Hermite Series: Common Functions in a Basis of Hermite Polynomials


This document notes down expressions for a few common functions in terms of Hermite polynomials. There are clear symmetries between certain special functions, i.e. \(\cos\) and \(\sin\), but interestingly a similar symmetry is formed between the Gaussian and the product of a Gaussian and the imaginary error function. The method of finding these expansions comes primarily from the Hermite transform. An approximate form for the \(\log\) function is given, but this may have complex branching problems.


Hermite Series Couplings: We can write a few common functions in a basis of Hermite polynomials \[\begin{aligned} e^{ax} = \sum_{n=0}^\infty \frac{a^n e^{a^2/4}}{2^n n!}H_n(x)\\ e^{x-\frac{1}{4}} = \sum_{n=0}^\infty \frac{H_n(x)}{2^n n!}\\ e^{-x^2} = \sum_{n=0}^\infty \frac{(-1)^n(2n-1)!!}{\sqrt{2}2^{2n}(2n)!}H_{2n}(x) \\ e^{-x^2} = \text{Re}\left[\sum_{n=0}^\infty \frac{i^n(n-1)!!}{\sqrt{2}2^{n}(n)!}H_{n}(x) \right] \\ ?(x) = \text{Im}\left[\sum_{n=0}^\infty \frac{i^n(n-1)!!}{\sqrt{2}2^{n}(n)!}H_{n}(x) \right] \\ \cos(ax) = \text{Re}\left[\sum_{n=0}^\infty \frac{a^ni^n}{e^{a^2/4}2^{n}n!}H_{n}(x)\right]\\ \sin(ax) = \text{Im}\left[\sum_{n=0}^\infty \frac{a^ni^n}{e^{a^2/4}2^{n}n!}H_{n}(x)\right]\\ x^n = \frac{n!}{2^n} \sum_{m=0}^{\lfloor \tfrac{n}{2} \rfloor} \frac{1}{m!(n-2m)! } ~H_{n-2m}(x)\end{aligned}\] where \(H_n(x)\) is a Hermite polynomial of order \(n\). I want to find standard forms such that a function can be written \[f(x) = \sum_{n=0}^\infty c_n H_n(x)\] for some constants \(c_n\). It is interesting to see that there is a function marked as \(?(x)\) above, which is made of the imaginary parts of the expansion of \(e^{-x^2}\). I would like to know what it is. We can see that \[\begin{aligned} ?(0)=0\\ ?(\infty)=0\end{aligned}\] and it has a single maximum just greater than \(x=1\). It seems we can write the Taylor expansion of this function as \[?(x) = \sqrt{2} x - \frac{2\sqrt{2}}{3}x^3 + \frac{4 \sqrt{2}}{15} x^5 - \frac{16 \sqrt{2}}{21} x^7 + \cdots \\\] we can see then it looks like \[\begin{aligned} \frac{?(x)}{\sqrt{2}} = \sum_{n=1}^\infty \frac{(-1)^{n+1} 2^{n-1}}{(2n-1)!!} x^{2n-1} \\ ?(x) = \sqrt{\frac{\pi}{2}}\text{Erfi}(x)e^{-x^2}\end{aligned}\]

For the logarithm, by experimental techniques we can formally write something like \[\begin{aligned} \log(x) = \frac{i}{2}\sqrt{\pi}(i\gamma + \pi + i\log(4)) + \sum_{n>0} \frac{(-1)^n (2n-2)! i \sqrt{\pi}}{2^{2n-1}(2n-1)!}H_{2n-1}(x) + 2\frac{(-1)^{n-1}4^{n-1}(n-1)!}{2^{2n}(2n)!}H_{2n}(x)\end{aligned}\] but it’s not clear how it converges, perhaps something slightly wrong with this. Other functions use hypergeometric expressions as coefficients, for example

\[\cosh(\sqrt{x}) = \sum_{n=0}^\infty \frac{_0F_2(;\frac{2n+1}{4},\frac{2n+3}{4};\frac{1}{256})}{2^n (2n)!}H_n(x)\]

by setting \(x=0\), this gives us some sums for unity\[\sum_{n=0}^\infty \frac{_0F_2(;\frac{2n+1}{4},\frac{2n+3}{4};\frac{1}{256})}{2^n (2n)!}\frac{2^n \sqrt{\pi}}{\Gamma(\frac{1-n}{2})} = 1\] for the exponential expression \(e^{x-1/4}\) given above we get \[\sum_{n=0}^\infty \frac{e^{1/4}\sqrt{\pi}}{n!\Gamma(\frac{1-n}{2})} = 1\] also \[\sum_{n=0}^\infty \frac{4^n \pi}{(2n)!\Gamma((1-n)/2)\Gamma(n+1/2)} \,_0F_4(;1/4 + n/2, 1/4+n/2,3/4+n/2,3/4+n/2;1/64) = 1\] and \[\frac{1}{\pi^{5/2}}=\sum_{n=0}^\infty \frac{2^{1-4n}}{(2n)!\Gamma(\frac{1}{2}-n)} \,_0\tilde{F}_4\left(;n+\frac{1}{4},n+\frac{1}{4},n+\frac{3}{4},n+\frac{3}{4};1/64\right)\] It seems that in general we have the result \[D_{2n-1} e^{-x^2} = \sum_{k=0}^\infty \frac{(-1)^{n+k}(2n+2k)!}{(2k+1)!(n+k)!}x^{2k+1} = \frac{(-1)^n x (2n)!}{n!} \;_1F_1\left(n+\frac{1}{2},\frac{3}{2},-x^2\right)\]

Nice Relationship/Transform

Using related functions above seems to make in interesting transform kernel, however, there does not seem to be a one to one mapping between transformed functions and input functions. Some results are given below: \[\int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}J_0(x)}{\sqrt{8\pi a}} = I_0(a)\] \[\int_{-\infty}^ \infty \frac{e^{-a-\frac{x^2}{8a}}I_0(x)}{\sqrt{8\pi a}} = J_0(a)\] \[\int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}e^x}{\sqrt{8\pi a}} = e^{3a}\] \[\int_{-\infty}^ \infty \frac{e^{-a-\frac{x^2}{8a}}e^x}{\sqrt{8\pi a}} = e^a\] \[\int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}\cos(x)}{\sqrt{8\pi a}} = e^{-a}\] \[\int_{-\infty}^ \infty \frac{e^{-a-\frac{x^2}{8a}}\cos(x)}{\sqrt{8\pi a}} = e^{-3a}\] \[\int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}\sin(x)}{\sqrt{8\pi a}} = 0\] \[\int_{-\infty}^ \infty \frac{e^{-a-\frac{x^2}{8a}}\sin(x)}{\sqrt{8\pi a}} = 0\] \[\int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}J_2(x)}{\sqrt{8\pi a}} = I_1(a)\] \[\int_{-\infty}^ \infty \frac{e^{a-\frac{x^2}{8a}}J_4(x)}{\sqrt{8\pi a}} = I_2(a)\]

The Integral Transform Relationship

We can write the Hermite polynomials as \[\begin{aligned} H_k(s)=\frac{2^{k+1}}{\sqrt{\pi}}\int_0^\infty e^{-x^2}x^k \cos(2sx-\frac{\pi k}{2}) \; dx \\ H_k(s)=\frac{2^k e^{ik\pi/2}}{\sqrt{\pi}}\int_0^\infty e^{-x^2}x^ke^{-2isx} \; dx + \frac{2^k e^{-ik\pi/2}}{\sqrt{\pi}}\int_0^\infty e^{-x^2}x^ke^{2isx} \; dx \\ H_k(s)=\frac{2^k i^k}{\sqrt{\pi}}\int_0^\infty e^{-x^2}x^ke^{-2isx} \; dx + \frac{2^k i^{-k}}{\sqrt{\pi}}\int_0^\infty e^{-x^2}x^ke^{2isx} \; dx \\ H_k(s)=\frac{2^k i^k}{\sqrt{\pi}}\left(\int_0^\infty e^{-x^2}x^ke^{-2isx} \; dx + (-1)^k\int_0^\infty e^{-x^2}x^ke^{2isx} \; dx\right)\end{aligned}\] as from the Wolfram functions website. Here the \(\cos\) term in effect swaps between \(\cos,\sin,-\cos,-\sin\) due to the phase shift. If we want two functions to be connected in the following manner \[f(x) = \sum_{k=0}^\infty a_k x^k \\ F(s) = \sum_{k=0}^\infty \frac{\sqrt{\pi}a_k}{2^{k+1}} H_k(s)\] then we have the transform \[\begin{aligned} F(s) = \int_0^\infty K(x,s)f(x) \; dx\\ F(s) = \int_0^\infty \sum_{k=0}^\infty a_k x^k K(x,s) \; dx\\ F(s) = \sum_{k=0}^\infty a_k \int_0^\infty x^k K(x,s) \; dx\end{aligned}\] if we set \[K_k(x,s) = e^{-x^2}\cos(2sx-\frac{\pi k}{2})\] then \[\begin{aligned} F(s) = \sum_{k=0}^\infty \frac{2^{k+1}a_k}{\sqrt{\pi}} \int_0^\infty x^k e^{-x^2}\cos(2sx-\frac{\pi k}{2}) \; dx\\ F(s) = \sum_{k=0}^\infty \frac{\sqrt{\pi}a_k}{2^{k+1}} H_k(s)\\\end{aligned}\] but the dependence on \(k\) disrupts this kernel. We can see that due to the periodicity of \(\cos\)...