Calculate second and third iterations Let $$L_\mu(x) = \mu x(1-x)$$
Calculate $$L_\mu^2(x)$$ and $$L_\mu^3(x)$$
I understand how this works when given f(x) but I don't understand how to do this with mu in the function. 
 A: Note that, for every specific value $\mu$, the definition defines a function $L_{\mu}:\mathbb R\rightarrow\mathbb R.$ The notation $L_{\mu}^2$ is the second iterate of the function $L_{\mu}$ - that is,
$$L_{\mu}^2(x)=L_{\mu}(L_{\mu}(x))$$
where you are doing the same thing you would with $f^2$ except doing it to $L_{\mu}$ instead. So, for instance, you could get
$$L_{\mu}^2(x)=L_{\mu}(\mu x(1-x))=\mu[\mu x(1-x)](1-[\mu x(1-x)])$$
as one valid representation of this, where the bracketed expressions are just $L_{\mu}(x)$ - although you might reasonably wish to simplify or expand this expression.
Generally, you might think of subscripts as having higher precedence than superscripts in this context: $L_{\mu}^n$ means $(L_{\mu})^n$. You can also think about this as if $\mu$ were variable; if we define
$$f(x)=\mu x(1-x)$$
your usual method for calculating $f^2(x)$ would work. The notation $L_{\mu}$ merely gives a name to the family of functions of this form one could get for various $\mu$.
A: We are given that $\,L_\mu(x) := \mu x(1-x).\,$
We have that
$$ x_1 := L_\mu(x) = \mu x(1-x), \tag{1}$$
$$ x_2 := L_\mu(x_1) = \mu x_1(1-x_1), \tag{2} $$
$$ x_3 := L_\mu(x_2) = \mu x_2(1-x_2) \tag{3} $$
are the first three iterates.
Substitute $\,x_2\,$ from equation $(2)$
to get
$$ x_3 = \mu  (\mu x_1(1-x_1)) (1-\mu x_1(1-x_1)). \tag{4}$$
Substitute $\,x_1\,$ from equation $(1)$ into this to get
$$ x_3 = \mu (\mu (\mu x(1-x))(1-\mu x(1-x))\\
 (1- \mu (\mu x(1-x))(1-(\mu x(1-x)))). \tag{5}$$
Expand $\,x_3\,$ into a polynomial in $\,x\,$ to get
$$ x_3 = \mu^3 x - (\mu^3+\mu^4+\mu^5)x^2 +\cdots + 4\mu^7 x^7 - \mu^7 x^8. \tag{6} $$
