Timeline for Heuristic argument for the Riemann Hypothesis
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2019 at 13:42 | comment | added | Will Sawin | @user1728 If you follow the link, he's talking about the explicit formula for the sum of Mobius, not the explicit formula for prime counting. So I tried to answer for the Mobius function. I agree with everything you said about the primes/von Mangoldt and logarithmic derivatives. | |
| Sep 4, 2019 at 13:39 | comment | added | user1728 | @WillSawin do you mean the residue of $-(\zeta'(s)/\zeta(s))(x^s/s)$? Logarithmic derivatives of $\zeta(s)$ (or $L$-functions) are what naturally occur in standard explicit formulas, and the function $\zeta'(s)/\zeta(s)$ (like any logarithmic derivative) has at worst simple poles, with the residue of $\zeta'(s)/\zeta(s)$ at each number in $\mathbf C$ being the order of vanishing (positive, negative, or zero) of $\zeta(s)$ at that number, so logarithmic derivatives of meromorphic functions never have residue $0$ at a pole. | |
| Sep 4, 2019 at 3:47 | comment | added | Will Sawin | @Michael Yes to both questions. It's not the residue of $\zeta$ which matters, but the residue of $x^s/\zeta(s)$, which can only be zero for all $x$ if $\zeta$ is nonvanishing there. | |
| Sep 3, 2019 at 20:33 | comment | added | Michael | @Pace Nielsen, thanks for the nice link. Could you elaborate whether the argument in the link "If Riemann was right, there will be no zeros for Re(σ)>1/2, and hence we could choose σ to be as little as 1/2+ε" applies to zeros with multiplicity? If zeta function has zeros off the 1/2 line such that the residue of 1/zeta is zero, would that have an impact on prime distribution? | |
| Sep 1, 2019 at 16:21 | comment | added | R.. GitHub STOP HELPING ICE | @ChanBae: When P and Q are logical propositions, it may be a poor choice of wording to talk about "causation", but I don't think its use qualifies as a correlation/causation fallacy like Gerald seems to have implied. | |
| Sep 1, 2019 at 16:19 | comment | added | R.. GitHub STOP HELPING ICE | This borders on being a link-only answer, no? | |
| Sep 1, 2019 at 15:35 | comment | added | Timothy Chow | To elaborate a bit on the comment by GH from MO, each zero off the line yields an irregularity that makes itself felt asymptotically. It is not the case that asymptotically, we can ignore a finite number of zeros. Of course if the zero is very close to the line then it might not be detectable numerically until you've gone a long way out. | |
| Sep 1, 2019 at 5:56 | comment | added | Solveit | @Gerald Edgar, is there a principled definition of "P causes Q" when P and Q are logical propositions? | |
| Aug 31, 2019 at 13:02 | comment | added | Gerald Edgar | "Cause", eh? I have to warn students: "If P then Q" does not mean "P causes Q". | |
| Aug 31, 2019 at 7:55 | comment | added | GH from MO | @Mark: A single zeta zero with real part greater than $1/2$ would already cause big irregularities in the distribution of prime numbers. | |
| Aug 31, 2019 at 7:50 | comment | added | Mark Schultz-Wu | If one had all but finitely many zeros on the critical line would this still hold? | |
| Aug 31, 2019 at 7:39 | history | answered | Pace Nielsen | CC BY-SA 4.0 |