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06/12/2022, 03:33 AM
(This post was last modified: 06/23/2022, 08:24 AM by Gottfried.
Edit Reason: typo in subject
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Does anyone know of a patterns for the terms of the Taylor series for tetration? (With the Kneser method.) Like does anyone know of any formula for the Taylor series for e4^^x? How could you find the 65536th term of the Taylor series, without having to differentiate 65536 times?
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(06/12/2022, 03:33 AM)Catullus Wrote: Does anyone know of a patterns for the terms of the Taylor series for tetration? (With the Kneser method.) Like does anyone know of any formula for the Taylor series for e4^^x? How could you find the 65536th term of the Taylor series, without having to differentiate 65536 times?
Hey Catullus....
I believe Fatou.gp has a manner of calculating taylor series! And you don't have to differentiate. Sheldon has a built in protocol. I've never used it, so I can't speak for its efficacy. I'm not sure if you'd be able to go that far out efficiently though...
This is also a reason I don't like sheldon's program for polynomial data. Every program I write will accept \(\text{sexp}(3+z)\) and spit out a polynomial to desired precision in \(z\), effectively solving the Taylor polynomial at \(3\). Unfortunately, Sheldon uses matrices and root finders built into his functions, so the values as input are required to be static (you can't plug in a polynomial into his functions, like I'd be able to).
The trouble is, of course, that Kneser is a slow process on its own. And there is no "convenient" formula for the polynomial. I had tried to write a program using the beta method which would estimate Kneser (and hence solve polynomial data and take polynomial arguments), but as I could never 100% prove the process works, I abandoned it. Plus, the code would probably run unreasonably slow compared to fatou.gp (god that program is fast).
Not sure what else to tell you. There's no convenient recursive formula for taylor terms, especially for Kneser.
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06/12/2022, 05:30 AM
(06/12/2022, 04:24 AM)JmsNxn Wrote: (06/12/2022, 03:33 AM)Catullus Wrote: Does anyone know of a patterns for the terms of the Taylor series for tetration? (With the Kneser method.) Like does anyone know of any formula for the Taylor series for e4^^x? How could you find the 65536th term of the Taylor series, without having to differentiate 65536 times?
Hey Catullus....
I believe Fatou.gp has a manner of calculating taylor series! And you don't have to differentiate. Sheldon has a built in protocol. I've never used it, so I can't speak for its efficacy. I'm not sure if you'd be able to go that far out efficiently though...
This is also a reason I don't like sheldon's program for polynomial data. Every program I write will accept \(\text{sexp}(3+z)\) and spit out a polynomial to desired precision in \(z\), effectively solving the Taylor polynomial at \(3\). Unfortunately, Sheldon uses matrices and root finders built into his functions, so the values as input are required to be static (you can't plug in a polynomial into his functions, like I'd be able to).
The trouble is, of course, that Kneser is a slow process on its own. And there is no "convenient" formula for the polynomial. I had tried to write a program using the beta method which would estimate Kneser (and hence solve polynomial data and take polynomial arguments), but as I could never 100% prove the process works, I abandoned it. Plus, the code would probably run unreasonably slow compared to fatou.gp (god that program is fast).
Not sure what else to tell you. There's no convenient recursive formula for taylor terms, especially for Kneser. What about for fraction iterations of exponentials?
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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(06/12/2022, 05:30 AM)Catullus Wrote: (06/12/2022, 04:24 AM)JmsNxn Wrote: (06/12/2022, 03:33 AM)Catullus Wrote: Does anyone know of a patterns for the terms of the Taylor series for tetration? (With the Kneser method.) Like does anyone know of any formula for the Taylor series for e4^^x? How could you find the 65536th term of the Taylor series, without having to differentiate 65536 times?
Hey Catullus....
I believe Fatou.gp has a manner of calculating taylor series! And you don't have to differentiate. Sheldon has a built in protocol. I've never used it, so I can't speak for its efficacy. I'm not sure if you'd be able to go that far out efficiently though...
This is also a reason I don't like sheldon's program for polynomial data. Every program I write will accept \(\text{sexp}(3+z)\) and spit out a polynomial to desired precision in \(z\), effectively solving the Taylor polynomial at \(3\). Unfortunately, Sheldon uses matrices and root finders built into his functions, so the values as input are required to be static (you can't plug in a polynomial into his functions, like I'd be able to).
The trouble is, of course, that Kneser is a slow process on its own. And there is no "convenient" formula for the polynomial. I had tried to write a program using the beta method which would estimate Kneser (and hence solve polynomial data and take polynomial arguments), but as I could never 100% prove the process works, I abandoned it. Plus, the code would probably run unreasonably slow compared to fatou.gp (god that program is fast).
Not sure what else to tell you. There's no convenient recursive formula for taylor terms, especially for Kneser. What about for fraction iterations of exponentials?
Using Kneser using fatou.gp I am totally unsure. I know how to find fraction iterations of exponentials in Shellthron as efficient as possible, using either beta or Schroder. But for \(e\) and using Kneser. I do not know. I'm sure Sheldon has something built in which might find that. I don't know though. Interesting question though.
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06/12/2022, 06:04 AM
(06/12/2022, 05:53 AM)JmsNxn Wrote: Using Kneser using fatou.gp I am totally unsure. I know how to find fraction iterations of exponentials in Shellthron as efficient as possible, using either beta or Schroder. But for \(e\) and using Kneser. I do not know. I'm sure Sheldon has something built in which might find that. I don't know though. Interesting question though. What about for other bases?
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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See tetration combinatorics for more information. The example derives \( D^4f^n(z) \). The following works for all smooth iterated functions, so it applies to tetration and the hyperoperators.
Enumerate the total partition of 65536 (not computationally practical), evaluate each enumeration and add the terms together.
Daniel
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06/12/2022, 06:33 AM
(This post was last modified: 06/12/2022, 06:37 AM by JmsNxn.)
(06/12/2022, 06:17 AM)Daniel Wrote: See tetration combinatorics for more information. The example derives \( D^4f^n(z) \). The following works for all smooth iterated functions, so it applies to tetration and the hyperoperators.
Enumerate the total partition of 65536 (not computationally practical), evaluate each enumeration and add the terms together.
Hey, Daniel. Catullus was asking about Kneser. This does not solve Kneser. This only solves the geometric solution about a Fixed point (Schroder iteration).
To Catullus:
For ShellThron region. I can calculate arbitrary Taylor series of \(b\uparrow \uparrow z\) of \(z\) about any point \(z_0\). I can do this using the beta method (which solves for varying periodic solutions) and for the standard Schroder solution. Though I've never released code from the Schroder equation, I can do it. Then if you ask for the taylor series with 500 terms about 3, set the series precision to 500, then you'd just run sexp(3+z) for whatever \(b\) is. But this only works within the Shellthron region. And it is much slower than fatou.gp. Be prepared to wait. But it will produce the taylor expansion to desired requirement.
The beta method will do this faster, but it will produce a different taylor series which is \(2\pi i / \lambda\) periodic, where \(\lambda\) is now a free variable as opposed to being fixed like with Schroder. And the Schroder iteration exists as a boundary value of the beta method. So you can't use the beta method to get Schroder unfortunately.
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(06/12/2022, 06:33 AM)JmsNxn Wrote: (06/12/2022, 06:17 AM)Daniel Wrote: See tetration combinatorics for more information. The example derives \( D^4f^n(z) \). The following works for all smooth iterated functions, so it applies to tetration and the hyperoperators.
Enumerate the total partition of 65536 (not computationally practical), evaluate each enumeration and add the terms together.
Hey, Daniel. Catullus was asking about Kneser. This does not solve Kneser. This only solves the geometric solution about a Fixed point (Schroder iteration). JmsNxn the technique in question works for all fixed points except super attracting.
Daniel
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06/12/2022, 07:32 AM
(This post was last modified: 06/12/2022, 07:33 AM by JmsNxn.)
(06/12/2022, 06:50 AM)Daniel Wrote: (06/12/2022, 06:33 AM)JmsNxn Wrote: (06/12/2022, 06:17 AM)Daniel Wrote: See tetration combinatorics for more information. The example derives \( D^4f^n(z) \). The following works for all smooth iterated functions, so it applies to tetration and the hyperoperators.
Enumerate the total partition of 65536 (not computationally practical), evaluate each enumeration and add the terms together.
Hey, Daniel. Catullus was asking about Kneser. This does not solve Kneser. This only solves the geometric solution about a Fixed point (Schroder iteration). JmsNxn the technique in question works for all fixed points except super attracting.
Well aware of that Daniel. That's not Kneser. That's just Schroder iteration about a fixed point. Kneser is far more involved. You are not solving a theta mapping...
