Operators: An algebraic synthesis

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Operators: An algebraic synthesis

Table of contents :
Preface
1. Sets; order
2. Relations; functions
3. Binary compositions; groups
4. Generalised addition and multiplication
5. An algebra of functions
6. Operators
7. Applications
8. Sequences
9. Summation of series
10. Linear difference equations
11. Linear differential equations
12. Infinite series of operators
13. Answers and solution notes

Citation preview

OPERATORS

OPERATORS An algebraic synthesis

D. R. DICKINSON, M.A., Ph.D. (CANTAB.)

MACMILLAN London - Melbourne - Toronto

ST. MARTIN’S PRESS New York

© D. R. DICKINSON 1967 First printed 1967 Library of Congress Catalog Number 67—19401

MACMILLAN AND COMPANY LIMITED Little Essex Street London W.C.2 also Bombay Calcutta Madras Melbourne THE MACMILLAN COMPANY or CANADA LIMITED 70 Bond Street Toronto 2 ST. MARTIN’S PRESS INC 175 fifth Avenue New York NY 10010

Made and printed in Great Britain by William Clowes and Sons, Liniited, London and Beocles

PREFACE I have made an attempt in this book to develop a theory of operators from the concepts of abstract algebra. Although the level is roughly that of university entrance, I would hope that the book could

be read with profit in the last school year. Equally well I hope that the book may give some pleasure to more mature mathematicians. The subject offers a challenge to precise habits of thought and I feel that I have myself learned a great deal in studying it. I derived considerable enjoyment from compiling the examples and exercises, particularly those in Chapters 7, 8 and 12, and I would urge the

reader to try his hand at constructing his own identities. I should certainly welcome any novel results for possible inclusion (with due acknowledgement) in a later edition. Mainly to help the beginner, the fundamentals of abstract algebra have been explained in the first four chapters. The more mature reader may well begin at Chapter 5 and refer to the earlier work as

required; the terminology of abstract algebra has changed so much over the last few years that he may find it advisable to refer to the index for the current meanings ascribed to certain terms, such as the range of a function. In Chapter 5 I have developed, in more detail than is usual, an

algebra of functions. This is indispensable for a proper study of operators and it soon becomes clear that there are, for our purpose,

serious shortcomings in the traditional manner of writing functiontheory. The general theory of operators is expounded in Chapter 6. Whilst it is rash to claim originality in mathematics, as far as I am aware the precise theoretical role of the Operator 0 has not before been studied in detail. Also the work on quasi-linear and normal operators is new to me. The remainder of the book is devoted to applications of translation, differencing and differentiating operators. These applications are mainly of academic interest, but some important formulae of

numerical analysis are given in Exercise 12.2(6).. As far as I am aware, the formulae given in 7.3 for successive difl'erences of a product are new. I have pleasure in recording my gratitude to Mr. A. J. Moakes of

vi

PREFACE

St. Paul’s School and to Dr. M. J. Zuckermann. Both these gentlemen

read the manuscript with great care and offered valuable criticism. I am also most grateful to three former pupils, T. W. Kismet, A. G.

Smith and R. P. A. Welch for their industry in checking the answers and compiling some of the exercises.

Bristol, 1967

D. R. D.

CONTENTS

N—‘O‘O

u—ou—n—a

axiom-hun—

Preface Sets; order

Applications Sequences Summation of series Linear difference equations Linear differential equations Infinite series of operators

13 26 41 57 73 96 119 135 148 178 195

Answers and solution notes

213

Index

243

Index of notation

245

Relations; functions Binary compositions; groups Generalised addition and multiplication An algebra of functions Operators

1 SETS; ORDER 1.1 Types In mathematics we are concerned with statements which ‘make sense’. More formally we say that a statement is significant ifi‘T it is either true or false. At first sight it might appear that every statement must of necessity be either true or false. However, if we

consider the statement This statement isfalse, we find that the assumption that the statement is true leads to the conclusion that it is false, and vice versa. Thus the statement is not a significant one, but is a meaningless collection of words. As a less subtle illustration, consider the statement

7‘} is odd. This statement is not significant because parity is not an attribute of rational numbers. It is now apparent that we must exercise some care in making statements about things whose properties we investigate; in other

words we must be careful to ask ourselves the right kind of questions about them. We say that two things are of the same type for the pur-

pose of some investigation ifl‘ every significant statement we wish to make about either of them may also be made significantly about the other.

A significant statement made about particular named things is called a proposition. Thus statements made in arithmetic are propositions. For example, 2 + 3 = 5,

2 < 5

are true propositions, and

i + i = t.

ix’+y”.

24

OPERATORS

When we here replace x and y by real numbers we obtain a proposition. This proposition is a value of the propositional function P defined by writing

P = {((x,y), (x + y)"‘ > x2 + y“) | (x, y) e 11’}Thus if we write Let P (x, y) be the statement (x + y)“ > x2 + y2

then we are indicating that we are using P as a name for a particular propositional function whose arguments are, by implication, ordered pairs of real numbers, and whose domain is R”. We show by a simple example how clarity and conciseness can be achieved in proofs using mathematical induction by introducing propositional functions. Example 2.1 Prove that if n e N and n 2 1 then

i(2r—l)=n’. r-1

Proof.

Write fln) = 2”: (2r — l),

g(n) = n’,

and let P(n) be the statement

f(n) = g(n)Thenf(l) = l, g(l) = l, and so P0) is true. Now assume that P(n) is true. On this hypothesis, fin + l) = 3:07 — l) =f(n) + (2n + l)

= g(n) + (2n + 1) = (n + l)“ = g(n + 1). Thus

P(n) => P(n + 1).

It now follows by induction that P(n) is true whenever n e N and n 2 l.

RELATIONS; FUNCTIONS

Exercise 2.7

25

Prove that ifn e N then n5 -— n_ is divisible by 5.

Note. We may formally define a theorem as a propositional function all of whose values are true propositions. In practice we write down only the enunciation of a theorem. This is an open statement containing variables (which may be words or symbols) and we imply that this statement is always true, so that whenever we replace the variables by admissible values a true proposition results. Unless it is implied by the context, the domain of a theorem must be included in the enunciation.

2+

3 BINARY COMPOSITIONS; GROUPS 3.1 Binary compositions Let 6’ be a concourse, and let S be a non-null subset of o". In this

chapter we investigate laws which associate a unique element of 6 with each ordered pair of elements of S. We call any such law a binary composition over S. For example, addition and multiplication are binary compositions over each of the concourses R, Z and N, and

so over any non-null subsets thereof. Subtraction is a binary composition over R and Z, but not over N. Considering the concourse R,

division is a binary composition over the subset {xeRIx aé 0}. As an example outside arithmetic, we observe that union and inter-

section are binary compositions over a concourse of sets of elements of a common type. Let us use x a: y for the element of 6’ which some binary composition over S associates with the ordered pair (x, y) of S“. Here a: is a variable, which in specific cases may be replaced by well-defined signs such as +, x , —, +, U and n. It increases our flexibility of

expression if we agree to use * in isolation as a pronoun representing any binary composition. More formally, we may define a: as a functional variable which maps S2 into 6’ by writing

* = {((x,y),v)eS”6’|v = My}Thus in studying binary compositions we are, in efi'ect, studying func-

tions of two variables. We use the notation x at y in place of *(x, y) because it is more convenient for our purpose and serves to emphasise the line of inquiry we have in mind. We say that a: is closed over S, or that S is closed under all, ifi‘ x a: y e S

whenever (x, y) e S“. For example, addition is closed over Z, but not over any subset of Z whose elements are all odd integers. (It should be mentioned that there are important binary compositions correlating elements of different types. In the theory of 26

z

BINARY COMPOSITIONS; GROUPS

27

vectors, for example, multiplication of a vector [x, y, 2] by a real number k yields the vector [kx, ky, kz], and scalar multiplication of two vectors [x1, y,, 21] and [x2, ya, 22] yields the real number xlxg +y1y3 + 2122. We are not however concerned with such binary

compositions here.) Exercise 3.1 Give examples of subsets of R which are not closed under (1) addition, (2) subtraction, (3) multiplication, (4) division. 3.2 Commutafivity and associativity We say that a binary composition a: is commutative for elements

x, y of 6 ifl' x as y and y a: x both exist and are equal. Also we say that * is commutative over S, or that S is commutative under *, ifi'

x*y=y*x

whenever (x, y) e 52. We say that a: is associative for elements x, y, z of 6, in this order,

ifl‘ (x a: y) * z and x a: (y a: 2) both exist and are equal. Here we must postulate the existence of (x * y) at: z and x * (y a: 2) even when (x, y, z) e 83, since we have not postulated that S is closed under 4:. Suppose, for example, that our concourse is Z and that S is the set of

all odd integers. If we take x at y to mean fix + y) then * is a binary composition over S, but there are elements x, y, z of S for which (xx: y) a: z and x * (ya: 2) do not exist. By contrast, whenever (x, y, z) e 53 neither x + y nor y + z are elements of S, yet nevertheless addition is associative for x, y, 2. Let us define x a: y * z to mean x a: (y * 2) whenever this exists. Then to assert that a: is associative for x, y, z is to postulate that

(x * y) a: z and x a: y a: 2 both exist and are equal. Let us also define w*x*y*z to mean w*(x*y*z) whenever this exists, and so on for any number of applications of *. More formally, if n is a natural number greater than 2, we define x1 :1: x2 *J_* x, to mean

x1 * (x2 *J_*x,,) whenever this exists. Here we presuppose that x2 *J_* x,, has already been defined. This process is called definition by recursion. We say that a: is associative over S, or that S is associative under

4:, ifi' * is associative for x, y, z wherever (x, y, z) 6 S3. When studying associativity we are chiefly concerned with cases in which S is closed under *. Closure ensures that if (x, y, z) e S3 then (x * y) at: z and x * (y a: 2) both exist and are elements of S. Examples of binary

28

OPERATORS

compositions which are closed, associative and commutative are addition and multiplication over R and union and intersection over the set of all subsets of a concourse. On the other hand, subtraction

is closed but is neither associative nor commutative over R. If a: is closed and associative over S we may insert and remove brackets at will in continued applications of * to elements of S, just

as we do in continued sums and products in arithmetic. If a: is also commutative over S we may alter at will the order of the elements of

S in any continued application of *. We usually gloss over the details, but the reader may find it instructive to supply them for himself occasionally. We give here two examples showing in detail how associativity is used. Example

3.1

If * is closed and associative over S, and

(w, x, y, z) 6 S4, prove that (w*x)*(y*z) = w*(x*y)*z. Proof.

Applying the associativity of a: to w, x and y 4: z,

m*w*o*a=w¥e*0*m w*(x*y*z) = w*x*y*z. Also

w*(x*y)*z=w*((x*y)*z)

=w*(x*y*z) = w*x*y*z. Thus Example 3.2

(w*x)*(y*z)=w*(x*y)*z. If a: is closed, associative and commutative over

S, and (x, y, z) e 83, prove that x*y*z=z*y*x. Proof:

x*y*z=x‘*(y*z) =x*(z*y) =(x*z)*y

=c*o*y =za-(xary) =z*(y*x) =z*y*x.

BINARY COMPOSITIONS; GROUPS

29

x*y*z=x*z*y =z*x*y =z=ry*x.

Morebriefly,

Algebraical manipulation of binary compositions is most fruitful when they are closed and associative. There are, however, important

instances of closed associative binary compositions which are not commutative. In particular, when we come to define multiplication of operators we shall find that this is a closed associative binary com-

position but is not always commutative. . There is a tendency for a newcomer to this work to feel that commutativity confers associativity. A counter-example will dispose of this misconception. For (x, y) e R“, write

x * y = l (x + y)Then clearly R is closed and commutative under *. Now, if

(x. y, 2) e R3, (x*y)*z = i:(x+y + 22')

and

x*(y*z)=i(2x+y+z).

Thus, in this case,

O*»*z=x*0*a ifi' Exercise 3.2 form

x=z. In the functional notation, commutativity takes the *(x9 y) = *0, X).

Write down the corresponding form for associativity. Exercise 3.3 For (x, y) e R“, let x u: y = x + y — xy, so that at is closed over R. Prove that an: is then commutative and associative over R.

Exercise 3.4 If :1: is closed and associative over S and (x, y) e S“, and ifx a: y = y a: x, prove that

x*x*y*y*y=y*y*y*x*x.

30 3.3

OPERATORS Neutral elements

In arithmetic 0 has the property that, as we may put it loosely, any . number is unchanged when 0 is added to it; in symbols,

x+0=x

whenever xeR. Moreover 0 is the only real number with this property, so that if (a e R, and x+w=x

whenever x e R, then at = 0. The number 1 has the corresponding property for multiplication; if k e R, and kx=x whenever x e R, then k = 1. We call 0 and l the neutral elements of R

under addition and multiplication respectively; some authors call them identity elements.

Now let a: be a binary composition over a set S. As we shall not postulate that * is necessarily commutative over S, in searching for neutral elements we must distinguish two cases:

(1) Ifi‘ i is an element of S such that i 4: x = x

whenever x e S then we call i a left-hand neutral element of S under *. (2) 111'j is an element of S such that x *j = x wheneVer x e S then we callj a right-hand neutral element ofS under *. As an example of a set which possesses no neutral element under a binary composition we cite the set of all even integers. Under multiplication this set is closed, associative and commutative, but possesses no neutral element. Provided S is not commutative under *, it is possible for a left-

hand neutral element to exist when there is no right-hand neutral element, and vice versa. For example, under subtraction in R, 0 is a right-hand neutral element but there is no left-hand neutral element.

Moreover, S may have more than one left-hand or right-hand neutral elements under ar, as the following example shows:

BINARY COMPOSITIONS; GROUPS

31

Example 3.3 Let S be a set consisting of two distinct elements a and )3, and write a*a=oe, fi*cz=a,

a*fl=fi, ,3*fi=fi.

Then both a and B are left-hand neutral elements of S under *, but

neither is a right-hand neutral element. > Note that S is closed and associative under *; whenever (x, y, z) e S", (x*y)*z=z,

x*(y*z)=z.

Theorem 3.1 If, under a binary composition *, a set S possesses both a left-hand neutral element i and a right-hand neutral elementj then j = i. Also iis the only left-hand neutral element and the only right-hand neutral element. Proof.

Since i is a left-hand neutral element,

j = i*j. Since j is a right-hand neutral element, i = i t j.

Therefore

j = i.

Moreover, if k is a left-hand neutral element then, by the same argument, j = k. Thus k = i, so that i is the only left-hand neutral element. Likewise iis the only right-hand neutral element.

By virtue of Theorem 3.1, to postulate the existence of both lefthand and right-hand neutral elements is equivalent to postulating the existence of an element i of S such that i* x = x * i = x whenever x e S. Then iis the only left-hand neutral element and the

only right-hand neutral element, and we may refer to i as the (unique) neutral element of S under an. Exercise 3.5 For subsets of a concourse 6’, is there a neutral element under (1) union, (2) intersection?

3.4 Inverse elements If x is a real number then there is a unique real number, which we call the negative of x and denote by — x, such that x+ (—x) =0.

32

OPERATORS

Also, provided x aé 0, there is a unique real number, which we call the reciprocal of x and denote by x‘ 1, such that

x‘lx = 1.

Note that, although subtraction is not commutative over R, by regarding x - y as x + (—y) we may use the commutativity of addition to write x — y = -y + x. Likewise division is not commutative, but if y aé 0 we may write

xy‘1 = y‘lx.

We call — x and x" 1 the inverses ofx under addition and multiplication respectively. We now give a more general definition of inverse elements. Let an be a binary composition over a set S, and suppose that S possesses a neutral element i under an. Let x e S. Then any element x1 of S such that x1*x=i

is called a left-hand inverse of x under an. Also any element x2 of S such that x*x2=i

is called a right-hand inverse ofx under It. Theorem 3.2 Suppose that, under a binary composition at, a set S is associative and possesses a neutral element 1'. Let x be an element of S which has both a left-hand inverse x1 and a right-hand inverse x2. Then x2 = x1. Also x1 is the only left-hand inverse of x, and is the only right-hand inverse of x.

Praofl

We are given that x1*x=i,

Hence

x*x2=i.

x2 = i an x2

= (x1 at x) * x2. Since S is associative under *,

(x1*x)*x2 =x1*(x*x2) =x1*i =x1.

BINARY COMPOSITIONS; GROUPS Thus

33

x2 = x1.

Moreover, if x3 is a left-hand inverse of x then, by the same argument, x2 = x3. Therefore x3 = x1, so that x1 is the only left-hand inverse of

x. Likewise x1 is the only right-hand inverse of x. Note that this proof makes no appeal to closure.

By virtue of Theorem 3.2, provided S is associative and possesses a neutral element 1' under at, to postulate the existence of both a lefthand and a right-hand inverse of an element x of S is equivalent to postulating the existence of an element x of S such that ' i*x=x*x=i.

Then 2': is the only left-hand inverse of x and the only right-hand inverse of x, and we may refer to a? as the inverse of x under *. Note that iis its own inverse. Also if x has inverse x then x has inverse x, so that f = x. If x and y are elements of S which both have inverses under *,

and if S is closed and associative under *, then (cf. Example 3.1, p. 28)

Likewise

(xaty)*(}7*3'c) =x*(y*y)*x =x*(i*x) =x*x=i. ()7 *3?)* (x4: y): i,

Thus x * y has an inverse under *, and x*y=y*x.

Exercise 3.6 Which integers of Z have inverses under multiplication ? For real numbers a and c we may regard a — c as being defined to mean a + (— c). Alternatively we may regard the problem of evaluating a — c as that of finding a real number x such that c + x = a.

For given a and c, this equation for x has unique solution x=a-c. 2.

34

OPERATORS

When we come to generalise this result for binary compositions which are not necessarily commutative, we must take care to dis-

tinguish between solutions of the two equations c 4: x = a,

Theorem 3.3

x * C = a.

Suppose that, under a binary composition *, a set S

is closed and associative and possesses a neutral element i. Let a and c be elements of S, and suppose that c has an inverse c' in S under *, Then, for elements x, y of 5,1" (1) c*x = a¢x = c'*a;

(2) y*c= a¢y=ata Proof:

We give the proof of (l) and leave that of (2) for the reader.

Inthefirstplaee, ifx = 6* athenxe Sand

c * x = c at (c‘ a: a) = (c * c') a: a = ita = a. Thus

x=c'*a=>c*x=a.

Now suppose, conversely, that c*x=a. Then

6*(c*x)=c'*a.

Now

é*(c*x)=(c'*c)*x =i*x =x.

Thus

c*x=a=>x=c'*a.

Note that, in particular, c a: x = c ¢> x = i. 3.5

Groups

Let a: be a binary composition over a set S. Then we say that S is a group under * ifi‘ the following conditions are satisfied:

(1) S is closed and associative under *. (2) S possesses a neutral element under *. (3) Each element of S has an inverse under *. When these conditions are satisfied S is called a commutative group 1‘ For propositions p and q, p q means p => q and q => p, Le. 11 '" q.

BINARY COMPOSITIONS; GROUPS

35

or a non-commutative group under * according as S is or is not commutative under 4:. If S is a group under * then, by Theorem 3.1, the neutral element of S is unique. Also, by Theorem 3.2, each element of S has a unique inverse. We give here some familiar examples of commutative groups: (1) Z under addition. (2) R under addition. (3) The set of all rational numbers under addition; this set is often called Q.

(4) {x e R | x > 0} under multiplication. (5) {x e Q | x aé 0} under multiplication. The reader should be able to add to this list. In 3.6 we study in detail a non-commutative group. Exercise 3.7 Is the set of all irrational real numbers a group under addition 7 The theorem which follows is sometimes useful for testing whether a given set is a group under a given binary composition. The hypotheses of the theorem may be taken as an alternative definition of a group. Theorem 3.4 Let S be a non-null set which is closed and associative under a binary composition * and suppose that, whenever (a, c) e S”, there are elements x, y of S such that c at x = a,

y a: c = a.

Then S is a group under *. Proof: Let (a, c) 6 82. Then, by hypothesis, there are elements i, j of S such that a * i = a,

j a: c = c.

Also there are elements 6, c' of S such that d * a = j,

c a: 6 = i.

Using the associativity of S, j*i=d*a*i=d*a=j, andalso

j*i=j*c*é=c*c'=i,

OPERATORS

36

Thereforej = i. It follows that iis unique, and is independent of the choice of a and c. Thus iis a neutral element of S under *.

Since

d*a=j,

c*c'=i,

every element of S has both a left-hand and a right-hand inverse

under an: and so, by Theorem 3.2, every element of S has a unique inverse under *. Thus S is a group under *. It now follows from Theorem 3.3 that in S, for given a and c, the

equations c a: x = a,

y a: c = a

.x=c'*a,

y=a*c'.

have unique solutions

Exercise 3.8 For subsets A, B of a concourse 6’ let A A B be the subset which consists of all the elements of a? that belong to just one of A and B. Prove that the set of all subsets of 6’ is a commutative

group under A. If, for subsets of of,

AAX= B, express X in terms of A and B. (A A B is called the symmetric difi'erence of A and B.) 3.6

Permutation Groups

There are six permutations of l, 2, 3. For convenience we give them letters as names, and write i = (l! 2! 3)!

P = (391: 2),

q = (2: 3:1),

1' = (3: 2: l),

S = (2’ 1’ 3):

t = (11 3: 2)‘

Let us call the set of these six permutations Sa. We shall define a

binary composition over Sa under which Se is a non-commutative

group. We call our binary composition multiplication, and we say that $3 is a permutation group under this multiplication; the alternative term symmetry group is sometimes used. For a geometrical realisation of this group in terms of transformations of an equilateral triangle, the reader is referred to The Core of Mathematics by A. J. Moakes (Macmillan), pp. 30—33. Let x and y be permutations of l, 2, 3. Then we define the product

BINARY COMPOSITIONS; GROUPS

37

xy as follows: Ifthe first component of x is [C (so that k is 1, 2 or 3)

we take the kth component ofy as the first component ofxy. We then repeat this process to obtain the second and third components of xy. Thus xy is a permutation of l, 2, 3, and so S3 is closed under multiplication.

We can express the rule for evaluating xy more concisely by writing x = (x1: 7‘2) X3),

y = 01: y2! J’s)-

xy = (yxp J’xz, yam)-

Then

For example,

ps = (3, l, 2)(2, l, 3) = (3, 2, l) = r,

and

sp = (2, l, 3)(3, 1, 2) = (l, 3, 2) = 1‘.

Thus ps 76 sp, and so multiplication is not commutative over SS. We now give the complete multiplication table, which the reader should verify: (Second factor) zpqrs t Hawa'uu.

(First factor)

t

s r q

1.’z

By inspection of the table we see that, under multiplication, SS has i as unique neutral permutation; since we are calling our binary composition multiplication, we call i the unit permutation. Also each permutation has a unique inverse under multiplication. We here use the term reciprocal in place of inverse and denote the reciprocal of a permutation x by x~ 1, so that we write

1'4 = 1',

if]l = q,

q‘1 =p,

r’1 = r,

s‘1 = s,

t‘1 = .

To prove that S3 is a group under multiplication it remains to prove that multiplication is assoCiative over $3. Let x, y, 2 be permutations of l, 2, 3, and write x = (x1, x2, 3‘8)!

y = 0’1. J’as y3):

Z = (21» 22: 23)-

38

OPERATORS 0002 = (yxp yxa’ yxal: 22: 23)

Then

= (211.9”, 21123, 23.13)

t1 = 2,“,

Now write

Then

t2 = 2,2,

t3 = 2,3.

3602) = (x1, x2, xa)(t1, t2, t3)

= (tan: txga 1x3)-

Clearly

t,,1 = 2m,

Thus

t,,2 = 2,“,

t,a = 2,”.

x(yz) = (xy)z.

This completes the proof that S3 is a group under multiplication. We observe that, in the body of our multiplication table, each row

and each column consists of the six permutations i, p, q, r, .9, tin some order; such a configuration is sometimes called a Latin square. By Theorem 3.4, this Latin-square property, together with associativity, provide an alternative proof that S:, is a group under multiplication. For example, by inspection ofthe second row ofthe table, the equation

px = t has unique solution x = r; alternatively,

x=p'1t=qt=r. Likewise yp = t ifi‘y = s. The theory which we have developed for S3 is readily extended to the set S,l of permutations of l, 2, . . ., n, where n e N and n 2 1. If

x and y are such permutations, and x=(x19x2s-'-,xn)’

y=(yl:y2’---,yn)r

we define xy by writing xy = (yxp yxaa - - -: ya.)-

Then evidently the multiplication table for S, is a Latin square. Also associativity can be established by the method we have used for 83. ' Exercise 3.9

For S3, verify by direct calculation that

(l) (.0103 = PODS), Exercise 3.10

(2) (")4 = r(Sq)-

Name the non-null proper subsets of Sa which are

BINARY COMPOSITIONS; GROUPS

39

groups under multiplication; we call each of these a proper subgroup of 8;; under multiplication.

Exercise 3.11

Solve, where possible, the following equations

in $3:

(1) rx = t.

(2) xr = t.

(3) sx = p.

(4) xs =1)-

(5) qx = p-

(6) xq = p-

(7) xx = p. (10) xrx = r. (13) xxt = t.

(8) xx = r. (11) q = i. (14) xr = sx.

(9) xrx = t. (12) q = q. (15) xr = px.

Exercise 3.12 For permutations (x1, x3, x3) and (yl, ya, ya) of 1, 2, 3, define (x1, x2, x3) * 0’1, J’a: J’s)

to be the permutation which, for m = l, 2, 3, has y,,I as its xmth

component. Write out the composition table for *. Is Ss a group under It? Exercise 3.13 Let M2 be the set of all 2 x 2 matrices whose elements are real numbers, and let R2 be the subset of non-singular 2 x 2matrices.

(1) Prove that M3 is a commutative group under addition. (2) Prove that R2 is a non-commutative group under multiplication. Is R2 a group under addition? [IfX and Y are 2 x 2 matrices, and

X= (x11 x21

x12),

Y= J’u J’m),

3‘22

21

.722

then X + Y and XY are defined by writing X + Y = (3511 + J’n

3‘12 + ym)

3‘21 + J’al

and

XY = (x11y11 + x12y21 x21J’11 + x22y21

3‘22 + Y22

x11J’12 + x1aJ’22). x21J'12 + xa’m

The determinant of X is the number 3‘11"” — 35217512:

40

OPERATORS

sometimes denoted by det X. We say that X is non-singular iff det X 75 0.] Exercise 3.14 Under a binary composition *, a set S is closed and associative and possesses a left-hand neutral element i. Corresponding to each element x of S there is an element a? of S such that i * x = i. Prove that S is a group under *.

4 GENERALIS ED ADDITION AND MULTIPLICATION 4.1 Postulates In Chapter 3 we studied the consequences of certain premises regarding a single binary composition. In arithmetic we have two closed,

commutative and associative binary compositions, addition and multiplication, which are interrelated: multiplication is distributive over addition, so that, if x, y, z are all natural numbers, or integers, or rational numbers, or real numbers,

x(y+z)=xy+xz. Our purpose now is to investigate in abstract form more general cases of two binary compositions, one of which is distributive over the other. We suppose that these binary compositions are both

associative but we do not insist that they are necessarily commutative. This being so, little headway can be made without two forms of

distributivity. These we shall state presently. To avoid tedious repetition we postulate throughout this chapter

that S is a non-null set which is closed and associative under each of two binary compositions. We call these compositions addition and multiplication, and if (x, y) e 82 we denote the unique elements of S which they associate with (x, y) by x + y and x x y respectively. As

in elementary algebra, we often write x. y or xy in place of x x y. If (x, y, z) e 83 then, as in 3.2, we take x + y + z to mean x + (y + z), and xyz to mean x(yz). Since, by hypothesis, addition and multiplication are associative over S, (x+y)+z=x+y+z,

(xy)z=xyz.

We further postulate that W + z) = xy + xz

and

(y+z)x=yx+zx,

so that what we may call pre-multiplication and post-multiplication 41

42

OPERATORS

are both distributive over addition. These two forms of distributivity

are equivalent ifi' both addition and multiplication are commutative over S.

It must be clearly understood that in the present context we are using the words addition and multiplication, and the signs + and x , as variables. Any statement we make concerning them is an open statement which does not become significant until, in a particular context, we have ascribed definite meanings to them. For example, if we take S to be the set of all subsets of a concourse 6’ then all the axioms we have so far postulated are satisfied if we take union as addition and intersection as multiplication; see 1.4. (It should be noted that, in this

case, if x and y are subsets of (6” then xy denotes x n y as well as the cartesian product of x and y. One cannot, at times, easily avoid having two diverse meanings for the same symbolism, but in practice it is usually clear which meaning is intended.) Alternatively our

axioms are all satisfied if we take intersection as addition and union as multiplication. In future, when it is clear that we intend certain symbols to

represent elements of S, we shall not always state this explicitly. Exercise 4.1 (1) Expand w(x + y + z). (2) Expand (w + x)(y + z) in two different ways. 4.2

Zeros and unifies

When we postulate that S possesses a neutral element under addition we call this element the zero element, or the zero, of S, and denote it by 0. Then, whenever x e S,

x + 0 = 0 + x = x. In particular,

0 + 0 = 0.

When we postulate that S possesses a neutral element under multiplication we call this element the unit element, or the unity, of S, and denote it by 1. Then, whenever x e S, x1 = 1x = x. In particular,

1 x 1 = 1.

Note that, in a general investigation, 0 and l are variables; they

are perhaps best described as parameters. If S possesses both a zero andaunitythen 0 x1=1x 0 = 0.

GENERALISED ADDITION AND MULTIPLICATION

43

The following exercise shows however that, unlike in arithmetic, there may be an element x of S such that 0x 95 0.

Exercise 4.2 Suppose that S consists of three distinct elements a, B, y, and that addition and multiplication are closed commutative

binary compositions over S, defined by the following tables:

X06137 flufifi 70:37 Prove that over S addition and multiplication are both associative, and that multiplication is distributive over addition. Show that S possesses a zero and a unity, and that there is an

element x of S such that 0x aé 0. 4.3 Negatives and reciprocals Suppose that an element x of S has an inverse under addition; this implies that S possesses a zero. Then we call this inverse the negative of x and denote it by —x. We usually write y — x in place of

y + (—x). Thus we write x — x = -—x + x = 0. Note that —x then also has a negative, and

—(—x) = x. If x has an inverse under multiplication, so that S possesses a unity, we call this inverse the reciprocal of x and denote it by x'l. Then xx‘1 = x'lx = 1.

Also

(x'1)‘1 = x.

By a result proved on p. 33, if —x and — y both exist then so does —(x + y), and —(x + y) = -y —x. Also if x“1 and y‘1 both exist then so does (xy)'1, and

000'1 = y'lx‘l-

44

OPERATORS

The following theorem is somewhat remarkable: Theorem 4.1 S is a set with the structure postulated in 4.1. If S possesses a zero and a unity, and the elements x, y both have negatives, then addition is commutative for x and y.

Proof: Using the distributivity over addition of both postmultiplication and pre-multiplication, (x+y)(1 + l)=x(l+l)+y(l +1) =x+x+y+fi cf. Exercise 4.1(2), p. 42. Also

(x+y)(l + l)=(x+y)l +(x+y)1 = x + y + x + y. Therefore

x+x+y+y=x+y+x+y.

Pre-adding —x and post-adding —-y to both sides,

(—x+x)+(x+y)+(y—y)=(—x+x)+(y+x)+(y—y). Since

—x+x=0,

it follows that

y—y=0,

x + y = y + x.

4.4 Residues In the theory of numbers, problems on divisibility deal with the remainders when natural numbers are divided by some natural number p greater than 1. The number p is called the modulus of division, and the set of all the p possible remainders is called the set of residues modulo p. Let us name this set N,” so that N,={reN|0 2 prove that the continued sum of n functions of 9', each of which is f, is the function n x f, and that their continued product is the functionf“. Exercise 5.2

Discuss the laws of indices for real functions of a

single real variable. 5.5

Composite functions

We now return to a consideration of the notation for the product of

two functions. We confine our attention to real functions of a single real variable, as it is with such functions that we are mainly concerned. Let us denote the set of all such functions by .931. For

functions f, g of 9?, we sometimes wish to interpret fg as f x g. For

AN ALGEBRA or FUNCTIONS

63

example, we may wish to interpret 2 cos x not only as the product of the numbers 2 and cos x but also as the value onto which the function 2 x cos maps the argument x. Unfortunately fg is traditionally used

to denote a composite function, often called a function of a function. For example, log sin x is taken to mean log (sin x) whereas on occasion we wish to take it to mean (log x) x (sin x). Note also that

if 2 is interpreted as a function then, according to the composite function convention, 2 cos x would mean 2(cos x), which is the

number 2. While we often retain the multiplication sign for emphasis, let us agree to break with tradition and always use fg to mean f x g. This

convention is particularly desirable in the study of operators. To denote a composite function let us use f o g for the function which maps each admissible argument x onto the valuef(g(x)). For example, we take 2(tan + l) to mean the product of the functions 2 and tan + l, and we use 2 0 (tan + l) to denote a composite function. This latter function is, of course, a constant function whose value

is always 2 and whose domain is 9 tan. There is one other source of possible ambiguity that is worth mentioning. This arises from the custom of using the same kind of enclosing brackets when writing argument as those used to indicate that an addition is to be performed before a multiplication. For example, consider the expression 2(tan x + 1). .Here we should normally mean the product of the numbers 2 and tan x + 1, assuming tan x to exist. However, if2 is interpreted as a function and (tan x + l)

as an argument of this function, then the value ofthe expression is the number 2. In cases such as this there is no traditional method for discriminating between the use of the symbols of arithmetic as numerals and as names of functions. It seems that the best we can do is to agree that these symbols shall represent numbers when otherwise there would be doubt as to their interpretation. While this does not seem a very satisfactory compromise it is always clear in practice which meaning is intended. For example, according to the convention that fx andfix) have the same meaning, if 2 denotes a function then 2x is the number 2. However, we should naturally take 2x to mean the

product of the numbers 2 and x if nothing were stated to the contrary. The difficulties in notation which we have been discussing un-

doubtedly arise from the fact that as, over the years, conventions in the writing of mathematics slowly developed, the need for an algebra of. functions was not fully realised.

64

OPERATORS Exercise 5.3

(l) Prove that o is associative over .921.

(2) Is 0 commutative over 91 ? (3) Has .921 a neutral element under 0 ? (4) Is .921 a group under 0 ? 5.6

Identity functions

Among real functions of a single real variable there is one important function which unfortunately has no agreed abbreviated name. This is the function which maps each argument onto itself. This function is called an identityfunction. We propose to give it the abbreviated name :9, and suggest that this be read ide. Thus

0={(x,x)lR}, and 15x = x whenever x e R. If we write, for example, f(x)=ax+b then f denotes the functional variable (11’ + b, where a and b now

represent constant functions. A long-standing convention for the use of x allows us to speak of ax + b, without explanation of the symbols employed, when we mean the value (a? + b)x, where a and b are constant functions.

The concept of an identity function applies whenever A = V. For example, when we study complex functions of a single complex variable we take 0 ={(z,z)|zeC}. For functions of two variables, with A = V“, we define the first

andsecond identityfunctions, :91 and :92, by writing

191 = {((x, y), x) I (x, y) 6 V2} and

192 = {((x. y), y) | (x, y) 6 Va}-

Thus, whenever (x, y) e V“, 01(x, y) = x and 192(x, y) = y. This notation can immediately be extended to functions of any number of variables.

Exercise 5.4 The following equalities are true whenever both sides are significant. State corresponding results for functions, using

AN ALGEBRA OF FUNCTIONS

65

the inclusion sign C where necessary. Suppose that an argument is represented by (x, y, z) in (2), by (x, y) in (4), and by x in the remaining equalities.

(1) (x+l)’=x3+2x+l.

(2)x(y+z)=xy+xz.

x2—1_ (3)x_l—x+l.

(4) xa+y3 x+y _ —xa

(5) sinzx + cosax = 1.

(6) secax = 1 + tan“ x.

(7) coseczx — cotzx = 1.

(8) sin'lx + cos‘lx = int.

xy+y"’.

(9) tan"1 x + tan-1(l/x) = in sgn x. (10) (l — x“‘)“2 = cos (sin'1 x).

(11) (i ; :)1/a = tan (‘3 cos"1 x).

(12) (1— x) x

5.7

112

_ = tan (sm' 1 x“).

Derivatives

Letfbe a function of3%. Then the derivative offis that function of 91 which maps an argument x onto the real number

lim f(X) -f(x) x-vx

X — x

whenever this limit exists. This derivative is traditionally denoted by '. Note that we have here another case of duplication of notation,

forfis a set of ordered pairs and in set-theory a prime is used to indicate the complement of a set; see Exercise 1.4 on p. 8. However, in the present context, there is little danger of confusion. Note also that

.

9f’ S 9f-

To define higher derivatives offwe first denote the derivativef’ by f‘1’, and call it the first derivative off. We then define f(’0, the nth derivative of f, inductively by writing f"0 = {f"“1)}“’ whenever n e N and n 2 2. It is also convenient to define f‘°’ to be f. As the reader is doubtless aware, we sometimes write f' for f‘2’, f”' for f‘3’ and f(lv) for f“)-

Suppose that Then 3.

f(x) = x2 + 3 sin x + 2. f’(x) = 2x + 3 cos x.

OPERATORS

66

We can insert particular numerical values for x in these statements and obtain correct results. For example,

f(0) = 2.

f’(0) = 3-

Now the symbol D is often used to denote ‘differentiation with respect to x’. No difliculty arises if we use Df as a compound symbol, with precisely the same meaning asf’. Then, in our example, Df(x) = 2x + 3 cos x, and we may still write

Df(0) = 3. However, it is common practice to write D(x2+3sinx+2)=2x+3cosx.

(1)

If we here replace x by 0 we obtain

D0) = 3,

(2)

an erroneous result. The explanation is that, in the present context, the use of the symbol D is only significant in the composite form DfZ It is not until we come to study operators that we learn to attach a meaning to the symbol D in isolation. It therefore emerges that we must interpret the left-hand side of (l) as meaning D092 + 3 x sin + 2)(x).

(3)

When a dependent variable y is introduced, so that we write y = x2 + 3 sin x + 2 and

Dy=2x+3cosx,

we must again interpret Dy as meaning the expression labelled (3). In contrast to (2), the correct statement D2 = 0

is an equality between two functions. This statement is, however, often taken to mean D2(x) = 0, an equality between two numbers. We mentioned above the phrase ‘difi‘erentiation with respect to x’. Evidently here x is not a pronumeral. Now, iff and g are functions

AN ALGBBRA 0F FUNCTIONS

67

of 931 we maydefine the function D,f, the derivative off withrespect to g, as that function of 9?, which maps an argument (x) onto the real number

. f(X) — f(x) 3333: goo—— g(—x) whenever this limit exists. We may then interpret the derivative of x2 + 3 sin x + 2 with respect to x as

D9092 + 3 x sin + 2)(x).

The same interpretation is to be given to the expression D,,(x2 + 3 sinx + 2). Likewise means

D,,(x2 + a sin x + b) D0093 + a x sin + b)(x),

where a and b denote constant functions and D00?“ + a x sin + b) is a functional variable. There is no difliculty in giving a correct account of diferentials on the lines we have indicated. With a function f of 9?, and a. real parameter h, usually called an increment, we may associate a functional

variable (1,,fsuch that 9d,.f = Qf’ and

dnf(x) = f’(X)h whenever x 6 9f. Thus d,f = f’ x h. In particular, (1,0 = h.

(4)

Hence if h aé 0, as is always assumed, I _ dhf(x). f (x) '- dh'xx)

(5)

It is customary to suppress any explicit reference to the increment h and to write (if in place of d,f. It must, however, be remembered that

dfdoes not become significant until fhas been replaced by the name of some particular function and h by some numeral. In place of (4) and (5) it is customary, though misleading, to write

dx=h,

f(x)= ”(i)-

OPERATORS

68

We now consider briefly real functions of two real variables. Let us denote the concourse of all such functions by .922. Wheneverf e 9?, we define the partial derivatives f1 and 1",, to be such that, for each admissible argument (x, y),

f1(x, y) = 11331 W —x and

f2(x, y) = $33 f__(x’ YY) : ;(x, J’)_

We also use D1f and D2f to denote f1 and f2 respectively. The expression D1(x2 + 2xy + 1), (6) for example, must then be taken to mean D109? + 2 x :91 x 192 + 1)(x, y).

In expressions such as (6), D1 is often replaced by D,‘ or am. It is perhaps worth remarking, in connexion with this notation, that it is only by a long-standing agreement that we denote an argument by (x, y) rather than by (y, x). We could equally well take

3:70? +2xy+ 1) to mean

D2093 + 2 x #2 x 61 + 1)(y, x).

We'have already called attention to this ambiguity on p. 22.

Exercise 5.5 (1) Iffe .922, define the differential offwith respect to an ordered pair (h, k) of increments. (2) If f, g, G are functions of .422 we define (D,f)G, the partial derivative offwith respect to g keeping G constant, by writing (Dgf)0(xs y) =

fix, Y) — f(x. y) «312a,, g‘_“_—o 0, (fo (6* + c))(X) exists if X e [x — 8, x + 8]. When this condition is satisfied, and X aé x,

0°09 + c))(X) - (foo? + c))(x) = f(X + c) -f(x + 6) (X+c)—(x+c)

X-x

From the definition of a derivative it follows that (fo (:9 + c))’(x) exists ifi' (x + c) 6 9f’, and then

(f' ° (:9 + 0))(x) =f’(x + 6)Thus, whenever f e .921,

(f° (19 + c))’ =f’°(t9 + c)The following exercise is intended merely to provide practice in the use of unfamiliar notation. The reader may find it profitable to construct further examples. Exercise 5.6 Compute the following derivatives. Here we use the convention that fg means f x g. (l) (193 + 319')'_

(2) (19.3I2Y.

(3) {'90 _ 20)1]3}1.

72

OPERATORS

O}.

In particular Xf may be a null function when f is not null. For example, if x e R,

D[{(x, 1) | x is rational} U {(x, - l) | x is irrational}] is null; so also is

D{(0. 0)}On the other hand an operator may map a null function onto a nonnull function. For example, if 11 is the operator which maps every function of 5" onto the constant function 1 then, in particular, 11¢ = 1.

Incidentally, for reasons which will appear, we do not call 11 a con-

stant operator and we do not use the symbol 1 as a name for 11. Exercise 6.1 Construct further examples of operators on .9331, using set notation. For each operator X state whetherfand tave a common domain whenever f e .921.

6.2 An algebra of operators Let .97 be a concourse of functions with a common range which is a field under some laws of addition and multiplication. For convenience we need a symbol to denote the concourse of operators on .97 ; let us use 0 for this purpose.

OPERATORS

75

Let A and B be operators of 0. Then we define A + B to be the operator of (D which mapsfonto Af + Bfwheneverfe .7. Thus

A+B={(f.Af+Bf)lf€:9’}, and addition is a closed binary composition over 0. According to this definition,

B + A = {(f, Bf+ Amfear}. Since addition of our functions is commutative over 9, it follows that '

A+B=B+A. If C is also an operator of (D then, using the associativity of addition over 3", we easily deduced that (A+B)+C=A+(B+C);

the reader should write out a proof, using set notation. Thus 0 is commutative and associative under addition. Consequently in a continued sum of operators we may omit brackets without ambiguity arising and alter the order of the operators at Will. Turning now to multiplication, we define AB to be the operator of (0 which maps f onto A(Bf) whenever fe .97, so that if Bf = fl then ABf = Afl. Thus

_AB = {(f, A(Bf)) IfE-V}. and multiplication is a closed binary composition over (9. For example, iffe .921 then

Thus

DDf = Df’ = f”. DD = {02m 1 fe .421}.

Ifg is a function of .97 then we also use the symbol g to denote the operator which maps f onto g x f whenever fe .97, and we call g a functional operator. Provided we agree that, when g denotes a function, gf denotes the product g x f we observe, as a matter of consistency, that gfdenotes the same function whether we interpret g as a

function or as an operator. Also if g1 and g, are functions of .9” then the operator which is given the name g1 + g, the Symbol for the sum of the functions g1 and g2, is the sum of the operators g1 and g2; for, whenever f e 9',

(g1 + g2)f = g1f + 32f

76

OPERATORS

whether g1 and g, are interpreted as functions or as operators. Likewise no ambiguity arises in the use ofglga both for a function and for

an operator. If a is a constant function of 9' then we call the operator which we name a a constant operator. In particular we have the constant operators 0 and l, where

0={(f;0 xf)|fe.97,06.9"}

and

l={(f,f)lf65‘}-

Note once more that we must guard against replacing 0f by 0, since the domain of Ofis that off; see 5.3. For this reason, if X e 0,

X+0¢X unless f and Xfhave a common domain wheneverfe .97. In general, the most we can say is that

m+w wheneverfe .97. It would thus be misleading to call 0 the zero operator. On the other hand the operator 1 is a genuine unity, for evidently X1 = 1X = X whenever X e 0. Exercise 6.2

Is X + 0 S X whenever X e (0?

Exercise 6.3 Construct an example to show that (X + 0)f may be null when neither f nor Xf is null.

We now prove that multiplication is associative over our concourse 0 of operators. Suppose that (A, B, C) e 03 and that fe .97", and write Cf = fl. Then

(AB)Cf = ABf: = A(Bfl) and

A(BC)f = A(BCf) = A(Bfi).

Thus

(AB)Cf = A(BC)f

whenever f e .97, and so

(AB)C = A(BC). Consequently in a continued product of operators we may omit brackets without ambiguity. Also, if n EN and n > 2, it is meaningful

OPERATORS

77

to speak of the continued product of n A’s. We denote this operator by A”. Also we define A1 to be A and A° to be the operator 1. Then, whenever (m, n) e N“, AmAn = AnAm ___. Am+n.

Multiplication is not commutative over (9. To justify this assertion

we need only produce one pair of operators for which multiplication is not commutative. As in 6.1, write

I1 = {(f, l) lfeé", 15.97}. Alsowrite

Io={(f,0)|fe.97,0637}.

Then, whenever fe .97, I1Iof= I10 = 1,

whereas

1011f = 101 = 0.

Since 1 ye 0, it follows that Illa aé 1011.

The reader may prefer a more practical example. Let us try again. Write

S = {(flf’) ”63"}Then, whenever fe .57,

S(— 1)f = S(-f) = f’, whereas

(—l)Sf= —(Sf) = —f3.

Now it is possible thatf2 = —f2 wheneverfe .97 ? The answer is yes! For suppose that the common range of our functions is the field {0, 1}. Then —0 = 0 and —1 = 1, from which it follows that

f2 = --f” wheneverfefl'. ' We now consider some examples of non-commutative multiplication of operators on 91. Evidently, for such operators, S(— 1) 75 (— l)S. Also, whenever f e .931,

SDf = Sf’ = (f’)’, whereas

DSf = s,

sothat

DSfafxf';

see 5.8. Hence

SD 9e DS.

78

OPERATORS

As another example,

fiDf = :9 x f’, D19f2f+29 xf’;

whereas

in more familiar notation, whenever x 6 9f,

f(x) = xf’(x), whereas

Dxf(x) = f(x) + xf’(x).

Thus

19D aé D19.

Exercise 6.4 Construct examples of pairs of operators for which multiplication is (l) commutative; (2) non-commutative.

We next consider the distributive laws. Suppose that A, B, C are operators of 0. Then, whenever f e .7, (B + C)Af = (B + C)(Af)

= MN) + C(Af) = BAf + CAf = (BA + CA)f. It follows that (B + C)A = BA + CA.

In particular,

(1 + 1)A = 1A + 1A,

2A = A + A.

so that

It is a simple matter to induce that if n e N and n 2 2 then the continued sum of n A’s is nA. When we come to consider the distributive law

A(B+C)=AB+AC we find that this law is not always true. For example, using the squaring operator S defined above,

S(B + C)f= (Bf+ Cf)” = (Bf)2 +(Cf)"1 + 2 X (Bf) x (CD. whereas (SB + SC)f = (Bf)2 + (CD2. Thus, in general,

S(B + C) aé SB + SC.

OPERATORS

79

Note in particular that, in general, S(l + l) 75 S + S, so that

$2 aé ZS.

We have already shown that, in general,

S(--1) sé (-l)SThus we cannot in general use —A to denote both A(— 1) and (— 1)A. We choose to use —A to denote (—1)A. Then B — A denotes B + (-1)A. Exercise 6.5 Give an example of an operator X such that X0 95 0X. 6.3 Linear operators Our concourse (9 of operators includes an important subset, closed under addition and multiplication, over which the distributive

law A(B + C) = AB + AC is valid. This is the subset of linear operators, defined as follows:

An operator X of0 is said to be linear W,for all constant operators a and b of 0,

X(af + bg) = aXf + n whenever f and g are functions of.5". To avoid tedious repetition, in the enunciations and proofs of the theorems which follow it is to be understood that all operators mentioned belong to 0 and all functions to 37". Theorem 6.1 If X is a linear operator and a is a constant operator then Xa = aX. Proof.

Taking g = f and b = 0 in our definition of linearity, X(af+ 0]) = aXf+ 0X]:

Since afand 0fhave a common domain, we are justified in writing af+0f=af2

80

OPERATORS

Likewise

aXf + OXf = aXf.

Thus

X(af) = aXf.

By definition,

X(af) = Xaf.

Hence

Xaf = aXf

whenever f e .97. Thus Xa = aX,

so that multiplication is commutative for a linear operator and a constant operator. In particular, if X is linear,

X0 = OK, a result which is not true for operators in general; see Exercise 6.5. However, X0 aé 0 unless, whenever f e .97, the functions f and Xf have a common domain. If X is linear, X(— l) = (—1)X.

We have previously chosen to use —X to denote (— l)X. We now see that, if X is linear, X( — l) = — X. Theorem 6.2 Every functional operator is linear. Proof.

Let h be a functional operator. Since, over .97, multiplica-

tion is commutative and is distributive over addition,

h(af + bg) = ahf + bhg whenever a, b are constants and f, g are functions of .97. Thus h is

linear. In particular, every constant operator is linear. Theorem 6.3

If X, B, C are operators, and X is linear, then

X(B + C) = XB + -XC.

Proof.

'Letfbe a function of .97. Then, by definition, (B +’ C)f= Bf+ Cf.

OPERATORS Hence

81

X(B + C)f = X(Bf + Cf).

Since X is linear,

X(Bf + Cf) = X(Bf) + X(Cf). By the definitions of addition and multiplication of operators, (XB + XC)f= XBf+ XCf = X(Bf) + X(Cf). Thus

X(B + C)f = (XB + XC)f

whenever f e .97, and so

X(B+C)=XB+XC. Theorem 6.4 If X and Y are linear operators then X + Y is linear.

Proof. Let a, b be censtants and letf, g be functions of .97. Then, by the definition of X + Y, (X + Y)(af + bg) = X(af + bg) + Y(af + bg). Since X and Y are linear, it follows that

(X + Y)(af + bg) =. (aXf + n) + (a + n). Now addition is commutative and associative over .9". Hence (X + Y)(af + bg) = (aXf + a) + (n + n). The constant a is a linear operator, and so

aXf+ a= a(Xf+ Yf). Thus

aXf + a = a(X + Y)f.

Likewise

n + n = b(X + Y)g.

Hence

(X + Y)(af+ bg) = a(X + Y)f+ b(X + We

and so X + Y is linear. Theorem 6.5 If X and Y are linear operators then XY is linear.

82 Proof.

OPERATORS Let a, b be constants and letf; g be functions of 9'. Then,

using in turn the linearity of Y and X, XY(af + bg) = X{Y(af + bg)} = X(a + n)

= aX(Yf) + bX(Yg)

= a(XY)f + b(XY)gThus XY is linear.

Theorems 6.4 and 6.5 justify the assertion made earlier that the subset of linear operators of 0 is closed under addition and multiplication. 6.4 The operators E, A and V We now introduce some linear operators on 921 with which we shall be mainly concerned in later work. We denote by E the operator which maps f onto f o (6‘ + 1) whenever fe .921. Thus

E = {(fif°(" + 1))lf6931}Note that f and Ef do not necessarily have a common domain; eEfifi‘x +169f,andthen

Ef(x) =f(x + 1)We may call E a translation, since it' has the effect of moving the graph off a distance —1 parallel to the x-axis. Theorem 6.6 E is a linear operator. Proof: ' hit a, b be constants and let f, g be functions of .921. Then

the functions F)(af + bg) and aEf + bEg have a common domain, namely the intersection 9Ef n 9Eg. For any argument x of this domain,

E(af + bg)(x) = (af + bg)(x + 1) = af(x + l) + bg(x + l) = aEf(x) + bEg(x) = (aEf + bEg)(x). Thus E is linear. Exercise 6.6 Prove that, whenever (f, g) e .923,

-Efg = (E!) x (E3)-

cannon-s

83

.If X is an operator on 921 then, wheneverf e 921, -EXf = Xfo (a9 + l).

(1)

It is not necessarily true that EX = XE, even when X is linear. For example, the functional operator 6‘ is linear and

Eflf=(a9+l)x Ef, whereas

flEf = 19 x Ef;

in more familiar notation, ifi' x 6 9E];

Exflx) = (x + l)f(x + l),

xEflx) = xf(x + 1).

Taking X = E in (l),

E2f= Ef°(1’ +1)=f°(0 + 2) and evidently, by induction, if n e N then E"f = f o (15‘ + n). For any pronumeral c we define the operational variable E‘ by writing

E‘ = {(flf° (9 + 6)) If6 991}Then, replacing l by c in the proof that E is linear, we see that E”. is linear, Also if c, and c, are pronumerals then, wheneverfe 931, E‘1E°2f = E”1fo (19‘ + c2) =f° (:9 + Cl + 0;) = E‘1+°af. Thus

E°1E°a = E°1 +‘a,

and our definition of Ec is consistent with the law of addition for indices. Note also that our definition is consistent with the requirement that E° should be the unit operator 1. We proved in 5.10 that, whenever f 6 Q1,

(f°(t’ + C))' =f’ °(9 + C) so that, in our present notation,

DE‘f = E‘Df.

Thus multiplication is commutative for the operator D and E“. Two further useful operators are the forward dzflerencing operator A and the backward dtferencing operator V (nabla), where

A=E—1,

V=1—E-1.

84

OPERATORS

The more general difl‘erencing operational variables A, and V,” where

A,=Eh—1,

Vh=l—E"',

are used in numerical analysis, usually with h representing a ‘small increment’. Note that VI! =

—A—h‘

Note also that, wheneverfe .921, 9A,,f E 9fand 9%,f E 9]: 6.5

Polynomial operators

Suppose that A is an operator on .97, that n e N, that a0, a1, . . . , an are constants, and that mo, m1, . . ., m, are natural numbers. Then we

use the term polynomial operator in A for any operator which is either of the form aoAmoalAml . . .anAmn

or is a continued sum of operators of this form. For example, 3A3 + 2A22A + v'3A + A—i— + 5 is a polynomial operator in A. More generally suppose that A1, A2, . . ., A, are operators on .9? and, for 0 < r < n, let B, represent any one of A1, A2, . . ., Ak. Then we use the term polynomial operator in A1, A2, . . ., Ak for any

operator which is either of the form aoBgoalBTl. . .anBfin

or is a continued sum of operators of this form. Our definition of a polynomial operator depends for its validity on the associativity of addition over 0. Since addition is also commutative over 0 the order of the terms in a polynomial operator may be altered at will. Provided that every operator concerned is linear, and multiplication is commutative for any two of these operators, polynomial operators may be multiplied together in the same way as in elementary algebra, except that it is not always permissible to suppress a term in which the operator 0 appears. When multiplica-

tion is not necessarily commutative, care is needed. For example, if A and B are linear operators we may assert that (A+B)2=A2+2AB+B3 only if we know that BA = AB; cf. Exercise 4.1(2) on p. 42.

ornnArons

85

As a simple example of the way in which linear polynomial operators are used, we have

A’=(E—1)’=E3-—2E+l. Therefore, in particular,

1 Afi—(E 2__

NW “3° 1 a_=_

am” a

Hen“

2_

21=:+1)'91

_.

1_ 1 ,1_;. Es-m’ E5‘a+2 1 1 1 1 A5=a""+"1"3‘ ‘ew+ 1)’ 1 1 = 2

(19+ 1)(&+2)+s(a+ 1)

2

1

.9(a+ 1)(.9+2)

2

1

0(a+1)(9+2)=a+2‘a+1+3'

Equating the values of these two functions for an argument x of their common domain it follows that, provided x aé — 2, — l, 0, 2

l

2

1

x(x+l)(x+2)_x+2—x+1+§'

(I)

It is usual, though incorrect, to replace the identity function 1‘} by the pronumeral x throughout the above working. For example, 1

l

x

x+l

is written in place of the correct statement 1

l

E 5 (x) - m (x), here, of course, it is quite correct to replace the right-hand side by l/(x + 1). As with the corresponding use of the operator D, no harm is done by using expressions such as [Kl/x), provided their meanings are correctly understood. We next discuss an example involving the operator 0. We have

A+l=(E—1)+l

=E+(-l+l) =E+Q

86

OPERATORS - -.

Inparticular,

l 0 A5+5=Eé +5.

Therefore, provided x aé — l, 0,

l

l

l

''

—x(x+l)+9_c=x+l'

(2)

It is seen that here, in contrast to (l), the two expressions equated do not have a common set of admissible values of x. When two ex; pressions involving a pronumeral x are equated the unwary reader might suppose that it is implied, unless anything is stated to the contrary, that the expressions have a common set of admissible values of x. If no restriction on the value of x is stated it would be more satisfactory to write, in place of (2), l

x(x+l)

1

x

l

0

x+1+x'

Regrettably perhaps, we are not always as meticulous as we might be, and make statements such as (2) without mentioning any restriction on x. Thus we suppress the operator 0 and write

1 1 (A+ 1);_n;, from which we deduce that

without further cement.

Exercise!” Showthat 1—A=2—E 1—A== 0+2E_— E2, and

A2'+2A+1=E=+0E+o.

Deduce identities involving. the pronumeral x by operating in each case on the functions (1) a9“, (2) 1/19.

OPERATORS . Exercise 6.8

87

Use operators to prove that if x e R,‘ andx '# —3,

"-2, '- 19 0!

l _ 2 + l (x+2)(x+3) (x+ l)(x+2) x(x+ l) 6

‘ x(" x' + 1')'(x" + 2')—(x + 3)‘ Exercise 6.9 Are the following statements correct? (l)E(l—A)=l—A“, 6.6

(2)(l+A)(l—A)=l—A2.

Quasi-linear and normal operatorsT

In 5.8 we showed that there are functionsfand g of .931 such that

D(f+ g) aé Df+ D3Thus D is not a linear operator. The most we can say in general is that if a, b are constants andf, g are functions of all, then

D(af + bg) 2 a + n. It is of some theoretical interest to investigate the consequences .of modifying our definition of linearity in the light of this result. Let us make the following definition: An operator X of 0 is said to be quasi-linear ifi; for all constant operators a and b of (9, X(af+ bg) 2 aXf+ n whenever f and g arefunctions of .97.

It is a simple exercise to modify the proofs of Theorems 6.1, 6.3 and 6.4 to obtain proofs of the theorems which follow. The proof of Theorem 6.9 depends on the following lemma:

Lemma 6-1 Iffi, gufat 82 are functions of .97 such that f1 2 g1 andfa. 2 32 thenfi +f2 2 g1 + 32Proof.

The function g1 + g, has domain 9g1 n 9g,. Since

f1 2 g1 and fa 2 ga, therefore Qfl 2 9371 and 9}; 2 9g. Hence ifa e 9(g1 + g,) then are 9(f1 +f2). Alsofla = glut andf,,a = gaa, andso(f1 +f,)a = (g; +g,)a.Thusf1 +f, 2 g1 + g,. 1' If desired a start my be made on Chapter 7 before reading the remainder of this chapter. ' ‘

88

OPERATORS

Theorem 6.7 If X is a quasi-linear operator and a is a constant operator then, whenever f e .97,

Xaf2 aXf. Note.

If a is a non-zero constant then Da = aD. It is not true,

however, that D0 = 0D. As shown in 5.8 we may have 9f = R while Df is the null function. In this case DOf is the zero function while 0Dfis null. Exercise 6.10 Theorem 6.8

Is it true that D + 0 = D? If X, B, C are operators and X is quasi-linear then,

whenever fe 5", X(B + C)f2 (XB + XC)f. Theorem 6.9 If X and Y are quasi-linear operators then X + Y is quasi-linear. A difliculty arises when we attempt to modify the proof ofTheorem 6.5 to show that if X and Y are quasi-linear then so is XY. With the notation used in Theorem 6.5, write

f1 = Y(af + bg),

f, = a'+ n.

Then, if Y is quasi-linear, f1 2 f2. However it does not follow that Xfl 2 s, and so we are unable to complete the proof unless we know more about X. Let us introduce the following definition: An operator X of (U is said to be normal If Xfl 2 s whenever f1

andf2 arefunctions of.9" such that f1 2 f2, so that iffla = fBa whenever a: é s then Xflfl = sfl whenever B 6 9X];

A11 operators used in practice are normal; in particular, E“ and D are normal. We now prove that the subset of normal operators of 0 is closed under addition and multiplication.

Theorem 6.10 If X and Y are normal operators then so are X + Y and XY. Proof. Suppose that f1 and f; are functions of .57 such that f; .3 f2. Then.Xf1.2 s and Yfl 2 Yfa. Hence, by Lemma 6.1,

(X + Y)f1 2 (X + Y1f2.ThusX + Yisnormal.

OPERATORS

89

If: Yf1'= Yfg then XYf1 = XYf2, and if Yf, :2 Yfa then XYf1 a XYfa. Thus XY is normal.

The proof of Theorem 6.5is easily modified to obtain a proof of the following theorem:

Theorem 6.11 If X and Y are quasi-linear operators, and X is normal, then XY is quasi-linear.

We now see that the subset of normal quasi-linear operators of 0 is closed under addition and multiplication. The fact that some important operators are not linear, but are only quasi-linear, is no great practical handicap. The position is best clarified by a simple example. Suppose that X is an operator, and write P=(X+1)(X+8)+(X+2)(X+7),

Q=2X2+18X+22, and

R=(X+3)(X+14)+(X+5)(X—4).

Then

P=X(X+8)+l(X+8)+X(X+7)+2(X+7)

=X(X+8)+X(X+7)+3X+22. Likewise R = X(X + 14) + X(X — 4) + 8X + 22.

If X is linear then P = Q and R = Q, and so P = R. Suppose, however, that all we know is that X is quasi-linear. Then all we can

say is that, for every relevant function f, Pf 2 Qfand Rf 2 Qf; to prove these statements we use Theorem 6.8 and an obvious extension of Lemma 6.1 to three pairs of functions. Thus Pfot = Rfa whenever

a e 9Qf. Whenever f e 9%, 9f 2 9Df 2 9D2f. It. follows that if we take X = D in our example then, wheneverfe .931, the functions P]; Qj, Rf have a common domain, namely 9Di‘f. Thus in this case we are justified in writing P = Q = R. Note also that, even if X is linear, in

general we must write

(X+1)(X—l)=X2+0X— 1, but that we may justifiably write (D + 1)(D"— l)‘= D2'—_ 1.

We noted in 6.4 that-multiplication is commutative for Band D.

4+

9.0

OPERATORS

We'nowprove that if m e N then multiplication is commutative for E“ and B“. Let P(m) be the statement

E°D"' = D"'E".

Then P(0) is trivially true and P(l) is true by what we have already proved in 5.10. Now assume that P(m) is true. On this hypothesis, whenever fe 911, ED“ 1f = E‘Dm = D’”E°Df = DmDE‘f = Dm+ 1E7:

Hence P(m) : P(m + 1). It follows by induction that P(m) is true whenever m e N. Since D is quasi-linear and normal it follows by induction from Theorems 6.10 and 6.11 that D'" is quasi-linear and normal. Hence, if fe 921 and a0, a1, co, c1 are constants,

D’"(aoE”o + a1E°1)f 2 DmaoEcof + Dma1E°1f. DmaoE°0f 2 aoDmE‘of,

Now

equality holding if (but not only if) an aé 0. Also DmalE‘If 2 alEclf.

Since D’"E°o = E‘oD’" and DE”: = EczD'" it follows that D'"(aoE°o + a1E°1)f 2 aoECODmf + a1E°1DmfZ

It is now a simple matter to induce that if neN and, for 0 S r s n, a,and cr are constants then

D'" i a,E”r 2 Z a,E°rD"'f. r=0

r=o

In particular

DMAgf 2 A:D"'f

and

DMVL‘f 2 Vg‘Dmf.

Since E"°D"' = D'"E"° and EMf 2 (E‘ + 0)’ff it follows from the normality of D’" that E"‘D"‘f 2 D'"(E" + 0)'ffl Thus Likewise

E"D"'f 2 D"'(Ac + 1W: E'“D"'f 2 D"‘(l — V,)'ff.

OPERATORS

9.1

6.7 Operators on functions of two variables Suppose thatf; p, q are functions of .922, and let f o (p, q) denote the function of_ 9?, which maps an argument (x, y) onto the value flp(x, y), q(x, y)) whenever this is defined. Then (x, y) 6 9f o (p, q) iff

(x. y) e 9p n 93 and (p(x, y), q(x, y)) 6 9fNow let c1 and c2 be constants. Then we define the translation

operators E‘i‘ and E? by writing

E? = {0513091 + 61, 192)) If6 ~92}

and

E? = {(fif° (19,, '92 + 62)) IfE 932}-

Thus, whenever (x, y) 5 QB?f,

E'i‘f(x, y) =f(x + 61, y) and, whenever (x, y) 6 9E?f,

E3706: 3’) = f(x, y + c,). Evidently E? and E? are linear and normal. Also the functions E? Efi’f and E? E?f have a common domain, and whenever (x, y) is an argument of this domain,

E‘i‘Eé’flx. y) = Efi’Ei‘flx. y) =f(x + 61. y + Ca)Thus multiplication is commutative for E? and E3“. We define the differencing operators A, 1-f-‘2 and Vcl 4,, by writing

Am, = E§1E3= — 1,

Vm, = 1 — Ef“E;“'.

Thus

Ac1.c,f(x,y) =f(x + 61,); + 6:) —f(x, y)

and

Vc,.c,f(x. y) =f(x, y) —f(x ~ c1.y — 6:)

whenever the right-t sides are defined. We may write E1 for E} and 13,; for E;. Also, when it is clear that

(x, y), or possibly (y, x), represents an argument, we may, for example, write E,,(x2 + xy)

to mean either

E109? + 191 x 090:, y)

01'

E209; + '92 x 1,003 x)-

Then

E,,(x2 + xy) = (x + 1)2 + (x + 1)y.

92

OPERATORS

On occasion, Axf(x, y) is used to denote A1l0f(x, y) so that, for example,

A,(x1 + xy) = (x +1)2 + (x + l)y — (x2 + xy). We mentioned the symbols D1 and D2 on p. 68 in connexion

with partial differentiation. We can now define these symbols as operators by writing

and

D1 ={(f,f1)|fE-%} D2 = {(flfa) IfE 9732}-

Evidently D1 and D2 are quasi-linear and normal. Also multiplication is commutative for either of D1, D2 and either of E21, E21. However, multiplication is not commutative for D1 and D,, for it is not always true that D1D2f and D2D1f are the same function; see,

for example, the author’s Calculus, Volume II (Harrap), Examples 94, p. 36. It is usual to denote D1D2 by D31, and DQD1 by D1,. Exercise 6.11

Show that, whenever f e 9%,

Ac,.c,f 2 (AME;a + Ao.c,)f and

Ac,.c,f 2 (AME? + Ac,.o)f-

Write down corresponding results for Vc1 3,f.

6.8 Inverses of operators An operator X is a many-to-one relation, and so the inverse relation X'1 is one-to-many. This inverse relation may not itself be an operator for one of two reasons. In the first place if X is not one-toone then X'1 is not many-to-one, and so is not an operator. In the

second place we have stipulated that the domain of an operator is the whole of the appropriate concourse .9" offunctions. Thus, even if X is one-to-one, X" 1 is not an operator unless X is a mapping of 31' onto itself. The inverse of any constant operator a other than 0 is an operator,

and we may consistently write a‘1 for the operator l/a. Note that strictly we should not use a‘1 for thefunction l/a; see p. 62. If a is a constant, the inverses of E‘, 1‘ and E; are operators; for example,

(15‘) ' 1 = E" c. On the other hand the relation D ' 1 is not an operator. Given a function f of .931, if there is 'a function F such that DF = f

OPERATORS

93

then there is an infinity of such functions. Furthermore, the problem of finding F such that DF = fmay have no solution. (For the reader who is interested and knows sufiicient functiontheory we outline a proof that if

and

f(x) = 1

when x is rational

f(x) = —1

when x is irrational

then there is no function F such that DF = f. The proof is by reductio ad absurdum. Suppose that F exists. Let a be any rational real

number, and let b be any irrational real number such that a < b. Then there is a real number £ such that a S é < b and F(x) < F(f) whenevera s x < b. Since F’(a) > 0, g ;é a. Since F’(b) < 0, E 9e b. Hence a < E < b. It follows that F’(§) = 0. We now have a contradiction, since f(E) 95 0. We have, in fact, provedlthat we cannot have F’(x) = f(x) fora S x < b, however small b — amay be.) Although D51 is not an operator, the symbolism D'1fis sometimes used to denote an antiderivative, or primitive, of f; that is a

function F such that DF = f. For example, we may write D-1191I2=§193l2 + C,

where c represents a constant; in more familiar notation, D—1x1l2 = gxsm + c.

We are here using D' 1:91” as a functional variable. In practice, when seeking antiderivatives of a function f, it is always tacitly assumed that we are concerned only with functions having domain 9]: It is only on this understanding that the above statements are valid. For let F be the function defined by writing

and

F(x) = ix”

(x 2 0),

F(x) = x”

(x < 0 and x rational),

F(x) = —x"’

(x < 0 and xirrational).

Then DF =91”. This solution is in fact a better one theoretically than 3:193”, since 0 is an argument of DF but not of D363”; see 5.9.

Thus when we write D—1191l2 _.__ $9312 + C

we are tacitly modifying the‘definition of a derivative.

94

OPERATORS

Exercise 6.12 :Criticisethe statement

Exercise 6.13

Construct a function F of .431 such that 9F = R

and F’ = {(0, 0)}.

The relation A;1 is not an operator. Here again, for c aé 0, the symbolism Ac‘lf is used on occasion as a functional variable to denote a function F such that AP = f. Now if A,F1 = f and Ac = f then, since A, is linear, Ac(F1 "' F2) = AcFl "‘ AcFa = 0 X f

However, this does not ensure that F1 — F2 is a constant function,

as the following exercise reveals: Exercise 6.14 Construct a function F of 9?, such that 9F = R, F is not a constant function, and AF = 0.

A polynomial of .921 is afunction of the form n

2 419”", where a0, a1, . . . , a" are constant functions. Provided do ye 0 we call

n the degree of this polynomial. In particular a non-zero constant function is a polynomial of degree 0. The zero function is also a polynomial, but we do not assign a degree to this function; some authors describe the zero function as being of degree —oo. A non-null function which is a quotient of two polynomials is called a rationalfunction. Clearly if F is a rational function then so is AcF. On the other hand, given a rational functionf there may be no function F such that AP = f. For example, there is no function F such that AP = 1/3. For suppose that F is such a function. Then F(c) — F(O) is not defined, and so either 0 or c is not an argument of F. Hence either —c or c is not an argument of ACE Since —c and c

are both arguments of 1/19 if c aé 0, we now have a contradiction.

Theorem 6.12 If e aé 0 and F1, F, are rational functions of 91’, such that A,F1 = AcF, then there is a constant function k such that

Fl—Fask.

OPERATORS

Proof: write

95

Clearly F1 — F, is a rational function. Hence we may P F1_F2=6,

where P and Q are polynomials and Q aé 0. Suppose without loss of generality that c > 0 and choose xo so that x 6 9E and x e 9F, whenever x 2 x0. Then, whenever n is positive and integral,

_

P(xo) _ n-1

1’05 + nc)

- A.,F,,)(xo + rc) _ 0_ Q06: + nc) _ 605—0) — ”2° (AcFl Hence there is an infinity of values of x for which the polynomial expression

Q(xo)P(X) - P(xo)Q(x) is 0 and so, by a well-known result, this expression is 0 whenever x e R. Therefore, writing k '= P(xo)/Q(xo),

F1(x) - F2(x) = k whenever x e 9F1 and x e 9F2. Thus

F1 — F; S k. Note that F1 and F, need not have a common domain. For

example, if

2.92

1

1

F1‘—02—1"a—"1‘19—+1+2 2.9 and

then

1

1

o

F":9($-1)‘.9—1".9+1+5

_;+;. ,=1__1_ AF1=AF :9- 0—1 19+2 0+1

7 APPLICATIONS 7.1

Rational functions

In the previous chapter we showed that, for all admissible real x,

A l: x

_ ___1_,

A21: ___1___.

x(x+l)

x

x(x+1)(x+2)

We now obtain more general results of this nature for products of the form m-1

l

_-_, H ,=0x+a+rc

where m is positive and integral. Our choice of letters doubtless leads the reader to believe that we intend x to be regarded as an independent variable, and 'a, c, m as

parameters. This is in fact the case. We regard the above product as the value onto which the functional variable

1—I19‘+al+rc maps an admissible argument x. It is, however, only because

mathematicians have been conditioned into habitually using certain letters as independent variables and certain other letters as parameters that our intention is clear. We could, for instance, intend

(x, a) to represent an argument of the functional variable 111-1

1

11 :91 + :92 + rc' Let us write m—l

l

W) = H, m' This makes it clear that we are considering our product as representing a value of the functional variable q,,,, where

c =nl9+a+r 7-0 96

APPLICATIONS

97

We omit reference to a and c in our abbreviated name qm on the

grounds that our notation should be no more elaborate than suits our convenience. The desirability of making reference to m will shortly become apparent. Now Aeq‘) = (E6 _ 1)qm(x) "1-1

1

m—1

l

= r=0 1—Ix+c+a+rc- r=0 Hx+a+rc

Thus

= (x + a)qm+1(x) - (x + a +m0)q..+1(x)AJAX) = -mcq...+1(x)-

So far we have not stated any restrictions on x. The result we have obtained is in fact valid provided x 9E —(a + rc) for 0 < r S m. In future we generally omit any reference to such restrictions. We could alternatively write Acqm = _ mcqm + 1:

an equality between two functional variables. This implies that whenever we replace a, c and m by pronumerals we obtain an equality between two functions, so that these functions have a common domain. This last result can be proved directly by replacing x by :9 in the products used in proving that

Acqm(x) = —mcqm+1(x)It is now a simple matter to induce that, if n is positive and integral, 15—1

AMA") = n (m + S)(_c)nqm+n(x)' 8=0

Thus

A:q,,,(x) = W (- c)"qm+..(x)-

This last result remains true when n = 0, even when c = 0, provided

we interpret 0° as 1. In particular, taking m = l, n

l

A"'x+a

=

y _

n

"

The reader is once again reminded that

A: x +1 a 4*

l _._.

n.( c)flx+a+rc

98

OPERATORS

means

A2 .9 +1-a(x)

Since A.c = — Vc, on replacing c by —c in the formula we have proved for Azq,"(x) we deduce that n

Ill-1

m+n—1

(m + n _1)!

1

V‘flx-Isa—rc

=

_

(m-l)!

(c)

n

1

—__.

Q x+a—rc

The reader is advised to write out an independent proof of this result. In particular,

l

n

Vcx+a

=

I...

7|



l

—.

n'(c),1:“[,x+a—rc

In the examples which follow, n and r represent non-negative integers, with 0 s r < n. The symbolism n r is used to denote the coeflicient of 19’ in the binomial expansion of (l + :9)”, so that n _ n! (r) _ r!(n — r)!In place of the product 111 ,=0 x +1 rc for example, we sometimes write 1 x(x + c)(x + 2c)- - -(x + nc) It must then be understood that this latter expression stands for 1/): when n = 0, for l/{x(x + c)} when n = l, and for 1/{x(x + c)(x + 2c)} when n = 2. Example 7.1

For c -,é 0, express 1 x(x + c)(x + 2c)- - -(x + nc)

in partial fractions.

APPLICATIONS

99-

1 1 ,, x(x + c)(x + 2c)- - -(x + nc) = Wide) 3: =n—!lc" (1— E9)“;

, n

0,1

=n!lc’I ,Zo(— )(r)E J_c

= n!lc'I ,2, (—)r(:) x -Il- rc Example 7.2 Sum the series

2:0“))(rrl')(x-r)(x—r:1Xx-r-2) "

rn

1

;o(_) (r) (x -— r)(x — r —1)(x — r - 2)

,2“) WWW?) _,.

1

=‘1‘E‘)m ,, 1 ‘ V —x(x — l)(—x — 2) (~)"(n + 2)!

_

_2x(x— l)---(x—n-2) Example 7.3 Prove that

,

20‘ )()(r+1)= Proof.

1

1

+12,”

Write

S= 2 1. Then clearly ALP... is a polynomial of degree m — 1. Also if a is a constant function then Aca = 0. It follows by induction that if n is integral andn > m thenAn = 0, and likewise VEP... = 0.

These results remain true if c = 0, when they are trivial. We imposed the condition c 5e 0 because we have not assigned a degree to the zero function; see 6.8.

Example 7.8 If m and n are integral, and 0 s m < n, prove that

i (—)*(’;)(r + 1y» = o. r-O

First proof 7|

2 (—)'(’;)(r + x)" = Z (—),(nr)E'x"'

= (—A)"xm = 0.

APPLICATIONS

' 107

In particular, setting x = l,

2" (—)'(n)(r + I)“ = o.

r=0

Secondproof.

' 1'

By induction on m,

i (—)r(’;)(r + 1)f =(D19)l"(l —..a)n(x). r- O

" Now if 0 < p < n then, by induction on p, (l - 19)"? is a factor of (D3)”(l — :9)". Hence

2 (—)'(’,‘)(r + I)“ = (Dam — 19m) = 0. We now study some polynomials of .921 of a particular form; For positive integral m, write m-1

pm=H(z9'+a+rc), r=0

where a, c, m are parameters.

Also write Po = 1-

Then, for non-negative integral m, pm is a functional variable which

represents a polynOmia'l Of .931 of degree m. For m > 1,

Even: (I — from = filo? +»a + rc) —fi1(:9_- c + a + rc) r=0

7'0

= {:9 + a + (m - l)c}p,,._1 - (:9 - C + “Pm-1Thus

cm = cm-l-

This result is also true for m = 1, since c1=(z9+a)-(0—c+a)

= c = lcpo.

It is now a simple matter to induce that if in and n are integral and non-negative, and n s m, then

I Vi-‘Pm = W C”Pm—n-

(1)

' 108

OPERATORS

In particular,

Vg'pm = m!c"'.

Also if n > m then V: m = 0. Next write

11-0 = 1 m-1

and

nm=H(19+a—rc)

(m>0).

r=o

Then, since V.c = —Ac, on replacing c by —c in (1) we deduce that n A27," = (m. m! — n)! c "M Exnnlple 7.9

(n < M)-

Sum the series

7: (—)'(r + l)(’;)(x + l — r)(x + 2 — r)- - -(x + n — r). Denote the sum of this series by S(x), and write

F = 2 (—)'(r + 1)(n)E". f=0

Then

80:) = F (x + 1)(x + 2)- - -(x + n)-

For1cvm(V:-'g). 1' = 0

We can obtain another expansion for A'c'fg as follows. We have

Acfg = (E‘YXAcg) + (A0103Interchanging f and g,

Acfg = (Acg)f + (E°g)(Acf)Now let us write

H. = E“ + 1, so that

Then

H,,. = A + 2.

ZAcfg = (HJXAeg) + (AJXHcg).

Applying this formula to each of the products on the right,

ZaAfifg = {(HcfcAc-g) + (AcD(H.Acg)} + {(HcAJXAcg) + (AAJXHcgB

= (H3f)(A§g) + 2(HcAcfXHeAcg) + (AEf)(H§g)The pattern is now clear, and the reader should have no difficulty in

proving the following theorem by induction. Theorem 7.2 If n is non-negative and integral,

2mm: = i (’r')(H:-'Am(HzA:-'g) r-o

and

2W2)”: =' i (Z)(H'.-.'vzf)(H'..vz-rg). r-o

:- Exerdse 7.4 (1)12:e that, if-n >1, A:xg(x) = (x + nc)A:g(x) + mA2‘1g(x).

APPLICATIONS

115

Obtain the corresponding formula for A2x3g(x) if‘n 2 2. Verify your results when .g(x) = l/x.(2) Prove that

1 (—)"n! " 1 A”(1+x)='=(1+x)(2+x)---(n.+1+x),;,r+l+x; cf. Example 7.3 on p. 99.

(3) Writing

u,(x) = $ an+1

prove that, for n 2 1,

u”(17) - un-1(x) = 56 — nDeduoe that

V"x"+1 = Kn + 1)!(2x - n);

cf. Example 7.11 on p.111. Example 7.12

The Gamma function 1" is a function of 931

defined in the first place by writing

I‘(x) = r tx‘le‘.‘ dt, 0

this definition being .efl'eetive ifi‘ x > 0. Using integratibn by parts, ifi‘ x > 0,

I‘tx + 1) ; r t"e"dt = t"(-e“)]: — J: xt"'1(—e“) dt = 0 + xI‘(x).

Thus

I‘(x + 1) = xI‘(x).

I‘(x) may now be defined for all non-integral negative values of x by taking l"(x) = 31; I‘(x + l), firstfor —l < x < 0,thenfor —2 < x < -—l,andsoon.

116

OPERATORS

We now have

EP(x) = x1"(x), provided only that x is not a negative integer. Hence E2F(x) = (x + l)I‘(x + 1)

= (x + l)xI‘(X), E3I‘(x) = (x + 2)(x + 1)I‘(x + l) = (x + 2)(x + 1)xP(x), and so on. If we write P005) = 1 1—1

end

p,(x) = H (x + s)

(r 2 1)

s=0

then evidently, by induction, if n is non-negative and integral and x is not a negative integer, E”l"(x) = pn(x)P(x). Also

A"I‘(x) = (E — l)"I‘(x)

= 2 (—)'(’;)En-'r(x) = :(—)'(;‘)p..-,(x)r(x). r

Write

0

P, = 20 (-')’(';)pn—r-

Then P,l is a polynomial of degree n, and A"I‘(x) = Pn(x)I‘(x). Now

Pn+1(x)I‘(x) = A"+1I‘(x) = AP,,I‘(x) =’ ,(x + l).xI‘(x) — Pn(x)I‘(x).

Also I‘(x) aé 0. Therefore

Pn+1(x) = xPn'(x + 1) - PM)-

(1)

Our proof of this'equality is valid provided x is not a negative integer.

APPLICATIONS

117

However, since the equality is one between two polynomial expressions, it is in fact valid for all x. This remark applies to other equalities we establish, and we make no further explicit reference to it. Setting x = 0 in (l),

Pn+1(0) = - 1.0)Now

Po(x) = po(x) = 1.

It follows by induction that

P..(0) = (-1)"For r > 1, A'x = 0. Therefore, using Theorem 7.2 with c = l and writing H for H1, ifn 2 l,

2"A"I‘(x + 1) = 2"A"xI‘(x) = (H"x)A"I‘(x) + n(H"‘1Ax)HA"'1I‘(x). Now

H”x = (2 + A)”x = 2"x + n2"‘1,

and

H"'1Ax = (2 + A)"'1l = 2"”.

Also

HA"'1I‘(x) = (E + 1)P,,_1I‘(x)

= n-1(x + l)1‘(x + 1) + Pn—1(X)I‘(x) =

u._1(x + l).xI‘(x) + Pn_1(x)I‘(x).

Thus 2"A"I‘(x + l) = (2"x + n2“'1)P,.(x)I‘(x)

+ n2""{xP.—1(x + 1) + Pn—1(x)}I‘(x)Now

2"A"I‘(x + 1) = 2"A"EI‘(x)

= 2"EA"I‘(x) = 2”P,.(x + l)I‘(x + 1) = 2"P,,(x + 1).xI‘(x). Therefore

(2"x + "2”“)Pn(x) + n2""{xPn-1(x + 1) + Pn-1(x)} = 2"xP,.(x + l), and so

(2x + n)Pn(X) + nxPu—ibc + l) + "Pa—10¢) = 2xPn(x + 1)From (1), 2xP,, x + l) = 2P,,+1(x) + 2P,(x).

(2)

118

OPERATORS

Also, using (1) with n replaced by n — 1, if n 2 l, nxPn_1(x + l) = nP,,(x) + nP,,_1(x).

It follows from (2) that, for n 2 l, P,,+1(x) = (x -|.- n — 1)P,.(x) + nP,.-1(x). Replacing n by n + l,

P..+2(x) = (x + n)P..+1(x) + (n + ’1)P..(x) for n 2 0.

Exeroise 7.5

(1) Compute Pn(x) for n = o, 1, 2, 3, 4.

(2) Prove that P,.(x + l) = nP,,_1(x + 1) + P,.(x) (n 2 l). (3) Prove that xP,.(x + l) = (x + n)P,,(x) + nP,._1(x) (n 2 1). (4) Provo that Pn(_1) = (—)"(n + l).

8 SEQUENCES ' 8.1 Terminology The word sequence is used to denote a function whose arguments are either natural numbers or integers. (A function whose arguments are ordered pairs of natural numbers or integers is called a double sequence, and so on.) A sequence is said to befinite or infinite according as its domain is a finite or an infinite set. The values of a sequence may be ofany type; for example, we may consider sequences of functions or sequences of intervals. However, when the word sequence is used without qualification it is always understood that the values are numbers of some type.

In this chapter we study sequences of real numbers. We choose to take our arguments to be integers. We may then regard any one of our sequences as a function of .931. (Strictly speaking, since an integral real number is of a difi‘erent type from the integer with the same name, we should observe that we can set up a one-to-one mapping of

the set of all real sequences onto the subset of .911 which contains those functions whose arguments are integral real numbers.)

Let u be a functional variable representing a sequence. Then it is usual to denote the value onto which u maps an argument r by u,, rather than by u(r). For the present we suppose that u is an infinite sequence and that u, is defined ifl‘ r is a positive integer, and we call u, the rth term of u. It is sometimes expedient to display the first few terms, and perhaps the rth term, of u in the form ulruasuasn'run'0'

and to use the word sequence informally (and inaccurately) for this array of numbers. 8.2 Applications of~operat0rs to sequences If u is a sequence with domain {reZ|r_> l} 119

OPERATORS

120

then the functions Eu and Au are sequences with the same domain as u. Consequently

(1 + A)u = Eu.

(E + 0)u = Eu, In the first place,

“3 = EH1.

Using brackets to make our intention clear, it must be stressed that

Eul means (E101, and not E(u1). We could only give a meaning to E(u1) by taking ul as a functional variable denoting a constant sequence, and then E(u1) would be the same constant sequence. Like-

wise ua = E142 = E2141,

and so on. It follows by induction that, if n is a positive integer, u" = E”'1u1.

Note that here we must regard n as a parameter and not as an independent variable. Unfortunately there is no agreed letter which is used exclusively as an independent variable when studying sequences. Because of this, misconceptions can arise if care is not taken. Using p as independent variable, and n as a parameter, it is correct to write

nan—1 = Emu,It is perhaps an aid to clarity if we consider the formula “1; = Endui

to be obtained by setting p = 1. Exercise 8.1 What is wrong with the following argument? Suppose that u2 = ul. Then Eua = Eul, so that us = uz.

Theorem 8.1

Nu, = (—)'- : (—)r(’r‘)u,,,,. r-o

Proof.

Nu, = (—)"(l — E)"u,,

SEQUENCES

Thus

12]

Mu, = (—)n 2 (—)'(’,‘)u.... 1|

r=0

In this proof we may regard p as an independent variable and n as a parameter. We have not explicitly stated that p and n are integral, with p 2 l and n 2 0. Here and elsewhere we take such conditions as being implied by the context. Setting p = 1,

Nu. = (on: (-)'(’;)u.... Thus

A141 = —u1 + 142, A2141 = ul — 2142 + u3, Aau1 = —u1 + 3u2 — 3143 + u4,

and so on. Example 8.1

For c aé 0, express 1

x(x + c)(x + 2c)- - -(x + nc) in partial fractions. 1 u, -— m'

. Write

Then

Nu,

_

(—)”n!c”

— {x + (p — l)c}{x +pc}{x +(p + l)c}---{x +(p + n — l)c}

Hence 1

(_)n A"u1

l " , n l nlc" 2:0 (—) (r) x + rc' We proved this result by a difi‘erent method in Example 7.1 on p. 98. Note that in the present proof x is a parameter. The notation l

n—

A x+(p—l)c

5+

122

OPERATORS

is clearly ambiguous unless we disclose which of x, p, c is to be re-

garded as the independent variable. However, if we use this notation our intention can be made clear by the effective, even if inaccurate,

instruction: Let A operate onp. Example 8.2 Sum the series

1_%+x(_x2?__l)_...+(_)nx(x_-l)'+"l(!";"+_l). Denote the sum ofthis series by $1 and, forl < p < n + 1, write u,=(n—p+l)!(x+l)x(x—1)---(x-p+2), the factor x + 1 being introduced to make ul of the same form as ug, . . ., an”. Then

(x + 1)s1 = "1,2(—)'(’,')u,+1 = (31%. For 1 < p < n,

um = (n —p)!(x + l)x(x - 1)---(x -p +1), and so

Au, = (n ~p)!(x +1)X(x — l)-'-(x -p + 2)

> 1,

vr-l = "r(b + dr),

and

12T — v,_1 = (a — b)u,.

Therefore

2 u, =

l

3.5 "

Here

(0,, — 1:0).

_1.3.5...(2n—1) —2.4.6...(2n)""

.3.5.. “' ‘ 2.4.6...(2r)

SUMMATION 0F SERIES

145

so that, with the above notation, a = l, b = 0 and d, = 2.9. Using

the method described above, rather than merely quoting the result, let us write ' yo =

and

l

_ l.3.5...(2r —1)(2r +1)

Ur —— W

(r ? l).

_ l.3.5...(2r - 1)

F" ’ > 1’

”"1 ‘ 2—.4.6. . .(2_r - 2)’

and

v, — v,_1 = {(2r + l) — 2r}u, = 11,.

A180

vl—vo=%—l=%=u1.

1:

Hence

2 u, = 11,, — v0 r=1

_1.3.5...(2n —1)(2n + l) _1 2.4.6...(2n) °

Otherwise:

u, = (—)'( 7%)-

Hence, if Ix] < l,

(1 — x)’“” =1 + Z urx'. r=1

Equating coeflicients of x” in the expansions of the two sides of the

identity

(1 — x)-1(1 — x)-“= = (1 — 20-3/2, it follows that

1+ Eu, = (—)"(“f) 3.5...(2n +1)_ 2.4. . .(2n)

Thus

" _3.5...(2n+1)__ ,Zfl‘__2.4...(2n) 1'

146

OPERATORS

Example 9.8 Sum the series l_fi+x(x2: l)

___+(_)nx(x—1)...(x—n+ 1)_

This series has n + 1 terms and, for r > 1,

(—x + r _ 2) . )...1 _ (—x)(—x +1r_

ur—

Since this formula does not apply when r = 1, we begin by finding n+1

14,. r=2

Write

1.71 = —x

and

v,=u,(—x+r—l)

(r>l).

Then, for r > 1, vr—1 = (r _ 1)ur9

and

12T - v,_1 = —xu,.

Hence, if x 75 0, n+1

1 Mar = - £02,.“ — v1) — - Evn+1 -1-

Also u1 = 1. Thus _

r21 “1 — (

_ n

)

(x —

n!

This result remains true when x = 0. The present method is more direct than those used previously for illustrative purposes to sum the same series; see Example 7.10 on p. 109, and Example 8.2 on p. 122. We do not propose to enlarge upon the results discussed in this chapter in any further detail, as they are incidental to our interests and there is already available an extensive amount of literature on this subject.

SUMMATION OF SERIES

Exercise 9.5

147

Sum the following series:

1.3

1.3...(2n—l)

(1) 2—.4+2.4.6+”'+2.4...(2n+2)

2

2.5

2.5...(3n—1)

(2) 3+7+"'+ '3.6...(3n) ' 2

n

(3) E + 1.3.5 +"‘+1.3...(2n +'1)‘ + c) + a(a + c)(a + 2c) (4) g+ a(a +--b(b + c) b(b + c)(b + 2c) +

a(a + c). ..{a + (n - l)c}

b(b + c). .'.{b + (n '—"'1)'c}’

wherebaéa+c. l

l!

2!

(5) x+l+(x+l)(x+2)+(x+l)(x+2)(x+3)+' (n-l)' +(x + l)(x+ 2). .(x+n)’ where x as 0; cf. Exercise 7.1 (14). x

x(x—1)

‘6’ l'y—fi+ 1 and neither a0 nor a,c is 0. Any sequence which satisfies

an equation of this form is called a recurring sequence oforder k and the equation is called the scale of relation of the sequence. For simplicity we study, in the first place, scales of relation of orders 1 and 2. Consider the difference equation, of order 1,

aouwi + alun = 0, where neither an nor a1 is 0. Writing a = -a1/ao, this equation is

satisfied ifl' “n+1 = omn-

Evidently, by induction,

an = Aa"'1, where A is a parameter which is usually called an arbitrary constant. We call this the general solution of the difference equation. For any particular solution, A = u1. Next consider the difference equation, of order 2,

aoum + a114,.“ + aaun = 0, where neither do nor a2 is 0. We can find numbers a and fl, neither of

which is zero, such that

_ _ fl,

a + B "'

a0

“fl

= ‘2. do!

in our general theory a and )8 are complex. Then our difl'erence equation may be written either as

ao(E - BXE - 0014,. = 0 or as

ao(E — a)(E — fl)u,. = 0.

LINEAR DIFFERENCE EQUATIONS

151

It follows from the first of these forms that our difference equation is satisfied if

u» Aa’H, where A is an arbitrary constant, since then

(B — a)u,, = 0. Likewise, from the second form, a solution is

u.. = 3“, where B is an arbitrary constant. Since the operator aoE2 + a1E + a2 is linear, it follows that our difference equation is satisfied if

u. = Aa"'1 + BB"‘1. Now suppose that a sé )3. Then, for prescribed values of ul and 14,, the equations A+B=u1 and

Aa+Bfl=u2

are satisfied simultaneouslyifi‘ _ "2 "‘ Bu}.

A——a—,3’

_ ““1 "‘ "a.

B_ a—fi

Now if 141 and u, are assigned definite values our difference equation

has a unique solution It follows that if 0: ye )3 then, for sequences of complex numbers, the difl'erence equation

00“”: + dilly-+1 + as“; = 0 has general solution

u.. = Aer-1 + Ban-1, where A and B are arbitrary constants. When the first term of u is called an instead of u; it is convenient to write Pain place of A and Q3 in place of B. We then have un=Pau+Qpn,

where P and Q are arbitrary constants; here u“ is the (n + l)th term of u.

152

OPERATORS

Example 10.1

Solve the difference equation ”n+2 '— 5“n+1 + 611" = 0

(n 21),

given that u1 = l and u2 = 4.

Here

(E2 — 5E + 6)un = 0,

so that

(E — 2)(E — 3)u,, = 0.

Hence

un = A.2"‘1 + B.3”‘1,

where

A + B = 1

and

2A + 33 = 4.

Therefore A = — l, B = 2, and the required solution is u” = 2.3"’1 — 2"“. Example 10.2 Solve the difference equation

“n+1 = u" + u.._1 (n 2 2). given that 141 = l and u2 = 1. Notice that it makes no material difi‘erenee whether three consecutive terms of u are called u", un+ 1, an” or Ila—1, u", an“. The

condition n 2 2 indicates that the first term is to be called ul. (Ea—E-1)u,,=0,

Here

so that

{E — 15(1 + V5)}{E — £1 — v5)}un = 0.

Hence

u,, = A(l—+2fi)n—1 + B(1_T‘/5)n—l,

where

A + B = 1

and Therefore

and

£1 + V5)A + £1 — V5)B = 1. A = Lil-71355:

B = — 1—217‘555,

u, = Fi/Tsw + v5)" — (1 — van}.

The nth term of u is called the nth Fibonacci number. Using the binomial theorem, 1

[(n+1)l2] r—i

n

“"‘2n-1 21 5 (2r—1)’

LINEAR DIFFERENCE EQUATIONS

153

where [t] denotes the greatest integer which does not exceed t. Although irrational real numbers have been used to obtain this solution, it is valid if un is taken to be an integer or a natural number.

This is because each of the sets N and {n E Z I n 2 0} is isomorphic with the set of non-negative integral real numbers. If we wish to evaluate a particular Fibonacci number it is simpler to appeal directly to the difference equation, rather than use either of the expressions obtained for u“. This procedure also has the advantage that all the Fibonacci numbers up to the one required are calculated at the same time. Thus u3=1+1=2,

u4=2+1=3,

u5=3+2=5,

and so on.

Example 10.3

Solve the difi‘erence equation

a“: - 4un+1 + 7”»: 0 (n> 0), giventhatuo =landu1 = 3.

Here

{(E — 2)2 + 3}u,. = 0.

Considering in the first place sequences of complex numbers we may factorise the operator and write the given difi‘erence equation in the form (E — 2 — iV3)(E — 2 + iV3)u,. = 0. Hence the general solution is u,l = P(2 + i1/3)'I + Q(2 — iv3)”, where P and Q are arbitrary constants. Denote the modulus and amplitude of the complex number 2 + iv3 by p and 9 respectively. Then

P = x/{22 + (V3)? = V7Also -1r 1):

given that u; = l and ua = —1.

Here

(4E2 + 4E + 1)u,, = 0,

so that

(2E + l)"’u,. = 0.

Hence

u,I = {A + B(n — 1)}(—§)“'1,

where and

A = 1 —'}(A + B) = —1,

so that B = 1. Therefore

u» = n(-ir)“'1Exercise 10.1 Obtain general solutions of the following difference equations:

(1) um: — 6a..” + 814.. = 0-

(2) "n+2 + um = 2a,.-

(3) an” = un-

(4) um. = 4a,.-

(5) 2“n+2 — 5“n+1 + 2“» = 0

(7) un+2 + un = 0(9) um; = 2(u..+1 - an)

(6) “n+2 + 2“n+1 = sun

(8) “n+2 + 3a.. = 0(10) 2un+2 + 214m + an = 0-

(ll) ”n+2 — 2“n+1 + 4“» = 0- (12) "n+2 + “n+1 + “n = o-

(13) “n+2 — 6am +1011» = 0- (14) 2un+2 + 2un+1 + 3a.. = 0(15) “n+2 — 4“n+1 + 4“» = 0- (16) “n+2 + 211M: + “a = 0Solve the following difi‘erence equations:

(17)“n+2—3un+1+2un=0(n> l),u1=3,.u3=5

(18) un+2 = un+1 + 611.. (n> 0), uo = 5 111 = 0 (l9) uu+2 + u" = 411m (n > 1). u1 = l “2 = 2 (20) 2un+2 + 2“n+1 = “n

(n> 0), “0 = 1,141 = ‘1-

(21) u,,+2 + 2u,,+1 + 2u = 0 (n 2 0), an = 0, u1 =1. (22) u,” = 6un+1 — 25u,, (n 2 1), u1 = 2, u2 = 18. (23)un+1+2un+6un—1=0(n>l)’ “0:2 “i=3(24)un+1_2un+un1=o(n1)’u0=_l

“i=1

(25) un+2 — 6u,,+1 + 911,. = 0 (n21), u1 = 2, u3= 12. (26) 4a,,” + 1214,.” + 914,. = 0 (n20), u0 = -1, u; = —6.

158

OPERATORS

(27) 4u..+2 — 4u..+1 + an = 0 (n21), u1 = 4, u2 = 2. (28) un+2-2u,,+1 cos 0 + u” = 0 (n21), u1 = 2coso, u2 = Zcos20.

(29) Suppose that, for n 2 0, ”n+2 _ (‘z + fl)un+1 + “pun = 09

where neither 0: nor B is zero. Writing ”11. = "n+1 — “um

show that

12,, = W00.

Deduce that, for n 2 1, u” _ a

1:21.01," 1 _ lflr. = 3 n_

Hence prove that, for n2 0, if o: #13 then _ ul — auo — _ _l3_“oan

"" ‘ —T a—

a — fl ’3"

and if a = [3 then = {05140 + n(u1 — auo)}a""1.

(30) A circular disc'15 marked off into n sectors, where n2 2, and each sector is coloured either red, green or blue, no two adjacent

sectors being of the same colour. If an is the number of ways of doing this prove that 3-2" = “n+1 + u,"

and deduce that “n+2 — “n+1 — 214a = 0-

Hence prove that



u, = 2,. + 2(-1)n.

10.3 Scales of relation of order greater than two . We now discuss briefly the general theory of scales of relation. We first consider the difi'erence equation (E _ “Vila = or

where a 75 0 and k 2 1. As for the case k = 2, we define a sequence

v by writing

14.. = “'10..-

LINEAR DIFFERENCE EQUATIONS

159

Then, forO S r < k, E’s-run = an+k-r-1Ek-rvm

Hence

(E — a)”u,,

2k: (-)'(I:)a'E""u..

r=0

= a'H'k‘l i (_)r(I:)Ek—rvn r-o

= an+k—1(E _ l)”v,.

= an+k-1Akvn_

Thus our difference equation is satisfied iff Nu, = 0,

so that v is either a zero sequence or a polynomial sequence of degree at most k — 1. We may thus write the general solution in either of the forms

u,l = ¢"'1{A1 + A2(n — 1)+ A3(n — l)(n — 2) + - -+ Ak(n — l)(n — 2)- - -(n — k + 1)} or un=

{Po+P1n+P2n(n- l)+---

+ Pk_1'n(n — 1)- - -(n — k + 2)}, where A1, . . ., A," Po, . . ., Pk_1 are arbitrary constants. If the first

term of u is called ul then it is a simple matter to express A1, . . ., A,‘ in terms of 11,, . . ., uk and 0:. Likewise if the first term of u is called an then Po, . . ., Pk_1 can easily be expressed in terms of uo, . . ., 14,.-1 and a.

Theoretically, for sequences of complex numbers, any scale of relation may be expressed in the form (10(E - a1)"1(E — ¢2)ka . . .(E _ a,.)"mu,. = 0,

where a1, a2, . . ., a". are all different, and not 0, and k1, k2, . . ., km are positive integers. A solution of this difference equation is obtained on equating u" to the sum ofm parts, each ofthe form obtained above. This is, moreover, the most general form of solution. To prove that this is the case we must show that the k1 + k2 + - - - + k,,.

arbitrary constants appearing in the solution can be expressed in terms 0fa1,a2,...,am and thefirstkl + k2 +---+ kmterms ofu.

A proof is given in the next article.

160

OPERATORS

Suppose that we require, in real form, the part of the general solution corresponding to a repeated factor in the operator of the

form

{(E - A)“ + #2}", where /\ and p, are real and p. aé 0. The contribution to the general

solution arising from this factor can, in the first place, be expressed in the form

(A + I'M)”Xn + (A - 110W», where X and Y are either zero sequences on polynomial sequences of degree at most h — 1. Writing, as before, A + ip. = p(cos 9 + isin 9),

and setting X+Y=U,

i(X—Y)=V,

the above expression takes the form, p"(U,, cos n0 + V, sin n0). Example 10.5

Solve the difi'erence equation

2un+2 + 3un+1 — 814.. + 3mm = 0 (n 2 1)

Here

(2E3 + 3E2 — 8E + 3)u,,_1 = 0,

so that

(E — l)(2E — 003 + 3)u,._1 = 0.

Hence

“a = P + Q01)" + R(__3)n,

where P, Q, R are arbitrary constants. It is left to the reader to verify that P = i(‘3Uo + 5111 + 2143),

Q = #6110 -' 2'41 " "2),

and

R = 513010 - 3111 + 2142).

Example 10.6 Solve the difi'erence equation “n+3 = 3un+a — 4“,: giventhatul = 4,143 = 4andu3 =18.

(n > 1):

LINEAR DIFFERENCE EQUATIONS

Here

(E3 — 3E2 + 4)u,I = 0,

so that

(E + l)(E — 2)’u,. = 0.

Hence

161

un = A(—l)"'1 + 2"'1{B + C(n — 1)},

where and

A + B = 4,

(l)

—A+2B+2C=4,

(2)

A + 4B + SC = 18.

(3)

Now (2) and (3) are satisfied ifl' A = §(10 — 40), Then

B = fill — 5C).

A+B—4=3—3C,

and so (1) is satisfied iffC = 1. Hence A = 2, B = 2, and u,, = 2(—1)”-1 + 2"'1(n + 1). Example 10.7 Solve the difference equation “n+4 = 2u3+3 _ 3un+2 + 2un+1 _ up;

Here

(E4 — 2E3 + 3E3 — 213 + 1)u,, = 0,

so that

Now where

(n > 1).

(E2 — E + l)2un = 0.

Ea—E+l=(E-a)(E—fl), a=cos§vr+isin§m

B=oos§1r—isin§1r.

Hence u“ = {A + B(n — 1)} cos $(n — 1)1r + {P + Q(n - 1)} sin fin — l)1r, where A, B, P, Q are arbitrary constants. Also “1 =

and

A,

u; = &A + &B + h/3P + sh/3Q, us = —‘}A — B + h/SP + V3Q, u4 = —A — 3B,

from which it follows that A = u1,

= —%(u1 + “4):

and 6*

P = %v3(—5u1 + 12u2 — 6143 + 4114), Q = §\/3(u1 — 2u2 + 2143 — u;).

162

OPERATORS

Exercise 10.2 Solve the following difference equations: (1) “n+s=7un+1"6un

(n> 0), u°=liu1=2 “2:24-

(2) “n+3 " “n+1 = 2(“n+2" um)

(n>1): 111:2,

“2:3,

“3:11.

(n ? 0),

(3)115“; = “n+2 + “n+1 " u"

“0:1, “1:2, “2:4.

(4) "n+3 = 3“n+2 — 3“n+1 + "n

(n> 1), “1:1, “2:3, ”3:6.

(5) “n+3 = “n+2 _ “n+1 + ”n

(n> 1), 141:], “2:1, “3:2.

(6)un+3=un

(n> l),u1=l, “2—01‘3-0

(7)u,,+3+8u,,=0

(n>0),u°=1,,.u1=1u3=l

(3) "n+2=2“n—4un-1

(n>1):uo—3 u1=1u2=2.

(9)un+4=un (n>0),uo=7,,,u1=7u2=9u3=l. (10)u,.+4+4u,,=0

(n> 1),u1—2 112—2 ua—O u4—4

(ll)un+4+2un+2+u,.=0 (n> 1), u1=l, u2=l,,u3=0u4=0. (12) un+4 = 4(u,.+3 — 214,.” + 211,,” — un)

(n> 0),

“0—4111—3142—2

(13) "n+4 = 4un+3 _ 6“n+2 + 4“n+1 — “n.

“3:1.

(n> 1):

u1=l, u2=2,u3=5,u4=16. (14) “n+4 = 4(un+3 _ 4“n+1 + 4“?!)

(n> 0):

“0:5, “1:0, “2:36, u3=64.

10.4

Recurring power series

Suppose that u is a recurring sequence, with first term 14;, satisfying a scale of relation (aoEk + alE’"1 +- -+ ak)u,, = 0 (n> l), where k 2 1 and neither an nor ak is 0. Let x be a pronumeral, which

in our general theory represents a complex number, and write

g(x) = aoul + (aou, + a1u1)x +- - - + (aou,c + alukd + - - - + ak_1u1x"")

a0 + a1x+---+akx"

LINEAR DIFFERENCE EQUATIONS

163

If

aoEk '1' “IE,"-1 + ' ' ' + a): = 4003 " a1)k1(E — aa)"=- ' '(E — “my", where 0:1, 0:3, . . ., on". are all different, and not 0, and k1, k2, . . ., k," are positive integers, then

an + alx + - - - + akx" = ao(1 — a1x)"1(l - wax)”=- - -(1 - “mx)k“' Hence g(x)m can be expressed in partial fractions in the form

b(r, 1) b(r 2) 1— +(—1—«1,202+

b(nkr) +——(1_a,x)n.}’

where b(r, l), b(r, 2). . ., b(r, k,) are constants. Provided |x| < 1/|a,| for 1 $ r < m, we may use the binomial theorem to expand g(x) in a power series. Let

g(x) = n=1 Z u.’.x"'1Equating coeflicients of powers of x in the identity D

(a0 + alx + - ~ - + akx") Z u,’.x"'1 1I=1

= 00141 + (00112 + ”1141)" + + (“0% + aluk—l + ' ° ' + ak-Iulxk_1): we have

aoui = aoul, aoué + alui = aoug + alul,

aoullg + 011412-]. + ' ' ' + “lg—114,1 = aouk + alike—1 + ' ' ' + (Ila—1111.

Also, forn 2 l, “0:4,a + “Iuy'Hk—I + ' ' ' + 01:“; = 0-

It follows thatun =u,, for n> >1. Thus, provided |x| < 1/|a,| for l < r s m,

g(x) = 21 um”For this reason g is called the generatingfunction of u. The series

2 um”.

n-il.

164

OPERATORS

which we have proved to be convergent for sufiiciently small values of |x| is called a recurringpower series. We can now obtain u,I from the expression for g(x) in partial fractions. In the first place, the coefficient of x"'1 in the expansion of (l — ux)'1 is a"'1. If s is an integer greater than 1, denote the coeficient of .x"'1 in the expression of (l — ax)" by cm. Then

cm = land, forn > 1, _ n_ s(s+1)---(s+n—2)

”M ‘ °‘ 1

(n — 1)!"

'

For n 2 s + l, c

= an_1s(s +1)---(n +1).n(n + l)- - -(n + s — 2)

""

(s —1)!s(s + l)- - -(n — 1)

For 1 < n g s — l,

c

_ afl_1n(n + l)- - -(s — l).s(s +1)---(n + s — 2)

"" _

(n -1)ln(n +1)---(s —- 1)

In either case, n_1n(n +1)- --(n + s - 2) This last formula is also valid when n = l and when n = s, so that it is valid for n 2 1. It now follows that

u” = Zar‘1{b(r,l)+ b(r,2)n +- --

+ 170, kr) __n(n + l)(k; .3133,“ — 2) ' Allowing for the difference in notation, we have now proved that the

general solution of our scale of relation takes the form stated in the previous article. By way of illustration, consider the difl'erence equation

ao(E - o:)(E — BM = 0, where a and )3 are unequal, and not zero. Here

_M

g(x) ‘ 1 (1 — axx1 — Bx) =

1 ("a " Bur _ "a - 05141). a — B l —- ax l — fix

LINEAR DIFFERENCE EQUATIONS

165

Therefore

_1 un—a_fi

{(142 — 5111)!”—1 — (“a " W03" '1}-

Next consider the difference equation ao(E — a)3u,. = 0, where as ye 0. Here

g(x) = u———1 +001: :93:q It is not necessary to express g(x) in partial fractions to obtain its expansion in ascending powers of x. For |x| < l/Ial, using the expansion of (l — ax)”, g(x) = {ul + (an — 2au1)x} 2 na”'1x"'1. Therefore, for n 2 1,

u,I = ulna?”1 + (u, — 2au1)(n — 1)a"" = a”"2{(n — 1)u2 — (n — 2)au1}. We now consider the application of generating functions to the difference equations solved in Examples 10.5, 10.6 and 10.7.

Example 10.8

Solve the difi‘erence equation

214,,” + 3un+1 — Sun + 3u,._1 = 0 (n 21). Here

2(E — 1)(E — &)(E + 3)u,._1 = 0

and, since the first term of u is uo,

g(x) =

2110 + (2u1 + 3uo)x + (2142 + 3111 — 8uo)x’

Write

2(1 _ x)(l — ixXl + 3x)

g(x)=%+fi+lfsx' _ 2110 + (2111 + 3%) + (211; + 3u1 — Sun)

The“

P‘

2(1 — 9(1 + 3)

= £(—3uo + 5u1 + 2142).

166

OPERATORS

Likewise Q = %(3u0 — 2“1 — "2),

For |x| < 11;,

R = fiuo _ 3111 + 2113).

g(x) = i unx".

Hence, as in Example 10.5,

un = P + Q6)" + R(—3)"Example 10.9 Solve the difference equation

u..+3 = 314,.” - 4a.. (n > 1). given that u1 = 4, uz = 4 and us =18.

Here

(E + 1)(E — 2)’u,l = 0

and, for |x| < 1}, g(x)=4+4x+18x2+u4x3 +---. Therefore Thus

(1 — 3x + 4x3)g(x) = 4 — 8x + 6x”. g(x)

_ 4—8Jc+6x2 _(l+x)(1—4x+4x2)

__2_1—4x+4x2 +_2i_

—l+x

_ 2 + 1 + 1 —1+x l—h (L—hP Hence, as in Example 10.6, 14,. = 2(_1)n-1 + 211—1 + n.2n—1.

Example 10.10 Solve the difi‘erence equation u,” = 2un+3 — 3un+2 + 2u,,+1 — an

(n 2 1).

Here

flfl= u1 + (“a —' 2u1)x + (us '— 2112 + 31.41).)Ca + (”4 — 2113 + 3112 — 2110x3-

O—x+flP

LINEAR DIFFERENCE EQUATIONS

167

We can avoid the labour of expressing g(x) in partial fractions. Since l—x+x’=(l—ax)(l —fix), where

a=cos§w+isin§m

B=cos§w—isin§n,

we can expand (1 — x + x3)'2 in a power series iff |x| < 1. Write

m: ,2 1"" Also writep, = 0forr = —3, —2, —1. Then, forn 2 l,

“n = uiPn—i + (“2 “ 2"1)Pn—2 + (“a -' 2142 + 3“1)Pn—s

+ (u4 — 2u3 + 3u2 — 2u1)p,._;.

l

_

1

a (1—x+x2)2—a—B(1—°‘x

NOW

_

I3 1—flx):

and so

1 (l — x + x2)2

= («c-3 p)*{(1 fax)” + (1 flex? ‘ «2:313 (1 fax ' 1 flex» Also

and

a — B = 1'13,

a' + [3* = 2 cos §r1r,

«3 = l,

a' - [3' = 2i sin in.

It follows that, for r 2 0

p, = — g(r + 1) cos v}(r + 2)” + $1/3 sin 4}(r + I)". This formula is also correct for r = —3, —2, — 1.

While this method does not give such a compact result as that obtained in Example 10.7, less calculation is involved here. Example 10.11

For n> >1, sum the series fl

2 rzx"1. r=1

Denote the sum of this series by Sn(x). Now if u, = r2 then V314, = 0, so that

r2—3(r—l)"’+3(r—2)2—(r-.—3)3=0

168

OPERATORS

Hence, multiplying Sn(x) by (1 - x)3, n+1

(1 — x)3s,,(x)= Z r2x"1 — 3 Z (r — 1)2x'- 1 i=2 n+2

n+3

+3Z(r—2)3x" 1— 2(r—3)2xr- 1 r-4

= 2 fix"1 — 3{ Z (r — l)2x"3l + nzx’} r=1

r=1

+ 3{ i (r - 2)2xr—1 _ 1 + (n __ l)2xn + nzxn+1} f=1

—{i(r—3)’x"1—4—x+(n—2)3x" r=1 + (n _ l)2xn+1 + n2xn+2}

= 1+ x — (n + l)2x" + (2n2 + 2n — 1)x"+1 - nax’HQ. Thus, provided x 9!: l, Sn(x)=

Also

1+x— n+12x"+ 2n2+2n—lx"+1—n“x"+2 (

)

(§_x)a

)

S,,(l) = %n(n + l)(2n + 1).

It is well known that if |x| < 1 then nkx" —> 0 as n —>oo. It follows that if |x| < 1then CD

2

1_

2 ”r Note.

1+3:

(—l-x)3

Ifu is a polynomial sequence of degree m, and x at: 1, the

series 1:

2w

r=1

may be summed by multiplying by (l — x)“ + 1. Exercise 10.3 Use generating functions to solve the following difference equations: (1) an” — Sun.” + 6a,, = 0 (2) 2“n+2 + 5“n+1 + 314,, = 0

(n 21) (n 9 0)

LINEAR DIFFERENCE EQUATIONS

169

(3) 414..” = “n (n>1-) (4) “n+2 = “n+1 + “11.

(5) 4un+2 + "71— = 0

(">1”)

(n>0)-1)

(6) um: = um — un (n> (7) "n+2 " 2”n+1 + “11:0

(71> 0-)

(8) u..+2 + 414m + 414.. = 0 (n> 0) (9) “n+3 + 2“n+2 = “n+1 + 2%

(n > 1)-

(10) "n+3 '_ 21(n+2 = 4(un+1 _ 21411)

(n > 0)

(ll) “n+3 + 6un+2 + 121411“ + 811:1 = 0

(12) “n+3 = “n

(n 9 1)-

(n> 0-)

[Exercises 10.] and 10.2 may be used for further practice] Sum the following series for n 2 1:

(13) 2 (r + l)x"1.

(14) 2 (2r — l)x"1.

1|

(15))2 r(r + l)x'”1.

(l6) 2 er-I.

1:1

10.5

r-1

The general case

We now consider the general linear difference equation of order k, with constant coefficients,

(aoE’c + a1E"'1 + - - - + noun = an (n > 1),

(l)

where q is a prescribed sequence. Suppose that we can find some particular sequence Q such that (1) is satisfied when u“ = Q". .Write

v = u — Q.

Then, since the operator in (l) is linear, (l) is satisfied ifi'

(aoEk + (11E"‘1 +

+ ak)vn = 0 (n> 1)

(2)

We can then theoretically determine :1,1 in terms of ul, uz, . . ., u,c and

the known values of Q1, Q2, . . . , Qk. Denote the general solution of (2), in which 141, 14,, . . ., uk are arbitrary constants, by v“ = V".

Then (1) has general solution 14,, = V, + Q...

170

OPERATORS

In practice we usually begin by solving the reduced equation, as it is called, (aoE’F + a,E"'1 + - - - + £1,914,I = 0 (n 2 l). Denote the general solution, containing k arbitrary constants, of this difference equation by u" = U,,.

Then U is called the complementary sequence. If we can find a particular solution of (1), say “u = QM

it follows that (1) has general solution “it = Un + Qu-

It is not always possible to obtain this solution in compact form without the use of summation signs. For example, if 1

“n+1 —un= n—+l

(">0

then, forn > 1,

un=u1+%+%+-~+%, and there is no compact form for 14,. We can always find, in compact form, a particular solution of (aoEk + (1113":—1 + ' ' ° + “13% = 41: when q satisfies a known difference equation of the form

(boEm + blE""1 + - - - + bm)qn = 0The method is best illustrated by examples. Note that if q,, is the sum of two or more terms, say N

4n = Z qnm r=1

and, for l < r < N,

u,, = Q”,r

is a particular solution of (aoEk + alEk—l + ' ' ' + ak)un = 4n.”

LINEAR DIFFERENCE EQUATIONS then

171

“n = i QIM' ra 1

is a particular solution of (“0131‘ + alEk_1 + ' ' ‘ + “13“» = 41:-

Example 10.12 Solve the difl‘erence equation un+2 — an“ — 614,I = 2.4" + 5.3"

(n 2 0),

given that uo = u1 = 1. Here

(E2 — E — 6)u,, = 2.4’I + 5.3",

so that

(E + 2)(E — 3)u,. = 2.4" + 5.3".

The general solution of the reduced equation is u,I = P(-2)” + Q.3", where P and Q are arbitrary constants. Now consider the difference equation

(E + 2)(E — 3)un = 2.4".

(1)

If c,. = 2.4“ then (E — 4)c,, = 0, a result which we may conveniently

(though inaccurately) record by writing (E — 4)2.4" = 0.

It follows that if (1) is satisfied then (B — 4)(E + 2)(E — 3)u,, = 0. Hence

14,, = A.4" + B(—2)" + C.3".

Here, however, A is not arbitrary. For a particular solution of (1) we write u" = A.4".

Then we require A(4n” — 4"+1 — 6.4'?) = 2.4”, which is satisfied ifl A(16-4—6)=2, so that

A = 5—.

OPERATORS

172

Next consider the difference equation

(E + 2)(E — 3)u,, = 5.3".

(2)

(E — 3)5.3" = 0,

Since

if (2) is satisfied then

(E - 3)(E + 2)(E - 3)u.. = 0. so that

(E + 2)(E — 3)9u,, = 0.

Hence

u,‘ = L(—2)" + (Mn + N)3".

Here M is not arbitrary. For a particular solution of (2) we write u,| = Mn.3".

Then we require M{(n + 2)3"+’ — (n + 1)3"+1 -— 6n.3"} = 5.3",

which is satisfied ifl' M(18 — 3) = 5, so that

M = g.

We now have, for the general solution of the given difference equation,

u,l = P(—2)” + Q.3" + £4" + n.3"). We also require P + Q + v} = l and

—2P+3Q+-}=l.

Hence P =_ g, Q = 0, and the required solution is

u,l = §{4" + n.3" — (—2)"+1}. Example 10.13 Solve the difl‘erence equation u“; — 411,,” + 411,, = n33"

Here

(n 2 1).

(E — 2)’u,, = n23",

and the general solution of the reduced equation is

u” = {A + B(n — 1)}2n'1,

where A and B are arbitrary constants.

(1)

LINEAR DIFFERENCE EQUATIONS

Now

173

(E - 3)3n“.3" = 0

and so, if (1) is satisfied,

(E — 3)3(E — 2)"‘u,, = 0. For a particular solution of (l) we choose u so that (E — 3)3u,, = 0. Then

u” = (ana + bn + c)3".

This is a solution of (1) ill, for n 2 l,

9{a(n2+4n+4)+b(n+2)+c} —12{a(n2 + 711 + l) + b(n + 1) + c}

+ 4(ana + bn + c) = n”. Equating coeflicients, we require a=l,12a+b=0, Thus

a=l,

24a+6b+c==0.

b=—12,

c=48.

Hence the given recurrence relation has general solution u,, = {A + B(n — 1)}2"‘1 + (n2 — 12n + 48)3". Here

A=u1—lll,

B=iu2—u1—15.

Example 10.14 Solve the difi‘erence equation "n+3 — “n+2 — “n+1 + 14.. = n’

(n 2 0).

Here

(Ea — Ea — E + l)u,, = n”,

so that

(E + l)(E — 1)2u,l = n“.

The general solution of the reduced equation is

un=P(-l)"+Q+Rn, where P, Q, R are arbitrary constants. Now

(E — l)"n2 = 0 and so, if the given difl‘erence equation is satisfied,

(E +1)(E — 0%,, = 0.

174

OPERATORS

For a particular solution we choose u so that

(E - 1W» = 0, and write

un = an(n — 1)(n — 2)(n — 3) + bn(n — 1)(n — 2) + cn(n' — 1); note that, from the form of the complementary sequence, it would

be pointless to add (dn + e) on the right. Then (E — l)’u,. = 4.3an(n — l) + 3.2bn + 2c, and

(E + 1)(E — l)’un = 12a{(n + l)n + n(n — 1)} +6b{(n + 1) + n} + 2c(l + 1)

= 24an’ + 6b(2n + 1) + 4c. Hence we require

a = 51;,

b = 0,

c = 0.

Thus the given difi‘erence equation has general solution un = P(—l)" + Q + Rn + 211n(n — 1)(n - 2)(n — 3). Here

P = ““0 " 2111 + “2),

Q = H3140 + 2111 — “2),

R = fl“: — “0)-

Example 10.15 Solve the simultaneous difl‘erenoe equations "n+2 + “n + ”n+2 + 2011 = n

and

“n+2 — 3“n+1 — 2“,, + ”n+2 _ 2”n+1 = 3n + l,

where n > 1. Here and

(E3 + l)u,. + (E2 + 2)v,, = n (E2 — 3E — 2)un + (E2 — 2E)v,l = 3n + 1.

We begin by assuming that these equations have solutions. Having obtained the only possible form of solution it must then be verified that the given equations are satisfied by this solution. Eliminating

the sequence 0,



{(132 - ZEXE2 + 1) - (E2 + 2)(132 - 3E - 2)}ua

= (E2 — 213»: — (E’ + 2)(3n + 1).

LINEAR DIFFERENCE EQUATIONS

175

Thus

(E3 + E2 + 4E + 4)u,. = {(n + 2) — 2(n + 1)} - {(3n + 7) + 2(3n + 1)}, so that

(E + 1)(E2 + 4)u,I = — lOn — 9. This difi'erence equation has general solution u“ = A(—1)"'1 + 2"'1{B cos fin — D11 + Csin fin — l)1r} — n, where A, B, C are arbitrary constants.

We can now determine 0,, without introducing any more arbitrary constants. For we can find polynomial operators P(E) and Q(E) such that

(E2 + 2)P(E) + (E2 — 2E)Q(E) is a constant operator other than 0. By successive division, as in the usual H.C.E. process,

E”—2E=(E3+2)—(2E+2), and

E2+2=(2E+2)(-}E—‘1r)+3.

Therefore 3 = (E2 + 2) + flE — l){(E2 — 2E) —- (E3 + 2)}, and so

6 = (—E + 3)(E2 + 2) + (E — 1)(E2 — 2E).

Hence

{(—E + 3)(E" + l) + (E — 1)(Ea — 3E — 2)}u,I + 612,, = (—E + 3)n + (E — 1)(3n + 1). Thus

60,, = {—(n +1)+ 3n} + {(3n + 4) — (3n +1)}+(Ea — 5)u,. = 2n + 2 — 4A(—l)"'1 + 2"+1 {B cos fin + 1)1r + C sin -}(n + l)1r} — 5.2"'1{B cosfln — 1)1r + Csinfln — l)1r} — {(n + 2) — 5n}.

Therefore

0,. = n — fi-A(—l)"'l — 3.2""{B cosfln — I)” + Csinvfln — D17}.

176

OPERATORS

It is left to the reader to check the solution we have obtained. It is instructive to do this by showing that the given difference equations are satisfied whenever

(E3+E’+4E+4)u,.=—10n—9 and

69,.=2n+2+(E“—5)u,,.

Exercise 10.4 Obtain general solutions of the following difference equations: (1) “n+1 "' 214,. = l-

(2) “n+1 + 214,. = 4n_

(3) um + u» =1+(-l)"(5) um - 2a.. = n-2"“-

(4) 2am — u» = n(6) um + un = COSlmr-

(7) u“; — 4u,l = 3" - 3.

(8) an” + 314,, = sin inn.

(9) "n+2 + “n = cos elm"-

(10) un+2 — 214m + un = n”-

(11) um + 3un+2 — 4a.. = n-3"- (12) “n+3 + 3un+2 - 414,. = nSolve the following difi'erence equations:

(l3)Tu,.+1 + an = 3-2” (n> 2), “2 = 6 (l4) u,1+1— u,.=n+l

(n2 l),u1=1.

(15) u“, — 4u,, =na (n2 0),uo =1, u1 = 3. (16)un+2+u,.= 2"+1 (n21),u1=l, u1=l.

(17) “n+3 - 6un+2 + llun+1 - Gun = 2" (n> 1) u1=1, u2=2, 143:6.

(18)u,.+3— u,1=l

(n2 0),uo=2,u1=l, u1=l.

Solve the following simultaneous difl'erence equations, supposing them to be valid for n 2 1: (l9) 1),. = u,1+1 + 2a”, 2u,, = 1.2,. + 014,1 +1, u2 = 2. (20) u,1+-1 -— 12,. = 4, vn+1 — u” = 4n, u1 = 0, v1 = 9.

(21) u. + v,. + vu+1 = 3, un + 3un+1 + ”n + 3vn+2 = 2"(22)un+1 '_ “13 = ”n+1 — ”n +15 “n+1 _ 211,, + ”n+1 + ”n = 7‘,

u1 = 0, u; = 2.

(23) “H2" 2un+1 + 2“» — ”n+2 + 20n+1 — 41),, = 2"”

= 9. u,.+1 — 3a,, — v,.+1 + 40,, = 9n, u1=10, v1— (24) un+1 + Bun + 4014.1— 412,, = 2. 3"“— 10, “n+1—3un+vn+2_vn=8n9 “1:0: u2=4, 91= (25) u,1 = 0,,” + 20,, + 8, v1 = un+2 + 211,, + 4sini-mr, u1=l, u3=3, v1=v1=1. 1' Cf. Exercise 10.1 (30).

LINEAR DIFFERENCE EQUATIONS

177

(26) ”n+3 — 3vn+2 _ l’n+1 + 301; _ “n+2 + 9“,, = 9_2n+1’

”n+3 — 2vn+2 — ”n+1 + 20,. + un+2 - 4a.. = 4.3““, ul =13, “2 = 30, v; = 6, ”2 =17.

(27) A vessel A contains a volume 01 of water and a vessel B contains a volume 173 of alcohol. A volume v from A is stirred into B

and a volume 0 from B is then stirred into A. After this process has been carried out n times what is the volume of alcohol in A? (Neglect any change in the total volume of liquid.) (28) At a dance there are n married couples. If u” is the number of ways in which each man can partner another man’s wife prove that, for n 2 3, “n = (n '_ lXun-l + urn—2)-

Deduce that, for n 2 2,

un - nun-1 = (“-1)" andthat

I l 1 uu=n!{-2—!—§-i+---+(—)”n—!}-

ll LINEAR DIFFERENTIAL EQUATIONS 11.1 Introduction A linear dtfierential equation of order k, with constant coefi‘icients, is a functional equation of the form (aoDk + al_1 +"'+ ak)F =1;

where k 2 1, a0, a1, . . ., ak are constants with ac aé 0, f is a pre-

scribed function and F represents a function which is to be determined. We have already noted on p. 89 that we may suppress or insert in the operator terms with zero coeflicient. We are primarily

concerned with difi‘erential equations for-real functions of a real variable but, as in our discussion of linear difference equations, our

general theory is facilitated by supposing our functions to be complex functions of a real variable. For the calculus of such functions the reader is referred to the author’s Calculus, Volume II, Chapter

25. In particular we require the exponential function. If x is real and a = A + in, where A and p. are real, then by definition exp ax = e""(cos ,u.x + isin 11.x). Also

Dexpax = «exp ax.

We shall usually write e" in place of exp ax. We are then assigning to e“, which is not uniquely determined, its principal value. When solving differential equations of the type under discussion it is tacitly assumed that we are concerned only with functions for which 9F = 9f. We have already discussed in 6.8 certain theoretical difl‘iculties concerning domains which arise in solving linear differential equations of the form

DF=f In some cases, unless we modify the definition of a derivative, we can only find functions such that

DFCfl 178

LINEAR DIFFERENTIAL EQUATIONS

179

In this chapter, however, we are only concerned with cases in Which 9f = R and our difl'erential equation has genuine solutions, so that 9F = R. In practice it is convenient to equate values onto which the two sides of our equation map an unspecified argument x, say, and to refer to

(aoDk + al'1 + - - - + ak)F(x) =f(x) as a difl‘erential equation. Furthermore an independent variable y, say, is usually used for F(x), and D'y or :7, Is written in place of

D'F(x), so that our typical equation appears either as

(aoDk + alD"-1 + - - - + ak)y = f(x) k

oras

k- 1y

aogxi+a1:Xk—_-y —‘i+" -+a,,y=f(x).

Linear difi'erential equations with constant coeflicients may be solved by methods which are analogous to those used to solve linear difl‘erence equations with constant coefficients. We describe in this chapter a method which depends upon factorisation of the operator. A second method, which in so far as it involves partial fractions is

analogous to the use of generating functions, employs Laplace transforms. For an account of this method the reader is referred to the author’s Calculus, Volume II, Chapter 27.

11.2 Equations of reduced form We begin by studying the linear differential equation (aoDk + al-1 + ' ‘ ' + ak)y = 0.

In our general theory we take x to be a real variable and y a complex variable. We may then suppose our equation to be written in the form

ao(D - «1)"I(D - «2)k2- - -(D - am)”my = 0, where k1, k2, . . ., k,,. are positive integers and a1, a2, . . ., a,” are all

different. There is one important difference from the corresponding theory of difi‘er‘ence equations. We do not stipulate that (1,, 9E 0, so that one ofal, a2, . . ., a," may be 0.

Consider first the equation

(D - a)? = 0-

(1)

180

OPERATORS

Multiplying by the integrating factor e'“", which is never zero, (1) is satisfied ifi‘ e_¢x(D — a)” = 05

that is

D(e""‘y = 0;

the reader is reminded that here the left-hand side is an abbreviation ’

for

D{GXP ° (- 0:19)}F(x)It follows that (l) is satisfied ifl‘ e—uxy = A,

where A is an arbitrary constant. Thus the general solution or complete primitive, as it is sometimes called, is y = Ae". This solution is valid for all values of x, so that

F = A exp 0 (a0). Notethaty = Awhenx = 0.

Next consider the equation

(D - «XD - 3))! = 0-

(2)

By what we have already proved this equation is satisfied ifl‘

(D — my = A'e“. where A’ is an arbitrary constant. Multiplying by the integrating factor e"" it follows that (2) is satisfied ifl‘ D(e""y) = A’e‘“"””.

(3)

Two cases now arise:

(i) Suppose that a aé ,3. Then (3) is satisfied ifl' I

e—fix

=

e(¢-fi)x + B,

where B is an arbitrary constant. Replacing A'/(a — B) by A, it

follows that (2) has general solution y = Ae” + Be“.

(4)

It is of course clear, as in the corresponding theory for difference

equations, that (4) gives us a solution of (2). The purpose of the above proof is to demonstrate that every solution is of this form.

LINEAR DIFFERENTIAL EQUATIONS

181

(ii) Suppose that a = ,3. Then (3) becomes D(e'“"y = A’. Therefore

e'“"y = A’x + B,

where B is an arbitrary constant. Replacing A’ by A, it follows that the equation

(D - are = has general solution y = (Ax + B)e‘"‘.

(5)

In practice we often require the solution of a linear differential equation of order k for which y, Dy, . . ., Dk'ly have assigned values when x = 0. To avoid repetition We denote these values by yo, y1, . . ., yk.1 respectively. It is left to the reader to show that (4) is satisfied ifl'

A _ y1—__fi_yo a—fl

- h, B = 0%a—p

and that (5) is satisfied ifi‘ B =J’o,

Example 11.1

A=}’1"°‘yo-

Solve the differential equation

2%};+dx Q= 0, given that yo = 0,y1 = 1. Here

(2D2 + D)y = 0,

so that

D(2D + l)y = 0.

Since e°" = 1, the general solution is y = A + Be'm. Then

We require

and

Dy == —}Be"‘”.

A + B =0

-§B = 1.

Hence B = -2, A = 2, and the required solution is y = 2(1 — e'””).

OPERATORS

182

Example 11.2 Solve the differential equation

g—ZAZ—:+(A2+u’)y=0, where Aandpare real and}; ye 0.

Here

{(D - A)2 + 1"}y == 0.

sothat

(D—A—ip)(D—A+ip.)y=0.

Hence the general solution is y = Ae(7\+lu)x + Bent—(10x,

that is y = e""{A(cos ,ux + i sin [1-36) + B(cos ax — isin px)}.

Setting

A + B = P,

i(A — B) = Q,

the general solution takes the form

y = e""(P cos ax + Q sin px). Since the calculus of real numbers is isomorphic with that of real complex numbers, this solution is valid for real functions of a real variable. Now Dy = e""{A(P cos px + Q sin px) + u.(—P sin [ix + Q cos px)}. Therefore and

so that

P = yo AP + F'Q = y1:

Q' = (y; - mm.

In particular, taking A = 0, if [L aé 0 the equation

%; + #2? = 0

has general solution

y = yo cos M + (J’i/I‘) sin I‘xWe now consider the differential equation

(D - 06)"? = 0. where m is a positive integer. Write

-z = e‘“"y.

(6)

LINEAR DIFFERENTIAL EQUATIONS

Then

Dz = e‘“"Dy + (—ae"‘"‘)y.

Thus

Dz = e'“"(D — a)y.

183

Replacing y by (D — a)y, it follows that

Daz = e‘“"(D — ac)a , and clearly, by induction,

D’"z = e""(D — a)"'y. Hence (6) is satisfied ifl' D'”z = 0,

so that

2: A0 + A1x+---+A,,._1x""1,

where A0, A1, . . ., Am.1 are arbitrary constants. Thus (6) has

general solution y = e“"(Ao + Alx + ‘ ' ‘ + Am_1x”"1).

For 0 < r S m —- 1 denote the value of D'z when x = 0 by 2,. Then Ar = zr/r!

We may express 2, in terms of yo, y1,.. ., y, and a. By Leibnitz’s theorem, 1‘

U: = 2 (;)(D'-'e-¢~)(D'y). s=0

Hence

T r 7—3 y,. (S)(-a) 2, = 3;)

We next tackle the differential equation

(D - a)“(D - 13)")! = 0,

(7)

where m and n are positive integers and 0: ye 3. By what we have just proved, (7) is satisfied ifi' (D " 3)"? = €“(Amo + Amlx + ' ' ' + Amm-lxm—l)’

where A”, A“, . . . , AM- 1 are arbitrary constants. These constants

can be expressed in terms of a and the values when x = 0 of (D — fl)" , (D — B)"+1y,. . ., (D - ,B"+""1y, and so can be expressed in terms of a, )3, yo, yl, . . ., y,,+,,,_1. Now

(D — fl)"y = e‘xD"(e"“y)-

184

OPERATORS

Hence (7) is satisfied ifl‘ D"(e""“y) = e‘“"”‘(A,.,o + Amlx + - ~ - + A,.,,,._1x’”'1).

Using repeated integration by parts it follows that (7) is satisfied ifi‘, say, Dn—1(e-flxy = ear-3):: x (An—1.0 + Arr-1.1x + ' ' ' + An—1.m-1xm'1) + Bn—l' Here An_1,o,A,,_1,1, .. .,A,._1_,,,_1 can be expressed in terms of a, p: Arno: Amla - - -: Arum—1: and SO in terms 0f“: p: J’o, yle ' - u yn+m-1'

Also B,,_1 can be expressed in terms of An-” and the value when

x = 0 of D“'1(e""y , and so in terms of a, )3, yo, yl, . . ., yn_,,,+1. Repeating this process a further It — 1 times, it follows that (7) has

general solution

e""y = em-B)x(Ao + Alx + - - - + A,,._1x""1) + (BO + l + ° ' ' + Bn—Ixndl)!

or y = e“"(Ao + A135 + ‘ ' '+ Am-1xm_1) + e‘”(Bo + l + - - - + B,,_1x“‘1). Here A0, A1, . . ., A,,._1, Bo, Bl, . . ., B".1 are arbitrary constants which can be expressed in terms of a, )3, yo, yl, . . ., y,,+,,,_1. This method of solution can be extended, by induction, to find the

general solution of a0(D — a1)k1(D _ aa)"2- "(D — army‘s-y = 0.

Writing

k=k1+k2+---+km,

the general solution contains k arbitrary constants which can be expressed in terms of a1, a2, . . ., am, yo, yl, . . ., yk_1. The reader should find no difficulty in showing that if A and p. are real, and

p aé 0, the contribution to the general solution arising from a factor

{03 — A)“ + W in the operator can be put in the form e""{(ao + alx + - - - + a,_1x"1) cos px + (b0 + bx + - - - + b,_1x"1) sin px}.

LINEAR DIFFERENTIAL EQUATIONS

Exercise 11.1 tial equations:

185

Obtain general solutions of the following difi‘eren-

(l):—:{=y.

(2)2— :J’HZy——2y=0.

(3)2dZ—1,’= 3‘1—5;

(4)d—y+4Z—fc+4y=o.

(5)9d—J’—6%+

—o.

(7)Z:—':+22;—y=0.

(6)::T 2':=+4y

o.

(8)2—y— g+y= o.

(9)d—y+2%+5y=0. (10)Z—;;— Zy+3y= o.

(11)—— 2—1— 6y=0. (12)%—2j7§— j—f—c+sy=o.. (13)Zx—{+jfc=(14)g+y=o. (15)d—”— g—x’;=o.

(16)diy—4y=o.

(17):?{+=4y

(13)—— Z—x§+y=o.

(19%

o.

4%:

(20)— +4 dx,+4y=o.

Solve the following differential equations: d2 (21)d7.:=4y9

yo=0,

y1=l.

d_y+ dy (22)dx2 +dx=0

(23):?:+=y

y0=19

Y1=1-

0 yo=0 y1=l-

d _(24)%—;az+9y=0, yo=0, y1=l. d 4ddi+5y=0, yo=l, y1=0. (25)Ey+4 7+

186

OPERATORS

d d yo=1, y1=2. (26)d-x—J,’-2d—:+5y=o, d2 d (27)2—— Eh l+=o yo=o y1=1 y2=o. d3 d2 d y) yo=2, y1=1, y2=2. 9(d§;+ dx——Z= (28)d—x%+ (294% =27(j§— y) yo=5, y1=o, y2=9. d3 yo=1, y1=1, y2=—5. (30)E{=y. 11.3 The general case We now consider the general linear differential equation of order k, with constant coeflicients, (aoDk + al_1 + ' ' ' + 0k)? = f(x)’

(1)

where f is a prescribed function. For simplicity we suppose that 9f = R. Suppose that we can find some particular function Q, with domain R, such that (l) is satisfied when y = Q(x). Write

z = y - Q(x)Then, since the operator in (l) is quasi-linear and we are studying functions with common domain R, (l) is satisfied iff (aoDk + al'1 + - - - + ak)z = 0.

(2)

We can then theoretically determine 2 in terms of yo, yl, . . ., y,,_1 and the known values Q(O), Q’(0), . . ., Q‘k‘1’(0). Denote the general solution of (2), in which yo, yl, . . ., yk_1 are arbitrary constants, by

z = P(x). Then (1) has general solution

y = P00 + Q(x)As in the case of difference equations we usually begin by solving the reduced equation (aoD" + al'1 + - - - + ak)y = 0.

LINEAR DIFFERENTIAL EQUATIONS

187

Denote the general solution, containing k arbitrary constants, of this difi'erential equation by

y = R(x)-

Then R is called the complementaryfunction. If we can find a particular solution of (1), say y = QC”): it follows that (1) has general solution

y = R(x) + Q(x). Example 11.3 Solve the differential equation (1’2 .

d

d7}; — (T: —- 6y = 2e” + 5e“, given that yo = y1 = 1. Here

(D + 2)(D — 3)y = 2e“ + 5e”,

and the general solution of the reduced equation is y = Ae'” + Re“, where A and B are arbitrary constants. Now consider the equation (D + 2)(D - 3)y = 2e“. Since

(1)

(D —- 4)2e*" = 0,

if (1) is satisfied then

(D - 4)(D + 2)(D — 3).v = 0Hence

y = ae‘“ + be"”‘ + cc“.

Here, however, a is not arbitrary. For a particular solution of (l) we write y = ae‘”. Then we require a(l6 — 4 - 6)e"‘ = 2e“, which is satisfied ifl‘ a = i. Next consider the equation

(D + 2)(D — 3)y = 5ea"-

(2)

188

OPERATORS

Since if (2) is satisfied then

(D — 3)5e3" = 0,

(D..— 3)(D + 2)(D — 3)y = 0, so that

(D + 2)(D — 3)“y = 0.

Hence

y = pe'z" + (qx + r)e3".

Here q is not arbitrary. For a particular solution of (2) we write y = qxea". Then

ye's" = qx,

and

(D — 3)ye'3" = q.

Therefore

(D — 3)y = qes",

and

(D + 2)(D — 3)y = qea"(3 + 2) =

e”.

Thus (2) is satisfied ifl‘ q = 1. We now have, for the general solution of the given differential equation,

y = Ae'” + Be” + vie” + xea". For this solution,

Dy = —2Ae'2" + 3Be3" + gen + (1 + 3x)e"*. Thus we require

A+B+§=1 and

-2A+3B+%=1.

Hence A = f, B = 0, and the required solution is

y = §(2e"'"" + e“ + 3xe3"). Example 11.4 Solve the differential equation

day dy _+2fi The reduced equation is

{(13 +1)2 + 2}y = 0.

LINEAR DIFFERENTIAL EQUATIONS

189

Hence, for the complementary function,

y = e'”(Acos 1/2x + B sin v2x), where A and B are arbitrary constants.

Dax“ = 0.

Now

Hence, if the given equation is satisfied,

D=’(D2 + 2D + 3)y = 0. For a particular solution we choose y so that D3y = 0,

y=ax3+bx+c.

andwrite

Then

Dy=2ax+b

and

Day = 2a.

Hence we require 2a+2(2ax+b)+3(ax2+bx+c)=x2. Equating coeflicients,

3a=l, 4a'+3b=o, 2a+2b+3c=0. Hence

a=§, b=—%, c=—2%.

Therefore the given difl‘erential equation has general solution y = e'“(A cos 1/2x + B sin v2x) + 517(9x2 — 12x + 2). It is left to the reader to verify that A=J’o—2_2'Ta

B='}\/2()’0+J’1—é'g‘)-

Example 11.5 Solve the differential equation 2

g; + y = x sin x.

The general solution of the reduced equation is y = Acosx + Bsinx,

'where A and B are arbitrary constants. If x is real, x sin x is the imaginary part of am“. Consider then the equation

(D2 + 1)}! = ace”.

(1)

190

OPERATORS

Since

(D - i)2xe"‘ = ‘0,

if (1) is satisfied then:

(D - 1')‘(D2 + 1).v = 0. (D + i)(D — 0"}: = 0.

so that

Accordingly, for a particular solution of (1), we write

y = (ax2 + bx)e"‘; note that there is no need to write y = (ax2 + bx + c)e"‘, since

y = ce‘x

is a solution of the reduced form of (1). If then

y = (ax2 + bx)e“‘ (D — i)y = e‘”D(ax2 + bx)

= (2ax + b)e"‘, and

(D + i)(D — i)y = {2a + 2i(2ax + b)}e"‘.

Hence (1) is satisfied if

4ia = 1, so that

a = —&i,

2a + 2ib = 0, b = &.

Thus (1) is satisfied if y = 9;(-ix2 + x)(cosx + isinx). Selecting the imaginary part on the right, it follows that the given differential equation has general solution y = (A — ix2)cosx + (B + ix)sinx.

HereA=yo,B =y1. Example 11.6 Solve the differential equation 2

Z7Z—y=xe"cos2x.

LINEAR DIFFERENTIAL EQUATIONS

191

The general solution of the reduced equation is y = Ae" + Be'”, where A and B are arbitrary constants. If x is real, xe" cos 2x is the real part of xe‘“"’”. Consider the

equation (D2 — l)y = xe‘l‘m’”.

(D — l — 2i)3xe(1+at)x = 0,

Since

if (I) is satisfied then (D - l — 2i)"‘(Da -1)y = 0. Accordingly, for a particular solution of (l), we write

y = (ax + b)e‘1+2”“. Dy = {a + (1 + 2i)(ax + b)}e 1, sum the series 11-1

2 (k +. r)x,,

r=0

r

where k is a non-negative integer. Treating 'x as a parameter and introducing an integral variable p, write

“'1 k + p + r ,

S,-— 2( r=0

p+r

)x.

198

OPERATORS

Then the sum of the given series is So; note that this is the sum of the first n terms of the binomial expansion for (l — x)‘“‘+1’. Now write k+

u" = ( p p). n-1

Then

S, = Z x'E'u,, r=0

and so

S, z (1 — x“E")(l — xE)‘1u,.

(1) For x 56 l, we write

(I — xE)'1u, z (1 — x - xA)'1u, l—x,_o(l—

)'

_ k+p+1_ k+p)

Now

Au"_(iv+l)

k+p)

+1



:1

and, by induction,

,

_ k+p

A“”_(iv+r) for r> 0. Thus A'u, = Oforr > k, and S,

n"

x'

k+p

~1—x(1‘xE)Z(1—x) _‘_'(p+r)' r=o

Making a fresh start,

(1 — xE)S, = (1 — x"E")u, so that, if x aé l,

(

x

xA) .

Hence 2:" + 1 {I - WHA’H-l} , —

k x)Ar(1- anfilp.

,,_ Now

A"+IS,, = Z x’E’A"+1u, = o. 7-0

INFINITE SERIES- OF OPERATORS

199

It follows that our tentative expression for SD is correct, so that

s=1—_x§o{(’;i’i)-x"(’;ifii’:)}u—f'x—yHence we have, for the sum of the given series,

8° =0 - waif“ ) ”(1; 13)}“1 " x)” A13

o

" k 7 20 (r)x (1

_

x)” —r = {(1

_

x) + x}k =

Thus, ifx 9’: l,

l

k+n ,

_,

5°: W? _ xao (n+ r)x(1_ x)" } (2) Now suppose that x = 1. Then Sp z (1 -— E")(—A)‘1u,,. This suggests that

s, z —(1-—E")(k +9) ”—1 In fact,

S, =

”f:

E u,

=2:Em — n(’; ii’) 1::" — ”(11: :11’) as predicted. Thus S _(k+p+n)_ (l;+p)



Since

+n—-1

( k1) = 0

we have, for the sum of the given series when x = l,

30 = (nk t’l’).

200

OPERATORS

This is a well-known result, usually proved by showing that So is the coeflicient of x""1 in the binomial expansion of (l — x)"“2, using the identity

_x)_l‘Tc_+a

2 x — "+1 f-O —x1) (1

(I x] 0}; see 5.9. (l — 30)(1 — 2:9)‘1". (4) iii-1’30 — 0r“. —«}{(9 —' l)/t9}1”(0 — 1)", with domain {x l x < Oorx > 1}. —i0'1"(fl — l)'°"“, with domain {x | x > 1}. —4 sin a 2:9. (8) 2 cos a :93 — 403(sin o 0’). 19*(2 sin — 20 cos — 0’ sin). (10) sin cos’ (2 cos’ — 3 sin”). 2709 + 1) exp”. (12) sin“1 + 00 — oar”.

(13) - W’ with domain {x | x > 1}. (14) 1—533 (15) (I9 8811)” = {(x, 1) I x > 0} U {(x, - 1) l x < 0}(16) (00)” = {(x, 0) l x 96 0}. (17) 1/0. (18) WW?” = {(x, l/x) l x > 0}. (19) cat. (20) tan secl” sec'“a = {(x, tan x) | sec x 2 1}. .(21) logsec tan + 01’30‘3” sec. (22) 61196-119 = {(x, 1)] x > 0}. 6.2 No. X + 0 E X requires that each ordered pair of X + 0 is an ordered pair of X, and this is so iff Xf + 0f = Xfwhenever f e 5'. 6.3

Let X = {(fifo (—19)) Ife 3,}. Then (X + 0X0 — 1)1”is null.

6.5

11, defined in 6.1, is such an operator. So also is D.

6.6 Efg and (Bf) x (Eg) have common domain QEfn QEg. For any argument x of this domain,

Efflx) =f£(x + 1) = f(x + 1)g(x + 1) = {(131) X (Eg)}(X).

6.7 1—A=1—(E—1)=l-—E+l=2—E. 1—A’=l—(E‘—2E+1)='l—E’+2E—l=0+2E—E’. A=+2A+1=(A+1)==(E+o==E=+0E+o. (1)x=—(2x+1)=2x=—(x+1)=, x3-2=2(x+l)‘—(x+2)’, 2+2(2x+l)+x’=(x+2)’. _1_. 1 2 1 2

0

2

l

E‘x(x+1)(x+—2)'=7c+x'+1'x+2’

4-; 1_;+_°_+2.x x(x+l)(x+2) x(x+l) x_x+2 x+1 6.8

LJ'LS. = (E2 '— 2E + l)x(—-x-1—_._D=Aax(Tl——_l_n=k.fi.s.

6.9

(1) No.

E(l—A)=E(2-E)=2E-E';l—A’=0+2E'é-'E’.

(2) Yes. Inthe first place, (1 +A)(1 —A) a: 11+ oA—A’.

Answaas AND SOLUTION NOTES

221

We'may delete the term OAon the grounds that, whenever f e an,

'

any; 9A}: Otherwise:

(I +A)(l -—A)=(E+0)(2—E)=0+2E—E’=l—A’. 6.10

Yes. Whenever f e 91;, 9f '2 9Df and so Df + 0f = Df.

6.11 Suppose that (x, y), (x + c1, y), (x, y + 0;), (x + c1,y +1 ca) are all arguments off. Then

f(x + c1..v + 02) -f(x,y) = {f(x + any + Ca) -f(x.y + 62)} _ ”(xi J’ + c2) _ [(x9 1’»-

The first result now follows; note that for (x, y) to be an argument of Ann] it is not necessary for (x + (:1, y) and (x, y + ca) to be arguments of f. The second result is proved in like manner. Also ve1.eaf2 (van) '1' v0.1:3E1-c1)f;

6.12

It would be better to write 1 _ l D 1?: -;+c1(x>0),

6.13

vc1,.¢.‘af-2 (V0.ea + V¢1_0E3_°3)fi

l _ l D 1;§= —;+Cg (x 0 as n —>oo, it follows that

.131; S,.(x) = 2(1 — V + V’)x2 = 2(xa — 2x + 3). (2) Write

Su(x) = 3:1 (1»)’(x — r)3

(n 2 l).

v=0

Then

(1 - iE")Sn(X) = {1 - (iE‘1)"}x3-

Now 1 — «1~E‘1 = §(l + iV) and (l — {V + iV’ — §V3)(1 + iV) = 1 — I—I‘V‘. It follows, as in (1), that 3131; S,.(x) = g0 — 5V + iV’ — §V°)x3 = i(4x3 — 6x2 + 12x — ll). (3) Write

S..(x) = ”:1 (r + 1)(~})'(x — r)2

(n 2 1).

i=0

Then

(1 — iE'1)’S,.(x) = {1 — (n + “GE-1)" + n(§E‘1)"+1}x3.

Thus

*0 + V)’Sn(x) = x2 - (n + l)(*})"(x - n)2 + n(i)"“(x - n - l)“Now

(1 — 2V + 3V')(1 + V)2 = 1 + 4V3 + 3V4.

It follows, as in (1), that :13; S,.(x) = 4(1 — 2V + 3V’)x' = 4(xa — 4x + 8).

ANSWERS AND SOLUTION NOTES n-1

(4) Write not) = Z (r + l)t'(x - r — 1)2

235

(n as 1).

1-0

Then

(1 — tE'1)’F..(x) = {l — (n + l)(1‘E'1)’I + n(tE'1)"+1}(x — 1)”. Thus

{(1 - t) + tV}’F..(x) = (x — 1)= - (n + l)t"(x — n — 1)"2 + nt"”(x — n — 2)=. NOW

{(1 — t)a — 2(1 — t)tV + 3t’V3}{(1 — t) + IV}2 = (l — t)‘ + 40 — t‘);°V3 + 3t‘V‘.

As in Example 12.], if It] < 1 then (1 — I)‘ liglnFfix) = {(1 —- t)a — 2(1 — t)tV + 313V’}(x - l)”. The stated result follows on setting t = x.

0+1):

(5) Write s,.(x) = “1:

,.,. WT?) 0'? ”-

where x is not 0 or a negative integer. Then

s,(x) =

7.0

(r + 1)(—A)' i

(1 + A)=s..(x) = {1 — (n + l)(—A)" + n(—A)"+1}Jl‘ NIH

Hence

1

x(1+;)(1+g)...(1+ "—3;

on simplification. Provided x > 0,

1 1 1 (1+2)(1+ 3)"(1+n+1)>x(i+§+'”+n—+1)' As )1 ->00,

1 l + 5+

1 '+ n—+_l ->°0.

Hence if x > 0 then E’Sn(x) —> 1/x as n—>oo. Therefore, provided x > 2, S..(x) —> l/(x — 2) as n —>00. The given series is not convergent if x < 2; if x < 1 the (r + l)th term does not tend to 0 as r->oo, and if 1 < x s 2 the (r + 1)th term is not less

than l/(r + 2).

(6) Write u.(x) = (— i}?— i r. v,(x)= (9m (r 2 o),

236

OPERATORS

where x is not 0 or a negative'integer. Then, as r —>oo, u,+1(x)/u,(x) —> 0 and v” 1(x)/v,(x) —> 0. Hence, by d’Alembert’s ratio test,

2 u,(x).

2 0.0:)

7-0

7-0

are absolutely convergent. Denote the sums of these infinite series by U(x), V(x). Also, for n 2 1, write 1| - 1

ll - 1

U.(x) = Z u,(x),

v,.(x) = 2; v.(x).

r-o

r-

'Il-l.

n-

1

Then

U.(x) = Z (—lE)' ,1: mo = o (—w;

New

(1 + new» = {1 — (—iE)”} i

and

1 + in = :(l + iA).

Hence 0-1

1

{1 — (—iA)"}Uu(x) = s 2 (—wu — (—im7-0

Thus

{1 - (-lA)"}Un(x) = H1 - (-lE)"}Vn(x)-

We prove that, as n—>oo, (—iA)"U..(x) —>0 and (—iE)"V,.(x)—>0. It then follows that U(x) = &V(x). Let m be the least positive integer such that x + m > 1, and take n > m. Then

'6)“ If (_ é), (x + r)(x + r +n:)- --(x + r + n)

n, we may write

S(x) = (—r 2 (-)' % V'f(x). 7-0

'

and this series is absolutely convergent. It follows that

eS(x) = Z 0.0:). where

m) = (—)- 2 fl (4-31! V'f(x) = (-)",1!E"f(x) = $0 — 1 — x)(r — 2 — x)---(r — n — x).(2) Write

S(x)= 2 if“ L,

this series being absolutely convergent by d’Alembert’s ratio test. Then

108-} - xS(x) = - log (1 - 4}) — xS(x)

1,)

p

=2 21-1—4 —:(§E) 1.

Hence

(1 — iE){los-% — xS(x)} = % x+l l

-

Now 1 - Q-E = 4}(l — iA). It follows that, if n is a positive integer,

+ 1 {1 -(11rA)"}{log§ - xS(x)}= i 2 GA ,— 1 x As n—>oo,(i-A)"{log§- — xS(x)}->0; cf. the solution to Exercise 12.1 (6). Hence

"’

2 (’) 7-1

,- (r — 1)!

2'

1

(x + l)(x + 2)---(x + r)

is convergent, with sum {log 1 — xS(x)}.

ANSWERS AND SOLUTION NOTES

239

(3) Let Q(m) be the statement

c. = f: (—)r-I}sr... r- 1

Then Q(l) is true. Assume that Q(m) is true. On this hypothesis, since

"n+1 = (6 - mc)1r,,., 121(—)r-1%A;Wn. CDWM+1 = C‘fl'm + (a — me)

Using Theorem 7.1 on p. 113, with f = :9 — mc, m+1

m+1

Z (—)'-1}A:«..+1 = 21 (—)'-1§{(-$ — mc — remznm + rcA2'11rn}

1:1

"I

= cw.» + (a — me) 2 (—)'-1}Azw.., 1-1

on simplification, since AWL“. = 0. Thus Q(m) => Q(m + 1). Hence, by induction, Q(m) is true for m 2 1. Since A51, = 0 for r > m, if] < m s n

c. = 2 (-)"1%A51r,..

then

f-1

There are constants no, a1, . . . , a“ such that ”-1

P = 2 am.-. + a... -o

I- 1

Hence

cDP = 2 a.cD1r,._, -1

= 2 a. i (-)'-1§Arm-. = i (—)'-1§A;P. 3-0

7-1

751

(4) By definition,

{log (1 + A»; = 2 (—)r-1}Av 1.1: 7'1

whenever this series is convergent. Then

1_ "

(r- 1)!

-x{1°8(1+ A»; ‘ 2, (x H»: + 2)---(x_. + r) Also

—JcDl = x. x

Forn >1, write

_ '

(r — 1)!

3"") “ .216: + l)(x + 2)- T(x + r‘)



240

OPERATORS

where x is not a negative integer. Then, provided x s6 0, 1

n!

,

5"") = I: " x(x + l)---(x + n)’ see Exercise 7.1 (14). Provided x > 0, S,.(x)—> 1/x as n—>oo; see Exercise 12.1 (5), to which this result is equivalent. -

1

(5) —{(l + A)'1log(l + A)}x—+1

-

.-1

I-

.

.2, 3(_A)x_1 +

p

z: (_A) ‘

Mud

n a

Z Z 0, $9: + l)(x + 2)-- -(x + n) > x 2:_. n+

and

3—1

—-'-(x+l)(x+2)-. -(x+n+1)l>xa

(n + 1).

7.1

Ell“

3-15 p

'

ANSWERS AND .SOLUTION NOTES n+ Now

l-

l

l

n+119

8=flf=1E-§(i-Zl;)

241

lu+1‘l' —5

ZIP

2 1 ->00 as n —->oo,

Since

I-1

it follows that if x > 0 then S..(x) —> l/JI:a as n —>oo. The given series is not convergent if x s 0; if x < —l the rth term does not tendtoOasr—>oo, andif —l < x s Otherthtermisnotlessthan l

1

l

n10 + §+"-'+7)' (6) (i) m“. = 1 + is?» and [‘0 = 1. Hence u»f(x) '9 (1 + W)“’f(x)(ii) From 6). f(x) #4 m0 + M“. - ThsiT‘flx). (iii) 8,.f(x) a? e"””—e"‘””. _ (iv) From (iii), t(x) § (8,. — 55MB“ —fi1,-3-h5D‘)f(x). First approximation: t(x) é! 8,.f('x)._ Second approximation: t(x) $4 (8,, — fiSflflx). Third approximation: t(x) #5 {8,I — 31:02 - 3.1138: - “1583}flx). (v) From (iv), 8,,f(x) $ (1 — 31:83 + 3218i)'1hD/'(x). (vi) From (iii) and (v),

M00 as 41.0 — is: + 1418:)(1 + as: — flawflx).

(vii) From (i) and (v),

unwed e: (1 + m — 7313:)(1 + 31:8: — gimme). Note that [93,. = 41:52,. + 0. If Df = F then

I": fix) dx = M0) a: mu - 1458mm) *9 m0 — l—‘zh'D'MFm as required. Also I; F(x) dx = 8M(0) é! 2(1 + £85," — fia-Sfi)hF(0)

5? {2 + KEN" — E""’)a - MD‘MHO) as 41MB“ + 4 + E“)F(0) — gigh'D‘Fw), as required. 12.3

(1) u, = A + (B + n)2""1.

(2) 14.. = A(—2)” + 4}.2n-3(n° - 3n2 + 5n + B). (3) u, = A(—2)" + B + (9n‘ — 30n° + 211:2 + 410/108.

OPERATORS

242

(4) u,| = A + i(-—)"(na — 2n '+ B). (5) u,| = §.2"‘a(n° — 3n2 +' An + B). (6) u” = Asinimr + (B — *n)cosi~mr. (7) u,I = A cosjmr — Kn“ — 2n + B) sin hm.

(8) u,I = 2""{A cos &mr r01” — 2n + B) sinimr}. (9) u. = Acos in + B sin in + (n’ —, 2n — 7)coss}mr — 4(n —- l)sin-}mr.

(10) 14,. = A(—1)'I + B + Cn + 2"(9na — '84:: + 230)]27. (1]) u" = (—)""1(A + Bu + Cu2 + 40n° — 15n‘ + 2n‘)/120. (12) u“ = A + B cos §mr + Csin firm + n‘(2h — 3)(n — 3) sin” i-mr.

11-4. Write

pan) = 2 c,Dm-'. v-o

Whenever x e 9D" 'f,

D""e""f(x) = "2" (m 3' r)a'e"‘D'”""’f(x) n-o

v

=' Wm + a)""f(x). Hence, whenever x e QD'ff,

P(D)e“"f(x) = e" 2 cm. + comm) 'IO

= e"P(D + a)f(x). 12.5 (1) y = Ac” + B — 384x — 96xa — 16xa — 2x‘ — *x“. (2) y = e'-"(Ae"”‘ + Bed") + e'3*(x° + 6x2 + 21 x+ 36). ' (3)y = e“‘(A cos x + B sin x) + e"{(2x — 1) cos x + (2x — 2) sin x}/l6. (4)y = ie“{(xa + A) sinx + (x + B) cos x}. (5)y = AeT“.+-Be“ + Ce" + (10x +13)cos2x + (10x - 8) sin 2x. (6) y = e”(A cos 2x + B sin 2x) + e’“(25x3 — 30xa — 6x + C). (7)y = e‘“(A + Bx) + e"(P + Qx — 30x2 + 15x3 — 5x4 + x‘). (8)y = e“'{(4x’ + 18x” + 15x + A) cos x + (4x3 — 15x + B) sin x} + e"(Pcosx + Qsinx),

INDEX Abmssmu VALUE, 3 Antiderivative, 93

Functional operator, 75 variable, 20

Argument, 19

Associativity, 27 Averaging operator, 204 BINARY comosmon, 26 Binomial coefficient, 98, 131 Boolean algebra, 213 CARTESIAN PRODUCT, 10 Central differencing operator, 204 Closed interval, 5 Closure, 26 Co-domain, 16

Complement, 8 Complementary function, 187

GENERATING FUNCTION, 163 Genesis of a relation, 14 Geometric progression, 128 Group, 34 IDEMPOTENT LAW, 8

Identity element, 30 function, 64 Inclusion, 6 Independent variable, 20 Integral element, 50 Intersection, 6 Into (mapping), 19 Inverse element, 32

relation, ’16 _

sequence, 170

Complete primitive, 180 Commutative group, 34 Commutativity, 27 Component, 8

Composite function, 62, 70 Concourse, 5 Constant, 23 function, 59 operator, 76

Converse relation, 16 DEFINITION m: RBCURSION, 27 De Morgan's laws, 8 Dependent variable, 20

Isom'orphism, -150 LAPLACE TRANSFORM, 179 Latin square, 38 Left-hand derivative, 70 Leibnitz’s theorem, 112 Linear operator, 79 MANY-ro-MANY nELA'rION, l7

Many-to-one relation, 17 Mapping, l9 Mod, 72 NEGATIVE

OF

A

wucnon,

Differential, 67 Distributivity, 41 Division ring, 52 Domain, 16

of an element, 31, 43 Neutral element, 30 Non-commutative group, 35 Non-elementary sequence, 144 Normal operator, 88

ELEMENT, 3

Null function, 57 relation, 15

Derivative, 65—71

Elementary function, 136 sequence, 136 FEnMAr’s THEOREM, 48 Fibonacci number, 152 Field, 52 Full relation, 15 Function, 19

set, 5 ONE-TO-MANY RELATION, 17 One-to-one relation, 17 Onto (mapping), 19 Open interval, 5 sentence, 2 Operator, 73

243

60

244

INDEX SCALE or RELATION, 150

Order, 8 Ordered n-tuple, 9

Sequence, 119

Set, 3 variable, 3

PARAMFI'ER, 23 Partial derivative, 68 Permutation group, 36 Place-holder, 3 _ _ Polynomial function, 94,106 operator, 84 Primitive, 9_3_ _ Pronumeral, 3 _ Proper subset, 6 Proposition, 1 Propositional function, 23

Ssh. 72

Skew field, 52 Subgroup, 39 Subset, 6 Successive difl'erences Of a product, 112

Mom, 25

QUASI-FIELD, 52 ‘ Quasi-linear operator, 87 Quaternion, 53

Translation operator, 82, 91

Twe. l

43

set, 5

Unity, 42

VARIABLE, 3 _ Value, 19

Relation, l4

Residue, 44 Right-t derivative, 70 Ring, 48

--

UNION, 6 Unitary ring, 49 Unit element, 42 function, 59.. permutation, 37

RANGE, 19

Rational element, 53 function, 94, 96 sequence, 140 Reciprocal of an element, 32, 37, of a function, 61 Recurring power series, 164 sequence, 150 . Reduced equation, 170, 186 Related variables, 13

'

Sum-sequence, 125 Symmetric difference, 36 Symmetry group,l36

Zeno, 42 element, 42 function, 59

co

r—aAI-M

961' I—l/‘fi-I ..' A u-yu-h “Hi“

INDEX OF NOTATION

E 82 U!

E”, 83 Ah, VII: 84

t

E1, Efia, 91

U-

E1: E2, Ex, 91 mug;

N.” (332:2.

m

A,V,83

Acid-'2- Vex-ca. 91

A,, 92

D21, Du. 92 “H 0‘

L?

u

Q!

d,f, ay; dx, 67 D1, D2, 68, 92

ar;57 52b 62 9h,68 0,74 6,64 ("),98,131 r D, 66, 73 Do, D“, 67

D“, 92

A; 1, 94 8:» P4» 204 C, E, :9 2: 6

'u, n, 6

|.4

an, 26

a, 63, 91 =>, 17 4», 34 z, 195