Prove that the Hermite polynomials satisfy the Hermite equation.

Question

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!

Answer 1

$$\cdot e^{-(x-z)^2}=(\sum_{n=0}^{\infty} \frac{z^n}{n !}\frac{d^n}{d z^n}e^{-(x-z)^2})|_{z=0}$$ 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 $g(x,z)$

Answer 2

$$g(x,z) = e^{-z^2+2zx} = \sum_{n=0}^{\infty}H_n(x)\frac{z^n}{n!}$$ $$\implies \frac{\partial g}{\partial x}=2ze^{-z^2+2zx}=\sum_{n=0}^{\infty}H_n(x)\frac{2z^{n+1}}{n!}=\sum_{n=0}^{\infty}H_n'(x)\frac{z^n}{n!}$$ Now equate coefficients of equal powers of $z$ in the last equality of the above expression. Doing that, one has $$H_n'(x)=2nH_{n-1}(x)$$
Similarly $$ \frac{\partial g}{\partial z}=(2x-2z)e^{-z^2+2zx}=\sum_{n=0}^{\infty}H_n(x)\frac{(2x-2z)z^{n}}{n!}=\sum_{n=0}^{\infty}H_n(x)\frac{z^{n-1}}{(n-1)!}$$Equating coefficients as previously, one has $$H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)$$ Using these recursions, its easy to verify that $H_n(x)$ satisfies the DE $$H_n''(x)-2xH_n'(x) + 2nH_n(x) = 0.$$