DIFFERENTIAL METHOD.

392. The Differential Method is the process of finding any term of a regular series, or the sum of any number of terms, by means of the successive differences of the terms.

393. To find any term of a series by the differential method.

If we subtract each term of a series from the next succeeding term, we shall obtain a new series called the first order of differences. If we subtract each term of this new series from the succeeding term, we shall obtain a series called the second order of differences; and so on.

Let a, b, c, d, e, . . . . represent a regular series, the successive terms being formed according to any fixed law. We will write the given terms in a vertical column, and proceed by actual subtraction to form the several orders of differences, placing each order in a separate column, and each difference at the right of the subtrahend. The result is as follows:

The quantity which stands first in any column, though a polynomial, is called the first term of the order of differences which designates the column.

Let d1, d2, dз, d, etc., represent the first terms of the first, second, third, fourth, etc., orders of differences. Then we shall have

d1 = ba,

d2 = c2b+a,

dd3c + 3b — a,

de4d6c-4b+a, etc.

By transposition, we may obtain, after making the necessary substitutions,

a = a,

b = a + dr

c = a + 2d1 + do

d = a + 3d1 + 3d2+ d ̧,

e = a + 4d1 + 6d2 + 4d3 + d, etc.

These equations express the values of a, b, c, d, e, etc., in terms of a, d, d, da, d, etc. The coefficients in the second members are formed according to the binomial formula; and we observe that the coefficients of the second power of a binomial are found in the third equation, the coefficients of the third power in the fourth equation, and so on.

Hence, if we let T2+1 denote the (n + 1)th term of the given series,

then we shall have

a, b, c, d, e,

And by substituting n-1 for n, we shall obtain a formula for the nth term of the given series; thus,

Tr=a+(n-1) d1+

(n−1) (n−2) d2+

2

2

3

394. To find the sum of any number of terms of a series, by the differential method.

And denote the sum of n terms by S. We are to find the values of S in functions of

a, d, da, d, etc.

Let us assume the auxiliary series,

0, a, a + b, a + b + c, a + b + c + d, . . . .

(2).

It is obvious that the (n + 1)th term of this series is the same as the sum of n terms of the given series (1), and may be placed equal to S. Now let

d', d'2, d'з, d'1, etc.,

represent the first terms of the successive orders of differences in the auxiliary series (2). Then by formula (m), we have

But if we proceed to form the first, second, third, etc., orders of differences for the auxiliary series (2), we shall have

Hence, by substitution in equation (n), we have

395. The use of formulas (4) and (B) may be illustrated by the following examples:

1. Find the 12th term of the series, 1, 5, 15, 35, 70, 126, etc. We first form the successive orders of differences, as follows:

Substituting these values in (4), and reducing the terms, we ebtain

T12 = 1 +44 + 330 + 660 + 3301365, Ans.

12

The series is broken off at the fifth term, because the subsequent differences are all zero.

2. Sum the series 1, 3, 6, 10, 15, 21, etc., to n terms.

By forming the successive orders of differences, as in the last example, we shall obtain

a = 1, d1 = 2, d1 = 1, d2 = 0.

Whence, by formula (B),

all the terms after the third becoming zero. By performing the indicated operations, adding the results, and then factoring, we have

1. Find the 9th term of the series 1, 4, 8, 13, 19, etc.

Ans. 53.

2. Find the 15th term of the series 1, 4, 10, 20, 35, etc.

Ans. 680.

3. Find the 8th and 9th terms of the series 1, 6, 21, 56, 126, 251, 456, etc. Ans. 771 and 1231.

4. Find the 20th term of the series 1, 8, 27, 64, 125, etc.

Ans. 8000.

5. Find the nth term of the series 1, 3, 6, 10, 15, 21, etc.

6. Find the nth term of the series 1, 4, 10, 20, 35, etc.

7. Find the nth term of the series 1, 5, 15, 35, 70, 126, etc.

8. Sum the series 1, 3, 6, 10, 15, 21, etc., to 20 terms.

Ans. 1540.

9. Sum the series 1, 5, 14, 30, 55, 91, etc., to 12 terms.

Ans. 2366.

10. Sum the series 1, 4, 13, 37, 85, 166, etc., to 10 terms.

Ans. 2755.

11. Sum the series 1.2, 2.3, 3.4, 4.5, 5.6, etc., to n terms.

12. Sum the series 1.2.3, 2.3.4, 3.4.5, 4.5.6, etc., to n terms. n (n + 1) (n + 2) (n + 3) ̧

Ans.

4

13. Sum the series 12, 22, 32, 42, 52, etc., to n terms.

14. Sum the series 13, 23, 33, 43, 53, etc., to n terms.

Ans.

15. Sum the series 14, 24, 34, 44, 54, etc., to n terms.

(n2 + n)2

n5 n1 Ans. + + 5 2 3 30

16. Sum the series (m + 1), 2 (m + 2), 3 (m + 3), 4 (m + 4), etc., to n terms. n (n + 1) (1 + 2n + 3m) 6

Ans.

INTERPOLATION.

396. Interpolation is the process of introducing between two consecutive terms of a series, a term or terms which shall conform to the law of the series. It is of great use in the construction of mathematical tables, and in the calculations of Astronomy.

397. The interpolation of terms in a series is effected by the differential method. In any series, the value of a term which has n terms before it, is expressed by formula (m) (393), which is n (n−1) (n−2) 2. 3

T1+n = a + nd1 + n (n−1) d2 +

1

2

If in this formula we make n a fraction, then the resulting equation will give the value of a term intermediate between two of the given terms, and related to the others by the law of the series.

If n is less than unity, the intermediate term will lie between the first and second of the given terms; if n is greater than 1 and less than 2, the intermediate term will lie between the second and third of the given terms; and so on.