Timeline for Regularized sums of Mobius sequence
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 2, 2015 at 18:14 | history | edited | Lucia | CC BY-SA 3.0 |
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| Mar 2, 2015 at 16:23 | comment | added | GH from MO | @JamesPropp: From Lucia's response it follows that $M(x)$ is the inverse Mellin transform of $\Gamma(s)/\zeta(s)$. Hence, by the residue theorem, $M(x)$ can be approximated well by a large finite sum of the sum of residues of $x^{-s}\Gamma(s)/\zeta(s)$ at the zeta-zeros. So I don't think there is an asymptotic expansion of $M(x)$, it is more like an infinite sum of winding terms each of size $x^{-1/2}$ and its fluctuations are nontrivial to analyze. This is a similar quantity as $\pi(x)-\mathrm{li}(x)$ or $\psi(x)-x$ in prime number theory. It would be interesting to explain your $-2$ though. | |
| Mar 2, 2015 at 14:42 | comment | added | James Propp | Is there some sort of asymptotic expansion for $M(x)$ in which (a) the constant term is $-2$, (b) the other terms correspond to roots of $\zeta$ on the critical line, and (c) the constant term dominates the other terms when $x >> 10^{-10}$? That would explain why it misleadingly seems to be converging to $-2$, and would reinforce Lucia's point that the zeroes are the reason for the non-convergence. | |
| Mar 2, 2015 at 10:13 | comment | added | GH from MO | @JamesPropp: I think Lucia answered all your questions. In particular, the convergence of $\sum\mu(n)/n^s$ for $s>1/2$ is equivalent to the Riemann Hypothesis, while the series is known to diverge for $s\leq 1/2$. See mathoverflow.net/questions/164874/… | |
| Mar 2, 2015 at 10:05 | comment | added | GH from MO | @JamesPropp: By Stirling's formula, $\Gamma(1/2+it)$ is much like $\sqrt{2\pi}e^{-\pi t/2}$, so one can check the $10^{-10}$ order even with a simple calculator. | |
| Mar 2, 2015 at 0:28 | comment | added | James Propp | For anyone who's interested in knowing just how small $\Gamma(1/2+14.1\dots i)$ is, Mathematica reports that its real and imaginary parts are on the order of $10^{-10}$. | |
| Mar 2, 2015 at 0:25 | comment | added | James Propp | @Lucia: I think your response settles my first two questions, but what about the third? I know that the Dirichlet series converges when Re $s > 1$, but I don't know whether the full domain of convergence includes some values of $s$ with Re $s \leq 1$ (such as real numbers strictly between 0 and 1). | |
| Mar 1, 2015 at 23:38 | comment | added | Lucia | @JoeSilverman: I added a clarification above. | |
| Mar 1, 2015 at 23:37 | history | edited | Lucia | CC BY-SA 3.0 |
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| Mar 1, 2015 at 23:33 | comment | added | Joe Silverman | I'm sure I'm missing something, but why does $\int_0^\infty x^s\,\frac{dx}{x}$ converge "a priori" for Re$(s)>1$? For, say, $s=2$, certainly $\int_0^\infty x\,dx$ doesn't converge. | |
| Mar 1, 2015 at 22:39 | history | answered | Lucia | CC BY-SA 3.0 |