Hermite Series: Common Functions in a Basis of Hermite Polynomials
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 e^{ax} = ^\infty }{2^n n!}H_n(x)\\ e^{x-{4}} = ^\infty {2^n n!}\\ e^{-x^2} = ^\infty {2^{2n}(2n)!}H_{2n}(x) \\ e^{-x^2} = \left[^\infty {2^{n}(n)!}H_{n}(x) \right] \\ ?(x) = \left[^\infty {2^{n}(n)!}H_{n}(x) \right] \\ \cos(ax) = \left[^\infty {e^{a^2/4}2^{n}n!}H_{n}(x)\right]\\ \sin(ax) = \left[^\infty {e^{a^2/4}2^{n}n!}H_{n}(x)\right]\\ x^n = {2^n} ^{\lfloor {2} \rfloor} {m!(n-2m)! } ~H_{n-2m}(x) where Hn(x) is a Hermite polynomial of order n. I want to find standard forms such that a function can be written f(x) = ^\infty c_n H_n(x) for some constants cn. 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−x². I would like to know what it is. We can see that ?(0)=0\\ ?(\infty)=0 and it has a single maximum just greater than x = 1. It seems we can write the Taylor expansion of this function as ?(x) = x - }{3}x^3 + }{15} x^5 - }{21} x^7 + \cdots \\ we can see then it looks like {} = ^\infty 2^{n-1}}{(2n-1)!!} x^{2n-1} \\ ?(x) = {2}}(x)e^{-x^2} For the logarithm, by experimental techniques we can formally write something like \log(x) = {2}(i\gamma + \pi + i\log(4)) + }{2^{2n-1}(2n-1)!}H_{2n-1}(x) + 24^{n-1}(n-1)!}{2^{2n}(2n)!}H_{2n}(x) but it’s not clear how it converges, perhaps something slightly wrong with this. Other functions use hypergeometric expressions as coefficients, for example \cosh() = ^\infty {4},{4};{256})}{2^n (2n)!}H_n(x) by setting x = 0, this gives us some sums for unity ^\infty {4},{4};{256})}{2^n (2n)!}}{\Gamma({2})} = 1 for the exponential expression ex − 1/4 given above we get ^\infty }{n!\Gamma({2})} = 1 also ^\infty {(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 {\pi^{5/2}}=^\infty }{(2n)!\Gamma({2}-n)} \,_0_4\left(;n+{4},n+{4},n+{4},n+{4};1/64\right) It seems that in general we have the result D_{2n-1} e^{-x^2} = ^\infty (2n+2k)!}{(2k+1)!(n+k)!}x^{2k+1} = {n!} \;_1F_1\left(n+{2},{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: ^ \infty {8a}}J_0(x)}{} = I_0(a) ^ \infty {8a}}I_0(x)}{} = J_0(a) ^ \infty {8a}}e^x}{} = e^{3a} ^ \infty {8a}}e^x}{} = e^a ^ \infty {8a}}\cos(x)}{} = e^{-a} ^ \infty {8a}}\cos(x)}{} = e^{-3a} ^ \infty {8a}}\sin(x)}{} = 0 ^ \infty {8a}}\sin(x)}{} = 0 ^ \infty {8a}}J_2(x)}{} = I_1(a) ^ \infty {8a}}J_4(x)}{} = I_2(a)
Fields of Study & Subjects
Mathematical functions and polynomials
Mathematical Analysis and Transform Methods
Mathematical and Computational Methods
Publication details
- Publication date
- 13 March 2026
- Language
- EN
- Document type
- Preprint