Timeline for Which polynomial's roots are its coefficients?
Current License: CC BY-SA 3.0
23 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Sep 5, 2015 at 1:36 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Smaller image.
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| Sep 4, 2015 at 23:51 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Sep 4, 2015 at 13:01 | comment | added | Lubin | Again as an outsider, I wonder whether it’s of any use or significance that, going in the direction I suggested, the square of the Jacobian determinant is the discriminant of the polynomial (maybe up to sign). | |
| Sep 3, 2015 at 21:33 | answer | added | Robert Israel | timeline score: 23 | |
| Sep 3, 2015 at 21:06 | comment | added | asmeurer | This leads to the question of how ordering the coefficients that way affects the roots. Which polynomials have inverse images under this map? | |
| Sep 3, 2015 at 20:43 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Typo.
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| Sep 3, 2015 at 19:41 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Sep 3, 2015 at 19:32 | answer | added | Richard Stanley | timeline score: 33 | |
| Sep 3, 2015 at 18:26 | comment | added | Martin M. W. | Lubin's suggestion gives a continuous map, which is nice. The process outlined here can't be made continuous (e.g. to define it for $z^2 + e^{i\theta}$ you essentially have to choose branches of the square root function for the coefficients). On its own, that seems unnatural, but it makes sense as picking a branch of the inverse of Lubin's map. Igor's idea of looking at all permutations then corresponds to taking the full inverse. | |
| Sep 3, 2015 at 17:49 | comment | added | Joseph O'Rourke | @Lubin (or Igor): Feel free to pose a new, related question. | |
| Sep 3, 2015 at 17:18 | answer | added | Lubin | timeline score: 8 | |
| Sep 3, 2015 at 17:02 | comment | added | Igor Rivin | @Lubin You got my vote... | |
| Sep 3, 2015 at 16:46 | comment | added | Lubin | As a complete outsider, I wonder whether it would make better sense to go in the opposite direction, from roots to coefficients, giving you the map defined by taking $n$ indeterminates and evaluating them at the $n$ symmetric functions. | |
| Sep 3, 2015 at 16:45 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
LaTeX spacing.
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| Sep 3, 2015 at 16:38 | comment | added | Joseph O'Rourke | @IgorRivin: Goofiness is in the eye of the beholder :-). Indeed I did initially consider arbitrary permutations of the roots, but decided to ask a more specific question. I do think the arbitrary permutations question is interesting, but it doesn't lead to an obvious iteration. | |
| Sep 3, 2015 at 16:04 | answer | added | wythagoras | timeline score: 11 | |
| Sep 3, 2015 at 15:54 | comment | added | Igor Rivin | Do you really want the coefficients in some goofy sorting order. A permutation of the roots seems much more natural. | |
| Sep 3, 2015 at 15:19 | answer | added | Per Alexandersson | timeline score: 10 | |
| Sep 3, 2015 at 15:08 | comment | added | Joseph O'Rourke | @BenoîtKloeckner: Good point. Changed. Still not ideal, but ... | |
| Sep 3, 2015 at 15:08 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Sep 3, 2015 at 15:07 | comment | added | Benoît Kloeckner | The notation $P'$ is somewhat unfortunate. | |
| Sep 3, 2015 at 15:00 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |