DIFFERENTIAL METHOD — Suppl. 3rd (1801)

DIFFERENTIAL METHOD, is the art of working with the differences of quantities. By this method any term of a series may be found from the several orders of differences being given; or vice versa, any difference may be found from having the terms of the series given: it likewise shews how to find the sum of such a series. And it gives rules to find by interpolation any intermediate term, which is not expressed in the series, by having its place or position given.

When any series of quantities is proposed, take the first term from the second, the second from the third, the third from the fourth, &c. then all these remainders make a new series, called the first order of differences. In this new series take the first term from the second, the second from the third, the third from the fourth, &c. as before; and these remainders make another series, called the second order of differences. In like manner, in this series, take the first term from the second, the second from the third, &c.; and these will make a series called the third order of differences; and after this manner you may proceed as far as you will. Thus in the following proposition A, b, c, d, e, &c. is the series; B, B¹, B², B³, &c. the first order of differences; C, C¹, C², &c. the second order of differences; D, D¹, &c. the third order; E, &c. the fourth order, and so on. But the first terms of these several orders of differences, as B, C, D, E, &c. are those that are principally made use of in calculations by this method.

PROP. I. If there be any series, A, b, c, d, e, &c. and if there be taken the first differences B, B¹, B², &c. the second differences C, C¹, C², &c. the third differences D, D¹, D², &c. and so on.

Then if T stand for the first term of the nth differences, $\pm T = A - nb + n \times \frac{n-1}{2} c - n \times \frac{n-1}{2} d + n \times \frac{n-1}{3} \times \frac{n-2}{2} \times \frac{n-3}{4} e - \dots$ that is, + T, when n is even, and - T when n is odd.

The several orders of differences being taken as before directed, will stand thus. Then,

\begin{array}{ccccccc} A & B & C & D & E & & \\ b & B^1 & C^1 & D^1 & & & \\ c & B^2 & C^2 & D^2 & & & \\ d & B^3 & C^3 & & & & \\ e & B^4 & & & & & \end{array}

series A, b, c, d, e, &c.
1st diff. $b-A$ , $c-b$ , $d-c$ , $e-d$ , &c.
2d diff. $c-2b+A$ , $d-2c+b$ , $e-2d+c$ , &c.
3d diff. $d-3c+3b-A$ , $e-3d+3c-b$ , &c.
4th diff. $e-4d+6c-4b+A$ , &c.

That is, $B=b-A$ , $C=c-2b+A$ , $D=d-3c+3b-A$ , $E=e-4d+6c-4b+A$ , &c. or $-B=A-b$ , $-C=A-2b+c$ , $-D=A-3b+3c-d$ , $-E=A-4b+6c-4d+e$ , &c. where, putting T successively equal to B, C, D, E, &c. and n = 1, 2, 3, 4, &c. the prop. will be evident.

Cor. Hence
A = A, the first term.
B = $-A+b$ , the first difference.
C = $-A+2b-c$ , the 2d difference.
D = $-A+3b-3c+d$ , the 3d difference.
E = $-A+4b-6c+4d-e$ , the 4th difference.
F = $-A+5b-10c+10d-5e+f$ , the 5th difference, &c.

PROP. II. If A, b, c, d, e, &c. be any series, and there be taken B, C, D, E, &c. the first of the several orders of differences;

Then, the nth term of the series will be $= A + \frac{n-1}{1} B + \frac{n-1}{1} \times \frac{n-2}{2} C + \frac{n-1}{1} \times \frac{n-2}{2} \times \frac{n-3}{3} D + \frac{n-1}{1} \times \frac{n-2}{2} \times \frac{n-3}{3} \times \frac{n-4}{4} E + \dots$

For from the equations in the last Prop. viz. $B=b-A$ , $C=c-2b+A$ , &c. we have, by transposing, $b=A+B$ , $c=A+2b+C=-A+2A+2B+C$ (expunging b); that is, $c=A+2B+C$ , $d=A-3b+3c+d=A-3A-3B+3A+6B+3C+D$ (expunging b and c); that is, $d=A+3B+3C+D$ . Also $e=-A+4b-6c+4d+E$ (expunging b, c, d) $-A+4A+4B-6A-12B-6C+4A+12B+12C+4D+E$ ; that is, $e=A+4B+6C+4D+E$ , &c.

Then putting A, b, c, d, &c. for the nth term, and n successively = 1, 2, 3, 4, &c. the series will be evident.

Cor. 1. If d, d¹, d², &c. be the first of the first, second, third order, &c. of differences; then

The nth term of the series A, b, c, d, &c. will be $= A + \frac{n-1}{1} d + \frac{n-1}{1} \times \frac{n-2}{2} d^1 + \frac{n-1}{1} \times \frac{n-2}{2} \times \frac{n-3}{3} d^2 + \frac{n-1}{1} \times \frac{n-2}{2} \times \frac{n-3}{3} \times \frac{n-4}{4} d^3 + \dots$

For $B=d$ , $C=d^1$ , $D=d^2$ , &c. And the coefficients are the uncles of the n-1th power.

Cor. 2. Hence also it follows, that any term of a given series may be accurately determined, if the differences of any order happen at last to be equal.

Cor. 3. Hence
A = A, the first term.
b = A+B, the 2d term.
c = A+2B+C, the 3d term.
d = A+3B+3C+D, the 4th term.
e = A+4B+6C+4D+E, the 5th term.
f = A+5B+10C+10D+5E+F, the 6th term.
g = A+6B+15C+20D+15E+6F+G, the 7th term, &c.

PROP. 3. If a, b, c, d, e, &c. be any series, and d¹, d², &c. the first of the several orders of differences; then

The sum of n terms of the series is $= na + n \times \frac{n-1}{2} d + n \times \frac{n-1}{2} \times \frac{n-2}{3} d^1 + n \times \frac{n-1}{2} \times \frac{n-2}{3} \times \frac{n-3}{4} d^2 + \dots$

Digges II
Diophanus.

$\times \frac{n-3}{4} d^n + n \times \frac{n-1}{2} \times \frac{n-2}{3} \times \frac{n-3}{4} \times \frac{n-4}{5} d^{n-1} + \dots$

For in the series of quantities,
$o, a, a+b, a+b+c, a+b+c+d, \dots$
1st diff. are $a, b, c, d, \dots$
2d diff. $d, d^2, d^3, \dots$
3d diff. $d^2, d^3, \dots$
4th diff. $d^3, \dots$

Therefore (by Cor. 1. Prop. II.) the $n+1$ th term of the series, $o, a, a+b, a+b+c, a+b+c+d, \dots$ or the $n$ th term of the series, $a, a+b, a+b+c, a+b+c+d, \dots$ is $= o + na + n \times \frac{n-1}{2} d + n \times \frac{n-1}{2} \times \frac{n-2}{3} d^2 + \dots$ . But the $n$ th term of the series $a, a+b, a+b+c, \dots$ is the sum of $n$ terms of the series, $a, b, c, d, \dots$ and therefore equal to $na + n \times \frac{n-1}{2} d + n \times \frac{n-1}{2} \times \frac{n-2}{3} d^2 + \dots$ .

For a fuller account of this method, and its application to curves, we refer the reader to Emerson's works, from which these three propositions are taken.