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 May 11th, 2016, 11:09 PM #1 Newbie   Joined: Oct 2014 From: China Posts: 12 Thanks: 2 Coefficient of bivariate polynomial as a determinant of matrix Given $$\begin{bmatrix} a\\ b\\ c\\ d\\ \end{bmatrix}=\begin{bmatrix} a_0t^3+a_1st^2+a_2s^2t+a_3s^3\\ a_4t^2+a_5st+a_6s^2\\ a_7t+a_8s\\ a_9\\ \end{bmatrix}$$ the following equation holds: $$-\frac{1}{d} \begin{vmatrix} d&c&b&a&0\\ 0&d&c&b&a\\ 3d&2c&b&0&0\\ 0&3d&2c&b&0\\ 0&0&3d&2c&b\\ \end{vmatrix} = \frac{1}{d}\begin{vmatrix} cd&2bd&3d\\ 2bd&3ad+bc&2c\\ 3ad&2ac&b\\ \end{vmatrix}\\ =\left(a_4^2a_7^2-4a_0a_7^3-4a_4^3a_9+18a_0a_4a_7a_9-27a_0^2a_9^2 \right)t^6 \\ +\left(2a_4a_5a_7^2-4a_1a_7^3+2a_4^2a_7a_8-12a_0a_7^2a_8-12a_4^2a_5a_9+18a_1a_4a_7a_9+18a_0a_5a_7a_9+18a_0a _4a_8a_9-54a_0a_1a_9^2 \right)st^5+\\ +(...)s^2t^4+(...)s^3t^3+(...)s^4t^2+(...)s^5t+(.. .)s^6$$ The resulting bivariate polynomial looks ugly, however, I find the coefficient of $t^6$ equals (that of $s^6$ has the same form): $$-\frac{1}{a_9}\begin{vmatrix} a_9&a_7&a_4&a_0&0\\ 0&a_9&a_7&a_4&a_0\\ 3a_9&2a_7&a_4&0&0\\ 0&3a_9&2a_7&a_4&0\\ 0&0&3a_9&2a_7&a_4\\ \end{vmatrix}$$ Thus I guess the coefficient of $st^5$, $s^2t^4$, etc could also be written as a determinant of matrix. But I tried rewrite that of $st^5$ in Mathematica for days in vain. Look forward for any clues. Last edited by whitegreen; May 11th, 2016 at 11:27 PM. Tags bivariate, coefficient, determinant, matrix, polynomial Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post panky Algebra 3 January 25th, 2016 11:08 AM wortel Linear Algebra 1 January 6th, 2016 09:12 AM PhizKid Calculus 5 October 3rd, 2012 11:12 AM davedave Algebra 1 November 10th, 2009 12:26 AM bull-roarer Algebra 2 May 23rd, 2009 02:44 PM

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