08/03/2022, 06:43 AM
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:
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:
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.
\[
b^z = e^i + i(z-e^i) + O(z^2)\\
\]
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*IAnd 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.
