Factorizations and Representations of Binary Polynomial Recurrences by Matrix Methods

In this paper we derive factorizations and representations of a polynomial analogue of an arbitrary binary sequence by matrix methods. It generalizes various results on Fibonacci, Lucas, Chebyshev and Morgan–Voyce polynomials. 1. Introduction In [10], the divisibility properties of the Fibonacci polynomial sequence ffn (x)g was studied. The Fibonacci polynomial sequence is de…ned by the recursion fn+2 (x) = xfn+1 (x) + fn (x) ; f0 (x) = 0; f1 (x) = 1: Five years later, Hoggatt and Long [2] considered the general Fibonacci type polynomial sequence fun (x; y)g of two variables. This sequence is de…ned by (1.1) un+2 (x; y) = xun+1 (x; y) + yun (x; y) where u0 (x; y) = 0 and u1 (x; y) = 1: The authors of [2, 10] found the roots of these polynomials and then obtained the factorizations of their polynomials. In [2], the authors found that un (x; y) = y (n 1)=2 n 1 Y k=1 x p y 2i cos k n : Further, in [1], the authors use the relationships between the determinants of certain tridiagonal matrices and the Fibonacci and Lucas numbers, and then by matrix methods, they obtained the factorizations and representations of these sequences. The factorization of Fibonacci numbers was initially proposed in [6], and the factorization of Lucas numbers was obtained in [11]. The (companion) generalized Lucas polynomial sequence vn (x; y) is de…ned by vn+2 (x; y) = xvn+1 (x; y) + yvn (x; y) where v0 (x; y) = 2 and v1 (x; y) = 1: Recently, in [5], the binary sequential analogues of the generalized Fibonacci and Lucas polynomial sequences was considered and factorizations and representations of these sequences was obtained. These sequences are de…ned by Un+1 = AUn +BUn 1 Vn+1 = AVn +BVn 1 2000 Mathematics Subject Classi…cation. 11B37, 11C20. Key words and phrases. Second order recurrences, factorization, Chebyshev polynomials, tridiagonal matrix, matrix methods. 1 2 EMRAH KILIC AND PANTELIMON ST 1⁄4 ANIC 1⁄4 A where U0 = 0; U1 = 1 and V0 = 2; V1 = A; respectively. Also, in [4], we gave the more general factorizations of second order linear recurrences fUng and fVng with indices in arithmetic progressions. Furthermore, we obtained the factorization of these general sequences by the matrix methods considering how these recurrences are related to the determinants of certain tridiagonal matrices. As it can be seen from the above mentioned results, the most general cases of these polynomial and binary sequences have not been studied. Consequently, we de…ne fAn (a; b; p; q) (x)g (we shall often drop the argument (a; b; p; q) and simply write fAn(x)g) to be a polynomial sequence satisfying An+1(x) = p(x)An(x) + q(x)An 1(x); and initial conditions A0 = a(x); A1 = b(x), where a; b; p; q are polynomials of an indeterminate x with real coe¢ cients. For easy notation, we shall sometimes write An; p; q; a; b for An(x); p (x), q (x), a (x) and b (x), respectively. We display some special cases of the sequence fAng in the following table. p(x) q(x) a(x) b(x) An (a; b; p; q) (x) Polynomial Type 2x 1 1 x Tn (x) 1st kind Chebyshev 2x 1 1 2x Un (x) 2nd kind Chebyshev x+ 1 1 1 x+ 1 bn (x) Morgan-Voyce x y 0 1 un (x; y) generalized Fibonacci x y 2 1 vn (x; y) generalized Lucas 2x 1 1 2x+ 1 Wn (x) 4th kind Chebyshev 2x 1 1 2x 1 Vn (x) 3rd kind Chebyshev Table 1 In Section 2 we present a recurrence, and de…ne a tridiagonal matrix whose determinant is precisely An (a; b; p; q) (x), with n in an arithmetic progression, n c (mod k). In our main Section 3 we derive the factorization and representations of the sequence fAn (a; b; p; q)g (under some assumptions), by matrix methods, thus generalizing some results of [1]–[5] and others. As consequences, we obtain the factorizations for the Chebyshev’s and generalized Lucas polynomials, among others. 2. A recurrence for An (a; b; p; q) (x), where n c (mod k) We start this section with the Binet formulas of both positively and negatively indexed terms of the sequence fAng, namely (2.1) A n = b a n + a b n where (2.2) = p+ p p2 + 4q 2 ; = p p p2 + 4q 2 : First, we prove the following lemma. Lemma 1. For k > 0; n > 0, the sequence fAng satis…es the following recursion (2.3) Ap(n+1; k;c) = y kAp(n; k;c) z kAp(n 1; k;c) where y k = k + , z k = q k and p(n; k; c) = nk + c (c constant). FACTORIZATIONS AND REPRESENTATIONS OF BINARY POLYNOMIAL RECURRENCES 3 Proof. From the de…nition of the sequence fAng, considering the case A0 = 2 and A1 = p; we write y k = A k (2; p; p; q) = k + k where and are given by (2.2). Further, for the positive case, we note that yk = Ak (2; p; p; q) = k + : Since the positively and negatively indexed terms cases are similar, we only consider the positively indexed terms case. By the Binet formula of the sequence fAng and since zk = q = ( ) ; ykAp(n;k;c) zkAp(n 1;k;c) = k + k h b a kn+c + a b kn+c i ( ) h b a k(n 1)+c + a b k(n 1)+c i = 1 ((b a ) k(n+1)+c + (a b) k(n+1)+c +(b a ) kn+c k + (a b) kn+c k (b a ) kn+c k (a b) kn+c ) = b a k(n+1)+c + a b k(n+1)+c = Ap(n+1;k;c); which proves the lemma. Now we present a relationship between the terms Ap( (n+1);k;c) and the determinant of a certain tridiagonal matrix. De…ne the n n tridiagonal matrix Mn;k as shown: (2.4) Mn; k = 2666664 Ap(1; k;c) Ap(0; k;c) p z k p z k y k p z k p z k y k . . . . . . . . . pz k p z k y k 3777775 : Theorem 1. For n > 1, we have detMn; k = Ap(n; k;c): Proof. As before, we only consider the positively indexed terms. The other case can be similarly derived. Expanding detMn;k using the cofactor expansion of a determinant with respect to the last column, we obtain detMn;k = yk detMn 1;k zk detMn 2;k: Replacing n = 1 in equation (2.3), we obtain Ap(2;k;c) = ykAp(1;k;c) zkAp(0;k;c) and detM2;k = ykAp(1;k;c) zkAp(0;k;c), and so, detM2;k = Ap(2;k;c). Obviously, detM1;k = Ap(1;k;c). Since the recurrence relations (and initial conditions) of detMn;k and the sequence Ap(n;k;c) are the same, the conclusion follows from Lemma 1. 4 EMRAH KILIC AND PANTELIMON ST 1⁄4 ANIC 1⁄4 A 3. Factorizations for An (a; b; p; q) (x) Now we investigate all possible cases of our main general considerations and then we give their factorizations and representations. However, a few special cases could be treated separately. For compactness, we do not consider all these cases, but we shall point out, whenever appropriate some hints in treating those cases. First, we consider the case Ap(1; k;c) = y k and Ap(0; k;c) = 1: Under these assumptions, we label the matrix Mn; k and the sequence Ap( n;k;c) byM (1) n; k and A (1) p(n; k;c); respectively. By Theorem 1, the matrixM (1) n; k takes the following form: M (1) n; k = 266664 y k p z k p z k p z k . . . . . . . . . pz k p z k y k 377775 and we have that (3.1) detM (1) n; k = A (1) p(n; k;c): Let Q1 be the following (n n) tridiagonal Toeplitz matrix Q1 = 266664 0 1 1 0 . . . . . . . . . 1 1 0 377775 : The characteristic equation of the matrix Q1 satis…es the following recursion, for n > 2 fn ( ) = fn 1 ( ) fn 2 ( ) ; where f1 ( ) = and f2 ( ) = 2 1: Taking 2x; the family ffn ( )g is reduced to the family fUn (x)g (Chebyshev polynomial of the second kind). From [9, 8, 7], the zeros of the Chebyshev polynomials are known, and so, the eigenvalues of Q1 are of the form: (3.2) r = 2 cos r n+1 ; r = 1; 2; : : : ; n: We can also write M (1) n; k = y kIn + p z kQ1, where In is the n n unit matrix. Theorem 2. Let Ap(1; k;c) = y k and Ap(0; k;c) = 1. Then for n > 1; A (1) p(n; k;c) = n Y r=1 y k 2 p z k cos r n+1 : Proof. Assume that r’s are the eigenvalues of the matrix Q1 with respect to the eigenvectors wr: Since M (1) n; kwr = y kIn + p z kQ1 wr = y kInwr + p z kQ1wr = y k + p z k r wr; FACTORIZATIONS AND REPRESENTATIONS OF BINARY POLYNOMIAL RECURRENCES 5 the y k + p z k r ’s are the eigenvalues of the matrix M (1) n; k with respect to the eigenvectors wr: Thus, by (3.2) and (3.1), we have the conclusion: detM (1) n; k = A (1) p(n; k;c) = n Y r=1 y k + p z k r = n Y r=1 y k 2 p z k cos r n+1 : Corollary 1. For n > 1; A (1) p(n; k;c) = 8><>>: y k n=2 Q r=1 y k 4z k cos r n+1 if n is even, (n 1)=2 Q r=1 y k 4z k cos r n+1 if n is odd. Proof. Since cos (k =n) = cos ((n k) =n) for 1 k n=2; the conclusion follows from Theorem 2. Now we give some applications of Theorem 2 in the following corollaries. Corollary 2. Let Un be nth Chebyshev polynomial of the second kind. For n; k > 1; then (3.3) U (n+1)k 1 = 2 U k 1 n Y r=1 T k cos r n+1 : where Tn is nth Chebyshev polynomial of the …rst kind. Proof. When p = 2x; q = 1; then the …rst coe¢ cient y k = 2T k, where Tk is the kth Chebyshev polynomials of the …rst kind (see Table 1). According to the …rst case, to satisfy A p(1; k;c) = y k = 2T k and A (1) p(0; k;c) = 1; we shall choose a = 1 and b = 2x. We see that for c = 1; A (1) p(1; k; 1) = U 2k 1 U k 1 = 2T k and A (1) p(0; k; 1) = U k 1 U k 1 = 1 and in general A (1) p(n; k; 1) = U (n+1)k 1 U k 1 : Then by Theorem 2, we get for yk = 2Tk and zk = 1 (3.4) A p(n; k;c) = U (n+1)k 1 U k 1 = n Y r=1 2T k 2 cos r n+1 : The required equation (3.3) follows immediately. Corollary 3. For n; k > 1; then U k(n+1) 1 = 8><>>: 2 U 2k 1 (n 1)=2 Q r=1 h T 2 k cos r n+1 i if n is odd,