TITLE: Iterative roots of sine
QUESTION [5 upvotes]: Is there an analytical function $f(z)$ such that $f(f(z)) = \sin(z)$? More generally, an analytical function such that f applied $n$ times to $z$ gives $\sin(z)$?
Is there a general theory for answering this question for functions besides $\sin(z)$?

REPLY [5 votes]: No, not analytic. On the real line it is $C^\infty$ and piecewise analytic. see https://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765 In the complex plane, the thing cannot even be defined in a neighborhood of the origin, as essential use is made of the logarithm. The final result is readily modified to give $n$-th iterative roots here, same conditions. 
There is an entire discipline devoted to this. After the traditional names, Fatou, Leau, Julia, the modern names are Baker and Ecalle. I put a number of items as pdf's at BAKER. I would like, someday, to see a translation of Ecalle's 1973 thesis.  
Now, given a specific $x$ with $x_1 = \sin x$ and $ x_{n+1} = \sin x_n,$ we may take
$$ \alpha(x) = \lim_{n \rightarrow \infty} \; \; \; \left( \frac{3}{x_n^2}  \; + \; \frac{6 \log x_n}{5}  \; + \; \frac{79  x_n^2}{1050}   \; + \; \frac{29  x_n^4}{2625}  \; - \; n \right).$$
Note that $\alpha$ actually is defined on  $ 0 < x < \pi$ with
$\alpha(\pi - x) = \alpha(x),$ but the symmetry also means that the inverse function $\alpha^{-1}$ returns to the interval  $ 0 < x \leq \frac{\pi}{2}.$ Meanwhile, $ \; \alpha(\sin x) = 1 + \alpha(x).$
Define
$$ f(x)  = \alpha^{-1} \left( \frac{1}{2} + \alpha(x)   \right) $$
Then $$  f(f(x)) = \sin x. $$
Note that $\alpha$ blows up at the origin. So, for $0 < x < \frac{1}{10},$ you might as well use the first half dozen terms of the asymptotic expansion I called $g$ in the MO question.
EDIT, MONDAY Aug 26:  it may help to point out that $\alpha$ is holomorphic in an open rhombus with opposite vertices at $0$ and $\pi,$ longer diagonal along the real axis, and $60^\circ$ angle at these two vertices. The rhombus is made up of two equilateral triangles. I have no idea how much larger a maximal domain of holomorphicity would be.