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$$ 1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^3-13\left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^3+\cdots $$

I went on evaluating the above series and encountered that solving $\displaystyle \sum_{n\ge 0}\left(\binom{2n}{n}\right)^3x^n$ would suffice.

But how do we make a generating function for the third power of a central binomial coefficient using the fact $\displaystyle \sum_{n\ge 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$

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  • $\begingroup$ The generating function is not algebraic. It can be expressed using a hypergeometric function: see OEIS sequence A002897. But that hypergeometric function is unlikely to help you. $\endgroup$ – Robert Israel Jan 13 '17 at 5:23
  • $\begingroup$ The "power 2" generating function is in terms of elliptic integral and this one as you say in terms of hypergeometric functions, so there isn't any closed form I guess. But Ramanujan did find a value to the series I posted above. It was like $\frac{2}{\pi}$ as per my memory. $\endgroup$ – Aditya Narayan Sharma Jan 13 '17 at 5:27
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This is not an answer but just a result obtained using a CAS.

Let $$f_k=\sum_{n=0}^\infty \binom{2 n}{n}^kx^n$$ The following expressions have been obtained $$f_1=\frac{1}{\sqrt{1-4 x}}$$ $$f_2=\frac{2 }{\pi }K(16 x)$$ $$f_3=\frac{4 }{\pi ^2}K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2$$ $$f_4=\, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1;256 x\right)$$ $$f_5=\, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;1 024 x\right)$$ where appear, for $k=2,3$, the complete elliptic integrals of the first kind and, for $k>3$, the generalized hypergeometric functions

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  • $\begingroup$ "+1" , Thanks for the additional info , Is there any proof of these 5 over the web ? $\endgroup$ – Aditya Narayan Sharma Jan 13 '17 at 5:48
  • $\begingroup$ @AdityaNarayanSharma. I did not search for that. After your comment, just out of curiosity, I gave a CAs to eat ! Cheers. $\endgroup$ – Claude Leibovici Jan 13 '17 at 5:50
  • $\begingroup$ In the third case , the argument inside K is squared right ? And not the entire K-function ? $\endgroup$ – Aditya Narayan Sharma Jan 13 '17 at 5:53
  • $\begingroup$ @AdityaNarayanSharma. It is the entire function which is squared. $\endgroup$ – Claude Leibovici Jan 13 '17 at 8:10
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    $\begingroup$ Rewriting both my comment and your answer in terms of $~x=\bigg[\dfrac{\sin(2a)}{2^k}\bigg]^2,~$ with $~|a|<\dfrac\pi4,~$ we have $$F_0~=~\sec^2(2a),\quad F_1~=~\sec(2a),\quad F_2~=~\dfrac2\pi~K\Big(\sin(2a)\Big),\quad F_3~=~\bigg[\dfrac2\pi~K(\sin a)\bigg]^2.$$ Note: I've used the traditional modulus notation $K(k)$ for the argument of elliptic integrals, instead of the Mathematica one in terms the parameter $m=k^2.~$ $\endgroup$ – Lucian Jan 13 '17 at 9:33
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This is one of the simplest and famous series given by Ramanujan and it's value is $2/\pi$. Unfortunately Ramanujan's technique requires a reasonable amount of effort to understand. I have presented the proof for this series and it's friend $$\frac{4}{\pi}= 1+\frac{7}{4}\left(\frac{1}{2}\right)^{3}+\frac{13}{4^{2}}\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}+\dots$$ in this post.

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