Determining Floating Point Square Root

Question

How do I determine the square root of a floating point number? Is the Newton-Raphson method a good way? I have no hardware square root either. I also have no hardware divide

Answer 1

A good approximation to square root on the range [1,4) is

def sqrt(x):
  y = x*-0.000267
  y = x*(0.004686+y)
  y = x*(-0.034810+y)
  y = x*(0.144780+y)
  y = x*(-0.387893+y)
  y = x*(0.958108+y)
  return y+0.315413

Normalise your floating point number so the mantissa is in the range [1,4), use the above algorithm on it, and then divide the exponent by 2. No floating point divisions anywhere.

With the same CPU time budget you can probably do much better, but that seems like a good starting point.