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Imagine there are numbers $a_1,\ldots,a_n \in \mathbb R$ and you want to know whether they are all positive. You cannot access the numbers themselves, but you can choose any symmetric polynomials you like to evaluate at $(a_1,\ldots,a_n)$. Actually, suppose you don't even get the results of those evaluations, but only the signs.

What is a good family $s_{i,n}$ of symmetric polynomials that allows you to do this?

To restate the problem, we want (for each $n=1,2,...$) a family $s_{1,n},s_{2,n},\ldots,s_{k,n}$ of symmetric polynomials in the variables $x_1,\ldots,x_n$ with the property that these are equivalent:

  • $a_j > 0$ for $j=1,\ldots,n$.
  • $s_{i,n}(a_1,\ldots,a_n) > 0$ for $i=1,\ldots,k$.

For extra credit, forget about $a_1,\ldots,a_n$ being real, and replace $a_j > 0$ with $\mathfrak{R}(a_j) > 0$.

How do I know this is even possible? It's because a related question has been treated extensively, namely that of determining, given the coefficients of a polynomial, whether all of the roots of the polynomial are in the right (or left or whatever) half-plane. Since the coefficients are (the elementary) symmetric polynomials applied to the roots of the polynomial, results like the Routh–Hurwitz stability criterion, which is usually expressed in terms of the signs of certain polynomials in the coefficients, can be rewritten in terms of symmetric polynomials applied to the roots.

But what if you don't care about going through the coefficients? What if you are happy enough to write down symmetric polynomials in terms of the roots themselves? Then what is the “best” or “most natural” family of symmetric polynomials that will do the job?

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    $\begingroup$ I'm not sure I understand your question correctly, so forgive me if this is completely off base. But the fundamental theorem of symmetric polynomial says that every symmetric polynomial can be expressed as a polynomial in the elementary symmetric polynomials, so you're not going to get anything that can't also be obtained through the coefficients. $\endgroup$ Feb 12, 2022 at 5:39
  • $\begingroup$ @JonathanLove: That's definitely not off-base. The thing is that the only way I know of to get a family that works is to start with something that was obtained in terms of the coefficients, and therefore is probably not the "simplest" possible. I put that term in quotes (and see also the last sentence of the question) because I'm not sure exactly what I mean by "best" or "most natural" or "simplest". I just suspect there is some nicer family of polynomials than what you could get by starting with polynomials in the coefficients. Of course, I'm not sure what I mean by "nicer" either. :) $\endgroup$ Feb 12, 2022 at 14:29

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The elementary symmetric functions $1=e_0,e_1,\dots,e_n$ will suffice. Clearly $e_j>0$ if each $a_i>0$. Conversely, let $P(x)= \prod_{i=1}^n(x+a_i)$. If some $a_i<0$ and all $e_j>0$ then $0=P(-a_i)=\sum_j e_j (-a_i)^{n-j}>0$, a contradiction.

Incidentally, if we assume only that $a_i\in\mathbb{C}$ we can still determine whether each $a_i$ is a positive real number using $n-1$ additional symmetric functions. See Gantmacher, Matrix Theory, vol. 2, Chapter 15, Section 9.1.

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  • $\begingroup$ Thanks, that is nice! In Gantmacher, Section 9.1 concerns the problem of counting the number of distinct real roots of a polynomial. Is this definitely the section you had in mind? In particular, I actually did find the answer I was looking for in Section 7.1, so this was helpful regardless. $\endgroup$ Feb 15, 2022 at 3:03
  • $\begingroup$ @LouisDeaett: I don't have access to Gantmacher to check. But 9.1 could be relevant from what you said since we can first check whether all the roots are real and then check whether they are all negative. $\endgroup$ Feb 16, 2022 at 16:49

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