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(06/09/2022, 05:58 PM)bo198214 Wrote: I think it is based on regular iteration of the formal power series of sqrt(2)^x at fixed point 2.
If you look at https://www.researchgate.net/publication..._tetration
chapter 2.3.1 under "Regular Citeration" you have a similar formula, though the infinite sum is not sorted by powers of s (in your case s=(ln2)^x ) but by powers of z, while in your case z=0 (at the regular iteration of f, not h). I guess after sorting it by powers of s you would arrive at the given formula (and know the further terms).
But this really involves quite some understanding and calculation, would be great if someone has some time to explicit it here.
I'm confused, isn't this just the Fourier expansion of \(F(x) = \sqrt{2} \uparrow \uparrow x\)?
$$
F(x) = 2 + \sum_{k=1}^\infty c_k \log(2)^{kx}\\
$$
At least, this second order expansion agrees as I look at it (you can just massage this to make the formal power series regular iteration method). This isn't much more than:
$$
\Psi^{1}(\log(2)^x\Psi(1))\\
$$
For the Schroder function \(\Psi\). Am I missing something?
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(06/10/2022, 11:27 PM)JmsNxn Wrote: (06/09/2022, 05:58 PM)bo198214 Wrote: I think it is based on regular iteration of the formal power series of sqrt(2)^x at fixed point 2.
If you look at https://www.researchgate.net/publication..._tetration
chapter 2.3.1 under "Regular Citeration" you have a similar formula, though the infinite sum is not sorted by powers of s (in your case s=(ln2)^x ) but by powers of z, while in your case z=0 (at the regular iteration of f, not h). I guess after sorting it by powers of s you would arrive at the given formula (and know the further terms).
But this really involves quite some understanding and calculation, would be great if someone has some time to explicit it here.
I'm confused, isn't this just the Fourier expansion of \(F(x) = \sqrt{2} \uparrow \uparrow x\)?
$$
F(x) = 2 + \sum_{k=1}^\infty c_k \log(2)^{kx}\\
$$
At least, this second order expansion agrees as I look at it (you can just massage this to make the formal power series regular iteration method). This isn't much more than:
$$
\Psi^{1}(\log(2)^x\Psi(1))\\
$$
For the Schroder function \(\Psi\). Am I missing something?
No, you are not missing anything. So can you calculate some more series coefficients, and show Catullus how its done?
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07/03/2022, 07:45 AM
(This post was last modified: 07/03/2022, 07:57 AM by JmsNxn.)
(07/02/2022, 10:37 AM)bo198214 Wrote: No, you are not missing anything. So can you calculate some more series coefficients, and show Catullus how its done?
Okay, so I can't generate the terms algebraically.
But each coefficient of \(z^k\) is the term \(c_k\) in \(c_k \log(2)^{kz}\) in the sum I wrote above.
This is done through a recursive protocol built into my program beta.gp. It's just a redundant amount of code meant to run the Schroder iteration, so it doesn't use any of the math of the beta method. I just installed this code into the program.
So this is \(A = \Psi(1)\) and the function output \(\Psi^{1}(A*z)\) for a variable \(z\). But it gives the coefficients \(c_k\).
Code: /*I've already initialized the base value as b=log(2)/2, and the beta polynomials, once you do that you can initialize the regular iteration, then you can run the code I'm writing*/
A = Sch_reg(1)
%39 = 0.63209866105082925035545064599078086279940391183279
Inv_Sch_reg(A*(z+O(z^50)))
%42 = 2.0000000000000000000000000000000000000000000000000  0.63209866105082925035545064599078086279940391183279*z  0.22563428568113651641314342556106075219383197400950*z^2  0.085408173026959759449911692849916118411760709773437*z^3  0.033577116075467402923000166342379529173359883175780*z^4  0.013567533990220214760849485909118650891001851681932*z^5  0.0055992068394587932789325712909644497530942417856943*z^6  0.0023500328878460392750960673121964884395590643434851*z^7  0.0010000364723451153067684842123773408264572502952945*z^8  0.00043048070830406712985249949329081051437808143735115*z^9  0.00018711645867148671128315207150288305360049496894112*z^10  8.2011402174512465373062128600968758671644383924212 E5*z^11  3.6202764736003317855243892589155487225622153451634 E5*z^12  1.6080724216503292699172049221643433258042551459311 E5*z^13  7.1816950016398788838199208354206545023038257355745 E6*z^14  3.2227189833782963297158720669992862252425048625638 E6*z^15  1.4522898416077888233911830908546717007993161862868 E6*z^16  6.5692689018586711099226950941511558913933794341841 E7*z^17  2.9815414068417090524975645714217900937721032122339 E7*z^18  1.3573017645273591657543672816286683203861576883508 E7*z^19  6.1957773072014452526396199289152844991250580056410 E8*z^20  2.8352284888659532737042040264288626168300288619249 E8*z^21  1.3003388847415089292648838389653585663335687232426 E8*z^22  5.9760834258357606554760750999794690256238177818190 E9*z^23  2.7516550155850334088613811741460857471873424553901 E9*z^24  1.2691789634394696392683795976593671457297353819964 E9*z^25  5.8633397392766035162414167673743212399309565423312 E10*z^26  2.7127434800812277698835765166438900199344726802836 E10*z^27  1.2568039997676115316158876879469486097115505664483 E10*z^28  5.8301536471094220748723560631635569554785296853437 E11*z^29  2.7077507782993934168201247406928073117302713586960 E11*z^30  1.2589830287685641289138737053231258637289280194117 E11*z^31  5.[+++]
At the amount of recursion I ran, slash series precision, digit precisionwe're probably accurate to at least 2530 digits. Might have more errors as we go further out.
Either way, setting \(z= \log(2)^x\) creates the asymptotic expansion Catullus is talking about, but it's actually just a Taylor expansion/Fourier expansion.
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07/03/2022, 09:20 AM
(This post was last modified: 07/03/2022, 09:28 AM by Catullus.)
(07/03/2022, 07:45 AM)JmsNxn Wrote: Code: /*I've already initialized the base value as b=log(2)/2, and the beta polynomials, once you do that you can initialize the regular iteration, then you can run the code I'm writing*/
A = Sch_reg(1)
%39 = 0.63209866105082925035545064599078086279940391183279
Inv_Sch_reg(A*(z+O(z^50)))
%42 = 2.0000000000000000000000000000000000000000000000000  0.63209866105082925035545064599078086279940391183279*z  0.22563428568113651641314342556106075219383197400950*z^2  0.085408173026959759449911692849916118411760709773437*z^3  0.033577116075467402923000166342379529173359883175780*z^4  0.013567533990220214760849485909118650891001851681932*z^5  0.0055992068394587932789325712909644497530942417856943*z^6  0.0023500328878460392750960673121964884395590643434851*z^7  0.0010000364723451153067684842123773408264572502952945*z^8  0.00043048070830406712985249949329081051437808143735115*z^9  0.00018711645867148671128315207150288305360049496894112*z^10  8.2011402174512465373062128600968758671644383924212 E5*z^11  3.6202764736003317855243892589155487225622153451634 E5*z^12  1.6080724216503292699172049221643433258042551459311 E5*z^13  7.1816950016398788838199208354206545023038257355745 E6*z^14  3.2227189833782963297158720669992862252425048625638 E6*z^15  1.4522898416077888233911830908546717007993161862868 E6*z^16  6.5692689018586711099226950941511558913933794341841 E7*z^17  2.9815414068417090524975645714217900937721032122339 E7*z^18  1.3573017645273591657543672816286683203861576883508 E7*z^19  6.1957773072014452526396199289152844991250580056410 E8*z^20  2.8352284888659532737042040264288626168300288619249 E8*z^21  1.3003388847415089292648838389653585663335687232426 E8*z^22  5.9760834258357606554760750999794690256238177818190 E9*z^23  2.7516550155850334088613811741460857471873424553901 E9*z^24  1.2691789634394696392683795976593671457297353819964 E9*z^25  5.8633397392766035162414167673743212399309565423312 E10*z^26  2.7127434800812277698835765166438900199344726802836 E10*z^27  1.2568039997676115316158876879469486097115505664483 E10*z^28  5.8301536471094220748723560631635569554785296853437 E11*z^29  2.7077507782993934168201247406928073117302713586960 E11*z^30  1.2589830287685641289138737053231258637289280194117 E11*z^31  5.[+++]
That has Big O of z to the power of thirty two, not z to the power of fifty.
Also, do you know of any closed forms of the terms after the one where z is raised to the power of 2?
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(07/03/2022, 09:20 AM)Catullus Wrote: (07/03/2022, 07:45 AM)JmsNxn Wrote: Code: /*I've already initialized the base value as b=log(2)/2, and the beta polynomials, once you do that you can initialize the regular iteration, then you can run the code I'm writing*/
Also, do you know of any closed forms of the terms after the one where z is raised to the power of 2?
Since I see that "regular" and "Schroeder" has been mentioned, that evaluation of mine might match your query. (They show polynomials for the coefficients of the power serie)
Gottfried Helms, Kassel
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07/04/2022, 11:19 PM
(This post was last modified: 07/04/2022, 11:21 PM by Catullus.)
What if the base is not the square root of two? For a real base, There is one a series like that for tetrations of a.
Also where in the complex plane does it have a series like that? For a real base it is when it is between one and eta non inclusive. But what about for a non real base?
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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07/04/2022, 11:55 PM
(This post was last modified: 07/04/2022, 11:56 PM by JmsNxn.)
(07/04/2022, 11:19 PM)Catullus Wrote: What if the base is not the square root of two? For a real base, There is one a series like that for tetrations of a.
Also where in the complex plane does it have a series like that? For a real base it is when it is between one and eta non inclusive. But what about for a non real base?
This method only works for \(a\) in the shell thron region. Then it'll work the same. It is holomorphic for at least \(z < \rho\) for some \(\rho\), too lazy to calculate it. But it should be about \(1\).
I do not know any more closed form expressions.
You can do a similar procedure for the boundary values of the Shellthron region using the beta method. For example, you can construct:
$$
F_\lambda(z)
$$
Such that:
$$
\begin{align}
F(0) &= 1\\
F(z+2 \pi i / \lambda) &= F(z)\\
F(z+1) &= \eta^{F(z)}\\
\Re \lambda &> 0\\
\end{align}
$$
Because of this, there exists a Fourier series (it's more chaotic than the above cases), but does look like:
$$
F_\lambda(z) = \sum_{k=\infty}^\infty c_k(\lambda) e^{\lambda k z}
$$
So we can make arbitrary periodic tetrations for neutral fixed points like the one \(\eta^z\) has.
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07/05/2022, 01:19 AM
(This post was last modified: 07/11/2022, 11:00 PM by Catullus.)
Thank you for helping.
This might be useful for calculating integer tetrations.
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ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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07/05/2022, 01:29 AM
(This post was last modified: 07/05/2022, 01:29 AM by JmsNxn.)
Oh I forgot to add I just cut off the terms at O(z^31), I used O(z^50) so that we'd at least be accurate for 30 or so terms. I can make all 50 terms if you'd like, but you can do it yourself using beta.gp. It's a little long winded though. beta.gp isn't intended for this, I just added it to calculate the elliptic functions spawned by the beta method.
