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 A Holomorphic Function Asymptotic to Tetration JmsNxn Long Time Fellow Posts: 509 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. « Next Oldest | Next Newest »

 Messages In This Thread A Holomorphic Function Asymptotic to Tetration - by JmsNxn - 03/23/2021, 12:21 AM RE: A Holomorphic Function Asymptotic to Tetration - by JmsNxn - 03/23/2021, 08:11 AM RE: A Holomorphic Function Asymptotic to Tetration - by JmsNxn - 03/24/2021, 09:58 PM

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