Prove identity concerning successive derivative of $e^{x^2/2a}$ 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 $x=0$.
 A: Use the fact(Hermite Polynomial) that
$$
e^{i2yx+y^2}=\sum_{k=0}^\infty \frac{H_k(x)}{k!}(iy)^k
$$
with $\sqrt{2a}y=x$ and then $x=0$, you'll get 
$$
\begin{equation}
\left[\frac{d^{2n}}{dx^{2n}}e^{x^2/2a}\right]_{x=0}
=\frac1{2^na^n}\left[\frac{d^{2n}}{dy^{2n}}e^{y^2}\right]_{x=0}=\frac1{2^na^n}\left[\sum_{k=0}^\infty \frac{(-i)^{2n}H_{2n+k}(0)}{k!}(iy)^k\right]_{y=0}=(-1)^{n}\frac{H_{2n}(0)}{2^na^n}
\end{equation}
$$
Now $H_{2n}(0)$ are the Hermite Numbers which are given as
$$
    H_n = \begin{cases} 0, & \mbox{if }n\mbox{ is odd} \\ (-1)^{n/2} 2^{n/2} (n-1)!! , & \mbox{if }n\mbox{ is even} \end{cases} 
$$
so $\displaystyle(-1)^{n}\frac{H_{2n}(0)}{2^na^n}=(-1)^{n}\frac{(-1)^{n} 2^{n} (2n-1)!!}{2^na^n} = \frac{(2n-1)!!}{a^n}$.
We need one more thing: A recurrence relation:
$$
H_{n+1}(x)=2xH_{n}(x)-2nH_{n-1}(x) \Rightarrow H_{n}(0)=-2(n-1)H_{n-2}(0) 
$$
Starting with $H_0=1$ we get $H_{2n}=(-1)^n2^n(2n-1)!!$
A: 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 
$$
\begin{align}
\left[\frac{d^{2n}}{dx^{2n}}e^{x^2/2a}\right]_{x=0} &= \left[\frac{d^{2n}}{dx^{2n}} \left(\dfrac{x^{2n}}{2^n a^n n!}\right)\right]_{x=0} \\
&= \frac{(2n)!}{2^n a^n n!} \\
&= \frac{(2(n)(2(n-1))(2(n-2) \cdots 2) (2n-1)(2n-3) \cdots 1)}{2^n a^n n!} \\
&= \frac{2^n n! (2n-1)!!}{2^n a^n n!} \\
&= \frac{(2n-1)!!}{a^n}.
\end{align}
$$
