square_root_newton_method.txt

/*
 Please write a function that accepts a floating number and returns its square-root. You may not use built-in square root function from your language. However, basic operators like addition, subtraction, multiplication are allowed. Please take into consideration the floating precision.
*/

#include <iostream>
#include "math.h"
using namespace std;

int main(){
    cout<<"Input an float number: ";
    float num;
    cin>>num;
    if(num<0){
        cout<<"wrong input."<<endl;
        return -1;
    }
    else if(num==0){
        cout<<"sqrt is 0"<<endl;
        return 0;
    }
    
    /*We will use Newton's method.*/
    float epsilon = 0.0000001;
    float x1=10;
    while(fabs(x1*x1-num)>epsilon){
    x1 = x1-(x1*x1-num)/(2*x1);
    }
    cout<<"The result is: "<<x1<<endl;
    return 0;
}

REMARK:
1. To use abs, you need to #include <cmath>. To use fabs, you need to #include <math.h>
2. We used Newton's method. On the numerical math textbook(author:Burden, Numerical Analysis 9th edition), in the chaptor2(page67, chaptor2 solutions of equations in one variable), he lists 

Newton's method works as:Given an equation f(p)=0, we want to get p. Using Taylor series, we have
f(p)= f(x0)+f'(x0)(p-x0)+...
where x0 is the first guess satisfing abs(p-x0)is small.
NOTE: f(p)=0 BECAUSE p IS THE SOLUTION OF THE EQUATION f(p)=0. Then we get:
p = x0 - f(x0)/f'(x0); that is
x1 = x0 - f(x0)/f'(x0)             (*)

For our problem, we need to calculate the square root a number. So the equation is x^2=num,i.e.
f(x) = 0 where f(x)=x^2-num;

Then replace f(x) into the equation (*) above, we get
x1 = x0 - (x0^2-num)/(2*x0) = 1/2 * (x0 + num/x0)