實驗目的

（1）通過對二分法與牛頓迭代法做程式設計練習和上機運算，進一步體會二分法和牛頓法的不同。
（2）編寫割線迭代法的程式，求非線性方程的解，並於牛頓迭代法作比較。

實驗內容

1、用牛頓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、編寫割線法程式求解第一問的方程

/* 牛頓迭代法的程式碼實現
 * 數值分析
 * 計科 1604 王宇晨 10430416414
 */

#include<iostream>
#include<string>
#include<cmath>
using
 
 namespace std;

const double e = 2.718281818284;
const double eps = 1e-6;

//求導數 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; //記錄迴圈的次數
    while(((a-b)>d) || ((a-b)<-1*d)){
        //cout<< a <<endl;
        a = b;
        b = f1(a);
        k ++;
        if(k>100){
            cout
 
<<"迭代失敗！（函式不收斂）"<<endl;
            return ;
        }
    }
    cout<< b <<endl;
    return;
}
void NewtonIterationMethod2(double x,double d){
    double a = x;
    double b = f2(a);
    int k=0; //記錄迴圈的次數
    while(((a-b)>d) || ((a-b)<-1*d)){
        //cout<< a <<endl;
        a = b;
        b = f2(a);
        k ++;
        if(k>100){
            cout<<"迭代失敗！（函式不收斂）"<<endl;
            return ;
        }
    }
    cout<< b <<endl;
    return;
}
void NewtonIterationMethod3(double x,double d){
    double a = x;
    double b = f3(a);
    int k=0; //記錄迴圈的次數
    while(((a-b)>d) || ((a-b)<-1*d)){
        //cout<< a <<endl;
        a = b;
        b = f3(a);
        k ++;
        if(k>100){
            cout<<"迭代失敗！（函式不收斂）"<<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<< "請輸入初始值x0：";
    double x,d = eps;
    cin>> x ;
    cout<< "牛頓迭代法" <<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
//割線法
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)   //第一問
{
    return x*x-exp(x);
}
float f1(float x)  //第二問
{
    return x*exp(x)-1;
}
float f2(float x)  //第三問
{
    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;
}

實驗結果：