Show series representation of orthogonal polynomials 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. 
 A: 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
