Question

I am using the code below, its basically babylonian method, how can i modify it so that it returns correct value for numbers that are b/w 0 and 1. moreover can anyone tell me how can i find cubic root of a number

int main(){
    double n;
    cin>>n;
    double e= 0.0000001;
    double x = n;
    double r = 1;
    while(x-r >0.0000001){
        x = (x+r)/2;
        r = n/x;
    }
    printf("%f",x);
    return 0;
}

Answer 1

I. Don’t use the difference, but its absolut value. E.g.,

  while((x-r >0.0000001)||(r-x>0.0000001)){ ...

II. Use the following iteration for y=x^(1/n): y’=((n-1)*y^n + x)/(n*y^(n-1))

EDIT

II. describes the iteration step for a n-th root of x. For the cube root, use

y’ = (2y^3 +x)/(3y^2)

The approach bases on Newton’s method, see http://en.wikipedia.org/wiki/Newton%27s_method