(For easier discussion, I suggest you to read the introduction of Schroder's equation and the section on 'Conjugacy' of iterated function, in case you are not familiar with these topics.)
Let $f(x)=\frac12\arctan x$, and $f_n(x)$ be the $n$th iteration of $f$.
Let us reduce functional iteration to multiplication: if we can solve the corresponding Schroder's equation
$$\Psi(f(x))=s\Psi(x)$$
then it is well known (and also straightforward) that
$$f_n(x)=\Psi^{-1}(f'(a)^n\cdot\Psi(x))$$ where $a$ is a fixed point of $f$.
For the moment, let us focus on $\Psi(f(x))=s\Psi(x)$.
Clearly, in our case, $a=0$, and $s=f'(a)=\frac12$.
For $a = 0$, if $h$ is analytic on the unit disk, fixes $0$, and $0 < |h′(0)| < 1$, then Gabriel Koenigs showed in 1884 that there is an analytic (non-trivial) $\Psi$ satisfying Schröder's equation $\Psi(h(x))=s\Psi(x)$.
Thus, $\Psi$ is analytic.
A few more observations:
- $\Psi(0)=0$.
- $\Psi'(0)$ is up to our choice, since if a function $\psi$ is a solution to the Schroder's equation, then so is $k\cdot \psi$ for any constant $k$. For convenience, set $\Psi'(0)=1$.
- All other Taylor series coefficients of $\Psi$ are then uniquely determined, and can be found recursively. (The method will be illustrated below.)
- By Lagrange inversion theorem, $\Psi$ is invertible in a neighbourhood of $0$, and $\Psi^{-1}(z)=0+\frac1{\Psi'(0)}z+o(z)\implies \Psi^{-1}(z)\sim z\quad(z\to 0)$.
- Therefore, $f_n(x)=\Psi^{-1}(f'(a)^n\cdot\Psi(x))=\Psi^{-1}(2^{-n}\Psi(x))\sim 2^{-n}\Psi(x)$ as $n\to\infty$.
Hence, for the limit the OP wanted to evaluate, $$\ell:=\lim_{n\to\infty}2^nf_n(x_0)=\Psi(x_0)$$
We shall now determine all the Taylor series coefficients of $\Psi(x)$ (valid only for $|x|<1$), since it can be assumed $0\le x_0<1$.
Obviously, $\Psi$ is an odd function. Let
$$\Psi(x)=x+\sum^\infty_{k=2}\phi_{2k-1} x^{2k-1}$$
The basic idea is to repeatedly differentiate both sides of $\Psi(f(x))=s\Psi(x)$ and substitute in $x=0$, then recursively solve for the coefficients.
For example, differentiating both sides three times and substitute in $x=0$, we obtain
$$-\Psi'(0)+\frac18\Psi'''(0)=\frac12\Psi'''(0)\implies\phi_3=-\frac49$$
Slightly modifying the notations of our respectable MSE user @Sangchul Lee, for $\lambda=(\lambda_1,\lambda_2,\cdots,\lambda_n)$ a $n$-tuple of non-negative integers:
- write $\lambda \vdash n$ if $\sum^n_{i=1}(2i-1)\lambda_i=2n-1$.
- write $|\lambda| = \sum_{i=1}^{n} \lambda_i$.
- define the tuple factorial as $\lambda !=\frac{|\lambda|!}{\lambda_1!\cdot\lambda_2!\cdots\lambda_n !}$.
I will state, without proof, Faà di Bruno's formula for odd inner function:
$$(\Psi\circ f)^{(2n-1)}=(2n-1)!\sum_{\lambda \vdash n}\lambda!\cdot\phi_{|\lambda|}\prod^n_{i=1}\left(\frac{f^{(2i-1)}(0)}{(2i-1)!}\right)^{\lambda_i}$$
$$\implies \frac12\phi_{2n-1}=\sum_{\lambda \vdash n}\lambda!\cdot\phi_{|\lambda|}\prod^n_{i=1}\left(\frac{(-1)^{i+1}}{2(2i-1)}\right)^{\lambda_i}$$
Further simplifications lead to the final result:
$$\ell=\Psi(x_0)=\sum^\infty_{k=1}\phi_{2k-1} x_0^{2k-1} \qquad{\text{where}}\qquad \phi_1=1$$
$$\phi_{2n-1}=\frac{(-1)^{n}}{2^{-1}-2^{1-2n}}\sum_{\substack{\lambda \vdash n \\ \lambda_1\ne 2n-1}}\phi_{|\lambda|}\frac{\lambda! (-1)^{(|\lambda|+1)/2}}{2^{|\lambda|}}\prod^n_{i=1}\frac1{(2i-1)^{\lambda_i}}$$
Yeah, I know it’s ugly. But that’s the best we can obtain.
If anyone have a nice math software, please help me calculate the first few Taylor coefficients.
