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 The bounded analytic semiHyper-operators JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/06/2016, 06:30 PM Hello everyone, I am happy to announce I think I have a constructive proof of our beloved semi-operators. Namely, the extension of hyper-operators to complex numbers. In such a sense I'll reiterate what is already known, and what I've found. Taking $\alpha \in [1, \eta]$ $\Re(z) > 0$ then by initially having $\alpha \uparrow^0 x = \alpha \cdot z$ and recursively defining $g_n(w) = \sum_{k=0}^\infty \alpha \uparrow^n \alpha \uparrow^n...(k+1\,times)...\uparrow^n \alpha \frac{w^k}{k!}$ $\alpha \uparrow^{n+1} z = \frac{d^{z-1}}{dw^{z-1}}|_{w=0}g_n$ where $\frac{d^{z-1}}{dw^{z-1}}|_{w=0} f = \frac{1}{\Gamma(1-z)}(\sum_{k=0}^\infty f^{(k)}(0)\frac{(-1)^k}{k!(k+1-z)} + \int_1^\infty f(-w)w^{-z}\,dw)$ then $\alpha \uparrow^{n} (\alpha \uparrow^{n+1} z) = \alpha \uparrow^{n+1} (z+1)$ $\alpha \uparrow^{n} 1 = \alpha$ therein defining tetration, pentation, hexation, septation,... etc These functions have monotone growth on the real positive line and are bounded for $\Re(z) > 0$ Now, I've taken the following two variable function $\vartheta(w,u) = \sum_{n=0}^\infty \sum_{k=0}^\infty \alpha \uparrow^{n+1}(k+1) \frac{w^ku^n}{k!n!}$ and I've shown $F(s,z) = \frac{d^{z-1}}{dw^{z-1}}\frac{d^{s-1}}{du^{s-1}}|_{u=0}|_{w=0}\vartheta$ is a well defined function for $\Re(s) > 0, \Re(z) > 0$. Where in further, trivially $F(n,z) = \alpha \uparrow^n z$ and much less trivially $F(s,F(s+1,z)) = F(s+1,z+1)$ which when written in hyper-operator notation $F(s,z) = \alpha \uparrow^s z$ $\alpha \uparrow^s (\alpha \uparrow^{s+1} z) = \alpha \uparrow^{s+1} (z+1)$ The function $\alpha \uparrow^t x \in \mathbb{R}^+$ for $t, x \in \mathbb{R}^+$ and is monotonely growing in $x$ I am finalizing the paper now, I am just fixing a few wordings and typos and polishing it up. The interesting thing, which I find curious and funny, is that no living computer on this god given earth has a chance of calculating this function, it is most definitely one of the most processor heavy functions ever. Sadly this means calculating or graphing the function is next to impossible. Meaning we dont get to look at the pretty picture these semi operators define. This also means there's no numerical evidence to support the conclusion. Furthermore we cannot even check the result with numerical evidence. I'm hoping there's a better way to define this function--an alternative expression--however I've had no luck so far. I'll post the paper up most likely in two weeks. I just thought I'd post a little hurrah before so. « Next Oldest | Next Newest »

 Messages In This Thread The bounded analytic semiHyper-operators - by JmsNxn - 05/06/2016, 06:30 PM RE: The bounded analytic semiHyper-operators - by MphLee - 05/23/2016, 07:25 PM RE: The bounded analytic semiHyper-operators - by JmsNxn - 05/27/2016, 04:03 AM

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