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




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^((k1)/2) for k >= 2, which leads to x  p_k/q_k < 2^(1/2k). > > 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 ?
Your confusion may be about the meanings 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).
> According to your post > 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.



