I'm trying to prove that the norm of the Hermite polynomials (physicist's version) equals $ 2^n n! $. I stumbled upon this answer and I don't understand parts of the proof. First of all, it seems implied by the author that:
$$ \frac{d^n}{dx^n}H_n(x) = 2^n n!$$
But I've failed to prove it myself, using the definition (physicists' convention) and by recursion.
Alternatively, I tried to use the recursion formula:
$$ H_n'(x) = 2x H_n(x) – H_{n+1}(x)$$
And calculate the norm similarly to how the proof for the general $n \ne m$ goes:
$$ \int |H_n(x)|^2 e^{-x^2}dx = \int H_{n-1}(x) H_n'(x)e^{-x^2}dx = \int H_{n-1}(x)(2 x H_n(x) – H_{n+1}(x))e^{-x^2}dx = \int H_{n-1}(x)2 x H_n(x)e^{-x^2}dx$$
Were in the last equality, I used the fact that $H_{n+1}(x)$ and $H_n(x)$ are orthogonal.
I don't know how to prove that
$$ \int H_{n-1}(x)2 x H_n(x)e^{-x^2}dx = 2n \int |H_{n-1}(x)|^2 e^{-x^2}dx$$
I read a comment to that answer which I don't understand, and that can prove the last missing piece:
$$ H_n'(x) = 2n H_{n-1}(x) $$
Why is this true? Isn't it contradictory to the known recursion formula?
Best Answer
It seems that you are happy to use the following definition of the Physicist's Hermite polynomials: $$ H_{n+1} = 2x H_n - H_n' , \quad H_0 \equiv 1 \tag{$D$}$$ The induction hypothesis we use is $$P(n): \forall n\ge 1,\quad H'_n=2nH_{n-1}\tag{$I$}.$$ Since $H_1(x) = 2x$, the base case is trivial: $H_1'=2\times 1\times H_0$. Inductively \begin{align} H_{n+1}' &\overset{(D)}=(2x H_n - H_n')' \\ &= 2H_n + 2xH_n' - H_{n}’’ \\ &\overset{(I)}= 2H_n+2x(2nH_{n-1}) -2n H'_{n-1} \\ &= 2H_n + 2n(2xH_{n-1} - H’_{n-1}) \\ &\overset{(D)}= 2H_n + 2n H_n \\ &= 2(n+1) H_{n} \end{align} which is what we wanted. One easily computes the norm from this and the calculation in the linked answer that shows $\int |H_n|^2 e^{-x^2}dx = \int H_{n-1} H_n' e^{-x^2}dx$.
Also, its not contradictory because the defining recursive formula is made up of different terms. It would be bad if we instead proved $H_{n+1} = ax H_n +bH_n'$ for some constants $a\neq2$ and $b\neq-1$, but this is not the case.