Timeline for What can be proved about the Ramanujan conjecture using elementary means?
Current License: CC BY-SA 3.0
5 events
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| Mar 30, 2013 at 14:55 | comment | added | Garth Payne | @Will Sawin. @Douglas Zare. You can get exponent $\frac{23}{2}$ by taking into account that most of the coefficients are 0. Let $T_i(n)$ bound the number of coefficients in $c_{i,0},\ldots,c_{i,n}$ that are nonzero. Then $|\tau_a(n)| \le B_1 \ldots B_m T_2(n) \ldots T_{m}(n)$. ($T_1$ doesn't figure into it because the choice of $t_1$ is determined by the choices of the other $t_i$.) When the $S_i$ are as above, we can take $T_i(n)=c\sqrt n$ for $i=1,\ldots 24$ and $T_{25}(n)=1$. | |
| Mar 30, 2013 at 2:04 | comment | added | Garth Payne | @Will Sawin. I'm not sure what you're asking. For Gowers' weak form of the Ramanujan Conjecture I'm taking $S_1=S_2=\cdots=S_{24}=\prod_{n=1}^\infty(1-q^n)=1-q-q^2+q^5+q^7-\cdots$ and $S_{25}=q$. | |
| Mar 29, 2013 at 23:31 | comment | added | Will Sawin | So here the exponent is 24? | |
| Mar 29, 2013 at 23:29 | history | edited | Garth Payne | CC BY-SA 3.0 |
inserted "infinite"
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| Mar 29, 2013 at 22:30 | history | answered | Garth Payne | CC BY-SA 3.0 |