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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

k

T

nk



k

and X

k

U

nk



k

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

n

and U

n

admit 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) .

(2)

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)

k

2

n−1−2k

n n − k

n − k k

 X

n−2k

, (3)

U

n

=

bn/2c

X

k=0

(−1)

k

2

n−2k

n − k k

 X

n−2k

. (4)

These relations allows to state that, for any n ≥ 0, T

n

and U

n

belong to E

n

[X] , (3) and (4) are their decompositions in the canonical basis B

n

:= X

n−2k



0≤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

(3)

The main goal of this paper is to prove that the families T

n

:= X

k

T

n−k



k

and U

n

:= X

k

U

n−k



k

for n − 2 bn/2c ≤ k ≤ n − bn/2c constitute two other basis of E

n

[X] (Theorem 1) for which T

n

and U

n

admit remarkableness integer coordinates.

Our first result is the following

Theorem 1. For any n ≥ 0, T

n

and U

n

are two basis of E

n

[X] .

So we can decompose T

2n+1

and U

2n+1

(resp. T

2n

and U

2n

) over each of the basis T

2n+1

and U

2n+1

(resp. T

2n

and U

2n

). Decompositions of T

2n

over T

2n

and U

2n

over U

2n

are trivial, it remains to examine the six decompositions:

1. The decomposition of T

2n+1

over T

2n+1

and U

2n+1

over U

2n+1

in Theorem 2.

2. The decomposition of T

2n

over U

2n

and U

2n

over T

2n

in Theorem 3.

3. The decomposition of T

2n+1

over U

2n+1

and U

2n+1

over T

2n+1

in 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,k

X

k

, (5)

(X − 1) (1 − 2X)

n

= X

k≥0

β

n,k

X

k

, (6)

(2X − 1)

n

+ 2 

1 + (2X − 1) + · · · + (2X − 1)

n−1



= X

k≥0

γ

n,k

X

k

, (7)

we deduce then, for (n, k) ∈ N

2

, the following relations α

n,k

= (−1)

k+1

2

k

n

k



,

(8)

(4)

β

n,k

= α

n,k

− α

n,k−1

, (9)

γ

n,k

= (−1)

n+1

α

n,k

+ 2

n−1

X

j=0

(−1)

j+1

α

j,k

. (10)

Theorem 2. For each n ≥ 0, we have

T

2n+1

=

n+1

X

k=1

β

n,k

X

k

T

2n+1−k

, (11)

U

2n+1

=

n+1

X

k=1

α

n+1,k

X

k

U

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,k

X

k

T

2n−k

. (14)

Theorem 4. For each n ≥ 0, we have

T

2n+1

=

n+1

X

k=1

n+1,k

− δ

k,1

) X

k

U

2n+1−k

, (15)

U

2n+1

=

n+1

X

k=1

n,k−1

+ β

n,k

) X

k

T

2n+1−k

, (16)

where δ

i,j

denotes the Kronecker symbol.

(5)

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,k

and γ

n,k

for 0 ≤ n ≤ 4

n α

n,0

α

n,1

α

n,2

α

n,3

α

n,4

0 −1

1 −1 2

2 −1 4 −4

3 −1 6 −12 8

4 −1 8 −24 32 −16

(6)

n β

n,0

β

n,1

β

n,2

β

n,3

β

n,4

β

n,5

0 −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,0

γ

n,1

γ

n,2

γ

n,3

γ

n,4

0 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

0

T

3

= 3XT

2

−2X

2

T

1

T

5

= 5XT

4

−8X

2

T

3

+4X

3

T

2

T

7

= 7XT

6

−18X

2

T

5

+20X

3

T

4

−8X

4

T

3

T

9

= 9XT

8

−32X

2

T

7

+56X

3

T

6

−48X

4

T

5

+16X

5

T

4

U

1

= 2XU

0

U

3

= 4XU

2

−4X

2

U

1

U

5

= 6XU

4

−12X

2

U

3

+8X

3

U

2

U

7

= 8XU

6

−24X

2

U

5

+32X

3

U

4

−16X

4

U

3

(7)

• Using Theorem 3

T

0

= U

0

T

2

= U

2

−XU

1

T

4

= U

4

−XU

3

T

6

= U

6

−XU

5

T

8

= U

8

−XU

7

U

0

= T

0

U

2

= T

2

+2XT

1

U

4

= T

4

+0XT

3

+4X

2

T

2

U

6

= T

6

+2XT

5

−4X

2

T

4

+8X

3

T

3

U

8

= T

8

+0XT

7

+8X

2

T

6

−16X

3

T

5

+16X

4

T

4

• Using Theorem 4

T

1

= XU

0

T

3

= 3XU

2

−4X

2

U

1

T

5

= 5XU

4

−12X

2

U

3

+8X

3

U

2

T

7

= 7XU

6

−24X

2

U

5

+32X

3

U

4

−16X

4

U

3

T

9

= 9XU

8

−40X

2

U

7

+80X

3

U

6

−80X

4

U

5

+32X

5

U

4

U

1

= 2XT

0

U

3

= 4XT

2

U

5

= 6XT

4

−8X

2

T

3

+8X

3

T

2

U

7

= 8XT

6

−16X

2

T

5

+16X

3

T

4

U

9

= 10XT

8

−32X

2

T

7

+64X

3

T

6

−64X

4

T

5

+32X

5

T

4

(8)

2. Proofs of Theorems 2.1. Proof of Theorem 1

T

n

and U

n

are 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

n

and U

n

relatively to the canonical basis B

n

of 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)/2

and 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−1

T

2m+1−k

and W

k(m)

= (2X)

k−1

U

2m+1−k

. Notice that V

k(m)

and W

k(m)

are polynomials of E

2m

[X] with dominant coefficient 2

2m−1

and 2

2m

respectively. 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

B2m



V

1(m)

, V

2(m)

, . . . , V

m+1(m)

 and

D

m

= det

B2m



W

1(m)

, W

2(m)

, . . . , W

m+1(m)

 . We have

m

= det

B2m



V

1(m)

, V

2(m)

− V

1(m)

, V

3(m)

− V

2(m)

, . . . , V

m+1(m)

− V

m(m)



= det

B2m

 V

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

,

(9)

and similarly, we obtain D

m

= 2

2m

D

m−1

= 2

2m+(2m−2)+···+2

D

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+r

X

r

T

2m

, X

r+1

T

2m−1

, . . . , X

r+m

T

m



= det

B2m

(T

2m

, XT

2m−1

, . . . , X

m

T

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]

N

defined 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

.

(10)

Using relations (5) , (6) and (7) , we have also

A

m

=

m

X

k=0

α

m,k

X

k

E

m−k

,

B

m

=

m+1

X

k=0

β

m,k

X

k

E

m+1−k

,

C

m

=

m

X

k=0

γ

m,k

X

k

E

m−k

.

Lemma 6. For any integer n, we have

(a) (2X − E)

n

T

m

= T

m−n

and (2X − E)

n

U

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

n

k=1

T

2k

(n ≥ 0) .

P roof.

(a) For m ≥ 1, one has (2X − E) T

m

= 2XT

m

− T

m+1

= T

m−1

and (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≥1

satisfies 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.

(11)

(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−2

from (2) .

(d) For n ≥ 0, one has U

2n

= U

0

+ P

n

k=1

(U

2k

− U

2k−2

) = 1 + 2 P

n k=1

T

2k

from (c).

P roof of T heorem 2. Relations (11) and (12) are respectively equivalent to

B

n

T

n

= 0 and A

n+1

U

n

= 0, for n ≥ 0.

These last relations follows from Lemma 6 (a). We have, for any integer n ≥ 0

B

n

T

n

= (−1)

n

(X − E) (2X − E)

n

T

n

= (−1)

n

(X − E) T

0

= (−1)

n

(XT

0

− T

1

) = 0, and

A

n+1

U

n

= (−1)

n+1

(2X − E) (2X − E)

n

U

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

n

T

n

, for n ≥ 0. One has indeed,

for any n ≥ 0

(12)

C

n

T

n

= 

(2X − E)

n

+ 2 P

n

k=1

E

k

(2X − E)

n−k

 T

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,k

X

k

U

2n+1−k

− XU

2n

(from (12) )

=

n+1

X

k=1

n+1,k

− δ

k,1

) X

k

U

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,k

X

k+1

T

2n−k

+

n+1

X

k=1

β

n,k

X

k

T

2n+1−k

, (from (11) and (14) )

=

n+1

X

k=1

n,k−1

+ β

n,k

) X

k

T

2n+1−k

.

(13)

References

[1] H. Belbachir and F. Bencherif, Linear recurrent sequences and powers of a square matrix, Integers 6 (A12) (2006), 1–17.

[2] E. Lucas, Th´eorie des Nombres, Ghautier-Villars, Paris 1891.

[3] T.J. Rivlin, Chebyshev Polynomials: From Approximation Theory to Algebra and Number Theory, second edition, Wiley Interscience 1990.

Received 30 April 2007

Revised 24 July 2007

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