## Homework 9

There is no programming questions in this homework. Everyone can submit up to 3 times.

#### Q9.1 \(\quad\) Transition matrix and cycle counting (Exercise 18.5)

Given the transition graph at the bottom of this page. You can easily get the transition matrix: \[ T = \begin{bmatrix} a & b \\ c & 0 \end{bmatrix} \] As is known from Chapter 18 formula (18.6), the trace \(tr T^n\) counts the number of periodic orbits of length \(n\), namely, \(N_n\), in this system. What is \(N_3\) ,the number of periodic orbits with period 3 given \(a=1\), \(b=2\), \(c=3\) ? ( please input an integer. )#### Q9.2 \(\quad\) Transition matrix and cycle counting -- countinued

From formulae (18.1) and (18.4), you know that the topological entropy indicates the average growth rate of the number of perodic orbits versus the length of the orbit, and it is given by the largest eigenvalue of the transition matrix. For the same system as in Q9.1 with the same parameters, what is the topological entropy ? (at least 3 significant digits should be accurate)#### Q9.3 \(\quad\) Alphabet \(\{a, b, c\}\) pruned \(\_ab\_\) (exercise 18.16)

Suppose there are three nodes \(\{a, b, c\}\) in the transition graph. There are transitions between each two nodes (same or different) except transition from \(a\) to \(b\). So basically, the transition graph is \[ T = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \] Let us try to find the spectral determinant by formula (18.13). \[ \begin{align*} \det(1-zT) = & 1 - (t_a + t_b + t_c + t_{ac} + t_{bc} + t_{acb}) \\ & + (t_at_b + t_bt_c + t_at_c + t_at_{bc} + t_bt_{ac}) \\ & - t_at_bt_c \\ = & 1 - (3z+2z^2+z^3) + (3z^2+2z^3) - z^3 \\ = & 1- 3z + z^2 \end{align*} \] The above expression is arranged in the same way as formula (18.13): we first enumerate all single non-intersecting cycles, and then enumerate all non-intersecting cycle pairs. Last, the non-intersecting cycle triples are enumerated. Note, 4 non-intersecting cycles do not exist simultaneously inside this system, so the above expression is complete. Also pay attention to how the pruning rule is implemented in it. \(t_{abc}\) and \(t_{acb}\) are two different orbits, but \(t_{abc}\) is pruned while \(t_{acb}\) is kept.Similarly, if both \(\_ab\_\) and \(\_ac\_\) are pruned in the system, you can write the spectral determinant in the same way. What is the coefficient of term \(z\) ? (answer should be an integer)

#### Q9.4 \(\quad\) Alphabet \(\{a, b, c\}\) pruned \(\_ab\_\) -- continued

If the pruning rule is finite and transition matrix is known, we can use formula (18.13) to write down the spectral determinant of this system, from which, topological entropy and the number of periodic orbits are all known, just as we did in Q9.3. But when the transition graph is infinite, we need to turn to topological zeta function (18.16) to valuate spectral determinant. But it does not mean that topological zeta function can only be applied to infinite graphs, but it also works for finite graphs like the one in Q9.3. Let us see how it works. \[ \begin{align*} \det(1-zT) = & \prod_p(1-t_p) \\ = & (1-t_a)(1-t_b)(1-t_c)(1-t_{ac})(1-t_{bc}) \\ & (1-t_{acb})(1-t_{aac})(1-t_{bbc})(1-t_{cca})(1-t_{ccb}) \\ = & 1 - (t_a + t_b + t_c) + (t_at_b + t_bt_c + t_at_c - t_{ac} - t_{bc}) \\ & + ( - t_at_bt_c + t_at_{bc} + t_at_{ac} + t_bt_{bc} + t_bt_{ac} + t_ct_{bc} + t_ct_{ac} \\ & - t_{acb} - t_{aac} - t_{bbc} - t_{cca} - t_{ccb}) \\ & \\ = & 1 - 3z + (3z^2-2z^2) + (-z^3+6z^3-5z^3) \\ = & 1- 3z + z^2 \end{align*} \] In the above expression, we have omitted periodic orbits whose period is larger than 3, and also we only keep terms up \(z^3\) in the expansion. We are sure that the higher order terms are all canceled out as discussed in section 18.6. So actually, the topological zeta function works as well as formula (18.13) for finite graphs. Note the difference between these two approaches is that we only count non-intersecting cycles in formula (18.13), but this restriction is not applied to topological zeta function, as shown in the above expansion. In this sense, topological zeta function seems more convenient.Similarly, if both \(\_ab\_\) and \(\_ac\_\) are pruned in the system, you can write the topological zeta function in the same way. What is the coefficient of term \(z^2\) ? (Note the coefficient of \(z^2\), not the coefficient of \(z\) as in Q9.3)

#### Q9.5 \(\quad\) Counting periodic orbit (Example 18.8, 18.9)

In chapter 18, we concentrate almost on one question: how to count periodic orbits ? Formulae (18.7) and (18.24) reveal the relation between the number of periodic orbits and the topological zeta function . In deriving these formulae, we rely heavily on two identities: \[ \ln det = tr \ln \] and \[ \ln(1-x) = -(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots) \] Now, let us turn to the third identity which is important for us to understand formula (18.34): \[ \frac{1}{1-x} = 1 + x + x^2 + \cdots \] Let's turn to example 18.8, the information about the number of periodic orbits is coded in \( \sum_{n=1} N_n z^n = \frac{Nz}{1-Nz}\). It is tempting to set \(z=1\) to get the total number of periodic orbits inside these system, but it turns out to be \(\frac{N}{1-N}\). What ? The problem is that here \(z\) serves as a book tracking variable, which will be set to 1 in the end. But when it appears in the denominator, it is not appropriate to set it to 1 immediately, we need to treat it as a Taylor series. For example, in this case, \[ \sum_{n=1}N_nz^n = Nz(1+Nz+(Nz)^2 + (Nz)^3 + \cdots) \,. \] Therefore, we obtain \(N_n = N^n\) which makes sense.Now turn to example 18.9. You are given the topological zeta function \(1/\zeta = 1 - z - 2z^4 + z^8\), what is the number of periodic orbits with period 15, namely, what is \(N_{15}\) ? ( hint: you can use Mathematica to verify your result if it is available )

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