解析立方方程的功能 [英] Function to solve cubic equation analytically

问题描述

我需要解析和实数求解三次方程(a x ^ 3 + b x ^ 2 + c * x + d = 0)，最好是纯JavaScript(无库) .由于可能会有1到3个根，所以我认为数字数组是一种合理的结果类型.

I need to solve a cubic equation (ax^3 + bx^2 + c*x + d = 0) analytically and in real numbers, preferably in pure javascript (no libs). As there could be 1 to 3 roots, I think an array of numbers is a reasonable result type.

P.S.在下面提供了我自己的解决方案，希望它会有用.

P.S. Provided my own solution below, hope it'll be useful.

推荐答案

在这里.包括处理退化案件.主要算法主要来自维基百科文章.

Here you go. Includes handling degenerate cases. Main algorithm is mostly from 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;
}

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