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I've found the following hypergeometric function value by numerical observation. The identity matches at least for $100$ digits.

$${_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right) \stackrel{?}{=} \frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$$

Or using a Pfaff transformation in an equivalent form

$$81^{1/6} \cdot {_2F_1}\left(\begin{array}c\tfrac16,\tfrac16\\\tfrac56\end{array}\middle|\,-80\right) \stackrel{?}{=} \frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$$

How could we prove it?


Other related problem: How could we prove that

$${_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right) \stackrel{?}{=} {_2F_1}\left(\begin{array}c\tfrac12,\tfrac56\\\tfrac12\end{array}\middle|\,\frac{4}{9}\right) = {_1F_0}\left(\begin{array}c\tfrac56\\\ - \,\end{array}\middle|\,\frac{4}{9}\right)$$

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    $\begingroup$ Is there a reason you are interested in this particular hypergeometric series? If you give some background, you will likely attract more attention to this problem. By the way, it's Pfaff, with an f. $\endgroup$ – A l'Maeaux Oct 6 '14 at 23:27
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    $\begingroup$ A related question. $\endgroup$ – Lucian Oct 6 '14 at 23:30
  • $\begingroup$ @Lucian Yes, the inspiration of the problem was Vladimir's question. This is an equivalent problem. $\endgroup$ – user153012 Oct 6 '14 at 23:42
  • $\begingroup$ @Daniel While I tried to solve Vladimir's question what Lucian linked as a related problem, I have stucked at this point. $\endgroup$ – user153012 Oct 7 '14 at 0:27
  • $\begingroup$ Maybe your problem can be answered using the technique in this paper: Transformations of Algebraic Gauss Hypergeometric Functions by Raimundas Vidunas $\endgroup$ – Anastasiya-Romanova 秀 Oct 15 '14 at 9:55
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My question is related to this question by Vladimir. Because it is already proved that $${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}} = \frac{3^{\small3/2}}{2^{\small4/3}\,5^{\small5/6}\,\pi }\Gamma^3\!\!\left(\tfrac13\right),$$

we have the answer for my question too, since according to Maple

$${\large\int}_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt5\,x}\ \left(1-x^2\right)^{\small2/3}} = \frac{2}{9} \frac{\sqrt[3]{4}\,\sqrt[3]{3}\,\pi^2\,{_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right)}{\Gamma^3 \left( \frac{2}{3} \right)}.$$

The second part of the question $-$ the part under the line $-$ is still open.

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  • $\begingroup$ Instead of finding the closed form of the hypergeometric function, why wouldn't we transform the hypergeometric function in RHS to LHS? All we need is some clever substitutions. +1 ≧◠◡◠≦✌ $\endgroup$ – Anastasiya-Romanova 秀 Oct 15 '14 at 13:55
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(This may not be complete, but generalizes the observation for context.) The OP's answer mentioned Reshetnikov's integral which can be expressed as, $$\frac{1}{48^{1/4}\,K(k_3)}\,\int_{-1}^1\frac{dx}{\sqrt[3]{9+4\sqrt{5}\,x}\,\left(1-x^2\right)^{\small2/3} } =\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-4\big)= \frac{3}{5^{5/6}} $$ so it will be fruitful to find a transformation between Reshetnikov's use of $\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};z_1\big)$ and the OP's use of $\,_2F_1\big(\tfrac{1}{6},\tfrac{2}{3};\tfrac{5}{6};z_2\big)$. Employing well-known transformations, we get,

$$G(y)=\,_2F_1\big(\tfrac{1}{3}, \tfrac{1}{3};\tfrac{5}{6}; -y\big) = \big(\tfrac1{1+2y}\big)^{1/3}\,_2F_1\Big(\tfrac{1}{6}, \tfrac{2}{3};\tfrac{5}{6}; \tfrac{(1+2y)^2-1}{(1+2y)^2}\Big)$$

It is conjectured that if $y$ are certain algebraic numbers, than $G(y)$ is also an algebraic number. Specializing the RHS, define as in this post,

$$H(\tau)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\text{where}\;\frac1{\delta_4}-1=\frac1{27}\left(\tfrac{\eta\big((\tau+1)/3\big)}{\eta(\tau)}\right)^{12}$$ with Dedekind eta function $\eta(\tau)$. If $\tau = \frac{1+N\sqrt{-3}}2$ for some integer $N>1$, then the conjecture implies both $\delta_4$ and $H(\delta_4)$ are algebraic numbers. For example, if we use $\tau = \frac{1+5\sqrt{-3}}2$ and $\tau = \frac{1+7\sqrt{-3}}2$, respectively, then we recover your, $$H\big(\tfrac{1+\color{blue}5\sqrt{-3}}2\big)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;\tfrac{80}{81}\big) = \tfrac3{\color{blue}5}\,(9\sqrt5)^{1/3}$$ as well as, $$H\big(\tfrac{1+\color{blue}7\sqrt{-3}}2\big)=\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;\tfrac{3024}{3025}\big) = \tfrac4{\color{blue}7}\,55^{1/3}$$ and so on.

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The Tito Piezas answer is related to this parameterization for ${}_2F_1(\frac{1}{6},\frac{2}{3};\frac{5}{6};z)$ in terms of modular forms: $$ {}_2F_1\left(\frac{1}{6},\frac{2}{3};\frac{5}{6}; \frac{j_{3A}}{j_{3A}-108}\right) = \eta(\tau)\; \eta(3\tau)\;(j_{3A}-108)^{1/6}\; \Gamma\left(\frac{2}{3}\right)^3 \frac{(-6i\tau+\sqrt{3}-3i)}{2^{7/3}\pi} \tag{1}$$ where $\eta$ is the Dedekind eta function and $j_{3A}$ is $$ j_{3A} = \left(\frac{\eta(\tau)^6}{\eta(3\tau)^6}+ \frac{3^3 \eta(3\tau)^6}{\eta(\tau)^6}\right)^2 $$ (Choosing the proper branches of the sixth root and the ${}_2F_1$).

Take $\tau = \frac{1}{2} + i \frac{5}{6}\sqrt{3}$ to get the required result from special values of $\eta$...

$$ \eta\left(\frac{1}{2} + i \frac{5}{6}\sqrt{3}\right) =2^{-3/2} 3^{3/8} 5^{-5/12} (1+\sqrt5)^{1/2} \pi^{-1} \Gamma\left(\frac{1}{3}\right)^{3/2} (-1)^{1/24} $$ and something similar for $\eta(3\tau)$, so $$ j_{3A}(\tau) = -8640,\qquad \frac{j_{3A}(\tau)}{j_{3A}(\tau)-108} = \frac{80}{81} $$ In $(1)$, the product of $\Gamma(1/3)$ and $\Gamma(2/3)$ simplify to something algebraic together with a $\pi$, and the powers of $\pi$ cancel.

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