@Gene: Actually, Newton's Method for sqrt(a), where a > 0, also
sometimes called Heron's Method, converges for every initial guess x_0
> 0. This is not true generally for Newton's Method, but it is true
for Newton's Method applied to f(x) = x^2 - a.
Dave
On Jan 15, 5:39 pm, Gene <gene.ress...@gmail.com> wrote:
> 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
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