Timeline for Why could Mertens not prove the prime number theorem?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2016 at 9:59 | history | edited | Nilotpal Kanti Sinha |
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| Feb 29, 2016 at 23:17 | comment | added | reuns | @Gjergji : I think Mertens could prove $M(x) = o(x)$ and with the fact that $\zeta(s)$ is meromorphic (for $Re(s) > 1-\epsilon$ is enough) this is enough to imply the PNT | |
| Jun 2, 2015 at 10:51 | answer | added | Vesselin Dimitrov | timeline score: 12 | |
| May 15, 2015 at 5:36 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 3.0 |
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| S Mar 30, 2015 at 22:35 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Improve wording of title
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| Mar 30, 2015 at 22:16 | review | Suggested edits | |||
| S Mar 30, 2015 at 22:35 | |||||
| May 6, 2012 at 7:02 | vote | accept | Nilotpal Kanti Sinha | ||
| May 4, 2012 at 7:08 | comment | added | Gjergji Zaimi | Incidentally, Mertens did give a proof of $\zeta(1+it)=0\implies \zeta(1+2it)=\infty$ based on his 3-4-1 inequality (this was after the proofs of Hadamard and de la Vallee Poussin). So in some sense he could prove PNT. :) | |
| May 4, 2012 at 6:27 | answer | added | Nilotpal Kanti Sinha | timeline score: 9 | |
| May 3, 2012 at 10:49 | comment | added | Nilotpal Kanti Sinha | @David, Thanks for the correction, I had mistyped. | |
| May 3, 2012 at 10:48 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 3.0 |
added 4 characters in body
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| May 3, 2012 at 2:38 | comment | added | GH from MO | @Greg: See my response for a precise estimate. | |
| May 3, 2012 at 2:34 | answer | added | GH from MO | timeline score: 21 | |
| May 2, 2012 at 17:47 | comment | added | Greg Martin | Indeed, I suspect that standard techniques (Landau's theorem) prove that the error term in Mertens's theorem is $\Omega(x^{-1/2})$; perhaps Littlewood's technique might even give $\Omega_\pm(x^{-1/2}\log\log\log x)$. | |
| May 2, 2012 at 12:59 | comment | added | David E Speyer | The error in Merten's theorem is $O(1/\log x)$, not $O(1/x)$. I suspect that $O(1/x)$ is not true. | |
| May 2, 2012 at 12:53 | history | edited | Gerry Myerson | CC BY-SA 3.0 |
apostrophe catastrophe
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| May 2, 2012 at 12:36 | answer | added | David E Speyer | timeline score: 84 | |
| May 2, 2012 at 10:03 | history | asked | Nilotpal Kanti Sinha | CC BY-SA 3.0 |