Re: Re: Please help with a Hypergeometric2F1 problem...

*To*: mathgroup at smc.vnet.net*Subject*: [mg23302] Re: [mg23252] Re: Please help with a Hypergeometric2F1 problem...*From*: BobHanlon at aol.com*Date*: Sun, 30 Apr 2000 21:13:53 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

expr = 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; D[expr, x] // FullSimplify // PowerExpand // FullSimplify (x - 1)^(1/3)*x^(2/3) Would need to verify conditions for which PowerExpand operation is valid. In a message dated 4/29/2000 10:36:41 PM, bruck at math.usc.edu writes: >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. > Bob BobHanlon at aol.com