If we have a iterative equation defined as
for $i=1:n$ $$a=a(4-a),$$ I need help finding the general formula for this in terms of n. I know some obvious ones like, $$n=1\quad\quad a=4a-a^2$$ $$n=2 \quad\quad a=4(a(4-a))-(a(4-a))^2.$$ Any help is greatly appreciated.
$$a_{n+1}=a_n(4-a_n)$$ $$a_{n+1}=4a_n-a_n^2$$ $$-a_{n+1}=-4a_n+a_n^2$$ $$4-a_{n+1}=4-4a_n+a_n^2=(2-a_n)^2$$ $$2-a_{n+1}=(2-a_n)^2-2$$
So we get $$2-a_{n+1}=(2-a_n)^2-2=\{(2-a_{n-1})^2-2\}^2-2= \ldots =[\ldots \{(2-a_1)^2-2\}^2 \dots \text{(n brackets)}\ldots ]^2 -2 $$
If $a_1$ is given, then the recursive formula is this: $$a_{n+1}=4-[\ldots \{(2-a_1)^2-2\}^2 \dots \text{(n brackets)}\ldots ]^2 $$
However, if $a_0$ is given instead of $a_1$, we will have to go one step further.