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On Jul 20, 11:33 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
> On Jul 20, 10:10 am, cliclic...@freenet.de wrote:
>
>
> > Vladimir Bondarenko schrieb:
>
> > > 4 (2 + Sqrt[3])^(1/4)/3 *
> > > Hypergeometric2F1[-1/4, 3/4, 7/4, (2 - Sqrt[3])/4]
>
> > > - (2 + Sqrt[3])^(1/4)/3 *
> > > Hypergeometric2F1[3/4, 3/4, 7/4, (2 - Sqrt[3])/4]
>
> > > + ((2 + Sqrt[3])^(1/4) *
> > > Hypergeometric2F1[3/4, 3/4, 7/4, (2 - Sqrt[3])/4])/(2 Sqrt[3])
>
> > > - (2*(2 + Sqrt[3])^(1/4) *
> > > Hypergeometric2F1[-1/4, 3/4, 7/4, (2 - Sqrt[3])/4])/Sqrt[3]
>
> > > ?
>
> > On Derive 6.10:
>
> > " a library function as discussed last year ... "
>
> > F21_aux2(a,b,c,z):=GAMMA(c)/(GAMMA(b)*GAMMA(c-b))*'INT(t^(b-1)*(~
> > 1-t)^(c-b-1)*(1-t*z)^(-a),t,0,1)
>
> > F21_aux1(a,b,c,z):=IF(RE(b)>0 AND RE(c)>RE(b),F21_aux2(a,b,c,z),~
> > ((c-a+1)*F21_aux1(a,b+1,c+2,z)+a*(c-(c-b)*z)*F21_aux1(a+1,b+1,c+~
> > 2,z)/c)/(c+1))
>
> > F21(a,b,c,z):=IF(ABS(2*RE(b)-RE(c))<ABS(2*RE(a)-RE(c)),F21_aux1(~
> > a,b,c,z),F21_aux1(b,a,c,z),'F21(a,b,c,z))
>
> > " the challenge ... "
>
> > 4*(2+SQRT(3))^(1/4)/3*F21(-1/4,3/4,7/4,(2-SQRT(3))/4)-(2+SQRT(3)~
> > )^(1/4)/3*F21(3/4,3/4,7/4,(2-SQRT(3))/4)+(2+SQRT(3))^(1/4)*F21(3~
> > /4,3/4,7/4,(2-SQRT(3))/4)/(2*SQRT(3))-2*(2+SQRT(3))^(1/4)*F21(-1~
> > /4,3/4,7/4,(2-SQRT(3))/4)/SQRT(3)
>
> > 2^(3/4)*SQRT(3*SQRT(3)-5)*INT(t^(3/4-1)*(1-t)^(7/4-3/4-1)*(1-t*(~
> > 1/2-SQRT(3)/4))^(1/4),t,0,1)/4-2^(3/4)*SQRT(3*SQRT(3)-5)*INT(t^(~
> > 3/4-1)*(1-t)^(7/4-3/4-1)*(1-t*(1/2-SQRT(3)/4))^(-3/4),t,0,1)/16
>
> > SQRT(3)/4-1/4
>
> > The reduction takes 86 + 1 steps, the 86th step fails in single-step
> > mode.
>
> > Martin.
>
> > (I had to remove two groups because aioe.org accepts three at most.)
>
> Martin, you showed me again what I felt for two decades. That Derive
> is built in a very powerful way.
>
> Now my question is if someone could show how to get to this nice
> answer using the built-in Maple or Mathematica operations?
>
> Vladimir
Can do this using a formula from functions.wolfram.com*.
http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/17/01/02/0001/
In[5]:= e1 = (4*(2 + Sqrt[3])^(1/4)*
Hypergeometric2F1[-1/4, 3/4, 7/4, (2 - Sqrt[3])/4])/
3 - (2*(2 + Sqrt[3])^(1/4)*
Hypergeometric2F1[-1/4, 3/4, 7/4, (2 - Sqrt[3])/4])/
Sqrt[3] - ((2 + Sqrt[3])^(1/4)*
Hypergeometric2F1[3/4, 3/4, 7/4, (2 - Sqrt[3])/4])/
3 + ((2 + Sqrt[3])^(1/4)*
Hypergeometric2F1[3/4, 3/4, 7/4, (2 - Sqrt[3])/4])/(2*Sqrt[3]);
In[6]:= e2 = FullSimplify[e1]
Out[6]= (1/(6*(26 + 15*Sqrt[3])^(
1/4)))*(4*Hypergeometric2F1[-(1/4), 3/4, 7/4, 1/4*(2 - Sqrt[3])] -
Hypergeometric2F1[3/4, 3/4, 7/4, 1/4*(2 - Sqrt[3])])
We'll define:
zz = 1/4*(2 - Sqrt[3]);
rul = {a -> 3/4, b -> 3/4, c -> 7/4, z -> zz};
Now extract the hypergeometrics as "variables".
e3 = e2 /. Hypergeometric2F1 -> hHypergeometric2F1;
vars = Cases[e3, HoldPattern[hHypergeometric2F1[___]], Infinity];
Next define a replacement we can use for algebraic substitution of
those hypergeometrics.
In[29]:= rul2 =
Expand[4*((a - c) Hypergeometric2F1[a - 1, b, c,
z] + (c - 2 a + (a - b) z) Hypergeometric2F1[a, b, c,
z]) /. rul] -
Expand[a (z - 1) Hypergeometric2F1[a + 1, b, c, z] /. rul] /.
Hypergeometric2F1 -> hHypergeometric2F1
Out[29]= 3/(8 (1 + 1/4 (-2 + Sqrt[3]))^(3/4)) + (3 Sqrt[3])/(
16 (1 + 1/4 (-2 + Sqrt[3]))^(3/4)) -
4 hHypergeometric2F1[-(1/4), 3/4, 7/4, 1/4 (2 - Sqrt[3])] +
hHypergeometric2F1[3/4, 3/4, 7/4, 1/4 (2 - Sqrt[3])]
WIth this we substitute and again simplify.
In[33]:= FullSimplify[PolynomialReduce[e3, rul2, vars][[2]]]
Out[33]= 1/16 (-1 + Sqrt[3])
Daniel Lichtblau
Wolfram Research
* I am not an actual hypergeometer, though sometimes I portray one on
television.
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