newton _070460.cpp

用牛顿迭代法来求解函数近似解

#include<iostream.h>
#include<iomanip.h>
#include<math.h>

double newton(double z,double eps)
{   int i=0;
    double x=0;
	while(i<=1000000 && fabs(x-z)>=eps)
	{
		x=z;
		z=x-((x*x*x/3.0)-x)/(x*x-1.0);
		i++;
	}
	if(i>100000)
		cout<<"次迭代格式不收敛!"<<endl;
	else return z;
}


void main()
{
	double y,eps;
    int w;

    cout<<"确定当X0在区间(-a,a)内取值，迭代序列收敛于根X=0时，尽可能大的a的值"<<endl;
	cout<<"测试结果如下,观察即可知道尽可能大的a值为0.77:"<<endl;
	double x=-0.8;
	while (x<=0.8&&x>=-0.8)
	{
		cout<<"x0="<<setw(4)<<x<<":"<<setw(8)<<"x="<<newton(x,0.0000001)<<endl;
		x+=0.01;
	}
	cout<<endl;


    do{
       cout<<"代入其它的初值为:"<<endl;
	   cin>>y;
	   cout<<"允许误差为： "<<endl;
       cin>>eps;
       cout<<"迭代求得的根："<<newton(y,eps)<<endl<<endl;
	   cout<<"继续请按1，退出请按0"<<endl;
	   cin>>w;
	   }
	while(w);

}