wikipedia has the following series expansion for hermite polynomials,
namely:
$$\exp \left\{xt-\frac{t^2}{2}\right\} = \sum_{n=0}^\infty {\mathit{He}}_n(x) \frac {t^n}{n!}.$$
Does anybody see how this can be shown. I tried rearranging a few terms, but did not get far.
Define the generating function by $$ F(x, t) = \sum_{n = 0}^{\infty} He_n(x) \frac{t^n}{n!}. $$ The backward shift operator of the Hermite polynomials is $\frac{d}{dx} He_n(x) = n He_{n-1}(x)$ and the three term recurrence relation is $$ He_{n+1}(x) = x He_n(x) - n He_{n-1}(x). $$ The generating function satisfies two first order differential equations. The first one is \begin{equation} \begin{split} \frac{d}{dx} F(x, t) = \sum_{n=1}^{\infty} He_{n-1}(x) \frac{t^n}{(n-1)!} = t F(x, t), \end{split} \end{equation} and the second one is \begin{equation} \begin{split} \frac{d}{dt} F(x, t) &= \sum_{n=1}^{\infty} He_n(x) \frac{t^{n-1}}{(n-1)!} = \sum_{n=0}^{\infty} He_{n+1}(x) \frac{t^n}{n!} \\ &= \sum_{n=0}^{\infty} (x He_n(x) - n He_{n-1}(x)) \frac{t^n}{n!} = x F(x, t) - t F(x, t), \end{split} \end{equation} where we assume that $He_n = 0$ if $n < 0$. The solution, using some initial values, to these two second order differential equations is $F(x, t) = \exp(xt - \frac{t^2}{2})$.
Best, Noud