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Given a finite sequence of real vectors $x_1, x_2, \dots, x_m \in \mathbb{R}^d$, how do I construct a polynomial $p \colon \mathbb{R}^d \to \mathbb{R}$, such that $p(x) = 0$ if $x \in \{ x_1, \dots, x_m \}$ and $p(x) < 0$ otherwise? Is that even possible? I only know how to construct a polynomial which has roots at $x_1, x_2, \dots, x_m$, but behaves »arbitrarily« elsewhere, e.g. by defining $$ p(x) = \left(\sum_i x_i - x_{1i}\right) \cdot \left(\sum_i x_i - x_{2i}\right) \cdots \left(\sum_i x_i - x_{mi}\right). $$

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    $\begingroup$ sure, $-\prod_k (x-x_k)^2$ $\endgroup$ – mathreadler May 15 at 12:45
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For example $P(x)=-\prod\limits_{i=1}^m (x-x_i)^2$

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  • $\begingroup$ Heh you were fast. $\endgroup$ – mathreadler May 15 at 12:45
  • $\begingroup$ @lisyarus: are you sure? $\endgroup$ – TonyK May 15 at 12:52
  • $\begingroup$ @lisyarus A sum wouldn't work... In that case how would you guarantee that $p(x_i)=0, \forall i$? Take just two points , say $-1, 1$. Would you say that $p(x)=-(x+1)^2-(x-1)^2$ vanishes at $x =-1,1$? $\endgroup$ – PierreCarre May 15 at 12:52
  • $\begingroup$ @lisyarus Maybe we just delete all the comments? $\endgroup$ – PierreCarre May 15 at 12:57

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