Timeline for Can we get good rational approximations in all residue classes?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Feb 19 at 16:51 | answer | added | user522439 | timeline score: 1 | |
| Jun 13, 2014 at 2:00 | answer | added | Steven Stadnicki | timeline score: 12 | |
| Jun 10, 2014 at 21:17 | comment | added | Joseph Vandehey | Although this answers a somewhat different question: it is known that almost all $\alpha$ satisfy a `Dirichlet-style' result as you mention. In fact, we know good asymptotics for how often $n$ should fall into a given arithmetic progression. See, for example, Ch.4 of Glyn Harman's Metric Number Theory. | |
| Jun 10, 2014 at 8:21 | answer | added | Douglas Zare | timeline score: 12 | |
| Jun 10, 2014 at 6:54 | vote | accept | Steven Stadnicki | ||
| Jun 10, 2014 at 5:17 | comment | added | duje | You may check this paper fq.math.ca/Scanned/34-1/elsner.pdf and references given there. | |
| Jun 10, 2014 at 4:37 | answer | added | Lucia | timeline score: 24 | |
| Jun 10, 2014 at 4:25 | comment | added | Steven Stadnicki | Ahh, mea culpa - in my head I actually have a slightly different version of this question; essentially, one with an arbitrary, possibly even $d$-dependent constant of approximation. (I only actually need the $a=1$ case for what I'm trying to prove, but was curious about the general result.) | |
| Jun 10, 2014 at 4:15 | comment | added | Douglas Zare | I don't see the reduction to the $a=1$ case if you allow unreduced fractions. If there are arbitrarily large coefficients in the simple continued fraction then there are very good approximations, and you can scale those up to produce good approximations in more residue classes, but for some irrationals (a set of measure $0$) there aren't arbitrarily large coefficients. | |
| Jun 10, 2014 at 4:09 | comment | added | Douglas Zare | Fibonacci numbers take few values mod $1597$, so I would guess that for $d=1597$, most values of $a$ produce bad approximations to the golden ratio. | |
| Jun 10, 2014 at 4:08 | comment | added | Steven Stadnicki | @DouglasZare Ideally, but it's not essential for my purposes (obviously if $m$ and $n$ don't have to be coprime then the question essentially reduces to the $a=1$ case) | |
| Jun 10, 2014 at 3:14 | comment | added | Douglas Zare | Do you want $m$ and $n$ to be coprime? | |
| Jun 10, 2014 at 2:35 | comment | added | Steven Stadnicki | @ChristianRemling More specifically, the convergents of that continued fraction - but it's possible to have 'good' approximations that aren't convergents, and it's certainly not true that convergents have to be in all residue classes. (For instance, the convergents for $\sqrt{2}$ have no $n\equiv 3\bmod 4$) | |
| Jun 10, 2014 at 2:31 | comment | added | Christian Remling | The best rational approximations are the continued fractions of $\alpha$. | |
| Jun 10, 2014 at 2:25 | history | asked | Steven Stadnicki | CC BY-SA 3.0 |