Page 40 - Foundations Of Differential Calculus

1. On Finite Differences 23
Furthermore, let

1
y = .
(x + nω)(x +(n +1) ω)
Since

I 1
y = ,
(x +(n +1) ω)(x +(n +2) ω)
we have

−2ω
∆y = ,
(x + nω)(x +(n +1) ω)(x +(n +2) ω)
and it follows that
1
Σ
(x + nω)(x +(n +1) ω)(x +(n +2) ω)
1 1
= − · .
2ω (x + nω)(x +(n +1) ω)
In a similar way we have

1
Σ
(x + nω)(x +(n +1) ω)(x +(n +2) ω)(x +(n +3) ω)
1 1
= − · .
3ω (x + nω)(x +(n +1) ω)(x +(n +2) ω)

35. We should observe this method carefully, since sums of differences of
this kind cannot be found by the previous method. If the difference has a
numerator or the denominator has factors that do not form an arithmetic
progression, then the safest method for finding sums is to express the frac-
tion as the sum of partial fractions. Although we may not be able to find
the sum of an individual fraction, it may be possible to consider them in
pairs. We have only to see whether it may be possible to use the formula
1 1 1
Σ − Σ = .
x +(n +1) ω x + nω x + nω
Although neither of these sums is known, still their difference is known.

36. In these cases the problem is reduced to finding the partial fractions,
5
and this is treated at length in a previous book. In order that we may see
its usefulness for finding sums, we will consider some examples.

5 Introduction, Book I, Chapter II.