Timeline for Is there real or complex analytic function whose positive real zeros are the primes?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2022 at 16:25 | comment | added | TravorLZH | @joro That infinite product expansion follows from Hadamard's factorization theorem, which allows you to construct entire functions with prescribed zeros. | |
| Oct 19, 2022 at 10:56 | history | became hot network question | |||
| Oct 18, 2022 at 21:29 | answer | added | juan | timeline score: 16 | |
| Oct 18, 2022 at 15:25 | comment | added | joro | @PietroMajer Has your function from the comment been studied? Experimentally it is heavily positive biased. | |
| Oct 18, 2022 at 13:36 | history | edited | joro | CC BY-SA 4.0 |
added 8 characters in body
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| Oct 18, 2022 at 12:36 | comment | added | joro | @PietroMajer Yes, I meant this. I asked why such closed form exists, if we don't have other constraints on $f$. | |
| Oct 18, 2022 at 12:31 | comment | added | Pietro Majer | Can you clarify the question? Certainly for positive real $x$ this $P(x)$ vanishes only if $x$ is an integer and $x$ divides $(x-1)!+1$, which by Wilson's theorem is true exactly for primes $x$ (as said in the linked question) | |
| Oct 18, 2022 at 12:19 | comment | added | joro | @PietroMajer Thanks. If we drop the analytic constraint, how do you explain the elementary form of $P(x)$? | |
| Oct 18, 2022 at 12:15 | comment | added | Pietro Majer | An entire function whose zeros are the primes should be $\prod_{n\ge1}\big(1-\frac z{p_n}\big)\exp\big(\frac z{p_n}\big)$, but I don't think it is related to any elementary function. | |
| Oct 18, 2022 at 11:44 | history | asked | joro | CC BY-SA 4.0 |