How do you solve $EllipticE [0.5,-600]$ analytically with series expansion method?

Question

How do you solve $EllipticE [0.5,-600]$ analytically with series expansion method?

I suspect a modulus transformation should be used inorder to set $0 <k <1$ where k is -600 in the example.

Any help would be appreciated

Answer 1

When $k$ is large, using Wolfram notation, and for for more simplicity letting $t=\frac 1{ \sqrt{k}}$ and consider $$E\left(\left.\frac{1}{2}\right|-k\right)= E\left(\frac{1}{2}|-\frac{1}{t^2}\right)$$ a series expansion is $$\frac{2 \sin ^2\left(\frac{1}{4}\right)}{t}+\frac{1}{4} t \left(-2 \log \left(t \left(2+\sqrt{2 (1+\cos (1))}\right)\right)+1+\log (64)+2 \log \left(\sin \left(\frac{1}{2}\right)\right)\right)+$$ $$\frac{1}{64} t^3 \left(4 \log \left(\frac{1}{4} t \left(2+\sqrt{2 (1+\cos (1))}\right)\right)+3-2 \log (2-2 \cos (1))+2 \sqrt{2 (1+\cos (1))} \csc ^2\left(\frac{1}{2}\right)\right)+O\left(t^5\right)$$

Truncated to these terms, for $k=600$, this would give $3.074528758$ while the "exact" value would be $3.074534459$.

Computing the coefficients and making them rational would give $$\frac{237}{1936 t}+\frac{1}{4} t \left(\frac{2416}{655}-2 \log \left(\frac{15813 }{4211}t\right)\right)+\frac{1}{64} t^3 \left(4 \log \left(\frac{3635 }{3872}t\right)+\frac{50914}{2761}\right)+O\left(t^5\right)$$

For the more general case, have a look here.