Question

c++题目 不知道怎么弄 1. 编程序 用迭代法求a的立方根,求立方根的迭代公式为 ::Xi+1=（2*Xi）/3+a/(3*Xi*X

1. 编程序 用迭代法求a的立方根,求立方根的迭代公式为 ::Xi+1=（2*Xi）/3+a/(3*Xi*Xi)
假定X的初值为a,迭代到|Xi+1-Xi|<10的-5次方为止. 显示a=3,27的值,并通过调用pow(a,1.0/3)函数加以验证.。

Answer 1

#include <stdio.h>
#include <math.h>

void main(void)
{
double a, fX, fXplus, fDiff, fRes;

printf("Enter a number: ");
scanf("%lf", &a);
fX = a;
fRes = pow(a, 1.0/3);
do
{
fXplus = fX * 2/3 + a / (3 * fX * fX);
fDiff = fXplus - fX;
fX = fXplus;
}
while((fDiff <= -0.00001) || (fDiff >= 0.00001));
printf("\nThe cube root of %lf is %lf.\n", a, fX);
printf("The result of function is %lf.\n", fRes);
}

Answer 2

有谁能给每行添加注解吗？

Answer 3

#include <iostream>
using namespace std;
#include <math.h>
void main()
{
double a,b,X,Y,z;
cout<<"Enter a number:";
cin>>a;
X= a;
Y= pow(a, 1.0/3);
do
{
b = X * 2/3 + a / (3 * X * X);
z= b-X;
X=b;
}
while((z<= -0.00001) || (z >= 0.00001));
cout<<"The cube root of a is:"<< a<<"\n";
cout<<"The result of function is:" <<Y<<"\n";

}

Answer 4

没有说X1=多少么？