Variation on Hermite Generating Function

Question

I am having trouble using the Hermite generating function to determine $e^{t^2}\cos(2xt)$. I know the generating function is $e^{2tx-t^2}=\sum_{n=0}^\infty (-1)^n \frac{t^n}{n!}H_n(x)$ but can't seem to get anywhere. Can anyone help direct me on this?

Answer 1

One may replace $t$ by $it$, getting $$ e^{2itx+t^2}=\sum_{n=0}^\infty (-1)^ni^n\frac{t^n}{n!}H_n(x) $$ then one may take the real part: $$ e^{t^2}\cos2xt=\sum_{n=0}^\infty \frac{t^{2n}}{(2n)!}H_{2n}(x) $$ where we have assumed that $t,x$ are real numbers.