Oh you're definitely right, Tommy!
I'm still screwing this up like crazy. So I thought I'd make some clarifications. This is actually something I remember Sheldon talking about. The function I just defined is not actually what we want. After careful numerical evaluations, it's garbage. It's a super cool function, and it's close to what we want, but not exactly.
Let's assume we have constructed Kneser's super logarithm \(\text{slog}_K(z) = P(z)\). Then, we're going to define a holomorphic function, with inverse, which "switches between Kneser and Beta". The function \(\beta\) is holomorphic (the tetration it induces through iterated logarithms is not holomorphic, but \(\beta\) itself is). We will call this function:
\[
\chi(z) = \beta(P(z))\\
\]
The function:
\[
\beta(z+1) = \frac{e^{\beta(z)}}{1+e^{-z}}\\
\]
Which is certainly analytic, except at the points \(j + (2k+1)\pi i\) for \(j,k\in\mathbb{Z}\) and \(j \ge 1\). This function satisfies:
\[
\chi(\text{tet}_K(z)) = \beta(z)\\
\]
So long as we stay in the strip \(|\Im(z)| < \pi\). Here's a large graph of \(\chi\) for \(|\Re(z)|<10\) and \(|\Im(z)|<10\):
This function is absolutely beautiful in the left half plane \(\Re(z) < 0\); it is also \(2 \pi i\) periodic, so we can assume that:
\[
\chi(z) = \beta(-2) + \sum_{j=1}^\infty a_j e^{jz}\\
\]
For some coefficients \(a_j\). This is confirmed also, by graphing \(\chi(\log(z))\) which in theory should now be holomorphic for \(|z| < 1\). Which is confirmed by this graph (ignore the small branch cut at zero, it's an artifact of using \(\log\) here).
\[
\chi(\log(z)) = \beta(-2) + \sum_{j=1}^\infty a_j z^j\\
\]
I've been grueling trying to find the coefficients \(a_j\) without referring to Kneser's slogarithm \(P\). If we know \(P\) it's rather trivial to find \(a_j\). But my hypothesis at this moment, is that we don't need the super logarithm \(P\) to find \(\chi\).
This is the really cool part!
\[
\chi(e^z) = \frac{e^{\chi(z)}}{1+e^{-P(z)}}\\
\]
So that \(\chi\) has a fair amount of invariance under \(z \mapsto e^z\). Now, as we take \(\Re(z) \to -\infty\):
\[
\chi(0) = \frac{e^{\beta(-2)}}{1+e^{2}}
\]
And if we want:
\[
\chi(x + O(x^2)) = \frac{e^{\beta(-2)}}{1+e^2} + \lambda x + O(x^2)\\
\]
Where we should be able to recover \(\lambda\), without needing the super logarithm. The exact value is:
\[
\begin{align}
\lambda &= e^{-z} \frac{d}{dz}\frac{e^{\chi(z)}}{1+e^{-P(z)}} \Big{|}_{z=-\infty}\\
&= \lim_{z\to -\infty} e^{-z}\left(\chi'(z)\chi(e^z) + \frac{e^{-P(z)}P'(z) \chi(e^z)}{1+e^{-P(z)}}\right)\\
&= \lim_{z\to -\infty} e^{-z}\chi'(z)\chi(e^z)\\
&= a_1\cdot \chi(0)\\
\end{align}
\]
Because \(P'(z) \to 0\) as \(z \to -\infty\) faster than any power of an exponential. The super logarithm's derivative has insanely fast decay at \(-\infty\); certainly exponential, and certainly estimating it as \(e^{-e^{-z}}\) is a rather modest estimate.
But then, we can also find \(a_1\), which solves \(\lambda\), the value I am finding is \(a_1 = \beta'(-2)\cdot P'(0)\). This is found by just writing:
\[
\frac{d}{dz}\Big{|}_{z=0} \beta(P(z)-1) = \beta'(-2)P'(0)\\
\]
The essential philosophy I'm thinking, is that we can use \(\beta\) and basic transformations to get \(a_j\), using \(\beta\)--with zero mention of \(P\). Then if we invert \(\chi\), we have the entirely expected:
\[
\chi^{-1}(\beta(z)) = \text{tet}_K(z)\\
\]
This would, instead of mapping the level sets to themselves, map the behaviour on the level set. I forgot to do some substitutions earlier. I apologize.
This is still super up in the air, Tommy! I'm not trying to claim anything too strong yet! I'm mostly just floating ideas around! Don't scorch me! I'm still just as confused as you! But the numbers are seeming plausible!
We are mapping:
\[
\chi(\exp^{\circ n}(0)) = \beta(n-1)\\
\]
And then filling in the blanks, without using knowledge of Kneser... That's the goal at least!
Another quick way of thinking of this, is to write:
\[
\begin{align}
\chi(z) &= \beta(-2) + \sum_{j=1}^\infty a_j e^{-jz}\,\,\text{for}\,\,\Re(z) < 0\\
\chi(z) &= \beta(-1) + \sum_{j=1}^\infty b_j z^j\,\,\text{for}\,\,|z| < L\\
\end{align}
\]
And we find \(a_j\) as we find \(b_j\) etc etc, and we don't need the super logarithm to do this. We just need \(\beta\)....
Honestly any comments are greatly welcomed!
Sincere, Regards, James---go easy on me, Tommy!
I'm still screwing this up like crazy. So I thought I'd make some clarifications. This is actually something I remember Sheldon talking about. The function I just defined is not actually what we want. After careful numerical evaluations, it's garbage. It's a super cool function, and it's close to what we want, but not exactly.
Let's assume we have constructed Kneser's super logarithm \(\text{slog}_K(z) = P(z)\). Then, we're going to define a holomorphic function, with inverse, which "switches between Kneser and Beta". The function \(\beta\) is holomorphic (the tetration it induces through iterated logarithms is not holomorphic, but \(\beta\) itself is). We will call this function:
\[
\chi(z) = \beta(P(z))\\
\]
The function:
\[
\beta(z+1) = \frac{e^{\beta(z)}}{1+e^{-z}}\\
\]
Which is certainly analytic, except at the points \(j + (2k+1)\pi i\) for \(j,k\in\mathbb{Z}\) and \(j \ge 1\). This function satisfies:
\[
\chi(\text{tet}_K(z)) = \beta(z)\\
\]
So long as we stay in the strip \(|\Im(z)| < \pi\). Here's a large graph of \(\chi\) for \(|\Re(z)|<10\) and \(|\Im(z)|<10\):
This function is absolutely beautiful in the left half plane \(\Re(z) < 0\); it is also \(2 \pi i\) periodic, so we can assume that:
\[
\chi(z) = \beta(-2) + \sum_{j=1}^\infty a_j e^{jz}\\
\]
For some coefficients \(a_j\). This is confirmed also, by graphing \(\chi(\log(z))\) which in theory should now be holomorphic for \(|z| < 1\). Which is confirmed by this graph (ignore the small branch cut at zero, it's an artifact of using \(\log\) here).
\[
\chi(\log(z)) = \beta(-2) + \sum_{j=1}^\infty a_j z^j\\
\]
I've been grueling trying to find the coefficients \(a_j\) without referring to Kneser's slogarithm \(P\). If we know \(P\) it's rather trivial to find \(a_j\). But my hypothesis at this moment, is that we don't need the super logarithm \(P\) to find \(\chi\).
This is the really cool part!
\[
\chi(e^z) = \frac{e^{\chi(z)}}{1+e^{-P(z)}}\\
\]
So that \(\chi\) has a fair amount of invariance under \(z \mapsto e^z\). Now, as we take \(\Re(z) \to -\infty\):
\[
\chi(0) = \frac{e^{\beta(-2)}}{1+e^{2}}
\]
And if we want:
\[
\chi(x + O(x^2)) = \frac{e^{\beta(-2)}}{1+e^2} + \lambda x + O(x^2)\\
\]
Where we should be able to recover \(\lambda\), without needing the super logarithm. The exact value is:
\[
\begin{align}
\lambda &= e^{-z} \frac{d}{dz}\frac{e^{\chi(z)}}{1+e^{-P(z)}} \Big{|}_{z=-\infty}\\
&= \lim_{z\to -\infty} e^{-z}\left(\chi'(z)\chi(e^z) + \frac{e^{-P(z)}P'(z) \chi(e^z)}{1+e^{-P(z)}}\right)\\
&= \lim_{z\to -\infty} e^{-z}\chi'(z)\chi(e^z)\\
&= a_1\cdot \chi(0)\\
\end{align}
\]
Because \(P'(z) \to 0\) as \(z \to -\infty\) faster than any power of an exponential. The super logarithm's derivative has insanely fast decay at \(-\infty\); certainly exponential, and certainly estimating it as \(e^{-e^{-z}}\) is a rather modest estimate.
But then, we can also find \(a_1\), which solves \(\lambda\), the value I am finding is \(a_1 = \beta'(-2)\cdot P'(0)\). This is found by just writing:
\[
\frac{d}{dz}\Big{|}_{z=0} \beta(P(z)-1) = \beta'(-2)P'(0)\\
\]
The essential philosophy I'm thinking, is that we can use \(\beta\) and basic transformations to get \(a_j\), using \(\beta\)--with zero mention of \(P\). Then if we invert \(\chi\), we have the entirely expected:
\[
\chi^{-1}(\beta(z)) = \text{tet}_K(z)\\
\]
This would, instead of mapping the level sets to themselves, map the behaviour on the level set. I forgot to do some substitutions earlier. I apologize.
This is still super up in the air, Tommy! I'm not trying to claim anything too strong yet! I'm mostly just floating ideas around! Don't scorch me! I'm still just as confused as you! But the numbers are seeming plausible!
We are mapping:
\[
\chi(\exp^{\circ n}(0)) = \beta(n-1)\\
\]
And then filling in the blanks, without using knowledge of Kneser... That's the goal at least!
Another quick way of thinking of this, is to write:
\[
\begin{align}
\chi(z) &= \beta(-2) + \sum_{j=1}^\infty a_j e^{-jz}\,\,\text{for}\,\,\Re(z) < 0\\
\chi(z) &= \beta(-1) + \sum_{j=1}^\infty b_j z^j\,\,\text{for}\,\,|z| < L\\
\end{align}
\]
And we find \(a_j\) as we find \(b_j\) etc etc, and we don't need the super logarithm to do this. We just need \(\beta\)....
Honestly any comments are greatly welcomed!
Sincere, Regards, James---go easy on me, Tommy!
