Timeline for Mertens formulas aren't enough for prime number theorem
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 7, 2020 at 22:30 | history | edited | user627482 | CC BY-SA 4.0 |
added 8 characters in body
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| Nov 7, 2020 at 19:42 | answer | added | reuns | timeline score: 3 | |
| Nov 7, 2020 at 19:32 | comment | added | reuns | Take $a_n\in 0,1$ such that $\sum_{n\in x} a_n=\frac{x}{\ln x}(1+\frac14 \cos(\ln x))+O(1)$ then do a partial summation to find the asymptotic of $\sum_{n\le x} \frac{a_n}{n}$. And you meant $\sum_{p\le x} \frac1p= \ln\ln x + C + O(1/\ln x)$. | |
| Nov 7, 2020 at 14:03 | comment | added | user627482 | Okay, now it's a bit more clear to me, thanks for your help! I've to think about it and then I'll try to write down your ideas. | |
| Nov 7, 2020 at 12:38 | comment | added | JoshuaZ | @user627482 Take the primes P. Then, for all the primes between $9*(10^k)$ and $10^{k+1}$, if the prime is less closer to 9*(10^k), replace it with the next available integer below $9*(10^k)$, and if the prime is closer to $10^{k+1}$ replace with the next available integer above $10^{k+1}$. This set will obey Mertens theorem by David's argument, but the ratio of $\Pi(x)/ (x/\log x)$ will not have a limit. Does this work for your purposes? (I haven't checked that the error term is precisely of the order you want.) | |
| Nov 7, 2020 at 12:28 | comment | added | user627482 | @Dror_Speiser: yes, if the error term is the same, it would be perfect | |
| Nov 7, 2020 at 12:28 | comment | added | user627482 | @JoshuaZ: I don't see how that example gives such a set. | |
| Nov 7, 2020 at 12:27 | review | Close votes | |||
| Nov 12, 2020 at 3:01 | |||||
| Nov 7, 2020 at 12:20 | comment | added | Dror Speiser | Should the error term be $1/log$? | |
| Nov 7, 2020 at 12:19 | comment | added | JoshuaZ | Does David's example using powers of ten not lead to such a set? | |
| Nov 7, 2020 at 12:11 | comment | added | user627482 | No, actually I linked that post because the questions are related, but there they only present reasoning about why Mertens isn't enough for pnt. What I need is an explicit subset of $\mathbb{N}$ of elements $a$ that gives a counterexample for Mertens>pnt. | |
| Nov 7, 2020 at 12:07 | comment | added | JoshuaZ | Does this answer your question? Why could Mertens not prove the prime number theorem? | |
| Nov 7, 2020 at 12:07 | comment | added | JoshuaZ | Duplicate of mathoverflow.net/questions/95743/… . | |
| Nov 7, 2020 at 11:55 | review | First posts | |||
| Nov 7, 2020 at 13:16 | |||||
| Nov 7, 2020 at 11:50 | history | asked | user627482 | CC BY-SA 4.0 |