Timeline for Improved upper bound for second moment of reduced residues modulo $q$?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 14, 2020 at 13:07 | comment | added | user45947 | Thanks for the effort! I'd be happy to accept that as an answer if you'd want to write it up as such. | |
| May 14, 2020 at 12:29 | comment | added | Ofir Gorodetsky | A heuristic for using $\{ \alpha \} (1-\{\alpha\}) \le \alpha$ is that $r^2/\phi(r)^2$ being large is correlated with $r$ being large which is correlated with $\{ h/r \}(1-\{h/r \})$ being of order $h/r$. | |
| May 14, 2020 at 12:27 | comment | added | Ofir Gorodetsky | Unfortunately, if you just use the fact that $\{ \alpha\}(1-\{\alpha\})$ is bounded, you do not obtain a superior upper bound. Indeed, assuming $q$ is squarefree (as we may), both $\mu(r)^2$ and $\prod_{p\mid q, \, p \nmid r} p(p-2)/(p-1)^2$ are $\asymp 1$, and your idea leads to an upper bound of order $qP^2 \sum_{r \mid q, \, r> 1} r^2/\phi(r)^2$, which is much larger qualitatively than $qPh$ for $q=\prod_{p \le x} p$ with $x \ge \log^2 h$... | |
| May 14, 2020 at 11:28 | comment | added | user45947 | Thanks for pointing to an important clarification! I've edited the question to focus on the case when $q=\prod_{p\leq h^C} p$ and $C<1$, as in this case the lower bound should be $M_2(q;h)\geq O(qhP^2)$ (is it a better or more correct way to write this inequality?). | |
| May 14, 2020 at 11:22 | history | edited | user45947 | CC BY-SA 4.0 |
Clarification of question
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| May 14, 2020 at 9:27 | comment | added | Ofir Gorodetsky | At least for some families of $q$, the upper bound is already optimal in the sense that there is a matching lower bound of the same order magnitude; this follows from the general lower bound given at the end of your linked question. To be concrete, taking $q = \prod_{p \le h^C} p$ for sufficiently large $C$, we have $M_2(q;h) \asymp qhP$. | |
| May 14, 2020 at 8:01 | history | asked | user45947 | CC BY-SA 4.0 |