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A last factorization

asked 2023-05-01 11:39:43 +0100

Cyrille gravatar image

Sorry for all, but one more time I fail to go to the last factorization. The following code

varx=var('x y') 
paramu=var('A α β U')
paramuc=tuple(list(paramu)+[U])
paramb=var('R p_x p_y')
varl=var('λ')
varg=tuple(list(varx)+[λ])
assume(x>=0)
assume(y>=0)
assume(p_x>0)
assume(p_y>0)
assume(A>0)
assume(α>0)
assume(α<1)
assume(β>0)
assume(β<1)
def Cobb_Douglas(x, y, A, α, β) :
    return A*x^α*y^β

def expense(x,y,p_x,p_y) :
    return p_x*x+ p_y*y

L(x,y,λ) = Cobb_Douglas(x,y,A, α, β) -λ*(expense(x,y,p_x,p_y)-R)
M=det(L.hessian()(x,y,λ)).full_simplify().factor().collect_common_factors()
show(M)

image description

But it is obvious that we can factor it in such a way to finaly obtain

$-(p_y^2 y^2 \alpha (\alpha-1)+ p_x^2 x^2 \beta (\beta-1) - 2 p_x p_y x y \alpha \beta) Ax^{\alpha-2}y^{\beta-2}$

Under this last factorization, it is obvious that according to the assumptions this last expression should be signed but which ever the question I ask with bool() the answer is False.

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answered 2023-05-01 20:29:14 +0100

achrzesz gravatar image

Restricting calculations to the last factorization

x,y,α,β=var('x,y,α,β')
P.<U,V> = PolynomialRing(SR, 'U,V')
hes0=-(V^2*y^2*α^2 - 2*U*V*x*y*α*β + U^2*x^2*β^2 - V^2*y^2*α - U^2*x^2*β)
hes0

(-x^2*β^2 + x^2*β)*U^2 + 2*x*y*α*β*U*V + (-y^2*α^2 + y^2*α)*V^2

Q = QuadraticForm.from_polynomial(hes0);
mG=Q.Gram_matrix();
mm=mG.apply_map(factor);mm

[-x^2*(β - 1)*β        x*y*α*β]
[       x*y*α*β -y^2*(α - 1)*α]

mm.det().factor()

-x^2*y^2*(α + β - 1)*α*β

one can observe that α,β in (0,1) doesn't imply definitness

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Asked: 2023-05-01 11:39:43 +0100

Seen: 73 times

Last updated: May 01 '23