Timeline for Optimality of the Riemann Hypothesis
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2021 at 14:29 | history | edited | Uzu Lim | CC BY-SA 4.0 |
added 21 characters in body
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| Apr 24, 2021 at 9:32 | vote | accept | Uzu Lim | ||
| Apr 24, 2021 at 1:43 | comment | added | KConrad | If you seek a bound on $\pi(x) - {\rm Li}(x)$, Monach and Montgomery conjectured that $\varliminf_{x \rightarrow \infty} (\pi(x)-{\rm Li}(x))/(\sqrt{x}(\log\log\log x)^2/\log x) = -1/(2\pi)$ and $\varlimsup_{x \rightarrow \infty} (\pi(x)-{\rm Li}(x))/(\sqrt{x}(\log\log\log x)^2/\log x) = 1/(2\pi)$. If that's true, then $\pi(x) - {\rm Li}(x) = O(\sqrt{x}(\log\log\log x)^2/\log x) = o(\sqrt{x})$, so in particular $\pi(x) - {\rm Li}(x) = O(\sqrt{x})$. That nobody has proved RH implies a better bound than $\pi(x) - {\rm Li}(x) = O(\sqrt{x}\log x)$ probably just reflects inadequate techniques. | |
| Apr 24, 2021 at 1:35 | comment | added | KConrad | That nobody has proved RH implies a sharper error term on $\psi(x) - x$ for all $x \geq 2$ than $O(\sqrt{x}(\log x)^2)$ could be considered more a matter of inadequate techniques than of the error term $O(\sqrt{x}(\log x)^2)$ being believed to be genuinely optimal as far as the factor $(\log x)^2$ factor is concerned. For instance, Gallagher showed RH implies $\psi(x) - x = O(\sqrt{x}(\log\log x)^2)$ outside a subset $E$ of $[3,\infty)$ with finite logarithmic measure, meaning the exceptions to such a bound are a subset $E$ such that $\int_E dx/x < \infty$. Perhaps $E$ is empty. | |
| Apr 24, 2021 at 1:31 | comment | added | KConrad | Oops, I means $O(\sqrt{x}(\log x)^2)$. Anyway, $\psi(x) - x \not= O(\sqrt{x})$ since Littlewood showed in 1914 that $\varliminf_{x \to \infty} (\psi(x) - x)/(\sqrt{x}\log\log\log x)< 0$ and $\varlimsup_{x \to \infty} (\psi(x) - x)/(\sqrt{x}\log\log\log x) > 0$. Monach and Montgomery conjectured that $\varliminf_{x \to \infty} (\psi(x) - x)/(\sqrt{x}(\log\log\log)^2) = -1/(2\pi)$ and $\varlimsup_{x \to \infty} (\psi(x) - x)/(\sqrt{x}(\log\log\log x)^2) = 1/(2\pi)$, which if true means the "correct" term to multiply by $\sqrt{x}$ in the bound on $\psi(x) - x$ is $(\log\log\log x)^2$. | |
| Apr 24, 2021 at 1:23 | comment | added | KConrad | The main procedural error you are making is using the explicit formula in its full form with no error term. You wind up wanting to estimate $|\sum_\rho x^\rho/\rho|$. However, that series (which sums over the nontrivial zeros $\rho$ with multiplicity) does not converge absolutely and there is no easy way to to estimate it directly. You need to use a truncated explicit formula, summing over nontrivial zeros $\rho$ with $|{\rm Im}(\rho)| \leq T$ and adding an error term on the right side that depends on $x$ and $T$. When $T = x$ you'll get the standard error term $O(\sqrt{x}\log x)$. | |
| Apr 23, 2021 at 17:44 | history | became hot network question | |||
| Apr 23, 2021 at 10:35 | comment | added | GH from MO | @OfirGorodetsky: You beat me by 3 minutes :-) | |
| Apr 23, 2021 at 10:23 | vote | accept | Uzu Lim | ||
| Apr 23, 2021 at 10:23 | |||||
| Apr 23, 2021 at 10:11 | comment | added | Uzu Lim | Thanks. I'll read the reference, but do you know whether the Littlewood's theorem translates to an impossibility of a power bound of $|\pi(x) - Li(x)| = O(x^\alpha)$ with $\alpha \leq \frac12$? | |
| Apr 23, 2021 at 10:10 | answer | added | GH from MO | timeline score: 17 | |
| Apr 23, 2021 at 10:07 | comment | added | Ofir Gorodetsky | To follow-up on Wojowu's comment: a concrete reference is Montgomery and Vaughan's book "Multiplicative Number Theory I". Littlewood's oscillation theorem is given there as Theorem 15.11 (page 478). Also see their Corollary 15.4 (weaker version of Littlewood's theorem). | |
| Apr 23, 2021 at 9:59 | comment | added | Wojowu | It's not quite known to be the best, but it is know that it cannot be much better than that. Look up Littlewood's oscillation theorem. | |
| Apr 23, 2021 at 9:40 | history | asked | Uzu Lim | CC BY-SA 4.0 |