The Fibonacci sequence modulo π, chaos and some rational recursive equations

J. Math. Anal. Appl. 310 (2005) 506 517 www.elsevier.com/locate/jmaa The Fibonacci sequence modulo π chaos and some rational recursive equations Mohamed Ben H. Rhouma Department of Mathematics and Statistics Sultan Qaboos University PO Box 36 P.C. 123 Al-khod Muscat Oman Received 29 January 2005 Available online 25 March 2005 Submitted by G. Ladas Abstract In this paper we provide the closed form solution to the inter-related equations y n+2 = y ny n+1 1 and z n+2 = (z n+1 + z n ) mod π. y n + y n+1 Both of these equations were suggested as open problems in the book by Kocic and Ladas [V.L. Kocic G. Ladas Global Behavior of Nonlinear Difference Equations of High Order with Applications Kluwer Academic Dordrecht 1993]. We also give the closed form solution to the equations x n+1 = x nx n 2 + a and x n+1 = x n 1x n 2 + a x n + x n 2 x n 1 + x n 2 studied by X. Li and D. Zhu [X. Li D. Zhu Two rational recursive sequences Comput. Math. Appl. 47 (2004) 1487 1494]. 2005 Elsevier Inc. All rights reserved. Keywords: Difference equations; Attractivity; Stability; Rational recursive sequences; Fibonacci sequence; Chaos E-mail address: rhouma@squ.edu.om. 0022-247X/$ see front matter 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2005.02.038

M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 507 1. Introduction Rational recursive sequences have been the focus of extensive research especially since the appearance of the monograph by Kocic and Ladas [4]. The more recent monograph by Kulenovic and Ladas [5] is exclusively dedicated to rational recursive equations and contains a long list of open problems and conjectures. The following open problem was suggested in [4]. Open Problem [4 6.11.1 p. 175]. The secant method for finding 1: Study the asymptotic behavior and the periodic nature of solutions of the difference equation y n+2 = y ny n+1 1 n= 0 1... (1) y n + y n+1 where y 0 and y 1 are arbitrary numbers such that y n exists for all n. The change of variables y n = cot z n leads to the equation cot z n+2 = cot z n cot z n+1 1 cot z n + cot z n+1 and so if the values of z n are restricted to the interval (0π) we obtain the Fibonacci sequence modulo π namely z n+2 = (z n + z n+1 ) mod π. (2) The study of this equation is also listed in [4] as: Open Problem [4 6.11.2 p. 175]. The Fibonacci sequence modulo π: Assume z 0 and z 1 are arbitrary numbers in (0π). Study the asymptotic behavior and the periodic character of solutions of the difference equation (2). As explained in [4] a possible derivation of Eq. (1) comes from using Newton s method for solving the equation f(x)= x 2 + 1 = 0 with the sequence x n+1 = x n f(x n) f (x n ) and replacing the derivative f (x n ) by the approximation f(x n ) f(x n 1 ). x n x n 1 Actually the intersection between Fibonacci numbers and Newton s method is not restricted to this example. In fact Gill and Miller [3] produced the subsequence {F 2 n +1/F 2 n} using Newton s method to solve the equation x 2 x 1 = 0 with initial approximation x 0 = 1. In the same note [3] they also proved that the secant method with initial approximations x 0 = 1 and x 1 = 2 produces the sequence {F Fn +1/F Fn }. A more recent paper [9] also extended some old results [1] about (±1) n /(F n F n+1 ) also using Newton s method. Recently Li and Zhu [6] have studied similar looking equations namely

508 M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 x n+3 = a + x n+2x n x n+2 + x n n= 0 1 2... (3) x n+3 = a + x n+1x n x n+1 + x n n= 0 1 2... (4) where a [0 ) and the initial conditions x 0 x 1 and x 2 are positive. For both Eqs. (3) and (4) it was found that the positive equilibrium of the equation is globally asymptotically stable. In a recent paper [2] we have obtained closed form solutions for several equations some of which having similar numerators and denominators to those of recursions (1) (3) and (4). In particular the solutions to the following equations x n+2 = x n+1x n + 1 x n+1 + x n (5) x n+2 = x n+1x n 1 x n+1 x n (6) x n+2 = x n+1x n 1 x n+1 x n (7) x n+2 = x n+1x n + x n x n+1 x n (8) x n+2 = x n+1 x n x n+1 x n + 1 (9) x n+2 = x n+1 x n x n+1 x n + 1 (10) as well as their good sets were given explicitly. For example when [( ) x0 1 Fn 2 ( ) x1 + 1 Fn 1 ] ( 1) n 1 n 1 x 0 + 1 x 1 1 and (x 0 x 1 )(x0 2 1)(x2 1 1) 0 solutions to (6) exist and are given by x n = 1 + [( x 0 1) Fn 2 ( x1 +1) Fn 1 ] ( 1) n x 0 +1 x 1 1 1 [( x 0 1) Fn 2 ( x1 +1) Fn 1 ] ( 1) n (11) x 0 +1 x 1 1 where F n denote the nth term in the Fibonacci sequence. On the other hand all solutions to (5) are periodic of prime period 3 while all solutions to (7) are periodic of prime period 6. In this paper we provide: (1) The closed form solutions to Eqs. (1) (3) and (4) as well as their asymptotic behavior. (2) Results related to Eq. (2) from our discussion of the solution to Eq. (1). In particular we establish that the Fibonacci sequence modulo π is chaotic and give the exact number of periodic points of period p. (3) A new set of solvable rational recursive equations.

M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 509 The rest of the paper is organized as follows: in the next section we provide a few preliminary lemmas. In Section 3 we find and discuss the closed form solution of Eq. (1) along with results and comments about (2). In Section 4 we give the closed form solution to both (3) and (4) and explain why their solutions always converge to a fixed value. Finally a few generalizations are made in Section 5. 2. Preliminary results We begin with the following elementary lemma. Lemma 1. Let b x y and z be arbitrary complex numbers such that b(x + y)(x + b)(y + b) 0. If then z b z + b = ( x b x + b )( ) y b (12) y + b z = xy + b2 x + y. (13) The following lemmas give the solution to a few nonlinear sequences of the form x n+1 = αx p n k xq n l. Lemma 2. Let F n be the Fibonacci sequence defined by F 0 = F 1 = 1 and F n+2 = F n+1 + F n. The sequence x n+2 = αx n+1 x n with initial conditions x 0 and x 1 has the general form x n = α ( 1+F n) x F n 2 0 x F n 1 1 for all integers n 2. (14) The proof of Lemma 2 can be easily obtained by induction. Lemma 3. Let λ 1 λ 2 and λ 2 be the roots of the polynomial λ 3 λ 1 = 0 and (abc)be the solution to the linear system (15) a + b + c = 0 aλ 1 + bλ 2 + c λ 2 = 1 aλ 2 1 + bλ2 2 + c λ 2 2 = 0. (16) Then the sequence x n+3 = x n+1 x n with the initial conditions x 0 x 1 and x 2 has the general form x n = x A n 2 0 x A n 1 xa n 1 2 (17)

510 M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 where A n is given by A n = aλ n 1 + bλn 2 + c λ n 2. (18) Proof. It is easy to establish that given initial conditions x 0 x 1 and x 2 the general term of x n can be written in the form x n = x C n 0 xa n 1 xb n 2 where A n B n C n satisfy the linear recursion Y n+3 = Y n+1 + Y n. Moreover C 0 = B 2 = A 1 = 1 and C 1 = C 2 = B 0 = B 1 = A 0 = A 2 = 0. The values of λ 1 λ 2 and λ 2 are the roots the polynomial λ 3 λ 1 = 0 which is the characteristic polynomial of (19). The rest of the proof follows from straightforward computations. Lemma 4. Let µ 1 µ 2 and µ 2 be the roots of the polynomial (19) µ 3 µ 2 1 = 0 and let (abc)be the solution to the linear system (20) a + b + c = 0 aµ 1 + bµ 2 + c µ 2 = 0 aµ 2 1 + bµ2 2 + c µ2 2 = 1. (21) Then the sequence x n+3 = x n+2 x n with the initial conditions x 0 x 1 and x 2 has the general form x n = x B n 1 0 x B n 2 1 x B n 2 (22) where B n is given by B n = aµ n 1 + bµn 2 + c µn 2. (23) The proof of Lemma 4 is identical to that of Lemma 3 and thus will be omitted. 3. Equation (1) and the Fibonacci sequence mod π First notice that by Lemma 1 a solution to (1) also satisfies the equation ( )( ) y n+2 i y n+2 + i = yn+1 i yn i. (24) y n+1 + i y n + i By virtue of Lemma 2 we obtain that y n i y n + i = ( y0 i y 0 + i ) Fn 2 ( y1 i y 1 + i ) Fn 1.

M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 511 Thus setting up e iθ 0 = y 0 i y 0 + i and e iθ 1 = y 1 i y 1 + i (25) or equivalently θ 0 = 2arccot y 0 and θ 1 = 2arccot y 1 (26) we have y n i y n + i = ei(f n 2θ 0 +F n 1 θ 1 ) and y n = i 1 + ei(f n 2θ 0 +F n 1 θ 1 ) 1 e i(f n 2θ 0 +F n 1 θ 1 ). (27) Theorem 1. The solution to Eq. (1) exists for all integers n 0 if and only if (F n 2 θ 0 + F n 1 θ 1 ) mod 2π 0 for all n 0. (28) Moreover when it exists the solution to (1) is given by ( ) Fn 2 θ 0 + F n 1 θ 1 y n = cot 2 = cot(f n 2 arccoty 0 + F n 1 arccoty 1 ). (29) In addition (i) If θ 0 and θ 1 are both rational multiples of π then either {y n } diverges in finitely many steps or y n is periodic. (ii) If θ 0 is a rational multiple of π and θ 1 is not (or vice-versa) then {y n } is aperiodic and does exist for all integers. Proof. Both (28) and (29) follow straight from (27). On the other hand if θ 0 = (m 0 π/n 0 ) and θ 1 = (m 1 π/n 1 ) then studying the sequence (F n 2 θ 0 + F n 1 θ 1 ) mod 2π is equivalent to studying the sequence G n = (m 0 n 1 F n 2 + m 1 n 0 F n 1 ) mod 2n 0 n 1 (30) which is bound to be periodic. (See for example [7] and references therein.) If G n contains zero then the sequence {y n } will diverge. Otherwise the sequence {y n } will be periodic. To illustrate item (i) of Theorem 1 the initial values y 0 = 1 + 2 and y 1 = 0 do correspond to values of θ 0 and θ 1 that are rational multiples of π and yield a periodic sequence of prime period 12. On the other hand when y 0 = 0 and y 1 = 3

512 M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 the sequence {y n } only exist for n 7. Consider now T π the square with corners (0 0) (0π) (π π) and (π 0) in the plane and the matrix A = ( 0 1 1 1 ). Using the definitions and results of [8 Chapter 5] we have that the matrix A induces a hyperbolic Anosov toral automorphism defined on T π namely the map L A : T π T π defined by L A (x y) = (y x + y mod π). (32) The following relation between the nth power of the matrix A and the Fibonacci sequence is well known: ( ) A n Fn 2 F = n 1. F n 1 F n Also known is the fact that the matrix A 2 induces the cat map one of the simplest examples of chaos in two dimensions. Perhaps less known is that the map L A induced by the matrix A is itself chaotic. The following theorem is just a translation of the fact that L A is chaotic in terms of the Fibonacci sequence mod π. Theorem 2. (i) The set of periodic points of the Fibonacci sequence modulo π is dense in T π. Moreover for any given integer n 1 the number of n-periodic points on the Torus T π is given by (31) F n 2 + F n 1 + ( 1) n 1. (33) (ii) Given any two open neighborhoods U and V in T π there exist initial conditions (z 0 z 1 ) U and an integer n 0 such that the nth iterate (z n z n+1 ) of the Fibonacci sequence modulo π belongs to the set V. Proof. Items (i) and (ii) are simply the density of periodic points and the transitivity properties of chaotic systems. As for the number of periodic solutions of period n it can be obtained by solving the equation A n (z 0 z 1 ) T = (z 0 z 1 ) T mod π and using the identity F 2 n F n 1F n+1 = ( 1) n. In terms of the sequence (1) Theorem 2 implies that if F and P are respectively the forbidden set and the set of initial conditions of periodic solutions of (1) then F P is dense in R 2. Theorem 2 also implies transitivity of the sequence (1) in R 2.

M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 513 4. Equations of the form x n+1 = (x n k x n l + 1)/(x n k + x n l ) When a>0 and k and l are distinct nonnegative integers all recursions of the form y n+2 = y n k+1y n l+1 + a (34) y n k+1 + y n l+1 can be transformed to the equation x n+2 = x n k+1x n l+1 + 1 (35) x n k+1 + x n l+1 by the change of variables y n = ax n. In this section we will present results for the case k = 0 and l = 1. With minor changes all results do apply to other cases. In particular Eqs. (3) and (4) studied in [6] can be treated in a very similar way. Consider the equation x n+2 = x nx n+1 + 1. (36) x n + x n+1 By virtues of Lemmas 1 and 2 we have that ( ) ( ) xn 1 x0 1 Fn 2 ( ) x1 1 Fn 1 =. (37) x n + 1 x 0 + 1 x 1 + 1 Thus the expression of x n is easily obtainable whenever (x 0 1) F n 2 (x 1 1) F n 1 (x 0 + 1) F n 2 (x 1 + 1) F n 1 to yield the closed form x n = (x 0 + 1) F n 2(x 1 + 1) F n 1 + (x 0 1) F n 2(x 1 1) F n 1 (x 0 + 1) F n 2(x 1 + 1) F n 1 (x 0 1) F n 2(x 1 1) F. (38) n 1 Notice also that using Eq. (37) we can write x n 1 x n + 1 = x 0 1 F n 2 [ F n 1 ] x 1 1 F Fn 2 n 2 x 0 + 1 x 1 + 1. (39) Since the ratio F n 1 /F n 2 (1 + 5 )/2 asn we can conclude that if 1+ 5 x 0 1 x 1 1 2 x 0 + 1 x 1 + 1 > 1 then the sequence {x n } 1. Similarly if 1+ 5 x 0 1 x 1 1 2 x 0 + 1 x 1 + 1 < 1 then the sequence {x n } 1asn.

514 M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 An immediate consequence of this fact is that if x 0 and x 1 are both positive then the sequence {x n } 1asn. A very similar explanation works for Eqs. (3) and (4). In fact the solutions to Eqs. (3) and (4) with a = 1 are respectively given by x n = 1 + ( x 0 1) An 2 ( x1 1) An ( x2 1) An 1 x 0 +1 x 1 +1 x 2 +1 1 ( x 0 1) An 2 ( x1 1) An ( x2 1) An 1 (40) x 0 +1 x 1 +1 x 2 +1 and x n = 1 + ( x 0 1) Bn 1 ( x1 1) Bn 2 ( x2 1) Bn x 0 +1 x 1 +1 x 2 +1 1 ( x 0 1) Bn 1 ( x1 1) Bn 2 ( x2 1) Bn (41) x 0 +1 x 1 +1 x 2 +1 where A n and B n are respectively given by Lemmas 3 and 4. Since Li and Zhu [6] considered positive initial conditions and that both sequences {A n } and {B n } converge to + as n we can easily see that {x n } 1 for both Eqs. (3) and (4). In fact in light of Eqs. (40) and (41) it is also easy to see that the sequence {x n } will converge to 1 for all initial conditions x 0 x 1 x 2 ( 0). Moreover obtaining the good sets of (40) and (41) and more detailed rules of convergence is done in a very similar way as for Eq. (36). The following theorem sums up the discussion of this section. Theorem 3. Let k>l>0 be two integers and consider Eq. (35) with initial conditions x 0 x 1...x k. There exist sequences {A (i) n } for i = 0...kall converging to as n such that ki=0 (x i + 1) A(i) n x n = ki=0 (x i + 1) A(i) n + k i=0 (x i 1) A(i) n k i=0 (x i 1) A(i) n (42) as long as the denominator in (42) does not vanish. Moreover if all initial conditions are positive then {x n } 1 and if all initial conditions are negative then {x n } 1. 5. Generalizations to other rational recursive equations In this section we illustrate how to obtain closed form solutions for some special rational recursive equations. To this purpose consider the equation x n+k+1 = α k i=0 x p i n+i n= 0 1 2... (43) with initial conditions x 0 x 1...x k. For simplicity we will consider only integer powers p i. As in Lemmas 2 4 by computing the first few items of the sequence {x n } we can easily obtain the general form of the solution namely x n = α B n k i=0 x A(i) n i (44)

M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 515 where the sequences {A (i) n } satisfy the linear equation A (i) n+k+1 = k j=0 p j A (i) n+j (45) with initial conditions A (i) i = 1 and A (i) j = 0 for all 0 i j k. The sequence {B n } also satisfies the linear equation k B n+k+1 = 1 + p j B n+j B 0 = B 1 = =B k = 0. (46) j=0 Appropriate changes of variables do lead to several interesting cases of which we only choose the following few. 5.1. A first order equation We start with the elementary first order equation x n+1 = αxn 2 (47) with initial condition x 0. It is easy to establish that the general term of x n is given by x n = α 2n 1 x (2n ) 0. (48) Immediately we can see that the closed form of the first order two-parameter recursive equation y n+1 = αyn 2 + 2αay n + a(aα 1) (49) is given by y n = α 2n 1 (y 0 + a) (2n) a. (50) In particular when αa = 1 we obtain that the closed form to y n+1 = αyn 2 + 2y n is given by ( y n = α 2n 1 y 0 + 1 ) (2 n ) 1 α α. With the change of variables x n = (y n a)/(y n + a) we can also deduce from (50) the closed form solutions to the 2-parameter equation y n+1 = (1 + α)ay2 n + 2(1 α)a2 y n + (1 + α)a 3 (1 α)yn 2 + 2(1 + α)ay n + (1 α)a 2 (51) and its special cases when α =±1 y n+1 = y2 n + a2 2y n y n+1 = 2a2 y n yn 2 +. a2 (52) (53)

516 M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 In fact the solution to (51) is given by y n = a (y 0 + a) (2n) + α 2n 1 (y 0 a) (2n ) (y 0 + a) (2n) α 2n 1 (y 0 a) (2n ) (54) whenever the denominator in (54) does not vanish for all n 0. 5.2. Some high order equations Let l>1 be an integer a be an arbitrary real number and x 0 a x 1 a... x l a be the initial conditions of the rational recursive equation x n+l+1 = a x n+l x n n= 0 1 2... (55) x n + a Using the change of variables x n = a(y n 1) Eq. (55) reduces to the simpler form y n+l+1 = y n+l n= 0 1 2... y n which is well defined whenever the product l y i 0. i=0 Our first conclusion then is that Eq. (55) is well defined if and only if l (x i + a) 0. i=0 Second whenever the polynomial λ l+1 λ l + 1 is solvable Eq. (55) is also solvable explicitly. For the sake of this discussion we will just concentrate on the general behavior of Eq. (55) for large values of l. Using Corollary 1.2.1 p. 6 and Theorem 1.3.6 p. 12 in [4] we can easily establish that unless x 0 = x 1 = =x l = 0 as n grows large all solutions to Eq. (55) will alternate between terms converging to zero and others diverging to. The second and last equation we consider is the rational recursive equation x n+1 x n x n+2 = n= 0 1 2... (56) x n+1 + x n + 1 with initial conditions x 0 and x 1. Setting y n = x n /(x n + 1) we obtain that y n+2 = y n+1 y n. Thus by Lemma 2 and some simple algebra we obtain that x n = x F n 2 0 x F n 1 1 (57) (1 + x 0 ) F n 2(1 + x 1 ) F n 1 x F n 2 0 x F n 1 1 whenever the initial conditions x 0 and x 1 are chosen such that the denominator never vanishes. We would like to end this paper by stressing that despite the fact that we can generate several closed form solutions for some interesting rational recursive equations the set of

M. Ben H. Rhouma / J. Math. Anal. Appl. 310 (2005) 506 517 517 solvable equations remains very small. In fact to our knowledge amongst the nontrivial equations of the form x n+2 = αx n+1 + βx n + γ ax n+1 + bx n + c only the following special equations have closed form solutions: x n+2 = αx n+1 x n x n+2 = αx n x n+1 x n+2 = αx n+1 βx n + αβ β 2 x n + β x n+2 = αx n βx n+1 + αβ β 2. (58) x n+1 + β Acknowledgment I thank my colleague Dr. Samir Siksek for his helpful comments on and around the Fibonacci sequence. References [1] B.A. Brousseau Summation of infinite Fibonacci series Fibonacci Quart. 7 (1969) 143 168. [2] M. Ben H. Rhouma G. Miller Closed form solutions of some rational recursive sequences of second order Comm. Appl. Nonlinear Anal. 12 (2005) 41 58. [3] J. Gill G. Miller Newton s method and ratios of Fibonacci numbers Fibonacci Quart. 19 (1981) 1 4. [4] V.L. Kocic G. Ladas Global Behaviour of Nonlinear Difference Equations of Higher Order with Applications Kluwer Academic Dordrecht 1993. [5] M.R.S. Kulenovic G. Ladas Dynamics of Second Order Rational Differential Equations Chapman & Hall CRC Boca Raton FL 2001. [6] X. Li D. Zhu Two rational recursive sequences Comput. Math. Appl. 47 (2004) 1487 1494. [7] M. Renault The Fibonacci sequence under various moduli Masters Thesis Wake Forest University May 1996. [8] S.N. Elaydi Discrete Chaos Chapman & Hall CRC Boca Raton FL 2000. [9] T. Taka-aki Algebraic independence of Fibonacci reciprocal sums associated with Newton s method Tsukuba J. Math. 27 (2003) 375 387.