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MathGroup Archive 2000

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Re: Please help with a Hypergeometric2F1 problem...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg23252] Re: Please help with a Hypergeometric2F1 problem...
  • From: Ronald Bruck <bruck at math.usc.edu>
  • Date: Sat, 29 Apr 2000 22:04:57 -0400 (EDT)
  • Organization: Univ of Southern California
  • References: <8e3b5h$kom@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

In article <8e3b5h$kom at smc.vnet.net>, zeno at magicnet.net wrote:

> I symbolically integrated the function..(x^2*(x-1))^(1/3) with respect to
> x.
> 
> There is in the answer... Hypergeometric2F1[2/3,2/3,5/3,x]
> 
> I can do nothing more with that..it just returns it. A Hypergeometric2F1
> with different parameters like Hypergeometric2F1[2,2,5,x] gives an 
> answer. I
> am using version 3. Is Mathematica unable to compute it?
> 
> I can get the Integral with out the Hypergeometric function on the 
> TI-92+,
> (it gives the answer in a different for using Tan, etc.) but I still 
> would
> like to work with the Mathematica answer.
> 

zeno e-mailed me the TI-92+ solution, which translated to Mathematica is

  Log[(((x - 1)^(1/3) - x^(1/3))^2/((x*(x - 1))^(1/3) + (x - 1)^(2/3) + 
              x^(2/3)))]/18 - 
  Sqrt[3]*ArcTan[Sqrt[3]*(2*(x - 1)^(1/3) + x^(1/3)) /(3*x^(1/3))]/9 + 
  x^(2/3)*(x - 1)^(4/3)/2 + x^(2/3)*(x - 1)^(1/3)/3

(I've removed a couple of Abs from the expression, since Mathematica 
doesn't like to differentiate these; as phrased, it's valid for x >= 1).

I am **impressed**.  While I am unable to coax Mathematica to 
differentiate this with respect to x and simplify the derivative to the 
original expression, when I plot the difference on [1,2] all I get is 
the typical roundoff noise.  It seems to be correct.

--Ron Bruck

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