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Trying to get Kneser from beta; the modular argument
#1
Hey, everyone!

As I'm refocusing my mathematics back to ODEs and PDEs, and the study of the compositional integral--I'm going to be much less of a regular here. As to that, I'm going to make some threads as an info dump on the fringe concepts of the \(\beta\) method. This is a result largely created from Tommy's work on Tommy's method using infinite composition. As I don't have the energy to dive deeper into tetration; and all of the beta method is flushed out. I thought I'd write what I consider my alternative formula for Kneser.

This result is inspired by Tommy's use of the gaussian rather than the logistic function. I sort of combined the best of both worlds. We get gaussian level decay, but we maintain much of the algebra of the \(\beta\) method. To recall,

$$
\begin{align}
\beta_\lambda(s) &= \Omega_{j=1}^\infty \frac{e^{\mu z}}{1+e^{-\lambda(s-j)}}\,\bullet z\\
\beta_\lambda(s+1) &= \frac{e^{\mu\beta_\lambda(s)}}{1+e^{-\lambda s}}\\
\end{align}
$$

Despite the work I've done trying to separate the \(\beta\) method from Kneser's method--the deeper Sheldon and I dug, the more the \(\beta\) method was non-holomorphic. But, it's still very possible to derive Kneser from the \(\beta\) function. This begins by making the change of variables:

$$
\begin{align}
\lambda &\mapsto \lambda+s\\
\end{align}
$$

This is done incredibly carefully, so that:

$$
\begin{align}
\beta(s) &= \Omega_{j=1}^\infty \frac{e^{\mu z}}{1+e^{-(\lambda+s)(s-j)}}\,\bullet z\\
\beta(s) &= \beta_{\lambda+s}(s)\\
\end{align}
$$

This means we are using a Gaussian in the asymptotic of \(s\), but we're using a geometric convergence of the infinite composition. This voids much of the problems with Tommy's method. Now we'll move \(\lambda\) as we move \(s\) so that \(\lambda(s+1) = \lambda(s) - 1\). This means that:

$$
\lambda(s) + s = \varphi(s)\\
$$

Where \(\varphi\) is holomorphic on \(\Im(s) > 0\) and is \(1\)-periodic; sending to \(\Re(z) > 0\). If we call \(\phi(s) =i\varphi(s)\), then \(\phi(s)\) is a modular function:

$$
\begin{align}
\mathbb{H} &= \{s \in \mathbb{C}\,:\,\Im(s) >0\}\\
\phi(s) &: \mathbb{H} \to \mathbb{H}\,\,\text{bijectively on a vertical strip width 1}\\
\phi(s+1) &= \phi(s)\\
\end{align}
$$

From here, everything is set up perfectly:

$$
\begin{align}
\beta(s+1) &= \frac{e^{\mu \beta(s)}}{1+e^{-(\lambda + s)s}}\\
\beta(s+1) &= \frac{e^{\mu \beta(s)}}{1+e^{i\phi(s)s}}\\
\end{align}
$$

Hidden within this, \(\beta(s)\) is only holomorphic for \(\Im(s) > 0\). We can construct a similar \(\beta(s)\) for \(\Im(s) < 0\), and both of these \(\beta\)'s agree on \(\mathbb{R}\). This effectively constructs a Gaussian type \(\beta\) function that is much easier to work with, then when we use a Tommy style Gaussian. We have a good amount of algebra at our disposal.

This means that:

$$
\begin{align}
\beta(s) &= \Omega_{j=1}^\infty \frac{e^{\mu z}}{1+e^{i\phi(s)(s-j)}}\,\bullet z \,\,\text{for}\,\,\Im(s) > 0\\
\beta(s) &= \Omega_{j=1}^\infty \frac{e^{\mu z}}{1+e^{-i\phi(s^*)^*(s-j)}}\,\bullet z \,\,\text{for}\,\,\Im(s) < 0\\
\end{align}
$$

From here, one can argue much more directly, when \(\mu > 1/e\), that Kneser can be found through iterated logarithms. This works essentially in the same manner I had originally argued; but the algebra is much cleaner now. This leads us to discussing \(\log(z)\) for \(\Im(z) > 0\) or \(\Im(z) < 0\), in distinct cases. Where we then paste these two functions together on the real line--where the \(\beta\) function agrees perfectly. From this point, it's a cake walk.

If we call \(\text{tet}_K(s)\) Kneser's tetration for \(\Im(s) > 0\) (\(\mu > 1/e,\,b>e^{1/e}\)), where \(\text{tet}_K(s) \to L\) as \(|s|\to\infty\) while \(\arg(s) \ge \pi/2\) (Left-half of the upper half-plane limits to a Fixed Point; Kneser's Fixed Point), then iterated logarithms on \(\beta\) behave identically in this limit. This is again, very attuned to what Tommy was talking about with the Gaussian method. I think he just missed a couple steps to making it perfect.  This is to mean, if I write:

$$
\tau^n(s) = \ln(1+\frac{\tau^{n-1}(s+1)}{\beta(s+1)})/\mu - \ln(1+e^{-\phi(s)s})/\mu\\
$$

And,

$$
\beta(s+1) + \tau^{n-1}(s+1) = \exp(\mu(\beta(s) + \tau^n(s)))\\
$$

Then, for \(F(s) = \beta(s) + \tau(s)\), we get the same asymptotic \(F(s) \to L\) as \(|s|\to\infty\) while \(\arg(s) \ge \pi/2\). Remember, we are referring solely to \(\mu > 1/e\) or base \(b > e^{1/e}\). This is the prime example of Kneser. To add to everything,

But...........

We now have access to everything from the library of work to do with modular functions. The function \(\phi\) is a modular function; any candidate for \(\phi\) must be a modular function. This brings us back to Kneser in a strange way; there exists one modular function \(\phi\); which allows for the construction of tetration... and that tetration is Kneser.

As I don't want to ramble. I am saying, there is a unique modular function \(\phi:\mathbb{H}\to\mathbb{H}\) such that:

$$
\beta(s) + \tau(s) = \text{Kneser's Tetration}\\
$$

Regards, James
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Trying to get Kneser from beta; the modular argument - by JmsNxn - 03/18/2022, 06:26 AM

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