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