What is the composition of a skewed tent map with itself? Given the skewed tent map:
$F(x) = 
     \begin{cases}
       \text{$\mu x$,} &\quad\text{$x\in [0,\frac{1}{1+s}]$} \\          
       \text{$\frac{\mu}{s}(1-x)$,} &\quad\text{$x\in [\frac{1}{1+s},1]$}            
     \end{cases}$
I want to find $F^2(x)$. I'm not really sure how to go about this; do I apply $F(x)$ again to the two parts of $F(x)$? 
Based on my notes for the non-skewed tent map I have something like this:
$F^2(x) = 
     \begin{cases}
       \text{$\mu^2 x$,} &\quad\text{$x\in [0,\frac{1}{\mu(1+s)}]$} \\
       \text{$\frac{\mu}{s}(1-\mu x) $,} &\quad\text{$x\in [\frac{1}{\mu(1+s)},\frac{1}{1+s}]$}  \\
       \text{$\frac{\mu}{s}(1-\frac{\mu}{s}(1-x)) x$,} &\quad\text{$x\in [\frac{1}{1+s},1-\frac{s}{\mu(1+s)}]$}\\        
       \text{$\frac{\mu^2}{s}(1-x)$,} &\quad\text{$x\in [1-\frac{s}{\mu(1+s)},1]$}            
     \end{cases}$
But I don't really understand how to get here; and I'm not even sure it's right as if $\mu\lt1$ then $\frac{1}{\mu(1+s)}\gt1-\frac{s}{\mu(1+s)}$ which means the intervals are all wrong...
 A: First of all, you are right that the formula for $F^2(x)$ does not make sense when $\mu < 1$. I'm going to assume that somewhere in your notes is an assumption that $\mu > 1$, and go from there. 
So now we have a formula
$$F^2(x) = F(F(x)) = 
\begin{cases}
\mu F(x) & F(x) \in [0,\frac{1}{1+s}] \\
\frac{\mu}{s} (1-F(x)) & F(x) \in [\frac{1}{1+s},1]
\end{cases}
$$
We would like to rewrite this without the expression "$F(x)$" anywhere on the right hand side. To do this, we break into cases.
Case 1: $x \in [0,\frac{1}{1+s}]$ and so $F(x) = \mu x$. We therefore have equivalent statements as follows:
$$F(x) \in \left[0,\frac{1}{1+s}\right] \quad \iff \quad 0 \le F(x) \le \frac{1}{1+s} \quad \iff \quad 0 \le \mu x \le \frac{1}{1+s}
$$
$$\iff 0 \le x \le \frac{1}{\mu(1+s)}
$$
Similarly,
$$F(x) \in \left[\frac{1}{1+s},1\right] \iff \frac{1}{\mu(1+s)} \le x \le \frac{1}{1+s}
$$
From this we obtain two lines of the desired formula for $F^2(x)$, namely
$$F^2(x) = \mu F(x) = \mu^2 x \quad\text{if}\quad x \in \left[0,\frac{1}{\mu(1+s)}\right]
$$
and
$$F^2(x) = \frac{\mu}{s}(1-F(x)) = \frac {\mu}{s}(1- \mu x) \quad\text{if}\quad x \in \left[\frac{1}{\mu(1+s)},\frac{1}{1+s}\right]
$$
I'll leave the formulation and proof of Case 2 to you.
