We’d like to understand how you use our websites in order to improve them. Register your interest.

# Vieta-Pell and Vieta-Pell-Lucas polynomials

## Abstract

In the present paper, we introduce the recurrence relation of Vieta-Pell and Vieta-Pell-Lucas polynomials. We obtain the Binet form and generating functions of Vieta-Pell and Vieta-Pell-Lucas polynomials and define their associated sequences. Moreover, we present some differentiation rules and finite summation formulas.

MSC:11C08, 11B39.

## 1 Introduction

Andre-Jeannin  introduced a class of polynomials $U n (p,q;x)$ defined by

$U n (p,q;x)=(x+p) U n − 1 (p,q;x)−q U n − 2 (p,q;x),n≥2,$

with the initial values $U 0 (p,q;x)=0$ and $U 1 (p,q;x)=1$.

Vieta-Lucas polynomials were studied as Vieta polynomials by Robbins . Vieta-Fibonacci and Vieta-Lucas polynomials are defined by

$V n ( x ) = x V n − 1 ( x ) − V n − 2 ( x ) , n ≥ 2 , v n ( x ) = x v n − 1 ( x ) − v n − 2 ( x ) , n ≥ 2 ,$

respectively, where $V 0 (x)=0$, $V 1 (x)=1$ and $v 0 (x)=2$, $v 1 (x)=x$ . The recursive properties of Vieta-Fibonacci and Vieta-Lucas polynomials were given by Horadam .

For $p=0$ and $q=1$, Vieta-Fibonacci polynomials are a special case of the polynomials $U n (p,q;x)$ in . Further, $U n , m (p,q;x)$ in  for $p=0$, $q=1$, $m=2$ gives Vieta-Fibonacci polynomials.

Chebyshev polynomials are a sequence of orthogonal polynomials which can be defined recursively. Recall that the n th Chebyshev polynomials of the first kind and second kind are denoted by $T n (x)$ and $U n (x)$, respectively.

It is well known that the Chebyshev polynomials of the first kind and second kind are closely related to Vieta-Fibonacci and Vieta-Lucas polynomials. So, in  Vitula and Slota redefined Vieta polynomials as modified Chebyshev polynomials. The related features of Vieta and Chebyshev polynomials are given as

$V n ( x ) = U n ( 1 2 x )  , v n ( x ) = 2 T n ( 1 2 x ) ( see [2, 6] ) .$

For $|x|>1$, we consider $t n (x)$ and $s n (x)$ polynomials by the following recurrence relations:

$t n ( x ) = 2 x t n − 1 ( x ) − t n − 2 ( x ) , n ≥ 2 , s n ( x ) = 2 x s n − 1 ( x ) − s n − 2 ( x ) , n ≥ 2 ,$

where $t 0 (x)=0$, $t 1 (x)=1$ and $s 0 (x)=2$, $s 1 (x)=2x$. We call $t n (x)$ the n th Vieta-Pell polynomial and $s n (x)$ the n th Vieta-Pell-Lucas polynomial.

The relations below are obvious

$s n ( x ) = 2 T n ( x ) , t n + 1 ( x ) = U n ( x ) .$

The first few terms of $t n (x)$ and $s n (x)$ are as follows:

$t 2 ( x ) = 2 x , s 2 ( x ) = 4 x 2 − 2 , t 3 ( x ) = 4 x 2 − 1 , s 3 ( x ) = 8 x 3 − 6 x , t 4 ( x ) = 8 x 3 − 4 x , s 4 ( x ) = 16 x 4 − 16 x 2 + 2 , t 5 ( x ) = 16 x 4 − 12 x 2 + 1 , s 5 ( x ) = 32 x 5 − 40 x 3 + 10 x , t 6 ( x ) = 32 x 5 − 32 x 3 + 6 x , s 6 ( x ) = 64 x 6 − 96 x 4 + 36 x 2 − 2 , t 7 ( x ) = 64 x 6 − 80 x 4 + 24 x 2 − 1 , s 7 ( x ) = 128 x 7 − 224 x 5 + 112 x 3 − 14 x .$

The aim of this paper is to determine the recursive key features of Vieta-Pell and Vieta-Pell-Lucas polynomials. In conjunction with these properties, we examine their interrelations and define their associated sequences. Furthermore, we present some differentiation rules and summation formulas.

## 2 Main results

Some fundamental recursive properties of Vieta-Pell and Vieta Pell-Lucas polynomials are given in this section.

### Characteristic equation

Vieta-Pell and Vieta-Pell-Lucas polynomials have the following characteristic equation:

$λ 2 −2xλ+1=0$

with the roots α and β

$α = x + x 2 − 1 , β = x − x 2 − 1 .$

Also, α and β satisfy the following equations:

$α + β = 2 x , α β = 1 , α − β = Δ = 2 x 2 − 1 .$
(1)

### Binet form

By appropriate procedure, we can easily find the Binet forms as

$t n (x)= α n − β n α − β ,$
(2)
$s n (x)= α n + β n .$
(3)

### Generating function

Vieta-Pell and Vieta-Pell-Lucas polynomials can be defined by the following generating functions:

$∑ n = 0 ∞ t n ( x ) y n = y ( 1 − 2 x y + y 2 ) − 1 , ∑ n = 0 ∞ s n ( x ) y n = ( 2 − 2 x y ) ( 1 − 2 x y + y 2 ) − 1 .$

### Negative subscript

We can also extend the definition of $t n (x)$ and $s n (x)$ to the negative index

$t − n ( x ) = − t n ( x ) , s − n ( x ) = s n ( x ) .$

### Simson formulas

$t n + 1 ( x ) t n − 1 ( x ) − t n 2 ( x ) = − 1 , s n + 1 ( x ) s n − 1 ( x ) − s n 2 ( x ) = 4 ( x 2 − 1 ) .$

We arrange the first ten coefficients of $t n (x)$ in Table 1. Let $T(n,j)$ denote the element in row n and column j, where $j⩾0$, $n⩾1$. As seen from the Table 1, it is obvious that

$T(n,0)=2T(n−2,0)+T(n−1,0)$

can be written like the coefficients of Pell polynomials in . Moreover,

$∑ j = 0 ⌊ n − 1 2 ⌋ T(n,j)=n.$

For example, for $n=8$ we can find

$∑ j = 0 3 T(8,j)=T(8,0)+T(8,1)+T(8,2)+T(8,3)=8.$

Let $P n$ denote the n th Pell number, so we have

$∑ j = 1 ⌊ n − 1 2 ⌋ |T(n,j)|= P n .$

### 2.1 Interrelations of $t n (x)$ and $s n (x)$

Most of the equations below can be obtained by using the Binet form and convenient routine operations

$t n + 1 (x)− t n − 1 (x)= s n (x)=2x t n (x)−2 t n − 1 (x),$
(4)
$s n + 1 ( x ) − s n − 1 ( x ) = 4 ( x 2 − 1 ) t n ( x ) , t n ( x ) s n ( x ) = t 2 n ( x ) , s n 2 ( x ) + 4 ( x 2 − 1 ) t n 2 ( x ) = 2 s 2 n ( x ) , s n 2 ( x ) − 4 ( x 2 − 1 ) t n 2 ( x ) = 4 , t n + 1 2 ( x ) − t n 2 ( x ) = t 2 n + 1 ( x ) , s n + 1 2 ( x ) − s n 2 ( x ) = 4 ( x 2 − 1 ) t 2 n + 1 ( x ) , s n + 1 2 ( x ) + s n 2 ( x ) = 2 x s 2 n + 1 ( x ) + 4 , t n + 1 2 ( x ) − t n − 1 2 ( x ) = 2 x t 2 n ( x ) , s n ( x ) s n + 1 ( x ) − 4 ( x 2 − 1 ) t n ( x ) t n + 1 ( x ) = 4 x , s n ( x ) s n + 1 ( x ) + 4 ( x 2 − 1 ) t n ( x ) t n + 1 ( x ) = 2 s 2 n + 1 ( x ) , s 4 n ( x ) − 2 = 4 ( x 2 − 1 ) t 2 n 2 ( x ) , t n + 1 ( x ) − x t n ( x ) = 1 2 s n ( x ) , 2 s n + 1 ( x ) − 2 x s n ( x ) = 4 ( x 2 − 1 ) t n ( x ) .$
(5)

Proposition 1 $s n (2 x 2 −1)− s n 2 (x)=−2$.

Proof Consider the expression $s n (2 x 2 −1)$. Then α, β, Δ are replaced by $α ∗$, $β ∗$, $Δ ∗$, respectively. So, $α ∗ = α 2$, $β ∗ = β 2$, $Δ ∗ =2xΔ$ and by using the Binet form, the proof is completed. □

### 2.2 Associated sequences

Definition 1 The k th associated sequences ${ t n ( k ) (x)}$ and ${ s n ( k ) (x)}$ of ${ t n (x)}$ and ${ s n (x)}$ are defined by, respectively ($k≥1$)

$t n ( k ) (x)= t n + 1 ( k − 1 ) (x)− t n − 1 ( k − 1 ) (x),$
(6)
$s n ( k ) (x)= s n + 1 ( k − 1 ) (x)− s n − 1 ( k − 1 ) (x),$
(7)

where $t n ( 0 ) (x)= t n (x)$ and $s n ( 0 ) (x)= s n (x)$.

Presently,

$t n ( 1 ) ( x ) = s n ( x ) ( by (4) ) , s n ( 1 ) ( x ) = Δ 2 t n ( x ) ( by (5) )$

are the members of the first associated sequences ${ t n ( 1 ) (x)}$ and ${ s n ( 1 ) (x)}$. If (6) and (7) are applied repeatedly, the results emerge

$t n 2 j ( x ) = s n ( 2 j − 1 ) ( x ) = Δ 2 j t n ( x ) , t n ( 2 j − 1 ) ( x ) = s n ( 2 j − 2 ) ( x ) = Δ 2 j − 2 s n ( x ) .$

### Some special values of $t n (x)$ and $s n (x)$

${ t n ( − x ) = ( − 1 ) n + 1 t n ( x ) , t n ( 1 ) = n , t n ( − 1 ) = ( − 1 ) n + 1 n , t 2 n ( 0 ) = 0 , t 2 n − 1 ( 0 ) = ( − 1 ) n + 1 , { s n ( 1 ) = 2 , s n ( − 1 ) = 2 ( − 1 ) n , s 2 n ( 0 ) = 2 ( − 1 ) n , s 2 n − 1 ( 0 ) = 0 .$

### 2.3 Differentiation formulas

$d s n ( x ) d x = 2 n t n ( x ) , d t n ( x ) d x = n s n ( x ) − 2 x t n ( x ) 2 ( x 2 − 1 ) , d 2 s n ( x ) d x 2 = n ( n s n ( x ) − 2 x t n ( x ) x 2 − 1 ) .$

Since the derivation function of $s n (x)$ is a polynomial, all of the derivatives must exist for all real numbers. Thus, we can give the following formulas.

Proposition 2

$d 2 s n ( x ) d x 2 | x = 1 = 2 3 ( n 4 − n 2 ) , d 2 s n ( x ) d x 2 | x = − 1 = 2 3 ( − 1 ) n − 1 ( n 2 − n 4 ) .$

Proof If we take the limit on $s n ″ (x)=n( n s n ( x ) − 2 x t n ( x ) x 2 − 1 )$, we have the numerical value of $s n ″$ at $x=1$ and $x=−1$.

$s n ″ (1)= n 2 lim x → 1 n s n ( x ) − 2 x t n ( x ) ( x − 1 ) .$

Since $s n (1)=2$, $t n (1)=n$, apply L’Hôpital’s rule:

$s n ″ ( 1 ) = n 2 lim x → 1 d d x ( n s n ( x ) − 2 x t n ( x ) ) d d x ( x − 1 ) = n 2 lim x → 1 d d x ( n s n ( x ) − 2 x t n ( x ) ) = n 2 lim x → 1 ( 2 n 2 t n ( x ) − 2 t n ( x ) − 2 x d d x t n ( x ) ) = n 2 ( 2 n 2 t n ( 1 ) − 2 t n ( 1 ) − 2 lim x → 1 d d x t n ( x ) ) = n 2 ( 2 n 3 − 2 n ) − 1 2 lim x → 1 d d x ( 2 n t n ( x ) ) = n 4 − n 2 − 1 2 s n ″ ( 1 ) .$

So, the proof for $x=−1$ is similar. □

### 2.4 Some summation formulas

In this section we deal with the matrix

$V= [ 2 x − 1 1 0 ] .$

By induction, we have

$V m = [ t m + 1 ( x ) − t m ( x ) t m ( x ) − t m − 1 ( x ) ] .$
(8)

So, the matrix V generates Vieta-Pell and Vieta-Pell-Lucas polynomials. Hence,

$[ t m + 1 ( x ) t m ( x ) ] = V m [ 1 0 ]$
(9)

and from (1.10) in , we get

$t m (x)= [ 1 0 ] V m − 1 [ 1 0 ] .$
(10)

It is known that

$V m + n = V m V n .$
(11)

From (8) and (11), the elementary formulas for $t n (x)$ are obvious

$t m + n + 1 ( x ) = t m + 1 ( x ) t n + 1 ( x ) − t m ( x ) t n ( x ) , t m + n ( x ) = t m + 1 ( x ) t n ( x ) − t m ( x ) t n − 1 ( x ) , t m + n − 1 ( x ) = t m ( x ) t n ( x ) − t m − 1 ( x ) t n − 1 ( x ) .$

If we use the matrix technique for summation in , we get the first finite summation as follows.

Proposition 3

1. (i)

$∑ n = 1 m t n (x)= t m + 1 ( x ) − t m ( x ) − 1 2 ( x − 1 )$,

2. (ii)

$∑ n = 1 m s n (x)= s m + 1 ( x ) − s m ( x ) + 2 − 2 x 2 ( x − 1 )$.

Proof (i) Let the matrix A,

$A=I+V+ V 2 +⋯+ V n − 2 + V n − 1$

be the series of matrices. Then we have

$VA=V+ V 2 +⋯+ V n − 1 + V n .$

Hence,

$( V − I ) A = V n − I , A = ( V − I ) − 1 ( V n − I ) A = 1 2 ( x − 1 ) [ t n + 1 ( x ) − t n ( x ) − 1 t n − 1 ( x ) − t n ( x ) + 1 t n ( x ) − t n − 1 ( x ) − 1 t n − 2 ( x ) − t n − 1 ( x ) + 2 x − 1 ] , ∑ n = 1 m t n ( x ) = [ 1 0 ] A [ 1 0 ] ( by (10) ) ∑ n = 1 m t n ( x ) = 1 2 ( x − 1 ) [ t m + 1 ( x ) − t m ( x ) − 1 t m − 1 ( x ) − t m ( x ) + 1 ] [ 1 0 ] ∑ n = 1 m t n ( x ) = 1 2 ( x − 1 ) [ t m + 1 ( x ) − t m ( x ) − 1 ] .$

Thus, the proof is completed.

1. (ii)
$∑ n = 1 m s n ( x ) = ∑ n = 1 m ( α n + β n ) ( by (3) ) = α ( 1 − α m 1 − α ) + β ( 1 − β m 1 − β ) = ( α + β ) − 2 α β − ( α m + 1 + β m + 1 ) + α β ( α m + β m ) 2 − 2 x = 2 x − 2 − ( α m + 1 + β m + 1 ) + α m + β m 2 ( 1 − x ) ( by (1) ) = s m + 1 ( x ) − s m ( x ) + 2 − 2 x 2 ( x − 1 ) ( by (3) ) .$

This completes the proof. □

Theorem 1 Let V be a square matrix such that $V 2 =2xV−I$. Then, for all $n∈ Z +$,

$V n = t n (x)V− t n − 1 (x)I,$

where $t n (x)$ is the nth Vieta-Pell polynomial and I is a unit matrix.

Proof The proof is obvious from induction. □

## References

1. 1.

Andre-Jeannin R: A note on general class of polynomials. Fibonacci Q. 1994, 32(5):445-454.

2. 2.

Robbins N: Vieta’s triangular array and a related family of polynomials. Int. J. Math. Math. Sci. 1991, 14: 239-244. 10.1155/S0161171291000261

3. 3.

Horadam AF: Vieta polynomials. Fibonacci Q. 2002, 40(3):223-232.

4. 4.

Djordjevic GB: Some properties of a class of polynomials. Mat. Vesn. 1997, 49: 265-271.

5. 5.

Vitula R, Slota D: On modified Chebyshev polynomials. J. Math. Anal. Appl. 2006, 324: 321-343. 10.1016/j.jmaa.2005.12.020

6. 6.

Jacobsthal E: Über vertauschbare polynome. Math. Z. 1955, 63: 244-276.

7. 7.

Halici S: On the Pell polynomials. Appl. Math. Sci. 2011, 5(37):1833-1838.

8. 8.

Mahon JM, Horadam AF: Matrix and other summation techniques for Pell polynomials. Fibonacci Q. 1986, 24(4):290-309.

## Acknowledgements

The authors thank the referees for their valuable suggestions, which improved the standard of the paper.

## Author information

Authors

### Corresponding author

Correspondence to Feyza Yalcin.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The authors declare that the research was realized in collaboration with the same responsibility and contributions. Both authors read and approved the final manuscript.

## Rights and permissions

Reprints and Permissions

Tasci, D., Yalcin, F. Vieta-Pell and Vieta-Pell-Lucas polynomials. Adv Differ Equ 2013, 224 (2013). https://doi.org/10.1186/1687-1847-2013-224 