First some comments on the previous comments: Let $theta=[0,a_0,a_1,ldots]$ have convergents $p_i/q_i$. Then the connection between the $a_i$ and how well $p_i/q_i$ approximates $theta$ is given by the following inequality:
$$frac{1}{q_k^2(a_{k+1}+2)} < |a-p_k/q_k| le frac{1}{q_k^2a_{k+1}}.$$
This is taken from Khintchin's book on continued fractions, Page 36 (with a typo corrected.)
This is why Roland's statement is true: For example you can make $theta:=[0,a_0,a_1,ldots]$ close to the number $2/3=[0,1,2]$, by taking for $theta$ something like $[0,1,2,10^{100},ldots]$.
As for Roth's theorem, if for some $epsilon>0$ and for infinitely many $k$ it holds that $a_{k+1}ge q_k^{epsilon}$, then by the above double inequality, Roth's theorem is violated, and $theta$ can't be algebraic. But consider $[0,1,2,3,ldots]$. Here the $q_k$ are bounded below by the Fibonacci sequence and the $a_k$ grow linearly, so there is no problem with Roth's Theorem. For all I can see (SEE THE COMMENTS BELOW!!), $[0,1,2,3,ldots]$ might be an algebraic number, although that might be ruled out by some of Lang's conjectural strengthenings of Roth.
No comments:
Post a Comment