Solution
= Solution
The result follows by induction. It is true for $n=1$. If
$$
F^n(x)=2^nx\pmod1,
$$
then
$$
F^{n+1}(x)
=F(F^n(x))
=2(2^nx\pmod1)\pmod1
=2^{n+1}x\pmod1.
$$
Therefore
$$
\boxed{F^n(x)=2^nx\pmod1}.
$$
= Solution
The result follows by induction. It is true for $n=1$. If
$$
F^n(x)=2^nx\pmod1,
$$
then
$$
F^{n+1}(x)
=F(F^n(x))
=2(2^nx\pmod1)\pmod1
=2^{n+1}x\pmod1.
$$
Therefore
$$
\boxed{F^n(x)=2^nx\pmod1}.
$$