# Thread: help solving orthogonal polynomial

1. ## help solving orthogonal polynomial

I managed to do the first part, stuck in the part circled. Any help will be appreciated, thanks.

2. Use the Rodrigues formula plus the following identity:

$\displaystyle H_{n+1}(x)=2xH_{n}(x)-2nH_{n-1}(x).$

Does that give you some ideas?

$\displaystyle \displaystyle e^{2xu - u^2} = \sum_{n=0}^{\infty} \dfrac{u^n}{n!}H_n(x)$

Then differentiating wrt x gives

$\displaystyle \displaystyle 2u e^{2xu - u^2} = \sum_{n=0}^{\infty} \dfrac{u^n}{n!}H'_n(x)$

so using the generating function again gives

$\displaystyle \displaystyle 2u \sum_{n=0}^{\infty} \dfrac{u^n}{n!}H_n(x) = \sum_{n=0}^{\infty} \dfrac{u^n}{n!}H'_n(x)$

or

$\displaystyle \displaystyle \sum_{n=0}^{\infty} \dfrac{u^{n+1}}{n!}2H_n(x) = \sum_{n=0}^{\infty} \dfrac{u^n}{n!}H'_n(x)$

Now, shifting indicies gives

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \dfrac{u^{n}}{(n-1)!}2H_{n-1}(x) = \sum_{n=0}^{\infty} \dfrac{u^n}{n!}H'_n(x)$

or

$\displaystyle \displaystyle \sum_{n=1}^{\infty} \dfrac{u^{n}}{n!}2nH_n(x) = \sum_{n=0}^{\infty} \dfrac{u^n}{n!}H'_n(x)$.

Thus,

$\displaystyle \displaystyle \sum_{n=0}^{\infty} \dfrac{u^{n}}{n!}2nH_n(x) = \sum_{n=0}^{\infty} \dfrac{u^n}{n!}H'_n(x)$

since the first term of the series on the left is zero. Equating coefficients of u gives your result.