Computing the $k$th iteration of the logistic map in one step

Question

Suppose we have the logistic map with parameter 4, that is take $$ f(x) = 4x(1-x). $$ For any point in $x_0 \in [0, 1] \cap \mathbb{Q}$, does there exists an equation I can use to compute $f^k(x_0)$ for any $k$? That is, if $k$ is very large, say $10^{10}$, and I want to know $f^{10^{10}}(x_0)$, is there a way I can quickly compute that value? The only way I know about is through iteration, or constructing the polynomial by taking $10^{10}$ compositions of $f$ (which gets you a polynomial of enormous degree). Is there a faster way, and if not, is there some mathematical reason for this?

Answer 1

For this parameter, and only for this, you can parametrize $x_0=\sin^2\theta_0$. Then it is easy to check that $$x_k=\sin^2(2^k\theta_0).$$ Now it remains to compute the reduced angle of $2^{10^{10}}\theta_0$ for the sine evaluation in sufficient accuracy, which requires to know $\pi$ with the same accuracy of about $3·10^9$ decimal digits.