# 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 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.

1. 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.
2. 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”

• wdjoyner says:

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