Timeline for Counting square-free numbers smoothly
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2013 at 15:48 | answer | added | H A Helfgott | timeline score: 2 | |
| Jan 29, 2013 at 15:12 | comment | added | H A Helfgott | Yes, Greg - I was being careless. | |
| Jan 29, 2013 at 2:02 | comment | added | Greg Martin | When I do the inclusion-exclusion, I get a main term of $2x/\zeta(2)$ rather than $x/\zeta(2)$ - each of the ranges $n\le x$ and $n>x$ seems to contribute an $x/\zeta(2)$. Do you think that's right? | |
| Jan 28, 2013 at 21:50 | comment | added | H A Helfgott | And thanks to Barry for the reference. | |
| Jan 28, 2013 at 21:48 | comment | added | H A Helfgott | Micah - I think you are right in principle, but I am not sure that an explicit constant has been worked out in that case. Does the smoothing help at all, at least numerically? | |
| Jan 28, 2013 at 21:47 | comment | added | H A Helfgott | Er, yes, $1/x$ was a typo for $1/x^2$. | |
| Jan 28, 2013 at 21:46 | history | edited | H A Helfgott | CC BY-SA 3.0 |
blush
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| Jan 28, 2013 at 19:34 | comment | added | Barry Cipra | FWIW, there's an update to the Cohen-Dress paper (by Cohen, Dress, and El Marraki) available through ams.org/mathscinet-getitem?mr=2357309 | |
| Jan 28, 2013 at 18:31 | comment | added | anon | Perhaps we want f(t)=0 in the first case so the sum is always finite? | |
| Jan 28, 2013 at 16:48 | comment | added | Gerhard Paseman | I'm with km. Indeed, for any set of integers n of positive density, I see the desired sum over that set as infinite. Gerhard "Ask Me About Unbounded Confusion" Paseman, 2013.01.28 | |
| Jan 28, 2013 at 15:25 | comment | added | Micah Milinovich | Since the generating function for square-free numbers is $\zeta(s)/\zeta(2s)$, I believe you can get an error of $o(\sqrt{x})$ by using the classical zero-free region for the zeta-function (even without smoothing the sum). To get an error like $O(x^{1/2-\delta})$ for some $\delta>0$ seems more or less equivalent to a quasi-Riemann hypothesis. | |
| Jan 28, 2013 at 15:24 | comment | added | user25235 | There must be some typo since the sum in the question is $+ \infty$. | |
| Jan 28, 2013 at 14:53 | history | asked | H A Helfgott | CC BY-SA 3.0 |