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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?
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The assumption here is that : the solution is less than equal to 1/2 => once you apply the first f, it becomes more than 1/2, so you then apply the second f. You can try to assume it is more than 1/2, and then you would use the "other way round". And the restraint is to ensure that the solution you compute (given by the $2a/(1+4a^2))$ is indeed less than 1/2. –  nonlinearism Jun 28 '13 at 6:06

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