Re: Hypergeometric2F1

*To*: mathgroup at smc.vnet.net*Subject*: [mg93157] Re: Hypergeometric2F1*From*: Roland Farnzius <roland.franzius at uos.de>*Date*: Wed, 29 Oct 2008 05:49:24 -0500 (EST)*References*: <ge6nfi$li8$1@smc.vnet.net>

Artur wrote: > Who know which Mathematica procedure to use to find such a,b,c that > ArcCosh[2]/ArcCosh[2-x]==Hypergeometric2F1[a,b,c,x] for {x,-Infinity,1} If you want a solution to hold exactly for all x it does not seem to work. coeff = (List @@ Series[ArcCosh[2]/ArcCosh[2 - x] - Hypergeometric2F1[a, b, c, x], {x, 0, 3}][[3]]) gives the coefficient list of the difference from the Tylor series in x at 0: {-((a*b)/c) + 1/(Sqrt[3]*ArcCosh[2]), -((a*(1 + a)*b*(1 + b))/(2*c*(1 + c))) + 1/(3*ArcCosh[2]^2) + 1/(3*Sqrt[3]*ArcCosh[2]), -((a*(1 + a)*(2 + a)*b*(1 + b)*(2 + b))/(6*c*(1 + c)*(2 + c))) + 1/(3*Sqrt[3]*ArcCosh[2]^3) + 2/(9*ArcCosh[2]^2) + 1/(6*Sqrt[3]*ArcCosh[2])} Two numerical solutions of the parameters {a,b,c} can be obtained sol = NSolve[(0 == #) & /@ coeff, {a, b, c}, 36] But the following coefficients in the Taylor expansion are not zero N[(List @@ Series[ArcCosh[2]/ArcCosh[2 - x] - Hypergeometric2F1[a, b, c, x], {x, 0, 7}][[3]])] /. sol // Simplify // Chop {{0, 0, 0, -0.0004795487345876648, -0.0010026916113038686, -0.0014561420339331, -0.0018271816030282495}, {0, 0, 0, -0.0004795487345877203, -0.0010026916113039241, -0.0014561420339331, -0.0018271816030282495}} -- Roland Franzius