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Rank-Wise Approximations of Hyper-Operations
#41
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
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#42
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
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#43
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.
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#44
(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
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#45
Question 
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
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#46
(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.
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#47
Question 
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
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