Timeline for On the equation $\zeta(s) = F(s)+F(s+1)$
Current License: CC BY-SA 4.0
26 events
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| Aug 29, 2022 at 9:06 | comment | added | Gui | @DanRomik Of course it's not the only way for $\zeta$ to vanish but I assume the intuition that $F$ share the same zeros as $\zeta$ would be the only way to say something meaningful about the zeros of $\zeta$. But it is of course possible that I'm wrong. | |
| Aug 29, 2022 at 1:09 | comment | added | Dan Romik | @Gui your other claim about $1-\rho$, $2-\rho$ and $1+\rho$ being zeros if $\rho$ is a zero also seems unjustified and probably wrong. | |
| Aug 29, 2022 at 1:06 | comment | added | Dan Romik | @Gui I still don’t see why $F(s)$ should have infinitely many zeros on the lines $\operatorname{Re}(s)=1/2, 3/2$. Are you implicitly assuming that the only way for $\zeta(s)$ to be 0 is if both $F(s)$ and $F(s+1)$ are 0? | |
| Aug 29, 2022 at 0:35 | history | edited | Gui | CC BY-SA 4.0 |
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| S Aug 29, 2022 at 0:29 | history | suggested | Buzz | CC BY-SA 4.0 |
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| Aug 29, 2022 at 0:27 | review | Suggested edits | |||
| S Aug 29, 2022 at 0:29 | |||||
| Aug 28, 2022 at 23:57 | history | edited | Gui | CC BY-SA 4.0 |
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| Aug 28, 2022 at 23:50 | comment | added | Gui | They are not, I edited my post in order to avoid any bad speculation. | |
| Aug 28, 2022 at 23:49 | history | edited | Gui | CC BY-SA 4.0 |
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| Aug 28, 2022 at 23:49 | comment | added | Dan Romik | Can you justify the claims you make in the last two paragraphs of your answer (current edit?)? I don’t see your basis for believing such statements to be true, and I’d guess that they are not correct. | |
| Aug 28, 2022 at 23:47 | history | edited | Gui | CC BY-SA 4.0 |
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| Aug 28, 2022 at 23:41 | history | edited | Gui | CC BY-SA 4.0 |
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| S Aug 28, 2022 at 23:30 | review | First answers | |||
| Aug 29, 2022 at 0:44 | |||||
| S Aug 28, 2022 at 23:30 | history | edited | Gui | CC BY-SA 4.0 |
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| Aug 28, 2022 at 23:20 | history | edited | Gui | CC BY-SA 4.0 |
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| Aug 28, 2022 at 23:07 | comment | added | Gui | It is, I'm not used to Latex sorry! | |
| Aug 28, 2022 at 23:06 | comment | added | paul garrett | @Guigui, wouldn't $>=$ be better for an inequality...? Or, just use MathJax/(La)TeX? :) | |
| Aug 28, 2022 at 23:01 | history | edited | Gui | CC BY-SA 4.0 |
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| Aug 28, 2022 at 22:50 | comment | added | Gui | But the statement about the zero of $F$ remains valid if your equation holds for at least $re(s)=>1/2$. If not, I'm not sure how you can deduce anything for the zero of $\zeta$ starting from $F$ (the $=>$ means greater or equal). | |
| S Aug 28, 2022 at 22:38 | review | First answers | |||
| Aug 28, 2022 at 23:26 | |||||
| S Aug 28, 2022 at 22:38 | history | edited | Gui | CC BY-SA 4.0 |
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| Aug 28, 2022 at 22:36 | comment | added | Gui | My functional equation was pure speculation. The main idea is that you should try to study $Q(x)$ in order to find some identity like the one for $\psi$. If such identity exist (you could start with Poisson formula) you might have a proof for the functional equation of $F$ that looks like the proof for the functional equation of $\zeta$. | |
| Aug 28, 2022 at 18:18 | comment | added | Dan Romik | Interesting, thanks. How does your suggested functional equation $q(s-1)F(s)=q(1-s)F(2-s)$ relate to the equation $q(s)(F(s)+F(s+1))=q(1-s)(F(1-s)+F(2-s))$? In any case, I tested your equation numerically and it does not seem to be satisfied unfortunately. | |
| Aug 28, 2022 at 14:02 | history | edited | Gui | CC BY-SA 4.0 |
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| S Aug 28, 2022 at 13:50 | review | First answers | |||
| Aug 28, 2022 at 15:15 | |||||
| S Aug 28, 2022 at 13:50 | history | answered | Gui | CC BY-SA 4.0 |