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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 4:42 PM

> Niklaus wrote:

> Say we need error less than 10^-50 ie correct value until the 50th
> decimal point then we get n=50 and it gives k as
> 50*log2(10) + 0.5 ~= 166, it says
> 166th convergent gives a good estimate.
> where as the other phi gives 50/log10(phi) ~= 239
> convergent gives a good estimate.

> 239th convergent is more accurate than 166th but it is too much of
> unnecessary work. i mean 166th iteself gives a good approximation.
> Aren't we overcalculating or i'm overlooking something.

> i could find them accurate until 50th place by the 102nd usually
> within 2*nth convergent fraction.

I have this: log_b(x) means log base b of x.
But log_b(x) = log(x)/log(b) in any base.
So, if k = n log_phi(10) will do, where phi = (1+sqrt(5))/2,
then, using base ten, k = n log(10)/log(phi)
= n / (log(1+sqrt(5)) - log(2)) = n / (0.20898764...)

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