09/11/2022, 11:02 AM
So just for Tommy
Normally it is clear that if you have two Abel functions \(\alpha_+\) and \(\alpha_-\) of \(f\) that are defined on a suitable region where you can take the inverse and which includes \(z\) and \(z+1\) that for \(\tau=\alpha_-\circ \alpha^{\circ -1}_+\):
\[ \tau(z+1)=\alpha_-(\alpha^{\circ -1}_+(z+1))=\alpha_-(f(\alpha^{\circ -1}_+(z)))=\alpha_-(\alpha^{\circ -1}_+(z))+1=\tau(z)+1 \]
Hence \(\theta(z)=\tau(z)-z\) is a 1-periodic function, and you can express \(\alpha_-(z)=\alpha_+(z+\theta(z))\)
In case of two adjacent petals the above conditions are satisfied and we can compute \(\tau\).
Lets do that for \(f(x)=e^x-1\):
We calculate the inverse Abel function as before:
\[ \alpha_+^{\circ -1}(t) = f^{\circ t}_+(z_0) \]
The Abel function can be defined similarly:
\begin{align}
\alpha_-(z)=\lim_{n\to\infty} \alpha_{20}(f^{\circ n}(z)) - n
\end{align}
where
\begin{align}
\alpha_{20}(z)=\frac{1}{3}\log(-z)-\frac{2}{x} - \frac{1}{36}x + \frac{1}{540}x^{2} + \frac{1}{7776}x^{3} - \frac{71}{435456}x^{4} + \frac{8759}{163296000}x^{5} + \frac{31}{20995200}x^{6} - \frac{183311}{16460236800}x^{7} + \frac{23721961}{6207860736000}x^{8} + \dots
\end{align}
So with these ingredients we can calculate \(\tau(t)=\alpha_-(\alpha_+^{\circ -1}(t))=\alpha_-( f^{\circ t}_+(z_0))\) where I choose \(z_0=3i\) and let t run from 0 to 2 and show the curve \(\tau(t)\) in the complex plane:
(09/10/2022, 12:08 PM)tommy1729 Wrote: Just like exp(x) - 1 has 2 solutions that fundamentally disagree ( no 1 periodic function unites them )I show how the two Abel functions of \(e^x-1\) are connected by a 1-periodic function.
Normally it is clear that if you have two Abel functions \(\alpha_+\) and \(\alpha_-\) of \(f\) that are defined on a suitable region where you can take the inverse and which includes \(z\) and \(z+1\) that for \(\tau=\alpha_-\circ \alpha^{\circ -1}_+\):
\[ \tau(z+1)=\alpha_-(\alpha^{\circ -1}_+(z+1))=\alpha_-(f(\alpha^{\circ -1}_+(z)))=\alpha_-(\alpha^{\circ -1}_+(z))+1=\tau(z)+1 \]
Hence \(\theta(z)=\tau(z)-z\) is a 1-periodic function, and you can express \(\alpha_-(z)=\alpha_+(z+\theta(z))\)
In case of two adjacent petals the above conditions are satisfied and we can compute \(\tau\).
Lets do that for \(f(x)=e^x-1\):
We calculate the inverse Abel function as before:
(08/31/2022, 09:47 PM)bo198214 Wrote: \begin{align}only that we change the \(\frac{1}{2}\) to \(t\).
f^{\circ \frac{1}{2}}_+(z) &= \lim_{n\to\infty} f^{\circ n}(h_{20}(f^{\circ -n }(z)))
\end{align}
where \(h_{20}\) the formal (non-converging) powerseries of the half iterate truncated to 20 (though 2, i.e. \(z+\frac{1}{4}z^2\) would be sufficient, imho).
\[ \alpha_+^{\circ -1}(t) = f^{\circ t}_+(z_0) \]
The Abel function can be defined similarly:
\begin{align}
\alpha_-(z)=\lim_{n\to\infty} \alpha_{20}(f^{\circ n}(z)) - n
\end{align}
where
\begin{align}
\alpha_{20}(z)=\frac{1}{3}\log(-z)-\frac{2}{x} - \frac{1}{36}x + \frac{1}{540}x^{2} + \frac{1}{7776}x^{3} - \frac{71}{435456}x^{4} + \frac{8759}{163296000}x^{5} + \frac{31}{20995200}x^{6} - \frac{183311}{16460236800}x^{7} + \frac{23721961}{6207860736000}x^{8} + \dots
\end{align}
So with these ingredients we can calculate \(\tau(t)=\alpha_-(\alpha_+^{\circ -1}(t))=\alpha_-( f^{\circ t}_+(z_0))\) where I choose \(z_0=3i\) and let t run from 0 to 2 and show the curve \(\tau(t)\) in the complex plane: