| American Citizen on Fri, 28 Mar 2025 04:35:40 +0100 |
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| question on converting general cubic equations to Weierstrass format |
For example, exploring body cuboids, I have the general cubic equation (where a,b is found from a Pythagorean ratio r, such as r = 3/4, i.e. a=3 and b=4 where a is the numerator of r and b is the denominator of r.
B(a,b) = 0 * X^3 - 4*a*b * X^2*Y + 2*(a^2-b^2) * X*Y^2 + 0 * Y^3 + 0 * X^2*Z + 2*(b^2+2*a*b-a^2) * X*Y*Z + 0 * Y^2*Z + 0 * X*Z^2 + 0 * Y*Z^2 + 0 * Z^3
I would like to find the Weierstrass normal form for this equation, i.e [a1,a2,a3,a4,a6] where y^2 + a1*x*y + a3*y = x^3 + a2*x^2 + a4*x + a6
For the edge and face cuboids I also have E(a,b) = a*v*(u*u+1) - b*u*(v*v+1) = 0 F(a,b) = (a^2+b^2)*u*(v^2-1) - 2*a*b*(u^2-1)*v where a,b are given parameters and u,v are the general variables.I am seeking the Weierstrass normal form for all 3 equations, B(a,b), E(a,b) and F(a,b)
Randall