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
New York San Francisco Washington, D.C. Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto
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)($)= <-#<E> = --& = -&
The fact that ( c ) and ( d ) are equal illustrates the associative law of multiplication.
+ q) = 4(3 - Q) = A($-&) 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,
x2-x-12
<
0
<
-3.
(~-4)(z+3) < 0
or
This last inequality holds only in the following cases.
+
Case 1: z - 4 > 0 and z 3 < 0, i.e. x both greater than 4 and less than -3.
>4
and x
This is impossible since x cannot be
+
x - 4 < 0 and x 3 > 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+
. * *
<
[CHAP. 1
1 for all positive integers n > 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 <
where n is any positive integer.
0, [ a+ Ya 1 1 2.
52.
Prove that for all real a
53.
Show t h a t in Schwarz's inequality (Problem 13) the equality holds if and only if a, = kb,, p = 1,2,3 , . , n where k is any constant.
54.
If al, at, a3 a r e positive, prove tha t &(a, as
f
..
+ + as) 2 v
a x .
CHAP. 11
NUMBERS
17
EXPONENTS, ROOTS and LOGARITHMS 55.
Evaluate
9 logm (A),
(a) 4*Og9*, (b)
( d ) 3-0
(e)
logg S,
(-Q)4/3
- (-27)-L/J.
Ans. ( a ) 64, ( b ) 7/4, (c) 60,000, ( d ) 1/25, (e) -7/144 56.
Prove
57.
Prove
(a) log, M N = log, blogba
M
+ log, N, (b)
log, M = r log, M
indicating restrictions if any.
= a giving restrictions if any.
COUNTABILITY 58. (a) Prove t h a t there is a one to one correspondence between the points of the interval 0 5 2 5 1 and -6 5 x S -3. (b) W h at is the cardinal number of the sets in (a)? Ans. (b) C , the cardinal number of the continuum. 59.
(a) Prove t h a t the set of all rational numbers is countable. (b) W h a t is the cardinal number of the set Ans. (b) in (a)?
60.
Prove t h a t the set of (a) all real numbers, (b) all irrational numbers is non-countable.
61.
The intersection of two sets A and B, denoted by A n B or AB, is the set consisting of all elements belonging to both A and B . Prove t h a t if A and B a r e countable, so is their intersection.
62.
Prove t h a t a countable set of countable sets is countable.
so
63. Prove t h a t the cardinal number of the set of points inside a square is equal to the cardinal number of the set of points on (a)one side, (b) all four sides. ( c ) W ha t is the cardinal number in this case? Ans. (c) C ( d ) Does a corresponding result hold for a cube?
LIMIT POINTS. BOUNDS. WEIERSTRASS-BOLZANO THEOREM (a) Is the set bounded? (b) Does the set 64. Given the set of numbers l , l . l , .9,1.01, .99,1.001, .999, . . have a 1.u.b. and g.l.b.? If so, determine them. (c) Does the set have any limit points? If so, determine them. ( d ) Is the set a closed set? Ans. (a) Yes (b) 1.u.b. = 1.1, g.1.b. = .9 (c) 1 (d) Yes
..
65. Given th e set -.9, .9, -.99, .99, -.999, .999 answer t h e questions of Problem 64.
Ans. ( a ) Yes
(b) 1.u.b. = 1, g.1.b. = -1
(c) 1, -1
(d) No
66.
Give a n example of a set which has ( a ) 3 limit points, (b) no limit points.
67.
(a) Prove t h a t every point of the interval 0 < x < 1 is a limit point. (b) Are there any limit points which do not belong to the set in (a)? Justify your answer.
68. Let S be the set of all rational numbers in ( 0 , l ) having denominator 2", n = 1 , 2 , 3 , have an y limit points? (b) Is S closed? 69.
. . ..
(a)Does S
(a) Give a n example of a set which h as limit points but which is not bounded. (b) Does this contradict the Weierstrass-Bolzano theorem? Explain.
ALGEBRAIC and TRANSCENDENTAL NUMBERS 70. Prove t h a t (a)
fi- fi
+ fi
( b ) @+ fi
fi+Jh'
a r e algebraic numbers.
71. Prove t h a t the set of transcendental numbers in ( 0 , l ) is not countable. 72.
Prove t h a t every rational number is algebraic but every irrational number is not necessarily algebraic.
COMPLEX NUMBERS. POLAR FORM 73. Perform each of the indicated operations: (a)2(5 - 3i) - 3(-2
+-
10
Ans. ( a ) 1 - 4i,
1-i
10
( b ) -9 - 46i,
(c)
y-
#i, ( d ) -1,
(e)
E,
+ i ) + S(i - 3), (f)
y-
8i
w
( b ) (3 - 2i)3, (c)
b
3
4i
74. If z1 and zz a r e complex numbers, prove 75.
[CHAP. 1
NUMBERS
18
Prove
(a)
1x1
+
5 1x11
76. Find all solutions of
+ IxzI,
( b ) 1x1
I:I
(a) - =
+ xz +
X S ~
I4 -, 1x21
5 1x11
+ lzzl + IxsI,
= 0.
2x'-3xS-7x~-8x+6
= Iz1Ia giving a ny restrictions.
(b)
Ans. 3,
(c) 1x1- 2 2 1 h
+, -1 + i
(XI/
- Izzl.
77. Let x1 and zu be represented by points P I and Pz in the Argand diagram. Construct lines OPI and OPS,
+
where 0 is the origin. Show t h a t X I 22 can be represented by the point PJ, where OPS is t h e diagonal of a parallelogram having sides OP1 and OPZ. This is called the parallelogram law of addition of complex numbers. Because of this and other properties, complex numbers can be considered as vectora in two dimensions.
78.
Interpret geometrically the inequalities of Problem 76.
79.
Express in polar form (a)3 f i + 3 i , (b) -2-2i, (c) 1 A m . (a)6 cisa/6 (b) 2 f i cis 5 ~ / 4 (c) 2 cis 6 ~ / 3 ( d ) 6 cis 0
-fii,
80. Evaluate (a) COS 26O
+5di,
A m . (a) -5fi
+ i sin 26O)][6(cosl l O o + i sin 110°)], (b) -2i
( d ) 6, (e) -6i. (e) 5 cis 3n/2
12 cis 16O ( b ) (3 cis 44O)(2 cis 62O) '
81. Determine all the indicated roots and represent them graphically: (a)( 4 6 + 4\/2i)'IS, (b) (-l)lI5, (c) i)'", ( d ) P4.
(fi-
(a) 2 cis 15O, 2 cis 136O, 2 cis 266O (b) cis 36O, cis 108O, cis 180' = -1, cis 252O, cis 324O (c) cis l l O o , cis 230°, cis 350' ( d ) cis 22.6O, cis 112.6O, cis 202.6', cis 292.6O
Ans.
fi
82.
Prove t h a t -1
83. If zI =
fi
~
+ ai is a n algebraic number.
cis@, and x z = Interpret geometrically.
pa
p,
cis @2, prove (a)
ZIZZ
=
plpz
+
cis
(b)
Zi/Ze
=
(pl/pz)
cis
- $J.
MATHEMATICAL INDUCTION Prove each of the following.
+
+ +
1 + 3 5 ... (2n-1) = no 1 1 1 1 - - n 85. - + - + - + ... 1.3 3-6 5-7 (2n - 1)(2n 1) 2n 1 a ( a d ) (a 2d) . [a (n - l)d] = &n[2a (n- l)d] 86. 1 1 1 n(n 3) +-+-+... 87. 1-2.3 2-3-4 3-4-6 + n(n l)(n 2) - 4(n l)(n 2)
84.
+ + + +
+
+
+
+ +
+
+
+
+
- 1) + ar + arz + . - . + urn-' = a(r* r-1 ' r # l lS+ 2S + 33 + . . . + = fn2(n+ 1)2 6 + (4n - 1)6"+' l(5) + 2(5)O + 3(5)3 + ,.. + n(6)"-' = 16
+
+
+
88. a 89. 90.
713
+
91. xZn--l 92. 93.
(cos @
4
is divisible by x
+ i sin @)"
= cos n+
+ cosx + cos2x + ... +
94. sin x
+ sin 22 +
95.
.
+
for n = 1,2,3,
+ i sin n@.
... .
Can this be proved if n is a rational number?
+
sin (n &)x , x#O,+2n,+4a, cosnx = 2 sin &x
+ sin nx
=
cos &x - cos (n
2 sin &x
...
+ &)x , x + 0, -e2a, *4r, .
+
*
.
n! n(n - l)(n - 2)...(n - r 1) - nCn-r. Here p ! = p ( p - l ) . . . l a nd O! r! r! (n - r ) ! n(n - 1) is defined as 1. This is called the binomial theorem. The coefficients %CO=1, ,,C1=n, nCs =- 2! '
where
nCr
=
. . . , n C n = 1 a r e called the binomial coeficients. ,C, is also written
(3. I
.
NUMBERS
CHAP. 11
19
MISCELLANEOUS PROBLEMS 96. Express each of the following integers (scale of 10) in the scale of notation indicated: (a)87 (two), A m . (a) 1010111, ( b ) 2101, (c) 2338 (b) 64 (three), (c) 1736 (nine). Check each answer. 97. If a number is 144 in the scale of 5, what is the number in the scale of (a) 2, (b) 8?
A m . ( a ) 110001, ( b ) 61
98.
Prove that every rational number p l q between 0 and 1 can be expressed in the form a1 + a + 2 = 9
2
2’
an + ... ... + 2”
where the a’s can be determined uniquely as 0’s or 1’s and where the process may or may not terminate. The representation 0 . a l a s . . . a , . . , is then called the binaw form of the rational number. [Hint: Multiply both sides successively by 2 and consider remainders.]
8 in the scale of (a) 2, (b) 3, (c) 8, ( d ) 10. A m . (a) 0.1010101.. ,, ( b ) 0.2 or 0.2000 (c) 0.6252.. ., (d) 0.6666..
99. Express
...,
.
100. A number in the scale of 2 is 11.01001. What is the number in the scale of 10. 101. In what scale of notation is 3 4 - 4 = 12?
Ans. 3.28125
Ans. 5
102. In the scale of 12, two additional symbols t and e must be used to designate the “digits” ten and
eleven respectively. Using these symbols, represent the integer 5110 (scale of ten) in the scale of 12. Ans. 2 e S t
103. Find a rational number whose decimal expansion is 1.636363..
..
Ans. 18/11
104. A number in the scale of 10 consists of six digits. If the last digit is removed and placed before the first digit, the new number is one-third as large. Find the original number. A m . 428571
105. Show that the rational numbers form a field. 106. Using as axioms the relations 1-9 on Page 2, prove that (c) (-2)(-3) = 6. ( U ) (-3)(0) = 0, ( b ) (-2)(+3) =-6,
+
107. ( a ) If z is a rational number whose square is less than 2, show that x (2-zsL)/10 is a larger such number. (b) If 5 is a rational number whose square is greater than 2, find in terms of z a smaller rational number whose square is greater than 2.
Chapter 2 Functions, limits and Continuity FUNCTIONS A function is a rule which establishes a correspondence between two sets. For our present purposes we consider sets of real numbers. If to each value which a variable x can assume there corresponds one or more values of a variable y, we call y a function of x and write y=f(x), y = G(z), . . . the letters f , G , . . . symbolizing the function while f(a), G(a), . . . denote the value of the function a t x =a. The set of values which x can assume is called the domain of definition or simply domain of the function; x is called the independent variable and y the dependent variable. If only one value of y corresponds to each value of x in the domain of definition, the function is called single-valued, If more than one value of y corresponds to some values of x, the function is called multiple-valued. Since a multiple-valued function can be considered as a collection of single-valued functions, we shall assume functions to be single-valued unless otherwise indicated. Examples: 1. If to each number in -1 5 x 5 1 we associate a number y given by x', then the correspondence between x and x' defines a function f which is single-valued. The domain of f is -1 d x 5 1. The value of f at x is given by y = f ( x ) = x'. For example, f(-1) = (-1)' = 1 is the value of the function a t x = -1. 2. With each time t after the year 1800 we can associate a value P for the population
of the United States. The correspondence between P and t defines a single-valued function, say F, and we can write P = F ( t ) .
3. If y' = x where x > 0 then to each x there correspond two values of y. Hence y is a double-valued function of 2. We can consider this as two single-valued functions f and g where f ( x ) = fi and g ( x ) =
-6.
Note that although a function is often defined by means of a formula as in Examples 1 and 3, it does not have to be, as seen in Example 2. For convenience we shall often speak of the function f ( x ) rather than the function f whose value a t x is f ( x ) . The distinction should however be kept in mind.
GRAPH of a FUNCTION The graph of a function defined by y = f(x) is a pictorial representation of the function and can be obtained by locating on a rectangular coordinate system the points defined by the number pairs (z,y)or [ x , f ( x ) ] .
BOUNDED FUNCTIONS If there is a constant M such that f ( x ) 5 M for all x in an interval (or other set of numbers), we say that f ( x ) is bounded above in the interval (or the set) and call M an upper bound of the function. 20
CHAP. 21
FUNCTIONS, LIMITS AND CONTINUITY
21
If a constant m exists such that f ( x ) Z m for all x in an interval, we say that f ( x ) is bounded below in the interval and call m a lower bound. If m 5 f ( x )5 M in an interval, we call f ( x ) bounded. Frequently, when we wish to indicate that a function is bounded we shall write If(x)l < P.
+
Examples: 1. f ( z ) = 3 z is bounded in -1 5 z d 1. An upper bound is 4 (or any number greater than 4). A lower bound is 2 (or any number less than 2). 2. f ( z ) = l/z is not bounded in 0 < z < 4 since by choosing z sufficiently close to zero, f ( z ) can be made as large a s we wish, so that there is no upper bound. However, a lower bound is given by $ (or any number less than &).
If f ( x ) has an upper bound it has a least upper bound (1.u.b.); if it has a lower bound it has a greatest lower bound (g.1.b.). (See Chapter 1 for these definitions.)
MONOTONIC FUNCTIONS A function is called monotonic increasing in an interval if for any two points $1 and x2 in the interval such that x1< x2, f(xl) 5 f ( x 2 ) . If f ( x 1 ) < f ( x 2 ) the function is called s tric tl y increasing. Similarly if f ( x 1 )2 f ( x 2 ) whenever x1< x2, then f ( x ) is monotonic decreasing; while if f(s1)> f ( x 2 ) , it is strictly decreasing. INVERSE FUNCTIONS. PRINCIPAL VALUES If y is a function of x, denoted by f ( z ) , then x is a function of y, denoted by x = f-'(y), called the inverse function. Interchange of x and y leads to consideration of y=f-'(x). If f ( x ) is single-valued, f - * ( x ) may be multiple-valued in which case it can be considered as a colltktion of single-valued functions each of which is called a branch. It is often convenient to choose one of these branches, called the principal branch, and denote it by f - l ( x ) . In such case the value of the inverse function is called the principal value. Example: The function y = sin x leads to consideration of = sin-' x which is multiple-valued, since for each x in -1 P x 5 1 there are many values of y. By restricting sin-l x to be such that - ~ / 2 S s i n - * x d a/2, for example, the function becomes single-valued. In such case the principal value of sin-1(-&) = -n/6.
MAXIMA and MINIMA If xo is a point of an interval such that f ( x > < f ( x o ) [or f ( x > > f(xo)] for all other x in the interval, then f ( x ) is said to have an absolute maximum [or absolute minimum] in the interval at x = x o of magnitude f ( x o ) . If this is true only for x in some deleted 6 neighborhood ofxo where 6 > 0 [i.e. for all x such that 0 < I x - x o I < 61, then f ( x ) is said to have a relative maximum (or relative minimum) at xo. TYPES of FUNCTIONS 1. Polynomial functions have the form
+
+
+
= aoxn a1xn-' ... an-lx + an (1) where ao, . . . , a n are constants and n is a positive integer called the degree of the polynomial if a0 z 0. The fundamental theorem o f algebra states that every polynomial equation f ( x ) = O has a t least one root. From this we can show that if the degree is n the equation has exactly n roots (counting a repeated root of multiplicity r as r roots). f(x)
22
FUNCTIONS, LIMITS AND CONTINUITY
[CHAP. 2
2. Algebraic functions are functions y = f(x) satisfying an equation of the form
+
+
Po(x)l/n + Pl(x)I/"-' + . * * P,-l(X) $4 pn(x) = 0 (2) where PO($), . . .,pn(x) are polynomials in x. If the function can be expressed as the quotient of two polynomials, i.e. P ( x ) / Q ( x ) where P(x) and Q ( x ) are polynomials, it is called a rationaZ algebraic function; otherwise it is an irrational algebraic function.
3. Transcendental functions are functions which are not algebraic, i.e. do not satisfy equations of the form (2).
Note the analogy with real numbers, polynomials corresponding to integers, rational functions to rational numbers, etc. SPECIAL TRANSCENDENTAL FUNCTIONS The following are sometimes called elementary transcendental functions. 1. Exponential function: f ( z ) = ax, a # 0 , l . For properties, see Page 3. 2 Logarithmic function: f ( z ) = logax, a z 0 , l . This and the exponential function are inverse functions. If a = e = 2.71828. . ., called the natural base of logarithms, we write f ( z ) = logex = lnx, called the natura2 logarithm of x. For properties, see Page 3.
3. Trigonometric functions: 1 1 - cos x sin x 1 sinx, cosx, tanx = cos - cscz = - sec$ =-cos , cot x = -- x' sin x ' t a n x sinx The variable x is generally expressed in radians ( X radians = 180'). For real values of x, sinx and cosx lie between -1 and 1 inclusive. The following are some properties of these functions.
1 + tan2z = sec2x sin2$ + cos2x = 1 sin (xk y ) = sinx cosy f cosx siny cos ( x k y ) = cosx cosy T sinx siny tanx 2 tang tan (X * Y) = 1 7 tan x tan g
+
1 cot2x = csc2x sin (-x) = - sinx cos (-x) = cos x tan (-5) = - tan x
The following is a list of the inverse trigonometric functions and their principal values.
4. Inverse trigonometric functions.
(a) y = sin-'x, ( - 4 2 S y S 4 ) (a) y = cos-'2, (0 5 y 5 X ) ( c ) y = tan-'x, (--r/2 < y < d 2 )
(d) y = csc-'x = sin-1l/x,
(-d2 5 g Id 2 ) ( e ) y = sec-'$ = ~ o s - ~ l l x , (0 5 2/ IX ) ( f ) y = cot-'x = d 2 - tan-'%, (0 < y < X )
5. Hyperbolic functions are defined in terms of exponential functions as follows. 1 0 eZ - e-= (a) sinhx = 2 1 0 1 L ex e-= ( e ) sechx = - - ex e-= (b) coshx = cosh x 2 sinhx - e Z - e - Z coshx - e Z + e - = ( f ) cothx = ( c ) tanhx = - sinhx - e X - e - " ex e-= cosh x
+
+
+
The following are some properties of these functions. 1- tanh2x = sech2x coth2x - 1 = csch2x cosh2x - sinh2x = 1
FUNCTIONS, LIMITS AND CONTINUITY
CHAP. 21
sinh (-x) = - sinh x cosh(-$) = cosh% tanh(-x) = -tanhx
sinh (x 2 y) = sinh x coshy 2 coshx sinhy cosh (x y) = cosh x cosh y f sinh x sinh y tanhx tanhy tanh (' y, = 1 f tanh x tanh y
*
23
*
6. Inverse hyperbolic functions. If x = sinhy then y = sinh-lx is the inverse hyperbolic sine of x. The following list gives the principal values of the inverse hyperbolic functions in terms of natural logarithms and the domains for which they are real.
(a) sinh-l x = In (x
+ d m ) ,all x
( d ) csch-'x =
(b) cosh-lx = In(x
+ d m ) ,x I 1
( e ) sech-lx =
(a), 1x1 < 1
x = +In 1 - x
(c) tanh-l
LIMITS of FUNCTIONS Let f ( x ) be defined and single-valued for all values of x near x = x o with the possible exception of X = X O itself (i.e. in a deleted 8 neighborhood of XO). We say that the number l is the limit of f(x)us x approaches xo and write lim f ( x ) = I if for any positive number c 2 4 2 0
(however small) we can find some positive number 8 (usually depending on C) such that If($) - lI < c whenever 0 < lx - x01 < 8. In such case we also say that f ( x ) approaches l as x approaches $0 and write f ( x )-* l as x -* XO. In words this means essentially that we can make the absolute value of the difference between f(x)and 1 as small as we wish by choosing x sufficiently close to XO, i.e. by choosing the, difference in absolute value between x and x o sufficiently small (but not zero, i.e. we exclude x = XO). Example:
Let f ( z ) =
I
" if x 2 2 0 if x = 2 '
Then as x gets closer to 2 (i.e. x approaches 2), f ( x ) gets
closer to 4. We thus suspect that lim f(x) = 4. To prove this we must see whether the above definition of limit (with I = 4) is satisfied. For this proof see Problem 10. Note that lim f ( x ) # f ( 2 ) , i.e. the limit of f(x) as x+ 2 is not the same as the =I))
r-+0
value of f ( x ) at x = 2 since f(2) = O by definition. The limit would in fact be 4 even if f ( x ) were not defined at x = 2,
When the limit of a function exists it is unique, i.e. i t is the only one (see Prob. 17).
RIGHT and LEFT HAND LIMITS In the definition of limit no restriction was made as to how x should approach XO. It is sometimes found convenient to restrict this approach. Considering x and xo as points on the real axis where $0 is fixed and x is moving, then x can approach xo from the right or from the left. We indicate these respective approaches by writing x+ XO+ and x + xo-. If lirn f ( x ) = ZI and lirn f ( x ) = 12, we call I 1 and k respectively the right and 2-+20+
+
2-20-
left hand limits of f ( x ) a t xo and denote them by f(xo+) or f(zo 0) and f(xo-) or f(z0 - 0). The Q, 8 definitions of limit of f(x) as x -* XO+ or x -* $0- are the same as those for x -* xo except for the fact that values of x are restricted to x > 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
can find a positive number 6 such that f ( x ) < -42 whenever 0 < Iz- $01 < 6. Analogous remarks apply in case x xo+ or x + XO-. Frequently we wish to examine the behavior of a function as x increases or decreases without bound. In such cases it is customary to write x+ +m (or 00) or x+ -00 respectively. We say that lim f ( x ) = I, or f ( x ) + l as x+ +a, if for any positive number c we +
X+
+oo
can find a positive number N (depending on L in general) such that x > 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 <9, f - ' ( x ) is double-valued. This is seen graphically from the fact t h a t every vertical line to the left of P and the right of A B meets the graph in two points.
+
The functions 6 d E and 6 - d G represent the two branches of f-'(z). The point where the two branches meet (or have the same value) is sometimes called a branch point, in this case at x=9, y = 6 .
5.
+
(a) Prove that g(x) = 5 d E is strictly decreasing in 0 5 x 5 9. (b) Is it monotonic decreasing in this interval? ( c ) Does g(x) have a single-valued inverse? g(z) is strictly decreasing if g ( x J >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
<
We must show that given any r > 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
<
I1 --21 +:-Ifs
e
> 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 <
E
2 < 1, then +1 <
E
when
+
elf" 1
2
1
>-,e'/">--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
<-. 1
rn
required result follows.
16. Present a geometric proof that
sine lim- 1. 8
e-o
Construct a circle with center at 0 and radius OA = OD = 1, as in Fig. 2-10 below. Choose point B on OA extended and point C on OD so t h a t lines BD and A C a r e perpendicular to OD. It is geometrically evident t h a t
< 4 s i n e cos e < h e <
Area of triangle OAC i.e. Dividing by
3 s in e, COS^
or
cose
As e + 0, cos e
+
Area of sector OAD
<
Area of triangle OBD
tan e
e 1 < sine < COS e sin e < 1 < e
cos e
e = 1. 1 and i t follows t h a t lirn sin e-0
e
Fig. 2-10
THEOREMS on LIMITS 17. If lirn f ( x ) exists, prove that it must be unique. z-+2 0
We must show t h a t if lirn f ( z )= ZI and =-+
20
lirn f(x) = 12, then Z I = Is. 0
3 "0
By hypothesis, given any c > 0 we can find 6 If(x)-LI I f ( x )-191
<
42 C d2
>0
when when
such t h a t 0 0
< IZ-ZO~ < < Ix-x~l <
6
6
[CHAP. 2
FUNCTIONS, LIMITS AND CONTINUITY
32
Then by inequality 2, Page 3, i.e.
(I1
+
IZr - Za1 = 111 - f ( z ) f ( ~-)121 5 [ Z i - f(z)l + If(%) - 191 < e/2 + a/2 = e - ZgI is less than any positive number e (however small) and so must be zero. Thus ZI = Zg.
18. If lirn g(x) = B 2-20
that there exists 8 > 0 such that (g(z)( > +(B( for 0 < Ix-x50( < 8
z 0, prove
< iIB1
Since lim g(z) = B , we can find 6 > 0 such that Ig(z)-BI ro
I+
Writing B = B - g(s) i.e.
IBI
<
+IB[
+ Ig(x)l,
+ g(z),
we have
+ lg(z>l <
IBI 5 IB-g(z)l from which Ig(s)l
> 0 we
(a) We must show that for any I[f(x)+g(x)]
-
>
< 1% - xo[ <
for 0
6.
+ IQ(4I
*lBl
can And 6 > 0 such that
(A+B)J<
0
when
e
<
<
Ix-xol
8
Using inequality 2, Page 3, we have
I [ f ( x )+ d~)] - ( A + B)I
= I [ f ( x )- A ]
> 0 we can If(z) -AI < 4 2 Ig(s)-Bl < 4 2
By hypothesis, given
c
Then from (I), (a) and (S), I[f( x) g(z)] - ( A B ) [
+
+
<
+
[ d z )- Bll
zs If(%)-AI + IdZ) - BI
(1)
find 61 > 0 and 6s > 0 such that when when
+ €12
e/2
0 0
=
< 1% - Z O ~ < < Ix-zo[ <
6, 8,
when
0
c
<
Ix-sol
<
6
where 6 is chosen as the smaller of 6, and 6,.
Since lim f ( x ) = A, we can find 8, such that If(z)-Al I-, ro
A
- 1 < f ( z )< A + 1, Since
0
for 0
< 1x-zol < a,,
lim g(z) = B, given
at3 so
c>O
lirn f(z) = A , given
=+I@
c
we can find S , > O
> 0 we can And
such that
6, > 0 such that
<
Ig(z)-B[
If(%) - AI <
< Ix - xol < 8,.
< P
~ f ( z ) g ( z-) ABI
(c)
< 1%-
ZO~
<6
&j
+ ( 1 ~ 1+ 1)
where 6 is the smaller of 8 , , 6 , , 6 ,
We must show that for any
By hypothesis, given
c
c
2(p;+1) -
>0
we can find 6, > 0 such that
Jg(x)- B J
< +B%
when
0
e
and the proof is complete.
> 0 we can find 6 > 0 such that
< J x- ZO] < 6,
c/W e
2(1BI + 1)
Using these in (4), we have
for 0
i.e.
f(x) is bounded, i.e. If(z)l< P where P is a positive constant.
< 1 % - 5 0 1 < 8,. Since
0
so that
<1
for for
CHAP. 21
33
FUNCTIONS, LIMITS AND CONTINUITY By Problem 18, since lim g(x) = B # 0, we can find 6,>0 such t h a t 0 3 20
Ig(x)l
> +IBI
0
when
Then if 6 is the smaller of 6, and
< ~ X - X O <~ 6,
a, we can write
and the required result is proved. ( d ) From parts (b) and ( c ) ,
This can also be proved directly (see Problem 69). The above results can also be proved in the cases x + XO+, x -* 20-, x + a, x + - a.
Note: In the proof of (a)we have used the results If(x)- AI < 4 2 and - BI < d2, so t h a t the final result would come out to be If(x) g ( x ) - ( A B)I < e. Of course the proof would be j u s t as valid if we had used 2c (or any other positive multiple of C) in place of C. A similar remark holds for the proofs of (b), ( c ) and (d).
+
+
20. Evaluate each of the following, using theorems on limits. (a) lim a
03
(xa- 6s
+ 4)
+ lims (-62) + lims 4 (I:? x)(!$ X) + (lii-6)(!li X) + lii4 (2)(2) + (-6)(2) + 4 = -4
= lirn xa
=
=+a
=3
=l,
In practice the intermediate steps are omitted.
+ 3 ) ( 2-~ 1) x'f32-2
(X
(c)
-
lirn, (x I4
-
liml - (2'
z 3 1
lim 2 ~ ' - 3 ~ ' + 1 ++a0 6x'+ xa-3x
-
lim 2
0 4
m
lirn 6
0 3w
by Problem 19.
sin x lim - =
+ 3)
sin x lim - *
liml (22 - 1)
-
+ 3x - 2) 0 4
- -2*(-3) -
-4
-
5
--
-1
2
3 . 1
+ +
+ lim 71 1 -3 lirn - + il~ -g x -3 lim -
x'
2400x
I
-
2 6
3
= 4 w
fi =
lirn
0 3 O +
sin x
-*
x
limfi
0 4 O +
= 1 . 0 = 0.
Note that in (c), ( d ) and (e) if we use the theorems on limits indiscriminately we obtain the so called indeterminate f o m m / a and O/O. To avoid such predicaments, note t h a t in each case the form of the limit is suitably modified. For other methods of evaluating limits, see Chapter 4.
CONTINUITY 21. Prove that f ( x ) = x2 is continuous at
x =2.
Method 1: By Problem 10, lirn f(x) = f(2) = 4 and so f(x) is continuous at x = 2 . 0 3I
Method 2: We must show that given any e > 0, we can find 6 > 0 (depending on e) such t h a t If(x) - f(2)I = 1x'-41 < c when Ix-21 < 6. The proof patterns that given in Problem 10.
34
FUNCTIONS, LIMITS AND CONTINUITY
[CHAP. 2
is not continuous at x=o.
22. (a) Prove that f ( x ) =
x=o define f ( 0 ) so that f ( x ) is continuous at x = O? ( a ) From Problem 13, lirn f ( z )= 0. tinuous at Z = O . z40
( b ) Can one re-
But this limit is not equal to f(0) = 6, so t h a t f ( z ) is discon-
( b ) By redefining f ( s )so that f ( 0 )= 0, the function becomes continuous. Because the function can be made continuous at a point simply by redefining the function at the point, we call the point a removable diacontinuity.
23. Is the function f ( x ) =
+ +
2x4- 6xs x2 3 continuous at x = I ? x-1
f(1) does not exist, so that f ( z ) is not continuous at x = 1 . By redefining f ( x ) so that f(1) = lim f ( z ) = -8 (see Problem l l ) , i t becomes continuous at z = 1, i.e. z = 1 is a removable discontinuity.
04.1
24. Prove that if f ( x ) and g(x) are continuous at x = x o , so also are (a) f(x)+g(x),
These results follow at once from the proofs given in Problem 19 by taking A = f ( z 0 )and B = g(z0) and rewriting 0 < Iz - zol < 6 as 1% - xol < 6, i.e. including x = XO.
25. Prove that f ( x ) = x is continuous at any point
X=XO.
We must show that, given any e > 0 , we can find 6 > 0 such that I f ( x ) - f ( z o ) l when Iz-z~l < 6. By choosing 6 = C the result follows at once.
= lz-zol
<
c
26. Prove that f ( x ) = Zx3+x is continuous at any point X = X O . Since x is continuous a t any point z = zo (Problem 26) so also is z x = za, z ' z = x', 2z3 and finally 2x3 continuous.
+
5,
using the theorem (Problem 24) t h a t sums and products of continuous functions are
27. Prove that if f ( x ) =
d-
If xo is any point such that lim
z45+
d
a
= 0 = f ( 6 ) and
for 5 5 x 5 9 , then f ( x ) is continuous in this interval. s = d s = f ( x 0 ) . Also, 6 < Z O < 9, then lirn f ( z ) = lim d 20 z-t 20 lim
+-to-
d s=
Here we have used the result that
at zo. An
e,
lirn
z-t zo
0 4
2 = f(9). Thus the result follows.
=
4 7
im f(z) =
d m
if f ( x ) is continuous
Z 4 2 0
6 proof, directly from the definition, can also be employed.
28. For what values of x in the domain of definition is each of the following functions continuous. (4 f ( 4 = Ans. all z except z = *1 (where the denominator is zero) A m . all z
2-x
x+x
X
X
-
If x > O , f ( z )= - = 0. If z
35
FUNCTIONS, LIMITS AND CONTINUITY
CHAP. 21
As in
U),f ( z ) is continuous
for z
Then since
it follows that f ( z ) is continuous (from the left) at z=O. Thus f ( z )is continuous for all ~ $ 0 i.e. , everywhere in its domain of definition. X
( h ) f ( z ) = z csc z = sin z'
(9
... .
A m . all z except 0, fr, %?a, * 3 ~ , X
Since lim z cscz = lim - = 1 = f ( O ) , we see that f ( z ) is =+o sin z Z+O continuous for all z except fr, *2r, f 3 r , . [compare (h)].
= z cscz, f ( 0 ) = 1.
f(2)
..
UNIFORM CONTINUITY 29. Prove that f ( x ) = x2 is uniformly continuous in 0 < x
< 1.
Method 1, using definition.
We must show that given any > 0 we can find 6 > 0 such that where 6 depends only on t and not on xo where 0 < zo< 1. If z and zo are any points in 0 < z < 1, then Iz~-zool = Iz+zoI
Thus if
IZ-ZO~
12
- zol < 6 it follows that
< 6,
12-201
<
- z:I < 26.
p+111z-xol
12%
where 6 depends only on
c
and not on
- ztl < E
12%
=
when
1% - zol < 6,
2IZ-Zol
Choosing 6 = c/2, we see that 12%- XXI < c when Hence f ( z ) = s * is uniformly continuous in XO.
O<x
Method 2: The function f ( z ) = 2%is continuous in the closed interval 0 5 z S 1. Hence by the theorem on Page 26 it is uniformly continuous in 0 5 z 5 1 and thus in 0 < z < 1.
30. Prove that f ( x ) = l/x is not uniformly continuous in 0 < x
< 1.
Method 1: Suppose that f ( z ) is uniformly continuous in the given interval. Then for any e > O we should ~ 6 for all z and zo be able to find 6, say between 0 and 1, such that If(z) -f(zo)l < c when 12 - Z O < in the interval. Let z = 6 and xo = However,
;1
l + E '
Then
-$I = lk--fl1 +
IZ-ZO~ =
-
>
6 e
<
6.
(since O<S
Thus we have a contradiction and it follows that f ( z ) = 1/z cannot be uniformly continuous in O
Method 2:
+
Let zo and zo 6 be any two points in (0,l). Then
I f ( z o ) - f ( z o + 6 ) ) = I 1g - f i1I
= zo(zo+8) 8
can be made larger than any positive number by choosing zo sufficiently close to 0. Hence the function cannot be uniformly continuous.
36
[CHAP. 2
FUNCTIONS, LIMITS AND CONTINUITY
MISCELLANEOUS PROBLEMS 31. If y=f(z) is continuous at x = x o , and x=g(y) is continuous at y=yo where yo=f(xo), prove that x = g { f ( s ) } is continuous at X = X O . Let h ( z )= g { f ( z ) } . Since by hypothesis f ( z ) and g(y) are continuous at zo and we have lim f(z) = f ( lim z) = f ( z 0 ) 2-b
Lo
2-b LO
lim g(y) = g( lim Y)
Y+Y@
Then
lim h ( z ) = 2-b
20
respectively,
Y-bYo
= ~ ( Y o )= g { f ( z o ) }
}
lim g { f ( z ) } = g lim f ( z ) {=+ao
s-bso
which proves that h ( z ) = g { f ( z ) } is continuous at z=zo.
= g { f ( z o ) ) = h(zo)
32. Prove Theorem 8, Page 26.
+
Suppose that f ( a )< 0 and f ( b ) > 0. Since f ( z )is continuous there must be a n interval (a,a h), h > 0, for which f ( z )< 0. The set of points (a,a h) has an upper bound and so has a least upper bound which we call c. Then f ( c )S 0. Now we cannot have f ( c )< 0, because if f ( c ) were negative we would be able to find an interval about c (including values greater than c) for which f ( z )< 0; but since c is the least upper bound this is impossible, and so we must have f ( c )= 0 as required.
+
If f ( a )> 0 and f ( b ) < 0, a similar argument can be used.
33. (a)Given f(x) = 2x3- 3x2+ 7x - 10, evaluate f(1) and f ( 2 ) . (b) Prove that f ( x ) = 0 for some real number x such that 1< x < 2. ( c ) Show how to calculate the value of x in (b). f(1) = 2(1)a- 3(1)’
+ 7(1) - 10 = -4,
f(2) = 2(2)’ - 3(2)’
+ 7(2) - 10 = 8.
If f ( z )is continuous in a 5 3 S b and if f ( a ) and f ( b ) have opposite signs, then there is a value of z between a and b such that f ( z )= 0 (Problem 32). To apply this theorem we need only realize that the given polynomial is continuous in 1 S z S 2, since we have already shown in (a)that f(1) < 0 and f ( 2 ) > 0. Thus there e z b t s a number c between 1 and 2 such that f ( c )= 0.
+
f(l.6) = 2(1.6)’- 3(1.6)* 7(1.6) - 10 = 0.6. Then applying the theorem of (b) again, we see that the required root lies between 1 and 1.6 and is “most likely” closer to 1.6 than to 1, since f(l.6) = 0.6 has a value closer to 0 than f(1) = -4 (this is not always a valid conclusion but is worth pursuing in practice). Thus we consider z = 1.4. Since f(1.4) = 2(1.4)a- 3(1.4)* 7(1.4) - 10 = -0.692, we conclude that there is a root between 1.4 and 1.6 which is most likely closer to 1.6 than to 1.4.
+
Continuing in this manner, we find that the root is 1.46 to 2 decimal places.
34. Prove Theorem 10, Page 26. Given any
c
> 0, we
can find z such that M - f ( z ) <
1 1 Then M - f ( % ) > TS so that
~
M - f (5)
c
by definition of the 1.u.b.
M.
is not bounded and hence cannot be continuous in view
of Theorem 4, Page 26. However if we suppose that f ( z ) # M , then since M - f ( z ) is continuous, by hypothesis, we must have
also - f (4
f ( z )= M for at least one value of
continuous. In view of this contradiction, we must have
z in the interval.
Similarly we can show that there exists an x in the interval such that f ( s )= m (Problem 93).
FUNCTIONS, LIMITS AND CONTINUITY
CHAP. 21
37
Supplementary Problems FUNCTIONS 35. Give the domain of definition for which each of the following functions is real and single-valued:
+
+
(a) d ( 3 - 2)(22 4), ( b ) (x - 2)/(xZ - 4), (c) & G X , (d) log10 (XS - 32' - 4 2 12). A m . (a) -2 d x 5 3 , ( b ) all z # '2, (c) 2m1/3 S z S (Zm l ) 1 / 3 , m = 0, +1, k 2 , . . ., (d) x > 3 , - 2 < x < 2 . 36. If
+
3 x + 1 , x Z 2 , find: (a) ' j ( - ' ) f(x) = x-2
- 2f(o) 6
+
3f(5);
(c) f ( 2 x -3);
( b ) {f(-&)}';
f(x) = 2x2, 0 < x d 2, find (a)the 1.u.b. and (b) the g.1.b. of f ( x ) . Determine whether f ( z )attains its 1.u.b. and g.1.b. Ans. (a) 8, ( b ) 0
37. If
38. Construct a graph for each of the following functions.
( a ) f(x) = 1x1, -3 ( b ) f ( z )= 2-1"1
x '
sx s 3
(f)
-2Sz52
- [%I where [x] = greatest integer 7
5x
(g) f ( x ) = coshx
+
39. Construct graphs for (a) z ' l d y'/bS = 1, ( b ) xz/a' - y2/b2 = 1, ( c ) y'= Zpx, and ( d ) y = 2az - z*, where a,b, p are given constants. If y = f ( x ) in each of these cases, is f ( x ) single-valued? 40.
(b) Show graphically why (a)From the graph of y = cosz construct the graph of y = cos-'S. cos-' x is a multiple-valued function. Indicate possible choices of a principal value of cos-' x. (c) Using the choice in (b), find cos-' ( 1 / 2 )- cos-' (-1/2). Does the value of this depend on the choice? Explain.
41. Work parts (a)and (b) of Problem 40 for
42.
(a)y = sec-' x, (b) y = cot-lx.
Given the graph for y = f ( x ) , show how to obtain the graph for y = f ( a x given constants. Illustrate the procedure by obtaining the graphs of (a)y = cos 32, (b) y = sin (6% 1/3), ( c ) y = tan (1/6 - 2x).
+ b),
where a and b are
+
43.
Construct graphs for (a)y = e-IZI, (b) y = In 1x1, (c) '
= e-Is1 sin x.
44. Using the conventional principal values on Pages 22 and 23, evaluate:
(a) sin-'
(-fi/~)
( b ) tan-' ( 1 ) - tan-' (-1) (c) cot-' ( l / f i )- cot-' ( - - 1 / l b ) ( d ) cosh-lfi ( e ) ,-coth-'(25/7)
cf) sin-' z + cos-'x, -1 d z 5 1
( g ) sin-' (cos Zx), 0 5 z 5 d 2
(h) sin-' (cos Zx), n/2 5 x 5 31/2 (i) tanh (csch-' 3 x ) , x # 0 ( j ) cos (2 tan-' x')
45. Evaluate (a)cos { I sinh (In 2)}, (b) cosh-' {coth (In 3 ) ) .
Ans. (a)-fi/Z,
(b) In 2
FUNCTIONS, LIMITS AND CONTINUITY
38 46.
+
(a) Prove that tan-' z cot-' z = n/2 if the conventional principal values on Page 22 a r e taken. (b) Is t a n - ' z tan-' (Uz) = n/2 also? Explain.
+
+
If f ( z ) = tan-lz, prove t h a t f ( z ) f(y) = f $a: ; ( 48. Prove that tan-' a - tan-' b = cot-' b - cot-' a. 47.
49.
[CHAP. 2
Prove the identities: (a)1 - tanh'z = sech'z, (d) tanh &x = (sinhz)/(l
discussing the case zy = 1.
(b) sin 3s = 3 sin z - 4 sins z, (c) cos 31; = 4 cos'z - 3 cosz, (e) In lcscz - c o t s ( = In [tan 8x1.
+ coshz),
50. Find the relative and absolute maxima and minima o f (a) f ( z ) = (sin%)/%, f ( 0 ) = 1; (b) f ( z ) = (sin' x)/z', f ( 0 )= 1. Discuss the cases when f ( 0 ) is undefined or f ( 0 ) is defined but # 1.
LIMITS 51.
Evaluate the following limits, first by using the definition and then using theorems on limits. 'x - 4 1 (2 h)' - 16 ( a ) lim (z'- 3z+2), (b) lim (d) lim-fi-' (e) lim 9 (c) lirn h =+a = + - I 22-6' r-el 2-2' r+4 4-x ' h-0
+
52. Let
f(z)
=
{
32-1, x < O 0, x=O 22+6, x > 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 <
-
-.
85. If a is a constant, prove that f ( z ) = l/z’ is (a)continuous in a < z < 00 if a h 0, (b) uniformly continuous in a < z < if a > 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
<
e
N (depending
or
> 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
when 2 Icl
<
e
>0
there is a number
i.e. np
Q,
>
>
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) <
c
for all n > 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-
<
111
lun-
n4m
n+m
We must show that for any e > 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
<
+
+c
+e
=
>0
c
ZI = IS.
+ bn) = A + B .
such that
+ (bn-B)[
+ bn) - (A + B)I <
I(&
lun-A(
d
e
for all
+ Ibn-BI
(1)
find NI and NP such that
[un-A[ Ib,- BI
+
< <
foralln>NI for all n > 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 <
&e
+
=
&e
for all n > 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
#
e
+ /AI
+ IAI
for all n >
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 <
< 81BI
+
for all n > 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
<
++
Qe
=
c
! a n ( Iba-B(
S PIbn-Bj e>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\
<
E.
<
Since
c
< ~/3,
jbn-anl
c/3,
lan-~l
<
c/3
(3)
is any positive number, we must have b - a = 0 o r a = b .
23. Prove the Weierstrass-Bolzano theorem (see Page 5 ) . Suppose the given bounded infinite set is contained in the finite interval [a,b ] . Divide this interval into two equal intervals. Then at least one of these, denoted by [ a ~ , bcontains ~] infinitely many points. Dividing [al,bl] into two equal intervals we obtain another interval, say [a,,bz], containing infinitely many points. Continuing this process we obtain a set of intervals [ a n , b n ] , n = 1 , 2 , 3 , . . . , each interval contained in the preceding one and such t h a t
bi
- ai
. . .,
( b - ~ ) / 2 , bp - a2 = ( b i - ~ 1 ) / 2= ( b - ~ ) / 2 ~ ,
b, - a, = ( b - a)/2"
lim ( b , - a n ) = 0.
from which we see t h a t
n-+ 30
This set of nested intervals, by Problem 22, corresponds to a real number which represents a limit point and so proves the theorem.
CAUCHY'S CONVERGENCE CRITERION 24. Prove Cauchy's convergence criterion as stated on Page 43. Necessity. such t h a t
Suppose the sequence Izt,-Z\
<
{U,,} converges
e/2 f o r all p > N
to 1.
and
=
I(2tP - I )
+ ( 1 - U,)]
5
<
luq-Zl
Then for both p > N and q > N , we have 12cp - U,\
Then given a n y E > O , we can find N
IZL,
- I1
e/2 f o r all q > N
+ 11 - uql
< E/2
+ cj2
=
E
Sufficiency. Suppose Iu,- u,I < e f o r all p , q > N and any e > 0 . Then all the numbers U N , . . . lie in a finite interval, i.e. the set is bounded and infinite. Hence by the Weierstrass-Bolzano theorem there is at least one limit point, say a. ?LS+I,
If a is the only limit point, we have the desired proof and lim Suppose there a r e two distinct limit points, say a and b , and suppose b > a (see Fig. 3-1). By definition of limit points, we have
< ( b - a ) / 3 for infinitely many values of - bl < ( b - a ) / 3 f o r infinitely many values of
lily Izt,
Then since Ib-a\
aj
b --a
= (b-U,)
+ (u,-up) + ( u , - u ) , + I1c,-uql + lu,-al
(1)
q
(8)
U,,=
a.
+b-a,
*La, 3 a
3
,
I
-
.
6
Fig. 3-1 (8)
> e
p
we have
= b - a IIb-u,l
Using ( 1 ) and (2) in (S),we see t h a t Iup- u,I thus contradicting the hypothesis t h a t jup - u,I < one limit point and the theorem is proved.
n+m
( b - a)/3 f o r infinitely many values of p and q, f o r p , q > N and any e > 0. Hence there is only
CHAP. 31
51
SEQUENCES
INFINITE SERIES 25. Prove that the infinite series (sometimes called the geometric series) = arn-1 a + ar + a p +
2
r=l
(a)converges to al(1- r ) if Irl< 1, (b) diverges if Irl2 1. S, = a
Let Then
=
TSn
Subtract,
ar
- ar"
U(1- P) = -
s, If 1 . 1
( b ) If
1.1
+ aZ + ... + urn-' + urn
= a
(1-y)Sn
or (a)
+ ar + a 9 + ... + urn-1 1-r
u(1-rr") CL by Problem 7. < 1, limm Sn = nlim 7 - +m 1 r 1-r > 1, lirn Sn does not exist {see Problem 44). m
( 1 3
26. Prove that if a series converges, its nth term must necessarily approach zero. Since S n = u l + u l + . * . + , , Sn-l = u l + u r + . . . + u , - ~ we have un = Sn-Sn--l. If the series converges to S, then lim us
n-em
lim (Sn - Sn-1)
=
1-1
27. Prove that the series Method 1: lim (-1)" n-, m
#
=
n+m
lim S, - lim Sn-l
(1-m
n 4m
+ 1 - 1 + 1 - 1 + ... =
n=l
= S-S
= 0
(-l)n-l diverges.
0, in fact it doesn't exist. Then by Problem 26 the series cannot converge, i.e.
it diverges.
Method 2: The sequence of partial $urns is 1, 1-1, 1-1+1, 1-1+1-1, Since this sequence has no limit, the series diverges.
...
i.e. l , O , l , O , l , O , l ,
MISCELLANEOUS PROBLEMS 28. If limu, = I, prove that lim n-,
n+m
n
Since lim
nr+m
U,
= 0, we can choose P
+U,
n
m
vl+V2+**'+vn
so that
Ul+U2+
-
= 1.
...+UP
Vl+V2+
+
vp+1+ VP+2
n
+ - . . + vn
n
so that IvnI < e/2
After choosing P we can choose N so that for n > N
for n > P .
Then
> P,
Then using (2)and (S), (1) becomes J v ~ + v ~* *+* + v n l n
thus proving the required result.
<
5 2
+
2
=
o
for n > N
... .
52
SEQUENCES
[CHAP. 3
+ + n2)lIn= 1.
29. Prove that lim (1 n Let
new
+ + n')'/" = 1 +
(1 n
Hence lim
U:
n + 00
Un
where
= 0 and lirn
a+ w
U,,=
Un
2 0.
Now by the binomial theorem,
Thus
0.
lirn ( l + n + n * ) ' l n
n+ w
= lim (l+un) = 1. n+ca
an = 0 for all constants a. n.
30. Prove that lim7 n+w
The result follows if we can prove that Let
[2lal
+ 11,
Un=
i.e.
lal" n!
.
Then 5 = UN+1 UN
1
c 2'
Multiplying these inequalities yields
!$ = n-
0
(see Problem 39). We can assume a # 0.
If n is large enough, say n > 2la[, and if we call N =
n the greatest integer 4 214 %-I
lirn
n 4 w
+ 1,
UN+S
then
1
< 2'
% < UN
(+)"-,
-< 1
.*.'
or
un < ( + ) " - N ~ ~ .
Since lim (&)"-" = 0 (using Problem 7), it follows that
lim
n e 00
31. The expression
a1
+
1
a2
...
+-as+
n-
indicated briefly by
1
2
un-1
U,,=
0.
00
a1
1 1 +- .. ., a2 + + a3
where
. * '
are positive integers, is an example of a continued fraction. Its value is 1 1 when this limit defined as the limit of the sequence al, al - a1 a2 1' . * . a2 + -
a1,a2,
+
+
as
exists, and the continued fraction is said to converge to this limit. The successive terms of the sequence are called the successive convergents of the continued fraction. In case the constants al,u2, . . . repeat after some point, the continued fraction is called recurring. Given the recurring continued fraction 1 1 1 2 --2+ 2+ 2+ . . * (a) Find the first ten convergents and guess a t a possible limit. (b) Assuming that the limit exists, find its value.
+
(a)
The first convergent The second convergent
= 2 = 2
The third convergent
=
The fourth convergent
=
The fifthconvergent
= 2
+ 1/2 = 6/2 = 2.6 1 1 - 12 2 + - 2 + 112 = 2 + = - 6 - 2 + 1 29 - 2.4166.. . 12 1 +29/12
70 = 2.4137..
= 2.4
.
29 Similarly, we find for the sixth through tenth convergents respectively the values
169 408 986 = 2.4140. . ., -= 2.4142. . ., -= 2.4142. 70 169 408
2378 . ., - 2.4142. . ., 986
6741 = 2.4142..
2378
,
From the results it is reasonable to guess that the required limit accurate to four decimal places is 2.4142.
CHAP. 31
53
SEQUENCES
It is of interest to note that if PJQn and Pn+l/Qn+l are the nth and (n respectively, then the (n 2)nd convergent is 2Pn+1 PQnn Pn+l - 2Qn+i
+
+ 1)st convergents
+
Qn+*
For the general result in the case of any continued fraction, see Problem 75(a).
+ 1/%. Thus 'x - 22 - 1 = 0 + 6.This agrees with the
( b ) Assume the limit to be given by x. Then clearly we must have x = 2 or x = 1f fi. Since the limit cannot be negative, it must be 1 guess in (a),since \/zi = 1.4142 approximately.
Note that this continued fraction can be defined by the recursion formula
+ z = 2 + l/x
Un+l
and if lim U-+
m
Un
= x exists, this yields
= 2
ui = 2
l/Un,
as above.
Supplementary Problems SEQUENCES 32. Write the first four terms of each of the following sequences:
33. Find a possible nth term for the sequences whose first 5 terms are indicated and find the 6th term: -1 3 -5 7 -9 (a)- - p p 11'14'17'
...
(-1)"(2n - 1) A m . (a) (3n 2)
(b)
+
... (c) Q , 0 , $ , 0 , + ,... (n + 3) .1- (-l)n
(b) l , O , l , O , l , 1- (-1)s 2
(cl
2
{U,} where U ~ + P= U n + l + Un and u1 = 1, UP = 1. (a)Find the first 6 terms of the sequence. (b) Show that the nth term is given by U,, = (an- b n ) / 6 where a = #l+fi), b = i(1-6). Ans. (a) 1,1,2,3,5,8
34. The Fibonacci sequence is the sequence
LIMITS of SEQUENCES 35. Using the definition of limit, prove that: 4 - 2 n - -2 (a) lim - - 3, (b) lim 2-''6 = 1, nr)m 3n+2 n-b 00
(c) lim n-b0O
n4+l -
nf
OQ,
sinn n
(d) lim - - 0. n+w
[CHAP. 3
SEQUENCES
54
+
- I
36.
Find the least positive integer N such that (3n 2)/(n - 1) 3 (b) c = .001, ( c ) C = .0001. AnS. (a)602, (b) 6002, ( c ) 60,002
37.
Using the definition of limit, prove that lim (2n - 1)/(3n
38.
Prove that lirn (-1)"n
I
U300
U300
40. If
lim a 3 00
U,,=
for all n > N if (a) c = .Ol,
e
cannot be
4.
does not exist.
*-+a
39. Prove that if lim
+ 4)
<
[Unl = 0
then lim un = 0. Is the converse true? .-boo
I, prove that (a) \im n-00
where p is a positive integer,
CU,,
= cl where
c is any constant, (b) lim U:
fi = fl, ZZO. *-em
Give a direct proof that lirn aJb,, = A/B if lirn a,, = A and lim b, = B # 0.
42.
Prove that (a) lim 3l/" = 1, (b) lim (3)'" = 1, ( c ) lirn
n+ a
n-+ 03
"3a
U 3 00
U300
that lirn r" =
"-+00
(d) lim
41.
> 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,
<
.. .
If
Sn
= u1+ u s + ... +&, prove that lim S,/n =
?p
U400
= 1, (b) lim (u+n)p'" = 1 where a and p are constants. n-+ 03
1, prove that lirn
Un
-4 00
=
0.
prove that lim n*a" = 0 where the constant p > O . n-+ 00
2"nf
n" = 0.
64.
Prove that lirn
65.
Prove that lirn n sin l l n = 1. a400
66. If {U,} is the Fibonacci sequence (Problem 34)' prove that lirn 67.
= Q(1+ &).
+
Prove that the sequence U n = (1 l/n)"+', n = 1,2,3, . . . is a monotonic decreasing sequence whose limit is e. [Hint: Show that uJUn-1 S 1.1
68. If a,, 2 69. If
lUnl
bn
5
for all n > N and lirn an = A, lim bn = B , prove that A 2 B . n-
IVnl
and hm
V,
0400
00
= 0, prove that lirn ~
I-+ 00
71.
Un+JUn
O*aO
Prove that [&,bn], where fining the number e.
an
= (1
+ l/n).
4
0
Un 0
= 0.
and b, = (l+l/n)"+', is a set of nested intervals de-
72. Prove t h a t every bounded monotonic (increasing or decreasing) sequence has a limit. 73. Verify the values of each of the following continued fractions.
(a) 3
1 1 + 2+3f + 51 " . =
+(3 + 6 6 )
56 74.
SEQUENCES
6, fi, and
Express (a)174/251, ( b ) (c) 1 1 1 1 1 1 1 Ans. (a) - - - - - 1+2+3+1+5+1+%
[CHAP, 3
(d) 3.14169 as continued fractions. (c) 2
1 1 1 1 +--2+ 4+ 2+ 4+"'
1 1 1 1 1 1 (d)3+---7+ 15+ 1+ 25+ 5 4
1 1 1 1 ( b ) l + ---1+ 2+ 1+ 2+"'
[Hint: In (b) add and subtract the greatest integer less than fi (namely 1) to obtain 1 1 = 1 + fi = l + ( f i - l ) = 1 + l / ( f i - 1) (fi 1)/2 Then add and subtract the greatest integer in (fi 1)/2 (namely 1) to obtain 1 1 = 1+(fl+ 1)/2 = 1 1)/2 = 1 2/@ - 1) 6+ 1 Then add and subtract the greatest integer in fi+ 1 (namely 2) to obtain 1 1 = 2 + f i + l = Z+(fi-l) = 2 + 1 / 0 5 - 1) (fi 1)/2 after which repetition occurs.]
+
+ (6-
+
+
+
75. Given the continued fraction
a1
+1 1 1 as+ aa+ a4+
,
u,,>O,
whose nth convergent is Pn/Qn, prove
each of the following and illustrate by means of examples.
+
+
n = hPn-1 Pn-4, Qn = anQn-1 Qn-1 ( b ) P n Q n - 1 - P n - l Q n = (-I)"-' (c) The successive convergents are alternately less than and greater than the continued fraction. ( d ) The convergents of odd order are less than the continued fraction but a r e increasing; the convergents of even order are greater than the continued fraction but are decreasing. (e) The continued fraction always converges. (U) P
76.
(a)Prove t h a t if PJQn and Pn+dQn+l are two successive convergents to the continued fraction in 1 1 (b) Find the first convergent to fi which is acProblem 75, then
A m . (b) 26/15
curate to two decimal places. 77.
+
Let {U,,} be a sequence such t h a t U n + a = a U s + 1 b U, where a and b a r e constants. This is called a second order difference equation for U,,. (a)Assuming a solution of the form Un = rn where T is a constant, prove that r must satisfy the equation f - ar - b = 0. (b) Use (a) to show t h a t a solution of the difference equation (called a general solution) is U,, = Art" Br,", where A and B are arbitrary = constants and rl and r~ are the two solutions of P - ar - b = 0 assumed different. (c) In case r ~ r~ in (b), show that a (general) solution is U, = ( A Bn)r;.
+
+
+
78. Solve the following difference equations subject to the given conditions: (a) Un+t = % + I Un, U'= 1, i 4un, U' = 2, U S= 8. 1 (compare Prob. 34); ( b ) %+a = 2un+l+ 3un, U I = 3, U S= 6 ; (c) Un+t = 4 U ~ +~a
+ (-l)n-l (c) U,, = n . 2 " (a)Prove that the nth convergent to the continued fraction 1 + - - ... 1+ 1+ ( b ) U,, = 2(3)"-'
Ans. (a) Same as in Prob. 34,
79.
[Hint: Use Prob. 34.1 (b) By taking the limits as n +
00
in (a), find the value of the continued fraction.
80. Work Problems 73(a) - (d) by first finding the nth convergent.
is
Chapter
4
Derivatives DEFINITION of a DERIVATIVE Let f ( x ) be defined a t any point xo in (a,b). The derivative of f ( z ) at x = xo is defined as if this limit exists. The derivative can also be defined in various other equivalent ways; for example,
A function is called differentiablea t a point x = xo if i t has a derivative a t this point, i.e. if f ’ ( x 0 ) exists, If f ( x ) is differentiable a t x = $0 it must be continuous there. However, the converse is not necessarily true (see Problems 3 and 4). RIGHT and LEFT HAND DERIVATIVES The right hand derivative of f ( x ) at x = $0 is defined as f; (50) = lim f ( x 0 + h) - f ( x 0 ) h h+O+ if this limit exists. Note that in this case 12 (= Ax) is restricted only to positive values as it approaches zero. Similarly, the left hand derivative of f ( x ) a t x = xo is defined as fi(x0)
=
- ?(So) lim f(xo + h) h
h+O-
(4)
if this limit exists. In this case h is restricted to negative values as i t approaches zero. A function f ( x ) has a derivative a t x = xo if and only if f: (SO) = f: (SO).
DIFFERENTIABILITY in an INTERVAL If a function has a derivative a t all points of an interval, it is said to be differentiable in the interval. In particular if f ( x ) is defined in the closed interval a d x 5 b, i.e. [a,b ] , then f ( x ) is differentiable in the interval if and only if f ’ ( x 0 ) exists for each xo such that a < xb < b and if f; (a) and f I ( b ) both exist. If a function has a continuous derivative, i t is sometimes called continuously differentiable . SECTIONAL DIFFERENTIABILITY A function is called sectionally or piecewise differentiable or sectiodly or piecewise smooth in an interval a S x S b if f’(s) is sectionally continuous. An example of a sectionally continuous function is shown graphically on Page 26. 67
58
DERIVATIVES
[CHAP. 4
GRAPHICAL INTERPRETATION of the DERIVATIVE Let the graph of y = f ( x ) be represented by the curve APQB shown in Fig. 4-1 below. The difference quotient
is the slope of the secant line joining points P and Q of the curve. As A X + 0, this secant line approaches the tangent line PS to the curve a t the point P. Then lim
A%*
~ ( Zi OAz)
Ax
0
-~(xo)
-
I
SR
= tana
is the slope of the tangent line to the curve a t the point P.
I
= *I-’
I
-
I I
t a
r
XI
I
I
I
x ~ + A
M
2
b
11
1 I
xo
2
Fig. 4-2
Fig. 4-1
An equation for the tangent line to the
given by
y -
f(x0)
=
ci rve y = f(x) a t the point where f ’ ( X 0 ) (z--0)
x =XO i (7)
The fact that a function can be continuous at a point and yet not be differentiable there is shown graphically in Fig. 4-2. In this case there are two tangent lines a t P represented by PM and PN. The slopes of these tangent lines are f: (XO) and f; (XO) respectively.
DIFFERENTIALS Let AZ = dx be an increment given to x. Then is called the increment in y = f ( x ) . If f(x) is continuous and has a continuous first derivative in an interval, then where
E+
0 as Ax+ 0. The expression
59
DERIVATIVES
CHAP. 41
= f’(x)dx (10) is called the d i f e r e n t i a l of y o r f ( x ) or the principal p a r t of Ay. Note that A y # d y in general. However if ~ x = d xis small, then d y is a close approximation of A y (see Problem 11). The quantity d x , called the differential of x, and d y need not be small. Because of the definitions (8) and (IO), we often write dy
It is emphasized that d x and d y are not the limits of AX and A y as A x + 0, since these limits are zero whereas d x and d y are not necessarily zero. Instead, given d x we determine d y from (lO), i.e. d y is a dependent variable determined from the independent variable d x for a given x. Geometrically, d y is represented in Fig. 4-1 above, for the particular value X = X O , by the line segment SR, whereas A y is represented by QR. RULES for DIFFERENTIATION If f , g and h are differentiable functions, the following differentiation rules are valid.
d 3. d x { C f ( z ) } =
6.
d C&f(x)
If y = f ( u ) where
U
C f ’ ( x ) where C is any constant
=
= g ( x ) , then
= g ( u ) and v = h(x), then dy du dv d y -- -*-.dx du dv dx
Similarly if y = f ( u ) where
U
The results (12) and (13) are often called chain rules for differentiation of composite functions.
7. If y = f ( x ) , then x = f-l(y);
and d y l d x and d x l d y are related by dll -- - 1 dx dxldy
8. If x = f ( t ) and y = g ( t ) , then d y - dyldt d x - dxldt
-
P(t)
- g’(t)
Similar rules can be formulated for differentials. For example, d { f ( x ) g ( x ) } = d f ( x ) d g ( x ) = f’(x)d x g’(x) d x = { f ’ ( x )+ g’(x)}dx d { f ( x )g ( N = f ( x ) d g ( 4 + g ( x ) = ( f ( 4g ’ ( 4 + g ( x )f ’ ( x ) > 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<-l
d 6. - cotu dx
du = - csc2u dx
d 21. tanhu dx
d 7. z s e c u
du = secutanudx
d du 22. Cothu = - c s c h 2 ~ dx dx
du = -cscucotu- d x
d du 23. - sech U = - sech U tanh U dx dx
8.
d
GCSCU
logs e d u a > 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<<
(16)
Rolle's theorem is a special case of this where f ( a )= f ( b )= 0. The result (16) can be written in various alternative forms; for example, if x and $0 are in (a$), then
f(x) = fpo)
+ f'(5)(x-xo)
5 between xo and x
(27')
We can also write (16) with b = a + h , in which case 5 = a + 6 h , where
o<e
The theorem of the mean is also called the law of the mean. 3. Cauchy's generalized theorem of the mean. If f ( x ) and g(x) are continuous in [a,b] and differentiable in (a$), then there exists a point 5 in ( a , b ) such that
a<<
(18)
where we assume g(a)z g ( b ) and f ' ( x ) ,g'(x) are not simultaneously zero. Note that the special case g ( x ) = x yields (16). 4. Taylor's theorem of the mean. If f ( n ) ( x )is continuous in [a,b] and differentiable in (a,b), then there exists a point 5 in (a,b) such that ( b - a)2 + ... + (b-a)" + Rn (19) f ( b ) = f(a) + f'(4 ( b - 4 + 2!
f'w
n!
where R n , called the remainder, can be written in either of the following forms:
Lagrange's form:
R,
=
f ( n + l ) ( t )( b
-a)"+'
a<[
( n + l)!
(20)
Cauchy's form: See Problems 26,81-84. In both forms the values of 5 'are different in general. The result (19) can be written in various alternative forms; for example, if x and xo are in (a,b), then using Lagrange's form of the remainder Rn, we have for 5 between xo and x, f ( " ) ( X O ) (x - $0)" f ( " + 1 ) ( 5 ) (x - xo)"+' f ( x ) = f ( x 0 ) f ' ( x o ) ( x - x o ) + ... + (22) n! (n 1)! This is often called a Taylor series f o r f ( x ) with a remainder and is used to approximate f(x) by a polynomial, in which case €3, is the error term. If lim Rn = 0 in (19), the infinite series obtained is called a Taylor series f o r f ( z ) n-, about x=xo. If xo=O, the series is called a MacZuurin series. Such series, called power series, generally converge f o r all values of x in some interval, called the interval of convergence, and diverge for all x outside this interval (see Chapter 11 for further discussion). In referring to Taylor's theorem of the mean, the Lagrange form of the remainder will be assumed unless otherwise stated.
+
00
+
+
62
[CHAP.4
DERIVATIVES
SPECIAL EXPANSIONS The following are some important expansions. The remainder Rn in each case can be obtained by using either (20) or (21). x2 xg X” e2 = I + x + - + - + . . * +,!fRn 1. 2! 3! x3
(-1)n-l~2n-l
4!
+ 7! + + ..* + 6!
23
x4
x5
2.
sin x
= x - - +3 !- - 5 !
3.
cosx
=
,4.
5.
x2 + -x4- 1 -2! x2
In ( 1 + x ) = x - - 2
tan-lx
x7
*..
36
+ x - -4 + ...
- - - x5 + . . .x7 = x - - +xs 3 6 7
+
+
(2n- 1) !
+ Rn
(-l)n-l X2n-2 (2n- 2) !
+ Rn
(-l)n-lxn
+ Rn
(-W- x2n-
+ Rn
n
2n-1
In 1-3, lirn R, = 0 for all x. In 4, lirn R, = 0 for -1 < x S1. In 5, lirn R, = 0 for n+oo
n 4 00
n 4 00
Further discussion of such expansions is given in Chapter 11.
-1sx51.
L’HOSPITAL’S RULES If lim f(x) = A and lirn g(x) = B where A and B are either both zero or both in2-20
2-20
’(%) is often called an indeterminate of the form 0/0 o r o o / a respectively, finite, lim 2 ‘ * q g(x) although such terminology is somewhat misleading since there is usually nothing indeterminate involved. The following theorems, called L’Hospital’s rules, facilitate evaluation of such limits. 1. If f(z) and g(x) are differentiable in the interval (a, b) except possibly at a point xo in this interval, and if g ’ ( x ) # 0 for X # X O , then
whenever the limit on the right can be found. In case f’(x) and g’(x) satisfy the same conditions as f ( x ) and g(x) given above, the process can be repeated. 2. If lim f(x) = 00 and lim g(x) = 00, the result (23)is also valid. 2-
20
2-
20
and to cases where xo = a or xo = b These can be extended to cases where x + or ---a, in which only one sided limits, such as x-) a+ or x + b-, are involved. Limits represented by the so-called indeterminate forms 0.00, WO, Oo, 1“ and 00 - * can be evaluated on replacing them by equivalent limits for which the above rules are applicable (see Problems 33-36). Sometimes evaluation of such limits is facilitated by using Taylor’s theorem of the mean, as in Problems 32 and 36.
APPLICATIONS 1. Maxima and minima. Suppose that at x = x o , f(x) satisfies the conditions f’(X0) = f”(x0) = ..- - f(2p-1)(x~)= 0 and f(2p)(x~) f 0 for some positive integer p (usually p = 1). Then (a ) f(x) has a relative maximum at x = xo if f ( 2 P ) ( ~ ~ )< 0 (b) f(x) has a relative minimum at x = xo if f(2p)(x~) >0
(24)
CHAP. 41
63
DERIVATIVES
See Problem 39. In practice, to find the relative maxima and minima of a differentiable function f ( x ) we solve the equation f ’ ( x ) = 0 to obtain the critical points x o and then use (24). Graphically the necessary condition f’(xo) = 0 follows, since a t a relative maximum or minimum point x = x o the tangent line to y = f ( x ) must be parallel to the x axis.
2. Rates of change. We can interpret dyldx = f ’ ( x ) as the rate of change of y = f ( x ) with respect to x. If f ’ ( x 0 ) > 0 , then y is increasing at X = X O ; if ~’(zo) <0, then y is decreasing at X = X O . 3. Velocity and acceleration. If s is the instantaneous displacement of a particle from a point 0 on a line a t time t then dsldt is its instantaneous velocity and d2s/dt2is its instantaneous acceleration at time t .
Solved Problems DERIVATIVES
x 1. Let f ( x ) = -t 3-x’
3. Evaluate f’(2) from the definition.
#
Note: By using rules of elementary calculus, we find
at all points x where the derivative exists. Putting x = 2 , we find f’(2)= 6 . Although such rules a r e often useful, one must be careful not to apply them indiscriminately (see Problem 6).
2.
Let f ( x ) = d
m .
Evaluate f’(5) from the definition.
lim 4 G X - 3 . d m h + 3 h d m h 3
+
h-0
94-2h-9 = lim = hlirn +o h ( d m h 3) Ir+O
+
2
d w + 3
1 - -
& (2%- 1 ) =
d rules of elementary calculus, we find f ’ ( z )= (2x - 1 ) l l a = i ( 2 x - 1)-ll2 dx (2%- l)-’/’. Then f ’ ( 6 ) = 9-’/* = &. b
3. If f ( z ) has a derivative at
X=XO,
prove that f ( x ) must be continuous at
+ h) - f(so) =
f(zo
f(%o + h,
h
- f(zo).
h,
h#0
lim f ( x o
+ h)
=
since ~ ’ ( z oexists ) by hypothesis. Thus lirn
h-0
f(x0
+ h) - ~ ( x o )=
showing t h a t f ( z ) is continuous at x = X O .
0
or
h e 0
f(z0)
X=XO.
64
4.
[CHAP. 4
DERIVATIVES
Let f ( x ) = {;,sin l/x, xx =#o0 Is f ( x ) continuous at x = O ? (b) Does f ( x ) have a derivative at x = O ? By Problem 22(b) of Chapter 2, f ( z ) is continuous at x= 0. f’(0)
lim f ( h )- f(0) = lim h sin l l h - 0 = lim sin 1 = lim f ( 0 + h)h - f ( 0 ) = h+O h h+O h h-0 h h+O
which does not exist. This example shows that even though a function is continuous at a point, it need not have a derivative at the point, i.e. the converse of the theorem in Problem 3 is not necessarily true. It is possible to construct a function which is continuous at every point of an interval but has a derivative nowhere.
f(x) =
{12sin l/x, x +
0 x=o
(b) Is f’(s) continuous at x = O ?
Is f ( x ) differentiable at x = O ?
by Problem 13, Chapter 2. Then f(s)has a derivative (is differentiable) at x= 0 and its value is 0. From elementary calculus differentiation rules, if z P 0,
+
’>
Since lim f ’ ( z ) = lii(-cos ; 22 sin - does not exist (because lim cos 1/z does not *+ 0 X =+O exist), f ’ ( z )cannot be continuous at x = 0 in spite of the fact that f’(0) exists.
1
This shows that we cannot calculate f’(0)in this case by simply calculating f ’ ( z ) and putting
z=O, as is frequently supposed in elementary calculus. It is only when the derivative of a function is continuous at a point that this procedure gives the right answer. This happens to be
true for most functions arising in elementary calculus.
6.
definition of the derivative of f ( x ) at x = x o . f ( z ) has a derivative f’(z0) at x = 20 if, given any c > 0, we can find 6 > 0 such that
Present an “c,S”
If(”+
h, - f(20) h
- f’(z0)) <
e
when
0 < ]hJ< 6
RIGHT and LEFT HAND DERIVATIVES 7. Let f ( x ) = 1x1. (a) Calculate the right hand derivative of f ( z ) at x = O . (b) Calculate the left hand derivative of f ( x ) at x = 0. ( c ) Does f ( x ) have a derivative at x = O ? (d) Illustrate the conclusions in (a), (b) and ( c ) from a graph.
since Ihl = h for h > 0.
since Ihl = -h for h < 0.
CHAP. 41 (c)
65
DERIVATIVES
No. The derivative at 0 does not exist if the right and left hand derivatives are unequal.
(d) The required graph is shown in the adjoining Fig. 4-3. Note that the slopes of the lines y = x and y = - x are 1 and -1 respectively, representing the right and left hand derivatives at x = O . However, the derivative at x = O does not exist.
Fig. 4-3
8. Prove that f ( x ) = x2 is differentiable in 0 5 x 5 1. Let zo be any value such that 0
< 50 < 1.
Then
At the end point x = O ,
At the end point z = 1,
Then f ( z ) is differentiable in 0 5 x S 1. We may write f ' ( x ) = 22 for any z in this interval. It is customary to write f: (0) = f ' ( 0 ) and fL (1)= f'(1) in this case.
9.
Find an equation for the tangent line to y = x2 at the point where (a) x = 1/3, ( b ) x =I. (a) From Problem 8, y
(b)
f'(z0)
- f(x0) =
As in part ( a ) ,
y
= 2x0 so that f'(1/3) = 2/3. Then the equation of the tangent line is
f'(x0)
(x - XO)
or
y
-9
- f(1) = f'(1) (z- 1) or
= p(x-+), y
i.e.
y
=
- 1 = 2(x - l), i.e.
3%-
y
= 22 - 1.
-
Z'
DIFFERENTIALS 10. If y = f ( x ) = (U)
Ay
+
x3 - 6x, find (a) Ay, (b) dy,
+
( c ) Ay - dy.
= f ( z A%) - f ( x ) = {(z +A%)' - 6 ( ~A%)} - {z' - 6%)
=
+ (A%)' - 62 + (A%)'
4- ~ x ' A x 4- 35(Az)* = ( 3 ~-* 6)Ax 4- 3x(Ax)' 2'
AS
4- 62
( b ) dy = principal part of Ay = (3x*-6)Ax = (3z2-6)dx, since by definition Ax =dx.
Note that f ' ( x ) = 3z* - 6 and dy = (3x*- 6)dz, i.e. d y l d z = 3x' - 6. It must be emphasized that dy and dx are not necessarily small.
+
(c) From (a) and (b), Ay - dy = 3x(A%)* (A%)' = EAX, where Note that
E+
0 as Ax+ 0, i.e.
* Ax
+
E
= 3xAx
+
0 as Ax+ 0. Hence Ay - dy is an infinitesimal
of higher order than Ax (see Problem 92).
In case Ax is small, d y and Ay are approximately equal.
11. Evaluate
@ approximately
by use of differentials.
-
If Ax is small, Ay = f(x 4- A%) - f ( z ) = f'(z)Ax approximately. - fi ix-'/'Ax (where EJ denotes approximately equal to). Let f ( z ) = fi. Then 9-
66
DERIVATIVES
-
[CHAP. 4
If x = 27 and A x = -2, we have
- fi
7Then
= 3-2/27
-3
i.e.
&(27)-2's(-2),
-
-2/27
or 2.926.
It is interesting to observe that (2.926)a=26.06, so that the approximation is fairly good.
DIFFERENTIATION RULES. DIFFERENTIATION of SPECIAL FUNCTIONS 12. Prove the formula
are differentiable. By definition,
lim f ( z
Are0
+ Ax)
+
Ax
Another method : Let U = f ( x ) , v = g(x). Then A u = f ( x 4- A x ) - f ( x ) and AV = g(x u + A u , g ( x + A x ) = v + AV. Thus --
d
dz
lim
(U
-k A ~ ) ( -k v AV) - uv
=
Ax
h-0
+ v-+--AV Ax Ax
,
)
lim
+ A x ) - g(x),
i.e. f(x
+Ax) =
UAV 4- V A U 4- A u A v Ax
As+
0
=
U-
dv dx
+
v-d u dx
where it is noted that AV -P 0 as A x * 0, since v is supposed differentiable and thus continuous.
13. If y = f ( u ) where U=&), differentiable.
2 20% =
prove that
Let x be given an increment A x Z O . Then as a consequence
A y respectively, where
Ay
= f ( u + A u ) - f(u),
U
assuming that f and g are and y take on increments A u and
= g(x+Ax) -
AU
g(2)
(1)
Note that as A x + 0, A y + 0 and A u + 0. If A ~ Z Olet , us write
e
=
* Au
- d y so that du Ay
= &Au du
e+O
as Au+O and
+ CAU
If A u = O for values of A x , then ( I ) shows that A y = O for these values of Ax. For such cases, we define c = O . It follows that in both cases, A u # 0 or A u = 0, (2)holds. Dividing (2) by A x # 0 and taking the limit as A x + 0, we have
GZ dx
=
lim
Are0
AX
=
lim(&e
&-+O
d u AX
+
Au ArdOAX
= d y . lim - + lim du
A-o
€
0
Au lim A-oAX
67
DERIVATIVES
CHAP. 41
d
d = cosx and -(cosx) = -sinx, derive the formulas dx 1 d = sec2x, (b) ~ ( s i n - l x )=
14. Given d<(sinx)
(a)
d
tans)
d (a) &(tan%)
dF2'
d cos x dx (sin x)
=
-
d sin x dx (cos x)
cosa x
-
(cosx)(cosx) - (sins)(- sin%)
=
cosa x
1
COSL x
-
sec%x
( b ) If y = sin-lx, then x = sing. Taking the derivative with respect to
2,
We have supposed here that the principal value -a/2 5 sin-lx Ia/2, is chosen so that cosy = d w rather than cosy =
cosy is positive, thus accounting for our writing +:
d
m
.
d log, - (a> 0, a# l), where 15. Derive the formula z ( 1 o g a u ) = U
function of x.
Consider y = f(u) = logau.
dx
U
is a differentiable
By definition,
Since the logarithm is a continuous function, this can be written
by Problem 19, Chapter 3, with x = u/Au. d log,e du = Then by Problem 13, -(log.u) dx U dx'
-
16. Calculate dyldx if
(a)xy3 - 3x2 = xy
+ 5,
(b) ea
+ ylnx
= cos2x.
( a ) Differentiate with respect to x, considering y as a function of s. (We sometimes say that y is an implicit function of x, since we cannot solve explicitly for y in terms of 2.) Then
d dx
-bys)
d d -dx ( 3 ~ ~=) dx (xy)
where y' = dy/dx. (b)
d
(ern)
+
d
(6)
or
+
(x)(3yay') (y3)(l)- 6 x = (x)(y')
+
Solving, y' = (62- ys y)/(3xyg- 5).
d +(y In x) dx
Solving,
=
d dx (cos 2x1, y'
=
+ y) + + (In x)y' - 22 sin2x + xyeq + y eq(xy'
x*eW
= -2 sin 22.
+ x In x
17. If y = cosh(x2-3x+1), find (a) dgldx, (b) d2y/dx2. (a) Let y = cosh U, where U = xa - 32 1. Then dyldu = sinh U, duldx =
(b)
2
* 2.2 dx
=
-( )
+ (y)(l) + 0
=
+
= (sinhu)(2x - 3) = (2x - 3) sinh (5% - 32
d d y = &(sinhug) = sinhu-d f u dx dx dx' (cosh u)(2x - 3)* = 2 sinh (2' - 3s = (sinh u)(2)
+
22 - 3, and
+ 1)
+ coshu( %)* + 1) + (22 - 3)' cosh (x9- 32 + 1)
68
DERIVATIVES
18. If x2y
+ y3 = 2,
[CHAP. 4
find (a) y’, (b) y” a t the point ( 1 , l ) .
+ 2xy + 3y2y’ = 0 and 1 y ‘ = - -- 2 w - -a t (1,l) x2 + 3y’ 2 ~ y+’ 2 ~ ) (~zY)(+ ~ z6yy’) ”( - 2SY ) -- - (x*+ 3 y * ) ( 2(d dx x’+33y8 + 3y’)*
(a) Differentiating with respect to x, x’y’
( b ) 24’’
=
d
=
&If)
Substituting x = 1,
= 1 and
y’
= -8, we find
= -8.
y’’
MEAN VALUE THEOREMS 19. Prove Rolle’s theorem. Case 1:
0 in [ a , b ] . Then f ’ ( x ) = 0 for all z in ( a , b ) .
f(x) Case 2 :
f ( s )f 0 in [a,b]. Since f ( z )is continuous there are points at which f ( z )attains its maximum and minimum values, denoted by M and m respectively (see Problem 34, Chapter 2). Since f(x) f 0, a t least one of the values M,m is not zero. Suppose, for example, M Z 0 and that f ( t ) = M If(z) (see Fig. 4-4). For this case, f ( E h) 4 f ( E ) .
If h > 0, then f(E + h,
h
+
- f(O 5
0 and
If h < 0 , then f(E + h, - f ( t ) 2 0 and h
+ -f(€)
lim f ( € h)
h
h-0-
Fig. 4-4
2 0
But by hypothesis f ( x ) has a derivative at all points in (a,b). Then the right hand derivative ( I ) must be equal to the left hand derivative (2). This can happen only if they are both equal to zero, in which case f ’ ( [ ) = 0 as required. A similar argument can be used in case M = 0 and m # 0.
20. (a) Prove the theorem of the mean.
theorem.
(a) Define
(b) Give a geometric interpretation of this
F ( x ) = f ( x ) - f(a) - ( z - a ) f ( bb)- af ( a )
.
Then F ( a ) = 0 and F(b) = 0. Also, if f ( x ) satisfies the conditions on continuity and differentiability specified in Rolle’s theorem, then F ( x ) satisfies them also. Then applying Rolle’s theorem to the function F ( x ) , we obtain
F’(€) = f’(E)
-
f(bl
= 0, a <
or
f’(t) = f ( b ) - f ( a ) , b-a
a
( b ) Let curve ACB in Fig. 4-6 represent the graph of
f ( z ) .Geometrically it appears that there is a point between a and b where the tangent line to this curve at point C is parallel to the chord AB.
= f’([). Slope of chord A B = f ( b )- f(a) b-a ‘ Slope of tangent line
Then
is such that
f’(0 = f ( bb)- af (a).
It is interesting to note that the function F ( z ) of part ( a ) represents the difference in ordinates of curve ACB and line A B at any point z in (a, b).
a
€
Fig. 4-5
b
CHAP. 41
69
DERIVATIVES
21. Verify the theorem of the mean for f ( x ) = 2x2-7x+10, a = 2 , b = 5 . f ( 2 ) = 4, f ( 6 ) = 26, f’(1) = 41- 7. Then the theorem of the mean states that (26 - 4)/(6 - 2) or ( = 3.6. Since 2 < 1< 6, the theorem is verified.
4E- 7 =
22. If f ’ ( x ) = O at all points of the interval (a,b),prove that f ( x ) must be a constant in
the interval. Let x1 < xt be any two different points in
(a,b). By the theorem of the mean, for
f ( 4-f ( 4 5 9
-
XI
< ( < xr,
= f’(() = 0
21
Thus f(xl) = f ( x r ) = constant. From this i t follows that if two functions have the same derivative at all points of (a,b), the functions can only differ by a constant.
23. If f ’ ( x ) > O at all points of the interval ( a , b ) , prove that f ( x ) is strictly increasing. Let x1< zpbe any two different points in (a,b). By the theorem of the mean, for
f(xd - f(xd x,
Then f(x2)> XI) for
24. (a) Prove that
x2
< tan-lb
- tan-%
+< tan-’- 34 25
7r3
(a) Let f(z) = tan-lx. the mean,
< E < zr,
= f’(1) > 0
> XI, and so f(x) is strictly increasing.
b-a 4
(b) Show that
- 21
21
< b-a
if a
< b.
7 r 1 <+4 6’
Since f ’ ( x ) = l/(l+~*) and f’(1) = l/(l+(*), tan-’ b - tan-’ a - -1 b-a 1 €2
+
Since ( > a , 1/(1 4-1’)
we have by the theorem of
a<[
< 1/(1 +a’). Since ( < b, 1/(1 + E r ) > 1/(1+ bZ). Then tan-’ b - tan-’ a < -1 +1 a’ l < b-a 1
+ b’
and the required result follows on multiplying by b - a. ( b ) Let b = 4/3 and a = 1 in the result of part (a). Then since tan-’ 1 = ~ / 4 , we have 3 T 3 < tan-1-4 - tan-11 < g1 or 4 25 < tan-’- 4 < 1 -1 26
+
3
25. Prove Cauchy’s generalized theorem of the mean. Consider G(s) = f(x) - f ( a ) - a { g ( z ) - g ( a ) } , where
3
4
+6
(Y is a constant. Then G ( x ) satisfies the conditions of Rolle’s theorem, provided f(x) and g(x) satisfy the continuity and differentiability conditions of Rolle’s theorem and if G(a) = G(b) = 0. Both latter conditions a r e satisfied if the con-
Applying Rolle’s theorem, G’(1)= 0 for a <
< b,
we have
70
DERIVATIVES
[CHAP. 4
TAYLOR’S THEOREM of the MEAN 26. Prove Taylor’s theorem of the mean with the Lagrange remainder for the case n = 1. We must show t h a t f(b) =
f ( 4 + f’(4 ( b - 4 + 2 f”(o !( b - a)*
a <E < b
(1)
Consider the function
H ( z ) = f ( b ) - f ( x ) - f ’ ( ~( b) - x ) - ( b - x ) ’ A where A is an undetermined constant. Motivation for forming H ( x ) is attained by replacing a by x in (I) and transposing all terms to the right. From (2)’ H ( b ) = 0. To obtain H(a) = 0, we must choose
Assuming f ( x ) and f ’ ( z )satisfy the continuity and differentiability conditions of Rolle’s theorem, then H ( s ) also satisfies the conditions of Rolle’s theorem; hence there is a value E between a and b such that H ’ ( 0 = 0. From (z),H ’ ( z ) = - f ” ( z ) ( b - x ) 2(b - $ ) A and El’([) = - f ” ( ( ) ( b - 4) 2(b - ( ) A = 0 for A = f ” ( [ ) / 2 ! (since [ # b). Substituting this value of A into (3) and solving for f ( b ) yields the required result (I). Generalizations to n > 1 follow by similar reasoning.
+
+
27. (a) Prove that
+
sinx = sina (cosa)(x-a) where 5 is between a and x.
(sin a)(x- a)2 - (cos t ) ( x2! 3!
Use part (a) to evaluate sin 51” and estimate the error made. Let f ( x ) = sin x. Then f ’ ( x ) = cos x, f”(x) = -sin x, f ” ’ ( x ) = -cos x, so that f ’ ( a )= cos a , f ” ( a )= - sin a, f ” ’ ( ( ) = - cos (.
f ( a )= sin a,
Then substituting into Taylor’s formula with n = 2, i.e.,
where
(
is between a and x, the required result is obtained.
Let x = 51” = 51~/180radians and a = 45” = 46~/180radians.
Then x - a = ~ / 3 0radians.
Thus from part (a), since sin46” = cos46’ = f i / 2 ,
1-
()(r/30)’1 < ‘ 3( !Z y30 .0002. 3! Hence the sum of the first three terms, namely 0.777, is accurate to 3 decimal places.
The absolute value of the error term
(COS
If more accuracy is desired, more terms in Taylor’s formula should be taken.
+
Prove that ez = 1 + x x2/2! + x3/3! Prove that e is an irrational number.
+ .
Let f ( z )= e”. Then all derivatives of f ( x ) are equal to e=. If a = 0, then f(0)= f ’ ( 0 )= = er, so that Taylor’s formula becomes
f(”’(0) = e0 = 1 while f(”+l)([)
where E is between 0 and x.
=
DERIVATIVES
CHAP. 41 ,p+1
Let
Rn
xn+l
x > 0, 1R.I < 7 ez and lim IRnI = 0 (see Prob. 30, Chap. 3).
er. If
=
Thus for all x, i.e.
--Q)
71
(n+ 1).
<x <
ez
n em
lim IRnI = 0 and we may write
Q),
U+
= 1
m
+ x + x2/2! + x3/3! +
i.e. the series converges for all x. in lowest terms where p and q ( b ) From (I),letting x = 1 and assuming that e is a rational number 23 9 are positive integers, we have e = 2 = 1 + 1 + & + - +1. . . + - + - 1 er (3) 0<(<1 ! l 3! n! (n+1)! Choose n > q and multiply both sides of ( I ) by n! Then n!e
= n!f
nf
n!
= n ! + n ! + 2 ! + - 3! +...+1+-
er n+l
(4)
+
Now et/(n 1) is a number between 0 and 1, while every other term in (4) is a positive integer. Thus assuming that e is rational, we arrive at the contradiction that an integer is equal to a non-integer. Hence e must be irrational.
L’HOSPITAL’S RULE 29. Prove L’Hospital’s rule for the case of the “indeterminate forms” (U) O/O, (b) a / m * (a) We shall suppose that f ( x ) and g(x) are differentiable in a < x < b and f(xo) = 0, SO) = 0, where U < xo < b. By Cauchy’s generalized theorem of the mean (Problem 25)’
Then lim =-+=o+
fo g(4
=
lim -=o+
(0
=
lim -=o+
f:(4 = L
g
(2)
since as x 3 xo+, ( + ZO+. Modification of the above procedure can be used to establish the result if
x (b)
3 a,
2 3 50-,
x + ZO,
x + -W.
We suppose that f ( z )and g(x) are differentiable in a < x < b, and lim f ( z ) = Q), lim g(z) = OQ Z-+Z0+ r-+zo+ where a < xo < b. Assume XI is such that a < $0 < x < x1 < b. By Cauchy’s generalized theorem of the mean,
Hence
from which we see that
Let us now suppose that
lim
-=o+
f’o = L (4 0
and write (I) as
72
DERIVATIVES
[CHAP. 4
We can choose ZI so close to xo that lf’(()/g’(() - L I C lim (I - g(21)’g(2)) 1 - f(zl)/f(z)
= 1
c.
since lirn f ( z ) = fo
“4
“4”0+
Keeping and
+
lim
0 4 “0
21 fixed,
g(x)
=
we see that 00
Then taking the limit as z + ZO+ on both sides of (a), we see that, as required,
Appropriate modifications of the above procedure establish the remlt if z +
x + a,
XO-,
z -P
20,
X’-W.
30. Evaluate (a) lim 240
e* - 1 7
1
(JbI)X!
+ COSnX
x2
(c) lim
- 2x + 1p
x-b~+
In cos3x In cos 2x
All of these have the “indeterminate form” O/O. (a) lim-
1
6”2
“40
= lim-2e’” = 2 1
“40
Note: Here L’Hospital’s rule is applied twice, since the first application again yields the “indeterminate form” 0/0 and the conditions for L’Hospital’s rule are satisfied once more.
(’)
lim
r 4 0 +
In cos32 In cos 22
lim
.+0+
(- 3 sin 32)/(cos 32) (-- 2 sin 2x)/(cos 22)
sin 32
=
(;).(;)
=
lim 3 sin 32 cos 22 2 sin 22 cos 32
++o+
3 cos 22
3 cos32
= 4 9
Note that in the fourth step we have taken advantage of a theorem on limits to simplify further c alcul ation S.
3x2-x+6
31. Evaluate
(U)
(’)
( b ) lim
;2
In tan 2x
(a) 24ao lirn 5x2 + 6~ - 3 9 (c) In tan3x Z-bW All of these have or can be arranged to have the “indeterminate form” 32’-~+6 62-1 6 3 52,+62-3 - lim - = lim - - s+Oo 102 6 =-+a 10 - 6
ii\
lim
o+ot
x2e-zp
-/00.
+
In tan22 In tan 3x -
lim (2 sec’ 2x)/(tan 22) (3 sec* 3z)/(tan 32)
Z40+
tan 3s
2 sec’22
32. Evaluate
lim 2 sec*22 tan 32 3 sec’ 32 tan 22
r 4 O t
sinx - tan-lx !. x21n(l . +5)
Although L’Hospital’s rule is applicable here, its successive use becomes involved. Using Taylor’s theorem of the mean, however, the limit is obtained quickly and easily. We use the results (see Page 62) sin2 =
2 - -
za
3!
,
Pz5 6!
’
tan-lz
=
2 - -
2’
+ Qx5,
2’ ln(l+z) = x - -
+ Rxa
DERIVATIVES
CHAP. 41
73
Then the required limit equals
33. Evaluate
lim x21nx.
x+o+
l/x -X' = lim lim - = o llx' I 4 0 + --2/xs r+O+ 2 The given limit has the "indeterminate form" 0 Q). In the second step the form is altered so as to give the indeterminate form Q)/Q) and L'Hospital's rule is then applied. lim %*In% =
=+O+
34. Find lim 240
lim
r+o+
(COSZ)~/~*.
Since lirn cosx = 1 and lim l / x * = Q),the limit takes the "indeterminate form" 1". =+O 2 4 0 Let F(x) = ( C O S X ) ' / ~ . Then In F ( x ) = (In cos x ) / x * to which L'Hospital's rule can be applied. We have In cos x lim -
t+O
a?
Thus lirn lnP(x) = -4. =+O Then
But since the logarithm is a continuous function, lim In F ( x ) = In (lim F ( x ) ) . S+O
Z-bO
35. If F ( x ) = (e32 - 5 ~ ) l / ~find , (a) lim F(x) and (b) lirn F ( x ) . zr,m
The respective indeterminate forms in (a)and (b) are Let G ( x ) = lnF(x) =
In (esS- 6 x ) X
.
Then
x+Q
a o and
1".
lirn G ( x ) and lim G ( x ) assume the indeterminate
o-b m
0-b 0
forms a / ~ and ) 0/0 respectively, and L'Hospital's rule applies. We have
Then, as in Problem 34, lirn (esz- 6x)'/" = es. =-b 00
36. Evaluate lim(
zr,~
sin2$ -
);
.
This has the indeterminate form
00
'x - sin'x - Q).By writing the limit as =+o lim 'x sin' x ' it is seen that
L'Hospital's rule is applicable. However, this process proves laborious. Two methods of procedure are possible. Method 1:
since lim 2-0
The required limit can be written
'X ' L
sin x
=
= 1. Now by successive applications of L'Hospital's rule,
74
DERIVATIVES 1im x* - s in2 x
t+o
‘ 2
- lim 22 t+
0
[CHAP. 4
- lim 22
- 2 sin x cosx 4x3
- sin 22 4x3
2 4 0
4 s i n 2 2 = lim- 8 cos22 1 = lim 2 - 2 c o s 2 x - lim - -
12xa
t+O
Method 2:
2 4 ~
Z+O
2 4 0
24
3
Using Taylor’s theorem, we have 1im
=+O
xz - sin’x x* sin*x
+ Px’)’ + Px’)*
= lim x2 - ,(x - 2’/6 Z-O
x*(&- x3/6
+ terms involving x” and higher + terms involving and higher lirn 1/3 + terms involving x* and higher = 1 + terms involving x2 and higher
= lim x4/3 =-o
=
x4
2”
0-0
1 3
MISCELLANEOUS PROBLEMS 37. If x = g ( t ) and 3 = f ( t ) are twice differentiable, find (a)dyldx, (b) d2yldx2. ( a ) Letting primes denote derivatives with respect to t, we have
dz/ dx
38. Let f ( x ) = Je-”’*’
1O,
2;.
- dz//dt
-
f’o
-
if g’(t) # 0
g’(t)
Prove that (a) f’(0) = 0, ( b ) f ” ( 0 ) = 0.
If h = l/u,using L’Hospital’s rule this limit equals
Similarly, replacing h + Of by h -+ 0f: (0) = f ( 0 ) = 0, and so f ’ ( 0 )= 0.
and
U
+
00
by
U
+
-a,
we find f l ( 0 ) = 0.
Thus
by successive applications of L’Hospital’s rule. Similarly, f y (0) = 0 and so f ” ( 0 ) = 0. In general, f ‘ “ ’ ( 0 ) = 0 fo r n = 1,2,3, . .
.
(see Problem 89).
39. Let f ( x ) be such that f ( I V ) ( x )exists in a 5 x 5 b, and suppose that ~ ’ ( x o=) ~ ” ( x o= ) ~ ” ’ ( x o= ) 0 where a < xo < b. Prove that f ( s ) has a relative maximum or minimum a t xo according as f ( * v ) ( x< ~ )0 or > 0 respectively. By Taylor’s theorem of the mean, if ( is between xo and x, then
CHAP. 41
75
DERIVATIVES
If f’”’(xa) > 0, then for all x in a deleted 6 neighborhood of xo we have f ( x ) > f ( x o ) , 80 that f ( x ) has a relative minimum value at x =no. Similarly if fclv)(xa)< 0, then for all x in a deleted 6 neighborhood of xo we have f ( x ) < f(zo), so that f ( x ) has a relative maximum value a t x = no.
40. Find the length of the longest ladder which can be carried around the corner of a
corridor, whose dimensions are indicated in the figure below, if it is assumed that the ladder is carried parallel to the floor. The length of the longest ladder is the same as the shortest straight line segment A B [Fig. 4-61 which touches both outer walls and the corner formed by the inner walls.
As seen from Fig. 4-6, the length of the ladder A B is
= a sec8
L
+ b csce
L is a minimum when dLlde = a sec8 t a n e - b csce cote i.e.,
a sin’e = b cosse
so t h at
L
= a sec8
+
ta n e
or
=
b csc8
= 0
=
kb-I
b2/s)s/2
Although i t is geometrically evident tha t this gives the minimum length, we can prove this analytically by showing tha t d2Llde2fo r 8 = t a n - l m is positive (see Problem 88).
Fig. 4-6
Supplementary Problems DERIVATIVES 41.
Use the definition to compute th e derivatives of each of the following functions at the indicated point:
-
(U) (3% 4)/(2z
+ 3), x = 1;
;:
A m . (4 17/25, ( b ) 2, (4 42.
Let f ( z )
{ =
+
( b ) z3- 32% 2x - 5, z = 2; (c)
t , (43
sin l/x, z # 0
x=O’
6,= 4; 2
(d)
$-,
x = 2.
Prove t h a t (a) f ( z )is continuous at z = 0, (b) f ( s )has a derivative
at z = 0, (c) f ’ ( z ) is continuous at z = 0. 43. Let f ( z ) = tive at z=O. 44.
# Determine whether f ( x ) (a) is continuous at z = 0, (b) has a derivax=O‘ Ans. (a) Yes; ( b ) Yes, 0
xe-1‘z2’
1 O*
Give a n alternative proof of the theorem in Problem 3, Page 63, using “ r , 6 definitions”.
45. If f ( z )= e+, show t h a t
f’(z0)
= ezo depends on the result lim (eh- l)/h = 1. he0
46. Use t h e results lim (sin h)/h = 1, lim (1 -cos h)/h = 0 to prove t h a t if f(z) = sin z, ~ ’ ( x o=) coszo. he0
h+O
76
DERIVATIVES
[CHAP. 4
RIGHT and LEFT HAND DERIVATIVES 47.
Let f ( x ) = 51x1. (a)Calculate the right hand derivative of f(s)at z=O. (b) Calculate the left hand derivative of f ( x ) at x = 0. (c) Does f ( x ) have a derivative at z = O? ( d ) Illustrate the conclusions in (a),(b) and (c) from a graph. A m . (a)0; (b) 0; ( c ) Yes, 0
48.
Discuss the (a) continuity and (b) differentiability of f(x) = positive number. What happens in case p is any real number?
2s-3, O S z S 2 2< 5 4
49.
z p
Discuss the (a) continuity
%a--,
sin l/x, f ( 0 ) = 0, where p is any
and
(b) differentiability of f ( z ) in
05x54. 50. Prove t h a t the derivative of f ( z ) at z = 50 exists if and only if f: 51.
= fL ( S O ) .
(20)
+
(a) Prove t h a t f(x) = x3 - xz 6z - 6 is differentiable in a 9 z 5 b, where a and b are any constants. (b) Find equations for the tangent lines to the curve y = x3 - 2'4- 6z - 6 at x = 0 and x = 1. Illustrate by means of a graph. ( c ) Determine the point of intersection of the tangent lines in (b). ( d ) Find f'(s), f " ( x ) , f"'(z), f"")(s), A m . ( b ) = 6 % - 6 , 2/ = 6%-7; (c) (1,-1); ( d ) 3sz-22+6, 6%-2, 6, 0, 0, 0, . . .
.. . .
52. Explain clearly the difference between (a) f:
(SO) and
f'(xo+), (b) f:
(20)
and ~'(zo-).
53. If f ( z ) = ~'1x1,discuss the existence of successive derivatives of f ( z ) at x
= 0.
DIFFERENTIALS 54.
55. If f ( x ) = xa
+
32, find (a)Ay, (b) d y , ( c ) AylAx, ( d ) dyldx and ( e ) (Ay - dy)/Ax, if x = 1 and Ax = .01. Ans. (a).0601, ( b ) .06, (c) 6.01, ( d ) 6, (e) .01 (a) sin 31°, (b) In (1.12),
56. Using differentials, compute approximate values for each of the following: Ans. (a)0.616, ( b ) 0.12, (c) 2.0126 (c)
z.
57.
If y = sin x, evaluate ( a ) Ay, (b) d y . (c) Prove that (Ay - dy)/Ax + 0 as Ax + 0.
DIFFERENTIATION RULES and SPECIAL FUNCTIONS 58.
Prove: (cl
59.
(a)
d
& {g}
-
= dx f ( 4
- f ( x )g ' ( 4 ,
9 ( 4Y ( 4
d + dx d4,
d ( b ) ~ { f b-)g(4)
=
d
d
dx f ( 4 - z g ( 4 ,
+ 0.
&)
[Q<X)l'
+
d d (a)-{ x 3 In (xz- 2% 4- 6 ) ) at x = 1, (b) ds {sinz(32 n/6)} at x = 0. dx Ans. (a) 3 In 4, (b) $fi
Evaluate
60. Derive the formulas:
d (c) -tanh U = sech' dx 61.
d
+ &)>
{f(x)
(a) U
du
a" =
where dx
d (a)- tan-'%, dx use of principal values.
Compute
d
(b)
U
U"
du In a - a > 0, a # 1; dx '
(b)
d
cscu - -csc
U
du. cot U -, dx
is a differentiable function of x.
d dz csc-' x,
d
( c ) - sinh-'
dx
d x, ( d ) coth-' x, paying attention to the dx
CHAP. 41 62. If y = x=, compute dyldx.
[Hint Take logarithms before differentiating].
+
find dy/dx at x = 0. 63. If y = {In (3% 2)}sin-1(22+*5), 64.
77
DERIVATIVES
If y = f ( u ) ,where differentiable.
U
= g(v) and v = h(x), prove that
A m . x"(1
(& + 2-)(lnInfiIn 2 2 = 2 2 assuming f , Ans.
+ In x).
2)716
$ 0
g and
h are
65. Calculate (a)dyldx and (b) d2y/dx* if xy - l n y = 1. AnS. (a)y2/(l - xy), ( b ) (3y3- 2xy4)/(l- ~ y provided ) ~ xy # 1
+
66. If y = tan x, prove that g"' = 2(1+ yz)(l 3y').
x = sec t and y = tan t, evaluate (a)dyldx, ( b ) #y/dxe, (c) dsy/dxs, at t = a/4. Ans. (a) fi, ( b ) -1, (c) 3 f i
67. If
68.
Prove that
= dx'
zy,
-*/dv'(
stating precise conditions under which it holds.
69. Establish formulas (a)7, (b) 18 and (c) 27, on Page 60.
MEAN VALUE THEOREMS 70. Let f(x) = 1 - (z- l ) % I 3 , 0 5 x 5 2. (a)Construct the graph of f(x). (b) Explain why Rolle's theorem is not applicable to this function, i.e. there is no value [ for which f ' ( [ ) = 0, 0 < [ < 2. 71. Verify Rolle's theorem for f(x) 72.
= x2(1- x ) ~ ,0 5 x 5 1.
Prove that between any two real roots of er sin x = 1 there is at least one real root of er cos x = -1. [Hint Apply Rolle's theorem to the function e-" - sin 2.1
73. (a)If 0 < a < b, prove that (1 - a/b) < In b/a < @/a- 1) (b) Use the result of (a)to show that Q < ln1.2 <
6.
74.
Prove that
(a/6
+ 6/15)
<
sin-l.6
<
+
( ~ / 6 1/8) by using the theorem of the mean.
75. Prove the statement in the last paragraph of Problem 20(b). 76.
(a)If f ' ( z )S 0 at all points of (a,b), prove that f(x) is monotonic decreasing in (a, b). (b) Under what conditions is f ( x ) strictly decreasing in ( a , b ) ?
77.
(a)Prove that (sin z)/x is strictly decreasing in (0, ~ / 2 ) .(b) Prove that 0 S sin x 5 2 x / ~for 0 S x S a/2.
78.
(a)Prove that
sin b - sin a = cot[, where [ is between a and b. cos a - cos b (b) By placing a = 0 and b = x in (a),show that [ = x/2. Does the result hold if x < O?
TAYLOR'S THEOREM of the MEAN 79. (a)Prove that
In (1 +x)
= x -
In (1.1) and estimate the accuracy.
2' + -xa32
x4 XS 0 < < x. 4+6(1+,
(b) Use (a)to evaluate
Ans. ( b ) 0.09631 with an error less than 2 10-'.
80.
Evaluate (a)cos64O, (b) tan-'0.2, (c) coshl, (d) e-O*s to 3 decimal places. Ans. (a) 0.438, ( b ) 0.197, (c) 1.643, (d) 0.741
81.
Prove Taylor's theorem of the mean with the Lagrange remainder for (a)n = 2, ( b ) n = 3, (c) n = any positive integer.
78 82.
DERIVATIVES
[CHAP. 4
Derive the result @I), Page 61, for the Cauchy form of the remainder. [ H i n t Apply Rolle's theorem to
( b - 1' " - ( b - z)A - f ' ( 4 ( b - 4 - f " ( 4 2!( b - X I 3 - ... - f'"'(') n!
H(z) = f(b) - f(4
where A is chosen so that H ( a ) = O . ]
83. Prove that the Lagrange and Cauchy forms of the remainder in Taylor's theorem can be written in
the forms
+
hn+lp+'(a eh) (n l)!
+
h n + l ( l- e)rf(x+l)(u+eh) n!
and
respectively, where h = b - a and 0 < e < 1. 84.
By writing (b - %)*Ain place of the last term ( b - - ) A in the hint of Problem 82, obtain Taylor's theorem with (a) the Lagrange remainder, (b) the Cauchy remainder, using suitable special values for p.
L'HOSPITAL'S RULE 85. Evaluate each of the following limits. 2 - sin x (a)lim (e) 0+0+ lim x31nx
zs
r+o
% ! I
(c)
lim t-b
+
e2z - 2ef 1 cos 32 - 2 cos 22 cos 5
l+
(2'
+
( f ) lim (3"- P ) / x *+o
(i) =-bO lim (l/x
- csc x)
( j ) lim z~~~~
=-+ m
=*O
00
(n)l i m sin ( 7 x)
=+0
1/z2
Z+O
(8) lim (1- 3 / ~ ) ~ =(k) lirn (Us2- cota x)
- 1)tan nz/2
x+3 s)
(m) =+ lim x ln(
(0)
+ + eb)'IZ
lim (x ef =* 00
(d) lirn x3e-2E f+ m
MISCELLANEOUS PROBLEMS 86. Prove that
87. If
A f ( z ) = f(z
ln(l+x)
+Ax) - f(z),
< 1
if 0 < x < 1 .
(a) prove t h a t A { A f ( z ) } = A2f(x) = f ( x 4- 2 A s ) - 2 f ( z 4- A s ) -k f ( x ) ,
(b) derive an expression for Anf(x) where n is any positive integer, f ( " ' ( z ) if this limit exists. 88. Complete the analytic proof mentioned at the end of Problem 40.
89.
90.
91.
(c) show that
-
lirn A U f ( x ) A-O
(As)'
-* .
(a) If f ( z ) is the function of Problem 38, prove t h a t f'"'(0) = 0 f o r n = 1,2,3, . . . (b) Write Taylor's series with a remainder for this function and prove t h a t f ( z ) = R,. (c) Explain why R, cannot approach zero a s n-, and discuss the consequences of this.
Find the relative maxima and minima of f ( z ) = XI, x > 0. A m . f ( x ) has a relative minimum when x = e - ' .
A I and particle 11 respectively travels with (seeconstant adjoining velocities Fig. 4-7). v1 Show and vathat in mediums in order to go from point P to point Q in the least time, i t must follow path PAQ where A is such that (sin e,)/(sin e,) = w
h
Medium I
vdocity 0 1
Medium I1
Fig. 1-7
CHAP. 41
79
DERIVATIVES
p, we say t h a t a is an infinitesimal of higher order (or the same order) if lim a / p = 0 (or lim alp = 1 Z 0 ) . Prove that as 2-0, (a)sin22x and (1- cos 3%) a r e infinitesimals of the same order, (b) (x' - sin'%) is a n infinitesimal of higher order than {z - In (1 z) - 1 cos z}.
92. A variable a is called an infinitesimal if it has zero as a limit. Given two infinitesimals a and
+
93. Why can we not use L'Hospital's rule to prove t h a t 94. Can we use L'Hospital's
Explain.
+
= 0 (see Prob. 91, Chap. 2) ?
lim Z/::::
z* 0
rule t o evaluate the limit of the sequence un = n'e-"', n = 1,2,3, . . . ?
95. If a is an approximate root of f ( x ) = 0 , show t h a t a better approximation, in general, is given by
a - f(a)/f'(a) (Newton's method). [ H i n t Assume the actual root is a h, so that f ( a h) = 0. f ( a h) = f(a) h f'(a) approximately.]
+
+
+
+
Then use the fact that for small h,
96. Using successive applications of Problem 95, obtain the positive root of (b) 6 s i n z = 4% to 3 decimal places. Ans. (a)3.268, ( b ) 1.131 97. If
D denotes the operator dldx so that D y D"(uv) = (D"u)v
where nCr = )(: 98.
Prove t h a t
(a) x 3 - 2 x 2 - 2 x -
7 = 0,
dyldx while D k y 3 dkyldxk, prove Leibnitz's formula
+ nC1(Dn-'U)(DV)+ nC s ( D n- s U ) ( D * V ) +
..*
+ nCr(Df-'u)(DrV) +
**.
+ UD"V
are the binomial coefficients (see Problem 95, Chapter 1).
dn (x2sin z) = dzn
+
{z2 - n(n - 1)) sin (z n d 2 )
-
+
2ns cos (z n ~ / 2 ) .
f'(z0)= f"(z0) = ... = f('")(zo) = 0 but f(2n+')(x~) # 0, discuss the behavior of f ( x ) in the neighborhood of x = XO. The point zo in such case is often called a point of inflection.
99. If
100. Let f ( z )be twice differentiable in (a,b) and suppose that f'(a) = f ' ( b ) = 0.
at least one point E in (a,b) such that If"(4)l
2
involving velocity and acceleration of a particle.
4 (b - U)' { f ( b )- f ( a ) } .
Prove that there exists
Give a physical interpretation
Chapter 5 Integrals DEFINITION of a DEFINITE INTEGRAL The concept of a definite integral is often motivated by consideration of the area bounded by the curve y = f ( x ) , the x axis and the ordinates erected at x = a and x = b (see Fig. 5-1). However, the definition can be given without appealing to geometry.
Subdivide the interval a 5 x Ib into n sub-intervals by means of the points . . , $ , , - I chosen arbitrarily. In each of the new intervals (a,xl), ( x l , ~ ~.). ,., (Xn-1, b) choose points tl,t2,. . . ,tn arbitrarily. Form the sum x1,x2,.
f(t1)('1 - a) + f ( t 2 ) ($2 - $1) + f ( t 3 ) ($3 - '2) + * * + f ( t n ) ( b - x n - 1 ) By writing xo=a, x n = b and xk- Xk-1 = Axk, this can be written *
(1)
Geometrically this sum represents the total area of all rectangles in the above figure. We now let the number of subdivisions n increase in such a way that each Axk+O. If as a result the sum ( I ) or (2) approaches a limit which does not depend on the mode of subdivision, we denote this limit by
which is called the definite integral of f ( x ) between a and b. In this symbol f ( x ) d x is often called the integrund, and [a,b] is called the range of integration. We sometimes call a and b the limits of integration, a being the lower limit of integration and b the upper limit. 80
INTEGRALS
CHAP. 61
81
The limit (3)exists whenever f ( x ) is continuous (or sectionally continuous) in a 5 x S b (see Problem 35). When this limit exists we say that f ( x ) is Riemann integrable or simply integrable in [a, b]. Geometrically the value of this definite integral represents the area bounded by the curve y = f ( x ) , the x axis and the ordinates at x = a and x = b only if f ( x )2 0. If f ( x ) is sometimes positive and sometimes negative, the definite integral represents the algebraic sum of the areas above and below the x axis, treating areas above the x axis as positive and areas below the x axis as negative.
MEASURE ZERO A set of points on the x axis is said to have meusure zero if the sum of the lengths of intervals enclosing all the points can be made arbitrary small (less than any given positive number e). We can show (see Problem 6) that any countable set of points on the real axis has measure zero. In particular, the set of rational numbers which is countable (see Problems 17 and 59, Chapter l), has measure zero. An important theorem in the theory of Riemann integration is the following: Theorem. If f ( x ) is bounded in [a,b],then a necessary and sufficient condition for the existence of
lb
f ( x ) d x is that the set of discontinuities of f ( x ) have measure zero.
PROPERTIES of DEFINITE INTEGRALS If f ( x ) and g(x) are integrable in [a,b] then 1.
ib
{ f ( x )* g(x)} d x
2. S b A f ( x ) d x
lbf(z) l ib
b
=
dx
= A
a
f ( x ) dx
= ff(z)dx
3. i b f ( x ) d x
g ( Z ) dx
where A is any constant
+ Jbf(z)dz
provided f ( x ) is integrable in [a, c] and [c, b ] . 4.
J ba f ( x ) d z
6.
If in a 5 x S b, m S f ( x ) d M where m and M are constants, then
=
-iaf(x)dx
m(b-a)
S
lbf(z)dx S
M(b-a)
7. If in a 5 x 5 b, f ( x )5 g(x) then
MEAN VALUE THEOREMS for INTEGRALS 1. First mean value theorem. If f ( x )is continuous in [a, b], there is a point 5 in (a, b) such that (4)
82
[CHAP. 5
INTEGRALS
2. Generalized first mean value theorem. If f ( x ) and g(x) are continuous in [a,b],and g ( x ) does not change sign in the interval, then there is a point 5 in (a, b ) such that
This reduces to (4) if g ( x ) = l . 3. Bonnet’s second mean value theorem. If f ( x ) and g(x) are continuous in [a,b] and if g(x) is a positive monotonic decreasing function, then there is a point 5 in (a,b) such that f ( x ) g ( x )d x = s ( a ) f ( 4 0% (6)
s’
L’
If g(x) is a positive monotonic increasing function, then there is a point 5 in ( a , b ) such that
(7) 4. Generalized second mean value theorem. If f ( x ) and g(x) are continuous in [a,b] and if g(x) is monotonic increasing or monotonic decreasing and is not necessarily always positive as in 3, there is a point in ( a , b ) such that
This result holds also if we replace continuity by integrability.
INDEFINITE INTEGRALS If f ( x ) is given, then any function F ( x ) such that F’(x) = f ( x ) is called an indefinite integral or anti-derivative of f ( x ) . Clearly if F ( x ) is a n indefinite integral of f ( x ) , so also is F ( x ) + c where c is any constant since [ F ( x )+ c ] ’ = F’(x) = f ( x ) . Thus all indefinite integrals differ by a constant. We often use the symbol Jf(z)dz to denote any indefinite integral of f (x). Example:
If F’(x) = x 2 , then F ( x ) = J x 2 d x = x3/3 tive of x 2 .
+c
is an indefinite integral or anti-deriva-
FUNDAMENTAL THEOREM of INTEGRAL CALCULUS If f ( x ) is continuous in [a, b] and F ( x ) is any function such that F’(x) = f ( x ) [i.e. F ( x ) is an indefinite integral or antiderivative of &)I, then
fi(z)dx
=
F(b)
-
F(a)
(9)
This important theorem enables us to calculate definite integrals without direct use of the definition whenever a n indefinite integral is known. Example:
To calculate
x 2 dx, we note t h a t P ( x ) = x2, F ( x ) = x3/3 J2XZdX
= F(2) - F(1) =
+ c,
($ +c)-(;+c)
Since c disappears anyway, it is convenient to write more simply
and we have 7
= 3
CHAP. 61
INTEGRALS
83
DEFINITE INTEGRALS with VARIABLE LIMITS of INTEGRATION An indefinite integral can be expressed as a definite integral with variable upper limit by writing J f(x)ds = J f(x)dx c a It follows that
+
Since a definite integral depends only on the limits of integration, we can use any f ( x )d x = f ( t )d t = J b f ( u )du, variable as the symbol of integration. For example, (ll), etc. For this reason the variable is often called a d u m m y variable. We can ;rite f or instance, as -&J)(t)dt = f(4
lb
The result can be generalized to the case where the lower and upper integration limits are variable. Thus we have
Example:
sin X~ d(x*)
dt
sin x d(s) -
7 d x - p x -dx
2 sin xe - sin x X
CHANGE of VARIABLE of INTEGRATION
S
If a determination of
f ( x ) d x is not immediately obvious in terms of elementary
functions, useful results may be obtained by changing the variable form x to t according to the transformation x = g ( t ) . The fundamental theorem enabling us to do this is summarized in the statement f ( x )d x = f { g ( t ) >g'(t) d t (1-6)
J
J
where after obtaining the indefinite integral on the right we replace t by its value in terms of x, i.e. t = g - l ( x ) , assumed single-valued. This result is analogous to the chain rule for differentiation (see Page 59). The corresponding theorem for definite integrals is Jb
f ( x )d x
=
s."
f { s ( t )g'(t) > dt
(15)
where g ( a ) = a and g ( p ) = b, i.e. a = g - l ( a ) , p = g - l ( b ) . This result is certainly valid if f ( x ) is continuous in [a,b] and if g ( t ) is continuous and has a continuous derivatiye in
astsp.
INTEGRALS of SPECIAL FUNCTIONS The following results can be demonstrated by differentiating both sides to produce an identity. In each case an arbitrary constant c (which has been omitted here) should be added. 1. s u n d u =
3.
s
Un+l
nS1
~
n#-1
4.
5.
sinu d u = - cos u
6.
s s s
cosudu
= sin U
t a n u d u = In lsecul = - In lcosul cotu d u = In lsinul
84 7. 8. 9. 10.
s s s s s
INTEGRALS
+
secu du
In lsecu tanu] = In ltan (u/2 d4)I
cscu du
= In lcscu - cotu) = In ltanu/2[
=
+
23. 24.
csc2udu = -cotu = sec U
12.
= - cscu
16. 17. 18. 19. 20.
1
s s s
or - cos-'
a
cothu du
= In (sinhul
sech u du
= tan-' (sinh U)
1 U 1 - -tan-'; or --cot-'; a a
- coth-I (cosh U)
32.
sd-du
33.
Je"usinbudu
s
34.
=
: d m
a
U
= uv - S v d u
or
+-sin-'a2 2
U
a
eau(asin bu - b cos bu) a2 b2 eau(acos bu b sin bu) eaucos bu du = a2 b2
+ + +
=
SPECIAL METHODS of INTEGRATION 1. Integration by parts. Judv
U -
31.
= In coshu
=
U
-
30.
tanhudu
u du
sin-'
29.
coshudu = sinhu
S csch
= - csch u
28.
= coshu
sinhudu
csch u coth u du
27.
14. S e u d u = eu 15.
= - sech u
26.
au - a>0, a # l = In a
Jaudu
sech u tanh U du
25.
11. f s e c u t a n u d u
13.
= tanh u
sech2u du
22. S c s c h 2 u d u = -cothu
sec2u du = tan u
cscu cotu du
s s s
21.
[CHAP. 6
s
= f(x)g(x) -
f(x)g'(x) dx
f'(x)g(x) dx
where U = f ( x ) and v = g(x). The corresponding result for definite integrals over the interval [a,b] is certainly valid if f(z) and g ( x ) are continuous and have continuous derivatives in [a,b]. See Problems 18-20. 2. Partial fractions. Any rational function ()' where P(x) and Q(x) are polynomials,
Q(x) with the degree of P ( x ) less than that of Q(x), can be written as the sum of A Ax+B rational functions having the form where r = 1,2,3, . . . (ax + b)" (ax2 bx c)' which can always be integrated in terms of elementary functions.
+ +
Example 1: Example 2:
3 ~ - 2 (42 - 3 ) ( 2 ~ 6)' (2'
6%'-
+ 2+2
+ 2 2 + 4)"x
- - A 4x-3 - 1) ,
+-( 2 ~B i - 6 ) ~ Ax+B
+ 2x + 4)'
(2%
+
2 '
D
C +-
+ + +-
Cx+D 22 4
2
E
-1
The constants, A , B , C, etc., can be found by clearing of fractions and equating coefficients of like powers of 2 on both sides of the equation or by using special methods (see Problem 21).
CHAP. 61
INTEGRALS
85
3. Rational functions of sin x and cos x can always be integrated in terms of elementary functions by the substitution tan x/2 = U (see Problem 22). 4. Special devices depending on the particular form of the integrand are often employed (see Problems 23 and 24).
IMPROPER INTEGRALS If the range of integration [a, b] is not finite or if f ( x ) is not defined or not bounded at one or more points of [a, b], then the integral of f ( x ) over this range is called an imp r o p e r integral. By use of appropriate limiting operations, we may define the integrals in such cases.
Since this limit does not exist we say that the integral diverges (i.e. does not converge).
For further examples, see Problems 33, 78-80. For further discussion of improper integrals, see Chapter 12.
NUMERICAL METHODS for EVALUATING DEFINITE INTEGRALS Numerical methods for evaluating definite integrals are available in case the integrals cannot be evaluated exactly. The following special numerical methods are based on subdividing the interval [a, b] into n equal parts of length A X = (b -a)/% For simplicity we denote f ( a k Ax) = f ( x k ) by ?/k, where k = 0,1,2,. . .,n. The symbol means “approximately equal”. In general, the approximation improves as n increases.
+
-
1. Rectangular rule.
The geometric interpretation is evident from the figure on Page 80. 2. Trapezoidal rule.
This is obtained by taking the mean of the approximations in (16). Geometrically this replaces the curve y = f ( z ) by a set of approximating line segments.
This is obtained by dividing [a,b] into an even number of equal intervals, (i.e. n is even) and approximating f ( x ) by a quadratic through 3 successive points corresponding to $0, XI, XZ; XI, x2, x3; . . .; ~ n - 2 ,Xn-1, Xn. Geometrically this replaces the curve 2/ = f(x) by a set of approximating parabolic arcs. 4. Taylor’s theorem of the mean can sometimes be used, as in Problem 26.
INTEGRALS
86
[CHAP. 6
APPLICATIONS The use of the integral as a limit of a sum enables us to solve many physical or geometrical problems such as determination of areas, volumes, arc lengths, moments of inertia, centroids, etc.
Solved Problems DEFINITION of a DEFINITE INTEGRAL 1. If f ( x ) is continuous in [u,b] prove that
Since f(x) is continuous, the limit exists independent of the mode of subdivision (see Problem 36). Choose the subdivision of [a,b] into n equal parts of equal length Ax = ( b - a)/n (see Fig. 5-1, Page 80). Let tk = a k(b-a)/n, k = 1,2,.. .,n. Then
+
2.
Express limL
2 f(g)
as a definite integral.
n k=1 Let a = 0, b = 1 in Problem 1. Then n+m
3. (a) Express
3
0
definite integral.
x 2 d x as a limit of a sum, and use the result to evaluate the given (b) Interpret the result geometrically.
(a) If f(x) = xa, then f(k/n) = (k/n)' = kl/na. Thus by Problem 2,
This can be written, using Problem 29 of Chapter 1,
- lim n(n+ W n + 1) = 6n3
lim (1
*-+00
+ l/n)(2 + l/n) 6
1 - -
3 which is the required limit. Note: By using the fundamental theorem of the calculus, we observe t h a t (xs/3)); = 13/3- OS/3 = 1/3. n+ 00
(b) The area bounded by the curve y = xz, the x axis and the line x = 1 is equal to
1
+
n+2
+
* * *
The required limit can be written
using Problem 2 and the fundamental theorem of the calculus.
9.
x1
x'dx =
87
INTEGRALS
CHAP. 61
'I
5. Prove that nlimsing ewn
+ sin-2tn + . . . + sin
1 - cost n
Let a = 0, b = t, f ( x ) = s i n x in Problem 1. Then
and so
1 n-l kt lim - 2 sin-
n+w
using the fact that
n
n
k=l
=
sint lim - - 0.
n-w
1 - cost t
n
MEASURE ZERO 6. Prove that a countable point set has measure zero.
. ..
and suppose t h a t intervals of lengths less than Let the point set be denoted by 21,xz, xs, x4, respectively enclose the points, where e is any positive number. Then the sum 42, 4 4 , d8, 416, = e (let a = 4 2 and T = in Problem of the lengths of the intervals is less than e/2 e/4 4 8 26(a) of Chapter 3), showing that the set has measure zero.
...
+
+ + + -
PROPERTIES of DEFINITE INTEGRALS 7. (a)If f(x) is continuous in [a,b] and m 5 f(s) 5 M where m and M are constants, prove that m(b-a) 5 i b f ( z ) d x S M(b-a) (b) Interpret the result of (a) geometrically. We have mAXk
MAXk
f(&)AXk
k = 1,2,.. . , n
Summing from k = 1 to n and using the fact that
i t follows that
I
m(b-U) Taking the limit as n-, the required result.
00
5
k=l
f([k)AXk
5 M(b-U)
and each A x k + O yields
Assume f ( x ) h 0 and continuous in [a, b] with graph shown in the adjoining Fig. 5-2. It is geometrically evident that Area ABCD 5 Area under y = f ( x ) 5 Area ABEF i.e., m(b-~)5 f(x)dx S M ( ~ - u )
lb
A similar interpretation can be made if the restriction f ( x ) h 0 is removed. The result also holds if f ( x ) is sectionally continuous in [ a , b ] .
-
a
b-a
Fig. 5-2
b L
88
INTEGRALS
8. Prove that
[Jbf(z)dxl 5 i b l f ( s ) l d x
[CHAP. 6
if a < b .
By inequality 2, Page 3,
Taking the limit as n
9.
-+
00
2a
Prove that
and each Axk
3
0, we have the required result.
sin nx -
Then
-,&I
!iylx m sin nx
2a
= 0, and so the required result follows.
MEAN VALUE THEOREMS for INTEGRALS If f ( x ) is continuous in [a, b ] , prove that there is a point Jb
in (a, b ) such that
= ( b - a) f (t )
f(x)
Interpret the result of (a) geometrically. Since f(x) is continuous in [a,b], we can And constants m and by Problem 7,
M such that
m 5 f(z) S
M. Then
Since f(x) is continuous it takes on all values between m and M (see Chapter 2, Problems 34, 93). In particular there must be a value 4 such that
The required result follows on multiplying by b-a. If f(x) 2 0 with graph as shown in the figure of Problem 7 ( b ) , we can interpret
L’
f(x) dz as
the shaded area under the curve y = f ( x ) . Geometrically this area should equal that of a rectangle with base b - a and height f(r) for some value 6 between a and b.
FUNDAMENTAL THEOREM of INTEGRAL CALCULUS 11. If
F(x) =
r‘
f ( t )d t
where f ( x ) is continuous in [a, b ] , prove that F’(x) = f(z). = f(r)
[
between x and x
+h
by the first mean value theorem for integrals (Problem 10). Then if x is any point interior to [a,b],
since f is continuous. If x = a or x = b, we use right or left hand limits respectively and the result holds in these cases as well.
CHAP. 61
89
INTEGRALS
12. Prove the fundamental theorem of the integral calculus. By Problem 11, if F ( x ) is any function whose derivative is f(z), we can write
F(z) = l ' f ( t ) dt
Ib
+c
where c is any constant (see last line of Problem 22, Chapter 4). Since F(a)= c, it follows that
F(b) =
f ( t ) dt
+ F(a)
F(z) =
13. If f ( ~ )is continuous in [a,bJ,,prove that
or
xb
f ( t ) dt
Jrf(t)dt
= F(b) - F(a).
is continuous in [a,b ] .
If x is any point interior to [a,b ] , then as in Problem 11, h lim 4 0 F ( z + h) - F(x) = h-+O lim h f ( t ) = 0 and F ( x ) is continuous. If z = a and x = b, we use right and left hand limits respectively to show that F ( z ) is continuous at x = a and s = b .
Another method : By Problem 11 and Problem 3, Chapter 4, it follows that F'(z) exists and so F ( s ) must be continuous.
CHANGE of VARIABLES and SPECIAL METHODS of INTEGRATION 14. Prove the result (14), Page 83, for changing the variable of integration. F ( x ) = x r f ( x ) dx
Let
G(t)
and
=
l*
f { g ( t ) } g'(t) dt,
where s = g(t).
dF = f ( z ) d x , dG = f { g ( t ) } g ' ( t ) d t .
Then
Since dx = g ' ( t ) d t , it follows that f ( z ) d z = f { g ( t ) } g ' ( t )d t so that d F ( s ) = dG(t), from which F ( x ) = G(t) c. Now when x = a, t = a or F(a) = G(a) c. But F(a) = G(a) = 0, so that c=O. Hence F(x) = G ( t ) . Since x = b when t = p , we have
+
+
as required.
s
15. Evaluate:
( x + 2 ) sin (x2+4x-6)dx
(c)
s' d ( ~ -1
(d) Method 1: Let xL 4% - 6 =
dx +2)(3 - X)
2-"tanh2l-"&x
(e)
J*'&
s, 0
(f)
x sin-'
diz?
x2
x dx X2+X+1
=
+ 4) dx = du, (z+ 2) ds = 3 du and the integral becomes - g1 c o s u + c = - ~1 cog(x*+4s-6) + c
(x 2) sin (x* 4s - 6) d x =
sin (x2 42 - 6) d(x2 4 2 - 6) = - 4 cos (xa 4 2 - 6)
+
U.
Then (22
kssinudzl Method 2:
+
Let l n x =
+
U.
+
+
+
Then (dz)lx = du and the integral becomes
c o t u d u = lnlsinul
+c
= ln)sin(lnz)I
+c
dx
+c
90
[CHAP. 6
INTEGRALS (c)
Method 1: dx
d(x
dx
+ 2)(3- x)
dx
- sin-l.2
dx
=
S d6 - (2' - X)
=
426/4
- (x - 9)'
+ sin-'.6
Method 2: Let x - 4 = by Problem 14,
U
as in Method 1. Now when x = - l ,
-
-2s 2 In2
4s u d u l/&
x sin-lx' dx
JG2
dx 16. Show that 1 2 ( x 2 - % + 4 ) 3 / 2 U
=
w
= &us
=
-In cosh 2*-= 2 In2
=
2xdx
+c
21n2'
2xdx
diTF
= &(sin-'xs)'
&(sin-1xa)'lr6
=
&(sin-'
so that the integral becomes
+
c
and the integral becomes
+c 9)'
= ' P
144 '
1 6'
-
dx Let x - 1 = 6 t a n U, dx = fisec'u [(x - l)a 313/' = tan-lO = 0; when x = 2, U = tan-' 1 / 6 = ~ / 6 . Then the integral becomes
Write the integral as x = 1,
l2
d
-
= -du
= du and 2-'dz
tanhudu
Let sin-'2' = U. Then du =
Thus
~ 1 6
17. Determine
fisec'u
Je2
du
-
Thus
sin-'.2 4- sin-' .6
(d) Let 2*-== U. Then -2'-"(In2)dx
(e)
and when x = l , U = + .
U=-$;
$=I6
+
du.
When
a
fisec2u du
-
[3 sec2u]~/' -
1 cosu du = - s i n u 3
I,
=I6
-
1 6'
&.
Let In x = y, (dx)lx = dy. When x = e, y = 1; when x = e', g =2.
Then the integral becomes
CHAP. 61
91
INTEGRALS
18. Find J x n In x dx if (a) n z -1, ( b ) n = -1. (U) Use integration by parts, letting U = In x, dv = xndx,
so that du
+
= ( d x ) / x , v = xnil/(n 1).
Then
J
Jx - ' l n x
(b)
=
dx
19. Find S 3 - d ~ . Let d W i = y, 2%+ 1 = 9%. Then dx = y dy and the integral becomes .f 3r y dy. Integrate by parts, letting U = y, dv = 3s dy; then du = dy, 2) = 3Y/(ln 3), and we have
= sudv
s3U-ydy
= yl n3' 3
= uv - s v d u
+
20. Find J ' x In (x 3 ) dx.
+
Let U = In (z 3), dv = xdx.
s
x In (x
+ 3) dz
v = Then du = dx x+3'
= -1n ( x + 3) X3
+ 3)
21. Determine
s
.
-
3y + (ln3)'
-
c
)dx
'
= - - 4 1 n 4 + 291 n 3 4
6-x
(x - 3)(2x + 5 ) dx*
6-x
(Z
- 3 ) ( 2+ ~ 6)
- -A
X-3
+ -2 ~B + 6 '
Method 1: To determine the constants A and B, multiply both sides by (x- 3)(2x
= A(2s+6)
6-x
+ B(x-3)
Since this is an identity, SA - 3B = 6, 2A
6-x 3)(2x
In3
f{$-3x+gln(x+3)
Use the method of partial fractions. Let
.f (x-
-y -3s
Hence on integrating by parts,
2
~ ' x l n ( x + 3 dx )
Then
X'
= -1n(x+3) X'
2
= -ln(x X' 2
=
s&dy
+ 6)
1 ' 3
dz
or
+ B = -1
6-2
= SA
+ 6)
- 3B
to obtain
+ (2A+B)x
and A =3/11, B = -17/11.
+ S-dz
= ~3 l n I x - 3 ] - -ln12%+61 17
2x+6
(11
Then
22
-4- c
Method 2: Substitute suitable values for z in the identity (1). For example, letting x = 3 and x = -6/2 in ( I ) , we find at once A = 3/11, B = -17/11.
22. Evaluate
S
+
/qu
dx cosx by using the substitution tanx12 = U.
From Fig. 6-3 we see that sin d 2 =
U
JG-x cos x/2
1
=-
dTTG
1 Fig. 5-3
92
[CHAP. 6
INTEGRALS Then
cosx
Also
du =
1 - U'I
= cosax/2 - sinax/2 = 1 +U"
4 secsx/2 dx
or
2 du 1+tCs*
dx = 2 cosax/2 du = -
- & tan" u/2
Thus the integral becomes
23. Evaluate
i= +
Let x = a-y.
i.e. I = n'/2
c0s2z
sinx
+c
=
4 tan-'(*
tan x/2)
+ c.
dx.
Then
- I or I = u'I4. nl2
x
dx = 4'
24. Prove that
from which 21 = a/2 and I = a/4. The same method can be used to prove that for all real values of m,
1 -..
sin" x sinms cosmz dx
+
a -
= 4
(see Problem 94). Note: This problem and Problem 23 show that some definite integrals can be evaluated without first finding the corresponding indefinite integrals.
NUMERICAL METHODS for EVALUATING DEFINITE INTEGRALS 25. Evaluate
approximately, using (a) the trapezoidal rule, (b) Simpson's
rule, where the interval [0, I] is divided into n = 4 equal parts.
+
Let f ( x ) = 1/(1 x'). Using the notation on Page 85, we find Ax = (b -a)/% = (1 - 0)/4 = 0.25. Then keeping 4 decimal places, we have: yo = f(0)= 1.0000, y~ = f(0.26) = 0.9412, ga= f(0.60) = 0.8000, = f(0.75) = 0.6400, l/4 = f(1) = 0.5000, ( a ) The trapezoidal rule gives
Ax 2 {go + 22/1+
+ +
2 ~ s 2 ~ s y4}
0.26 = (1.0000
2
+ 2(0.9412) + 2(0.8000) + 2(0.6400) + 0.500}
= 0.7828.
CHAP. 51
INTEGRALS
93
( b ) Simpson's rule gives Ax 7 {YO + 4161 + 2 ~ + 2 4y3 + y4)
=
0 25 % {l.OOOO + 4(0.9412)+ 2(0.8000) + 4(0.6400) + 0.5000)
= 0.7854.
The true value is ~ / = 4 0.7854.
J1
26. (a) Evaluate
e z 2 d x approximately by using Taylor's theorem of the mean and
( b ) estimate the maximum error. As in Problem 28, Chapter 4, we find
Then replacing x by x2, ex2 = 1 + x2 +
x4
xs
x8
x'Oe6
g + g + z-gj+
O<[<X2
Integrating from 0 to 1, xlezzdx
= il(l + x2 =
1 +,+
+
dx
+15.12! + -7-31
+
9-41 + E
E =
(where the error E = 1.4618
+E
Thus the maximum error is less than 0.0021, and so the value of the integral is 1.46 accurate to two decimal places. By using more terms in Taylor's theorem, better accuracy can be attained.
APPLICATIONS 27. Find the (a)area and ( b ) moment of inertia about the y axis of the region in the xy plane bounded by y = 4 - x 2 and the x axis. (a) Subdivide the region into rectangles as in the figure on Page 80. A typical rectangle is shown in the adjoining Fig. 5-4. Then
( b ) Assuming unit density, the moment of inertia about the y axis of the typical rectangle shown above is S i f([k) AXk. Then Required moment of inertia
= lim
zl
n
X [if(&)Axk
nrtm k=l
n
=
128 15
E;(4 - E;) Axk
94
INTEGRALS
[CHAP. 6
28. Find the length of arc of the parabola y = x2 from Required arc length
=
1’dl +
(dy/dx)* d x
=
x = 0 to x = I.
1’d
w dx
29. Find the volume generated by revolving the region of Problem 27 about the x axis. Required volume = Iirn m+ao
I
2 r y t ~ x k= k=l
r
$-: (4 - xa)ad x
= 612~/15.
MISCELLANEOUS PROBLEMS 30. If f ( x ) and g(x) are continuous in [u,b], prove Schwurz’s inequality f o r integrub: b
We have
for all real values of A.
Hence by Problem 13 of Chapter 1, using ( I ) with
At = s.’{g(z)>”x,
B2 =
Jb
{f(x)>*dx,
c
=
Jb
f ( x ) d z )d x
we find Ca 5 A’B’, which gives the required result.
31. Prove the second mean value theorem of equation (8),Page 82, under the assumption of the existence and continuity of g’(x) in [ u , b ] in addition to the other assumptions. Let F ( z ) =
iz
f(t) dt.
Then on integrating by parts,
Cwe I : g(x) is monotonic increasing, i.e. g’(z) B 0.
Case 2: g(s) is monotonic decreasing, i.e. g ’ ( x ) S 0.
The proof is similar to that in Case 1.
CHAP. 61
95
INTEGRALS
32. (a) If fcn+l)(x)is continuous in [a,b], prove that for x in [a,b], f"(4( 5 - 4 ' + . . . + (a)(5- a)* + +J (x- t)" f ( 4 = f ( 4+ f'(4(-4 + 2! n! 2
f'")
f ( * + 1)
( t )dt
(b) Use (a) to obtain the Lagrange and Cauchy forms of the remainder in Taylor's theorem (see Page 61). (a) We use mathematical induction (see Chapter 1). The result holds for n = 0 since
f ( x ) = f ( 4+ J " f ' ( t ) dt = f ( 4+ f ( t )1:
= f ( 4 + f ( d- f ( 4
Assume that it holds for n = k. Then integrating by parts, using
we find
which shows that the result holds for n = k (b)
+ 1. Thus it holds for all integers
n
2 0.
We have by the generalized first mean value theorem for integrals (see Page 82),
f
l ' F ( t ) G(t)dt = F([)
G(t)dt
[
between a and x
0
F ( t ) = f ( * + ' ) ( t )G(t) , =
Letting
q,
we obtain
n.
giving the Lugrange f o r m of the remainder, [equation (ZO), Page 611, with b replaced by x.
giving the Cuuchy f o r m of the remainder, [equation (Zl),Page 611, with b replaced by x.
33. Prove that
lim
M-POO
+
XM&
=
+
8' P
+
+
4 = 2' 4s' 4 - 42' = (s' 2)' - (22)' = (2' We have 2' According to the method of partial fractions, assume Ax+B Cx+D - 1- 'x 4 z9+2x+2 + + - 2 2 + 2
Then so that
+
+
+ 2 + 2x)(x9+ 2 - 2x).
+
+
+
1 = (A+C)X' (B-ZA++C+D)X' ( ~ A - ~ B + ~ C + ~ D2B ) X 20 A + C = 0, B - U + 2 C + D = 0 , 2 A - - B + 2 C + 2 D = 0, 2 B + 2 D = 1
Solving simultaneously, A = 4, B = &, C = -6, D = i. Thus
-is
dx
+ +1
(x 1)'
= -1l n ( x a + 2 x + 2 ) 16
dx
+
dx $,f(x+l)'+l
+ -81t a n - ' ( x + l )
8
x-l dz (x-l)*+l
- 11 6l n ( x P - 2 z + 2 )
++s
+ -tan-'(x1 8
dx (x - 1)' 1)
+1
+C
96
INTEGRALS
We denote this limit by
[CHAP. 6
x*--& ,
called an improper integral o f the first kind. Such integrals are
considered further in Chapter 12. See also Problem 78.
34. Evaluate lim
fi sin t3 dt X'
z+a
The conditions of L'Hospital's rule are satisfied, so that the required limit is
--"a (4x5) dx
35. Prove that if f ( z ) is continuous in [a,b] then t
Let
U
=
2 f ( E k ) Axk,
k=l
ib
f(z)dx exists.
using the notation of Page 80.
Since f ( z ) is continuous we can find
numbers Mk and mk representing the 1.u.b. and g.1.b. of f ( x ) in the interval m 5 f(x) IMk. We then have
where m and
M are the g.1.b. and 1.u.b. of f ( x ) in [a,b ] . The sums
lower and u p p e r
8um8
respectively.
8
[%k-l,Zk],
i.e. such that
and S a r e sometimes called the
Now choose a second mode of subdivision of [a,b] and consider the corresponding lower and upper sums denoted by 8' and S' respectively. We must have 8'
5
s
and
s'
2
(2)
8
To prove this we choose a third mode of subdivision obtained by using the division points of both the first and second modes of subdivision and consider the corresponding lower and upper sums, denoted by t and T respectively. By Problem 89, we have 8 5 t t T T S '
and
8 ' 6 t l T I S
(3)
which proves (2). From ( 2 ) it is also clear that as the number of subdivisions is increased, the upper sums are monotonic decreasing and the lower sums are monotonic increasing. Since according to (1) these ~ u m s are also bounded, it follows that they have limiting values which we shall call s' and _S respectively. By Problem 90, B S S. In order to prove that the integral exists, we must show that ii = _S. Since f ( x ) is continuous in the closed interval [a,b], it is uniformly continuous. Then given any can take each Axk so small that MI, - mk < d ( b - a). It follows that
> 0, we
Now S-8 = ( S - 8 ) 4- @-a) 4- (9- 8 ) and it follows that each term in parentheses is positive and so is less than c by (4). In particular, since ,S - B is a definite number it must be zero, i.e. = 8'. Thus the limits of the upper and lower sums are equal and the proof is complete.
s
97
INTEGRALS
CHAP. 61
Supplementary Problems DEFINITION of a DEFINITE INTEGRAL x3dx as a limit of a sum. (b) Use the result of (a)to evaluate the given definite integral. (c) Interpret the result geometrically. 37.
Using the definition, evaluate (a) r
38. Prove that 39. Prove that
lim
i-
n
+n + ...
ns+ 1 '
lim
1P
A m . (b) (xa-4x)dx.
+&}
>
A m . (a)8, ( b ) 9
= 4' U
ns++'
+ 2 P + 3P + ... + np
U 4 00
40. Using the definition, prove that
xb
er d x
- -.:1>if
p>-l.
= eb - ea.
41. Work Problem 6 directly, using Problem 94 of Chapter 1.
42.
Prove that
lim
43.
Prove that
lim
1 ++ ... +
{d?-TT &%F
2 n k=lns+ Vx' U
n400
tan-'x if 5
XZO.
PROPERTIES of DEFINITE INTEGRALS 44.
Prove (a) Property 2, (b) Property 3 on Page 81.
45.
If f ( x ) is integrable in (a,c) and (c, b), prove that
lb 1'f ( 4 + lb lb xb f ( z )dz
=
d~
46. If f ( x ) and g ( x ) are integrable in [a,b] and f ( x ) 5 g(z), prove that 47.
Prove that
48.
Prove that
49.
Prove that
f(z)
f ( z )dz.
g b ) ch.
5
1 - cosx 1 x z / r for 0 S x 5 d 2 .
IJ
5 In 2 for all n.
-dx( f i e~- "
sind x
xS
126. U l
MEAN VALUE THEOREMS for INTEGRALS 50. Prove the result (5), Page 82. [Hint If m 5 f(x) S M, then m g(z) S f ( x ) g ( x ) S M g(x). Now integrate and apply Property 7, Page 82.1 51. Prove that there exist values
tl and Et in
0 S 2: S 1 such that
52. Prove that there is a value E in 0 S x S U such that
s
i u e - r c o s x dx = sin[.
lam, (d) s
CHANGE of VARIABLES and SPECIAL METHODS of INTEGRATION 53. Evaluate:
(e)
(a)
J:m. dx
xae*inzscosx'dx,
Am. (a)Qesinzs
(b) J ' tan-' m d tt ,
+ c,
(c)
&
(b) rL/32, (c) d3, (d) - 2 coth
6 + c,
(e)
*dU,
6
In 3.
98
INTEGRALS
[CHAP. 5
+ c.
54.
(a)
55. Prove that
S=
(b)Sd=du
-C
+ 41-
+aaIn I U
+ +aasin-'ula + c,
+ud=
x dx
56.
iud-
=
u2k a2du
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
using the transformation x = l / y .
= a12. Thus a / 2 = 0.
100
[CHAP. 6
INTEGRALS
1 if x is irrational is not Riemann integrable in [0,1]. 0 if x is rational [Hint: In (Z), Page 80, let [k, k = 1,2,3,. . , n be first rational and then irrational points of subdivision and examine the lower and upper sums of Problem 36.1 f(x) =
88. Prove that
.
89. Prove the result (3) of Problem 36. [Hint: First consider the effect of only one additional point of
subdivision.]
90.
In Problem 36, prove that ii 5
S.
[Hint Assume the contrary and obtain a contradiction.]
91. If f ( x ) is sectionally continuous in [a, b], prove that
xb
f(x) d x exists. [Hint Enclose each point of
discontinuity in an interval, noting that the sum of the lengths of such intervals can be made arbitrarily small. Then consider the difference between the upper and lower sums.] 92. If
{
f(x) =
93. Evaluate
22
O<X
3 x =1 , find 6s-1 l < ~ < 2
ls{x
- [XI
+ +} dx
(a)Prove that (b) Prove that
1 2~
+
95. Prove that
dx tan,x
1 0.8
96. Show that 97. Show that
F
d
xdx
dx
=
=
for all real values of m.
r.
exists.
dx x
= 0.4872 approximately.
-
Am. 9
where [x] denotes the greatest integer less than o r equal to x. Inter-
sin"'x
=Ia
Interpret the result graphically.
Am. 3
pret the result graphically. 94.
f f ( x )dx.
Pa
Chapter 6 Partial Derivatives FUNCTIONS of TWO or MORE VARIABLES A variable x is said to be a function of two variables x and y if for each given pair ( x , y ) we can determine one or more values of x . This definition is in keeping with the general definition of function as a correspondence between two sets (see Page 20). Here the two sets are (1) a set of number pairs (x,y) [represented geometrically by a two dimensional point set in the xy plane] and (2) a set of real numbers represented by the variable x . We use the notation f(x,y), F(x,y), etc., to denote the value of the function at (x,y) and write x = f ( x ,y), ,x = F(x, y), etc. We shall also sometimes use the notation x = x(x, y) although it should be understood that in this case x is used in two senses, namely as a function and as a variable. Example: If f(x, y) = xz
+ 2y3, then
f(3, -1) = (3)2
+ 2(--1)3 = 7.
The concept is easily extended. Thus w = F ( x , y , x ) denotes the value of a function at (x,y , x ) [a point in 3 dimensional space], etc.
DEPENDENT and INDEPENDENT VARIABLES. DOMAIN of a FUNCTION If x = F(x, y), we call x a dependent variable and x and y the independent variables. The function is called single-vdued if only one value of x corresponds to each pair (x,y) for which the function is defined. If there is more than one value of x , the function is multiple-valued and can be considered as a collection of single-valued functions. Hence we shall restrict ourselves to single-valued functions, unless otherwise indicated. The set of values (points), (x,y) for which a function is defined is called the domain of definition or simply domain of the function. Example: If z = dl - ( x 2 + y 2 ) , the domain for which z is real consists of the set of points ( 2 ,y) such that x2 y2 5 1, i.e. the set of points inside and on a circle in the zy plane having center at (0,O) and radius 1.
+
THREE DIMENSIONAL RECTANGULAR COORDINATE SYSTEMS A three dimensional rectangular coordinate system obtained by constructing 3 mutually perpendicular axes (the x, y and x axes) intersecting in point 0 (the origin) forms a natural extension of the usual xy plane for representing functions of two variables graphically. A point in 3 dimensions is represented by the triplet (x,y , x ) called coordinutes of the point. In this coordinate system x = f ( x ,y ) [or F ( x ,y, x ) = 01 represents a surface, in general. Example: The set of points ( x , y , x ) such that x = dl - (x2+ y2) comprises the surface of a hemisphere of radius 1 and center at (O,O,O).
For functions of more than two variables such geometric interpretation fails, although the terminology is still employed. For example, ( x , y , x , w ) is a point in 4 dimensional space, and w = f ( x ,y, x ) [or F ( x ,36, x, w)= 01 represents a hypersurface in 4 dimensions; thus x 2 + y 2 + x 2 + w 2= a2 represents a hypersphere in 4 dimensions with radius a > 0 and center a t (O,O,O,0). 101
102
PARTIAL DERIVATIVES
[CHAP. 6
NEIGHBORHOODS The set of all points (x,y) such that I x - x ~ l < 6, Iy-yol < 6 where 6 > 0, is called a rectangular 6 neighborhood of (x0,yO); the set 0 < I x - x ~ l < 6, 0 < Iy-yo/ < 6 which excludes (xo,yo) is called a rectangular deleted 6 neighborhood of ($0, yo). Similar remarks ~ is a circular 6 neighcan be made for other neighborhoods, e.g. (x- ~ 0 +) (y~- y ~ <) S2 borhood of ($0, yo). A point (xo,y0) is called a limit point, accumulation point or cluster point of a point set S if every deleted 6 neighborhood of (x0,yo) contains points of S. As in the case of one dimensional point sets, every bounded infinite set has at least one limit point (the Weirstrass-Bolzano theorem, see Pages 5 and 50). A set containing all its limit points is called a closed set. REGIONS if there exists A point P belonging to a point set S is called an interior point of a deleted 6 neighborhood of P all of whose points belong to S. A point P not belonging to S is called an exterior point of S if there exists a deleted 6 neighborhood of P all of whose points do not belong to S. A point P is called a boundary point of S if every deleted 8 neighborhood o f P contains points belonging to S and also points not belonging to S. If any two points of a set S can be joined by a path consisting of a finite number of broken line segments all of whose points belong to S, then S is called a connected set. A region is a connected set which consists of interior points or interior and boundary points. A closed region is a region containing all its boundary points. An open region consists only of interior points. Examples of some regions are shown graphically in Figures 6-l(a),( b ) and ( c ) below. The rectangular region of Fig. 6-l(a),including the boundary, represents the set of points a 5 x 5 b, c 5 y 5 d which is a natural extension of the closed interval a 5 x 2 b for one dimension. The set a < x < b, c < y < d corresponds to the boundary being excluded. In the regions of Figures 6-l(a)and 6 - l ( b ) , any simple closed curve (one which does not intersect itself anywhere) lying inside the region can be shrunk to a point which also lies in the region. Such regions are called simply-connected regions. In Fig. 6 - l ( c ) however, a simple closed curve ABCD surrounding one of the “holes” in the region cannot be shrunk to a point without leaving the region. Such regions are called multiplyconnected regions.
Fig. 6-1
CHAP. 61
PARTIAL DERIVATIVES
103
, any positive number we can find some positive number 6 [depending on L and ( x o , ~ o )in general] such that If(x,y)-Zl < L whenever 0 < Ix-xol < 6 and 0 < 1y-y01 < 6.
If desired we can use the deleted circular neighborhood 0 instead of the deleted rectangular neighborhood.
< (x- Z O )+~ (y - y
~ <) B2 ~
Example : gets closer to 3(1)(2) = 6 and we suspect that lirn f ( z , y ) = 6. To prove this we must r-1 U-2
show that the above definition of limit with Z=6 is satisfied. Such a proof can be supplied by a method similar to t h a t of Problem 4. Note that lirn f ( x , y ) f f ( l , 2 ) since f(1,2) = 0. The limit would in fact be 6 2 4 1
8 4 2
even if f ( z , g ) were not defined at (1,2). Thus the existence of the limit of f(z,y) as (z,y) + (20,go) is in no way dependent on the existence of a value of f(z,y) at (zo,yo).
Note that in order for
(x,
lim
(x,I/)
(+
YO)
f(x,y) to exist, it must have the same value re-
gardless of the approach of y) to ($0, yo). It follows that if two different approaches give different values, the limit cannot exist (see Problem 7). This implies, as in the case of functions of one variable, that if a limit exists it is unique. The concept of one-sided limits for functions of one variable is easily extended to functions of more than one variable. Example 1: lim tan-' (ylx) = a/2, r-O+ v+1
lirn tan-' (ylz) = -a/2
r-0Y-1
Example 2: lirn tan-'(ylx) r e 0
does not exist, as is clear from the fact that the two different ap-
U-+'
proaches of Example 1 give different results.
In general the theorems on limits, concepts of infinity, etc., for functions of one variable (see Page 24) apply as well, with appropriate modifications, to functions of two or more variables.
ITERATED LIMITS The iterated limits
lim
X-zQ
{
}
lim f(x,y)
u4yO
and
.-..I,-,.} lim
lim f ( x , y )
, [also denoted by
lim lirn f ( x ,y) and lim lim f ( x ,y) respectively] are not necessarily equal. 2-q)
1I-yo
II-Yo
z-xo
Although
they must be equal if lim f(z, y) is to exist, their equality does not guarantee the existence 24x0 of this last limit. y-y0 Example:
( %-y>
If f(z,y) = x:--2/ then lim lim - = lirn (1) = 1 and x+y' 3 4 0 y 4 0 S+Y r e 0 lirn (-1) = -1. Thus the iterated limits are not equal and so lirn f ( z , y) cannot exist.
U 4 0
r-0
U40
CONTINUITY Let f ( x , y ) be defined in a 8 neighborhood of (x0,yO) [i.e. f(x,y) must be defined at ( z o , ~ oas ) well as near it]. We say that f(x,y) is continuous at (x0,yo) if for any positive number L we can find some positive number 6 [depending on L and (x0,yo) in general] such that y) - f ( x 0 , go) < whenever 1s - 501 < 6 and Iy - y01 < 6. Note that three conditions must be satisfied in order that f(z,y) be continuous at (zo,~o).
I
104
[CHAP. 6
PARTIAL DERIVATIVES
-
lim
1.
( x ,U )
( X O ,Y o )
f(x,y) = I,
i.e. f ( x , g ) is defined at
f(xo,yo)must exist,
2.
i.e. the limit exists as (x,y)-P ( x o , ~ o ) (XO,VO)
3. l = f(x0,yo) If desired we can write this in the suggestive form
lim f ( x ,y) = f(lim 3, lim ?I)2-30
Y-YO
x-20
Y*Yo
f(z,y) is not continuous at (1,2). If we redefine the function so that f ( z , y ) (2, y)
= (1,2), then the function is continuous at (1,2).
= 6 for
If a function is not continuous at a point (xo,yo), i t is said to be discontinuous at (x0,yo) which is then called a point of discontinuity. If, as in the above example, it is
possible so to redefine the value of a function at a point of discontinuity that the new function is continuous, we say that the point is a removable discontinuitg of the old of the xy plane if it is confunction. A function is said to be continuous in a region tinuous at every point of %. Many of the theorems on continuity for functions of a single variable can, with suitable modification, be extended to functions of two o r more variables.
UNIFORM CONTINUITY In the definition of continuity of f(x,g) at ( X O , Z / ~ ) 8, depends on L and also ( x o , ~ oin ) general. If in a region % we can find a 8 which depends only on L but not on any particular point (xo,go)in % [i.e. the same 8 will work for all points in %], then f ( x , g ) is said to be uniformly continuous in %. A s in the case of functions of one variable, it can be proved that a function which is continuous in a closed and bounded region is uniformly continuous in the region. PARTIAL DERIVATIVES The ordinary derivative of a function of several variables with respect to one of the independent variables, keeping all other independent variables constant, is called the partial derivative of the function with respect to - the variable. Partial derivatives of f ( x ,y) with respect to x and y are denoted by f,,(x,g),
$Iz] -
af ax [or
fZ, f z ( x , y ) ,
$IY]
9
and
$ [or fv,
respectively, the latter notations being used when i t is needed to emphasize A
which variables are held constant. By definition,
when these limits exist. often indicated by Example:
) The derivatives evaluated at the particular point ( x o , ~ o are
$l~zo,uo~
= f&o,
WO)
and
2 1 ay
= fy(xo,go) respectively.
(ZoBuo)
+
If f ( x ,y) = 2z3 3xy*, then f Z = af/ax = 62.9 -I-3y4 and fz(1,2) = 6(1)*+ 3(2)L= 18, f,(1,2) = 6(1)(2) = 12.
fY
= df/dz/ = 6xy. Also,
If a function f has continuous partial derivatives dflax, aflag in a region, then f must be continuous in the region. However, the existence of these partial derivatives alone is not enough to guarantee the continuity of f (see Problem 9).
PARTIAL DERIVATIVES
CHAP. 61
105
HIGHER ORDER PARTIAL DERIVATIVES If f ( x , y ) has partial derivatives a t each point ( x , y ) in a region, then aflax and aflay are themselves functions of x and y which may also have partial derivatives. These second derivatives are denoted by
If f z y and f v z are continuous, then f m = f v z and the order of differentiation is immaterial; otherwise they may not be equal (see Problems 13 and 43). Example:
If f(z, y) = 22’ 4- 3xy* (see preceding example), then fw = 122, fur = 62, fY = 6y = fit. In such case fu(1,2) = 12, fYu(l,2)= 6, f a ( l , 2 ) = fBx(1,2)= 12.
In a similar manner higher order derivatives are defined. For example, @f a X 2 ay is the derivative of f taken once with respect to g and twice with respect to x.
fvxx
DIFFERENTIALS Let Ax = d x and AY = d y be increments given to x and y respectively. Then = f(x+AX,y+A?j) - f(x,?/) = A f (8) is called the increment in x = f ( x ,g). If f ( x ,y) has continuous first partial derivatives in a region, then A2
A2
where c1 pression
= -afA x
af + c1Ax + c,Ay = Gax d x + - dazy + c l d x + c,dy = A f (4) +-AY ax ay al/ and c2 approach zero as Ax and A y approach zero (see Problem 14). The ex-
is called the total differential or simply d ife r e n t i a l of x or f , or the principal part of Ax or A f . Note that Ax Z dx in general. However, if Ax = dx and A y = d y are “small”, then dx is a close approximation of Ax (see Problem 15). The quantities d x and diy, called differentials of x and y respectively, need not be small. If f is such that A f (or Ax) can be expressed in the form (4) where c1 and c2 approach zero as Ax and A y approach zero, we call f differentiable a t ( x , y ) . The mere existence of f i and f v does not in itself guarantee differentiability; however, continuity of f 5 and ftr does (although this condition happens to be slightly stronger than necessary). In case f x and f v are continuous in a region q,we shall say that f is continuously diflerentiable in T.
THEOREMS on DIFFERENTIALS In the following we shall assume that all functions have continuous first partial derivatives in a region q,i.e. the functions are continuously differentiable in T. 1. If
=
f ( x 1 ,X Z ,
. . .,X n ) ,
then
regardless of whether the variables X I , X Z , . . .,X n are independent or dependent on other variables (see Problem 20). This is a generalization of the result (5). In ( 6 ) we often use x in place of f .
PARTIAL DERIVATIVES
106 2. If XI,
~(XI,XZ,
. . .,Xn
x2,
. . .,xn)
= c, a constant, then df = 0. cannot all be independent variables.
3. The expression P(x, y) dx
+ Q(x, y) dy
tial of f(x,y) if and only if exact differential. 4.
[CHAP.6
=
+
Note that in this caae
or briefly Pdx
Pdx
z.In such case +
+ Q dy
is the differen-
+ Qdy
is called an
or briefly P d x + aP a& a& - aR Q dy R dz is the differential of f ( x ,y, z) if and only if - - - - - ay ax' ax ay' aR--- a' In such case P d x Qdy R d z is called an exact diflerential. ax az' The expression
P(x,y,z)dx
+
Q(x,y,z)dy
+
R(x,y,z)dz
+
Proofs of Theorems 3 and 4 are best supplied by methods of later chapters (see Chapter 10, Problems 13 and 30).
DIFFERENTIATION of COMPOSITE FUNCTIONS Let x = f(x,y) where x = g ( r , s ) , y = h(r,s) so that z is a function of r and s. Then
then
In general, if
U
= F(x1,
E -ark
If in particular
XI,X~,
. . .,G)
au axl axl ark
. . .,xn
where
au- ax2 + + -ax2 ark
XI
=
fl(r1,
. . .,rp), . . .,
du dXn ... + dxn a r k
depend on only one variable
au dxl du - -- ds axl ds
au dxt +-+ axe ds
xn = f,,(rl, . . . , r p ) ,
k = 1,2, . . . , p 8,
then
... + aU dxn
(9)
axn ds
I -
These results, often called chain rules, are useful in transforming derivatives from one set of variables to another. Higher derivatives are obtained by repeated application of the chain rules.
EULER'S THEOREM on HOMOGENEOUS FUNCTIONS A function F(xl,xZ, . . .,xn) is called homogeneous of degree p if, for all values of the parameter x and some constant p, we have the identity
Example:
+
F ( x , y) = x4 Z2y3 - Sy' is homogeneous of degree 4, since F ( b , Ay) = (A%)' 2(hx)(Ay)' - S(Xy)' = A'(%' 22y3 - Sy') = X'F(z, y)
+
+
Euler's theorem on homogeneous functions states that if F(x1, XZ, geneous of degree p then (see Problem 25)
. . .,Xn)
is homo-
CHAP. 61
107
PARTIAL DERIVATIVES
IMPLICIT FUNCTIONS In general, an equation such as F ( x , p , x ) = 0 defines one variable, say x, as a function of the other two variables x and 9. Then x is sometimes called an implicit function of x and y, as distinguished from a so-called expZicit function f , where x = f ( x , y ) , which is such that F [ x , y , f ( x , g ) ]= 0. Differentiation of implicit functions is not difficult provided the dependent and independent variables are kept clearly in mind. JACOBIANS If F(u,v) and G(u,v) are differentiable in a region, the Jacobian determinant, or briefly the Jacobian, of F and G with respect to U and v is the second order functional determinant defined by
FO
(7)
Gv Similarly, the third order determinant
is called the Jacobian of F , G and H with respect to made.
U, v
and w . Extensions are easily
PARTIAL DERIVATIVES USING JACOBIANS Jacobians often prove useful in obtaining partial derivatives of implicit functions. Thus for example, given the simultaneous e(pations
=0 and v as functions of x and y. In this case, we have (see
F(X,Y,U,V)
we may, in general, consider Problem 31)
U
a(F, G )
a(F, G ) a --. --~ ( v J )
ax
a(F,G)’
a(%
= 0,
-au= - ag
U)
a(~&
a(F,G)’
a(% v )
G(~,Y,U,V)
av -_ - -
a(F,G )
d(F,G)
a(u4
a(F,G) ’
-av - -- - a(% ?I)
a(% v )
a(F,G ) a(% U )
The ideas are easily extended. Thus if we consider the simultaneous equations = 0, G ( w , w , x , ? I ) = 0, H ( ~ , V , W , X , Y )= 0 we may, for example, consider U, v and w as functions of x and g. In this case, F(U,V,W,X,Y)
- -- -
d(F,G, H ) a(x,v , w )
a(F,G , H) aw -- -
v , Y)
a(F, G, H) ’ a2/ a(F,G, H ) a(% v, w ) a(% v , w) with similar results for the remaining partial derivatives (see Problem 33).
ax
108
PARTIAL DERIVATIVES
[CHAP. 6
THEOREMS on JACOBIANS In the following we assume that all functions are continuously differentiable. 1. A necessary and sufficient condition that the equations F(u,v,x,y,x) = 0, a(F,G) is G(u,v,x,y,x)= 0 can be solved €or U and v (for example) is that a(% v ) not identically zero in a region Similar results are valid for m equations in n variables, where m < n . 2.
If x and y are functions of
U
and U while
U
and v are functions of r and s, then (9)
This is an example of a chain rule for Jacobians. These ideas are capable of generalization (see Problems 114 and 116, for example). 3. If
U
= f ( x , y ) and v = g ( x , y ) , then a necessary and sufficient condition that a
a(% v ) functional relation of the form +(u,v)= 0 exists between U and v is that a($, Y) be identically zero. Similar results hold for n functions of n variables.
Further discussion of Jacobians appears in Chapter 7 where vector interpretations are employed.
TRANSFORMATIONS The set of equations defines, in general, a transformation or mapping which establishes a correspondence between points in the uv and x y planes. If to each point in the uv plane there corresponds one and only one point in the x y plane, and conversely, we speak of a one to one transformation or mapping. This will be so if F and G are continuously differentiable with Jacobian not identically zero in a region. In such case (which we shall assume unless otherwise stated) equations (10) are said to define a continuouslg diflerentiable transformation or mapping. Under the transformation (10) a closed region % of the xy plane is, in general, mapped into a closed region %’ of the uv plane. Then if AA,, and AA,, denote respectively the areas of these regions, we can show that where lim denotes the limit as AA,, (or AA,,) approaches zero. The Jacobian on the right of (11) is often called the Jacobian of the transformation (10). If we solve (10)for U and v in terms of x and y, we obtain the transformation U = f ( x , y ) , v = g ( x , y ) often called the inverse transformation corresponding to (10).
a(% V ) The Jacobians a(z,y) and a@, v ) of these transformations are reciprocals of each other (see Problem 45). Hence if one Jacobian is different from zero in a region, so also is the other. The above ideas can be extended to transformations in three or higher dimensions. We shall deal further with these topics in Chapter 7, where use is made of the simplicity of vector notation and interpretation.
PARTIAL DERIVATIVES
CHAP. 61
109
CURVILINEAR COORDINATES If ( x , y ) are the rectangular coordinates of a point in the x y plane, we can think of (u,v) as also specifying coordinates of the same point, since by knowing (u,v) we can determine ( x , y ) from (10). The coordinates (u,v) are called curvilinear coordinates of the point. Example:
The polar coordinates ( p , +) of a point correspond to the case U = p , v = +. In this case the transformation equations (10) are z = p cos +, y = p sin +.
For curvilinear coordinates in higher dimensional spaces, see Chapter 7.
MEAN VALUE THEOREMS 1. First mean value theorem. If f ( z , y ) is continuous in a closed region and if the first partial derivatives exist in the open region (i.e. excluding boundary points), then f(xo+h,2/o+k) - f ( ~ o , 2 / 0 )= hfz(~o+Oh,?/o+Ok)+ kfu(xo+Oh,go+Ok)
0 < 6 < 1 (12)
This is sometimes written in a form in which h = A$ = x - xo and k = Ay = g - WO.
2. Taylor's theorem of the mean. If all the nth partial derivatives of f ( x ,y) are continuous in a closed region and if the (n 1)st partial derivatives exist in the open region, then
+
f ( ~ o + h , y ~ + k= )
P(XO,YO)
i a + ( h & + I c $ ) f ( x ~ , y ~ +) s(h% +k$)k~,y~)+
...
where R n , the remainder after n terms, is given by
and where we use the operator notation
f(x0,2/0)
E
+
E
hfz(Xo,Yo)
h2fut ($0, go)
+ 2hk
+
fzu ($0,
kf~(X0,1/0)
YO) + k2fuu (XO,YO)
(15)
etc., where we expand ( h a k d ) . formally by the binomial theorem. ax a?/ Equation (IS) is sometimes written in a form where h = As = x - xo and k = Ay = y-WO. Note that (12) is a special case of (Is)where n = 0 . In case lim Rn = 0 for all ( x , y ) in a region, the result can be used to obtain an n-+w infinite series expansion of f(z, y) in powers of x - xo and y - yo convergent in this region, called the region of convergence. This series is called a Tuylor series in 2 variables. Extensions to 3 or more variables can be made.
110
[CHAP. 6
PARTIAL DERIVATIVES
Solved Problems FUNCTIONS and GRAPHS 1. If f(x,y) = x3-2xy+3y2, find: k#0. (a) f(-2,3)
= (-2)3 - 2(-2)(3)
( b ) f($;)
=
+ 3(3)2 =
(;>”-2 ( 3 ( 3 +
(c) f(x’ y+k) - f ( x 9)’
k
(b) f
(a) f(-2,3); -8
3()s
+ 12 + 27
= 31
= 1-4 + 2 x3
xy
y’
+ k) + 3(y + k)2] - [x3- 22y + 3y2]} 1 ( x 3 - 2 ~ 2/ 2kx + 3y2+ 6ku + 3k’ - + 2 ~ 2/ 3~’) 1 -(-2kx + 6kv + 3k2) = - 2 ~+ 62/ + 3k. k
= k{[x3 - 2x(y = =
X’
Give the domain of definition for which each of the following functions are defined and real, and indicate this domain graphically. (a)f ( x ,y) = In { (16 - x2 - y2)(x2 y2- 4))
+
The function is defined and real for all points (x, y) such that
(16-x2-y2)(x2+y2-4)
>
0,
i.e.
4
<
se+g2
<
16
which is the required domain of definition. This point set consists of all points interior to the circle of radius 4 with center a t the origin and exterior to the circle of radius 2 with center a t the origin, as in the figure. The corresponding region, shown shaded in Fig. 6-2 below, is an open region.
Fig. 6-2
( b ) f ( x , y ) = l/6
- (2%+3Y)
Fig. 6-3
+
The function is defined and real for all points (2,y) such that 2s 3y 5 6, which is the required domain of definition. The corresponding (unbounded) region of the xy plane is shown shaded in Fig. 6-3 above.
3. Sketch and name the surface in 3 dimensional space represented by each of the following. (a)2x+4y+3x = 12. Trace on xy plane ( x = O ) is the straight line 3 + 2 y = 6, x = O . Trace on yz plane (x=O) is the straight line 4 ~ 4 - 3=~12, z=O. Trace on xx plane (y=O) is the straight line 22+3x = 12, y=O. These are represented by AB, BC and AC in Fig. 6-4. The surface is a plane intersecting the x, y a n d x axesin the points A(6,0,0), B(O,3,0), C(O,O, 4). The lengths OA = 6, OB = 3, Oc = 4 are called the x, y and x intercepts respectively.
Fig. 6-4
CHAP. 61
PARTIAL DERIVATIVES
111
x2 -+ 5 = 1, x = 0. u2 b
Trace on xy plane ( z = 0) is the ellipse
Trace on yx plane (x = 0) is the hyperbola
y2 b2
x2
-- - =
c2
1, x = 0.
x2 2 2 Trace on xx plane (y = 0) is the hyperbola -- ~2 = 1, y = 0. U2
Trace on any plane x = p parallel to the xy plane is the ellipse
ay1
+p W ) X2
+
Y2
+p W )
by1
= 1
Fig. 6-5
As lpl increases from zero, the elliptic cross-section increases in size. The surface is a hyperboloid of one sheet (see Fig. 6-5).
LIMITS and CONTINUITY 4. Prove that lim (x2+2y) = 5. 2-1 Y-+2
Method 1, using definition of limit. We must show that given any E > O , we can find 6 > O such that 0
<
Ix-11
< 6, 0 < Iy-21 < 6. < Ix-11 < 6 and 0 <
If 0 excluding x = 1, y = 2. Thus
1- 26
+ 6'
5-46+S2
<
<
x2
<
x2+2y
(y-21
+ + S2
1 26
<
<
6, then
and
5+46+S2
4 - 26
1-6
<
x
<
<
<
4
+ 26.
2y
<
-4s+s2
or
1+6
Ix2+2y-51
and 2 - 6
<
e
when
y
<
24-6,
<
Adding,
x2+2y-5
<
46+S2
Now if 6 5 1, it certainly follows that -56 < x 2 + 2y- 5 < 56, i.e. Ix2+ 2y- 51 < 56 whenever 0 < [ x - 11 < 6, 0 < Iy - 21 < 6 . Then choosing 56 = e, i.e. 6 = J5 (or 6 = 1, whichever is smaller), it follows that Ix2 2y - 51 < e when 0 < Ix - 11 < 6, 0 < Iy-21 < 6, i.e. lim (x2+ 2y) = 5.
+
2-1 Y e 2
Method 2, using theorems on limits. lim ( x2 + 2y ) 1
2-
Y+2
5. Prove that
lirn 2y
2 41
=
1
+4
= 5
Y+2
f(x,y) = x 2 + 2 y is continuous at (1,2). 2 4 1 Y 4 2
Then lirn f(x, y) = f ( l , 2 ) .%+I Y-2
6.
+
Y e 2
By Problem 4, lim f(x, y) = 5.
any
= 2-1 lirn x2
+
Also, f ( l, 2) = l2 2(2) = 5.
and the function is continuous at (1,2).
Alternatively we can show, in much the same manner as in the first method of Prob. 4, that given we can find 6 > 0 such that \f(%,y)-f(l,2)1 < e when 1%-11 < 6, 1y-2) < 6 .
e>O
Determine whether (a) has a limit as
f (x, y) =
x + 1 and
x2 + 2Y,
(x, Y) + (192) Y) = (112) ($9
y + 2, ( b ) is continuous a t (1,Z).
112
PARTIAL DERIVATIVES (a)
(b)
By Problem 4, it follows that value at (1,Z).
[CHAP. 6
lirn f(z,y) = 6, since the h i t has nothing to do with the
r-1
U-+'
Since lim f(x,y) = 6 and f ( l , 2 ) = 0, it follows that lirn f(z,y) # f(l,2). r-+ 1 U-+'
r-1
Hence the function
U-+'
is d&continuous at (1,Z).
7. Investigate the continuity of Let x + 0 and y + O in such a way that y = m x (a line in the sy plane). Then along this line, lim-
r-0 1 4 0
+
- -l1-mm' +ms
xs-maxa - lim x'(1- m') lim - r-0 x'+m'x' - r-+o x'(l+ m')
'x - y' 'X
'/1
Since the limit of the function depends on the manner of approach to (0,O) (i.e. the slope m of the line), the function cannot be continuous at (0,O).
Another method: Since lim r-+o
{
lim
">
~-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,
G = hkf,,(xo+e,h,yo+e,k) G = hkf,,(xo+e,h,po+e4k)
+ elk)
= f,(m
0 in ( 9 ) we have, since
fry
and
+ e&, fvz
YO+ e,k)
are assumed continuous at
YO),
(ZO,
(xo,Yo) = fY (Zo, Yo)
flfz
as required. For an example where this fails
to hold, see Problem 43.
DIFFERENTIALS 14. Let f ( x , p ) have continuous first partial derivatives in a region Prove that Af
where
c1
and
of the x y plane.
= f ( X + A X , y + A z / ) - f ( X , y ) = f z A X 4- f u A y 4- c i A X 4-
approach zero as A x and Ay approach zero.
Applying the theorem of the mean for functions of one variable (see Page 61), we have (1)
Af
{ f ( & Y + AY) - f ( % Y)> = { f ( x -I-A% I/ + A?/) - f ( x , Y -I-AY)) = Ax f a ( % e,Ax, y 4- Ay) 0 < 8, < 1, 0 < 8, AY fu(x,Y @,Ay)
+
+
+
<1
CHAP. 61
PARTIAL DERIVATIVES
115
Since, by hypothesis, fi and fu are continuous, it follows that f=(x+@,As,1/+&/) = f=(X,Y)
where
e1
+
0,
Thus
e2 4 0
Af
as Ax
= fzAx
+
+
fY(%Y+4A.y) = fr(x,li)
€1,
+ f yAy 4-
df = f = d x
€2
0 and Ay+O. e1 Ax
+
e,
Ay
as required.
Defining Ax=dx, Ay=dy, we have Af = f t d x We call
+
+ f Y d y the diferential
+ f Y d y+ e,dx + e,dy.
of f (or z ) or the principal part of Af (or Az).
15. If x = f(x, y) = x2y - 3y, find (a) AZ, (b) dx. (c) Determine Az and dx if x = 4, g = 3, AX = -0.01, A# = 0.02. (d) How might you determine f(5.12,6.85) without direct computation ? Sohtion: (4 = =
f(x
+ Ax, + Ay) - f(z, Y)
+
AY) - 3(y +Ay)} - {Z'Y - 31/} = 2 ~ A% y 4- (x' - 3) AY 4- (Ax)'~-k 2~ AS AY 4- (Ax)'A~
{(x 4- A x ) ' ( ~
Y
L
J
Y
(4
(B) The sum (A) is the principal part of Az and is the differential of z, i.e. dz. Thus (b)
dz
Another method: (c)
AZ
dz
dz
+ az + -dy al/
+
= ~ x ~ A x(x' - 3) Ag = 2 ~ y d Z (x'az = -dx
ax
= 2xydx
3) dg
+ (xa-3)dy
+
= f ( -k~Ax, Ag) - f(x, U) = f(4 - 0.01, 3 0.02) - f(4,3) = {(3.99)'(3.02) - 3(3.02)} - {(4)*(3)- 3(3)} = 0.018702
= 2xy d s
+ (d- 3) dv
= 2(4)(3)(-0.01)
+ (4'-
3)(0.02) = 0.02
Note that in this case Az and dz are approximately equal, due to the fact that Ax=dx and Ay = dy are sufficiently small.
+
+
+
+
(d) We must find f(x Ax, y Ay) when x As = 6.12 and y Ay = 6.86. We can accomplish this by choosing x = 6, Ax = 0.12, y = 7, Ay = -0.16. Since Ax and Ay are small, we use the fact that f(x Ax, y 4- Ay) = f(x, y) A z is approximately equal to f(x, y) dz, i.e. z dz.
+
Now
+
+
z = f ( x , y ) = f(6,7) = (6)'(7)
dz = 2xy dx
+ (x*- 3)dy
+
= 2(6)(7)(0.12) (6' - 3)(--0.16) = 6.1.
Then the required value is 164+6.1 = 169.1 approximately. direct computation is 169.01864.
16. (a) Let
+
- 3(7) = 164
U = x2e~/x.Find dU. (b) Show that (3x2y-2g2)dx
The value obtained by
+ (x3-4x2/+6y2)dy can
be written as an exact differential of a function +(x,g) and find this function.
Then
116
PARTIAL DERIVATIVES
[CHAP. 6
From (l), integrating with respect to x keeping y constant, we have + = xsy - 2xy' F(y)
+
where F(y) is the "constant" of integration. Substituting this into (2) yields
xs - 4 x y
+ F'(y)
= xs - 4xy
+ 6y*
from which
Hence the required function is # = xsy - 2xy'
+ 2ys + c,
F'(y) = 6y', i.e. P(y) = 2ys
+c
where c is an arbitrary constant.
Note that by Theorem 3, Page106, the existence of such a function is guaranteed, since if P = 3x*y -2y' and Q = x s - 4 x y + 6y*, then aP/ay = 32'-4y = aQ/ax identically. If aP/dy # aQ/& this function would not exist and the given expression would not be an exact differential. Method 2: ( 3 2 ' ~- 2 ~ ' d) x
+
+ (d - 4 2 +~6y2)d2/
+
+ +
= d(zsy- 2 ~ y ' 2 ~ '
Then the required function is xsy - 2xy'
+
+
= ( 3 ~ ' yd x X' d g ) - ( 2 ~ d' z 4 ~ d y ) 69%d v = d(x'Y) - d ( 2 ~ y ' ) d(2y') = d(Xs'l/-2xy2+221/')
+ 2y' + c.
C)
This method, called the grouping method, is based on one's ability to recognize exact differential combinations and is less direct than Method 1. Naturally, before attempting to apply any method, one should determine whether the given expression is an exact differential by using Theorem 3, Page 106. See last paragraph of Method 1.
DIFFERENTIATION of COMPOSITE FUNCTIONS 17. Let z = f ( x , y) and x = +(t),y = +(t)where f , 4, + are assumed differentiable. Prove that az d x az d y dx -
dt
ax dt + -ay- d t
=
Using the results of Problem 14, we have A' d z = lim = lim ~ Z A X a' ~y dt At-0 At At-0 since as A t + O we have A x + O , A y + O , e 1 + 0 ,
A% At
+*-)0,
"}
'At
Ax -+& At dt'
az d x = -
*+ At
dt
dt'
18. If z = ex? x = t cost, y = t sin t, compute dzldt a t t = ~ / 2 . dz - a' d x -ax- d+t - - ay a' dd yt = (y'em*)(- t sin t + cos t ) + (2xyezvl)(t cos t dt
+
At t = a / 2 , z = 0, y = a/2. Then
az d y
dx d t -I-dy
+ sin t ) .
= (aS/4)(--a/2) (0)(1) = --aa/8. d t t=7r/* Another method. Substitute x and y to obtain z = eB sin*t and then differentiate. 'Oat
19. If x = f ( x , y ) where x = &,v) az ax - az ag = (a) ax au ay a ~ )
+--
{:E
and
I/
= +(U,v), prove that
( a ) From Problem 14, assuming the differentiability of f,+,+, we have a' - lim AE = lim + a2 AY + Ax au
~ u - 0 AU
Att-0
'A u
ay AU
( b ) The result is proved as in (a) by replacing Au by AV and letting AV
20. Prove that
az
dz = ax dx
+ -daxy
even if x and
I/
+
0.
are dependent variables.
Suppose x and y depend on three variables u,v,w, for example. Then ( 1 ) dx
= xudu
+ x u d v + xrodw
(2) d y
= yudu
+ y v d v + ywdw
CHAP. 61
PARTIAL DERIVATIVES
Thus
2,
dx
+
Z,
dy
+
+
117
+
= x y y u ) du (Zrxv z,yv) dv = z,du 4- z v d v xWdw = dz (ZIXU
+
+ (zzxw
z , U w ) dw
using obvious generalizations of Prob. 19.
21. If
T = x3 - xy + g3, x = COS +, y = sin +, find (a) aT/ap, (b) aT/a+. a T ax aTay aT -- - = (3x2- y)(cos +) + (3y2- x)(sin #) aP a% ap + ay ap This may also be worked by direct substitution of x and y in T.
+
U = x sin g/x where x = 3 9 28, y = 4 r - 2s3, x = 2P - 3s2, find (a) aUlar, ( b ) au/as. a u ax auay auaz aU - -
22. If
(a)
ar
+
ax a r
=
(b)
au as ’
{
+
ay a r
( x cos
$(-
az ar
‘7cos 5 + 2
-
--
-
a u ax -
=
{
+
ax as
( x cos
+ {( z cos 9(:)}(4)
-$)}(6r)
auay ay as
23. If x = p c o s + , y = p s i n + ,
X
az as
+ {( z
3)}(2)
- -2yz cog ?r - @? x* x x
:)(4r)
aug
+
$)(-
+ 4r sin
cos
+ (sin
cog
E X
+ (sin
cos :)(:)}(-6s’)
$)(-68)
6s gin 1/
show that
X
(gJ+(%I (gJ+~(qav]. =
1
Using the subscript notation f o r partial derivatives, we have Vp = V,x, V4 = V,z4 Dividing both sides of (2) by
p,
+ V,yp + Vry4
we have 1 -V4 = - V , s i n # P
Then from (1) and (S), we have 1 V; -pV$ = (V, cos 9
+
+
= Vrcos,p V, sin+ = V,(-p sin+) V,(pcos#)
+ V, sin #)% +
+
+ V,cos+ (- V , sin
+
+ V, cos +)*
=
V:
+
24. Show that x = f(x2y), where f is differentiable, satisfies x(ax/ax) = Zy(ax/dy). Let x’y = U. Then x = f(u). Thus az au az - - ax au ax - f’(u)*2~11, Then
az az x- = f’(u)*2xay, 2 y =~ f’(u).Zx*y ax
a~ au az = au - = fyu).x’ ay ay ax az and so x- = 2y-. all ax
118
[CHAP. 6
PARTIAL DERIVATIVES
Another method : We have
dz
= f’(x’y)d(x’y) = f’(z*y)(2xydx
+ x’dy).
+ az/
az
az --U.
= -dz
Also,
dz
Then
az = 2xyf’(z’y),
az
as
ay
x g =
Elimination of f’(z3y) yields
= x’f’(x2y).
az
2y -
ay
.
25. If for all values of the parameter
and for some constant p, F(hx,Xy) = P F ( x , y ) identically, where F is assumed differentiable, prove that x(dF/dx) y(dF/dy) = pF. Let hz=u, x y = v .
+
Then F ( u , v ) = x’F(z,y)
The derivative with respect to
(1)
of the left side of (1) is
The derivative with respect to X of the right side of ( I ) is pXP-lF. Then
Letting A = 1 in (2),so that
U
26. If F(x,g ~ )= x4y2sin-’y/x,
+
= x, v = y, we have %@Flax) y(aF/ay) = pF.
show that
z(dF/dx)
+ y(dF/dy) = 6F.
Since F ( x z ,Ay) = (Xs)‘(Xy)’ sin-’ XyIXx = X6x4yasin-’ y/x = As F ( x , y), the result follows from Problem 26 with p = 6 . It can of course also be shown by direct differentiation.
+
27. Prove that Y = f(a: +at) g(x - at) satisfies d2Y/dt2= u2(d2Y/dz2),where f and g are assumed to be a t least twice differentiable and U is any constant.
CHAP. 61
28. If
119
PARTIAL DERIVATIVES
x = 2 r - s and
g = r + 2 s , find
Solving x = 2r-s,
y
a2U in terms of derivatives with respect to r and s. ag ax
= r+28 for r and 8 : r = (2%+y)/6, s = ( 2 y - x ) / 6 . ar/dy = 1/6, dslay = 2/6. Hence we have
Then W d x = 2/6, ds/dx = -1/6,
assuming U has continuous second partial derivatives.
IMPLICIT FUNCTIONS and JACOBIANS 29. If U = x3y,find d U / d t if (I) x5 y = t, ( 2 ) x2 g3 = t2.
+
+
Equations ( 1 ) and ( 2 ) define x and y as (implicit) functions of t. Then differentiating with respect to t, we have ( 8 ) 6x4(dx/dt) d y / d t = 1 (4) 2 x ( d x / d t ) 3ya(dy/dt) = 2 t
+
+
Solving (3) and ( 4 ) simultaneously for d x l d t and d y l d t , 1 d x -44 . = 3Ya 3ys-2t dt 1 6 ~ ' ~ ' 2~ - ' dt
I it I
I E4 I 2't
-
10x4t-222 1 6 ~ -~ 22 ~ '
of the $9 30. If F(x, y, X) = 0 defines x as an implicit function of x and g in a region plane, prove that (a)a d d x = - F J F , and (b) azlag = - F,/F,, where F, # 0. Since z is a function of x and y,
az
= -dx
dz
ax
az + -dy. all
Since x and y a r e independent, we have aF dF az aF dFaz ( I ) = + - = 0 ( 2 ) - + - = 0 az ax ay a2 ay from which the required results are obtained. If desired, equations ( I ) and (2) can be written directly.
31. If F(x,y,u,v) = 0 and G(x,;y,u,v)= 0, find (a) auldx, ( b ) duldy, ( c ) W a x , (d) dv/dy. The two equations in general define the dependent variables U and v as (implicit) functions of the independent variables x and y. Using the subscript notation, we have (1) d F ( 2 ) dG
Also, since
U
= Fxdx = Gzdx
+ F # d y + F u d u + F,dv + GWdW + Gudu + Gvdv
= 0 = 0
and u are functions of x and y, (8) d u
= uIdx
+ u,dy
(-6)
du
= vrdx
+ v,dy.
120
PARTIAL DERIVATIVES Substituting ( 8 ) and (4) in (I) and (2) yields (5) (6)
dF dG
=
(Fr (Gs
+ Fuus + Fuvs )dx + + G u ~ +t G vvr )dx +
(Fy (G,
[CHAP. 6
+ F s h + Fvv,)dy + G u h + Guv,)dl/
= =
0 0
Since x and y are independent, the coefficients of dx and dy in ( 5 ) and ( 6 ) a r e zero. Hence we obtain FWuz f Fvvz = -Fa G,vr = -Gz Gruz
+
Solving (7)and ( 8 ) gives
U,
F and G with respect t o
U
I
I
Fu Fv , denoted by G, Gv and v and is supposed # 0.
The functional determinant
or a(% v )
J
(s),
is the Jacobian of
Note that it is possible to devise mnemonic rules for writing at once the required partial derivatives in terms of Jacobians (see also Problem 33).
32. If u2-v = 3x + y and u-2v2 = x -2y, find
(U)&lax,
( b ) avlax, (c) aulay, ( d ) dvldy.
Method 1: Differentiate the given equations with respect to x , considering U and v as functions of z and y. Then au (2) au - 4 v -av = 1 ( I ) 2u - "ax2 1 = 3 ax ax
ax
Solving,
av - -2u - 3 1 - ~ U V ' ax 1 -~ U
a u- -I-12~ f3X
V
Differentiating with respect to y, we have
Solving,
a u- - - 2- 4v -
a?/
1-auv
-1 1-auv.
av = -4u
ay
We have, of course, assumed t h a t 1- 8uv # 0.
Method 2: The given equations are F = ua- v - 32 - y = 0, G = Problem 31,
U-
2vs- x
+ 2y
1 - 12v 1 -~ U provided 1 - 8uv # 0. Similarly the other partial derivatives are obtained.
33. If F(u,V , w,x,y) = 0, G(u,v,w,x,y) = 0, H(u, v,w,x,y) = 0, find
= 0. Then by
V
CHAP. 61
121
PARTIAL DERIVATIVES
From 3 equations in 6 variables, we can (theoretically at least) determine 3 variables in terms of the remaining 2. Thus 3 variables are dependent and 2 are independent. If we were asked to determine avlay, we would know that v is a dependent variable and y is a n independent variable, but would not know the remaining independent variable. However, the particular notation
ay =
serves to indicate
that we are to obtain av/az/ keeping x constant, i.e. x is the other independent variable. (a) Differentiating the given equations with respect to y, keeping
x constant, gives
F, = 0 (2) G,u,+ G v v , + G w w , + G, = 0 (3) H , u , + H , v , + H , w , + H , = 0
( I ) F,u,+ F,u,+ F,w,+
Solving simultaneously for
U,,
we have
I
I Fa
F W
Fw
I F,
F.
Fw I
I
H, Hv H, I
Equations (I), (2) and (3) can also be obtained by using differentials as in Problem 31. The Jacobian method is very suggestive for writing results immediately, as seen in this problem and Problem 31. Thus observe that in calculating
the result is the negative of the quotient of ay t two Jacobians, the numerator containing the independent variable y, the denominator containing the dependent variable v in the same relative positions. Using this scheme, we have
a22
- - 3z2+ x (3Z2 - x)'
34. If z 3 - x z - g = 0, prove that - 8% a?.l
'
Differentiating with respect to x , keeping y constant and remembering that z is the dependent variable depending on the independent variables z and y, we find
Differentiating with respect to y, keeping x constant, we find
-
1
=
0
and
(2)
1 av - 32'-x
Differentiating (2)with respect to x and using (I),we have
The result can also be obtained by differentiating (I) with respect to y and using (2).
U = f ( x ,g) and v = g(x,g), where f and g are continuously differentiable in some region T. Prove that a necessary and sufficient condition that there exists a functional relation between U and v of the form +(%,U) = 0 is the vanishing of the Jacobian,
35. Let
i.e.
a@, 24
= 0 identically,
122
PARTIAL DERIVATIVES
[CHAP. 6
Necessity. We have to prove that if the functional relation +(u,v)= 0 exists, then the Jacobian
e(u,v)= a(%,Y)
0 identically, To do this, we note that
= Then
(+uUr
(1)
Now
+ +,dv = # , ( u r d x + u y d Y ) + ~ , , ( v z d x + v v d ~ ) + vr) d x + (+,, + +, d~ = 0 + = 0 (2) + = 0
d+ = +udU
and
+,
vy)
+uUr
+"U2
+OVY
+U%
cannot be identically zero since if they were, there would be no functional relation,
- - - 0 identically.
contrary to hypothesis. Hence it follows from ( I ) and (2)that
Sufficiency. We have to prove t h a t if the Jacobian tional relation between U and U , i.e. +(u,v)= 0. U
$4)
= 0 identically, then there exists
a func-
Let us first suppose that both U+ = 0 and uu= 0. In this case the Jacobian is identically zero and is a constant c1, so that the trivial functional relation U = C I is obtained.
Let us now assume that we do not have both uz = 0 and uy= 0; for definiteness, assume uz# 0. We may then, according to Theorem 1, Page 108, solve for x in the equation U = f(z,y) to obtain 2 = F ( u , y ) , from which it follows that
('1
= f{F(u,y)J y>
From (.!?), u z F r = 1 and urFy
But by hypothesis
A
Y)
+ uu = 0
or
= g { F ( u , y ) , Y>
(2)
(5) F ,
=
-U&.
uZvy- u,v,
=
Using this, (4)becomes
= 0 identically,
so t h a t ( 6 ) becomes
d v = v Z F udu. This means essentially that referring to (2), av/dg = 0 which means that v is not dependent on y but depends only on U, i.e. v is a function of U , which is the same as saying that the functional relation +(U, U ) = 0 exists.
'+'
36. (a) If U = - and v = t a n - ' x + 1-xw (b) Are U and v functionally related? If so, find the relationship.
( b ) By Problem 36, since the Jacobian is identically zero in a region, there must be a functional relationship between U and U . This is seen to be tan v = U , i.e. +(u,v) = U - tan v = 0. We can show this directly by solving for x (say) in one of the equations and then substituting in the t a n - l y we find t a n - l x = v - tan-'y and so other. Thus, for example, from v = tan-'$
+
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 <1
CHAP. 61
125
PARTIAL DERIVATIVES
+ ht, y = yo + kt, then F ( t ) = f ( z ,y), so that by Problem 17, F’(t) = f=(dx/dt)+ f i ( d y / d t ) = hfi + k f v and F’(e) = h f , ( x o + eh, yo+ ek) + k f , ( z o + eh, yo + ek) If x = xo
where 0
< e < 1. f(x0
where 0
<e <1
Thus (2) becomes
= h f , (xo+ eh, go+ ek)
+ h, YO+ k) - f(xo, y0)
+k
fy
+
(xo eh, yo
+ ek)
(9)
as required.
Note that (2), which is analogous to (2) of Problem 14 where h = A x , has the advantage of being more symmetric (and also more useful), since only a single number e is involved.
41. Prove Taylor’s theorem of the mean for functions of two variables. Let F ( t ) = of one variable,
+ ht, + kt) as in Problem 40. ‘By Taylor’s theorem of the mean for functions 0 F(0) + F’(0)t + 2 ! F”(0) t2 ... + Fn ( )t” + O<e
~ ( X O
and if t = l ,
F(n+1)
(n)
=
F(t)
I/O
+
o
t
,
,
l
_-
From Problem 40, where we have used the symbolic operator notation. Similarly,
=
h2fn
from which
F”(0) =
+
+
Zhkf,
h*ficic(xo,YO)
+
k2fyr
2hk f U ( x 0 ,
1/01
+
a
ke fyY(xo,YO) = (
h
+ ~k aay ) ’
f(z0,
YO)
since the second partial derivatives of f are supposed continuous. In like manner we can verify (by mathematical induction) that f o r all positive integers n, F(n)(0)
where 0 < e
< 1.
+ k$)nf(zo,yo),
= (h&
F(n+l)(e)
Substituting these in (2), the required result follows.
42. Expand x 2 y + 3 y - 2
in powers of x - 1
and y+2.
Use Taylor’s theorem of the mean with h =
f ( ~y), =
=
+ 31/- 2,
fic
= ~ Z Yf ,r =
5’
+ 3,
ftt
x--0,
= 2y,
fty
k =
= 2x9
y - y ~ , where x 0 = 1 , y0=-2. fyr
= 0,
= 0,
fr-
frty
= 2,
ffyy
Then
= 0 , fyyy = O
All higher derivatives are zero. Thus f ( 1 , -2) = -10,
f z ( 1 , -2) = -4, fy(1, -2) = 4, fZr(1, -2) = -4, f r u ( 1 , -2) = 2, fyy(1, -2) = 0 ficicr(l,-2) = 0, ficIy(1,-2) = 2, fqy(19-2) = 0, f y y y ( l p - 2 ) = 0
By Taylor’s theorem, f ( x , Y)
= f(1, -2)
+
1
+ h f 4 , -2) + k f d , -2) + g1p ’ f d l ,-2) + 2hk fty(1, -2) + k’ fvy(1, -2)) {h3 ficicr(l,-2)
+ 3h9k fuel (1, -2) + 3hk2fty, (1, -2) +
k3 fry#
(1,-2))
+
R~
where R3 is the remainder and is zero in this case. Substituting the values of the derivatives obtained above, we find
x*I/
+ 3y - 2
=
-10
-
4(x- 1)
+ 4(y+2) - 2 ( z - 1)* + 2 ( x -
as can be verified directly in this case by algebraic processes.
l)(y+2)
+ (z- 1)2(y+2)
126
PARTIAL DERIVATIVES
[CHAP. 6
MISCELLANEOUS PROBLEMS 43. Let f ( 2 , Y ) =
1
)-(yx
X2
+ y2
(x,y) + (0,O)
Then
Note that
-
fzY
#
fyz
at (0,O). See Problem 13.
-44. Show that under the transformation a2v iav 1 !f!iIg2 = 0 becomes ap2 p2
x = p cos +,
+ - -a p
+
a2v
g = p sin
+
the equation
anv
+
= 0.
We have
DifFerentiate x = p cos c, y = p sin tions of x and g 1
=
Solving simultaneously,
+
- p s i n + - $# ax
ax
cos 9,
az
x,
o =
c o s + -aP,
-a P _ -
-
with respect to
remembering that p
as
=
Solving simultaneously, -~
a4 +
- p s i n + - 841
c o s + -aP,
ay
and
+
are func-
aP sin 9 032
3! = --sin 9 ax
P
Similarly, differentiate with respect to y. Then 0
+
as cos 9 I
p
1
=
p
a+
cos 9 al/
+ sin 9 9 aY !4!
Then from (1) and
c o s + -aV aP
sin+ aV --P
a+
PARTIAL DERIVATIVES
CHAP. 61
127
Hence
=
aV - sin#aV dp
aP
dP
a v
a+ - ass
aLV cos%#-
+
cos+-
a+
P*
aPs
$kos+-aP
-P a+
a1 a+
Adding (7) and (8) we find, as required,
a v - COS+ aV -P a+ a#
8%
aPa+
P
+
2 sin + COS + aLV
spa+
P
a'V + a'V a ~ 9
si;+ azy)( -a@
aP
2 sin + cos + a V
2 sin @ cos # aV
2 sin + COS+
agv a" - - sin*+ -
+
8%
sin # aV sin + aZV +-- -)(cos+) P I a+ P spa+
P9
Similarly,
a+
(-sin+dp+
aP9
ay'
P
av
+ which simplifies to
--)
d(eos+-
+
P
aP'
P aP
PL
aP
cosL+ aV --
= a v + 1av
ay9
sinL+ a V
+ -sine - +@-aV P
+
aP
+
si;+)
COS* @
a+L aLV
PL
p)v = PL a+%
(7)
(8)
0.
-a(U9)' = I provided do # 0, and interpret geometrically. a@, v) Y) a(% v ) xr
w,
2.
Yr Y.
I=l
+ xwvr yuur + yavr XUUI
XUU.
+ xwv.
yuu.+ ywv.
using a theorem on multiplication of determinants (see Problem 116). We have assumed here, of course, the existence of the partial derivatives involved.
The equations x = f(u,v), y = g(u,v) defines a transformation between points (x,y) in the xy plane and points (u,v) in the uv plane. The inverse transformation is given by U = +(z,y), v = $(x,y). The result obtained states that the Jacobians of these transformations are reciprocals of each other.
46. Show that F ( q ,z- 2x) = 0 satisfies under suitable conditions the equation z(dz/dz) z/(dx/@)
= 2x. What are these conditions?
Let u = x y , v = (11
dF
2-
2x. Then F ( u , v ) = 0 and
= Fudu
+ Fadv
+
= Fu(xdy+yd~) Fw(dz-2d~) = 0
Taking x as dependent variable and x and y as independent variables, we have dz = x+ dx Then substituting in (1),we find (yFu
+ Fox, - 2) ds +
(XF,+ F w z , ) d y =
Hence we have, since x and y are independent,
(a) yF,
+ Fuzz - 2
= 0
(3)
XF,
0
+ Fwz, = 0
+ zydg.
PARTIAL DERIVATIVES
128
[CHAP. 6
Solve for F , in (3) and substitute in (2). Then we obtain the required result xzt-yzr = 22 upon dividing by F,, (supposed not equal to zero). The result will certainly be valid if we assume that F(u,v) is continuously differentiable and that F,,# 0.
Supplementary Problems FUNCTIONS and GRAPHS 41.
If f ( % I I )
=
An8. (a) -4,
22 + y find G,
(a)f(1,-3),
11 (b) 6(3h 6) ' (')
+
+
( b ) f(2 + h, 3,h - f ( 2 9 3, , (c) f ( x y, xy).
+ +
2 s 2y xy 1 - x'y - xy'
48.
49.
Give the domain of definition for which each of the following functions are defined and real, and indicate this domain graphically.
50. (a)What is the domain of definition for which
x+y+2-1 x2 y* 2' 1
+ + -
f(x,y,x) =
(b) Indicate this domain graphically. A m . (a) x + y + x S 1, s s + y * + z p < 1 and x + y + z Z 1, x x + y a + z a > 1.
is defined and real?
51. Sketch and name the surface in 3 dimensional space represented by each of the following.
(a)3 2 + 22 = 12, ( b ) 42 = x*+y', (c)
2
= z'- 4y',
(d) x'+ 2' = y', ( e ) x p + y p +xp = 16, ( f ) ~ * - 4 y ' - 4 ~ * = 36,
= 2Y,
( 8 ) %'+Y'
( h ) 2 = z+y,
(i)
y'
= 42,
(i) X* + y'
+
+
2%- 4% 4- 6y 22 - 2 = 0. A m . ( a ) plane, ( b ) paraboloid of revolution, ( c ) hyperbolic paraboloid, ( d ) right circular cone, ( e ) sphere, ( f ) hyperboloid of two sheets, ( g ) right circular cylinder, (h) plane, (9 parabolic cylinder, (i) sphere, center at (2,-3,-1) and radius 4.
52.
Construct a graph of the region bounded by x'+ya = a' and x'++'
= a', where a is a constant.
53. Describe graphically the set of points (z,y,x) such t h a t (a)s * + y ' + z ' = 1, x'+y* = 22; ( b ) 2 * + y ' < 2 < x+y.
54.
The level curve8 for a function z = f(x,y) are curves in the zy plane defined by f ( z , y ) = c, where c is any constant. They provide a way of representing the function graphically. Similarly, the level surfaces of w = f ( z ,y,x) a r e the surfaces in a rectangular (xyz) coordinate system defined by f ( x , y , z ) = c, where c is any constant. Describe and graph the level curves and surfaces for each of the following functions: (a) f ( x ,y) = In (xp+ys - l), ( b ) f(z,y) = 4xy, ( c ) f ( z ,y) = tan-'y/(x l), ( d ) f ( z ,y) = x''' y"', ( e ) f(x,y, x ) = x ' 4- 4y' 4-16x*, ( f ) sin (z x)/(l - y).
+
+
+
CHAP. 61
129
PARTIAL DERIVATIVES
LIMITS and CONTINUITY 55.
Prove that (a)
ip4(3%-2y)
= 14 and (b)
lirn
(zy-3x+4)
(r,v ) - + ( 2 , 1 )
= 0 by using the definition.
U-+ - 1
56. If lirn f ( x , y) = A and lirn g(z,y) = B , where lim denotes limit as (z,y)+ (zo,yo),
(a)lim { f @ , U) + &, Y)} = A
57.
+B,
d g(z, Y)> = AB.
( b ) lim
prove t h a t
Under what conditions is the limit of the quotient of two functions equal to the quotient of their limits? Prove your answer.
58. Evaluate each of the following limits where they exist.
(b) lirn
2 4 0
32 - 2y
(4 !$
22 - 3y
~
+
z sin (z' y2) xs+y'
Ans. (a) 4, ( b ) does not exist, ( c ) 8fi,
2x-g (f) lirn r-0
2 '
+ y'
sin-l (zy- 2) (h) !,'"c tan - (3xy - 6)
(d) 0, (e) 0, (f) does not exist, ( g ) 0, ( h ) 113
59. Formulate a definition of limit for functions of (a)3, (b) n variables. 60. Does
+
42 y - 32 22 - 61/ 22 a s (z,y, z ) + (0, 0,O)
lirn
+
exist?
Justify your answer.
61. Investigate the continuity of each of the following functions a t the indicated points:
(4z' + 8'; ( 2 0 , YO).
(b)
m; X
+
(0,O). (c) (z* y*) sin
1
if
(2,
Y) f
0 if (z,21)
= (0,O); (0,O).
Ans. (a)continuous, ( b ) discontinuous, (c) continuous 62.
Using the definition, prove that f(2,y) = z y + 6 x is continuous at (a)(1,2), (b) ( s o , ~ o ) .
63. Prove that the function of Problem 62 is uniformly continuous in the square region defined by
oszdl,o~yYll.
PARTIAL DERIVATIVES 64. If f(z,y)
=
z,
find (a)df/dz and (b) dfldy a t (2,-1) from the definition and verify your answer
by differentiation rules.
Ans. (a) -2,
( b ) -4
65.
66. Investigate
lirn
(r,y)-+(O,O)
fi(z,y)
for the function in the preceding problem and explain why this limit
(if it exists) is or is not equal to f 2 ( 0 , 0 ) . 67.
68.
+
(a)Prove by direct differentiation t h a t z = zy t a n (ylz) satisfies the equation z(az/dz) y(az/dy) = 22 if (%,U) # (0,O). (b) Discuss part (a)for all other points (z,y)assuming z = 0 a t (0,O).
69. Verify that f a = fur for the functions (a) (2s - y)/(z
+
y), (b) z t a n s y and (c) cosh (y indicating possible exceptional points and investigate these points.
70.
Show that z = In {(z- a)'
+ (y - b)'}
+
satisfies d s z / d ~ * agz/i3yL = 0 except at (a,b).
+ cos a!),
PARTIAL DERIVATIVES
130 71.
Show t h a t z = x cos ( y l x ) x=o.
+ tan (ylx)
satisfies
Z'Z+~
[CHAP. 6
+ 2syz,, + y'z,,,
= 0 except at points for which
72. Show th at if w =
aw
aw aw a*w ax2 9-ay z- a2 = 0, (b) XfIndicate possible exceptional points.
+
(a)s -a+s
DIFFERENTIALS
a*w a2w + y2 + z2 allZ azz
2 x a2w ~ a +q 2xz-+ax az a2w
+
2yz
- =
ay az
0.
+
73. If z = x3 - xy 3y2, compute (a)Az and (b) dz where s = 5 , y = 4, A s = -0.2, A y = 0.1. Explain why Az and dz a r e approximately equal. ( c ) Find Az and dz if x = 5, y = 4, Ax = -2, A y = 1. Ans. (a)-11.658, ( b ) -12.3, ( c ) Az=-66, d z = -123. 74.
+ 2(2.1)'
5
Compute d(3.8)' Ans. 2.01
approximately, using differentials.
+
(a) F ( x , y) = x3y - 4xy2 8y3, (b) G(z, y, z ) = 8xy2z3- 3x2yz, ( c ) F ( z , y) = xy' In ( y l x ) . Ans. (a) (3s*y - 4y2)dx (x' - 8xy 24y') dy (b) (8y2z3- Gxyz) dx (16xyz3 - 32'2) dy (24xy*z*- 32'16) dz ( c ) {y' In ( y l s ) - y'} dx {2xy In (y/x) xy} dy
75. Find d F and dG if
+
+
+
+
+
+
+
V d U , (b) d(U/V) = ( V d U - UdV)/V', (c) d(ln U) = ( d U ) / U , 76. Prove t h a t (a) d(UV) = U d V (d) &(tan-' V) = (dV)/(l+ VZ) where U and V are differentiable functions of two or more variables. 77.
Determine whether each of the following a r e exact differentials of a function and if so, find the function. (a) (2x3' 3y cos 32) dx ( 2 2 ' ~ sin 3%)d y (b) (6xy - yz) dz (2xeY- 2 ' ) dy (c) (2' - 3y)dx (12yz - 32)dy 3xz2dz Ans. (a) x 2 v 2 y sin32 c, (b) not exact, ( c ) 2z3 4ys - 3sy c
+
+ +
+
+ + +
+
+
+
DIFFERENTIATION of COMPOSITE FUNCTIONS 78. (a) If U(x,y,z) = 2 x L - y z + x z 2 , x = 2 s i n t , y = t * - t + l , z = 3e-*, find dUldt at t = 0 . (b) If H(x, y) = sin (32 - y), z3 2y = 2ts, s - y* = t Z + 3t, find d H l d t .
+
Ans. (a) 24, ( 6 )
(36t21/+ 12t +6x54-+ 2 92'
+
79. If F ( x , ~ )= (2% ~ ) / ( y- 2 4 , 2 = 2 ~ - 3 ~ , ( d ) d2F/av2, (e) aLFIduav, where U = 2, v = 1. 80. If 81.
(U) aF/au, (b) aFlav, (c) a2Flauz, An8. (a)7 , ( b ) -14, ( c ) 21, (d) 112, (e) -49
+
U = x 2 F ( y l s ) , show t h a t under suitable restrictions on F , z(dU/az) y(dU/dy) = 2U.
If x =
U COS& -
v sin (Y and y =
U
(avlax)' 82.
= u + ~ v ,find
sin + v cosa, + (aV/aY)2 = (Y
where a is a constant, show t h a t (av/au)'
+ (avlav)'
Show th a t if x = p cos 9, y = p sin 9, the equations
83. Use Problem 82 to show th at under the transformation x = p cos +, y = p sin +, the equation a2u 1 au 1 azu = 0 = 0 becomes
-+a22
ag
+P aP- +>c aP' P 9
CHAP. 61
131
PARTIAL DERIVATIVES
IMPLICIT FUNCTIONS and JACOBIANS 84. If F ( x , y ) = 0, prove that d y / d x = -F,/F,. 85. Find (a)dyldx and ( b ) #y/dxa if x 3 + y 8 - 3 x y = 0. A m . (U) (2/ - z2)/(y2- x), ( b ) - ~ x Y / ( Y ' - x)' 86. If x u a + v = y3, 2 y u - x v 3
87. If U
au
= 42, find (a)(b) 8X
+
vs - 3XULVZ 4
av
= f ( x , U), v = g(x,y) are differentiable, prove that
A m . (a) 6x*uve4- 2 y
au
a% du +
' (b)
2xuz
+ 3ya
3 2 z ~ v4-a y
= 1. Explain clearly which
variables are considered independent in each partial derivative. ay ar
88. If f ( x , y , r , s ) = 0, g(x,y,r,s) = 0, prove that
ay as +a8 a X
= 0, explaining which variables are
independent. What notation could you use to indicate the independent variables considered? d% -
89. If F(x,y) = 0, show that 90.
dxa
-
-
F,,Fi - 2F,F+Fr Fy"
+ F,FZ
Evaluate -if F ( u , v ) = 3 u 2 - u v , G(u,v) = 2uvZ+v8.
91. If F = x + 3 y e - x 8 ,
G = 2x2yx, and H = 2 z z - x y ,
+
+
Ans. 2 4 u b - t 16uv'-3v3 A m . 10
evaluate a ( F 9 G J H )at (1,-1,O). a(x, YJ
U = sin-lx sin-ly and v = x i y d m , determine whether there is a functional relationship between U and U , and if so find it.
92. If 93.
If F = x y + y z + z x , G = x Z + y * + z a , and H = x + y + z , determine whether there is a functional Am. H Z - G - 2 F = 0 relationship connecting F , G , and H , and if so find it.
94.
(a) If x = f ( u , v , w ) , y = g ( u , v , w ) , and z = h(u,v,w), prove that
w) v , w ) a(%, Y,
= 1 provided
a(x'yJz) # 0. (b) Give an interpretation of the result of (a) in terms of transformations.
a(% v , w ) 95.
If f ( x , y , z ) = 0 and g(x,y,z) = 0, show that
- d-x - -- '(fJ
g)
dy
a(f,g )
-
dx
w, g )
a(%, U)
a(z, $1 d(y, 2) giving conditions under which the result is valid. 96. If x + y Z = U, y + z L =
U,
s, (4
z + x L = w , find (a) G a% , ( b ) asx
assuming that the equations
define x, y and x as twice differentiable functions of U, v and w . 16yaz - 8x2 - 32xzys 16x'y - 8yz - 32xzza 1 Ans. (a)(1 8 x y @ ' (1 8 2 ~ 2 ) ~ 1 8xyx ' ( b )
+
97.
+
+
State and prove a theorem similar to that in Problem 36, for the case where U = f ( x , y , z ) , v = g ( x , y , z ) , = h(x,y,z).
w
TRANSFORMATIONS, CURVILINEAR COORDINATES 98.
Given the transformation x = 2 u + v , y = U - 3 v . (a)Sketch the region !I(' of the uv plane into which the region of the xy plane bounded by z = 0, x = 1,y = 0, y = 1 is mapped under the transformation. (b) Compute
A m . ( b ) -7 99.
(c) Compare the result of (b) with the ratios of the areas of 9( and
+
+
l('.
(a)Prove that under a linear transformution x = a ~ u a i v , y = b l u bzv (a1bs - ai bl # 0) lines and circles in the xy plane are mapped respectively into lines and circles in the uv plane. (b) Compute the Jacobian J of the transformation and discuss the significance of J=O.
132
[CHAP. 6
PARTIAL DERIVATIVES
100. Given x = cos U cosh U , y = sin U sinh U. (a) Show that in general the coordinate curves U = a and U = b in
t h e u u plane a r e mapped into hyperbolas a n d ellipses, respectively, in t h e xy plane.
+ cos2U sinh2v, ( c ) (sin2 101. Given the transformation z = 21c + 3v - w , y = A m . (6) sin2j c cosh2v
T ' of the z
NVW
zi
+ cos2
cosh2v
U
( 6 ) Com-
sinh' v)-l
+
+
ZL - 2v w , z = 2u - 2v w. (a) Sketch the region space into which the region 5J( of the zyz space bounded by z = 0,z = 8, y = 0, y = 4, z = 0,
= 6 is mapped. (6) Compute
of 9( and 9('.
Ans. (6) 1
ao. ( c ) Compare the result of (6) with the ratios of the volumes a(% v,w )
102. Given the spherical coordinate transformation x = r sin e cos $, y = T sin 8 sin 9, z = T cos 8, where T 2 0, 0 5 e Z 7;, 0 S $ < 27. Describe the coordinate surfaces ( a ) r = a, ( 6 ) 8 = 6 , and ( c ) $ = c , where a , 6 , c a r e any constants. Ans. (a)spheres, (6) cones, ( c ) planes 103. ( a ) Verify t h a t for the spherical coordinate transformation of Problem 102,
(6) Discuss the case where J = 0.
J =A a(z ?I z, - r* sin 8, a(r, 8 , $)
MEAN VALUE THEOREMS %+U x+y-2 0 < e < 1, where x > 0 , y > 0. 104. Prove t h a t In - 2 + B ( X + y - 2) ' 2
f(x,y) = sin XI/ in powers of x - 1 and y - SR, up to and including second degree terms. Ans. 1 - &r2(z - 1)' - +(z - 1)(y - &(y -
105. Expand
AT)
106. Expand f ( x , y) = y2/z3 in powers of
write the remainder. Ans. 1 - 3 ( x - 1) - 2(y - i w x - 1)3[1 - 8(y
4,)'
x - 1 and y
+ 1,
up to and including second degree terms and
+ 1) + 6 ( 2 - 1)' + 6 ( x - l ) ( y + 1) + (Y + 1)' + 1)y + 12(x - i ) y y + i ) [ i + e(x
-
[ I + e(x -
where 0 < e < 1.
1)][1 - e(y
+ 111 + 3(x -
l)(y
+
U2[1 + e b
-
1)12
i)16
107. Prove the first mean value theorem f o r functions of 3 variables. 108. Generalize and prove Taylor's theorem of the mean for functions of 3 variables.
These results a r e useful in thermodynamics, where P , V , T correspond to pressure, volume and temperature of a physical system.
+
110. Show t h a t F(x/y,z/y) = 0 satisfies %(Wax) y(az/dy) 111. Show t h a t F ( x
+ y - z, x 2 + y')
= z.
= 0 satisfies x(az/ay) - y(az/az) = x av
112. If x = f(tc,v) and 1~ = g(z(,v), prove t h a t - = - - - where J
ax
113. If
J au
x = f(zc,v),y = g ( u , v ) , z = h(u,v) and F ( x , y , z ) = 0, prove t h a t
-
U.
a(x, Y) = a(?(, U)* ~
CHAP. 61
133
PARTIAL DERIVATIVES
I 1' I I 1
I
a b bg a f + b h , thus establishing the rule for the = ce+dg c f + d h c d product of two second order determinants referred to in Problem 45. (b) Generalize the result of (a) to determinants of order 3,4, . . ..
115. (a) Prove that
116. If x, y and z are functions of U, v and w , while U, v and w are functions of
T,
s and t, prove that
117. If D, and D , denote the operators d/ax and slay respectively, show that if the Taylor series for f ( x h, y k) exists it can be written in the form
+
+
f(x
+ h, y +k ) .
118. Given the equations F ~ ( Z.I.,znr,y1,. ,. .,yn)
conditions on
= ebD=+kDv f ( z , y)
= 0 where
i = 1,2,.. .,n.
Prove that under suitable
Fj,
119. (a)If F(x,y) is homogeneous of degree 2, prove that
+
aaF aSF + x' 2xy a ~ 9 ax ay
yad'Ii.
ay'
= 2F.
( b ) Illustrate by using the special case F(z, y) = xa In (ylz).
Note that the result can be written in operator form, using D, d/dx and D, = d/dy, as ( x D z + y D s ) ' F = 2F. [Hint Differentiate both sides of equation ( I ) , Problem 25, twice with respect to A.]
.
120. Generalize the result of Problem 119 as follows. If F(ZI,XL, . .,2,) is homogeneous of degree p, then
for any positive integer r, if D,
= msj,
( E I D , ~ + ~ S D , + +xnDq,)'F 121. (a) Let x and y be determined from
transformation the equation
$+ 3 = o
U
= P(P-1).**(p-r++)F
and v according to x + i y = (u+iv)*. Prove that under this is transformed into
az$ -
as+ = +-
aul av9 (b) Is the result in (a) true if s + iy = F ( u + i v ) ? Prove your statements.
o
Chapter
7
Vectors
/ /
VECTORS and SCALARS There are quantities in physics characterized by both magnitude and direction, such as displacement, velocity, force and acceleration. To describe such quantities, we introduce the concept of a vector as a directed line segment 8 from one point P called the initial point to another point Q called the terminal point. We denote vectors by bold - P faced letters or letters with an arrow over them. Thus PQ is denoted by A or A as in Fig. 7-1, The magnitude or + Fig. 7-1 length of the vector is then denoted by IPQI, PQ, IAI or Other quantities in physics are characterized by magnitude only, such as mass, length and temperature. Such quantities are often called scalars to distinguish them from vectors, but it must be emphasized that apart from units such as feet, degrees, etc., they are nothing more than real numbers. We can thus denote them by ordinary letters as usual.
1x1.
VECTOR ALGEBRA The operations of addition, subtraction and multiplication familiar in the algebra of numbers are, with suitable definition, capable of extension to an algebra of vectors. The following definitions are fundamental. 1. Two vectors A and B are equal if they have the same magnitude and direction regardless of their initial points. Thus A = B in Fig. 7-1 above.
/ /
2. A vector having direction opposite to that of vector
A but with the same magnitude is denoted by -A
Fig. 7-2
[see Fig. 7-21.
3. The sum or resultant of vectors A and B of Fig. 7-3(a) below is a vector C formed by placing the initial point of B on the terminal point of A and joining the initial point of A to the terminal point of B [see Fig. 7-3(b) below]. The sum C is written C = A + B. The definition here is equivalent to the parallelogram law for vector addition as indicated in Fig. 7-3(c) below.
Fig. 7-3
134
CHAP. 71
135
VECTORS
Extensions to sums of more than two vectors are immediate. For example, Fig. 7-4 below shows how to obtain the sum or resultant E of the vectors A, B, C and D.
Fig. 7-4
4. The diference of vectors A and B, represented by A - B , is that vector C which added to B gives A. Equivalently, A - B may be defined as A+(-B). If A = B , then A - B is defined as the nuZE or zero vector and is represented by the symbol 0. This has a magnitude of zero but its direction is not defined. 5. Multiplication of a vector A by a scalar m produces a vector mA with magnitude Iml times the magnitude of A and direction the same as or opposite to that of A according as m is positive o r negative. If m = 0 , mA=0, the null vector.
LAWS of VECTOR ALGEBRA If A, B and C are vectors, and m and n are scalars, then 1. A + B = B + A 2. A + ( B + C ) = ( A + B ) + C 3. m(nA) = (mn)A = n(mA) 4. (m+n)A = m A + n A 5. m(A+B) = m A + m B
Commutative Law for Addition Associative Law for Addition Associative Law for Multiplication Distributive Law Distributive Law
Note that in these laws only multiplication of a vector by one or more scalars is defined. On Pages 136 and 137 we define products of vectors.
UNIT VECTORS Unit vectors are vectors having unit length. If A is any vector with length A>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 O<)E~ whenever Iu-uol < 6. This is equivalent to the statement lim A(u)= A(%). U-%
2. The derivative of A(u) is defined as
dA du -
provided this limit exists.
lim
+
A(u Au) - A(u) AU
Au*O
+
+
In case A(u) = Al(u)i Az(u)j As(u)k; then
dA dA2 dAsk - -d Ai i+ - j + du du du du Higher derivatives such as d2A/du2,etc., can be similarly defined. 3. If
A ( x ,y , x ) = A l ( x , y , x)i
+ A ~ ( xy ,,x)j + As($,y , x)k,
then
is the differential of A. 4. Derivatives of products obey rules similar to those for scalar functions.
when cross products are involved the order may be important. are:
a
(b) 5 ( A * B ) =
(c) za( A X B )
=
A *d-B + dA * B , ay aB A X - +
ax
ay aA
-a2X B
GEOMETRIC INTERPRETATION of a VECTOR DERIVATIVE If r is the vector joining the origin 0 of a coordinate system and the point (x,y, x), then specification of the vector function r(u) defines x, y and x as functions of U. As U changes, the terminal point of r describes a space curve (see Fig. 7-8) having parametric equations x = $(U), 9 = y(u), x = x(u). If the parameter U is the arc length s measured from some fixed point on the curve, then dr
- = T
ds
(9)
is a unit vector in the direction of the tangent to the curve and is called the unit tangent vector. If '1c is the time t, then Fig. 7-8
is the velocity with which the terminal point
However, Some examples
140
VECTORS
[CHAP. 7
of r describes the curve. We have from which we see that the magnitude of v is v = dsldt. Similarly, d2r dt2 - a is the acceleration with which the terminal point of r describes the curve. cepts have important applications in mechanics and differentid gemetW.
These con-
GRADIENT, DIVERGENCE and CURL (del) defined by Consider the vector operator
v
a i -a + j -a ax ay Then if +(x,y,z) and A(x,y,z) have continuous first partial derivatives in a region (a condition which is in many cases stronger than necessary), we can define the following.
V
E
+
1. Gradient. The gradient of
kz
+ is defined by
An interesting interpretation is that if +(x,y,z) = c is the equation of a surface, then V+ is a normal to this surface (see Problem 36).
2. Divergence. The divergence of A is defined by divA = V - A = ( i , a+ j ~ + ka - & ) * ( A I i + A 2 j + A 3 k ) aAl - -
ax
+-aA2 +-aA3 ay az
3. Curl. The curl of A is defined by
curlA
=
v xA
=
=
(a
i -ax
+ j -aya + k Ga ) x
(Ali
+ A2j + Aak)
ilG a A2
=
("
3
-
Note that in the expansion of the determinant, the operators alax, a&, precede AI, A2, As.
alaz must
CHAP. '71
141
VECTORS
FORMULAS INVOLVING v If the partial derivatives of A, B, U and V are assumed to exist, then
+ +
+
V(U+V) = VU VV or grad(U+V) = gradu gradV *(A+B) = v * A + v - B or div(A+B) = divA divB or c u r l ( A + B ) = curlA curlB x(A+B) = v x A + v x B *(UA) = ( V U ) * A U ( V * A ) X (UA) = (VU) X A U(V XA) * ( A X B ) = B * ( V X A )- A * ( V X B ) X ( A X B ) = ( B * V ) A - B ( V * A ) - ( A * V ) B -I- A ( V * B ) 8. V ( A * B ) = ( B * V ) A ( A * V ) B B X ( V X A ) AX(vxB) a2U a2U a2U 9. v ( v U ) = v 2 U = - + - + 7 is called the Laplacian of U ax2 ay2 dz 1.
+
+
+
and
v x (vU)
v2
=
+
+
a2
+7 + ay
a
+
~
a2
2
a2
a22
is called the Laplacian operator.
The curl of the gradient of U is zero. 11. (V x A) = 0. The divergence of the curl of A is zero. 12. V X ( V X A ) = V ( V * A ) - V2A 10.
= 0.
v
VECTOR INTERPRETATION of JACOBIANS. ORTHOGONAL CURVILINEAR COORDINATES. The transformation equations x = f(ui,u2,~3), y = g(ui,U2,u3), z = h(ui,U2,u3) (17) [where we assume that f , g , h are continuous, have continuous partial derivatives and have a single-valued inverse] establish a one to one correspondence between points in an xyz and ulu2u3 rectangular coordinate system. In vector notation the transformation ( I 7) can be written
+
+
r = xi yj xk = f(ul,u~,u3)i + g(ul,uz,u3)j A point P in Fig. 7-9 can then be defined not only by rectangular coordinates ( x , y,z) but by coordi) well. We call ( U I , U Z , U ~the ) nates ( U I , U Z , U ~as curvilinear coordinates of the point. If US and 243 are constant, then as u1 varies, r describes a curve which we call the u1 coordinate curve. Similarly we define the u2 and u3 coordinate curves through P . From (129,we have
+ h(ul,US,u3)k
(18)
The vector drldul is tangent to the u1 coordinate curve at P . If el is a unit vector at P in this direction, we can write drldul = hle1 where hl = I drldul I. Similarly we can write drldu2 = h2e2 and drldu3 = h3e3, where h2 = 1 arldu2 I and h3 = I drlau3 I respectively. Then (19) can be written
+
+
dr = hldulel h2du2e2 h3du3e3 The quantities hl, h2, h3 are sometimes called scale factors.
(20)
142
[CHAP. 7
VECTORS
If el, e2, e3 are mutually perpendicular a t any point P, the curvilinear coordinates are called orthogonal. In such case the element of arc length ds is given by
+
+
ds2 = dr-dr = h; du: hg dui hi dui (21) and corresponds to the square of the length of the diagonal in the above parallelepiped. Also, in the case of orthogonal coordinates the volume of the parallelepiped is given by d V = I (hidui el) (h2du2 e2) X (h3du3 e3) 1 which can be written as
=
hi h2 h3 dul du2 du3
(22)
, us) is the Jacobian of the transformation. where d(x, y, x ) l d ( u ~u2, It is clear that when the Jacobian vanishes there is no parallelepiped and explains geometrically the significance of the vanishing of a Jacobian as treated in Chapter 6.
GRADIENT, DIVERGENCE, CURL and LAPLACIAN in ORTHOGONAL CURVILINEAR COORDINATES If cp is a scalar function and A = Ale1 A2e2 A3e3 a vector function of orthogonal curvilinear coordinates U I ,U Z ,u3, we have the following results. 1 dcp 1 a+ 1 dcp 1. vcp = gradcp = --er +--e2 +--e3 hi dui h2 du2 h3 au3
+
+
-
3.
v xA
4.
v2cp
=
=
curlA
1
d a 1 d = - - - hih2h3 due dua dui hiAi h2A2 h3A3
Laplacian of
=
These reduce to the usual expressions in rectangular coordinates if we replace (u1,u2, 243) by (x,y, x ) , in which case er, e2 and e3 are replaced by i, j and k and hl = h2 = h3 = 1.
SPECIAL CURVILINEAR COORDINATES 1. Cylindrical Coordinates ( p , +, 2). See Fig. 7-10. Transforrnation equations: x = pcos+, y = psin+, x = x where p 2 0, 0 2 + < Z X , - 00 < x < W . Scale factors: hl = 1, h:!= p , h3 = 1 Element of arc length: ds2 = dp2 4- ~ ~ d +dz2~ + Element of volume: dV = p d p d + d x L uplacian:
a2U + -I -d + U - - 1 d2U a p ~ ap p 2 a42
a2U +
d ~ 2
CHAP. 71
143
VECTORS
Note that corresponding results can be obtained for polar coordinates in the plane by omitting x dependence. In such case f o r example, ds2 = dp2 + p2d+2, while the element of volume is replaced by the element of area, d A = pdpd+. 2. Spherical Coordinates ( r , 0, +). See Fig. 7-11.
Transformation equations: x = r sin 8 cos +, y = r sin 8 sin +, x = r cos 8 where r 2 0, 0 5 8 5 7T, 0 5 + < 2 r . Scale factors: h~= 1, h2 = r , h3 = r sin 8 Element of arc length: ds2 = dr2
+ r2do2 + r2sin28 d+2
Jaco bian: Fig. 7-11
Element of volume: d V = r2 sin 8 drdOd+ Laplacian: Other types of coordinate systems are possible.
Solved Problems VECTOR ALGEBRA 1. Show that addition of vectors is commutative, i.e. A O P + P Q = OQ O R + R Q = OQ
and Then A + B = B + A.
or or
+B
= B
+ A.
See Fig. 7-12 below.
A + B = C, B + A = C.
Q
0
0 U
Fig. 7-12
Fig. 7-13
2. Show that the addition of vectors is associative, i.e. A Fig. 7-13 above. O P + P Q = OQ = ( A + B )
+
Since we have
A
+ ( B + C)
and
OP P R = OR = D, O Q + QR = OR = D,
= (A+B)
+ C.
+ (B + C) = (A+ B) + C.
See
P Q + Q R = PR = ( B + C )
+ + +
i.e. A (B C) = D i.e. ( A + B ) C = D
Extensions of the results of Problems 1 and 2 show that the order of addition of any number of vectors is immaterial.
VECTORS
144
[CHAP. 7
3. An automobile travels 3 miles due north, then 5 miles northeast as shown in Fig, 7-14. Represent these displacements graphically and determine the resultant displacement (a) graphically, (b) analytically. Vector OP or A represents displacement of 3 mi due north, Vector PQ or B represents displacement of 6 mi northeast. Vector OQ or C represents the resultant displacement or sum of vectors A and B, i.e. C = A + B . This is the triangle law of vector addition. The resultant vector OQ can also be obtained by constructing the diagonal of the parallelogram OPQR having vectors OP = A and OR (equal to vector PQ or B) as sides. This is the parallelogram law of vector addition. Graphical Determination of Resultant. Lay off the 1 mile unit on vector OQ to find the magnitude 7.4 mi (approximately). Angle EOQ = 61.6O, using a protractor. Then vector OQ has magnitude 7.4 mi and direction 61.6O north of east.
w-jk-E Unit = 1 mile
S
Analgtical Determination of Resultant. From triangle OPQ,denoting the magnitudes of A, B, C by A , B, C,we have by the law of cosines
0 = A'
t
B2 - ~ A B C OLOPQ S = 3'
+ 6'
m.7-14
- 2(3)(6)~ 0 ~ 1 3=6 ~34
+ 16fi
= 66.21
and C = 7.43 (approximately). By the law of sines, sin LOQP =
A -C sin LOQP sin LOPQ
Then
*
A sin LOPQ - -3(0.707) - o.2855 C 7.43
Thus vector OQ has magnitude 7.43 mi and direction (46O
4.
Prove that if a and b are non-collinear, then xa
+
LOQP = 16O35'
and
+ 16O36') = 61O36' north of east.
+ vb = 0
implies x = y = 0.
Suppose x # 0. Then xa yb = 0 implies xa = -yb or a = -(y/x)b, i.e. a and b must be parallel to the same line (collinear), contrary to hypothesis. Thus x = 0; then yb = 0, from which y = 0.
5. If x1a + y1b = x2a + yzb, where a and b are non-collinear, then z l a + y t b = xza+ynb can be written xla
+ ylb - (xza + yrb)
Hence by Problem 4,
21
- zz = 0,
= 0
or
yi - yz = 0 or
(21
2 1
- 22)a
+ (g1- y2)b
XI = x2
and y1=
= 0
= XZ, y1= y,.
Extensions are possible (see Problem 49).
6.
Prove that the diagonals of a parallelogram bisect each other. Let ABCD be the given parallelogram with diagonals intersecting a t P as shown in Fig. 7-16. Since B D f a = b, BD = b - a . Since AC = a
+ b,
C
Then BP = x(b -a).
+ b). + + (y - z)b.
AP = y(a
But AB = A P + P B = AP-BP, i.e. a = y(a b) - x(b - a) = (z y)a
+
Since a and b are non-collinear we have by Problem 6, x + y = 1 and y - x = 0, i.e. z = y = & and P is the midpoint of both diagonals.
A Fig. 7-15
y2.
CHAP. 71
VECTORS
145
7. Prove that the line joining the midpoints of two sides of a triangle is parallel to the
third side and has half its length.
From Fig. 7-16, AC+ CB = AB or b + a = c. Let D E = d be the line joining the midpoints of sides AC and CB. Then d = DC+CE = + b + + a = +(b+a) = 4 c
C
A
Thus d is parallel to c and has half its length.
B
Fin. 7-16
8. Prove that the magnitude A of the vector A = Ali+Azj+Ask is A = dA;+A:+Ag. See Fig. 7-17. By the Pythagorean theorem, (@)2
+ (QP)2
= (03)2
where @ denotes the magnitude of vector OP, etc. = (~&)2. Similarly, (03)~ Then (=)O = ( O R ) 2 (RQ)2 (QP)2 or A2 = A : + A : + A ; , i.e. A = d A ; + A : + A ; .
+ +
+
9. Determine the vector having initial point P(x1, y1, XI)
and terminal point Q ( x z , ~ ~ ,and x z ) find its magnitude. See Fig. 7-18. The position vector of P is rl = x l i The position vector of Q is r2 = x2i r l + P Q = rz or PQ = r2 - rl = (x2i y2j = ( X Z - x1)i
+ ylj + xlk.
+ j + x2k. y2
+ + x2k) - (xli + y1 j + xlk)
+
Magnitude of PQ = P &
(y2
- y1)j
+
- x1)k.
(22
+
+
= l/(x2 - X1)2 (y2 - y1)2 ( 2 2 Note that this is the distance between points P and Q .
Fig. 7-18
The DOT or SCALAR PRODUCT 10. Prove that the projection of A on B is equal to A . b, where b is a unit vector in the direction of B.
E
Through the initial and terminal points of A pass planes perpendicular to B at G and H respectively as in the adjacent Fig. 7-19; then Projection of A on B = G H = EF = A cos e = A . b
11. Prove A * ( B + C ) = A - B
I I I
I I I
I
I I
G
H
B
Fig. 7-19
+ A-C.
Let a be a unit vector in the direction of A; then Projection of (B
+ C) on A
= projection of B on A
(B+C)*a =
+ projection of C on A Boa + C - a
Multiplying by A, and
+
(B+C)*Aa = B*Aa C-Aa (B+C)*A = B * A + C * A
Then by the commutative law for dot products,
A*(B+C) = A * B + A * C and the distributive law is valid.
I
+ C)
I
I (B I
I
I
I
I
I
I
I
I I
I I
E
F
I
I
I I
1
G A
Fig. 7-20
F
[CHAP. 7
VECTORS
146 12. Provethat
(A+B)*(C+D) = A.C
+ A-D + B*C + B*D. +
+
ByProblem11, ( A + B ) * ( C + D ) = A * ( C + D ) B * ( C + D ) = A * C A * D The ordinary laws of algebra are valid for dot products where the operations are defined.
+ B * C+ B*D.
13. Evaluate each of the following. (a) i - i = lil lil (b) i * k = lil Ikl (c) k - j = Ikl Ijl (d) jo(2i-33jfk) (e) ( 2 i - j ) * ( 3 i + k )
cos 'O = (1)(1)(1)= 1 cos 90° = (1)(1)(0) = 0 cos 90' = (1)(1)(0) = 0 = 2 j * i - 3 j * j + j * k = 0 - 3 + O = -3 = 2i*(3i+k) - j * ( 3 i + k ) = 6i.i 2i.k - 3 j - i - j * k = 6
+
+0 -0 -0
= 6
14. If A = Ali+AZj+Ask and B = Bli+BZj+Bak, prove that A * B = A I B I + A ~ B z + A3 B3.
+
+ + + + + + +
= (Ali Azj + A s k ) (Bli Bzj B3k) = Ali.(Bli+Bzj+B3k) Azj*(Bli+Bzj+Bak) A s k * (Bli+Bzj+Bak) = A1Bli.i A1Bzi.j AlB3i.k A2Blj.i AzBsj*j A ~ B 3 j . k A3Blk.i A3Bzk-j A3B3k.k = AiBi AzBz A3B3 since i i = j j = k k = 1 and all other dot products are zero. A B
+
.
15. If
+
+
+
A = Ali+Azj+Ask,
+
Then A = I / C A
+
show that A = d T A = f A T + A g + A , 2 .
A - A = (A)(A) cosOo = A2. Then A = d T A . Also, A A = (Ali A2 j Ask) (Ali A2 j A3li) = (Ai)(Ai) (Az)(Az) i(A3)(A3) = A: by Problem 14, taking B =A.
+
+
+
+
= dA:
+ A: + A:
+
+
+ A: + A:
is the magnitude o f A.
Sometimes A *A is written A'.
The CROSS or VECTOR PRODUCT 16. Prove A X B = - B x A .
A X B = C has magnitude A B sin e and direction such that A, B and C form a right-handed system [Fig. 7-21 (a)above]. B X A = D has magnitude BA sin e and direction such t h a t B, A and D form a right-handed system [Fig. 7-21(b) above]. Then D has the same magnitude as C but is opposite in direction, i.e. C = -D or A X B = -B X A. The commutative law for cross products is not valid.
CHAP. 71
147
VECTORS
17. Provethat A x ( B + C ) = A X B + A X C f o r the case where A is perpendicular to B and also to C. Since A is perpendicular to B, AX B is a vector perpendicular to the plane of A and B and having magnitude A B sin90° = A B or magnitude of AB. This is equivalent to multiplying vector B by A and rotating the resultant vector through 90° to the position shown in Fig. 7-22. Similarly, A X C is the vector obtained by multiplying C by A and rotating the resultant vector through 90° to the position shown. In like manner, A X (B C) is the vector obtained by multiplying B C by A and rotating the resultant vector through 90° to the position shown. Since A X ( B + C) is the diagonal of the parallelogram with A X B and AX C as sides, wehave A X ( B + C ) = A X B + A X C .
+
+
Fig. 7-22
+
AxC 18. Provethat A x ( B + C ) = A x B in the general case where A, B and C are non-coplanar. See Fig. 7-23. Resolve B into two component vectors, one perpendicular to A and the other parallel to A, and denote them by B, and B I Irespectively. Then B = B, BII. If e is the angle between A and B, then B , = B sin 8. Thus the magnitude of AX B, is A B sin 8, the same as the magnitude of A X B. Also, the direction of AXB, is the same a s the direction of A X B. Hence A X B, = A X B. Similarly if C is resolved into two component vectors C I I and C,, parallel and perpendicular respectively to A, then A X C, = A X C. Also, since B C = B, BII C, CII = (B,
+
+
+ + +
Fig. 7-23
+ C,) + (BII+ C I I ) it follows that
A X (B,+ C,) = A X ( B + C) Now B, and C, are vectors perpendicular to A and so by Problem 17, A X (B,+ C,) AX(B+C)
Then
+
= A X B, A X C, = AXB+AXC
+
and the distributive law holds. Multiplying by -1, using Problem 16, this becomes (B C) X A = BX A C X A. Note that the order of factors in cross products is important. The usual laws of algebra apply only if proper order is maintained.
+
19. If A = A l i + A z j + A s k and B = B l i + B 2 j + B s k , provethat A x B = AxB
= (Ali+Azj+Ask)
X
(Bli+Bzj+Bsk)
+
+
= Ali X (Bli+Bzj+B3k) Azj X ( B l i + B * j + B s k ) A3k X ( B l i + B z j + B 3 k ) = A l B l i X i AlB2i X j t A1B3i X k A2Blj X i A2B2j X j AzB3j X k
+
=
+
+
+ A3Blk X i + A3BZk X j + A3B3k X k (AzB3-AsBz)i + ( A ~ B I - A I B ~+) ~(A1Bz-AzBl)k
+
=
148
[CHAP. 7
VECTORS
20. If A = 3 i - j + 2 k and B = 2 i + 3 j - k , i
j
2
3
k
2
-1
=
find A X B . 3
3 -1
2
+ 7j + I l k
-5i
21. Prove that the area of a parallelogram with
11
sides A and B is IAXBI.
Area of parallelogram = h IBI = IAl sine IBI = IAXBI.
J
e
Note that the area of the triangle with sides A and B = gIAXBI.
B
Fig. 7-24
22. Find the area of the triangle with vertices at P(2,3,5), Q(4,2,-1), R(3,6,4). PQ = (4 -2)i PR = (3 -2)i
Area of triangle
=
+ (2 - 3)j + (-1 - 5)k = 2i - j - 6k + (6 - 3)j + (4 - 5)k = i + 3j - k
-& I PQ X PR I
Il
i 2
=
-& 1
=
+./(19)2
j -1 3
3 1 (2i - j - 6k) X
= k -6 -11
I
=
(i -I- 3j - k) I
&I19i-4j+7kI
+ (-4)2 + (7)2
=
**
TRIPLE PRODUCTS 23. Show that A (B x C) is in absolute value equal to the volume of a parallelepiped with sides A, B and C. Let n be a unit normal to parallelogram I, having the direction of B X C, and let h be the height of the terminal point of A above the parallelogram I. Volume of parallelepiped
Fig. 7-25
= (height h)(area of parallelogram I) = (A n)(IB X Cl) = A*{IBXCIn} = A * ( B X C ) A n
If A, B and C do not form a right-handed system,
24. If
<0
and the volume = I A
(B X C)
A = A l i + A ~ j + A s k , B = Bli+B2j+B3k7 C = C l i + C ~ j + C 3 k show that A1 A2 A3 A *( B x C ) = Bi B2 B3
I
c1
c 2
c 3
I
A * (BX C) = A *
= (Ali+Azj+Ask)
= Ai(BzC3-BaC2)
[(BzCS-B3C2)i
+ ( B Q C I - B I C ~+) ~(BiC2-BzCl)kl
+ Az(B3Ci-BiC3) + A Q ( B ~ C ~ - B ~ =C ~ )
A1 Az A3
I.
CHAP. 71
149
VECTORS
25. Find the volume of a parallelepiped with sides A = 3i - j, B = j By Problems 23 and 24, volume of parallelepiped
A1 Az A3
A * (BX C) =
By Problem 24:
,
=
IA
=
1-201
(B X C) =
C = i + 5j + 4k.
+ 2k,
I
=
I
3 -1
0
1
4
5
20.
(AXB)*C = C*(AXB) =
Ci Ca Cs A1 As As Bi Bz B3
Fig. 7-26
x-x1 y-y1 xz - XI yz - y1 X3-xi y3-yi
or, using Problem 24,
2-21
xz
- xi
=
0
23-21
28. Find a n equation for the plane passing through the points PI@,1,-2), P3(2, -1,l).
P2(-1,
2,4),
The position vectors of PI,Pz,P3 and any point P ( x , y, x ) on the plane are respectively r1
= 3i
+ j - 2k,
r z = -i
+ 2j + 4k,
r3
= 2i - j
+ k,
r = xi
+ y j + xk
Then PP1 = r - rl, PzPl = r 2 - r1, P3P1 = r3 - r1, all lie in the required plane and so the required equation is ( r - r l ) (rz-r1) X (r3 -rl) = 0, i.e.,
+ (y - l ) j + ( x + 2)k) {-4i + j + 6k) X (4- 2j + 3k) = 0 + (y - l ) j + ( x + 2)k) (15 + 6j + 9k) = 0 15(x-3) + 6(y- 1)+ 9 ( x + 2) = 0 or 5% + 2y + 32 = 11
{(x - 3)i {(x - 3)i
Another method:
By Problem 27, the required equation is
x-3 -1-3 2-3
y-1 2-1 -1-1
x+2 4+2 14-2
=
0
or
5%
+ 2y 4-32
= 11
150
VECTORS
29. If A = i + j , B = 2 i - 3 j + k ,
C = 4j-3k,
= i-j-6k.
(b)
1 i 23 1 i 3;
BXC =
k
[CHAP. 7
find
Then ( A X B ) X C =
= 6i+6j+8k.
1
i
#
(U)
dr dt,
d
( t 3 + 2t)i
d +3 (-3 e-9 j +
A t t = 0, drldt = 2i (b)
From (a), (drldtl =
(')
dsr dt' -
L ) dt(* dt
At t = 0 ,
(d) From
(c),
d
i j k 1 1 0 6 6 8
dr
(2 sin 6t)k =
(3tx+ 2)i
+ 6j + 10k
d(Z)*+ (6)'+
= z{(3t2 d
= Zfi
(10)* =
+ 2)i +
= 23i+3j+4k.
= 8 i - 8j + k.
1 ~ 1 , =,
(b)
t = 0 and give a possible physical significance. dr dt =
1;
A X (B X C).
DERIVATIVES 30. If r = (tS+2t)i- 3e-2fj+ 2 sin5t k, find (a)
k
j
l;
Then A X ( B X C ) =
It follows that, in general, (A X B) X C
1
(AxB)x C , (b) A x (BxC).
(U)
6e-2tj
+
d2r
at
(d)
+ 6e-*'j + 10 cos6t k
at t
10 cos6t k}
(c)
= 0.
= 6ti - 12e-"j - 60 s i n & k
dsrldt2 = -12j.
jdsrldt') = 12
at t = 0 .
If t represents time, these represent respectively the velocity, magnitude of the velocity, acceleration and magnitude of the acceleration at t = 0 of a particle moving along the space curve z = t3 2 4 y = -3e-s' , z = 2 sin6t.
+
31. Prove that tions of U. Method 1:
d
dB dA (A B) = A B, where A and B are differentiable funcdu +
d z(A*B)
= =
Method 2:
Let
lim (A+AA)
AueO
lim A*AB
(B+AB) - A - B Au
+ AA*B + AA*AB
Awe0
A = Ali+Aej+Ask,
+
Au
B = B l i + B z j +B3k.
a2
32. If +(x,8, x ) = x%x and A = 3x2yi pz2j - xxk, find -(+A)
+
Then
at the point (1,-2, -1).
= (z'yz)(3z2yi yz8j - zxk) = 3x'y'xi 4- zay*zaj- s'yz*k a ($A) = a (3z4ypzi+ z2&'j - z'yz'k) = 3dy2i 4- 3z2&2j - 2z3yzk az
$A
CHAP. 71
VECTORS
a
as
= - (3x'y2i
-($A) ay az
JY
+ 3x2y2z2~ - 2x3yzk)
If x = 1, y = - 2 ,
z = -1, this becomes
33. If A = x2 sin yi
+ x2 cosy j - xy2k,
151
+ 6x2yzLj- 2x'zk
= 6x'yi
- 12j + 2k.
-12i
find dA.
Method 1: JA = ax
2x s i n y i
y4k,
+
+
x2 c o s y i - z' s i n g j - 2xyk,
= =
= =
d ( r 2s i n y ) i (2x s i n y dx
+
Method 2: dA
aA = ay
JA aA -dx fldy -dz ax ay az (x2 c o s y i - z* s i n y j (2x s i n v i - y2k)dx (22 sin y d x xz c o s 9 dy)i (2%cosy dz -
=
dA
-
+
+
+ d(z2c o s y ) j - d(xgP)k + x2 cosy dy)i + (2%cosy dz
2'
JA = Jz
= *i+*j+c*k
= 2xyz3i (b)
V*A
ax
+ xZz3j+ 3x2yz2k
ay
-
- z2 s i n y dy)j - (y'dz
+
=
+ (22 cosy j ) dz (y'dx + 2xy dy)k
2xyk) dy sin g dy)j
GRADIENT, DIVERGENCE and CURL 34. If + = x2yx3 and A = xxi - y2j 2x2yk, find (a)v+, ( b ) (+A),(4 curl (+A)* (a) V#
22 c o s y j
v *A,(c) v X A, (d)div
= -J( x z y x 3 ) i + - ( Jx P y z 3 ) j + - ( xaz y z ' ) k
az
Jx
JY
= (iG J + j day + k ~ ) * ( x z i - y y P j + 2 x z y k ) J = (xz)
ax
a (+) +all
a + -(2z29) a2
=
2
-
2g
X ( x z i - y 2 j +2x*yk)
(d) div ($A)
(e)
curl ($A)
= V *(x3yz'i - zpy3z3j + 2x4y2z3k) a a ( - x ~ ~ ~+z ~%(2x4gz3) a) = - (x3yz4) + -
=
V (+A)
=
ax JY 3x22/z4- 3 ~ ~ 26x4yPz2 / ~ ~ ~
=
+
V
X
($A)
=
+ 2x4y2z3k)
V X (x3yz4i - x2y3z3j
k x3yz4 - ~ 2 y 3 2 ~ 2 ~ 4 ~ 2 ~ 3
=
(4x'yz3
+ 3x2y3z2)i+
+ 2xydy)k
(4x3yz3 - 8x31/'z3)j - (2xy3x3
+ z'z')~
a2
152
VECTORS
36. Prove that constant.
[CHAP. 7
V+ is a vector perpendicular to the surface
Let r = xi
+(s,y,z) = c, where c is a
+ y j + zk be the position vector to any point P ( z ,y, z ) on the surface. + dy j + dz k lies in the plane tangent to the surface at P. But
Then dr = dx i
i.e. V$ dr = 0 so that V $ is perpendicular to dr and therefore to the surface.
37. Find a unit normal to the surface
2x2
+ 492 - 5x2 = -10
By Problem 36, a vector normal to the surface is V(2x2+4yz-6xa)
= 4xi
+ 4zj + (4y-
Then a unit normal to the surface a t P is Another unit normal to the surface at P is
38. If
+
= 2x2y - xx3, find (a) ~4 and (b)
=
4y
10x)k = 12i
+ 8j - 24k
12i+8j-24k d(12)' (8)' (-24)' - 3i 2j - 6k
+
+ + 7
at the point P(3,-1,2). a t (3,-1,2)
- 3i + 2 j - 6k 7
.
'
v2+.
- 6x2
Another method:
=
4y
-
6x2
39. Prove div curl A = 0. divcurlA
= V (V XA)
=
V
= o assuming that A has continuous second partial derivatives so t h a t the order of differentiation is immaterial.
CHAP. 71
VECTORS
153
JACOBIANS and CURVILINEAR COORDINATES 40. Find ds2 in (a)cylindrical and (b) spherical coordinates and determine the scale factors. (a) Method 1:
x =
=
dx
dsz =
Then
p
cos +,
y = p sin +, dy = p cos
+ cos+ dp, d x 2 + dgZ + dn2 = sin 9 d+
-p
(-
+
+
p
sin + d+
+
(p
x
=z
+ d$
+ sin+ dp,
+ cos+ d p ) %
cos+ d+
+
+ s i n + dp)2 + + hr(dn)z
dz = dz (dd2
= (dp)2 pZ(d+)' (dz)' = h:(dp)' h,9(d+)' and h~= h, = 1, hz = h4 = p , ha = h, = 1 are the scale factors. Method 2:
The position vector is r = p cos + i dr ar dr dr = -dp -d+ -dz
+ a+ + (cos+i + sin$j)dp +
aP
= =
Thus
ds2
(b) Then
=
Then
a2
+ pcos+j)d+ + k d n + p cos + d+)j + k d z (cos + dp - p sin + d$)2 + (sin + dp + p cos + d+)' + (dp)' + pZ(d+)' + (dz)'
(cos+ dp dr- dr
= =
p
sin + d+)i
(-psin+i (sin + dp
+
( d ~ ) ~
y = r sine sin $, z = r cos e x = r sin e cos +, = - r sin e sin + d+ r cos 8 cos 9 de i- s i n e cos # d r r cose sin + de sin e sin $ d r dg = r sin e cos+ d+ cos e d r dx = - r sin e de
dx
+
and
+ p sin 4 j + xk.
( d ~ )=~ ( d x ) z
+
+ ( d ~ +) (dn)z ~
+
+
= (dr)2
+ r"(de)z + r" sinze (d+)'
hl = h, = 1, hz = he = r, hs = h, = r sin 8.
The scale factors are
41. Find the volume element dV in (a) cylindrical and (b) spherical coordinates and sketch. The volume element in orthogonal curvilinear coordinates dV
=
( a ) In cylindrical coordinates,
hi hz h3 dui dun d ~ 3
=
I
UI,UZ,U S is
dui du2 dus
u2 = +, u 3 = z, hi = 1, h2 = p, h3 = 1 [see Problem 4O(a)]. = ( l ) ( p ) ( l ) dp d+ d z = p dp d+ d z
U I = p,
dV
1
Then
This can also be observed directly from Fig. 7-27(a) below.
(b) Volume element in spherical coordinates.
(a) Volume element in cylindrical coordinates.
Fig. 7-27
154
[CHAP. 7
VECTORS ( b ) In spherical coordinates,
UI
= r , u2 = 8,
US
= 9,hl = 1, h2 = T , hs = r sin e [see Problem 4 0 ( b ) ] . Then
= (l)(r)(r sin e) dr de d+
dV
= 73 sin e dr de d+
This can also be observed directly from Fig. 7-27(b) above.
42. Express in cylindrical coordinates: (a) grad a, (b) div Let w = p , U S = # , Page 142. Then
US=Z,
A, (c)
v2+.
h l = 1, h S = p , hs= 1 [see Problem 4O(a)] in the results 1, 2 and 4 on
where el, e2, es are the unit vectors in the directions of increasing p, $, z respectively.
where A = Alel+Azev+Ases.
MISCELLANEOUS PROBLEMS 43. Prove that grad f ( r ) = 7 r(r)r, where r = d m and f’(r) = df/dr is assumed to exist. a a a grad f ( r ) = V f ( r ) = - f ( r ) i -t - f ( r ) i + ~ f b k ‘) ax
= Another method:
f’(r)zi
+
f’(r):j
all
+
f’(r)$k =
W(xi
+ y j + zk)
= U
T
r
In orthogonal curvilinear coordinates U I ,US,U S , we have
If in particular we use spherical coordinates, we have U I= r , U I = 8 , U S = 9. Then letting of r alone, the last two terms on the right of (1) are zero. Hence we have, on observing that el = r/r and h l = 1, the result
+ = f ( r ) , a function
v f(r) 44. (a) Find the Laplacian of equation v2+= 0.
=
af(d T -11 dr r
=
f’m T
+ = f ( r ) . (b) Prove that + = l/r
(2)
is a solution of Laplace’s
CHAP. 71
155
VECTORS
By Problem 36, assuming that f ( r ) has continuous second partad derivatives, we nave
Laplacian of cp
Another method:
=
va+ = v *(vcp) = v
*{Tr}
In spherical coordinates, we have
If U = f ( r ) ,the last two terms on the right a r e zero and we find
( b ) From the result in part (a),we have
showing that l l r is a solution of Laplace’s equation.
45. A particle moves along a space curve r = r(t), where t is the time measured from some initial time. If v = Idr/dtl = ds/dt is the magnitude of the velocity of the particle (s is the arc length along the space curve measured from the initial position), prove that the acceleration a of the particle is given by a
dv
V2
= ZT
+-N P
where T and N are unit tangent and normal vectors to the space curve and - 1/2 P
The velocity of the particle is given by v a = - dv
dt
= zd( v T )
dv
= -dtT
f
= WT.
Then the acceleration is given by
dv d!r = -T dt dt
v-
d8 + v dT -ds dt
dv = -T dt
+
dT v*-
Since T has unit magnitude, we have T O T= 1. Then differentiating with respect to T * E + g * T = 0, d8 d8
m = 0 2T*d8
or
(2)
da
8,
T o dT = 0
da
from which it follows that dT/d8 is perpendicular to T . Denoting by N the unit vector in the direction of dTld8, and called the principal normal to the space curve, we have
VECTORS where
K
[CHAP. 7
is the magnitude of dT/da. Now since T = dr/da [see equation ( 9 ) , Page 1391, we have
Defining p = l/~,(2) becomes dT/ds = N/p. a
Thus from ( 1 ) we have, as required,
dv dt
= -T+-N
V2
P
The components dvldt and v'lp in the direction of T and N are called the tangential and normul components of the acceleration, the latter being sometimes called the centripetd acceleration. The quantities p and K are respectively the radius of curvature and curvature of the space curve.
Supplementary Problems VECTOR ALGEBRA 46. Given any two vectors A and B, illustrate geometrically the equality 4A
+ 3(B - A)
= A
+ 3B.
47. A man travels 26 miles northeast, 16 miles due east and 10 miles due south. By using an appropriate
scale determine graphically (a) how f a r and (b) in what direction he is from his starting position. Am. 33.6 miles, 13.2O north of east Is it possible to determine the answer analytically?
48. If A and B are any two non-zero vectors which do not have the same direction, prove that mA is a vector lying in the plane determined by A and B.
+ nB
49. If A, B and C are non-coplanar vectors (vectors which do not all lie in the same plane) and XI A ~1 B 21C = X P A yl B za C, prove that necessarily XI = XS, y~ = y ~ ,21 = zt.
+
50.
+
+
+
Let ABCD be any quadrilateral and points P , Q , R and S the midpoints of successive sides. Prove (a)that PQRS is a parallelogram and (b) that the perimeter of PQRS is equal to the sum of the lengths of the diagonals of ABCD.
51. Prove that the medians of a triangle intersect at a point which is a trisection point of each median. 52. Find a unit vector in the direction of the resultant of vectors Ans. (6i - 2j 7 k ) / 6 9 C = 3i - 2j 4k.
+
+
The DOT or SCALAR PRODUCT 53. Evaluate I ( A f B) (A--)( if A = 2 i - 3j
.
54.
+ 6k
A = 2 i - j +k,
and B = 3 i + j -2k.
B = i+j+2k,
Ane. 24
Prove the law of cosines for a triangle. [Hint Take the sides as A , B , C where C = A - B . use C C = ( A - B ) (A-B).]
55. Find a so that 2i - 3 j
56. If A = 2 i + j direction of B.
+ k,
+ 6k
and 31
B = i - 2j Ans. 1713
+ 2k
+ a j -2k and
are perpendicular.
C = 3 i - 4 j + 2k,
Am.
Q
Then
= -4/3
find the projection of A + C in the
CHAP. 71
157
VECTORS
57. A triangle has vertices at A(2,3, l), B(-l,l,2), C(1,-2,3). Find (a)the length of the median drawn from B to side AC and (b) the acute angle which this median makes with side BC. Ans. (a) &*6, (b) c 0 s - ~ m / 1 4 58. Prove that the diagonals of a rhombus are perpendicular to each other. 59. Prove that the vector (AB
+ BA)/(A+ B )
represents the bisector of the angle between A and B.
The CROSS or VECTOR PRODUCT 60. If
A = 2i-jfk
and B = i + 2 j - 3 k ,
find 1(2A+B) X (A-2B)I.
61. Find a unit vector perpendicular to the plane of the vectors
A m . *(2j
+ k)/fi
A = 3i - 2j
A m . 26fi
+ 4k
and B = i
+ j - 2k.
62. If A X B = A X C, does B = C necessarily?
63. Find the area of the triangle with vertices (2, -3, l ) , (1, -1,2), (-1,2,3).
Am.
64. Find the shortest distance from the point (3,2,1) to the plane determined by
46
(l,l,O), ( 3 , - l , l ) ,
Ana. 2
(-l,0,2).
TRIPLE PRODUCTS 65. If A = 2 i + j -3k, B = i - 2 j + k , C = - i + j -4k, find (a) A ( B X C), ( b ) C ( A X B), (c) A X (BX C), (d) (A XB) X C. A m . (a) 20, ( b ) 20, (c) 8i-19j-k, (cl) 261-16j-lOk 66. Prove that
(a) A (BX C ) = B (C X A ) = C (AX B) ( b ) A X (B X C) = B(A C) - C(A*B).
67. Find an equation for the plane passing through (2, -1, -2), (-1,2, -3), (4,1,0).
Ans. 2 % + 2 / - 3 ~= 9
68. Find the volume of the tetrahedron with vertices at (Z,l,l), (1,-1,2), (O,l,-1), (l,-Z,l). 69. Prove that
(AXB) (CXD)
+ ( B X C)
(AXD)
+ (CX A)
DERIVATIVES
Am.
4
( B X D ) = 0.
+
70. A particle moves along the space curve r = e-* cos t i e-* sin t j 1- e-t k. Find the magnitude Ans. (a)f i e - * , (b) fie-' of the (a)velocity and (b) acceleration at any time t . 71.
Prove that
d
(A X B) = A X
dB
dA + &U X B
where A and B are differentiable functions of
U.
72. Find a unit vector tangent to the space curve x = t, y = ts, z = tJ at the point where t = 1.
+ 2j + 3k)/fl4 = a cos ot + b sin at,
Ans. (i 73. If r
scalar, prove that
where a and b are any constant noncollinear vectors and dr d'r (a)r X - = o(a X b), ( b ) dt' war = 0. dt
+
A = xai-yj +xzk, B = y i + z j -xyzk ( b ) d[A (B X C)] at the point (1, -1,2).
74. If
75. If
R = z'yi - 2y'zj
+ zy'z*k,
find
and
Am.
1s *I X
ay'
+
C = i - y j z'zk, find ( a ) 8j, ( b ) 8 d z
(a) -4i
+
at the point (2,1, -2).
o
is a constant
a ( A X B)
Ans. 1 6 6
and
VECTORS
[CHAP. 7
GRADIENT, DIVERGENCE and CURL 76. If U , V, A, B have continuous partial derivatives prove that: (a) V ( U + V ) = V U + VV, ( b ) V * ( A + B ) = V * A V * B , (c) V X ( A + B ) = V X A
+
+ +
+
+
77.
If @ = xy yz zx and A = x2yi y2zj z'zk, find (a)A V@, ( b ) at the point (3, -1,2). Ans. (a) 25, ( b ) 2, (c) 56i - 30j 4- 47k
78.
Show that V X(Tlr) = 0 where r = x i + y j
79.
Prove: (a)V
80.
Prove that curl grad U = O ,
X
(UA) = (VU) X A
+ zk
+ U(V XA),
(b) V
83. ( a ) Prove that Problem 82.
- yzj + (z+ 22)k, find curl curl A. V
X
(V
X
( A X B ) = B * (V XA) - A * (V XB).
stating appropriate conditions on U.
*
A = 3xz'i
A and (c) (V+)X A
and r = .1
81. Find a unit normal to the surface z2y - 2x2 4- 2y22' Ans. (3i -t 4 j - 6k)/61
82. If
cV
+ VXB.
A) = - V'A f
V(V A).
= 10 at the point (2,1, -1). An8. -6xi
+ (62-
l)k
(b) Verify the result in
(U)
if A is given as in
JACOBIANS and CURVILINEAR COORDINATES
85. Express
(a) grad 9, (b) div A, (c) V2+ in spherical coordinates.
a+
Ans. ( a ) are1
86.
f
1 a+ --e~ r ae
+
1 a+ r sin e en
The transformation from rectangular to parabolic cylindrical coordinates is defined by the equations 2. (a) Prove that the system is orthogonal. (b) Find ds' and the scale factors. (c) Find the Jacobian of the transformation and the volume element. dz', hl = hz = h3 = 1 Ans. ( b ) ds' = (u2+w2)du2 (U'+ v2) dw2 (c) U' v2, (U' w*) du dw dz z = &(u2- w2), y = uw, z =
+
87. Write (a) V2+ and (b) div A A m . (a) V2+
=
+
+
+
d w ,
in parabolic cylindrical coordinates.
-(% 1
+
2)+
ag+
88. Prove that for orthogonal curvilinear coordinates,
+
+
[Hint: Let V+ = ale1 use2 uses and use the fact t h a t d+ = V+ dr must be the same in both rectangular and the curvilinear coordinates.] 89. Give a vector interpretation to the theorem in Problem 36 of Chapter 6.
VECTORS
CHAP. 71
MISCELLANEOUS PROBLEMS 90. If A is a differentiable function of
U
159
and IA(u)I = 1, prove that d A / d u is perpendicular to A.
91. Prove formulas 6, 7 and 8 on Page 141.
+
+ a r e polar coordinates and A , B, n are any constants, prove that U = pn(Acos n#J B sin n+) satisfies Laplace’s equation.
92. If p and
=
93. If
2 cos8
+
3 sin3@c o s + ,
find
vpv.
Anss 6 sin 8 cos + (4 - 6 sin%) r4
T2
(a)the cylindrical coordinate p, (b) the spherical coordinate r , ( c ) the spherical coordinate 8 which satisfies Laplace’s equation. B In (csc 8 - cot 8 ) where A and B are any constants. A m . (a)A B In p , (b) A B / r , (c) A
94. Find the most general function of
+
95.
+
+
Let T and N denote respectively the unit tangent vector and unit pmhcipal normal vector to a space curve r = r(u), where r(u) is assumed differentiable. Define a vector B = T X N called the unit b i n o m a l vector to the space curve. Prove that dT dN = rB - KT d s = KN9 ds = --N, ds
*
These are called the Frenet-Serret formulas and are of fundamental importance in differential geometry. In these formulas K is called the curvature, r is called the torsion; and the reciprocals of these, p = 1 / K and U = 117, are called the radius of curvature and radius of torsion respectively. 96. (a) Prove that the radius of curvature a t any point of the plane curve y
I
differentiable, is given by
P
=
= f ( z ) , z = 0 where f ( x ) is
(1 +yy,’~)3/z~
(b) Find the radius of curvature at the point ( ~ / 2 , 1 , 0 ) of the curve y = sin x, z = 0.
Ans. ( b ) 2 f i 97.
Prove that the acceleration of a particle along a space curve is given respectively in (a)cylindrical, (b) spherical coordinates by
(+ -
-
r i ) sins @)er
+
6.- p 2 ) e p + ( p g + 2i$)e, (TY + 24; - r i ) sin e cos 8)ee +
+ ?ex (2;
sin 8
+ 2 4 4 cos 8 + r;
sin e)e4
where dots denote time derivatives and ep,e,,e,,e,,ee,e4 are unit vectors in the directions of increasing p , +,.z, r , 8 , + respectively. 98. Let E and H be two vectors assumed to have continuous partial derivatives (of second order a t least)
with respect to position and time. Suppose further that E and H satisfy the equations 1 aE V * E = 0, V * H = 0, V X E = --V X H = -c at prove that E and H satisfy the equation
it:,
[The vectors E and H are called e l e c t r k and magnetic field vectors in electromagnetic theory. Equations ( I ) are a special case of Maxwell’s equations. The result ( 2 ) led Maxwell to the conclusion that light was an electromagnetic phenomena. The constant c is the velocity of light.] 99. Use the relations in Problem 98 to show that
a
-at{ * ( E s + H’)}
+
cV*(EXH) =
0
100. Let A I ,A z ,A3 be the components of vector A in a n zyz rectangular coordinate system with unit vectors il, in, i 3 (the usual i, j, k vectors), and A’I,A’%, A’J the components of A in an x’y’z’ rectangular coordinate
system which has the same origin as the zyz system but is rotated with respect to it and has the unit vectors i’l, i’2, i’3. Prove that the following relations (often called invariance relations) must hold:
where iL
in = lmn.
An = 11wA;
+ L A : + 13nA:
n = 1,2,3
160
VECTORS
[CHAP. 7
101. If A is the vector of Problem 100, prove that the divergence of A, i.e. V * A , is an invariant (often called a scalar invariant), i.e. prove that
The results of this and the preceding problem express an obvious requirement that physical quantities must not depend on coordinate systems in which they are observed. Such ideas when generalized lead to an important subject called tensor anulysis which is basic to the theory of relativity. 102. Prove that
( a ) A B, ( b ) A X B, ( c ) V
X
A are invariant under the transformation of Problem 100.
103. If U I , US,U S are orthogonal curvilinear coordinates, prove that
and give the significance of these in terms of Jacobians. 104. Use the axiomatic approach to vectors to prove relation (8) on P8ge 138. 105. A set of n vectors
..
AI,A,, .,A, is called linearly dependent if there exists a set of scalars ... c,A, = 0 identically, otherwise the set is cl, CI, . . . cm not all zero such that CIAI c z An A 2 = i+j-2k, AS = called linearly independent. (a) Prove that the vectors AI = 2 i - 3 j + S k , (b) Prove that any four 3 dimensional vectors are linearly 3 i - 7 j + 12k are linearly dependent. dependent. ( c ) Prove that a necessary and sufficient condition that the vectors AI = a l i bl j clk, A2 = asi b~j crk, AS = asi bs j csk be linearly independent is that AI At X As # 0. Give a geometrical interpretation of this.
+ +
+
+ +
+ +
+ +
106. A complex number can be defined as a n ordered pair ( a , b ) of real numbers a and b subject to certain
rules of operation for addition and multiplication. ( a ) What are these rules? (b) How can the rules in (a) be used to define subtraction and division? (c) Explain why complex numbers can be considered as two-dimensional vectors. ( d ) Describe similarities and differences between various operations involving complex numbers and the vectors considered in this chapter.
Chapter 8 Applications of Partial Derivatives APPLICATIONS to GEOMETRY 1. Tangent Plane to a Surface. Let F ( x ,y, z ) = 0 be the equation of a surface S such as shown in Fig. 8-1. We shall assume that F,and all other Eunctions in this chapter, is continuously differentiable unless otherwise indicated. Suppose we wish to find the equation of a tangent plane to S a t the point P(x,, yo, zo). A vector normal to S a t this point is No= V F J , , the subscript P indicating that the gradient is to be evaluated a t the point m o , Yo,
20).
If ro and r are the vectors drawn respectively from 0 to P(xo,ZJO, 20) and &($, Y,2 ) on the plane, the equation of the plane is
VFI, = 0 since r - ro is perpendicular to NO. In rectangular form this is (r -ro)
No = (r -ro)
(I)
Fig. 8-1
In case the equation of the surface is given in orthogonal curvilinear coordinates in the form F(ul,u~.,us)= 0, the equation of the tangent plane can be obtained using the result on Page 142 for the gradient in these coordinates. See Problem 4. 2. Normal Line to a Surface. Suppose we require equations for the normal line to the surface S at P(xo,YO,20). If we now let r be the vector drawn from 0 in Fig. 8-1 to any point (2,y, x ) on the normal NO,we see that r-ro is collinear with NOand so the required condition is (r--0)
x NO =
(r--0)
X
VFI,
=
0
(4
In rectangular form this is
Setting each of these ratios equal to a parameter (such as t or U) and solving for x, y and z yields the parametric equations of the normal line. The equations for the normal line can also be written when the equation of the surface is expressed in orthogonal curvilinear coordinates.
161
162
APPLICATIONS OF PARTIAL DERIVATIVES
[CHAP. 8
3. Tangent Line to a Curve. Let the parametric equations of curve C of Fig. 8-2 be x = f(u), y = g(u), x = h(u) where we shall suppose, unless otherwise indicated, that f , g and h are continuously differentiable. We wish to find equations f o r the tangent line to C a t the point P(xo,y, 20) where U = UO. If R = f(u)i g(u)j h(u)k, a vector tan-
+
+
gent to C a t the point P is given by TO=
;1$
If ro and r denote the vectors drawn respectively from 0 to P(xo,YO,20) and Q ( x ,y, x ) on the tangent line, then since r - ro is collinear with TOwe have (r - ro) x TO = (1: - ro) x - = 0 ( 5 )
ZIP
In rectangular form this becomes x - -0 --- Y - Y o
2 - 20 - h’(uo) The parametric form is obtained by setting each ratio equal to U. If the curve C is given as the intersection of two surfaces with equations F(x,y, Z ) = 0 and G ( x ,y, x ) = 0, the corresponding equations of the tangent line are
f ’(uo)
g’(u0)
Note that the determinants in (7) are Jacobians. A similar result can be found when the surfaces are given in terms of orthogonal curvilinear coordinates. 4. Normal Plane to a Curve.
Suppose we wish to find an equation for the normal plane to curve C a t P(xo,YO,20) of Fig. 8-2 (i.e. the plane perpendicular t o the tangent line to C a t this point). Letting r be the vector from 0 t o any point ( x , y , x ) on this plane, it follows that r -ro is perpendicular to TO. Then the required equation is In rectangular form this becomes when the curve has parametric equations x = f ( u ) , y = g ( u ) , z=h(u), and
when the curve is defined by F ( x ,y, x ) = 0, G(x,y, x ) = 0. 5. Envelopes. If +(x,y, a) = 0 is a one parameter family of curves in the xy plane, there may be
a curve E which is tangent a t each point to some member of the family and such that
163
APPLICATIONS OF' PARTIAL DERIVATIVES
CHAP. 81
each member of the family is tangent to E . If E exists, its equation can be found by solving simultaneously the equations +(x,y,a) = 0, +&,Y,a) = 0 (11) and E is called the envelope of the family. The result can be extended to determine the envelope of a one parameter family of surfaces +(x,y, x, a). This envelope can be found from +(X,Y,X,a)
= 0,
+,(X,Y,Z,a) = 0
(12)
Extensions to two (or more) parameter families can be made.
DIRECTIONAL DERIVATIVES Suppose F ( x , y , x ) is defined at a point (s,y,x) on a given space curve C. Let F(x + A s , y +Ay, x + Ax) be the value of the function at a neighboring point on C and let A s denote the length of arc of the curve between those points. Then AF F(x A$, 2j 4- Ay, 2 AZ) - F ( Z , y, 2 ) lim - = lim (18) AS As-0
+
As
+
As+O
if it exists, is called the directional derivative of F a t the point (x,y,x) along the curve C and is given by In vector form this can be written = vF.T (15) ds from which it follows that the directional derivative is given by the component of V F in the direction of the tangent to C. The maximum value of the directional derivative is given by IvFI. These maxima occur in directions normal to the surfaces F(x, y, x ) = c (where c is any constant) which are sometimes called equipotential surfaces o r level surfaces.
ds
DIFFERENTIATION UNDER the INTEGRAL SIGN P%
where u1 and u2 may depend on the parameter
a.
Then
for a S a 5 b, if f ( x ,a ) and af/aa a re continuous in both x and a in some region of the x a plane including u15 x 5 u2, a Ia S b and if U I and 2.42 are continuous and have continuous derivatives for a iQ Ib. In case U I and U P are constants, the last two terms of (17) are zero. The result (1r), called Leibnitx's rule, is often useful in evaluating definite integrals (see Problems 15, 29).
INTEGRATION UNDER the INTEGRAL SIGN If +(R) is defined by (16) and f ( x , a ) is continuous in x and al S x 5 242, a S x S b, then if u1 and u2 are constants, Xb+(a)da =
lb{
a
in a region including
i r f ( x . o ) d z ) d o = J:{Jbf(x,a)da}dx
(18)
164
APPLICATIONS OF PARTIAL DERIVATIVES
[CHAP. 8
The result is known as interchange of the order o f integration or integration under the
integral sign.
MAXIMA and MINIMA A point (xo,y0)is called a relative maximum point or relative minimum point of f ( x , y ) respectively according as ~ ( X + O h, YO + k) < f(xo,V O ) or f ( x o+ h, z/o + k) > f(xo,YO) for all h and k such that 0 < lhl < 6 , 0 < lkl < 6 where 6 is a sufficiently small positive number. A necessary condition that a differentiable function f ( x , y) have a relative maximum or minimum is df = o -af _ - 0, ax dY If ( x o . g 0 )is a point (called a critical point) satisfying equations (19) and if A is de-
2.
( X O , 90) is
a relative minimum point if A > 0 and
> 0 (or
-
3 1
)
ay2 ( x u . y o ) > o 3. (xo,yo) is neither a relative maximum o r minimum point if A < 0. If A < 0, ( X o , Yo) is sometimes called a saddle point. dx'
(Xo.Yo)
4. no information is obtained if A = 0 (in such case further investigation is necessary).
METHOD of LAGRANGE MULTIPLIERS for MAXIMA and MINIMA A method for obtaining the relative maximum or minimum values of a function F ( x , y , z ) subject to a constraint condition +(x,y,x) = 0, consists of the formation of the auxiliary function G ( W J ) = F(T,Y,Z) + X+(X,Y,4 (21) subject to the conditions aG aG aG = 0, - = 0, =o az ax which are necessary conditions for a relative maximum or minimum. The parameter A, which is independent of x , y, x , is called a Lagrange multiplier. The method can be generalized. If we wish to find the relative maximum or minimum values of a function F(x1,X Z , X S , . . . ,x,J subject to the constraint conditions + , ( X I , . . .,xn) = 0, + p ( ~ ~. ., ., rn)= 0, . . ., + k ( ~ l , . . ., x,) = 0, we form the auxiliary function subject to the (necessary) conditions aG aGaG E 0 - 0, - = 0, . . ., -ax dX2 dxn where A,, A,, . . .,A,, which are independent of X I , x2, . . ., x,,, are the Lagrange niultipliers.
APPLICATIONS to ERRORS The theory of differentials can be applied to obtain errors in a function of x , y , z , etc., when the errors in x , y , x , etc., are known. See Problem 28.
APPLICATIONS O F PARTIAL DERIVATIVES
CHAP. 81
165
Solved Problems TANGENT PLANE and NORMAL LINE to a SURFACE 1. Find equations for the (a) tangent plane and (b) normal line to the surface x2yx 3y2 = 2xx2-82 a t the point (1,2,-I).
+
(a) The equation of the surface is
F = z a y z + 3 y a - 2 z z * + 8 x = 0. A normal to the surface at
(1,2,-1) is
NO =
VFl(i,a,-,)
=
(2zyz
=
-6i
+ + 6y)j + + 14k
- 2z')i
+
(5%
llj
(z'y - 4x2
+ 8)kl(1,2,
-1)
Referring to Fig. 8-1, Page 161: The vector from 0 to any point (z,y,z) on the tangent plane is r = z i + y j +zk. The vector from 0 to the point (1,2, -1) on the tangent plane is r o = i 2j - k.
The vector r - ro = (z- l ) i perpendicular to NO.
+
+ (y - 2)j + ( z + l)k
Then the required equation is
+
lies in the tangent plane and is thus
+ + l)k}
i.e. {(z - 1)i (y - 2)j ( z (r - ro) NO = 0 -6(z - 1) 11(g - 2) 14(x 1) = 0 or
+
+
+
+ + +
{-6i l l j 14k} = 0 6% - l l y - 142 2 = 0
( b ) Let r = z i + y j + z k be the vector from 0 to any point ( x , y, z ) of the normal NO. The vector from 0 to the point (1,2, -1) on the normal is ro = i 2 j - k. The vector r - ro = (x - 1)i (y - 2)j ( z l ) k is collinear with NO. Then
+
+
XNO
(r--0)
= 0
i.e.
I
+
-6
14
11
+
I
which is equivalent to the equations I
11(z-1) = -6(g-2),
14(y-2) = l l ( z + l),
14(x- 1) = -6(z+ 1)
These can be written as often called the standard form for the equations of a line. By setting each of these ratios equal to the parameter t, we have 5
= 1 - 6t,
y
=2
+ llt,
z = 14t - 1
called the parametric equationa for the line.
2.
In what point does the normal line of Problem l(b) meet the plane x+3y-22
= IO?
Substituting the parametric equations of Problem l(b), we have
+ 3(2 + l l t ) - Z(14t - 1) = 10 = 2 + l l t = -9, z = 1 4 t - 1 = -16
1 - 6t Then z = 1- 6 t = 7, y
or
t =,-1
and the required point is (7,-9, -16).
+ +
3. Show that the surface x2 - 2yx y3 = 4 is perpendicular to any member of the family of surfaces x 2 1 = (2 -4a)y2 ax2 a t the point of intersection (1,-1,2).
+
Let the equations of the two surfaces be written in the form Then
+ ys - 4 = 0 and 2xi + (3g2-22x)j - Zyk,
F = xa - 2yz VF =
G = xz
+ 1 - (2-4a)y'-
a2
V G = 2xi - Z(2-44a)pj - Zazk
= 0
166
APPLICATIONS OF PARTIAL DERIVATIVES
[CHAP. 8
Thus the normals to the two surfaces at (1, -1,2) are given by NI
= 2i
-j
+ 2k,
Nz = 2i
Since Nt Ns = (2)(2) - 2(2 - 4a) - (2)(4a) all a, and so the required result follows.
4.
f
+ 2(2 -4a)j
- 4ak
0, it follows that NI and Na are perpendicular for
The equation of a surface is given in spherical coordinates by F(r,6 , + ) = 0, where we suppose that F is continuously differentiable. (a) Find an equation for the tangent plane to the surface a t the point (ro,8,, +o). (b) Find an equation for the tangent plane to the surface r = 4 cos 8 a t the point (2@, d 4 , 3d4). ( c ) Find a set of equations for the normal line to the surface in (b) a t the indicated point. (a) The gradient of 9 in orthogonal curvilinear coordinates is
(see Pages 141, 142).
In spherical coordinates ut = r , U S= 8, U S= +, hl = 1, h1= r , ha = r sin 8 and r = zi zk = r sin e cos +i r sin 8 sin +j r cos ek. Then sin 8 sin + j cose k et = sine c o s + i cos e sin + j - sin e k er = COB 8 cos + i es = - s i n + i cos+ j and aF 1 dF 1 aF V F = -et --el --es dr r ae r sin e a+
+
+
+
+ +
+ yj +
+
+
(1)
+
As on Page 161, the required equation is (r - ro) V F l p = 0. Now substituting ( I ) in (Z), we have
+ +
Denoting the expressions in braces by A, B, C respectively so that VFI, = A i Bj Ck, B ( y - PO) C(z - 20) = 0. This can be written we see that the required equation is A(z - ZO) in spherical coordinates by using the transformation equations for z,y and z in these coordinates.
+
(b)
We have Since
T
-
4 cos8
= 0.
=2fi,
8,
= u/4,
+, = 3n/4,
F = TO
Then
+
dF/dr = 1, aF/& = 4 sine, aF/a+ = 0.
we have from part (a), V F l p = Ai
+ B j + Ck = -i + j.
From the transformation equations the given point has rectangular coordinates (and 50 r - TO = (z f i ) i (y - f i ) j (z - 2)k.
+
+
+
+
a)
The required equation of the plane is thus - (z+fi) (y= 0 or y In spherical coordinates this becomes r sin e sin + - r sin e cos 9 = 2 f i .
+ +
fi,fi,2),
-z
= 2fi.
In rectangular coordinates the equation r = 4 cos e becomes xa y' (z- 2)' = 4 and the tangent plane can be determined from this as in Problem 1. In other cases, however, it may not be so easy to obtain the equation in rectangular form, and in such cases the method of part (a) is simpler to use.
APPLICATIONS O F PARTIAL DERIVATIVES
CHAP. 81 (c)
167
The equations of the normal line can be represented by
the significance of the right hand member being that the line lies in the plane z = 2 . required line is given by ---1
- I/-*
,
or
z=O
Thus the
x+y=O,z=O
TANGENT LINE and NORMAL PLANE to a CURVE 5. Find equations for the (a)tangent line and (b) normal plane to the curve x = t - cost, y = 3 sin2t, x = 1 cos3t a t the point where t = ir.
+
+
(a) The vector from origin 0 (see Fig. 8-2, Page 162) to any point of curve C is R = (t - cos t)i (3 sin 2t)j (1 cos 3t)k. Then a vector tangent to C at the point where t = +T is
+
+ +
+ 3j + k.
The vector from 0 to the point where t = &r is ro =
The vector from 0 to any point (x, y, z ) on the tangent line is r = xi Then equation is
= (z-&r)i
r-ro
+ (y-3)j + (2-1)k i
(r--0)
X
To = 0,
i.e.
Z - ~ T
1
and the required equations are = 3-24 x = 3t+1.
y
2
+ vj + xk.
is collinear with TO,so that the required
j y-3 - 2
3 = 2/-3 = z-1 2 -2 3 X-
+ +
+
3
or in parametric form x = 2t
+i ~ ,
( b ) Let r = xi yj zk be the vector from 0 to any point (2, y, x ) of the normal plane. The vector from 0 to the point where t = &r is ro = &i 3j k. The vector r - ro = (z- &r)i (y - 3)j (z - l ) k lies in the normal plane and hence is perpendicular to To. Then the required equation is ( r - r o ) * T o = 0 or 2(x- 81.) - 2(y-3) 3(2- 1) = 0.
+ +
+
+
+
6.
Find equations for the (a) tangent line and (b) normal plane to the curve 2 x 2 - x2y = 3 a t the point (1,-1,l).
y2x = -2,
3x2y
+
(a) The equations of the surfaces intersecting in the curve are
F = 3x4y
+ 9% + 2
G = 2x2 - x22/ - 3 = 0
= 0,
The normals to each surface at the point P(1,-1,l) are respectively NI = VFI, NZ = VGI,
+
= 6xyi (3x2+22/2)j -t yek = -6i = (22-2xy)i - xej 22k = 4i - j
+
Then a tangent vector to the curve at P is TO = NI XNZ = ( - 6 i + j + k )
X (4i- j + 2k)
= 3i
+j +k + 2k
+ 16j + 2k
Thus, as in Problem 5(a),the tangent line is given by i.e.,
(r - ro) X TO = 0 or 2 1 y + l 2-1 ----3 16
((5
- 1)i
or
+ (y + 1)j + ( x - l)k} X (3i + 16j + Zk}
x=1+3t,
y=16t-l,
= 0
z = 2 t + 1
( b ) As in Problem 5 ( b ) the normal plane is given by
(r - ro) TO = 0 i.e.,
3(x - 1)
or
((5
- 1)i + (y + l ) j + ( x - l)k} (3i + 16j + 2k) = 0
+ 16(y + 1) + 2(x - 1) =
0
or
3x
+ 16y + 22
= -11
The results in (a)and (b) can also be obtained by using equations (7) and (10)respectively on Page 162.
168 7.
APPLICATIONS O F PARTIAL DERIVATIVES
[CHAP. 8
Establish equation (IO),Page 162. Suppose the curve is defined by the intersection of two surfaces whose equations are F ( x , y, z) = 0, G(x,y,z) = 0 where we assume F and G continuously differentiable. The normals to each surface at point P are given respectively by NI= VFI, and NS= VGlp. Then a tangent vector to the curve at P is TO = NIX Nr = V F ( , X VGI,. Thus the equation of the normal plane is (r - ro) TO = 0. Now
To
=
VFI,
X
VG1,
=
{(Fzi
+ F Y j+ F,k) X (Gzi + G,j + G,k)} ,1
and so the required equation is
ENVELOPES 8. Prove that the envelope of the family + ( x , ~ , c Y=) 0, if it exists, can be obtained by solving simultaneously the equations = 0 and 4, = 0. Assume parametric equations of the envelope to be x = f(a), y = g(a). Then +(f(a),g(a), a) = 0
+
identically, and so upon differentiating with respect to a [assuming t h a t #, f and g have continuous derivatives], we have #,f'(a) + ##'(a) + 46, = 0 (1) The slope of any member of the family @ ( x , y , a )= 0 at (x,y) is given by $,dx
9=I The .slope of d,
+ #,dy
= 0 or
the envelope at (x,y) is dy - ___ dy/da - - Then a t any point where the dx ddda f'(a)' envelope and a member of the family are tangent, we must have
Comparing (2)with ( I ) we see that da = 0 and the required result follows.
9.
(a) Find the envelope of the family x sin CY geometrically.
+ y cos
CY
= 1. (b) Illustrate the results
( a ) By Problem 8 the envelope, if i t exists, is obtained by
+
solving simultaneously the equations +(x,y,a) = x s i n a y s i n a - 1 = 0 and # , ( x , y , a ) = x c o s a - y coscu = 0. From these equations we find x = sina, = cosa or x 2 + y ' = 1.
\
X /
( b ) The given family is a family of straight lines, some members of which are indicated in Fig. 8-3. The envelope is the circle x'+y* = 1.
Fig. 8-3
10. Find the envelope of the family of surfaces x = 2ax-a22/. By a generalization of Problem 8 the required envelope, if it exists, is obtained by solving simultaneously the equations (1)
+
= 2ax - a2y - x = 0
and
(2) $,
= 2 x - 2ay = 0
From (2)(Y = d y . Then substitution in ( I ) yields x5 = yz, the required envelope.
CHAP. 81
169
APPLICATIONS OF PARTIAL DERIVATIVES
11. Find the envelope of the two parameter family of surfaces x = ~ ~ x + p y - a / ? . The envelope of the family F ( x , y , x , a , P ) = 0, if it exists, is obtained by eliminating a: and p between the equations F = 0, F , = 0, F , = 0 (see Problem 43). Now
F = Then
Fa = - x + P
= 0,
z-a~-Py+ap
F, = - y + a
= 0,
= 0
p = x, = y and we have x = xy. (Y
DIRECTIONAL DERIVATIVES 12. Find the directional derivative of F = x2yx3 along the curve x = e-u, y = 2 s i n u x = U - cos U at the point P where U = 0. The point P corresponding to
U
= O is ( l , l , - 1 ) .
V F = 2xyx'i
Then
+ z2zsj -t- 3z2yz2k
= -2i - j
+ 3k
at P
A tangent vector to the curve is d -dr _ - -{e-"i (2 s i n u du du - -e i 2 cosuj
+
-U
+ l ) j + (U - cosu)k} + (1 + sinu)k = -i + 2j + k
+
and the unit tangent vector in this direction is TO = Then Directional derivative
=
V F TO =
(-2i
+ 1,
atP
-i+2j+k
fi
- j + 3k)
(-i +$+ ") =
=
-
fi
Since this is positive, F is increasing in this direction.
Lfi. 2
13. Prove that the greatest rate of change of F, i.e. the maximum directional derivative,
takes place in the direction of, and has the magnitude of, the vector V F .
dr This projection is a maximum ds ds * when VF and drlds have the same direction. Then the maximum value of dFlds takes place in the direction of V F ,and the magnitude is (VF(.
E ds
= V F - is the projection of V F in the direction
dr
14. (a) Find the directional derivative of U = 2x3y-3y2x at P(l,2,-1) in a direction
toward Q(3,-1,5). ( b ) In what direction from P is the directional derivative a maximum? ( c ) What is the magnitude of the maximum directional derivative? (a) V U = 6zzyi
+ (2%'- 6yz)j - 3y*k =
12i
+ 14j - 12k
The vector from P to Q
= (3-1)i
The unit vector from P to Q
= T =
Then Directional derivative at P
+ (-1
a t P. -2)j
2i - 3j
+
+
[ 5 - (-l)]k
+ 6k + (6)'
d(2)* (-3)'
= 2i - 3j
- 2i - 3j + 6k -
7
= (12i + 14j - 12k)
+ 6k.
. 90 7
i.e. U is decreasing in this direction. ( b ) From Problem 13, the directional derivative is a maximum in the direction 12i
(c) From Problem 13, the value of the maximum directional derivative is d 1 4 4 + 1 9 6 + 1 4 4 = 22.
+ 14j - 12k.
(12i+ 14j -12k( =
170
APPLICATIONS O F PARTIAL DERIVATIVES
[CHAP. 8
DIFFERENTIATION UNDER the INTEGRAL SIGN 15. Prove Leibnitz's rule for differentiating under the integral sign.
s., 'U
Let A$
$(a)
=
=
(a)
(a) f ( z ,a)
$(a+Aa)
-
$(a)
dz.
Then
=
1
u,(a+Aa)
f(z, a
(a+Aa)
=
U1 ( 0 )
f(x,a+Aa)dx
(a+Aa)
,"a (a)
f ( ~a +, A a ) d z
U1 (a)
-
ll + i,,.) (a)
-
(a) f ( x ,a)
U2
(a+Ac~)
f(x,a
+ Aa) d z
(a) J - a )
f ( x , a) d x
l,,,) 'U
=
+J
+ Aa) d x
(a)
[f(% a + A 4 - f @ , 41 d x +
u2 ( a+ ha)
f(x,a
+ Aa) dx
-
f ( x , a -k Aa) dz
By the mean value theorems for integrals, we have
Taking the limit as Acr derivatives, we obtain
16. If
+(a) =
+
0, making use of the fact that the functions are assumed to have continuous
I7 a'sin
a2
dx, find
where
+'(a)
a#
0.
By Leibnitz's rule,
-
17. If
f a-cosx dx
2 sin as - sin 'a fa'cosax d x + a a
da
sin ax
7r
-& 5 q '
+
2 sinas
sin2
a
a
a>l
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' <
Iff1>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
(8)
A 3 d z
where A I ,A 2 and A 3 are functions of X, y and x. Other types of line integrals, depending on particular curves, can be defined. For example, if A s k denotes the arc length along curve C in the above figure between points ( X k , Y k ) and ( X k t 1, Y k t I), then
is called the line integral of U(z,y) along curve C. dimensions are possible.
195
Extensions to three (or higher)
196
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
[CHAP. 10
VECTOR NOTATION for LINE INTEGRALS It is often convenient to express a line integral in vector form as a n aid in physical or geometric understanding as well as for brevity of notation. For example, we can express the line integral (3) in the form LAldx
+ A2dy + Asdx
=
1
(Ali
+
+
+ Azj + A s k )
(dxi + d y j
+ dxk)
(5)
+
where A = Ali + Azj Ask and dr = d x i d y j dx k . The line integral (2) is a special case of this with x = O . If at each point (x,y,x) we associate a force F acting on an object (i.e. if a force field is defined), then LF-dr represents physically the total work done in moving the object along the curve C.
EVALUATION of LINE INTEGRALS If the equation of a curve C in the plane x = O is given as y=f(x), the line integral (2)is evaluated by placing g = f ( x ) , d y = f ’ ( x )dx in the integrand to obtain the definite integral which is then evaluated in the usual manner. Similarly if C is given as x = g(y), then d x = g’(y) dy and the line integral becomes
If C is given in parametric form x = +(t),y = +(t), the line integral becomes
where tl and t 2 denote the values of t corresponding to points A and B respectively. Combinations of the above methods may be used in the evaluation. Similar methods are used for evaluating line integrals along space curves.
PROPERTIES of LINE INTEGRALS Line integrals have properties which are analogous to those of ordinary integrals. For example: 1.
S,
P ( x ,Y) dx
s
(a2, bz)
2.
(ol,bl)
Pdx
+
Q ( x , Y) d~
+ Qdy
=
S, P ( ~ , Ydx) + S Q(x, Y)d2/
(a,. b,)
=
- J a s . b,)
Pdx
+ Qdy
Thus reversal of the path of integration changes the sign of the line integral.
CHAP. 101
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
Pdx
+ Qdy
s
(as. b t )
=
( ~ 1 b, l )
Pdx
+
Qdy
+
s
(UZ,
bz)
197
P d x i- Q d y
( ~ 3 b3) ,
where (U&) is another point on C. Similar properties hold for line integrals in space.
SIMPLE CLOSED CURVES. SIMPLY and MULTIPLY-CONNECTED REGIONS A simple closed curve is a closed curve which does not intersect itself anywhere. Mathematically, a curve in the x y plane is defined by the parametric equations x = + ( t ) , y = $(t) where + and $ are single-valued and continuous in an interval t 1 5 t 5 t 2 . If +@I) = &) and $ ( t l ) = I/&), the curve is said to be closed. If +(U) = + ( U ) and $(U)= $(U) only when U = v (except in the special case where U = tl and v = t 2 ) , the curve is closed and does not intersect itself and so is a simple closed curve. We shall also assume, unless otherwise stated, that + and + are piecewise differentiable in t~5 t 5 tz. If a plane region has the property that any closed curve in i t can be continuously shrunk to a point without leaving the region, then the region is called simply-connected, otherwise it is called multiply-connected (see Page 102 of Chapter 6). As the parameter t varies from tl to t ~the , plane curve is described in a certain sense or direction. For curves in the x y plane, we arbitrarily describe this direction as positive or negative according as a person traversing the curve in this direction with his head pointing in the positive x direction has the region enclosed by the curve always toward his left or right respectively. If we look down upon a simple closed curve in the x y plane, this amounts to saying that traversal of the curve in the counterclockwise direction is taken as positive while traversal in the clockwise direction is taken as negative.
GREEN’S THEOREM in the PLANE Let P , Q, aPlay, aQlax be single-valued and continuous in a simply-connected region bounded by a simple closed curve C. Then $Pdr where
Jc
i-
Qdy
=
s$(p$.w
is used to emphasize that C is closed and that it is described in the positive
direction. This theorem is also true for regions bounded by two or more closed curves (i.e. multiply-connected regions). See Problem 10.
CONDITIONS for a LINE INTEGRAL to be INDEPENDENT of the PATH Theorem 1. A necessary and sufficient condition for Pdx Q d y to be independent of the path C joining any two given points in a region is that in
s,
+
where it is supposed that these partial derivatives are continuous in T .
198
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
[CHAP. 10
+ +
The condition (11) is also the condition that P d x Qdy is an exact differential, i.e. that there exists a function +(x,y) such that P d x Q dy = d+. In such case if the end points of curve C are (XI, yl) and (x2, y2), the value of the line integral is given by
s
( 1 2 . Y2 1
(Xl,Y1)
Pdx
+ Qdy
(X2,YZ)
=
(XI,Y1)
d+ =
In particular if (11) holds and C is closed, we have
5Pdx
+ Qdy
XI
=
$4~2,112)
-
+(xi,~i)
(12)
= x2, y l = y2 and 0
For proofs and related theorems, see Problems 11-13. The results in Theorem 1 can be extended to line integrals in space. Thus we have
Theorem 2. A necessary and sufficient condition for
A1dx
+ A2dy + A3dZ
to be inde-
pendent of the path C joining any two given points in a region !l( is that in !l(
where it is supposed that these partial derivatives are continuous in !l(. The results can be expressed concisely in terms of vectors. If A = A1i the line integral can be written
s,
+ A2j + Ask,
A * d r and condition (14)is equivalent to the con-
dition v X A = 0. If A represents a force field F which acts on an object, the result is equivalent to the statement that the work done in moving the object from one point x A = 0. to another is independent of the path joining the two points if and only if Such a force field is often called conservative.
v
v
The condition ( 1 4 ) [or the equivalent condition x A = 01 is also the condition that A1 dx A2 dy A3 dx [or A . dr] is an exact differential, i.e. that there exists a function +(x,y, x ) such that A1dx A2dy A3dx = d+. In such case if the endpoints of curve C are (xl,y1, xl) and (x2, y2, a),the value of the line integral is given by
+
+
s
(X2,Y2,%2)
(XI, Y 1, %I)
+
+
s
( Z 2 ’ Y2, % 2 )
Aedr
=
In particular if C is closed and
d+
=
+(x2, y2,x2)
- +(XI, y1,xl)
( 5 1 ’ Y 1 , %I)
vxA
= 0, we have
5A*dr = 0
SURFACE INTEGRALS Let S be a two-sided surface having projection ?i: on the xy plane as in the adjoining Fig. 10-2. Assume that an equation for S is x = f(x, y), where f is single-valued and continuous for all x and y in ?i:. Divide ?i: into n subregions of area AA,, p = 1,2, . . .,n, and erect a vertical column on each of these subregions to intersect s in an area A s p . Fig. 10-2
(15)
CHAP. 101
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
199
Let + ( x , y, x ) be single-valued and continuous at all points of S. Form the sum
in such a way where (tp,vp,S,) is some point of AS,. If the limit of this sum as ?a+ + 0 exists, the resulting limit is called the surface i n t e g r d of +(x,y, x ) that each over S and is designated by
JJ +@,
Y 9
2)
dS
(18)
S
Since AS, = lsecY,I AA, approximately, where yP is the angle between the normal line to S and the positive x axis, the limit of the sum (17)can be written
Then assuming that x = f ( x , y) has continuous (or sectionally continuous) derivatives in T,(19) can be written in rectangular form as
In case the equation for S is given as F ( x , y , x ) = 0, (21) can also be written
The results (21) or (22) can be used to evaluate (18). In the above we have assumed that S is such that any line parallel to the x axis intersects S in only one point. In case S is not of this type, we can usually subdivide S into surfaces S I , S ~. ., . which are of this type. Then the surface integral over S is defined as the sum of the surface integrals over S&, . . . . The results stated hold when S is projected on to a region of the xy plane. I n some cases it is better to project S on to the yx or xx planes. For such cases (18) can be evaluated by appropriately modifying (21) and (22).
The DIVERGENCE THEOREM Let S be a closed surface bounding a region of volume V . Choose the outward drawn normal to the surface as the positive normal and assume that p, Y are the angles which this normal makes with the positive x, y and x axes respectively. Then if A I , A2 and A3 are continuous and have continuous partial derivatives in the region (Y,
which can also be written
200
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
In vector form with A = A l i + A z j +&k be simply written as
[CHAP. 10
and n = c o s a i + c o s p j + c o s y k , these can
In words this theorem, called the divergence theorem or Green’s theorem in space, states that the surface integral of the normal component of a vector A taken over a closed surface is equal to the integral of the divergence of A taken over the volume enclosed by the surface.
STOKES’ THEOREM Let S be an open, two-sided surface bounded by a closed non-intersecting curve C (simple closed curve). Consider a directed line normal to S as positive if i t is on one side of S, and negative if it is on the other side of S. The choice of which side is positive is arbitrary but should be decided upon in advance. Call the direction or sense of C positive if an observer, walking on the boundary of S with his head pointing in the direction of the positive normal, has the surface on his left. Then if A I , A s , Aare ~ single-valued, continuous, and have continuous first partial derivatives in a region of space including S, we have
+ In vector form with A = Ali simply expressed as
(g-COS^ + ($-
+ A2 j + Ask
and n = cosai
$)cosY]dS
+ cosp j + cosy k,
this is
In words this theorem, called Stokes’ theorem, states that the line integral of the tangential component of a vector A taken around a simple closed curve C is equal to the surface integral of the normal component of the curl of A taken over any surface S having C as a boundary. Note that if, as a special case v x A = 0 in (27),we obtain the result ( I 6 ) .
Solved Problems LINE INTEGRALS
f(0.1) ( x 2 - y ) d x + ( y 2 + x ) d y along (1,2)
1. Evaluate
(a) a straight line from ( 0 , l ) to (1,2),
(b) straight lines from ( 0 , l ) to (1, I) and then from ( 1 , l ) to (1,2), ( c ) the parabola
x = t , y = t2+1.
(a) An equation for the line joining ( 0 , l ) and (1,2)in the zy plane is y = z the line integral equals
+
{z*- (Z 1)) dz
+ {(x + 1)’+
2)
dx
=
x1
+ 1.
Then dy = d x and
( 2 2 * + 2 x ) d x = 6/3
CHAP. 101
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
s'
201
( b ) Along the straight line from (0,l) to (1,l), y = 1, d y = 0 and the line integral equals
Lo
+ (1+ z)(O)
1)dx
(2'-
=
(z'- 1)dx = -2/3
0
Along the straight line from (1,l) to (1,2), z = 1, d x = O and the line integral equals
J:'
1'
+ (U'+ 1)d y = Then the required value = -2/3 + 10/3 = 8/3. (c)
= 10/3
(y'+ 1)d y
Since t = 0 a t (0,l) and t = 1 at (1,2), the line integral equals (t'
2.
(1- y)(O)
- (t'
+ 1))d t + ((t' + 1)' + t} 2t d t
=
(2t'
+ 4ts + 2t' + 2t - 1)dt
&
+
+
= 2
If A = (3x2- 6gx)i + (2g 3xx)j (1- 4xgx2)k, evaluate A *dr from (0, 0,O) to (1,1,1) along the following paths C: (a) x = t, g = t 2 , 2 = t 3 . ( b ) the straight lines from (O,O, 0) to (O,O, I), then to ( O , l , l),and then to (1,1,1). (c) the straight line joining (0,0,O) and (1,1,l).
1
A dr
=
((3%'- 6yz)i
s,
=
(32*-6yx)ds
+ (2y + 3xz)j + (1- 4zyz')k)
(dsi
+ dg j + d z k )
+ (2y+3sz)dy + (1-4xy z')dx
(a) If x = t , 2/= t*, z = t3, points (O,O,O) and (l,l,l)correspond to t = O and t = l respectively. Then
=
A dr
(3t'(3t'
6(t')(ts)} d t
- 6t')
Another method : Along C, A = (3t*-6ts)i dr = (i+2tj+3tSk)dt. Then
1
A *d r
(b)
+ (4t3 + 6t') dt + (3t2- 12t") d t
+ (2ta+3t4)j+ (1-4t')k
l1
=
dt
+ (2t' + 3(t)(ts)}d ( P ) + ( 1 - 4(t)(tL)(ts)2}d(ta)
(3t'- 6t') dt
and
=
2
r = z i + y j + z k = t i + t'j+ t%,
+ (4ta + 66') dt + (3t'-
12t") dt
=
2
Along the straight line from (O,O, 0) to (O,O, l), z = 0, y = 0, dx = 0, dy = 0 while x varies from 0 to 1. Then the integral over this p a r t of the path is {3(0)' - S(O)(z)}O
+ (2(O) + 3(0)(~)}0+ (I - 4(0)(O)(~')}dz
=
L:.
dz
= 1
Along the straight line from (O,O, 1) to (0,1, l), x = 0, z = 1, dz = 0, dz = 0 while y varies from 0 to 1. Then the integral over this part of the path is
s.l,
{3(0)' - W W O
+
+ 3(0)(1)) d2/ + (1- 4(0)(Y)(1)'}0
=
Jl0
2ydy
= 1
Along the straight line from (0, 1,l)to (1,1, l), y = 1, z = 1, dy = 0, dz = 0 while x varies from 0 to 1. Then the integral over this p a r t of the path is (32' - 6(1)(1)}dx
Adding ,
+ (2(1) + 3~(1)}0+ (1- 4~(1)(1)*}0 -
LA*dr = 1
+1-5
= -3.
= -5
202
[CHAP. 10
L INE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS (c)
The straight line joining (O,O, 0) and ( l , l ,1) is given in parametric form by x = t , g = t , x = t . Then A dr = (3t2- 6t2)dt (2t 3t2) dt (1 - 4t4)d t = 6/5
l:o
+
+
+
3. Find the work done in moving a particle once around an ellipse C in the xy plane, if the ellipse has center a t the origin with semi-major and semi-minor axes 4 and 3 respectively, as indicated in Fig. 10-3,and if the force field is given by
= (3~-42/+22)i
F
+ (4x+2y-3x2)j + (2xz-4y2+x3)k
+
In the plane z = 0, F = (32 - 4y)i (42 dr = dx i dy j so t h a t the work done is
$ F dr
+
=
1
((32 - 4g)i
= $(32--4y)dx
+ 2y)j - 4yZk and
+ (42 + 2y)j - 4y2k}
(dxi
U1
r = d+yj = 4coati+3sintj
+ dy j)
+ (42+2y)dy
Choose the parametric equations of the ellipse as x = 4 cos t ,
y
= 3 sin t where t varies from 0 to 2n (see Fig. 10-3). Then the
Fig. 10-3
line integral equals
l:
{3(4 cos t ) - 4(3 sin?)}{- 4 sin t } dt -
i:
(48 - 30 sin t c os t) d t
+
=
{4(4 cos t )
+ 2(3 sin t)}{3 cos t } d t
(48t - 16 einZt)li" =
96n
In traversing C we have chosen the counterclockwise direction indicated in Fig. 10-3. We call this the positive direction, o r say t h a t C ha s been traversed in the positive sense. If C were traversed in the clockwise (negative) direction the value of the integral would be - 96a.
4.
Evaluate
3y
Since ds =
ds along the curve
d
w
from x = 3 to x = 24.
C given by y = 2 f i
= d m d x = d m d x , we have
GREEN'S THEOREM in the PLANE 5. Prove Green's theorem in the plane if C is a closed curve which has the property that any straight line parallel to the coordinate axes cuts C in a t most two points. Let the equations of the curves AEB and AFB (see adjoining Fig. 10-4) be y = YI(x) and y = Y2(z) respectively. If 9( is the region bounded by C, we have
Y
A I
0
b
a
I
Fig. 10-4
=
-Jb
P(x, Y , )dx -
ia
P(x,Y,)dx
=
-$ P dx
X
CHAP. 101
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
203
Then Similarly let the equations of curves E A F and EBF be
2
= XI(Y)and z = x2(11) respectively.
Then
6.
5
Verify Green’s theorem in the plane for
(2xy-x2)dx
+ (x+y2)dy
where C is the closed curve of the region bounded by y = x2 and y2= x. The plane curves y = x 2 and y 2 = x intersect at (0,O) and ( 1 , l ) . The positive direction in traversing
C is as shown in Fig. 10-5.
Along y = x 2 , the line integral equals
-CO
{(2x)(x2)- xz} dx
+ {x + (z~)~} d(x2)
Along y2 = x the line integral equals (2(yz)(y) (y’))”} d ( y 2 )
+ {y2+ 8’) d y
=
Then the required line integral = 7 / 6 - 17/15 = 1/30.
Hence Green’s theorem is verified.
7. Extend the proof of Green’s theorem in the plane given in Problem 5 to the curves C for which lines parallel to the coordinate axes may cut C in more than two points. Consider a closed curve C such as shown in the adjoining Fig. 10-6, in which lines parallel to the axes may meet C in more than two points. By constructing and line ST the region is divided into two regions “Ip, which are of the type considered in Problem 5 and for which Green’s theorem applies, i.e.,
=
s.’
(22’
+ x2 + 2x5)dx
= 7/6
+ 2y2) d y
= -17/15
(4y4 - 221’
204
[CHAP. 10
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
(1)
s
Pdx
+ QdzJ
= Jf($-$)dxdy,
STUS
(2)
s
Pdx
+ Qdy
=
%.
SVTS
T 1
)
I f ( g -” -’ \ex
Adding the left hand sides of ( I ) and (Z),we have, omitting the integrand P d x
ay
+ Q dy
dxdy
in each
case, STUS
SVTS
using the fact that
ST
TUS
s=-s
ST
TS
SVT
TUS
SVT
TS
Adding the right hand sides of ( I ) and (2),omitting the integrand, where % consists of regions ql and q2. Pdx
Then
+ Qdy
=
ss (2
TUSVT
- z ) d x dy
ss Tl
TUSVT
+
Jf
=
*2
ss 9
and the theorem is proved.
s
A region % such as considered here and in Problem 5, f o r which any closed curve lying in % can be continuously shrunk to a point without leaving l(,is called a simpZy-connected region. A region which is not simply-connected is called multiply-connected. We have shown here t h a t Green’s theorem in the plane applies to simply-connected regions bounded by closed curves. In Problem 10 the theorem is extended to multiply-connected regions. For more complicated simply-connected regions it may be necessary to construct more lines, such as ST, to establish the theorem.
8. Show that the area bounded by a simple closed curve C is given by Q In Green’s theorem, put P = -y, Q = x . Then
where A is the required area. Thus A =
9.
3
t
f
C
x d y - ydx.
x d y - y dx.
Find the area of the ellipse x = U cos 8, y = b sin 8 . Area
= =
-&cJ(
x dy - y dx
=
&i2T+ a,b(cos2e
“J,
sin2e) de
( a cos e)(b cos e) de - ( b sin e)(- a sine) de
=
8
i2= ab de
=
rab
10. Show that Green’s theorem in the plane is also such valid for a multiply-connected region as shown in Fig. 10-7. The shaded region %, shown in the figure, is multiply-connected since not every closed curve lying in q can be shrunk to a point without leaving q,as is observed by considering a curve surrounding DEFGD for example. The boundary of %, which consists of the exterior boundary A H J K L A and the interior boundary DEFGD, is to be traversed in the positive direction, so that a person traveling in this direction always has the region on his left. It is seen that the positive directions are those indicated in the adjoining figure.
Fig. 10-7
In order to establish the theorem, construct a line, such as A D , called a cross-cut, connecting the exterior and interior boundaries. The region bounded by ADEFGDA L K J H A is simply-connected, and so Green’s theorem is valid. Then
CHAP. 101
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
205
=
Pdx+Qdy
9
ADEFGDALK JHA
But the integral on the left, leaving out the integrand, is equal to
since
ID l; = -
Thus if C1 is the curve ALKJHA, CO is the curve DEFGD and C is the boundary
of % consisting of C1 and CO (traversed in the positive directions), then
I + I /.
/.
=
and so
INDEPENDENCE of the PATH 11. Let P(x,y) and Q ( x , y ) be continuous and have continuous first partial derivatives at each point of a simply connected region T. Prove that a necessary and sufficient condition that
=
$Pdx+Qdy
aP/@ = a&/& identically in
around every closed path C in
0
is that
T.
Sufficiency. Suppose dP/dz/ = dQ/dz. Then by Green’s theorem, $Pdx+
=
Qdy
Suppose
5
P dx
+ Q dy
0
9
where 9( is the region bounded by C. Necessity.
=
s s ( g - g ) d z d y
= 0 around every closed path C in % and that dP/dy # dQ/dx at
some point of %. In particular suppose dP/dy - aQ/ax > 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
Al(s+eAz,y,z)
(Z,U,Z)
where we have applied the law of the mean for integrals.
= AI.
Taking the limit of both sides as Ax 4 0 gives Similarly we can show that
= As, a&x
d+/dy
= AS.
Thus A = A l i + A ¶ j + A s k = 2 a%i + * ay j + 3 k a2
= V#*
+
+
30. (a) Prove that a necessary and sufficient condition that A ~ d x Atdg Asdz = d+, an exact differential, is that x A = 0 where A = A l i A2 j Ask. ( b ) Show that in such case,
+
v
(rp, ~ 2 . x p )
(Zl.YI.Z1)
(a) Necessity.
Aldx If
(za,us,r*)
+ Atdy + A&
Aldx
+
=
+ Apdy + A3dz
(r1,v1,q)
d+ =
= d+ = *dx ax
-
+(~2,1/2,~2)
+
+ *dz, az
+(xI,~~,zI)
then
Then by differentiating we have, assuming continuity of the partial derivatives,
which is precisely the condition V X A
Another method:
from which V
X
Sufficiency. If V
If
Atdz
A = V X
X
=
0.
+ A ~ d g+ Asdz
= d+, then
V + = 0.
A = 0, then by Problem 29, A = V + and
A l d x + A ¶ d y + A s d z = A * d r = V + * d r = *dx+*dy+*dz a% ay az (r,U.
( b ) From part ( a ) ,
2)
+(x,y,z) =
AI dz (a, b. c)
+ Asdy + Asdz.
= d+
CHAP. 101
217
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
F = (2xz3+ 6y)i + (6%- 2yz)j + (3x2x2- y2)k is a conservative force Fe dr where C is any path from (I,-1, I) to (2,1, -1). (c) Give field. (b) Evaluate
31. (a) Prove that
s,
a physical interpretation of the results. ,
n
A force field F is conservative if the line integral
Jc
F dr is independent of the path C joining
any two points. A necessary and sufficient condition that F be conservative is that V
I V
Since here
X
'a
F =
k
j
+
I
aza
a aY
2 % ~6y~ 6%- 21/2 3~'z' - 'V
Method 1: By Problem 30, F dr = ( 2 d ferential d+, where + is such that
From these we obtain respectively
+
= x*zs
+ Gxy + f ~ ( y , z )
These are consistent if
f1
(y, 2)
1
= 0,
X
F = 0.
F is conservative.
+ 6y) dx + (6%- 2yz) dy + (32'2' - y*) dz
is an exact dif-
+ f ~ ( z , z ) + = xgza- y*z + f ~ ( z , g ) + c, f ~ ( xz), = xsza+ c, fs(x, y) = 6xy + c in which case
9 = 6sy - y'z
= -y'z
9 = xLzS+6xy-yZz+c. Thus by Problem 30,
J
(1, - 1 . 1 )
Alternatively we may notice by inspection that
+ 3x%* dz) + (6y d x + 6%d y ) - (2yz dy + 9' dz) d ( h 8 ) + 4 6 x 9 ) - d ( 2 / * ~ ) = d(x*zs + 6 % -~ 1/% +
F dr = (2xz'dx
= from which
+
C)
is determined.
Method 2: Since the integral is independent of the path, we can choose any path to evaluate it; in particular we can choose the path consisting of straight lines from (1,-1,l) to (2,-l,l), then to (2,1,1) and then to (2,1,-1). The result is f
l
(2x-6)dx
+
(12-2V)dV
+
l::
(12z*-l)dz
= 16
where the first integral is obtained from the line integral by placing y = -1, z = 1, dy = 0, dz = 0; the second integral by placing x = 2, z = 1, dz = 0, dz = 0; and the third integral by placing x = 2 , y = l , dz=0, dy=0. Physically
F * d r represents the work done in moving an object from (1,-1,l) to (2,1, -1)
along C . In a conservative force field the work done is independent of the path C joining these points.
MISCELLANEOUS PROBLEMS 32. (a) If x = f(u,w), y = g(u,w ) defines a transformation which maps a region xy plane into a region T' of the uw plane, prove that
of the
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
218
[CHAP. 10
(b) Interpret geometrically the result in (a). (a) If C (assumed to be a simple closed curve) is the boundary of q,then by Problem 8, s s d x d y = i i x d y - ydx 9 Under the given transformation the integral on the right of ( I ) becomes
where C' is the mapping of C in the uv plane (we suppose the mapping to be such t h a t C' is a simple closed curve also). By Green's theorem if q' is the region in the uv plane bounded by C', the right side of (2) equals
where we have inserted absolute value signs so a s to ensure t h a t the result is non-negative as is s s d x d y . 9 In general, we can show (see Problem 83) t h a t
(b)
JJdx d y
JJ1-
and
d u d v represent the area of region 9' tangular coordinates, the second in curvilinear coordinates. %
33. Let F =
(a) V X F
-gi
+ xj .
22+g2
=
(a) Calculate
i
j
k
a% -
a ay
a az
+
v x F.
= 0
'+(, the first expressed in rec-
F dr around any closed
(b) Evaluate
in any region excluding (0,O).
+
Y S 0 'x "a 2' y' -y dx
+x dy .
Let x = p cos 9, y = p sin #,
Then dx = - p sin
and so
d+
+ dp cos #, + y*
-ydxfxdy 39
dy
=
p
where ( p , +) are polar coordinates.
cos # d#
+ dp sin +
= d+ = d
For a closed curve ABCDA [see Fig. 10-18(a)below] surrounding the origin, # = 0 at A and
+=2n
after a complete circuit back to A. In this case the line integral equals
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
CHAP. 101
219
(b)
Fig. 10-18 For a closed curve PQRSP [see Fig.lO-l8(b) above] not surrounding the origin, # = #o at P and # = $0 after a complete circuit back to P. In this case the line integral equals Since F = Pi+ Qj, V X F = 0 is equivalent to dP/dy = dQ/dx X
2'
+
ckb = 0.
and the results would
seem to contradict those of Problem 11. However, no contradiction exists since P = Q=-
i:
and
do not have continuous derivatives throughout any region including (O,O), and this
was assumed in Problem 11.
34. If div
A denotes the divergence of div A
a vector field
=
lim
AV+O
A at
a point P , show that
$JA*ndS AV
where AV is the volume enclosed by the surface ASand the limit is obtained by shrinking A V to the point P. By the divergence theorem, div A dV = A n dS
ss
ss
AV
AS
By the mean-value theorem for integrals, the left side can be written drA SSsdV
-
-
= divA AV
AV
where div A is some value intermediate between the maximum and minimum of div A throughout AV. Then JJ A * n dS As div A = AV
-
Taking the limit as AV --* 0 such that P is always interior to AV, div A approaches the value div A at point P;hence JJ A * n dS AS div A = lim AV-0 AV This result can be taken as a starting point for defining the divergence of A, and from i t all the properties may be derived including proof of the divergence theorem. We can also use this to extend the concept of divergence to coordinate systems other than rectangular (see Page 142). Physically,
(JJJA As
n dS)/AV
represents the flux or net outflow per unit volume of the vector
A from the surface AS. If div A is positive in the neighborhood of a point P it means that the outflow from P is positive and we call P a source. Similarly, if div A is negative in the neighborhood of P the outflow is really an inflow and P is called a sink. If in a region there are no sources or sinks, then div A = 0 and we call A a solenoidat vector field.
220
[CHAP. 10
L INE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
Supplementary Problems
il,ll +
LINE INTEGRALS
(4.2)
35.
Evaluate
y) dx
(x
+ (y - x) dy
along (a)the parabola y* = x, (b) a straight line, ( c ) straight
+ + 1, y = t* + 1.
lines from ( 1 , l ) to (1,2) and then to (4,2), ( d ) the curve x = 2tZ t A m . ( a ) 34/3, ( b ) 11, (c) 14, ( d ) 32/3 36.
Evaluate
$ (22 -
y
+ 4) dx + (5y + 3x - 6)dy
around a triangle in the xy plane with vertices at
An8. 12
(0, 0), (3,0), ( 3 , 2 ) traversed in a counterclockwise direction. 37.
Evaluate the line integral in the preceding problem around a circle of radius 4 with center at (0,O). An8. 64a
38. (a) If
F = (x2- y2)i
+ 2xyj,
evaluate
s,
F * d r along the curve C in the zy plane given by
y = x 4 - x from the point (1,O) to (2,2). (b) Interpret physically the result obtained. Ans. (a) 124/15
39.
Evaluate
( 2 x + y ) d 8 , where C is the curve in the xy plane given by x 2 + y 2 = 25 and
If
F = (32 - 2y)i
+ (y + 2z)j - x2k,
evaluate
x
F *dr from (O,O,O)
is the
Ans. 15
ar c length parameter, from the point (3,4) to (4,3) along the shortest path. 40.
8
to ( l , l ,l), where C is a path
consisting of (a)the curve x = t, y = t Z , z = t3, (b) a straight line joining these points, ( c ) the straight lines from (0’0, 0) to (0’1, 0), then to (0’1, 1) and then to ( l , l , l), (d) the curve x = zz, z = y2. An8. ( a ) 23/15, ( b ) 5/3, ( c ) 0, ( d ) 13/15 41.
If T is the unit tangent vector to a curve C (plane or space curve) and F is a given force field, prove th at under appropriate conditions
F*dr =
F * T ds
where
is the a rc length parameter.
8
Interpret the result physically and geometrically.
GREEN’S THEOREM in the PLANE. INDEPENDENCE of the PATH 42. Verify Green’s theorem in the plane for
9;,
+ (y3-2zy)dy
(z’-zya)dx
where C is a square with
Am. common value = 8
vertices at (0, 0), (2,0), (2,2), (0,2) and counterclockwise orientation. 43.
Evaluate the line integrals of (a)Problem 36 and ( 6 ) Problem 37 by Green’s theorem.
44.
(a) Let C be any simple closed curve bounding a region having area A. Prove t h a t if a r e constants, ( a l x a t y +as) dx ( b l x b t y b3) dy = (bl - az)A
f
+
+
+
C be zero?
45.
Find the area bounded by the hypocycloid x2I3+ y213 = a*/’. [Hint: Parametric equations a re x = a cos3 t , y = a sin3t , 0 S t S 2a.1
46.
If x = p cos +, y = p sin cp, prove t h a t
4 $ x dy - y dx
f
(x3- s t y ) dx
=
(a)Prove th a t
~~~~~)
(2xy - y4
+ 3) dx + (xz- 4xy3)d y
( 2 , l ) . (b) Evaluate the integral in ( a ) .
Ans. (b) 5
3
pzdcp
+ zy*dy,
region enclosed by the circles x 2 + y 2 = 4 and x P + y 2 = 16. 48.
02,
as,b l , bz, b3
+
(b) Under what conditions will the line integral around any path
47. Verify Green’s theorem in the plane for
UI,
where
Ans. ( b )
a 2
= bl
An8. 3na’/8 and interpret.
C is the boundary of the
An8. common value = 1 2 0 ~
is independent of the path joining (1,O) and
CHAP. 101 49.
Evaluate
221
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
1
( 2 % ~-' y' cos x) d z
+
(1 - 2y sin z
+ 3xayy')d y
2%= ay'
along the parabola
from
Ans. af/4
(($0) to (a/2,1).
50. Evaluate the line integral in the preceding problem around a parallelogram with vertices at (O,O), (3,0), (5,2), (292). Am. 0
+
+ + + + +
+
G = (22' zy - 2y') d z (3%' 2xy) d y is not an exact differential. (b) Prove that eu/"Glz is an exact differential of 9 and find 9. (c) Find a solution of the differential equation (2%' ZY - 2 ~ ' )d x (3x2 2 % ~d y) = 0. (x ' 2x9) c, (c) x 2 + 2xg+ ce-y/' = 0 Ans. ( b ) 9 = eY/*
51. (a) Prove that
+
SURFACE INTEGRALS 52.
(a)Evaluate
ss
(x'+y*) d S , where S is the surface of the cone z' = 3(x'+y')
S
bounded by z = O
Ans. (a) 9a
and z = 3. (b) Interpret physically the result in (a).
53. Determine the surface area of the plane 2 x + y + 2 2 = 16 cut off by (a) z=O, y=O, x = 2 , y = 3 , ( b ) z = 0 , y = 0 and x * + y ' = 64. A m . (a)9, ( b ) 24a
54. Find the surface area of the paraboloid 22 = x * + y ' which is outside the cone z = Ans. j a ( 5 6 - 1) 55, Find the area of the surface of the cone z' = 3(xL+2/*) cut out by the paraboloid z
Ans. 6~
56.
Find the surface area of the region common to the intersecting cylinders Ans. 16a'
2'
+ y'
d m .
= x'+
= 'a and x'
y'.
+ z'
= a'.
+
' + y * Z' = a' contained within the cone z t a n a = 57. (a)Obtain the surface area of the sphere x 0 < a < a / 2 . (b) Use the result in (a)to find the surface area of a hemisphere. (c) Explain why formally placing a = P in the result of (a)yields the total surface area of a sphere. Ans. (a) 2aa'(l - cos a), (b) 2aa' (consider the limit as a + ~ / 2 )
d w ,
58. Determine the moment of inertia of the surface of a sphere of radius a about a point on the surface. A m . 2Ma', where mass M = 4na'u Assume a constant density U. 59.
(a) Find the centroid of the surface of the sphere z'+y*+z' = a' contained within the cone d m , 0 < < ~ / 2 . (b) From the result in (a) obtain the centroid of the surface of a hemisphere. Ans. (a) &a(l cosa), (b) a/2 z tana =
+
The DIVERGENCE THEOREM
+ +
+
y'j - (z 3y)k taken over the region bounded 60. Verify the divergence theorem for A = (2x9 z)i by 2 x + 2 y + z = 6, x = O ; y=O, z = O . A m . common value = 27
61.
Evaluate
ss ss
Fa n d S ,
S
bounded by z = 4 -U*, 62. Evaluate
where
F =
(2'
- x)i - xyj + 3zk
x = 0, x = 3 and the xy plane.
+
A n dS, where A = (2% 3z)i - (xz
8
s$
sphere having center at (3,-1,2) and radius 3. 63. Determine the value of
8
x dydz
and S is the surface of the region
Ans. 16
+ y)j + (y' + 2x)k
and S is the surface of the
Am. 108~
+ydzdz + zdxdy,
where S is the surface of the region
bounded by the cylinder z*-+y' = 9 and the planes z = O and z = 3 , Ans. 81u theorem, (b) directly.
(a) by using the divergence
222
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
f S 4 x z d y d z - y*dzdx
64. Evaluate
+ yzdx dy,
[CHAP. 10
where S is the surface of the cube bounded by x = 0,
S
y = 0, z = 0, x = 1, y = 1, z = 1, (a)directly, (b) by Green’s theorem in space (divergence theorem). Ans. 312
ss
(V X A) n dS = 0 for any closed surface S.
65. Prove that 66. Prove that
jJ
n d S = 0, where n is the outward drawn normal to any closed surface S .
S
67.
If n is the unit outward drawn normal to any closed surface S bounding the region
V, prove that
SSdivndV = S V
STOKES’ THEOREM 68. Verify Stokes’ theorem for A = 2yi+3xj-zSk, where S is the upper half surface of the sphere A m . common value = 9a x * + y ’ + z * = 9 and C is its boundary.
+
+
A = (y z)i - zzj y*k, where S is the surface of the region in the first octant bounded by 2 x + z = 6 and y = 2 which is not included in the (U) xy plane, (b) plane y = 2, (c) plane 2% x = 6 and C is the corresponding boundary. An8. The common value is (a) -6, (b) -9, (c) -18
69. Verify Stokes’ theorem for
ss
70. Evaluate
8
cone z = 2 71.
+
( V X A) n dS, where A = (x - z)i
d x -
+ (x*+ yz)j - 3xy’k
and S is the surface of the
A m . 12n
above the xy plane.
If V is a region bounded by a closed surface S and B = V X A, prove that S S B . n d S = 0.
72. (a)Prove that F = (2sy
F = V+. A m . (b)
( c ) Evaluate
+
+ 3)i + (z’- 4z)j - 4yk
s,
F dr, where
= xXy-4yz+3x+constant,
S
y * + z*
+ such
that
C is any path from (3, -1’2) to (2,1, -1). (c) 6
73. Let C be any path joining any point on the sphere z 9 + y * + z s =
x’+
(b) Find
is a conservative force field.
U’
to any point on the sphere
= b*. Show that if F = W r , where r = z i + y j +zk, then
1
Fodr = b S - a s .
/.
74.
r
F dr if F = f ( r ) r , where f ( r ) is assumed to be continuous.
In Problem 73 evaluate
An8.
b
rf(r)dr
J o
75. Determine whether there is a function + such that (a)F = (xz- y)i (say z3)j (3xz*- zy)k.
+
+
+ +
F = V+t where:
( b ) F = 2xe-’i (cosz - z*ee-’)j - y sin zk. If so, find it. y cosz 4- constant An8. (a)+ does not exist. (b) + = x’e-’
+
76. Solve the differential equation (z3- 4xy) dx An8. zz’- 2x*y 4- 31/* z = constant
+ (6y - 2xP)dy + (322’ + 1) dz = 0.
MISCELLANEOUS PROBLEMS 77. Prove that a necessary and sufficient condition that closed path C in a region two, at least) is that
+
(where
$X d y - aUdx az/
be zero around every simple
U is continuous and has continuous partial derivatives of order
78. Verify Green’s theorem for a multiply-connected region containing two “holes” (see Problem 10).
CHAP. 101 79. If P d s
LINE INTEGRALS, SURFACE INTEGRALS, INTEGRAL THEOREMS
+
223
+
Q d y is not an exact differential but p ( P d x Q d y ) is an exact differential where p is some function of x and y, then p is called a n integrating factor. (a) Prove that if F and G a r e functions of x done, then ( F y G ) dx d y has a n integrating factor p which is a function of z alone and find p. What must be assumed about F and G? ( b ) Use (a) to find solutions of the differential equation xy' = 2% 3y. Ant?. (a)p = eJcc+)dt ( b ) y = c x a - s , where c is any constant
+ +
80.
+
+ y' + ( z - a)' = as contained within the paraboloid
Find the surface area of the sphere x'
A m . 2ra
z
= x'
+ .'y
81. If f ( r ) is a continuously differentiable function of r = J / x z + y z + 2, prove that
82. Prove that
JJV
X
(+n)dS = 0
where
+
is any continuously differentiable scalar function of
8
position and n is a unit outward drawn normal to a closed surface S . 83. Establish equation (3), Problem 32, by using Green's theorem in the plane. [Hint: Let the closed region % in the xy plane have boundary C and suppose that under the transformation s = f ( u , v ) ,y = g(u,v) these are transformed into T' and C' in the uv plane respectively. First prove that
Jf
F ( z , y) d z dy =
Q ( x , y) d y
9 from sign this last integral is equal to
theorem to transform this into
ss 9'
where aQ/ay
l,
= F(x,y).
Q [ f ( u ,U), g(u, v ) ] [ g d u
Then show that apart
+ $d v ] .
Finally use Green's
F [ f ( u U), , g(u, U)]I d o . l d u dv. v)
84. If 2 = f(u,U,w ) , y = g(u,U ,w ) , z = h(u,U ,w ) defines a transformation which maps a region % of xyz space into a region q' of uvw space, prove using Stokes' theorem that
fs
fs$
[HI
F ( x , y, z ) dx d y dz = G(u,U,w ) du dv dw 9 %' where G(u,U ,w ) F [ f ( u U, , w ) , g(u, U,w ) , h(u,U,w ) ] . State sufficient conditions under which the result is valid. See Problem 83. 85. (a) Show that in general the equation r = r(u,v) geometrically represents a surface. (b) Discuss the geometric significance of U = C I ,V = C I where c1 and ca are constants. (c) Prove that the element of arc length on this surface is given by
d8'
= Edus
+ 2 F d u d v + Gdv'
86. (a) Referring to Problem 85, show that the element of surface area is given by dS = J / w d u d v .
( b ) Deduce from (a)that the area of a surface r = r(u, U) is [Hint: Use the fact that
X
$1
=
d(2
au ( A X B ) ( C X D ) = ( A * C ) ( B * D )- ( A * D ) ( B * C ) .
+
JJd
w
d
u dv.
S
X
*) (* *) av au av X
and then use the identity
+
r = a sin U cos v i a sin u sin v j a cos U, 0 5 U 5 'R, 0 5 v < 27 represents a sphere of radius a. (b) Use Problem 86 to show that the surface area of this sphere is 4na'.
87. (a) Prove that
88. Use the result of Problem 34 to obtain divA in (a)cylindrical and (b) spherical coordinates. Page 142.
See
-
Chapter
~
II
-
infinite Series CONVERGENCE and OIVERGENCE of INFINITE SERIES Consider the infinite series
2
=
Un
?%=l
U1
+
+ U3 +
. . a
Let the sequence of partial sums of the series be S1, SZ,S3, . . . where SI= U I , SO = ~1 + ~ 2 , Ss = Ui uz + ~ 3 , . . ., Sn = U1 + U2 + *
+
*
+
Un
(2)
If this sequence is convergent, i.e. if there exists a number S such that lim S,, = S, the n+oo
series ( I ) is called conurngent and S is called i t s sum. If lim S, does not exist, Vne series n+m
is called divergent. (Compare Chap. 3, Page 43). We shall sometimes abbreviate the infinite series (1) by 2% and shall-refer to U,, as the nth term of the series. Example I: “=12” 2 “ 1 = -+-+?+ 1 - 2’1 1 .... Here S, = s u m o f f i r s t n t e r m s = 1 - (see Prob. 26, 2 2” Chap, 3). Then since lim = 1, the series is convergent and has sum S = 1. U+CG
Example 2:
Q
(-l)--l = 1 - 1
a=l
Hence lirn S,
+1-1+ .
Here S , = 0 or 1 according as n is even o r odd.
does not exist and the series is divergent.
n-m
FUNDAMENTAL FACTS CONCERNING INFINITE SERIES 1. If Bun converges, then lirn un = 0 (see Prob. 26, Chap. 3). The converse hown+oo
ever is not necessarily true, i.e. if lim un = 0, Bun may or may not converge. n 4 m
fi; fol!ows that if the nth term of a series does not approach zero the series is divergent
.
2.
Multidication of each term of a series by a constant different from zero does not affect the convergence or divergence.
3.
Removal (or addition) of a finite number of terms from (or to) a series does not. affect the convergence or divergence.
SPECIAL SERIES 1. Geometricseries
Ca
n=l
ar”-l = a
+ ar + ar2-+ . . .,
a 1-r a(1- rn) terms is Sn = (see Prob. 25, Chap. 3). 1-r
where a and r are constants,
converges to S-= - if Irl
224
~
226
INFINITE SERIES
CHAP.111
+ +- +
1 1 1 where p is a constant, converges for n = 1~ 2p 3p . . . p > 1 and diverges for p 5 1. The series with p = 1 is called the harmonic series.
2. The p series
O01
n=l
L;;
TESTS for CONVERGENCE and DIVERGENCE of SERIES of CONSTANTS 1. Comparison test for series of non-negative terms. Convergence. Let V n 2 0 for all n > N and suppose that B V , converges. Then if OSunSv, for all n > N , Bun also converges. Note that n > N means from some term onward. Often, N = 1. 1
Since - - and 2"+ 1 - 2"
Example:
1 1 22; converges, 2-2n
+
also converges.
Divergence. Let V n 2 0 for all n.> N and suppose that Zvn diverges. if un 2 V n for all n > N, Bun also diverges, 1 = 1 Example: Since -> - and 2 - diverges, n=t n Inn n
Then
5 Inn also diverges.
n=2
2. Quotient test for series of non-negative terms. (U) If U n 2 0
and
V n 20
Un
lim - = A # O or
and if
n-+m
Vn
00,
then 2% and
BVn
either
both converge or both diverge. (b) If A = 0 in (a) and 2 V n converges, then Bun converges. ( c ) If A = 00 in (a)and B V n diverges, then Zu, diverges. This test is related to the comparison test and is often a very useful alternative to it. In particular, taking V n = l/nP, we have from known facts about the p series the
Theorem 1. Let lirn n P U n = A. n-00
Then
(i) Bun converges if p > 1 and A is finite (ii) Zun diverges if p 5 1 and A # O (A may be infinite). Examples :
1.
n 2-4nSconverges since 2
2.
n 2-vIn m
lim n9
n-00
n - 1 -4n - 2 4'
J
Inn diverges since lim nl/L (n 1)lI2 U 4 00
+
QO.
3. Integral test for series of non-negative terms. If f ( x ) is positive, continuous and monotonic decreasing for x 2 N and is such that f ( n )= un,n = N , N+1, N+2, . . ., then Zun converges or diverges ac-
cording as
xz
f ( z )dx = lirn Me-
lM
f ( z )dx
converges or diverges.
In particular
we may have N = 1, as is often true in practice. Example:
1 2" 7 n
n=1
converges since
lim
M+w
lM$
= lirn (1 M - c 00
$) exists.
4. Alternating series test. An alternating series is one whose successive terms are alternately positive and negative.
226
INFINITE SERIES
[CHAP.11
An alternating series converges if the following two conditions are satisfied (see Problem 15). ( a ) Iuntl(5 1u.l for n 1 1 ( b ) limu, = o (or IimIu,,l = 0) n+
Example:
1 -4
For the series
Iu,~
= ;p1
+ 8 - $ + & - -.. -- 5 kT2 n
=n+l'
IUn+ll
n+m
w
Then for n Z 1 ,
IUn+ll
5
we have
an = (-1)n-l
n
'
lim 1u.l = 0. Hence
\U,,\.Also
U-+=
the series converges.
Theorem 2. The numerical error made in stopping at any particular term of a convergent alternating series which satisfies conditions (a) and (b) is Iess than the absolute value of the next term. Example: If we stop a t the 4th term of the series 1 - 3 + Q - t + -& - -.-,the error made is less = 0.2.
than
5.
Absolute and conditional convergence. The series Bun is called absolutely convergent if BIu.1 converges. If Bun converges but BIu.1 diverges, then %,, is called conditionaZly convergent.
Theorem 3. If BIu.1 converges, then Bun converges. is convergent (see Prob. 17). Example 1:
1 2 + +1 + 31 - . ' . is absolutely convergent and thus convergent, since
-1+ - -1- - 1 1a
22
32
the series of absolute values
Example 2:
In words, an absolutely convergent series
6+ & + f + -$+ -
' ' '+
I--+---
... converges,but 1 , 2 3 4 5 . is conditionally convergent. 4
+
1 1 1 +-+-+-+ 2 3 4
' . converges.
... diverges. Thus 1
1 1 --+-2 3
Any of the tests used for series with non-negative terms can be used to test for absolute convergence. 6. Ratio test. Let
I
lim - = L. Untl
n+wl
a
(a) converges (absolutely) if
(b) diverges if L > 1. If L = 1 the test fails.
Then the series 2un
L <1
7. The nth root test. Let l i m m = L. n+-
Then the series 2%
(a)converges (absolutely) if L < 1 (b) diverges if L > 1. If L = 1 the test fails. 8. Raabe's test.
( lui:ll)
Let lim n 1- - = L. Then the series Bun n+ w
(a) converges (absolutely) if L > 1 (b) diverges or converges conditionally if L
< 1.
CHAP. 111
INFINITE SERIES
221
If L = 1 the test fails. This test is often used when the ratio test fails. 9. Gauss’ test. If series Bun
Untl
-
where 1--+n n2’ Cn
1Cl.
< P for all n > N, then the
(a) converges (absolutely) if L > l ( b ) diverges or converges conditionally if L 5 1. This test is often used when Raabe’s test fails.
THEOREMS on ABSOLUTELY CONVERGENT SERIES Theorem 4. The terms of an absolutely convergent series can be rearranged in any order, and all such rearranged series will converge to the same sum. However, if the terms of a conditionally convergent series are suitably rearranged, the resulting series may diverge or converge to any desired sum (see Problem 76). Theorem 5. The sum, difference and product of two absolutely convergent series is absolutely convergent. The operations can be performed as for finite series.
INFINITE SEQUENCES and SERIES of FUNCTIONS. UNIFORM CONVERGENCE
.
Let {U&)}, n = 1,2,3,. . be a sequence of functions defined in [a, b]. The sequence is said to converge to F(x), or to have the limit F ( x ) in [a,b], if for each c > 0 and each x in [a,b] we can find N>O such that lun(x)- F ( z ) I < c for all n > N . In such case we write lim U,@) = F(x). The number N may depend on x as well as C. If i t depends n4w only on c and not on x, the sequence is said to converge to F(z) uniformly in [a,b] or to be uniformly convergent in [a,b]. The infinite series of functions
is said to be convergent in [a,b] if the sequence of partial sums {Sfl(z)},n = 1,2,3,. . ., where S,(x)= u 1 ( x )+U&) . . . +U,($), is convergent in [a, b]. In such case we write lim S,,(x) = S ( x ) and call S ( x ) the sum of the series.
+
n 4 EO
It follows that 2u,,(x) converges to S ( x ) in [alb] if for each c > O and each x in [a,b] we can find N > O such that IS,(s)-S(s)I < c for all n > N . If N depends only on c and not on x, the series is called uniformly convergent in [a,b]. Since S ( x )- S f l ( x )= R,,(x), the remainder after n terms, we can equivalently say that 2 U n ( x ) is uniformly convergent in [a, b] if for each c > 0 we can find N depending on c but not on x such that IRn(x)J< c for all n > N and all x in [a,b].
228
INFINITE SERIES
[CHAP.11
These definitions can be modified to include other intervals besides a 5 x 5 b, such as a < x < b, etc. The domain of convergence (absolute or uniform) of a series is the set of values of x for which the series of functions converges (absolutely or uniformly).
SPECIAL TESTS for UNIFORM CONVERGENCE of SERIES 1. Weierstrass M test. If a sequence of positive constants Ml,Mz,Ma, found such that in some interval (a) Iun(x)I 5 Mn
( b ) BMn converges
...
can be
n = 1,2,3,...
then 2un(x) is uniformly and absolutely convergent in the interval. Example:
5 n2 is uniformly and absolutely convergent in [0,2a] since x-$converges. ,,=I
This test supplies a sufficient but not a necessary condition for uniform convergence, i.e. a series may be uniformly convergent even when the test cannot be made to apply. One may be led because of this test to believe that uniformly convergent series must be absolutely convergent, and conversely. However, the two properties are independent, i.e. a series can be uniformly convergent without being absolutely convergent, and conversely. See Problems 30, 123. 2. Dirichlet’s test. Suppose that (a) the sequence {a,} is a monotonic decreasing sequence of positive constants having limit zero,
( b ) there exists a constant P such that for ~ 5 x 3 1
IuI(~)+u~(x)+ Then the series
alul(x)
. . . +un(x)I < P
+ a2U2(x) +
for all n > N . =
is uniformly convergent in a 5 x 5 b.
zanun(x)
n=l
THEOREMS on UNIFORMLY CONVERGENT SERIES If an infinite series of functions is uniformly convergent, i t has many of the properties possessed by sums of finite series of functions, as indicated in the following theorems. Theorem 6. If { U n ( X ) } , n = 1,2,3, . . . are continuous in [a, b] and if Bu,,(x) converges uniformly to the sum S ( x ) in [a, b ] , then S ( x ) is continuous in [a, b ] . Briefly, this states that a uniformly convergent series of continuous functions is a continuous function. This result is often used to demonstrate that a given series is not uniformly convergent by showing that the sum function S ( x ) is discontinuous a t some point (see Problem 30).
229
INFINITE SERIES
CHAP. 111
In particular if xo is in [a,b],then the theorem states that
where we use right or left hand limits in case xo is an endpoint of
[U,
Theorem 7. If {U&)}, n = 1,2,3, . . ., are continuous in [a,b] and if &,,(x) to the sum S ( x ) in [a,b], then
b].
converges uniformly
or Briefly, a uniformly convergent series of continuous functions can be integrated term by term. Theorem 8. If {un(z)}, n = 1,2,3, . . ., are continuous and have continuous derivatives in [a,b] and if Bun(%)converges to S ( x ) while Bu&) is uniformly convergent in [a,b], then in [a, b]
S'(x) = or
2.:(x)
n=l
This shows conditions under which a series can be differentiated term by term. Theorems similar to the above can be formulated for sequences. For example, if {U&)}, n = 1,2,3, . . . is uniformly convergent in [a,b], then b
l i m l un(s)dx =
n4~1
1 b
limu,(x)dx
n4-
which is the analog of Theorem 7.
POWER SERIES. A series having the form a0
+ alx + a2x2 +
...
(9)
where ao,al,a2, . . . are constants, is called a POW ries in x. It is often onvenient to abbreviate the series (9) as Ha,#. In general a power series converges for 1x1 < R and diverges for 1x1 > R, where the constant R is called the rudius of convergence of the series. For [xI=R, the series may or may not converge. The interval 1x1 < R or -R < x < R, with possible inclusion of endpoints, is called the interval o f convergence of the series. Although the ratio test is often successful in obtaining this interval, it may fail and in such cases other tests may be used (see Prob. 22). The two special cases R = O and R=.o can arise. In the first case the series converges only for z=O; in the second case it converges for all x, sometimes written -00 < x < 00 (see Problem 25). When we speak of a convergent power series we shall assume, unless otherwise indicated, that R > 0. Similar remarks hold for a power series of the form ( 9 ) , where x is replaced by (x- a).
230
INFINITE SERIES
[CHAP. 11
THEOREMS on POWER SERIES Theorem 9. A power series converges uniformly and absolutely in any interval which lies entirely within its interval of convergence. Theorem 10. A power series can be differentiated or integrated term by term over any interval lying entirely within the interval of convergence. Also, the sum of a convergent power series is continuous in any interval lying entirely within its interval of convergence. This follows a t once from Theorem 9 and the theorems on uniformly convergent series on Pages 228 and 229. The results can be extended to include endpoints of the interval of convergence by the following theorems. Theorem 11. A bel's theorem. When a power series converges up to and including an endpoint of its interval of convergence, the interval of uniform convergence also extends so f a r as to include this endpoint. See Problem 42. Theorem 12. Abel's limit theorem. If
2 anxn converges a t x = xo which may be an interior point or an endpoint of
n=O
the interval of convergence, then
If xo is an endpoint we must use x-, XO+ or x + $0- in (10)according as xo is a left or right hand endpoint. This follows a t once from Theorem 11 and Theorem 6 on the continuity of sums of uniformly convergent series.
OPERATIONS with POWER SERIES In the following theorems we assume that all power series are convergent in some interval. Theorem 13. Two power series can be added or subtracted term by term for each value of x common to their intervals of convergence. Theorem 14. Two power series, for example CnXn where
2
fl=O
+
03
n=O
anxn and
+
03
n=O
bnz", can be multiplied to obtain
+
+
= aobn albn-1 azbn-2 anbo the result being valid for each x within the common interval of convergence. Cn
Theorem 15. If the power series
6
.
.
(14
OD
n=O
anxn is divided by the power series 2 b n P where bo # 0, the
quotient can be written as a power series which converges for sufficiently small values of x.
Theorem 16. If y =
231
INFINITE SERIES
CHAP. 111
2
n=O
anP,
then by substituting x =
9 bnyn
we can obtain the coefficients
n=O
b n in terms of an. This process is often called reversion of series.
EXPANSION of FUNCTIONS in POWER SERIES Suppose that f(x) and its derivatives f'(x),f"(x), . . .,f(")(x) exist and are continuous ) in the open interval a < x < b. in the closed interval a S x 5 b and that f ( n + l ) ( ~exists Then as we have seen in previous chapters (see Pages 61 and 95) where R,, the remainder, is given in either of the forms (Lagrange's form) R, = f ( n + ') (Is) (x - a)n+1 (n 1)!
+
Rn
=
f'n
'n ! (')
(x - 5)" (x - a)
(Cauchy's form)
(14)
where 5, which lies between a and x, is in general different in the two forms. As n changes, 5 also changes in general. If for all x and 5 in [a,b] we have lim Rn = 0, n4 then (12) can be written 00
which is called the Taylor series or expunsion of f(x). In case a = 0 , it is often called the Maclaurin series o r expunsion of f(x). For problems involving such expansions see also Chapter 6. One might be tempted to believe that if all derivatives of f(x) exist at x = a , the expansion (15) would be valid. This however is not necessarily the case, for although one can then formally obtain the series on the right of (l5),the resulting series may not converge to f(x). For an example of this see Problem 104. Precise conditions under which the series converges to f(x) are best obtained by means of the theory of functions of a complex variable. See Chapter 17. SOME IMPORTANT POWER SERIES The following series, convergent to the given function in the indicated intervals, are frequently employed in practice. x2n- 1 x7 + . , .( -1)-1 - x - - +x3- - x5 + ... -oo<x
x2
x3
+ x - z+ x7 + -+ 7 +
= x -2 x3 x5 l+x 5. i I n l z l = x +K 4.
x4
InIl+xl
6,
tan-lx
= x -
7.
(l+X)P
= 1
23
-
x5 - 7 x7 + +5
...(-1)n-I-
Xn
n
+ ...
+2n-1 + X2n-1
a..
x2n
...(-l)n-I-
+ px + *(*-l)x2 2! +
-1
2n-1
-1<xSl
...
-l<x
+ ...
-1SxSl
... + p ( p - - l ) .n. .!( p - n + l )
3"
+ ...
INFINITE SERIES
232
[CHAP. 11
This is the binomial series. (a) If p is a positive integer or zero, the series terminates.
(b) If p > 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
and
1 S 2nS- 1
n=l
Inn 2ns- 1
1 we have na'
*
5
2n'- 1 - ns
=1
n8'
" 1 Then the given series converges, since 2 - converges. u p 1 nr
a
Let
U,,
and vn be positive. If
un = constant A # 0, lirn -
n+m
prove that Bun converges
Vn
or diverges according as B V n converges or diverges. By hypothesis, given c > 0 we can choose an integer N such that - A < c for all n > N. Then for n = N + l , N + 2 , ... -c < -urn A < c or ( A - c)vn < un < ( A + €)U,, (2) Urn
I? I
Summing from N + 1 to
~1
(more precisely from N + l to M and then letting M ( A-e)
2 v,,
N+1
s
5 w,,
N+1
s
( A +e)
5
N+1
vn
-+
a),
236
INFINITE SERIES
[CHAP. 11
There is no loss in generality in assuming A - c > 0. Then from the right hand inequality of (Z), Bun converges when Bv, does. From the left hand inequality of (a), 2% diverges when HV, does. For the cases A = 0 or A = ~ 1 ,see Problem 62.
9. Test for convergence:
+ 2n
( a ) For large n, 4 n * - n + 3
ns
sl
4n2-n+3 n3 2n
(a)
Inn
cQ
+
4 4n'=-. ns n
is approximately
Taking
U,,=
4n'-n'3
+ 2n
ns
and v n = -4 n'
Un
lim - = 1. vn Since Hvn = 4H l l n diverges, Bun also diverges by Problem 8.
we have
n 4 m
Note that the purpose of considering the behavior of un for large n is to obtain an appropriate comparison series vn. In the above we could just as well have taken v n = l / n . Another method :
(b)
lim n (
n 4 m
For large n, un = n
Then by Theorem 1, Page 225, the series diverges.
fi is approximately
vn = n -- 1
2ns- 1
Since converges.
+3
lim n s / S ( L ) 4
m
n'
5
=
=
lim ns(&T) U
U
2 vn
lim 3 = 1 and vn
n 4 w
Another method:
(c)
+
4n'Fp2') = 4.
40
2ns
1 1 - 22 n'
1 5.
2n"
converges (p series with p = 2), the given series
Then by Theorem 1, Page 226, the series converges. In n
lim n*/*
(by L'Hospital's rule or otherwise).
n+m
Then by Theorem 1 with p = 3l2, the series converges. n, n = --, 1 but nothing can In n be deduced Note that the method of Problem 6(a) yields n"+3 < since 2 l l n diverges.
10. Examine for convergence: (a) (a)
lim n'e-"
m+ m
2 e-"',
n=1
(b)
n=l
sin3(
k).
= 0 (by L'Hospital's rule or otherwise). Then by Theorem 1 with p = 2 , the series
converges. ( b ) For large n, sin ( l l n ) is approximately l l n . This leads to consideration of
from which we deduce, by Theorem 1 with p = 3, that the given series converges.
INTEGRAL TEST 11. Establish the integral test (see Page 225). We perform the proof taking N = 1. Modifications are easily made if N
> 1.
From the monotonicity of f ( x ) , we have un+l = f ( n + 1 ) S
Integrating from x = n to x = n
f ( x ) 5 f(n) = un
+ 1, using Property 7 , Page 81,
S.
n+l
un+l 5
f(x)dx 5
un
n = 1,2,3,
n = 1 , 2 , 3 , .. .
...
237
INFINITE SERIES
CHAP. 111 Summing from n = 1 to M
- 1,
UZ+U3+
. * *
+UM
5 l'f(x)dx
5
UI
+ uz + +
uM-1
If f ( x ) is strictly decreasing, the equality signs in (I) can be omitted. If
exists and is equal to S, we see from the left hand inequality in ( I ) that
lim J " f ( x ) d x
M403
+ U S+ - - .+ U,
uz
If lim M-tm
is monotonic increasing and bounded above by S, so that 2% converges.
f f ( x )d x
is unbounded, we see from the right hand inequality in (I) that Bu, diverges.
Thus the proof is complete.
12. Illustrate geometrically the proof in
Problem 11.
+ +- +
Geometrically, uz u3 - U, is the total area of the rectangles shown shaded in Fig. 11-1, while u1+ 242 -. u,-~ is the total area of the rectangles which are shaded and non-shaded.
+- +
The area under the curve g = f ( x ) from x = 1 to x = M is intermediate in value between the two areas given above, thus illustrating the result ( 1 ) of Problem 11.
iM f f
l-
x l - ~ M M1-P-1 x-pdx = - where p # 1. l-P1 1-P M1-P - 1 If p < 1, lim ____ = 00, so that the integral and thus the series diverges. M+CQ 1-$3
(a) Consider
=
If p > 1, lim
M e w
If p = 1,
~
~
M1-P - 1 - - so that the integral and thus the series converges. 1-p p-1'
lMf 1$ =
M
= In M and lim 1nM = M 4 m
00,
so that the integral and thus the
series diverges. Thus the series converges if p >.land diverges if p d 1.
M
dx
= lim In (In x ) ] : = M-+w
{In (In M)
- In (In 2)) =
00
and the series diverges.
Note that when the series converges, the value of the corresponding integral is not (in general) the same as the sum of the series. However, the approximate sum of a series can often be obtained quite accurately by using integrals. See Problem 70.
238
[CHAP. 11
INFINITE SERIES
1 17r < n "= 1 < -+ -. n2+1 2 4
14. Prove that
From Problem 11 it follows that
i.e.
3 na+ 1 <1 . 4. < 2-n a + 1 '
,,=%
Since
1 2" < na + 1
z,
from which 4 4
5 na+l
<
& to each
we obtain, on adding
as required. 1 a 5na+l <-+-. 2 4
side,
nri
The required result is therefore proved.
ALTERNATING SERIES 15. Given the alternating series a1 - a2 a3 - a4 . . . where 0 5 &+ 1 5 a, and where lim a,, = 0. Prove that (a) the series converges, (b) the error made in stopping a t
+
+
n-t o
any term is not greater than the absolute value of the next term. (a) The sum of the series to 2M terms is
s,, = =
(a1 -as)
+ (us - a 4 + + (uaM-l- uaM) - * .
- ( c c ~ - u -~ )( & - U S ) -
a1
- (uSM-~
. * *
- arm
- agM--)
Since the quantities in parentheses are non-negative, we have
2 0,
SaM
S S
s 4
5 S S d SR 5
*. *
5 SZM 5
a1
Therefore { S S Mis } a bounded monotonic increasing sequence and thus has limit S.
+
Also, Szm+i = SUC aaM+l.Since lirn SUC= S and lim lim
n-t oo
U,,=
M-too
0), it follows that
?loo s%Y+1=
lim Sau Me00
+ lim
M+oo
= 0 (for, by hypothesis,
U,,+,
= S
00
+0
= S.
Thus the partial sums of the series approach the limit S and the series converges. ( b ) The error made in stopping after 2M terms is (aSM+l
- a2M+S) +
(%M+3
- %M+4) +
and is thus non-negative and less than or equal to
=
aSM+l,
Similarly, the error made in stopping after 2M -a%M+S
+ (aSM+9 -
u3M+4)
+
*"
=
which is non-positive and greater than -a-
16. (a) Prove that the series
-(%M+L
n=l
(-1)n+l
- (%M+S - % M + 3 ) -
" *
the first term which is omitted.
+ 1 terms is - aSM+3)
-
(%+4
- a2M+s) -
* * *
+ a.
2converges. 2n- 1 m
a!lM+l
(b) Find the maximum error made in
approximating the sum by the first 8 terms and the first 9 terms of the series. ( c ) How many terms of the series are needed in order to obtain an error which does not exceed ,001in absolute value? 1 1 1 = - Since -I- and since lim -= 0, it follows by Problem 16(a) 2n 1 . 2n+1 2n-1 2n-1 that the series converges. IUn+l(
+
CHAP. 111
239
INFINITE SERIES
( b ) Use the results of Problem 16(b). Then the first 8 terms give 1 and the error is positive and does not exceed &.
-Q +Q -3+Q -
+ + + fr +
+
+&-
+
and the error is Similarly, the first 9 terms are 1 - Q Qf3 - fs negative and greater than or equal to -h, i.e. the error does not exceed fg in absolute value.
(c) The absolute value of the error made in stopping after M terms is less than 1/(2M+1). To obtain the desired accuracy we must have 1/(2M+1) 5 .001, from which M1499.5. Thus a t least 500 terms are needed.
ABSOLUTE and CONDITIONAL CONVERGENCE 17. Prove that an absolutely convergent series is convergent. Given that 2
IZG.~
Let S M = U I+US
converges, we must show that Bun converges.
+ ... + and TM = lull + + ... + /U,]. Then + + ].PI) + + (U, + I u , ~ ) SM + TM = (UI+ + 2lu2l + + 2 1 u M l 1 ~ 2 1
U,
IUtl)
(UZ
2lUll
'*'
Since B Iu-1 converges and since U,+ lunl 2 0, for n = 1,2,3,
+
.. .,
+
it follows that SM T M is a
bounded monotonic increasing sequence, and so lim (SM TM)exists. M+W
Also, since lim TM exists (since the series is absolutely convergent by hypothesis), m+m
must also exist and the result is proved.
18. Investigate the convergence of the series
sinfi
sinfi sin fi -231'2 + 33/2- . . . *
Since each term is in absolute value less than or equal to the corresponding term of the series 1 1 1 ..., which converges, it follows that the given series is absolutely convergent and 1312 2312 3312
+
+ -+
hence convergent by Problem 17.
19. Examine for convergence and absolute convergence:
n which is divergent by Problem 13(b). Hence the given 2O=7 (a) The series of absolute values is % i n+ 1 series is not absolutely convergent. However, if a, =
n lu-l = and ne + 1
anti
n all n 2 1, and also lim a, = lim 7 = 0. n 4w n 4 w n + 1
=
Iun+11
= (n+ n 1)2 +
+
,
then
Un+l5an
for
Hence by Problem 16 the series converges.
Since the series converges but is not absolutely convergent, it is conditionally convergent. (b)
The series of absolute values is
nz2m. "
1
M
By the integral test, this series converges or diverges according as or does not exist.
dx
exists
[CHAP. 11
INFINITE SERIES
240 If
U
= lnx,
J*
=
J$ = - - +1c
= --+c.1
In x
U
dx
and the integral exists. Thus the
series converges. Then
0O
2 ,,=a
(-1)n-l
-converges absolutely and thus converges. n ln'n
Another method:
1 1 = 0, it follows, by Prob. 16(a), that Since ( n + 1) Ins (n 1) <= n lnsn and lim n InZn the given alternating series converges. To examine its absolute convergence we must proceed as above.
+
(c)
Since limu, # 0 where us = n+ 00
(-'La
n-1
lim 1u.l = lirn
that lim un# 0, it suffices to show that 1 -
the given series cannot be convergent.
2",
I-00 )
00
2"
n
# 0.
To show
This can be accomplished
by L'Hospital's rule or other methods [see Prob. 21(b)].
RATIO TEST 20. Establish the ratio test for convergence. Consider first the series
if
+ + U S+ -..
UI
u 2
where each term is non-negative. We must prove that
lim - - L < 1, then necessarily Bun converges. un+t
1-00
Un
By hypothesis, we can choose an integer N so large that for all n 2 N, L < r < l . Then UN+l
etc. By addition, U,+,
< < <
+ U,+, +
ruN rUN-kl ruN-k2
... <
<
Similarly we can prove that if
uN(r+r'+rS+ .-.)
I I
lirn - = L un Un+i
< 1.
Since 0 < 1, the series converges.
< 1,
+ IGS/+
l u ~ l +... . Then by the
then Eu, converges (absolutely).
= L > 1 the series Pu, diverges, while if
= L = 1 the ratio test fails [see Prob. 21(c)].
21. Investigate the convergence (a) Here un = n4e-na. Then
where
rSuN
In case the series has terms with mixed signs, we consider lull n+w
r
<
and so the given series converges by the comparison test, since 0 C r
above proof and Problem 17, it follows that if
<
(U,+ ,/U,,)
CHAP. 111
241
INFINITE SERIES
Since 2 > 1, the series diverges. Compare Prob. 19(c).
(c)
Here
U,,
(-l)n-ln . Then = n' 1
+
(n+1)(nS+1) = 1 ,+= (n' 2n 2)n
+ +
and the ratio test fails. By using other tests [see Prob. 19(a)],the series is seen to be convergent.
MISCELLANEOUS TESTS 22. Test for convergence 1 + 2r (c) . . r = 4/3.
+ + + 2rS + r"'+
2++
Iu~' I
...
Here the ratio test is inapplicable, since - = 2lrl or even. However, using the nth root test, we have
7m
m = {rn Then lim
n 4 00
Thus if Irl
= Irl
= fi Irl = I4
where
91rl
(U)
r = 2/3, ( b ) r = -2/3,
depending on whether n is odd or
if n is odd if n is even
(since lirn 2l" = 1). n-tm
< 1 the series converges, and if
Irl
>1
the series diverges.
Hence the series converges for cases (a)and (b), and diverges in case (c).
23. Test for convergence
3.6
The ratio t a t fails since lim n ( 1 0-
00
-
1 ~ 1 )=
+("""J+ 3.609 lim ("">' (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
CQ.
This limit is infinite if x # a. Then the series converges only for z = a.
Thus the series converges for /x- 11 The test fails for
1% - 11 = 2,
For x = 3 the series becomes zero.
<
1 2 and diverges for Is - 1
i.e. x
" n 3 3n~
,,=I
For x = -1 the series becomes approach zero.
O0
,,=I
Then the series converges only for
- 1 = +2 1
(-'3nInn
>
2.
or x = 3 and z = -1.
which diverges since the nth term does not approach
____
1
which also diverges since the nth term does not
1% - 11 < 2,
i.e. -2
< x -1 <
2 or -1
< z < 3.
243
INFINITE SERIES
CHAP. 111
26. For what values of
x does
(97.
U"==
(U) 1
Then
5-
*I
Then the aeries converges if
<
Ix-1
i f - x + 2 = I, i.e. x = -4.
I
fi > 1,
1, diverges if
12-1
and the test fails
12-11
If x = 1 the series diverges. If x = - 2 the series converges. If x = -3 the series is
OD
,=I
-which
Thus the series converges for
'+' 1, x=-+
-c
1 (x+n)(x+n-1) we see that if x # 0,-1,-2,. .,-n, U1
4- us 4-
*.*
+
.
U1
=
and x=-2,
12-11
= 1, where
The ratio test fails since that
s, =
converges.
2n - 1
-
U,
i.e. for x ~ - +
1 = (x + n)(x n - 1)' However, noting
+
--
1
x+n-1
1
x+n
1 1 (:---A)+(--a) +
. * *
+
1 (x+n-l
-2) x+n
= -1 - - 1 x
and lim S, = 1/x, provided x *4m
x+n
.. .
# 0,-1,-2,-3!
Then the series converges for all x except x = 0,-1,-2,-3,
UNIFORM CONVERGENCE 27. Find the domain of convergence of (1- x ) + x(1- x) Method 1: Sum of first n terms = Sn(Z) = (1 - X) ~ ( 1 -X) = 1- x + x - 2 8 +
+
.. .,
and its sum is 1/x.
+ x2(1 - x) + ... .
+ ~ ' ( 1 -X) + 2:
= 1-xx"
+ xs-'(l-
- '5 + ... + xn-1 - 2'
X)
If 1x1 < 1, lirn S,(x)= lim (1-xn) = 1. .+a0
If 1x1 > 1, lirn
a+ m
Sn(X)
does not exist.
If x = 1, S,(x)= 0 and lirn S,(x)= 0. U 4 a0
If x = -1, S,(x)= 1 - (-1)" and lirn S,(x)does not exist. 0 4 O D
Thus the series converges for 1x1 < 1 and x = 1, i.e. for -1 Method 2, using the ratio test. The series converges if x = 1. If x # 1 and
U,, =
< x S 1.
x"-' (1 - x), then
1 I
lirn - = lim 1x1. U,+'
*400
0 3 Q
Thus the series converges if 1x1 < 1,diverges if 1x1 > 1. The test fails if 1x1 = 1. If x = 1 the series converges; if x = -1 the series diverges. Then the series converges for -1 < x S 1.
244
INFINITE SERIES
[CHAP.11
28. Investigate the uniform convergence of the series of Problem 27 in the interval -4 < x < 3, (b) -3 5 z S 9, ( c ) -.99 5 x 5.99, (d) -1 < x < 1, (e) 0 S x <2.
-*
By Problem 27, S.(Z) = 1- z", S(z) = lim Sn(x) = 1 if n+m in this interval. We have Remainder after n terms on
e,
< x < i; thus the series converges
= R,(x) = S ( x ) - S,(x) = 1 - (1-2") = xn
The series is uniformly convergeitt in the interval if given any but not on z,such that IRI(x)I < c for all n > N. Now
IRn(z)l = 1x.l
= lzl" <
e
when
n l n 1x1
< In€
o
> 0 we can find N
or n
dependent
In t
>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
0
In Problem 34 it is shown that if a series is uniformly convergent in an interval, the sum function S(z) must be continuous in the interval. It follows that if the sum function is not continuous in an interval, the series cannot be uniformly convergent. This fact is often used to demonstrate the non-uniform convergence of a series (or sequence).
30. Investigate the uniform convergence of x2
+
X2
X2 + (m + . . . + (1+ x2)n + . . ' .
Suppose z # 0. Then the series is a geometric series with ratio 1/(1+x') Prob. 26, Chap. 3). -1 ic = l+zz S(x) = 1 - 1/(1+x*)
X2
whose sum is (see
CHAP. 111
245
INFINITE SERIES
If x = 0 the sum of the first n terms is Sn(0) = 0; hence S(0) = lim Sn(0) = 0. U-,
lim S(x) = 1 # S(O), S(x) is discontinuous at x=O.
Since
t-+ 0
03
Then by Problem 34, the series
cannot be uniformly convergeiit in any interval which includes x = 0, although it is (absolutely) convergent in any interval. However, i t is uniformly convergent in any interval which excludes x = 0. This can also be shown directly (see Problem 89).
WEIERSTRASS M TEST 31. Prove the Weierstrass M test, i.e. if lUn(x)l S Mn, n = 1'2'3, . . ., where Mn are positive constants such that BMn converges, then Bun(x) is uniformly (and absolutely) convergent.
+
The remainder of the series Bu,(x) after n terms is R,(x) = un+l(x) u,+z(x)
+ .
Now
+
But Mn+1+ Mn+3 - - . can be made less than c by choosing n > 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 <
r/3
whenever lhl
<6
Since the series, by hypothesis, is uniformly convergent, we can choose
IRn(z)I
for lhl
< 6,
<
43
I Rn(z + h) I <
and
c/3
(2)
N
so that
for n > N
and so the continuity is established.
35. Prove Theorem 7, Page 229. If a function is continuous in [a,b ] , its integral exists. Then since S(z), S,(z) and Rn(z) are continuous,
lb
S(z)
=
lb
Sn(z)dz
+
lb
Rm(z) dz
To prove the theorem we must show that
can be made arbitrarily small by choosing n large enough. This, however, follows at once, since by the uniform convergence of the series we can make 1Rn(z)1 < e / ( b - a) for n > N independent of z
241
INFINITE SERIES
CHAP. 111
36. Prove Theorem 8, Page 229. Let g ( x ) =
n=l
u:(x).
Since, by hypothesis, this series converges uniformly in [ a , b ] , we can
integrate term by term (by Problem 36) to obtain
i'
because, by hypothesis,
2 ~ ' u : ( zd x)
=
g(2)dx
5 u n ( x ) converges to S ( x ) in
g ( x ) d x = S ( x ) - S(u) then shows that g(z) = S'(x), which
Differentiating both sides of proves the theorem.
37. Let S,(x)= nxe-nz', n = 1,2,3, . . .) 0 S x 5 1. lim
n-a
3 { u n ( z ) - un(a))
U==l
[U,b ] .
n=l
(a) Determine whether
=
1=1
fSn(x)dx
=
0
i1
lirn Sn(x) dx.
n 4 m
(b) Explain the result in (a). (a) i l S u ( z ) d x
= xlnxe-Jdx
= -+e-nrP(t = * ( l - e - " ) .
Then
S ( x ) = lim S , ( x ) = lim nze-& = 0, whether x = O or O < z S l . *-+a
U-+ 00
i'S(z)dz It follows that the integral sign.
lirn
*+a
l*
S,(x) dx #
Then,
= 0
1'
lim Sn(z) d z , i.e. the limit cannot be taken under
n-+w
( b ) The reason for the result in (a) is that although the sequence Sn(z) converges to 0, it does not converge uniformlg to 0. To show this, observe that the function nxe-& has a maximum at x = l/& (by the usual rules of elementary calculus), the value of this maximum being a n e - l p . Hence arr n 3 0 0 , Sn(x) cannot be made arbitrarily small for a2Z z and so cannot converge uniformly to 0 .
38. Let f(x) =
nx . 2 sin n3
Prove that
,,=1
x7
f ( x )dx = 2
Then by the Weierstrass all x , in particular 0 5 x S
7, and
gl(zn1
1)4
M test the series is uniformly convergent for
can be integrated term by term. Thus
248
INFINITE SERIES
[CHAP. 11
POWER SERIES 39. Prove that both the power series
01
and the corresponding series of derivatives
n=O
9 nu,,xn-l have the same radius of convergence.
n=O
Let R > O be the radius of convergence of CCLX". Let O < 1x01 < R .
'
Then, as in Problem 33,
we can choose N so that Janl < - for n > N .
Id*
Thus the terms of the series
B (nanxn-'I = 2 n ICG.~ IxI"-'
corresponding terms of the series 2 n-'-"zl
Ixol"
can for n > N be made less than
which converges, by the ratio test, for
Hence Pnu,,x"-' converges absolutely for all points xo (no matter how cbse If, however, converge.
1x1 > R,
lim
an%"
P 0 and thus
(1400
IXO~
151
< 1x01 < R .
is to R).
lim nanx"-' # 0, so that Bnanx"-' does not
n-* m
Thus R is the radius of convergence of B naax"-'. Note t h a t the series of derivatives may or may not converge for values of x such that 1x1 = R.
40. Illustrate Problem 39 by using the series
*
x" -
n=ln2*3n'
so that the series converges for 1x1 < 3. A t x = +3 the series also converges, so that the interval of convergence is -3 5 x 5 3.
The series of derivatives is
By Problem 26(a) this has the interval of convergence -3 5 x < 3. The two series have the same radius of convergence, i.e. R = 3 , although they do not have the same interval of convergence. Note t h a t the result of Problem 39 can also be proved by the ratio test if this test is applicable. The proof given there, however, applies even when the test is not applicable, a s in the series of Problem 22.
41. Prove that in any interval within its interval of convergence a power series
a continuous function, say f ( z ) , (b) can be integrated term by term to yield the integral of f ( x ) , ( c ) can be differentiated term by term to yield the derivative of f ( x ) . (U) represents
We consider the power series Ba,x", although analogous results hold for Ban(x - a)". (a) This follows from Problems 33, 34 and the fact that each term an%"of the series is continuous.
( b ) This follows from Problems 33, 36 and the fact that each term an%"of the series is continuous and thus integrable. (c)
From Problem 39, the series of derivatives of a power series always converges within the interval of convergence of the original power series and therefore is uniformly convergent within this interval. Thus the required result follows from Problems 33 and 36.
If a power series converges a t one (or both) end points of the interval of convergence, i t is possible to establish (a)and (b) to include the end point (or end points). See Problem 42.
CHAP. 111
249
INFINITE SERIES
42. Prove Abel's theorem that if a power series converges at an end point of its interval
of convergence, then the interval of uniform convergence includes this end point. 00
For simplicity in the proof, we assume the power series to be
2
with the end point of
k=O
its interval of convergence at x = 1, so that the series surely converges for 0 5 z 4 1. Then we must show that the series converges uniformly in this interval. Let Rn = an U ~ + I U ~ + S Rn(x) = auxn an+1zn+l u , + ~ x * + * *",
+
+
+
+
To prove the required result we must show that given any
c
+
+
> 0, we can find N such that IRn(X)l < c
for all n > N, where N is independent of the particular z in 0 d z d 1. Now
+
+
(Rm+r - R n + ~ ) x ~ + ' * * . Rn(x) = (Rn - Rn+l)P (Rn+1- Rn+S)z*+' i= Ruzn Rn+1(zn+'- 2.) R , + S ( X ~+ ~xn+') = z"{Rn - (l--)(Rn+l R n + t Z Rn+Jx% * * * ) } Hence for 0 S x < 1,
+
lRn(z)I 5 lRkl
Since
2 a k
< ~ / 2 for
+
+
l ~ n l
+
(l-X)(lRn+11
+
+
+
+ IRn+sIs + IRn+s(z' +
converges by hypothesis, it follows that given all k B n . Then for n > 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<
x5 -+-+ x7 . . . 6
7
COB 1c
= sin 22.
00.
-l<x
Prove that the formal Taylor series about x = 0 corresponding to f ( z ) x=o exists but that it does not converge to the given function for any x Z 0 . [Hint See Problem 38, Chapter 4.1
104. Let
257
INFINITE SERIES
CHAP. 111 105. Prove that
-
l+x) (4 l n1( + x
for -1
<x < 1
MISCELLANEOUS PROBLEMS 106. Prove that the series for Jp(x) converges (a) for all x , (b) absolutely and uniformly in any finite interval. d (a) ~ { J O ( Z )=}
107. Prove that (c)
JptI(x)
- JI(s),
(b)
d
=
{xPJp(x))
zPJp-l
(x),
- Jp-l(x).
= %Jp(x)
108. Assuming that the result of Problem 107(c) holds for p = 0, -1, -2, (U)
J-i(x) =
109. Prove that
- Jl(x),
eHz(t-lIL)
( b ) J - s ( x ) = J s ( x ) , (c) J-,(x) = (-l)rJn(x),
=
5
p = -02
[Hint: Write the left side as 110. Prove that
circle lzl = 1, 111. (a) If
m
n=l
. . ., prove that n
e"'Is e-*Int,
expand and use Problem 108.1
x" (n+l 1)z" m is absolutely and uniformly convergent m
2 b,x" r=l
-I
at all points within and on the
for all x in the common interval of convergence 1x1 < R where R > 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
132. Prove that 133
U,,),
n=l
OD
n=l
'>
places and compare with the true value.
..
134 Prove that the series 1 + 0 - 1 + 1 + 0 - 1 + 1 + 0 - 1 +
is C - 1
summable to zero.
135 Prove that the Cdsaro method of summability is regular. [Hint: See Prob. 28, Chap. 3.1
CHAP. 111
INFINITE SERIES
+ + 3x* + 4sa+ . + mn-'+ - -
136. Prove that the series 1 22 137. A series
ca
00
(a) 2 (-l)"(n n=O
5
(b) n=O
+ 1) +
2
x m y d u
(b) Use (a) to prove that
141. If
0 0 0 0
(ms+ ,p)p
-.n - 1
* anx"" *=z
for 1x1 < 1.
00
lim
z-1-
2 anx"
exists. Prove that
n=O
~
where p is a constant, converges or diverges according
1 +2! 2' - -1 - 2 2 - 3! c
a
-
_ _ -d
~1 x
f(x) has an asymptotic expansion given by
expansion
5)'
l)(n i-2, is Abel summable to 1/8.
> 1or p S 1respectively.
139. (a) Prove that
converges to 1/(1-
is Abel summable to 1/4 and
138. Prove that the double series as p
if S =
2 a, is called Abel sumrnable to S
,=O
269
2'
+
... ( - 1 ) n - I
2" (n
- 1) ! +
+x-2+ ... ~ xs 2'
acr 2-", prove that ,=L
(-l)"n! l c a s d u .
J
r
f(z)
dx has the asymptotic
Chapter
I2
Improper Integrals DEFINITION of an IMPROPER INTEGRAL The integral
ib
f ( x ) d x is called an improper integral if
(1) a = -QO or b = QO or both, i.e. one or both integration limits is infinite, (2) f ( x ) is unbounded at one or more points of a 5 x 5 b. Such points are called singutarities of f(z). Integrals corresponding to (1)and ( 2 ) are called improper integrals of the first and second kinds respectively. Integrals with both conditions (1) and ( 2 ) are called improper integrals of the third kind. Example 1: Example 2:
1i4
sin x' dx is an improper integral of the first kind. is an improper integral of the second kind.
00
ZcZz is an improper integral of the third kind.
Example 3: Example 4:
l1
%dz
sin x is a proper integral since lim -- 1. ++o+ x
In this chapter we formulate tests for convergence or divergence of improper integrals. It will be found that such tests and proofs of theorems bear close analogy to convergence and divergence tests and corresponding theorems for infinite series (see Chapter 11).
IMPROPER INTEGRALS of the FIRST KIND Let f ( x ) be bounded and integrable in every finite interval a d x 5 b. Then we define
The integral on the left is called convergent or divergent according as the limit on the right does or does not exist. series
W
U,, n=l
where
U,,
lm lb
Note that
= f ( n ) , while
f ( z ) dx bears close analogy to the infinite
f ( x ) cix corresponds to the partial sums of such
infinite series. We often write M in place of b in ( I ) . Similarly, we define J:f(s)dz
=
lim i b f ( x ) d s
a+--
260
261
IMPROPER INTEGRALS
CHAP.121
and call the integral on the left convergent or divergent according as the limit on the right does or does not exist. Example 1:
xme = lim i bx' g = XS
s-:
b+m
f
Example 2: s:mcos x dx = l i m m exist,
lim
b-toc
(1- k)
cos x dx
= 1 so thatJ*$
converges to 1.
lim -00 (sin U - sin a). Since this limit does not
=
a+
cos x dx is divergent.
In like manner, we define
where zo is a real number, and call the integral convergent or divergent according as the integrals on the right converge or not as in definitions ( I ) and (2).
SPECIAL IMPROPER INTEGRALS of the FIRST KIND
1. Geometric or exponential integral x m e - t z d z , where t is a constant, converges if t > 0 and diverges if t S 0. Note the analogy with the geometric series if r = e - f so that e-tz = p. 2. The p integral of the first kind
lm$,
where p is a constant and a > 0, con-
verges if p > l and diverges if p S 1 . Compare with the p series.
CONVERGENCE TESTS for IMPROPER INTEGRALS of the FIRST KIND The following tests are given for cases where an integration limit is a. Similar tests exist where an integration limit is -a (a change of variable x = - g then makes the integration limit a). Unless otherwise specified we shall assume that f ( x ) is continuous and thus integrable in every finite interval a S x Ib. 1. Comparison test for integrals with non-negative integrands. (a) Convergence. Let g ( x ) Z O for all xZa, and suppose that
verges.
Then if O S f ( x ) S g ( x ) for all x z a ,
Example: Since
1 1 = e'" e"+ 1 - ez
and
e-z dx converges,
( b ) Divergence. Let g ( x ) 2 0 for all X Z U ,and suppose that
1 Example: Since - > 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
<
[a1,a2]
if for each c > 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 <
when z 1 N. Thus for z 2 N, we have A -c d f (2) d A g(z)
Then
( A - C)
+
or
c
g ( 2 ) dx
( A - E ) g(z) S f ( z ) S ( A
lb f(z)dz
S
There is no loss of generality in choosing A - c If
J r g(z) n dz
(A
+
e)
lb g(z) d z
0.
converges, then by the inequality on the right of (1), lim
b-
If
>
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 =
=
$(a)
lim - e - * * b-t
m
L0
-
lim 1 - e - a b = 1 if a > 0 . b+*
Thus the integral converges t o 1 for all a > 0. Method 1, using definition.
I lu
The integral converges uniformly to 1 in a 2 a l > 0 if for each r > O we can find N, depending on
but not on a, such t h a t
z
11
Since
-
lU
ae+Zdz
I
l -
ae-azdxl
for all U > N.
= 11 - (1 - e - f f u )I = e-*" 5 e-alu
<
e
for U
>1 In 1 = N, a1
the result follows. Method 2, using the Weierstrass M test. Since lim xp ae-"" = 0 for a 2 a, > 0, we can choose lae-azl t+ 00
say
2: 2x0.
1
Taking M ( z ) = - and noting t h a t XP
integral is uniformly convergent to 1 for a 2 a, > 0.
1;$
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) <{cosax)
=
lim
1 - e-(r-a)M 8-a
M - , ~
lrn e - 8 r c o s a x dx
=
8
8'
-k U'
-
-
8-U
ifs>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 <{sin ax}
8 ~ { c o s a x }= x{sinax} = s2 as' 8' U' '
+
The above fomza2 method can be justified using methods of Chapter 17.
U
+
280
IMPROPER INTEGRALS
[CHAP. 12
35. Prove that (a) x { Y ’ ( z ) }= soC{Y(z)}- Y(o), ( b ) ~ { Y ” ( x )=} s2-C‘{y(~)} - S y ( 0 ) - y’(0)
under suitable conditions on Y ( x ) .
(a) By definition,
4{Y’(z)} =
=
im
=
lim {e-aaY(x)lr
+
e-’” Y’(x)dx
H*oo
lim
f e-’= Y’(z)d x
81M } M*m
e-aZY(z)dz
= 8 ~ w e - a z Y ( Z ) dX Y(0) = 84{Y(X)}- Y ( 0 ) assuming that
8
is such that lim e - ’ * Y ( M ) = 0. M400
(b)
Let U ( x ) = Y ’ ( z ) . Then by part (a), 4{ U’(x)} = 8 ~ { U ( X ) } U(0). Thus 84{Y‘(S)}- Y‘(0) =
4{Y”(Z)} =
8[84{Y(Z)} - Y(o)]- Y’(0)
= 8’4{Y(Z)}- 8 Y(0)- Y’(0)
36. Solve the differential equation
+
Y”(x) Y ( x ) = x, Y(0)= 0, Y’(0)= 2.
Take the Laplace transform of both sides of the given differential equation. Then by Problem 36,
+
and
4c(Y”(x)+ Y ( z ) > = 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)<{l/G), (b) K{cosh ax}, (c) <{(sin x ) l x } .
A m * (4
m,
(b)
8 > 0
&,
8
> lal
(c) tan"
76. (a)If < { F ( x ) } = f ( 8 ) , prove that <{e"F(s)}
( b ) (8 77. (a) If
and use the fact
diverges.
y+flx*'dx
73. (a) Prove that
dx
+
= f ( 8 - a), ( b ) Evaluate
z{e" sin
bs}.
8>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
<-I
+
Am. (b)
(m, "-'
{ f ( s ) g ( s ) } = 4'' ( f ( s ) }
8 > 0
+ 4"
{g(8)},
stating any restrictions.
79. Solve using Laplace transforms, the following differential equations subject to the given conditions.
+
+
+
+
(a) Y"(x) 3Y'(x) 2Y(z) = 0 ; Y ( 0 )= 3, Y'(0)= 0 ( b ) Y"(x)- Y'(x) = x ; Y(0)= 2, Y'(0)= -3 (c) Y"(x) 2Y'(x) 2 Y ( x ) = 4 ; Y(0)= 0, Y'(0)= 0
Ans. (a) Y ( x ) = 6e-'80.
3e-**, ( b ) Y ( z ) = 4 - 2 e " - a x s - x , ( c ) Y ( x ) = 1 - e-'(sinx
+ cosx)
Prove that < { F ( x ) } exists if F ( x ) is sectionally continuous in every finite interval [ O , b ] where b > 0 and if F ( z ) is of exponential order a s x 4 00, i.e. there exists a constant a such that le-u*F(z)l < P (a constant) for all z > b.
284 81.
If
f(8)
= x { F ( x ) } and g( s ) = X { G ( x ) } , prove t h a t
H(z) =
ir
f(8)
pint:
Write
f(8)g ( 8 )
=
{iM
e-” F ( u )du}{
lim
M-bCQ
g(8)
= - ( { H ( x ) } where
F ( u )G ( x - U)du
is called the convolution of F and G , written F*G.
l’
e-’” G(v) dv
F(u)G(v)du dv 82.
[CHAP. 12
IMPROPER INTEGRALS
and then let
U
1
+ v = t.
(b) Solve Y ” ( z )iY ( x ) = R ( z ) , Y(0)= Y’(0)= 0.
(a) Find 4-l
+J
fi*
(c) Solve the integral equation
Y(x) = x
Y ( u )sin (x - U ) du.
[Hint: Use Problem 81.1
0
Ans. (a)&(sinx - xcosx), (b) Y ( s ) =
R(u) sin (x - U ) du, (c) Y ( z ) = x -t x3/6
83. Let f ( z ) ,g(z) and g’(z) be continuous in every finite interval a d x S b and suppose t h a t g’(x) 4 0.
dx is bounded for all s Z a and lim g(z) = 0. =
(a)Prove that
-
=+0
jr g ’ ( x ) h(s)dx.
(b) Prove that the integral on the right, and hence the integral on the left, is convergent. The result
is t h a t under the given conditions on f ( z ) and g ( x ) , called A bel’s integral test. [ H i n t For (a),consider
lim
b-bm
lb
1
x iw
Use Problem 83 to prove that (a)
85. (a) Given that
evaluate
1
f ( s )g(x) dx after replacing f(x) by h’(z) and integrating by parts.
For (b), first prove that if Ih(z)l < H (a constant), then let b + -.I 84.
f(s)g(z) dx converges and is sometimes
sin x d z and ( b )
sin x2 dx = l * c o s x2 d z =
xa
2
4
i b g ’ ( z )h(z)dx
sin zpdz, p
d H ( g ( u ) - g ( b ) ) ; and then
> 1,
converge.
[see Problems 27 and 68(a), Chapter 131,
+
sin (x’ $) dx dy
(b) Explain why the method of Problem 31 cannot be used to evaluate the multiple integral in (a). Aria. ~ / 4
Chapter 13 Gamma and Beta Functions GAMMA FUNCTION The gamma function denoted by r(n) is defined by
r(n> = l m x n - l e - z d x which is convergent for n > 0 (see Problem 18, Chapter 12). A recurrence formula for the gamma function is r ( n + l ) = nr(n) (2) where r ( l ) = l (see Problem 1). From (Z), r(n) can be determined for all n>O when the values for 15 n < 2 (or any other interval of unit length) are known (see table below). In particular if n is a positive integer, then n = 1,2,3,. . .
r(n+l) = n!
(3)
For this reason r(n) is sometimes called the f a c t o r i d function. r(5) = 4! = 12. Examples: r(2) = l ! = 1, I’(6) = 5 ! = 120, r(3)
2!
It can be shown (Problem 4) that
r(8 = I F
(4)
The recurrence relation ( 2 ) is a difference equation which has (I) as a solution. By taking ( I ) as the definition of r(n) for n>0, we can generalize the gamma function to n
+
See Problem 7, for example. The process is called analytic continuation.
TABLE of VALUES and GRAPH of the GAMMA. FUNCTION n
r(n)
1.00 1.10 1.20 1.30 1.40 1.50 1.60 1.70 1.80 1.90 2.00
1.0000 0.9514 0.9182 0.8975 0.8873 0.8862 0.8935 0.9086 0.9314 0.9618 1.0000
Fig. 13-1
285
286
G A M M A A N D BETA FUNCTIONS
[CHAP.13
ASYMPTOTIC FORMULA for r(n) If n is large, the computational difficulties inherent in a calculation of parent. A useful result in such case is supplied by the relation
r(n) are ap-
+
r(n 1) = .\/2mznn e-n ee/12(n+l) o < e < 1 (6) For most practical purposes the last factor, which is very close to 1 for large n, can be omitted. If n is an integer, we can write
-
-
n! +nnne-n (7) where means “is approximately equal to for large n”. This is sometimes called Stirling’s factorial approximation or asymptotic formula f o r n !
MISCELLANEOUS RESULTS INVOLVING the GAMMA FUNCTION T
r(x)r(l-x) = sin x x In particular if x = &, r(+)= 6.
O<x
(8)
+
222-1 r ( x )r(x 3) = f i r ( 2 x )
(9)
This is called the duplication formula for the gamma function.
The result ( 9 ) is a special case of (10) with m = 2.
2) m
-z,m}
This is an infinite product representation for the gamma function. Euler’s constant (see Problem 49, Chapter 11).
where n(x, k) is sometimes called Gauss’
7
The constant
y
is
function.
{
1 r ( z + 1 ) = e x 2 e - Z I+-+-12x
1 288x2
139
5 1 8 4 0 ~+~* * * }
This is called Stirling’s usgmptotic series for the gamma function. The series in braces is an asymptotic series (see Problem 20).
r’p)
=
J e-zlnx dx 0
= -7
where y is Euler’s constant.
The BETA FUNCTION denoted by B ( m , n ) is defined by B(m,n) =
I’
zm-’ (1- z ) ~ - dx ’
which is convergent for m > 0, n > 0. See Problem 16, Chapter 12.
(14)
CHAP. 131
287
GAMMA AND BETA FUNCTIONS
The beta function is connected with the gamma function according to the relation
See Problem 11. Many integrals can be evaluated in terms of beta or gamma functions. results are
Two useful
valid for m > 0 and n > 0 [see Problems (10) and ( 1 4 and
See Problem 17.
DIRICHLET INTEGRALS If V denotes the closed region in the first octant bounded by the surface
y:(
($ +(iy+
= 1 and the coordinate planes, then if all the constants are positive,
Integrals of this type are called Dirichlet integrals and are often useful in evaluating multiple integrals (see Problem 21).
Solved Problems The GAMMA FUNCTION 1. Prove: (a)r(n+ 1) = nr(n), (U)
r(n+ 1)
= lmxne-rdx f
( b ) r(1) = l = e - . d x Put n = 1,2,3,
=
. ..
lim f e - = d x
n*oo
in r(n
= lim (l-e'Y) Y * W
= 1.
+ 1) = fir(%).Then
r(2) = lr(1) = 1, r(3) = 2r(2) = 2.1 = 21, r(4) = 3r(s)= 3.21 = 81 In general,
r(fi+ 1) = n ! if
n is a positive integer.
288
[CHAP. 13
GAMMA AND BETA FUNCTIONS
2. Evaluate each of the following.
3. Evaluate each integral. (U)
I m s a e - = d x = F(4) = 31
(b)
J m
Let 22 = y. Then the integral becomes
z6e-'*dz.
4. Prove that
= 6
r(Q) = 6. z - ~ "e - = d x .
2
iw
Letting x = U' thk integral becomes
e-U' d u = 2
($) =
6
using Problem 31, Chapter 12
5. Evaluate each integral. (a)
I*
f i e - # dg.
Letting
ys = x ,
the integral becomes
becomes
6. Evaluate
I*
xm e-at" dx
where m,n,a are positive constants.
Letting uxn=y, the integral becomes
7. Evaluate (a) r(-1/2), ( b ) r(-5/2).
-. n+
We use the generalization to negative values defined by r ( n ) = r(n 1)
(b)
289
GAMMA AND BETA FUNCTIONS
CHAP. 131
Letting n = -3/2, Then
8. Prove that
r(-3/2)
=
!k!.!?d
-312
r(-3/2) r(-6/2) = -6/2
J1xm(~nx
*
- -3/2
i, using (a). - 'f 3
-5: 6'
(-l)"n! ) ~ = x (m 1)"+' where n is a positive integer and m > -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
2 r(p) COS ( p d 2 ) ’
We have
fi
Then
-d ~U P~x)W ~ p u P - l e - m c o s x d w d x
-
l
W
(11
- %I&dU where we have reversed the order of integration and used Problem 22, Chapter 12. Letting u s = v in the last integral, we have by Problem 17
ilkq)‘P-l)”
l*&dU
=
dv
=
+l)d2
U
2 sin (p
-
U
2 COSpd2
(9)
Substitution of (a) in (1) yields the required result.
27. Evaluate
JW
cos x2 dx.
Letting z * = y , the integral becomes Problem 26.
f i W - cos d yy
vii
= A( 2 2 r(+) UCOS p/4) = + \ 1 ; 7 2
by
This integral and the corresponding one for the sine [see Problem 6S(a)] are called Fremel integrals.
296
GAMMA AND BETA FUNCTIONS
CHAP. 131
Supplementary Problems The GAMMA FUNCTION 28. Evaluate
(a)
r(7) 2 r(4) r(3)
'
r($i;iy),
(c) r(l/2) r(3/2) r(6/2).
(b)
A m . (a)SO, ( b ) 16/106, (c) #aUa 29. Evduate
(a) 1 w x 4 e - s d z , (b) S m x 6 e - & & ,
(c) l w z ' e - a d dx.
Am.
(a) 24,
(b) 243 80 ' (')
1 6 6
30.
31.
Show that
32. Prove that
fi
r(n) =
$,
=
imcdt
i'
8>0. n > 0.
(In ;>"-'dz,
33.
Am. (a)24, ( b ) -3/128, 34. Evaluate
(U) r(-7/2),
35. Prove that
(c)
+r(#
(b) 1?(-1/3).
lim r(z) =
where m = 0 , 1 , 2 , 3 , .
-m
t-,
A m . (a) (166)/106, ( b ) -3r(2/3)
J
r(-m
A*
Prove that
r'(1) =
+ Q)
(-1)- 2" 6 ... (2m - 1) is a negative number (it is equal to -y, where y = 0.677216..
36. Prove that if m is a positive integer, 37.
..
e-" In z dx
=
is called Euler's constant as in Problem 49, Chapter 11).
The BETA FUNCTION 38. Evaluate (a)B(3,6), (b) B(3/2,2), (c) B(1/3,2/3).
1'
39. Find (a)
Am.
x4 sa
(U)1/60,
40. Evaluate
(a)
41.
Prove that
42.
Evaluate
(a)
43. Evaluate
(a)
1'{m I*-
dz, ( b )
~'(1-
dz, (c)
( b ) r/2, (c) 3a uaf1(4- u )du,~ (~b )
od-
1
(dx l/G=P*
(4
(c) 2 d 6
z*)v'dz.
Am.
(U) 1
2 ~ ,( b )
P
0
dlt
-
{IW4)}* 4a&
-*
PIS
44. Prove that
A m . (a)1/106, ( b ) 4/16,
sin'e cos'
iP
8
de, (b) r c o s 6 8 de.
sin'8 de, (b)
f"
COS'
8 sin'e de.
Ans. (a)3d256, ( b ) 6a/8
Am. (a) 16/16, ( b ) 8/105
.
296
GAMMA AND BETA FUNCTIONS
46.
Prove that
47.
Prove that
2P
> 0.
where a, b
3fiarfab1j8
[CHAP. 13
2n
[Hint Differentiate with respect to b in Problem 46.1 48. Use the method of Problem 31, Chapter 12, to justify the procedure used in Problem 11.
DIRICHLET INTEGRALS 49. Find the mass of the region in the zy plane bounded by U = fi. A N . a/24 50.
Find the mass of the region bounded by the ellipsoid
y' +3 +2 a' 2'
2'
aabck (U' Am. -
square of the distance from its center.
+
z y = 1, z = 0, y = 0 if the density is
30
= 1 if the density varies as the
+ 6' + c'),
k = constant of proportionality
+ gala+ z'Ia = 1.
51. Find the volume of the region bounded by
Am.
zrIa
4a/35
+ +
52. Find the centroid of the region in the A r s t octant bounded by xel8 A m . d = # = E = 21/128
53. Show that the volume of the region bounded by
z*+ym+zm
= 1.
@'I8
=
54. Show that the centroid of the region in the first octant bounded by
U",
where m > 0 ,
+ + z*
zm y*
is given by
is given by
= a", where m > 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
x c o s d dx.
Ans. (a)
2 r(p) sin (pa/2) ' (b) Discuss the cases p = 0 and p = 1.
68. Evaluate
(U) Jm
sin z ' dz, (b)
Jw
a.]
0
- -u'cSCpacotplr,
69. Prove that 70. Show that
71. Let
Jp(z) =
+a, (b)
7r
3 f i r(1/3)
O
16 (-l)n
n=O
+ p + 1)'
(z/2)p+1n
n ! r(n
a generalization of the Bessel function (16), Page 232, to the case
+ +
where p may not be a positive integer. Prove that Jp(z)satisfies the equation zly" zy' z sinz, (b) J - I / s ( ~ = ) 72. Referring to Problem 71, show th at ( U ) JI/B(z)= K results of Problem 107(b) and ( 6 ) of Page 267 are valid for all p .
(2'
-pl)y = 0.
cosx,
(c)
73.
If a > 0, b > 0 and 4a.c > bl, prove that
74.
Obtain (13)on Page 286 from the result (4) of Problem 20. [Hint: Expand eOs'OG) . .. in a power series and replace the lower limit of the integral by
75.
+
Obtain the result (15) on Page 286.
the
-00.1
Chapter 14 Fourier Series PERIODIC FUNCTIONS A function f ( x ) is said to have a period T or to be periodic with period T if for all x , f ( x T ) = f ( x ) , where T is a positive constant. The least value of T > 0 is called the least period or simply the period of f ( x ) . Example 1 : The function sin x has periods 2a,4a,67, . . ., since sin (z+ 2a), sin (z+ 4a), sin (z+ 6a),
+
...
all equal sin x.
However, 2a is the least period or the period of sin z.
Example 2: The period of sinnz or cosnz, where n is a positive integer, is 2aln. Example 3: The period of tanx is
P.
Example 4: A constant has any positive number as period.
Other examples of periodic functions are shown in the graphs of Figures 14-1(u), ( b ) and (c) below.
Fig. 14-1
FOURIER SERIES Let f ( x )be defined in the interval (-L,L)and outside of this interval by f ( x 2L)= f ( x ) , i.e. assume that f ( x ) has the period 2L. The Fourier series or Fourier expansion corresponding to f ( x ) is given by.
+
where the Fourier coeficients an and
{ from
l
bn
are nnX
L
an = ZJ-, f ( x ) c o s ~ d x
S_:
bn =
nnx f ( x ) s i n L CEX
n = 0,1,2,.
.
If f ( x ) has the period 2L, the coefficients a, and b, can be determined equivalently
1
an
=
zc(:
c+2L
$l
nnx f ( x ) cos -d x
L
(3)
c+2L
bn =
f ( x )s i n F Lc i x
where c is any real number. In the special case c = 4,(3)becomes (2). 298
299
FOURIER SERIES
CHAP. 141
To determine a0 in (I), we use (%) or (3) with n=0.
&Jl
see that
a0
=
EX, l
L
f ( z ) dx.
For example, from (2) we
Note that the constant term in (1) is equal to
$=
f ( s ) d z , which is the mean of f(x) over a period.
If L = T , the series (I) and the coefficients (2) or (8) are particularly simple. The function in this case has the period ZA.
DIRICHLET CONDITIONS Suppose that (1) f ( z ) is defined and single-valued except possibly a t a finite number of points in (-L,L) with period 2L (2) f ( x ) is periodic outside (-L,L) (3) f ( z ) and f ’ ( x ) are sectionally continuous in (-L,L). Then the series (I) with coefficients (2)or (3) converges to (U)
f(x) if x is a point of continuity i
point of discontinuity
+
Here f(x 0) and f ( x - 0) are the right and left hand limits of f ( z ) at x and represent lim f ( x + ~ )and lim f ( x - ~ ) respectively. For, a proof see Problems 18-23. €40+
€WO+
The conditions (l),(2) and (3) imposed on f ( z ) are suficient but not necessary, and are generally satisfied in practice. There are at present no known necessary and sufficient conditions for convergence of Fourier series. It is of interest that continuity of f ( s ) does not done insure convergence of a Fourier series.
ODD and EVEN FUNCTIONS A function f(s)is called odd if f(-z) = -f(x). are odd functions.
Thus 9,x 5 - 3 9 + 2 x , sinx, tan32
A function f ( x ) is called even if f(-x) = f ( x ) . Thus x4, 2 9 -4x2 are even functions.
+ 6 , cosx, eZ + e-=
The functions portrayed graphically in Figures 14-l(a) and 14-l(b) are odd and even respectively, but that of Fig. 14-l(c) is neither odd nor even. In the Fourier series corresponding to an odd function, only sine terms can be present. In the Fourier series corresponding to an even function, only cosine terms (and possibly a constant which we shall consider a cosine term) can be present.
HALF RANGE FOURIER SINE or COSINE SERIES A half range Fourier sine or cosine series is a series in which only sine terms or only cosine terms are present respectively. When a half range series corresponding to a given function is desired, the function is generally defined in the interval (0, L) [which thus accounting for the name half range] and then the is half of the interval (-L,L),
FOURIER SERIES
300
[CHAP. 14
function is specified as odd or even, so that it is clearly defined in the other half of the interval, namely (-L,O). In such case, we have nnx an = 0, b,, = - J L f ( x ) sin - dx for half range sine series L (4) L nnx f ( x )cos -dx for half range cosine series b n = 0, an =
E
i
1
L
PARSEVAL’S IDENTITY states that
if a,, and b n are the Fourier coefficients corresponding to f ( x ) and if f ( z ) satisfies the Dirichlet conditions.
DIFFERENTIATION and INTEGRATION of FOURIER SERIES Differentiation and integration of Fourier series can be justified by using the theorems on Pages 228 and 229 which hold for series in general. It must be emphasized, however, that those theorems provide sufficient conditions and are not necessary. The following theorem for integration is especially useful. Theorem: The Fourier series corresponding to f ( x ) may be integrated term by term from a to x , and the resulting series will converge uniformly to l x f ( x ) d x provided that f ( x ) is sectionally continuous in 4 5 x 5 L and both a and x are in this interval.
COMPLEX NOTATION for FOURIER SERIES Using Euler’s identities, eie
= cos8
+ i sin 8 ,
tie = cos 8 - i sin 8
(6)
where i = c l (see Problem 48, Chap. 11, Page 251), the Fourier series for f ( z ) can be written as f ( x ) = n = - m Cn einuz/L (7) where Cn = 2 2L J L- L f ( x ) e - f n W x f L d x (8)
9
In writing t h t equality (7),we are supposing that the Dirichlet conditions are satisfied and further that f ( x ) is continuous at x . If f ( x ) is discontinuous at x , the left side of (7) f ( x + 0) + f ( x - 0) should be replaced by 2
BOUNDARY-VALUE PROBLEMS Boundary-value problems seek to determine solutions of partial differential equations satisfying certain prescribed conditions called boundary conditions. Some of these problems can be solved by use of Fourier series (see Problem 24).
CHAP. 141
301
FOURIER SERIES
ORTHOGONAL FUNCTIONS Two vectors A and B are called orthogonal (perpendicular) if A. B = 0 or A d 1 A2B2+A3B3 = 0, where A = Ali+ Az j+ ASk and B = Bli+ Bz j+ Bs k. Although not geometrically or physically evident, these ideas can be generalized to include vectors with more than three components. In particular we can think of a function, say A @ ) ,as being a vector with an infinity of components (i.e. an infinite dimensional vector), the value of each component being specified by substituting a particular value of x in some interval (a$). It is natural in such case to define two functions, A($) and B(x), as or thogml in (a,b) if
+
JbA(z)B(z)dz = 0
(9)
A vector A is called a unit vector or normalized vector if its magnitude is unity, i.e. if A *A = A2 = 1. Extending the concept, we say that the function A($) is normal or normalized in (a,b ) if P b J {A(~)}~= dz 1 a
From the above it is clear that we can consider a set of functions {+&)},
k = 1,2,3, . . .,
In such case, each member of the set is orthogonal to every other member of the set and is also normalized. We call such a set of functions an orthonormal set. The equations (11) and (12) can be summarized by writing
lb +m
where
am,,
(x)+n (x)dx =
called Kronecker’s symbol, is defined as 0 if m f n and 1 if m = n .
Just as any vector r in 3 dimensions can be expanded in a set of mutually orthogonal unit vectors i, j, k in the form r = c d c 2 j c3k, so we consider the possibility of expanding a function f ( x ) in a set of orthonormal functions, i.e.,
+ +
Such series, which are generalizations of Fourier series, are of great interest and utility both from theoretical and applied viewpoints.
FOURIER SERIES
302
[CHAP. 14
Solved Problems FOURIER SERIES 1. Graph each of the following functions.
f (4
I
t-
Period
-t
piS. 14-2
Since the period is 10, that portion of the graph in -6 < x < 6 (indicated heavy in Fig. 14-2 above) is extended periodically outside this range (indicated dashed). Note that f ( x ) is not defined at x = 0,6, -6,10, -10,15, -16, etc. These values are the discontinuities of f ( x ) .
s i nx
OSxSr
Period = 27
p4.143 Refer to Fig. 14-3 above. Note that f ( x ) is defined for all x and is continuous everywhere.
0 1 0
osx<2 25x<4 45x<6
Period = 6
- -
f (4
Period
I
0
2
--
I . -
4
8
6
1
0
1
I
8
1
1
4
X
Fig. 144
Refer to Fig. 14-4 above. Note that f ( x ) is defined for all x and is discontinuous at x = 22, 24,*8, k10, +14, . .
..
2.
kxx
Prove
l:
s_1
= 0
knx sinxdx
=
krx cosLdz
=
L kr
L krz -sinTl kr
kn
= -L
if k = l , 2 , 3 ,....
cos k r
L cos(-ka) +kr
L L ka sinkn - kn sin (-kr)
= 0
= 0
FOURIER SERIES
CHAP. 141
3. Prove
lL
(a)
mxx nrx cos-pos-dx L
=
mrx nnx s i n L cos L dx
=
s-L L
(b)
L
303
mxx nnx sin-pinzdx
0
=
mfn
L m=n
0
where m and n can assume any of the values 1,2,3, . . . (a) From trigonometry: cos A cos B cos ( A B ) } .
+
=
+{cos ( A
- B ) + cos ( A + B ) } ,
sin A ain B = +{cos ( A - B ) -
Then, if m Z n, by Problem 2,
Similarly if m # n,
If m = n , we have
m = n = 0 these integrals a r e equal
Note that if (b)
to 2L and 0 respectively.
We have sinA cosB = +{sin (A - B) + sin (A + B)}.
Then by Problem 2, if m # n ,
If m = n ,
=
sin mpz cos-n m dz
L
s-:sinT 2nuz dz
= 0
The results of parts (a)and (b) remain valid even when the limits of integration -L, L are
replaced by c, c
4.
A
If the series
+ 2L
+
respectively.
cos
n=l
nnx
show that for n = 1,2,3, . . .,
(U)
+ b, sin L
Multiplying f(z) by
COS
1;
=
mnz - and integrating from -L mUZ
f ( z ) cos- L
dz
=
A
1;
=
an
a,,,L
=
bn
sin-
L
to L, using Problem 3, we have dz
5{
n=1
+
n=l
cos
+ Thus
+
A
converges uniformly to f ( x ) in (4 L), ,
~
(8) - m ~m c nuz ~ ~
~
+
c
obn s
ifm#O
IL
mrz f ( z ) cosdz
L
if m = 1,2,3,. . .
~
d
x
304
FOURIER S E R I E S and integrating from - L to L, using Problem 3, we have
( b ) Multiplying ( 1 ) by s in= L max f ( x ) sin -d x
=
L
A
1:
sin-
max
L
5
+
{ a n
n=1
= Thus (c)
bmL
=
b,
[CHAP. 14
dx
1;
+
sin m 7 n x cos nax dx L
1;
bn
ax sin max sin ndx L
if m = 1 , 2 , 3 , . . .
f ( x ) sin-m m d x
Integration of (1) from -L to L, using Problem 2 , gives J-Lf(x)dx
= 2AL
or
=
A
Putting m = 0 in the result of part (a),we find
a0
LJL
2L
f(x)dx
-L
l L = - J-L f(z)d x L
The above results also hold when the integration limits
and so A = 5. 2
-L,L a r e replaced by c , c + 2L.
Note t h a t in all p arts above, interchange of summation and integration is valid because the series is assumed to converge uniformly to f(z) in ( - L , L ) . Even when this assumption is not warranted, th e coefficients urnand b, as obtained above a r e called Fourier coeficients corresponding to f ( x ) ,and the corresponding series with these values of a,,,and b m is called the Fourier series corresponding to f ( x ) . An important problem in this case is to investigate conditions under which this series actually converges to f ( x ) . Sufficient conditions for this convergence a r e the Dirichlet conditions established in Problems 18 through 23 below.
5.
(a) Find the Fourier coefficients corresponding to the function
( b ) Write the corresponding Fourier series.
at x = -5, x = 0 and x = 5 in order that the Fourier series will converge to f ( x ) for -5Sx5;5?
( c ) How should f ( x ) be defined
The graph of f ( z )is shown in Fig. 14-6 below.
4
-
1-1
-15
-10
-5
5
I 10
I 15
X
Fig. 14-5 (a) Period = 2L = 10 and L = 5 . Choose the interval c to c
+ 2L
as -6 to 6, so t h a t c = -5. Then
CHAP. 141
305
FOURIER S E R I E S c+2L
=
bn
;{ J-1
=
nax f ( x ) sindx
(0) sin?
+
dx
1;
=
L
f ( x ) sin-n a x d x
l ’ ( 3 ) sin?
6
l’
sin nax dx 6
=
dx}
-COS~T)
3(1
na
The corresponding Fourier series is
% 2
+
2 (a,,c o nsa x7 + b, sin
3
n=l
+
nax 5 3(1 -nac o s n a ) sin 6
n=l
Since f ( x ) satisfies the Dirichlet conditions, we can say that the series converges to f ( x ) at all points of continuity and to f ( x iO) 4-’(’- O) at points of discontinuity. A t z = -6, 0 and 6, which are 2 points of discontinuity, the series converges to ( 3 + 0 ) / 2 = 3/2 as seen from the graph. If we redefine f(x) as follows, 312 ~ z - 6 0 -6<~<0 f(%) = 3/2 x=O Period = 10 3 O<x<6 3/2 x=6
I
then the series will converge to f ( x ) for -6 4 x S 6 .
6.
Expand f ( x ) = x2, 0 < x < 27 in a Fourier series if (a)the period is 27, (b) the period is not specified. (a) The graph of f ( x ) with period 27 is shown in Fig. 14-6 below.
f (4
I -6a
0
/
/
b,
/
/
I -4a
0
/
/
/
/
/
/
/
/
/
/
I /
;$’”
f ( x ) sin-n a x dx
L
/ / /
0
I
0
-2a
/
/
/
0
f ( x ) cos-n m
=
/
I
if+2LL dx
=
a,
/
/
=
0
2P
1
L
X
0
I
I
4a
6a
xs sin n x d x
U
n 4a2
/
/
xs cos n x d x
U
=
/
/
/
FOURIER SERIES
306 This is valid for 0 < x
<2
[CHAP. 14
~ .At x = 0 and x = 2n the series converges to 2 2 .
( b ) If the period is not specified, the Fourier series cannot be determined uniquely in general.
1 1 1 F +g+ -+ -.. -- 7r26' 32 4a' " 4 6 reduces to -
7. Using the results of Problem 6, prove that At
2
= 0 the Fourier series of Problem
3
+
2F
By the Dirichlet conditions, the series converges at x = 0 to #O Then
4n9 4 + n=l2" = 3 n'
5 n'
and so
29,
=
+ 4 ~ ' ) = 2a'.
at6 '
ODD and EVEN FUNCTIONS. HALF RANGE FOURIER SERIES 8. Classify each of the following functions according as they are even, odd, o r neither even nor odd.
From Fig. 14-7below it is seen that f ( - x ) = -f(z),
so that the function is odd.
Fig. 14-7
cosx
o<x<7r
Period = ZT
From Fig. 14-8 below it is seen that the function is neither even nor odd.
Fig.14-8
( c ) f ( x ) = x(10 - x), 0 < x
< 10, Period = 10.
From Fig. 14-9 below the function is seen to be even.
f (4 I \
\
'
/
\
/ I
- 10
/
/
/
/
0-\
\
\
\
\
\
\
\
0
Fig. 14-9
/ I
I
I
10
/
/
/ 2
FOURIER SERIES
CHAP. 141 9.
307
Show that an even function can have no sine terms in its Fourier expansion. Method 1: No sine terms appear if b, = 0, n = 1,2,3, ba
21;
=
. .. . To show this, let us write
=
f ( z ) sin nnx dz
f(z) s i n E d z
L
+
f ( z ) sin-nnz dz
L
(11
If we make the transformation z = -U in the first integral on the right of (I), we obtain
$ J-:f(z)
nnx
sin L dz
=
iL
=
-;IL
f(-u) sin
(-7 )du nnu
=
f ( u )sin-du L
lL - iL
=
-
f(-u) sin
npu L du
(2)
f ( z ) sin nrz -dz
L
where we have used the fact that for an even function f(-U) = f ( u ) and in the last step that the dummy variable of integration U can be replaced by any other symbol, in particular z. Thus from (Z), using (a), we have
Method 2:
Assume
f(z)
= 2 2
+
n=l
If f ( z ) is even, f(-z) = f(z). Hence
5 bnsin-nnz 0, L
and so
i.e.
n-1
f(z)
+ 3 a , c onnz sx
=
n=1
and no sine terms appear. In a similar manner we can show that an odd function has no cosine terms (or constant term) in its Fourier expansion.
10. If f ( x ) is even, show that
(4
&r
=
(a) an =
f(z) cos
nnz
dz
Letting x = - u ,
dx h S-:f(x) cos L nuz
=
JL
=
$
iL
nrx
f ( s ) cos-
s-1
L dx, ( b ) b n = 0 .
nnz f ( z )cosdx
L
+; JL
nux f(x) cosdx
i"
(7 du ) =
f(-u) cos -nm
L
f ( u )cos-nbu du L
since by definition of an even function f(-u) = f(u). Then a,
=
ii
L
f(u) c o s F d u
L
+;
nnz
J L f(z) C O S T
dz
=
J L f ( z ) cos naz dz
( b ) This follows by Method 1 of Problem 9.
11. Expand f(x) = sinx, 0 < x
<x,
in a Fourier cosine series.
A Fourier series consisting of cosine terms alone is obtained only for an even function. Hence we extend the definition of f(z) so that it becomes even (dashed part of Fig. 14-10 below). With this extension, f ( z ) is then defined in an interval of length 2 ~ .Taking the period as ZU, we have 2 L = 2 a so that L = a .
FOURIER S E R I E S
308
[CHAP.14
Mg. 14-10
By Problem 10, b, = 0 and a,
=
El
L
L
(sin(x+nx)
U
1 - cos (n n+l
=
+ sin (x-nz)} +
n-1
if
~ ( n' 1)
a0
sin x cos n z dx dx
+ 1 ) ~ cos (n - l ) u -
- - 2(1 + cos nu)
For n = 0,
lT
=
f(x) cos-dx nuz
+
!IT
s i n z dx
'>
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<2
A method commonly employed in practice is to assume the existence of a solution of the partial differential equation having the particular form U ( z ,t ) = X ( z ) T(t), where X ( x ) and T ( t ) are functions of cc and t respectively, which we shall try to determine. For this reason the method is often called the method of separation of variables. Substitution in the differential equation yields
where we have written X and T in place of X ( z ) and T ( t ) . Equation (I?)can be written as
Since one side depends only on t and the other only on z, and since z and t are independent variables, it is clear that each side must be a constant c. In Problem 47 we see that if c 2 0, a solution satisfying the given boundary conditions cannot exist. Let us thus assume that c is a negative constant which we write as -Aa. Then from (3)we obtain two ordinary differential equations
$ + 3h'T
= 0,
+ h'X
=
0
(4)
whose solutions are respectively
A solution is given by the product of X and T which can be written
U ( z ,t ) = where A and B are constants.
PSt ( A cos xz
+ B sin xz)
(6)
FOURIER SERIES
314
[CHAP. 14
We now seek to determine A and B so that ( 6 ) satisfies the given boundary conditions. To satisfy the condition U ( 0 , t ) = 0, we must have
e-sA't(A) = 80
o
A =o
or
(7)
that ( 6 ) becomes
U(%,t ) = ~ e - 3 ~ sin ' t ~z To satisfy the condition U(2, t) = 0, we must then have
Be-""
sin2A
= 0
(9)
Since B = O makes the solution (8) identically zero, we avoid this choice and instead take sin2X = 0,
i.e.
- mn 2A = mu or A --
where m = 0, kl,a2, . . . .
(10 )
2
Substitution in (8) now shows that a solution satisfying the first two boundary conditions is
U($, t ) = B,e-Sm'04t/4sin E 2
(11)
where we have replaced B by B,, indicating that different constants can be used for different values of m. If we now attempt to satisfy the last boundary condition U(%, 0) = x , 0 < x < 2, we And it to be impossible using (11). However, upon recognizing the fact that sum8 of solutions having the form (11) are also solutions (called the principle of superposition), we are led to the possible solution
From the condition U ( x ,0) = x, 0 < x < 2, we see, on placing t = 0, that (12) becomes
This, however, is equivalent to the problem of expanding the function f(z) = x for O C x < 2 into a -4
sine series. The solution to this is given in Problem 12(a), from which we see that B , = -cos mr mn so that (12) becomes
which is a fomtal solution. To check that (14)is actually a solution, we must show that i t satisfies the partial differential equation and the boundary conditions. The proof consists in justification of term by term differentiation and use of limiting procedures for infinite series and may be accomplished by methods of Chapter 11. The boundary value problem considered here has an interpretation in the theory of heat con-
au
duction. The equation - = k-
a'u
is the equation for heat conduction in a thin rod or wire located
at a%) on the x axis between x = 0 and x = L if the surface of the wire is insulated so that heat cannot enter
or escape. U ( x , t) is the temperature at any place x in the rod a t time t. The constant k = K/sp (where K is the thermal conductivity, 8 is the specific heat, and p is the density of the conducting material) is called the diffusivity. The boundary conditions U(0,t ) = 0 and U(L, t) = 0 indicate that the end temperatures of the rod are kept a t zero units for all time t > 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<3
-3c~C0 cosna) 7L'
Period 6
nm cos 4
6(COSna - 1) cosn~x 3
% sin nu
30. In each part of Problem 29, tell where the discontinuities of f ( x ) are located and to what value the series converges at these discontinuities. Am. (a)z = 0, *2, =t4, . . ; 0 (b) no discontinuities (c) x = 0,*10,*20, ; 20 (d) x = *3,*9,*16, ;3
.
...
...
31. Expand
-
=
f(s)
2-6
+ s1 c
Am. g{cosf U '
4CxC8 3ux
in a Fourier series of period 8.
sax + 1cos+ 4
o s ~
32. (a)Expand f ( x ) = cos x , 0 < x C U, in a Fourier sine series. (b) How should f ( x ) be defined a t 1~ = 0 and x = U so that the series will converge to f ( x ) for 0 d x S U? 8 " n sin2nx ( b ) f ( 0 ) = f(d = 0 A M * (U) 4+ 1
;zl
33. (a)Expand in a Fourier series f ( x ) = cos x , 0 < x < U if the period is a; and (b) compare with the result of Problem 32, explaining the similarities and differences if any. Ans. Answer is the same as in Problem 32.
f(x) =
34. Expand
I".
8-x
4 OC c xX<<48
32 1 nr nux Am. (a)23 sin- sinO0
' U
n=1
2
35. Prove that for 0 S x d U,
8
cos2x (-i+
(a) x(7r-x)
=
6 -
(b) z(r-z)
=
T8 (sinx F +
U '
in a series of (a)sines, ( b ) cosines. (b)
5 ( 2 cos n d 2 - cos nu - 1)
' U
n'
n=t
cos6x + ..> +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
2 + 2(T-;;isin2x
-+sinx
x coax
Use (a)to show that for xsinx
-U
= 12(7- --
- Asingx
+ 3 4. 5 4 4 2 -
5 x S -a,
= 1
-
+coax
cos22 1.3
-
cos3x 2.4
cos42 +3.6
...)
...)
differentiating the result of Problem 36(b), prove that for 0 S z 5 I,
=
p 2
-
$(COSX P 1'
COS~X
3'
COS~X +5 '+
**.>
PARSEVAL'S IDENTITY 41. By using Problem 36 and Parseval's identity, show that (a)
42.
Show that
1 + 1i.31
1
+
1 x" = n4
r=l
1
" (a) 3
44. Showthat
1 1 1'.2'.g' + m'+
n=l
P4
IT4
-
+
1 = (2~96 '
43. Show that
-
1 = U" (b) " 946 n = l no
90
-. 16
[Hint Use Problem 11.1
U'-a
-
* 1 ( b ) ]E (2n - 1)" =
1
+
-.
960' P6
- 4u'-39
... -
16
BOUNDARY-VALUE PROBLEMS
45. (a)Solve
au = 2 $$subject to the conditions at
U(0,t ) = 0, U(4, t ) = 0, U ( z ,0) = 3 sin ux
- 2 sin 6 ~ 2 ,
where O < x < 4 , t > O . (b) Give a possible physical interpretation of the problem and solution. (a) ~ ( x , t )= Qe-'P*' s i n r x - ~ e - sin ~ liax. ' ~
A=.
46. Solve
Tz
dt = -
subject to the conditions
U(0,t) = 0, U(6, t) = 0, U(x,0 ) =
and interpret physically.
47. Show that if each side of equation (3),Page 313, is a constant c where c h 0, then there is no solution
satisfying the boundary-value problem.
319
FOURIER SERIES
CHAP.141
as. A flexible string of length U is tightly stretched between points z = 0 and 2 = w on
the z axis, its ends Axed at these points. When set into small transverse vibration the displacement Y(z,t) from the x
apY = a'axis of any point x at time t is given by where ' a = T/p, T = tension, at a axo unit length.
p
= mass per
(a)Find a solution of this equation (sometimes called the wave equation) with a'=4 which satisfies t) = 0,Y ( z ,0) = 0.1 sin z 0.01 sin 42, Yt(2, 0) = 0 for 0 C z C r, t > 0. the conditions Y(0,t) = 0,Y(u, (b) Interpret physically the boundary conditions in (a)and the solution. A m . ( U) Y(z, t) = 0.1 sinz cos2t 0.01 sin42 cos8t
+
+
= 9-a'Y subject to the conditions Y(0,t) = 0, Y(2, t) = 0, 13XL where 0 < z C 2, t > 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
= -1- - + -2a- a d - 12
2a a'-
2 '
a'
2a
- 3'
+ ...
1,
- 1- - - 2a -
+
a'-12
a
(c)
[CHAP. 14
2a
-
+ ...
2a
From (a),(b) and Problem 17, Chapter 13, complete the proof of the fact that
= P O
-a)
QO,
1 = l-x+z'-z*+ l+s
... .]
60. Let r be any three dimensional vector. Show that
(a)(r i)' i(re j)' 5 (r)',
( b ) (r *i)'
+ (r 1)' + (r*k)' = 3
and discuss these with reference to Besael's inequality and Parseval's identity. 61. If {+,,(z)},fi = 1,2,3, a minimum whem
.. .
is orthonormal in (a,b), prove that ca
=
lb
Discuss the relevance of this result to Fourier series.
+,(z) dz
rb
Ja
{f(z)
-
5
n=l
Ca+a(s))'d~
is
Chapter 15 Fourier Integrals The FOURIER INTEGRAL Let us assume the following conditions on f ( x ) : 1. f ( x ) satisfies the Dirichlet conditions (Page 299) in every finite interval (-LJ). 2.
Jm
-W
If(x)ldx converges, i.e. f ( x ) is absolutely integrable in
(-a,00).
Then Fourier’s integral theorem states that f(x) =
/.w
J,
{A(&)C O S ~ Z
A(a) = 5 J T
where B(a) =
1 7r
+ B(a)sinax) da
(1)
f ( ~COS ) ax d s
-m
lw f ( x ) sin ax dx
The result ( I ) holds if x is a point of continuity of f ( x ) . If x is a point of discontinuity, we must replace f ( x ) by f ( x +O) if ( x - O) as in the case of Fourier series. 2
Note that
the above conditions are sufficient but not necessary. The similarity of (I)and (2) with corresponding results for Fourier series is apparent. The right hand side of (I) is sometimes called a Fourier integral expansion of f(z).
EQUIVALENT FORMS of FOURIER’S INTEGRAL THEOREM Fourier’s integral theorem can also be written in the forms f(u)COS a($ - U)du da
-
(3)
& s_”,lm eLa(u-z)du da f(u)
where it is understood that if f ( x ) is not continuous at x the left side must be replaced by
f ( x + 0) + f ( x - 0) 2
These results can be simplified somewhat if f ( z ) is either an odd or an even function, and we have if f(x) is even f(x) = R cos ax dcy f(u)cos du du (5)
qW im
f ( x ) = gT f k n a x da i w f ( as)i n a d u
321
if f ( z ) is odd
(6)
322
FOURIER INTEGRALS
[CHAP. 16
FOURIER TRANSFORMS From (4) it follows that if
1 F(a) = -
(7)
then The function F(a) is called the Fourier transform of f ( x ) and is sometimes written F(a) = 3 { f ( x ) } . The function f ( x ) is the inverse Fourier transform of F(a) and is written f ( 4 = 3-’{F(a))* Note: The constants preceding the integral signs in (7) and (8) were here taken as However, they can be any constants different from zero so long as their equal to U-. product is l/%. The above is called the symmetric form. If f ( x ) is an even function, equation ( 5 ) yields
.-
J
(9)
and we call F c ( a ) and f ( x ) Fozcrier cosine transforms of each other. If f ( x ) is an odd function, equation (6) yields
and we call F+) and f ( x ) Fourier sine transforms of each other.
PARSEVAL’S IDENTITIES for FOURIER INTEGRALS If F,(cY)and Gs(a)are Fourier sine transforms of f ( x ) and g ( x ) respectively, then
Similarly if F&) and Gc(a)are Fourier cosine transforms of f ( z ) and g(s), then Fc(a)Gc(a)da
Ja
=
s , ’ f ( x ) g ( 4 dx
In the special case where f ( x ) = g(x), (11)and (12) become respectively
The above relations are known as Purseval’s identities for integrals. Similar relations hold for general Fourier transforms. Thus if F(a) and G(a) are Fourier transforms of f(x) and g(x) respectively, we can prove that
CHAP. 161
323
FOURIER INTEGRALS
(15)
where the bar signifies the complex conjugate obtained by replacing i by -i,
See Prob. 30.
The CONVOLUTION THEOREM If F(a) and G(a) are the Fourier transforms of f ( x ) and g ( z ) respectively, then
s_".F((Y)G(a)e-iaz da
=
If we define the convotution, denoted by f
lm
f ( u )s(z - U ) du
*g,
of the functions f and g to be
then (16)can be written
or in words, the Fourier transform of the convolution of two functions is equal to the product of their Fourier transforms. This is called the convolution theorem for Fourier transforms.
Solved Problems The FOURIER INTEGRAL and FOURIER TRANSFORMS
1. (a) Find the Fourier transform of f ( x ) =
{l lxl> < aa (b) Graph f ( x ) and its Fourier transform for a = 3.
Ix'
.
(a) The Fourier transform of f ( z )is
For a = 0, we obtain F(a) = &a. ( b ) The graphs of f ( z )and F(a) for a = 3 are shown in Figures 16-1 and 16-2 respectively.
Fig. 15-1
Fig. 15-2
324 2.
FOURIER INTEGRALS (U) Use
[CHAP. 16
sin aa COS a$
the reerult of Problem 1 to evaluate
da
(b) Deduce the value of (a) From Fourier's integral theorem, if
F(a) =
1l:f(u)P'du
F(a) e - - d a
then
I& Then from Problem 1,
The left side of (1) is equal to
The integrand in the second integral of (2) is odd and so the integrsl is zero. Then from (a), we have
(1) and
(b)
If z=O and a = 1 in the result of (a), we have or sin a da = U
laT& = U
sin a
2
since the integrand is even.
(a)
If f(u) is even, f(u) cosxu is even and f(u) sin Xu is odd. right of (1) is zero and the result can be written F(a)
(b)
=
2lm *La * f ( u )COS a U du
=
Then the second integral on the f ( u )cos au du
From (a), F(-a) = F(u) so that #'(a) is an even function. Then by using a proof exactly analogous to that in (a), the required result follows. A similar result holds for odd functions and can be obtained by replacing the coaine by the sine.
4.
Solve the integral equation Let
l m f ( z )cos as dx
lm f ( x ) cos a s dx
f(z)
= F(a) and choose F(a) = F(a) C O S U Z da
=
1-a
= ( 0
=
osaS1
*I*$
a>l
. ThenbyProb.3,
(1- a ) C O S U Z &
= ~ ~ * ~ l - a ~ c o s n=x & 2(1 - cos 5 ) U
OSaSl a>l
ux'
FOURIER INTEGRALS
CHAP. 161
5.
du =
Use Problem 4 to show that
326
7r 2'
As obtained in Problem 4,
Taking the limit as
a+O+,
we find
1-
But this integral can be written as
dz
which becomes
l-9
du
on letting
z = 2u, so that the required result follows.
6.
COS aX
Show that
7r
-e-=,
2
x 20.
Let f ( z )= e-" in the Fourier integral theorem
But by Problem 22, Chapter 12, we have
e-" coaau du
=
d+l'
Then
PARSEVAL'S IDENTITY 7. Verify Parseval's identity for Fourier integrals for the Fourier transforms of Prob. 1.
This is equivalent to
or i.e., By letting used to find
aa = U
la
and using Problem 6, it is seen that this is correct. The method can also. be du directly.
326
FOURIER INTEGRALS
[CHAP. 16
PROOF of the FOURIER INTEGRAL THEOREM 8. Present a heuristic demonstration of Fourier's integral theorem by use of a limiting form of Fourier series. Let
Then by substitution (see Problem 21, Chapter 14)'
If we assume that
L+
00,
1:
If(u)ldu converges, the first term on the right of (8) approaches zero as
while the remaining part appears to approach lim L L
L-+OO
$
f(u)c o s F ( u - z)du
-oo
r=l
(3)
This lsst step is not rigorous and makes the demonstration heuristic. Calling AU = P/L, (8) can be written
=
f(z)
But the limit (4)is equal to
=
f(z)
=
I"F(u)du
5 Aa F(nAu)
lim
Aa-0
1 P
r-1
loo&
J:af(u) cosa(w-x)
du
which is Fourier's integral formula. This demonstration serves only to provide a possihle result. To be rigarOus, we start with the integrd f da f ( u )cos a(u - z)d z P
I* 1:
and examine the convergence. This method is considered in Problems 9-12.
9.
Prove that: ((I)
(a) alirn -oo
Let av = y. Then
xLe
0
lr
sin ~ Z V dv =
2)
lim i L v d v =
a-*-
,'%
aL
=
-du 2/
lr
2.
sin y
29, Chap. 12.
( b ) Let uv = -y.
Then
sin uv dv
=
iFp
aL
9
d
y
U
= 2'
10. Riemann's theorem states that if F ( x ) is sectionally continuous in (a,b), then
!I
1 b
F ( z ) sinaxdx
= 0
with a similar result for the cosine (see Problem 31). Use this to prove that
by Fbvb.
FOURIER INTEGRALS
CHAP. 1/51
327
where f ( x ) and f’(x) are assumed sectionally continuous in (0, L) and (-L, 0) respectively. (a) Using Problem 9(a),i t is seen that a proof of the given result amounts to proving that sin av lim J L {f(z + v ) - f ( x 0)) 2, dv = 0
+
a-+w
This follows at once from Riemann’s theorem, because F ( v ) = continuous in ( 0 , L ) since
:ly+ F ( v )
f(s+ v ) - f(z + 0) is sectionally V
exists and f ( x ) is sectionally continuous.
( b ) A proof of this is analogous to that in part (a)if we make use of Problem 9(b).
11. If f ( x ) satisfies the additional condition that
(a) ~ i mJwf(x a-m
+
sin av dv + v) 7
rw
If(x)I dx converges, prove that 0
= i f ( x 0), ( b ) lim a--
--m
f ( x+ U )
sin uv
dv 7
= Tf(x-0). 2
We have
Subtracting,
Denoting the integrals in (3) by Z, ZI, ZZ and ZS respectively, we have I = ZI
IZl Now
lZZl
Also
1131
5
IZll
+
IZZl
+
+ ZZ + Zs
so t h a t
lZSl
(4)
sin av dv both converge, we can choose L so 1 rge th t 1ZZ1 5 €/3,
IZ31 5 4 3 . Also, we can choose a so large that IZll 5 43. Then from (4) we have IZl sufficiently large, so t h a t the required result follows.
<e
for a and L
This result follows by reasoning exactly analogous to t h a t in p a r t (a).
---:o
12. Prove Fourier’s integral formula where f ( z ) satisfies the conditions stated on Page 321. We must prove that Since
1
J-:f(u)
strass test that
1
cos a ( z - U ) du
Ilf(u)
I
f(u) cos a ( x - U ) du da
2
J-: If(u)l du
=
f(z
+ 0) + f ( z- 0) 2
which converges, it follows by the Weir-
cos a(z - U ) du converges absolutely and uniformly for all a. Thus we can
reverse the order of integration to obtain
328
FOURIER INTEGRALS
where we have let
U
=x
[CHAP. 16
+v.
Letting L + a, we see by Problem 11 that the given integral converges to f(x required.
+ 0) + f ( x - 0) 2
as
MISCELLANEOUS PROBLEMS 13. Solve - - - subject to the conditions
ax2
at
U ( 0 , t )= 0, U(x,O) =
U ( x ,t ) is bounded where x > 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
<1
dx.
o s x < 1 find the (a) Fourier sine transform, (b) Fourier cosine transform of f ( x ) . X Z l In each case obtain the graph of f ( x ) and its transform.
18. If
f(x) =
1
Ans. (a) '$(l
(b)
$*. r
a
19. (a) Find the Fouriepr sine transform of e-",
x sinrnx
X-zTTi- d x =
(b) Show that
ge-"',
x B 0. m
>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
tB2
lSt<2
and verify the solution by direct substitution. A m . Y(x) = (2
+2
COS
x
- 4 COS 2z)/az
PARSEVAL'S IDENTITY 21. Evaluate
dx
xs d x
by use of Parseval's identity.
[Hint Use the Fourier sine and cosine transforms of e-=, x
> 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'
.
(4),Page 321, from equation (S), Page 321.
29. Prove the result (18),Page 323. [Hint If
F(u) =
E1 J-,f(w)"
etau du
F(a) G(a) = Now make the transformation u + v =
G(Q)
and
& J-: J-:
=
1
-
J-@g(v) elaOdv,
d a ( u + "f(u) ) g(V) dU dv
2.1
30. (a)If F(a) and G(a) are the Fourier transforms of f ( z )and g(z) respectively, prove that J-=j(a)
Go da
=
where the bar signifies the complex conjugate. (b) From (a)obtain the results (11)- (14), Page 322. 31.
Prove Riemann's theorem (see Problem 10).
Sp,f(dgo d z
then
Chapter 16 Elliptic Integrals The INCOMPLETE ELLIPTIC INTEGRAL of the FIRST KIND is defined aa U
= F(k,+) =
l*VI-
d0
ICB sin20
O
where + is the amplitude of F(k,+) or U, written + = am U , and k is its M d w , written k = m o d u . The integral is also called Legendre’s form for the eltiptic integral of the fiT8t I%ind. If + = d Z the integral is called the complete integral of the first kind and is denoted by K ( k ) or simply K . For all purposes it will be assumed that k is a given constant.
The INCOMPLETE ELLIPTIC INTEGRAL of the SECOND KIND is defined by E(k,+) = ~ * ~ l - V s i n ’ B d B O < k < l also called Legendre’s form for the elliptic integrd of the second kind. If + = d Z the integral is called the comptete elliptic i n t e g d of the second kind and is denoted by E(k) or simply E . This integral arises in the determination of the length of arc of an ellipse and supplies a reason for use of the term elliptic integral.
The INCOMPLETE ELLIPTIC INTEGRAL of the THIRD KIND is defined by
W?%+) =
6
d0
O
(8)
(1 + n s i n 2 8 ) d 1- k2sintJ8 also called Legendre’s form f o r the elliptic integral of the third k i d . Here n is a constant assumed different from zero since if n = 0, (8) reduces to ( I ) . If +=& the integral is called the complete etliptic integral of the third kind.
JACOBI’S FORMS for the ELLIPTIC INTEGRALS If the transformation v = sin 8 is made in the Legendre forms of the elliptic integrals above, we obtain the following integrals with x = sin+
El(k,x) =
ix,/1-h2 1-VB
331
dv
332
ELLIPTIC INTEGRALS
[CHAP.16
called Jacobi's forms for the elliptic integrals of first, second and third kinds respectively. These are complete elliptic integrals if x = 1.
INTEGRALS REDUCIBLE to ELLIPTIC TYPE If R(x,y) is a rational algebraic function of x and y, i.e. the quotient of two polynomials in x and y, then R(x,?4 d x (7)
+
J
can be evaluated in terms of the usual elementary functions algebraic, trigonometric, inverse trigonometric, exponential and logarithmic) if g = ux b or g = duxg bx c, where a, b, c are given constants. If y = d a x s + b x 2 + c x + d or y = ~ a x 4 + b x s + c x 2 + d x + ewhere a,b,c,d,e are given constants, (7) can be evaluated in terms of elliptic integrals of first, second or third kinds or for special cases in terms of elementary functions. If 1/ = d m where P(x) is a polynomial of degree greater than four, (7) may be integrated with the aid of hyper-elliptic functions.
+
+ +
JACOBI'S ELLIPTIC FUNCTIONS The upper limit x in the Jacobi form for the elliptic integral of the first kind is related to the upper limit + in the Legendre form by x = sin +. Since 9 = am U, it follows that x = sin (am U). Thus we are led to define the elliptic functions x = sin (am U) - sn U
dF2 d m
= cos(amu)
- cn U
= d1-k2sn2u = dnu - X- - - sn U = tnu j/-cn U
which have many important properties analogous to those of trigonometric functions as presented in the problems. It is also possible to define inverse elliptic functions; for example, if x = s n u then U = sn-l x . Note that U depends on k. To emphasize this dependence we sometimes write U = sn-' ( x , k ) or U = sn-l x,mod k.
LANDEN'S TRANSFORMATION By use of the transformation sin 2+1 tan+ = or k sin+ = sin@+,-+) k C082+, called Landen's transformation, we can show that
+
1dl -&
l"
d+l - -2 1i-k f1 - k; sin29, k2sin2+ where kl = 2fi (see Problem 61). This can be written l+k n 0
333
ELLIPTIC INTEGRALS
CHAP.161
It is seen that k < k l < 1. By successive applications of Landen's transformation, a sequence of moduli kn, n = 1,2,3,. . . is obtained such that k < kl < k2 < . . . < 1 and we can prove that lim kn = 1. From this it follows that
By using this result, F ( k , + ) can be computed. In practice it is possible to achieve accurate
results after only a few applications of the transformation.
Solved Problems ELLIPTIC INTEGRALS 1. Prove that if 0 < k < 1.
By the binomial theorem,
Let z = k1 sins 8. Then by the uniform convergence of the series, we can integrate term by term from 0 to d 2 to obtain
1 dl WS
de
1
'k s i d e - 21(1 -
+
+ z1k a s i n S e + m1 . 3k
(+>'.+ ( g ) ' k 4
4s i d e
+(
E
+ E2 . k4 -66 s i n 6 e + ) l k 6 +
...
. a . }
using Problem 15, Chapter 13.
2. Evaluate
&
Awl2
to 3 decimal places by first expressing the integral as an
ra
elliptic integral. Let z = a/2 - y and the integral becomes Let cosy = cos' dy
U.
=
Then -sin y dy = 2 cos U sin u du
v-
-
qu -q-
=
fi~fi,d2)
fi in the result of
and
2 cos U sin u du
-
du 2 - sin'u
du
Hence
Substituting k =
&*
- 2 cos U sin U du
or
fi~4).
2 cos U du
dd 1 - Q sin'u
Problem 1 yields the value 2.622 for the integral.
ELLIPTIC INTEGRALS
334
[CHAP. 16
Another method :
= sin 9. Then f i / 2 cosg/2 d y = cos 9 d+, v 2 cos+ d+
=
dy
3. Evaluate
and the integral becomes
41 - 9 sin'+
s,"" d
s""d G '
G dx
fiJ*/' d9 41 - 9 sin'
=
Im(4)
in terms of elliptic integrals.
Let c o s x = cos'u as in Problem 2. Then dx
q-
0
=
4.
Evaluate
2fi
2 cosu
s"" 0
dl - + sin' u d u -
lTf2 dl + IT''dl +
4(1 - cos'x) d x
Letting x = d 2 - y, the integral becomes
6 - 4 sin'y d y
5.
Express Let
where
dl - 4 sin2U du dZ-ZFii =
+ = sin-'
=
lufa \/6
lmia -
fl
41
du
uta
sinay d y
4 cos'x d x .
=
6E ( m 6 )
in terms of incomplete elliptic integrals where 0 S x S d 6 .
2 sinu = sin +.
(2 sin 2).
=
du
6
4 sin2xdx.
The integral equals
cos'u
=n
CHAP. 161
6.
Show that
sd F z &
- (coszx / 2 - sin*x/2) = 3 - 2 cos*x/2
Letting x / 2 = n / 2 - u elliptic type.
From Problem 6,
since from x / 2 = n/2 - U
8.
which is of
du
-
it is seen that
U
= n/2 and n/4 respectively when z = 0 and a / 2 . Then
Find the length of arc of the curve y = sinx, 0 S x ST. Arc length
9.
du -2 fi VI - Q sinsu
2 - ~ { F ( ~ , T /- ZF )( w , ~ / 4 ) } .
=
I
and the given integral can be written
in this last integral, it becomes
dx
Prove that
335
is reducible to integrals of elliptic type.
dx
2 - cosx = 2
7.
ELLIPTIC INTEGRALS
=
fdl
+ (dy/dx)*dx
=
=
x ' q m d x
21T/zq-dx
Find the length of arc of the ellipse x = a sin +, y = b cos +, a > b > 0. Arc length
=
4 1
=
4 1
a/¶
R/*
d (dx)' + (dy)*
daz -
=
4
iw" ++ +
( a L - bZ)sinZ+d+
d a Zcosz
=
4a i
T I z
bz sinz d+
41 - eL sinL+ d+
where eL = (a*- b*)/a*= c * / d is the square of the eccentricity of the ellipse, The result can be written as 4aE(e,n/2) or 4aE(e).
10. Express
l'dl +
d+ in terms of elliptic integrals. k2 sin2+
Let k sin + = tan U. Then k cos + d+ = secLU du and
and
du
IU4 k L
- (k'+ 1)sinsu
Now proceed as in Problem 6. Let U
du
-
1
d
m sin U
l=
= k sin x. Then dx
-
41 -
cos U du = k cos x dx 1
')
dkZ- (k'+ 1)sinLu d m F( &l ' d1 - [kz/(ke+l)] sin*x Upon retracing steps it is observed that the upper limit x in the last integral is related to the d 1 sin+ upper limit # in the original integral by x = sin" 1 kz sinz+
+
336
ELLIPTIC INTEGRALS
[CHAP. 16
11. Evaluate each of the following in terms of elliptic integrals.
.
dx
Letting x = 2 sin 8, the integral becomes
Letting e = ~ / 2 $, this last integral becomes
.
dx
Let d z 3 = U or x = 3
+ us.
Then the integral becomes
du
Letting
U
= tang, this integral becomes
dx
where P 3 ( z ) is a third degree polynomial with real. zeros, can be integral by the method indicated.
12. Evaluate Let
U
du
lm
+
i ( u 2- 1)(u2 3)
= sec e and the integral becomes
13. Show how to evaluate
dx in terms of elliptic integrals. d(x- l)(x - 2)(x - 3 ) ( x - 4)
* +
Make the fractional linear transfomnation x = ct correspond to t = 0, 1, respectively. QO
d
and choose a, b , c , d so that x = 1 , 2 , 3
CHAP. 161
337
ELLIPTIC INTEGRALS
This leads to 1 =
ab ,
a+b a 3t+1 2 = - 3 = - from which a = 3d, b = d , c = d so that x = c+d' c t+ l.
Using this transformation the given integral becomes Then if t =U', we obtain In general
Jdm dx
2
du
Jd(u' - l)(u' + 3)
dt d t ( t - l)(t
+ 3)
and the method of Problem 12 is applicable.
where P,(s) is a fourth degree polynomial having real zeros, can be
transformed into an elliptic integral by the method indicated. Similar procedures are available if some o r all zeros are complex (see Problem 14; the method employed there can also be used in this problem).
14. Evaluate
s
dx .\/(x2- 2x 10)(x2 x
+ + 7)
+
in terms of elliptic integrals.
Here the polynomial under the radical has no real zeros, so that the method of Problem 13 is inapplicable. We proceed as follows. Let z = y + a , where a is a constant. Then the integral becomes
I
=
J
Choose a so that the constant terms of each quadratic are equal, i.e. a'or a = 1 . Then
I
J
=
2a
+ 10 = 'a + a + 7
dy
fly'
+ 9)(yL+ 3y + 9)
Let y = pu, where p is a positive constant. Then
Choose p so that the coefficient af U' in each quadratic is equal to the constant term, i.e. p' = 9 or p = 3 . Then du I d(u' l)(u' U 1)
J
Let
U
2 dt m, d u = (1 - t)'
= l+t
+
+ +
Then
*
I
=
fiJ d(t'+
dt l)(t'+ 3)
which is reducible to elliptic type by substituting t = tan 8 as in Problem l l ( b ) .
15. Express
dx .\/(z2- 2x + 10)(z2- 2x
+ 7)
as a n elliptic integral.
If we let x = y + a as in Problem 14, we are led to the condition a a - 2 a + 10 = a'-2a+ 7 which is impossible. In this case however, completing the square in each quadratic suggests the transformation x - 1 = y and the integral becomes
which is reducible to standard form by letting y = jb tan 8.
338
ELLIPTIC INTEGRALS
la
16. Evaluate
du
+
[CHAP. 16
+
(3u2 1)d(u2- 1)(u2 3)
Letting
U
= sec 8, the integral becomes (3
+
COS'
e) - 3
where the second integral is the complete elliptic integral of the third kind.
dx in tern8 of elliptic integrals. x \/(x - l)(x - 2)(x - 3)(x - 4)
17. Show how to evaluate
Using the same transformation a s in Problem 13, the integral becomes Now let t=u' to obtain 2
+
(U*
+ 1) d u
(3u2 1) q(u' - l)(u'
+ 3)
'
-
#3u*
(324'
+ 1)d(u' - l)(u' + 3 ) du +
du
(3u'
+ 1) d(u' - l)(u' + 3)
Letting x - 1 = l / y , the integral becomes
for which the method of Problem l l ( c ) is applicable.
ELLIPTIC FUNCTIONS 18. Prove
d du
d du
(a) - (snu) = cn U dnu, ( b ) - ( m u )
By definition, if
U
=
J'd1-z
Singe
since
* d+
=
1
dl - k4 sin'$
and so
& ! du
=
= - s n u dnu.
then
snu = sin + = sin (amu),
c n u = cos$ = cos (amu)
dl - kL sina+
= dnu.
( t + 1) d t + 1) dt(t- l ) ( t + 3 )
du
d(u* - l)(u'+ 3) . .
Then proceed as in Problems 12 and 16.
Another method:
+ 1) + Q
(3t
CHAP. 161
ELLIPTIC INTEGRALS
339
d
19. Prove -(dn U) = - k2 sn u cn U.
du
1 d = -(1 k' sn'u)-l/' -(-
2
- k'
=
k* sn'u)
du
2(sn U)
d
=
(sn U)
- 'k
2 dnu
- 0
2 snu cnu dnu
-k'snucnu
20. Prove (a) sn (-U) = - sn U, (b) cn (-U) = cn U, ( c ) dn (-U) = dn U, (d)tn (-U) = - tn U. (a) Let
U
=
o - ks sin* e d1 - P sin*e sn U = sin +, sn v = sin (-#) = - sin + = - sn U. Now letting e = -$ in the second integral, it becomes v =
41
Then
-
Thus sn (-U) = - sn U. (b)
Since cn U =
d w ,
(c)
Since dn U =
ql - k'
sn (-U) = (d) tn(-U) = cn (-U)
cn (-U) = \/1 - SIP (-U) = (
sn'u,
dn (-U) = ( 1
sn u = - cn u
u/2
21. Show that if K =
d0
(a) sn (U 2K)= - sn U, (b) cn (U tn u.
1
*+U
(a) Consider
de d1 - k' sin'e
8
=
U+$,
Q+U
(-U)
=
-
-U.
u = cn U.
dl - ks snsu
= dnu.
, then
+ 2 K ) = - cn U, -
The first integral on the right Letting
- k' an'
m
sin'$
--U.
dl - k2sin28
+
dl - k'
o
+
de dl - ks sin'e PI'
= 2 1
+
( c ) dn (U 2 K ) = dn U, (d) tn (U 2 K ) = *+U
de
dl - k'
sin'e
de
dl - k'sin'e
+
= 2K.
the second integral on the right becomes
de
1dl 6
d+
k' sin'+
-
U.
= u+2K or am(u+2K) = + + T side sn (u+2K) = s i n ( + + a ) = -sin+ = - - n u
Then
dl - k'
and
tn
(U+
2K) = s n ( u + 2 K ) - s n u c n ( u + 2 ~ )- GE - t n u
22. Prove that s n u and c nu have periods equal to 4K,while d n u and t n u have periods 2K. Replacing
U
by
U
+ 2K in Problem 21,
sn(u+4K) = -sn(u+2K) cn(u+4K) = - c n ( u + 2 K ) and the proof is complete.
= -(-snu) = -(-mu)
= snu = cnu
340
ELLIPTIC INTEGRALS
[CHAP. 16
From Problem 21(c) it is seen that d n u and t n u have period 2K. sn U
It can be shown that the elliptic functions have other periods which are complex. For example, has periods 4K and ZiK', cn U has periods 4K and 2K 4- ZiK', dn U has periods 2K and 4X', where
For this reason elliptic functions are sometimes known as doubly-periodia functions.
23. Prove that
(a)
d sn-l (x,k) dx
0
1 d ( 1 - x2)(1- k2XZ)
=
dv d(1 - vZ)(1 - k2v2)
= sn-l(x,k)
= F(k, sin-%).
(a) We shall write sn-lx in place of sn'l(2, k), the dependence on the modulus k being understood. By Problem 18, if x = s n u , then
dx -- -d( m u ) du du Hence
$Dd-
= cnudnu =
du -dx
d (sn-12)
=
= d(l-xa)(l-kV) 1
d(1- d)(l-k'z')
(b) Integrating the result in (a)from 0 to x , we have, since z = sin + where dv
Jo
d(1- v ' ) ( l - kavr)
-
sn-lz
+ = amu,
= F(k,+) = P(k, sin'lz)
Note the similarity with the result for trigonometric functions (corresponding to k = 0): dv = sin-'
x.
The elliptic functions a r e generalizations of trigonometric functions.
MISCELLANEOUS PROBLEMS
d
7
24. Prove that
F(m ~,/ 2 ) =
* 3). rO)
From the properties of the gamma function,
once.
But from Problem 2, this integral equals
25. Determine the period length 1.
T of
8
f i F ( m ,4) and
so the required result follows at
simple pendulum of
A simple pendulum consists of a mass m attached to a rigid rod OP of negligible mass and length I (see Fig. 16-1). Assuming the rod to be suspended from a Axed point 0,the differential equation of motion of mass m is given by m l c = -mgsine or d% - - - sIi n e dtr dt' 1 de @e = dp = d P d e dP the Letting - = p , dt dt3 dt de d t 'de' equation becomes p dPz = - f sine. Integrating,
Fig. 16-1
CHAP. 161
ELLIPTIC INTEGRALS
341
If the pendulum makes an angle e = 8, > 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
and diverges if lzl > 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
f l+xdx. ZP-1
Since x = O is a branch point,
choose C as the contour of Fig. 17-12 where A B and GH are actually coincident with the x axis but are shown separated for visual purposes. The integrand has the pole x = - 1
lying within C.
Residue at x = -1 = er' is
Jc I t 2
or, omitting the integrand,
s+s+s+s
AB
BDEFG
GH
HJA
=
We thus have
where we have to use z = xez+ for the integral along G H , since the argument of x is increased by 2a in going around the circle BDEFG.
368
FUNCTIONS OF Taking the limit a8 r - 0 and R + zero, we And
00
A COMPLEX VARIABLE
[CHAP. 17
and noting that the second and fourth integrals approach
or M)
that
Supplementary Problems FUNCTIONS, LIMITS, CONTINUITY 39. Describe the locus represented by (a)1% 2 - 3il = 6, ( b ) lz 21 = 21%- 11, (c) Construct a figure in each case. Ans. (a)Circle (z 2)' (y- 3)' = 26, center (-2,3), radius 6. (b) Circle (z - 2)' y' = 4, center (2,0), radius 2. (c) Branch of hyperbola zr/Q- yV16 = 1, where z L 3.
+
+
1% + 61 - 1% - 61 = 6.
+ + +
40. Determine the region in the z plane represented by each of the following: Iz+3l < 10. (a)1%-2+il 2 4, ( b ) IzI 1 3 , 0 S a r g z S z , (c) 12-31 4 Construct a figure in each case. (y 1)' = 16. A m . (a)Boundary and exterior of circle (z - 2)' (b) Region in the first quadrant bounded by z'+y' = 9, the z axis and the line y = z. (c) Interior of ellipse z'/26+y1/16 = 1.
+
+ +
41.
+
Express each function in the form u(x,y) iv(z, y), where U and v are real. (a)$+2iz, (b) z/(3+z), (c) @,' (d) l n ( l + z ) . A m . (U) U = z a - 3 ~ y ' - 2 ~ , v = 3 ~ ' y - - ~ + 2 ~
(a) U
42. Prove that
= + I n ((1+z)'
+ y*),
(a) lim z* = z;, ( b ) f ( z ) = 2' 2 4 20
'
+
v = tan-' 1 + 2 2ka, k=O,*l,*2,
...
is continuous at z = zo directly from the definition.
43. (a)If Z = O is any root of zs= 1 different from 1, prove that all the roots are 1,w,w',oa,w4. (b) Show that 1 + w + o ' + w a + w 4 = 0. (c) Generalize the results in ( a ) and (b) to the equation z" = 1. DERIVATIVES, CAUCHY-RIEMANN EQUATIONS 44.
(a)If w = f ( z ) = z
+ ;,1
dw
directly from the definition. find (b) For what finite values of z is f ( z ) non-analytic? Ans. (a)1 - 1/2,( b ) z = O
CHAP. 171
369
FUNCTIONS OF A COMPLEX VARIABLE
45. Given the function w =z4. (a)Find real functions U and v such that w = u + i u . (b) Show that the Cauchy-Riemann equations hold at all points in the finite z plane. (c) Prove that U and v are harmonic A m . (a) U = z4-6&ys+y4, v = 4&- 4zya (d) 4za functions. (d) Determine dwldz. 46. Prove that f ( z ) = z 121 is not analytic anywhere. 47. Prove that f ( z )
=
is analytic in any region not including z = 2 .
48. If the imaginary part of an analytic function is 2 4 1 - y), determine (a)the real part, (b) the function. A m . (a) y' - z' - 2y c, ( b ) 2iz - z1 c, where c is real
+
+
49. Construct an analytic function f ( z ) whose real part is e"(zcosy A m . ze-.+ 1
+ ysiny)
and for which f ( 0 ) = 1.
50. Prove that there is no analytic function whose imaginary part is d - 2 y .
and f ( l + Q= -3i.
51. Find f ( z ) such that f ' ( z ) = 42-3
Ans. f(z) = 225-32+3-4i
INTEGRALS, CAUCHY'S THEOREM, CAUCHY'S INTEGRAL FORMULAS
+ 3) dz:
52. Evaluate f+1(22
+
(a)along the path 2 = 2t 1, y = 4 f - t - 2 0 S t 5 1. (b) along the straight line joining 1 - 2i and 3 i. (c) along straight lines from 1 - 2i to 1 i and then to 3 i. A m . 17 19i in all cases
+
53. Evaluate
+
+
l
+
+
(2- z 2) dz, where C is the upper half of the circle Izl = 1 traversed in the positive
A m . -14/3
sense. 54. Evaluate
$&,
where C is the circle (a)Izl = 2, (b) 12-31 = 2.
A M . (a) 0 , ( b ) S d 2
z' $ + - 1) where C is: (a) square with vertices at -1 - i, -1 + i, -3 + i, -3 (b) the circle lz + il = 3; (c) the circle lzl = fi. A m . (a) - 8 d 3 ( b ) -2ai (c) 2 d 3
55. Evaluate
56. Evaluate
(2
cos lrz (a)f x d z ,
Ans. (a) -2ai 57.
dz,
2)(2
a
(b)
$-
8+z
dz
i;
where C is any simple closed curve enclosing z = 1 .
C
( b ) &e/3
Prove Cauchy's integral formulas. [Hint Use the definition of derivative and then apply mathematical induction.]
SERIES and SINGULARITIES 58. For what values of z does each series converge?
Ans. (a)all z
( b ) Iz-il
(c)
z = -1 * i
zn 5is (a)absolutely convergent, (b) uniformly convergent for d 1. n(n + 1) Prove that the series 5 converges uniformly within any circle of radius R such that 2,,
59. Prove that the series
60.
C 1
lz+il
2.
n=1
12)
370 61.
FUNCTIONS O F A COMPLEX VARIABLE Locate in the finite z plane all the singularities, if any, of each function and name them:
A m . ( a ) z = -8, pole of order 4
( d ) z = 0, essential singularity
( b ) z = 1, simple pole; z = -2, double pole (c) Simple poles z
62.
[CHAP. 17
= -1
-t
( e ) z = ~ / 3 removable , singularity
( f ) z = 2 2 4 double poles
i
Find Laurent series about the indicated singularity for each of the following functions, naming the singularity in each case. Indicate the region of convergence of each series. cos z z--a ; z=a
( b ) z'e"/z;
Ans. (a) -- 1
4- --
(a)
z-n
2!
2-77
( b ) z2 - z 1
(4
z=O
(z-n)* 4!
(z -
2%
...,
(Z-7d5
+
6!
1 1 1 1 +-+-+2! 3 ! z 4!z2 6!z3
+ - + -7- 16(z- 1)
(')
9 ( z - 1) 266
9 64
+3) ; = 1
+
.,
simple pole, all z # a essential singularity, all
...,
double pole, 0
< lz-
Z Z O
1 < 4
RESIDUES and the RESIDUE THEOREM 63. Determine the residues of each function at its poles:
Am.
(U)
~ = 2 ;7/4,
( b ) z = O ; 8/25, 64.
Find the residue of e" tan z at the simple pole z = 3 d 2 .
f 66.
( c ) z = 2; @2e't (d) z=i; 0, 2 z - i ; 0
~ z - 2 ;1/4 ~ = - 6 ; -8/26
(2
+ l)(z + 3 ) '
where C is a simple closed curve enclosing all the poles.
A m . -8ni
If C is a simple closed curve enclosing z = Ai, show that
=
$-dz 67.
A m . -esrtlr
itsin t
If f ( z ) = P(z)/Q(z), where P(z) and Q(z) are polynomials such that the degree of P(z) is at least two less than the degree of Q ( z ) , prove that
f ( z ) d z = 0, where C encloses all the poles of f ( z ) .
EVALUATION of DEFINITE INTEGRALS Use contour integration to verify each of the following 73-
J-: PT
2r
75-
"*
1
dx (x2+ l)~(xX+4) de
-
de (2+cose)~
-
9"
2n
- -4 n 6 9
side de 6 - 4coSe
= E 8
FUNCTIONS OF A COMPLEX VARIABLE
CHAP. 171 cosne de
78.
S.
2T
79.
-
de
(a
+ bcose)S
2aan l-as,
- (2a' + b2)n ((&' -
80.
x sin22 xL+4 dx
= 4
81.
cos 2nx x4+4 d x
= 8
n = 0 , 1 , 2 , 3 ,..., O < a < l
a
J
'Ibl
re-=
sin' x xs dx
84.
85.
dx
=
11-1-
sin x x(x'+ 1)' d x
ae-'
82.
86.
U
2 cosh (a /2).
[Hint: Consider
at (-R, 0), (R, 0), @,a), ( - R , a ) . Then let R
MISCELLANEOUS PROBLEMS
+ i v(p,+),
87. If z = pet@ and f ( z ) = u(p,@) Riemann equations are
371
=
~ ( 2-e 3)
4 e
= .Z 2
sins x 3a xJ dx = 8
f & dz,
where C is a rectangle with vertices
+ 00.1
+ are polar coordinates,
where p and
show t h a t the Cauchy-
88. If w = f ( z ) , where f ( z ) is analytic, defines a transformation from the z plane to the w plane where z = x + i y and w = u + i v , prove that the Jacobian of the transformation is given by
89.
90.
91.
Let F ( x , y) be transformed to G(u,v ) by the transformation w = f ( z ) . Show that if aSG aeG then a t all points where f ' ( z ) # O , = 0.
+
a'F a'F += axs
ayL
0,
+
az b Show t h a t by the bilinear transformation w = - where a d - bc # 0 , circles in the z plane are cz+d' transformed into circles of the w plane. If f ( z ) is analytic inside and on the circle Iz - a1 = R, prove Cauchy's inequality, namely, n!M If'"'(a)l 5 -
R"
where If(z)l S M on the circle.
[Hint: Use Cauchy's integral formulas.]
92. Let C1 and C2 be concentric circles having center a and radii
+
If a h is any point in the annular region bounded by prove Laurent's theorem that f(a+h) =
rl
and
r2
respectively, where r l < ra.
CIand C2, and f ( z ) is analytic in this region,
5 anhn
--m
where
C being any closed curve in the angular region surrounding CI. LHint: Write
f ( a + h)
two different ways.
1
=
2ni 1 fa
f (2) dz - (a h) +
z - (U+ h)
and expand
-a -h
in
372
FUNCTIONS OF A COMPLEX VARIABLE
93. Find a Laurent series expansion for the function f ( z ) = ( z + l;z
and diverges elsewhere.
+
[CHAP. 17
2) which converges for 1< IzI
<2
-1
+ 94. Let
f*e-.'F(t) dt
2 0
=
f(8)
where
f(8)
'*.
is a given rational function with numerator of degree less
than that of the denominator. If C is a simple closed curve enclosing all the poles of f ( s ) , we can show that F(t) = e"f(z) dz = sum of residuea of e" f ( z ) at its poles
&f
[Note that f ( 8 ) is the Laplace tramform of F ( t ) , and F ( t ) is the inverse Laplace transform of (see Chapter 12). Extensions to other functions f ( 8 ) are possible.] A m . (a)cos t , (b) i e - t sin 2t, (c) & Qte" teat, (d)+(sin t - t cos t )
+
+
f(8)
INDEX Abel, integral test of, 284 summability, 269 theorems of, 230, 249 Absolute convergence, of integrals, 262, 266, 270, 271 of series, 226, 233, 239, 240 theorems on, 227, 239 Absolute maximum or minimum, 2 1 (see also Maxima and minima) Absolute value, 3 of complex numbers, 6 Acceleration, 63, 140 centripetal, 166 in cylindrical and spherical coordinates, 169 normal and tangential components of, 166 Accumulation, point of, 6, 102 (see also Limit points) Addition, 1 associative law of, 2, 7 commutative law of, 2 formulas for elliptic functions, 341, 342, 344 (see also Elliptic functions) of complex numbers, 6, 12 of vectors, 134, 144 Aerodynamics, 363 Aleph-null, 4 Algebra, fundamental theorem of, 21 of complex numbers, 6, 12, 13 of vectors, 134, 136, 143-146 Algebraic functions, 22 Algebraic numbers, 6, 12 countability of, 12 Alternating series, 225, 226, 238, 239 convergence test for, 226, 226 error in computations using, 226, 238, 239 Amplitude, 6 of elliptic integrals, 331 Analytic continuation, of gamma function, 286 Analytic functions, 346 Analytic part, of a Laurent series, 348 Anti-derivatives, 82 Approximations (see also Numerical methods) by partial sums of Fourier series, 310 by use of differentials, 66, 66, 116 least square, 176 to irrational numbers, 8 using Newton’s method, 79 using Taylor’s theorem, 70, 86, 93 Arc length, 94 element, 142, 143, 163 Area, 80, 93 expressed as a line integral, 204 of a n ellipse, 179 of a parallelogram, 137, 148
373
Argand diagram, 6 Argument, 6 Arithmetic mean, 9 Associative law, 2, 7 for vectors, 136, 143 Asymptotic series or expansions, 233, 234, 262, 263, 269
for gamma function, 286, 292 Axiomatic foundations, of complex numbers, 6 of real numbers, 3 of vector analysis, 138 Axis, real, 2 x , y and z, 110 Base, of logarithms, 3 Bernoulli numbers, 268 Bernoulli’s inequality, 14 Bessel functions, 232, 260, 267, 297 Bessel’s differential equation, 232, 260 Bessel’s inequality, 310, 320 Beta functions, 273, 286, 287, 289-292 relation, to gamma functions, 287, 290 Bilinear transformation, 371 (see also Fractional linear transformation) Binary scale, 16, 19 system, 1 Binomial coefficients, 18 series, 231, 232 theorem, 18 Bolzano-Weierstrass theorem (see Weierstrass-Bolzano theorem) Bonnet’s mean value theorem, 82 Boundary conditions, 300 Boundary point, 102 Boundary-value problems, and Fourier integrals, 328 and Fourier series, 300, 313, 314 in heat conduction, 313, 314, 328 in vibration of strings, 319 separation of variables method for solving, 313
Bounded functions, 20, 21 sequences, 42, 47-49 sets, 6 Bounds, lower and upper, 6, 11, 12 Box product, 137 Branches of a function, 2 1 Branch line, 367 Branch point, 28, 348, 367 Briggsian system of logarithms, 3 Calculus, fundamental theorem of integral, 82, 88, 89
374 Cardinal number of the continuum, 4 Cardioid, 99 Catenary, 98 Cauchy principal value, 263, 272 Cauchy-Riemann equations, 346, 361-363 derivation of, 362 in polar form, 371 Cauchy ' 8 , convergence criterion, 43, 60 form of remainder in Taylor's theorem, 61, 96, 231 generalized theorem of the mean, 61, 69 inequality, 371 integral formulas, 347, 363-366 theorem, 346, 347, 363-366 Centripetal acceleration, 166 CCsaro summability, 233, 262 Chain rules, 69, 106 for Jacobians, 108 Circle of convergence, 232 Class, 1 (see also Sets) Closed interval, 4 region, 102 set, 6, 11, 12, 102 Closure law or property, 1 Cluster point, 6, 102 (see also Limit points) Collection, 1 (see also Sets) Commutative law, 2 for dot products, 136 f o r vectors, 136, 143 Comparison test, for integrals, 261, 264, 268 for series, 226, 236, 236 Complementary modulus, 343 Completeness, of an orthonormal set, 310 Complex numbers, 6, 12, 13 absolute value of, 6 amplitude of, 6 argument of, 6 as ordered pairs of real numbers, 6 as vectors, 18 axiomatic foundations of, 6 conjugate of, 6 equality of, 6 modulus of, 6 operations with, 6, 12, 13 polar form of, 6, 7, 13 real and imaginary parts of, 6 roots of, 7, 13 Complex plane, 6 Complex variable, 345 (see also Functions of a complex variable) Components, of a vector, 136 Composite functions, 26 continuity of, 36 differentiation of, 69, 106, 116-119 Conditional convergence, of integrals, 262, 266, 270, 271 of series, 226 Conductivity, thermal, 314 Conformal mapping or transformation, 366 (see also Transformations) Conjugate, complex, 6
INDEX Connected region, 197 set, 102 simply-, 102, 197, 204 Conservative field, 198 Constants, 4 Constraints, 164 Continued fraction, 62, 63, 66, 66 convergence of, 62, 63 convergents of, 62, 63 recurring, 62 Continuity, 20-40, 103, 104, 111, 112, 346, 360, 361 and differentiability, 67, 63, 64, 105, 113 definition of, 24, 26 in a n interval, 26 in a region, 104 of an infinite series of functions, 228, 229, 246 of functions of a complex variable, 345, 360, 361 of integrals, 89, 266 of vector functions, 139 right and left hand, 26 sectional or piecewise, 26 theorems on, 26, 26 uniform, 26, 104 Continuous (see Continuity) diff erentiability, 67, 106 Continuously differentiable functions, 67, 105 Continuum, cardinality of, 4 Contour integration, 349 Convergence, absolute (see Absolute convergence) circle of, 232 conditional (see Conditional convergence) criterion of Cauchy, 43, 60 domain of, 228 interval of, 61 of continued fractions, 62, 53 of Fourier integrals (see Fourier's integral theorem) of Fourier series, 299, 311-313 of improper integrals (see Improper integrals) of infinite series (see Infinite series) of series of constants, 234, 235 radius of, 229, 232 region of, 109 uniform (see Uniform convergence) Convergent (see also Convergence) integrals, 260-266 (see d s o Improper integrals) of continued fraction, 62, 63 sequences, 41, 227 (sec also Sequences) series, 43 (see also Infinite series) Convolution theorem, for Fourier transforms, 323 for Laplace transforms, 284 Coordinate curve, 141 Coordinates, curvilinear, 109 (see also Curvilinear coordinates) cylindrical, 142, 163, 164 hyperbolic, 186 polar, 6
INDEX Coordinates (cont.) rectangular, 6, 101, 141 spherical, 143, 163, 164 Correspondence, 2, 10, 20, 41, 101, 141 one to one, 2, 10 Countability, 4, 10, 11 of algebraic numbers, 12 of rational numbers, 10, 11 Countable set, 4, 10, 11 measure of a, 81, 87 Critical points, 63 Cross products, 137, 146-148 proof of distributive law for, 147 Curl, 140, 151, 152 in curvilinear coordinates, 142 Curvature, radius of, 166, 169 Curve, coordinate, 141 simple closed, 102, 197, 204 space, 139 Curvilinear coordinates, 109 curl, divergence, gradient and Laplacian in, 142 Jacobians and, 141, 142 multiple integrals in, 181, 182,217, 218 orthogonal, 142 special, 142, 143 transformations and, 123, 124, 141 vectors and, 141, 142 Cut (see Dedekind cut) Cycloid, 99 Cylindrical coordinates, 142, 153, 164 arc length element in, 142, 163 divergence in, 164 gradient in, 164 Laplacian in, 142, 164 multiple integrals in, 189 parabolic, 168 volume element in, 142, 163 Decimal representation of real numbers, 1 Decimals, recurring, 1 Decreasing functions, 21, 26 monotonic, 21 strictly, 21, 26 Decreasing sequences, monotonic and strictly, 42 Dedekind cuts, 4, 16 Definite integrals, 80, 81, 86, 87 (see also Integrals) change of variable in, 83, 89-92 definition of, 80, 81 mean value theorems for, 81, 82, 88 numerical methods f o r evaluating, 86, 92, 93 properties of, 81, 87, 88 theorem for existence of, 81 with variable limits, 83, 163, 170, 266 Degree, of a polynomial equation, 6 of homogeneous functions, 106 Del (V),140 formulas involving, 141 in curl, gradient and divergence, 140 Deleted neighborhood, 6, 102 De Moivre’s theorem, 7, 13 Denominator, 1
375 Dense, everywhere, 2 Denumerable set (sec Countable set) Dependent variable, 20, 101 Derivatives, 67-79,101-133 (see also Differentiation) chain rules for, 69, 106 continuity and, 67, 63, 64, 106, 113 definition of, 57, 104 directional, 163, 169 evaluation of, 63, 64 graphical interpretation of, 68 higher order, 60, 105 of elliptic functions, 338-340 of functions of a complex variable, 346, 361-363 of infinite series of functions, 229, 247 of special functions, 60, 66-68 of vector functions, 139, 160, 161 partial (see Partial derivatives) right and left hand, 67, 64, 66 rules for finding, 69, 66-68 table of, 60 Determinant, for cross product, 137 for curl, 140 for scalar triple product, 137 Jacobian (see Jacobian) Dextral system, 136 Difference equations, 66, 286 Differentiability, 67, 106 and continuity, 67, 63, 64 continuous, 67 sectional or piecewise, F7 Differential equation, Bessel’s, 232, 260 Gauss’, 232 solution of, by Laplace transforms, 267, 280 Differential geometry, 140, 169 Differentials, 68, 69, 66, 66, 106, 114-116 approximations by use of, 66, 66, 116 exact, 106, 116, 116, 198 geometric interpretation of, 69 of functions of one variable, 6 of functions of several variables, 106 of vector functions, 139 total, 105 Differentiation (see also Derivatives) of Fourier series, 300, 311 rules for, 69, 66-68 under the integral sign, 163, 170, 266 Diffusivity, 314 Directed line segments, 134 Directional derivatives, 163, 169 Dirichlet conditions, 299 integrals, 287, 292, 293 Dirichlet’s test, for integrals, 266 for series, 228, 268 Discontinuities, 26, 104 removable, 34, 104 Distance between points, 146 Distributive law, 2 for cross products, 137 for dot products, 136
INDEX Divergence, 140, 161 in curvilinear coordinates, 142 in cylindrical coordinates, 164 of improper integrals, 260-266 (see also Improper integrals) of infinite series (see Infinite series) Divergence theorem, 199, 200, 210-213 proof of, 210, 211 Divergent integrals, 260-266 sequences, 41 (see &o Sequences) series, 43 (see &o Series) Division, 1 by zero, 8 of complex numbers, 7, 12 Domain, of a function, 20, 101 of convergence, 228 Dot products, 136, 137, 145, 146 commutative law for, 136 distributive law for, 136 laws for, 136, 137 Double series, 233 Doubly-periodic functions, 340 Dummy variable, 83 Duplication formula for gamma function, 286, 293, 294
3 proof of irrationality of, 70, 71 Electric field vector, 169 Electromagnetic theory, 169 Elementary transcendental functions, 22, 23 of a complex variable, 346, 346 Elements, of a set, 1 Ellipse, 99 area of, 179 length of arc of, 336 Elliptic functions, 332, 338-340 (see also Elliptic integrals) addition formulas for, 341, 342, 344 derivatives of, 338-340 hyper-, 332 inverse, 332 Jacobi’s, 332 periods of, 339, 340 Elliptic integrals, 331-344 (see also El!iptic functions) complementary modulus of, 343 complete and incomplete, 331 Jacobi’s forms for, 331, 332 Landen’s transformation for, 332, 333 Legendre’s forms for, 331 modulus of, 331 of the first kind, 331-338 of the second kind, 331-338 of the third kind, 331, 332, 338 Empty set, 1 Envelopes, 162, 163, 168, 169 Equality, of complex numbers, 6 of vectors, 134 Equations, difference, 66, 286 differential (see Differential equation) integral, 324, 328, 329
6,
Equations (cont.) polynomial, 6, 21 Equipotential surfaces, 163 Errors, applications to, 164, 174 in computing sums of alternating series, 226, 238, 239 mean square, 310 Essential singularity, 348, 368 Euler’s, constant, 261, 286, 296 formulas or identities, 7, 261 theorem on homogeneous functions, 106 Even functions, 299, 306-309 Everywhere dense set, 2 Evolute, 176 Exact differentials, 106, 116, 116, 198 (see also Differentials) Expansion of functions, in Fourier series (see Fourier series) in power series, 231 Expansions (see Series) Explicit functions, 107 Exponential function, 22 order, 283 Exponents, 3, 10 Factorial function (see Gamma functions) Fibonacci sequence, 63, 66 Field, 2 conservative, 198 scalar, 138 vector, 138 Fluid mechanics, 363 Fourier coefficients, 298 expansion (see Fourier series) Fourier integrals, 321-330 (see also Fourier transforms) convergence of (see Fourier’s integral theorem) solution of boundary-value problems by, 328 Fourier series, 298-320 complex notation for, 300 convergence of, 299, 311-313 differentiation and integration of, 300, 311 Dirichlet conditions for convergence of, 299 half range, 299, 300, 306-309 Parseval’s identity for, 300, 309, 310 solution of boundary-value problems by, 300, 313, 314 Fourier’s integral theorem, 321, 323-326 heuristic demonstration of, 326 proof of, 326-328 Fourier transforms, 322-326 (see also Fourier integrals) convolution theorem for, 323 inverse, 322 Parselval’s identities for, 322, 323, 325 symmetric form for, 322 Fractional linear transformation, 336 (see also Bilinear transformation) Fractions, 1 continued (see Continued fractions) Frenet-Serret formulas, 159 Fresnel integrals, 294
INDEX Frullani’s integral, 282 Functional determinant 107, 120 (see also Jacobians) Functional notation, 20, 101 Functions, 20-40,101, 107, 346 algebraic, 22 Bessel, 232, 260, 267, 297 beta (see Beta functions) bounded, 20, 21 branches of, 21 composite (see Composite functions) continuity of (see Continuity) decreasing, 21, 26 definition of, 20, 101 derivatives of (see Derivatives) differential of (see Differentials) domain of, 20, 101 double-valued, 367 doubly-periodic, 340 elementary transcendental, 22, 23 elliptic (see Elliptic functions) even, 299, 306-309 explicit and implicit, 107 gamma (see Gamma functions) harmonic, 346 hyperbolic, 22, 23 hypergeometric, 232, 267 increasing, 21, 26 inverse (see Inverse functions) limits of (see Limits of functions) maxima and minima of (see Maxima and minima) monotonic, 21 multiple-valued (see Multiple-valued function) normalized, 301 odd, 299, 306-309 of a complex variable (see Functions of a complex variable) of a function (see Composite function) of several variables, 101, 110, 111 orthogonal, 301, 314, 316 orthononnal, 301 periodic, 298 polynomial, 21 sequences and series of, 227, 228, 232, 242, 243 single-valued, 20, 101, 346 staircase or step, 28 transcendental, 22, 23 types of, 21, 22 value of, 20 vector (see Vector functions) Functions of a complex variable, 346-372 analytic, 346 Cauchy-Riemann equations and (see Cauchy-Riemann equations) continuity of, 346, 360, 361 definition of, 346 derivatives of, 346, 361-363 elementary, 346, 346 imaginary part of, 346, 362, 363 integrals of, 346, 363-366
377 Functions of a complex variable (cont.) Jacobians and, 371 Laplace transforms and, 372 limits of, 346, 360, 361 line integrals and, 346 multiple-valued, 346 poles of, 347, 348 real part of, 346, 362, 363 residue theorem for (see Residue theorem) series of, 347, 367-360 single-valued, 346 singular points of, 347 Fundamental theorem, of algebra, 21 of integral calculus, 82, 88, 89 Gamma functions, 274, 286-297 analytic continuation of, 286 asymptotic formulas for, 286 duplication formula for, 286, 293, 294 infinite product for, 286 recurrence formula for, 286, 287 Stirling’s formulas and asymptotic series for, 286, 292 table and graph of, 286 Gauss’, differential equation, 232 r function, 286 test, 227, 241 Generalized theorem of the mean, 61, 69 Geometric integral, 261 mean, 9 series, 61, 224 G.1.b (see Greatest lower bound) Gradient, 140, 161, 162 in curvilinear coordinates, 142 in cylindrical coordinates, 164 Graph, of a function of one variable, 20 of a function of two variables, 110, 111 Greater than, 2 Greatest limit (see Limit superior) Greatest lower bound, 6 of a function, 21 of a sequence, 42, 43, 49 Green’s theorem in the plane, 197, 202-206 in space (see Divergence theorem) Grouping method, for exact differentials, 116 Gyration, radius of, 189, 190 Half range Fourier sine or cosine series, 299, 300, 306-309 Harmonic functions, 346 series, 226 Heat conduction equation, 313, 314 solution of, by Fourier integrals, 328 solution of, by Fourier series, 313, 314 Homogeneous functions, Euler’s theorem on, 106 Hyperbolic coordinates, 186 Hyperbolic functions, 22, 23 inverse, 23 Hyperboloid of one sheet, 111 Hyperelliptic functions, 332
378 Hypergeometric function or series, 232, 267 Hypersphere, 101 Hypersurface, 101 Hypocycloid, 207 area bounded by, 220 Identity, with respect to addition and multiplication, 2 Image or mapping, 108, 366 Imaginary part, of a complex number, 6 of functions of a complex variable, 346, 362, 353 Imaginary unit, 6 Implicit functions, 107 and Jacobians, 119-123 Improper integrals, 86, 96, 99, 260-284 absolute and conditional convergence of, 262, 266, 270, 271 comparison test for, 261, 264, 268 containing a parameter, 266 definition of, 260 of the first kind, 260-262, 268-270 of the second kind, 260, 263-266,267, 272,273 of the third kind, 260, 266, 273, 274 quotient test for, 262, 264, 268 uniform convergence of, 266, 266, 274, 276 Weierstrass M test for, 266, 274-279 Increasing functions, 21, 26 monotonic, 21 strictly, 21, 26 Increasing sequences, monotonic and strictly, 42 Indefinite integrals, 82 (see also Integrals) Independence of the path, 197, 198, 206-207, 216 Independent variable, 20, 101 Indeterminate foims, 33, 62, 71-74 L’Hospital’s rules f o r (see L’Hospital’s rules) Taylor’s theorem and, 62, 72-74 Induction, mathematical, 7, 14 Inequalities, 2, 9, 10 Inequality, Bernoulli’s, 14 Bessel’s, 310, 320 Cauchy’s, 371 Schwarz’s, 9, 16, 94 Inferior limit (see Limit inferior) Infinite, countably, 4 dimensional vectors, 301 interval, 4 Infinite product, 233, 261, 268 f o r gamma function, 286 for sin z, 316, 316 Wallis’, 316 Infinite series, 43, 61, 224-269 (see also Series) absolute convergence of, 226, 233, 239, 240 comparison test for, 226, 235, 236 conditional convergence of, 226 convergence testa for, 226-227 functions defined by, 232 Gauss’ test for, 227, 241 integral test for, 226, 236-238 nth root test for, 226 of complex terms, 232
INDEX Infinite series (cont.) of functions, 227, 228, 232, 242, 243 partial sums of, 43, 224 quotient test for, 226 Raabe’s test for, 226, 227, 241 ratio test for, 226, 240, 241 rearrangement of terms in, 227, 266 special, 224 uniform convergence of, 227, 228 (see also Uniform convergence) Weierstrass M test for, 228, 246, 246 Infinitesimal, 66, 79 Infinity, 24, 42 Inflection, point of, 79 Initial point, of a vector, 134 Integers, positive and negative, 1 Integrable, 81 Integral calculus, fundamental theorem of, 82, 88, 89 Integral equations, 324, 328, 329 Integral formulas of Cauchy, 347, 363-366 Integrals, 80-100, 180-223, 260-297, 321-344, 346, 347, 349, 363-366, 360-368 (see also Integration) definite, 80, 81 (see also Definite integrals) Dirichlet, 287, 292, 293 double, 180 elliptic (see Elliptic integrals) evaluation of, 267, 276-278, 349, 362-366 Fresnel, 294 Frullani’s, 282 improper (see Improper integrals) indefinite, 82 iterated, 180, 181 line (see Line integrals) mean value theorems for, 81, 82 multiple (see Multiple integrals) of functions of a complex variable, 346, 363-366 of infinite series of functions, 229, 246 of special functions, 83, 84 Schwarz’s inequality for, 94 table of, 83, 84 transformations of, 83, 89-92, 181, 182, 188-191 uniform convergence of, 266, 266, 274, 276 Integral test f o r infinite series, 226, 236-238 Integrand, 80 Integrating factor, 223 Integration, applications of, 86, 93, 94 (see also Integrals) by parts, 84, 98 contour, 349 interchange of order of, 181 limits of, 80 of Fourier series, 300, 311 of special functions, 83, 84 partial fractions used in, 84, 91, 96 range of, 80 special methods of, 84, 86, 89-92 under integral sign, 163, 164, 171 Intercepts, 110 Interior point, 102 Intermediate value theorem, 26 Intersection of sets, 11
379
INDEX Intervals, closed, 4 continuity in, 25 infinite, 4 nested, 43, 50 of convergence, 61 open, 4 unbounded, 4 Invariance relations, 159, 160 Invariant, scalar, 160 Inverse elliptic functions, 332 Fourier transforms, 322 (see also Fourier transforms) Laplace transforms, 280, 372 (see also Laplace transforms) Inverse functions, 21 continuity of, 26 hyperbolic, 23 trigonometric, 22 Inverse, of addition and multiplication, 1, 2 Irrational algebraic functions, 22 Irrationality of fl,proof of, 8 Irrational numbers, 1, 2, 8, 9 approximations to, 8 definition of, 1 (see also Dedekind cut) Isolated singularity, 347 Iterated integrals, 180, 181 limits, 103 Jacobian determinant (see Jacobians) Jacobians, 107, 119-123, 141, 142, 153, 154 chain rules for, 108 curvilinear coordinates and, 141, 142 functions of a complex variable and, 371 implicit functions and, 119-123 multiple integrals and, 181 of transformations, 108, 142 partial derivatives using, 107 theorems on, 108 vector interpretation of, 141 Jacobi’s elliptic functions, 332 addition formulas for, 341, 342, 344 Jacobi’s forms f o r elliptic integrals, 331, 332 Kronecker’s symbol, 301 Lagrange multipliers, 164, 172-174 Lagrange’s form of the remainder, in Taylor series, 61, 95, 231 Landen’s transformation, 332, 333 Laplace’s equation, 113 (see also Laplacian operator) Laplace transforms, 267, 279, 280, 284, 372 convolution theorem for, 284 inverse, 280, 372 relation of functions of a complex variable to, 372 table of, 267 use of, in solving differential equations, 267, 280 Laplacian operator, 141 (see also Laplace’s equation)
Laplacian operator (cont.) in curvilinear coordinates, 142 in cylindrical coordinates, 142, 164 in spherical coordinates, 143 Laurent’s series, 348, 369, 360 theorem, 371 Least limit (sec Limit inferior) Least square approximations, 175 Least upper bound, 5 of functions, 21 of sequences, 42, 43, 49 Left hand continuity, 25 derivatives, 67, 64, 65 limits, 23 Legendre’s forms for elliptic integrals, 331 Leibnitz’s formula for nth derivative of a product, 79 rule for differentiating under the integral sign, 163, 170, 266 Lemniscate, 99 Length, of a vector, 134 Less than, 2 Level curves and surfaces, 128, 163 L’Hospital’s rules, 62, 71-74 proofs of, 71, 72 Limit inferior, 43, 49 Limit points, 5, 11, 12, 102 Weierstrass-Bolzano theorem on (see Weierstrass-Bolzano theorem) Limits of functions, 20-40, 102, 103, 111, 112, 345, 350, 351 definition of, 23 iterated, 180, 181 of a complex variable, 345, 350, 361 proofs of theorems on, 31-33 right and left hand, 23 special, 24 theorems on, 24 Limits of integration, 80 Limits of sequences, 41, 44, 45, 227 definition of, 41 of functions, 227 theorems on, 41, 42, 45-47 Limits of vector functions, 139 Limit superior, 43, 49 Linear dependence of vectors, 160 Linear transformations, 131 fractional (see Fractional linear transformation) Line integrals, 196-198, 200-202 evaluation of, 196 independence of path of, 197, 198, 206-207, 215 properties of, 196, 197 relation of, to functions of a complex variable, 346 vector notation for, 196 Line, normal (see Normal line) tangent (see Tangent line) Logarithms, 3, 10, 351 as multiple-valued functions, 351 base of, 3
380 Lower bound, 5, 11, 12 of functions, 21 of sequences, 42 Lower limit (see Limit inferior) L.u.b. (see Least upper bound) Maclaurin series, 61, 231 Magnetic field vector, 169 Magnitude, of a vector, 134 Many-valued function (see Multiple-valued function) Mappings, 108 (aee also Transformations) conformal, 366 Mathematical induction, 7, 14 Maxima and minima, 21, 62, 63, 74, 76, 164, 171-174 absolute, 21 Lagrange’s multiplier method for, 164, 172-174 of functions of one variable, 62, 63, 74, 75 of functions of several variables, 164, 171-174 relative, 21 Taylor’s theorem and, 74, 75, 171, 172 Maximum (see Maxima and minima) Maxwell’s equations, 159 Mean, arithmetic, 9 geometric, 9 Mean square error, 310 Mean, theorem or law of the, 61, 69 proof of, 68 Mean value theorems, f o r derivatives, 61, 68-71, 109, 124, 125 f o r integrals, 81, 82, 88, 94 Measure zero, 81, 87 Mechanics, 140 fluid, 363 Members, of a set, 1 Minimum (see Maxima and minima) Modulus, complementary, 343 of a complex number, 6 of elliptic integrals, 331 Moebius strip, 210 Moment of inertia, 93 polar, 182, 186 Monotonic functions, 21 Monotonic sequences, 42, 47-49 fundamental theorem on, 42 Multiple integrals, 180-194 improper, 267 in curvilinear coordinates, 181, 182, 217, 218 in cylindrical cool dinates, 189 in spherical coordinates, 189 Jacobians and, 181 transformations of, 181, 182 Multiple-valued functions, 20, 21, 101, 345 logarithm as a, 351 Multiplication, 1 associative law of, 7 involving vectors, 136 of complex numbers, 7, 12 Multiply-connected regions, 102
INDEX Napierian system of logarithms, 3 Natural base of logarithms, 3 Natural numbers, 1 Negative integers, 1 numbers, 1, 2 Neighborhoods, 5, 102 Nested intervals, 43, 50 Newton’s method, 79 Normal component of acceleration, 156 Normalized vectors and functions, 301 Normal line, parametric equations for, 161 principal, 166, 169 to a surface, 140, 152, 161, 165-167 Normal plane, 162, 167, 168 nth root test, 226 Null set, 1 vector, 136 Number, cardinal, 4 Numbers, 1-19 algebraic (see Algebraic numbers) Bernoulli, 268 complex (sec Complex numbers) irrational (see Irrational numbers) natural, 1 negative, 1, 2 operations with, 2, 6-8, 12, 13 positive, 1, 2 rational (see Rational numbers) real (see Real numbers) roots of, 3 transcendental, 6, 12 Numerator, 1 Numerical methods (see also Approximations) for evaluating definite integrals, 86, 92, 93 Odd functions, 299, 306-309 Open interval, 4 region, 102, 110 Operations, with complex numbers, 6, 12, 13 with power series, 230, 231 with real numbers, 2, 7, 8 Ordered pairs of real numbers, 6 triplets of real numbers, 138 Order, exponential, 283 of derivatives, 60 of poles, 347, 348 Orientable surface, 210 Origin, of a coordinate system, 101 Orthogonal curvilinear coordinates (see Curvilinear coordinates) Orthogonal families, 352, 353 functions, 301, 314, 315 Orthonormal functions, 301 Pappus’ theorem, 194 Parabola, 27 Parabolic cylindrical coordinates, 158 Parallelepiped, volume of, 137, 148, 149 Parallelogram, area of, 137, 148 law, 18, 134, 144
381
INDEX Parametric equations, of line, 165 of normal line, 161 of space curve, 139 Parseval’s identity, f o r Fourier integrals, 322, 323, 325 f o r Fourier series, 300, 309, 310 Partial derivatives, 101-133 applications of, 161-179 definition of, 104 evaluation of, 112-114 higher order, 106 notations for, 104 order of differentiation of, 105 using Jacobians, 107 Partial fractions, 84, 91, 96 Partial sums of infinite series, 43, 224 Pendulum, period of a simple, 340, 341 Period, of a function, 298 of a simple pendulum, 340, 341 of elliptic functions, 339, 340 Piecewise continuous, 26 differentiable, 67 p integrals, 261, 263 Plane, complex, 6 Plane, equation of, 149 normal to a curve (see Normal plane) tangent to a surface (see Tangent plane) Point, boundary, 102 branch, 28, 348, 367 cluster, 6, 102 (see also Limit points) critical, 63 interior, 102 limit (see Limit points) neighborhood of, 6, 102 of accumulation, 5 (see also Limit points) singular (see Singular points) Point set, one dimensional, 4 two dimensional, 101 Polar coordinates, 6 Polar form, of complex numbers, 6, 7, 13 Poles, 347, 348 defined from a Laurent series, 348 of infinite order, 348 residues at, 348 Polynomial equations, 6, 21 degree of, 6 Polynomial functions, 21 Position vector, 136 Positive definite quadratic form, 179 Positive direction, 197 normal, 199 Positive integers, 1 numbers, 1, 2 Potential, velocity, 363 Power series, 61, 229, 248-250 Abel’s theorem on, 230 expansion of functions in, 231 operations with, 230, 231 radius of convergence of, 229 special, 231, 232 theorems on, 230
Power series (cont.) uniform convergence of, 230 Prime, relatively, 8 Principal branch, of a function, 21 of a logarithm, 351 Principal normal, to a space curve, 155, 159 Principal part, 69, 105 of a Laurent series, 348 Principal value, of functions, 21, 22 of integrals (see Cauchy principal value) of inverse hyperbolic functions, 23 of inverse trigonometric functions, 22 of logarithms, 351 Product, 1 box, 137 cross or vector (see Cross products) dot or scalar (see Dot products) infinite (see Infinite product) nth derivative of, 79 triple (see Triple products) Wallis’, 316 Projection, of a vector, 195 p series, 225 .Quadratic equation, solutions of, 13 Quadratic form, 179 Quotient, 1 Quotient test, for integrals, 262, 264, 268 f o r series, 225, 235, 236 Raabe’s test, 226, 227, 241 Radius of convergence, 229, 232 of curvature, 156, 159 of gyration, 189, 190 of torsion, 159 Radius vector, 136 Range, of integration, 80 Rates of change, 63 Rational algebraic functions, 22 Rational numbers, 1, 8, 9 countability of, 10, 11 Ratio test, 226, 240, 241 proof of, 240 Real axis, 2 Real numbers, 1 (see aZso Numbers) absolute value of, 3 axiomatic foundations of, 3 decimal representation of, 1 geometric representation of, 2 inequalities for (see Inequality) non-countability of, 11 operations with, 2, 7, 8 ordered pairs and triplets of, 6, 138 roots of, 3, 10 Real part, of a complex number, 6 of functions of a complex variable, 345, 352, 353 Rectangular component vectors, 136 Rectangular coordinates, 6, 101, 141
INDEX Rectangular neighborhood, 102 rule for integration, 86 Recurring continued fraction, 62 decimal, 1 Region, 102 closed, 102 connected, 197 multiply-connected, 102 of convergence, 109 open, 102, 110 simply-connected, 102, 197, 204 Regular summability, 233, 268 Relativity, theory of, 160 Removable discontinuity, 34, 104 singularity, 348, 368 Residues, 348, 360-362 Residue theorem, 348, 349, 360-362 evaluation of integrals by, 349, 362-366 proof of, 361 Resultant of vectors, 134, 144 Reversion of series, 231 Riemann integrable, 81 Riemann’s theorem, 311, 326, 327 Right hand continuity, 26 derivatives, 67, 64, 66 limits, 23 Right handed rectangular coordinate system, 136, 136 Rolle’s theorem, 61 proof of, 68 Roots, of complex numbers, 7, 13 of real numbers, 3, 10 Roots of equations, 21 computations of, 36 Newton’s method for finding, 79 Saddle points, 172 Scalar, 134 field, 138 invariant, 160 product (see Dot products) triple product, 137, 138 Scale factors, 141 Scale of two (see Binary scale) Schwards inequality, for integrals, 94 f o r real numbers, 9, 16 Section (see Dedekind cut) Sectional continuity, 26 differentiability, 67 integration and, 81 Separation of variables, in boundary-value problems, 313 Sequence, Fibonacci, 63, 66 Sequences, 41-66,227 bounded, monotonic, 42, 47-49 convergent and divergent, 41, 227 decreasing, 42 definition of, 41 finite and infinite, increasing, 42
Sequences (cont.) limits of, 41, 227 (see also Limits of sequences) of functions, 227 terms of, 41 uniform convergence of, 227 Series (see also Infinite series) alternating (see Alternating series) asymptotic (see Asymptotic series) binomial, 231, 232 double, 233 geometric, 61, 224 harmonic, 226 Laurent’s, 348, 369, 360, 371 Maclaurin, 61, 231 of functions of a complex variable, 367-360 p-, 226 partial sums of, 43, 224 power (see Power series) reversion of, 231 special, 62 sum of, 43, 224 Taylor (see Taylor series) telescoping, 234 terms of, 224 test for integrals, 262 Sets, 1 bounded, 6 closed, 6, 11, 12 connected, 102 countable or denumerable (see Countable set) elements of, 1 everywhere dense, 2 intersection of, 11 orthonormal, 301 point, 4, 101 union of, 11 Simple closed curves, 102, 197, 204 Simple poles, 347 Simply connected region, 102, 197, 204 Simpson’s rule, 86, 92, 93 Single-valued function, 20, 101, 346 Singular points or singularities, 124, 260, 347, 367-360 defined from Laurent series, 348 essential, 348, 368 isolated, 347 removable, 348, 368 Sink, 219 Slope, 68 Smooth function (see Sectional differentiability) Solenoidal vector fields, 219 Source, 219 Space curve, 139 Specific heat, 314 Spherical coordinates, 143, 163, 164 arc length element in, 143, 163 divergence in, 168 gradient in, 168 Laplacian in, 143, 168 multiple integrals in, 189 volume element in, 143, 163, 164 Staircase or step function, 28 Stirling’s asymptotic formula and series, 286, 292
383
INDEX Stokes’ theorem, 200, 2 3-217 proof of, 213, 214 Stream function, 353 Subset, 1 Subtraction, 1 of complex numbers, 12 of vectors, 134 Sum, 1 of series, 43, 224 of vectors, 134, 144 partial, 43, 224 Summability, 233, 252, 258 Abel, 269 Chsaro, 233, 252 regular, 233, 258 Superior limit (see Limit superior) Superposition, principle of, 314 Surface, 101 equipotential, 163 level, 128, 163 normal line to (see Normal line) orientable, 210 tangent plane to (see Tangent plane) Surface integrals, 198, 199, 207-210 Tangential component of acceleration, 156 Tangent line, to a coordinate curve, 141 to a curve, 58, 162, 167, 168 Tangent plane, 161, 165-167 in curvilinear coordinates, 166, 167 Tangent vector, 139, 159 Taylor series, in one variable, 61, 231 (see also Taylor’s theorem of th e mean) in several variables, 109 of functions of a complex variable, 347 uniqueness of, 257 Taylor’s theorem of the mean, 61, 109, 124, 125 (see also Taylor series) approximations using, 70 f o r functions of one variable, 61 f o r functions of several variables, 109, 124, 125 in approximate integration, 85, 93 indeterminate forms and, 62, 72-74 proof of, 70, 125, 358 remainder in, 61, 95, 231 Telescoping series, 234 Tensor analysis, 160 Term, of a sequence, 41 of a series, 224 Terminal point of a vector, 134 Thermal conductivity, 314 Thermodynamics, 132 Torsion, radius of, 159 Total differential, 105 (see also Differentials) Trace, on a plane, 111 Transcendental functions, 22, 23 numbers, 5, 12 Transformations, 108, 123, 124 and curvilinear coordinates, 123, 124, 141 bilinear o r fractional linear, 336, 371 conformal, 366 Jacobians of, 108, 142
Transformations (cont.) Landen’s, 332, 333 linear, 131 of integrals, 83, 89-92, 181, 182, 188-191 Transforms (see Fourier transforms and Laplace transforms) Transitivity, law of, 2 Trigonometric functions, 22 derivatives of, 60 integrals of, 83, 84 inverse, 22 Triple integrals, 181, 186-188 transformation of, 182, 188-191 Triple products, scalar, 137, 138 vector, 137 Unbounded interval, 4 Uniform continuity, 26, 104 Uniform convergence, 227, 228, 243-245 of integrals, 265, 266, 274, 275 of power series, 230 of sequences, 227 of series, 227, 228 tests for integrals, 266 tests for series, 228 theorems for integrals, 266 theorems for series, 228, 229, 246, 247 Weierstrass M test for (see Weierstrass M test) Union of sets, 11 Unit tangent vector, 139 Unit vectors, 135, 136, 301 infinite dimensional, 301 rectangular, 135, 136 Upper bound, 5 of functions, 20, 21 of sequences, 42 Upper limit (see Limit superior) Variable, 4, 20 change of, in differentiation, 59, 106 change of, in integration, 83, 89-92, 181, 182 complex, 345 (see also Functions of a complex variable) dependent and independent, 20, 101 dummy, 83 limits of integration, 83, 163, 170, 266 Vector algebra, 134, 135, 143-145 Vector analysis (see Vectors) Vector field, 138 solenoidal, 219 Vector functions, 138 limits, continuity and derivatives of, 139, 150, 151 Vector product (see Cross products) Vectors, 18, 134-160 algebra of, 134, 135, 143-145 axiomatic foundations for, 138 complex numbers as, 18 components of, 136 curvilinear coordinates and, 141, 142 equality of, 134 infinite dimensional, 301
384 Vectors (cont.) Jacobians interpreted in terms of, 141 length or magnitude of, 134 normalized, 301 null, 136 position, 136 radius, 136 resultant or sum of, 134, 144 tangent, 139, 169 unit, 136, 136, 301 Vector triple product, 137 Velocity, 63, 139 of light, 169 potential, 363 Vibrating string, equation of, 319 Volume, 94 element of, 142, 143, 163, 164 of parallelepiped, 137, 148, 149 Wallis’ product, 316 Wave equation, 319
INDEX Weierstrass-Bolzano theorem, 6, 11, 12, 42, 43, SO, 102
proof of, 60 Weierstrass M test, for integrals, 266, 274-279 for series, 228, 246, 246 Work, as a line integral, 196 2
axis, 101 intercept, 110
g
axis, 101 intercept, 110
z axis, 101
intercept, 110 Zero, 1 division by, 8 measure, 81, 87 vector, 136