Now I realized that \( f_n \) has a lot of singularities except for \( n=1,2 \) where it is a linear function.
A singularity can only occur if the logarithm of 0 is taken.
For \( f_1 \) no singularity occurs because \( 0 \) is not in the image of \( \exp \). And we can simplify it to:
\( f_1(z)=\log(\exp_\eta(z)) = \log(\eta) z = e^{-1}z \)
Because \( f_1(z)=0 \) iff \( z=0 \) the only singularity can occur when \( \exp_\eta(z)=0 \) which is again not possible. So \( f_2 \) will also not have any singularities. It can indeed be written as a affine function.
\( f_2(z)=\log(e^{-1}(\exp_\eta(z)))=-1+e^{-1}z \)
For \( f_3 \)
\( f_3(z)=\log(-1+e^{-1}\exp_\eta(z)) \)
we determine the zeros of \( f_2(z)=0 \):
\( -1+e^{-1}z=0\leftrightarrow z=e \)
So whenever \( \exp_\eta(z)=e \), then there is a singularity of \( f_3 \), this is for:
\( \exp_\eta (z)=\exp(e^{-1}z)=e \leftrightarrow z=e+2\pi i k e \), \( k\in\mathbb{Z} \).
Thatswhy \( f_3 \) has a singularity at each \( e+2\pi i k e \).
Particularly at \( e \) which restricts the domain of definition of \( f_3 \) to \( (e,\infty) \).
Let us generalize this some more.
Let now \( f_n=\log_a^{[n]}\circ \exp_b^{[n]} \) and \( f=\lim_{n\to\infty} f_n \).
\( \log_a^{[n]} \) has singularities exactly if \( \log_a^{[m]}(x)=0 \) for some \( m\le n-1 \). As we assume that we take the real logarithm of real numbers (otherwise singularities could be avoided by choosing non-real branches).
In other words \( \log_a^{[n]} \) has singularities at \( A_n:=\{0,1,a,\dots,\exp_a^{[n-1]}(0)\} \).
These values are reached by \( \exp_b^{[n]}(z) \) at \( \exp_b^{[-n]}(A_n) \)
where \( \exp_b^{[-1]}(X):=\{\log_b(z)+\frac{2\pi i k}{\ln(b)}: z\in X, k\in\mathbb{Z}\} \).
However \( f_n \) does not need to have singularities at all of these values, as some \( \log_a \) may take a different branch \( \frac{2\pi i k}{\ln(a)} \) instead of returning 0 (or one of \( \exp_a^{[m]} \)) if the argument of \( f_n \) was non-real.
So if we always take the primary logarithm \( \log_b \) we obtain a possible set of singularities which lies inside the set \( B_n:=\log_b^{[n]}(\mathbb{C}) \). As the primary logarithm \( \log_b \) maps \( \mathbb{C} \) bijectively to \( \{z:-\pi/\ln(b) < \Im(z) \le \pi/\ln(b)\} \) we conclude that \( \exp_b^{[n]} \) is bijective on \( B_n \). More importantly any path \( \gamma \) to \( z \) from \( B_n \) in the upper halfplane will be mapped to \( \exp_b^{[n]}(\gamma) \) in the upper halfplane, i.e. it will not wind around 0.
Hence by our construction we must take the primary logarithm \( \log_a \) of the point \( \exp_b^{[n]}(z) \), \( z\in B_n \). But the primary logarithm \( \log_a \) lies again in the upper halfplane and so on, that means we must always take the primary logarithm \( \log_a \) of \( \log_a^{[m]}(\exp_b^{[n]}(z)) \), \( m\le n-1 \).
Particularly this is true singular choices of \( z\in B_n \), they *must* yield singularities \( \log_a \) can not escape to some other branch. Hence
Proposition. Every element of the set \( S_n := \log_b^{[n]}\(\{a,\dots,\exp_a^{[n-2]}(1)\}\) \) is a singularity of \( f_n \) (which is defined via path-continuation).
Where we consider \( \log_b \) to be the primary branch.
(0 and 1 are excluded because they have no logarithm for \( n\ge 2 \).)
These singularities are not isolated but they are branch points.
So depending how the path to a point winds around these singularites we get different results of the \( f_n \).
So we have to restrict ourselves to a simply connected neighborhood of the real axis where no singularities exist, there we have a unique continuation.
The interesting question is now how these singularities are distributed in the limit case.
Do some singularities converge to the real axis?
A singularity can only occur if the logarithm of 0 is taken.
For \( f_1 \) no singularity occurs because \( 0 \) is not in the image of \( \exp \). And we can simplify it to:
\( f_1(z)=\log(\exp_\eta(z)) = \log(\eta) z = e^{-1}z \)
Because \( f_1(z)=0 \) iff \( z=0 \) the only singularity can occur when \( \exp_\eta(z)=0 \) which is again not possible. So \( f_2 \) will also not have any singularities. It can indeed be written as a affine function.
\( f_2(z)=\log(e^{-1}(\exp_\eta(z)))=-1+e^{-1}z \)
For \( f_3 \)
\( f_3(z)=\log(-1+e^{-1}\exp_\eta(z)) \)
we determine the zeros of \( f_2(z)=0 \):
\( -1+e^{-1}z=0\leftrightarrow z=e \)
So whenever \( \exp_\eta(z)=e \), then there is a singularity of \( f_3 \), this is for:
\( \exp_\eta (z)=\exp(e^{-1}z)=e \leftrightarrow z=e+2\pi i k e \), \( k\in\mathbb{Z} \).
Thatswhy \( f_3 \) has a singularity at each \( e+2\pi i k e \).
Particularly at \( e \) which restricts the domain of definition of \( f_3 \) to \( (e,\infty) \).
Let us generalize this some more.
Let now \( f_n=\log_a^{[n]}\circ \exp_b^{[n]} \) and \( f=\lim_{n\to\infty} f_n \).
\( \log_a^{[n]} \) has singularities exactly if \( \log_a^{[m]}(x)=0 \) for some \( m\le n-1 \). As we assume that we take the real logarithm of real numbers (otherwise singularities could be avoided by choosing non-real branches).
In other words \( \log_a^{[n]} \) has singularities at \( A_n:=\{0,1,a,\dots,\exp_a^{[n-1]}(0)\} \).
These values are reached by \( \exp_b^{[n]}(z) \) at \( \exp_b^{[-n]}(A_n) \)
where \( \exp_b^{[-1]}(X):=\{\log_b(z)+\frac{2\pi i k}{\ln(b)}: z\in X, k\in\mathbb{Z}\} \).
However \( f_n \) does not need to have singularities at all of these values, as some \( \log_a \) may take a different branch \( \frac{2\pi i k}{\ln(a)} \) instead of returning 0 (or one of \( \exp_a^{[m]} \)) if the argument of \( f_n \) was non-real.
So if we always take the primary logarithm \( \log_b \) we obtain a possible set of singularities which lies inside the set \( B_n:=\log_b^{[n]}(\mathbb{C}) \). As the primary logarithm \( \log_b \) maps \( \mathbb{C} \) bijectively to \( \{z:-\pi/\ln(b) < \Im(z) \le \pi/\ln(b)\} \) we conclude that \( \exp_b^{[n]} \) is bijective on \( B_n \). More importantly any path \( \gamma \) to \( z \) from \( B_n \) in the upper halfplane will be mapped to \( \exp_b^{[n]}(\gamma) \) in the upper halfplane, i.e. it will not wind around 0.
Hence by our construction we must take the primary logarithm \( \log_a \) of the point \( \exp_b^{[n]}(z) \), \( z\in B_n \). But the primary logarithm \( \log_a \) lies again in the upper halfplane and so on, that means we must always take the primary logarithm \( \log_a \) of \( \log_a^{[m]}(\exp_b^{[n]}(z)) \), \( m\le n-1 \).
Particularly this is true singular choices of \( z\in B_n \), they *must* yield singularities \( \log_a \) can not escape to some other branch. Hence
Proposition. Every element of the set \( S_n := \log_b^{[n]}\(\{a,\dots,\exp_a^{[n-2]}(1)\}\) \) is a singularity of \( f_n \) (which is defined via path-continuation).
Where we consider \( \log_b \) to be the primary branch.
(0 and 1 are excluded because they have no logarithm for \( n\ge 2 \).)
These singularities are not isolated but they are branch points.
So depending how the path to a point winds around these singularites we get different results of the \( f_n \).
So we have to restrict ourselves to a simply connected neighborhood of the real axis where no singularities exist, there we have a unique continuation.
The interesting question is now how these singularities are distributed in the limit case.
Do some singularities converge to the real axis?