A different approach to the base-change method JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 03/17/2021, 11:15 PM Hey, everyone. I made a post last night, but it really wasn't what I wanted, so I deleted it today and decided to write it out better this time. I'll begin by constructing tetration base $\eta = e^{\frac{1}{e}}$ and proving some properties about it, and then branch off into how I envision a base-change method looking. This is intended as a manner to use hyper-operations base $\eta$ to construct hyper-operations base $e$. Start with the function $\eta^z$ which has a neutral fixed point at $e$. There is an attractive petal $\mathcal{P}$ which includes the line $(-\infty,e)$. We can define an Abel function on this petal $\alpha : \mathcal{P} \to \mathbb{C}$ such that, $\alpha(\eta^z) = \alpha(z) + 1\\$ Let, $\mathcal{U} = \{|z-e| < \delta\} \cap \mathcal{P}$ be a neighborhood of the point $e$ in the attracting petal $\mathcal{P}$. The function $\alpha$ is univalent on $\mathcal{U}$ (for a small enough $\delta$) in this neighborhood. Further, as $z \to e$ in this neighborhood $\Re(\alpha(z)) \to \infty$--which corresponds to iterating $\eta^z$ which converts to right translations. This allows us to define a function, $\eta^{\circ s}(z) : \mathbb{C}_{\Re(s) > 0} \times \mathcal{U} \to \mathcal{U}\\$ Which can be written, $\eta^{\circ s}(z) = \alpha^{-1}(\alpha(z) + s)\\$ Upon which, as $|s|\to\infty$ while $0 \le \arg(s) < \pi/2$ we get that $\eta^{\circ s}(z) \to e$; as this relates to right translations which eventually approach the fixed point $e$ and approaches in the manner $\mathcal{O}(\frac{1}{\sqrt{\Re(s)}})$. What we want now, is to know what happens as $\Im(s) \to \pm\infty$; upon which it can't grow faster than $\mathcal{O}(\Im(s))$ by nature of Abel functions on neutral fixed points (this is an argument from John Milnor and the manner of constructing $\alpha$), where at infinity it looks like the identity function through a substitution. So now we are safe and have the following bounds (which are fairly weak from what we can say, but all we need), $|\eta^{\circ s}(z)| \le C e^{\tau|\Im(s)| + \rho|\Re(s)|}\\$ For, $0 < \tau < \pi/2$ and $C,\rho > 0$--which is again, a very weak bound. And now we enter the inverse mellin transform. Let $0 < \sigma < 1$ then, $f(x) = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(s) \eta^{\circ 1-s}(z) x^{-s}\,ds\\$ The poles of this integrand in the left half plane are at $s = -n$ and their residues are, $\text{Res}(s=-n;\Gamma(s) \eta^{\circ 1-s}(z) x^{-s}) = \eta^{\circ n+1}(z) \frac{(-x)^n}{n!}\\$ Since we have the exponential bounds of $\eta^{\circ s}$ and that, $\Gamma(s) \sim \sqrt{2\pi}s^{s-\frac{1}{2}}e^{-s}\,\,\text{as}\,\,|s|\to\infty\,\,\text{while}\,\,|\arg(s)| < \pi\\$ By a standard exercise in contour integration, we get, $f(x) = \sum_{n=0}^\infty \eta^{\circ n+1}(z) \frac{(-x)^n}{n!}\\$ And also, by Mellin's inversion theorem, we know that, $\int_0^\infty |f(x)|x^{\sigma - 1}\,dx < \infty\\ \int_0^\infty f(x)x^{s-1}\,dx = \eta^{\circ 1-s}(z)\Gamma(s)\\$ Since $f$ is entire, we can analyticially continue this expression by breaking the integral into $\int_0^1 + \int_1^\infty$--and we make the substitution $1-s \mapsto s$ so that, $\Gamma(1-s) \eta^{\circ s}(z) = \sum_{n=0}^\infty \eta^{\circ n+1}(z)\frac{(-1)^n}{n!(n+1-s)} + \int_1^\infty (\sum_{n=0}^\infty\eta^{\circ n+1}(z) \frac{(-x)^n}{n!})x^{-s}\,dx\\$ Which provides a formula for $\eta^{\circ s}$ depending solely on the natural iterates $\eta^{\circ n+1}$. Now since, there exists $K$ such that $\eta^{\circ K}(1) \in \mathcal{U}$, which implies the function $\eta^{\circ s}(1)$ is holomorphic for $\Re(s) > K$.  And now we extend this function to its maximal domain $\mathbb{C}/(-\infty,-2]$. To do this, we go by induction. Assume that $\eta^{\circ s}(1)$ is holomorphic for $\Re(s) > K$ and $\Im(s) > 0$. Assume that, $\eta^{\circ s_0}(1) = \eta^{\circ s_1}(1)\\$ Then certainly for all $k \in \mathbb{N}$, $\eta^{\circ s_0 + k} = \eta^{\circ s_1 + k}$ Take $N$ such that $f_{01}(s) = \eta^{\circ s_{01} + s + N}$ is holomorphic for $\Re(s) > 0$. Then, for all $n \in \mathbb{N}$, $f_0(n) = f_1(n)\\$ Now the function $f_0 - f_1$ is exponentially bounded as above. And therefore it's inverse mellin transform is, $\sum_{n=0}^\infty (f_0(n+1) - f_1(n+1)) \frac{(-x)^n}{n!} = 0\\$ But, by the invertibility of the mellin transform, when we take the mellin transform we must get that it is zero while also equaling $f_0 - f_1$. And therefore, $f_0 - f_1 = 0\\ f_0 = f_1\\$ This then implies that $\eta^{\circ s_0 + s + N} = \eta^{\circ s_1 + s + N}$ for all $s$. This implies that $s_1 - s_0$ is a period of $\eta^{\circ s}$; but $\eta^{\circ s}$ has no period, therefore $s_0 = s_1$. Therefore, since $\eta^{\circ -1}(1) = 0$ there can be no other point where $\eta^{\circ s}(1) = 0$. Therefore the logarithm of $\eta^{\circ s}$ is always defined for $\Im(s) > 0$. A similar analysis works for $\Im(s) < 0$. Therefore, $\eta^{\circ s}(1)\,\,\text{is holomorphic for}\,\,\mathbb{C}/(-\infty,-2]\\$ Further, on the attractive petal $\mathcal{P}$ we have that, $\eta^{\circ s} : \mathbb{C}/(-\infty,-2] \to \mathcal{P}\,\,\text{bijectively}\\$ This implies a couple thing. First of all, if $z \in \mathcal{P}$ then $z+ 2\pi i e \not \in \mathcal{P}$ because $\eta^{z} = \eta^{z+2\pi i e}$ and this is impossible by injectivity of tetration. We have a holomorphic slog function which is very well behaved, $\text{slog} : \mathcal{P} \to \mathbb{C}/(-\infty,-2]$--which is equally as bijective. When we use $f(z) = e\log(z)$ which is the functional inverse of $\eta^z$ , we know that $f$ is repelling on the petal $\mathcal{P}$ and tends to infinity. This implies that $\eta^{\circ s}(z) \to \infty$ while $|s|\to\infty$ for $0 < |\arg(-s)| < \pi/2$. Now here is where we start for constructing a basechange function. Typically as I've seen attempts, we would use the unbounded branch of $\eta$ (the chi function, I believe we called it)--I want to use the bounded tetration to construct unbounded tetration. Let's try to look for a function $\varphi$ such that, $\varphi : (-\infty,e)\to(-\infty,\infty)\,\,\text{bijectively}\\ \varphi: \mathcal{P} \to \mathbb{C}\\ \varphi(\infty \pm it) = L^{\pm}\,\,\text{where}\,\,e^{L^{\pm}} = L^{\pm}\,\,\text{and}\,\,\overline{L^{\pm}} = L^{\mp}\\ \varphi(0) = 0\\ \overline{\varphi(z)} = \varphi(\overline{z})\\$ So we really only care about $\varphi$ for $\Im(s) \ge 0, s\not \in (e,\infty)$. Now, it's obvious there are many functions which satisfy these properties; of which the goal I have in mind, is to sequentially approximate $\varphi$ with $\varphi_n$ in which, $e^{\varphi(z)} = \varphi(\eta^z)\\$ Now the existence of these functions are ABSOLUTELY guaranteed. The example I have in mind is to use Kneser's tetration, $\text{tet}_{K}(z)$ and simply take, $\varphi(z) = \text{tet}_{K}(\text{slog}_{\eta}(z))\\$ But, as the reader may guess, this is cheating. We use Kneser's solution to get $\varphi$, which is putting the cart before the horse really. But I'm wondering if we can use what we know about Kneser's tetration, to attempt to construct $\varphi$ independently. Usually the constructive manner of creating this function is to use iterated logs. This won't work in this case because we are using the bounded branch of $\eta$'s iteration. And largely, by this point, I believe the iterated log method has become sort of hopeless to produce holomorphy. Instead, we want to look at the behaviour of $\varphi$ as $|z| \to \infty$ for $0 < |\arg(-z)| < \pi/2$ where $\varphi(z) \to L^{\pm}$ a complex fixed point, and use this to approximate way off in the left half plane. I am not certain how to do this, but I'm experimenting with certain processes. For example, $\varphi_n(z) = \exp^{\circ n}(L^{\pm} + f(\text{slog}_\eta(z) - n)}\\$ satisfies, $e^{\varphi_n(z)} = \varphi_{n+1}(\eta^z)\\$ For a function $f$. If we were able to choose this $f$ properly so that, convergence is guaranteed and the properties of $\varphi$ are satisfied for each $\varphi_n$, we may have an approach as solving these equations. Finding a function $f$ will be the difficult part here. The alternative way I think we might be able to do this; rather than dealing directly with iterated exponentials, we look at the Inverse Schroder function of $e^z$. $\Psi^{-1}(z) : \mathbb{C} \to \mathbb{C}/\{0\}\\$ Where, $\varphi(z) = \Psi^{-1}(e^{L\text{slog}_\eta(z) + \theta(\text{slog}_{\eta}(z))})\\$ Where $\theta$ is a $1$-periodic function. Let's let this be holomorphic for $\Im(z) > 0$. This function will satisfy the identity, $\theta(\text{slog}_\eta(\eta^z)) = \theta(\text{slog}_\eta(z))$ and, $e^{\varphi(z)} = e^{\Psi^{-1}(e^{L\text{slog}_\eta(z) + \theta(\text{slog}_{\eta}(z))})}\\ =\Psi^{-1}(e^{L\text{slog}_\eta(z) +L+ \theta(\text{slog}_{\eta}(z))})\\ = \Psi^{-1}(e^{L(\text{slog}_\eta(z)+1) + \theta(\text{slog}_{\eta}(z))})\\ =\Psi^{-1}(e^{L\text{slog}_\eta(\eta^z) + \theta(\text{slog}_{\eta}(\eta^z))})\\ =\varphi(\eta^z)\\$ This function will pretty much satisfy all our requirements of $\varphi$ except for the real-valued requirement, unless we choose $\theta$ properly. Where, spoiler alert, Kneser's choice is the right choice. Now, we have restricted our attention to the function $h(z) = \theta(\text{slog}(z))$; but now we are asking for a function $h$ which is invariant under the operator $\mathcal{C}_\eta : f(z) \mapsto f(\eta^z)$. Which is $h(\eta^z) = h(z)$ for $\Im(z) > 0$. And I wonder if we can find a suitable function $F$ by using a sequence of functions $h_n$. Now, the main reason I chose to write all this out, is a question about hyper-operators. Does there exist a tetration base $\eta_1$ such that, $\eta_1(z) : \mathbb{C}/(-\infty,-2] \to \mathbb{C}\\$ Such that, $\eta_1(x_0) = x_0\\ \eta_1'(x_0) = 1\\ \eta_1^{\circ n}(1) \to x_0\,\,\text{as}\,\,n\to\infty\\$ If there is, we can construct a pentation function which satisfies pretty much all the above properties that $\eta^{\circ s}$ satisfy--except we change the domains of holomorphy to suit pentation. But the mellin transform identity will still hold, and therefore so will the injectivity (which is crucial), non-periodicity will occur, so the inverse function will be unbounded; when using periodic functions the inverse function will be restricted to a horizontal strip the length of periodicity (why we didn't use $\sqrt{2}$ and we used $\eta$). All in all, I'm not quite too sure if this is doable. But upon which we'd have the function, $\text{tet}(s) = \varphi(\eta^{\circ s}(1))$ Any comments, suggested reading, questions regarding my construction, would be greatly appreciated, Regards, James « Next Oldest | Next Newest »

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