# On the Sums of Powers of Chebyshev Polynomials and Their Divisible Properties

Authors: Lan Zhang; Wenpeng Zhang
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

The main purpose of this paper is using mathematical induction and the Girard and Waring formula to study a problem involving the sums of powers of the Chebyshev polynomials and prove some divisible properties. We obtained two interesting congruence results involving Fibonacci numbers and Lucas numbers as some applications of our theorem.


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## Body

## 1. Discussion and Conclusions

For any integern≥0, the first-kind Chebyshev polynomials {Tn(x)} and the second-kind Chebyshev polynomials {Un(x)} are defined as T0(x)=1, T1(x)=x, U0(x)=1, U1(x)=2x and Tn+2(x)=2xTn+1(x)-Tn(x), Un+2(x)=2xUn+1(x)-Un(x) for all n≥0. If we write α=x+x2-1 and β=x-x2-1, then we have (1)Tnx=12αn+βn,Unx=1α-βαn+1-βn+1∀n≥0.Since Chebyshev polynomials occupy very important position in the theory and application of mathematics, many scholars have studied their various properties and obtained a series of important results. See [1–11]. For example, Li [1] proved some identities involving power sums of Tn(x) and Un(x). She obtained some divisibility properties involving Chebyshev polynomials as some applications of these results. Ma and Lv [2] studied the computational problem of the reciprocal sums of Chebyshev polynomials and obtained some identities. Some theoretical results related to Chebyshev polynomials can be found in Ma and Zhang [3], Cesarano [4], Lee and Wong [5], Bhrawy and others (see [6–9]), and Wang and Zhang [10]. Bircan and Pommerenke [11] also obtained many important applications of the Chebyshev polynomials.In this paper, we will focus on the problem involving the sums of powers of Chebyshev polynomials. These contents not only are widely used in combinatorial mathematics, but also have important theoretical significance for the study of Chebyshev polynomials themselves. Here we will use mathematical induction and the Girard and Waring formula (see [12, 13]) to prove some interesting divisible properties for Chebyshev polynomials. That is, we will prove the following.Theorem 1.
Letn and h be any positive integers; then we have the congruence (2)U0xU2x⋯U2nx∑m=1hU2m-12n+1x≡0modT2h+1x-xx2-12.Taking x=i/2 and i2=-1 in Un(x) and Tn(x), then we have (3)Uni2=in51+52n+1-1-52n+1=in·Fn+1,Tni2=in21+52n+1-52n=in2·Ln, where Fn and Ln are Fibonacci numbers and Lucas numbers, respectively.There are many very interesting and important results related to Fibonacci numbers and Lucas numbers; some of them can be found in Yi and Zhang [14], Ozeki [15], Prodinger [16], Melham [17], and Wang and Zhang [18].We can also deduce interesting congruence properties involving Fibonacci numbers and Lucas numbers as an application of our theorem. That is, we have the following corollaries.Corollary 2.
For any positive integersn and h, we have the congruence (4)F1F3F5⋯F2n+1∑m=12h-1mF2m2n+1≡0modL4h+1-152.Corollary 3.
For any positive integersn and h, we have the congruence (5)F1F3F5⋯F2n+1∑m=12h-1-1mF2m2n+1≡0modL4h-1+152.


## 2. Several Simple Lemmas

To complete the proof of our main result, we need several simple lemmas. First we have the following.Lemma 4.
For any positive integerh, we have (6)x2-1,T2h+1x-x=x2-1,x2-1,T2h+1x-xx2-1=1.Proof.
Ifh=1, then T2h+1(x)=T3(x)=4x3-3x and T3(x)-x=4xx2-1. It is clear that x2-1,T3(x)-x=x2-1 and (7)x2-1,T3x-xx2-1=x2-1,4x=1. So without loss of generality we can assume that h>1. Then from the definition of Tn(x) and binomial theorem we have the identity(8)T2h+1x=12∑k=02h+12h+1kxkx2-12h+1-k/2+12∑k=02h+12h+1kxk-12h+1-kx2-12h+1-k/2=∑k=0h2h+12k+1x2k+1x2-1h-k. From (8), we have the polynomial congruence (9)T2h+1x=∑k=0h2h+12k+1x2k+1x2-1h-k≡x2h+1≡xx2-1+1h≡xmodx2-1 or (10)T2h+1x-x≡0modx2-1, which implies that(11)x2-1,T2h+1x-x=x2-1. On the other hand, from (8) we also have (12)T2h+1x=∑k=0h2h+12k+1x2k+1x2-1h-k≡x2h+1+h2h+1x2h-1x2-1≡xx2-1+1h+h2h+1xx2-1+1h-1x2-1≡x+hxx2-1+h2h+1xx2-1modx2-12, which implies that(13)x2-1,T2h+1x-xx2-1=x2-1,2hh+1x=1.It is clear that Lemma 4 follows from (11) and (13).Lemma 5.
Letn and h be any positive integers; then we have the polynomial congruence (14)4xT2h+12n+1x+U2n-2x-U2n+2x≡0modT2h+1x-x.Proof.
For any fixed positive integerh, we prove this polynomial congruence by complete induction for positive integer n. Note that U0(x)=1, U1(x)=2x, U2(x)=4x2-1, U4(x)=4x2-12-4x2=16x4-12x2+1, and 4T2h+13(x)=T3(2h+1)(x)+3T2h+1(x), so if n=1, then we have (15)4xT2h+12n+1x+U2n-2x-U2n+2x=4xT32h+1x+U0x-U4x=4xT32h+1x+1-16x4+12x2-1=4x4T2h+13x-3T2h+1x-16x2+12x2≡4x4x3-3x-16x2+12x2≡0modT2h+1x-x. That is to say, Lemma 5 is true for n=1.
Ifn=2, then note that the identities (16)U6x-U2x=64x4-80x4+20x2,4T2h+13x=T32h+1x+3T2h+1x,16T2h+15x=T32h+1x+5T32h+1x+10T2h+1x; we have the polynomial congruence (17)4xT2h+12n+1x+U2n-2x-U2n+2x=4xT52h+1x+U2x-U6x=4x16T2h+15x-5T32h+1x-10T2h+1x-64x4-80x4+20x2≡4x16x5-54T2h+13x-3T2h+1x-10T2h+1x-64x4-80x4+20x2=4x16x5-54x3-3x-10x-64x4-80x4+20x2≡0modT2h+1x-x. So Lemma 5 is true for n=2.
Suppose Lemma5 is true for all integers n=1,2,…,k. That is,(18)4xT2h+12n+1x+U2n-2x-U2n+2x≡0modT2h+1x-x for all 1≤n≤k.
Then, forn=k+1≥3, note that the identities (19)2T22h+1xT2h+12k+1x=T2h+12k+3x+T2h+12k-1x,U2k+4x=4x2-1U2k+2x-2xU2k+1x=4x2-1U2k+2x-U2k+2x-U2kx,T22h+1x=2T2h+12x-1≡2x2-1modT2h+1x-x; from inductive assumption (18), we have (20)4xT2h+12n+1x+U2n-2x-U2n+2x=4xT2h+12k+3x+U2kx-U2k+4x=8xT22h+1xT2h+12k+1x-4xT2h+12k-1+U2kx-U2k+4x≡8x2x2-1T2h+12k+1x-4xT2h+12k-1+4x2-1U2k-2x-U2k-2x-U2k-4x-4x2-1U2k+2x+U2k+2x+U2kx=4x2-14xT2h+12k+1+U2k-2x-U2k+2x-4xT2h+12k+1x+U2k-2x-U2k+2x-4xT2h+12k-1x+U2k-4x-U2kx≡0modT2h+1x-x. Now Lemma 5 follows from complete induction.Lemma 6.
For any integersh≥1 and n≥0, we have the congruence (21)U0xU2x⋯U2nx∑m=1hU2m2n+1-1x-2n+1U2m-1x≡0modT2h+1x-xx2-12.Proof.
For integern≥0, note that the identities αβ=1, Tn(x)=1/2αn+βn,(22)∑m=1hU2m2n+1-1x=12x2-1∑m=1hα2m2n+1-β2m2n+1=12x2-1α22n+1α2h2n+1-1α22n+1-1-β22n+1β2h2n+1-1β22n+1-1=12x2-1α2h+12n+1-α2n+1α2n+1-β2n+1+β2h+12n+1-β2n+1α2n+1-β2n+1=12x2-1U2nxT2h+12n+1x-T2n+1x,∑m=1hU2m-1x=12x2-1T2h+1x-x. It is clear that from (22) we know that, to prove Lemma 6, we need only to prove the polynomial congruence(23)T2h+12n+1x-T2n+1x-2n+1U2nxT2h+1x-x≡0modT2h+1x-x2. Now we use complete induction to prove (23). It is clear that polynomial congruence (23) is true for n=0. If n=1, then note that U2(x)=4x2-1, T3(x)=4x3-3x, and T3(2h+1)(x)=4T2h+13(x)-3T2h+1(x); we have (24)T2h+12n+1x-T2n+1x-2n+1U2nxT2h+1x-x=T32h+1x-T3x-34x2-1T2h+1x-x=4T2h+13x-3T2h+1x-4x3-3x-34x2-1T2h+1x-x=4T2h+1x-x2T2h+1x+2x≡0modT2h+1x-x2. So congruence (23) is true for n=1. Suppose (23) is true for all integers n=0,1,2,…,k. Namely,(25)T2h+12n+1x-T2n+1x-2n+1U2nxT2h+1x-x≡0modT2h+1x-x2 for all n=0,1,…,k.
Then, forn=k+1≥2, note that the identities (26)2T22h+1xT2h+12k+1x=T2h+12k+3x+T2h+12k-1x,T22h+1x=2T2h+12x-1=2T2h+1x-x+x2-1≡4xT2h+1x-2x2-1modT2h+1x-x2; from inductive assumption (25) and Lemma 5, we have (27)T2h+12n+1x-T2n+1x-2n+1U2nxT2h+1x-x=T2h+12k+3x-T2k+3x-2k+3U2k+2xT2h+1x-x=2T22h+1xT2h+12k+1x-T2h+12k-1x-T2k+3x-2k+3U2k+2xT2h+1x-x≡8xT2h+1x-4x2-2T2h+12k+1x-T2h+12k-1x-4x2-1T2k+1x+T2k+1x+T2k-1x-2U2k+2xT2h+1x-x-2k+14x2-1U2kx-U2kx-U2k-2xT2h+1x-x=2T2h+1x-x4xT2h+12k+1x+U2k-2x-U2k+2x+4x2-2T2h+12k+1x-T2k+1x-2k+1U2kxT2h+1x-x-T2h+12k-1x-T2k-1x-2k-1U2k-2xT2h+1x-x≡0modT2h+1x-x2. This proves Lemma 6 by complete induction.Lemma 7.
For all nonnegative integersu and real numbers X and Y, we have the identity (28)Xu+Yu=∑k=0u/2-1kuu-ku-kkX+Yu-2kXYk, where [x] denotes the greatest integer ≤x.Proof.
This formula is given by Girard and Waring, which can be found in [12] or [13].


## 3. Proof of the Theorem

In this section, we will prove our theorem by mathematical induction. TakingX=α2m, Y=-β2m, and u=2n+1 in Lemma 7, note that XY=-1; from the definition of Un(x) and binomial theorem, we have(29)U2m2n+1-1x=12x2-1α2n2n+1-β2m2n+1=∑k=0n-1k2n+12n+1-k2n+1-kk4n-kx2-1n-kU2m-12n+1-2kx-1k=∑k=0n2n+12n+1-k2n+1-kk4n-kx2-1n-kU2m-12n+1-2kx. For any integer h≥1, from (29) we have the identity(30)∑m=1hU2m2n+1-1x-2n+1U2m-1x=∑k=0n-12n+12n+1-k2n+1-kk4n-kx2-1n-k∑m=1hU2m-12n+1-2kx. Taking n=1 in (30), we have the identity(31)∑m=1hU6m-1x-3U2m-1x=4x2-1∑m=1hU2m-13x. From Lemma 4, we know that x2-1,T2h+1(x)-x/x2-1=1, so applying Lemma 6 and (31) we can deduce that(32)U0xU2x∑m=1hU2m-13x≡0modT2h+1x-xx2-12. That is to say, the theorem is true for n=1.Suppose that the theorem is true for all integersn=1,2,…,s. That is,(33)U0xU2x⋯U2nx∑m=1hU2m-12n+1x≡0modT2h+1x-xx2-12 for all integers 1≤n≤s.Then, forn=s+1, from (29) we have(34)∑m=1hU2m2s+3-1x-2s+3U2m-1x=∑k=0s2s+32s+3-k2s+3-kk4s+1-kx2-1s+1-k∑m=1hU2m-12s+3-2kx=∑k=1s2s+32s+3-k2s+3-kk4s+1-kx2-1s+1-k∑m=1hU2m-12s+3-2kx+4s+1x2-1s+1∑m=1hU2m-12s+3x. From Lemma 6, we have(35)U0xU2x⋯U2s+2x∑m=1hU2m2s+3-1x-2s+3U2m-1x≡0modT2h+1x-xx2-12. Applying inductive assumption (33), we have(36)U0xU2x⋯U2sx∑k=1s2s+32s+3-k2s+3-kk×4s+1-kx2-1s+1-k∑m=1hU2m-12s+3-2kx≡0modT2h+1x-xx2-12. Combining (34), (35), (36), and Lemma 6, we can deduce the polynomial congruence(37)U0xU2x⋯U2s+2x4s+1x2-1s+1∑m=1hU2m-12s+3x≡0modT2h+1x-xx2-12. From Lemma 4, we know that x2-1,T2h+1(x)-x/x2-1=1, so from (37) we may immediately deduce the polynomial congruence (38)U0xU2x⋯U2s+2x∑m=1hU2m-12s+3x≡0modT2h+1x-xx2-12. This completes the proof of our theorem by mathematical induction.



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