For repelling fixed points, we can compute the Abel function of the inverse function which has attracting fixed points:
\( \beta(f^{-1}(x))=\beta(x)+1 \)
and then is clear, that \( \alpha(x)=-\beta(x) \) is an Abel function of the original function \( f \), because
\( -\beta(f(x))=-\beta(x)+1 \) iff \( \beta(x)=\beta(f(x))+1 \) iff \( \beta(f^{-1}(y))=\beta(y)+1 \).
So for repelling fixed points \( a \) we get the formula:
\( \alpha_{b,a}(x)=\lim_{n\to\infty} n-\log_{1/\log(a)}(a-\log_b^{\circ n}(x)) \)
which works for arbitrary repelling complex fixed points and arbitrary \( b>1 \) as long as we chose a branch of the involved logarithm such that \( \log_b^{\circ n}\to a \).
For computing the regular super logarithm however we face a major problem with repelling fixed points: we can not compute directly \( \alpha_{b,a}({^nb}) \), as \( \log_b^{\circ n+2}({^nb})=-\infty \). This presents a problem because for the rslog we have to compute \( \alpha_{b,a}(1) \) to normalize the values.
The good news however is that \( \lim_{x\to {^nb}} \alpha_{b,a}(x) \) seems to always exists. So the regular super logarithm is then:
\( \text{rslog}_{b,a}(x)=\alpha_{b,a}(x)-\lim_{\xi\to 1}\alpha_{b,a}(\xi) \) for \( x\neq {^nb},n\in\mathbb{N}_0 \) and
\( \text{rslog}_{b,a}(x)=\lim_{\xi\to x}\alpha_{b,a}(x)-\lim_{\xi\to 1}\alpha_{b,a}(\xi) \) otherwise.
Following the idea of Jay to add the regular iteration at conjugate fixed points (and my idea to divide by 2 to get an Abel function again) let us consider
\( \alpha_{b,a}^\ast(x)=\frac{\alpha_{b,a}(x)+\alpha_{b,\overline{a}}(x)}{2} \)
where \( a \) is a fixed point in the upper halfplane.
Proposition: \( \alpha_{b,\overline{a}}(x)=\overline{\alpha_{b,a}(x)} \) for \( x\in\mathbb{R} \), \( x\neq {^nb} \). Particularly this implies that \( \alpha_{b,a}^\ast(x)=\Re(\alpha_{b,a}(x))=\Re(\alpha_{b,\overline{a}}(x)) \)
Note, that we define \( \alpha_{b,a}^\ast \) merely on the real axis, because this is the intersection of the domain of definition of \( \alpha_{b,a} \) (upper halfplane) and \( \alpha_{b,\overline{a} \) (lower halfplane).
Proof:
The first question that appears is: Which branch of the logarithm converges to \( \overline{a} \). While the usual logarithm is defined to yield imaginary values \( -\pi<y\le\pi \), for the lower primary fixed point we use the logarithm that yields imaginary values \( -\pi\le y<\pi \) denote this by \( \log^\ast \). For the non-primary fixed points \( \log(z)+2\pi i k \) is appropriate for the \( k+1 \)th upper fixed point and \( \log^\ast(z)-2\pi i k \) is appropriate for the conjugated fixed point (though \( \log(z)-2\pi i k \) is also ok, for \( k>0 \)).
We first verify that \( \log(\overline{z})=\overline{\log^\ast(z)} \) and hence \( \log(\overline{z})+2\pi i k=\overline{\log^\ast(z)-2\pi i k} \).
\( \begin{align*}
\log(\overline{z})&=\log(\overline{x+iy})=\ln(x-iy)\\
&=\log(r(\cos(\varphi)-i\sin(\varphi))) & \text{let} -\pi< -\varphi\le\pi\\
&=\ln( r)+\log(\cos(-\varphi)+\sin(-\varphi)) & -\pi\le \varphi<\pi\\
&=\ln( r)+\log(e^{-i\varphi})=\ln( r)-i\varphi=\overline{\ln( r)+i\varphi}=\overline{\log^\ast(x+iy)}=\overline{\log^\ast(z)}\end{align*} \).
A further consequence is that \( \log_{\overline{c}}(\overline{z})=\overline{\log_c(z)} \).
The rest is then easily established, let \( a \) be the \( k+1 \)th fixed point in the upper half plane:
\( \alpha_{b,\overline{a}}(x)=\lim_{n\to\infty} n-\log_{1/\log(\overline{a})}(\overline{a}-\left(\log^\ast_b-\frac{2\pi i k}{\ln(b)}\right)^{\circ n}(x))=\overline{\lim_{n\to\infty} n-\log_{1/\log(a)}\left(\log_b+\frac{2\pi i k}{\ln(b)}\right)^{\circ n}(x))}=\overline{\alpha_{b,a}(x)} \).
\( \beta(f^{-1}(x))=\beta(x)+1 \)
and then is clear, that \( \alpha(x)=-\beta(x) \) is an Abel function of the original function \( f \), because
\( -\beta(f(x))=-\beta(x)+1 \) iff \( \beta(x)=\beta(f(x))+1 \) iff \( \beta(f^{-1}(y))=\beta(y)+1 \).
So for repelling fixed points \( a \) we get the formula:
\( \alpha_{b,a}(x)=\lim_{n\to\infty} n-\log_{1/\log(a)}(a-\log_b^{\circ n}(x)) \)
which works for arbitrary repelling complex fixed points and arbitrary \( b>1 \) as long as we chose a branch of the involved logarithm such that \( \log_b^{\circ n}\to a \).
For computing the regular super logarithm however we face a major problem with repelling fixed points: we can not compute directly \( \alpha_{b,a}({^nb}) \), as \( \log_b^{\circ n+2}({^nb})=-\infty \). This presents a problem because for the rslog we have to compute \( \alpha_{b,a}(1) \) to normalize the values.
The good news however is that \( \lim_{x\to {^nb}} \alpha_{b,a}(x) \) seems to always exists. So the regular super logarithm is then:
\( \text{rslog}_{b,a}(x)=\alpha_{b,a}(x)-\lim_{\xi\to 1}\alpha_{b,a}(\xi) \) for \( x\neq {^nb},n\in\mathbb{N}_0 \) and
\( \text{rslog}_{b,a}(x)=\lim_{\xi\to x}\alpha_{b,a}(x)-\lim_{\xi\to 1}\alpha_{b,a}(\xi) \) otherwise.
Following the idea of Jay to add the regular iteration at conjugate fixed points (and my idea to divide by 2 to get an Abel function again) let us consider
\( \alpha_{b,a}^\ast(x)=\frac{\alpha_{b,a}(x)+\alpha_{b,\overline{a}}(x)}{2} \)
where \( a \) is a fixed point in the upper halfplane.
Proposition: \( \alpha_{b,\overline{a}}(x)=\overline{\alpha_{b,a}(x)} \) for \( x\in\mathbb{R} \), \( x\neq {^nb} \). Particularly this implies that \( \alpha_{b,a}^\ast(x)=\Re(\alpha_{b,a}(x))=\Re(\alpha_{b,\overline{a}}(x)) \)
Note, that we define \( \alpha_{b,a}^\ast \) merely on the real axis, because this is the intersection of the domain of definition of \( \alpha_{b,a} \) (upper halfplane) and \( \alpha_{b,\overline{a} \) (lower halfplane).
Proof:
The first question that appears is: Which branch of the logarithm converges to \( \overline{a} \). While the usual logarithm is defined to yield imaginary values \( -\pi<y\le\pi \), for the lower primary fixed point we use the logarithm that yields imaginary values \( -\pi\le y<\pi \) denote this by \( \log^\ast \). For the non-primary fixed points \( \log(z)+2\pi i k \) is appropriate for the \( k+1 \)th upper fixed point and \( \log^\ast(z)-2\pi i k \) is appropriate for the conjugated fixed point (though \( \log(z)-2\pi i k \) is also ok, for \( k>0 \)).
We first verify that \( \log(\overline{z})=\overline{\log^\ast(z)} \) and hence \( \log(\overline{z})+2\pi i k=\overline{\log^\ast(z)-2\pi i k} \).
\( \begin{align*}
\log(\overline{z})&=\log(\overline{x+iy})=\ln(x-iy)\\
&=\log(r(\cos(\varphi)-i\sin(\varphi))) & \text{let} -\pi< -\varphi\le\pi\\
&=\ln( r)+\log(\cos(-\varphi)+\sin(-\varphi)) & -\pi\le \varphi<\pi\\
&=\ln( r)+\log(e^{-i\varphi})=\ln( r)-i\varphi=\overline{\ln( r)+i\varphi}=\overline{\log^\ast(x+iy)}=\overline{\log^\ast(z)}\end{align*} \).
A further consequence is that \( \log_{\overline{c}}(\overline{z})=\overline{\log_c(z)} \).
The rest is then easily established, let \( a \) be the \( k+1 \)th fixed point in the upper half plane:
\( \alpha_{b,\overline{a}}(x)=\lim_{n\to\infty} n-\log_{1/\log(\overline{a})}(\overline{a}-\left(\log^\ast_b-\frac{2\pi i k}{\ln(b)}\right)^{\circ n}(x))=\overline{\lim_{n\to\infty} n-\log_{1/\log(a)}\left(\log_b+\frac{2\pi i k}{\ln(b)}\right)^{\circ n}(x))}=\overline{\alpha_{b,a}(x)} \).
