Periodic points of the doubling map
= Periodic points of the doubling map
The fixed points of $F^n$ are
$$
x=\frac{j}{2^n-1},
\qquad
j=0,\ldots,2^n-2.
$$
Thus $F^n$ has $2^n-1$ fixed points.
= Periodic points of the doubling map
The fixed points of $F^n$ are
$$
x=\frac{j}{2^n-1},
\qquad
j=0,\ldots,2^n-2.
$$
Thus $F^n$ has $2^n-1$ fixed points.