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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 [1] 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),n2,

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 [2]. 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 [3]. The recursive properties of Vieta-Fibonacci and Vieta-Lucas polynomials were given by Horadam [3].

For p=0 and q=1, Vieta-Fibonacci polynomials are a special case of the polynomials U n (p,q;x) in [1]. Further, U n , m (p,q;x) in [4] 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 [5] 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 ) [3] , 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 j0, n1. As seen from the Table 1, it is obvious that

T(n,0)=2T(n2,0)+T(n1,0)

can be written like the coefficients of Pell polynomials in [7]. 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 .
Table 1 The first ten coefficients of t n (x)

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 (k1)

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 [8], 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 [8], 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 =2xVI. 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. □

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Acknowledgements

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

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Correspondence to Feyza Yalcin.

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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

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Keywords

  • Vieta-Pell
  • Vieta-Pell-Lucas polynomials