Editing Elliptic integral (section)

===Inverting the period ratio===
Here, we use the complete elliptic integral of the first kind with the ''parameter'' <math>m</math> instead, because the squaring function introduces problems when inverting in the complex plane. So let
:<math>K[m]=\int_0^{\pi/2}\dfrac{d\theta}{\sqrt{1-m\sin^2\theta}}</math>
and let
:<math>\theta_2(\tau)=2e^{\pi i\tau/4}\sum_{n=0}^\infty q^{n(n+1)},\quad q=e^{\pi i\tau},\, \operatorname{Im}\tau >0,</math>
:<math>\theta_3(\tau)=1+2\sum_{n=1}^\infty q^{n^2},\quad q=e^{\pi i\tau},\,\operatorname{Im}\tau >0</math>
be the [[theta function]]s.

The equation
:<math>\tau=i\frac{K[1-m]}{K[m]}</math>
can then be solved (provided that a solution <math>m</math> exists) by
:<math>m=\frac{\theta_2(\tau)^4}{\theta_3(\tau)^4}</math>
which is in fact the [[modular lambda function]].

For the purposes of computation, the error analysis is given by<ref>{{cite web |url=https://fungrim.org/topic/Approximations_of_Jacobi_theta_functions/ |title=Approximations of Jacobi theta functions |last= |first= |date= |website=The Mathematical Functions Grimoire |publisher=Fredrik Johansson |access-date=August 29, 2024}}</ref>
:<math>\left|{e}^{-\pi i \tau / 4} \theta_{2}\!\left(\tau\right) - 2\sum_{n=0}^{N - 1} {q}^{n \left(n + 1\right)}\right| \le \begin{cases} \frac{2 {\left|q\right|}^{N \left(N + 1\right)}}{1 - \left|q\right|^{2N+1}}, & \left|q\right|^{2N+1} < 1\\\infty, & \text{otherwise}\\ \end{cases}\;</math>
:<math>\left|\theta_{3}\!\left(\tau\right) - \left(1+2\sum_{n=1}^{N - 1} {q}^{n^2}\right)\right| \le \begin{cases} \frac{2 {\left|q\right|}^{N^2}}{1 - \left|q\right|^{2N+1}}, & \left|q\right|^{2N+1} < 1\\\infty, & \text{otherwise}\\ \end{cases}\;</math>
where <math>N\in\mathbb{Z}_{\ge 1}</math> and <math>\operatorname{Im}\tau >0</math>.

Also
:<math>K[m]=\frac{\pi}{2}\theta_3(\tau )^2,\quad \tau=i\frac{K[1-m]}{K[m]}</math>
where <math>m\in\mathbb{C}\setminus\{0,1\}</math>.