Timeline for Precise asymptotic of diophantine approximation
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2016 at 9:55 | comment | added | Douglas Zare | @Siminore: That's not a particularly special condition on a good approximation to $\xi$. For convergents, it means that the index $n$ is even, since those are the approximations slightly below $\xi$ while the odd convergents are slightly above $\xi$. There are infinitely many even indices. | |
| Apr 24, 2016 at 9:12 | comment | added | Siminore | Okay, but I don't think this is enough. To reverse your argument, it seems to me that $p$ must be something special: the integer part of $q \xi$. | |
| Apr 24, 2016 at 8:52 | comment | added | Douglas Zare | @Siminore: Since there are infinitely many normalized errors in an interval, there must be limit points. | |
| Apr 24, 2016 at 8:29 | comment | added | Siminore | According to my edit, my question seems to be equivalent to the following: given $a$ and $b$ arbitrarily close to some (positive) number $x$, do there exist infinitely many integers $m$ and $n$ such that $$a \leq n^2 \left( \xi - \frac{m}{n} \right)$$ and $$n \left( \xi - \frac{m}{n} \right) < \frac{1}{2}?$$ Hence the question is related to your normalized error, and - roughly - I am asking if this error can converge to some positive number. | |
| Apr 24, 2016 at 3:34 | history | answered | Douglas Zare | CC BY-SA 3.0 |