Editing Jacobi elliptic functions (section)

===Using elliptic integrals===
Equivalently, Jacobi's elliptic functions can be defined in terms of the [[theta function]]s.<ref>{{cite book |last1=Whittaker |first1=Edmund Taylor |authorlink1=Edmund T. Whittaker |last2=Watson |first2=George Neville |authorlink2=George N. Watson |date= 1927 |page=492 |edition=4th |title=A Course of Modern Analysis |title-link=A Course of Modern Analysis |publisher= Cambridge University Press}}</ref> With <math>z,\tau\in\mathbb{C}</math> such that <math>\operatorname{Im}\tau >0</math>, let

:<math>\theta_1(z|\tau)=\displaystyle\sum_{n=-\infty}^\infty (-1)^{n-\frac12}e^{(2n+1)iz+\pi i\tau\left(n+\frac12\right)^2},</math>
:<math>\theta_2(z|\tau)=\displaystyle\sum_{n=-\infty}^\infty e^{(2n+1)iz+\pi i\tau \left(n+\frac12\right)^2},</math>
:<math>\theta_3(z|\tau)=\displaystyle\sum_{n=-\infty}^\infty e^{2niz+\pi i\tau n^2},</math>
:<math>\theta_4(z|\tau)=\displaystyle\sum_{n=-\infty}^\infty (-1)^n e^{2niz+\pi i\tau n^2}</math>
and let <math>\theta_2(\tau)=\theta_2(0|\tau)</math>, <math>\theta_3(\tau)=\theta_3(0|\tau)</math>, <math>\theta_4(\tau)=\theta_4(0|\tau)</math>. Then with <math>K=K(m)</math>, <math>K'=K(1-m)</math>, <math>\zeta=\pi u/(2K)</math> and <math>\tau=iK'/K</math>,

:<math>\begin{align}\operatorname{sn}(u,m)&=\frac{\theta_3(\tau)\theta_1(\zeta|\tau)}{\theta_2(\tau)\theta_4(\zeta|\tau)},\\
\operatorname{cn}(u,m)&=\frac{\theta_4(\tau)\theta_2(\zeta|\tau)}{\theta_2(\tau)\theta_4(\zeta|\tau)},\\
\operatorname{dn}(u,m)&=\frac{\theta_4(\tau)\theta_3(\zeta|\tau)}{\theta_3(\tau)\theta_4(\zeta|\tau)}.\end{align}</math>

The Jacobi zn function can be expressed by theta functions as well:
:<math>\begin{align}\operatorname{zn}(u,m)&=\frac{\pi}{2K}\frac{\theta_{4}'(\zeta|\tau)}{\theta_{4}(\zeta|\tau)}\\ &=\frac{\pi}{2K}\frac{\theta_{3}'(\zeta|\tau)}{\theta_{3}(\zeta|\tau)}+m\frac{\operatorname{sn}(u,m)\operatorname{cn}(u,m)}{\operatorname{dn}(u,m)}\\
&=\frac{\pi}{2K}\frac{\theta_{2}'(\zeta|\tau)}{\theta_{2}(\zeta|\tau)}+\frac{\operatorname{dn}(u,m)\operatorname{sn}(u,m)}{\operatorname{cn}(u,m)}\\
&=\frac{\pi}{2K}\frac{\theta_{1}'(\zeta|\tau)}{\theta_{1}(\zeta|\tau)}-\frac{\operatorname{cn}(u,m)\operatorname{dn}(u,m)}{\operatorname{sn}(u,m)}\end{align}</math>
where <math>'</math> denotes the partial derivative with respect to the first variable.