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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?

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    $\begingroup$ 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. $\endgroup$ – Philipp Lampe Oct 29 '18 at 17:18
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    $\begingroup$ According to your definition, if $f$ is not identically $0$ you must have $n$ even, because $f(x) = f(-(-x)) = (-x)^n f(-1/x) = (-x)^n (1/x)^n f(x)$. $\endgroup$ – Robert Israel Oct 29 '18 at 18:18
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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 $\{i,-i\}$.

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  • $\begingroup$ $n$ stands for the degree. By $\mathbb{C}^{\times}$ you mean non-zero complex number or set of all unit elements? $\endgroup$ – SuperMario Oct 29 '18 at 21:35
  • $\begingroup$ @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). $\endgroup$ – François Brunault Oct 29 '18 at 22:46
  • $\begingroup$ Do you think if the coefficients are even or odd the roots the number of roots lying on the unit circle will change? $\endgroup$ – SuperMario Oct 29 '18 at 23:08
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    $\begingroup$ Coefficients even or odd? Were we meant to assume the coefficients are integers? This was never stated. $\endgroup$ – Gerry Myerson Oct 30 '18 at 4:15
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    $\begingroup$ @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). $\endgroup$ – François Brunault Oct 30 '18 at 7:01

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