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Topic: square root continued fraction
Replies: 9   Last Post: Sep 21, 2008 5:22 PM

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 Nat Silver Posts: 2,082 Registered: 12/6/04
Re: square root continued fraction
Posted: Sep 20, 2008 10:25 AM

Niklaus wrote:
> Robert Israel wrote:

> > For general results, see e.g. Khinchin, "Continued Fractions".
> > We have the inequality
> > 1/(q_k q_{k+1}) > |x - p_k/q_k| > 1/(q_k (q_k + q_{k+1}))
> > To answer your question we need lower bounds on q_k. An easy one is
> > q_k >= 2^((k-1)/2) for k >= 2, which leads to |x - p_k/q_k| < 2^(1/2-k).
> > Thus you are guaranteed accuracy to within 10^(-n) if you take
> > k >= 1/2 + n log_2(10). Actually the lower bound can be improved to
> > q_k >= F_(k+1) (the (k+1)'th Fibonacci number), so approximately
> > k = n log_phi(10) will do, where phi = (1+sqrt(5))/2.

> i'm little confused by your results. For n=100 (say for the 100th
> continued fraction of sqrt(2))
> gives accurate results upto which place ?

of "k" and "n" in Prof. Israel's post.

I think p_k/q_k is the rational number
that results after k steps that generate
k integers of the continued fraction
expansion of sqrt(d).

If one needs a certain accuracy, say an error
less than 10^-50, one takes n = 50 in the
formula k >= 1/2 + n log_2(10) to find k
(a number of steps that gives a sufficiently
small error).

> 1) 1/2 + 100*log_2(10) which is 332
> 2) 100/log10(phi) which is 478

> Am i wrong ?

> My results were something like it gives around 60 places. Maybe
> something wrong with my calcs.

Date Subject Author
9/19/08 niklaus@gmail.com
9/19/08 amzoti
9/19/08 Robert Israel
9/21/08 Robert Israel
9/20/08 niklaus@gmail.com
9/20/08 Nat Silver
9/20/08 niklaus@gmail.com
9/20/08 Nat Silver
9/21/08 Robert Israel
9/20/08 gerry@math.mq.edu.au