Consider the following functions:
$X_{n+1} = 2aX_na$ on $0 \leq X_n \leq \dfrac{1}{2}$
$X_{n+1} = -2aX_n + 2a$ on $\dfrac{1}{2} \leq X_n \leq$
We want to find if there are period 2 solutions for this triangular shaped 'object'. What my teacher did and what I don't understand now is:
- Normally if you want period 2 solutions you have to solve $X=X_{n+2}$, so $x= f(f(x))$. However, what my teacher I believe is combine the functions:
$x = -2a \cdot 2ax + 2a$
.. $x=\dfrac{2a}{1+4a^2}$
Why do we have to fill in the first function in the second (or vice versa)
Doesn't doing it the other way around yield different answers? ($x=2a(-2aX_n + 2a$))?
He then continued on with his answer $\dfrac{2a}{1+4a^2}$, and did the following:
$$0 < \dfrac{2a}{1+4a^2} < \dfrac{1}{2}$$
- Why do we have to add this restraint?