# Some Identities Involving the Derivative of the First Kind Chebyshev Polynomials

Authors: Tingting Wang; Han Zhang
Journal: Mathematical Problems in Engineering
 (2015)

---

## Abstract

We use the combinatorial method and algebraic manipulations to obtain several interesting identities involving the power sums of the derivative of the first kind Chebyshev polynomials. This solved an open problem proposed by Li (2015).


---

## Body

## 1. Introduction

For any integern
≥
0, the famous Chebyshev polynomials of the first and second kind T
n
(
x
) and U
n
(
x
) are defined as follows: (1)
T
n
x
=
n
2
∑
k
=
0
n
/
2
-
1
k
n
-
k
-
1
!
k
!
n
-
2
k
!
2
x
n
-
2
k
,
U
n
x
=
∑
k
=
0
n
/
2
-
1
k
n
-
k
!
k
!
n
-
2
k
!
2
x
n
-
2
k
,where n
/
2 denotes the greatest integer ≤
n
/
2.It is clear thatT
n
(
x
) and U
n
(
x
) are the second-order linear recurrence polynomials; they satisfy the recurrence formulae:T
0
(
x
)
=
1, T
1
(
x
)
=
x, and T
n
+
1
(
x
)
=
2
x
T
n
(
x
)
-
T
n
-
1
(
x
) for all n
≥
1,
U
0
(
x
)
=
1, U
1
(
x
)
=
2
x, and U
n
+
1
(
x
)
=
2
x
U
n
(
x
)
-
U
n
-
1
(
x
) for all n
≥
1.The general term formulae ofT
n
(
x
) and U
n
(
x
) are (2)
T
n
x
=
1
2
x
+
x
2
-
1
n
+
x
-
x
2
-
1
n
,
U
n
x
=
1
2
x
2
-
1
x
+
x
2
-
1
n
+
1
-
x
-
x
2
-
1
n
+
1
.Both T
n
(
x
) and U
n
(
x
) are orthogonal polynomials. That is, (3)
∫
-
1
1
T
m
x
T
n
x
1
-
x
2
d
x
=
0
,
if
m
≠
n
;
π
2
,
if
m
=
n
>
0
;
π
,
if
m
=
n
=
0
,
∫
-
1
1
1
-
x
2
U
m
x
U
n
x
d
x
=
0
,
if
m
≠
n
;
π
2
,
if
m
=
n
.About the other various properties of the Chebyshev polynomials, some authors had studied them and obtained many interesting conclusions. For example, Ma and Zhang [1], Wang and Han [2], Cesarano [3], and Lee and Wong [4] proved a series of identities involving Chebyshev polynomials. Bhrawy and others (see [5–10]) obtained many important applications of the Chebyshev polynomials.Very recently, Li [11] proved some identities involving power sums of T
n
(
x
) and U
n
(
x
). That is, for any positive integers h and n, one has the identities:(a)
(4)
∑
m
=
0
h
T
2
m
+
1
2
n
+
1
x
=
1
2
2
n
+
1
∑
k
=
0
n
2
n
+
1
n
-
k
U
2
2
k
+
1
h
+
1
-
1
x
U
2
k
x
,
(b)
(5)
∑
m
=
0
h
U
2
m
2
n
+
1
x
=
1
4
n
x
2
-
1
n
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
U
2
k
+
1
h
+
1
-
1
2
x
U
2
k
x
,
(c)
(6)
∑
m
=
1
h
T
2
m
2
n
+
1
x
=
1
2
2
n
+
1
∑
k
=
0
n
2
n
+
1
n
-
k
U
2
k
+
1
2
h
+
1
-
1
x
-
U
2
k
x
U
2
k
x
,
(d)
(7)
∑
m
=
1
h
U
2
m
-
1
2
n
+
1
x
=
1
2
2
n
+
1
x
2
-
1
n
+
1
·
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
T
2
k
+
1
2
h
+
1
x
-
T
2
k
+
1
x
U
2
k
x
.As for some applications of these results, Xiaoxue Li obtained some divisibility properties involving Chebyshev polynomials. At the same time, she also proposed the following open problem.Does there exist 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)? That is to say, does there exist an exact expression for the summations(8)
∑
m
=
1
h
T
m
′
x
n
or
∑
m
=
1
h
∫
0
x
T
m
y
d
y
nin terms of the Chebyshev polynomials of the first kind?Does there exist an exact expression for∑
m
=
1
h
‍
T
m
n
(
x
) in terms of T
m
′
(
x
) or ∫
0
x
‍
T
m
(
y
)
d
y where T
m
′
(
x
) denotes the derivative of T
m
(
x
) with respect to x?In this paper, as a note of  [11], we give some identities involving the derivative of the first kind Chebyshev polynomials. That is, we will prove the following.Theorem 1.
For any integersh
≥
1 and n
≥
0, one has the identity (9)
∑
m
=
1
h
T
2
m
′
x
2
n
+
1
=
∑
i
=
1
n
+
1
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
·
R
i
,
h
U
2
i
2
k
+
1
-
1
x
+
S
i
,
h
U
2
i
-
2
2
k
+
1
-
1
x
4
i
x
2
-
1
n
+
i
U
2
k
2
i
x
+
∑
i
=
1
n
+
1
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
·
H
i
,
h
U
2
i
+
2
h
2
k
+
1
-
1
x
+
K
i
,
h
U
2
i
-
2
+
2
h
2
k
+
1
-
1
x
4
i
x
2
-
1
n
+
i
U
2
k
2
i
x
,where R
(
i
,
h
), S
(
i
,
h
), H
(
i
,
h
), and K
(
i
,
h
) are computable constants.Theorem 2.
For any positive integersh and n, one has the identity (10)
∑
m
=
1
h
T
m
′
x
2
n
=
-
1
n
4
n
x
2
-
1
n
·
2
n
!
n
!
2
·
∑
m
=
1
h
m
2
n
+
∑
i
=
1
n
+
1
∑
k
=
1
n
2
n
n
-
k
-
1
n
-
k
·
r
i
,
h
T
2
k
i
x
+
s
i
,
h
T
2
k
i
-
1
x
4
n
+
i
x
2
-
1
n
+
i
U
k
-
1
2
i
x
+
∑
i
=
1
n
+
1
∑
k
=
1
n
2
n
n
-
k
-
1
n
-
k
·
p
i
,
h
T
2
k
h
+
i
x
+
q
i
,
h
T
2
k
h
+
i
-
1
x
4
n
+
i
x
2
-
1
n
+
i
U
k
-
1
2
i
x
,where r
(
i
,
h
), s
(
i
,
h
), p
(
i
,
h
), and q
(
i
,
h
) are computable constants.For some specialn, from Theorem 1 with n
=
0 and Theorem 2 with n
=
1, we can also deduce the following two corollaries.Corollary 3.
For any positive integerh, one has the identity (11)
∑
m
=
1
h
T
2
m
′
x
=
h
U
2
h
+
1
x
-
h
+
1
U
2
h
-
1
x
2
x
2
-
1
.Corollary 4.
For any positive integerh, one has the identity (12)
∑
m
=
1
h
T
m
′
x
2
=
-
h
h
+
1
2
h
+
1
12
x
2
-
1
+
h
2
-
2
T
2
h
+
2
x
-
h
+
1
2
T
2
h
x
8
x
2
-
1
2
+
T
2
h
+
4
x
-
T
2
h
+
2
x
16
x
2
-
1
3
.


## 2. Two Simple Lemmas

In this section, we will give two simple lemmas, which are necessary in the proofs of our theorems. First we expressT
m
′
(
x
)
2
n and T
m
′
(
x
)
2
n
+
1 in terms of T
k
(
x
) and U
k
(
x
). That is, we have the following.Lemma 1.
For any positive integersm and n, one has the identities (13)
T
m
′
x
2
n
=
1
4
n
x
2
-
1
n
-
1
n
2
n
!
n
!
2
m
2
n
+
2
∑
k
=
1
n
2
n
n
-
k
-
1
n
-
k
m
2
n
T
2
k
m
x
,
T
m
′
x
2
n
+
1
=
1
4
n
x
2
-
1
n
·
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
m
2
n
+
1
U
m
2
k
+
1
-
1
x
.Proof.
For any positive integern and real number x
≠
0 and 1, by using the familiar binomial expansion we have (14)
y
-
1
y
2
n
=
-
1
n
2
n
!
n
!
2
+
∑
k
=
1
n
2
n
n
-
k
-
1
n
-
k
y
2
k
+
1
y
2
k
,
y
-
1
y
2
n
+
1
=
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
y
2
k
+
1
-
1
y
2
k
+
1
.
Now we takey
=
x
+
x
2
-
1
m in (14); note that 1
/
y
=
x
-
x
2
-
1
m; from the definition of T
n
(
x
) and U
n
(
x
) we have the identities (15)
U
m
-
1
2
n
x
=
1
4
n
x
2
-
1
n
-
1
n
2
n
!
n
!
2
+
2
∑
k
=
1
n
2
n
n
-
k
-
1
n
-
k
T
2
k
m
x
,
U
m
-
1
2
n
+
1
x
=
1
4
n
x
2
-
1
n
·
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
U
m
2
k
+
1
-
1
x
.Note that T
n
′
(
x
)
=
n
U
n
-
1
(
x
); from (15) we may immediately deduce (16)
T
m
′
x
2
n
=
1
4
n
x
2
-
1
n
-
1
n
2
n
!
n
!
2
m
2
n
+
2
∑
k
=
1
n
2
n
n
-
k
-
1
n
-
k
m
2
n
T
2
k
m
x
,
T
m
′
x
2
n
+
1
=
1
4
n
x
2
-
1
n
·
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
m
2
n
+
1
U
m
2
k
+
1
-
1
x
.This proves Lemma 1.Lemma 2.
Leth and k be two positive integers. Then for any x
≠
0 and 1, we have the recurrence formula: (I)
(17)
∑
m
=
1
h
m
k
x
m
=
x
1
-
x
+
x
1
-
x
∑
i
=
1
k
-
1
k
i
∑
m
=
1
h
m
i
x
m
-
h
+
1
k
x
h
+
1
1
-
x
.There exist some computable constants C
(
r
,
s
) and D
(
r
,
s
) such that (II)
(18)
∑
m
=
1
h
m
k
x
m
=
∑
i
=
1
k
+
1
C
i
,
h
·
x
i
1
-
x
i
+
∑
i
=
1
k
+
1
D
i
,
h
·
x
h
+
i
1
-
x
i
.In particular for k
=
1 and 2, we have the identities: (A)
(19)
∑
m
=
1
h
m
x
m
=
x
-
h
+
1
x
h
+
1
+
h
x
h
+
2
1
-
x
2
;
(B)
(20)
∑
m
=
1
h
m
2
x
m
=
x
+
x
2
1
-
x
3
-
h
+
1
2
x
h
+
1
1
-
x
-
2
h
+
3
x
h
+
2
1
-
x
2
-
2
x
h
+
3
1
-
x
3
.Proof.
For any positive integerk, let f
(
x
,
k
,
h
)
=
∑
m
=
1
h
‍
m
k
x
m; it is clear that (21)
1
-
x
f
x
,
k
,
h
=
∑
m
=
1
h
m
k
x
m
-
∑
m
=
1
h
m
k
x
m
+
1
=
∑
m
=
1
h
m
k
x
m
-
∑
m
=
1
h
-
1
m
k
x
m
+
1
-
h
k
x
h
+
1
=
x
+
∑
m
=
1
h
-
1
m
+
1
k
x
m
+
1
-
∑
m
=
1
h
-
1
m
k
x
m
+
1
-
h
k
x
h
+
1
=
x
+
∑
m
=
1
h
-
1
m
+
1
k
-
m
k
x
m
+
1
-
h
k
x
h
+
1
.Note that the binomial expansion is as follows: (22)
m
+
1
k
-
m
k
=
∑
i
=
0
k
-
1
k
i
m
i
,and from (21) we may immediately deduce the identity: (23)
f
x
,
k
,
h
=
∑
m
=
1
h
m
k
x
m
=
x
1
-
x
+
x
1
-
x
∑
i
=
0
k
-
1
k
i
∑
m
=
1
h
-
1
m
i
x
m
-
h
k
x
h
+
1
1
-
x
=
x
1
-
x
+
x
1
-
x
∑
i
=
0
k
-
1
k
i
∑
m
=
1
h
m
i
x
m
-
x
1
-
x
∑
i
=
0
k
-
1
k
i
h
i
x
h
-
h
k
x
h
+
1
1
-
x
=
x
1
-
x
+
x
1
-
x
∑
i
=
0
k
-
1
k
i
∑
m
=
1
h
m
i
x
m
-
x
h
+
1
1
-
x
h
+
1
k
-
h
k
-
h
k
x
h
+
1
1
-
x
=
x
1
-
x
+
x
1
-
x
∑
i
=
0
k
-
1
k
i
∑
m
=
1
h
m
i
x
m
-
h
+
1
k
x
h
+
1
1
-
x
=
x
1
-
x
+
x
1
-
x
∑
i
=
0
k
-
1
k
i
f
x
,
i
,
h
-
h
+
1
k
x
h
+
1
1
-
x
.This proves the recurrence formula (I).
It is easy to prove (II) by recurrence formula (I) and the complete mathematical induction.
From the recurrence formula (I) and noting that the identity(24)
∑
m
=
1
h
x
m
=
x
-
x
h
+
1
1
-
xwe can also deduce (25)
∑
m
=
1
h
m
x
m
=
x
1
-
x
+
x
1
-
x
x
-
x
h
+
1
1
-
x
-
h
+
1
x
h
+
1
1
-
x
=
x
-
h
+
1
x
h
+
1
+
h
x
h
+
2
1
-
x
2
.This proves formula (A).
Identity (B) follows from (24), (A), and the recurrence formula (I) with k
=
2. This proves Lemma 2.Some Note. The first part of Lemma 2 obtained an interesting recurrence formula for the computation of the summation ∑
m
=
1
h
‍
m
k
x
m. But if positive integer k is large enough, then the computation of the recurrence formula is more complex, and so we have not given the exact constants C
(
i
,
h
) and D
(
i
,
h
) in formula (II).


## 3. Proofs of the Theorems

In this section, we will complete the proofs of our theorems. First we prove Theorem1. Let α
=
x
+
x
2
-
1, β
=
x
-
x
2
-
1; then note that α
·
β
=
1, 1
-
α
2
(
2
k
+
1
)
=
-
α
2
k
+
1
α
2
k
+
1
-
β
2
k
+
1; from the definition of U
n
(
x
) and the second formula of Lemma 1 we have (26)
∑
m
=
1
h
T
2
m
′
x
2
n
+
1
=
1
4
n
x
2
-
1
n
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
·
∑
m
=
1
h
2
m
2
n
+
1
U
2
m
2
k
+
1
-
1
x
=
1
x
2
-
1
n
+
1
/
2
·
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
·
∑
m
=
1
h
m
2
n
+
1
α
2
m
2
k
+
1
-
m
2
n
+
1
β
2
m
2
k
+
1
.Applying formula (II) of Lemma 2 we have (27)
∑
m
=
1
h
m
2
n
+
1
α
2
m
2
k
+
1
-
m
2
n
+
1
β
2
m
2
k
+
1
=
∑
i
=
1
2
n
+
2
C
i
,
h
·
α
2
i
2
k
+
1
1
-
α
2
2
k
+
1
i
+
∑
i
=
1
2
n
+
2
D
i
,
h
·
α
2
2
k
+
1
h
+
i
1
-
α
2
2
k
+
1
i
-
∑
i
=
1
2
n
+
2
C
i
,
h
·
β
2
i
2
k
+
1
1
-
β
2
2
k
+
1
i
-
∑
i
=
1
2
n
+
2
D
i
,
h
·
β
2
2
k
+
1
h
+
i
1
-
β
2
2
k
+
1
i
=
∑
i
=
1
2
n
+
2
C
i
,
h
·
-
1
i
α
i
2
k
+
1
-
β
i
2
k
+
1
α
2
k
+
1
-
β
2
k
+
1
i
+
∑
i
=
1
2
n
+
2
D
i
,
h
·
-
1
i
α
i
2
k
+
1
+
2
h
2
k
+
1
-
β
i
2
k
+
1
+
2
h
2
k
+
1
α
2
k
+
1
-
β
2
k
+
1
i
.Note that 2
U
2
k
(
x
)
T
(
2
i
-
1
)
(
2
k
+
1
)
(
x
)
=
U
2
i
(
2
k
+
1
)
-
1
(
x
)
-
U
(
2
i
-
2
)
(
2
k
+
1
)
-
1
(
x
); from the definitions of U
n
(
x
) and T
n
(
x
) we have (28)
∑
i
=
1
2
n
+
2
C
i
,
h
·
-
1
i
α
i
2
k
+
1
-
β
i
2
k
+
1
α
2
k
+
1
-
β
2
k
+
1
i
=
∑
i
=
1
n
+
1
C
2
i
,
h
x
2
-
1
i
-
1
/
2
·
2
U
2
i
2
k
+
1
-
1
x
4
i
U
2
k
2
i
x
-
C
2
i
-
1
,
h
x
2
-
1
i
-
1
/
2
·
T
2
i
-
1
2
k
+
1
x
4
i
-
1
U
2
k
2
i
-
1
x
=
∑
i
=
1
n
+
1
C
2
i
,
h
x
2
-
1
i
-
1
/
2
·
2
U
2
i
2
k
+
1
-
1
x
4
i
U
2
k
2
i
x
-
∑
i
=
1
n
+
1
C
2
i
-
1
,
h
x
2
-
1
i
-
1
/
2
·
2
U
2
i
2
k
+
1
-
1
x
-
2
U
2
i
-
2
2
k
+
1
-
1
x
4
i
U
2
k
2
i
x
,
(29)
∑
i
=
1
2
n
+
2
D
i
,
h
·
-
1
i
α
i
2
k
+
1
+
2
h
2
k
+
1
-
β
i
2
k
+
1
+
2
h
2
k
+
1
α
2
k
+
1
-
β
2
k
+
1
i
=
∑
i
=
1
n
+
1
D
2
i
,
h
x
2
-
1
i
-
1
/
2
·
2
U
2
i
+
2
h
2
k
+
1
-
1
x
4
i
U
2
k
2
i
x
-
∑
i
=
1
n
+
1
D
2
i
-
1
,
h
x
2
-
1
i
-
1
/
2
·
2
U
2
i
+
2
h
2
k
+
1
-
1
x
-
2
U
2
i
-
2
+
2
h
2
k
+
1
-
1
x
4
i
U
2
k
2
i
x
.Combining (27)–(29) we may deduce that (30)
∑
m
=
1
h
m
2
n
+
1
α
2
m
2
k
+
1
-
m
2
n
+
1
β
2
m
2
k
+
1
=
∑
i
=
1
n
+
1
R
i
,
h
U
2
i
2
k
+
1
-
1
x
+
S
i
,
h
U
2
i
-
2
2
k
+
1
-
1
x
4
i
x
2
-
1
i
-
1
/
2
U
2
k
2
i
x
+
∑
i
=
1
n
+
1
H
i
,
h
U
2
i
+
2
h
2
k
+
1
-
1
x
+
K
i
,
h
U
2
i
-
2
+
2
h
2
k
+
1
-
1
x
4
i
x
2
-
1
i
-
1
/
2
U
2
k
2
i
x
.From (26) and (30) we have the identity (31)
∑
m
=
1
h
T
2
m
′
x
2
n
+
1
=
∑
i
=
1
n
+
1
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
·
R
i
,
h
U
2
i
2
k
+
1
-
1
x
+
S
i
,
h
U
2
i
-
2
2
k
+
1
-
1
x
4
i
x
2
-
1
n
+
i
U
2
k
2
i
x
+
∑
i
=
1
n
+
1
∑
k
=
0
n
2
n
+
1
n
-
k
-
1
n
-
k
·
H
i
,
h
U
2
i
+
2
h
2
k
+
1
-
1
x
+
K
i
,
h
U
2
i
-
2
+
2
h
2
k
+
1
-
1
x
4
i
x
2
-
1
n
+
i
U
2
k
2
i
x
.This proves Theorem 1.Now we prove Theorem2. From the first identity of Lemma 1 we have (32)
∑
m
=
1
h
T
m
′
x
2
n
=
-
1
n
4
n
x
2
-
1
n
2
n
!
n
!
2
∑
m
=
1
h
m
2
n
+
2
∑
k
=
1
n
2
n
n
-
k
-
1
k
∑
m
=
1
h
m
2
n
T
2
k
m
x
.From (II) of Lemma 2 and the method of proving (30) we have (33)
2
∑
m
=
1
h
m
2
n
T
2
k
m
x
=
∑
m
=
1
h
m
2
n
α
2
k
m
+
β
2
k
m
=
∑
i
=
1
2
n
+
1
C
i
,
h
·
α
2
k
i
1
-
α
2
k
i
+
∑
i
=
1
2
n
+
1
D
i
,
h
·
α
2
k
h
+
i
1
-
α
2
k
i
+
∑
i
=
1
2
n
+
1
C
i
,
h
·
β
2
k
i
1
-
β
2
k
i
+
∑
i
=
1
2
n
+
1
D
i
,
h
·
β
2
k
h
+
i
1
-
β
2
k
i
=
∑
i
=
1
n
2
C
2
i
,
h
T
2
i
k
x
4
i
x
2
-
1
i
U
k
-
1
2
i
x
-
∑
i
=
1
n
+
1
2
C
2
i
-
1
,
h
T
2
i
k
x
-
T
2
i
-
1
k
x
4
i
x
2
-
1
i
U
k
-
1
2
i
x
+
∑
i
=
1
n
2
D
2
i
,
h
T
2
k
h
+
i
x
4
i
x
2
-
1
i
U
k
-
1
2
i
x
-
∑
i
=
1
n
+
1
2
D
2
i
-
1
,
h
T
2
k
h
+
i
x
-
T
2
k
h
+
i
-
1
x
4
i
x
2
-
1
i
U
k
-
1
2
i
x
.Combining (32) and (33) we may immediately deduce (34)
∑
m
=
1
h
T
m
′
x
2
n
=
-
1
n
4
n
x
2
-
1
n
·
2
n
!
n
!
2
·
∑
m
=
1
h
m
2
n
+
∑
i
=
1
n
+
1
∑
k
=
1
n
2
n
n
-
k
-
1
n
-
k
·
r
i
,
h
T
2
k
i
x
+
s
i
,
h
T
2
k
i
-
1
x
4
n
+
i
x
2
-
1
n
+
i
U
k
-
1
2
i
x
+
∑
i
=
1
n
+
1
∑
k
=
1
n
2
n
n
-
k
-
1
n
-
k
·
p
i
,
h
T
2
k
h
+
i
x
+
q
i
,
h
T
2
k
h
+
i
-
1
x
4
n
+
i
x
2
-
1
n
+
i
U
k
-
1
2
i
x
.This proves Theorem 2.Now we prove Corollary3. Applying the second formula of Lemma 1 with n
=
0 and formula (A) of Lemma 2 we have (35)
∑
m
=
1
h
T
2
m
′
x
=
∑
m
=
1
h
2
m
U
2
m
-
1
x
=
1
x
2
-
1
·
∑
m
=
1
h
m
α
2
m
-
β
2
m
=
1
x
2
-
1
α
2
-
h
+
1
α
2
h
+
2
+
h
α
2
h
+
4
1
-
α
2
2
-
β
2
-
h
+
1
β
2
h
+
2
+
h
β
2
h
+
4
1
-
β
2
2
=
h
U
2
h
+
1
x
-
h
+
1
U
2
h
-
1
x
2
x
2
-
1
.This proves Corollary 3.Similarly, applying the first formula of Lemma1 with n
=
1 and formula (B) of Lemma 2 we have (36)
∑
m
=
1
h
T
m
′
x
2
=
1
4
x
2
-
1
-
2
∑
m
=
1
h
m
2
+
2
∑
m
=
1
h
m
2
T
2
m
x
=
1
4
x
2
-
1
-
h
h
+
1
2
h
+
1
3
+
∑
m
=
1
h
m
2
α
2
m
+
β
2
m
.Note that (37)
∑
m
=
1
h
m
2
α
2
m
+
β
2
m
=
α
2
+
α
4
1
-
α
2
3
-
h
+
1
2
α
2
h
+
2
1
-
α
2
-
2
h
+
3
α
2
h
+
4
1
-
α
2
2
-
2
α
2
h
+
6
1
-
α
2
3
+
β
2
+
β
4
1
-
β
2
3
-
h
+
1
2
β
2
h
+
2
1
-
β
2
-
2
h
+
3
β
2
h
+
4
1
-
β
2
2
-
2
β
2
h
+
6
1
-
β
2
3
=
h
+
1
2
α
2
h
+
4
-
α
2
h
+
2
-
2
h
+
3
α
2
h
+
4
1
-
α
2
2
+
2
α
2
h
+
8
-
α
2
h
+
6
1
-
α
2
4
+
h
+
1
2
β
2
h
+
4
-
β
2
h
+
2
-
2
h
+
3
β
2
h
+
4
1
-
β
2
2
+
2
β
2
h
+
8
-
β
2
h
+
6
1
-
β
2
4
=
h
2
-
2
T
2
h
+
2
x
-
h
+
1
2
T
2
h
x
2
x
2
-
1
+
T
2
h
+
4
x
-
T
2
h
+
2
x
4
x
2
-
1
2
.Combining (36) and (37) we have the identity (38)
∑
m
=
1
h
T
m
′
x
2
=
-
h
h
+
1
2
h
+
1
12
x
2
-
1
+
h
2
-
2
T
2
h
+
2
x
-
h
+
1
2
T
2
h
x
8
x
2
-
1
2
+
T
2
h
+
4
x
-
T
2
h
+
2
x
16
x
2
-
1
3
.This completes the proof of all results.Some Comments. In our theorems, we can give the exact expressions for all constants R
(
i
,
h
), S
(
i
,
h
), r
(
i
,
h
), and s
(
i
,
h
), if k is a small positive integer. If k is large enough, then we can only give an exact computational method for these constants (because of the reason of Lemma 2), but the computation is more complex, and so we have not obtained the exact expression for R
(
i
,
h
), S
(
i
,
h
), r
(
i
,
h
), and s
(
i
,
h
), 1
≤
i
≤
h.



---

*Source: 146313-2015-07-01.xml*