Exercise 3

Law (2)

The dynamical law (2) is:

\[N(n+1) = N(n) - 1\]

It is very similiar to the law (1), in differs only in direction of time: the state transistions are executed in reverse order. The law is deterministic and reversible.

Law (3)

\[N(n+1)=N(n)+2\] With law (3), the next state over time is always 2 greater than the previous one. There are two possible paths through the space of states, depending on where you start. If you start in an even numbered state, the path consists of all even numbered states, and if you start in an odd numbered state, you stay only in those states.

The law is deterministic and reversible.

Law (4)

Law (4) is given as: \[ N(n+1) = N(n)^2 \] Again, the path through the space of states depends on the initial condition.

If we start in state 0, which means \(N(n=0)=0\), we stay in this state forever. It isn’t a dynamical law at all. The same holds true if we start in state 1, \(N(n=0)=1\).

If we start in state 2 (\(N(n=0)=2\)), we end up in state 4. If we start in state -2, we also end up in state 4. This violates the rules for valid dynamical laws, as it is not reversible.

Law (5):

\[N(n+1) = -1^{N(n)} N(n)\]

It is exactly equivalent to the law \(N(n+1) =-N(n)\).

If you start in state 0, the state is never left. It’s not a dynamical law.

If you start in any other state, you flip state with every transition between \(N\) and \(-N\). There is an infinite number of these two-state-paths in the space of states, but the law is nevertheless deterministic and reversible.

Perhaps the law (5) should actually read: \[N(n+1) = (-1)^{N(n)} N(n)\]

Now, if you start in an even numbered state, you stay there for ever, it’s no dynamcial law.

If you start in an odd numbered state, you flip state with every transition. This law is valid.