Schaum's Outline of Calculus of Finite Differences and Difference Equations

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CALCULUS Of FINITE DIFFERENCES AND DIFFERENCE EQUATIONS MURRAY R. SPIEGEL

The perfect aid for better grades Covers all course fundamentals and supplements any class text

Teaches effective problem-solving Features fully worked problems

Ideal for independent study

THE ORIGINAL AND MOST POPULAR COLLEGE COURSE SERIES AROUND THE WORLD

SCHAUM'S OUTLINE OF

THEORY AND PROBLEMS OF

Calculus of FINITE DIFFERENCES and

DIFFERENCE EQUATIONS

MURRAY R. SPIEGEL, Ph.D. Formw Profesror and Clba .wa Mathematics Deport rt

Rmsd. tPolpteclbdcIre Hartford Graduate Cutter

SCHAUM'S OUTLINE SERIES McGraw-Hill. Inc. New York St. Louis San Francisco Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto

Copyright © 1971 by McGraw-Hill, Inc. All rights reserved. Printed in the United States of America. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. 07-060218-2

.2345678910 SH SH 75432106

Preface In recent years there has been an increasing interest in the calculus of finite differences and difference equations. There are several reasons for this. First, the advent of high

speed computers has led to a need for fundamental knowledge of this subject. Second, there are numerous applications of the subject to fields ranging from engineering, physics and chemistry to actuarial science, economics, psychology, biology, probability and statistics.

Third, the mathematical theory is of interest in itself especially in view of the remarkable analogy of the theory to that of differential and integral calculus and differential equations. This book is designed to be used either as a textbook for a formal course in the calculus of finite differences and difference equations or as a comprehensive supplement to all current standard texts. It should also be of considerable value to those taking courses in which difference methods are employed or to those interested in the field for self-study. Each chapter begins with a clear statement of pertinent definitions, principles and theorems, together with illustrative and other descriptive material. 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 formulas are included among the solved problems. The large number of supplementary problems with answers serves as a complete review of the material of each chapter. Topics covered include the difference calculus, the sum calculus and difference equations [analogous to differential calculus, integral calculus and differential equations respectively] together with many applications. Considerably more material has been included here than can be covered in most first 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 Nicola Monti, David Beckwith and Henry Hayden for their splendid cooperation. M. R. SPIEGEL

Rensselaer Polytechnic Institute November 1970

CONTENTS Page

Chapter

1

THE DIFFERENCE CALCULUS

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

1

........

32

Operators. Some Definitions Involving Operators. Algebra of Operators. The Difference Operator. The Translation or Shifting Operator. The Derivative Operator. The Differential Operator. Relationship Between Difference, Derivative and Differential Operators. General Rules of Differentiation. Derivatives of Special Functions. General Rules of the Difference Calculus. Factorial Polynomials. Stirling Numbers. Generalized Factorial Functions. Differences of Special Functions. Taylor Series. Taylor Series in Operator Form. The Gregory-Newton Formula. Leibnitz's Rule. Other Difference Operators.

Chapter

2

APPLICATIONS OF THE DIFFERENCE CALCULUS

Subscript Notation. Difference Tables. Differences of Polynomials. GregoryNewton Formula in Subscript Notation. General Term of a Sequence or Series. Interpolation and Extrapolation. Central Difference Tables. Generalized Interpolation Formulas. Zig-Zag Paths and Lozenge Diagrams. Lagrange's Interpolation Formula. Tables with Missing Entries. Divided Differences. Newton's Divided Difference Interpolation Formula. Inverse Interpolation. Approximate Differentiation.

Chapter

3

THE SUM CALCULUS .........................................

79

The Integral Operator. General Rules of Integration. Integrals of Special Functions. Definite Integrals. Fundamental Theorem of Integral Calculus. Some Important Properties of Definite Integrals. Some Important Theorems of Integral Calculus. The Sum Operator. General Rules of Summation. Summations of Special Functions. Definite Sums and the Fundamental Theorem of Sum Calculus. Differentiation and Integration of Sums. Theorems on Summation Using the Subscript Notation. Abel's Transformation. Operator Methods for Summation. Summation of Series. The Gamma Function. Bernoulli Numbers and Polynomials. Important Properties of Bernoulli Numbers and Polynomials. Euler Numbers and Polynomials. Important Properties of Euler Numbers and Polynomials.

Chapter

4

APPLICATIONS OF THE SUM CALCULUS .................. 121 Some Special Methods for Exact Summation of Series. Series of Constants. Power Series. Approximate Integration. Error Terms in Approximate Inte-

gration Formulas. Gregory's Formula for Approximate Integration. The Euler-Maclaurin Formula. Error Term in Euler-Maclaurin Formula. Stirling's Formula for n!

CONTENTS Chapter

5

DIFFERENCE EQUATIONS

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

Page 150

Differential Equations. Definition of a Difference Equation. Order of a Difference Equation. Solution, General Solution and Particular Solution of a Difference Equation. Differential Equations as Limits of Difference Equations. Use of the Subscript Notation. Linear Difference Equations. Homogeneous Linear Difference Equations. Homogeneous Linear Difference Equations with Constant Coefficients. Linearly Independent Solutions. Solution of the Nonhomogeneous or Complete Equation. Methods of Finding Particular Solutions. Method of Undetermined Coefficients. Special Operator Methods. Method of Variation of Parameters. Method of Reduction of Order. Method of Generating Functions. Linear Difference Equations with Variable Coefficients. SturmLiouville Difference Equations. Nonlinear Difference Equations. Simultaneous Difference Equations. Mixed Difference Equations. Partial Difference Equations.

Chapter

6

APPLICATIONS OF DIFFERENCE EQUATIONS ............ 199 Formulation of Problems Involving Difference Equations. Applications to Vibrating Systems. Applications to Electrical Networks. Applications to Beams. Applications to Collisions. Applications to Probability. The Fibonacci Numbers. Miscellaneous Applications.

..........

Appendix A

Stirling Numbers of the First Kind sk

Appendix B

Stirling Numbers of the Second Kind Sk ................................. 233

Appendix C

Bernoulli Numbers .................................................. 234

Appendix D

Bernoulli Polynomials ............................................... 235

Appendix E

Euler Numbers .................................................... 236

Appendix F

Euler Polynomials .................................................. 237

Appendix G

Fibonacci Numbers .................................................

232

238

ANSWERS TO SUPPLEMENTARY PROBLEMS .............. 239 INDEX

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

255

Chapter 1 The Difference Calculus

OPERATORS

We are often concerned in mathematics with performing various operations such as squaring, cubing, adding, taking square roots, etc. Associated with these are operators, which can be denoted by letters of the alphabet, indicating the nature of the operation to be performed. The object on which the operation is to be performed, or on which the operator is to act, is called the operand. Example 1. If C or [

]3 is the cubing operator and x is the operand then Cx or [

]3x represents the cube of x, i.e. x3.

Example 2.

If D or d is the derivative operator and the operand is the function of x given by f (x) = 2x4 - 3x2 + 5 x

then

D f (x)

= D(2x4 - 3x2 + 5) = dx (2x4 - 3x2 + 5)

= 8x3 - 6x

Example 3.

If J or f (

) dx is the integral operator then

7(2x4 - 3x2 + 5)

= Ff (

) dx

I (2x4 - 3x2 + 5) =

f

(2x4 - 3x2 + 5) dx

2x5-x3+5x+c 5

where c is an arbitrary constant. Example 4. The doubling operator can be represented by the ordinary symbol 2.

Thus

2(2x4 - 3x2 + 5) = 4x4 - 6x2 + 10

It is assumed, unless otherwise stated, that the class of operands acted upon by a given

operator is suitably restricted so that the results of the operation will have meaning. Thus for example with the operator D we would restrict ourselves to the set or class of differentiable functions, i.e. functions whose derivatives exist.

Note that if A is an operator and f is the operand then the result of the operation is indicated by Af. For the purposes of this book f will generally be some function belonging to a particular class of functions. SOME DEFINITIONS INVOLVING OPERATORS 1.

Equality of operators. Two operators are said to be equal, and we write A = B or B = A, if and only if for an arbitrary function f we have Af = Bf.

2.

The identity or unit operator. If for arbitrary f we have If = f then I is called the identity or unit operator. For all practical purposes we can and will use 1 instead of I.

3.

The null or zero operator. If for arbitrary f we have Of = 0 then 0 is called the null or zero operator. For all practical purposes we can and will use 0 instead of O. 1

2

THE DIFFERENCE CALCULUS

4. Sum and difference of operators.

[CHAP. 1

We define

(A+B)f = Af + Bf,

(A - B)f = Af - Bf

(1)

and refer to the operators A + B and A - B respectively as the sum and difference of operators A and B. Example 5. 5.

(C + D)x2 = Cx2 + Dx2 = x6 + 2x,

(C - D)x2 = Cx2 - Dx2 = x6 - 2x

Product of operators. We define (A B) f = (AB) f = A(Bf) (2) and refer to the operator AB or A B as the product of operators A and B. If A = B we denote AA or A A as A2. Example 6.

(CD)x2 = C(Dx2) = C(2x) = 8x3,

C2x2 = C(Cx2) = C(x6) = x18

6.

Linear operators. If operator A has the property that for arbitrary functions f and g and an arbitrary constant a A(f + g) = A f + Ag, A(af) = aAf (3) then A is called a linear operator. If an operator is not a linear operator it is called a non-linear operator. See Problem 1.3.

7.

Inverse operators. If A and B are operators such that A(Bf) = f for an arbitrary

function f, i.e. (AR) f = f or AB =1 or AB = 1, then we say that B is an inverse of A and write B = A-1= 1/A. Equivalently A-1 f = g if and only if Ag = f. ALGEBRA OF OPERATORS We will be able to manipulate operators in the same manner as we manipulate algebraic quantities if the following laws of algebra hold for these operators. Here A, B, C denote any operators. I-1. A+B=B+A Commutative law for sums Associative law for sums A + (B + C) = (A + B) + C 1-2. Commutative law for products 1-3. AB = BA Associative law for products A(BC) = (AB)C 1-4. Distributive law 1-5. A(B + C) = AB + AC Special care must be taken in manipulating operators if these do not apply. If they do

apply we can prove that other well-known rules of algebra also hold, for example the index law or law of exponents AmA" = Am+" where Am denotes repeated application of operator A m times. THE DIFFERENCE OPERATOR Given a function f (x) we define an operator 0, called the difference operator, by Af(x) = f(x + h) - f(x)

(4)

where h is some given number usually positive and called the difference interval or diferencing interval. If in particular f (x) = x we have

Ox = (x + h) - x = h

or

h = Ox

(5)

Successive differences can also be taken. For example

A2f(x) = o[of(x)] = A[f(x+h)- f(x)] = f(x+2h) - 2f(x+h) + f(x)

(6)

catAP.1J

THE DIFFERENCE CALCULUS

3

We call As the second order difference operator or difference operator of order 2. In general we define the nth order difference operator by Anf(x) = A[An-i f(x)] (7)

THE TRANSLATION OR SHIFTING OPERATOR We define the translation or shifting operator E by Ef(x) = f(x + h) By applying the operator twice we have

(8)

E2f(x) = E[Ef(x)] = E[f(x+h)] = f(x+2h) In general if n is any integer [or in fact, any real number], we define Enf (x) = f (x + nh)

(9)

We can show [see Problem 1.10] that operators E and A are related by

E-1

or

E=1+a

(10)

using 1 instead of the unit operator I. THE DERIVATIVE OPERATOR From (4) and (5) we have Af(x)

_

f(x + h) - f(x)

h

Ax

where we can consider the operator acting on f (x) to be A/Ax or A/h. The first order derivative or briefly first derivative or simply derivative of f (x) is defined as the limit of the quotient in (11) as h or Ax approaches zero and is denoted by Df (x)

=

f'(x)

=

lim Af (x)

Ax-.o Ax

=

lim h-.o

f (x + h) - f (x) h

(12)

if the limit exists. The operation of taking derivatives is called differentiation and D is the derivative or differentiation operator. The second derivative or derivative of order two is defined as the derivative of the first derivative, assuming it exists, and is denoted by D2f(x) = D[Df(x)1 = f"(x) We can prove that the second derivative is given by f(x + 2h) - 2f(x+h) + f(x) D2f(x) = lim A2f(x) = lim Ax-+o (Ax)2

h-.o

h2

(13)

(14)

and can in fact take this as a definition of the second derivative. Higher ordered derivatives can be obtained similarly.

THE DIFFERENTIAL OPERATOR The differential of first order or briefly first differential or simply differential of a function f (x) is defined by

df(x) = fl(x)Ax = f'(x)h

In particular if f (x) = x we have dx = Ax = h so that (15) becomes df (x) = f'(x) dx = f'(x)h or df (x) = D f (x) dx = hDf (x)

(15)

(16)

4

THE DIFFERENCE CALCULUS

[CHAP. I

We call d the differential operator. The second order differential of f (x) can be defined as d2f(x) = f"(x)(ox)2 = f"(x)(dx)2 (17) and higher ordered differentials are defined similarly. Note that df (x), dx = ax = h, d2f (x), (dx)2 = (Ox)2 = h2 are numbers which are not necessarily zero and not necessarily small. It follows from (16) and (17) that

fl(x) = dfdx (x) = Df(x), ... (18) f"(x) = d2fdx(2) = D2f(x), where in the denominator of (18) we have written (dx)2 as dx2 as indicated by custom or convention. It follows that we can consider the operator equivalence

D= Similarly we shall write

D2 = d2

d

dx2'

YX

(ox)2 f(x)

as

(19)

ox2 AX)

RELATIONSHIP BETWEEN DIFFERENCE, DERIVATIVE AND DIFFERENTIAL OPERATORS From (16) we see that the relationship between the derivative operator D and the differential operator d is

D = dx =

h

or

d = hD

(20)

Similarly from (12) and (16) we see that the relationship between the difference operator A and the derivative operator D is A D = AX-0 lim = lim = dx d (21) Ax n-.o h

with analogous relationships among higher ordered operators. Because of the close relationship of the difference operator 0 with the operators D and d as evidenced by the above we would feel that it should be possible to develop a difference calculus or calculus of differences analogous to differential calculus which would give the results of the latter in the special case where h or Ax approaches zero. This is in fact the case as we shall see and since h is taken as some given constant, called finite, as opposed to a variable approaching zero, called infinitesimal, we refer to such a calculus as the calculus of finite differences. To recognize the analogy more clearly we will first review briefly some of the results of differential calculus.

GENERAL RULES OF DIFFERENTIATION It is assumed that the student is already familiar with the elementary rules for differentiation of functions. In the following we list some of the most important ones. II-1.

D[f(x) + g(x)]

D f (x) + Dg(x)

11-2.

D [-f (x)]

aD f (x)

11-3.

D[f(x)g(x)]

II -4 .

D

11-5.

D[f(x)]m

[f(x)] g(x)

a = constant f(x)Dg(x) + g(x) Df (x) g(x) Df (x) - f (x) Dg (x) [g(x)]2

_ m[f(x)]--1Df(x)

m = constant

CHAP. 1] 11-6.

THE DIFFERENCE CALCULUS

D[f(g(x))]

5

= D[f(u)] where u = g(x) df , du = f'(g(x)) g'(x)

du dx

DILf Dg where D. = d/du The above results can equivalently be written in terms of d rather than D. Thus for example 11-3 becomes

d[f(x) g(x)] = f(x) dg(x) + g(x) df(x) = [f(x) g'(x) + g(x) f'(x)]dx DERIVATIVES OF SPECIAL FUNCTIONS In the following we list derivatives of some of the more common functions. The constant e = 2.71828... is the natural base of logarithms and we write loge x as In x, called the natural logarithm. All letters besides x denote given constants. III-1.

D [c]

111-2. 111-3.

D [x-] D [(px + q)m]

mp(px + q)--l

111-4.

D[bx]

bx In b

111-5.

D [e-]

111-6.

111-7.

D[sin rx] D[cos rx]

re'x r cos rx

111-8.

D [In x]

111-9.

D [logb x]

0 mxm-1

-r sin rx b>0, b =A1

x

These results can also be written in terms of d rather than D. Thus for example 111-2 becomes

d(xm) = mxri-1 dx

GENERAL RULES OF THE DIFFERENCE CALCULUS The following results involving A bear close resemblances

to the results II-1 to 11-4 on

page 4 . IV-1. IV-2. IV-3.

o[af (x)]

of (x) + Og(x) aif(x) a = constant

0[f (x) g(x)]

f (x) Ag(x) + g(x + h) A f (x)

AV (x) + g(x)]

g(x) of (x) + f (x + h) zg(x)

f(x) og(x) + g(X)of(x) + of(x) t.g(x) IV -4 .

A

x)

g(x) of (x) - f(x) Ag(x) g(x) g(x + h)

I g(x)

Note that if we divide by Ax and let Ax - 0 the above results become those of II-1 to 11-4. FACTORIAL FUNCTIONS Formula 111-2 states that Dxm = mxii-1. In an effort to obtain an analogous formula involving A we write Oxm _ (x + h)m - xAX

-

h

THE DIFFERENCE CALCULUS

6

[CHAP. 1

which however does not resemble the formula for De'. To achieve a resemblance we introduce the factorial function defined by x(m) = x(x - h)(x - 2h) . . . (x - [m -1]h) m = 1, 2, 3, ... (22) consisting of m factors. The name factorial arises because in the special case x = m, h = 1, we have m(m) = m(m - 1)(m - 2) 2. 1 = m!, i.e. factorial m. If m = 0, we define x(m) = 1, i.e. .

x(0) = 1 For negative integers we define [see Problem 1.18(a)] x(-m)

_

I

(x + h)(x + 2h)

-

(x + mh) Note that as h - 0, x(m) - xm, x(-m) -* x-m. .

(! 1

(x + mh)(m)

m = 1-,2,3,

...

(24)

Using (22), (23), (24) it follows that [see Problems 1.17 and 1.18(b)] for all integers in Ax(m)

mx(m-1)

Ox

or

AX(m) = mx(m-1)h

2.

5)

in perfect analogy to Dxm = and d(xm) = mxm-1 dx respectively. We can also define x(m) for nonintegral values of m in terms of the gamma function mxm-1

[see page 103].

FACTORIAL POLYNOMIALS From (22) we find on putting m = 1, 2, 3, .. . X(1) = x X(2) = x2

- xh

x(l) = x3 - 3x2h + 2xh2 X(41 = x' - 6x3h + 11x2h2 - 6xh3 X(51 = xs - 10x4h + 35x3h2 - 50x2h3 + 24xh4 etc. If p is any positive integer, we define a factorial polynomial of degree p as aox(p) + alx(p-1) +

(26)

+ ap

where ao 0, a1, . . ., a1, are constants. Using (26) we see that a factorial polynomial of degree p can be expressed uniquely as an ordinary polynomial of degree p. Conversely any ordinary polynomial of degree p can be expressed uniquely as a factorial polynomial of degree p. This can be accomplished by noting that

x = x(1) + x(1)h + 3x (2)h + x(1)h2

(27)

+ 7x(3)h + 6x(2)h2 + x(1)h3 + 15x 14)h + 25x (31h2 + 10x(2)h3 + x(1)h4

etc. Another method for converting an ordinary polynomial into a factorial polynomial

uses synthetic division [see Problems 1.23-1.25].

STIRLING NUMBERS Any of the equations (26) can be written as x(n) _

n k=1

skxkhn-k

(28)

THE DIFFERENCE CALCULUS

CHAP. 1]

7

where the coefficients sk are called Stirling numbers of the first kind. A recursion formula for these numbers is n+ n n Sk

where we define

sn = 1, sk = 0

for

= Sk 1 - nsk

k < 0, k

n+1

(29)

where n > 0

(30)

Similarly any of the equations (27) can be written as n

xn =

Skx(k)hn-k

(31)

k=1

where the coefficients Sk are called Stirling numbers of the second kind. A recursion formula for these numbers is Sk+1 = Sk-1 + kSk

(32)

where we define S" n= 1, Sk = O

for k ' 0, k ? n + 1

where n > O

(33)

In obtaining properties of the Stirling numbers results are greatly simplified by choosing h = 1 in (28) and (31). For tables of Stirling numbers of the first and second kinds see pages 232 and 233. GENERALIZED FACTORIAL FUNCTIONS The generalized factorial function corresponding to any function f(x) is defined by = f(x) f (x - h) f (x - 2h) ... f(x - [m -1] h) (34) m = 1, 2, 3, ... [f (x) ] (m) [f(x)](-m)

=

1

f(x+h) f(x+2h) f(x+mh) [f(x)](a)

=

m = 1 , 2, 3, ...

(35)

1

(36)

The special case f (x) = x yields (22), (24) and (23) respectively. It should be noted that as h - 0, [f(x)] (m) - [f (x)]m and [f(x)] (-m) - [ f (x)] -m.

DIFFERENCES OF SPECIAL FUNCTIONS In the following we list difference formulas for special functions analogous to the differentiation formulas on page 5. = 0 V-1. A[c] V-2. V-3. V-4. V-5. V-6. V-7. V-8. V-9.

= 0[(px + q)(m)] =

o[x(m)]

mx(m-1)h

A[bx]

mph(px + bx(bh -1)

1[e'] A[sin rx] A[cos rx] A[ln x] 0 [logb x]

q)(m-1)

= = erx(erh - 1) = 2 sin (rh/2) cos r(x + h/2) _ -2 sin (rh/2) sin r(x + h/2) = In (1 + h/x) = loge (1 + h/x) -

Note that if we divide each of these results by ox = h and take the limit as Ox or h

approaches zero we arrive at the corresponding formulas for derivatives given on page 5. For example V-5 gives lerh - 1\ , [erx] d (e-) = erx = lim lim h ) = rerx = dx k-,0 Ax-+0 Ox

-

D(erx)

THE DIFFERENCE CALCULUS

8

[CHAP. 1

TAYLOR SERIES From the calculus we know that if all the derivatives of f (x) up to order n + 1 at least exist at a point a of an interval, then there is a number 77 between a and any point x of the interval such that f(n) (a) (x f"(a)(x - a)2 - a)n f (x) = f (a) + f'(a) (x - a) + + Rn 2! + (37) + n! where the remainder Rn is given by f(n+l)(,q) (x - a)n+1

Rn

(38)

(n + 1)!

This is often called Taylor's theorem or Taylor's formula with a remainder. The case where n = 0 is often called the law of the mean or mean value theorem for derivatives. If lira Rn = 0 for all x in an interval, then fl . f" a) 2x! - a)2 + ... (39) f (x) = f(a) + f, (a) (x -a) + 00

is called the Taylor series of f (x) about x = a and converges in the interval. We can also write (39) with x = a + h so that f"(a)h2 + ... f(a + h) = f (a) + f'(a)h + (40) The special case where a = 0 is often called a Maclaurin series. Some important special series together with their intervals of convergence [i.e. the values of x for which the series converges] are as follows. ez

1.

=

1+x+2i+Zi+ x3

sinx =

X-

3.

cosx

x2 x4 1 - 2i + 4i

=

3!

+

x7

x5

2.

0 for some value between a and b. Then it follows since f (x) is continuous that it attains its maximum value somewhere between a and b, say at V. Consider now 4f(x) Ax

--

f(n + h) - f(n) h

where we choose h = Ax so small that -1 + h is between a and b. Since f (-q) is a maximum value, it follows that Af(x)/Ax r' 0 for h < 0 and &f(x)/Ax : 0 for h > 0. Then taking the limit as It - 0 through positive values of It, we have f'(n) < 0, while if the limit is taken through negative values of It, we have f'(n) i' 0. Thus f'(n) = 0. A similar proof holds if f(x) < 0 for some value between a and b.

CHAP. 1]

1.45.

THE DIFFERENCE CALCULUS

27

(a) Prove the mean value theorem for derivatives: If f (z) is continuous in a < x < b and has a derivative in a < x < b, then there is at least one value I between

a and b such that

f(b) - f(a)

b-a

(b) Use (a) to show that for any value of x such that a < x < b, f(x) = f(a) + (x - a)f'(7) where 0 is between a and x.

(c) Give a geometric interpretation of the result. (a) Consider the function

- f(a) F(x) = f(x) - f(a) - (x -a) f(b)b-a

(1)

From this we see that F(a) = 0, F(b) = 0 and that F(x) satisfies the conditions of Rolle's theorem [Problem 1.44]. Then there is at least one point p between a and b such that F'(,) = 0.

But from (1)

so that

- f(a) F'(x) = F(X) - f(b)b-a

(2)

=0 F'(n) = f'(n) - f(b)b -1(a) -a MO

or

f (b

-

a(a)

(3)

(b) Replacing b by x in (3) we find as required f (z) = f (a) + (x - a)f'(n)

(4)

where , is between z and a.

Note that (4) is a special case of Taylor's series with a remainder for n = 1. The general Taylor series can be proved by extensions of this method.

(c) The theorem can be illustrated geometrically with reference to Fig. 1-2 where it is geometrically evident that

there is at least one point R on the

curve y = f (z) where the tangent line PRQ is parallel to the secant line ACB. Since the slope of the tangent line at R is f'(,) and the slope of the secant line

is [f(b) - f(a)]/(b - a), the result follows. It is of interest to note that F(x) in equation (1) represents geometrically RC of Fig. 1-2. In the case where f(a) = 0, f(b) = 0, RC represents the maximum value of f (x) in the interval

a- a' b.

THE DIFFERENCE CALCULUS

28

Supplementary Problems

OPERATORS 1.46.

[CHAP. 1

Let of = [

]2

following.

be the squaring operator and D the derivative operator. Determine each of the

(a) ?f(1 + V)

(e)

(b) (2ef + 3D)(x2 - x) (c) ?JD(3x + 2) (d) D,?f(3x + 2)

(g)

(.J2 + 2,?f - 3)(2x -1) (f) (D + 2)(1- 3)x2

(z)

x3D3f(x + 1)

(j) (xeJ - 2f x)D,,f x2

f - 3)(D + 2)x2 (h) (xD)3ef (x + 1) (

1.47.

(a) Prove that the operator j of Problem 1.46 is a nonlinear operator. (b) Explain the significance of @f-1 and determine whether it always exists and is unique.

1.48.

Prove that

1.49.

Let,?f be the squaring operator and a be any real number. (a) Explain the meaning of the operators a2f and 2f a. (b) Do the operators a and 05 obey the commutative law? Illustrate by an example.

1.50.

Is the operator (xD)4 the same as the operator x4D4? Justify your answer.

1.51.

C2 where of and C are the squaring and cubing operators respectively.

Prove that (a) D2x - xD2 = 2D, (b) D3x - xD3 = 3D2. Obtain a generalization of these results and

prove it.

THE DIFFERENCE AND TRANSLATION OPERATORS 1.52.

Find each of the following. (a) A(2x - 1)2 (e) (A + 1)2(x + 1)2 (b) E(3 5x --4 ) (f) (xE2 + 2xE + 1)x2 (c) A2(2x2 - 5x)

(d) 3E2(x2 + 1)

(2E - 1)(3A + 2)x2 (j) (xAE)2x2 (1)

(g) A2E3x (h) (3A + 2)(2E -1)x2

1.53.

Determine whether (a) (E - 2)(A + 3) _ (A + 3)(E - 2), (b) (E - 2x)(A + 3x) = (A + 3x)(E - 2) and discuss the significance of the results.

1.54.

Prove that E is a linear operator.

1.55.

Determine whether (a) A2 and (b) E2 are linear operators. Do your conclusions apply to A" and E'"? Explain.

1.56.

Verify directly that A3 = (E - 1)3 = E3 - 3E2 + 3E - 1.

1.57.

Prove formulas (a) V-7, (b) V-8, (c) V-9 on page 7.

1.58.

Prove that the commutative law for the operators D and A holds (a) with respect to addition, (b) with respect to multiplication.

1.59.

(a) Does the associative law with respect to multiplication hold D, A and E? (b) Does the commutative law with respect to multiplication hold for D and E?

1.60.

Show that

(a) (b)

A [x(2 - x)] = D[x(2 - x)] = 2(1 - x) lim AX-0 AZ

lim A2 [x(2 - x)] =

Ax-.o Axe

d2 [x(2 x2

- x)] = -2

directly from the definition. 1.61.

Find (a) d(x3 - 3x2 + 2x - 1), (b) d2(3x2 + 2x - 5).

1.62.

Prove that D [ f (x) + g(x)] = D f (x) + Dg(x) giving restrictions if any.

1.63.

Prove that

D Lg(x)J = g(x) Df([g(x)]2

Dg(x)

giving restrictions if any.

CHAP. 1]

1.64.

THE DIFFERENCE CALCULUS A

Prove that lim

Ox-.0 Ax

29

[cos rx] -_ -r sin rx by using the fact that lim

sin 9

o-+0

1.

9

1.65.

Prove that

1.66.

Prove equation (14), page 3, giving suitable restrictions.

1.67.

Obtain a relationship similar to that of equation (14), page 3, between D3f(x) and A3f(x)/Ax3.

lim

AX-0 Ax

[bx] = bx In b stating assumptions made.

FACTORIAL FUNCTIONS 1.68. Find (a) A (3x(5) + 5x(4) - 7x(2) + 3x(1) + 6),), (b)

Ax

(X(-3)

-

3x(-2)),

A

1.69.

Find

1.70.

Express each of the following as factorial polynomials for h = 1 and for h

(a) A2(2x(-3) - 3x(-2) + 4x(2)), (b)

x3

x(-2)

x(2) + 2

(x(4) + x(-4)). 1.

(a) 3x2-5x+2, (b) 2x4+5x2-4x+7. A2

1.71.

Find

1.72.

Express each of the following as a product of suitable factors using the indicated values of h. (a) (2x - 1) (4) if h = 2, (b) (3x + 5) (3) if h = 1, (c) (4x - 5)(-2) if h = 1, (d) (5x + 2)-(4) if h = 2.

1.73.

Write each of the following as a factorial function.

(a) Ox (x4 - 2x2 + 5x - 3), (b) Ox2 (x4 - 2x2 + 5x - 3).

(a) (3x - 2)(3x + 5)(3x + 12),

(b) (2 + 2x)(5 + 2x)(8 + 2x)(11 + 2x)

2)(1

(c)

x(x +

x + 4) '

(d

(2x - 1)(2x + 3)(2x + 7)(2x + 11)

1.74.

Prove that Ox (px + q)(m) = mp(px + q)(m-1) for (a) m = 0, 1, 2, ..., (b) m = -1, -2,-3,

1.75.

Express in terms of factorial functions

x2-1

(a) (x + 2)(x + 4)(x-F-6) '

... .

2x+1 (b) (2x + 3)(2x + 5)(2x + 9)

STIRLING NUMBERS 1.76.

Obtain Stirling numbers of the first kind sk for n, k = 1, 2, 3 by using the recursion formula (29), page 7.

1.77.

Obtain Stirling numbers of the second kind Sk for n, k = 1, 2, 3 using the recursion formula (32), page 7.

1.78.

Find Sk for k = 1, 2, ..., 6 by using the method of Problem 1.25.

1.79.

Prove that

+ 6n = 0 1 + 82 + and illustrate by referring to the table of Stirling numbers of the first kind in Appendix A, page 232.

1.80.

Prove that (a) Si - 82 + 83 (b)

+ (-1)n-1sn = (-1)a-1n! 18i + 1s21 + ... + Isnl = nl

THE GREGORY-NEWTON FORMULA AND TAYLOR SERIES 1.81.

Express each of the following as factorial polynomials for the cases h = 1 and h the Gregory-Newton formula.

(a) 3x2 - 5x + 2, (b) 2x4 + 5x2 - 4x + 7.

1 by using

THE DIFFERENCE CALCULUS

30 1.82.

1.83.

[CHAP. 1

Under what conditions is the remainder R,,, in (45), page 9, equal to zero?

Show how to generalize Problem 1.35 by obtaining the Gregory-Newton formula for the case where h

1, a=0.

1.84.

Prove that An[aoxn + alxn-1 + by using the Gregory-Newton formula.

1.85.

Obtain the Taylor series expansion for f(x) from the Gregory-Newton formula by using a limiting

+ an] = n! aohn

procedure. 1.86.

Obtain the Taylor series expansions

(a) 1,

(b) 2,

(c)

3 and (d) 4 on page 8 and verify the

intervals of convergence in each case. LEIBNITZ'S RULE 1.87.

Use Leibnitz's rule to find A3(x2.2x) if h = 1.

1.88.

Find An(xax).

1.89.

Find Ari(x2ax).

1.90.

Obtain Leibnitz's rule for derivatives from Leibnitz's rule for differences by using an appropriate limiting procedure.

OTHER DIFFERENCE OPERATORS 1.91.

If f (x) = 2x2 + 3x - 5 find (a) V f (x), (b) S f (x), (c) V 1f (x),

1.92.

Evaluate (V2 - 3 V S + 282)(x2 + 2x).

1.93.

Prove that

1.94.

Determine whether the operators V and 8 are commutative.

1.95.

Demonstrate the operator equivalence E = C2 +

1.96.

Prove that VA = AV = 82.

1.97..

Is it true that

1.98.

Show that

1.99.

Determine whether (a) M and (b) µ commutes with A, D and E.

1.100.

Show that

(d) 82f(x)

(a) V2 = (AE-1)2 = A2E-2, (b) V n = AnE-n.

2

(a)

ox m a

Sx

= 4Y, dx'

(b)

lim Sny ax-.o Sxn

2

1 + 4

-

dny ? dxn

Explain.

(a) M = j(1 + E) _ E - JA, (b) µ = M/E112.

(a) A = µ8+- 82, (b) A2m+1 = Em[1182m+1+48M+21.

MISCELLANEOUS PROBLEMS 1.101.

(a) If A and B are any operators show that (A - B) (A + B) = A2 - B2 + AB - BA. (b) Under what conditions will it be true that (A - B)(A + B) = A2 - B27 (c) Illustrate the results of parts (a) and (b) by considering (A2 - D2)x2 and (A - D)(A + D)x2.

1.102.

Prove that

(a)

A sin (px + q) = 2 sin 2h sin [px + q + .(ph + ir)]

(b)

A cos (px + q) = 2 sin

2 cos [px + q + , (ph +,,,)]

CHAP. 1]

1.103.

1.104.

THE DIFFERENCE CALCULUS

31

m

Prove that

[2 sin 2-] sin [px + q + 2 (ph + 7r) jJ

(a)

Am sin (px + q)

(b)

Am cos (px + q) =

rL2 sin 2

r

cos Lpx + q +

J

2

(ph + Tr)]

Use Problem 1.103 to show that

2

m

dm

d dx,n sin x = sin x + mw

(a)

_

n

aim cos x

(b)

x +

cos

C

n+1

2 7r)

1.105.

Verify that

1.106.

(a) Show that A tan x = 1

1.107.

h (a) Show that A tan-1 x = tan-1x2 + hx + 1 and (b) deduce that dx tan-1 x

r +

-

d tan-1xa

dx

1.108.

(r +It 1) se

and thus complete the proof of Problem 1.11, Method 2.

(r + 1

tan h tan x tan h and (b) deduce that dx tan x = sect X. 2

x2 + a2.

(a) Show that A sin-1 x = (x + h) 1 --x2 - x 1 - -(x+ h)2.

(b) Deduce from (a) that dx sin-1 x =

and dx sin-1

11 x2

1.109.

1.110.

1.111.

and

a

Prove that for h = 1

Sk =

Prove that Prove that

sk = n) Dnx(k) x=0'

(a)

(-1)k k

k

(-1)P O

k/ p

pn

=

a Sk =

(b)

a21 x2

Akxn k)

x=0

and illustrate by using the table on page 233.

(a) E[f(x) g(x)] = [Ef(x)] [Eg(x)]

(b) E[f(x)]n = [r''f(x)]n Em[fl(x)...fn(x)] (c) =

[E-f1(x)]...[Emfn(x)]

1.112.

Show that the index law for factorial functions, i.e. X(m)X(n) = x(m+n)

1.113.

Prove that

An[f(x) g(x)]

= ik=0 (-1)k+n

1t

does not hold.

f(x + kh) g(x + kh)

k

and discuss the relationship with Leibnitz's rule. 1.114.

Prove that

n! = It, - n(n-1)n + 1.115.

Show that

n(n - 1)

2!

3+

(n-2)n -

h4D4

n(n - 1)(n - 2)

3!

(a)

A = hD + h2D2 2! +

(b)

A2 = h2D2 +h3D3

(c)

A3 = h3D3 + 3h4D4 + 5h5D5 + 3h6D6 + 903h7D7 + 2

h3D3

4!

+

h5D45

y1jh4D4 +

4

(n-3)n + ...

31h360 6D6

+ 4

+ ...

2520

1.116.

Find (a) A2(3x3 - 2x2 + 4x - 6), (b) A3(x2 + x)2 by using Problem 1.115 and compare results by direct evaluation.

1.117.

If U(t) is the function defined by (5) of Problem 1.33, prove that U(n+1)(t) = 0 for at least one value t = n between the smallest and largest of x, x0, x1, ..., xn. [Hint. Apply Rolle's theorem successively.]

Chapter 2 Applications of the Difference Calculus

SUBSCRIPT NOTATION Suppose that in f (x) we make the transformation x = a + kh from the variable x to the variable k. Let us use the notation y = f (x) and yk = f(a + kh)

(1)

It follows that [see Problem 2.1] DYk = yk+1 - yk,

and so as on page 3

Eyk = yk+1

(2)

0=E-1 or E=1+0

(3)

Using this subscript notation it is clear that a unit change in the subscript k actually corresponds to a change of h in the argument x of f (x) and conversely. In addition all of the basic rules of the difference calculus obtained in Chapter 1 can be written with subscript notation. Thus for example formula IV-4, page 5, becomes 0 (Yk) Zk

=

()

Zk oyk - Ilk AZk Zk Zk+1

where yk = f(a + kh), Zk = g(a + kh).

Also since x = a + kh becomes x = k if a = 0 and h = 1, results involving x remain valid if we replace x by k and put h = 1. Thus for example equations (22) and (25), page 6, become respectively (5) kcm) = k(k -1)(k - 2) ... (k - m + 1), ok(m) = mk(m-1'

It should be noted that k need not be an integer. In fact k is a variable which is discrete or continuous according as x is. An important special case with which we shall be mainly concerned arises however if k = 0, 1, 2, ... so that the variable x is equally spaced, i.e. x = a, a + h, a + 2h, .... In such case we have yo = f(a), yl = f(a + h), y2 = f(a + 2h), ... .

DIFFERENCE TABLES A table such as that shown in Fig. 2-1 below which gives successive differences of y = f(x) for x = a, a + h, a + 2h, ..., i.e. Ay,o2y,o3y, ..., is called a difference table. Note that the entries in each column after the second are located between two successive entries of the preceding column and are equal to the differences between these entries. Thus for example 02y1 in the fourth column is between Dye and oy2 of the third column and Similarly 03 y2 = O2y3 - 02'y2.

d2yi = oy2 - oy1.

32

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

y

X

Ay

03y

22y

33

o4y

A5y

yo=f(a)

a

A7fo

a+h

711 = f (a + h)

A2yo

Ay,

a+ 2h

A3Yo

p4yo

A2y1

712 = f(a + 2h) AY2

a + 3h

713 = f(a+ 3h)

a + 4h

y4 = f(a+4h)

a + 5h

715 = f(a+ 5h)

A5Yo

037 f1

,4y1

02712

Iy3

03712 22713

1Y4

Fig. 2-1

Example 1.

The difference table corresponding to y = f(x) = x3 for x = 1, 2, ..., 6 is as follows. x

y = f (X) = x3

1

1

2

8

Ay

02y

I3y

A471

7 12 19 3

27

6

18 37

4

24

64 61

5

0 6

125

0

6

30

91 6

216

Fig. 2-2

The first entry in each column beyond the second is called the leading difference for that column. In the table of Fig. 2-1 the leading differences for the successive columns are Oyo, o2yo, A3yo, .... The leading differences in the table of Fig. 2-2 are 7, 12, 6, 0. It is often desirable also to include the first entry of the second column called a leading difference of order zero. It is of interest that a difference table is completely determined when only one entry in each column beyond the first is known [see Problem 2.4].

DIFFERENCES OF POLYNOMIALS It will be noticed that for f (x) = x3 the difference table of Example 1 indicates that the third differences are all constant, i.e. 6, and the fourth differences [and thus all higher differences] are all zero. The result is a special case of the following theorem already proved in Problem 1.41, page 25. Theorem 2-1. If f (x) is a polynomial of degree n, then o"f(x) is a constant and A" "f (x),

0"+2f (x), ... are all zero.

34

APPLICATIONS OF THE DIFFERENCE CALCULUS

[CHAP. 2

GREGORY-NEWTON FORMULA IN SUBSCRIPT NOTATION If we put x = a + kh in the Gregory-Newton formula (44) on page 9, it becomes + &2f(2 k(2) + ... _+ On f(a) k(n) + Rn f(a + kh) = f(a) + n! i where the remainder is given by Af(1'kaw

.

(6)

hn+lf(n+1)(,0 k(n+1)

Rn =

(7)

(n + 1)!

the quantity 9 being between a and a + kh and where k(1) = k,

k(2) = k(k - 1),

k(3) = k(k - 1)(k - 2),

In subscript notation (6) can be written as

If An+ 1yo, On+2yo,

=

yo +

avok(1)

&2y°k(2)

+ ... +

An

k(n)

+ R. n! ... are all zero, then R. - 0 and yk is a polynomial of degree n in k.

1!k

1!

+

2!

y0

(8)

GENERAL TERM OF A SEQUENCE OR SERIES The Gregory-Newton formula is often useful in finding the general law of formation of terms in a sequence or series [see Problems 2.8 and 2.9]. INTERPOLATION AND EXTRAPOLATION Often in practice we are given a table showing values of y or f (x) corresponding to various values of x as indicated in the table of Fig. 2-3. Z

xo

x1

...

XP 1/

Yo F 1!1

...

1/y

Fig. 2-3

< xp. We assume that the values of x are increasing, i.e. xo < xl < An important practical problem involves obtaining values of y [usually approximate] corresponding to certain values of x which are not in the table. It is assumed of course that we can justify seeking such values, i.e. we suspect some underlying law of formation which may be mathematical or physical in nature.

Finding the [approximate] value of y corresponding to an untabulated value of x between xo and x, is often called interpolation and can be thought of as a "reading between the lines of the table". The process of obtaining the [approximate] value of y corresponding to a value of x which is either less than xo or greater than xi,, i.e. lies outside the table, is often called extrapolation. If x represents the time, this can involve a problem in prediction or forecasting. Suppose that the x values are equally spaced and the nth differences of y or f (x) as obtained from the table can be considered as small or zero for some value of n. Then we can obtain a suitable interpolation or extrapolation formula in the form of an approximating polynomial by using the Gregory-Newton formula. See Problem 2.11. If the x values are not equally spaced we can use the Lagrange interpolation formula [see page 38]. Because formulas for interpolation can also in general be used for extrapolation we shall refer to such formulas collectively as interpolation formulas.

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

35

CENTRAL DIFFERENCE TABLES In the table of Fig. 2-1 it was assumed that the first value of x was x = a, the second x = a + h, and so on. We could however have extended the table backwards by considering x = a - h, a - 2h, .... By doing this we obtain the table of Fig. 2-4 which we call a central difference table. x

y

a - 3h

21-3

a - 2h

Y-2

Ay

A2y

037!

047!

A5y

A6y

Ay-3

A2y-3 A3y-3

071-2

a-h

21-1

A3Y-2

AY-1 a

A2y-1

Yo

AY,

a + 2h

ABy-3 A5y- 2

A4y-1

A2Yo

211

0571-3

0471-2

A3Y-1

Ay0

a+h

A4y-3

0271-2

A3yo A2y1

212

Ay2

a+3h

213

Fig. 2-4

The table of Fig. 2-4 can also be written equivalently in terms of central differences as shown in Fig. 2-5. X

y

a - 3h

y-3

a - 2h

y-2

8y

82y

83y

Boy

85y

86y

8y-5/2 82y-2 821-3/2

a-h

s3y-3/2

s2y-1

21-1 821-112

a

82y0

210

S3y3/2

8213/2

a + 2h

a+3h

86210

85211/2

84y1

82y1

211

8521-1/2 84y0

83y1/2

8211/2

a + h

soy-1 8321-1/2

82712

212

8215/2 713

Fig. 2-5

Note that the entries of this table can be related to those in the table of Fig. 2-4 by simply using the operator equivalence 8 = AE-112 on page 9. Thus for example, 83y-3/2 = (AE-1/2)3y_3/2 = A3E-3/2Y-3/2 = A3Y-3

Other tables can be made using the backward difference operator V.

APPLICATIONS OF THE DIFFERENCE CALCULUS

36

[CHAP. 2

GENERALIZED INTERPOLATION FORMULAS In using the Gregory-Newton formula (6) or (8) for interpolation, greater accuracy is

obtained when x or a + kh is near the beginning of the table rather than near the middle or end of the table. This is to be expected since in this formula use is made of the leading differences Af (a), A2f (a),

... .

To obtain more accuracy near the middle or end of the table we need to find interpolation

formulas which use differences near the middle or end of the table. The following are formulas which do this. All of these formulas are exact and give the same result when the function is a polynomial. If it is not a polynomial, a remainder or error term can be added. The results can also be expressed in terms of the operators V or 8 of page 9. For the purposes of completeness and reference we include in the list the result (8). 1.

Gregory-Newton forward difference interpolation formula.

2.

+ 1 2yo

!

+ A3yo

_

= yo + Ay-1

k(1)

2

1! +

k(1)

=

A Y-2

(k + 1)(2)

+ A3y-3

2!

k(2)

yo +Ayo 1 + A y-1 2, + A y-1 3

2

k(1)

yo +

yk

(k + 2)(3)

+

3!

AY-1 1 + A 2y-1

(k + 1)(2)

2!

(k + 1)(3)

4 + A y-2

3!

+ A y-2 3

(k + 1)(3)

3!

(k + 1)(4)

4!

4

+ A y-2

(k + 2)(4)

4!

. .

+

Stirling's interpolation formula. Yk

=

+

yo

1

k(1)

(Ay-1+ Ayo)

+ A21_1

1)(2) [1(k(2)+(k+ 2!

(k

+ 2 (A3y-z + A3y-1) 5.

+

!

Gauss' interpolation formulas. Yk

4.

k(3)

k(2)

k(l)

Gregory-Newton backward difference interpolation formula. ilk

3.

yo + Ayo

=

Yk

2!

2 )(3)

3

+ .. .

Bessel's interpolation formula. ilk

=

1

(yo+yl) + A yo r12/ka) 1+ k(z)

1

+

2

(A2y 1+ A2yo)

2

(k -1)(1) 1!

+A3y 1

[1 ((k + 1)c3) 3!

kc3)\

+ 3 /)

+

ZIG-ZAG PATHS AND LOZENGE DIAGRAMS There exists a simple technique for not only writing down all of the above interpolation formulas but developing others as well. To accomplish this we express the central difference

table of Fig. 2-4 in the form of a diagram called a lozenge diagram as shown in Fig. 2-6 below. A path from left to right such as indicated by the heavy solid line in Fig. 2-6 or the heavy dashed line is called a zig-zag path.

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

Y-z 1

1

-,Yo

j

(k + 2)(1)/1!

Qty-3

(k+3)(3)/31

37

A4Y-4

\ /\ \ AY-2

(k+2)(2)/2!

Q3y-3

( + 1)(1)/11

Qty-2

(k + 2)(3)/3,

AY-1

(k+1)(2)/2!

Q3y_z

k(')/l !

1)(3)/3!

(k+3)(4)/4! Q4

y -3

(k+2)(4)/4! Q4y

-- 1 - - - - Q3lo - - -- - k(2)/2! - - - Q3y_1 - - - (k+ 1)(4)/4!

y1 \ 1

(k - 1)(1)1 - AY,

(k - 2) (1) /1!

1

j y3

Qy2

ti

c

3)(1)/11

Q2yo

1)(2)2

Q2111

(k

2)(z)2

A21,

2

k(3)/31

Q4y_1

Q3yo

k(4)/4!

-1)(3)3! MY,

(k

2)

Q47lo

1)(4)/4!

Q4y,

Fig. 2-6

The following rules are observed in obtaining an interpolation formula. 1. A term is to be added whenever any column containing differences is crossed from left to right. The first column in Fig. 2-6 is to be considered as containing differences of order zero. 2. If the path enters a difference column from the left with positive slope [as for example from yo to Ay-, in Fig. 2-6], the term which is to be added is equal to the product of the difference and the coefficient which is indicated directly below the difference.

3. If the path enters a difference column from the left with negative slope [as for example from oy-1 to A2y-1 in Fig. 2-6], the term which is to be added is equal to the

product of the difference and the coefficient which is indicated directly above the difference.

4. If the path enters a difference column with zero slope [as for example from 1 to Dyo in Fig. 2-6], the term which is to be added is equal to the product of the difference and the arithmetic mean of the coefficients directly above and below the difference.

5. If the path crosses with zero slope a column between two differences [as for example in the path from oyo to O3y-1 in Fig. 2-6], the term which is to be added is equal to the product of the coefficient between the two differences and the arithmetic mean of the two differences.

6. A reversal of path changes the sign of the corresponding term to be added.

APPLICATIONS OF THE DIFFERENCE CALCULUS

38

[CHAP. 2

LAGRANGE'S INTERPOLATION FORMULA

In case the table of Fig. 2-1 either has nonequally spaced values of x or if the nth differences of y are not small or zero the above interpolation formulas cannot be used. In such case we can use the formula (x-x1)(x-x2).. .(x-xp) (x-x0)(x-x2). ..(x-xp) Yo

. (x0 - xp) +

.

(x0 - x1)(x0 - x2) .

y1

(x1 - xo)(xl - x2) .

.

(x1 - xp)

(x-x0)(x-x1)...(x-X,_1)

+ yP (XP - x0)(xP - xl) . .

. (XP - XP-1) + which is called Lagrange's interpolation f ormula [although it was known to Euler].

The result holds whether the values xo, x1, . . ., xP are equally spaced or not and whether nth differences are small or not. Note that y is a polynomial in x of degree p.

TABLES WITH MISSING ENTRIES Suppose that in a table the x values are equally spaced, i.e. x = a, a + h, a + 2h, .. . and that all the corresponding entries f (x), except for a few missing ones, are given. Then there are various methods by which these missing entries can be found as shown in Problems 2.28-2.30.

DIVIDED DIFFERENCES If a table does not have equally spaced values of x, it is convenient to introduce the idea of divided differences. Assuming that the values of x are xo, x1, x2, ... and that the function is f (x) we define successive divided differences by f(x0, x1)

f(x1)

=

f(xo, x1, x2)

- f (xo) x1-x0

=

f(x0, x1, x2, x3)

(9)

f (xi, x2) - f(xo, XI)

(10)

X2 - xo

f (XI, x2, x3) - f (x0, x1, x2) X3 - XO

etc. They are called divided differences of orders 1, 2, 3, etc. We can represent these differences in a divided difference table as in Fig. 2-7. xo

f (xo)

x1

f(x1)

f(xo, x1, x2)

x2

f (X2)

A XI, x2, x3)

f (x0, x1, x2, x3) f (x0, x1, x2, x3, x4)

f(x1, x2, x3, x4) X3

&3)

f (X2, x3, x4) f (X3, x4)

X4

f (X4)

Fig. 2-7

Various other notations are used for divided differences. For example f (xo, x1, x2) is sometimes denoted by [xo, x1, x2] or A[xo, xl, x2].

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

39

The following are some important results regarding divided differences.

1. The divided differences are symmetric. For example f (X01 XI) = f (XI, x0),

f (x0, x1, x2) = f(xi, x0, x2) = f(X2, x0, x1),

etc.

2. If f (x) is a polynomial of degree n then f (x, xo) is a polynomial of degree n - 1, f (x, xo, xi) a polynomial of degree n - 2, .... Thus f (x, xo, x1, ... , xn-I) is a constant and f (x, xo, x1, . . ., xn) = 0.

3. Divided differences can be expressed in terms of first differences. For example, f(xo) f (x2) A XI) (x0 -' x1)(x0 - X2) + (x1 - x0)(X1 - x2) + (X2 - x0)(x2 - XI)

f (xo, X1, x2)

NEWTON'S DIVIDED DIFFERENCE INTERPOLATION FORMULA From the above results we obtain f (X) = f (xo) + (x - x0) f (xo, x1) + (x - xo)(x - xi)f(xo, X1, x2)

+ where

+(x-xo)(x-x1). (x-xn-I)f(xo,x1, ...,xn +R

R=

f(n)( 1

(12)

n

(x - Xk)

n i,

(13)

k

(x - xk) = (x-xo)(x-x1)...(x-xn)

(14)

and is some value between the smallest and largest of x, xo, xI, ... , xn. The result (12) can be written as AX) = pn(x) + R (15)

where pn(x) is a polynomial in x of degree n often called an interpolating polynomial.

INVERSE INTERPOLATION We have assumed up to now that the values of x in a table were specified and that we wanted to interpolate [or extrapolate] for a value of y = f (x) corresponding to a value of x not present in the table. It may happen however that we want a value of x corresponding to a value y = f (x) which is not in the table. The process of finding such a value of x is often called inverse interpolation and a solution for x in terms of y is often designated by x = f -1(y) where f -1 is called the inverse function. In this case the Lagrange interpolation formula or Newton's divided difference formula are especially useful. See Problem 2.38 and 2.39. APPROXIMATE DIFFERENTIATION From the formula ehD = 1 + A [equation (43), page 8] we obtain the operator equivalence 1 D = -hln(1+A) =

/o2+334+ ...1

h(

by formally expanding In (1 + A) using the series on page 8.

(16)

/

By using (16) we can obtain approximate derivatives of a function from a difference Higher derivatives can be obtained by finding series expansions for D2, D3, .. . from (16). We find for example table.

40

APPLICATIONS OF THE DIFFERENCE CALCULUS

[CHAP. 2

...l

(17)

D2 =

/L2(A2-13+12 A4

D3 =

1 (A.,

K3

_5

46-

6

- 32 04 + 74 i5 - ... )

(18)

Use can also be made of different interpolation formulas to find approximate derivatives.

In these cases a remainder term can be included enabling us to estimate the accuracy.

Solved Problems SUBSCRIPT NOTATION 2.1. If y = f (x) and ilk = f (a + kh), show that Ayk = yk+l - Yk

(a)

(b)

Eyk = yk+1

(c)

E = I+ A

Since ilk = f (a + kh) we have (b)

Dyk = Af(a+kh) = f(a + kh + h) - f(a + kh) = ilk+1 - ilk Eyk = f(a + kh) = f(a + kh + h) = ilk+1

(c)

From parts (a) and (b), ilk + Ayk = Eyk or (1 + A)yk = Eyk

(a)

Then since ilk is arbitrary we have E = 1 + A.

2.2.

Prove that if ilk = f (a + kh), Zk = g(a + kh) then (a) (a)

f (yk + Zk) = Dyk + AZk

(b)

AYkZk = ykOZk + Zk+lAyk

A(yk + Zk) = A[f (a + kh) + g(a + kh)]

[f (a + kh + h) + g(a + kh + h)] - [f (a + kh) + g(a + kh)] [ilk + l + zk + 1] - [yk + zk]

Ayk+Azk (b)

A(ykzk) = A[f (a + kh) g(a + kh)]

= f(a+kh+h)g(a+kh+h) - f(a+kh)g(a+kh) = yk+1 zk+1 - Ykzk = yk(zk+1 - zk) + zk+1(Yk+1 - yk) = YkAzk + zk+lAyk

DIFFERENCE TABLES 2.3. Let f (x) = 2x2 - 7x + 9 where x = 3, 5, 7, 9, 11. Set up a table for the differences of f (x). The required table is shown in Fig. 2-8 below. Note that in column 2 the various values of f (x) are computed from f (x) = 2x2 - 7x + 9. Thus for example f (5) = 2(5)2 - 7(5) + 9 = 24, etc.

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

x

Ay

y = 1(x)

3

6

5

24

7

58

A2y

A3y

41

A4y

18 16 34

0

16

0

50

108

9

16

0

66 11

174

Fig. 2-8

From this table we see that the second differences are constant and equal to 16 while the

higher differences are zero illustrating Theorem 2-1, page 33.

2.4.

Find the numerical values corresponding to the letters in the following difference table where it is assumed that there is equal spacing of the independent variable x. y

A B

Ay

A2y

Asy

-1

M

J

H

-3

N

K

C

D E

Asy

F G

3

A4y

-2

P

-5

4

L I

Fig. 2-9

Starting at the right end it is clear that P - (-3) = -5 or P = -8. Similarly we find in

succession the equations 4 - N = P or N = 12, N - M = -3 or M = 15, J - (-1) = M or J = 14,

K-J=N or K=26, L-K=4 or L=30, I-(-2)=L or I=28, -2-H=K or H=-28, H-G=J or G=-42, G-F=-1 or F=-41, 3-B=G or B=45, B-A=F or A=86,

C-3H or C=-25, D-C=-2 or D=-27, E-DI or E=1.

The final difference table with the required numbers replacing letters is as follows: y 86 45 3

-25

Ay

A2y

-41 -42

-1

-28

Asy

Asy

15

-3

14

-5

12

-8

26

-2 -27

A4y

4

30

28 1

Fig. 2-10

The problem illustrates the fact that a difference table is completely determined when one entry in each column is known. From the manner in which the letters were obtained it is clear that they can each have only one possible value, i.e. the difference table is unique.

APPLICATIONS OF THE DIFFERENCE CALCULUS

42

2.5.

[CHAP. 2

By using only the difference table of Problem 2.3 [and not f (x) = 2x2 - 7x + 9] find the values of y = f(x) for x = 13, 21. We must extend the difference table of Problem 2.3 as shown in Fig. 2-11. To do this we first note that the second differences are all constant and equal to 16. Thus we add the entries 16 in this column as shown in Fig. 2-11. We then fill in the remaining entries of the table.

From the table it is seen that the values of y corresponding to x = 13,15,17,19, 21 are 256,

354, 468, 598, 744 respectively.

x

A2y

Ay

3

6

5

24

18

16 34

7

58

16 50

108

9

16 66

174

11

16 82

256

13

16 98

15

354

16

114 17

468

16 130

19

598

16

146 21

744

16

Fig. 2-11

THE GREGORY-NEWTON FORMULA AND APPLICATIONS 2.6.

Prove that (kh)(n) = hnk(n) for n = 1,2,3,.... Since (kh)(1) = kh = hk(1), the result is true for n = 1. Since (kh)2 = (kh)(kh - h) = h2k(k - 1) = h2k(2), the result is true for n = 2. Using mathematical induction [Problem 2.67] we can prove for all n

(kh)(n) _ (kh) (kh - h) ... (kh - nh + h) = hnk(k -1) ... (k - n + 1) = hnk(n)

2.7.

Show that if we put x = a + kh in the Gregory-Newton formula given by equations (44) and (45) on page 9 we obtain (6) and (7) or (8) on page 34. From equations (44) and (45) on page 9 we have

f(x) = f(a)

&f(a) (x -a)")

Ax

&2f(a) (x - a)(2)

+

1!

Rn =

where

Axe

21

+ ... +

Anf(a (x - a)(n)

n!

Axn

+ Rn

(1)

f(n+1)(17) (x - a)(n+1) (n + 1) 1

Then putting x = a + kh in (1) we find using Problem 2.6 and Ax = h

f(a+kh) = f(a) + = f(a) +

Af(a)

h

(kh)(1) + A2f(a) (kh)(2) +

1!

Af(a)k(1) 1!

+

2!

h2

A2f(a)k(2)

2!

+

... +

... +

Anf(a) (kh)(n) + R n hn n!

Anf(a)k(n) n1

R.

(2)

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

where

Rn =

f(n+1)(7j)(kh)(n+1)

Since f (a + kh) = yk, f (a) = yo, Af (a) = Ayo,

2.8.

_

(n+1)!

,

43

(n+1)!

(2) yields the required result.

The first 5 terms of a sequence are given by 2, 7, 16, 35, 70. (a) Find the general term of the sequence. (b) What is the tenth term of the sequence? (c) What assumptions are you making in these cases?

(a) Let us represent the sequence by u0, u1i u2, ... where the general term is uk. The given terms can be represented in the table of Fig. 2-12. k

0

1

2

3

4

Uk

2

7

16

35

70

Fig. 2-12

The difference table corresponding to this is shown in Fig. 2-13. Auk

Uk

A2Uk

A4uk

ASUk

2 5

7

4 9

6

16

10

0

19

6

35

16 35

70

Fig. 2-13

Then by the Gregory-Newton formula with Uk replacing yk, we have since uo = 2, Auo = 5,

A2uo = 4, Asuo = 6, A4uo = 0 Uk

= u° +

Au°k(1) 1!

= 2 + 5k +

+

A3uok(s)

A2uok(2)

2!

4k(k - 1)

+

+

3!

+

41

6k(k - 1)(k - 2)

2

6

ks-k2+5k+2 It should be noted that if the first 5 terms of the sequence are represented by u1, u2. ..., US rather than u0, u1i ..., u4, we can accomplish this by replacing k by k -1 in the above polynomial to obtain

uk = (k-1)$-(k-1)2+6(k-1)+2 = k3-4k2+10k-5

(b) The 10th term is obtained by putting k=9 and we find uo = 9$ - 92 + 5(9) + 2 = 695

(c)

In obtaining the above results we are assuming that there exists some law of formation of the terms in the sequence and that we can find this law on the basis of limited information provided by the first 5 terms, namely that the fourth and higher order differences are zero. Theoretically any law of formation is possible [see Problem 2.9].

APPLICATIONS OF THE DIFFERENCE CALCULUS

44 2.9.

.

[CHAP.2

Are there other formulas for the general term of the sequence in Problem 2.8? Explain. In obtaining the general term for the sequence in Problem 2.8 we assumed first the existence of some underlying law or formula giving these terms. Second we assumed that the function of k describing this law was such that the fourth and higher differences were all zero. A natural consequence of this was that the function was a third degree polynomial in k which we found by use of the Gregory-Newton formula. In such case the general term not only exists but is unique. By using other assumptions we can obtain many other formulas. For example if we write as general term

Uk = k3-k2+5k+2+ k(k-1)(k-2)(k-3)(k-4)

(1)

we obtain all the data in the table of Problem 2.8. On putting k = 9 however we would obtain for the 10th term the value ug = 15,815 differing from that of Problem 2.8. The formula (1) does not however have the fourth and higher differences equal to zero.

From these remarks it is clear that one must give careful consideration as to the meaning of the three dots in writing a sequence such as 2, 7,16, 35, 70, ... .

2.10.

Find a polynomial which fits the data in the following table. x

3

5

7

9

11

y

6

24

58

108

174

Method 1. The difference table corresponding to this is given in Fig. 2-8, page 41.

In this case a = 3,

h = 2 and we can use the Gregory-Newton formula (8) on page 34. We have for the leading differences

Then

1!k

Yo = 6, Ayo = 18, A21Jo = 16, 12!2)

= 6 + 18k(1) +

A3Y0 = 0,

041!0 = 0,

+ 0 = 6 + 18k + 8k(k-1)

...

8k2 + 10k + 6

To obtain this in the form y = f (x) we can write it equivalently as f (x) = f (a + kh) = f(3 + 2k) = 8k2 + 10k + 6

Thus using x = 3 + 2k, i.e. k = (x - 3)/2, we find f (x) = 8[(x - 3)/2] 2 + 10[(x - 3)/2] + 6 = 2(x - 3)2 + 5(x - 3) + 6 = 2x2 - 7x + 9

That this is correct is seen from the fact that the entries in the table of Fig. 2-8 were actually obtained by using f (x) = 2x2 - 7x + 9 [see Problem 2.3]. Method 2.

Letting a = 3, h = Ax = 2 in the Gregory-Newton formula (6), page 34, we have f(x)

(x-a)(1) + A2f(a) (x-a)(2) + ... = f(a) + Af(a) Axe 21 1! Ax

6 + 18 (x - 3) (1) + 16 (x - 3)(2) 2

1!

22

2!

-

6 + 9(x - 3) + 2(x - 3)(x - 3 - 2)

2x2-7x+9 using the fact that (x - a)(2) = (x - a) (z - a - h). It should be noted that there are other polynomials of higher degree which also fit the data [see Problem 2.12]. Consequently it would have been more precise to ask for the polynomial of least degree which fits the data.

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

45

INTERPOLATION AND EXTRAPOLATION 2.11.

Use the table of Problem 2.10 to find the value of y corresponding to (a) x = 8, (b) x = 5.5, (c) x = 15, (d) x = 0. (e) What assumptions are you making in these cases? Method 1.

From Method 1 of Problem 2.10 we have Yk = 8k2 + 10k + 6.

To find the value of k corres-

ponding to x = 8 we use x = a + kh with a = 3, h = 2 to obtain k = (x - a)/h = (8 - 3)/2 = 2.5. Thus the required value of y corresponding to x = 8 is Y2.5 = 8(2.5)2 + 10(2.5) + 6 = 81

In a similar manner the values of k corresponding to x = 5.5, x = 15, x = 0 are (5.5 - 3)/2, Then the required values of y corresponding

(15 - 3)/2, (0 - 3)/2 or 1.25, 6, -1.5 respectively. to x = 5.5, 15 and 0 are respectively

Y1.25 = 8(1.25)2 + 10(1.26) + 6 = 31,

y8 = 8(6)2 + 10(6) + 6 = 354

y_1.5 = 8(-1.5)2 + 10(-1.5) + 6 = 9 Method 2.

From either Method 1 or Method 2 of Problem 2.10 we have

y = f (x) = 2x2 - 7x + 9 Then the required values are

f(8)

= 2(8)2 - 7(8) + 9 = 81

f(5.5) = 2(5.5)2 - 7(5.5) + 9 = 31

f(15) = 2(15)2 - 7(16) + 9 = 354

f (0)

= 2(0)2 - 7(0) + 9 = 9

Note that in parts (a) and (b) we are interpolating, i.e. finding values within the table, while in parts (c) and (d) we are extrapolating, i.e. finding values outside the table. (e)

2.12.

In using the method we are assuming that there exists some underlying law which the data follows and that we can find this law on the basis of the limited information supplied in the table which suggests that all third and higher differences are actually zero so that the data is fitted by a polynomial of second degree. It is possible however that even if an underlying law exists, it is not unique. The analogy with finding general terms of sequences or series as discussed in Problem 2.9 is apparent.

(a) Give an example of a polynomial of degree higher than two which fits the data of Problem 2.3. (b) Discuss the relationship of this to problems of interpolation and extrapolation. (a) One example of a polynomial fitting the data of Problem 2.3 is obtained by using F(x) = 2x2 - 7x + 9 + (x - 3)(x - 5)(x - 7)(x - 9)(x - 11) (1) which is a polynomial of degree 5. For x = 3, 5, 7, 9, 11 the values agree with those of the table or equivalently those obtained from f(x) = 2x2 - 7x + 9. Other examples can easily be made up, for example the polynomial of degree 11 given by (2) F1(x) = 2x2 - 7x + 9 + 3(x - 3)2(x - 5)(x - 7)4(x - 9)(x - 11)3 For an example of a function which fits the data but is not a polynomial we can consider (3) F2(x) = 2x2 - 7x + 9 + (x - 3)(x - 5)(x - 7)(x - 9)(x 11)e-x

(b)

Since there are many examples of functions which fit the data, it is clear that any interpolation formula based on data from a table will not be unique. Thus for example if we put x = 8 in (1) of part (a), we find the value corresponding to it is 126 rather than 81 as in Problem 2.11. Uniqueness is obtained only when we restrict ourselves in some way as for example requiring a polynomial of smallest degree which fits the data.

46 2.13.

APPLICATIONS OF THE DIFFERENCE CALCULUS

[CHAP. 2

The table of Fig. 2-14 gives the values of sin x from x = 25° to x = 30° in steps of 1°. (a) Using interpolation find sin 28°24' and (b) compare with the exact value.

sin x

25°

26°

27°

28°

29°

300

0.42262

0.43837

0.45399

0.469.47

0.48481

0.50000

Fig. 2-14

To avoid decimals consider f (X) = 105 sin x and let x = a + kh where a = 25° and h = 1°. Then we can write yk = f (a + kh) = 105 sin (25° + k 1°). The difference table is given in Fig. 2-15. In this table k = 0 corresponds to x = 25°, k = 1 to 26°, etc. k

11k

0

42,262

1

43,837

QYk

&2Yk

1,575

-13

1,562 2

45,399

3

46,947

4

48,481

5

50,000

-14

1,548

-14 -15

1,534

03Yk

-1 0

-1

1,519

Fig. 2-15

It should be noted that the third differences are very small in comparison with the Ilk and for all practical purposes can be taken as zero. When x = 28°24' = 28.40, we have k = (28.4° - 25°)/1° = 3.4. Thus by the Gregory-Newton formula

y3.4 = 105 sin (28° 24') 02yok(2)

4yok(1)

+

= 110+

2!

+... (-13)(3.2'(3.4 - 1) +

= 42,262 + (1,575)(3.4) +

= 47,564

This yields a value of sin 28° 24' = 0.47564. For an estimate of the error see Problem 2.14. (b) The exact value to 5 decimal places is 0.47562. Thus the absolute error made is 0.00002 and the percent error is 0.004/0. 2.14.

Estimate the error term in Problem 2.13. The error is given by the remainder (7), page 34, with n = 2 R2

h3f"'(n) k(3)

3.

(1)

To evaluate this note that in formulas of calculus involving trigonometric functions, angles must be expressed in radians rather than degrees. This is accomplished by using 1° = r,/180 radians. If x is in radians we have f(x) = 105 sin x, f'(x) = 1.05 cos x, f"(x) _ -105 sin x, f"'(x) _ --105 cos x so that (1) becomes

R2 = -105

rr

13 (cos 7))(3.4)(3.4 - 1)(3.4 - 2) 3!

180/

(2)

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

47

Since the largest value of cos n is 1 - sine 25° _ 0.90631 we see from (2) that R2 = -1 approximately, i.e. the best value we can get for sin 28°24' is 0.47563. The fact that this is still not correct is due to rounding errors in the given table.

CENTRAL DIFFERENCE TABLES AND INTERPOLATION FORMULAS 2.15. (a) Construct a central difference table corresponding to the data of Problem 2.13 by choosing a = 28°. (b) Use Stirling's interpolation formula to work Problem 2.13 and (c) compare with the exact value and the value obtained from the GregoryNewton formula. (a) The central difference table is shown in Fig. 2-16.

k

Yk

Ayk

-3 -2 -1

43,837

0

46,947

A2Yk

A3yk

-13 -14

-1

42,262 1,575 45,399

1,562 1,548 1,534

48,481

1

1,519 2

0

-14

-1

-15

50,000

Fig. 2-16

(b) Here k = 0 corresponds to 28°, k = 1 corresponds to 29°, k = -1 corresponds to 26°, etc. The value of k corresponding to 28.4° is k = (28.4° -28°)/1° = 0.4.

From the table we see that yo = 46,947, Ay-1 = 1,548,

Ayo = 1,534,

A2y_i = -14, A3y_2 = -1, A3y_I = 0

Then by Stirling's formula 110.4 = 105 sin (28° 24') 1

1 f k(2)

WD

110 +

2(Ay-I+AY0)

1! + Azy

(1,548 + 1,534) (Z

46,947 +

1

2 ( 2! + k+ (2!1)(2) \ 1 + ... /J

((0.4)2

0.6) + (1.2'0.4))1 +

(-14) C2

2

= 47,562

Thus sin 28° 24' = 0.47562. (c)

The result obtained in (b) is correct to 5 decimal places. The additional accuracy of Stirling's formula over the Gregory-Newton formula is explained by the fact that 28°24' = 28.4° occurs in the central part of the table. Thus the Stirling formula, which uses the differences near the center, is expected to be more accurate than the Gregory-Newton formula, which uses differences near the beginning of the table.

APPLICATIONS OF THE DIFFERENCE CALCULUS

48 2.16.

[CHAP.2

(a) Explain why you might expect Bessel's interpolation formula to. give a better approximation to sin 28°24' than Stirling's formula. (b) Can you confirm your expectation given in part (a) ? (a)

Since the differences used in Bessel's interpolation formula lie to one side of the center it might be expected that since 28.4° lies to the same side, a better approximation would be obtained.

(b) From the table of Fig. 2-16, yo = 46,947,

y1 = 48,481,

Ayo = 1,534,

A2y_1 = -14,

A2yo = -15

Thus Bessel's interpolation formula gives 71k

or

yo.4

=(7l0 + y1) + Ayo =

kcl)

DI

(k - 1)(1)

(1!! +

k(2)

1

+(4271_1 + + A2yo) 2 + .. .

)J

1!

(46,947 + 48,481) + 1,534[1(0.4 - 0.6)] + J(-14 - 15)

0.6)] +

[LO .4)

2

= 47,562

Thus we obtain sin 28° 24' = 0.47562. Although Bessel's formula might have been expected to give a better approximation than Stirling's formula we cannot confirm the expectation in this case since the result obtained from Stirling's formula is already accurate to 5 decimal places.

2.17.

Derive the Gregory-Newton backward difference formula by (a) using symbolic operator methods, (b) using any other method. (a) We have since E - A = 1 E Yk

= (1- AE-1)-kyo

k

E - A) Yo

C1 + (-k)(-AE-1) +

=

11 + kAE-1 +

k(k2

= yo + kA y-1

i kc1)

2!

3!

1) A2E-2 + k(k + 3)' k + 2) A3E-3 + ...

k(k + 1)(k + 2) + 1) A,Y-3 -3 A2y-2 + 3! 2!

+

7l0+Ay-1

(-k)(-k -1)(-AE-1)2 + (-k)(-k -1)(-k - 2)(-AE-1)3

k(2)

+0271-22! + A3y-3

yo

+

k(3) 31

where we have used the binomial theorem.

(b) Assume Then

Yk

= A0 + A1k(') + A2(k + 1)(2) + A3(k + 2)(3) + ... + An(k + n - 1)(11)

+ nAn(k+n- 1)(n-1) Ayk = Al + 2A2(k + 1)(1) + 3A3(k+2)(2) + + n(n -1)An(k + n - 1)(n-2) A2yk = 2A2 + 6A3(k + 2)(1) + .................................................................

Anyk = n! An

Putting k = 0 in the expression for

Yk,

k = -1 in Ayk, k = -2 in A2yk, ..". we find A2y-2

Ao = yo, and so

Al = Ay-1,

A2 =

ku1

Yk = Yo + All-1 1! +

A2y-2

2!

,

...,

An =

Any-n n!

(k+1)(2) 2! + ... + Any_n (k+n-1)(n) n!

CHAP. 2]

2.18.

APPLICATIONS OF THE DIFFERENCE CALCULUS

49

Derive the Gauss interpolation formulas from the Gregory-Newton formula. The Gregory-Newton formula is Yk = Y0 + A71o 1(F) + &2Yo 2 i, + &3Yo k3 i 1 + .. .

Now from the generalized difference table of Fig. 2-4, page 35, we have A2Y0 - A2y_l = &311-1

A221o = A2y_l + A3y-1

or

Similarly, ,&311o - A3y-1 = A4y-1,

&411-1 - &411-2 = &511-2

from which A3y0 = A3y-1 + &411-1 = A3y-1 + A4y-2 + A511-2

Then by substituting in the Gregory-Newton formula we find

Yk =

k(2)

kc1 110

+ A710 li + k(l)

110 + &1/0 11

Yo

+ Ayo

kc1

kc2>

+ A2y_l 2i + A31!_1 + A2y_l

k(2)

+ A31!_1

21

kc3

+ (&311-1+&411-2+A511-2)

(A2y_,+A3y-1)

k(2)

+

2!

k(3)

3

(k + 1)(3

3!

+ A4y_2

+ &4y_2 (k

k(s)

3!

+ 1)(4

4!

3 + ... k(4)

+_4!

1+

+ ...

The second Gauss formula can be derived in a similar manner [see Problem 2.94].

2.19.

Derive Stirling's interpolation formula. From the Gauss formulas we have kc1

11k

= Y0 + A11o 1 + ' 21l-1

11k

= 110 + AY-1

k(2) 2

k(D i

+ &211-1

+ &37l-1

(k+l)(2) 21

(k + 1)(3)

+

3

+ &311-2

(k+1)(3) 31

+

Then taking the arithmetic mean of these two results we have

Yk = y0 + - (Ay-1 + Ayo)

k(l) !

1

+ &2y- i

[_2

kc2)

(2

+

(k + 1)(2) 2!

) -

2.20. Show how to express the Gregory-Newton backward difference interpolation formula in terms of the backward difference operator V. By Problems 1.39(a), page 25, and 1.93, page 30, we have V2 = A2E-2, ..., Vn = AnE-n V = AE-1, Thus

n = 1, 2, 3, .. .

AnE-nf(a) = Vnf(a) or equivalently

AnE-nyo = Vnyo Using this in (8), page 34, it becomes 01)

11k

= 110 + Vyo 1 + V211o

(k + 1) (2)

2!

+ V3Yo

(k + 2) (3)

3!

+ ...

APPLICATIONS OF THE DIFFERENCE CALCULUS

50

[CHAP. 2

ZIGZAG PATHS AND LOZENGE DIAGRAMS 2.21. Demonstrate the rules for writing interpolation formulas by obtaining (a) Bessel's formula, (b) Gauss' second formula on page 36. (a) We follow the path shown dashed in Fig. 2-5, page 37. Since the path starts with 1 in the first column of the difference table, the first term of the interpolation formula by rule 5, page 37, is 1 multiplied by the mean of the numbers yo and yl above and below 1, i.e. -(yo + yl)

Similarly by rule 4, page 37, since the path goes through Ayo the next term to be added is Ayo multiplied by the mean of the coefficients directly above and below, i.e. Ay 1 /k(1) + (k -1)(1)\ 021\ 1! 1!

Again using rule 5 since the path passes through the coefficient k(2)/2!, the next term to be added is k(2)/2! multiplied by the mean of the differences above and below the coefficient, i.e. k(2) 1

2 t!

2 (A2y _I + A27lo)

Thus continuing in this manner we find the required formula (k -1)(1) 1 k(1) (r+ -1!

Yk = .-(yo + yl) + Ayo 2 (b)

k(2)

+

(A2y-1 + A21fo) 21 + .. .

In this case we follow the zig-zag path indicated by the solid lines of Fig. 2-5, page 37. By rule 2 if we assume that the path enters yo with negative slope, the first term is yo multiplied by the coefficient 1 directly above, i.e. the first term is Yo ' 1 = I/o The same term is obtained if we assume that the path enters yo with positive or zero slope. By rule 3 since the path enters Ay-1 with positive slope, the second term is Ay-1 multiplied by the coefficient directly below, i.e. AY-1

k(l) -f -I

By rule 2 since the path enters A2y-1 with negative slope, the third term is A2y_1 multiplied by the coefficient directly above, i.e. k + 1) (2)

A2y-1 (2!

Continuing in this manner we obtain the required formula

yk = 2.22.

k(1)

I/O + Ay-1 11 + A2y-1

(k + 1)(2) 21

+ Asy-2

(k + 1)(s) 3!

+ ...

Let P denote a zig-zag path around any closed lozenge or cell in Fig. 2-5, page 37, i.e. the path begins and ends on the same difference. Prove that according to the rules on page 37 the contribution is equal to zero. A general cell or lozenge is shown in Fig. 2-17. (k - r)(n-1)

A' Yr-1

(n-1)1

-L 28 An-lyr (k-r+ 1)(n-1) (n - 1)!

(

k -M;) (n)

(k - r +1)(n+1) (n + 1)! 2 ='PathAf+lyr-1

(k - r)(n+l) Any,

Fig. 2-17

(n+1)!

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

51

By rule 6 a reversal of path changes the sign of the corresponding term added. Thus we need only show that the contributions corresponding to the three paths from An-1yr to On+Iyr-I in Fig. 2-17 are all equal. Now if path 1 is chosen the contribution is

= Qn

Cl

7lr-1

(k - r) (n) + An+1

n!

1!r-D

(k - r + 1)(n+ 1)

(1)

(n+ 1)!

Similarly if paths 2 and 3 are chosen the contributions are respectively

n Aft1!r) y,) C2 =( 71r-1

+

n!

n+D 7lr-121

(k-r)(n+1)

+ (n+l)! ) (n + 1)! ((k-r+1)n+D

(2)

C3 = O r(knf)(n) + On+Dyr-1(k(n+1)!D)

(3)

From (1) and (3) we see that C1

-

C3 =

(k-r)(n) n!

(0"vr-1 - Any,) + On + 171r- 1

(kn')(n) (-An +Iyr-1)

- (k

)Cn)

Qn+171r-1

n!

(k

On+Dyr-1

+ On+ Yr-1

_ (k-r)(n+l) -(n+1) r + 1) (n + 1) !

(n + 1) !

)

(kn,)(n) /(k-r-1) 1k-r-n)1

(k ni)(n)

= 0 so that C1 = C3. Also we note that the mean of C1 and C3 is

n+iyr_121 (k-r+l)(n+D) (k-r)(n+1) + (n } 1)! ) (n + 1)! Thus C,=C2=C3 and the required result is proved.

(CI+C3) _ - (k-r)(n)1 (Onvr-1+O"?1r) n!

C2

2.23. Prove that the sum of the contributions around any closed zig-zag path in the lozenge

diagram is zero. We shall prove the result for a typical closed path such as JKLMNPQRJ in Fig. 2-18. This path encloses 3 cells I, II, III.

Q

Fig. 2-18

If we denote the contribution corresponding to path JK, for example, by CJK and the total contribution around cell I, for example, by C1, then

CI+CII+ CIII

(CJK + CKN + CNR + CRJ) + (CKL + CLM + CMN + CNK) + (CRN + CNP + CPA + CQR) CJK + CKL + CLM + CMN + CNP + CPA + CQR + CRJ

+ (CKN + CNK) + (CRN + CNR) CJKLMNPARJ

since CKN = -CNK and CRN = -CNR. lows that CJKLMNPARJ = 0.

Thus since CI = 0, C11 = 0, C111 = 0 by Problem 2.22, it fol-

APPLICATIONS OF THE DIFFERENCE CALCULUS

52 2.24.

[CHAP. 2

Prove that if two zig-zag paths end at the same place in the lozenge diagram then the interpolation formulas corresponding to them are identical. Consider for example the two zig-zag paths indicated dashed and solid in Fig. 2-19. two paths end at the same place but begin at different places.

These

(k-r)(1) Yr

1!

(k-r-1)(1) 1!

AYr+1

Fig. 2-19

The total contribution to the interpolation formula corresponding to the dashed path is given by

Yr+A1lr

(k

r)

1!

-AYr

(k

1!

+R

(1)

when R is the remaining part of the formula. Similarly the total contribution to the interpolation formula corresponding to the solid path is (k - r - 1)(1)) 1 C(k - r)c1) + -Dyr (k - r1 !- 1)(" + R (2) J(Yr + Yr+ 1) + Ayr 2 it 1! Note that the minus signs in (1) and (2) are used because of rule 6, page 37. To show that the results (1) and (2) are actually equal we need only show that their difference is zero. We find this difference to be (k r)! 1AY.

1!

- JAY,.

(8)

11

Since yr+1 = yr + Ayr, (3) can be written 12y r

r 1+ (k-r-1)(1) -(k-r)(1)l 1

1!

A7lr[1 + k - r - 1 - (k - r)]

=0

so that the required result is proved.

2.25.

Prove the validity of all the interpolation formulas on page 36. Since all the formulas can be obtained from any one of them, it follows by Problem 2.24 that they are all valid if we can prove just one of them. However we have already proved one of them, namely the Gregory-Newton formula on page 36. This proves that all of them are valid. Note that Problems 2.18 and 2.19 illustrate how some of these formulas can be obtained from the Gregory-Newton formula.

LAGRANGE'S INTERPOLATION FORMULA 2.26. Prove Lagrange's interpolation formula on page 38. Since there are p + 1 values of y corresponding to p + 1 values of x in the table of Fig. 2-1 on

page 32, we can fit the data by a polynomial of degree p. Choose this polynomial to be y = a0(x - x1)(x

-x2)...(x - xp) + a1(x - x0)(x - x2) ...(x - xp)

+ ... +ap(x-xo)(x-x1)...(x-xp-1)

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

53

Then putting x = xo, y = yo we have yo = ao(xo - xl)(xo - x2) ... (xo - xP)

Yo

ao =

or

(xo - x1)(xo -

x2)...(xo - x0)

Similarly putting x = x1, y = yj we find yi = al(xl - xo)(xl - x2) ... (xl - xP) with corresponding values for a2, a3...., aP. obtain Lagrange's formula.

2.27.

Y1

a1 =

or

(x1- xo)(xl - x2) ... (XI - XP)

Putting these values in the assumed polynomial we

(a) Work Problem 2.10 by using Lagrange's formula and thus (b) solve Problem 2.11. (a)

Substituting the values from the table of Fig. 2-13 into Lagrange's formula we obtain _ y

(x - 5)(x - 7)(x - 9)(x - 11)

6

(x - 3)(x - 7)(x - 9)(x - 11)

(3 - 5)(3 - 7)(3 - 9)(3 - 11) +

+ 58

24 (5

- 3)(5 - 7)(5 - 9)(5 - 11)

(x - 3)(x - 5)(x - 9)(x - 11)

+ 108

(7 - 3)(7 - 5)(7 - 9)(7 - 11) + 174

(x - 3)(x - 5)(x - 7)(x - 11)

(9 - 3)(9 - 5)(9 - 7)(9 -11)

(x - 3)(x - 5)(x - 7)(x - 9)

(11 - 3)(11 - 5)(11 - 7)(11 - 9)

Performing the algebra we find y = 2x2 - 7x + 9 which is a polynomial of degree two and not four as we might have expected.

(b) The values of y for x = 8, 5.5, 15, 0 can be obtained by substituting in the Lagrange interpolation formula of part (a).

TABLES WITH MISSING ENTRIES 2.28. Find the value of u3 missing in the following table. k

0

1

2

uk

3

8

15

3

4

47

Fig. 2-20

Method 1.

In this problem we cannot construct a difference table. However from the four pairs of values (k, uk) which are known we would be able to fit a polynomial of degree 3. This leads us to assume that the fourth differences are zero, i.e. A4uo = 0. Thus we have

(E -1)4uo = (E4 - 4E3 + 6E2 - 4E + 1)uo = 0 which becomes u4 - 4u3 + 6u2 - 4u1 + uo = 0

Using the values from the table we then have 47 - 4u3 + 6(15) -4(8) + 3 = 0

or

u3 = 27

Method 2.

Since there are 4 values of k for which Uk is known, we can fit the data by a polynomial

of degree 3 which involves 4 unknown coefficients. Let us therefore assume that

Uk = A0k3+A1k2+A2k+A3

APPLICATIONS OF THE DIFFERENCE CALCULUS

54

[CHAP. 2

Then putting k = 0, 1, 2, 4 in succession we obtain the equations

A3 = 3 Al + A2 + A3 = 8 8A0+ 4A1+2A2+A3 = 15 A0 +

64A0 + 16A1 + 4A2 + A3 = 47 Solving simultaneously we find A0

Al=-i-, A2 = 5, A33

,

Uk = Jk3 - --k2 + 5k + 3

and so

Putting k = 3 we then find u3 = 27. Method 3.

By Lagrange's formula we have Uk

(k - 1)(k - 2)(k - 4)

(k - 0)(k - 2)(k - 4)

u0 (0 - 1)(0 - 2)(0 - 4)

+ ul (1 - 0)(1 - 2)(1 - 4)

+ u2

(k - 0)(k - 1)(k - 4) (2 - 0)(2 - 1)(2

4)

+ u4 (k - 0)(k - 1)(k - 2) (4-O)(4-1)(4--2)

- 8 (k - 1)(k - 2)(k - 4) + 3 (k)(k - 2)(k - 4) - 4b (k)(k - 1)(k - 4) + 24 (k)(k - 1)(k - 2)

Then putting k = 3 we find u3 27. Note that Method 1 has the advantage in that we need not obtain the polynomial. 2.29. Find loglo 7 and loglo 11 from the following table. x

6

AX) = log10 x

0.77815

7

J

8

9

10

0.90309

0.95424

1.00000

11

12

1.07918

Fig. 2-21

Since there are 5 known pairs of values we could fit a ,polynomial of degree 4 to the data. We are thus led to assume that the 5th differences of f(x) = log10 x are zero. If we let x = a + kh and uk = f (a + kh), then taking a = 6 and h = 1 the given table can be written in the k notation as follows. k

0

uk = log10(6 + k)

0.77815

1

2

3

4

0.90309

0.95424

1.00000

5

6 1.07918

Fig. 2-22

From A5u0 = 0 and A5u1 = 0, i.e. (E -1)5u0 = 0 and (E -1)5u1 = 0, we are led to the equations (E5 - 5E4 + 10E3 -10E2 + 5E -1)u1 = 0 (E5 - 5E4 + 10E3 -10E2 + 5E -1)u0 = 0, u5 - 5u4 + 10u3 - 10u2 + 5u1 - u0 = 0 or

U6 - 5u5 + 10u4 - 10u3 + 5u2 - u1 = 0

(1)

Putting in the values of u0, u2, u3, u4 and us from the table we find

5u1 + u5 = 5.26615 u1 + 6u5 = 6.05223

(2)

Solving simultaneously we find u5 = 1og1011 = 1.04146 u1 = log10 7 = 0.84494, These should be compargd with the exact values to 5 decimal places given by loglo 7 = 0.84510, log1011 = 1.04139 so that the errors made in each case are less than 0.02% and 0.01% respectively.

CHAP. 2]

2.30.

APPLICATIONS OF THE DIFFERENCE CALCULUS

55

The amount Ak of an annuity [see Problem 2.144] after k years is given in the following table. Determine the amounts of the annuity for the missing years corresponding to k = 4 and 5. k

3

Ak

3.1216

FT

6

6.6330 __ __

9

12

10.5828

15.0258

Fig. 2-23

Let A3 and E3 respectfully denote the difference and shifting operators for the 3 year intervals while Al and El denote the corresponding operators for 1 year intervals. Then we have

E1 = 1+Al

E3 = 1+A3,

Uk+3 = E3uk = Eluk

Since

E3 = E3

we have

or

1 + A3 = (1 + A1)3

Al = (1 + A3)1/3 - 1

Thus

(1)

Using the binomial theorem we then have 1 (1/3)(-2/3) (1/3)(-2/3)(-5/3) A3 + ... 2, Al = A3 + A3 + 3

= 31A3 -

1

9

A3

5

+ 81 A3 + .. .

Since there are only four pairs of values in the table we can assume that third differences A are constant. Thus we can omit fourth and higher differences in the expansions. We thus have Al = 3 A3 - A3 + si A3 (2) 29

2

Al

2

3

(3)

= 9 A3 - 27 A3 Ai

=

27

(4)

A3

The difference table for the 3 year intervals is shown in Fig. 2-24. k 3

3.1216

6

6.6330

3

2

A3uk

Uk

A3uk

A3 uk

3.5114 0.4384

0.0548

3.9498 9

10.5828

12

15.0258

0.4932 4.4430

Fig. 2-24

Using (2), (3) and (4) we then have 1

1

5

1 1 2 + 81 A3u3 = 3 (3.5114) - 9 (0.4384) + Sl (0.0548) A1u3 = 3A3u3 - 9A3u3

= 1.12513827

Al u3 =

A3 u3 - 27 A3u3

(0.4384) - 27 (0.0548)

= 9

9

= 0.04465185 3

Alu3

(0.0548)

= 0.00202963

= 27 By using the above leading differences we can now complete the difference table for 1 year intervals as shown in Fig. 2-25 below. 27 A3u3

56

APPLICATIONS OF THE DIFFERENCE CALCULUS

k

Uk

3

3.1216

4

4.2467

5

5.4165

[CHAP. 2

Diuk

Qluk

Diuk

1.12513827 0.04465185

1.16979012

0.00202963

0.04668148 1.21647160

6.6330

6

Fig. 2-25

From this table we see that U4 = 4.2467,

us = 5.4165

By proceeding in a similar manner we can find u7, u8, u10, u11.

DIVIDED DIFFERENCES 2.31. Construct a divided difference table corresponding to the data in the following table. x

2

4

5

6

8

10

1(x)

10

96

196

350

868

1746

Fig. 2-26

The divided difference table is as follows x

f(x)

2

10

4

96

43

19

100 5

2

27

196

0

2

154 6

35

350

0

259 8

2

45

868 439

10

1746

Fig. 2-27 In the third column the entry 43 is obtained from

96 - 10 4

-

2

, 100 is obtained from

196 - 96 5 - 4

'

etc.

etc.

- 100 In the fourth column the entry 19 is obtained from 100 - 43 , 27 is obtained from 154 6-4

5 --2

In the fifth column the entries are obtained from

equal to 2. 2.32.

Prove that

(a)

27- 19 6 - 2

35 - 27 45 - 35 8 - 4 and 10 - 5 and are all

f (x) = f (xo) + (x - xo)f (x, xo)

(b) f(x, xo) = f(xo, x1) + (x - xl)f (x, xo, xl)

(C) A XI x0, x1) = f(xo, x1, x2) + (x - x2)f(x, x0, x1, x2) (a) By definition,

f (x, x0) =

Ax)

- f (x0) x-x0

so that

f(x) = f(xo) + (x - x0)f(x, XO)

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

(b) By definition,

f (x, x0, xi) = f(x, x0) - f(xo, xi)

57

so that

X - xi

Ax, x0) = f (xo, xi) + (x - xi)f (x, x0, xi)

(c) By definition,

A XI xo, xi, x2) =

f(x, xo, xl) - f(xo, x1, x2) X

- x2

so that

f(x, xo, xi) = f(xo, xi, x2) + (x - x2)f(x, xo, xv x2) Generalizations can easily be made. 2.33.

Prove that

f (x) = f (xo) + (x - xo) f (xo, xi) + (x - xo)(x - xi)f(xo, xi, x2) + R

R = (x - xo)(x - xl)(x - x2)f(x, xo, xi, x2)

where

Using the results of Problem 2.32(a), (b) and (c) we have

f(x) = f(xo) + (x-xo)f(x,xo) = f(xo) + (x - xo)[f(xo, xl) + (x - xi)f(x, x0, xi)] = f(xo) + (x - xo)f(xo, xi) + (x - xo)(x - xi)f(x, X0, xi) = 1(x0) + (x - xo)f(xa, xi) + (x - xo)(x - xi)[f(xo, x1, x2) + (x - x2)f(x, xo, xi, x2)]

= 1(x0) + (x - xo)f(xo, xi) + (x - xo)(x - xl)f(xo, xi, x2) + (x - xo)(x - xi)(x - x2)f(x, xo, x1, which is equivalent to the required result.

2.34.

x2)

Prove Newton's divided difference interpolation formula f (X) = f (xo) + (x - xo)f (xo, xi) + (x - xo)(x - xl)f(xo, xl, x2)

+ . .. + (x - xo) ... (x - xn-i)f (xo, . .. , xn) + R(x)

R(x) = (x - xo)(x - xi)

where

.

. (x - xn)f (x, xo, xi, ..., xn)

We shall use the principle of mathematical induction. Assume that the formula is true for n = lc, i.e.

f(x) = f(xo) + (x - xo)f(xo, x1) + ... + (x - xo)... (x - xk-i)f(xo, ..., xk) + (x - xo) (x - x1) ... (x - xk)f (x, xo, xi, ... , xk)

(1)

Now by definition as in Problem 2.32 we have

fx, xo, xi, ..., xk+i) = so that

AZ, xo, ... , xk) - A X0, xi, ... , xk+ i) X

(2)

Xk+1

f(x, x0, x1, ..., xk) = f(xo, xi, ..., xk+l) + (x - xk+l)f(x, xo, xi, ..., xk+l)

(3)

Using this in (1) we find

f(x) = f(x0) + (x-xo)f(xo,xi) + ... + (x-xo)...(x-xk)f(xo, ...,xk+l) + (x - xo) (x - x)

. (x - Xk + 1)A X, xo,

xi, ... , xk+ l)

Thus assuming that it is true for n = k we have proved it true for n = k + 1. But since it is true for n = 1 [see Problem 2.32], it is true for n = 2 and thus for n = 3 and so on for all positive integers. 2.35.

Prove that the remainder in Problem 2.34.can be written as fn + l)(q)

= (x - xo)(x - xi) . . . (x - xn) (n + 1)! where 7 is a number between the smallest and largest of the values x, xo, xi, ... , x,,. R(x)

By Problem 2.34 the remainder is R(x) = (x - x0)(x - xi) ... (x - x,)f (x, x0, x1, ... , xn)

(1)

APPLICATIONS OF THE DIFFERENCE CALCULUS

58

[CHAP. 2

Since R =0 at x = x0, x1, ...,x.. , we know by Problem 1.117, page 31, that the (n + 1)th derivative

R(n+i)(x) is zero for some value ,, i.e. R(n+1)(71) = 0

(E)

where , is a number between the smallest and largest of the values x, x0, xl, ..., x,. Now by taking the (n + 1)th derivative of f (z) in Newton's divided difference formula, we have

f(n+i)(x) = n! f(xo, xl, ..., x,) + R(n+1)(x) Then letting x = 11 in (8) and using (2) we find

(s)

f(n+1)(y)

f(xo,xl,...,x,) =

(4)

(n+1)!

Thus (1) becomes

f(n+i)(,) R(x) = (x - xo)(x - xl)

(x - xn)

(5)

(n + 1)!

2.36. Use Newton's divided difference formula to express f(x) as a polynomial in x for the data of Problem 2.31. From Problem 2.31 we have

xo=2, x1=4, x2=5,

x3 = 6,

x4=8, x5=10

Also,

f(xo) = 10, f(xo, x1) = 43, f(xo, x1, x2) = 19, Then using Problem 2.33 we have

f(xo, x1, x2, x3) = 2, f(xo, x1, x2, x3, x4) = 0

f(x) = 10 + (x - 2)(43) + (x - 2)(x - 4)(19) + (x - 2)(x - 4)(x - 5)(2) + (x - 2)(x - 4)(x - 5)(x - 6)(0)

= 2x3-3x2+5x-4 2.37.

Solve the equation X3-3z -4 = 0. Consider y = f(x) = x3 - 3x - 4. We are looking for the values of x for which y = f(x) = 0. When x = 2, y = -2 and when x = 3, y = 14. It follows that x is between 2 and 3 since y changes sign from -2 to +14. Also it seems reasonable from these results that the value of x is closer to 2 than to 3. In view of these observations we set up a table of values for x and y as shown in Fig. 2-28. We can think of x as a function of y, i.e. x = g(y). x

2.0

2.1

2.2

2.3

2.4

y

-2.0000

-1.0390

0.0480

1.2670

2.6240

Fig. 2-28

We now set up the following table of divided differences. y

x=g(y)

-2.0000

2.0

-1.0390

-

0.10406 2.1

0.09200 0.0480

2.2

1.2670

2.3

2.6240

2.4

0.08203 0.07369

Fig. 2-29

-0.00589 -0.00432 -0.00324

0.00048 0.00029

APPLICATIONS OF THE DIFFERENCE CALCULUS

CHAP. 2]

59

By Newton's divided difference formula, g(y) = 9(yo) + (y - 7!0)9(7!0, yl) + (y

y1)9(y0, yl, y2) (1)

+ (y - yo)(y - Y1)(Y - y2)g(yo, yl, y2, y3) +

Since the required value of x is that for which y = 0, we put y = 0 in (1) to obtain x = 9(0) = 9(y0) - y09(yo, y1) + (y0)(y1)9(y0, y1, y2) - y0y1y29(y0, y1, y2, y3) + " '

(f)

Then letting yo = -1.0390 so that g(y0) = 2.1 and yl = 0.0480, y2 = 1.2670 we have g(yo, yj) _ 0.09200, g(y0, y1, y2) = -0.00432, 9(y0, y1, y2, y3) = 0.00029. Thus (2) becomes x = 2.1 + (1.0390)(0.09200) + (-1.0390)(0.0480)(-0.00432) - (-1.0390)(0.0480)(1.2670)(0.00029) = 2.1 + 0.09559 + 0.00022 + 0.00002

= 2.19583

As a check we note that when x = 2.19583 we have x3 - 3x - 4 = 0.00008 so that the value is quite accurate. To find the remaining roots of the equation x3 - 3x - 4 = 0 we use the fact that x - 2.19583 must be a factor of x3 - 3x - 4 and find by division that the other factor is x2 + 2.19583x + 1.82167. Setting this factor equal to zero we find that the other two roots are complex numbers given by -1.09792 i- 0.78502i.

INVERSE INTERPOLATION 2.38. Using the data in the table of Problem 2.10 find the value of x corresponding to y = f (x) = 100 by using (a) Lagrange's formula, (b) Newton's divided difference formula and (c) the Gregory-Newton forward difference formula. (a) Interchanging the roles of x and y in Lagrange's formula we have

x = 3 (y - 24)(y - 58)(y - 108)(y - 174) + 5 (6 - 24) (6 - 58)(6 - 108) (6 - 174)

(y - 6)(y - 58)(y - 108)(y - 174) (24 - 6)(24 - 58)(24 - 108)(24 - 174)

(y - 6)(y - 24)(y - 58)(y - 174) 6)(y - 24)(y - 108)(y - 174) + 9 (108 - 6)(108 - 24)(108 - 58)(108 - 174) + 7 (58(y--6)(58 - 24)(58 - 108)(58 -174)

+ 11

(y - 6)(y - 24)(y - 58)(y -108)

(174 - 6)(174 - 24)(174 - 58)(174 - 108)

Then putting y = 100 we find x = 8.656. (b) From the table of Problem 2.10 we obtain the following divided difference table. y

x

174

11

108

9

0.030303

-0.00008360 0.040000

58

7

-0.00022410

0.058824 24

0.000000937

-0.000000040

0.000007661

-0.00100552

5

0.111111 6

3

Assuming the fourth order difference to be negligible we have by Newton's divided difference formula x = 11 + (y - 174)(0.030303) + (y - 174)(y - 108)(-0.00008360) + (y -174)(y - 108)(y - 58)(0.000000937)

Putting y = 100 we find x = 8.628.

APPLICATIONS OF THE DIFFERENCE CALCULUS

60 (c)

[CHAP. 2

Since the values of x are equally spaced we can use the Gregory-Newton formula to find an interpolating formula for f (x). As in Problem 2.11 we find f (x) = 2x2 - 7x + 9. If f (x) = 100, i.e. 2x2 - 7x + 9 = 100, then 2x2 - 7x - 91 = 0 and by solving this equation we find x = 7 -! 777 _ 7 -'-- 27.8747 4

4

Since only the positive square root is significant for our purposes, we use it to find x = 8.718. This can in fact be taken as the true value of x since the table of Problem 2.10 was actually constructed by using the function f (x) = 2x2 - 7x + 9.

2.39. Work Problem 2.37 by using inverse interpolation involving Lagrange's formula. We use the table of Fig. 2-28, page 58. Then by Lagrange's interpolation formula we have (y + 1.0390)(y - 0.0480)(y - 1.2670)(y - 2.6240) x = 2.0 (-2.0000 + 1.0390)(-2.0000 - 0.0480)(-2.0000 - 1.2670)(-2.0000 - 2.6240)

+ 2.1

(y + 2.0000)(y - 0.0480)(y -1.2670)(y - 2.6240) (-1.0390 + 2.0000)(-1.0390 - 0.0480)(-1.0390 - 1.2670)(-1.0390

+ 2.2

(y + 2.0000)(y + 1.0390)(y - 1.2670)(y - 2.6240) (0.0480 + 2.0000)(0.0480 + 1.0390)(0.0480 - 1.2670)(0.0480 - 2.6240)

2.6240)

(y + 2.0000)(y + 1.0390)(y - 0.0480)(y - 2.6240) + 23 (1.2670 + 2.0000)(1.2670 + 1.0390)(1.2670 - 0.0480)(1.2670 - 2.6240)

+ 2.4

(y + 2.0000)(y + 1.0390)(y - 0.0480)(y - 1.2670) 0.0480)(2.6240 - 1.2670)

(2.6240 + 2.0000)(2.6240 + 1.0390)(2.6240

Since the required value of x is the one for which y = 0, we obtain on putting y = 0 in the above result x = 2.19583 in agreement with that of Problem 2.37.

APPROXIMATE DIFFERENTIATION 2.40. Establish equation (16), page 39, (a) by symbolic operator methods and (b) by using the Gregory-Newton formula. (a)

In using symbolic operator methods we proceed formally. We start with

ehD = 1+A [equation (43), page 8]. Taking logarithms this yields

hD = In (1 +A)

D=

or

h

In (1 + A)

Then using the formal Taylor series on page 8 we find

2 +. 3 4_+... 2

D = hln(1+A) _ (b)

4

3

By the Gregory-Newton formula,

f(a+kh) = f(a) + kAf(a) +

k(k

21

1) A2f(a)

+ k(k - 3)1 k - 2)

Aaf (a) +

k(k - 1)(k

i 2)(k - 3) Aaf(a) + .. .

Then differentiating with respect to k,

hf'(a + kh) = Af(a) + 2k2!

1 A2f(a)

+ 3k2 -36k + 2 A3f(a)

+ 40 - 18k2 +

22k - 6 A4f(a) + ...

4!

Putting k = 0 we have hf'(a)

= Af(a) -

AZf(a) 2

+

Aa3 3

-

A4

4

+ ...

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

61

i.e.

+3344 ...)f(a)

Df(a) = Since f(a) is arbitrary we have 2

D=

4

3

2+

A

4+

...1

If we take into account the remainder term in the Gregory-Newton formula, then we can find a corresponding remainder term for the derivative. 2.41.

Given the table of values for the function f (x) in Fig. 2-30, find the approximate value of f'(0.25). x

0.25

0.35

0.45

0.55

0.65

0.75

AX)

0.24740

0.34290

0.43497

0.52269

0.60518

0.68164

Fig. 2-30 We consider the difference table for 105f(x) so as to avoid decimals.

This table is as follows.

The difference columns are headed A, A2, ... in place of 105Af (x),105A2f (x), ... for brevity. A

105f(x)

A3

A2

A4

A5

24,740 9550

-343

34,290 9207

-92

-435

43,497 8772

-523

52,269 8249

4 8

-80

-603

60,518

4

-88

7646

68,164

Fig. 2-31

Using the formula of Problem 2.40 we have (A D[105f(x)]

=h

- 22 + 3 - 44 ..}.....1 [105f(x)]

and neglecting differences of order 5 and higher we find for x = 0.25

1 L9550 - (-342 3) + (-92) -

105f'(0.25)

4(4)1

= 96,898

or f'(0.25) = 0.96898.

The table of values are those for f(x) = sin x where x is in radians. From this we see f'(x) = cos x. By way of comparison the true value is f'(0.25) = cos 0.25 = 0.96891. 2.42. Use the table of Fig. 2-31 to find f"(0.25). From Problem 2.40,

D=

2

3

4

(A-T+T-T }

so that on squaring we find D2

= h2 (A2-A3+12A4+ ..

that

APPLICATIONS OF THE DIFFERENCE CALCULUS

62 Thus

=

D2f (x)

[CHAP. 2

-(A2-A3+ 204 + ... )f (x)

22

From this it follows that 105f"(0.25)

(-343+92 +3.7)

1

(0.1)2

or f"(0.25) = -0.25470.

-25,470

The exact value is f"(0.25) = -sin 0.25 = -0.24740.

2.43. Referring to the table of Fig. 2-31 find f'(0.40).

From the Gregory-Newton formula we have k(

f(a+kh) = f(a) + kef(a) + 2! 1) o2f(a) + k(k- 3!k-2) o3f(a) + ... Differentiation with respect to k yields hf'(a + kh)

= of(a) +

2k2'

3k2 - 6k + 2 03f(a) +

102f(a) +

+ 4k3 - 18k2 + 22k - 6 4!

3!

Letting a = 0.25, h = 0.1 and a + kh = 0.40 so that k = 1.5 we find using the difference table of Problem 2.41

= 011 [9550 + (-343) - z1(-92) + 0(4)]

105f'(0.40)

f'(0.40) = 0.92108

The true value is f'(0.40) = cos 0.40 = 0.92106.

MISCELLANEOUS PROBLEMS 2.44. The population of a particular country during various years is given in the following table. Obtain an estimate for the population in the year (a) 1980, (b) 1955 and (c) 1920. Year

1930

1940

1950

1960

1970

Population in millions

1.0

1.2

1.6

2.8

5.4

Fig. 2-32 Since the years given are Let us denote the year by x and the population in millions by P. equidistant it is convenient to associate the years 1930, 1940, 1950, 1960, 1970 with the numbers

This is equivalent to writing x = a + kh where a = 1930 and

k = 0, 1, 2, 3, 4 respectively. h = 10 so that

x = 1930 + l0k

or

k -

x - 1930 (1)

10

Using this we can rewrite the given table as follows. k

0

1

2

3

4

P

1.0

1.2

1.6

2.8

5.4

Fig. 2-33

The difference table corresponding to this is as follows.

CHAP. 2]

APPLICATIONS OF THE DIFFERENCE CALCULUS

k

P

0

1.0

1

1.2

OP

02P

03P

63

D4P

0.2 0.2

0.4

2

0.6

1.6

0.8

0

1.2

3

0.6

2.8

1.4

2.6

4

5.4

Fig. 2-34

We shall assume that the third differences are all constant [equal to 0.6] and that the fourth

differences are zero. Then using the Gregory-Newton formula we find

P = =

1.0 + (0.2)k(') +

0.2(2) + 0.63! (3)

1.0 + 0.2k + 0.1k(k - 1) + 0.1k(k - 1)(k - 2)

= llo (k3 - 2k2 + 3k + 10)

(a) From (1) we see that the year 1980 corresponds to k = 5 so that

P=

1 oo

[53 - 2(5)2 + 3(5)+ 10]

Thus we estimate that in 1980 the population will be 10 million.

(b) From (1) we see that the year 1955 corresponds to k = 1955 - 1930 2.5

10

Then

P=

11o[(2.5)3-2(2.5)2+3(2.5)+10] = 2.0625

Thus to two significant figures we would estimate that the population in 1955 was 2.1 million. (c)

The value of k corresponding to the year 1920 is from (1)

k =

1920 - 1930

= -1

10

Then

P = i o [(-1)3-2(-1)2+3(-l)+10] = 0.4 Thus we estimate that the population in 1920 was 0.4 million. 2.45.

Suppose that it is necessary to determine some quantity y at the time when it is a maximum or minimum. Due to error we may not be able to do this exactly. However we observe that at times t1, t2, t3 near the time for maximum or minimum, the values are given respectively by y1, y2, y3. Prove that the time at which the quantity is a maximum or minimum is given by y1(t2 - t3) + y2(t3 - t1 + y3(t1 - t2 2[yl(t2 - t3) + y2(t3 - t1) + y3(t1 - t2)]

By Lagrange's formula the value of y at time t is given by (t - t1)(t - t3) + (t - t2) (t - t3) (t - t1)(t - t2) + y2 y3 (1) (t3 - t1R 3 - t2) (tl - t2i' 1 - t3) (t2 - t1)(t2 - t3) The maximum or minimum of y occurs where dy/dt = 0 [assuming it to be a relative maximum or minimum]. Then taking the derivation of (1) with respect to t, setting it equal to zero and solving for t we find the required value given above.

Y=

111

APPLICATIONS OF THE DIFFERENCE CALCULUS

64 2.46.

[CHAP. 2

If an error is made in one of the entries of a table show how the various differences are affected. Suppose that the value yo + E is given in a table in place of the correct value yo so that is the error. Then the difference table can be constructed as in Fig. 2-35. Ay-4 0271-4

Ay-3

0371-4

A2Y-3

AY-2

A2Y-2 + E

Ay_1+

,14Y-3 - 4E

A3y-2 - 3E

A2y-1 - 2E

Ayo - E

A4Y-2 + 6E

A3y-1+3E

A2yo + E AY,

A4y-4 + E

A3Y-3 + C

A3yo - E

,&2711

A4y-1-4E ,&4710 + E

A3Y1

AY2 A2Y2

Ay3

Fig. 2-35

Note that the error is propagated into the higher ordered differences as indicated by the shading in Fig. 2-35. It is of interest to note that the successive coefficients of e in each column alternate in sign and are given numerically by the binomial coefficients. Also the largest absolute error occurring in an even difference column lies on the same horizontal line in which yo occurs. 2.47.

x f(x)

In reading a table the entries given in Fig. 2-36 were obtained. (a) Show that an error was made in one of the entries and (b) correct the error. 1

2

4

3

7

6

5

8

9

10

11

12

.195520 .289418 .379426 .464642 .544281 .617356 .683327 .741471 .791207 .832039 .863558 .886450

Fig. 2-36

(a) The difference table omitting decimal points is as follows. 106f(x)

A

A2

A3

6,4

AS

196,520 93,898 289,418

90,008

379,426 85,216 464,642

79,639 544,281

73,075

617,356 65,971

683,327 58,144 741,471

49,736 791,207 40,832 832,039 31,519

863,558

-3890

-4792

-5577 -6564 -7104 -7827

-8408 -8904 -9313

-9627

-902 217

-785

-987

-202

-419 1268 649

447

-540

-723

-183

-630

955 325

142

-57

-581 85

-496 -409

21,892 885,450

Fig. 2-37

-382 59

2

87

6 8

95

-314

-1279

CHAP. 2]

APPLICATIONS OF TIE DIFFERENCE CALCULUS

65

The violent oscillations of sign which occur near the beginning to the column of fourth differences A4 and the practically constant values toward the end of this column indicate an

error near the beginning of the table. This is further supported in the column of sixth

differences Ao. In view of the fact that the largest absolute error in an even difference column is in the same horizontal line as the entry in which the error occurs, it would seem that the error should occur in the entry corresponding to -1279 which is largest numerically. It thus appears that the error should be in the entry 0.544281 corresponding to x = 5.

(b) Denoting the entry which is in possible error, i.e. 0.544281, by yo + e where a is the error, it is

clear from comparison with the difference table of Problem 2.46 that A4y_4+E = 217 A4Y-3 - 4E = -202 A4y-2 + 6e =

447

A4y_1-4e = -183 = 142 A4yo + Assuming that the correct fourth differences should all be constant, i.e. A4y_4 = A4y-3 = A4y-2 = A4y_l = A4y-0, we obtain by subtracting the first two equations, the second and third equation, etc. the equations 5E = 419 or e = 84, 10e = 649 or e = 65, 10E = 630 or e = 63, 5E = 325 or e = 65 Because most of these indicate the approximate value e = 65 [the value e = 84 seems out of line] we shall consider this as the most probable value. Taking into account the decimal point, this leads to e = 0.000065 and so since yo + 0.544281, the true value should be close to yo = 0.544281 - 0.000065 = 0.544216.

It is of interest to note that the true value was actually 0.544218. It appears that in copying the table the 1 and the 8 in this true value must have been interchanged. In Fig. 2-38 the difference table taking into account this error is shown.

106f(X)

A

A2

As

A4

A5

195,520 93,898

-3890

289,418

-902

90,008

-4792

379,426

54

-5640

464,642

50

-798

79,576

-6438

544,218

-7167

-7827 -8408 -8904

6

2

87

-409

40,832

-9313

832,039

1

85

-496

49,736 791,207

10 79

-581

58,144 741,471

0 69

-660

65,971

683,327

19 69

-729

73,138

617,356

-4

-848

85,216

-2 85

-314

31,519

-9627

863,558 21,892

885,450

Fig. 2-38

The changes of sign occurring in the fifth differences could be due to rounding errors as well as the fact that the function is not actually a polynomial.

APPLICATIONS OF THE DIFFERENCE CALCULUS

66

[CHAP. 2

2.48. Using the following table find the value of y corresponding to x = 0.6. x

0

0.1

0.2

0.3

0.4

0.5

y

1.250

1.610

2.173

3.069

4.539

7.029

Fig. 2-39

On constructing a difference table from the given data we obtain that shown in Fig. 2-40 where we use 103y in place of y to avoid decimals. 103y

0

1250

0.1

1610

A4

A3

A2

x

A5

360

203 130

563

111

333

2173

0.2

896

241

1470

446

1020

4539

0.4

94

205

574

3069

0.3

2490

7029

0.5

Fig. 2-40

Letting a = 0, h = Ax = 0.1 in the Gregory-Newton formula (6) on page 34 we have

103y = 1250 +

3060

x(1) +

130 2

0 1)2 x(2) + 3! (0.1)3

x0

(21)

and we can show [see Problem 3.32] that (22)

r(p + 1) = pr(p)

The recursion formula (22) can be used to define r(p) for p -- 0 [see Problem 3.34(c)]. Some important properties of the gamma function are as follows. 1.

If p is a positive integer, r(p+1)=p! r(21)=v---

2.

r(r) r(1 - r) =

3.

4.

,

r'(1) =

e-t In t dt

0 < r < 1

sin r

-y = -0.5772156649 .

fo

where ,y is called Euler's constant.

If m is an integer we can write [see Problem 3.35]

h-r(h+1)

(23)

r(h-m+1) Since (23) has meaning for all values of m we shall take it as a definition of xlm' for all M.

86

THE SUM CALCULUS

From (23) it follows that

mx(m-1)h

ox(m) =

(24) x(m+i)

m _ (m + 1)h thus generalizing the results (25), page 6, and IV-2, page 83. If m = -1, we obtain [see Problem 3.39] and

[CHAP. 3

X(M)

-1

hr (h + 1)

(25)

(26)

The function on the right of (26) is sometimes called the digamma function and is denoted by *(x).

The results (23) through (26) can also be written in the "k notation" with k replacing

x and h=1.

BERNOULLI NUMBERS AND POLYNOMIALS Various sequences of functions f o (x), fl (X), f2 (x), ... have the property that Dfn(x) = fn-1(x) (27) where in (27), as well as what follows, we shall assume that n = 1, 2, 3, .... If we consider the function f. (x) to be a polynomial of degree n then we can show that n n fn (x) = cn + (n x 1) f + ... + c..-ix + cn (28) 1

where co, c1, ..., cn are constants independent of n [see Problem 3.40].

To specify the polynomials, i.e. to determine co, ci, . . ., cn, further conditions need to be given. Different conditions will result in different sequences of polynomials. Let /3n(x) be a sequence of polynomials satisfying (27), i.e.

D/3n(x) = fl,-,(x)

(29)

and the added condition (n

0/3n(x)

1)

n

(30)

where we take h = 1 and define 0! = 1. Conditions (29) and (30) enable us to specify completely the polynomial 8,,(x). We find in fact that the first few polynomials are /30(x) = 1, (31(x) = x -1, a2(x) = (31) - jx + , 133(x) = Jx3 - jx2 + See Problem 3.41. jx2

It should be noted that once (3.(x) is found, we can use (29) to find ,6 1(x), 8_2(x), , by successive differentiations. This is particularly useful since it is possible to obtain /3n(x)

in terms of Stirling numbers of the second kind by the formula _ 1 d n nx(k+1) fln(x) n! dx Y Sk k + 1

(32)

See Problem 3.44.

We define the Bernoulli polynomials as = n ! /3n(x) fl .(x)

so that from (32) Bn(x)

d dx

(33)

tx(k+l) Sk

k -+1

(34)

THE SUM CALCULUS

CHAP. 3]

87

The first few Bernoulli polynomials are given by Bo(x) = 1, Bj(x) = x - -, B2(x) = x2 - x + J,

B3(x) = xa - 2x2 + sx (35) We define the Bernoulli numbers, denoted by B,,, as the values of the Bernoulli polynomials with x = 0, i.e.

B = B.(0)

(36)

The first few Bernoulli numbers are given by Bo = 1, B1 = B2 = 5, B3 = 0, B4 = - slo (37) For tables of Bernoulli numbers and polynomials see Appendixes C and D, pages 234 2,

and 235.

IMPORTANT PROPERTIES OF BERNOULLI NUMBERS AND POLYNOMIALS In the following we list some of the important and interesting properties of Bernoulli numbers and polynomials. 1.

=

B,,(x)

xn +

+ ... +

+ ()xt_2B2

()xi-1Bi

(n)B.

This can be expressed formally by Bn(x) = (x + B)n where after expanding, using the binomial formula, Bk is replaced by Bk. It is a recurrence formula which can be used to obtain the Bernoulli polynomials. 1) - B,,(x) = nxi--1

2.

Bn (x) = nBn-1 (x)

3. 4.

B,, = Bn(0) = Bn(1),

5.

1 ±- (n )B1 + (2)B2 +

n = 2,3,4,...

0

Ben--1

... + (n n

1)Bn-1

=

0

This can be expressed formally as (1 + B)n - B" = 0 where after expanding, using the binomial fomula, Bk is replaced by Bk. This recurrence formula can be used to obtain the Bernoulli numbers. Bn(x) = (-1)nBn(1- x)

6.

=i

text 7.

e

1

Bnnx) to

n=0

=

1 + Bi(x) t +

B2(x) t2

+ .. .

This is often called the generating function for the Bernoulli polynomials and can be used to define them [see. Problems 3.51 and 3.53].

t 8.

et

1

=

n=0

Bn nt

2

=

1 + Bit + 2t + .. .

This is called the generating function for the Bernoulli numbers and can be used to obtain these numbers [see Problem 3.52]. x 9.

coth

x 2

=

jr + 2T + 31 +

10.

1 + B2x2 2! + + (n -

B4x4

1)T

+ Bsxo 6!

4!

-

(

+ .. .

n+B)r+1 - B'

1

r + 1

where r=1,2,3,... and where the right side is to be expanded and then Bk replaced by Bk.

88

THE SUM CALCULUS

[CHAP. 3

1)n-1B2n(27r)2n 11.

12n + 22n + '3& +

12. If n = 1, 2, 3,

n=1,2,3,...

2 (2n) !

..., then =

B2,,(x)

sin 2k7rx

_ 2(-1)n(2n -1) !

B2n-1(x)

2(-1)n-1(2n) !

k=1

(2k7r)2n-1

"'

2k7rx k-1 (2k7r)2n

EULER NUMBERS AND POLYNOMIALS The Euler polynomials E,,(x) are defined by equation (27), i.e. DE,,(x)

En-1(x)

(38)

together with the added condition

n

MEn(x)

or

!

I [En(x + 1) + En(x)] = n

(39)

where M is the averaging operator of page 10 and h = 1. The first few Euler polynomials are given by Eo(x) = 1, E1(x) = x - 1, E2(x) = jx2 - fix, E3(x) = Jx3 -4 x2 + The Euler numbers are defined as En = 2nn ! En(d)

(140)

(41)

and the first few Euler numbers are given by

Eo = 1, E1 = 0, E2 = -1, E3 = 0, E4 = 5,

Es = 0

(42)

For tables of Euler numbers and polynomials see Appendixes E and F, pages 236 and 237.

IMPORTANT PROPERTIES OF EULER NUMBERS AND POLYNOMIALS In the following we list some of the important and interesting properties of Euler numbers and polynomials. E,(x)

1.

where

en +

1

_

elxn + (n -1) ! + ... + en-1x + en

eoxn n!

1

el

e0

2Ln! + (n- 1)!+

e2

.

en-17

The first few values of ek are eo = 1,

e1 = -1,

e2 = 0,

e3 = 24 I-,

2.

En (x) = E,,-,(x)

3.

E,,(x) = (-1)nEn(1- x)

4.

En(0) + En(1) = 0

5.

En(0) + En(-1) =

1

e4 = 0,

2(-J) n

e5

- 240

for n = 1, 2, 3, .. . for n = 1, 2, 3, .. .

6.

E2. = 22tt(2n) ! E2n(j) = 0

for n = 1, 2, 3, .. .

7.

E2.(0) = E2n(1) = e2. = 0

for n = 1, 2, 3, .. .

CHAP. 3]

THE SUM CALCULUS 2exc

8.

e°

+1

-

89

1 + E1(x)t + E2(x)t2 +

j Ek(x)tk =

k=0

This is often called the generating function for the Euler polynomials and can be used to define them. 61 x6 + + 5x4 9. + 720 sec x = I IE2klx2k = 1 + 1x2 2 24 (2k)!

k=O

17

k

tan x

10.

=

=

k=i k !

5 x°

x + 3x3 + 15 xs +

283562

+

Tk = I2kk! ekl

where

are called tangent numbers and are integers. 11.

121+1

-

(_22n n

321+1

+

521+1

12. If n = 1, 2, 3, ..., then for 0 '

x

n = 0, 1, 2, .. .

1,

4(

=

E2n-1(x) E2n(x)

2n+1

+2(2n) !

k2. co(2k

2n

Lrsin

7r

4( -1)n

_

1) )mix

+

=O

2-

(2k + 1)7rx (2k + 1)2n+1

°°

,r2n+1

Solved Problems THE INTEGRAL OPERATOR AND RULES OF INTEGRATION 3.1. Find each of the following: (c) D-1 - 3 sin 2x) (b) D (a) D-1(2x2 - 5x + 4)

r(

1(4e-3x

L.

(a)

D-1(2x2 -5x+4) = f (2x2 - 5x + 4) dx =33s

(b)

D-1(4e-3x - 3 sin 2x)

f (4e-3x - 3 sin 2x) dx

-3x\

4(e 3J

- 3(- cos 2x ) +

- a e-3x + (c)

D-1

rL (Nrx- -3)2

f(x-6\+9 x3

J

- 622 + 4x +

2

- cos

2x + c

dx

_ J('x-6x1/2+9dx x3/2 =

Jf (x-1/2 \ -x+ 6

9x-3/21

/ dx

1/2

22 - 6Inx + 9x 1/2 + 1

= 2x1/2 - 6 In x -

C

18x-112 + c

C

c

3)21

THE SUM CALCULUS

90 3.2.

[CHAP. 3

Prove the formula 1-3 on page 79 for integration by parts. We have by the rules of diffejentiation D[f(x) g(x)] = f(x) Dg(x) + g(x) Df(x) or

f(x) Dg(x) = D[f(x) g(x)] - g(x) Df(x) Operating with D-1 on both sides and remembering that D-1 is a linear operator, we have D-1[f(x) Dg(x)] = D`1D[f(x) g(x)] - D-1[g(x) Df(x)]

3.3.

-

51(x) Dg(x) dx = f (x) g(x)

or

Find

(a) f x cos 4x dx,

(b)

f g(x) D/(x) dx

f In x dx.

(a) We use integration by parts here. Let

f(x) = x, so

th at

Dg(x) = cos 4x

D (x) = 1 ,

g( x )

=

sin 4x 4

where in finding g(x) we omit the arbitrary constant. Then using the formula for integration by parts,

f x cos 4x dx = x

sin 4x 4

(b) Let f(x) = In x, Dg(x) = 1 so that

)

-f

Df(x) = x,

sin 4x 4

dx =

x sin 4x 4

cos 4x 16

+

+

c

g(x) = x

Then using integration by parts we have

f In x

3.4.

Find

(a) f x20y dx,

(b)

dx = x In x-f x l dx

= x In x - x

f x4 sin 2x dx.

(a) In this case we must apply integration by parts twice. We obtain f x2e3x dx

/ \

= (x2)1 3) - f (2x)

3x)

dx

=

(x2)

(3) - [ (2x) (9) -

_

(x2)

(3)

f

(2)

dx]

- (2x) (e9x) + (2) (._)

apart from the additive constant of integration. In this result we make the following observations. The first factors in each of the last two terms, i.e. 2x and 2, are obtained by taking successive derivatives of x2 [the first factor of the first term]. (ii) The second factors in each of the last two terms, i.e. e3x/9 and e3x/27, are obtained by (i)

taking successive integrals of e3x/3 [the second factor of the first term].

(iii) The signs of the terms alternate +, -, +. The above observations are quite general and can be used to write down results easily in cases where many integrations by parts might have to be performed. A proof of this method which we shall refer to as generalized integration by parts is easily formulated [see Problem 3.143].

CHAP. 3]

THE SUM CALCULUS

91

(b) Using the generalized integration by parts outlined in part (a) we find the result 5x4 sin 2x dx = (x4) C-cos 2x1 (4x3) /-sin 2x 1(12x2 (cOs2x)

J

)

(\

- (24x) (sin sx) + (24)

l

3.5.

Find

J

(a)

5e'1dx,

(b)

(1 x gl cos xx ,

(c)

(' 1n2x

)

(-cos 2x1 + c 32

)

+ 4) dx.

(a) Make the substitution

x - cos x = t so that x - cos x = t2 and (1 + sin x) dx = 2t dt on taking the differential of both sides. Then the given integral becomes (' (1 + sin x) dx = (' 2t dt = 2t + c = 2 x - cos x + c t J x- cos x

(b)

Let

3

f

J

2x - 1 = t so that 2z-1=0, x = 4 (t3 + 1), dx = t2 dt. Then the given integral

becomes

f A2--1 dx

= f et 2 t2 dt =

2J

t2et dt

3 [(t2)(et) - (2t)(et) + (2)(et)] + c 2

Z et(t2 -

(c)

2t + 2) + c = 2 e3 1 [(2x -1)2/3 - 2(2x - 1)1/3 + 2] + c

Let In (x + 4) = t so that x + 4 = et, x = et - 4, dx = et dt. Then the given integral becomes

f

1n2 (x + 4)

x+4

dx

(' t2(et dt)

J

=

et

f t2dt

(ax + b)m+1 &6.

Prove that

f (ax + b)m dx

(m + 1)a In (ax + b)

=

a

3ln3(x+4) +

3+ m

c

-1

m=-1

Let t = ax + b so that dx = dt/a. Then the required integral becomes tm+1

f (ax + b)m dx = 1 ( tmn dt =

aJ

0n+ 1)a in ,5

lnt a

Replacing t by ax + b yields the required result.

-1

m=-1

DEFINITE INTEGRALS AND THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS 2

3.7.

(a) Evaluate f (x2 + 1)2 dx and (b) give a geometric interpretation. 1

(a) We first find the corresponding indefinite integral as follows:

= f (x2 + 1)2 dx =

D -1(x2 + 1)2

=

X5

f (x4 + 2x2 + 1) dx =

Jf x4 dx + 2 f x2 dx + f dx

+233+x

where we can omit the constant of integration. Thus by the fundamental theorem of integral calculus,

f2

s

(x2 + 1)2 dx 1

2

3

T

-3

178

= 11.867

15

11

V (approx.)

92

THE SUM CALCULUS

[CHAP. 3

(b) The value 178/15 = 11.867 (approx.) represents the area [shaded in Fig. 3-21 bounded by the curve y = (x2 + 1)2, the x axis and the ordinates at x = 1 and x = 2.

y

1

Fig. 3-2

zr/2

3.8.

Find

(a) )

sin 6x dx,

fo

2

i2 (a)

J0 2

(b)

Si

6

(x) (e--s3x)

xe-3x dx

x2 - 2x + 2 '

0

cos 6x a/2

sin 6x dx

dx

2

(c) f

xe-3x dx,

(b) .f

Io

-- 6

cos 37

cos 0

+

dx

Method 1.

1

1

1

- 6+ 6 - 3

- (1) (e 93x) I2

(2) (e 3) - (e-6) - r(1) (e-3) (c)

6

- (!)] 93= ee-a - 9e-s

dx

J x2-2x+2 - J (x-1)2+1

Let x-1 = tan t. Then dx = sec2 t dt and so dx

(x -1)2 + 1

- ` sect t dt

J tang t+ 1

f dt

rsec2 t dt J sec2 t

t + c = tan-i (x -1) +

,Thus Co 2

dx

x2-2x+2

tan-' (x - 1)

2 o

=

tan-' (1) - tan-1 (-1) = 4 -

r- !0

It 2

Method 2.

As in Method 1 we let x -1 = tan t but we note that when x = 0, tan t = -1, t = -a14 and when x = 2, tan t = 1, t 7r/4. Then

(2

dx

2

dx

,lx_ox2-2x+2 - ,f _0(x-1)2+1

('4

sec2 t dt

Jt=_It/4tan2t+1

dt = t = -7r/4

Note that all of these can be interpreted geometrically as an area.

THE SUM OPERATOR AND RULES OF SUMMATION 3.9.

Prove that if ox Fi (x) = f (x) and ox F2 (x) = f (X), where C(x + h) = C(x).

then

F,(x) - F2(x) = C(x)

CHAP. 3]

THE SUM CALCULUS

93

By subtraction we have

xFl(x) - AxF2(x) = f(x) - f(x) = 0 or

Ax

[Fl (x) - F2(x)] = 0

i.e.

A[Fl (x) - F2 (x)] = 0

If we write Fl (x) - F2 (x) = C(x) then AC(x) = 0 or C(x + h) - C(x) = 0,

C(x + h) = C(x)

i.e.

We refer to C(x) as a periodic constant with period h. We have proved that if two functions have the same difference, then they can differ by at most an arbitrary periodic constant. The result is analogous to the theorem of integral calculus which states that two functions which have the same derivative can differ at most by an arbitrary constant. 3.10.

Prove that A-1 is a linear operator, i.e. (a) A-1[f (x) + g(x)] = A-'f (x) + A-lg(x) and (b) A-1[af (x)] = a0-lf (x) where f (x) and g(x) are any functions and a is any constant. (a) By definition if AF(x) = f(x) and AG(x) = g(x) then F(x) = A-lf(x), G(x) = A-lg(x) Also by adding AF(x) = f (x) and AG(x) = g(x) we have A[F(x) + G(x)] = f(x) + g(x) so that by definition F(x) + G(x) = A-1 [f (x) + g(x)]

Using (1) in (2) we find as required A-1[f (x) +g(x)] = A-lf(x) + A-19'(x)

(1)

(2)

(3)

(b) For any constant a we have

A[aF(x)] = a0F(x) = of (x) Thus from the first equation in (1) A-1[af(x)] = aF(x) = aA-1[f(x)]

3.11.

If we define F(x) = o-1[hf(x)] = I f(x)h +C(x) as in (8), page 82, prove that (a) Am = 1 or Y. = A-1, (b) A-1[f (x)] = Y, f (x) + Ci(x) where Ci(x) is an arbitrary periodic con-

stant, and (c) ox)

f (x) Ox + C(x).

f (x)

(a) By operating with A on

F(x) = A-1[hf(x)] _

f(x)h + C(x)

(1)

we obtain

AF(x) = hf(x) = A Y, f(x)h + AC(x)

(2)

since A is a linear operator. Now by Problem 3.9, AC(x) = 0 so that (2) becomes

hf(x) = AIf(x)h But since f (x) is arbitrary and h

0 it follows from (3) that

A=1 (b)

(3)

or

= A-1

It follows from this and Problem 3.10 that Y. is a linear operator. Dividing equation (1) by h using the fact that 7. and A-1 are linear operators, we have A-1[f(x)]

MAX) + Ch x)

_

where C(x) and thus Cl(x) = C(x)/h are periodic constants.

f(x) + C1(x)

THE SUM CALCULUS

94 (c)

[CHAP. 3

This follows at once from (1) on noting that 1

= h1 f(x) =

A-1[hf(x)]

(W

f(x)

and replacing h by Ax.

3.12.

Prove that

(a) I [f (x) + g(x)] _ I f (x) + I g(x),

(b) I of (x) = a I f (x).

These follow at once from the fact that Y = A-1 and Problem 3.10.

3.13.

Prove the formula for summation by parts, i.e. I f (x) Ag(x)

= f (x) g(x) -

g(x + h) Af (x)

We have AV (x) g(x)] = Ax) Ag(x) + g(x + h) Af (x)

f (x) Ag(x) = A[f(x) g(x)] - g(x + h) Af(x)

or

Operating on both sides with A-1 yields A-1[f(x) Ag(x)] = f(x) g(x) - A-1 [g(x + h) Af(x)]

or since A-1 = M,

I f(x) Ag(x) = f(x) g(x) - I g(x + h) Af(x)

(m+1)

3.14.

Prove

(a)

-1,

x(m) = (m -+1) h' m

sin rx =

(b)

_ cos

Then if m # -1, x(m+1) A

A-lx(m) =

x(m+l) (m + 1)h x(m+1)

I x(m)

or

= x(m)

(m+1)h

(m + 1)h

apart from an additive periodic constant. (b) From V-7, page 7, A[cos rx]

-2 sin 2 sin r (x + 2)

Replacing x by x - 4h, we have A[cos r(x - ,-h)]

= -2 sin 2 sin rx

Thus on dividing by -2 sin 2 , we have cos r(x - .h) A [ -2 sin jrh

]

= sin rx

1

2 sin\ Jrh

(a) We have from V-2, page 7, or (24), page 86, on replacing m by m + 1 Ax(m+i) = (m+1)x(m)h

(

)

CHAP. 3]

THE SUM CALCULUS

95

so that r(x - i-h) = - cos2 sin 4rh

A-1 sin rx or

- 4-h) sin rx = - cos2r(x sin jh

apart from an additive periodic constant.

3.15. Use Problem 3.14(b) to obtain r sin rx dx. From Problem 3.14(b) we have on multiplying by h - 4-h) - h cos2 sinr(x4-rh

h sin rx = Letting h - 0 this yields

f hcosr(x-}-h) J sinrx dx = h?.o - 2 sin4-rh -rh

cos r(x - 4-h)

lim h=o

r

sin 4,,h

[,,,n-cos r(x - h) ] h lim Jrh

r

h=o

LLh-+o sin 2rh

cos rx

r

The arbitrary periodic constant C(x) becomes an arbitrary constant c which should be added to the

indefinite integral. The result illustrates Theorem 3-1, page 83.

3.16.

(a) Find A-1(xax) = I xax, a

1, and (b) check your answer.

(a) We have from Problem 3.13 A-l[f(x) og(x)]

= f(x) g(x) - A-1[g(x + h) of(x)]

Let f (x) = x and Ag(x) = ax. Then Af (x) = Ax = h and g(x)

=

A-lax =

ax

ah - 1

Thus

= x . ahax 1 - A-1 Cahx+hl h]

A-l[xax]

_

hah

xax

ah - I A -

ah - 1 xax

_

ah - 1

(b) Check:

xax

hax+h_

Cah - 1

(ah - 1)2]

_

-

[

(x + h)ax+h

¢h - 1

hax+h (ah - 1) 2

hax+2h

(ah - 1)2]

xax

- [ah - 1

hax+h (ah - 1)2

(ah - 1)(x + h)ax+h - hax+sh - (ah - 1)(xax) + hax+h

= xax

(ah - 1)2

THE SUM CALCULUS

96

[CHAP. 3

DEFINITE SUMS AND THE FUNDAMENTAL THEOREM OF SUM CALCULUS 3.17. Prove Theorem 3-2, the fundamental theorem of sum calculus, i.e. a+(n-1)h a Y'

a+nh

f(x)

I-1F

! (x)

=

+

a

If we assume that F(x) and f (x) are related by

= f(x)

AF(x)

AF(x) = hf(x)

or

Ax

(1)

F(x + h) - F(x) = hf(x) Then by putting x = a, a + h, ..., a + (n -1)h successively in (2) we obtain the equations F(a + h) - F(a) = hf(a) we have

(2)

F(a+2h) -F(a+h) = hf(a+h) F(a + nh) - F[a + (n -1)h] = hf [a + (n -1)h] Thus by addition we have

F(a + nh) - F(a) = h{f(a) + f (a + h) + a + (n-I)h f(x)

or

= F( a + n

a

+ f[a + (n - 1)h]}

- F(a) - F(x) la+nh h

h

(3)

a

where the sum is taken from x = a to a + (n - 1)h in steps of h.

But from the second equation in (1) it follows that

ALF(x)J = /(x)

Fhx) = A-If(x)

or

(4)

and the required result is proved on using this in (3).

3.18.

Evaluate each of the following using the fundamental theorem of sum calculus and verify directly 21r

7

8x(3) if h = 3

(a)

Y, cos 5x if h = 7r/2

(c)

0 19

6

(b) I [3x(2) - 8x(1) + 10] if h = 2

(d)

2

(a) Since

calculus

x(-3) if h= 4

3

8x(3) = 8

X(4)

2x(4)

x(3) = 8 4-3

we have by the fundamental theorem of sum

3

using a=1 and h=3, 7

8x(3)

_

2x(4) 10 3

_ 2.10(4) _ 2.1(4) 3

3

I1

2(10)(7)(4)(1)

_ 2(1)(-2)(-5)(-8)

240

3

3 7

Check:

8x(3) = 8[1(3) + 4(3) + 7(3)] = 8[(1)(-2)(-5) + (4)(1)(-2) + (7)(4)(1)] = 240 x3

=

[3x(2) - 8x(1) + 10]

(b)

3

(3)

,

2-X(2)2

2-

8

+ 10

x 2

=

x(3) 2

- 2x(2) + 5x

Then 6

8

X(3)

Y, [3x(2) - 8x(1) + 10]

2x(2) + 5x

2

2

2

x(8)(6)(4)

x

2

- 2(8)(6) + 5(8) 1 - Lr(2)(0)(-2) - 2(2)(0) + 5(2) 2

= 30

CHAP. 3]

THE SUM CALCULUS

97

Check : 6

I, [3x(2) - 8x(1) + 10] = [3(2)(0) - 8(2) + 10] + [3(4)(2) - 8(4) + 10] + [3(6)(4) - 8(6) + 10] = 30 (c)

By IV-7, page 83, with r = 5, h = 7x/2 we have sin 5[x - (7x/4)]

A-1 cos 5x = Thus

2 sin (57x/4)

2a

sin 5[x - (7x/4)]

0

2 sin (57x/4)

Y, cos 5x =

sin (-57r/4) 2 sin (57x/4)

2>r

I0 cos 5x = (d) Since

0

_

sin (457x/4) 2 sin (577./4)

Check:

57x/2

cos 0 + cos (57x/2) + cos (57x) + cos (157x/2) + cos (107x)

(-2)

A-1[x(-3)] _ (x2)(4)

X(-2)

-8

3

we have

-8 x(-2)

19

1 x(-3)

23

8 [(23)(-2) - (3)(-2)]

3

_1

19x(_3)

=

+

1

3

+

1

7-11-15

_

1

(23)(27)

8

Check:

=1

25 4347

1

(3)(7)

+

1

11.15.19

1

15-19-23

_

25 4347

DIFFERENTIATION AND INTEGRATION OF SUMS 3.19. Find A-1[xerx] by using Theorem 3-4, page 84. We have from IV-5, page 83, A-1[erx]

=

erx

erh - 1

Then by differentiating both sides with respect to r [which corresponds to a in Theorem 3-4] we find

A-1 [8r or

erx]

r a

=

(erhe-

1/

(erh - 1)(xerx) - (erx)(herh) (erh - 1)2

A-1 [xerx]

(x - h)er(x+h) - xerx (erh - 1)2

This can be checked by referring to an alternative method of obtaining the result given in Problem 3.16(a). From that problem we find on putting a = e7. A-1[xerx]

her(x+h)

xerx

_ erh - 1 - (erh - 1)2 -

(x - h)er(x+h) - xerx (erh - 1)2

SUMMATION OF SERIES

3.20. Sum the series 1.3 - 5 + 3- 5-7 + 5.7.9 +

to n terms.

Since the nth term of the series is (2n - 1)(2n + 1)(2n + 3), the series can be written as 2n+3 , 5

x(3)

where h = 2

THE SUM CALCULUS

98

[CHAP. 3

Then by the fundamental theorem of sum calculus, 2n+3 1 x(3)

2n+5

x(4) 2n+5

5

4-2,

= A-1x(3)

5

_

(2n + 5)(2n + 3)(2n + 1)(2n -1)

(5)(3)(1)(-1)

8

8

(2n + 5)(2n + 3)(2n + 1)(2n -1) + 15 8

and (b) 22+52 +8 2 + 112 + . . . ton terms.

3.21. Sum the series (a) 12 +2 2 +31+4 2 +

Since we wish to find 1 x2 we need to express x2 as a factorial polynomial. This is found from

x2 = x(2) + hx(1)

(a) Using h = 1 the series is given by ccn

L X2

n = G' [x(2) + x(1)]

x(3)

n+1 _ 0-1[x(2) + x(1)]

3

L

-

1

2

n(n + 1)(2n + 1)

2

3

2

r(')(°)(-1) +(1)(O)

= r(n+1)(n)(n-1) + (n +1)(-)1 2 3 L (n + 1)(n)(n - 1) + (n + 1)(n)

(2) n+1

3

1

1

6 Then we obtain

As a possible check we can try a particular value of n, say n = 3.

12 + 22 + 32 =

= 14

(2)(3)(7)

6

which is correct.

(b) The nth term is (3n - 1)2. Thus using h = 3 the series can be represented by 3n-1

3n-1

22 x2

3n+2

r(3n + 2) (3n -1)(3n - 4) + (3n + 2)(3n -1)1 2

9

Check:

- r(2)(-1)(-4) + (2)(-1)1 L

2

Let n = 3. Then (3)(6.32+3.3-1) = 93 22 + 52 + 82 = 2

which is correct.

3.22. Sum the series 1.3.5 + 3.5.7 + 5.7.9 + ... ton terms. The nt h term o f

the

1

series Ss (2n - 1)(2n + 1)(2n + 3)

Now since (x-h)(-3)

=

1 x(x + h)(x + 2h)

it follows that if we let h = 2, the required sum of the series is

3n+2 2

_ (3n + 2)(3n -1)(6n + 1) +2 _ n(6n2 + 3n -1) 18

x(2)

x(3)

3.3 + 32.3

I [x(2) + 3x(1)] = A-1[x(2) + 3x(1)] 12

9

2

CHAP. 3]

THE SUM CALCULUS 2n-1

2n-1

(x - h)(-s)

(x - 2)(-s)

99 2n+1

A-1(x - 2)(-3)

11

(x-2)(-2) 2n+1

[(2n-1)(-2)j

(-2)(2) it 1

(2n + 1)(2n + 3)]

4 1

__

Putting n = 2 we find

1

1

4

(1)(3)

-4

1

- 4(2n + 1)(2n + 3)

12

Check:

-[

(-1)(-2) L

-4

1.

__

-

1.31

2

+ 3.5.7

.5

.7.9+

323. Sum the series 1.3 b + 3.5.7 +

1

4(5)(7)

- l8

which is correct.

. . .

In this case we are to sum the series of Problem 3.22 to infinitely many terms. The required sum using the result of Problem 3.22 is lim

1

12

- 4(2n+1)(2n+3)] 1

3.24. Sum the series I + cos B + cos 20 +

__

1

12

+ cos nO.

Using h = e the series is given by no

z+ , cos x= 2+1 0-1 cos x Ie(n+1)9 = 2+ 1

1 + sin (n+,-)e 2

2 sin --e

3.25. Sum the series 114 +2 15 + 316 + .

. .

sin (x - 4-e) (n+1)e 2 sin 4-e

e

sin 4-e

sin (n+--)e

2 sin --e

2 sin --e

to n terms.

Using h = 1 the sum of the series to n terms is given by n

1

fx(x + 3) To find this we attempt to express the general term as a sum of factorials. We first write 1

__

x(x + 3)

_

(x + 1)(x + 2)

(x + 1)(x + 2)

x(x + 1)(x + 2)(x + 3)

(1)

(x + 3)(4)

We then seek to determine constants A0, A1i A2, ... such that the numerator is the sum of factorials in the form (x + 1)(x + 2) = Ao + A1(x + 3)(1) + A2(x + 3)(2) + Ag(x + 3)(3) + .. .

Since the left side of (2) is of degree 2, it is clear that A3, ... must be zero. Thus

(x+1)(x+2) = Ao+A1(x+3)(1)+A2(x+3)(2) = Ao+A1(x+3)+A2(x+3)(x+2) From this identity we find Ao = 2, Al = -2, A2 = 1. Thus (1) becomes _ 2 - 2(x + 3) + (x + 3)(x + 2) 1 x(x + 1)(x + 2)(x + 3)

x(x + 3)

_

2

x(x + 1)(x + 2)(x + 3)

_

2

1

x(x + 1)(x + 2) + x(x + 1)

2(z - 1)(-4) - 2(x - 1)(-3) + (x - 1)(-2)

(2)

THE SUM CALCULUS

100

[CHAP. 3

Thus apart from an additive constant,

2(x -1)(-3) _ 2(x - 1)(-2) + (x -1)(-1)

1

-3

-1[x(x+3)

-2

-1

Then by the fundamental theorem of sum calculus n

11

1

f 2(x - 1)(-3)

x(x+3)

5l

-3

- 2(.-J)(-2) -2

-1)(-1)1 fIn+1

-1

+

1

- 3 n(-3) + n(-2) - n(-1) + 3 (0)(-3) - (0)(-2) + (0)(-1) 2

_

11

_

1

_

1

n -+1 +

18

_

1

3(n + 1)(n + 2)(n + 3) + (n + 1)(n

(n + 1)(n + 2)

1

2

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

2)

-

1

1

(1)(2) + 1

2 3(n + 1)(n + 2)(n + 3)

3.26. Sum the series 14+15+166+ From Problem 3.25 by letting n -

it follows that the required sum is 11/18.

SUMMATION USING SUBSCRIPT NOTATION

3.27. Sum the series 1.3.5 + 3.5.7 + 5.7.9 +

ton terms by using the subscript or

k notation.

The kth term of the series is Yk = (2k - 1)(2k + 1)(2k + 3) = (2k + 3)(3) where the difference interval h = 1 [as is always the case for the k notation]. Then the sum of the series is n

I (2k + 3) (3)

k=1

(2k + 3)(4) n+1

_

4.2

8 5(4)

(2n + 5) (4) 8

1

(2n + 5)(2n + 3)(2n + 1)(2n - 1) _ (5)(3)(1)(-1) 8

8

(2n + 5) (2n + 3) (2n + 1) (2n - 1) + 15 8

to n terms by using the subscript or k notation.

3.28. Sum the series 22 +5 2 + 82 +

The kth term of the series is Ilk = (3k - 1)2. Then the sum of the series is n

I (3k -1)2 = k=1

n

n

Y, (9k2 - 6k + 1)

k=1

_ 1 [9(k(2) + k(1)) - 6k(1) + 1] k=1

_ I [9k(2) + 3k(1) + 1] = k=1

93 + 32 - +k

n+1

)

1

_ [3(n + 1)(3) +

(n + 1)(2) + n + 1] - [3(1)(3) + (1)(2) + 1] 2

a

[3(n + 1)(n)(n - 1) + 8 (n + 1)(n) + (n + 1)] - [(3)(1)(0)(-1) + n(6n2 + 3n - 1) 2

2

(1)(0) + 1]

CHAP. 3]

THE SUM CALCULUS

101

ABEL'S TRANSFORMATION 3.29. Prove Abel's transformation.

nn+1k=1I vk - k=1 I IAUk p=1 Vp

=

uk7Jk k=1

Using summation by parts and the fundamental theorem of sum calculus we have a + (n-1)h

a+nh

=

f(x) og(x)

-

f(x) g(x) a

a + (n-1)h

I

g(x+h)of(x)

a

(1)

where the summation is taken in steps of h.

Putting a = 1, h = 1, x = k and writing f(k) = fk, g(k) = gk, (1) becomes n

n

I fkAgk = fn+lgn+1 - f1g1

k=1

I gk+lAfk

(2)

k=1

Now let fk = Uk and Ogk = Vk. Then gk+1- gk = Vk and so on summing from k = 1 to n - 1 n-1 n-1

we find

= gn - g1 = 1 Vk

I (gk+1 - gk)

k=1

n-1

i.e.

gn = 91 + IVk 1

From this we have

k

gk+1 = g1 + P=1 I VP Then (2) becomes

r91 k=1

ukvk

n

- u1g1 - k1

+ k1 vk =

un+1 n

= untl 1 Vk n

un+1 k=1

since

Ir

L

&uk

p=1

k=1

+ p=1 vpJ n

vp I

+ un+1g1 - u1g1 -

LAuk

u1g1

glAuk

p=1 VPJ n

rz

un+1g1

k=1

k

n

vk -

[gi

k

tt

k= 1

k

Duk

k=1

g10uk

untlgl

- u1g1 - g1

un+1g1

- u1g1 - gl k=1

un+Sgl

- u1g1 - g1(un+1 - ul)

Auk k=1 n

(uk+1 - uk)

OPERATOR METHODS OF SUMMATION 1 3.30. Prove that if R A-1[NkP(k)]

=

NkP(k)

8202 RO Rk _ R-1L1-/3-1+(R-1)2

/3303

(R-1)3

+

...

apart from an arbitrary additive constant. We have for any F(k)

DakF(k) = 8k+1F(k + 1) - pkF(k) = pk+1EF(k) - /3kF(k) = ak(RE - 1)F(k)

J

P(k)

THE SUM CALCULUS

102

[CHAP. 3

Let

(,RE -1)F(k) = P(k)

F(k) =

or

1

1 P(k)

RE

Then

ApkF(k) = RkP(k) Thus

0-1[,akP(k)]

= BkF(k) = Rk/El R

k

ak

- a -11 Pk

P(k)

(3(1 +A) - 1

1P(k)

p303.

p2A2

QD

.8-1 + (,6-1)2

(3 - 1 L1

1

[Ro/(a -1)]

(,a - 1)3

P(k)

+ ... P(k)

apart from an arbitrary additive constant and assuming $ 1. Since A-1$kP(k) = I RkP(k) the required result follows. Note that if P(k) is any polynomial the series terminates. n

3.31. Use Problem 3.30 to find I k - 2k. k=1

We have by Problem 3.30 with R = 2, P(k) = k 0-1[k.2k] = 2k[1-2o+402-...]k

I=

= 2k(k-2)

Then by the fundamental theorem of sum calculus n

I k2" = 0-1[k 2k]

2k(k - 2)

k=1

k=1

n+1 k=1

= 2n+l(n-1) + 2

THE GAMMA FUNCTION 3.32. Prove that r(p + 1) = pr(p). Integrating by parts we have for p > 0 M

f tne-t dt = M-.w

r(p + 1) =

lim

o

r

IM

lim

M_.

- fM M

lim

M-+w P

f

(-e-t)(pty-1)

0

C-Mpe-M + p

f

tP-1e-t dt]

0

tp-1e-t dt

pr(p)

3.33.

Prove that r(2) = V7r. We have

f,

r(2) =

t-1/2e-tdt = 2

e-x2 dx

0

on letting t = x2. Then {r(1)12 =

' e-x2 dx) f ' e-1/2 dy/ ' ' 4 f f e-(x2+12) dx dy (25

(2

0

0

0

{1)

o

where the integration is taken over the first quadrant of the xy plane. We can however equivalently perform this integration by using polar coordinates (p, 0)

Fig. 3-3

dt]

CHAP. 3]

THE SUM CALCULUS

103

rather than rectangular coordinates (x, y). To do this we note that the element of area in the xy plane shown shaded in Fig. 3-3 is (p do) dp = p dp 4. Also since x = p cos 0, y = p sin 0, we have x2 + y2 = p2 so that the result (1) with dx dy replaced by p dp do becomes

(r(j))2

v/2

=4

f f e-P2 p dp do d=0

(2)

P=0

The limits for p and 0 in (2) are determined from the fact that if we fix 0, p goes from 0 to CO and then we vary 0 from 0 to 7r/2. The integral in (2) is equal to

f

7/2

a/2

O

do = 4

=4J If Thus {r(,}))2 = T and since r(j) > 0 we must have r(4) _ . 4

a- P2 p

d'=0

3.34. Find (a)

C P=O

d¢

--e- P2

J0

dpJ

0

f J0=0 r/2

Ido

(a) r(5), (b) r(7/2), (c) r(-1/2).

Since r(p + 1) = pr(p) we have on putting p = 4, r(5) = 4r(4). Similarly on putting p = 3, r(4) = 3r(3). Continuing, we find r(5) = 4r(4) = 4.3x(3) = 4. 3.2r(2) = 4.3.2 lr(1). But r(1) = J . e-t dt = -e-t = 1 0

0

Thus

r(5) =4-3-2-1 = 4!

In general if p is any positive integer, r(p + 1) = p ! (b) Using the recursion formula we find

r(2) = 2r(2) = 2 2r(2) = 2 (c)

_

5

2

2 2 -_ 15 3

g

Assuming that the gamma function is defined for all values of p by assuming that it satisfies the recursion formula r(p + 1) = pr(p) for all p, we have on putting p = -1/2

rO _ (-j)r(-j) 3.35.

2 r()

r(-) = -2r(j) = -2 f

or

Prove that if m is an integer then hmr (h X(m)

+ 1)

r(h-m+1)

If m is a positive integer then

x(m) = x(x -h)(x - 2h)

(x -mh + h)

hm(h)(hf -1)(z-2) ... (3-m+1)

(1)

Now r(p + 1) = pr(p) = p(p - 1)r(p - 1) = p(p - 1)(p - 2)r(p - 2) = p(p - 1)(p - 2)...(p - m + 1)r(p - m + 1)

Thus r(p + 1)

(p-m+1)

r(p-m+1) = p(p-1)(p-2) Using(2)with p = x/h together with (1) we have

(2)

\

hmr (h-x.+1)

x(m) _

(3)

r(h-m+1)

Similarly we can prove that if m is a negative integer the result holds. In general we shall take the result as defining x(m) for all values of m. See Problem 3.109.

104

THE SUM CALCULUS

[CHAP. 3

3.36. Assuming that the result obtained in Problem 3.35 is the defining equation for x(m' for all m, prove that AX(m) = mx(m-3'h. We have hmJ' h

+ 1)

Lr(Fm+1)J hmr (x±h+i) h r x h - m+i) hmr(kx + 2)

r(-m+2)

r(h=m+1)

x

hm(+1Jr(h+1)

hmr(1+1)

-m+1)r(-m+1)

r(h-m+1)

(?III

h+

-m+1 mhmr( +1

r 3.37. Show that

x('") =

= mx(m-1)h,

-m+2)

x(m+1)

From Problem 3.36 Ox(m+1) _ (m + 1)x(m)h. A

3.38.

x(m+1)

(m + 1)h

= x(m)

-1.

for all m

(m + 1)h

or

Then if m-1,

(m + 1)h '

i.e.

x(m) =

x(m+1)

(m+1)h

m

Let the digamma function be given by mi(x)

=

dx In r (h + 1)

Prove that ni(x) = 1/(x + h). Since a and D are commutative with respect to multiplication [see Problem 1.58, page 28], 0*(x) OD In r(x/h + 1) = D O In r(x/h + 1)

=

Drlnr(x , h+1)

= Dln

/x

S

- inr(+1)1

= Dln(h+1) _

-1

CHAP. 3]

3.39.

THE SUM CALCULUS

105

r'(x+1)

Prove that I x(-1)

.x+h hrh+1

From Problem 3.38 we have since 1

x +1 h = x(-l),

1 x+h = A-lx0

=

0

(b) 14 + 24 + 34 +

Prove that Ben+1(j) = 0.

EULER NUMBERS AND POLYNOMIALS 3.119. Find (a) e4, e5; (b) E4(x), E5(x); (c) E4, E5. 3.120.

Prove that

(a)

3.121.

Show that

(a) e7 =

e,, 17 40, 320 '

0,

(b)

(b) es

(c)

-es.

-31 = 362 880-

MISCELLANEOUS PROBLEMS 3.122.

Determine whether the operators (a) D and D-1, (b) A and A-1 are commutative.

116 3123.

3.124.

THE SUM CALCULUS

i2

Prove that (a)

A-1 sin (px + q)

(b)

A-1 cos (px + q)

[CHAP.3

sin (px + q - jph - 17r) 2 sin i-ph

cos (px + q - -ph - Jr) sin -kph

Let A-1f(x) = f1(x), A-1f1(x) = f2 W,

...

.

Prove that

l f(x) g(x) = A-1[f(x) g(x)l = f1(x)g(x) - f2(x+h)Ag(x) + f3(x+2h)A2g(x) + (-1)n-1fn(x + nh - h) An-1g(x) + (-1)nA-1{fn(x+nh)Ang(x)} which is called generalized summation by parts. 3.125.

Use Problem 3.124 to find A-1[x3.2-] if h = 1.

3.126.

Find I k3 cos 2k.

3.127.

Show that

3.128.

Show that if sin a # 0 sine a + sine 2a +

n

k=1

3.129.

Show that

3.130.

Prove that

22.4

12

n2 .4n-1

32.42

2.3 + 3.4 + 4.6 +

n

1

_n

+ sine na

2

(n - 1)4n

6 + 3n+6

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

sin na cos (n + 1)a 2 sin a

(a2n2 - 2an2 - tan + n2 + 2n + a + 1)an+1 - (a2 + a)

k2 . ak

(a - 1)3

k=1

(-1)k (kl uk

k-n

(-1)fAnuo

O

3.131.

Use Problem 3.130 to show that (a)

(-1)k(n)

1 k 2k+1

k=o

(b)

k=O

n (-1)k+n k

=

2nn!

xn = n!

3.132.

Evaluate 15 + 25 + 35 +

3.133.

Show that

+ n5.

sine + sin 29 + sin 40 + sin 8e + 3.134.

3.135.

to n terms = cot (e/2) - cot 2n-19

(a)

sec2 x tan h Show that A-1 1 - tan x tan h L

(b)

Use (a) to deduce that f sec2 x dx = tan x + c.

(a)

Show that

(b)

Use (a) to deduce that

A-1 [tan-1

_ tan x + c.

tan-1 x + c. x2 + hx + 1] =

f dx x2+1

= tan-1 x + c.

CHAP. 3] 3.136.

THE SUM CALCULUS

117

(a) Show that

1 tan-1 (k2+k+l) = tan-1(n + 1)

k=1

4

(b) Use (a) to show that 1

i tan-1

k2 + k +

k=1

4

3.137.

Are there any theorems for integrals corresponding to the Theorems 3-3, 3-4 and 3-5 on page 84? Explain.

3.138.

(a) Show that formally 0-1

__

1

_

1

__

1 -E -

0

-(1+E+E2+

)

(b) Use (a) to obtain the formal result o-1f(x)

I f(u) u=x

where the sum on the right is taken for u = x, x + h, x + 2h, (c)

Is the result in (b) valid? Illustrate by considering the case where f (x) = with h = 2. x(x + 2)

3.139.

Prove Theorem 3-1, page 83.

3.140.

Prove (a) Theorem 3-3 and (b) Theorem 3-4, page 84.

[Hint.

Let

G(a) = fb(a) F(x, a) dx

and consider

G(a + Aa) - G(a) Aa

(a)

3.141.

... .

Prove that r'(1) =

f

e-x In x dx = -y where y is Euler's constant [see page 85].

0

3.142.

Use Problem 3.141 to prove that

++3+ 1

r'(n + 1)

+n1

r(n+1) +

3.143.

Supply a proof of the method of generalized integration by parts [see Problem 3.4].

3.144.

The beta function is defined as B(m, n)

(a) Show that B(m, n) = 2 f

=

f

1

xm-1(1 -

x)n-1 dx

0

ri2 sin2m-1 a

cos211-1 o do.

0

(b) Show that B(m, n) = B(n, m). 3.145.

Show that the beta function of Problem 3.144 is related to the gamma function by B(m, n)

= r(m) r(n) r(m + n)

ff

x2m-1y2n-1e-(zQ+i2) dx dy. Use Problem 3.106 to show that r(m) r(n) = 4 0 0 transform to polar coordinates and use the result of Problem 3.144.]

[Hint.

3146.

Evaluate (a)

f

0

1

y2

1- y2 dy,

(b)

f

0

n/2

sin4 0 cost a do,

(c)

f

0

rr/2

tan e do.

Then

118

3.147.

THE SUM CALCULUS Show that

sin2n x dx

J

zr/2

sin2n+1 x

°

Show that

2n 2n - 2 dx = n-I-1-2n-1

AM-1(1) _ (-1)nB(m, n) n

2 3

if h=1.

3.149.

Prove property 6, page 87, i.e. B.(x) _ (-1)nBn(1- X)-

3.150.

Prove that B3 = B5 = B7 = are equal to zero.

3.151.

Show that

3.152.

Show that cot x

x 2

= 1 x

coth

= 0, i.e. all Bernoulli numbers with odd subscript greater than 1 -

x 2

e-x/2)

+ = 2x (ex/2 ex12 - e-x/2

B4x4

f2 X2

2! + 4! + .. .

4

1 1 x-3x-46x

tan x

=1

22B+24Bx3-2sBxs+... 2x s

- 2i

1

(b)

r

2*2

Use the fact that 1 = f e-nx dx. 1 n J o

[Hint.

(a)

1

2n-2

2n

0

(b) f 3.148.

2n-1 2n-3

ir/2

(a)

[CHAP. 3

3

22(22 -

1) B2x

945

-

xs

4725

x7

24(24-1)

B4x3 +

4!

x + 3x3 + 16x5 + 3 bx7 + ... [Hint.

For (a) use Problem 3.151 with x replaced by ix. For (b) use the identity tan x = cot x

2 cot 2x.] 3.153.

[Hint. 3.154.

csc x =

Show that

(-1)k-1(22k - 2)B2kx2k-1 (2k)!

kk=o

Use the identity csc x = cot x + tan (x/2).]

Let f (x) be periodic with period equal to 1 and suppose that f (x) and f'(x) are bounded and have a finite number of discontinuities. Then f(x) has a Fourier series expansion of the form

where

AX) = `1

ak = 2 J

°+

2

k=1

(ak cos 2k7rx + bk sin 2k7rx)

bk = 2

f (x) cos 2k7rx dx,

f

1

f (x) sin 2kirx dx

0

0

(a) If AX) _ (32.(x) [see page 88] show that 2(-1)n-1

ao = 0, so that

bk = 0

ak =

(2kr-)2n

P2n(x)

= 2(-1)n-1k=1 1

k = 1, 2, .. .

cos 2krrx (2kxr)2n

(b) Use the result in (a) to obtain the second result of property 12, page 88.

3.155.

Prove the first result of property 12, page 88. [Hint. property 12 already obtained in Problem 3.154(b).]

3.156.

Prove property 11, page 88.

Take the derivative of the second result of

CHAP. 31

3.157.

THE SUM CALCULUS

Show that

(a)

12+22+22+

(b)

14 + 24 + 24 +

119

2

=

64 90

3.158.

Show that

(a)

12

-

22_

+ 32

2

+

42

12

1 1 1 +... 1 + 3444 () 1424

(b)

3.159.

=

720

- 23 + 23 - 73 + 3 (b) () 5I - + b575 +... = 1536

Show that

(a)

13

1

5ir5

1

1

35

3.160.

Prove (a) property 4 and (b) property 6 on page 88 by using the generating function.

3.161.

Show that

3.162.

(a) Show that the Bernoulli numbers can be found from B = n (-1)kk!

1 - 1 B,

B,,(-)

n

2n-1

=1 k + 1

n

n

Sk

where Sk are Stirling numbers of the second kind.

(b) Use the result in (a) to obtain the first few Bernoulli numbers. 3.163.

Prove that (a)

In sin x = In x -x26 - -

(b)

In cos x

x6

(-1)k-122k(22k - 1)B2kx2k

2

12

45

2k(2k)!

B,2 +

L4

4!

B-

1-

3.165.

Prove that

B,, = 4n Jf00

3.166.

Prove that 1

__

e2 + 1

L Hint.

e2x1

1

1

_

=

4

(22

(ex 1

-7/2 < x < 7r/2

0.

- 1 dx.

-1)B2x

(24

2!

2

0

- a)" + R

,/ between a and x

APPLICATIONS OF THE SUM CALCULUS and integrating from x = a to

f

a+h

a+h

f (x) dx =

a

a 1

`R =

1

x = a + h we have

f(a) dx +

f

133

A f(a) fa+h h

(x -a) dx +

(1)

a+h

(2)

f"(o)(x - a)(2) dx

a

Performing the integrations in (1) we find

Ja

h

f(x) dx =

[f(a) + f(a+ h)] + `R

Now to estimate the remainder or error R we note that (x - a)(2) = (x - a)(x - a - h) does not

change sign for x between a and a + h. We can thus apply the mean-value theorem [see page 81] to (2) to obtain a+h

`R = I f"(E) f

(x - a)(x - a - h) dx = -12

a

where Z is also between a and a + h. This yields the required error term.

4.19.

Find the error term in Simpson's one-third rule. If we proceed as in Problem 4.18 the error term will be given by a+2h 3!

f

(n)(x - a)

ax

a

However (x - a)(3) _ (x - a)(x - a - h)(x - a - 2h) changes sign between x = a and x = a + 2h so that we cannot apply the mean-value theorem. We must thus proceed in another manner. One approach is to write the error as

fa+2h

f(x) dx -

a

[f(a) + 4f(a+ h) + f(a+2h)]

(1)

3

To simplify details by introducing a symmetry to this we shall choose a = -h so that the error (1) is given by `R(h) = f(x) dx - 3 [f(-h) + 4f(0) + f(h)] (2)

f

h

We now consider h as a parameter and differentiate with respect to it. We find using Leibnitz's rule on page 81 `R'(h)

= f(h) + f(-h) - 3 [-f'(-h) + f'(h)] - ' [f(-h) + 4f(0) + f(h)]

= ,f(h) + If(-h) - If(0) + 3 [f'(-h) - f'(h)] `R"(h)

_ I f'(h) - If'(-h) +

[-f"(-h) - f"(h)] + [f'(-h) - f'(h)] a

3

kf'(h) - kf'(-h) - 3 [f"(-h) + f "(h)] IR ."(h)

_

--

f"(h) + .-f"(-h) - [-f ".(-h) + f"(h)] - k[f"(-h) + f"(h)] 3

_h [f...(h) - f".(-h)] 3

Thus

qz(0) = k,(0) = 'R"(0) = ' "'(0) = 0

(3)

If we now assume that f(IV)(x) is continuous we can apply the mean-value theorem for derivatives [see page 8] to obtain cR..'(h)

h [f... (h) - f...(-h)]

23 2h2 [fcrvl(n)]

n between -h and h

APPLICATIONS OF THE SUM CALCULUS

134 Now considering

9Z...(h)

= G(h)

[CHAP. 4

232 f(IV)(o)

(4)

we can show by integrating three times using conditions (3) that [see Problem 4.89] h

`R(h)

=

3

f u2(h-u)2f(IV)(7)) du

(5)

Since u2(h - u)2 does not change sign between u = 0 and u = h we can now apply the mean-value

theorem for integrals in (5) to obtain

- h5 f(IV)(0

h

`R(h)

= -if(IV)(E) f u2(h - u)2 du

90

0

which is the required error term.

GREGORY'S FORMULA FOR APPROXIMATE INTEGRATION

- 1)) (\

fa+nh

4.20.

f (x) dx = h E,nIn E

Show that

f (a)'

We have by definition

D-If(x) = fa+nh

so that

f f(x) dx

1(x) dx = D If(x)

a

(1)

a+nh (2)

a

But since E = ehD we formally have D =

In E so that h

f(x) =

D-If(x)

In

Ax)

Then (2) can be written as

J

a+nh a

a+nh

f (x) dx = InhE f (x)

In E

a

h

4.21.

Show that

fa+nh

f (x) dx

[f (a + nh) - f (a)]

(,nE1) f(a) h In E

_

[f(a + nh) - f(a)]

1n E

(yn - yo)

where Yk = f (a + kh). From Problem 4.20 we have, using Yk = f (a + kh) so that yo = f (a), yn = f (a + nh),

fa+nh

=

h(EnE1) f(a) _

f (x) dx a

4.22. Show that 1

In(1+x)

_

- x(1+2x-2x2

19

1

+ 24

3

720

We have [see page 8] 2

In

3

4

5

5

3 4

+ 160 x5 -

l

CHAP. 4]

APPLICATIONS OF THE SUM CALCULUS

135

Then by ordinary long division we obtain 1

x2

x4

x3

2+3

-1

1

x+

12x+241

-19

x2

+163

X4

720x3

0

1x 1+x1x2+ 1x319x4+ 3 xs2 12 24 720 160 4.23.

Obtain Gregory's formula (20), page 124, for approximate integration. From Problem 4.21 we have

f

a+nh

(In

f (x) dx = h

-

f(,)

= InEE [f(. + nh) - f (a)]

nE1

a

E = 1+0,

Now

so that

= InE (yn - 7l0)

E-1 = 1- V In K = -ln(1- V)

In E = In (1 + 0),

Then (1) can be written

fa+nh a

-1 h[ln(1-V)Yn

f(x)dx

- In(1+0)YoI 1

(4)

Now using Problem 4.22 we find

-1 In (1 --V)

7!n

=

_

-

1

1V

-

1 V2 24

12

2

=

1

1

- 720 19 Vs -

3 V4

160

1 (+;+23+4 12

24

720

-

...

11n

160

By subtraction we then have

-1

1n (1 - V) pn

_

1

In (1 + A) p0

1

=

V 71n -

- 2 (y,, + YO)

1

A

1

Y0

_24 (V271.+,&2710) - 19 (V371n-03y0)

-

160 (V4?!n

12

(Vyn - 0710)

047la)

Also since V = E-1A = AE-1 [see Problem 1.39, page 25] we have 1

1

V on

0

1

1

= L,,_1,& 71n

710

0

yo =

Eyn - 71o

_ (En+1 -1)Yo = (En+l - 1

E-1 ) Yo

D

= (1+ E + E2 + - - - + En)y0 = Yo + Y1 + 712 + ' - ' + Yn Thus (4) becomes

fa+nh

[710+2y1+2712+ ... +2yn-1+y,]

f(x) dx = a

2 -12

(V Y. - Ay0) - 4 (V 2yn + A2y0) 720 M71n - 03710)

1

(V 4yn + 04710) -

4.24. Work Problem 4.12 by using Gregory's formula. Putting a = 0, n = 6, h = 1/6 and f (X) = 1/(1 + x2) in Gregory's formula it becomes

136

APPLICATIONS OF THE SUM CALCULUS

[CHAP. 4

J0

1

+2ys+yb]

126 [yo+2y1+2yz+

1-x2

-12 (V y6 - AYo) 19(1/6) 720

2/4

(V2y6 + A2yo)

(V3y6-A3y8 )

- 3(1/6) (V4y8+A4yo) 160

The ordinates y0, yl, ..., y6 can be taken from the table of Problem 4.12. To obtain the various differences we form the following difference table where the factor 108 has been used so as to avoid decimals. k

yk X 108

0

100000000

1

97297297

2

90000000

3

80000000

4

69230769

5

59016393

6

50000000

Ayk X 108

A2yk X 108

-2702703

-4594594

-7297297

-2702703

-10000000 -769231 -10769231

+554855

A3yk X 106

A4yk X 106

1891891 41581

1933472

-609386 1324086

-680958

-10214376

643128

+1197983 -9016393

Fig. 4-5

From this table we see that

Vu6 = AE-1y6 = Ays =

-0.09016393,

V2y6 = A2E-2y6 = A2y4 =

0.01197983,

A2yo = -0.04594594

0.00643128,

A3yo =

0.01933472

= -0.00680958,

A4yo =

0.00041581

V3y6 =

A3E-3y6

A3y3 =

V4y6 =

A4E-4y6

A4Y2

AlIo

= -0.02702703

Thus we have

(' 1 0

dx 1+ x2 =

0.78424077 -12) (-0.0631369)

(/) 24

(-0.03396611) -

19(1/6) (-0.01290344) - 3(1/6) (-0.00639377) 160

720

0.78424077 + 0.00087690 + 0.00023588 + 0.00005675 + 0.00001998 0.78543028

It should be noted that the value 0.78424077 in the above is that obtained using the trapezoidal Thus we can think of the remaining terms in Gregory's formula as supplying correction terms to the trapezoidal rule. It is seen that these correction terms provide a considerable improvement since the value obtained above has only an error of 0.00423% whereas that obtained without the correction terms is 0.147%. It should be noted however that Simpson's one-third rule is still the victor. rule.

THE EULER-MACLAURIN FORMULA 4.25. Show that A-1f

(x)

ehD1 1! (x) hD f(x)

2 f(x)

_I

AX) [k=o Bk(hD)k] k!

+

12

hDf(x)

720 h3D3f (x) + .

.

CHAP. 4]

APPLICATIONS OF THE SUM CALCULUS

137

We have since A _ E - 1 = ehD - 1 [see page 8]

o-1f(x) = o f(x) _

hD

1

= hD ehD - 1 f(x) Then using property 8, page 87, with t = hD this becomes B hD)k A-1f(x)

hD [ k

f (x)

kki o

B1hD hD

[Bo +

4.26.

B2h2D2

+

1

2f

hD Ax) on

f(x)

ehDi

+

2!

(x) +

12

hDf(x)

B3h3D3 3!

+

B4h4D4

720 h3D3f

4!

+ ...l f(x)

(x) + .. .

... .

using the values for B0, B1, B2, B31

Derive the Euler-Maclaurin formula (21) on page 124. By the fundamental theorem of sum calculus we have [see page 83]

I

a+ (n- 1)h

f(x)

a

= o-1f(x)

a+nh (1) a

where the summation in (1) is to be taken from a to a + (n - 1)h in steps of h. of Problem 4.25 we then have a + (n-1)h a

a+nh

Ax) _ TD f(x)

a

- 1 f(x)

a+nh

a+nh

+

a

1 1

which can be written as

+

hJ

1

h3 f"(x)

20

a

a+nh

1

+ f (a + (n -1)h) =

f (a) + f (a + h) +

hf'(x)

Using the results a+nh a

f(,) dx - 2 [ f (a + nh) - f (a)]

1 [f'(a + nh) - f'(a)] - 720 [f"'(a+nh) - f... (a)] + ...

199

4.27.

Find

1

k=100

k

by using the Euler-Maclaurin formula.

Let f (x) = 1/x, a = 100, a + (n - 1)h = 199, h = 1 in the result of Problem 4.26 so that n = 100. Then f'(x) _ -1/x2, f"(x) = 2x-3, f"'(x) _ -6x-4 so that 199

200 dx 100 x

1

k-100 k

1

1

2

720

[

1

1

[200

100

1

+ 12

(200)4 + (100)4]

(200)2 + (100)2

+

In 2 + -(0.005) + x(0.000075) + --0.693147 + 0.002500 + 0.00000625 + 0.695653

to 6 decimal places. In Problem 4.32 we shall discover how accurate the result is.

ERROR TERM IN THE EULER-MACLAURIN SERIES 4.28. Let 0p (x) = B9 (x) - Bp, p = 0, 1, 2, 3, ... , where Bp (x) are the Bernoulli polynomials and Bp are the Bernoulli numbers. Show that (a)

(b)

0p( x) = pBp-1(x)

op (x) _

0p+1(x) P + 1.

- Bp

(c)

¢p(0) = 0

(d)

op(1) =

P=1 fo P>1 1

138

APPLICATIONS OF THE SUM CALCULUS

[CHAP. 4

(a) We have using property 3, page 87, 0n(x) _ B'p(x) = pBp-1(x) (b)

We have since op+i(x) = Bp+1(x) -Bp+1

op+l(x) = Bp+1(x) = (p + 1)Bp(x) _ + 1)[Op(x) + Bp] p+l(x) - B

Then (c)

p (x)

_

p

p + 1

Since by definition Bp(0) = B. we have

¢p(0) = Bp(0) - Bp = 0 01(x) = B1(x) - B1 = x - 4- - (-4-) = x Thus 01(1) = 1 If P> 1 then by property 4, page 87,

(d) We have

(1)

¢p(1) = Bp(1) - Bp = 0 Then from (1) and (2) we see that

p=1 p>i

1

=

0p(1)

4.29.

(2)

0

Show that a +h

S

f(x) dx

= 2 [f (a) + f (a + h)] +

2 f 4(u) f"(a + hu) du

Letting x = a + hu, using the fact that p2(u) = u2 - u, and integrating by parts we have

J

f (x) dx = h f f(a+hu)du a

0

h f 1 f(- + hu) d(2u -1) 2

0

1

h

[(2u -1) f (a + hu)

o

-h

f

1

(2u - 1)f'(a + hu) dul

0

[f (a) + f (a + h)] - 22 f (2u - 1)f'(a + hu) du

2

0

f

[f (a) + f (a + h)] - 2

2 h

1

f'(a + hu) d02(u)

0

[f (a) + f (a + h)] -

h2

1

f'(a + hu)c2(u)

2 h

1

f 02(u)f"(a + hu) du o + hs 0

1

L3 f 02(u)f"(a+hu) du [f(a) + f(a+h)] + 2i 0

4.30.

Show that (a)

2

f

1

02(u) f"(a + hu) du

22h2 [f'(a

+

+ h) - f'(a)]

hs

1

44(u) f('V' (a + hu) du

4I

(b)

fo

h' f 1$4(u)f (lv) (a + hu) du = - B4h4 [f,11(a + h) - f111(a)] +

hf 7

0B(u) f(vl) (a + hu) du

o

(a) From Problem 4.28(b) with p = 2 we have 02(u) = -B2 + }03'(u)

and

o3(u) _ jo4(u)

CHAP. 4]

APPLICATIONS OF THE SUM CALCULUS

139

Then s

/ 1

1

If f 02(u)f"(a + hu) du

=

h2 3

0

J

[-B2 + j0s(u)]f"(a + hu) du 0

B2h.s r1

2 f"(a J + hu) du +

hs

1

f 03(u)f"(a+hu)du

0

0 1

-B2 2 f'(a + hu)

10

+ 3s [o.(u)f"(a + hu) 0

- h"'(a fo&)f+ hu) dul a

B2h2

- 2 ! [f'(a + h) - f'(a)] -

h4 fl 0

B2h2

- 2 ! [f'(a + h) - f'(a)] - 4 f 1

04(u)f,,,(a + hu) du

!

0

B2h2

2!

1

[f'(a + h) - f'(a)] - 4 !

[04(u)f,,,(a + hu)

-h

B2h2

2 [f'(a + h) - f'(a)] + 4h5i

f

0

f

04(u)f crv>(a + hu) du]

0

1

04(u)f(1v> (a + hu) du

0

(b) From Problem 4.28(b) with p = 4 we have

04(u) = -B4 + jos(u)

and

05(u) = jos(u)

Then h5

1

4! f ep4(u)f(v)(a+hu)du = 0

h5

[-B4 + 4, fl 0 B41 s

-

fl B2 h4

*0s,(u)]fIly>(a+

f(lv)(a

hu) du

+ hu) du + 5 f 1 05(u)f Ilv)(a + hu) du 0

f,,,(a+ hu) II + b [o5(u)f(1v)(a+ hu)

f

-h

1

0

1

o5(u)f (v)(a + hu) du]

o

- 4 (a[f+ h) - f,,,(a)] h) - f"'(a)] - s f 'e(u)f (v) (a + hu) du B4h4

h6

B4 h4

g

1

f 05(u)f(v)(a + hu) du 0

B4 [ f",(a + h) - f,,,(a)]

1

0

4

1

- s6 [os(u)f cv>(a + hu)

0

1

- h f 00(u)f(vn(a+hu)du] 0

_ B4 4 [f,,,(a+h)

-

f"'(a)]

1

7

+ 6j f 0o(u)fcvn(a+hu)du 0

4.31.

Obtain the Euler-Maclaurin series with a remainder for the case where n = 1 [see equation (24), page 125]. From Problems 4.29 and 4.30 we have fa+h f (x) dx a

2 [f(a) + f (a + h)]

-

B2 h2

[f'(a + h) - f'(a)] - B44 [f"(a + h) - f,,,(a)] +

L7

1

f 00(u)f(VI)(a+ hu) du 0

which is the required result. This result can of course be generalized and leads to that on page 125.

140 4.32.

APPLICATIONS OF THE SUM CALCULUS

[CHAP. 4

Estimate the error made in the approximation to the sum in Problem 4.27. Using f (z) = 11x, a = 100, h = 1, n = 100, m = 3 in the result of Problem 4.27 we can

write it as

1

1

k 00 k

1

1

x - 2 [200

J100

+2[

100

(200)2 + (100)2]

720

1(200)4 + (100)4]

-R

where

R =

1

99

1

720 f [Be(,) - Be] L0 (p + 100 6! + u)7] du 0

f 1 [u0 - 3u5 + 5 u4 - }u2] 0

99

1

99 1z)=0 (p + 100 + u)7

] du

Now by the mean-value theorem for integrals we have rr9

R

1

1

3u5 + u4 - - u2) du Lr-=o (p + 100 + E)7] ,Jo (u0 42 v=0 (p +

Thus the absolute value of the error is JRI

_

1

1

99

42 v1o (p+100+C)7

But since 0< f < 1, it follows that each term in the series on the right is less than 1/(100)7 and so the sum of the first 100 terms of the series is less than 100/(100)7 so that ]RI

1


0. Then (35) states that the eigenfunctions {v' r k O. ,k) where m = 1, ... , N are mutually orthogonal. The terminology used is an extension to N dimensions of the idea of orthogonality in 3 dimensions where two vectors A = Ali + A2j + A3k, B = Bli + B2j + B3k are orthogonal or perpendicular if 3

A1B1 + A2B2 + A3B3 = 0

i.e. k=1

AkBk = 0

(36)

The extension to N dimensions is obtained on replacing the upper limit 3 in the sum by N thus yielding (35). In case (35) holds we also say that the functions {¢m k} are orthogonal with respect to the weight or density rk. Given a function Fk it is often possible to obtain an expansion in a series of eigenfunc-

tions. We find that

N

I Cm Om, k 1

(37)

where the coefficients cm are given by N

Y,

cm

=

rk Fk Om, k

k=1

m=1,...,N

N

k=1

(38)

rk'rm,k

If we suitably normalize the functions 0m k so that N

k=1

rk 0m, k = 1

(39)

then the coefficients (38) are simplified since the denominator is equal to 1. The results (35) and (39) can be restated as r0m,k0n,k

0 min fl m = n

(40)

In such case we say that the set { V k m k} is an orthonormal set. NONLINEAR DIFFERENCE EQUATIONS An important class of nonlinear difference equations can be solved by applying suitable transformations which change them into linear difference equations. See Problems 5.43 and 5.44. SIMULTANEOUS DIFFERENCE EQUATIONS

If two or more difference equations are given with the same number of unknown functions we can solve such equations simultaneously by using a procedure which eliminates all but one of the unknowns. See for example Problem 5.45. MIXED DIFFERENCE EQUATIONS In addition to differences which may occur in equations there may also be derivatives or integrals. In such case we refer to the equations as differential-difference equations, integral-difference equations, etc. These equations can sometimes be solved by various special techniques. See Problems 5.46 and 5.47 for example.

160

DIFFERENCE EQUATIONS

[CHAP. 5

PARTIAL DIFFERENCE EQUATIONS Up to now we have been concerned with difference equations involving unknown functions of one variable. These are often called ordinary difference equations in contrast to difference equations involving unknown functions of two or more variables which are called partial difference equations.

To consider such equations we must generalize the concepts of difference operators to functions of two or more variables. To do this we can consider for example a function f (x, y) and introduce two difference operators Al, 02 defined so that if h = Ox, 1 = Ay are given,

Alf (x, y) = f(x + h, y) - f (x, y),

o2f(x, y) = f(x, y + l) - f(x, y)

(41)

These are called partial difference operators. Similarly we define partial translation operators E1, E2 so that E1f(x, y) = f (x + h, y), E2f(x, y) = f (X, y + 1) (42) These are related to partial derivatives considered in elementary calculus. It is clear that

E1 = 1 +A,,

E2 = 1 + 02

(43)

We can also define powers of E1, E2, Ol, 02

If we use the subscript notation zk,,n = f (a + kh, b + ml)

(44)

it follows for example that Elzk,m =

zk+1, m,

E2zk, m

zk,m+l

(45)

We can now consider partial difference equations such as for example (El - 3E1E2 + 2E2)zk,m = 0

(46)

zk+2, m - 3Zk+1, m+l + 2Zk,m+2 = 0

(47)

which can also be written Any function satisfying such an equation is called a solution of the equation.

The general linear partial difference equation in two variables can be written as O(E1, E2)zk,m = Rk,m

(48)

where (p(E1, E2) is a polynomial in E1 and E2 of degree n. A solution of (48) which contains n arbitrary functions is called a general solution. The general solution of (48) with the right side replaced by zero is called the complementary solution. Any solution which satisfies the complete equation (48) is called a particular solution. As in the one variable case we have the following fundamental theorem. Theorem 5-4:

The general solution of (48) is the sum of its complementary solution and any particular solution.

Various methods are available for finding complementary solutions as in the one variable case. See Problem 5.55.

As might be expected from the one variable case it is possible to express partial differential equations as limits of partial difference equations and to obtain their solutions in this manner. See Problem 5.56.

CHAP. 5]

DIFFERENCE EQUATIONS

161

Solved Problems DIFFERENTIAL EQUATIONS 5.1.

Show that the general solution of the differential equation dx - 3 dx + 2y = 4x2

is y = clex + c2e2x + 2x2 + 6x + 7. We have

y = clex + c2e2x + 2x2 + 6x + 7

(1)

dx = clex + 2c2e2x + 4x + 6

(2)

d2y

dx2 = clex + 4c2e2i + 4 Then

d2y dx2

3 dx + 2y =

(3)

(clex + 4c2e2x + 4) - 3(clex + 2c2e2x + 4x + 6)

+ 2(clex + c2e2x + 2x2 + 6x + 7)

= 4x2

Since the given differential equation is of order 2 and the solution has 2 arbitrary constants it is the general solution.

5.2.

Find the particular solution of the differential equation in Problem 5.1 such that y(O) = 4, y'(0) = -3. From (1) and (2) of Problem 5.1 we have

y(0) = c1 + c2 + 7 = 4 y'(0) = cl + 202 + 6 = -3 so that cl = 3, c2 = -6. Then the required particular solution is

or

c1 + C2 = -3

or

cl + 2c2 = -9

y = 3em - 6e2x + 2x2 + 6x + 7

DIFFERENCE EQUATIONS 5.3.

Show that the difference equation oz - 3 ox + 2y = 4x(2) can be written as f (x + 2h) - (3h + 2) f (x + h) + (2h2 + 3h + 1)f(x) = 4h2x(2)

where

y = f (x).

Writing Ax = h and y = f(x) the difference equation can be written as o2f(x) _ 3 Af(x) + 2f (x) = 4x(2) h2

or

f (x + 2h) - 2f (x + h) + f(x)

(1)

h

_ 3 [f(x + h) -Ax) h

+ 2f (x)

=

4x(2)

Multiplying by h2 and simplifying we obtain

f(x+2h) - (3h + 2)f(x + h) + (2h2 + 3h + 1)f (x) = 4h2x(2) 5.4.

(2)

Show that the general solution of the difference equation in Problem 5.3 is given by y = cl(x)(1 + 2h)x1h + c2(x)(1 + h)x/h + 2x2 + (6 - 2h)x + 7

(1)

where ci(x) and c2(x) have periods equal to h. We have since c1(x + h) = cl(x), c2(x + h) = c2(x),

y=

cl(x)(1 + 2h)x/h + c2(x)(1 + h)x/h + 2x2 + (6 - 2h)x + 7

(2)

162

DIFFERENCE EQUATIONS Ay Ax

clhx)

[(1 + 2h)(z+h)/h - (1 + 2h)x/h] +

Czhx)

[CHAP.5

[(1 + h)(x+h)/h - (1 + h)x/h]

+ 2[(x + h)2 - x2] + (6 - 2h)[(x + h) - x] h 2c1(x)(1 + 2h)x/h + c2(x)(1 + h)x/h + 4x + 6

Ax

Similarly,

Axz

= 4c 1( x)(1

+ 2h)z/h

c2(x)(1 + h)x/h + 4

Then A2Y

[4c1(x)(1 + 2h)x/h + c2(x)(1 + h)x/h + 4]

Axe

- 3[2c1(x)(1 + 2h)x/h + C2(x)(1 + h)x/h + 4x + 6] + 2[cl(x)(1 + 2h)x/h + c2(x)(1 + h)x/h + 2x2 + (6 - 2h)x + 7]

4x2 - 4hx = 4x(x - h) =

4x(2)

Since the difference equation (1) has the difference between the largest and smallest arguments divided by h equal to (x + 2h) - x

-2

h

it follows that the order is 2. Then since the solution (1) has two arbitrary independent periodic constants, it is the general solution.

5.5.

Show that h2 ox + 2h ox + y = 2x(3' is not a second order difference equation. We can write the equation if y = f(x) as h2

or

Ff(x+2h) - 2f(x+h) + f(x) L

+ 2h[f(x+h) - f(x)1j + f(x)

=

2x(3)

f(x + 2h) = 2x(3)

This equation involves only one argument and in fact is equivalent to

f(x) = 2(x-2h)(3) which is not really a difference equation. The result shows that we cannot determine the order of a difference equation by simply looking at the largest value of n in Any/Axn. Because of this we must define the order as given on page 151.

DIFFERENTIAL EQUATIONS AS LIMITS OF DIFFERENCE EQUATIONS 5.6. Show that the limit of the difference equation [see Problem 5.3]

ox - 3 ox + 2y =

4x(2'

as Ax or h approaches zero is [see Problem 5.1]

dx - 3dx + 2y = This follows at

once since

lim

AX-0

and

4x2

zy3Ay+2y Axe

lim 4x(2) h-.0

Ax

=

d?y dx2

- 3dydx + 2y

= lim 4x(x - h) = 4x2 h-.0

DIFFERENCE EQUATIONS

CHAP. 5]

5.7.

163

Show that the limit as Ax or h approaches zero of the solution of the difference equation in Problem 5.6 is the solution of the differential equation in that problem. By Problem 5.4 the general solution of the difference equation in Problem 5.6 is, if h = Ax, y = c1(x)(1 + 2h)--/h + c2(x)(1 + h)x/h + 2x2 + (6 - 2h)x + 7

(1)

From the calculus we have lim

C1 + 1

n

ft-.m

= e = 2.71828...

(2)

= 2.71828...

(3)

or, if n = a/h where a is some given constant,

/

a/h

=

lim (1 +h a-

h-,0

\

Equivalently (3) can be written

/

lim 1 1 + a-)

e

1/h (4)

h-+0 \ /

lim (1 + ha-)

or

xm

= ex/a

(5)

Using (5) with a = 1/2 we find

lim (1 + 2h)x/h = e2x

(6)

lim (1 + h)x/h = ex

(7)

h-,0

Using (5) with a = 1 we find

h-+0

lim c1(x) Also since cl(x + h) = c1(x), c2(x + h) = c2(x) it follows that h-,0 where cl and c2 are constants, provided that these limits exist.

C1,

lim C2(x) = C2

h-+0

Then the limit of the solution (1) as h - 0 is y = c1e2x + c2ex + 2x2 + 6x + 7 which is the general solution of the differential equation of Problem 5.6 [see Problem 5.1].

USE OF THE SUBSCRIPT NOTATION 5.8. Write the difference equation of Problem 5.3 with subscript or k notation. We have by the definition of the subscript notation

Vk = f(x),

yk+1 = f(x+h),

Yk+2

= f(x+2h)

Also using x = kh, x(2) = x(x - h) = kh(kh - h) = h2k(k - 1) = h2k(2)

Then the equation f(x+2h) - (3h + 2)f(x + h) + (2h2 + 3h + 1)f (x) = 4x(2) becomes

Yk+2- (3h+2)yk+1+ (2h2+3h+1)yk =

4h2k(2)

or using the operator E E2yk - (3h + 2)Eyk + (2h2 + 3h + 1)yk = 4h2kc2>

which can be written

[E2 - (3h + 2)E + (2h2 + 3h + 1)] yk = 4h2k(2)

164 5.9.

DIFFERENCE EQUATIONS

[CHAP. 5

Write the difference equation f(x + 4h) + 2f(x + 3h) - 4f(x + 2h) + 3f(x + h) - 6f(x) = 0

in the subscript or k notation. We use Yk = f(x), Yk+1 = f(x+h), Yk+2 = f(x+2h), Yk+3 = f(x+3h), Yk+4 = f(x+4h) Then the difference equation is (E4 + 2E3 - 4E2 + 3E - 6)yk = 0

LINEARLY INDEPENDENT FUNCTIONS 5.10. Determine which of the following sets of functions are linearly dependent and which are linearly independent. (a) 2k, 2k+3

(b) 2k, 2k+3 4k

(c) 2k, 4k

(d) 2k, k . 2k, k2 -2 k.

(a) Consider A,2k + A22k+3 which can be written as

A12k + A2.23.2k = (A, + 8A2)2k

(1)

Since we can find Al and A2 not both zero such that (1) is zero [as for example A2 = -1, A, = 8] it follows that the functions 2k, 2k + 3 are linearly dependent. (b)

Consider A,2k + A22k+3 + A34k which can be written as (A1 + 8A2)2k + A34k

(2)

Then since we can find the 3 constants A1i A2, A3 not all zero such that (2) is zero [for example A2 = -1, Al = 8, A3 = 0] it follows that the functions 2k, 2k+3, 4k are linearly dependent.

In general if a set of functions is linearly dependent it remains linearly dependent when one or more functions is added to the set. (c)

Consider A12k + A24k where A1, A2 are constants.

2k(A1 + A22k) = 0

or

This will be zero if and only if

Al + A22k = 0

However this last equation cannot be true for all k unless A,=A2=0. Thus it follows that 2k and 4k are linearly independent. (d)

Consider A12k + A2k . 2k + A3k2.2k where A1i A2, A3 are constants. This will be zero if and only if 2k(A1 + A2k +A3k2) = 0 or Al + A2k + A3k2 = 0

But this last equation cannot be true for all k unless Al = A2 = A3 = 0. Thus the functions are linearly independent.

5.11. Work Problem 5.10 by using Theorem 5-1, page 154. (a) In this case f 1(k) = 2k, f2 (k) = 2k+3 and the Casorati is given by 20

23

1

8

21

24

2

16

Then by Theorem 5-1 the functions 2k, 2k+s are linearly dependent. (b)

In this case f, (k) = 2k,

f2 (k) = 2k + 3,

20

23

40

21

24

41

22

25

42

=

f3 (k) = 4k and the Casorati is

1

8

1

2

16

4

4

32

16

=

8

1

1

2

2

4

4

4

16

1

0

CHAP. 5]

DIFFERENCE EQUATIONS

165

where we have taken the factor 8 from elements of the second column in the determinant and noted that since the first two columns are identical the determinant is zero. Thus by Theorem 5-1 the functions 2k, 2k+3, 4k are linearly dependent. (c)

In this case f1 (k) = 2k, f2(k) = 4k and the Casorati is 20

40

1

1

21

41

2

4

=

2

Then by Theorem 5-1 the functions 2k, 4k are linearly independent. (d)

In this case f 1(k) = 2k, f 2 (k) = k 2k, f 3 (k) = k2. 2k and the Caso rati is 20 21 22

0-20 1.21 2.22

02. 20

1

0

0

12. 21

2

2

2

22. 22

4

=

2

2

8

16

1

=

16

16

8

Thus the functions 2k, k 2k, k2 - 2k are linearly independent.

5.12.

Prove Theorem 5-1, page 154. We prove the theorem for the case of 3 functions. The general case can be proved similarly. By definition the functions f1(k), f2(k), f3(k) are linearly independent if and only if the equation

A,f,(k) + A2f2(k) + A3f3(k) = 0 identically holds only when Al = 0, A2 = 0, A3 = 0. Putting k = 0, 1, 2 in (1)

(1)

Alf,(0) + A2f2(0) + A3f3(0) = 0 Alfl(1) + A2f2(1) + A3f3(1) = 0 Alf,(2) + A2f2(2) + A3f3(2) = 0

(2)

Now (2) will have the sole solution Al = 0, A2 = 0, A3 = 0 if and only if the determinant fl(0)

f2(0)

f3(0)

fl(1)

f2(1)

f3(1)

fi(2)

f2(2)

f3(2)

0

(3)

which is the required result.

HOMOGENEOUS LINEAR DIFFERENCE EQUATIONS WITH CONSTANT COEFFICIENTS 5.13. (a) Find linearly independent solutions of the difference equation 8yk = 0 and (b) thus write the general solution. (a)

Let Yk = rk in the difference equation.

Then it becomes

rk+2 - 6rk+1 + 8rk = 0 Dividing by rk [assuming r the auxiliary equation

Yk+2 - 62,k+1 +

or

rk(r2 - 6r + 8) = 0

0 which otherwise leads to the trivial solution Yk = 0] we obtain

r2 - 6r + 8 = 0

i.e.

(r -2)(r - 4) = 0

or

r = 2,4

Thus two solutions are 2k and 4k. By Problem 5.11(c) these are linearly independent solutions.

Note that the difference equation can be written as

(E2 - 6E + 8)yk = 0 and that the auxiliary equation can immediately be written down from this on formally replacing E by r in the operator E2 - 6E + 8 and setting the result equal to zero.

166

DIFFERENCE EQUATIONS

[CHAP. 5

(b) Using Theorem 5-2, page 154, i.e. the superposition principle, the general solution is yk =

cj2k + c24k

where cl and c2 are arbitrary constants.

5.14.

(a) Find that particular solution of yk+2 - 6yk+1 + 8yk = 0 such that yo = 3, y1 = 2. (b) Find y5.

(a) From Problem 5.13 the general solution of the difference equation is yk = c12k + C24k

Then from yo = 3, y1 = 2 we have on putting k = 0 and k = 1 respectively 3 = c1 + c2i 2 = 2c1 + 4c2 from which c1 = 5, c2 = -2. Thus the required particular solution is

yk = 5.2k-2.4k (b)

Letting k = 5 we find y5 = 5 - 25 - 2.45 = 5(32) - 2(1024) = -1888 The value of y5 can also be found directly from the difference equation [see Problem 5.84].

5.15.

Solve the difference equation yk+2 - 2yk+1 + 5yk = 0. The difference equation can be written

(E2 - 2E + 5)yk = 0 and the auxiliary equation obtained by letting yk = rk in this equation is

r2 - 2r + 5 = 0 from which

r=

2-*

4-20 _

2 -h 4i

= 1 ± 2i

2

2

It follows that solutions are (1 + 2i)k and (1 - 2i)k which can be shown to be linearly independent [see Problem 5.74]. Since in polar form

1 + 2i

=VC1+?

1 - 2i = where

1

_

it

=

(cos B + i sin o)

=

i

=

(cos 9 - i sin e)

_

2

cos e =

2

,

vF5

eie

e-te

sin e = 2

(1)

it follows that the two linearly independent solutions are (/ eie)k = (f )kekie = 5k/2ekie = 5k/2(cos ke + i sin ke)

(f e-ie)k = (Vr5)ke-kie = 5k/2e-kie = 5k/2(cos ke - i sin ke) Then the general solution can be written as yk =

5k/2(c1 cos ke + c2 sin ke)

(2)

CHAP. 5]

5.16.

DIFFERENCE EQUATIONS

167

Find the particular solution of the difference equation of Problem 5.15 satisfying the conditions yo = 0, yl = 1. Since yo = 0, y1 = 1 we have on putting k = 0, k = 1 respectively in the general solution (2) of Problem 5.15 0 = cl,

from which

1 = 51/2(cl cos e + c2 sine)

c1 = 0,

1

c2 =

NF5 sin e

Yk =

Thus

5k/2

sin ke =

5(k-1)/2

NF5 sine

sin ke sin 6

Note that 9 can be found from (1) of Problem 5.15.

5.17.

(a) Solve the difference equation yk+2 - 4yk+1 + 4yk = 0 and (b) find the solution such that yo = 1, yl = 3. (a) The difference equation can be written as

(E2 - 4E + 4)Yk = 0

and the auxiliary equation is

r2 - 4r + 4 = 0 or (r - 2)2 = 0 so that r = 2, 2, i.e. 2 is a repeated root. It follows that Yk = rk = 2k is a solution. However we need another solution which is linearly independent. To find it let Yk = 2kvk in the given difference equation. Then it' becomes or

Vk+2 - 2Vk+1 + Vk = 0

Thus

and

Vk= C1 + c2k

Yk =

A2Vk

= 0

2k(c1 + c2k)

which gives the required general solution.

(b)

Putting k = 0 and k = 1 we find

1 = c1,

3 = 2(c1 + c2)

Thus Cl = 1, c2 = 1/2, and the required solution is

Uk = 2k

5.18.

(k + 2)2k-1

(1 + 2

(a) Solve the difference equation yk+3 + Yk+2 - yk+1 - Yk = 0 and (b) find the solution which satisfies the conditions yo = 2, y1= -1, y2 = 3. (a) The difference equation can be written as (E3 + E2 - E - 1)yk = 0 and the auxiliary equation is

r3+r2-r-1 = 0 This can be written as r2(r + 1) - (r + 1) = 0,

(r + 1) (r2 - 1) = 0,

or

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

so that the roots are r = -1, -1, 1. Corresponding to the two repeated roots -1, -1 we have the solution C,(-1)k + C2k(-1)k = Corresponding to the root 1 we have the solution C3 (1)k = c3. Then by the principle of superposition the general solution is (-1)k(cl + c2k).

Yk = (-1)k(Cl + C2k) + C3 (b)

Putting k = 0, 1, 2 respectively we find 2 = Cl + C3, -1 = -Cl - C2 + C3, from which C1 = 4 , c2 = 4, C3 = 4.

3 = c1 + 202 + C3

Thus the required solution is

Yk =

3

+

(

4

)k (2k + 5)

168 5.19.

DIFFERENCE EQUATIONS

[CHAP. 5

Find the general solution corresponding to a linear homogeneous difference equation

if the roots of the corresponding auxiliary equation are given byY3,,-21 + 3, -2, -3, 3, 4.

Vr3

2

z,-3,

Corresponding to the repeated roots 3, 3, 3 we have the solution 3k(cl + c2k + c3k2).

Corresponding to the complex roots - 2 ±

cos 0 = - 2 ,

sin e = 2 or

2

i [which have polar form cos 0 ± i sin 0 where

0 = 120° = 27r/3 radians] we have the solution

c4 cos ke + c5 sin ke =

c4 cos

23k

+ c5 sin 23k

Corresponding to the repeated roots -3, -3 we have the solution (-3)k(e6 + c7k). Corresponding to the single root -2 we have the solution c6(-2)k. Corresponding to the single root 4 we have the solution c94k. Then the required general solution is by the principle of superposition given by Yk = 3k(c1 + c2k + c3k2) + c4 cos 3k + c5 sin 23k + (-3)k(c6 + CA) + c6(-2)k + c94k

METHOD OF UNDETERMINED COEFFICIENTS 5.20.

Solve

Yk+2 - 6yk+1 + 8Yk = 3k2 + 2 - 5.3k.

The general solution of the homogeneous equation [also called complementary or reduced

equation] is by Problem 5.13

Ilk = C12k + c24k

(1)

Corresponding to the polynomial 3k2 + 2 on the right hand side of the difference equation we assume as trial solution A1k2 + A2k + A3

(2)

since none of these terms occur in the complementary solution (1)

Corresponding to the term -5. 3k on the right hand side of the difference equation we assume the trial solution . A43k

(3)

Thus corresponding to the right hand side we assume the trial solution [or particular solution]

Yk = A1k2+A2k+A3+A43k Substituting in the given difference equation we find Ilk+2 - 6yk+1 + 8Yk = A1(k + 2)2 + A2(k + 2) + A3 + A43k+2 - 6[A1(k + 1)2 + A2(k + 1) + A3 + A43k+1] + 8[A1k2 + A2k + A3 + A43k]

= 3A1k2+(3A2-8A1)k+3A3-4A2-2A1-A43k Since this must equal the right hand side of the given difference equation we have

3A 1k2 + (3A2- 8A1)k + 3A3 - 4A2 - 2A1 - A43k = 3k2 + 2 - 5.3k Equating coefficients of like terms we have

3A1 = 3, Thus

3A2 - 8A1 = 0,

Al = 1,

3A3 - 4A2 - 2A1 = 2,

A2 = 8/3,

A3 = 44/9,

A4 = 5

and the particular solution is

k2 +3k +

s4+5.3k

Adding this to the complementary solution (1) we find llk

c12k+c24k+k2 +3k+

44+5.3k

A4 = 5

CHAP. 5]

5.21.

DIFFERENCE EQUATIONS

Solve

169

yk+2 - 4yk+1 + 4Yk = 3.2k + 5.4k.

The complementary solution is by Problem 5.17 c12k + c2k 2k

(1)

Corresponding to the term 3.2k on the right side of the given difference equation we would normally assume the trial solution A12k. This term however is in the complementary solution (1) so that we multiply by k to obtain the trial solution Alk 2k. But this also is in the complementary solution (1). Then we multiply by k again to obtain the trial solution A1k2.2k. Since this is not in the complementary solution (1) it is the appropriate trial solution to use. Corresponding to the term 5.4k we can assume the trial solution A24k since this is not in the complementary solution (1).

The trial solution or particular solution is thus given by

Yk = A1k2.2k + A24k Substituting this into the given difference equation we find Yk+2 - 49lk+1

+ 4Yk = A1(k + 2)2.2k+2 + A24k+2 - 4[A1(k + 1)2.2k+1 + A24k+1] + 4[A1k2.2k + A24k] 8A12k + 4A24k

8A12k + 4A24k = 3.2k + 5.4k

Thus we must have

Al=1,

from which

A2=I

Then the required solution is Yk

Yk = c12k + C2k 2k +

or

5.22.

c12k + c2k 2k + Jk2.2k + I.4k 3k2-2k-3 +

5-4k-1

Solve yk+3 - 3yk+2 + 3yk+1 - yk = 24(k+2). The complementary equation is

(E3 - 3E2 + 3E -1)yk = 0 and the complementary solution is thus

or

(E -1)3yk = 0

Yk = cl + c2k + c3k2

Corresponding to the term 24k + 48 on the right of the given equation we would normally use the trial solution A1k+A2. Since these terms occur in the complementary solution we multiply by a power of k which is just sufficient to insure that none of these terms will appear. This is k3 so that the trial solution is Yk = A 1k4 + A2k3. Substituting this into the given difference equation we find

24A1k + 36A1 + 6A2 = 24k + 48

so that Al = 1, A2 = 2. Thus the required solution is Yk = c1 + c2k + c3k2 + k4 + 2k3

SPECIAL OPERATOR METHODS a1En-1 + .. . + an. (a) Prove that 0(E)pk = 4,(f3)Rk. (b) Prove that 5.23. Let ¢(E) = aoEn + k 1 if a particular solution of the equation ¢(E)yk = Rk is given by 4,(E) Rk = 0(R)

R)R)

o.

(a) We have R/jk = Rk+1 = /3.13k, E2Qk = pk+2 = 132./3k,

...

Enjjk = ek+n = pn. Rk

170

DIFFERENCE EQUATIONS

[CHAP. 5

Thus

(E)Rk = (a0En + a1Ett-1 + ... + a, )Rk (ao/;n + a1pn-1 + ... + a, )/3k = O(/;)ak (b)

Since o(E)fjk = 0(R)Rk from part (a) it follows that

(E) pk 5.24.

0(R)

00)

0

Prove that 1/0(E) is a linear operator. Let

E) Rl(k)= U1,

E) R2(k) = U2

(1)

Then by definition if R1(k) and R2(k) are any functions

o(E)U1 = RI(k), 0(E)U2 = R2(k) and since O(E) is a linear operator it follows that o(E) [U1 + U2] = R1(k) + R2(k)

so that

1 [RI(k)1- R2(k))

O(E)

= U1 + U2

or using (1)

(E) [R1(k) + R2(k)]

= o(E) R1(k) + o(E) R2(k)

(2)

Also if a is any constant and R(k) is any function let

_ 10( EaR(k) )

U

(3)

Then

c(E)U = aR(k) Dividing by a * 0 we have since Sb(E) is a linear operator U

O(E)

a

= R(k)

or

O(E)

R(k) =

(4)

a

From (2) and (3) we have O(E)

aR(k)

= a (E) R(k)

(5)

The results (2) and (5) prove that 1/¢(E) is a linear operator. 5.25.

Solve the difference equation yk+2 - 2yk+1 + 5yk = 2.3k - 4.7k. The equation can be written as (E2 - 2E + 5)Yk = 2 3k - 4. 7k. a particular solution is given by 1

E2-2E+5(2.3k-4.7k)

3k_4. - 4-

2

21 4

By Problems 5.23 and 5.24 E2-2E+57k

32-2(1

3)+53k - 4

72-2(7)+57k

. 3k- 1 .7k 10

Since the complementary solution is as given in equation (2) of Problem 5.15 the required general solution is

Ilk

= 5k12(cl cos ke + c2 sin ke) +

where a is determined from equation (1) of Problem 5.15.

3k -

. 7k

DIFFERENCE EQUATIONS

CHAP. 5]

5.26.

Let O(E) = aoE" + a,En-1 +

171

Prove that

+ an.

0(E) [akF(k)] = pkp(pE)F(k) We have

Ef3kF(k)

= ek+IF(k+ 1) = /3k[REF(k)]

E2/3kF(k) = pk+2F(k + 2) = 8k[p2E2F(k)]

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

En/3kF(k)

Then

= pk+nF(k + n) = pk[P"E"F(k)]

0(E)[pkF(k)] = (aoEn+a1E"-1+ ... +an)[,9kF(k)] a1(RE)n-1 + ... + a, ]F(k)

= Rk[ao(l3E)" +

= Qk,(PE)F(k) 5.27.

Prove that O(E) f3kF(k) = qk 0(1E) F(k). Then by Problem

F(k) so that ¢(,@E) G(k) = F(k).

Let G(k) _

5.26

O(E)[RkG(k)] = Rko(RE) G(k) = RkF(k) Thus O(E)

5.28.

Find

/3kF(k) = akG(k) = qk 00E) F(k)

E2-4E+42k

The method of Problem 5.23 fails since if O(E) = E2 - 4E + 4 then 0(2) = 0. However we can use the method of Problem 5.27 with F(k) = 1, /3 = 2 to obtain 1

o(E)

2k = 2k =

2k

1

o(2E)

5.29.

Find

= 2k

1

(2E)2 - 4(2E) + 4 (1) = 2k

1

4(1 +,&)2 - 8(1 + p) + 4 (1)

2k k(2)

4

(1)

_

= 2k 4p2 (1)

2k

k(k - 1)2k

8

2

E2 - 6E + 8

1

4E2-8E+4(1)

(3k2 + 2).

We use the method in entry 3 of the table on page 156. Putting E = 1 + A we have 1 (3k2 + 2) _ Ez - 6E+8

1

(1 + A)2 - 6(1 + A) + 8

(3kz + 2)

1 2 (3k2 + 2) 3-4p+p

1

1

31- (*p - }pz) (3k2 + 2) [1+(

-

J,&2)2

[3k2 + 2] + $p[3k(z) + 3k"+ 2] + 2z p2[3k(2) + 3k(l) + 2] k[3k2 + 2] + $[6k(" + 3] + z-; [6]

k2+$k+v

172

DIFFERENCE EQUATIONS

[CHAP. 5

5.30. Work Problem 5.21 by operator methods. The equation can be written (E2 - 4E + 4)yk = 3-2k+5-4k. Then by Problem 5.23 and

5.28 a particular solution is given by

E2-4E+4( 3.2k+5.4k) = 3. E2-4E+4( 2k) 1

k(k - 1)2k +

+5

k

E2-4E+4) (4

5 4k 42-4(4) + 4

= Ik(k -1)2k + I.4k Since the complementary solution is C12k + C2k 2k the general solution is

Yk = C12k + C2k 2k + Ik(k - 1)2k + j . 4k This can be written

Yk = C12k + (C2-- )k.2k + sk22k + .4k and is the same as that obtained in equation (2) or (3) of Problem 5.21 if we take

c2=C2-I

c1 = C1,

METHOD OF VARIATION OF PARAMETERS 5.31. Solve yk+2 - 5yk+1 + 6Yk = k2 by the method of variation of parameters. The complementary solution, i.e. the general solution of Yk + 2 - 51!k+ 1 + 6yk = 0, is c12k + c23k.

Then according to the method of variation of parameters we assume that the general solution of the complete difference equation is 7!k = K12k + K23k

(1)

where K1 and K2 are functions of k to be determined. Now

AYk = K12k + K22.3k + 2k+' K1 + 3k+1AK2

(2)

Since two conditions are needed to determine K1 and K2, one of which is that (1) satisfy the given difference equation, we are free to impose one arbitrary condition. We shall choose this condition to be that the sum of the last two terms in (2) is identically zero, i.e.

so that (2) becomes

2k+10K1 + 3k+10K2 = 0

(3)

Ayk = K12k + K22.3k

(4)

A2Yk = K12k + K24.3k + 2k+1AK1 + 2. 3k+10K2

(5)

From (4) we find

Now the given equation can be written as (E2 - 5E + 6)Yk = k2

(6)

(02 - 30 + 2)yk = k2

(7)

or since E =I+ A

Using (1), (4) and (5) in (7) we find after simplifying 2k+1pK1 + 2. 3k+1pK2 =

k2

We must thus determine K1 and K2 from the equations (3) and (8), i.e. = 2k+1oK1 + 3k+10K2 2k+1oK1 + 2.3k+1pK2 = k2 0

(8)

(9)

Note that these equations could actually be written down directly by using equations (24) on page 157.

CHAP. 5]

DIFFERENCE EQUATIONS

173

From (9) we find 0

3k+1

k2

2.3k+1

2k+1 2k+1

-k2

3k+1 2 . 3k+1

AK,

2k+1'

2k+1

0

2k+1

k2

k2

2k+1

3k+1

3k+1

2k+1

2-3k+1

Then K1

= - 1 (2k+k2 1) '

K2 = 0-1 1

L-2

3k+1

These can be obtained by using the methods of Chapter 4 or the operator method of Problem 5.27. Using the latter we have, omitting the arbitrary additive constant,

K1 ()k (c) =

(k2)

_

E

_( )k

k_z

1

IE-1 2

-(j)k E 1 2 (k2) ()k 1 1 Q (k2) (4)k(1 + 0 + 02 + ... )(k(2) + k(1))

=

(.)k(k(2) + k(') + 2k(l) + 1 + 2)

=

(j)k(k2 + 2k + 3)

Thus taking into account an arbitrary constant c1,

K1 =

(k2 + 2k + 3) + c1 2k

Similarly we find

K2 = -213k(k2+k+1) + C2 Then using these in (1) we find the required solution

Ilk = c12k+c23k+jk2+8k+j

METHOD OF REDUCTION OF ORDER 5.32. Solve yk+1- ryk = Rk where r is a constant. Method 1, using variation of parameters. The complementary or homogeneous equation corresponding to the given equation is

yk+1-ryk = 0

(1)

From this we find

Yk = ryk-1,

Ilk-1 = rllk-2,

,

Y2 = ry1,

so that the solution of (1) is

Yk = rky0

or

where we take co = yo as the arbitrary constant.

yk = cork

4l1 = ryo

DIFFERENCE EQUATIONS

174

[CHAP. 5

We now replace co by a function of k denoted by K(k), i.e.

Yk = K(k)rk (2) and seek to determine K(k) so that the given equation is satisfied. Substitution of (2) in the given equation yields K(k + 1)rk+1 - K(k)rk+l = Rk or dividing by rk+1,

K(k + 1) - K(k) = OK =

(1

Rk -l f/rk+l/

K

Thus

Rk rk+1

and so from (2) Rk \)

Ilk

= rk0-1 rk+l/

-

rk

k-1 R P P=1 rP+1

+ Cr k

Method 2, multiplying by a suitable summation factor.

Multiplying the given equation by 1/rk+l it can be written as Yk+1

Ilk

rk+1

rk

Rk rk+1

Thus

or

Ilk

rk

(Yk) = rk

or

Yk

I Rk

rk

1\ rk+l

-1

rk

CRk

rktl /

Rk rk+1

k-1R P I

+ Cr k P-1 rP+l

where c is an arbitrary constant. The factor 1/rk+l used is called a summation factor. 5.33.

Solve yk+2 - 5yk+1 + 6Yk = k2 by the method of reduction of order. Write the equation as (E - 3)(E - 2)yk = V. Letting Zk = (E - 2)Yk we have

(E - 3)zk = k2 Then by Problem 5.32 the solution is 2k =

3k-1

(k2) + o13k = c13k - -(k2 + k + 1)

(E - 2)yk = c13k - j(k2 + k + 1)

Thus

Applying the method of Problem 5.32 again the solution is f ci3k - j(k2 + k + 1)\ Ilk

= 2k0-1 = j k2 +

+ C22k

+1 3k+1

k + $ + C22k + C33k

We can use the method to find particular solutions by omitting the arbitrary constants.

METHOD OF GENERATING FUNCTIONS 5.34. Solve the difference equation yk+2-3Yk+1+2yk = 0

if yo=2, y1=3

by the method of generating functions. Let

G(t) _ Ik=0 Yktk. Multiplying the given difference equation by tk and summing from

k = 0 to - we have

k=0

cc Yk+2tk - 3 k=0 Ilk+itk + 2 k=0

yktk = 0

CHAP. 5]

DIFFERENCE EQUATIONS

175

(y2+y3t+y4t2+...) - 3(111+y2t+y3t2+ ...) + 2(110+y1t+y2t2+

= 0

This can be written in terms of the generating function G(t) as G(t) - y2o -

yet

- 3 (G(t)_t

yo) + 2G(t)

=

0

(1)

Putting yo = 2, yl = 3 in (1) and solving for G(t) we obtain G(t)

2 - 3t 1 - 3t + 2t2

=

2 - 3t

=

(1 - t)(1 - 2t)

Writing this in terms of partial fractions we find G(t)

=

1 1

-

t + 1 1 2t

Thus

I yktk k=0

1 tk +

k=0

k=0

(2t)k

=

k=0

(1 + 2k) tk

= 1 (1 + 2k)tk k=0

and so

Yk = 1 + 2k

(2)

which can be verified as the required solution.

LINEAR DIFFERENCE EQUATIONS WITH VARIABLE COEFFICIENTS 5.35. Solve the first order linear difference equation with variable coefficients given by yk+1 - Akyk = Rk or (E - Ak)yk = Rk. Method 1, using variation of parameters. The complementary or homogeneous equation corresponding to the given equation is Yk+1 - AkYk = 0

(1)

From this we have

Yk = Ak-lyk-1,

..,

11k-1 = Ak-2yk-2,

Y2 = Alyl

so that the solution of (1) is 11k = Ak-lAk-2* . A1111 Yk = c1A1A2...Ak-1

or

where we take

c1 = 111 as the arbitrary constant.

We now replace cl by a function of k denoted by K(k), i.e. Yk =

K(k)AlA2...Ak-1

(2)

and seek to determine K(k) so that the given equation is satisfied. Substitution of (2) in the equation yields K(k + 1)A1A2...Ak - AkK(k)A1A2...Ak-1 Rk

or dividing by A1A2 Ak Rk

K(k + 1) - K(k) = AK = Rk

K = A-1

Thus

A1A2...Ak

A1A2..

J

and so from (1) we find the required solution Rk

Yk

A1A2...Ak-lA-1 A1Az...Ak

RP

k-1

A1A2...Ak-1P=1 I AA 2 1

where c is an arbitrary constant.

.

,A

+ cA1A2...Ak-1 p

given

176

DIFFERENCE EQUATIONS

[CHAP. 5

Method 2, multiplying by a suitable summation factor.

Multiplying the given equation by 1/A1A2 Ak it can be written yk+1

Rk

Yk

=

A1A2...Ak - A1A2...Ak_1 yk

or

Then

Yk

=

AlA2...Ak

A1A2...Ak-1o-1

Yk

or

Rk

=

A1A2...Ak-1

A1A2..Ak-1

k-1

A1A2...Ak

Rk

A1A2...Ak)

R p

A A . ..A

=1

1

2

+ cA1A2...Ak-1 p

We call 1/A1A2 . Ak used to multiply the given equation the summation factor.

5.36. Show that the equation yk+2 - (k + 2)yk+1 + kyk = k can be written in the factored form (E - Ak)(E - Bk)yk = k. The given equation is in operator form

[E2 - (k + 2)E + k] yk = k Let us try to determine Ak and Bk so that (1) is the same as

(1)

(E - Ak)(E - Bk)yk = k Now the left side of (2) can be written as

(2)

(E-Ak)(yk+1-Bkyk) = yk+2-Bk+1yk+1 -Ak(yk+i-Bkyk) = yk+2 - (Ak+Bk+l)yk+1 + AkBkyk Comparing with the left side of the given equation (1) we have

Ak + Bk+1 = k + 2, AkBk = k (3) From the second equation of (3) we are led to try either Ak = k, Bk = 1 or Ak = 1, Bk = k. Since the second set, i.e. Ak = 1, Bk = k, satisfies the first equation of (3) the given equation can be written as (E -1)(E - k)yk = k 5.37.

Solve

yk+2 - (k + 2)yk+ 1 + kyk = k, y1= 0, y2 = 1.

By Problem 5.36 the equation can be written as

(E -1)(E - k)yk = k Let zk = (E - k)yk.

Then (1) becomes

which has the solution

(E - 1)zk = k

zk = A -1k = p 1kc1)

(2)

=

k(2)

+ c1

2

k(2)

(E - k)yk = 2 + cl

Then

(1)

= . -k(k -1) + e1

By Problem 5.35 this has the solution

/2k(k- 1) + c1 1

Yk

I

k-1 / .p(p - 1) + c1/ or

Yk =

(k - 1)!

I\

p

p=l (

k-1)! k-1 p(p 2

p=1

p!

1)

(k-1)!p

l

zk(k-1) + cl k!

+ c2(k - 1)t

k-1 + c1(k - 1)! Y 1 + c2(k - 1)! P=1 P.

(3)

CHAP. 5]

DIFFERENCE EQUATIONS

177

From y1 = 0, y2 = 1 and the difference equation we find, if k =1, y3 - 3Y2 + Y1 = 1 so that y3 = 4. If we put k = 2 and k = 3 in (3), we find using 0! = 1

C1+c2 = 1, 1+3c1+2C2 = 4 from which c1 = 1, c2 = 0.

Thus the required solution is

k1 1 _ (k- 1)! k-+1p(p-1) + (k-1)! 2

p!

P=1

P=1 p!

This can be written for k > 3 as Yk

5.38.

Solve

(k + 1)yk+2 - (3k +2)yk+1 + (2k -1)yk = 0.

The equation can be written as (k + 1)t2yk - koyk - 2yk = 0

Assume a solution of the form

I

yk = where we take

(1)

cP k(P)

(2)

c, = 0 for p = -1, -2, -3, ...

(3)

From (2) we obtain

Dyk = A2Yk

P=-

P=-W

pcpk(P-1)

(4)

p(p - 1)cpk(P-2)

(5)

Substituting (2), (4) and (5) into (1), it becomes

j p(p - 1)cpkk(P-2) + j p(p -

1)c17k(P-2) - j pckk(P-1) - Y.2cpk(P) = 0

(6)

where we have omitted the limits of summation. On using the result kk(m) = k(m+l) + mk(m)

(6) can be written -1)cp[k(P-1) + (p - 2)k(P-2)] + I p(p

p(p

-

- 1)cpk(P-2)

Y. pc,[k(P) + (p -1)k(P-1)] -

2cpk(P)

=0

This in turn becomes on collecting coefficients of k(P)

I {(p + 2)(p + 1)2cp+2 - (p + 2)cp}k(P) = 0

(7)

Since (7) is an identity each coefficient must be zero, i.e.

(p + 2)(p + 1)2cP+2 - (p + 2)cp = 0

Using (3) it follows that we must have p = 0, 1, 2.... so that (8) becomes on dividing by p + 2 (p + 1)2Cp+2 - Cp = 0

22C3 - C1 = 0,

32C4 - C2 = 0,

0

(9)

Putting p = 0, 1, 2, ... in (9) we find 12C2 - Co = 0,

(8)

4205 - C3 = 0,

DIFFERENCE EQUATIONS

178 Then

Cl

CO

C2 = 12 ,

C2

C3 = T21

14 = g2

[CHAP. 5

C3

CO

12.32'

cs

42

Cl

22-421

and so the required solution (2) becomes k(2)

7!k

=

oo

+ 12 +

k(4)

k(e)

12.32 + 12 . 32. 52 + kcsl

.. .

kc5

k+ c1 and using the fact that v1 = 2 we find c1 =

vk =

2

and the required solution is

ilk

5.44.

Solve

2

yk+2yk

_

k(k - 1) + Cl

(3)

2

Thus

.

k(k - 1)

(2)

I

1

2=

k2 - k + l 2

2

- k2-k+1

(4)

(5)

= yktl if y, = 1, y2 = 2. 3

Taking logarithms of the given difference equation it becomes

In yk+2 - 3 In yk+l + 2 In yk = 0 Letting

Vk = In ?1k

the equation (1) can be written as a linear difference equation v1 = 0, v2 = In 2 Vk+2 - 3Vk+1 + 2Vk = 0,

(3)

The general solution to the difference equation in (3) is Vk = C1 + C22k

(4)

DIFFERENCE EQUATIONS

182

Using the conditions in (3) we find c1 = -In 2, c2 =

[CHAP. 5

In 2 so that

-

vk = (ln 2)(2k-1- 1)

(5)

The required solution of the given difference equation is thus

In Y/k = (ln 2)(2k-1 -1) or

e(ln2)(2k-1 - 1)

?1k

22k 1 - 1

which can be written

yk

(6)

SIMULTANEOUS DIf FERENCE EQUATIONS 5.45.

- 3y

Solve the system L

k+1k

+ zk+1 k

5zkk

= 4k subject to the conditions y1 = 2, zl = 0.

Write the system of equations as (E - 3)yk + Zk = k

(1)

3yk + (E - 5)zk = 4k

Operating on the first equation in (1) with E - 5 and leaving the second equation alone the system (1) can be written

(E-5)(E-3)yk+ (E-5)zk = 1-4k1

J}

3Yk + (E - 5)zk = 4k

(2)

Subtracting the equations in (2) yields

(E2-8E+12)yk = 1 - 4k - 4k

(3)

The general solution of (3) found by any of the usual methods is

yk = c12k + c26k + 41

k

. 4k

-

19 25

(4)

and so from the first equation in (1) c12k - 3c26k -

4

4k

5

k

(5)

T5_

Putting k = 1 in (4) and (5) we have since y1 = 2, 1 = 0 2c1 + 602 - 64, 25

74

2c1 - 18c2 = 25

(6)

from which c1 = 133/100, c2 = -1/60. Thus the required solution is 7!k

=

zk

=

1332k-

2+

L3-3 -2k

100

16k+4k_1-4k-19

co

100

25

5

1

6k-

20

5 k- 34 25

4k-1 - 3

MIXED DIFFERENCE EQUATIONS 5.46. Solve the differential-difference equation yk+1\t) = Yk(t), Putting k = 0 we find

yp(t) = t,

2

yi (t) = yo(t) = t Then since y1(0) = 1, we have c1

or

y1(t) = 2 + cl

1 so that y1(t)

yk(0) = k

=

t2

1 + 2

(7)

(8)

CHAP. 5]

DIFFERENCE EQUATIONS

183

Putting k = 1 we find y2(t) = y1(t) = 1 +

t2

i12(t) = t+3 +c2

or

Then since y2(0) = 2, c2 = 2 so that y2(t)

=2+t+

5

Continuing in this manner we find

3+2t+2iz

y3(t)

=

Y4(t)

= 4 + 3t +

Y5(t)

= 5+4t+

+Ti-

+ t33+ 3t2

t5 i

33+4

+

+6

and in general

y"(t)

+ ... + t"-1 = n + (n -1)t + (n -2!2)t2 + (n3)t3 3! (n-1)!

+

t"+1

(n+1)!

which is the required solution.

5.47.

Solve Problem 5.46 by using the method of generating functions. Let the generating function of yk(t) be

= Ik=0 yk(t)Sk

Y(8, t)

(1)

Then from the difference equation,

j yk+1 (t)Sk = Lk=0yk(t)sk

(2)

k=0

which can be written in terms of (1) as

sat {Y(8, t) - y0(t)} = or since yo(t) = t

ay at

Y(8, t)

(3)

- By = 1

(4)

Putting t = 0 in (1) and using Yk(O) = k yields

=

Y(s, 0)

Y, ksk

(5)

k=0

Equation (4) can be written as

sat In (sY + 1) = 1 so that Using (5) and (6) we find

Y(s, t)

Y(s, t)

s

s 1 S

In (sY + 1) = st + c1

or

cent - 1

_

(6)

8

{est(1+ics11) - 1 } k=O f °° (st)" / 1

y0 p! ` `'

(st)P

P=1 p!

+j

ksk+1)

k=0

+P=O IY k=0

/

-

}

ksP+k+1tpl

p!

Jy

(7)

184

DIFFERENCE EQUATIONS

[CHAP. 5

Since the coefficient of sn in this expansion is yn(t) we find it to be 1!n(t)

=

to+l

+n

Y.

(n + 1) !

(n - p)tp (8)

p!

This is obtained by taking the term corresponding to p = n + 1 in the first summation of (7) and p + k + 1 = n +.1 in the second summation of (7). The required solution (8) with n replaced by k agrees with that of Problem 5.46.

PARTIAL DIFFERENCE EQUATIONS 5.48. Write the partial difference equation u(x + h, y) - 4u(x, y + 1) = 0 the operators E1, E2 and (b) in terms of subscript notation. (a)

(a)

in terms of

Since E1u(x, y) = u(x +h, y), E2u(x, y) = u(x, y + 1), the given equation can be written as

E1u - 4E2u = 0

(E1- 4E2)u = 0

or

(1)

where we have denoted u(x, y) briefly by u. (b) In subscript notation

u(x + h, y) = uk+ 1, m,

u(x, y + l)

u(x, y) = uk,m

uk, m+ 1,

and the given equation can be written as

uk+l,m - 4Uk, m+1 = 0

5.49.

or

(E1- 4E2)uk,m = 0

(2)

Solve the difference equation of Problem 5.48. We can write equation (2) of Problem 5.48 as Eluk,m = 4E2uk,m

(1)

Considering m as fixed (1) can be thought of as a first order linear difference equation with coefficients in the single variable k. Calling uk,,n = Uk, 4E2 = a the equation becomes

EUk = aUk A solution of (2) is

Uk = akC

where C is an arbitrary constant as far as k is concerned but may depend on m, i.e. it is an arbitrary function of in.

We can thus write (3) as Uk = akCm

(4)

Restoring the former notation, i.e. Uk = uk,m, a = 4E2, (4) becomes uk,m = (4E2)kCm

But this can be written as uk,m =

4kE2k

= 4kCm+k

(5)

To obtain this in terms of x and y we use the fact that x = a + kh, y = b + ml.

Then (5)

becomes u(x, y)

_ 4(x-a)/hC x h- a + y

b

which can be written u(x, y)

_

(6)

where H is an arbitrary function.

We can show that (6) satisfies the equation of Problem 5.48. arbitrary function it is the general solution.

Then since it involves one

CHAP. 5]

DIFFERENCE EQUATIONS

185

5.50. Work Problem 5.49 by finding solutions of the form akg- where k, µ are arbitrary constants.

Substituting the assumed solution uk, m = xkµm in the given difference equation (2) of Problem 5.48 we find Xk+1µm - 4xkµm+1 = 0 Xor

4µ = 0

on dividing by Xkµm 0 0.

From a = 4µ it follows that a solution is given by Xkµm = (4µ)kµm = 4kµk+m

Since sums of these solutions over any values of µ are also solutions, we are led to the solutions 4kµk+m = 4k µk +m = 4kF(k + m) lA

{A

where F is an arbitrary function. Thus we are led to the same result as given in Problem 5.49 where the arbitrary function F(k + m) = Ck+m 5.51.

Solve u(x + 1, y) - 4u(x, y + 1) = 0, u(0, y) = y2. The difference equation is the same as that of Problem 5.48 with h = 1, 1 = 1. general solution is from equation (6) of Problem 5.49,

Thus the

u(x, y) = 4xH(x + y)

(1)

Using the boundary condition u(0, y) = y2 we have from (1)

u(0, y) = H(y) = y2 H(x + y) = (x + y)2

Thus and the required solution is

u(x,y) = 4x(x + y)2

5.52.

Solve

(2)

u(x + 1, y) - 2u(x, y + 1) = 3u(x, y).

We can write the equation in subscript notation as E1uk,m - 2E2uk.m = 3uk,m

or

(E1 - 2E2- 3)uk,m = 0

Rewriting it as E1uk, m = (2E2 + 3)uk,,m we can consider it as a first order linear difference equation with constant coefficients as in Prob-

lem 5.49 having solution uk, m = (2E2 + 3)kCm

This can be written as uk,m = 3k(1 + IE2)kC,m

or on expanding by the r[binomial theorem,

uk,m = 3k L1 + k(1E2) +

=

k(2 -

1) (E2)2 +

k(k - 3)' 1)(k - 2) (JE2)3 + ... C.

3k[Cm+IkCm+1+2k(k-1)Cm+2+

sk(k-1)(k-2)Cm+3+...]

To obtain the result in terms of x and y we can use x = a + kh, y = b + ml, choosing h = 1,

1=1, a = 0, b =0 so that k = x, m=y. Then writing C,m = H(y) we obtain u(x, y) = 3x[H(y) + IxH(y + 1) + 9x(x - 1)H(y + 2) +

81-

x(x - 1)(x - 2)H(y + 3) +

This contains one arbitrary function H(y) and is the general solution.

]

DIFFERENCE EQUATIONS

186

5.53.

[CHAP. 5

if uk,o = 0 for k > 0, uo.,n = 1 for m > 0 and p, q are given positive constants such that p + q = 1. Solve Uk,m = PUk-l,m + quk.m-1

In operator notation the given equation can be written

(PEI1 + qE2 1 - i)uk,m = 0 This can also be written as

PEi 1uk,m = (1 - qR 1)uk,m

Eluk,m = 1 - qE P 2 1 uk, m Considering this last equation as one in which m is fixed the solution is or

P k i-gE21) C.

Pk(1- gE21)-kCm

= Pk [I+ k(gE21) + k(k2 1) (qE2 1)2 +

uk.m = Pk Ic-

or

kgCm-1 +

k(k + 1)

2

k(k + 3)! k + 2) (gE21)3 + ...

q Cm-z +

2!

k(k + 1)(k + 2) 3

C.

ggC+n-s + .. .

Now if m = 0, k > 0, this becomes

0 = Pk [Co + kgC_1 + k(k + 1) q2C_ 2

2!

from which

+ ... 1

Co=0, C-1=0, C-2=0,

If k = 0, m > 0, it becomes

m>0

Cm = 1, Then for all k > 0, in > 0,

uk,m = 5.54.

Pk[1+kq+k(k2+1)g2+... +

k(k+1)...(k+m-2)qm-1 (m-1)1

Solve Problem 5.53 by the method of generating functions. Let

Gk(t) = uk, 0 + uk.1 t + uk, 2 t2 + ...

= Im=0uk, m tm

(1)

Then multiplying the given difference equation by t'n and summing from m = 1 to *0 we have m=1

uk,mtm = P Y. uk-l,mtm + q m=1 Uk,m.-ltm m=1

This can be written in terms of the generating function (1) as Gk(t) - uk.0 = P[Gk-1(t) - Uk-1,0] + gtGk(t) Since uk,0 = 0 and uk-1,0 = 0 for k z 1 and uo, 0 = 0, (2) becomes Gk(t) = pGk-1(t) + gtGk(t) Gk(t)

or

=

This equation has solution

Gk(t) =

1

P

/

qt

(2)

(3)

Gk-1(t)

(b)

1k

1 1 -!-q t / Go(t)

and since

Go(t) = uo, 0 + uo,1 t + uo, 2 t2 + ...

/ = t + t2 + t3 + ...

(5)

t = 1-t

(5) becomes Gk(t)

-

Pkt

(1 - gt)k(1 - t)

(6)

CHAP. 5]

DIFFERENCE EQUATIONS

We can write this as

187

Gk(t) = pk(1- qt)-b 1 t t

and expand into a power series in t to obtain

Gk(t) = pk [1 + kqt + k(2+ 1) g2t2 + k(k + 11)(k + 2) q t3] [t + t2 + is + ... ] The coefficient of tm in this product is then given by

[1+kq+1t1

uk.m =

k(k + 1(k m - 2)

qz + ... +

(m -1)1 which is the required solution in agreement with that of Problem 5.53. 5.55.

2!

qm-ll

J

Solve u(x + 1,y) - 4u(x, y + 1) = 6xzy + 4. By Problem 5.49 the equation with the right hand side replaced by zero, i.e. the homogeneous equation

u(x + 1, y) - 4u(x, y + 1) = 0

(E1- 4E2)u(x, y) = 0

or

has the solution 4xH(x + y)

where H is an arbitrary function.

Then to solve the complete equation we need only find a par-

ticular solution. Method 1, using undetermined coefficients.

Assume a particular solution having the form u(x, y) = A1x2y + A2xy + As X2 + A4x + A5Y + As

where A1, ...,A6 are constants. Substitution in the complete equation yields -3Alx2y + (2A1- 3A2)xy - (4A1 + 3A3)x2 + (2A3 - 3A4 - 4A2)x

+(A1+A2-3A5)y+A3+A4-4A5-3A6 = 6x2y+4 Then equating coefficients of corresponding terms

-3A1 = 6, 2A1 - 3A2 = 0, 4A1 + 3A3 = 0, 2A3 - 3A4 - 4A2 = 0, Al + A2 - 3A5 = 0,

A3+A4-4A5-3A6 = 4 so that

Al = -2, A2

4

As =

8 , 3

A4 =

32 9

A5 = - 10 9 , A6 =

,

20 9

Then the required solution is

y+9

u(x, y) = 4xH(x + y) - 2x2y - xy + 3 xz + s2 x -

(1)

9

3 Method 2, using operators.

Since the complete equation is (El - 4E2)u(x, y) = 6x2y + 4 a particuar solution is, using

E1=1+4 E2=1+A2,

1 1 (6xzy + 4) _ -3+O1-402 E1-4E2

(6x2y + 4)

- 3 1 - (A1 - 402)/3 (6xzy + 4) 1

-3

1

[1 + O1

-2xzy

402 3

+

(

44z)z

1

+ ...

(6xzy + 4)

9

-3xy+3x2. 92x- 9y+ 9°

and so the required solution is the same as that obtained in equation (1) of Method 1.

188

DIFFERENCE EQUATIONS

[CHAP. 5

MISCELLANEOUS PROBLEMS 5.56. Solve the partial differential equation au

+ 2 ay

=

u(x, 0) = x3 - 3 sin x

0,

The given partial differential equation is the limit as h -+ 0, 1 -> 0 of u(x + h, yh - u(x, y)

+2

- u(x, y)

u(x, y + l

0

i

which can be written in operator and subscript notation as (

lE1+ 2hE2-l- 2h)uk,m = 0

or

E1uk, m

1+ 2

=

Z

-1 E2 u k,m Z

This can be solved as in Problem 5.49 to yield

1 + 2h

uk, m

or in terms of x and y as

- 2h E2

k

Cm

(l+_E2)H()

(1)

In order to obtain meaningful results as h-0, 1-0 choose 1+2h/1=0 so that (1) becomes u(x, y)

= H (Y-2x) = J(2x - y)

where J is an arbitrary function. The required general solution is thus u(x, y) = J(2x - y)

u(x, 0) = J(2x) = x3 - 3 sin x

Since

J(x) _ jx3 - 3 sin (x/2)

we have Then from (2) we have u(x , y)

5.57.

(2)

_ *(2x - y)3 - 3 sin C\ 2x2- y1)

Solve the mixed difference equation

u(x + 1, y) - u(x, y)

= 2y

The given equation can be written as

E1u - u = 2D2u E1u = (1 + 2D2)u

or

Then treating 1 + 2D2 as a constant we find as in Problem 5.49 u(x, y) = (1 + 2D2)XH(y)

(3)

where H(y) is an arbitrary function of y. Expanding by the binomial theorem we then find u(x, y)

=

Ll + (1 ) 2D2 + (2 ) (2D2)2 + ... + (m) (2D2)m +

= H(y) + 2 l 1J H'(y) + 22

...l H(y)

(x) H"(y) + ... + 2m (ml H(-)(y) + ...

Then if x is a positive integer the series terminates and we have

u(x,y) = H(y) + 2 l 1 J H'(y) + 22 (2) H"(y) + ... + 2x (:) H(x)(y)

(4)

CHAP. 5]

5.58.

DIFFERENCE EQUATIONS

189

Solve the preceding problem subject to u(0, y) = y3. From (4) of Problem 5.57 we see that

u(0, y) = H(y) = v3 Thus

H'(y) = 3y2,

H"(y) = 6y,

H,,,(y) = 6,

H(1v)(y)

= 0,

.

and so u(x, v)

= y3 + 6 (1 ) y2 + 24 (2) y + 48 (X) = y3 + 6xy2 + 12x(x - 1)y + 8x(x - 1)(x - 2)

5.59.

Find the general solution of Uk,m = puk+l,m-1 + qUk-1,m+1 where p and q are given constants such that p + q = 1.

Method 1.

Assume a solution of the form uk,m = hkNm

Then by substitution in the given difference equation we find after dividing by Akµm

pA2 - X + gji2 = 0 from which

-

µ±

µ2 - 4pq 2

µ ± µ 1- 4pq

2p

2

--

µ,

µq

p

Thus solutions are given by µk , µm = uk+m and (µq/p)kµm = ,k+m(q/p)k.

Since sums of such solutions are also solutions, it follows that F(k + m) and (q/p)kG(k + m) are solutions and the general solution is uk,,n = F(k + m) + (q/p)kG(k + m)

(1)

where F and G are arbitrary functions. Method 2.

It is noted that the sum of the subscripts in the given difference equation is a constant. If we denote this by c we have k + m = c. The equation is thus given by

uk,c-k = PUk+l,c-(k+1) + quk-1,c-(k-1)

(2)

uk,c-k = vk

(3)

Vk = PVk+l + qvk-1

(4)

If we let (2) becomes

which involves only one subscript and can be solved by the usual methods. We find Vk = C1 + c2(q/p)k

(5)

Now the constants cl and c2 can be considered as arbitrary functions of the constant c = k + m:

cl = F(k + m), c2 = G(k + m). Thus we obtain from (3) and (5) Uk,m = F(k + m) + G(k + m)(q/p)k

190

DIFFERENCE EQUATIONS

[CHAP. 5

Supplementary Problems DIFFERENTIAL EQUATIONS AND RELATIONSHIPS TO DIFFERENCE EQUATIONS 5.60.

(a) Show that the general solution of the differential equation dy ,4X

is (b) 5.61.

y = xz

y=cex-x2-2x-2.

Find that particular solution to the equation in (a) which satisfies the condition y(0) = 3.

(a) Show that the equation

Ay Ax

y -

X(2)

[compare Problem 5.60] where y = f(x), Ax = h can be written as f (x + h) - (1 + h) f (x) = hx2 - hex (b) Show that the general solution of the equation in (a) is y(x) = f(x) = C(x)(1 + h)xih - x2 + (h - 2)x - 2

where C(x + h) = C(x). (c)

Find that particular solution for which y(0) = 3.

5.62.

Explain clearly the relationship between the difference equation and the differential equation of Problems 5.60 and- 5.61 discussing in particular the limiting case as h -> 0.

5.63.

(a) Show that the general solution of d2y dx2

is (b)

+y=

3x2 - 5x + 4

y = c1cosx+c2sinx+3x2-5x-2.

Find that particular solution to the equation in (a) which satisfies the conditions y(0) = 2, y'(0) = 0.

5.64.

(a) Show that the equation

A2

0x2 +y = 3X(2)-5X(l) + 4

can be written as f(x+2h) - 2f(x+h) + (1+h2)f(x) = 3h2x2 - (3h3 + 5h2)x + 4h2 where y = f (x). (b) Show that the general solution of the equation in (a) is y = f (Z) = C1(x)(1 + ih)xih + C2(x)(1- ih)xih + 3x2 - (3h + 5)x - 2 where C1(x) and C2(x) are periodic constants with period h.

5.65.

(a) Explain clearly the relationships between Problems 5.63 and 5.64, in particular the limiting case as h -> 0. (b) What are the boundary conditions for the difference equation in Problem 5.64 which correspond to those for the differential equation in Problem 5.63? (c) Obtain the particular solution of Problem 5.64 corresponding to the boundary conditions just found in (b), and explain the relationship between this and the particular solution of Problem 5.63.

THE SUBSCRIPT NOTATION FOR DIFFERENCE EQUATIONS 5.66.

Write the difference equation of (a) Problem 5.61 and (b) Problem 5.64 with subscript notation.

5.67.

Write each of the following in subscript notation. (a) f(x + 3h) - 3f(x + 2h) + 3f(x + h) - f(x) = x(3) - 2x(2) (b) f(x + 4h) + f(x) = cosx

CHAP. 5] 5.68.

DIFFERENCE EQUATIONS

191

Change each of the following equations to one involving the "x notation". (a)

211k+2 - 3yk+1 + Yk = 0

(b)

yk+3+4yk+2-5Yk+1+2Yk = k2-4k+1

LINEARLY INDEPENDENT FUNCTIONS 5.69.

Determine which of the following sets of functions are linearly dependent and which are linearly

independent. (a)

2k, k - 3

(e)

(b) k+4, k-2, 2k+1 (c)

k2 - 2k, 3k, 4

(d) 2k, 22k, 23k 5.70. 5.71.

1, k, k2, k3, k4

(f) k2 - 3k + 2, k2+5k-4, 2k2, 3k (g) 2k cos ke, 2k sin ke (h) k2, (k + 1)2, (k + 2)2, (k + 3)2

Work Problem 5.69 using Theorem 5-1, page 154.

Show that if 0 is added to any linearly independent set of functions then the new set is linearly dependent.

5.72.

Prove Theorem 5-1, page 154, for any number of functions.

5.73.

Prove Theorem 5-2, page 155.

5.74.

If a 0 b show that ak and bk are linearly independent.

HOMOGENEOUS LINEAR DIFFERENCE EQUATIONS WITH CONSTANT COEFFICIENTS 5.75.

Find the solution of each of the following subject to given conditions (if any). Yk+2+5yk+1+6yk = 0 (f) yk+2+2yk+1+2yk-0, YO =01 y1=-1

(a) (b)

Yk+2 - 3yk+ 1 + 2yk - O, 110 - 1, 111 = 2

(c)

3yk+2 + 8yk+ 1 + 4yk = 0

(g) Ilk+2 + 16yk = O, Yo = 0, y1 = 1 (h) 411k+2 + 25yk = O

(d)

Yk+2-9Yk-0, y0=2, 111=-1

(1.)

Yk+2-6Yk+1+9yk = 0

(9)

4Yk+2+411k+1+Yk = 0

(e) 2yk+1 + 3Yk = 0, 110 = 4 5.76.

Solve

Yk+3 - 6Yk+2 + 11yk+1 - 6yk = 0

5.77.

Solve

Yk+4 + Ilk = 0

5.78.

Solve

4Yk+4 - 25Yk = 0

5.79.

Solve

Yk+3 - 8Yk '

5.80.

Solve

Ilk+4 + 12yk+2 - 64Yk = 0.

5.81.

Solve

Ilk+3 - 3Yk+2 + 4Yk+1 - 2yk = 0

subject to the conditions y0=0, y1=1, Y2=1-

0

5.82.

Find the general solution corresponding to a linear homogeneous difference equation if the roots of the corresponding auxiliary equation are given by 2, -2, -3 ± 4i, -2, -2, 4, -2, 2.

5.83.

What is the difference equation of Problem 5.82?

5.84.

Work Problem 5.14(b) directly by using the difference equation.

METHOD OF UNDETERMINED COEFFICIENTS 5.85.

Solve each of the following subject to given conditions if any.

(a) Yk+2 - 3Yk+1 + 2yk = 4k (b) Ilk+2 - 4Yk+1 + 4yk = k2 (c) 4v+ 2 + Ilk = 2k2 - 5k + 3 (d) (e)

Ilk+1-Yk = k, Y0 = 1 Yk+2+2Yk+1+Ilk = k+2k

(f) Ilk+2 - 811k+1 + 16yk = 3 . 4k, y0 = 0, y1 = 0 (g) 2Yk+2 - 3Yk+1 + Ilk = k2 - 4k + 3 (h) Ilk+2 + 4Yk = 5(-3)k + 10k (i)

Ilk+2-1lk = 1/3k

192

DIFFERENCE EQUATIONS

5.86.

Solve

yk+1 - ayk = R for all values of the constants a and 6.

5.87.

Solve

Ilk+2 + yk = 4 cos 2k.

5.88.

Solve

yk+3 - 6yk+2 + 11Yk+1 - 6yk = 4k + 3-2k - 5k.

5.89.

Solve

Yk+4 - 16yk = k2 - 5k + 2 - 4. 3k.

5.90.

Solve

Yk+3 + Yk = 2k cos 3k.

5.91.

Solve

3yk+4 - 4yk+2 + Yk = 3k(k - 1).

[CHAP. 5

SPECIAL OPERATOR METHODS 5.92.

Solve the following difference equations by operator methods. Yk+2 - 3yk+1 + 2yk = 4-3k - 2. 5k Ilk+2 + 4yk+1 + 4yk

5.93.

= k2 - 3k + 5

(f)

3yk+2 - 8yk+1 - 3yk = 3k - 2k + 1 Yk+2 - 8yk+1 + 16yk = 3 . 4k - 5(-2)k 2yk+2 - 2yk+1 + Yk = k3 - 4k + 5k Yk+2 - 4yk+1 + 4yk = k2. 2k

(g)

Yk+3 - 3yk+2 + 3yk+1 - Yk = 2k + 4 - 3k-1

(h)

Yk+4 - 16yk = 3k + 2k + 3k

Show that

O(E) cos ke = Re

(a)

ik9

e

iB

(b)

O(E) sin ke = Im

where Re denotes "real part of" and Im denotes "imaginary part of". 5.94.

Use Problem 5.93 to solve Yk+2 + 4Yk+1 - 12Yk = 5 cos (Irk/3).

5.95.

(a)

Determine E

1

a R(k).

-

1 1 1 1 (b) Show that E2 - 6E + 8 2 E - 4 E - 2 i.e. the method of partial fractions can be used for operators. (c) Show how the results of (a) and (b) can be used to solve the equation

1

Yk+2 - 6yk+1 + 8yk = k2 + 3k 5.96.

Show how operator methods can be used to solve (E -1)2(E + 2)(E - 3)yk = k2 - 3k + 5 + 3k

5.97.

Solve

5.98.

Solve Problem 5.85(a)-(i) by operator methods.

5.99.

Solve (a) Problem 5.87, (b) Problem 5.88, (c) Problem 5.89, (d) Problem 5.90 and (e) Problem 5.91 by operator methods.

(E2 + 1)2Yk = 2k - 3.

METHOD OF VARIATION OF PARAMETERS 5.100.

Solve each of the following difference equations by using the method of variation of parameters. (a)

Yk+2 - 5yk+1 + 6yk = k2

(b)

Yk+2-2yk+l+yk = 3+k+4k

(C)

4yk+2 - 4yk+1 + Yk = 3/2k

(d)

Yk+2 + Yk = 1/k 5yk+2 - 3yk+1 - 2yk = 3k + (-2)k Yk+2 - Yk+1 + Yk = 1/k!

/(e)

()

CHAP. 5]

DIFFERENCE EQUATIONS

193

5.101.

Solve Problem 5.85(a)-(i) by using variation of parameters.

5.102.

Solve Problem 5.92(a)-(h) by using variation of parameters.

5.103.

Solve 1/k+3 - 69lk+2 + 11Uk+1 - 611k = 5.2k + k2 by using variation of parameters.

5.104.

Verify that the determinant of the coefficients (24) on page 157 is equal to the Casorati of the system [see (15), page 154]. Discuss the significance of the case where the determinant is or is not zero.

METHOD OF GENERATING FUNCTIONS 5.105.

Verify each of the following generating functions. m

(a)

1t = ko

(b)

1

(c)

(d) 5.106.

tk

(-1)kt2k+1 k=0 (2k + 1) !

(e)

sin t

(1+t)n = k 0 k)tk

(f)

cost =

et = ko k -In (1- t) _

(g)

(1

(h)

tan-1 t =

tk+1

k=0k+1

(_1)kt2k k=O

1 t)2 = kI

(2k)!

ktk-1 (-1)kt2kt 1

=0 2k+1

Solve each of the following by the method of generating functions.

(d)

11k+2 - 6Yk+1 + 8Yk = 0, to = 0, Ui = 2 2Yk+1 - 1!k = 1, Yo = 0, 9k+2 + 4yk+1 + 4Yk = 0, Yo = 1, 911 = 0 6Uk+2 - 5Uk+1 + Yk = 0

(e)

Yk+2+211k+1+1!k = 0,

(a) (b)

(c)

yo = 2, 911 = -1 Ilk+3-3yk+2 + 3yk+1 - yk = 0 1/k+2+2Uk+1+Yk = 1+k 4Yk+2 - 4Yk+1 + Yk = 2-k, yo = 0, 1!1 = 1

(f) (9) (h) 5.107.

Solve

5108.

Show how to obtain the generating function of the sequence (a) Uk = k, (b) Uk = k2, (c) Uk = k3. [Hint. Use Problem 5.105(a) and differentiate both sides with respect to t.]

5.109.

If G(t) and H(t) are the generating functions of Uk and Vk respectively show that G(t)H(t) is the k generating function of uk vk = p-0 uk-P vP

Uk+2 + 3yk+1 - 4Yk =

k,

ho = 0, 111 = 0-

which is called the convolution of Uk and Vk. 5.110.

Show that (a) (b)

uk V k = vk uk uk (Vk + Wk) = ukVk + ukWk

(c) uk (vk Wk) = (4Vk)*wk What laws of algebra does the convolution satisfy?

LINEAR DIFFERENCE EQUATIONS WITH VARIABLE COEFFICIENTS 5.111. Solve each of the following. (d) Uk+1 + 4kyk = k!, Ui = 1 (a) ?1k + 1 - kyk = k, 711 = 0 k91k+1 + 211k = 1 (e) kUk+l - Uk = 2k, 111 = 0 (b) (c)

Yk+1 + kyk =

1c 2)

DIFFERENCE EQUATIONS

194

[CHAP. 5

5.112.

Solve

Yk+2 - (k + 1)yk+1 + kyk = k,

5.113.

Solve

yk+2 - (2k + 1)yk+1 + k2Yk = 0.

5.114.

Work Problem 5.113 if the right side is replaced by 2k, using variation of parameters.

5.115.

Solve

(k + 3)yk+2 - 3(k + 2)yk+l + (2k + 1)yk = 0-

5.116.

Solve

(k -1)yk+2 - 2kyk+1 + 1(k + 1)yk = 0-

5.117.

Let Yk = 0 be any solution of the difference equation

y1 = 0, Y2 = 1-

Yk+2 + akyk+1 + bkyk = 0

Show that this second order linear difference equation can be reduced to a first order linear difference equation by means of the substitution Yk = YkUk and thus obtain the solution. 5.118.

Show that if Yk is a solution of the equation yk+2 + akyk+l + bkyk = 0

then the substitution Ilk = Ykuk can be used to solve the equation yk+2 + akyk+l + bkyk = Rk 5.119.

Use Problems 5.117 and 5.118'to solve the equations

(1 - k)yk+1 - 2yk = 0 [Hint. A solution of (a) is Yk = k.] (a)

kyk+2

(b)

kyk+2 + (1- k)yk+1 - 2yk = 1

5.120.

Discuss the relationship of the methods of Problems 5.117 and 5.118 with the method of variation of parameters.

5.121.

Solve

Yk+2 +

5.122.

Solve

ak+2yk+2 - (aak+i + bk+l)yk+l + abkyk

(a+k+2)yk+1 + k+1yk = 0 0

where a is a given constant.

5.123.

Solve yk+2+kyk = k, yo = 0, yi = 1 by the method of generating functions.

5.124.

Solve Problem 5.116 by the method of generating functions.

5.125.

(a) Solve Problem 5.37 by making the substitution yk = (k - 1)! Vk. (b) Show that k inl Yk = -

e.

STURM-LIOUVILLE DIFFERENCE EQUATIONS 5.126.

Find eigenvalues and eigenfunctions for the Sturm-Liouville system y1 = ye, yN+1 = yN 2yk-1 + Xyk = 0,

5.127.

Work Problem 5.126 if the boundary conditions are yi = yo, yN+1 = 0-

5.128.

Write the orthogonality conditions for the eigenfunctions of

(a) Problem 5.126 _

and

(b) Prob-

lem 5.127. 5.129.

(a) Show that if the eigenfunctions of a Sturm-Liouville system are not necessarily real then the orthogonality of page 159 must be rewritten as N

_

I rkcm.kcbn,k = 0 k=1 (b)

m#n

If Fk = 7. am(m,k determine the coefficients c,,, for this case.

DIFFERENCE EQUATIONS

CHAP. 5] 5.130.

195

Can any second order linear difference equation akA2yk + bkAyk + aklk = 0

be written as a Sturm-Liouville difference equation?

Justify your answer.

5.131.

Prove the results (a) (10), (b) (11) and (c) (12) on page 181 directly without using Sturm-Liouville theory.

5.132.

Explain how you could solve the equation A2Yk-1 + Xyk = k if yo = 0, YN+1 = 0. relationship with Sturm-Liouville theory.

Discuss the

NONLINEAR DIFFERENCE EQUATIONS 5133. Solve yk+yk+1 = Ykyk+1. [Hint. Let 1/yk = Vk.] 5.134.

Solve ykyk+1 = 2yk + 1 by letting Yk = c + (1/Vk) and choosing c.

5.135.

Solve Ykyk+lyk+2 = Yk+Yk+1+Yk+2

5.136.

Solve Ikyk+1Yk+2 = K(yk+yk+1+yk+2) where K is a given constant.

5.137.

Solve yk+1 = 311k, Y0=1-

[Hint.

5.138.

Solve

(b) Yk+1

5.139.

Solve yk = 1/yk-1yk+1-

5.140.

2 0. Solve yk2 +1 - 5yk+1 Yk + 6yk =

5141.

Solve Yk+1 = Y i

5142.

(a)

5.143.

(a)

(a) Yk+2 = Ykyk+I

Solve Yk+ i =

[Hint.

Let Ilk = tan Uk.]

Take logarithms.]

= 2yk -1.

Nk

2 + yk, y0=0

Solve Yk+1 = 2 (Yk+W-)

k,

and

yo = a

(b)

find lim yk k-+oa

and

(b)

find kllim yk.

SIMULTANEOUS DIFFERENCE EQUATIONS 5.144.

Solve

f2uk+1 + Vk+1 = Uk + 3Vk Sl

nk+1 + Vk+1 = uk + Vk

Yk+1 - kzk = 0

5145.

Solve

5146.

Solve the system of equations

5.147.

Solve Problem 5.146 if 1 - p + q = 0.

5.148.

Solve

zk+1 - 4kyk = 0

uk+ 1 = puk + 4vk vk+1 = (1 - p)uk + (1 - q)vk

if 1 - p + q # 0.

f 2uk+1 + vk+2 = 16vk uk+2 - 2vk+1 = 421k

MIXED DIFFERENCE EQUATIONS 5.149. 5.150.

Solve yk+1(t) = yk(t) subject to the conditions yo(t) = 1, Yk(O) = 1-

(a) Determine lim yk(t) for Problem 5.149. (b) Could this limit have been found without solving k-+oo the equation? Explain.

196 5.151.

5152.

DIFFERENCE EQUATIONS

[CHAP. 5

(a) Solve yk(t) = Ilk+1(t) subject to the conditions yo(t) = 1, yk(0) = 1. (c) Discuss the relationship of this problem with Problems 5.149 and 5.150.

(b) Find lira yk(t).

(a) Solve yk+1(t) + yk(t) = 0 subject to the conditions yo(t) = 1, yo (t) = 0, yk(0) = 1. cuss the limiting case of (a) as k -), ao,

k-,w

(b) Dis-

PARTIAL DIFFERENCE EQUATIONS 5.153.

Solve

u(x + 1, y) + 2u(x, y + 1) = 0.

5.154.

Solve

u(x + 1, y) - 2u(x, y + 1) = 0, u(0, y) = 3y + 2.

5.155.

Solve

u(x, y + 1) + 2u(y, x + 1) = 3xy - 4x + 2.

5.156.

Solve

u(x + h, y) - 3u(x, y + k) = x + y.

5.157.

Solve

(E1E2 - 2)(E1- E2)u = 0-

5.158.

Solve

uk+l,m+2 - Uk,m = 2k-m,

5.159.

Solve

uk+l,mtl - uk+l,m =' uk

5.160.

Solve

uk+2,m+2 - 3uk+l,m+1 + 2uk,m = 0.

5.161.

Solve

uk+l,m + uk,m+l = 0.

5.162.

Solve

uk,m = uk-l,m-1 + uk+1,m+1

5.163.

Solve

uk+1(y) - uk(y) + 3

m

a8yuk(7l)

if

= 0,

uk 0 ={1 0

k>0 k =0

u0(y) = 3e-21j'3.

MISCELLANEOUS PROBLEMS 5164.

Show that the general solution of

is given by

yk+2 - 2(COS a)yk+l + yk = Rk k RP sin (k - P)a

I1

Yk

sin a

where it is assumed that a # 0, ±lr, -21r, ... . 5.165.

Obtain the general solution in Problem 5.164 in the case a = 0, -tea, ±2a, ... .

5.166.

Let yk+2 + 3yk+1 + 2Yk = 6.

5.167.

(a)

Express the differential equation

(b)

as a limiting case of a difference equation having differencing interval equal to h. Solve the difference equation of (a) by assuming a factorial series expansion. Use the limiting form of (b) as h -+ 0 to obtain the solution of the differential equation in (a).

(c)

5.168.

Prove that if lim Yk exists then it must be equal to 1. k- w

y" - xy' - 2y = 0

(a) Solve the difference equation

+ P(x)y = Q(x) (b) By finding the limit of the results in (a) as Ax = h - 0 show how to solve the differential Ax

equation

dx + P(x)y = Q(x)

CHAP. 5] 5.169.

(a)

DIFFERENCE EQUATIONS Solve the difference equation

z

Axz

197

3y = x + 2

- 4 Ax -f-

(b) Use an appropriate limiting procedure in (a) to solve the differential equation zy

dx+3y = x+2

dX2

5.170.

Obtain the solution of the differential equation e-x

dx2+2dx+y by using difference equation methods. 5.171.

denote a one-parameter family of curves. The differential equation of the family is obtained by finding dyldx from f (x, y, c) = 0 and then eliminating c. (a) Find the differential equation of the family y = cxz and discuss the geometric significance. (b) Explain how you might find the difference equation of a family and illustrate by using y = cxz. (c) Give a geometric interpretation of the results in (b). Let f(x, y, c) = 0

5.172. Work Problem 5.171 for y = cx + c2. 5.173.

Generalize Problem 5.171 to two-parameter families of curves A XI y, c1, c2) = 0 and illustrate by using y = clan + c2bx where a and b are given constants.

5.174.

(a) Show that two solutions of

y = X AY +

mix J 2

are y = cx + c2 and y =C2- Ix2 + AM - jch(-1)xi4./ (b)

5.175.

Discuss the relationship of your results in (a) with those of Problem 5.172.

(a) Show that two solutions of the differential equation y

dy

xdx +

dy

z

(TX-)

are y = cx + c2 and y = -x2/4. (b) By obtaining the graph of the solutions in (a) discuss the relationship of these solutions. (c) 5.176.

Explain the relationship between the solutions in part (a) and Problem 5.172.

The difference equation Uk = k1uk + F(t uk) or in "x notation"

Ax +

y (a)

Xd

Ox (41Y)

+ F (dx

Solve Clairaut's differential equation by differentiating both sides of it with respect to x. Illustrate by using the equation (dy)z dy Y = xdx +

dx

and explain the relationship with Problem 5.175. (b)

5.177.

Is there an analogous method of obtaining the solution of Clairaut's difference equation? cuss with reference to Problem 5.174.

Solve

Yk+1 + (-2)kyk = 0.

Dis-

DIFFERENCE EQUATIONS

198 5.178.

Solve

5.179.

Show that the nonlinear equation

[CHAP. 5

Ykyk+l + 1 = 2k(yk+1 - yk)

Yk+lyk + Akyk + Bkyk+1

Ck

can be reduced to a linear equation by the substitution

Yk =

vk+1

- Ak

vk

Illustrate by solving Problem 5.134. 5.180.

(a) Show that if all a2, a3 are constants then the equation Ilk+3 + al/3kyk+2 + a2QkPk-lyk+1 + a3$kPk-lPk-2Yk = Rk can be reduced to one with constant coefficients by means of the substitution 9192... /3k-3uk Yk = (b) What is the corresponding result for first and second order equations? What is the result for nth order equations?

a

by letting

b + Yk =

uk+1

5.181.

Solve

5.182.

Solve

YkYk+1 = a/(k + 1).

5.183.

Solve

uk+3,m - 3uk+2,m+1 + 3uk+1,m+2 - uk,m+3 = 0-

Solve

uk.m+1 = uk-1,m - muk,m

where uo,m = 0

for

m = 1, 2, 3, ... and uk,m = 0

where

uo,.m = 0

for

m = 1, 2, 3, ...

5.184.

Yk+1 = b + Yk

Uk

for k > m. 5.185.

Solve

uk,m+1 = Uk-1,m + kuk,m

5.186.

Solve

(a) ax au +2 ay au = 4,

5.187.

Solve

ykyk+1 +

5.188.

Solve

Y(x) +

5.189.

Solve

Y(x) -

5.190.

Solve

kuk = 1 + ul + U2 +

5.191.

Solve the partial differential equations

5.192.

and

uk,m = 0

for k>m.

+x-y (b) 2 au = au ay ax

(1 - yk)(1-Ilk+1) = cos (47r/k)

Y(x + 1) 1!

+

Y(x + 2) 2!

+

(a)

2 ax - 3 ay =

(b)

ax + 3 a-uy

Y(x + 2) 2!

+

Y(x+4)

-

4!

=

by using difference equation methods.

subject to the condition yl = 0.

X.

x2-4x+1.

+ uk_2,

u1 = 1, u2 = 1/2.

u(0, y) = 712 + e-v

0,

= x - 2y,

u(x, 0) = x2 + x + 1

Solve the partial differential equations 2

2

(a)

ax2 - 3 ax u

+ 2 a2 =

(6)

ax

- 4a

= x + y,

0,

u(x, 0) = x2, u(0, y) = y

u(x, 0) = 0, u(0, y) = y

Chapter 6 Applications of Difference Equations FORMULATION OF PROBLEMS INVOLVING DIFFERENCE EQUATIONS Various problems arising in mathematics, physics, engineering and other sciences can

be formulated by the use of difference equations.

In this chapter we shall consider

applications to such fields as mechanics, electricity, probability and others.

APPLICATIONS TO VIBRATING SYSTEMS There are many applications of difference equations to the mechanics of vibrating systems. As one example suppose that a string of negligible mass is stretched between two fixed points P and Q and is loaded at equal intervals h by N particles of equal mass m as indicated in Fig. 6-1. Suppose further that the particles are set into motion so that they vibrate transversely in a plane, i.e. in a direction perpendicular to PQ. Then the equation

of motion of the kth particle is, assuming that the vibrations are small and that no external forces are present, given by

m d2yk = dt2

h (y k-i

- 2y k + yk+i )

(1)

Fig. 6-1

where yk denotes the displacement of the kth particle from PQ and T is the tension in the string assumed constant. Assuming that yk = Ak cos (wt + a) so that the particles all vibrate with frequency 27r/w we obtain the difference equation

Ak+i +

(m2h - 2\t Ak + Ak-i =

0

(2)

T

This leads to a set of natural frequencies for the string given by n = 1, 2, ... , N

(3)

mh sin 2(N + 1) = Each of these corresponds to a particular mode of vibration sometimes called natural modes or principal modes. Any complex vibration is a combination of these modes and involves more than one of* the natural frequencies. fn

In a similar manner we can formulate and solve other problems involving vibrating systems. See Problems 6.36 and 6.39. 199

200

APPLICATIONS OF DIFFERENCE EQUATIONS

[CHAP. 6

APPLICATIONS TO ELECTRICAL NETWORKS In Fig. 6-2, a set. of N capacitors or condensers having capacitance or capacity C and N resistors having resistance R are connected by electrical wires as shown to a generator supplying voltage V. The problem is to find the current and voltage through any of the resistors. C

C

R

C

JR

C

C

R

C

R

R

R

Fig. 6-2

The formulation of this problem and other problems of a similar nature involving various combinations of capacitors, resistors or inductors is based on Kirchhoff's two laws. These laws can be stated as follows. Kirchhoff's laws. 1. The algebraic sum of the currents at any junction point of a network is zero. 2. The algebraic sum of the voltage drops around any closed loop is zero.

-

If I denotes the current, Q denotes the charge on a capacitor and t denotes time then we have:

Voltage drop across a resistor of resistance R = IR Voltage drop across a capacitor of capacitance C = Q Voltage drop across an inductor of inductance L = L dt APPLICATIONS TO BEAMS

Suppose that we have a continuous uniform beam which is simply supported

at points which are equidistant as indicated in Fig. 6-3. If there is no load

h

h

acting between the supports we can show

that the bending moments at these successive supports satisfy the equation Mk-1 + 4Mk + Mk+1 = 0

Fig. 6-3

(4)

sometimes called the three moment equation. The boundary conditions to be used together with this equation depend on the particular physical situation. See Problems 6.5 and 6.6. For cases where there is a load acting between supports equation (4) is modified by having a suitable term on the right side. APPLICATIONS TO COLLISIONS Suppose that two objects, such as spheres, have a collision. If we assume that the spheres are smooth the forces due to the impact are exerted along the common normal to the spheres, i.e. along the line passing through their centers.

APPLICATIONS OF DIFFERENCE EQUATIONS

CHAP. 6]

201

In solving problems involving collisions we make use of Newton's collision rule which can be stated as follows.

Newton's collision rule.

If V12 and v12 are the relative velocities of the first sphere with respect to the second before and after impact respectively then V12 = -Evil

(5)

where E, called the coefficient of restitution, is generally taken as

a constant between 0 and 1. If c = 0 the collision is called perfectly inelastic while if

1 the collision is called perfectly

elastic.

Another important principle which is used is the principle of conservation of momentum. The momentum of an object is defined as the mass multiplied by the velocity and the principle can be stated as follows. Conservation of momentum. The total momentum of a system before a collision is equal to the total momentum after the collision. Some applications which lead to difference equations are given in Problems 6.7, 6.8 and 6.53.

APPLICATIONS TO PROBABILITY

If we denote two events by Al and A2 then the probabilities of the events are indicated by P(A1) and P(A2) respectively. It is convenient to denote the event that either Al or A2 occurs or both occur by Al + A2. Thus P(A1 + A2) is the probability that at least one of the events A1, A2 occurs.

We denote the event that both Ai and A2 occur by A1A2 so that P(A1A2) is the probability that both Ai and A2 occur. We denote by P(A2 I A1) the probability that A2 occurs given that A, has occurred. This is often called the conditional probability.

An important result is that P(A1A2) = P(A1)P(A2I Al)

(6)

In words this states that the probability of both Ai and A2 occurring is the same as the probability that Al occurs multiplied by the probability that A2 occurs given that Al is known to have occurred. Now it may happen that P(A2 I Ai) = P(A2). In such case the fact that A, has occurred is irrelevant and we say that the events are independent. Then (6) becomes P(A1A2) = P(A1)P(A2)

A1, A2 independent

(7)

We also have P(A1 + A2) = P(A1) + P(A2) - P(AlA2)

(8)

i.e. the probability that at least one of the events occurs is equal to the probability that Ai occurs plus the probability that A2 occurs minus the probability that both occur. It may happen that both Al and A2 cannot occur simultaneously. In such case we say that Al and A2 are mutually exclusive events and P(A1A2) = 0. Then (8) becomes P(A1 + A2) = P(A1) + P(A2)

A,, A2 mutually exclusive

(9)

202

APPLICATIONS OF DIFFERENCE EQUATIONS

[CHAP. 6

The ideas presented above can be generalized to any number of events.

Many problems in probability lead to difference equations. for example.

See Problems 6.9-6.12

THE FIBONACCI NUMBERS The Fibonacci numbers Fk, k = 0, 1, 2, ..., are defined by the equation Fk = Fk-1 + Fk-2

(10)

Fo = 0, F1 = 1

where

(11)

It follows that the numbers are members of the sequence 0, 1, 1, 2, 3, 5, 8, 13, 21,

...

(12)

where each number after the second is the sum of the two preceding ones. Fibonacci numbers see Appendix G, page 238.

For a table of

We can show [see Problem 6.13] that the kth Fibonacci number in the sequence (12) is given by

Fk =

i

2V5-/k

L1

- \1

2-vF 5/k]

k = 0, 1, 2....

(18)

These numbers have many interesting and remarkable properties which are considered in the problems.

MISCELLANEOUS APPLICATIONS Difference equations can be used in solving various special problems. Among these are

Evaluation of integrals. See Problem 6.15. 2. Evaluation of determinants. See Problem 6.16. 3. Problems involving principal and interest. See Problems 6.17 and 6.18. 4. Numerical solution of differential equations. See Problems 6.19-6.24. 1.

Solved Problems APPLICATIONS TO VIBRATING SYSTEMS 6.1. A string of negligible mass is stretched between two points P and Q and is loaded at equal intervals h by N particles of equal mass m as indicated in Fig. 6-1, page 199. The particles are then set into motion so that they vibrate transversely. Assuming that the displacements are small compared with h find (a) the equations of motion and (b) natural frequencies of the system. (a) Let Yk denote the displacement of the kth particle from PQ. Let us call r the tension in the string which we shall take as constant.

CHAP. 6]

APPLICATIONS OF DIFFERENCE EQUATIONS

203

The vertical force acting on the kth particle due to the (k - 1)th particle is

-r sine = -

r(7!k

yk-1)

(1)

(vk - yk-1)2 T -h2

and since we assume that the displacements are small compared with h, we can replace the result in (1) by r(11k - yk-1) h to a high degree of approximation.

Similarly the vertical force on the kth particle due to the (k + 1)th particle a high degree of approximation by T(yk -hyk+1)

The total vertical force on the kth particle is thus T(yk - yk-1) h

-

r(yk - yk+1) h

r = h(Yk-l-2yk+yk+1)

(4)

Assuming that there are no other forces acting on the kth particle we see that by Newton's law the equation of motion is d2yk r (5) m dt2 = h (yk-1 - 2yk + yk+1) If there are other forces acting, such as gravity for example, we must take them into account.

(b) To find the natural frequencies assume that yk = Ak cos (wt + a)

(6)

i.e. assume that all the particles vibrate with the same frequency w/2?r. Then (5) becomes

-1riw2Ak =(Ak-1-2Ak+Ak+l)

(7)

which can be written as

h - 2) Ak + Ak-1 = 0

Ak+1 + C

(8)

Now we can consider the points P and Q as fixed particles whose displacements are zero. Thus we have

A0 = 0, AN+1 = 0

(9)

Letting Ak = rk in (8) we find that

r2+ rh2r+1 = 0 (2 mw2h2 r = 2 (2 -

(10)

T

or

4}

We can simplify (11) if we let

2 - mm2h r so that

= 2 cos e

(12)

r = cosy ± isine

In this case we find that the general solution is Ak = c1 cos ke + c2 sin ke

(14)

Using the first of conditions (9) we find c1 = 0 so that

Ak = c2 sin ke

(15)

APPLICATIONS OF DIFFERENCE EQUATIONS

204

[CHAP. 6

From the second of conditions (9) we must have

c2 sin (N + 1)e = 0 or since 02

(16)

0,

n = 1,2, ..

(17)

n7r 2CosN+1

(18)

Thus (12) yields '2h 2

from which W2 =

2T

mh

n7r -LT Cl - cos N+1) = mh

sln2

n7r

2(N + 1)

(19)

Denoting the N different values of 1,2 corresponding to n = 1, 2, 3, ..., N by ml, w21 w3, ..., wN it then follows that 4z wn = Nmh sin 2(Nnrr+ 1)

n = 1, 2,..., N

(20)

Thus the natural frequencies are fn

6.2.

r4z

1

=

sin

mh

na

2(N+1)

n=1,2,...,N

(21)

Find the displacement of the kth particle of the string in Problem 6.1. Let us consider the particular mode of vibration in which the frequency corresponds to ton, i.e. the nth mode of vibration. In this case we see from Problem 6.1 that the displacement of the kth particle is given by (1) yk,n = Ak,nCOs((0nt+an)

Ak,n = cn sin N + 1

where

(2)

the additional subscript n being used for w, a and c to emphasize that the result is true for the nth mode of vibration.

Now since the difference equation is linear, sums of solutions are also solutions so that we have

for the general displacement of the kth particle N Yk

::--

I c,, sin N n1

cos (tint + a.,,)

(8)

Since this satisfies the boundary conditions at the ends of the string, i.e. yo = 0, yN+1 = 0, it represents the required displacement. The 2N constants cn, an, n = 1, . . ., N, can be found if we specify the positions and velocities of the N particles at the time t = 0. 6.3.

Find the natural frequencies of the string of Problems 6.1 and 6.2 if the number of particles N becomes infinite while the total mass M and length L of the string remain constant. The total mass M and length L of the string are given by

M = Nm,

L = (N+1)h

(1)

Then the natural frequencies are given by

fn = or

In =

1 27r

nit 4TN(N + 1) sin 2(N + 1) ML

ML

N(N + 1) sin 2(N + 1)

(2)

(3)

CHAP. 6]

APPLICATIONS OF DIFFERENCE EQUATIONS

205

Now for any particular value of n we have lim

N-*

nr N(N + 1) sin 2(N+1)

= nr 2

[by using for example the fact that for small angles e, sin a is approximately equal to e]. the required natural frequencies are given by

fn=

n = 1, 2, .. .

ML

211

Thus

(4)

If we let p denote the density in this case, then

M

L = p or M = pL

(5)

and (4) becomes

fn =

2L

n=1,2,...

p

(6)

APPLICATIONS TO ELECTRICAL NETWORKS 6.4. An electrical network has the form shown in Fig. 6-4 where R1, R2 and R are given constant resistances and V is the voltage supplied by a generator. (a) Use Kirchhoff's laws to express the relationship among the various currents. (b) By solving the equations obtained in (a) find an expression for the current in any of the loops.

Fig. 6-4

(a) For the first loop, I1R1 + (I1- I2)R2 - V = 0

(1)

IkR1 + (Ik - Ik+1)R2 - (Ik-1- Ik)R2 = 0

(2)

For the kth loop, For the (N + 1)th loop, IN+1R1 - (IN-IN+1)R2 = 0

(3)

(b) Equation (2) can be written as

Ik+1 -

(2+)Ik + Ik-1

0

(4)

2

Letting Ik = rk in (4) it can be written as

/

r2 - (2 +

RR1

2

r+1=0

(5)

so that

r =

2 + R1/R2 _

(2 + R1/R2)2-4

2

R1

= 1 + 2R2

1

Ri

R2 + 4R2

206

APPLICATIONS OF DIFFERENCE EQUATIONS

[CHAP. 6

Then the general solution of (4) is Ik = clr1 + c2r2

(6)

2

where

rl

1 + RIZ +

RZ

r2 = 1 +

+ 812,

R,

From (1),

-

I z

Rl + R lz

2

2

2

4R 2

II(R! + R2) - V R,

(7)

(8)

From (3),

(R+R2)IN+I (9)

R2

Now if we put k = 2 and k = N in (6) we have 12 = c1r2 + c2r2 IN = c1r1 + c2r2

from which I2r2 - INr2 cl

so that In

=

2 N' rlr2 - r2r1

-

( I2r2 - INr2 \

INri - I2ri

2N

2NrIr2

2N r2r1

e2

rn1

2 N - r2rl 2 N rlr2

+

2 Ni

(r1INrr2

- 12ri 2N - r2r /

rn2

where rl, r2 are given in (7) and I2, IN are given by (10) and (11).

APPLICATIONS TO BEAMS 6.5. Derive the equation of three moments (4) on page 200. Assume that the spacing between supports is h and consider three successive supports as shown in Fig. 6-5. Choosing origin 0 at the center support and letting x be the

_zS

Mn

n-1

distance from this origin we find that the bending moment at x is given by

Fig. 6-5

Mn + (Mn - M.n-1) h M(x)

Mn+ I

=

x _0 (1)

x

Mn + (Mn+1- Mn) h

0x-5- h

where Mn _ 1, Mn, Mn+ I are the bending moments at the (n - 1)th, nth and (n + 1)th supports.

Now we know from the theory of beams that the curve of deflection of a beam is y(x) where

YIy" = M(x)

(2)

and YI, called the flexural rigidity of the beam, is assumed to be constant. Here Y is Young's modulus of elasticity and I is the moment of inertia about an axis through the centroid. Integrating (2) using (1) we find

Mnx + (Mn - M.-,) 2h + c1 YIy'

=

-h.x_0 (3)

2

Mnx + (Mn + 1 - M11) 2h + c2

Now since y' must be continuous at x = 0, it follows that cl

h

c2 = c.

CHAP. 6]

APPLICATIONS OF DIFFERENCE EQUATIONS

207

Integrating (3) we have similarly Mnx2

-

YIy

2

x3

M2x2

+

cx

-h

x

0

Mn) 63 -1- cx

0

x

h

(Mn - Mn-1) 6h

(4)

+- (Mn+

Since the deflection y is zero at x = -h, 0, h we find from (4) Mnh2

(Mn-Mn_1)

2

h2

- ch

6

Mnh2 h2 +- (Mn+1 _ Mn) 6 2

0

(5)

+ Ch = 0

(6)

Addition of these equations leads to the result Mn-1 + 4Mn + Mn+1 = 0

(7)

as required.

6.6.

Suppose that the beam of page 200 is of infinite extent on the right and that at the distance h to the left of the first support there is a load W which acts on the beam [see Fig. 6-6]. Prove that the bending moment at the nth support is given by Mn = (-1)n+1 Wh(2 - ,,,F3)n W h

I Fig. 6-6

The bending moment at the distance It to the left of the first support is given by

MO = -Wh

(1)

Also at infinity the bending moment should be zero so that

lim Mn = 0

(2)

Mn-1+ 4Mn+ Mn+1 = 0

(3)

Now from the difference equation

we have on making the substitution Mn = rn

rn-1-I-4rn+rn+1 = 0

r=

-4 ---

or

16 - 4

r2-i-4r+1 = 0

(4)

- -2 +

2

Then the general solution of (3) is

Mn =

c1(-2 -f- V )n + c2(-2 -

Now since -2 + - is a number between -1 and 0 while -2 lows that condition (2) will be satisfied if and only if c2 = 0.

)n

(5)

is approximately -3.732, it folThen the solution (5) becomes

Mn = c1(-2 +f 13-)-

(6)

Also from (1) we have c1 = --Wh so that

Mn = -Wh(-2 +

)n = (-1)n+1 Wh(2 - V )n

(7)

APPLICATIONS OF DIFFERENCE EQUATIONS

208

[CHAP. 6

APPLICATIONS TO COLLISIONS 6.7. Two billiard balls of equal mass m lie on a billiards table with their line of centers perpendicular to two sides of the table. One of the balls is hit so that it travels with constant speed S to the other ball. Set up equations for the speeds of the balls before the kth impact occurs assuming that the table is smooth and the sides are perfectly elastic. Denote by Uk and Vk the speeds of the balls before the kth impact occurs. that Uk refers to the ball which is hit initially we have

Then assuming

U1 = S, Vl = 0 (1) Since the sides are perfectly elastic, the speeds of the balls after the kth impact are Uk+1 and Vk+1 respectively. Then the total momentum before and after the kth impact are given by mUk +mVk and mUk+1- mVk+1 respectively. Thus by the conservation of momentum we have

mUk + mVk = mUk+1 - mVk+1

Uk + Vk = Uk+1 - Vk+1

or

(2)

In addition we have by Newton's collision rule

Uk+1 + Vk+1 = -e(Uk - Vk) where e is the coefficient of restitution between the two billiard balls.

(3)

The required equations are given by (2) and (3) which are to be solved subject to the conditions (1).

6.8.

Determine the speeds of the balls of Problem 6.7 before the kth impact. Writing the equations (2) and (3) of Problem 6.7 in terms of the shifting operator E we obtain

(E-1)Uk-(E+1)Vk = 0

(1)

(E + e) Uk + (E - e)Vk = 0

(2)

Then operating on (1) with E - e, on (2) with E + 1 and adding we obtain (E2 + e) Uk = 0

(3)

Letting Uk = rk, (3) yields

r2 + e = 0,

r = ±iYC =

so that the general solution of (3) is

Uk = ek/2(Aeaki/2 + Be-arks/2) =

k/2

(e1 cos

2+

c2 sin 2

(4)

By eliminating Vk+1 between equations (2) and (3) of Problem 6.7 we find

(e+1)Vk = 2Uk+1+(e-1)Uk

(5)

Then using (4) Vk

ek/2 r Tk rk L{(e - 1)c2 - 2 c1} sin 2 + {(e - 1)cl + 2f c2} cos 2

(6)

+1

From conditions (1) of Problem 6.7 we have (Cl

COS 7 +

=S

c2sin2\

(e-1)C2-2Ve- c1 = 0

so that

(e -1)S Cl

2e

c2

=

(7)

(8)

S e

(9)

CHAP. 6]

APPLICATIONS OF DIFFERENCE EQUATIONS

209

Using these in (4) and (6) we obtain the required results

Uk = .-Se(k-2)/2 [(e - 1) cos

k + 2 I sin 2

(10)

Vk = .Se(k-2)/2(e + 1) cos 2

(11)

APPLICATIONS TO PROBABILITY 6.9. A and B play a game. In each step of the game A can win a penny from B with probability p while B can win a penny from A with probability q, where p and q are positive constants such that p + q = 1. We assume that no tie can occur. Sup-

pose that the game is started with A and B having a and b pennies respectively, the total being a + b = N, and that the game ends when one or the other has M pennies. Formulate (a) a difference equation and (b) boundary conditions for the probability that A wins the game. (a)

Let uk be the probability that A wins the game when he has k pennies. one of the following two mutually exclusive events occurs: 1. On the next step A wins a penny and then wins the game.

Now A will win if

2. On the next step A loses a penny and then wins the game. The probability of the first event occurring is given by puk+1

(1)

since if A wins on the next step [for which the probability is p] he has k + 1 pennies and can then win the game with probability uk+1

Similarly the probability of the second event occurring is given by quk-1

(2)

since if A loses on the next step [for which the probability is q] he has k - 1 pennies and can then win the game with probability uk_1. It follows that the probability of A winning the game is the sum of the probabilities

(1) and (2), i.e.

Uk = puk+1 + quk-1

(3)

(b) To determine boundary conditions for Uk we note that A wins the game when he has M pennies, i.e.

U"

(4)

Similarly A loses the game when B has M pennies and thus A has N - M pennies, i.e.

UN-M = 0 6.10.

(5)

Obtain the probability that A wins the game of Problem 6.9 if (a) p q, (b) p = q =. We must solve the difference equation (3) of Problem 6.9 subject to the conditions (4) and (5). To do this we proceed according to the usual methods of Chapter 5 by letting Uk = rk in equation (3) of Problem 6.9. Then we obtain

pr2-r+q = 0

from which

r = (a)

1± 1-4pq _ 2p

1 -!

(1)

1-4p+4p2 _ 1±:(1-2p) 2p

2p

=

1, qlp

(2)

If p # q, the roots of (1) are 1 and q/p so that Uk = C1 + c2(q/p)k

(3)

210

APPLICATIONS OF DIFFERENCE EQUATIONS

[CHAP. 6

Using the condition (4) of Problem 6.9, i.e. um = 1, we have 1 = cl + c2(q/p)M

(4)

Similarly using condition (5) of Problem 6.9, i.e. uN_M = 0, we have

0 = cl +

(5)

c2(q/p)N-M

Then solving (4) and (5) simultaneously we obtain -(q/p)N-M

c

I

-

(q/p)M -

e2

(q/p)N-M '

(6)

(q/p)M -

(q/p)N-M

and so from (3) we have (q/p)k - (q/p)N_M (q/p)M -

Uk

(7)

(q/p)N-M

Since A starts with k = a pennies, the probability that he will win the game is Ua

(q/p)a -

(q/p)N-M

(8)

(q/p)N_M

(q/p)M -

while the probability that he loses the game is

1 - ua = (b)

(q/p)M - (q/p)a (q/p)M -

(9)

(q/p)N-M

If p = q = -k, the roots of (1) are 1, 1, i.e. they are repeated. In this case the solution of equation (3) of Problem 6.9 is Uk = cg + c4k

(10)

Then using um = 1, uN_M = 0, we obtain c3 + c4(N - M) = 0

cs + c4M = 1,

M-N °s = 2M-N' Substituting in (10)

_

uk =

_

1

14 - 2M-N

(11) (12)

M-N+k 2M-N

Thus the probability that A wins the game is

_

ua =

M-N+a

(14)

2M-N

while the probability that he loses the game is

M-a 1-u6 = 2M-N 6.11.

(15)

Suppose that in Problem 6.9 the game is won when either A or B wins all the money,

i.e. M = N. Find the probability of A being "ruined", i.e. losing all his money if (a) p (a)

q, (b) p = q = 1/2.

Putting M = N in equation (9) of Problem 6.10 we find for the probability of A being ruined

1 -ua (b)

(q/p)N - (q/p)a (q/p)N - 1

(q/p)a+b - (q/p)a (q/p)a+b - 1

(1)

Putting M = N in equation (14) of Problem 6.10 we find for the probability of A being ruined b 1-ua = N-a = a+b N

2

CHAP. 6]

APPLICATIONS OF DIFFERENCE EQUATIONS

211

The case where p = q = 1/2 is sometimes called a "fair game" since A and B then have equal chances of getting a point at each step. From (2) we note that even in a fair game the chance of A being ruined is proportional to the initial capital of B. It follows as a consequence that if B has many more pennies than A initially then A has a very great chance of being ruined.

This is illustrated by events which take place at many gambling casinos where a person is competing in a game [which often is not even fair] against an "infinitely rich" adversary.

6.12. A and B play a game. A needs k points in order to win the game while B needs m points to win. If the probability of A getting one point is p while the probability of B getting one point is q, where p and q are positive constants such that p + q = 1, find the probability that A wins the game. Let uk,,,, denote the probability that A wins the game. The event that A wins the game can be realized if one of the following two possible mutually exclusive events occurs:

1. A wins the next point and then wins the game. 2. A loses the next point and then wins the game. The probability of the first event occurring is given by puk-1,m

(1)

since if A wins the next point [for which the probability is p] he needs only k - 1 points to win

the game [for which the probability is uk_1,m].

Similarly the probability of the second event occurring is quk,m-1

(2)

since if A loses the next point [for which the probability is q] he still needs k points to win the game while B needs only m-1 points to win [for which the probability is uk,m_1]. It follows that the probability of A winning the game is the sum of the probabilities (1) and (2), i.e.

uk,m = puk-1,m + quk,m-1

(3)

To determine boundary conditions for uk,m we note that if k = 0 while m > 0, i.e. if A needs no more points to win, then he has already won. Thus

uo,m = 1 Also if k > 0 while m = 0, then B has already won so that uk, o = 0

k>0

(5)

The solution of the difference equation (3) subject to conditions (4) and (5) has already been found [see Problems 5.53 and 5.54, page 186] and the result is

uk,m = pk [1 + kq + k(k i 1) q2 + ... +

k(k + 1).(..(k jm - 2) qm-1

FIBONACCI NUMBERS 6.13.

Prove that the kth Fibonacci number is given by

Fk =

k

1

CV1 +

1

kJ

We must solve the difference equation

subject to the conditions

Fk = Fk-1 + Fk-2

(1)

Fo = 0,

(2)

F1 = 1

212

APPLICATIONS OF DIFFERENCE EQUATIONS

[CHAP. 6

Letting Fk = rk in (1) we obtain the auxiliary equation

r2-r-1 = 0

r=

or

-L

(3)

2

Then the general solution is

+C2 (l_V\k 2

Fk - ci/1+ 2 \\

(4)

From the conditions (2) we find

(1+ V5 )C1+(1-f

C1+C2 = 0, Solving (5) we find

1

Cl =

)c2 =

2

(5)

1

C2 =

,

(6)

xF5

XF5

. -and so from (4) we have V5- )k

Fk = 2

L

6.14.

Prove that

l

-

(1 - / )k] (7)

J

2

lim Fk+1 = 1 +

k-..

Fk

2

We have using Problem 6.13

[(1+V'5)./2}k+1 - [(1- f )/2]k+1

Fk+1 Fk

[(1 + NF5)/2]k _ [(1 _.)/2]k

[(1 + NF5)/2]k+1{1 - [(1- /5)/(1 + I )]k+1) [(1 + f )/2]k{1 - [(1- NF5 )/(1 + NF5)]kl

=

J)/(1 + f )]k+1) [(1 +

)/2]

{1 - [(1- V5-)/(1 + XF5)1k)

Then since 1(1- f )/(1 + V) I < 1 we have + F5

lim Fk+1

k-... Fk

2

This limit, which is equal to 1.618033988... , is often called the golden mean. It is supposed to represent the ratio of the sides of a rectangle which is "most pleasing" to the eye.

MISCELLANEOUS APPLICATIONS 6.15.

Evaluate the integral Sk

=

(' cos k9 - cos ka

f ,Jo

cos 0 - cos of

de

Let a, b, c be arbitrary constants and consider

k+1 + bSk

C`Sk-1 =

a[cos (k + 1)e - Cos(k + 1)a]

O

COO B - COs a

do

" b[cos ke - cos ka] 0

Cos e - Cos a

+

7 c[cos U

do

(k -1)e - cos (k -1)a] cos O - cos a

do

("r [(a + c) cos e + b] cos ke + [(c - a) sin e] sin ke do Jo Cos 0 - Cos a

CHAP. 6]

APPLICATIONS OF DIFFERENCE EQUATIONS

213

If we now choose a, b and c so that

c = a, the last integral becomes

b = -2a cos a

f cos ko de = 0 Ir

2a

if k = 1, 2, 3, .. .

0

It thus follows that Sk+1 - 2(cos a)Sk + Sk-1 = 0

k = 1, 2, 3,

...

(1)

From the definition of Sk we have

So = 0,

S1 = Q

(2)

We must thus solve the difference equation (1) subject to the conditions (2).

According to the usual methods for solving difference equations we let Sk = rk in (1) and

obtain

rk+1 - 2(cos a)rk + rk-1 = 0

r2 - 2(cos a)r + 1 = 0

or

so that

r=

2cosa!

2

4costa-4

= cos a --* i sin a

Then by de Moivre's theorem,

rk = cos ka --* i sin ka so that the general solution is

Sk = Cl cos ka + c2 sin ka

(S)

Using the conditions (2) we have

c1 = 0,

c1 cos a + c2 sin a = a

(4)

so that c2 = 7r/sin a and thus zr sin ka

Sk =

sin a

(5)

which is the required value of the integral.

6.16.

Find the value of the determinant b

a

0

0

0

a

b

a

0

0

0

a

b

a

0

0

0

0

0

0

0

0

0

0

0

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

0

0

0

0

0

0

0

0

0

a

b

a

D

a

b

Denote the value of the kth order determinant by Ak. Then by the elementary rules for evaluating determinants using the elements of the first row we have

Ak = bAk-1 - a2Ak-2 or Ak - bAk-1 + a2Ak-2 = 0 Letting Ak = rk we obtain rk - brk-1 + a2rk-2 = 0 or r2 - br + a2 = 0 from which r

bi '-

b2-4a2 2

(1)

(2)

214

APPLICATIONS OF DIFFERENCE EQUATIONS

[CHAP. 6

It is convenient to let

b = 2a cos e

(8)

r = a cos e ± ai sin 6

(4)

so that (2) becomes

Thus since rk = ak (cos ke ± i sin ke) we can write the general solution as Ak = ak[cl cos ke + c2 sin ke]

(5)

Now if k = 1, the determinant of order 1 has the value A 1 = b, while if k = 2, the determinant of order 2 has the value

=

A2

b

al

a

b

Putting k = 2 in (5) we find that

A2-bA1+a2A0 = 0 from which we see on putting Al = b, A2 = b2 - a2 that A0 = 1. (1) subject to the conditions

It follows that we can solve

Al=b

A0 = 1,

(8)

These values together with (5) lead to

C1 = 1,

02

=

b - a cos a a sln B

cos e

Mn e

(7)

using (3) so that (5) becomes Ak

=

ak [cos ke + sin ke cos e 1

=

sin B

ak sin (k + 1)e sin e

(8)

Using the fact that 6 = cos-1 (b/2a)

(9)

we can write the value of the determinant in the form A k

-

Ak =

or

The above results hold for

Jbi < 21a1.

ak sin [(k + 1) cos-1 (b/2a)] sin [cos-1 (b/2a)]

(10)

2ak+1 sin [(k+ 1) cos-1 (b/2a)] 4a2 - b2

In case IbI > 21a1 we must replace cos-1 by cosh-1.

For the case Jbl = 21al see Problem 6.97.

6.17. A man borrows an amount of money A. He is to pay back the loan in equal sums S to be paid periodically [such as every month, every 3 months, etc.]. This payment includes one part which is interest on the loan and the other part is used to reduce the principal which is outstanding. If the interest rate is i per payment period, (a) set up a difference equation for the principal Pk which is outstanding after the kth payment and (b) solve for Pk. (a)

Since the principal outstanding after the kth payment is Pk, it follows that the principal out-

standing after the (k+1)st payment is Pk+1 Now this last principal to be paid is the principal outstanding from the kth payment plus the interest due on this payment for the period minus the sum S paid for this period.

Thus

Pk+1 = Pk + 1Pk - S or

(1)

Pk+1 - (1 + i)Pk = -S

The outstanding principal when k = 0 is the total debt A so that we have PO

=A

(2)

CHAP. 6] (b)

APPLICATIONS OF DIFFERENCE EQUATIONS

215

Solving the difference equation (1) we find Pk = C(1 + 2)k + S/i .

and using condition (2) we find c = A - S/i so that

Pk = (A \ - S (1 +,)k + 6.18.

+ 2)k - 1 A(1 + i)k - S r(1 LL

i

2

2

(3)

In Problem 6.17 determine what sum S is to be paid back each period so as to pay back the debt in exactly n payments. We must determine S so that Pf, = 0. From Problem 6.17 this is seen to be equivalent to A(1 + 2)n - S

jj

L(1 + i)n -

=0

S = A 1-f - (1 i+ i)-n]

or

(1)

(2)

The process of paying back a debt according to (2) and Problem 6.17 is often called amortization. The factor a-n-, i

_ 1-(1+i)-n

(3) i is often called the amortization factor and is tabulated for various values of n and i. In terms of this we can write (2) as

S=

A (4)

a> 0 is given by N

1/k

nkrr cos (w,,t + an) _ = I n=1 c,,, sin N+1

N

Y, (an cos wnt + b,, sin wnt) sin

n=1

N+1

where an, b or cn and an are determined from N

an = c, cos an bn = 6.34.

-cn sin an = N-2 +1

N k=

7!k(0) sin

=1

k7r

N71 + 1

(a) Show that if the (N + 1)th mass of the string of Problem 6.1 is free while the first mass is fixed, then the frequency of the lowest mode of vibration is 11 = (b)

(a)

mh sin 2N71 1

What are the frequencies of the higher modes of vibration?

[Hint. 6.35.

= N Z-- 1 I yk(0) sin Nnk7r +1 k=1

If the force on the (N + 1)th mass is zero then YEN =

Give a physical interpretation of the boundary conditions aN+lyN+1 = 0

in Problem 6.1.

(b) Obtain the natural frequencies of the system for this case.

a0y0 + ajh i = 0,

aNYN +

CHAP. 6] 6.36.

APPLICATIONS OF DIFFERENCE EQUATIONS

223

An axis carries N equally spaced discs which are capable of torsional vibrations relative to each other [see Fig. 6-13]. If we denote by ek the angular displacement of the kth disc relative to the first disc show that d2ek

o(9k+1-20k+Bk-1)

I dt2

where I is the constant moment of inertia of each disc about the axis and a is the torsion constant.

Fig. 6-13 6.37.

(a) Using appropriate boundary conditions obtain the natural frequencies for the system of Problem 6.36. (b)

Explain how you would find the motion for the kth disc by knowing the state of the system

(c)

Compare with results for the vibrating system of Problem 6.1.

at time t=0.

6.38.

Work Problem 6.37 if the first disc is fixed while the last is free.

6.39.

A frictionless horizontal table AB has on it N identical objects of mass m connected by identical springs having spring constant K as indicated in Fig. 6-14. The system is set into vibration by moving one or more masses along the table and then releasing them.

k h VUVU '

m

m

k

' VVVV "°

I" VVVV '°'l

B

A Fig. 6-14

(a) Letting Xk denote the displacement of the kth object, set up a difference equation. (b) (c) 6.40.

Find the natural frequencies of the system. Describe the analogies with the vibrating string of Problem 6.1.

Suppose that the string of Problem 6.1 with the attached particles is rotated about the axis with uniform angular speed R. (a) Show that if Yk is the displacement of the kth particle from the axis, then Yk+1 e 2 C1 -

2MS2 2

Th Yk + Yk-1 = 0

(b) Show that the critical speeds in case m22h/4r < 1 are given by

Stn =

mh sin 2(N + 1)

n = 1, 2, ..., N

and discuss the physical significance of these. 6.41.

Discuss the limiting case of Problem 6.40 if N -- o in such a way that the total mass and length remain finite.

6.42.

Explain how you would find a solution to the partial differential equation in Problem 6.30 subject to appropriate boundary conditions by applying a limiting procedure to the results obtained in Problem 6.2.

APPLICATIONS OF DIFFERENCE EQUATIONS

224

[CHAP. 6

APPLICATIONS TO ELECTRICAL NETWORKS 6.43.

An electrical network consists of N loops with capacitors having constant capacitance C1, C2 in each loop as indicated in Fig. 6-15. An alternating voltage given by VN = K sin at where K and , are constants is applied across ANBN. If it is required that the voltage across AOB0 be Vo = 0, show that the voltage Vk across AkBk is given by

Vk _ K sink µk

sinh µN sin wt

where

cosh-1

µ

C'

41

1 +

C 2C

rl

CI

BN-I

AN-1

AN

Bo

11

Ao

Fig. 6-15 6.44.

Work Problem 6.43 if the capacitors C, and C2 are replaced by inductors with constant inductance LI, L2-

6.45.

Work Problem 6.43 if the capacitors Cl are replaced by resistors having constant resistance R1.

6.46.

Can Problem 6.43 be worked if it is required that the voltage across A,B0 be

Vo > 0?

Explain.

APPLICATIONS TO BEAMS 6.47.

Suppose that in Problem 6.6 the beam is of finite length Nh. Show that the bending moment at the nth support is given by rr1+1-N - r2+1-N1 Mn = -WhL rl-N - rl-N

where ri=-2+f, r2=-2-f.

1

6.48.

Show that as N -> - the result in Problem 6.47 reduces to that of Problem 6.6.

6.49.

In Fig. 6-16 AB represents a continuous uniform beam on which there is a triangular load distribution where the total load on the beam is W. Suppose that the beam is simply supported at the N + 1 points PO, Pi, ... , PN at distances h apart. (a) Show that the three moment equation can be written as -Whk Mk_i + 4Mk + Mk+1 N2 where the bending moments at the ends A and B are zero, i.e. Mo = 0, MN = 0. the bending moment at the kth support is given by Mk

=

WhL 6N (-1)

(2+f

+ f )N - (2 - f )N

llllllllllllllllllllllll

0 Fig. 6-16 6.50.

Does lim Mk exist in Problem 6.49?

N-.

Explain.

kl

N

(b) Show that

CHAP. 6]

APPLICATIONS OF DIFFERENCE EQUATIONS

225

APPLICATIONS TO COLLISIONS 6.51.

Show that the speeds of the balls in Problem 6.7 follow a periodic pattern and determine this period.

6.52.

(a) Work Problem 6.7 if the two masses are given by m1 and m2 where m1 m2. (b) Determine the values of ml/m2 for which the speeds will be periodic and determine the periods in such cases. (c) Discuss the motion for the case m1 = 2M2-

6.53.

A billiard ball is at rest on a horizontal frictionless billiards table which is assumed to have the shape of a square of side l and whose sides are assumed to be perfectly elastic. The ball is hit so that it travels with constant speed S and so that the direction in which it travels makes the angle e with the side. Determine the successive points at which the ball hits the sides.

6.54.

What additional complexities arise in Problem 6.53 if the billiard table has the shape of a rectangle?

Explain.

APPLICATIONS TO PROBABILITY 6.55.

Suppose that in Problem 6.9 we let uk,,,n be the probability of A winning the game when he has k pennies while B has m pennies.

(a) Show that the difference equation and boundary conditions for uk,m are given by uk,m = PUk+1, m-1 + quk-1, m+1

UM, N-M = 1, (b)

6.56.

UN-M, M = 0

Solve the difference equation in (a) subject to the conditions and reconcile your results with those given in Problem 6.9. [Hint. See Problem 5.59, page 189.]

Suppose that in a single toss of a coin the probabilities of heads or tails are given by p and q = 1 - p respectively. Let uk,,n denote the probability that exactly k heads will appear in n tosses of the coin.

(a) Show that

uk + 1, n + 1 = PUk,n + quk + 1, n 1

where

0

k=0 k>0

(b) By solving the problem in (a) In pkgn-k k

This is called Bernoulli's probability formula for repeated trials. 6.57.

Suppose that A and B play a game as in Problem 6.9 and that A starts out with 10 pennies (a) Show that if p = q = 1/2, i.e. the game is fair, then the probability that A will lose all of his money, i.e. A will be "ruined", is 0.9. (b) Show that if p = 0.4, q = 0.6 then the probability that A will be "ruined" is 0.983. (c) Discuss the significance of the result in (b) if A represents a "gambler" and B represents the "bank" at a while B starts out with 90 pennies.

gambling casino. "gambler"? 6.58.

(d) What is the significance of the result in (b) if A is the "bank" and B is the

Suppose that in Problem 6.9 p > q, i.e. the probability of A's winning is greater than that of B. Show that if B has much more capital than A then the probability of A escaping "ruin" if he starts out with capital a is approximately 1 - (q/p)a

What is this probability if p = 0.6, q = 0.4 and (a) a = 5, (b) a = 100?

226

APPLICATIONS OF DIFFERENCE EQUATIONS

6.59.

Let Dk denote the expected duration of the game in Problem 6.9 when the capital of A is k. (a) Show that Dk satisfies the equation

Dk = pDk+l + qDk-1 + 1,

(b) Show that if p (c)

Dk =

q,

k

q-p

-

Do = 0,

[CHAP. 6

Da+b = 0

a + b r (q/p)k

q- p (q/p)a+b Show that if p = q = 1/2, Dk = k(a + b - k) and thus that if A and B start out with

initial capitals a and b respectively the expected duration is ab. 6.60.

Work Problems 6.9-6.11 in case a tie can occur at any stage with constant probability e > 0 so

that p+q+e=1.

THE FIBONACCI NUMBERS 6.61.

Let Fk denote the kth Fibonacci number and let Rk = Fk/Fk+1

(a) Show that (b) Show that

Rk =

1 1

Rk_1 ,

Ro

= 0. Rk

=

i.e. a continued fraction with k divisions. 6.62.

(a) Solve the difference equation of part (a) in Problem 6.61 for Rk.

-1 and thus obtain the limiting value of the infinite continued (b) Show that slim Rk = 2 fraction in part (b) of Problem 6.61. 6.63.

If Fk denotes the kth Fibonacci number prove that z Fk+1Fk-1-F'k = (-1)k

6.64.

Generalize Problem 6.63 by showing that

Fk+mFk-m - Fk = 6.65.

k=1,2,3,...

k = m, m + 1, m + 2, .. .

(-1)k+m-1F2m

A planted seed produces a flower with one seed at the end of the first year and a flower with two seeds at the end of two years and each year thereafter. Assuming that each seed is planted as soon as it is produced show that the number of flowers at the end of k years is the Fibonacci number Fk.

6.66.

Find the number of flowers at the end of k years in Problem 6.65 if a planted seed produces a flower with four seeds at the end of the first year and a flower with ten seeds at the end of two years and each year thereafter.

6.67.

Generalize Problems 6.65 and 6.66 by replacing "four seeds" and "ten seeds" by "a seeds" and "b seeds" respectively. What restriction if any must there be on a and b?

6.68.

Find the sum of the series

6.694

Show that

Fk

'

Fk+z where Fk is the kth Fibonacci number. k=0 Fk+1Fk+9

= 2k 1[(1)

+

(k)

5+

(6) 52 +

]

CHAP. 6]

APPLICATIONS OF DIFFERENCE EQUATIONS

227

MISCELLANEOUS PROBLEMS 6.70.

In Pascal's triangle [see Fig. 6-17] each number is obtained by taking the sum of the numbers which lie immediately to the left and right of it in the preceding row. Thus for example in the sixth row of Fig. 6-17 the number 10 is the sum of the numbers 6 and 4 of the fifth row. 1 1

2

1

3

1

4

1

1

6

1

3

1

6

10

5

1

1

4

10 20

15

1

1

5

15

6

1

Fig. 6-17

(a) Show that if uk,,n denotes the kth number in the mth row, then nk+l,m + uk,m

uk+1, m+1 (b)

where

uk,0 =

r1 k> O

0 k=0

Solve the difference equation of (a) subject to the given conditions and thus show that

uk,m =

(k)

i.e. the numbers represent the binomial coefficients. 6.71.

The Chebyshev polynomials Tk(x) are defined by the equation

Tk+l (x) - 2xTk(x) + Tk-1(x) = 0

To(x) = 1,

where

T1(x) = x

(a) Obtain the Chebyshev polynomials T2(x), T3(x), T4(x). (b) By solving the difference equation subject to the boundary conditions show that

Tk(x) = cos (k cos-1 x) 6.72.

The Bessel function Jk(x) of order k can be defined by (ke - x sin e) do Jk(x) = 1 f rcos 0

Show that Jk(x) satisfies the difference equation xJk+1(x) - 2kJk(x) + Jk-1(x) = 0 6.73.

Show that the Bessel functions of Problem 6.72 have the property that

Jk(x) = 4[Jk-1(x) - Jk+l(x)] 6.74.

The Laplace transform of a function F(x) is denoted by

.C{F(x)} = f(s) = f oe-s: F(x) dx 0

6.75.

(a)

Show that .C{J0(x)} =

(b)

Show that .C{Jk(x)} _

1

82+1 [

s2 + 1 - s]k s2+1

Suppose that a principal P is deposited in a bank and kept there. If the interest rate per period is given by i, (a) set up a difference equation for the total amount Ak in the bank after k periods and (b) find Ak.

APPLICATIONS OF DIFFERENCE EQUATIONS

228 6.76.

[CHAP. 6

A man deposits a sum S in a bank at equal periods. If the interest rate per period is i, (a) set up

a difference equation for the amount A. in the bank after n periods and (b) show that $r(1+iTn-1l

An =

= Ssiij t

Such a set of equal periodic payments is called an annuity.

....

6.77.

Work Problem 6.76 if the successive deposits are given by S, S + a, S + 2a, S + 3a, the significance of the cases (a) a > 0, (b) a < 0.

6.78.

In a particular town the population increases at a rate a% per year [as measured at the end of each year] while there is a loss of /3 individuals per year. Show that the population after k years is given by l0 POa/il\1+lUU)k+

Pk =

Discuss

lUa /3

where P0 is the initial population. 6.79.

Does the result of Problem 6.78 hold if (a) a < 0, /3 < 0, Interpret each of the cases.

6.80.

Let Sk denote the sum of the kth powers of the n roots of the equation

(b) a > 0, e < 0,

(c) a < 0, /3 > 0?

rn + alrn-1 + a2rn-2 + ... + an = 0 (a) Show that S1 + a1 = 0, S2 + a1S2 + 2a2 = 0,

...,

Sn + a1Sn-1 + ... + an-1S1 + nan = 0

(b) Show that if k > n Sk + alSk-1 + a2Sk-2 + ... + anSk_n = 0

6.81.

Obtain the generating function for the sequence Sk of Problem 6.80.

6.82.

A box contains n identical slips of paper marked with the numbers 1, 2, ..., n respectively. The slips are chosen at random from the box. We say that there is a coincidence if the number M is picked at the Mth drawing. Let uk denote the number of permutations of k numbers in which there are no coincidences. (a) Prove that the number of permutations in which there is 1 coincidence is (k) uk_1. (b) Prove that the number of permutations in which there are m coincidences

is (m) Uk_,n where m = 0, 1, ..., k. 6.83.

(a)

kk k

Prove that Uk in Problem 6.82 satisfies the equation

uk +

(k)

+

()Uk_2 + ... +

(k1)ui + ()0

-' k!'

u0=1

(b) By solving the difference equation in (a) prove that

un = n. Lit (c)

gt+4t

+ nt

Show that the probability of getting no coincidences in the drawing of the n slips from the box of Problem 6.82 is given by 1

_ 1+

2!

3!

(-1)n n!

(d) Show that for large values of n the probability in (c) is very nearly

1-1e (e)

= 0.63212...

Discuss how large n must be taken in (d) so that the probability differs from 1- 1/e by less than 0.01.

APPLICATIONS OF DIFFERENCE EQUATIONS

CHAP. 6] 6.84.

229

An inefficient secretary puts n letters at random into n differently addressed envelopes. Show that the probability of getting at least one letter into its correct envelope is given by 1 - 1 + 1 - 1 + ... _!. (-1)n+l 2!

3!

if

n!

and thus show that for large n this probability is about 1/e = 0.36787... . Can you explain from a qualitative point of view why this probability does not change much with an increase of n as for example from 100 to 200? 6.85.

Find the probability that at least two people present at a meeting will have their birth dates the same [although with different ages].

6.86.

A group of players plays a game in which every time a player wins he gets m times the amount of his stake but if he loses he gets to play again. Let uk denote the stake which a player must make in the kth game so that he gets back not only all that he has lost previously but also a profit P. k (a) Show that Uk satisfies the sum equation

muk = P + M Up p=1

(b)

Show that the sum equation of (a) can be written as the difference equation M uk = 0 uk+1 m - 1

(c)

Using appropriate boundary conditions show that uk =

6.87.

6.88.

State some limitations of the "sure profit" deal in Problem 6.86 illustrating the results by taking

m=2.

A box contains slips of paper on which are written the numbers 1, 2, 3, .. ., n [compare Problem The slips are drawn one at a time from the box at random and are not replaced. Let uk,,n denote the number of permutations [out of the total of n! permutations] in which there is no coincidence in k given drawings. (a) Show that Uk,O = k! uk,m = uk-1,m - uk-1,m-1, (b) Show that the solution to the difference equation of part (a) is

6.821.

uk,m =

()m! -

C(m-1)! +

((m-2)!

+ (-1)k

0!

where 0 ! = 1. Show that the probability that there will be no coincidence in the k drawings is uk,,,,/m!. (d) Show that for k = m the probability obtained in (c) reduces to that obtained in Problem 6.82. 6.89.

6.90.

A and B play a game with a deck of n cards which are faced downwards. A asks B to name a card. Then A shows B the card and all the cards above it in the deck and all these cards are removed from the deck. A then asks B to name another card and the process is repeated over and over again until there are no cards left. If B does not name any of the top cards then A wins the game, otherwise B wins the game. Find the probability that A wins the game.

If the deck of Problem 6.89 has 52 cards, determine whether the game is fair if B offers

A

2 to 1 odds. 6.91.

6.92.

A sequence has its first two terms given by 0 and 1 respectively. Each of the later terms is obtained by taking the arithmetic mean of the previous two terms. (a) Find the general term of the sequence and (b) find the limit of the sequence. An urn has b blue marbles and r red marbles. Marbles are taken from the urn one at a time and each one is returned following the next marble taken. Determine the probability that the kth marble taken from the urn is blue if it is known that the nth marble taken is blue.

APPLICATIONS OF DIFFERENCE EQUATIONS

230

[CHAP. 6

6.93.

In the preceding problem what is the probability if nothing is known about the nth marble taken?

6.94.

(a) Show that the Stirling numbers of the second kind satisfy the equation where

uk+l,m+1 = (k+ 1)uk+1,m + uk,m

3-

(b) Use (a) to show that (-kk-1

[() 1- (2) 2m + ()3m k! = I and verify this for some special values of k and m. 2lk m

nk, 0 =

... + (-1)k-1

()icm1

6.95.

There are three jugs which have wine in them. Half of the wine in the first jug is poured into the second jug. Then one-third of the wine in the second jug is poured into the third jug. Finally one-fifth of the wine in the third jug is poured back into the first jug. Show that after many repetitions of this process the first jug contains about one-fourth of the total amount of wine.

6.96.

Generalize Problem 6.95.

6.97.

Work Problem 6.16 if (a)

6.98.

(a) Show that Newton's method [Problem 2.169, page 78] is equivalent to the difference equation

(b) IbI = 21a1.

IbI > 21a1,

f (xn) xn+1 - xn - f,(x)

x0 = a

n

(b) Can you find conditions under which lim xn exists? 6.99.

Let

xn+1 = 21 xn + x

\

n

Justify your conclusions.

) , xn >O, xo = a, A > 0

(a) Prove that lim xn exists and is equal to VA. n-.ao

(b) Show how the result of (a) can be used to find

to six decimal places.

analogous to that of Problem 6.99.

6.100.

Can you formulate a method for finding

6.101.

Work Problem 6.20 by using h = 0.05 and discuss the accuracy.

6.102.

(a) Solve the differential equation

dy dx

= x+y,

y(0) = 1

for y(0.5) using h = 0.1 by using the Euler method and compare with the exact value.

(b) Work part (a) using h = 0.05. 6.103.

Work Problem 6.22 by using h = 0.05 and discuss the accuracy.

6104.

Given the differential equation

ddx = xb1/2

y(1) = I

(a) Use the Euler method to find y(1.5) and compare with the exact value.

(b) Find y(1.5) using the modified Euler method. 6.105.

(a)

Obtain from the equation dx = f (z, y) the approximate "predictor" formula Ylk+1 = ?1k-1 + 3 (7fk-1 + 4y' + yk'+l)

(b)

Explain how the result in (a) can be used to solve Problem 6.20 and compare the accuracy with the modified Euler method.

CHAP. 6] 6.106.

APPLICATIONS OF DIFFERENCE EQUATIONS

231

(a) Explain how the method of Taylor series can be used to solve a differential equation.

(b) Illus-

trate the method of (a) by working Problem 6.20. [Hint. If y' = x + y, then by successive differentiations y" = 1 + y', y"` = y", .... Now use the Taylor series y(x) = y(a) + (x - a)y'(a) + (x - a)' 2 y"(a) + ....] 6.107.

Work (a) Problem 6.102, (b) Problem 6.104 by the method of Taylor series.

6.108.

Explain the advantages and disadvantages of (a) the Euler method, (b) the modified Euler method, (c) the Taylor series method for solving differential equations.

6.109.

(a) Explain how the Euler method can be used to solve the second order differential equation d2y dx2

= f x , y, dy) dJ

(b) Use the method of (a) to solve d2y

dx2 = X + y,

y(0) = 1,

y'(0) = 0

for y(0.5) and compare with the exact value. 6.110.

Show how the differential equation of Problem 6.109(a) can be solved using the modified Euler method and illustrate by solving the differential equation of Problem 6.109(b).

6.111.

Work Problem 6.109(b) by using the Taylor series method.

6.112.

Explain how you would solve numerically a system of differential equations. Illustrate your procedure by finding the approximate value of x(0.5) and y(0.5) given the equations

x - dt = t,

dt + y = et,

x(0) = 0, y(0) = 0

6.113.

and compare with the exact value. Work Problem 6.24 by choosing a grid which is subdivided into 16 squares.

6.114.

Explain how you could use the results of Problem 6.24 or 6.113 to obtain the steady-state temperature in a square plate whose faces are insulated if three of its sides are maintained at 0°C while the fourth side is maintained at 100°C.

6.115.

Explain how you could use the results of Problem 6.24 or 6.113 to obtain the steady-state temperature in a square plate if the respective sides are kept at 20°C, 40°C, 60°C and 80°C.

6.116.

Solve the equation axe + ay2 = 0 for the plane region shown shaded in Fig. 6-18 with the indicated boundary conditions for U.

6.117.

The equation a2U ax2

(0,2)

(1,2)

U=0 a2U + ay2

p(x, y)

.U=0

where p(x, y) is some given function of x and y is called Poisson's equation. Explain how you would solve Poisson's equation approximately for the case where p(x, y) = 100/(1 + x2 + y2) if the region and boundary conditions are the same as those of Prob-

(2,1)

U=0 (0,0)

U=1

(2,0)

Fig. 6-18

lem 6.23. 6.118.

U= O

Explain how you would solve approximately Laplace's equation a2U

a2U

in three dimensions given by

a2U

- 0 inside a cube whose edge is of unit length if U = U(x, y, z) is equal to 1 on one face and 0 on the remaining faces. 5x2 + ay2 + a22

6.119.

Explain how you would find an approximate solution to the boundary-value problem

au _ at

a2U ax2

U(0, t) = 0,

U(1, t) = 0, U(x, 0) = 1

Stirling Numbers of the First Kind s;

Appendix A

o CV

0 L-

o 10

GV

M

CO

00

o N 00

rao

CO

00 C']

to

o 10

LO

1,14

CO

00

N 10

LIZ

r-q

110

N

10

in

Co

CV CD

N

,

N N

IM

c I

CV

C13

CV CO

to

CO

x

C

CO

to

1fJ

CO

C CO CO

,-1 T-4

in

,O

C) to

C

10 CV

o co

0 co

Co

10

c'i

10

C11

r-i

M

to 00

CV CV

IM

M

N

CD

00

L ^

CV

M

CV CO

M

r-4 C0 ,--1

CO

,

CO

rl

10

N CV

N

c

C

0

00 c9

C

N

°

C

C N c'i N

r-

to

O

00 10

w N 10

CO

C

10

00 GO

06 r1

,--I

o rl

N

r O -

r-q

C11

.c

N

01

r.l

C)

C>

CC']

O

N

Cl

0 00

e

cli

co

N

00

10

C

to

N

O

06 10

00

10 CI

N

t

10 C6 10

N

c

M

N rl C

ri 10

o

0

CO

00

00 cq

C

ori

0

CC

0 06 0 00

CO

10 GV

0 00 1.6

C CO

cq

CO

10

CO

232

N

00

r-I

r1

Cl

Stirling Numbers of the Second Kind

Appendix B

N

10

o N

C 00 Cu

-

CD CO

00

O

O

N

00

10

00

LC

10

00

to

00 00

N

000

m 00

to

N

N

Cl

N

Cl

00

Ca

C6 CO

r-q

o

o C>

r1 X0

1 Cl

o CO Cv

0

a r-;

o

1

° CO

o N ri

N N

N

o ,-.

N

eN

,ri

to

r-I

10

10

Cl

o

M

o CO

M 10

C

w

10

c

CO

C0

10

CO

C

00

01

Cl

CO

N

.

i

CO

1.0

Cl

N

LO

7-4

Cl

,-I

N

G9

233

N

00

M

CO

o

CI

O

,--

Cl 10

1 00 N

0

Cq

Appendix C

... are equal to zero.

All values B3, B5, B7,

=

Bernoulli Numbers

1

-0.500000000..

=

-0.0333333333..

30 1

B6

=

42 1

5 66

Blo

691 2,730

B12

7

B14

6

3,617 510

B16

43,867 798

B13

854,513 138

=

B30

1.1666666666...

=

-7.092156862745098039215...

= 54.971177944862155388...

-529.1242424242...

6

=

=

236,364,091 2,730 8,553,103

B26

0.075757575...

=

330

B22

.

-0.2531135531135...

174,611

B20

B28

=

.

0.02389523895...

-0.0333333333..

30

B24

0.166666666...

1

B4

.

6,192.1231884057898550...

-86,580.2531135531135..

=

.

1,425,517.16666...

- 23,749,461,029 870

-27,298,231.067816091954.. .

8,615,841,276,005 14,322

601,580,873.90064236838..

234

.

Appendix D

Bernou

B1(x)

=x

B2(x)

= x2 - x +

B3(x)

= XS ®2x2 + x

2

B4(x)

x4-2x3+ x2° 30 1

B5(z)

xs - 5x4 + 5x3 2 3

B6(x) B7(x) B8(x)

-1 6

= x6 - 3x5 + 2x4 - 2x2 +

42

x7-2x8+2x5-6x3+6x = x8 - 4x7 +

14 xe 3

- 3 x4 + 3x2

30

B9(x)

x9 - 2x8 + 6x7 -

B,o(x)

X10 _5x9+ 2x8-?xo+5x4-2x2+66

235

5x6 + 2x3 -

10-

Appendix E

Euler Numbers

All values El, E3, ... are equal to zero.

Eo = 1 E2

= -1

E4 = 5

E6 =-61 E3 = 1,385 Eio = -50,521 E12 = 2,702,765

E14 = -199,360,981 Eas = 19,391,512,145

E18 = -2,404,879,675,441 E20

= 370,371,188,237,525

E22

= -69,348,874,393,137,901

E24

= 15,514,534,163,557,086,905

236

Appendix F

E2(x) =

Euler Polynomials

2

x2 - x 2

x3-4x2+24

E3(x) = 1

E4(x)

24x4

12

x3

237

+ 24 x

Appendix G

Fibonacci Numbers

Fo = 0

F26 = 121,393

F1 = 1

F27 = 196,418

F2 = 1

F28 = 317,811

F3 = 2

F29 = 514,229

F4 = 3

F3o = 832,040

F5 = 5

F31 = 1,346,269

Fs = 8

F32 = 2,178,309

F7 = 13

F33 = 3,524,578

Fs = 21

F34 = 5,702,887

Fo = 34

F35 = 9,227,465

Flo = 55

F36 = 14,930,352

F11 = 89

F37 = 24,157,817

F12 = 144

F38 = 39,088,169

F13 = 233

F39 = 63,245,986

F14 = 377

F41 = 102,334,155

F15 = 610

F41 = 165,580,141

F16 = 987

F42 = 267,914,296

F17 = 1,597

F43 = 433,494,437

F18 = 2,584

F44 = 701,408,733

F19 = 4,181

F45 = 1,134,903,170

F20 = 6,765

F46 = 1,836,311,903

F21 = 10,946

F47 = 2,971,215,073

F22 = 17,711

F48 = 4,807,526,976

F23 = 28,657

F49 = 7,778,742,049

F24 = 46,368

Fso = 12,586,269,025

F25 = 75,025 238

Answers to Supplementary Problems CHAPTER 1 1.46.

1.52.

(a)

1 + 2V + x

(e)

16x4 - 32x3 + 32x2 - 22x + 6

(i)

0

(b) 2x4 - 4x3 + 2x2 + 6x - 3

(f)

2x4 + 4x3 - 6x2 - 6x

(j)

16x7(1- x)

(c)

(g)

4x4 + 8x3 - 2x2 - 6x

9

(d) 6(3x+2)

(h) 2x(4x + 1)

(a) 8xh + 4h2 - 2h

(e)

(x + 2h + 1)2

(i)

2x2 + 14xh + 19h2

(f)

3x3 + 8x2h + 6xh2 + x2

(j)

4x2h2 + 9xh3

0

+ 5h - 4

(b)

3 5x

(c)

4h2

(g)

(d)

3x2 + 12xh + 12h2 + 3

(h) 2x2 + 14xh + 19h2

1.53.

(a) Yes

(b) No

1.55.

(a) Yes

(b) Yes

1.59.

(a) Yes

(b) No

1.61.

(a) (3x2 - 6x + 2) dx

1.68.

(a) h(15x(4) + 20x(3) - 14x(1) + 3)

1.69.

(a) h2(24x(-5) - 18x(-4) + 8)

1.70.

(a)

3x(2) - 2x(1) + 2 and 3x(2) + (3h - 5)x(') + 2

(b)

2x(4) + 12x(3) + 19x + 3x(1) + 7 and 2x(4) + 12hx(3) + (14h2 + 5)x(2) + (2h3 + 5h - 4)x(l) + 7

(a)

4x3 + 6x2h - 12xh2 - 4x + 25h3 - 2h + 5

(b)

12x2 + 24xh + 14h2 - 4

1.71.

1.72.

(c) Yes

(b) 6(dx)2

(b) -3x(-4) + 6x(-3)

(b) 24x - 120x(-7)

(a) (2x-1)(2x-5)(2x-9)(2x-13)

1 (d\

(5x + 12)(5x + 22)(5x + 32)(5x + 42)

(b) (3x+ 5)(3x + 2)(3x - 1) 1 (c)

1.73.

(4x - 1)(4x + 3)

(a) (3x - 2) (3), h = 7/3

(c)

(2x + 2)(4), h = 3/2

(d)

(b)

(c) h(x(1> - x(-3))

(x - 2)(-3), h = 2 (2x - 5)(-4), h = 2 239

ANSWERS TO SUPPLEMENTARY PROBLEMS

240

1.76.

31 = 1, si = -1, s3 = 2, s2 = 0, s2 = 1, s2 = -3, s3 = 0, s3 = 0, s3 = 1

1.77.

Si = 1, S1 = 1, Si = 1, S2 = 0, S2 = 1, S2 = 3, S3 = 0, S3 = 0, S3 = 1

1.78.

Si = 1, S2 = 31, S3 = 90, S4 = 65, Ss = 15, SB = 1

1.81.

(a)

3x(2) - 2x + 2 and 3x(2) + (3h - 5)x(') + 2

(b)

2x(4) + 12x(3) + 19x(2) + 3x(1) + 7 and 2x(4) + 12hx(3) + (14h2 + 5)x(2) + (2h3 + 5h - 4)x(1) + 7

1.87.

2x(x2 + 12x + 30)

1.88.

(ah - 1)nxax + nh(ah - 1)n-lax+h

1.89.

(ah - 1)nx2ax + 2nh(ah - 1)n-lxax+h + nh2(nah - 1)(ah - 1)n-2ax+h

1.91.

(a) 4xh - 2h2 + 3h

1.92.

0

1.94.

Yes

1.97.

Yes

1.116.

(a) 18h2x - 4h2 + 18h3

(b) 4xh + 3h

(c) 4h2

(d) 4h2

(b) 24h3x + 12h3 + 36h4

CHAPTER 2 2.52.

Yk + nLyk + n( 2! 1) A2Yk + ... + Onyk

(a)

Yk+n =

(b)

Anyk = Yk+n - nyk+n-1 +

n (n

-1)

2!

Yk+n-2 -

+

(-1)n

yk

Ilk = k(3) - 02) - kc1 - 3

nyk = 3k(2) - 2k(1) - 1 = 3k2 - 5k - 1 A2yk = 6k(1) - 2 = 6k - 2 (d)

O3yk = 6

(e)

A4yk = 0

2.56.

60k2 + 108k + 54

2.58.

1489, 2053, 2707, 3451, 4285, 5209

2.59.

(b) 2366, 3994, 6230, 9170, 12910

2.61.

5

2.63.

A=4, B=O, C=-2, D=4, E=4, F=-1, G=-5, H=2, J=3, K=-1, L=0

2.64.

(a) 6

(b) 7

(c) 5

ANSWERS TO SUPPLEMENTARY PROBLEMS 2.68.

(a)

Ilk = 3k2 - 2k + 4

2.69.

(a)

Ilk = k3 - k2 + 21- k - 4

2.71.

y = x3- 4x2+5x+1

2.72.

y = *(x3 - 5x2 + 7x + 21)

2.74.

Exact value = 0.70711,

2.76.

(a) Exact value = 0.71911

(b) Exact value = 0.73983

2.77.

(a) Exact value = 0.87798

(b) Exact value = 0.89259

2.79.

(a) Exact value = 0.93042

(b) Exact value = 0.71080

Ilk =

(b)

s

k(k + 1)(k + 2)

(b) 651

Computed value = 0.71202

Computed value = 0.90399 2.80.

241

Computed value = 0.70341

(a) Exact value = 0.49715

(b) 1.75 X 10-4

Computed value = 0.49706 2.81.

(a) Computed value = 0.49715

2.82.

Exact values

2.83.

Exact value = 0.92753

2.84.

Exact values

2.85.

(b) Computed value = 0.71911

2.86.

Computed values

(a) 0.87798, 0.89725

(b) 0.49715

2.87.

Computed values

(a) 0.71940, 0.73983

(b) 0.87800, 0.89726

(c) 0.49715

2.88.

Computed values

(a) 0.71915, 0.73983

(b) 0.87798, 0.89726

(c) 0.49715

2.92.

Computed value = 0.92752

2.93.

Computed values

(b) 0.87798, 0.89720

(c) 0.49715

2.98.

(a) y = x3 - 4x2 + 5x + 1

2.99.

(a) y = 2x3 - 3x2 + 5x + 2

2.100.

(a)

(a) 1.27507

(a) 0.30788

(b) 1.50833

(b) 0.62705

(a) 0.71912, 0.73983

4n3 - 27n2 + 59n - 35

(c) 0.84270

(b) 0.49715

(b) 44 (b)

7, 85

(c) 202

(d) 874

(d) 0.92756

ANSWERS TO SUPPLEMENTARY PROBLEMS

242 2.101.

(a) sin x = 2.80656(x/7r)4 - 6.33780(x/7r)3 + 0.21172(x/7r)2 + 3.12777(x/ir)

(b) Computed values and exact values are given respectively by 0.25859, 0.25882; 0.96612, 0.96593; 0.64725, 0.64723 2.103.

15

2.104.

106, 244

2.105.

Exact values are 0.25882, 0.96593

2.106.

(a) 44, 202

2.107.

6.2, 7.6, 11, 17

2.108.

8, 14, 22, 32

2.111.

Exact values are 0.97531, 0.95123, 0.92774

2.113.

(b)

f(x) = 3x2 - lix + 15

2114.

(a)

f(x) = x3 - 20x + 40

2121.

(b) 1.32472

2.122.

1.36881, -1.68440 ± 3.43133i

2.123.

2.90416, 0.04792 ± 1.31125i

2.124.

0.5671

2.125.

0, 4.9651

2.126.

Exact value = 3.67589

2.127.

2.2854%

2.128.

3.45

2.129.

Exact value = 0.2181

2.131.

Exact value = 0.08352

2.132.

Exact values are

(a) 0.4540

(b) -0.8910

2.133.

Exact values are

(a) 1.2214

(b) 1.2214

2.134.

Exact values are

(a) -0.31683

(b) -0.29821

2.135.

Exact values are

(a) 0.91230

(b) 0.72984

2.138.

A=6, B=8, C=6, D=3, E=2, F=-2, G=-4, H=-4, J=-2, K=-3

(b) 874

(c)

19, 159

(b) 65, 840

(c) -0.66236 ± 0.56228i

ANSWERS TO SUPPLEMENTARY PROBLEMS 2.140.

Exact value = 3.80295

2.141.

Exact values are

2.142.

11.72 years

2.143.

(a) 0.9385

2.144.

Exact value = 22.5630

2.145.

2.285%

2.146.

(a) 38.8

2.148.

Exact values are 0.74556, 0.32788, 0.18432, 0.13165, 0.08828, 0.05261

2.150.

0.97834, 0.66190, 0.47590, 0.34600, 0.24857, 0.17295, 0.11329, 0.06604, 0.02883 f (X)

=3

2.151.

(a) 1.7024

(b) 2.5751

(b) 0.6796

(b) 29.4

2xa+x+1

2.157.

8k(k + 1)(2k + 1)

2.158.

(a) 8k(k + 1)

2.159.

-k(6k2 + 3k - 1)

2.160.

(a) 2k2 - 5k + 3 + 2k

2.165.

k = 10

2.166.

All values of k

2.168.

At x = 6, f (x) = 0.68748 should be 0.68784

(b) 1k2(k + 1)2

(b) 1177

CHAPTER 3 3,5

3.55.

(a) b

4

2

2+2-x+c

(b) I x3/2 - 9 x413 + C (c)

3.57.

8(2 sin 3x + cos 3x) + c

(a) -*e-3i(x + J) + c (b)

2 sin x - x cos x + c

(d) 2e2x -

je-4x + c

(e)

4 1n (x - 2) + c

U)

In 2

2x

+c

Jx2(ln x - J) + c (d) jx3 ln3 x - 8x3 In x + 7x3 + c (c)

243

244 3.58.

ANSWERS TO SUPPLEMENTARY PROBLEMS

(a) -x sin 2x - -x2 cos 2x + j cos 2x + c (b) *e2x(4x3 - 6x2 + 6x - 3) + c

3.59.

(a) In (ex + sin x) + c

(d) - tan-1( x

(b) 2e(V-x - 1) + c

6 cos V + 6I sin T - 33x2 cos T +

(c)

3.65.

(e)

2

2) -I-

-1

2(2x - 3) + c

c

x sin r(x - -h) h cos rx 2 sin -rh + 4 sine --rh -x cos r(x - jh) h sin rx 2 sin - rh + 4 sine jrh

(a)

(b)

3.66.

(a) 768

3.67.

1 12

(b) 0

-

(c) 1

(d) 660

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

3.68.

(n + 1)2n+1 - 2n+2 + 2

3.72.

(a)

x sin r(x - -h) h cos rx 2 sin -rh + 4 sine , rh -x cos r(x - -h) h sin rx 2 sin jrh + 4 sine - rh

(b)

(b) 4n(3n - 1)

3.74.

(a) , n(n + 1)

3.75.

(a) *n(n + 1)(n + 2)

(c)

jn(4n2 + 6n - 1)

(b) in(n + 1)(n + 2)(n + 3)

(d)

n(3n2 + 6n + 1)

n

1

3.76.

(a)

3.78.

(n-1)2n+1+2

3.81.

(a) 1

3.82.

-&n(n + 1) (n + 2)(3n + 5)

3.83.

211+1(n2 - 2n + 3) - 6

3.84.

*n(2n - 1)(2n + 1)

3.85.

(a)

n+

1

(b) .

(b)

(c)

2n + 1

1-1

2

(b)

(n + 1)(n + 2)

(c) 24

(d)

1

6(3n + 1)(3n + 4)

(d)

1 -

4

2(n + 1)(n + 2)

ANSWERS TO SUPPLEMENTARY PROBLEMS 2n + 5

5

(a)

3.89.

112n(n + 1)(n + 2)(3n + 1)

3.90.

(a)

12

2(n+2)(n+3)' 12

5n2 + n

(a)

3.93.

(a)

23

12

(b)

+ 10 (2n + 7)(2n + 5)(2n + 3)(2n + 1)(2n - 1)

6(2n1+ 1)(279+ 3)

"

(b)

(b)

1

90

-

1

6(2n + 1)(2n + 3)(2n + 5)

211 4

sn(n+1)(2n+1) a a)3 [1 + a - (n + 1)2an + (2n2 + 2n - 1)an+1 - n2an+2]

3.96.

(1

3.102.

(a) 6

(b)4

3.103.

(a) 3

(b) 4

3.104.

1P(j)

3.105.

s

3.108.

fr'(x+n+1)12 r(x+n+1) J

(c) -15

- 1'"(x+n+1) -

3.110.

(a)

nn+1)

1'(x

n In general,

-

1

n k=1 (x + k)r

Q4(x) =

x4 24

-

x3 12

+

x2

24

-

Jr'(x+1)}2 P(x+1)

r'(x + n + 1)

(1 - 1) ! dxr`1

P(x + n + 1)

1

x5 120

20

Rs(x) =

-1/30, 1/42, -1/30

3.114.

(a)

3.115.

3B0(x) - 4B1(x) + B2(x) + 2B3(x) + 5B4(x)

3.119.

(a)

e4 = 0, e5 = -1/240

(b)

E4(x) = 24 1 ro4 -12 1x3 + 24 1 x, Es(x') =

(c)

E4 = 5, Es = 0

n2(n + 1)2

P"(x+1)

+ r(x+1)

(-1)r-1 dr-1

3.111.

3.125.

2n2 + 14n + 23

4(n+2)(n+3)(n+4)(n+5)' 4-80-

2 - n2 2

3.94.

3.95.

21

-

(b) 480

5

12(n + 2)(n+ 3)

3.91.

23

5

3.86.

245

-

x4 48

+

+ 1) - ox P(x + 1)

x3 72

-_

x 720

(b)

(b) 30 n(n + 1)(2n + 1)(3n2 + 3n - 1)

2T(x3 - 6x2 + 18x - 26)

1 x 120

s

1

4

1

48 x+ 48 x

21 240

1/30, 0

ANSWERS TO SUPPLEMENTARY PROBLEMS

246 3.126.

(n + 1)3 cos 2(n + 1) - (3n2 + 9n + 7) cos 2(n + 2) + 6(n + 2) cos 2(n + 3) - 6 cos 2(n + 4) - cos 2 + 7 cos 4 - 12 cos 6 + 6 cos 8

3.132.

i2n2(2n4 + 6n3 + 5n2 - 1)

3.146.

(a) it/4

3.162.

Bl = -1/2, B2 = 1/6, B3 = 0, B4 = -1/30, B5 = 0, B6 = 1/42

(c) 7r//

(b) Tr/32

CHAPTER 4 4.40.

(a) n2

(d) a n(n + 1)(n + 2)

(b) *n(2n -1)(2n + 1)

(e)

(c)

jn(6n2 + 3n - 1)

jn2(n + 1)2

x+1 (1- x)3

4.42.

[1 + x - (n + 1)2xn + (2n2 + 2n - 1)xn+1 - n2xn+2J/(1 - x)3

4.43.

9/32

4.44.

(a)

3 + 6x - x2

x(x + 1)

x (b)

(1 - x)3

(c)

4.45.

359/500

4.46.

5e - 1

4.48.

(x4 - 7x2) sin x + (x - 6x3) cos x

4.49.

Ix cosh

4.50.

x 3n(3

- x)3

+

(h) An(3n3 + 46n2 + 255n + 596)

(f) J(2n - 1)(2n + 1)(2n + 3)(2n + 5) + 1$5

4.41.

(1 - x)2

(g) Ii2n(n + 1)(n + 2)(3n + 17)

(1 - x)3

IV-x (1 + x) sinh '

[3n+2 + 3n+ lx - 9(n + 1)2xn + 3(2n2 + 2n - 1)xn+1 - n2xn+2]

4.59.

Exact value = 1.09861

4.61.

Exact value = 1

4.62.

0.63212

ANSWERS TO SUPPLEMENTARY PROBLEMS 4.64.

(a) sn(n + 1)(2n + 1)

(c)

(b) in2(n + 1)2

(d) 30 n(n + 1)(2n + 1) (3n2 + 3n -1)

4.67.

y = 0.577215

4.70.

1.20206

4.78.

Exact value = 3,628,800

4.79.

2100/5 2a - 4.036 X 1028

4.83.

xe-7(2 - x)

4.84.

(a)

-n(n + l)(n + 2)

6n + 11 0 - 12(2n + 1)(2n + 3)(2n + 5)

1811

4.86.

1.092

4.88.

(a) 1.0888

4.94.

1(-1)n-ln(n + 1)

4.95.

2"(n2-2n+3) - 3

4.99.

(a) 1.09861

(b) 0.785398

1811

(b)

0

(c) 0.63212

CHAPTER 5 5.60.

(b) y = Sex - x2 - 2x - 2

5.61.

y(x) = 5(1 + h)xih - x2 + (h - 2)x - 2

5.63.

y = 2 cos x + 5 sin x + 3x2 - 5x - 2

5.66.

(a) yk+1 - (h+1)yk = h3k(2) (b) Yk+2 - 2yk+1 + (1 + h2)yk = 3h4k2 - (3h3 + 5h2)hk + 4h2

5.67.

(a) Yk+3 -' 3yk+2 + 3yk+1 - Yk = h3k(3) - 2h2k(2)

(b) yk+4 + yk = coshk 5.68.

(a) 2f(x+2h) - 3f(x+h) + f(x) = 0 (b) f(x+3h) + 4f(x+2h) - 5f(x+h) + 2f(x) = k2 - 4h + 1

5.69.

(a), (c), (d), (e), (f), (g) linearly independent; (b), (h) linearly dependent

247

ANSWERS TO SUPPLEMENTARY PROBLEMS

248

Yk = cl(-3)k + C2(-2)k Yk = 2k

Yk = cl(-2/3)k + c2(-2)k Yk = 5(3)k + 6 (-3)k Yk =

(f)

yk =

(9)

Yk =

(h)

yk = yk = 3k(cl + c2k) yk = (-1/2)k(cl + c2k)

2k-1 sin (k7r12)

(5/2)k[cl sin (k7r/2) + c2 cos (kvr/2)]

(i) (7)

4(-3/2)k.

2k/2 cos (37rk/4)

5.76.

yk = 5.2k-2.3k-3

5.77.

Yk =

Cl COS

5.78.

yk

cl(5/2)k12 + c2(-5/2)k/2 + (5/2)k/2 (C3 COS

5.79.

Yk

2k

5.80.

yk

5.81.

yk =

5.82.

47r

+ C2 sin

k4

(cl + C2 COS 23k

+ c3 COS

+

= cl2k + c2(-2)k + 4k

Yk =

+ c4 sin 3

2r

4

+ C4 sin 2-'r

)

C3 sin 23k

\

(c2 cos 34k

Cl + 2k/2

34ar

c3 COS

k2

+ c4 sin 2

+ c3 sin 34k

)

(Cl + c2k)2k + (c3 + c4k + csk2 + c6k3)(-2)k

+ C74k + 5k( C8 cos ko + cs sin ko)

where 8 = tan-1 (-4/3) 5.83.

Yk+9 + 6yk+8 + 5yk+7 - 136yk+6 - 484yk+5 + 464yk+4 + 3376yk+3

+ 1792yk+2 - 6336yk+1 - 6400yk = 0 (E - 2)2(E 4- 2)4(E - 4)(E2 + 6E + 25)yk = 0

or 5.85.

. 4k

= Cl + c22k +

(a)

yk

(b)

Yk = 2k(cl + c2k) + k2 + 4k + 8

(c)

yk = 2

k

(C3 COS

k7r

2

k7r + C2 sin 2+

(e)

yk = 2k2 - 2k + 1 yk = (-1)k(Cl + C2k) + .-k -

(f)

yk

(9)

yk = cl + c22-k +

(h)

Yk

(d)

=

'ffk(k - 1) . 4k-2

2k (Ci

k3

cost'r +

_

57k

371

25 + 12-5-

+ 1-2k

- 9k2 + 127k 2

3

6

c2 sin 2'r)

+ 13 (-3)k + 2k - 5

J

9. 3-k

=

2k2

f cak + /3/(1 - a)

a

1c+/3k

a=1

1

5.86.

yk

5.87.

yk = 2[cos 2k + cos (2k - 4)]/(1 + cos 4)

ANSWERS TO SUPPLEMENTARY PROBLEMS 5.88.

Uk = Cl + c22k + c33k + 2k + k2 - k . 2k - 2 45k

5.89.

yk = c2k + c2(-2)k + 2k

(03

cos k7 + c4

sink27r

2

\

l

-

67k

k2

249

422

lb + 225 - 33 75

-

4. 3k

5.90.

2k Uk = cl(-1)k + c2 Cos. 2 + c3 sin 3 + 65 + 16 [8 cos (3k - 9) + cos 3k] cos 9

5.91.

11k

5.92.

(a)

Uk

Cl + c22k + 2 . 3k - 6 .5k

(b)

Yk

(-2)k(cl +c2k) +

(c)

Yk

(Ci

(d)

Uk

4k(01 + c2k) + 32 k2-4k -

(e)

Yk =

(J)k

(f)

Yk

2k(cl + c2k) +

(9)

Yk

Cl + c2k + c3k2 + *k3 + i2k4 -

(h)

Yk

c12k + c2(-2)k + 2k (03

5.94.

Yk

5.95.

(0)

= Cl 3-k/2 + C2(-3)-k/2 + C3(-1)k

Cl cos 4-

+ 2k

48

k

c2 sin 4T

3

6

(-2)k

+ k3 - 6k2 + 2k + 18 +

(d)

5k

41

(k4 - 4k3 + 5k2 - 2k)

Cos

k7r

4

.3k-1

+ C4 sin

kir 4

_

k

_

4

75 +

k 2k 64

3k

+

65

3

Yk = cl2k + 024k - 3k +

k2 3

+9+

238

cl+c2k+c3(-2)k+ c4+60 3k Yk

52 11 ,041

9

-

36

_

7

87 +

+ 30) 3k + c2(-3)k +

k 208 C

1 / Irk Irk = 01(-6)k + c22k + 258 l 25. sin 3 - 105 cos

5.96.

5.97.

k2 7

C4 + 3k

65

=

4

U

3

2+ 54

77k 2

18

(cl + c2k) cos--- + (c3 + CA sin 2 + }-k - 7

Nk

c12k+ c23k + jk2 + Ik -

Yk

Cl+c2k+k2+Wk3+ 9.4k

Yk

(Cl + c2k + k2)2-k

Yk

7

2

(e)

Yk

3k 1 Cl + C2(,)k - 598 + 14 + 2-4(-2)k

(f)

vk

Cl cos 3 + c2 sin 3

k = C1cos k + C2sin2 + ==0 k+2r r

= Cl +

296k + k2 2

5.103.

Yk

5.106.

(a) Yk = 2k(2k - 1)

+ k3

kr 1 yr=o (k-r-2) 2

+

+ (c2 - k)2k + c33k

6

(e)

Yk = 2(-1)k(1- k)

(b)

Uk = 1 - 1/2k

(f) Yk = Cl + C2k + C3k2

(C)

Uk = 2k(1 - k)

(9)

Yk = (-1)k(C1 + C2k) + *(2k - 1)

(h)

Uk = 2-k -1(k2 + 3k)

(d) Yk = 013-k + 022-k

sin

r(r + l) 3

ANSWERS TO SUPPLEMENTARY PROBLEMS

250

=

5107.

Yk

5108.

(a)

5111.

(a)

1

5

1)r4r-1 1r!] + 15 - b1 (-4)k IIL (k-r)! + k

1

(

I ktk

t

=

k=0

k2tk = W + 1)

(b)

(1 - t)2

4k!

(1 - t)3

k=O

(c)

k=0

k3tk

t(t2 + 4t + 1) (1 - t)4

0

e(k -1) ! - I (k -1)(-T)

Yk

T=0

_

m

c(-2)k

(-1)rk(r)

(b)

Yk

(k - 1) ! + T-wo

(C)

Vk

c(-1)k(k -1)! + k - 1 + 1 (-1)r+ 1(k -1)(-r)

(d)

Vk = -k(k -1) ! [1 + (-4)k-1]

(e)

Uk

2r+1 m

T=0

1 Ik [2r+ir! -2r+k(k+r-l)!] (k-1) ! r=0

k-1 Yk

(1+e)

5.113.

Uk

A(k-1)!

5.114.

Uk = (k -

5.115.

Yk

5.112.

ci

r=1

1)!

k-1

(r - 1) ! -s=1,

k-1 1

- + B(k-1)!

r=1 r

k-i [is_i21.

+

+

s r=1 r!

s=i

C

(s -1)(-*> r=0

A] + B(k-1)! s

k(4)

k(2)

C

+ c2 k(l) +

kcs_

k(5)

k(3)

5.116.

Yk = c1(3/2)k(k - 3) + c2(2)-k(k + 1)

5.119.

(a)

Vk -

(b)

Yk = c(k + 1) - 1 +

5122. k -

c( k + 1 )

+

113+1.32.5+1.32.52.7...

kc7>

+

\

24+2.42.6+2.42.62.8... J

k-i r(-2) r + ( k+1) r=o v (r + 2)! k- r(-2)r (k + 1)

r=O

ak+r-11 //bk-1 bk-2 +1 (!k bk+1)]} + C2f k+r)(bk+r-1/ ... \ ak / ak-i) ... {1 + r=i L ak+r

Ciak bk

5.133.

Uk = 2/[1 + c(-1)k]

5.134.

Yk = 1 - NF2 + 2//[1 + c(2V - 3)-k]

5.135.

Yk

tan

5.136.

Yk

Kli2 tan I Cl cos 23k + c2 sin 23k]

5.137.

Yk =

e3-k

23k Cl [cos

+ c2 sin 23k]

(b1'

a2

AKRw i gym. 5.138.

(a) 1. (b)

=`i`t+c

TtARY PROBLEMS

e s = 4(1 + NF5), R = - (1- V5-)

!It = cos (c 2t)

5.139.

yk = e

5.140.

Yk = c13k or Yk = 022k

5.141.

(a) Yk = 2 cos (7r/2k+1)

5.144.

Uk = cl(-2)k + C2,

5.145.

Yk = }(k - 1) ! [c12k+1 + c2(-2)k+1],

5146.

Uk = {q(uo + vo) + [(1 - p)uo - gvo](p - q)k}l(1 - p + q)

c zk)

(b) 2

vk = -el(-2)k + 2e2 Zk = (k - 1) ! [c12k + c2(-2)k]

Vk = {(1- p)(uo + vo) + [qvo - (1- p)uo] (p - q)k}/(1 - p + q) 5.147.

Uk = uo + (u0 + v0)gk,

5.148.

Uk = (Cl + C2k)(V o )k + (c3 + c4k)(-r8 )k

Vk = v0 - (u0 + vo)gk

Vk = (2c1 + 602 + 2c2k)(NF8)k-1 + (2c3 + 604 +

2c4k)(-f )k-1

5.149.

t2 t3 t+k yk(t) = 1+t+ 2!+ 3!+...-ki

5.150.

(a) et

5.153.

u(x, y)

= (-2)xF(x + y)

5.154.

u(x, y)

= 2x(3x + 3y + 2)

5.155.

u(x, y)

= (-2)-xF(x + y) + xy - $ x - ly + 2

5.156.

u(x, y)

3x/hF (! + k)

5.157.

u(x, y)

2xihF+(

5.158.

uk.m = F(2k-m) -

5.159.

uk.m =

5.160.

uk,m = 2kF(k - m) + G(k - m) + 2mH(k - m)

5.161.

uk.m = F(k - m)

5.162.

Ztk.m = F(k - m) cos 3 + G(k - m) sin 3 + H(k - m) cos 3 + J(k - m) sin 3

- Ix - ly - 1h + lk

I -k) + G (

+ k/

2k-m+l

(k)

251

252

ANSWERS TO SUPPLEMENTARY PROBLEMS

5.163.

uk(y) = 3k+1e-2y/3

5.167.

(c)

y=

co

5.169.

(a)

y

el(1 + h)x/h + c2(1 + 3h)x/h + 3 + 9

(b)

y

clex + C2e3x + 3

5170.

y

5.177.

Yk

5.178.

yk

5.181.

Yk

5.182.

Yk

1+x+

x3 x4 1*31.3.5+...1 + xe +cl(x+2

10

+

e-x cl+c2x+ 002) r (-1)½('-- 1) 2%k(k-1) Jll(-1)%k21 k(k-1)

tan

k odd k even

Ic + Y, tan-1 (1/2k)]

clak+1 + c2 /3k+1

2.4.6 (k-1)ac

. , (3 = I [b i Vb2 _+4a ]

where

- b

Clerk + C2/jk

k odd

k even

yk

5.183.

uk, m = F(k + m) + kG(k + m) + k2H(k + m)

5.184.

uk,m = Sm =

5.185.

uk,m = Sm = Stirling numbers of the second kind

5.187.

Yk = cos l 47r

Stirling numbers of the first kind

/

k-11

\\\

r=1 r

= (x -1)/e

5.188.

Y(x)

5.189.

1 + 2x tan 1 Y(x) = x2 - 4x +cos 1

5.190.

Uk

5.191.

(a)

u(x,y) = J(3x + 2y)2 + e- (3x+i1)/2

(b)

u(x,y)

(a)

u(x, y)

(b)

u(x, y)

5.192.

1

1

2-

1

+ if - ...

x2+x

+ (-1)k-1

5y2

18

k!

xy + 1 3

y 3

= x2 + y + jxy y11

(- 8]

+

CHAPTER 6 k

6.62.

(a)

2

1

11

ak+1

where

a=

1-y 1+v

xs

x7

+2.4+2.4.6+...1

ANSWERS TO SUPPLEMENTARY PROBLEMS

{(4 +

1

6.66.

2

)k - (4 -

)k)

19

6.68.

2

6.71.

(a)

6.75.

(a) Ak+1 = Ak + iAk

6.76.

An+1 - (1+2)An = S

6.77.

An+1 - (1 + i)A,, = S + na,

6.89.

T2(x) = 2x2 - 1, T3(x) = 6x3 - 3x, T4(x) = 8x4 - 8x2 + 1

(b) Ak = P(1 + i)k

- 3!1 + 4!1 - ... +

1

2!

An = a 2Sz [(1 + i)n - 1] - ai

(-1)n n!

6.91.

(a) 3 + 3.2n-1

6.92.

b + r + (b + r)(b + r

6.93.

b b+r

6.102.

(a)

Exact value = 1.7974

6.109.

(b)

Exact value = 1.1487

6.112.

Exact values: x(0.5) = 0.645845 y(0.5) = 0.023425

6.113.

U=0

(b) 3

(-1)k-nr

b

U=0

-1)k-n

1

2

3

4

5

6

7

8

9

U=1

U1 = 0.071428571, U2 = 0.098214285,

U=0

U3 = U1, U4 = 0.1875, U5 = 0.2500, U6 = U4, U7 = 0.428571428, U8 = 0.526785714, U9 = U7

253

INDEX Abel's transformation, 85 proof of, 101 Actuarial tables, 75 Algebra of operators, 2 Amortization, 215 Amortization factor, 215 Annuities, 55, 75, 228 Anti-derivative operator, 79 Approximate differentiation, 39, 40, 60-62 Approximate integration, 122-124, 129-136

Coincidences, problem of, 228, 229 Collisions, applications to, 200, 201, 208, 209 Commutative laws, 2, 13 Complementary equation, 168 [see also Homogeneous equation] Complementary function or solution, 155, 160 Complete equation, 153, 155 Complex numbers, conjugate, 153 polar form of, 153, 154

Complex roots of auxiliary equation, 153 Conditional probability, 201 Conservation of momentum, 201 Continued fraction, 226 Convergence, intervals of, 8

error terms in, 132-134

Gregory's formula for, 124, 134-137 Arbitrary periodic constant, 82, 93 Area, 80, 92 Associative laws, 2

Convolution, 193 Corrector methods, 218, 230 Cube roots, 11, 75 Cubing operator, 1, 10, 11 Currents, 200

Auxiliary equation, 153, 165-168 Averaging operators, 10, 88, 108

Backward difference operator, 9, 24, 25, 124 Beams, applications to, 200, 206, 207 Bending moments, 200, 206, 207 Bernoulli numbers, 86-88, 105-108, 234 generating function for, 87, 110 Bernoulli polynomials, 86-88, 105-108, 235 generating function for, 87, 109, 110 Bernoulli's probability formula for repeated trials, 225 Bessel functions, of order k, 227 of zero order, 75 Bessel's interpolation formula, 36, 48, 50 Beta function, 117 Billiards, 208 Binomial coefficients, 9, 13, 227 Boundary conditions, 150, 151 Boundary-value problems, involving difference equations, 151 involving differential equations, 150 numerical solution of, 219-221 [see also Numerical solution] Sturm-Liouville, 158, 178-181

De Moivre's theorem, 154 Definite integrals, 80, 81, 91, 92 approximate values of [see Approximate integration] Definite sums, 83, 84, 96, 97 Density or weight, 159 Dependent events, 201 Derivative operator, 1, 3, 10, 14, 15 relationship of to difference and differential

operators, 4

Derivatives, 3 Leibnitz's rule for, 9, 24 mean-value theorem for, 8, 27 partial, 81, 160 of special functions, 5 Determinants, evaluation of, 202, 213, 214 Difference calculus, 1-31 applications of, 32-78

general rules of, 5 Difference or differencing interval, 2 Difference equations, 150-198 applications of, 199-231 boundary-value problems involving, 151 with constant coefficients, 152 definition of, 150, 151 formulation of problems involving, 199 linear, 152 mixed, 159, 182, 183, 188, 189 nonlinear, 152, 159, 181, 182 order of, 151, 152

Calculus of differences, 4 [see also Difference calculus] Capacitance or capacity, 200 Capacitors or condensers, 200 Casorati, 54, 157, 164, 165 Central difference operator, 9, 24, 25

Central difference tables, 35, 47-49

Characteristic functions, 158 [see also Eigenfunctions] Characteristic values, 158 [see also Eigenvalues] Chebyshev polynomials, 227 Clairaut's differential and difference equations, 197 Coefficient of restitution, 201, 208

partial [see Partial difference equations] series solutions of, 158 simultaneous, 159, 182 Difference operator, 2, 3, 11-14 backward, 9, 24, 25, 124 central, 9, 24, 25 255

INDEX

256

Difference operator (cont.)

forward, 9 relationship of to differential and derivative operators, 4 second and nth order, 3 Difference and sum of operators, 2 Difference tables, 23, 32, 33, 40-42 central, 35, 47-49 Differences, calculus of, 4 central, 9, 24, 25, 35, 47-49 leading, 33 Leibnitz's rule for, 9, 24 of polynomials, 33 of special functions, 7 of zero, 26 Differentiable functions, 1, 10

Differential, 3, 4 Differential calculus, 4 Differential-difference equations, 159, 182 Differential equations, 150, 161 boundary-value problems involving, 150 as limits of difference equations, 151, 152,

Events, dependent and independent, 201 mutually exclusive, 201 Everett's formula, 77 Expected duration of a game, 226 Exponents, laws of, 2 Factorial functions, 5, 6, 16-20 generalized, 7 Factorial polynomials, 6 Factorization of operators, 158, 176 Fair game, 211 Families of curves, one- and two-parameter, 197 Fibonacci numbers, 202, 211, 212, 238 table of, 238 Finite differences, 4 Flexural rigidity, 206 Fluid flow, 221 Forecasting, 34 Forward difference operator, 9 Fourier series, 118 Frequency, 199, 202-205

162, 163

numerical solution of, 202, 215-221 order of, 150

partial [see Partial differential equations] Differential operator, 3, 4, 14, 15 relationship of to difference and derivative operators, 4 Differentiation, approximate, 39, 40, 60-62 general rules of, 4, 5 of an integral, 81, 84, 111 of sums, 84, 97 Diagamma function, 86, 104, 105 Distributive law, 2 Divided differences, 38, 39, 56-59 Doubling operator, 1

Gamma function, 6, 70, 76, 85, 86, 102-105 Gauss' interpolation formulas, 36, 50 derivation of, 49 General solution, of a difference equation, 151, 161, 162 of a differential equation, 150, 161 of a partial difference equation, 160 Generalized interpolation formulas, 36 Generating function, for Bernoulli numbers, 87, 110 for Bernoulli polynomials, 87, 109, 110 for Euler polynomials, 89 method for solving difference equations, 157, 158, 174, 175, 183, 184, 186, 187

Gravity, vibrating string under, 221 e, 5

Eigenfunctions, 158, 178, 179 expansions in series of, 159, 179, 180 mutually orthogonal, 159, 178 Eigenvalues, 158, 178, 179 reality of, 158, 179 Elastic collisions, 201 Electrical networks, applications to, 200, 205, 206

Equality of operators, 1 Error function, 70 Error term, in approximate integration formulas, 123, 132-134 in Euler-Maclaurin formula, 125, 137-140 Euler-Maclaurin formula, 124, 136, 137 error term in, 125, 137-140

evaluation of integrals using, 124 Euler method, 216 modified, 217

Euler numbers, 88, 89, 108, 109, 236 Euler polynomials, 88, 89, 108, 109, 237 generating function for, 89 Euler transformations, 148 Euler's constant, 85, 117

Gregory-Newton backward difference formula, 36, 48, 49

derivation of, 48 Gregory-Newton formula, 9, 21-24, 36, 42-47

inverse interpolation by means of, 59, 60 with a remainder, 9, 22, 23 in subscript notation, 34

used in approximate integration, 122, 129, 130

Gregory-Newton forward difference formula [see Gregory-Newton formula]

Gregory's formula for approximate integration, 124, 134-137 Grid, 219 Heat flow, 221

-

[see also Steady-state temperature] Homogeneous linear difference equations, 153 with constant coefficients, 153, 165-168 in factored form, 153

Identity or unit operator, 1 Indefinite integrals, 79 Independent events, 201 Index law, 2

INDEX

Inductance and inductors, 200 Induction, mathematical, 14, 42 Inelastic collisions, 201 Integral calculus, fundamental theorem of, 81, 91, 92 important theorems of, 82 Integral-difference equations, 159 Integral operator, 1, 79, 89-91

Integrals, 79-81 definite [see Definite integrals] evaluation of, 202, 212, 213 indefinite, 79 mean-value theorem for, 81 of special functions, 80 Integrating factor, 158, 216 Integration, 79 approximate [see Approximate integration] constant of, 79 by parts [see Integration by parts] by substitution, 79 of sums, 84 Integration by parts, 79, 90, 91 generalized, 90, 91, 117 Interest, 75 problems involving, 202, 214, 215 Interpolating polynomial, 39 uniqueness of, 44, 45 Interpolation and extrapolation, 34, 45-47 Interpolation formula, Bessel's, 36, 48, 50 Gauss', 36, 49, 50 generalized, 36 Gregory-Newton backward difference, 36, 48, 49 Gregory-Newton forward difference [see Gregory-Newton formula] Lagrange's [see Lagrange's interpolation formula] Stirling's, 36, 47-49 Intervals of convergence, 8 Inverse function, 39 Inverse interpolation, 39, 59, 60 Inverse operator, 2, 11, 79, 156 Iteration, method of, 220, 221 Kirchhoff's laws, 200, 205

Lagrange's interpolation formula, 34, 38, 52, 53 inverse interpolation by means of, 59, 60 proof of, 52, 53 Laplace transforms, 227 Laplace's equation, in three dimensions, 231 in two dimensions, 221 Law of the mean for derivatives, 8 Leading differences, 33 Leibnitz's rule, for derivatives, 9, 24 for differences, 9, 24 for differentiation of an integral, 81, 84, 111 Limits, 3, 15 Linear difference equations, 152 Linear operators, 2, 12, 93, 170 Linearly dependent and independent functions, 154, 164, 165

257 Linearly dependent and independent

solutions, 154, 155, 165, 166

Logarithm, natural, 5 Lozenge diagrams, 36, 37, 50-52

Maclaurin series, 8 Mathematical induction, 14, 42 Mean-value theorem for derivatives, 8, 27 proof of, 27 Mean-value theorem for integrals, 81 Mechanics, applications to, 199 Mesh, 219

Mixed difference equations, 159, 182, 183, 188, 189

Modes, natural or principal, 199 Modes of vibration, 199 Moment equation, 200, 206, 207 Moments, bending, 200, 206, 207 Momentum, conservation of, 201 Montmort's formula, 121 Mutually exclusive events, 201

Natural base of logarithms, 5 Natural frequencies, 199, 202-205 Natural logarithm, 5 Natural modes, 199 Network, electrical, 200, 205, 206 Newton-Gregory formula [see Gregory-Newton formula] Newton's collision rule, 201 Newton's divided difference interpolation formula, 39 inverse interpolation by means of, 59, 60 proof of, 57, 58 Newton's method, 78, 230 Noncommutative operators, 10 Nonhomogeneous equation, 153, 155 solution of, 155 Nonlinear difference equations, 152, 159, 181, 182 Nonlinear operator, 11 Nontrivial and trivial solutions, 158, 165, 180 Normalization, 159 Numerical solutions, of boundary-value problems, 219-221 of differential equations, 202, 215-221 Operand, 1 Operations, 1

Operator, 1 [see also Operators] anti-derivative, 79 cubing, 1, 10, 11 derivative [see Derivative operator] difference [see Difference operator] differential [see Differential operator] doubling, 1

identity or unit, 1 integral, 1, 79, 89-91

null or zero, 1 squaring, 2 translation or shifting, 3, 11-14, 160

INDEX

258

Operator methods, for solving difference equations, 156, 169-172, 187

for summation, 85, 101, 102 symbolic [see Symbolic operator methods] Operators, 1, 10, 11 [see also Operator] algebra of, 2 associative laws for, 2 averaging, 10, 88, 108 definitions involving, 1, 2 distributive law for, 2 equality of, 1 factorization of, 158, 176 inverse, 2, 11, 79, 156 linear, 2, 12, 93, 170 noncommutative, 10 nonlinear, 2 partial difference, 160 product of, 2 sum and difference of, 2 Order, of a difference equation, 151, 152 of a differential equation, 150 reduction of, 157, 173, 174 Orthogonal functions, 159, 178 Orthonormal set, 159

Roots, of auxiliary equation, 153, 154 of equations, 58, 59 Newton's method for finding, 78

Sequences and series, general terms of, 34, 43, 44 [see also Series] uniqueness of, 43, 44

Series, 8 of constants, 121, 126 of eigenfunctions, 159, 179, 180 intervals of convergence for, 8 Maclaurin, 8 power, 121, 122, 127-129 summation of, 97-100, 121, 142, 143 [see also Summation] Taylor [see Taylor series] Series solutions of difference equations, 158 Shifting or translation operator, 3, 11-14, 160 Simply supported beams, 200, 206, 207 Simpson's one-third rule, 122 error term in, 123, 133, 134 proof of, 131 Simpson's three-eighths rule, 123, 132 error term in, 123 proof of, 132

Partial derivative, 81, 160 Partial difference equations, 160, 184-187 solutions of, 160 Partial difference operators, 160 Partial differential equations, 160, 188 numerical solutions of, 218-221 Partial fractions, 192 Partial translation operators, 160 Particular solutions, of a difference equation, 151, 166, 167 of a differential equation, 150, 161 methods of finding, 155 Pascal's triangle, 227 Payment period, 214 Periodic constant, 82, 93 Points, problem of, 211 Poisson's equation, 231 Polar coordinates, 102, 117 Potential theory, 221 Power series, 121, 122, 127-129 Prediction or forecasting, 34 Predictor-corrector methods, 218, 230 Principal, 75, 214 Principal modes, 199 Probability, applications to, 201, 202, 209-211 conditional, 201 Product of operators, 2

Real roots of auxiliary equation, 153 Rectangular coordinates, 103 Reduced equation, 153, 168 [see also Homogeneous equation] Reduction of order, 157, 173, 174 Repeated trials, probability formula for, 225 Resistance and resistor, 200 Restitution, coefficient of, 201, 208 Rolle's theorem, 26

Simultaneous difference equations, 159, 182 Solution, of a difference equation, 150 of a differential equation, 150

nontrivial and trivial, 158, 165, 180 Springs, vibrations of, 223 Squaring operator, 28 Steady-state temperature in a plate, 221, 231 Step by step method, 216 Stirling numbers, 6, 7, 20, 21 of the first kind, 7, 20, 232 recursion formulas for, 7, 20, 21 of the second kind, 7, 20, 21, 86, 106, 230, 233

Stirling's formula for )r;, 125, 140-142 proof of, 141

Stirling's interpolation formula, 36, 47, 48 derivation of, 49 String, under gravity, 221 rotating, 223 tension in, 202

vibrations of, 199, 202-205 Sturm-Liouville boundary-value problem, 158 Sturm-Liouville difference equations, 158, 178-181

Subscript notation, 32, 40, 84 use of in difference equations, 152, 163, 164 Sum calculus, 79-120 applications of, 121-149 fundamental theorem of, 83, 84, 96-98, 100 Sum and difference of operators, 2 Sum operator, 82, 92-95 Summation, 82 general rules of, 82, 92-95 operator methods for, 85, 101, 102 by parts [see Summation by parts] of series, 97-100, 121, 142, 143 of special functions, 83 theorems using subscript notation, 84 using subscript notation, 100

INDEX Summation factor, 158, 174, 176 Summation by parts, 82, 149 generalized, 116

proof of, 94 Sums, definite, 83, 84, 96, 97 indefinite, 82 Superposition principle, 154, 166, 168 Symbolic operator methods, 8 differentiation formulas obtained by, 60, 61 Gregory-Newton backward difference formula obtained by, 48 Gregory-Newton formula obtained by, 24

Synthetic division, 6, 18 Tables, central difference, 35, 47-49

divided difference, 38

location of errors in, 64, 65 with missing entries, 38, 53-56 Tangent numbers, 89, 120 Taylor series, 8 in operator form, 8 solution of differential equations by, 231 Taylor's theorem or formula, 8 Temperature, steady-state, 221, 231 Tension, in a string, 202 Three-moment equation, 200 Torsion constant, 223 Torsional vibrations, 223 Translation or shifting operator, 3, 11-14, 160 Trapezoidal rule, 122, 123, 130 error term in, 123, 132, 133 proof of, 130

259

Trial solution, 155, 156 Trivial and nontrivial solutions, 158, 165, 180 Undetermined coefficients, method of, 155, 156, 168, 169, 187

Variable coefficients, linear difference equations with, 157, 158, 175-178 Variation of parameters, 156-158, 172, 173, 175 solution of first-order linear difference equations by, 175 Vectors, 159, 180 Vibrating systems, applications to, 199, 202-205 Vibration, 199 frequencies of, 199, 202-205 modes of, 199 of springs, 223

of a string [see String] torsional, 223

Wallis' product formula, 141, 149 Wave equation, 222 Weddle's rule, 123, 143, 144 error term in, 123 proof of, 143 Weight or density, 159 Wronskian, 154 Young's modulus of elasticity, 206 Zig-zag paths, 36, 37, 50-52

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