On a unified approach to homogeneous second-order linear difference equations with constant coefficients and some applications

Issam Kaddoura; Bassam Mourad

doi:10.1515/spma-2023-0115

Article Open Access

On a unified approach to homogeneous second-order linear difference equations with constant coefficients and some applications

Issam Kaddoura and Bassam Mourad

Published/Copyright: June 6, 2024

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From the journal Special Matrices Volume 12 Issue 1

Abstract

In this article, we establish a new closed formula for the solution of homogeneous second-order linear difference equations with constant coefficients by using matrix theory. This, in turn, gives new closed formulas concerning all sequences of this type such as the Fibonacci and Lucas sequences. Next, we show the main advantage of our formula which is based on the fact that future calculations rely on the previous ones and this is true from any desired starting point. As applications, we show that Binet’s formula, in this case, is valid for negative integers as well. Finally, new summation formulas relating elements of such sequences. As a conclusion, we present a formula for the sum of squares of Chebychev polynomials of the first and second kind.

Keywords: difference equations; Fibonacci sequence; Lucas sequence; Pell sequence; Pell-Lucas sequence; Jacobsthal sequence; Jacobsthal-Lucas sequence; Chebychev polynomials; Fibonacci polynomials; Lucas polynomials; Generalized Binet’s formula

MSC 2010: 11B39; 65Q30

1 Introduction

For any ( n , z ) ∈ Z × C , where Z is the usual ring of integers and C is the field of complex numbers, we call a generalized linear second-order recurrent sequence, any sequence which is given by the following second-order linear difference equation:

(1) R n + 1 ( z ) = f ( z ) R n ( z ) + g ( z ) R n − 1 ( z ) , R 0 ( z ) = h ( z ) , R 1 ( z ) = k ( z ) ,

where f , g , h , and k are any complex functions^[1]. Without loss of generality, we may assume that f ( z ) ≠ 0 and g ( z ) ≠ 0 since otherwise we obtain one trivial case for f ( z ) = g ( z ) = 0 , and two other cases that are easy to handle; one corresponds to f ( z ) = 0 , g ( z ) ≠ 0 , and the other is associated with f ( z ) ≠ 0 , g ( z ) = 0 . In addition, for convenience, we may sometimes use the notation: R n = R n ( z ) = R ( n , f , g , h , k ) to refer to the nth term of such a sequence, and we omit the argument from the functional notation when there is no ambiguity; so f , for example, will mean f ( z ) . Our intention here is to study this sequence assuming that f , g , h , and k are any complex functions.

Certainly, there is extensive work in the literature concerning linear second-order difference equations and their applications [8,9]. However, only a few deals with finding a general framework under which all these sequences come together under one theory [1,2,5,10,11]. Our main objective in this note is to deal with this issue subject to the extra condition that n is any integer in Z . A different approach to the very same problem appears in the following papers [3,4,6].

This article is organized as follows. In Section 2, we present a new closed formula solution for the sequence defined by (1), which, in turn, gives new closed formulas for many sequences of this type such as Fibonacci, Lucas, Pell, Pell-Lucas, Jacobsthal, and Jacobsthal-Lucas number sequences as well as Tchebychev, Fibonacci, and Lucas polynomials. Next, we show the main advantage of our formula, which is based on the fact that future calculations rely on the previous ones, and this is true from any desired starting point. Section 3 deals with showing that Binet’s formula in this case is valid for negative integers as well. Finally, new summation formulas relating elements of the sequence { R n } n are presented in Section 4. As a conclusion, we present a formula for the sum of squares of Chebychev polynomials of the first and second kind.

2 Main results

We shall start by fixing some notations. The determinant of a square matrix X will be denoted by ∣ X ∣ . For our purposes, we next introduce the following four matrices, which will be used throughout this article, and they constitute the essential foundations of our main results. Define

A = 2 g h + f 2 h − f k 2 k g − f h g 2 k − f h 2 g h + f k , B = 0 g 1 f , C = 2 g f g f f 2 + 2 g , and D = − f 2 g 2 f ,

where f , g , h , and k are as above. Now, a simple check shows that matrices A , B , C , and D are pairwise commuting.

As a result, we have the following useful lemma that relates matrices A , B , C , and D to only R n and R n − 1 so that it gives the interesting consequence that solving a second-order linear homogeneous difference equation that has constant coefficients and which is coupled with initial conditions is essentially equivalent to solving a non-homogeneous linear first-order difference equation.

Lemma 2.1

Let matrices A, B, C, and D be defined as above and let R n be the general term of the sequence defined by (1). Then, for any integer n, we have

(2) A B n = C R n + D g R n − 1 .

Proof

Using induction, we first do the proof for the case n ≥ 1 . It is clear that (2) is true for n = 1 since A B = C R 1 + D g R 0 as an inspection shows that

2 g h + f 2 h − f k 2 k g − f h g 2 k − f h 2 g h + f k 0 g 1 f = 2 g f g f f 2 + 2 g k + − f 2 g 2 f g h .

Now, suppose that (2) is true for n , then we can write

2 g h + f 2 h − f k 2 k g − f h g 2 k − f h 2 g h + f k 0 g 1 f n = 2 g f g f f 2 + 2 g R n + − f 2 g 2 f g R n − 1 .

Multiplying both sides of this last equation to the right by B = 0 g 1 f , we obtain the following equality:

A B n + 1 = ( C R n + D g R n − 1 ) B = f g R n + 2 g 2 R n − 1 g ( 2 g + f 2 ) R n + f g 2 R n − 1 ( f 2 + 2 g ) R n + f g R n − 1 f ( f 2 + 3 g ) R n + g ( 2 g + f 2 ) R n − 1 = 2 g f g f f 2 + 2 g ( f R n + g R n − 1 ) + − f 2 g 2 f g R n = C R n + 1 + D g R n .

Next, for the case n = 0 , it is easy to see that the sequence defined by (1) becomes R − 1 = g − 1 ( k − f h ) . Therefore, we can easily see that A = C h + D ( k − f h ) so that (2) is also valid for n = 0 . Moreover, as B − 1 = − f g 1 1 g 0 , then a simple check shows that A B − 1 = C R − 1 + D g R − 2 so that (2) is also true for n = − 1 as well. Finally, the proof for the case n < − 1 can also be done by induction in a similar manner as in the first case, and the proof is complete.□

Next, note that an inspection shows that the following determinant formulas are valid:

∣ A ∣ = ( f 2 + 4 g ) ( g h 2 − k 2 + f h k )

and

∣ C R n + D g R n − 1 ∣ = g ( f 2 + 4 g ) ( R n 2 − g R n − 1 2 − f R n R n − 1 ) .

As a conclusion, we have the following lemma that also appears as Theorem 5 in [11] for z real, albeit arrived at by different means.

Lemma 2.2

Let R n , f , g , h , and k be as given above. Then, it holds that

(3) R n 2 − g R n − 1 2 − f R n R n − 1 = ( − g h 2 + k 2 − f h k ) ( − g ) n − 1 ,

for all integers n.

Proof

We divide the proof into two cases.

Case 1 : ̲ If f 2 + 4 g = 0 , then we obtain that R n + 1 = f R n − f 2 4 R n − 1 . From this last formula, we can easily find that R 2 = 2 ( 2 k − f h ) + f h 2 2 f and R 3 = 3 ( 2 k − f h ) + f h 2 3 f 2 and hence by induction we can easily show that R n = n ( 2 k − f h ) + f h 2 n f n − 1 for n ≥ 2 . With this in mind, we next use these formulas of R n , R n − 1 together with g = − f 2 4 to prove that (3) is valid. That in turn means that we need to prove that R n 2 − g R n − 1 2 − f R n R n − 1 − ( − g h 2 + k 2 − f h k ) ( − g ) n − 1 = 0 . Indeed, an inspection shows that R n 2 − g R n − 1 2 − f R n R n − 1 − ( − g h 2 + k 2 − f h k ) ( − g ) n − 1 = ( − 2 k + f h ) 2 f 2 n − f 2 n 2 2 n f 2 = 0 , and thus (3) holds.

Case 2 : ̲ If f 2 + 4 g ≠ 0 , then taking determinants of both sides of (2), we obtain ∣ A B n ∣ = ∣ C R n + D g R n − 1 ∣ or

2 g h + f 2 h − f k 2 g k − f g h 2 k − f h 2 g h + f k 0 g 1 f n = 2 g f g f f 2 + 2 g R n + − f 2 g 2 f g R n − 1 .

Using the fact that ∣ X Y ∣ = ∣ X ∣ ∣ Y ∣ for any square matrices of the same size, we obtain

( f 2 + 4 g ) ( g h 2 − k 2 + f h k ) ( − g ) n = g ( f 2 + 4 g ) ( R n 2 − g R n − 1 2 − f R n R n − 1 ) .

Therefore, (3) is valid.□

As a consequence, we are now able to present the following result.

Theorem 2.3

For any integer n and for any complex functions f ≠ 0 and g ≠ 0 , the general term R n of the recurrence R n + 1 = f R n + g R n − 1 , with seeds R 0 = h , R 1 = k , satisfies one of the following:

If f 2 + 4 g ≠ 0 and g h 2 − k 2 + f h k ≠ 0 , then
(4) R n 2 = ( − g ) n ( g h 2 − k 2 + f h k ) ( f 2 + 4 g ) ∣ M + ( − g ) n B − 2 n ∣ ,
where
M = g h 2 + f 2 h 2 + k 2 − 2 f h k g h 2 − k 2 + f h k g h 2 k − f h g h 2 − k 2 + f h k h 2 k − f h g h 2 − k 2 + f h k g h 2 + k 2 g h 2 − k 2 + f h k a n d B = 0 g 1 f .
If f 2 + 4 g = 0 then we obtain R n + 1 = f R n − f 2 4 R n − 1 , which implies that
(5) R n = n ( 2 k − f h ) + f h 2 n f n − 1 .
If, in addition, g h 2 − k 2 + f h k = 0 , then R n = h f 2 n .
If f 2 + 4 g ≠ 0 and g h 2 − k 2 + f h k = 0 , then we obtain the following closed formula: R n = k n h n − 1 .

Proof

Case 1 : First notice that the right-hand side of (4) can be rewritten as

RHS = ( − g ) n ( g h 2 − k 2 + f h k ) ( f 2 + 4 g ) ∣ B − n ∣ ∣ M B n + ( − g ) n B − n ∣ = g h 2 − k 2 + f h k ( f 2 + 4 g ) ∣ M B n + ( − g ) n B − n ∣ ( since ( − g ) n ∣ B − n ∣ = 1 ) .

Now, from equality (2), we know that B n = A − 1 ( C R n + D g R n − 1 ) . Finding A − 1 and multiplying, we obtain

B n = g h R n − k R n − 1 g h 2 − k 2 + f h k g − k R n + f h R n + g h R n − 1 g h 2 − k 2 + f h k − k R n + f h R n + g h R n − 1 g h 2 − k 2 + f h k g h R n − f k R n + f 2 h R n − g k R n − 1 + f g h R n − 1 g h 2 − k 2 + f h k .

As a result, we obtain

M B n = g h R n + k R n − 1 − f h R n − 1 g h 2 − k 2 + f h k g k R n + g h R n − 1 g h 2 − k 2 + f h k k R n + g h R n − 1 g h 2 − k 2 + f h k g h R n + f k R n + g k R n − 1 g h 2 − k 2 + f h k = 1 g h 2 − k 2 + f h k g ( h R n + k R n − 1 − f h R n − 1 ) g ( k R n + g h R n − 1 ) k R n + g h R n − 1 g h R n + f k R n + g k R n − 1 .

Similarly, using Lemma 2.2, we can write

( − g ) n B − n = ( − g ) n ( C R n + D g R n − 1 ) − 1 A = ( − g ) n g h R n − f k R n + f 2 h R n − g k R n − 1 + f g h R n − 1 g ( − g R n − 1 2 + R n 2 − f R n R n − 1 ) − k R n + f h R n + g h R n − 1 g R n − 1 2 − R n 2 + f R n R n − 1 − k R n + f h R n + g h R n − 1 g ( g R n − 1 2 − R n 2 + f R n R n − 1 ) h R n − k R n − 1 − g R n − 1 2 + R n 2 − f R n R n − 1 = g ( g h 2 − k 2 + f h k ) ( g h − f k + f 2 h ) R n − ( g k − f g h ) R n − 1 g ( k − f h ) R n − g h R n − 1 ( k − f h ) R n − g h R n − 1 g h R n − k R n − 1 .

Substituting these two expressions of B n and ( − g ) n B − n , we obtain an expression that involves a determinant of sum of two matrices. More explicitly,

RHS = ( g h 2 − k 2 + f h k ) ( f 2 + 4 g ) ( g h 2 − k 2 + f h k ) 2 g ( h R n + k R n − 1 − f h R n − 1 ) g ( k R n + g h R n − 1 ) k R n + g h R n − 1 g h R n + f k R n + g k R n − 1 + g g h R n − f k R n + f 2 h R n − g k R n − 1 + f g h R n − 1 g k R n − f h R n − g h R n − 1 k R n − f h R n − g h R n − 1 g h R n − k R n − 1 .

After arranging terms and simplifying by g h 2 − k 2 + f h k , we obtain

RHS = 1 ( f 2 + 4 g ) ( g h 2 − k 2 + f h k ) R n ( f 2 h + 2 g h − f k ) g R n ( 2 k − f h ) R n ( 2 k − f h ) R n ( 2 g h + f k ) = R n 2 .

This completes the proof of the first case.

Case 2 : (5) can be easily obtained by a direct substitution. Now, if g h 2 − k 2 + f h k = 0 , then solving for k , we obtain k = f h 2 . Substituting in (5), the proof can be easily completed.

Case 3 : Observe first that g h 2 − k 2 + f h k = 0 , which implies that k = h f ± f 2 + 4 g 2 . Moreover, (3) implies that

R n 2 − g R n − 1 2 − f R n R n − 1 = ( − g h 2 + k 2 − f h k ) ( − g ) n − 1 = 0 .

As a result, we obtain R n = f ± f 2 + 4 g 2 R n − 1 , and hence R n = k h R n − 1 . Thus, R n = k n h n − 1 . □

Remark

It is worthy to note that formulas in Cases 2 and 3 seem to follow from elementary results on linear difference equations, see [6] for similar formulas.

An immediate consequence of the preceding theorem is the following corollary which gives new closed formulas for the following sequences; see, for example, [7–9] and references therein for more details on such sequences.

Corollary 2.4

The following new closed formulas hold.

If f = x , g = 1 , h = 0 , and k = 1 , we obtain the Fibonacci polynomials:
f n ( x ) = ( − 1 ) n + 1 4 + x 2 − 1 0 0 − 1 + − ( 1 + x 2 ) x x − 1 n .
For f = 1 , g = 1 , h = 0 , and k = 1 , the Fibonacci numbers are given by:
F n = ( − 1 ) n + 1 5 − 1 0 0 − 1 + − 2 1 1 − 1 n .
If f = x , g = 1 , h = 2 , and k = x , we obtain the Lucas polynomials:
l n ( x ) = ( − 1 ) n 1 0 0 1 + − ( 1 + x 2 ) x x − 1 n .
For f = 1 , g = 1 , h = 2 , and k = 1 , the Lucas numbers are given by
L n = ( − 1 ) n 1 0 0 1 + − 2 1 1 − 1 n .
If f = 1 , g = 2 x , h = 1 , and k = 1 , we obtain the Jacobsthal polynomials:
J n ( x ) = ( − 1 ) n ( 2 x ) n + 1 1 + 8 x 1 1 1 2 x 1 + 1 2 x + − 1 2 x − 1 1 1 2 x − 1 n .
If f = 1 , g = 2 , h = 0 , and k = 1 , we obtain the Jacobsthal numbers:
J n = ( − 2 ) n − 9 − 1 0 0 − 1 + − 3 ∕ 2 1 1 ∕ 2 − 1 n .
For f = 1 , g = 2 x , h = 2 , and k = 1 , we obtain the Jacobsthal-Lucas polynomials:
j n ( x ) = ( − 2 x ) n 1 0 0 1 + − 1 2 x − 1 1 1 2 x − 1 n .
For f = 1 , g = 2 , h = 2 , and k = 1 , we obtain the Jacobsthal-Lucas numbers:
j n = ( − 2 ) n 1 0 0 1 + − 3 ∕ 2 1 1 ∕ 2 − 1 n .
If f = 2 , g = 1 , h = 2 , and k = 2 , we obtain the Pell-Lucas numbers:
Q n = ( − 1 ) n 1 0 0 1 + − 5 2 2 − 1 n .
For f = 2 , g = 1 , h = 0 , and k = 1 , we obtain the Pell numbers:
P n = ( − 1 ) n + 1 8 − 1 0 0 − 1 + − 5 2 2 − 1 n .
For f = 2 x , g = − 1 , h = 1 , and k = x , we obtain the Chebychev polynomial of the first kind:
T n ( x ) = 1 4 1 0 0 1 + 4 x 2 − 1 2 x − 2 x − 1 n .

3 Generalized Binet formula

The main objective in this section is to show that Binet’s formula for the difference equation (1) is also valid for negative integers. But first, we need to introduce some notations. Let P be the matrix which is defined by

P ≔ 1 1 f + f 2 + 4 g 2 g f − f 2 + 4 g 2 g .

Now, it is easy to check that the inverse P − 1 of P is given by

P − 1 = − f + f 2 + 4 g 2 f 2 + 4 g g f 2 + 4 g f + f 2 + 4 g 2 f 2 + 4 g − g f 2 + 4 g .

In addition, P diagonalizes all matrices A , B , C , and D . More explicitly, an inspection shows that

P − 1 B P = 1 2 ( f + f 2 + 4 g ) 0 0 1 2 ( f − f 2 + 4 g ) ,
P − 1 D P = f 2 + 4 g 0 0 − f 2 + 4 g ,
P − 1 C P = 1 2 ( f 2 + 4 g + f 2 + 4 g ) 0 0 1 2 ( f 2 + 4 g − f 2 + 4 g ) ,
P − 1 A P = 1 2 ( h ( f 2 + 4 g ) + ( 2 k − f h ) f 2 + 4 g ) 0 0 1 2 ( h ( f 2 + 4 g ) − ( 2 k − f h ) f 2 + 4 g ) .

As a result, we have the following conclusion that ensures the validity of Binet’s formula for negative integers.

Corollary 3.1

For any integer n, it holds that

R n = h 2 + 2 k − f h 2 f 2 + 4 g f + f 2 + 4 g 2 n + h 2 − 2 k − f h 2 f 2 + 4 g f − f 2 + 4 g 2 n .

Proof

From (2), we can easily obtain that P − 1 A P P − 1 B n P = P − 1 C P R n + P − 1 D P g R n − 1 . After substitution and equating terms in the preceding matrix equation, we obtain the following system:

1 2 ( h ( f 2 + 4 g ) + ( 2 k − f h ) f 2 + 4 g ) f + f 2 + 4 g 2 n = R n 2 ( f 2 + 4 g + f f 2 + 4 g ) + g R n − 1 2 f 2 + 4 g 1 2 ( h ( f 2 + 4 g ) − ( 2 k − f h ) f 2 + 4 g ) f − f 2 + 4 g 2 n = R n 2 ( f 2 + 4 g − f f 2 + 4 g ) − g R n − 1 2 f 2 + 4 g .

The proof can be easily achieved by adding the two equations in the preceding system.□

Next, we shall present one of the main advantages of our formula, which shows that future calculations rely on the previous ones and this is true from any desired starting point.

Theorem 3.2

For any integer n with f ≠ 0 and g ≠ 0 , the general term R n of the recurrence R n + 1 = f R n + g R n − 1 , with seeds R 0 = h , R 1 = k , satisfies the following functional relations:

R n + m ( R 0 , R 1 ) = R m ( R n , R n + 1 ) = R n ( R m , R m + 1 ) .

Proof

By substituting the value of

R m = h 2 + 2 k − f h 2 f 2 + 4 g f + f 2 + 4 g 2 m + h 2 − 2 k − f h 2 f 2 + 4 g f − f 2 + 4 g 2 m

and that of

R m + 1 = h 2 + 2 k − f h 2 f 2 + 4 g f + f 2 + 4 g 2 m + 1 + h 2 − 2 k − f h 2 f 2 + 4 g f − f 2 + 4 g 2 m + 1

in the following formula of R n ( R m , R m + 1 ) , we obtain

R n ( R m , R m + 1 ) = R m 2 + 2 R m + 1 − f R m 2 f 2 + 4 g f + f 2 + 4 g 2 n + R m 2 − 2 R m + 1 − f R m 2 f 2 + 4 g f − f 2 + 4 g 2 n = h 2 + 2 k − f h 2 f 2 + 4 g f + f 2 + 4 g 2 m + n + h 2 − 2 k − f h 2 f 2 + 4 g f − f 2 + 4 g 2 m + n = R n + m ( h , k ) ,

which is the required result.□

4 Summation relations satisfied by the elements R i of sequence (1)

In this section, we present new summation relations that the elements R i of sequence (1) satisfy.

We start with the following result which can be proven by using only the definition of sequence (1).

Theorem 4.1

For any integer n and for any complex functions f , g , h , and k , the following two equations, concerning the elements R i of sequence (1), are valid:

( f 2 − g 2 + 2 g − 1 ) ∑ i = 1 n R i 2 = ( R n + 1 2 − k 2 ) − g 2 ( R n 2 − h 2 ) + 2 S g , f ( f 2 − g 2 + 2 g − 1 ) ∑ i = 1 n R i R i − 1 = ( 1 − g ) ( R n + 1 2 − k 2 ) + g ( g − 1 + f 2 ) ( R n 2 − h 2 ) + S ( 1 − f 2 − g 2 ) ,

where

S = ( k 2 − g h 2 − f h k ) 1 − ( − g ) n 1 + g if g ≠ − 1 n ( k 2 − g h 2 − f h k ) if g = − 1 .

Proof

It is easy to see that

(6) ∑ i = 1 n R i + 1 2 = ∑ i = 1 n ( f R i + g R i − 1 ) 2 = f 2 ∑ i = 1 n R i 2 + g 2 ∑ i = 1 n R i − 1 2 + 2 f g ∑ i = 1 n R i R i − 1 .

Now, if we let x n = ∑ i = 1 n R i + 1 2 and y n = ∑ i = 1 n R i R i − 1 , then it is easy to see that we can write ∑ i = 1 n R i 2 = x n + k 2 − R n + 1 2 and ∑ i = 1 n R i − 1 2 = x n + h 2 + k 2 − R n 2 − R n + 1 2 . Substituting these in (6) and then rearranging the terms, we obtain

(7) x n ( 1 − f 2 − g 2 ) − 2 f g y n = f 2 ( k 2 − R n + 1 2 ) + g 2 ( h 2 + k 2 − R n 2 − R n + 1 2 ) .

Making use of (3) of Lemma 2.2 in (6), we obtain

∑ i = 1 n R i 2 − g ∑ i = 1 n R i − 1 2 − f ∑ i = 1 n R i R i − 1 = ∑ i = 1 n ( k 2 − g h 2 − f h k ) ( − g ) i − 1 .

After arranging terms, we obtain

(8) ( 1 − g ) x n − f y n = S − ( k 2 − R n + 1 2 ) + g ( h 2 + k 2 − R n 2 − R n + 1 2 ) ,

where

S = ∑ i = 1 n ( k 2 − g h 2 − f h k ) ( − g ) i − 1 = ( k 2 − g h 2 − f h k ) 1 − ( − g ) n 1 + g if g ≠ − 1 n ( k 2 − g h 2 − f h k ) if g = − 1 .

Solving system (7) and (8) for x n and y n , we then obtain two equations where the first involves x n and is given by

( f − g + 1 ) ( f + g − 1 ) x n = ( f 2 + 2 g − g 2 ) ( R n + 1 2 − k 2 ) − g 2 ( R n 2 − h 2 ) + 2 S g ,

and the second equation involves only y n and is given by

f ( f − g + 1 ) ( f + g − 1 ) y n = ( 1 − g ) ( R n + 1 2 − k 2 ) + g ( g − 1 + f 2 ) ( R n 2 − h 2 ) + S ( 1 − f 2 − g 2 ) .

This completes the proof.□

Of particular interest is the application of the preceding theorem on Chebychev polynomials of the first and second kind. First, recall that the Chebychev polynomials of the first kind T n ( x ) are defined by the following recurrence relation:

T n + 1 ( x ) = 2 x T n ( x ) − T n − 1 ( x ) , T 0 ( x ) = 1 , T 1 ( x ) = x ,

and the Chebychev polynomials of the second kind U n ( x ) are defined by the following recurrence relation:

U n + 1 ( x ) = 2 x U n ( x ) − U n − 1 ( x ) , U 0 ( x ) = 1 , U 1 ( x ) = 2 x .

By applying the previous theorem, we obtain the following conclusion.

Corollary 4.2

For Chebychev polynomials of the first kind T n ( x ) , we have the following summation formulas:

∑ i = 1 n T i 2 = ( T n + 1 2 − T n 2 ) − ( 1 − 2 n ) ( x 2 − 1 ) 4 ( x 2 − 1 ) ,

∑ i = 1 n T i T i − 1 = T n + 1 2 + ( 1 − 2 x 2 ) T n 2 − 2 n x 2 ( 1 − x 2 ) + x 2 − 1 4 x ( x − 1 ) ( x + 1 ) .

Similarly, for Chebychev polynomials of the second kind U n ( x ) , we have the following summation formulas:

∑ i = 0 n U i 2 = U n + 1 2 − U n 2 − 2 n − 3 4 ( x 2 − 1 ) , ∑ i = 1 n U i − 1 U i = U n + 1 2 + ( 1 − 2 x 2 ) U n 2 − 2 ( n + 1 ) x 2 − 1 4 x ( x − 1 ) ( x + 1 ) .

Another different relation involving products of elements of sequence (1), is given in the next theorem (for similar results of this kind see [11]).

Theorem 4.3

For any integers n and i in Z , the elements R n of sequence (1) satisfy the following:

(9) ( k 2 − f k h − g h 2 ) R i + n + ( f 2 h − f k + g h ) R i R n + h g 2 R i − 1 R n − 1 + ( f h − k ) g ( R n R i − 1 + R i R n − 1 ) = 0 .

Proof

Since A and B commute, for any integers i and n , we obtain the matrix equation A B n A B i = A 2 B n + i . By (2), this last equation can be rewritten as

( C R n + D g R n − 1 ) ( C R i + D g R i − 1 ) = A ( C R n + i + D g R n + i − 1 ) .

From this matrix equation, we obtain the following two equations:

g ( 4 g + f 2 ) ( R i R n − h R n + i − k R n + i − 1 + g R i − 1 R n − 1 + f h R n + i − 1 ) = 0

g ( 4 g + f 2 ) ( − k R n + i + g R n R i − 1 + g R i R n − 1 + f R i R n − g h R n + i − 1 ) = 0 .

Since 4 g + f 2 ≠ 0 , we conclude the following two equations:

g h ( R i R n − h R n + i + g R i − 1 R n − 1 + ( f h − k ) R n + i − 1 ) = 0 ,

( f h − k ) ( − k R n + i + g R n R i − 1 + g R i R n − 1 + f R i R n − g h R n + i − 1 ) = 0 .

Finding the value of ( f h − k ) R n + i − 1 from the first equation and substituting it in the second equation, we obtain (9).□

Acknowledgements

The authors sincerely thank the handling editor and the reviewers for their constructive reports and for their valuable suggestions and many useful comments. Issam Kaddoura is supported by Lebanese International University.

Author contributions: Both authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results and approved the final version of the manuscript. The authors confirm contribution to the paper as follows: study conception and design: I. Kaddoura, B. Mourad; analysis and interpretation of results: I. Kaddoura, B. Mourad; draft manuscript preparation: I. Kaddoura, B. Mourad.
Conflict of interest: Authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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Received: 2023-07-04

Revised: 2024-01-26

Accepted: 2024-02-20

Published Online: 2024-06-06

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/spma-2023-0115

Keywords for this article

difference equations; Fibonacci sequence; Lucas sequence; Pell sequence; Pell-Lucas sequence; Jacobsthal sequence; Jacobsthal-Lucas sequence; Chebychev polynomials; Fibonacci polynomials; Lucas polynomials; Generalized Binet’s formula

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