Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2014, Article ID 387675, 6 pages http://dx.doi.org/10.1155/2014/387675
Research Article Complex Factorizations of the Lucas Sequences via Matrix Methods Honglin Wu Department of Mathematics, Shanghai Business School, Shanghai 201400, China Correspondence should be addressed to Honglin Wu;
[email protected] Received 9 October 2013; Accepted 27 November 2013; Published 2 January 2014 Academic Editor: Mehmet Sezer Copyright © 2014 Honglin Wu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Firstly, we show a connection between the first Lucas sequence and the determinants of some tridiagonal matrices. Secondly, we derive the complex factorizations of the first Lucas sequence by computing those determinants with the help of Chebyshev polynomials of the second kind. Furthermore, we also obtain the complex factorizations of the second Lucas sequence by the similar matrix method using Chebyshev polynomials of the first kind.
1. Introduction Given two nonzero integers 𝑃 and 𝑄 satisfying 𝑃2 − 4𝑄 ≠ 0. The first Lucas sequence {𝑈𝑛 (𝑃, 𝑄)}𝑛≥0 and the second Lucas sequence {𝑉𝑛 (𝑃, 𝑄)}𝑛≥0 are defined by the recurrence relations 𝑈0 (𝑃, 𝑄) = 0, 𝑈1 (𝑃, 𝑄) = 1, 𝑈𝑛 (𝑃, 𝑄) = 𝑃𝑈𝑛−1 (𝑃, 𝑄) − 𝑄𝑈𝑛−2 (𝑃, 𝑄) ,
(1) 𝑛 ≥ 2,
𝑉𝑛 (𝑃, 𝑄) = 𝑃𝑉𝑛−1 (𝑃, 𝑄) − 𝑄𝑉𝑛−2 (𝑃, 𝑄) ,
𝑛−1
𝐹𝑛 = ∏ (1 − 2i cos 𝑘=1
𝑉0 (𝑃, 𝑄) = 2, 𝑉1 (𝑃, 𝑄) = 𝑃,
There is a long tradition of using matrix methods to study Lucas sequences [2–5]. In 2003, Cahill et al. obtained complex factorizations for Fibonacci numbers and Lucas numbers by using the determinants of two slightly different sequences of tridiagonal matrices [2]. They used the 𝑛 × 𝑛 tridiagonal matrix 𝑀(𝑛) with entries 𝑚𝑘,𝑘 = 1 (1 ≤ 𝑘 ≤ 𝑛), and 𝑚𝑘−1,𝑘 = 𝑚𝑘,𝑘−1 = i (2 ≤ 𝑘 ≤ 𝑛) to prove that
(2) 𝑛 ≥ 2,
respectively. By assigning 𝑃 and 𝑄 some special values, we will see some well-known Lucas sequences, which are important historically and for their own sake. The numbers {𝑈𝑛 (1, −1)}𝑛≥0 are called the Fibonacci numbers while the numbers {𝑉𝑛 (1, −1)}𝑛≥0 are called the Lucas numbers, the numbers {𝑈𝑛 (2, −1)}𝑛≥0 and {𝑉𝑛 (2, −1)}𝑛≥0 are the Pell numbers and the Pell-Lucas numbers, {𝑈𝑛 (1, −2)}𝑛≥0 and {𝑉𝑛 (1, −2)}𝑛≥0 are Jacobsthal numbers and Jacobsthal-Lucas numbers, respectively, {𝑈𝑛 (3, 2)}𝑛≥0 are Mersenne numbers, and so on.
𝜋𝑘 ), 𝑛
𝑛 ≥ 2,
(3)
where 𝐹𝑛 is the 𝑛th Fibonacci number and i = √−1, and also used the 𝑛 × 𝑛 tridiagonal matrix 𝑆(𝑛) with entries 𝑠1,1 = 1/2, 𝑠𝑘,𝑘 = 1 (2 ≤ 𝑘 ≤ 𝑛) and 𝑠𝑘−1,𝑘 = 𝑠𝑘,𝑘−1 = i (2 ≤ 𝑘 ≤ 𝑛) to prove that 𝑛
𝐿 𝑛 = ∏ (1 − 2i cos 𝑘=1
𝜋 (𝑘 − (1/2)) ), 𝑛
𝑛 ≥ 1,
(4)
where 𝐿 𝑛 is the 𝑛th Lucas number and i = √−1. In 2011, Burcu Bozkurt et al. obtained the complex factorization of the second Lucas sequence {𝑉𝑛 (𝑃, −1)}𝑛≥0 [4]: 𝑛
𝑉𝑛 = ∏ (𝑃 − 2i cos 𝑘=1
𝜋 (𝑘 − (1/2)) ), 𝑛
𝑛 ≥ 1,
(5)
by using the 𝑛 × 𝑛 tridiagonal matrix 𝐷(𝑛) with entries 𝑑1,1 = 𝑃/2, 𝑑𝑘,𝑘 = 𝑃 (2 ≤ 𝑘 ≤ 𝑛), and 𝑑𝑘−1,𝑘 = 𝑑𝑘,𝑘−1 = i (2 ≤ 𝑘 ≤ 𝑛), where i = √−1.
2
Journal of Applied Mathematics
In this study, number theory and linear algebra (with the help of orthogonal polynomials) are similarly intertwined to yield the complex factorizations of the first and second Lucas sequences. Now we give the following lemma which will be needed later. Lemma 1 ([2]). Let {𝐻(𝑛), 𝑛 = 1, 2, . . .} be a sequence of tridiagonal matrices of the form ℎ1,1 ℎ1,2 ( ( ( 𝐻 (𝑛) = ( ( ( (
Proof. According to Lemma 1, successive determinants of 𝐴(𝑛) are given by the recursive formula: |𝐴 (𝑛)| = 𝑃 |𝐴 (𝑛 − 1)| − 𝑄 |𝐴 (𝑛 − 2)| ,
ℎ3,2 ℎ3,3
d
d
d ℎ𝑛,𝑛−1
(
) ) ) ). ) ℎ𝑛−1,𝑛 ) )
for 𝑛 ≥ 2.
𝑈𝑛 (𝑃, 𝑄) = |𝐴 (𝑛 − 1)|
(12)
ℎ𝑛,𝑛
(6)
)
(7) 𝑛 ≥ 3.
Note that 𝑈0 (𝑃, 𝑄) = 0 = (𝑎 − 𝑏 )/(𝑎 − 𝑏) and 𝑈1 (𝑃, 𝑄) = 1 = (𝑎1 − 𝑏1 )/(𝑎 − 𝑏). In order to prove (10), it is sufficient to prove 𝑎𝑛 − 𝑏𝑛 (14) for 𝑛 ≥ 2 𝑎−𝑏 by (9). Now we prove (14) by induction on 𝑛. When 𝑛 = 2, 3, we have |𝐴 (1)| = 𝑃 = 𝑎 + 𝑏 =
First of all, we introduce the tridiagonal matrix sequence {𝐴(𝑛), 𝑛 = 1, 2, . . .} and express the first Lucas sequence 𝑈𝑛 (𝑃, 𝑄) by the determinant of 𝐴(𝑛 − 1) for 𝑛 ≥ 2. Then we use this connection to prove the explicit formula for 𝑈𝑛 (𝑃, 𝑄) which is a generalization of the Binet Form for 𝑈𝑛 (𝑃, −1). Theorem 2. Let {𝐴(𝑛), 𝑛 = 1, 2, . . .} be a sequence of tridiagonal matrices of the form 𝑃 𝛼 𝛽 𝑃 𝛼
) 𝛽 𝑃 d ) ) ) ) d d 𝛼) )
(8)
𝑎𝑛 − 𝑏𝑛 𝑎−𝑏
for 𝑛 ≥ 0,
𝑎3 − 𝑏3 . 𝑎−𝑏 (15)
= (𝑎 + 𝑏) |𝐴 (𝑘 − 1)| − 𝑎𝑏 |𝐴 (𝑘 − 2)| = (𝑎 + 𝑏)
𝑎𝑘 − 𝑏𝑘 𝑎𝑘−1 − 𝑏𝑘−1 − 𝑎𝑏 𝑎−𝑏 𝑎−𝑏
(16)
𝑎𝑘+1 − 𝑎𝑏𝑘 + 𝑎𝑘 𝑏 − 𝑏𝑘+1 − 𝑎𝑘 𝑏 + 𝑎𝑏𝑘 = 𝑎−𝑏 𝑎𝑘+1 − 𝑏𝑘+1 . 𝑎−𝑏 Hence, (14) is proved. =
Now we give two complex factorizations and one explicit formula of the first Lucas sequence by using the determinants of tridiagonal matrices and Chebyshev polynomials of the second kind.
𝑛−1
with 𝛼𝛽 = 𝑄, where 𝑃 and 𝑄 are nonzero integers satisfying 𝑃2 − 4𝑄 ≠ 0. Then for 𝑛 ≥ 2,
|𝐴 (2)| = 𝑃2 − 𝑄 = (𝑎 + 𝑏)2 − 𝑎𝑏 = 𝑎2 + 𝑎𝑏 + 𝑏2 =
Theorem 3. The first Lucas sequence {𝑈𝑛 (𝑃, 𝑄)}𝑛≥0 satisfies
𝛽 𝑃 )
𝑈𝑛 (𝑃, 𝑄) = |𝐴 (𝑛 − 1)|
𝑎2 − 𝑏2 , 𝑎−𝑏
Next we assume (14) holds for 3 ≤ 𝑛 ≤ 𝑘. Then when 𝑛 = 𝑘+1, we have |𝐴 (𝑘)| = 𝑃 |𝐴 (𝑘 − 1)| − 𝑄 |𝐴 (𝑘 − 2)|
2. Complex Factorizations of the First Lucas Sequence
(
(13)
0
|𝐴 (𝑛 − 1)| =
|𝐻 (𝑛)| = ℎ𝑛,𝑛 |𝐻 (𝑛 − 1)| − ℎ𝑛−1,𝑛 ℎ𝑛,𝑛−1 |𝐻 (𝑛 − 2)| ,
𝑎𝑏 = 𝑄. 0
|𝐻 (2)| = ℎ1,1 ℎ2,2 − ℎ1,2 ℎ2,1 ,
𝑈𝑛 (𝑃, 𝑄) =
(11)
Clearly, this is also the first Lucas sequence, starting with 𝑈2 (𝑃, 𝑄). Therefore,
𝑎 + 𝑏 = 𝑃,
Then the successive determinants of 𝐻(𝑛) are given by the recursive formula:
( ( ( 𝐴 (𝑛) = ( ( ( (
𝑛 ≥ 3.
Since 𝑎 and 𝑏 are roots of 𝑥2 − 𝑃𝑥 + 𝑄 = 0,
ℎ2,1 ℎ2,2 ℎ2,3
|𝐻 (1)| = ℎ1,1 ,
|𝐴 (2)| = 𝑃2 − 𝑄,
|𝐴 (1)| = 𝑃,
(9) (10)
where |⋅| denotes the determinant 𝑎 and 𝑏 are roots of 𝑥2 −𝑃𝑥+ 𝑄 = 0.
𝑈𝑛 (𝑃, 𝑄) = ∏ (𝑃 + 2√𝑄 cos 𝑘=1
𝑛−1
𝑈𝑛 (𝑃, 𝑄) = ∏ (𝑃 − 2√𝑄 cos 𝑘=1
𝑛−1 sin (𝑛
𝑈𝑛 (𝑃, 𝑄) = (√𝑄)
𝑘𝜋 ), 𝑛
𝑛 ≥ 2,
(17)
𝑘𝜋 ), 𝑛
𝑛 ≥ 2,
(18)
arccos (𝑃/2√𝑄)) , sin ( arccos (𝑃/2√𝑄))
𝑛 ≥ 1, (19)
where √𝑄 ≡ √−𝑄 i and i = √−1 if 𝑄 is a negative integer.
Journal of Applied Mathematics
3
Proof. Choosing 𝛼 = 𝛽 = √𝑄 in {𝐴(𝑛), 𝑛 = 1, 2, . . .}, where √𝑄 represents √−𝑄i when 𝑄 is negative, we obtain a sequence of tridiagonal matrices {𝐵(𝑛), 𝑛 = 1, 2, . . .} of the form
( ( ( 𝐵 (𝑛) = ( ( ( (
𝑃
√𝑄
√𝑄
𝑃
√𝑄
√𝑄
𝑃 d
) ) ) ). ) d √𝑄) ) d
√𝑄
(20)
for 𝑛 ≥ 1
1 0 1 ( ) ( 1 0 d ) ( ) ). 𝐺 (𝑛) = ( ( ) ( ) d d 1 ( )
(21)
(22)
)
Note that 𝐵(𝑛) = 𝑃𝐼𝑛 + √𝑄𝐺(𝑛), where 𝐼𝑛 is the 𝑛 × 𝑛 identity matrix. Let 𝜆 𝑘 , 𝑘 = 1, 2, . . . , 𝑛, be the eigenvalues of 𝐺(𝑛) with associated eigenvectors 𝑥𝑘 . Then for each 1 ≤ 𝑗 ≤ 𝑛, we have 𝐵 (𝑛) 𝑥𝑗 = (𝑃𝐼𝑛 + √𝑄𝐺 (𝑛)) 𝑥𝑗
= 𝑃𝑥𝑗 + √𝑄𝜆 𝑗 𝑥𝑗
𝑝𝑛 (𝜆) = 𝜆𝑝𝑛−1 (𝜆) − 𝑝𝑛−2 (𝜆) ,
𝑛 ≥ 3.
(26)
This family of characteristic polynomials can be transformed into another family {𝑈𝑛 (𝑥), 𝑛 = 1, 2, . . .} by the transformation 𝜆 ≡ 2𝑥 as follows: 𝑈2 (𝑥) = 4𝑥2 − 1,
𝑈1 (𝑥) = 2𝑥,
𝑈𝑛 (𝑥) = 2𝑥𝑈𝑛−1 (𝑥) − 𝑈𝑛−2 (𝑥) ,
𝑛 ≥ 3,
(27)
and this family {𝑈𝑛 (𝑥), 𝑛 = 1, 2, . . .} is the set of Chebyshev polynomials of the second kind. It is well known [6] that defining 𝑥 ≡ cos 𝜃 allows the Chebyshev polynomials of the second kind to be written as follows: 𝑈𝑛 (𝑥) =
sin [(𝑛 + 1) 𝜃] . sin 𝜃
𝜆 𝑘 = 2 cos
𝑘𝜋 , 𝑛+1
(28)
𝑘 = 1, 2, . . . , 𝑛.
𝑛
𝑘=1
(23)
(24)
𝑘𝜋 ), 𝑛+1
𝑛 ≥ 1,
(30)
which is equivalent to (17). Moreover, by taking 𝑘 = 𝑛 + 1 − 𝑡, we get 𝑛
Therefore, 𝑃 + √𝑄𝜆 𝑘 , 𝑘 = 1, 2, . . . , 𝑛 are the eigenvalues of 𝐵(𝑛). Hence,
(29)
Combining (21), (24), and (29), we finally have 𝑈𝑛+1 (𝑃, 𝑄) = ∏ (𝑃 + 2√𝑄 cos
= (𝑃 + √𝑄𝜆 𝑗 ) 𝑥𝑗 .
𝑛 ≥ 1.
𝑝2 (𝜆) = 𝜆2 − 1,
From (28), we can easily see that the roots of 𝑈𝑛 (𝑥) = 0 are 𝜃𝑘 = 𝑘𝜋/(𝑛 + 1), 𝑘 = 1, 2, . . . , 𝑛, or equally, 𝑥𝑘 = cos 𝜃𝑘 = cos(𝑘𝜋/(𝑛 + 1)), 𝑘 = 1, 2, . . . , 𝑛. So we get the eigenvalues of 𝐺(𝑛) as follows:
1 0
= 𝑃𝐼𝑛 𝑥𝑗 + √𝑄𝐺 (𝑛) 𝑥𝑗
(25)
)
𝑝1 (𝜆) = 𝜆,
0 1
𝑘=1
) ) ) ); ) d d −1) )
−1 𝜆 d
we use Lemma 1 to obtain a recursive formula for the characteristic polynomials of {𝐺(𝑛), 𝑛 = 1, 2, . . .}:
by Theorem 2. Considering that the determinant of a matrix can be found by taking the product of its eigenvalues, we will compute the spectrum of 𝐵(𝑛) in order to find an alternate formulation for |𝐵(𝑛)|. We now introduce another sequence of tridiagonal matrices {𝐺(𝑛), 𝑛 = 1, 2, . . .}, where 𝐺(𝑛) is the 𝑛×𝑛 tridiagonal matrix with 𝑔𝑘,𝑘 = 0 (1 ≤ 𝑘 ≤ 𝑛) and 𝑔𝑘−1,𝑘 = 𝑔𝑘,𝑘−1 = 1 (2 ≤ 𝑘 ≤ 𝑛). That is,
|𝐵 (𝑛)| = ∏ (𝑃 + √𝑄𝜆 𝑘 ) ,
−1 𝜆 −1
(
Then
𝑛
( ( ( 𝜆𝐼𝑛 − 𝐺 (𝑛) = ( ( ( (
−1 𝜆
)
(
𝜆 −1
𝑃
(
𝑈𝑛+1 (𝑃, 𝑄) = |𝐵 (𝑛)|
Next we compute 𝜆 𝑘 ’s by recalling that each 𝜆 𝑘 is a zero of the characteristic polynomial 𝑝𝑛 (𝜆) = |𝜆𝐼𝑛 − 𝐺(𝑛)|. Notice that
𝑈𝑛+1 (𝑃, 𝑄) = ∏ (𝑃 + 2√𝑄 cos (𝜋 − 𝑡=1 𝑛
𝑡𝜋 )) 𝑛+1
𝑡𝜋 = ∏ (𝑃 − 2√𝑄 cos ), 𝑛+1 𝑡=1 which is identical to (18).
(31) 𝑛 ≥ 1,
4
Journal of Applied Mathematics
From (25), we can think of Chebyshev polynomials of the second kind as being generated by determinants of successive matrices of the form 2𝑥 −1 ) ) ) ), ) d d −1) )
−1 2𝑥 d
(32)
)
where 𝐻(𝑛, 𝑥) is 𝑛 × 𝑛. Note that 𝐵(𝑛) = −√𝑄𝐻(𝑛, −𝑃/2√𝑄). So we obtain 𝑛 𝑛 𝑃 𝑃 ) = (−√𝑄) 𝑈𝑛 (− ). |𝐵 (𝑛)| = (−√𝑄) 𝐻 (𝑛, − 2√𝑄 2√𝑄 (33) Since the Chebyshev polynomial of the second kind 𝑈𝑛 (𝑥) is an even or odd function, involving only even or odd powers of 𝑥, as 𝑛 is even or odd, the right side of (33) can be represented as 𝑛
(−√𝑄) 𝑈𝑛 (−
𝑛 𝑃 𝑃 ) = (√𝑄) 𝑈𝑛 ( ). 2√𝑄 2√𝑄
+ 1) arccos (𝑃/2√𝑄)) , sin (arccos (𝑃/2√𝑄))
𝑛 ≥ 1, (35)
which is identical to (19) since it also holds for 𝑛 = 0: 0 sin (arccos (𝑃/2√𝑄))
sin (arccos (𝑃/2√𝑄))
.
(36)
3. Complex Factorizations of the Second Lucas Sequence
Theorem 5. The second Lucas sequence {𝑉𝑛 (𝑃, 𝑄)}𝑛≥0 satisfies
𝑘=1
(𝑘 − (1/2)) 𝜋 ), 𝑛
𝑛 ≥ 1, (37)
𝑛
𝑉𝑛 (𝑃, 𝑄) = ∏ (𝑃 − 2√𝑄 cos 𝑘=1
(𝑘 − (1/2)) 𝜋 ), 𝑛
√𝑄
(40)
𝑃 )
(
where √𝑄 represents √−𝑄i when 𝑄 is negative. Then 𝑉𝑛 (𝑃, 𝑄) = 2 |𝐹 (𝑛)|
(41)
|𝐹 (𝑛)| =
1 (𝐼𝑛 + 𝑒1 𝑒1𝑇 ) 𝐹 (𝑛) , 2
(42)
where 𝐼𝑛 is the 𝑛 × 𝑛 identity matrix and 𝑒𝑗 is the 𝑗th column of 𝐼𝑛 . Moreover, we can express the right-hand side of (42) as follows:
where 𝐺(𝑛) is the 𝑛 × 𝑛 matrix defined in Theorem 3. Let 𝜇𝑘 , 𝑘 = 1, 2, . . . , 𝑛 be the eigenvalues of 𝐺(𝑛) + 𝑒1 𝑒2𝑇 with associated eigenvectors 𝑦𝑘 . Then for each 1 ≤ 𝑗 ≤ 𝑛, we get (𝑃𝐼𝑛 + √𝑄 (𝐺 (𝑛) + 𝑒1 𝑒2𝑇 )) 𝑦𝑗
The complex factorization of {𝑉𝑛 (𝑃, −1)}𝑛≥0 was derived in [4], and we generalize the formula to {𝑉𝑛 (𝑃, 𝑄)}𝑛≥0 for any integer 𝑄 in the following theorem.
𝑛
) (√𝑄 𝑃 √𝑄 ) ( ) ( ) ( ) ( √𝑄 𝑃 d ), ( 𝐹 (𝑛) = ( ) ) ( ) ( ) ( √𝑄 d d ) (
1 1 (𝐼 + 𝑒 𝑒𝑇 ) 𝐹 (𝑛) = 𝑃𝐼𝑛 + √𝑄 (𝐺 (𝑛) + 𝑒1 𝑒2𝑇 ) , (43) 2 2 𝑛 1 1
Remark 4. For 𝑃 = 1, 𝑄 = −1, (18) and (19) are exactly the formulas for Fibonacci numbers in [2].
𝑉𝑛 (𝑃, 𝑄) = ∏ (𝑃 + 2√𝑄 cos
√𝑄
by Theorem 1 of [4]. Unlike the derivation in Theorem 3, we will not compute the spectrum of 𝐹(𝑛) directly. Instead, since |𝐼𝑛 + 𝑒1 𝑒1𝑇 | = 2, we have
𝑛 sin ((𝑛
𝑈1 (𝑃, 𝑄) = 1 = (√𝑄)
(39)
(34)
Combining (21), (33), (34), and (28), we derive 𝑈𝑛+1 (𝑃, 𝑄) = (√𝑄)
𝑛 ≥ 1,
Proof. In order to obtain (37), we introduce a sequence of tridiagonal matrices {𝐹(𝑛), 𝑛 = 1, 2, . . .} of the form 𝑃 2
−1 2𝑥 (
𝑃 )) , 2√𝑄
where √𝑄 ≡ √−𝑄 i and i = √−1 if 𝑄 is a negative integer.
−1 2𝑥 −1
( ( ( 𝐻 (𝑛, 𝑥) = ( ( ( (
𝑛
𝑉𝑛 (𝑃, 𝑄) = 2(√𝑄) cos (𝑛 arccos (
𝑛 ≥ 1, (38)
= 𝑃𝐼𝑛 𝑦𝑗 + √𝑄 (𝐺 (𝑛) + 𝑒1 𝑒2𝑇 ) 𝑦𝑗 (44) = 𝑃𝑦𝑗 + √𝑄𝜇𝑗 𝑦𝑗 = (𝑃 + √𝑄𝜇𝑗 ) 𝑦𝑗 . Therefore, 𝑃 + √𝑄𝜇𝑘 , 𝑘 = 1, 2, . . . , 𝑛, are the eigenvalues of 𝑃𝐼𝑛 + √𝑄(𝐺(𝑛) + 𝑒1 𝑒2𝑇 ). Hence, 𝑛 (𝐼𝑛 + 𝑒1 𝑒1𝑇 ) 𝐹 (𝑛) = ∏ (𝑃 + √𝑄𝜇𝑘 ) , 𝑘=1
𝑛 ≥ 1.
(45)
Journal of Applied Mathematics
5
Next we compute 𝜇𝑘 ’s by recalling that each 𝜇𝑘 is a zero of the characteristic polynomial 𝑞𝑛 (𝜇) = |𝜇𝐼𝑛 − (𝐺(𝑛) + 𝑒1 𝑒2𝑇 )|. Since |(𝐼𝑛 − (1/2)𝑒1 𝑒1𝑇 )| = 1/2, we can represent the characteristic polynomial as follows: 1 𝑞𝑛 (𝜇) = 2 (𝐼𝑛 − 𝑒1 𝑒1𝑇 ) (𝜇𝐼𝑛 − (𝐺 (𝑛) + 𝑒1 𝑒2𝑇 )) . (46) 2 Notice that 1 (𝐼𝑛 − 𝑒1 𝑒1𝑇 ) (𝜇𝐼𝑛 − (𝐺 (𝑛) + 𝑒1 𝑒2𝑇 )) 2 𝜇 −1 2 (−1 𝜇 −1 ) ( ) ( ) −1 𝜇 d =( ). ( ) ( ) d d −1
𝜇 , 2
𝑞2 (𝜇) =
(48)
𝑛 ≥ 3.
This family of characteristic polynomials can be transformed into another family {𝑇𝑛 (𝑥), 𝑛 = 1, 2, . . .} by taking 𝜇 = 2𝑥: 𝑇1 (𝑥) = 𝑥,
𝑇2 (𝑥) = 2𝑥2 − 1,
𝑇𝑛 (𝑥) = 2𝑥𝑇𝑛−1 (𝑥) − 𝑇𝑛−2 (𝑥) ,
(49)
𝑛 ≥ 3.
In fact, this family {𝑇𝑛 (𝑥), 𝑛 = 1, 2, . . .} is the set of Chebyshev polynomials of the first kind. It is well known [6] that defining 𝑥 ≡ cos 𝜃 allows the Chebyshev polynomials of the first kind to be written as 𝑇𝑛 (𝑥) = cos 𝑛𝜃.
(50)
It is easy to see that the roots of 𝑇𝑛 (𝑥) = 0 are given by 𝜃𝑘 = (𝑘 − (1/2))𝜋/𝑛, 𝑘 = 1, 2, . . . , 𝑛, or 𝑥𝑘 = cos 𝜃𝑘 = cos(𝑘 − (1/2))𝜋/𝑛, 𝑘 = 1, 2, . . . , 𝑛. Thus, we derive (𝑘 − (1/2)) 𝜋 𝜇𝑘 = 2 cos , 𝑘 = 1, 2, . . . , 𝑛. 𝑛 Combining (41), (42), (45), and (51), we finally have 𝑛
𝑉𝑛 (𝑃, 𝑄) = ∏ (𝑃 + 2√𝑄 cos 𝑘=1
(𝑘 − (1/2)) 𝜋 ), 𝑛
(51)
𝑛 ≥ 1. (52)
Furthermore, by taking 𝑘 = 𝑛 + 1 − 𝑡, we obtain 𝑛
𝑉𝑛 (𝑃, 𝑄) = ∏ (𝑃 + 2√𝑄 cos (𝜋 − 𝑡=1 𝑛
= ∏ (𝑃 − 2√𝑄 cos 𝑡=1
where 𝐾(𝑛, 𝑥) is 𝑛 × 𝑛. Note that 𝐹(𝑛) = −√𝑄𝐾(𝑛, −𝑃/2√𝑄). So we get 𝑛 𝑛 𝑃 𝑃 ) = (−√𝑄) 𝑇𝑛 (− ). |𝐹 (𝑛)| = (−√𝑄) 𝐾 (𝑛, − 2√𝑄 2√𝑄 (55) Since the Chebyshev polynomial of the first kind 𝑇𝑛 (𝑥) is an even or odd function, involving only even or odd powers of 𝑥, according as 𝑛 is even or odd, the right side of (55) can be represented as 𝑛
(−√𝑄) 𝑇𝑛 (−
𝑛 𝑃 𝑃 ) = (√𝑄) 𝑇𝑛 ( ). 2√𝑄 2√𝑄
𝑛 ≥ 1.
(56)
Combining (41), (55), (56), and (50), we derive 𝑛
𝑉𝑛 (𝑃, 𝑄) = 2(√𝑄) cos (𝑛 arccos (
𝑃 )) , 2√𝑄
𝑛 ≥ 1. (57)
Equation (39) is completed. Remark 6. For 𝑃 = 1, 𝑄 = −1, (38) and (39) are exactly the formulas for Lucas numbers in [2]. Remark 7. For 𝑄 = −1, (38) and (39) are exactly the formulas for {𝑉𝑛 (𝑃, −1)} in [4].
4. Examples Finally, we give some examples. Example 1. Choosing 𝑃 = 2, 𝑄 = −1, we can get the complex factorization of Pell numbers by (17). Let 𝑃𝑛 be the 𝑛th Pell number and i = √−1. Then the fourth Pell number is 3
𝑃4 = 𝑈4 (2, −1) = ∏ (2 + 2i cos
𝑘𝜋 ) 4
2𝜋 𝜋 = (2 + 2i cos ) (2 + 2i cos ) 4 4
(53) Equations (37) and (38) are proved.
(54)
)
𝑘=1
(𝑡 − (1/2)) 𝜋 )) 𝑛
(𝑡 − (1/2)) 𝜋 ), 𝑛
−1 2𝑥 −1 ( ) ( ) −1 2𝑥 d ( ) ), 𝐾 (𝑛, 𝑥) = ( ( ) ( d d −1) ( ) (
(47)
𝜇2 − 1, 2
𝑞𝑛 (𝜇) = 𝜇𝑞𝑛−1 (𝜇) − 𝑞𝑛−2 (𝜇) ,
𝑥 −1
−1 2𝑥
−1 𝜇 ( ) We use Lemma 1 to obtain a recursive formula for the characteristic polynomials: 𝑞1 (𝜇) =
Now we think of Chebyshev polynomials of the first kind from (47) as being generated by determinants of successive matrices of the form
× (2 + 2i cos
3𝜋 ) 4
= (2 + √2i) (2 + 0) (2 − √2i) = 12.
(58)
6
Journal of Applied Mathematics
Example 2. Choosing 𝑃 = 1, 𝑄 = −2, we can get the complex factorization of Jacobsthal numbers by (18). Let 𝐽𝑛 be the 𝑛th Jacobsthal number and i = √−1. Then the third Jacobsthal number is 2
𝐽3 = 𝑈3 (1, −2) = ∏ (1 − 2√2i cos 𝑘=1
𝑘𝜋 ) 3
𝜋 2𝜋 = (1 − 2√2i cos ) (1 − 2√2i cos ) 3 3
(59)
= (1 − √2i) (1 + √2i) = 3. Example 3. Choosing 𝑃 = 2, 𝑄 = −1, we can get the complex factorization of Pell-Lucas numbers by (37). Let 𝑝𝑛 be the 𝑛th Pell-Lucas number and i = √−1. Then the third Pell-Lucas number is 3
𝑝3 = 𝑉3 (2, −1) = ∏ (2 + 2i cos 𝑘=1
(𝑘 − (1/2)) 𝜋 ) 3
3𝜋 5𝜋 𝜋 = (2 + 2i cos ) (2 + 2i cos ) (2 + 2i cos ) 6 6 6
(60)
= (2 + √3i) (2 + 0) (2 − √3i) = 14. Example 4. Choosing 𝑃 = 1, 𝑄 = −2, we can get the complex factorization of Jacobsthal-Lucas numbers by (38). Let 𝑗𝑛 be the 𝑛th Jacobsthal-Lucas number and i = √−1. Then the third Jacobsthal-Lucas number is 3
𝑗3 = 𝑉3 (1, −2) = ∏ (1 − 2√2i cos 𝑘=1
(𝑘 − (1/2)) 𝜋 ) 3
𝜋 3𝜋 = (1 − 2√2i cos ) (1 − 2√2i cos ) 6 6 × (1 − 2√2i cos
5𝜋 ) 6
(61)
= (1 − √6i) (1 − 0) (1 + √6i) = 7.
5. Conclusion We use the tridiagonal matrices 𝐵(𝑛) and 𝐹(𝑛) to obtain the complex factorizations of the first and second Lucas sequences, respectively. It is possible to find other suitable matrices for different factorizations of Lucas sequences.
Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this article.
Acknowledgments The author’s research is supported by Excellent Young Teacher Foundation of Shanghai (sxy10004) and Foundation of Shanghai Business School (A-0201-00-050-205). The author also appreciates the editors and the anonymous referees for their helpful suggestions and comments.
References [1] P. Ribenboim, My Numbers, My Friends, Popular Lectures on Number Theory, Springer, Berlin, Germany, 2000. [2] N. D. Cahill, J. R. D’Errico, and J. P. Spence, “Complex factorizations of the Fibonacci and Lucas numbers,” The Fibonacci Quarterly, vol. 41, no. 1, pp. 13–19, 2003. [3] N. D. Cahill and D. A. Narayan, “Fibonacci and Lucas numbers as tridiagonal matrix determinants,” The Fibonacci Quarterly, vol. 42, no. 3, pp. 216–221, 2004. [4] S¸. Burcu Bozkurt, F. Yılmaz, and D. Bozkurt, “On the complex factorization of the Lucas sequence,” Applied Mathematics Letters, vol. 24, no. 8, pp. 1317–1321, 2011. [5] J. Feng, “Fibonacci identities via the determinant of tridiagonal matrix,” Applied Mathematics and Computation, vol. 217, no. 12, pp. 5978–5981, 2011. [6] J. C. Mason and D. C. Handscomb, Chebyshev Polynomials, CRC Press, 2003.
Advances in
Operations Research Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Advances in
Decision Sciences Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Journal of
Applied Mathematics
Algebra
Hindawi Publishing Corporation http://www.hindawi.com
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Journal of
Probability and Statistics Volume 2014
The Scientific World Journal Hindawi Publishing Corporation http://www.hindawi.com
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
International Journal of
Differential Equations Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Volume 2014
Submit your manuscripts at http://www.hindawi.com International Journal of
Advances in
Combinatorics Hindawi Publishing Corporation http://www.hindawi.com
Mathematical Physics Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Journal of
Complex Analysis Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
International Journal of Mathematics and Mathematical Sciences
Mathematical Problems in Engineering
Journal of
Mathematics Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Discrete Mathematics
Journal of
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Discrete Dynamics in Nature and Society
Journal of
Function Spaces Hindawi Publishing Corporation http://www.hindawi.com
Abstract and Applied Analysis
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
International Journal of
Journal of
Stochastic Analysis
Optimization
Hindawi Publishing Corporation http://www.hindawi.com
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Volume 2014
