# Are 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. $$a_1 = \sum_{i=1}^n x_i > 0\\ a_2 =\sum_{(i>j)} x_i x_j > 0\\ a_3 =\sum_{(i>j>k)} x_i x_j x_k> 0\\ \dots \\ a_n =\prod_{i=1}^n x_i > 0$$ where the sums go over all possible indicated pairs, triples, ... , $n$-tupels.

Question: Are these conditions sufficient to show that all $x_i>0$?

I observed that this is linked to roots of polynomials. Let $a_0=1$. We have the identity (Vieta): $$p(x) = \prod_{i=1}^n (x + x_i) = \sum_{k=0}^n a_k x^{n-k}$$ The roots of the polynomial $p(x)$ are given by $-x_i$, so if all roots are negative, then all $x_i$ are positive and all Vieta's coefficients are positive. The question asks for the other way.

For $n=2$, it is easy to show that it's true. For higher $n$, it will become cumbersome / impossible to give an analytic solution of the roots, since it is known that no such analytic solutions exist for $n >4$.

Before I start trying to produce (numerical) counterexamples, I ask if there is an already existing answer to this question.

• If all $a_k$ are positive then $\sum_{k=0}^n {n\choose k} a_k x^{n-k}$ cannot have a positive root. Am I overlooking or misunderstanding something? – Martin R Mar 28 '18 at 7:39
• @MartinR Your statement would be the (affirming) answer to my question. So: why can there be no positive root? – Andreas Mar 28 '18 at 7:43
• If $p(x)$ has (real) positive coefficients then $x > 0$ implies $p(x) >0$, trivially. – Martin R Mar 28 '18 at 7:44
• @MartinR So this gives a proof by contradiction, right? – Andreas Mar 28 '18 at 7:47
• There should be no $\binom{n}{k}$ in the Vieta identity, should it? – user Mar 28 '18 at 7:57

## 2 Answers

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, strictly positive coefficients.

If $x_i \le 0$ for some $i$ then $$0 = p(-x_i) = a_n + \sum_{k=0}^{n-1} a_k (-x_i)^{n-k} > 0$$ gives a contradiction.

So yes, if $x_1, \ldots, x_n$ are given real numbers and all their elementary symmetric polynomials $a_k$ are positive, then all $x_i$ are necessarily positive.

If the $x_i$ are complex numbers and the $a_k$ are positive real numbers then one can only conclude that the $x_i$ are not zero or negative real numbers.

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 none of the roots can be zero (since the constant term is nonzero).

Moreover, none of the roots can be negative either, since if $x_1<0$ and $n$ is even, then substituting $x_1$ for $x$, all terms on the $\text{LHS}$ of the equation would be positive, and if $x_1<0$ and $n$ is odd, then substituting $x_1$ for $x$, all terms on the $\text{LHS}$ of the equation would be negative.

Hence, if any of $x_1,...,x_n$ are real, they must be positive.

• For odd $n$, none of the roots can be negative either. – Martin R Mar 28 '18 at 8:23
• Thanks (it was a momentary blind spot), but now it's clear. I've edited it accordingly. – quasi Mar 28 '18 at 8:40