In article <firstname.lastname@example.org>, "David W. Cantrell" <DWCantrell@sigmaxi.org> writes: >email@example.com wrote: >> In article <muqqag9mwbv2@legacy>, >> David Powers <firstname.lastname@example.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. >> >.... >>... >> 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*m-4)*E(phi,m)^5 >> +1/5040*m*(493*m^2-284*m+16)*E(phi,m)^7 >> +1/362880*m*(37369*m^3-31224*m^2+4944*m-64)*E(phi,m)^9 ... > >Yes. That's the same series I posted last week. See my Out, in which z >is the same as your E(phi,m). > >Of course, the coefficients of the polynomials in m are obtainable by >reversion of series, as we both did. But does anyone know of a simpler way >of calculating them? (I may submit the sequence of coefficients to the OEIS >in the near future.)
The direct approach to get the Taylor series of the inverse Elliptical Integral phi(E,m) of the second kind without knowing the Taylor series of the function E(phi,m) in advance is "implicit derivation". One starts from dE/dphi= sqrt(1-m*sin^2 phi), inverts it dphi/dE= 1/sqrt(1-m*sin^2 phi), inserts phi=E=0 into the right hand side to get d phi/dE (at phi=0)=1 One builds second, third etc derivatives of the equation w.r.t E and inserts the results of the previous steps iteratively into the right hand sides. d^2phi/dE^2= m sin phi cos phi/(1-m sin^2 phi)^(3/2) dphi / dE -> 0 (with dphi/dE=1, phi=0) d^3phi/dE^3= m [ (cos^2 phi -sin^2 phi)/(1-m sin^2 phi)^(3/2) -3/2*...] * d phi/dE + m sin phi cos phi /( 1-m sin^2 phi)^3/2 d^2 phi/dE^2 -> m (because phi=0, d phi/dE=1, d^2phi/dE^2=0) Finally, with E=0, d phi/dE=1, d phi^2/dE^2=0, dphi^3/dE^3=m one builds the power series phi(E)=E+mE^3/3!... etc. There is no real advantage in this case compared to inverting the power series.