Answer by quasi for Are positive Vieta's equations sufficient for negative...
Based on your definition of $a_1,...,a_n$, each of $x_1,...,x_n$ is a root of$$x^n -a_1 x^{n-1} + a_2 x^{n-2} + \cdots + (-1)^na_n=0$$Note the alternating signs.Now suppose $a_1,...,a_n > 0$.Then...
View ArticleAnswer by Martin R for Are positive Vieta's equations sufficient for negative...
Your $a_k$ are the “elementary symmetric polynomials” of $x_1, \ldots, x_n$. If all $a_k$ are positive then$$ p(x) = \prod_{i=1}^n (x + x_i) = \sum_{k=0}^n a_k x^{n-k}$$is a polynomial with real,...
View ArticleAre positive Vieta's equations sufficient for negative roots?
Let $n$ (unknown) real numbers $x_i$ be given. Suppose all Vieta's coefficient equations are positive, i.e.$$a_1 = \sum_{i=1}^n x_i > 0\\a_2 =\sum_{(i>j)} x_i x_j > 0\\a_3...
View Article
