showing $\frac{H_{2n}(0)}{(2n)!} = \frac{(-1)^n}{n!}$

Question

How do I show that, given the Hermite polynomials $H_n(x)\equiv (-1)^n e^{x^2}\frac{d^n}{dx^n}(e^{-x^2})$, $$\frac{H_{2n}(0)}{(2n)!} = \frac{(-1)^n}{n!}?$$

Answer 1

Taylor expansion:

$$e^{-x^2} = \sum \frac{(-1)^n}{n!}x^{2n} \qquad \implies\qquad \frac{d^{2n}}{dx^{2n}}(e^{-x^2})\mid_{x=0} = (-1)^n \frac{(2n)!}{n!}$$

Then $$H_{2n}(0) = (-1)^{2n} e^0 \times (-1)^n \frac{(2n)!}{n!} = (-1)^n \frac{(2n)!}{n!}$$

Answer 2

Use the generating function:

$$e^{2 x t-t^2} = \sum_{n=0}^{\infty} H_n(x) \frac{t^n}{n!}$$

Note that

$$e^{-t^2} = \sum_{n=0}^{\infty} (-1)^n \frac{t^{2 n}}{n!}$$

Compare this with the result of the generating function at $x=0$:

$$e^{-t^2} = \sum_{n=0}^{\infty} H_n(0) \frac{t^n}{n!}$$

Note that $H_n(0)=0$ when $n$ is odd. Thus

$$\frac{H_{2 n}(0)}{(2 n)!} = \frac{(-1)^n }{n!}$$

ADDENDUM

You can also use the recurrence

$$H_{n+1}(x) = 2 x H_n(x) - 2 n H_{n-1}(x)$$

to get

$$H_{2 n}(0) = -2 (2 n-1) H_{2(n-1)}(0) = (-2)^n (2 n-1) (2 n-3)\cdots (1)$$

or

$$H_{2 n}(0) = (-1)^n \frac{(2 n)!}{n!}$$