How do I show that, given the Hermite polynomials $H_n(x)\equiv (-1)^n e^{x^2}\frac{d^n}{dx^n}(e^{-x^2})$, $$\frac{H_{2n}(0)}{(2n)!} = \frac{(-1)^n}{n!}?$$
Asked
Active
Viewed 70 times
1
-
1How have you defined the Hermite polynomials? What can you assume? – not all wrong Aug 28 '13 at 23:51
-
@sharkos $H_n(x) = (-1)^n e^{x^2}\frac{d^n}{dx^n}(e^{-x^2})$ – hasExams Aug 28 '13 at 23:53
2 Answers
4
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 $$H_{2n}(0) = (-1)^{2n} e^0 \times (-1)^n \frac{(2n)!}{n!} = (-1)^n \frac{(2n)!}{n!}$$
not all wrong
- 16,178
- 2
- 35
- 57
3
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
$$H_{2 n}(0) = (-1)^n \frac{(2 n)!}{n!}$$
Ron Gordon
- 138,521
-
Is there any other way? I am doing the opposite ... finding the generating function. I have to show that if $f(x,t)$ is generatig function then $f(x,t) = k(t)e^{2xt}$ where $k(t) = e^{-x^2}$ – hasExams Aug 28 '13 at 23:54
-
@hasExams: sorry, I don't read minds. Yes, I am sure there is another way. – Ron Gordon Aug 28 '13 at 23:55
-