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.
