25
$\begingroup$

Working on this conjecture, I found its corollary, which is also supported by numeric calculations up to at least $10^5$ decimal digits: $$K\left(\frac{\sqrt{2-\sqrt3}}2\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\Gamma\left(\frac13\right)}{4\ \sqrt[4]3\ \sqrt\pi},$$ where $K(x)$ is the complete elliptic integral of the 1st kind. I did not find this specific value at MathWorld, Wolfram Functions Site, Wikipedia or DLMF.

Is it a known value?

$\endgroup$
  • 5
    $\begingroup$ I'm always enjoying your integral calculations and conjectures! Again, a nice discovery! (+1) $\endgroup$ – Sangchul Lee May 28 '13 at 2:22
  • $\begingroup$ Added some plain text to the title, the dropdown menus don't work otherwise. $\endgroup$ – zyx May 28 '13 at 3:54
  • $\begingroup$ Does the corresponding elliptic curve have CM? $\endgroup$ – Bruno Joyal May 28 '13 at 5:47
19
$\begingroup$

See here: http://mathworld.wolfram.com/EllipticIntegralSingularValue.html and also here: http://mathworld.wolfram.com/EllipticIntegralSingularValuek3.html

Your value is actually $$ \sin \frac\pi{12} = \frac{\sqrt{2-\sqrt3}}{2}, $$ and according to MathWorld, it is known as the third singular value $k_3$. It satisfies $$ K(\sqrt{1-k_3^2}) = \sqrt{3}K(k_3) $$ and $$ K(k_3) = \frac{\sqrt{\pi}\Gamma(1/6)}{2\cdot 3^{3/4}\Gamma(2/3)}. $$ Mathematica says that the two closed forms are equal.

$\endgroup$

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.