Timeline for Möbius square root function: existence of multiplicative and bounded function
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20 at 14:03 | history | edited | Alexei Entin | CC BY-SA 4.0 |
added 123 characters in body
|
| Aug 20 at 13:55 | history | edited | Alexei Entin | CC BY-SA 4.0 |
added 25 characters in body
|
| Aug 20 at 13:53 | comment | added | Alexei Entin | I agree, my reduction doesn't work. I've updated my answer. | |
| Aug 20 at 13:48 | history | edited | Alexei Entin | CC BY-SA 4.0 |
added 381 characters in body
|
| Aug 20 at 13:43 | history | edited | Alexei Entin | CC BY-SA 4.0 |
added 381 characters in body
|
| Aug 20 at 12:29 | vote | accept | Virgile Dine | ||
| Aug 20 at 10:59 | comment | added | Will Sawin | @VirgileDine The reduction does not work, but the method of proof easily transfers. One only needs the analogue of Proposition 2.1 for arbitrary $S^1$-valued multiplicative functions, because the squareroot of Möbius is not pretentious as that would force Möbius itself to be pretentious. But one just has to copy the proof of Proposition 2.1 and insert absolute values and complex conjugates in appropriate places to establish this. | |
| Aug 20 at 10:40 | comment | added | Virgile Dine | I agree with @EmilJeřábek that you can't easily reduce the problem this way. | |
| Aug 20 at 10:38 | comment | added | Emil Jeřábek | Explicitly: let $f$ be a square root of $\mu$ as in the question, let $g$ be the multiplicative function such that $f(p)=\pm i\implies g(p)=\pm1$ for prime $p$, and for $j=0,1,2,3$, let $S_j(x)\in\mathbb Z$ be the sum of $g(n)$ over square-free $n\le x$ with $j\pmod4$ prime factors. Then the partial sums of $g$ are $\sum_jS_j(x)$, and the partial sums of $f$ are $\sum_ji^jS_j(x)=(S_0(x)-S_2(x))+i(S_1(x)-S_3(x))$. Thus, the assumption that $f$ has bounded sums means that $|S_0(x)-S_2(x)|$ and $|S_1(x)-S_3(x)|$ are bounded. How does this imply that $|(S_0(x)+S_2(x))+(S_1(x)+S_3(x))|$ is bounded? | |
| Aug 20 at 9:49 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
include a proper reference
|
| Aug 20 at 9:44 | comment | added | Emil Jeřábek | I don’t understand how the reduction works. If I replace each $\pm i$ with $\pm1$, and make it multiplicative, the sign of the function will be negated for square-free numbers with $2\pmod4$ prime factors, but stay the same for square-free numbers with $0\pmod4$ prime factors. Thus I do not see why this should preserve the property of bounded partial sums. | |
| Aug 20 at 9:34 | history | answered | Alexei Entin | CC BY-SA 4.0 |