Does an inverse exist for discrete summation?

Question

Is an inverse function defined for discrete summation (sigma)(finite or infinite), in a way that the differential exists as an inverse for integration?
For example if f(x) = sigma(n from 0 to infinity) ( x^n)
What would be its inverse?

Answer 1

Well, [n=0, ∞]∑x^n = 1/(1-x), so if f(x)=[n=0, ∞]∑x^n, then f^(-1)(x) = 1 - 1/x

Of course, I doubt you actually mean to ask whether that particular function has an inverse (since the free variable doesn't even appear in the limits of summation), but rather, whether there is a way to reverse the operation of summation in a manner analogous to how differentiation reverses integration. Indeed there is. Given any function F(x) defined on the integers, there exists a function f(x) such that F(b)-F(a) = [x=a, b-1]∑f(x) (provided b≥a). This function is simply f(x) = F(x+1) - f(x). (This is kind of obvious when you think about it).

Answer 2

[n=0,∞]∑[(x^n)/(n!)]= exp(x) so [n=2,∞]∑[(x^n)/(n!)] = exp(x) -1-x then x can be calcute by numerical methodes such as newton-raphson.