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 Alternative manners of expressing Kneser JmsNxn Long Time Fellow Posts: 571 Threads: 95 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. « Next Oldest | Next Newest »

 Messages In This Thread Alternative manners of expressing Kneser - by JmsNxn - 03/18/2021, 02:27 PM RE: Alternative manners of expressing Kneser - by JmsNxn - 03/19/2021, 01:02 AM

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