3 ms·

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 rescali

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.

dataflow 2y ago
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.
mppm 2y ago
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?
- fxj 2y ago
  Yes the dirac comb for example. Actually there are infinitely many. https://en.wikipedia.org/wiki/Dirac_comb https://en.wikipedia.org/wiki/Dirac_comb and for other: http://www.systems.caltech.edu/dsp/ppv/papers/journal08post/PPVIETEeigenFT.pdf http://www.systems.caltech.edu/dsp/ppv/papers/journal08post/...
  - 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 ago
      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.
- perihelions 2y ago
  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.
  - kkylin 2y ago
    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.
    - deleted 2y ago
      [deleted]
JoshTriplett 2y ago
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.
- deleted 2y ago
  [deleted]

dataflow 2y ago

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.

mppm 2y ago

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?

fxj 2y ago
Yes the dirac comb for example. Actually there are infinitely many. https://en.wikipedia.org/wiki/Dirac_comb https://en.wikipedia.org/wiki/Dirac_comb and for other: http://www.systems.caltech.edu/dsp/ppv/papers/journal08post/PPVIETEeigenFT.pdf http://www.systems.caltech.edu/dsp/ppv/papers/journal08post/...
- 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 ago
    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.
perihelions 2y ago
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.
- kkylin 2y ago
  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.
  - deleted 2y ago
    [deleted]

JoshTriplett 2y ago

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.

deleted 2y ago
[deleted]