Timeline for A determinant of perfect square polynomials
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 19, 2023 at 16:22 | vote | accept | T. Amdeberhan | ||
| Feb 13, 2022 at 16:23 | comment | added | T. Amdeberhan | @SamHopkins: I think all determinants/Pfaffians in the paper (you mentioned) involves the Vandermonde determinant, not the "cyclic type" in the question. | |
| Feb 13, 2022 at 16:21 | comment | added | T. Amdeberhan | @darijgrinberg: that's a nice "cousin". Note: $\det(x_{\min\{i,j\}})=(-1)^{m-1}x_1(x_1-x_2)\cdots(x_{m-1}-x_m)$. | |
| Feb 13, 2022 at 12:53 | answer | added | LeechLattice | timeline score: 6 | |
| Feb 13, 2022 at 2:27 | history | edited | T. Amdeberhan | CC BY-SA 4.0 |
added 6 characters in body
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| Feb 13, 2022 at 2:26 | comment | added | T. Amdeberhan | I'm sorry but there was a typo: $x_n$ should be $x_{2n}$, in the definition on the matrix. | |
| Feb 13, 2022 at 0:33 | comment | added | Sam Hopkins | Seems similar to various determinants and Pfaffians considered by Okada in math.kobe-u.ac.jp/publications/rlm18/9.pdf. | |
| Feb 12, 2022 at 23:45 | comment | added | darij grinberg | Also, the true "little brother" of this claim is not the Vandermonde determinant, but the known fact that $\det\left(\left(x_{\min\left\{i,j\right\}}\right)_{1\leq i,j\leq m}\right) = x_1 \left(x_1 - x_2\right) \left(x_2 - x_3\right) \cdots \left(x_{m-1}-x_m\right)$ for any $m \geq 1$. | |
| Feb 12, 2022 at 23:35 | comment | added | darij grinberg | This is reminiscent of (4.2) in Donald E. Knuth, Overlapping Pfaffians, Electronic Journal of Combinatorics 32 (1996), issue 2. | |
| Feb 12, 2022 at 23:31 | comment | added | darij grinberg | Note also that the matrix is antisymmetric, so it is not surprising that its determinant is a square. The question is why its Pfaffian is this nice. | |
| Feb 12, 2022 at 23:31 | comment | added | darij grinberg | Apparently, you can remove the "$-x_n$" part and the determinant becomes $x_1^2 \left(x_1-x_2\right)^2 \left(x_2-x_3\right)^2 \cdots \left(x_{2n-1}-x_{2n}\right)^2$. This should generalize the corrected version of your question. | |
| Feb 12, 2022 at 20:33 | history | asked | T. Amdeberhan | CC BY-SA 4.0 |