# How to solve $\Sigma f(k,n)=g(n)$?

Let $x$ be a real number. Consider $\Delta f(x)=f(x)-f(x-1)$ and $\Sigma\Delta f(x)=\Delta\Sigma f(x) = f(x)$. The operator $\Delta$ and $\Sigma$ are eachothers inverse.

Example : $\Sigma$ $x=(x^2+x)/2$

It is well known that for every entire function $a(x)$ , both $\Delta$ $a(x)$ and $\Sigma$ $a(x)$ are defined.

I would like to extend this idea with the problem below.

How to solve $\Sigma_{k=1}^{x} f(k,x)=g(x)$ when $g(x)$ (entire) is given and we want to find a meromorphic $f(x,k)$?

A specific example is this : $\Sigma_{k=1}^{x} \binom{k}{x}=2^x$ where $2^x$ was given. I write $\Delta_{1}^{x}$ $2^x= \binom{k}{x}$. The binomium is evaluated by using the Gamma function and hence we have our meromorphic solution.

I am somewhat familiar with trinomials (to solve the case $g(x) = 3^x$) and such but I think there must be a nicer way to solve this.

( with "this" I mean expanding $g(x)$ as $a 2^x + b 3^x + c 4^x ...$ and then using multinomials )

Also I wonder about the uniqueness of a solution.

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 In retrospect , particularly intresting would be a nontrivial solution $f$ for $g(x) =$ $Constant$. With nontrivial I mean e.g. exclude $f=0$ if $g=0$. – mick Feb 12 at 22:39