Eigenvalues of Fibonacci stochastic adding machine A. Messaoudi, D. Smania 0-0
Let N ∈ N k ε i ( N )2 i = ε k ( N ) . . . ε 0 ( N ) � N = i =0 where ε i ( N ) = 0 or 1 for all i. It is known that there exists an algorithm that computes the digits of N + 1. Ex: 1011 + 1 = 1100 1
0 1 1 0 1 11 1 1100 This algorithm can be described by by the following manner: c − 1 ( N + 1) = 1, ε i ( N + 1) = ε i ( N ) + c i − 1 ( N + 1) mod (2) c i ( N + 1) = [ ε i ( N ) + c i − 1 ( N + 1) ] . 2 What happens if the machine dos not work. P.R. Killeen and T.J. Taylor [KT] consider fallible adding machine by the following: 2
They consider the algorithm: ε i ( N + 1) = ε i ( N ) + e i ( N ) c i − 1 ( N + 1) mod (2) c i ( N + 1) = [ ε i ( N ) + e i ( N ) c i − 1 ( N + 1) ] , 2 where e i ( N ) = 1 with probability p e i ( N ) = 0 with probability 1 − p is an independent, identically distributed family of random variables. Hence the transition graph is: 3
1-p 1-p 1-p 1-p 1-p 1-p 1-p 1-p 2 2 4 p p 3 p p p p p p 1 2 3 5 7 0 4 6 p (1-p) p (1-p) p (1-p) p (1-p) 2 2 (1-p) p (1-p) p 3 p (1-p) Figure 1: Transition graph of adding machine in base 2 The transition operator P associated to the transition graph is: P = 4
1 − p p 0 0 0 0 0 0 . . . p 2 p (1 − p ) 1 − p 0 0 0 0 0 . . . 0 0 1 − p p 0 0 0 0 . . . p 2 (1 − p ) p 3 0 p (1 − p ) 1 − p 0 0 0 . . . 0 0 0 0 1 − p p 0 0 . . . p 2 0 0 0 0 p (1 − p ) 1 − p 0 . . . 0 0 0 0 0 0 1 − p p . . . p 3 (1 − p ) p 2 (1 − p ) 0 0 0 0 p (1 − p ) 1 − p . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
In [KT], P.R. Killen and J. Taylor study the spectrum of the operator P . They prove that the spectrum of P in l ∞ ( N ) is connected to the Julia set of f where f : C �→ C defined by: f ( z ) = ( z − (1 − p )) 2 /p 2 . In particular the set of eigenvalues E satisfies E = { z ∈ C , f n ( z ) bounded } 1 Fibonacci base F 0 = 1 , F 1 = 2 , F n = F n − 1 + F n − 2 ∀ n ≥ 0 . k � N = ε i ( N ) F i = ε k ( N ) . . . ε 0 ( N ) i =0 6
where ε i = 0 , 1 , ε i ε i +1 � = 11 , ∀ 0 ≤ i ≤ k ( N ) − 1 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , ... 51 = 34 + 13 + 3 + 1 = F 7 + F 5 + F 2 + F 0 = 10100101 It is known (see Frougny) that the addition of 1 in base ( F n ) n ≥ 0 is given by a finite transductor. How to construct this transductor 7
10 / 00 00 / 01 I T 001 / 010 x / x 101 / 000 Figure 2: Transductor of Fibonacci adding machine Ex: N = 10010 8
10010 1 10011 = 10100 If N = 10010 then N = ( I, 10 / 00 , I ) ( I, 00 / 01 , T ) ( T, 1 , T ) . Hence N + 1 = 10100 . Now we define the stochastic adding machine by the following manner 9
( 10 / 10, 1 - p ) ( 10 / 00, p ) ( 00 / 00, 1 - p ) ( 00 / 01, p ) I T ( 001 / 010, p ) ( 001 / 001, 1 - p ) ( 101 / 000, p ) ( x / x, 1 ) ( 101 / 101, 1 - p ) Figure 3: Transductor of Fibonacci fallible adding machine 10
If we have a path ( p 0 , ( a 0 /b 0 , t 0 ) , p 1 ) . . . ( p n , ( a n /b n , t n ) , p n +1 ) where p 0 = I and p n +1 = T, then we say that the number a n . . . a 0 transitions to the number b n . . . b 0 and we can remark that the probability of transition is t 0 t 1 . . . t n . Example: 10101 transitions to 100000 with probability p 3 and 10101 transitions to 10000 with probability p (1 − p ) . The transition graph: 11
1-p 1-p 1-p 1-p 1-p 1-p 1-p 1-p 1-p 1-p 1-p 1-p 1-p 1-p 3 3 p p p p p p p 2 p 2 p p 2 p p p 3 10 0 1 9 13 8 11 12 4 5 6 2 7 p (1-p) p (1-p) p (1-p) p (1-p) p (1-p) 2 p (1-p) 2 p (1-p) Figure 4: Transition graph of Fibonacci fallible adding machine The transition operator P is: 12
1 − p p 0 0 0 0 0 0 0 0 1 − p p 0 0 0 0 0 0 p 2 p (1 − p ) 0 1 − p 0 0 0 0 0 0 0 0 1 − p p 0 0 0 0 p 2 p (1 − p ) 0 0 0 1 − p 0 0 0 0 0 0 0 0 1 − p p 0 0 P = 0 0 0 0 0 0 1 − p p 0 p 2 (1 − p ) p 3 0 0 0 0 p (1 − p ) 0 1 − p 0 0 0 0 0 0 0 0 1 − p 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 p (1 − p ) 0 0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
Remark 1.1 The diagonal region of the transition matrix is formed 1 − p p 0 by two blocs A and B where A = and 0 1 − p p p (1 − p ) 0 1 − p 1 − p p B = . We can prove that the sequence of 0 1 − p occurences of A and B is given by the fixed point of Fibonacci substitution. 14
Figure 5: 1 /p = 1 . 1 15
Figure 6: 1 /p = 1 . 5 16
Figure 7: 1 /p = 1 . 6 17
Figure 8: 1 /p = 1 . 61 18
Figure 9: 1 /p = 1 . 8 19
Figure 10: 1 /p = 1 . 9 Theorem 1 Let f : C 2 �→ C 2 be the function defined by: f ( x, y ) = ( 1 p 2 ( x − 1 + p )( y − 1 + p ) , x ) . Then the point spectrum of P, S ( P ) in l ∞ ( N ) is contained in the set E p = { λ ∈ C | ( λ 1 , λ ) ∈ J f } 20
where J f is the filled Julia set of f and λ 1 = 1 − p + (1 − λ − p ) 2 . p Moreover The set E p satisfies the following topological properties. 1. C \ E p is a connected set. √ 2. If 0 < p < 1 /β where β = 1+ 5 , then E p is a disconnected set . 2 3. When p converges to 1 then E p is a connected set . Conjecture S ( P ) = E p Idea of Proof. Let λ be an eigenvalue of P associated to the eigenvector v = ( v i ) i ≥ 0 . Since P i,i + k = 0 for all k ≥ 2 and i , we can prove by induction on k that v k = q k ( a, λ ) v 0 , ∀ k ∈ N (1) 21
where a = 1 /p and q k ( a, λ ) ∈ C . We can also prove that for all integer n ≥ 2 , q F n = aq F n − 1 q F n − 2 − ( a − 1) , ∀ n ≥ 2 . where q F 0 = q 1 = − 1 − λ − p p and q F 1 = q 2 = (1 − λ − p ) 2 . p 2 Let g ( x, y ) = ( axy − ( a − 1) , x ) . We have ( q F k , q F k − 1 ) = g ( q F k − 1 , q F k − 2 ) = · · · = g k − 1 ( q F 1 , q F 0 ) . (2) 22
We have λ ∈ S ( P ) ⇔ v n bounded ⇔ q n bounded ⇒ q F n bounded ⇒ ( q F n , q F n − 1 ) bounded . Since ( q F n , q F n − 1 ) = g n − 1 ( q F 1 , q F 0 ) = h − 1 f n − 1 h ( q F 1 , q F 0 ) = h − 1 f n − 1 ( λ 1 , λ ) Where h is the C 2 map defined by: h ( x, y ) = ( ax − ( a − 1) , ay − ( a − 1)) . It follows S ( P ) ⊂ { λ ∈ C , f n ( λ 1 , λ ) bounded } = J ( f ) . � To give the exact value of S ( P ) , we need the following lemma. 23
Lemma 1 for all 0 < k < F n − 1 , we have q F n + k = q F n q k . In particular for each n ∈ N with n = F n 1 + · · · + F n k (Fibonacci representation), we have q n = q F n 1 . . . q F nk . � Remark 1.2 We have S ( P ) = { λ ∈ C , q n ( λ ) is bounded for all n ∈ N } and E p = { λ ∈ C , q F n ( λ ) is bounded for all n ∈ N } . By doing many computations, we conjecture that S ( P ) = E p . Properties of E p 1.1 Let ( f n ) n ≥ 0 be the function sequence defined by: f 0 ( z ) = z, f 1 ( z ) = z 2 , f n ( z ) = af n − 1 ( z ) f n − 2 ( z ) − ( a − 1) , ∀ n ≥ 2 , 24
Let K = K p = { z ∈ C , f n ( z ) is bounded } = { z ∈ C , g n ( z 2 , z )) bounded } . It is easy to see that E p = ψ ( K ) where ψ ( z ) = pz + 1 − p, for all z ∈ C . Then we will study properties of K Proposition 1 There exists a real number R > 1 such that if that | f k ( z ) | > R for some integer k, then the sequence ( f n ( z )) n ≥ 0 is unbounded. With this, we obtain that 25
+ ∞ � f − 1 K = n D (0 , R ) where D (0 , R ) n =0 Working more, we can prove that for all n f − 1 n +1 D (0 , R ) ⊂ f − 1 n D (0 , R ) � C \ K is a connected set . Since + ∞ � f − 1 K = n D (0 , R ) n =0 , then + ∞ � C \ f − 1 C \ K = n D (0 , R ) . n =0 26
We have for all n, f − 1 n D (0 , R ) is a connected set. Since for every n D (0 , R ) ⊂ C \ f − 1 n, C \ f − 1 n +1 D (0 , R ) , we deduce that C \ K is a connected set. � K is disconnected if p > 1 /β Remark 1.3 This work extends to all Parry sequences ( F n ) n ≥ 0 given by the relation F n + d = a 1 F n + d − 1 + · · · + a d F n ∀ n ≥ 0 , and with initial conditions (Parry conditions) F 0 = 1 , F n = a 1 F n − 1 + · · · + a n F 0 + 1 ∀ 0 ≤ n < d, where a i , 1 ≤ i ≤ d are non-negative integers which satisfy the 27
relations a j a j +1 . . . a d ≤ lex a 1 a 2 . . . a d − j +1 for 2 ≤ j ≤ d. References [Be] J. Berstel , Transductions and context-free languages , Teubner, 1979. [Ei] S. Eilenberg , Automata, languages, and machines , Volume 1, Academic Press, New York and London (1974) [FS] C. Frougny, B. Solomyak , Finite beta-expansions , Ergodic Theory Dynam. Systems 12 (1992), 713-723. [Fr] C. Frougny , Syst` emes de num´ eration lin´ eaires et automates finis , PHd thesis, Universit´ e Paris 7, Papport LITP 89-69, 1989. 28
Recommend
More recommend
