Until now we were doing all our constructions \(f^{\circ t}(x)=\alpha^{-1}(t+\alpha(x))\) in the parabolic case with \(\alpha\) having a pole (i.e. a rather simple singularity) as \(\tan(x)\) has at \(\frac{\pi}{2}\) or \(\cot(x)\) has at 0.
However according to Ecalle starting with the julia function or logit j one can obtain \(\alpha(x)=\int \frac{{\rm d}x}{j(x)}\), and hence it has to be of the general form:
\[ c\log(x) + a_{-m}x^{-m} + ... + a_{-1}x^{-1} + a_0 + a_1 x + a_2 x^2 + ... \]
And the criterion that \(f^{\circ t}\) is analytic is only that the logit j converges.
So I was looking what one could construct with a \(\log\) and came up with this scenario:
\begin{align}
\alpha(x) &= \log(x) + \frac{1}{x}\\
\alpha'(x) &= \frac{1}{x} - \frac{1}{x^2}\\
j(x) &= \frac{1}{\alpha'(x)} = \frac{1}{\frac{1}{x} - \frac{1}{x^2}} = \frac{x^2}{x-1}
\end{align}
So the logit j is analytic at 0, which according to Ecalle causes \(f^{\circ t}\) to be analytic at 0 for all t too. How does \(f^{\circ t}\) look like? First we have to find the inverse of \(\alpha\):
\begin{align}
\log(x)+\frac{1}{x} &= y\\
xe^{\frac{1}{x}} &= e^y\\
-\frac{1}{x}e^{-\frac{1}{x}} &=- e^{-y}\\
-\frac{1}{x} &= W(-e^{-y})\\
x&=\frac{-1}{W(-e^{-y})}
\end{align}
Just need to choose the right branch of the LambertW (0 for x>0 and -1 for x<0 in \(f^{\circ t}(x)\)). So we have
\[ f^{\circ t}(x) = \frac{-1}{W(-e^{-(t+\log(x)+\frac{1}{x})})} \]
Nobody would believe that this gourd is analytic at 0, but according to Ecalle it is!
However according to Ecalle starting with the julia function or logit j one can obtain \(\alpha(x)=\int \frac{{\rm d}x}{j(x)}\), and hence it has to be of the general form:
\[ c\log(x) + a_{-m}x^{-m} + ... + a_{-1}x^{-1} + a_0 + a_1 x + a_2 x^2 + ... \]
And the criterion that \(f^{\circ t}\) is analytic is only that the logit j converges.
So I was looking what one could construct with a \(\log\) and came up with this scenario:
\begin{align}
\alpha(x) &= \log(x) + \frac{1}{x}\\
\alpha'(x) &= \frac{1}{x} - \frac{1}{x^2}\\
j(x) &= \frac{1}{\alpha'(x)} = \frac{1}{\frac{1}{x} - \frac{1}{x^2}} = \frac{x^2}{x-1}
\end{align}
So the logit j is analytic at 0, which according to Ecalle causes \(f^{\circ t}\) to be analytic at 0 for all t too. How does \(f^{\circ t}\) look like? First we have to find the inverse of \(\alpha\):
\begin{align}
\log(x)+\frac{1}{x} &= y\\
xe^{\frac{1}{x}} &= e^y\\
-\frac{1}{x}e^{-\frac{1}{x}} &=- e^{-y}\\
-\frac{1}{x} &= W(-e^{-y})\\
x&=\frac{-1}{W(-e^{-y})}
\end{align}
Just need to choose the right branch of the LambertW (0 for x>0 and -1 for x<0 in \(f^{\circ t}(x)\)). So we have
\[ f^{\circ t}(x) = \frac{-1}{W(-e^{-(t+\log(x)+\frac{1}{x})})} \]
Nobody would believe that this gourd is analytic at 0, but according to Ecalle it is!
