# Reference request for anti-palindromic polynomials.

I have come across a lot of papers that are written about the palindromic polynomials, however, I am recently interested in polynomials satisfying $$f(-x) = x^nf(1/x)$$ for $$n\geq 1$$ and for all $$x\in \mathbb{R}$$ except $$0.$$ Is there any reference where the roots of such polynomials are studied?

• Is $n$ the degree of $f$? In the literature, authors typically use the functional equation $-f(x)=x^nf(x^{-1})$ to define antipalindromic polynomials. Oct 29, 2018 at 17:18
• According to your definition, if $f$ is not identically $0$ you must have $n$ even, because $f(x) = f(-(-x)) = (-x)^n f(-1/x) = (-x)^n (1/x)^n f(x)$. Oct 29, 2018 at 18:18

Assuming $$n=\deg f$$, it is easy to see that there is no such polynomial if $$n$$ is odd. So let us assume $$n$$ is even from now on. We may also assume $$f$$ is monic.

Let $$S$$ denote the multiset of roots of $$f$$ in $$\mathbb{C}$$ (note that $$\# S$$ is even). Then $$f$$ satisfies your condition if and only if $$\prod_{\alpha \in S} (x+\alpha) = \prod_{\alpha \in S} (1-\alpha x) = \bigl(\prod_{\alpha \in S} \alpha\bigr) \cdot \prod_{\alpha \in S} (x-1/\alpha).$$ This is equivalent to $$\prod_{\alpha \in S} \alpha = 1$$ and $$S$$ is invariant under the involution $$\iota : z \mapsto -1/z$$ (note that $$S$$ is contained in $$\mathbb{C}^\times$$). Now $$\iota$$ has 2 fixed points, namely $$i$$ and $$-i$$. If $$S$$ contains $$\{i,-i\}$$ then $$f$$ is divisible by $$x^2+1$$, and since $$x^2+1$$ satisfies your condition, the polynomial $$f(x)/(x^2+1)$$ also does. Dividing by $$x^2+1$$ as many times as needed, we may assume that $$S$$ does not contain $$\{i,-i\}$$. There are three cases:

1. $$S \cap \{i,-i\}=\emptyset$$. In this case $$f(x) = \prod_{j=1}^{n/2} (x-z_j)(x+1/z_j)$$ for some non necessarily distinct $$z_j \in \mathbb{C}^\times \backslash \{\pm i\}$$. Since the product of the roots is equal to $$1$$, we must have $$n \equiv 0 \textrm{ mod } 4$$.
2. $$S$$ contains $$i$$. Then the multiplicity of $$i$$ as a root of $$f$$ is even, so we may write $$f(x) = \prod_{j=1}^{n/2} (x-z_j)(x+1/z_j)$$ for some non necessarily distinct $$z_j \in \mathbb{C}^\times \backslash \{-i\}$$, and again $$n \equiv 0 \textrm{ mod } 4$$.
3. $$S$$ contains $$-i$$. Similarly $$f(x) = \prod_{j=1}^{n/2} (x-z_j)(x+1/z_j)$$ with $$z_j \in \mathbb{C}^\times \backslash \{i\}$$ and $$n \equiv 0 \textrm{ mod } 4$$.

In conclusion, the polynomials $$f$$ satisfying your condition are those of the form $$\lambda (x^2+1)^h \prod_{j=1}^{2m} (x-z_j)(x+1/z_j)$$ for some non necessarily distinct $$z_j \in \mathbb{C}^\times$$ and $$\lambda \in \mathbb{C}^\times$$, $$h \in \{0,1\}$$. Equivalently, a polynomial $$f$$ satisfies your condition if and only if its multiset of roots is the union of an even number of multisets $$\{z,-1/z\}$$ with $$z \in \mathbb{C}^\times$$ plus possibly one copy of $$\{i,-i\}$$.

• $n$ stands for the degree. By $\mathbb{C}^{\times}$ you mean non-zero complex number or set of all unit elements? Oct 29, 2018 at 21:35
• @SuperMario By $\mathbb{C}^\times$ I mean the multiplicative group of all nonzero complex numbers. I should also have mentioned that in the case $h=0$ (so $n \equiv 0 \textrm{ mod } 4$), the polynomial $g(x)=f(ix)$ is reciprocal, and there are many results on roots of such polynomials (lying within, on, or outside the unit circle). Oct 29, 2018 at 22:46
• Do you think if the coefficients are even or odd the roots the number of roots lying on the unit circle will change? Oct 29, 2018 at 23:08
• Coefficients even or odd? Were we meant to assume the coefficients are integers? This was never stated. Oct 30, 2018 at 4:15
• @SuperMario If $f$ has integral coefficients then there will be of course additional conditions on the roots. If you are interested with the roots on the unit circle, I encourage you to ask a new question (but make it precise enough). Oct 30, 2018 at 7:01