I encountered this problem in my assignment.
$S: [0,1]\to [0,1] $
$$ S(x_n)=\begin{cases} 3x_n&0\leq x_n<\dfrac{1}{4}\\ x_n+\dfrac{1}{2}&\dfrac{1}{4}\leq x_n<\dfrac{1}{2}\\ 3-4x_n&x_n\in[\dfrac{1}{2},\dfrac{3}{4})\\ 2x_n-3/2&x_n\in[\dfrac{3}{4},1) \end{cases} $$ Show it has periodic points of all periods, or disprove this.
I have written out the shift patterns:
Let $\displaystyle x=\sum_{n=1}^{+\infty}\dfrac{a_n}{2^n}$, where $a_n\in\{0,1\}$, then
If $a_1=a_2=0$, then $x\in[0,\dfrac{1}{4}]$, therefore $$ S(x)=S[0,0,a_3,a_4,\cdots]=[0,a_3,a_4+a_3,a_5+a_4,\cdots] $$
If $a_1=0,a_2=1$, then $x\in[\dfrac{1}{4},\dfrac{1}{2}]$, therefore $$ S(x)=S[0,1,a_3,a_4,\cdots]=[1,1,a_3,a_4,\cdots] $$
If $a_1=1,a_2=0$, then $x\in[\dfrac{1}{2},\dfrac{3}{4}]$, therefore $$ S(x)=S[1,0,a_3,\cdots]=[1-a_3,1-a_4,\cdots] $$
If $a_1=1,a_2=1$, then $x\in[\dfrac{3}{4},1]$, therefore $$ S(x)=S[1,1,a_3,a_4,\cdots]=[0,a_3,a_4,a_5,\cdots] $$
The fixed points of $S(x)$ are $x_0=\bar 0$, $x_1=\dfrac{3}{5}={0.1\overline{0011}}$. 2 period point $x_2=\overline{10}$.
But for period larger than $3$, I cannot find a point of that period, could someone tell me how to proceed?