The following two functions
f(u)=-u/2+min_symbolic(0,u) + min_symbolic(u,2*u) and
g(u)=-u/2+min_symbolic(u, 3*u)
are mathematically equivalent, but the following symbolic integrals give
different results.
F(x)=integral(f(u)*u*exp(x*u),u,-1,1)
FF(y)=F(y)
G(x)=integral(g(u)*u*exp(x*u),u,-1,
The following Sage code
var('u')
integrate(-1/8*(u - 2*min_symbolic(-1/2*u - 1/2, 1/2*u - 1/2) - 1)^2 +
1/2*(min_symbolic(0, u) + 1)^2,u,-1,1)
produces 1/2 as a (correct) result with SageMath 8.3 (running with Software
Environment Ubuntu 18.04 @ 2018-08-27 in CoCalc).
With SageMath 9.3 (ru
Yes I am aware of it and also use it. What I need is a step by step Fourier
Motzkin decomposition. It's for teaching reasons. I want also be able to draw
the projections implied by it from 3D to 2D and 1D. Then I plot it in Asymptote
(because on html it has latex label even in 3D, which is not t
Just checking - are you aware of Sage's functionality to deal with
constructing of polyhedra by linear inequalities and equations? It seems
you are trying to do the same, but in a rather inefficient way.
On Tue, 15 Jun 2021, 09:58 cyrille.piate...@univ-orleans.fr, <
cyrille.piate...@univ-orleans.f
The following code write a system of inequations
x = vector(SR, SR.var('x_', 7))
A = random_matrix(SR,7,7)
b = random_vector(SR,7)
o = zero_vector(SR,7)
Z=[SR(A[i]*x-b[i])<=SR(o[i]) for i in range(A.nrows())]
SR(x[1]).variables()[0]
Sol=[solve(SR(Z[i]),SR(x[2]).variables()[0])[1] for i in r
