Timeline for Generalizations of the Euler–Maclaurin Summation Formula
Current License: CC BY-SA 4.0
12 events
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| Jul 27, 2023 at 18:19 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
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| Jul 26, 2023 at 23:38 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
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| Jul 26, 2023 at 23:22 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
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| Jul 26, 2023 at 23:15 | comment | added | Sidharth Ghoshal | @CalebBriggs see the General Theory section, I just added. It's all elementary stuff but it gets a little bit abstract. The pole argument is a one-liner that happens at the end. Your original intuition might be right but we need to find a reason WHY the pole of $\frac{1}{1-e^x}$ at $0$ implies the pole of $\zeta$ at $s=1$ | |
| Jul 26, 2023 at 23:15 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
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| Jul 22, 2023 at 22:36 | comment | added | Caleb Briggs | @SidharthGhoshal I'm excited to see the write-up! My argument is only a heuristic based on divergent series, but it relies on the fact that $\frac{d^z}{dx^z} \zeta(x) \big|_{x = 0}$ doesn't have any poles. So it would be great to see a rigorous argument that relies only at looking at $\zeta(0)$. | |
| Jul 22, 2023 at 21:53 | comment | added | Sidharth Ghoshal | @CalebBriggs sorry for the delay, yes you are correct, I was trying to find a counterexample and then found an argument (which I’ll write up later) that basically shows that because $\zeta(0)$ is not a pole, that no such correction exists | |
| Jul 6, 2023 at 20:12 | comment | added | Caleb Briggs | @SidharthGhoshal If I'm not mistaken, there should be no correction term for the log series. The correction term for the E-M formula comes from the fact that the zeta function has a pole (the correction term is exactly recovered by evaluating this pole). There should be no pole in your $\eta$ function. Hence, I would not expect a correction term. | |
| Jun 29, 2023 at 19:46 | history | edited | Sidharth Ghoshal | CC BY-SA 4.0 |
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| Jun 27, 2023 at 5:20 | comment | added | Sidharth Ghoshal | I would be surprised if they were the same to a given order. For one, we are comparing derivatives to the $x \frac{d}{dx}$ operator and vice versa and its not obvious to me that the rate of convergence of the $\log$ series is the same as the corresponding power series. There probably is still some kind of connection between the two representations to relate convergence rates on one side to another but at least to me it's not clear. Thanks for taking the time to read this post :) | |
| Jun 27, 2023 at 5:09 | comment | added | user196574 | I find this very interesting. Is it obvious that for a function with a log series and with a regular power series that the two Euler-Maclaurin formulae to a given order are not the same? I think your conjecture would make a good math overflow question | |
| Jun 27, 2023 at 3:34 | history | answered | Sidharth Ghoshal | CC BY-SA 4.0 |