# Some Identities Involving the Reciprocal Sums of One-Kind Chebyshev Polynomials

Authors: Yuankui Ma; Xingxing Lv
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

We use the elementary and analytic methods and the properties of Chebyshev polynomials to study the computational problem of the reciprocal sums of one-kind Chebyshev polynomials and give several interesting identities for them. At the same time, we also give a general computational method for this kind of reciprocal sums.


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## Body

## 1. Introduction

It is well known that Chebyshev polynomials of the first and second kindTnx and Un(x) are defined as follows: T0(x)=1, T1(x)=x, and the recursion formula Tn+1(x)=2xTn(x)-Tn-1(x) for all integers n≥1. U0(x)=1, U1(x)=2x, and the recursion formula Un+1(x)=2xUn(x)-Un-1(x) for all integers n≥1. The generation functions of these polynomials are (1)1-xt1-2xt+t2=∑n=1∞Tnxtn,x<1,t<1,11-2xt+t2=∑n=1∞Unxtn,x<1,t<1.The general term formulae of Tn(x) and Un(x) are expressed as(2)Tnx=12x+x2-1n+x-x2-1n,Unx=12x2-1x+x2-1n+1-x-x2-1n+1.If we take x=cos⁡θ, then (3)Tncos⁡θ=cos⁡nθ,Uncos⁡θ=sin⁡n+1θsin⁡θ.Since all these definitions and properties of Chebyshev polynomials can be found in any handbook of mathematics, there is no need to list the source everywhere.Recently, some authors studied the properties of Chebyshev polynomials and obtained many interesting conclusions. For example, Li [1] obtained some identities involving power sums of Tn(x) and Un(x). As some applications of these results, she obtained some divisibility properties involving Chebyshev polynomials. At the same time, she also proposed the following open problem.Whether there exists an exact expression for the derivative or integral of the Chebyshev polynomials of the first kind in terms of the Chebyshev polynomials of the first kind (and vice-versa) is the question.Wang and Zhang [2] and Zhang and Wang [3] partly solved these problems. Some theoretical results related to Chebyshev polynomials can be found in Ma and Zhang [4], Cesarano [5], Babusci et al. [6–8], Lee and Wong [9], and Wang and Han [10]. Doha and others [11–14] and Bircan and Pommerenke [15] also obtained many important applications of the Chebyshev polynomials.In this paper, we consider the computational problem of the reciprocal sums of Chebyshev polynomials. That is, letq and k be positive integers with q≥3, for any real number x; if Tn(x)≠0, then we consider the computational problem of the summations(4)∑a=1q-11Ta2kx,∑a=1q-11Ua-12kx.Although there are many results related to Chebyshev polynomials, it seems that none had studied the computational problem of (4). The main reason may be that a computational formula does not exist. But for some special real number x, we can really get the precise value of (4). In this paper, we will illustrate this point. That is, we will use the elementary and analytic methods and the properties of Chebyshev polynomials to prove the following results.Theorem 1.
Letq be an integer with q≥3. Then for any integer h with (h,q)=1, one has the identities (5)∑a=1q-11Ua-12cos⁡πh/q=sin2πh/q3q2-1;∑a=1q-11Ua-14cos⁡πh/q=sin4πh/q45q2-1q2+11;∑a=1q-11Ua-16cos⁡πh/q=sin6πh/q945q2-12q2-11q2+17.Theorem 2.
Letq be an odd number with q≥3. Then for any integer h with (h,q)=1, one has the identities (6)∑a=1q-11Ta2cos⁡πh/q=q2-1;∑a=1q-11Ta4cos⁡πh/q=13q2-1q2+3;∑a=1q-11Ta6cos⁡πh/q=115q2-12q4+7q2-363.Some Notes. First in Theorem 2, we must limit q as an odd number. Otherwise, if q is an even number, then a=1/2·q is an integer, 1≤a≤q-1 and cos⁡πa/q=cos⁡π/2=0. Therefore, the fraction 1/Ta2kcos⁡πa/q is meaningless.Second, for any positive integerk and x=cos⁡πh/q with (h,q)=1, we can give an computational formula for (4). Of course, the calculation is very complicated when k is larger. So we do not give a general conclusion for (4), only give an efficient calculating method. In fact if we use computer MatLab program, and by means of recursive method in Lemma 4, we can also obtain all precise values of (4) for any positive integer k.


## 2. Several Lemmas

To complete the proofs of our theorems, we need following lemmas. First we haveLemma 3.
Letq>3 be an integer. Then for variable s with 0<s<1 and function fπs=ln⁡sin⁡πs, one has the identity (7)∑a=1q-1f2kπaq=-1k·22k-1·B2kk·q2k-1,where f(n)(s) denotes the n-order derivative of f(s), B2k is Bernoulli numbers.Proof.
For any real numbers, from Corollary 6 (Section 3, Chapter 5) in [16] we have the identity(8)sin⁡πs=πs∏n=1∞1-s2n2.Then from (8), the definition and properties of derivative we have (9)πf′πs=1s+∑n=1∞1n+s-1n-s,πf′′πs=-1s2-∑n=1∞1n+s2+1n-s2.Generally, for any positive integer k, we have(10)π2kf2kπs=-2k-1!s2k-∑n=1∞2k-1!n+s2k+2k-1!n-s2k.Now taking s=a/q in (10), and summation for all 1≤a≤q-1 and noting the definition and properties of complete residue system modq (see [16]) we have(11)π2k∑a=1q-1f2kπaq=-2k-1!∑a=1q-1q2ka2k-∑a=1q-1∑n=1∞2k-1!n+a/q2k+2k-1!n-a/q2k=-q2k∑a=1q-1∑n=0∞2k-1!qn+a2k-2k-1!∑a=1q-1q2kq-a2k-q2k∑a=1q-1∑n=2∞2k-1!qn-a2k=-2q2k∑a=1q-1∑n=0∞2k-1!qn+a2k=-22k-1!∑n=1∞q2kn2k-∑n=1∞1n2k=-22k-1!q2k-1ζ2k.Note that Riemann ζ-function ζ(2k)=∑n=1∞‍1/n2k=(-1)k+1(2π)2kB2k/22k! (see [17], Theorem  12.17). Then from (11) we have (12)π2k∑a=1q-1f2kπaq=-22k-1!q2k-1·-1k+12π2kB2k22k!=-1k·2π2k·B2k2k·q2k-1.This proves Lemma 3.Lemma 4.
Letf(s)=ln⁡sin⁡(πs), α=α(s)=cot⁡πs. Then one has (13)f′′s=-π21+α2;f4s=-π46α4+8α2+2,f6s=-π6120α6+240α4+136α2+16.Proof.
Noting that the identity1+cot2s=1/sin2s, from the definition and properties of derivative we have f′(s)=πcot⁡πs=πα and (14)f′′s=-π2sinπs=-π21+cot2πs=-π21+α2.This proves the first formula of Lemma 4.
Similarly, we have(15)f3s=-2π2αα′=2π3α+2π3α3,f4s=2π3α′+6π3α2α′=-π46α4+8α2+2.This is the second formula of Lemma 4.
It is easy to prove that(16)f6s=-π6120α6+240α4+136α2+16.This completes the proof of Lemma 4.Lemma 5.
Letq be an integer with q≥3. Then for any integer h with (h,q)=1, one has the identities (17)∑a=1q-11sin2πha/q=13q-1q+1;∑a=1q-11sin4πha/q=145q2-1q2+11;∑a=1q-11sin6πha/q=1945q2-1q2+172q2-11.Proof.
Since(h,q)=1, if a pass through a complete residue system modq, then ha also pass through a complete residue system modq. Therefore, without loss of generality we can assume that h=1. Noting that B2=1/6 and 1+cot2(x)=1/sin2(x), from Lemmas 3 and 4 with k=1 we have (18)-∑a=1q-11+cot2πaq=-∑a=1q-11sin2πa/q=-13q2-1or(19)∑a=1q-11sin2πa/q=13q-1q+1.
Similarly, noting thatB4=-1/30, from Lemmas 3 and 4 with k=2 and applying (19) we have (20)-∑a=1q-16cot4⁡πaq+8cot2⁡πaq+2=-215q4-1,which implies that (21)∑a=1q-161sin2πa/q-12+81sin2πa/q-1+2=215q4-1or(22)∑a=1q-11sin4πa/q=145q2-1q2+11.Noting that B6=1/42, from Lemmas 3 and 4 with k=3, applying (19) and (22) we have (23)-∑a=1q-1120cot6⁡πaq+240cot4⁡πaq+136cot2⁡πaq+16=-1663q6-1or (24)∑a=1q-1151sin2πa/q-13+301sin2πa/q-12+171sin2πa/q-1+2=263q6-1,which implies that(25)∑a=1q-11sin6πa/q=1945q2-1q2+172q2-11.Now Lemma 5 follows from (19), (22), and (25).
In fact, by using Lemma4 and the method of proving Lemma 5 we can give a computational formula for (26)∑a=1q-11sin2kπa/qwith all positive integer k. Here just in order to meet the demands of main results we only calculated k=1,2, and 3.


## 3. Proofs of the Theorems

In this section, we shall complete the proofs of our theorems. First we prove Theorem1. For any integer q≥3, taking x=cos⁡πh/q with (h,q)=1, from (3) we have(27)∑a=1q-11Ua-12kx=∑a=1q-11Ua-12kcosπh/q=sin2kπhq∑a=1q-11sin2kπha/q.Now Theorem 1 follows from (27) and Lemma 5 with k=1,2, and 3.To prove Theorem2, we note that, for any odd number q≥3, if a pass through a complete residue system modq, then 2a also pass through a complete residue system modq. So from the properties of trigonometric functions we have(28)∑a=1q-11sin2πa/q=∑a=1q-11sin22πa/q=∑a=1q-1sin2πa/q+cos2πa/q4sin2⁡πa/qcos2πa/q=14∑a=1q-11sin2πa/q+14∑a=1q-11cos2πa/q.From (3), (28), and Lemma 5 we may immediately deduce that(29)∑a=1q-11Ta2cos⁡πh/q=∑a=1q-11cos2πha/q=∑a=1q-13sin2πa/q=q2-1.Similarly, we also have(30)∑a=1q-11sin4πa/q=∑a=1q-11sin42πa/q=∑a=1q-1sin2πa/q+cos2πa/q216sin4⁡πa/qcos4πa/q=116∑a=1q-11sin4πa/q+116∑a=1q-11cos4πa/q+12∑a=1q-11sin22πa/q.So from (30) and Lemma 5 we have(31)∑a=1q-11cos4πa/q=∑a=1q-115sin4πa/q-∑a=1q-18sin2πa/q=13q2-1q2+3.Combining (3) and (31) we have the identity(32)∑a=1q-11Ta4cos⁡πh/q=∑a=1q-11cos4πha/q=13q2-1q2+3.From the method of proving (30) we also have(33)∑a=1q-11sin6πa/q=∑a=1q-11sin62πa/q=∑a=1q-1sin2πa/q+cos2πa/q364sin6⁡πa/qcos6πa/q=164∑a=1q-11sin6πa/q+164∑a=1q-11cos6πa/q+34∑a=1q-11sin42πa/q.From (3), Lemma 5, and (33) we can deduce that(34)∑a=1q-11Ta6cos⁡πh/q=∑a=1q-11cos6⁡πha/q=∑a=1q-163sin6⁡πha/q-∑a=1q-148sin4⁡πha/q=q2-12q4+7q2-36315.Now Theorem 2 follows from (29), (32), and (34).This completes all proofs of our results.



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*Source: 4194579-2017-07-11.xml*