# Asymptotic expression for the complete Elliptic integral of the first kind

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?

• Welcome to Mathematics Stack Exchange Joe! Take the short tour to see how how to get the most from your time here. – dantopa Feb 28 at 2:46
• Can you provide a link to the correct version? Thank you. – HarryT Feb 28 at 11:45
• Graphically, the expression given on Wikipedia, with $\frac{\pi}{16}$ seems to be a better approximation! – HarryT Feb 28 at 11:54

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

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

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

• Thank you for answering. Where does the first equality come from? Do you have any recommended reading for this subject? – HarryT Feb 28 at 3:13
• Do you mean the first series ? It is the binomial series $(1+x)^{-1/2} = \sum_{r=0}^\infty {-1/2 \choose r} x^r$ – reuns Feb 28 at 3:14
• I see, thank you. Do you know where I can find more examples, perhaps for the complete elliptic integral of the second kind? – HarryT Feb 28 at 3:22
• For the second kind it works exactly the same way with ${1/2 \choose r}$ – reuns Feb 28 at 3:25
• How would I determine the coefficients $c_{r}$? – HarryT Feb 28 at 11:05

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 $$(1-k^2)K(k)=\frac{\pi }{2} (1-k^2)+\pi \sum_{n=1}^p a_n k^{2n}$$

• This is fantastic, thank you. Could you provide a link to the paper? – HarryT Feb 28 at 10:19
• @JoePenn. Fantastic is not the word I should use : it is a reasonable approximation, that's all. Concerning the paper, it was published around 40 years ago in the proceedings of I do not remember which conference (problem of age !). The only thing I kept was the formula I wrote (I do not know where are the higher order models we built). If you want better approximations, I could do it. If you are seriously interested, let me know. With the current machinary, it would be easy (even for you). But again, I do not refuse to do it. Cheers :-) – Claude Leibovici Feb 28 at 10:49
• No problem, thank you for the help. There is no need for a better approximation, thank you! – HarryT Feb 28 at 11:15
• Sorry to bother you again, but do you have a source for the first expression you gave? Thanks! – HarryT Feb 28 at 19:08
• @JoePenn wolframalpha.com/input/… – Claude Leibovici Mar 1 at 4:34