Timeline for Riemann–Von Mangoldt formula
Current License: CC BY-SA 4.0
39 events
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| Jun 15, 2023 at 17:13 | comment | added | Steven Clark | For example, see this WolframAlpha plot. | |
| Jun 15, 2023 at 17:02 | comment | added | Steven Clark | Exactly which formula did you plot? If you plot it further out I believe you'll find that it does not evaluate correctly (e.g. in the range $\left\{T,\frac{282.4}{2 \pi},\frac{282.5}{2 \pi}\right\}$ which is in the neighborhood of Gram point $g_{126}$ and zeta zero $ρ_{127}$ where the first violation of Gram's law occurs). | |
| Jul 15, 2022 at 21:44 | vote | accept | Jesse Elliott | ||
| Jul 15, 2022 at 21:44 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Jul 15, 2022 at 21:40 | history | undeleted | Jesse Elliott | ||
| Jul 15, 2022 at 21:39 | history | deleted | Jesse Elliott | via Vote | |
| Nov 17, 2021 at 22:15 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 17, 2021 at 22:09 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 7, 2021 at 0:13 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 6, 2021 at 23:45 | comment | added | Jesse Elliott | Oh, use the fact that the imaginary part of $\operatorname{Log}\zeta(1/2+iT)$ is differentiable between such ordinates. | |
| Nov 6, 2021 at 23:25 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 6, 2021 at 9:37 | comment | added | Anurag Sahay | If there's no zero in between the horizontal lines, there's no obstruction to shifting the defining contour around. In particular, that means the branch of the logarithm you're on doesn't change. | |
| Nov 6, 2021 at 9:17 | comment | added | Jesse Elliott | That makes sense as a definition. Thank you. Why is $S(T)$ continuous between such ordinates? | |
| Nov 6, 2021 at 9:10 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 6, 2021 at 7:53 | comment | added | Anurag Sahay | I might be remembering incorrectly, but $S(T)$ is typically defined by $\frac{S(T^+) + S(T^-)}{2}$ whenever $T$ is the ordinate of a zero (including those that are not on the critical line). Hopefully this answers your question about how the argument jumps around zeroes (as it should)? | |
| Nov 6, 2021 at 6:07 | comment | added | Jesse Elliott | OK, edited. I need to add an integer correction term $R(T)$ that is $O(\log T)$, and then the result is $\frac{1}{\pi}S(T)$. I still don't know how $\arg \zeta(x+iT)$ varies "continuously" as $x+iT$ passes over a potential zero of $\zeta(s)$ off the critical line (as $x$ passes from $2$ to $1/2$). Do you add $\pi$ or subtract $\pi$? Also, you have to divide those large $S(T)$ values by $\pi$ in my formula--they are just over $\pi$, which means the $R(T)$ values are probably $\pm 1$ in those cases, but Mathematica can't calculate out that far to confirm. I'd love to see a graph of $R(T)$. | |
| Nov 6, 2021 at 5:33 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 6, 2021 at 5:15 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 6, 2021 at 5:08 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 6, 2021 at 4:41 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 6, 2021 at 4:06 | comment | added | Lucia | If I understand you correctly, you would be maintaining that $|S(t)| \le 1$ --- $S(t)$ is $1/\pi$ times the argument as usually defined in the literature. The growth of this quantity is notoriously slow. But look at arxiv.org/pdf/1607.00709.pdf which gives examples of large values of $S(t)$: at the moment the numerical values are only about $3.3$ or so. Of course it is well understood that this gets arbitrarily large. | |
| Nov 6, 2021 at 3:14 | comment | added | Jesse Elliott | E.g, what's the smallest value of $T$ for which the arguments are different? | |
| Nov 6, 2021 at 3:12 | comment | added | Jesse Elliott | Or, as noted on p. 98, "or, equivalently, by continuous horizontal movement from $\infty+ iT$ to $1/2 + iT$, starting with the value $0$." What do you do if you hit a potential zero off the critical line? Please give me an example where it is not the principal value of $\arg \zeta(1/2+iT)$ where $T$ is not an imaginary part of a zero of $\zeta(s)$, and provide a reference that $\Delta_L \arg \zeta(s)$ is unbounded off those horizontal lines. A charge of "nonsense" requires a reference or a proof. | |
| Nov 6, 2021 at 3:02 | comment | added | Lucia | The argument is defined to be zero at $s=2$. Now consider the line joining $2$ to $2+iT$ and define the argument so that it is a continuous function on that line. Finally go from $2+iT$ to $1/2+iT$ on the horizontal line, again defining the argument so that it is continuous on that line segment. What you get will not in general be simply the principal value of $Arg \zeta(1/2+iT)$. Why don't you answer my question on what you meant? The argument defined in the correct way (as given here) is not a bounded function. | |
| Nov 6, 2021 at 2:47 | comment | added | Jesse Elliott | Please answer my question and write out an exact formula for $\Delta_L \operatorname{arg} \zeta(s)$, and please stop being disrespectful. I'm trying to figure out what Davenport means by his notation. | |
| Nov 6, 2021 at 2:33 | comment | added | Lucia | Just to understand what you're saying: are you claiming that the difference between $N(2\pi T)$ and $\theta(2\pi T)/\pi$ is bounded? That would be nonsense. | |
| Nov 6, 2021 at 2:21 | comment | added | Jesse Elliott | By the way, the black curve I plotted is exactly the formula for $N(2\pi T)$ I gave, and it checks out to as far as I can compute in Mathematica. And here is the non-arxiv version of the paper: iopscience.iop.org/article/10.1209/0295-5075/102/10006 | |
| Nov 6, 2021 at 1:45 | comment | added | Jesse Elliott | $\Delta_C \arg$ is not in Davenport's index of notation, and I can't find it defined anywhere. So when you say "read Davenport carefully," I say, either show me where he defines the notation or tell me how it is defined. I've never seen it anywhere else. | |
| Nov 6, 2021 at 1:35 | comment | added | Jesse Elliott | E.g., please write out an exact formula for $\Delta_L \arg \zeta(s)$, where $L$ is the line from $+\infty+iT$ to $1/2+iT$. | |
| Nov 6, 2021 at 1:33 | comment | added | Jesse Elliott | Obviously, the argument needs to jump up by $\pi$ for every zero ordinate on the critical line, but it also has to jump at potential zeros off of the critical line; so at least the Riemann Hypothesis should imply that my statement is correct, as I doubt that $\zeta(s+iT)$ has any winding numbers as $s \to 1/2$ from $\infty$ if $T$ is not an imaginary part of a zero of $\zeta(s)$. Is this wrong? I will continue to be mystified by Davenport's explanation until someone shows me the precise definition of $\Delta_C \arg$.... | |
| Nov 6, 2021 at 0:18 | comment | added | Jesse Elliott | Davenport says also by continuous variation from $+\infty+iT$ to $1/2+iT$, which should be the same as $\operatorname{Arg}\zeta(1/2+iT)$ as long as $T$ isn't the imaginary part of a zero of $\zeta(T)$. Can you give a counterexample to this claim? Also, Davenport doesn't mention the Riemann-Siegel theta function $\theta(t)$, so obviously he isn't the last word on the subject. Riemann never proved the asymptotic for $N(T)$, but that didn't detract from its being true. Obviously there's a lot more going on here than is noted in Davenport. | |
| Nov 6, 2021 at 0:14 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 6, 2021 at 0:03 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 5, 2021 at 23:45 | comment | added | Lucia | The argument here is defined by continuous variation along straight lines from $2$ to $2+iT$ to $1/2+it$. I don't know why you call this the principal branch. I would not rely on the arxiv paper by physicists, but please read Davenport carefully. | |
| Nov 5, 2021 at 23:38 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 5, 2021 at 23:33 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 5, 2021 at 23:07 | history | edited | Jesse Elliott | CC BY-SA 4.0 |
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| Nov 2, 2021 at 10:44 | vote | accept | Jesse Elliott | ||
| Nov 6, 2021 at 2:13 | |||||
| Nov 2, 2021 at 10:43 | history | answered | Jesse Elliott | CC BY-SA 4.0 |