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Problem Related to Gauss's Hypergeometric Function

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Shobhit Site Admin
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If \(\displaystyle p=\frac{\sqrt{6\sqrt{3}-9}-1}{2}\), then prove that

\(\displaystyle _2F_1 \left(\frac{1}{2},\frac{1}{2};1;p^3 \frac{2+p}{1+2p} \right)=\frac{\sqrt{\pi}}{\Gamma^2 \left(\frac{3}{4}\right)\sqrt{6\sqrt{3}-9}}\)

:lol:


Posts: 38
Location: India, West Bengal
ETYUCAN wrote:
If \(\displaystyle p=\frac{\sqrt{6\sqrt{3}-9}-1}{2}\), then prove that

\(\displaystyle _2F_1 \left(\frac{1}{2},\frac{1}{2};1;p^3 \frac{2+p}{1+2p} \right)=\frac{\sqrt{\pi}}{\Gamma^2 \left(\frac{3}{4}\right)\sqrt{6\sqrt{3}-9}}\)

:lol:


Post away the answer, E.

Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

mathbalarka wrote:
Post away the answer, E.


Actually, I don't know the answer myself. :)

I thought you guys might have some ideas on it.


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