Timeline for An apparently simple question (behaviour at infinity of a power series)
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 16, 2014 at 18:56 | comment | added | Joni Teräväinen | In that case, the remainder formula for the Taylor series tells that one can take $c_n=a_0+a_1\log \log n+...+a_n(\log \log n)^n$ (for example), unless the derivatives of $S(x)$ grow too fast. I wouldn't expect a nice formula without any further assumptions. | |
| May 16, 2014 at 15:38 | comment | added | Josh | Indeed,I expect that for every $n\in\mathbb{N}$, the $n$-element $c_n$ must be a function of $a_0,a_1,\ldots,a_n$. That's why I mentioned convolution. | |
| May 16, 2014 at 13:10 | history | answered | Joni Teräväinen | CC BY-SA 3.0 |