2 ms·
A question I hadn't even thought to ask, thanks. So, basically, the eigenfunctions of the Fourier transform are Hermite polynomials times a Gaussian [0] [1].
by abetusk 2y ago
A question I hadn't even thought to ask, thanks.
So, basically, the eigenfunctions of the Fourier transform are Hermite polynomials times a Gaussian [0] [1].
[0] https://math.stackexchange.com/questions/728670/functions-that-are-their-own-fourier-transform https://math.stackexchange.com/questions/728670/functions-th...
[1] https://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions_as_eigenfunctions_of_the_Fourier_transform https://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_fu...
- perihelions 2y agoAs well as the linear combinations (including infinite sums!) of Hermite functions with the same eigenvalue under the Fourier transform. (Those eigenvalues are infinitely degenerate). You could express sech(x) as such a sum.