Yes, that is perfectly possible.
Let's say we write \(b^{z} = \exp_b(z)\) for \(b \approx e\), and we discuss \(L(b)\) as the fixed point as a function of \(\exp_b(z)\).
There is a neighborhood where the iterated exponential is indistinguishable for each \(b\approx e\)--at least, topologically. We are just talking about moving \(L\) while we iterate. At least for the Schroder case--these are isomorphic objects until you hit the boundary of Shell-Thron. In many senses, the behaviour of \(\exp_2(z)\) is indistinguishable topologically from \(\exp(z)\).
What your response suggests, is that you are tracing a path across the fixed point \(L(b)\), and finding a holomorphic solution. There are countably infinite solutions to this though. You are choosing one, which I agree is very natural, but there are many many more--which satisfy the uniqueness you are arguing for.
\[
\exp^{\circ s}_b(z)\,\,\text{for}\,\,|z-L(b)| < \delta\\
\]
While:
\[
\exp^{\circ s}_b(L(b)) = L(b)\\
\]
This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.
There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.
This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea.
Let's say we write \(b^{z} = \exp_b(z)\) for \(b \approx e\), and we discuss \(L(b)\) as the fixed point as a function of \(\exp_b(z)\).
There is a neighborhood where the iterated exponential is indistinguishable for each \(b\approx e\)--at least, topologically. We are just talking about moving \(L\) while we iterate. At least for the Schroder case--these are isomorphic objects until you hit the boundary of Shell-Thron. In many senses, the behaviour of \(\exp_2(z)\) is indistinguishable topologically from \(\exp(z)\).
What your response suggests, is that you are tracing a path across the fixed point \(L(b)\), and finding a holomorphic solution. There are countably infinite solutions to this though. You are choosing one, which I agree is very natural, but there are many many more--which satisfy the uniqueness you are arguing for.
\[
\exp^{\circ s}_b(z)\,\,\text{for}\,\,|z-L(b)| < \delta\\
\]
While:
\[
\exp^{\circ s}_b(L(b)) = L(b)\\
\]
This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.
There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.
This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea.
