Inverse super-composition

[sheldonison]

For a fixed point of zero, with a fixed point multiplier of 2, the general solution for the Abel function generated at the fixed point of zero is:
\( f(z) = 2x + \sum_{n=2}^{\infty}a_n z^n \)

\( \alpha(z) = \log_2(S(z))\;\;\; \) This is the Abel function for f(z) \( \;\alpha(f(z)) = \alpha(z)+1 \)

where S(z) is the formal Schröder equation solution;
\( S(f(z)) = 2\cdot S(z)\;\;\; S(z)=z+\sum_{n=2}^{\infty}b_n z^n\;\; \)
This is sometimes called Koenig's solution. It can be modified to work with any fixed point multiplier of k, |k|<>1. Using pari-gp one can easily write a program to generate the formal power series for S(x) given f(x).

I almost undestand it.
Okey, than what is the a[n] and b[n] in the sum formula?
I feel we are closer then ever before.
Could you show me this way with another example, please?
For instance, let us invastigate it: \( cos ^o ^N (x) = sin(x) \). (And cos' fixed point is ~0.739.) The question is that what N is and how I can calculate it.

\( L \approx 0.73908513322; \)
\( f(x) = \cos(x+L)-L = \cos(L)\cos(x)-\sin(L)\sin(x)-L \)
\( f(x) \approx -0.67361202918x-0.36954256661x^2+0.11226867153x^3... \)
I make no attempt to address your question, partly because I doubt your question has an answer.

Using f(x) as an example to explore the Schröder equation probably isn't a good first example either because the multiplier at the fixed point, which is the a1 coefficient of f(x), is negative. Therefore it is no longer possible for the fractional iterates of the function \( f^{[\circ x] \) to be real valued. So the Taylor series of the Schröder equation for f(x) isn't relevant either, but it follows the definition from my previous post. Here are the first couple of terms.
\( {S(x)\approx x+ 0.32779313060x^2-0.74862437767x^3... \)