Is it known that the extrema of $H_n(x)\exp(-\frac{x^2}{2})$ are the zeros of $H_{n+1}(x)$?

Question

The extrema of $H_n(x)\exp(-\frac{x^2}{2})$ can be calculated $$ \frac{d}{dx}H_n(x)\exp(-\frac{x^2}{2})=\left[nH_{n-1}(x) - H_n(x)H_1(x)\right]\exp(-\frac{x^2}{2}) $$ the second term can be expanded using the Hermite linearization formula $$ H_n(x)H_1(x) = \sum_{k=0}^1 {1 \choose k}{n \choose k}k!H_{n+1-2k}(x) $$ with result $$ \frac{d}{dx}H_n(x)\exp(-\frac{x^2}{2})=\left[nH_{n-1}(x) - H_{n+1}(x) - nH_{n-1}(x)\right]\exp(-\frac{x^2}{2})\\ = -H_{n+1}(x)\exp(-\frac{x^2}{2}) $$ Since $\exp(-\frac{x^2}{2}) >0$ the extrema of $H_n(x)\exp(-\frac{x^2}{2})$ are precisely the zeros of $H_{n+1}(x)$.

Is this a known result? It is quite straight-forward but I haven't seen it anywhere.

Answer 1

This is well known, see Equation 18.9.26 in the https://dlmf.nist.gov/18.9 .