4
$\begingroup$

While I was working on this question by @Vladimir Reshetnikov, I've found the following relations between Gaussian hypergeometric function values and the Baxter constant:

$$\begin{align}{_2F_1}\left(\begin{array}c\tfrac13,\tfrac13\\1\end{array}\middle|\,-1\right) &\stackrel{?}{=} \frac{1}{2^{\small2/3}}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac23,\tfrac23\\1\end{array}\middle|\,-1\right) &\stackrel{?}{=} \frac{1}{2}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac13,\tfrac13\\1\end{array}\middle|\,\frac19\right) &\stackrel{?}{=} \frac{1}{\sqrt[3]{3}}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac23,\tfrac23\\1\end{array}\middle|\,\frac19\right) &\stackrel{?}{=} \frac{\sqrt[3]{3}}{2}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac13,\tfrac13\\1\end{array}\middle|\,9\right) &\stackrel{?}{=} \frac{3-i\sqrt3}{6}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac23,\tfrac23\\1\end{array}\middle|\,9\right) &\stackrel{?}{=} -\frac{i}{2\sqrt3}\,C^2_\text{B4CC},\\ \end{align}$$

where ${_2F_1}$ is the Gaussian hypergeometric function, and

$$ C^2_\text{B4CC} = \frac{3}{4\pi^2}\Gamma^3\left(\tfrac{1}{3}\right) \approx 1.460998486206318358158873117846059697\dots $$

is Baxter's four-coloring constant.

The first two identity are known, but with the last four relations I've never met before.

How could we prove these identities?

In this paper, there is another connection between a hypergeometric value and Baxter constant.

$\endgroup$
  • $\begingroup$ Your third formula is the case $a=\frac13$ of this $\endgroup$ – nospoon Aug 21 '15 at 19:16
4
+100
$\begingroup$
  1. To prove the third identity, it suffices to take the limit $z\to -\frac12$ of the Ramanujan's cubic transformation. It yields $$_2F_1\left(\frac13,\frac23;1;-\frac18\right)=\lim_{z\to-\frac12^+} \frac{_2F_1\left(\frac13,\frac23;1;1-\left(\frac{1-z}{1+2z}\right)^3\right)}{1+2z}=\frac23\,C_{\mathrm{B4CC}}^2,\tag{1}$$ where the limit is evaluated using the connection formulae expressing $_2F_1(\ldots;z)$ as a combination of two $_2F_1(\ldots;z^{-1})$. Combining (1) with the well-known transformation $$_2F_1\left(a,b;c;z\right)=\left(1-z\right)^{-a}{}_2F_1\left(a,c-b;c;\frac{z}{z-1}\right),$$ we get $$_2F_1\left(\frac13,\frac13;1;\frac19\right)=\left(1-\frac19\right)^{-\frac13}{}_2F_1\left(\frac13,\frac23;1;-\frac18\right)=\frac{C_{\mathrm{B4CC}}^2}{\sqrt[3]{3}}.\tag{2}$$

  2. The fourth identity immediately follows from (2) and the transformation $$_2F_1\left(a,b;c;z\right)=\left(1-z\right)^{c-a-b}{}_2F_1\left(c-a,c-b;c;z\right).\tag{3}$$

  3. The fifth and sixth identity are meaningless in the present form: you have to specify the on which side of the branch cut $[1;\infty)$ we evaluate $_2F_1$. In any case the corresponding hypergeometric functions are related by the same transformation (3) - it suffices to correctly take into account the branching of $(1-z)^{c-a-b}$. Therefore only one of these identities is independent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.