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

$$U_n(p,q;x)=(x+p)U_{n-1}(p,q;x)-q{U}_{n-2}(p,q;x),n\ge 2,$$

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 [2]. Vieta-Fibonacci and Vieta-Lucas polynomials are defined by

$$\begin{array}{c}{V}_n(x)=x{V}_{n-1}(x)-V_{n-2}(x),n\ge 2,\hfill \\ v_n(x)=x{v}_{n-1}(x)-v_{n-2}(x),n\ge 2,\hfill \end{array}$$

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

$$\begin{array}{c}{V}_n(x)=U_n\left(\frac{1}{2}x\right)\text{[3]},\hfill \\ v_n(x)=2T_n\left(\frac{1}{2}x\right)\left(\text{see [2, 6]}\right).\hfill \end{array}$$

For $|x|>1$, we consider $t_n(x)$ and $s_n(x)$ polynomials by the following recurrence relations:

$$\begin{array}{c}{t}_n(x)=2x{t}_{n-1}(x)-t_{n-2}(x),n\ge 2,\hfill \\ s_n(x)=2x{s}_{n-1}(x)-s_{n-2}(x),n\ge 2,\hfill \end{array}$$

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

$$\begin{array}{c}{s}_n(x)=2T_n(x),\hfill \\ t_{n+1}(x)=U_n(x).\hfill \end{array}$$

The first few terms of $t_n(x)$ and $s_n(x)$ are as follows:

$$\begin{array}{ll}{t}_2(x)=2x,& s_2(x)=4x^2-2,\\ t_3(x)=4x^2-1,& s_3(x)=8x^3-6x,\\ t_4(x)=8x^3-4x,& s_4(x)=16x^4-16x^2+2,\\ t_5(x)=16x^4-12x^2+1,& s_5(x)=32x^5-40x^3+10x,\\ t_6(x)=32x^5-32x^3+6x,& s_6(x)=64x^6-96x^4+36x^2-2,\\ t_7(x)=64x^6-80x^4+24x^2-1,& s_7(x)=128x^7-224x^5+112x^3-14x.\end{array}$$

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:

$${\lambda }^2-2x\lambda +1=0$$

with the roots α and β

$$\begin{array}{c}\alpha =x+\sqrt{x^2-1},\hfill \\ \beta =x-\sqrt{x^2-1.}\hfill \end{array}$$

Also, α and β satisfy the following equations:

$$\begin{array}{c}\alpha +\beta =2x,\hfill \\ \alpha \beta =1,\hfill \\ \alpha -\beta =\Delta =2\sqrt{x^2-1}.\hfill \end{array}$$

(1)

Binet form

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

$$t_n(x)=\frac{{\alpha }^n-{\beta }^n}{\alpha -\beta },$$

(2)

$$s_n(x)={\alpha }^n+{\beta }^n.$$

(3)

Generating function

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

$$\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}{t}_n(x)y^n=y{(1-2xy+y^2)}^{-1},\hfill \\ \sum _{n=0}^{\mathrm{\infty }}{s}_n(x)y^n=(2-2xy){(1-2xy+y^2)}^{-1}.\hfill \end{array}$$

Negative subscript

We can also extend the definition of $t_n(x)$ and $s_n(x)$ to the negative index

$$\begin{array}{c}{t}_{-n}(x)=-t_n(x),\hfill \\ s_{-n}(x)=s_n(x).\hfill \end{array}$$

Simson formulas

$$\begin{array}{c}{t}_{n+1}(x)t_{n-1}(x)-t_n^2(x)=-1,\hfill \\ s_{n+1}(x)s_{n-1}(x)-s_n^2(x)=4(x^2-1).\hfill \end{array}$$

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 [7]. Moreover,

$$\sum _{j=0}^{\lfloor \frac{n-1}{2}\rfloor }T(n,j)=n.$$

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

$$\sum _{j=0}^{3}T(8,j)=T(8,0)+T(8,1)+T(8,2)+T(8,3)=8.$$

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

$$\sum _{j=1}^{\lfloor \frac{n-1}{2}\rfloor }|T(n,j)|=P_n.$$

Table 1 The first ten coefficients of ${\mathit{t}}_{\mathit{n}}\mathbf{(}\mathit{x}\mathbf{)}$

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

$$t_{n+1}(x)-t_{n-1}(x)=s_n(x)=2x{t}_n(x)-2t_{n-1}(x),$$

(4)

$$\begin{array}{c}{s}_{n+1}(x)-s_{n-1}(x)=4(x^2-1)t_n(x),\hfill \\ t_n(x)s_n(x)=t_{2n}(x),\hfill \\ s_n^2(x)+4(x^2-1)t_n^2(x)=2s_{2n}(x),\hfill \\ s_n^2(x)-4(x^2-1)t_n^2(x)=4,\hfill \\ t_{n+1}^2(x)-t_n^2(x)=t_{2n+1}(x),\hfill \\ s_{n+1}^2(x)-s_n^2(x)=4(x^2-1)t_{2n+1}(x),\hfill \\ s_{n+1}^2(x)+s_n^2(x)=2x{s}_{2n+1}(x)+4,\hfill \\ t_{n+1}^2(x)-t_{n-1}^2(x)=2x{t}_{2n}(x),\hfill \\ s_n(x)s_{n+1}(x)-4(x^2-1)t_n(x)t_{n+1}(x)=4x,\hfill \\ s_n(x)s_{n+1}(x)+4(x^2-1)t_n(x)t_{n+1}(x)=2s_{2n+1}(x),\hfill \\ s_{4n}(x)-2=4(x^2-1)t_{2n}^2(x),\hfill \\ t_{n+1}(x)-x{t}_n(x)=\frac{1}{2}{s}_n(x),\hfill \\ 2s_{n+1}(x)-2x{s}_n(x)=4(x^2-1)t_n(x).\hfill \end{array}$$

(5)

Proposition 1 $s_n(2x^2-1)-s_n^2(x)=-2$.

Proof Consider the expression $s_n(2x^2-1)$. Then α, β, Δ are replaced by ${\alpha }^{\ast }$, ${\beta }^{\ast }$, ${\Delta }^{\ast }$, respectively. So, ${\alpha }^{\ast }={\alpha }^2$, ${\beta }^{\ast }={\beta }^2$, ${\Delta }^{\ast }=2x\Delta$ 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 ($k\ge 1$)

$$t_n^{(k)}(x)=t_{n+1}^{(k-1)}(x)-t_{n-1}^{(k-1)}(x),$$

(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,

$$\begin{array}{c}{t}_n^{(1)}(x)=s_n(x)\left(\text{by (4)}\right),\hfill \\ s_n^{(1)}(x)={\Delta }^2{t}_n(x)\left(\text{by (5)}\right)\hfill \end{array}$$

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

$$\begin{array}{c}{t}_n^{2j}(x)=s_n^{(2j-1)}(x)={\Delta }^{2j}{t}_n(x),\hfill \\ t_n^{(2j-1)}(x)=s_n^{(2j-2)}(x)={\Delta }^{2j-2}{s}_n(x).\hfill \end{array}$$

Some special values of $t_n(x)$ and $s_n(x)$

$$\begin{array}{c}\{\begin{array}{l}{t}_n(-x)={(-1)}^{n+1}{t}_n(x),\\ t_n(1)=n,\\ t_n(-1)={(-1)}^{n+1}n,\\ t_{2n}(0)=0,\\ t_{2n-1}(0)={(-1)}^{n+1},\end{array}\hfill \\ \{\begin{array}{l}{s}_n(1)=2,\\ s_n(-1)=2{(-1)}^n,\\ s_{2n}(0)=2{(-1)}^n,\\ s_{2n-1}(0)=0.\end{array}\hfill \end{array}$$

2.3 Differentiation formulas

$$\begin{array}{c}\frac{d{s}_n(x)}{dx}=2n{t}_n(x),\hfill \\ \frac{d{t}_n(x)}{dx}=\frac{n{s}_n(x)-2x{t}_n(x)}{2(x^2-1)},\hfill \\ \frac{d^2{s}_n(x)}{d{x}^2}=n\left(\frac{n{s}_n(x)-2x{t}_n(x)}{x^2-1}\right).\hfill \end{array}$$

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

$$\begin{array}{c}\frac{d^2{s}_n(x)}{d{x}^2}{|}_{{}_{x=1}}=\frac{2}{3}(n^4-n^2),\hfill \\ \frac{d^2{s}_n(x)}{d{x}^2}{|}_{{}_{x=-1}}=\frac{2}{3}{(-1)}^{n-1}(n^2-n^4).\hfill \end{array}$$

Proof If we take the limit on $s_n^{″}(x)=n(\frac{n{s}_n(x)-2x{t}_n(x)}{x^2-1})$, we have the numerical value of $s_n^{″}$ at $x=1$ and $x=-1$.

$$s_n^{″}(1)=\frac{n}{2}\underset{x\to 1}{lim}\frac{n{s}_n(x)-2x{t}_n(x)}{(x-1)}.$$

Since $s_n(1)=2$, $t_n(1)=n$, apply L’Hôpital’s rule:

$$\begin{array}{rcl}{s}_n^{″}(1)& =& \frac{n}{2}\underset{x\to 1}{lim}\frac{\frac{d}{dx}(n{s}_n(x)-2x{t}_n(x))}{\frac{d}{dx}(x-1)}\\ =& \frac{n}{2}\underset{x\to 1}{lim}\frac{d}{dx}(n{s}_n(x)-2x{t}_n(x))\\ =& \frac{n}{2}\underset{x\to 1}{lim}(2n^2{t}_n(x)-2t_n(x)-2x\frac{d}{dx}{t}_n(x))\\ =& \frac{n}{2}(2n^2{t}_n(1)-2t_n(1)-2\underset{x\to 1}{lim}\frac{d}{dx}{t}_n(x))\\ =& \frac{n}{2}(2n^3-2n)-\frac{1}{2}\underset{x\to 1}{lim}\frac{d}{dx}(2n{t}_n(x))\\ =& n^4-n^2-\frac{1}{2}{s}_n^{″}(1).\end{array}$$

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

2.4 Some summation formulas

In this section we deal with the matrix

$$\mathbf{V}=\left[\begin{array}{cc}2x& -1\\ 1& 0\end{array}\right].$$

By induction, we have

$${\mathbf{V}}^m=\left[\begin{array}{cc}{t}_{m+1}(x)& -t_m(x)\\ t_m(x)& -t_{m-1}(x)\end{array}\right].$$

(8)

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

$$\left[\begin{array}{c}{t}_{m+1}(x)\\ t_m(x)\end{array}\right]={\mathbf{V}}^m\left[\begin{array}{c}1\\ 0\end{array}\right]$$

(9)

and from (1.10) in [8], we get

$$t_m(x)=\left[\begin{array}{cc}1& 0\end{array}\right]{\mathbf{V}}^{m-1}\left[\begin{array}{c}1\\ 0\end{array}\right].$$

(10)

It is known that

$${\mathbf{V}}^{m+n}={\mathbf{V}}^m{\mathbf{V}}^n.$$

(11)

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

$$\begin{array}{c}{t}_{m+n+1}(x)=t_{m+1}(x)t_{n+1}(x)-t_m(x)t_n(x),\hfill \\ t_{m+n}(x)=t_{m+1}(x)t_n(x)-t_m(x)t_{n-1}(x),\hfill \\ t_{m+n-1}(x)=t_m(x)t_n(x)-t_{m-1}(x)t_{n-1}(x).\hfill \end{array}$$

If we use the matrix technique for summation in [8], we get the first finite summation as follows.

Proposition 3

• (i) ${\sum }_{n=1}^m{t}_n(x)=\frac{t_{m+1}(x)-t_m(x)-1}{2(x-1)}$,

• (ii) ${\sum }_{n=1}^m{s}_n(x)=\frac{s_{m+1}(x)-s_m(x)+2-2x}{2(x-1)}$.

Proof (i) Let the matrix A,

$$\mathbf{A}=\mathbf{I}+\mathbf{V}+{\mathbf{V}}^2+\cdots +{\mathbf{V}}^{n-2}+{\mathbf{V}}^{n-1}$$

be the series of matrices. Then we have

$$\mathbf{VA}=\mathbf{V}+{\mathbf{V}}^2+\cdots +{\mathbf{V}}^{n-1}+{\mathbf{V}}^n.$$

Hence,

$$\begin{array}{r}(\mathbf{V}-\mathbf{I})\mathbf{A}={\mathbf{V}}^n-\mathbf{I},\\ \mathbf{A}={(\mathbf{V}-\mathbf{I})}^{-1}({\mathbf{V}}^n-\mathbf{I})\\ \phantom{\mathbf{A}}=\frac{1}{2(x-1)}\left[\begin{array}{cc}{t}_{n+1}(x)-t_n(x)-1& t_{n-1}(x)-t_n(x)+1\\ t_n(x)-t_{n-1}(x)-1& t_{n-2}(x)-t_{n-1}(x)+2x-1\end{array}\right],\\ \sum _{n=1}^m{t}_n(x)=\left[\begin{array}{cc}1& 0\end{array}\right]\mathbf{A}\left[\begin{array}{c}1\\ 0\end{array}\right]\left(\text{by (10)}\right)\\ \phantom{\sum _{n=1}^m{t}_n(x)}=\frac{1}{2(x-1)}\left[\begin{array}{cc}{t}_{m+1}(x)-t_m(x)-1& t_{m-1}(x)-t_m(x)+1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \phantom{\sum _{n=1}^m{t}_n(x)}=\frac{1}{2(x-1)}[t_{m+1}(x)-t_m(x)-1].\end{array}$$

Thus, the proof is completed.

• (ii)
  $$\begin{array}{rcl}\sum _{n=1}^m{s}_n(x)& =& \sum _{n=1}^m({\alpha }^n+{\beta }^n)\left(\text{by (3)}\right)\\ =& \alpha \left(\frac{1-{\alpha }^m}{1-\alpha }\right)+\beta \left(\frac{1-{\beta }^m}{1-\beta }\right)\\ =& \frac{(\alpha +\beta )-2\alpha \beta -({\alpha }^{m+1}+{\beta }^{m+1})+\alpha \beta ({\alpha }^m+{\beta }^m)}{2-2x}\\ =& \frac{2x-2-({\alpha }^{m+1}+{\beta }^{m+1})+{\alpha }^m+{\beta }^m}{2(1-x)}\left(\text{by (1)}\right)\\ =& \frac{s_{m+1}(x)-s_m(x)+2-2x}{2(x-1)}\left(\text{by (3)}\right).\end{array}$$

This completes the proof. □

Theorem 1 Let V be a square matrix such that ${\mathbf{V}}^2=2x\mathbf{V}-\mathbf{I}$. Then, for all $n\in {\mathbb{Z}}^{+}$,

$${\mathbf{V}}^n=t_n(x)\mathbf{V}-t_{n-1}(x)\mathbf{I},$$

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

Proof The proof is obvious from induction. □