First of all I have to withdraw my assertion. It was based on an error in my computations 
And anyway to good to be true 
A powerseries in the usual sense, i.e.
\( f(x)=\sum_{n=0}^\infty f_n x^n \)
(developed at 0).
If \( f \) is analytic then you get the coefficients via
\( f_n = \frac{f^{(n)}(0)}{n!} \).
If you develop \( f \) at the point \( a \) (given that \( a \) is in the radius of convergence)
\( f(x)=\sum_{n=0}^\infty \alpha_n (x-a)^n \) you can compute the coefficients \( \alpha_n \) via:
\( \alpha_n = (f\circ \tau_a)_n = \sum_{m=n}^\infty \left(m\\n\right) f_m a^{m-n} \)
You can verify this by explicitely expanding \( (f\circ \tau_a)(x)=f(x+a)=\sum_{n=0}^\infty f_n (x+a)^n \) and regather the powers of \( x \).
By \( \tau_a^{-1}\circ f\circ \tau_a \) you move the fixed point \( a \) to 0. And if you have the fixed point at 0 (which means nothing else than \( f_0=0 \)), then you can do the usual regular iteration. For example if you want to compute \( g=f^{\circ 1/2} \) you simply need to solve the equation system:
\( (g\circ g)_n = f_n \)
Here I showed how to solve this equation for \( f_1>0,f_1\neq 1 \) (hyperbolic case).
If \( f_1>0 \) then there is exactly one solution \( g \) (without involving limits but via a recurrrence formula) to this equation system.
Similarely for each other \( m \) in \( g^{\circ m}=f \).
So the fractional iterations of a (formal) power series are uniquely determined (whether the obtained powerseries has a radius of convergence at all is another question). And this can be continuously extended to real iteration counts. You can also get the real iteration count \( g=f^{\circ t} \) by a finite formula, namely by solving the equation system (by recurrence):
\( f\circ g=g\circ f \) with the starting condition \( g_1={f_1}^t \).
In the complex numbers there are more than one regular solution, as \( {f_1}^t \) has a whole set of solutions.
And anyway to good to be true jaydfox Wrote:And I'm not entirely sure what you mean by "power series".
A powerseries in the usual sense, i.e.
\( f(x)=\sum_{n=0}^\infty f_n x^n \)
(developed at 0).
If \( f \) is analytic then you get the coefficients via
\( f_n = \frac{f^{(n)}(0)}{n!} \).
If you develop \( f \) at the point \( a \) (given that \( a \) is in the radius of convergence)
\( f(x)=\sum_{n=0}^\infty \alpha_n (x-a)^n \) you can compute the coefficients \( \alpha_n \) via:
\( \alpha_n = (f\circ \tau_a)_n = \sum_{m=n}^\infty \left(m\\n\right) f_m a^{m-n} \)
You can verify this by explicitely expanding \( (f\circ \tau_a)(x)=f(x+a)=\sum_{n=0}^\infty f_n (x+a)^n \) and regather the powers of \( x \).
By \( \tau_a^{-1}\circ f\circ \tau_a \) you move the fixed point \( a \) to 0. And if you have the fixed point at 0 (which means nothing else than \( f_0=0 \)), then you can do the usual regular iteration. For example if you want to compute \( g=f^{\circ 1/2} \) you simply need to solve the equation system:
\( (g\circ g)_n = f_n \)
Here I showed how to solve this equation for \( f_1>0,f_1\neq 1 \) (hyperbolic case).
If \( f_1>0 \) then there is exactly one solution \( g \) (without involving limits but via a recurrrence formula) to this equation system.
Similarely for each other \( m \) in \( g^{\circ m}=f \).
So the fractional iterations of a (formal) power series are uniquely determined (whether the obtained powerseries has a radius of convergence at all is another question). And this can be continuously extended to real iteration counts. You can also get the real iteration count \( g=f^{\circ t} \) by a finite formula, namely by solving the equation system (by recurrence):
\( f\circ g=g\circ f \) with the starting condition \( g_1={f_1}^t \).
In the complex numbers there are more than one regular solution, as \( {f_1}^t \) has a whole set of solutions.
