Simplifying: $\sum_{n=0}^\infty\sum _{i=0}^n \left(\frac{a}2\right)^{2n}\frac1{i!n!(n-i)!}\cdots$ 
I'm trying to simplify the following infinite sum:
$$\sum_{n=0}^\infty\sum_{i=0}^n \left(\frac{a}2\right)^{2n}\frac1{i!n! (n-i)!}\frac{\partial^{2n}f(x,y)}{\partial x^{2i}\, \partial y^{2(n-i)}},\ \ a,x,y\in\Bbb R$$
Where $f:\Bbb{R}\to\Bbb{R}$ is a sufficiently well-behaved function.


The only thing I've thought about is to use a binomial coefficient, i.e.
$$\sum_{n=0}^\infty\frac{\left(\frac{a}2\right)^{2n}}{(n!)^2}\sum_{i=0}^n \binom{n}i\frac{\partial^{2n}f(x,y)}{\partial x^{2i}\, \partial y^{2(n-i)}}$$
Maybe the substitution $m=n-i$ will make the sum more symmetric? However, I don"t know how to change the limits of summation in that case.
I'm interested to hear what general techniques people use when dealing with such infinite sums.

EDIT
I suspect that this sum can be written as an infinite sum over 'diagonal terms'. If we construct a matrix for the above elements, say $A_{i,j}$, then this sum should be possible to be written as $\sum_{k=0}^\infty$. such that $k=n+i$.
Alternatively, writing
$$\sum_{n=0}^\infty \frac{{\left(\frac{a}2\right)}^{2n}}{(n!)^2}x^n=I_0(a\sqrt{x})$$
with $I_0$ the modified Bessel function of first kind, maybe one can 'hide' the summation completely?

From 
$$\sum_{n=0}^\infty\frac{\left(\frac{a}2\right)^{2n}}{(n!)^2}\sum_{i=0}^n \binom{n}i\frac{\partial^{2n}f}{\partial x^{2i}\, \partial y^{2(n-i)}}$$
We can identify
$$\Delta^n(f)=\sum_{i=0}^n \binom{n}i\frac{\partial^{2n}f}{\partial x^{2i}\, \partial y^{2(n-i)}}$$
where $\Delta^n$ is the Laplacian composed $n$ times.
$$\sum_{n=0}^\infty\frac{\left(\frac{a}2\right)^{2n}}{(n!)^2}\Delta^n(f)=\left(\sum_{n=0}^\infty\frac{\left(\frac{a}2\right)^{2n}}{(n!)^2}\Delta^n\right)f$$
We can treat the term in brackets as the definition of a certain Differential operator through Taylor expansion.
Writing
$$\sum_{n=0}^\infty \frac{{\left(\frac{a}2\right)}^{2n}}{(n!)^2}x^n=I_0(a\sqrt{x})$$
with $I_0$ the modified Bessel function of first kind, we could write the following form:
$$\left(\sum_{n=0}^\infty\frac{\left(\frac{a}2\right)^{2n}}{(n!)^2}\Delta^n\right)f=I_0(a\sqrt{\Delta})(f)$$
