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I need help finding the value of $${_2F_1}\left(\frac12,\frac12;\ 1;\ \frac{1}{9} \right)\tag1$$

Not in terms of elliptic functions.


I've tried many methods on this and have reduced it down to many forms, none of which have provided any help to me, however I think they may be of some use here.

We have the identity $${_2F_1}\left(a,1-a;1;\ z\right)= P_{-a}(1-2z)$$ Which leads to $(1)$ being equal to $$\frac{2 \operatorname{K}({1/9})}{\pi}^*$$

Where $\operatorname{K}(x)$ is the Complete Elliptic Integral of the First Kind. This is technically a closed form, but I would prefer to find a closed form not in terms of elliptic integrals - since elliptic integrals can be expressed in the form ${_2F_1}\left(\frac12,\frac12;\ 1;\ x^2\right)$ this closed form feels awful circular.

It also seems to be equal to $${_2F_1}\left(\frac14,\frac34;\ 1;\ 9/25 \right)\frac{3}{\sqrt{10}}$$

and many other equivalent but seemingly useless forms that can be computed using (1) (2)

I've gotten tantalizingly close but haven't been able to crack it. What is the closed form for this hypergeometric function, if there is one?


*Also, this is probably a very stupid question, but on both Wolfram Mathworld and Wikipedia they claim ${_2F_1}\left(\frac12,\frac12;\ 1;\ k^2\right)=\operatorname{K}(k)$ which means $(1)$ should be equal to $\operatorname{K}(1/3)$ but Wolfram Alpha and Mathematica both seem to say ${_2F_1}\left(\frac12,\frac12;\ 1;\ k\right)=\operatorname{K}(k)$

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    $\begingroup$ Note that Mathematica uses the parameter convention for its elliptic integrals, and not the usual modulus. So to achieve $K(k)$ you'd type EllipticK[k^2]. $\endgroup$ – nospoon Aug 10 '16 at 20:38
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This is not an answer but some hints which rather indicate that there is no simple closed formula for this expression.

We can find an expression of ${_2F_1}\left(\frac{1}{2},\frac{1}{2};1;\frac{1}{9} \right)$ in terms of the central binomial coefficient $\binom{2n}{n}$. Using D. Knuths notation of rising factorial $x^{\overline{n}}=x(x+1)\cdots(x+n-1)$ and the notation for double factorials $$(2n)!=(2n)!!(2n-1)!!=2^nn!(2n-1)!!$$ we obtain \begin{align*} \left(\frac{1}{2}\right)^{\overline{n}}&=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdots\left(\frac{1}{2}+(n-1)\right) =\frac{(2n-1)!!}{2^n}=\frac{(2n)!}{2^{2n}n!}\\ \\ 1^{\overline{n}}&=1\cdot2\cdot3\cdots (1+(n-1))=n! \end{align*}

$$ $$

It follows \begin{align*} {_2F_1}\left(\frac{1}{2},\frac{1}{2};1;\frac{1}{9} \right) &=\sum_{n=0}^\infty\frac{\left(\frac{1}{2}\right)^{\overline{n}}\left(\frac{1}{2}\right)^{\overline{n}}}{1^{\overline{n}}n!} \left(\frac{1}{9}\right)^n\\ &=\sum_{n=0}^\infty\frac{(2n)!}{2^{2n}n!}\cdot\frac{(2n)!}{2^{2n}n!}\cdot\frac{1}{n!^2}\left(\frac{1}{9}\right)^n\\ &=\sum_{n=0}^\infty\binom{2n}{n}^2\frac{1}{144^n}\tag{1} \end{align*} Wolfram alpha evaluates the series with the square of the central binomial coefficient (1) as \begin{align*} \sum_{n=0}^\infty\binom{2n}{n}^2\frac{1}{144^n}=\frac{2}{\pi}K\left(\frac{1}{9}\right)\tag{2} \end{align*} with $K$ the complete elliptic integral of the first kind with parameter $k^2=\frac{1}{9}$.

$$ $$

I did some recherche on (1) and found two related results.

  • In the paper Central binomial coefficients by S.Finch we find on page $5$ a bunch of formulas of series containing the central binomial coefficients and precisely one with $\binom{2n}{n}^2$ as numerator. \begin{align*} \sum_{n=0}^\infty\binom{2n}{n}^2\frac{1}{2^{4n}(2n+1)}=\frac{4G}{\pi}\tag{3} \end{align*} with $G$ Catalan's constant. Note that Catalan's constant $G$ is \begin{align*} G=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}=\frac{1}{2}\int_0^1K(k)\,dk \end{align*} with $K(k)=\int_0^\frac{\pi}{2}\frac{dt}{\sqrt{1-k^2\sin^2t}}$. So, there is no simple closed formula for (3).

Another interesting paper with series containing $\binom{2n}{n}^2$ in the numerator is

  • A certain series associated with Catalan's constant by V. Adamchik. He considers the formula \begin{align*} S(r)=\sum_{n=0}^\infty\frac{\binom{2n}{n}^2}{(n+r)16^n}=\frac{1}{r}{_3F_2}\left(\frac{1}{2},\frac{1}{2},r;1,r+1;1\right) \end{align*} Starting point is formula (11) in his paper which introduces a function $g(x,r)$ defined as \begin{align*} g(r,x)=x^{r-1}\sum_{n=0}^\infty\binom{2n}{n}^2\frac{x^n}{16^n}=\frac{2}{\pi}x^{r-1}K(x)\qquad\qquad |x|<1 \end{align*} Note that $g\left(1,\frac{1}{9}\right)$ is our formula (2). He continues integrating $g(r,x)$ in order to reach after some further investigations and calculations $S(r)$ for certain $r$. Although it was not obvious for me, but maybe his proof could be adapted to attack formula (2).

To your last question: I'm astonished, too. I would also expect $K\left(\frac{1}{3}\right)$ according to the Wolfram page describing the Complete elliptic integral of the first kind. This notation seems to be quite common. We can read e.g. in section 3.2 of Special functions (Encyclopedia of Mathematics and its Applications) by G.E. Andrews et al. \begin{align*} K(k)=\frac{\pi}{2} {_2F_1}\left(\frac{1}{2},\frac{1}{2};1;k^2\right) \end{align*} When evaluating the LHS of (2) with Wolfram Alpha it returns the RHS and states as comment: $K(m)$ is the complete elliptic integral of the first kind with parameter $m=k^2$.

This rather cryptic note from WA indicates some weird notation convention.

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  • $\begingroup$ @TreFox: Many thanks for granting the bounty! :-) $\endgroup$ – Markus Scheuer Aug 19 '16 at 22:49
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You can express it in terms of arithmetic–geometric mean: $$ {_2F_1}\left(\frac12,\frac12;\ 1;\ \frac{1}{9} \right) = \frac{1}{\operatorname{agm}\left(1,\frac{2\sqrt{2}}{3}\right)}. $$

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    $\begingroup$ Good answer. For any practical purpose this is the best way to compute the value of this number $\endgroup$ – Yuriy S Aug 27 '16 at 22:07
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    $\begingroup$ I agree. That can be written as $\frac{3}{\text{AGM}(2,4)}$, too. $\endgroup$ – Jack D'Aurizio Feb 5 '17 at 22:47
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In case this is still interesting I have found that $$ {_2F_1}\left(\frac12,\frac12;\ 1;\ \frac{1}{9} \right)\tag1 =\frac{3}{\pi}\int_0^1 \frac{1}{\sqrt{9x-13x^2+8x^3-4x^4}}\;dx $$ or alternatively from your representation $${_2F_1}\left(\frac14,\frac34;\ 1;\ 9/25 \right)\frac{3}{\sqrt{10}}$$ we have $$ {_2F_1}\left(\frac12,\frac12;\ 1;\ \frac{1}{9} \right)\tag1 =\frac{3\sqrt{5}}{2\pi}\int_0^1 \frac{1}{(1-x)^{1/4}x^{3/4}\sqrt{25-36x+36x^2}}\;dx $$

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