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 $G_0(x)=-a$, $G_1(x)=x-a$

$$G_{n+2}^{(k)}(x)=x^k{G}_{n+1}^{(k)}(x)+G_n^{(k)}(x).$$

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

MSC:11B39, 11B37.

1 Introduction

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

$$F_{n+2}(x)=x{F}_{n+1}(x)+F_n(x);F_1(x)=1,F_2(x)=x$$

(1)

are called Fibonacci polynomials and denoted by $F_n(x)$, [1]. The Fibonacci-type polynomials $G_n(x)$, $n\ge 0$, are defined by

$$G_{n+2}(x)=x{G}_{n+1}(x)+G_n(x),$$

(2)

where $G_0(x)$ and $G_1(x)$ are seed polynomials. There are several studies about the properties of zeros of polynomials $G_n(x)$. 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 $G_n(x)$ with the initial values $G_0(x)=-1$, $G_1(x)=x-1$. In [4], the authors generalized Moore’s result for these polynomials. They considered the initial values $G_0(x)=a$, $G_1(x)=x+b$, 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:

$$G_{n+2}^{(k)}(x)=x^k{G}_{n+1}^{(k)}(x)+G_n^{(k)}(x),n\ge 0,$$

(3)

where k is a positive integer number. The initial values of the recursive relation (3) are $G_{{}_0(k)}^{}(x)=-1$ and $G_{{}_1(k)}^{}(x)=x-1$. 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 $G_n^{(k)}(x)$

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

$$G_{n+2}^{(k)}(x)=x^k{G}_{n+1}^{(k)}(x)+G_n^{(k)}(x);G_0^{(k)}(x)=-a;G_1^{(k)}(x)=x-a.$$

(4)

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

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

and

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

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

$$G_n^{(k)}(x)=A(x){\alpha }^n(x)+B(x){\beta }^n(x),$$

(5)

where

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

(6)

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

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

(7)

Proof Let $H_r^{(k)}(x,t)$ be the generating function for polynomials $G_{n+r}^{(k)}(x)$. So, we write

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

(8)

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

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

and

$$t^2{H}_r^{(k)}(x,t)=G_r^{(k)}(x)t^2+G_{r+1}^{(k)}(x)t^3+G_{r+2}^{(k)}(x)t^4+\cdots +G_{r+n-2}^{(k)}(x)t^n+\cdots .$$

The last two equations give us the following equation:

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

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

$$H_r^{(k)}(x,t)-x^{k}t{H}_r^{(k)}(x,t)-t^2{H}_r^{(k)}(x,t)=G_r^{(k)}(x)t^0+(G_{r+1}^{(k)}(x)-x^k{G}_r^{(k)}(x))t,$$

i.e.,

$$H_r^k(x,t)=\{\begin{array}{cc}\frac{G_r^{(k)}(x)+G_{r-1}^{(k)}(x)t}{1-x^{k}t-t^2},\hfill & r=1,2,3,\dots ,\hfill \\ \frac{t(a{x}^k+x-a)}{1-x^{k}t-t^2},\hfill & r=0.\hfill \end{array}$$

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

$$G_{n-1}^{(k)}(x)G_{n+1}^{(k)}(x)-{[G_n^{(k)}(x)]}^2={(-1)}^{n-1}[A(x)B(x)],$$

(9)

where

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

and

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

In the following propositions, we give some sums formulas related to polynomials $G_n^{(k)}(x)$.

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

$$H_0^{(k)}(x,1)-H_{N+1}^{(k)}(x,1)=\sum _{r=0}^N{G}_r^{(k)}(x)=\frac{2a-x-a{x}^k+G_{N+1}^{(k)}(x)+G_N^{(k)}(x)}{x^k}.$$

(10)

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

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

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

(11)

and

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

(12)

Proof From the Binet formula, we can write

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

(13)

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

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

and

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

into equation (13), then we can get

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

If we rearrange the last equation, then we have

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

By taking aid of the Binet formula, we can write

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

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

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

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

3 Asymptotic behaviors of the maximal roots for polynomials $G_n^{(k)}(x)$

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

$$G_{n+2}^{(2)}(x)=x^2{G}_{n+1}^{(2)}(x)+G_n^{(2)}(x),$$

(14)

where

$$G_0^{(2)}(x)=-a,G_1^{(2)}(x)=x-a$$

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 $f(x)>0$ for all $x>r$. Conversely, if $f(x)>0$ for all $x\ge t$, then $r<t$. If $f(s)<0$ , then $s<r$ [2].

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

Proof Some of polynomials $G_n^{(2)}(x)$ are as follows:

$$\begin{array}{c}{G}_2^{(2)}(x)=x^3-a{x}^2-a,\hfill \\ G_3^{(2)}(x)=x^5-a{x}^4-a{x}^2+x-a,\hfill \\ G_4^{(2)}(x)=x^7-a{x}^6-a{x}^4+2x^3-2a{x}^2-a,\hfill \\ ⋮\hfill \end{array}$$

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

$$\begin{array}{c}{G}_1^{(2)}(a)=0,\hfill \\ G_2^{(2)}(a)=-a<0,\hfill \\ G_3^{(2)}(a)=-a^3=a^2{G}_2^{(2)}(a)<0,\hfill \\ G_4^{(2)}(a)=-a^5-a\le -a^5=a^2{G}_3^{(2)}(a)<0,\hfill \\ ⋮\hfill \end{array}$$

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

$$G_{k+1}^{(2)}(a)=a^2{G}_k^{(2)}(a)+G_{k-1}^{(2)}(a)<0.$$

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

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

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

Proof Firstly, we consider polynomials $G_n^{(2)}(x)$ with odd indices. By a direct computation, we get $G_3^{(2)}(a)=-a^3<0$, $g_3>a$, $a=g_1$. Assume that $g_1<g_3<g_5<\cdots <g_{2k-3}<g_{2k-1}$. We can write $G_{2k-3}^{(2)}(g_{2k-1})>0$. Thus, it can be easily seen that

$$G_{n+k}^{(2)}(g_n)={(-1)}^{k+1}{G}_{n-k}^{(2)}(g_n).$$

(15)

By using equation (15), we can write

$$G_{2k+1}^{(2)}(g_{2k-1})=G_{(2k-1)+2}^{(2)}(g_{2k-1})=-G_{(2k-1)-2}^{(2)}(g_{2k-1})=-G_{2k-3}^{(2)}(g_{2k-1}).$$

(16)

So, from equation (16) we write

$$G_{2k+1}^{(2)}(g_{2k-1})<0.$$

(17)

Therefore, polynomials $G_{2k+1}^{(2)}(x)$ must have a root greater than $g_{2k-1}$. So, we get

$$g_{2k+1}>g_{2k-1}.$$

(18)

After that we consider polynomials $G_n^{(2)}(x)$ with even indices. From the recursive relation (14), we can obtain

$$G_{2k+1}^{(2)}(g_{2k-1})=g_{2k-1}^2{G}_{2k}^{(2)}(g_{2k-1})+G_{2k-1}^{(2)}(g_{2k-1}).$$

(19)

Since $G_{2k-1}^{(2)}(g_{2k-1})=0$, by using Lemma 3.1, we can get $G_{2k+1}^{(2)}(g_{2k-1})<0$. Thus, we get

$$g_{2k-1}<g_{2k}.$$

(20)

Again, by using Lemma 3.1, we can write

$$G_{2k-1}^{(2)}(g_{2k})>0.$$

(21)

From the recursive relation (14), we can write

$$G_{2k}^{(2)}(g_{2k})=g_{2k}^2{G}_{2k-1}^{(2)}(g_{2k})+G_{2k-2}^{(2)}(g_{2k}).$$

(22)

From equation (22), we can get

$$-g_{2k}^2{G}_{2k-1}^{(2)}(g_{2k})=G_{2k-2}^{(2)}(g_{2k})<0.$$

So, we have $g_{2k}<g_{2k-2}$. Thus, $\{g_{2n-1}\}$ is a monotonically increasing sequence and bounded above by the number $a+1$. Similarly, $\{g_{2n}\}$ 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 $G_{2n-1}^{(2)}(x)$ and $G_{2n}^{(2)}(x)$, the sequences $\{g_{2n-1}\}$ and $\{g_{2n}\}$ converge to the following number ζ:

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

(23)

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

$$\underset{n\to \mathrm{\infty }}{lim}{\alpha }^n(x)=+\mathrm{\infty };\underset{n\to \mathrm{\infty }}{lim}{\beta }^n(x)=0.$$

(24)

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

$$A(g_{2k-1}){\alpha }^{2k-1}(g_{2k-1})+B(g_{2k-1}){\beta }^{2k-1}(g_{2k-1})=0.$$

(25)

And from equation (25) we write

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

(26)

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

$$\underset{k\to \mathrm{\infty }}{lim}A(g_{2k-1})=A(g_{\mathrm{odd}})=0.$$

(27)

From Binet formula (5), we have

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

(28)

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

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

That is,

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

(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

$$a<\zeta <a+1.$$

(30)

Now, we give a proposition for the maximal real roots of $G_n^{(k)}(x)$ without proof.

Proposition 3.6 The maximal real roots of $G_n^{(k)}(x)$ provide the following equation:

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

(31)

where the numbers $g=g_n=g_n(k)$ are the maximal real roots of $G_n^{(k)}(x)$, that is,

$$a{g}_n=a^2-g_n^{2-k}+2a{g}_n^{1-k},$$

(32)

which implies

$$\frac{a}{1+a{(a+1)}^{k-1}}<g_n(k)-a<\frac{a}{1+a^k},$$

whenever $k>2$.

$$\frac{a-1}{a(a+1)}<\frac{2a-g_n(2)}{a{g}_n(2)}=g_n(2)-a<\frac{1}{a+1}$$

and

$$\underset{k\to \mathrm{\infty }}{lim}{g}_n(k)=a,$$

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 □