I'm trying to solve the inequality 18bcd−4b3d+b2c2−4c3−27d2>0 with b,c,d∈R and c<0. The goal is to obtain an expression for d.
My guess would be to use solve_ineq() and perhaps assume() for the additional condition, but I can't figure out how. In case of a symbolic equation I'd use
b,c,d = var('b,c,d')
Delta = (18*b*c*d - 4*b^3*d + b^2*c^2 - 4*c^3 - 27*d^2 == 0)
solve(Delta,d)What version is this? I get with 6.8beta8
sage: b,c,d = var('b,c,d')
sage: Delta = (18*b*c*d - 4*b^3*d + b^2*c^2 - 4*c^3 - 27*d^2 == 0)
sage: solve(Delta,d)
[d == -2/27*b^3 + 1/3*b*c + 2/27*(b^2 - 3*c)^(3/2), d == -2/27*b^3 + 1/3*b*c - 2/27*(b^2 - 3*c)^(3/2)]which is fine.
Your question is unclear: do you want to solve the inequality, or do you want to express d in terms of the other variables, knowing f(a,b,c,d)>0 and c<0?
In the first case, you have a polynomial expression f(a,b,c,d) which you want to be positive. The usual way to solve that is to understand where the expression is zero.
Since f is continuous, it has a fixed sign on each connected component of the complement of this zero locus, so all that remains to understand is on which components f is positive.
In the second case, calling Δ the fixed value of f(a,b,c,d), notice that the equation f(a,b,c,d)=Δ can be rewritten as A∗d2+B∗d+C=0, where A, B, C are expressions in a, b, c, d, Δ.
In the first case your solution is fine:
var('b,c,d')
assume(c<0)
solve(18bcd - 4b^3d + b^2c^2 - 4c^3 - 27d^2==0,d)
But I would like to know if there is a way to have a solution without "trying to understand" where the sign change. I think is the main point to use a math software. Furthermore it'll be pretty annoying if there is anything like that implemented already as with Mathematica is a one line command
Reduce[{b^2 c^2-4 c^3-4 b^3 d+18 b c d-27 d^2>0,c<0},d]
Asked: 10 years ago
Seen: 2,580 times
Last updated: Jul 15 '15
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