Thread: Square Root Recursive Problem

I am trying to write a function to express this but I keep confusing myself. Maybe I am thinking about it too hard?

4.5 is halfway between 1 and 8. 4.52 = 20.25, so the square root of 8 must be less than 4.5.
2.75 is halfway between 1 and 4.5. 2.752 = 7.5625, so the square root of 8 must be greater than 2.75.
3.625 is halfway between 2.75 and 4.5. 3.6252 = 13.140625, so the square root of 8 must be less than 3.625.
(and so on…)
Continuing this process long enough reveals that the square root of 8 is about 2.828.

My code is messed up somewhere because I keep getting back the same number that I scanned in. This is what I have so far:

Code:

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

double recursive_square (double n);

int main (void)
{
    double num = 0;
    double result = 0;
    
    
    printf("Enter a number to compute the square root of:\n");
    scanf("%lf", &num);
    
    result = recursive_square(num);
    
    printf("The square root of %lf is about %.3lf \n", num, result);
    
    system ("PAUSE");
    return 0;   
}

double recursive_square (double n)
{
    double tempval = n/2;
    
    do
    {
        // If x2 is bigger than n, then we know that the square 
        //root of n must be between 1 and x,
        if ((tempval*tempval) > n)
        {
            tempval /= 2;
        }
        //otherwise it's between x and n
        if (tempval * tempval < n)
        {
            tempval += (n - tempval);
        }
    } while (((n - tempval) >= .001) || ((tempval - n) >= .001));
    
    return tempval;
}

A better way to revise your estimate is to use the value half way between tempval and n/tempval