Editing Hermite polynomials (section)

== Hermite polynomial expansion ==
Similar to Taylor expansion, some functions are expressible as an infinite sum of Hermite polynomials. Specifically, if <math>\int e^{-x^2}f(x)^2 dx < \infty</math>, then it has an expansion in the physicist's Hermite polynomials.<ref>{{Cite web |title=MATHEMATICA tutorial, part 2.5: Hermite expansion |url=https://www.cfm.brown.edu/people/dobrush/am34/Mathematica/ch5/hermite.html |access-date=2023-12-24 |website=www.cfm.brown.edu}}</ref>

Given such <math>f</math>, the partial sums of the Hermite expansion of <math>f</math> converges to in the <math>L^p</math> norm if and only if <math>4 / 3<p<4</math>.<ref>{{Cite journal |last1=Askey |first1=Richard |last2=Wainger |first2=Stephen |date=1965 |title=Mean Convergence of Expansions in Laguerre and Hermite Series |url=https://www.jstor.org/stable/2373069 |journal=American Journal of Mathematics |volume=87 |issue=3 |pages=695–708 |doi=10.2307/2373069 |jstor=2373069 |issn=0002-9327}}</ref><math display="block">x^n = \frac{n!}{2^n} \,\sum_{k= 0}^{\left\lfloor n/2 \right\rfloor} \frac{1}{k! \,(n-2k)!} \, H_{n-2k} (x) = n! \sum_{k= 0}^{\left\lfloor n/2 \right\rfloor} \frac{1}{k! \,2^k \,(n-2k)!} \, \operatorname{He}_{n-2k} (x) , \qquad n \in \mathbb{Z}_{+} .
</math><math display="block">e^{ax} = e^{a^2 /4} \sum_{n\ge 0} \frac{a^n}{n! \,2^n} \, H_n (x) , \qquad a\in \mathbb{C}, \quad x\in \mathbb{R} .</math><math display="block">e^{-a^2 x^2} = \sum_{n\ge 0} \frac{(-1)^n a^{2n}}{n! \left( 1 + a^2 \right)^{n + 1/2} 2^{2n}}\, H_{2n} (x) .</math><math display="block">\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} ~dt=\frac{1}{\sqrt{2 \pi}} \sum_{k \geq 0} \frac{(-1)^k}{k !(2 k+1) 2^{3 k}} H_{2 k}(x) .</math><math display="block">\cosh (2x) = e \sum_{k\ge 0} \frac{1}{(2k)!}\, H_{2k} (x) , \qquad \sinh (2x) = e \sum_{k\ge 0} \frac{1}{(2k+1)!} \, H_{2k+1} (x) .</math><math display="block">\cos (x) = e^{-1/4} \,\sum_{k\ge 0} \frac{(-1)^k}{2^{2k} \, (2k)!} \, H_{2k} (x) 
\quad \sin (x) = e^{-1/4} \,\sum_{k\ge 0} \frac{(-1)^k}{2^{2k+1} \, (2k+1)!} \, H_{2k+1} (x) </math>