參考書目：《數(shù)值分析》 

y=x*x-a求零點 用零點附近的切線求x軸的交點，在以交點做切線進而不斷迭代來近似拋物線的零點 由于該函數(shù)在零點附近是收斂的(二階收斂)所以能不斷迭代

//main函數(shù)入口

#include <stdio.h>#include <math.h>double sqrt1(double a,int tot);double sqrt2(double a,int tot);double sqrt3(double a,int tot);int main(void){double i;int j;PRintf("input number to be sqrt:/n");scanf("%lf %d",&i,&j);printf("sqrt number is /n%.9f /n%.9f /n%.9f",sqrt1(i,j),sqrt2(i,j),sqrt3(i,1E-10));return 0;}double sqrt1(double a,int tot)  //直接傳入迭代次數(shù){double x,y;int i;x=a/2;for(i=0;i<tot;i++){y=(x+a/x)/2;x=y;}return y;}double sqrt2(double a,int tot) 
#include <stdio.h>#include <math.h>double sqrt1(double a,int tot);double sqrt2(double a,int tot);double sqrt3(double a,int tot);int main(void){	double i;	int j;	printf("input number to be sqrt:/n");	scanf("%lf %d",&i,&j);	printf("sqrt number is /n%.9f /n%.9f /n%.9f",sqrt1(i,j),sqrt2(i,j),sqrt3(i,1E-10));	return 0;}double sqrt1(double a,int tot)//傳入二分迭代次數(shù){	double x,y;	int i;	x=a/2;	for(i=0;i<tot;i++)	{		y=(x+a/x)/2;		x=y;	}	return y;}double sqrt2(double a,int tot){	int i;	double low,mid,high;	low=0;	high=a>1?a:1;	mid=(low+high)/2;	for(i=0;mid*mid!=a&&i<tot;i++)//如果直接相等退出循環(huán)	{		if(mid*mid<a)		{			low=mid;			mid=(low+high)/2;		}		else		{			high=mid;			mid=(low+high)/2;		}	}	return mid;}double sqrt3(double a,int err){	double x,y,temp;	temp=a/2;	do{		x=temp; //temp保留中間結(jié)果 方便x,y的比較		temp=(x+a/x)/2;		y=temp;	}while(fabs(x-y)>err);	return y;}
測試結(jié)果：二分求根收斂速度遠小于牛頓迭代}