Timeline for Equidistribution of $\{\sqrt{p}: p \text{ primes }\}$ modulo 1
Current License: CC BY-SA 4.0
11 events
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| Jul 28, 2020 at 21:17 | history | edited | Kyle Yip | CC BY-SA 4.0 |
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| Jul 28, 2020 at 21:11 | history | edited | Kyle Yip | CC BY-SA 4.0 |
added 24 characters in body
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| Jul 28, 2020 at 19:45 | comment | added | user44143 | For the first 100,000 primes, the first digit after the decimal point is distributed as $$\{\{0, 9904\}, \{1, 10060\}, \{2, 10042\}, \{3, 9941\}, \{4, 10059\},\\ \{5, 9967\}, \{6, 10069\}, \{7, 9973\}, \{8, 10065\}, \{9, 9920\}\}$$ according to Mathematica with 10 Mod[Sqrt[Prime[Range[10^5]]], 1] // Floor // Tally // Sort | |
| Jul 28, 2020 at 18:21 | history | edited | Kyle Yip | CC BY-SA 4.0 |
edited body
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| Jul 28, 2020 at 18:20 | comment | added | Kyle Yip | Oh yes, that's a typo. I have corrected that. | |
| Jul 28, 2020 at 17:00 | comment | added | mathworker21 | I still don't see how you got the bound $x^{1/2}(bc)^{-1}+bc$. Should the $\ll$ be $\ll_\alpha$? | |
| Jul 28, 2020 at 15:47 | comment | added | Kyle Yip | For fixed $b,c$, $$|\sum_{bcl \leq x} e(\alpha \sqrt{bcl})| \ll x^{1/2} (bc)^{-1} \alpha^{-1}+ \alpha^2 bc$$ follows easily from lemma 1. The sum is over $l \leq x/bc$, and the "$\alpha$" in lemma 1 will be replaced $\alpha \sqrt{bc}$. | |
| Jul 28, 2020 at 13:04 | comment | added | mathworker21 | i don't see how $y^2z^2$ comes from the $\alpha^2$ term from lemma 1. specifically, I don't see how you got $x^{1/2} (bc)^{-1}+bc$. | |
| Jul 28, 2020 at 5:43 | history | edited | Kyle Yip |
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| Jul 28, 2020 at 5:42 | review | First posts | |||
| Jul 28, 2020 at 5:49 | |||||
| Jul 28, 2020 at 5:37 | history | asked | Kyle Yip | CC BY-SA 4.0 |