find square root with Newton using integers-CodePudding

In this program, I'm trying to calculate the square root with the newton equation 1/2(x a/x) just using integers. So if I repeat this equation at least 10 times, it should divide the number by 1000 and give an approximate value of the square root of a/1000. This is the code:

int main (){
    int32_t a, x;  //integers a and x
    float root;

     do{
    scanf ("%d", &a);   // get value of a from the user
    
    
    if (a < 1000 ){   // if chosen number is less than 1000 the programm ends.
    break;
    }
    
    x = ((float) a / 1000);   //starting value of x is a / 1000;
    for (int i = 0; i < 50;i  )
    {

        root = ((float) x * (float) x   a/1000) / ((float)2*x);  // convert int to float //through casting
        x = (float)root;    // refresh the value of x to be the root of the last value.


    }
    printf ("%f\n", (float)root);
     }while (1);

    return 0;

}

so if I calculate the square root of 2000, it should give back the square root of 2(1.414..), but it just gives an approximate value: 1.50000 How can I correct this using integers and casting them with float? thanks

CodePudding user response：

The iterates of x = (x a / x) / 2 for a = 2000000 and x0 = 1000 (all integer variables) are 1500, 1416 and 1414. Then 200000000 gives 14142 and so on.

CodePudding user response：

    #include <stdlib.h>
    #include <stdio.h>
    
    int main (int argc, char * *argv, char * *envp) {
        int32_t a, x;  //integers a and x
        float root;
    
        do {
            scanf ("%d", &a);   // get value of a from the user
            if (a < 1000) {   // if chosen number is less than 1000 the program ends.
                break;
            }
            x = (int32_t)((float) a / 1000.0f);   //starting value of x is a / 1000;
            for (int i = 0; i < 1000; i  ) {
// Fixed formula based on (x a/x)/2
                root = ((float)x   (((float)a) / (float)x)) / 2.0f;
//Below, wrong old formula
                //root = ((float) x * (float) x   a / 1000) / ((float) 2 * x);  // convert int to float //through casting
                x = (int32_t) root;    // refresh the value of x to be the root of the last value.
            }
            printf ("%f\n", root);
        } while (1);
    
        return (EXIT_SUCCESS);
    }