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...
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.