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 The bounded analytic semiHyper-operators JmsNxn Ultimate Fellow Posts: 754 Threads: 104 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. MphLee Long Time Fellow Posts: 270 Threads: 23 Joined: May 2013 05/23/2016, 07:25 PM (This post was last modified: 05/23/2016, 08:58 PM by MphLee.) I already saw it on Arxiv! I'm pretty updated on your arxiv page (stalker mode ;P). I find all of this extremely exciting... I can't even imagine what is the role of this function, that I guess we agree must be a very special one in the sea of new functions that your iteration method can produce, in the universe and physics... I mean... what the hell it stands for? what does it computes? History tell us that almost every mathematical concept has a role in our understanding of the universe even if sometimes it needed decades or centuries before the correct link was found. I also think that this is, as usual, only the tip of the iceberg: this is only the analytic view on the subject. But we know that complex analysis is deeply linked with a lot of other interesting fields... from geometry to number theory! Think of the historical jump in complexity from integer/rational/irrational algebra to analysis/real and complex exponentiation. And apply it to tetration/all the hyperoperations. Are we maybe on the verge of a new mathematical revolution? Ok, this last paragraph is a bit more extreeme: a lot of fantasy, sorry. But I'm still very excited. But back on the topic: I was not able to finish your last two papers (too difficult for me so I'm slower at reading) So forgive me if I ask you something that appears there. Now you have some kind of operator that maps functions to their iteration (using your notation). $\uparrow:H\to H$ (where $H\subset {\mathbb C}^S$ is a set of complex-valued function on an addition-closed subset $S$ of $\mathbb C$) Some serious mathematician would ask if you have started to compare it with other known methods, when they match and when they do not. But I'm not so serious or professional so I would like to ask you: have you some ideas/intuition on the behavior of this map $\uparrow:H\to H$ and its dynamics in general? Is it injective? Has it fixed points? These bounded analytic semiHyper-operators are just an orbit (a flow) of this map. $\{\uparrow^{\circ s}[\times]\}_{s\in S\subset {\mathbb C}}$ __________ Last note: from the algebraic point of view, this may sound unexpected to me but obvious to the expert, the flow is just the "manifestation" of some more basic "fundamental blocks/bricks": See the following, I believe that this progression is quite remarkable and speaks for itself: Monoid/group action $\phi:E\times X\to X$ - $(E,+)$ abelian monoid (or group) - $\phi (a+b,\chi)=\phi (a,\phi (b,\chi))$ (or equivalently $\phi' : (E,+) \to ({X^X},\circ)$ is an homomorphism where $\phi':a \mapsto \phi (a,-)$ concept of "evaluation" ${\diamond}:F\times X\to X$ - $(F,*)$ monoid (or group ) - $(f*g) \diamond \chi=f\diamond (g\diamond \chi)$ (or equivalently $\diamond' : (F,*) \to ({X^X},\circ)$ is an homomorphism where $\diamond' :f \mapsto (f\diamond -)$ concept of "Iteration" ${\triangle}:E\times F\to F$ - $(E,+,\cdot)$ ring (or field ) - $(F,*)$ monoid (or group ) - $(a\cdot b) \triangle f=a \triangle (b \triangle f)$ or equivalently $\triangle' : (E,\cdot) \to ({F^F},\circ)$ is an homomorphism where $\triangle' :a \mapsto (e \triangle -)$ - $(a+b)\triangle f=(a\triangle f) *(b\triangle f)$ or equivalently $\triangle ({_},f) : (E,+) \to (F,*)$ is an homomorphism Is it known that the flows $\phi:E\times X\to X$ are definable by iteration and evaluation... $\phi(e,\chi)=(e\triangle f)\diamond \chi$ but the fact that all those concepts are structurally and algebraically very similar... are kind of group actions or "enriched" actions (such as module structures or vector space structures) what changes are just the domains it's mind-blowing. At this point we may ask for $E\subseteq X$ (I.E. we extend the domain of iteration to the domain of evaluation like for example from natural iteration of complex function to complex iteration of complex functions) and for a new external operation ${\nabla}:X\times F\to F$ satisfying $(\chi \nabla f)\diamond e=(e \triangle f)\diamond \chi$ (the main example is $F_\chi (e)=F^{\circ e}(\ch)$) Here is where we hget the superfunction and where we get an external operation that is structurally very different from the previous. Do the problems with iteration comes from this algebraic loss of "regular structure"? in fact the rank arise as iteration of ${\nabla}$ on the right( with this notation): ${\rm ra}:{\mathbb N}\times F\to F$ ${\rm ra}_\chi(0,f):=f$ ${\rm ra}_\chi(n+1,f):=\chi \nabla{\rm ra}_\chi(n,f)$ Now it gets interesting: (05/06/2016, 06:30 PM)JmsNxn Wrote: $F(s,z) = \frac{d^{z-1}}{dw^{z-1}}\frac{d^{s-1}}{du^{s-1}}|_{u=0}|_{w=0}\vartheta$ 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 $F(s,F(s+1,z)) = F(s+1,z+1)$ by definition of nabla ${\rm ra}_\chi(n+1,f)\diamond \xi=[\xi \triangle{\rm ra}_\chi(n,f)]\diamond \chi$ apply $[{\rm ra}_\chi(n,f)]\diamond$ to both side using the evaluation $[{\rm ra}_\chi(n,f)]\diamond[{\rm ra}_\chi(n+1,f)\diamond \xi]=[{\rm ra}_\chi(n,f)]\diamond[[\xi \triangle{\rm ra}_\chi(n,f)]\diamond \chi]$ Applyng the algebraic properties of these operation we get $[{\rm ra}_\chi(n,f)*{\rm ra}_\chi(n+1,f)] \diamond \xi=[(1+\xi) \triangle{\rm ra}_\chi(n,f)]\diamond \chi$ where 1 is the multiplicative unit of the ring/field of exponents, we then apply the definition of nabla and of "rank-exponentiation" again $[{\rm ra}_\chi(n,f)*{\rm ra}_\chi(n+1,f)] \diamond \xi=[\chi \nabla{\rm ra}_\chi(n,f)]\diamond (1+\xi)$ $[{\rm ra}_\chi(n,f)*{\rm ra}_\chi(n+1,f)] \diamond \xi={\rm ra}_\chi(n+1,f)\diamond (1+\xi)$ If the link is not clear fix the value of $f\in F$ and $\chi\in X$ and define a new function ${\mathcal F}:{\mathbb N}\times X\to X$, like we did for the flow: ${\mathcal F}(n,\xi):={\rm ra}_\chi(n,f)\diamond \xi$ and we get ${\mathcal F}(n,{\mathcal F}(n+1,\xi))={\mathcal F}(n+1,1+\xi)$ This is just to point that much of the equations are out there embedded in the intrinsic algebraic structure... and can be derived algebraically without analysis. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Ultimate Fellow Posts: 754 Threads: 104 Joined: Dec 2010 05/27/2016, 04:03 AM Hey Mphlee. I corrected some of the mistakes I made on my old arxiv page. They were rather small (the same result holds), the new version is also being edited by my professor, and just some double checking being done by him. And to your question on the super function operator, I have added a little bit about this. If $H$ is a set in which all its elements $\phi(z)$ are holomorphic for $\Re(z) > 0$, take the real positive line to itself, have a fix point $x_0 \in \mathbb{R}^+$ such that $0<\phi'(x_0) < 1$ and $\phi^{\circ n}(x) \to x_0$ for all $x\in[0,x_0]$ then.. There is a semi group of operators $S$ such that $S$ acts on $H$. Where more beneficially, the semi group is isomorphic to $\{\mathbb{R}^+,+\}$ This semi group being $\uparrow^t$ for $t \ge 0$ where $\uparrow^t \uparrow^{s} = \uparrow^{t+s}$ and if $F_t(x) = \uparrow^t \phi(x)$ then $F_t(F_{t+1}(x)) = F_{t+1}(x+1)$ (05/23/2016, 07:25 PM)MphLee Wrote: I would like to ask you: have you some ideas/intuition on the behavior of this map $\uparrow:H\to H$ and its dynamics in general? I have no idea to be honest. This is where I'm looking. The analysis is extensively simplified using ramanujan's master theorem. and some key points arise. I have a few well thought out points but nothing too expansive. Quote: Is it injective? $\alpha \uparrow^t z$ is injective in $t \ge 0$ and $0 < \Im(z) < \ell_t$ (where $\ell_t$ is the imaginary period in $z$. The operator $\uparrow^t$ is injective as well on the function space $H$ as defined above. Quote:Has it fixed points? The only fixed point of $\uparrow$ I can think of are constant functions. Since $\uparrow \phi(z) = \phi^{\circ z}(1)$ and if $\phi(z) = \alpha$ then surely $\phi^{\circ z}(1)= \alpha$ It always follows that $\uparrow^n \phi(x) \to \phi(1)$ though. This is exemplified by $\alpha \uparrow^n x \to \alpha$ Catullus Fellow Posts: 144 Threads: 31 Joined: Jun 2022   06/29/2022, 10:37 PM (05/27/2016, 04:03 AM)JmsNxn Wrote: Quote:Has it fixed points? The only fixed point of $\uparrow$ I can think of are constant functions. Since $\uparrow \phi(z) = \phi^{\circ z}(1)$What about the successor function? Sincerely: Catullus JmsNxn Ultimate Fellow Posts: 754 Threads: 104 Joined: Dec 2010 06/29/2022, 11:46 PM (06/29/2022, 10:37 PM)Catullus Wrote: (05/27/2016, 04:03 AM)JmsNxn Wrote: Quote:Has it fixed points? The only fixed point of $\uparrow$ I can think of are constant functions. Since $\uparrow \phi(z) = \phi^{\circ z}(1)$What about the successor function? Oh yes it is a fixed point in the general sense, but not in the space I'm concerning myself. So yes, the successor function is a fixed point. But the successor function isn't bounded in the right half plane. I kind of implicitly meant this when I wrote that. You are right though. The space in question is $$|\phi(z)| \le M$$ for $$\Re(z) \ge 0$$. I apologize for the confusion. « Next Oldest | Next Newest »

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