Timeline for A hypergeometric series for $\sqrt3\pi$ with converging rate $1/9$
Current License: CC BY-SA 4.0
12 events
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| Oct 21, 2023 at 11:45 | comment | added | Agno | @ho boon san, maybe you know this already, but an impressive inverse symbolic calculator is Ries: mrob.com/pub/ries/index.html It is written in C and produces results (in terms of algebraic equations) very fast. | |
| Oct 21, 2023 at 11:09 | comment | added | ho boon suan | Finally, I note that there is a lot of sophisticated hypergeometric-identity proving going on at Math StackExchange, with many interesting methods presented; see for example math.stackexchange.com/q/3736208 and math.stackexchange.com/q/2123298 and related posts. | |
| Oct 21, 2023 at 11:09 | comment | added | ho boon suan | … So, you could compute many $_5F_4$ values with integer or half-integer parametric excess at various rational entries with small denominators, perhaps out to a hundred decimal places or so, then you could use an inverse symbolic calculator to guess a nice closed form in terms of common irrational numbers such as square roots of natural numbers, $e$, and $\pi$. I think in the literature on experimental mathematics there should be discussions of exhaustive searches of certain input values for interesting hypergeometric identities. | |
| Oct 21, 2023 at 11:09 | comment | added | ho boon suan | @SidharthGhoshal I don't know how the author found these identities, but I can suggest a possible way using a computer: The first identity is an identity for a generalized hypergeometric series $_5F_4$; namely, it states that $_5F_4(3/4,1,1,5/4,13/8; 5/8,3/2,3/2,3/2; 1/9)=\pi\sqrt3/5$. Also, the parametric excess of a hypergeometric is the sum of its bottom parameters, minus the sum of its top parameters, and is an important quantity in the study of hypergeometric series, and it happens to be $-1/2$ in this case… | |
| Oct 21, 2023 at 0:58 | comment | added | Zhi-Wei Sun | Agno, $2\sqrt3 Gi/3$ coincides with the right-hand side of $(2)$. | |
| Oct 20, 2023 at 23:25 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
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| Oct 20, 2023 at 23:24 | comment | added | Zhi-Wei Sun | Deyi, thank you for pointing out the typo. | |
| Oct 20, 2023 at 17:26 | comment | added | Deyi Chen | (4) should be $$\sum_{k=1}^{(p-1)/2}\frac{(8k-3)\binom{4k}{2k}}{k(\color{red}{4}k-1)9^k\binom{2k}k^2}\equiv-\frac5{36}pB_{p-2}\left(\frac13\right)\pmod{p^2}.$$ | |
| Oct 20, 2023 at 12:17 | comment | added | Agno | Equation (2) seems to have the closed form $\frac{2\,\sqrt{3}\,Gi}{3}$, where $Gi$ is Gieseking's constant ( mathworld.wolfram.com/GiesekingsConstant.html ). | |
| Oct 20, 2023 at 1:47 | comment | added | Sidharth Ghoshal | How did you find this?? | |
| Oct 20, 2023 at 1:36 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
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| Oct 20, 2023 at 1:07 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |