ON SOME PROPERTIES OF CHEBYSHEV POLYNOMIALS
Hac` ene Belbachir and Farid Bencherif USTHB, Faculty of Mathematics,
P.O.Box 32, El Alia, 16111, Algiers, Algeria e-mail: hbelbachir@usthb.dz
or hacenebelbachir@gmail.com e-mail: fbencherif@usthb.dz
or fbencherif@gmail.com
Abstract
Letting T
n(resp. U
n) be the n-th Chebyshev polynomials of the first (resp. second) kind, we prove that the sequences X
kT
n−kk
and X
kU
n−kk
for n − 2 bn/2c ≤ k ≤ n − bn/2c are two basis of the Q-vectorial space E
n[X] formed by the polynomials of Q [X]
having the same parity as n and of degree ≤ n. Also T
nand U
nadmit remarkableness integer coordinates on each of the two basis.
Keywords: Chebyshev polynomials, integer coordinates.
2000 Mathematics Subject Classification: 11B83, 33C45, 11B39.
1. Introduction and main results
For each integer n ≥ 0, T
n= T
n(X) and U
n= U
n(X) denote the unique polynomials, with integer coefficients, satisfying
cos nx = T
n(cos x) and sin((n + 1)x) = sin x U
n(cos x), (x ∈ R) .
The well known Simpson’s formulae [2], for all n ∈ N and x ∈ R : cos nx = 2 cos x cos (n − 1) x − cos (n − 2) x sin (n + 1) x = 2 cos x sin nx − sin (n − 1) x give the recurrence relations
T
n= 2XT
n−1− T
n−2, with T
0= 1, T
1= X, (1)
U
n= 2XU
n−1− U
n−2, with U
0= 1, U
1= 2X.
(2)
We deduce (see [1]), for n ≥ 1, the following relations
T
n=
bn/2c
X
k=0
(−1)
k2
n−1−2kn n − k
n − k k
X
n−2k, (3)
U
n=
bn/2c
X
k=0
(−1)
k2
n−2kn − k k
X
n−2k. (4)
These relations allows to state that, for any n ≥ 0, T
nand U
nbelong to E
n[X] , (3) and (4) are their decompositions in the canonical basis B
n:= X
n−2k0≤k≤bn/2c
For instance, the first fifth values of thus polynomials are
T
0= 1 U
0= 1
T
1= X U
1= 2X
T
2= 2X
2− 1 U
2= 4X
2− 1 T
3= 4X
3− 3X U
3= 8X
3− 4X
T
4= 8X
4− 8X
2+ 1 U
4= 16X
4− 12X
2+ 1
The main goal of this paper is to prove that the families T
n:= X
kT
n−kk
and U
n:= X
kU
n−kk
for n − 2 bn/2c ≤ k ≤ n − bn/2c constitute two other basis of E
n[X] (Theorem 1) for which T
nand U
nadmit remarkableness integer coordinates.
Our first result is the following
Theorem 1. For any n ≥ 0, T
nand U
nare two basis of E
n[X] .
So we can decompose T
2n+1and U
2n+1(resp. T
2nand U
2n) over each of the basis T
2n+1and U
2n+1(resp. T
2nand U
2n). Decompositions of T
2nover T
2nand U
2nover U
2nare trivial, it remains to examine the six decompositions:
1. The decomposition of T
2n+1over T
2n+1and U
2n+1over U
2n+1in Theorem 2.
2. The decomposition of T
2nover U
2nand U
2nover T
2nin Theorem 3.
3. The decomposition of T
2n+1over U
2n+1and U
2n+1over T
2n+1in Theorem 4.
Let us define the families of integers (α
n,k) , (β
n,k) and (γ
n,k) by the following equalities
− (1 − 2X)
n= X
k≥0
α
n,kX
k, (5)
(X − 1) (1 − 2X)
n= X
k≥0
β
n,kX
k, (6)
(2X − 1)
n+ 2
1 + (2X − 1) + · · · + (2X − 1)
n−1= X
k≥0
γ
n,kX
k, (7)
we deduce then, for (n, k) ∈ N
2, the following relations α
n,k= (−1)
k+12
kn
k
,
(8)
β
n,k= α
n,k− α
n,k−1, (9)
γ
n,k= (−1)
n+1α
n,k+ 2
n−1
X
j=0
(−1)
j+1α
j,k. (10)
Theorem 2. For each n ≥ 0, we have
T
2n+1=
n+1
X
k=1
β
n,kX
kT
2n+1−k, (11)
U
2n+1=
n+1
X
k=1
α
n+1,kX
kU
2n+1−k. (12)
Theorem 3. For each n ≥ 0, we have
T
2n= U
2n− XU
2n−1, n ≥ 1, (13)
U
2n=
n
X
k=0
γ
n,kX
kT
2n−k. (14)
Theorem 4. For each n ≥ 0, we have
T
2n+1=
n+1
X
k=1
(α
n+1,k− δ
k,1) X
kU
2n+1−k, (15)
U
2n+1=
n+1
X
k=1
(γ
n,k−1+ β
n,k) X
kT
2n+1−k, (16)
where δ
i,jdenotes the Kronecker symbol.
The sequences of integers (α
n,k) , (β
n,k) and (γ
n,k) satisfy the following recurrence relation
α
n,0= −1, for n ≥ 0, α
0,k= 0, for k ≥ 1,
α
n,k= α
n−1,k− 2α
n−1,k−1, for n, k ≥ 1,
β
n,0= −1, for n ≥ 0,
β
0,1= 1 and β
0,k= 0, for k ≥ 2, β
n,k= β
n−1,k− 2β
n−1,k−1, for n, k ≥ 1,
γ
n,0= 1, for n ≥ 0, γ
0,k= 0, for k ≥ 1,
γ
n,k= −γ
n−1,k+ 2γ
n−1,k−1, for n, k ≥ 1.
The following tables give the values of α
n,k, β
n,kand γ
n,kfor 0 ≤ n ≤ 4
n α
n,0α
n,1α
n,2α
n,3α
n,40 −1
1 −1 2
2 −1 4 −4
3 −1 6 −12 8
4 −1 8 −24 32 −16
n β
n,0β
n,1β
n,2β
n,3β
n,4β
n,50 −1 1
1 −1 3 −2
2 −1 5 −8 4
3 −1 7 −18 20 −8
4 −1 9 −32 56 −48 16
n γ
n,0γ
n,1γ
n,2γ
n,3γ
n,40 1
1 1 2
2 1 0 4
3 1 2 −4 8
4 1 0 8 −16 16
Notice that α
n,k= γ
n,k= 0 for k > n and β
n,k= 0 for k > n + 1.
According to these tables, one obtains
• Using Theorem 2 T
1= XT
0T
3= 3XT
2−2X
2T
1T
5= 5XT
4−8X
2T
3+4X
3T
2T
7= 7XT
6−18X
2T
5+20X
3T
4−8X
4T
3T
9= 9XT
8−32X
2T
7+56X
3T
6−48X
4T
5+16X
5T
4U
1= 2XU
0U
3= 4XU
2−4X
2U
1U
5= 6XU
4−12X
2U
3+8X
3U
2U
7= 8XU
6−24X
2U
5+32X
3U
4−16X
4U
3• Using Theorem 3
T
0= U
0T
2= U
2−XU
1T
4= U
4−XU
3T
6= U
6−XU
5T
8= U
8−XU
7U
0= T
0U
2= T
2+2XT
1U
4= T
4+0XT
3+4X
2T
2U
6= T
6+2XT
5−4X
2T
4+8X
3T
3U
8= T
8+0XT
7+8X
2T
6−16X
3T
5+16X
4T
4• Using Theorem 4
T
1= XU
0T
3= 3XU
2−4X
2U
1T
5= 5XU
4−12X
2U
3+8X
3U
2T
7= 7XU
6−24X
2U
5+32X
3U
4−16X
4U
3T
9= 9XU
8−40X
2U
7+80X
3U
6−80X
4U
5+32X
5U
4U
1= 2XT
0U
3= 4XT
2U
5= 6XT
4−8X
2T
3+8X
3T
2U
7= 8XT
6−16X
2T
5+16X
3T
4U
9= 10XT
8−32X
2T
7+64X
3T
6−64X
4T
5+32X
5T
42. Proofs of Theorems 2.1. Proof of Theorem 1
T
nand U
nare two families of polynomials of E
n[X] with cardT
n= cardU
n= dim E
n[X] = bn/2c + 1. Using the following Lemma, we prove that the determinant of T
nand U
nrelatively to the canonical basis B
nof E
n[X] are not zero. Theorem 1 follows.
Lemma 5. For any integer n ≥ 0, by setting m = bn/2c , we have det
Bn(T
n) = 2
m(m−1)/2and det
Bn(U
n) = 2
m(m+1)/2.
P roof. For any integer m ≥ 0 and for 1 ≤ k ≤ m + 1, set V
k(m)= (2X)
k−1T
2m+1−kand W
k(m)= (2X)
k−1U
2m+1−k. Notice that V
k(m)and W
k(m)are polynomials of E
2m[X] with dominant coefficient 2
2m−1and 2
2mrespectively. Using the recurrence equations (1) and (2), we obtain for m ≥ 1
V
k+1(m)− V
k(m)= V
k(m−1)and W
k+1(m)− W
k(m)= W
k(m−1).
Let ∆
m= det
B2mV
1(m), V
2(m), . . . , V
m+1(m)and
D
m= det
B2mW
1(m), W
2(m), . . . , W
m+1(m). We have
∆
m= det
B2mV
1(m), V
2(m)− V
1(m), V
3(m)− V
2(m), . . . , V
m+1(m)− V
m(m)= det
B2mV
1(m), V
1(m−1), V
2(m−1), . . . , V
m(m−1)= 2
2m−1∆
m−1= 2
(2m−1)+(2m−3)+···+1∆
0= 2
m2,
and similarly, we obtain D
m= 2
2mD
m−1= 2
2m+(2m−2)+···+2D
0= 2
m(m+1). For n = 2m + r, with m = bn/2c and r ∈ {0, 1} , we have
det
Bn(T
n) = det
B2m+rX
rT
2m, X
r+1T
2m−1, . . . , X
r+mT
m= det
B2m(T
2m, XT
2m−1, . . . , X
mT
m)
= 2
−(1+2+···+m)∆
m= 2
m(m−1)/2,
and similarly det
Bn(U
n) = 2
−(1+2+···+m)D
m= 2
m(m+1)/2.
2.2. Proof of Theorems 2, 3 and 4
Let us denote E denote the shift operator on Q [X]
Ndefined by
E ((W
n)
n) = (W
n+1)
n,
or in a more simple form
EW
n= W
n+1, (n ≥ 0) .
For any m ≥ 0, define the operators A
m= − (E − 2X)
m, B
m= (X − E) (E − 2X)
m,
C
m= (2X − E)
m+ 2
m
X
k=1
E
k(2X − E)
m−k.
Using relations (5) , (6) and (7) , we have also
A
m=
m
X
k=0
α
m,kX
kE
m−k,
B
m=
m+1
X
k=0
β
m,kX
kE
m+1−k,
C
m=
m
X
k=0
γ
m,kX
kE
m−k.
Lemma 6. For any integer n, we have
(a) (2X − E)
nT
m= T
m−nand (2X − E)
nU
m= U
m−n, (m ≥ n ≥ 0) . (b) T
n= U
n− XU
n−1(n ≥ 1) .
(c) 2T
n= U
n− U
n−2(n ≥ 2) . (d) U
2n= 1 + 2 P
nk=1
T
2k(n ≥ 0) .
P roof.
(a) For m ≥ 1, one has (2X − E) T
m= 2XT
m− T
m+1= T
m−1and (2X − E) U
m= 2XU
m− U
m+1= U
m−1. We conclude by induction.
(b) Letting for n ≥ 1, W
n= T
n− U
n+ XU
n−1. The sequence (W
n)
n≥1satisfies the following relation
W
n= 2XW
n−1− W
n−2, (n ≥ 3) , with W
1= W
2= 0,
leading to W
n= 0, for n ≥ 1.
(c) For n ≥ 2, one has 2T
n= U
n+ (U
n− 2XU
n−1) from (b) and thus 2T
n= U
n− U
n−2from (2) .
(d) For n ≥ 0, one has U
2n= U
0+ P
nk=1
(U
2k− U
2k−2) = 1 + 2 P
n k=1T
2kfrom (c).
P roof of T heorem 2. Relations (11) and (12) are respectively equivalent to
B
nT
n= 0 and A
n+1U
n= 0, for n ≥ 0.
These last relations follows from Lemma 6 (a). We have, for any integer n ≥ 0
B
nT
n= (−1)
n(X − E) (2X − E)
nT
n= (−1)
n(X − E) T
0= (−1)
n(XT
0− T
1) = 0, and
A
n+1U
n= (−1)
n+1(2X − E) (2X − E)
nU
n= (−1)
n+1(2X − E) U
0= (−1)
n+1(2XU
0− U
1) = 0.
P roof of T heorem 3. Relation (13) follows from Lemma 6 (b).
Relation (14) is equivalent to U
2n= C
nT
n, for n ≥ 0. One has indeed,
for any n ≥ 0
C
nT
n=
(2X − E)
n+ 2 P
nk=1
E
k(2X − E)
n−kT
n= T
0+ 2
n
X
k=1
T
2k= U
2n, (from (d) of Lemma 6).
P roof of T heorem 4. For any n ≥ 0, we have
T
2n+1= U
2n+1− XU
2n, (from (b) of Lemma 6)
=
n+1
X
k=1
α
n+1,kX
kU
2n+1−k− XU
2n(from (12) )
=
n+1
X
k=1
(α
n+1,k− δ
k,1) X
kU
2n+1−k.
We have also, for any n ≥ 0
U
2n+1= XU
2n+ T
2n+1, (from (b) of Lemma 6)
=
n
X
k=0
γ
n,kX
k+1T
2n−k+
n+1
X
k=1
β
n,kX
kT
2n+1−k, (from (11) and (14) )
=
n+1
X
k=1