closed form for arbitrary derivatives of gaussian function

Question

What is the closed form expression for the $n$th order derivative of the Gaussian function? $$ \frac{d^n}{dx^n}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}}$$ Wikipedia says that it can be expressed in terms of the Gaussian times Hermite polynomials, but it does not illuminate any further.

Answer 1

The relation

Indeed, via Rodrigues' formula, the Hermite polynomials are $$H_n(x) = (-1)^n \, e^{x^2} \, \frac{d^n}{dx^n} e^{-x^2} \; , $$ for $n \in \mathbb{N}$ including zero. And by changing variables $x \to \frac{x-\mu}{\sigma\sqrt{2}}$ ones get $$ H_n\left(\frac{x-\mu}{\sigma\sqrt{2}}\right) = (-1)^n \, e^{\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} \, 2^{\frac{n}{2}} \, \sigma^n \, \frac{d^n}{dx^n} e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} \; , $$ and the answer for your question is $$ \frac{d^n}{dx^n} e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} = (-1)^n \left( \frac{1}{2 \sigma^2} \right)^{\frac{n}{2}} \, e^{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}} \, H_n\left(\frac{x-\mu}{\sigma\sqrt{2}}\right) \; . $$

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The Hermite polynomials

The above expression is useful only if you know the polynomial that you are interested in. For n = 0, ... , 10 there is a table here and some polynomials are \begin{align} H_0(x) &= 1 \; , \\ H_1(x) &= 2x \; , \\ H_2(x) &= 4x^2 - 2 \; ,\\ & \; \; \vdots \end{align} and the polynomial $H_n(x)$ is of order $n$ and the polynomial has the same parity (odd or even) of $n$. The general expression for this polynomials is not trivial and normally one use its property instead of its explicitly expression. If you desire the explicitly expression I copy it from the wiki and it is \begin{align} H_{2n}(x) \; & = \; (2n)! \, \sum_{l=0}^{n} \frac{(-1)^{n-l}}{(2l)!(n-l)!} \, (2x)^{2l} \; ,\\ H_{2n+1}(x) \; & = \; (2n+1)! \, \sum_{l=0}^{n} \frac{(-1)^{n-l}}{(2l+1)!(n-l)!} \, (2x)^{2l+1} \; ,\\ \end{align} where I separate the case for odd and even polynomials for ease notation.

Answer 2

For this, I will find what Wikipedia https://en.wikipedia.org/wiki/Hermite_polynomials calls the "probabilist's Hermite polynomials", which differ from the "physicist's" version by the factor of $\frac{1}{2}$ in the exponent.

There are actually multiple equivalent definitions. The first one given by the article actually solves your problem by definition! There are other equivalent versions: defined by orthogonality conditions, or through a generating function, or through the Hermite differential equation.

First of all, change variables so that $u = \frac{x - \mu}{\sigma}$. That way $\frac{d}{dx} = \frac{1}{\sigma} \cdot \frac{d}{du}$. Also let $k = \frac{1}{2\pi \sigma^2}$.

Let $$G = k\cdot e^{\frac{-u^2}{2}}$$ denote that Gaussian function $G(u)$, and $G^{(n)}$ denote its $n$th derivative with respect to $u$.

$G^{(0)} = (-1)^0 \cdot 1 \cdot G$

$G^{(1)} = -uG = (-1)^1 \cdot u \cdot G$

$G^{(2)} = -G + u^2G = (-1)^2 \cdot (u^2 - 1)\cdot G $

$G^{(3)} = (2u)G + (u^2-1)(-u)G = (-1)^3 \cdot (u^3 - 3 u)\cdot G$

The first few Hermite polynomials are $1, u, (u^2 -1), (u^3 - 3u),...$ To get the physicist version, multiply by $2^n$.