Timeline for Friedlander-Iwaniec Flipping moduli
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2018 at 19:43 | comment | added | literature-searcher | At the end of the day, you have to bound $$\sum_{c|r}\sum_{m|c} \Bigl({1\over\sqrt{cm}}+{m\over c^{3/2}}\Bigr)$$ in terms of $r$, and while I'm not sure why they just don't say (e.g) this is $\tau_3(r)$ via lazily turning the summand into $1$, the asserted bound of $\tau_2(r)\log R$ seems correct. | |
| Dec 18, 2018 at 18:40 | comment | added | Greg Martin | However, I see what you mean about the sums over $c$. Since $r_1$ and $r_2$ are close to $R$, their gcd $c$ should take values from $1$ up to $R$ish, meaning that $\sum_c c^{-1/2}$ looks to be about $R^{1/2}$ ... and a quick look through that section of the paper doesn't find anything to contradict this. So, I'm not being very helpful here, other than to say that I too don't see how this step goes yet. | |
| Dec 18, 2018 at 18:39 | comment | added | Greg Martin | Thank you for asking your question in such helpful detail! It's even slightly worse than you posit, because the $\ll$ bound for $V_{cm}(f,g)$ means that $\mu(m)$ must be replaced by $|\mu(m)|$ (hence effectively by $1$) when summing over $m$. Still, that part of the sum should be covered by the $\tau(r)$ in the eventual bound. | |
| Dec 18, 2018 at 17:10 | review | First posts | |||
| Dec 18, 2018 at 17:18 | |||||
| Dec 18, 2018 at 17:05 | history | asked | user133643 | CC BY-SA 4.0 |