A polynomial is called reciprocal if its coefficients read forwards are the same as those read backward - there is an obvious translation of reciprocal polynomials into cosine polynomials (since a term like $a_i x^{n + k}+ a_i x^{n-k}$ can be written like $a_i \exp(i n t) \cos 2\pi i k t,$ under the obvious substitution. It is known that a random reciprocal polynomial with i.i.d. Gaussian coefficients has 58% (= $1/\sqrt{3}$) of its roots on the unit circle, and experiments with i.i.d. uniform integer coefficients between -1000 and 1000, and degree 600 show that the smallest number of unimodular roots in 1000 trials is 290 (which is a lot).

It has been shown (Borwein, Peter; Erdélyi, Tamás Lower bounds for the number of zeros of cosine polynomials in the period: a problem of Littlewood. Acta Arith. 128 (2007), no. 4, 377–384.) that a family of reciprocal polynomials with bounded coefficients and growing degree will have more and more unimodular roots. So, the conclusion is that a reciprocal polynomial with no unimodular roots and of high degree will have large coefficients. But how large? And what are natural ways of producing these non-unimodular guys? (this question is strongly connected to Littlewood's conjectures, thus the harmonic analysis tag).

PS: Of course, only irreducible polynomials are really of interest. And the coefficients are integers, which, sadly, has remained unsaid above.

  • $\begingroup$ "Irreducible": over what field? $\endgroup$ – Laurent Moret-Bailly Oct 7 '14 at 17:38
  • $\begingroup$ @LaurentMoret-Bailly I was intending $\mathbb{Q}$... $\endgroup$ – Igor Rivin Oct 7 '14 at 19:18
  • $\begingroup$ If you are intending irreducibility over the rationals, are you intending integer coefficients? $\endgroup$ – Gerry Myerson Oct 7 '14 at 22:47
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    $\begingroup$ @GerryMyerson Yes, indeed. $\endgroup$ – Igor Rivin Oct 7 '14 at 23:09
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    $\begingroup$ I think you're misreading the result of that paper. For example, the reciprocal polynomial $x^{2n} + 3 x^n + 1$ has no unimodular roots, and I think it is irreducible for all $n$. $\endgroup$ – Robert Israel Oct 7 '14 at 23:47

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