I'm looking for a closed form for this, not sure if it exists:

Question

kainui · Answer

$$f(x,a)=\sum_{n=0}^x \frac{(-a)^n}{n!}$$

kainui · Answer

Basically the way I was thinking of handling it was to just write it as $$f(x,a)=\sum_{n=0}^x \frac{(-a)^n}{n!}=e^{-a}-\sum_{n=x+1}^\infty \frac{(-a)^n}{n!}$$
and then say for large x, the sum can be thrown away but I don't like that very much.

imqwerty · Answer

closed form?

kainui · Answer

Yeah, so if I plug in a value of x, I don't want to have to add up all the terms form n=0 to n=x haha.

ganeshie8 · Answer

you're looking for a more closed form of
$$\Gamma(x,a)=(x-1)!e^{-a}\sum\limits_{n=0}^{x-1}\dfrac{a^k}{n!}$$

?

ganeshie8 · Answer

$$\Gamma(a,x) = \int\limits_{x}^{\infty} t^{a-1}e^{-t}\,dt$$

ganeshie8 · Answer

https://en.wikipedia.org/wiki/Incomplete_gamma_function

ganeshie8 · Answer

not sure how that is any better than the partial sum that you have...

ikram002p · Answer

what do u mean by from n=0 to x ? what is x ?

kainui · Answer

I don't think that's what I'm looking for, I guess I should clarify that I am mostly interested in x being an integer greater than 0.

ikram002p · Answer

and a can be anything ?

ikram002p · Answer

are u messing for Fourier ?

kainui · Answer

$$f(x,a)=\sum_{n=0}^x \frac{(-a)^n}{n!}$$

So for example:

$$f(3,a)=\sum_{n=0}^3 \frac{(-a)^n}{n!} = 1 - a + \frac{a^2}{2}-\frac{a^3}{6}$$

a can be anything, real or complex, but I think it's best to keep it to positive integer for now but it really doesn't matter.

kainui · Answer

Specifically I'm looking at solving this recurrence relation:

$$f(x+1)+(x+1)f(x) =a^x$$

kainui · Answer

Oh well, I already found the solution and it's not too difficult to plug it in and see that it does indeed satisfy the equation.

$$f(x)=(-1)^x x! \left(f(0) + \sum_{n=0}^x \frac{(-a)^n}{n!} \right)$$

ganeshie8 · Answer

$f(x) = a^{x-1} - xf(x-1) \~\= a^{x-1} - x(a^{x-2}-(x-1)f(x-2))\~\
=a^{x-1}-xa^{x-2}+x(x-1)f(x-2)\~\
 = \cdots $

ganeshie8 · Answer

Okay I think I see how you got that weird sum :)

kainui · Answer

Yeah I discovered that for recurrence relations of this form with p(x), q(x) and f(0) given:

$$f(x+1)+p(x)f(x)=q(x)$$

Has solution:

$$f(x)=(-1)^x\prod_{n<x} p(n) \left( f(0)+\sum_{n=0}^{x-1}\frac{(-1)^{n+1}}{\prod_{k \le n} p(k)}q(n) \right)$$

kainui · Answer

Anyways I'm just sorta trying to test it out on random stuff to make sure it works, kinda weird and fun haha. The problem is I don't know many closed form functions that look like these, which is why I picked specifically $p(x)=x+1$ and $q(x)=a^x$ because it's what we started with here