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Jun 24, 2020 at 17:43 comment added Nick S Functions vanishing at infinity fall into the class of null weakly almsot periodic functions. These are exactly the weakly almost periodic functions with vanishing Fourier--Bohr coefficients. In particular, in the sense of Besicovitch or Weyl almsot periodicity, they are the same as the 0 function. So Besicovitch or Weyl almost periodicity will definitely NOT give you the coefficients. @YemonChoi
Apr 9, 2020 at 0:43 comment added Alexandre Eremenko @Alessandro Zunino: On any finite interval, you can approximate any reasonable function by a partial sum of its Fourier series. Or a series of exponentials provided that they are complete on this interval.
Apr 8, 2020 at 21:06 comment added Yemon Choi It would be helpful if, in the body of the question, you explain which notion of "almost periodic function" you are hoping to use. There are several variants in the literature, but my impression is that they all exclude non-zero $C_0$ functions as remarked by @AlexandreEremenko
Apr 8, 2020 at 18:30 history became hot network question
Apr 8, 2020 at 14:11 comment added Igor Khavkine Alessandro, do you need this interval to be "large" in any way? If not, fix your favorite interval, then compute the usual Fourier series on that interval.
Apr 8, 2020 at 13:43 history edited DrManhattan CC BY-SA 4.0
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Apr 8, 2020 at 13:39 comment added DrManhattan I want to thank both and @IgorKhavkine and Alexandre Eremenko for their very useful comments. Indeed, the generalized Fourier series I am looking for does not need to represent the function $J_0(x)$ over all $\mathbb R$. It is enough that the series approximates well the Bessel function in an interval centered around $x=0$ (let's say for a few "quasi-periods").
Apr 8, 2020 at 13:35 history edited DrManhattan CC BY-SA 4.0
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Apr 8, 2020 at 13:29 comment added Igor Khavkine I think this question could make sense in the following way. Is there an almost periodic function that asymptotically approximates $x^{1/2} J_0(x)$ with better error than $O(1/x)$, which is already accomplished by $(\sin x + \cos x)/\sqrt{\pi}$? If not, could this error be improved on some interval that doesn't extend all the way to infinity, but still gets asymptotically large?
Apr 8, 2020 at 13:16 comment added Alexandre Eremenko Almost periodic function cannot tend to zero as $x\to\infty$. The series you wrote never tends to zero as $z\to\infty$ on the real line. And Bessel functions do.
Apr 8, 2020 at 11:58 history edited DrManhattan CC BY-SA 4.0
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Apr 8, 2020 at 11:52 history edited DrManhattan CC BY-SA 4.0
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Apr 8, 2020 at 11:26 history edited YCor CC BY-SA 4.0
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Apr 8, 2020 at 11:21 answer added Gerald Edgar timeline score: 4
Apr 8, 2020 at 11:15 review Close votes
Apr 9, 2020 at 11:11
Apr 8, 2020 at 11:08 comment added DrManhattan @GeraldEdgar is any expansion in discrete "frequencies" possible, for a Bessel function? Even if it is not exactly a Fourier series.
Apr 8, 2020 at 11:04 comment added Gerald Edgar $J_{1/2}$ is a trigonometric function. But (as you note) $J_0$ decays to $0$, so it is not of the form you wrote.
Apr 8, 2020 at 11:00 history edited DrManhattan CC BY-SA 4.0
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Apr 8, 2020 at 11:00 comment added DrManhattan @CarloBeenakker since I suspect that Bessel functions are almost-periodic, I would be satisfied even with an approximated relation.
Apr 8, 2020 at 10:56 comment added Carlo Beenakker how would that equation make sense, since the right-hand-side is periodic, while the left-hand-side is not?
Apr 8, 2020 at 10:30 review First posts
Apr 8, 2020 at 10:55
Apr 8, 2020 at 10:29 history asked DrManhattan CC BY-SA 4.0