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?
 A: 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
$$\frac{\operatorname{d}}{\operatorname{d}x} H_n(x)=2 n H_{n-1}(x)$$
A: $H_n'(x) = 2n H_{n-1}(x)$ can be proved by induction.
The base case $H_1'(x) = 2H_0(x)$ is straightforward to verify.
Now, assume $H_k'(x) = 2k H_{k-1}(x)$ and we will show $H_{k+1}'(x) = 2(k+1) H_{k}(x)$.
Taking derivatives on both sides of the recurrence relation $H_{k+1}(x) = 2xH_k(x)-H_k'(x)$, one has
$$ H_{k+1}'(x) = 2H_k(x) + 2xH_k'(x) - H_{k}''(x). $$
Using the inductive hypothesis $H_k'(x) = 2k H_{k-1}(x)$, we have
\begin{aligned}
H_{k+1}'(x) 
&= 2H_k(x) + 4kxH_{k-1}(x) - 2kH_{k-1}'(x) \\
& = 2H_k(x) + 2k[2xH_{k-1}(x) - H_{k-1}'(x)] \\
&= 2H_k(x) + 2kH_k(x) \\
&= 2(k+1)H_k(x).
\end{aligned}
