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CHAPTER II

FINITE DIFFERENCES
DEFINITIONS AND FUNDAMENTAL FORMULAE

1. In most mathematical operations there are two classes of quantities. One class consists of those quantities which have the same value throughout the operation, and the other of quantities which may take different values. The first class are constants and the second are variables. If for example throughout a particular investigation y = 5, then wherever y occurs we may substitute the value 5 and y is said to be constant. If however y = x + 2, then to any particular value of x there corresponds a different value of y. In this example, if x may take up any value that we care to give it, then x is called an independent variable. On the other hand, y will vary according to the value that we assign to x and is said to be a function of x, or simply a dependent variable. A function of x is generally expressed in either of the following notations: f (x),

F (x), _q(x), ... or u_x, v_x, U_x, .... There may be more than one

independent variable on which the value of the function depends. Suppose that y = x sin a + z cos /3, where x, z, a, _/3 all vary: then x, z, a, _/3 are the independent variables and y may be written as f (x, z, a, /3) or ^ux,.0

A rational integral function is a simple form of function depending upon one variable.

y = a + bx + cx² + dx³ + ... + kxⁿ is a rational integral function of the nth degree in x, where a, b, c, d, ... k are constants and the indices are positive integers, n being the greatest.

It should be noted that for any one value of x in such a function there is one and only one value of y.
An alternative name for a rational integral function is a parabolic function. When represented graphically the curve
y=a+bx+cx2+... is said to be of the parabolic form.

TABLE OF DIFFERENCES 23
2. Consider the function y = u,, = 1 + x + x². It is quite easy to obtain the value of y corresponding to any value of x by substituting that value of x on the right-hand side of the equation. For example

x 0 I 2 3 4 5 ⁶7 ⁸Y ^I3 7 ¹3 ²¹3¹ 43 57 73

It will be noticed that for successive integral values of x in the above table the values of y follow a certain definite law. If from each value of y the previous value of y be subtracted, we obtain a new set of figures :

(a) 2 4 6 8 10 12 14 16

and if the subtraction be performed on these figures in the same way the new differences are

(13) 2 2 2 2 2 2 2

The sequence of 2^'S in ((3) is not a mere coincidence : it will be shown later that when y has the value supposed all the terms in (/3) have the same value, 2, however far the series extends.
This leads us to another method of obtaining values of y. Suppose that we write down the original table in a different form, and include in the table the two sets of figures (a) and (/3) thus :

^x(^u)    ([³)
8    73
We can now find any further value of y by extending the columns (a) and (/3). We must however work from (3) to (a) and then to

24    FINITE DIFFERENCES

y instead of from y to (a) and then to (S) as has already been done. For example, to obtain the value of y when x has the value 9, i.e. to obtain u₉, a new 2 must be inserted in the _(S)column: the new value in the (a) column will be 16 + 2 = 18, and the required value of y will be 73 + 18 = 91. To find u₁₀the process is continued. Any value of y corresponding to an integral value of x can be obtained in a similar manner.

The above is a particular instance of a far more general set of operations. We have used the simplest possible numerical values of x, namely the natural numbers, and we have evolved our example from a known quadratic function y = ux = 1 + x + x². As a general rule the form of the function is not known and the given values of x are not necessarily consecutive integers.

4. Now suppose that instead of numerical values of x differing by

unity we have the following consecutive values of x:

a+h,

a+2h,

a+3h,...,

where the values of

differ by a quantity

instead of by unity. Then if the function be still

the values of

corresponding to the above values of x will be

^Zi

a+h,

^Zt

a+2h,

^Zt

a+3h

In order to form a column similar to column (a) above we shall

have to write down

²¹

a+h —

^Zi

¹¹

a+2h —

^Zt

a+h,

^Z1

a+3h —

¹¹

a+2h

These are the

first differences

of the function

and are denoted by

Alta,

ua+h,

^Du

a+2h, ...

where 0 is not a quantity but a symbol representing an " operation."

Column

_(S)

being the differences of column (a), will be

(

^Z1

a+21t —

a+h) — (

a+h — ua),

(

a+3h —

a+2h) — (

^Z1

a+2h —

^Zt

a+h),

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

or, more shortly,

^Du

a+h —

^OZt

a+2h —

AZta+h r

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

DIFFERENCE TABLE 25 These are the second differences of u_x and are denoted by

^L2ua, ^02ua+h, A2ua+2h, • • •,
where, it must be emphasized, the symbol A²does not represent the square of a quantity but denotes the repetition of an operation. Similarly, third, fourth, ... nth differences are denoted by
^A3ua, A4ua, ... ^Onua^.

Before forming a difference table similar to that in para. 2, it is convenient to introduce alternative names for x and y in our equation y = u_x. Where our ultimate object is to obtain numerical values of x or y, the independent variable is often termed the argument, and the corresponding value of y the entry.

In a table of logarithms the number itself is the argument and the

logarithm the entry. The converse holds in a table of antilogarithms,

where the logarithm is the argument. Similarly in a table of sin

a, a is

the argument and the sine the entry, whereas

a is

the entry in a table of

sin_'

Our new difference table is therefore

FirstSecondThird Argument Entry differences differences differences

Aua

a + h

a+h

^A2u

Dua+h

_ua+2h

_+h

^Du

a+2h

^A3u

a+h

a + 3h

a+3h

^A2u

a+2h

^Du

a+sh

^A3u

a+2h

a + 4h

a+4h

^A2u

a+3h

^Du

a+4h

The first term in the table

)

is called the

leading term,

and the differences which stand at the head of the respective columns, namely,u

ua,

ua, ...,

are called the

leading differences.

Although we have expressed the terms in the difference table by the use of A symbols, it is quite easy to obtain any difference in terms of the function alone.

For example,

is the difference between

_a+h

and

ua,

A2ua+h —

^A2u

26 FINITE DIFFERENCES
Again, L1²1La is the difference between Dua_+h and Du_a, or
^A211.^=L1ua+h^—Dua,
and as Alt_a = ^ua+h — ^ua'
we have ,3ua = O^zu_a+h— c,² ua
(L1ua+2h — ^Lua+h) — ('~^ua+h — '⁶'^ua) = /Aua+2h — 2L1 ^ua+h + 0ua
(^ua+3h — ¹ⁱa+2h) — ²(^ua+2h — ^Ua+h) + (^Ua+h — ^ua) = ^ua+3h — 3^ua+2h + 3^ua+h — ^Ua^.8. It is a simple matter to construct a difference table from a given set of data.
Consider the following examples :
Example 1.
Construct a difference table from the following values, where y is a function of x:

x I	y I		o2y_.	o3y
2	8		12
		19		6
3	²7		18
		37		6
4	⁶4		24
		61		6
5	¹²5		3⁰
		91		6
6	216		3⁶
		127
7	343

Example 2.
Show that, in the following table of annuity-values, third differences are practically constant:

Argument x	Entry a_x	.la_x	D~ax	Wa_x
35	14'298
		— '¹54
3⁶	¹4'¹44		— •004
		— •158		+ •001
37	13^.986		— •003
		— •161		•000
3⁸	¹3'⁸²5		— •003
		— .164		+•001
39	13^.661		— •002
		— •166		+ -ow
4⁰	¹3'495		— •001
		— •167
41	13^.328

THE SYMBOL E 27

It will be observed that in Ex. 1 third differences are invariably the same. In the second example, however, third differences are not quite constant, although the error in assuming them to be so is very small.
The difference in the two examples lies in the fact that, while the first function is y = x³, the table of annuity-values from which the data in the second example have been taken does not conform to a mathelaw.

Example 3.
Assuming third differences constant, find the values of u₂ and u₃ from the data :
x    4    5    6    7    8
u_x    _'35•88    1^.71    2^.90    4^.51
Construct the difference table from the given values, and fill in the vacant spaces in the O³u_,, column with the constant third difference, thus :
x    u_r    dux    _O2ux^03ux
2    •Os
•11

3	+ •o6		•r8
		•29		•o6
4	+ '35		'²4
		'53		•o6
5	+ •88		'3⁰
		•83		•o6
6	+ 1^.71		^.36
		1^.19		•o6
7	+ 2^.90		•42
		I^.61
8	+ 4^.51

9. Now it has been stated above that a convenient method for expressing the difference between two successive values of a function _ua+hand u_ais by the symbol A prefixed to u_a, so that Au_a = u_a+h— u_a. It will be seen therefore that to find Au_a we perform two operations : we change u_ato Ua+h and subtract u_afrom it. The new function _ua+hresulting from the first of these operais denoted symbolically by Eu_a, and the double operation may be written

Au_a=Eua — u_a.

This gives Eu_a = u_a+ Au_a.
Eu_a may therefore otherwise be expressed as the sum of u_aand its first difference.

28 FINITE DIFFERENCES

Suppose that in either of the above relations the u_a which occurs in each of the terms be omitted. Then we can state that the two operations denoted by ^"E^" and "z " are connected by the symbolic equation

Ei+0.

It must be distinctly understood that we have not "factorized out^" u_a in the relation Eu_a = u_a + Du_a, and that we must relate the symbols to the functions on which they operate. If, therefore, we were using the equivalence 0 E — i, and we operated on the function sin x, it would be wrong to say that A sin x = E sin x — i. The correct statement is A sin x = E sin x — sin x. Since we are dealing with symbols we cannot increase or decrease either of them by unity, and on forming the algebraic or trigonometrical identity the function must be included in all three terms. In other words, in the identity E i + 0 the r is a symbol of operation just as are E and 0, and its meaning is that the function on which it operates is to be taken once without alteration.

In the same way as p2 denotes, when operating on a function, the difference of the difference of the function, i.e. the second difference, so E² denotes the operation of repeating E. That is to say

ux = E.

Eux

= Eu

_x+h

x+zn,

^'3u

x+3h,

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

and, generally,

x+ah

Care must be taken not to confuse the expression

with

(Eu

)

For example,

)

(x + zh)

4hx + 411

but

(Ex)

(x +

2hx + h

It is evident that the first difference of a function of the form cx, where c is a constant, is constant : for Ocx = c (x + h) — cx = ch, which is constant.

Let us consider the effect of differencing a function of x of higher degree than the first.

DIFFERENCES OF A RATIONAL INTEGRAL FUNCTION 29 Example 4.
Difference successively the functions (i) y = bx² and (ii) y = ax³.

(i)tXbx² = b (x + h)² — bx² = zbhx + bh²,

(zbhx

)

zbh

(x + h) +

—

zbhx —

zbh

and since

zbh

is constant all

further differences will be zero.

(ii)Dax³ = a (x + h)³ — ax³ = 3ahx² + 3ah²x + ah³,

= 6ah

6ah

and

6ah

further differences vanishing.

Collating the above results, we have that

the first differences of functions of the form

are constant,

,,,,

,>>>

^ax3

It follows therefore that third differences of ax³ + bx² + cx + d are constant, for before we reach the third differences the terms bx², cx and d will have been eliminated.

12. The above considerations lead us to the following important proposition:

be a rational integral function of the nth degree in x, then the nth difference of the function is constant.

Let the function be

ⁿ

ⁱ

+ ... +

then

Dux=a(x+h)'n+b(x+h)n-l+c(x+h)'i-2+...+s

—

ⁿ

— bx

^n—1

—

^n—2

—

...

—

anx

ⁿ

^n—2

^n—3

+ ... +

where

c', ... r'

are coefficients not involving

Similarly,

^A2u, = an (n — I) x^n—2h²+ b^"x^n—3+ C^"x^n—4+ ... + _Q'"_,

and so on.

Each time that we difference we lower the degree of the function by unity. After differencing n times no terms after the first will appear, and we shall be left with

ⁿ

=an(n —

I) (n—2)

(n — 3) ... 2 .

ⁿ

an!h

ⁿ

which is independent of x and is therefore constant.

As a corollary we may note that

ⁿ}

a property of a

the second the third

30 FINITE DIFFERENCES
rational integral function of the nth degree which is of value in the practical application of the work.

The converse proposition is of importance : if the (n+ I)th difference of a function is zero, the function is a rational integral function of the nth degree.

It should be remembered that we are dealing here with a particular form of function. Should the function be other than a rational integral function the nth difference will not vanish how-ever great n may be. Thus, we have

Example 5.

Find the nth difference of

= e

^x+h

—

= e

—

1),

—

₁

)

(ex+h — ex) = ex

(eh —

₁

Similarly,

A3ex =

(eh —

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

Ultimately

Anex = e

—

)

ⁿ

, which is still a function of

and is

therefore not constant.

Although it has been said that the symbols A and E are in no sense algebraic quantities, our definitions, namely that Aⁿ denotes the operation of differencing the function n times, and that Eⁿdenotes the operation of obtaining a new function when the arguis increased by n unit differences, enable us to apply to these symbols the ordinary algebraic laws. For example,

A (ux + uy) =

x+h

+ uy+h — u

—

x+h — u

+ u,,+h — uy,

which is

+ Au

This relation is exactly similar to the ordinary algebraic identity 3

3y.

The three simple algebraic laws are the laws of (i) distribution, (ii) commutation, (iii) indices.

(i)

A(ux+vx+wx+...)

(

x+h

x+h + •—) —

(ux + vx + wx + ...)

(

x+h —

x) +

(

x+h — vx) + (

x+h —

=Aux+Avx+Aw

+....

Similarly,

E(ux+vx+wx+...)=Eux+Evx+Ewx+....

APPLICATION OF THE LAW OF INDICES TO L AND E 31

(ii)The symbols 0 and E are commutative in their operation as regards constants. For if c be a constant,

Acur = cur+h — cur = ^c(us+h — ur) = cLur, and Ecu_s = ^cur+h = cEu_r .

(iii)The application of indices to the symbols 0 and E may be shown thus :

If m be a positive integer, then

represents the operation of differencing

times.

= (AAAA ...

times)

ⁿ

)

(AA AA ...

times)

(00Az ... m

times)

= (MMA0 ...

times)

= Qm+nu

x•

Similarly,

x+mh,

(

^Emu

^Enu

x+mh

x+mh+nh

Em+nu

15. In connection with the law of indices we must be careful to define A^m, Oⁿ, E^m, ... when m and n are not positive integers. So far, the symbols ^Omand E^m are intelligible only when we can actually perform the operations defined above and obtain the values of the new functions. We have not yet defined these symbols when the indices are negative. Consider for example the symbol 0-¹. Since we have assumed that the symbol 0 obeys the ordinary algebraic laws, ^A—1must be such that A (A-^lu_x) givesi.e. i.e. u_r.Let in be a positive integer. Then we define _i—muxas a function such that if it be operated on by Om the result will be Q^m-^mu_x, i.e. us. In the same way we have a meaning for E^—mus, namely, that E^m operating on E-^mu_r produces us. But if in be a positive integer, E^m operating on u_x__mhproduces _{ur_mh+mh,}i.e. us. Therefore the same result is obtained by operating with E^m on E-^mu_x as on u_x__mh. In other words just as E^mu_s gives _ur+mhso E-^mu_r gives ^ur_mh^.The symbols E and 0 may be manipulated in a manner similar to algebraic quantities provided that it is always remembered that they are operators and that they have no actual values. There are, however, two important points in which algebraic precedent cannot be safely followed. These are :

32 FINITE DIFFERENCES
(i) Operators are not commutative with regard to variables. E.g., L (uxvx) does not as a rule equal u_s Av_.,.

(z) It is fundamental in algebra that if a function vanishes, then one of its factors must vanish. It is not true that if the result of a series of operations on ux is equivalent to o.u_x (i.e. zero), then some one of the operations on ux must produce o . ux. For example, if x² = o, then x = o ; it does not necessarily follow, however, that if 02 o, then 0 = o.
In many problems it is convenient to use operators alone and to omit the functions on which they operate. Where this practice is followed the sign = (is equivalent to) should be adopted in place of = (equals). Thus, Eu_x = (I + 0) ux, but E (I + 0).
For further information on the difficulties connected with the use of operators the student may refer to J.S.S. vol. II, pp. 237 et seq. (S. H. Alison).

16. Proceeding from the definition of differencing, it has been shown that
^ux+h = ux + Otlx ,
^ux+2h = ^ux+h + ^Dux+h
⁼ux+Aux+0(ux+Aux)
= u_x + 20ux + ^02ux , ^ux+3h = ^ux+2h + ^Dux+2h
= ux + 20ux + 0²ux + 0 (ux + 20ux + 0²u_x) = ux + 3^0ux + 30²ux + 0³ux.

The coefficients of the various terms in these expansions are the coefficients of x in the expansions of (I + x), (I + x)², (I + x)³ by the binomial theorem. If we assume, for positive integral values of n, that the general relation between _ux+nhand ux and its differences follows the same law, we can prove the truth of the assumption by the method of mathematical induction.

Assume therefore that
^ux+nh = ux + n₁0ux + 1₂0²ux + ... + nr/^rux + ... + Oⁿu_x is true for the value n.
Then, since ^ux+(n+l) h = ^ux+nh + ^Dux+nh

OPERATORS A AND E
we have
ⁿx+(n+l)h = Us + n_iAux + n₂ X²ux + ... + n_rtX^ru_x + ... + Oⁿnx
+ 0 (ux + n₁0nx + n,A²ux + ... + nr A^rux + ... + ^iXnux) = ^ux + Aux (n_l + I) + ,X ^tux (n₂ + n_l) + .. .
+ Orux (nr + ⁿr-1) + ... + Qn+ln_x
But n_T + ⁿr-1 = (ⁿ+ ^I)r,
ⁿx+(n+1) h = Ux + (n + 1)₁ Au_x + (n + ¹)2 ^A2ux + .. .
+(n+ I)rO^rux+ ... +O^n}1nx,
which is of the same form in (n + I) as was the original expression in n.
Therefore if the assumption is true for n it is true for n + I. But the theorem holds when n = I, 2, 3.
Therefore it is true when n = 4, 5, ... and for all positive integral values.
Therefore, for positive integral values of n,

ⁿx+nh = Ux + n,Aux + n₂0²ux + n₃A³ux + ... + nrt^rnx + ... + Oⁿnx.

17. When the relation between the operator 0 and E was discussed it was stated that our definition of these operations enables us to apply the ordinary algebraic laws to these symbols. We may therefore use the equivalent relation

E-(I+0),

and if we operate on the function ux we shall have
ⁿx+nh = Eⁿux = (1 + L,)ⁿ ux
=(1+n₁0+n₂L1²+...+nrL^r+...+Aⁿ)ux.
If we introduce the fact that the symbols follow the algebraic distributive law, we may write
ⁿx+nh = nx + n₁0ux + n₂0²nx + ... + n4^ru_x + ... + Aⁿnx, . which is the relation proved above for positive integral values of n.

This result is true whatever the form of the function so long as n is a positive integer. If n be other than a positive integer we cannot adopt the binomial expansion without further investigation. For the purposes of this chapter it will be sufficient to assume that

the relation Eⁿ = (i + .,)ⁿ= I + n₁ + n2 A²+ n₃t³ + ... holds
without restriction. The question of the validity of the expansion will be discussed at a later stage (see Chap. III).

34 FINITE DIFFERENCES

18. We are now in a position to state that if n+ I consecutive values of a rational integral function of the nth degree are given, then, by the method of finite differences, we can obtain the actual function in the form

ux=uo+x_l.,ua+x₂0²uo+...+x_"L1"uo,

where x x(x—I)...(x—r+1)
T=
r ! .
or u_x=A+Bx₁+Cx₂+... +Kx_",
where the coefficients A, B, C, ... K are obtained by inspection of a table of differences.

Now if we are given n + I corresponding values of x and u_x it does not immediately follow that u₂ is a rational integral function of the nth degree.

For example, suppose that we have the following data:
X o I 2 3 4 5
ux I 4 9 16 25 36
Since six values are given there are the following possibilities:

(i)they may be actually given as values of the function (I + x)²;

of the second degree in x, and it may be required to find the function ;

(iii)

they may be given as values of a rational integral function

of the nth degree, where n is less than 6, and it may be required to

find the function ;

(iv)they may be given as values of a rational integral function without any information as to degree ;

The answer to (ii) and (iii) is obviously

= (1 +

The answer to (iv) is

= (I

x (x —

I) ...

(x — 5) A

where

is a rational integral function of x which does not become infinite at any of the points o,

I, 2,

3, 4, 5

6 x (x — I) ... (x — 5) A_x will then obviously vanish for these values.

The answer to (v) is the same as to (iv) except that A, need not

be a rational integral function.

It is of importance to realise that we can always find a value for

EXPANSION FOR A"ttx 35

A_x which will make the function u_x —(1 + x)² + A_xi_s agree with any additional value whatever. For example, if _u1.5=19'75the function u_x = (I + x)² + z⁹xs will agree with the given values and also with the additional value which has been inserted at the point x = 4'5.
Conversely we can say that whatever be the complete form of the function of which the six given values are samples, the value at any other point is the value of the function (I + x)² at that point with an error A_x x₆. Whether the value is a good approximation or not depends on the magnitude of A_x, and we may or may not have reason to suppose that A_x is so small that it can be neglected. It should be understood that we are not at liberty to say that (I + x)² gives an approximate value at the point in question unless we can give such a reason.

If instead of writing Eⁿ- (1 + A)ⁿ and expanding this by the binomial theorem, we write An - (E — 1)ⁿ and expand, a new series is obtained :

ⁿ

ux = (E —

I)"

= [En —

₁

^n-1

₂

^a-2

—

₃

^n-3

+ ... + (—

nrE

^n-r

L+...+(—

ⁿ

x+nh —

x+(n-1)

ⁿ

¹¹

x+(n-2) h — • • •

+ (— )r

ⁿ

x+(n-r)h + ... + (—

ⁿ

ux.

Just as the relation established in para.

enables us to obtain the value of

_ux+nh

in terms of

and its leading differences, so the above relation gives any required difference of the function

in terms of successive values of the function.

A few simple illustrations of the use of these two formulae are given below.

Example 6.

Find

₆

, given

= —

3, u

₁

₂

= 8,

₃

12;

third differences

being constant.

The leading differences are easily found to be

Duo

= 9;

A2110

= — 7;

₀

=9.

₆

⁶

₀

= (I +

6A + 15.,

+ 20:,

)

Ito

= up + 6A:

₀

+ 2o,

=—3+54—105+180=126.

Note.

There is no need to continue the expansion beyond third differ

as further differences are zero.

36 FINITE DIFFERENCES
Example 7.
Find u_, given u₄ = o, u_; = 3, us = 9; second differences being constant.
Here the initial term of the known series is u₄ , so that in order to find u₂ we must use the relation
u₂ = u4_₂ = E^—2u₄ = (I + 0)2 u₄ = (I — 2L + 3X2) 11₄, as far as second differences.

u₂= u₄ — 2,u₄ + 3^02u4
^=0—6+9⁼3,

since Du_o = 3 and / ²u₄= 3.
Example 8.
From the following values of ux, calculate A⁵uo:
U₆= 3, u₁ = 12, u₂ = 81, 11₃ = 200, U₄= 100, U₅= 8.
Since we require one value only of A⁵ u_x, we do not need to form a difference table, but may write at once
0⁵11₀ = (E — 1)5 ^uo
_ (E⁵ — 5E4 + I0E³ — I0E² + 5E — I) uo

= E⁵u, — 5^E4uo + IoE³uo — IoE²uo + 5Euo — uo

= U₅— 5^U4 + 10U₃ — I0U: + 5u₁ — Uo
= 755^.
Note that before we can find the fifth difference six terms of the series must be given.
21. Separation of symbols.

In obtaining the value of Eⁿu_x in terms of u_x and its differwe have used the symbolic relation E _ (1 + 0) and have expanded (I + A)' by the binomial theorem without introducing the function ux until the last stage. This method, in which in fact ux is omitted from both sides of the identity, is known as the method of separation of symbols, and enables many relations involving u_x and differences of u_x to be readily established.

Example 9.
Show that uo+u_l+u₂+...+un
=(n+ _I)Iuo+(n+1)₂Duo+(n+ 1)₃A²uo+...+Aⁿzio.

SEPARATION OF SYMBOLS

uo+111+u2+... +un
= uo + Euo + E²uo + ... + Eⁿuo =(I+E+E2+...+En)110
En+l — I
E —I ¹¹⁰En+1 — I

A ^uo
(1+z)' — I

uo
= _j[1 ± (n + I )₁J + (n + 1 ) ₂+( n+ I )₃0³... + O^n+lI ] u₀
_ [(n + I), + (n + 1)2 A + (n + 1)3 A2 + ... + AnJ ^u0
= (n + I), ^u0 + (n + 1)₂ Duo + (n + 1)₃ 0²uo + ... + Qn uo.
Example 10.
Prove by the method of separation of symbols that

u_x = ux_₁ + 1 u_{x_2}+ 0²11,_3 + 03ux_4 + ... + Q^n—lnx_n + Oⁿux _n
ux ^—,nux, = ^ux — ^:1nE—nux = {1 — CE.~ⁿ} ^21x= En_EnOⁿ_ux

I En — An
=_En~_E_z_X}u_x, sinceE — O-I,

_ E-ⁿ (Eⁿ-¹ + DEⁿ-² + A²E^n—3+ ... + An—1) u_x = (^E—1+ AE^—2 + A²E^—3 + ... + An-'E-n) u_x = 11_{r_1}+ ^Dux—2 + 0²ux_₃ + ... + ,^n_lux_n.
u_x= 11_{x_1}+ Atl_{x_2} + A2ux_3 + ... + ,^n—lux_n + Oⁿu_{x_} _n.
Since this is true for all values of n we have the convenient formulae u_x = u_{x_1}+ Lux__I (which is otherwise evident),
ux = 11,._1 + Dux_₂ + L ²ux—2 ,

u_x = 1[a._1 + Du_{x_2}+ 0²u_{x_3}+ ^03ux—3 ,
and so on.
Example 11.

Obtain a formula based on u_n similar to that given by the relation
E^xuo = (¹+ 0)^x uo.

38 FINITE DIFFERENCES

It will be found that this is an ordinary formula which could be obtained by using the values in the reverse order u_n , u_{n_l}, u_{tt_2}, ... u_fl . There is as much justification for using one order as the other. It should be noticed that the same numerical values appear in the table of differences, but that they are in the reverse order with a change of sign for the odd differences.

Example 12.
Find the value of

~x^m — ~0²x^m + 2.4 A'x'" — a.3.6 Q4xm + ... to m terms. 4 2.4.0
Since ^Qm+lxmand higher differences of x^m are zero, the sum of the series to m terms is the same as the sum to infinity.
Omitting the function x^m, and working on symbols alone, we have ¹-3_,y_ 1-3.5 A4+...=L.(I- o+1-3A2—03+...) 2.4 2.4.6 ₍₁2.4 2.4.6 \

_ A²- ³~ 0³+..₎
3•

0 (I +v)-i=0E-1.
The value of the given series is therefore

z E^-Ix^m= A (x — )^m= (x + 1)^m — (x — 1\m_, if the interval of differencing be taken as unity. 2l

Further examples of the application of the method of separation of symbols to the operators A and E and to other operators will be found in Chapters vii and VIII.

22. Factorial notation.

For convenience in working it is often useful to use a notation for the product of m factors of which the first is x and the successive factors decrease