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Given the complete elliptic integral of the first kind $K(k)$ for the modulus $k$,

can the elliptic function $$\text{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\bigg|\frac{1}{2}\right)$$ be expressed in closed form?

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    $\begingroup$ of course. You are at a finite order point. $\endgroup$ – Start wearing purple Aug 24 '15 at 13:39
  • $\begingroup$ @Startwearingpurple: it would be really interesting to see that comment expanded into an answer. $\endgroup$ – Jack D'Aurizio Aug 24 '15 at 13:40
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    $\begingroup$ Perhaps tell us which convention you are using. (I.e., is that last $1/2$ supposed to be $k$ or $m$?) Is this 0.473058826656122429 ? $\endgroup$ – GEdgar Aug 24 '15 at 13:41
  • $\begingroup$ @GEdgar It could be $m$ $\endgroup$ – Nicco Aug 24 '15 at 13:44
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To explain how an algebraic equation for $Z:=\operatorname{cn} \left(\frac{2K}{3}|m\right)$ can be obtained, notice that

  1. The usual parity properties imply that $$\operatorname{cn} \left(\frac{4K}{3}\biggl|\,m\right)=\operatorname{cn} \left(2K-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} \left(-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} \left(\frac{2K}{3}\biggl|\,m\right)=-Z.$$

  2. On the other hand the doubling formula implies that $$\operatorname{cn} \left(\frac{4K}{3}\biggl|\,m\right)=\frac{1-2\operatorname{sn}^2 \left(\frac{2K}{3}|m\right)+m\operatorname{sn}^4 \left(\frac{2K}{3}|m\right)}{1-m\operatorname{sn}^4 \left(\frac{2K}{3}|m\right)}=\frac{2Z^2-1+m\left(1-Z^2\right)^2}{1-m\left(1-Z^2\right)^2}.$$

Thus $Z$ satisfies the equation $$\frac{2Z^2-1+m\left(1-Z^2\right)^2}{1-m\left(1-Z^2\right)^2}=-Z.$$ Although naively it is of $5$th order, it has an obvious root $Z=-1$, so we can deduce from it a $4$th order equation $$\boxed{\quad m\left(Z^2-1\right)\left(Z-1\right)^2-2Z+1=0\quad}$$

In particular, the relevant solution for $m=k^2=\frac12$ is given by $$Z=\frac{1-\sqrt2\cdot\sqrt[4]{3}+\sqrt3}{2}\approx 0.435421.$$

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  • $\begingroup$ @ Start wearing purple:thank you very much ,your answer is very nice. $\endgroup$ – Nicco Aug 24 '15 at 14:36
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    $\begingroup$ And for the $k=1/2$ version, take $m=1/4$ in this equation. $\endgroup$ – GEdgar Aug 24 '15 at 14:38
  • $\begingroup$ @GEdgar This Wolfram convention always causes a lot of confusion to me. $\endgroup$ – Start wearing purple Aug 24 '15 at 14:40
  • $\begingroup$ @Startwearingpurple Same here. $\endgroup$ – user153012 Aug 25 '15 at 13:27
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Assuming $1/2$ is $k$, as in Maple ... Maple JacobiCN(2/3*EllipticK(1/2),1/2) evaluates to $0.473058826656122429170671314726$. From ISC we find that this is a solution of $$ Z^4-2Z^3-6Z+3=0 $$ which may be written $$ {\frac { \left( 1+2\,\sqrt [3]{6} \right) ^{3/4}+\sqrt [4]{1+2\, \sqrt [3]{6}}-\sqrt {-2\,\sqrt [3]{6}\sqrt {1+2\,\sqrt [3]{6}}+2\, \sqrt {1+2\,\sqrt [3]{6}}+14}}{2\sqrt [4]{1+2\,\sqrt [3]{6}}}} $$

On the other hand, if $1/2$ is $m=k^2$, then in Maple we want JacobiCN(2/3*EllipticK(1/sqrt(2)),1/sqrt(2)) which evaluates to $0.43542054468233904782250442376$. This is a solution of $-1+2Z+2Z^3-Z^4=0$.

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  • $\begingroup$ GEdgar wonderful.now generalizing this case to any $k$ ,is it possible to find an explicit formula? $\endgroup$ – Nicco Aug 24 '15 at 14:01
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    $\begingroup$ @Nicco Yes. Express $\operatorname{cn} (3u|k)$ in terms of $\operatorname{cn} (u|k)$ using addition formulas and then use this to obtain an algebraic equation for $\operatorname{cn} (2K/3|k)$. $\endgroup$ – Start wearing purple Aug 24 '15 at 14:05

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