Suppose we have the tent map, $T_\mu(x) = \mu \min(x, 1-x)$ with parameter $\mu = 2$.
Then note that if we pick an $x_0 \in [0, 1]\cap \mathbb{Q}$, so that if $x_0 = p/q$ with $p, q$ co-prime and $q$ an odd integer, then the orbit $\mathcal{O}(x_0)$ under $T_2$ is necessarily periodic with the period bounded above by $\lceil q/2\rceil$. In particular, all elements of the orbit $\mathcal{O}(x_0)$ are of form $a/q$, where after the first iteration $a$ is necessarily an even number.
This means that on a computer we can exactly compute such an orbit without a loss of precision. My quesiton is, then, given such a setup, can one compute $T_2^n(x_0)$ in a single step without iterating (for any $n$), with the same perfect precision that you would get if you iterated?
If you see the related question here, a commentor shows how you can compute $T^n_2(x_0) = x_n$ by $$ x_n = \frac{2}{\pi}\arcsin (|\sin(2^{n-1}\pi x_0)|). $$ But of course one can no longer compute this quantity with perfect precision as before. So do we have a similar formula where in our special case with $\mu=2$ and $x_0 = p/q$ where we can explicitly compute $x_n$ without loss of precision?