Timeline for Proof of a Fourier pair with Bessel functions?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 17, 2015 at 14:36 | comment | added | Pavel | I was surprised to see Mathematica 10.2 cannot compute this integral. | |
| Aug 15, 2015 at 10:22 | comment | added | Pavel | Many thanks to @Terry Tao. Computing the FT of a 4-dim sphere in two ways does give the desired relation. Using another dimension gives the relation for other c. They are related by dim=2c+c. Pavel | |
| S Aug 10, 2015 at 9:19 | history | bounty ended | Igor Khavkine | ||
| S Aug 10, 2015 at 9:19 | history | notice removed | Igor Khavkine | ||
| Aug 8, 2015 at 15:38 | answer | added | Johannes Trost | timeline score: 2 | |
| Aug 4, 2015 at 21:37 | comment | added | Suvrit | The integral representation $J_\nu(z) = \frac{(z/2)^\nu}{\Gamma(\nu+\frac12)\sqrt{\pi}}\int_{-1}^1 e^{izt}(1-t^2)^{\nu-\frac12}dt$ may be useful it seems.... | |
| S Aug 4, 2015 at 20:51 | history | bounty started | Igor Khavkine | ||
| S Aug 4, 2015 at 20:51 | history | notice added | Igor Khavkine | Draw attention | |
| S Aug 4, 2015 at 11:32 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Added tag.
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| Aug 4, 2015 at 11:00 | review | Suggested edits | |||
| S Aug 4, 2015 at 11:32 | |||||
| Aug 3, 2015 at 4:49 | comment | added | Terry Tao | Actually, I think the $c=1$ case might be obtainable by computing the Fourier transform of surface measure on the sphere $\{ (z,x) \in {\bf R}^4 \times {\bf R}: |z|^2 + x^2 = a^2 \}$ at $(b,0,0,0,y)$ in two ways: (i) by first taking Fourier transform in the z variable, and then in the x variable; (ii) by using spherical symmetry to replace $(b,0,0,0,y)$ with $(0,0,0,0,\sqrt{b^2+y^2})$ and then using cylindrical coordinates. | |
| Aug 3, 2015 at 4:38 | comment | added | Terry Tao | Another approach would be to multiply both $f$ and $\hat f$ by $t^c$ and sum over natural number $c$, using the generating function for the Bessel function; this should reduce matters (formally at least) to a simpler identity, at least for the case of natural number $c$ which is what the OP wants. | |
| Aug 3, 2015 at 4:33 | comment | added | Terry Tao | Perhaps the OP may wish to share some initial attempts to resolve the problem? It looks likely that one can rescale one of $a$ or $b$ to equal $1$, although this only achieves a modest simplification. One approach would be to use the Bessel equation to work out the ODE that $f$ and the claimed value of $\hat f$ satisfy; if these Fourier transform to each other, and if one can verify suitable boundary conditions at infinity, one should be done. | |
| Aug 2, 2015 at 18:10 | comment | added | Igor Khavkine | The question is asking for a proof of what seems like a non-trivial identity. Those voting to close because the question is "too trivial" are invited to make that triviality manifest by at least pointing to a semblance of an answer. | |
| Aug 2, 2015 at 15:47 | history | edited | Joonas Ilmavirta | CC BY-SA 3.0 |
Reformatted and retagged the question to make it work better at MSE.
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| Aug 2, 2015 at 10:09 | review | Close votes | |||
| Aug 3, 2015 at 17:13 | |||||
| Aug 2, 2015 at 9:24 | review | First posts | |||
| Aug 2, 2015 at 9:53 | |||||
| Aug 2, 2015 at 9:24 | history | asked | Pavel | CC BY-SA 3.0 |