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Chapter 13. Counting
Solution 13.1: A transition matrix for 3-disk pinball. a) As the disk is convex, the transition to itself is forbidden. Therefore, the Markov diagram is
1 2
3
with the corresponding transition matrix
T =
Note that T^2 = T + 2. Suppose that Tn^ = anT + bn, then
Tn+1^ = anT^2 + bnT = (an + bn)T + 2an.
So an+1 = an + bn , bn+1 = 2an with a 1 = 1 , b 1 = 0.
b) From a) we have an+1 = an + 2an− 1. Suppose that an ∝ λn. Then λ^2 = λ + 2. Solving this equation and using the initial condition for n = 1, we obtain the general formula
an =
(2n^ − (−1)n) ,
bn =
(2n−^1 + (−1)n).
c) T has eigenvalue 2 and − 1 (degeneracy 2 ). So the topological entropy is ln 2, the
same as in the case of the binary symbolic dynamics. (Yueheng Lan)
Solution 13.2: Sum of Aij is like a trace. Suppose that Aφk = λkφk , where λk , φk are eigenvalues and eigenvectors, respectively. Expressing the vector v = (1, 1 , · · · , 1)t^ in terms of the eigenvectors φk, i.e., v = Σkdkφk , we have
Γn = Σij [An]ij = vtAnv = ΣkvtAndkφk = Σkdkλnk (vtφk) = Σkckλnk ,
where ck = (vtφk)dk are constants.
a) As tr An^ = Σkλnk , it is easy to see that both tr An^ and Γn are dominated by the largest eigenvalue λ 0. That is
ln |tr An| ln |Γn|
n ln |λ 0 | + ln |Σk( λ λk 0 )n| n ln |λ 0 | + ln |Σkdk( λ λk 0 )n|
→ 1 as n → ∞.
soluCount - 8oct2003 boyscout version11.9.2, Aug 21 2007
b) The nonleading eigenvalues do not need to be distinct, as the ratio in a) is controlled by the largest eigenvalues only.
(Yueheng Lan)
Solution 13.4: Transition matrix and cycle counting. a) According to the definition of Tij , the transition matrix is
T =
a c b 0
b) All walks of length three 0000 , 0001 , 0010 , 0100 , 0101 , 1000 , 1001 , 1010 (four sym- bols!) with weights aaa, aac, acb, cba, cbc, baa, bac, bcb. Let’s calculate T^3 ,
T^3 =
a^3 + 2abc a^2 c + bc^2 a^2 b + b^2 c abc
There are altogether 8 terms, corresponding exactly to the terms in all the walks.
c) Let’s look at the following equality
Tnij = Σk 1 ,k 2 ,···,kn− 1 Tik 1 Tk 1 k 2 · · · Tkn− 1 j.
Every term in the sum is a possible path from i to j, though the weight could be zero. The summation is over all possible intermediate points (n − 1 of them). So, Tnij gives the total weight (probability or number) of all the walks from i to j in n steps.
d) We take a = b = c = 1 to just count the number of possible walks in n steps. This is the crudest description of the dynamics. Taking a, b, c as transition probabilities would give a more detailed description. The eigenvlues of T is (1 ±
5)/ 2 , so we get
N (n) ∝ ( 1+
√ 5 2 )
n.
e) The topological entropy is then ln 1+
√ 5 2.^ (Yueheng Lan)
Solution 13.6: “Golden mean” pruned map. It is easy to write the transition matrix T
T =
The eigenvalues are (1 ±
5)/ 2. The number of periodic orbits of length n is the trace
Tn^ =
5)n^ + (1 −
5)n 2 n^
(Yueheng Lan)
Solution 13.5: 3-disk prime cycle counting. The formula for arbitrary length
cycles is derived in sect. 13.4.
boyscout version11.9.2, Aug 21 2007 soluCount - 8oct
Solution 13.11: Whence M¨obius function? Written out f (n) line-by-line for a few values of n, (13.38) yields
f (1) = g(1) f (2) = g(2) + g(1) f (3) = g(3) + g(1) f (4) = g(4) + g(2) + g(1) · · · f (6) = g(6) + g(3) + g(2) + g(1) · · · (S.46)
Now invert recursively this infinite tower of equations to obtain
g(1) = f (1) g(2) = f (2) − f (1) g(3) = f (3) − f (1) g(4) = f (4) − [f (2) − f (1)] − f (1) = f (4) − f (2) · · · g(6) = f (6) − [f (3) − f (1)] − [f (2) − f (1)] − f (1) · · ·
We see that f (n) contributes with factor − 1 if n prime, and not at all if n contains
a prime factor to a higher power. This is precisely the raison d’etre for the M¨obius
function, with whose help the inverse of (13.38) can be written as the M¨obius inversion formula [29.29] (13.39).
boyscout version11.9.2, Aug 21 2007 soluCount - 8oct
