Let $f(x)$ and $g(x)$ be infinitely differentiable. Is it possible to simplify the following infinite sum?
$$S=fg+\frac{df}{dx}\frac{dg}{dx}+\frac{1}{2!}\frac{d^2f}{dx^2}\frac{d^2g}{dx^2}+\frac{1}{3!}\frac{d^3f}{dx^3}\frac{d^3g}{dx^3}+\cdots \;=\;\sum_{n=0}^{\infty}\frac{1}{n!}\frac{d^nf}{dx^n}\frac{d^ng}{dx^n} \tag{1}$$
The expression above bears some resemblance with the following sum:
$S_{f}=f+\frac{df}{dx}+\frac{1}{2!}\frac{d^2f}{dx^2}+\frac{1}{3!}\frac{d^3f}{dx^3}+... \tag{2}=f(x+1)$
I feel it may be possible to compute $(1)$ using the Fourier transform. Although it is quite evident that $S_f$ can be obtained from the Taylor series, using the Fourier transform
$\mathcal{F}\{S_f\}=\mathcal{F}\{f\}+\mathcal{F}\{df/dx\}+\frac{1}{2!}\mathcal{F}\{d^2f/dx^2\}+...=(1+i\omega+\frac{1}{2!}(i\omega)^2+...)\mathcal{F}\{f\}=e^{i\omega}\mathcal{F}\{f\}\tag{3}$
therefore, $S_f=\mathcal{F}^{-1}\{e^{i\omega}\mathcal{F}\{f\}\}=f(x+1)$.
If I use the same approach for $S$, assuming that $\mathcal{F}$ and $\mathcal{G}$ are the Fourier transform of $f(x)$ and $g(x)$ respectively, using the convolution theorem, I get:
$S=\mathcal{F}^{-1}\{\mathcal{F}*\mathcal{G}+i\omega\mathcal{F}*i\omega \mathcal{G}+\frac{1}{2!}(i\omega)^2\mathcal{F}*(i\omega)^2 \mathcal{G}+...\} \tag{4}$
where $*$ is the convolution operator. Any hint how can I proceed further?