用牛顿迭代法求方程f(x)=x^6-x-1=0在区间【1,2】内的实根，要求｜f(x(k))｜<10^-8 用C语言编写此编程

答案：#include<stdio.h>
#include<math.h>
#define eps  1e-8
void main()
{
	double a=1,b=2;
	double t,t0,f0,f00,m,n;
	t0=(a+b)/2;
	m=pow(t0,5);
	n=pow(t0,4);
	f0=6*m-1;
	f00=30*n;
	t=t0-f0/f00;
	while(fabs(t-t0)>eps)
	{
		t0=t;
		m=pow(t0,5);
	    n=pow(t0,4);
		f0=6*m-1;
	    f00=30*n;
	    t=t0-f0/f00;
		printf("t0=%12.10lf,t=%12.10lf\n",t0,t);
	}
	printf("用Newton切线法得：%12.10lf\n",t);
}
结果为：
t0=1.2065843621,t=0.9809945654
t0=0.9809945654,t=0.8207881793
t0=0.8207881793,t=0.7300742137
t0=0.7300742137,t=0.7013898132
t0=0.7013898132,t=0.6988457773
t0=0.6988457773,t=0.6988271198
t0=0.6988271198,t=0.6988271188
用Newton切线法得：0.6988271188
Press any key to continue

其他：#define _CRT_SECURE_NO_WARNINGS
#include "stdio.h"
#include "math.h"
#define e0 1e-8

void main()
 {
 double x0,x1;
   x0=0;
 x1=1;
 do
    {
    x0=x1;
     x1=x0-(x^6-x-1)/(6*x^5-1);
    }
 while(fabs(x1-x0)>=e0);
 printf("方程的根是%f\n",x1);
  } //牛顿法的迭代序列：x(n+1)=x(n)－f(x(n))/f'(x(n))。
//f(x)=x^6-x-1;
//f'(x)=6*x^5-1;

#include<stdio.h>
#include <math.h>  
void main()
{  
 double x,x0,f,f1;  
 x0=1.5;
 while(1)
 {
  f=pow(x0,6)-x-1;
  f1=6*pow(x0,5)-1;
  x=x0-f/f1;
  if(fabs(x-x0)<1e-8) break;
  x0=x;
 }
 printf("x=%f\n",x);    //x=1.134724138
}

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