Timeline for Are these continued fractions for the "tails" of $\zeta(3)$ and of the Catalan constant known?
Current License: CC BY-SA 4.0
14 events
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| Oct 14, 2023 at 19:28 | comment | added | Wolfgang | @HenriCohen Nice findings! In fact you have my first one in section 11.1 in your paper where you say "Bauer–Muir acceleration leads to the family $ζ(3) = ((2n − 1)(n^2 − n + 2k^2 +2k + 1), −n^6)$" (for the $A(n)$, your $n$ is $n+1$ in my notation). Similarly the Bauer–Muir transformation of your first family in section 2.1 is my conjecture for the tail of $\eta(2)$. Feel free to post it as an answer. | |
| Sep 3, 2023 at 14:15 | answer | added | Nanhui Lee | timeline score: 1 | |
| Apr 25, 2023 at 21:06 | comment | added | Henri Cohen | please have a look at my new preprint arXiv:2304.11727 | |
| Jul 23, 2022 at 11:31 | history | edited | Wolfgang | CC BY-SA 4.0 |
added corresponding conjectures for half-integers!
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| Jul 5, 2022 at 14:35 | history | edited | Wolfgang | CC BY-SA 4.0 |
added Gradshteyn and Ryzhik integrals
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| Jun 22, 2022 at 16:15 | history | edited | Wolfgang | CC BY-SA 4.0 |
added similar CF for tail of eta(2)
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| Jun 22, 2022 at 7:37 | comment | added | Wolfgang | In fact, for even $k$, it is still (twice) the tail with the appropriate sign $$\frac{1}{f(x^2+(x-1)^2+k^2-k,-x^4)}=2 \sum_{j=k}^{\infty}\frac{(-1)^{j}}{j^2} .$$ | |
| Jun 21, 2022 at 20:43 | comment | added | Wolfgang | @HenriCohen I have looked again at the CF you mentioned in your first comment. And I made the fascinating discovery that it corresponds to the tail of the alternating sum $\eta(2)=\frac{\pi^2}{12}$ rather than to $\zeta(2)$, because we have in fact (supposedly) for odd $k$ $$\frac{1}{f(x^2+(x-1)^2+k^2-k,-x^4)}=\zeta(2)-2\sum_{j=1}^{k-1}\frac{(-1)^{j+1}}{j^2}=2 \sum_{j=k}^{\infty}\frac{(-1)^{j+1}}{j^2} ,$$ while for even $k$ the RHS would be negative. | |
| Jun 21, 2022 at 7:16 | comment | added | Wolfgang | @HenriCohen Note that at the end of my question, I had mentioned that these CFs do not have an accelerated convergence, so in fact, the exposé of your colleagues doesn't address them. But it does provide a proof for the closed form of a different CF, which had so far been considered unproven! | |
| Jun 20, 2022 at 15:02 | comment | added | Wolfgang | @HenriCohen Great, I managed to find it. But I see that in the development of $S_n^\prime$ on the bottom of p.23-17, $n$ is fixed while $k$ varies, likewise on p. 23-19. This is just the converse of mine, but certainly there should be a very close relationship anyway. | |
| Jun 19, 2022 at 21:33 | comment | added | Henri Cohen | Sorry, I just found the reference: STDN 1979-1980 by Batut-Olivier, Expose 23, the formula you mention for $\zeta(3)$ and mine for $\zeta(2)$ are on page 23-17 and 23-19 (S=Seminaire instead of Journal). | |
| Jun 19, 2022 at 20:39 | comment | added | Wolfgang | @HenriCohen Interesting! I couldn't locate such a paper on the JTNB site or digizeitschriften.de. Do you have more details? For $\zeta(2)$, I may have failed to find anything because I didn't use high enough precision. It seems like $f(x^2+(x-1)^2+k^2-k,-x^4)$ only yields closed forms for odd $k$, but the pattern is still unclear. Putting $$g(k):=\frac{1}{f(x^2+(x-1)^2+k^2-k,-x^4)}-\left(\zeta(2)-\sum_{j=1}^{k-2}\frac{1}{j^2}\right),$$ I found $g(3)=\frac12, g(5)= \frac{17}{72}, g(7)= \frac{659}{3600}, g(9)= \frac{42137}{352800}$ etc. | |
| Jun 19, 2022 at 13:39 | comment | added | Henri Cohen | These CF are well known and are part of Ap\'ery's work on $\zeta(3)$ although I am not sure he published them. I have a paper in JTNB around 1979 with C. Batut and M. Olivier which probably have them. The ones for $\zeta(2)$ are similar, $f(x^2+(x-1)^2+k^2-k,-x^4)$ if I am not mistaken. | |
| Jun 18, 2022 at 20:47 | history | asked | Wolfgang | CC BY-SA 4.0 |