Solving for argument of complete elliptic integral of first kind

Question

I have the following equation to be solved for $m$

$\frac{K(1-m)}{K(m)} = a$

where $a$ is known value, $K(m)$ is the complete elliptic integral of the first kind. $K(m)$ can be expressed as an infinite series which I found on Wikipedia https://en.wikipedia.org/wiki/Elliptic_integral

I don't know how to go about solving this equation.

Answer 1

If

$$ f(m) := \frac{K(1-m)}{K(m)}, \tag{1}$$ then $$ f(m) = -\frac{\log(q(m))}{\pi} \tag{2}$$ where $\, q(m) \equiv q(k^2) \,$ is the Jacobi elliptic nome $\,q\,$ defined as

$$ q(m) := e^{-\pi K(1-m)/K(m)} = e^{-\pi f(m)}. \tag{3} $$

Given $\, f(m), \,$ to find $\, m, \,$ use equation $\,(3)\,$ to compute $\, q(m). \,$ Then $$ m = \frac{ \theta_2(0,q)^4} {\theta_3(0,q)^4} = 16q - 128q^2 + 704q^3 +\dots \tag{4} $$ where $\,\theta_2, \, \theta_3\,$ are Jacobi theta functions. TheOEIS sequence A115977 has more information about the $q$ series expansion of $m$.

As a numerical example, assume $\,f(m) = 1.6.\,$ Then $$ \theta_2 \approx 0.569243 , \, \theta_3 \approx 1.013123 , \, q \approx 0.0065614, \\ 16q - 128q^2 + 740q^3 \approx 0.0996652, \, m \approx 0.0996653.$$

Answer 2

This is a problem I worked years ago and I give you some old results of mine.

As said in answer and comments, you will need some numerical methods and, for sure, a guess of the solution.

If we use a very limited series expansion built around $m=0$, we have $$\frac{K(1-m)}{K(m)}=\frac 1 {2 \pi} \left(8\log(2)-2\log(m)-m \right)+O\left(m^2\right)$$ from which $$m=2 W\left(8 e^{-\pi a}\right)\tag 1$$ where appears Lambert function. This is quite good for $a >1$ as shown below $$\left( \begin{array}{ccc} a & (1) & \text{exact} \\ 1.0 & 0.530366 & 0.500000 \\ 1.1 & 0.411170 & 0.396910 \\ 1.2 & 0.315098 & 0.308663 \\ 1.3 & 0.239066 & 0.236256 \\ 1.4 & 0.179861 & 0.178666 \\ 1.5 & 0.134392 & 0.133894 \\ 1.6 & 0.099869 & 0.099665 \\ 1.7 & 0.073898 & 0.073816 \\ 1.8 & 0.054501 & 0.054468 \\ 1.9 & 0.040096 & 0.040083 \\ 2.0 & 0.029442 & 0.029437 \end{array} \right)$$

Edit

If you are only concerned by the case $a < 1$, we can build the $[3,3]$ Padé approximant built at $m=\frac 12$,we should get $$\frac{K(1-m)}{K(m)}=\frac{1+b_1 \left(m-\frac 12\right)+b_2\left(m-\frac 12\right)^2+b_3\left(m-\frac 12\right)^3}{1-b_1 \left(m-\frac 12\right)+b_2\left(m-\frac 12\right)^2-b_3\left(m-\frac 12\right)^3}$$ where $$b_1=\frac{\Gamma \left(-\frac{1}{4}\right) \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)^2}\qquad b_2=-\frac{32}{15}\qquad b_3=\frac{1275 \pi \Gamma \left(\frac{3}{4}\right)}{2048 \sqrt{2} \Gamma \left(\frac{9}{4}\right)^2 \Gamma \left(\frac{13}{4}\right)}$$ and solve a cubic equation in $\left(m-\frac 12\right)$. This seems to give acceptable results for $0.5 \leq a \leq 1$.