Contents - ChaosBook

CHAPTER 1. OVERTURE 16We could now proceed to estimate the location of the leading singularity ofΓ(z) from finite truncations of (1.7) by methods such as Padé approximants. However,as we shall now show, it pays to first perform a simple resummation thatconverts this divergence into a zero of a related function.1.5.2 Dynamical zeta functionIf a trajectory retraces a prime cycle r times, its expanding eigenvalue is Λ r p .Aprime cycle p is a single traversal of the orbit; its label is a non-repeating symbolstring of n p symbols. There is only one prime cycle for each cyclic permutationclass. For example, p = 0011 = 1001 = 1100 = 0110 is prime, but 0101 = 01is not. By the chain rule for derivatives the stability of a cycle is the same [exercise 13.5]everywhere along the orbit, so each prime cycle of length n p contributes n p terms[section 4.5]to the sum (1.7). Hence (1.7) can be rewritten asΓ(z) =∑∞∑n pp r=1( zn p) r ∑=|Λ p |pn p t p1 − t p, t p = zn p|Λ p |(1.8)where the index p runs through all distinct prime cycles. Note that we have resummedthe contribution of the cycle p to all times, so truncating the summationup to given p is not a finite time n ≤ n p approximation, but an asymptotic, infinitetime estimate based by approximating stabilities of all cycles by a finite number ofthe shortest cycles and their repeats. The n p z n pfactors in (1.8) suggest rewritingthe sum as a derivativeΓ(z) = −z d ∑ln(1 − t p ) .dzpHence Γ(z) is a logarithmic derivative of the infinite product∏1/ζ(z) = (1 − t p ) , t p = zn p|Λ p |p. (1.9)This function is called the dynamical zeta function, in analogy to the Riemannzeta function, which motivates the ‘zeta’ in its definition as 1/ζ(z). This is theprototype formula of periodic orbit theory. The zero of 1/ζ(z) is a pole of Γ(z),and the problem of estimating the asymptotic escape rates from finite n sums suchas (1.2) is now reduced to a study of the zeros of the dynamical zeta function(1.9). The escape rate is related by (1.6) to a divergence of Γ(z), and Γ(z) diverges [section 20.1]whenever 1/ζ(z) has a zero.[section 17.4]Easy, you say: “Zeros of (1.9) can be read off the formula, a zeroz p = |Λ p | 1/n pfor each term in the product. What’s the problem?” Dead wrong!intro - 13jun2008.tex