Timeline for A reformulation of the Riemann Hypothesis
Current License: CC BY-SA 3.0
8 events
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| Jul 18, 2013 at 9:23 | history | edited | Subhajit Jana | CC BY-SA 3.0 |
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| Jul 10, 2013 at 22:27 | answer | added | Will Jagy | timeline score: 6 | |
| Jul 8, 2013 at 21:57 | comment | added | Subhajit Jana | I think that $|R(x,N)|\leq\tau(N)$ is a crude bound, as you are losing 'cancellation opportunity' by change of sign of $\mu(n)$. $|R(n,N)|=O(2^{\omega(N)})=O(2^{\log N})$ | |
| Jul 8, 2013 at 21:42 | comment | added | v08ltu | It seems to me that $|R(x,N)|\le \tau(N)$ by the first expression. So the $n^2$ in the denominator dominates, making only small $n$ relevant in $K(N)$. But for those, you can take $N$ a product of primes up to some point $P$, then $R(x,N)=x\prod_{p\le P}(1-1/p)-1$ for $x\le P$, which is essentially $x{e^{-\gamma}\over\log P}$ (say for $x\ge P/2$ and $P$ large). Dividing by $x$ and squaring in the $K(N)$ sum, truncating this sum at $P$, you have $K(N)\ge \sum_{x\le P} e^{-2\gamma}{1\over (\log P)^2}=e^{-2\gamma}{P\over (\log P)^2}$, so at least on these special $N$ the $K(N)$ must be big. | |
| Jul 8, 2013 at 19:49 | history | edited | Subhajit Jana | CC BY-SA 3.0 |
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| Jul 8, 2013 at 13:39 | history | edited | Subhajit Jana |
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| Jul 8, 2013 at 13:18 | review | First posts | |||
| Jul 8, 2013 at 13:28 | |||||
| Jul 8, 2013 at 13:00 | history | asked | Subhajit Jana | CC BY-SA 3.0 |