数值分析之牛顿法多项式求根

很朴素的方法，如果在区间[a,b]内有根，那么f(a)*f(b)<0

不难得到以下代码：

 

#include <iostream>

#include <memory>



using namespace std;



double f(int m, double c [], double x)

{

	int i;

	double p = c[m];



	for (i = m; i > 0; i--)

		p = p*x + c[i - 1];

	return p;

}



int newton(double x0, double *r,

	double c [], double cp [], int n,

	double a, double b, double eps)

{

	int MAX_ITERATION = 1000;

	int i = 1;

	double x1, x2, fp, eps2 = eps / 10.0;



	x1 = x0;

	while (i < MAX_ITERATION) {

		x2 = f(n, c, x1);

		fp = f(n - 1, cp, x1);

		if ((fabs(fp)<0.000000001) && (fabs(x2)>1.0))

			return 0;

		x2 = x1 - x2 / fp;

		if (fabs(x1 - x2) < eps2) {

			if (x2<a || x2>b)

				return 0;

			*r = x2;

			return 1;

		}

		x1 = x2;

		i++;

	}

	return 0;

}



double Polynomial_Root(double c [], int n, double a, double b, double eps)

{

	double *cp;

	int i;

	double root;



	cp = (double *) calloc(n, sizeof(double) );

	for (i = n - 1; i >= 0; i--) {

		cp[i] = (i + 1)*c[i + 1];

	}

	if (a > b) {

		root = a; a = b; b = root;

	}

	if ((!newton(a, &root, c, cp, n, a, b, eps)) &&

		(!newton(b, &root, c, cp, n, a, b, eps)))

		newton((a + b)*0.5, &root, c, cp, n, a, b, eps);

	free(cp);

	if (fabs(root) < eps)

		return fabs(root);

	else

		return root;

}



int main()

{

	double c[2] = { 1,2 };

	int n = 1, a = -5, b = 5;

	double eps = 1e-8;

	double root = Polynomial_Root(c, n, a, b, eps);

	cout << root << endl;

}