我需要通过分析和实数求解一个三次方程(a x ^ 3 + b x ^ 2 + c * x + d = 0),最好是纯javascript(无libs) 。由于可能有1到3个根,我认为数组是合理的结果类型。
P.S。在下面提供了我自己的解决方案,希望它有用。
答案 0 :(得分:17)
你走了。包括处理退化案例。主算法主要来自wikipedia article。
function cuberoot(x) {
var y = Math.pow(Math.abs(x), 1/3);
return x < 0 ? -y : y;
}
function solveCubic(a, b, c, d) {
if (Math.abs(a) < 1e-8) { // Quadratic case, ax^2+bx+c=0
a = b; b = c; c = d;
if (Math.abs(a) < 1e-8) { // Linear case, ax+b=0
a = b; b = c;
if (Math.abs(a) < 1e-8) // Degenerate case
return [];
return [-b/a];
}
var D = b*b - 4*a*c;
if (Math.abs(D) < 1e-8)
return [-b/(2*a)];
else if (D > 0)
return [(-b+Math.sqrt(D))/(2*a), (-b-Math.sqrt(D))/(2*a)];
return [];
}
// Convert to depressed cubic t^3+pt+q = 0 (subst x = t - b/3a)
var p = (3*a*c - b*b)/(3*a*a);
var q = (2*b*b*b - 9*a*b*c + 27*a*a*d)/(27*a*a*a);
var roots;
if (Math.abs(p) < 1e-8) { // p = 0 -> t^3 = -q -> t = -q^1/3
roots = [cuberoot(-q)];
} else if (Math.abs(q) < 1e-8) { // q = 0 -> t^3 + pt = 0 -> t(t^2+p)=0
roots = [0].concat(p < 0 ? [Math.sqrt(-p), -Math.sqrt(-p)] : []);
} else {
var D = q*q/4 + p*p*p/27;
if (Math.abs(D) < 1e-8) { // D = 0 -> two roots
roots = [-1.5*q/p, 3*q/p];
} else if (D > 0) { // Only one real root
var u = cuberoot(-q/2 - Math.sqrt(D));
roots = [u - p/(3*u)];
} else { // D < 0, three roots, but needs to use complex numbers/trigonometric solution
var u = 2*Math.sqrt(-p/3);
var t = Math.acos(3*q/p/u)/3; // D < 0 implies p < 0 and acos argument in [-1..1]
var k = 2*Math.PI/3;
roots = [u*Math.cos(t), u*Math.cos(t-k), u*Math.cos(t-2*k)];
}
}
// Convert back from depressed cubic
for (var i = 0; i < roots.length; i++)
roots[i] -= b/(3*a);
return roots;
}