Infinite Sum of Hermite Polynomials

Question

I'm trying to find a closed form for this sum: $$\sum_{n=0}^{\infty}\frac{1}{n!}H_{2n}(x)c^{2n}$$ where $H_m(x)$ are the Hermite polynomials according to the physicists' convention. I have found several formulas that come close to being the right form. For example, $$\sum_{n=0}^\infty\frac{1}{(n/2)!}H_n(x)w^n=(4w^2+2xw+1)(4w^2+1)^{-3/2}\exp{\left(\frac{4w^2x^2}{4w^2+1}\right)},$$ but that doesn't quite fit. Any suggestions as to how I might do this?

Answer 1

I think you're closer than you might expect. We have $$ H_n(-x)=(-1)^nH_n(x), $$ so adding the formula you've provided to itself with $x \mapsto -x$ gives $$ \sum_{n=0}^{\infty} \frac{1}{(n/2)!}H_n(x)w^n + \frac{1}{(n/2)!}H_n(-x)w^n = \sum_{n = 0}^{\infty} \frac{1}{(n/2)!}H_n(x)w^n(1 + (-1)^n), $$ which gives double the even terms you desire. Doing the same thing on the right should give the closed form you desire.