Timeline for A limit arising from Mellin Inversion: How to compute a specific term of an asymptotic series?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 8, 2023 at 21:40 | vote | accept | Sidharth Ghoshal | ||
| Jul 8, 2023 at 21:40 | comment | added | Sidharth Ghoshal | i have come to the conclusion that my question was just bad. | |
| Jun 30, 2023 at 22:59 | comment | added | Sidharth Ghoshal | I'm sorry for the delay. As I started writing it up I realized there were things I wasn't paying attention to and I want to take some time to organize my thoughts carefully. I will post an update sometime next week | |
| Jun 30, 2023 at 18:49 | comment | added | Iosif Pinelis | @SidharthGhoshal : Can you explain in detail the relation of your procedure with the residue in question? Also, what "will be periodic"? | |
| Jun 30, 2023 at 18:38 | comment | added | Sidharth Ghoshal | For $f = 2^n$ this is equal to $\ln(2)$ numerically, which tells us that the Asymptotic series of $\sum_{n=0}^{\infty} e^{-\frac{2^n}{e^x}} = \frac{1}{\ln(2)}x + ... $ | |
| Jun 30, 2023 at 18:38 | comment | added | Sidharth Ghoshal | So I described a procedure which seems like it works at least for $f(n)=c^n$. The procedure is to consider: $\lim_{x \rightarrow \infty} \frac{d}{dx} \left[ e^{- \frac{f(n)}{e^x}} \right]$ observe that in the limit this is equal to a sum of exponentials, that experimentally appears periodic for $f = 2^n$ and $f = n!$ (but I cannot prove that it will be periodic in all cases). In the case that it looks periodic we can extract the constant term from the fourier series that it converges to. | |
| Jun 30, 2023 at 17:12 | comment | added | Iosif Pinelis | @SidharthGhoshal : If you now want $\text{Res}\,F(s)|_{s=0}$ for a certain $F$, instead of $\lim_{s\to0+}sF(s)$, how do you define $\text{Res}(F(s)|_{s=0}$, given that $F(s)=\infty$ for $s\le0$, so that $F$ is not meromorphic in any neighborhood of $0$? | |
| Jun 29, 2023 at 19:57 | comment | added | Sidharth Ghoshal | Sorry for that glitch, I corrected the question. I appreciate you taking the time to answer the original post and apologize for any time wasted. | |
| Jun 29, 2023 at 19:56 | comment | added | Sidharth Ghoshal | you know what you are right, im being hasty here, I should've just used the residue instead of the limit | |
| Jun 29, 2023 at 19:52 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |