用牛顿法解方程xe(x) -1=0
程序流程图如下：

//方程求根的迭代法——牛顿迭代法
/*************Analysis**********
*1.初值x0,精度e,迭代次数N
*2.牛顿法解方程f(x)=x*e**x-1
*3.取初值为0.5
********************************/
#include<iostream>
#include<math.h>
using namespace std;
//原函数
double f(double x)
{return x*exp(x)-1;
}
//导数
double derivative(double x)
{return (1+x)*exp(x);
}
int main()
{double x0,x1;double e;//precisionunsigned N=0;unsigned k=0;cout<<"----------NewtonIteration-----------"<<endl;cout<<"Please input initial value x0:"<<endl;cin>>x0;cout<<"导数："<<derivative(x0)<<endl;cout<<"Please input precision e:"<<endl;cin>>e;cout<<"Please input iteration times N:"<<endl;cin>>N;k=1;while(true){if(derivative(x0)!=0){//迭代关系式x1=x0-f(x0)/derivative(x0);cout<<"x1:"<<x1<<endl;if(fabs(x1-x0)<e){cout<<"x1:"<<x1<<endl;break;}else{if(k==N){cout<<"Sorry,the iteratting failed..."<<endl;break;}else{k=k+1;//cout<<"k:"<<k<<endl;x0=x1;cout<<"x0:"<<x0<<endl;}}}else{cout<<"The derivative of equation at x0 is 0 "<<endl;break;}}cout<<"Hello,Boker..."<<endl;system("pause");return 0;
}

运行结果如下：