Timeline for Is there any Menelaus-type theorem for polynomials?
Current License: CC BY-SA 4.0
12 events
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Oct 15, 2018 at 20:19 | answer | added | David E Speyer | timeline score: 3 | |
Oct 9, 2018 at 16:04 | comment | added | MR_BD | @DavidESpeyer I want to prove that this $x_{12}$ is only dependent on other $x_{ij}$. i.e. There exist $x^*=x_{12}$ such that for every polynomials satisfying $P_i(x_{ij})=P_j(x_{ij})$ we have $P_1(x^*)=P_2(x^*)$ | |
Oct 9, 2018 at 16:02 | comment | added | MR_BD | @AlexandreEremenko I was considering real polynomials. | |
Oct 9, 2018 at 14:43 | comment | added | Alexandre Eremenko | Are your polynomials and points $x_{i,j}$ real or complex? Or this is a question about arbitrary field? | |
Oct 9, 2018 at 14:17 | comment | added | David E Speyer | There will always exist $x_{12}$ such that $P_1(x_{12}) = P_2(x_{12})$ (unless $P_1(x) - P_2(x)$ is a nonzero constant. Is the question to give a formula for $x_{12}$ in terms of the other $x_{ij}$? | |
Oct 9, 2018 at 14:15 | comment | added | MR_BD | @StanleySnelson Of course there is plenty degree of freedom. And the question is to find some $x_{ij}$ such that there exist $x_{12}$ which is independent of the other parameters. | |
Oct 9, 2018 at 14:14 | history | edited | MR_BD | CC BY-SA 4.0 |
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Oct 9, 2018 at 14:11 | comment | added | MR_BD | @LSpice To be easier, you may assume they all are distinct. But I think it is still true with less assumption | |
Oct 9, 2018 at 13:36 | history | edited | Chris Godsil |
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Oct 9, 2018 at 10:59 | comment | added | user126920 | When $n=1$, there are generically 6 points of intersection between the 4 lines, and you specify 5 of them. In general, there are $n\binom{n+3}{2}$ points of intersection, but you only specify $\binom{n+3}{2} - 1$ of them, so the problem seems very underdetermined. | |
Oct 9, 2018 at 10:14 | comment | added | LSpice | I don't understand "there are distinct (it is not necessary that all of them be distinct) numbers". Are they distinct or not? Or do you perhaps mean just to hypothesise that they are not all the same? | |
Oct 9, 2018 at 9:21 | history | asked | MR_BD | CC BY-SA 4.0 |