I'm struggling with this question, which says "given the generating funcion $g(x,z)=e^{-z^2 + 2xz} = \sum_{n=0}^{\infty}H_n(x) \frac{z^n}{n!}$ prove that the Hermite polynomials satisfy the Hermite equation".
So far I tried using the Rodrigues' expresion (which I already derived from the generating function, and it's the most explicit formula I could find for $H_n$) and tried to plug it into the Hermite equation, but I wasn't succesful.
Is this the way to work it out? Any ideas?
Thanks a lot!
Best Answer
$$\cdot e^{-(x-z)^2}=(\sum_{n=0}^{\infty} \frac{z^n}{n !}\frac{d^n}{d z^n}e^{-(x-z)^2})|_{z=0}$$ Make the substitution $y=x-z$ $$\frac{d^n}{d z^n}e^{-(z-x)^2})|_{z=0}=(-1)^n\frac{d^n}{d y^n}e^{-y^2})|_{y=x}= e^{-x^2} H_n(x) \dots (\text{Rodrigue's Formula})$$
$$\rightarrow \cdot e^{-(x-z)^2}=e^{-x^2} (\sum_{n=0}^{\infty} \frac{z^n}{n !}H_n (x))$$ Multiply both sides by $e^{x^2}$ to get $g(x,z)$