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 A Holomorphic Function Asymptotic to Tetration JmsNxn Long Time Fellow Posts: 507 Threads: 89 Joined: Dec 2010 03/23/2021, 12:21 AM (This post was last modified: 03/23/2021, 05:32 AM by JmsNxn.) Hey Everyone, After the stunning defeat of my function $\phi$ when trying to construct holomorphic tetration, I've gone back to the drawing board. And instead of producing Tetration, I've focused on producing an asymptotic function to tetration. Or rather, functions which satisfy an asymptotic relationship similar to the functional equation of tetration. Enter in the function, $ \beta(s) = \Omega_{j=1}^\infty \frac{e^z}{e^{j-s} + 1}\,\bullet z\\$ Which is holomorphic for $\mathbb{C} / P$ where $P = \{j + (2 k+1)\pi i \,|\,j \in \mathbb{N}\,j\ge 1, k \in \mathbb{Z}\}$. Of which $\beta$ satisfies the identity, $ \beta(s+1) = \frac{e^{\beta(s)}}{e^{-s} + 1}\\ \frac{\beta(s+1)}{e^{\beta(s)}} = 1 + \mathcal{O}(e^{-s})\\$ Wherein, this relationship will also look like, $ \log \beta(s+1) = \beta(s) + \mathcal{O}(e^{-s})\\$ I have a proof the iterated log procedure will work on the real-line, as it does with $\phi$. I also can't see this producing as many errors in the complex plane, because the functional equation is eventually tetration. This would mean, $ F_n(s) = \log^{\circ n} \beta(s+n)\\$ May have a better chance at converging. Plus the differential relationship will be better satisfied (I think; at least better than where it wasn't with $\phi$), which is, $ \beta'(s+1) \approx \beta'(s)e^{\beta(s)} + \mathcal{O}(e^{-s})\\ \frac{\beta'(s+1)}{\beta'(s) e^{\beta(s)}} = 1 + \mathcal{O}(e^{-s})\\$ And somewhat similarly for higher order derivatives. The trouble with $\beta$ being it isn't entire. And it has poles which will show up in the limit formula, unless we're careful. I do have hope though that this might construct a function tetration function in some domain in the complex plane (think $|\Im(s)| < \pi, s \not \in (-\infty,-2]$). Anyway, I think this guy might actually work. I'm not going to get ahead of myself though; at the moment all I can do is produce a $C^1$ proof on the real line, but this seems far more hopeful. As it has to solve the functional equation minus an error which drops off exponentially. Regards, James PS: As a manner of avoiding the poles and avoiding the periodic problem, it may make sense to take a function, $ \lambda(s) = \sum_{k=0}^\infty a_ke^{2\pi i k s}\\ \lambda:\mathbb{R} \to \mathbb{R}\\ \lambda(s+1) = \lambda(s)\\$ And assuming we choose $\lambda$ such that it satisfies $\Re(\lambda) >\delta > 0$ for $s$ in some domain (thinking a domain $\mathcal{S}$ which may look like $\Im(s) \ge 0$), then, $ \beta_\lambda(s) = \Omega_{j=1}^\infty \frac{e^z}{e^{\lambda(s)(j-s)} + 1}\,\bullet z\\$ Still satisfies the functional equation/asymptotic relationship (though slightly weaker), $ \beta_{\lambda}(s+1) = \frac{e^{\beta_\lambda(s)}}{e^{-\lambda s} + 1}\\ \log(\beta_{\lambda}(s+1)) = \beta_\lambda(s) + \mathcal{O}(e^{-\lambda s}) = \beta_\lambda(s) + \mathcal{O}(e^{-\delta \Re(s)})$ Then through logarithms, again, try to define this tetration on $\mathbb{C}/(-\infty,-2]$. I'd have to play with this more though.... Anyways. To clarify, I'll use the mobius map as a starting point, $ \mu(z) = \frac{i(1+z)}{1-z} : \mathbb{D} \to \mathbb{H}\\$ Where $\mathbb{D},\,\mathbb{H}$ are the unit disk and the upper half plane, respectively. The boundary of $\mathbb{D}$ is sent to the real line including the point at infinity. If we take, $\lambda(s) = \frac{i(1+e^{2\pi i s})}{1 - e^{2\pi i s}}$ Then $\lambda : \mathbb{H} \to \mathbb{H}$, where on the real line, it hits the boundary values of $\mathbb{D}$ which includes the anomalous points $\lambda(j) = \infty$ and $\lambda(j+1/2) = 0$--which we'll some how have to work around. But otherwise, in the complex plane, we should get a well behaved $\beta_\lambda(s)$ which has an asymptotic relationship to tetration at infinity (wherever $\Re(\lambda(s)) > 0$). Where now the singularities are almost voided because they'll begin to appear less sporadically as the real argument grows (in the complex plane), and it's still real valued; minusing the most anomalous points $\lambda(j),\lambda(j + 1/2)$. Taking the conjugate shouldn't pose a problem. Upon which, after taking our iterated logs, we're reduced to showing the singularities are removable at $j, j+1/2$--which is saying a lot, lol. And then, what we want is definitely not $\mu(e^{2\pi i z})$; that would be far too easy. We want a NICE enough Riemann mapping that takes $\varphi(s) : \mathbb{D} \to \{z \in \mathbb{C}\,|\, \Re(z) > 0,\,\,0 < \Im(z) < \pi\}$ so that, $\varphi(e^{2\pi i s}) = \lambda(s) : \mathbb{H} \to \{z \in \mathbb{C}\,|\, \Re(z) > 0,\,\,0 < \Im(z) < \pi\}$ Or at least, something close enough to this situation; or something similarly well controlled. Even $\varphi(s): \mathbb{D} \to \{z \in \mathbb{C} \, | \, 0<\arg(z) < \theta\}$ might have a hope of working. So long as we choose our Riemann mapping to be real valued, and well enough for the log process to work, $\Re(\lambda) > 0$; I think we may have a chance in hell. For instance, the function $\lambda(s) = \sqrt[4]{\mu(e^{2\pi i s})} : \mathbb{H} \to \{z\,:\,0 < \arg(z) < \pi/4\}$ may be a hopeful solution; but what I want is something like this, without being this exactly. EDIT2: Last edit, I swear; I'm not going to touch this anymore. I've already done like 15 edits. It may be better to use $\lambda(s+1) = \lambda(s) + \mathcal{O}(\frac{1}{\sqrt{s}})$, and just go along with the log construction. Because in this case we can still have, $ \log(\beta_\lambda(s+1)) = \beta_{\lambda}(s) + \mathcal{O}(e^{-\lambda(s)s})\\$ But the exponential drop off may be more like $\mathcal{O}(e^{-\sqrt{s}})$ in this case. PPS: A BIG thank you goes to Tommy for this one. He didn't give up on how to construct tetration in this manner; and I think we might be able to do it with a switched up difference relation than $\phi$, largely because of Tommy's weird choices for $\phi$. He was definitely right when he said $\log(\phi(s+1)) = \phi(s) + s$ is the wrong asymptotic solution. And his talking points are the main reason I think $\log(\beta(s+1)) = \beta(s) + \mathcal{O}(e^{-\lambda s})$ might work. JmsNxn Long Time Fellow Posts: 507 Threads: 89 Joined: Dec 2010 03/23/2021, 08:11 AM (This post was last modified: 03/23/2021, 09:27 AM by JmsNxn.) Hey, everyone. I'm going to run through a proof of uniform convergence of $\log^{\circ n} (\beta_\lambda(s+n))$; to get a feel for what it might look like. I might make some errors, but this is looking like it might work. Let's start by considering $\lambda \in \mathbb{C}$ and $s \in \mathbb{C}$ as variables; assume that $\Re(\lambda) > \delta$ and $|\lambda(j-s) - (2k+1)\pi i| > \delta$ for some $\delta > 0$; and $j,k \in \mathbb{Z}$ with $j \ge 1$. We'll call this set $\mathbb{L}$. As such, $(s,\lambda) \in \mathbb{L}$. We have a holomorphic function for $(s,\lambda) \in \mathbb{L}$ given by, $ \beta_\lambda(s) = \Omega_{j=1}^\infty \frac{e^z}{e^{\lambda(j-s)} + 1}\,\bullet z\\$ This satisfies the functional equation, $ \beta_\lambda(s+1) = \frac{e^{\beta_\lambda(s)}}{e^{-\lambda s} + 1}\\$ And the way we've constructed $\mathbb{L}$ is such that for any compact subset $\mathcal{U} \subset \mathbb{L}$ we must have, $ \sum_{j=1}^\infty ||\frac{1}{e^{\lambda(j-s)} + 1}||_{(s,\lambda) \in \mathcal{U}} < \infty\\$ This is all that's needed to prove holomorphy of $\beta_\lambda(s)$ by the same proof which constructed $\phi$. In which, $ \log \beta_{\lambda}(s+1) = \beta_\lambda(s) + \mathcal{O}(e^{-\delta \Re(s)})\\$ And just as well by Taylor's theorem, $ \log(\beta_\lambda(s+1) + \mathcal{O}(e^{-\delta \Re(s)})) = \beta_\lambda(s) + \mathcal{O}(e^{-\delta \Re(s)})\\$ Which holds as $\Re(s) \to \infty$ by construction. So in defining a sequence of convergents $\tau_\lambda^{n}(s)$ where, $ \tau_\lambda^n(s) = \log(\beta_\lambda(s+1) + \tau_\lambda^{n-1}(s+1)) - \beta(s)\\$ As $\Re(s)\to \infty$ for $(s,\lambda) \in \mathbb{L}$, we get by induction $\tau_\lambda^n(s) = \mathcal{O}(e^{-\delta \Re(s)}) \to 0$ as $\Re(s) \to \infty$. We may expect a swift enough convergence in this manner. Then, $ \beta_\lambda(s+1) + \tau_\lambda^n(s+1) = e^{\beta(s) + \tau_\lambda^{n+1}(s)}\\$ To which, $ |\tau_\lambda^n(s) - \tau_\lambda^m(s)| \le Ae^{-\delta \Re(s)}\\$ So at infinity we have a normality condition, in which $\tau_\lambda^n(s)$ is bounded for large enough $n > N$. Where convergence should follow for $(s,\lambda) \in \mathbb{L}$ when $\Re(s) > X$.  And here is where we need a Riemann mapping on $\lambda$... Please, I need help at this point of the proof. But I think I have it. JmsNxn Long Time Fellow Posts: 507 Threads: 89 Joined: Dec 2010 03/24/2021, 09:58 PM (This post was last modified: 03/25/2021, 02:54 AM by JmsNxn.) Start by taking the function $\beta_\lambda$, and define a sequence of functions $\tau_\lambda^n(s)$ such that, $ \beta_\lambda(s) + \tau_\lambda^{n+1}(s) = \log(\beta_\lambda(s+1) + \tau_\lambda^n(s+1))\\$ We aren't too concerned about where $\tau_n$ is holomorphic at the moment. We know that, $ \tau_\lambda^0(s) = 0\\ \tau_\lambda^1(s) = \sum_{k=1}^\infty \frac{(-1)^k}{k} e^{-k\lambda s}\\$ Assume that, $ \tau_\lambda^n(s) = \sum_{k=1}^\infty c_{nk}(\lambda) e^{-k\lambda s}\\$ And we are reduced to the equation, $ \tau_\lambda^{n+1}(s) = \sum_{k=1}^\infty c_{(n+1)k}(\lambda) e^{-k\lambda s} = \log(\beta_\lambda(s+1) + \sum_{k=1}^\infty c_{nk}(\lambda)e^{-k\lambda} e^{-k\lambda s}) - \beta_\lambda(s)\\$ Now we can calculate $c_{(n+1)k}(\lambda)$ inductively on $k$. Assuming we know $c_{(n+1)j}(\lambda)$ for $1 \le j \le k$; then we can find $c_{(n+1)(k+1)}(\lambda)$ by, $ \lim_{\Re(s) \to \infty} \frac{\tau_\lambda^{n+1}(s) - \sum_{j=1}^k c_{(n+1)j}(\lambda) e^{-j\lambda s}}{e^{(k+1)\lambda s}} = c_{(n+1)(k+1)}(\lambda)\\$ Of which this limit necessarily exists; and constructs a sequence $c_{(n+1)k}(\lambda)$. And now our problem is reduced to solving $\lim_{n\to\infty} c_{nk}(\lambda)$. We are also given the quick relation, $ \tau^{n+1}_\lambda(s) = \log(\beta_\lambda(s+1) + \tau^n_\lambda(s+1)) - \beta_\lambda(s)\\ = -\log(1+e^{-\lambda s}) + \log(1 + e^{-\beta_\lambda(s)}(e^{-\lambda s} + 1) \tau^n(s+1))\\ =\sum_{k=1}^\infty \frac{(-1)^k}{k}e^{-k\lambda s} + \log(1+e^{-\beta_\lambda(s)}(e^{-\lambda s} +1)\tau^{(n)}(s+1))\\$ Then since we know that $\tau^{n+1}_\lambda(s) \to 0$ we know that, $ e^{-\beta_\lambda(s)}(e^{-\lambda s} +1)\tau_\lambda^{n}(s+1) \to 0\\$ Because, $e^{-\beta_\lambda(s)}(e^{-\lambda s} +1) = \frac{1}{\beta(s+1)}$, this implies that, $ \frac{\tau^n_\lambda (s)}{\beta_\lambda(s)} \to 0\\$ This gives us hope we can devise a normality condition on $\tau^n_\lambda(s)$ as it controls how fast $\beta(s)$ can dip to zero; which is the exact problem we had with $\phi$. This can be more aptly written, $ \frac{1}{\beta(s)} = o(e^{\lambda s})\\$ But we can say something stronger. We know that $\tau^n_\lambda(s) = \mathcal{O}(e^{-\lambda s})$ and this means that, $ \frac{\tau^n_\lambda (s)}{\beta_\lambda(s)} = \mathcal{O}(e^{-\lambda s})\\$ As this is the term in the logarithm, and in order for $\tau^{n+1}_\lambda(s)$ to tend to zero exponentially the term in the logarithm has to as well. Therefore we can say that, $ \frac{1}{\beta(s)} = \mathcal{O}(h(s))\\$ Where for any $\delta > 0$ we have $e^{-\delta s} h(s) \to 0$ as $\Re(s) \to \infty$. Which tells us that, $ |\beta(s)| \ge \frac{A}{h(s)}$ Which informs us asymptotically how $\beta(s)$ approaches zero at infinity... slower than any exponential. This should help us with the limit; in the end, and making sure we stay away from zero when defining our function $F_n(s) = \log^{\circ n} \beta(s+n)$. As to the second part, when we actually go about and take this limit we want $\lambda$ to be a function of $s$. And in this sense, we want $\beta_{\lambda(s)}(s) : \{s \in \mathbb{C} \,:\, |\arg(s)| < \theta\} \to \mathbb{C}$ where  $(s,\lambda(s)) \in \mathbb{L}$ and $\lambda : \mathbb{R}^+ \to \mathbb{R}^+$. This would mean for every imaginary part of $s$ there is a large enough $N$ in which $F_n(s)$ exists for $n > N$. This is where we'll need a clever Riemann mapping on $\lambda$ over some simply connected domain $\mathbb{D} \subset \mathbb{L}$; which I'm not sure how to do; especially as this requires a two variable ideation. Forget I said anything about periodic functions, I don't think that's necessary here. I'm going to keep plugging and playing with this, but the more I fiddle the more this has none of the problems $\phi$ has. Regards, James. EDIT: Some even more evidence this is doable. Let $\lambda(s)$ be an implicitly defined function such that, $ \log \beta_{\lambda(s+1)}(s+1) - \log \beta_{\lambda(s)}(s+1) = \log(1+ e^{-\lambda(s) s})\\ \log \beta_{\lambda(s+1)}(s+1) - \beta_{\lambda(s)}(s) + \log(1+e^{-\lambda(s)s}) = \log(1+ e^{-\lambda(s) s})\\ \log \beta_{\lambda(s+1)}(s+1) = \beta_{\lambda(s)}(s)\\$ Which would require an implicit solution $\lambda(s)$; hinting that the logarithm trick might work for correctly chosen $\lambda$; as it's idempotent in this case. « Next Oldest | Next Newest »

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