Prove the following identity of Gamma : $\frac{\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})}=\frac{(1+\sqrt{2})\Gamma(\frac{5}{8})}{\Gamma(\frac{1}{8})}$

Question

Prove $$\frac{\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})}=\frac{(1+\sqrt{2})\Gamma(\frac{5}{8})}{\Gamma(\frac{1}{8})}$$

We know that :

$\Gamma(x+\frac{1}{2})=\frac{(2x)!\sqrt{π}}{4^{x}x!}$

So : $\Gamma(\frac{7}{8})=\Gamma(\frac{1}{2}+\frac{3}{8})$ !!

Use the reflection formula for $\Gamma$. – metamorphy Jun 29 '19 at 15:43 — metamorphy, Jun 29 '19 at 15:43

Robert Z · Answer 1 · 2019-06-29T16:09:59.463

4

By the reflection formula, if $z\notin\mathbb{Z}$, $$\Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \dfrac \pi {\sin \left({\pi z}\right)}$$ so you can evaluate $\Gamma(\frac{1}{8})\Gamma(\frac{7}{8})$ and $\Gamma(\frac{3}{8}) \Gamma(\frac{5}{8})$ and verify that $$\frac{\Gamma(\frac{1}{8})\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})\Gamma(\frac{5}{8})}=1+\sqrt{2}.$$

P.S. See also Evaluate $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$ or the Silver ratio (due to user 493905).

edited Jun 29 '19 at 16:09

answered Jun 29 '19 at 15:46

Robert Z

145,942

$1+ \sqrt 2$ = $\delta_s$ – user 493905 Jun 29 '19 at 16:03
What is $\delta_s$? – Robert Z Jun 29 '19 at 16:07
1

$\delta_s$ is the Silver Ratio (a Metallic Number). – user 493905 Jun 29 '19 at 16:09
1

@user493905 Thanks for the reference! – Robert Z Jun 29 '19 at 16:12

Prove the following identity of Gamma : $\frac{\Gamma(\frac{7}{8})}{\Gamma(\frac{3}{8})}=\frac{(1+\sqrt{2})\Gamma(\frac{5}{8})}{\Gamma(\frac{1}{8})}$

1 Answers1