# On Chebyshev Polynomials, Fibonacci Polynomials, and Their Derivatives

Authors: Yang Li
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We study the relationship of the Chebyshev polynomials,
Fibonacci polynomials, and theirrth derivatives. We get the formulas for the rth derivatives of Chebyshev polynomials being represented by Chebyshev polynomials
and Fibonacci polynomials. At last, we get several identities about the Fibonacci
numbers and Lucas numbers.


---

## Body

## 1. Introduction

As we know, the Chebyshev polynomials and Fibonacci polynomials are usually defined as follows: the first kind of Chebyshev polynomials isT
n
+
2
(
x
)
=
2
x
T
n
+
1
(
x
)
-
T
n
(
x
) and n
≥
0, with the initial values T
0
(
x
)
=
1 and T
1
(
x
)
=
x; the second kind of Chebyshev polynomials is U
n
+
2
(
x
)
=
2
x
U
n
+
1
(
x
)
-
U
n
(
x
) and n
≥
0, with the initial values U
0
(
x
)
=
1 and U
1
(
x
)
=
2
x; the Fibonacci polynomials are F
n
+
2
(
x
)
=
x
F
n
+
1
(
x
)
+
F
n
(
x
) and n
≥
0 with the initial values F
0
(
x
)
=
0 and F
1
(
x
)
=
1. From the second-order linear recurrence sequences, we have
(1)
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
]
,
F
n
(
x
)
=
1
2
n
x
2
+
4
×
[
(
x
+
x
2
+
4
)
n
-
(
x
-
x
2
+
4
)
n
]
.
These polynomials play a very important role in the study of the theory and application of mathematics and they are closely related to the famous Fibonacci numbers {
F
n
} and Lucas numbers {
L
n
} which are defined by the second-order linear recurrence sequences
(2)
F
n
+
2
=
F
n
+
1
+
F
n
,
L
n
+
2
=
L
n
+
1
+
L
n
,
where n
≥
0, F
0
=
0, F
1
=
1, L
0
=
2, and L
1
=
1. Therefore, many authors have investigated these polynomials and got many properties and corollaries. For example, Wu and Zhang [1] have obtained the general formulas involving F
n
(
x
)
(3)
⌊
(
∑
k
=
n
∞
‍
1
F
k
(
x
)
)
-
1
⌋
=
{
F
n
(
x
)
-
F
n
-
1
(
x
)
,
if
n
is
even
,
n
≥
2
,
F
n
(
x
)
-
F
n
-
1
(
x
)
-
1
,
if
n
is
odd
,
n
≥
1
,
⌊
(
∑
k
=
n
∞
‍
1
F
k
2
(
x
)
)
-
1
⌋
=
{
x
F
n
(
x
)
F
n
-
1
(
x
)
-
1
,
if
n
is
even
,
n
≥
2
,
x
F
n
(
x
)
F
n
-
1
(
x
)
,
if
n
is
odd
,
n
≥
1
,
where x is any positive integer. Wu and Yang [2] studied Chebyshev polynomials and got a lot of properties.Recently, several authors also studied the derivatives of these polynomials. For example, Zhang [3] used the rth derivatives of Chebyshev polynomials to solve some calculating problems of the general summations. Falcón and Plaza [4–6] presented many formulas and relations between Fibonacci polynomials and their derivatives. This fact allows them to present a family of integer sequences in a new and direct way.In this paper, we combine Sergio Falcón and Wenpeng Zhang's ideas. Then we obtain the following theorems and corollaries. These results strengthen the connections of two kinds of polynomials. They are also helpful in dealing with some calculating problems of the general summations or studying some integer sequences.Theorem 1.
For any positive integers,n and r, one has the following formulas:
(4)
T
2
n
(
2
r
-
1
)
(
x
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
n
s
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
k
-
r
-
s
+
1
)
!
(
k
+
s
+
1
-
r
)
!
×
U
2
s
-
1
(
x
)
,
T
2
n
+
1
(
2
r
)
(
x
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
(
2
n
+
1
)
s
(
n
+
k
)
!
(
n
-
k
)
!
(
k
+
1
+
s
-
r
)
!
(
k
+
1
-
s
-
r
)
!
×
U
2
s
-
1
(
x
)
,
T
2
n
+
1
(
2
r
-
1
)
(
x
)
=
∑
s
=
0
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
-
2
(
2
n
+
1
)
(
2
s
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
k
+
s
+
2
-
r
)
!
(
k
-
s
-
r
+
1
)
!
×
U
2
s
(
x
)
,
T
2
n
(
2
r
)
(
x
)
=
∑
s
=
0
n
-
r
‍
∑
k
=
0
n
‍
(
-
1
)
n
-
k
·
2
2
r
·
(
n
+
k
-
1
)
!
·
(
2
s
+
1
)
·
n
(
n
-
k
)
!
(
k
-
s
-
r
)
!
(
k
+
s
-
r
+
1
)
!
×
U
2
s
(
x
)
,
where T
n
(
r
)
(
x
) denotes the rth derivative of T
n
(
x
) with respect to x.Theorem 2.
For any positive integers,n and r, one has the following formulas:
(5)
T
2
n
+
1
(
2
r
-
1
)
(
x
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
-
1
(
2
n
+
1
)
(
n
+
k
)
!
T
2
s
(
x
)
(
n
-
k
)
!
(
s
+
k
+
1
-
r
)
!
(
k
+
1
-
s
-
r
)
!
+
∑
k
=
r
n
‍
(
-
1
)
n
-
k
·
(
2
n
+
1
)
·
(
n
+
k
)
!
2
2
-
2
r
(
n
-
k
)
!
(
k
+
1
-
r
)
!
(
k
+
1
-
r
)
!
,
T
2
n
(
2
r
)
(
x
)
=
∑
s
=
1
n
-
r
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
·
2
2
r
+
1
·
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
k
-
r
+
s
)
!
(
k
-
r
-
s
)
!
T
2
s
(
x
)
+
∑
k
=
r
n
‍
(
-
1
)
n
-
k
·
2
2
r
·
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
k
-
r
)
!
(
k
-
r
)
!
,
T
2
n
+
1
(
2
r
)
(
x
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
(
n
+
k
)
!
(
2
n
+
1
)
T
2
s
-
1
(
x
)
(
n
-
k
)
!
(
s
+
k
-
r
)
!
(
k
-
r
-
s
+
1
)
!
,
T
2
n
(
2
r
-
1
)
(
x
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
n
(
n
+
k
-
1
)
!
T
2
s
-
1
(
x
)
(
n
-
k
)
!
(
k
+
s
-
r
)
!
(
k
-
r
-
s
+
1
)
!
.Theorem 3.
For any positive integersn and r one has the following formulas:
(6)
T
2
n
+
1
(
2
r
-
1
)
(
x
)
=
∑
s
=
1
n
-
r
+
2
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
r
-
s
2
2
k
+
r
-
1
(
2
s
-
1
)
(
2
n
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
s
+
k
-
r
+
1
)
!
(
k
-
r
-
s
+
2
)
!
×
F
2
s
-
1
(
x
)
,
T
2
n
(
2
r
)
(
x
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
r
-
s
+
1
·
2
2
k
+
r
(
2
s
n
-
n
)
(
n
+
k
-
1
)
!
(
k
+
s
-
r
)
!
(
k
-
r
-
s
+
1
)
!
(
n
-
k
)
!
×
F
2
s
-
1
(
x
)
,
T
2
n
(
2
r
-
1
)
(
x
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
0
n
‍
(
-
1
)
n
-
r
-
s
+
1
2
2
k
+
r
(
n
+
k
-
1
)
!
s
n
(
n
-
k
)
!
(
k
-
r
-
s
+
1
)
!
(
k
+
s
-
r
+
1
)
!
×
F
2
s
(
x
)
,
T
2
n
+
1
(
2
r
)
(
x
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
r
-
s
+
1
2
2
k
+
r
+
1
(
2
s
n
+
s
)
(
n
+
k
)
!
(
n
-
k
)
!
(
s
+
k
-
r
+
1
)
!
(
k
-
r
-
s
+
1
)
!
×
F
2
s
(
x
)
.Corollary 4.
For any positive integersn, m, and r, one has the following identities:
(7)
T
2
n
(
2
r
-
1
)
(
i
m
L
m
2
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
s
m
+
n
-
k
+
m
·
i
m
·
2
2
r
n
s
(
n
+
k
-
1
)
!
F
2
s
m
(
n
-
k
)
!
(
k
-
r
-
s
+
1
)
!
(
k
+
s
+
1
-
r
)
!
F
m
,
T
2
n
+
1
(
2
r
)
(
i
m
L
m
2
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
s
m
+
n
+
m
-
k
2
2
r
(
2
n
+
1
)
i
m
s
(
n
+
k
)
!
F
2
m
s
(
n
-
k
)
!
(
k
+
1
+
s
-
r
)
!
(
k
+
1
-
s
-
r
)
!
F
m
,
T
2
n
+
1
(
2
r
-
1
)
(
i
m
L
m
2
)
=
∑
s
=
0
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
s
m
+
n
2
2
r
-
2
(
2
n
+
1
)
(
2
s
+
1
)
(
n
+
k
)
!
F
m
(
2
s
+
1
)
(
-
1
)
k
(
n
-
k
)
!
(
k
+
s
+
2
-
r
)
!
(
k
-
s
-
r
+
1
)
!
F
m
,
T
2
n
(
2
r
)
(
i
m
L
m
2
)
=
∑
s
=
0
n
-
r
‍
∑
k
=
0
n
‍
(
-
1
)
s
m
+
n
-
k
·
2
2
r
(
n
+
k
-
1
)
!
(
2
s
+
1
)
n
F
m
(
2
s
+
1
)
(
n
-
k
)
!
(
k
-
s
-
r
)
!
(
k
+
s
-
r
+
1
)
!
F
m
,
where i denotes the square root of -
1.Corollary 5.
For any positive integersn, m, and r, one has the following identities:
(8)
T
2
n
+
1
(
2
r
-
1
)
(
i
m
L
m
2
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
s
m
+
n
-
k
2
2
r
-
2
(
2
n
+
1
)
(
n
+
k
)
!
L
2
s
m
(
n
-
k
)
!
(
s
+
k
+
1
-
r
)
!
(
k
+
1
-
s
-
r
)
!
+
∑
k
=
r
n
‍
(
-
1
)
n
-
k
·
(
2
n
+
1
)
·
(
n
+
k
)
!
2
2
-
2
r
(
n
-
k
)
!
(
k
+
1
-
r
)
!
(
k
+
1
-
r
)
!
,
T
2
n
(
2
r
)
(
i
m
L
m
2
)
=
∑
s
=
1
n
-
r
‍
∑
k
=
r
n
‍
(
-
1
)
s
m
+
n
-
k
·
2
2
r
·
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
k
-
r
+
s
)
!
(
k
-
r
-
s
)
!
L
2
s
m
+
∑
k
=
r
n
‍
(
-
1
)
n
-
k
·
2
2
r
·
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
k
-
r
)
!
(
k
-
r
)
!
,
T
2
n
+
1
(
2
r
)
(
i
m
L
m
2
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
s
m
+
n
-
k
2
2
r
-
1
(
n
+
k
)
!
(
2
n
+
1
)
L
2
s
m
-
m
i
m
(
n
-
k
)
!
(
s
+
k
-
r
)
!
(
k
-
r
-
s
+
1
)
!
,
T
2
n
(
2
r
-
1
)
(
i
m
L
m
2
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
s
m
+
n
-
k
2
2
r
-
1
n
(
n
+
k
-
1
)
!
L
2
s
m
-
m
i
m
(
n
-
k
)
!
(
k
+
s
-
r
)
!
(
k
-
r
-
s
+
1
)
!
.


## 2. Some Lemmas

Lemma 6.
For any nonnegative integersm and n, one has the following identities:
(9)
∫
-
1
1
‍
T
m
(
x
)
T
n
(
x
)
1
-
x
2
d
x
=
{
0
,
m
≠
n
,
π
2
,
m
=
n
>
0
,
π
,
m
=
n
=
0
,
(10)
∫
-
1
1
‍
U
m
(
x
)
U
n
(
x
)
1
-
x
2
d
x
=
{
0
,
m
≠
n
,
π
2
,
m
=
n
,
(11)
T
n
(
cos
⁡
θ
)
=
cos
⁡
n
θ
,
(12)
U
n
(
cos
⁡
θ
)
=
sin
(
n
+
1
)
θ
sin
θ
.Proof.
See [7].Lemma 7.
For any positive integersm and n, one has the following identities:
(13)
U
n
(
i
2
)
=
i
n
F
n
+
1
,
T
n
(
i
2
)
=
i
n
2
L
n
+
1
.Proof.
See [3].Lemma 8.
For any positive integern, one has
(14)
T
n
(
T
m
(
x
)
)
=
T
n
m
(
x
)
,
U
n
(
T
m
(
x
)
)
=
U
m
(
n
+
1
)
-
1
(
x
)
U
m
-
1
(
x
)
.Proof.
See [3].Lemma 9.
For any positive integersn and r, one has
(15)
T
2
n
(
r
)
(
x
)
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
2
2
k
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
2
k
-
r
)
!
x
2
k
-
r
,
T
2
n
+
1
(
r
)
(
x
)
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
2
2
k
(
2
n
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
2
k
+
1
-
r
)
!
x
2
k
+
1
-
r
.Proof.
From Theorem 2 of [2], we can get the following result easily:
(16)
T
2
n
(
x
)
=
∑
k
=
0
n
‍
(
-
1
)
n
-
k
×
[
2
2
k
(
n
+
k
n
-
k
)
-
2
2
k
-
1
(
n
+
k
-
1
2
k
-
1
)
]
x
2
k
=
∑
k
=
0
n
‍
(
-
1
)
n
-
k
·
2
2
k
-
1
×
[
2
(
n
+
k
)
!
(
2
k
)
!
(
n
-
k
)
!
-
(
n
+
k
-
1
)
!
(
2
k
-
1
)
!
(
n
-
k
)
!
]
x
2
k
=
∑
k
=
0
n
‍
n
(
-
1
)
n
-
k
2
2
k
(
n
+
k
-
1
)
!
(
2
k
)
!
(
n
-
k
)
!
x
2
k
.
From Theorem 2 of [2], we know
(17)
T
2
n
+
1
(
x
)
=
∑
k
=
0
n
‍
(
-
1
)
n
-
k
×
[
2
2
k
+
1
(
n
+
k
+
1
n
-
k
)
-
2
2
k
(
n
+
k
2
k
)
]
x
2
k
+
1
.
In the similar way, we can get the following result easily:
(18)
T
2
n
+
1
(
x
)
=
∑
k
=
0
n
‍
(
-
1
)
n
-
k
(
2
n
+
1
)
2
2
k
(
n
+
k
)
!
(
2
k
+
1
)
!
(
n
-
k
)
!
x
2
k
+
1
.
If we derive both sides of the above properties rth times, we will get
(19)
T
2
n
(
r
)
(
x
)
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
2
2
k
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
2
k
-
r
)
!
x
2
k
-
r
,
T
2
n
+
1
(
r
)
(
x
)
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
2
2
k
(
2
n
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
2
k
+
1
-
r
)
!
x
2
k
+
1
-
r
.
This proves Lemma 9.Lemma 10.
For any positive integersn and r, let
(20)
T
2
n
(
r
)
(
x
)
=
∑
s
=
0
+
∞
‍
a
2
n
,
r
,
s
U
s
(
x
)
,
T
2
n
+
1
(
r
)
(
x
)
=
∑
s
=
0
+
∞
‍
a
2
n
+
1
,
r
,
s
U
s
(
x
)
,
where T
n
(
r
)
(
x
) denotes the rth derivative of T
n
(
x
) with respect to x. Then one can get
(21)
a
2
n
,
r
,
s
=
{
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
2
2
k
+
1
n
(
n
+
k
-
1
)
!
(
s
+
1
)
(
n
-
k
)
!
(
2
k
-
s
-
r
)
!
!
(
2
k
+
s
-
r
+
2
)
!
!
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
s
-
r
i
s
e
v
e
n
,
0
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
o
t
h
e
r
w
i
s
e
,
(22)
a
2
n
+
1
,
r
,
s
=
{
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
·
2
2
k
+
1
·
(
2
n
+
1
)
·
(
s
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
2
k
+
s
+
3
-
r
)
!
!
(
2
k
+
1
-
s
-
r
)
!
!
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
s
-
r
i
s
o
d
d
,
0
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
o
t
h
e
r
w
i
s
e
.Proof.
To begin with, we multiply1
-
x
2
U
m
(
x
) to both sides of the following identity:
(23)
T
n
(
r
)
(
x
)
=
∑
s
=
0
+
∞
‍
a
n
r
s
U
s
(
x
)
,
and then integrate it from -
1 to 1. Applying property (10), we can get
(24)
∫
-
1
1
‍
1
-
x
2
T
n
(
r
)
(
x
)
U
m
(
x
)
d
x
=
∑
s
=
0
∞
‍
∫
-
1
1
‍
a
n
,
r
,
s
1
-
x
2
U
m
(
x
)
U
s
(
x
)
d
x
=
π
2
a
n
,
r
,
m
and then we have
(25)
a
n
r
m
=
2
π
∫
-
1
1
‍
1
-
x
2
T
n
(
r
)
(
x
)
U
m
(
x
)
d
x
.
We define
(26)
w
n
k
=
2
π
∫
0
π
‍
cos
⁡
n
θ
sin
(
k
+
1
)
θ
sin
θ
d
θ
.
From [8], we know
(27)
w
n
k
=
{
2
(
k
+
1
)
n
!
(
n
+
k
+
2
)
!
!
(
n
-
k
)
!
!
,
k
+
n
is
even
,
n
≥
k
,
0
,
otherwise
,
where n and k are any nonnegative integers. Let x
=
cos
⁡
θ; then we can get the following identity by applying property (10):
(28)
a
n
r
m
=
2
π
∫
0
π
‍
T
n
(
r
)
(
cos
⁡
θ
)
U
m
(
cos
⁡
θ
)
sin
2
θ
d
θ
=
2
π
∫
0
π
‍
T
n
(
r
)
(
cos
⁡
θ
)
sin
(
m
+
1
)
θ
sin
θ
d
θ
.
According to Lemma 9 and property (27), we have
(29)
a
2
n
,
r
,
m
=
2
π
∫
0
π
‍
T
2
n
(
r
)
(
cos
⁡
θ
)
sin
⁡
(
m
+
1
)
θ
sin
θ
d
θ
=
2
π
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
·
n
·
(
n
+
k
-
1
)
!
(
-
1
)
n
-
k
(
n
-
k
)
!
(
2
k
-
r
)
!
×
∫
0
π
‍
cos
⁡
2
k
-
r
θ
sin
⁡
(
m
+
1
)
θ
sin
θ
d
θ
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
·
n
·
(
n
+
k
-
1
)
!
(
-
1
)
n
-
k
(
n
-
k
)
!
(
2
k
-
r
)
!
w
2
k
-
r
,
m
.
Then we have a
2
n
,
r
,
m
=
0 if m
-
r is odd. If m
-
r is even, we have
(30)
a
2
n
,
r
,
m
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
·
n
·
(
n
+
k
-
1
)
!
(
-
1
)
n
-
k
(
n
-
k
)
!
(
2
k
-
r
)
!
·
2
(
m
+
1
)
(
2
k
-
r
)
!
(
2
k
-
m
-
r
)
!
!
(
2
k
+
m
+
2
-
r
)
!
!
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
2
2
k
+
1
n
(
n
+
k
-
1
)
!
(
m
+
1
)
(
n
-
k
)
!
(
2
k
-
m
-
r
)
!
!
(
2
k
+
m
-
r
+
2
)
!
!
.
This proves property (21). In the similar way, we have a
2
n
+
1
,
r
,
m
=
0 if m
-
r is even. If m
-
r is odd, we have
(31)
a
2
n
+
1
,
r
,
m
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
2
2
k
(
2
n
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
2
k
+
1
-
r
)
!
w
2
k
+
1
-
r
,
m
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
·
2
2
k
+
1
·
(
2
n
+
1
)
·
(
m
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
2
k
+
m
+
3
-
r
)
!
!
(
2
k
+
1
-
m
-
r
)
!
!
.
That is property (22). This proves Lemma 10.Lemma 11.
For any positive integersn and r, let
(32)
T
2
n
(
r
)
(
x
)
=
1
2
b
2
n
,
r
,
0
T
0
(
x
)
+
∑
s
=
1
+
∞
‍
b
2
n
,
r
,
s
T
s
(
x
)
,
T
2
n
+
1
(
r
)
(
x
)
=
1
2
b
2
n
+
1
,
r
,
0
T
0
(
x
)
+
∑
s
=
1
+
∞
‍
b
2
n
+
1
,
r
,
s
T
s
(
x
)
.
Then one can get
(33)
b
2
n
,
r
,
s
=
{
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
+
1
(
-
1
)
n
-
k
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
2
k
-
r
+
s
)
!
!
(
2
k
-
r
-
s
)
!
!
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
s
-
r
i
s
e
v
e
n
,
0
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
o
t
h
e
r
w
i
s
e
,
(34)
b
2
n
+
1
,
r
,
s
=
{
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
·
2
2
k
+
1
·
(
2
n
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
s
+
2
k
+
1
-
r
)
!
!
(
2
k
+
1
-
r
-
s
)
!
!
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
s
-
r
i
s
o
d
d
,
0
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
o
t
h
e
r
w
i
s
e
.Proof.
In order to prove property (22) we must multiply T
m
(
x
)
/
1
-
x
2 to both sides of the following identity:
(35)
T
n
(
r
)
(
x
)
=
1
2
b
n
,
r
,
0
T
0
(
x
)
+
∑
s
=
1
+
∞
‍
b
n
,
r
,
s
T
s
(
x
)
,
and then integrate it from -
1 to 1. Applying property (9) we can get
(36)
∫
-
1
1
‍
T
n
(
r
)
(
x
)
T
m
(
x
)
1
-
x
2
d
x
=
∫
-
1
1
‍
b
n
r
0
T
0
(
x
)
T
m
(
x
)
2
1
-
x
2
d
x
+
∑
k
=
1
∞
‍
∫
-
1
1
‍
b
n
r
k
T
m
(
x
)
T
n
(
x
)
1
-
x
2
d
x
=
π
2
b
n
r
m
and then we have
(37)
b
n
r
m
=
2
π
∫
-
1
1
‍
T
n
(
r
)
(
x
)
T
m
(
x
)
1
-
x
2
d
x
.
We define
(38)
q
n
k
=
2
π
∫
0
π
‍
cos
⁡
n
θ
cos
⁡
k
θ
d
θ
.
From [8], we know
(39)
q
n
k
=
{
2
n
!
(
n
+
k
)
!
!
(
n
-
k
)
!
!
,
k
+
n
is
even
,
n
≥
k
,
0
,
otherwise
,
where n and k are any nonnegative integers. Let x
=
cos
⁡
θ; then we can get the following identity by applying property (10):
(40)
b
n
r
m
=
2
π
∫
0
π
‍
T
m
(
cos
⁡
θ
)
T
n
(
r
)
(
cos
⁡
θ
)
sin
θ
sin
θ
d
θ
=
2
π
∫
0
π
‍
T
m
(
cos
⁡
θ
)
T
n
(
r
)
(
cos
⁡
θ
)
d
θ
.
According to Lemma 9 and property (39), we have
(41)
b
2
n
,
r
,
m
=
2
π
∫
0
π
‍
T
2
n
(
r
)
(
cos
⁡
θ
)
cos
⁡
m
θ
d
θ
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
n
(
n
+
k
-
1
)
!
q
2
k
-
r
,
m
(
-
1
)
n
-
k
(
n
-
k
)
!
(
2
k
-
r
)
!
,
so we have b
2
n
,
r
,
m
=
0 if m
-
r is odd. If m
-
r is even, we have
(42)
b
2
n
,
r
,
m
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
2
k
-
r
)
!
·
2
(
-
1
)
n
-
k
(
2
k
-
r
)
!
(
2
k
-
r
+
m
)
!
!
(
2
k
-
r
-
m
)
!
!
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
+
1
(
-
1
)
n
-
k
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
2
k
-
r
+
m
)
!
!
(
2
k
-
r
-
m
)
!
!
.
This proves property (33). In the similar way we have b
2
n
+
1
,
r
,
m
=
0 if m
-
r is even. If m
-
r is odd, we have
(43)
b
2
n
+
1
,
r
,
m
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
(
2
n
+
1
)
(
n
+
k
)
!
q
2
k
+
1
-
r
,
m
(
-
1
)
n
-
k
(
n
-
k
)
!
(
2
k
+
1
-
r
)
!
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
-
k
·
2
2
k
+
1
·
(
2
n
+
1
)
(
n
+
k
)
!
(
n
-
k
)
!
(
m
+
2
k
+
1
-
r
)
!
!
(
2
k
+
1
-
r
-
m
)
!
!
.
That is property (34). This proves Lemma 11.Lemma 12.
For any positive integersm and n, one has the following identities:
(44)
F
n
(
2
i
cos
⁡
θ
)
=
i
n
+
3
sin
n
θ
sin
θ
,
(45)
∫
-
1
1
‍
x
2
+
4
F
m
(
x
)
F
n
(
x
)
d
x
=
{
2
i
2
m
-
1
π
,
m
=
n
>
0
,
0
,
o
t
h
e
r
w
i
s
e
.Proof.
As we know,(46)
F
n
(
x
)
=
1
2
n
x
2
+
4
×
[
(
x
+
x
2
+
4
)
n
-
(
x
-
x
2
+
4
)
n
]
.
Let x
=
2
i
cos
⁡
θ; then we have
(47)
F
n
(
2
i
cos
⁡
θ
)
=
1
2
sin
θ
[
(
i
cos
⁡
θ
+
sin
θ
)
n
-
(
i
cos
⁡
θ
-
sin
θ
)
n
]
=
1
2
sin
θ
(
i
n
e
-
i
n
θ
-
i
n
e
i
n
θ
)
=
i
n
2
sin
θ
(
cos
⁡
n
θ
-
i
sin
n
θ
-
cos
⁡
n
θ
-
i
sin
n
θ
)
=
i
n
+
3
sin
n
θ
sin
θ
.
This proves property (44). Let x
=
2
i
cos
⁡
θ in the following identity:
(48)
A
=
∫
-
2
i
2
i
‍
x
2
+
4
F
m
(
x
)
F
n
(
x
)
d
x
;
then we can get
(49)
A
=
∫
0
π
‍
2
sin
θ
F
n
(
2
i
cos
⁡
θ
)
F
m
(
2
i
cos
⁡
θ
)
2
i
sin
θ
d
θ
=
∫
0
π
‍
4
i
sin
2
θ
i
n
+
3
sin
n
θ
sin
θ
i
m
+
3
sin
m
θ
sin
θ
d
θ
=
4
i
n
+
m
-
1
∫
0
π
‍
sin
n
θ
sin
m
θ
d
θ
=
2
i
n
+
m
-
1
∫
0
π
‍
cos
⁡
(
n
-
m
)
θ
-
cos
⁡
(
m
+
n
)
θ
d
θ
.
Then we can get property (45). This proves Lemma 12.Lemma 13.
For any positive integern, let
(50)
T
2
n
(
r
)
(
x
)
=
∑
s
=
1
+
∞
‍
c
n
,
r
,
s
F
s
(
x
)
,
T
2
n
+
1
(
r
)
(
x
)
=
∑
s
=
1
+
∞
‍
c
n
,
r
,
s
F
s
(
x
)
;
then we can get
(51)
c
2
n
,
r
,
s
=
{
∑
k
=
⌈
r
/
2
⌉
n
‍
2
4
k
-
r
+
1
n
s
(
n
+
k
-
1
)
!
i
1
-
r
-
s
(
-
1
)
n
(
n
-
k
)
!
(
2
k
-
r
-
s
+
1
)
!
!
(
2
k
-
r
+
s
+
1
)
!
!
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
s
-
r
i
s
o
d
d
,
0
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
o
t
h
e
r
w
i
s
e
,
c
2
n
+
1
,
r
,
s
=
{
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
2
4
k
+
2
-
r
(
2
s
n
+
s
)
i
2
-
r
-
s
(
n
+
k
)
!
(
n
-
k
)
!
(
2
k
+
s
-
r
+
2
)
!
!
(
2
k
-
r
-
s
+
2
)
!
!
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
s
-
r
i
s
e
v
e
n
,
0
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
o
t
h
e
r
w
i
s
e
.Proof.
At first, we multiplyx
2
+
4
F
m
(
x
) to both sides of the following identity:
(52)
T
n
(
r
)
(
x
)
=
∑
s
=
1
+
∞
‍
c
n
,
r
,
s
F
s
(
x
)
,
and then integrate it from -
2
i to 2
i; we can get the following identity by applying Lemma 12, where m is any positive integer. Consider
(53)
∫
-
2
i
2
i
‍
x
2
+
4
F
m
(
x
)
T
n
(
r
)
(
x
)
d
x
=
∑
s
=
1
∞
‍
∫
-
2
i
2
i
‍
c
n
,
r
,
s
x
2
+
4
F
s
(
x
)
F
m
(
x
)
d
x
=
2
i
2
m
-
1
π
c
n
,
r
,
m
;
then we have
(54)
c
n
,
r
,
m
=
(
-
i
)
2
m
-
1
2
π
∫
-
2
i
2
i
‍
x
2
+
4
F
m
(
x
)
T
n
(
r
)
(
x
)
d
x
.
Let x
=
cos
⁡
θ; then we can get the following identity by applying Lemma 12:
(55)
c
n
,
r
,
m
=
(
-
i
)
2
m
-
1
2
π
×
∫
0
π
‍
T
n
(
r
)
(
2
i
cos
⁡
θ
)
F
m
(
2
i
cos
⁡
θ
)
4
i
sin
2
θ
d
θ
=
i
2
m
+
2
2
π
∫
0
π
‍
T
n
(
r
)
(
2
i
cos
⁡
θ
)
·
i
m
+
3
sin
m
θ
sin
θ
·
(
2
sin
θ
)
2
d
θ
=
2
i
3
m
+
1
π
∫
0
π
‍
T
n
(
r
)
(
2
i
cos
⁡
θ
)
sin
m
θ
sin
θ
d
θ
.
According to property (27), we have
(56)
c
2
n
,
r
,
m
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
2
k
+
1
·
n
i
3
m
+
1
(
n
+
k
-
1
)
!
(
-
1
)
n
-
k
(
n
-
k
)
!
(
2
k
-
r
)
!
π
×
∫
0
π
‍
(
2
i
cos
⁡
θ
)
2
k
-
r
sin
m
θ
sin
θ
d
θ
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
4
k
+
1
n
(
n
+
k
-
1
)
!
i
1
-
r
-
m
(
-
1
)
n
(
n
-
k
)
!
(
2
k
-
r
)
!
2
r
π
×
∫
0
π
‍
cos
⁡
2
k
-
r
θ
sin
m
θ
sin
θ
d
θ
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
4
k
-
r
n
(
n
+
k
-
1
)
!
i
1
-
r
-
m
(
-
1
)
n
(
n
-
k
)
!
(
2
k
-
r
)
!
w
2
k
-
r
,
m
-
1
,
so we have c
2
n
,
r
,
m
=
0 if m
-
r is even. If m
-
r is odd, we can get
(57)
c
2
n
,
r
,
m
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
4
k
-
r
n
(
n
+
k
-
1
)
!
i
1
-
r
-
m
(
-
1
)
n
(
n
-
k
)
!
(
2
k
-
r
)
!
×
2
m
(
2
k
-
r
)
!
(
2
k
-
r
-
m
+
1
)
!
!
(
2
k
-
r
+
m
+
1
)
!
!
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
4
k
-
r
+
1
·
n
m
(
n
+
k
-
1
)
!
i
1
-
r
-
m
(
-
1
)
n
(
n
-
k
)
!
(
2
k
-
r
-
m
+
1
)
!
!
(
2
k
-
r
+
m
+
1
)
!
!
.
In the similar way, we have c
2
n
+
1
,
r
,
m
=
0 if m
-
r is odd. If m
-
r is even, we can get
(58)
c
2
n
+
1
,
r
,
m
=
∑
k
=
⌈
r
/
2
⌉
n
‍
2
4
k
+
1
-
r
(
2
n
+
1
)
i
2
-
r
-
m
(
n
+
k
)
!
(
-
1
)
n
(
n
-
k
)
!
(
2
k
+
1
-
r
)
!
w
2
k
+
1
-
r
,
m
-
1
=
∑
k
=
⌈
r
/
2
⌉
n
‍
(
-
1
)
n
2
4
k
+
2
-
r
(
2
m
n
+
m
)
i
2
-
r
-
m
(
n
+
k
)
!
(
n
-
k
)
!
(
2
k
+
m
-
r
+
2
)
!
!
(
2
k
-
r
-
m
+
2
)
!
!
.
This proves Lemma 13.


## 3. Proof of the Theorems and Corollaries

In this section, we will prove our theorems and corollaries. First of all, we can prove all the theorems from Lemmas10, 11, and 13 easily. Then we prove our corollaries.Proof of Corollary4.
Letx
=
T
m
(
x
) in Theorem 1. We can get the following properties from Lemma 8:
(59)
T
2
n
(
2
r
-
1
)
(
T
m
(
x
)
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
n
s
(
n
+
k
-
1
)
!
U
2
s
m
-
1
(
x
)
(
n
-
k
)
!
(
k
-
r
-
s
+
1
)
!
(
k
+
s
+
1
-
r
)
!
U
m
-
1
(
x
)
,
T
2
n
+
1
(
2
r
)
(
T
m
(
x
)
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
(
2
n
+
1
)
s
(
n
+
k
)
!
U
2
s
m
-
1
(
x
)
(
n
-
k
)
!
(
k
+
1
+
s
-
r
)
!
(
k
+
1
-
s
-
r
)
!
U
m
-
1
(
x
)
,
T
2
n
+
1
(
2
r
-
1
)
(
T
m
(
x
)
)
=

∑
s
=
0
n
+
1
-
r
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
-
2
(
2
n

+

1
)
(
2
s

+

1
)
(
n

+

k
)
!
U
m
(
2
s
+
1
)
-
1

(
x
)
(
n
-
k
)
!
(
k
+
s
+
2
-
r
)
!
(
k
-
s
-
r
+
1
)
!
U
m
-
1
(
x
)
,
T
2
n
(
2
r
)
(
T
m
(
x
)
)
=
∑
s
=
0
n
-
r
‍
∑
k
=
0
n
‍
(
-
1
)
n
-
k
·
2
2
r
(
n
+
k
-
1
)
!
(
2
s
+
1
)
n
U
m
(
2
s
+
1
)
-
1
(
x
)
(
n
-
k
)
!
(
k
-
s
-
r
)
!
(
k
+
s
-
r
+
1
)
!
U
m
-
1
(
x
)
.
Then, taking x
=
i
/
2 in the above identities, according to Lemma 7, we can get Corollary 4.Proof of Corollary5.
Letx
=
T
m
(
x
) in Theorem 2. We can get the following properties from Lemma 8:
(60)
T
2
n
+
1
(
2
r
-
1
)
(
T
m
(
x
)
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
-
1
(
2
n
+
1
)
(
n
+
k
)
!
T
2
s
m
(
x
)
(
n
-
k
)
!
(
s
+
k
+
1
-
r
)
!
(
k
+
1
-
s
-
r
)
!
+
∑
k
=
r
n
‍
(
-
1
)
n
-
k
·
(
2
n
+
1
)
·
(
n
+
k
)
!
2
2
-
2
r
(
n
-
k
)
!
(
k
+
1
-
r
)
!
(
k
+
1
-
r
)
!
,
T
2
n
(
2
r
)
(
T
m
(
x
)
)
=
∑
s
=
1
n
-
r
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
·
2
2
r
+
1
·
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
k
-
r
+
s
)
!
(
k
-
r
-
s
)
!
T
2
s
m
(
x
)
+
∑
k
=
r
n
‍
(
-
1
)
n
-
k
·
2
2
r
·
n
(
n
+
k
-
1
)
!
(
n
-
k
)
!
(
k
-
r
)
!
(
k
-
r
)
!
,
T
2
n
+
1
(
2
r
)
(
T
m
(
x
)
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
(
n
+
k
)
!
(
2
n
+
1
)
T
2
s
m
-
m
(
x
)
(
n
-
k
)
!
(
s
+
k
-
r
)
!
(
k
-
r
-
s
+
1
)
!
,
T
2
n
(
2
r
-
1
)
(
T
m
(
x
)
)
=
∑
s
=
1
n
-
r
+
1
‍
∑
k
=
r
n
‍
(
-
1
)
n
-
k
2
2
r
n
(
n
+
k
-
1
)
!
T
2
s
m
-
m
(
x
)
(
n
-
k
)
!
(
k
+
s
-
r
)
!
(
k
-
r
-
s
+
1
)
!
.
Then, taking x
=
i
/
2 in the above identities, according to Lemma 7, we can get Corollary 5.



---

*Source: 451953-2014-03-13.xml*