Eric Nitardy
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 Mar 15 asked Estimating the Primarithm Mar 13 accepted Linear Algebra of Symmetric Sums: a converse question Mar 13 comment Linear Algebra of Symmetric Sums: a converse question I found a straightforward way to calculate the constant term from the from the powers sums, but I have not found a good way to calculate the discriminant from the power sums. Is the only way to calculate the all the coefficients of the polynomial and then find the determinant of the Sylvester matrix? Mar 12 comment Do equal mean and equal moment imply equal distribution? @user7815: This paper http://ramanujan.math.trinity.edu/wtrench/research/papers/RP-112.pdf by William F. Trench appears relevant. Mar 12 revised Linear Algebra of Symmetric Sums: a converse question added 864 characters in body; added 14 characters in body Mar 12 comment Linear Algebra of Symmetric Sums: a converse question Thank you for your answer. I think that you have resolved the "some of $z_j$'s are zero" issue. There is an algebra error in your $n=2$ non-distinct discussion. The criteria should be $$\frac{1}{2}\mathcal{Z_1}^2=\mathcal{Z_2}.$$ Mar 12 comment Center of gravity of a self intersecting irregular polygon @mixkat: Given your description of the problem, finding the centroid of the polygon your group forms seems like overkill. Treat each person like a point mass and find the center of mass of your collection of points as: $$\left(\frac{\sum_{i=0}^n x_i}{n}, \frac{\sum_{i=0}^n y_i}{n}\right).$$ Mar 12 comment Linear Algebra of Symmetric Sums: a converse question My question is: given only the $\mathcal{Z_k}$'s and that $x=(1,1, \dots, 1)$ can I tell if the matrix $M$ is singular or not? Mar 11 asked Linear Algebra of Symmetric Sums: a converse question Mar 10 revised From baby Hartshorne: showing the exterior of a circle is segment connected Fixed an erroneous statement in the second proof, clarified the third proof, and added an illustration.; added 6 characters in body Mar 10 comment Equidecomposability of a Cube into 6 Trirectangular Tetrahedra Just making sure everyone is aware: the easiest generalization of the 2-dimesional splitting of a square into two right triangles comes out of noting that you can split the square region ($0y$ and $y>x$. This generalizes to 3 dimensions by splitting the cubic region (\$0