How to implement Sqrt(double k) efficiently?

Question

Hussein · Accepted Answer

double sqrt2(double num) {
	double i = 0, j = num, mid;
	
	if(j  0.00001) {
		if(mid*mid > num) j = mid;
		else i = mid;
		
		mid = (i+j)/2.0;
	}
	
	return mid;
}

xy · Answer

binary search is not a good approach, too slow. compare to that newton's method is simple and fast.

double sqrt_newton(double v) {
  double x, nx = 1;
  while (abs(nx - x) > 1e-9) {
    x = nx;
    nx = (v / x + x) / 2;
  }
  return nx;
}

Eskimo · Answer

You could also use Newton's Method. It might be a bit (just a bit) less efficient, but it will work with every function, not just sqrt.
en.wikipedia.org/wiki/Newton's_method

Anoniem · Answer

Binary search, start with reasonable range and shrinking the range until precision level is satisfied

Bruno · Answer

i came up with a iterative algorithm bu having problem with the precision. for example, i get to this solution 2.236068 instead of 2.23606798 for sqrt(5) using double numbers, any ideas why?

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