How do you find the Inverse of Elliptic Integral of Second Kind when modulus is large

Question

So I tried to take the inverse of EllipticE when modulus is large, in Mathematica, but the solution gives wrong answer.

InverseSeries[Series[EllipticE[x, -k], {x, 0, 12}, {k, Infinity, 1}],y] = InverseFunction[y,k]

For example, I tried EllipticE[0.5,-9.9] = 0.656 where x:0.5 , k:-9.9, y:0.656

But InverseFunction[y,k] is not equal to 0.5. Am I not correctly taking the inverse of the function? I need a general form of an equation for the inverse of EllipticE. Polynomial approximation is also fine. The approximation should definitely work around when x-->0 and k-->-infinity. So for the above example, the approximation function result should yield to 0.5 when y=0.656 and k=-9.9. I need to code this function in MCU, so I need an analytical approximation.

Caner

Answer 1

You can always do this if you are not tied to a series:

f = InverseFunction[EllipticE[#1, -9.9] &]

f[.656025]
(*0.5*)

or

g = InverseFunction[Function[{x, y}, EllipticE[x, y]], 1, 2]

g[.656025, -9.9]
(*0.5*)

This series works for smaller x

Clear[k]

f[x_, k_] = InverseSeries[Series[EllipticE[x, k], {x, 0, 20}]] // Simplify // Normal

k = -9.9;

f[.1, k]
(*0.0984504*)

EllipticE[%, k] // Chop
(*0.1*)

k = -50;

f[.1, k]
(*0.0935462*)

EllipticE[%, k] // Chop
(*0.0999678*)

The series still blows up for an x value as high as 0.656, so I guess it is a numerical issue and will probably require many more terms in the series with much higher precision than is practical. For smaller values of x this seems to give a decent approximation.