research article some identities involving chebyshev...

Research ArticleSome Identities Involving Chebyshev Polynomials

Xiaoxue Li

School of Mathematics, Northwest University, Xi’an, Shaanxi 710127, China

Correspondence should be addressed to Xiaoxue Li; lxx20072012@163.com

Received 18 December 2014; Accepted 18 March 2015

Academic Editor: Fabio Tramontana

Copyright © 2015 Xiaoxue Li. This is an open access article distributed under the Creative Commons Attribution License, whichpermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The main purpose of this paper is using the combinatorial method and algebraic manipulations to study some sums of powers ofChebyshev polynomials and give several interesting identities. As some applications of these results, we obtained several divisibilityproperties involving Chebyshev polynomials.

1. Introduction

For any integer 𝑛 ≥ 0, the famous Chebyshev polynomialsof the first and second kind 𝑇

𝑛(𝑥) and 𝑈

𝑛(𝑥) are defined as

follows:

𝑇0(𝑥) = 1, 𝑇

1(𝑥) = 𝑥, and 𝑇

𝑛+1(𝑥) = 2𝑥𝑇

𝑛(𝑥) −

𝑇𝑛−1(𝑥) for all 𝑛 ≥ 1;

𝑈0(𝑥) = 1, 𝑈

1(𝑥) = 2𝑥, and 𝑈

𝑛+1(𝑥) = 2𝑥𝑈

𝑛(𝑥) −

𝑈𝑛−1(𝑥) for all 𝑛 ≥ 1.

It is clear that these polynomials are the second-orderlinear recurrence polynomial; they satisfy the computationalformulae:

𝑇𝑛(𝑥) =

1

2

[(𝑥 + √𝑥2− 1)

𝑛

+ (𝑥 − √𝑥2− 1)

𝑛

] ,

𝑈𝑛(𝑥) =

1

2√𝑥2− 1

⋅ [(𝑥 + √𝑥2− 1)

𝑛+1

− (𝑥 − √𝑥2− 1)

𝑛+1

] .

(1)

About the elementary properties of Chebyshev polyno-mials and related second-order linear recurrences, manyauthors had studied them and obtained a series of interestingconclusions. For example, some of the theoretical results canbe found in [1–4], and other some important applications ofthe Chebyshev polynomials can also be found in [5–10].

Recently, several authors studied the sums of powersof Fibonacci numbers {𝐹

𝑛} and Lucas numbers {𝐿

𝑛}, and

obtained a series of important identities; see [11–13]. At thesame time, Melham [13] also proposed the following twoconjectures.

Conjecture 1. Let𝑚 ≥ 1 be an integer. Then the sum

𝐿1𝐿3𝐿5⋅ ⋅ ⋅ 𝐿2𝑚+1

𝑛

∑

𝑘=1

𝐹2𝑚+1

2𝑘(2)

can be expressed as (𝐹2𝑛+1

− 1)2

𝑃2𝑚−1

(𝐹2𝑛+1

), where 𝑃2𝑚−1

(𝑥)

is a polynomial of degree 2𝑚 − 1 with integer coefficients.

Conjecture 2. Let𝑚 ≥ 0 be an integer. Then the sum

𝐿1𝐿3𝐿5⋅ ⋅ ⋅ 𝐿2𝑚+1

𝑛

∑

𝑘=1

𝐿2𝑚+1

2𝑘(3)

can be expressed as (𝐿2𝑛+1

− 1)𝑄2𝑚(𝐿2𝑛+1

), where 𝑄2𝑚(𝑥) is a

polynomial of degree 2𝑚 with integer coefficients.

Wang and Zhang [14] solved the Conjecture 2 completelyand made some substantial progress for the Conjecture 1.

The main purpose of this paper is using the algebraicmanipulations to obtain some identities involving Chebyshevpolynomials of the first and second kind 𝑇

𝑛(𝑥) and𝑈

𝑛(𝑥). As

some applications, we give three interesting corollaries. Thatis, we will prove the following two results.

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015, Article ID 950695, 5 pageshttp://dx.doi.org/10.1155/2015/950695

2 Mathematical Problems in Engineering

Theorem 3. For any positive integers ℎ and 𝑛, we have theidentities

(a)

ℎ

∑

𝑚=0

𝑇2𝑛+1

2𝑚+1(𝑥) =

1

22𝑛+1

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)

𝑈2(2𝑘+1)(ℎ+1)−1

(𝑥)

𝑈2𝑘(𝑥)

; (4)

(b)

ℎ

∑

𝑚=0

𝑈2𝑛+1

2𝑚(𝑥) =

1

4𝑛(𝑥2− 1)𝑛

⋅

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

) (−1)𝑛−𝑘

𝑈2

(2𝑘+1)(ℎ+1)−1(𝑥)

𝑈2𝑘(𝑥)

.

(5)

Theorem 4. For any positive integers ℎ and 𝑛, we have theidentities

(A)

ℎ

∑

𝑚=1

𝑇2𝑛+1

2𝑚(𝑥)

=

1

22𝑛+1

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)

𝑈(2𝑘+1)(2ℎ+1)−1

(𝑥) − 𝑈2𝑘(𝑥)

𝑈2𝑘(𝑥)

;

(6)

(B)

ℎ

∑

𝑚=1

𝑈2𝑛+1

2𝑚−1(𝑥)

=

1

22𝑛+1

(𝑥2− 1)𝑛+1

⋅

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

) (−1)𝑛−𝑘

𝑇(2𝑘+1)(2ℎ+1)

(𝑥) − 𝑇2𝑘+1

(𝑥)

𝑈2𝑘(𝑥)

.

(7)

The benefit of these identities is that it can transformthe complex sums of powers of Chebyshev polynomials thatbecome relatively simple linear sums of Chebyshev polyno-mials. This can simplify the calculation problems related tothe sums of powers of Chebyshev polynomials.

Whether there exists an exact expression for the deriva-tive or integral of the Chebyshev polynomials of the first kindin terms of the Chebyshev polynomials of the first kind (andvice-versa) is an open problem. we will be looking for somenew methods to further research.

Note that 𝑇𝑛(cos 𝜃) = cos(𝑛𝜃) and 𝑈

𝑛(cos 𝜃) = sin((𝑛 +

1)𝜃)/ sin 𝜃; it is clear that from Theorems 3 and 4 we candeduce some identities involving sin𝑥 and cos𝑥. On theother hand, we can also obtain some divisibility propertiesinvolving Chebyshev polynomials. That is, we have thefollowing.

Corollary 5. Let ℎ ≥ 1 and 𝑛 ≥ 0 be two integers. Then thesum

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

ℎ

∑

𝑚=0

𝑇2𝑛+1

2𝑚+1(𝑥) (8)

can be divided by polynomials 𝑈2ℎ+1

(𝑥).The sum

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

ℎ

∑

𝑚=0

𝑈2𝑛+1

2𝑚(𝑥) (9)

can be divided by polynomials 𝑈2ℎ(𝑥).

Corollary 6. Let ℎ ≥ 1 and 𝑛 ≥ 0 be two integers. Then thesum

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

ℎ

∑

𝑚=1

𝑇2𝑛+1

2𝑚(𝑥) (10)

can be expressed as (𝑈2ℎ(𝑥) − 1)𝑃

2𝑛(𝑥, 𝑇2ℎ+1

(𝑥)), where𝑃2𝑛(𝑥, 𝑦) is an integer coefficients polynomial of two variables

with degree 2𝑛 of 𝑦.

Corollary 7. Let ℎ ≥ 1 and 𝑛 ≥ 0 be two integers. Then thesum

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

ℎ

∑

𝑚=1

𝑈2𝑛+1

2𝑚−1(𝑥) (11)

can be expressed as (𝑇2ℎ+1

(𝑥) − 𝑥)𝑄2𝑛(𝑥, 𝑇2ℎ+1

(𝑥)), where𝑄2𝑛(𝑥, 𝑦) is an integer coefficients polynomial of two variables

with degree 2𝑛 of 𝑦.

2. Proof of the Theorems

In this section, we will use the algebraic manipulationsto complete the proof of our theorems. First we proveTheorem 3. In fact, for any positive integer 𝑛 and real number𝑦 = 0, by using the familiar binomial expansion

(𝑦 +

1

𝑦

)

𝑛

=

𝑛

∑

𝑘=0

(

𝑛

𝑘

)𝑦𝑛−2𝑘 (12)

we may get

(𝑦 +

1

𝑦

)

2𝑛+1

=

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)(𝑦2𝑘+1

+

1

𝑦2𝑘+1

) ,

(𝑦 −

1

𝑦

)

2𝑛+1

=

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

) (−1)𝑛−𝑘

(𝑦2𝑘+1

−

1

𝑦2𝑘+1

) .

(13)

Nowwe take 𝑦 = (𝑥+√𝑥2 − 1)2𝑚+1 in (13); then note that1/𝑦 = (𝑥 − √𝑥

2− 1)2𝑚+1; from the definitions of 𝑇

𝑛(𝑥) and

𝑈𝑛(𝑥), we may immediately deduce the identities

𝑇2𝑛+1

2𝑚+1(𝑥) =

1

4𝑛

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)𝑇(2𝑚+1)(2𝑘+1)

(𝑥) , (14)

Mathematical Problems in Engineering 3

𝑈2𝑛+1

2𝑚(𝑥) =

1

4𝑛(𝑥2− 1)𝑛

⋅

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

) (−1)𝑛−𝑘

𝑈4𝑚𝑘+2𝑚+2𝑘

(𝑥) .

(15)

If we take𝑦 = (𝑥+√𝑥2 − 1)2𝑚 in (13), thenwe can also deducethe identities

𝑇2𝑛+1

2𝑚(𝑥) =

1

4𝑛

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)𝑇2𝑚(2𝑘+1)

(𝑥) , (16)

𝑈2𝑛+1

2𝑚−1(𝑥) =

1

4𝑛(𝑥2− 1)𝑛

⋅

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

) (−1)𝑛−𝑘

𝑈2𝑚(2𝑘+1)−1

(𝑥) .

(17)

Let 𝛼 = 𝑥+√𝑥2 − 1 and 𝛽 = 1/𝛼 = 𝑥−√𝑥2 − 1. Then for anyinteger ℎ > 0, note that 𝛼𝛽 = 1; from (14) and the definitionsof 𝑈𝑛(𝑥) and 𝑇

𝑛(𝑥) we have

ℎ

∑

𝑚=0

𝑇2𝑛+1

2𝑚+1(𝑥)

=

1

4𝑛

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)

ℎ

∑

𝑚=0

𝑇(2𝑚+1)(2𝑘+1)

(𝑥)

=

1

2 ⋅ 4𝑛

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)[

𝛼2𝑘+1

(𝛼2(2𝑘+1)(ℎ+1)

− 1)

𝛼2(2𝑘+1)

− 1

+

𝛽2𝑘+1

(𝛽2(2𝑘+1)(ℎ+1)

− 1)

𝛽2(2𝑘+1)

− 1

]

=

1

2 ⋅ 4𝑛

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)

𝛼4𝑘ℎ+4𝑘+2ℎ+2

− 𝛽4𝑘ℎ+4𝑘+2ℎ+2

𝛼2𝑘+1

− 𝛽2𝑘+1

=

1

22𝑛+1

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)

𝑈4ℎ𝑘+4𝑘+2ℎ+1

(𝑥)

𝑈2𝑘(𝑥)

.

(18)

This proves the identity (a) of Theorem 3.Similarly, from formula (15) we can deduce the identity

(b) of Theorem 3.Now we proveTheorem 4. From (16) we haveℎ

∑

𝑚=1

𝑇2𝑛+1

2𝑚(𝑥)

=

1

4𝑛

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)

ℎ

∑

𝑚=1

𝑇2𝑚(2𝑘+1)

(𝑥)

=

1

22𝑛+1

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)[

𝛼2(2𝑘+1)

(𝛼2ℎ(2𝑘+1)

− 1)

𝛼2(2𝑘+1)

− 1

+

𝛽2(2𝑘+1)

(𝛽2ℎ(2𝑘+1)

− 1)

𝛽2(2𝑘+1)

− 1

]

=

1

22𝑛+1

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)[

𝛼(2𝑘+1)(2ℎ+1)

− 𝛼2𝑘+1

𝛼2𝑘+1

− 𝛽2𝑘+1

−

𝛽(2𝑘+1)(2ℎ+1)

− 𝛽2𝑘+1

𝛼2𝑘+1

− 𝛽2𝑘+1

]

=

1

22𝑛+1

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)

𝑈(2𝑘+1)(2ℎ+1)−1

(𝑥) − 𝑈2𝑘(𝑥)

𝑈2𝑘(𝑥)

.

(19)

This proves the identity (A) of Theorem 4.Similarly, from formula (17) we can also deduce the

identity (B) of Theorem 4.Now we use Theorem 3 to prove Corollary 5. From

the properties of Chebyshev polynomials we know that𝑇𝑛(𝑇𝑚(𝑥)) = 𝑇

𝑚𝑛(𝑥) and𝑈

𝑛(𝑇𝑚(𝑥)) = 𝑈

𝑚(𝑛+1)−1(𝑥)/𝑈

𝑚−1(𝑥);

from (a) of Theorem 3 we may immediately deduce that

ℎ

∑

𝑚=0

𝑇2𝑛+1

2𝑚+1(𝑥)

=

1

22𝑛+1

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

)

𝑈2ℎ+1

(𝑥) ⋅ 𝑈2𝑘(𝑇2ℎ+2

(𝑥))

𝑈2𝑘(𝑥)

.

(20)

That is, the power sum

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

ℎ

∑

𝑚=0

𝑇2𝑛+1

2𝑚+1(𝑥) (21)

can be divided by polynomial 𝑈2ℎ+1

(𝑥).Similarly, note that 𝑈

2

(2𝑘+1)(ℎ+1)−1(𝑥) = 𝑈

2

ℎ(𝑥) ⋅

𝑈2

2𝑘(𝑇ℎ+1(𝑥)); from (b) of Theorem 3 we know that 𝑈2

ℎ(𝑥)

divide the power sum

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

ℎ

∑

𝑚=0

𝑈2𝑛+1

2𝑚(𝑥) . (22)

This proves Corollary 5.Now we use (B) of Theorem 4 to prove Corollary 7. It is

clear that if 𝐹(𝑥) is an integer coefficients polynomial withvariable 𝑥, then, for any polynomials 𝑎(𝑥) and 𝑏(𝑥), we have𝑎(𝑥) − 𝑏(𝑥) divides 𝐹(𝑎(𝑥)) − 𝐹(𝑏(𝑥)). From those propertiesand noting that the identity 𝑇

2𝑘+1(𝑇2ℎ+1

(𝑥)) = 𝑇(2ℎ+1)(2𝑘+1)

(𝑥)

we can deduce that

(𝑇2ℎ+1

(𝑥) − 𝑥) | 𝑇2𝑘+1

(𝑇2ℎ+1

(𝑥)) − 𝑇2𝑘+1

(𝑥)

= 𝑇(2ℎ+1)(2𝑘+1)

(𝑥) − 𝑇2𝑘+1

(𝑥) .

(23)

From (23) and (B) of Theorem 4 we can deduce that

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

ℎ

∑

𝑚=1

𝑈2𝑛+1

2𝑚−1(𝑥)

=

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

22𝑛+1

(𝑥2− 1)𝑛+1

4 Mathematical Problems in Engineering

⋅ (

𝑛

∑

𝑘=0

(

2𝑛 + 1

𝑛 − 𝑘

) (−1)𝑛−𝑘

𝑇(2𝑘+1)(2ℎ+1)

(𝑥) − 𝑇2𝑘+1

(𝑥)

𝑈2𝑘(𝑥)

)

= (𝑇2ℎ+1

(𝑥) − 𝑥)𝑄2𝑛(𝑥, 𝑇2ℎ+1

(𝑥)) ,

(24)

where 𝑄2𝑛(𝑥, 𝑦) is an integer coefficients polynomial of two

variables with degree 2𝑛 of 𝑦. This proves Corollary 7.Finally, we prove Corollary 6. We first prove that

(𝑈2ℎ(𝑥) − 1) | (𝑈

(2𝑘+1)(2ℎ+1)−1(𝑥) − 𝑈

2𝑘(𝑥)) (25)

for all integers 𝑘 ≥ 0. It is clear that if 𝑘 = 0, then the con-clusion is correct. So without loss of generality we can assume𝑘 ≥ 1. Note that the identity 𝑈

(2𝑘+1)(2ℎ+1)−1(𝑥) = 𝑈

2ℎ(𝑥) ⋅

𝑈2𝑘(𝑇2ℎ+1

(𝑥)), so we have

𝑈(2𝑘+1)(2ℎ+1)−1

(𝑥) − 𝑈2𝑘(𝑥) = (𝑈

2ℎ(𝑥) − 1) ⋅ 𝑈

2𝑘(𝑇2ℎ+1

(𝑥))

+ 𝑈2𝑘(𝑇2ℎ+1

(𝑥)) − 𝑈2𝑘(𝑥) .

(26)

Therefore, to prove (25), we only need to prove that

(𝑈2ℎ(𝑥) − 1) | (𝑈

2𝑘(𝑇2ℎ+1

(𝑥)) − 𝑈2𝑘(𝑥)) . (27)

Since 𝑈2𝑘(𝑥) is an even function, from (𝑎 − 𝑏) | (𝑈

2𝑘(𝑎)−

𝑈2𝑘(𝑏)) we can deduce that (𝑇

2ℎ+1(𝑥) ± 𝑥) | (𝑈

2𝑘(𝑇2ℎ+1

(𝑥))−

𝑈2𝑘(𝑥)); note that the coprime relations (𝑇

2ℎ+1(𝑥) + 𝑥,

𝑇2ℎ+1

(𝑥) − 𝑥) = 𝑥, so from the properties of polynomials weknow that

(𝑇2

2ℎ+1(𝑥) − 𝑥

2

) | 𝑥 ⋅ (𝑈2𝑘(𝑇2ℎ+1

(𝑥)) − 𝑈2𝑘(𝑥)) . (28)

Since (𝑈2ℎ(𝑥) − 1, 𝑥) = 1, from (28) we know that to prove

(27), we only need to prove

(𝑈2ℎ(𝑥) − 1) | (𝑇

2

2ℎ+1(𝑥) − 𝑥

2

) . (29)

Note that the identity

𝑇2

2ℎ+1(𝑥) − 𝑥

2

= (𝑥𝑈2ℎ(𝑥) − 𝑈

2ℎ−1(𝑥))2

− 𝑥2

= 𝑥2

𝑈2

2ℎ(𝑥) − 2𝑥𝑈

2ℎ(𝑥)𝑈2ℎ−1

(𝑥) + 𝑈2

2ℎ−1(𝑥) − 𝑥

2

= 𝑥2

(𝑈2

2ℎ(𝑥) − 1) − 𝑈

2ℎ−1(𝑥)𝑈2ℎ+1

(𝑥)

= 𝑥2

(𝑈2

2ℎ(𝑥) − 1) −

1

4 (𝑥2− 1)

⋅ [(𝛼2ℎ+2

− 𝛽2ℎ+2

) ⋅ (𝛼2ℎ

− 𝛽2ℎ

)]

= 𝑥2

(𝑈2

2ℎ(𝑥) − 1) −

1

4 (𝑥2− 1)

⋅ [(𝛼2ℎ+1

− 𝛽2ℎ+1

)

2

− (𝛼 − 𝛽)2

]

= 𝑥2

(𝑈2

2ℎ(𝑥) − 1) − 𝑈

2

2ℎ(𝑥) + 𝑈

2

0(𝑥)

= (𝑥2

− 1) ⋅ (𝑈2

2ℎ(𝑥) − 1) .

(30)

From this identity we may immediately deduce (29). That is,(25) is correct.

Applying (25) and (A) ofTheorem 4 we can easily deducethat the sum

𝑈0(𝑥)𝑈2(𝑥)𝑈4(𝑥) ⋅ ⋅ ⋅ 𝑈

2𝑛(𝑥)

ℎ

∑

𝑚=1

𝑇2𝑛+1

2𝑚(𝑥) (31)

can be expressed as (𝑈2ℎ(𝑥)−1)𝑃

2𝑛(𝑥, 𝑇2ℎ+1

(𝑥)), where 𝑃2𝑛(𝑥,

𝑦) is an integer coefficients polynomial of two variables withdegree 2𝑛 of 𝑦.

This completes the proofs of our all results.

Conflict of Interests

The author declares that there is no conflict of interestsregarding the publication of this paper.

Acknowledgments

The author would like to thank the referees for their veryhelpful and detailed comments, which have significantlyimproved the presentation of this paper. This work is sup-ported by the P. S. F. (2014JM1009) and N. S. F. (11371291).

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