Newton迭代法解非线性方程组的C语言实现

#include<stdio.h>
#include<math.h>
 
#define tol -3   /*精度*/
/*--------------------
*方程组及一阶导数
--------------------*/
double f1(double x, double y){
	return x*x + y*y -1;
}
 
double f2(double x, double y){
	return x*x*x -y;
}
 
double f1_x(double x, double y){
	return 2*x;
}

double f1_y(double x, double y){
	return 2*y;
}

double f2_x(double x, double y){
	return 3*x*x;
}

double f2_y(double x, double y){
	return -1;
}
 
int main(void){
	unsigned int c_times = 1;
	double x_1,y_1,x_0,y_0;
	double n_tol=1;
	x_0 = 0.8;
	y_0 = 0.6;
	n_tol = pow((double)10,tol);  /*误差界限*/
 
	/*牛顿迭代*/
	while(1){
		x_1 = x_0 + (-f1(x_0,y_0)*f2_y(x_0,y_0) + f2(x_0,y_0)*f1_y(x_0,y_0))/(f1_x(x_0,y_0)*f2_y(x_0,y_0) - f2_x(x_0,y_0)*f1_y(x_0,y_0));
		y_1 = y_0 + (-f1(x_0,y_0)*f2_x(x_0,y_0) + f2(x_0,y_0)*f1_x(x_0,y_0))/(f1_y(x_0,y_0)*f2_x(x_0,y_0) - f2_y(x_0,y_0)*f1_x(x_0,y_0));
		printf("迭代次数：%d\t X的值：%-.20