Page 38 - Foundations Of Differential Calculus

1. On Finite Differences 21
30. It is clear that if n is not a positive integer, then the expression for the
sum is going to be an infinite series, nor can it be expressed in finite form.
Furthermore, here we should note that not all powers of x with exponents
n−2 n−4 n−6
less than n occur. All of the terms x , x , x ,... are lacking, that
n
is, they have coefficients equal to zero, although the second term, x ,does
1
not follow this law, since it has coefficient − .If n is negative or a fraction,
2
then this sum can be expressed as an infinite series with the sole exception
that n cannot be −1, since in that case the term
n+1
x
(n +1) ω
would be infinite, since n + 1 = 0. Hence, if n = −2, then
1 1 1 1 ω 1 ω 3 1 ω 5 3 ω 7
Σ = C − − − · + · − · + ·
x 2 ωx 2x 2 2 3x 3 6 5x 5 6 7x 7 10 9x 9

5 ω 9 691 ω 11 35 ω 13 3617 ω 15
− · + · − · + · − ··· .
6 11x 11 210 13x 13 2 15x 15 30 17x 17

31. If a given difference is any power of x, then its sum, or the function
from which it came, can be given. However, if the given difference is of
some other form, so that it cannot be expressed in parts that are powers of
x, then the sum may be very difficult, and frequently impossible, to find,
unless by chance it is clear that it came from some function. For this reason
it is useful to investigate the difference of many functions and carefully to
note them, so that when this difference is given, its sum or the function
from which it came can be immediately given. In the meantime, the method
of infinite series will supply many rules whose use will marvelously aid in
finding sums.
32. Frequently, it is easier to find the sum if the given difference can be
expressed as a product of linear factors that form an arithmetic progression
whose difference is ω. Suppose the given function is (x + ω)(x +2ω). Since
when we substitute x + ω for x we obtain (x +2ω)(x +3ω), then the dif-
ference will be 2ω (x +2ω). Hence, going backwards, if the given difference
is 2ω (x +2ω), then its sum is (x + ω)(x +2ω). From this it follows that
1
Σ(x +2ω)= (x + ω)(x +2ω) .
2ω
Similarly, if the given function is (x + nω)(x +(n +1) ω), since its differ-
ence is 2ω (x +(n +1) ω), we have
1
Σ(x +(n +1) ω)= (x + nω)(x +(n +1) ω) ,
2ω