【计算方法】实验一 非线性方程求根数值解法

实验目的

（1）通过对二分法与牛顿迭代法做编程练习和上机运算，进一步体会二分法和牛顿法的不同。
（2）编写割线迭代法的程序，求非线性方程的解，并于牛顿迭代法作比较。

实验内容

1、用牛顿迭代法求下列方程的根
(1) x^2-e^x=0
(2) 〖xe〗^x-1=0
(3) lgx+x-2=0
2、编写割线法程序求解第一问的方程

/* 牛顿迭代法的代码实现* 数值分析* 计科 1604 王宇晨 10430416414*/#include<iostream>
#include<string>
#include<cmath>
using namespace std;const double e = 2.718281818284;
const double eps = 1e-6;//求导数 X = x - f(x)/f'(x)
double f1(double x){double a = pow(e,x);return x-(x*x-a)/(2*x-a);
}
double f2(double x){double a = pow(e,x);return x-(x*a-1)/(a*x+a);
}
double f3(double x){double a = x*log(10);return x-(log10(x) + x - 2)/( 1/a + 1);
}
void NewtonIterationMethod1(double x,double d){double a = x;double b = f1(a);int k=0; //记录循环的次数while(((a-b)>d) || ((a-b)<-1*d)){//cout<< a <<endl;a = b;b = f1(a);k ++;if(k>100){cout<<"迭代失败！（函数不收敛）"<<endl;return ;}}cout<< b <<endl;return;
}
void NewtonIterationMethod2(double x,double d){double a = x;double b = f2(a);int k=0; //记录循环的次数while(((a-b)>d) || ((a-b)<-1*d)){//cout<< a <<endl;a = b;b = f2(a);k ++;if(k>100){cout<<"迭代失败！（函数不收敛）"<<endl;return ;}}cout<< b <<endl;return;
}
void NewtonIterationMethod3(double x,double d){double a = x;double b = f3(a);int k=0; //记录循环的次数while(((a-b)>d) || ((a-b)<-1*d)){//cout<< a <<endl;a = b;b = f3(a);k ++;if(k>100){cout<<"迭代失败！（函数不收敛）"<<endl;return ;}}cout<< b <<endl;return;
}
//二分法
double check1(double x){double a = pow(e,x);return x*x-a;
}
double check2(double x){double a = pow(e,x);return a*x - 1;
}
double check3(double x){return log10(x) + x - 2;
}
void BinarySearch1(double d){double st = -1000,ed = 1000;double mid = 0.0f;while(ed - st > d){mid = st + (ed - st)/2;if(check1(mid) > 0){st = mid;}else if(check1(mid) < 0){ed = mid;}}cout<< mid <<endl;
}
void BinarySearch2(double d){double st = -1000,ed = 1000;double mid = 0.0f;while(ed - st > d){mid = st + (ed - st)/2;if(check2(mid) < 0){st = mid;}else if(check2(mid) > 0){ed = mid;}}cout<< mid <<endl;
}
void BinarySearch3(double d){double st = -1000,ed = 1000;double mid = 0.0f;while(ed - st > d){mid = st + (ed - st)/2;if(check3(mid) < 0){st = mid;}else if(check3(mid) > 0){ed = mid;}}cout<< mid <<endl;
}
int main(){cout<< "请输入初始值x0：";double x,d = eps;cin>> x ;cout<< "牛顿迭代法" <<endl;NewtonIterationMethod1(x,d);NewtonIterationMethod2(x,d);NewtonIterationMethod3(x,d);cout<< "二分法" <<endl;BinarySearch1(d);BinarySearch2(d);BinarySearch3(d);return 0;
}

#include<stdio.h>
#include<math.h>
#define eta 1e-6
//割线法
float ge(float(*f)(float),float x1,float x0)
{float x2,d;int k=0;do{k++;x2=x1-((*f)(x1)*(x1-x0))/((*f)(x1)-(*f)(x0));d=x1-x0;x0=x1;x1=x2;printf("x(%d)=%f\n",k,x0);}while(fabs(d)>eps&&fabs((*f)(x1))>eta);return x1;
}float f(float x)   //第一问
{return x*x-exp(x);
}
float f1(float x)  //第二问
{return x*exp(x)-1;
}
float f2(float x)  //第三问
{return log10(x)+x-2;
}int main()
{float x1,x0,y0,y1,y2;printf("please insert x1,x0\n");scanf("%f,%f",&x1,&x0);printf("x(0)=%f\n",x0);y0=ge(f,x1,x0);y1=ge(f1,x1,x0);y2=ge(f2,x1,x0);printf("one answer is %6f\n",y0);printf("two answer is %6f\n",y1);printf("three answer is %6f\n",y2);return 0;
}

实验结果：