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Using a numerical search on my computer I discovered the following inequality: $$\left|\,{_2F_1}\left(\frac14,\frac34;\,\frac23;\,\frac13\right)-\rho\,\right|<10^{-20000},\tag1$$ where $\rho$ is the positive root of the polynomial equation $$12\,\rho^8-12\,\rho^4-8\,\rho^2-1=0,\tag2$$ that can be expressed in radicals: $$\rho=\frac1{\sqrt{\sqrt{\frac4{\sqrt{2-\sqrt[3]{4\vphantom{\large1}}}}+\sqrt[3]{4}+4}-\sqrt{2-\sqrt[3]4}-2}}.\tag3$$ Based on this inequality I conjecture that the actual difference is the exact zero, i.e. $$\color{#808080}{_2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\rho}.\tag4$$ I looked up in DLMF and MathWorld, but did not find a known special value with exactly these parameters. It also appears that CAS like Maple or Mathematica do not know this identity.


Could you please suggest any ideas how to prove the conjecture $(4)$?


Update: I can propose even more general conjecture: $$\color{#808080}{27\,(x-1)^2\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^8+18\,(x-1)\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^4-8\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;x\right)^2=1}$$

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    $\begingroup$ Really spectacular! $\endgroup$
    – Peter
    Dec 1, 2013 at 22:10
  • $\begingroup$ I would like to check it in PARI, but I do not know how the 2F1-function can be calculated there. Any ideas ? $\endgroup$
    – Peter
    Dec 1, 2013 at 22:13
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    $\begingroup$ I am wondering how you ensured an error less than $10^{-20000}$(?!*>+/@) $\endgroup$
    – user17762
    Dec 1, 2013 at 22:51
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    $\begingroup$ @user17762 Strictly speaking, I did not prove this boundary rigorously. I evaluated the difference using Mathematica with a precision higher than $10^{-20000}$ (it took quite long time) and the result was not distinguishable from 0. Mathematica documentation claims the result of a numeric calculation always contains only provably correct digits. $\endgroup$
    – HWᅠ
    Dec 1, 2013 at 22:58
  • $\begingroup$ This is really beautiful ! Congratulations. $\endgroup$ Dec 4, 2013 at 7:53

2 Answers 2

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Let's start with this Pfaff transformation for $\,a=\frac 14,b=\frac34,c=\frac23$ : $$\tag{1}_2F_1\left(a,b\;;c\;;z\right)=(1-z)^{-a}\;_2F_1\left(a,c-b\;;c\;;\frac z{z-1}\right)$$

The 'Darboux evaluation' $(42)$ of Vidunas' "Transformations of algebraic Gauss hypergeometric functions" is : $$\tag{2}_2F_1\left(\frac14,-\frac 1{12};\,\frac23;\,\frac {x(x+4)^3}{4(2x-1)^3}\right)=(1-2x)^{-1/4}$$

Solving $\,\displaystyle\frac {x(x+4)^3}{4(2x-1)^3}=\frac z{z-1}\,$ gives : $$\tag{3}z=\frac{x(x+4)^3}{(x^2-10x-2)^2} $$ that we will use as : $$\tag{4}z-1=\frac {4(2x-1)^3}{(x^2-10x-2)^2}$$

while $(1)$ and $(2)$ return : $$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[(z-1)(2x-1)\right]^{-1/4}$$

so that :

$$_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)=\left[\frac{4\,(2x-1)^4}{(x^2-10x-2)^2}\right]^{-1/4}$$ and (up to a minus sign) : $$\tag{5}_2F_1\left(\frac14,\frac34;\,\frac23;\,z\right)^2=-\frac{x^2-10x-2}{2\,(2x-1)^2}$$ and indeed the substitution of $(z-1)$ and $_2F_1()^2$ with $(4)$ and $(5)$ in your formula gives : $$27\,(z-1)^2\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^8+18\,(z-1)\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^4-8\cdot{_2F_1}\left(\tfrac14,\tfrac34;\tfrac23;z\right)^2=1$$

Many other formulae of this kind may be deduced using Vidunas' paper.

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    $\begingroup$ I was about to bet that I should find you here ! Cheers. $\endgroup$ Dec 4, 2013 at 10:54
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    $\begingroup$ Yes @Claude I am fond of special functions ! Cheers, $\endgroup$ Dec 4, 2013 at 17:39
  • $\begingroup$ Dear Raymond - Outstanding. Beautifully resolves the "Update" conjecture. I was wondering if it also addresses the initial conjecture(4)? I.e., does the update come out of (4), or are they distinct? Thanks and best regards, $\endgroup$
    – user12802
    Feb 17, 2014 at 18:18
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    $\begingroup$ Thanks @96Tears! It doesn't give the explicit formula for $\rho$ (i.e. the initial $(3)$) but it provides the equation $(2)$ which is merely the specific case $z=\dfrac 13$ of the update formula (if you remember that $\displaystyle _2F_1\left(\frac14,\frac34;\,\frac23;\,\frac13\right)=\rho$). Cheers, $\endgroup$ Feb 17, 2014 at 21:21
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The January issue of the Notices has an article by Frits Beukers. There is a criterion to decide whether certain hypergeometric functions are algebraic. Checking these numbers in Theorem 2 of that article, we find that $$ {}_2F_1\left(\frac{1}{4},\frac{3}{4};\frac{2}{3};z\right) $$ is an algebraic function of $z$.$^*$ He also mentions H. A. Schwarz's list of 1873; I did not check, but this is probably in it.

$^*$ Of course Raymond found (much more than) that in his solution.

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