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

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

Board index **‹** Special Functions **‹** Problem Related to Gauss's Hypergeometric Function
## Problem Related to Gauss's Hypergeometric Function

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

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

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

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

Post away the answer, E.

Actually, I don't know the answer myself.

I thought you guys might have some ideas on it.

**Moderator:** Shobhit

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

ETYUCAN wrote:

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

Post away the answer, E.

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.

3 posts
• Page **1** of **1**