Hermite Polynomials

1. Apr 5, 2006

No Name Required

Im stuck on this question :(

The Hermite polynomials can be defined through

$$\displaystyle{F(x,h) = \sum^{\infty}_{n = 0} \frac{h^n}{n!}H_n(x)}$$

Prove that the $$H_n$$ satisfy the hermite equation

$$\displaystyle{H''_n(x) - 2xH'_n(x) + 2nH_n(x) = 0}$$

Using

$$\displaystyle{\sum^{\infty}_{n = 0} \frac{h^n}{n!}nH_n(x) = h\frac{\partial}{\partial h}F(x,h)}$$

Can someone give me a bit of a push in the right direction?

2. Apr 5, 2006

dextercioby

Who's $F(x,h)$ ...?

IIRC, it has to have some exponential, right...?

Daniel.

3. Apr 5, 2006

HallsofIvy

Staff Emeritus
What's wrong with "just do it!"? Just go ahead and differentiate that sum, term by term, (your hint essentially says that you can do that), plug into the differential equation and see what happens.