# Abstract

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.

# Main

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 $$\cos$$...