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Let $S$ be a finite subset of the positive integers. Define $N_S(x) = 1-(1-x)\sum_{j\in S}x^j$. Assume that $N_S(x)$ is symmetric, i.e., $x^dN_S(1/x)=N_S(x)$, where $d=\deg N_S(x)$. It seems that $N_S(x)$ tends to have many zeros of absolute value 1. As two random examples, if $S=\{1,2,7,8,9,10,13\}$ then $N_S(x)$ is irreducible and has ten such zeros, and if $S=\{2,3,4,6,8,12,13\}$ then $N_S(x)$ is irreducible and has twelve such zeros. Is there a reason for this? I don't know whether the symmetry is relevant or whether $\sum_{j\in S}x^j$ can be replaced by a more general polynomial.

Addendum. Of the $32$ sets $S$ with max$(S)=11$ for which $N_S(x)$ is symmetric, $16$ of them have eight zeros of $N_S(x)$ on the unit circle, $6$ of them have ten zeros, and $10$ of them have twelve zeros (and are hence a product of cyclotomic polynomials by a theorem of Kronecker).

Addendum 2. Let $F_n(x)=\prod_S N_S(x)$, where $S$ ranges over all subsets of the positive integers with maximum element $n=2m+1$ for which $N_S(x)$ is symmetric. Can some analytic technique be used to estimate the number of zeros of $F_n(x)$ on the unit circle (and hence the average number of zeros of $N_S(x)$ on the unit circle)?

Here is some related data. Let $g(n)$ be the number of zeros of $F_n(x)$ on the unit circle. Then $$ (g(1),g(3),g(5),\dots,g(33)) = (2,8,22,54,126,308, 660, 1538, 3350, 7368, 15904, 34764, 73480, 158424, 336256, 712958, 1502306). $$ It looks like the relative number of zeros of $N_S$ on the unit circle is slowly decreasing at an irregular rate as $n$ increases. For $n=33$ it is $\frac{1502306}{33\cdot 2^{16}}= 0.6946\cdots$.

Richard Stanley
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    symmetry is certainly relevant: if $a$ is a root with $|a|=1$, then $1/a$ is also a root, and without symmetry $N_s(x)$ and $x^dN_s(1/x)$ are normally coprime – Fedor Petrov Jan 08 '24 at 21:58
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    If you remove $13$ from your first example of $S$, then none of the roots are unit magnitude. Of course, in this case the polynomial is not symmetric (palindromic). I agree with Fedor. – T. Amdeberhan Jan 08 '24 at 22:02
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    This interesting problem prompted me to this question: if $P\in\mathbb{R}[x]$ has all roots on the unit circle $Gamma$ then $P$ is symmetric or anti-symmetric. Your polynomials are $N_S\in\mathbb{R}[x]$. If your anticipation holds of having the majority roots on $\Gamma$ then is there a density argument (in the sense, polynomials $P$ with roots on $\Gamma$ are dense in polynomials set $\mathbb{R}[x]$ for symmetric/anti-symmetric)? – T. Amdeberhan Jan 09 '24 at 05:31
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    Still, it can happen that only one third of the roots have absolute value $1$. For $S={1, 2, 3, 5, 6, 7, 8, 13, 17}$ the resulting degree $18$ polynomial has only $6$ such roots. – Peter Mueller Jan 09 '24 at 13:14
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    ... and even for $S={1, 2, 3, 5, 6, 7, 9, 10, 11, 12, 17, 21, 25}$ there are only $6$ such roots. – Peter Mueller Jan 09 '24 at 14:16
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    @T.Amdeberhan: if $P$ has no roots on the unit circle, then the same is true for all polynomials in a sufficiently small neighborhood of $P$. Thus polynomials with a least one root on the unit circle cannot be dense. – Richard Stanley Jan 09 '24 at 19:07
  • If I remember correctly, questions of a similar kind are the subject of Schur-Cohn theory, see e. g. chapter X in Morris Marden, Geometry of polynomials, chapter 7 in Sheil-Small, Complex polynomials or chapter 11.5 in Rahman/Scheisser, Analytic theory of polynomials. I never worked through that stuff though, it might not be (easily) applicable to this particular case. – thomashennecke Jan 29 '24 at 21:44

2 Answers2

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I'm not sure to what extent it's relevant, but your second example is a Salem polynomial (or more accurately, the minimal polynomial of a Salem number), i.e., it is a symmetric monic integer polynomial (of degree at least 4) with exactly one root outside the unit circle. I'm not sure if there are "generalized $k$-Salem polynomials" with exactly $k$ roots outside the unit circle, but your first example would be a $2$-Salem polynomial. It's also amusing how the same objects seem to acquire multiple names. Your symmetric polynomials are often called reciprocal polynomials, and also sometimes palindromic polynomials.

Joe Silverman
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The following might be relevant.

Proposition. Let $f(x)=a_dx^d+\cdots +a_1x+a_0\in \mathbb{R}[x]$ be a reciprocal polynomial of even degree $d=2n$. If $|a_n|\le |a_j|$ for some $j\ne n$ then $f$ has at least two complex zeros of modulus $1$.

We keep the notation and assumptions of the proposition, and $S^1$ denotes the unit circle in $\mathbb{C}$. Also, we define the rational function $g(x)=f(x)/x^n$ and we write $e(t)=e^{2\pi i t}$.

Lemma. $g(z)\in \mathbb{R}$ for every $z\in S^1$.

Proof of the lemma. Let $z\in S^1$. As $f$ is reciprocal we have $z^df(1/z)=f(z)$, hence $$ \begin{aligned} \overline{g(z)}&= g(\overline{z})=g(1/z)=z^nf(1/z)\\ &= z^d f(1/z)/z^n =f(z)/z^n=g(z). \end{aligned}$$

Let us now prove the proposition, by contradiction. Suppose that $f(x)$ has at most one zero in $S^1$, then so does $g(x)$ and therefore it cannot change sign on $S^1$ (it takes real values there by the lemma and it is continuous.) It follows that $$ \left|\int_0^1 g(e(t)){\rm d}t\right| = \int_{0}^1 |g(e(t))|{\rm d}t. $$ Let $I$ be the integral on the left and $J$ the one on the right, so that the equation is $|I|=J$. We note that $$ I=\int_0^1 f(e(t))e(-nt){\rm d}t = a_n $$ hence, $|I|= |a_n|$. On the other hand, since $|e(\theta)|=1$ for all real $\theta$ we see that $$ \begin{aligned} J&=\int_0^1 |f(e(t))e(-nt)|{\rm d}t\\ &=\int_0^1 |f(e(t))e(-jt)|{\rm d}t\\ &\ge \left|\int_0^1 f(e(t))e(-jt){\rm d}t\right| =|a_j| \end{aligned}$$ where $j\ne n$ satisfies $|a_n|\le |a_j|$. Hence, all the previous inequalities are actually equalities and in particular $$ \int_0^1 |f(e(t))e(-jt)|{\rm d}t= \left|\int_0^1 f(e(t))e(-jt){\rm d}t\right|. $$ By the equality condition in the triangle inequality, there is certain number $w\in S^1$ such that $w\cdot f(e(t))e(-jt)$ is real and non-negative for all real $t$. We deduce that for each $z\in S^1$ $$ \begin{aligned} w\cdot f(z)/z^j&=\overline{w}\cdot f(\overline{z})/\overline{z}^j\\ & = \overline{w}\cdot z^jf(1/z)\\ &=\overline{w}z^{j-d}\cdot f(z). \end{aligned}$$ Thus, for all $z\in S^1$ with at most one exception (more precisely, where $f$ does not vanish) we have $z^{2j-d}=w^2$. As $w$ is fixed and $2j\ne d$, we reach a contradiction.

Pasten
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