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#include<stdio.h>
#include<math.h>
//显示中文
#define _CRT_SECURE_NO_WARNINGS
#include <windows.h>//用于函数SetConsoleOutputCP(65001);更改cmd编码为utf8
#include<iostream>
using namespace std;
#define phi01(x) 20/((x)*(x)+2*(x)+10)//迭代函数
#define phi02(x) (20-2*((x)*(x))-((x)*(x)*(x)))/10
#define f(x) ((x)*(x)*(x))+2*((x)*(x))+10*(x)-20
#define f1(x) 3*((x)*(x))+4*(x)+10
void test01()
{
int n=1,N;
double x,x0,del;
printf("x0 = 1");
//scanf("%f",&x0);
x0 = 1;
printf("\ndel = 0.000000001");
//scanf("%f",&del);
del = 0.000000001;
printf("\nN = 100");
//scanf("%d",&N);
N = 100;
printf("\n k x(k)");
printf("\n %2d %f",0,x0);
while (n<N)
{
x = phi01(x0);
if(fabs(x-x0)<del)
{
printf("\n\n 迭代法式1的近似解 = %.9lf \n",x);
return;
}
printf("\n %2d %.9lf",n,x0);
n = n+1;
x0 = x;
}
printf("\n\n%d次迭代后未达到精度要求.\n",N);
}
void test02()
{
int n=1,N;
double x,x0,del;
printf("x0 = 1");
//scanf("%f",&x0);
x0 = 1;
printf("\ndel = 0.000000001");
//scanf("%f",&del);
del = 0.000000001;
printf("\nN = 20");
//scanf("%d",&N);
N = 20;
printf("\n k x(k)");
printf("\n %2d %f",0,x0);
while (n<N)
{
x = phi02(x0);
if(fabs(x-x0)<del)
{
printf("\n\n 迭代法式2的近似解 = %.9lf \n",x);
return;
}
printf("\n %2d %.9lf",n,x0);
n = n+1;
x0 = x;
}
printf("\n\n%d次迭代后未达到精度要求.\n",N);
}
void Aitken01()
{
int n=1,N;
double x0,x1,x2,x3,del;
printf("x0 = 1");
//scanf("%f",&x0);
x0 = 1;
printf("\ndel = 0.000000001");
//scanf("%f",&del);
del = 0.000000001;
printf("\nN = 100");
//scanf("%d",&N);
N = 100;
printf("\n k x(k)");
printf("\n %2d %f",0,x0);
while (n<N)
{
x1 = phi01(x0);
x2 = phi01(x1);
x3 = x2-(((x2-x1)*(x2-x1))/(x2-2*x1+x0));
if(fabs(x3-x0)<del)
{
printf("\n\n 迭代法式1的Aitken近似解 = %.9lf \n",x3);
return;
}
printf("\n n = %2d x1 = %.9lf",n,x3);
n = n+1;
x0 = x3;
}
printf("\n\n%d次迭代后未达到精度要求.\n",N);
}
void Aitken02()
{
int n=1,N;
double x0,x1,x2,x3,del;
printf("x0 = 1");
//scanf("%f",&x0);
x0 = 1;
printf("\ndel = 0.000000001");
//scanf("%f",&del);
del = 0.000000001;
printf("\nN = 100");
//scanf("%d",&N);
N = 100;
printf("\n k x(k)");
printf("\n %2d %f",0,x0);
while (n<N)
{
x1 = phi02(x0);
x2 = phi02(x1);
x3 = x2-(((x2-x1)*(x2-x1))/(x2-2*x1+x0));
if(fabs(x3-x0)<del)
{
printf("\n\n 迭代法式2的Aitken近似解 = %.9lf \n",x3);
return;
}
printf("\n n = %2d x1 = %.9lf",n,x3);
n = n+1;
x0 = x3;
}
printf("\n\n%d次迭代后未达到精度要求.\n",N);
}
void Newton()
{
int n=1,N;
double x,x0,del;
printf("x0 = 1");
//scanf("%f",&x0);
x0 = 1;
printf("\ndel = 0.000000001");
//scanf("%f",&del);
del = 0.000000001;
printf("\nN = 100");
//scanf("%d",&N);
N = 100;
printf("\n k x(k)");
printf("\n %2d %f",0,x0);
while (n<N)
{
float f_1 = f(x0);
float f_2 = f1(x0);
float d = f_1/f_2;
x =x0 - d;
if(fabs(x-x0)<del)
{
printf("\n\n Newton迭代法的近似解 = %.9lf \n",x);
return;
}
printf("\n %2d %.9lf",n,x);
n = n+1;
x0 = x;
}
printf("\n\n%d次迭代后未达到精度要求.\n",N);
}
int main()
{
//显示中文
SetConsoleOutputCP(65001);
test01();
cout<<endl;
test02();
cout<<endl;
Aitken01();
cout<<endl;
Aitken02();
cout<<endl;
Newton();
cout<<endl;
system("pause");
return 0;
}
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