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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
Rodrigo de Azevedo
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Stefan Smith
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2 Answers2

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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".

Emily
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  • I see. This will be hard to do with arbitrary coefficients, if you have no other criteria. The best you'll get is a framework for showing that, say, $P(t)$ has all positive coefficients. In fact, that's not hard, but in some sense that's still a "numerical" approach. – Emily Aug 06 '13 at 20:59
  • The main issue here is that you're asking for "I want to show that all permissible polynomials have this desired property." In general, this is not true, because it is easy to construct a polynomial that doesn't have that property. However, any counterexample that can be constructed generally might not be permissible under your "rules." But since we don't know what those rules are, we can't give you a general proof. – Emily Aug 06 '13 at 21:01
  • I edited my question to indicate exactly what the rules were. I would be interested in lesser-known tools for finding signs of polynomials that might work. There might be some things in http://en.wikipedia.org/wiki/Properties_of_polynomial_roots, for example. – Stefan Smith Aug 06 '13 at 21:23
  • I'll think on it. It might be a lot easier to show that the determinants of the individual matrices have the same sign, however. – Emily Aug 06 '13 at 21:26
  • @StefanSmith You said that the resulting polynomial is the determinant of the product of a Cauchy matrix (call it $C$) and a diagonal matrix (call it $D$). Then $\det M = \det CD = \det C \det D$. $\det D$ is easy to calculate (the product of its entries) and $\det C$ has an explicit form. – Emily Aug 06 '13 at 22:07
  • unfortunately, except for the $z=1$ case, which simplifies, $M(z)$ is the sum, not the product, of two matrices, of which one is a Cauchy matrix and the other is the product of a Cauchy matrix and a diagonal matrix. – Stefan Smith Aug 06 '13 at 22:25
  • I think it should be clarified that $P(t) \neq 0$ for all $t > 0$ is not sufficient to have $P'(t) \neq 0$ for all $t > 0$. The Gauss-Lucas theorem is irrelevant to your answer; it does not imply that $P'(t)$ has no positive real roots. – Antonio Vargas Aug 06 '13 at 23:46
  • @AntonioVargas That is true, which I see now that the conditions have been updated. Certainly we could have $P(t)$ with roots with positive real components, and then $P'(t)$ could have a positive real root. Prior to the clarification, I thought that it was known that no roots have positive real components, so this answer is no longer relevant! – Emily Aug 07 '13 at 00:37
  • Oh, I didn't know! In that case I apologize for the brusqueness of my last comment. Your answer is, of course, completely correct in that context. – Antonio Vargas Aug 07 '13 at 01:01
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$P(t+1)$ 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 $P(t+1)$)

Hagen von Eitzen
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  • You are misunderstanding something somewhere. If all the nonzero coefficients of all the powers of $t$ in the expansion of $P(t+1)$ are positive, then $P(t+1)$ is positive for all $t \geq 0$, therefore $P(z)$ is positive for all $z \geq 1$. In all my experiments, the coefficients of all the powers of $t$ in the expansion of $P(t+1)$ are positive, which is a sufficient, not necessary, condition for $P(z)$ to be positive for all $z \geq 1$. I only need to show $P(z) > 0$ for $z \geq 1$. Regarding your suggestion, I am looking for an analytic proof. – Stefan Smith Aug 06 '13 at 20:01
  • Also, I do have an analytic proof that $P(1) > 0$, but I have not proven anything about any other derivatives, except for the highest (such as the 20th and the 19th in my example above). – Stefan Smith Aug 06 '13 at 20:03