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 Rank-Wise Approximations of Hyper-Operations JmsNxn Ultimate Fellow Posts: 935 Threads: 111 Joined: Dec 2010 06/27/2022, 11:57 PM That's a very deep question, Catullus. I honestly don't know the answer to it. But to point out, when you use $$1 \le \alpha \le \eta$$, you have a very fine growth in the hyper-operations. In that, they tend to $$\alpha$$ for $$\Re(z) > 0$$. So doing this in this interval is about the closest I ever got. Where, as I've written before, I'll write again. $$\vartheta(w,u) = \sum_{n=0}^\infty\sum_{k=0}^\infty \alpha \uparrow^{n+2} (k+1)\frac{u^nw^k}{n!k!}\\$$ If we differentiate repeatedly in $$w$$, we get: $$\frac{d^j}{dw^j} \vartheta(w,u) \approx e^w\sum_{n=0}^\infty \omega_{n+2} \frac{u^n}{n!}\\$$ Where $$\omega_{n+2} = \alpha \uparrow^{n+2} \infty$$, which are the fixed points of $$\alpha \uparrow^{n+1} z$$. And if we differentiate repeatedly in $$u$$, we get: $$\frac{d^l}{du^l} \vartheta(w,u) \approx \alpha e^{w}e^{u}\\$$ Now as asymptotic limits, both of these things are differintegrable. Whereby: $$\frac{d^{s}}{du^{s}} \frac{d^{z}}{dw^z}\Big{|}_{u=0,w=0} \vartheta(w,u) = \alpha \uparrow^{s+2} z+1\\$$ This will satisfy the functional equation, but only if you can turn this heuristic into a proof. Which, I was unable to do. But I was only off by a few lemmas. Not sure what else you can do here... Your question is very fucking hard, don't expect an honest and easy answer to it. Regards, James tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 06/28/2022, 03:03 PM Maybe this is a dumb remark but .. What if 2 fixpoints have the same Taylor polynomial when the fixpoints are recentered at 0 ? So around fixpoint A we get a X + b X^2  + … ( recenter ) And the same polynomial around Fixpoint B. The the fixpoint formulas would agree around the 2 fixpoints. Another fixpoint in the middle might destroy the connection. Otherwise I see little objection ? Ofcourse you might need a periodic theta adjustment to get desirable properties such as real to real  ( kneser for instance )  Or even to make them analytically connected. But what is stopping us to unite 3 fixpoints if they also locally have the same Taylor ? —-  Regards  Tommy1729 JmsNxn Ultimate Fellow Posts: 935 Threads: 111 Joined: Dec 2010 06/30/2022, 12:05 AM Ah unfortunately that's not possible in the general sense, it's very restrictive. Assume that: $$f(z+c) - c = \sum_{j=1}^\infty a_j z^j\\$$ And $$f(z+b) - b = \sum_{j=1}^\infty a_j z^j\\$$ Then: $$f(z) = f(z+b-c) -b+c\\$$ Then if $$\mu = b-c$$ we're restricted to functions such that: $$f(z+\mu) = f(z) + \mu\\$$ Which implies that $$f(z) - z$$ is $$\mu$$-periodic. tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 07/01/2022, 10:01 PM (06/30/2022, 12:05 AM)JmsNxn Wrote: Ah unfortunately that's not possible in the general sense, it's very restrictive. Assume that: $$f(z+c) - c = \sum_{j=1}^\infty a_j z^j\\$$ And $$f(z+b) - b = \sum_{j=1}^\infty a_j z^j\\$$ Then: $$f(z) = f(z+b-c) -b+c\\$$ Then if $$\mu = b-c$$ we're restricted to functions such that: $$f(z+\mu) = f(z) + \mu\\$$ Which implies that $$f(z) - z$$ is $$\mu$$-periodic. wow this resonates so hard with the new thread i started before i read this ! regards tommy1729 Catullus Fellow Posts: 205 Threads: 46 Joined: Jun 2022   07/05/2022, 01:49 AM If you used similar uniqueness criterion to the one I proposed for tetration‚ but for higher hyper-operations than for tetration‚ when in the complex plane would a circulated to the infinity converge? ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus JmsNxn Ultimate Fellow Posts: 935 Threads: 111 Joined: Dec 2010 07/05/2022, 02:32 AM (07/05/2022, 01:49 AM)Catullus Wrote: If you used similar uniqueness criterion to the one I proposed for tetration‚ but for higher hyper-operations than for tetration‚ when in the complex plane would a circulated to the infinity converge? I'd wager a conjecture is in order. There exists a domain $$a \in \mathcal{D}$$ such that: $$a \uparrow^\infty z\\$$ is finite. I'd propose that solely (under any iteration protocol) this domain is $$\mathcal{D} \supseteq \mathcal{S}$$ for the Shell-thron region $$\mathcal{S}$$. This would be based on the fact that I know $$1 \le \alpha \le \eta$$ converges for $$\Re(z) > 0$$. As we sort of move and iterate more exotic areas of $$a \in \mathcal{S}$$, the domain $$\Re(z) > 0$$ moves in some manner. I do not know how. But, bounded iterations ellicit bounded iterations. So since bounded iterations of the exponential exist in the shell-thron region, and these iterations ellicit bounded iterations themself. We can expect that the Shell thron region $$a \in \mathcal{S}$$ always has a value $$z$$ such that: $$a \uparrow^\infty z = a\\$$ This is all I'm willing to talk on the matter. When the simple answer is I don't know, and I don't think anyone knows. I think this would be worthy of the greats if you could answer this question entirely. Catullus Fellow Posts: 205 Threads: 46 Joined: Jun 2022   07/05/2022, 08:25 AM (This post was last modified: 07/09/2022, 07:01 AM by Catullus.) What about i circulated to the infinity? Does that converge? If so, then what does it converge to? ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Please remember to stay hydrated. Sincerely: Catullus « Next Oldest | Next Newest »

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