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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 $x=0$.

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Is $a$ a constant or a variable? – 000 Nov 26 '12 at 21:49
Have you tried induction? – glebovg Nov 26 '12 at 21:51
@glebovg, that's what I was thinking. – 000 Nov 26 '12 at 21:52
@glebovg, I think induction goes nowhere on this problem. Unfortunate, that. – 000 Nov 26 '12 at 22:37
up vote 4 down vote accepted

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 $H_{2n}=(-1)^n2^n(2n-1)!!$

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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 $$ \begin{align} \left[\frac{d^{2n}}{dx^{2n}}e^{x^2/2a}\right]_{x=0} &= \left[\frac{d^{2n}}{dx^{2n}} \left(\dfrac{x^{2n}}{2^n a^n n!}\right)\right]_{x=0} \\ &= \frac{(2n)!}{2^n a^n n!} \\ &= \frac{(2(n)(2(n-1))(2(n-2) \cdots 2) (2n-1)(2n-3) \cdots 1)}{2^n a^n n!} \\ &= \frac{2^n n! (2n-1)!!}{2^n a^n n!} \\ &= \frac{(2n-1)!!}{a^n}. \end{align} $$

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I like this answer because it uses a fairly simple tool. ("a fairly simple tool" meaning something that a Calculus student would typically understand.) – 000 Nov 26 '12 at 23:03

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