To explain this further.
Assume you have a function \( H(s) \) holomorphic way off in the right half plane (with singularities, or branch cuts, or what ever)--and non-zero. And you have the identity,
\(
\log H(s+1) - H(s) = |s|^{-\rho}\,\,\text{as}\,\,\Re(s) \to \infty\,\,\rho > 1\\
\)
Then,
\(
H(s+1) = e^{H(s) + \epsilon}\\
\)
For a function \( \epsilon \sim |s|^{-\rho} \) as \( \Re(s) \to \infty \). Now I'm going to cite Milnor, but it's actually proven by two other authors--it escapes my mind at the moment. But the accumulation points of \( \{c_n\}_{n=0}^\infty = \{\exp^{\circ n}(z)\}_{n=0}^\infty \), for almost all \( z\in\mathbb{C} \) is the sequence \( \{0,1,e,e^e,...,\} \) (the orbit of the exponential at zero). And therefore the sequence diverges to infinity (eventually). Think, if \( \exp^{\circ n_k}(z) \approx 1 \) then \( \exp^{\circ n_k}(z+\delta) = 1 \)--and this divergence is guaranteed.
Returning to our situation, this means that \( H(s) \to \infty \) as \( \Re(s) \to \infty \)--a.e. To prove this is a tad subtle, but I'll lay it out,
\(
\mathcal{N}_\epsilon^s = \{z \in \mathbb{C}\,|\,|z-H(s)| < 2\epsilon\}\\
\)
By our definition,
\(
H(s+1) \in \exp(\mathcal{N}_\epsilon^s)\\
\)
And further, \( H(s+1) \in \mathcal{N}_\epsilon^{s+1} \). But, for a small enough \( \epsilon_1 \) then,
\(
\mathcal{N}_{\epsilon_1}^{s+1} \subset \exp(\mathcal{N}_\epsilon^s)\\
\)
And we can find a decreasing sequence of \( \epsilon_n \to 0 \) in which,
\(
\mathcal{N}_{\epsilon_n}^{s+n} \subset \exp^{\circ n}(\mathcal{N}_\epsilon^s)\\
\)
Therefore, \( H(s+n) \in \exp^{\circ n}(\mathcal{N}_\epsilon^s) \)--but this diverges almost everywhere. Therefore \( \lim_{\Re(s) \to \infty} H(s) = \infty \) (where we mean this almost everywhere, but this is still an effective statement).
And now, we describe the sequence of convergents, for \( \tau^0 = 0 \),
\(
\tau^n(s) = \log(H(s+1) + \tau^{n-1}(s+1)) - H(s)\\
= \epsilon(s) + \log(1 + \frac{\tau^{n-1}(s+1)}{H(s+1)})\\
\)
For a compact set \( \mathcal{K} \) such that \( \Re(s) > K \) and \( |\Im(s)| \le T \); we can put a bound on \( |1/H(s+1)| \le 1/2 \) (which is where we need \( H \) to be non-zero). This is because for large enough \( \Re(s) > K \)--this eventually goes to zero (almost everywhere). We'd have to exclude potential anomalous points; but almost everywhere this works.
Then since \( \epsilon(s) \sim |s|^{-\rho} \); by induction, \( \tau^n(s) = \mathcal{O}(|s|^{-\rho}) \) as \( \Re(s) \to \infty \). But even better,
\(
|\tau^{n+1}(s) - \tau^n(s)| \le |\log(1+ \frac{\tau^n(s+1)}{H(s+1)}) - \log(1+ \frac{\tau^{n-1}(s+1)}{H(s+1)})| \\
\le (1+\delta)|\frac{\tau^n(s+1)}{H(s+1)} - \frac{\tau^n(s+1)}{H(s+1)}|\,\,\,\text{by a Lipschitz argument}\\
\le q|\tau^n(s+1) - \tau^n(s+1)|\,\,\text{for}\,\,0<q<1\\
\le q^n|\tau^1(s+n) - \tau^0(s+n)|\\
\le Mq^n|s+n|^{-\rho}\\
\)
This is a summable sequence, so that,
\(
|\tau_n - \tau_m| \le \sum_{j=n}^m \frac{Cq^n}{n^\rho}\\
\)
And this can get as small as you want, and the compact set argument works fine. Recall that \( \rho > 1 \) (the series converges when \( q=1 \)).
This guarantees a tetration WAY WAY off in the right half plane. And the pullback is the easy part. The hard part is getting a function \( H \) that works relatively well. The \( H \) I chose is very convoluted--but it works. It has to work, for the same argument as above.
\(
H(s) = \beta(s) = \Omega_{j=1}^\infty \frac{e^z}{e^{\displaystyle \frac{j-s}{\sqrt{1+s}}} + 1}\,\bullet z\\
\)
I chose this function, SPECIFICALLY because the pull back will be easy. There will be no singularities. There will be no zeroes. There will be nothing but complex behaviour, controlled.
Please note, I simplified the arguments from my paper. It's a bit more difficult in reality. But pretty much exactly the same. I have not used the change of variables \( s = -\log(w)/\lambda \); which I used tremendously in the paper. That's because this is a simplistic viewpoint. You need to make the change of variables somewhere down the line.
I'd like to say, your looking for the coefficients \( a_n \) are exactly what I was trying to do. But, I focused on solving the equation at \( \infty \). Which, through a certain variable change, could be mapped to a fixed point at zero, rather than infinity. And then, we're just talking about geometrically converging functions about a fixed point. Yada yada...
Assume you have a function \( H(s) \) holomorphic way off in the right half plane (with singularities, or branch cuts, or what ever)--and non-zero. And you have the identity,
\(
\log H(s+1) - H(s) = |s|^{-\rho}\,\,\text{as}\,\,\Re(s) \to \infty\,\,\rho > 1\\
\)
Then,
\(
H(s+1) = e^{H(s) + \epsilon}\\
\)
For a function \( \epsilon \sim |s|^{-\rho} \) as \( \Re(s) \to \infty \). Now I'm going to cite Milnor, but it's actually proven by two other authors--it escapes my mind at the moment. But the accumulation points of \( \{c_n\}_{n=0}^\infty = \{\exp^{\circ n}(z)\}_{n=0}^\infty \), for almost all \( z\in\mathbb{C} \) is the sequence \( \{0,1,e,e^e,...,\} \) (the orbit of the exponential at zero). And therefore the sequence diverges to infinity (eventually). Think, if \( \exp^{\circ n_k}(z) \approx 1 \) then \( \exp^{\circ n_k}(z+\delta) = 1 \)--and this divergence is guaranteed.
Returning to our situation, this means that \( H(s) \to \infty \) as \( \Re(s) \to \infty \)--a.e. To prove this is a tad subtle, but I'll lay it out,
\(
\mathcal{N}_\epsilon^s = \{z \in \mathbb{C}\,|\,|z-H(s)| < 2\epsilon\}\\
\)
By our definition,
\(
H(s+1) \in \exp(\mathcal{N}_\epsilon^s)\\
\)
And further, \( H(s+1) \in \mathcal{N}_\epsilon^{s+1} \). But, for a small enough \( \epsilon_1 \) then,
\(
\mathcal{N}_{\epsilon_1}^{s+1} \subset \exp(\mathcal{N}_\epsilon^s)\\
\)
And we can find a decreasing sequence of \( \epsilon_n \to 0 \) in which,
\(
\mathcal{N}_{\epsilon_n}^{s+n} \subset \exp^{\circ n}(\mathcal{N}_\epsilon^s)\\
\)
Therefore, \( H(s+n) \in \exp^{\circ n}(\mathcal{N}_\epsilon^s) \)--but this diverges almost everywhere. Therefore \( \lim_{\Re(s) \to \infty} H(s) = \infty \) (where we mean this almost everywhere, but this is still an effective statement).
And now, we describe the sequence of convergents, for \( \tau^0 = 0 \),
\(
\tau^n(s) = \log(H(s+1) + \tau^{n-1}(s+1)) - H(s)\\
= \epsilon(s) + \log(1 + \frac{\tau^{n-1}(s+1)}{H(s+1)})\\
\)
For a compact set \( \mathcal{K} \) such that \( \Re(s) > K \) and \( |\Im(s)| \le T \); we can put a bound on \( |1/H(s+1)| \le 1/2 \) (which is where we need \( H \) to be non-zero). This is because for large enough \( \Re(s) > K \)--this eventually goes to zero (almost everywhere). We'd have to exclude potential anomalous points; but almost everywhere this works.
Then since \( \epsilon(s) \sim |s|^{-\rho} \); by induction, \( \tau^n(s) = \mathcal{O}(|s|^{-\rho}) \) as \( \Re(s) \to \infty \). But even better,
\(
|\tau^{n+1}(s) - \tau^n(s)| \le |\log(1+ \frac{\tau^n(s+1)}{H(s+1)}) - \log(1+ \frac{\tau^{n-1}(s+1)}{H(s+1)})| \\
\le (1+\delta)|\frac{\tau^n(s+1)}{H(s+1)} - \frac{\tau^n(s+1)}{H(s+1)}|\,\,\,\text{by a Lipschitz argument}\\
\le q|\tau^n(s+1) - \tau^n(s+1)|\,\,\text{for}\,\,0<q<1\\
\le q^n|\tau^1(s+n) - \tau^0(s+n)|\\
\le Mq^n|s+n|^{-\rho}\\
\)
This is a summable sequence, so that,
\(
|\tau_n - \tau_m| \le \sum_{j=n}^m \frac{Cq^n}{n^\rho}\\
\)
And this can get as small as you want, and the compact set argument works fine. Recall that \( \rho > 1 \) (the series converges when \( q=1 \)).
This guarantees a tetration WAY WAY off in the right half plane. And the pullback is the easy part. The hard part is getting a function \( H \) that works relatively well. The \( H \) I chose is very convoluted--but it works. It has to work, for the same argument as above.
\(
H(s) = \beta(s) = \Omega_{j=1}^\infty \frac{e^z}{e^{\displaystyle \frac{j-s}{\sqrt{1+s}}} + 1}\,\bullet z\\
\)
I chose this function, SPECIFICALLY because the pull back will be easy. There will be no singularities. There will be no zeroes. There will be nothing but complex behaviour, controlled.
Please note, I simplified the arguments from my paper. It's a bit more difficult in reality. But pretty much exactly the same. I have not used the change of variables \( s = -\log(w)/\lambda \); which I used tremendously in the paper. That's because this is a simplistic viewpoint. You need to make the change of variables somewhere down the line.
I'd like to say, your looking for the coefficients \( a_n \) are exactly what I was trying to do. But, I focused on solving the equation at \( \infty \). Which, through a certain variable change, could be mapped to a fixed point at zero, rather than infinity. And then, we're just talking about geometrically converging functions about a fixed point. Yada yada...