Asymptotic expression for the complete Elliptic integral of the first kind

Question

On the Wikipedia page Elliptic integral it states, that the complete elliptic integral of the first kind has asymptotic expression $$K(k) = \frac{\pi}{2}+\frac{\pi}{8}\left(\frac{k^2}{1-k^2}\right)-\frac{\pi}{16}\left(\frac{k^4}{1-k^2}\right)+...$$ but without a reference to where it came from. Can someone provide a source? Is it possible to obtain a similar expression for the complete elliptic integral of the second kind?

Answer 1

$$K(k)=\int_0^\tfrac{\pi}{2} \frac{d\theta}{\sqrt{1-k\sin^2\theta}} = \sum_{r=0}^\infty {-1/2 \choose r} \int_0^\tfrac{\pi}{2} (-k \sin^2 \theta)^r d\theta$$

$$(1-k^2)(K(k)-K(0)) = \sum_{r=1}^\infty c_r k^r$$

$$K(k) = K(0)+\sum_{r=1}^\infty c_r \frac{k^r}{1-k^2}$$

Answer 2

In fact, what is given as an asymptotic expression in the Wikipedia page seems to be resulting from a curve fit by analogy with the series expansion (as @reuns explained) built around $k=0$ of $$(1-k^2)\left(K(k)-\frac \pi 2\right)=\frac{\pi }{8} k^2-\frac{7 \pi }{128} k^4-\frac{11 \pi }{512}k^6-\frac{375 \pi }{32768}k^8-\frac{931 \pi }{131072}k^{10}+O\left(k^{12}\right)$$ which is not fantastic at all.

Long time ago, I made similar things for approximating $K(k)$ with a sufficient accuracy for the range $0 \leq k \leq 0.95$ using as a model $$K(k)=\frac \pi 2 +\pi \sum_{n=1}^p a_n\frac{k^{2n}}{1-k^2}$$

Using $p=4$, the results are highly significant $(R^2 > 0.999999)$; they are given in the following table $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a_1 & +0.134857 & 0.001166 & \{+0.132540,+0.137174\} \\ a_2 & -0.132812 & 0.006139 & \{-0.145007,-0.120618\} \\ a_3 & +0.162040 & 0.010247 & \{+0.141685,+0.182395\} \\ a_4 & -0.151968 & 0.005431 & \{-0.162756,-0.141179\} \\ \end{array}$$

To make the results nicer in a paper published by my research team, the coefficients were rationalized and the result was given as $$K(k)\approx \frac{\pi }{2}+\frac{53 \pi }{393 }\frac{ k^2}{ \left(1-k^2\right)}-\frac{17 \pi }{128 }\frac{ k^4}{ \left(1-k^2\right)}+\frac{35 \pi }{216 }\frac{ k^6}{ \left(1-k^2\right)}-\frac{31 \pi }{204}\frac{ k^8}{\left(1-k^2\right)}$$

In fact,we built higher order models for better accuracy.

Thinking more about it, if I had to repeat it today, what I should probably use as a model is $$(1-k^2)K(k)=\frac{\pi }{2} (1-k^2)+\pi \sum_{n=1}^p a_n k^{2n}$$