SCHAUM’S OUTLINE OF
THEORY ANB PROBLEMS
OF
Calculus
FINITE
ENCES
and
DIFFERENCE EQUAT
e
w
BY
MURRAY R. SPIEGEL, Ph.D.
Professor of Mathematics
Rensselaer Polytechnic Znstitute
SCHAUM’S OUTLINE SERIES
McGRAW-HILL BOOM COMPANY
New York, St. Louis, San Francisco, Dusseldorf, Johannesburg, Kuala Lumpur, London, Mexico,
Montreal, New Delhi, Panama, Rio de Janeiro, Singapore, Sydney, and Toronto
Copyright @ 1971 by McGraw-Hill, Inc. All rights reserved. Printed in the
United States of America. No p a r t 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
5 6 7 8 9 10 11 12 13 14 15 SH SH 8 7 6 5 4 3 2 1
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 s.ubject. 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 f o r 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 t o 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
Operators. Some Definitions Involving Operators. Algebra of Operators.
The Difference Operator. The Translation o r 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
........
32
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 f o r 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. E r r o r Terms in Approximate Integration Formulas. Gregory’s Formula for Approximate Integration. The
Euler-Maclaurin Formula, E r r o r Term in Euler-Maclaurin Formula. Stirling’s
Formula f o r n !
CONTENTS
Page
Chapter
5
DIFFERENCE EQUATIONS
..................................
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. Line:rly 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 s;
Appendix
B
Stirling Numbers of the Second Kind ,YE..
Appendix
c
Bernoulli Numbers
Appendix
D
Bernoulli Polynomials
Appendix
E
Euler Numbers
Appendix
F
Appendix
G
.................................
232
...............................
233
...................................................
...............................................
234
235
....................................................
236
Euler Polynomials
..................................................
237
Fibonacci Numbers
.................................................
238
ANSWERS TO SUPPLEMENTARY PROBLEMS . . . . . . . . . . . . . . 239
INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
255
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 o r [
13
is the cubing operator and x is the operand then Cx or [
13s represents the cube of x, i.e. x3.
Example 2.
d
If D or is the derivative operator and the operand is the function of x given by f ( x ) = 2x4 - 3x2
d
x
then
Df(x) = D(2x4-3x2f5)
Example 3.
If
4 or
s
~ ( 2 -~3~4
(
d
= -((2x4-3$+
dx
5 ) = 8x3
+5
- 6%
) d x is the integral operator then
+5)
=
[J (
)ds](22'-3x2+5)
=
(2&-3x2+5)ds
= -2x5
5
53
+ 5s + c
where c is a n arbitrary constant.
Example 4.
The doubling operator can be represented by the ordinary symbol 2.
2(2x4 - 3x2
+ 5) =
4x4
- 6x2 + 10
Thus
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 f u n c t i o n s , 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 t o be equal, and we write A = B or
B = A, if and only if for an arbitrary function f we have Af = B f .
2. The identity or unit operator. If for arbitrary f we have I f = f then I is called the
i d e n t i t y 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 0.
1
2
THE DIFFERENCE CALCULUS
4. Sum and difference of operators.
[CHAP. 1
We define
+
( A - B ) f = A f - Bf
(A+B)f= A f Bf,
and refer t o the operators A + B and A - B respectively as the s u m and difference of
operators A and B.
Example 5.
(C
+ D)xz
= Cx2
+ Dxz
=
x6
+ 2x,
(C - D)x* = Cx2 - 0 x 2 =
x6
- 2re
5. Product of operators. We define
(A*B)f= (AB)f = A ( B f )
(2)
and refer t o the operator AB or A B as the product of operators A and B. If A = B
we denote A A or A . A as A2.
Example 6.
(CD)xZ = C(Dx2) = C ( 2 x ) = 8x3,
C2x2 = C ( C x z ) = C ( x 6 ) =
%la
6. Linear operators. If operator A has the property that for arbitrary functions f and g
and an arbitrary constant CY
+
A ( f + g ) = A f A s , A(cuf) = 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 ( B f )= f for an arbitrary
function f, i.e. ( A B ) f= f or A B = I or AB = 1, then we say that B is an inverse of
A and write B = A-I = 1/A. Equivalently A - ' f = g if and only if Ag = f .
ALGEBRA OF OPERATORS
We will be able t o manipulate operators
quantities if the following laws of algebra
any operators.
1-1.
A+B = B+A
1-2.
A + ( B -I-C ) = ( A + B) + C
1-3.
AB = B A
1-4.
A(BC) = (AB)C
1-5.
A ( B + C ) = A B + AC
in the same manner as we manipulate algebraic
hold f o r these operators. Here A , B, C denote
Commutative law for sums
Associative law for sums
Commutative law for products
Associative law for products
Distributive law
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
inde:c latw o r law o f exponents AqnAn= Am+nwhere Am denotes repeated application of
operator A m times.
THE DIFFERENCE OPERATOR
Given a function f ( x ) we define an operator 3, called the difference operator, by
A f ( x ) = f ( x + h) - f ( x )
(4
where h is some given number usually positive and called the difference interval or
di.fei*encing interval. If in particular f ( x ) = x we have
a x = ( x + h ) - x = h or h
= Ax
(5)
Successive differences can also be taken. For example
Azf(Z)
+
= A [ A f ( x ) ] = A [ f ( x h) - f ( x ) ] = f ( x -k 2h) - 2 f ( X -k h )
+ f(x)
(6)
CHAP. 11
THE D I F F E R E N C E CALCULUS
3
We call A2 the second order difference operator or difference operator o f order 2 . In general we define the 92th order difference operator by
A n f ( x ) = A[An-’f(x)]
(7)
THE TRANSLATION OR SHIFTING OPERATOR
We define the translation or s h i f t i n g operator E by
Ef(x)= f ( x + h )
By applying the operator twice we have
+
E2f(lr) = E [ E f ( x ) l = E [ f ( x h)] = f ( x + Z h )
In general if n is any integer [or in fact, any real number], we define
E”f(x) = f ( x + n h )
We can show [see Problem 1.101 that operators E and A are related by
or
A = E - 1
E = l + A
using 1 instead of the unit operator I .
THE DERIVATIVE OPERATOR
From ( 4 ) and (5) we have
A f ( x-)
-
-
f ( x+ h )- f ( x )
h
AX
where we can consider the operator acting on f ( x ) to be AIAX o r Alh. The first order
derivative or briefly first derivative o r simply derivative of f ( x ) is defined as the limit
of the quotient in (11) as h o r Ax approaches zero and is denoted by
if the limit exists. The operation of taking derivatives is called differen,tiation and D is
the derivative or differentiation operator.
The second derivative or derivative of order t w o is defined as the derivative of the
first derivative, assuming it exists, and is denoted by
D”(x) = D [ D f ( X ) ]= f”(x)
(63)
We can prove that, the second derivative is given by
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 diflerential of a
function f ( x )is defined by
(15 )
d f ( x ) = f ’ ( x )A x = f’(z)h
In particular if f ( x )= x we have d x = Ax = h so that (15) becomes
d f ( x ) = f ’ ( x ) d x = f’(x)h
or
df(x) = D f ( x ) d x = h D f ( x )
(16 )
4
T H E D I F F E R E N C E CALCULUS
[CHAP. 1
We call d the differential operator. The second order differential of f ( x ) can be defined as
d 2 f ( x ) = f ” ( ~ ) ( A x ) ~= f ” ( x ) ( d x ) 2
(17)
and higher ordered differentials are defined similarly.
= h2 are numbers which are not
Note that d f ( x ) , d x =ax = h, d2f(x), (dx)2=
necessarily zero and not necessarily small.
It follows from (16) and (17) that
where in the denominator of (18) we have written (dx)2as dx2 as indicated by custom or
convention. I t follows that we can consider the operator equivalence
Similarly we shall write
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 = - d= - d
or
d = hD
dx
h
Similarly from (12) and (16) we see that the relationship between the difference operator A
and the derivative operator D is
with analogous relationships among higher ordered operators.
Because of the close relationship of the difference operator A 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 o f 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
o f 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 f o r differentiation of functions. In the following we list some of the most important ones.
m[f(x)]“-l D f ( x )
m = constant
CHAP. 11
T H E D I F F E R E N C E CALCULUS
=
-
Duf Dg
where
5
D, = dldu
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 )d g ( x ) + g ( x )d f ( x ) =
If(%)
g'(x) + g ( x )f ' ( x ) l d x
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 log, x as In x , called
the natural logarithm. All letters besides x denote given constants.
111-1.
111-2.
111-3.
111-4.
111-5.
111-6.
111-7.
111-8.
D[c]
D[x"]
D [ ( p x q)"]
D[bs]
D[erz]
D[sin Y X ]
D[cos r x ]
D[ln x ]
111-9.
D[log, X ]
+
= o
-
-
mxm-l
m p ( p x + q)m-l
b"1nb
rers
r cos r x
-r sin r x
=
llx
=
=
-
-
-
-logb e
b>0, b # 1
X
These results can also be written in terms of d rather than D. Thus for example 111-2
d(x") = m x m - ' d x
becomes
G E N E R A L R U L E S O F T H E D I F F E R E N C E CALCULUS
The following results involving A bear close resemblances to the results 11-1 t o 11-4 on
page 4.
A [ f ( x )+ g ( x ) ] = A f ( x ) + Ag(x)
IV-1.
A[a:f(X)]
= a:af(x)
a: = constant
IV-2.
A [ f ( x )g ( x ) ]
= f ( x )A s ( % ) + g ( x + h) A f ( x )
IV-3.
= g ( x ) A f ( x ) -k f ( x + h ) A g ( x )
= f ( x ) A g ( X ) + g ( x ) A f ( x ) -l
Af(x)As(x)
Note' that if we divide by Ax and let
AX
+
0 the above results become those of 11-1 to 11-4.
FACTORIAL FUNCTIONS
d
Formula 111-2 states that Dxm = mxm-l. In an effort t o obtain an analogous formula
involving A we write
(X
h)" - xrn
AXm
_ _ AX
+
h
6
T H E DIFFERENCE CALCULUS
[CHAP. 1
which however does not resemble the formula for Dxm. To achieve a resemblance we
introduce the factorial function defined by
= x ( x - h ) ( x - 2h) . ( x - [m- l ] 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- l ) ( m- 2 ) . . a 2 1 = m !, i.e. factorial m.
If m = 0, we define x ( " ) = 1, i .e.
x(0) = 1
(23)
X(m)
a
-
For negative integers we define [see Problem l . l 8 ( u ) ]
1
m = 1,2,3, . . .
( x + h ) ( x 2 h1 ) . . ( x mh) - ( x mh)(m)
Note that a s h + 0, x c m )+ xm, x ( - ~+). x - ~ .
Using ( Z Z ) , (23), (24) it follows that [see Problems 1.17 and 1.18(b)] f o r all integers m
X(-m)
=
+
+
+
in perfect analogy to Dxm = mxm-I and d(xm)= mxm-l d x respectively.
We can also define x ( " ) for nonintegral values of m in terms of the gamma function
[see page 1031.
FACTORIAL POLYNOMIALS
From (22)we find on putting m = 1,2,3,
x(1) = X
...
x(2) = x2 - x h
3x'h + 2xh2
x ( ~=
) x4 - 6x3h
1 l ~ ' h ' - 6xh3
x ( ~ =) x 5 - 10x4h 35x3h2- 50x2h3+ 24xh4
x(') =
X' -
+
+
I
etc. If p is any positive integer, we define a factorial polynomial of degree p as
a0x(p)+ a,x(P-l) + . . +a,
where a0 # 0, U I , . . . ,a, are constants. Using ( 2 6 ) 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 2 = x ( 2 ) + x'l'h
I
x 3 = x ( 3 ) + 3x(2)h+ x(i)hz
$4 = ~ ( 4 +
) 7x(3)h+ 6 ~ ( 2 ) h+
2 x(l)h3
1
+
x 5 = x ( ~+) 1 5 ~ ' ~ ) 25xc3)h2
h
+ l O ~ ( ~+ )x (hl )~h 4
etc. Another method for converting an ordinary polynomial into a factorial polynomial
uses synthetic division [see Problems 1.23-1.251.
STIRLING NUMBERS
Any of the equations ( 2 6 ) can be written as
CHAP. 11
7
THE DIFFERENCE CALCULUS
where the coefficients s: are called Stirling numbers of the first kind. A recursion formula
for these numbers is
s;+l = s:-~ -ns:
(29)
where we define
s,"= 1, s: = 0
k S 0, k 2 n
for
+1
where n > 0
(30)
Similarly any of the equations ( 2 7 ) can be written as
where the coefficients S: are called Stirling numbers o f the second kind.
formula for these numbers is
Slt" = 8:-1 + k S :
where we define
SE = 1, Slt = 0 for k d 0, k 2 n+ 1 where n > 0
A recursion
(32)
(33)
In obtaining properties of the Stirling numbers results are greatly simplified by
choosing h = 1 in (28) and ( 3 1 ) .
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
If (x)]'"'
=
f ( x )f (X - h) f ( X
- 2h) *
* *
f ( z- [m - l]h)
m = 1,2,3, . . .
(34)
The special case f ( x )= x yields (as),(2.4) and (223) respectively. It should be noted that as
h + 0, [ f ( ~ ) ] ( +
~ ) [f(x)]" and [f(x)](-")
+ [ f (43
DIFFERENCES OF SPECIAL FUNCTIONS
In the following we list difference formulas for special functions analogous to the
differentiation formulas on page 5 .
v-1.
A[c]
- 0
v-2.
A[Xcm)]
- mx(m-1) h
V-3.
V-4.
V-5.
V-6.
v-7.
V-8.
v-9.
+
A[(pX + q ) c m ) ]= mph(px q)'m-"
A[b"]
- bx(bh- 1)
- er5(erh- 1)
A[erx]
- 2 sin (rh/2)cos ~ ( xh/2)
A[sin rx]
A[cos TX]
- -2 sin (rh/2)sin r(x h/2)
A[lnx]
- In (1 h / s )
A[lOgb X]
- b g b (1 h / x )
+
+
+
+
Note that if we divide each of these results by AX = h and take the limit as AX or h
approaches zero we arrive a t the corresponding formulas for derivatives given on page 5.
For example V-5 gives
8
THE DIFFERENCE CALCULUS
[CHAP. 1
TAYLOR SERIES
From the calculus we know that if all the derivatives of f(x) up t o order n 1 a t 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
?’(a) (x - a ) 2 + . , . + fy
a) (x - a).
+ Rn (37)
f(x) = f ( 4 + ?(a) (2 - a ) +
2!
n!
+
where the remainder Rn is given by
f(n+l)(v)(x - U)”+l
(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.
Rn
=
+
If lim Rn = 0 for all x in an interval. then
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
+
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.
1.
x7
x3
x5
+ - +
5!
7!
3!
2.
sin%
1
x
3.
cosx
=
x6
x2
x4
I - - + - - - +
6!
2!
4!
4.
In(l+x)
=
x - 2
-
x2
x3
x4
+--+
3
4
...
--m<x<m
*..
--m<x<m
TAYLOR SERIES IN OPERATOR FORM
If in (40) we replace a by x we have
f(x+h)
= f(x)
+
+
f’(x)h
- l < x l l
* * *
f”(X)h2 +
,
.,
which can formally be written in terms of operators as
E~(x) =
[ l + h D h2D2
+ T +h3D3
m
+
. * * ] f ( x ) = ehDf(x)
using the series for ehDobtained from the result 1 above on replacing x by hD. This leads
to the operator equivalence
E = ehD
(42)
or using ( l o ) ,page 3,
A = ehD- 1
or
ehD = 1 + A
(43)
Such formal series of operators often prove lucrative in obtaining many important results
and the methods involved are often called symbolic operator methods.
CHAP. 11
9
THE DIFFERENCE CALCULUS
THE GREGORY-NEWTON FORMULA
As may be expected there exists a formula in the difference calculus which is analogous
t o the Taylor series of differential calculus. The formula is called the Gregory-Newton
formula and one of the ways in which it can be written is
which is analogous t o ( 3 7 ) . The remainder R, is given by
where 17 lies between a and x. If lim Rn = 0 for all x in an interval, then (44) can be
n-t m
written as an infinite series which converges in the interval.
If f(x) is any polynomial of degree n, the remainder Rn = 0 for all x.
LEIBNITZ'S RULE
In the calculus there exists a formula for the nth derivative of the product of two
functions f ( x ) and g(x)known as Leibnitx's rule. This states that
qfs)
=
( f ) ( D " g )+ ( ; ) ( D f ) ( D n - W
+ (;)(DZf)(D9 + .*.+
(;)(w(s)
where for brevity we have written f and g for f ( x ) and g ( x ) and where
+
n!
n(n- 1).. .(n- r 1) 0!=1
=
r!
r!(n-r)!'
are the binomial coefficients,i.e. the coefficients in the expansion of (1 x),.
(:)
+
An analogous formula exists for differences and is given by
A n ( f g ) = ( f ) ( A n g )+ ( : ) ( A f ) ( A n - l E g ) 4- ( ~ ) ( A z f ) ( A n - z E ' g+
)
* ' .
+
(:)
(A"f)(EV)
which we shall refer to 'as Leibnitx's rule f o r differences. See Problem 1.36.
OTHER DIFFERENCE OPERATORS
Various other operators are sometimes used in the difference calculus although these
can be expressed in terms of the fundamental operators A and E. Two such operators are
v and 6 defined by
(49)
Sf(X)
= f(x+i) -
f(%-k)
(50)
v the backward differenceoperator [in contrast with A which is then called the
forward difference operator] and 6 the central difference operator. These are related to
A and E by
We call
See Problems 1.38 and 1.39.
10
T H E D I F F E R E N C E CALCULUS
[CHAP. 1
Two other operators which are sometimes used are called averaging operators and are
denoted by M and p respectively. They are defined by the equations
M f ( x ) = H f ( x + h) + f (43
;[++;)
=
pf(x)
(53)
+ f (x-S)]
Some relationships of these operators to the other operators are' given by
M = *(E+1) =
~
1+*A
(55)
= +(E1/2+E-l/2
)
(56)
See Problem 1.40.
Solved Problems
OPERATORS
1.1. If C is the cubing operator and D is the derivative operator, determine each of the
following.
+
( e ) (C 0 ) ( 4 x+ 2 )
( f ) ( D -i-C ) ( ~ X2 )
( 8 ) (Cz- 4C l ) ( F l )
(a) C ( 2 x + l )
( b ) D(42 - 5x3)
+
(c) CD(2x - 3 )
+
+
+
+4)
(h) (D2- 2 0 1)(3x2- 5~
(i) (C - 2 ) ( 0 3)xZ
( j ) (xCxD)x3
( d ) DC(2x - 3 )
+ 1) = (2%+ 1)3
(c) CD(2x - 3 ) = C * 2 = 23 = 8
( b ) D(4x - 5x3) = 4 - 15x2
( d ) D C ( 2 x - 3 ) = D(22 -3)3 = 3 ( 2 ~ - 3 ) 2 * 2 6 ( 2 ~ - 3 ) 2
(a) c(2x
+
+
Another method. DC(2x - 3) = D(2x - 3)3 = D[8x3- 36x2 542 - 271 = 24x2 - 72x 54
Note that from (c) and (d) i t is seen that the result of operating with CD is not the same as
operating with DC, i.e. CD # DC so that C and D are noncommutative with respect to multiplication.
(e) (C
( f ) (D
+ D)(4x+ 2)
+ C ) ( 4 x-t 2)
= C(4x
+ 2) + D(4x + 2)
= D(4x -t 2)
+ C(4x + 2)
+ 2)s + 4
+ (4%+
= (4x
= 4
The results ( e ) and ( f ) illustrate the commutative law o f addition f o r operators C and D,
i.e. C-t D = D C.
+
( 9 ) (Cz - 4C
(h)
(02
+ 1)(=1)
= C Z ( e 1 ) - 4C(=l)
+l
3
( g 1 ) = ( x - 1)3- 4(x - 1 ) -t&=i
+ 4 ) - ZD(3x2- 5 2 + 4 ) + l ( 3 ~ 2 -5~ + 4)
+ 3 6 - 5%+ 4 = 3x2 - 1 7 +~ 20
(C - 2)[Dx2+ 3x21 = ( C - 2)(2x + 3%')
C(2x + 3x2) - 2(2x + 3x2) = (2%+ 3x2)3 - 2(2x + 3x2)
- 2 0 + 1 ) ( 3 ~-2 52 + 4 ) =
D 2 ( 3 ~ 2- 5,
= 6 - 2 ( 6~ 5)
(i) ( C - 2)(U
+ 3 ) ~ 2=
=
( j ) (xCxD)x3 = xCx(Dx3) = sCx(3x') = xC(3x3) = ~ ( 2 7 x 9 )= 27x10
1.2.
I f D is the derivative operator and f is any differentiable function, prove the operator
equivalence D x - X D = I = 1, i.e. the unit o r identity operator.
We have
(Dx-xD)f
= Dxf - xDf
d
df
df
= -dx
( ( x j ) - x-dx = x-+
dx
f -
Then for any differentiable f , ( D x - xD)f = If = 1 o r D x - X D = Z = 1.
xL'df
dx = f
= If
CHAP. 11
1.3.
T H E D I F F E R E N C E CALCULUS
11
Prove that (a) the cubing operator C is a nonlinear operator while ( b ) the derivative
operator D is a linear operator.
(a) If
f and g a r e any functions we have
+
+
+
+
C ( f 9) = ( f s)3, C f cg = f 3 g3
Then in general since ( f g ) 3 f f 3 g 3 i t follows that C ( f 4- g) ;f Cf
be a linear operator, i.e. C is a n o n l i n e a r operator.
+
+
+ Cg
so that C cannot
( b ) If f and g a r e any differentiable functions, we have
Also if a is a n y constant, then
Then D is a linear operator.
1.4.
D ( a f ) = a -ddfx
= aDf
(a) If C denotes the cubing operator, explain what is meant by the inverse operator
C-'. ( b ) Does C-' exist always? (c) Is C-l unique? Illustrate by considering C-l(f3).
is that operator such t h a t C B f ( x )= f(x) o r CB = I = 1, the unit
or identity operator. Equivalently C - l f ( x ) = g ( x ) if and only if C g ( x ) = f ( x ) o r [ g ( x ) ] 3= f(z),
i.e. g(x)is a cube r o o t of f ( x ) . Thus C-l is the o p e r a t i o n o f taking cube r o o t s . As a particular
case C-l(8) is the result of taking cube roots of 8.
( b ) If f(x) is real, then C - l [ f ( x ) ] willalways exist but may o r may not be real. For example if
C-l(8) is denoted by x then x3 = 8, i.e. (x - 2 ) ( x 2 2 x 4) = 0 and x = 2, -1 fi i so that
there a r e both real and complex cube roots. The only case where C - l f ( x ) is always real when
f ( x ) is real is if f(x) = 0.
If f ( x ) is complex, C - l f ( x ) always exists.
( c ) It is clear from ( b ) t h a t C - 1 f ( x ) is not always unique since f o r example C-l(8) has three different values. However if f ( x ) is real and if we restrict ourselves to real values only, then C - l f ( z )
will be unique. Thus if we restrict ourselves to real values we have for example C-l(8) = 2 .
( a ) By definition B = C - l
+ +
*
THE DIFFERENCE AND TRANSLATION OPERATORS
1.5. Find each of the following.
(a) 4 2 x 2 3 X )
( d ) E3(3X - 2 )
( 9 ) (A 1)(2A- 1)(x2 + 2x + 1)
'
(h) (E- 2)(E - 1)(2"Ih+ X)
( b ) E(4x - X 2 )
( e ) (2A24-A - 1 ) ( X 2 + 23: 1)
(C) A2(X3- X 2 )
( f ) (E2- 3E + 2)(2"Ih+ X)
+
(a) A(2&
+ 3%)
( b ) E ( 4 x - x2)
(c) A2(x3 - 22)
(d) E 3 ( 3 ~ 2)
+
+ +
+ 2/22 + 3 h
+ +
+
= 4 ( x + h) - (x + h)2 = 4 2 + 4 h - x2 - 2 h z - h2
= A [ { ( %+ h)3 - (X + h)2}- (x3- xz)]
= A [ A ( d - &)I
= A[3x2h + 3xh2 + h3 - 2hx - h2]
= [ 3 ( x + h)2h + 3 ( x + h)h2 -t h3 - 2 h ( z + h) - h2] - [3x2h+ 3xh3 + h3 - 2hx - h2]
= 6h% + 6h3 - 2hZ
E2[3(x+ h) - 21 = E .E [ 3 ( x+ h) - 21
= EZE(3x -2 )
= [ 2 ( ~ h)2 3 ( x + h)]- [222 + 3x1
= 2x2 4 h z 2h2 t 3x 3h - 2x2 - 3% = 4hx
= E[3(x+2h)-Z]
( e ) (2A2
+
+ A - I)(&+ 2% + 1) =
Note t h a t we use 1 in 2Az
= 3(x+3h) - 2
+
+ +
+ +
+ +
+ +
+
+
+ + +
+ + +
+ +
+
+
+ +
+
+
+ +
2A2(x2 2% f 1) A(x2 2% f 1) - 1(x2 2 2 1)
= ~ A [ { ( x h)2 2 ( x f h) f 1}- { X 2 2% I}]
[{(x h)2 t 2 ( x h) l}- {x2 2%-I l}]- 5 2 - 2% - 1
= 2 A [ 2 h 2 h2 -k 2h] [2h% hz 2/21 - X 2 - 2% - 1
= 2 [ { 2 h ( x h) h2 2 h } - {Zhs h2 2 h } ]
2 h x i- h2 2 h - x2 - 2s - 1
= 5/22 2 h x 2 h - 2 2 - 2 x - 1
+ A - 1 instead of the unit o r identity operator I.
12
THE D I F F E R E N C E CALCULUS
(f) (E2- 3E
+ 2)(2z/h + X)
+
[CHAP. 1
+
+ +
= E2(25/h
X) - 3E(2z/h X)
2(2”/h X)
h x + h) + 2 ( ~ + h ) / +
h 22
= (2(X+2h)/h + x + 2h) - 3 ( 2 ( ~ t h ) / +
+ x + 2h) - 3(2
= (4 2xlh
2x/h
+ x + h) + 2
2z/h
+ 22
= -h
(9) (A
+ 1)[2A(x2+ 2%4-1) - l(x2 + 2%+ I)]
= (A + 1)[2{(x-th)24-2(2 + h) 1)- 3(x2 + 22 + l)]
= (A + 1)[4hx + 2 h 2 + 4h 2x - 11
= A [ 4 h ~+ 2h2 + 4h - x2 - 2%- 11 + [4hz+ 2h2 + 4h - x2 - 2 2 - 11
= 4h(x + h) + 2h2 + 4h - (x+
- 2(x + h) - 1
=
+ 2hx + 2 h - - 2x - 1
+ 1)(2A - 1)(x2+ 2x + 1) =
(A
~
2
-
h)2
5/22
22
Note t h a t this result is the same a s t h a t of ( e ) and illustrates the operator equivalence
1)(2A - 1) zz 2A2 A - 1.
(A
+
+
+
+
+
(h) (E - 2)(E - 1)(2z/h x) = (E - 2)[E(25’h x) - (2z/h x)]
= (E - 2)[(2(%+h)/h x h) - (2Z’h
= (E - 2)[2X/h h]
+
= E[2Z/h
X I
+ +
+ x)]
+ h] - 2[25/h + h]
2(zth)/h
+ h - 2(z+h)/h - 2 h
= -h
Note t h a t this result i s the same as that of (f) and illustrates the operator equivalence
(E-2)(E-l)
= Ez-3E+2.
CHAP. 11
1.8.
THE DIFFERENCE CALCULUS
Prove formulas (a) V-4 ( b ) V-5 (c) V-6, page 7.
(4
A[bZ] = b Z t h - b x =
(b)
A[eTZ] = eT(Zth) - eTZ =
(4
A[sinrx] = sinr(x
13
&(bh-1)
eTZ(eT'h-
1)
+ h) -
on using the trigonometric formula
sin el - sin e2 = 2
with el = r ( x
1.9.
+ h) and
e2 = r x .
Prove that AE = EA, i.e. the operators A and E are commutative with respect to
multiplication.
We have for arbitrary f(z)
A E f ( x ) = A [ E f ( x ) ] = A [ f ( ~ + h ) ]= f ( x + 2 h )
+
E ~ f ( x=
) E [ ~ ( z~ t) f ( ~ ) ]= f ( x
Then A E f ( x ) = E A f ( x ) , i.e. A E = E A .
-f(z+h)
+ 2 h ) - f(x + h)
1.10.
We have for arbitrary f(z)
Thus A = E
-1
A f b ) = f(x -th) - f ( s ) = E f b ) - f ( x )
(E- l)f(z)
where we use 1 instead of the unit or identity operator I .
We have for arbitrary f(x)
+
E f ( x ) = f(z+ h) = fb) [fb+
h) -fb)] = fb)+ A f ( x ) = ( 1 + A ) f ( x )
Thus E = 1 + A . Note that this illustrates the fact that we can treat operators A and E as
ordinary algebraic quantities. In particular we can transpose the 1 in A = E - 1 to obtain
E=l+A.
We have f o r arbitrary f(s)
Azf(s) = A[Af(s)] =
Thus A2 = ( E - 1 ) 2 = E2 - 2 E
+ 1.
-
1.11. Obtain a generalization of Problem 1.10 f o r an where n is any positive integer.
Method 1.
Usinn the fact that A and E can be manipulated as ordinary algebraic quantities we have by
where
(:>
n!
= r ! ( n- r ) !
are the binomial coefficients.
By operating on f(x) the result (I) is equivalent t o
~ n f ( x )= f ( x
+ nh) -
f(x
+ (n- 1)h) +
f(x
+ (n - 2)h) -
* *
+ (-l)nf(s)
14
THE DIFFERENCE CALCULUS
Method 2.
Assume that
~ n f ( x ) = f(x + n h )
-
f(x
[CHAP. 1
+ (n- l)h) +
+ (-l)nf(x)
is true for a positive integer n, i.e. assume the operator equivalence
Then operating on both sides of ( 2 ) with A we have
+ (n+ 1)h) - f ( x + n h ) -
A n t l f ( z ) = f(x
+
+ n h ) - f(x + (n- l ) h ) ]
+ (-l)"+'f(x)
(:)
or since [see Problem 1.1051
An+'f(x) = f ( x
[f(x
... + ( - l ) " [ f ( x + h) - f(41
- ...
we have
(3
(A)= (Z)
+
n+l
+ (n+ 1)h) - n + l
- ... + ( - l ) " + l f ( x )
(8)
It follows that if ( 2 ) is true then (3)is true, i.e. if the result is true for n i t is also true for n+ 1.
Now since ( 2 ) is true for n = 1 [because then ( 2 ) reduces to A f ( z ) = f(x h) -f(x)], it follows
that it is also true for n = 2 and thus for n = 3 and so on. I t is thus true for all positive integer
values of n. The method of proof given here is called mathematical induction.
+
1.12.
Prove that (a) E - l f ( x ) = f ( x - h), (b) E-nf(x) = f(x - nh) for any integer n.
+
(a) B y definition if E - l f ( x ) = g(x), then E g ( s ) = f(z), i.e. g ( z h) = f(x) or g(s) = ffz - h)
on replacing x by x - h. Thus E - l f ( x ) = f(z - h).
( b ) Case 1. If n is a negative integer or zero, let n = -m where m is a positive integer or zero.
Then
Case 2.
E m f ( x ) = f ( s + m h ) , i.e. E - n f ( x ) = f ( s - n h )
If n is a positive integer, then by definition if
E-nf(x) = g(x) then E n g ( x ) = f(z) or g ( x + n h ) = f(x)
Then replacing x by x - nh we have
E-nf(s) = g(z) = f ( s - n h )
Enf(.x) = f ( x + n h )
In general we shall define
for all real numbers n.
THE DERIVATIVE AND DIFFERENTIAL OPERATORS
1.13.
A
Show that
lim -((zx2
(a)
A x r ~Ax
(b)
AX+O
+ 3x) = -axd( 2 x 2 +
3x) = 4x
a2
= 6x - 2
lim 4 ( x 3 - x 2 ) = 7(x3-x2)
dx
Ax
+3
directly from the definition.
(a) From Problem 1.5(a) we have since h = Ax
A
-(2x2+32)
Ax
Then
A
lim -(2x2+32)
AX
Ax-rO
=
4hx
+ 2h2 + 3h
h
= lim ( h x + 2 h + 3 )
hrO
=
42
4x
+
+3
2h
+
d
= ;EjE(2x2+ 3s)
CHAP. 11
THE DIFFERENCE CALCULUS
16
( b ) From Problem l.S(c) we have since h = Ax and h2 = AX)^
+
= 6 h z ~ 6h3 - 2hz = 6 ~ + 6 h - 2
h2
A2
-(x3-x2)
Ax2
Then
lim
AZHO
= lim (6x + 6h-2)
- A2
(x3-x2)
Ax2
The results illustrate the fact that ~
An = dn
lim dxn.
Axn
bz-ro
The notation D =
speaking we should write
= 6x
h-0
An
as
Axn
-2
= ddz
22(x3-~2)
f= (~dnfx ( )x ) .Note the operator equivalence
~ An
o ~
d
d2
z,
D2 = a,
. . ., D n = dn
can also be used.
dxn
An and
(Ax)"
dn a s
d"
However
(dx)n *
Strictly
by custom we leave off the
parentheses in the denominator,
1.14. Find (a) d(2x2+3x), ( b ) d 2 ( z 3 - x 2 ) .
(u) By definition, using
dx = Ax = h we have
d
d ( 2 ~ 2 + 3 ~=) - ( 2 ~ 2 + 3 ~dx
)
dx
(b)
(42
+ 3) dx
+
= ( 4 ~ 3)h
B y definition,
dZ(x3
1.15.
=
- $2) =
- x 2)(dx)z =
-C
P
( ~ 3
dx2
(62 - 2 ) ( d ~ ) 2 = ( 6 s -2)hZ
Prove that D [ f ( x )g(x)] = f(x) D g ( x ) + g(x)Df(x).
From Problem 1.7(u) we haye on dividing by h = Ax
A
Ax [f(49(4]
=
f(4A
9(4
A
+ 9@ + h)1
:f(4
Then taking the limit as h = Ax+O we have
d
[t(x,9(x)]
d
= f ( 4 ;i;Ed4
+ 8(4 d f ( 4
which gives the required result on writing d / d x = D.
sin 8
d
1.16. Use Problem 1.8(c) and the fact that lim 7
= 1 to prove that ds sin T X = T cos z.
8-0
From Problem 1.8(c) we have since Ax = h
A
[sin rx]
Ax
2 sin (rh/2) cos r(x
h
+ h/2)
= r*l*cosrx
- r cosrx
We have used here the theorem familiar from the calculus that the iimit of a product of functions is
equal to the product of the limits of the functions whenever these limits exist.
16
THE DIFFERENCE CALCULUS
[CHAP. 1
FACTORIAL FUNCTIONS
A
1.17. If x C m )= x(z - h)(x - 2h) * . (x - [m - llh), m = 1, 2, 3, . . ., prove that -xCm) =
AX
mx(m-l)
or equivalently A d r n ) = mx(m-l)hwhere h = A x .
We have
Az(lil) = (z + h)(Vl) - x ( m )
L1
= ( z + h ) ( x ) ( z - h ) . . . ( ~[-m - 2 ] h ) - ~ ( x - h ) ( x - 2 h ) . . . ( x - - [ m - l ] h )
= x(x - h ) . * .(x - [ m- 2]h){(x h ) - (x - [ m- ljh)}
+
-
mx(m-l)h
Note t h a t f o r m = 1 this formally reduces to A x ( 1 ) = z(O)h
However since Ax(” = Ax = h we a r e led to define d o ) = 1.
1.18. (a) Motivate the definition of x ( - ~ )m, = 1 , 2 , 3 , . . . , given on page 6 . ( b ) . Using the
defihition on page 6 prove that
...
( a ) Fr o m the definition of z(m) f o r “m = 1 , 2 , 3 ,
Z(m)
= z(z - h) *
x(m+l)
so t h a t
we have
. .( 5 - [ m - l ] h )
= z(z-h)...(x-mh)
= x(m)(x-mh)
z(m+l)
If now we formally p u t m = -1 in this last result, we a re led t o
X ( 0 ) = x(-l)(x.+ h)
Using d o ) = 1, from Problem 1.17 we a r e thus led t o define
= I
x(-1)
Similarly, putting m = -2
2-h
in (1) we find
=
2(-1)
x(--2)($
+ 2h)
so t h a t we a r e led t o define
x(-1)
1
= -(x h)(x 2h)
x 2h
Proceeding i n this way we a r e thus led to define
1
x(-m)
=
m = 1,2,3,
(X
+
+
x(-2)
+
+ h ) ( z+ 2h). - . (z + mh)
as on page 6.
( b ) We have
Ax(-rn)
=
(%+
-
1
1
( x + 2 h ) ( z + 3 h ) . . . ( z +[ m + l ] h ) ( x + h ) ( x + 2 h ) . . . ( x + m h )
-
-
h)(-m) -
%(-m)
1
1’
-
1
[
+ (m+l)h a + h
-mh
(x + h)(x + 2h). (x + mh)(x+ [m+ 11h)
( x + 2 h ) * . ’ ( z + m h )x
-mx(-m-l)h
Ax(-m)
-
o r equivalently
-
-
-mx(
- m - 1)
Ax
Note t h a t the results of this problem and Problem 1.17 enable us t o write for all integers m
Ax(m)
__ = m z ( m - l )
Ax
or
~ x ( m )=
mx(m-l)h
CHAP. 11
THE DIFFERENCE CALCULUS
17
1.19. If m is any integer and c is any constant prove that
( b ) Since AX = h we have from part (a)
A[cx(m)] = cAx(m) = mchx(m-1)
Method 2.
Since A and thus A f A x are linear operators we have
and the required result follows a t once from Problems 1.17 and 1.18.
A
A
ax
[ 3 ~ ( ~(b)
) ] ,A [ ~ O X ( ~ ) ](c)
, ax [ 5 ~ ( - ~ )(d)
] , A [ - ~ z ( - ~ ) ] expressing all results
1.20. Find (a)
in terms of h.
From the results of Problem 1.19 we have
= 4 -3253) = l2x(3) = 1 2 4 %- h)(x -2h)
(a) f-[3x(4)]
( b ) ~ [ 1 0 ~ ( 3 )=1 3 . 1 0 ~ ( 2 ) h = 3ohx(2) = 3 0 h ~ ( x - - )
1.21. Find (a)
A
+
( 2 ~ '-~3d2)
)
x - 4), ( b )
- 2 ~ '-~5 d) - l ) 1.
By Problem 1.19 and the fact that A and thus AlAx are linear operators we have
(a)
A
(2d4) - 3 d 2 )+ x - 4) =
Ax
8d3) - 62")
A ( 3 ~ - 2- 2x(2) - 5xc-l))
=
(6) Ax
(-62(-3)
+1
- &(l) f 5x(-Z))h
+
1.23. Express 2x3 - 3x2 5%- 4 as a factorial polynomial in which the difference interval is h.
Method 1.
From page 6 we have
2 x 3 - 3~ + 52
-4 =
+ 3 x ( 2 ) h+ .(1)-2) - 3 ( ~ ( 2+) ~ ( 1 ) h+) 5 ~ ( 1-) 4
+ (6h - 3)2(2) 4- (2h2- 3h + 5)x(') - 4
2(q3)
= 2x(3)
18
T H E D I F F E R E N C E CALCULUS
Method 2.
Write
2x3 - 3 x 2
+
+ 5x - 4
[CHAP. 1
+
+
= A , d 3 ) Alx(2) A,x(l) A ,
= A,.(% - h ) ( z- 2h) A,%(%- h)
+
+ AZx + A3
+ ( A , - 3A0h)x2 + (2Aoh2- A , h + A,)% + A ,
= A,x3
where A , , A , , A,, A , a r e constants to be determined. Equating coefficients of like powers of x we
have
A , = 2 , A1 - 3Aoh
-3,
2Aoh2- A l h A , = 5 , A3 = -4
+
A , = 2,
from which
Thus
2x3 - 3x2
A , = 6h - 3,
+ 5%- 4
A2 = 2 h 2 - 3 h
+ 5,
A , = -4
= 2 x ( 3 ) + (6h - 3 ) x ( 2 ) + (Zhz - 3h
+ 5 ) x ( 1 )- 4
Method 3.
A s in Method 2 we have
2x3
Let x = 0.
- 3x2
+ 5x - 4
Then A , = -4
+ 5%
2x3 - 3x2
Then dividing by
%
= A,%(%- h)(x - 2h)
+ A,%(%- h) + A,% + A3
= A,x(x - h)(x - 2 h )
$-
so t h a t
we find
2x2 - 3%
+5
= A,(%- h ) ( x- 2h)
A l x ( x - h)
+ A , ( %- IL) - A ,
Now let x = h. Then A , = 2h2 - 3 1 ~ + 5 so t h a t
Z(x2 - h2) - 3 ( 2 - h) = Ao(x - h ) ( x - 2h)
Then dividing by x - h we find
2(x
+ h) - 3
+ A2x
= A,(%- 2 h )
+ A,(%- h)
+ A,
Letting x = 2h we then find A , = 6 h - 3 so t h a t 2x - 4h = A,(%- 2 h ) and A , = 2.
obtain the same result as in Method 1.
Thus we
Method 4.
A s suggested by Method 3 we see t h a t A , is the remainder on dividing 2x3 - 3x2 5%- 4 by x
yielding a first quotient, A , is the remainder on dividing the first quotient by x - h yielding a
second quotient, A , is t h e remainder on dividing the second quotient by x - 2 h yielding a third
quotient and finally A , is the remainder on dividing the third quotient by x - 3 h yielding a quotient
which should be zero. The results can be illustrated as follows
+
+ 5x
X
)2%3 - 3x2
x - h
)2x2-3x+5
- 4
- 4
C-
F i r s t remainder
C-
Second remainder
~~
x - 2h) 22
+ (2h-3 ) +
+5
2h2 - 3h
x - 3 h k
6h
-
3
2
Third remainder
-
C-
Fourth remainder
The required coefficients a r e given by these remainders reading upwards.
2x3 - 3x2
+ 5x - 4
= 2 x ( 3 ) + (6h - 3 ) 2 ( 2 )
Thus we have
+ (2h2 - 3h + 5 ) x ( 1 ) - 4
Method 5.
Since the quotients and remainders can be found by use of synthetic divisiolt in which only
coefficients of the various powers of x a r e used we can a rra nge the computation of Method 4 in
the following form
h
1
2
-3
5
I
2h
~
2h
2
(2>
2h4;3
1
2 h 2 - 3h
j
)@
-
(m)
where t h e required coefficients a r e shown encircled.
tipliers to be used at each stage.
The numbers a t the extreme left indicate mul-
CHAP. 11
19
T H E D I F F E R E N C E CALCULUS
+
1.24. Express 2x3 - 3x2 5%- 4 as a fakorial polynomial in which the differencing interval h = 1.
I n this case a n y of the methods of Problem 1.23 can of course be used and the results a r e
greatly simplified because h is replaced by 1. For example Method 5 which is the simplest method
reduces to the following
2
4
3
2 x ( x - 1)(x
2x3 - 6x2
2x3 - 3x2
- 2) + 3x(x - 1) + 4% - 4
+ 4x + 3x2 - 3% + 4x - 4
+ 5x - 4
1.25. Express x 4 + x - 2 as a factorial polynomial in which h = 1.
We must be careful in using Method 5 of Problem 1.23 to introduce zero coefficients where
needed. Thus we have the following
1
1
0
0
1
1
1
1
2
1
6
3
1
3
3
7
+ -2
=
-
2
-
Thus
x4
Check.
+ 6 ~ ( 3 +' 7 ~ ( "+ ZX"'
+~
d4)
2
+
~ ( 3 7x(2)
)
+ 2 x c ~) 2
- 2 ) ( -~ 3) + 62(x - 1 ) (-~2) +
- 1)+ 2 x - 2
+ 11x2 - 6%+ 6x3 - 18x2 + 12%+ 7x2 - 7 x + 22 - 2
- 2 = X(X - l)(x
IL.(~)
= x4 - 6x3
~ X ( X
= x4+x-2
1.26. Show that if m = 1 , 2 , 3 , . , .
(ax b)'") = (ax b)(ax b - ah)(ax b - 2ah). * (ax 3. b - mah
+
+
+
+
The result follows from definition ( 3 4 ) , page 7, on letting f(x) = ax
f(x - h) =
U(X
-h)
+b
= ax 4- b - ah,
f ( x - [m - l ] h ) = ~ ( -x [m- l]h)
+b
f ( x - 2 h ) = a(%- 2 h )
=
rtx
+b -~
+b
+b
so that
= ux
[ ml ] h = ax
+ah)
+ b - Zah, . . .,
+ b - mah 4-ah
1.27. Show that if m = 1,2,3, . . .
(ax
+ b)(-m)
=
1
(ax + b + ah)(ax + b + 2ah). (ax + b + maG
+
The result follows from definition ( 3 5 ) , page 7, on letting f ( x ) = ax b so t h a t
b
f ( x + h ) = a ( x + h ) b = a x b ah, f ( x + 2h) = a(%+2h) b = a x
+
+ +
f ( x + m h ) = a(x +mh)
+
+b
= ax + b
+ mah
+ + Zah,
. . .,
20
THE DIFFERENCE CALCULUS
(a) Letting h
= 1, a = 2, b = 9 and
(2s
( b ) Letting h = 1, a= 3, b = -5
m
= 3 in Problem 1.26 we have
+ 9)(3) =
and m
(3s - 5)(-4'
[CHAP. 1
(2x
+ 9)(2s + 7)(2x + 5 )
= 4 in Problem J.27 we have
=
1
(3s - 2)(3s
+ 1 ) ( 3f~ 4)(3s 4-7)
STIRLING NUMBERS
1.29. Derive the recursion formula (29), page 7 , for Stirling numbers of the first kind.
Using h = 1 in (28),page 6, we can write the result as
where we take
sz
=0
for k 5 0, k 2 n + l
where n
>0
since the series (1) is actually finite and
s;
= 1
since both sides of (1)are polynomials of degree n.
Replacing n by n
+ 1 in
(I) we have
$(n+l)
=
sjt+lxk
k=-x
Now if h = 1 we have
= X(n)(s-n)
X(n+l)
Then substituting (I) and (a) into ( 4 ) we have
Equating coefficients of xk we find as required
s;+'
= s : - ~ -ns:
1.30. Derive the recursion formula (32),page 7 , for Stirling numbers of the second kind.
where we use ( J I ) , page 7, with h = 1. Since both sides are polynomials of degree n we have
S;=O
for k S 0, k 2 n S 1 where n > 0
so that the series (1)is actually finite.
From ( I ) we have on replacing n by n -t 1
(2)
CHAP. 13
T H E D I F F E R E N C E CALCULUS
Since
~ n +=
l X, e x ,
21
we have from (1) and (4.)
But
Thus
Equating coefficients of
d k )
we find as required
SE+l = S;-l+kS;
1.31. Show how the method of Problem 1.25 can be used to obtain Stirling numbers of the
second kind by using the polynomial f(x) = x4.
We use the same method as t h a t of Problem 1.25 as indicated in the following
1
1
0
0
0
1
1
1
lo
1
1
It follows t h a t
%4
=
%(4)
6
+ 6%(3) + 72(2) +
and the Sticling numbers a r e 1, 6, 7, 1.
THE GREGORY-NEWTON FORMULA AND TAYLOR SERIES
1.32. Prove the Gregory-Newton formula (44), page 9, for the case where f(x) is a polynomial and a = 0.
If f(z)is a polynomial of degree n we can write it as a factorial polynomial, i.e.
f(x) = A ,
Then
+ A , z ( l ) + A2x(2)+ A3x(3)+
* *
+ Anx(n)
(1)
22
THE DIFFERENCE CALCULUS
[CHAP. 1
Putting x = 0 in the above equations we find
A , = f(O),
...,
A, =
A , = --
which we can agree to write as
Using these in (1) we obtain the required formula for a = 0.
If we use the same method above with
general formula (44) on page 9 is obtained.
replaced by x
2
-a
and then put z = a the more
1.33. Prove the Gregory-Newton formula with a remainder given by (44) and (QS),page 9.
Let p,(x) denote the polynomial of degree n as determined in Problem 1.32. Writing
+ a p - - l + *..+ a,
p,(s) = a,@
+
.
we see that it has n 1 coefficients a,,a,,. .,a,. It follows that if f(z) is some given function
we can determine these n 1 coefficients uniquely in terms of the values of f(z)at n 1 different
values of 2, say xo,x,, . . .,x,. We shall suppose that this is done. In such case
+
+
f ( z o ) = P,(zo), f(xJ = P,(s~),
.
.
.
.
f(%)
.
.
.
.
.
=
.
.
I
.
.
.
.
.
P,(Xn)
.
.
.
..
f(4 =
* 3
Pn(xn)
(1)
.
+ g(x,)
Then using (1) we see that
dxo) = 0,
= 0,
. . ., dx,)
It thus follows that unless g ( 2 ) is identically zero [in which case
have f ( z )= p,(z)J we must have
g(2)
f(2) is
a polynomial and we will
= K ( x )(x- x,)(z - E l ) . ' *(x- Z,)
Thus from (Z),
f(x) =
= 0
Pn(z)
+ K ( x )(x - X O ) ( X - $0)
*
*
($1
( z - 2,)
(4)
To obtain K ( s ) in terms of f(x) let us consider the function
U ( t ) = f(t) - p , ( t ) - K ( z ) ( t - x,)(t - 5 , ) ' * ' ( t- an)
(5)
.
+
It follows from (4) that this equation has the n 2 roots t = 2,x,, z,,. .,a,. Then by Problem
1.117 the ( n + 1)st derivative of U ( t ) , i.e. U(n+')(t),is zero for at least one value, say t = 7,
between the smallest and largest of 2, xo,xl, . . ,x,. But from ( 5 )
.
+
U(n+"(t) = f ( " + I ) ( t ) - (n l)!K ( s )
since p(;fl)(x) = 0. Putting t = 7 in (6) and setting it equal to zero i t thus follows that
(6)
THE DIFFERENCE CALCULUS
CHAP. 11
Since we can choose any values for
$0
= a,
$0, xl,
.. .,x,, let us choose
= a+h,
51
23
52
...,
= a+2h,
x, = a + n h
Then
which is Gregory-Newton's formula with a remainder.
1.34. Work Problem 1.24 by using the Gregory-Newton formula.
For any polynomial f(x) of degree n we have
or using Ax = h = 1,
f(x) = f(0)
+ AfdO)x(l) + A2f(0)
x(2)
+
2!
Now f(z) = 2x3 - 3 x 2
+ 5z - 4
f(o) =
Then
. . . + TA"f(0)
r n ( n )
so that
f(1) = 0, f(2) = 10, f(3) = 38, f(4) = 96
-4,
Af(o) = f(1) - f(0) =
- f(1) = 10,
Af(3) = f(4) - f(3) = 58
Af(2) = f(3) - f(2) = 28
Af(1) = f(2)
4,
From these we find
- Af(0) =
A2f(o) = Af(1)
A2f(l) = Af(2) - Af(1) = 18,
6,
Azf(2) = Af(3) - Af(2) = 30
Similarly,
A3f(0) = Azf(l)
- Azf(0) = 12,
and finally
A3f(l) = A2f(2) - Azf(1) = 12
~ 4 f ( o )= ~ 3 f ( i ) ~ 3 j ( o )=
o
From these we see that
f(0) = -4,
and so (I) becomes
Af(0) = 4, Azf(0) = 6,
223 - 322
+ 6s - 4
=
-4
A3f(0) = 12, A4f(0) = 0
+ 4x(1) +
3s(2)
+ 2d3)
in agreement with the result of Problem 1.24.
X
0
1
2
3
4
f (4
Af(4
A2f (2)
A3f (4
A4f (2)
-4
0
10
38
96
4
10
28
58
6
18
30
12
12
0
P
24
T H E D I F F E R E N C E CALCULUS
[CHAP. 1
1.35. Show how to arrive a t the Gregory-Newton formula for the case where h = 1, u = 0
by using symbolic operator methods.
W e have f o r any value of n [see Problem 1.121
(11
Enf(u) = f ( u + n h )
Then in particular if we choose u = 0, n = x, h = 1, this becomes
EZf(0) =
f(4
(3)
Using E = 1 + A in (2) and expanding formally by the binomial theorem, ( 2 ) becomes
f(z) = E x f ( 0 ) = (1
+ A)"f(O)
The result is the same as the infinite series obtained from (4.4), page 9, with h = Ax = 1 and a = 0.
Extensions to the case where h f 1, u f 0 can also be obtained symbolically [see Problem 1.831.
LEIBNITZ'S RULE
1.36. Prove Leibnitz's rule for the nth difference of a product of functions.
Let us define operators E l and E , which operate only on f(z)and g(2) respectively, i.e.
.
E , [ f ( x )g(x)] = f(x
+
+ h) dx),
+
E,[f(X)g b ) ] = f(x) Q(x
+ h)
Then E I E z [ f ( xg(z)]
)
= f(x h) g(x h) = E [ f ( x g(x)]
)
so t h a t E = E,E,. Associate the difference
operators Al, A, with E l , E , respectively, i.e. E l = 1 Al, E , = 1 A,. Then
A = E
- 1 = EIE, - 1 =
and so
A"[fg]
=
(A2
+
(1 +A1)E,- 1 = Ez
+A1Edn[fg]
1.37. Find A"[xzua"].
Let f = x2, g = ax.
Then by Leibnitz's rule we have
OTHER DIFFERENCE OPERATORS
1.38. Find (a) v ( x 2 + 2 x ) , ( b ) S ( x 2 + 2 x ) .
(a) V(x2 + 2%) = [x2 + 2x1 - [(2- h)2 + 2(2 - h)]
= x 2 + 2 x - [,Z-2hx+h2+22-2h]
= 2hs
+ 2h - h2
+
+ AIEz - 1 =
A2
+ PIE,
CHAP. 11
T H E D I F F E R E N C E CALCULUS
(b)
= [(X+;)'+2(x+;)]
6(X2+2X)
- [(X-;)2+2(x-;)]
+ + rh2 +2%+
= (9 hx
1.39. Prove that
(a)
11
)-
+ 2h
= 2hx
25
( x 2 - K x + - +k 22 x - h
4
)
v = AE-' = E - ~ A= 1- E-1
( b ) 8 = ,73112 - E-112 = ~ E - l / 2
= VE112
(a) Given a n y function f ( x ) we have
vf(%)= f(x) - f ( ~ - h )= A f ( x - h ) = AE-'f(%)
so t h a t V = AE-L Similarly
vf(X)=
so t h a t V = E - l A .
+
f(z)- f(z- h) = E - l [ f ( x h) - f(x)] = E-lAf(x)
Finally
V f ( X ) = f(x) - f ( x - h) = f(x)
= (1- E - ' ) f ( x )
E-If(%)
so t h a t V = 1- E-1.
= f
Sf($)
(b)
(. +);
so t h a t 6 = E 1 / 2 - E - 1 / 2 .
6f(x) =
so t h a t 6 = A E d 1 I 2 .
Sf(%)
so t h a t 6
1.40.
Find
-f
(. - );
= E1/2f(z) - E-'/Zf(x) =
Similarly
f ( ~ + k )- f ( ~ - k )
Finally
= f(x
+
k) -
f(x -
= VE1/2.
a)
= Af(2-a)
= Of (x
(a) M ( 4 x 2- 8x), ( b ) p(4x2- 8%).
= 4[4(% h)2 - 8 ( ~ h) 42' - 8x1
+
( a ) M ( 4 x 2 - 82)
(E1/2-E-m)f(X)
+ +
4x2
=
+ );
A.?C-1/2f(~)
= VE1/2f(x)
+ 4 h +~ 2h2 - 8% - 4h
MISCELLANEOUS PROBLEMS
1.41. If f(x) = aoxn ulxn-l+ - * +a,, i.e. a polynomial of degree n, prove that
(a) A n f ( x ) = n!aoh"
( b ) A n t l f ( % ) = 0, An"f(x) = 0, , . .
+
Method 1.
W e have
A~(z)
+
+
+
+
= [ao(. 4- h)n u,(z+
a,] = [aonxn-lh terms involving xn-2, 5%--3,. . .7
+
C L ~ X ~f - ~
+ a,]
It follows t h a t if A operates on a polynomial of degree n the result is a polynomial of degree n - 1.
From this we see t h a t An!(%)
must be independent of x, i.e. a constant, and so Antlf(x) = 0,
A n * 2 f ( ~=
) 0, . . ., which proves ( b ) . To find the constant value of Ant(%) note that we need only
consider the term of highest degree.
Thus we have
+
h)n-'h
. * ] - [aonxn-lh
= ao.[(.
hp-1- z n - l l h
* '.
= a,?&[(%
- l)xn-Zh
.* ' ] h ...
= aon(n- 1)Xn-Zhz
.* *
Azf(x) = [ U o n ( x
+
+
+
+
+
+.
'
'1
26
T H E D I F F E R E N C E CALCULUS
[CHAP. 1
Proceeding in this manner we see finally t h a t
= n!aohn
A f l f ( x ) = a,n(n-l)(n-2)...(l)hn
For another method see Problem 1.84.
Method 2.
Since every polynomial of degree n can be written as a factorial polynomial of degree n, we
have
~ 0 % " al~n-1
* . a, = boxfn) b r ~ ( n - 1 )
b,
+
Equating coefficients of
xn
+
+
+
+
on both sides we find bo = a,.
+
The required result then follows since
1.42. Prove that
We have
Now from Problem 1.41 with h = 1 we have f o r
T
>n
while if r = n,
Thus the required result follows.
1.43. Show that
(a) ArOn = 0 if r
> n,
( b ) AflOfl= n !
( a ) Putting x = 0 in (2) of Problem 1.42 we have f o r
(a)
Putting x = 0 i n (3) of Problem 1.42 we have f o r
T
> n,
ArOn
= 0.
= n,
Anon
= n!
We call ATOn the differences of zero.
1.44. Prove Rolle's theorem: If f ( x ) is continuous in a S x S b , has a derivative in
a < x < b , and if f(a) = 0, f ( b )= 0, then there is a t least one value 7 between a
and b such that f'(7) = 0.
W e assume t h a t f(x) is not identically zero since i n such case the result is immediate. Suppose
then t h a t f(x) > 0 f o r some value between a and b. Then i t follows since f(x) i s continuous t h a t
i t attains its maximum value somewhere between a and b, say at q. Consider now
afo =
f(7
+ h) - f ( 7 )
h
Ax
+
where we choose h = Ax so small t h a t 7 h is between a and b. Since f(7) is a maximum value,
i t follows t h a t Af(2)lA.x 2 0 f o r h < 0 and Af(x)/Ax 5 0 f o r h > 0. Then taking the limit a s
h + 0 through positive values of h, we have f'(7) S 0, while if the limit is taken through negative
values of h, we have f ' ( 7 ) 2 0. Thus f ' ( 7 ) = 0. A similar proof holds if [(z) < 0 for some value
between a and b.
CHAP. 11
1.45.
27
T H E D I F F E R E N C E CALCULUS
(a) Prove the mean value theorem for derivatives: If f(x) is continuous in a 5 x S b
and has a derivative in a < x < b, then there is at least one value TJ between
a and b such that
( b ) Use (a) to show that for any value of x such that a d x 5 b,
where
7)
is between a and x.
f(x) = f(a) 3- (3 - a)f’(rl)
(c) Give a geometric interpretation of the result.
(a,) Consider the function
F ( x ) = f(’)
-
f(a) - (X-a)
f(b)-f(a)
b-a
(11
From this we see t h a t F ( a ) = 0, F ( b ) = 0 and t h a t F ( x ) satisfies the conditions of Rolle’s
theorem [Problem 1-44]. Then there is at least one point 7 between a and b such t h a t F’(q) = 0.
B u t from (1)
(b) - f ( 4
F’(x) = f’(x) - f~(2)
b-a
so t h a t
or
( b ) Replacing b by x in (3) we find as required
where q is between x and a.
f(x) =
f ( 4+ (x - a)f’(v)
(4)
Note t h a t ( 4 ) i s a special case of Taylor’s series with a remainder f o r 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 i t is geometrically evident t h a t
there is at least one point R on the
curve y = f(x) where the tangent line
PRQ is parallel to the secant line ACB.
Since the slope of the tangent line at R
is f’(v) and the slope of the secant line
is [ f ( b )-!(a)]/@ - a), the result follows. It is of interest to note t h a t F ( x )
in equation (1) represents geometrically RC of Fig. 1-2. I n the case where
f ( a ) = 0, f ( b ) = 0, RC represents the
maximum value of f(x) in the interval
a S x 5 b .
A
a
X
7
Fig. 1-2
b
28
T H E D I F F E R E N C E CALCULUS
[CHAP. 1
Supplementary Problems
OPERATORS
1.46.
Let d = [
following,
12
be the squaring operator and D the derivative operator.
(4 6(1+Vz)
(e)
(b) ( 2 d
(f) (D
+ 3 D ) ( ~-2 X)
(4 d W 3 X + 2)
( 6 2
+ 2 d - 3)(2x - 1)
(i) x3D3&(2
+ 2 ) ( d- 3 ) ~ '
(I.)
(
.
d -d
+ 1)
W d X 2
+
(9) (@J- 3)(D 21x2
(h) (xD)3&(2 1)
+
(4 D d ( 3 x + 2 )
1.47.
of Problem 1.46 is a nonlinear operator.
( a ) Prove t h a t the operator
of d-l and determine whether i t always exists and is unique.
1.48.
Prove t h a t @ = C2 where
1.49.
Determine each of the
d and
( b ) Explain the significance
C a r e the squaring and cubing operators respectively.
Let eJ be the squaring operator and a: be any real number. (a) Explain the meaning of the operators
obey the commutative law? Illustrate by a n example.
and &a. ( b ) Do the operators CY and
ad
1.50.
Is the operator ( x D ) ~
the same as the operator x4D4? Justify your answer.
1.51.
Prove t h a t (a) D2x - xD2 = 2 0 , (b)
prove it.
032
- 5 0 3 = 302.
Obtain a generalization of these results and
THE DIFFERENCE AND TRANSLATION OPERATORS
1.52. Find each of the following.
+ 1)2(x+
( i ) (2E - 1)(3A+ 2 ) ~ z
(f) (xE2 + 2xE + 1)x2
(j) ( x A E ) ~ ~ ~
(9) A2E3x
(h) (3A + 2)(2E - l)s2
(a) (E - 2)(A + 3) = (A + 3)(E - Z), (b) (E - %)(A + 3%)= (A + 3x)(E - 2) and
(a) 4 ( 2 ~ 1)2
(b) E ( & F - z )
(c) A2(2x2-5x)
(e) (A
+
(d) 3E2(~2 1)
1.53.
Determine whether
discuss the significance of the results.
1.54.
Prove t h a t E is a linear operator.
1.55.
Determine whether (a) A2 and (b) Ez a r e linear operators. Do your conclusions apply to An and
Explain.
En?
= (E - l ) 3 = E3 - 3E2
+ 3E - 1.
1.56.
Verify directly t h a t
1.57.
Prove formulas (a) V-7, ( b ) V-8, (c) V-9 on page 7.
1.58.
Prove t h a t the commutative law f o r 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 f o r D and E?
1.60.
Show t h a t
(a) lim
AZ+O
(b)
A3
A
[x(Z- x)] =
AX
A2
lim - [ ~ ( 2 - x ) ]
Ax-0
Ax2
D[%(2- %)I = 2(1 - X)
d2
= -[x(2-x)]
dx2
= -2
directly from the definition.
+ 2%- l), (b) d2(3X2+ 2%- 5).
D [ f ( x )+ g(x)] = Df(x) + D g ( x ) giving restrictions if
1.61.
Find (a) 4 x 3 - 3x2
1.62.
Prove t h a t
any.
CHAP. 11
1.64,
T H E D I F F E R E N C E CALCULUS
Prove t h a t
29
A
sin e
lim [cos r x ] = --r sin rx by using the fact t h a t lim 7
= 1.
Ax
6-0
Ar-tO
1.65.
Prove t h a t
A
[br]
Ax
lim
AZ+O
=
bx
In b stating assumptions made.
1.66.
Prove equation ( I h ) , page 3, giving suitable restrictions.
1.67.
Obtain a relationship similar to t h a t of equation ( U ) ,page 3, between D3f(x) and A3f(x)/Ax3.
1.70.
Express each of the following as factorial polynomials for h = 1 and for h f 1.
(a) 3x2 - 5% 2, ( b ) 2x4 5x2 - 4% 7.
+
+
A
+
+ 5x - 3),
A2
(x4- 2x2 + 5x - 3).
Ax2
1.71.
Find
1.72.
Express each of the following as a product of suitable factors using the indicated values of h.
(a) (2s - 1)(4)
if h = 2, (b) (32 5)(3) if h = 1, (c) (4x - 5)c-Z) if h = 1, (d) (5x 2)-(4) if h = 2.
(a)
(24
- 2x2
(b)
+
+
1.73.
Write each of the following as a factorial function.
+ 5 ) ( 3+~ 12), (b) (2 + 2 x ) ( 5 + 2 ~ ) (+8 22)(11+ 2%)
1
1
x(z + 2)(z + 4) ' (d) (2%- 1)(2x + 3)(2z + 7 ) ( 2 x+ 11)
A
Prove t h a t - (px + q ) ( m ) = mp(px + q)(m-l) for (a) m = 0,1,2, . . . , ( b ) m = -1, -2, -3, . .
Ax
21c + 1
22 - 1
Express in terms of factorial functions (a) (s + 2)(x + 4)(x + 6) '
(2%+ 3)(22 + 5)(22 + 9)
(a) (3%- 2)(3x
1.74.
1.75.
STIRLING NUMBERS
Obtain Stirling numbers of the first kind
1.76.
page 7.
s t for
n, k = 1 , 2 , 3
by using the recursion formula (2%
1.77.
Obtain Stirling numbers of the second kind SE f o r n, k = 1,2,3 using the recursion formula (a@,
page 7.
1.78.
Find S i for k = 1,2,
1.79.
Prove t h a t
. . ., 6
by using the method of Problem 1.25.
s;+s;+"'+s;
= 0
and illustrate by referring to the table of Stirling numbers of the first kind in Appendix A,
page 232.
1.80.
Prove t h a t
(a)
(b)
8;
- 8; + 53" - . + (-1)n-h;
* *
+ Is;/ +
*.'
+
=
(-l)n-bt!
= n!
T H E GREGORY-NEWTON FORMULA AND TAYLOR SERIES
1.81.
Express each of the following as factorial polynomials for the cases h = 1 and h f 1 by using
the Gregory-Newton formula. (a) 3x2 - 5 s 2, (b) 2x4 5x2 - 4 s 7.
+
+
+
30
T H E D I F F E R E N C E CALCULUS
[CHAP. 1
1.82.
Under what conditions is the remainder R, in ( 4 5 ) , page 9, equal to zero?
1.83.
Show how to generalize Problem 1.35 by obtaining the Gregory-Newton formula for the case where
h P 1 , UPO.
1.84.
Prove t h a t
An[aOxn
+ U1xn-' +
* *
+ a,]
=
YL!
aohn
by using the Gregory-Newton formula.
1.85.
Obtain the Taylor series expansion f o r f ( x ) from the Gregory-Newton formula by using a limiting
procedure.
1.86.
Obtain the Taylor series expansions
intervals of convergence in each case.
( a ) 1, ( b ) 2, ( c ) 3 and (13) 4 on page 8 and verify the
LEIBNITZ'S RULE
1.87. Use Leibnitz's rule to find A3(x2 2.) if h = 1.
1.88.
Find An(xa").
1.89.
Find An(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 3 x - 5 find ( a ) V f ( x ) , ( b ) S f ( % ) , (c) V 2 f ( x ) , ( d ) S Z f ( x ) .
+
+ 2S*)(x2+ 2 x ) .
1.92.
Evaluate ( V 2 -3VS
1.93.
Prove t h a t
1.94.
Determine whether the operators V and S a r e commutative.
1.95.
Demonstrate the operator equivalence E =
1.96.
Prove t h a t V A = A V = S2.
1.97.
Is it true t h a t
1.98.
Show t h a t
1.99.
Determine whether ( a ) M and ( b ) p commutes with A, D and E.
1.100. Show t h a t
(a)
V2
= ( ~ E - 1 ) 2= A2E-2, ( b ) V n = AnE-n.
( a ) lim "= dy
6x-toSx
(a) M = +(l
( a ) A = pS
dx'
( b ) lim
+ E ) = E - +A,
+
@2,
(i+ d-)'.
( b ) A2mt1
3 =
6 x - ~ SX"
Explain.
7
dxn'
( b ) p = M/E1/2.
+
= Em[pSZm+l J2- S m + 2
I.
MISCELLANEOUS PROBLEMS
1.101. (a) If A and B a r e any operators show t h a t ( A - B ) ( A B ) = A2 - B2 A B - B A . ( b ) Under what
conditions will it be true t h a t ( A - B ) ( A + B ) = A2- B2? ( c ) Illustrate the results of parts ( a )
and ( b ) by considering (A2 - D 2 ) x 2 and (A - D ) ( A D)x2.
+
+
+
1.102. Prove t h a t
(a)
A sin ( p x
+ q)
= 2 sin- 2 sin [ p x
(b)
A cos ( p x
+ q)
= 2 s i nPhF cos [ p x + q
ph
q f $ ( p h 4-T ) ]
+ $ ( p h + TI]
CHAP. 11
THE DIFFERENCE CALCULUS
1.103. Prove that
(a)
Am sin (px
+ q)
= r2
(b)
Am cos (px
+ q)
+ q + 2m (ph +
sin r p x
L
31
T)]
L
J
=
px 4-q
+ m,(ph+
1
T)
1.104. Use Problem 1.103 t o show that
(a)
dm
dxm
sinx
= sin
(b)
1.105.
dm
~
C
O
X
=
S
cos
and thus complete the proof of Problem 1.11, Method 2.
1.106. ( a ) Show that A tan x =
sec2 x tan h
1 - tan x tan h
1.107. ( a ) Show that A tan-1x = tan-lx2
d
X
a
dx tan-1- a = x2 u2'
+
and
h
hx
+
( b ) deduce that -dt a n s
dx
and
( b ) deduce that
= sec2x.
d
tan-1 x =
+
1.108. ( a ) Show that A sin-1 x = (x
d
sin-lx =
dx
( b ) Deduce from (a)that
1.109. Prove that for h = l
1.110. Prove that
5': =
(a)
k!
+ h) d
8:
5
p=o
and
m - x d l - (X + h)2.
1
d
and -sin-'dx
~
1
= n
-Dnx(k)I
!
(-l)p(E)
x2 + 1
pn
x=o
1
=-
X
a
d m '
1
'
( b ) S: = a A k x n ! z = O
and illustrate by using the table on page 233.
1.111.
1.112. Show that the index law for factorial functions, i.e. x ( ~ ) x (=~ x)( m + n ) , does not hold.
1.113.
Prove that
and discuss the relationship with Leibnitz's rule.
1.114. Prove that
n! = nn
1.115. Show that
(a) A
- n(n-1)"
= hD
n(n - 1)
+ 2!(n-2)"
-
n(n - l)(n
3!
- 2) (n-3)"
+
a
'
*
h2D2 I h3D3 h4D4 . . .
+x
3! +4!
+
h5D5 31h6Ds . . .
+ hh4D4 + 7
4-360
3h4D4 5h5D5 3hsDs 903h7D7+
h3D3 + 2 +-+-+4
4
2520
(b)
~2
= h2Dz + h3D3
(c)
A3
=
1.116. Find (a) A2(3x3-2x2+ 4x-6),
direct evaluation.
+
.,
( b ) A3(x2+ x)2 by using Problem 1.115 and compare results by
1.117. If U ( t ) is the function defined by ( 5 ) of Problem 1.33, prove that U ( " + l ) ( t )= 0 for at least one
[Hint. Apply Rolle's theorem
value t = 9 between the smallest and largest of x, xo, xl, . . .,x,.
successively.]
1
Chapter 2
I
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)
(4
It follows that [see Problem 2.11
'
Ayk = Ykti-yk,
Eyk =
a=E-l
E = l + A
Yktl
and so as on page 3
or
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
Also since x = a + k h becomes x = k if a = 0
valid if we replace x by k and put h = 1. Thus for
example equations (22) and (25), page 6, become respectively
= ~,k(~-')
k C m )= k ( k - l ) ( k - 2) * * * ( k - W t 4-I),
(5)
It should be noted that k need not be an integer. In fact k is a variable which is discrete
or continuous according a5 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), y1 = f(a + h), yz = f(a 2h), . . . .
+
DIFFERENCE TABLES
A table such as that shown in Fig. 2-1 below which gives successive differences of
24 = f ( x ) for x = a, a + h , a+%, . . ,, i.e. Ay,A2y,A3y, . . ., is called a diflerence tabZe.
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 A2y1 in the fourth column is between Ay1 and Ay2 of the third column and
A2y1 = Ayz - Ay1. Similarly A3y2 = a2y3- A2yz.
32
CHAP. 21
APPLICATIONS OF THE DIFFERENCE CALCULUS
X
Y
AY
A2Y
A3y
33
A4Y
A5Y
U
a+h
+ 2h
a + 3h
a + 4h
a
a
t 5h
Fig. 2-1
Example 1.
The difference table corresponding t o y = f(x) = 2 3 for x = 1 , 2 , . . ., 6 is as follows.
X
y
= f(x) = 2 3
1
1
2
8
AY
*2Y
A3Y
A4Y
7
12
19
3
27
6
18
0
37
4
64
6
24
0
61
5
125
6
30
91
6
216
Fig. 2-2
The first entry in each column beyond the second is called the leading difference f o r
that column. In the table of Fig. 2-1 the leading differences for the successive columns are
A y ~ , h ~ y o , ~. .~ .y .~ ,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 diflerence of o r d e r 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 2-41.
DIFFERENCES OF POLYNOMIALS
It will be noticed that for f(z) = 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
~ . + ~ f ( x.).,. are all zero.
Anf(x)
is a constant and A n t l f ( x ) ,
34
APPLICATIONS O F T H E D I F F E R E N C E 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
+
f(a+kh) =
f(a) +
A f ( a )k ( l )
1!
where the remainder is given by
R,
the quantity
7
.
1
being between a and a
k“’ =
I t ,
=
Azfja)k(2)
2!
+
+
...
+
A n f ( a )W n )+
Rn
n!
(6)
hn+ I f ( n + I ) ( 9 ) k(n+1)
(n+ 1)!
+ kh and where
k(2) = k ( k - l),
k‘3’ = k j k - l ) ( k - 2),
...
In subscript notation (6) can be written as
If An+lyo,Ant2y0, . . . are all zero, then R n = 0 and y k is a polynomial of degree n in k .
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 2.91.
INTERPOLATION AND EXTRAPOLATION
Often in practice we are given a table showing values of y or f ( x ) corresponding t o
various values of x as indicated in the table of Fig. 2-3.
X
xo
X1
...
XP
Y
Yo
Yl
...
YP
Fig. 2-3
We assume that the values of x are increasing, i.e. xo < $1 <
< xP.
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 t o an untabulated value of 2
b e t w e e n xo and x p 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 xp,i.e. lies outside the table,
is often called extrapolation. If x represents the time, this can involve a problem in
prediction o r 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 o r 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 f o r m u l a
[see page 381.
Because formulas for interpolation can also in general be used for extrapolation we
shall refer t o such formulas collectively as interpolation f o r m u l a s .
CHAP. 2:
APPLICATIONS O F 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 diflerence table.
+
Y
X
a - 3h
Y-3
- 2h
76-2
AY
A2Y
A3Y
A4Y
A5Y
A6Y
AY-3
a
A%-3
AY-2
a-h
Y-1
A3Y-3
A2Y
-2
a
Yo
A'Y-1
+ 2h
a + 3h
a
A'y - 3
A*Y-~
AOY - 3
A3y -
AYO
a+h
A4Y - 3
A3y - 2
AY-1
A4y-
A2Yo
Y1
AY 1
A5y-,
ASY 0
A2Y 1
Y2
AY2
Y3
The table of Fig. 2-4 can also be written equivalently in terms of central differences as
shown in Fig. 2-5.
~
X
a
- 3h
Y-3
a
- 2h
Y-2
6Y -5/2
sY-3/2
a-h
Y-1
SY-1/2
a
Yo
8Y1/2
a+h
+ 2h
a + 3h
a
Yl
sy3/2
YZ
6Y.5/,
Y3
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 6 = AE-l12 on page 9. Thus for example,
S3y,3/2
=
(AE-1'2)3y-3/2
= A 3 E - 3 / 2y - 3 / 2 =
n3y-3
Other tables can be made using the backward difference operator V.
'
36
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[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 A f ( a ) ,A z f ( 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 adaed. The results can also be expressed in terms of the operators V or 6 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. Gregory-Newton backward difference interpolation formula.
3. Gauss' interpolation formulas.
4.
Stirling's interpolation formula.
5. Bessel's interpolation formula.
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. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
.
,
Y-2
(k
PLY-3
I
1,
\
+ 3)'"/3!
/
37
A4Y--4
'
(k - l)(2)/2!
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 t o 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-1 in Fig. 2-61, 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 Ay-1 to A2y-l in Fig. 2-61, 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
A ~ in
O Fig. 2-61, the term which is to be added is equal to the product of the differ-
ence 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 A ~ to
O ~ ~ y in
- 1Fig. 2-61, 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.
38
APPLICATIONS O F T H E DIFFERENCE CALCULUS
[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
Y =
(x- Xl)(X - xz)*
yo(xo-xl)(xo-xz)'
* *
(x- x,)
.'(XO--X,)
+
... +
(x-xo)(x-x2)~~'(x-x,)
+
YP
y1 (x1- xo)(x1-
x2) *
.
a
(x1- x p )
(x-xo)(x-xl).'.(x-x,--l)
(x, - xo)(xp - Xl) (x, - xp-1)
*
which is called Lagrange's interpolation formula [although it was known t o Euler].
The result holds whether the values XO,21, . . . , x, are equally spaced o r 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, XI, XZ, . . . and that the
function is f (x) we define successive divided differences by
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.
Various other notations are used for divided differences.
sometimes denoted by [xo,XI, x2] or A[xo, XI, $21.
For example f(xo, xi, $2) is
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
39
The following are some important results regarding divided differences.
1. The divided differences are symmetric. For example
=
f(x0, $1)
f(x1,XO),
f(xo,$1,
$2)
=
f ( x 1 ,XO,$2)
= f ( x z ,XO,XI),
etc.
2. If f ( x ) is a polynomial of degree n then f(x, $0) is a polynomial of degree n - 1,
f(x, XO,xl) a polynomial of degree n - 2, . . . . Thus f(x,xo,XI, .,G-I) is a constant
and f(x,XO, 21, . . ., X n ) = 0.
..
3. Divided differences can be expressed in terms of first differences.
f (4
(xo- x1)(xo - $2)
=
f ( x 0 ,x1, $2)
+
($1
f (xd
- xo)(x1- xz)
+
For example,
f (4
(xz- xo)(x2- Xl)
NEWTON'S DIVIDED DIFFERENCE INTERPOLATION FORMULA
From the above results we obtain
f(x) =
f(x0)
+ (x- xo)f(xo,X l ) + (x- xo)(x- x1)f(xo,
+ + (X - - O ) ( X - - X ~ ) . . . ( X - - ~ - ~ ) ~ ( X ~ , X ~ ,
$1, x2)
*
a
*
...,~ n S) R
(Id)
where
n (x
n
- Xk)
=
(x- xo)(x - $1) ' * (x- x n )
k=O
and 7 is some value between the smallest and largest of x,XO,XI,
can be written as
f ( 4 = pn(x) + R
. . .,xn.
The result (12)
(15)
where p3,(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 - l ( y ) where f-' 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 81 we obtain the operator equivalence
D
=
1
hln(l+A)
=
A'
f -A3
- - + A4
3
4
...
by formally expanding In (1+ A ) using the series on page 8.
By using (16) we can obtain approximate derivatives of a function from a difference
table. Higher derivatives can be obtained by finding series expansions for D2, D3, . . .
from (16): We find for example
40
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
Use can also be made of different interpolation formulas t o 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 y k = f ( a kh), show that
+
(a)
Since
Yk
=
yk+l-yk
+ kh)
we have
Ayk
= f(a
(b)
E y k
=
yk+l
=
(a) A u k = A f ( a + k h ) = f ( a + k h + h ) - f ( a + k h )
= f(U+kh) = f ( a + k h f h ) =
(b)
EYk
(c)
From parts ( a ) and ( b ) ,
Then since
Yk
Yk+AYk
is arbitrary we have E = 1
(c)
E = l+a
Ykti-yk
Uktl
= E'&
Or
(1
+ A)Yk = E l 4 k
+ A.
DIFFERENCE TABLES
2.3. Let f(x) = 2x2- 7%+ 9 where x = 3,5,7,9,11. Set up a table for the differences
of f ( 4 .
The required table is shown in Fig, 2-8 below. Note t h a t in column 2 the various values of f ( s )
a r e computed from f(x) = 239 - 7x 9. Thus f o r example f(5) = 2(5)2 - 7(5) 9 = 24, etc.
+
+
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
X
ld
3
6
5
24
7
AY
11
16
34
174
0
0
0
16
66
A42/
'
16
50
108
A3Y
A2Y
18
58
9
2.4.
=f(4
41
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.
$4
A
B
AY
F
-1
G
D
E
A4Y
H
-3
N
K
-2
P
4
L
Z
A5Y
M
J
3
C
A3Y
A%
-5
Fig. 2-9
Similarly we find in
Starting at the right end i t is clear t h a t P - (-3) = -5 or P = -8.
succession the equations 4 - N = P or N = 12, N - M = -3 o r M = 15, J - (-1) = M or J = 14,
or Z=28, - 2 - H = K
or H = - 2 8 ,
K - J = N or K = 2 6 , L - K = 4 or L = 3 0 , Z - ( - 2 ) = L
G-F=-1
or F = - 4 1 ,
3 - B = G or B = 4 5 , B - A = F or A = 8 6 ,
H - G = J or G = - 4 2 ,
C - 3 = H or C = - 2 5 ,
D-C=-2
or D=-27,
E - D = I or E = l .
The final difference table with the required numbers replacipg letters is as follows:
Y
86
45
3
-25
AY
-41
-42
-28
A3Y
A22/
-1
28
A52/
I
15
14
12
26
-b
-8
4
-2
-27
A416
30
1
Fig. 2-10
The problem illustrates the f a c t that a difference table is completely determined when one
entry in each column is known. From the manner in which the letters were obtained i t is clean
that they can each have only one possible value, i.e. the difference table is unique.
42
2.5.
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
By using only the difference table of Problem 2.3 [and not f ( x )= 2 x 2 - 7 x + 9 ] find
the values of 3 = f ( x ) f o r x = 13,15,17,19,21.
W e must extend the difference table of Problem 2.3 as shown in Fig. 2-11. To do this we first
note t h a t the second differences a r e 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 i t is seen t h a t the values of y corresponding to x = 13,15,17,19,21 are 256,
354, 468, 598, 744 respectively.
THE GREGORY-NEWTON FORMULA AND APPLICATIONS
2.6. Prove that ( k h ) ( n=
) h"k(") for n = 1,2,3, . . .
.
Since ( k h ) ( l )= k h = h k ( l ) , the result is true f o r n = 1.
Since (kh)2 = (kh)(kh- h) = h?k(k - 1)= h2k(2), the result is true for n = 2.
Using mathematical induction [Problem 2.671 we can prove f o r all n
(kh)(n) = (kh)(kh-h)...(kh-nF,fh)
2.7.
= hnk(k--l)...(k-n+l)
= hnk(n)
+
Show that if we put x = a k h 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) +
Ma)
(3
-a)(')
11
R,
where
Then putting x = Q
+ kh
+ -A2f(a)
Ax2
(Z
=
- a)(')
+
.. .
f(ntl)(ll)
(n
(x-
I A"f(a) ( x - a)',)
Axn
2!
n!
+ Rn
a)(ntl)
+ 1) !
in ( 1 ) we find using Problem 2.6 and A x = h
f(a+kh) = f(a)
= f(a)
Af(a) ( k h ) ( l ) A2f(a)(kh)(2)
+ ~+
h
l ! + -h2
2!
A f ( a ) k ( l ) A2f(u)k(2) . , .
+-+
2!
l!
+
+
A"f(a) ( k h ) ( n )
+ ~+ Rn
hn
n!
Anf(a)k(n)
n!
+ Rn
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
43
where
Since f ( a
2.8.
+ kh) = gk,f(a)= yo, Af(a) = Avo, . . .,
( 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 uo,u1,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.
35
70
6
9
16
10
19
0
6
16
35
I
Fig. 2-13
Then by the Gregory-Newton formula with uk replacing yk,we have since uo= 2, Auo = 5,
A2uO= 4, A3u0 = 6, A4u0 = 0
Auok(l)
A2uok(2) A3uOk(3) A4uok(*4)
uk = u0
3!
4!
2!
l!
+
+-+-
+-
2+5k+-
= k3 - k2
4k(k - 1)
2
+
6k(k - l)(k - 2)
6
+ 6k + 2
It should be noted t h a t if the first 5 terms of the sequence a r e represented by ul, u2, . . .,u5
rather than uo,ul, . . .,u4, we can accomplish this by replacing k by k - 1 in the above polynomial to obtain
Uk = (k - 1)s - (k - 1)2
5(k - I) -I- 2 = k3 - 4k2 10k - 5
+
+
( b ) The 10th term i s obtained by putting k = 9 and we find
ug
= g3 - 92
+ 5(9) + 2
= 695
(c) I n obtaining the above results we a r e assuming t h a t there exists some law of formation of
the terms in the sequence and t h a t we can find this law on the basis of limited information
provided by the first 6 terms, namely t h a t the fourth and higher order differences a r e zero.
Theoretically a n y law of formation is possible [see Problem 2.91.
44
2.9.
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
Are there other formulas f o r the general term of the sequence in Problem 2.8?
Explain.
I n obtaining the general term for the sequence in Problem 2.8 we assumed first the existence
of some underlying law o r formula giving these terms. Second we assumed t h a t the function
of k describing this law was such t h a t the fourth and higher differences were all zero. A natural
consequence of this was t h a t the function was a third degree polynomial in k which we found by
use of the Gregory-Newton formula. I n 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
(11
uk = k3 - k2 5k 2 + k ( k - l ) ( k - 2 ) ( k - 3 ) ( k - 4)
+ +
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 t h a t of Problem 2.8. The formula (1)
does not however have the fourth and higher differences equal to zero.
From these remarks i t is clear t h a t one must give careful consideration as to the meaning
of the three dots in writing a sequence such as 2 , 7 , 1 6 , 3 5 , 7 0 , . . .
.
).
2.10. Find a polynomial which fits the data in the following table.
24
58
Method 1.
The difference table corresponding to this is given in Fig. 2-8, page 41. I n this case a = 3,
h = 2 and we can use the Gregory-Newton formula (8) on page 34. We have for the leading
differences
yo = 6, Avo = 18, A'yo = 16, A3y0 = 0, A4yo = 0, ...
Then
yk
= 6
+ 18k(l) + 16k(2) + 0
= 6
+ 18k + 8k(k-1)
= 8k2
+
10k
+6
To obtain this in the form y = f(x) we can write i t equivalently as
f ( x ) = f ( a + k h ) = f ( 3 + 2 k ) = 8k2
Thus using x = 3
+ 2k,
+ 10k + 6
i.e. k = (x - 3)/2, we find
f(x) = 8 [ ( x- 3)/2]2
+ lo[($ - 3)/2] + 6
= Z(X - 3)2
+5
( -~ 3)
+6
= 2x2 - 7x
+9
T h a t 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 2.31.
+
Method 2.
Letting a = 3, h = Ax = 2 in the Gregory-Newton formula (6), page 34, we have
= 6
+ 9(x-3) + 2 ( 2 - 3 ) ( ~ - 3 - 2 )
= 2 x 2 - 72
+9
using the fact t h a t (x - a ) ( 2 )= (x - a)(%- a - h).
It should be noted t h a t there a r e other polynomials of higher degree which also fit the data
[see Problem 2.121. Consequently i t would have been more precise to ask f o r the polynomial
of least degree which fits the data.
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
45
INTERPOLATION AND EXTRAPOLATION
2.11. Use the table of Problem 2.10 to find the value of y corresponding t o (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 corresponding to x = 8 we use x = a k h with a = 3, h = 2 to obtain k = (x - a ) / h = ( 8 - 3)/2 = 2.6.
Thus the required value of y corresponding to x = 8 is
+
+
+
+ lO(2.5) + 6
= 81
I n a similar manner the values of k corresponding t o x = 5.5, x = 15, x = 0 a r e (6.6 - 3)/2,
~ 2 . 5=
8(2.5)’
(15 - 3)/2, (0 - 3)/2 or 1.25, 6, -1.5 respectively.
to x = 5.5, 15 and 0 a r e respectively
~1.25
= 8(1.25)2
+ lO(1.25) + 6
= 31,
= 8(-1.5)2
1/-1.5
Then the required values of y corresponding
8(6)2
+ 10(-1.5) + 6
+ lO(6)
6 = 354
= 9
Method 2.
From either Method 1 or Method 2 of Problem 2.10 we have
y
Then the required values a r e
f(8)
= f(x) = 2x2 - 7%
+9
+ 9 = 81
+ 9 = 31
= 2(15)2 - “(15) + 9 = 354
= 2(0)2 - 7(0) + 9 = 9
= 2(8)2 - 7(8)
f(5.5) = 2(5.5)2 - 7(5.5)
f(15)
f(0)
Note t h a t in parts (a) and ( b ) we a r e interpolating, i.e. finding values within the table, while
in parts (c) and ( d ) we a r e extrapolating, i.e. finding values outside the table.
(e)
*
I n using the method we a r e assuming t h a t there exists some underlying law which the data
follows and t h a t we can find this law on the basis of the limited information supplied in the
table which suggests t h a t all third and higher differences a r e actually zero so t h a t the data
is fitted by a polynomial of second degree. It is possible however t h a t even if a n underlying
law exists, i t is not unique. The analogy with finding general terms of sequences or series
as discussed in Problem 2.9 is apparent.
2.12. (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
+ +
(1)
F ( x ) = 2x2 - 7 x 9 (x - 3)(x - 5)(x - 7)(x - 9)(x - 11)
For
x
=
3
,
5
,
7
,
9
,
1
1
the
values
agree
with
those
of
the
which is a polynomial of degree 5.
table or equivalently those obtained from f ( z ) = 2x2 - 7x 9.
+
Other examples can easily be made up, f o r example the polynomial of degree 11 given by
F,($1 = 2x2 - 7%+ 9 + 3(x - 3)2(x - 5)(x - 7)4(2 - 9)(2 - 11)3
(2)
F o r a n example of a function which fits the data but is not a polynomial we can consider
F,(x)= 2x2 - 7x + 9 + (x- 3)(x - 5)(x - 7)(x - 9)(x - 1l)e-5
(4
( b ) Since there a r e many examples of functions which fit the data, i t is clear t h a t any interpolation formula based on data from a table will not be unique. Thus f o r example if we put x = 8
in (1)of p a r t ( a ) , we find the value corresponding to i t 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
APPLICATIONS O F THE DIFFERENCE CALCULUS
[CHAP. 2
2.13. The table of Fig. 2-14 gives the values of sinx from x = 25" t o x = 30" in steps
of 1". (a) Using interpolation find sin 28' 24' and ( b ) compare with the exact value.
X
25"
26"
27"
28O
29"
30"
sin x
0.42262
0.43837
0.45399
0.46947
0.48481
0.50000
To avoid decimals consider f(z)= lo5 sin x and let x = a+ kh where a = 25" and h = lo.
Then we can write v k = f ( a kh) = 105 sin (25O k lo). The difference table is given in
Fig. 2-15. In this table k = 0 corresponds t o x = 25", k = 1 to 26O, etc.
+
+
1
42,262
0
1,575
43,837
~1
I
-13
-1
1,562
2
45,399
3
46,947
4
48,481
5
50,000
-14
0
1,548
-14
-1
1,534
-15
1,519
Fig. 2-15
It should be noted that the third differences are very small in comparison with the
practical purposes can be taken a s zero.
When x = 28'24'
formula
= 28.4O, we have k = (28.4' - 25')/1'
1/k
and for all
= 3.4. Thus by the Gregory-Newton
y3,4 = 105 sin (28' 24')
+ (1,575)(3.4)+ (-13)(3.42!)(3.4- 1) .. .
+
= 42,262
= 47,564
This yields a value of
sin 28O 24' = 0.47564. For an estimate of the error see Problem 2.14.
( b ) The exact value t o 5 decimal places is 0.47562. Thus the absolute error made is 0.00002
and the percent error is 0.004%.
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"'(q) 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 lo = 1;/180
radians. If x is in radians we have
f(x) = 105 sinx, f'(x) = 105 cosx, f " ( z ) = -105 sinx, f " ' ( x ) = -10s cos x
so that (1) becomes
(COS
R,
= -105(~)
7)(3.4)(3.4- 1)(3.4- 2)
3!
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
47
Since the largest value of cos q is dl - sin225O = 0.90631 we see from (2) that R, = -1
approximately, i.e. the best value we can get for sin28O24' is 0.47563. The fact t h a t 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.
~~
-3
42,262
-2
43,837
-1
1,575
45,399
0
46.947
~,~
~-
1
48,481
2
50,000
1,562
1.548
-13
-1
-14
0
-14
-1
1,534
1,519
-15
Fig. 2-16
( b ) Here k = 0 corresponds to 28O, k = 1 corresponds to 29O,
The value of k corresponding to 28.4O is k = (28.4O -28')/1'
k = -1 corresponds to 2 6 O , etc.
= 0.4.
From the table we see t h a t
YO
= 46,947, A u - 1 = 1,548, Ayo = 1,534, A'y-1 = -14,
A 3 ~ - 2 = -1,
A3y-1 = 0
Then by Stirling's formula
yo,4 = lo5 sin (28'24')
= 47,662
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 28O.24' = 28.4'
occurs in the central p a r t 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.
48
[CHAP.2
APPLICATIONS O F T H E DIFFERENCE CALCULUS
2.16. (a) Explain why you might expect Bessel's interpolation formula to give a better
approximation t o 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 i t
might be expected t h a t since 28.4O lies to the same side, a better approximation would be
obtained.
( b ) From the table of Fig. 2-16,
yo
= 46,947,
~1
= 48,481,
hyo = 1,534,
= -14,
Azy-1
A21Jo
= -15
Thus Bessel's interpolation f o r m u b gives
or
yo,4
= +(46,947
+ 48,481) + 1,534[+(0.4 - 0.6)] + &(-14 - 15)
L 2 1 . ..
(0.4)(--0.6)
+
= 47,562
Thus we obtain sin 28O24' = 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
=
= [l
(&)210
k
=
+ (-k)(-AE-I) + (-k)(-k
(1 - AE-l)--kyo
- l)(-AE-1)2
+ (-k)(-k
- 1)(-k - Z)(-AE-1)3
2!
3!
where we have used the binomial theorem.
( b ) Assume
Then
+ A , k ( l ) + Az(k + 1)(2) + A 3 ( k + 2)(3) + + A,(k + n - l ) ( n )
= A , + 2A2(k+
+ 3 A 3 ( k + 2)tz) + - - - + nA,(k+ n - 2A2 + 6A3(k + 2)(1) + . + n(n- 1)A,(k + n - 1)'n-Z)
-
yk = A ,
Ayk
Azuk
9
* *
l)(n-l)
l)(l)
...............................................
..............
I . . ,
A"yk = % ! A ,
Putting k = 0 in the expression f o r yk, k = -1 in Auk, k = -2
in A2yk,
.. .
we find
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
CHAP. 21
49
2.18. Derive the Gauss interpolation formulas from the Gregory-Newton formula.
The Gregory-Newton formula is
yk
k(1)
= yo
Avo-
l!
+
k'2)
A2yy,-
2!
k(3)
+ A 3 yO +
3!
..*
Now from the generalized difference table of Fig. 2-4, page 35, we have
A2yo - A 2 y - 1
= A3Y-1
or
A2yo = A 2 y - l
4- A3Y-1
Similarly,
A3yo-A3$/-l
= A4y-1,
from which
+ A4yVl
A3y0 = A32/-1
=
A4u-1-A42/-2
= A3y-1
A5Y-2
+ A4yP2 + A5y-2
Then by substituting in the Gregory-Newton formula we find
The second Gauss formula can be derived in a similar manner [see Problem 2. )4].
2.19. Derive Stirling's interpolation formula.
From the Gauss formulas we have
Then taking the arithmetic mean of these two results we have
2.20. Show how t o express the Gregory-Newton backward difference interpolation formula
in terms of the backward difference operator V.
By Problems 1.39(u), page 25, and 1.93, page 30, we have
V = AE-1,
Thus
AnE-nf(u)
V2
= A2E-2,
=
Vnf(a)
or equivalently
AnE-nyo
Using this in (8), page 34, it becomes
...,
Vn = AnE-"
n = 1,2,3,
== V n y o
...
50
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
ZIG-ZAG PATHS AND LOZENGE DIAGRAMS
2.21. Demonstrate the rules f o r writing interpolation formulas by obtaining (a) Bessel's
formula, ( b ) Gauss' second formula on page 36.
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 Ago the next term to be added
is Ay, multiplied by the mean of the coefficients directly above and below, i.e.
Again using rule 5 since the path passes through the coefficient k ( 2 ) / 2 ! , the next term t o
be added is k ( 2 ) / 2 ! multiplied by the mean of the differences above and below the coefficient, i.e.
Thus continuing in this manner we find the required formula
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 t h a t 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
Y o ' l = Yo
The same term is obtained if we assume t h a t the path enters yo with positive o r zero slope.
By rule 3 since the path enters Ay-l with positive slope, the second term is Ay-1 multiplied
by the coefficient directly below, i.e.
k(l)
Ay-1By rule 2 since the path enters A2yP1 with negative slope, the third term is A2y-1 multiplied
by the coefficient directly above, i.e.
( k 4-1)(2)
A ' Y - 1 ~
2!
Continuing in this manner we obtain the required formula
2.22. Let P denote a zig-zag path around any closed lozenge o r 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.
Fig. 2-17
CHAP. 21
51
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
By rule 6 a reversal of path changes the sign of the corresponding term added, Thus we need only
show t h a t the contributions corresponding to the three paths from An-'y,. to Antly,.-l
in Fig. 2-17
a r e all equal.
Now if path 1 is chosen the contribution is
Similarly if paths 2 and 3 a r e chosen the contributions are respectively
From (1) and (3) we see t h a t
c1 - c3
=
(k- ~ ) ( n )
(An?./,.-l-Any,.)
n!
+ An+l?./,.-l
(k-r + l)(n+l)
(n 1)!
+
- (k -r ) ( n + l )
(n
+ 1 )!
= o
so t h a t C1 = C3. Also we note t h a t the mean of C1 and C3 is
Thus C, = C p = C3 and the required result is proved.
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.
path encloses 3 cells I, 11, 111.
This
L
Fig. 2-18
If we denote the contribution corresponding to path J K , for example, by C j K and the total
contribution around cell I, for example, by CI, then
CI + CII
+ CHI
+ C X N + Ch'R + C R J )
+ ( C K L + C L M + C M N + Ch7K)
=
(cJK
=
+ (CRN+ C N P+ CPQ+ CQR)
CJK + CKL+ CLM+ CMN+ CNP + CPQ+ CQR+ CRJ
+ (CKN + CNK) + (CRN + CNR)
= CJKLMNPQRJ
since CKN= -CNK and CRN = -CNR.
lows t h a t CJKLMNPQRJ
= 0.
Thus since C, = 0, CIr = 0,
= 0 by Problem 2.22, i t fol-
52
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
224. Prove that if two zig-zag paths end a t 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 a t the same place but begin at different places.
These
( k - ?.)(I)
yr \
\
l!
\
Fig. 2-19
The total contribution to the interpolation formula corresponding to the dashed path is given by
when R is the remaining p a r t of the formula.
Similarly the total contribution to the interpolation formula corresponding to the solid path is
-
Aur
( k - v - 1)(1)
+ R
1!
(2)
Note t h a t the minus signs in ( 1 ) and ( 2 ) a r e used because of rule 6, page 37.
To show t h a t the results (1)and ( 2 ) are actually equal we need only show t h a t their difference
is zero. W e find this difference to be
Since y r t l = y r +Aur, (3) can be written
so t h a t 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, i t follows by Problem 2.24 that
they a r e 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 t h a t all of them a r e valid.
Note t h a t 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 a r e p 1 values of g 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
u = ao(x- xl)(x - x2). * * (x- xp) + a l ( z - x o ) ( 5 - x2)' ' * (x - xp)
+
* * *
+ Up(% - x o ) ( x- X1)'
*
'(X - x p - J
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
53
Then putting x = xo, y = yo we have
yo
= ao(xo-x,)(xo-~2)~~~(~O
o r - ~ pa.) =
Yo
(xo- x1)(xo - x 2 ) ' * * (xo - x p )
Similarly putting x = xl, y = y1 we find
Y1
y1 = a l ( x l - ~ O ) ( ~ 1 - ~ 2 ) ~ ~ ~ or
( ~ 1 -al~ p=)
(XI
- xo)(x1- xz). * '(xl-
xp)
with corresponding values for a2,a3. . . , , up, Putting these values in the assumed polynomial we
obtain Lagrange's formula.
2.27.
( 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 ) ( 2- 9)(x - 11)
(3 - 5)(3 - 7)(3 - 9)(3 - 11)
+
+
(x - 3)(x - 7)(x- 9)(x - 11)
24 (5 - 3)(5 - 7)(5- 9)(5- 11)
(x - 3)(x --5)(x - 9)(x - 11)
(x - 3)(x - 5)(x - 7)(x - 11)
58 (7 - 3)(7 - 5)(7 - 9)(7 - 11) + log(9 - 3)(9 - 5)(9 - 7)(9 - 11)
+
(x- 3)(x - 5)(x - 7)(x - 9)
174 (11- 3)(11 - 5)(11- 7)(11- 9)
Performing the algebra we find y = 2x2 - 7x
four as we might have expected.
+9
which is a polynomial of degree two and not
( b ) The values of y for 2 = 8,5.5,15,0 can be obtained by substituting in the Lagrange interpolation formula of p a r t (a).
TABLES WITH MISSING ENTRIES
2.28. Find the value of
u3
missing in the following table.
Fig. 2-20
Method 1.
In this problem we cannot construct a difference table. However from the four pairs of values
(k,u,) which a r e known we would be able to fit a polynomial of degree 3. This leads us to assume
t h a t the fourth differences a r e zero, i.e. A4u0 = 0. Thus we have
(E - l)4u0 = (E4- 4E3
which becomes
~q
+ 6Ez - 4 E + l ) u 0 =
+
-4 ~ 36 ~2 4Ul
+ uo =
0
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 a r e 4 values of k for which uk i s known, we can fit the data by a polynomial
of degree 3 which involves 4 unknown coefficients. Let us therefore assume t h a t
uk
= A0k3
+ Alk2 + A,k + A ,
54
APPLICATIONS OF THE DIFFERENCE CALCULUS
[CHAP. 2
Then putting k = 0, 1 , 2 , 4 in succession we obtain the equations
A,+
AI+
A3 =
A,+A3 =
3
8
= 15
A, = 47
8Ao+ 4A,+2A,+A3
64A,
+ 16A1 + 4Az +
Solving simultaneously we find
A, = 1,
A, = -12, Az = 5, A3 = 3
a
and so
Uk = 8k3 - Qkz
5k f 3
+
Putting k = 3 we then find u3 = 27.
Method 3.
By Lagrange's formula we have
uk
=
(k- l)(k - 2)(k - 4)
uo (0 - 1)(0 - 2)(0 - 4)
+
(k - O)(k - 2)(k - 4)
u1 (1- 0)(1- 2)(1- 4)
(k - O)(k - l ) ( k - 2)
(k - O)(k - l ) ( k - 4)
- 0)(2 - 1)(2 - 4) u4 (4 - 0)(4 - 1)(4 - 2)
8
15
= - 83 (k - l)(k - 2)(k - 4) 3
(k)(k- 2)(k- 4) - ;?- (k)(k - l)(k - 4)
+
u2 (2
+
+ ;47i;i(k)(k - l)(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.
7
5
6
f ( x ) = loglo x
0.77815
k
Uk
+
0.77815
9
0.90309
0.95424
1
0
= log10 (6 k)
8
I
10
~
11
1.00000
2
3
4
0.90309
0.95424
1.00000
12
1.07918
5
6
1.07918
Fig. 2-22
From A5u0 = 0 and A5u, = 0, i.e. (E - l ) 5 u ~= 0 and (E - l ) b , = 0, we are led to the equations
(E5- 5E4
+
lOE3
- 10E2 + 5E - l ) ~ ,=
+
(E5- 5E4 + lOE3 - 10E2 5 E - 1)ul = 0
0,
or
Putting in the values of uo,u2,us, u4 and u6 from the table we find
5U, 4U,
~5
= 5.26615
+ 5 ~ 5=
6.05223
Solving simultaneously we find
U,
= loglo7 = 0.84494,
~5
= log10 11 = 1.04146
These should be comparcd with the exact values to 5 decimal places given by loglo 7 = 0.84510,
loglo 11 = 1.04139 so that the errors made in each case are less than 0.02% and 0.01% respectively.
CHAP. 21
55
APPLICATIONS OF THE DIFFERENCE CALCULUS
2.30. The amount Ak of an annuity [see Problem 2.1441 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
I
Ak
3
6
9
12
3.1216
6.6330
10.5828
15.0258
Fig. 2-23
Let A3 and E 3 respectfully denote the difference and shifting operators for the 3 year intervals
while A1 and E l denote the corresponding operators for 1 year intervals. Then we have
E3 = 1 + A , ,
Since
Uk+3
=
El = 1 + A 1
E3Uk
=
E;Uk
E 3 = EY
we have
or
1 r'
Thus
A1
= (I+AI)~
A3
= (1+A3)1'3 - 1
(11
Using the binomial theorem we then have
.. .
(1/3)(-2/3)(-5/3)~:
+
= 1A3 - s1A i + sl
5 A: +
3!
Since there are only four pairs of values in the table we can assume that third differences A
constant, Thus we can omit fourth and higher differences in the expansions.
1
1
5
A, = ? A 3 - -A; 4- EA:
We thus have
9
= 1
+
+
k
A3Uk
uk
I
.
are
(8)
A; uk
uk
II
3.1216
3.5114
6.6330
0.4384
3.9498
10.5828
0.0548
0.4932
4.4430
Using (Z),(3) and ( 4 ) we then have
1
Alu3 = l3A 3 u 3 - -Aiu3
9
5
+ a Aiu3
1
1
= -(3.5114)
3
- 9 (0.4384)
+
5
(0.0548)
= 1.12513827
A:U~
2
27
= $Aiu3 - -A:u3
1
2
= -(0.4384)
9
- ~(0.0548)
= 0.04465185
~ : u 3=
1
-27
A;u3
1
= -(0.0548)
27
= 0.00202963
By using the above leading differences we can now complete the difference table for 1 year
intervals a s shown in Fig. 2-25 below.
56
[CHAP. 2
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
k
uk
3
3.1216
4
4.2467
5
1.12513827
0.04465185
1.16979012
5.4165
6
4Uk
A? Uk
*lUk
6.6330
0.00202963
0.04668148
1.21647160
From this table we see t h a t
uq
= 4.2467,
US
= 5.4165
By proceeding in a similar manner we can find u,,
us,ul0,u l l .
X
f(x)
1
,
2
2
4
5
6
8
10
10
96
196
350
868
1746
1
10
4
43
96
5
196
6
350
8
868
10
1746
19
100
154
259
27
0
35
0
45
439
Fig. 2-27
In the third column the entry 43 is obtained from
etc.
is obtained from
In the fourth column the entry 19 is obtained from ____
loo- 43 , 27 is obtained from
In the fifth column the entries a r e obtained from
equal to 2.
Prove that
196 - 96
5-4
,
154 - 100 ,
5-2
etc.
2.32.
- 10 , 100
4--2
96
(a)
f(x) = f(xo)+ (x- xo)f(x, $ 0 )
( b ) f(x, xo) =
(c)
(a) By definition,
m,8-4 and 4510--355 and a r e all
2 7 - 19 35 - 2 7
f(x, xo,X l )
f(x,x0) =
(x - % l ) f ( X , Xot X l )
= f ( x 0 ,x1, xz) + (X - x z ) f ( x ,xo,x1, xz)
f ( x 0 ,X l )
f(x) -
mo)
x - xo
-t
so that
~
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
57
2.34. Prove Newton's divided difference interpolation formula
f(x) = f ( x 0 )+ (x- xo)f(lro,X l ) + (x - xo)(x- x1)f(xo, x1, xz)
(x- xo) * ' (x- xn-l)f(xo, . . ., x,) R ( x )
+
where
a
+
+
R(x) = (x - xo)(x- X l ) *
* *
(x- xn)f(x,xo,x1, . . . ' x,)
Assume t h a t the formula is true f o r
We shall use the principle of mathematical induction.
n = lc, i.e.
f(x) = f(x0)
(x- ~ k - l l ) f ( x O , . . ., Xk)
+ (5 - xo)f(xo, + + (x + (x - xO)(x - x1) (x- xk)f(x, x0, x1, , zk)
%O) ' * *
* ' '
* * *
* * *
(1)
Now by definition as in Problem 2.32 we have
f(x, $0,
so t h a t
,
%1t * * * $ k + 1)
. . ., % k )
f(x,xo, xi,
=
= f(x, x0, . *
f ( x 0 , $1,
. . ., x k
f
Using this in (1) we find
f(x) = f(%o)
+ (x - %o)f(xo, +
Xi)
'
*
1)
* t
xk)
- f(x0, 51,
+ (x -
+ (x -
* *
*, Z k + l )
(2)
x - xk+l
2,)
%k
+
' ' '
x0, %1, *
(x- % k ) f ( Z o ,
-k (x- xo)(x - E l ) ' ' '(x - x k t l ) f ( x , xot $1,
* * * t
x k t 1)
* * f
t
($1
zk + 1)
zk+l)
+
Thus assuming t h a t it is true f o r n = k we have proved it true for n = k 1. But since it is
true f o r n = 1 [see Problem 2.321, i t is true f o r n = 2 and thus for n = 3 an& so on for all
positive integers.
2.35. Prove that the remainder in Problem 2.34 can be written as
R(x) =
(x-xXo)(X-xXI)...(X--Xn)
fcn+Y?)
(n
+ 1)!
where 7 is a number between the smallest and largest of the values x,xo,xi,
By Problem 2.34 the remainder is
R ( z ) = (x - xo)(x - x1) *
.( 5 - x,)f(x, xo,x1, . . .,x,)
. . .,Xn.
(1)
60
APPLICATIONS O F THE DIFFERENCE CALCULUS
[CHAP. 2
( c ) Since the values of z are equally spaced we can use the Gregory-Newton formula to find an
interpolating formula for f(s). As in Problem 2.11 we find f(x) = 2x2 - 7x
2x2 - 72
If f(x) = 100, i.e.
equation we find
+ 9 = 100,
6
x =
7
2x2 - 7 2 - 91 = 0
then
-
‘-c
4
7
+ 9.
and by solving this
27.8747
4
Since only the positive square root is significant for our purposes, we use i t 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 - 7% 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.
x =
(y
2*o(-2.0000
+
+
+
+
Then by Lagrange’s interpolation formula we have
+ 1.0390)(16- 0.0480)(~- 1.2670)(~- 2.6240)
+ 1.0390)(-2.0000 - 0.0480)(-2.0000 - 1.2670)(-2.0000 - 2.6240)
(y + 2.0000)(~
- 0.0480)(~- 1.2670)(~- 2.6240)
+ 2.0000)(-1.0390 - 0.0480)(--1.0390 - 1.2670)(--1.0390 - 2.6240)
(y + 2.0000)(~+ 1.0390)(~
- 1.2670)(1/- 2.6240)
2*2 (0.0480 + 2.0000)(0.0480 + 1.0390)(0.0480 - 1.2670)(0.0480 - 2.6240)
(y + 2.0000)(~
+ 1.0390)(~- 0 . 0 4 8 0 ) ( -~ 2.6240)
2*3(1.2670 + 2.0000)(1.2670 + 1.0390)(1.2670 - 0.0480)(1.2670 - 2.6240)
(y + 2.0000)(~+ 1.0390)(~
- 0.0480)(2/ - 1.2670)
2*4 (2.6240 + 2.0000)(2.6240 + 1.0390)(2.6240 - 0.0480)(2.6240 - 1.2670)
2*1 (-1.0390
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.
ehD = 1
We start with
+A
[equation ( 4 4 , page 81. Taking logarithms this yields
Then using the formal Taylor series on page 8 we find
q
1
D = ~In(1-l-A) = h
A2
A3
A--+---+..-
2
A4
3
4
( b ) By the Gregory-Newton formula,
f ( a + kh) = f(a)
+ k A f ( a ) + k(k - 1)A2f ( a )
- 2)
+ k(k - l)(k
Asf(a) +
3!
k(k - l ) ( k - 2)(k - 3)
A4f(a)
4!
Then differentiating with respect to k,
hf’(a+ kh)
= Af(a)
3k2 - 6k + 2
2k - 1
A3f (4
+ --gA2f(a) +
3!
+ 4k3 - 18k24!+ 22k - 6 A4f(a) 4~
* *
*
*
CHAP. 21
APPLICATIONS O F THE DIFFERENCE CALCULUS
61
Since !(a) is arbitrary we have
D
= - A--
A2
hl (
2
+ -A3- - +A4
3
4
...
If we take into account the remainder term in the Gregory-Newton formula, then we can
find a corresponding remainder term for the derivative.
X
f(z)
0.25
0.35
0.24740
0.34290
.
0.45
0.55
0.65
0.75
0.43497
0.52269
0.60518
0.68164
We consider the difference table for lOjf(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),. . . f o r brevity.
105f(4
A
24,740
9550
34,290
-88
-523
8249
60,518
-92
-435
8772
52,269
A4
A5
-343
9207
43,497
A3
A2
4
8
4
-80
-603
7646
68,164
Fig. 2-31
Using the formula of Problem 2.40 we have
’
A2
D[105f(x)] =
(A
-
A3
A4
+3-7+
and neglecting differences of order 5 and higher we find for x = 0.25
o r f’(0.25) = 0.96898.
The table of values are those f o r f ( x ) = sinx where x is in radians. From this we see that
f’(z)= 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 . 2 5 ) .
From Problem 2.40,
so that on squaring we find
62
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
Thus
=
D2f(x)
...)f( x)
L ( A 2 - A 3 + u A 41 +2
h2
From this i t follows t h a t
=
105f”(0.25)
1
(-343
+ 92 + 3.7)
= -25,470
o r f”(0.25) = -0.25470.
.
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
f(a + k h ) = f(a) + k A f ( a ) +
Differentiation with respect to k yields
2k - 1
hf’(a+kh)
= Af(a)
+
k(k - 1)
)(k
2! A2f(a) + k(k - l3!
+
3k2 - 6k
3!
- 2)
+ 2 A3f(a) +
A3f(a)
+
’
+ 4k3 - 18k24!+ 22k - 6
+
Letting a = 0.25, h = 0.1 and a k h = 0.40 so t h a t k = 1.5 we find using the difference table of
Problem 2.41
1
0(4)]
105f’(0.40) = - [9550 (-343) - &(-92)
0.1
+
+
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
Let us denote the year by x and the population in millions by P. Since the years given a r e
equidistant i t is convenient to associate the years 1930, 1940, 1950, 1960, 1970 with the numbers
k = 0, 1 , 2 , 3 , 4 respectively. This is equivalent to writing x = a kh where a = 1930 and
h = 10 so t h a t
- 1930
x = 1 9 3 0 + 10k
or
k =
(1
10
+
Using this we can rewrite the given table as follows.
I
;
Fig. 2-33
The difference table corresponding to this is as follows.
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
AP
I
A2P
A3P
0.6
1.4
I
63
I
I
Fig. 2-34
We shall assume t h a t the third differences a r e all constant [equal to 0.61 and t..at the fourth
differences a r e zero. Then using the Gregory-Newton formula we find
P
= 1.0
0 2k(2) +
+ (0.2)k(’) + 2!
+
+ O.lk(k - 1) +
+ 3k + 10)
= 1.0
0.212
= A ( k 3 - 2k2
(a)
0 6k(3)
3!
O.lk(k - l)(k - 2)
-
From (1) we see t h a t the year 1980 corresponds to k = 5 so t h a t
P = &[53-2(5)2+
3(5)+ 101
Thus we estimate t h a t in 1980 the population will be 10 million.
(b) From ( 1 ) we see t h a t the year 1955 corresponds to
1955 - 1930 = 2.5
10
k =
Then
P = &[(2.5)3
- 2(2.5)2 + 3(2.5) + 101 =
2.0625
Thus to two significant figures we would estimate t h a t the population in 1955 was 2.1 million.
( c ) The value of k corresponding to the year 1920 is from (1)
k =
Then
P =
1920 - 1930
10
- 2(-1)2
&[(-l)3
= -1
+ 3(-1) + 101 =
0.4
Thus we estimate t h a t the population in 1920 was 0.4 million.
,2.45. Suppose that it is necessary to determine some quantity y a t the time when it is a
maximum or minimum. Due t o error we may not be able t o do this exactly. However we observe that at times t1,t~,t3near the time for maximum or minimum, the
values are given respectively by yl, y ~y3.
,
Prove that the time a t which the quantity
is a maximum or minimum is given by
+
y1(tz” - ti) Y2(ti - t l )
2[yl(t2 - t3) f yZ(t3 - tl)
+ y3(t:: - ti)
+ Y3(tl
t2)]
By Lagrange’s formula the value of y at time t is given by
I d =
( t - t2)(t - t3)
t l - t3)
?dl(tl - M
( t - tl)(t - t z )
( t - M t - t3)
+
y2 (t2 - t,)(t2- t3)
+
y3
(t3
- tl)(t3 - t 2 )
(11
The maximum or minimum of y occurs where dyldt = 0 [assuming it to be a relative maximum or
minimum]. Then taking the derivation of (1) with respect to t , setting i t equal to zero and solving
for t we find the required value given above.
64
2.46.
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[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 t h a t
is the error. Then the difference table can be constructed as in Fig. 2-35.
e
Y-4
AY-4
A2Y -4
76-3
AY-3
Y-2
AY-2
7d- 1
AY-1-k
A2y-2
E
YO+€
A2&1
AYO - E
A2yo
Yt
AY 1
YZ
+
43y-3
E
- 2e
+
+
A4y-4
E
A3y-1
E
+
E
A4y-3 - 4~
A3y-z - 3s
+ 3~
A3y0 - E
A2Y 1
AY2
Y3
A3Y -4
4211 - 3
A4y-2
+ 6~
A42/ - 1 - 4~
~
4
+y
E
~
A3Y 1
A2Y 2
AY 3
Y4
Fig. 2-35
Note t h a t the error is propagated into the higher ordered differences as indicated by the shading
i n Fig. 2-35. It is of interest to note t h a t the successive coefficients of e in each column alternate
in sign and a r e given numerically by the binomial coefficients. Also the largest absolute error
occurring in a n even difference column lies on the same horizontal line in which yo occurs.
2.47. 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.
X
1
2
3
4
5
6
7
9
8
11
10
12
f(x) .195520 .289418 .379426 .464642 .544281 .617356, .683327 .741471 .791207 A32039 363558 .885450
Fig. 2-36
(a) The difference table omitting decimal points is as follows.
195,520
289,418
93,898
90,008
379,426
464,642
544,281
617,356
683,327
741,471
85,216
79,639
-4792
-5577
-6564
73,075
-7104
65,971
58,144
49,736
-7827
-8408
-8904
791,207
832,039
-3890
40,832
-9313
31,519
-902
-785
-987
-540
-723
-202
447
-183
142
-581
-496
-409
85
87
95
-314
-9627
863,558
217
21,892
885,450
Fig. 2-37
-419
649
-630
325
-57
2
8
1268
-1279
955
-382
69
6,
CHAP. 21
65
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
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 A6. I n view of the fact t h a t the largest absolute error in an even difference column
i s in the same horizontal line as the entry in which the error occurs, i t would seem that the
error should occur in the entry corresponding to -1279 which i s largest numerically, It thus
appears t h a t the error should be in the entry 0.544281 corresponding to z = 5.
+
( b ) Denoting the entry which is in possible error, i.e. 0.544281, by yo E where
clear from comparison with the difference table of Problem 2.46 t h a t
E
is the error, it is
= 217
A4y-3-4e = -202
A4y-z+6e = 447
A4y-1 - 4~ = -183
A4y0+€
= 142
A4y--4+e
Assuming t h a t the correct fourth differences should all be constant, i.e. A4y-4 = A4yds =
= A4g-1 = A42/-0, we obtain by subtracting the first two equations, the second and
third equation, etc. the equations
A42/-2
5s = 419 or
E
= 84,
1 0=
~ 649 o r
E
= 65,
1 0=
~ 630 or
E
= 63,
5e = 325 or
Because most of these indicate the approximate value B = 65 [the value
of line] we shall consider this as the most probable value.
e=
E
= 65
84 seems out
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.
+
E
=
It is of interest to note t h a t the true value was actually 0.544218. It appears t h a t in
copying the table the 1 and the 8 i n this true value must have been interchanged.
I n Fig. 2-38 the difference table taking into account this error is shown.
106f(x)
A
~-
A2
A3
~~
A4
~
~~
A5
~-
195,520
93,898
-3890
289,418
-902
90,008
54
-4792
379,426
50
-5640
464,642
69
-6438
-729
73,138
-660
65,971
-581
58,144
-496
49,736
-2
-409
40,832
85
-9313
832,039
2
87
-8904
791,207
6
85
-8408
741,471
10
79
-7827
683,327
0
69
-7167
617,356
19
-798
79,576
544,218
-4
-848
85,216
-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 t h a t the function is not actually a polynomial.
66
2.48.
APPLICATIONS O F THE D I F F E R E N C E CALCULUS
[CHAP. 2
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 t h a t shown in Fig. 2-40
where we use lO3y in place of y to avoid decimals.
X
103y
0
1250
0.1
1610
A
2173
0.3
3069
0.4
4539
0.5
7029
A5
360
203
563
0.2
A4
A3
A2
130
111
333
896
574
1470
241
205
94
446
1020
2490
Fig. 2-40
Letting a = 0, h = AX = 0.1 in the Gregory-Newton formula (6) on page 34 we have
y = 1.250
or
+ 3.60~(~’
+ 10.15~(2)I- 21.67~(3)+ 46.25~(4)+ 78.33~(5)
Then the value of y corresponding to x = 0.6 is
y = 1.250
+ 3.60(0.6)i10.15(0.6)(0.5)+ 21.67(0.6)(0.5)(0.4)
+ 46.25(0.6)(0.5)(0.4)(0.3)
= 11.283
2.49.
+ 78.33(0.6)(0.5)(0.4)(0.3)(0.2)
(a) Work Problem 2.48 by first obtaining logloy in place of y.
(a) to arrive a t a formula for y.
(a) From the table of Fig.
( b ) Use the result in
2-39 we obtain the following table.
X
0
0.1
0.2
0.3
0.4
0.5
log,, y
0.0969
0.2068
0.3371
0.4870
0.6570
0.8469
X
104 log,, Y
0
969
0.1
2068
0.2
3371
0.3
4870
0.4
6570
0.5
5469
A
A2
A3
A4
A5
1099
204
-8
1303
1499
1700
196
201
199
1899
Fig. 2-42
-5
-2
3
0
3
CHAP. 21
APPLICATIONS O F THE DIFFERENCE CALCULUS
67
Then we find by the Gregory-Newton formula
loglo y = 0.0969
or
+ 1.099d1) + 1.02d2) - 0.1333d3) + 0.1250~(4)
(1)
Putting x = 0.6 we then obtain
log10 Y = 0.0969
+ 1.099(0.6) + 1.02(0.6)(0.5) - 0.1333(0.6)(0.5)(0.4) + 0.1250(0.6)(0.5)(0.4)(0.3)
= 1.0508
y = 101.0508 = 11.24
Thus
(b) The result (1) can be written as
1.099%
logloy = 0.0969
+
+ 1 . 0 2 ~ -( ~0.1) - 0 . 1 3 3 3 ~ -( ~O.l)(x - 0.2)
+ 0.1250x(~- O.l)(x - 0.2)(2 - 0.3)
To a high degree of approximation this can be written a s
loglo y = 0.0969
Thus
+ 1.099%+ 1 . 0 2 ~ ( -2 0.1)
= 0.0969
+ 1 . 0 2 +~ ~0.9972
y = 100.0969+ 1.0&+0.997x
- 10o~ogsg, 101.02x2+0.997x
- 1.250. 1ol.02z2+0.997x
(2)
The values corresponding to the table of Fig. 2-39 were in fact obtained by using the
formula
x=+x
y = 1.250.10
(3)
Thus the exact value corresponding t o x = 0.6 is 11.40. It is of interest that the value
obtained in Problem 2.48 is closer t o this true value than that obtained in part ( a ) using
logarithms. This can be accounted for by realizing that in the process of taking logarithms
and antilogarithms we introduce small errors each time. Further evidence of this is seen by
comparing ( 2 ) and (3). By recognizing that such errors do occur however we could be led
from ( 2 ) t o (3) on replacing 1.02 by 1 and 0.997 by 1. Then (3) could be recognized a s the
true formula by in fact putting x = 0, 0.1, . . .,0.5 and comparing with the corresponding
values of the original table.
Supplementary Problems
2.52.
Generalize the results of Problem 2.51.
2.54.
If yk = k3- 4k2+2k
(d)
( e ) A41/k*
- 3 ( a ) express
yk as a factorial polynomial and find (b)
Auk,
(c)
68
2.55.
2.56.
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
If
denotes differentiation with respect to k , show that
Dk
+
+ +
+
+ $0: + g D i +
...
3
(a)
Ak
(b)
A:
= Dk f +Di &Dk f &D;
= 0:
Df
&Dt
&Dt
(c)
At
= Df
+
* * a
...
Illustrate Problem 2.55 by finding Af(k5 - 3k4 f 2k3) and check your answer.
DIFFERENCE TABLES
2.57. Let f(x) = 5x2+ 3%- 7 where x = 2,5,8,11,14.
2.58.
[CHAP. 2
Set up a difference table f o r f(x).
By using only the difference table obtained in Problem 2.57, find the values of
y = f(x) for
x = 17,20,23,26,29,32.
+
2.59.
Let f(x) = 2x3 - 8x2 6%- 10 where x = 2,4,6,8,10. (a) Set u p a difference table for f ( x ) and
( b ) deduce the values of f(x) f o r x = 12,14,16,18,20 from this difference table.
2.60.
( a ) Suppose t h a t in Problem 2.57 the diRerence table had been set up f o r values x = 2,5,8,11.
Would the resulting table enable us to obtain the values of y = f(x) f o r x = 14,17,20,23? Explain.
(b) Answer p a r t (a) if the difference table had been set up f o r values x = 2 , 5 , 8 and if the values
f o r x = 11,14,17,20 were wanted.
2.61.
Suppose t h a t i t is desired t o obtain the values of f(x) = 3x4 - 2x2 7 x 5 for x = 3,7,11,15,
1 9 , . .,43. W h a t is the minimum number of values which we must obtain directly from f(x) so
t h a t from the resulting difference table we can determine the remaining values of f(x)?
2.62.
Generalize the results of Problem 2.61 to any polynomial f(x) and any set of equidistant values of x.
2.63.
Find the numerical values corresponding to the letters in the following difference table where it
i s assumed t h a t there is equal spacing of the independent variable x.
+ +
.
Y
-1
n
AY
F
G
A2Y
4
A3Y
A4Y
K
I
Fig. 2-43
2.64.
A difference table h a s one entry in the column Aoy. Determine how many entries there would be
in the column (a) Ay, (b) y, ( c ) A2y.
2.65.
Generalize the result of Problem 2.64 by determining how many entries there would be in the
column Amy if there were a entries in the column Any.
THE GREGORY-NEWTON FORMULA AND APPLICATIONS
2.66. Prove t h a t ( a ) ( k h ) ( 3 )= h3k(3), (b) ( k ! ~ ) ( ~=) h4kC4).
2.67.
Prove t h a t ( k h ) ( n )= hnk(n) by using mathematical induction and thus complete the proof in
Problem 2.6.
2.68.
Find a formula f o r the general term uk of a sequence whose first few terms a r e (a) 5,12,25,44,
(b) 1,5,14,30,55. W h a t assumptions must you make? Are the results unique? Explain.
CHAP. 21
69
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
2.6%
Given a sequence whose first four terms a r e 3,7,14,30. (a) Can you obtain a formula for the
general term of the sequence? ( b ) Can you find the 10th term of this sequence? State clearly
the assumptions made and whether you feel t h a t results obtained a r e unique.
2.70.
Suppose t h a t in Problem 2.69 the next three terms of the sequence a r e given as 61,113,192.
effect if any would this additional information have i n your answer to t h a t problem?
2.71.
Find g as a polynomial function in x which fits the data in the folldwing table and state clearly
the assumptions which you make in determining this polynomial.
X
1
2
3
4
5
6
Y
3
3
7
21
51
103
What
Fig. 2-44
2.72.
Suppose t h a t the values of x corresponding to those of y in the table of Fig. 2-44 a r e 1,3,5,7,9,11.
Explain how you might obtain y a s a polynomial function in x from the answer to Problem 2.71.
Can you generalize the result of your conclusions?
X
O0
30 O
60 O
90'
sin x
0.00000
0.50000
0.86603
1.00000
2.75.
Estimate the error term in Problem 2.74.
2.76.
Use the following table to find (a) loglo 5.2374, ( b ) loglo 5.4933.
X
5.1
5.2
5.3
5.4
5.5
log,, x
0.70757
0.71600
0.72428
0.73239
0.74036
X
60 O
61'
62'
63 O
64 '
sin x
0.86603
0.87462
0.88295
0.89101
0.89879
2.78.
Estimate the error terms in (a) Problem 2.76 and (b) Problem 2.77.
2.79.
Use the table of Problem 2.74 to find (a) sin68'30',
values.
X
3.12
log10 x
0.49415
3.13
j 0.49554
(b) sin45'18'
65'
0.90631
and compare with the exact
3.14
3.15
3.16
0.49693
0.49831
0.49969
70
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
2.81.
Determine if there is any difference to your answer for Problem 2.80 if we add t o the table the
values of loglo x for x = 3.11 and x = 3.17 given by 0.49276 and 0.50106 respectively.
2.82.
Given the following table find (a) e0.243, ( b ) e0.411.
I
e5
0.2
0.1
X
1
1
1.10517
0.3
I
1.22140
0.4
1.34986
0.5
I
1.49182
~
I
1.64872
Fig. 2-49
2.83.
The gamma function denoted by r ( x ) is given for various values of x in the following table.
Find r(1.1673).
1
z
I
1.15
1
1.16
1
1.17
I
1.18
I
1.19
I
I
r(x)
1
0.93304
I
0.92980
I
0.92670
1
0.92373
1
0.92088
I
Fig. 2-50
X
0.00
0.25
0.50
0.75
erf (x)
0.00000
0.27633
0.52049
0.71116
CENTRAL DIFFERENCE TABLES AND INTERPOLATION FORMULAS
2.85.
(a) Construct a central difference table corresponding to the data of Problem 2.76 by choosing
(a) Use Stirling’s formula to work Problem 2.76(a) and compare with that obtained
a = 5.2.
by use of the Gregory-Newton forward difference formula.
2.86.
Work (a) Problem 2.77, ( b ) Problem 2.80 by using Stirling’s formula.
2.87.
Work (a) Problem 2.76, ( b ) Problem 2.77, ( c ) Problem 2.80 by using the Gregory-Newton backward difference formula.
2.88.
Work (a) Problem 2.76,
formula.
2.89.
Show how a central difference table can be expressed using the backward difference operator
2.90.
Write the interpolation formulas on page 36 using the “x notation” in place of the “k notation”.
2.91.
Show how to write Bessel’s interpolation formula using (a) central differences and ( b ) backward
differences.
2.92.
Work Problem 2.83 by using Stirling’s formula with a = 1.17 and compare with that obtained by
using the Gregory-Newton forward difference formula.
2.93.
Use Gauss’ interpolation formula to work
( d ) Problem 2.83.
2.94.
Derive the second Gauss formula and thus complete Problem 2.18.
( b ) Problem 2.77,
( c ) Problem 2.80
by using Bessel’s interpolation
v, v2, v3, . . . .
(a) Problem 2.76, ( b ) Problem 2.77,
(c)
Problem 2.80,
CHAP. 21
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
71
ZIG-ZAG PATHS AND LOZENGE DIAGRAMS
2.95. Demonstrate the rules f o r obtaining interpolation formulas [see page 371 by obtaining (a) the
Gregory-Newton forward difference formula, ( b ) the Gregory-Newton backward difference formula,
(c) Gauss' first formula on page 36, ( d ) Stirling's formula.
Indicate the zig-zag path in the
lozenge diagram which corresponds to each formula.
2.96.
By choosing a n appropriate zig-zag path in Fig. 2-6, page 37, obtain an interpolation formula
differing from those on page 36.
2.97.
Explain whether you could find t h e error term in (a) the Gauss formulas and ( b ) the Stirling
formula by knowing the error in the Gregory-Newton forward difference formula.
LAGRANGE'S INTERPOLATION FORMULA
2.98.
Work (a) Problem 2.71, (b) Problem 2.80 by using Lagrange's interpolation formula.
2.99.
( a ) Find a polynomial in x which fits the d a t a in the following table and use the result to find
the values of y corresponding to (b) x = 3, (c) x = 5, ( d ) x = 8.
X
1
2
4
6
7
Y
6
16
102
376
576
X
0 (0')
n/6 (30')
n/4 (45')
n/3 (SO')
n/2 (90')
sin x
0.00000
0.50000
0.70711
0.86603
1.00000
2.102, Show t h a t Lagrange's interpolation formula reduces t o the Gregory-Newton forward difference
formula in the case where the x values a r e equally spaced.
TABLES WITH MISSING ENTRIES
2.103. Find the value u2 missing from the following table.
Fig. 2-54
2.104. Find the values zij and ug missing from the following table.
k
II
4
I
Uk
I
72
5
7
6
I
I
I
I
I
146
I
Fig. 2-55
192
9
8
I
I
I
302
I
I
72
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
2.105.
( a ) Use the data of Problem 2.101 to find the^ missing entries for sin x corresponding to x = 16O
and x = 75O and compare with the exact values. ( b ) Is there any difference between the answers
obtained from the method of p a r t ( a ) and those using the method of Problem 2.101(b)? Explain.
2.106.
( a ) Find the values of z/ corresponding to x = 3 and x = 5 in the table of Problem 2.99 using
Method 1 of Problem 2.28. ( b ) Can you use the method of p a r t (a)to find the value of y corresponding to x = 8 ? If so what is this value?
2.107.
Find the values
ztl,
Z L ~ Z, L ~ Z, L ~from
the following table.
1
Fig. 2-56
2.108.
Find the values
Z L ~ Z, L ~ uj,
,
us
from the following table.
k
0
UIC
2
5
10
15
20
92
212
382
Fig. 2-57
2.109.
A sequence is given by U , , , U ~ , U ~ , U ~ , U ~in
, U which
~
all values but u3 a r e known.
fourth differences a r e assumed constant then
- 0 . 5 ~ 1 UZ
0 . 5 ~ 4- 0 . 1 ~ 5
Z L ~= 0 . 1 ~ 0
Show t h a t if
+ +
2.110.
The following table shows the population of the United States f o r the years 1930 to 1970 at intervals of 10 years. Use the table t o find the population f o r the years (a) 1980 and ( b ) 1962, 1964,
1966 and 1968.
1930
Year
I
Population of
United States (millions)
I
122.8
1940
1
131.7
1950
1
151.1
1960
1
179.3
1970
I
205.2
I
2.111.
X
0
0.1
0.2
0.3
0.4
e-X
1.00000
0.90484
0.81873
0.74082
0.67032
2.112.
DIVIDED DIFFERENCES
2.113.
( a ) Construct a divided difference table corresponding to the data in the following table.
3
X
P(x)
I
9
I
5
8
10
11
35
119
205
257
.~
CHAP. 21
2.114.
73
APPLICATIONS O F THE DIFFERENCE CAJCULUS
( a ) Find a polynomial in 3: which fits the data in the following table and ( b ) obtain values of f(z)
corresponding t o x = 5 and x = 10.
X
0
2
3
6
7
8
f (2)
40
8
7
136
243
392
Fig. 2-61
2.115. Work Problem 2.28 using Newton’s divided difference interpolation formula.
2.116.
Show how to derive Newton’s divided difference formula by assuming a n expansion of the form
f(x) =
A0
+ A,(%- xO) + A,(%- x ~ ) (-x $1) +
and then determining the constants A,, Al, A,,
2.117.
2.118.
2.119.
* * ‘
+ A,($ - x ~ ) (-x 21).
*
(X
- x,)
. . ., A , .
Obtain the Gregory-Newton formula as a special case of Newton’s divided difference formula.
Prove that the value of f(z,,xl, . . .,2,) remains the same regardless of the arrangement of
f ( z 0 ,xl, . .,x,) = f(z,, xo,zl,
. .,x , - ~ ) , etc.
.
xo,zl, . . .,x,, i.e.
.
Prove that the divided differences of f(x)
+ g(z) are equal to the sum of
the divided differences of
f(4and dx).
2.120.
Prove that the divided differences of a constant times f(z) is equal to the constant times the divided
differences of f(x).
2.121.
( a ) Show that there is a root of z3- x - 1-= 0 between z = 1 and z = 2.
in (a). ( c ) Determine the remaining roots of the equation.
2.122.
Solve the equation x3
2.123
Determine the roots of
2.124.
Solve the equation e - 2 = x.
2.125.
Solve the equation x = 5 ( 1 - e - 2 ) .
( b ) Find the root
+ 2x2 + 10%- 20 = 0.
x3
- 3x2 + 2x - 5 = 0.
[Hint. Let
g
= f(x) = e-5 - 2.3
INVERSE INTERPOLATION
Use the data in Problem 2.71 to find the value of x corresponding to g = f(z)= 15 by using
( a ) Lagrange’s formula, ( b ) Newton’s divided difference formula and (c) the Gregory-Newton
forward difference formula. Compare your results with the exact value.
2.126.
Interest rate
l&%
2%
23%
3%
Amount of annuity
23.1237
24.2974
25.5447
26.8704
74
2.128.
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
Given the data in the following table find the value of x which corresponds to y = 10.
1
15
11
14
12
Fig. 2-63
2.129.
The following table gives values of f(x) f o r certain values of x. (a) Find the value of x which
corresponds to f ( x ) = 2.2000. (b) Using the fact t h a t the entries of the table were actually
2) where x is in radians, determine the exact value of x corobtained from f(x) = 10 sin &(x
responding to f ( x ) = 2.2000 and thus determine the accuracy of the result obtained in (a).
+
I
x
I
1
0.1
2.0846
f(x)
I
I
0.2
I
I
0.3
I
2.1823
I
0.4
I
2.2798
I
0.5
I
2.3770
I
I
2.4740
APPROXIMATE DIFFERENTIATION
2.131. ( a ) Use the table of Problem 2.76 to find the derivative of logl0z f o r x = 5.2 and ( b ) compare
with the exact value.
2.132.
Use the table of Problem 2.77 to find (a) the first derivative and ( b ) the second derivative of sin x
corresponding to x = 63O and compare with the exact values.
2.133.
Find the ( a ) first derivative and ( b ) second derivative of e2 at x = 0.2 using the table of Problem
2.82 and compare with the exact values.
2.134.
Find the derivative of the gamma function of Problem 2.83 f o r (a) x = 1.16, (b) 1.174.
2.135.
Find the derivative of t h e error function of Problem 2.84 f o r (a) x = 0.25, (b) x = 0.45.
2.136.
Show t h a t
(a) D =
2.137.
Show t h a t
(a) D2 =
V
+ 21 V2 + 31 V3 + ;1iV 4 +
Vz
* * *
1
6
+ V3 + 1V4 + - V j +
12
6
MISCELLANEOUS PROBLEMS
( a ) Is i t possible to determine unique numerical values corresponding to the letters in the following
difference table where i t is assumed t h a t there i s equal spacing of the independent variable X ?
( b ) Does your answer to ( a ) conflict with the remark made at the end of Problem 2.3?
2.138.
-1
H
J
1
Fig. 2-65
K
2
5
CHAP. 21
2.139.
APPLICATIONS O F THE DIFFERENCE CALCULUS
75
Can you formulate a rule which tells the minimum number of entries in a difference table which
must be known before all entries can be uniquely determined?
2.140.
2.141.
X
50
52
54
56
58
60
&
3.6840
3.7325
3.7798
3.8259
3.8709
3.9149
If a unit amount of money [the principal] is deposited a t 6% interest compounded semi-annually
and not withdrawn, the total amount A , accumulated a t the end of k years is given in the following
table. Determine the amount accumulated after ( a ) 9 years, ( b ) 16 years.
k
6
8
10
12
14
Ak
1.4258
1.6047
1.8061
2.0328
2.2879
Fig. 2-67
2.142.
Referring t o Problem 2.141 after how many years would one expect the money to double?
2.143.
X
0
0.2
0.4
0.6
0.8
1.0
J,(x)
1.0000
0.9900
0.9604
0.9120
0.8463
0.7652
r
2, 44.
2.145.
A ur t amount of money deposited a t the end of each year for n years a t an interest rate of r%
compounded annually is called an annuity. The following table shows the amount of an annuity
accumulated after 20 years a t various interest rates. Use the table t o determine the amount of
an annuity after 20 years if the interest rate is 1&%.
Interest rate r
1
1+
2
28
3
Amount of annuity
22.0190
23.1237
24.2974
25.5447
26.8704
A man takes out an annuity [see Problem 2.1441 with an insurance company who agrees t o pay
back after 20 years an amount equal to 25 times the yearly payment. Determine the interest rate
compounded annually which is involved.
2.146.
s
10
20
30
40
50
60
70
80
f(z)
50.3
42.8
35.6
29.4
22.3
15.5
10.9
6.8
76
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
[CHAP. 2
2.147.
(a) Give a possible explanation f o r the fact t h a t the differences in the table of Problem 2.146 tend to
decrease and then increase. ( b ) Does the result produce an error in interpolation? Explain.
( c ) How might more accurate results be obtained?
2.148.
The following table gives f o r various values of x corresponding values of l n r ( x ) where r(x) is
the gamma function [see page 701. From the known fact t h a t
r ( x ) r ( l-2)
x
= -
sin XT
obtain the values of In r(x) corresponding t o the missing values.
Fig. 2-71
2.149
From the theory of the gamma function we have the result
22=-1r(x) r ( x
+ 4) =
6132%)
Can this be used in Problem 2.148 to obtain greater accuracy in the values obtained f o r r(x)?
Explain.
2.150.
Use Problems 2.148 and 2.149 to find values of l n r ( x ) f o r x = 0.1,0.2,
. . .,0.9.
X
0.1
0.2
0.3
0.4
0.5
0.6
f(x)
0.3423
0.4080
0.5083
0.6616
0.9000
1.2790
2.152.
Complete the solution to Problem 2.45 by actually setting the derivative of (1) with respect to t
equal to zero.
2.153.
Suppose t h a t in Problem 2.45 the times in seconds a r e given by t , = 7.2, t, = 8.1, t3 = 8.4 and
t h a t the quantity y at these times has values 15.6, 15.9, 15.7 respectively. Find the time at which
y is a maximum and the value of this maximum.
2.154.
A t times t , , t,, . . ., t, measured from some fixed time as origin an object is observed to be at positions yl,y2, . . , , y n on a straight line measured from some fixed position as origin. If ~d i s the
position of the object at time t show t h a t
2.155.
Show t h a t Stirling's interpolation formula can be written in the form
yk
=
yo
+ +(Ay-,+Ayo)k + A2Y- ' 2k2!
+
-
- ')
~ 4 y k2(k2
- ~
4!
+
3(A3Y-2+A3Y-1)-
(n-l),]
+
-
[kz - (n - 1)2]
(272 - 1) !
+ + ( ~ 2 n - l y - ~+ ~ z n - l y - , , +
2)...[kz+ A2ny-n k z ( k z - l 2 ) ( k 2 - 2 (2n)
!
2.156.
k(k2 - 1)
3!
k(k2 - 12)(k2- 22)
...
Referring to Problem 2.44 what estimate of the population would you give f o r the year 1910?
Explain the significance of your result and give possible limitations on the reliability of interpolation and extrapolation.
CHAP. 21
2.157.
APPLICATIONS O F THE D I F F E R E N C E CALCULUS
77
Find the sum of the series
2.158.
+ 22 + 3* + + k2
[Hint. Find a general term for the sequence 12, l2+ 22, 12 + 22 +
. . . .]
Find the sum of the series (a) 1 + 2 + 3 + . + k, (b) l3+ Z3 + i. . + k3.
2.159.
Find the sum of the first k terms of the series
12
32,
*
22
33
*
+ 5 2 + + 112 +
82
*
* * *
2.160.
( a ) Find a formula f o r the general term of the sequence 2,5,14,31,60,109,194,347, . . . and ( b ) use
the result to find the 10th term.
2.161.
Prove t h a t
2k = 1
2.162.
Let
=
2.163.
Prove Everett's formula
@k,O
k(1) k(2) k(3)
++++ ..
l!
2!
3!
1 k=O
0 k f O '
I
.,
for k = 0 , 1 , 2 ,
Show that f o r k = 0,1,2,
.
f
An advantage of this formula is t h a t only even order differences a r e present.
2.164.
Prove t h a t
76k
2.165.
=
k(k - 1 ) . * ( k- n
(n- 1)!
+ 1)
[k
1 -
(" ') k - n + (";
Yn-2
2
+
1)
F
;;
Given the following table find the positive value of k such t h a t
ykt3 =
f
3(76k+2-76k)
I
I
I
18
I
I
Fig. 2-73
2.166.
Given the following table find the values of k such that
Fig. 2-74
2.167.
Referring to Problem 2.110(a) estimate the population for the year 1920 and compare with the
actual population which was 105.7 million. ( b ) Estimate the population for the years 1910, 1900
and 1890 and compare with the actual populations which were 92.0, 76.0 and 62.9 million respectively. ( c ) Estimate the population for the years 1990 and 2000. Discuss the reliability of your
predictions.
78
APPLICATIONS O F T H E D I F F E R E N C E CALCULUS
X
2
1
3
4
5
6
7
8
[CHAP. 2
9
10
f(x) 0.28685 0.36771 0.44824 0.52844 0.60831 0.68748 0.76705 0.84592 0.92444 1.00263
2.169.
Suppose t h a t x = xo is a n approximation to the root of the equation f(x) = 0.
appropriate conditions a better approximation is given by
Show that under
xo and determine these conditions. The method of obtaining roots is called Newton’s method.
[Hint. Assume t h a t the exact root is xo h so t h a t f ( x o + h) = 0. Then use the fact that
f ( x o It) = f ( x o ) hf’(x,) approximately.]
+
2.170.
+
+
Use the method of Problem 2.169 to work (a)Problem 2.37, ( b ) Problem 2.122, (c) Problem 2.123,
( d ) Problem 2.124, ( e ) Problem 2.125.
1
THE INTEGRAL OPERATOR
We have seen on page 2 that if A and B are operators such that ( A B ) f= f , i.e. A B = 1,
then B = A-I which is called the inverse operator corresponding to A or briefly the
inverse of A .
In order to see the significance of D-' where D is the derivative operator we assume
that f(x) and F ( x ) are such that
F(x) = D-'f(x)
(1)
Then by definition if we apply the operator D to both sides of (1)we obtain
D [ F ( x ) ]= D [ D - ' f ( x ) ] = ( D D - ' ) f ( x ) = f(x)
(2)
Thus F ( x ) is that function whose derivative is f(x). But from the integral calculus we
know that
F(x) = J-f(x)dx
c
(3)
+
where c is the constant o f integration. Thus we have
D-lf(x)
=
Jf(x)dx
+
c
(4.)
We can interpret D-' or 1 / D as an integral operator or anti-derivative operator.
D-l
J ( )dx
=
(5)
The integral in ( 3 ) o r ( 4 ) is called an indefinite integral of f(x) and as we have noted all
such indefinite integrals can differ only by an arbitrary constant. The process of finding
indefinite integrals is called integration.
GENERAL RULES OF INTEGRATION
I t is assumed that the student is already familiar with the elementary rules for
integration of functions. In the following we list the most important ones. In all cases
we omit the constant of integration which should be added.
1-1.
1-2.
1-3.
J [f(x)+ g ( x ) Jd x
J L Y f ( X ) dx
s
f(x) D g ( x )dx
=
=
=
J f ( x ) dx
+
J f(x) dx
f(x) g ( x )
-
J g ( x ) dx
s
O(
= constant
g ( x )D f ( x )d x
This is often called integration b y parts.
1-4.
f(x) d x
=
ddt)
f [ + ( t ),-dt]
where x = d t )
This is often called integration b y substitution.
79
80
THE SUM CALCULUS
[CHAP. 3
It should be noted that these results could have been written using D-I.
example, 1-1 and 1-2 become respectively
D-'[f(x) g(x)] = D-'f(x) D - ' ~ ( x ) , D-'[d(x)] = ao-'f(x)
Thus, for
+
+
Note that these indicate that D-I is a linear operator [see page 21.
INTEGRALS OF SPECIAL FUNCTIONS
In the following we list integrals of some of the more common functions corresponding
to those on page 5. In all cases we omit the constant of integration which should be added.
11-2.
r
m
J
11-3.
s
s
-1
m = -1
(ax+ b)m+'
( a x + b)"dx
=
+
(m 1)a
In ( u x b )
+
U
11-5.
f
xmdx
-
-bx
m#-1
m = -1
b>0, b P 1
In b
riO
erxdx
r#O
-
cos rx d x
11-7.
sin r x
r#O
r
DEFINITE INTEGRALS
Let f(x) be defined in an interval a 5 x 5 b. Divide the interval into n equal parts
of length h = A X = ( b - a)/n. Then the definite integral of f ( x ) between x = a and x = b
is defined as
f ( x )dx =
Jb
+ f [ a+ (n- 1)hl)
+ + + f ( a + 2h) +
lim it{f(a) f ( a h)
h-0
ern-m
if this limit exists.
If the graph of y = f ( x ) is the curve C of Fig. 3-1,then this limit represents geometrically the area bounded by the curve c, the x axis and the ordinates a t x = a and x = b.
In this figure Xk = U kh, X k t i = U -k (k -k 1)h and y k = f ( X k ) .
+
I
a
xk
Fig. 3-1
xk+l
b
CHAP. 31
81
T H E S U M CALCULUS
In a definite integral the variable of integration does not matter and so any symbol
can be used, Thus we can write
lb
b
f(u)du =
_Jb.’f(x)dx =
f ( t ) d t , etc.
For this reason the variable is often called a dummg variable.
FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS
d
If f ( x ) = D F ( x ) = - F ( x )
dx
that
f(x)dx =
then the f u n d a m e n t a l theoyem o f integral calculus states
’ d
,,F(z)
b
dx = F ( x ) I a = F ( b ) - F(a)
This can also be written as
=
F(x$
=
F(b) - F(u)
Example 1.
SOME IMPORTANT PROPERTIES OF DEFINITE INTEGRALS
2.
lb
f(x) dx
=
1‘
f(x) dx
+
1’
f(x) dx
where c is between a and b
or
SOME IMPORTANT THEOREMS OF INTEGRAL CALCULUS
Two theorems which will be found useful are the following.
1, Mean-value theorem for integrals. If f(x) is continuous in an interval a 5 x d b and
g(x) does not change sign in the interval, then there is a number t such that
_Jb.’ f ( x )g(x)d x
=
f ( t ) J bg(x)dx
t between a and b
2. Leibnitz’s rule for differentiation of an integral.
where U ( N ) , b(a) are assumed to be differentiable functions of a and F ( x , a ) and dFlaa:
are assumed to be continuous functions of x. Note that aF‘laN is the partial derivative
of F with respect to i.e. the ordinary derivative with respect t o N , when x is considered
to be constant.
(Y,
82
THE SUM CALCULUS
[CHAP. 3
THE SUM OPERATOR
Because of the analogy of D and A [or more precisely.A/Ax = A/h], it is natural t o seek
the significance of the operator A-' which we conjecture would have properties analogous
t o those of D-I.
To discover the analogy we suppose that F ( x ) and f(x) are such that
A
-F(x)
Ax
= f(x)
or
AF(x) = hf(x)
(6)
Then by definition A-' is that operator such that
F ( x ) = A-'[hf(x)]
(7)
Now we can show [see Problem 3.101 that A - I is a linear operator and also that if two
functions have the same difference, they can differ a t most by an arbitrary function C ( x )
such that C ( x h) = C ( x ) , i.e. an arbitrary periodic function with period h. Since this
is analogous to the arbitrary constant of integral calculus we often refer t o it as an
arbitrary periodic constant or briefly periodic constant.
+
From ( 7 ) we are led by analogy with formulas ( 3 ) and (4) of the integral calculus to
define
a-'[f(x)l =
or
c f(x) + Cdx)
(9)
where CI(X)is also a periodic constant. It is of interest that (9) can be written as
=
c f(x)Ax + C ( x )
(10)
which displays the remarkable analogy with
D-'f (x) =
J-f(x)dx
+c
(14
We shall call B the sum operator and B f(x) [or Z f(x) Ax = h B f(x)] the indefinite sum
of f(x). Note that because of (9) and (10) we can consider A-' and B or A l A x and Z ( ) A x
as inverse operators, i.e. A B = 1 o r B = A-'. See Problem 3.11. The process of finding
sums is called summation.
GENERAL RULES OF SUMMATION
The following formulas bear close resemblance to the rules of integration on page 79.
In all cases we omit the periodic constant of summation which should be added.
111-1.
111-2.
111-3.
c [ f ( 4+g(x)]
c
c f ( x )u ( x )
af(X)
=
=
=
c f ( x ) + c g(x)
c f(x)
= constant
f(x) Q(4- c g(x + h)
a
(Y
Af(4
This is often called sumnzation b y parts.
Note that these formulas could have been written using A-1 instead of 8. The first
two represent the statement that B o r A-' is a linear operator.
CHAP. 31
THE SUM CALCULUS
83
SUMMATIONS OF SPECIAL FUNCTIONS
In the following we list sums of some of the more common functions. The results are
analogous t o those on page 80. In all cases we omit the periodic constant which should
be added.
IV-1.
c
ff
ffX
h
-
X(m+l)
-
+
-
m#-1
( m 1)h
For the case where m = -1 or is not an integer we use the gamma function
[see page 861.
IV-4.
c bx
IV-5.
2 erx
IV-6.
c sinrx
-
- cos Y ( X - 3h)
2 sin drh
sin r ( x - Ah)
2 sinfrrh
in place of H. The analogy of the above
These results can also be written with
with corresponding integrals is clear if we multiply the sums by h = Ax. For example,
IV-2 becomes
IV-7.
2 cosrx
-
which corresponds to
with h = Ax, xcm)and H corresponding to dx, xm and $ respectively.
This relationship between the sum calculus and integral calculus is expressed in the
following important theorem.
r
1
n
Theorem 3-1.
DEFINITE SUMS AND THE FUNDAMENTAL THEOREM OF SUM CALCULUS
The definite sum of f(x) from x = a t o x = a + (n- 1)h in steps of h is denoted by
a
+ (n-llh
Ca
f(x) = f ( a )
+ f ( a + h) + f ( a+ 2h) + - + f[a +
* *
The following theorem analogous to the fundamental theorem of integral calculus on
page 81 is of great importance and is called the fundamental theorem of the sum calculus.
Theorem 3-2 [fundamental theorem of sum calculus].
a
+ (n-l)h
84
THE SUM CALCULUS
[CHAP. 3
C
where it is emphasized once again that the sum on the left is taken from x = a to
of h. It should also be noted that the right side of (15) means that
the indefinite sum is to be found as a function of x and we are then to substitute the upper
limit a + n h and lower limit a and subtract the results. In such case the added arbitrary
periodic constant is subtracted out as in the case of definite integrals.
x = a + (n-- 1)h in steps
Example 2.
If a = 1, h = 2, n = 3, then from (15) and IV-2 we have
+
+
This checks since the left side is (1)(-1) (3)(1) (5)(3) = 17.
It is of interest t o note that if we multiply both sides of (15) by h and then let h + 0
and n + so that nh = b - a, we obtain the fundamental theorem of integral calculus.
DIFFERENTIATION AND INTEGRATION OF SUMS
The following theorems are sometimes useful.
Theorem 3-3. If A-'F(X)
= BF(x) = G(x) then
4--1(g)
2 -d F
=
Theorem 3-4. If a-'F(x,a) = B F ( x , a ) =
=
dG
dx
dx
then
G(x,~t)
Theorem 3-4 is analogous t o Leibnitz's rule for differentiating an integral [see page 811.
Similar results hold for integrals. For example we have
Theorem 3-5.
THEOREMS ON SUMMATION USING THE SUBSCRIPT NOTATION
As in the case of the difference calculus all of the above results for the sum calculus
can be expressed in terms of the subscript notation yk = f ( a kh). For example 111-3 on
page 82 becomes using Zk = g ( a + kh)
+
c
zjicaZk
=
YkZk
-
c
Zk+lAyk
(16)
The periodic constant of summation in this case becomes an arbitrary constant.
Similarly formulas IV-1 through IV-7 can be written with x replaced by k and h = 1.
For example, IV-2 becomes
k(m+l)
2 k(m) = mi-1
(17)
m+l
The fundamental theorem of sum calculus can in subscript notation be written, using the
particular value a = 1, as
n+l
$ y k
= A-%k
k=l
I
It is seen that the subscript notation has some obvious advantages since h does not
ap p ea F.
CHAP. 31
THE SUM CALCULUS
85
ABEL'S TRANSFORMATION
The following result, called Abel's transformation, is sometimes useful.
n
Ukvk
Untl
-
vk
k
[Aukpz
(19)
k=l
k=l
k=l
5
OPERATOR METHODS FOR SUMMATION
Various operator methods are available which can be used to obtain summations quickly.
One useful result valid for p # 1 is given by
1
1
BkP(k) = A-'[pkP(k)] = -pkP(k)
= ---pkP(k)
A
E-1
=
pk-
1
,BE - 1 P ( k )
where c is an arbitrary constant. The series terminates if P ( k ) is a polynomial.
SUMMATION OF SERIES
The above results are very useful in obtaining sums of various series [see Problems
3.20-3.281. Further applications involving the important topic of summation of series are
given in Chapter 4.
THE GAMMA FUNCTION
) the case where m is not an integer we make
In order to generalize the factorial x ( ~ to
use of the gamma function. This function is defined by
r(p) =
dt
tP-le-t
=
lim
M-tol
xM
tp-'ect dt
p
>0
(21)
and we can show [see Problem 3.321 that
r ( p + 1) = W ( P )
(22)
The recursion formula (22) can be used to define r(p) for p 5 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(+) = fi
2.
x
r(r)r(l-r) = sin r?:
3.
e - t l n t d t = -7 = -0.5772156649..
r'(1) = Jffi
4.
O<r<l
.
0
where
y
is called Euler's constant.
If m is an integer we can write [see Problem 3.351
X(m)
=
(i+ 1)
hmr
Since ( 2 3 ) has meaning for all values of m we shall take it as a definition of
for all m.
x ( ~ )
86
THE SUM CALCULUS
[CHAP. 3
and
thus generalizing the results ( 2 5 ) , page 6, and IV-2, page 83.
If m = -1, we obtain [see Problem 3.391
The function on the right of (26) is sometimes called the digamma function and is denoted
by qx).
The results ( 2 3 ) 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 fo(x),f~(x),
f z ( x ) , . . . have the property that
Dfn(x) = fn-l(x)
(27)
If
we
conwhere in (27), as well as what follows, we shall assume that n = 1 , 2 , 3 , . , . .
sider the function f n ( x )to be a polynomial of degree n then we can show that
coxn
c1xn-1
fn(x)
= - +-+
. . . C n - I X + Cn
n!
(n-l)!
where CO, CI, . . ,en are constants independent of n [see Problem 3.401.
+
.
To specify the polynomials, i.e. to determine CO, CI, . . .,Cn, further conditions need t o be
given. Different conditions will result in different sequences of polynomials.
Let p,(x) be a sequence of polynomials satisfying (27), i.e.
OP,(x) = P,&)
and the added condition
p - 1
@n(x)
=
(n- 1) !
where we take h = 1 and define O ! = 1, Conditions (29) and (30) enable us to specify
completely the polynomial p,(x). We find in fact that the first few polynomials are
Po@) = 1, /3,(x) = x - 4, &(x) =
- &x
+ A,
&(x) =
Qx3
- *x2
+ &x
(31)
See Problem 3.41.
It should be noted that once p,(x) is found, we can use (29) t o find P,-,(x), Pn-2(x),. . . ,
by successive differentiations. This is particularly useful since it is possible to obtain P,,(x)
in terms of Stirling numbers of the second kind by the formula
See Problem 3.44.
We define the Bernoulli polynomials as
B,(4 = n!P,(x)
so that from (32)
(33)
THE SUM CALCULUS
CHAP. 31
87
The first few Bernoulli polynomials are given by
Bo(x) = 1, Bi(x) = x - 6, B z ( x ) = X'
-x
+ 9,
B ~ ( x=) x3 - $ x 2 + 4%
(35)
We define the Bernoulli numbers, denoted by Bn, as the values of the Bernoulli polynomials with x = 0, i.e.
Bn = Bn(0)
(36)
The first few Bernoulli numbers are given by
Bo = 1, B1 = -4, Bz = 9, B3 = 0 , Bq = -&j
(37)
For tables of Bernoulli numbers and polynomials see Appendixes C and D, pages 234
and 235.
IMPORTANT PROPERTIES OF BERNOULLI NUMBERS AND POLYNOMIALS
In the following we list some of the important and interesting propyties of Bernoulli
numbers and polynomials.
1.
B,(x)
=
xn
+
(Y)xn-lBl
+
+
(~)X,-~BZ
*
+
(:>Bn
+
This can be expressed formally by B,(x) = (x B)" where after expanding, using
the binomial formula, Bk is replaced by Bk. It is a recurrence formula which can be
used t o obtain the Bernoulli polynomials.
2.
B,(x
+ 1) - B,(x) = nxn-l
3.
B;(x) = nB,-i(x)
4.
B, = B,(O) = B n ( l ) , B2n-l = 0
n = 2 , 3 , 4 , .. .
5.
+
This can be expressed formally as (1 B)" - Bn = 0 where after expanding, using
the binomial fomda, Bk is replaced by Bk. This recurrence formula can be used to
obtain the Bernoulli numbers.
7.
This is often called the generating function for the Bernoulli polynomials and can
be used t o define them [see Problems 3.51 and 3.531.
8.
This is called the generating function for the Bernoulli numbers and can be used t o
obtain these numbers [see Problem 3.521.
X
X
-cothZ
2
9.
10.
1'
+
where r = 1 , 2 , 3 , . . ,
by Bk.
=
1
++ B ~ x ' + -B4x4
4!
B6X6 +
6!
...
(n+ B)'+' - Brtl
r+l
and where the right side is to be expanded and then Bk replaced
2'
+
3'
+
* * *
+
(n-1)'
=
88
THE SUM CALCULUS
-1+ - +1- + . 1. .
11.
12n
22"
( -l)n-1B2n(2x)2n
2 (2n)!
=
2(-1)"(2n - I ) !
=
n = 1,2,3, . . .
-
32n
12. If n = 1,2,3, ., . . , then
Bz,-l(x)
[CHAP. 3
ffi
k=l
sin2kii-x
(2kT)'"-'
EULER NUMBERS AND POLYNOMIALS
The Euler polynomials E,(x) are defined by equation (27),i.e.
DEn(x) = E,-I(x)
together with the added condition
Xn
MEn(X) = n!
or
+[E,(x+1)
Xn
+ E,(x)] = a
(39)
where M is the averaging operator of page 10 and h = 1.
The first few Euler polynomials are given by
+
E o ( x ) = 1, El(%)= x - 4, E ~ ( x =
) ax2 - +x, E ~ ( x =
) +x3-
+g
The Euler numbers are defined as
En = 2"n! En(+)
and the first few Euler numbers are given by
(40)
(41)
Eo = 1, El = 0, Ez = -1, E B = 0, E4 = 5, Eg = 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 n ( ~ )=
1.
'oxn
n!
+ -( n - l ) ! + . . .
+ en
+
where
The first few values of e k are
eo = 1, el = --&,e2 = 0,
= &,
e3
2.
E ; ( x ) = En-l(x)
3.
E,(x) = (-l)nE,(l- x )
4.
E,(O)
+ E,(1)
+ En(-l)
=
e4
= 0
2(-1)"
n!
5.
E,(O)
6.
Epn = 2*"(2n)!Ezn(&)= 0
7.
= 0,
1
e5
= -m
f o r n = 1,2,3, . . .
for n = 1,2,3,
. ..
f o r n = 1,2,3, .
..
THE SUM CALCULUS
CHAP. 31
89
8.
This is often called the generating function f o r the Euler polynomials and can be used
t o define them.
secx =
iE2klX2k
= 1 + -1- 2 + -x4
5
+ -x6
61
+ ...
9.
k=O
(2k)!
2
24
720
2
10.
tanx
=
2 -TkXk
k!
-
x
+ 31 +
-x3
-2x 5
15
where
Tk = / z k k !e k l
are called tangent numbers and are integers.
k=l
+
17 x
315
62 x 9 + . . .
' i2835
11.
12. If n = 1 , 2 , 3 , . . . , thenfor O d x d l ,
Solved Problems
THE INTEGRAL OPERATOR AND RULES OF INTEGRATION
3.1. Find each of the following:
( a ) D-'(2x2 - 5% + 4)
( b ) D-1(4e-3" - 3 sin 2 x )
(a)
0-'(222-
=
S
(b)
D-1(4e-3" - 3 sin 2x)
=
5 2 + 4)
(222-5%
r
+ 4) dx
2x3
--
112
(4e-3" - 3 sin 2x) dz
61nx
+ 9-
5 % ~
+ 4 x + c
= 3 -2
J
$1/2
( c ) Ddl[
x-1/2
+ c
90
3.3.
THE SUM CALCULUS
Find
(a)
[CHAP. 3
1
s x cos4x dx, ( b )
In x dx.
( a ) We use integration by parts here.
Let
Dg(x) = cos4x
f(x) = x,
D f ( x ) = 1,
so t h a t
g(x) =
sin 42
4
where in finding g(x) we omit the arbitrary constant.
Then using the formula f o r integration by parts,
S
+(
x cos4x dx = x-
sin44x)
- Jsin442 cis =
x sin4x
4
+-
COS~X
16
+ C
( b ) Let f ( x ) = l n x , D g ( x ) = 1 so t h a t
Df(4 =
;,1
dx) =
Then using integration by parts we have
s
3.4.
Find
(a) sx2e3'.dx,
lnxdx
= xlnx-
= xlnx-x
x4 sin2x dx.
(b)
( a ) I n this case we must apply integration by parts twice. We obtain
s
x2e3x dx = (x2)
(7) - S (7)
-[
($) (2x)
=
(~2)(?)
=
(%2)($)
(2%)
-( 2 4 3
dx
s ( 2 ) ($)dx]
+ (2)($)
a p a r t from the additive constant of integration.
I n this result we make the following observations.
(i)
The Erst factors in each of the last two terms, i.e. 2%and 2, a r e 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. e35/9 and e3x/27, a r e obtained by
taking successive integrals of e3"/3 [the second factor of the first term].
+, +.
The above observations a r e quite general and
-,
(iii) The signs of the terms alternate
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 p a r t s is easily formulated [see Problem 3.1431.
THE SUM CALCULUS
CHAP. 31
91
( b ) Using the generalized integration by parts outlined in part (a) we find the result
S
= (x4)
3c4 sin22 dx
3.5.
Find
(a)
x) dx
J (1v +x sin
- cos x
f
(24z)(y)
(4
( b ) Je\"dx,
+ (24) (9)
+c
J In2(x++44)dx.
- cosx = t so that x - c o s x = t z and (1
on taking the differential of both sides. Then the given integral becomes
(a) Make the substitution q x
S(l+sinx)dx
v x - cosx
sy
=
= 2 t + c = 2q-f
c
+
3
( b ) Let dbecomes
= t so that 22 - 1 = t3, x = 8(t3 l), dz = Qtzdt. Then the given integral
zs
= Setitzdt = 2
$ek=i&
=
[(tZ)(et)
+ 4) = t
Let t = ax
+b
Replacing t by ax
so that x
t2etdt
- (it)(et)+ (2)(et)] +
= iet(t2-2t+2)
(c) Let In (x
+ sinx) dx = 2t dt
+ 4 = et,
so that dx = dtla.
c
3
+c
= 32 e ~ [ ( ~ s - l ) 2 / 3 - 2 ( 2 x - l ) 1 / 3 + 2 1
x = et - 4, dx = et dt.
+c
Then the given integral becomes
Then the required integral becomes
+ b yields the required result.
DEFINITE INTEGRALS AND THE FUNDAMENTAL THEOREM OF
INTEGRAL CALCULUS
2
3.7.
(a) Evaluate
+
( x2 1)2dx and ( b ) give a geometric interpretation.
(a) We first find the corresponding indefinite integral a s follows:
D-1(x2+1)2
= J(x2+l)zdx
= J(x4+2x2+l)dx
= sz4dx
+ 2Jx2dx + J d s
where we can omit the constant of integration. Thus by the fundamental theorem of integral
calculus,
%5
2x3
(&+1)2dx = x
= (y+u+2)
- (&+Q+l)
3
5
3
+
_ =
- 178
15
+
1
11.867 (approx.)
92
THE SUM CALCULUS
[CHAP. 3
( b ) The value 178/16 = 11.867 (approx.) represents the area [shaded in Fig. 3-21 bounded by the
the x axis and the ordinates a t x = 1 and x = 2.
curve ld = (x2+
X
Fig. 3-2
3.8.
Find
(a)
SnI2 6x dx,
sin
iz -%
( b ) J 2~ e - ~ ~ d(C)x ,
0
(c)
s
Method 1.
Let x
dx
x2
x2
+
2.
dx
- 22 + 2
- 1 = tan t. Then d x = see2 t d t and so
= S d t = t
+c
= tan-'(x-1)
+ c,
Thus
Method 2.
As in Method 1 we let x - 1 = tan t but we note that when x = 0, tan t = -1, t = -d4
and when x = 2, tan t = 1, t = id4 Then
dx
-
dx
JVl4
t = -7r/4
+1
sec2t d t
tan2 t
=
JTi4
at =
t = --a/4
77
-
2
Note that all of these can be interpreted geometrically as an area.
THE SUM OPERATOR AND RULES OF SUMMATION
3.9.
A
A
Prove that if ~ F I ( x=) f(x) and z F z ( x ) = f(x), then Fl(x) - Fz(x) =
+
where C(x h) = C ( x ) .
c(x)
CHAP. 31
T H E SUM CALCULUS
93
By subtraction we have
A
-[Fl(x)
-F2(x)] = 0
Ax
or
AIFl(x)-F,(x)]= 0
i.e.
If we write Fl(z)
- F 2 ( 2 )= C ( x ) then
AC(x) = 0
or
C ( x + h) - c(z) = 0,
i.e.
C(z
+ h) =
C(x)
We refer to C ( x ) as a periodic constant with period h. We have proved t h a t if two functions
have the same difference, then they can differ by at most a n arbitrary periodic constant, The
result i s analogous to the theorem of integral calculus which states t h a t two functions which have
the same derivative can differ at most by a n arbitrary constant.
3.10. Prove that A-' is a linear operator, i.e. (a) A-'[f(x) 4- g ( x ) ]= A-'f(x) + A-'g(x) and
( b ) a-'[.f(x)] = .A-lf(x) where f(x) and g(x) are any functions and (Y is any constant.
( a ) By definition if AF(x)
= f(x) and AG@) = g(x) then
F ( x ) = A-lf(x),
G(x) = A-'g(z)
(1)
Also by adding AF(z) = f(z) and 'AG(x) = g(x) we have
A[F(z)
so that by definition
F(x)
Using (1) in ( 2 ) we find as required
A-l[f(x)
( b ) F o r a n y constant
(Y
+ G(z)l
= f(x) + g ( 4
+G(4
= A-'[f(x)
+ g(x)I
+ g(x)]
= A-lf(x)
+ A-lg(x)
we have
A[aF(%)] = aAF(x) =
(Y~(x)
Thus from the first equation in (1)
A-'[af(%)]
= &(x)
= aA-l[f(~)]
+
3.11. If we define P ( x ) = a - l [ h f ( x )=
] Bf(x)h C ( x )as in (S), Page 82, prove that (a) A X = 1
(
b
)
A-'[f(x)]
=
B
f(x) Cl(x) where CI(X) is an arbitrary periodic conor B = Ad',
+
stant, and (c)
(&)-I
f(x) = 2 f(x) A x -I- c(X).
By operating with A on
+
F ( x ) = A - l [ h f ( z ) ] = x f ( x ) h c(o)
we obtain
AF(x) = h f ( z ) = A x f ( x ) h-I- Ac(x)
since A is a linear operator.
Now by Problem 3.9, AC(x) = 0 so that ( 2 ) becomes
hfb) = A x f ( d h
B u t since f(x) is arbitrary and h # 0 i t follows from ( 8 ) t h a t
AX
= 1
or
2
=
~ - 1
It follows from this and Problem 3.10 t h a t B is a linear operator.
Dividing equation (1) by h using the fact t h a t B and
A-1
a r e linear operators, we have
where C ( x ) and thus Cl(x) = C ( x ) / h a r e periodic constants.
[CHAP. 3
THE SUM CALCULUS
94
(c) This follows a t once from (1) on noting that
and replacing h by Ax.
3.12. Prove that
(a)
B [f(x) +g(x)] = Bf(x) + z g ( X ) , ( b ) B d ( X ) = a zf(x).
= A--l and Problem 3.10.
These follow a t once from the fact that
3.13. Prove the formula f o r summation by parts, i.e.
x'"+l'
3.14. Prove
(a) B d m ) =
(m+l)F,'
m z -1,
( b ) B sinrx =
- cos r ( x - gh)
2 sin &rh
(a) We have from V-2, page 7, or (24), page 86, on replacing rn by m
Ax(m+1) = (m 4-l ) x ( m ) h
Then if m # -1,
i.e.
or
apart from an additive periodic constant.
( b ) From V-7, page 7,
AIcosrm]
= -2
Replacing x by x - i h , we have
rh
A[cos r ( x - *h)] = -2 sin- sin r x
2
rh
Thus on dividing by -2 sin 2 , we have
cos r ( x - i h )
-2 sin 3rh
1-
sin rx
+1
CHAP. 31
THE SUM CALCULUS
95
so that
- +h)
- cos2 r(x
sin &rh
=
A-1 sinrs
or
2sinrx
- cos2rs( izn-a h+h)
=
L
apart from an additive periodic constant.
3.15.
Use Problem 3.14(b) to obtain
S sinrx dx.
From Problem 3.14(b) we have on multiplying by h
2hsinrx
= -
h cos T ( Z - &h)
2 sin &rh
Letting h + 0 this yields
S
s i n m dx
= lim -
h cos r(5 - &h)
= lim
h+O
=
[
[
-cos r(x -
lim
")I[ 3
@)I[ ]
2 sin&&
h-rO
irh
sin i r h
-cos r(x -
r
h+O
lim sin +rh
h-0
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-'(xax) =
B xu2, a # 1, and ( b ) check your answer.
(a) We have from Problem 3.13
A - l [ f ( ~ A) d z ) ] = f ( z ) &)
Let f(z)= x and A&)
= ax. Then
Af(z)=Ax
g(2)
=
A-laZ
- A-'[ds
-I-h) A f ( s ) ]
= h and
ax
= ah
-1
Thus
-
xax
ah
-1
hax+h
(ah- 1)2
( b ) Check:
(ah- 1)(x
+ h ) a Z + h - haz+2h - ( a h - l ) ( x a s ) + has+h
(ah
xax
- 1)2
96
[CHAP.3
THE S U M CALCULUS
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-llh
f ( x ) = a-Y(a)la+nh
c
a
If we assume that F ( z ) and f(x) are related by
AF(x)
- - f(z)
Ax
we have
Then by putting x = a, a
or
= hf(x)
AF(X)
F ( z + h) - F ( x ) = h f ( z )
+ h, . . .,a + (n - 1)h successively in (2) we obtain the
(1)
(8)
equations
F ( a + h) - F ( a ) = hf(a)
2h) - F ( a h) = h f ( a h)
+
+
+
F ( a + nh) - F [ a+ ( n - l)h] = h f [ a+ (n - l)h]
F(a
...........................................
Thus by addition we have
+ + + - + f [ a+ (.n - l)h]}
+
F ( a nh) - F(a) = h { f ( a ) f ( a h)
*
*
or
where the sum is taken from x = a t o a
+ (n-
1)h in steps of h.
But from the second equation in (1) it follows that
and the required result is proved on using this in (8).
3.18.
Evaluate each of the following using the fundamental theorem of sum calculus and
verify directly
2 8 ~ ' if~ )h = 3
(c)
0
1
6
(b)
2 [ 3 ~ ( ~ ) - 8 +x (10~1) if
19
h =2
2
(a) Since 2 8 d 3 )
c cos 5x if h = d 2
if h = 4
(d) c
277
7
(a)
3
~ ( 4 )
20(4)
= 8 2 x ( 3 ) = 8 4 . 3 = 3 we have by the fundamental theorem of sum
calculus using a = 1 and It = 3,
97
THE SUM CALCULUS
CHAP. 31
Check:
5
(c)
+ 101 + [3(4)(2)- 8(4) + 101 + [3(6)(4)- 8(6) + 101 =
22 [ 3 d 2 )- 8 d 1 ) + 101 =
[3(2)(0) - 8(2)
By IV-7, page 83, with
= 5, h = a/2 we have
T
A-1
C O S ~ Z
Check:
2a
20 cos5a:
Check:
19
Xs(-3)
= COSO
=
3
+
sin 512 - ( ~ / 4 ) ]
2 sin ( 5 d 4 )
=
=
sin 5[2 - ( d 4 ) ] 5 ~ 1 2
2 sin ( 5 d 4 ) o
-
sin (45d4)
2 sin (5r/4)
2Tl
20 cos5s
Thus
-
+
sin (-5d4)
=
2 sin ( 5 d 4 )
COS(~~T/~) COS(~T)
1
3.7011
+
___
1
7-11.15
30
+ ~ 0 ~ ( 1 5 a / 2+) c o s ( l 0 r )
1
1
1 1 * 1 5 * 1 9+ 1 5 - 1 9 . 2 3
~
+
= 1
- -25
- 4347
DIFFERENTIATION AND INTEGRATION OF SUMS
3.19. Find A-l[xerz] by using Theorem 3-4, page 84.
We have from IV-5, page 83,
~-l[erz]
erz
= erh
-1
Then by differentiating both sides with respect t o r [which corresponds to a in Theorem 3-41 we
find
or
- l)(zerz) - (e")(herh)
(erh - 1)2
( x - h ) e r ( z + h ) - se'z
(erh - 1)2
=
A-I[zerz]
(erh
This can be checked by referring t o an alternative method of obtaining the result given in
Problem 3.16(a). From that problem we find on putting a = er
A-~[x~?Z]
= erh - 1
her(%+
(erh - 1)2
h)
($ - h ) e r ( z + h ) - Z e r z
SUMMATION OF SERIES
3.20. Sum the series 1 . 3 - 5 + 3.507
(erh
- 1)2
+ 5 0 7 . 9 + to n terms.
Since the nth term of the series is (2%- 1)(2n+ 1)(2n+ 3), the series can be written as
2n.e 3
2
5
x(3)
where h = 2
98
THE SUM CALCULUS
[CHAP. 3
Then by the fundamental theorem of sum calculus,
-
(2n
-
(2n
+ 5)(2n + 3)(2n + 1 ) ( 2 ~ -1) - (5)(3)(1)(-1)
8
8
+ 5)(2n + 3)(2n + I)(2n - 1)+ 15
8
+ + + +
+ + + +
3.21. Sum the series (a) l2 22 32 42
and ( b ) 22 5' 82 112 . t o n terms.
Since we wish to find X 2 2 we need to express x 2 as a factorial polynomial. This is found from
3 2 = X ( 2 ) + hxcl)
As a possible check we can t r y a particular value of n, say n = 3.
12
+ 22 + 32
which is correct.
Then we obtain
= 14
= (2)(3)(7)
6
~
( b ) The nth term is (3%- 1)2. Thus using h = 3 the series can be represented by
=c
+
(3% 2)(3n - 1)(3n - 4)
9
Check:
+
(3% 2)(3n - l)(6n
18
- l)]
+ (3%+ 2)(3n
2
+ 1)4-2
n(6d
-
+ 6 2 + 82
(3)(6 ~3~
=
9
+ 3%- 1)
2
Let n = 3. Then
22
- p2)(-1)(-4)
+ 3 . 3 - 1)
= 93
2
which is correct.
3.22. Sum the series
1
103.5
~
...
+-----+-----+
1
1
3.5.7
to n terms.
5.709
1
The n t h term of the series is (2n - 1)(2n + l)(2n
+
3)
.
Now since
(%-h)(-3)
= -1
+
+
x(z h)(x 2h)
i t follows t h a t if we let h = 2, the required sum of the series is
I (2)(-1)]
2
CHAP. 31
T H E SUM CALCULUS
L-
=
12
Check:
99
1
4(2n
+ l)(2n + 3)
1 - 4(5)(7)
1
1 . 31. 5 + 3 . 510 7 - 12
- 105 which is correct.
Putting n = 2 we find
1.3
1.
5+ 3.5.7
1 + 5.7.9
1 + - .
3.23. Sum the series
*
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
ne w
[L12
4(2n
+ + cos 0 + cos 20 +
3.24. Sum the series
]=.
1
+ 1)(2n+ 3)
. + cos no.
Using h = e the series is given by
sin ( n + + ) e
2 sinie
i1 +
=
1
1
1
3.25. Sum the series - - 1.4 2-5 3.6
+
+
+ -.
- sin (n+ +)e
- -s i n i s
2 sin48
2 sinis
to n terms.
Using h = 1 the sum of the series to n terms is given by
To find this we attempt to express the general term as a sum of factorials. We first write
- 1
x(2
-
+ 3)
+
*
+ 2)
= A,
. ..
+ 1)(x+ 2)
= A,
(2
3)(4)
such that the numerator is the sum of fac-
+ A,(%+ 3)(1) + A2(x+ 3)(2) + A3(x + 3 ) ( 3 ) +
Since the left side of ( 2 ) is of degree 2, i t is clear that A,,
(5
+
+ 1)(2 + 2)(2 + 3)
We then seek to determine constants A,, A,, A,,
torials in the form
(x l ) ( m
- (x + l)(x + 2)
(2+1)(X+2)
x(x
. , . must be zero.
+ A1(x + 3)(') + A2(2 + 3)(*) =
A,
* * *
Thus
+ A1(x + 3) + A2(5 + 3 ) ( +~ 2)
From this identity we find A , = 2, A l = -2, A2 = 1. Thus (1) becomes
1
- -
x(x
+ 3)
-
2
- 2(x + 3) + (x + 3)(x + 2)
+ l ) ( x + 2)(x + 3)
2
2
+- 1+ 1)
x(z + l)(x + 2)
x(2 + 1)(2+ 2)(x + 3)
x(2
-
= 2(2-11)(-4) - q x - i y - 3 )
x(2
+
($- 1 ) ( - 2 )
(2)
100
THE SUM CALCULUS
[CHAP. 3
Thus apart from an additive constant,
L
Then by the fundamental theorem of sum calculus
{
2(2 - 1)(-3)
-3
n
2
- 1 =
1 2(2+3)
- 11
- 18
3.26.
1
Sum the series 1 4
- 2(2 - 1 ) ( - 2 )
-2
1
1
n
+1
(5
+
(n+ l)(n+2)
1
++ 7+
2.5 3 6
- 1)(-1)
-1
+
1
* *
-
2
3(n+l)(n+2)(n$3)
.
From Problem 3.26 by letting n + m 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 + *
k notation.
- . t o n terms by using the subscript or
+
The kth term of the series is ’& = (2k - 1)(2k 1)(2k 4-3) = (2k
interval h = 1 [as is always the case for the k notation].
+ 3)(3)
where the difference
Then the sum of the series is
- (2n + 5)(2n + 3)(2n + 1)(2n - 1) - (6)(3)(1)(-1)
8
8
- 1) + 1 6
- (2n + 6)(2n + 3)(2n + l)(%
8
3.28. Sum the series 22
+ 52 + 82 + - -
The kth term of the series is
n
(3k - 1 ) 2 =
x (9k2-
k=l
k=l
-
uk
n
n(6n2
6k
to n terms by using the subscript or k notation.
= (3k - 1)2. Then the sum of the series is
+ 1)
+ 3n - 1)
2
*
n
=
2
k=l
+
[9(k(2) k(1)) - 6k(”
+ 11
CHAP. 31
T H E SUM CALCULUS
101
ABEL'S TRANSFORMATION
3.29. Prove Abel's transformation.
Using summation by parts and the fundamental theorem of sum calculus we have
a t (n-l)h
2
atnh
f(x)Ag(z)
= . f(z)g(z)
(1
-
t (n-l)h
(2 )
g(x+h)Af(z)
where the summation is taken in steps of h.
Putting a = 1, h = 1,
Now let
we find
fk
2
= k and writing f(k) = fk, g(k) = g k , (1) becomes
= u k and A g k = v k . Then
gktl
- g k = wk and so on summing from k = 1 t o
i.e.
From this we have
Then ( 2 ) becomes
= o
OPERATOR METHODS OF SUMMATION
3.30. Prove that if p # 1
apart from an arbitrary additive constant.
We have for any F ( k )
ApkF(k) =
pk+lF(k
+ 1)-
/3kF(k)
=
/3k+lEF(k)
- /3kF(k) =
pk(/3E
- l)F(k)
n
-1
102
THE SUM CALCULUS
Let
(PE-l)F(k) = P(k)
or
F(k) =
[CHAP. 3
1
PE - 1 P ( k )
Then
APkF(k) = pkP(k)
Thus
-
1
pk
P(1i-A)
- 1 P(k)
=
apart from an arbitrary additive constant and assuming P Z 1. Since A-'pkP(k) = 2 pkP(k) the
required result follows. Note that if P ( k ) is any polynomial the series terminates.
3.31. Use Problem 3.30 t o find
f:k ~ 2 ~ .
k=l
We have by Problem 3.30 with p = 2, P ( k ) = k
A-l[k*2k] =
2 k*2k =
2k[l-2A+4A2--*-]k
= 2k(k-2)
Then by the fundamental theorem of sum calculus
THE GAMMA FUNCTION
+
3.32. Prove that r(p 1)= p r ( p ) .
Integrating by parts we have for p
3.33. Prove that
r(4)= 6.
We have
on letting t = 22.
r(4) =
>0
tw
t-lIZe-tdt
= 2 t w k-2' d s
Then
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 , $)
Fig. 3-3
CHAP. 31
103
THE SUM CALCULUS
rather than rectangular coordinates (sly). To do this we note that the element of area in the
xy plane shown shaded in Fig. 3-3 is ( p dqj) dp = p dp d+. Also since x = p cos
y = p sin +, we
have x2 y2 = p2 so that the result (1) with dx dy replaced by p dp d+ becomes
+
+f
= 4
{r(+)}2
sir'Srn
'
e-P2
(2)
p dp d+
p=o
@=0
The limits for p and qj in ( 2 ) are determined from the fact that if we fix
and then we vary qj from 0 t o ~ 1 2 .
+f p goes from 0 to
m
The integral in (2) is equal t o
Thus
3.34. Find
{r(+)}2
=n
and since
r(3)> 0
r(4)= G.
we must have
(a) r(5), ( b ) r(7/2), (c) 1?(-1/2).
+
Since r ( p 1) = p r ( p ) we have on putting p = 4, r(5) = 4r(4). Similarly on putting p = 3,
r(4) = 3r(3). Continuing, we find 135) = 4134) = 4 3133) = 4 3 2132) = 4 3 2 lr(1).
But
J r ne - t d t
r(1) =
=
= 1
-e-tiw
0
0
Thus r(5) = 4 . 3 . 2 . 1 = 4 !
In general if p is any positive integer, r(p
+ 1) = p !
Using the recursion formula we find
7
5
5
r(z)= z r ( p )=
5
3
3
zr(5)=
:*
:*
+r(&)=
+fi
=
1 5 6
8
Assuming that the gamma function is defined f o r all values of p by assuming that i t satisfies
the recursion formula r(p 1) = pI'(p) for all p , we have on putting p = -1/2
r(&)= (-&)r(-+) or r(-+) = -2134) = -2fi
+
3.35.
Prove that if m is a n integer then
%(m)
h m r ( i+ 1)
=
r
(i- m + I)
If m is a positive integer then
Z(m)
= hm(;)(;
Now
+ h)
. ..(; - m + 1)
= x(x - h)(x - 2h) * * * (x - mh
r(p+ 1) = p r ( p ) =
- 1)(
;-2)
p ( p - l ) r ( p - 1) = p ( p - l ) ( p - 2 ) r ( p - 2 )
= p ( p - l ) ( p - 2)- ' ( p - m
+ l ) r ( p- m + 1)
=
***
Thus
+
r(p 1)
p ( p - m+ 1)
= P(P - 1)b- 2). * '(P- m
+ 1)
(2)
Using (2) with p = x/h together with ( 1 ) we have
hmr
x(m)
(f + 1)
=
(3)
r(;-m+l)
Similarly we can prove that if m is a negative integer the result holds. In general we shall take
for all values of m. See Problem 3.109.
the result as defining drn)
104
THE SUM CALCULUS
[CHAP. 3
3.36. Assuming that the result obtained in Problem 3.35 is the defining equation for xCm)
f o r all m, prove that A x ( m ) = mx(m-l)h,
We have
3.37. Show that
Ex(")=
X(m+l)
(m
+ 1)h
for all m # -1.
From Problem 3.36 A z ( m + l ) = (m -t l)xcm)h.
Then if m # -1,
3.38. Let the digamma function be given by
hr(i
Prove that A q ( x ) = l/(x
+ h).
+ 1)
Since A and D a r e commutative with respect to multiplication [see Problem 1.58, page 281,
A+(x) = AD In r ( x / h 1) = DA In r ( x / h 1)
+
+
CHAP. 31
THE SUM CALCULUS
3.39. Prove that
x+h =
From Problem 3.38 we have since
=
1
A-1-
+ 1)
r'(;
1
2 x(-l) =
105
h r ( i + 1)'
-- x(-l),
x+h
A-1~(--1)
= *(z) =
x+h
i.e.
BERNOULLI NUMBERS AND POLYNOMIALS
3.40. Prove that if fn(x) is a polynomial of degree n having the property that
f n - l ( x ) , then fn(x) must have the form
coxn
c1xn-1
fn(x)
= ' * *
Cn-1X
Cn
n!
(n-l)!
where CO, CI, . . .,cn are constants independent of n.
+-
Suppose that
fn(x) = Ao(n)xn
+
+ Al(n)x*-l +
+
Dfn(Z)
+
+ An-l(n)x + A,(n)
* *
Then by using D f n ( x ) = fn-l(x) we have
nAo(n)xn-l
+ (n- l)A,(n)xn-2 +
* * *
= Ao(n - 1)xn-1
+ A,-,(n)
+ A,(% - 1)xn-2 + - + An-l(n - 1)
* *
Since this must be an identity, it follows that
%Ao(%)= Ao(n - l), (n - l)Al(n) = Al(n- l),
From the first of these we have
A&)
=
. . .,
An-l(n) = An-l(n- 1)
A,(% - 1)
n
Then replacing n by n - 1 successively we find
A,(%) =
A,(% - 1)
n
- A,@-2)
n(n - 1)
-
=
..I
AOP)
n(n - 1). *
CO
- e
l
n!
where co = Ao(0) is a constant independent of n.
Similarly from the second we have
AlW
=
A l b - 1) n-1
A1h-2)
(n - l)(n- 2)
-
...
=
A l(0)
C1
(n - l)(n - 2). * *1
(n - 1)!
where c1 = A,(O) is a constant independent of n.
By proceeding in this manner the required result is obtained.
3.41. Find p,(x) for n = 0, 1,2,3.
By definition [see (SO), page 86, with n = 4 ] Ap4(x) = x3/3!.
A - 1 - 23 = -4-1x3
1
3!
3!
Thus with h = 1
=
106
THE SUM CALCULUS
3.42. Obtain the Bernoulli polynomials B,(x) f o r
92
[CHAP. 3
= 0,1,2,3.
Since by definition B,(x) = n ! B,(x), we have using Problem 3.41
B ~ ( x=
) O ! B ~ ( x=
) 1
B ~ ( x=
) l! B , ( x ) = z - 3
B ~ ( x=
) 2! B ~ ( x=
)
22
B ~ ( x=
) 3! B,(x) =
~3
-x +6
- 4x2 + Q X
3.43. Obtain the Bernoulli numbers B, f o r n = 0, 1 , 2 , 3 .
Since by definition B , = B,(O), we obtain from Problem 3.42
Bo = 1,
B1 =
-4 ,
B2 =
4,
B3 = 0
Prove that
where 5'; are the Stirling numbers of the second kind.
Use this result to solve Problems 3.41, 3.42 and 3.43.
From definition (SO), page 86, we have
P,(x)
Replacing n by n
=
A - 1 - %n-l
(n - 1) !
+ 1 and using (31),page 7, with
--
- ~ -1
1$n-l
(n - 1)!
h = 1 we have
Putting n = 3 in the result of (a)we have
L
-
1x3
-
A52
+ &X
in agreement with Problem 3.41. From these we can determine P 2 ( x ) ,
that problem.
PI(%) and Po(%) as in
We can from these also determine the Bernoulli polynomials and Bernoulli numbers a s
in Problems 3.42 and 3.43.
CHAP. 31
3.45.
THE SUM CALCULUS
107
(a) If
prove that for n = 2,3, . . .
( 6 ) If B n are the Bernoulli numbers prove that they must satisfy the recurrence
formula (1 B)" - Bn= 0 where, after formally expanding by the binomial
theorem, Bk is replaced by B k .
+
Use ( b ) to obtain the first few Bernoulli numbers.
We have for n = 1 , 2 , 3 , . . .
APn(4
b&"
blAXn--l
= n! + -
+
* * *
+ bfl-lAs
Then putting x = 0 in ( I ) and using the fact that
AX" Ix=o = [ ( x
we find
APn(0)
=
b0
7
+
+
- zn] I x = o =
bl
+
+
n=2,3,
n=l
...
1
bfl-i
Also since
we have on putting x = 0
APnW
=
0
1
and the required result follows from ( 2 ) and (3).
The Bernoulli numbers are given by
B , = B,(O) = n!B,(O) = n! b,
Thus
where O! = 1.
Then using ( 4 ) in the result of part (a) we have
Multiplying by n! ( 5 ) becomes since Bo = 1
which is the same a s expanding (1
Putting n = 2,3,4,
...
+ B)n- Bn = 0
formally and then replacing Bk by Bk.
in ( 5 ) we find
I + ( Y ) B 1 + ( ; ) B 2 = 0,
1+3Bl+3Bz = 0
or Bz =
Q
l + ( ~ ) B 1 + ( ~ ) B ~ + ( ~ ) B 3 = 0
1+
, 4Bl+6Bz+4B3=0
etc.
or
B,=O
108
T H E SUM CALCULUS
[CHAP. 3
3.46. Prove that for r = 1,2,3, . . .
+
- Brtl
r+l
where the right side is to be expanded formally using the binomial theorem and then
replacing Bk by B k .
1'
+ 2' + 3' + . . . +
(n-1)'
From equation (SO), page 86, we have on putting n = r
Then using n = r
(n
=
B)'+l
+1
+ 1 in equation ( 3 4 , page 86, and property 1, page 87,
= r!PTtl(x) =
A-lx'
~
(x+B)++'
r+l
B,+,(s) r+l
Thus by the fundamental theorem of sum calculus,
n-1
B
'x
=
x=l
(x + B )
r
+
;+'I:
(n
+ B)'+'
- BVfl
r + l
=
+ + + . . + n2.
3.47. Use Problem 3.46 to evaluate l 2 22 32
*
Putting r = 2 in the result of Problem 3.46 we have
12+22+32+
+(n-1)2
=
+ B)3 - B3
3
Q ( n 3 + 3n2B1+ 3nB2 + B3 - B3)
$(n3f 3n2B1 + h B 2 )
=
in3 - + n 2 +
=
Qn(n- 1)(2n- 1)
=
=
Then replacing n by n
+ 1 we have
12 + 22 + 32 +
*
*
+ n2
(n
ln
6
+
+
= Qn(n 1 ) ( 2 n 1)
Note t h a t this agrees with the result of Problem 3.21(a).
E U L E R NUMBERS AND POLYNOMIALS
3.48. Show that the Euler polynomials are given by
where the constants eo, . . .,en are given for n = 1,2,3,
...
by
and eo = 1.
From Problem 3.40 and definition (38) on page 88 i t follows t h a t
Then using definition ( 3 9 ) on page 88 together with ( 1 ) i t is seen t h a t we must have
Xn
n!
where
e0
-M(xn)
n!
+
el
-M(xn-1)
(n - 1)!
+
+ endlM(x) + en
M is the averaging operator with h = 1 which is a linear operator.
T H E SUM CALCULUS
CHAP. 31
109
If n = 0, ( 2 ) becomes
M E o ( x ) = 1 = eo
while ( 1 ) yields f o r n = 0
EOb) = eo
We thus conclude from (3) and ( 4 ) t h a t
E ~ ( x=
) 1
If n = 1 , 2 , 3 , . . . we have by definition of M
M(xn) =
*[(X
+ 11%+ Xn]
so t h a t on putting x = 0
M(sn) Ix=o
=
4
Thus on putting x = 0 in (2) we obtain for n = 1,2,3, . .
as required.
3.49. (a) Find
eo,el,e2,e3
and ( b ) obtain the first four Euler polynomials.
(a) W e already know from Problem 3.48 t h a t eo = 1. Then putting n = 1,2,3 successively in the
result ( 6 ) of Problem 3.48 we have
From this we find
el =
-*,
e2 = 0,
e3 = 1
24
( b ) Using p a r t ( a ) we obtain from (1) of Problem 3.48
x2
2!
x
2
E ~ ( x=
) - - -,
X3
X2
1
E ~ ( x=
) -- 3! 2.2!+24
For a table of Euler polynomials see page 237.
3.50.
Obtain the first four Euler numbers.
By definition the Euler numbers
given by
5 , = 2nn!En(4)
Then from t h e results of Problem 3.49(b) we have
Eo = 1,
El =‘O,
Ez = -1,
E3 = 0
For a table of Euler numbers see page 236.
MISCELLANEOUS PROBLEMS
3.51. Prove that the generating function for the Bernoulli polynomials is given by
If we let G ( x , t ) denote the generating function, then
using the fact t h a t &(x)
= n ! P,(x).
110
THE SUM CALCULUS
[CHAP. 3
Now by definition
Xn-l
=
APn(Z)
~
(n - 1) !
Then multiplying by tn and summing over n we have
Using ( I ) , we can write ( 2 ) as
AG(x, t ) = text
G ( x ,t ) = A-'(teZt)
or
so that
G ( x ,t )
Taking the limit of (3) as t
+
=
text
et - 1 +
(3)'
0 we find since
lim G ( x , t ) = 1,
t*O
that c = 0.
o
lim t - r o et - 1
Thus we have as required
This generating function can be used t o define the Bernoulli polynomials from which all other
properties are obtained. See for example Problem 3.53.
3.52.
Prove that the generating function for the Bernoulli numbers is given by
t
=
Bntn
et - 1
n=O n !
%--.
This follows a t once from Problem 3.51 on putting x = 0 and noting that Bn = Bn(0). The
generating function can be used t o obtain all other properties of the Bernoulli numbers.
3.53.
Use the generating function of Problem 3.51 t o define the Bernoulli polynomials.
(a) Find the polynomial Bn(x) for n = 0, 1,2,3. ( b ) Prove that BL(x) = nBn-i(x).
(a) We have
Then multiplying both sides by
we have
Dividing both sides by t and using the expansion
ext = 1
this becomes
xt + X*tZ
X3t3 +
++-
l!
2!
3!
THE SUM CALCULUS
CHAP. 31
1
+ -xt +
l!
X2t2
~
+ -+
X3t3
2!
.**
3!
111
= Bob) + [ s + y ] t
Then equating coefficients of corresponding powers of t we have
From these we find
Bo(z) = 1,
B,(z) = x
- 4,
B,(x) =
x2
- x + 4,
B3(X) =
x3
Ib) Differentiating both sides of
with respect t o x we have
-t2e"tet - 1
-
5 BA(x)tn
n=O
But this is the same as
Bn(x)tn
t n=O x
T
=
n!
Bh(x)
i--
tn
n=O
n!
i.e.
or
since BA(x) = 0.
Equating coefficients of tn on both sides we thus find
-Bn-l(x)
-
(n - 1)!
so that
3.54.
Find
2
-
__
BA(4
n!
BA(x) = nBn-l(x)
!%?E
dx.
X
We have by Leibnitz's rule on page 81
cos xa dx
-
4 sin a4
2 sina3
+ 3 sina4 - ~
- 3 sin a3
a
- 8 x 2 + 42, * .
.
112
T H E SUM CALCULUS
[CHAP. 3
Supplementary Problems
THE INTEGRAL OPERATOR AND RULES OF INTEGRATION
3.55.
Find each of the following:
( a ) 0 - 1 ( 3 ~ 4- 2x3
+ x - 1)
(c) 0-1(4 cos 3x
( b ) D - 1 ( 2 6 - - 3%)
( d ) 0-1(4e25
-2
sin 3%)
+ 3e-4")
3.56.
Prove ( a ) formula 1-1 and ( b ) formula 1-2 on page 79.
3.57.
Find
(a)
3.58.
Find
(a)
3.59.
Find
s
s
~ e - dx,
~ z (b)
$2
sin2x d x ,
s
(b)
(x sin x
+ cos x) dx,
(c)
(f)
s
D-W)
x In x dx, (d)
x2
ln2x d x .
f x3e2"dx.
c
THE SUM OPERATOR AND RULES OF SUMMATION
sin r ( x - $h)
3.60.
Prove t h a t A - ' [ c o s r x ] = 2 C O S ~=
2 sin + r h '
3.61.
Use Problem 3.60 to obtain
3.62.
Find
3.63.
( a ) Find A-1[x2ax] =
3.64.
Prove t h a t
3.65.
Find
A-l[xez] =
(a)
S
cos rx
dx.
z xe".
z xzax,
a # 1,
and
( b ) check your answer.
c o s m ) , ( b ) A-'(x sinrx).
DEFINITE SUMS AND THE FUNDAMENTAL THEOREM OF SUM CALCULUS
Evaluate each of the following using the fundamental theorem of sum calculus and verify directly.
3.66.
8
(a)
2
2d4)
2
71
if h = 2
(c)
[3x(2)- 2 x ( 3 ) ] if h = 1
1
if h = r/3
0
10
4
(b)
2 cos 4x
n
3.67.
Find
2
~ ( - 3 ) if
1
3.68.
Find
2x0
(d)
22 6
~ ( - if
~ )h = 4
h = 1.
n
2 if~ h = 1.
1
a
3.69.
Show t h a t
+ (n-llh
z
$2
=
na2
+ +n(n - l)h(2a + h) + Qn(n- l)(n - 2)hz
x=a
DIFFERENTIATION AND INTEGRATION OF SUMS
Verify t h a t Theorem 3-3, page 84, is t r u e for
3.70.
CHAP. 31
T H E SUM CALCULUS
113
3.71.
Use Theorem 3-4, page 84, to obtain IV-7 from IV-6, page 83.
3.72.
Find
3.73.
Prove (a) Theorem 3-3, (b) Theorem 3-4, (c) Theorem 3-5, on page 84.
(a)
xx
(b)
COSTX
2 x sinrx
SUMMATION OF SERIES
3.74.
Sum the series (a) 1 2
+ +3+ -
3.75.
+ + 7 + 10 +
to n terms, (b) 1 4
* *
(b)
+ 2.3 + 3.4 +
1 . 2 0 3 + 2 . 3 0 4 + 304.5 +
(c) 1 . 3
.**
*
a
(d) 2.5
*
Sum each of the following series to n terms.
1
1
1
1
(a)
* * *
2.3
3.4
(')
1.4.7
1
1
1
(d) i
-&
(b) E + E + f i +
=+-+-+
..'
3.77.
* *
to n terms.
Sum each of the following series to n terms.
(a) 1 . 2
3.76.
by using Theorem 3-4, page 84.
(a) Show that
xn zk-k = 2 - n + 2 and thus
2n
k=l
+ 3.5 + 5.7 +
3- 5 . 8 + 8.11 +
+-4 * 71* 1 0 7 * 110 * 1 3 + ...
1
1
+2 - 3 . 4 + -+
3.405
+
*"
( b ) show that
-1 + - +
2 - +3- + .4. .
2
22
23
+ 2-22 + 3
=
24
+ 4.24 +
3.78.
Find the sum of the series 1* 2 l
3.79.
Show that
sin e
3.80.
Show that
13
3.81.
Find the sum to infinity of each of the series in Problem 3.76.
3.82.
Sum the series 1 22
3.83.
Sum the series
3.84.
Sum the series
3.85.
(a) Sum the series
29
+ sin 2e + sin 36 + - + sin n6
*
+ 23 + 33 +
+2
32
+ n3
*
+3
42
+
*
*
* * a
+
n*2n.
+
cos (e/2) - cos ( n #e
2 sin (e/2)
=
= (1 + 2 + 3 +
2
- - - + n)2
=
nz(n
t o n terms.
n
x k2
k=l
12
Zk.
+ 32 + 52 + - *
1-2.3
and ( b ) find the limit a s n +=
3.86.
to n terms.
+-2 . 340 4 + -3+. 47- 0+5. ' " 4 0105 . 6
m.
Sum each of the following series to n terms and to infinitely many terms.
1
1
1
1
+ - + 1- + * . . 1
.'.
(a) - 2.4.6 3 . 5 . 7
5.709
2-4
3'5
5'7
+
+
+
SUMMATION USING SUBSCRIPT NOTATION
Use the subscript notation to sum the series of Problem 3.75.
3.87.
3.88.
to n terms
Use the subscript notation to sum the series of Problem 3.76.
+ 1)2
4
114
THE SUM CALCULUS
n
3.89.
Sum the series
k*(k
k=l
3.90.
3.91.
Sum the series
2k-1
( k 2)(k
+
W
+ 4)
(b)
2k - 1
(k
k?1
+ l ) ( k + 2)(k + 3)
Sum the following series to n terms.
+ 3 - 5 - 7 . 9 + 5 * 7 * 9 * 1 1+ ...
1
1
+ ...
3 * 5 * 7 * 9+ 5 - 7 * 9 . 1 1
1
1.3.5.7
+
Show t h a t
(a)
kak
=
(n+l)an+l
a-1
k=O
3.93.
+ 1).
kZ1
(a) 1 . 3 - 5 - 7
3.92.
[CHAP. 3
-
an+2
-a
a Z 1
( a - 1)2
Sum the series
1'3.5
+ -3+. 56-0 7
5'7'9
+ *
( a ) to n terms.
( b ) to infinity.
ABEL'S TRANSFORMATION
Use Abel's transformation to sum the series
3.94.
kzl5.
"
k
n
3.95.
Sum the series
k2 by using Abel's transformation.
k=l
n
3.96.
Sum the series
2
k=l
k2ak'
a f 1, by using Abel's transformation.
OPERATOR METHODS OF SUMMATION
Work Problem 3.92 by using the result of Problem 3.30.
3.97.
3.98.
3.99.
Show t h a t
A-l[rk
sin ak ] =
rk+l
A-l[rk
cos ak] =
rk+l
+
cos a ( k - 1)- y k cos ak
r2 - 2r cos a
1
+
Problem 3.98 to show t h a t if jr/ < 1
5
rk
sinak
=
k=l
?'+
2
3.100.
sin a( k - 1) - r k sin ak
r2 - 2r cosa
1
5 rkcosak
k=l
r2
=
r sin a
- 2r cos a
+1
1 - r2
2(r2 - 2r cos a
+ 1)
Show t h a t if p # 1 is a constant and h # 1,
thus generalizing the result of Problem 3.30.
3.101.
Use Problem 3.100 to find A - l [ x a z ] .
Compare with Problem 3.16.
*
CHAP. 31
T H E SUM CALCULUS
115
THE GAMMA FUNCTION
(a) I’(4), jb) r(5/2), (c) r(-5/2).
3.102.
Find
3.103.
Evaluate
3.104.
Evaluate
im
3.105.
Evaluate
A’
3.106.
Prove t h a t
3.107.
Prove t h a t
3.108.
Use Problem 3.107 to express in terms of gamma functions
xe-x3 dx.
x2 ln2 x dx.
1
1
(X+1)2
+
1
+
*
I
’
(x+ n)2
-
+
and generalize.
3.109.
Show t h a t equation (3) of Problem 3.35 holds for all values of
In.
BERNOULLI NUMBERS AND POLYNOMIALS
3.110. ( a ) Find & ( x ) for n = 4 and 5 and thus obtain ( b ) the Bernoulli polynomials and (c) the Ber’
noulli numbers f o r n = 4 and 5.
3.111.
Obtain the Bernoulli numbers B4, B6,B B without first finding the Bernoulli polynomials.
3.112.
Prove t h a t B i ( x ) = nB,-,(x)
3.113.
Prove t h a t B , = B,(O) = B,(l).
3.114.
Evaluate
3.115.
Expand
3.117.
Prove t h a t
3.118.
Prove t h a t 132,+1(+) = 0.
without using the generating function.
+ 23 + 33 + - + ( b ) 1 4 + 24 + 34 + +
5x4 - 8x3 + 3x2 - 4x + 5 in a series of Bernoulli polynomials.
( a ) 13
ilBn(X)dX
* *
=
{
1
0
123,
* * *
n4.
n=O
n>O’
EULER NUMBERS AND POLYNOMIALS
3.119. Find ( a ) e4, e,; ( b ) E 4 ( x ) ,E&); (c) E,, E,.
4.120.
Prove t h a t
(a)E,(O) = en, ( b ) E,(O)
3.121.
Show t h a t
17
(a)e7 = 40,320 ’ ( b )
+ E,(1) = 0,
(c) E,(1) = -en*
-31
= 362,880 *
MISCELLANEOUS PROBLEMS
3.122. Determine whether the operators (a)D and D-1, ( b ) A and
A-1
a r e commutative.
116
3.123.
THE SUM CALCULUS
Prove that
(a)
A - l sin(pz
(b)
A-1
+ q)
=
cos(px+ q) =
[CHAP. 3
+
sin (PX q - +ph - 4%)
2 sin Bph
cos (PX
+
4 - Jph
2 sin -$ph
- -&T)
3.125.
Use Problem 3.124 to find A-l[x3 2x] if h = 1.
3.126.
Find
3.127.
Show that
3.128.
Show that if sin (Y # 0
n
2
k=l
k3 cos2k.
21. 23
~
sin2a
$
22'4
3.4
~
W*ak =
Show that
3.130.
Prove that
3.131.
Use Problem 3.130 t o show that
3.132.
Evaluate
3.133.
Show that
3.134.
3.135.
15
I
...
n2.4n-1
(n+l)(n+2)
+
+ sin22a + -.. + sinzna
3.129.
k=l
32'42
4,
+ 25 + 35 + - - +
*
(aW
(b)
Use (a) to deduce that
(a)
Show that
(b)
Use (a) to
cos (n
2 sin (I
+1)(~
(a - 1)s
725.
sec2x
tan tan
tanh
Show that A-l[
n - sin na
= 2
(?~-1)4"
s + x-n-
- 2an2 - 2an + n2 + 2n + a + l ) a n + l - (a2 + a )
1
1
1 +
1
+++sin B
sm2e
sin 4e
sin88
(a)
1
=
S
1
= tan
9
t o n terms =
*
+ c.
sec2x dx = tan x
+ c.
= tan-' s + c.
cot (e/2)
- cot 2n-le
THE SUM CALCULUS
CHAP. 31
3.136.
117
( a ) Show that
5
k=l tan-1 (kZ
+ ‘k + 1)
3.137.
Are there any theorems for integrals corresponding t o the Theorems 3-3, 3-4 and 3-5 on page 84?
Explain.
3.138.
( a ) Show that formally
=
A-1
= -- 1
1-E
A
- -(1+ E + E 2 +
* *
*)
( b ) Use ( a ) t o obtain the formal result
f(x) = f(u)
u=x
where the sum on the right is taken for u = x, x + h , x + 2 h ,
A-l
(0)
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 ( 6 ) Theorem 3-4, page 84.
i(al
b(a)
[Hint.
3.141.
... .
Let
G(a) =
Prove that I?’@) =
F ( x ,a ) d x
Ja
and consider G(a + A @ ) - G(a)
ha
e--5 In x d x = -y
where y is Euler’s constant [see page 851.
0
3.142.
Use Problem 3.141 t o prove that
1
I+,+,+
1
. . . + -n1
+
+
r’(n 1)
= ____
r ( n 1) + Y
3.143.
Supply a proof of the method of generalized integration by parts [see Problem 3.41.
3.144.
The beta function is defined as
( a ) Show that B(m, n) = 2
( 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
w
[Hint.
Use Problem 3.106 t o show that r ( m )r(n) = 4
0
1
w
x 2 ~ - l y 2 n - 1 e - ~ x 2 t Y 2d1x dy.
0
transform to polar coordinates and use the result of Problem 3.144.1
3.146.
Evaluate
(a)
i’
*dyl
( b ) Jn/2sin4e
0
cos2e
d s , (c)
&7’zG
de.
Then
118
3.148.
THE SUM CALCULUS
Show that
c
Hint.
= (-l)nB(m,n)
Am-l(i)
In. =
Use the fact that
if h = 1.
1%
e-nsdx.]
'
3.149.
Prove property 6, page 87, i.e. B,(x) = (-l)nBn(l - x).
3.150.
Prove that B3 = B5 = B7 =
are equal t o zero.
3.151.
Show that
3.152.
Show that
5
= 0, i.e. all Bernoulli numbers with odd subscript greater than 1
ex12
X
-coth2
2
2(
= g
- -1- - x 1
--x3-1
x
(b)
tans =
ex/2
24
+ -B4x3
4!
22
(a) cotx = - - -B2x
x
2!
3
)
5 - -
= 1
1
4725
15
+
3.154.
...
...
...
replaced by ix. For ( b ) use the identity tan x = cot x
E
i
cscx =
Use the identity csc x = cot x
+ tan (x/2).]
-
- 2)B2k x2k-1
(-1)k-l(22k
k=O
[Hint.
4!
315
[Hint. For (a) use Problem 3.151 with
2 cot ex.]
Show that
2!
- ...
$7
4!
+ -x3
1 + -2x 5 + -x7
17
+ -B-~+x-' + B4~4
+ ...
26
- -B6x5
6!
2 x
945
45
3
3.153.
+ e-x/2
- e-x/2
22(22 - 1)
2!
B ~ x- 24(24- 1) BqX3 +
= x
[CHAP. 3
(2k) !
Let f(x) be periodic with period equal t o 1 and suppose that f(x) and f'(x) are bounded and have a
finite number of discontinuities. Then f(x) has a Fourier series expansion of the form
f(X) =
ao2 +
(Uk
COS2knx
k=l
+
bk
Sin2kn-X)
1
where
(a) If f(x)
ak
= 2
f(X) COS2knX dX,
bk
= 2 J1f(E) S i n 2 k T ~dX
= PZn(x) [see page 881 show that
a,, = 0,
ak
=
2(-1)n-1
(2kn)Zn
'
bk
= 0
k=1,2,
...
so that
(b) Use the result in ( a ) to obtain the second result of property 12, page 88.
3.155.
Prove the first result of property 12, page 88. [Hint.
property 12 already obtained in Problem 3.154(b).]
3.156.
Prove property 11, page 88.
Take the derivative of the second result of
T H E SUM CALCULUS
CHAP. 31
3.157.
1
119
1 1
772
12+$+s+
= 6
Show t h a t
'.*
1
1
1
774
( b ) -+-+t34+
*.. = 14
24
90
3.158.
3.159.
I + -1- - +1. . .
(a) -- -
Show t h a t
32
42
1
( b ) -1- - +1- - - +
14
24 34
1
44
12
22
1 33
1
(a) 13 - -
Show t h a t
(6)
3.161.
Show t h a t
3.162.
(a)
73
- - 32
* * *
1- -1+ 1 - -1+
15
35 55 7 5
Prove ( a ) property 4 and
12
7774
= 720
...
1
1
+ ---+
53 73
3.160.
= 772
...
5775
- -
-
1536
property 6 on page 88 by using the generating function.
B,(&) =
Show t h a t the Bernoulli numbers can be found from
(-l)kk!
B, =
k + l
k=l
where S; a r e Stirling numbers of the second kind.
2
"
Use the result in ( a ) to obtain the first few Bernoulli numbers.
3.163.
Prove t h a t
x2
x4
+ -B4
4!
-
x6
- -- -(a) In s i n x = I n x - 6
180
2835
772
3.164.
Prove t h a t
1 - -B2
2!
3.165.
Prove t h a t
B,
3.166.
Prove that
774
cc
- 1ex 1
+
... - (-1)k-'B2k(2%)2k
2k(2k)!
-
...
O<x<n
= 0.
*.*
32,-1
= 4 n l m d x .
- _1 2
(-
1
(22- 1)Bzx
2!
1
-
-).I
(24 - 1 1 ~ ~ x 3
4!
... -
(22k - 1 ) B 2 k $ 2 k - 1
(2k) !
-
...
1
Hint. e2x - 1 = 2 e x - 1 - e x + 1
3.167.
Prove t h a t the only zeros of the polynomial B,,(x) - B2, in the interval 0 S x 5 1 a r e x = 0
and x = 1.
.
3.168.
(a) Obtain the generating function 8, page 89, for the Euler polynomials and ( b ) use i t to find the
first four polynomials.
3.169.
(a) Obtain the generating function f o r the Euler numbers and ( b ) use i t to find the first four
numbers.
3.170.
Prove properties
3.171.
Prove formulas
(a) 3 ( b ) 6 ( c ) 7
(a) 9
on page 88.
(b) 10 on page 89.
120
THE SUM CALCULUS
3.172.
Derive the results 12 on page 89.
3.173.
Obtain property 11 on page 89.
3.174.
(a) Show that
[CHAP. 3
where S t are Stirling numbers of the second kind.
( b ) Use ( a ) to obtain a relationship between the Euler and Stirling numbers and from this find the
first four Euler numbers.
3.175.
-1
ex
1
+
( a ) Prove that
- i{Eo(r3)+ E,(O)x + E,(O)xZ
+
a
'
*}
( b ) Use part ( a ) and Problem 3.166 to show that
EO(O) = 1,
3.176.
E2k(0)
= 0,
EZk-l(O)
=
- 2 ( 2 * k - l)&
(2k.i
k = 1,2,.
.
(a) Show that the first few t a n g e n t n u m b e r s [see page 891 are given by
TI = I, T2 = 0, T, = 2, T, = 0, T, = 16, T, = 0, T, = 272, T, = 0, T B= 7936
( b ) Use ( a ) to show that
tanx
= x
+ 2 . g + 16.-5 ! +
53
x5
X7
272.5
+ 7936.91x9 +
9 . .
Y
Applications
the Sum Calc
SOME SPECIAL METHODS FOR EXACT SUMMATION OF SERIES
In Chapter 3 we saw how the fundamental theorem of the sum calculus could be
used to find the exact sum of certain series. We shall consider, in the following, various
special methods which can be used in the exact summation of series.
SERIES OF CONSTANTS
Suppose the series to be summed is
In particular the series can be summed exactly if uk is any polynomial in k. See Problems
4.1 and 4.2.
POWER SERIES
A power series in x is defined as a series having the form
Suppose that in this series we can write
in k and where z)k is such that
= U k v k where Uk can be written as a polynomial
ak
v(X)
=
2
(4)
z)kXk
k=O
i.e. the series on the right of (4) can be summedexactly t o V ( X ) . Then we can show that
[see Problem 4.31
3
=
akXk
V(XE)Uo =
V(X
k=O
+ XA)UO
x*V”(x)
(5)
2 ! A2Uo 3- * * *
where primes denote derivatives of V ( X ) . Since Uk is a polynomial, the series on the
right of ( 5 ) terminates and yields the exact sum of the series on the left.
The following are some special cases.
=
V(X)Uo
+ --j-j-
c xk =
sc
I.
z)k
= 1.
In this case V ( X )=
2
k=O
akXk
=
5
k=O
k=O
UkXk
-
1 - X
uo
1-x
+ x4uo
(1-x)~
This is sometimes called Montmort’s formula.
121
and we find that
+
~
X2A2uo +
(1-~)3
..
122
2.
APPLICATIONS O F THE S U M CALCULUS
Vk
= U k ! . In this case V ( X )=
*
[CHAP. 4
Xk
- = ex and we find that
k=O k !
Note that we take O! = 1 by definition.
For other special cases see Problems 4.37 and 4.47.
The power series method can also be used to sum various series of constants by letting
x be equal to particular constants. See for example Problems 4.6 and 4.8.
APPROXIMATE INTEGRATION
Finite difference methods are valuable in obtaining formulas from which we can compute the aproximate value of a definite integral. One important method uses the GregoryNewton formula [page 91 and yields the result [Problem 4.91
+ llOn - 90) A”@)
a3f(a)+ n(6n3 - 45n2
120
+
9
.
.]
(8)
The following are some important special cases.
1. Trapezoidal rule.
I‘
f ( x )d x
=
h
2 [f(a) + f(a + h)1
(9)
This is obtained by letting n = 1 and neglecting differences of order 2 and higher
[see Problem 4.101.
By using (9) repeatedly we can arrive at the extended trapezoidal rule
which holds for any positive integer N . If we let
(10)can be written as
76k
+
= f(a kh) denote the ordinates,
2. Simpson’s one-third rule.
This is obtained by letting n = 2 and neglecting differences of order 4 and higher.
It is of interest that third order differences do not enter in this case. See Problem
4.13.
By using (12) repeatedly we can arrive a t the extended Simpson’s one-third rule
which holds for any even positive integer N . In ordinate notation ( 1 3 ) can be written as
h
f ( X ) d% = 3 [yo 4Yi 232 4Ys -k * * * f 4Yiv-1 -k Y N ]
(14)
latNh +
+
+
CHAP. 41
APPLICATIONS OF THE SUM CALCULUS
123
3. Simpson’s three-eighths rule.
This is obtained by letting n = 3 and neglecting differences of order 4 and higher.
See Problem 4.16.
By using (15) repeatedly we can arrive a t an extended Simpson’s three-eighths rule.
4. Weddle’s rule,
+ 5 { f ( a+ h) + f(a + 5h)) + 6 f ( a+ 3h)l
which can also be written as
a+6h
3h
f(x) d x = 10[(YO + Y Z + Y4 + YS)
+ ~ ( Y+I95) f
(16)
(17)
6 ~ 3 1
This is obtained by letting n = 6 and replacing the coefficient 41/140 of asf(a) by
42/140 = 3/10 and neglecting differences of order 7 and higher. See Problem 4.38.
By using (16) or (17)repeatedly we can arrive a t an extended Weddle’s rule [see
Problem 4.911.
In practice to obtain an approximate value for the definite integral
(18)
we subdivide the interval from a to b into N subintervals where N is some appropriate
positive integer and then use one of the above rules.
ERROR TERMS IN APPROXIMATE INTEGRATION FORMULAS
If we let Ii denote the error term to be added to the right hand side of any of the
approximate integration formulas given above, then we can find 5Y in terms of h and the
derivatives of f ( x ) . The results are given in the following table.
Name of rule
Trapezoidal
I
I
Error term
h3
--f”([),
12
Simpson’s one-third
Simpson’s three-eighths
Weddle’s
[ between
[
a and a
+h
between a and a
+ 2h
3h5 f ( * V ) ( [ ) , 5 between a and a
-~
[h4ei(a)
140
+ 1400
E-f(8)([)],
I
+ 3h
E between a and a
+ 6h
For a specified value of h it is in general true that Simpson’s one-third rule and Weddle’s rule lead to greatest accuracy followed in order of decreasing accuracy by Simpson’s
three-eighths rule and trapezoidal rule.
124
APPLICATIONS OF THE SUM CALCULUS
[CHAP. 4
GREGORY’S FORMULA FOR APPROXIMATE INTEGRATION
We can express a definite integral in operator form as
(See Problem 4.20.) By using this we obtain Gregory’s formula f o r approximate inteyration given by
a+nh
h
f(z)dx = 2 [yo i- 2Y1+ 2y2 + * 2yn-1+ yn]
+
+
where the ordinates are given by zjk = f(a k h ) , k = 0, 1, . . . ,n, and where V = A E - ~is
the backward difference operator [see page 91. In practice the series is terminated a t or before terms involving the nth differences so that ordinates outside the range of integration
are not involved. See Problem 4.23.
THE EULER-MACLAURIN FORMULA
The Euler-Maclaurin formula, which is one of the most important formulas of the calculus of finite differences, provides a relationship connecting series and integrals. The
formula states that
n--1
1
f ( x )d x - 3 [ f ( a+ nh) - f(a)]
C f ( a + kh) =
k=O
+h [?(a nh) - ?(a)] - h3 [?”(a + nh) - ?”(a)]
12
wo
+
and in this form is useful in evaluation of series involving f(x) when the integral of f ( x )
can be found.
Conversely the formula provides an approximate method for evaluating definite integrals in the form
a+nh
h
f ( x )d x = 3 [yo 2y1+ 2y2 * * * + 2yn-1+ yn]
+
S,
I
+
h2 [y! - y;]
+h4 [y;’
12
- YY’]
720
= f(a+ kh), 2 ~ 6= f’(a), U! = f’(a +nh), etc.
We can also express the Euler-Maclaurin formula in terms of differences rather than
derivatives [see Problem 4.1011.
where we have used the notation
!/k
ERROR TERM IN EULER-RIIACLAURIN FORMULA
It is possible to express the Euler-Maclaurin formula in terms of Bernoulli numbers
and polynomials including an expression for the remainder or error term.
APPLICATIONS O F T H E SUM CALCULUS
CHAP. 41
125
To do this we first show that
+
+
+
Then replacing a by a h, a 2h, , . ., a (n - l ) h and adding, we obtain the Euler-Maclaurin formula with a remainder given by
lQ"'f(x) dx
=
+ 2 f ( a+ h) + . + 2 f ( a+ nh - h) + f ( a+ nh)]
h [f(a)
2
z1rn
m-l
-
Bzph2p
[f(Zp-l)(a
+ nh) - f ( Z ~ - l ) ( a )+
] R
(24
where the remainder R is given by
This can be used in connection with evaluation of series by writing it as
The result (26) is of course equivalent to (21) except that the error term R has been
introduced.
STIRLING'S FORMULA FOR n !
An important formula which can be obtained by using the Euler-Maclaurin formula
is that for n ! , called Stirling's formula and given by
When n is not an integer the result can also be used if the left side is replaced by the
gamma function r ( n 1) [see page 851. The series in (27) is an asymptotic series and is
useful for large values of n [actually there is good accuracy even if n = 10 as shown in
Problem 4.781.
+
1
APPLICATIONS O F THE SUM CALCULUS
126
[CHAP. 4
Solved Problems
SERIES OF CONSTANTS
4.1. Prove that
+ u1 + uz +
Uo
+
* . *
+
= nu0
Un-1
n(n- 1)
2!
AUo
4-
n(n- l)(n- 2)
3!
4-
A2Uo
".
We have from the definition of the operator E = 1 + A
UO
+
~1
+ +
~2
.**
+ un-l
=
+ Euo + E*uO +
UO
*.*
+ En-1~0
= ( 1 + E + E 2 + *.*+En-l)uo
- -En - 1
E-lUo
4.2.
Sum the series l2+ 22+ 32+ *
+ A)n - 1
-
(1
=
nuo
A
UO
- 2)
A*UO +
+ n(n - 1)Auo 4- n(n- l)(n
3!
''I
- . to n terms.
Comparing with the series of Problem 4.1 we have
uo = 12,
u1 = 22,
u2
= 32,
. . .,
u,-~ =
n2
Then we can set up the following difference table.
0
1
1
4
2
9
3
16
4
25
3
2
5
0
2
7
0
0
2
9
Fig. 4-1
It is clear from this table that
uo = 1,
Auo = 3,
A%,
= 2,
which we could also obtain by observing that
uk
k = 0.
,
A3u0 = 0,
= (k
+
A4u0
Auk = 2k
= 0,
+ 3,
...
'I
A2uk= 2 and putting
Thus using Problem 4.1 the required sum is
- n(n+ 1)(2n+ 1)
6
in agreement with Problem 3.21(a), page 98.
Note that the method of Problem 4.1 is especially useful when the kth term can be expressed
a s a polynomial in k so that differences are ultimately zero.
CHAP. 41
APPLICATIONS O F THE SUM CALCULUS
127
POWER SERIES
+ XA)Uo
= V(X
where we have used the Taylor series expansion
+
V ( x h)
with h = xA,
Note that although the above has been proved without any restriction on u k an important
case where the series can be summed exactly occurs when u k is a polynomial in k so that higher
differences are ultimately zero.
Put
wk
= 1 in the results of Problem 4.3. Then
m
1
+x+
22
+
23
+
* ’
. = - 1
1 -x,
Thus
and so from Problem 4.3 we have
m
4.5.
Sum the series
+
( k + l ) ( k 2)xk = 1.2
k=O
Comparing with Problem 4.4 we have
ing difference table.
uk
+ 2.3% + 3 *4x2+ - . .
*
+
= (k l)(k
+ 2).
Then we can construct the follow-
128
APPLICATIONS O F T H E SUM CALCULUS
I
[CHAP. 4
I
Fig. 4-2
From this table i t is clear t h a t
uo = 2, Auo = 4, A2uo = 2,
A3zt0
= 0, A4u0 = 0,
These could also have been obtained by letting k = 0 in the results
2k f 4, A 2 U k = 2,
..
..
z6k
...
= (k
+ l ) ( k+ 2),
Auk
=
Thus we have from Problem 4.4
m
3
( k t1 ) ( k + 2 ) x k
4x
=
2x2
k=O
-
2(1 - x ) f~ 4 ~ ( 1 -X)
(1- x)3
-
___ 2
(1- x)3
+ 2x2
It should be noted t h a t the series converges only if !xi< 1.
4.6.
Sum the series 1 . 2
-
2.3 3.4 4.5
5.6
- + 7- 53
i
54
5
5
. * * .
This series is a special case of the series of ProbIem 4.5 in which we let x = -1/5.
sum of t h e series i s
4.7.
Prove that
1
Put
2ik
= l / k ! in the results of Problem 4.3.
Then
Thus
V'(x) =
and so from Problem 4.3 we haxie
ex,
V"(x) = ex,
.. .
Then the
CHAP. 41
4.8.
APPLICATIONS O F T H E SUM CALCULUS
Sum the series
22
+
52
-
1!4
129
112 + *..
+ 2 82! 4 - + 3!43
The series is a special case with x = 114 of
11'
+ -l5 2x! + -2!822 9 + --3
3!
22
+ ..-
or
Thus we can obtain the sum by putting
difference table i n Fig. 4-3 we find
uo = 4,
uk
Auo = 21,
= (3k+2)2 in the result of Problem 4.7.
A2uo = 18,
A3u0 = 0,
From the
. ..
21
25
64
121
196
39
57
75
18
0
18
0
18
Fig. 4-3
Thus
and putting x = 114 the required sum is seen to be 157&16.
APPROXIMATE INTEGRATION
4.9. Prove that
n(2n- 3 )
+
A2f( a )
+
n(6n3- 45n2 110n - 90)A 4 f ( 4
n(n - 2)2
24
A3ff(a)+
120
By the Gregory-Newton formula (44) on page 9 we have
Using Ax = h and the definition of ( z f(z) = f(a)
+ ?(x-a)
this can be written
+A s f ( a ) ( x - a)(%- a - h)
2 ! hz
A 3 f ( a ) ( x- a ) ( x -a - h)(x - a - 2h)
. ..
+ -~
3!
+
h3
Integrating from x = a to a
+ nh
we then have
+ .
*
.]
130
APPLICATIONS O F THE SUM CALCULUS
[CHAP. 4
+-
+
~ / ~ ) ~ a t( x"- ah) ( x - a - h ) ( x - a - 2 h ) ( x - u - 3 h ) d x
n(n- 2)2
+ A,3(af)
+
~ ( 6 -~4 35 ~ 2 110%- 90)
A4f(a)
120
+
+
*
4.10. Prove the trapezoidal rule
Put n = 1 in the result of Problem 4.9.
two a r e zero we have
Then assuming that differences of order higher than
lath
h [ f ( a )+ ?pV(a)I
= h [ f ( a )+ & { f ( a+ h) - f(a)}I
f b )d x =
= 5h [fb)+ f ( a + h)]
4.11. Use Problem 4.10to obtain the extended trapezoidal r u l e
From the results of Problem 4.10 we have
It
= 5
[!(a
+ h) + f(a+ 2h)]
...............................................
atNh
h
f ( x ) d x = z [ f ( a+ ( N - 1)h) + f ( a +- N h ) ]
at(N-1)h
Then by addition
l a t N b f ( x )d x
or using ordinates
1
4.12. (a) F i n d l
?dk = f ( a
dx
h
2
+
+ + - + 2 f ( a+ ( N - 1)h) + f ( a + Nh)!
= - [ f ( a ) 2 f ( a h)
* *
+ kh),
by the extended trapezoidal rule where the interval of integration
is subdivided into 6 equal parts.
( b ) Compare the result. with the exact value.
(a) We have a = 0, a + N h = 1 so that since N = 6 , h = 1/6. The work of computation can be
arranged in the following table.
131
APPLICATIONS O F THE SUM CALCULUS
CHAP. 41
x
0
116
216
316
416
1
516
f ( x ) = 1/(1+ 6)1.00000000 0.97297297 0.90000000 0.80000000 0.69230769 0.59016393 0.50000000
Then applying the extended trapezoidal rule of Problem 4.11 we have
i*+
1
x2
3
11 + 2(0.97297297 + 0.90000000 4- 0.80000000 + 0.69230769 + 0.59016393)
2
=
+ 0.50000000]
= 0.78424077
( b ) The exact value is
i1
dx
i-j-~ = tan-1%
1
= tan-ll
10
= 4 = 0.785398163..
,
Thus the trapezoidal rule for this number of subdivisions yields a result which is accurate t o
two decimal places. The percent error is 0.147%.
4.13. Prove Simpson’s one-third rule
Then assuming that differences of order 4 and higher
P u t n = 2 in the result of Problem 4.9.
are zero we obtain
+ Af(a) + &A2f(a)]
+ { f ( a+ h) - f(a)> + + 2h) - 2 f b + h) + f(a)>l
3 [ f ( a )+ 4f(a + h) + f ( a + 2h)]
l‘+zh
f ( x )dx = 2h[f(a)
= 2h[f(a)
=
4.14. Use Problem 4.13 to obtain the extended Simpson’s one-third rule
where it is assumed that N is even.
From the results of Problem 4.13 we have
J‘+zhf(m)
J-==;;
dx =
4
+
[f(a) 4f(a
+
h-[ f ( a 2h)
f(x) dx = j
+ h) + f ( a + 2 4 1
+ 4f(a + 3h) + f ( a+ 4h)]
....................
h
3
[ f ( a+ (N- 2)h) + 4f(a + (N- 1)h) + f ( a + WI
.............................................
a
S
+ NIL
f(x) dx =
af(N-2)h
Then by addition
J-=,+,
f(x) dx =
Using ordinates yk = f ( a
+ 4f(a + h) + 2f(a + 2h) + 4f(a + 3h) +
[f(a)
+ kh)
* *
+f(a
this can be written
h
f(z)dz = - [yo
3
+ 4Y1+ 2y2 + 4u3 +
- .. + ~ Y N - I+ YNI
Nh)]
132
4.15.
APPLICATIONS OF THE SUM CALCULUS
[CHAP. 4
(a) Work Problem 4.12 by using the extended Simpson’s one-third rule and ( b ) compare your result with the exact value and that obtained from the trapezoidal rule.
( a ) We can apply the rule since N = 6 [which is a multiple of 31. It is convenient to write the
rule in the form
=
3
+ 0.50000000 + 4(0.97297297 + 0.80000000 + 0.59016393)
+ 2(0.90000000 + 0.69230769)l
[l.OOOOOOOO
= 0.78539794
where we have used the value h = 1/6 and the table of values obtained in Problem 4.12 which
is applicable here.
( b ) The result is seen to be accurate to 6 decimal places on comparison with the exact value
0.78539816.. , a s obtained in Problem 4.12(b). The percent error is 0.00028%. It is clear that
the result is f a r more accurate than that obtained using the trapezoidal rule.
4.16.
Prove Simpson’s three-eighths rule
[ f ( a )+ 3f(a + h) + 3 f ( a+ 2h) + f ( a+ 3h)]
Jat3hf(z)dz =
P u t n = 3 in Problem 4.9 and neglect differences of order 4 and higher.
at3h
f ( x )dx
Then
+
+ &Asf(a)]
3h [f(a)+ $Cf(a + h) - f ( a ) >+ f { f ( a+ 2h) - 2f(a + h) + f ( 4 1
+ & { f ( a+ 3h) - 3f(a + 2h) + 3f(a + h) - f(a)>]
3h
[f(a)+ 3f(a + h) + 3f(a + 2h) + f ( a + 3h)]
= 3h [ f ( u )+ %Af(a) fA2f(a)
=
=
atNh
We can obtain an extended Simpson’s three-eighths .rule for the evaluation of
N is a multiple of 3 [see Problem 4.571.
4.17.
f(x)dx when
(a) Work Problem 4.12 by the extended Simpson’s three-eighths rule and ( b ) compare with the result obtained from the one-third rule.
(a) Using Problem 4.1 and also the table of Problem 4.12 we have
=
+ 0.50000000)
+ 3(0.97297297 + 0.90000000 + 0.69230769 + 0.69016393) + 2(0.80000000)]
3O
[(l.OOOOOOOO
8
= 0.78539586
( b ) The percent error is 0.00293% which is about 1 0 times the error made using Simpson’s onethird rule.
ERROR TERMS IN APPROXIMATE INTEGRATION FORMULAS
4.18. Find the error term in the trapezoidal rule.
The Gregory-Newton formula with a remainder after 2 terms is
f(x)
where
R =
= f(a)
+ *(z-u)(l)
f”(v)(x - u p )
2!
7 between
+R
a and x
CHAP. 41
APPLICATIONS O F T H E SUM CALCULUS
133
Putting Ax = h aild integrating from x = a to x = a + h we have
a’h
f(x) dz
=
f ( a )dx
I’
where
a
5y
= 1
2!
a
+
+-
11
I”(?)(%--a
p dx
Performing the integrations in (1) we find
[-”
f(x)
=
dz
2h [ f ( a ) + f ( a + h)] + T
Now to estimate the remainder or error T( we note that (x - a ) ( z )= (x- a)(%- a - h) does not
change sign f o r x between a and a h. We can thus apply the mean-value theorem [see page 811
t o ( 2 ) to obtain
9 = $if’([) 4 7 j / T T 0 j 0 . 0 % T ( [ 0 8 . 1 p 4 ) T 5 T j T 1 . e r r o r f 0 T c 1 2 . 3 0 2 / T 4 7 j
+
/TT00 1
APPLICATIONS O F THE SUM CALCULUS
134
Now considering
[CHAP. 4
--2h2
f(IV’(7)
3
q”’(h) = G (h) =
we can show by integrating three times using conditions ( 9 ) that [see Problem 4.891
q(h) =
h
-
u2(h- w)2f(IV)(q)du
(5)
0
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
which is the required error term.
GREGORY’S FORMULA FOR APPROXIMATE INTEGRATION
4.20. Show that x a t n h f ( x )dx = h En- f@).
(&)
We have by definition
D-’f(z)
so that
=
f(x) dz
latnh
Datnh
1
But since E = e h D we formally have D = - 1 n E
h
4.21. Show that
atnh
so that
h
1nE
+
where gk = f ( a kh).
From Problem 4.20 we have, using
vk
= f(a-k kh)
SO
that yo = f(a),yn = f ( a inh),
4.22. Show that
1
In (1+ x)
X
We have [see page 81
In(1fx)
= z - 22
2
53 - 5
$5
+-4 + -
3
4
5
APPLICATIONS O F THE SUM CALCULUS
CHAP. 41
135
Then by ordinary long division we obtain
- x1+ 2 - -1" + - "12 - 12
1
24
19 3
720x
3
+ 160
-x4
-
*..
4.23. Obtain Gregory's formula (LO),page 124, f o r approximate integration.
From Problem 4.21 we have
Now
so t h a t
Then (1) can be written
Now using Problem 4.22 we find
By subtraction we then have
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 i t becomes
APPLICATIONS O F THE SUM CALCULUS
136
[CHAP. 4
The ordinates yo, yl, . . .,
can be taken from the table of Problem 4.12. To obtain the
various differences we form the following difference table where the factor 10' has been used so as
t o avoid decimals,
k
0
100000000
1
97297297
2
3
4
5
6
-2702703
-7297297
90000000
-10000000
80000000
-10214376
59016393
-2702703
-769231
-10769231
69230769
-4594594
+554855
1891891
1933472
1324086
41581
-609386
-680958
643128
$1197983
-9016393
50000000
Fig. 4-5
From this table we see that
VY0
V2y0
= A E - ~ Y=~Ayb = -0.09016393,
= A 2 E - 2 ~=~ A2y4 = 0.01197983,
V3y0 = A 3E-3g6
V4y6
= A3y3 =
0.00643128,
= A4E-4y0 = A4y2 = -0.00680958,
Ayo
A2yo
= -0.02702703
= -0.04594594
A3Uo =
0.01933472
=
0.00041581
A416,-,
Thus we have
s,'&
-
0.78424077
--(1'6)
24
-
(-0.0631369)
(-0.03396611)
= 0.78424077
= 0.78543028
-720
3(1'6) (-0.00639377)
19(1'
6) (-0.01290344) - 160
+ 0.00087690 + 0.00023588 + 0.00005675 + 0.00001998
I t should.be noted that the value 0.78424077 in the above is that obtained using the trapezoidal
rule, 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%. I t should be noted however that Simpson's one-third rule
is still the victor.
THE EULER-MACLAURIN FORMULA
CHAP. 41
APPLICATIONS O F THE SUM CALCULUS
137
We have since A = E - 1 = e h D - 1 [see page 81
Then using property 8, page 87, with t = h D this becomes
- 51 f ( x ) + 12
1 h D f ( 2 ) - 720
1 h3D3f(~+
)
= -f(x)
h1D
on using the values f o r Bo,B,, B,, B3, . . . .
4.26. Derive the Euler-Maclaurin formula (21) on page 124.
By the fundamental theorem of sum calculus we have [see page 831
a t (n-llh
atnh
f(x) = A - l f b ) I
where the summation in ( I ) is to be taken from a t o a
of Problem 4.25 we then have
a .t ( n - l ) h
afnh
(1)
+ ( n- l ) h in steps
atnh
-
of h.
Using the results
-h3f“‘(~)
+
...
which can be written a s
+ + +
f(a) f ( a h)
* * *
+ f ( a+ (n- 1 ) h )
=
a+nh
[
1
f ( 4 dx - 2 [f(a+ n h ) - f b ) ]
It
h3
+ 12
[ f ’ (+
~ nh) - f’(a)] - - [f’”(a + nh) - f”’(a)] +
720
199
4.27. Find
2
*
9
1
-
by using the Euler-Maclaurin formula.
k=100
Let
n
= 100.
f(x) = l/x, a = 100, a + (n- l ) h = 199, h = 1 in the result of Problem 4.26 so that
Then f ’ ( x ) = -1/x2, f”(z) = 2%-3, f”’(x) = -62-4
so t h a t
6
-&[-m+&]
+
* ’ *
+ i ( 0 . 0 0 5 ) + &(0.000075) +
0.693147 + 0.002500 + 0.00000625 +
= In2
=
= 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 +l,(x) = B,(x)- Bp, p = 0 , 1 , 2 , 3 , . . ., where B p ( z )are the Bernoulli polynomials
and Bp are the Bernoulli numbers. Show that
(4 9 ; w
=
?q-m
(c)
9,(0) = 0
138
APPLICATIONS OF THE SUM CALCULUS
[CHAP. 4
( a ) We have using property 3, page 87,
3
+b(4 = BL(4 = PB,-l(%)
( b ) We have since +,tl(x) =B,+l(x)-Bp+l
+
(P + W,(4 = (P+ 1)[$fJp(4 Bp]
+:,+A4
= B;+l(x) =
c:, + 1(x)
Then
= P + 1 - BP
c p ( 4
( c ) Since by definition BJO) = B,
we have
+,(O) = Bp(0)- B, = 0
+1(x) = B,(x) - B , = x - - (-*)
*
(d) We have
+1(1) = 1
Thus
If p
= z
> 1 then by property
4, page 87,
- B,
= B,(1)
+,(l)
= 0
Then from (1) and (2) we see that
4.29.
1
=
$fJp(l)
p = l
p>1
0
Show that
Letting x = a
+ hu,
using the fact that +z(u)= u2 - u, and integrating by parts we have
f(x) d x = h
i1+
f ( a hu) du
=
$il
+
=
;[(Zu -
f(a hu) d(2u - 1)
lo
1
+
l ) f ( a hu)
-
h
1
+ +
h
= [ f ( a ) f ( a h)] 2
+ +
= h
[ f ( a ) f ( a h)] 2
J
7
1
(2u - l ) f ’ ( a +hu) d u ]
+
(2u - l ) f ’ ( a hu) du
0
+
f’(a hu) d+,(u)
J1
0
4.30. Show that
(a)
(b)
$J1+z(u)f”(a+ hu)du
1’
=
Bzh2
--
2 ! [?(a
+ h) - f’(a)l
+
+$
+
+,(u)f(IV)(a hu) du =
+,(u)f(IV)(a hu)du
B4h4
T
[?”(a + h) - f”’(a)]
l1
+ h6!
‘
+6(U)f‘v”(a
+ hu)du
( a ) From Problem 4.28(b) with p = 2 we have
qjZ(u)= -B2
+ &&(u)
and
cdu) = &kW
CHAP. 41
APPLICATIONS O F THE SUM CALCULUS
139
Then
i1
+
B4h4
;y +5(u)f(v)(a+ hzc)d u
- --41 [f”’(u+ h) - f”’(a)] - -h
+;(u)f(v)(a hu) du
i1
-
4.31.
1
he
--B4h4 [f”’(a+h) - f”’(a)] - zs
&(u)f‘v’(a+ hu) du
4!
0
Obtain the Euler-Maclaurin series with a remainder f o r the case where n = 1
[see equation (a+$),page 1251.
From Problems 4.29 and 4.30 we have
+5
I’
+
+8(u)f‘””(a hzt) d u
which is t h e required result. This result can of course be generalized and leads to that on page 125.
140
APPLICATIONS O F T H E SUM CALCULUS
[CHAP. 4
4.32. Estimate the error made in the approximation to the sum in Problem 4.27.
Using f(x) = 1/x, a = 100, h = 1, n = 100, rn = 3 in the result of Problem 4.27 we can
write i t as
1 - R
6
where
]du
Now by the mean-value theorem for integrals we have
Thus the absolute value of the error i s
.
99
But since O < 5 < 1, i t follows t h a t each term in the series on the right is less than
so the sum of the first 100 terms of the series is less than l O O / ( l O O ) 7 so t h a t
l/(lOO)7
and
The fact t h a t the error i s so small shows t h a t not only is the result 0.695653 accurate to 6
decimal places but t h a t we could have obtained accuracy to 12 decimal places at least by simply
obtaining more decimal places i n Problem 4.27. If we do this we find
199
2 -
= 0.693147180559945
k=100
+ 0.0025 + 0.00000625 + 0.000000000078125
= 0.695653430638
accurate to all 12 decimal places.
By taking only one more term in the Euler-Maclaurin formula i t is possible to obtain even
more decimal place accuracy [see Problem 4.751.
This problem serves to illustrate further the importance of being able to find a n error term
since one then can determine how many figures in a calculation a r e accurate.
STIRLING'S FORMULA FOR n !
4.33. Prove that
c + ( n + i ) ~ n n- n
Inn!
1
1
+ 12n - 36Qn2
I
where c is a constant independent of n.
Let f(x) = l n x , h = 1 in the Euler-Maclaurin formula.
Y
Then
+
*..
APPLICATIONS O F THE SUM CALCULUS
CHAP. 41
141
From Problem 3.3(b), page 90,
latn
8
lnxdx
= xlnx - x
In a
Thus we have on adding
+
In ( a
+ n)
= ( a + n ) In ( a + n ) - ( a + n )
to both sides of equation ( I ) and using equation ( 2 )
Assuming that a and n are both positive integers and replacing a
this result we have
+ n by n, i.e. n by n - a, in
which can be written
Inn!
= c
+ (n++)lnn - n + 12n
1
+
1
where c is a constant independent of n.
4.34. Prove that in the result of Problem 4.33 the constant
c =&
ln2r.
We make use of Wallis' product [see Problem 4.1001 which states that
lim
12'-
2 2 4 4 6 6
1.3-3.5.5.7.
Now we can write
9
-
(2n) (2%)
n
- 2
(2%- 1) ' (2% 1)
* *
*
- . - ( 2 n ) (2%)
....( 2 n - 1) (2%+ 1)
2 2-4 4 6 6
1.3.3.5.5.7.
*
+
-
2 9 n !)4
+
(2% 1 ) [ ( 2 n ) ! ] z
It follows that
or
lim [ 4 n I n 2
n-+m
+ 4 I n n ! - In ( 2 n + 1)- 2 In ( 2 n ) ! ] =
In' 2
Using the result of Problem 4.33 to find In n ! and In (2%)! and substituting in ( 2 ) we have
4.35.
Prove Stirling's formula f o r n ! .
Using c = 4 In 2a in Problem 4.33 we have
Inn!
or
i.e.
= qln2n
! =
n!
(2)
1
+ (n++)Inn - n + 12n
e% In2rre(n t M ) In n e-ne1/12n
-,
1
'
+
142
4.36.
APPLICATIONS O F T H E SUM CALCULUS
[CHAP. 4
(a) Evaluate 20! by using Stirling's formula and ( b ) compare with the true value.
(a) Method 1.
From Problem 4.35, equation ( 2 ) ,
20! =
-6-q
(220 1020)(~-zo)(1.004167)
- (11.21000)(1.048576 X 1026)(2.061096 X 10-9)(1.004167)
- 2.43282 X 1018
Method 2.
From Problem 4.35, equation ( I ) ,
ln20!
41
1
+In20 - 20 + - - * .
2
240
0.91894 + -(2.99574)
41
- 20 + 0.0041667 2
0.91894 + 61.41267 - 20 + 0.0041667
-21 In27
=
=
=
*.*
= 42.33578
Then
log,, 20! = 0.434294 In 20! = 18.38618
and so
20! = 2.433 X 10'8
( b ) The exact result is 2,432,902,008,176,640,000 and thus the percent error of the result in Method 1
of p a r t (a) is less than 0.0033%.
MISCELLANEOUS PROBLEMS
4.37. Sum the series
22x2
x+- 1
Consider the series
uox
This can be written
uox
+-2
32x3
42x4
+
$3
u2x3
u3x4
+ u1x2
++
1 +2
3
~
...
.*.
U ~ XE~uOX~
~
+ E++ E~uOX~
-+
1
2
3
* . *
= [ I lx i
E +
x2E2
- +x3E3
-+...
3
2
Using series 4, page 8, this can be written as
[-In (1- xE)]uox =
{-In (1- x [ l f A])}uox = {-In (1 - x
= -u0x In (1- x)
x~AuO
+1- x
- xA)}uox
1 X ~ A ~ U O 1 x4A3Uo +
+
5(1--.)2
+
2-
...
CHAP. 41
APPLICATIONS OF THE SUM CALCULUS
u,
n=O
1
n = l
4
n=2
9
n=3
16
n=4
25
Au,
Azu,
143
A~u,
3
2
5
0
2
7
0
2
9
c
~~
Fig. 4-6
Thus
u,
= 1, Au, = 3,
= 2, A3u0 = 0,
A%,
A4u0 = 0,
.
Then the sum of the series is
-x In (1 - x)
3x2
++1-x
23
(1-X)Z
-
3x2 - 2x3
(1- x p
-
x ln(1-x)
4.38. Prove Weddle’s rule [equation (16),page 1231.
P u t n = 6 in Problem 4.9 and neglect differences of order 6 and higher. In doing this we
need the terms involving A5f(a) and Aef(a) which are not given in Problem 4.9 but are presented
in Problem 4.60. We then find
a+6h
.(:
[
f(x) dx = 6h f ( a )
+ 3Af(a) + 29A 2 f ( a )+ 4A3f(a)
41
11
41
+ ~0
A4f(a) + %A5f(a) + 840 A6f(a)]
6f(a)
+ 18Af(a) + 27Azf(a) + 24A3f(a)
Now if we change the coefficient 41/140 to 42/140 = 3/10 the error made is quite small.
make this change we obtain the approximation
r a t e h f ( ~ ) d r = hr6f(a)
L
=
+ 18Af(a) + 27AZf(a) + 2 4 ~ 3 f ( a )
+ + + f(a+ 4h) + f ( a+ 6h))
+ 5{f(a + h) + f ( a+ 5h)3 + 6f(a f3h)]
[ { f ( a ) f ( a 2h)
f(x) dx when N is a multiple of 6 [see Problem 4.911.
The result can be extended to evaluate
4.39.
If we
(a) Work Problem 4.12 by Weddle’s rule and ( b ) compare the result with the exact
value and that obtained by Simpson’s one-third rule and Simpson’s three-eighths rule.
( a ) We can use the table of Problem 4.12 since N = 6.
So1&
=
Then using Problem 4.38 we have
3O
[(l.OOOOOOOO + 0.90000000 + 0.69230769 + 0.50000000)
10
+ 5(0.97297297 + 0.59016393) + 6(0.80000000)]
= 0.78539961
144
APPLICATIONS O F T H E SUM CALCULUS
[CHAP. 4
Percent error
Rule
0.147%
Trapezoidal
Simpson's one-third
Simpson's three-eighths
Weddle's
0.00028%
0.00293%
0.00185%
Supplementary Problems
SERIES OF CONSTANTS
Sum each of the following series to n terms by using the method of Problem 4.1, page 126.
4.40.
(a) 1
3
5
7
* ' .
(e) 1 3
23
33
43
(b) 1 2
32
52
72
.*.
(f) 1 . 3 . 5
305.7
5.7.9
(c) 22
52
82
112
* *
(9) 1 . 2 9 5
2.306
3.4-7
+ + + +
4.41.
+ + + +
+ + + +
+
+
+
+ + + +
+
+
+
(d) 2 . 4 + 4.6 + 6.8 +
(h) 3 . 5 0 1 0 + 4 . 6 0 1 2 + 5 - 7 - 1 4 +
Sum the series l2+ 22% + 32x2 + 42x3 +
.
4.42.
Can the series of Problem 4.41 be summed to n terms?
4.43.
Sum the series
**.
* *
12
22
-+
3
POWER SERIES
4.44.
Sum the series
(a) x
2x2
3x3
(b)
(c)
42
32
32
--33
52
62
+-+
34
35
+ + + 4x4 +
+ 22x2 + 32x3 + 42x4 +
1 9 3 + 3.623 + 5'7x2 + 709x3 +
a
'
* * .
'
''*
32
4.45.
Sum the series
2.5
32 -
4.46.
Sum the series
12
5
+
4.47.
If
5.8
34
-
32
8.11
36
72
+ 352 + a
+
* * -
11.14 I
38
*
.
and
show t h a t
= A ( % )s i n x
+ B ( x ) cosx
...
Explain.
.
*
a
*
APPLICATIONS O F T H E SUM CALCULUS
CHAP. 41
33x3
5325
73x7
+++
4!
6!
.
4.48.
Sum the series
13x - 2!
4.49.
Sum the series
-12x
+ - +22x2
l!
3!
4.50.
Can the power series method of page 121 be used f o r the finite series
32x3
5!
~
42x4 I
7!
*
...
1 3 + 22x2
__ + 32x3
- + 42x4
-+
3
32
33
34
4.51.
145
...
+ -n2xn
3n
Can the sum of the first n terms of the series of Problem 4.48 be found by using the power series
method of page 121? Explain.
x3$
APPROXIMATE INTEGRATION
4.52.
Find
by using the extended trapezoidal rule where the interval of integration i s subdivided
into (a) 6 (b) 10 equal parts. Show t h a t the true value is I n 3 = 1.09861 approx. and compare
this with the values obtained in ( a ) and ( b ) .
4.53.
Discuss how you would improve the accuracy in Problem 4.52,
4.54.
Work Problem 4.52 by using Simpson’s one-third rule and discuss the accuracy.
4.55.
Show t h a t Simpson’s one-third rule is exact if f ( x ) is a polynomial of degree 3.
metrical significance of this?
4.56.
(a) Work Problem 4.52(a) by using Simpson’s three-eighths rule and discuss the accuracy.
you work Problem 4.52(b) by Simpson’s rule? Explain.
4.57.
Derive the extended Simpson’s three-eighths rule.
4.58.
Explain the difficulty involved in attempting to find the approximate value of
without actually evaluating the integral.
Can you resolve the difficulty i n p a r t (a)?
4.59.
Prove the formula
+
where yk = f ( a kh).
Use ( a ) to obtain an approximate value f o r
and compare with the exact value.
4.60.
Show t h a t the next two terms in the series f o r
{ a i n h f ( 2 ) dx
given in equation (8),page 122, a r e
n(Zn4 - 24x3 105n2 - 200% 144)
A 5 f( a )
1440
+
+
4725n2 + 7672n - 5040)
Asf(a)
+ n(12ns - 210124 + 1 4 2 8 ~ -360,480
and use this to complete the proof in Problem 4.38.
What is a geo-
( b ) Can
146
APPLICATIONS O F T H E SUM CALCULUS
[CHAP. 4
T/2
4.61.
4.63.
Find a n approximate value of
sin x dx and compare with the exact value.
X
0
0.2
0.4
0.6
0.8
1.0
f(x)
1.00000
0.81873
0.67032
0.54881
0.44933
0.36788
Discuss what steps you would use to work Problem 4.62 if the values corresponding to (a) x = 0.4
( b ) x = 0.4 and x = 0.8 were not present in the table.
THE EULER-MACLAURIN FORMULA
4.64.
Use the Euler-Maclaurin formula t o find the sums of the series
(c)
+ 22 + 32 + +
1 3 + 23 + 33 + . . - $- n3
1 ' 2 + 2 . 3 + 3 . 4 + ." + n ( n f 1 )
(d)
14
(a)
(b)
. * '
12
n2
+ 24 + 34 +
4.65.
Use the Euler-Maclaurin formula to work Problem 3.46, page 108.
4.66.
Show t h a t
1
1
+ -12 + 31 t ... + -n1
= y
1
1
+ Inn + -f
2n
12n2
1
Eon4
where y is Euler's constant.
4.67.
Use Problem 4.66 to obtain an approximate value f o r y.
4.69.
Show t h a t
4.70.
Sum the series
formula.
1
13
1
1
+i+
23
33
*
to 5 decimal place accuracy by using the Euler-Maclaurin
ERROR TERM I N THE EULER-M$CLAURIN FORMULA
4.71.
Show t h a t
4.72.
Obtain the Euler-Maclaurin formula with a remainder for the case where m = 5 .
CHAP. 41
APPLICATIONS O F THE SUM CALCULUS
147
4.73.
(a) Use mathematical induction to extend the results of Problems 4.71 and 4.72 and thus ( b ) prove
result (23), page 125.
4.74.
Prove t h e result ( 2 5 ) , page 125.
4.75.
(a) Show how to improve the accuracy of the result in Problem 4.32 by taking one more term in
the Euler-Maclaurin formula. ( b ) Determine the accuracy of the result obtained in p a r t (a).
4.76.
Prove t h a t if f(*P-”(x) does not change sign in the interval from a to a, nh then the remainder R
given in the Euler-Maclaurin formula ( 2 4 ) , page 125, has the property t h a t
4.77.
Work ProbIem 4.70 by taking into account the error term in the Euler-Maclaurin formula.
+
STIRLING’S FORMULA FOR n !
Obtain l o ! from Stirling’s formula and compare with the exact value.
4.78.
4.79.
4.80..
Calculate the approximate value of the binomial coefficient
Use Stirling’s formula to show t h a t
fi =
(?;)
*
2?n(n ! ) 2
lim
~
(2?z)!fi
n+o
and show t h a t the result i s equivalent t o Wallis’ product.
4.81.
4.82.
Show t h a t f o r large values of n
(-Y2)
1
=
Generalize the result of Problem 4.35 to obtain
n!
MISCELLANEOUS PROBLEMS
(12
l)x
4.83.
Sum the series
l!
+
~
4.84.
Sum the series
4.85.
Show t h a t
= *nne-n
-
(22
+ 2)x2
+
(32
2!
+ 3)23 - ...
3!
1
1
1
1.3
.7 + 3.5
.9 + m
1
+
1 + x + x 2 + - - * + x n = n
*
3
n(n- 1)
*
(a)to ?z terms and ( b ) to infinity.
T (x-1)
+
4.86.
Find a n approximate value f o r L ’ & .
4.87.
Obtain a n approximate value for the integral
+ n(n - 13!) ( n- 2) (x-
1)2
+
* “
+ (x - 1 ) n
t T ” x cot x dx
and compare with the exact value given by
4.88.
-6.
In 2 = -1.0888.
., .
( a ) Use integration by p a r t s and the result of Problem 4.87 to find
TI2
In s i n e de
( b ) Explain why numerical methods a r e not immediately applicable to the integral in (a).
148
APPLICATIONS O F THE SUM CALCULUS
[CHAP. 4
4.89.
Obtain equation ( 5 ) of Problem 4.19 and thus complete the proof in t h a t problem.
4.90.
Show t h a t t h e seventh differences in Weddle's rule drop out and explain the significance of this.
4.91.
Derive Weddle's extended rule.
4.92.
Show t h a t
These a r e sometimes called Euler transformations and a r e often useful f o r speeding up the convergence of slowly convergent series.
4.93.
Show how Problem 4.92 can be used to find a n approximate value f o r the series
(b)
-1 - -1 + -1 - -1 + -1 1
3
5
7
9
..
and compare with the exact values given by In 2 and 714 respectively.
4.94.
Sum the series
12
-
4.95.
Sum the series
12
+2
4.96.
If x # -1, -2, -3,
+ 32 - 4 2 +
22
...
22
x+l
Show t h a t
+8
32
+
.
42
(-1)n-1n2.
+ 16
+ --
52
to n terms.
F+l ) ( x + 2 )
1
~
(n
2!
l!
+
+
+
1)2
where
4.98.
9
show t h a t
1
4.97.
+4
-
+
(x+ l ) ( x + 2 ) ( x + 3 )
+ 2)2 +, ...
1
~
(n
O
S
+
+ 1 +-2 ( n +
S
A
6(n +
...
X
1
= -
n
R
1
= -
1)2
+ R
l)3
Let f(x) have the values yo,y1,y2 a t x = a, a + h, a + 2h respectively.
( a ) Obtain an interpolating polynomial f o r f ( x ) .
( b ) Use the interpolating polynomial i n ( a ) to obtain the approximate result
s,
a+h
(c)
+
h
-[5f(a) 8f(a
f(x) dx = 12
+ h) - f ( a+ 2h)]
Use the interpolating polynomial in (a) to obtain the approximate result
1""
f ( x ) dx =
+
+ + 2f(a + 3h)]
[2f(a h) - f ( a 2h)
(d) Show t h a t the formula in ( c ) is exact when f(x) is a polynomial of degree 3 but that the
formula in ( b ) is not exact in such case.
4.99.
Use the results of Problem 4.98(b) and ( c ) to obtain approximate values of
and discuss the accuracy.
APPLICATIONS O F T H E SUM CALCULUS
CHAP. 41
4.100.
149
( a ) Show t h a t
( b ) By taking the limit as n
+
m in
( a ) obtain Wallis' product formula of Problem 4.34.
1
T/2
[
Hint. For (a) see Problem 3.147, page 118, while for ( b ) first prove t h a t 1 S
sinznx dx
7,2
d
sin2nfl x dx
1+&
2n ]
4.101.
Obtain the Euler-Maclaurin formnla in terms of differences instead of derivatives and give an example of its use.
4.102.
Let S,(s) =
n- 1
2
k=O
akxk.
(a) Show t h a t on using summation by parts
( b ) Generalize the result of ( a ) by showing t h a t after p summations by parts
4.103.
( a ) By letting
n
+
m in Problem 4.102(b) show t h a t
( b ) Discuss the case where p
--3
m in p a r t
( a ) . Under what conditions will i t be true that
5 (-)1--x
S ( x ) = l- x1k = O
Compare with Problem 4.4.
k
Aka,
DIFFERENTIAL EQUATIONS
In previous chapters we have examined the remarkable analogies which exist between
the difference and sum calculus and the differential and integral calculus respectively. It
should thus come as no surprise that corresponding t o the theory of differential equations
there is a theory of difference equations.
To develop this analogy we recall that a differential equation is a relationship of the
form
between x and the derivatives of an unknown function y = f ( x ) . If we can solve for
dnyldxn in ( 1 ) we obtain
and we call the order n of this highest derivative the order of the differential equation.
A solution of a differential equation is any function which satisfies the equation, i.e., reduces it to an identity when substituted into the equation. A general solution is one which
involves exactly n arbitrary independent constants. A particular solution is a solution
obtained from the general solution by assigning particular values t o these constants.
Example 1.
The equation
y
= cle5 4- c2e2S
*
+
+
- 3 dx 2y = 4x2 is a differential equation of order two. A general solution is
dx2
2x2 6x 7 [see Problem 5.11 and a particular solution is y = 2eZ - 6e2Z f 2s2 60 4- 7.
+ +
+
To determine the n arbitrary constants we often prescribe n independent conditions
called boundary conditions for the function y . The problem of determining solutions of the
differential equation subject to these boundary conditions is called a boundary-value probl e m involving differential equations. In many cases differential equations arise from
mathematical o r physical problems and the boundary conditions arise naturally in the
formulation of such problems.
DEFINITION OF A DIFFERENCE EQUATION
By analogy with differential equations we shall define a difference equation as a relationship of the form
(3)
150
CHAP. 51
151
DIFFERENCE EQUATIONS
Since A x = h [and since by custom or convention (Ax)” = hP is written Axp when appearing in the denominator] we see that the difference equation ( 3 ) is a relationship connecting
x,y, a y , a2y, . . .,a n y or equivalently x, f(x), f(x h), f(x 2h), . ., f(x nh) since y = f(x).
Thus we can define a difference equation as a relationship of the form
+
+
+
.
G(x, f(x), f(x + h),f(x + 2h), . . . f(x + nh)) = 0
(4
where f(x) = y is the unknown function.
ORDER OF A DIFFERENCE EQUATION
From the analogy with differential equations it might be suspected that we would
define the order of a difference equation as the order of the highest difference present,
i.e. n in (3) or (4). However this leads to some difficulties [see Problem 5.51 and so instead
we define the order to be the difference between the largest and smallest arguments for
the function f involved divided by h. Thus in (4) if both f(x) and f ( x nh) appear explicitly
the order is [(x+ nh) - x ] / h= n. However if f(x) does not appear but f ( x h) and f(x nh)
both appear then the order is [(x nh) - (x h)]/h= n - 1.
+
+
+
+
+
SOLUTION, GENERAL SOLUTION AND PARTICULAR
SOLUTION OF A DIFFERENCE EQUATION
A solution of a difference equation is [as in the case of differential equations] any
function which satisfies the equation. A general solution of a difference equation of order
n is a solution which involves n arbitrary periodic constants [as on page 821. A particular
solution is a solution obtained from the- general solution by’ assigning par-ticular periodic
constants.
Example 2.
The equation
*Ax2
3@+
Ax
f(x
2y = 4 ~ ( ~ which
),
can be written as
+ 2h) - (3h + 2)f(x + h) + (2h2 + 3h + l)f(X) = 4h2,tz)
where y = f(x), is a difference equation of order 2. A general solution is given by
y = f(x) = Cl(X)(l+ 2h)X/h
+ c2(x)(l+ h ) x / h + 2x2 + (6 - 2h)x + 7
as can be verified by substitution [see Problem 5.41. A particular solution is found f o r example by letting
cl(x) = 2
+ 4h sin ( 2 ~ x l h ) ,
cz(x) = 5
+ h2[1- cos (28Xlh)I
both of which have periods h.
To determine the n arbitrary periodic constants of an nth order difference equation
we can prescribe n independent boundary conditions for the unknown function y . The
problem -of determining solutions to the difference equation subject to these boundary conditions is called a boundary-value problem involving difference equations. As in the case
of differential equations these may arise from the nature of a mathematical or physical
problem.
DIFFERENTIAL EQUATIONS AS LIMITS OF DIFFERENCE EQUATIONS
I t will be noted that the limit of the difference equation of Example 2 as Ax = h approaches zero is the differential equation of Example 1. Also the limit of the solution of
the difference equation as a x = h approaches zero is the solution of the differential
equation [see Problems 5.6 and 5.71.
152
D I F F E R E N C E EQUATIONS
[CHAP. 6
In general we can express a differential equation as the limit of some corresponding
difference equation.
USE OF THE SUBSCRIPT NOTATION
If the subscript notation is used, the difference equation ( 4 ) with x = a
f ( a kh) can be written as
H ( k , y k , y k + l , . . ., y k f n ) = 0
+
If we are given the n values zjo,y~,y~,. . .,yn -1 then we can determine
Example 3.
The difference equation
f(x
+ 3h) - 6f(x + 2h) + 6f(x + h) + 3 f ( ~ )=
yk
+ kh,
gk =
(5)
for k=n,n+l,. . .
.
0
can be written in subscript notation a s
Ykt3
-bk+2 +6 ~ k ++
1 3Yk =
0
Then given yo = 1, y, = -2, y2 = 0 we find on putting k = O,l, , . . the values y3 = 9, y4 = 61,
. .. .
The difference equation ( 5 ) can also be written in terms of the operator E , i.e.
H ( k ,y k ,
Eyk, E2yk,
. . ., E n y k )
= 0
(6)
Example 4.
The difference equation of Example 3 can be written a s
(E'-6E2+6E+3)~k
= 0
LINEAR DIFFERENCE EQUATIONS
An important class of difference equations which arises in practice is a linear difference equation. A linear difference equation of order n is a difference equation having the
form
ao(k)E"yk a l ( k ) E " - l y k f ' * ' an(k)yk = R ( k )
(7)
+
+
where ao(k)# 0. This equation can also be written as
[ao(k)E"+ al(k)En-' +
- + an(k)]yk = R ( k )
*
where the linear operator + ( E )is given by
+(E) = ao(k)E"
+ al(k)En-l +
* * *
+ an(k)
A particularly important case arises when ao(k),a ~ ( k ). ,. . ,an(k) are all constants [i.e.
independent of k ] and in such case we refer to ( 7 ) ,(8) or (9) as a linear difference equation
of order n with constant coefficients.
A difference equation which does not have the form ( 7 ) ,(8) or (9),i.e. which is not linear,
is called a nonlinear diference equation.
Example 5.
The equation of Example 4 is a linear difference equation with constant coefficients.
Example 6.
The equation [(2k
variable] coefficients.
Example 7.
The equation
+ 1)E2- 3 k E 4-
Ykyk+l
4]yk
= 4 k 2 - 3k is a linear difference equation with nonconstant [i.e.
= yE-l is a nonlinear difference equation.
CHAP. 51
D I F F E R E N C E EQUATIONS
153
HOMOGENEOUS LINEAR DIFFERENCE EQUATIONS
If the right side of (8) is replaced by zero we obtain
(uOE" + aiEn-l +
+ an)yk = 0
(11)
or
$(E)gk = 0
(12)
which is called the homogeneous equation corresponding to ( 9 ) . It is also called the
reduced equation and equation ( 9 ) is called the complete equation o r nonhomogeneous
equation.
*
*
Example 8.
The difference equation of Example 4 is a homogeneous linear equation.
Example 9.
The homogeneous equation corresponding to the complete equation of Example 6 is
[ ( 2 k 1)E' - 3kE 4]yk = 0
+
+
It turns out that in order for us to be able to solve the complete equation (9) we first
need to solve the homogeneous or reduced equation (12) [see Theorem 5-3 below].
Since the general theory of linear difference equations is in many ways similar to that
of linear difference equations with constant coefficients we shall consider this simplified
case first.
HOMOGENEOUS LINEAR DIFFERENCE EQUATIONS
WITH CONSTANT COEFFICIENTS
Assume that y k = rk is a solution of (11) where we suppose that
stants. Then on substitution we obtain
U O P + ~ a l ~ n + k+
- l. - - + anrk = 0
or
(norn a1P-l
* * *
an)? zz 0
It follows that r k is a solution of (9) if T is a solution of
aorn alTn-1
* * '
a, = 0
+
+
+
+
+
+
+
U O , al,
. . . ,an are con-
(13)
which is called the auxiliary equation and can be written +(r)= 0. This equation has n
roots which we take to be TI, r2, . . .,rn which may or may not be different. In such case
(11)can be written in f a c t o r e d form as
aO(E-?"l)(E-r2).'. . ( E - T n ) p k - 0
(1-4)
Solutions of (14) depend on the nature of the roots rl, . . . ,r,. The following cases can
arise.
Case 1. Roots are all real and distinct.
In this case r : , ~ : ,. . .,rk are all solutions of (11). Since constant multiples of these
solutions are also solutions and since sums of solutions are solutions [because +(E) is a
linear operator] we have the solution
* + cnrk
y k = clrt + c2r,k +
Case 2. Some of the roots are complex numbers.
If ao, al, . . .,an are assumed to be real it follows that complex roots if they occur must
be conjugate complex numbers, i.e. if ,+pi is a root then so also is ,-pi where a and p
and (.-pi)" are solutions and so we have the solution
are real. In this case
Kl((Y p i ) k K2(a - p i ) k
where K1, K2 are constants. To get this in real form we write the complex numbers in
polar form, i.e.
- pi = p(cos 6 - i sin 6)
pi = p(cos 6 i sin e),
+
(Y
+
+
+
(Y
156
DIFFERENCE EQUATIONS
[CHAP. 5
The only requirement which must be met t o guarantee success of the method is that no
term of the trial solution can appear in the complementary function. If any term of the
trial solution does happen to be in the complementary function then the entire trial
solution corresponding to this term must be multiplied by a positive integer power of k
which is just large enough so that no term of the new trial solution will appear in the
complementary function. The process is illustrated in Problems 5.20-5.22.
SPECIAL OPERATOR METHODS
We have seen that any linear difference equation can be written as
#(E)%= R ( k )
called the inverse of + ( E )by the relationship
(18)
Let us define the operator 1/@)
1
---R(k)
= U
where +(E)U = R ( k )
(19)
$(E)
We shall assume that U does not have any arbitrary constants and so is the particular
solution of the equation (18). We can show that l/+(E)
is a linear operator [see Problem
5.241.
The introduction of such inverse operdtors leads t o some special operator methods
which are very useful and easy to apply in case R ( k ) takes on special forms as indicated
in the following table. In this table P ( k ) denotes as usual a polynomial of degree m.
For illustrations of the methods see Problems 5.23-5.30.
~ ( pf) 0, otherwise use method 4.
1.
2.
1
-sinuk or -coscuk
+(E)
dE)
Write cosak =
3.
eik
1
P(k) =
+(I+
+(El
+ e-tk
9
~
A)
sinak =
P ( k ) = (9,
eiak
- e-iak
2
+ blA +
and use 1.
+ bmAm +
* *
*)P(k)
where the expansion need be carried out only as f a r a s Am since
A m + I P ( k ) = 0.
1
pkP(k) =
P(k). Then use method 3.
dPE)
The result also holds for any function F ( k ) in place of P ( k ) .
pk-
I
Fig. 5-2
METHOD OF VARIATION OF PARAMETERS
In this method we first find the complementary solution of the given equation
+(E)yk =
(20)
R k
i.e. the general solution of
+(E)’h= 0
in the form
yk =
+ czuz+
CIU~
* * *
+
CnUn
CHAP. 61
157
D I F F E R E N C E EQUATIONS
We then replace the arbitrary constants
and seek to determine these so that
f/k
c1,
= Klul
. . .,Cn by functions of k denoted by Ki, Kz, . . .,Kn
CZ,
+ K ~ U+Z
* *
'
-tKnun
(28)
satisfies (20). Since n conditions are needed to find K I ,K z , . . .,Kn and one of these is that
(20) be satisfied there remain n - 1 arbitrary conditions which can be imposed. These are
Usually chosen so as to simplify as much as possible the expressions for Ayk,A2yk, , . . . The
equations obtained in this way are
+
+
ul&
(AU~)AK,
+
u 2 ~ K 2 . - .+
( A U ~ ) A K ~* * *
+
UnAKn = 0
(AUn)AKn = O
+
..............................................
+
(An-'Ul)AK1 + ( A ~ - ~ u ~ ) A *K *~ + (An-'Un)AKn = O
(hn-lul)aK1+ (An-%2)AKz * * * (An-lUn)AKn = R k
+
1
(24)
which can be solved for a K 1 , a K z , . . ., AKn and thus K1, Kz, . . .,Kn from which (23) can be
found. For an illustration of the method see Problem 5.31. It should be noted that the
determinant of the coefficients of (2.4) is equal to the Casorati of the system [see (15)]which
is different from zero if ~1,242, . . .,U n of (23) are linearly independent. See Problem 5.104.
METHOD OF REDUCTION OF ORDER
If an nth order equation can be written as
( E - Tl)(E- r z ) *
then on letting zk = ( E - r2) *
-
* *
(E- Tn)f/k =
( E - rn)yk we are led to the first order equation
=
(E-Ti)Zk
which has the solution
zk
where
CI is
=
(25)
R k
($)
(26)
R k
k-l
=
T:
P=l
RP
rptl -/-
Clr:
1
an arbitrary constant. By continuing in this manner (25) can be solved.
METHOD OF GENERATING FUNCTIONS
The generating function f o r yk is defined as
5 yktk
G(t) =
k=O
By using this a linear difference equation can be solved. See Problem 5.34.
LINEAR DIFFERENCE EQUATIONS WITH VARIABLE COEFFICIENTS
Up to now we have considered linear difference equations with constant coefficients by
using the fundamental Theorem 5-3, page 155. Since this theorem also applies t o linear
difference equations with variable coefficients it can also be used t o solve such equations.
Any linear first order equation with variable coefficients can be written as
y k f i - Akyk =
R k
or
(E-Ak)Yk =
and can always be solved. The solution is given by
R k
(29)
158
DIFFERENCE EQUATIONS
[CHAP. 5
where c is an arbitrary constant. This solution can be obtained by multiplying both sides
of (29) by the summation factor uk = 1 / A l A 2 . ' A k - 1 analogous to the integrating factor
f o r first order differential equations. See Problem 5.35.
Second or higher order linear equations with variable coefficients cannot always be solved
exactly. In such cases special methods are used. Among the most common methods are
the following.
1. Factorization of the operator.
tion as
In this method we attempt to write the difference equa(31)
by finding A k , B k , . . ., u
lems 5.36 and 5.37.
k
and then using the method of reduction of order. See Prob-
2. Variation of parameters. This can be used when the complementary solution can be
found. See Problem 5.114.
3. Generating functions.
4.
Series solutions.
See Problem 5.124.
In this method we assume a solution in the form of a factorial series
(32)
where c, = 0 for p
< 0. See Problem 5.38.
STURM-LIOUVILLE DIFFERENCE EQUATIONS
A Sturm-Liouville difference equation is one which has the form
A(Pk-dyk-1)
f
(qk
+
hrk)yk
= 0
(33)
where 1 5 k 5 N - 1 and where h is independent of k .
We shall associate with (33) the boundary conditions
= 0,
(34)
ayNY, + @ N + I Y N + l = 0
where a,, al,aN,a N t l are given constants.
We seek nontrivial solutions [i.e. solutions which are not identically zero] of the equation
(33) subject to conditions (34). The problem o f finding such solutions is called a SturmLiouville boundary-value problem. The equations (33) and conditions (3.4) are called a
a,Y,
+ a1Y1
Sturm-Liouville szjstem.
We can show that in general there will be nontrivial solutions only for certain values
of the parameter h. These values are called characteristic values or eigenvalues. The
solutions corresponding to these are then called characteristic functions or eigenfunctions.
In general there will be a discrete set of eigenvalues and eigenfunctions. If Am and An
are two different eigenvalues we can denote the corresponding eigenfunctions by +7n, and
+n,k respectively.
In case p k , q k , r k are real we can show that the eigenvalues are also real [see Problem 5.401. If the eigenfunctions are real we can also show that
c
N
k=l
rk+m,k+n,k
=
'
if
(35)
CHAP. 51
’
159
DIFFERENCE EQUATIONS
In general we can take r k > 0. Then (35) states that the eigenfunctions {fi+m,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 = A l i +
A2j
A&, B = Bli B 2 j f B& are orthogonal or perpendicular if
+
+
AIBIf AZBZ+ AJ33 = 0
3
i.e.
2 AkBk
= 0
(36)
k=l
The extension to N dimensions is obtained on replacing the upper limit 3 in the sum by
N thus yielding ( 3 5 ) .
In case ( 3 5 ) holds we also say that the functions
to the weight o r density r k .
{+m,k}
are orthogonal with respect
Given a function Fk it is often possible to obtain an expansion in a series of eigenfunctions. We find that
N
Fk
=
C
m=l
(37)
cm+m,k
where the coefficients cm are given by
N
2
k = l rkFk+m,k
Cm
=
m=l,
N
2 ‘k+h,
k=l
If we suitably normalize the functions
+m,k
...,N
k
so that
N
then the coefficients (38) are simplified since the denominator is equal to 1. The results
(35) and (39) can be restated as
N
k=i
0 mPn
1 m=n
In such case we say that the set { f i + mis, an
k }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
o r integrals. In such case we refer to the equations as differential-differenceequations,
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 o r 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 AI, AZ defined so that if h = Ax, 1 = Ay are
given,
A f ( x , 9 ) = f ( x + h, y) - f ( x ,Y),
A z f ( X , y ) = f ( x ,Y + 1) - f ( x ,V )
(41)
These are called partial difference operators. Similarly we define partial translation operators El, Ez so that
E l f ( %Y)
, = f ( x+ h, Y ) , Ezf(x,Y ) = f ( x ,Y + 1)
(42)
These are related t o partia2 derivatives considered in elementary calculus. I t is clear that
Ez = 1 + A z
El = 1 + A i ,
(43)
We can also define powers of El, Ez, A I , Az.
If we use the subscript notation
Zk,m
+
= f ( a kh, b
+ WZ)
(44)
it follows for example that
E1Zk,m
=
zk+l,m,
EzZk,m
=
Zk,m+l
(45)
We can now consider partial difference equations such as for example
which can also be written
Zkt2,rn
- 3Zk+l,m+l
+
2Zk,mt2
= 0
(47)
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
+(E1,Ez)Zk,m
=
Rk,m
(48)
where +(El,E z ) is a polynomial in El and EZ of degree n. A solution of (48) which contains n arbitrary functions is called a general solu,tion. 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. 51
DIFFERENCE EQUATIONS
161
Solved Problems
DIFFERENTIAL EQUATIONS
5.1. Show that the genera1 soIution of the differential equation
is y = cleX c2eZx 2x2 6% 7.
+ + +
+
We have
+ c2eZs + 2x2 + 62 + 7
clex + 2c2eZX+ 42 + 6
= clez
y
-d’ _ dx
d2y
-
Then
+
4x2
(1)
(2)
+
= (cles -I- 4c2eZX 4) - 3(c1eX 2c2e22-I- 4x
3& -I- 2y
dx
dx2
G - 3 dx
9+2y =
+ 6)
+ 2(cle2 + c2e2X+ 2x2 + 60 + 7)
= 4x2
Since the given differentia1 equation is of order 2 and the solution has 2 arbitrary constants i t is the
general solution.
Find the particular solution of the differential equation in Problem 5.1 such that
y(0) = 4, y’(0) = -3.
From (1) and ( 2 ) of Problem 6.1 we have
so that c1 = 3,
c2
y(0)
=
y’(0)
=
+ c2 + 7 = 4
c1 + 2c2 + 6 = -3
or
c1
or
+ c2 = -3
c1 + 2c2 = -9
c1
= -6.
Then the required particular solution is
y
= 3eX - 6e25
+ 2x2 + 6x + 7
DIFFERENCE EQUATIONS
A2Y
3 AY ~ 2y += 4 ~ ‘ ~
can
) be written as
5.3. Show that the difference equation 3-
f(x + 2h) - (3h + 2)f(x + h) + (2h2+ 3h + l)f(x) = 4h2d2) where y = f(x).
Writing Ax = h and y = f(x) the difference equation can be written a s
WE)
hz
or
f(x
Multiplying by
F
+
2f(x)
= 4x(2)
+ 2h) - 2f(x
+ h) + f(x) h2
h2
and simplifying we obtain
f(z
5.4.
3
+ 2h) - (3h + 2)f(x + h) + (2h2 + 31%+ l)f(x) = 4h2x(’)
(2)
Show that the general solution of the difference equation in Problem 5.3 is given by
y = cl(x)(l 2h)”Ih C Z ( X ) ( ~ h)“Ih+ 2%’ (6 - 2h)x 7
(1)
+
where cl(x) and
CZ(X)
+
+
+
+
have periods equal to h.
+ h) = el(%), cz(x + h) = c2(x),
= cl(z)(l + 2h)z’h + cZ(x)(l + h)z’h + 2x2 + (6 - 2h)x + 7
We have since cl(z
2/
(2)
162
DIFFERENCE EQUATIONS
-A”
-Axz
Similarly,
4c l(a)(l+ 2h)zlh
[CHAP. 6
+ cz(z)(l+ h ) z / h + 4
Then
A’YAxz
AY
3~
+ 2y
= [4C,(Z)(I
-
+
+ c , ( z ) ( ~+ h ) z l h + 41
+ 2h)’lh + c , ( z ) ( ~+
+ 2[~1(2)(1+ 2h)‘” +
2h)s/h
3[201(2)(1
h)s/h
+ 4 2 + 61
+ h ) z / h + 2x2 + (6 - 2h)z + 71
CZ(Z)(~
= 422 - 4hz = 4 2 ( 2 - h )
= 4x(2)
Since the difference equation ( I ) has the difference between the largest and smallest arguments
divided by h equal to
(z 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 h A2Y
2 s 2h-AY
Ax
+y
= 2 ~ ‘ is
~ )not a second order difference equation.
We can write the equation if g = f(x) as
f(x +2h) = 2 ~ ( 3 3
or
This equation involves only one argument and in fact is equivalent t o
= 2(x--22h)(3)
which is not really a difference equation.
The result shows that we cannot determine the order of a difference equation by simply looking
a t the largest value of n in Any/Axn. Because of this we must define the order a s given on page 151.
DIFFERENTIAL EQUATIONS AS LIMITS OF DIFFERENCE EQUATIONS
5.6. Show that the limit of the difference equation [see Problem 5.31
A2YAX’
3 ,AY
x
+ 2y
as A x o r h approaches zero is [see Problem 5.11
This follows a t once since
and
=
4X(Z)
CHAP. 51
5.7.
163
DIFFERENCE EQUATIONS
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 = Cl(X)(I 2h)'Ih
+ 02(x)(l+ h)'Ih + 2x2 + (6 - 2 h ) ~+ 7
(11
From the calculus we have
(2)
or, if n = alh where a is some given constant,
e
= 2.71828.. .
Esuivalently (3) can be written
or
Using ( 5 ) with a = '2 we fin'
Using ( 5 ) with a = 1 we find
+
+
Also since cl(x h) = cl(x), c2(x h) = c2(x) i t follows that lim cl(z) = cl,
h-ro
where c1 and c2 are constants, provided that these limits exist.
lim c2(x) = c2
h-ro
Then the limit of the solution (1) as h + 0 is
y =
cle2S
+ c2e5 + 2x2 + 62 + 7
which is the general solution of the differential equation of Problem 5.6 [see Problem 5.11.
USE OF THE SUBSCRIPT NOTATION
5.8. Write the difference equation of Problem 5.3 with subscript o r k notation.
We have by the definition of the subscript notation
Also using
2
= kh,
Then the equation
becomes
or using the operator
which can be written
164
5.9.
DIFFERENCE EQUATIONS
[CHAP. 6
Write the difference equation
f(x + 4h) + 2f(x + 3h) - 4f(x + 2h) + 3 f ( x + h) - 6f(x) = 0
in the subscript or k notation.
We use
gk
= f(z),
gk+l
= f(x + h),
Then the difference equation is
(E4
= f(s + 2h),
76lc+2
= f(z f 3h),
gk+3
+ 2E3 - 4E2 + 3E - 6)yk
?dk+4
=
f
(
%
'
4h)
= 0
LINEARLY INDEPENDENT FUNCTIONS
5.10. Determine which of the following sets of functions are linearly dependent and which
are linearly independent.
(a) 25 2 k + 3
(q 25 2k+3,4k
(c) 2 k , 4k
( d ) Z k , k*Zk,k 2 - Z k .
+
(a) Consider A12k AzZk+3which can be written as
A,2k
+ A2
23
2k = (A,
+ 8A2)Zk
(11
Since we can find A , and A , not both zero such that (1)is zero [as for example A, = -1, A , = 81
it follows that the functions 2k, Z k + , are linearly dependent.
+
( b ) Consider A12k 4-A2Zk+3 A34k which can be written a s
(A,
+ 8A2)2k + A34k
(2)
Then since we can find the 3 constants A , , A,, A, not all zero such that ( 2 ) is zero [for example
A , = -1, A l = 8, A, = 01 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 t o the set.
+
(c) Consider AIZk A24k where A , , A, are constants. This will be zero if and only if
2k(A, +A22k) = 0
or
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
only if
+ A&
+ A3k2*2k where A,, A,, A, are constants. This will be zero if and
2k(A1 + A & + A,@) = 0 or A , + A2k + A3k2 = 0
2k
But this last equation cannot be true for all k unless A l = A , = A3 = 0.
are linearly independent.
5.11.
Thus the functions
Work Problem 5.10 by 'using Theorem 5-1, page 154.
(a) In this case fl(k) = 2k, f 2 ( k ) = 2kt3 and the Casorati is given by
Then by Theorem 5-1 the functions Zk, Z k + 3 are linearly dependent.
( b ) In this case f , ( k ) = 2k, f 2 ( k )= 2k+3, f 3 ( k ) = 4k and the Casorati is
\
20 23 40
2 1 2 4 4 1
22 25 42
1
=
8
1
2 1 6
4 32
4
16
=
8
1
2
4
1
2
4
1
4
16
=
o
CHAP. 61
165
D I F F E R E N C E EQUATIONS
where we have taken the factor 8 from elements of the second column in the determinant and
noted t h a t since the first two columns a r e identical the determinant is zero,
Thus by Theorem 6-1 the functions 2k, 2k+3, 4k a r e linearly dependent.
( c ) I n this case
f,(k) = zk, f z ( k ) = 4k and the Casorati is
1
20
21
22
0 '20
02 20
1.21
2'22
12-21
22-22
1
1
2
4
=
0
0
2
21
8 1 6
=
1:
11;
=
16
Thus t h e functions 2k, k 2k, k2 2k a r e 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 f l ( k ) , f 2 ( k ) ,f 3 ( k ) a r e linearly independent if and only if the equation
A i f i ( k ) + Azf2W)
+ A3f3(k) =
identically holds only when A , = 0, A, = 0 , A , = 0.
(11
0
Putting k = 0, 1,2 in (1)
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.
rk+2
- 6rkfl
ykt2
-6
+
~ k t i
Then i t becomes
+ 8rk
= 0
or
rk(r2-
6r+ 8 ) = 0
Dividing by r k [assuming r # 0 which otherwise leads to the trivial solution yk = 01 we obtain
the auxiliary equation
r2
- 6r + 8 =
i.e.
0
(r - 2 ) ( r - 4 ) = 0
or
r = 2,4
Thus two solutions a r e 2k and 4k. By Problem S.ll(c) these a r e linearly independent solutions.
Note t h a t the difference equation can be written as
(E2- 6E
8)yk = 0
and t h a t the auxiliary equation can immediately be written down from this on formally replacing E by r in the operator Ez - 6E 8 and setting the result equal t o zero.
+
166
D I F F E R E N C E EQUATIONS
[CHAP. 5
(b) Using Theorem 5-2, page 154, i.e. the superposition principle, the general solution is
yk =
+
C1zk
C2'Ik
where c1 and c2 a r e arbitrary constants.
5.14. (a) Find that particular solution of
(b) Find
ykt2
- 6ykti
+ 8yk = 0
such that yo = 3, yl = 2.
y5.
(a) From Problem 5.13 the general solution of the difference equation is
yk =
+ E2dk
C1zk
Then from yo = 3, y, = 2 we have on putting k = 0 and k = 1 respectively
3 =
from which c1 = 5,
c2
C1
+
C2,
2 = 2C1 f 4C2
= -2.
Thus the required particular solution is
yk = 5 9 2 k - 2 . 4 k
( b ) Letting k = 5 we find
765
= 5 - 2 5 - 2.45 = 5(32) - 2(1024) = -1888
The value of y5 can also be found directly from the difference equation [see Problem 5.841.
5.15.
Solve the difference equation
YktZ
- 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 i s
r2-2r+5
= 0
-
2"4i
2
from which
r =
2.-dF%
2
= 1'2i
+
It follows t h a t solutions a r e (1 2i)k and (1- 2i)k which can be shown to be linearly independent
[see Problem 5.741. Since in polar form
Then the general solution can be written as
yk = 5k/2(c1 cos k s
+ c2 sinks)
CHAP. 51
5.16.
D I F F E R E N C E EQUATIONS
167
Find the particular solution of the difference equation of Problem 5.15 satisfying
the conditions yo = 0, 21 = 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 = el,
1 = 51/2(c1 cos e c2 sine)
+
from which
= 0,
c1
1
cp =
I
_
_
fi sin e
Thus
Note t h a t e can be found from (1) of Problem 5.15.
5.17.
(a) Solve the difference equation
such that yo = 1, y1 = 3.
Ykt2
+
- 4y1,+1 4yk = 0 and ( b ) find the solution
+
( a ) The difference equation can be written as ( E 2- 4E 4)yk = 0 and the auxiliary equation is
rz - 4r 4 = 0 or (Y - 2)2 = 0 so t h a t r = 2,2, i.e. 2 is a repeated root.
+
It follows t h a t Yk = rk = 2k is a solution. However we need another solution which is
linearly independent. To find i t let yk = 2kvk in the given difference equation. Then i t
becomes
or
A2wk = 0
vk+2-2vk+l.+vk = 0
Thus
vk=
C1
and
f Cpk
z/k
-
2k(C1
-tC,k)
which gives the required general solution.
( b ) Putting k = 0 and k = 1 we find
1 =
3 = 2(c,
c1,
+
c2)
Thus c1 = 1, c2 = 1/2, and the required solution is
5.18.
+
(a) Solve the difference equation y k 7 3 y k + 2 - yktl- yk = 0 and ( b ) find the solution which satisfies the conditions yo = 2, y1 = -1, 3 2 = 3.
(a) The difference equation can be written as (E3
+ Ez -E - l)yk = 0
13+rz--r-1=
and the auxiliary equation is
0
This can be written a s
r * ( r + 1) - ( r + 1) = 0,
(Y+ l ) ( r 2 -
1) = 0, or
(r
+ l ) ( r +l ) ( r - 1) =
0
so t h a t the roots a r e -r = -1, -1,l.
=
Corresponding to the two repeated roots -1, -1 we have the solution ~ ~ ( -4-1c2k(
)~
Then
by
the
Corresponding
to
the
root
1
we
have
the
solution
c
,
(
l
)
k
=
c3.
( - l ) k ( ~c&).
l
principle of superposition the general solution is
+
~k
+
= ( - l ) k ( ~ i cnk)
+
~3
( b ) Putting k = 0, 1 , 2 respectively we find
2 =
c1
+ CQ,
-1 =
from which c1 = $, c2 = 4, c3 = f.
yk
-c1
- c2 + c3,
3
=
c1
Thus the required solution is
= 3 + (-l)k(2k+5)
4
4
f 2c,
+
c3
168
5.19.
D I F F E R E N C E EQUATIONS
[CHAP. 5
Find the general solution corresponding to a linear homogeneous difference equation
if the roots of the corresponding auxiliary equation are given by 3, - 1f1
-C -i,
-3,
2
2
3, -2, -3, 3, 4.
+ + c3k2).
Corresponding to the repeated roots 3 , 3 , 3 we have the solution 3k(c1 c2k
2
Corresponding to the complex roots - - f -i [which have polar form cos e
2
1 , sin e = - o r e = 120° = 2 ~ 1 3radians] we have the solution
cos e = - i
2
2nk
c4 cos ke
c5 sin ke = c4 c o s y
c5 sin- 3
6
+
f
i s i n e where
+
Corresponding to the repeated roots -3, -3 we have the solution (-3)k(C6
Corresponding to the single root -2 we have the solution c8(--2)k,
+qk).
Corresponding to the single root 4 we have the solution cg4k.
Then the required general solution is by the principle of superposition given by
l/k
+
= sk(c1 c2k+ c3k2)
+ c4 cos-2nk3 +
c5 sin
2nk
3
+ (-3)k(C6+ c7k) + C8(-2)' + c04'
METHOD OF UNDETERMINED COEFFICIENTS
5.20. Solve y k + 2 - 6 y k t 1 + 8yk = 3k2 + 2 - 5 ' 3 k .
The general solution of the homogeneous equation [also called complementary or reduced
equation] is by Problem 5.13
u k = C1zk
C2qk
(11
+
+
Corresponding t o 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 ( I )
Corresponding to the term -5
the trial solution
3k on the right hand side of the difference equation we assume
A43k
(8)
Thus corresponding to the right hand side we assume the trial solution [or particular solution]
+ A2k + A3 + A43k
yk = A1k'
Substituting in the given difference equation we find
uk+2
- 6uk+l
81/k
1
=
+ + A3 A43k32
+ + A2(k + 1) + A3 + A43k+1]
+ 8[Alkz + A2k + A3 + A43k]
3Alk2 + ( 3 A 2 - 8 A l ) k + 3A3 - 4A2 - 2 A 1 - A43'
Al(k
2)2 A2(k 2)
- 6[A1(k 1)'
Since this must equal the right hand side of the given difference equation we have
3Alk2
+ (3A2 - 8 A l ) k + 3A3 - 4A2 - 2 A 1 -
A43k = 3k2
+2 -5
3k
Equating coefficients of like terms we have
3A1 = 3,
Thus
A1
3.42
- 8A1 = 0,
= 1, A , = 813,
and the particular solution is
k2
3A3
- 4Az - 2A1 = 2,
A3
4419,
A4
=
A4 = 5
5
44
+ g8 k + s
+ 6.3'
Adding this t o the complementary solution ( I ) we find
yk
= c12k
44
+ c24k + k2 + -k83 + +503~
9
CHAP. 51
D I F F E R E N C E EQUATIONS
169
t.
5.21. Solve
Yk+2-4yk+l$4yk
= 3~2~+5*4~.
The complementary solution is by Problem 5.17
+ c2k.2k
c12k
(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 t h a t 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 Alk2*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 = Alk2* 2k
+ Az4k
Substituting this into the given difference equation we find
- 41/k+l + 4yk = A,(k+2)2*2k+' +
- 4[A1(k + 1)2* 2 k f l
yk+2
+ A,4k+l]
+ 4[A1k2*2k + A24k]
+ 4A24k
= 3*2k + 5*4k
= 8A12k
Thus we must have
5.22. Solve
9k+3
8A,2k
- 3ykt2
+ 3Yk+l -
yk
+ 4A24k
= 24(k+2).
The complementary equation is
(E3-3E2+3E-l)gk
= 0
and the complementary solution is thus
yk
=
C1
+
or
(E- l)3yk = 0
+ c2k + ~ 3 k '
Corresponding to the term 24k 48 on the right of the given equation we would normally use
the trial solution Alk Az. Since these terms occur in the complementary solution we multiply by
a power of k which is j u s t sufficient to insure t h a t none of these terms will appear. This is k3 so
t h a t the trial solution is yk = Alk4+A2k3. Substituting this into the given difference equation
we find
24Aik 36A1+ 6Az = 24k 48
+
+
so t h a t A, = 1, A, = 2.
+
Thus the required solution is
l/k
= c1 + c2k
SPECIAL OPERATOR METHODS
5.23. Let +(E)= aoEn+ alEn-' + . . . + a,.
+ c3k2 + k4 + 2k3
(a) Prove that
+(E)pk= +(p)pk. (b) Prove that
Pk if
a particular solution of the equation +(E)yk = pk is given by
+(P) + 0.
( a ) W e have
Epk =
pk+l
= p.pk,
E2pk
=
pk+2
= pZ.pk,
...,
Enpk
=
hkm
=
pk+n
= pn.pk
170
D I F F E R E N C E EQUATIONS
Thus
+
+
@ ( E ) p k = (aoEn alEn-l
-
(a,pn -I- alp"-l
[CHAP. 5
+ an),@
+ an)pk =
*
+
@(p)pk
( b ) Since @(E)pk
= #(B)pk from p a r t ( a ) i t follows t h a t
1
mPk= m
5.24.
if G(p)
pk
+o
Prove t h a t 1/+(E)is a linear operator.
Let
1
---Rp(lc)
dE)
u,,
-1_R,(k)=
4E')
=
u,
Then by definition if R,(k) and R 2 ( k )a r e any functions
@ ( E ) U l = R , (k ),
= R,(k)
@)U2
and since @ ( E is
) a linear operator it follows t h a t
d E ) [U,+ U,] = R,(k) + R,(k)
so t h a t
1
or using (1)
Also if
01
1
u, +
4 R,(k)] =
UZ
1
+ --RR,(k)
dE)
1
[R,(k)+R,(k)] = --RR,(k)
dE)
is a n y constant and R ( k ) is any function let
Then
Dividing by
(Y
# 0 we have since +(E)is
a linear operator
U
or
= R(k)
U
L R ( k ) = @(E)
From (2) and (3) we have
1
-cuR(k)
dE)
1
= cu-R(k)
#m)
The results ( 2 ) and ( 5 ) prove t h a t l / @ ( Eis
) a linear operator.
5.25.
Solve the difference equation
v kt2
- 2y, + 1
The equation can be written as (E2- 2 E
a particular solution is given by
E2 - 2 E
+ 5 (2.3k-4.7k)
+ 5yk = 2
+ 5)yk = 2
3k - 4 * Yk.
3k - 4 7k. By Problems 5.23 and 5.24
1
= 2 ' E 2 - 21E f 5 3 k - 4 4 E 2 - 2 E + 5 7 k
1
1
3k - 4 -~
7' - 2(7) 5 7 k
32 - 2 ( 3 ) 5
-1. 3 k - - - - 1 71 k
4
10
Since the complementary solution is as given in equation ( 2 ) of Problem 5.15 the required general
solution is
yk = S k / 2 ( c 1 cos ke + c2 sin ke)
3k - 2
10 7k
-
I
+
+
where 8 is determined from equation (1) of Problem 5.15.
+
DIFFERENCE EQUATIONS
CHAP. 51
171
Then
5.27.
1
-pkF(k)
Prove that
=
O m
pk-
~ ( k =) I
F ( k ) so that
dPE)
Let
1
+(pE)F(lc)*
@(RE)G ( k ) = F ( k ) .
Then by Problem 5.26
Thus
5.28.
E2 - 4E +
Find
2k.
+
The method of Problem 5.23 fails since if g ( E ) = E z - 4 E 4 then @) = 0. However we can
use the method of Problem 5.27 with F ( k ) = 1, p = 2 to obtain
1 2 k
= 2k-
dE)
5.29.
d2E)
1
E2 - 6E
Find
(1) = 2 k 1
(2E)Z - 4 ( 2 E )
+ 4 (l) =
1
2k 4Ez - 8 E + 4 ( l )
+
(3k2 2).
+
We use the method in entry 3 of the table on page 156. Putting E = 1 + A we have
E2
- 6E
+ 8 (3k2+2)
=
1
_(1 -I- A)2 - 6 ( 1 -I- A )
-
1
3 - 44
I
+ 8 (3k2 + 2 )
+ 4 2 (3k2 + 2 )
1
- -
1
(3k2
3 1 - ( $ A - 442)
+ 2)
+ ($A-&AZ) + ( $ A - + A 2 ) 2 + * * . ] ( 3 k 2 + 2 )
&[I -I- % A + ?A2 + * ] ( 3 k 2+ 2 )
+[3k2 + 21 + @[3k(2) + 3k(1) + 21 + g A 2 [ 3 k ( 2 )+ 3k(1) + 21
= *[I
=
=
+ + 8-61
= 5 [ 3 k z + 21 t $ [ 6 k ( l ) 31
= kZ+$k+V
172
5.30.
DIFFERENCE EQUATIONS
Work Problem 5.21 by operator methods.
The equation can be written (E2- 4E + 4)yk =
6.28 a particular solution is given by
3
(3 2k 5 4 k )
E2 - 4E
4
+
+
[CHAP. 5
3 2k 4- 5 4k.
= 3 * E 2 - 41E + 4 f2k) +
Then by Problem 5.23 and
E2
- 4 E + 4 (4k)
5
+ 42 - 4(4)
+ 4 4k
#k(k - 1 ) 2 k + :4k
= $k(k-l)2k
=
Since the complementary solution is C12k
+ C2k
2k the general solution is
+
# k ( k - 1)2k
yk
= C1zk 4 Czk 2k
yk
= C1zk f ( C ~ - # ) k * 2+~ 9k22k
This can be written
+$
4k
+ Q*4k
and is the same a s that obtained in equation ( 2 ) or (3) of Problem 5.21 if we take c1 = C1,
cg
= c, - #*
METHOD OF VARIATION OF PARAMETERS
5.31. Solve y k + 2 - 5 ? / k + l + 6yk = k2 by the method of variation of parameters.
The complementary solution, i.e. the general solution of yk+2 - 5yk+l + 6Yk = 0, is c1zk + C
Z ~ ~ .
Then according to the method of variation of parameters we assume that the general solution of
the complete difference equation is
yk = K12k f Kg3k
(1)
where K, and K2 are functions of k to be determined.
Since two conditions are needed t o determine K, 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.
2kflAK1 +3k+lAKZ = 0
so that
( 2 ) becomes
AYk
- Kl2k
-
(3)
+ K22.3'
Now the given equation can be written as
5E
+ 6)yk
= k2
(A2734
+ 2)yk
= kz
(E2-
or since
E = 1+ A
.
(6)
(7)
Using (I), ( 4 ) and ( 5 ) in (7) we find after simplifying
2k+1AK1
We must thus determine
+ 2*3k+lAK2 =
K1and K2 from the equations
2kt1AK1
2k+lAK1
k2
(3) and (8),i.e.
+ 3k+lAK2
+ 2*3k+lAK2 =
(9)
k2
Note that these equations could actually be written down directly by using equations ( 2 4 ) on page 157.
CHAP. 51
D I F F E R E N C E EQUATIONS
173
Then
These can be obtained by using t h e methods of Chapter 4 or the operator method of Problem 6.27.
Using the latter we have, omitting the arbitrary additive constant,
K1 =
-:(&)
=
- AE (- &1 ) k ( F )
Thus taking into account a n arbitrary constant cl,
+ c1
K1 = $(kz+2k+3)
Similarly we find
= --
K2
2 3k
+ c2
(k2+k+1)
Then using these in (1) we find the required solution
vk
=
C1zk
+
C23k
+ &k2 + $ k
Q
METHOD OF REDUCTION OF ORDER
5.32. Solve y k t 1 - ryk = Rk where r is a constant.
Method 1, using variation of parameters.
The complementary o r homogeneous equation corresponding to the given equation is
Yk+l
- ryk
=
From this we find
yk
=
‘rVk-1,
yk-1
so t h a t the solution of (1) is
76k
=
=
...
rZ/k-zr
rkyO
1
or
where we take co = yo as the arbitrary constant.
yk
Y2
=
= ryll
cork
111
= ‘Yo
174
DIFFERENCE EQUATIONS
[CHAP. 6
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 l ) r k + l - K ( k ) r k + l = Rk
+
or dividing by
rk+l,
K ( k + 1)
Thus
-
K
K ( k ) = AK
= A-1
=
Rk
yk+l
(&)
and so from (2)
Method 2, multiplying by a suitable summation factor.
Multiplying the given equation by llrk+l it can be written as
Thus
or
where c is an arbitrary constant, The factor
5.33.
Solve
ykt2
llrk+1
used is called a summation factor.
- 5yk+l+ 6yk = k2 by the method of reduction of order.
Write the equation as ( E - 3 ) ( E - 2)yk = k2. Letting
(E-3)Zk
Zk
= ( E - 2)yk we have
= k2
Then by Problem 5.32 the solution is
( E - 2)yk = C13k - +(k2 k
Thus
+I)
Applying the method of Problem 5.32 again the solution is
= &k'
+ 4 k + 8 + ~22' + ~33k
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+l
+2yk = 0
if yO=2, y 1 = 3
by the method of generating functions.
m
Let G ( t )
=
x
k=O
2/ktk.
Multiplying the given difference equation by tk and summing from
CHAP. 51
DIFFERENCE EQUATIONS
175
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)(l- 2 t )
Writing this in terms of partial fractions we find
G(t)
1 + 2
1-2t
1-22t
=
5
Thus
5
=
tk
5
+
k=O
5 (1
=
Vktk
(2t)k
k=O
=
(1+Zk)tk
k=O
+2k)tk
k=O
k=O
and so
yk
= 1
+ zk
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
Z / k + i - Ak'&
= R k O r ( E-Ak)Z/k = R k .
Method 1, using variation of parameters.
The complementary or homogeneous equation corresponding to the given equation is
?dk+l-Ak?dk
(11
= 0
From this we have
Yk
=
Ak-1%-1,
=
yk-1
Ak-2yk-2,
9
yz = A l y l
so that the solution of (1) is
?dk
or
Yk
=
Ak-iAk-2.
=
*
'A~yi
~1A1A2.**Ak-1
where we take c1 = yl as the arbitrary constant.
We now replace c1 by a function of k denoted by K ( k ) , i.e.
~k
= K(k)A,Az** ' A k - 1
(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 ) A I A 2 ** ' A k - A k K ( k ) A , A , . * A k - 1 = R k
+
Thus
and so from (1)we find the required solution
?dk
=
= AlA2*
where c is an arbitrary constant.
176
D I F F E R E N C E EQUATIONS
[CHAP. 6
Method 2, multiplying by a suitable summation factor.
Multiplying the given equation by l I A , A Z - * * A i, t can be written
or
Then
or
* * A kused to multiply the given equation the summation factor.
w e call 1/AlA2*
5.36.
Show that the equation y k + 2
form (E- A k ) ( E - B k ) Y k = k.
-(k
+
2)yk+l+
k y k
= k can be written in the factored
The given equation is in operator form
+
[E2- ( k 2)E
+
k]yk
= k
Let us t r y to determine Ak and Bk so t h a t (2) is the same as
-B k ) Y k = k
(ENow t h e left side of ( 2 ) can be written a s
=
=
(E-Ak)(Yk+l-BkYk)
?dk+Z - B k + l Y k + l
yk+2
-Ak(yk+l-BkYk)
- ( A k + Bk+l)?dk+l
+AkBkYk
Comparing with the left side of the given equation (1) we have
= k+2,
A k + B k + l
AkBk
(4
= k
From the second equation of (8) we a r e led to t r y either A k = k , B k = 1 or A k = 1, B k = k.
Since the second set, i.e. A k = 1, Bk = k , satisfies the first equation of (3) the given equation Can
be written as
( E - 1)(E- k ) Y k = k
5.37.
Sohe
YZ=1.
Yl=o,
Yk+?-(ki-2)Yk+l+kyk=k,
By Problem 5.36 the equation can be written as
( E - 1)(E- k ) y k = k
Let xk = ( E - k ) y k .
Then (2) becomes
(E-l)~k
which h a s the solution
Then
zk =
~ - 1 k
(E-k)yk
By Problem 5.35 this has the solution
or
=
=
k(2)
2
= k
k(2)
A-lk(1)
+
C1
= 2 +
= +k(k-1)
01
+
C1
CHAP. 51
177
DIFFERENCE EQUATIONS
From yl = 0, y2 = 1 and the difference equation we find, if k = 1, y3 - 32/2+ g1 = 1
If we put k = 2 and k = 3 in (S),we find using O! = 1
SO
that
y3 = 4.
C1
from which c1 = 1, c2 = 0.
+ C2
= 1,
1
+ 3C1 + 2C2
= 4
Thus the required solution is
+
2
p=1
=
2/k
k-1
1
2p=1P!
(k-l)!
This can be written for k > 3 as
gk
5-38, Solve
=
Wql
(k
+
I-'+
2
+ l + -1+ F1+
.'.
l)yk+Z
- (3k+ 2 ) g k + 1
+ (2k- 1)yk
2!
+ (k-l)!
(k -3)!
(k - l ) !
= 0.
The equation can be written as
(k
+ 1)A22/k- kAYk - 22/k
Assume a solution of the form
= 0
4
2
p =-m
2/k
Cpk(p)
where we take
cp = 0
for p = -1, -2, -3,
From (2)we obtain
x
...
m
-
p=--oo
pcpk(P--l)
Substituting (Z),( 4 ) and ( 5 ) into (I), i t becomes
2 p ( p - 1)Cpkk(P--2) + 2 p ( p - l)Cpk(P--P) - 2 pc,kk(p-') -
2~,k(p) = 0
(6)
where we have omitted the limits of summation. On using the result
kk(rn) = k(rnt1) + mk(rn)
(6) can be written
2 p ( p - l)cp[k(P-')+ ( p - 2)k(P-2)] + 2 p ( p - 1)Cpk(p-2)
- 2 pc,[k(P) + ( p - l)k(p-1)] - 2 2cPk(p) =
0
This in turn becomes on collecting coefficients of k(p)
2 { ( p + 2)(p+ 1)2cp+, - (P + 2 ) c p W )
= 0
(7)
Since (7) is an identity each coefficient must be zero, i.e.
(P + 2 ) ( P + W
p + 2
Using (3) it follows that we must have p = 0,1,2,
- (P+ 2)cp = 0
. ..
so that (8) becomes on dividing by p
(p+1)2cp+2- cp = 0
Putting p = 0,1,2,
...
+2 P 0
(9)
in (9) we find
12cz - co = 0,
22c3 - c1 = 0,
32c,
- c* =
0,
42c, - cs
= 0,
. ..
r.
178
DIFFERENCE EQUATIONS
[CHAP. 5
STURM-LIOUVILLE DIFFERENCE EQUATIONS
5.39. Prove that the eigenfunctions +m,k, +n,k belonging to two different eigenvalues Am,
An satisfy the orthogonality condition
N
c
rk+m,k+n,k
k=l
=
Since Am, @m,k and An, @n,k are the corresponding eigenvalues and eigenfunctions they must by
their definition satisfy the equations
A ( P k - 1 A@m,k- 1) + ( q k + Amrk)@m, k =
(1)
+ ( q k + Anrk)@n,k
A(pk-lA@n.k-l)
=
(2)
for k = 1, , . .,N .
If we multiply these equations by
(Am
- An)rk @m,k @n,k
@n,k
=
and
@m,k
respectively and then subtract we find
@m,k A [ p k - IA@n, k- 11
- @n,k A b k -
(3)
1‘@m, k- 11
Summing from k = 1 t o N then yields
Using summation by parts this can be written
N+l
N
(Am--n)
3
k = l rk@m,k@n,k
=
Pk-l[@m,kA@n,k-l
=
PN[@m, N + 1A@n,N
=
PN[@m, N@n,N
- @n,kA@m.k-l]/k=l
- @n,N + 1A@m, NI
- Po[Cm,lACn,o - @ n , ~ A @ m , o l
+ 1 - @n,N@m,N + 11
- Po[@,, o@n,1 - Gn, 0, Cm, 11
From the boundary conditions (34), page 158, which must be satisfied by
aO@rn,O
+ al@m,l
“O@n,O f a l @ n , l
= 0,
= 0,
@m,k
(4)
and
@n,k
+ UN+l@rn,N+l =
aN@n,N + a N + l @ n , N f l =
aN@m,N
By eliminating a0, a l , ah,,a N + lfrom these equations we see that they are equivalent t o
‘$rn,O@n,l
- @n,O@m,l = 0,
Using these in ( 4 ) we thus have
@m,N@n,N+l - @n,N@m,N+l
N
(Am-
Ax,)
2
k=l
rk@m,k@n,k
and if Am # An we must have as required
N
2
k=l
rk@m,k’&,k
=
= 0
=
we have
CHAP. 51
5.40.
179
D I F F E R E N C E EQUATIONS
Prove that if p k , q k , r k in equation (33), page 158, are real then any eigenvalues of
the Sturm-Liouville system must also be real.
Suppose that Am is a n eigenvalue which may be complex and that
eigenfunction which may be complex. Then by definition,
+ ( q k + Amrk)@m,k
A(Pk- lA@rn,k-l)
@m,k
is the corresponding
= 0
(11
Taking the complex conjugate of this equation [denoted by a bar over the letter] we have
+
A(pk-lA$m,k-l)
using the fact t h a t
Pk-1,
qk, T k
+Lrk)zm,k
(qk
= 0
B y proceeding as in Problem 5.39 multiplying the first equation
&,k
(2)
a r e real.
the second equation by
&k,
and then subtracting, we find [compare equation (3) of Problem 5.391
-
N
x
k=l
rk@m,k@m,k
= 0
N
= o
or
Since r k > 0, the sum in (3) cannot be zero and so we must have X,
Am i s real.
-
- Am = 0
-
or A, = ,,A
i.e.
N
5.41.
Let
F k
c,&,~
=
where
are the eigenfunctions of the Sturm-Liouville
+m,k
m=l
system on page 158 assumed to be real.
(a) Show that formally
N
c
k = l rkFk+m,k
-
Cm
m = 1,.. . , N
N
( b ) What limitations if any are there in the expansion obtained in part (a)?
N
then multiplying both sides by
rk@n,k
and summing from k = 1 to N we have
N
N
3
k=l
rkFk@n,k
=
N
x
2 Cmrk@m,k'h,k
m=l
k=l
N
=
2
m = l cm{k$l
rk@m,k@n,k
N
where we have used the fact t h a t
N
3 rk@m,k@n,k =
k=l
by Problem 5.39.
We thus have
mS;n
N
cn
=
k=l
rkFk&,k
N
3 rk@?t,k
k=l
which i s the same as the required result on replacing n by m.
For the case where the eigenfunctions may be complex see Problem 5.129.
180
DIFFERENCE EQUATIONS
(a)
[CHAP. 5
In order to understand the limitations for the expansion in (a) it is useful to use the analogy
of vectors in 3 dimensions a s indicated on page 159. In 3 dimensions i, j, k represent unit
vectors. If we wish to expand an arbitrary 3-dimensional vector in terms of i, j, k we seek
constants A,, A2, As such that
A = Ali+Aj+Ask
Al = A - i ,
We find
A, = A * j ,
A, = A * k
as the solution of this problem.
Now if the only vectors given t o us were i and j [with k not present] it would not be
possible to expand the general 3-dimensional vector A in terms of i and j, i.e. we could not have
A = A,i+Azj
We say in such case that the set of unit vectors i and j are not a complete set since we have
in fact omitted k.
In an analogous manner i t is necessary for us to have a complete set of eigenfunctions
in order to be able to write the expansion in (a).
5.42.
Illustrate the results of Problems 5.39-5.41 by considering the Sturm-Liouville system
A’Yk-1
+
= 0,
XYk
YO = 0,
= 0
YN+1
The system is a special case of equations (38) and ( 3 4 ) on page 168 with
pk-1
= 1, qk = 0,
Tk
= 1,
“0
= 1,
= 0,
“1
“N
= 0,
“Ni-1
=1
The equation can be written
?dk+l
Letting
yk
= rk we find that
+ yk-1
- (2-1)yk
= 0
- (2 - 1)r + 1 = 0
2 - h -C q ( 2 - k p - 4
=
r2
so that
r
2
It is convenient to write
=
2-A
so that (2) becomes
= cos B
r
2cose
* i sine
Then the general solution of the equation is
yk
=
01 COS
ke
+
C2
sin kO
(5)
From the first boundary condition yo = 0 we have c1 = 0 so that
yk
= c2 sin ke
From the second boundary condition y N t l = 0 we have since
C,
Z0
sin ( N - t 1)s = 0
(7)
From this we h a d
( N + l ) e = nr
or
6
=
nn
N S 1
n = 1,2,3,.
where we have omitted n = 0 since this leads to the trivial solution yk = 0 and we have omitted
= -1, -2, -3, . since these lead to essentially the same solutions as n = 1 , 2 , 3 , . . . From (3)
and ( 8 ) we see that
n
..
n = 1,2,3, ...
CHAP. 61
181
DIFFERENCE EQUATIONS
Since these values start to repeat after n = N , there will be only N different values given by
= 4 sin2-
A,,
n7;
2(N 1)
n = 1,2, . . .
+
These represent the eigenvalues of the Sturm-Liouville system. The corresponding eigenfunctions
are obtained from (6) a s
c2 sin
kna
N+1
From the orthogonality property of Problem 5.39 we would have
N
B
k=l
sin
krn%
kna
sin= 0
N + 1
N S 1
m#n
~
which can in fact be shown directly [see Problem 5.131(a)].
Similarly if we are given a function Fk we can obtain the expansion
N
where
krnp
c, sin N + l
B
m=l
=
=
c,
These can also be shown directly [see Problem 5.131(b),(c)].
NONLINEAR DIFFERENCE EQUATIONS
5.43. Solve y,,, - y, ky,+,y, = 0,
y1 = 2.
+
Dividing by
Ykl/k+l,
the equation can be written as
1
1
Yk+l
Then letting
=
k
Yk
= wk the difference equation to be solved is
= k,
'vk+l-wk
v1 =
4
The solution to the linear difference equation in (2) is
and using the fact that w1 =
4 we find
Wk
=
c1
= &.
Thus
-k (+k2-- 1)
k2-kkl
2
1 =
2
and the required solution is
2
= kZ-k+1
5.44.
Solve y,+,yi = yE+l if
y1 = 1, y z = 2.
Taking logarithms of the given difference equation it becomes
lnyk+2
- 3 l n y k + 1 + 21nYk =
Letting
Vk
=
0
h g k
the equation (1) can be written a s a linear difference equation
wk+2
- 32)k+l
+ 2vk
= 0,
211
The general solution t o the difference equation in (3) is
wk
=
c1
+
CZZk
= 0,
VUg
= In 2
182
DIFFERENCE EQUATIONS
[CHAP. 5
Using the conditions in (3) we find c1 = -In 2, c2 = 4 In 2 so that
= (In 2)(2k--1- 1)
'vk
The required solution of the given difference equation is thus
In yk = (In 2)(2k--1- 1)
(ln2)W-l
or
yk = e
which can be written
yk = 2
- 1)
1
pk-1-
SIMULTANEOUS DIFFERENCE EQUATIONS
5.45.
Solve the system
38,
= k
- 3yk
8 k + l +'k
+
-
52, =
subject to the conditions y1 = 2,z, = 0.
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 - 4k
($1
3yk f (E-5)Zk = 4k
Subtracting the equations in ( 2 ) yields
(E2
- 8E + 12)yk = 1 - 4k
- 4k
The general solution of (3) found by any of the usual methods is
=
uk
19
+ ~26k+ 1 dk - -k45 - 26
C1zk
and so from the first equation in (1)
zk = c12k
-
Putting k = 1 in ( 4 ) and ( 5 ) we have since
+
2 ~ 1 6 ~ 2=
3c26k
111
- -41
= 2,
64
25 '
z1
4k
34
- -36 k - 26
=0
201 - 18c2 =
74
25
from which c1 = 133/100, c2 = -1/60.
Thus the required solution is
19
- -54 k - 25
133
-2k
100
1
- -6k
+
133
zk = -2k
100
1
+ -6k
34
- 4k-l - -3k - -
uk =
60
20
4k-l
5
25
MIXED DIFFERENCE EQUATIONS
5.46. Solve the diff erential-diff erence equation
y ; + ~ ( ~=
)
= t>
yk(t)t
Z/k(O)
=
Putting k = 0 we find
y;(t)
=
yo(t)
= t
Then since ~ ~ (=0 1,
) we have c1 = 1 so that
or
yl(t) =
Ftz + c1
CHAP. 51
DIFFERENCE EQUATIONS
183
Putting k = 1 we find
d(t) =
yl(t) = 1
t2
+2
or
y2(t)
= t
+ 3t3 + c2
Then since yz(0) = 2, c2 = 2 so that
yz(t) = 2
t3
+t+3!
Continuing in this manner we find
y3(t) = 3
+ 2t + +
2!
y4(t) = 4
2tz
t5
+ 3t + ++2!
3!
5!
y5(t)
t4
t2
t3
= 5 + 4 t + -3t2+ 3 +2t3+ + gt 4
t6
2!
and in general
y,(t)
= n
(n- 2)t2 + (n- 3)t3 + ...
tn-1
+ (n-l)t + +-+2!
3!
(n-l)!
which is the required solution.
5.47.
Solve Problem 5,46 by using the method of generating functions.
Let the generating function of y k ( t ) be
00
3
y ( s , t, =
Yk(t)sk
k=O
Then from the difference equation,
which can be written in terms of (1) a s
;;
o r since yo(t) = t
--
SY
= 1
Putting t = 0 in ( 1 ) and using y k ( 0 ) = k yields
Lo
Y(s,O) =
3
k=O
ksk
Equation ( 4 ) can be written as
i a
--In
s at
( s Y + 1) = 1
or
l n ( s Y + 1) = s t
ceSt
-1
Y ( s ,t ) = -
so that
6
Using ( 5 ) and (6) we find
Y ( s ,t ) =
A{&
8
(1
+
k=O k s k + l )
-
l}
+ c1
tntl
(n+l)!
184
DIFFERENCE EQUATIONS
Since the coefficfent of
8n
[CHAP. 6
in this expansion is y n ( t ) we find i t t o be
(8)
This i s obtained by taking the term corresponding to p = n
p k 1 = n 1 in the second summation of (7).
+ +
+
+ 1 in the first summation of
(7) and
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(z, y I ) = 0
the operators El, E2 and ( b ) in terms of subscript notation.
5.49.
Solve the difference equation of Problem 5.48.
We can write equation
(a) in terms of
( 2 ) of Problem 6.48 as
E1uk,m
=
(11
4EZUk,m
Considering m as fixed (1)can be thought of as a first order linear difference equation with constant
coefficients in the single variable k. Calling U k , m = u k , 4E2 = a the equation becomes
Euk = auk
(8)
U k = akC
(8)
A solution of (2) is
where C is a n arbitrary constant a s f a r a s k is concerned but may depend on m, i.e. it is an
arbitrary function of m. We can thus write ( 8 ) a s
uk
Restoring the former notation, i.e.
uk
= akCm
(4)
= uk,m, a = 4 E 2 , ( 4 ) becomes
Uk,m
kc,
=
But this can be written as
Uk,m
= 4kE:Cm = 4 k C m + k
(5)
To obtain this in terms of x and y we use the fact that x = a
becomes
u(x,y)
=
4(z--a)lhC
(
where
H is an arbitrary function.
y =b
+ ml.
Then ( 5 )
c+y-6)
h
which can be written
u(s,y)
+ kh,
= 4*/hH f'
(h
1
J
We can show that (6) satisfies the equation of Problem 5.48.
arbitrary function it is the general solution.
(6)
Then since it involves one
CHAP. 51
5.50.
DIFFERENCE EQUATIONS
Work Problem 5.49 by finding solutions of the form
constants.
186
Akpm
where A,
p
are arbitrary
Substituting the assumed solution
Uk,m = Akam
in the given difference equation ( 2 ) of Problem 5.48 we find
),k+lpm-
= 0
4),kpm+l
A - 4p = 0
or
on dividing by hkpm # 0.
From A = 4p it follows that a solution is given by
Akpm
= (4p)kpm = 4kpk+m
Since sums o f these solutions over any values of p are also solutions, we are led t o the solutions
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) = 4 W ( z + y )
(1)
Using the boundary condition u ( 0 , y) = y2 we have from (1)
U(0,Y) = H(y) =
?d2
H ( x + y ) = (x+y)2
Thus
and the required solution is
u ( x , y ) = 4=(x+y)2
5.52.
+
Solve u(x 1, y) - 2u(x, y
(8)
+ 1) = 3u(x,y).
We can write the equation in subscript notation as
EIUk,m-
~ E Z U =
~ 3Uk,m
, ~
or
(Ei-2E2-3)uk,m
= 0
Rewriting it as
ElUk,m
=
(2EZ+
3)Uk.m
we can consider i t as a first order linear difference equation with constant coefficients as in Problem 5.49 having solution
u k , m = (2Ez
3)kCm
+
This can be written a s
Uk,m
= 3k(l++Ez)kCm
or on expanding by the binomial theorem,
+ $kCm+l + $ k ( k - l ) C m + z + & k ( k - l ) ( k - 2 ) C m + , +
To obtain the result in terms of x and y we can use x = a + k h , y = b + ml,
= 3k[C,
1 = 1, a = 0, b = 0 so that k = x, m = y.
U ( X , 2/)
= 35[H(y)
a * * ]
choosing h = 1,
Then writing Cm = H ( y ) we obtain
+ Q x H ( y + 1) + $X(X - 1)H(y+ 2 ) + &X(X
-l
This contains one arbitrary function H ( y ) and is the general solution.
) (- 2~) H ( y
+3) +
* *
a ]
188
DIFFERENCE EQUATIONS
[CHAP. 5
MISCELLANEOUS PROBLEMS
5.56. Solve the partial differential equation
dU
dU
dX
dY
-+2-
0,
=
U ( X ,0)
The given partial differential equation is the limit as h
u(x+ h,
= x3 - 3 sinx
-+
0, 2
-+
0 of
- u(z, 7d)
h
which can be written in operator and subscript notation a s
(ZE,+ 2 h E 2 - 1 - 2 h ) u k , , = 0
(
E ~ . L I=
; , ~ 14--- - E 2
or
)
Uk,m
This can be solved a s in Problem 5.49 t o yield
or in terms of x and g a s
(1
u(x,y) =
+ 2h
~ - T
2h E ~'Ih) H ( Y )
In order to obtain meaningful results as h
0, I
-+
+
0 choose 1 2h/l = 0 so that (1) becomes
= H(+ y - 2x
u(x,y)
where J is an arbitrary function.
-+
= 4 2 %- y)
The required general solution is thus
u(x,v) = J ( 2 x - v )
u ( ~0),
Since
J(x) =
we have
Then from (2) we have
&x3
x3
- 3 sin x
- 3 sin (xl2)
x - Y\
= 3 2 % -y)s - 3 sin ( 2-
u(x,y)
5.57.
= J(2x) =
\
2
)
Solve the mixed difference equation
+
u(x 1, y) - u(x, y)
=
dU
2-
dY
The given equation can be written as
= 2DZu
E1u-u
(1)
E,u = ( 1 + 2 D 2 ) u
or
(2)
Then treating l + 2 D 2 as a constant we find a s in Problem 5.49
= ( 1 + 2D,)"H(v)
2@,2/)
where H ( y ) is an arbitrary function of y.
(3)
Expanding by the binomial theorem we then find
Then if x is a positive integer the series terminates and we have
u(x,u) = H ( y )
+2
(3
H'(y)
+
22
(i)
H"(y)
+ - + 2"
* *
H(")(y)
(4)
CHAP. 61
DIFFERENCE EQUATIONS
189
5.58. Solve the preceding problem subject to u(0,y) = y3.
From ( 4 ) of Problem 6.67 we see that
U(0,Y) = H(y) = Y3
Thus
H’(y) = 3y2, H”(y) = 6y,
H”’(y) = 6, H(Iv)(y) = 0,
,
and so
5.59.
Find the general solution of
=
uk,m
pUkt1,m-1
+
qUk-l,m+l
+ q = 1.
where p and q are given constants such that p
Method 1.
Assume a solution of the form
=
%k,m
Xkpm
Then by substitution in the given difference equation we find after dividing by
px2
- x p + qp2
A k p
= 0
from which
x =
P
*
m
-
P*Pdi=Gl
2P
=
2
p,clQ
P
Thus solutions are given by p k * pm = p k t m and ( p q / p ) k p m = p k t m ( q / p ) k . Since sums of such
solutions are also solutions, i t folIows that F ( k + m ) and (q/p)lCG(k+m)are solutions and the
general solution is
%k,m = F ( k
m) ( q / p ) k G ( 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.
denote this by c we have k m = c. The equation is thus given by
+
Uk,c-k
=
PUktl.c-(ktl)
f
=
vk
If we
(2)
qUk-ll,c-(k-l)
If we let
Uk,c-k
(2) becomes
vk
=
p v k t 1 f qvk- 1
which involves only one subscript and can be solved by the usual methods.
vk
= c1
We find
+ CZ(q/I))‘
Now the constants c1 and c2 can be considered as arbitrary functions of the constant c = k
c1 = F ( k + m ) , c2 = G ( k + m ) . Thus we obtain from (8)and ( 5 )
Uk,m
+ + G(k + m)(q/p)k
= F ( k m)
(5)
+ m:
190
D I F F E R E N C E EQUATIONS
[CHAP. 5
Supplementary Problems
DIFFERENTIAL EQUATIONS AND RELATIONSHIPS TO DIFFERENCE EQUATIONS
5.60.
( a ) Show t h a t the general solution o f the differential equation
is y = c e ~ - x 2 - 2 x - 2 .
( b ) Find t h a t particular solution to the equation in ( a ) which satisfies the condition y(0) = 3.
5.61.
( a ) Show t h a t the equation
[compare Problem 5.601 where y = f(x), Ax = h can be written as
f(x
+ h) - (1 + h)f(x) =
hx2
( b ) Show t h a t the general solution of the equation in (a) is
y(x) = f(x) = C(x)(l hp’h - x2
where C(x
(c)
+
+ h) = C(x).
- h2x
+ ( h- 2)x - 2
Find t h a t particular solution f o r which y(0) = 3.
5.62.
Explain clearly the relationship between the difference equation and the differential equation o f
Problems 5.60 and- 5.61 discussing in particular the limiting case a s h + 0.
5.63.
( a ) Show t h a t the general solution of
3x2
i s y = c1 cos x
-
5x
+4
+ c2 sin x + 3x2 - 5%- 2.
( b ) Find t h a t particular solution to the equation in ( a ) which satisfies the conditions y(0) = 2,
Y’(0) = 0.
5.64.
(a) Show t h a t the equation
can be written as
f(x
“16 +
Ax2
16
= 3x(2) - 5x(1) + 4
+ 2h) - 2 f ( x+ h) + (1 + h2)f(X) =
3h222 - (3h3
+ 5h2)x + 4h2
where y = f(x).
( b ) Show t h a t the general solution o f the equation in ( a ) is
y = f(x) = Cl(X)(l+ ih)”/” C,(x)(l- ih)”/h
+
+ 3x2 - (3h + 5)x - 2
where Cl(x) and C2(x) a r e 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. (21) What a r e the boundary conditions f o r 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
Write the difference equation of ( a ) Problem 5.61 and ( b ) Problem 5.64 with subscript notation.
5.66.
5.67.
Write each of the following in subscript notation.
f(x 3h) - 3f(x 2h) 3f(x h) - f(x) = x(3) - 2392)
(a)
(b)
+
f(x+4h)
+
+
+ f(x) = cosx
+
CHAP. 51
5.68.
DIFFERENCE EQUATIONS
191
Change each of the following equations to one involving the “ x notation”.
(a)
(b)
2 Y k t 2 -3ldk+l+yk
= 0
yk+3 + 4Ykt2 - 52/k+l 211; = 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 ) 1, k, kz, k3, k4
( b ) k 4, k - 2, 2k 1
(f) k2 - 3k 2, k2 5k - 4, 2k2, 3k
( 0 ) k2 - 2k, 3k, 4
( 8 ) 2k cos ke, 2k sin ks
( d ) 2 5 225 23k
( h ) k2, (k 1 ) 2 , (k 2)2, (k 3)2
+
+
+
+
+
+
+
5.70.
Work Problem 5.69 using Theorem 5-1, page 154.
5.71.
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
5.82.
Find the general solution corresponding t o 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.
# b
show that
ak
and bk are linearly independent.
*
P
192
[CHAP. 5
DIFFERENCE EQUATIONS
5.86.
Solve
5.93.
Show that
&+I
- a u k = p for all values of the constants
coske
dE)
(a)
= Re
(b)
(Y
and p.
sinke = Im
{-I
where Re denotes “real part of” and Im denotes “imaginary part of”.
5.94.
Use Problem 5.93 t o solve yk+2
5.95.
(a) Determine E R ( k ) .
+ 4uktl - 122/k =
5 cos ( d 3 ) .
1
1
( b ) Show that E2- 6 E + 8
used for operators.
1
1
=L
i.e. the method of partial fractions can be
2 (
E y
4 -E
-2 ,
( c ) Show how the results of (a) and ( b ) can be used t o solve the equation
2/k+2
- 61/k+l + 816k = k2 + 3k
5.96.
Show how operator methods can be used t o solve
(E 1)2(E 2)(E - 3)uk = kz
5.98.
Solve Problem 5.86(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.
-
+
- 3k f 5 + 3k
METHOD OF VARIATION OF PARAMETERS
Solve each of the following difference equations by using the method of variation of parameters.
5.100.
(a)
uk+2-
(b)
vk+2
(c)
41dkt2-41dkt1
(d)
u k t 2 f ?dk
(e)
5Yk+2 - 3 Y k t l
(f)
Yk+2-1/ktl+16k
b k t l f 6?dk
- 2yk+l f
= k2
= 3
uk
+ uk
+ k + 4k
= 3/2k
=
- 2uk = 3k
= l/k!
+ (-2)k
CHAP. 51
193
DIFFERENCE EQUATIONS
5.101.
Solve Problem 6.86(a)-(i) by using variation of parameters.
5.102.
Solve Problem 5.92(a)-(It) by using variation of parameters.
5.103.
Solve
5.104.
Verify t h a t the determinant of the coefficients (24) on page 157 is equal to the Cmorati of the
system [see (Is), page 1541. Discuss the significance of the case where the determinant is or i s
not zero.
l/k + 3
- 676k + 2 + 1176, + 1 - 676k = 5 2k + kz by using variation of parameters.
METHOD OF GENERATING FUNCTIONS
5.105. Verify each of the following generating functions.
5.106.
Solve each of the following by t h e method of generating functions.
(a)
(b)
(c)
5.108.
+
(e)
676k+Z - 5yk+l + 76k =
76k+2
21/k+l f 76k = 0,
(f)
76k+3-376k+2
(d)
5.107.
- 62/k+i 8 Y k = 0, Yo = 0,
276k+l - 76k = 1, 760 = 0,
Y k t 2 + 476k+l
476k = 0, 760 = 1,
76kt2
+
+ 276k+l
(9)
Ykt2
(h)
476k+2 - 476k+l
Solve
gk+2
760
+ 376k+1 - Yk
+ 76k = +
+ 76k
= 2-',
+ 376k+1- 476k
=
= 2,
761
=2
761
=0
761
= -1
=
760
1
z,
= 0,
760
= 0,
761
=
761
= O.
Show how to obtain the generating function of the sequence (a) uk = k, ( b ) uk = k2,
Use Problem 6.105(a) and differentiate both sides with respect to t.]
(C)
uk = k3.
[Hint.
5.109.
If G(t) and H ( t ) a r e the generating functions of uk and vk respectively show t h a t G(t)H(t) is the
generating function of
k
u:vk =
Uk-pvp
g=O
which is called the convolution of uk and
Vp
194
D I F F E R E N C E EQUATIONS
[CHAP. 5
5.112.
5.113.
5.114.
Work Problem 5.113 if the right side is replaced by 2k, using variation of parameters.
5.115.
5.116.
5.117.
Let Yk
f;
0 be a n y solution of the difference equation
Y k + 2 + akyk+l + bkyk = 0
Show t h a t 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 t h a t if Yk is a solution of the equation
+
Y k + 2 + ak?dk+l
bkyk = 0
then the substitution yk = Ykuk can be used to solve the equation
Yk+2
5.119.
+ a k Y k + i + bkYk
= Rk
Use Problems 5.117 and 5.118 to solve the equations
( a ) kyk+2 (1- k)Yk+l - 276k = 0
[Hint. A solution of (a) is yk=k.]
(b)
h k + 2
f (l--k)Yk+l
- 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
5.122.
Solve
where
CY
ak+2ykt2 - (aak+l f bk+1)2/k+l+
i s a given constant.
+ kyk = k, yo = 0,
abkYk
=
yl = 1 by the method of generating functions.
5.123.
Solve yk+z
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- I)!'uk.
(b) Show t h a t lim yk = # e .
k-co
STURM-LIOUVILLE DIFFERENCE EQUATIONS
5.126. Find eigenvalues and eigenfunctions f o r the Sturm-Liouville system
A'Yk-1
+
= 0,
?i&
Y l = YO!
YN+l
=
UN
5.127.
Work Problem 5.126 if the boundary conditions a r e y1 = YO, yN+1 = 0.
5.128.
Write the orthogonality conditions f o r the eigenfunctions of
lem 5.127.
5.129.
(a) Show t h a t if the eigenfunctions of a Sturm-Liouville system a r e not necessarily real then the
(a) Problem 5.126
orthogonality of page 159 must be rewritten as
N
2 rk?m,k@n,k
k=l
(b) If
F k
= 2
Cm@m,k
=
determine the coefficients c,
0
mZn
f o r this case.
and
(b) Prob-
CHAP. 51
5.130.
D I F F E R E N C E EQUATIONS
195
Can any second order linear difference equation
+
+
akh2yk bkAyk Ckyk = 0
be written as a Sturm-Liouville difference equation? Justify your answer.
5.131.
Prove the results (a) ( l o ) , ( b ) ( 2 1 ) and (c) (12) on page 181 directly without using Sturm-Liouville theory.
5.132.
Explain how you could solve the equation A2yk-1
relationship with Sturm-Liouville theory.
5.136.
Solve. yk yk + 1 yk + 2 = K(Yk
+ yk + + yk +
1
2)
+
Xldk
= k if YO = 0,
Y N + ~=
0. Discuss the
where K is a given constant.
SIMULTANEOUS DIFFERENCE EQUATIONS
5.146.
Solve the system of equations
5.147.
Solve Problem 5.146 if 1- p
5.148.
Solve
2ukt.l
uk+2
+
vk+2 =
- zwk+1
=
uk+l = P U k + qwk
% + 1 = (1(1- q)vk
+
if 1- p
+ q # 0.
+ q = 0.
16vk
4uk
MIXED DIFFERENCE EQUATIONS
5.149. Solve ~ ; + ~ =
( tg )k ( t ) subject to the conditions y ~ ( t =
) 1, Yk(0) = 1.
5.150.
(a) Determine lim yk(t) f o r Problem 5.149,
k-rm
the equation?
Explain.
( 6 ) Could this limit have been found without solving
196
5.151.
D I F F E R E N C E EQUATIONS
y k ( t ) = yk+l(t) subject to the conditions yo($)= 1, &(o) = 1.
Discuss the relationship of this problem with Problems 5.149 and 5.150.
(a) Solve
(c)
5.152.
[CHAP. 5
( b ) Find lim yk(t).
k r m
+
( a ) Solve y:+,(t)
&(t) = 0 subject to the conditions y & t ) = 1, yi ( t )= 0, yk(0) = 1.
cuss the limiting case of ( a ) as k + m.
( 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)
5.155.
Solve
u(x, y
5.156.
Solve
+ 1) + 2u(y, x + I) = 3xy - 4x + 2.
u ( x + h, y) - 3u(x, y + k ) = x + y.
5.157.
Solve
( E l E 2- 2)(E1- E2)u = 0.
5.158.
Solve
Uk+l,m+z
5.159.
Sohe
ukfl,m+l - uk+l,m
- uk,m =
= 0, u ( 0 , y ) = 3y
+ 2.
2k-m.
-
if
uk,m
uk,O =
0
1
k>O
k=O’
t
MISCELLANEOUS PROBLEMS
5.164. Show t h a t the general solution of
yk+2 - 2(cos a)yk+l
is given by
yk
where i t is assumed t h a t a
5.165.
5.166.
5.167.
# 0,
ktn,
= Rk
k
+ c2sin ka + 2
*2n,. . . .
c1cos ka
=
+ yk
R , sin ( k - p ) a
sine!
p=1
Obtain the general solution in Problem 6.164 in the case a = 0, kz-, &2a,
Uki.2
+ 3?4k+,+ 2yk = 6.
. .. .
Prove t h a t if lim l/k exists then i t must be equal to 1.
Express the differential equation
k-kw
y”
- xu’-
2y = 0
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).
5.168.
Solve the difference equation
+
!
!
I
! P(x)y = Q(x)
Ax
By finding the limit of the results in (a) as Ax = h + 0 show how to solve the differential
equation
P(x)y = Q(x)
dx
*+
CHAP. 51
5.169.
D I F F E R E N C E EQUATIONS
197
( a ) Solve the difference equation
6216-4i%+3y
= x + 2
Ax2
Ax
( b ) Use an appropriate limiting procedure in ( a ) to solve the differential equation
dzll
dx2
5.170.
4a
dx + 3y
+2
Obtain the solution of the differential equation
dzv + 2 d y +
dxz
by using difference equation methods.
5.171.
= x
y
dx
= e-3
Let f ( x , y , c) = 0 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 = 0x2 and discuss the geometric significance.
( b ) Explain how you might find the difference equation of a family and illustrate by using y = cx2.
( c ) Give a geometric interpretation of the results in (b).
5.172. Work Problem 6.171 for y = cx
+ cz.
5.173.
Generalize Problem 5.171 to two-parameter families of curves f ( x ,y, cl, c2) = 0 and illustrate by
using y = claz c2bx where a and b a r e given constants.
5.174.
( a ) Show t h a t two solutions of
+
(2)
2
a r e y = cx
+ c2
and y
y
= x&
Ax+
= c2 - i x z + &h2 - +ch(-l)*/h.
( b ) Discuss the relationship of your results in (a) with those of Problem 5.172,
5.175.
( a ) Show t h a t two solutions of the differential equation
( b ) By obtaining the graph of the solutions in ( a ) discuss the relationship of these solutions,
(c) Explain the relationship between the solutions in p a r t ( a ) and Problem 5.172.
5.176.
The difference equation
uk
=khk
+F(Auk)
y
or in “ x notation”
= x z
+ F($
is called Clairaut’s diference equation by analogy with Clairaut’s differential equation
(a\ Solve Clairaut’s differential equation by differentiating both sides of i t with respect to x.
Illustrate by using the equation
2
>
,
and explain the relationship with Problem 6.175.
( b ) Is there a n analogous method of obtaining the solution of Clairaut’s difference equation?
cuss with reference to Problem 6.174.
Dis-
198
5.179.
DIFFERENCE EQUATIONS
[CHAP. 6
Show that the nonlinear equation
yktlvk
+ Akyk + BkZ/ktl
=
c k
can be reduced to a linear equation by the substitution
liktl
= -
yk
Ak
lik
Illustrate by solving Problem 6.134.
5.180. (a) Show that if q, a2, as are constants then the equation
yk+3
+ alPkl/k+2 + a2PkPk-lvk+l + a3PkPk-lPk-Zyk
=
Rk
can be reduced to one with constant coefficients by means of the substitution
z/k
=
P1P2' ' *Pk-Suk
( b ) What is the corresponding result for first and second order equations?
What is the result for
nth order equations?
5.184.
Solve
for k
5.186.
Solve
5.187.
Solve
Uk,m+l
> m.
au
(a) ax
- muk,m
au
+2=
ay
+ ((1
Y(x)
5.188. Solve
-
where u
au
( b ) 2= aY
ax
4,
-yE)(1-
~ = , 0 ~ for
+x -y
y ! + ~= cos (4a/k)
Y(xf1)+ Y
(x+2)
..,
+l!
2!
+
=
Solve
5.191.
Solve the partial differential equations
X.
1.
+
Solve the partial differential equations
(4
(b)
8%
- 3-axau
+ 2-au2
a2U
2u
- 4- aa@
= x
ax2
a2U
= 0,
+ 21,
U(X,O)
= $2,
and
subject to the condition y1 = 0.
( a ) 2 -au - 3-au = 0,
u(O,z/) = y2+ e-v
ax
ay
au
au
( b ) - 3= x - 2y,
u(x,O) = $ 2 + x + 1
ax
ay
5.192.
...
Uk,m
by using difference equation methods.
= x'-4x+
5.189.
m = 1, 2, 3,
U(0,Y)
u ( x , O ) = 0, u ( 0 , ~=)
= 76
= 0
Applicutims of
S
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 a t equal intervals h by N particles of equal mass rn 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
rn
Fig. 6-1
where ?Jk denotes the displacement of the kth particle from PQ and T is the tension in the
string assumed constant.
Assuming that ?jk = A k cos (,t a ) so that the particles all vibrate with frequency
2 ~ we/ obtain
~
the difference equation
+
0
This leads to a set of natural frequencies for the string given by
Each of these corresponds to a particular mode of vibration sometimes called natural
principal modes. Any complex vibration is a combination of these modes and
involves more than one of the natural frequencies.
In a similar manner we can formulate and solve other problems involving vibrating systems. See Problems 6.36 and 6.39.
modes o r
199
200
APPLICATIONS O F D I F F E R E N C E 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 t o a generator
supplying voltage V . The problem is t o find the current and voltage through any of the
resistors.
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.
1. The algebraic sum of the currents a t 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
Kirchhoff’s laws.
c
dl
Voltage drop across an inductor of inductance L = L z
APPLICATIONS TO BEAMS
Suppose that we have a continuous
uniform beam which is simply supported
a t points which are equidistant as indicated in Fig. 6-3. If there is no load
acting between the supports we can show
that the bending moments at these successive supports satisfy the equation
Mk-1
+ 4Mk +
h
_I_
1-
h
Fig. 6-3
Mktl
= 0
(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 6.6 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.
CHAP. 61
APPLICATIONS O F DIFFERENCE EQUATIONS
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 V I Z and V I ; are the relative velocities of the first sphere with
respect to the second before and after impact respectively then
viz = -EVIL
(5)
called the coefficient o f restitution, is generally taken as
where C,
a constant between 0 and 1. If = 0 the collision is called
perfectly inelastic while if E = 1 the collision is called perfectly
elastic.
Another important principle which is used is the principle o f conservation o f 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 A1 and A Z then the probabilities of the events are indicated by P(A1) and P(A2) respectively. It is convenient to denote the event that either A1
or A Z occurs o r b o t h occur by A1 Az. Thus P(A1 -tAz) is the probability that a t least one
of the events A I ,A2 occurs.
+
We denote the event that both A1 and A Z occur by A1A2 so that P(AiA2) is the probability that both A1 and A Z occur.
We denote by P(AzI A1)the probability that Az occurs g i v e n t h a t A1 h a s occurred. This
is often called the conditional probabilitp.
An important result is that
P(AiAz) = P(Ai)P(Az1 Ai)
In words this states that the probability of both A1 and A Z occurring is the same as the
probability that A1 occurs multiplied by the probability that A Z occurs given that A1 is
known to have occurred.
Now i t may happen that P(Az1 AI) = P(A2). In such case the fact that A1 has occurred
is irrelevant and we say that the events are independent. Then (6) becomes
P(A1Ar) = P(A1)P(Az)
We also have
+
A I ,AZindependent
(7)
+
P(Ai Az) = P(Ai) P(A2) - P(A1-42)
i.e. the probability that a t least one of the events occurs is equal to the probability that
A 1 occurs plus the probability that A z occurs minus the probability that both occur.
It may happen that both A1 and A2 cannot occur simultaneously. In such case we
say that A1 and A2 are mutually exclusive events and P(AiA2)= 0. Then (8) becomes
A I ,A2 mutually exclusive
P(A1 + A 4 = P(A1) + P(Az)
(9)
202
APPLICATIONS O F 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-I
Fo = 0,
where
4-Fk--2
(10)
Fi = 1
(11)
It follows that the numbers are members of the sequence
0 , 1 , 1 , 2 , 3 , 5 , 8 , 1 3 , 2 1 , ...
where each number after the second is the sum of the two preceding ones. For a table of
Fibonacci numbers see Appendix G, page 238.
We can show [see Problem 6.131 that the kth Fibonacci number in the sequence (12)is
These numbers have many interesting and remarkable properties which are considered in the problems.
MISCELLANEOUS APPLICATIONS
Difference equations cail be used in solving various special problems. Among these are
1. 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.
Solved Problems
APPLICATIONS TO VIBRATING SYSTEMS
6.1. A string of negligible mass is stretched between two points P and Q and is loaded a t
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
string which we shall take as constant.
T
the tension in the
APPLICATIONS O F DIFFERENCE EQUATIONS
CHAP. 61
203
The vertical force acting on the kth particle due to the ( k - 11th particle is
(1
and since we assume that the displacements are small compared with h, we can replace the
result in ( I ) by
- T(Yk - Yk-1)
h
to a high degree of approximation.
Similarly the vertical force on the kth particle due t o the ( k
a high degree of approximation by
I
r(Yk
_
+ 1)th particle is given t o
- Yk i1)
(3)
h
The total vertical force on the kth particle is thus
Assuming that there are no other forces acting on the kth particle we see that by Newton's law
the equation of motion is
k! -?2d
=
dt2
r
(5)
2Yk f Y k i 1)
k(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 (Ot -I-CY)
(6)
i.e. assume that all the particles vibrate with the same frequency d 2 r . Then (5) becomes
Now we can consider the points P and Q a s fixed particles whose displacements are zero.
Thus we have
A0
0, AN+l = 0
(9)
Letting A k = ~k in (8) we find that
(E+)Y+l
r2+
= 0
(10 )
or
We can simplify (11) if we let
2
so that
r
- 2cose
- -mw2h
7
= cose
I
i sine
In this case we find that the general solution is
Ak = c1 coske
+ c2 sin ke
Using the first of conditions (9) we find c1 = 0 so that
A k
= C2 sinks
.
(13)
204
APPLICATIONS O F D I F F E R E N C E EQUATIONS
[CHAP. 6
From the second of conditions (9) we must have
or since c2 # 0,
Thus (12) yields
from which
Denoting the N different values of
then follows t h a t
w9
corresponding to n = 1,2,3,
= f i s i n n7rm
w,
n=1,2,
. . .,N
by
w1,02,w3,
...,N
. . ., WN i t
(20)
Thus the natural frequencies a r e
f,
6.2.
=
'6
nr
sin- 2(LV 1)
2Ti
n = 1 , 2 , . . .,N
+-
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 w,,
i.e. the n t h mode of vibration. In this case we see from Problem 6.1 t h a t the displacement of
the kth particle is given by
Yk,n = A k , n
(Writ an)
(1)
+
where
nl;
c, sin N+1
Ak,n
the additional subscript n being used for
mode of vibration.
W,
(2)
a and c to emphasize t h a t the result is true for the nth
Now since the difference equation is linear, sums of solutions a r e also solutions so that we have
f o r the general displacement of the kth particle
N
yk =
2
n=t
+
nki;
c, sin -cos (ant a,)
N + l
(3)
Since this satisfies the boundary conditions at the ends of the string, i.e. yo = 0, Y N + =
~ 0, it
represents the required displacement.
The 2 N constants c,, a,, 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 a r e given by
= Nm,
L = (Ar+
~ ) J L
Then the natural frequencies a r e given by
or
f,
=
-L
2s
sin
n7r
CHAP. 61
APPLICATIONS O F DIFFERENCE EQUATIONS
205
Now for any particular value of n we have
[by using for example the fact that f o r small angles
the required natural frequencies are given by
f
:&
=
-
L
8,
sine is approximately equal t o el.
...
n=1,2,
Thus
(4)
If we let p denote the density in this case, then
-: -_ p
M = ~ L
or
(5)
and ( 4 ) becomes
n=1,2,
...
APPLICATIONS TO ELECTRICAL NETWORKS
6.4. An electrical network has the form shown in Fig. 6-4 where RI, RZ and R are given
constant resistances and V is the voltage supplied by a generator. (a) Use Kirchhoff's laws to e x p e s s 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,
ZlRl
For the kth loop,
+ (ZI-
22)Rz
+ (Ik-Ik+l)R2-
-V
=
0
(Ik-l-Ik)R2
For the ( N + 1)th loop,
INtlRl-
( b ) Equation
Letting
(a) can
Ik
be written a s
= r k in ( 4 ) it can be written a s
(zN-zNt1)R2
= 0
= 0
206
APPLICATIONS O F D I F F E R E N C E EQUATIONS
[CHAP. 6
Then the general solution of ( 4 ) is
I!, =
clr:
+ c2rt
where
From ( I ) ,
From (3),
Now if we put k = 2 and k = N in (6) we have
I, =
c l r : -I-
I N = clr:
c2ri
+ c2r:
from which
c1
=
z2rf
rIrf
- zNri
- r,"rT '
c2
= INrT - Z2ry
-rirr
ryr:
so t h a t
, a r e given by (10)and ( 1 1 ) .
where r l , r2 a r e given in (7) and I,, Z
APPLICATIONS TO BEAMS
6.5. Derive the equation of three moments
( 4 ) on page 200.
IZI
9:
h
Assume t h a t 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
distance from this origin we find t h a t the
bending moment a t x is given by
M,
M(x)
=
M,
where M,-l,
M,,, M,+
h
-I
M"
Fig. 6-5
+ (M,- M,-.l)
+ (M,+l-M,);
-h 5 x 5 0
05x5h
a r e the bending moments at the (n - l ) t h , nth and (n
+ 1)th supports.
Now we know from the theory of beams t h a t the curve of deflection of a beam is y(x) where
YZy" = M(x)
(2)
and YZ, 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 a n axis through the centroid.
r
Integrating (2) using ( 1 ) we find
YZy'
=
M,x
M,x
+ ( M n - M n - l ) X2c + ~1
+ (M,+1-Mn)c +
X2
c2
-h S x 5 0
(8
0
sx
Now since y' must be continuous a t x = 0, i t follows t h a t c1 = c2 = c.
5h
CHAP. 61
APPLICATIONS O F D I F F E R E N C E EQUATIONS
207
Integrating (3) we have similarly
Since the deflection g is zero at x = -h, 0, h we find froni ( 4 )
Addition of these equations leads to the result
M,-I
as required.
6.6.
+ 4M, + Mn+l
= 0
(7)
Suppose that the beam of page 200 is of infinite extent on the right and that a t the
distance h to the left of the first support there is a load W which acts on the beam
[see Fig. 6-61. Prove that the bending moment at the nth support is given by
Mn = (-I),+' Wh(2- $3 )"
Fig. 6-6
The bending moment at the distance h to the left of the first support is given by
(11
M, = -Wh
Also at infinity the bending moment should be zero so that
lim M , = 0
n-+m
Now from the difference equation
M,-1
+ 4M, + Mn+l
= 0
we have on making the substitution M , = r n
= 0
rn-1+4r,+m+l
or
r2+
4r+ 1 = 0
(4)
i.e.
Then the general solution of (3) is
M, = ~,(-2
+ fl)"+ c2(-2
+
- fi )"
Now since -2
fi is a number between -1 and 0 while -2
lows t h a t condition (2)will be satisfied if and only if c2 = 0.
M , = el(-2
Also from (1) we have c1 = -Wh
- fi is
approximately -3.732, i t folThen the solution ( 5 ) becomes
+ fi),
(6)
so t h a t
M , = -Wh(-2
+
fi)'L
= (-l)n+lWh(2 - 'v6)'
(7)
208
APPLICATIONS O F D I F F E R E N C E EQUATIONS
[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 f o r the speeds of the balls before
the kth impact occurs assuming that the table is smooth and the sides are perfectly
elastic.
that
Denote by Uk and V k the speeds of the balls before the kth impact occurs.
u k refers to the ball which is hit initially we have
s,
u1 =
Then assuming
= 0
v1
(1)
Since the sides a r e perfectly elastic, the speeds of the balls after the kth impact a r e U k t 1
Vk+l respectively.
Then the total momentum before and after the kth impact a r e given by
m u k mVk and m u k + - mVk+ respectively. Thus by the conservation of momentum we have
and
+
+ mvk
=
mUk+l
Uk f V k
=
Uk+l
mUk
or
- mVk+l
- Vk+l
(2)
I n addition we have by Newton’s collision rule
=
Uk+l f Vk+l
where
E
(3)
-E(Uk-Vk)
is the coefficient of restitution between the two billiard balls.
The required equations a r e given by (2) and (3) which a r e 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 ) u k
- (E
(EfE)Uk
+
= 0
(2)
+ 1 and adding we obtain
(E2+e)Uk
Uk
(1)
(E-E)Vk
Then operating on ( 1 ) with E - E , on (2) with E
Letting
1)Vk = O
= 0
(3)
= r k , (3)yields
= 0, r = & i f i =
r2+,
fie..i/2
so t h a t the general solution of (3)i s
Uk =
By eliminating
Vk+l
,k/2(AerkI/2
+ Be-rkil2)
=
&/2
(
c1 cos- T k
2
+ c2 sin -
(4)
between equations (2)and (3) of Problem 6.7 we find
(Ef1)Vk
=
2Uk+l
+
(5)
(E-1)Uk
Then using ( 4 )
vk
=
E + 1
[{(e
- 1)c2 - 2~
ak
cl) sin 2
+ { ( e - 1)cl + 2
6 c2> cos -
(6)
From conditions ( 1 ) of Problem 6.7 we have
so t h a t
c1
+ c2 sin -;)
\
/
fi (cl
=
cos 2
?r
- 1)s
2E
(E
c2
=
s
s
= -
fi
(7)
(9)
CHAP. 61
APPLICATIONS O F D I F F E R E N C E EQUATIONS
209
Using these in ( 4 ) and (6) we obtain the required results
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. Suppose 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 o r the other has M
pennies. Formulate (a) a difference equation and ( b ) boundary conditions for the
probability that A wins the game.
+
+
Let u k be the probability t h a t A wins the game when he has k pennies.
one of the following two mutually exclusive events occurs:
Now A will win if
1. On the next step A wins a penny and then wins the game.
2. On the next step A loses a penny and then wins the game.
The probability of the first event occurring is given by
(1)
puk+ 1
since if A wins on the next step [for which the probability is p ] he has k
then win the game with probability u k + l .
+ 1 pennies and can
Similarly the probability of the second event occurring is given by
(2)
quk-l
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 U k - 1 .
It follows t h a t the probability of A winning the game is the sum of the probabilities
( 1 ) and ( 2 ) , i.e.
uk
=
I)uk+l
+ qZLk-1
(3)
To determine boundary conditions for uk we note t h a t A wins the game when he has M
pennies, i.e.
uM = 1
(4)
Similarly A loses the game when B has M pennies and thus A has N - M pennies, i.e.
uN-M
= 0
(5)
6.10. Obtain the probability that A wins the game of Problem 6.9 if (a) p i
q, ( b ) p = q = 4.
W e must solve the difference equation ( 3 ) of Problem 6.9 subject t o the conditions ( 4 ) and ( 5 ) .
To do this we proceed according to the usual methods of Chapter 5 by letting u k = y k in equation
(8) of Problem 6.9. Then we obtain
prz--.+q
= 0
(1)
from which
r
( a ) If p
=
1kd2P
-
1kd1-4p+4p2
213
# q, the roots of ( 1 ) a r e 1 and
-
q l p so t h a t
uk
=
c1
C2(q/p)k
1 2 (1-2P)
2P
= 1, q / p
(2)
210
APPLICATIONS OF DIFFERENCE EQUATIONS
[CHAP. G
Using the condition ( 4 ) of Problem 6.9, i.e. uM = 1, we have
1 =
+ cz(q/p)M
c1
~ = 0,
- we
~ have
Similarly using condition (5) of Problem 6.9, i.e. u
0
=
c1
+ cz(df3)N-M
Then solving ( 4 ) and (5) simultaneously we obtain
and so from (3) we have
Since A starts with k = a pennies, the probability that he will win the game is
while the probability that he loses the game is
( b ) If p = q = 8, the roots of (1) are 1, 1, i.e. they are repeated.
equation (3) of Problem 6.9 is
Uk = c3
c4k
+
Then using uM= 1, u
(10)
~ = 0,
- we
~ obtain
~3
i.e.
In this case the solution of
+ c4M
= 1,
CS
+ c ~ ( N- M )
M - N
2M-N'
= 0
-- 1
cs = ___
c4-
2M-N
Substituting in (10)
uk
=
M-N+k
2M-N
Thus the probability that A wins the game is
u, =
M - N + a
2M-N
while the probability that he loses the game is
1-u,
M-a
= 2M-N
~
6.11. 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 f 4, ( b ) p = Q = 1/2.
( a ) Putting M = N in equation (9) of Problem 6.10 we find for the probability of A being ruined
( b ) Putting M = N in equation (14) of Problem 6.10 we find for the probability of A being ruined
1-u,
6
N-a - = N
a f b
(2)
CHAP. 61
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 t o the initial capital of B. I t 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 a t 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 t o 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 u k , m 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
(11
PUk--l,m
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 ~ l ~ - ~ , ~ ] .
Similarly the probability of the second event occurring is
(2)
qUk,m--l
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 t o win [for which the probability is U k , m - l ] .
It follows that the probability of A winning the game is the sum of the probabilities ( 1 ) and
(a), i.e.
Uk,m
=
Puk--l,m
+ qUk,m--l
To determine boundary conditions f o r U k , , , we note that if k = 0 while m
no more points to win, then he has already won. Thus
ug,m
Also if k
>0
= 1
m> 0
(8)
> 0, i.e. if A needs
(4)
while m = 0, then B has already won so that
uk,o
= 0
k
>0
(5)
The solution of the difference equation ( 8) subject t o conditions ( 4 ) and ( 5 ) has already been
found [see Problems 5.53 and 5.54, page 1861 and the result is
FIBONACCI NUMBERS
6.13. Prove that the kth Fibonacci number is given by
We must solve the difference equation
Fk
subject to the conditions
=
Fk-1
F” = 0 ,
+
Fk--2
F1 = 1
212
APPLICATIONS O F DIFFERENCE EQUATIONS
Letting
F k
[CHAP. 6
= rk in (1) we obtain the auxiliary equation
Then the general solution is
From the conditions (2)we find
= 0,
c,+c,
(l+fi)C~+(l-fi)c,
Solving ( 5 ) we find
c1 =
and so from ( 4 ) we have
6.14.
Provethat
lim
k-tco
Fk+l
Fk
=
c2
6’
L[(T)k
1+fi
fiL
=
F k
1
-
-
= 2
--1
fi
(!5.d!)k]
= l+fi
2
-
We have using Problem 6.13
Then since
I (1- fi )/(1+fi) I < 1 we have
This limit, which is equal to 1.618033988.. . , is often called the golden mean. I t is supposed to
represent the ratio of the sides of a rectangle which is “most pleasing” t o the eye.
MISCELLANEOUS APPLICATIONS
6.15. Evaluate the integral
sk
=
cl:
= cos kd - cos ka
cose - cosa do
Let a, b, c be arbitrary constants and consider
ask+l
+ bSk +
Csk-1
=
7 U[COS
(k
+ 1)e - cos ( k + 1)a] de
cos e
+
- cos a
f b[cos ke - cos ka]
do
cos e - cos (Y
[(a
+ c ) cos e
c[cos ( k - 1)s - cos
+
i
+
+
i7
cos e
(k - l)a]
- cos a
de
b] cos ke
[(c - a) sin 81 sin ke
de
cos e - cos a
CHAP. 61
APPLICATIONS O F DIFFERENCE EQUATIONS
213
If we now choose a, b and c so that
b = -2acosa
c = a,
the last integral becomes
2a l r c o s k e de = 0
if k = 1,2,3,
...
It thus follows that
Sk+i-2(COSCX)Sk+Sk-l
From the definition of
Sk
k = l , 2 , 3 ,...
= 0
(1)
we have
so =
0,
s1
=
(2)
P
We must thus solve the difference equation ( I ) subject to the conditions (2).
According t o the usual methods for solving difference equations we let S k = r k in (1) and
obtain
r k + l - ~ ( C O Sa ) r k
rk-1 = 0
or TZ - 2(cosa)r 1 = 0
+
+
so that
2 cosa -r- d4 cos2a
2
r =
-4
-
cos a
f
i sin
Then by de Moivre's theorem,
= coska
rk
f
so that the general solution is
Sk
= c1 cos ka
Using the conditions ( 2 ) we have
c1
so that c., = &ina
= 0, c1 cosa
isinka
+ c2 sin ka
+ c2 sincr
=
T
and thus
which is the required value of the integral.
6.16. Find the value of the determinant
b a O O O
a b a O O
O a b a O
...
...
0 0 0
0 0 0
0 0 0
...
..............................
0 0 0 0 0 ... a b
0
0
0
0
0
...
a
O 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
- U'Ak-2
or
Ak
- bAk-1
+ U'Ak-2
= 0
Letting Ak = r k we obtain
yk
- b r k - 1 + a2yk-z
from which
r =
= 0
or
r2
b * d m
2
- br +
= 0
(11
214
APPLICATIONS O F D I F F E R E N C E EQUATIONS
’
It i s convenient to let
SO
t h a t ( 2 ) becomes
vk
= ak(cos ke
* i sin ke)
* ai sine
= a cose
T
Thus since
[CHAP. 6
we can write the general solution as
Ak =
U k [ C 1 COS
ke
+
C2 Sin ke]
(5)
Now if k = 1, the determinant of order 1 has the value A , = b, while if k = 2, the determinant of order 2 has the value
Putting k = 2 in ( 5 ) we find t h a t
Rz
- bA1+
&A0 = 0
from which we see on putting A , = b, A2 = b2- a2 t h a t A . = 1. It follows t h a t we can solve
(I) subject to the conditions
A0 = 1, A1 = b
.
(6)
These values together with ( 5 ) lead to
6,
= 1,
c2
= b - u cos 6 - -cos 6
using (3) so t h a t ( 5 ) becorn&
Ak
=
ak [eoske
cOsel
+
=
sine
(7)
sin e
a sin e
+
ak sin (k 1)e
sin 8
Using the fact t h a t
e = cos-1(b/2a)
(9)
we can write the value of the determinant in the form
a k sin [ ( k 1)cos-1 (b/2a)]
Ak = sin [cos-1 (b/2a)]
+
Ak
or
The above results hold f o r Ibl
=
2&+1
sin [(k
+ 1)cos-1 ( b / 2 a ) ]
dzFT
< 21~1.
I n case jbl
> 21al
we must replace cos-1 by cosh-1.
F o r the case 161 = 21aj 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 f o r Pk.
( a ) Since the principal outstanding after the kth payment is Pk, i t follows t h a t the principal outstanding after the ( k + 1)st payment is Pk+,. 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+l = Pk
or
+ ipk - s
Pk+l-(I+i)Pk =
(1)
-s
The outstanding principal when k = 0 is the total debt A so that we have
Po = A
(2)
CHAP. 61
(a)
APPLICATIONS O F DIFFERENCE EQUATIONS
Solving the difference equation ( 1 ) we find
Pk
and using condition (2) we find c = A
6.18.
215
+
= c(1 +i)k S/i
- S / i so
that
In Problem 6.17 determine what sum S is t o be paid back each period so as to pay
back the debt in exactly n payments.
We must determine S so that P, = 0.
From Problem 6.17 this is seen to be equivalent t o
A ( l + i ) n - S [ (1
+ i)"
]
-1
r
= 0
1
S
or
The process of paying back a debt according t o (2) and Problem 6.17 is often called amortization.
The factor
- 1 - (1 + i ) p
"qi
-
(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) a s
6.19.
Given the differential equation
Show that if y(xk)= yk, y'(xk) = y; = f ( x k ,yk) then the differential equation is approximated by the difference equation
Yk+l
=
Yk
+ hy;
=
Yk
+ hf(xkJ
'Yk)
where h = Ax = x,,, - xk.
From the differential equation we have on integrating from x = x k t o
corresponding values of y are y k and y k + l respectively
%k+l
or
If we consider f ( x , y) t o be approximately constant between
as required.
x k and x k +
1,
( 2 ) becomes
for which the
216
6.20.
APPLICATIONS O F D I F F E R E N C E EQUATIONS
[CHAP. 6
(a) Use the method of Problem 6.19, sometimes called the Euler method, to find the
value of y corresponding t o x = 0.5 f o r the differential equation
dy
dx
= 2x
+ y,
4
v(0) = 1
( b ) Compare the result obtained in (a) with the exact value.
(a) W e have f o r initial conditions xo = 0, yo = 1. Then by Problem 6.19 with k = O , l , 2, . . .
we find
2/1 = Yo $- hfb0,YO)
762 = ?dl
hLf(X1,Yl)
(11
Y 3 = Y,
hf(X2,Yz)
etc.
+
+
We can use a convenient value f o r h, f o r example h = 0.1, and construct the table of
Fig. 6-7 using equations (1) with f(Xk, yk) = 2xk yk.
+
yk + 1 = Yk + h2/i
k
xk
2/k
76d = f(xk, yk)
0
0
1.0000
1.0000
1.1000
1
0.1
1.1000
1.3000
1.2300
0.2
1.2300
1.6300
1.3930
2
,
3
0.3
1.3930
1.9930
1.5923
4
0.4
1.5923
2.3923
1.8315
5
0.5
1.8315
Fig. 6-7
From the table we can see t h a t the value of y corresponding to x = 0.5 is 1.8315.
( b ) The differential equation can be written as
_
dy dx
= 2x
y
a first order linear differential equation having integrating factor e
by e-X we have
Thus on integrating,
e-xy
or
=
s
2xe-Xdx
y
= -2xe-5
= -2x - 2
Since Y = 1 when x = 0, we find c = 3.
-
2e-5
’
( - I ) dZ
(2)
= e-5. Multiplying
+c
+ cex
(3)
Thus the required solution is
y = 3e5- 2x
-2
(4)
Putting x = 0.5 we have for the required value of y
y = 3eo.5 - 2(0.5)
-2 =
3(1.6487) - 3 = 1.9461
(5)
It is seen t h a t the result i n (a) is not too accurate since the percent error is about 64%.
Better results can be obtained by using smaller values of h f o r example h = 0.05 [see
Problem 6.1011.
This method, also called the step by step method, is important because of its simplicity
and also theoretical value.
There a r e of course many methods which provide greater
accuracy. One of these which is a modification of the Euler method presented here is
given in Problems 6.21 and 6.22. Other methods a r e given in the supplementary problems.
CHAP. 61
APPLICATIONS OF D I F F E R E N C E EQUATIONS
217
6.21. Given the differential equation of Problem 6.19 show that a better approximation is
given by the difference equation
=
Yk+l
Yk
+ W(Y,'+ Y;+J
Using equation ( 2 ) of Problem 6.19 we have
Instead of assuming t h a t f(x, y) is practically constant over the range from
trapezoidal rule to approximate the integral in (1). We thus obtain
$k
to Xk+
we use the
6.22. Use the method of Problem 6.21, sometimes called the modified Euler method, to work
Problem 6.20.
The initial values of x and y a r e xo = 0, yo = 1 and a r e entered in the table of Fig. 6-7. I n
order to start the procedure using Problem 6.21 we must first find a n approximate value of y
corresponding to x = 0.1, assuming t h a t we choose h = 0.1 as in Problem 6.20. F o r this purpose we use the ordinary Euler method of Problem 6.20 and find
Yl = Y(O.1) = Yo + hf(xo,YrJ) = Yo
= 1.0000 0.1(1.0000)
+ hYd
+
= 1.1000
This is entered in the table of Fig. 6-8 under Approximation I.
5
I
1.9247
0.5
I
I
2.9247
I
1.9471
2.9471
I
I
1.9482
I
2.9482
1.9483
I
Fig. 6-8
From the differential equation we now find
YI = f ( % Y l )
= 2x1 + Yl
= 2(0.1) 1.1000
+
= 1.3000
which i s also entered under Approximation I in Fig. 6-8.
From the modified Euler formula of Problem 6.21 we have on putting k
Y1
Y;)
= Yo -i- &h(ldA+= 1,0000 3- +(0.1)(1.0000
0
+ 1.3000)
= 1.1150
This better approximation to yl is entered in the table of Fig. 6-8 under Approximation 11.
218
APPLICATIONS O F D I F F E R E N C E EQUATIONS
[CHAP. 6
Repeating the process we then find
Y; = f(X1,Yl) = 2x1 + Y 1
= 2(0.1) 1.1150
+
= 1.3150
and so Approximation I11 f o r y, i s given by
Y, = Yo + -p(Yi+Y;)
= 1,0000 ~(0.1)(1.0000 1.3150)
+
+
= 1.1153
which yields
I/; =
= 2x1 + Y l
= 2(0.1) i- 1.1158
f(X1,Yl)
= 1.3158
We then find
?dl
= I40 + $h(Yi+ YI)
= 1.0000 +(0.1)(1.0000
+
+ 1.3158)
= 1.1158
Since this gives the same value as t h a t obtained above for y1 the desired accuracy to four decimal
places i s achieved.
We now proceed in the same manner to find y(0.2) = y8.
responding to x1 = 0.1 we have for Approximation I
Y2
= Y1
+ hf(x,,ld,)
Using the value y1 = 1.1158 cor-
= Y l + hYf
= 1.1158 O.l(l.3158)
= 1.2474
+
Y; = f(X2,Y2) = 2x2 + I42
= 2(0.2) 1.2474
and
+
= 1.6474
By continuing the process over and over again we finally obtain the required value y5 = ~ ( 0 . 5 )=
1.9483 as indicated in the table. This compares favorably with the exact value 1.9461 and is
accurate to two decimal places.
It will be noticed t h a t in t h e modified Euler method the equation
Yktl
is used to “predict” the value of yk+1.
=
+
Yk
&h(YL+?dLtl)
The differential equation
Yk = f(xk,Yk)
Because of this we often refer to the method as a “preMany such methods a r e available which yield better accuracy. See f o r
is used to “correct” the value of ykfl.
dictor-corrector” method.
example Problem 6.105.
6.23. Show how the boundary-value problem
U(x,O) = 1, U ( x , l ) = 0, U(0,y) = 0,
’ can be approximated by a difference equation.
U(1, y) = 0
It is convenient to approximate the derivatives by using central differences. If we let 8, and
the central difference operator and shifting operator with respect to x, then the partial
Ex denote
derivative operator with respect to x is given by
where the symbol
U we have
-
- - -h _
a
D, = ax
denotes “corresponds to”.
8,
-
El/Z
X
- E-l/2
z
h
Thus by squaring the operators and operating on
CHAP. 61
APPLICATIONS O F D I F F E R E N C E EQUATIONS
= -1( E x - 2
hz
=
219
+ E L ~ ) U ( y)S ,
h21 [ U ( x+ h, Y) - 2Wx, Y) + U ( x - h, Y)]
Similarly using operators with respect to y we have
It follows by addition t h a t
Thus the equation
is replaced by
+
+
+
+ +
+
u (2 h, Y) U ( x- h, y) U ( x , y h) V ( x ,y - h) - 4U(x, y) = 0
or
U ( x , Y) = A [ U ( x+ h, Y)
U ( x - h, Y) U(X, y h) U ( x , y - h)]
which is the required difference equation,
The boundary conditions associated with this difference equation a r e left the same.
6.24.
+
+ +
(11
Show how to obtain an approximate solution to the boundary-value problem of
Problem 6.23.
We must find U ( X , ? Jat
) all points inside the square 0 5 x 5 1, 0 5 y 5 1. Using a suitable
value of N and h such t h a t hrh = 1, we subdivide the square into smaller squares of sides
h = 1/N as in Fig. 6-9. The result is often called a g r i d o r mesh. Instead of finding U ( X y)
, at
all points inside the square we find the values of U ( x ,y) a t the points of the grid such as A , B ,
in Fig. 6-9. To accomplish this we make use of the difference equation ( 1 ) of Problem 6.23
which is, as we have seen, an approximation to the partial differential equation. By referring
to Fig. 6-10, which represents a n enlargement of a p a r t of Fig. 6-9, we see t h a t this difference
. ..
states t h a t the value of U at P is the arithmetic mean of the values at the neighboring points
By using this difference equation together with the values of U on the boundary
of t h e square as shown in Fig. 6-9, we can obtain approximate values of U at all points in the
grid.
A , B, C and D.
Y
I
I
L'ix. 1) = 0
B
C(0,y) = 0
Fig. 6-9
Fig. 6-10
220
APPLICATIONS O F DIFFERENCE EQUATIONS
In order to illustrate the procedure we shall use
the grid of Fig. 6-11 in which we have chosen for
the purposes of simplification h = 1/3. In this case
there are four points in the grid denoted by 1 , 2 , 3 , 4
of Fig. 6-11. We wish to find the values of U a t
U3,U4 respectively.
these points denoted by Ul, U2,
u=o
By applying equation (1) we see that U1, U2, U3,
U4 must satisfy the equations
u1 = & ( O + O + U 2 + U 3 )
U, = & ( U l + 0 4 0 + U 4 )
u3 = Q ( O + U 1 + U 4 + 1 )
U, = $ ( U 3 + U 2 + O + l )
[CHAP. 6
u=o
U = l
(2)
Fig. 6-11
The first equation for example is obtained by realizing that U1,i.e. the value of U at point 1, is
the arithmetic mean of the values of U at the neighboring points, i.e. O,O, U,, U3 the zeros being
the values of U at points on the boundary of the square.
The values U1, U,, U3 and U4 can be found by solving simultaneously the above system of
equations. For a grid with a large number of subdivisions this procedure is often less desirable
than a method called the method of iteration which can take advantage of machine calculators.
In this method of iteration we first assume a trial set of solutions. The simplest is to take
We refer to this as Trial Number 1 or the first trial as recorded in Fig, 6-12.
U1 = 0, U , = 0, U3 = 0, U, = 0.
Using the values of U1, U2,U3,U4 from the Trial 1 in equation (2) we then calculate the new
values
U1 = 0, UZ = 0, U3 = 0.2500, U4 = 0.2500
These are entered in Fig. 6-12 for Trial 2.
Trial
Number
1
Ul
u,
u3
u
4
0
0
0
0
Fig. 6-12
CHAP. 61
APPLICATIONS O F D I F F E R E N C E EQUATIONS
221
OP Ul, U,,U3,U , from the Trial 2 in equations
(2) we then calculate the
Using the values
new values
U ; = 0,0625,
Ug = 0.0625,
U3
= 0.3125,
U, = 0.3125
These a r e entered in Fig. 6-12 for Trial 3.
By continuing in this manner the remaining entries in Fig. 6-12 a r e obtained. The process
is continued until the values foi. U,, U,, U s , U4 no longer change as indicated by the last two lines.
We see t h a t the values a r e
U1
0.1250,
U , = 0.1250,
U3
= 0.3750,
U4
= 0.3750
As a check the values can be substituted back into equations (2) and will be found to satisfy
them.
The equation
is called Laplace’s equation in two dimensions. This equation arises in various problems on heat
flow, potential theory and fluid flow. I n problems on heat flow for example, U = U(x,y) represents the steady-state temperature in a plane region such as a thin plate whose faces a r e
insulated so t h a t heat cannot enter o r escape. I n the problem above, the region is a square, one
of whose sides is maintained at temperature 1 while the remaining sides a r e maintained at temperature 0. In such case the values of U represent the steady-state temperatures at various
points in the square.
Supplementary Problems
APPLICATIONS TO VIBRATING SYSTEMS
6.25.
Show t h a t if we assume that the tension of the string in Problem 6.1 is constant but t h a t the
displacements a r e not necessarily small, then the equation of motion f o r the kth particle is
given by
m- dtz
-
where
6.26.
Show t h a t if
l(yk - ~lc-l)/IzI < 1 the equation of Problem 6.25 can be approximated by
d2yk
m- dt2
6.27.
r
= -h( ( ~ k - : - 2 ~ k + ~ k + 1 )
+ 2h3
-(
--I-[(~k-~yii-:)~
~ k t i - ~ k ) ~ I
Show t h a t if the string of Problem 6.1 vibrates in a vertical plane so t h a t gravity must be taken
into account, then assuming t h a t vibrations a r e small the equation of the kth particle is given by
where g is the acceleration due to gravity.
6.28.
By letting ifk = u,+F(lc) in the equation of Problem 6.27 and choosing F ( k ) appropriately, show
t h a t t h e equation can be reduced t o
and discuss the significance of the result.
222
6.29.
APPLICATIONS O F D I F F E R E N C E EQUATIONS
[CHAP. 6
Suppose t h a t the string of Problem 6.27 fixed at points P and Q [see Fig. 6-11 is left to hang
vertically under the influence of gravity. Show t h a t the displacement yk of the kth particle is
given by
yk = -mg
k = 1,2, . . .,N
27h X k ( L - Xk)
where xk = kh i s the distance of the kth particle from P.
6.30.
Suppose t h a t in the string of Problem 6.1 the number of particles N becomes infinite, in such
a way t h a t the total mass and length a r e constant. Show t h a t the difference equation in this
limiting case becomes the wave equation
where T is the tension and p is the density [mass per unit length] of the string, both assumed
constant.
6.31.
W h a t does the equation of Problem 6.30 become if the vibration takes place in a vertical plane
and gravity is taken into account?
6.32.
Suppose t h a t in Problem 6.1 the tension in the string is not constant but t h a t in the portion of
the string immediately to the right of particle k the tension is r k . Show t h a t the equation
of motion of the kth particle is
m-
d2yk
d t2
__
-
1
hA(Tk-l&k-1)
and write corresponding boundary conditions.
(a) By letting yk = A!, cos w t in the equation of Problem 6.32 show t h a t we obtain the SturmLiouville difference equation
A(Tk-iAz4k-i)
+ WdhAk
= 0
( b ) Discuss the theory of the solutions of this equation from the viewpoint of eigenvalues and
eigenfunctions a s presented in Chapter 5.
(c) Illustrate the results of this theory by considering the special case where the tension is
constant.
6.33.
Suppose t h a t at time t = 0 the position and velocity of the kth particle of Problem 6.1 a r e given
Show t h a t the position of the kth particle at any time t > 0 is given by
by y k ( 0 ) and &(O).
N
N
nkr
nkr
2 (a, cos w,t b, sin writ) sin yk =
c, sin K
1
cos (ant a,) = n=l
N + l
nzl
+
a r e determined from
where a,, b, or c, and
b,
6.34.
+
= --en sin an
(a) Show that, if the ( N i . 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
( b ) W h a t a r e the frequencies of the higher modes of vibration?
[Hint. If the force on the ( N + 1)th mass is zero then yx = YN+~.]
6.35.
(a) Give a physical interpretation
aAT+l?dN+l
= 0
of the boundary conditions
in Problem 6.1.
( b ) Obtain t h e natural frequencies of the system for this case.
CUOYO
+ alY1
= 0,
axY.v
CHAP. 61
6.36.
APPLICATIONS O F D I F F E R E N C E EQUATIONS
223
An axis carries N equally spaced discs which a r e 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 t h a t
d2ek
I - dt2 = " ( ' k t l - 2 8 k
+ @k--1)
where Z is the constant moment of inertia of each disc about the axis and
u
is the torsion constant.
Fig. 6-13
6.37.
appropriate boundary conditions obtain the natural frequencies for the system of
Problem 6.36.
(a,) Using
( b ) Explain how you would find the motion f o r the kth disc by knowing the state of the system
at time t = 0.
(c)
Compare with results for the vibrating system of Problem 6.1.
6.38.
Work Problem 6.37 if the first disc is fixed while the last is free.
6.39.
A frictionless horizontal table A B 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 o r more masses along the table and then releasing them.
k
k
m
m
0
.
0
+-)qyp+
.
B
A
Fig. 6-14
( a ) Letting % k denote the displacement of the kth object, set up a difference equation.
( b ) Find the natural frequencies of the system.
(c) Describe the analogies with t h e vibrating string of Problem' 6.1.
6.40.
Suppose that the string of Problem 6.1 with the attached particles is rotated about the axis with
uniform angular speed a.
(a) Show t h a t if yk is the displacement of the kth particle from the axis, then
y k t l
-2
(1
-3)
Yk
-
( b ) Show t h a t the critical speeds in case m i P h f 4 r
$2,
=
<1
nr
sin mh
2(N + 1)
4- y k - 1
= 0
a r e given by
n=l,2,
...
and discuss the physical significance of these.
6.41.
Discuss the limiting case of Problem 6.40 if N
remain finite.
6.42.
Explain how you would find a solution t o the partial differential equation in Problem 6.30 subject
to appropriate boundary conditions by applying a limiting procedure to the results obtained in
Problem 6.2.
+
5)
in such a way that the total mass and length
224
APPLICATIONS O F D I F F E R E N C E EQUATIONS
[CHAP. 6
APPLICATIONS TO ELECTRICAL NETWQRKS
6.43.
An electrical network consist,s of N loops with capacitors having constant capacitance C1, C2
in each loop as indicated in Fig. 6-15. An alternating voltage given by V N= K sin at where
K and o a r e constants is applied across A$,.
If i t is required t h a t the voltage across A&
= 0, show t h a t the voltage Vk across AkBk is given by
be
v,
sinh ,uk
Vk = K sinhfiN sin at
where
Fig. 6-15
6.44.
Work Problem 6.43 if t h e capacitors C1 and C, a r e replaced by inductors with constant inductance
Ll, Lz.
6.45.
Work Problem 6.43 if the capacitors C, a r e replaced by resistors having constant resistance Rl.
6.46.
Can Problem 6.43 be worked if i t is required t h a t the voltage across A$,
APPLICATIONS TO BEAMS
Suppose that in Problem 6.6 the beam is of finite length NF,.
at the n t h support is given by
6.47.
Tntl--N
where rl = -2
+ fi, r2 = -2
M,
= -Wh[
- g3.
V o > O?
Explain.
Show t h a t the bending moment
-1
I
~ 11 - N -
be
ntl--N
r2
1-N
1
6.48.
Show t h a t as N
6.49.
I n Fig. 6-16 A B represents a continuous uniform beam on which there is a triangular load distribution where the total load on the beam is W . Suppose t h a t the beam is simply supported at
the W + 1 points Po,P,, . . .,P, at distances h apart. ( a ) Show t h a t the three moment equation
can be written as
WIL
=
Mk-1
4fi!f,
Ml,+l
N2
+ m
the result in Problem 6.47 reduces to t h a t of Problem 6.6.
+
+
where the bending moments a t the ends A and B a r e zero, i.e. iM, = 0, M v = 0.
the bending moment at the kth support is given by
( b ) Show that
B
Fig. 6-16
6.50.
Does lim M k exist in Problem 6.49?
N-+m
Explain.
CHAP. 61
APPLICATIONS O F D I F F E R E N C E EQUATIONS
225
APPLICATIONS TO COLLISIONS
6.51.
Show t h a t 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 a r e given by m, and m2 where ml f m2. ( b ) Determine
the values of m,lm2 f o r which the speeds will be periodic and determine the periods in such cases.
( c ) Discuss the motion for the case ml = 2m2.
6.53.
A billiard ball is a t rest on a horizontal frictionless billiards table which is assumed to have the
shape of a square of side 1 and whose sides a r e assumed to be perfectly elastic. The ball is hit
so t h a t i t travels with constant speed S and so t h a t the direction in which it travels makes the
angle e with the side. Determine the successive points a t 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
Suppose that in Problem 6.9 we let
k pennies while B has m pennies.
6.55.
Uk,m
be the probability of A winning the game when he has
( a ) Show t h a t t h e difference equation and boundary conditions f o r
Uk,m
= Puk+l,m-l
~M,N-M
= 1,
Uk,m
a r e given by
+ qUk-l,mtl
UN-M.M
= 0
( b ) 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 189.1
6.56.
Suppose t h a t i n a single toss of a coin the probabilities of heads o r tails a r e given by p and
Let Uk,,$ denote the probability t h a t exactly k heads will appear in 12
q = 1- p respectively.
tosses of the coin.
( a ) Show t h a t
where
i:
uk, 0
k=O
k>O
This is called Bernoulli’s probability formula for repeated trials.
6.57.
658.
Suppose t h a t A and B play a game a s in Problem 6.9 and that A starts out with 10 pennies
while B starts out with 90 pennies. ( a ) Show that if p = q = 1/2, i.e. the game is fair, then the
probability t h a t A will lose all of his money, i.e. A will be “ruined”, is 0.9. ( b ) Show t h a t 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
gambling casino. ( d ) What is the significance of the result in ( b ) if A is the “bank” and B is the
“gambler” ?
Suppose t h a t in Problem 6.9 p > q, i.e. the probability of A’s winning is greater than t h a t of
Show t h a t if B has much more capital than A then the probability of A escaping “ruin” if
he starts out with capital a is approximately
B.
1
W h a t is this probability if
- (alp)“
p = 0.6, q = 0.4 and ( a ) a = 5, ( b ) a = loo?
226
6.59.
APPLICATIONS O F D I F F E R E N C E EQUATIONS
[CHAP. 6
Let Dk denote the expected duration of the game in Problem 6.9 when the capital of A is k.
( a ) Show t h a t D , satisfies the equation
Dk = p D k + I + q D k - 1 + 1 ,
( b ) Show t h a t if p # q ,
=
Dk
DO = 0,
Da+b
= 0
k
--
(c) Show t h a t if p = q = 112, D k= k ( a + b - k) and thus t h a t 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 a t any stage with constant probability
t h a t p q E = 1.
+ +
THE FIBONACCI NUMBERS
Let F , denote the kth Fibonacci number and let
6.61.
1
Show t h a t
Rk = 1 t Rk-1 , R, =
Show t h a t
Rk
Rk
0.
1
=
i.e. a continued fraction with k divisions.
1
l + -1
+* -11
Solve the difference equation of p a r t ( a ) in Problem 6.61 for
Show t h a t lim
k-r
01
Rk =
5-2
2
> 0 so
= FkIF,+l.
1 +
6.62.
B
Rk.
and thus obtain the limiting value of the infinite continued
fraction in p a r t ( b ) of Problem 6.61.
6.63.
6.64.
If
F k denotes the kth Fibonacci number prove t h a t
Fk+lFk,-I - F t = ( - l ) k
k = 1,2,3,
...
Generalize Problem 6.63 by showing t h a t
F k t , n F k - m - F ; = (-l)k+m-IF;
k = m,m+l,m+2,
...
6.65.
A planted seed produces a flower with one seed a t the end of the first year and a flower with
two seeds at the end of two years and each year thereafter. Assuming t h a t each seed is planted
as soon as i t is produced show t h a t the number of flowers at the end of k years is the Fibonacci
number F k .
6.66.
Find t h e 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. W h a t restriction if any must there be on a and b ?
6.68.
Find the sum of the series
6.69.
Show that
-
2
k=O
Fkt2
where F , is the kth Fibonacci number.
FkilFkt3
CHAP. 61
APPLICATIONS O F 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 f o r 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
1
1
2
1
1
1
1
3
3
4
1
6
4
1
1 6 1 0 1 0 6
1
6
16 20 16 6
1
Fig. 6-17
( a ) Show that if U,k,m denotes the kth number in the mth row, then
Uktl,m+l
=
+ Uk,m
uk+l,m
where
uk,O
=
1 k>O
0 k=O
( b ) Solve the difference equation of ( a ) subject to the given conditions and thus show that
uk,m
(3
=
i.e. the numbers represent the binomial coefficients.
6.71.
The Chebyshev polynomials Tk(x) are defined by the equation
- 2%Tk(X)f T k - i ( $ )
To(%)= 1, TI(%)=
= 0
Tk+I(x)
where
x
( a ) Obtain the Chebyshev polynomials T&), T3(x),T&).
( b ) By solving the difference equation subject to the boundary conditions show that
T k (8)
6.72.
=
COS
(k
COS-l
X)
The Bessel function Jk(X) of order k can be defined by
Show that Jk(x) satisfies the difference equation
dk+i(Z)
6.73.
=
&[Jk-l(X)
- Jk+l(X)]
The Laplace transform of a function F ( x ) is denoted by
dl{F(x))
(a)
6.75.
= 0
Show that the Bessel functions of Problem 6.72 have the property that
JL(X)
6.74.
- ZkJk(x) f J k - i ( X )
Show that <{Jo(x)> =
= f(4 =
[
O0e--ss
F ( x )dx
A
@Ti'
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 A k in the bank after k periods
and ( b ) find A k .
228
6.76.
APPLICATIONS O F DIFFERENCE EQUATIONS
[CHAP. 6
A man deposits a sum S in a bank a t 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
]
An = S [ (1+i)"
i -1
= Ssqi
Such a set of equal periodic psyments is called an annuity.
+ a, S + 2a, S f 3a, . . . .
6.77.
Work Problem 6.76 if the successive deposits are given by S, S
the significance of the cases ( a ) a > 0, ( b ) a < 0.
6.78.
In a particular town the population increases a t a rate a% per year [as measured a t the end of
each year] while there is a loss of p individuals per year. Show that the population after k years
is given by
p, = ( P o - T ) ( l + & ) k +
. .
where Po is the initial population.
Discuss
9
6.79.
Does the result of Problem 6.78 hold if (a) a:
Interpret each of the cases.
6.80.
Let
S k
(a) Show that
S1
+ al = 0,
( b ) Show that if k
Sz
>n
+
alSk-1
Sk
. .. + a ,
S,
+ azSk-2 +
Obtain the generating function for the sequence
> 0, p < 0,
(c) a
< 0, /3 > O?
* ' *
= 0
+
+ an-,S1 + nu,
+
+
anSk-n
=0
= 0
of Problem 6.80.
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 u k 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
is
6.83.
+ a2rn-Z +
+ all'n-l
+ alSz + 2a2 = 0, . ..,
Sk
6.82.
(b) a
denote the sum of the kth powers of the n roots of the equation
p
6.81.
< 0, p < 0,
(3
(:)
\
uk-1.
( b ) Prove that the number of permutations in which there are m coincidences
I
Uk-m
where m = o , ~ , . ..,k.
Prove that u, in Problem 6.82 satisfies the equation
By solving the difference equation in (a) prove that
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
2!
1 + ... + -3!
n!
Show that for large values of n the probability in (c) is very nearly
1-
= 0.63212..
.
Discuss how large n must be taken in ( d ) so that the probability differs from 1- l / e by less
than 0.01.
APPLICATIONS O F DIFFERENCE EQUATIONS
CHAP. 61
6.84.
An inefficient secretary puts n letters a t random into n differently addressed envelopes.
that the probability of getting a t least one letter into its correct envelope is given by
1 - -1 +
1 - -1 + 2!
3!
4!
Show
... +-( - 1 )nn +! l
and thus show that for large n this probability is about l l 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 a t 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 ~, times the amount
of his stake but if he loses he gets t o play again. Let u k denote the stake w%ch 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 .
( a ) Show that
uk
satisfies the sum equation
muk
k
= P
+ 2 up
p=1
( b ) Show that the sum equation of (a) can be written as the difference equation
(c)
Using appropriate boundary conditions show that
k
6.87.
State some limitations of the “sure profit” deal in Problem 6.86 illustrating the results by taking
m = 2.
6.88.
A box contains slips of paper on which are written the numbers 1, 2, 3, ., n [compare Problem
6.821. The slips are drawn one at a time from the box a t random and are not replaced. Let
‘lLk,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
u k , m = u k - - l , m - uk--l,m--li
Uk,o = k!
..
( b ) Show that the solution t o the difference equation of part ( a ) is
(c)
Show that the probability that there will be no coincidence in the k drawings is
Uk,mh!.
( d ) Show that for k = m the probability obtained in (c) reduces to that obtained in Problem 6.82.
6.89.
A and B play a game with a deck of n cards which are faced downwards. A asks B t o name a
card. Then A shows B the card and all the cards above i t 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.
6.90.
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.
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.
6.92.
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 i t is known that the nth marble taken is blue.
230
APPLICATIONS O F DIFFERENCE EQUATIONS
[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
( b ) Use ( a ) to show that
and verify this for some special values of k and m.
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.
6.98.
( a ) Show that Newton’s method [Problem 2.169, page 781 is equivalent t o the difference equation
Xnt1
-
x,
-
f (En)
f’(xn) ’
( b ) Can you find conditions under which lim x, exists?
n+-
xg = a
Justify your conclusions.
6.99.
( a ) Prove that lim x, exists and is equal t o
It-m
fl.
( b ) Show how the result of (a) can be used to find
6.100.
Can you formulate a method for finding
6.101.
Work Problem 6.20 by using h
6.102.
( a ) Solve the differential equation
fi t o six decimal places.
9
fi
analogous t o that
of Problem 6.99.
= 0.06 and discuss the accuracy.
dy =
dx
x
+ y,
y(0)
= 1
for y(0.6) using h = 0.1 by using the Euler method and compare with the exact value.
( b ) Work p a r t (a) using h = 0.05.
6.103.
Work Problem 6.22 by using h = 0.05 and discuss the accuracy.
6.104.
Given the differential equation
dy =
dx
( a ) Use the Euler method t o
x y q
y(1) = 1
find ~ ( 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(x, V ) the approximate “predictor” formula
yk+l
= yk-1
h
+ -(Y;-1+4Y{+y{+1)
3
( b ) Explain how the result in ( a ) can be used t o solve Problem 6.20 and compare the accuracy with
the modified Euler method.
CHAP. 61
APPLICATIONS O F DIFFERENCE- EQUATIONS
231
6.106. (a) Explain how the method of Taylor series can be used to solve a differential equation. ( b ) Illustrate 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(z) = y(a) (z - &’(a)
(x - a)2y”(a) +
2!
6.107.
. .. *I
+
+
.
+
+
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 t o solve the second order differential equation
2
=
f(X,Y,$)
( b ) Use the method of (a)t o solve
d2y = x
dx2
+ 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
and compare with the exact value.
6.113. 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 t o obtain the steady-state
temperature in a square plate whose faces are insulated if three of its sides are maintained a t O°C
while the fourth side is maintained a t 100OC.
6.115. Explain how you could use the results of Problem 6.24 or 6.113 t o obtain the steady-state temperature in a square plate if the respective sides are kept a t 2OoC, 4OoC, 6OoC and 80OC.
6.116.
Solve the equation
u = 0
$$+ a2
for the plane
a Y2
region shown shaded in Fig. 6-18 with the indicated
boundary conditions for U.
6.117. The equation
a2u
aw =
+ax2
ay2
p(x ’ y)
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+ ~2 y2) if the region and
boundary conditions are the same a s those of Problem 6.23.
+
Fig. 6-18
6.118. Explain how you would solve approximately Laplace’s equation in three dimensions given by
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.
6.119. Explain how you would find an approximate solution t o the boundary-value problem
at
=
a&’
U(0, t ) = 0,
U(1,t) = 0,
U(z,O) = 1
Appendix A
2
7
rl
rl
4
w
W
I
0
rl
Q,
W
?t
W
I d ‘
u3
*
*
hl
*
‘%
0
rl
I
0
N
2
w
0
0
2
3*
I
--?
1
3
2
5
4
6
7
8
9
10
1
1
2
1
1
3
1
3
1
4
1
7
6
1
5
1
15
25
10
1
6
1
31
90
65
15
1
7
1
63
301
350
140
21
8
1
127
966
1,701
1,050
266
28
9
1
255
3,025
7,770
6,951
2,646
462
36
10
1
511
9,330
34,105
42,525
22,827
5,880
750
45
1
11
1
1,023
28,501
145,750
246,730
179,487
63,987
11,880
1,155
55
12
1
2,047
86,526
611,501
1,379,400
1,323,652
627,396
159,027
22,275
1,705
K
J
03
03
--
1
1
1
Appendix C
All values B3,BE,,B?,
Bo
=
Bernoulli Numbers
. . , are equal to zero.
1
-0.500000000. .
.
.
0.166666666..
-0.0333333333...
..
0.02389523895.
-0.0333333333...
5
BlO = 66
zI'
=
--691
2,730
0.075757575..
.
-0.2531135531135. .
.
B14
7
= 6
Bl5
=
--3,617
510
-7.092156862745098039215...
Bis
=
43,867
798
54.971177944862155388...
Bzo
=
--174,611
-529.1242424242. . .
Bzz =
B24
=
B
2
5
=
Bz8
=
B
3
0
=
330
854,513
138
- 236,364,091
2,730
8,553,103
6
- 23,749,461,029
870
,
8,615,841,276,005
14,322
~
1.1666666666.. .
6,192.1231884057898550.. .
-86,580.2531135531135.. .
.
1,425,517.16666..
-27,298,231.067816091954.. .
.
601,580,873.90064236838.
.
234
Appendix D
Bernoulli Polynomials
Bo(x) = 1
B&)
= x
- 31
+ g1
Bz(x) =
x2
-x
B3(x) =
29
- -323 2
+ -21x
B5(4 =
x5
5
- -x4
2
5
+ -x3
3
B7(X) =
27
7
- -39
+ 27
7
- -x3
6
1
+ -x
6
B&)
=
23
- 4x7
+ -14
x6
3
7
- -34
3
+ 23
B&)
=
x9
'2
-2
21
+ 6x7 - -x6
+ 2x3 - --x103
5
Bl0(X) =
x10
2
-
5x9
1
6
--x
/-
-x5
15
+ -x*
2
235
- 7x6
-552
1
-30
3
5
+ 5x4 - -xz
+2
66
Appendix E
All values El, Es,
Euler Numbers
. . . are equal t o zero.
Eo = 1
E2
= -1
E4 = 5
Es = -61
Es = 1,385
Elo = -50,521
E
l
2
= 2,702,765
E
l
4
= -199,360,981
El6
= 19,391,512,145
E
l
8
= -2,404,879,675,441
Ezo = 370,371,188,237,525
E22
= -69,348,874,393,137,901
-
E24 = 15,514,534,163,557,086,905
236
Appendix F
PolynomiaIs
Eo(x) = 1
1
El@) = x - 2
Ez(x) =
-39
1
2
1
-2x
E3(x) =
-x3
1
6
1
-4x
E&)
1
= -x4
24
E&)
=
1
120
-x5
1
- -12
x3
+ -241
+ -241x
1
1
1
- -x4
+ -x2
-48
237
48
240
Appendix G
Fibonacci Numbers
F26
= 121,393
F27
= 196,418
Fzs = 317,811
F29
= 514,229
F30
= 832,040
Fsi = 1,346,269
F32
= 2,178,309
F33
= 3,524,578
F34
= 5,702,887
F35
= 9,227,465
Fs6 =
14,930,352
F37
= 24,157,817
F3s
= 39,088,169
F39
= 63,245,986
F40
= 102,334,155
F41
= 165,580,141
F42
= 267,914,296
F43
= 433,494,437
F u = 701,408,733
238
F45
= 1,134,903,170
F46
= 1,836,311,903
F47
= 2,971,216,073
F48
= 4,807,526,976
F49
= 7,778,742,049
F50
= 12,586,269,025
Answers to Supplementary Problems
CHAPTER 1
1.46.
(4 9
- 32x3 + 32x2 - 22% + 6
+ 4x3 - 6x2 - 6%
(g) 4x4 + 8x3 - 2x2 - 6%
( d ) 6(3x+ 2)
(h) 2x(4x+ 1)
(a) 1
+2fi +x
- 4x3 + 222 + 6%- 3
(b) 2x4
1.52.
(a) 8 x h
+ 4h2 - 2h
(b) d 5 x + 5 h - 4
+ 12xh + 12h2 + 3
(a)Yes
( b ) No
1.55.
(a)Yes
(b) Yes
1.59.
(a) Yes
(b) No
1.61.
(a)(3x2- 6%
1.71.
(a) 4x3
1.72.
+ + 1)s
3x3 + 8x2h + 6xh2 +
x2
+ 2) dx
(c) Yes
(b) 6 ( d ~ ) 2
+ 6 ~ 2 h- 12xh2 - 4 3 + 25h3 - 2h + 5
12x2 + 2 4 ~ h+ 14h2 - 4
(a) (2%- 1)(2x- 5)(2$ - 9)(2x - 13)
+
1
(d) (5x
+
1
(4%- 1)(4x
+ 12)(5x + 22)(5x + 32)(5x + 42)
+ 3)
(a) (3%- 2)(3), h = 7/3
(c) (x-2)(-3),
(b) (2x+2)(4), h = 3/2
(a)
E, = 2
(23- 5)(-4), h = 2
239
+ 14xh + 19h2
(j) 4 x W
+ 14xh + 19h2
(h) 2x2
~ 1)
( b ) (3% 5 ) ( 3 ~ 2 ) ( 3 -
1.73.
(i) 2x2
(g) 0
1.53.
(b)
(f)
4h2
( d ) 3x2
(i) 0
( j ) 1627(1- X)
(e) (x 2h
3
(c)
( e ) 16x4
(f) 2x4
+ 9xh3
ANSWERS TO SUPPLEMENTARY PROBLEMS
240
s;=2,
s;=o,
s;=1,
5;=-3,
s+1,
s:=-1,
1.77.
s:=1,
S ? = l , S 13= 1 , s z1 = o , s ; = 1 ,
1.78.
5'; = 1, Si = 31, Ss6 = 90, S46 = 65, S," = 15, S: = 1
1.81.
(a) 3 x ( 2 ) - 2 d l )
1.76.
+2
s;=o,
s;=3, s;=o,
s;=o,
s;=1
s 32 = 0 ,
s 33 = 1
1.87.
+ (3h - 5)d1) + 2
(b) 2 d 4 ) -t l2x(3) + 19x(2) + 3 d 1 ) + 7 and 2 d 4 ) + 1 2 h d 3 ) + (14h2 + 5 ) d 2 ) -t (2h3 + 5h - 4)d1)4- 7
25(22 + 12%+ 30)
1.88.
(ah - 1)nxU"
1.89.
(ah- l)nx2(15
1.91.
(a) 4xh
1.92.
0
1.94.
Yes
1.97.
Yes
1.116.
((I) 18h%
and 3 d 2 )
+ nh((Ih- l)n--l(Ir+h
+ 2nh(ah - l)n--1zu5+h + nhZ(nah - I)(&
- 2/22 + 3h
- 4h2 + 18
(b) 4xh
3
+ 3h
(b) 24h3x
CHAPTER 2
2.54.
2.56.
2.58.
1489,2053, 2707, 3451, 4285, 5209
2.59.
(b) 2366, 3994, 6230, 9170,12910
2.61.
2.63.
2.64.
(c) 4h2
- l)n--2uz+h
( d ) 4/22
+ l2h3 + 36h
ANSWERS TO SUPPLEMENTARY PROBLEMS
+4
2.68.
( a ) Zlk =
2.69.
( a ) qc = k3 - 9 k 2
9-
2.71.
y
2.72.
y = &(x3-55X2+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
=
3k* - 2k
x3 -
4x2
(b)
Uk
+ a2 k - 4
= Q k ( k + l ) ( k + 2)
( b ) 651
+ 5%+ 1
Computed value = 0.71202
Computed value = 0.90399
2.80.
241
( a ) Exact value = 0.49715
Computed value = 0.70341
( 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
=
~3
( a ) 1.27507
(a) 0.30788
- 4x2
(b) 1.50833
(b) 0.62705
( a ) 0.71912, 0.73983
+ 5%+ 1
(b) 0.49715
2.99.
2.100.
(a) 4n3 - 27n2
+ 59% - 35
(c) 0.84270
(b) 7, 85
( d ) 0.92756
242
2.101.
ANSWERS TO SUPPLEMENTARY PROBLEMS
+
+
(a) sin x = 2 . 8 0 6 5 6 ( x / ~ ) ~ 6 . 3 3 7 8 0 ( ~ / ~ ) 3 0 . 2 1 1 7 2 ( x / ~ ) ~3.12777(x/~)
( 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, 1 7
2.108.
8, 14, 22, 32
2.111.
Exact values are 0.97531, 0.95123, 0.92774
2.113.
( b ) f(x) = 3x2 - 11% 1 5
2.114.
(a) f ( x ) =
2.121.
(b) 1.32472
2.122.
1.36881, -1.68440
2.123.
2.90416, 0.04792
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,
( b ) 874
+
x3
- 20%
+ 40
(c) -0.66236
k
(c) 19, 159
(b) 65, 840
2
0.56228i
3.43133i
* 1.311252'
G=-4,
Hz-4,
Jz-2,
K=-3
ANSWERS TO SUPPLEMENTARY PROBLEMS
2.140.
Exact value = 3.80295
2.141.
Exact values a r e
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.OG604, 0.02883
2.151.
f(%)
2.157.
Qk(k
2.158.
( a ) &k(k+ 1)
2.159.
ik(6k2
2.160.
( a ) 2k2 - 6 k
2.165.
k = 10
2.166.
All values of k
2.168.
At x = 6, f(x) = 0.68748 should be 0.68784
(a) 1.7024
(b) 2.5751
( b ) 0.6796
(b) 29.4
=p*+r+1
+ 1)(2k+ 1)
( b ) & V ( k + 1)2
+ 3k - 1)
+ 3 -+ 2h-
( b ) 1177
CHAPTER 3
3.55.
(a) 3x5
5
;+d
-x+c
2
4 ~ 3 1 2- 2
$ X 4 '3
(b) 8
+c
(4 $(Z sin 3% + cos 3%)+ c
3.57.
+ 9) + c
2 sin x - x cos x + c
( d ) 2eZS - $3-42
(e) 4 In (x - 2)
(f)
2"
+c
+c
f0
ixZ(1nx
- 4) + c
( a ) -4e-3r(x
(c)
(b)
(d) + 9 1 n s z - Q x 3 l n x + & x 3 + ~
243
244
3.58.
ANSWERS TO SUPPLEMENTARY PROBLEMS
(a)
(b)
3.59.
4%sin 2x - 3x2 cos 2% + & cos 2% + c
&ezz(4x3- 6x2 + 6% - 3) + c
+ sinx) + c
2efi(fi-l) + c
6 cos $ + 6% sin
(a) In (e"
(b)
(c)
x sin r(x - +h)
3-65.
(a)
-1
(e)
- 3@
2 sin +rh
3.67.
1
-2(n
12
3.72.
(4
+
h sin rx
4 sin2 ?ph
1
+ 2)(n+ 3 )
x sin r(x - i h )
sin $rh
+
-x cos r(x - $h)
('1
+c
h cos rx
4 sin2 i r h
2 sin i r h
-x cos r(x - &h)
(b)
cos
2 sin &rh
+
h cos rx
4 sin2 +rh
h sin rx
4 sin2 &rh
B
3.74.
( a ) @ ( n + 1)
3.75.
(a) gn(n
(c)
(b)
( d ) n(3n2 6%
(b) i n ( 3 n - 1)
+ l)(n + 2)
*n(n + l)(n + 2)(n+ 3)
3.76.
3.78.
3.81.
3.82.
3.83.
3.84.
3.85.
+
4n(2n - 1)(272 1)
hn(4nz
+ 6%- 1)
+ + 1)
zqG-=3j+ C
A N S W E R S TO SUPPLEMENTARY PROBLEMS
5
t5
2n
23
&n(n -t. l ) ( n
3.91.
(a)y
3’93’
( a ) 6(2n+ 1)(2n+ 3)
3.94“
n+2
2 - 2n
3.95.
3n(n
3.104.
&r(Q)
3.105.
8
21
2n2
+ 14% + 23
23
480 - 4(n + 2)(n + 3)(n+ 4)(n + 5 ) ’ A80
+ lo1 (2n + 7)(2n + 6)(2n + 3)(2n + 1)(2n- 1)
11x2
+ 19n
1
(b)
1
90 -
6(2n
+ 1)(2n+ 3)(2n+ 5)
11
(b)
5
+ 1)(2n+ 1)
-=
In general,
3.111.
-1130, 1/42, -lit30
3.114.
( a ) ;in2(n
3.115.
3Bo(2)- 4B,(z)
3.119.
(a)
1
~
-
* -253
xz
1
= x
24 - 720’
24 12 + -
&(x)
3.110. ( a )
+
(b)
1)2
r’(x+n+ 1)
r ( x n 1)
+ +
= 120
85
’
r(x + 1)
x4 + -$8
x
48
72 720
(b)
1
30
n(n + 1)(2n+ 1)(3n2+ 3n - 1)
+ B 2 ( x )+ 2B3(x)+ 5B4(x)
e4 = 0, e5 = -1/240
( b ) E4(x) =
(c)
(b)
+ 2)(3n+ 1)
3.89.
3.125.
5
12 - 2(n + 2)(n + 3) ’ 12
3’86’
245
1
$24
- 12x3
E , = 5, E, = 0
P ( x 3 - Gx2
+ 18x - 26)
+ 241 x, E 5 ( x )= 1120x
1
5
48 x4
--
1
+ -xz
48
1
-240
-1/30, 0
246
3.126.
ANSWERS TO SUPPLEMENTARY PROBLEMS
+
(n 1)s cos 2(n
+ 1) - (3n2 + 9%+ 7) cos 2(n + 2)
+ 6(n+2)
+ 7 cos 4
C O S ~ ( ~ 3)
+
-
-
6 C O S ~ (4)~ +- c o s 2
1 2 cos 6
+ 6 cos 8
+ + 5x2 - 1)
3.132.
&n2(2n4 6n3
3.146.
(a) r / 4
(c) r/fi
(b) r/32
CHAPTER 4
4.40.
+ + 2)
(e) )+(n + 1)2
(f) &(2n - l)(2n + 1)(2n+ 3)(2n + 5) + 9
( d ) Qn(n l ) ( n
(a) n2
(b) Qn(2n - 1)(2n
an(6n2
(c)
+ 1)
+ 3n - 1)
x + l
(1- x)3
4.41.
__-
4.43.
9/32
4.44.
(a) ___
4.45.
3591500
4.46.
5e
4.48.
(x4 - 7x2) sin x
4.49.
$x cosh v%
4.59.
Exact value = 1.09861
4.61.
Exact value = 1
4.62.
0.63212
x(x
X
+ 1)
pTp
(1- x ) 2
-1
+ (x - 6x3) cos x
+ f e ( 1 - k x ) sinh fi
3 4 6% - x2
(1- 4 3
+ l ) ( n+ 2)(3n + 17)
Qn(3n3+ 46n2 + 255% + 596)
(9) &n(n
(h)
ANSWERS TO SUPPLEMENTARY PROBLEMS
4.64.
+
+ 1)2
( a ) Qn(n 1)(2n-f- 1)
( c ) +n(n
(a)
(d)
&22(n
4.67.
y = 0.577215
4.70.
1.20206
4.78.
Exact value = 3,628,800
4.79.
2100/5fi
4.83.
xe-s(2
4.84.
-
1028
- x)
11
(a)
4.036 X
&R(R
+ l ) ( n+ 2)
+ 1)(2n + 1)(3n2+ 3 n - 1)
180 -
+
6n 11
12(2n+ 1 ) ( 2 n + 3 ) ( 2 n + 5 )
4.86.
1.092
4.88.
(a) 1.0888
4.94.
+(-l)n-ln(n
4.95.
29n2 - 2n
4.99.
( a ) 1.09861
11
180
+ 1)
+ 3) - 3
( b ) 0.785398
(c) 0.63212
CHAPTER 5
5.60.
(b) y = 5e5 - x2 - 2%- 2
5.61.
Y(X) = 5(1
5.63.
y
5.68.
(a) 2 f ( x + 2 h )
5.69.
+ h ) z / h - $2 + (h-
= 2 cos x
2)s - 2
+ 5 sin x + 3x2 - 5 x - 2
+ f(x) = 0
( b ) f ( x + 3 h ) + 4f(x + 2h) - 5 f ( x + h ) + 2j(x)
- 3 f ( x + h)
x2
= j-p
- 4 x ~ 1+
( a ) , (c), ( d ) , ( e ) , (f), (9)linearly independent; ( b ) , ( h ) linearly dependent
247
248
ANSWERS TO SUPPLEMENTARY PROBLEMS
ka
4
5.77.
2/lc
= c1 cos-
5.79.
Yk
= 2k c1
5.81.
yk
= c1
5.82.
Z/k
= (cl
+ c2 sin-kn4 + c3 cos-3kn
+ c4 sin-3kn
4
4
+ c2 cos-2 3r k + c3 sin- 3
+ Zkl2
+ c,k)2k -I- (c, + c,k + ~ 5 k 24-Cgk3)(-2)k
+ c,4k + 5 k ( c , cos ke + cg sin ke)
where e = tan-l(-4/3)
kn
5.87.
yk
=
~ [ C O S2k
+ cos (2k - 4)]/(1 4- cos 4)
249
ANSWERS TO SUPPLEMENTARY PROBLEMS
+ ~ 2 ( - 2 )+~ 2k
5.89.
yk
= c12k
5.90.
yk
= cl(-l)k
=
k2 + 67k --
+ c4 sin- 2
15
225
422 - 4*3k
3375
65
2k
[8 cos (3k - 9) + cos 3k]
+ c2 c o kr
s i + c3 sin-kir +
3
3
65 + 16 cos9
+ c22k + 258
5.94.
yk
5.95.
(c)
ylc =
(b)
(c)
yk = C,
c,k
yk = (el +'c&+
(d)
yk = c1 cos- 2
~1(-6)k
C12k
+ C,4k
+
k7i
-
3k
irk
25fi s i n 3
-
105 cos-
38
k2
8k
+27
3 + 9 +-
+ k2 + $k3 + 1 * 4 k
9
3k2)zpk
+ c2 sin-k7i2 +
(-1)"
3k + 2~
,.=O
(f)
uk
= c1 COS-kn
3
+ c p sin-k7i3
1
sin
-(r
+ 1)
3
1
250
ANSWERS TO SUPPLEMENTARY PROBLEMS
241
+ c(-l)k]
1-
fi +
2fi/[l
2-k
tan c1 cosL
-
3
+ ~(2fS-
3)-k]
+ c2 sin-2tik
3
1
ANSWERS TO SUPPLEMENTARY PROBLEMS
5.150.
(a) et
5.153.
Z C ( X , y)
5.158.
uk,,, = F ( 2 k - m )
5.160.
uk,m
5.161.
uk,nz = F ( k - m )
5.162.
7k
uk,m = F ( k - m) cos 2
3
+
= ( - 2 ) x F ( ~ 21)
= 2kF(k-m)
-
2k-mt1
+ G ( k - m ) + 2,,H(k-m)
rrk
nm
rrm
+ H ( k - m ) cos + J ( k - m ) sin 3
+ G ( k - m ) sin 3
3
251
252
5.167.
ANSWERS TO SUPPLEMENTARY PROBLEMS
(c)
=
y
CO
1 . 3 5 . * * ( k -1)
2 4 6.m.k~
Yk
x7
x4
k even
+ k G ( k + m ) + k2H(k+m)
5.183.
Uk,m
= F(k+m)
5.184.
uk,m
=
:
8
= Stirling numbers of the first kind
5.185.
Uk,m
=
sk
= Stirling numbers of the second kind
5.188.
Y(x) = (x - l ) / e
5.189.
Y(x) =
5.190.
uk
5.191.
(a)
u(x, y)
= A(3x
5.192.
(a)
U(X,Y)
=
- 4%
x2
l!
+ 1 -t- 2 x tan 1
cos 1
-
2!
1
+3!
x2
-
k -1
... +-( - l )k!
+ 2y)2 +
e-(32*2~)/2
+ y + QXY
CHAPTER 6
6.62.
(a)
fl-1
(
71-- aa kk + l )
where
a =
1-fi
-
l + f i
A N S W E R S TO S U P P L E M E N T A R Y PROBLEMS
6.68.
2
6.71.
( a ) Tg(5)= 2x2-11,
6.75.
( a ) A k + l = Ak f iAk
6.76.
A,+l
6.77.
A,+1 - (1 i ) A , =
6.89.
- -
-
6.91.
(a)
2
+
6.92.
b +
b+r
6.93.
-
6.102.
(a)
Exact value = 1.7974
6.104.
(a)
Exact value = = 1.72265625
256
6.109.
(b)
Exact value = 1.1487
6.112.
Exact values: x(0.5) = 0.645845
( b ) A , = P(1+ i ) k
- (1+ i ) A , = S
+
1
2!
T~(x)
= 6 ~ 3 - 3 3 ~ ,T ~ ( x
=)8 x 4 - 8 x 2 + 1
S
+ nu,
a
+ si
A , = -[(1+i)"
i2
an
- 11 - T
1
3!
b
bSr
441
~ ( 0 . 5 )= 0.023425
u=o
6.113.
U1 = 0.071428571, Uz = 0.098214285,
u=o
u=o
Us = U1, U4 10.1875, Us = 0.2500,
UB = Uq, U , = 0.428571428,
U8
U = l
= 0.526785714, Ug =
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
error terms in, 132-134
Gregory’s formula for, 124, 134-137
Arbitrary periodic constant, 82, 93
Area, 80, 92
Associative laws, 2
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 f o r
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
Calculus of differences, 4
[see also Difference calculus]
Capacitance or capacity, 200
Capacitors o r 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
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
Convolution, 193
Corrector methods, 218, 230
Cube roots, 11, 75
Cubing operator, 1, 10, 11
Currents, 200
De Moivre’s theorem, 154
Definite integrals, 80, 81, 91, 92
approximate values of
[see Approximate integration]
Definite sums, 83, 84, 96, 97
Density o r 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
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
INDEX
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,
162, 163
numerical solution of, 202, 215-221
order of, 150
partial [serz Partial differential equations]
Differential operator, 3, 4, 14, 1 5
relationship of to difference and derivative
operators, 4
Differentiation, approximate, 39, 40, 60-62
general rules of, 4, 5
of a n integral, 81, 84, I l l
of sums, 84, 97
Diagamma function, 86, 104, 105
Distributive law, 2
Divided differences, 38, 39, 66-69
Doubling operator, 1
e, 5
Eigenfunctions, 158, 178, 179
expansions in series of, 169, 179, 180
mutually orthogonal, 169, 178
Eigenvalues, 158, 178, 179
reality of, 158, 179
Elastic collisions, 201
Electrical networks, applications to, 200,
205, 206
Equality of operators, 1
E r r o r function, 70
E r r o r 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
modificd, 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
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
F a i r game, 211
Families of curves, one- and
two-parameter, 197
Fibonacci numbers, 202, 211, 212, 238
table of, 238
Finite differences, 4
Flrxural rigidity, 206
Fluid flow, 221
Forecasting, 34
Forward difference operator, 9
Fourier series, 118
Frequency, 199, 202-205
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 hterpolation formulas, 36
Generating function, for Bernoulli
numbers, 87, 110
for Bernoulli polynomials, 87, 109, 110
for Euler polynomials, 89
method for solving difference equations,
167, 158, 174, 175, 183, 184, 186, 187
Gravity, vibrating string under, 221
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-Ncwton forward difference formula
[sce 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 coeficients, 153, 165-168
in factored form, 153
Identity or unit operator, 1
Indefinite integrals, 79
lndependent 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 p a r t s [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
i n two dimensions, 221
Law of the mean f o r derivatives, 8
Leading differences, 33
Leibnitz’s rule, f o r derivatives, 9, 24
f o r differences, 9, 24
f o r differentiation of a n integral, 81, 84, 111
Limits, 3, 1 5
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 f o r 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, na tura l o r 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
N a tura l base of logarithms, 5
N a tura l frequencies, 199, 202-205
N a tura l logarithm, 5
N a tura l 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, 1 0
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 [ s e e Difference operator]
differential [see Differential operator]
doubling, 1
.identity o r unit, 1
integral, 1, 79, 89-91
null o r zero, 1
squaring, 2
translation o r shifting, 3, 11-14, 160
INDEX
Operator methods, f o r solving difference
equations, 156, 169-172, 187
f o r 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
Pa r t i a l derivative, 81, 160
Pa r t i a l difference equations, 160, 184-187
solutions of, 160
Pa r t i a l difference operators, 160
P a r t i a l differential equations, 160, 188
numerical solutions of, 218-221
Partial fractions, 192
P a r t i a l 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 o r 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
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
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 f o r n ! , 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 o r shifting operator, 3, 11-14, 160
Trapezoidal rule, 122, 123, 130
error term in, 123, 132, 133
proof of, 130
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
Yoling’s modulus of elasticity, 206
Zig-zag paths, 36, 37, 50-52
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