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I have spent some time using gp-pari. There is, of course, a formal power series solution to $ f(f(x)) = \sin x.$ It is displayed, below, identified by the symbol $g$ because I am not entirely sure whether it is a function of anything. On the other hand, should the coefficients continue to (by and large) decrease, this suggests a nonzero radius of convergence. If the radius of convergence is nonzero, then inside that, not only is a function defined and, you know, analytic, but the functional equation is satisfied. Indeed, all that is necessary is radius of convergence strictly larger than $\frac{\pi}{2}$ owing to certain symmetries. For instance, given my polynomial $g,$ it seems we have $g=1$ at about $x \approx 1.14.$ Then we seem to have a local maximum at $x =\frac{\pi}{2},$ and apparently there $g \approx 1.14,$ strictly larger than 1 which is an important point. So everything would fall into place with large enough nonzero radius of convergence.

$$ \begin{array}{lll} g & = & x - \frac{x^3 }{ 12} - \frac{x^5 }{ 160} - \frac{53 x^7 }{ 40320} - \frac{23 x^9 }{71680} - \frac{92713 x^{11}}{1277337600} - \\\ & & \\\ & & \frac{742031 x^{13} }{79705866240} + \frac{594673187 x^{15} }{167382319104000} + \frac{329366540401 x^{17} }{91055981592576000} + \\\ & & \\\ & & \frac{104491760828591 x^{19} }{62282291409321984000} + \frac{1508486324285153 x^{21} }{4024394214140805120000} + \cdots \end{array} $$

Note that the polynomial $g$ is smaller than $x$ but larger that $\sin x,$ for, say, $0 < x \leq \frac{\pi}{2}.$

So, that is the question, does the formal power series beginning with $g$ converge anywhere other than $x = 0$?

EDIT: note that the terms after the initial $x$ itself have all turned out to be $$ \frac{a_{2 k + 3} x^{2 k + 3} }{2^k ( 2 k + 4)!} $$ where each $a_{2 k + 3}$ is an integer. This much seems provable, although I have not tried yet.

EDIT, Friday 12 November. It now seems really unlikely that this particular problem gives an analytic answer. I suspect that the answer is $C^\infty$ and piecewise analytic, with failure of analyticity at only the points "parabolic" where the derivative has absolute value as large as 1, those points being $0,\pi, 2 \pi, \ldots.$ However, we need the anchor point at the fixpoint 0, otherwise how to begin? And I do think the power series will serve as an asyptotic expansion around 0.

Given the problem with the size of the derivative, now I am hoping for great things, and an obviously periodic and analytic solution, to the easier variant $f(f(x)) = g(x) = (1/2) \sin x.$ I would like both a nice power series and a nice answer by methods summing iterates $ g^{[k]}(x),$ which for the moment is an entirely mysterious method to me, but attractive for periodic target functions as periodicity would be automatic.

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It might help if you wrote the recurrence relation for the coefficients of $g$ so people could think about how to solve it without having to rederive it. –  Warren Schudy Nov 10 '10 at 22:43
    
Let $g=\sum_{i = 0}^{\infty} a_i x^{2i+1}$. Do you have a conjecture of the asymptotic behavior of the coefficients? For example do you suppose $a_i = \Theta(c^i}$ for some constant $c$? A tabulation of $\ln a_i$ for $0 \le i \le 30$ might help one make such a guess. –  Warren Schudy Nov 10 '10 at 22:59
    
Hi Will, are you saying that you had trouble quickly computing terms beyond those shown in your question? –  Warren Schudy Nov 10 '10 at 23:03
7  
Let $\sin^{\langle k\rangle}(x)$ denote the composition of $\sin x$ with itself $k$ times. Write $\sin^{\langle k\rangle}(x) =\sum_{n\geq 1} \varphi_n(k)x^n/n!$. Then $\varphi_n(k)$ is a polynomial in $k$, and $f(x)=\sum_{n\geq 1} \varphi_n(1/2)x^n/n!$. Thus it might be interesting to look at the polynomial $\varphi_n(k)$. See Exercise 5.52 of Enumerative Combinatorics, vol. 2. Part (c) of this exercise is concerned with the formal power series $h(x)$ satisfying $h(h(x))=e^x-1$ and seems to behave similarly to $f(x)$. –  Richard Stanley Nov 11 '10 at 4:01
2  
Regarding my previous comment, here are the polynomials $(2n+1)!n!\varphi_{2n+1}(k)$ for $0\le n\le 6$: $$ 1 $$ $$ -k $$ $$ 10k^2-8k $$ $$ -350k^3+672k^2-32k $$ $$ 29400k^4-95424k^3+102912k^2-36864k $$ $$ -4851000k^5+22915200k^4-40187840k^3+30666240k^2-8542720k $$ $$ 1387386000k^6-8772603840k^5+21909888000k^4-26678446080k^3 $$ $$ \ \ \ \ \ \ \ +15602895360k^2-3449118720k $$ –  Richard Stanley Nov 11 '10 at 20:09

7 Answers 7

This is also a comment. There's another reasonably efficient way to do this sort of computation. Let $L$ be the linear operator on formal power series defined by $L(g) = g(\sin x)$. (Instead of $\sin x$ we could use any formal power series starting with $x$.) Let $I$ be the identity operator, and let $\Delta= L-I$. Then $\Delta$ kills the lowest degree term of its argument, so any infinite sum $\sum_n a_n \Delta^n(g)$ converges as a formal power series. If $\alpha$ is a nonnegative integer then $$L^\alpha(g) = (I+\Delta)^\alpha(g) = \sum_i \binom{\alpha}{i}\Delta^i(g).$$ The coefficient of $x^n$ on the right is a polynomial in $\alpha$ and thus makes sense for any $\alpha$, so we can define $L^\alpha$ for any $\alpha$ by this formula; and we will always have $L^\alpha\circ L^\beta= L^{\alpha+\beta}$. So $f(x) = L^{1/2}(x)$ satisfies $f(f(x)) = \sin x$. Using this approach we can easily compute the coefficients of $f(x)$ up to $x^{100}$ in a few seconds in Maple (though I don't claim that this approach is more efficient than Kevin O'Bryant's).

It might be pointed out this this approach is closely related to the representation of composition of power series as matrix multiplication.

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Yes, I'm at Brandeis. Calculations suggest that the infinite series for $f(x)=L^{1/2}(x)$ given above converges for all real $x$, and the radius of convergence of the power series is something like $\pi/2$ but maybe a little smaller. You could probably prove convergence by getting a bound on $\Delta^i(x)$ as defined above, though I haven't tried to do this. There is a list of references on fractional iteration at reglos.de/lars/ffx.html, and a possibly relevant paper is G.Labelle, Sur l'inversion et l'itération continue des séries formelles. European J. Combin. 1 (1980), 113–138. –  Ira Gessel Nov 11 '10 at 6:23
1  
The same idea is explained in mathoverflow.net/questions/17605/how-to-solve-ffx-cosx/…. –  Ira Gessel Nov 11 '10 at 6:36

EDIT, September 2014: I wrote to Prof. Ecalle, it turns out (as I had hoped) that the fractional iterates constructed by the recipe below really do come out $C^\infty,$ including a growth bound, in terms of $n,$ on the $n$-th derivatives at $0.$ The key word phrase is Gevrey Class. Also, I recently put a better exposition and example of the technique at http://math.stackexchange.com/questions/911818/how-to-get-fx-if-we-know-ffx-x2x/912324#912324

ORIGINAL: The correct answer to this belongs to the peculiar world of complex dynamics. See John Milnor, Dynamics in One Complex Variable.

First, an example. Begin with $f(z) = \frac{z}{1 + z},$ which has derivative 1 at $z=0$ but, along the positive real axis, is slightly less than $x$ when $x > 0.$ We want to find a Fatou coordinate, which Milnor (page 107) denotes $\alpha,$ that is infinite at $0$ and otherwise solves what is usually called the Abel functional equation, $$ \alpha(f(z)) = \alpha(z) + 1.$$ There is only one holomorphic Fatou coordinate up to an additive constant. We take $$ \alpha(z)= \frac{1}{ z}.$$ To get fractional iterates $f_s(z)$ of $f(z),$ with real $0 \leq s \leq 1,$ we take $$ f_s (z) = \alpha^{-1} \left( s + \alpha(z) \right) $$ and finally $$f_s(z) = \frac{z}{1 + s z}.$$ The desired semigroup homomorphism holds, $$ f_s(f_t(z)) = f_{s + t}(z), $$ with $f_0(z) = z$ and $f_1(z) = f(z).$

Alright, the case of $\sin z$ emphasizing the positive real axis is not terribly different, as long as we restrict to the interval $ 0 < x \leq \frac{\pi}{2}.$ For any such $x,$ define $x_0 = x, \; x_1 = \sin x, \; x_2 = \sin \sin x,$ and in general $ x_{n+1} = \sin x_n.$ This sequence approaches 0, and in fact does so for any $z$ in a certain open set around the interval $ 0 < x \leq \frac{\pi}{2}$ that is called a petal.

Now, given a specific $x$ with $x_1 = \sin x$ and $ x_{n+1} = \sin x_n$ it is a result of Jean Ecalle at Orsay that we may take $$ \alpha(x) = \lim_{n \rightarrow \infty} \; \; \; \frac{3}{x_n^2} \; + \; \frac{6 \log x_n}{5} \; + \; \frac{79 x_n^2}{1050} \; + \; \frac{29 x_n^4}{2625} \; - \; n.$$

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 returns to the interval $ 0 < x \leq \frac{\pi}{2}.$

Before going on, the limit technique in the previous paragraph is given in pages 346-353 of Iterative Functional Equations by Marek Kuczma, Bogdan Choczewski, and Roman Ger. The solution is specifically Theorem 8.5.8 of subsection 8.5D, bottom of page 351 to top of page 353. Subsection 8.5A, pages 346-347, about Julia's equation, is part of the development.

As before, we define ( at least for $ 0 < x \leq \frac{\pi}{2}$) the parametrized interpolating functions, $$ f_s (x) = \alpha^{-1} \left( s + \alpha(x) \right) $$

In particular $$ f_{1/2} (x) = \alpha^{-1} \left( \frac{1}{2} + \alpha(x) \right) $$

I calculated all of this last night. First, by the kindness of Daniel Geisler, I have a pdf of the graph of this at:

http://zakuski.math.utsa.edu/~jagy/sine_half.pdf

Note that we use the evident symmetries $ f_{1/2} (-x) = - f_{1/2} (x)$ and $ f_{1/2} (\pi -x) = f_{1/2} (x)$

The result gives an interpolation of functions $f_s(x)$ ending at $ f_1(x)=\sin x$ but beginning at the continuous periodic sawtooth function, $x$ for $ -\frac{\pi}{2} \leq x \leq \frac{\pi}{2},$ then $\pi - x$ for $ \frac{\pi}{2} \leq x \leq \frac{3\pi}{2},$ continue with period $2 \pi.$ We do get $ f_s(f_t(z)) = f_{s + t}(z), $ plus the holomorphicity and symmetry of $\alpha$ show that $f_s(x)$ is analytic on the full open interval $ 0 < x < \pi.$

EDIT, TUTORIAL: Given some $z$ in the complex plane in the interior of the equilateral triangle with vertices at $0, \sqrt 3 + i, \sqrt 3 - i,$ take $z_0 = z, \; \; z_1 = \sin z, \; z_2 = \sin \sin z,$ in general $z_{n+1} = \sin z_n$ and $z_n = \sin^{[n]}(z).$ It does not take long to show that $z_n$ stays within the triangle, and that $z_n \rightarrow 0$ as $n \rightarrow \infty.$

Second, say $\alpha(z)$ is a true Fatou coordinate on the triangle, $\alpha(\sin z) = \alpha(z) + 1,$ although we do not know any specific value. Now, $\alpha(z_1) - 1 = \alpha(\sin z_0) - 1 = \alpha(z_0) + 1 - 1 = \alpha(z_0).$ Also $\alpha(z_2) - 2 = \alpha(\sin(z_1)) - 2 = \alpha(z_1) + 1 - 2 = \alpha(z_1) - 1 = \alpha(z_0).$ Induction, given $\alpha(z_n) - n = \alpha(z_0),$ we have $\alpha(z_{n+1}) - (n+1) = \alpha(\sin z_n) - n - 1 = \alpha(z_n) + 1 - n - 1 = \alpha(z_0).$

So, given $z_n = \sin^{[n]}(z),$ we have $\alpha(z_n) - n = \alpha(z).$

Third , let $L(z) = \frac{3}{z^2}+ \frac{6 \log z}{5} + \frac{79 z^2}{ 1050} + \frac{29 z^4}{2625}$. This is a sort of asymptotic expansion (at 0) for $\alpha(z),$ the error is $| L(z) - \alpha(z) | < c_6 |z|^6.$ It is unlikely that putting more terms on $L(z)$ leads to a convergent series, even in the triangle.

Fourth, given some $ z =z_0$ in the triangle. We know that $z_n \rightarrow 0$. So $| L(z_n) - \alpha(z_n) | < c_6 |z_n|^6.$ Or $| (L(z_n) - n ) - ( \alpha(z_n) - n) | < c_6 |z_n|^6 ,$ finally $$ | (L(z_n) - n ) - \alpha(z) | < c_6 |z_n|^6 .$$ Thus the limit being used is appropriate.

Fifth, there is a bootstrapping effect in use. We have no actual value for $\alpha(z),$ but we can write a formal power series for the solution of a Julia equation for $\lambda(z) = 1 / \alpha'(z),$ that is $\lambda(\sin z ) = \cos z \; \lambda(z).$ The formal power series for $\lambda(z)$ begins (KCG Theorem 8.5.1) with $- z^3 / 6,$ the first term in the power series of $\sin z$ after the initial $z.$ We write several more terms, $$\lambda(z) \asymp - \frac{z^3}{6} - \frac{z^5}{30} - \frac{41 z^7}{3780} - \frac{4 z^9}{945} \cdots.$$ We find the formal reciprocal, $$\frac{1}{\lambda(z)} = \alpha'(z) \asymp -\frac{6}{z^3} + \frac{6}{5 z} + \frac{79 z}{525} + \frac{116 z^3}{2625} + \frac{91543 z^5}{6063750}\cdots.$$ Finally we integrate term by term, $$\alpha(z) \asymp \frac{3}{z^2} + \frac{6 \log z }{5} + \frac{79 z^2}{1050} + \frac{29 z^4}{2625} + \frac{91543 z^6}{36382500}\cdots.$$ and truncate where we like, $$\alpha(z) = \frac{3}{z^2} + \frac{6 \log z }{5} + \frac{79 z^2}{1050} + \frac{29 z^4}{2625} + O(z^6)$$

Numerically, let me give some indication of what happens, in particular to emphasize $ f_{1/2} (\pi/2) = 1.140179\ldots.$

    x      alpha(x)      f(x)       f(f(x))     sin x       f(f(x))- sin x
1.570796   2.089608    1.140179    1.000000    1.000000      1.80442e-11
1.560796   2.089837    1.140095    0.999950    0.999950      1.11629e-09
1.550796   2.090525    1.139841    0.999800    0.999800      1.42091e-10
1.540796   2.091672    1.139419    0.999550    0.999550      3.71042e-10
1.530796   2.093279    1.138828    0.999200    0.999200      1.97844e-10
1.520796   2.095349    1.138070    0.998750    0.998750      -2.82238e-10
1.510796   2.097883    1.137144    0.998201    0.998201      -7.31867e-10
1.500796   2.100884    1.136052    0.997551    0.997551      -1.29813e-09
1.490796   2.104355    1.134794    0.996802    0.996802      -1.14504e-09
1.480796   2.108299    1.133372    0.995953    0.995953      9.09416e-11
1.470796   2.112721    1.131787    0.995004    0.995004      1.57743e-09
1.460796   2.117625    1.130040    0.993956    0.993956      5.63618e-10
1.450796   2.123017    1.128133    0.992809    0.992809      -3.00337e-10
1.440796   2.128902    1.126066    0.991562    0.991562      1.19926e-09
1.430796   2.135285    1.123843    0.990216    0.990216      2.46512e-09
1.420796   2.142174    1.121465    0.988771    0.988771      -2.4357e-10
1.410796   2.149577    1.118932    0.987227    0.987227      -1.01798e-10
1.400796   2.157500    1.116249    0.985585    0.985585      -1.72108e-10
1.390796   2.165952    1.113415    0.983844    0.983844      -2.31266e-10
1.380796   2.174942    1.110434    0.982004    0.982004      -4.08812e-10
1.370796   2.184481    1.107308    0.980067    0.980067      1.02334e-09
1.360796   2.194576    1.104038    0.978031    0.978031      3.59356e-10
1.350796   2.205241    1.100627    0.975897    0.975897      2.36773e-09
1.340796   2.216486    1.097077    0.973666    0.973666      -1.56162e-10
1.330796   2.228323    1.093390    0.971338    0.971338      -5.29822e-11
1.320796   2.240766    1.089569    0.968912    0.968912      8.31102e-10
1.310796   2.253827    1.085616    0.966390    0.966390      -2.91373e-10
1.300796   2.267522    1.081532    0.963771    0.963771      -5.45974e-10
1.290796   2.281865    1.077322    0.961055    0.961055      -1.43066e-10
1.280796   2.296873    1.072986    0.958244    0.958244      -1.58642e-10
1.270796   2.312562    1.068526    0.955336    0.955336      -3.14188e-10
1.260796   2.328950    1.063947    0.952334    0.952334      3.20439e-10
1.250796   2.346055    1.059248    0.949235    0.949235      4.32107e-10
1.240796   2.363898    1.054434    0.946042    0.946042      1.49412e-10
1.230796   2.382498    1.049505    0.942755    0.942755      3.42659e-10
1.220796   2.401878    1.044464    0.939373    0.939373      4.62813e-10
1.210796   2.422059    1.039314    0.935897    0.935897      3.63659e-11
1.200796   2.443066    1.034056    0.932327    0.932327      3.08511e-09
1.190796   2.464924    1.028693    0.928665    0.928665      -8.44918e-10
1.180796   2.487659    1.023226    0.924909    0.924909      6.32892e-10
1.170796   2.511298    1.017658    0.921061    0.921061      -1.80822e-09
1.160796   2.535871    1.011990    0.917121    0.917121      3.02818e-10
1.150796   2.561407    1.006225    0.913089    0.913089      -3.52346e-10
1.140796   2.587938    1.000365    0.908966    0.908966      9.35707e-10
1.130796   2.615498    0.994410    0.904752    0.904752      -2.54345e-10
1.120796   2.644121    0.988364    0.900447    0.900447      -6.20484e-10
1.110796   2.673845    0.982228    0.896052    0.896052      -7.91102e-10
1.100796   2.704708    0.976004    0.891568    0.891568      -1.62699e-09
1.090796   2.736749    0.969693    0.886995    0.886995      -5.2244e-10
1.080796   2.770013    0.963297    0.882333    0.882333      -8.63283e-10
1.070796   2.804543    0.956818    0.877583    0.877583      -2.85301e-10
1.060796   2.840386    0.950258    0.872745    0.872745      -1.30496e-10
1.050796   2.877592    0.943618    0.867819    0.867819      -2.82645e-10
1.040796   2.916212    0.936899    0.862807    0.862807      8.81083e-10
1.030796   2.956300    0.930104    0.857709    0.857709      -7.70554e-10
1.020796   2.997914    0.923233    0.852525    0.852525      1.0091e-09
1.010796   3.041114    0.916288    0.847255    0.847255      -4.96194e-10
1.000796   3.085963    0.909270    0.841901    0.841901      6.71018e-10
0.990796   3.132529    0.902182    0.836463    0.836463      -9.28187e-10
0.980796   3.180880    0.895023    0.830941    0.830941      -1.45774e-10
0.970796   3.231092    0.887796    0.825336    0.825336      1.26379e-09
0.960796   3.283242    0.880502    0.819648    0.819648      -1.84287e-10
0.950796   3.337412    0.873142    0.813878    0.813878      5.84829e-10
0.940796   3.393689    0.865718    0.808028    0.808028      -2.81364e-10
0.930796   3.452165    0.858230    0.802096    0.802096      -1.54149e-10
0.920796   3.512937    0.850679    0.796084    0.796084      -8.29982e-10
0.910796   3.576106    0.843068    0.789992    0.789992      3.00744e-10
0.900796   3.641781    0.835396    0.783822    0.783822      8.10903e-10
0.890796   3.710076    0.827666    0.777573    0.777573      -1.23505e-10
0.880796   3.781111    0.819878    0.771246    0.771246      5.31326e-10
0.870796   3.855015    0.812033    0.764842    0.764842      2.26584e-10
0.860796   3.931924    0.804132    0.758362    0.758362      3.97021e-10
0.850796   4.011981    0.796177    0.751806    0.751806      -7.84946e-10
0.840796   4.095339    0.788168    0.745174    0.745174      -3.03503e-10
0.830796   4.182159    0.780107    0.738469    0.738469      2.63202e-10
0.820796   4.272614    0.771994    0.731689    0.731689      -7.36693e-11
0.810796   4.366886    0.763830    0.724836    0.724836      -1.84604e-10
0.800796   4.465171    0.755616    0.717911    0.717911      3.22084e-10
0.790796   4.567674    0.747354    0.710914    0.710914      -2.93204e-10
0.780796   4.674617    0.739043    0.703845    0.703845      1.58448e-11
0.770796   4.786234    0.730686    0.696707    0.696707      -8.89497e-10
0.760796   4.902777    0.722282    0.689498    0.689498      2.40592e-10
0.750796   5.024513    0.713833    0.682221    0.682221      -3.11017e-10
0.740796   5.151728    0.705339    0.674876    0.674876      7.32554e-10
0.730796   5.284728    0.696801    0.667463    0.667463      -1.73919e-10
0.720796   5.423842    0.688221    0.659983    0.659983      -1.66422e-10
0.710796   5.569419    0.679599    0.652437    0.652437      5.99509e-10
0.700796   5.721838    0.670935    0.644827    0.644827      -2.45424e-10
0.690796   5.881501    0.662231    0.637151    0.637151      -6.29884e-10
0.680796   6.048843    0.653487    0.629412    0.629412      1.86262e-10
0.670796   6.224333    0.644704    0.621610    0.621610      -5.04285e-10
0.660796   6.408471    0.635883    0.613746    0.613746      -6.94697e-12
0.650796   6.601802    0.627025    0.605820    0.605820      -3.81152e-10
0.640796   6.804910    0.618129    0.597834    0.597834      4.10222e-10
0.630796   7.018428    0.609198    0.589788    0.589788      -1.91816e-10
0.620796   7.243040    0.600231    0.581683    0.581683      -4.90592e-10
0.610796   7.479486    0.591230    0.573520    0.573520      4.29742e-10
0.600796   7.728570    0.582195    0.565300    0.565300      -1.38719e-10
0.590796   7.991165    0.573126    0.557023    0.557023      -4.05081e-10
0.580796   8.268218    0.564025    0.548690    0.548690      -5.76379e-10
0.570796   8.560763    0.554892    0.540302    0.540302      1.49155e-10
0.560796   8.869925    0.545728    0.531861    0.531861      1.0459e-11
0.550796   9.196935    0.536533    0.523366    0.523366      -1.15537e-10
0.540796   9.543137    0.527308    0.514819    0.514819      -2.84462e-10
0.530796   9.910004    0.518054    0.506220    0.506220      6.24335e-11
0.520796   10.299155    0.508771    0.497571    0.497571      -9.24078e-12
0.510796   10.712365    0.499460    0.488872    0.488872      8.29491e-11
0.500796   11.151592    0.490122    0.480124    0.480124      3.31769e-10
0.490796   11.618996    0.480757    0.471328    0.471328      2.27307e-10
0.480796   12.116964    0.471366    0.462485    0.462485      3.06434e-10
0.470796   12.648140    0.461949    0.453596    0.453596      4.77846e-11
0.460796   13.215459    0.452507    0.444662    0.444662      1.53162e-10
0.450796   13.822186    0.443041    0.435682    0.435682      -2.87541e-10
0.440796   14.471963    0.433551    0.426660    0.426660      -5.20332e-11
0.430796   15.168860    0.424037    0.417595    0.417595      -8.17951e-11
0.420796   15.917436    0.414501    0.408487    0.408487      -4.6788e-10
0.410796   16.722816    0.404944    0.399340    0.399340      3.70729e-10
0.400796   17.590771    0.395364    0.390152    0.390152      -6.97547e-11
0.390796   18.527825    0.385764    0.380925    0.380925      -2.45522e-10
0.380796   19.541368    0.376143    0.371660    0.371660      4.09758e-10
0.370796   20.639804    0.366503    0.362358    0.362358      1.15221e-10
0.360796   21.832721    0.356843    0.353019    0.353019      -4.75977e-11
0.350796   23.131092    0.347165    0.343646    0.343646      -4.27696e-10
0.340796   24.547531    0.337468    0.334238    0.334238      2.12743e-10
0.330796   26.096586    0.327755    0.324796    0.324796      4.06133e-10
0.320796   27.795115    0.318024    0.315322    0.315322      -2.71476e-10
0.310796   29.662732    0.308276    0.305817    0.305817      -3.74988e-10
0.300796   31.722372    0.298513    0.296281    0.296281      -1.50491e-10
0.290796   34.000986    0.288734    0.286715    0.286715      2.17798e-11
0.280796   36.530413    0.278940    0.277121    0.277121      4.538e-10
0.270796   39.348484    0.269132    0.267499    0.267499      5.24261e-11
0.260796   42.500432    0.259311    0.257850    0.257850      7.03059e-11
0.250796   46.040690    0.249475    0.248175    0.248175      -1.83863e-10
0.240796   50.035239    0.239628    0.238476    0.238476      4.06119e-10
0.230796   54.564668    0.229768    0.228753    0.228753      -2.56253e-10
0.220796   59.728239    0.219896    0.219007    0.219007      -7.32657e-11
0.210796   65.649323    0.210013    0.209239    0.209239      3.43103e-11
0.200796   72.482783    0.200120    0.199450    0.199450      -1.20351e-10
0.190796   80.425131    0.190216    0.189641    0.189641      1.07544e-10
0.180796   89.728726    0.180303    0.179813    0.179813      9.93221e-11
0.170796   100.721954    0.170380    0.169967    0.169967      2.63903e-10
0.160796   113.838454    0.160449    0.160104    0.160104      6.74095e-10
0.150796   129.660347    0.150510    0.150225    0.150225      4.34057e-10
0.140796   148.983681    0.140563    0.140332    0.140332      -2.90965e-11
0.130796   172.920186    0.130610    0.130424    0.130424      4.02502e-10
0.120796   203.060297    0.120649    0.120503    0.120503      -1.85618e-11
0.110796   241.743576    0.110683    0.110570    0.110570      4.2044e-11
0.100796   292.525678    0.100711    0.100626    0.100626      -1.73504e-11
0.090796   361.023855    0.090734    0.090672    0.090672      2.88887e-10
0.080796   456.537044    0.080752    0.080708    0.080708      -2.90848e-10
0.070796   595.371955    0.070767    0.070737    0.070737      4.71103e-10
0.060796   808.285844    0.060778    0.060759    0.060759      -3.90636e-10
0.050796   1159.094719    0.050785    0.050774    0.050774      3.01403e-11
0.040796   1798.677124    0.040791    0.040785    0.040785      3.77092e-10
0.030796   3159.000053    0.030794    0.030791    0.030791      2.4813e-10
0.020796   6931.973789    0.020796    0.020795    0.020795      2.95307e-10
0.010796   25732.234731    0.010796    0.010796    0.010796      1.31774e-10
    x       alpha(x)        f(x)        f(f(x))     sin x       f(f(x))- sin x
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This is more a comment than an answer. The following Mathematica code gave the first 100 coefficients in 44 seconds.

Do[
   f[x_] = Sum[a[k] x^k, {k, 0, exp}];
   term1 = Coefficient[f[f[x]], x, exp];
   term2 = SeriesCoefficient[Sin[x], {x, 0, exp}];
   a[exp] = a[exp] /. First[FindInstance[term1 == term2, a[exp], Rationals]],
 {exp, 0, 100}]
Table[ a[k], {k, 0, 100}]

Here, $f(x) = \sum_{k=0}^\infty a_k x^k$. As expected, $a_{2k}=0$ for $0\leq k \leq 50$, and $a_{2k+1} (2k+2)! 2^{k-1}$ is an integer for $0\leq k \leq 49$.

Here's the list of $a_{2k+1} (2k+2)! 2^{k-1}$ for $0\leq k \leq 22$.

1, 
-2, 
-9, 
-212, 
-9315, 
-556278, 
-25971085, 
4757385496, 
2964298863609, 
1044917608285910, 
215713544372776879, 
-62932769961642167868, 
-98704332065950259333867, 
-30188592688651749114181790, 
58856949571932104601673308075, 
77375921970586388105168106822960, 
-72564223774641266435601127563343119, 
-334464255008553673036506122999946116946, 
-40744061094877107085401232437389280011673, 
2173769171456754713290183664020158569935376220, 
3467462783233757169265913185746537990884591231373,
-21502898790444864584967220140381964189431832253894982,
-93866159932956697746363373697973240405899859356681018397

And here is $\log(|a_k|)$ rounded to the nearest integer for odd $k$ between 0 and 200:

0, -2, -5, -7, -8, -10, -12, -13, -13, -13, -15, -16, -16, -18, -17, 
-18, -19, -18, -21, -18, -19, -19, -19, -19, -18, -20, -18, -19, -17, 
-18, -17, -16, -16, -15, -15, -14, -15, -13, -15, -11, -13, -10, -10, 
-8, -8, -7, -6, -5, -4, -4, -2, -2, 0, -1, 2, 1, 4, 2, 6, 4, 8, 8, 
10, 10, 13, 13, 15, 16, 17, 18, 20, 21, 22, 24, 25, 27, 27, 29, 30, 
32, 33, 35, 35, 38, 38, 41, 39, 44, 43, 47, 47, 50, 50, 53, 54, 57, 
57, 60, 61, 63

That looks to me like super-exponential growth.

share|improve this answer
    
Call the normalized integers $b_k=a_{2k+1}(2k+2)!2^{k-1}$. Then $b_{k+1}/b_k$ is typically quite small, but varies dramatically. $b_{39}/b_{38}$ is just over 360 000, while $b_{40}/b_{39}$ is only about 3500. There doesn't seem to be much structure in the ratios, but admittedly there isn't much data. –  Kevin O'Bryant Nov 11 '10 at 2:55
    
In the range of $0\leq k \leq 100$, the three largest values of $|a_k|$ are (in decreasing order) $a_1>a_3 > a_{99}$. That's right, the third largest coefficient is the 99th. –  Kevin O'Bryant Nov 11 '10 at 3:07
1  
I see now that Gottfried has already noted the apparently super-exponential growth of the coefficients. –  Kevin O'Bryant Nov 11 '10 at 22:42

Checking the numerators 53,23,92713 (ignoring signs) in the trusty OEIS leads to A048602. Which has references and comments Recursion exists for coefficients, but is too complicated to process without computer algebra system

If you try in the obvious way to compose g with itself when it goes up to $x^{23}$ then you will get terms up to $x^{529}$ all but one of which are useless. Maple has a power series package which allows composition and truncates all terms past the order you specify. I've never used it before now but it looks as if it might be pretty snappy.

update I've removed my terms because others calculated further by better methods. Kevin points out that the largest terms of the first 100 are $a_1=1,a_3=-0.083$ and $a_{99}=0.0231$. 100 seems like a reasonable place to stop, but Gottfreid went further. Unless you click the link to his plots you might miss that (according to him) $a_{255}>10^{48}$. I do think he is correct about the sizes. I thought maybe it was an artifact of calculation but my own modest calculations using Ira's lovely method agree with his (based on a plot) as far as I went which was up to :

[97, -0.011673], [99, 0.023144], [101, 0.83376e-1], [103, -.11914], [105, -.62229], [107, .60156], [109, 4.8816], [111, -2.6819], [113, -40.354], [115, 6.0469], [117, 351.82], [119, 88.156]

share|improve this answer
3  
Three cheers for OEIS! –  Warren Schudy Nov 10 '10 at 23:08
    
So do those references solve the problem? –  Warren Schudy Nov 10 '10 at 23:12
    
This does not solve the problem. The references there are three articles by the person who made the OEIS entry. Two of those papers are unpublished preprints whose titles seem quite unrelated to this problem (i.e.,"W. C. Yang, Polynomials are essentially integer partitions, preprint, 1999 "). –  J.C. Ottem Nov 11 '10 at 0:17
    
Probably A098932 is more informative. –  Gerald Edgar Nov 11 '10 at 2:12

If you create the Bell-matrix for the function f(x) = sin(x), say SI, then you can compute the matrix-logarithm of SI, say SIL = MLog(SI). Then a formal power of SI is SIP(h) = MExp(h*SIL) and the Bell-matrix for the height-dependend function sin_iter(x,h), which has polynomials in h for the coefficients at x. SI begins with

  1        .     .       .
  0        1     .       .
  0        0     1       .
  0     -1/6     0       1
  0        0  -1/3       0
  0    1/120     0    -1/2
  0        0  2/45       0
  0  -1/5040     0  13/120

where column 1 contains the coefficients for the power series sin(x), column 2 that for (sin(x))^2, column 0 that for (sin(x))^0 = 1 and similar for column k.

The matrix-logarithm SIL begins with

  0         .      .      .     .     .  .  . 
  0         0      .      .     .     .  .  . 
  0         0      0      .     .     .  .  . 
  0      -1/6      0      0     .     .  .  . 
  0         0   -1/3      0     0     .  .  . 
  0     -1/30      0   -1/2     0     0  .  . 
  0         0  -1/15      0  -2/3     0  0  . 
  0  -41/3780      0  -1/10     0  -5/6  0  0 

Here the column k is the k'th multiples of column 1

Then column 1 of SIP(h)= MExp(h*SIL) is

                              0
                              1
                              0
                         -1/6*h
                              0
                1/24*h^2-1/30*h
                              0
  -5/432*h^3+1/45*h^2-41/3780*h

and the function sin_iter is

$ siniter(x,h) = 1*x -1/6*h*x^3 + (1/24*h^2-1/30*h)*x^5 - (...)*x^7 + O(x^9) $

Inserting h=1/2 gives you the powerseries for the half-iterate.

Using 64 terms it looks as if the radius of convergence for h=1/2 will be 1, since the absolute values of the coefficients seem to stabilize to the interval $ \pm 1*E-7 $ but I'll look at this further later this day.

[Update]

using 256 terms there occurs a clear growthrate of the coefficients. Looking at the log of absolute values of that coefficients we get a rough impression. See here:

http://go.helms-net.de/math/images/sincoeff_c.png

These are the coefficients at x^123,x^125,x^127 and x^251, x^253, x^255:

c_123     -2156.72733764089915  // 4 digits
c_125     31313.42875545542423  // 5 digits
c_127     34859.64557727596911  // 5 digits 
...    
c_251       -35365220492708296140377087748804440170254492009.570  // 46 digits    
c_253     -1378449672866233726070664896135098313484573633108.4    // 48 digits    
c_255       987848122496441964413343332623221752473112662017.00   // 47 digits    

Differences of the logs are also quotients of the coefficients By the plot of the differences we get also a trend of logarithmic increase. (If the differences continue to increase then the radius of convergence of the powerseries is zero)

http://go.helms-net.de/math/images/sincoeff_d.png

[end update]

Pari/GP computes this pretty fast, it took,let say 5 seconds to handle 64-term-matrices.

share|improve this answer
    
On your plots you indicate that the non-zero coefficients alternate in sign. Actually the pattern starts out +------++++---++---++--++---++--++--++---++--++--++--++--+++ An entry in the OEIS oeis.org/A095883 indicates that the inverse function might have even terms 0 and odd terms positive. –  Aaron Meyerowitz Nov 11 '10 at 15:42
    
@Aaron - hmm, I had such an effect of apparently cyclic signs often. If I would understand more of fourier-analysis I'd like to apply such an analysis to that sequences of coefficients. If we assume some periodic effect, say we have 4 interleaved partial sequences then we get smoother curves. However - they are not eventually smooth. It looks, as if either more cyclic effects are overlaid or the period-length is dependent on the index and/or is irrational measured by the index. –  Gottfried Helms Nov 11 '10 at 19:47
2  
Pretty amazing. You should put in the body of your answer that $a_{255}>10^{48}$. That would be easy to overlook. I'd love to see the pattern of the signs as far as you have it. Can you push the calculations further? Your plot of the ratios is also quite a surprise. –  Aaron Meyerowitz Nov 11 '10 at 22:41
    
Sorry, the comment-function hangs, I've to format my comment as an answer. –  Gottfried Helms Nov 11 '10 at 23:47
    
@Will: I'm not sure I got you right. If your question is, whether the given powerseries of sin°0.5(x) can still be summed (as I've done it with $exp(x)-1 $ ) although its convergence-radius is (seems to be) zero, then yes, but again it needs Noerlund-summation, and I've just tried it for some small values of x (I'll provide a plot later; note that there was a plot given by Anixx in the related thread concerning the iteration of the cos(x)) If your question concerns the method I'll put it in another answer because comments allow more than this number of characters :-) –  Gottfried Helms Nov 12 '10 at 0:35

Another helping comment: There is a general statement about the convergence radius of fractional iterates developed at a fixed point with multiplier 1:

The set of values $\lambda$ for which the regular iteration formal powerseries $f^\lambda$ has non-zero convergence radius is either: (1) only $\lambda=0$ (2) the points $k\lambda_0$, $k\in\mathbb{Z}$, for one $\lambda_0\in\mathbb{C}$. Example $e^z-1$ with $\lambda_0=1$. (3) the whole complex plane. Example $\frac{z}{1-z}$.

This result is due to Écalle [1] and preliminary work of Baker [2]. In our case the original function $\sin(x)$ has non-zero convergence radius, and hence all its integer iterations too. So it may only occur case (2) with $\lambda_0=\frac{1}{n}$ for some integer $n$ or case (3). My conjecture is case (2) with $\lambda_0=1$, but the particular proof needs to be done, (like Baker did it for $e^x-1$)

[1] Écalle, J. (1973). Nature du groupe des ordres d’itération complexes d’une transformation holomorphe au voisinage d’un point fixe de multiplicateur 1. C. R. Acad. Sci., Paris, Sér. A, 276, 261–263.

[2] Baker, I. N. (1962). Permutable power series and regular iteration. J. Aust. Math. Soc., 2, 265–294.

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(This should go as a comment, but was impossible) @ Aaaron: I've uploaded a list of the first 128 nonzero coefficients, see:

http://go.helms-net.de/math/tables/sinxcoeffs.htm

Also here is a routine for Pari/GP to compute the sqrt of a lower triangular Bell-matrix (the matrix SI in my earlier answer) With this you can compute the powerseries for the half-iterate (by column 1 of sqrt of SI) in a second even if the matrix size is 256x256.

\\ square-root of a lower triangular Bell-matrix
\\ only implemented for operator/"Bell"-matrices for functions
\\ where f(x) = ax  + bx^2+ cx^3 + ... with a>0
\\
 trisqrt(m) = local(tmp, rs=rows(m), cs=cols(m), c);
  tmp=matrix(rs,cs,r,c,if(r==c,sqrt(m[r,r])));
  for(d=1,rs-1,
       for(r=d+1,rs,
          c=r-d;
          tmp[r,c]=(m[r,c]-sum(k=c+1,r-1, tmp[r,k]*tmp[k,c]) )/(tmp[c,c]+tmp[r,r])
          );
      );
 return(tmp);
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