Some identities involving the bi-periodic Fibonacci and Lucas polynomials

Tingting Du; Zhengang Wu; Tingting Du; Zhengang Wu

doi:10.3934/math.2023294

AIMS Mathematics

2023, Volume 8, Issue 3: 5838-5846. doi: 10.3934/math.2023294

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Research article

Some identities involving the bi-periodic Fibonacci and Lucas polynomials

Tingting Du ,
Zhengang Wu ^,

Northwest University, Xian, Shaanxi 710000, China

Received: 02 November 2022 Revised: 04 December 2022 Accepted: 09 December 2022 Published: 26 December 2022
MSC : 11B37, 11B39

In this paper, by using generating functions for the Chebyshev polynomials, we have obtained the convolution formulas involving the bi-periodic Fibonacci and Lucas polynomials.

Keywords:

Citation: Tingting Du, Zhengang Wu. Some identities involving the bi-periodic Fibonacci and Lucas polynomials[J]. AIMS Mathematics, 2023, 8(3): 5838-5846. doi: 10.3934/math.2023294

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Abstract

In this paper, by using generating functions for the Chebyshev polynomials, we have obtained the convolution formulas involving the bi-periodic Fibonacci and Lucas polynomials.

1. Introduction

For any real number $x$ , the Fibonacci polynomials $\left \{ F_{n}\left (x \right) \right \}$ and Lucas polynomials $\left \{ L_{n}\left (x \right) \right \}$ are defined by the recurrence relations as follows:

$\begin{equation*} F_{0}\left ( x \right ) =0, \quad F_{1}\left ( x \right ) =1, \quad F_{n}\left ( x \right ) =xF_{n-1}\left ( x \right ) +F_{n-2}\left ( x \right ) , \quad n\ge 2, \end{equation*}$

and

$\begin{equation*} L_{0}\left ( x \right ) =2, \quad L_{1}\left ( x \right ) =x, \quad L_{n}\left ( x \right ) =xL_{n-1}\left ( x \right ) +L_{n-2}\left ( x \right ) , \quad n\ge 2. \end{equation*}$

For $x=1$ , the Fibonacci and Lucas polynomials are well known, respectively, Fibonacci sequences $\left \{ F_{n} \right \}$ and Lucas sequences $\left \{ L_{n} \right \}$ . The various properties of $\left \{ F_{n}\left (x \right) \right \}$ and $\left \{ L_{n}\left (x \right) \right \}$ have been investigated by many authors; see ^[1,2,3,4,5]. In particular, in ^[6,7,8] the authors established a series of connection formulaes between Fibonacci polynomials, Lucas polynomials and Chebyshev polynomials.

In ^[9], Yi and Zhang considered the convolution involving the Fibonacci polynomials:

$\begin{equation*} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k}=n }F_{a_{1}+1 } \left ( x \right )\cdot F_{a_{2}+1 } \left ( x \right )\cdots F_{a_{k}+1 } \left ( x \right ), \end{equation*}$

where the summation is over all k-dimension nonnegative integer coordinates $\left (a_{1}, a_{2}, \cdots, a_{k} \right)$ such that $a_{1}+a_{2} +\cdots +a_{k}=n$ , and $k$ is any positive integer.

In ^[10], Zhang obtained a series of identities that consists of the Fibonacci and Lucas sequences, by using generating functions for the second kind Chebyshev polynomials $\left \{ U_{n}\left (x \right) \right \}$ and their partial derivatives to prove the following:

$\begin{equation*} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }F_{m\left (a_{1}+1 \right ) } \cdot F_{m\left (a_{2}+1 \right ) }\cdots F_{m\left (a_{k+1}+1 \right ) }=\left ( -i \right )^{mn}\frac{F_{m}^{k+1} }{2^{k} \cdot k! }U_{n+k}^{\left ( k \right ) }\left ( \frac{i^{m} }{2}L_{m} \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} &\sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n+k+1 }L_{ma_{1} } \cdot L_{ma_{2} }\cdots L_{ma_{k+1} }\\ &=\left ( -i \right )^{m\left ( n+k+1 \right ) }\frac{2}{k!}\sum\limits_{h=0}^{k+1}\left ( \frac{i^{m+2} }{2} L_{m} \right )^{h}\binom{k+1}{h}U_{n+2k+1-h}^{\left ( k \right ) }\left ( \frac{i^{m} }{2} L_{m} \right ), \end{split} \end{equation*}$

where $k, m$ are any positive integers, $n, a_{1}, a_{2}, \cdots, a_{k+1}$ are nonnegative integers, $i$ is the square root of $-1$ , $U^{\left (k \right) }\left (x \right)$ denotes the $k$ -order derivative of $U\left (x \right)$ for $x$ , and $\binom{k+1}{h}=\frac{\left (k+1 \right)! }{h!\left (k+1-h \right)! }$ .

In addition, in ^[11], the author introduced the bi-periodic Fibonacci polynomials $\left \{f_{n} \left (x \right) \right \}$ , defined by

$\begin{equation} f_{0}\left ( x \right ) =0, \quad f_{1}\left ( x \right )=1, \quad {f_{n}\left ( x \right ) =} \begin{cases} axf_{n-1}\left ( x \right ) +f_{n-2}\left ( x \right ) , \quad & if\, n \, is \, even ; \\ bxf_{n-1}\left ( x \right ) +f_{n-2}\left ( x \right ) , \quad & if\, n \, is \, odd, \end{cases} \quad n\ge 2, \end{equation}$

(1.1)

where $a, b$ are any nonzero real numbers, and $x$ is any real numbers. The bi-periodic Lucas polynomials $\left \{l_{n} \left (x \right) \right \}$ are defined by

$\begin{equation} l_{0}\left ( x \right ) =2, \quad l_{1}\left ( x \right ) =ax, \quad {l_{n}\left ( x \right ) =} \begin{cases} bxl_{n-1}\left ( x \right ) +l_{n-2}\left ( x \right ) , \quad & if\, n \, is \, even ; \\ axl_{n-1}\left ( x \right ) +l_{n-2}\left ( x \right ) , \quad & if\, n \, is \, odd, \end{cases} \quad n\ge 2, \end{equation}$

(1.2)

where $a, b$ are any nonzero real numbers, and $x$ is any real numbers. For $x=1$ , the bi-periodic Fibonacci and Lucas polynomials are well known, respectively, bi-periodic Fibonacci and Lucas sequences.

Recently, in ^[12], Komatsu and Ramirez considered the convolution identities of order $2$ , $3$ and $4$ for the bi-periodic Fibonacci sequences $\left \{f_{n} \right \}$ are given with binomial cofficients. Other works related to convolved sequences can be found in ^{[13,14,15,16,17,18,19,20]}.

In this paper inspired by ^[10], we use the generating function of the first kind Chebyshev polynomials $\left \{ T_{n}\left (x \right) \right \}$ , the second kind Chebyshev polynomials $\left \{ U_{n}\left (x \right) \right \}$ and their partial derivative to study the following two theorems:

Theorem 1. Let $\left \{ f_{n}\left (x \right) \right \}$ be bi-periodic Fibonacci polynomials defined by (1.1), for any positive integer $k$ , nonnegative integer $n, a_{1}, a_{2}, \cdots, a_{k+1}$ . Then, we have the identity

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }&\left ( \frac{b}{a} \right )^{\frac{\xi \left ( a_{1} \right )+\xi \left ( a_{2} \right )+\cdots + \xi \left ( a_{k+1} \right )}{2} }f_{a_{1}+1 } \left ( x\right )\cdot f_{a_{2}+1 }\left ( x\right )\cdots f_{a_{k+1}+1 }\left ( x\right ) \\ & =\frac{1}{k!}\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( n+k-h \right )!\cdot i^{2h} }{h!\left ( n-2h \right )! }\left ( \sqrt{ab} x \right )^{n-2h}, \end{split} \end{equation*}$

where $\left \lfloor z \right \rfloor$ denotes the floor function, the greatest integer less than or equal to $z$ , $\xi \left (n \right)=n-\left \lfloor \frac{n}{2} \right \rfloor$ is the parity function, and $i$ is the square root of $-1$ .

Theorem 2. Let $\left \{ l_{n}\left (x \right) \right \}$ be bi-periodic Lucas polynomials defined by (1.2), for any positive integer $r$ , nonnegative integer $m, a_{1}, a_{2}, \cdots, a_{r+1}$ . Then, we have the identity

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}=m }&\left ( \frac{b}{a} \right )^{\frac{\xi \left ( a_{1} \right )+\xi \left ( a_{2} \right )+\cdots + \xi \left ( a_{r+1} \ \ \ \right )}{2} }l_{a_{1} }\left ( x\right )\cdot l_{a_{2} }\left ( x\right )\cdots l_{a_{r+1} }\left ( x\right ) \\ & =\frac{2^{r+1} \left ( 1-xt \right )^{r+1}\left ( m+r \right ) }{\left ( 1-t^{2} \right ) \cdot r!}\sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( m+r-h-1 \right )! \cdot i^{2h}}{h!\left ( m-2h \right )! }\left ( \sqrt{ab} x \right )^{m-2h}, \end{split} \end{equation*}$

From these two theorems we may immediately deduce the following corollaries:

Corollary 1. Let $\left \{ f_{n} \right \}$ be bi-periodic Fibonacci sequences and $\left \{ l_{n} \right \}$ be bi-periodic Lucas sequences, for any positive integers $k, r$ , and any nonnegative integers $n, m$ . Then, we have the following:

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }&\left ( \frac{b}{a} \right )^{\frac{\xi \left ( a_{1} \right )+\xi \left ( a_{2} \right )+\cdots + \xi \left ( a_{k+1} \ \ \ \right )}{2} }f_{a_{1}+1 }\cdot f_{a_{2}+1 }\cdots f_{a_{k+1}\ \ \ +1 } \\ & =\frac{1}{k!}\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( n+k-h \right )!\cdot i^{2h} }{h!\left ( n-2h \right )! }\left ( \sqrt{ab} \right )^{n-2h}, \end{split} \end{equation*}$

and

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}\ \ \ =m }&\left ( \frac{b}{a} \right )^{\frac{\xi \left ( a_{1} \right )+\xi \left ( a_{2} \right )+\cdots + \xi \left ( a_{r+1} \ \ \ \right )}{2} }l_{a_{1} }\cdot l_{a_{2} }\cdots l_{a_{r+1} }\\ & =\frac{2^{r+1} \left ( 1-t \right )^{r+1}\left ( m+r \right ) }{\left ( 1-t^{2} \right ) \cdot r!}\sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( m+r-h-1 \right )! \cdot i^{2h}}{h!\left ( m-2h \right )! }\left ( \sqrt{ab} \right )^{m-2h}, \end{split} \end{equation*}$

Corollary 2. Let $\left \{ F_{n} \left (x \right) \right \}$ be Fibonacci polynomials and $\left \{ L_{n} \left (x \right) \right \}$ be Lucas polynomials, for any positive integers $k, r$ , and any nonnegative integers $n, m$ . Then, we have the following:

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }&F_{a_{1}+1 }\left ( x\right )\cdot F_{a_{2}+1 }\left ( x\right )\cdots F_{a_{k+1}+1 }\left ( x\right ) \\ &=\frac{1}{k!}\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( n+k-h \right )! \cdot i^{2h} }{h!\left ( n-2h \right )! } x ^{n-2h}, \end{split} \end{equation*}$

and

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}=m }L_{a_{1} }&\left ( x\right )\cdot L_{a_{2} }\left ( x\right )\cdots L_{a_{r+1} }\left ( x\right )\\ &=\frac{2^{r+1} \left ( 1-xt \right )^{r+1}\left ( m+r \right ) }{\left ( 1-t^{2} \right ) \cdot r!}\sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( m+r-h-1 \right )!\cdot i^{2h}}{h!\left ( m-2h \right )! } x^{m-2h}, \end{split} \end{equation*}$

where $\left \lfloor z \right \rfloor$ denotes the floor function, the greatest integer less than or equal to $z$ , and $i$ is the square root of $-1$ .

Corollary 3. Let $\left \{ F_{n} \right \}$ be Fibonacci sequences and $\left \{ L_{n} \right \}$ be Lucas sequences, for any positive integers $k, r$ and any nonnegative integers $n, m$ . Then, we have the following:

$\begin{equation*} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }F_{a_{1}+1 }\cdot F_{a_{2}+1 }\cdots F_{a_{k+1}+1 } =\frac{1}{k!}\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( n+k-h \right )!\cdot i^{2h} }{h!\left ( n-2h \right )! }, \end{equation*}$

and

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}=m }&L_{a_{1} }\cdot L_{a_{2} }\cdots L_{a_{r+1} }\\ &=\frac{2^{r+1} \left ( 1-t \right )^{r+1}\left ( m+r \right ) }{\left ( 1-t^{2} \right ) \cdot r!}\sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( m+r-h-1 \right )! \cdot i^{2h}}{h!\left ( m-2h \right )! }, \end{split} \end{equation*}$

where $\left \lfloor z \right \rfloor$ denotes the floor function, the greatest integer less than or equal to $z$ , and $i$ is the square root of $-1$ .

2. Auxiliary results

In this section, we shall give several lemmas which are necessary in the proofs of the theorems. First, we introduce Chebyshev polynomials $\left \{ T_{n} \left (x \right) \right \}$ and $\left \{ U_{n} \left (x \right) \right \}$ . For any positive integer $n$ , the first kind Chebyshev polynomials $\left \{ T_{n}\left (x \right) \right \}$ are defined by

$\begin{equation*} T_{0}\left ( x \right ) =1, \quad T_{1}\left ( x \right ) =x, \quad T_{n}\left ( x \right ) =2xT_{n-1}\left ( x \right ) -T_{n-2}\left ( x \right ) , \quad n\ge 2, \end{equation*}$

and the second kind Chebyshev polynomials $\left \{ U_{n}\left (x \right) \right \}$ are defined by

$\begin{equation*} U_{0}\left ( x \right ) =1, \quad U_{1}\left ( x \right ) =2x, \quad U_{n}\left ( x \right ) =2xU_{n-1}\left ( x \right ) -U_{n-2}\left ( x \right ) , \quad n\ge 2. \end{equation*}$

The generating function $F\left (t, x\right)$ of the polynomials $\left \{ T_{n}\left (x \right) \right \}$ is given by

$\begin{equation} F\left ( t, x\right )=\sum\limits_{n=0}^{\infty }T_{n}\left ( x \right )t^{n} = \frac{1-xt}{1-2xt+t^{2} }, \quad \left | x \right | < 1, \quad \left | t \right | < 1, \end{equation}$

(2.1)

and the generating function $G\left (t, x\right)$ of the polynomials $\left \{ U_{n} \left (x \right) \right \}$ is given by

$\begin{equation} G\left ( t, x\right )=\sum\limits_{n=0}^{\infty }U_{n}\left ( x \right )t^{n} = \frac{1}{1-2xt+t^{2} }, \quad \left | x \right | < 1, \quad \left | t \right | < 1. \end{equation}$

(2.2)

Their general expressions are

$\begin{equation} T_{n}\left ( x \right )=\frac{n}{2}\sum\limits_{h=0}^{\left \lfloor \frac{n }{2} \right \rfloor }\frac{\left ( -1 \right )^{h}\left ( n-h-1 \right )! }{h!\left ( n-2h \right )! }\left ( 2x \right ) ^{n-2h}, \quad \left | x \right | < 1, \end{equation}$

(2.3)

and

$\begin{equation} U_{n}\left ( x \right )=\sum\limits_{h=0}^{ \left \lfloor \frac{n }{2} \right \rfloor }\frac{\left ( -1 \right )^{h}\left ( n-h \right )! }{h!\left ( n-2h \right )! }\left ( 2x \right ) ^{n-2h}, \quad \left | x \right | < 1. \end{equation}$

(2.4)

Lemma 1. ^[11] Let $\left \{ f_{n} \left (x \right) \right \}$ be bi-periodic Fibonacci polynomials, $\left \{ l_{n} \left (x \right) \right \}$ be bi-periodic Lucas polynomials, $\left \{ T_{n} \left (x \right) \right \}$ be Chebyshev polynomials of the first kind and $\left \{ U_{n} \left (x \right) \right \}$ be Chebyshev polynomials of the second kind. For any positive integer $n$ , we have the following identities:

$\begin{equation} f_{n+1}\left ( x \right )=\left ( \frac{a}{b} \right ) ^{\frac{\xi \left ( n \right ) }{2} } i^{n}U_{n}\left ( \frac{\sqrt{ab} }{2i} x \right ), \end{equation}$

(2.5)

and

$\begin{equation} l_{n}\left ( x \right )=2\left ( \frac{a}{b} \right ) ^{\frac{\xi \left ( n \right ) }{2} } i^{n}T_{n}\left ( \frac{\sqrt{ab} }{2i} x \right ), \end{equation}$

(2.6)

where $i$ is the square root of $-1$ , and $\xi \left (n \right)=n-\left \lfloor \frac{n}{2} \right \rfloor$ is the parity function.

Lemma 2. Let $\left \{ U_{n} \left (x \right) \right \}$ be Chebyshev polynomials of the second kind, and for any positive integer $k$ and nonnegative integer $n$ , we have the identity:

$\begin{equation*} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }U_{a_{1} }\left ( x \right )\cdot U_{a_{2} }\left ( x \right )\cdots U_{a_{k+1} }\left ( x \right )=\frac{1}{k!}\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor }\left ( -1 \right )^{h}\frac{\left ( n+k-h \right )! }{h!\left ( n-2h \right )! } \left ( 2x \right )^{n-2h}. \end{equation*}$

Proof. Noting that the degree of $U_{n} \left (x \right)$ has degree $n$ and taking the partial derivative $\left (\frac{\partial }{\partial x} \right)^{k}$ on both sides of (2.2), we have

$\begin{equation*} \frac{\partial^{k} G\left ( t, x \right ) }{\partial x^{k} }=\frac{\left ( 2t \right ) ^{k}\cdot k! }{\left ( 1-2xt+t^{2} \right )^{k+1}}=\sum\limits_{n=0}^{\infty }U_{n+k}^{\left ( k \right ) }\left ( x \right )t^{n+k}, \end{equation*}$

where $U^{\left (k \right) }\left (x \right)$ denotes the $k$ -order derivative of $U_{n}\left (x \right)$ for $x$ . Then, we obtain that

$\begin{equation} \begin{split} &\sum\limits_{n=0}^{\infty }\left ( \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n } U_{a_{1}}\left ( x \right ) \cdot U_{a_{2}}\left ( x \right )\cdots U_{a_{k+1}}\left ( x \right ) \right )t^{n}= \left (\sum\limits_{n=0}^{\infty } U_{n}\left ( x \right )t^{n} \right ) ^{k+1} \\ &= \frac{1}{\left ( 1-2xt+t^{2} \right )^{k+1}}=\frac{1}{\left ( 2t \right ) ^{k}\cdot k! } \frac{\partial^{k} G\left ( t, x \right ) }{\partial x^{k} }=\frac{1}{2^{k} \cdot k!}\sum\limits_{n=0}^{\infty } U_{n+k}^{\left ( k \right ) } \left ( x \right )t^{n}. \end{split} \end{equation}$

(2.7)

Comparing the coefficients of $t^{n}$ on both sides of Eq (2.7), we obtain that

$\begin{equation} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }U_{a_{1} }\left ( x \right ) \cdot U_{a_{2} }\left ( x \right )\cdots U_{a_{k+1} }\left ( x \right )=\frac{1}{2^{k}\cdot k!}U_{n+k}^{\left ( k \right ) }\left ( x \right ). \end{equation}$

(2.8)

From (2.4), we can deduce the $k^{th}$ derivative of $U_{n+k}\left (x \right)$ ,

$\begin{equation} U_{n+k}^{\left ( k \right ) } \left ( x \right )=\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor } \left ( -1 \right ) ^{h} \binom{n+k-h}{h}\left ( n+k-2h \right )_{k}2^{n+k-2h}x^{n-2h}, \end{equation}$

(2.9)

where the falling factorial polynomials $\left (x \right)_{n}$ are given by

$\begin{equation} \left ( x \right )_{0}=1, \quad \left ( x \right ) _{n} =x\left ( x-1 \right ) \cdots \left ( x-n+1 \right ), \quad n\ge 1 . \end{equation}$

(2.10)

Then, combining (2.8) and (2.9), we obtain

$\begin{equation*} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }U_{a_{1} }\left ( x \right )\cdot U_{a_{2} }\left ( x \right )\cdots U_{a_{k+1} }\left ( x \right )=\frac{1}{k!}\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor }\left ( -1 \right )^{h}\frac{\left ( n+k-h \right )! }{h!\left ( n-2i \right )! } \left ( 2x \right )^{n-2h}. \end{equation*}$

This completes the proof of the Lemma.

Lemma 3. Let $\left \{ T_{n} \left (x \right) \right \}$ be Chebyshev polynomials of the first kind. For any positive integer $r$ and nonnegative integer $m$ , we have the identity

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}=m}&T_{a_{1} }\left ( x \right ) \cdot T_{a_{2} }\left ( x \right )\cdots T_{a_{r+1} }\left ( x \right )\\ &=\frac{\left ( 1-xt \right )^{r+1} \left ( m+r \right ) }{\left ( 1-t^{2} \right )\cdot r! } \sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right ) ^{h}\frac{\left ( m+r-h-1 \right )! }{h!\left ( m-2h \right )! } \left ( 2x \right )^{m-2h} . \end{split} \end{equation*}$

Proof. Noting that the degree of $T_{m} \left (x \right)$ has degree $m$ and taking the partial derivative $\left (\frac{\partial }{\partial x} \right)^{r}$ on both sides of (2.1), we have

$\begin{equation*} \frac{\partial^{r} F\left ( t, x \right ) }{\partial x^{r} }=\frac{\left ( t-t^{3} \right ) \left ( 2t \right )^{r-1}\cdot r! }{\left ( 1-2xt+t^{2} \right )^{r+1}}=\sum\limits_{m=0}^{\infty }T_{m+r}^{\left ( r \right ) }\left ( x \right )t^{m+r}, \end{equation*}$

where $T^{\left (r \right) }\left (x \right)$ denotes the $r$ -order derivative of $T_{m}\left (x \right)$ for $x$ . Then, we obtain that

$\begin{equation} \begin{split} &\sum\limits_{m=0}^{\infty }\left ( \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}=m } T_{a_{1}}\left ( x \right ) \cdot T_{a_{2}}\left ( x \right )\cdots T_{a_{r+1}}\left ( x \right ) \right )t^{m}\\ &= \left (\sum\limits_{m=0}^{\infty } T_{m}\left ( x \right )t^{m} \right ) ^{r+1} = \frac{\left (1-xt \right )^{r+1} }{\left ( 1-2xt+t^{2} \right )^{r+1}}\\ &=\frac{\left ( 1-xt \right )^{r+1} }{\left ( t-t^{3} \right ) \left ( 2t \right ) ^{r-1}\cdot r! }\cdot \frac{\partial^{r} F\left ( t, x \right ) }{\partial x^{r} } =\frac{\left ( 1-xt \right )^{r+1} }{\left ( 1-t^{2} \right ) 2 ^{r-1}\cdot r! } \sum\limits_{m=0}^{\infty }T_{m+r}^{\left ( r \right ) }\left ( x \right )t^{m}. \end{split} \end{equation}$

(2.11)

Comparing the coefficients of $t^{m}$ on both sides of equation (2.11), we obtain that

$\begin{equation} \sum\limits_{a_{1}+a_{1}+\cdots +a_{r+1}=m}T_{a_{1} }\left ( x \right ) \cdot T_{a_{2} }\left ( x \right )\cdots T_{a_{r+1} }\left ( x \right )=\frac{\left ( 1-xt \right )^{r+1} }{\left ( 1-t^{2} \right ) 2^{r-1} \cdot r!}T_{m+r}^{\left ( r \right ) }\left ( x \right ). \end{equation}$

(2.12)

From (2.3), we can deduce the $r^{th}$ derivative of $T_{m+r}\left (x \right)$ ,

$\begin{equation} T_{m+r}^{\left ( r \right ) } \left ( x \right )=\frac{m+r}{2} \sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right ) ^{h}\frac{1}{m+r-h} \binom{m+r-h}{h}2^{m+r-2h} \left ( m+r-2h \right )_{r}x^{m-2h}. \end{equation}$

(2.13)

Then, combining (2.12) and (2.13), where the condition $\left (x \right)_{n}$ is defined by (2.10), we obtain

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}=m}&T_{a_{1} }\left ( x \right ) \cdot T_{a_{2} }\left ( x \right )\cdots T_{a_{r+1} }\left ( x \right )\\ &=\frac{\left ( 1-xt \right )^{r+1} \left ( m+r \right ) }{\left ( 1-t^{2} \right ) r! } \sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right ) ^{h}\frac{\left ( m+r-h-1 \right )! }{h!\left ( m-2h \right )! } \left ( 2x \right )^{m-2h} . \end{split} \end{equation*}$

This completes the proof of the Lemma.

3. Proof of the theorems

Proof of Theorem 1. By Lemma 2, we obtain that

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }&U_{a_{1} } \left ( \frac{\sqrt{ab} }{2i}x \right ) \cdot U_{a_{2} } \left ( \frac{\sqrt{ab} }{2i}x \right )\cdots U_{a_{k+1} } \left ( \frac{\sqrt{ab} }{2i}x \right )\\ & =\frac{1}{k!}\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( n+k-h \right )! }{h!\left ( n-2h \right )! }\left ( \frac{\sqrt{ab} }{i} x \right )^{n-2h}. \end{split} \end{equation*}$

By (2.5) of Lemma 1, we obtain that

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=n }&\frac{f_{a_{1}+1 }\left ( x \right )\cdot f_{a_{2}+1 }\left ( x \right )\cdots f_{a_{k+1}+1 }\left ( x \right )}{\left ( \frac{a}{b} \right )^{\frac{\xi \left ( a_{1} \right )+\xi \left ( a_{2} \right )+\cdots +\xi \left ( a_{k+1} \right ) }{2} }\cdot i^{a_{1}+a_{2}+\cdots +a_{k+1} } }\\ & =\frac{1}{k!}\sum\limits_{h=0}^{\left \lfloor \frac{n}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( n+k-h \right )! }{h!\left ( n-2h \right )! }\left ( \frac{\sqrt{ab} }{i} x \right )^{n-2h}. \end{split} \end{equation*}$

Therefore, we obtain

This completes the proof of the Theorem.

Proof of Theorem 2. By Lemma 3, we obtain that

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}=m }&T_{a_{1} } \left ( \frac{\sqrt{ab} }{2i}x \right )\cdot T_{a_{2} } \left ( \frac{\sqrt{ab} }{2i}x \right )\cdots T_{a_{r+1} } \left ( \frac{\sqrt{ab} }{2i}x \right )\\ &=\frac{\left ( 1-xt \right )^{r+1}\left ( m+r \right ) }{\left ( 1-t^{2} \right ) r!}\sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( m+r-h-1 \right )! }{h!\left ( m-2h \right )! }\left ( \frac{\sqrt{ab} }{i} x \right )^{m-2h}. \end{split} \end{equation*}$

By (2.6) of Lemma 1, we obtain that

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{k+1}=m }&\frac{l_{a_{1} }\left ( x \right )\cdot l_{a_{2} }\left ( x \right )\cdots l_{a_{r+1}}\left ( x \right )}{2^{r+1}\left ( \frac{a}{b} \right )^{\frac{\xi \left ( a_{1} \right )+\xi \left ( a_{2} \right )+\cdots +\xi \left ( a_{r+1} \right ) }{2} }\cdot i^{a_{1}+a_{2}+\cdots +a_{r+1} } }\\ & =\frac{\left ( 1-xt \right )^{r+1}\left ( m+r \right ) }{\left ( 1-t^{2} \right ) \cdot r!}\sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( m+r-h-1 \right )! }{h!\left ( m-2h \right )! }\left ( \frac{\sqrt{ab} }{i} x \right )^{m-2h}. \end{split} \end{equation*}$

Therefore, we obtain

$\begin{equation*} \begin{split} \sum\limits_{a_{1}+a_{2}+\cdots +a_{r+1}=m }&\left ( \frac{b}{a} \right )^{\frac{\xi \left ( a_{1} \right )+\xi \left ( a_{2} \right )+\cdots + \xi \left ( a_{r+1} \right )}{2} }l_{a_{1} }\left ( x\right )\cdot l_{a_{2} }\left ( x\right )\cdots l_{a_{r+1} }\left ( x\right ) \\ & =\frac{2^{r+1} \left ( 1-xt \right )^{r+1}\left ( m+r \right ) }{\left ( 1-t^{2} \right ) \cdot r!}\sum\limits_{h=0}^{\left \lfloor \frac{m}{2} \right \rfloor } \left ( -1 \right )^{h}\frac{\left ( m+r-h-1 \right )! i^{2h}}{h!\left ( m-2h \right )! }\left ( \sqrt{ab} x \right )^{m-2h}. \end{split} \end{equation*}$

This completes the proof of the Theorem.

4. Conclusions

In this paper, by using generating functions for the Chebyshev polynomials, we have obtained the convolution formulas involving the bi-periodic Fibonacci and Lucas polynomials. In the past, scholars considered the convolution of linear recursive polynomials. In this paper, we extend the previous research to non-linear recursive polynomials. Specifically, we consider the convolution formula of bi-periodic recursive polynomials. Furthermore, we hope to consider extending the convolution formula of t-periodic recursive polynomials.

Acknowledgments

The authors express their gratitude to the referee for very helpful and detailed comments. Supported by the National Natural Science Foundation of China (Grant No. 11701448)

Conflicts of interest

All authors declare no conflicts of interest in this paper.

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AIMS Mathematics

Some identities involving the bi-periodic Fibonacci and Lucas polynomials

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Abstract

1. Introduction

2. Auxiliary results

3. Proof of the theorems

4. Conclusions

Acknowledgments

Conflicts of interest

References

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AIMS Mathematics

Some identities involving the bi-periodic Fibonacci and Lucas polynomials

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Abstract

1. Introduction

2. Auxiliary results

3. Proof of the theorems

4. Conclusions

Acknowledgments

Conflicts of interest

References

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