THE ASYMPTOTIC GROWTH RATE OF RANDOM FIBONACCI TYPE SEQUENCES Hei-Chi Chan Mathematical Sciences Program, University of Illinois at Springfield, Springfield, IL 62703-5407 email: chanhei-chi@uisedu Submitted November 2002-Final Revision July 2003 ABSTRACT Estimating the growth rate of random Fibonacci-type sequences is both challenging and fascinating In this paper, by using ergodic theory, we prove a new result in this area Let a denote an infinite sequence of natural numbers {a, a 2, } and define a random Fibonaccitype sequence by f = 0, f 0 =, a 0 = 0, and f k = 2 a k f k + 2 a k f k 2 for k Then, for almost all such infinite sequences a, we have n n ln f n = 30022988 INTRODUCTION It is well-known that the Fibonacci numbers, F n, are given by Binet s formula: F n = 5 + n 5 n 5 2 5 2 With, we can compute the asymptotic growth rate of the sequence {F k }, which is given by n n ln F + 5 n = ln = 0482 2 In the case of random Fibonacci type sequences, defined by with fixed f and f 2 f k = ak f k + bk f k 2, where ak and bk are random coefficients, the quest for the asymptotic growth rate will be much more difficult, if not impossible Recently, Viswanath made a surprising breakthrough in [26] He considered the random Fibonacci sequences defined by f = f 2 = and f k = ±f k ± f k 2, 2 where the signs are chosen independently and with equal probabilities He proved the following remarkable result 243
Theorem : Viswanath, 2000 The asymptotic growth rate of the random Fibonacci sequences defined in 2 is given by n n ln f n = ln398824 = 02397559 3 with probability For an interesting introduction, see Peterson s article [8] The hard part of Viswanath s work is the computation of a fractal measure That involves a large system of equations with dimension in the order of millions Since then, some authors [4, 29] have generalized this result in various directions Viswanath s method is not the only way to estimate the growth rate of random Fibonacci type sequences Write x [0, as x = + 2 a 2 a 2 + 2 a 3 +, 4 where a k = a k x {0,, 2, } see Section 2 Note that, in terms of the following compact notation which we will use in the rest of this paper the above can be written as A B + A 2 B 2 + x = 2 a A B + A 2 B 2 +, 5 + 2 a 2 Our main result is the following theorem + 2 a 3 + Theorem 2: For each x [0,, we associate with it an infinite sequence of natural numbers {a, a 2, } through 4 Consider the random Fibonacci type sequences, {f k }, defined by f = 0, f 0 =, a 0 = 0, and f k = 2 a k f k + 2 a k f k 2, 6 for k Then we have, for almost all x, n n ln f n = 30022988 7 In fact, this is a generalization of the following theorem, which is proved by Lévy []: Theorem 3: Levy, 929 For each x [0,, we associate with it an infinite sequence {b, b 2, } b k =, 2, through the regular continued fraction representation of x; ie, x = b + b 2 + 244
Consider the random Fibonacci type sequences, {Q k }, defined by Q = 0, Q 0 =, and for k Then we have, for almost all x, n n ln Q n = Q k = b k Q k + Q k 2 8 π2 2 ln 2 = 865690 9 See also Khintchin [8, 9] Later, the same theorem is proved using ergodic theory; eg, see [, 3, 3, 9, 2] See also Kac [7] We will prove Theorem 2 by following the same strategy that uses ergodic theory A question may be raised at this point: what is ergodic theory and why is it able to compute the asymptotic growth rate of random recurrences like that of 6 and 8? The concept of ergodicity was originated in physics in the nineteenth century Given a system that evolves dynamically, suppose one would like to study the behavior of a certain physical quantity P of this system The ergodic hypothesis asserts that, under certain conditions, the time average of P should be the same as the phase space average of P For example, consider a sealed box with a permeable partition that divides the box into two equal chambers Then the ergodic hypothesis asserts that an air molecule should spend half of its time in each chamber In the early part of the twentieth century, Birkhoff, von Neumann, Khintchin and others developed this into a mathematical theory which is now known as the ergodic theory For a physicist s introduction, see [24] Ergodic theory is closely related to the theory of chaos, see [2, 6, 6] For introductions in the form of book to ergodic theory, consult [, 3, 3, 7, 9, 23, 27] See also Young s lecture [30] Kac [7] gave a very readable account that connects the physicist s concept of the ergodic hypothesis and the mathematical foundation of ergodic theory With the above understood, we can look into the reason why ergodic theory can be used to compute the asymptotic growth rate of random recurrences Random sequences, like that of 6 or 8, can be thought of being generated by certain dynamical systems in a non-technical sense By standard arguments see Section 2 below, we can associate these sequences with continued fractions, like that of 4 This induces dynamics on continued fractions of which ergodic theory allows us to assert the asymptotic properties It should be remarked that Theorem 3 is hard to generalize: to compute explicitly the asymptotic growth rate, one has to know the analytical closed-form of the invariant measure see Section 2 below associated with the random recurrence In the case of Theorem 2, since we found the invariant measure involved, therefore we are able to compute the asymptotic growth rate of the f n defined in 6 The outline of this paper is as follows In Section 2, we fix some notations and introduce the generalized Gauss map associated with the random Fibonacci type sequence 6 Then in Section 3, we prove Theorem 2 The main ingredient of this proof is the ergodicity of the generalized Gauss map Since the proof of ergodicity is rather long and complicated, therefore, in order to minimize digression, we assume this property in Section 3 and prove it in Section 4 Section 5 is our concluding remarks 245
2 THE GENERALIZED GAUSS MAP The goal of this section is to define the generalized Gauss map In brief, this map is the machinery that generates random recurrences To proceed, let us introduce a certain continued fraction representation of all x [0, Lemma : For all x [0,, we have x = 2 a + 2 a 2 where a k {0,, 2, } Remark: One can think of the a k as the digits of x Proof: For any x [0,, we have 2 a + < x 2 a, where a {0,, 2, } This implies, for some p [0,, x = p 2 a + p 2 2 a = p 2 2 a Define x [0, by x = p/2 p, we can write x as x = 2 a + x Since x [0,, we can repeat the same iteration and obtain x = 2 a + 2 a 2 + = 2 a + [a, a 2, ], 0 + 2 a 2 + Remark: If x is irrational, then x has an infinite expansion in the form of 0 This continued fraction representation of x is closely related to the random Fibonacci type sequence in 6 First, we fix an x and this fixes a sequence of natural numbers, {a k }, through Lemma Note that, a k are functions of x; ie, a k = a k x Next, we define the following sequences: Definition : For all x = [a, a 2, ] [0,, define g k = 2 a k g k + 2 a k g k 2, k 2 f k = 2 a k f k + 2 a k f k 2, k where g 0 = 0, g =, f = 0 and f 0 = Note that f k and g k are integer-valued functions of x Also, the second equation of is the defining equation for the random Fibonacci type sequences in 6 All these machinery are tied up together by the following identity: for t [0, ], we have g k + t 2 a k g k f k + t 2 a = 2 a + 2 a 2 + + 2 a k k fk + t 2 246
This can be proved by standard induction arguments see eg, Niven s book [5] With these understood, we define the generalized Gauss map: Definition 2: For all x [0,, the generalized Gauss map T : [0, [0, is defined as follows: for x = 0, T 0 0; for x 0, we have, using the notation in 0, T x = T [a, a 2, a 3, ] [a 2, a 3, a 4, ] 3 There is a beautiful way to define T alternatively, as pointed out by the referee; see equation 37 and the last section in this paper One can think of T as a shift map, as it shifts the digits of x Note that the original Gauss map is defined as follows Write x [0, as a regular continued fraction; ie, x = b + b 2 + [b, b 2, ] G, where b k {, 2, } The Gauss map T G : [0, [0, is defined as follows: for x = 0, T G 0 0; for x 0, we have T G x = T G [b, b 2, b 3, ] G [b 2, b 3, b 4, ] G The following remarkable property of the generalized Gauss map 3 is the key to the proof of Theorem 2: Theorem 4: Given an integrable function f on the unit interval, then, for almost all x, we have n ft k x = n n 0 ρxfx dx 4 Here, ρx, the probability density of an invariant measure see 3, is given by ρ 0 ρx = x + x + 2, 5 where ρ 0 = ln 4/3 Note that, equation 4 is the same as saying that the time average of f ie, the lhs of 4 is the same as the phase space average of f ie, the rhs of 4 The normalization ρ 0 in 5 is chosen such that ρx dx = Note that ρx satisfies the 0 following equation: for x [0, ], we have ρx = 2 k + x 2 ρ 2 k + x This can be understood as the fact that ρx is the eigenfunction of eigenvalue of the Frobenius-Perron operator; eg, see [5, 22] This will be explored elsewhere The integrability in Theorem 4 is defined with respect to the Lebesgue measure, or an equivalent invariant measure the generalized Gauss measure, which will be introduced in Section 4 In the case of the original Gauss map, we have a similar theorem, with T replaced by T G, and ρx replaced by ρ G x = /x + Note that this Gauss map and its ergodicity is closely related to the Gauss-Kusmin-Lévy problem; see [, 5, 0, 3, 20, 22, 25] 247
With the generalized Gauss map 3, we can derive a formula cf 20 which will be useful in the subsequent sections Using the T map, we can write x as x = 2 a + T x = = 2 a + 2 a 2 + + 2 a n + T n x 6 Note that we have used the iteration procedure used in the proof of Lemma Equations 6 and 2 imply x = g nx + T n x 2 a nx g n x f n x + T n x 2 a nx f n x 7 For subsequent sections, we define the following: Definition 3: For n, A n x g n x + T n x 2 a nx g n x 8 B n x f n x + T n x 2 a nx f n x 9 Of course, A n x and B n x are simply the numerator and the denominator of the quotient in 7; ie, x = A n x/b n x Note that, in the same manner, we have T x = A n T x/b n T x This is because, the expansion of T x is obtained by removing the top level of x; ie, T x = 2 a 2 + 2 a 3 + + 2 a n + T n x, and there are n levels in this expansion In particular, we have B n T x = A n x 20 To see this, we compare the numerator in the first quotient and the numerator in the last quotients in the following equation: x = A nx B n x = 2 a + T x = 2 a + A n T x/b n T x = B n T x 2 a [An T x + B n T x] In order to minimize digression, we will assume and apply Theorem 4 in the next section We will come back for a proof of this theorem in Section 4 As we will see, Theorem 4 is a consequence of the ergodicity of the generalized Gauss map 3 PROOF OF THEOREM 2 In this section, following the strategy in chapter 24 of Schweiger s book [22], we prove Theorem 2 by applying Theorem 4 See also [, 2] 248
Step We need to establish the following formula n T k x = f n + T n x 2 a = n fn B n x 2 Indeed, write the product on the left by A k and B k defined in the last section: n T k x = A nx A n T x B n x B n T x AT n x B T n x Using 20 repeatedly, we have most of the factors canceled, except the denominator of the first factor ie B n x, and the numerator of the last factor ie A T n x This gives the desired result as A T n x = Step 2 Equation 2 and the fact that 0 T n x imply n T k x 22 2f n f n Note that, in the upper inequality, we have used 2 and the fact that, for k 0, f k > 0 this can be shown by induction For the lower inequality, we have used the fact that f n + T n x2 a n f n f n + 2 a n f n f n + 2 a n f n + 2 a n f n 2 = 2f n The log of inequality 22 will be useful in the next and final step: ln f n n 2 n n n ln T k x ln f n n 23 Step 3 Taking the it n in 23 and using Theorem 4 with fx = ln x, we have ae almost everywhere n n n ln f n = n n Since, for x [0, ], 0 ln T k x = ρ 0 x + x + 2 3 x + 2, the last integral in 24 can be shown to be bounded: 0 ln x x + x + 2 dx 0 0 ln x x + x + 2 ln x 3 x + dx = 5 2 2 249 dx 24
Numerical integration gives This completes the proof of Theorem 2 n n ln f n = 30022988 4 THE ERGODICITY OF THE GENERALIZED GAUSS MAP It is time to prove Theorem 4 To this end, all we need to do is to establish the ergodicity of the generalized Gauss map Then Theorem 4 follows by applying the ergodic theorem see [, 3, 3, 7, 23, 27] We will proceed as follows First, in Section 4, we introduce the concept of cylinders It is a special way to partition the unit interval so that the subsequent computations may be simplified Next, in Section 42, we introduce the generalized Gauss measure; cf the probability density ρx in 5 With these two major ingredients defined, we proceed to prove Theorem 4 in Section 43 4 Cylinders Fix non-negative integers a,, a k Let t [0, and define ψ a a k ψ a a k t 2 a + 2 a 2 + + 2 a k + t = g k + t 2 ak g k f k + t 2 a k fk 25 For the second equality, we have used 2 The cylinder or fundamental interval of rank k is defined by a a k = {ψ a a k t; t [0, } A crucial property is that the lengths of these cylinders shrink to zero: let λ denote the Lebesque measure and we have Lemma 2: λ a a k = 0 26 k Remarks: The implication of this lemma is as follows Let B k denote the σ-algebra generated by the cylinders of order k and F denote the σ-algebra of the Borel sets Then Lemma 2 implies B k = F; 27 k= ie, the class of all cylinders generates the σ-algebra F of Borel sets For a proof, see p 48 of [22] This fact allows us to reduce many calculations in the subsequent sections to be done in terms of cylinders We now turn to the proof of Lemma 2 Proof: All we need to do is to establish, for a constant, D, which is independent of k, In particular, this implies Lemma 2 Indeed, by direct computation, we can show that λ a a k = ψ a a k ψ a a k 0 = λ a a k D 2 k 28 250 k i= 2a i f k f k + 2 a k fk by k i= 2a i f 2 k 29
To obtain the desired bound, we consider first the case of an odd k To this end, since f k f k, which can be shown by induction, we have This implies f i = 2 a i f i + 2 a i f i 2 2 a i + 2 a i f i 2 f k f By combining 29 and 30, we obtain m=3,5,,k 2 a m + 2 a m 30 λ a a k 2a f 2 m=3,5,,k 2 a m 2 a m 2 a m + 2 a m 2 = 2a f 2 m=3,5,,k 2 + 2a m 2 a + 2am m 2 a m 2a f 2 m=3,5,,k 2 2 2 k = 2 2 k Note that f = 2 a and therefore the prefactor 2 a /f 2 = /f For the inequality in the third line, we set x = 2 a m /2 a m and used the fact that, for x > 0, x + x 2 This proves the case of k being odd The case of an even k can be shown in a similar manner Therefore we establish 28 42 The Generalized Gauss Measure In this section, we define the generalized Gauss measure and study some of its crucial properties Definition 4: The generalized Gauss measure is given by P A = ρ 0 where A F and ρ 0 is given in Theorem 4 By using the inequality A dx x + x + 2, 3 we obtain 6 x + x + 2, x [0, ] 2 λa 6 ln4/3 P A λa 2 ln4/3 32 25
Note that P and λ are absolutely continuous with respect to each other they have the same set of measure zero This implies that if the sequence a x, a 2 x, has a certain property ae with respect to P, it will also have the same property ae with respect to λ An important property of P is that it is preserved by the generalized Gauss map T measure preserving; ie, we have P T A = P A for every A F To establish this, we need Lemma 3: For t > 0, we have P T [0, t] = P [0, t] Proof: Let us set γ = /2 and note that T [0, t] = {x : 0 T x t} = [ γ k + t, γk With these understood, we proceed to compute P T [0, t]: P T [0, t] = ρ 0 γ k γ k +t dx x + x + 2 ] γ k + γ k+ + = ρ 0 ln γ k +t + ln γ k+ +t + = ρ 0 ln 2 +t + = ρ 0 [ ln 2 + ln ] t + t + 2 Note that the infinite sum is a telescoping sum and only the first term survives Next, we proceed to compute P [0, t]: P [0, t] = ρ 0 t 0 [ dx t + x + x + 2 = ρ 0 ln ln ] t + 2 2 This proves the lemma Remark: let F 0 be the sub-algebra of disjoint unions of intervals contained in [0, ] Lemma 3 implies that P T A = P A for every A F 0 By Proposition 2 in p 27 of [3], this can be extended to every A F This shows that T is measure-preserving 43 Proof of Theorem 4 The crux of the issue is the following theorem: Theorem 5: T is ergodic with respect to P Proof: To prove this, we follow standard procedure eg, see []: we need to show that if A F is such that T A = A and P A > 0, then P A = Fix a,, a k First, we show that, for all A F, we have Here, λab = λa B/λB 2 λa λ T k A a a k 33 252
To this end, we first consider the interval B = [x, y, where 0 x < y and we have λ T k B a a k = ψ a a k y ψ a a k x ψ a a k ψ a a k 0 Again, 25 has been used Since h, hx, y and therefore, 34 implies f k f k + 2 a k f k = y x f k + y 2 a k fk f k + x 2 a 34 k fk }{{} hx,y h, = f k + 2 a k f k f k f k + 2 ak f k + 2 ak f k 2 f k = 2, 2 λb λ T k B a a k The last inequality also holds if B is a disjoint union of intervals, and therefore, it is also true for any A F This proves 33 By 32 and 33, we have a similar inequality for P A: C P A P T k A a a k 35 3 Here, the constant C is given by ln4/3 To complete the proof of Theorem 5, we proceed as follows Suppose A is invariant under T ie, T A = A and P A > 0 Then inequality 35 implies P A C P A a a k ; ie, P A C P A a a k P a a k We multiply both sides by P a a k /P A and obtain P a a k C P a a k A This implies P E C P EA 36 for all finite disjoint unions E of cylinders; since the sets of cylinders generate F, 36 is also true for any E F Setting E = A c the complement of A, 36 implies P A c = 0 and so P A = This completes the proof of Theorem 5 Finally, Theorem 4 follows immediately as the consequence of Theorem 5 and the ergodic theorem 5 CONCLUDING REMARKS In this paper, we have applied ergodic theory in computing the asymptotic growth rate of random Fibonacci type sequences The key ingredients are the generalized Gauss map 3 and its ergodicity with respect to the generalized Gauss measure 3 The result obtained 253
here can be used to estimate the effectiveness of continued fractions in the form of 0 in the sense of Lochs [2] This will be addressed in a forthcoming paper By imitating Mayer [4] and Wirsing [28], one can derive not just the constant but also the rate of convergence to the constant Again, this will be addressed in a future paper Note added: The anonymous referee pointed out a beautiful generalization of the generalized Gauss map: Let k = 2, 3,, we have T x = klog k x mod 37 k Denote the density of the invariant measure by ρ k x Then, for x 0,, one can show that with C being a normalizing constant ρ k x = C + k xk + k x, as it is the eigenfunction of eigenvalue of the corresponding Frobenius-Perron operator; ie, it satisfies ρ k x = a=0 k k a + k x 2 ρ k k a + k x The situation discussed in this paper corresponds to k = 2 Note that 37 leads to a continued fraction expansion: following the notation in 5, this expansion will have B i = for i ; A = k a and A m = k k a m for m 2 It will be interesting to generalize Theorem 2 for the interval map 37 ACKNOWLEDGMENTS I would like to thank Chung-Hsien Sung and Doug Woken for discussion and encouragement Grateful acknowledgment is made of the very useful comments and the beautiful generalization see above of an anonymous referee I would also like to thank the participants of the Tenth International Conference on Fibonacci Numbers and Their Applications for their interest in the present work The conference was a very stimulating and enjoyable occasion REFERENCES [] P Billingsley Ergodic Theory and Information John Wiley & Sons, New York, 965 [2] R M Corless Continued Fractions and Chaos The Am Math Mon 99 992: 203-25 [3] I P Cornfeld, S V Fomin and Y G Sinai Ergodic Theory Springer-Verlag, New York, 982 [4] M Embree and L N Trefethen Growth and Decay of Random Fibonacci Sequences Proc Roy Soc Lond A 455 999: 247-2485 [5] M Iosifescu and S Grigorescu Dependence with Complete Connections and its Applications Cambridge University Press, New York, 990 [6] E A Jackson Perspectives of Nonlinear Dynamics Vol, Cambridge University Press, New York, 99 254
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