Timeline for Is Li(x) the best possible approximation to the prime-counting function?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 22, 2013 at 14:07 | answer | added | Jan-Christoph Schlage-Puchta | timeline score: 10 | |
| Jul 4, 2010 at 18:47 | comment | added | Bo Peng | There are well-known explicit formulas for the very-related Mangoldt function, involving zeros of the Riemann zeta function: math.ucsb.edu/~stopple/explicit.html en.wikipedia.org/wiki/Explicit_formula | |
| Jul 4, 2010 at 16:10 | answer | added | Will Jagy | timeline score: 29 | |
| Jul 4, 2010 at 15:00 | comment | added | Emerton | My understanding is that Riemann's suggested improvement (mentioned at the end of your question) is wrong, and that all the higher order terms $Li(x^{1/n})$, for $n > 1$, get swamped (for very large $x$) by the error terms coming from the non-trivial zeroes of the $\zeta$-functions. This is discussed in Edwards's book (and surely many other places too, but Edwards's book is my standard reference for $\zeta$ and the prime number theorem). | |
| Jul 4, 2010 at 13:06 | comment | added | Thomas Bloom | The form of the PNT in my comment should read $\pi(x)=Li(x)+O(xe^{c\sqrt{\log x}})$. | |
| Jul 4, 2010 at 13:04 | comment | added | Thomas Bloom | Not about the question, but your comment in the first paragraph - $\pi(x)=x/\log x+O(x/\log^2x)$, but the error term here is not $o(x/\log^2x)$, unlike the error term in the standard form of the PNT, $\pi(x)=\Li(x)+O(xe^{-\sqrt{x}))$. This makes precise what is meant by "$Li(x)$ is a better approximation than $x/\log x$". | |
| Jul 4, 2010 at 12:09 | history | asked | Sam Derbyshire | CC BY-SA 2.5 |