Why the beta-method is non-zero in the upper half plane JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 09/01/2021, 01:57 AM (This post was last modified: 09/01/2021, 09:31 AM by JmsNxn.) Hey, everyone! I thought I'd make a post explaining why the beta-method is non-zero in the upper half plane. I know that this may still be confusing, tommy seems unconvinced (and I don't blame him)--as the math is a little difficult. The important equivalence to remember is that, if a tetration has essential singularities, it'll have zeroes. And the beta-method has no essential singularities (if it has singularities they will be logarithmic), and we can show from that that there are no zeroes. First of all, Tommy is absolutely correct; if you have a holomorphic tetration on some domain, it cannot necessarily be extended ad infinitum to the left by taking logarithms. It will probably fail--especially in the neighborhood of the real line. The perfect example of this is actually within the steps we take to construct the beta-method. First of all, the function, $ F_\lambda(s) : \mathcal{U}_\lambda \to \mathbb{C}\\ \overline{\mathcal{U}_\lambda} \simeq \mathbb{C}/2\pi \mathbb{Z}\\$ Where there will be at least one branch cut. Now, when $\lambda \in \mathbb{R}^+$ and we restrict $|\Im(s)| < \pi / \lambda$, then the only branch cut that happens is at $s = -2$ which follows the real line. This is because, within this strip there are no essential singularities; but at $\Im(s) =\pi / \lambda$ we have a bunch of essential singularities at integer lengths. Here is where you will find zeroes too. Now it's important to remember that $s \mapsto s+1$ acts on $\mathcal{U}_\lambda$ for all $\lambda$; but in the real case, this acts on the strip $|\Im(s)| < \pi / \lambda$ perfectly; where this operation is "in parallel" so to say, with the shape of the cylinder. If I take $\lambda = 1+i$, then the strip without essential singularities will be slanted by the period $2 \pi i / 1+i$. But it has to have a period of order $2 \pi i / (1+i)$ and it must still allow $s \mapsto s+1$ to act on $\mathcal{U}_{1+i}$--so things can get a little whacky. Which means, we're getting a grid of essential singularities at different "slants" (so to speak); and this causes zeroes. To explain why the zeroes are happening is pretty straight forward. At an essential singularity, the tetration function attains the value $1$ INFINITELY often, and so, the logarithms (so to speak) have no choice but to make one of the $1$'s map to zero. This is to mean, if we have an accumulation of $1$'s near a point than the log of this will force a zero to happen somewhere in there, and then a branch cut once you apply log again. This is why in a lot of my slanted graphs you have zeroes and branch cuts popping out of the wood work. The essential singularities do not line up with the mapping $s \mapsto s+1$; and so an otherwise normal point, may be pulling back from a neighborhood of an essential singularity, and then it must hit a zero, and then a branch cut, all while being periodic. Chaos ensues. Now, I want to bring the original proof I have that the beta method is non zero into this discussion. But I may have underworded it. I got very excited because I knew this couldn't have any essential singularities, and had to behave nice. But I probably wrote it too fast, because the other parts of my paper were so much harder to justify, and I started playing a little fast and loose. So I appreciate tommy's questioning. I'll get to my original argument later, which honestly, is just a worse worded version of this post. So instead, I'll go slower and work on the sort of keystone lemma which holds everything together: Lemma 1: Suppose $y(z) : \mathbb{D} = \{z \in \mathbb{C}\,|\, |z| < 1\,\}\,\to\mathbb{C}$ is a holomorphic function with $y(0)= 0$, and $y(1^-) = 1$. Suppose, using the principal branch of the logarithm, $ \log(y(z)) : \mathbb{D}/(-1,0]\to\mathbb{C}\\$ Then, $ y : [0,1] \to [0,1]\\$ This lemma is basically saying that if a tetration function has a branch cut, and that branch cut is parallel to the real line, then the tetration must be real valued along the real line. This lemma really hides a lot of the legwork of why the beta-method is holomorphic in the upper half plane, and has no zeroes/branch cuts. And the key is that we can apply the principal branch of the logarithm and make sure the beta method has it's branch cut along the real line. We essentially choose this branch cut. But once you've made this choice, it implies that $y : [0,1] \to [0,1]$ which can't happen with the beta method, unless we're on the real line. So, in the upper half plane, it doesn't happen--by contradiction there are no zeroes. This is a very nuanced argument. So, I think it's safe to say, it's up in the air a bit. We should justify this more. And so, I'll justify it more. To do so, I'll continue with the toy model of $\lambda = 1 + i$ and $\lambda = 1$ which counts for all $\lambda \in \mathbb{R}^+$. The key to remember, is that in the above lemma, we have assumed that each $1$ is isolated. This happens for $\lambda = 1$ in the strip $\Im(s) < \pi$ (at least compactly). But for $\lambda = 1+i$ we do not get that each $1$ is isolated on the fundamental strip $\mathcal{U}_\lambda$. So the above lemma doesn't apply, because, there are an accumulation of singularities, which is caused by the askew essential singularities, not in parallel with $s \mapsto s+1$. This is to say, a zero is happening because $y(1^-) = 1$ and $y(0) = 0$ but $y$ is not holomorphic on $[0,1]$.... there's a singularity in there. We don't have this problem with the beta method. Only the periodic cases. With this, I'll restate my original proof. The beta-method, first of all, only maps the real line to the real line, and no where in the upper half plane does it map $t + ib \to \mathbb{R}^+$ while b is constant. So we get the benefit of $\lambda =1$ but on an unbounded domain. Further. There are absolutely no essential singularities in the original sector we prove convergence of the beta method. The function, $ \text{tet}_\beta: S_\theta \to \mathbb{C}\\$ where $S_\theta = \{s \in \mathbb{C}\,|\, |\arg(s)| < \theta\}$ for some $\theta<\pi/2$. In this sector, we have no essential singularities. In fact, it looks similarly to the case $\lambda \in \mathbb{R}^+$, excepting we're in a sector rather than a strip. And we can pull back for the same reasons we can pull back when $\lambda \in \mathbb{R}^+$. We will not get a cluster of 1s to throw off the logarithm. And we'll have no path parallel to $s \mapsto s+1$ in which we have a singularity. The first fact to note is that in this sector, if $\text{tet}_\beta(t+s_0) \in \mathbb{R}^+$ for $t \in (-\delta,\delta)$ then $\Im(s_0) = 0$.  This is because if it is, then as we increase t it stays real valued, but this asymptotically approaches $\beta(t+s_0)$ which is only real valued for $\Im(s_0) = 0$ (it may hit real values, but it cannot stay real valued on a line parallel to $\mathbb{R}^+$). So now if $\text{tet}_\beta(s_0) = 1$ then we know that $\text{tet}_\beta(s_0+t) \not \in \mathbb{R}^+$ for $t \neq 0,\, t \in (-\delta,\delta)$. If we assume that $\text{tet}_\beta(s_0 - 1) = 0$ then we can apply our lemma on $y(z) = \text{tet}_\beta(s_0+z-1)$ and $\log y(z) = \text{tet}_\beta(s_0+z-2)$. This will be the principle branch of the logarithm, and the branch cut can be made horizontal to $\mathbb{R}^+$--use the implicit function theorem. But then, this implies that $y : [0,1] \to [0,1]$ which is a contradiction. So that, $\text{tet}_\beta(s_0 - 1) \neq 0$. Now, it's important to remember this argument fails near an essential singularity, because we can find a 1 arbitrarily close to the singularity; and we can't apply this argument on the sequence of 1's to make sure each logarithm is non-zero. The word isn't quite "isolated 1", but it's something close to it. We cannot find a holomorphic logarithm for each 1--hence a break down, and a zero. And then, pulling back again, a logarithmic singularity coupled with a branch cut. And we can see this in our lemma. It is dependent on $y$ being holomorphic on $[0,1]$; this doesn't happen necessarily; unless the push forward has no singularities.  If we have a slanted grid, with a bunch of singularities; the pull back under the mapping $s \mapsto s+1$ gets arbitrarily close to a singularity. And eventually $y(1^-) = 1\,\,\text{and}\,\,y(0) = 0$ but $y\,\,\text{is not holomorphic for}\,\,[0,1]$. Here is two graphs explaining the $\lambda =1, 1+i$ difference. Here is $\lambda = 1+i$ over $-1 \le \Re(z) \le 3$ and $-1.5 \le \Im(z) \le 2$:     You can see the zeroes appearing almost arbitrary. But it's because somewhere forward we're getting arbitrarily close to an essential singularity. Remember that this function has a slant of the form $2 \pi i / 1+i$ (this is the period, so the almost cylinder of holomorphy is slanted as this period). Conversely, looking at when $\lambda = 1$ when $-3 \le \Re(s) \le 3$ and $|\Im(s)| \le 3$,     There are no zeroes in the strip. But we're getting closer and closer to zero near the essential singularities--because the above argument begins to fail. But it can only actually attain zero at $\Im(s) = \pi$, by some pullback near an essential singularity. This argument is very subtle. But the proof is sound. The essential trick is Lemma 1; and the fact we have a lot of control on how the branch cut will look--we can always make it parallel with the real line. Which is just an elementary application of the implicit function theorem. This is an argument that is tailored specifically for tetration too--it has little generality to other functions. Any comments, questions for clarification, are greatly appreciated. Regards, James Just for fun here is the graph of $\lambda = 1+3i$ which has a plethora of zeroes, because somewhere down the line of applying $s \mapsto s+1$ we have a singularity. For $-1 \le \Re(z) \le 1$ and $0 \le \Im(z) \le 1$--see the zeroes which come from pullbacks in neighborhoods of singularities.     Notice the prominent zero, has a branch cut between it and 1; $s \mapsto s+1$ has a singularity inbetween. Hence lemma 1 is unapplicable, hence a zero is possible. As to overexplain, heres a better graphic:     The only way a zero arises, is if we have this break of infinity==in this case just a branch cut; but still a singularity/discontinuity at least. If we don't have this break of infinity; lemma 1 applies and there is no zero.   And remember that:     Where we're pulling back in the neighborhood of an essential singularity.... I'm just trying to explain the subtleties of when you can use Lemma 1 and when you can't.... « Next Oldest | Next Newest »

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