05/23/2008, 09:24 AM
andydude Wrote:This must mean that:
\( \text{ilog}(f^{\circ t}) = t \text{ilog}(f) \)
so
\( f^{\circ t} = \text{ilog}^{-1}(t \text{ilog}(f)) \)
At first I thought this was either an Abel or Schroeder function, but it seems that it is neither.
The categories (non-mathematical) are different:
\( \text{ilog} \) maps a function to a function (or better a formal powerseries to a formal powerseries) while the Abel function maps values to values. And before writing \( \text{ilog}^{-1} \) you should assure that it is invertible, which stronly seems not to be the case.
To be shorter I adopt Ecalle's notation and write \( f_\ast \) for \( \text{ilog}(f) \) and write \( f^{\ast} \) for the regular Abel function of \( f \). Then the relation between them both is:
\( \frac{\partial f^\ast(x)}{\partial x}=1/f_\ast(x) \)
This can easily deduced by:
\( \begin{align*}
f^\ast(f^{\circ t}(x))&=t+f^\ast(x)\\
\frac{\partial f^\ast(f^{\circ t})}{\partial t}&=1\\
\frac{\partial f^\ast(x)}{\partial x} \frac{\partial f^{\circ t}}{\partial t}&=1\\
\end{align*}
\)
In particular \( 1/f_\ast \) is a meromorphic function i.e. can be expressed as
\( c_{-n}x^{-n}+\dots+c_{-1}x^{-1}+c_0+c_1x+c_2x^2+\dots \)
If you now integrate this expression \( x^{-1} \) becomes \( \log(x) \). As this is so important, Ecalle calls \( c_{-1} \)
the "residu iteratif" (which he introduces in his first theorem in "Theorie des Invariants Holomorphes") so that we have
\( f^\ast(z) = c_{-1} \log(z) + F(z) \) where \( F(z) \) is a meromorphic function. I think similar considerations can be found in Szerkes' paper too.
Quote:Because I've also heard Abel functions referred to as "iterational logarithm" before (I think Peter Walker calls them this).
Ecalle calls the Abel function of \( f \) "iterateur de \( f \)" and he calls \( f_\ast \) "logarithme iteratif de \( f \)".
