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Question: How to simplify this hypergeometric function?

Question:How to simplify this hypergeometric function?

Markiyan Hirnyk 7804 Maple 2016
convert simplify hypergeom elementary
January 04 2017
0

It is suggested  

hypergeom([1/3, 2/3], [3/2], (27/4)*z^2*(1-z)) = 1/z

if z > 1. Here is my try to prove that with Maple:


 

a := `assuming`([convert(hypergeom([1/3, 2/3], [3/2], (27/4)*z^2*(1-z)), elementary)], [z > 1])

-(1/((1/2)*(27*z^3-27*z^2+4)^(1/2)+(3/2)*z*(3*z-3)^(1/2))^(1/3)-1/((1/2)*(27*z^3-27*z^2+4)^(1/2)-(3/2)*z*(3*z-3)^(1/2))^(1/3))/(z*(3*z-3)^(1/2))

 (1)

b := `assuming`([simplify(a, symbolic)], [z >= 1])

2*(-(12*(3*z+1)^(1/2)*z-12*z*(3*z-3)^(1/2)-8*(3*z+1)^(1/2))^(1/3)+(12*(3*z+1)^(1/2)*z+12*z*(3*z-3)^(1/2)-8*(3*z+1)^(1/2))^(1/3))/((3*z-3)^(1/2)*(12*(3*z+1)^(1/2)*z+12*z*(3*z-3)^(1/2)-8*(3*z+1)^(1/2))^(1/3)*(12*(3*z+1)^(1/2)*z-12*z*(3*z-3)^(1/2)-8*(3*z+1)^(1/2))^(1/3)*z)

 (2)

plot(1/b, z = 1 .. 10)

  

simplify(diff(1/b, z), symbolic)

-48*(((3*z-2)*(3*z+1)^(1/2)+z*(3*z-3)^(1/2))*((12*z-8)*(3*z+1)^(1/2)-12*z*(3*z-3)^(1/2))^(1/3)+((12*z-8)*(3*z+1)^(1/2)+12*z*(3*z-3)^(1/2))^(1/3)*((-3*z+2)*(3*z+1)^(1/2)+z*(3*z-3)^(1/2)))/((3*z+1)^(1/2)*(3*z-3)^(1/2)*((12*z-8)*(3*z+1)^(1/2)+12*z*(3*z-3)^(1/2))^(2/3)*((12*z-8)*(3*z+1)^(1/2)-12*z*(3*z-3)^(1/2))^(2/3)*(((12*z-8)*(3*z+1)^(1/2)-12*z*(3*z-3)^(1/2))^(1/3)-((12*z-8)*(3*z+1)^(1/2)+12*z*(3*z-3)^(1/2))^(1/3))^2)

 (3)

``


 

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