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

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.

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.
In this paper, we consider the computational problem of the reciprocal sums of Chebyshev polynomials.That is, let  and  be positive integers with  ≥ 3, for any real number ; if   () ̸ = 0, then we consider the computational problem of the summations 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 , 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.Let  be an integer with  ≥ 3. Then for any integer ℎ with (ℎ, ) = 1, one has the identities Theorem 2. Let  be an odd number with  ≥ 3. Then for any integer ℎ with (ℎ, ) = 1, one has the identities Some Notes.First in Theorem 2, we must limit  as an odd number.Otherwise, if  is an even number, then  = (1/2) ⋅  is an integer, 1 ≤  ≤  − 1 and cos(/) = cos(/2) = 0. Therefore, the fraction 1/ 2  (cos(/)) is meaningless.Second, for any positive integer  and  = cos(ℎ/) with (ℎ, ) = 1, we can give an computational formula for (4).Of course, the calculation is very complicated when  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 .

Several Lemmas
To complete the proofs of our theorems, we need following lemmas.First we have Lemma 3. Let  > 3 be an integer.Then for variable  with 0 <  < 1 and function () = ln sin(), one has the identity where  () () denotes the -order derivative of (),  2 is Bernoulli numbers.
Similarly, we have This is the second formula of Lemma 4.
It is easy to prove that  (6) This completes the proof of Lemma 4.

Lemma 5.
Let  be an integer with  ≥ 3. Then for any integer ℎ with (ℎ, ) = 1, one has the identities Proof.Since (ℎ, ) = 1, if  pass through a complete residue system mod , then ℎ also pass through a complete residue system mod .Therefore, without loss of generality we can assume that ℎ = 1.Noting that  2 = 1/6 and 1 + cot 2 () = 1/sin 2 (), from Lemmas 3 and 4 with  = 1 we have or Similarly, noting that  4 = −1/30, from Lemmas 3 and 4 with  = 2 and applying (19) we have In fact, by using Lemma 4 and the method of proving Lemma 5 we can give a computational formula for with all positive integer .Here just in order to meet the demands of main results we only calculated  = 1, 2, and 3.
To prove Theorem 2, we note that, for any odd number  ≥ 3, if  pass through a complete residue system mod , then 2 also pass through a complete residue system mod .So from the properties of trigonometric functions we have Similarly, we also have (31)