Timeline for Sum of the sum-of-divisors function
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 24, 2012 at 18:09 | comment | added | user9072 | @Eric Naslund: Sorry, in case you found my comment odd. My motication was that I was worried, perhaps unnecessarily so, that somebody else reading your "in the display, the Omega plus/minus" (while in the display in the question there is only an Omega) could get away with the wrong idea that Omega and Omega plus/minus are synonymous. Regarding the question, it is, as indicated by the tag, a history question, on whether Gronwall (in that paper) actually proved what is claimed in Pétermann's paper. This is as documented by one other answers not immediate. So, to me the qu. makes sense. | |
| Jun 24, 2012 at 14:44 | comment | added | Eric Naslund | @quid: Well I was referring to the existing result on the error term. The OP's question is somewhat strange, and feels pointless if one understands Petermann's paper. The paper proves $\Omega_{-}$, and says $\Omega_{+}$ follows from somewhere else, but the methods actually prove both, that is $\Omega_{\pm}$. | |
| Jun 20, 2012 at 12:30 | comment | added | user9072 | @Eric Naslund: a nit-pick, there is no Omega plus/minus, but of course you are still right regarding the main issue. | |
| Jun 20, 2012 at 11:57 | comment | added | Eric Naslund | My first comment above needs an additional $O(x)$, and a factor of $x$ for the sum. That is, we have $$\frac{\pi^2}{12}x^2 -\sum_{n\leq x} \sigma(n)=x\sum_{n\leq x} \frac{1}{n} s\left(\frac{x}{n}\right) +O(x).$$ | |
| Jun 20, 2012 at 11:05 | comment | added | Eric Naslund | @Gerry: You say that "It seems to me that this gives a poorer error term than the one in your display." Note that in the display, the $\Omega_{\pm}$ means it is a lower bound on the maximal size of the error term, whereas the conclusion from the formula you write is an upper bound. | |
| Jun 20, 2012 at 11:01 | comment | added | Eric Naslund | In the definition of $\psi(x)$, the $x\sum_{n>x} \frac{1}{n^2}$ is superfluous since it can just be migrated into the $O(x)$ term appearing in the first line. Perhaps a better way to write this entire thing is $$\frac{\pi^2}{12} x^2 -\sum_{n\leq x } \sigma(n)=\sum_{n\leq x}\frac{1}{n}s\left(\frac{x}{n}\right)$$ where $s(\alpha )=\alpha-1/2$ is the sawtooth function. | |
| Jun 20, 2012 at 4:26 | history | answered | Gerry Myerson | CC BY-SA 3.0 |