Verifying a polynomial is positive on the half-line defined by $z \geq 1$

Question

Math people:

I am running experiments that produce polynomials $P(z)$ that, in every experiment I have run, are always positive on the half-line $\{z \geq 1\}$. I want to prove analytically that the polynomials that I generate using my rules must have this property. Unfortunately, the signs of the coefficients of the powers of $z$ in $P(z)$ usually change sign often (of course the leading coefficient is positive). In all of my experiments so far, all the nonzero coefficients of the powers of $t$ in $P(t+1)$ are always positive (which is certainly more than sufficient for $P$ to be positive on $\{z \geq 1\}$). I would like a way to verify that $P(z)$ is positive for $z \geq 1$ without having to multiply out $P(t+1)$. I would like to emphasize that I only need to show that $P(z)$ is positive for $z \geq 1$, I don't need to prove $P(t+1)$ has all positive coefficients. The fact that apparently $P(t+1)$ always has positive coefficients suggests that there may be a simple test for me to use to prove what I need.

Clarification: I do have analytic proof that $P(z=1)>0$ and the leading coefficient of $P(z)$ is positive. I do not have analytic proof that the coefficients of $P(t+1)$ are positive, but the polynomials I am generating in my experiment seem to have this strong property. Therefore I am hoping there is a simple test for signs of polynomials that will show that my polynomials are always positive on $\{z \geq 1\}$. I want an analytic, not a numerical tool.

UPDATE: Here is where the polynomials come from: let $n \geq 1$ and let $\mu_1, \mu_2, \ldots, \mu_n, \gamma_1,\gamma_2, \ldots, \gamma_n$ be distinct positive integers. For real $z$, define the $n$-by-$n$ matrix $M(z)$ by

$$M(z)_{i,j} = \frac{1}{\mu_i + \gamma_j} + \frac{z^{\gamma_j}}{\mu_i - \gamma_j},$$

and define the polynomial $P$ with rational coefficients by $P(z) = \operatorname{det}(M(z))$. I want to show $P(z)$ has the same, nonzero sign for all $z \geq 1$. As I note below, $P(1)$ is known to be nonzero. If $P(1) < 0$, just switch the top two rows of $M(z)$ for all $z$ and $P(1)$ becomes positive.

$M(1)$ is the product of a "Cauchy matrix'' (http://en.wikipedia.org/wiki/Cauchy_matrix) and a nonsingular diagonal matrix. There is a known exact formula for the determinant of a Cauchy matrix. For $z>>1$, $M(z)$ also resembles the product of a nonsingular diagonal matrix and a Cauchy matrix, and the determinant of $M(z)$ works out to have the same sign as that of $M(1)$ (which we have rigged to be positive).

The special case where all the $\mu$'s are bigger than all the $\gamma$'s has been solved.

Here is an example of $P(z)$:

    (1/1512)*z^20-(1/29700)*z^18-(11/33048)*z^15+(715/301644)*z^14

-(17879/7068600)*z^13-(13/430920)*z^12+(13/326700)*z^11-(65/6593076)*z^9+

(13/367200)*z^8-(30281/564074280)*z^7-(13/127234800)*z^6+(1/4740120)*z^5+

(1/161164080)*z^2-1/223373414880

and this is $P(t+1)$:

    (1/1512)*t^20+(5/378)*t^19+(1451/11550)*t^18+(13052/17325)*t^17+(443411/138600)*t^16+

(650561753/63617400)*t^15+(4800134477/188024760)*t^14+(1467084181/28779300)*t^13+

(99985934971/1208730600)*t^12+(6513521209/59093496)*t^11+(2573705693/21205800)*t^10+

(286764920/2590137)*t^9+(134656965781/1611640800)*t^8+(24368103409/470061900)*t^7+

(337565813/12927600)*t^6+(221604617/21205800)*t^5+(10773737/3357585)*t^4+

(215655973/302182650)*t^3+(5221684/50363775)*t^2+(394186/50363775)*t+

169/1438965

Answer 1

$P(t+1)$ is a polynomial with positive values for all positive $t$. A such it definitely need not have only positive coefficients, consider for example $P(t+1)=t^2-t+1$ (corresponding to $P(t)=t^2-3t+3$). So even the simple test of looking at coefficients of $P(t+1)$ fails.

My suggestion: First verify that $P(1)=\sum a_k>0$ and of course $a_n>0$. Numerically determine all roots of $P'(t)$ with $t>1$ and verify that $P(t)>0$ for these. (Of course this is even more laborious than computing the coefficients of $P(t+1)$)

Answer 2

Edit: The below assumed that we knew that all the roots of $P(t)$ had real part less than zero, and that $P(z)$ was a polynomial over $\mathbb{C}$. So it's no longer relevant, given the updated information. I'll try to edit in an answer that is correct.

---

Since the coefficients of $P(t)$ are all positive, then $P(t) > 0$ when $t > 0$. By the Gauss-Lucas theorem, the zeros of $P'(t)$ must lie in the convex hull of the zeros of $P(t)$. Therefore, $P'(t)$ has no zeros on the positive real line.

Let $z = t+1$. Then, $$\frac{dP}{dz} = \frac{dP}{dt}\frac{dt}{dz} = P'(z-1).$$ The convex hull containing the zeros of $P'(z)$ is just the convex hull containing the zeros of $P'(t)$ shifted over by $1$.

Therefore, $P'(z)$ has no zeros on the real line to the right of $z = 1$.

---

Note: this is just a hackneyed way of saying "P(z) is P(t) shifted to the right by 1".