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Obtain series expansions using Frobenius Method

Posted 7 years ago
POSTED BY: Brad Klee
4 Replies
Posted 7 years ago

Hi Daniel,

I think this looks okay. The other "hacker" way to do the calculation is:

Normal@Series[ x Exp[   FullSimplify@CoefficientList[
       Normal@  Series[- 2 Pi  /Sqrt[3] Hypergeometric2F1[1/3, 2/3, 1, 1 - x]
           /Hypergeometric2F1[1/3, 2/3, 1, x], {x, 0, 10}], Log[x]][[1]]], {x, 0, 3}]

Outputs:

Out[]:=(1/27)*x + (5 /243)*x^2 + (31/2187)* x^3

And the integerization is as above. There is another question as to whether the output should be a series with rational coefficients or lots of Gamma, PolyGamma whatever. I guess it doesn't matter if FullSimplify then works, nice. You are the expert, so I guess I will just leave it up to you from here, thanks though. This is much better than the original call.

--Brad

POSTED BY: Brad Klee

All four of the original examples will give more "expanded" expansions in future. Here is the second one, pared back for brevity.

Series[Exp[-2 Pi/Sqrt[3] Hypergeometric2F1[1/3, 2/3, 1, 1 - x]/Hypergeo
metric2F1[1/3, 2/3, 1, x]], {x, 0, 1}] // InputForm                             

(* Out[20]//InputForm= 
x^((2*Pi)/(Sqrt[3]*Gamma[1/3]*Gamma[2/3]))*
 SeriesData[x, 0, {E^((4*EulerGamma*Pi)/(Sqrt[3]*Gamma[1/3]*Gamma[2/3]) + 
     (2*Pi*PolyGamma[0, 1/3])/(Sqrt[3]*Gamma[1/3]*Gamma[2/3]) + 
     (2*Pi*PolyGamma[0, 2/3])/(Sqrt[3]*Gamma[1/3]*Gamma[2/3])), 
   (-4*E^((4*EulerGamma*Pi)/(Sqrt[3]*Gamma[1/3]*Gamma[2/3]) + 
       (2*Pi*PolyGamma[0, 1/3])/(Sqrt[3]*Gamma[1/3]*Gamma[2/3]) + 
       (2*Pi*PolyGamma[0, 2/3])/(Sqrt[3]*Gamma[1/3]*Gamma[2/3]))*Pi*
     (2 + PolyGamma[0, 1/3] + PolyGamma[0, 2/3] - PolyGamma[0, 4/3] - 
      PolyGamma[0, 5/3]))/(9*Sqrt[3]*Gamma[1/3]*Gamma[2/3])}, 0, 2, 1] *)
POSTED BY: Daniel Lichtblau
Posted 7 years ago
POSTED BY: Brad Klee
POSTED BY: Daniel Lichtblau
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