To find sqrt(a), this is equivalent to Newton's Method with f(x)=x^2 - a. Newton is: x_{i+1} = x_i + f'(x_i) / f'(x_i). So you have x_{i+1} = x_i + (x_i^2 - a) / (2 x_i) = (x + a/x) / 2, which is just what the Babylonian method says to do.
Newton's method roughly doubles the number of significant bits in the answer with every iteration _when it converges_. The problem is that it doesn't converge to every root. There's a huge literature on this, which I'll let you find yourself. On Jan 15, 10:22 pm, Ankur Garg <ankurga...@gmail.com> wrote: > Hello > > I was going through this link > > http://www.geeksforgeeks.org/archives/3187 > > Wonder what is the time complexity for this..?Can anyone explain > > > Regards > Ankur -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.