TITLE: Prove identity concerning successive derivative of $e^{x^2/2a}$
QUESTION [8 upvotes]: Prove the following identity:
\begin{equation}
\left[\frac{d^{2n}}{dx^{2n}}e^{x^2/2a}\right]_{x=0}=\frac{(2n-1)!!}{a^n}
\end{equation}
The final derivative must be evaluated at $x=0$.

REPLY [4 votes]: Use the Maclaurin series for $e^x$.  Then substitute $(x/\sqrt{2a})^2$ for $x$.  This yields
$$e^{x^2/2a} = \sum_{k=0}^{\infty} \frac{x^{2k}}{2^k a^k k!}.$$
Since subbing in $x=0$ at the end means we need only concern ourselves with the constant term after taking $2n$ derivatives, we have 
$$
\begin{align}
\left[\frac{d^{2n}}{dx^{2n}}e^{x^2/2a}\right]_{x=0} &= \left[\frac{d^{2n}}{dx^{2n}} \left(\dfrac{x^{2n}}{2^n a^n n!}\right)\right]_{x=0} \\
&= \frac{(2n)!}{2^n a^n n!} \\
&= \frac{(2(n)(2(n-1))(2(n-2) \cdots 2) (2n-1)(2n-3) \cdots 1)}{2^n a^n n!} \\
&= \frac{2^n n! (2n-1)!!}{2^n a^n n!} \\
&= \frac{(2n-1)!!}{a^n}.
\end{align}
$$