36
$\begingroup$

I discovered the following conjecture numerically, but have not been able to prove it yet: $$_2F_1\left(\frac13,\frac13;\frac56;-27\right)\stackrel{\color{#808080}?}=\frac47.\tag1$$ The equality holds with at least $10000$ decimal digits of precision. It can be written in equivalent forms in terms of definite integrals: $${\large\int}_0^1\frac{dx}{\sqrt{1-x}\ \sqrt[3]{x^2+(3x)^3}}\stackrel{\color{#808080}?}=\frac{\sqrt[3]4\,\sqrt3}{7\pi}\Gamma^3\!\!\left(\tfrac13\right),\tag2$$ or $${\large\int}_0^\pi\frac{d\phi}{\sqrt[3]{\sin\phi}\,\sqrt[3]{55+12\sqrt{21}\cos\phi}}\stackrel{\color{#808080}?}=\frac{\sqrt[3]4\,\sqrt3}{7\pi}\Gamma^3\!\!\left(\tfrac13\right).\tag3$$


Update: A several more equivalent forms: $$_2F_1\left(\frac13,\frac12;\frac56;\frac{27}{28}\right)\stackrel{\color{#808080}?}=\frac{2^{\small8/3}}{7^{\small2/3}}\tag4$$ $$\int_0^\infty\frac{dx}{\sqrt[3]{55+\cosh x}}\stackrel{\color{#808080}?}=\frac{\sqrt[3]2\,\sqrt3}{7\pi}\Gamma^3\!\!\left(\tfrac13\right)\tag5$$ $$C_{\small-1/3}^{\small(1/3)}(55)\stackrel{\color{#808080}?}=\frac{3}{7\pi^2}\Gamma^3\!\!\left(\tfrac13\right)\tag6$$ $$P_{\small-1/2}^{\small1/6}(55)\stackrel{\color{#808080}?}=\frac{\sqrt2\,\sqrt[4]3\,e^{\small-\pi\,i/12}}{7^{\small13/12}\,\pi^{\small3/2}}\Gamma^2\!\!\left(\tfrac13\right)\tag7$$ where $C_n^{(\lambda)}(x)$ is the Gegenbauer polynomial and $P_l^m(x)$ is the Legendre function of the first kind.


  • Please suggest ideas how to prove this conjecture.
  • What are other points where the function $_2F_1\left(\frac13,\frac13;\frac56;z\right)$ takes simple special values?
$\endgroup$
10
  • 1
    $\begingroup$ The most obvious suggestion that comes to mind is some set of contiguous identities combined with a cubic transformation...but that's decidedly not the same as me seeing how to follow that path. $\endgroup$ Jul 26, 2014 at 23:12
  • 1
    $\begingroup$ I suggest looking through the transformations in Section 15.3 of Abramowitz and Stegun, Handbook of Mathematical Functions, to see if any of them transform your quantity into another ${}_2F_1$ value where a closed form is known (as in Section 15.1). $\endgroup$ Jul 27, 2014 at 0:03
  • 1
    $\begingroup$ Another special value is apparently $_2F_1\left(\frac13,\frac13;\frac56;-4\right)\stackrel{\color{#808080}?}=3\cdot5^{-5/6}$. $\endgroup$ Jul 27, 2014 at 0:05
  • 1
    $\begingroup$ By changing the argument from $-27$ to $5$, we get one of the roots of $~5^5x^6+12^3=0$. $\endgroup$
    – Lucian
    Jul 27, 2014 at 0:06
  • 1
    $\begingroup$ Other special values are $x=-\dfrac13$ and $x=+\dfrac12$, for which the minimal polynomials are $9x^3-8$ and $16x^6-27$. $\endgroup$
    – Lucian
    Jul 27, 2014 at 0:43

3 Answers 3

26
+100
$\begingroup$

The conjecture is true, as are the other cases reported in the comments where $f(z) := {}_2F_1 \left( \frac13, \frac13; \frac56; z \right)$ takes algebraic values for special rational values of $z$. There are a few others obtained from the symmetry $z \leftrightarrow 1-z$ (these ${}_2F_1$ parameters correspond to a hyperbolic triangle group with index $6,6,\infty$ at $c=0,1,\infty$, so the $z=0$ and $z=1$ indices coincide); e.g. $f(-1/3) = 2 / 3^{2/3}$ pairs with $f(4/3) = 3^{-2/3} (5-\sqrt{-3})/2$. ($z=1/2$ pairs with itself, and the pair $f(-4)$ and $f(5)$ has been noted already; the OP's $f(-27) = -4/7$ pairs with $f(28) = \frac12 - \frac3{14} \sqrt{-3}$.) Somewhat more exotic are $$ f\big({-}4\sqrt{13}\,(4+\sqrt{13})^3\big) = \frac7{13\,U_{13}}\\ f\big({-}\sqrt{11}\,(U_{33})^{3/2}\big) = \frac{6}{11^{11/12}\, U_{33}^{1/4}}, $$ with fundamental units $U_{13}=\frac{3+\sqrt{13}}2,\;U_{33}=23+4\sqrt{33}$ and further values at algebraic conjugates and images under $z \leftrightarrow 1-z$.

In general, for $z<1$ the integral formula for $f(z)$ relates it with $$ \int_0^1 \frac{dx}{ \sqrt{1-x} \; x^{2/3} (1-zx)^{1/3} } $$ which is half of a "complete real period" for the holomorphic differential $dx/y$ on the curve $C_z : y^6 = (1-x)^3 x^4 (1-zx)^2$. This curve has genus $2$, but is in the special family of genus-$2$ curves with an automorphism of order $3$ (multiply $y$ by a cube root of unity), for which both real periods are multiples of the real period of a single elliptic curve $E_z$ (a.k.a. a complete elliptic integral). In general the resulting formula doesn't simplify further, but when $E_z$ has CM (complex multiplication) its periods can be expressed in terms of gamma functions. For $z = -27$ and the other special values listed above, not only does $E_z$ have CM but the CM ring is contained in ${\bf Z}[\rho]$ where $\rho = e^{2\pi i/3} = (-1+\sqrt{-3})/2$. Then the $\Gamma$ and $\pi$ factors of the period of $E_z$ exactly match those in the integral formula, leaving us with an algebraic value of $f(z)$. It turns out that the choice $z = -27$ makes $E_z$ a curve with complex multiplication by ${\bf Z}[7\rho]$. The others from the comments lead to ${\bf Z}[m\rho]$ with $m=1,2,3,5$, and the examples where $z$ is a quadratic irrationality come from ${\bf Z}[13\rho]$ and ${\bf Z}[11\rho]$.

One way to get from $C_z$ to $E_z$ is to start from the change of variable $u^3 = (1+cx)/x$, which gives $$ f(z) = \int_{\root 3 \of {1-z}}^\infty \frac{3u \, du}{\sqrt{(u^3+z)(u^3+z-1)}}. $$ and identifies $C_z$ with the hyperelliptic curve $v^2 = (u^3+z)(u^3+z-1)$. Now in general a curve $v^2 = u^6+Au^3+B^6$ has an involution $\iota$ taking $u$ to $B^2/u$, and the quotient by $\iota$ is an elliptic curve; we compute that this curve has $j$-invariant $$ j = 6912 \frac{(5+2r)^3}{(2-r)^3(2+r)} $$ where $A = rB^3$. (There are two choices of $\iota$, related by $v \leftrightarrow -v$, and thus two choices of $j$, related by $r \leftrightarrow -r$; but the corresponding elliptic curves are $3$-isogenous, so their periods are proportional.) In our case $r = A/B^3 = -(2z+1)/\sqrt{z^2+z}$ (in which the $z \leftrightarrow 1-z$ symmetry takes $r$ to $-r$). Taking $z=-27$ yields $j = -2^{15} 3^4 5^3 (52518123 \pm 11460394\sqrt{21})$, which are the $j$-invariants of the ${\bf Z}[7\rho]$ curves; working backwards from the $j$-invariants of the other ${\bf Z}[m\rho]$ curves we find the additional values of $z$ noted in the comments and earlier in this answer.

$\endgroup$
7
  • $\begingroup$ I like this approach very much, despite only understanding this algebraic curves machinery just enough to mostly follow it. I'm a bit puzzled, though: Your answer makes clear that the $z=-27$ should be algebraic, but I don't see where the fact of it being $4/7$ in particular is shown. $\endgroup$ Aug 4, 2014 at 17:48
  • 4
    $\begingroup$ Good question. Basically you follow the differential $dx / (\sqrt{1-x} \; x^{2/3} (1-zx)^{1/3})$ through the changes of variable corresponding to the maps from $C_z$ to $E_z = C_z/\iota$ and the 3-isogenous curve $C_z/-\iota$, and then use the $m$-isogeny from the ${\bf Z}[m\rho]$ curve to a ${\bf Z}[\rho]$ curve to relate the period to the Beta integral $\int_0^1 dt/(t-t^2)^{1/3}$ which cancels the transcendental factor $\Gamma^3(1/3)/\pi$. Once $m$ gets as large as $5$ or $7$, let alone $11$ or $13$, the actual formula for the $m$-isogeny can get complicated, but we know how to find it. $\endgroup$ Aug 5, 2014 at 1:49
  • 2
    $\begingroup$ @NoamD.Elkies I know this is an old answer, but how do you know which values of $j$ to look for? $\endgroup$
    – Kirill
    Oct 12, 2014 at 23:52
  • $\begingroup$ @NoamD.Elkies: Made some minor changes for aesthetics. I hope it's ok. $\endgroup$ Dec 4, 2016 at 4:02
  • $\begingroup$ Not sure what you changed and why . . . $\endgroup$ Dec 4, 2016 at 4:55
9
$\begingroup$

(This is more a comment than answer, but I couldn't get MathJax to properly show it in comments)

Here is a nice identity (equation (21) of this paper with $x=-1/7$): $$_2F_1 \left(a,a+\frac{1}{2};\frac{4a+5}{6};-\frac{1}{7}\right)=\left(\frac{7}{4}\right)^a {_2}F_1 \left(\frac{a}{3},\frac{a+1}{3};\frac{4a+5}{6};-27\right)$$

It's an example of a cubic transformation. Possibly, one can at this point use contiguous relations to make some progress.

$\endgroup$
1
  • $\begingroup$ Note that the obvious choice of $a=0$ is useless since both sides equal one in that case. $\endgroup$ Jul 27, 2014 at 0:29
4
$\begingroup$

Regarding your secondary question, by appealing to the classical j-function at defined arguments, it seems there are infinitely many algebraic numbers $z$ such that the $_2F_1$ evaluates to an algebraic number. Some examples, $$_2F_1\left(\frac13,\frac13;\frac56;-z_1\right)= \frac9{17} \big(833+324\cdot17^{1/3}-252\cdot17^{2/3}\big)^{1/6}$$ $$2F_1\left(\frac13,\frac13;\frac56;-z_2\right)= \frac{10}{3\cdot19} \big(2+2\cdot19^{1/3}-19^{2/3}\big)$$ where, $$z_1 =4\big(19894+7737\cdot17^{1/3}+3009\cdot17^{2/3}\big)$$ $$z_2 =\frac{1}{3}\big(1464289+548752\cdot19^{1/3}+205648\cdot19^{2/3}\big)$$ See also this post.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .