Polynomials with Nonnegative Coefficients Whose Zeros Have Modulus One

Evans, Ronald J.; Greene, Joseph P.

doi:10.1137/0522076

Cited by 6 publications

(5 citation statements)

References 3 publications

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“…We complete our argument by applying the following interesting nontrivial result due to Barnard et al [3,Theorem 1]. (Or see [4,Theorem 2.4.5] or [11,Theorem 4] ).…”

Section: The Higher-dimensional Casementioning

confidence: 91%

Maximal Exponents of K-Primitive Matrices: The Polyhedral Cone Case

Loewy,

Tam

2009

Preprint

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Let K be a proper (i.e., closed, pointed, full convex) cone in R n . An n × n matrix A is said to be K-primitive if there exists a positive integer k such that A k (K \ {0}) ⊆ int K; the least such k is referred to as the exponent of A and is denoted by γ(A). For a polyhedral cone K, the maximum value of γ(A), taken over all K-primitive matrices A, is denoted by γ(K). It is proved that for any positive integers m, n, 3 ≤ n ≤ m, the maximum value of γ(K), as K runs through all n-dimensional polyhedral cones with m extreme rays, equals (n − 1)(m − 1) + 1 when m is even or m and n are both odd, and is at least (n − 1)(m − 1) and at most (n−1)(m−1)+1 when m is odd and n is even. For the cases when m = n, m = n+1 or n = 3, the cones K and the corresponding K-primitive matrices A such that γ(K) and γ(A) attain the maximum value are identified up to respectively linear isomorphism and cone-equivalence modulo positive scalar multiplication.

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“…We complete our argument by applying the following interesting nontrivial result due to Barnard et al [3,Theorem 1]. (Or see [4,Theorem 2.4.5] or [11,Theorem 4] ).…”

Section: The Higher-dimensional Casementioning

confidence: 91%

Maximal Exponents of K-Primitive Matrices: The Polyhedral Cone Case

Loewy,

Tam

2009

Preprint

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“…Finally, it is in order to mention some works on factorizations with constrained coefficients. Indeed, intuitively, it would make sense that some progress on the conjecture would follow from a study of the divisors and the roots of 0-1-polynomials (Newman polynomials) [29,37,9,10,14,26,7,35] and of polynomials with nonnegative coefficients [3,11,12,27]. Variations on this theme include the study of polynomials with coefficients that belong to a finite set [6] or that are constrained by inequalities [5].…”

Section: Introductionmentioning

confidence: 99%

Progress on the unfair 0-1-polynomials conjecture using linear recurrences and numerical analysis

Ghidelli

2022

Preprint

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If the product of two monic polynomials with real nonnegative coefficients has all coefficients equal to 0 or 1, does it follow that all the coefficients of the two factors are also equal to 0 or 1? % Here is an equivalent formulation of this intriguing problem: is it possible to weigh unfairly a pair of dice so that the probabilities of every possible outcome (roll them and take the sum) were the same? If the two dice have six faces numbered 1 to 6, it is easy to show that the answer is no. But for general dice with finitely many faces, this is an open problem with no significant advancement since 1937. In this paper we examine, in some sense, the first infinite family of cases that cannot be treated with classical methods: the first die has three faces numbered with 0,2 and 5, while the second die is arbitrary. In other words, we examine factorizations of 0-1-polynomials with one factor equal to x5+a x2 +1 for some nonnegative a. We discover that this case may be solved (that is, necessarily a=0 or a=1) using the theory of linear recurrence sequences, computation of resultants, as well as analytic and numerical approximations. 2020 MSC Classification: Primary 12D05 , 60B15 , 60-08 , 11B39 , 13P15 , 26C10; Secondary 00A08 , 11C08 , 60A10 , 11C20 , 26A05 , 26D05 , 65H04.

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“…It has been recently shown by R. Evans and P. Montgomery [4] that in the special case when p is the polynomial P N defined in (1) that the corresponding reduced p t polynomials are all strictly unimodal (in their coefficients). Also, R. Evans and J. Greene [3] have shown for polynomials p of the form p(z) = (z M k − 1)/(z k − 1) that the corresponding reduced p t polynomials all have positive coefficients.…”

Section: Remarkmentioning

confidence: 99%