answer to this post has not yet been answered ...i.e abt complexity.
seems log(n) to me..
correct me if i am wrong.
On Mon, Jan 16, 2012 at 8:53 AM, Gene <gene.ress...@gmail.com> wrote:
> I'm sorry for not being specific. I meant it doesn't converge for all
> roots, e.g. cube root.
>
> On Jan 15, 10:18 pm, Dave <dave_and_da...@juno.com> wrote:
> > @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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