I want to derive the derivative recurrence relation for the Hermite polynomials straight from the Hermite differential equation. That is, I want to go from left to right in the following sequence without going through the generating function, Rodrigues formula, or any other representation of the Hermite polynomials.
$$H''_n(x)-2xH'_n(x)+2nH_n(x)=0 \longrightarrow H'_n=2nH_{n-1}$$
Every derivation of the above identity that I've seen has been from either the generating function or Rodrigues formula. I essentially would like to find the simplest method for getting from the Hermite differential equation to that recurrence relation.
Sorry if this sounds/looks trivial. I've been stuck on it for hours. My first instinct was to differentiate the equation, which gives
$$H'''_n(x)-2xH''_n(x)+2(n-1)H'_n(x)=0$$
From this, restricting ourselves to the finite-polynomial Hermite polynomials, I would say that $H'_n=\alpha H_{n-1}$, where $\alpha$ is some constant. However, I don't know how to then deduce that $\alpha=2n$. I've tried further differentiation, integration, substitution, putting it in Sturm-Liouville form, etc.
Best Answer
Starting with the equation $$ y''(x) - 2x y'(x) +2n y(x) = 0, $$ you can differentiate: $$ y'''-2y'-2xy''+2ny' = 0 \\ (y')''-2x(y')'+2(n-1)(y')=0. $$ In other words, $f=y'$ satisfies $$ f''-2xf'+2(n-1)f=0, $$ which is the Hermite equation with $2n$ replaced by $2n-1$. There are two linearly independent solutions of the Hermite equation, but only one solution is a polynomial. So you are correct that $H_n'=\alpha_n H_{n-1}$ must hold for some constant $\alpha_n$. The constant $\alpha_n$ depends on normalization, which means that $\alpha_n$ is not uniquely determined by the equation; that's why you're stuck at that point.