Exemplo n.º 1
0
        // f(x) = x^3 + cx + d = 0
        public static double[] Solve(double c, double d)
        {
            if (d < 0)
            {
                return(Solve(c, -d).Reverse().Select(x => - x).ToArray());
            }
            // 自明解
            if (d == 0 && c >= 0)
            {
                return new[] { 0D }
            }
            ;

            // 負の実数解
            var x1 = SolveNegative();

            // f(x) = (x - x_1) (x^2 + x_1 x + x_1^2 + c)
            return(QuadraticEquation0.Solve(1, x1, x1 * x1 + c).Prepend(x1).ToArray());

            double SolveNegative()
            {
                var f = CreateFunction(c, d);

                var f1 = CreateDerivative(c);
                var x0 = -1D;

                while (f(x0) > 0)
                {
                    x0 *= 2;
                }
                return(NewtonMethod.Solve(f, f1, x0));
            }
        }
    }
Exemplo n.º 2
0
        // f(x) = ax^3 + bx^2 + cx + d = 0
        public static double[] Solve(double a, double b, double c, double d)
        {
            if (a == 0)
            {
                throw new ArgumentException("The value must not be 0.", nameof(a));
            }
            if (a < 0)
            {
                return(Solve(-a, -b, -c, -d));
            }

            var f  = CreateFunction(a, b, c, d);
            var xc = -b / (3 * a);
            var yc = f(xc);

            if (yc < 0)
            {
                return(Solve(a, -b, c, -d).Reverse().Select(x => - x).ToArray());
            }

            var f1 = CreateDerivative(a, b, c);

            // 自明解
            if (yc == 0 && f1(xc) >= 0)
            {
                return new[] { xc }
            }
            ;

            // xc より小さい実数解
            var x1 = SolveNegative();
            var p  = a * x1 + b;
            var q  = p * x1 + c;

            return(QuadraticEquation0.Solve(a, p, q).Prepend(x1).ToArray());

            double SolveNegative()
            {
                var x0 = -1D;

                while (f(xc + x0) > 0)
                {
                    x0 *= 2;
                }
                return(NewtonMethod.Solve(f, f1, xc + x0));
            }
        }
    }