Items tagged with elementary

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Using convert(f,elsymfun) when f is a symmetric polynomial will write it in terms of elementary symmetric polynomials. For example, x^2+y^2 would become (x+y)^2-2x*y. For this command not to return an error, f must be symmetric, equivalently of type symmfunc(all indeterminants in expression).

I'd like to use this in a broader sense, when there are extra variables hanging around. For example, suppose f is a polynomial in a,b,c,x,y,z, and symmetric in x,y,z, so that type(f,symmfunc(x,y,z)) returns true. Then it is still possible to write this in terms of the elementary symmetric polynomials on x,y,z (with coefficients taken being rational polynomials in a,b,c).

For example, a+x^2+y^2 can be written as a+(x+y)^2-2x*y.

Is there a command available for this? Or is there a roundabout way to make Maple forget temporarily that a,b,c are indeterminants, before putting them back in after?

Thanks

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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