Timeline for Asymptotic propeties of Euler function
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 7, 2010 at 16:23 | vote | accept | Fedor Petrov | ||
| May 5, 2010 at 11:52 | comment | added | Fedor Petrov | Oh, thank you, now I see, why it may seem so! The little problem is that $$(1-e^{-u})^2(u^{-2}+1/6)$$ is less then 1 when $u\searrow 0$, while evaluation suggest that $\zeta(2)S(t)(1-t)^2$ is greater then 1 when $t\nearrow 1$. But probably this has some stupid explanation, like miscalculation somewhere :) | |
| May 4, 2010 at 19:13 | comment | added | David Hansen | (it seems I assumed the Riemann hypothesis there. :)) | |
| May 4, 2010 at 19:12 | comment | added | David Hansen | Push the contour to (-1/2) instead. You'll get residues coming from poles of the $\zeta$-function along the 1/2-line, but the first zero is at 1/2+14.1347I, and Gamma(1/2+14.1347I) is of size 10^{-10}. So the contribution of these residues will be invisible to the naked eye. The pole at $s=0$ has residue 1/6. So to the eye it seems that S(e^-u)=zeta(2)^-1 u^{-2} + 1/6 + o(1) as $u\to 0$, but in reality there are wiggly terms of such small magnitude the eye cannot see them. | |
| May 4, 2010 at 18:55 | comment | added | Fedor Petrov | Yes, thank you, but why this remainder seems to have fixed sign? It suffices to take $1000$ summands fo $t=0.98$ and 2000 for $t=0.99$ in $\sum \varphi(n)t^n$ to get value grater then \zeta(2)^{-1}(1-t)^{-2) , each next summand only increases the sum. Another sum, in which I am more interested, $u^2\sum_{n=1}^{\infty} -\varphi(n) \ln(1-e^{-nu})$ is greater then limit value $\zeta(3)/\zeta(2)$ when $u$ decreases to 0, while analogous sum for $n$ instead $\varphi(n)$ increases for $u\searrow 0$. | |
| May 4, 2010 at 17:23 | history | answered | David Hansen | CC BY-SA 2.5 |