A Note on Fibonacci-type Polynomials

We opt to study the convergence of maximal real roots of certain Fibonacci-type polynomials given by Gn = xGn−1 + Gn−2. The special cases k = 1 and k = 2 were done by Moore, and Zeleke and Molina, respectively. 1. Main Results In the sequel, P denotes the set of positive integers. The Fibonacci polynomials [2] are defined recursively by F0(x) = 1, F1(x) = x and Fn(x) = xFn−1(x) + Fn−2(x), n ≥ 2. Fact 1. Let n ≥ 1. Then the roots of Fn(x) are given by xk = 2i cos ( πk n + 1 ) , 1 ≤ k ≤ n. In particular a Fibonacci polynomial has no positive real roots. Proof. The Fibonacci polynomials are essentially Tchebycheff polynomials. This is well-known (see, for instance [2]). ! Let k ∈ P be fixed. Several authors ([3]-[7]) have investigated the so-called Fibonacci-type polynomials. In this note, we focus on a particular group of polynomials recursively defined by G n (x) =    −1, n = 0 x− 1, n = 1 xkG n−1(x) + G (k) n−2(x), n ≥ 2. When there is no confusion, we suppress the index k to write Gn for G (k) n (x). We list a few basic properties relevant to our work here. INTEGERS: 10 (2010) 14 Fact 2. For each k ∈ P, there is a rational generating function for Gn; namely, ∑ n≥0 G n (x)t n = (xk + x− 1)t− 1 1− xkt− t2 . Proof. The claim follows from the definition of Gn. ! Fact 3. The following relation holds G n (x) = G n−1Fn−1(x) + (−1)n−1 Fn−2(x) . Proof. We write the equivalent formulation G n (x) = det   x− 1 −1 0 0 . . . 0 0 0 −1 xk −1 0 . . . 0 0 0 0 1 xk −1 . . . 0 0 0 .. .. .. .. . . . .. .. .. 0 0 0 0 . . . xk −1 0 0 0 0 0 . . . 1 xk −1 0 0 0 0 . . . 0 1 xk   , then we apply Dodgson′s determinantal formula [1]. ! Fact 4. For a fixed k, let {g n }n∈P be the maximal real roots of {G n (x)}n. Then {g 2n }n and {g (k) 2n−1}n are decreasing and increasing sequences, respectively. Proof. First, each gn exists since Gn(0) = 1 < 0 and Gn(∞) = ∞. Assume x > 0. Invoking Fact 3 from above, twice, we find that F2n−3(x)G (k) 2n (x) = F2n−1(x k)G 2n−2(x) + x , F2n−2(x)G (k) 2n+1(x) = F2n(x k)G 2n−1(x)− x. From these equations and Fn(x) > 0 (see Fact 1), it is clear that G2n−2(x) > 0 implies G2n(x) > 0; also if G2n−2(x) = 0 then G2n(x) > 0. Thus g2n−2 > g2n. A similar argument shows g2n+1 > g2n−1. The proof is complete. ! Define the quantities α(x) = x + √ x2 + 4 2 , β(x) = x− √ x2 + 4 2 , p(x) = (x− 1) + β(xk) α(xk)− β(xk) , q(x) = (x− 1) + α(xk) α(xk)− β(xk) . INTEGERS: 10 (2010) 15 Fact 5. For n ≥ 0 and k ∈ P, we have the explicit formula G n (x) = p(x)α (x)− q(x)β(x). Proof. This is the standard generalized Binet formula. ! For each k ∈ P, we introduce another set of polynomials H(x) = x − xk−1 + x− 2. Fact 6. For each k ∈ P, the polynomial H(k)(x) has exactly one positive real root ξ(k) > 1. Proof. Since H(k)(x) = (x − 1)(xk−1 + 1) − 1 < 0, whenever 0 < x ≤ 1, there are no roots in the range 0 < x ≤ 1. On the other hand, H(k)(1) < 0,H(k)(∞) = ∞ and the derivative d dx H(x) = xk−1(k(x− 1) + 1) + 1 > 0 whenever x ∈ P, suggest there is only one positive root (necessarily greater than 1). ! Fact 7. If k is odd (even), then H(k)(x) has no (exactly one) negative real root. Proof. For k odd, H(k)(−x) = (−x− 1)(xk−1 +1)− 1 < 0. For k even, H(k)(−x) = xk + xk−1 − x − 2 changes sign only once. We now apply Descartes′ Rule to infer the claim. ! Now, we state and prove the main result of the present note. Theorem. Preserve the notations of Facts 4 and 6. Then, depending on the parity of n, the roots {g n }n converge from above or below so that g n → ξ(k) as n→∞. Note also ξ(k) → 1 as k →∞. Proof. For notational brevity, we suppress k and write gn and ξ. From Gn(gn) = 0 and Fact 5 above, we resolve p(gn) q(gn) = βn(gk n) αn(gk n) , or 2(gn − 1) + gk n − √ g2k n + 4 2(gn − 1) + gk n + √ g2k n + 4 = (−1) ( 1− 2gn gk n + √ g2k n + 4 )n . (1) INTEGERS: 10 (2010) 16 Using Gershgorin′s Circle theorem, it is easy to see that 1 ≤ gn ≤ 2. When combined with Fact 4, the monotonic sequences {g2n}n and {g2n−1}n converge to finite limits, say ξ+ and ξ− respectively. The right-hand side of (1) vanishes in the limit n→∞, thus 2(ξ − 1) + ξ − √ ξ2k + 4 = 0. Further simplification leads to H(k)(ξ) = ξk − ξk−1 + ξ − 2 = 0. From Fact 6, such a solution is unique. So, ξ+ = ξ− = ξ completes the proof. ! 2. Further Comments In this section, we discuss the bivariate Fibonacci polynomials, of the first kind (BFP1), defined as gn(x, y) = xgn−1(x, y) + ygn−2(x, y), g0(x, y) = x, g1(x, y) = y. If x = y = 1 then the resulting sequence is the Fibonacci numbers. The following is a generating function for the BFP1: ∑ n≥0 gn(x, y)t = x + (y − x2)t 1− xt− yt2 . It is also possible to give an explicit expression: