From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Approximation of sqrt(2)
Date: 21 Jan 2000 03:03:29 -0500
Newsgroups: sci.math
Summary: [missing]

In article <388765d5.27241797@news.cloud9.net>,
Chris <chriss@spammers.die> wrote:
:By repeatedly iterating the formula (1/2)(n + 2/n) (IE, taking the
:result of it and plugging it back into it), one can come up with an
:increasingly accurate approximation of the square root of two. 
:
:I have been able to figure out the proof that if one plugs in a value
:of n that can be expressed as sqrt(2)+c (0<c), it will result in a
:value less than sqrt(2)+c. I think, but I am not sure, that the value
:can also not be less than sqrt(2). I don't have enough knowledge of
:inequalities to prove this one way or another.
:
:Anyway, in order to check myself, to make certain that I am not
:screwing anything up, I wrote a perl script to repeat this over and
:over again. My initial value for n was 1, and I repeated it 10 times.
:Each time, I subtracted the computers version of sqrt(2) from the
:approximation I got after each iteration, resulting in the following:
[...]

Actually, this method (known to Heron centuries before Newton) has a
closed formula: 

if you approximate sqrt(a) (a>0) by x_old>0  and improve it by

   x_new = (x_old + a / x_old) / 2

then

(x_new - sqrt(a)) / (x_new + sqrt(a)) 

= ((x_old - sqrt(a))/(x_old + sqrt(a)))^2

 It follows that x_new is theoretically greater than sqrt(a), regardless
of where  x_old was, as long as it was different from sqrt(a).

 And it leads to a closed formula for the result of k iterative steps as a 
function of k, too.

 It was rather frustrating that Newton's method for cube and higher roots
does not have a closed formula for the k-th iterate. The a connection
between higher roots and fractals was discovered.

Cheers, ZVK(Slavek).