Math Forum / Math Software / Mathematica / October 2008

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}}

Hi,

FindInstance[
 ArcCosh[2]/ArcCosh[2 - x] == Hypergeometric2F1[a, b, c, x],
  {x, a, b, c}]

??

Regards
  Jens

It seems that you are wanting to determine a,b,c such that

ArcCosh[2]/ArcCosh[2-x]==Hypergeometric2F1[a,b,c,x]

would be an identity for x < 1. But that is not possible. What made you
think it would be possible?

David

On 10/29/08 at 5:50 AM, DWCantrell@sigmaxi.net (David W. Cantrell)
wrote:

Given

In[17]:= Hypergeometric2F1[a, b, c, 0]

Out[17]= 1

It appears there are infinitely many solutions. Is the result
returned by Mathematica for Hypergeometric2F1[a,b,c,x] incorrect
when x = 0? What am I missing here?