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I have the following conjecture, which is supported by numerical calculations up to at least $10^5$ decimal digits: $$4\times{_2F_1}\left(-\frac{1}{4},\frac{3}{4};\frac{7}{4};\frac{2-\sqrt{3}}{4}\right)-{_2F_1}\left(\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{2-\sqrt{3}}{4}\right)\,\stackrel?=\,\frac{3\sqrt[4]{2+\sqrt{3}}}{\sqrt{2}},$$ where $_2F_1$ denotes the hypergeometric function.

Can you suggest any ideas how to prove it?


The conjectural closed form was obtained using WolframAlpha query

ToRadicals[RootApproximant[2.94844576626425580599908814238570067699233]]
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  • $\begingroup$ I guess you occasionally discovered this while playing with your calculator? $\endgroup$ May 23, 2013 at 23:02

2 Answers 2

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This is the identity 15.5.12 from DLMF, with $a=-1/4$, $b=3/4$, $c=7/4$ and the special form $$ F(b,a,a,x)=(1-x)^{-b}. $$ Is this how you got your identity?

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    $\begingroup$ No, it all happened as I wrote in my comment to O.L.'s answer. I am far from being an expert in hypergeometric functions - actually, I just started to study them. Many things that are immediately obvious to you, may not be obvious for me at all. $\endgroup$ May 24, 2013 at 0:28
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    $\begingroup$ Okay. Generally what you want to do first with such identities, after trying Mathematica, for example, is to check if the Digital Library of Mathematical Functions has any identities that might help out, because all I did was look up the section on identities relating hypergeometric functions with different parameters. $\endgroup$
    – Kirill
    May 24, 2013 at 0:46
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Edited on request of moderator team

  1. Gauss hypergeometric function satisfies a linear transformation formula $$ (c-a) _2F_1(a-1,b,c,z)+(2a-c-az+bz){}_2F_1(a,b,c,z)+a(z-1){}_2F_1(a+1,b,c,z)=0.$$ Setting $a=b$ makes $z$ disappear from the 2nd prefactor, and one obtains $$(c-a) _2F_1(a-1,a,c,z)+(2a-c){}_2F_1(a,a,c,z)+a(z-1){}_2F_1(a+1,a,c,z)=0.\tag{1}$$

  2. Also, sometimes hypergeometric function simplifies to elementary functions. For example, $_2F_1(\alpha,\beta,\alpha,z)=(1-z)^{-\beta}$. Setting in this formula $\alpha=a+1$, $\beta=a$ and combining it with (1) with $c=a+1$, one obtains $$ _2F_1(a-1,a,a+1,z)+(a-1){}_2F_1(a,a,a+1,z)=a(1-z)^{1-a}.\tag{2}$$

  3. Setting in (2) $\displaystyle a=\frac34$ and $\displaystyle z=\frac{2-\sqrt{3}}{4}$ gives the quoted relation.

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  • $\begingroup$ You got the answer first. (My answer is the same, I hadn't seen yours when I posted.) $\endgroup$
    – Kirill
    May 24, 2013 at 0:18
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    $\begingroup$ @O.L. Thanks for your proof! Actually, the story was a bit different. I was doing some calculations in Mathematica and got a huge, clumsy expression (that I would not copy here) containing this linear combination of hypergeometric functions as a part. I tried FullSimplify command in Mathematica, but it could not do anything. Then I used RootApproximant to get a conjectural expression in radicals in case I was lucky and the expression was indeed algebraic. I got the result and then checked it up to $10^5$ decimal digits to get more confidence and then posted it here as a conjecture. $\endgroup$ May 24, 2013 at 0:20
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    $\begingroup$ Interesting thing to note: Mathematica will actually simplify this correctly if you differentiate the left hand side once w.r.t. $x$, then use FullSimplify, then integrate back. $\endgroup$
    – Kirill
    May 24, 2013 at 0:22
  • $\begingroup$ Well, it is a shame that Mathematica cannot simplify an expression in case it can be done several simple identities. $\endgroup$ May 24, 2013 at 0:31

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