Iterating at eta minor
#21
Here's a graph of \(\text{tet}_{\sqrt{2}}\) with period \(2 \pi i/\lambda = 2 \pi i/0.38\). This is just before it turns into regular iteration. The weak julia set is the bits of black/white, which will turn into nothing as we let \(\lambda \to 0.3665.... = - \log \log 2\). These are again, fractal lines and the such, a.e. this tetration converges (When we think of a lebesgue area measure).

   


For reference, this is what the regular iteration looks like:

   




I'd also like to add that for \(b = e\) and say, period \(2 \pi i\), that this will be SMOOTH on \((-2,\infty)\) and be a smooth bijection to \(\mathbb{R}\). In the complex plane we get a weird equivalent of "smooth". In the sense that we make an arbitrarily accurate approximation, but the taylor expansion is nowhere analytic.

When you let \(\lambda \to 0\) though, as you let \(2 \pi i/\lambda \to i \infty\), you get a much calmer feature in the taylor expansion. And by proxy, the Taylor series is no longer zero radius of convergence. This is sort of the reallly reallllllly hard idea, I have no idea how to justify. All I know is that the Taylor coefficients calm down a lot, and it starts to look like the Crescent iteration (taylor coefficient wise).
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#22
I did some more numerical tests about Parabolic points on the Shell thron boundary. And everything is behaving as expected. To begin, I'm only going to focus on \(b = e^{i/e^i}\) which has a fixed point at \(e^i\) and whose exponential looks like:




\[


b^z = e^i + i(z-e^i) + O(z^2)\\


\]



Then, we can construct the \(2 \pi i\) periodic beta function as usual:




\[


\beta(s) = \Omega_{j=1}^\infty \frac{e^{iz/e^i}}{1+e^{j-s}}\,\bullet z\\


\]




Which satisfies:





\[


\beta(s+1) = \frac{e^{i\beta(s)/e^i}}{1+e^{-s}}\\


\]




Then we can take the iterated log's to get:




\[


F(s) = \beta(s) + \tau(s)\\


\]



Which satisfies:




\[


e^{iF(s)/e^i} = F(s+1)\\


\]




Not sure if it is tetrational, but this works. Then, translating this into beta.gp protocols, we can expand a taylor series as:




Code:
Sexp_N(I+z)

%37 = (0.43009170191496182527819681445577872265895208294413 + 0.88847636341401428615623698692762553770452570705891*I) + (-0.069584068940860811257612615440706823489129811880706 - 0.18102770794170207293331437410597567940199544990193*I)*z + (0.15150881644369075786155539894156295098223943567963 - 0.047729122157401620231741125727642254960282726926393*I)*z^2 + (0.016424246298951876050611631105005345959276864009557 + 0.094758438404327937727395945653829229385887506786094*I)*z^3 + (-0.043729428921885622644137705211342133807458850848791 - 0.0010425898373530911964797969751961292558908919214905*I)*z^4 + (0.0066349128260833502921175328749154657039183935400590 - 0.024092604904287956387725954451807197757948657065673*I)*z^5 + (0.0076472133389980087683540143385632099940176662432575 + 0.0059315910895966384527804006995320089339429070819386*I)*z^6 + (-0.0043139598767842683166008043980190970196954093290590 + 0.0056296778112537799087140970529319621698005213804829*I)*z^7 + (-0.0010283086201463810463241227586071422304312755471997 - 0.0026555511603110758468921921464804363906880329307130*I)*z^8 + (0.0016442892348491365299774805161898912432413177184532 - 0.0011515960240409491585119246942565511047831313116251*I)*z^9 + (5.6445888954836777426098997249445269145412807914002 E-5 + 0.00097242821320896717809665568971346691510203520071688*I)*z^10 + (-0.00057157880516207629056853344712870315383636662670608 + 0.00013430728523423432405065744848835832033997741719931*I)*z^11 + (6.0298960484903285468731095058196078338293203346480 E-5 - 0.00033863684707853356815906927246585322070560229218733*I)*z^12 + (0.00018731567215036970760533417186454241908001387294504 + 3.8158748701786273429925176681313790447313664227573 E-5*I)*z^13 + (-5.3024172351565935395524896590314842125159683152928 E-5 + 0.00010818477809364977936133563581553376415665602019808*I)*z^14 + (-5.4815209521402094301909223104445137017889794830744 E-5 - 3.8879467761401215874337991208927969574261352617448 E-5*I)*z^15 + (2.99565929408229214882071399717[+++]



I chose a taylor series about \(z=i\) because it's away from the singular behaviour that is to the south of \(z=0\), so the Taylor series is calmer. By which we can test that this is still a valid Taylor expansion, by treating it as a polynomial \(P(z)\) by which \(P(1/2) = exp(iP(-1/2)/e^i)\) should be indistinguishable quantities. Which we can check:




Code:
A = Pol(%37,z)
....
subst(A,z,1/2) - exp(base*subst(A,z,-1/2))
%39 = 2.0998198604133409115 E-47 + 2.257127662184212178 E-44*I


And we're accurate to about 44 digits.





Here's a graph of \(beta(x) + tau(x)\) for \(x\in(0,30)\):




   




Here's a graph of \(\beta(x) + \tau(x)\) for \(x \in(0,5)\):




   




Additionally here is a complex plot of \(\beta(z)\) which is the asymptotic solution that is \(2 \pi i\) periodic. The orange area is the good well behaved area, the black and white essentially means we have more chaos, and the superfunction will be more poorly behaved here. This is over \(|\Im(z)| < 7.5\) and \(0 < \Re(z) < 15\).



   






Essentially everything behaves the same for any parabolic point.
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#23
So, I cut the rendering of my graph, because I was lazy, and it's been about 30 hours, but here's the result of the beta super function for base \(b = e^{i/e^i}\) and period \(2 \pi i\). The black and white areas, are again, the chaos. The small line of them, is where the essential singularities inherited from \(\beta\) appear. The bigger black and white areas are the actual chaos inherited from \(b^z\). And the nice orange areas are where we are holomorphic and unproblematic. We, again, are holomorphic almost everywhere (under an area measure); but there is a lot of chaos/branching such that it's difficult to compute and graph perfectly.

   

This is about \(0 \le \Re(z) \le 15\) and contains a full periodic strip (a strip of width \(2 \pi\)).
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