# Lyapunov exponent of the logistic map (Mathematica Code)

In a previous post I’d shown a way to get the Lyapunov exponent from the time series data of any map. In this quick tutorial, I’ll show you a cleaner way to get the Lyapunov exponent for the specific case of the logistic map, and then using a really short script in Mathematica, plot it against r.

First the mathematical analysis that simplifies the expression for the Lyapunov exponent for a map, and particularly the logistic map. This discussion follows this article.

Suppose the initial infinitesimal perturbation is $\delta x_{0}$. Then we have, for $n\rightarrow\infty$: $\left|\delta x_{n}\right|=\left|\delta x_{0}\right|e^{\lambda n}\Rightarrow e^{\lambda n}=\underset{\delta x_{0}\rightarrow0}{\lim}\left|\frac{\delta x_{n}}{\delta x_{0}}\right|=\left|\frac{dx_{n}}{dx_{0}}\right|$.

But we can write: $\frac{dx_{n}}{dx_{0}}=\frac{dx_{n}}{dx_{n-1}}\frac{dx_{n-1}}{dx_{n-2}}\ldots\frac{dx_{1}}{dx_{0}},$

where each $x_{i}=f(x_{i-1})$. So we have: $\frac{dx_{n}}{dx_{0}}=f'(x_{n-1})f'(x_{n-2})\ldots f'(x_{0}).$

Therefore, $\;e^{\lambda n}=\left|f'(x_{n-1})f'(x_{n-2})\ldots f'(x_{0})\right|$ $\Rightarrow\lambda(r;x_{0})=\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\ln\left|f'(x_{n-1})f'(x_{n-2})\ldots f'(x_{0})\right|$ $=\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\sum\limits_{k=0}^{n-1}\ln\left|f'(x_{k})\right|.$

For the logistic map, $f'(x)=r(1-2x)$.
So we have: $\lambda(r;x_{0})=\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\sum\limits_{k=0}^{n-1}\ln\left|r(1-2x_{k})\right|$.

We can put the above formula in a short Mathematica script to obtain $\lambda$ as a function of $r$ and plot it. The following is the code:

\[Lambda][r_] := Module[{f, l},
f[x_] := r x (1 - x);
l[x_] := Log[Abs[r (1 - 2 x)]];
Mean[l[NestList[f, 0.1, 1*^2]]]];
Plot[\[Lambda][r], {r, 0, 4}, PlotStyle -> Thickness[.0001],
AxesLabel -> {"r", "\[Lambda](r)"}]


And the following is the output: In the line that uses Nestlist, we specify the starting point of the trajectories. However, I noticed that the output does not depend on the starting point.

## 4 thoughts on “Lyapunov exponent of the logistic map (Mathematica Code)”

1. Galdino Vanlex

Hi, very good. But, and of system dynamics, i. e., flow?

Like

2. midhun

Is it l or f ? in the function

Like

3. syed zubair shah

if the function is sin(x) then can you draw it graphically its lyaponuv exponent?

Like