
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...)

