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We know that a periodic function (e.g. a trigonometric function) has the property

$$ f(x+n\Lambda)=f(x) \qquad n\in\mathbb Z $$

A Bessel function is not exactly periodic, because the value of the function roughly decreases after each oscillation. However, one could say that is not very far from being periodic. I would like to know if it is possible to express this almost periodicity of Bessel functions, generalizing the above formula.

Would it be possible to expand such an almost periodic function in a generalized Fourier series?

More in detail, is it legitimate to write the following relation?

$$ J_{0}(kz)\sim\sum_{m=-\infty}^{+\infty}\varphi_m e^{i\lambda_mz} $$

where the approximated expansion holds in an interval centered around $z=0$ and extends for a few quasi-periods.

If so, how is $\lambda_m$ calculated?

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  • $\begingroup$ how would that equation make sense, since the right-hand-side is periodic, while the left-hand-side is not? $\endgroup$ – Carlo Beenakker Apr 8 at 10:56
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    $\begingroup$ 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. $\endgroup$ – Alexandre Eremenko Apr 8 at 13:16
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    $\begingroup$ 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? $\endgroup$ – Igor Khavkine Apr 8 at 13:29
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    $\begingroup$ 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. $\endgroup$ – Igor Khavkine Apr 8 at 14:11
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    $\begingroup$ @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. $\endgroup$ – Alexandre Eremenko Apr 9 at 0:43
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Maple gave me this... $$ J_0(x) = \left( {\frac {\sin \left( x \right) }{\sqrt {\pi}}}+{\frac {\cos \left( x \right) }{\sqrt {\pi}}} \right) x^{-1/2}+ \left( -{\frac {\cos \left( x \right) }{8\sqrt {\pi}}}+{\frac {\sin \left( x \right) }{8\sqrt {\pi}}} \right) x^{-3/2} \\+ \left( -{\frac {9\,\sin \left( x \right) }{128\,\sqrt {\pi}}}-{ \frac {9\,\cos \left( x \right) }{128\,\sqrt {\pi}}} \right) x^{-5/2}+ \left( {\frac {75\,\cos \left( x \right) }{ 1024\,\sqrt {\pi}}}-{\frac {75\,\sin \left( x \right) }{1024\,\sqrt { \pi}}} \right) x^{-7/2} \\+ \left( {\frac {3675\, \sin \left( x \right) }{32768\,\sqrt {\pi}}}+{\frac {3675\,\cos \left( x \right) }{32768\,\sqrt {\pi}}} \right) x^{-9/2} \\+ \left( -{\frac {59535\,\cos \left( x \right) }{262144 \,\sqrt {\pi}}}+{\frac {59535\,\sin \left( x \right) }{262144\,\sqrt { \pi}}} \right) x^{-11/2}+o(x^{-11/2}) $$ as $x \to \infty$

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  • $\begingroup$ Thank you very much. Your Maple result looks nice, but I am looking for something like a Generalized Fourier series, as those of Besicovitch almost periodic functions (en.wikipedia.org/wiki/…). The multiplication term $x^{-1/2-n}$ looks like a big departure from an expansion in "discrete frequencies". $\endgroup$ – Alessandro Zunino Apr 8 at 11:48
  • $\begingroup$ @AlessandroZunino Besicovitch AP seems to be a red herring if you merely want an approximation in some interval around $x=0$. As pointed out above, AP functions in the sense of Bohr cannot tend to zero at infinity, and I think the same also applies to the Besicovitch $B_p$-spaces $\endgroup$ – Yemon Choi Apr 8 at 21:04

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