Timeline for Surprisingly long closed form for simple series
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 15, 2022 at 17:01 | comment | added | GH from MO | @joro My previous comment was directed to you. | |
| May 15, 2022 at 15:58 | comment | added | GH from MO | I don't know what you mean by "algebraic dependency". $f(A)=C$ means that $C$ is a linear combination of the participating logarithms of algebraic numbers, with coefficients taken from the Kummer field $\mathbb{Q}(e^{2\pi i/A},A^{1/A})$. I don't know if my approach works for the BPP formula, I don't have time to check it. My guess is that the answer is "yes". I mean, I don't think there are many other ways to evaluate this sum. Iosif Pinelis's approach leads to similar terms, I think. | |
| May 15, 2022 at 15:55 | comment | added | joro | So if it happen $f(A)=C$ where C is some named constant, you will find algebraic dependency between C and logarithms of algebraic numbers, right? Does your approach work for the similar BPP formula for pi: thatsmaths.com/2013/08/08/the-remarkable-bbp-formula | |
| May 15, 2022 at 15:37 | comment | added | LSpice | @GHfromMO, re, interestingly, ISO 31-11 prescribes $\mathrm e^t$ (but I don't like it either, and I agree the ‘standard’ is non-standard in the mathematics community). | |
| May 15, 2022 at 13:52 | comment | added | GH from MO | @MarkWildon Thanks. Your extra line was missing the summation over $m$, which I added now. I also changed $\mathrm{e}^t$ to $e^t$ as it is more customary that way (see e.g. en.wikipedia.org/wiki/Exponential_function). | |
| May 15, 2022 at 13:51 | history | edited | GH from MO | CC BY-SA 4.0 |
deleted 21 characters in body
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| May 15, 2022 at 13:47 | comment | added | GH from MO | @joro I thought it was clear what I wrote. The $m$-sum is a logarithm, hence the right-hand side is a sum of $A$ logarithms. For your sum you will need the left-hand side for $r=1$ and $r=A-1$, and get rid of the $n=0$ term. Hence $2A$ logarithms (and 2 fractions) will do. This works for every positive integer $A$, not just $A=11$. | |
| May 15, 2022 at 10:10 | comment | added | Mark Wildon | I added one more line in the middle while changing a few $\exp^{ix}$ to $\mathrm{e}^{ix}$. | |
| May 15, 2022 at 10:10 | history | edited | Mark Wildon | CC BY-SA 4.0 |
Corrected exp with superscript to e and added one more line.
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| May 15, 2022 at 10:01 | comment | added | joro | Thanks. What is closed form for the infinite sum? Can you compute by hands f(11)? | |
| May 15, 2022 at 9:43 | history | edited | GH from MO | CC BY-SA 4.0 |
added 1 character in body
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| May 15, 2022 at 9:37 | history | answered | GH from MO | CC BY-SA 4.0 |