对于一个函数f(x),x_0为函数零点的近似值,使用牛顿迭代法,有迭达公式x_{k+1} = x_k - \frac{f(x_k)}{f^{\prime}(x_k)} , k=0, 1,2...。当k逐渐增大时,x_{k+1}趋近与所求零点的精确值。
实现如下:
#include<iostream>
#include<string>
#include<cmath>
#define e 0.00001 //迭代精度要求
#define N 50 //最大迭达次数
using namespace std;
double f(double x) //目标函数
{
double c = sin(10 * x) - x;
double d = 10 * cos(10 * x) - 1;
return (c / d);
}
void newton(double x0)
{
double a = x0;
double x = x0 - f(x0);
int i = 0;
while (abs(x - a) > e)
{
cout << a << endl;
a = x;
i++;
x = x - f(x);
if (i > N)
{
cout << "Can't find zeros!" << endl;
return;
}
}
cout << a << " is a root" << endl;
}
int main()
{
double x0;
cin >> x0;
newton(x0);
return 0;
}
Comments NOTHING