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A note on the roots of some special type polynomials

Abstract

In this study, we investigate the polynomials for n2 and positive integers k and a positive real number a, with the initial values G 0 (x)=a, G 1 (x)=xa

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 n0, 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), n0, 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)=x1. 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),n0,
(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)=x1. 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 ( k ) n (x), for n0, 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)=xa.
(4)

The characteristic equation for (4) is t 2 x k t1=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 n0, 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 n0, 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 N0, 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 N0, 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)=xa

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 xt, then r<t. If f(s)<0 , then s<r [2].

Lemma 3.2 For n2, 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 k2, 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 nN. 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=2k1 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 a1, 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 [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<ζ<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:

g2a+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 nN.

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.

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The authors declare that they have no competing interests.

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

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Keywords

  • Fibonacci polynomials
  • Binet formula
  • generating function