"Expand $x^{2m}$ in a series of hermite polyniomials". We start with the definition of an orthogonal expansion:
$$f(x) = \sum_{0}^{\infty}\frac{<f,H_n>_wH_n}{||H_n||^2}$$
with the weight function $w = e^{-x^2}$
Now, since $x^{2m}$ is of degree $2m$, we need the hermite polyniomials of degree $2m$. $$<f,H_n>_w = <x^{2m},H_{2m}>_w = \int_{-\infty}^{\infty}x^{2m}H_{2m}e^{-x^2}dx$$ $$=\int_{-\infty}^{\infty}x^{2m}\frac{d^{2m}}{dx^{2m}} e^{-x^2}dx$$ and after $2m$ integration by parts $$=\int_{-\infty}^{\infty}\frac{d^{2m}}{dx^{2m}}x^{2m}e^{-x^2}dx = (2m)!\int_{-\infty}^{\infty}e^{-x^2}dx = (2m)!\sqrt{\pi}$$
Calculating the norm, by the same rule as above and as $\frac{d^{2m}}{dx^{2m}} H_{2m} = 2^{2m}(2m)!$
$$||H_{2m}||^2 = <H_{2m},H_{2m}> = 2^{2m}\sqrt{\pi}$$
Putting it all together we get
$$x^{2m} = \sum_{k=0}^{m}\frac{H_{2k}}{2^{2m}}$$
which is very much wrong. Problem is, I have no idea where the problem arises. I have tried many different approaches, and there is obviously something I am missing. Any hints and ideas on where to look for the problem would be appreciated!
edit: Adding the answer:
$$\frac{(2m)!}{2^{2m}}\sum_{k=0}^{m}\frac{H_{2k}}{(2k)!(m-k)!}$$
$x^{2m}$ is an even function, hence the expansion just depends on Hermite polynomials with even index. We have: $$ H_{2k}(x) = e^{x^2}\cdot\frac{d^{2k}}{dx^{2k}}e^{-x^2} \tag{1} $$ together with: $$ \int_{-\infty}^{+\infty} H_{2k}(x)^2 e^{-x^2}\,dx = 2^{2k}(2k)!\sqrt{\pi}.\tag{2} $$ We have: $$\langle x^{2m},H_{2k}(x)\rangle = \int_{-\infty}^{+\infty} x^{2m} H_{2k}(x) e^{-x^2}\,dx=\int_{-\infty}^{+\infty}x^{2m}\frac{d^{2k}}{dx^{2k}}e^{-x^2}\,dx \tag{3}$$ hence by applying integration by parts: $$\begin{eqnarray*}\langle x^{2m},H_{2k}(x)\rangle &=& \frac{(2m)!}{(2m-2k)!}\int_{-\infty}^{+\infty}x^{2m-2k}e^{-x^2}\,dx\\&=&\frac{(2m)!}{(2m-2k)!}\cdot\frac{(2m-2k-1)!!}{2^{m-k}}\sqrt{\pi}\tag{4}\end{eqnarray*} $$ so by $(2)$ and $(4)$ we have: $$\begin{eqnarray*} x^{2m} &=& \sum_{k=0}^{m}H_{2k}(x)\cdot\frac{(2m)!}{(2m-2k)!}\cdot\frac{(2m-2k-1)!!}{2^{m+k}(2k)!}\\&=&\sum_{k=0}^{m} H_{2k}(x)\cdot\frac{(2m)!}{(2m-2k)!!}\cdot\frac{1}{2^{m+k}(2k)!}\\&=&\sum_{k=0}^{m}H_{2k}(x)\cdot\frac{(2m)!}{(m-k)!(2k)! 2^{2m}}\tag{5}\end{eqnarray*}$$ as wanted.