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?
  • 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. $\endgroup$ – Semiclassical Jul 26 '14 at 23:12
  • $\begingroup$ 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). $\endgroup$ – Greg Martin Jul 27 '14 at 0:03
  • 1
    $\begingroup$ Another special value is apparently $_2F_1\left(\frac13,\frac13;\frac56;-4\right)\stackrel{\color{#808080}?}=3\cdot5^{-5/6}$. $\endgroup$ – Vladimir Reshetnikov Jul 27 '14 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 '14 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 '14 at 0:43

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.

  • $\begingroup$ 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. $\endgroup$ – Semiclassical Aug 4 '14 at 17:48
  • 3
    $\begingroup$ 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. $\endgroup$ – Noam D. Elkies Aug 5 '14 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 '14 at 23:52
  • $\begingroup$ @NoamD.Elkies: Made some minor changes for aesthetics. I hope it's ok. $\endgroup$ – Tito Piezas III Dec 4 '16 at 4:02
  • $\begingroup$ Not sure what you changed and why . . . $\endgroup$ – Noam D. Elkies Dec 4 '16 at 4:55

(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.

  • $\begingroup$ Note that the obvious choice of $a=0$ is useless since both sides equal one in that case. $\endgroup$ – Semiclassical Jul 27 '14 at 0:29

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.


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.