# A note on the roots of some special type polynomials

## Abstract

In this study, we investigate the polynomials for $n\ge 2$ and positive integers k and a positive real number a, with the initial values ${G}_{0}\left(x\right)=-a$, ${G}_{1}\left(x\right)=x-a$

${G}_{n+2}^{\left(k\right)}\left(x\right)={x}^{k}{G}_{n+1}^{\left(k\right)}\left(x\right)+{G}_{n}^{\left(k\right)}\left(x\right).$

We give some fundamental properties related to them. Also, we obtain asymptotic results for the roots of polynomials ${G}_{n}^{\left(k\right)}\left(x\right)$.

MSC:11B39, 11B37.

## 1 Introduction

The polynomials defined by Catalan, for $n\ge 0$, as follows

${F}_{n+2}\left(x\right)=x{F}_{n+1}\left(x\right)+{F}_{n}\left(x\right);\phantom{\rule{2em}{0ex}}{F}_{1}\left(x\right)=1,\phantom{\rule{2em}{0ex}}{F}_{2}\left(x\right)=x$
(1)

are called Fibonacci polynomials and denoted by ${F}_{n}\left(x\right)$, [1]. The Fibonacci-type polynomials ${G}_{n}\left(x\right)$, $n\ge 0$, are defined by

${G}_{n+2}\left(x\right)=x{G}_{n+1}\left(x\right)+{G}_{n}\left(x\right),$
(2)

where ${G}_{0}\left(x\right)$ and ${G}_{1}\left(x\right)$ are seed polynomials. There are several studies about the properties of zeros of polynomials ${G}_{n}\left(x\right)$. However, there are no general formulas for zeros of Fibonacci-type polynomials. In [2, 3], the authors studied the limiting behavior of the maximal real roots of polynomials ${G}_{n}\left(x\right)$ with the initial values ${G}_{0}\left(x\right)=-1$, ${G}_{1}\left(x\right)=x-1$. In [4], the authors generalized Moore’s result for these polynomials. They considered the initial values ${G}_{0}\left(x\right)=a$, ${G}_{1}\left(x\right)=x+b$, where a and b are integer numbers. In [5], the author determined the absolute values of complex zeros of these polynomials. In [6], Ricci studied this problem in the case $a=1$ and $b=1$. In [7], Tewodros investigated the convergence of maximal real roots of different Fibonacci-type polynomials given by the following relation:

${G}_{n+2}^{\left(k\right)}\left(x\right)={x}^{k}{G}_{n+1}^{\left(k\right)}\left(x\right)+{G}_{n}^{\left(k\right)}\left(x\right),\phantom{\rule{1em}{0ex}}n\ge 0,$
(3)

where k is a positive integer number. The initial values of the recursive relation (3) are ${G}_{{}_{0}\left(k\right)}^{}\left(x\right)=-1$ and ${G}_{{}_{1}\left(k\right)}^{}\left(x\right)=x-1$. In this study, firstly, we investigate some fundamental properties of Fibonacci-type polynomials. We give some combinatorial identities related to equation (3). Then, we investigate the limit of maximal real roots of these polynomials. We notice that Tewodros [7] studied a special case $a=1$ of the polynomials we investigate.

## 2 Some fundamental properties of polynomials ${G}_{n}^{\left(k\right)}\left(x\right)$

In this section, we give some fundamental properties of polynomials ${G}_{{}_{n}\left(k\right)}^{}\left(x\right)$, for $n\ge 0$, defined by the recursive formula as follows:

${G}_{n+2}^{\left(k\right)}\left(x\right)={x}^{k}{G}_{n+1}^{\left(k\right)}\left(x\right)+{G}_{n}^{\left(k\right)}\left(x\right);\phantom{\rule{2em}{0ex}}{G}_{0}^{\left(k\right)}\left(x\right)=-a;\phantom{\rule{2em}{0ex}}{G}_{1}^{\left(k\right)}\left(x\right)=x-a.$
(4)

The characteristic equation for (4) is ${t}^{2}-{x}^{k}t-1=0$ and its roots are

$\alpha \left(x\right)=\frac{{x}^{k}+\sqrt{{x}^{2k}+4}}{2}$

and

$\beta \left(x\right)=\frac{{x}^{k}-\sqrt{{x}^{2k}+4}}{2}.$

Note that $\alpha \left(x\right)\beta \left(x\right)=-1$, $\alpha \left(x\right)+\beta \left(x\right)={x}^{k}$ and $\alpha \left(x\right)-\beta \left(x\right)=\sqrt{{x}^{2k}+4}$. For relation (4), the Binet formula is

${G}_{n}^{\left(k\right)}\left(x\right)=A\left(x\right){\alpha }^{n}\left(x\right)+B\left(x\right){\beta }^{n}\left(x\right),$
(5)

where

$A\left(x\right)=\frac{2\left(x-a\right)+a{x}^{k}-a\sqrt{{x}^{2k}+4}}{2\sqrt{{x}^{2k}+4}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}B\left(x\right)=\frac{-2\left(x-a\right)-a{x}^{k}-a\sqrt{{x}^{2k}+4}}{2\sqrt{{x}^{2k}+4}}.$
(6)

Proposition 2.1 For $n\ge 0$, the generating function for polynomials ${G}_{n}^{\left(k\right)}\left(x\right)$ is

${H}_{r}^{\left(k\right)}\left(x,t\right)=\sum _{n\ge 0}{G}_{n+r}^{\left(k\right)}\left(x\right){t}^{n}=\left\{\begin{array}{cc}\frac{{G}_{r}^{\left(k\right)}\left(x\right)+{G}_{r-1}^{\left(k\right)}\left(x\right)t}{1-{x}^{k}t-{t}^{2}},\hfill & r=1,2,3,\dots ,\hfill \\ \frac{t\left(a{x}^{k}+x-a\right)}{1-{x}^{k}t-{t}^{2}},\hfill & r=0.\hfill \end{array}$
(7)

Proof Let ${H}_{r}^{\left(k\right)}\left(x,t\right)$ be the generating function for polynomials ${G}_{n+r}^{\left(k\right)}\left(x\right)$. So, we write

${H}_{r}^{\left(k\right)}\left(x,t\right)=\sum _{n\ge 0}{G}_{n+r}^{\left(k\right)}\left(x\right){t}^{n}.$
(8)

If we multiply both sides of equation (8) by ${x}^{k}t$ and ${t}^{2}$, respectively, then we can get

${x}^{k}t{H}_{r}^{\left(k\right)}\left(x,t\right)={x}^{k}{G}_{r}^{\left(k\right)}\left(x\right){t}^{1}+{x}^{k}{G}_{r+1}^{\left(k\right)}\left(x\right){t}^{2}+{x}^{k}{G}_{r+2}^{\left(k\right)}\left(x\right){t}^{3}+\cdots +{x}^{k}{G}_{r+n-1}^{\left(k\right)}\left(x\right){t}^{n}+\cdots$

and

${t}^{2}{H}_{r}^{\left(k\right)}\left(x,t\right)={G}_{r}^{\left(k\right)}\left(x\right){t}^{2}+{G}_{r+1}^{\left(k\right)}\left(x\right){t}^{3}+{G}_{r+2}^{\left(k\right)}\left(x\right){t}^{4}+\cdots +{G}_{r+n-2}^{\left(k\right)}\left(x\right){t}^{n}+\cdots .$

The last two equations give us the following equation:

$\begin{array}{rcl}{H}_{r}^{\left(k\right)}\left(x,t\right)-{x}^{k}t{H}_{r}^{\left(k\right)}\left(x,t\right)-{t}^{2}{H}_{r}^{\left(k\right)}\left(x,t\right)& =& {G}_{r}^{\left(k\right)}\left(x\right){t}^{0}+\left({G}_{r+1}^{\left(k\right)}\left(x\right)-{x}^{k}{G}_{r}^{\left(k\right)}\left(x\right)\right)t\\ +\left({G}_{r+2}^{\left(k\right)}\left(x\right)-{x}^{k}{G}_{r+1}^{\left(k\right)}\left(x\right)-{G}_{r}^{\left(k\right)}\left(x\right)\right){t}^{2}+\cdots \\ +\left({G}_{n+r}^{\left(k\right)}\left(x\right)-{x}^{k}{G}_{n+r-1}^{\left(k\right)}\left(x\right)-{G}_{n+r-2}^{\left(k\right)}\left(x\right)\right){t}^{n}+\cdots .\end{array}$

If we use the recurrence relation and simplify it, we write

${H}_{r}^{\left(k\right)}\left(x,t\right)-{x}^{k}t{H}_{r}^{\left(k\right)}\left(x,t\right)-{t}^{2}{H}_{r}^{\left(k\right)}\left(x,t\right)={G}_{r}^{\left(k\right)}\left(x\right){t}^{0}+\left({G}_{r+1}^{\left(k\right)}\left(x\right)-{x}^{k}{G}_{r}^{\left(k\right)}\left(x\right)\right)t,$

i.e.,

${H}_{r}^{k}\left(x,t\right)=\left\{\begin{array}{cc}\frac{{G}_{r}^{\left(k\right)}\left(x\right)+{G}_{r-1}^{\left(k\right)}\left(x\right)t}{1-{x}^{k}t-{t}^{2}},\hfill & r=1,2,3,\dots ,\hfill \\ \frac{t\left(a{x}^{k}+x-a\right)}{1-{x}^{k}t-{t}^{2}},\hfill & r=0.\hfill \end{array}$

Thus, the proof is completed. □

Let us give the well-known formula, which is called the Cassini-like formula, without proof.

Proposition 2.2 (Cassini-like)

For $n\ge 0$, we have

${G}_{n-1}^{\left(k\right)}\left(x\right){G}_{n+1}^{\left(k\right)}\left(x\right)-{\left[{G}_{n}^{\left(k\right)}\left(x\right)\right]}^{2}={\left(-1\right)}^{n-1}\left[A\left(x\right)B\left(x\right)\right],$
(9)

where

$A\left(x\right)=\frac{2\left(x-a\right)+a{x}^{k}-a\sqrt{{x}^{2k}+4}}{2\sqrt{{x}^{2k}+4}}$

and

$B\left(x\right)=\frac{-2\left(x-a\right)-a{x}^{k}-a\sqrt{{x}^{2k}+4}}{2\sqrt{{x}^{2k}+4}}.$

In the following propositions, we give some sums formulas related to polynomials ${G}_{n}^{\left(k\right)}\left(x\right)$.

Proposition 2.3 For $N\ge 0$, we have

${H}_{0}^{\left(k\right)}\left(x,1\right)-{H}_{N+1}^{\left(k\right)}\left(x,1\right)=\sum _{r=0}^{N}{G}_{r}^{\left(k\right)}\left(x\right)=\frac{2a-x-a{x}^{k}+{G}_{N+1}^{\left(k\right)}\left(x\right)+{G}_{N}^{\left(k\right)}\left(x\right)}{{x}^{k}}.$
(10)

Proof Proof of formula (10) follows now immediately from (7). □

Proposition 2.4 For $N\ge 0$, we have the following sum formulas:

$\sum _{r=0}^{N}{G}_{2r}^{\left(k\right)}\left(x\right)=\frac{-{x}^{k+1}-a{x}^{k}\left({x}^{k}-1\right)-{G}_{2N}^{\left(k\right)}\left(x\right)+{G}_{2N+2}^{\left(k\right)}\left(x\right)}{{x}^{2k}}$
(11)

and

$\sum _{r=0}^{N}{G}_{2r+1}^{\left(k\right)}\left(x\right)=\frac{a{x}^{k}-{G}_{2N+1}^{\left(k\right)}\left(x\right)+{G}_{2N+3}^{\left(k\right)}\left(x\right)}{{x}^{2k}}.$
(12)

Proof From the Binet formula, we can write

$\sum _{r=0}^{N}{G}_{2r}^{\left(k\right)}\left(x\right)=\left(\frac{x-a+a\beta }{\alpha -\beta }\right)\sum _{r=0}^{N}{\alpha }^{2r}-\left(\frac{x-a+a\alpha }{\alpha -\beta }\right)\sum _{r=0}^{N}{\beta }^{2r},$
(13)

where $\alpha =\alpha \left(x\right)$, $\beta =\beta \left(x\right)$. If we substitute the equations

$\sum _{r=0}^{N}{\alpha }^{2r}=\frac{1-{\alpha }^{2N+2}}{1-{\alpha }^{2}},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\sum _{r=0}^{N}{\beta }^{2r}=\frac{1-{\beta }^{2N+2}}{1-{\beta }^{2}}$

and

$\left(1-{\alpha }^{2}\right)\left(1-{\beta }^{2}\right)=-{x}^{2k}$

into equation (13), then we can get

$\sum _{r=0}^{N}{G}_{2r}^{\left(k\right)}\left(x\right)=\left(\frac{x-a+a\beta }{\alpha -\beta }\right)\frac{1-{\alpha }^{2N+2}}{1-{\alpha }^{2}}-\left(\frac{x-a+a\alpha }{\alpha -\beta }\right)\frac{1-{\beta }^{2N+2}}{1-{\beta }^{2}}.$

If we rearrange the last equation, then we have

$\begin{array}{rcl}\sum _{r=0}^{N}{G}_{2r}^{\left(k\right)}\left(x\right)& =& \frac{-a\left(\alpha -\beta \right)+\left(x-a\right)\left({\alpha }^{2}-{\beta }^{2}\right)+a\left({\alpha }^{3}-{\beta }^{3}\right)}{\left(1-{\alpha }^{2}\right)\left(1-{\beta }^{2}\right)\left(\alpha -\beta \right)}\\ -\frac{\left[\left(x-a+a\beta \right){\alpha }^{2N+2}-\left(x-a+a\alpha \right){\beta }^{2N+2}\right]}{\left(1-{\alpha }^{2}\right)\left(1-{\beta }^{2}\right)\left(\alpha -\beta \right)}\\ +\frac{{\alpha }^{2}{\beta }^{2}\left[\left(x-a+a\beta \right){\alpha }^{2N}-\left(x-a+a\alpha \right){\beta }^{2N}\right]}{\left(1-{\alpha }^{2}\right)\left(1-{\beta }^{2}\right)\left(\alpha -\beta \right)}.\end{array}$

By taking aid of the Binet formula, we can write

$\begin{array}{rcl}\sum _{r=0}^{N}{G}_{2r}^{\left(k\right)}\left(x\right)& =& \frac{-a+\left(x-a\right)\left(\alpha +\beta \right)+a\left({\alpha }^{2}+\alpha \beta +{\beta }^{2}\right)}{\left(1-{\alpha }^{2}\right)\left(1-{\beta }^{2}\right)}\\ +\frac{{G}_{2N}^{\left(k\right)}\left(x\right)}{\left(1-{\alpha }^{2}\right)\left(1-{\beta }^{2}\right)}-\frac{{G}_{2N+2}^{\left(k\right)}\left(x\right)}{\left(1-{\alpha }^{2}\right)\left(1-{\beta }^{2}\right)}.\end{array}$

If we substitute $\alpha +\beta ={x}^{k}$, $\alpha \beta =-1$, $\left(1-{\alpha }^{2}\right)\left(1-{\beta }^{2}\right)=-{x}^{2k}$ into the last equation, we obtain the following equation:

$\sum _{r=0}^{N}{G}_{2r}^{\left(k\right)}\left(x\right)=\frac{{x}^{k+1}-a{x}^{k}+a{x}^{2k}-{G}_{2N+2}^{\left(k\right)}\left(x\right)+{G}_{2N}^{\left(k\right)}\left(x\right)}{-{x}^{2k}}.$

Thus, the proof is completed. Similarly, the second part of the proposition can be seen. □

## 3 Asymptotic behaviors of the maximal roots for polynomials ${G}_{n}^{\left(k\right)}\left(x\right)$

In this section, firstly for $k=2$, we investigate the roots of polynomials ${G}_{n}^{\left(k\right)}\left(x\right)$. After that, we generalize the obtained results for all positive real numbers k. When $k=2$, we write

${G}_{n+2}^{\left(2\right)}\left(x\right)={x}^{2}{G}_{n+1}^{\left(2\right)}\left(x\right)+{G}_{n}^{\left(2\right)}\left(x\right),$
(14)

where

${G}_{0}^{\left(2\right)}\left(x\right)=-a,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{G}_{1}^{\left(2\right)}\left(x\right)=x-a$

and a is a positive real number. Now, we can give the following lemma to be used the later.

Lemma 3.1 If r is a maximal root of a function f with positive leading coefficient, then $f\left(x\right)>0$ for all $x>r$. Conversely, if $f\left(x\right)>0$ for all $x\ge t$, then $r. If $f\left(s\right)<0$ , then $s [2].

Lemma 3.2 For $n\ge 2$, ${G}_{n}^{\left(2\right)}\left(x\right)$ has at least one real root on the interval $\left(a,a+1\right)$ and ${g}_{n}\in \left(a,a+1\right)$, where ${g}_{n}$ is the maximal root of polynomial ${G}_{n}^{\left(2\right)}\left(x\right)$.

Proof Some of polynomials ${G}_{n}^{\left(2\right)}\left(x\right)$ are as follows:

$\begin{array}{c}{G}_{2}^{\left(2\right)}\left(x\right)={x}^{3}-a{x}^{2}-a,\hfill \\ {G}_{3}^{\left(2\right)}\left(x\right)={x}^{5}-a{x}^{4}-a{x}^{2}+x-a,\hfill \\ {G}_{4}^{\left(2\right)}\left(x\right)={x}^{7}-a{x}^{6}-a{x}^{4}+2{x}^{3}-2a{x}^{2}-a,\hfill \\ ⋮\hfill \end{array}$

Note that polynomials ${G}_{n}^{\left(2\right)}\left(x\right)$ are monic polynomials with degree n and constant term −a. If we write for $x=a$, then we have

$\begin{array}{c}{G}_{1}^{\left(2\right)}\left(a\right)=0,\hfill \\ {G}_{2}^{\left(2\right)}\left(a\right)=-a<0,\hfill \\ {G}_{3}^{\left(2\right)}\left(a\right)=-{a}^{3}={a}^{2}{G}_{2}^{\left(2\right)}\left(a\right)<0,\hfill \\ {G}_{4}^{\left(2\right)}\left(a\right)=-{a}^{5}-a\le -{a}^{5}={a}^{2}{G}_{3}^{\left(2\right)}\left(a\right)<0,\hfill \\ ⋮\hfill \end{array}$

For $k\ge 2$, if we suppose ${G}_{k}^{\left(2\right)}\left(a\right)\le {a}^{2}{G}_{k-1}^{\left(2\right)}\left(a\right)<0$, then by using the recursive relation (14), we get

${G}_{k+1}^{\left(2\right)}\left(a\right)={a}^{2}{G}_{k}^{\left(2\right)}\left(a\right)+{G}_{k-1}^{\left(2\right)}\left(a\right)<0.$

Thus, for $x=a$, we get ${G}_{n}^{\left(2\right)}\left(x\right)<0$. Similarly, when $x=a+1$, we have ${G}_{n}^{\left(2\right)}\left(x\right)>0$. Therefore, ${G}_{n}^{\left(2\right)}\left(x\right)$ has at least one real root on the interval $\left(a,a+1\right)$, and we write ${g}_{n}\in \left(a,a+1\right)$ for the maximal root of ${G}_{n}^{\left(2\right)}\left(x\right)$, which results easily from Lemma 3.1 and the recursive relation for ${G}_{n}^{\left(2\right)}\left(x\right)$. □

Let ${g}_{n}$ denote the maximal root of polynomial ${G}_{n}^{\left(2\right)}\left(x\right)$ for every $n\in \mathbb{N}$. Then we can give the following proposition to illustrate the monotonicity of $\left\{{g}_{2n-1}\right\}$ and $\left\{{g}_{2n}\right\}$.

Proposition 3.3 The sequence $\left\{{g}_{2n-1}\right\}$ is a monotonically increasing sequence and the sequence $\left\{{g}_{2n}\right\}$ is a monotonically decreasing sequence.

Proof Firstly, we consider polynomials ${G}_{n}^{\left(2\right)}\left(x\right)$ with odd indices. By a direct computation, we get ${G}_{3}^{\left(2\right)}\left(a\right)=-{a}^{3}<0$, ${g}_{3}>a$, $a={g}_{1}$. Assume that ${g}_{1}<{g}_{3}<{g}_{5}<\cdots <{g}_{2k-3}<{g}_{2k-1}$. We can write ${G}_{2k-3}^{\left(2\right)}\left({g}_{2k-1}\right)>0$. Thus, it can be easily seen that

${G}_{n+k}^{\left(2\right)}\left({g}_{n}\right)={\left(-1\right)}^{k+1}{G}_{n-k}^{\left(2\right)}\left({g}_{n}\right).$
(15)

By using equation (15), we can write

${G}_{2k+1}^{\left(2\right)}\left({g}_{2k-1}\right)={G}_{\left(2k-1\right)+2}^{\left(2\right)}\left({g}_{2k-1}\right)=-{G}_{\left(2k-1\right)-2}^{\left(2\right)}\left({g}_{2k-1}\right)=-{G}_{2k-3}^{\left(2\right)}\left({g}_{2k-1}\right).$
(16)

So, from equation (16) we write

${G}_{2k+1}^{\left(2\right)}\left({g}_{2k-1}\right)<0.$
(17)

Therefore, polynomials ${G}_{2k+1}^{\left(2\right)}\left(x\right)$ must have a root greater than ${g}_{2k-1}$. So, we get

${g}_{2k+1}>{g}_{2k-1}.$
(18)

After that we consider polynomials ${G}_{n}^{\left(2\right)}\left(x\right)$ with even indices. From the recursive relation (14), we can obtain

${G}_{2k+1}^{\left(2\right)}\left({g}_{2k-1}\right)={g}_{2k-1}^{2}{G}_{2k}^{\left(2\right)}\left({g}_{2k-1}\right)+{G}_{2k-1}^{\left(2\right)}\left({g}_{2k-1}\right).$
(19)

Since ${G}_{2k-1}^{\left(2\right)}\left({g}_{2k-1}\right)=0$, by using Lemma 3.1, we can get ${G}_{2k+1}^{\left(2\right)}\left({g}_{2k-1}\right)<0$. Thus, we get

${g}_{2k-1}<{g}_{2k}.$
(20)

Again, by using Lemma 3.1, we can write

${G}_{2k-1}^{\left(2\right)}\left({g}_{2k}\right)>0.$
(21)

From the recursive relation (14), we can write

${G}_{2k}^{\left(2\right)}\left({g}_{2k}\right)={g}_{2k}^{2}{G}_{2k-1}^{\left(2\right)}\left({g}_{2k}\right)+{G}_{2k-2}^{\left(2\right)}\left({g}_{2k}\right).$
(22)

From equation (22), we can get

$-{g}_{2k}^{2}{G}_{2k-1}^{\left(2\right)}\left({g}_{2k}\right)={G}_{2k-2}^{\left(2\right)}\left({g}_{2k}\right)<0.$

So, we have ${g}_{2k}<{g}_{2k-2}$. Thus, $\left\{{g}_{2n-1}\right\}$ is a monotonically increasing sequence and bounded above by the number $a+1$. Similarly, $\left\{{g}_{2n}\right\}$ is a monotonically decreasing sequence and bounded below by the number a. If we denote the ${lim}_{x\to \mathrm{\infty }}{g}_{2n-1}$ by ${g}_{\mathrm{odd}}$ and ${lim}_{x\to \mathrm{\infty }}{g}_{2n}$ by ${g}_{\mathrm{even}}$, then we can write ${g}_{\mathrm{odd}}={g}_{\mathrm{even}}$. □

Proposition 3.4 For polynomials ${G}_{2n-1}^{\left(2\right)}\left(x\right)$ and ${G}_{2n}^{\left(2\right)}\left(x\right)$, the sequences $\left\{{g}_{2n-1}\right\}$ and $\left\{{g}_{2n}\right\}$ converge to the following number ζ:

$\zeta =\frac{\sqrt{{\left(1-{a}^{2}\right)}^{2}+8{a}^{2}}-\left(1-{a}^{2}\right)}{2a}.$
(23)

Proof Using the Binet formula of relation (14), for all $\left[a,a+1\right]$, we can see that $\alpha \left(x\right)\ge \alpha \left(a\right)>1$ and $|\beta \left(x\right)|=\frac{1}{\alpha \left(x\right)}\le \frac{1}{\alpha \left(a\right)}$. Thus, we get

$\underset{n\to \mathrm{\infty }}{lim}{\alpha }^{n}\left(x\right)=+\mathrm{\infty };\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{\beta }^{n}\left(x\right)=0.$
(24)

If we write $n=2k-1$ and $x={g}_{2k-1}$ in equation (5), then we have

$A\left({g}_{2k-1}\right){\alpha }^{2k-1}\left({g}_{2k-1}\right)+B\left({g}_{2k-1}\right){\beta }^{2k-1}\left({g}_{2k-1}\right)=0.$
(25)

And from equation (25) we write

$A\left({g}_{2k-1}\right)=-B\left({g}_{2k-1}\right)\left(\frac{{\beta }^{2k-1}\left({g}_{2k-1}\right)}{{\alpha }^{2k-1}\left({g}_{2k-1}\right)}\right).$
(26)

$A\left(x\right)$ and $B\left(x\right)$ are continuous on the interval $\left[a,a+1\right]$, this implies that $|A\left(x\right)|$ and $|B\left(x\right)|$ are bounded below and above on $\left[a,a+1\right]$. So, since $a\ge 1$, we get

$\underset{k\to \mathrm{\infty }}{lim}A\left({g}_{2k-1}\right)=A\left({g}_{\mathrm{odd}}\right)=0.$
(27)

From Binet formula (5), we have

$\underset{k\to \mathrm{\infty }}{lim}{g}_{2k-1}=\frac{\sqrt{{\left(1-{a}^{2}\right)}^{2}+8{a}^{2}}-\left(1-{a}^{2}\right)}{2a}.$
(28)

Also, by the aid of similar discussion, if we take $n=2k$ and $x={g}_{2k}$, then we find that

$\underset{k\to \mathrm{\infty }}{lim}{g}_{2k}=\frac{\sqrt{{\left(1-{a}^{2}\right)}^{2}+8{a}^{2}}-\left(1-{a}^{2}\right)}{2a}.$

That is,

$\zeta =\frac{\sqrt{{\left(1-{a}^{2}\right)}^{2}+8{a}^{2}}-\left(1-{a}^{2}\right)}{2a}.$
(29)

Notice that if we take $a=1$ in equation (29), then our result coincides with the result of Tewodros [7]. □

For ζ numbers in equation (23), from Proposition 3.4 we can deduce the following result.

Corollary 3.5 For every positive integer a, we have

$a<\zeta
(30)

Now, we give a proposition for the maximal real roots of ${G}_{n}^{\left(k\right)}\left(x\right)$ without proof.

Proposition 3.6 The maximal real roots of ${G}_{n}^{\left(k\right)}\left(x\right)$ provide the following equation:

$g-2a+a{g}^{k}-{a}^{2}{g}^{k-1}=0,$
(31)

where the numbers $g={g}_{n}={g}_{n}\left(k\right)$ are the maximal real roots of ${G}_{n}^{\left(k\right)}\left(x\right)$, that is,

$a{g}_{n}={a}^{2}-{g}_{n}^{2-k}+2a{g}_{n}^{1-k},$
(32)

which implies

$\frac{a}{1+a{\left(a+1\right)}^{k-1}}<{g}_{n}\left(k\right)-a<\frac{a}{1+{a}^{k}},$

whenever $k>2$.

$\frac{a-1}{a\left(a+1\right)}<\frac{2a-{g}_{n}\left(2\right)}{a{g}_{n}\left(2\right)}={g}_{n}\left(2\right)-a<\frac{1}{a+1}$

and

$\underset{k\to \mathrm{\infty }}{lim}{g}_{n}\left(k\right)=a,$

whenever $a>1$ for every $n\in \mathbb{N}$.

Proof The proof can be easily seen as being similar to the proof of Proposition 3.4 □

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## Acknowledgements

The authors are very grateful to the referees for very helpful suggestions and comments about the paper which improved the presentation and its readability.

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Correspondence to Serpil Halıcı.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors completed the paper together. All authors read and approved the final manuscript.

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Halıcı, S., Akyüz, Z. A note on the roots of some special type polynomials. J Inequal Appl 2013, 466 (2013). https://doi.org/10.1186/1029-242X-2013-466