Even and odd Hermite polynomials

Question

I have to derive the following relation: $$H_{2n}(0)= (-1)^{n} \frac{(2n)! } {n! } $$ where $H_{n} (x) $ are Hermite polynomials. What I've done so far: 1) tried using this formula: : $H_{n}(x)= (-1)^{n} e^{x^2} \frac{d^n} {dx^n} (e^{-x^2} )$ but i couldn't find a relation for n-th derivation of $e^{-x^2}$. I even tried using Taylor series expansion for $e^{-x^2} $ but that didn't get me far. 2)tried using this recurrence formula for Hermite polynomials: $H_{n+1} = 2x H_n - 2nH_{n-1}$ knowing $H_0(x)=1$ and $H_1(x)=2x$. I solved the problem this way but I don't think it's an adequate way to do it. Does anyone have any ideas? Thanks!

Answer 1

Rodrigues' formula gives $$ H_{2n}(x) = e^{x^2}\frac{d^{2n}}{dx^{2n}}e^{-x^2}=\sum_{k\geq 0}\frac{x^{2k}}{k!}\cdot\sum_{k\geq n}\frac{x^{2k-2n}}{k!}(2n)!\binom{2k}{2n}(-1)^k $$ and by evaluating the RHS at $x=0$, the only relevant terms are the term associated to $k=0$ in the former series and the term associated to $k=n$ in the latter series, leading to $$ H_{2n}(0) = 1\cdot \frac{1}{n!}(2n)!\binom{2n}{2n}(-1)^n = (-1)^n\frac{(2n)!}{n!}.$$