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Topic: Inverting elliptic integrals
Replies: 6   Last Post: Aug 10, 2004 3:59 AM

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Richard Mathar

Posts: 15
Registered: 12/8/04
Re: Inverting elliptic integrals
Posted: Aug 9, 2004 5:19 AM
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In article <muqqag9mwbv2@legacy>,
David Powers <> writes:
>I need the inverse of the (incomplete) elliptic integral of the second
>kind. It is well defined but I can't track down a solution or code
>for it.

For a numerical representation of the inverse in terms of the angle phi,
where E(phi,m)=int(theta=0..phi) sqrt(1-m*sin^2 theta) dtheta is the
elliptic integral fo the second kind, one could expand E(phi,m) in a power
series around phi=0,

Emphi := phi
+1/9*(1/630*m-1/40*m^2+1/16*m^3-5/128*m^4)*phi^9 ...

The expansion coefficients in front of the order phi^(n+1) (n=2,4,6,...) are
[sum over k=2,4,..,n of U(k,n)*F(k,m)/k!]/(n+1)!
U(k,n) = (-1)^[(k+n)/2]/2^(k-n)*[sum l=0,1,...k of (-1)^l (l-k/2)^n*binom(k,l)]
F(k,m) = -m^(n/2)*[(k-1)!!]^2/(k-1) ,
with (k-1)!! = 1*3*5*7*...*(k-1) .

Then invert this as outlined in chapt 3.6.25 of the book edited by M Abramowitz
and I Stegun:

phi := E(phi,m)
+1/362880*m*(37369*m^3-31224*m^2+4944*m-64)*E(phi,m)^9 ...

Another efficient ansatz is a higher order Newton method, since the
derivatives of E(phi,m) with respect to phi are well known:
d E(phi,m)/d phi = sqrt(1-m*sin^2 phi)
This needs in addition a solid implementation of the original E(phi,m) itself.

Richard J. Mathar,

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