Page 39 - Foundations Of Differential Calculus

22 1. On Finite Differences
and
1
Σ(x + nω)= (x +(n − 1) ω)(x + nω) .
2ω

33. If the function is the product of several factors, such as

y =(x +(n − 1) ω)(x + nω)(x +(n +1) ω) ,
then since
I
y =(x + nω)(x +(n +1) ω)(x +(n +2) ω) ,
we have

∆y =3ω (x + nω)(x +(n +1) ω) .
It follows that
1
Σ(x + nω)(x +(n +1) ω)= (x +(n − 1) ω)(x + nω)(x +(n +1) ω) .
3ω
In the same way we find that
Σ(x + nω)(x +(n +1) ω)(x +(n +2) ω)

1
= (x +(n − 1) ω)(x + nω)(x +(n +1) ω)(x +(n +2) ω) .
4ω
Hence the law for finding sums is quite clear if the difference is the product
of several factors of this kind. Although these differences are polynomials,
still this method of finding their sums seems to be easier than the previous
method.
34. From this method the way is now clear to finding the sums of fractions.
Let the given fraction be
1
y = .
x + nω
Since
I 1
y = ,
x +(n +1) ω
we have
1 1 −ω
∆y = − = ,
x +(n +1) ω x + nω (x + nω)(x +(n +1) ω)
and it follows that
1 1 1
Σ = − · .
(x + nω)(x +(n +1) ω) ω x + nω