Timeline for Smooth functions for which $f(x)$ is rational if and only if $x$ is rational
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Oct 9, 2022 at 10:00 | comment | added | Somnium | So, if you fix enumeration for A, how you get enumeration for B? I am trying to construct explicit example. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 10, 2010 at 18:18 | history | edited | Pietro Majer | CC BY-SA 2.5 |
deleted 16 characters in body
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| Dec 10, 2010 at 18:10 | comment | added | Pietro Majer | yes, it was explicit in the lines above. Sorry, I though it was clear enough. | |
| Dec 10, 2010 at 14:04 | comment | added | Willie Wong | Ah, so you are implicitly using an enumeration of the two sets. Okay. | |
| Dec 10, 2010 at 14:03 | comment | added | Pietro Majer | that's no problem, "already settled" just means the first (approximatively) $n/2$ of $A$ plus their images, and the first $n/2$ of $B$ plus their pre-images. | |
| Dec 10, 2010 at 13:41 | comment | added | Willie Wong | @Pietro: I don't understand how you can run your procedure described as is in that other question here. You say: "inductively add only odd degree polynomials $p_n$, that vanish on the (finitely many) already settled points, and do not destroy the invertibility on $\mathbb{R}$, but if you start with $p_1(x) = qx$ for some rational $q$, then the "already settled points" is already in fact all the points that you want... | |
| Dec 10, 2010 at 13:15 | history | answered | Pietro Majer | CC BY-SA 2.5 |