Timeline for Orbits of rational functions
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2017 at 17:38 | comment | added | Noam D. Elkies | If not then it only takes finitely many rational values, so no infinite orbit is possible. | |
| Sep 4, 2017 at 17:25 | comment | added | Alexandre Eremenko | Why do you assume that the rational function has coefficients in $Q$? | |
| Sep 3, 2017 at 17:36 | comment | added | Noam D. Elkies | That interpretation seems to have a negative answer for reasons unrelated to what the OP has in mind. Given a computable function $f$ from ${\bf Z}_{\geq 0}$ to ${\bf Z}_{\geq 0}$, define another such function $a_f$ as follows: $a_f(n)=n$ if $f(m)\neq 0$ for all $m<n$, but $a_f(n) = n+1$ if $f(m)=0$ for some $m<n$. Then there exists a rational function taking each $a_f(n)$ to $a_f(n+1)$ iff $f(m) \neq 0$ for all $m$; and there's no algorithm that can decide that for every choice of $f$. | |
| Sep 3, 2017 at 17:10 | comment | added | Christian Remling | @NoamD.Elkies: In the computability theory context (perhaps) hinted at by the OP, the sequence could be given by the label of a Turing machine that on input $n$ prints the $n$th member of the sequence. The 'real question' could then be if there is another TM that takes this label as its input and outputs the correct answer to the original question. (Of course, now you can't be given an arbitrary sequence, only computable ones.) | |
| Sep 3, 2017 at 16:49 | comment | added | Noam D. Elkies | Not sure what the "real question" means as it stands: how can I be "given" an infinite sequence of rationals? | |
| Sep 3, 2017 at 16:44 | comment | added | Igor Rivin | True, but of course, the real question is whether you can tell. | |
| Sep 3, 2017 at 16:42 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |