Trying to get Kneser from beta; the modular argument JmsNxn Ultimate Fellow Posts: 993 Threads: 117 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 JmsNxn Ultimate Fellow Posts: 993 Threads: 117 Joined: Dec 2010 03/29/2022, 04:48 AM (This post was last modified: 03/29/2022, 06:12 AM by JmsNxn.) So, I reread my modular stuff, and I made a couple math typos in the above post, but the general idea is still there. Again, this is just undeveloped research. Take it all with a grain of salt. I'm doing an info dump.... As I realize I haven't explained what modular functions are, or do; I thought I'd remind/explain to everyone what I mean here. $$\phi(s) = \sum_{j=-k}^\infty a_k e^{2\pi i j s}$$ This series converges uniformly on compact subsets of $$\Im(s) > 0$$ upto some singularities; though you need to change $$a_k$$ (varying where you center the Laurent expansion), and also converges at least uniformly on $$[0,1]$$ up to two discontinuities (for us at least). The key to a modular function has nothing to do with this though. A modular function is invariant under a special group of automorphisms of $$\mathbb{H} = \{\Im(s) > 0\}$$. So not only is it periodic, it also satisfies: $$\phi(\frac{az+b}{cz+d}) = \phi(z)\\$$ Where $$ad-bc = 1$$ for $$a,b \in \mathbb{Z}$$. We can say this about our $$\phi$$ because it is invariant under the automorphism $$z \mapsto z+1$$; and not under any $$z \mapsto z+c$$ for $$c \in \mathbb{R}/\mathbb{Z}$$; This produces the first generator of the special group of automorphisms of $$\mathbb{H}$$, the modular $$\Gamma$$ group. To get the second generator is a bit more complicated, and relates deeper to work I did in the paper. If you can derive that $$\phi(-1/z) = \phi(z)$$, then $$\phi$$ must be modular... If I write $$h(s) = 1/\beta(s)$$, then: $$h(s+1) = e^{-1/h(s)}(1+e^{i\phi(s)s})$$--notice that $$h$$ uses the exact generator we need: $$-1/z$$. This caries over to tetration too, $$H = 1/\text{tet}$$, then $$H(s+1) = b^{-1/H(s)}$$. This is oddly very important, and just helps us get a hang that the modular aspect isn't that far out there, considering the second generator makes a direct appearance with the multiplicative inverse. To get there though, we have to analyze what we're asking of $$\phi$$. We must have that $$\phi(s) \to \infty$$ as $$\Im(s) \to \infty$$ (look at the expansion). Additionally, as we want to relate this to the beta method, we have to address that we have surjectivity to $$\widehat{\mathbb{C}}$$. First of all $$\beta(s) \to 0$$ as $$|s| \to \infty$$ for $$\arg(s) \ge \pi/2$$; and further we'd like $$\tau(s) \to L$$. (Remember, we are trying to reconstruct Kneser). Exactly as in Tommy's Gaussian method, we want as $$|s| \to \infty$$ while $$0\le \arg(s) \le \pi/2$$ that $$\beta(s+1) = e^{\beta(s)}$$. This means we want $$e^{i\phi(s)s} \to 0$$, but since here, $$\phi(s) \to i\infty$$ and $$i\phi(s)s \to - \infty$$, we're all good. So in this area $$\beta(s+1) \approx e^{\beta(s)}$$. But $$\phi:\mathbb{R}\cup\infty\to\mathbb{R}\cup\infty$$ as well (it must map the boundary to itself).  But additionally, this means it has poles on $$[0,1]$$, we can choose a pole such that $$\phi(\infty) \sim \phi(-0)$$, and this gets us $$\phi(-1/s) \sim \phi(s)$$ as $$|s| \to 0,\infty$$ for $$s \in \mathbb{H}$$. Ensure this for all the derivatives too, and $$\phi(s) = \phi(-1/s)$$ by virtue of holomorphy. This is enough for us to say that it fixes the automorphic group. The automorphic group $$\Gamma$$ is generated by $$z\mapsto z+1,\,\,z\mapsto-1/z$$. ...I might need a bit more, I'll have to reread all the modular books I've read but this is at least 90% there. There's probably something small you also need to derive that $$\phi$$ must be modular, but I can't think of much off the top of my head. Again, I'm just spitballing areas of research I haven't fully developed.... All in all, we've asked for a lot of the modular function $$\phi(s)$$, I'd bet money that the only one that works is unique (I know it's unique because Kneser is unique), and so there's some modular function that makes Kneser. This would mean that $$\beta$$ would have a flavor of modular functions; it would still admit singularities but the singularities would look like discontinuities between $$\beta(s+1) = e^{\beta(s)}$$ or $$\beta(s+1) = 0$$, based on how $$e^{i\phi(s)s} = 0,\infty$$ when $$\phi$$ has a singularity. Additionally you'd be able to talk fluidly about the transformation $$1/\beta(s)$$ is equivalent to $$z \mapsto -1/z$$ much more fluidly (again, this was a large part of the paper for describing the weak Julia set). This would mean we wouldn't get "singularities" in $$\beta$$, we would get essential singularities that look like $$\beta(s+1) = 0,\beta(s+1) = e^{\beta(s)}$$. These, in the end would become our values of kneser that equal the orbit $$\exp^{n}(0)$$. This shit is unbelievably fascinating. But I have so much ODE/PDE research I've put on hold for 2 years to look at tetration; I need a break, rofl. So, I'm going back into my dark little corner of studying Schrodinger's equation, and how it relates to infinite compositions. See you guys on the otherside !!! Regards, James JmsNxn Ultimate Fellow Posts: 993 Threads: 117 Joined: Dec 2010 03/29/2022, 06:34 AM It's important to remember; the correct manner of computing $$\tau$$ is a small little trick: $$\beta(s) = \exp^{\circ n}(\beta(s-n) + \tau^{n}(s-n))\\$$ This is the exact implicit manner that we'd be required to do to get Kneser. It's largely slow and, not very efficient. But mathematically is what we need to remember. Where we'll start to see $$\tau^{n}(s-n) \approx L$$ and $$\beta(s-n) \approx 0$$. « Next Oldest | Next Newest »

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