Tetration Asymptotic Series JmsNxn Ultimate Fellow Posts: 993 Threads: 117 Joined: Dec 2010 06/10/2022, 11:27 PM (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 C-iteration" 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? bo198214 Administrator Posts: 1,615 Threads: 101 Joined: Aug 2007 07/02/2022, 10:37 AM (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 C-iteration" 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? JmsNxn Ultimate Fellow Posts: 993 Threads: 117 Joined: Dec 2010 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 E-5*z^11 - 3.6202764736003317855243892589155487225622153451634 E-5*z^12 - 1.6080724216503292699172049221643433258042551459311 E-5*z^13 - 7.1816950016398788838199208354206545023038257355745 E-6*z^14 - 3.2227189833782963297158720669992862252425048625638 E-6*z^15 - 1.4522898416077888233911830908546717007993161862868 E-6*z^16 - 6.5692689018586711099226950941511558913933794341841 E-7*z^17 - 2.9815414068417090524975645714217900937721032122339 E-7*z^18 - 1.3573017645273591657543672816286683203861576883508 E-7*z^19 - 6.1957773072014452526396199289152844991250580056410 E-8*z^20 - 2.8352284888659532737042040264288626168300288619249 E-8*z^21 - 1.3003388847415089292648838389653585663335687232426 E-8*z^22 - 5.9760834258357606554760750999794690256238177818190 E-9*z^23 - 2.7516550155850334088613811741460857471873424553901 E-9*z^24 - 1.2691789634394696392683795976593671457297353819964 E-9*z^25 - 5.8633397392766035162414167673743212399309565423312 E-10*z^26 - 2.7127434800812277698835765166438900199344726802836 E-10*z^27 - 1.2568039997676115316158876879469486097115505664483 E-10*z^28 - 5.8301536471094220748723560631635569554785296853437 E-11*z^29 - 2.7077507782993934168201247406928073117302713586960 E-11*z^30 - 1.2589830287685641289138737053231258637289280194117 E-11*z^31 - 5.[+++] At the amount of recursion I ran, slash series precision, digit precision--we're probably accurate to at least 25-30 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. Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022 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 E-5*z^11 - 3.6202764736003317855243892589155487225622153451634 E-5*z^12 - 1.6080724216503292699172049221643433258042551459311 E-5*z^13 - 7.1816950016398788838199208354206545023038257355745 E-6*z^14 - 3.2227189833782963297158720669992862252425048625638 E-6*z^15 - 1.4522898416077888233911830908546717007993161862868 E-6*z^16 - 6.5692689018586711099226950941511558913933794341841 E-7*z^17 - 2.9815414068417090524975645714217900937721032122339 E-7*z^18 - 1.3573017645273591657543672816286683203861576883508 E-7*z^19 - 6.1957773072014452526396199289152844991250580056410 E-8*z^20 - 2.8352284888659532737042040264288626168300288619249 E-8*z^21 - 1.3003388847415089292648838389653585663335687232426 E-8*z^22 - 5.9760834258357606554760750999794690256238177818190 E-9*z^23 - 2.7516550155850334088613811741460857471873424553901 E-9*z^24 - 1.2691789634394696392683795976593671457297353819964 E-9*z^25 - 5.8633397392766035162414167673743212399309565423312 E-10*z^26 - 2.7127434800812277698835765166438900199344726802836 E-10*z^27 - 1.2568039997676115316158876879469486097115505664483 E-10*z^28 - 5.8301536471094220748723560631635569554785296853437 E-11*z^29 - 2.7077507782993934168201247406928073117302713586960 E-11*z^30 - 1.2589830287685641289138737053231258637289280194117 E-11*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? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ Gottfried Ultimate Fellow Posts: 880 Threads: 129 Joined: Aug 2007 07/03/2022, 10:18 AM (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 Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022   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? Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ JmsNxn Ultimate Fellow Posts: 993 Threads: 117 Joined: Dec 2010 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 Shell-thron 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. Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022   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. Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ JmsNxn Ultimate Fellow Posts: 993 Threads: 117 Joined: Dec 2010 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. « Next Oldest | Next Newest »

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