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# A note on the roots of some special type polynomials

## Abstract

In this study, we investigate the polynomials for $n≥2$ 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 $n≥0$, 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 $F n (x)$, . The Fibonacci-type polynomials $G n (x)$, $n≥0$, 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 , 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 , the author determined the absolute values of complex zeros of these polynomials. In , Ricci studied this problem in the case $a=1$ and $b=1$. In , Tewodros investigated the convergence of maximal real roots of different Fibonacci-type polynomials given by the following relation:

$G n + 2 ( k ) (x)= x k G n + 1 ( k ) (x)+ G n ( k ) (x),n≥0,$
(3)

where k is a positive integer number. The initial values of the recursive relation (3) are $G ( k ) 0 (x)=−1$ and $G ( k ) 1 (x)=x−1$. 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  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 $G ( k ) n (x)$, for $n≥0$, 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

$α(x)= x k + x 2 k + 4 2$

and

$β(x)= x k − x 2 k + 4 2 .$

Note that $α(x)β(x)=−1$, $α(x)+β(x)= x k$ and $α(x)−β(x)= x 2 k + 4$. For relation (4), the Binet formula is

$G n ( k ) (x)=A(x) α n (x)+B(x) β n (x),$
(5)

where

$A(x)= 2 ( x − a ) + a x k − a x 2 k + 4 2 x 2 k + 4 ,B(x)= − 2 ( x − a ) − a x k − a x 2 k + 4 2 x 2 k + 4 .$
(6)

Proposition 2.1 For $n≥0$, the generating function for polynomials $G n ( k ) (x)$ is

$H r ( k ) (x,t)= ∑ n ≥ 0 G n + r ( k ) (x) t n ={ G r ( k ) ( x ) + G r − 1 ( k ) ( x ) t 1 − x k t − t 2 , r = 1 , 2 , 3 , … , t ( a x k + x − a ) 1 − x k t − t 2 , r = 0 .$
(7)

Proof Let $H r ( k ) (x,t)$ be the generating function for polynomials $G n + r ( k ) (x)$. So, we write

$H r ( k ) (x,t)= ∑ n ≥ 0 G n + r ( k ) (x) t n .$
(8)

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

$x k t H r ( k ) (x,t)= x k G r ( k ) (x) t 1 + x k G r + 1 ( k ) (x) t 2 + x k G r + 2 ( k ) (x) t 3 +⋯+ x k G r + n − 1 ( k ) (x) t n +⋯$

and

$t 2 H r ( k ) (x,t)= G r ( k ) (x) t 2 + G r + 1 ( k ) (x) t 3 + G r + 2 ( k ) (x) t 4 +⋯+ G r + n − 2 ( k ) (x) t n +⋯.$

The last two equations give us the following equation:

$H r ( k ) ( x , t ) − x k t H r ( k ) ( x , t ) − t 2 H r ( k ) ( x , t ) = G r ( k ) ( x ) t 0 + ( G r + 1 ( k ) ( x ) − x k G r ( k ) ( x ) ) t + ( G r + 2 ( k ) ( x ) − x k G r + 1 ( k ) ( x ) − G r ( k ) ( x ) ) t 2 + ⋯ + ( G n + r ( k ) ( x ) − x k G n + r − 1 ( k ) ( x ) − G n + r − 2 ( k ) ( x ) ) t n + ⋯ .$

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.,

$H r k (x,t)={ G r ( k ) ( x ) + G r − 1 ( k ) ( x ) t 1 − x k t − t 2 , r = 1 , 2 , 3 , … , t ( a x k + x − a ) 1 − x k t − t 2 , r = 0 .$

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 $n≥0$, 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

$A(x)= 2 ( x − a ) + a x k − a x 2 k + 4 2 x 2 k + 4$

and

$B(x)= − 2 ( x − a ) − a x k − a x 2 k + 4 2 x 2 k + 4 .$

In the following propositions, we give some sums formulas related to polynomials $G n ( k ) (x)$.

Proposition 2.3 For $N≥0$, we have

$H 0 ( k ) (x,1)− H N + 1 ( k ) (x,1)= ∑ r = 0 N G r ( k ) (x)= 2 a − x − a x k + G N + 1 ( k ) ( x ) + G N ( k ) ( x ) x k .$
(10)

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

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

$∑ r = 0 N G 2 r ( k ) (x)= − x k + 1 − a x k ( x k − 1 ) − G 2 N ( k ) ( x ) + G 2 N + 2 ( k ) ( x ) x 2 k$
(11)

and

$∑ r = 0 N G 2 r + 1 ( k ) (x)= a x k − G 2 N + 1 ( k ) ( x ) + G 2 N + 3 ( k ) ( x ) x 2 k .$
(12)

Proof From the Binet formula, we can write

$∑ r = 0 N G 2 r ( k ) (x)= ( x − a + a β α − β ) ∑ r = 0 N α 2 r − ( x − a + a α α − β ) ∑ r = 0 N β 2 r ,$
(13)

where $α=α(x)$, $β=β(x)$. If we substitute the equations

$∑ r = 0 N α 2 r = 1 − α 2 N + 2 1 − α 2 , ∑ r = 0 N β 2 r = 1 − β 2 N + 2 1 − β 2$

and

$( 1 − α 2 ) ( 1 − β 2 ) =− x 2 k$

into equation (13), then we can get

$∑ r = 0 N G 2 r ( k ) (x)= ( x − a + a β α − β ) 1 − α 2 N + 2 1 − α 2 − ( x − a + a α α − β ) 1 − β 2 N + 2 1 − β 2 .$

If we rearrange the last equation, then we have

$∑ r = 0 N G 2 r ( k ) ( x ) = − a ( α − β ) + ( x − a ) ( α 2 − β 2 ) + a ( α 3 − β 3 ) ( 1 − α 2 ) ( 1 − β 2 ) ( α − β ) − [ ( x − a + a β ) α 2 N + 2 − ( x − a + a α ) β 2 N + 2 ] ( 1 − α 2 ) ( 1 − β 2 ) ( α − β ) + α 2 β 2 [ ( x − a + a β ) α 2 N − ( x − a + a α ) β 2 N ] ( 1 − α 2 ) ( 1 − β 2 ) ( α − β ) .$

By taking aid of the Binet formula, we can write

$∑ r = 0 N G 2 r ( k ) ( x ) = − a + ( x − a ) ( α + β ) + a ( α 2 + α β + β 2 ) ( 1 − α 2 ) ( 1 − β 2 ) + G 2 N ( k ) ( x ) ( 1 − α 2 ) ( 1 − β 2 ) − G 2 N + 2 ( k ) ( x ) ( 1 − α 2 ) ( 1 − β 2 ) .$

If we substitute $α+β= x k$, $αβ=−1$, $(1− α 2 )(1− β 2 )=− x 2 k$ into the last equation, we obtain the following equation:

$∑ r = 0 N G 2 r ( k ) (x)= x k + 1 − a x k + a x 2 k − G 2 N + 2 ( k ) ( x ) + G 2 N ( k ) ( x ) − x 2 k .$

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 $x≥t$, then $r. If $f(s)<0$ , then $s .

Lemma 3.2 For $n≥2$, $G n ( 2 ) (x)$ has at least one real root on the interval $(a,a+1)$ and $g n ∈(a,a+1)$, where $g n$ is the maximal root of polynomial $G n ( 2 ) (x)$.

Proof Some of polynomials $G n ( 2 ) (x)$ are as follows:

$G 2 ( 2 ) ( x ) = x 3 − a x 2 − a , G 3 ( 2 ) ( x ) = x 5 − a x 4 − a x 2 + x − a , G 4 ( 2 ) ( x ) = x 7 − a x 6 − a x 4 + 2 x 3 − 2 a x 2 − a , ⋮$

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

$G 1 ( 2 ) ( a ) = 0 , G 2 ( 2 ) ( a ) = − a < 0 , G 3 ( 2 ) ( a ) = − a 3 = a 2 G 2 ( 2 ) ( a ) < 0 , G 4 ( 2 ) ( a ) = − a 5 − a ≤ − a 5 = a 2 G 3 ( 2 ) ( a ) < 0 , ⋮$

For $k≥2$, if we suppose $G k ( 2 ) (a)≤ a 2 G k − 1 ( 2 ) (a)<0$, 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 $g n ∈(a,a+1)$ 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 $n∈N$. Then we can give the following proposition to illustrate the monotonicity of ${ g 2 n − 1 }$ and ${ g 2 n }$.

Proposition 3.3 The sequence ${ g 2 n − 1 }$ is a monotonically increasing sequence and the sequence ${ g 2 n }$ 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 $g 1 < g 3 < g 5 <⋯< g 2 k − 3 < g 2 k − 1$. We can write $G 2 k − 3 ( 2 ) ( g 2 k − 1 )>0$. 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

$G 2 k + 1 ( 2 ) ( g 2 k − 1 )= G ( 2 k − 1 ) + 2 ( 2 ) ( g 2 k − 1 )=− G ( 2 k − 1 ) − 2 ( 2 ) ( g 2 k − 1 )=− G 2 k − 3 ( 2 ) ( g 2 k − 1 ).$
(16)

So, from equation (16) we write

$G 2 k + 1 ( 2 ) ( g 2 k − 1 )<0.$
(17)

Therefore, polynomials $G 2 k + 1 ( 2 ) (x)$ must have a root greater than $g 2 k − 1$. So, we get

$g 2 k + 1 > g 2 k − 1 .$
(18)

After that we consider polynomials $G n ( 2 ) (x)$ with even indices. From the recursive relation (14), we can obtain

$G 2 k + 1 ( 2 ) ( g 2 k − 1 )= g 2 k − 1 2 G 2 k ( 2 ) ( g 2 k − 1 )+ G 2 k − 1 ( 2 ) ( g 2 k − 1 ).$
(19)

Since $G 2 k − 1 ( 2 ) ( g 2 k − 1 )=0$, by using Lemma 3.1, we can get $G 2 k + 1 ( 2 ) ( g 2 k − 1 )<0$. Thus, we get

$g 2 k − 1 < g 2 k .$
(20)

Again, by using Lemma 3.1, we can write

$G 2 k − 1 ( 2 ) ( g 2 k )>0.$
(21)

From the recursive relation (14), we can write

$G 2 k ( 2 ) ( g 2 k )= g 2 k 2 G 2 k − 1 ( 2 ) ( g 2 k )+ G 2 k − 2 ( 2 ) ( g 2 k ).$
(22)

From equation (22), we can get

$− g 2 k 2 G 2 k − 1 ( 2 ) ( g 2 k )= G 2 k − 2 ( 2 ) ( g 2 k )<0.$

So, we have $g 2 k < g 2 k − 2$. Thus, ${ g 2 n − 1 }$ is a monotonically increasing sequence and bounded above by the number $a+1$. Similarly, ${ g 2 n }$ is a monotonically decreasing sequence and bounded below by the number a. If we denote the $lim x → ∞ g 2 n − 1$ by $g odd$ and $lim x → ∞ g 2 n$ by $g even$, then we can write $g odd = g even$. □

Proposition 3.4 For polynomials $G 2 n − 1 ( 2 ) (x)$ and $G 2 n ( 2 ) (x)$, the sequences ${ g 2 n − 1 }$ and ${ g 2 n }$ converge to the following number ζ:

$ζ= ( 1 − a 2 ) 2 + 8 a 2 − ( 1 − a 2 ) 2 a .$
(23)

Proof Using the Binet formula of relation (14), for all $[a,a+1]$, we can see that $α(x)≥α(a)>1$ and $|β(x)|= 1 α ( x ) ≤ 1 α ( a )$. Thus, we get

$lim n → ∞ α n (x)=+∞; lim n → ∞ β n (x)=0.$
(24)

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

$A( g 2 k − 1 ) α 2 k − 1 ( g 2 k − 1 )+B( g 2 k − 1 ) β 2 k − 1 ( g 2 k − 1 )=0.$
(25)

And from equation (25) we write

$A( g 2 k − 1 )=−B( g 2 k − 1 ) ( β 2 k − 1 ( g 2 k − 1 ) α 2 k − 1 ( g 2 k − 1 ) ) .$
(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 $a≥1$, we get

$lim k → ∞ A( g 2 k − 1 )=A( g odd )=0.$
(27)

From Binet formula (5), we have

$lim k → ∞ g 2 k − 1 = ( 1 − a 2 ) 2 + 8 a 2 − ( 1 − a 2 ) 2 a .$
(28)

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

$lim k → ∞ g 2 k = ( 1 − a 2 ) 2 + 8 a 2 − ( 1 − a 2 ) 2 a .$

That is,

$ζ= ( 1 − a 2 ) 2 + 8 a 2 − ( 1 − a 2 ) 2 a .$
(29)

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

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<ζ
(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

$a 1 + a ( a + 1 ) k − 1 < g n (k)−a< a 1 + a k ,$

whenever $k>2$.

$a − 1 a ( a + 1 ) < 2 a − g n ( 2 ) a g n ( 2 ) = g n (2)−a< 1 a + 1$

and

$lim k → ∞ g n (k)=a,$

whenever $a>1$ for every $n∈N$.

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.

## Author information

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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• #### DOI

https://doi.org/10.1186/1029-242X-2013-466

### Keywords

• Fibonacci polynomials
• Binet formula
• generating function 