code5_1.cpp

NEWTON迭代法求非线性方程的根

#include<iostream>
#include<math.h>
#include<iomanip>
using namespace std;

#define MAX 100
#define e 1e-4
double coef[MAX];
double coef_orginal[MAX];
double x[MAX];

double f(double x,double t[])
{
	
	double temp=t[0];

	for(int i=1;i<=MAX;i++)
	{
		if(t[i]==99999)
			break;
		temp += t[i]*pow(x,i);
	}
	return temp;
}

void qiudao()
{
	int i;
	for(i=0;i<=MAX;i++)
	{
		if(coef[i] == 99999 )
			break;
	}
	i--;
	for(int j=1;j<=i;j++)
	{
		coef[j-1] = coef[j]*j;
	}
	coef[i] = 99999;
}

int Newton(double x0)
{
	int k;
	x[0] = x0;

	qiudao();
	for(k=1;k<=MAX;k++)
	{
		x[k] = x[k-1]-f(x[k-1],coef_orginal)/f(x[k-1],coef);
		if(x[k]-x[k-1]<e&&x[k]-x[k-1]>-e)
			break;
	}
	if(k==MAX)
		return -1;
	else
		return k;
}
int main()
{

	cout.precision(8);
	cout.setf(ios::fixed);

	int i;
	cout<<"请输入原函数的系数，按照低次向高次的顺序!并且在完成之后输入99999! 记得系数为0的输入0!"<<endl;
	for(i=0;i<=MAX;i++)
	{
		cin>>coef[i];
		coef_orginal[i] = coef[i];
		if(coef[i] == 99999)
			break;
	}
	int k;
	double x0;
	cout<<"请输入x0:"<<endl;
	cin>>x0;
	k = Newton(x0);
	double error[MAX];
	for(i=0;i<=k;i++)
	{
		double t1,t2,t3,temp;
		t1 = x[i]-0;
		t2 = sqrt(3)-x[i];
		t3 = -sqrt(3)-x[i];
		if(t1<0) t1 = -t1;
		if(t2<0) t2 = -t2;
		if(t3<0) t3 = -t3;
		if(t1<=t2)
			temp = t1;
		else temp = t2;
		if(temp>t3)
			temp = t3;
		error[i] = temp;
	}
	cout<<"用Newton法做出的结果为："<<endl;
	cout<<"x0"<<"              k"<<"             x[k]                   f(x)             error"<<endl;
	for(i=0;i<=k;i++)
	{
		cout<<x0<<"      "<<i<<"           "<<x[i]<<"            "<<f(x[i],coef)<<"       "<<error[i]<<endl;
	}
	cout<<"sqrt(3) = "<<sqrt(3)<<endl;

	return 0;
}