The Moment Generating Function for a Standard Normal $X\sim\mathcal{N}(0,1)$ looks like so:
$$M_X(s) = \text{exp}(\frac{s^2}{2})$$
Usually we are interested in finding the $n^{th}$ moment of $X$, so we end up taking the $n^{th}$ derivative of $M_X(s)$ with $s$ evaluated at zero. However I am more curious as to what the final form of the polynomial $\frac{d^n}{ds^n} M_X(s)$ looks like.
Through taking success derivatives we know our polynomial should be of the form:
$$p_n(s) = \sum_{i=0}^{n} a_{i}^{(n)}s^i\text{exp}(\frac{s^2}{2})$$
Where $p_n$ is the $n^{th}$ derivative of $M_X(s)$
From the relation $p_{n+1} = \frac{d}{ds} p_n$ we obtain the following recurrence relation:
$$a_i^{(n+1)} = (i+1)a_{i+1}^{(n)} + a_{i-1}^{(n)} \hspace{1cm} \text{ with } a_i^{(n)} = 0 \text{ for } i \notin [0,1,\ldots,n]$$
The initial condition would be $a_0^{(0)} = 1$
How can I go about solving this recurrence relation?
$e^{\frac {s^{2}} 2} =\sum_{n=0}^{\infty} \frac {(s^{2})^{n}} {2^{n} n!}$. For any convergent power series $\sum_{n=0}^{\infty} a_n s^{n}$ the n-th derivative at $0$ is simply $(n!)a_n$. Can you write down the values of the n-th derivatives now?
[ $\frac {d^{k}} {ds^{k}} M_X(s)=\sum_{n\geq k/2} \frac {(2n)(2n-1)\cdots (2n-k+1) s^{2n-k}} {2^{n} n!}].$