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From what I learned here, for evaluating complete elliptic integrals, using the binomial theorem is quite effective as it puts the elliptic integral in the form of. $$\sum_{n=1}^{\infty} some-constants\int_0^{\dfrac{\pi}{2}} \sin^{2n} \phi d\phi $$ which turns out to be easy to evaluate. Yet when it turns in to the form $$\sum_{n=1}^{\infty} some-constants\int_0^{\theta} \sin^{2n} \phi d\phi $$ which is quite common among elliptic integrals, I start to not know what to do. I resolved to go on with conducting a Taylor expansion where the derivatives(as I do not know how to evaluate the derivatives) I computed using Matlab. Yet what is the best way or the standard way(if there is such a way) to conduct/approximate these kinds of incomplete elliptic integrals? As a reference, I want am trying to compute $$ F(\dfrac{\theta + \gamma}{2}|\dfrac{2a}{a-1})$$

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  • $\begingroup$ @Rohan Thanks for the comment! But isn't there a way to at least approximate it? I really will prefer to be able to do that. $\endgroup$ – Isamu Isozaki Dec 28 '17 at 7:20
  • $\begingroup$ @Rohan wait, then how do we find values associated with the incomplete elliptic integrals? $\endgroup$ – Isamu Isozaki Dec 28 '17 at 7:52
  • $\begingroup$ @Rohan I'm sorry for my persistence, yet when I inputted (in WolframAlpha ) the elliptic integral EllipticF(0.4 | 0.5) it returns the value 0.405352. It also returned a graph when I inputted a random elliptic function. So, I think it is possible. $\endgroup$ – Isamu Isozaki Dec 28 '17 at 8:18
  • $\begingroup$ @Rohan I am sorry for any mistakes I am making. $\endgroup$ – Isamu Isozaki Dec 28 '17 at 8:21
  • $\begingroup$ Sorry sir, I am wrong. See under the section Connections within the group of incomplete elliptic integrals and with other function groups here. $\endgroup$ – Rohan Dec 28 '17 at 8:21
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If you want to compute $$y=F(x|k)$$ for small values of $k$ and $x$, you could use series expansion $$y=\sum_{n=1}^p a_n\, x^{2n-1}+O(x^{2p+1})$$ where the first coefficients would be $$a_1=1 \qquad a_2=\frac{k}{6}\qquad a_3=\frac{ k (9 k-4)}{120}\qquad a_4=\frac{k \left(225 k^2-180 k+16\right)}{5040}$$

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  • $\begingroup$ Thanks for the answer! Even if the values are not especially small, will the above trick still work? $\endgroup$ – Isamu Isozaki Dec 28 '17 at 10:27
  • $\begingroup$ I am sorry, a correction. "Will the trick above work?" is the correct wording $\endgroup$ – Isamu Isozaki Dec 28 '17 at 10:45

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