Roots of a Type of Generalized Quasi-Fibonacci Polynomials

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1 Roots of a Type of Generalzed Quas-Fbonacc Polynomals Danel Thompson July 8, 2013 Abstract Let a be a nonnegatve real number and defne a quas-fbonacc polynomal sequence by F1 a(x = a, F 2 a(x = x, and F n a (x = Fn 1 a (x + xf n 2 a (x for n 2. Let ra n denote the maxmum real root of Fn a. We prove for certan values of a that the sequence {r2n a } converges monotoncally to β a = a 2 + a from above and the sequence {r2n+1 a } converges monotoncally to β a from below. 1 Introducton Consder a Fbonacc type polynomal sequence defned recursvely by G 0 (x = α, G 1 (x = x + β, and for n 2 G n = xg n 1 + G n 2. Here α and β are ntegers wth α 0. The propertes of these polynomals have been studed extensvely: when α = 1 and β = 0, G n s the classcal Fbonaccc polynomal sequence. Lkewse, when α = 2 and β = 0, one gets the classcal Lucas polynomals. Hoggatt and Bcknell explctly fnd the zeros of these two sequences n [2]. Moore [4] and Prodnger [5] studed the asymptotc behavor of the maxmal roots when α = β = 1. Moore s results have been generalzed to more cases; n [6], Yu, Wang, and He study the case when α = β = a for all postve ntegers a, whle Mátyás [3] examnes the sequence for α = a 0 and β = ±a. More recently, Wang and He fully generalzed ther results to any ntegers α and β wth α 0. 1

2 Let a be a postve real number and defne the quas-fbonacc polynomal sequence F a n recursvely by F a 0 (x = a, F a 1 (x = x, and for n 2, F a n(x = F a n 1(x + xf a n 2(x. (1 In [1], B. Alberts studes ths sequence when a = 1 and fnds that the maxmum roots of even ndexed polynomals converge monotoncally to 2 from above, whle the maxmum roots of the odd ndexed polynomals converge monotoncally to 2 from below. In general, denote the maxmum root of Fn(x a by rn a. When no ambguty wll arse we wll omt the superscrpt a. The paper s organzed as follows: n secton 2 we present techncal results used to prove later theorems. In secton 3 we examne the exstence, boundedness, and monotonc behavor of the maxmum roots. In sectons 4 and 5 we study propertes of the dervatve needed to show convergence. Fnally, man results are provded n secton 6. 2 Techncal results Lemma 1. Denote the standard Fbonacc sequence by f 0 = 0, f 1 = 1, and for n 2, f n = f n 1 + f n 2. Then F n (1 = f n f n+1 (2 F n (0 = a (3 F n ( 1 { a, 1, 1, a, 1 + a, 1} (4 Proof. The proofs for Equatons (2 and (3 follow drectly from the ntal condtons and recurson. For Equaton (4, we notce that F 0 ( 1 = a, F 1 ( 1 = 1, F 2 ( 1 = 1, F 3 ( 1 = a, F 4 ( 1 = 1 + a, F 5 ( 1 = 1. An nductve argument shows ths pattern holds for all values of n. Lemma 2. Let β = a 2 + a. Then F n (β = ( a n+1. Proof. Ths s clear for F 0 and F 1. Suppose t holds through n 1. Then F n (β = F n 1 (β + β F n 2 (β = ( a n + β( a n 1 = ( a n 1 ( a + β = ( a n 1 (a 2 = ( a n+1. 2

3 Lemma 3. F n (x = (1 + 2xF n 2 (x x 2 F n 4 (x5 Proof. Ths follows drectly from manpulaton of the recursve formula. Theorem 4. F n (x = =0 [( n 1 ( n ] x (6 Proof. Notce that each sum s fnte snce for large enough, 1 > n and > n. Drect computaton verfes the formula for F 0 and F 1 ; suppose we have shown the theorem through F n 1. Then F n (x = F n 1 (x + xf n 2 (x [( = n 1 1 n 1 ] x + x = = =0 =0 =0 [( n 1 1 [( n 1 ( n 1 n ] x ] x + j=1 [( n 2 1 =0 [ (n 1 j j 2 ( n 2 ( n 1 j j 1 ] x ] x j + 0 Corollary. F 2n (x = n =0 [( 2n 1 ( 2n ] x (7 n+1 [( ] 2n + 1 2n + 1 F 2n+1 (x = x (8 1 =0 3 Postve real roots Lemma 5. If ( n 1 n > 0, then n (+1 (+1 1 n (+1 (+1 > 0. If ( n 1 n 0, then n ( 1 ( 1 1 n ( 1 ( 1 < 0. 3

4 Proof. Suppose ( ( n 1 n n > 0. Then = ( 1 n 2+1 n > a. So ( n (+1 (+1 1 ( n (+1 (+1 = ( + 1 n 2( > n > a Our concluson follows drectly from ths fact; the second result s proven smlarly. Theorem 6. F 2n+1 has exactly one nonnegatve real root. When n a, F 2n has no postve real roots, and when n > a, F 2n has exactly one nonnegatve real root. Proof. Frst notce that f a = 0 then the maxmal root of each F n s 0 for n 1. Suppose a > 0. We use Descartes Rule of Sgns. Equaton (8 shows that the constant term of F 2n+1 s a < 0 and the leadng coeffcent of F 2n+1 s 1 > 0. Thus, by Lemma 5, there must be exactly one change n sgn of the coeffcents of F 2n+1. Ths mples that the polynomal has exactly one real postve root. Smlarly, Equaton (7 shows the constant term of F 2n s a < 0. However, when n a, the leadng coeffcent of F 2n s n 0, so there are no sgn changes and therefore no postve real roots. When n > a, n > 0, so there must be exactly one sgn change and therefore exactly one postve real root. Corollary. Let r n denote the maxmal real root of the polynomal F n (f t exsts. Then for all n, r 2n+1 exsts and r 2n+1 < β. Smlarly, for all n > a, r 2n exsts and β < r 2n. Corollary. For all n and for all x (0, r 2n+1, F 2n+1 (x < 0; for all x > r 2n+1, F 2n+1 (x > 0. Smlarly, for all n > a, for all x (0, r 2n, F 2n (x < 0 and for all x > r 2n, F 2n (x > 0. 4

5 Theorem 7. For all n, Furthermore, when N > a and n > N, a = r 1 < r 3 < < r 2n+1 < < β. (9 β < < r 2n < < r 2N+2 < r 2N. (10 Proof. Clearly r 1 = a and F 3 (r 1 = F 2 (a < 0, so r 1 < r 3. Suppose we have shown Equaton (9 holds through r 2n 1. Then snce r 2n 3 < r 2n 1, Equaton (5 gves us F 2n+1 (r 2n 1 = (1 + 2r 2n 1 F 2n 1 (r 2n 1 r 2 2n 1F 2n 3 (r 2n 1 Thus, r 2n+1 > r 2n 1. The proof of (10 s smlar. = r 2 2n 1F 2n 3 (r 2n 1 < 0. 4 Behavor of dervatves For the remander of the paper, we wll use N a (or just N to denote an nteger satsfyng the followng four propertes: ( N a > a. ( r 2Na+1 a 2. ( For x β, F 2N a 1 (x > a2na 1 and F 2N a (x > a 2Na. (v For x > r 2Na+1, F 2N a+1 (x 1, F 2N a (x 1, and F 2N a+2 (x xf 2N a (x. Numercal evdence suggests that such an N a exsts for all a 0. For nstance, when 0 a 1, t can be shown drectly that N a = 2 satsfes the propertes. Lemma 8. Let x β. Then for all n 2N 1, F n(x > a n. 5

6 Proof. Snce N satsfes condton (, F 2N 1 (x > a2n 1 and F 2N (x > a 2N. Suppose we have shown the lemma holds for all ntegers between 2N 1 and 2n. Then snce F 2n 1 > 0 on [β,, F 2n 1 (x F 2n 1 (β, so F 2n+1(x = F 2n(x + xf 2n 1(x + F 2n 1 (x The proof for F 2n+1 follows smlarly. a 2n + β a 2n 1 + a 2n = a 2n a 2n > a 2n+1. Lemma 9. For all n > N and x > r 2n 1, F 2n+1(x x n N F 2n+2(x x n N+1 F 2n 1(x x n N 1 F 2N+1(x 1 and F 2n(x x n N F 2N(x 1. Proof. Let x > r 2n 1. Snce N satsfes condton (v, F 2N+1 (x 1, F 2N (x 1, and F 2N+2 (x xf 2N (x. Suppose we have shown the lemma holds through n 1. Let x r 2n 1 > r 2n 3. Then so F 2n+1(x = F 2n(x + xf 2n 1(x + F 2n 1 (x x n N + xf 2n 1(x + 0 xf 2n 1(x, F 2n+1(x x n N F 2n 1(x x. n N 1 Smlarly, snce F 2n(x xf 2n 2(x, Equaton (5 shows: F 2n+2(x = (1 + 2xF 2n(x x 2 F 2n 2(x + 2F 2n (x 2xF 2n 2 (x F 2n(x + (2x xf 2n(x + 2F 2n 1 (x x n N + xf 2n(x + 0 xf 2n(x, so F 2n+2(x x n N+1 F 2n(x x n N. 6

7 Remark Gven some n 0 that satsfes condton (, each n n 0 also satsfes condton (. Lkewse, f n 1 satsfes condton (v, each n n 1 also satsfes condton (v. Lemma 10. When x β, lm F 2n(x a 2n =. Proof. Let x β. Notce from Lemma 8 that when n > N and x β, F 2n(x a 2n Then F 2n 2(x = 1 ( F a 2n 2 a 2n 2n 1 (x + xf 2n 2(x + F 2n 2 (x 2 F 2n 2 1 ( a 2n 1 + (β 2 F a 2n 2(x 2n 1 2n 1 ( a a 2n 2 = 1 a 2n a = lm n =N+1 lm n =N+1 = lm [ F 2n (x a 2n = lm F 2n(x a 2n 1 a [ F 2 (x F ] 2 2(x a 2 F 2N (x a 2N a 2 2 Lemma 11. Let {x 2n+1 } be a sequence of numbers satsfyng x 2n+1 r 2n 1. Then F lm 2n+1(x 2n+1 =. a 2n+1 Proof. Snce N satsfes condton (, r 2N+1 a 2. Then for n > N, Lemma 9 and a smlar argument to the prevous proof show that when x > r 2n 1 r 2N+1 a 2, F 2n+1(x a 2n+1 F 2n 1(x 1 ( a 2n = 1 a 2n 1 a 2n+1 a 2N a. 2N+1 Thus, by an analogous argument to the prevous proof, we fnd F lm 2n(x =. a 2n 7 ]

8 5 Exstence of N a We wll now nvestgate the exstence of an N a whch satsfes the base cases for each of the proofs n the prevous sectons. The propertes and condtons mentoned below are lsted at the begnnng of secton 4. Notce that f n 0 satsfes any gven condton, then each n n 0 also satsfes that condton. Furthermore, drect computaton of N a for some values of a (see table below suggests that as a ncreases, the mnmum possble value of N a ncreases. Ths mples that the value N a would work for all values less than or equal to a; e.g., snce N 25 = 102 satsfes the condtons when a = 30, 102 should satsfy the condtons for all a [0, 30]. a n satsfyng ( n satsfyng ( n satsfyng (v N a Lemma 12. Suppose there exsts some N satsfyng condton (v for a gven value of a. Then there exsts k 0 such that N + k satsfes condton (. Proof. By Lemma 9, for all k 0 and x > r 2(N+k 1, F 2(N+k+1(x x k and F 2(N+k(x x k Let k be large enough such that Then for all x β, (a 2 + a k 1 (a 2 k 1 a 2N+1 and (a 2 + a k (a 2 k a 2N. F 2(N+k 1(x x k 1 β k 1 = (a 2 + a k 1 a 2(N+k 1. Lkewse, for all x β, Thus, N + k satsfes condton (. F 2(N+k(x x k (a 2 + a k a 2(N+k. 8

9 6 Man results Theorem 13. Let a be an nteger and n > 1. If n s odd or greater than 2a, then r n s rratonal. Proof. Suppose r s a ratonal root of F n. Then n reduced form, r = p q where p s a factor of a. So p a, whch means r a. However, from Theorem 7 we know r 1 = a, r 2n+1 > r 1, and when n 2 > a, r 2n > β > a. Theorem 14. The sequence r 2n converges to β monotoncally from above, and the sequence r 2n+1 converges to β monotoncally from below. Proof. Recall that F n (β = ( a n+1, r 2n > β for n > a, and r 2n+1 < β. By the Mean Value Theorem, for each n > a there exsts c 2n [β, r 2n ] such that F 2n (r 2n F 2n (β r 2n β = F 2n(c 2n. Then lm F F 2n (β 2n(c 2n = lm r 2n β lm (r a 2n 2n β = a lm F 2n(c 2n = 0 lm r 2n = β. By combnng a smlar argument and Lemma 11, we obtan our second result. 7 Acknowledgements I would lke to thank Mchgan State Unversty and the Lyman Brggs College for ther support of our REU, as well as my faculty mentor, Dr. Akllu Zeleke. I would also lke to thank hs graduate assstants, Justn Droba, Ran Satyam and Rchard Shadrach. Project sponsored by the Natonal Securty Agency under Grant Number H Project sponsored by the Natonal Scence Foundaton under Grant Number DMS

10 References [1] Alberts, B.: On the propertes of a quas-fbonacc polynomal sequence. June SURIEM [2] Hoggatt, Verner Jr. ; Bcknell, Marjore: Roots of Fbonacc polynomals. In: Fbonacc Quart. 11 (1973, Nr. 3, S ISSN [3] Mátyás, Ferenc: Bounds for the zeros of Fbonacc-lke polynomals. In: Acta Acad. Paedagog. Agrenss Sect. Mat. (N.S. 25 (1998, S (1999. ISSN [4] Moore, Gregory: The lmt of the golden numbers s 3. In: Fbonacc 2 Quart. 32 (1994, Nr. 3, S ISSN [5] Prodnger, Helmut: The asymptotc behavor of the golden numbers. In: Fbonacc Quart. 34 (1996, Nr. 3, S ISSN [6] Yu, Hongquan ; Wang, Y ; He, Mngfeng: On the lmt of generalzed golden numbers. In: Fbonacc Quart. 34 (1996, Nr. 4, S ISSN