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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 $\sin(z)$?

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You may find this MathOverflow post helpful. – Michael Albanese Aug 26 '13 at 1:18
@JohnD.Cook also, see… as was brought to my attention by Will Jagy in… – James S. Cook Aug 26 '13 at 1:22
I wonder if there is a way to numerically approximate such a function, assuming it exists. Let $f:(2\pi/n)Z_n \rightarrow (2\pi/n)Z_n$, for instance, with $f$ essentially an element in $Z_n^n$ and we just seek to minimize $\sum |\sin(2\pi k/N)-f(f(2 \pi k/N))|^2$. – nayrb Aug 26 '13 at 2:42
I've given some information on this in . It also contains a general approach(using Carleman/Bell-matrices and matrix-logarithm). However, power series for such half-iterates might have convergence-radius zero and thus cannot conventionally be used. Sometimes a divergent summation procedure can be meaningful to arrive at a valid result. – Gottfried Helms Aug 26 '13 at 19:57
up vote 4 down vote accepted

No, not analytic. On the real line it is $C^\infty$ and piecewise analytic. see 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.

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Google the phrase "What you have to do twice in order to sin" (in quotes, verbatim) and you'll find a paper about this.

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The Catholic Church's Council of Trent, meeting in the 1500s and issuing reactions against Protestantism, dogmatically decreed it to be a sin to get baptized more than once. I suspect it was a reaction to the Anabaptists. But that's not what that paper is about. – Michael Hardy Aug 26 '13 at 1:22
OK, it now appears that this was a talk that someone gave, but Google will lead you to some pages that mention both that talk and some papers dealing with the problem. – Michael Hardy Aug 26 '13 at 1:26
Google results don't look the same for everyone. – Ilmo Euro Aug 26 '13 at 4:56

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