Timeline for Why could Mertens not prove the prime number theorem?
Current License: CC BY-SA 3.0
9 events
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| May 6, 2012 at 7:02 | vote | accept | Nilotpal Kanti Sinha | ||
| May 4, 2012 at 17:13 | comment | added | David E Speyer | To connect Terry's comment and my post, having no primes between $9 \times 10^k$ and $10 \times 10^k$ is the same as having $p^{2 \pi i/\log 10}$ avoid a certain wedge in $\mathbb{C}$. I just find it easier to think "how would such and such sum behave if no primes started with $9$" then "how would it behave if $\mathrm{arg}(p^{it})$ was never near $\pi$?" | |
| May 4, 2012 at 6:54 | comment | added | Terry Tao | Or, one can use sieve theory to stop $p^{it}$ clustering near the negative axis too often (this is how the Erdos-Selberg elementary proof goes). | |
| May 4, 2012 at 6:53 | comment | added | Terry Tao | Indeed, one can view the known proofs of the prime number theorem as methodically isolating and then eliminating such "conspiracies" among the primes. In particular, by working either with the zeroes of the zeta function or with Selberg's explicit formula, one soon finds that the only conspiracy that can really cause trouble is if primes p hate having $p^{it}$ stray far from the negative axis, for some non-zero real $t$... but then $p^{2it}$ stays too close to the positive real axis, which one can show is in contradiction with the conditional convergence of $n^{2it}/n$. | |
| May 3, 2012 at 2:35 | comment | added | GH from MO | See also my response. | |
| May 2, 2012 at 16:39 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| May 2, 2012 at 14:42 | history | edited | Kevin O'Bryant | CC BY-SA 3.0 |
removed "=0" where it was incorrect
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| May 2, 2012 at 13:00 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| May 2, 2012 at 12:36 | history | answered | David E Speyer | CC BY-SA 3.0 |