On Sun, 2011-10-16 at 20:25 +0000, Adam Majewski wrote:
> Hi,
>
> I would like to find limit of Continued fraction [3,2,1000,1...]
> http://www.warwick.ac.uk/~masiay/Research/Siegel.html
. . .
> Can I find it's limit ?
I don't know if Maxima knows much about the algebra of continued
fractions, but it can be of some help hacking out the manipulation
details of a derivation. A most useful fact is that
[a1, a2, a3, ...] = a1 + 1/[a2, a3, ...]
provided the continued fraction converges. If we apply that three
times by hand to [3, 2, 1000, 1, 1, ...] we obtain
3 + 1/(2 + 1/(1000 + 1/[1, 1, ...]))
Now it is known that [1, 1, ...] converges to the Golden Ratio,
(1+sqrt(5))/2. So now we can use Maxima as follows:
(%i20) 3+(1/(2+(1/(1000+(1/((1+sqrt(5))/2))))));
1
(%o20) ---------------------- + 3
1
------------------ + 2
2
----------- + 1000
sqrt(5) + 1
(%i21) factor(%o13);
7003 sqrt(5) + 7017
(%o21) -------------------
2001 sqrt(5) + 2005
(%i22) %o21,numer;
(%o22) 3.499750279196346
You set a to [0, 3, 2, 1000, 1, 1, ...], which by our useful
fact must be the reciprocal of [3, 2, 1000, 1, 1, ...], and indeed the
reciprocal of 3.499750279196346 is 0.2857346725405882, which is what
your float(t) evaluates to, so we seem to get consistent results.
If all of the continued fractions for the rotation numbers exhibited
on the link you provided do end up repeating 1 forever then the method
I used above can be used to determine their limits as ratios of linear
expressions in sqrt(5).
--
Bill Wood