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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