Hermite polynomials recurrence relation

Hermite polynomials $H_n (x)$ can be obtained using the recurrence relation $$H_{n+1} (x)=2xH_n (x)-2nH_{n-1} (x).$$ To prove this, I started by calculating the first derivative of the Hermite's Rodrigues formula $H_n (x)=(-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2 } $. The process goes like this: $$ \frac{d}{dx}H_n (x)=(-1)^n 2xe^{x^2}\frac{d^n}{dx^n} e^{-x^2 }+(-1)^n e^{x^2} \frac{d^{n+1}}{dx^{n+1}}e^{-x^2 } $$ Rearranging the terms in the previous equation produces $$H_{n+1} (x)=2xH_n (x)-\frac{d}{dx}H_n (x)$$ or $$H_{n+1} (x)=2xH_n (x)-H_n'(x).$$

That's it. Im stuck here. I don't know how to show that $H_n' (x)=2nH_{n-1} (x)$. Can you help me?

Solution 1:

Let’s write $D = \frac{\operatorname{d}}{\operatorname{d}x}$, and take the derivative of $H_n$ $$ \begin{aligned}D H_n &= D \left( e^{x^2} D^n e^{-x^2} \right) \\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} D^n \left( - 2 x e^{-x^2} \right) \\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} \sum_{k=0}^n \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2} \\ & \left(\text{note that $D^k(-2x)=0$ for $k\ge 2$, so $\sum_{k=0}^n=\sum_{k=0}^1$}\right)\\ &= 2 x e^{x^2} D^n e^{-x^2} + e^{x^2} \sum_{k=0}^1 \binom{n}{k} D^k (-2 x) D^{n-k} e^{-x^2}\\ &= {{2 x e^{x^2} D^n e^{-x^2} }}+ e^{x^2} \left(\underbrace{\binom{n}{0}}_1\underbrace{D^0(-2x)}_{-2x}D^{n}e^{-x^2}+\underbrace{\binom{n}{1}}_n\underbrace{D^1(-2x)}_{-2}D^{n-1}e^{-x^2} \right) \\ &= \color{red}{{2 x e^{x^2} D^n e^{-x^2} }}+ e^{x^2} \left(\color{red}{-{2 x D^{n} e^{-x^2} }}- 2 n D^{n-1} e^{-x^2} \right) \end{aligned} $$ So we have the rather simple end result $$\frac{\operatorname{d}}{\operatorname{d}x} H_n(x)=2 n H_{n-1}(x)$$