I started with the the following four-variable function, $f(s,x,y,u)$, expressed as a summation involving the product of Hermite polynomials.
$f(s,x,y,u)=\sum_{n=0}^{\infty}\frac{H_{n}(x)H_{n}(y)}{(n-s)!}\big(\frac{u}{2}\big)^n \tag{1}$
I rewrote $(1)$ as follows.
$=u^s\frac{\partial^s }{\partial u^s}\sum_{n=0}^{\infty}\frac{H_{n}(x)H_{n}(y)}{n!}\big(\frac{u}{2}\big)^n \tag{2}$
Using the closed-form expression for the sum: $\sum_{n=0}^{\infty}\frac{H_{n}(x)H_{n}(y)}{n!}\big(\frac{u}{2}\big)^n$ given by the Mehler's formula, I wrote $(2)$ as follows.
$=u^s \frac{\partial^s }{\partial u^s}\exp{\big(-\frac{1}{2}\log{(1-u^2)}+\frac{2uxy}{1+u}-\frac{u^2(x-y)^2}{1-u^2}\big)} \tag{3}$
Now, using the Faà di Bruno's formula, I expressed the $s^{th}$ partial derivative with respect to $u$ in $(3)$ in terms of a summation involving incomplete Bell polynomials, $B_{s,k}$, as follows.
$=u^s e^{g(x,y,u)}\sum_{k=1}^{s}B_{s,k}\big(\frac{\partial}{\partial u}g(x,y,u),\frac{\partial^2}{\partial u^2}g(x,y,u),\dots,\frac{\partial^{s-k+1}}{\partial u^{s-k+1}}g(x,y,u)\big) \tag{4}$
where, $g(x,y,u)=-\frac{1}{2}\log{(1-u^2)}+\frac{2uxy}{1+u}-\frac{u^2(x-y)^2}{1-u^2}$.
I'm wondering how to further simplify the sum in $(4)$ to get a much simpler closed-form expression. I need to use the function $f(s,x,y,u)$ to perform further computations and hence I'm looking to get a simpler expression.
Alternatively, is there a more efficient way to compute the sum in $(1)$ compared to the approach I used?
Any ideas on this would be much appreciated. Thanks!
