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by ziofill 428 days ago
I know the article is about sec(x) but I want to share this tidbit about its cousin, the hyperbolic secant: sech(x) is its own Fourier transform (modulo rescalings). That’s right, exp(-x^2) is not the only one.
3 comments

Learned something new today, thank you!

If I understand correctly, the Hermite functions are the eigenfunctions of the Fourier Transform and thus all have this property -- with the Gaussian being a special case. But sech(x) is doubly interesting because it is not a Hermite function, though it can be represented as an infinite series thereof. Are there other well-behaved examples of this, or is sech(x) unique in that regard?

Yes the dirac comb for example. Actually there are infinitely many.

https://en.wikipedia.org/wiki/Dirac_comb

and for other:

http://www.systems.caltech.edu/dsp/ppv/papers/journal08post/...

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-th...

[1] https://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_fu...

As 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.
There has to be a link to the harmonic oscillator here. That's the Hamiltonian that's symmetric under exchange of position and momentum, and the Hermite functions are its eigenfunctions.
Indeed, the (quantum) harmonic oscillator Hamiltonian (with suitable scalings) commutes with the Fourier transform. Since the former has the Hermite functions as eigenbasis, the Hermite functions also form an eigenbasis for the latter.
It still involves e, though: sech(x) = 2 * e^x / (e^(2x) + 1)

Makes sense, given that the definition of e goes hand in hand with its property of e^x being its own integral and derivative.

The impulse train is another well-known one, though I suppose someone will chime in here to rebut that it's arguably not a function.