2
$\begingroup$

Is it possible to show that: \begin{align} \frac{\Gamma\left(\frac{1}{78}\right) \Gamma\left(\frac{29}{78}\right) \Gamma\left(\frac{35}{78}\right) \Gamma\left(\frac{53}{78}\right) \Gamma\left(\frac{55}{78}\right) \Gamma\left(\frac{61}{78}\right)}{\Gamma\left(\frac{2}{78}\right) \Gamma\left(\frac{28}{78}\right) \Gamma\left(\frac{32}{78}\right) \Gamma\left(\frac{44}{78}\right) \Gamma\left(\frac{58}{78}\right) \Gamma\left(\frac{70}{78}\right) } = \sqrt{3} \end{align}

There are other known ratios of Gamma functions, but as always there seems to be a product rule or trick to evaluate the ratios.

$\endgroup$
  • $\begingroup$ Well yes, else you wouldn't have that particular identity to post. Not to be condescending, but posting a reference, your own work, and motivation would be appreciated. $\endgroup$ – Zach466920 Jun 21 '15 at 16:00
  • 1
    $\begingroup$ $$\frac{\Gamma(1/34)\Gamma(9/34)\Gamma(13/34) \Gamma(15/34)\Gamma(19/34)\Gamma(21/34) \Gamma(25/34)\Gamma(33/34)}{\Gamma(3/34)\Gamma(5/34) \Gamma(7/34)\Gamma(11/34)\Gamma(23/34) \Gamma(27/34)\Gamma(29/34)\Gamma(31/34)} = 1 . $$ See Amer. Math. Monthly, November 2010, page 842 $\endgroup$ – Dietrich Burde Jun 21 '15 at 16:09
2
$\begingroup$

One method to find such formulas is the duplication formula $$ \Gamma(z) \Gamma(z+1/2) = 2^{1-2z} \sqrt{\pi} \Gamma(2z), $$ and apply it to nominator and denominator, e.g., to obtain $$ {\Gamma(1/8) \Gamma(5/8) \Gamma(6/8) \over \Gamma(2/8) \Gamma(3/8) \Gamma(7/8)} = \sqrt{2}. $$ There seem to be more advanced ideas in Deligne, P. Valeurs de fonctions L et périodes d'intégrales, Amer. Math. Soc., Providence, R.I., 1979.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.