Timeline for Precise asymptotic of diophantine approximation
Current License: CC BY-SA 3.0
12 events
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| Apr 24, 2016 at 21:41 | comment | added | GH from MO | @DouglasZare: We talk about different things. Of course the same limsup (or sup), when it is at most $1/3$, can be attained by continuum many (pairwise inequivalent) $\xi$'s. This harmonizes with my response. If the limsup (or sup) exceeds $1/3$, then $\xi$ is equivalent (in a certain sense) to $\sqrt{9-4m^{-2}}$ with $m$ a Markov number, and the limsup (or sup) is given in terms of this $m$, and there are countably many such $\xi$'s. | |
| Apr 24, 2016 at 19:21 | comment | added | Douglas Zare | Given $[a_0; a_1, a_2,...]$, we're looking at $\limsup_k a_k + [0;a_{k+1},a_{k+2},...] + [0;a_{k-1},a_{k-2},...,a_1]$, right? If we let $a_{n^2+\epsilon_n}=1$ and $a_n=2$ otherwise, where each $\epsilon_n \in \{0,1\}$, that gives uncountably many reals with lim sup efficiency of $2+1/\sqrt{2}+1/(1+\sqrt{2}) = 3.121$. Maybe why this is greater than $3$ is clear from some perspective I don't have yet. | |
| Apr 24, 2016 at 19:10 | comment | added | Douglas Zare | Thanks. I'm still missing something. For large enough elements of the spectrum (maybe only greater than the critical 4.5278?), I think you can have an uncountable collection of reals with the same lim sup of approximation efficiency. Say, choose blocks A and B, and any simple continued fraction built from an infinite string of A and B blocks such that each finite pattern occurs infinitely often should have the same lim sup. I think you are saying that something prevents that lim sup from being less than $3$, that only very special sequences occur then that can't be built that way. | |
| Apr 24, 2016 at 15:03 | comment | added | GH from MO | @DouglasZare: Note also that by an element in the Markov spectrum I meant the quadratic irrational $\sqrt{9-4m^{-2}}$, where $m$ is a Markov number. | |
| Apr 24, 2016 at 14:16 | comment | added | GH from MO | @DouglasZare: I think no, since $\mathrm{SL}_2(\mathbb{Z})$ is countable. Two $\xi$'s are equivalent if and only if their continued fraction expansions are the same apart from finitely many coefficients and a shift of the coefficients. | |
| Apr 24, 2016 at 4:14 | comment | added | Douglas Zare | Can't you have uncountably many $\xi$ corresponding to the same element of the Markov spectrum? | |
| Apr 23, 2016 at 18:10 | comment | added | Siminore | @GHfromMO It is very interesting. Can you please read my edit, where I try to be more precise about my question? | |
| Apr 23, 2016 at 16:53 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Apr 23, 2016 at 16:50 | comment | added | GH from MO | @Wojowu: Thanks, I will clarify my response. | |
| Apr 23, 2016 at 16:40 | comment | added | Wojowu | Per what OP said in the comments, they meant the double inequality to hold for infinitely many $p/q$, not that the left inequality holds for all $p/q$. However, your answer still contains neat results, so I advise you to keep it, but add how you understood OP's question. | |
| Apr 23, 2016 at 16:30 | history | edited | GH from MO | CC BY-SA 3.0 |
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| Apr 23, 2016 at 16:25 | history | answered | GH from MO | CC BY-SA 3.0 |