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 Trying to get Kneser from beta; the modular argument JmsNxn Ultimate Fellow Posts: 889 Threads: 110 Joined: Dec 2010 03/18/2022, 06:26 AM (This post was last modified: 03/29/2022, 03:32 AM by JmsNxn.) 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 « Next Oldest | Next Newest »

 Messages In This Thread Trying to get Kneser from beta; the modular argument - by JmsNxn - 03/18/2022, 06:26 AM RE: Trying to get Kneser from beta; the modular argument - by JmsNxn - 03/29/2022, 04:48 AM RE: Trying to get Kneser from beta; the modular argument - by JmsNxn - 03/29/2022, 06:34 AM

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