Zeros of polynomials with four-term recurrence

TL;DR

This paper characterizes when the zeros of polynomials satisfying a specific four-term recurrence are all real, providing explicit conditions on parameters and describing the density of zeros on a real interval.

Contribution

It establishes necessary and sufficient conditions for the real zeros of four-term recurrence polynomials and describes the density of their zeros on a specific interval.

Findings

Zeros are real under certain parameter conditions.

Explicit interval where zeros are dense.

Conditions depend on parameters b and c.

Abstract

For any real numbers $b,c\in\mathbb{R}$, we form the sequence of polynomials $\left\{ H_{m}(z)\right\} _{m=0}^{\infty}$ satisfying the four-term recurrence \[ H_{m}(z)+cH_{m-1}(z)+bH_{m-2}(z)+zH_{m-3}(z)=0,\qquad m\ge3, \] with the initial conditions $H_{0}(z)=1$, $H_{1}(z)=$$-c$, and $H_{2}(z)=-b+c^{2}$. We find necessary and sufficient conditions on $b$ and $c$ under which the zeros of $H_{m}(z)$ are real for all $m$, and provide an explicit real interval on which ${\displaystyle \bigcup_{m=0}^{\infty}\mathcal{Z}(H_{m})}$ is dense where $\mathcal{Z}(H_{m})$ is the set of zeros of $H_{m}(z)$.

Equations

$$H_m(z)+c{H}_{m-1}(z)+b{H}_{m-2}(z)+z{H}_{m-3}(z)=0,m\ge 3,$$

$$\sum _{k=0}^n{a}_k(z)H_{m-k}(z)=0,m\ge n,$$

$$H_m(z)+c{H}_{m-1}(z)+b{H}_{m-2}(z)+z{H}_{m-3}(z)$$

$$H_0(z)$$

$$H_1(z)$$

$$H_2(z)$$

$$c^3.\left(-\infty ,\frac{-2+9b/c^2-2\sqrt{(1-3b/c^2{)}^3}}{27}\right].$$

$$\sum _{m=0}^{\infty }{H}_m(z)t^m=\frac{1}{1+ct+b{t}^2+z{t}^3}.$$

$$\sum _{m=0}^{\infty }{H}_m(z)t^m=\frac{1}{1+t+a{t}^2+z{t}^3}$$

$$I_a=\left(-\infty ,\frac{-2+9a-2\sqrt{(1-3a{)}^3}}{27}\right]$$

$$\mathcal{Z}(H_m(z,a))\in I_a$$

$$\mathcal{Z}(H_m(z,a^{\ast }))\in I_{a^{\ast }}$$

$$\left|H_m(z^{\ast },a_n)\right|={\gamma }^{(n)}\prod _{k=1}^{\mathrm{deg}{H}_m(z,a_n)}\left|z^{\ast }-z_k^{(n)}\right|$$

$$H_m(z^{\ast },a^{\ast })=\underset{n\to \infty }{\lim }{H}_m(z^{\ast },a_n)\ne 0$$

$$(1+t+a{t}^2+z{t}^3)\sum _{m=0}^{\infty }{H}_m(z)t^m=1.$$

$$H_{m+3}(z)+H_{m+2}(z)+a{H}_{m+1}(z)+z{H}_m(z)=0$$

$$\sum _{m=0}^{\infty }{H}_m(z)t^m=\frac{1}{1+t+a{t}^2+z{t}^3}=\frac{1}{z(t-t_0)(t-t_1)(t-t_2)}.$$

$$\frac{1}{z(t-t_0)(t_0-t_1)(t_0-t_2)}+\frac{1}{z(t-t_1)(t_1-t_0)(t_1-t_2)}+\frac{1}{z(t-t_2)(t_2-t_0)(t_2-t_1)}$$

$$-\sum _{m=0}^{\infty }\frac{1}{z}\left(\frac{1}{(t_0-t_1)(t_0-t_2)t_0^{m+1}}+\frac{1}{(t_1-t_0)(t_1-t_2)t_1^{m+1}}+\frac{1}{(t_2-t_0)(t_2-t_1)t_2^{m+1}}\right)t^m.$$

$$\frac{1}{(t_0-t_1)(t_0-t_2)t_0^{m+1}}+\frac{1}{(t_1-t_0)(t_1-t_2)t_1^{m+1}}+\frac{1}{(t_2-t_0)(t_2-t_1)t_2^{m+1}}=0.$$

$$\frac{1}{(1-t_1/t_0)(1-t_2/t_0)}+\frac{1}{(t_1/t_0-1)(t_1/t_0-t_2/t_0)(t_1/t_0{)}^{m+1}}+\frac{1}{(t_2/t_0-1)(t_2/t_0-t_1/t_0)(t_2/t_0{)}^{m+1}}=0.$$

$$\frac{1}{(1-e^{2i\theta })(1-\zeta e^{i\theta })}+\frac{1}{(e^{2i\theta }-1)(e^{2i\theta }-\zeta e^{i\theta })(e^{2i\theta }{)}^{m+1}}+\frac{1}{(\zeta e^{i\theta }-1)(\zeta e^{i\theta }-e^{2i\theta })(\zeta e^{i\theta }{)}^{m+1}},$$

$$\frac{1}{e^{2i\theta }(-2i\sin \theta )(e^{-i\theta }-\zeta )}+\frac{1}{(2i\sin \theta )(e^{i\theta }-\zeta )(e^{2i\theta }{)}^{m+2}}+\frac{1}{(\zeta -e^{-i\theta })(\zeta -e^{i\theta })(\zeta {)}^{m+1}(e^{i\theta }{)}^{m+3}}.$$

$$0$$

$$=\frac{(\zeta -e^{i\theta })e^{i(m+1)\theta }-(\zeta -e^{-i\theta })e^{-i(m+1)\theta }}{2i\sin \theta }+\frac{1}{{\zeta }^{m+1}}$$

$$=\frac{\zeta (e^{i(m+1)\theta }-e^{-i(m+1)\theta })+e^{-i(m+2)\theta }-e^{i(m+2)\theta }}{2i\sin \theta }+\frac{1}{{\zeta }^{m+1}}$$

$$=\frac{\zeta (2i\sin (m+1)\theta )-2i\sin (m+2)\theta }{2i\sin \theta }+\frac{1}{{\zeta }^{m+1}}$$

$$=\frac{2i\zeta \sin (m+1)\theta -2i\sin (m+1)\theta \cos \theta -2i\cos (m+1)\theta \sin \theta }{2i\sin \theta }+\frac{1}{{\zeta }^{m+1}}$$

$$=\frac{(\zeta -\cos \theta )\sin (m+1)\theta }{\sin \theta }-\cos (m+1)\theta +\frac{1}{{\zeta }^{m+1}}.$$

$$t_0+t_1+t_2$$

$$t_0{t}_1+t_0{t}_2+t_1{t}_2$$

$$t_0{t}_1{t}_2$$

$$1+e^{2i\theta }+\zeta e^{i\theta }$$

$$e^{2i\theta }+\zeta e^{i\theta }+\zeta e^{3i\theta }$$

$$\zeta e^{3i\theta }$$

$$\frac{1+e^{2i\theta }+\zeta e^{i\theta }}{e^{2i\theta }+\zeta e^{i\theta }+\zeta e^{3i\theta }}$$

Full text

[TABLE]

with the initial conditions $H_{0}(z)=1$, H_{1}(z)=$$-c, and $H_{2}(z)=-b+c^{2}$. We find necessary and sufficient conditions on $b$ and $c$ under which the zeros of $H_{m}(z)$ are real for all $m$, and provide an explicit real interval on which ${\displaystyle\bigcup_{m=0}^{\infty}\mathcal{Z}(H_{m})}$ is dense where $\mathcal{Z}(H_{m})$ is the set of zeros of $H_{m}(z)$.

1. Introduction

Consider the sequence of polynomials $\left\{H_{m}(z)\right\}_{m=0}^{\infty}$ satisfying a finite recurrence

[TABLE]

where $a_{k}(z)$, $1\leq k\leq n$, are complex polynomials. With certain initial conditions, one may ask for the locations of the zeros of $H_{m}(z)$ on the complex plane. There are two common approaches: asymptotic location and exact location. The first direction asks for a curve where the zeros of $H_{m}(z)$ approach as $m\rightarrow\infty$, and the second direction aims at a curve where the zeros of $H_{m}(z)$ lie exactly on for all $m$ (or at least for all large $m$). Recent works in the first direction include [1, 2, 3, 5, 6]. Results in this first direction are useful to prove the necessary condition for the reality of zeros of $H_{m}(z)$ that we will see in Section 3.

It is not an easy problem to find an explicit curve (if such exists) where the zeros of all $H_{m}(z)$ lie for a general recurrence in (1.1). For three-term recurrences with degree two and appropriate initial conditions, the curve containing zeros is given in [10]. The corresponding curve for a three-term recurrence with degree $n$ is given in [11]. Among all possible curves containing the zeros of the $H_{m}(z)$s, the real line plays an important role. We say that a polynomial is hyperbolic if all of its zeros are real. There are a lot of recent works on hyperbolic polynomials and on linear operators preserving hyperbolicity of polynomials, see for example [4, 7]. In some cases, we could find the curve on the plane containing the zeros of a sequence of polynomials by claiming certain polynomials are hyperbolic, see for example [8].

The main result of this paper is the identification of necessary and sufficient conditions on $b,c\in\mathbb{R}$ under which the zeros of the sequence of polynomials $H_{m}(z)$ satisfying the recurrence

[TABLE]

are real. We use the convention that the zeros of the constant zero polynomial are real. Let $\mathcal{Z}(H_{m})$ denote the set of zeros of $H_{m}(z)$.

Theorem 1**.**

Suppose $b,c\in\mathbb{R}$, and let $\left\{H_{m}(z)\right\}_{m=0}^{\infty}$ be defined as in (1.2). The zeros of $H_{m}(z)$ are real for all $m$ if and only if one of the two conditions below holds

(i) $c=0$ and $b\geq 0$

*(ii) $c\neq 0$ and $-1\leq b/c^{2}\leq 1/3$.

Moreover, in the first case with $b>0$, ${\displaystyle\bigcup_{m=0}^{\infty}\mathcal{Z}(H_{m})}$ is dense on $(-\infty,\infty)$. In the second case, ${\displaystyle\bigcup_{m=0}^{\infty}\mathcal{Z}(H_{m})}$ is dense on the interval*

[TABLE]

Our paper is organized as follow. In Section 2, we prove the sufficient condition for the reality of the zeros of all $H_{m}(z)$ in the case $c\neq 0$. The case $c=0$ follows from similar arguments whose key differences will be mentioned in Section 3. Finally, in Section 4, we prove the necessary condition for the reality of the zeros of $H_{m}(z)$.

2. The case $c\neq 0$ and $-1\leq b/c^{2}\leq 1/3$

We write the sequence $H_{m}(z)$ in (1.2) using its generating function

[TABLE]

From (2.1), if we make the substitutions $t\rightarrow t/c$, $b/c^{2}\rightarrow a$, and $z/c^{3}\rightarrow z$, it suffices to prove the following form of the theorem.

Theorem 2**.**

Consider the sequence of polynomials $\left\{H_{m}(z)\right\}_{m=0}^{\infty}$ generated by

[TABLE]

where $a\in\mathbb{R}.$ If $-1\leq a\leq 1/3$ then the zeros of $H_{m}(z)$ lie on the real interval

[TABLE]

and ${\displaystyle\bigcup_{m=0}^{\infty}\mathcal{Z}(H_{m})}$ is dense on $I_{a}$.

We will see later that the density of the union of zeros on $I_{a}$ follows naturally from the proof that $\mathcal{Z}(H_{m})\subset I_{a}$ and thus we focus on proving this claim. We note that each value of $a\in[-1,1/3]$ generates a sequence of polynomials $H_{m}(z,a)$. The lemma below asserts that it suffices to prove that $\mathcal{Z}(H_{m}(z,a))\subset I_{a}$ for all $a$ in a dense subset of $[-1,1/3]$.

Lemma 3**.**

Let $S$ be a dense subset of $[-1,1/3]$, and let $m\in\mathbb{N}$ be fixed. If

[TABLE]

for all $a\in S$ then

[TABLE]

for all $a^{*}\in[-1,1/3]$.

Proof.

Let $a^{*}\in[-1,1/3]$ be given. By the density of $S$ in $[-1,1/3]$, we can find a sequence $\{a_{n}\}$ in $S$ such that $a_{n}\rightarrow a^{*}$. For any $z^{*}\notin I_{a^{*}}$, we will show that $H_{m}(z^{*},a^{*})\neq 0$. We note that the zeros of $H_{m}(z,a_{n})$ lie on the interval $I_{a_{n}}$ whose right endpoint approaches the right endpoint of $I_{a^{*}}$ as $n\rightarrow\infty$. If we let $z_{k}^{(n)}$, $1\leq k\leq\deg H_{m}(z,a_{n})$, be the zeros of $H_{m}(z,a_{n})$ then

[TABLE]

where $\gamma^{(n)}$ is the lead coefficient of $H_{m}(z,a_{n})$. We will see in the next lemma that $\deg H_{m}(z,a_{n})$ is at most $\left\lfloor m/3\right\rfloor$. From this finite product and the assumption that $z^{*}\notin I_{a}$, we conclude that there is a fixed (independent of $n$) $\delta>0$ so that $|H_{m}(z^{*},a_{n})|>\delta$, for all large $n$. Since $H_{m}(z^{*},a)$ is a polynomial in $a$ for any fixed $z^{*}$, we conclude that

[TABLE]

and the lemma follows. ∎

Lemma 3 allows us to ignore some special values of $a$. In particular, we may assume $a\neq 0$. In our main approach, we count the number of zeros of $H_{m}(z)$ on the interval $I_{a}$ in (2.3) and show that this number of zeros is at least the degree of $H_{m}(z)$. To count the number of zeros of $H_{m}(z)$ on $I_{a}$, we write $z=z(\theta)$ as a strictly increasing function of a variable $\theta$ on the interval $(2\pi/3,\pi)$. Then we construct a function $g_{m}(\theta)$ on $(2\pi/3,\pi)$ with the property that $\theta$ is a zero of $g_{m}(\theta)$ on $(2\pi/3,\pi)$ if and only if $z(\theta)$ is a zero of $H_{m}(z)$ on $I_{a}$. From this construction, we count the number of zeros of $g_{m}(\theta)$ on $(2\pi/3,\pi)$ which will be the same as the number of zeros of $H_{m}(z)$ on $I_{a}$ by the monotonicity of the function $z(\theta)$. We first obtain an upper bound for the degree of $H_{m}(z)$ and provide heuristic arguments for the formulas of $z(\theta)$ and $g_{m}(\theta)$.

Lemma 4**.**

The degree of the polynomial $H_{m}(z)$ defined by (2.2) is at most $\left\lfloor m/3\right\rfloor$.

Proof.

We rewrite (2.2) as

[TABLE]

By equating the coefficients in $t$ of both sides, we see that the sequence $\left\{H_{m}(z)\right\}_{m=0}^{\infty}$ satisfies the recurrence

[TABLE]

and the initial condition $H_{0}(z)\equiv 1$, $H_{1}(z)\equiv-1$, and $H_{2}(z)\equiv 1-a$. The lemma follows from induction. ∎

2.1. Heuristic arguments

We now provide heuristic arguments to motivate the formulas for two special functions $z(\theta)$ and $g_{m}(\theta)$ on $(2\pi/3,\pi)$. Let $t_{0}=t_{0}(z)$, $t_{1}=t_{1}(z)$, and $t_{2}=t_{2}(z)$ be the three zeros of the denominator $1+t+at^{2}+zt^{3}$. We will show rigorously in Section 2.2 that $t_{0}$, $t_{1}$, $t_{2}$ are nonzero and distinct with $t_{0}=\overline{t_{1}}$. We let $q=t_{1}/t_{0}=e^{2i\theta}$, $\theta\neq 0,\pi$. We have

[TABLE]

We apply partial fractions to rewrite the generating functions $1/z(t-t_{0})(t-t_{1})(t-t_{2})$ as

[TABLE]

which can be expanded as a series in $t$ below

[TABLE]

From this expression, we deduce that $z$ is a zero of $H_{m}(z)$ if and only if

[TABLE]

After multiplying the left side of (2.5) by $t_{0}^{m+3}$ we obtain the equality

[TABLE]

Setting $\zeta=t_{2}/t_{0}e^{i\theta}$, we rewrite the left side as

[TABLE]

or equivalently

[TABLE]

We multiply this expression by $(\zeta-e^{-i\theta})(\zeta-e^{i\theta})e^{i(m+3)\theta}$ and set the summation to [math] and rewrite (2.5) as

[TABLE]

The last expression will serve as the definition of $g_{m}(\theta)$.

We next provide a motivation for the specific form of $z(\theta)$. Since $t_{0}$, $t_{1}$, and $t_{2}$ are the zeros of $D(t,z)=1+t+at^{2}+zt^{3}$, they satisfy the three identities

[TABLE]

If we divide the first equation by $t_{0}$, the second by $t_{0}^{2}$, and the third by $t_{0}^{3}$ then these identities become

[TABLE]

We next divide the first identity by second, and the second by the third to obtain

[TABLE]

from which we deduce that

[TABLE]

This equation is equivalent to

[TABLE]

or simply

[TABLE]

Lemma 5**.**

For any $a\in[-1,1/3]$ and $\theta\in(2\pi/3,\pi)$, the zeros in $\zeta$ of the polynomial

[TABLE]

are real and distinct.

Proof.

We consider the discriminant of the above polynomial in $\zeta$:

[TABLE]

There are three possible cases depending on the value of $a$. If $1/4\leq a\leq 1/3$, the inequality $\Delta>0$ comes directly from $a\geq 4a-1>(1-4a)\cos^{2}\theta$. If $0\leq a<1/4$, the claim $\Delta>0$ is trivial since $1-4a>0$. Finally, if $a<0$, we have $\Delta\geq(1-4a)/4+a=1/4$. It follows that the zeros of (2.10) are real and distinct for any $a\in[-1,1/3]$ and $\theta\in(2\pi/3,\pi)$. ∎

To obtain the formula $z(\theta)$, we multiply both sides of (2.7) and (2.8)

[TABLE]

and divide by (2.9) to arrive at the resulting equation

[TABLE]

2.2. Rigorous proof

Motivated by Section 2.1, we will rigorously prove Theorem (2) in this Section 2.2. We start our proof of Theorem (2) by defining the function $\zeta(\theta)$ according to (2.10):

[TABLE]

which, from Lemma 5, is a real function on $(2\pi/3,\pi)$ with a possible vertical asymptote at

[TABLE]

when $1/4<a\leq 1/3$. However we note that the function $1/\zeta(\theta)$ is a real continuous function on $(2\pi/3,\pi)$.

Lemma 6**.**

Let $\zeta(\theta)$ be defined as in (2.12). Then $|\zeta(\theta)|>1$ for every $a\in(-1,1/3)$ and every $\theta\in(2\pi/3,\pi)$ with $1-4a\cos^{2}\theta\neq 0$.

Proof.

From (2.10), we note that $\zeta_{+}:=\zeta(\theta)$ and

[TABLE]

are the zeros of

[TABLE]

Note that

[TABLE]

If $\theta\in(2\pi/3,\pi)$ and $a\in(-1,1/3)$, this product is negative since

[TABLE]

Thus exactly one of the zeros of the quadratic function $f(\zeta)$ lies outside the interval $[-1,1]$. The claim follows from the fact that $|\zeta_{+}|>|\zeta_{-}|$. ∎

Although one can prove Lemma 6 for the extreme value $a=-1$ or $a=1/3$, that will not be necessary by Lemma 3. Next we define the real function $z(\theta)$ by

[TABLE]

on $(2\pi/3,\pi)$. From Lemma 6, $1+2\zeta\cos\theta\neq 0$ and so does $2\cos\theta+\zeta$ by (2.10). Dividing the numerator and the denominator by $\zeta^{2}(\theta)$ and combining with the fact that $1/\zeta(\theta)$ is continuous on $(2\pi/3,\pi)$, we conclude that the possible discontinuity of $z(\theta)$ in (2.13) is removable. Finally, motivated by (2.6), we define the function $g_{m}(\theta)$ by

[TABLE]

which has the same vertical asymptote as that of $\zeta(\theta)$ in (2.13) when $1/4<a\leq 1/3$.

From Lemma 6, we see that the sign of the function $g_{m}(\theta)$ alternates at values of $\theta$ where $\cos(m+1)\theta=\pm 1$. Thus by the Intermediate Value Theorem the function $g_{m}(\theta)$ has at least one root on each subinterval whose endpoints are the solution of $\cos(m+1)=\pm 1$. However, in the case $1/4\leq a\leq 1/3$, one of the subintervals will contain a vertical asymptote given in (2.13). The lemma below counts the number of zeros of $g_{m}(\theta)$ on such a subinterval.

Lemma 7**.**

Let $g_{m}(\theta)$ be defined as in (2.15). Suppose $1/4<a\leq 1/3$ and $m\geq 6$. Then there exists $h\in\mathbb{N}$ such that

[TABLE]

where $\lfloor 2(m+1)/3\rfloor+1\leq h-1<h\leq m+1$. Furthermore, as long as

[TABLE]

the function $g(\theta)$ has at least two zeros on the interval

[TABLE]

whenever $h$ is at most $m$, and at least one zero when $h$ is $m+1$.

Proof.

Suppose $a\in[1/4,1/3]$. Since the function $\cos^{-1}\left(-1/2\sqrt{x}\right)$ is decreasing on the interval $[1/4,1/3]$, we conclude that

[TABLE]

Then the existence of $h$ comes directly from the inequality

[TABLE]

when $m\geq 6$.

The vertical asympote at $\cos^{-1}(-1/2\sqrt{a})$ of $g_{m}(\theta)$ divides the interval $J_{h}$ in (2.17) into two subintervals. We will show that each subinterval contains at least a zero of $g_{m}(\theta)$ if $h\leq m$. In the case $h=m+1$, only the subinterval on the left contains at least zero of $g_{m}(\theta)$. We analyze these two subintervals in the two cases below.

We consider the first case when $\theta\in I_{h}$ and $\theta<\cos^{-1}(-1/2\sqrt{a})$. By Lemma 6 and (2.15) we see that the sign of $g_{m}(\theta_{h-1})$ is $(-1)^{h}$. We now show that the sign of $g_{m}(\theta)$ is $(-1)^{h-1}$ when $\theta\rightarrow\cos^{-1}(-1/2\sqrt{a})$. From (2.12), we observe that $\zeta(\theta)\rightarrow+\infty$ as $\theta\rightarrow\cos^{-1}(-1/2\sqrt{a})$. Since $\theta\in I_{h}$, the sign of $\sin(m+1)\theta$ is $(-1)^{h-1}$ and consequently the sign of $g_{m}(\theta)$ is $(-1)^{h-1}$ when $\theta\rightarrow\cos^{-1}(-1/2\sqrt{a})$ by (2.15). By the Intermediate Value Theorem, we obtain at least one zero of $g_{m}(\theta)$ in this case.

Next we consider the case when $\theta\in I_{h}$ and $\theta>\cos^{-1}(-1/2\sqrt{a})$. In this case the sign of $g_{m}(\theta_{h})$ is $(-1)^{h-1}$ if $h\leq m$ by Lemma 6. Since $\zeta(\theta)\rightarrow-\infty$ as $\theta\rightarrow\cos^{-1}(-1/2\sqrt{a})$ and the sign of $\sin(m+1)\theta$ is $(-1)^{h-1}$, the sign of $g_{m}(\theta)$ is $(-1)^{h}$ as $\theta\rightarrow\cos^{-1}(-1/2\sqrt{a})$. By the Intermediate Value Theorem, we obtain at least one zero of $g_{m}(\theta)$ in this case and $h\leq m$. ∎

Note that we may assume (2.16) by Lemma3. From the fact that the sign of $g_{m}(\theta)$ in (2.15) alternates when $\cos(m+1)\theta=\pm 1$, we can find a lower bound for the number of zeros of $g_{m}(\theta)$ on $(2\pi/3,\pi)$ by the Intermediate Value Theorem. We will relate the zeros of $g_{m}(\theta)$ to the zeros of $H_{m}(z)$ by (2.6). However to ensure that the partial fractions procedure preceding equation (2.6) is rigorous, we need the lemma below.

Lemma 8**.**

Let $\theta\in(0,\pi)$ such that $\theta\neq\cos^{-1}\left(-1/2\sqrt{a}\right)$ whenever $a>1/4$. The zeros in $t$ of $1+t+at^{2}+z(\theta)t^{3}$ are

[TABLE]

where $\zeta:=\zeta(\theta)$ is given in (2.12).

Proof.

We first note that

[TABLE]

where $\zeta$ is a root of the quadratic equation $(2\cos\theta+\zeta)\zeta-a(1+2\zeta\cos\theta)^{2}=0$. We apply the following identities

[TABLE]

and

[TABLE]

to conclude that $P(t_{0})=0$. Similarly, we have

[TABLE]

Finally

[TABLE]

∎

As a consequence of Lemma 8, if $\theta\in(2\pi/3,\pi)$, then the zeros of $1+t+at^{2}+z(\theta)t^{3}$ will be distinct and $t_{1}=\overline{t_{0}}$ since $\zeta\in\mathbb{R}$ by Lemma 5. Thus we can apply partial fractions given in the beginning of Section 2.1. From this partial fraction decomposition, we conclude that if $\theta$ is a zero of $g_{m}(\theta)$, then $z(\theta)$ will be a zero of $H_{m}(z)$. In fact, we claim that each distinct zero of $g_{m}(\theta)$ on $(2\pi/3,\pi)$ produces a distinct zero of $H_{m}(z)$ on $I_{a}$. This is the content of the following two lemmas.

Lemma 9**.**

Let $\zeta(\theta)$ be defined as in (2.12). The function $z(\theta)$ defined as in (2.14) is increasing on $\theta\in(2\pi/3,\pi)$.

Proof.

Lemma 8 gives

[TABLE]

We differentiate the three terms and obtain

[TABLE]

where

[TABLE]

If we set

[TABLE]

then $f(t_{0})=\overline{f(t_{1})}$, and consequently $f(t_{0})f(t_{1})\geq 0$. Thus (2.19) implies that

[TABLE]

After solving this equation for $dt_{0}$ and substituting it into (2.19), we obtain

[TABLE]

With $t_{0}=\tau e^{-i\theta}$, $\tau\in\mathbb{R}$, we have

[TABLE]

We now substitute $3=-3t_{0}-3at_{0}^{2}-3zt_{0}^{3}$ and have

[TABLE]

In the formula of $t_{0}$ in Lemma 8, we substitute $\tau=-1/\zeta-2\cos\theta$ and obtain

[TABLE]

We finish this lemma by showing that $-\cos\theta+a/\zeta+2a\cos\theta>0$, which implies $f(t_{0})=\overline{f(t_{1})}\neq 0$ and and the lemma will follow from (2.20). We expand and divide both sides of (2.10) by $\zeta$ to get

[TABLE]

or equivalently,

[TABLE]

Finally, using the definition of $\zeta$ in (2.12) and Lemma 5, we note that

[TABLE]

The proof is complete. ∎

Lemma 10**.**

The function $z(\theta)$ as defined in (2.14) maps the interval $(2\pi/3,\pi)$ onto the interior of $I_{a}$.

Proof.

Since $z(\theta)$ is a continuous increasing function on $(2\pi/3,\pi)$, we only need to evaluate the limits of $z(\theta)$ at the endpoints. Since $|\zeta|>1$ by Lemma 6, the formula of $\zeta(\theta)$ in (2.12) implies that $\zeta(\theta)\rightarrow 1^{+}$ as $\theta\rightarrow(2\pi/3)^{+}$. Consequently, (2.18) gives

[TABLE]

Finally, from the fact that

[TABLE]

and (2.14), we have

[TABLE]

where we obtain (2.21) by multiply and divide the fraction by $\left(-1+6a-\sqrt{1-3a}\right)\left(-1+2\sqrt{1-3a}\right)$. ∎

Before making final arguments to connect all results in this section, we check the sign of $g_{m}(\theta)$ at one of the endpoints.

Lemma 11**.**

If $-1\leq a<1/4$ then the sign of $g_{m}(\pi^{-})$ is $(-1)^{m}$.

Proof.

Since $-1\leq a<1/4$, one can check that

[TABLE]

The result then follows directly from (2.15) and the fact that

[TABLE]

∎

With all the lemmas at our disposal, we produce the final arguments to finish the proof of Theorem 2. We consider the function $g_{m}(\theta)$ at the points

[TABLE]

We note that the number of such values of $h$ is

[TABLE]

where the equality can be checked by considering $m$ congruent to $0,$1, or $2$ modulus $3$. From the formula of $g_{m}(\theta)$ in (2.15) and Lemma 6, the sign of $g_{m}(\theta_{h})$ is $(-1)^{h-1}$. By the Intermediate Value Theorem and Lemma 7, there are at least $\left\lfloor m/3\right\rfloor-1$ zeros of $g_{m}(\theta)$ on $(2\pi/3,\pi)$. In fact, we claim that there are at least $\left\lfloor m/3\right\rfloor$ zeros of $g_{m}(\theta)$ on $(2\pi/3,\pi)$. In the case $-1\leq a<1/4$, we obtain one more zero of $g_{m}(\theta)$ from Lemma 11. On the other hand, if $1/4<a\leq 1/3$, then we obtain another zero of $g_{m}(\theta)$ by Lemma 7. From Lemmas 9 and 10, we obtain at least $\left\lfloor m/3\right\rfloor$ zeros of $H_{m}(z)$ on $I_{a}$. Since the degree of $H_{m}(z)$ is at most $\left\lfloor m/3\right\rfloor$ by Lemma 4, all the zeros of $H_{m}(z)$ lie on $I_{a}$. Recall that we can ignore the case $a=1/4$ by Lemma 3. The density of ${\displaystyle\bigcup_{m=0}^{\infty}\mathcal{Z}(H_{m}(z))}$ on $I_{a}$ comes from the density of ${\displaystyle\bigcup_{m=0}^{\infty}\mathcal{Z}(g_{m}(\theta))}$ on $(2\pi/3,\pi)$ and from $z(\theta)$ being a continuous map.

3. The case $c=0$ and $b\geq 0$

It is trivial that if $c=0$ and $b=0$ then the zeros of $H_{m}(z)$ are real under the convention that the constant zero polynomial is hyperbolic. In the case $b>0$, we make a substitution $t\rightarrow t/\sqrt{b}$ and reduce the problem to the following theorem.

Theorem 12**.**

The zeros of the sequence of polynomials $H_{m}(z)$ generated by

[TABLE]

are real, and the set ${\displaystyle\bigcup_{m=0}^{\infty}\mathcal{Z}(H_{m})}$ is dense on $(-\infty,\infty)$.

The proof of Theorem 12 follows from a similar procedure seen in Section 2. We will point out some key differences. The following lemma comes directly from the recurrence relation

[TABLE]

and induction.

Lemma 13**.**

The degree of the polynomial $H_{m}(z)$ generated by (3.1) is at most

[TABLE]

We define the functions

[TABLE]

on the interval $(\pi/3,\pi/2)$.

The proof of the lemma below is similar to that of Lemma 8. We leave the detailed computations to the reader.

Lemma 14**.**

Suppose $\theta\in(\pi/3,\pi/2)$, $\zeta=\zeta(\theta)$, and $z=z(\theta)$ defined by (3.2). The three zeros of $1+t^{2}+z(\theta)t^{3}$ are

[TABLE]

[TABLE]

[TABLE]

Looking at $z^{\prime}(\theta)$, one can check that $z(\theta)$ is strictly decreasing on the interval $(\pi/3,\pi/2)$. By the partial fraction decomposition of

[TABLE]

similar to the previous section, we conclude that for each zero of $g_{m}(\theta)$ on the interval $(\pi/3,\pi/2)$ we obtain two zeros $\pm z(\theta)$ of $H_{m}(z)$. We can also check by induction that $z=0$ is a simple zero of $H_{m}(z)$ if $m$ is odd, and $z=0$ is not a zero of $H_{m}(z)$ when $m$ is even. The formula of $g_{m}(\theta)$ implies that the sign of this function alternates when $\cos(m+1)\theta=\pm 1$, that is,

[TABLE]

Since $g_{m}(\theta)$ is continuous on $(\pi/3,\pi/2)$, we may apply the Intermediate Value Theorem to compute the number of zeros of $g_{m}(\theta)$ on $(\pi/3,\pi/2)$ and correspond to the number of zeros of $H_{m}(z)$ on $(-\infty,\infty)$. We will see that this number is equal to the degree of $H_{m}(z)$ and Theorem 12 follows. We quickly summarize in the six cases below where $\theta^{*}$ denotes the smallest solution $(m+1)\theta=k\pi$ on the interval $(\pi/3,\pi/2)$.

1. (1)

$m\equiv 1\pmod{3}$ and $m$ is even: there are

[TABLE]

zeros of $g_{m}(\theta)$ on $(\pi/3,\pi/2)$, which gives $(m-4)/3$ zeros of $H_{m}(z)$ on $(-\infty,\infty)$. 2. (2)

$m\equiv 1\pmod{3}$ and $m$ is odd: there are

[TABLE]

zeros of $g_{m}(\theta)$ on $(\pi/3,\pi/2)$ which gives$(m-7)/3$ nonzero roots of $H_{m}(z)$. We add a simple zero $z=0$ and obtain $(m-4)/3$ zeros of $H_{m}(z)$ on $(-\infty,\infty)$. 3. (3)

$m\equiv 0\pmod{3}$ and $m$ is even: with the observation that $\lim_{\theta\rightarrow\pi/3}g_{m}(\theta)=-3<0$ and $g_{m}(\theta^{*})>0$ we obtain

[TABLE]

zeros of $g_{m}(\theta)$ on $(\pi/3,\pi/2)$, which gives $m/3$ zeros of $H_{m}(z)$ on $(-\infty,\infty)$. 4. (4)

$m\equiv 0\pmod{3}$ and $m$ is odd: with the observation that $\lim_{\theta\rightarrow\pi/3}g_{m}(\theta)=3>0$ and $g_{m}(\theta^{*})<0$ we obtain

[TABLE]

zeros of $g_{m}(\theta)$ on $(\pi/3,\pi/2)$, which gives $(m-3)/3$ nonzero roots of $H_{m}(z)$. We add a simple zero $z=0$ and obtain $m/3$ zeros of $H_{m}(z)$ on $(-\infty,\infty)$. 5. (5)

$m\equiv 2\pmod{3}$ and $m$ is even: with the observation that $g_{m}(\pi/3)=0$, $g_{m}^{\prime}(\pi/3)>0$, and $g_{m}(\theta^{*})<0$ we obtain

[TABLE]

zeros of $g_{m}(\theta)$ on $(\pi/3,\pi/2)$, which gives $(m-2)/3$ zeros of $H_{m}(z)$ on $(-\infty,\infty)$. 6. (6)

$m\equiv 2\pmod{3}$ and $m$ is odd: with the observation that $g_{m}(\pi/3)=0$, $g_{m}^{\prime}(\pi/3)<0$, and $g_{m}(\theta^{*})>0$ we obtain

[TABLE]

zeros of $g_{m}(\theta)$ on $(\pi/3,\pi/2)$, which gives $(m-5)/3$ nonzero roots of $H_{m}(z)$. We plus add simple zero $z=0$ and obtain $(m-2)/3$ zeros of $H_{m}(z)$ on $(-\infty,\infty)$.

In all cases above the number of zeros of $H_{m}(z)$ on $(-\infty,\infty)$ corresponds to the degree of $H_{m}(z)$ and Theorem 12 follows.

4. Necessary condition for the reality of zeros

To prove the necessary condition of Theorem 1, we first show that if $c=0$ and $b<0$ then not all polynomials $H_{m}(z)$ are hyperbolic. In fact, with the substitution $t$ by $it$, we conclude that all the zeros of $H_{m}(z)$ will be purely imaginary by Theorem 12.

It remains to consider the sequence $H_{m}(z)$ generated by

[TABLE]

and show that if $a\notin[-1,1/3]$ then there is an $m$ such that not all the zeros of $H_{m}(z)$ are real. In fact, we will show if $a\notin[-1,1/3]$, $H_{m}(z)$ is not hyperbolic for all large $m$. To prove this, let us introduce some definitions (discussed in [9]) related to the root distribution of a sequence of functions

[TABLE]

where $\alpha_{k}(z)$ and $\beta_{k}(z)$ are analytic in a domain $D$. We say an index $k$ dominant at $z$ if $|\beta_{k}(z)|\geq|\beta_{l}(z)|$ for all $l$ ($1\leq l\leq n$). Let

[TABLE]

Let $\liminf\mathcal{Z}(f_{m})$ be the set of all $z\in D$ such that every neighborhood $U$ of $z$ has a non-empty intersection with all but finitely many of the sets $\mathcal{Z}(f_{m})$. Let $\limsup\mathcal{Z}(f_{m})$ be the set of all $z\in D$ such that every neighborhood $U$ of $z$ has a non-empty intersection infinitely many of the sets $\mathcal{Z}(f_{m})$. We will need the following theorem from Sokal ([9, Theorem 1.5]).

Theorem 15**.**

Let $D$ be a domain in $\mathbb{C}$, and let $\alpha_{1},\ldots,\alpha_{n},\beta_{1},\ldots,\beta_{n}$ $(n\geq 2)$ be analytic function on $D$, none of which is identically zero. Let us further assume a ’no-degenerate-dominance’ condition: there do not exist indices $k\neq k^{\prime}$ such that $\beta_{k}\equiv\omega\beta_{k^{\prime}}$ for some constant $\omega$ with $|\omega|=1$ and such that $D_{k}$ $(=D_{k^{\prime}})$ has nonempty interior. For each integer $m\geq 0$, define $f_{m}$ by

[TABLE]

Then $\liminf Z(f_{m})=\limsup Z(f_{m})$, and a point $z$ lies in this set if and only if either

(i) there is a unique dominant index $k$ at $z$, and $\alpha_{k}(z)=0$, or

(ii) there a two or more dominant indices at $z$.

If $z^{*}\in\mathbb{C}$ such that the zeros in $t$ of $1+t+at^{2}+z^{*}t^{3}$ are distinct then by partial fractions given in (2.4) and Theorem 15, $z^{*}$ will belong to $\liminf\mathcal{Z}(H_{m})$ when the two smallest (in modulus) roots of $1+t+at^{2}+z^{*}t^{3}$ have the same modulus. We also note that $t_{0}(z)$, $t_{1}(z)$, and $t_{2}(z)$ are analytic in a neighborhood of $z^{*}$ by the Implicit Function Theorem. If we let $\omega=e^{2i\theta}$ then the ’no-degenerate-dominance’ condition in Theorem 15 comes directly from equations (2.14) and (2.12) since $\theta$ is a fixed constant (and thus $z$ is a fixed point which has empty interior).

Suppose $a\notin[-1,1/3]$. With the arguments in the previous paragraph, our main goal is to find a $z^{*}\notin\mathbb{R}$ so that the zeros of $1+t+at^{2}+z^{*}t^{3}$ are distinct and the two smallest (in modulus) zeros of this polynomial have the same modulus. If we can find such a point, then $z^{*}\in\liminf\mathcal{Z}(H_{m})=\limsup\mathcal{Z}(H_{m})$. This implies that on a small neighborhood of $z^{*}$ which does not intersect the real line, there is a non-real zero of $H_{m}(z)$ for all large $m$ by the definition of $\liminf\mathcal{Z}(H_{m})$. Our choice of $z^{*}=z(\theta^{*})$ comes from (2.14) for a special $\theta^{*}$. Unlike in Section 2, $\theta^{*}$ will not belong to $(2\pi/3,\pi)$ to ensure that $z^{*}\notin\mathbb{R}.$ In particular, we consider the two cases $a<-1$ and $a>3$.

Case $a<-1$.

We will pick $\theta^{*}\neq\pi/2$ but sufficiently close to $\pi/2$ on the left. Since from (2.12)

[TABLE]

we can pick $0<\theta^{*}<\pi/2$ sufficiently close to $\pi/2$ so that $\zeta:=\zeta(\theta^{*})\in\mathbb{C}\backslash\mathbb{R}$ and $\left|\zeta(\theta^{*})\right|>1$. By Lemma 8, we have $t_{2}=\zeta t_{0}e^{i\theta^{*}}$ and $t_{1}=t_{0}e^{2i\theta^{*}}$. The fact that $|\zeta|>1$ and $\theta^{*}\neq 0,\pi/2$ implies that the polynomial $1+t+at^{2}+z(\theta^{*})t^{3}$ have distinct zeros and not all its zeros are real. We will show that $z(\theta^{*})\notin\mathbb{R}$ by contradiction. In deed, if $z(\theta^{*})\in\mathbb{R}$ then the zeros of the polynomial $1+t+at^{2}+z(\theta^{*})t^{3}\in\mathbb{R}[t]$ satisfy $t_{0}=\overline{t_{1}}$ and

[TABLE]

This gives a contradiction because the first equation implies that $t_{0}e^{i\theta^{*}}\in\mathbb{R}$, while the second equation implies that $t_{0}e^{i\theta^{*}}\notin\mathbb{R}$ since $\zeta\notin\mathbb{R}$.

Case $a>1/3$.

We will pick $\theta^{*}$ so that $\cos\theta^{*}=\beta^{+}$ where $\beta=\sqrt{a/(4a-1)}<1$. The definition of $\zeta(\theta)$ in (2.12) gives

[TABLE]

where we can easily check that

[TABLE]

Thus if $\cos\theta^{*}=\beta^{+}$ then $0<\theta^{*}<\pi/2$, $|\zeta(\theta^{*}|>1$, and $|\zeta(\theta^{*})|\notin\mathbb{R}$ where the last statement comes from (2.12) and the inequality

[TABLE]

With $0<\theta^{*}<\pi/2$, $|\zeta(\theta^{*}|>1$, and $|\zeta(\theta^{*})|\notin\mathbb{R}$, we apply the same arguments given in the previous case to complete this section.

Acknowledgments

The authors would like to thank Professor T. Forgács for his careful review of the paper.