I was reconsidering the fractional iteration of the sine-function and remembering that the power series for the fractional iterates have convergence radius zero I looked at the variant of the sine with different bases, defined by $$ \sin_b(x) = (\exp(\beta x î) - \exp( - \beta x î))/(2 î) = (b^{xî}- b^{-xî})/(2î) $$ where $\beta:=\log b$.
Of course, $\sin_b(x) = \sin(\beta x)$ . For $0 \lt \beta \lt 1$ the fixpoint is zero and it seems that the power series for the fractional iterates have nonzero radius of convergence, as well as the power series for the Schroeder-function.

The problem which I face now occurs for $\beta=\pi/2$. Then the function $f(x)=\sin_b(x)$ has an attracting fixpoint at $t=1$ and I determine the formal power series for the Schroeder-function by $g(x) = f(x+t)-t $ getting something like $$\small g(x) = -1.23370055 x^2 + 0.253669508 x^4 - 0.0208634808 x^6 + 0.000919260275 x^8 - O(x^{10}) $$ $\qquad \qquad $ (which seems to be simply $\cos_b(x)-1 = \cos(\beta x)-1 $ )

Nicely the function has no constant term, but I don't know how to handle this case where also the linear term is zero.

So, how would I proceed here to arrive at a formal power series for the fractional iterate?

Surely I have an option to define a half-iterate, say $ \sin_b^{o0.5} (0.5)$, as a limit, using the observation, that the integer iterates converge to the fixpoint by the following rate. But I do not yet see, how I could use that observation for the generating of power series.

The limit-process which I apply goes this way:
let's denote the $h$'th iterate from some $x_0$ as $x_h$ (in this case we start at $x_0=0.5$). Then heuristically $$ \lim_{h\to \infty} {\log(t-x_{h+1}) \over \log(t-x_{h})} = 2 $$ Then I define the fractional iterate by the fractional power of $2$ for instance the half-iterate implicitely $$ \lim_{h\to \infty} {\log(t-x_{h+0.5}) \over \log(t-x_{h})} = 2^{0.5} $$ and with some approximation, using some small positive integer $h$, its evaluation $$\begin{array} {rl} \log(t-x_{h+0.5}) &= 2^{0.5}\log(t-x_{h}) \\ x_{h+0.5} &= t-\exp(2^{0.5}\log(t-x_{h})) \\ x_{0.5} &= \sin_b^{o-h} ( x_{h+0.5} ) \\ \end{array}$$ gives for the so-defined half-iterate $x_{0.5}$ for $h=11,12,13$ the following approximations: $$ h=11: x_{0.5} \approx 0.600735954803 \\ h=12: x_{0.5} \approx 0.600727116170 \\ h=13: x_{0.5} \approx 0.600722696772 $$ however seemingly needing exorbitant internal precision (in Pari/GP I needed more than $800$ digits precision and used then simply $1600$ digits precision)

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