Programming

// Implementation of Newton - Raphson method for
// determination square root od positive number 
// Date: 10. april 2011
// Author: Solaris_user 
// Solution for programming exercise 13

#include <stdio.h>

int main(void) {
    
    double tolerance, input, y, x = 1.0;
    
    // Ask user to enter some number to find square root 
    // if user enters negative number then re-ask for new input
    // input must be greater then numbers zero and one. 
    // also user must enter some tolerance which will controll loop 
    // test condition    

    printf("Newton - Raphson square root algorithm\n");
    
    do {
        printf("Enter some number to find squre root: ");
        scanf("%lf", &input);
        if (input <= 1) {
            printf("Can't find square root for input number\n");
            }
        } while (input < 1);
    
    do {
        printf("Enter desired tolerance ( < 0.01): ");
        scanf("%lf", &tolerance);
        if (tolerance >= 0.01) {
            printf("Too small tolerance\n");
            }
        } while (tolerance >= 0.01);
    
    // Newton - Raphson method: 
    // start with x = 1.0. Then set y = (x + input/2) / 2 . If x and y are not 
    // close enough then set x = y, stop when difference is small enough 
    // to be negligible. 
    
    do {
        y = (x + (input / 2) ) / 2;
        if (x > y ) {
            x = y;
            }
        } while ( (x - y) > tolerance);
    
    printf("Square root is %.7lf\n", y);    
    return 0;
}

Solaris:~/Desktop> ./a.out
Newton - Raphson square root algorithm
Enter some number to find squre root: 9
Enter desired tolerance ( < 0.01): 0.00000001
Square root is 2.7500000

    do {
        y = (x + (input / 2) ) / 2;
        if (x > y ) {
            x = y;
            }
        } while ( (x - y) > tolerance);

     do {
        y = (x + (input / ( double ) 2.0) ) / ( double ) 2.0;
        if (x > y ) {
            x = y;
            }
        } while ( (y - x) > tolerance);