Timeline for Possible regularisation for sum of function of primes
Current License: CC BY-SA 4.0
20 events
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| Feb 23, 2023 at 6:53 | history | edited | Zaza | CC BY-SA 4.0 |
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| Feb 6, 2023 at 13:46 | history | edited | Zaza | CC BY-SA 4.0 |
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| Jan 28, 2023 at 0:14 | history | edited | Zaza | CC BY-SA 4.0 |
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| Jan 14, 2023 at 8:13 | comment | added | Gottfried Helms | @SylvainJULIEN : like "Oil-eir" (second syllable might have a better representation; likely in WP there is a better one) | |
| Jan 14, 2023 at 6:12 | history | edited | Zaza | CC BY-SA 4.0 |
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| Dec 31, 2022 at 16:07 | history | edited | Zaza | CC BY-SA 4.0 |
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| Dec 30, 2022 at 9:06 | comment | added | Zaza | @KConrad Regularisation as in 'Zeta Function Regularisation'. | |
| Dec 30, 2022 at 5:40 | comment | added | Zaza | @SylvainJULIEN thank you for the comment. Tried to regularize by classical methods in literature but couldn't do it . ( Can't derive a unique numerical value) | |
| Dec 29, 2022 at 19:56 | comment | added | Sylvain JULIEN | Like the infinite sum of all the negative integers is regularized by $\zeta(-1)=-1/12$. Something along these lines may be possible seeing the expression in your first logarithm as a factor of a Euler product (not sure of how his name is pronounced in English, is the first sound in it vocalic or consonantic?). | |
| Dec 29, 2022 at 18:35 | comment | added | Zaza | @KConrad same meaning, sir! To assign a finite value to an infinite/divergent sum | |
| Dec 29, 2022 at 18:13 | comment | added | KConrad | What do you mean by the term “regularization”? I thought it meant “find a divergent expression so that the difference with that converges”. Maybe you mean something else. | |
| Dec 29, 2022 at 17:54 | comment | added | Zaza | @KConrad again thank you for the comment. I'm not that used to MathOverflow site so i don't know how it works. I thought the divergence is elementary for number theorist so i didn't gave clarification. | |
| Dec 29, 2022 at 17:54 | comment | added | Zaza | @KConrad again thank you for the comment. I'm not that used to MathOverflow site so i don't know how it works. I thought the divergence is elementary for number theorist so i didn't gave clarification! But Regularisation seems much more tough | |
| Dec 29, 2022 at 17:30 | comment | added | KConrad | I did not downvote. The only thing “wrong” in the question is that it lacks context: why are you interested in that strange-looking sum? If you want people to see previous posts for that context, include links to them in your question. | |
| Dec 29, 2022 at 17:27 | comment | added | Zaza | If something's wrong with the question, please mention. I couldn't understand downvotes. | |
| Dec 29, 2022 at 17:26 | comment | added | Zaza | @KConrad thank you for the comment. But tried doing the Regularisation by converting the sum exactly to sum of series of log Zeta derivatives.( You can see my previous posts ) . But I'm unable to get a value so asked again. | |
| Dec 29, 2022 at 16:25 | comment | added | KConrad | You didn't say how you know the sum diverges. As $x\to 0$, $-\log(1-x) \sim x$, so for large $p$, $-\log(1-1/\sqrt{ep})\sim 1/\sqrt{ep}$. Thus the $p$-th term in the sum, for large $p$, grows like $(1/\sqrt{ep})\log(p) = \log(p)/\sqrt{ep}$. So your sum over $p \leq x$ is asymptotic (as $x \to \infty$) to $\sum_{p \leq x}\log(p)/\sqrt{ep}$ if the 2nd series tends to $\infty$ with $x$, and it does: it can be shown that $\sum_{p \leq x} \log(p)/\sqrt{p} \sim 2\sqrt{x}$, so $\sum_{p \leq x} \log(p)/\sqrt{ep} \sim 2\sqrt{x}/\sqrt{e}$. Thus to regularize, at least subtract off $2\sqrt{x}/\sqrt{e}$. | |
| Dec 29, 2022 at 15:13 | history | edited | Zaza | CC BY-SA 4.0 |
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| Dec 29, 2022 at 7:00 | history | edited | YCor | CC BY-SA 4.0 |
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| Dec 29, 2022 at 6:51 | history | asked | Zaza | CC BY-SA 4.0 |