Hay varias relaciones de recurrencia que los polinomios de Hermite satisfacen. Acá obtengo una que es de particular utilidad para mostrar que $\displaystyle{\frac{d^nH_n(\zeta)}{d\zeta^n}=2^n\,n!}$ En la literatura pueden encontrarse muchas relaciones, incluso la que aquí muestro, sin embargo la siguiente forma de obtenerla es bastante interesante. Es necesario recordar la forma general de la regla del producto de Leibniz
$$\frac{d^n(fg)}{dx^n}=\sum_{k=0}^n\binom{n}{k}\,\frac{d^kf}{dx^k}\,\frac{d^{n-k}g}{dx^{n-k}}$$ Se tiene la definición por fórmula de Rodrigues de los polinomios de Hermite
$$H_n(\zeta)=(-1)^n\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}$$ entonces
\begin{align*}\frac{dH_n}{d\zeta}&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+\mathrm{e}^{\zeta^2}\frac{d^{n+1}\mathrm{e}^{-\zeta^2}}{d\zeta^{n+1}}\right]\\[0.1in]&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+\mathrm{e}^{\zeta^2}\frac{d^n}{d\zeta^n}\left(-2\zeta\mathrm{e}^{-\zeta^2}\right)\right]\\[0.1in]&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+\mathrm{e}^{\zeta^2}\sum_{k=0}^n\binom{n}{k}\frac{d^k(-2\zeta)}{d\zeta^k}\,\frac{d^{n-k}\left(\mathrm{e}^{-\zeta^2}\right)}{d\zeta^{n-k}}\right]\\[0.1in]&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+\mathrm{e}^{\zeta^2}\sum_{k=0}^1\binom{n}{k}\frac{d^k(-2\zeta)}{d\zeta^k}\,\frac{d^{n-k}\left(\mathrm{e}^{-\zeta^2}\right)}{d\zeta^{n-k}}\right]\\[0.1in]&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}-2\mathrm{e}^{\zeta^2}\left(\zeta\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+n\frac{d^{n-1}\mathrm{e}^{-\zeta^2}}{d\zeta^{n-1}}\right)\right]\\[0.1in]&=2n(-1)^{n-1}\mathrm{e}^{\zeta^2}\frac{d^{n-1}\mathrm{e}^{-\zeta^2}}{d\zeta^{n-1}}\\[0.1in]&=2n\,H_{n-1}\end{align*} de donde se sigue que
$$\frac{d^nH_n}{d\zeta^n}=2^n\,n!$$ como se quería mostrar.
$$\frac{d^n(fg)}{dx^n}=\sum_{k=0}^n\binom{n}{k}\,\frac{d^kf}{dx^k}\,\frac{d^{n-k}g}{dx^{n-k}}$$ Se tiene la definición por fórmula de Rodrigues de los polinomios de Hermite
$$H_n(\zeta)=(-1)^n\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}$$ entonces
\begin{align*}\frac{dH_n}{d\zeta}&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+\mathrm{e}^{\zeta^2}\frac{d^{n+1}\mathrm{e}^{-\zeta^2}}{d\zeta^{n+1}}\right]\\[0.1in]&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+\mathrm{e}^{\zeta^2}\frac{d^n}{d\zeta^n}\left(-2\zeta\mathrm{e}^{-\zeta^2}\right)\right]\\[0.1in]&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+\mathrm{e}^{\zeta^2}\sum_{k=0}^n\binom{n}{k}\frac{d^k(-2\zeta)}{d\zeta^k}\,\frac{d^{n-k}\left(\mathrm{e}^{-\zeta^2}\right)}{d\zeta^{n-k}}\right]\\[0.1in]&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+\mathrm{e}^{\zeta^2}\sum_{k=0}^1\binom{n}{k}\frac{d^k(-2\zeta)}{d\zeta^k}\,\frac{d^{n-k}\left(\mathrm{e}^{-\zeta^2}\right)}{d\zeta^{n-k}}\right]\\[0.1in]&=(-1)^n\left[2\zeta\mathrm{e}^{\zeta^2}\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}-2\mathrm{e}^{\zeta^2}\left(\zeta\frac{d^n\mathrm{e}^{-\zeta^2}}{d\zeta^n}+n\frac{d^{n-1}\mathrm{e}^{-\zeta^2}}{d\zeta^{n-1}}\right)\right]\\[0.1in]&=2n(-1)^{n-1}\mathrm{e}^{\zeta^2}\frac{d^{n-1}\mathrm{e}^{-\zeta^2}}{d\zeta^{n-1}}\\[0.1in]&=2n\,H_{n-1}\end{align*} de donde se sigue que
$$\frac{d^nH_n}{d\zeta^n}=2^n\,n!$$ como se quería mostrar.
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