Suppose I am working with polynomials $p(z)$ over $\mathbb{C}$, of some degree $N$, and where $a_n=1$. I would like to find a simple way to parameterize the set for which there exists a prescribed linear condition $C(r_1,r_2,\cdots r_N)$ on the roots $r_i$ of $p(z)$. For instance, we know by Vieta's formulas that if the condition is $\sum{r_i}=c$ for some $c \in \mathbb{C}$, then we simply have $a_{n-1}=c$. What about a general condition $\sum{\alpha_ir_i}=c$? perhaps some transformation of division of polynomials?
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2It's strange to impose such a condition that isn't symmetric in the roots, since you have to pick an order on the roots to check it. – Qiaochu Yuan Oct 09 '20 at 18:29
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Right. Let's put it this way: there exists an order on the roots that obeys this. – Amir Vaxman Oct 09 '20 at 18:31
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If the condition is that there exists some ordering of the roots with this property, this is equivalent to the vanishing of the polynomial
$$\prod_{\sigma \in S_n} \left( \sum \alpha_i r_{\sigma(i)} - c \right).$$
This expands out to some (complicated) sum of symmetric functions in the roots $r_i$ so it can be written as some (complicated) polynomial in the coefficients of $\prod (x - r_i)$ (the elementary symmetric polynomials in the $r_i$). For example, if $n = 2, c = 1, \alpha_1 = 1, \alpha_2 = 2$ this gives
$$(r_1 + 2r_2 - 1)(r_2 + 2r_1 - 1) = (2r_1^2 + 5r_1 r_2 + 2r_2^2) - (3r_1 + 3r_2) + 1 = (2e_1^2 + e_2) - 3e_1 + 1 = 0$$
where $e_1 = r_1 + r_2$ and $e_2 = r_1 r_2$.
Qiaochu Yuan
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