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Let $z=x+iy$ ($x,y$ real) be a complex root of $$S_r(z)=z^{4m}+rz^{4m-2}+r^{4m-3}z^2+r^{4m}$$ where $m\geq 2$ is an integer and $r\gt 1$ is a real number. Then, I conjecture that $(*) : R_r(y^2)=0$ where $$ R_r(y^2)=2y^{2m}-\bigg(mr^2+\frac{r}{2}\bigg)y^{2(m-1)}\\+\sum_{d=2}^{m} \small\Bigg(\frac{m}{4^{d-1}d}\binom{2m-(d+1)}{d-1}r^{2d}+\frac{m-1}{4^{d-1}(d-1)}\binom{2m-(d+2)}{d-2}r^{2d-1}\Bigg)(-1)^dy^{2(m-d)}. $$

Does anyone know how to prove (*) or find a counterexample ?

My thoughts: I have checked (*) for $2 \leq m \leq 15$. Numerical examples seem to indicate that $R_r(y^2)$ is in fact the minimal polynomial of $y$ whenever $x+iy$ is a root of $S_r$, which is remarkably simple - if we had replaced $S_r$ with a random irreducible polynomial of degree $4m$, different $y$'s might have different minimal polynomials, and those minimal polynomials would often have degree larger than the original $4m$ not smaller.

It is easy to see that all the roots of $S_r$ have modulus $r$. Indeed, this follows from $$S_r(re^{i\theta})=2r^{4m-1}e^{2im\theta}\bigg(r\cos((2m)\theta)+\cos((2m-2)\theta)\bigg),$$ (see Jyrki Lahtonen's second comment on this other question). So we have the promising-looking identity $x^2+y^2=r^2$.

Ewan Delanoy
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1 Answers1

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$\def\e{\mathrm{e}}\def\i{\mathrm{i}}\def\R{\mathbb{R}}\def\peq{\mathrel{\phantom{=}}{}}$This proof utilizes the given result that $S_r(z) = 0 \Longrightarrow |z| = r$ and\begin{gather*} S_r(r\e^{\i θ}) = 2r^{4m - 1} \e^{2\i mθ} (r\cos(2mθ) + \cos(2(m - 1)θ)). \quad \forall θ \in \R \end{gather*}

Lemma: If $\{T_n(x)\}$ are Chebyshev polynomials defined by$$ T_0(x) = 1, \quad T_1(x) = x, \quad T_{n + 1}(x) = 2x T_n(x) - T_{n - 1}(x), \quad \forall n \geqslant 1 $$ then$$ T_{2m}(x) = (-1)^m m \sum_{k = 0}^m \frac{(-1)^k}{2m - k} \binom{2m - k}{k} (4(1 - x^2))^{m - k}. $$

Proof: It is known that$$ T_{2m}(x) = m \sum_{k = 0}^m \frac{(-1)^k}{2m - k} \binom{2m - k}{k} (2x)^{2m - 2k} $$ and $T_{2m}(\cos θ) = \cos(2mθ)$ for $θ \in \R$ (See, e.g. here). Thus for $θ \in \left( 0, \dfrac{π}{2} \right)$,\begin{gather*} T_{2m}(\sin θ) = T_{2m}\left( \cos\left( θ - \frac{π}{2} \right) \right)\\ = \cos(2mθ - mπ) = (-1)^m \cos(2mθ) = (-1)^m T_{2m}(\cos θ), \end{gather*} which implies that for $x \in (0, 1)$,\begin{align*} &\peq T_{2m}(x) = (-1)^m T_{2m}(\sqrt{1 - x^2})\\ &= m \sum_{k = 0}^m \frac{(-1)^k}{2m - k} \binom{2m - k}{k} (2\sqrt{1 - x^2})^{2m - 2k}\\ &= m \sum_{k = 0}^m \frac{(-1)^k}{2m - k} \binom{2m - k}{k} (4(1 - x^2))^{m - k}. \end{align*} Note that both $T_{2m}(x)$ and $m \sum\limits_{k = 0}^m \dfrac{(-1)^k}{2m - k} \dbinom{2m - k}{k} (4(1 - x^2))^{m - k}$ are polynomials of $x$, therefore the above identity holds for all $x \in \R$. $\quad\square$

Now return to the question. Since\begin{gather*} \frac{1}{r^{2m - 1}} R_r(r^2 \sin^2 θ) = 2r (\sin^2 θ)^m - \left( mr + \frac{1}{2} \right) (\sin^2 θ)^{m - 1}\\ + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} \left( \frac{m}{k} \binom{2m - k - 1}{k - 1} r + \frac{m - 1}{k - 1} \binom{2m - k - 2}{k - 2} \right) (\sin^2 θ)^{m - k}, \end{gather*} it suffices to prove that if $r\cos(2mθ) + \cos(2(m - 1)θ) = 0$, then\begin{gather*} 2r (\sin^2 θ)^m - \left( mr + \frac{1}{2} \right) (\sin^2 θ)^{m - 1}\\ + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} \left( \frac{m}{k} \binom{2m - k - 1}{k - 1} r + \frac{m - 1}{k - 1} \binom{2m - k - 2}{k - 2} \right) (\sin^2 θ)^{m - k} = 0. \end{gather*} By the lemma,\begin{align*} &\peq 2 (\sin^2 θ)^m - m(\sin^2 θ)^{m - 1} + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} · \frac{m}{k} \binom{2m - k - 1}{k - 1} (\sin^2 θ)^{m - k}\\ &= 2 (\sin^2 θ)^m - m(\sin^2 θ)^{m - 1} + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} · \frac{m}{2m - k} \binom{2m - k}{k} (\sin^2 θ)^{m - k}\\ &= \sum_{k = 0}^m \frac{(-1)^k}{4^{k - 1}} · \frac{m}{2m - k} \binom{2m - k}{k} (\sin^2 θ)^{m - k}\\ &= \frac{1}{4^{m - 1}} · m \sum_{k = 0}^m \frac{(-1)^k}{2m - k} \binom{2m - k}{k} (4(1 - \cos^2 θ))^{m - k}\\ &= \frac{1}{4^{m - 1}} · (-1)^m T_{2m}(\cos θ) = \frac{(-1)^m}{4^{m - 1}} \cos(2mθ), \end{align*}\begin{align*} &\peq {-}\frac{1}{2} (\sin^2 θ)^{m - 1} + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} · \frac{m - 1}{k - 1} \binom{2m - k - 2}{k - 2} (\sin^2 θ)^{m - k}\\ &= -\frac{1}{2} (\sin^2 θ)^{m - 1} + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} · \frac{m - 1}{2m - k - 1} \binom{2m - k - 1}{k - 1} (\sin^2 θ)^{m - k}\\ &= \sum_{k = 1}^m \frac{(-1)^k}{4^{k - 1}} · \frac{m - 1}{2m - k - 1} \binom{2m - k - 1}{k - 1} (\sin^2 θ)^{m - k}\\ &= -\frac{1}{4^{m - 1}} · (m - 1) \sum_{k = 0}^{m - 1} \frac{(-1)^k}{2(m - 1) - k} \binom{2(m - 1) - k}{k} (4(1 - \cos^2 θ))^{(m - 1) - k}\\ &= -\frac{1}{4^{m - 1}} · (-1)^{m - 1} T_{2(m - 1)}(\cos θ) = \frac{(-1)^m}{4^{m - 1}} \cos(2(m - 1)θ), \end{align*} so\begin{align*} &\peq 2r (\sin^2 θ)^m - \left( mr + \frac{1}{2} \right) (\sin^2 θ)^{m - 1}\\ &\peq + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} \left( \frac{m}{k} \binom{2m - k - 1}{k - 1} r + \frac{m - 1}{k - 1} \binom{2m - k - 2}{k - 2} \right) (\sin^2 θ)^{m - k}\\ &= \left( 2 (\sin^2 θ)^m - m(\sin^2 θ)^{m - 1} + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} · \frac{m}{k} \binom{2m - k - 1}{k - 1} (\sin^2 θ)^{m - k} \right) r\\ &\peq + \left( -\frac{1}{2} (\sin^2 θ)^{m - 1} + \sum_{k = 2}^m \frac{(-1)^k}{4^{k - 1}} · \frac{m - 1}{k - 1} \binom{2m - k - 2}{k - 2} (\sin^2 θ)^{m - k} \right)\\ &= \frac{(-1)^m}{4^{m - 1}} \cos(2mθ) · r + \frac{(-1)^m}{4^{m - 1}} \cos(2(m - 1)θ)\\ &= \frac{(-1)^m}{4^{m - 1}} (r\cos(2mθ) + \cos(2(m - 1)θ)) = 0. \end{align*}

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