS OF A COMPLEX VARIABLE

CHAP. 171

Expressed in another way, ( 5 ) is equivalent to the statement that

ly

f ( x ) dx

has a

value independent o f the path joining XI and 22. Such integrals can be evaluated as These results are similar to corresponding results for F(x2) - F(x1) where F’(z) = f ( x ) . line integrals developed in Chapter 10. Example: Since f ( z ) = 22 is analytic everywhere, we have for any simple closed curve C i2zdz

Also,

= 0 = (1 +i)’ - (2iy = 2i

+4

CAUCHY’S INTEGRAL FORMULAS If f ( x ) is analytic within and on a simple closed curve C and a is any point interior to C , then where C is traversed in the positive (counterclockwise) sense. Also, the nth derivative of f ( x ) at z = a is given by

These are called Cauchy’s integral formulas. They are quite remarkable because they show that if the function f ( x ) is known on the closed curve C then i t is also known within C, and the various derivatives at points within C can be calculated. Thus if a function of a complex variable has a first derivative, it has all higher derivatives as well. This of course is not necessarily true for functions of real variables.

TAYLOR’S SERIES Let f ( x ) be analytic inside and on a circle having its center at x=a. Then for all points x in the circle we have the Taglor series representation of f ( x ) given by

See Problem 21.

SINGULAR POINTS A singular point of a function f ( z ) is a value of x at which f ( x ) fails to be analytic. If f ( x ) is analytic everywhere in some region except at an interior point x=a, we call x = a an isolated singularity of f ( x ) . Example: If f ( z ) = - then x = 3 is an isolated singularity of f(z). ( 2 - 3)s ’

POLES If

=’(’) ‘ ( a ) # 0, where +(x) is analytic everywhere in a region including ( 2 - a). ’ z = a , and if n is a positive integer, then f ( z ) has a n isolated singularity a t x = a which is called a pole of order n. If n = 1, the pole is often called a simple poZe; if n = 2 i t is called a double pole, etc. f(2)

348

FUNCTIONS OF A COMPLEX VARIABLE

[CHAP. 17

z

Example 1: f(z) = ( z - 3)1(% + 1) has two singularities: a pole of order 2 or double pole at z = 3,

and a pole of order 1 or simple pole at z=-1. Example 2: f(z) =

32-1

-

-

32

(z

-1

+ 29(z - 29

has two simple poles at x = *2i.

A function can have other types of singularities besides poles. For example, f ( x ) = fi sin x has a branch point at x = 0 (see Problem 37). The function f ( x ) = - has a singularity x sin x is finite, we call such a singularity a at x = O . However, due to the fact that lim2-0 2 removable singularity. LAURENT'S SERIES If f(z) has a pole of order n at x = a but is analytic a t every other point inside and on a circle C with center a t a, then ( x - a). f ( x ) is analytic at all points inside and on C and has a Taylor series about x = a so that

+ f@) = ( 2 - a). a-,

U-*+

1

(2 - f2)n-I

a-1 + ... + + 2-U

+ a1(x-a) + az(x-a)2 + ( 9 ) a0 + ( x - a) + a2(x - a)2+ . . . is called a0

a . .

This is called a Laurent series for f(x). The part a1 the analytic part, while the remainder consisting of inverse powers of x - a is called the

principal part. More generally, we refer to the series

k=-m

ak(x

- a)k as a Laurent series

where the terms with k < O constitute the principal part. A function which is analytic in a region bounded by two concentric circles having center at x = a can always be expanded into such a Laurent series (see Problem 92). It is possible to define various types of singularities of a function f ( x ) from its Laurent series. For example, when the principal part of a Laurent series has a finite number of terms and a-n # 0 while a-,,- 1, a-, - 2, . . . are all zero, then x = a is a pole of order n. If the principal part has infinitely many terms, x = a is called an essential singularity or sometimes a pole o f infinite order. 1 +1 + . . . has an essential singularity at z = 0. Example: The function e l l X = 1 + ; 2!

2'

RESIDUES The coefficients in ( 9 ) can be obtained in the customary manner by writing the coefficients for the Taylor series corresponding to ( x - a), f(x). In further developments, the coefficient a-I, called the residue of f(x) at the pole x = a, is of considerable importance. It can be found from the formula 1 dn-' a-1 = lim((x-~)~f(x)> r - r a (n- 1) ! dxn-1 where n is the order of the pole. For simple poles the calculation of the residue is of particular simplicity since it reduces to =

lim ( x - a ) f ( z ) z-ra

(14

RESIDUE THEOREM If f ( x ) is analytic in a region % except for a pole of order n at x = a and if C is any simple closed curve in % containing x=a, then f(z) has the form (9). Integrating ( 9 ) , using the fact that

FUNCTIONS OF A COMPLEX VARIABLE

CHAP. 171

349

(see Problem 13), it follows that

i i.e. the integral of f ( x ) around a closed path enclosing a single pole of f ( x ) is 2 ~ times the residue at the pole.

More generally, we have the following important except at a Theorem. If f ( x ) is analytic within and on the boundary C of a region finite number of poles a, b, c, . . . within q,having residues a-I, L 1 ,c-1, . . . respectively, then f ( x ) dx = Z~i(a-1 b-I c-1 . . .) (14)

5

+

+

+

i.e. the integral of f ( x ) is 2ri times the sum of the residues of f ( x ) at the poles enclosed by C. Cauchy's theorem and integral formulas are special cases of this result which we call the residue theorem.

EVALUATION of DEFINITE INTEGRALS The evaluation of various definite integrals can often be achieved by using the residue theorem together with a suitable function f ( x ) and a suitable path o r contour C, the choice of which may require great ingenuity. The following types are most common in practice.

1. J w F ( x ) dx, F ( x ) is an even function. 0

Consider

$ F(x)dx

along a contour C consisting of the line along the x axis

from -R to +R and the semi-circle above the x axis having this line as diameter. Then let R+ 00. See Problems 29, 30.

2.

x2T -

G(sin 8, cos 8) do, Let x = e?

G is a rational function of sin 8 and cos 8. x

- 2-1

2

Then sine = -, COS^ =-

2i

+

2-1

2

and dx = ieiede or do =

dx/ix. The given integral is equivalent to

F(x)dx where C is the unit circle 9;, with center at the origin. See Problems 31, 32. 3. l w F ( x ){ ~dx, F (~x ) is a rational ~ function. ~ ~ } Here we consider F(x)eimzdx where C is the same contour as that in 9;, Type 1. See Problem 34. 4. Miscellaneous integrals involving particular contours.

See Problems 35, 38.

350

[CHAP. 17

FUNCTIONS OF A COMPLEX VARIABLE

Solved Problems FUNCTIONS, LIMITS, CONTINUITY 1. Determine the locus represented by (c) ]x-33j 12-21 = 3, ( b ) Ix-21 = Ix+41,

+ Ix+31

= 10.

Method 1: 12-21 = 1 x + i y - 2 ( = Ix-2+iyl = d ( x - 2 ) 2 + y 2 = 3 or ( ~ - 2 ) ~ + = y ~ 9, a circle with center at (2,O) and radius 3. Method 2: Ix - 21 is the distance between the complex numbers x = x iy and 2 Oi. If this distance is always 3, the locus is a circle of radius 3 with center a t 2 Oi or (2,O).

+ d ( ~ - 2 )+~y2 = d(x+ 4)2 + y2.

+

+

+

Method 1: 1% iy- 21 = Ix+iy+41 or Squaring, we find x = -1, a straight line. Method 2: The locus is such that the distances from any point on it to (2,O) and (-4,0) are equal. Thus the locus is the perpendicular bisector of the line joining (2,O) and (-4,0), or x = - 1 . - 3)2 y' = Method 1: The locus is given by d(x- 3)2 y2 d(x 3)' g2 = 10 or I/(% 10 - d(x 3)2 y2. Squaring and simplifying, 25 3% = 5I/{x 3)2 g2. Squaring and

+ +

+ +

simplifying again yields

x2 + 218 25 16

+

+ +

+

+ +

= 1, an ellipse with semi-major and semi-minor axes of

lengths 5 and 4 respectively. Method 2: The locus is such that the sum of the distances from any point on it to (3,O) and (-3,0) is 10. Thus the locus is an ellipse whose foci are at (-3,0) and (3,O) and whose major axis has length 10.

2.

Determine the region in the x plane represented by each of the following. 1x1 < 1. Interior of a circle of radius 1. See Fig. 17-l(u) below.

1 < I2+2il 5 2. ( x + 2il is the distance from x to -2i, so that Ix+2il = 1 is a circle of radius 1 with center at -2i, i.e. (0, -2); and Ix 2i[ = 2 is a circle of radius 2 with center at -2i. Then 1 < I x 2il S 2 represents the region exterior to I x + 2il = 1 but interior to or on Ix+2iI = 2. See Fig. 17-l(b) below.

+

+

~ / 5 3 argx 5 d 2 . Note that argz = $, where x = pe'b. The required region is the infinite region bounded by the lines + = ~ / 3and $ = ~ / 2 ,including these lines. See Fig. 17-l(c) below.

3. Express each function in the form U($,y) (a) x3, ( b ) 1/(1-x ) , (c) e32, ( d ) In x . (U)

w = x3 = ( X

+

+

i ~ = ) x3 ~ 3x2(iy ) = x 3 - 3x32 i(3x2y - y3)

Then

Fig.17-1 (c)

Fig. 17-1 ( b )

Fig.17-1 (a)

+

+ i v ( x ,y),

where

U

+ 3 ~ ( i y+) ~( i ~=) x3~ + 3ix2y - 3

u(x,y) = x 3 - 3xy2, v(x,y) = 3x2y - y3.

and v are real: %- ~iy* ~

FUNCTIONS OF A COMPLEX VARIABLE

CHAP. 171

(b) w

=

1 1 1 - (x 3- iy) 1--2-iy

1 1-2 -

( d ) In z = In (pe'b) = lnp

+ ln(xe+y'),

=

U

- 1-x+iy - (1 - x)* y'

1-x+iy 1-x+iy

+ + = In i x *+ y* + i tan-'

351

+

ylx

and

v = tan-'y/x

Note t h a t In z is a multiple-valued function (in this case it is infinitely many-valued) since can be increased by any multiple of 2 ~ .The principal value of the logarithm is defined as that value f o r which 0 S d < 2~ and is called the principal branch of In z. $

4.

Prove

+ cos (x+ iy)

+ i cos x sinh y

(a) sin (x iy) = sin x cosh y

(b)

= cos x cosh y - i sin x sinh y.

+ i sin z,

eir = cos z

We use the relations

ek

sin z =

= sin x cosh y

2i

= cos z - i sin z, from which ,c +

'

=

cos

e-ix

2 -

+ i cos x sinh y + e-i(r+iY) 2 e-'"+y} = +{e-v(cos x ei(z+iy)

= cos(x+iy) =

Similarly, cos z

- e-iz

e-*"

+{eir-y (cos x )

+

+ i sin x) + ew (cos x - i sin x)}

(7) x) ( F =) x i(sin

cos

cosh y - i sin z sinhy

DERIVATIVES. CAUCHY-RIEMANN EQUATIONS 5.

d Prove that -8? dx

By definition,

where 8 is the conjugate of x, does not exist anywhere. d

-f(z) dz

=

jlzo f ( z

- f ( z ) if this limit exists independent of the manner in

Az which Az = Axi-iAy approaches zero. Then d ~2

=

-

-

lim Z - k A z - 5

AzeO

A2

+

lim x + i y + A ~ + i A y- x + & Ax 4- i Ay

Az- 0

AV-0

lim ~ - i i 2 / + A ~ - i A y - ( ~ - i i y ) = A=+ 0 Ax iAy Am-

0

+

If Ay = 0, the required limit is If Ax = 0, the required limit is

lim

Ax = 1.

lim

-iAy - -1.

lim AX - iAy

AZ+Q

Aw*O

AX 4- i Ay

hz+O AX

Am-0

$Ay

These two possible approaches show that the limit depends on the manner in which Az+O, so that the derivative does not exist; i.e. Z is non-analytic anywhere.

362 6.

[CHAP. 17

FUNCTIONS O F A COMPLEX VARIABLE (U) If

'+'

dw

find w = f(x) = 1-2' dz '

(a) Method 1:

dw

= -

lim

+ +

(4 Determine

1 (z Az) 1 - (z+Az)

-

- -l + z

Az

Az-b 0

a

(1 - 4

where w is non-analytic.

1-2

=

lim

At+o

2 (1 - Z - AZ)(l - 2 )

provided z # 1, independent of the manner in which Az + 0.

Method 2: The usual rules of differentiation apply provided z # 1. Thus by the quotient rule for differentiation, d d (1 - 2 ) -&(1 2 ) - (1 2 ) -(1 - 2 ) dz 2 - (1 - z)(l) - (1 z)(-l) (1 - 2)s (1 - 2)' (1 - 2 ) s

+

+

+

( b ) The function is analytic everywhere except at z = 1, where the derivative does not exist; i.e. the function is non-analytic a t z = 1.

+ iv(x,y)

7. Prove that a necessary condition for w = f ( x ) = u(x,y) du av -a region is that the Cauchy-Riemann equations 9~ ay' the region. Since f(z) = f ( x iy) = U(%,y) + i v ( x , y), we have f ( z 4- Az) = f[x 4- A X 4- i(y + Ay)] = u(x 4- Ax, y -tAy)

au -

ay

+

-k i v ( s

Then

to be analytic in

-? ax

be satisfied in

+ Ax, y + Ay)

If Ax=O, the required limit is

If the derivative is to exist, these two special limits must be equal, i.e.,

Conversely, we can prove that if the first partial derivatives of U and v with respect to x and y a r e continuous in a region, then the Cauchy-Riemann equations provide sufficient conditions for f ( z ) to be analytic.

8.

+

(U)If f ( x ) = u(x, y) i v(x,y) is analytic in a region T,prove that the one parameter families of curves u(x, y) = CI and v(x,y) = CZ are orthogonal families. (b) Illustrate by using f ( z ) = x2.

(a) Consider any two particular members of these families u ( x ,y) = %, v(x,y) = vo which intersect at the point ( x o , ~ o ) .

+ uydy = 0, we have dv = vr dx + vydg = 0, dx

Since du = u z d x Also since

dy = dx

= - -.

v8 VY

US -_

uy '

FUNCTIONS OF A COMPLEX VARIABLE

CHAP. 171

353

When evaluated at ( z o , ~ o )these , represent respectively the slopes of the two curves at this point of intersection. By the Cauchy-Riemann equations, u z = v y , uy=-vl, we have the product of the slopes at the point (z0,yo) equal to

(+)(p)

=

-1 X

so that any two members of the respective families are orthogonal, and thus the two families are orthogonal. ( b ) If f ( z ) = z’, then U = 2’- ys, v = 22y. The graphs of several members of 2’-y’ = CI, 2xy=Ce are shown in Fig. 17-2.

9.

Fig. 17-2

++

+

i$, where In aerodynamics and fluid mechanics, the functions and $ in f ( x ) = f ( x ) is analytic, are called the velocity potential and stream function respectively. If = z 2 + 4 x - y 2 + 2 y , (a) find $ and (b) find f ( x ) .

+

(U)

By the Cauchy-Riemann equations,

as

=

ay’

3 = -% !ay* as

a9 = 22 + 4 -

(I)

a#

(2)

Then

- 2y

-2

+ + F(z). + G(y).

Method 1: Integrating (I), 9 = 2sy 4y Integrating (2), 9 = 22y - 22

F(z) = -2s+c,

These a r e identical if 9 = 2xy+4y--x+c.

G(y) = 4 y + c

where c is any real constant.

Thus

Method 2: Integrating (I), 9 = 2 x y + 4 y + F ( s ) . Then substituting in (Z), 2y+F’(z) = 2 y - 2 F’(x) = -2 and F ( x ) = -22+c. Hence J , = 222(+4y-222+~. (b)

From (a), f ( z ) =

=

where y

=

+ 42 - y2 + 2y + i(2xy + 4y - 22 + c) + 2ky) + 4(2 + i ~ -) 2i(%4-ill) ic = 4- 42 - 2i2 +

++i~,= (%*

-

2%

Z*

CI is a pure imaginary constant. This can also be accomplished by noting that z = z + i y , Z = x - i y

Ci

=2 ’ The result is then obtained by substitution; the terms involving E drop out.

2-2 F .

so that

2

INTEGRALS, CAUCHY’S THEOREM, CAUCHY’S INTEGRAL FORMULAS 10. Evaluate

or

Jy

4i

x2 dx

(a) along the parabola x = t , y = t2 where 1 5 t i2, ( b ) along the straight line joining 1 i and 2 4i, ( c ) along straight lines from 1 i to 2 i and then to 2

+

We have z’dz

=

J

=

(2.4)

(1.1)

+

+

+ iy)’(dz + idy)

=

(za-g*)dx - 2xydy

+

(2.4)

(1.1)

+

(z

+ 4i.

J

(2,4)

(2’

(1,l) (2.4)

i s (1.11

- y2 + 2izy)(dz + idy)

2zydz

+ (d-y*)dy

Z+B

354

FUNCTIONS O F A COMPLEX VARIABLE

[CHAP. 17

Method 1: (a) The points (1,l)and (2,4) correspond to t = 1 and t = 2 respectively. Then the above line integrals become

is

+

{ ( t * - t') dt - 2(t)(t2)2tdt}

The line joining (1,l) and (2,4) has the equation we find

(b)

Xl +

From 1 i to 2

(c)

From 2

{2x(3~-2)dx

is' t=l

+ i \or (1,l)to (2,1)1,

+ i to 2 + 4i (4 + 329

4-1

=2-1 (x-1)

or

y

3

- 6i

= 32-2.

Then

+ [x2- ( 3 ~ - 2 2 ) ~ ] 3 d x } =

-_ 86 3

- 6i

= 1, dg = 0 and we have

y

[or (2,l) to (2,4)], x = 2, dx = 0 and we have

1=,-4ydy

Adding,

y -1

-86

=

- 2 ~ ( -32)3 ~ dx}

{ [x2- (32 - 2)2]dx

+

+ (t2- t4)(2t)dt}

{2(t)(t2)dt

t=l

+ (-

+

=

il=l(4-y2)dy

=

30 - 9;)

-y

-30

- 9i

- 6i.

Method 2: By the methods of Chapter 10 it is seen that the line integrals are independent of the path, thus accounting for the same values obtained in (a), (b) and ( c ) above. In such case the integral can be evaluated directly, a s for real variables, as follows:

11. (a) Prove Cauchy's theorem: If f ( x ) is analytic inside and on a simple closed curve C ,

then

$ f ( x ) d x = 0.

(b) Under these conditions prove that i : f ( z ) d z P I and Pz. =

$f(z)dz

=

$(u+iv)(dx+idy)

$s (-E-

is independent of the path joining

fudx-vdy

+

i $ vC d x + u d y

By Green's theorem (Chapter l O ) ,

=

iudx-vdy

$vdx+udy

$)dz:dy,

R

=

{J(g-g)dxdy

*

where (!5 is the region (simply-connected) bounded by C. au = av = au (Problem 7), and so the above integrals are Since f ( z ) is analytic, a s ay' ax ay zero. Then f ( z ) dz = 0, assuming f ' ( z ) [and thus the partial derivatives] to be continuous.

f

(b)

Consider any t w o paths joining points PI and Pt (see Fig. 17-3). By Cauchy's theorem,

S

f(z)dz

=

PI

0

Pl U F l

Then

f(z)dz

+

Pl'Pt

or

f PI *PI

f(z)dz

f(z)dz

=

f ( z )dz

=

0

P P 1

=

-

f P P l

f ( z ) dz

Fig. 17-3

PI BPI

i.e. the integral along PlAPr (path 1) = integral along PlBPr (path 2), and so the integral is independent of the path joining P1 and Pt. This explains the results of Problem 10, since f ( z ) = z'

is analytic.

FUNCTIONS O F A COMPLEX VARIABLE

CHAP. 171

355

12. If f ( x ) is analytic within and on the boundary of a region bounded by two closed curves CI and CB (see Fig. 17-4), prove that n

As in Fig. 17-4, construct line A B (called a crossCZ and a point on CI. By Cauchy's theorem (Problem l l ) ,

cut) connecting an y point on

S

f ( z )dz

=

0

Fig. 17-4

AQPABRSTBA

since f ( x ) is analytic within the region shaded and also on the boundary. Then

AQPA

AB

BRSTB

AQPA

BRSTB

BA

BTSRB

i.e. Note t h a t f ( x ) need not be analytic within curve CZ.

13. (a) Prove that

$&

=

1 { 20 ~ ififi nn == 2,3,4, ...

where C is a simple closed

curve bounding a region having x = a as interior point. ( b ) What is the value of the integral if n = 0, -1, -2, -3,

. ..?

(a) Let C1 be a circle of radius E having center at z = a (see Fig. 17-5). Since ( x - u ) - " is analytic within and on the boundary of the region bounded by C and C I , we have by Problem 12,

To evaluate this last intkgral, note t h a t on C1, lz- ul = e o r x - a = eeie and dx = ieeiede. The integral equals

If n = 1, the integral equals

ifT

Fig. 17-5

de = 2Ti.

( b ) F o r n = 0, -1, -2, . . . the integrand is 1, ( z - a), ( z - a)', . . . and is analytic everywhere inside C1, including z = a. Hence by Cauchy's theorem the integral is zero.

14. Evaluate

& zE3 -

where C is ( a ) the circle 1x1 = 1, ( b ) the circle Iz i-il = 4.

(a) Since z = 3 is not interior to 1x1

( b ) Since z = 3 is interior to Iz

= 1, the integral equals zero (Problem 11).

+ il = 4,

the integral equals 2ni (Problem 13).

356

FUNCTIONS OF A COMPLEX VARIABLE

[CHAP. 17

15. If f ( z ) is analytic inside and on a simple closed curve C, and a is any point within C, prove that Referring to Problem 12 and the figure of Problem 13, we have $ s a d z

=

dz

2-U

+

Letting z - a = re'e, the last integral becomes i is continuous. Hence

f(a

ce'@) de.

But since f ( z ) is analytic, i t

and the required result follows.

(a)

Since z = a lies within C, a

= r.

cos r = -1

by Problem 16 with f ( z ) = cos 2,

= -2ri.

Then $ sZ - dU z

2aie0 - !hie-l

=

= 2141 - e-l)

by Problem 15, since z = O and x = -1 are both interior to C.

17. Evaluate

$ 52"(;-"~~2 dx

where C is any simple closed curve enclosing x = 1.

Method 1: By Cauchy's integral formula,

f(")(a) =

2f

If n = 2 and f ( z ) = 6%'- 32+ 2, then f"(1) = 10.

+ 7(2- 1) + 4. Then 6(2 - 1)' + 7(2 - 1) + 4 dz

Method 2: 62'- 3 2 + 2 = 6 ( z - 1)'

= by Problem 13.

lOai

(z f(z)

Hence

dz.

357

FUNCTIONS O F A COMPLEX VARIABLE

CHAP. 171

SERIES and SINGULARITIES 18. For what values of x does each series converge?

By the ratio test the series converges if IzI

test fails.

However, the series of absolute values

51 converges. n'

n=:

2. If lzl = 2 the ratio

5 1n'2" ~ = 1 -& !

I=:

121= 2,

converges if

since

Thus the series converges (absolutely) for lzl 5 2, i.e. at all points inside and on the circle

1.1 = 2.

= o Then the series, which represents sinz, converges for all values of z.

lz-il 3

The series converges if lz-il

< 3,

and diverges if Iz-il

If I x - i l = 3, then z- i = 3ei@ and the series becomes the nth term does not approach zero a s n+ W .

.

> 3. m n=l

e"@. This series diverges since

Thus the series converges within the circle Iz - il = 3 but not on the boundary.

19. If

m

n=O

anxn is absolutely convergent for 1x1 S R, show that it is uniformly convergent

for these values of x . The definitions, theorems and proofs for series of complex numbers a r e analogous to those for real series. In this case we have

lanZnl

by the Weirstrass M test that

5 la,,lR" = M n . Since by hypothesis

5 a,z"

n=O

2

n=l

Mn converges, i t follows

converges uniformly for /zI5 R.

20. Locate in the finite x plane all the singularities, if any, of each function and name them. 2'

(z

+

l)s.

z = -1 is a pole of order 3.

+

2 2 -z 1 z = 4 is a pole of order 2 (double pole); z = i and z = 1 - 2i are (z- 4)*(z - i ) ( Z - 1 29 ' poles of order 1 (simple poles).

+

sin mz m # 0. Since n S + 2 z + 2 = 0 when z = - 2 + 9 4 - 8 - - --2+2izp 22 2 ' 2 2 can write 'z 22 2 = {z - (-1 i ) } { z- (-1 - i)} = (z 1 - i ) ( z 1 9.

+ +

+ +

+

The function has the two simple poles: z = -1

+

+i

+ +

and z = -1 - i .

-1

2

i, we

FUNCTIONS O F A COMPLEX VARIABLE

368 (a)

1 - cosz

-- 0,

singularity. However, since lim 1 - c o s z

z = 0 appears to be a

z

[CHAP. 17

r+O

z

it is a

removable singularity. Another method:

is a removable singularity.

1--

1

( z - 1)'

+

This is a Laurent series where the principal part has an infinite number of non-zero terms. Then z = 1 is an essential singularity.

This function has no finite singularity. However, letting z = l/u, we obtain ellY which has an essential singularity at U = 0. We conclude that z = QO is an essential singularity of e'. In general, to determine the nature of a possible singularity of f ( z ) at z = m, we let z = l/u and then examine the behavior of the new function at u=O.

21. If f ( z ) is analytic a t all points inside and on a circle of radius R with center a t a, and if a h is any point inside C , prove Taylor's theorem that

+

By Cauchy's integral formula (Problem 16), we have

By division,

-

1 z-a-h

-

1

( z - u ) [ ~- h / ( z - a ) ]

(2

-4

h

h'

(2- a)

(2- a)'

h"

( z - a)"

+

(z

- a). ( z - a - h)

Substituting ( 8 ) in (1) and using Cauchy's integral formulas, we have

R,

where

=

%.f

f (2) dz ( z - a)"+'( x - a - h)

and [z-aa( = R, so that by (4), Page 346, we have, since

Now when z is on C, 2nR is the length of C, 1R.I

As n + Q), 1R.I

+

0. Then Rn -+ 0 and the required result follows.

If f ( z ) is analytic in an annular region

rl d Iz-al

5 n , we can generalize the Taylor series

to a Laurent series (see Problem 92). In some cases, as shown in Problem 22, the Laurent series can

be obtained by use of known Taylor series.

FUNCTIONS OF A COMPLEX VARIABLE

CHAP. 171

369

22. Find Laurent series about the indicated singularity for each of the following functions. Name the singularity in each case and give the region of convergence of each series.

- -e.1)'

(a)

'

(2

Let z - 1 = U. Then z = 1 + U and

z = 1.

- 6"(2

- -,1

-

- 1)'

+U

US

- -

U '

+-2-1

e

6

(2-1)'

e e(z-1) +-+-+2!

e ( z - 1)'

+

41

3!

...

z = 1 i s a pole of order A?, or double pole. The series converges for all values of z # 1. (b) z

COS

1

-

'

2'

z=O.

zcosf

Z

z(l--

=

1 1 2!2'+412'-

G-2

+

..)

=

z - - + -1- - +

212

1 4!za

z = O is an essential singularity.

The series converges for all values of z # 0. (c)

sin z z--p; z=a.

Let

Then

Z - P = U .

sinz- - sin (u+P) -

-

U

2-U

and

Z = P + U

sin U --

-ua + us 3! 6!

1

=

U

U

z = P is a removable singularity.

The series converges for all values of z.

(d) ( z

+

Let z

1,"(+ 2) ; z = -1. +

+ 1 = U.

Then

U-1 - u(u+ 1) -

2

(Z+l)(Z+2)

U - 1 (1 - U

-

--

+

2 - 2u

+

-

--z + l

+2-

2(2+1)

U

2242 - 2u3

z = -1 is a pole of order 1 , or simple pole.

The series converges for values of z such that 0 (e)

-*

z(z

1

+ 2)d ' z = 0,-2.

Case 1 , z = 0.

Using the binomial theorem,

1 -- + 3 82

16

16

2

- -2'6

32

+ ...

z = 0 is a pole of order 1, or simple pole.

The series converges for 0

< IzI < 2.

+ U = - U'+

U4

- ...)

U

+

+

a . .

2(2+1)* - * . .

< Iz + 11 < 1.

1

6!z5

...

360

FUNCTIONS O F A COMPLEX VARIABLE Let z + 2 =

C u e 2, z = -2.

- 1z(z

+ 2)5

-

- 1 (U - 2)uS

-

Then

U.

1 -2u5(1- u/2)

z = -2 is a pole of order S.

The series converges for 0

[CHAP. 17

...}

-${l

-

< Iz + 2) < 2.

RESIDUES and the RESIDUE THEOREM 23. If f ( z ) is analytic everywhere inside and on a simple closed curve C except at x = u which is a pole of order n so that U-n a-n + 1 + . - * + U0 + al(2-U) + a2(2-a)2 + ...

+ f(2) = ( 2 - a).

where a-n (a)

f

(b) (U)

#

(2 -

0, prove that

f ( 2 ) d x = 27ria-1

=

1 dn-l lim --( ( 2 - ~ ) ~ f ( z ) > . %-+a ( n - l ) ! dP-'

By integration, we have on using Problem 13

=

2uiu-1

Since only the term involving ( b ) Multiplication by (2

(z-U)"

remains, we call a-1 the residue of f ( z ) at the pole

U-I

Z=U.

gives the Taylor series

- u).f(z) =

a-,

+

a-n+I(z-~a)

+

...

+

a-1(z-u)"-'

+

Taking the (n - 1)st derivative of both sides and letting z + a, we find dn-1

lim -{ ( z - u ) " f ( z ) } z-+o dP-'

=

(n-l)!u-l from which the required resuIt follows.

24. Determine the residues of each function at the indicated poles. 2'

(2

- 2 ) ( 2+ 1) ' *

z = 2,i,-i.

Residue at z = 2 is Residue at z = i is Residue a t z = -i is

These are simple poles. Then: lima

z-,

(2

- 2)

lim ( z - i ) 2 3i

m ;!,

(z

{ {

+ i)

}

Za

4 = S.

( z - 2)(2a + 1) (2

- 2)(22a- i ) ( 2 + i ) > = (i- 2)(2i)

(2

- 2 ) ( z - i ) ( z i)

Za

+

>-

2%

2%

- 1 - 2i -

(-i - 2)(-2i)

10

- -1 + 2i 10

FUNCTIONS O F A COMPLEX VARIABLE

CHAP. 171

(b)

i(xipx = 0,-2.

is a simple pole, z =--2

x=O

Residue at x = O is

lim x

230

*-

1 z(z +2)'

-

361

is a pole of order 3.

Then:

1

- 8'

Residue at x = -2 is

z(z

+ 2)s -

1 8'

Note that these residues can also be obtained from the coefficients of l / x and l/(z+2) in the respective Laurent series [see Problem 22(e)].

(4

zeXt x = 3, m;

a pole of order 2 or double pole. Then:

Residue is

{

lim dz ( x - 3)'

2-3

(2-3)'

d

= 2-9 lim -((xe") dZ = est

(d) cot x ; x = 57,

= lim (e" 2 33

+ xte")

+ 3te3t

a pole of order 1. Then:

Residue is

cos2 $I (2 - 5n) sin x

(-l)(-1) = 1 where we have used L'Hospital's rule, which can be shown applicable for functions of a complex variable.

25. If f ( x ) is analytic within and on a simple closed curve C except at a number of poles a, b, c, . . interior to C, prove that

.

9;,f(d dx

=

2 ~{sum i of residues of f ( x ) at poles a, b, c, etc.}

Refer to Fig. 17-6. By reasoning similar to that of Problem 12 (i.e. by constructing cross cuts from C to CI, CZ,Cs, etc.), we have

For pole a,

hence, as in Problem 23,

f ( x ) dx = 2 ~ i a - I .

Similarly for pole b,

so that Continuing in this manner, we see that

f f(x)dx

=

+ b-1 + ...)

27i(u-1

=

2 ~ i ( s u mof residues)

[CHAP. 17

FUNCTIONS OF A COMPLEX VARIABLE

362 26*

(2

ez dx where C is given by (U) 1x1 = 3/2, ( b ) 1x1 = 10. - l)(x + 3)2

Residue at simple pole x = 1 is

ez

lim

(2

x 41

- l)(x

+ 3)2} = ;

Residue at double pole x = -3 is

et (2

x 4 -3

-l)(x

+ 3)Z

= lim x4-3

( x - l ) e x - ex -

1)2

(2-

-

-5e-S 16

(a) Since zI = 3/2 encloses only the pole x = 1,

= 2~i($)

the required integral (b)

XI

Since

=

Tie 8

= 10 encloses both poles x = 1 and x = -3, the required integral

(& 5)=

= 2 ~ i

-

Ti(e - 5 e - s ) 8

EVALUATION of DEFINITE INTEGRALS 27. If

If(x)l 5

M Rk for

x = Reie, where k > 1 and

$El, f ( x ) dx

M

/.

are constants,

prove that

= 0

where r is the semi-circular arc of radius R shown in Fig. 17-7. By the result ( 4 ) , Page 346, we have

since the length of arc L = TR. Then

M

1

28. Show that for x = Reie, If(x)l 5 F , k > 1 if f ( x ) = 1 + ~ 4 * ~f 2

=~ e i e ,

~ f ( z ) l= 11

1

+ R4e4ieI

for example) so that M = 2 , k = 4 .

1 -5 IR4e4{f)1 R4- 1 -1

Note that we have made use of the inequality 1x1

29. Evaluate Consider

+

2 if R is large enough (say R > 2, -

I - R4

1 1x11 - lxzl with zi = R4e4(0 and

22

= 1.

&.

f&,

where C is the closed contour of Problem 27 consisting of the line from

-R to R and the-semi-circle I‘, traversed in the positive (counterclockwise) sense. Since x 4 + 1 = 0 when = #i14, e 3 T i / 4 9 e5?ri14 9 e7aa4, these are simple poles of l/(x4 poles eriJ4and 8ni/4 lie within C . Then using L’Hospital’s rule,

+ 1).

Only the

363

FUNCTIONS OF A COMPLEX VARIABLE

CHAP. 171

=

Residue at err/4

lim

z-, er414

{(z - eriI4

Thus

i.e.

Taking the limit of both sides of (2) as R +

x2dx

30* Show that

s_",(x2+ 1)2(x2+ 2x + 2)

The poles of

+ l)*(z*+ 22 + 2) ZL

(2'

and using the results of Problem 28, we have

OQ

77r - 50

enclosed by the contour C of Problem 27 are z = i of order 2

and z = - l + i of order 1. Residue at z = i is

lim -& fiz - 21' r-+t dz

Residueat x = - l + i

is

$ + %f:+

Then

-1

+4

(z+l--q

za dz l)"2 22

(2'

or

lim

z-,

+ zy(z - i)*(z* + 22 + 2) } 2

(2

(x2

+ + 2)

xadx l)L(X'

+ 22 + 2)

+

9i- 12

= TCiii-.

2'

(2'

+ I)+ + 1 - i ) ( z + 1 + 9

s,

26

26

78 - -

zLdz

(2'

3 - 4i - -

+ l)*(z' + 22 + 2)

60

7a -

60

Taking the limit as R + 00 and noting that the second integral approaches zero by Problem 27, we obtain the required result.

31. Evaluate

2n

5

do

+ 3sine' sine =

Let z = ei9. Then

x

2 1 ~

6

de

+ 3 sin 8

et9 - e-@ -2i - 2i

'- '-', dz = iei9 de = iz de

-

-

$6

dzliz

+ 3(q)

=

so that

2 dz f31'+10C-3

where C is the circle of unit radius with center a t the origin, as shown in Fig. 17-8 below.

364

[CHAP. 17

FUNCTIONS OF A COMPLEX VARIABLE

The poles of

32'

2

+ loix - 3 -loi

x =

are the simple poles f

d=iiGT% 6

-

-lOi*

8i

6

= -3i, -i/3.

Fig. 17-8

Only -i/3 lies inside C. Residue at 4 3 = Then

.f

-if3

2 dz

32'

2

+

+ lok -3

1

= x-, lim - i / 3 62 1oi = 4i by L'Hospital's rule.

lim

%+

= 2~i($)

P 2,

=

the required value.

Then

where C is the contour of Problem 31. The integrand has a pole of order 3 at x = 0 and a simple pole x = 4 within C. 21

Residue a t z = O is Residue at x = 8 is

33. If If(x)l S

M for R"

~ ~ (2 l)(x 2 - 2)

x = Reie, where k > 0 and M are constants, prove that

lim

R-+W

eim*f(x) dx = 0

where r is the semi-circular arc of the contour in Problem 27 and m is a positive constant. If x = Rei@, Then

A

lxn

etmxf ( z ) dx

=

ln

eimRe'e

eimR."f(Re'0) S e t @del

sT

f(Re'@)iRei@de. S

leimR2'f(Rei@) iRe4el do

0

365

FUNCTIONS O F A COMPLEX VARIABLE

CHAP. 171

Now sine 2 2 e / ~for 0 5 e 5 ~ / 2(see Problem 77, Chapter 4). than or equal to

Then the last integral is less

As R + 00 this approaches zero, since m and k are positive, and the required result is proved.

34. Show that Consider

xrn-

cos mx

f x"+l dx eimx

n.

dx = -e-m, 2

> 0.

where C is the contour of Problem 27.

The integrand has simple poles at z = 'i, Residue a t z = i is

m

lim

(z -i)(x

x+i

but only z = i lies within C.

+9

Then or i.e. and so 2 i Rcosmm xd r

+

l

g

d

z

=

re-m

Taking the limit as R -+ 00 and using Problem 33 to show that the integral around zero, we obtain the required result.

35. Show that

dx imF

r approaches

7r

= -

2'

The method of Problem 34 leads us to consider the integral of eh/x around the contour of Problem 27. However, since x = O lies on this path of integration and since we cannot integrate through a singularity, we modify that contour by indenting the path a t z = 0, as shown in Fig. 17-9, which we call contour C' or ABDEFGHJA. Since x = O is outside C',we have i , $ d z

Fig. 17-9

= 0

or

BDEFG

HJA

Replacing x by -x in the first integral and combining with the third integral, we find,

or

HJA

BDEFG

HJA

BDEFG

[CHAP. 17

FUNCTIONS OF A COMPLEX VARIABLE

366

Let T + O and R + 00. By Problem 33, the second integral on the right approaches zero. The first integral on the right approaches

since the limit can be taken under the integral sign. Then we have

MISCELLANEOUS PROBLEMS 36. Let w = x 2 define a transformation from the x plane (xy plane) to the w plane (UW plane). Consider a triangle in the x plane with vertices at A(2,1), B(4,1), C(4,3). (a) Show that the image or mupping of this triangle is a curvilinear triangle in the uw plane. ( b ) Find the angles of this curvilinear triangle and compare with those of the original triangle. ( a ) Since w = x2, we have ZL = x2- y2, v = 2xy a s the transformation equations. Then point A ( 2 , l ) in the xy plane maps into point A’(3,4) of the wv plane (see figures below). Similarly, points B and C map into points B’ and C’ respectively, The line segments AC,BC,AB of triangle ABC map respectively into parabolic segments A’C’, B’C’,A’B’ of curvilinear triangle A’B’C‘ with equations a s shown in Figures 17-10(a) and ( b ) .

( b ) The slope of the tangent to the curve v 2 = 4(1+

U)

The slope of the tangent to the curve u2 = 2v

a t ( 3 , 4 ) is

+1

m1 = dv

at (3,4) is

, , ,I:

Zl(3,4) =

rn4

dvI

= du

(3,4)

1 - - 2

- u = 3 .

-

Then the angle e between the two curves a t A’ is given by 3-* t a n e = m2-m1 = = 1, and e = ~ / 4 1 mime 1 (3)(*) Similarly we can show t h a t the angle between A’C’ and B’C’ is n/4, while the angle between A’B‘ and B’C’ is ~ / 2 . Therefore the angles of the curvilinear triangle are equal to the corresponding ones of the given triangle. In general, if w = f ( x ) is a transformation where f ( x ) is analytic, the angle between two curves in the x plane intersecting a t x = xo has the same magnitude and sense (orientation) as the angle between the images of the two curves, so long as f ’ ( x 0 ) # 0. This property is called the conformal property of analytic functions and for this reason the transformation w = f ( x ) is often called a conformal transformation or conformal mapping function.

+

+

FUNCTIONS O F A COMPLEX VARIABLE

CHAP. 171

367

37. Let w = fi define a transformation from the x plane to the w plane. A point moves counterclockwise along the circle 1x1 = 1. Show that when it has returned t o its starting position for the first time its image point has not yet returned, but that when it has returned for the second time its image point returns for the first time. Let .x = e'e. Then w = fi = ere/z. Let e = 0 correspond to the starting position. Then x = 1 and w = 1 [corresponding to A and P in Figures 17-ll(u) and (b)].

Fig. 17-11 ( b )

Fig. 17-11 (a)

When one complete revolution in the x plane has been made, e = 2a, x = 1 but w = efe/z= e'" = -1

so the image point has not yet returned to its starting position.

However, after two complete revolutions in the x plane have been made, w = eiela = e2* = 1 so the image point has returned for the first time.

e =4a, x = 1 and

It follows from the above that w is not a single-valued function of x but is a doubZe-valued function of z; i.e. given z, there are two values of w . If we wish to consider it a single-valued function, we must restrict 8. We can, for example, choose 0 5 e < 2a, although other possibilities exist. This represents one branch of the double-valued function 7 0 = fi. In continuing beyond this interval we are on the second branch, e.g. 2 r 5 e < 4a. The point x = 0 about which the rotation is taking place is called a brunch point. Equivalently, we can insure that f ( x ) = fi will be single-valued by agreeing not to cross the line O x , called a brunch line.

XP-1

38. Show that Consider

7r

dx = O