
Re: Inverting elliptic integrals
Posted:
Aug 9, 2004 5:19 AM


In article <muqqag9mwbv2@legacy>, David Powers <powers@computer.org> 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(1m*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 1/6*m*phi^3 +1/5*(1/6*m1/8*m^2)*phi^5 +1/7*(1/45*m+1/12*m^21/16*m^3)*phi^7 +1/9*(1/630*m1/40*m^2+1/16*m^35/128*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)! where U(k,n) = (1)^[(k+n)/2]/2^(kn)*[sum l=0,1,...k of (1)^l (lk/2)^n*binom(k,l)] and F(k,m) = m^(n/2)*[(k1)!!]^2/(k1) , with (k1)!! = 1*3*5*7*...*(k1) .
Then invert this as outlined in chapt 3.6.25 of the book edited by M Abramowitz and I Stegun:
phi := E(phi,m) +1/6*m*E(phi,m)^3 +1/120*m*(13*m4)*E(phi,m)^5 +1/5040*m*(493*m^2284*m+16)*E(phi,m)^7 +1/362880*m*(37369*m^331224*m^2+4944*m64)*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(1m*sin^2 phi) This needs in addition a solid implementation of the original E(phi,m) itself.
Richard J. Mathar, http://www.strw.leidenuniv.nl/~mathar

