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函数:用牛顿迭代法求方程的根
2022-07-06 09:24:00 【|光|】
要求
用牛顿迭代法求方程的根。方程为ax3+bx2+cx+d=0,系数由用户输入,求x在1附近的根。
,
代码
#include<math.h>
#define EPSILON 1E-6
/* * 函数f(x)=a*x**3+b*x**2+c*x+d */
double f(double a,double b,double c, double d, double x)
{
double t;
t=a*x*x*x+b*x*x+c*x+d;
return t;
}
/* * 函数f(x)=a*x**3+b*x**2+c*x+d的导函数 */
double derivatives(double a,double b,double c, double d, double x)
{
double t;
t=3*a*x*x+2*b*x+c;
return t;
}
/* * 在该函数中用迭代法求解方程在1附近的根,可以调用上面的f函数和f的导函数 */
double fun(double a,double b,double c,double d)
{
double x1,x=1;
do
{
x1=x;
x = x1 - f(a,b,c,d,x)/derivatives(a,b,c,d,x);
}
while(fabs(x1-x)>=1e-3);
return x;
}
测试
测试输入
10 5 -11 -3.28351
输出
方程在1附近的根为0.97
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