Timeline for Asymptotic behaviour of $K$-Bessel function in transition range
Current License: CC BY-SA 4.0
17 events
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| Sep 17, 2021 at 14:19 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Nov 5, 2014 at 14:52 | comment | added | GH from MO | @Kunnysan: I am glad that I could help! | |
| Nov 5, 2014 at 5:00 | comment | added | Subhajit Jana | Thank you very much for such an illuminating discussion! | |
| Nov 5, 2014 at 3:16 | comment | added | GH from MO | @Kunnysan: There are other techniques for general groups, but the resulting bounds are weaker than $(*)$ for $N$ small. See for example Theorem 3.2 in Iwaniec's book. In particular, this theorem implies that for $N>r$ we can remove $r^\epsilon$ in $(*)$. | |
| Nov 5, 2014 at 2:16 | comment | added | Subhajit Jana | Or in other words how do you estimate sum of Fourier coefficients for groups where you don't have nice Hecke relation like $\rho(n)=\rho(1)\lambda(n)$? | |
| Nov 5, 2014 at 0:41 | comment | added | Subhajit Jana | So the only way to find uniform estimate of partial sum of Fourier coefficients is to go through Hecke eigenvalue relation with Fourier coefficient? Is there any way to find the estimate directly, without using Ramanujan on average? | |
| Nov 5, 2014 at 0:15 | comment | added | GH from MO | @Kunnysan: One more small comment. In fact $(*)$ is as strong as Corollary 0.3 in Hoffstein-Lockhart, since for $N=2$ it yields $|\rho(1)|^2\ll e^{\pi r}r^\epsilon$. This also shows that without $r^\epsilon$ the bound $(*)$ would be too strong to be true. | |
| Nov 5, 2014 at 0:10 | comment | added | GH from MO | @Kunnysan: (8.15) says that the sum in $(*)$ is asymptotically an absolute constant times $e^{\pi r}N$. There is no explicit error term there, i.e. it does not tell us anything about the size of the sum for small $N$, say for $N<r$. In particular, (8.15) does not yield that the sum in $(*)$ is $\ll e^{\pi r}N$. Compare this with (8.17) in Iwaniec's book and plug in $X=|s_j|$ there. You will see that the error term in (8.17) exceeds the main term provided by the asymptotic formula. The value of $(*)$ is that it provides a strong uniform upper bound, even when $N$ is very small compared to $r$. | |
| Nov 4, 2014 at 21:44 | comment | added | Subhajit Jana | I am sorry, but I am really confused. Equation (8.15) of Iwaniec's 'Spectral Methods of Automorphic forms' does not have $r^\epsilon$ term. It is, in our language,$$N^2\mathrm{res}_{s=1} L(s,\phi\times\phi)$$ and the residue is of size $e^{\pi r}$. | |
| Nov 4, 2014 at 19:57 | vote | accept | Subhajit Jana | ||
| Nov 4, 2014 at 16:45 | comment | added | GH from MO | @Kunnysan: In $(*)$ we have $|\rho(n)|^2=|\rho(1)|^2.|\lambda(|n|)|^2$, where $\lambda(n)$ is the $n$-the Hecke eigenvalue. Hence $(*)$ follows from two bounds: one for $|\rho(1)|^2$, and one for $\sum_{n=1}^N |\lambda(n)|^2$. The bound for $|\rho(1)|^2$ is $e^{\pi r}r^\epsilon$, as follows from Corollary 0.3 in Hoffstein-Lockhart (note their different normalization of $\rho(n)$). The bound for $\sum_{n=1}^N |\lambda(n)|^2$ is $r^\epsilon N$, and this is due to Iwaniec. So there are two sources of $r^\epsilon$ in $(*)$. | |
| Nov 4, 2014 at 4:18 | comment | added | Subhajit Jana | Thank you very much for a detailed answer. One confusion is: how did you get $r^\epsilon$ term in (*)? Don't Rankin-Selberg and Hoffstein-Lockhart give $e^{\pi r}N$? | |
| Nov 4, 2014 at 2:39 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 4, 2014 at 2:32 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 4, 2014 at 2:23 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 4, 2014 at 2:17 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Nov 4, 2014 at 1:58 | history | answered | GH from MO | CC BY-SA 3.0 |