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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?

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  • $\begingroup$ Welcome to Mathematics Stack Exchange Joe! Take the short tour to see how how to get the most from your time here. $\endgroup$ – dantopa Feb 28 at 2:46
  • $\begingroup$ Can you provide a link to the correct version? Thank you. $\endgroup$ – HarryT Feb 28 at 11:45
  • $\begingroup$ Graphically, the expression given on Wikipedia, with $\frac{\pi}{16}$ seems to be a better approximation! $\endgroup$ – HarryT Feb 28 at 11:54
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$$K(k)=\int_0^\tfrac{\pi}{2} \frac{d\theta}{\sqrt{1-k\sin^2\theta}} = \sum_{r=0}^\infty {-1/2 \choose r} \int_0^\tfrac{\pi}{2} (-k \sin^2 \theta)^r d\theta$$

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

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

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

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  • $\begingroup$ This is fantastic, thank you. Could you provide a link to the paper? $\endgroup$ – HarryT Feb 28 at 10:19
  • $\begingroup$ @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 :-) $\endgroup$ – Claude Leibovici Feb 28 at 10:49
  • $\begingroup$ No problem, thank you for the help. There is no need for a better approximation, thank you! $\endgroup$ – HarryT Feb 28 at 11:15
  • $\begingroup$ Sorry to bother you again, but do you have a source for the first expression you gave? Thanks! $\endgroup$ – HarryT Feb 28 at 19:08
  • $\begingroup$ @JoePenn wolframalpha.com/input/… $\endgroup$ – Claude Leibovici Mar 1 at 4:34

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