Alternative manners of expressing Kneser JmsNxn Ultimate Fellow Posts: 1,064 Threads: 121 Joined: Dec 2010 03/18/2021, 02:27 PM (This post was last modified: 03/18/2021, 11:48 PM by JmsNxn.) So, I've been fiddling with Kneser's tetration, and I have been trying to think of different ways to express it. Let's write $\text{tet}_K$ for Kneser's tetration. Now depending on how it grows as $\Re(z) \to \infty$ we have different manners of expressing it. But there is one way I can say for sure. Let's take $\varphi(z) = i\sqrt{z}$ which maps the upper half plane of $\mathbb{C}$ to the upper left quadrant of $\mathbb{C}$. As $|z|\to\infty$ in the upper left quadrant $\text{tet}_K(z) \to L$ so long as we stay away from the real line. This implies the function $\text{tet}_K(i\sqrt{iz}) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}$ and this function satisfies $\text{tet}_K(i\sqrt{iz}) \to L$ as $|z|\to\infty$ in $\mathbb{C}_{\Re(z) > 0}$. And from this the Mellin transform is a viable option. Therefore, if we write, $ a_n = \text{tet}_K(i\sqrt{i(n+1)})\\$ Then, for $\Re(z) > 0$, $ \Gamma(1-z)\text{tet}_K(i\sqrt{iz}) = \sum_{n=0}^\infty a_{n} \frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty a_{n}\frac{(-x)^n}{n!})x^{-z}\,dx\\ $ Or if you prefer, for $\Im(z) > 0$ and $\Re(z) < 0$, $ \Gamma(1-iz^2) \text{tet}_K(z) = \sum_{n=0}^\infty a_{n} \frac{(-1)^n}{n!(n+1-iz^2)} + \int_1^\infty (\sum_{n=0}^\infty a_{n}\frac{(-x)^n}{n!})x^{-iz^2}\,dx\\$ This form gives us a way of representing Kneser's tetration using only the data points $\text{tet}_K(i\sqrt{i(n+1)})$. What I would really like to know is how fast Kneser's tetration grows as $\Re(z) \to \infty$ in the complex plane and how fast $\frac{1}{\text{tet}_K(z)}$ grows as $\Re(z) \to \infty$. Which would translate, how close does Kneser's tetration get to zero (which it must), and how fast does it do so. Ideally, I wonder if we can weaken this with a less intrusive function than $i\sqrt{iz}$. The best option being, simply using the data points $\frac{1}{\text{tet}_K(i(n+1))}$ to express tetration in the upper-half plane. This would require a careful analysis though. I believe this may be helpful, as we'd only have to approximate $a_n$ which for large $n$ should look something like, $ L + e^{iL\sqrt{i(n+1)}}\\$ Which has fairly fast convergence. It may also make sense to take a double sequence $a_{nk}$ where $\lim_{k\to\infty} a_{kn} = a_n$ and each $a_{nk}$ generates an exponential-like function which is Mellin Transformable as well. Think of using the partial sums of an exponential series which approximates $\text{tet}_K(z)$ as $|z|\to\infty$ in the upper left half quadrant; which will Mellin-transformable as well. Anywho, I'll let you guys know if I can think of a better kind of representation up this alley. Regards, James EDIT: For instance, if we write, $ \text{tet}_K(z) = \Psi^{-1}(e^{Lz}\theta(z))\\$ For a one-periodic function $\theta$ and $\Psi^{-1}$ the inverse Schroder function at a fixed point $L$ of $e^z$. Then, $ \text{tet}_K(z) = L + \sum_{j=1}^\infty \sum_{m=0}^\infty c_{mj} e^{Ljz + 2\pi i m z}\\$ If we truncate this series, then, $ F_k(z) = L + \sum_{j=1}^k \sum_{m=0}^k c_{mj} e^{Ljz + 2\pi i m z}\\$ Then calling $F_k(i\sqrt{iz}) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}$ and as $|z|\to\infty$ in this half plane $F_k(z) \to L$. So if we define, $ a_{nk} = L + \sum_{j=1}^k \sum_{m=0}^k c_{mj} e^{(iLj - 2\pi m)\sqrt{i(n+1)} }\\$ And we define, $ f_k(x) = \sum_{n=0}^\infty a_{nk} \frac{(-x)^n}{n!}\\$ Then, for $0 < \Re(z) < 1$, $ F_k(i\sqrt{i(1-z)})\Gamma(z) = \int_0^\infty f_k(x)x^{z-1}\,dx\\$ Whereupon, $ \lim_{p\to\infty} f_k^{(p)}(x) = Le^{-x}\\$ And for large values of $n > N$ we may be able to uncover an asymptotic relationship that is easier to derive than bruteforcing $c_{mj}$ using a Riemann Mapping theorem. EDIT2: As I remember people aren't so used to the mellin transform/fractional calculus approach as I am, this also generates a uniqueness condition. Let $h(z)$ be the function such that $i\sqrt{ih(z)} = i \sqrt{iz} + 1$, and assume there exists a function, $ G(z) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}\\$ Such that, $ |G(z)| \le C e^{\tau|\Im(z)| + \rho|\Re(z)|}\,\,\text{for}\,\,0 < \tau <\pi/2\,\,C,\rho > 0\\ G(n) = \text{tet}_K(i\sqrt{in})\\$ Then necessarily $e^{G(z)} = G(h(z))$ and $G = \text{tet}_K(i\sqrt{iz})$. We can also derive a uniqueness condition by ignoring the data points and requiring only that $G$ be bounded exponentially (as in the above), tends to the same fixed point, satisfies $e^{G(z)} = G(h(z))$, and only requiring that $G(b_n) = \text{tet}_K(i\sqrt{ib_n})$ for some sequence $b_n \to \infty$. Though this is a tad more complicated to derive. EDIT3: All in all, these transformations are intended as methods of understanding tetration through equations of the form, $ \vartheta_\lambda(w) = \sum_{n=0}^\infty e^{\lambda\sqrt{n+1}}\frac{w^n}{n!}\\ \frac{d^z}{dw^z} \vartheta_\lambda(w) = \sum_{n=0}^\infty e^{\lambda\sqrt{n+z+1}}\frac{w^n}{n!}\,\,\text{where}\,\,\\ \frac{d^z}{dw^z} \vartheta_\lambda(w)|_{w=0} = e^{\lambda \sqrt{z+1}}\\$ For $\lambda \in \mathbb{C}$. And summing over $\lambda_n$ with coefficients $c_n \in \mathbb{C}$ in an effort to approximating $G(h(z)) = e^{G(z)}$ with $\frac{d^z}{dw^z} \sum_n c_n \vartheta_{\lambda_n}(w) |_{w=0} = G(z)$. Whereby Kneser's solution, we know such sequences $c_n,\lambda_n$ exist. JmsNxn Ultimate Fellow Posts: 1,064 Threads: 121 Joined: Dec 2010 03/19/2021, 01:02 AM (This post was last modified: 03/19/2021, 01:08 AM by JmsNxn.) I think I may be on the way to answering this question. I've always wondered if Hilbert spaces could be useful for Tetration. This may be the first case I've ever seen it viable. I'll just post what I have here, and the rest of my thoughts I will compose in a PDF. This is an attempt at placing Kneser's tetration in a Hardy space, and deriving uniqueness/existence in this manner. Avoiding the Riemann mapping theorem only in technicality. I'll be brief here. Let's call $\mathcal{T}$ the space of holomorphic functions, $ h(z) = \sum_{n} c_n e^{\lambda_n \sqrt{z}}\,\,\text{for}\,\,z \in \mathbb{C}/\mathbb{R}^{-}\\$ For $\Re(\lambda_n) < 0$. Where the sum is infinite or finite; so long as we are guaranteed convergence. We'll define the inner-product as, for $0 < \sigma < \infty$, $ (f,g) = \lim_{T\to\infty} \frac{1}{2 T}\int_{\sigma-iT}^{\sigma + iT}f(z^2)\overline{g(z^2)}\,dz\\ $ This produces a hilbert space where, $ (e^{\lambda\sqrt{z}}, e^{\omega \sqrt{z}}) = A_{\lambda,\omega}\delta_{\Im(\lambda) = \Im(\omega)}\\$ Where if $\lambda = \omega$ we get $\delta = 1$, if $\Im(\lambda) = \Im(\omega)$ we get a finite real number; and otherwise $\delta = 0$. To those of you who may have fiddled with Hilbert Spaces, this is isomorphic to the space of almost periodic functions, if we look at it only in terms of rays for $z,\lambda$, this is precisely the space of almost periodic functions. This is a space, not exactly a familiar kind of Hilbert Space, but a non-separable Hilbert space. Nonetheless, we can now translate from one Hilbert space to the next. If we call, $\frac{d^z}{dw^z}\vartheta(w)|_{w=0} = f(z)$, where $\vartheta$ exists in a different $L^1$ space, specifically, $ \int_0^\infty |\vartheta(x)|x^{\sigma-1}\,dx < \infty\\$ Of which their exists an inner product $[\vartheta,\theta] = (f,g)$ (This can be explained better with a good understanding of Hilbert spaces; this is just a mapping between spaces). In this new space the inner product looks like, $ [\vartheta,\theta] = \int_0^\infty \vartheta(x)\overline{\theta(x)}\,d\mu\\$ For an appropriately chosen measure $\mu$ (where again, it requires some work to justify this, but it's something in this vector). This allows us to no longer talk about $f(z) = \text{tet}_K(i\sqrt{iz})$ as a function on its own, but rather talk about other functions which exist in a different Hilbert space, and approximating the solution in the different Hilbert space. I apologize if I'm being a bit top heavy in my analysis right now. I'm going to start a paper on this, mostly with the goal of finding a different manner of defining uniqueness and expressibility of Kneser's Tetration. Even if I have to use Kneser's tetration to do everything I'm going to do; I think it may be very important to view it as a Hilbert Space, and view the construction in a different Hilbert space. « Next Oldest | Next Newest »

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