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Research ArticleSome Identities Involving Chebyshev Polynomials
Xiaoxue Li
School of Mathematics, Northwest University, Xiβan, Shaanxi 710127, China
Correspondence should be addressed to Xiaoxue Li; [email protected]
Received 18 December 2014; Accepted 18 March 2015
Academic Editor: Fabio Tramontana
Copyright Β© 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 π β₯ 0, the famous Chebyshev polynomialsof the first and second kind π
π(π₯) and π
π(π₯) are defined as
follows:
π0(π₯) = 1, π
1(π₯) = π₯, and π
π+1(π₯) = 2π₯π
π(π₯) β
ππβ1(π₯) for all π β₯ 1;
π0(π₯) = 1, π
1(π₯) = 2π₯, and π
π+1(π₯) = 2π₯π
π(π₯) β
ππβ1(π₯) for all π β₯ 1.
It is clear that these polynomials are the second-orderlinear recurrence polynomial; they satisfy the computationalformulae:
ππ(π₯) =
1
2
[(π₯ + βπ₯2β 1)
π
+ (π₯ β βπ₯2β 1)
π
] ,
ππ(π₯) =
1
2βπ₯2β 1
β [(π₯ + βπ₯2β 1)
π+1
β (π₯ β βπ₯2β 1)
π+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 [1β4], and other some important applications ofthe Chebyshev polynomials can also be found in [5β10].
Recently, several authors studied the sums of powersof Fibonacci numbers {πΉ
π} and Lucas numbers {πΏ
π}, and
obtained a series of important identities; see [11β13]. At thesame time, Melham [13] also proposed the following twoconjectures.
Conjecture 1. Letπ β₯ 1 be an integer. Then the sum
πΏ1πΏ3πΏ5β β β πΏ2π+1
π
β
π=1
πΉ2π+1
2π(2)
can be expressed as (πΉ2π+1
β 1)2
π2πβ1
(πΉ2π+1
), where π2πβ1
(π₯)
is a polynomial of degree 2π β 1 with integer coefficients.
Conjecture 2. Letπ β₯ 0 be an integer. Then the sum
πΏ1πΏ3πΏ5β β β πΏ2π+1
π
β
π=1
πΏ2π+1
2π(3)
can be expressed as (πΏ2π+1
β 1)π2π(πΏ2π+1
), where π2π(π₯) is a
polynomial of degree 2π 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 π
π(π₯) andπ
π(π₯). 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 β and π, we have theidentities
(a)
β
β
π=0
π2π+1
2π+1(π₯) =
1
22π+1
π
β
π=0
(
2π + 1
π β π
)
π2(2π+1)(β+1)β1
(π₯)
π2π(π₯)
; (4)
(b)
β
β
π=0
π2π+1
2π(π₯) =
1
4π(π₯2β 1)π
β
π
β
π=0
(
2π + 1
π β π
) (β1)πβπ
π2
(2π+1)(β+1)β1(π₯)
π2π(π₯)
.
(5)
Theorem 4. For any positive integers β and π, we have theidentities
(A)
β
β
π=1
π2π+1
2π(π₯)
=
1
22π+1
π
β
π=0
(
2π + 1
π β π
)
π(2π+1)(2β+1)β1
(π₯) β π2π(π₯)
π2π(π₯)
;
(6)
(B)
β
β
π=1
π2π+1
2πβ1(π₯)
=
1
22π+1
(π₯2β 1)π+1
β
π
β
π=0
(
2π + 1
π β π
) (β1)πβπ
π(2π+1)(2β+1)
(π₯) β π2π+1
(π₯)
π2π(π₯)
.
(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 ππ(cos π) = cos(ππ) and π
π(cos π) = sin((π +
1)π)/ sin π; it is clear that from Theorems 3 and 4 we candeduce some identities involving sinπ₯ and cosπ₯. On theother hand, we can also obtain some divisibility propertiesinvolving Chebyshev polynomials. That is, we have thefollowing.
Corollary 5. Let β β₯ 1 and π β₯ 0 be two integers. Then thesum
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
β
β
π=0
π2π+1
2π+1(π₯) (8)
can be divided by polynomials π2β+1
(π₯).The sum
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
β
β
π=0
π2π+1
2π(π₯) (9)
can be divided by polynomials π2β(π₯).
Corollary 6. Let β β₯ 1 and π β₯ 0 be two integers. Then thesum
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
β
β
π=1
π2π+1
2π(π₯) (10)
can be expressed as (π2β(π₯) β 1)π
2π(π₯, π2β+1
(π₯)), whereπ2π(π₯, π¦) is an integer coefficients polynomial of two variables
with degree 2π of π¦.
Corollary 7. Let β β₯ 1 and π β₯ 0 be two integers. Then thesum
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
β
β
π=1
π2π+1
2πβ1(π₯) (11)
can be expressed as (π2β+1
(π₯) β π₯)π2π(π₯, π2β+1
(π₯)), whereπ2π(π₯, π¦) is an integer coefficients polynomial of two variables
with degree 2π of π¦.
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 π and real numberπ¦ = 0, by using the familiar binomial expansion
(π¦ +
1
π¦
)
π
=
π
β
π=0
(
π
π
)π¦πβ2π (12)
we may get
(π¦ +
1
π¦
)
2π+1
=
π
β
π=0
(
2π + 1
π β π
)(π¦2π+1
+
1
π¦2π+1
) ,
(π¦ β
1
π¦
)
2π+1
=
π
β
π=0
(
2π + 1
π β π
) (β1)πβπ
(π¦2π+1
β
1
π¦2π+1
) .
(13)
Nowwe take π¦ = (π₯+βπ₯2 β 1)2π+1 in (13); then note that1/π¦ = (π₯ β βπ₯
2β 1)2π+1; from the definitions of π
π(π₯) and
ππ(π₯), we may immediately deduce the identities
π2π+1
2π+1(π₯) =
1
4π
π
β
π=0
(
2π + 1
π β π
)π(2π+1)(2π+1)
(π₯) , (14)
Mathematical Problems in Engineering 3
π2π+1
2π(π₯) =
1
4π(π₯2β 1)π
β
π
β
π=0
(
2π + 1
π β π
) (β1)πβπ
π4ππ+2π+2π
(π₯) .
(15)
If we takeπ¦ = (π₯+βπ₯2 β 1)2π in (13), thenwe can also deducethe identities
π2π+1
2π(π₯) =
1
4π
π
β
π=0
(
2π + 1
π β π
)π2π(2π+1)
(π₯) , (16)
π2π+1
2πβ1(π₯) =
1
4π(π₯2β 1)π
β
π
β
π=0
(
2π + 1
π β π
) (β1)πβπ
π2π(2π+1)β1
(π₯) .
(17)
Let πΌ = π₯+βπ₯2 β 1 and π½ = 1/πΌ = π₯ββπ₯2 β 1. Then for anyinteger β > 0, note that πΌπ½ = 1; from (14) and the definitionsof ππ(π₯) and π
π(π₯) we have
β
β
π=0
π2π+1
2π+1(π₯)
=
1
4π
π
β
π=0
(
2π + 1
π β π
)
β
β
π=0
π(2π+1)(2π+1)
(π₯)
=
1
2 β 4π
π
β
π=0
(
2π + 1
π β π
)[
πΌ2π+1
(πΌ2(2π+1)(β+1)
β 1)
πΌ2(2π+1)
β 1
+
π½2π+1
(π½2(2π+1)(β+1)
β 1)
π½2(2π+1)
β 1
]
=
1
2 β 4π
π
β
π=0
(
2π + 1
π β π
)
πΌ4πβ+4π+2β+2
β π½4πβ+4π+2β+2
πΌ2π+1
β π½2π+1
=
1
22π+1
π
β
π=0
(
2π + 1
π β π
)
π4βπ+4π+2β+1
(π₯)
π2π(π₯)
.
(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 haveβ
β
π=1
π2π+1
2π(π₯)
=
1
4π
π
β
π=0
(
2π + 1
π β π
)
β
β
π=1
π2π(2π+1)
(π₯)
=
1
22π+1
π
β
π=0
(
2π + 1
π β π
)[
πΌ2(2π+1)
(πΌ2β(2π+1)
β 1)
πΌ2(2π+1)
β 1
+
π½2(2π+1)
(π½2β(2π+1)
β 1)
π½2(2π+1)
β 1
]
=
1
22π+1
π
β
π=0
(
2π + 1
π β π
)[
πΌ(2π+1)(2β+1)
β πΌ2π+1
πΌ2π+1
β π½2π+1
β
π½(2π+1)(2β+1)
β π½2π+1
πΌ2π+1
β π½2π+1
]
=
1
22π+1
π
β
π=0
(
2π + 1
π β π
)
π(2π+1)(2β+1)β1
(π₯) β π2π(π₯)
π2π(π₯)
.
(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 thatππ(ππ(π₯)) = π
ππ(π₯) andπ
π(ππ(π₯)) = π
π(π+1)β1(π₯)/π
πβ1(π₯);
from (a) of Theorem 3 we may immediately deduce that
β
β
π=0
π2π+1
2π+1(π₯)
=
1
22π+1
π
β
π=0
(
2π + 1
π β π
)
π2β+1
(π₯) β π2π(π2β+2
(π₯))
π2π(π₯)
.
(20)
That is, the power sum
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
β
β
π=0
π2π+1
2π+1(π₯) (21)
can be divided by polynomial π2β+1
(π₯).Similarly, note that π
2
(2π+1)(β+1)β1(π₯) = π
2
β(π₯) β
π2
2π(πβ+1(π₯)); from (b) of Theorem 3 we know that π2
β(π₯)
divide the power sum
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
β
β
π=0
π2π+1
2π(π₯) . (22)
This proves Corollary 5.Now we use (B) of Theorem 4 to prove Corollary 7. It is
clear that if πΉ(π₯) is an integer coefficients polynomial withvariable π₯, then, for any polynomials π(π₯) and π(π₯), we haveπ(π₯) β π(π₯) divides πΉ(π(π₯)) β πΉ(π(π₯)). From those propertiesand noting that the identity π
2π+1(π2β+1
(π₯)) = π(2β+1)(2π+1)
(π₯)
we can deduce that
(π2β+1
(π₯) β π₯) | π2π+1
(π2β+1
(π₯)) β π2π+1
(π₯)
= π(2β+1)(2π+1)
(π₯) β π2π+1
(π₯) .
(23)
From (23) and (B) of Theorem 4 we can deduce that
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
β
β
π=1
π2π+1
2πβ1(π₯)
=
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
22π+1
(π₯2β 1)π+1
4 Mathematical Problems in Engineering
β (
π
β
π=0
(
2π + 1
π β π
) (β1)πβπ
π(2π+1)(2β+1)
(π₯) β π2π+1
(π₯)
π2π(π₯)
)
= (π2β+1
(π₯) β π₯)π2π(π₯, π2β+1
(π₯)) ,
(24)
where π2π(π₯, π¦) is an integer coefficients polynomial of two
variables with degree 2π of π¦. This proves Corollary 7.Finally, we prove Corollary 6. We first prove that
(π2β(π₯) β 1) | (π
(2π+1)(2β+1)β1(π₯) β π
2π(π₯)) (25)
for all integers π β₯ 0. It is clear that if π = 0, then the con-clusion is correct. So without loss of generality we can assumeπ β₯ 1. Note that the identity π
(2π+1)(2β+1)β1(π₯) = π
2β(π₯) β
π2π(π2β+1
(π₯)), so we have
π(2π+1)(2β+1)β1
(π₯) β π2π(π₯) = (π
2β(π₯) β 1) β π
2π(π2β+1
(π₯))
+ π2π(π2β+1
(π₯)) β π2π(π₯) .
(26)
Therefore, to prove (25), we only need to prove that
(π2β(π₯) β 1) | (π
2π(π2β+1
(π₯)) β π2π(π₯)) . (27)
Since π2π(π₯) is an even function, from (π β π) | (π
2π(π)β
π2π(π)) we can deduce that (π
2β+1(π₯) Β± π₯) | (π
2π(π2β+1
(π₯))β
π2π(π₯)); note that the coprime relations (π
2β+1(π₯) + π₯,
π2β+1
(π₯) β π₯) = π₯, so from the properties of polynomials weknow that
(π2
2β+1(π₯) β π₯
2
) | π₯ β (π2π(π2β+1
(π₯)) β π2π(π₯)) . (28)
Since (π2β(π₯) β 1, π₯) = 1, from (28) we know that to prove
(27), we only need to prove
(π2β(π₯) β 1) | (π
2
2β+1(π₯) β π₯
2
) . (29)
Note that the identity
π2
2β+1(π₯) β π₯
2
= (π₯π2β(π₯) β π
2ββ1(π₯))2
β π₯2
= π₯2
π2
2β(π₯) β 2π₯π
2β(π₯)π2ββ1
(π₯) + π2
2ββ1(π₯) β π₯
2
= π₯2
(π2
2β(π₯) β 1) β π
2ββ1(π₯)π2β+1
(π₯)
= π₯2
(π2
2β(π₯) β 1) β
1
4 (π₯2β 1)
β [(πΌ2β+2
β π½2β+2
) β (πΌ2β
β π½2β
)]
= π₯2
(π2
2β(π₯) β 1) β
1
4 (π₯2β 1)
β [(πΌ2β+1
β π½2β+1
)
2
β (πΌ β π½)2
]
= π₯2
(π2
2β(π₯) β 1) β π
2
2β(π₯) + π
2
0(π₯)
= (π₯2
β 1) β (π2
2β(π₯) β 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
π0(π₯)π2(π₯)π4(π₯) β β β π
2π(π₯)
β
β
π=1
π2π+1
2π(π₯) (31)
can be expressed as (π2β(π₯)β1)π
2π(π₯, π2β+1
(π₯)), where π2π(π₯,
π¦) is an integer coefficients polynomial of two variables withdegree 2π of π¦.
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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