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I'm trying to find a closed form (in terms of simpler functions) for the following hypergeometric function with a complex argument: $$\mathcal{Q}=\,_2F_1\left(\frac12,\frac23;\,\frac32;\,\frac{8\,\sqrt{11}\,i-5}{27}\right).\tag1$$ I have a guess (supported by thousands of digits from numerical calculations) about its argument (phase), but no ideas about its absolute value yet: $$\arg\mathcal{Q}\approx0.168669236010871306727578153...\stackrel?=\arccos\left(\frac{12+\sqrt{33}}{18}\right)\tag2$$ $$|\mathcal{Q}|\approx0.915170225773196416688677425...\tag3$$

Can you suggest any ideas how to prove the conjecture $(2)$? Is there a closed form for the absolute value?


As suggested by gammatester in a comment, the conjecture $(2)$ is equivalent to $$\arg\,B\left(\frac{8\,\sqrt{11}\,i-5}{27};\,\frac12,\frac13\right)\stackrel?=\frac\pi3,\tag4$$ where $B$ denotes the incomplete beta function.


Also, it can be shown that another equivalent form of the conjecture $(2)$ is $$B\left(\frac19;\,\frac16,\frac13\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\,\Gamma\left(\frac13\right)}{2\,\sqrt\pi}.\tag5$$

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  • $\begingroup$ Let $x=\frac{-5+i8\sqrt{11}}{27}$ (assuming that the positive square-root is chosen). Then one can show that $27 x^2 + 10 x + 27=0$ and $|x|=1$. You can try using the formula for the hypergeometric function as a series and then using the quadratic formula to reduce every higher power of $x$ to one that is at most linear in $x$. $\endgroup$
    – suresh
    Commented Feb 24, 2014 at 1:43
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    $\begingroup$ I don't know if this really helps, but your hypergeometric function is related to the incomplete Beta function with $a=\frac{1}{2}, \; b=\frac{1}{3}$ $$B_x(a,b)= \frac{x^a}{a}F(a,1-b,a+1,x), $$ see functions.wolfram.com/06.19.26.0005.01 $\endgroup$ Commented Feb 24, 2014 at 13:47
  • $\begingroup$ @gammatester Thanks, it makes the conjecture much more nicely-looking. $\endgroup$ Commented Feb 24, 2014 at 18:09
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    $\begingroup$ The right hand side of (5) can be rewritten as $\frac12 B(\frac16,\frac13)$. $\endgroup$
    – Kirill
    Commented Feb 25, 2014 at 19:21
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    $\begingroup$ @VladimirReshetnikov: Care to shed some light regarding this question? :) $\endgroup$ Commented Dec 11, 2016 at 6:19

1 Answer 1

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Not sure how to transform conjecture $(2)$ to $(5)$. However, $(5)$ is true.

Let $\;\displaystyle t = \frac{1}{1+y^3}\;$, we can rewrite the integral $\;\displaystyle B\left(\frac19;\frac13,\frac16\right)\;$ as $$ \int_0^{1/9} t^{-5/6} (1-t)^{-2/3} dt = \int_\infty^2 (1+y^3)^{5/6}\left(\frac{1+y^3}{y^3}\right)^{2/3}\frac{-3y^2 dy}{(1+y^3)^2} = 3 \int_2^\infty \frac{dy}{\sqrt{1+y^3}} $$ Let $\omega = e^{\pi i/3}$ and $\mathbb{T} = \big\{\; m+n\omega : m, n \in \mathbb{Z}\;\big\}$ be the triangular lattice span by $1$ and $\omega$. Let $\wp(z)$ be the Weierstrass elliptic $\wp$ function with double poles on lattice $\mathbb{T}$:

$$\wp(z) = \frac{1}{z^2} + \sum_{\lambda \in \mathbb{T} \setminus \{ 0 \}} \left(\frac{1}{(z-\lambda)^2} - \frac{1}{\lambda^2}\right)$$

Let $\;\displaystyle\eta = \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)}{\sqrt{3\pi}}\;$, it is known that $\wp(z)$ satisfies a differential equation of the form:

$$\wp'(z)^2 = 4 \wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ where }\quad g_2 = 0 \;\text{ and }\;g_3 = \frac{\eta^6}{16}$$

Let $\;\displaystyle y(z) = -\frac{4}{\eta^2} \wp\left(\frac{iz}{\eta}\right)$. Using symmetry, it is not hard to see as $z$ varies from $0$ to $\frac{\eta}{\sqrt{3}}$ along the real axis, $y(z)$ remains real and positive, decreases from $\infty$ at $z = 0$ to $0$ at $z = \frac{\eta}{\sqrt{3}}$.

In terms of $z$, we have:

$$\frac{dy}{\sqrt{1+y^3}} = \frac{-\frac{4i}{\eta^3}\wp'\left(\frac{iz}{\eta}\right)dz}{ \sqrt{1-\frac{64}{\eta^6}\wp\left(\frac{iz}{\eta}\right)^3}} = \frac{-4i\wp'\left(\frac{iz}{\eta}\right)dz}{ \sqrt{16g_3-64\wp\left(\frac{iz}{\eta}\right)^3}} = - dz $$ From this, we get

$$B\left(\frac19;\frac13,\frac16\right) = -3 \left[ y^{-1}(\infty) - y^{-1}(2)\right] = 3y^{-1}(2)$$

This allow us to simplify conjecture $(5)$

$$ B\left(\frac19;\frac13,\frac16\right) \stackrel{?}{=} \frac{\Gamma\left(\frac13\right)\Gamma\left(\frac16\right)}{2\sqrt{\pi}} = \frac{\sqrt{3}}{2}\eta \iff y^{-1}(2) \stackrel{?}{=} \frac{\eta}{2\sqrt{3}} \iff \wp(\frac{i}{2\sqrt{3}}) \stackrel{?}{=} -\frac{\eta^2}{2} $$ Let $u = \frac{i}{2\sqrt{3}}$. Since $\wp(2u) = \wp\left(\frac{i}{\sqrt{3}}\right) = 0$, we can use the duplication formula for Weierstrass elliptic $\wp$ function to obtain

$$\begin{align} & 0 = \wp(2u) = \frac14\left(\frac{(6\wp(u)^2-\frac12 g_2)^2}{4\wp(u)^3-g_2\wp(u)-g_3}\right) -2\wp(u) \\ \iff & \wp(u)\left(\frac{\wp(u)^3 + 2g_3}{4\wp(u)^3-g_3}\right) = 0\\ \implies & \wp(u) = (-2g_3)^{1/3} = \left(-\frac{\eta^6}{8}\right)^{1/3} = -\frac{\eta^2}{2} \end{align}$$

i.e. the conjecture $(5)$ is true.

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  • $\begingroup$ Could you give a reference to understand how the value $g_3 = \eta^6/16$ is found? Thanks. $\endgroup$ Commented Mar 1, 2014 at 10:06
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    $\begingroup$ @EstebanCrespi Writing $g_3$ as $\frac{\eta^6}{16}$. In our case, it corresponds to half period $\frac12$. This implies the half period for the case $g_3 = 1$ will be equal to $\frac12 \left(\frac{\eta^6}{16}\right)^{1/6} = 2^{-5/3}\eta$. The last number is known as the $\omega_2$ constant and has value equal to $\frac{\Gamma(1/3)^3}{4\pi}$, you can use that to deduce the expression of $\eta$ I use. $\endgroup$ Commented Mar 1, 2014 at 18:42
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    $\begingroup$ Two questions: 1.) How do we know that we need the lattice with $\omega=\exp(i \pi/3)$, despite of the fact that it works out in the end? For example, the choice of $\wp(z)$ with $g_2=0, g_3=-1$ seems to be more natural at first glance 2.) How do we know that $\wp(i\eta/\sqrt3)=0$? Away from the case of square lattices finding solutions to $\wp(z)=0$ is highly non-trivial (See for example people.mpim-bonn.mpg.de/zagier/files/doi/10.1007/BF01453974/…) $\endgroup$
    – asgeige
    Commented Sep 10, 2020 at 16:38
  • $\begingroup$ @achillehui push $\endgroup$
    – asgeige
    Commented Oct 9, 2020 at 18:33

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