A curious conjecture about self-reciprocal polynomials

Let $p$ be a polynomial
$p(z) = a_0 + a_1z+\dots + a_nz^n\ \ \ \ \ \ a_i\in {\bf C},$
and let $p^*$ denote the reciprocal polynomial
$p^*(z) = a_n + a_{n-1}z+\dots + a_0z^n=z^np(1/z).$
We say $p$ is self-reciprocal if $p=p^*$ . For example,
$1+2z+3z^2+2z^3+z^4$
and
$1+2z+3z^2+3z^3+2z^4+z^5$
are self-reciprocal. Suppose
$p(z) = a_0 + a_1z+\dots + a_dz^d+a_{d+1}z^{d+1}+a_dz^{d}+\dots +a_{1}z^{2d+1}+a_0z^{2d+2},\ \ \ \ \ \ a_i\in {\bf C},$
or
$p(z) = a_0 + a_1z+\dots + a_dz^d+a_{d+1}z^{d+1}+a_{d+1}z^{d+2}+a_dz^{d+3}+\dots +a_{1}z^{2d+2}+a_0z^{2d+3},\ \ \ \ \ \ a_i\in {\bf C},$

The question is this: for which increasing sequences $a_0<a_1<\dots a_{d+1}$ do the polynomial roots $p(z)=0$ lie on the unit circle $|z|=1$ ?

Some examples:

This represents, when $d=1$ and $a_0=1$ and $a_1=1.1$ the largest $a_{d+1}$ which has this roots property is as in the plot.

This represents, when $d=1$ and $a_0=1$ and $a_1=1.2$ the largest $a_{d+1}$ which has this roots property is as in the plot.

This represents, when $d=1$ and $a_0=1$ and $a_1=1.3$ the largest $a_{d+1}$ which has this roots property is as in the plot.

This represents, when $d=1$ and $a_0=1$ and $a_1=1.4$ the largest $a_{d+1}$ which has this roots property is as in the plot.

This represents, when $d=1$ and $a_0=1$ and $a_1=1.5$ the largest $a_{d+1}$ which has this roots property is as in the plot.

Conjecture 1: Let $s:{\bf Z}_{>0}\to {\bf R}_{>0}$ be a ”slowly increasing” function.

Odd degree case.
If $g(z)= a_0 + a_1z + \dots + a_dz^d$ , where $a_i=s(i)$ , then the roots of $p(z)=g(z)+z^{d+1}g^*(z)$ all lie on the unit circle.
Even degree case.
The roots of
$p(z)=a_0 + a_1z + \dots + a_{d-1}z^{d-1} + a_{d}z^{d} + a_{d-1}z^{d+1} + \dots + a_{1}z^{2d-1}+a_0z^{2d}$
all lie on the unit circle.

The next conjecture gives an idea as to how fast a “slowly increasing” function can grow.

Conjecture 2:* Consider the even degree case only. The polynomial
$p(z)=a_0 + a_1z + \dots + a_{d-1}z^{d-1} + a_{d}z^{d} + a_{d-1}z^{d+1} + \dots + a_{1}z^{2d-1}+a_0z^{2d}$ ,
with $a_0=1$ and all $a_i>0$ , is symmetric increasing if and only if it can be written as a product of quadratics of the form $x^2-2\cos(\theta)x+1$ , where all but one of the factors satisfy $2\pi/4\leq \theta\leq 4\pi/3$ . One of the factors can be of the form $x^2+\alpha x+1$ , for some $\alpha \geq 0$ .

* It was pointed out to me by Els Withers that this conjecture is false.

2 thoughts on “A curious conjecture about self-reciprocal polynomials”

I don’t see the conjecture, but this paper seems quite relevant:

http://imamat.oxfordjournals.org/content/8/3/397.short

wdjoyner says:

2014/12/05 at 6:30 am

Thanks for the reference! I added a vaguely-worded conjecture, the best I have at this point.

Reply

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A curious conjecture about self-reciprocal polynomials

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