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Let $x_1, \ldots, x_n$ be variables, $e_n$ be the elementary symmetric polynomials. I will denote the discriminant by $$D_n(x_1, \ldots, x_n) = \prod_{i<j} (x_i - x_j)^2$$ And a generalized discriminant by $$D_k(x_1, \ldots, x_n) = \sum_{S \subset \{1, \ldots, n\}, |S| = k} ~~~\prod_{\{i,j\} \subset S} (x_i - x_j)^2$$

If $p$ is the monic polynomial with $x_i$ as roots, then it is well known that $$D_n(x_1, \ldots, x_n) = R(p,p') = \left| \begin{matrix} n & (n-1) e_1 & (n-2) e_2 & \ldots & 0& 0 & 0 \\ 1 & e_1 & e_2& \ldots & 0 & 0 &0\\ 0 & n & (n-1)e_1 & \ldots & 0 & 0 & 0\\ 0 & 1 & e_1 & \ldots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & \ldots & 0& 1 & e_1 & \ldots & e_n \\ 0 & \ldots & 0 & 0 & n & \ldots & e_{n-1} \end{matrix} \right|$$ (This matrix is formed by the row $(n , (n-1) e_1, \ldots, e_{n-1})$ and $(1, e_1, e_2, \ldots, e_n)$ alternating and moving over to the right once each time). \ \

After testing multiple examples, I have hypothesized that $D_k$ is the leading principal minor of this matrix of size $(2k+1) \times (2k+1)$. I'm sure this result must be known, so I was hoping for a proof or a reference.

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  • $\begingroup$ I think you mean $D_k$ corresponding to size $(2k-1)\times(2k-1)$ minor. $\endgroup$ – T. Amdeberhan Sep 14 '16 at 0:20
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I've seen your $D_k$ be called a subdiscriminant. Similarly the principal minors of the Sylvester matrix of two polynomials are the subresultants. The result you want is that subdiscriminants are (up to a constant, depending on the definition) equal to the corresponding subresultant of the polynomial and its derivative. A possible reference is the beginning of chapter 4 of "Algorithms in Real Algebraic Geometry" by S. Basu, R. Pollack, M-F. Roy. Your result is Proposition 4.28.

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