CHAPTER VII
FINITE DIFFERENCES. SUMMATION OR INTEGRATION
- Summation is the process of finding the sum of any number of terms of a given series. This can be accomplished either if the law of the series is known or if a sufficient number of terms is given to enable the law to be assumed. As will be shown in Chapter XX, if no mathematical law is apparent, methods can be applied by which the approximate sum of a series can be obtained.
Consider a function F(x) whose 1st difference is f (x). Then we have
- —F(0)=f(0)
—F(1)=f (1)
F(a) —F(a—1)=f(a—1) F(a+1)—F(a) =f(a)
F (n — 1) — F (n — 2) = f (n — 2) F(n) —F(n—1)=f(n— 1).
|
or |
Summing both sides, we obtain
F(n)—F(0)=f(0)+f(1)
F(a)—F(0)=f(0)+f(1) |
+...+f(n—1)
+...+f(a—1) |
|
or |
F(n)—F(a)=f(a)+f(a+1)+...+f(n—1) ...(1). |
It is clear, therefore, that the sum of any number of terms of a series of values of f (x) can be represented by the difference between two values of another function F (x) whose 1st difference is f (x). By analogy with the system of notation already adopted for ex-pressing orders of differences, the process of finding the function whose 1st difference is f (x) may be denoted by A—' f (x). It is
n-1
customary to express f (0) + f (1) + ... + f (n — 1) as E f (x), the
0
terms at which the summation is commenced and terminated (designated respectively the inferior and superior limits of sumbeing indicated in the manner shown.
46 FINITE DIFFERENCES
- The process of finding the value of F (x) is known as Finite Integration and F (x) is called the Finite Integral of f (x). Where the limits of summation are known we obtain by summation of f (x) the Definite Integral of f (x); if the limits of summation are not expressed we obtain merely the Indefinite Integral of f (x).
As stated above, in obtaining the indefinite integral of f (x) no point is specified at which the summation is to commence and, since an unknown number of terms of the series is included, it is necessary to include in the value of F (x) a constant term which is of unknown value.This constant vanishes in the case of definite integrals, since iff(x)=F(x)+c,n-1thenf (x) = [F (n) + c] — [F (0) + c]0= F (n) — F (0) = [0-'f (x)];.
- It is obviously always possible to find the first difference of any function, but it does not follow that every function can be integrated. The functions which can be integrated are limited in number and the process of integration rests largely on the ingenuity of the solver aided by such analogous forms as may be obtained by the formulas of finite differences.
Thus we haveA ax = (a — 1) ax, whence it is easily seen thatax =D ax a—1'and therefore, since the result of differencing aaxl is to give. ax, we havex%ax= a +0, a—1where c is the constant introduced by integration.The sum of the series ar + ar+' + ... + W.1n—' is at once obtained r+n-1 by finding the value of the definite integralax, which by 2rar+narIs equal to—, which agrees with the familiar result for
a—1 a—1the sum of a geometrical progression.
SUMMATION OR INTEGRATION 47
6. Similarly since by analogy
Ox (m) = 7nx(m–1)r 1x(m) = x(m+l)
+ c. m+1
- 7.The following table gives the values of some of the simpler integrals. They should be verified as an exercise by the student.
|
Function |
Indefinite Integral |
|
x |
x (x—1)+c |
|
2 |
|
ax |
axa—1+c |
|
x(') |
x(m+l) |
| m+l +0 |
|
x(-m) |
x(-mm-l)— (m—1)+o |
| |
(ax+ b) (ax -1 + b)(ax — m + b) |
|
(ax+b)(ax—1+b)...(ax—m+1+b) |
... |
| a(m+1)+c |
| |
(ax—1+b) (ax+b) ... (ax+m—1+b)+c |
|
(ax+b)(ax+1+b)...(ax+m— l+b) |
| a(m+1) |
| |
1 |
|
(ax+b) (ax+l+b)...(ax+m—1+b) |
—a(m—1)(ax+b)(ax+l+b)...(ax+m—2+b)+c |
| |
1 |
|
(ax+b)(ax— 1 +b)...(ax — m+ 1 +b) |
+c |
| —a (m— 1) (ax—1 +b) ... (ax—m+1+b) |
- 8.If the form of the function is unknown, a general formula for the sum of a series of values may be obtained as follows, since (x)=f(0)+xAf(0)+ x—(x 2!
— 1) .6,2f (0)x x—1x—2)+3! Osf(0)+.... Integrating both sides, we havex(x—1)x(x—1)(x—2) 0f(x) =c+ xf(0) + 2! Of (0) + 3!2f(0)+..., or, integrating between limits, we havenn(n—1)n(n—1)(n—2)f(x)=nf(0)+ Zi Af(0)+ 3) Ay(o) +...0
48 FINITE DIFFERENCES
or, more generally, a+n-1
f(x)=f(a)+f(a+1)+...+f(a+n—1)
a
=nf(a)+n(21 1)Af(a)+n(n— 31(n—2),2f(a)+...
(2).
It is instructive to obtain this result by the method of separation of symbols. For
f(a)=f(a),
f (a + 1) = (1 + A) f (a),
f(a+2)=(1+0)2f(a),
f(a+n— 1)=(1 +0)n 'f (a). Adding
a+n-1
f(a)={1+(1+A)+(1+0)2+...+(1+A)n—'}f(a)
a
(1+0)n—1 a) — (1+A)—1}(
=j1+n0+n(2, 1)02+...—1} f(a)
=~n+n(n—1)O+n(n—1)(n—2)02+...f(a) 2! 3! J
=nf(a)+n(2! 1)Af(a)+n(n— 3),(n—2) 2f(a)+....
9. An example will illustrate the use of the formula.
Example 1. Find the sum of the first n terms of the series whose
initial terms are 1, 8, 27, 64, 125.
We have
|
J (x) |
A |
A2 |
|
04 |
|
1 |
7 |
12 |
6 |
0 |
|
8 |
19 |
18 |
6 |
|
|
27 |
37 |
24 |
|
|
|
64 |
61 |
|
|
|
|
125 |
|
|
L |
|
SUMMATION OR INTEGRATION 49
Whence
n(n—1) n(n—1)(n—2)
of(x)=nx1+—21 x7+ 3~ x12
+n(n—1)(n—2)(n—3)x6 41
n(n3+2n'+n) n2(n+1)2 4 — 4
which agrees with the formula for the sum of the cubes of the natural numbers.
10. Where it is desired to integrate a function which is the product of two factors, the following device may often be utilised with advantage.
Let the function be y =uxvx.
Then Auxvx='tlx+lvx+1 uxvx
= ux+1 (vx+i — vx) + vx (ux+1 — uz) = ux+1 Avx + vx Aux.
Integrating both sides of the equation, we obtain uxvx=lux+,0vx+lvx Aux
or lvxAux= uxvx — lux+,Avx (3).
Thus, if the original function can be put in the form vx Aux, its integral can be made to depend upon that of ux+l0vx, and, if the latter is in a form which can be readily integrated, the value of an apparently intractable integral may often be obtained in this way.
Example 2. To find the value of Ixax.
ax
Since lax = a , we may write -1
Lax
Ixax = Ix - a—1'
Using the above formula (3) we get
x Dax = xax — ax+1 ~x
a—1 a—1 a—1
xax ax+1
— ! , since Ax = 1,
a—1 a—1
_ xax ax+1 a—1 (a—1)2+c.
E. T. B. Z. 4
50 FINITE DIFFERENCES
11. Sometimes it may be necessary to apply the formula more than once in order to reduce the integral by stages to a standard form. The process is illustrated in the following example.
Example 3. To find the value of 12zx3. Remembering that 02z = 2z, we may write
~2zx3 =x3022
= x3 2z -12z-" Ax3
=x32z z+'(3x'+3x+1).
It will be observed that in applying foriiiula (3), the degree of x, in the terms within the integral, has been reduced by unity. Proceeding as before we obtain
52z13x'22—s(3x'+3x+1)A2z+1
| = x322 — [23'+' (3x' + 3x + 1) — £2z+'0 (3x' |
+ 3x + 1)] |
| =2z(x3—6x'—6x—2)+ 2z+'(6x+6) |
|
| =22(x3—6x'—6x—2)+1 (6x + 6) A2z+2 |
|
| =2z (xt— 6x'— 6x—2)+ [2z+' (6x+6)— |
2z+3A (6x + 6)] |
| =2z(x3—6x'+18x+22)—1'22+3x6 |
|
=2z(x'—6x'+18x+22)—6 x 20+3+c =2z(x'—6x'+18x—26)+c.
The above process is analogous to that of "Integration by Parts," which is dealt with in the Integral Calculus, Chapter XVIII, S 6.