# Chaos Theory: Tent Map (Part 2)

Continuing from the previous post; the graph below shows the tent map for a range of $\mu$ values, here we’ll use initial condition 0.4.

Tent map for a range of values of $\mu$

For $\mu < 1$ the system converges to 0 for all initial conditions. If $\mu = 1$ then all initial conditions less than or equal to $\frac{1}{2}$ are fixed points of the system, otherwise for initial conditions $x_0 > \frac{1}{2}$ they converge to the fixed point $1 - x_0$. For example, for $x_0 = 0.93$ then it converges to the fixed point 0.07 as seen by the red points in the top right plot for $\mu = 1$ in the figure above.

For $\mu > 1$ the system has fixed points at 0 and the other at $\frac{\mu}{\mu + 1}$, we can show this mathematically using the tent map. For when $0 \leq x \leq \frac{1}{2}$ then the fixed point is when $\mu x = x$ which implies that $x = 0$ is a fixed point. For when $\frac{1}{2} \leq x \leq 1$ then the fixed point is when $\mu - \mu x = x$. Rearranging gives $\mu = x + \mu x = (\mu + 1) x$. It follows that the fixed point is at $\frac{\mu}{\mu + 1}$. However, both fixed points are unstable, we can show this mathematically by looking at the gradient of the fixed point. If the gradient is less than one then its said to be stable, if its greater than then its unstable.

Let $T(x^*)$ denote a fixed point of the tent map, then $T'(x^*)$ is the gradient. Given the fixed point 0, then $T'(x^*) = T'(0) = \mu$. For the fixed point $\frac{\mu}{\mu + 1}$, then $T'(x^*) = T'(\frac{\mu}{\mu + 1}) = \frac{1}{(\mu + 1)^2}$. Since $\mu > 1$ in both cases then both fixed points are unstable. In other words, for a value x near the given fixed point, it will diverge away from it rather than converge. For $\mu = 2$, the system maps the interval [0,1] onto itself, becoming chaotic.

The dynamics will be aperiodic for initial conditions that are irrational, and periodic when rational. One effect of this is that we cannot run long term simulations of the tent map on a computer because all numbers in a computer are rational. Since $x_n$ is expressed in binary notation, each successive iteration of the tent map will eventually hit 0 since the leftmost bit will always be removed (Ex 1). For irrational numbers there’s an infinite binary expansion and so will never go to 0 (Ex 2) [1].

(Ex 1) $\quad x_0 = 0.5322265625 = \frac{1}{2} + \frac{1}{32} + \frac{1}{1024} = 0.1000100001$.

(Ex 2) $\quad x_0 = \frac{\pi}{10}$.

In the next post we will (again) be exploring further the tent map, looking at computational limitations regarding the calculation of the tent map, bifurcation diagrams and histogram.

Sources:

[1] GLEICK, J. (1998). Chaos: the amazing science of the unpredictable. London, UK. Vintage.