Timeline for Möbius square root function: existence of multiplicative and bounded function
Current License: CC BY-SA 4.0
10 events
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| Aug 20 at 15:45 | history | became hot network question | |||
| Aug 20 at 14:46 | comment | added | Noam D. Elkies | Certainly multiplicativity does not imply boundedness. Given $x$, once you've chosen $\sqrt\mu(p)$ for each prime $p \leq x/2$, you have determined $\sum_{n\leq x} \sqrt\mu(n)$ except for the $\pi(x) - \pi(x/2) \sim x / (2 \log x)$ terms $\sqrt\mu(p)$ with $x/2 < p \leq x$, each of which you can choose to be either $+i$ or $-i$. So in the worst case the sum can grow at least as fast as a positive multiple of $x / \log x$. | |
| Aug 20 at 12:29 | vote | accept | Virgile Dine | ||
| Aug 20 at 10:45 | history | edited | Virgile Dine | CC BY-SA 4.0 |
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| Aug 20 at 9:37 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
grammar and typographical corrections
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| Aug 20 at 9:34 | answer | added | Alexei Entin | timeline score: 8 | |
| Aug 20 at 9:30 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Aug 20 at 8:32 | comment | added | GH from MO | Please use a high-level tag like "nt.number-theory". I added this tag now. I also removed the tag "algebraic-number-theory" as irrelevant. Regarding high-level tags, see meta.mathoverflow.net/q/1075 | |
| Aug 20 at 8:31 | history | edited | GH from MO |
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| Aug 20 at 7:44 | history | asked | Virgile Dine | CC BY-SA 4.0 |